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Laplace Transform – II Exercise Solved Problems

Kumar — Sat, 02 Sep 2023 10:55:09 +0000

Laplace Transform Exercise 2

1. If F(t) is continuous for all t≥0 and of exponential order a as t→∞ and if ⋅F^’ (t) is of class A, then prove that the Laplace transform of F'(t) exists when p>a and L[F'(t)]=pf(p)-F(0) where L[F(t)]=f(p).

Solution:

If F(t) is continuous for all t≥0 and of exponential order a as t→∞ and if ⋅F^’ (t) is of class A

L$\left[F^{\prime}(t)\right]=\int_0^{\infty} e^{-p t} F^{\prime}(t) d t=\int_0^{\infty} e^{-p t} d[F(t)]=\left[e^{-p t} F(t)\right]_0^{\infty}+\int_0^{\infty} p e^{-p t} F(t) d t$

= $0-F(0)+p \int_0^{\infty} e^{-p t} \cdot F(t) d t$ (because F(t) is exponential order)

∴ $L\left[F^{\prime}(t)\right]=p L[F(t)]-F(0)=p f(p)-F(0)$

2. Prove that $L\left[F^{\prime \prime}(t)\right]=p^2 f(p)-p f(0)-F^{\prime}(0)$.

Solution:

L$\left[F^{\prime \prime}(t)\right]=L\left[\left(F^{\prime}(t)\right)^{\prime}\right]=p L\left[F^{\prime}(t)\right]-F^{\prime}(0)=p \cdot[p L(F)-F(0)]-F^{\prime}(0)$

= $p^2 L F(t)-p f(0)-F^{\prime}(0)$

3. Prove that $L\left\{F^{(n)}(t)\right\}=p^n f(p)-p^{n-1} F(0)-p^{n-2} F^{\prime}(0) \cdots F^{(n-1)}(0)$.

Solution:

L$\left\{F^{\prime}(t)\right\}=\int_0^{\infty} e^{-p t} F^{\prime}(t) d t=e^{-p t} F(t)_0^{\infty}-\int_0^{\infty} F(t) e^{-p t}(-p) d t$

= $-F(0)+p \int_0^{\infty} e^{-p t} F(t) d t=p f(p)-F(0)$

Suppose $L\left\{F^{(m)}(t)\right\}=p^m f(p)-p^{m-1} F(0)-p^{m-2} F^{\prime}(0) \cdots F^{(m-1)}(0)$

⇒ $\left.L\left\{F^{(m+1)}(t)\right\}=\int_0^{\infty} e^{-p t} F^{(m+1)}(t) d t=e^{-p t} F^{(m)}(t)\right]_0^{\infty}-\int_0^{\infty} e^{-p t}(-p) F^{(m)}(t) d t$

= $-F^{(m)}(0)+p \int_0^{\infty} e^{-p t} F^{(m)}(t) d t=-F^{(m)}(0)+p L\left\{F^{(m)}(t)\right\}$

= $-F^{(m)}(0)+p^{m+1} f(p)-p^m F(0)-p^{m-1} F^{\prime}(0)-\cdots-p F^{(m-1)}(0)$

= $p^{m+1} f(p)-p^m F(0)-p^{m-1} F^{\prime}(0) \cdots-p F^{(m-1)}(0)-F^{(m)}(0)$

∴ If the statement is true for n=m then it is true for n=m+1

Hence by induction, $L\left\{F^{(n)}(t)\right\}=p^n f(p)-p^{n-1} F(0)-p^{n-2} F^{\prime}(0)-\cdots-F^{(n-1)}(0)$.

Advanced Laplace Transform Solved Examples Step-By-Step

4. State and prove the initial value theorem.

Solution:

Initial value theorem

Statement: If F(t) is peicewise continuous for all $t \geq 0$ and of exponential order as $t \rightarrow \infty$ or $F^{\prime}(t)$ is of class A, then Lt F(t) = $\underset{p \rightarrow 0}{\text{Lt}} p t\{f(t)\}=\underset{p \rightarrow \infty}{\text{Lt}}[p f(p)]$.

Proof: By the theorem on the Laplace transform of derivatives we have

L $\left\{F^{\prime}(t)\right\}=\int_0^{\infty} e^{-p t} F^{\prime}(t) d t=p L(F(t)]-F(0) \rightarrow(1)$

Since $F^{\prime}(t)$ is piecewise continuous and of exponential order, we have $\underset{p \rightarrow \infty}{L t} \int_0^{\infty} e^{-p t} F^{\prime}(t) d t=0$

(because $\underset{p \rightarrow \infty}{L t} \int_0^{\infty} e_0^{-p t} F^{\prime}(t) d t=\int_0^{\infty}\left(\underset{p \rightarrow \infty}{L t} e^{-p t}\right) F^{\prime}(t) d t=\int_0^{\infty}(0) F^{\prime}(t) d t=0$

Taking limit as $p \rightarrow \infty$ in (1), we have $0=\underset{p \rightarrow \infty}{\text{Lt}} p L\{F(t)\}-F(0)$

⇒ $F(0)=\underset{p \rightarrow \infty}{\text{Lt}} p \dot{L}\{F(t)\} \Rightarrow \underset{t \rightarrow 0}{\text{Lt}} F(t)=\underset{p \rightarrow \infty}{\text{Lt}} p L\{F(t)\}=\underset{p \rightarrow \infty}{\text{Lt}} p F(p)$

5. State and prove the final value theorem.

Solution:

Final value theorem

Statement: If F(t) is continuous for all $t \geq 0$ and of exponential order as $t \rightarrow \infty$ or $F^{\prime}(t)$ is of class A, then

Lt F(t) = $\underset{p \rightarrow 0}{\text{Lt}} p L\{F(t)\}=\underset{p \rightarrow 0}{\text{Lt}}[p f(p)]$.

Proof: By a theorem of Laplace transform on derivatives, we have

L$\left\{F^{\prime}(t)\right\}=\int_0^{\infty} e^{-p t} F^{\prime}(t) d t=L_{p \rightarrow 0} p L\{F(t)\}-F(0) \rightarrow(1)$

Taking limit as $p \rightarrow 0$ in (1), we have $\underset{p \rightarrow 0}{L} \int_0^{\infty} e^{-p t} F^{\prime}(t) d t=\underset{p \rightarrow 0}{L t} p L\{F(t)\}-F(0)$

⇒ $\int_0^{\infty}\left(L_{p \rightarrow 0}^{L t} e^{-p t}\right) F^{\prime}(t) d t=L_{p \rightarrow 0} p L\{F(t)\}-F(0)$

⇒ $\int_0^{\infty} F^{\prime}(t) d t=\mathrm{Lt}_{p \rightarrow 0} p L\{F(t)\}-F(0)$

⇒ $F[(t)]_0^{\infty}=\text{Lt}_{p \rightarrow 0} p L\{F(t)\}-F(0) \Rightarrow \underset{t \rightarrow \infty}{\\text{Lt}} F(t)-F(0)$

= $\underset{p \rightarrow 0}{\text{Lt}} p L\{F(t)\}-F(0)$

∴ $\mathrm{Lt}_{t \rightarrow \infty} F(t)=\underset{p \rightarrow 0}{L t} p L\{F(t)\}=\mathrm{Lt}_{p \rightarrow 0} p f(p)$

6. Find the Laplace transform of $e^{a t}$ using the theorem on transforms of derivatives.

Solution:

Let $F(t)=e^{a t} \text {. Then } F^{\prime}(t)=a e^{a t} \text { and } F(0)=1$

Now, $L\left\{F^{\prime}(t)\right\}=p L\{F(t)\}-F(0) \Rightarrow L\left\{a e^{a t}\right\}=p L\left\{e^{a t}\right\}-1$

⇒ $a L\left\{e^{a t}\right\}-p L\left\{e^{a t}\right\}=-1 \Rightarrow(a-p) L\left\{e^{a t}\right\}=-1 \Rightarrow L\left\{e^{a t}\right\}=\frac{-1}{a-p}=\frac{1}{p-a}$

Laplace Transform II Exercise Problems With Solutions

7. Find the Laplace transform of cos⁡ at using the theorem on transforms of derivatives.

Solution:

Let F(t)=cos a t.

Then $F^{\prime}(t)=-a \sin a t, F^{\prime \prime}(t)=-a^2 \cos a t$

∴ $L\left\{F^{\prime \prime}(t)\right\}=p^2 L\{F(t)\}-p F(0)-F^{\prime}(0) \rightarrow(1)$

Now $F(0)=\cos 0=1$ and $F^{\prime}(0)=-a \sin 0=0$.

Substituting in (1), we get $L\left\{-a^2 \cos a t\right\}=p^2 L\{\cos a t\}-p(1)-0$

⇒ $\left(p^2+a^2\right) L\{\cos a t\}=p \Rightarrow L\{\cos a t\}=\frac{p}{p^2+a^2}$

8. Find the Laplace transform of t sin⁡ at using the theorem on transforms of derivatives.

Solution:

Let F(t) = t sin at.

Then $F^{\prime}(t)=\sin a t+a t c o s$ at and $F^{\prime \prime}(t)=a \cos a t+a[\cos a t-a t \sin a t]=2 a \cos a t-a^2 t \sin a t$

Also $F(0)=0$ and $F^{\prime}(0)=0$

Now$L\left\{F^{\prime \prime}(t)\right\}=p^2-L\{F(t)\}-p F(0)-F^{\prime}(0)$

⇒ $L\left\{2 a \cos a t-a^2 t \sin a t\right\}=p^2 L\{t \sin a t\}-0-0$

⇒ $2 a L\{\cos a t\}-a^2 L\{t \sin a t\}-p^2 L\{t \sin a t\}=0$

⇒ $2 a L\{\cos a t\}-\left(a^2+p^2\right) L\{t \sin a t\}=0$

⇒ $\frac{2 a p}{p^2+a^2}=\left(a^2+p^2\right) L\{t \sin a t\} \Rightarrow L\{t \sin a t\}=\frac{2 a p}{\left(p^2+a^2\right)^2}$

9. If $L\left[2 \sqrt{\frac{t}{\pi}}\right]=\frac{1}{p^{3 / 2}}$, then show that $L\left\{\frac{1}{\sqrt{t \pi}}\right\}=\frac{1}{\sqrt{p}}$.

Solution:

Given

$L\left[2 \sqrt{\frac{t}{\pi}}\right]=\frac{1}{p^{3 / 2}}$

Let $F(t)=2 \sqrt{\frac{t}{\pi}} \text {. Then } F^{\prime}(t)=\frac{2}{\sqrt{\pi}} \cdot \frac{1}{2 \sqrt{t}}=\frac{1}{\sqrt{\pi t}} \text { and } F(0)=0$

Now, $L\left\{F^{\prime}(t)\right\}=p L\{F(t)\}-F(0) \Rightarrow L\left\{\frac{1}{\sqrt{\pi} t}\right\}=p L\left\{2 \sqrt{\frac{t}{\pi}}\right\}-0=p\left(\frac{1}{p^{3 / 2}}\right)=\frac{1}{\sqrt{p}}$

10. If $L[\sin \sqrt{t}]=\frac{\sqrt{\pi} e^{1 / 4 p}}{2 p^{3 / 2}}$, then find $L\left\{\frac{\cos \sqrt{t}}{\sqrt{t}}\right\}$.

Solution:

Given

$L[\sin \sqrt{t}]=\frac{\sqrt{\pi} e^{1 / 4 p}}{2 p^{3 / 2}}$

Let $F(t)=\sin \sqrt{t} \text {. Then } F^{\prime}(t)=\frac{1}{2 \sqrt{t}} \cos \sqrt{t} \text {. Also } F(0)=0$

Now, $L\left\{F^{\prime}(t)\right\}=p L\{F(t)\}-F(0) \Rightarrow L\left\{\frac{\cos \sqrt{t}}{2 \sqrt{t}}\right\}=p L\{\sin \sqrt{t}\}-0$

⇒ $\frac{1}{2} L\left\{\frac{\cos \sqrt{t}}{\sqrt{t}}\right\}=p \frac{\sqrt{\pi} e^{-1 /(4 p)}}{2 p^{3 / 2}}=\frac{\sqrt{\pi} e^{-1 /(4 p)}}{2 p^{1 / 2}} \Rightarrow L\left\{\frac{\cos \sqrt{t}}{\sqrt{t}}\right\}=\sqrt{\frac{\pi}{p}} e^{-1 /(4 p)}$

11. Find $L\left\{\frac{\cos \sqrt{t}}{\sqrt{t}}\right\}$.

Solution:

⇒ $\sin \sqrt{t}=\sqrt{t}-\frac{(\sqrt{t})^3}{3!}+\frac{(\sqrt{t})^5}{5!}-\frac{(\sqrt{t})^7}{7!}+\cdots=t^{1 / 2}-\frac{t^{3 / 2}}{3!}+\frac{t^{5 / 2}}{5!}-\frac{t^{7 / 2}}{7!}+\cdots$

∴ $L\{\sin \sqrt{t}\}=L\left\{t^{1 / 2}\right\}-\frac{1}{3!} L\left\{t^{3 / 2}\right\}+\frac{1}{5!} L\left\{t^{5 / 2}\right\}-\frac{1}{7!} L\left\{t^{7 / 2}\right\}+\cdots$

= $\frac{\Gamma\left(\frac{3}{2}\right)}{p^{3 / 2}}-\frac{1}{3!} \frac{\Gamma\left(\frac{5}{2}\right)}{p^{5 / 2}}$

+ $\frac{1}{5!} \frac{\Gamma\left(\frac{7}{2}\right)}{p^{7 / 2}}-\frac{1}{7!} \frac{\Gamma\left(\frac{9}{2}\right)}{p^{9 / 2}}+\cdots$

∴ $L\left\{t^n\right\}=\frac{\Gamma(n+1)}{p^{n+1}}$

= $\frac{\frac{1}{2} \Gamma\left(\frac{1}{2}\right)}{p^{3 / 2}}-\frac{1}{3!} \frac{\frac{3}{2} \cdot \frac{1}{2} \Gamma \cdot\left(\frac{1}{2}\right)}{p^{5 / 2}}+\frac{1}{5!} \frac{\frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \Gamma\left(\frac{1}{2}\right)}{p^{7 / 2}}$

–$\frac{1}{7!} \frac{\frac{7}{2} \cdot \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \Gamma\left(\frac{1}{2}\right)}{p^{9 / 2}}+\cdots$

= $\frac{\sqrt{\pi}}{2 p^{3 / 2}}\left[1-\frac{3}{2 \cdot 3!p}+\frac{5 \cdot 3}{2 \cdot 2 \cdot 5!p^2}-\frac{7 \cdot 5 \cdot 3}{2 \cdot 2 \cdot 2 \cdot 7!p^3}+\cdots\right]$

because $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$

= $\frac{\sqrt{\pi}}{2 p^{3 / 2}}\left[1-\frac{1}{2^2} p+\frac{1}{\left(2^2 p\right)^2 2!}-\frac{1}{\left(2^2 p\right)^3 3!}+\cdots\right]=\frac{\sqrt{\pi}}{2 p^{3 / 2}}\left(e^{\left.-1 / 2^2 p\right)}\right)=\frac{\sqrt{\pi}}{2 p^{3 / 2}} e^{-1 / 4 p} \text {. }$

Let $F(t)=\sin \sqrt{t}$. Then $F^{\prime}(t)=\frac{1}{2 \sqrt{t}} \cos \sqrt{t}$. Also F(0)=0

Now $L\left\{F^{\prime}(t)\right\}=p L\{F(t)\}-F(0) \Rightarrow L\left\{\frac{\cos \sqrt{t}}{2 \sqrt{t}}\right\}=p L\{\sin \sqrt{t}\}-0$

⇒ $\frac{1}{2} L\left\{\frac{\cos \sqrt{t}}{\sqrt{t}}\right\}=p \frac{\sqrt{\pi} e^{-1 /(4 p)}}{2 p^{3 / 2}}=\frac{\sqrt{\pi} e^{-1 /(4 p)}}{2 p^{1 / 2}}$

⇒ $L\left\{\frac{\cos \sqrt{t}}{\sqrt{t}}\right\}=\sqrt{\frac{\pi}{p}} e^{-1 /(4 p)}$

Solved Problems On Laplace Transform For Advanced Exercises

12. If F(t) is piecewise continuous and of exponential order, then show that the Laplace transform of $\int_0^t F(t) d t \text { is } \frac{1}{p} L[F(t)] \text { i.e., } L\left[\int_0^t F(t) d t\right]=\frac{1}{p} L[F(t)]$

Solution:

Given

F(t) is piecewise continuous and of exponential order

L$\left[\int_0^t F(t) d t\right]=\int_0^{\infty}\left[\int_0^t F(t) d t\right] e^{-p t} d t=\int_0^{\infty}\left[\int_0^t F(t) d t\right] d\left(\frac{e^{-p t}}{-p}\right)$

= $\left[\frac{e^{-p p t}}{-p} \int_0^t F(t) d t\right]+\frac{1}{p} \int_0^{\infty} e^{-p t} F(t) d t=\frac{1}{p} \int_0^{\infty} e^{-p t} F(t) d t$

∴ $L\left[\int_0^t F(t) d t\right]=\frac{1}{p} L[F(t)]$

13. Find the Laplace transform of $\int_0^t e^{-t} \cos t d t$.

Solution:

We know that $L\{\cos t\}=\frac{p}{p^2+1}$

L$\left\{e^{-t} \cos t\right\}=\frac{p+1}{(p+1)^2+1}=\frac{p+1}{p^2+2 p+2}$

∴ $L\left[\int_0^t e^{-t} \cos t d t\right]=\frac{1}{p} L\left(e^{-t} \cos t\right)=\frac{1}{p} \frac{p+1}{p^2+2 p+2}$

14. Find the Laplace transform of $\int_0^t e^{-t} \sinh t d t$

Solution:

We know that $L\{\sinh t\}=\frac{1}{p^2-1} \text {. }$

L$\left\{e^{-t} \sinh t\right\}=\frac{1}{(p+1)^2-1}=\frac{1}{p^2+2 p}$

∴ $L\left[\int_0^t e^{-t} \sinh t d t\right]=\frac{1}{p} L\left(e^{-t} \sinh t\right)=\frac{1}{p} \frac{1}{p^2+2 p}=\frac{1}{p^2(p+2)}$

15. Find the Laplace transform of $\int_0^t \int_0^t \cosh a t d t$.

Solution:

We know that $L\{\cosh a t\}=\frac{p}{p^2-a^2}$

L$\left[\int_0^t \cosh a t\right]=\frac{1}{p} f(p)=\frac{1}{p}\left(\frac{p}{p^2-a^2}\right)=\frac{1}{p^2-a^2} .$

∴ $\int_0^t \int_0^t \cosh a t d t=\frac{1}{p}\left(\frac{1}{p^2-a^2}\right)=\frac{1}{p\left(p^2-a^2\right)}$

Inverse Laplace Transform II Examples With Solutions

16. If f(t) is a function of class A and if L{f(t)}=f(p) prove that $L\{t f(t)\}=-f^{\prime}(p)$.

Solution:

Given

f(t) is a function of class A and if L{f(t)}=f(p)

f(p) = $L\{F(t)\}=\int_0^{\infty} e^{-p t} F(t) d t$

⇒ $\frac{d}{d p}[f(p)]=\frac{d}{d p} \int_0^{\infty} e^{-p t} F(t) d t=\int_0^{\infty} \frac{\partial}{\partial p}\left[e^{-p t} F(t)\right] d t=\int_0^{\infty} e^{-p t}(-t) F(t) d t$

= $-\int_0^{\infty} e^{-p t} t F(t) d t=-L\{t F(t)\}$

∴ $L\{t F(t)\}=-\frac{d}{d p}[f(p)]$

17. If L{F(t)}=f(p) then show that $L\left\{t^n F(t)\right\}=(-1)^n \frac{d^n}{d p^n}[f(p)], n=1,2,3 \ldots$.

Solution:

We prove $L\left\{t^n F(t)\right\}=(-1)^n \frac{d^n}{d p^n} f(p) \rightarrow(1)$ by Inductic on n.

f(p) = $\int_0^{\infty} e^{-p t} F(t) d t \Rightarrow \frac{d}{d p}[f(p)]=\frac{d}{d p} \int_0^{\infty} e^{-p t} F(t) d t=\int_0^{\infty} \frac{\partial}{\partial p}\left[e^{-p t} F(t)\right] d t$

= $\int_0^{\infty} e^{-p t}(-t) F(t) d t \stackrel{=}{=}-\int_0^{\infty} e^{-p t} t F(t) d t=-L\{t F(t)\}$

∴ $L\{t F(t)\}=-\frac{d}{d p}[f(p)] \Rightarrow(1)$ is true for n=1.

Assume that (1) is true for n=m.

Now $\frac{d}{d p}\left[L\left\{t^m F(t)\right\}\right]=(-1)^m \frac{d^{m+1}}{d p^{m+1}}[f(p)]$

⇒ $\frac{d}{d p}\left(\int_0^{\infty} e^{-p t} t^m F(t) d t\right)=(-1)^m \cdot \frac{d^{m+1}}{d p^{m+1}}[f(p)]$

⇒ $\int_0^{\infty} \frac{d}{d t}\left(-e^{-p t} t t^m F(t)\right) d t=(-1)^m \frac{d^{m+1}}{d p^{m+1}}[f(p)]$

⇒ $\int_0^{\infty} e^{-p t}\left(-t^{m+1}\right) F(t) d t=(-1)^m \frac{d^{m+1}}{d p^{m+1}}[f(p)]$

⇒ $-L\left\{t^{m+1} F(t)\right\}=(-1)^m \frac{d^{m+1}}{d p^{m+1}}[f(p)] \Rightarrow L\left\{t^{m+1} f(p)\right\}=(-1)^{m+1} \frac{d^{m+1}}{d p^{m+1}}[f(p)]$

Thus (1) is ture for n=m+1.

By mathematical induction $L\left\{t^n F(t)\right\}=(-1)^n \frac{d^n}{d t^n}[f(p)] \forall n \text {. }$

18. Find $L[t \sin a t]$.

Solution:

We have $L[\sin a t]=\frac{a}{p^2+a^2}=f(p) \text { (say) }$

∴ $L[t \sin a t]=–\frac{d}{d p}\left[f^{\prime}(p)\right]=-\frac{d}{d p}\left[\frac{a}{p^2+a^2}\right]=\frac{2 a p}{\left(p^2+a^2\right)^2}$

19. Evaluate $L\{t \cos a t\}$.

Solution:

Given

$L\{t \cos a t\}$

Since $L\{\cos a t\}=\frac{p}{p^2+a^2}=f(p)$

∴ $L\{t \cos a t\}=(-1) \frac{d}{d p}[f(p)]=(-1) \frac{d}{d p}\left\{\frac{p}{p^2+a^2}\right\}=(-1)\left[\frac{\left(p^2+a^2\right) 1-(2 p)}{\left(p^2+a^2\right)^2}\right]$

= $\frac{p^2-a^2}{\left(p^2+a^2\right)^2}$

Applications Of Laplace Transform Ii In Engineering And PhysicsApplications Of Laplace Transform Ii In Engineering And Physics

20. Find $L\{t \cos t\}$.

Solution:

L$\{\cos t\}=\frac{p}{p^2+1}=f(p)$

∴ $L\{t \cos t\}=(-1) \frac{d}{d p}[f(p)]=(-1) \frac{d}{d p}\left\{\frac{p}{p^2+1}\right\}=(-1)\left[\frac{\left(p^2+1\right) 1-p(2 p)}{\left(p^2+1\right)^2}\right]$

= $\frac{p^2-1}{\left(p^2+1\right)^2}$

21. Evaluate $L\{\sin a t-a t \cos a t\}$.

Solution:

Given

$L\{\sin a t-a t \cos a t\}$

L$\{\sin a t-a t \cos a t\}=L\{\sin a t\}-a L\{t \cos a t\}=\frac{a}{p^2+a^2}-\frac{a\left(p^2-a^2\right)}{\left(p^2+a^2\right)^2}$

= $\frac{a\left(p^2+a^2\right)+a^3-a p^2}{\left(p^2+a^2\right)^2}=\frac{2 a^3}{\left(p^2+a^2\right)^2} .$

22. Find the Laplace transform of $L\{t(3 \sin 2 t-2 \cos 2 t)\}$.

Solution:

L$\{3 \sin 2 t-2 \cos 2 t\}=3 L\{\sin 2 t\}-2 L\{\cos 2 t\}$

= $3 \cdot \frac{2}{p^2+4}-2 \cdot \frac{p}{p^2+4}=\frac{6-2 p}{p^2+4}$

L$\{t(3 \sin 2 t-2 \cos 2 t)\}=\frac{-d}{d p}\left\{\frac{6-2 p}{p^2+4}\right\}$

= $-\frac{\left[\left(p^2+4\right)(-2)-2 p(6-2 p)\right]}{\left(p^2+4\right)^2}=\frac{12 p+8-2 p^2}{\left(p^2+4\right)^2}$

Laplace Transform II Detailed Examples And Solutions

23. Find $L\left(t \sin ^2 t\right)$.

Solution:

L$\left(\sin ^2 t\right)=L\left[\frac{1-\cos 2 t}{2}\right]=\frac{1}{2}[L(1)-L(\cos 2 t)]=\frac{1}{2}\left[\frac{1}{p}-\frac{p}{p^2+4}\right]$

= $\frac{4}{2 p\left(p^2+4\right)}=\frac{2}{p\left(p^2+4\right)}$

∴ $L\left[t \sin ^2 t\right]=(-1) \frac{d}{d p}\left[\frac{2}{p\left(p^2+4\right)}\right]=-\left[\frac{p\left(p^2+4\right) \cdot 0-2 \cdot\left(3 p^2+4\right)}{\left[p\left(p^2+4\right)\right]^2}\right]=\frac{6 p^2+8}{p^2\left(p^2+4\right)^2}$

24. Evaluate $L[t \sin 3 t \cos 2 t]$.

Solution:

Given

$L[t \sin 3 t \cos 2 t]$

We have $L\{\sin 3 t \cos 2 t\}=\frac{1}{2} L\{2 \sin 3 t \cos 2 t\}=\frac{1}{2} L\{\sin 5 t+\sin t\}$

= $\frac{1}{2}\left[\frac{5}{p^2+25}+\frac{1}{p^2+1}\right]=f(p)$

∴ $L[t \sin 3 t \cos 2 t]=(-1) \frac{d}{d p}[f(p)]=-\frac{1}{2} \frac{d}{d p}\left[\frac{5}{p^2+25}+\frac{1}{p^2+1}\right]$

= $-\frac{1}{2}\left[\frac{-10 p}{\left(p^2+25\right)^2}-\frac{2 p}{\left(p^2+1\right)^2}\right]=\frac{5 p}{\left(p^2+25\right)^2}+\frac{p}{\left(p^2+1\right)^2}$

Worked Examples On Laplace Transform II For Differential Equations

25. Find $L\left\{t^2 \sin t\right\}$.

Solution:

L$\{\sin t\}=\frac{1}{p^2+1}$

L$L\left\{t^2 \sin t\right\}=(-1)^2 \frac{d^2}{d p^2}[L\{\sin t\}]=\frac{d^2}{d p^2}\left\{\frac{1}{p^2+1}\right\}=\frac{d}{d p}\left\{\frac{-1}{\left(p^2+1\right)^2}(2 p)\right\}$

= $-\frac{d}{d p}\left\{\frac{2 p}{\left(p^2+1\right)^2}\right\}=-\left[\frac{\left(p^2+1\right)^2 2-2 p 2\left(p^2+1\right) 2 p}{\left(p^2+1\right)^4}\right]$

= $-\left[\frac{2\left(p^2+1\right)-8 p^2}{\left(p^2+1\right)^3}\right]=\frac{6 p^2-2}{\left(p^2+1\right)^3}$

26. Find $L\left\{t^2 \cos t\right\}$.

Solution:

L$\{\cos t\}=\frac{p}{p^2+1}$

L$\left\{t^2 \cos t\right\}=(-1)^2 \frac{d^2}{d p^2} L\{\cos t\}=\frac{d^2}{d p^2}\left\{\frac{p}{p^2+1}\right\}=\frac{d}{d p}\left\{\frac{\left(p^2+1\right) 1-p(2 p)}{\left(p^2+1\right)^2}\right\}$

= $\frac{d}{d p}\left\{\frac{1-p^2}{\left(p^2+1\right)^2}\right\}=\frac{\left(p^2+1\right)^2(-2 p)-\left(1-p^2\right) 2\left(p^2+1\right) 2 p}{\left(p^2+1\right)^4}$

= $\frac{\left(p^2+1\right)(-2 p)-4 p\left(1-p^2\right)}{\left(p^2+1\right)^3}=\frac{2 p^3-6 p}{\left(p^2+1\right)^3} .$

27. Find $L\left\{t^2 \cos a t\right\}$.

Solution:

L$\left\{t^2 \cos a t\right\}=(-1)^2 \frac{d^2}{d p^2}[L\{\cos a t\}]=\frac{d^2}{d p^2}\left[\frac{p}{p^2+a^2}\right]=\frac{d}{d p}\left[\frac{\left(p^2+a^2\right) 1-p(2 p)}{\left(p^2+a^2\right)^2}\right]$

= $\frac{d}{d p}\left[\frac{a^2-p^2}{\left(p^2+a^2\right)^2}\right]=\frac{\left(p^2+a^2\right)^2(-2 p)-\left(a^2-p^2\right) 2\left(p^2+a^2\right) 2 p}{\left(p^2+a^2\right)^4}$

= $\frac{-2 p^3-2 p a^2-4 p \dot{a}^2+4 p^3}{\left(p^2+a^2\right)^3}=\frac{2 \dot{p}^3-6 p a^2}{\left(p^2+a^2\right)^3}$

28. Find $L\left[t^2 \cos 2 t\right]$.

Solution:

We know that $L[\cos 2 t]=\frac{p}{p^2+4}$

Hence, $L\left[t^2 \cos 2 t\right]=(-1)^2 \frac{d^2}{d p^2}\left[\frac{p}{p^2+4}\right]=\frac{d}{d p}\left[\frac{p^2+4-2 p^2}{\left(p^2+4\right)^2}\right]=\frac{d}{d p}\left[\frac{4-p^2}{\left(p^2+4\right)^2}\right]$

= $\frac{\left(p^2+4\right)^2(-2 p)-\left(4-p^2\right) 4 p\left(p^2+4\right)}{\left(p^2+4\right)^4}=\frac{-2 p\left(p^2+4\right)-4 p\left(4-p^2\right)}{\left(p^2+4\right)^3}$

= $\frac{-2 p\left(p^2+4+8-2 p^2\right)}{\left(p^2+4\right)^3}=\frac{2 p\left(p^2-12\right)}{\left(p^2+4\right)^3}$

Laplace Transform II Solved Problems In Control Systems

29. Find $L\left\{t^2 \cos 3 t\right\}$.

Solution:

Since, $ L\{\cos 3 t\}=\frac{p}{p^2+9}=f(p)$

∴ $L\left\{t^2 \cos 3 t\right\}=(-1)^2 \frac{d^2}{d p^2}[f(p)]=\frac{d^2}{d p^2}\left[\frac{p}{p^2+9}\right]=\frac{d}{d p}\left[\frac{9-p^2}{\left(p^2+9\right)^2}\right]$

= $\frac{\left(p^2+9\right)^2(-2 p)-\left(9-p^2\right) 2\left(p^2+9\right) 2 p}{\left(p^2+9\right)^4} \text {, using Quotient Rule }$

= $\frac{\left(p^2+9\right)\left[-2 p\left(p^2+9\right)-4 p\left(9-p^2\right)\right]}{\left(p^2+9\right)^4}=\frac{-2 p^3-18 p-36 p+4 p^3}{\left(p^2+9\right)^3}=\frac{2 p^3-54 p}{\left(p^2+9\right)^3} .$

30. Find $L\left\{t^2 \sin 2 t\right\}$.

Solution:

L$\left\{t^2 \sin 2 t\right\}=(-1)^2 \frac{d^2}{d p^2}[L(\sin 2 t)]=\frac{d}{d p}\left[\frac{d}{d p}\left\{\frac{2}{p^2+4}\right\}\right]$

= $2 \frac{d}{d p}\left[\frac{-1}{\left(p^2+4\right)^2} 2 p\right]=(-4) \frac{d}{d p}\left[\frac{p}{\left(p^2+4\right)^2}\right]=(-4) \frac{\left(p^2+4\right)^2-2 p\left(p^2+4\right) 2 p}{\left(p^2+4\right)^4}$

= $(-4) \frac{\left(p^2+4\right)\left[\left(p^2+4\right)-4 p^2\right]}{\left(p^2+4\right)^4}=(-4) \frac{4-3 p^2}{\left(p^2+4\right)^3}=\frac{4\left(3 p^2-4\right)}{\left(p^2+4\right)^3} \text {. }$

31. Find $L\left\{\left(t^2-3 t+2\right) \sin 3 t\right\}$.

Solution:

We have $L\{\sin 3 t\}=\frac{3}{p^2+9}=f(p)$.

Now $L\{t \sin 3 t\}=-\frac{d}{d p}\left(\frac{3}{p^2+9}\right)$

L$\left\{t^2 \sin 3 t\right\}=(-1)^2 \frac{d^2}{d p^2}\left(\frac{3}{p^2+9}\right)=\frac{d}{d p}\left[\frac{d}{d p}\left(\frac{3}{p^2+9}\right)\right]$

= $\frac{d}{d p}\left[3 \frac{-2 p}{\left(p^2+9\right)^2}\right]=\frac{d}{d p}\left[\frac{-6 p}{\left(p^2+9\right)^2}\right]$

= $-6\left[\frac{\left(p^2+9\right)^2(1)-p(4 p)\left(p^2+9\right)}{\left(p^2+9^2\right)^4}\right]=-6\left[\frac{\left(p^2+9\right)-4 p^2}{\left(p^2+9\right)^3}\right]=\frac{6\left(3 p^2-9\right)}{\left(p^2+9\right)^3}$

∴ L $\left\{\left(t^2-3 t+2\right) \sin 3 t\right\}=\frac{6\left(3 p^2-9\right)}{\left(p^2+9\right)^3}-\frac{18 p}{\left(p^2+9\right)^2}+\frac{6}{p^2+9}=\frac{6 p^4-18 p^3+126 p^2-162 p+432}{\left(p^2+9\right)^3}$

Complex Laplace Transform II Problems With Solutions

32. Find $L\left(t^2 \sin a t\right)$.

Solution:

L$\left\{t^2 \sin a t\right\}=(-1)^2 \frac{d^2}{d p^2}[L\{\sin a t\}]=\frac{d}{d p} \frac{d}{d p}\left\{\frac{a}{p^2+a^2}\right\}=a \frac{d}{d p}\left\{\frac{-2 p}{\left(p^2+a^2\right)^2}\right\}$

= $-2 a\left[\frac{\left(p^2+a^2\right)^2-2 p\left(p^2+a^2\right) 2 p}{\left(p^2+a^2\right)^4}\right]=-2 a\left[\frac{\left(p^2+a^2\right)\left(p^2+a^2-4 p^2\right)}{\left(p^2+a^2\right)^4}\right]$

= $-2 a\left[\frac{a^2-3 p^2}{\left(p^2+a^2\right)^3}\right]=\frac{2 a\left(3 p^2-a^2\right)}{\left(p^2+a^2\right)^3}$

33. Find the Laplace transform of $t^3 \cos a t$.

Solution:

L$\{\cos a t\}=\frac{p}{p^2+a^2}$

L$\left\{t^3 \cos a t\right\}=(-1)^3 \frac{d^3}{d p^3}\left(\frac{p}{p^2+a^2}\right)=-\frac{d^2}{d p^2}\left(\frac{\left(p^2+a^2\right) 1-p(2 p)}{\left(p^2+a^2\right)^2}\right)=\frac{d^2}{d p^2}\left(\frac{p^2-a^2}{\left(p^2+a^2\right)^2}\right)$

= $\frac{d}{d p}\left(\frac{\left(p^2+a^2\right)^2(2 p)-\left(p^2-a^2\right) 2\left(p^2+a^2\right) 2 p}{\left(p^2+a^2\right)^4}\right)=\frac{d}{d p}\left(\frac{2 p^3+2 p a^2-4 p^3+4 p a^2}{\left(p^2+a^2\right)^3}\right)$

= $\frac{d}{d p}\left(\frac{6 p a^2-2 p^3}{\left(p^2+a^2\right)^3}\right)=\frac{\left(p^2+a^2\right)^3\left(6 a^2-6 p^2\right)-\left(6 p a^2-2 p^3\right) 3\left(p^2+a^2\right)^2 2 p}{\left(p^2+a^2\right)^6}$

= $\frac{\left(p^2+a^2\right)\left(6 a^2-6 p^2\right)-6 p\left(6 p a^2-2 p^3\right)}{\left(p^2+a^2\right)^4}=\frac{6 p^2 a^2-6 p^4+6 a^4-6 p^2 a^2-36 p^2 a^2+12 p^4}{\left(p^2+a^2\right)^4}$

= $\frac{6 p^4-36 p^2 a^2+6 a^4}{\left(p^2+a^2\right)^4}=\frac{6\left(p^4-6 p^2 a^2+a^4\right)}{\left(p^2+a^2\right)^4}$

Second method: We know that $L\left(t^3\right)=\frac{3!}{p^4}=\frac{6}{p^4}$.

L$\left(t^3 \cos a t\right)=\text { R. P. } L\left(t^3 e^{a i i}\right)=\text { R. P. } \frac{6}{(p+a i)^4}=\text { R.P. } \frac{6(p-a i)^4}{\left(p^2+a^2\right)^4}=\frac{6\left(p^4-6 p^2 a^2+a^4\right)}{\left(p^2+a^2\right)^4}$

34. Find $L\left(t^3 \cos t\right)$.

Solution:

L$\{t \cos t\}=\frac{p}{p^2+1}$

L$\left\{t^3 \cos t\right\}=(-1)^2 \frac{d^3}{d p^2}\left\{\frac{p}{p^2+1}\right\}=-\frac{d^2}{d p^2}\left\{\frac{\left(p^2+1\right) 1-p(2 p)}{\left(p^2+1\right)^2}\right\}$

= $-\frac{d^2}{d p^2}\left\{\frac{p^2-1}{\left(p^2+1\right)^2}\right\}=\frac{d}{d p}\left\{\frac{\left(p^2+1\right)^2 \cdot 2 p-\left(p^2-1\right) 2\left(p^2+1\right) 2 p}{\left(p^2+1\right)^4}\right\}$

= $\frac{d}{d p}\left\{\frac{2 p^3+2 p-4 p^3+4 p}{(p+1)^3}\right\}=\frac{d}{d p}\left\{\frac{6 p-2 p^3}{\left(p^2+1\right)^3}\right\}=\frac{\left(p^2+1\right)^3\left(6-6 p^2\right)-\left(6 p-2 p^3\right) 3\left(p^2+1\right)^2(2 p)}{\left(p^2+1\right)^6}$

= $\frac{\left(p^2+1\right)\left(6-6 p^2\right)-6 p\left(6 p-2 p^3\right)}{\left(p^2+1\right)^4}=\frac{6 p^2+6-6 p^4-6 p^2-36 p^2+12 p^4}{\left(p^2+1\right)^4}$

= $\frac{6 p^4-30 p^2+6}{\left(p^2+1\right)^4}=\frac{6\left(p^4-5 p^2+1\right)}{\left(p^2+1\right)^4}$

35. Find the Laplace transform of $\left(1+t e^{-t}\right)^2$.

Solution:We have to $\left(1+t e^{-t}\right)^2=1+2 t e^{-t}+t^2 e^{-2 t}$

∴ $L\left\{\left(1+t e^{-t}\right)^2\right\}=L(1)+2 L\left\{t e^{-t}\right\}+L\left\{t^2 e^{-2 t}\right\}$

= $\frac{1}{p}+2(-1)+(-1) \frac{d}{d p}\left(\frac{1}{p+1}\right)+(-1)^2 \frac{d^2}{d p^2}\left(\frac{1}{p+2}\right)=\frac{1}{p}-\frac{2}{(p+1)^2}+\frac{2}{(p+2)^3}$

36. Find the Laplace transform of $t^2 e^{-2 t}$.

Solution:

L$\left\{e^{-2 t}\right\}=\frac{1}{p+2} \Rightarrow L\left\{t^2 e^{-2 t}\right\}=(-1)^2 \frac{d^2}{d p^2}\left(\frac{1}{p+2}\right)=\frac{d}{d p}\left[\frac{-1}{(p+2)^2}\right]=\frac{2}{(p+2)^3}$

37. If $L\left\{t^{1 / 2}\right\}=\frac{\sqrt{\pi}}{2 p^{3 / 2}}$ then find $L\left\{t^{-1 / 2}\right\}$.

Solution:

Given

$L\left\{t^{1 / 2}\right\}=\frac{\sqrt{\pi}}{2 p^{3 / 2}}$

Let $F(t)=t^{-1 / 2} \text { and } L[F(t)]=f(p) \text {. }$

We have $\left.L\{t F(t)\}=(-1) \frac{d}{d p}[f(p)] \Rightarrow L \mid t \cdot t^{-1 / 2}\right\}=(-1) \frac{d}{d p}[f(p)]$

⇒ $L\left\{t^{1 / 2}\right\}=(-1) \frac{d}{d p}[f(p)] \Rightarrow \frac{d}{d p}[f(p)]=-L\left\{t^{1 / 2}\right\}=-\frac{\sqrt{\pi}}{2 t^{3 / 2}}$

⇒ $L\left\{t^{-1 / 2}\right\}=f(p)=-\frac{\sqrt{\pi}}{2}\left[\frac{t^{-1 / 2}}{-1 / 2}\right]=\frac{\sqrt{\pi}}{\sqrt{t}}$

38. Find $L\left(t e^{-t} \sin t\right)$.

Solution:

We know that $L(\sin t)=\frac{1}{p^2+1} \Rightarrow L\left(e^{-t} \sin t\right)=\frac{1}{(p+1)^2+1}=\frac{1}{p^2+2 p+2}$

L$\left(t e^{-t} \sin t\right)=-\frac{d}{d p}\left(\frac{1}{p^2+2 p+2}\right)=\frac{2(p+1)}{\left(p^2+2 p+2\right)^2}$

39. Find the Laplace transform of the function $t e^{-t} \sin 2 t$.

Solution:

We have $L\{\sin 2 t\}=\frac{2}{p^2+2^2}=\frac{2}{p^2+4}=f(p)$

∴ $L\{t \sin 2 t\}=(-1) \frac{d}{d p}[f(p)]=(-1) \frac{d}{d p}\left[\frac{2}{p^2+4}\right]=(-2)\left[\frac{-1}{\left(p^2+4\right)^2} 2 p\right]=\frac{4 p}{\left(p^2+4\right)^2}$

Hence by First Shifting theorem, $L\left\{t e^{-t} \sin 2 t\right\}=\left[\frac{4 p}{\left(p^2+4\right)^2}\right]_{p \rightarrow p+1}=\frac{4(p+1)}{\left[(p+1)^2+4\right]^2}=\frac{4(p+1)}{\left(p^2+2 p+5\right)^2}$

40. Find $L\left[e^{-t} t \sin 3 t\right]$.

Solution:

We have $L\{\sin 3 t\}=\frac{3}{p^2+9}$

L$L\{t \sin 3 t\}=-\frac{d}{d p}[L\{\sin 3 t\}]=-\frac{d}{d p}\left[\frac{3}{p^2+9}\right]=\frac{6 p}{\left(p^2+9\right)^2}$

By first shifting theorem $L\left\{e^{-t} t \sin 3 t\right\}=\frac{6(p+1)}{\left[(p+1)^2+9\right]^2}=\frac{6(p+1)}{\left(p^2+2 p+10\right)^2} \text {. }$

41. Find the Laplace transform of $t e^{2 t} \sin 3 t$.

Solution:

Since, $ L\{\sin 3 t\}=\frac{3}{p^2+9} \text {. }$

∴ $L\left\{e^{2 t} \sin 3 t\right\}=\frac{3}{(p-2)^2+9}=\frac{3}{p^2-4 p+13}$

Hence $L\left\{t e^{2 t} \sin 3 t\right\}=(-1) \frac{d}{d p}\left(\frac{3}{p^2-4 p+13}\right)=(-3) \frac{d}{d p}\left[\left(p^2-4 p+13\right)^{-1}\right]$

= $(-3)(-1)\left(p^2-4 p+13\right)^{-2}(2 p-4)=\frac{6(p-2)}{\left(p^2-4 p+13\right)^2}$

42. Find $L\left\{t e^{3 t} \sin 2 t\right\}$.

Solution:

L$\{\sin 2 t\}=\frac{2}{p^2+4} L\left\{e^{3 t} \sin 2 t\right\}=\left(\frac{2}{p^2+4}\right)_{p \rightarrow p-3}=\frac{2}{(p-3)^2+4}=\frac{2}{p^2-6 p+13}$

Hence, $L\left\{t e^{3 t} \sin 2 t\right\}=(-1) \frac{d}{d p}\left(\frac{2}{p^2-6 p+13}\right)=(-2) \frac{-1}{\left(p^2-6 p+13\right)^2}(2 p-6)$

= $\frac{4(p-3)}{\left(p^2-6 p+13\right)^2}$

43. Find $L\left\{t e^{-2 t} \sin 3 t\right\}$.

Solution:

We have $L\{\sin 3 t\}=\frac{3}{p^2+9}$

L$\{t \sin 3 t\}=-\frac{d}{d p}[L\{\sin 3 t\}]=-\frac{d}{d p}\left[\frac{3}{p^2+9}\right]=\frac{6 p}{\left(p^2+9\right)^2}$

By first shifting theorem $L\left\{e^{-2 t} t \sin 3 t\right\}=\frac{6(p+2)}{\left[(p+2)^2+9\right]^2}=\frac{6(p+2)}{\left(p^2+4 p+13\right)^2}$

44. Find $L\left[t e^{-2 t} \cdot \cos 3 t\right]$.

Solution:

We have $L[\cos 3 t]=\frac{p}{p^2+3^2}$

∴ $L\left[e^{-2 t} \cos 3 t\right]=\frac{p^2+3^2}{(p+2)^2+3^2}=\frac{p+2}{p^2+4 p+13}$

∴ $L\left[t e^{-2 t} \cos 3 t\right]=-\frac{d}{d p}\left[\frac{p+2}{p^2+4 p+13}\right]=-\frac{\left(p^2+4 p+13\right) \cdot 1-(p+2)(2 p+4)}{\left(p^2+4 p+13\right)^2}$

= $-\frac{p^2+4 p+13-2 p^2-8 p-8}{\left(p^2+4 p+13\right)^2}=-\frac{-p^2-4 p+5}{\left(p^2+4 p+13\right)^2}=\frac{p^2+4 p-5}{\left(p^2+4 p+13\right)^2}$

45. Find $L\left\{t e^{2 t} \sin 3 t\right\}$.

Solution:

We have $L(\sin 3 t)=\frac{3}{p^2+9}$

∴ $L\left(e^{2 t} \sin 3 t\right)=\frac{3}{(p-2)^2+9}=\frac{3}{p^2-4 p+13}$

∴ $L\left[t e^{2 t} \sin 3 t\right]=-\frac{d}{d p}\left[\frac{3}{p^2-4 p+13}\right]=-\frac{-3(2 p-4)}{\left(p^2-4 p+13\right)^2}=\frac{6(p-2)}{\left(p^2-4 p+13\right)^2}$

46. Find $L\left\{t e^{a t} \sin b t\right\}$.

Solution:

L$\{\sin b t\}=\frac{b}{p^2+b^2} \Rightarrow L\left\{e^{a t} \sin b t\right\}=\left(\frac{b}{p^2+b^2}\right)_{p \rightarrow p-a}=\frac{b}{(p-a)^2+b^2}$

Hence $L\left\{t e^{a t} \sin b t\right\}=(-1) \frac{d}{d p}\left[\frac{b}{(p-a)^2+b^2}\right]=(-b) \frac{-1}{\left[(p-a)^2+b^2\right]^2} 2(p-a)$

= $\frac{2 b(p-a)}{\left[(p-a)^2+b^2\right]^2}$

47. Find $L\left\{t e^{-t} \cosh t\right\}$.

Solution:

L$\{\cosh t\}=\frac{p}{p^2-1} \Rightarrow L\{t \cosh t\}=(-1) \frac{d}{d p}\left[\frac{p}{\dot{p}^2-1}\right]=(-1)\left[\frac{\left(p^2-1\right) 1-p \cdot 2 p}{\left(p^2-1\right)^2}\right]=\frac{1+p^2}{\left(p^2-1\right)^2}$

By First Shifting theorem, $L\left\{e^{-t} t \cosh t\right\}=\left[\frac{1+p^2}{\left(p^2-1\right)^2}\right]_{p \rightarrow p+1}=\frac{1+(p+1)^2}{\left.\left[(p+1)^2-1\right]^2\right]}=\frac{p^2+2 p+2}{\left(p^2+2 p\right)^2}$

48. Find $L\left[t e^{-2 t} \cosh t\right]$.

Solution:

We know that, $L[\cosh t]=\frac{p}{p^2-1}$

L$\left[e^{-2 t} \cosh t\right]=\frac{p+2}{(p+2)^2-1}=\frac{p+2}{p^2+4 p+3}$

∴ $L\left[t e^{-2 t} \cosh t\right]=(-1) \frac{d}{d p}\left[\frac{p+2}{p^2+4 p+3}\right]=(-1) \frac{\left[\left(p^2+4 p+3\right) \cdot 1-(p+2)(2 p+4)\right]}{\left(p^2+4 p+3\right)^2}$

= $\frac{-\left[p^2+4 p+3-\left(2 p^2+8 p+8\right)\right]}{\left(p^2+4 p-3\right)^2}=\frac{p^2+4 p+5}{\left(p^2+4 p-3\right)^2}$

49. Find the Laplace transform of $t^2 e^{-2 t} \cos t$.

Solution:

We know that $L\{\cos t\}=\frac{p}{p^2+1}$

L$\left\{t^2 \cos t\right\}=\frac{d^2}{d p^2}\left\{\frac{p}{p^2+1}\right\}=\frac{d}{d p}\left\{\frac{1-p^2}{\left(p^2+1\right)^2}\right\}$

= $\frac{\left(p^2+1\right)^2(-2 p)-\left(1-p^2\right) 2\left(p^2+1\right) 2 p}{\left(p^2+1\right)^4}=\frac{2 p\left(p^2-3\right)}{\left(p^2+1\right)^3}$

L$\left\{t^2 e^{-2 t} \cos t\right\}=\frac{2(p+2)\left[(p+2)^2-3\right]}{\left[(p+2)^2+1\right]^3}=\frac{2(p+2) \cdot\left(p^2+4 p+1\right)}{\left(p^2+4 p+5\right)^3}$

50. Find the Laplace transform of $t^3 e^{2 t} \sin t$.

Solution:

L$\left\{t^3 e^{2 t} \sin t\right\}=\text { I.P } L\left\{t^3 e^{2 t} e^{i t}\right\}=\text { I.P } L\left\{t^3 e^{(2+i) t}\right\}$

L$\left\{t^3\right\}=\frac{6}{p^4} \Rightarrow L\left[t^3 e^{(2+i) t}\right]=\frac{6}{(p-2-i)^4}$

L$\left\{t^3 e^{2 t} \sin t\right\}=\text { I.P } \frac{6}{[(p-2)-i]^4}=\text { I.P } \frac{6[(p-2)+i]^4}{\left[(p-2)^2+1\right]^4}=\frac{6\left[4(p-2)^3-4(p-2)\right]}{\left(p^2-4 p+5\right)^4}$

= $\frac{24(p-2)\left(p^2-4 p+3\right)}{\left(p^2-4 p+5\right)^4}$

51. Find $L\left\{\int_0^t t e^{-t} \sin 4 t d t\right\}$

Solution:

L$\{\sin 4 t\}=\frac{4}{p^2+16} \Rightarrow L\left\{e^{-t} \sin 4 t\right\}=\left(\frac{4}{p^2+16}\right)_{p \rightarrow p+1}=\frac{4}{(p+1)^2+16}=\frac{4}{p^2+2 p+17}$

∴ $L\left[t e^{-t} \sin 4 t\right]=(-1) \frac{d}{d p}\left[\frac{4}{p^2+2 p+17}\right]=(-4)\left[-\frac{1}{\left(p^2+2 p+17\right)^2}(2 p+2)\right]$

= $\frac{8(p+1)}{\left(p^2+2 p+17\right)^2}=f(p), \text { say }$

$L\left\{\int_0^t t e^{-t} \sin 4 t d t\right\}=\frac{1}{p} f(p)=\frac{1}{p} \frac{8(p+1)}{\left(p^2+2 p+17\right)^2}=\frac{8(p+1)}{p\left(p^2+2 p+17\right)^2} \text {. }$

52. Find $L\left[\int_0^t t e^{-t} \sin 2 t d t\right]$.

Solution:

L$\{\sin 2 t\}=\frac{2}{p^2+4} \Rightarrow L\left\{e^{-t} \sin 2 t\right\}=\left(\frac{2}{p^2+4}\right)_{p \rightarrow p+1}=\frac{2}{(p+1)^2+4}=\frac{2}{p^2+2 p+5}$

∴ $L\left[t e^{-t} \sin 2 t\right]=(-1) \frac{d}{d p}\left[\frac{2}{p^2+2 p+5}\right]=(-2)\left[-\frac{1}{\left(p^2+2 p+5\right)^2}(2 p+2)\right]$

= $\frac{4(p+1)}{\left(p^2+2 p+5\right)^2}=f(p), \text { say }$

Hence, $L\left\{\int_0^t t e^{-t} \sin 2 t d t\right\}=\frac{1}{p} f(p)=\frac{1}{p} \frac{4(p+1)}{\left(p^2+2 p+5\right)^2}=\frac{4(p+1)}{p\left(p^2+2 p+5\right)^2}$

53. Show that $\int_0^{\infty} t^3 e^{-t} \sin t d t=0$.

Solution:

⇒ $\int_0^{\infty} t^3 e^{-t} \sin t d t=\int_0^{\infty} e^{-p t}\left(t^3 \sin t\right) d t=L\left\{t^3 \sin t\right\} \text { where } p=1$

L$\left\{t^3 \sin t\right\}=(-1)^3 \frac{d^3}{d p^3}[L\{\sin t\}]=-\frac{d^3}{d p^3}\left\{\frac{1}{p^2+1}\right\}=-\frac{d^2}{d p^2}\left[\frac{-2 p}{\left(p^2+1\right)^2}\right]$

⇒ $\left.=2 \frac{d^2}{d p^2}\left\{\frac{p}{\left(p^2+1\right)^2}\right\}=2 \frac{d}{d p}\left\{\frac{\left(p^2+1\right)^2-p 2\left(p^2+1\right) 2 p}{\left(p^2+1\right)^4}\right\}=2 \frac{d}{d p}\left\{\frac{\left(p^2+1\right)\left[p^2+1-4 p^2\right]}{\left(p^2+1\right)^4}\right\}\right\}$

= $2 \frac{d}{d p}\left\{\frac{1-3 p^2}{\left(p^2+1\right)^3}\right\}=2\left[\frac{\left(p^2+1\right)^3(-6 p)-\left(1-3 p^2\right) 3\left(p^2+1\right)^2 2 p}{\left(p^2+1\right)^6}\right]$

= $2\left[\frac{-6 p^3-6 p-6 p+18 p^2}{\left(p^2+1\right)^4}\right]=2\left[\frac{12 p^2-12 p}{\left(p^2+1\right)^4}\right]$

∴ $\int_0^{\infty} t^3 e^{-t} \sin t d t=0$

54. If L[F(t)]=f(p), then show that $L\left[\frac{F(t)}{t}\right]=\int_p^{\infty} f(p) d p$, provided the integral

Solution:

Given $f(p)=L[F(t)]=\int_0^{\infty} e^{-p t} F(t) d t$

Integrating both sides w.r.t ‘ p’ from p to ∞, we get $\int_p^{\infty} f(p) d p=\int_p^{\infty}\left[\int_0^{\infty} e^{-p t} F(t) d t\right]$

The order of integration in the double integral can be interchanged since ‘ p ‘ and ‘ t ‘ are independent variables.

∴ $\int_p^{\infty} f(p) d p=\int_0^{\infty} d t \int_p^{\infty} e^{-p t} F(t) d p=\int_0^{\infty} F(t) d t \int_p^{\infty} e^{-p t} d p=\int_0^{\infty} F(t) d t\left[\frac{e^{-p t}}{-t}\right]_p^{\infty}$

= $\int_0^{\infty} \frac{F(t)}{t} \cdot e^{-p t} d t=L\left[\frac{F(t)}{t}\right]$

55. Find the Laplace transform of $\frac{1-e^t}{t}$.

Solution:

We know that $L\left(1-e^{\prime}\right)=\frac{1}{p}-\frac{1}{p-1}$

∴ $\left.\left.L\left[\frac{1-e^t}{t}\right]=\int_p^{\infty}\left(\frac{1}{p}-\frac{1}{p-1}\right) d p=\log p-\log (p-1)\right]_p^{\infty}=\log \frac{p}{p-1}\right]_p^{\infty}$

= $0-\log \frac{p}{p-1}=\log \left(\frac{p-1}{p}\right)=\log \left(1-\frac{1}{p}\right) .$

56. Find $L\left\{\frac{e^{-a t}-e^{-b t}}{t}\right\}$.

Solution:

Let $F(t)=e^{-a t}-e^{-b t} \text {. Then } L\left\{e^{-a t}-e^{-b t}\right\}=\frac{1}{p+a}-\frac{1}{p+b}=f(p)$

L$\left\{\frac{e^{-a t}-e^{-b t}}{t}\right\}=L\left\{\frac{F(t)}{t}\right\}=\int_p^{\infty} f(p) d p=\int_p^{\infty}\left(\frac{1}{p+a}-\frac{1}{p+b}\right) d p$

= $\log (p+a)-\log (p+b)]_p^{\infty}=\log \left(\frac{p+a}{p+b}\right)_p^{\infty}=\log \left(\frac{p+b}{p+a}\right) .$

57. Find $L\left\{\frac{e^{a t}-e^{-b t}}{t}\right\}$

Solution:

Let $F(t)=e^{a t}-e^{-b t} \text { : Then } L\left\{e^{a t}-e^{-b t}\right\}=\frac{1}{p-a}-\frac{1}{p+b}=f(p)$

L$\left\{\frac{e^{a t}-e^{-b t}}{t}\right\}=L\left\{\frac{F(t)}{t}\right\}=\int_p^{\infty} f(p) d p=\int_p^{\infty}\left(\frac{1}{p-a}-\frac{1}{p+b}\right) d p$

= $\log (p-a)-\log (p+b)]_p^{\infty}=\log \left(\frac{p-a}{p+b}\right)_p^{\infty}=\log \left(\frac{p+b}{p-a}\right) .$

58. Evaluate $L\left[\frac{\sin t}{t}\right]$.

Solution:

Given

$L\left[\frac{\sin t}{t}\right]$.

L$(\sin t)=\frac{1}{p^2+1} \Rightarrow L\left\{\frac{\sin t}{t}\right\}=\int_p^{\infty} \frac{1}{p^2+1} d p=\ Tan^{-1} p_p^{\infty}$

= $\frac{\pi}{2}-\ Tan^{-1} p=\ Cot^{-1} p=\ Tan^{-1} \frac{1}{p}$

59. Evaluate $L\left[\frac{\sin a t}{t}\right]$.

Solution:

Given

$L\left[\frac{\sin a t}{t}\right]$

We know that $L(\sin a t)=\frac{a}{p^2+a^2}$

∴ $L\left(\frac{\sin a t}{t}\right)=\int_p^{\infty} \frac{a}{p^2+a^2} \dot{d p}=\left[\text{Tan}^{-1} \frac{p}{a}\right]_p^{\infty}=\frac{\pi}{2}-\text{Tan}^{-1} \frac{p}{a}=\text{Cot}^{-1} \frac{p}{a}$

60. Show that $L\left\{\frac{\cos a t}{t}\right\}$ does not exist.

Solution:

L$\{\cos a t\}=\frac{p}{p^2+a^2}=f(p)$

L$\left\{\frac{\cos a t}{t}\right\}=\int_p^{\infty} \frac{p}{p^2+a^2} d x=\frac{1}{2} \log \left(p^2+a^2\right)_p^{\infty}=\frac{1}{2}\left[\log \infty-\log \left(p^2+a^2\right)\right] \text {, does not exist. }$

61. Find $L\left\{\frac{\sinh a t}{t}\right\}$.

Solution:

We have $L\{\sinh (a t)\}=\frac{a}{p^2-a^2}=f(p)$

L$\left\{\frac{\sinh a t}{t}\right\}=\int_n^{\infty} \frac{a}{p^2-a^2} d p=a \frac{1}{2 a} \log \left|\frac{p-a}{p+a}\right|_p^{\infty}=\frac{1}{2} \log \left|\frac{1-a / p}{1+a / p}\right|_p^{\infty}$

= $\frac{1}{2}\left[\log \left(\frac{1-10}{1+0}\right)-\log \left(\frac{1-a / p}{1+a / p}\right)\right]=\frac{1}{2}\left[-\log \left(\frac{p-a}{p+a}\right)\right]=\frac{1}{2} \log \left(\frac{p+a}{p-a}\right)$

62. Show that $L\left\{\frac{\cosh (a t)}{t}\right\}$ does not exist.

Solution:

We have $L\{\cosh (a t)\}=\frac{p}{p^2-a^2}=f(p)$

L$\left\{\frac{\cosh a t}{t}\right\}=\int_p^{\infty} \frac{p}{p^2-a^2} d p=\frac{1}{2} \log \left|p^2-a^2\right|_p^{\infty}=\frac{1}{2}\left[\log (\infty)-\log \left(p^2-a^2\right)\right] .$

This does not exist.

63. Find the Laplace transform of $\frac{1-\cos t}{t}$.

Solution:

We know that $L[1-\cos t]=L(1)-L[\cos t]=\frac{1}{p}-\frac{p}{p^2+1}$

∴ $\left.\left.L\left[\frac{1-\cos t}{t}\right]=\int_p^{\infty}\left(\frac{1}{p}-\frac{p}{p^2+1}\right) d p=\log p-\frac{1}{2} \log \left(p^2+1\right)\right]_p^{\infty}=\log \frac{p}{\sqrt{p^2+1}}\right]_p^{\infty}$

= $0-\log \frac{p}{\sqrt{p^2+1}}=\frac{1}{2} \log \left(\frac{p^2+1}{p^2}\right)$

64. Evaluate $L\left\{\frac{1-\cos a t}{t}\right\}$.

Solution:

Given

$L\left\{\frac{1-\cos a t}{t}\right\}$

Since $L\{1-\cos a t\}=\frac{1}{p}-\frac{p}{p^2+a^2}$

∴ $L\left\{\frac{1-\cos a t}{t}\right\}=\int_p^{\infty}\left(\frac{1}{p}-\frac{p}{p^2+a^2}\right) d p=\left[\log p-\frac{1}{2} \log \left(p^2+a^2\right)\right]_p^{\infty}$

= $\frac{1}{2}\left[2 \log p-\log \left(p^2+a^2\right)\right]_p^{\infty}=\frac{1}{2}\left[\log \left(\frac{p^2}{p^2+a^2}\right)\right]_p^{\infty}=\frac{1}{2}\left[\log 1-\log \frac{p^2}{p^2+a^2}\right]$

= $\frac{1}{2}\left[\log \left(\frac{1}{1+\frac{a^2}{p^2}}\right)\right]_p^{\infty}=-\frac{1}{2} \log \left(\frac{p^2}{p^2+a^2}\right)=\log \left(\frac{p^2}{p^2+a^2}\right)^{-1 / 2}=\log \frac{\sqrt{p^2+a^2}}{p^2} .$

65. Find the Laplace transform of $\left[\frac{\cos t-\cos 2 t}{t}\right]$.

Solution:

We know that $L[\cos t-\cos 2 t]=\frac{p}{p^2+1}-\frac{p}{p^2+4}$

L$L\left[\frac{\cos t-\cos 2 t}{t}\right]=\int_p^{\infty}\left(\frac{p}{p^2+1}-\frac{p}{p^2+4}\right) d p=\left[\frac{1}{2} \log \left(p^2+1\right)-\frac{1}{2} \log \left(p^2+4\right)\right]$

= $\frac{1}{2}\left[\log \frac{\left(1+1 / p^2\right)}{\left(1+4 / p^2\right)}\right]_p^{\infty}=\frac{1}{2}\left[\log 1-\log \left(\frac{p^2+1}{p^2+4}\right)\right]=\frac{1}{2}\left(-\log \frac{p^2+1}{p^2+4}\right)=\frac{1}{2} \log \frac{p^2+4}{p^2+1}$

66. Find $L\left\{\frac{\cos 2 t-\cos 3 t}{t}\right\}$.

Solution:

We have $L\{\cos 2 t-\cos 3 t\}=L\{\cos 2 t\}-L\{\cos 3 t\}$ $=\frac{p}{p^2+4}-\frac{p}{p^2+9}=f(p), \text { say }$

∴ $L\left\{\frac{\cos 2 t-\cos 3 t}{t}\right\}=\int_p^{\infty} f(p) d p=\int_p^{\infty}\left(\frac{p}{p^2+4}-\frac{p}{p^{2+9}}\right) d p=\frac{1}{2} \int_p^{\infty}\left(\frac{2 p}{p^2+4}-\frac{2 p}{p^2+9}\right) d p$

= $\frac{1}{2}\left[\log \left(p^2+4\right)-\log \left(p^2+9\right)\right]=\frac{1}{2}\left[\log \left(\frac{p^2+4}{p^2+9}\right)\right]_p^{\infty}=\frac{1}{2}\left[\log \left(\frac{1+4 / p^2}{1+9 / p^2}\right)\right]_p^{\infty}$

= $\frac{1}{2}\left[\log \left(\frac{1+0}{1+0}\right)-\log \left(\frac{1+4 / p^2}{1+9 / p^2}\right)\right]=\frac{1}{2}\left[0-\log \left(\frac{p^2+4}{p^2+9}\right)\right]=\frac{1}{2} \log \left(\frac{p^2+9}{p^2+4}\right)$

= $\log \left(\frac{p^2+9}{p^2+4}\right)^{1 / 2} L\{\sin 3 t \cos t\}=L\left\{\frac{\sin 4 t+\sin 2 t}{2}\right\}=\frac{1}{2} \frac{1}{p^2+16}+\frac{1}{2} \frac{2}{p^2+4}$

= $\frac{2}{p^2+16}+\frac{1}{p^2+4}$

67. Find the Laplace transform of $\left[\frac{\cos a t-\cos b t}{t}\right]$.

Solution:

We have $L\{\cos a t-\cos b t\}=L\{\cos a t\}-L\{\cos b t\}=\frac{p}{p^2+a^2}-\frac{p}{p^2+b^2}=f(p)$, say

L$\left\{\frac{\cos a t-\cos b t}{t}\right\}=\int_p^{\infty} f(p) d p=\int_p^{\infty}\left(\frac{p}{p^2+a^2}-\frac{p}{p^2+b^2}\right) d p=\frac{1}{2} \int_p^{\infty}\left(\frac{2 p}{p^2+a^2}-\frac{2 p}{p^2+b^2}\right) d p$

= $\frac{1}{2}\left[\log \left(p^2+a^2\right)-\log \left(p^2+b^2\right)\right]=\frac{1}{2}\left[\log \left(\frac{p^2+a^2}{p^2+b^2}\right)\right]_p^{\infty}=\frac{1}{2}\left[\log \left(\frac{1+a^2 / p^2}{1+b^2 / p^2}\right)\right]_p^{\infty}$

= $\frac{1}{2}\left[\log \left(\frac{1+0}{1+0}\right)-\log \left(\frac{1+a^2 / p^2}{1+b^2 / p^2}\right)\right]=\frac{1}{2}\left[0-\log \left(\frac{p^2+a^2}{p^2+b^2}\right)\right]=\frac{1}{2} \log \left(\frac{p^2+b^2}{p^2+a^2}\right)$

68. Find $L\left\{\frac{\sin 3 t \cos t}{t}\right\}$

Solution:

⇒ $\left.L\left\{\frac{\sin 3 t \cos t}{t}\right\}=\int_p^{\infty}\left[\frac{2}{p^2+16}+\frac{1}{p^2+4}\right] d p=(2) \frac{1}{4} \ Tan^{-1} \frac{p}{4}+\frac{1}{2} \ Tan^{-1} \frac{p}{2}\right]_p^{\infty}$

= $\frac{1}{2}\left(\frac{\pi}{2}\right)-\frac{1}{2} \ Tan^{-1} \frac{p}{4}+\frac{1}{2}\left(\frac{\pi}{2}\right)-\frac{1}{2} \ Tan^{-1} \frac{p}{2}=\frac{\pi}{2}-\frac{1}{2} \ Tan^{-1} \frac{p}{4}-\frac{1}{2} \ Tan^{-1} \frac{p}{2} .$

69. Find $L\left[\frac{\cos 4 t \sin 2 t}{t}\right]$.

Solution:

L$\{\cos 4 t \sin 2 t\}=\frac{1}{2} L\{2 \cos 4 t \sin 2 t\}=\frac{1}{2} L\{\sin 6 t-\sin 2 t\}=\frac{1}{2}\left[\frac{6}{p^2+6^2}-\frac{2}{p^2+2^2}\right]$

= $\frac{3}{p^2+6^2}-\frac{1}{p^2+2^2}=f(p)$

∴ $L\left\{\frac{\cos 4 t \sin 2 t}{t}\right\}=\int_p^{\infty} f(p) d p=\int_p^{\infty}\left(\frac{3}{p^2+6^2}-\frac{1}{p^2+2^2}\right) d p$

= $\left[3 \frac{1}{6} \ Tan^{-1} \frac{p}{6}-\frac{1}{2} \ Tan^{-1} \frac{p}{2}\right]_0^{\infty}=\frac{1}{2}\left(\frac{\pi}{2}-\frac{\pi}{2}\right)-\frac{1}{2}\left(\ Tan^{-1} \frac{p}{6}-\ Tan^{-1} \frac{p}{2}\right)$

= $\frac{1}{2}\left(\ Tan^{-1} \cdot \frac{p}{2}-\ Tan^{-1} \frac{p}{6}\right) .$

70. Solve $L\left\{\frac{e^t-\cos t}{t}\right\}$.

Solution:L$\left\{e^t-\cos t\right\}=L\left\{e^t\right\}-L\{\cos t\}=\frac{1}{p-1}-\frac{p}{p^2+1}$

L$\left\{\frac{e^t-\cos t}{t}\right\}=\int_p^{\infty}\left[\frac{1}{p-1}-\frac{p}{p^2+1}\right] d p=\left[\log (p-1)-\frac{1}{2} \log \left(p^2+1\right)\right]_p^{\infty}$

= $\left[\log \left(\frac{p-1}{\sqrt{p^2+1}}\right)\right]_p^{\infty}=0-\log \left(\frac{p-1}{\sqrt{p^2+1}}\right)=\frac{1}{2} \log \left[\frac{p^2+1}{(p-1)^2}\right]$

71. Find $L\left[\frac{e^{-2 t} \sin 2 t}{t}\right]$.

Solution:

L$[\sin 2 t]=\frac{2}{s^2+2^2} \Rightarrow L\left[e^{-2 t} \sin 2 t\right]=\frac{2}{(s+2)^2+2^2}$

∴ $\left.L\left[\frac{e^{-2 t} \sin 2 t}{t}\right]=\int_p^{\infty} \frac{2}{(p+2)^2+2^2} d s=\ Tan^{-1} \frac{p+2}{2}\right]_p^{\infty}=\frac{\pi}{2}-\ Tan^{-1} \frac{p+2}{2}=\ Cot^{-1}\left(\frac{p+2}{2}\right)$

72. Find $L\left[\frac{e^{-3 t} \sin 2 t}{t}\right]$.

Solution:

L$\{\sin 2 t\}=\frac{2}{p^2+2^2} \Rightarrow L\left\{e^{-3 t} \sin 2 t\right\}=\left(\frac{2}{p^2+2^2}\right)_{p \rightarrow p+3}=\frac{2}{(p+3)^2+2^2}=f(p), \text { say }$

∴$L\left[\frac{e^{-3 t} \sin 2 t}{t}\right]=\int_p^{\infty} f(p) d p=\int_p^{\infty} \frac{2}{(p+3)^2+2^2} d p=2 \frac{1}{2}\left[\ Tan^{-1}\left(\frac{p+3}{2}\right)_p^{\infty}\right.$

= $\frac{\pi}{2}-\ Tan^{-1}\left(\frac{p+3}{2}\right)=\ Cot^{-1}\left(\frac{p+3}{2}\right)$

73. Evaluate $L\left[e^{-4 t} \frac{\sin 3 t}{t}\right]$.

Solution:

Given

$L\left[e^{-4 t} \frac{\sin 3 t}{t}\right]$

L$\{\sin 3 t\}=\frac{3}{p^2+3^2} \Rightarrow L\left\{e^{-4 t} \sin 3 t\right\}=\left(\frac{3}{p^2+3^2}\right)_{p \rightarrow p+4}=\frac{3}{(p+4)^2+3^2}=f(p) \text {, say }$

∴ $L\left[\frac{e^{-4 t} \sin 3 t}{t}\right]=\int_p^{\infty} f(p) d p=\int_p^{\infty} \frac{3}{(p+4)^2+3^2} d p=3 \frac{1}{3}\left[\ Tan^{-1}\left(\frac{p+4}{3}\right)_p^{\infty}\right.$

= $\frac{\pi}{2}-\ Tan^{-1}\left(\frac{p+4}{3}\right)=\ Cot^{-1}\left(\frac{p+4}{3}\right)$

74. Find $L\left\{\frac{1-\cos t}{t^2}\right\}$.

Solution:

We know that $L[1-\cos t]=L(1)-L[\cos t]=\frac{1}{p}-\frac{p}{p^2+1}$

∴ L$\left[\frac{1-\cos t}{t}\right]=\int_p^{\infty}\left(\frac{1}{p}-\frac{p}{p^2+1}\right) d p=\log p-\frac{1}{2} \log \left(p^2+1\right)_p^{\infty}=\log \frac{p}{\sqrt{p^2+1}}_p^{\infty}$

= $0-\log \frac{p}{\sqrt{p^2+1}}=\frac{1}{2} \log \left(\frac{p^2+1}{p^2}\right)$

∴ $L\left\{\frac{1-\cos t}{t^2}\right\}=\int_p^{\infty} \frac{1}{2} \log \left(\frac{p^2+1}{p^2}\right) d p=\frac{1}{2} \int_p^{\infty}\left[\log \left(p^2+1\right)-\log p^2\right] d p$

= $\frac{1}{2} \int_p^{\infty}\left[\log \left(p^2+1\right)-2 \log p\right] 1 d p$

= $\frac{1}{2}\left[\left\{\log \left(p^2+1\right)-2 \log p\right\}_p\right]_p^{\infty}-\frac{1}{2} \int_p^{\infty}\left(\frac{2 p}{p^2+1}-\frac{2}{p}\right) p d p$

= $\left[\frac{p}{2} \log \left(\frac{p^2+1}{p^2}\right)\right]_p^{\infty}+\int_p^{\infty} \frac{1}{p^2+1} d p=\left[\frac{p}{2} \log \left(1+\frac{1}{p^2}\right)\right]_p^{\infty}+\left(\text{Tan}^{-1} p\right)_p^{\infty}$

= $-\frac{p}{2} \log \left(1+\frac{1}{p^2}\right)+\left(\frac{\pi}{2}-\text{Tan}^{-1} p\right)=\text{Cot}^{-1} p-\frac{p}{2} \log \left(1+\frac{1}{p^2}\right)$

75. Find the Laplace transform of $\int_0^t \frac{\sin t}{t} d t$.

Solution:

L$(\sin t)=\frac{1}{p^2+1}$

∴ $\left.\Rightarrow L\left\{\frac{\sin t}{t}\right\}=\int_p^{\infty} \frac{1}{p^2+1} d p=\ Tan^{-1} p\right]_p^{\infty}=\frac{\pi}{2}-\ Tan^{-1} p=\ Cot^{-1} p=\ Tan^{-1} \frac{1}{p} .$

L$\left\{\int_0^t \frac{\sin t}{t}\right\}=\frac{1}{p} L\left\{\frac{\sin t}{t}\right\}=\frac{1}{p} \ Tan^{-1} \frac{1}{p}$

76. Find $L\left[\int_0^t \frac{1-e^{-t}}{t} d t\right]$.

Solution:

Let $F(t)=1-e^{-t} \text {. Then } f(p)=L\left\{1-e^{-t}\right\}=\frac{1}{p}-\frac{1}{p+1}$

∴ $L\left\{\frac{1-e^{-t}}{t}\right\}=\int_p^{\infty}\left(\frac{1}{p}-\frac{1}{p+1}\right) d p=[\log p=\log (p+1)]_p^{\infty}=\left[\log \left(\frac{p}{p+1}\right)\right]_p^{\infty}$

= $\left[\log \left(\frac{1}{1+1 / p}\right)\right]_p^{\infty}=\log \left(\frac{1}{1+0}\right)-\log \left(\frac{1}{1+1 / p}\right)=0-\log \left(\frac{p}{p+1}\right)=\log \left(\frac{p+1}{p}\right)=f(p)$

⇒ $L\left[\int_0^t \frac{1-e^{-t}}{t} d t\right]=\frac{1}{p} f(p)=\frac{1}{p} \log \left(\frac{p+1}{p}\right)=\frac{1}{p} \log \left(1+\frac{1}{p}\right)$

77. Find $L\left\{\int_0^t \frac{e^t \sin t}{t} d t\right\}$

Solution:

We know that $L\{\sin t\}=\frac{1}{p^2+1}$

∴ $L\left\{\frac{\sin t}{t}\right\}=\int_p^{\infty} \frac{1}{p^2+1} d p=\left(\tan ^{-1} p\right)_p^{\infty}=\tan ^{-1} \infty-\tan ^{-1} p=\frac{\pi}{2}-\tan ^{-1} p=\cot ^{-1} p$

By First Shifting theorem, $L\left\{e^t \frac{\sin t}{t}\right\}=\left(\cot ^{-1} p\right)_{p \rightarrow p-1}=\cot ^{-1}(p-1)=f(p)$

Using the theorem of L.T. of the integral. $L\left\{\int_0^t \frac{e^t \sin t}{t} d t\right\}=\frac{1}{p} f(p)=\frac{1}{p} \cot ^{-1}(p-1)$.

78. Find $L\left[e^{-3 t} \int_0^t \frac{\sin t}{t} d t\right]$.

Solution:

Since $L\{\sin t\}=\frac{1}{p^2+1}=f(p)$

∴ $L\left\{\frac{\sin t}{t}\right\}=\int_p^{\infty} f(p) d p=\int_p^{\infty} \frac{1}{p^2+1} d p=\left[\ Tan^{-1} p\right]_p^{\infty}=\frac{\pi}{2}-\ Tan^{-1} p=\text{Cot}^{-1} p$

Hence $L\left[\int_0^t \frac{\sin t}{t} d t\right]=\frac{1}{p} \ Cot^{-1} p$ By First Shifting theorem,

L$\left[e^{-3 t} \int_0^t \frac{\sin t}{t} d t\right]=\left(\frac{1}{p} \ Cot^{-1} p\right)_{p \rightarrow p+3}=\frac{1}{p+3} \ Cot^{-1}(p+3)$

79. Find the Laplace transform of $\int_0^t e^{-t} \cosh t d t$.

Solution:

We know that $L\{\cosh t\}=\frac{p}{p^2-1} \cdot \Rightarrow\left\{e^{-t} \cosh t\right\}=\frac{p+1}{(p+1)^2-1}=\frac{p+1}{p^2+2 p} \text {. }$

∴ $L\left[\int_0^t e^{-t} \cosh t d t\right]=\frac{1}{p} L\left(e^{-t} \cosh t\right)=\frac{1}{p} \frac{p+1}{p^2+2 p}=\frac{p+1}{p^2(p+2)}$

80. Evaluate $\int_0^{\infty} t e^{-3 t} d t$

Solution:

Given

$\int_0^{\infty} t e^{-3 t} d t$

⇒ $\int_0^{\infty} t e^{-3 t} d t=\int_0^{\infty} t e^{-p t} d t \text { where } p=3 . \text { But } \int_0^{\infty} t e^{-p t} d t=L\{t\}=\frac{1}{p^2}$

Putting p=3 , we get $\int_0^{\infty} t e^{-3 t} d t=\frac{1}{3^2}=\frac{1}{9} \text {. }$

81. Evaluate $\int_0^{\infty} e^{-4 t} \sin 3 t d t .$

Solution:

Given

$\int_0^{\infty} e^{-4 t} \sin 3 t d t .$

⇒ $\int_0^{\infty} e^{-4 t} \sin 3 t d t=\int_0^{\infty} e^{-p t} \sin 3 t d t=L\{\sin 3 t\} \text { where } p=4$

But $L\{\sin 3 t\}=\frac{3}{p^2+9} \Rightarrow \int_0^{\infty} e^{-p t} \sin 3 t d t=\frac{3}{p^2+9}$

Putting p=4, we get $\int_0^{\infty} e^{-4 t} \sin 3 t d t=\frac{3}{16+9}=\frac{3}{25} \text {. }$

82. Evaluate $\int_0^{\infty} t e^{-t} \sin t d t$.

Solution:

Given

⇒ $\int_0^{\infty} t e^{-t} \sin t d t=\int_0^{\infty} t e^{-p t} \sin t d t=\int_0^{\infty} e^{-p t}(t \sin t) d t=L\{t \sin t\} \text {, where } p=1 \text {. }$

But $L\{t \sin t\}=(-1) \frac{d}{d p}[L(\sin t)]=(-1) \frac{d}{d p}\left(\frac{1}{p^2+1}\right)=(-1) \frac{-1}{\left(p^2+1\right)^2} 2 p=\frac{2 p}{\left(p^2+1\right)^2}$

Putting p=1, we get $\int_0^{\infty} t e^{-t} \sin t d t=\frac{2}{(1+1)^2}=\frac{1}{2}$.

83. Evaluate $\int_0^{\infty} t e^{-2 t} \sin t d t$.

Solution:

Given

$\int_0^{\infty} t e^{-2 t} \sin t d t$

⇒ $\int_0^{\infty} t e^{-2 t} \sin t d t=\int_0^{\infty} e^{-p t} t \sin t d t \text { where } p=2$

But $\int_0^{\infty} e^{-p t} t \sin t d t=L\{t \sin t\}=(-1) \frac{d}{d p}\left(\frac{1}{p^2+1}\right)=\frac{2 p}{\left(p^2+1\right)^2}$

⇒ $\int_0^{\infty} e^{-p t} t \sin t d t=\frac{2 p}{\left(p^2+1\right)^2} \text {. Taking } p=2 \text {, we get } \int_0^{\infty} t e^{-2 t} \sin t d t=\frac{4}{25}$

84. Prove that $\int_0^1 t e^{-2 t} \cos t d t=\frac{3}{25} .$.

Solution:

⇒ $\int_0^{\infty} t e^{-2 t} \cos t d t=\int_0^{\infty} e^{-p t}(t \cos t) d t \text { where } p=2$

= $L\{t \cos t\}=(-1)^1 \frac{d}{d p}[L\{\cos t\}]=-\frac{d}{d p}\left\{\frac{p}{p^2+1}\right\}=-\left[\frac{\left(p^2+1\right) 1-p(2 p)}{\left(p^2+1\right)^2}\right]$

= $-\frac{1-p^2}{\left(p^2+1\right)^2}=\frac{p^2-1}{\left(p^2+1\right)^2}=\frac{3}{25}$

85. Show that $\int_0^{\infty} t e^{-3 t} \sin t d t=\frac{3}{50} .$.

Solution:

⇒ $\int_0^{\infty} t e^{-3 t} \sin t d t=\int_0^{\infty} e^{-p t} t \sin t d t \text { where } p=3 .$

But $\int_0^{\infty} e^{-p t} t \sin t d t=L\{t \sin t\}=(-1) \frac{d}{d p}\left(\frac{1}{p^2+1}\right)=\frac{2 p}{\left(p^2+1\right)^2}$

⇒ $\int_0^{\infty} e^{-p t} t \sin t d t=\frac{2 p}{\left(p^2+1\right)^2} \text {. Taking } p=3 \text {, we get } \int_0^{\infty} t e^{-3 t} \sin t d t=\frac{6}{100}=\frac{3}{50}$

86. Evaluate $\int_0^{\infty} t e^{-3 t} \cos 4 t d t$.

Solution:

Given

$\int_0^{\infty} t e^{-3 t} \cos 4 t d t$.

We know that $L[\cos 4 t]=\frac{p}{p^2+16}$

L$[t \cos 4 t]=-\frac{d}{d p} L[\cos 4 t]=-\frac{d}{d p}\left[\frac{p}{p^2+16}\right]=-\left[\frac{\left(p^2+16\right) 1-p(2 p)}{\left(p^2+16\right)^2}\right]=\frac{p^2-16}{\left(p^2+16\right)^2}$

∴ $\int_0^{\infty} t e^{-3 t} \cos 4 t d t=\int_0^{\infty} e^{-p t} t \cos 4 t d t=\int_0^{-p t} e^{-p}(t \cos 4 t) d t=L[t \cos 4 t] \text { where } p=3$

⇒ $\int_0^{\infty} e^{-3 t} t \cos 4 t=\frac{9-16}{(9+16)^2}=-\frac{7}{625}$

87. Evaluate $\int_0^{\infty} t e^{-4 t} \cos 2 t d t$.

Solution:

Given

$\int_0^{\infty} t e^{-4 t} \cos 2 t d t$

We know that $L_s[\cos 2 t]=\frac{p}{p^2+4}$

∴ $L[t \cos 2 t]=-\frac{d}{d p} L[\cos 2 t]=-\frac{d}{d p}\left[\frac{p}{p^2+4}\right]=-\left[\frac{\left(p^2+4\right) 1-p(2 p)}{\left(p^2+4\right)^2}\right]=\frac{p^2-4}{\left(p^2+4\right)^2}$

Now $\int_0^{\infty} t e^{-4 t} \cos 2 t d t=\int_0^{\infty} e^{-p t} t \cos 2 t d t=\int_0 e^{-p t}(t \cos 2 t) d t \doteq \ L[t \cos 2 t] \text { where } p=4$

∴ $\int_0^{\infty} e^{-4 t} t \cos 2 t=\frac{16-4}{(16+4)^2}=\frac{12}{400}=\frac{3}{100} .$

88. Show that $\int_0^{\infty} t^3 e^{-t} \sin t d t=0$.

Solution:

⇒ $\int_0^{\infty} t^3 e^{-t} \sin t d t=\int_0^{\infty} e^{-p t} t^3 \sin t d t$, where p=1

But $\int_0^{\infty} e^{-p t} t^3 \sin t d t=L\left\{t^3 \sin t\right\}$

Now $L\left\{t^3 \sin t\right\}=(-1)^3 \frac{d^3}{d s^3}[L\{\sin t\}]=(-1) \frac{d^3}{d p^3}\left(\frac{1}{p^2+1}\right)$

= $(-1) \frac{d^2}{d p^2}\left(\frac{-2 p}{\left(p^2+1\right)^2}\right)=2 \frac{d}{d p}\left(\frac{1-3 p^2}{\left(1+p^2\right)^3}\right)=\frac{-24 p\left(1-p^2\right)}{\left(1+p^2\right)^4}$

⇒ $\int_0^{\infty} e^{-p t} t^3 \sin t d t=\frac{-24 p\left(1-p^2\right)}{\left(1+p^2\right)^4}$

Putting p=1, we get $\int_0^{\infty} e^{-t} t^3 \sin t d t=\frac{24(1-1)}{(1+1)^4}=0$

89. Evaluate $\int_0^{\infty} \frac{\sin t}{t} d t$

Solution:

Given

$\int_0^{\infty} \frac{\sin t}{t} d t$

⇒ $\int_0^{\infty} \frac{\sin t}{t} d t=\int_0^{\infty} e^{-p t} \frac{\sin t}{t} d t=L\left[\frac{\sin t}{t}\right] \text { where } p=0$

L$(\sin t)=\frac{1}{p^2+1}=f(p) \Rightarrow L\left(\frac{\sin t}{t}\right)=\int_p^{\infty} f(p) d p=\int_p^{\infty} \frac{d p}{p^2+1}$

= $\left[\ Tan^{-1} p\right]_s^{\infty}=\frac{\pi}{2}-\ Tan^{-1} p$

Taking p=0, we get $\int_0^{\infty} \frac{\sin t}{t} d t=\frac{\pi}{2} \text {. }$

90. Evaluate $\int_0^{\infty} \frac{\sin 2 t}{t} d t$.

Solution:

Given integral is the Laplace transform of $\frac{\sin 2 t}{t} \text {. with } p=0$

∴ $\int_0^{\infty} e^{-p t} \frac{\sin 2 t}{t} d t=L\left\{\frac{\sin 2 t}{t}\right\}=\int_p^{\infty} L\{\sin 2 t\} d p=\int_p^{\infty} \frac{2}{p^2+4} d p$

= $2 \frac{1}{2}\left(\tan ^{-1} \frac{p}{2}\right)_p^{\infty}=\frac{\pi}{2}-\tan ^{-1} \frac{p}{2}$

Putting p=0, we get $\int_0^{\infty} \frac{\sin 2 t}{t} d t=\frac{\pi}{2} \text {. }$

91. Using Laplace transforms show that $\int_0^{\infty} \frac{\sin ^2 t}{t^2 \cdot} d t=\frac{\pi}{2}$.

Solution:

⇒ $\int_0^{\infty} \frac{\sin ^2 t}{t^2} d t=\int_0^{\infty} e^{-p t} \frac{\sin ^2 t}{t^2} d t \text { where } p=0$

= $L\left\{\frac{\sin ^2 t}{t^2}\right\}=\int_p^{\infty} L\left\{\frac{\sin ^2 t}{t}\right\} d p=\frac{1}{2} \int_0^{\infty} L\left\{\frac{1-\cos 2 t}{t}\right\} d p$

92. Evaluate $\int_0^{\infty} \frac{e^{-a t}-e^{-b t}}{t} d t$.

Solution:

Given

$\int_0^{\infty} \frac{e^{-a t}-e^{-b t}}{t} d t$

Let $F(t)=e^{-a t}-e^{-b t} \text {. Then } L\left\{e^{-a t}-e^{-b t}\right\}=\frac{1}{p+a}-\frac{1}{p+b}=f(p)$

L $\left\{\frac{e^{-a t}-e^{-b t}}{t}\right\}=L\left\{\frac{F(t)}{t}\right\}=\int_p^{\infty} f(p) d p=\int_p^{\infty}\left(\frac{1}{p+a}-\frac{1}{p+b}\right) d p$

= $\log (p+a)-\log (p+b)]_p^{\infty}=\log \left(\frac{p+a}{p+b}\right)_p^{\infty}=\log \left(\frac{p+b}{p+a}\right) .$

⇒ $\int_0^{\infty} e^{-p t}\left(\frac{e^{-a t}-e^{-b}}{t}\right) d t=\log \left(\frac{p+b}{p+a}\right)$

p=0, we get $\int_0^{\infty} \frac{e^{-a t}-e^{-b t}}{t} d t=\log \left(\frac{b}{a}\right)$

93. Evaluate $\int_0^{\infty} \frac{e^{-t}-e^{-3 t}}{t} d t$.

Solution:

Given

$\int_0^{\infty} \frac{e^{-t}-e^{-3 t}}{t} d t$

Let $F(t)=e^{-t}-e^{-3 t} \text {. }$

Now, $\int_0^{\infty} \frac{e^{-t}-e^{-3 t}}{t} d t=\int_0^{\infty} \frac{F(t)}{t} d t==\int_0^{\infty} e^{-p t} \frac{F(t)}{t} d t=L\left[\frac{F(t)}{t}\right] \text { where } p=0$

Then $L[F(t)]=L\left[e^{-t}-e^{-3 t}\right]=\frac{1}{p+1}-\frac{1}{p+3}=f(p) \text {. }$

∴ $L\left[\frac{F(t)}{t}\right]=\int_p^{\infty} f(p) d p=\int_p^{\infty}\left(\frac{1}{p+1}-\frac{1}{p+3}\right) d p=[\log (p+1)-\log (p+3)]_p^{\infty}$

= $\left[\log \left(\frac{1+1 / p}{1+3 / p}\right)\right]_p^{\infty}=\log 1-\log \left(\frac{p+1}{p+3}\right)=0-\log \frac{p+1}{p+3}=\log \left(\frac{p+3}{p+1}\right)$

Taking p=0, we get $\int_0^{\infty}\left(\frac{e^{-t}-e^{-3}}{t}\right) d t=\log \frac{3}{1}=\log 3 \text {. }$

94. Evaluate $\int_0^{\infty} \frac{e^{-3 t}-e^{-6 t}}{t} d t=\log 2$.

Solution:

Given

$\int_0^{\infty} \frac{e^{-3 t}-e^{-6 t}}{t} d t=\log 2$

Let $F(t)=e^{-3 t}-e^{-6 t} \text {. Then } L\left\{e^{-3 t}-e^{-6 t}\right\}=\frac{1}{p+3}-\frac{1}{p+6}=f(p)$

L$\left\{\frac{e^{-3 t}-e^{-6 t}}{t}\right\}=L\left\{\frac{F(t)}{t}\right\}=\int_p^{\infty} f(p) d p=\int_p^{\infty}\left(\frac{1}{p+3}-\frac{1}{p+6}\right) d p$

= $=\log (p+3)-\log (p+6)]_p^{\infty}=\log \left(\frac{p+3}{p+6}\right)_p^{\infty}=\log \left(\frac{p+6}{p+3}\right)$

⇒ $\int_0^{\infty} e^{-p t}\left(\frac{e^{-3 t}-e^{-6 t}}{t}\right) d t=\log \left(\frac{p+6}{p+3}\right)$

Putting p=0, we get $\int_0^{\infty} \frac{e^{-3 t}-e^{-6 t}}{t} d t=\log \left(\frac{6}{3}\right)=\log 2 \text {. }$

95. Evaluate $\int_0^{\infty} \frac{e^{-t}-e^{-2 t}}{t} d t$.

Solution:

Given

$\int_0^{\infty} \frac{e^{-t}-e^{-2 t}}{t} d t$

Let $F(t)=e^{-t}-e^{-2 t} \text {. Then } L\left\{e^{-t}-e^{-2 t}\right\}=\frac{1}{p+1}-\frac{1}{p+2}=f(p)$

⇒ $L\left\{\frac{e^{-t}-e^{-2 t}}{t}\right\}=L\left\{\frac{F(t)}{t}\right\}=\int_p^{\infty} f(p))^{\infty} d p=\int_p^{\infty}\left(\frac{1}{p+1}-\frac{1}{p+2}\right) d p$

= $\log (p+1)-\log (p+2)]_p^{\infty}=\log \left(\frac{p+1}{p+2}\right)_p^{\infty}=\log \left(\frac{p+2}{p+1}\right)$

⇒ $\int_0^{\infty} e^{-p t}\left(\frac{e^{-t}-e^{-2 t}}{t}\right) d t=\log \left(\frac{p+2}{p+1}\right)$

Putting p=0, we get $\int_0^{\infty} \frac{e^{-t}-e^{-2 t}}{t} d t=\log \left(\frac{2}{1}\right)=\log 2 \text {. }$

96. Using Laplace transform, evaluate $\int_0^{\infty} \frac{\cos a t-\cos b t}{t} d t$.

Solution:

⇒ $\int_0^{\infty} \frac{\cos a t-\cos b t}{t} d t=\int_0^{\infty} e^{-p t}\left(\frac{\cos a t-\cos b t}{t}\right) d t=L\left\{\frac{\cos a t-\cos b t}{t}\right\} \text { where } p=0$

We know that $L\{\cos a t-\cos b t\}=\frac{p}{p^2+a^2}-\frac{p}{p^2+b^2}$

∴ $L\left[\frac{\cos a t-\cos b t^2}{t}\right]=\int_p^{\infty}\left(\frac{p}{p^2+a^2}-\frac{p}{p^2+b^2}\right) d p=\frac{1}{2} \int_p^{\infty}\left(\frac{2 p}{p^2+a^2}-\frac{2 p}{p^2+b^2}\right) d p$

= $\frac{1}{2}\left[\log \left(p^2+a^2\right)-\log \left(p^2+b^2\right)\right]_p^{\infty}=\frac{1}{2}\left[\log \left(\frac{p^2+a^2}{p^2+b^2}\right)\right]_p^{-\infty}=\frac{1}{2}\left[\log \left(\frac{1+a^2 / p^2}{1+b^2 / p^2}\right)\right]_p^{\infty}$

= $\frac{1}{2}\left[\log 1-\log \left(\frac{1+a^2 / p^2}{1+b^2 / p^2}\right)\right]=\frac{1}{2}\left[0-\log \left(\frac{p^2+a^2}{p^2+b^2}\right)\right]=\frac{1}{2} \log \left(\frac{p^2+b^2}{p^2+a^2}\right)$

⇒ $\int_0^{\infty} e^{-p t}\left(\frac{\cos a t-\cos b t}{t}\right) d t=\frac{1}{2} \log \left(\frac{p^2+b^2}{p^2+a^2}\right) .$

Take p=0 Then $\int_0^{\infty} \frac{\cos a t-\cos b t}{t} d t=\frac{1}{2} \log \left(\frac{b}{2}\right)^2=\log \left(\frac{b}{a}\right)$

97. Using Laplace transform, evaluate $\int_0^{\infty} \frac{e^{-a t} \sin ^2 t}{t} d t$.

Solution:

⇒ $\int_0^{\infty} \frac{e^{-a t} \sin ^2 t}{t} d t=L\left[\frac{\sin ^2 t}{t}\right], \text { where } p=a$

Now, $L\left\{\frac{\sin ^2 t}{t}\right\}=L\left\{\frac{1-\cos 2 t}{2 t}\right\}=\frac{1}{2} L\left\{\frac{1-\cos 2 t}{t}\right\}=\frac{1}{2} \int_p^{\infty} L\{1-\cos 2 t\} d p$

= $\frac{1}{2} \int_p^{\infty}\left(\frac{1}{p}-\frac{p}{p^2+4}\right) d p=\frac{1}{2}\left[\log p-\frac{1}{2} \log \left(p^2+4\right)\right]_p^{\infty}$

= $\frac{1}{4}\left[\log \left(\frac{p^2}{p^2+4}\right)\right]_p^{\infty}=\frac{1}{4}\left[\log \left(\frac{p^2+4}{p^2}\right)\right]$

⇒ $\int_0^{\infty} e^{-p t} \frac{\sin ^2 t}{t} d t=\frac{1}{4} \log \left(\frac{p^2+4}{p^2}\right)$

Putting p=a, we get $\int_0^{\infty} \frac{e^{-a t} \sin ^2 t}{t} d t=\frac{1}{4} \log \left(\frac{a^4+4}{a^2}\right)$

98. Define error function.

Solution:

The error function is denoted by erf(t) and is defined } erf(t) = $\frac{2}{\sqrt{\pi}} \int_0^t e^{-x^2} d x$

99. Prove that $L[\ erf(\sqrt{t})]=\frac{1}{p \sqrt{p+1}}$ and hence deduce that

1) $L\{\ erf(2 \sqrt{t})\}=\frac{2}{p \sqrt{p+4}}$

2) $L[t \ erf(2 \sqrt{t})]=\frac{3 p+8}{p^2(p+4)^{3 / 2}}$

Solution:

⇒ $\ erf(\sqrt{t})=\frac{2}{\sqrt{\pi}} \int_0^{\sqrt{t}} e^{-x^2} d x=\frac{2}{\sqrt{\pi}} \int_0^{\sqrt{t}}\left(1-x^2+\frac{x^4}{2 !}-\frac{x^6}{3 !}+\ldots\right) d x$

= $\frac{2}{\sqrt{\pi}}\left[x-\frac{x^3}{3}+\frac{x^5}{5 \cdot 2 !}-\frac{u^7}{7 \cdot 3 !}+\ldots\right]_0^{\sqrt{t}}=\frac{2}{\sqrt{\pi}}\left[t^{1 / 2}-\frac{t^{3 / 2}}{3}+\frac{t^{5 / 2}}{5 \cdot 2 !}-\frac{t^{7 / 2}}{7 \cdot 3 !}+\ldots\right]$

∴ $L\{\ erf(\sqrt{t})\}=\frac{2}{\sqrt{\pi}}\left[L\left\{t^{1 / 2}\right\}-\frac{1}{3} L\left\{t^{3 / 2}\right\}+\frac{1}{5 \cdot 2 !} L\left\{t^{5 / 2}\right\}-\frac{1}{7 \cdot 3 !} L\left\{t^{7 / 2}\right\}+\ldots\right]$

⇒ $\frac{2}{\sqrt{\pi}}\left[\frac{\Gamma\left(\frac{3}{2}\right)}{p^{3 / 2}}-\frac{1}{3} \frac{\Gamma\left(\frac{5}{2}\right)}{p^{5 / 2}}+\frac{1}{5 \cdot 2 !} \frac{\Gamma\left(\frac{7}{2}\right)}{p^{7 / 2}}-\frac{1}{7 \cdot 3 !} \frac{\Gamma\left(\frac{9}{2}\right)}{p^{9 / 2}}+\ldots\right]$

= $\frac{2}{\sqrt{\pi}}[\frac{\sqrt{\pi}}{2} \cdot \frac{1}{p^{3 / 2}}-\frac{\sqrt{\pi}}{2} \cdot \frac{1}{2} \cdot \frac{1}{p^{5 / 2}}+\frac{\sqrt{\pi}}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{p^{7 / 2}}$

–$\frac{\pi}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{1}{p^{9 / 2}}+\ldots$

= $\frac{1}{p^{3 / 2}}\left[1-\frac{1}{2} \cdot \frac{1}{p}+\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{p^2}-\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{1}{p^3}+\ldots\right]$

= $\frac{1}{p^{3 / 2}}\left(\mathfrak{1}+\frac{1}{p}\right)^{-1 / 2}=\frac{1}{p \sqrt{p+1}}$

Thus $L\ erf(\sqrt{t})\}=\frac{1}{p \sqrt{p+1}} \text {. }$

By change of scale property,

L$[\text{erf}(2 \sqrt{t})]=L[\text{erf}(\sqrt{4 t})]=\frac{1}{4} \frac{1}{(p / 4) \sqrt{p / 4+1}}=\frac{2}{p \sqrt{p+4}}$

∴ $L[\ t erf(2 \sqrt{t})]=-\frac{d}{d p}\left[\frac{2}{p \sqrt{p+4}}\right]=\frac{3 p+8}{p^2(p+4)^{3 / 2}}$

100. Prove that $L\left\{e^{3 t} \ erf(\sqrt{t})\right\}=\frac{1}{(p-3) \sqrt{p-2}}$

Solution:

⇒ $\ erf(\sqrt{t})=\frac{2}{\sqrt{\pi}} \int_0^{\sqrt{t}} e^{-x^2} d x=\frac{2}{\sqrt{\pi}} \int_0^{\sqrt{t}}\left(1-x^2+\frac{x^4}{2 !}-\frac{x^6}{3 !}+\ldots\right) d x$

= $\frac{2}{\sqrt{\pi}}\left[\frac{\Gamma\left(\frac{3}{2}\right)}{p^{3 / 2}}-\frac{1}{3} \frac{\Gamma\left(\frac{5}{2}\right)}{p^{5 / 2}}+\frac{1}{5 \cdot 2 !} \frac{\Gamma\left(\frac{7}{2}\right)}{p^{7 / 2}}-\frac{1}{7 \cdot 3 !} \frac{\Gamma\left(\frac{9}{2}\right)}{p^{9 / 2}}+\ldots\right]$

= $\frac{2}{\sqrt{\pi}}\left[\frac{\sqrt{\pi}}{2} \cdot \frac{1}{p^{3 / 2}}-\frac{\sqrt{\pi}}{2} \cdot \frac{1}{2} \cdot \frac{1}{p^{5 / 2}}+\frac{\sqrt{\pi}}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{p^{7 / 2}}-\frac{\pi}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{1}{p^{9 / 2}}+\ldots\right]$

= $\frac{1}{p^{3 / 2}}\left[1-\frac{1}{2} \cdot \frac{1}{p}+\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{p^2}-\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{1}{p^3}+\ldots\right]=\frac{1}{p^{3 / 2}}\left(1+\frac{1}{p}\right)^{-1 / 2}=\frac{1}{p \sqrt{p+1}}$

Thus $L\{\ erf(\sqrt{t})\}=\frac{1}{p \sqrt{p+1}}$By first shifting theorem, $L\left\{e^{3 t} \text{erf}(\sqrt{t})\right\}=\frac{1}{(p-3) \sqrt{(p-3)+1}}=\frac{1}{\cdot(p-3) \sqrt{p-2}} \text {. }$

101. If $E_i(t)=\int_0^{\infty} \frac{e^{-u}}{u}$, then evaluate $L\left\{E_i(t)\right\}$.

Solution:

We have $E_i(t)=\int_t^{\infty} \frac{e^{-u}}{u} d u$

∴ $L\left\{E_i(t)\right\}=L\left\{\int_t^{\infty} \frac{e^{-u}}{u} d u\right\}=L\left\{\int_1^{\infty} \frac{e^{-t v}}{v} d v\right\} \text {, putting } u=t v \text { so that } d u=t d v$

= $\int_0^{\infty} e^{-p t}\left\{\int_1^{\infty} \frac{e^{-t v}}{v} d v\right\} d t=\int_1^{\infty} \frac{1}{v}\left\{\int_0^{\infty} e^{-p t} e^{-t v} d t\right\} d v \text {, changing the order of integration }$

= $\int_1^{\infty} \frac{1}{v}\left\{\int_0^{\infty} e^{-(p+v) t} d t\right\} d v=\int_1^{\infty} \frac{1}{v}\left[\frac{e^{-(p+v) t}}{-(p+t)}\right]_0^{\infty} d v=\int_1^{\infty} \frac{1}{v} \frac{1}{p+v} d v=\int_1^{\infty} \frac{1}{p}\left(\frac{1}{v}-\frac{1}{p+v}\right) d v$

= $\frac{1}{p}[\log v-\log (p+v)]=\frac{1}{p}\left[-\log \left(\frac{p}{v}+1\right)\right]_1^{\infty}=\frac{1}{p} \log (p+1) .$

102. Define the Bessel function.

Solution:

Bessel function

The Bessel Function of order n is denoted by $J_n(t)$ and defined as $J_n(t)=\frac{t^n}{2^n \Gamma(n+1)}\left[1-\frac{t^2}{2(2 n+2)}+\frac{t^4}{2 \cdot 4(2 n+2)(2 n+4)}-\ldots\right]$

= $\sum_{r=0}^{\infty} \frac{(-1)^r}{r ! \Gamma(n+r+1)}\left(\frac{t}{2}\right)^{n+2 r}$

103. Prove that $L\left[\ J_0(t)\right]=\frac{1}{\sqrt{1+p^2}}$

Solution:

⇒ $J_n(t)=\sum_{r=0}^{\infty} \frac{(-1) r}{r ! \Gamma(r+n+1)}\left(\frac{t}{2}\right)^{n+2 r}$

∴ $J_0(t)=\sum_{r=0}^{\infty} \frac{(-1)^5}{(r !)^2}\left(\frac{t}{2}\right)^{2 r}=1-\frac{t^2}{2^2}+\frac{t^4}{2^2 4^2}-\frac{t^6}{2^2 4^2 6^2}+\ldots$

∴ $L\left\{J_0(t)\right\}=L\{1\}-\frac{1}{2^2} \dot{L}\left\{t^2\right\}+\frac{1}{2^2 4^2} L\left\{t^4\right\}-\frac{1}{2^2 4^2 6^2} L\left\{t^6\right\}+\ldots$

= $\frac{1}{p}-\frac{1}{2^2} \frac{2 !}{p^3}+\frac{1}{2^2 4^2} \frac{4 !}{p^5}-\frac{!}{2^2 4^2 6^2} \frac{6 !}{p^7}+\ldots$

= $\frac{1}{p}\left[1-\frac{1}{2} \frac{1}{p^2}+\frac{1 \cdot 3}{2 \cdot 4}\left(\frac{1}{p^2}\right)^2-\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\left(\frac{1}{p^2}\right)^3+\ldots\right]$

= $\frac{1}{p}\left(1+\frac{1}{p^2}\right)^{-1 / 2}=\frac{1}{p \sqrt{\left(\frac{1+p^2}{p^2}\right)}}=\frac{1}{\sqrt{1+p^2}}$

104. Prove that $L\left[J_0(a t)\right]=\frac{1}{\sqrt{p^2+a^2}}$⋅

Solution:

Using change of scale of property, $L\left\{J_0(a t)\right\}=\frac{1}{a} f\left(\frac{p}{a}\right)=\frac{1}{a} \frac{1}{\sqrt{1+p^2 / a^2}}=\frac{1}{\sqrt{a^2+p^2}} \text {. }$

105. Prove that $L\left[t J_0(a t)\right]=\frac{p}{\left(p^2+a^2\right)^{3 / 2}}$.

Solution:

L$\left\{t J_0(a t)\right\}=-\frac{d}{d p} L\left\{J_0(a t)\right\}=-\frac{d}{d p}\left[\frac{1}{\sqrt{a^2+p^2}}\right]=\frac{p}{\left(p^2+a^2\right)^{3 / 2}}$

106. Prove that $L\left[e^{-a t} J_0(a t)\right]=\frac{1}{\sqrt{p^2+2 a p+2 a^2}}$.

Solution:

By using the First Shifting theorem, L$\left\{e^{-a t} J_0(a t)\right\}=\frac{1}{\sqrt{(p+a)^2+a^2}}=\frac{1}{\sqrt{p^2+2 a p+2 a^2}}$

107. Prove that $\int_0^{\infty} J_0(t) d t=1$.

Solution:

We have $L\left\{J_0(t)\right\}=\frac{1}{\sqrt{1+p^2}} \Rightarrow \int_0^{\infty} J_0(t) e^{-p t} d t=\frac{1}{\sqrt{1+p^2}}$

Putting p=0, we get $\int_0^{\infty} J_0(t) d t=1 \text {. }$

108. Prove that $L\left\{J_1(t)\right\}=1-\frac{p}{\sqrt{p^2+1}}$.

Solution:

We know that $J_0^{\prime}(t)=-J_1(t)$

From theorem on Laplace Transform of Derivative, we have L$\left\{F^{\prime}(t)\right\}=p L\{F(t)\}-F(0)$

∴ $L\left\{J_1(t)\right\}=L\left\{-J_0{ }^{\prime}(t)\right\}=-L\left\{J_0{ }^{\prime}(t)\right\}=-\left[p L\left\{J_0(t)\right\}-J_0(0)\right]$

= $-\left[p \frac{1}{\sqrt{p^2+1}}-1\right]=1-\frac{p}{\sqrt{p^2+1}}$

109. Prove that $L\left\{t J_1(t)\right\}=\frac{1}{\left(p^2+1\right)^{3 / 2}}$.

Solution:

L$\left\{t J_1(t)\right\}=-\frac{d}{d p} L\left\{J_1(t)\right\}=-\frac{d}{d p}\left[1-\frac{p}{\sqrt{p^2+1}}\right]=\frac{1}{\left(p^2+1\right)^{3 / 2}}$

110. Show that $L\left\{J_0(a \sqrt{t})\right\}=\frac{1}{p} e^{-a^2 / 4 p}$

Solution:

L$\left\{J_0(\sqrt{t})\right\}=L\left\{1-\frac{t}{2^2}+\frac{t^2}{2^2 4^2}-\frac{t^3}{2^2 4^2 6^2}+\ldots\right\}$

= $L\{1\}-\frac{1}{2^2} L\{t\}+\frac{1}{2^2 4^2} L\left\{t^2\right\}-\frac{1}{2^2 4^2 6^2} L\left\{t^3\right\}+\ldots$

= $\frac{1}{p}-\frac{1}{4}\left(\frac{1}{p^2}\right)+\frac{1}{2^2 4^2} \frac{2 !}{p^3}-\frac{1}{2^2 4^2 6^2} \frac{3 !}{p^4}+\ldots=\frac{1}{p}-\frac{1}{4 p^2}+\frac{1}{32 p^3}-\frac{1}{384 p^4}+\ldots$

= $\frac{1}{p}\left[1-\frac{1}{4 p}+\frac{1}{32 p^2}-\frac{1}{384 p^3}+\ldots\right]=\frac{1}{p} e^{-1 / 4 p}$

Using change of scale property, $L\left\{J_0(a \sqrt{t})\right\}=\frac{1}{a^2} \times \frac{a^2}{p} e^{-a^2 / 4 p}=\frac{1}{p} e^{-a^2 / 4 p} \text {. }$

111. Evaluate $L\left\{t^2 u(t-2)\right\}$

Solution:

Given

$L\left\{t^2 u(t-2)\right\}$

The unit step function } u(t-2) is defined by u(t-2)= $\begin{cases}0, & \text { if } t<2 \\ 1, & \text { if } t>2\end{cases}$

∴ $L\{u(t-2)\}=\int_0^{\infty} e^{-p t} u(t-2) d t=\int_0^2 e^{-p t} u(t-2) d t+\int_2^{\infty} e^{-p t} u(t-2) d t$

= $\int_0^2 e^{-p t} 0 d t+\int_2^{\infty} e^{-p t} 1 d t=0+\int_2^{\infty} e^{-p t} d t=\left(\frac{e^{-p t}}{-p}\right)_2^{\infty}=\frac{-1}{p}\left(0-e^{-2 p}\right)=\frac{e^{-2 p}}{p}$

Hence, $L\left\{t^2 u(t-2)\right\}=(-1)^2 \frac{d^2}{d p^2}[L\{u(t-2)\}]=\frac{d^2}{d p^2}\left\{\frac{e^{-2 p}}{p}\right\}$

= $\frac{d}{d p}\left\{\frac{p(-2) e^{-2 p}-e^{-2 p}}{p^2}\right\}=(-1) \frac{d}{d p}\left\{\frac{e^{-2 p}(2 p+1)}{p^2}\right\}$

= $\frac{(-1)}{p^4}\left[p^2\left\{(-2) e^{-2 p}(2 p+1)+2 e^{-2 p}\right\}-2 p e^{-2 p}(2 p+1)\right]$

= $\frac{2 e^{-2 p}}{p^4}\left[p^2(2 p+1-1)+p(2 p+1)\right]=\frac{2 e^{-2 p}}{p^4}\left[2 p^3+p(2 p+1)\right]=\frac{2 e^{-2 p}}{p^3}\left(1+2 p+2 p^2\right)$

112. Find $L\left\{e^{t-3} u(t-3)\right\}$.

Solution:

L$\{u(t-3)\}=\frac{e^{-3 p}}{p} \Rightarrow L\left\{e^{t-3} u(t-3)\right\}=L\left\{e^{-3} e^t u(t-3)\right\}=e^{-3} L\left\{e^t u(t-3)\right\}$

= $e^{-3} \frac{e^{-3(p-1)}}{p-1}=\frac{e^{-3} e^{-3 p} e^3}{p-1}=\frac{e^{-3 p}}{p-1}$

113. If F(t) is a periodic function with period T then show that $L[F(t)]=\frac{1}{1-e^{-p T}} \int_0^T e^{-p t} F(t) d t$.

Solution:

L[F(t)] = $\int_0^{\infty} e^{-p t} F(t) d t=\int_0^T e^{-p t} F(t) d t+\int_T^{2 T} e^{-p t} F(t) d t+\cdots \text { up to } \infty \rightarrow \text { (1) }$

Put t=u+T in the second integral.

Then d t=d u; when t=T, u=0 and t=2 T, u=T.

⇒ $\int_T^{2 T} e^{-p t} F(t) d t=\int_0^T e^{-p(u+T)} F(u+T) d u$

= $\int_0^T e^{-p T} \cdot e^{-p u} F(u) d u=e^{-p T} \int_0^T e^{-p u} F(u) d u$

By changing the dummy variable u to ‘ t’, we get $\int_T^{2 T} e^{-p t} F(t) d t=e^{-p T} \int_0^T e^{-p t} F(t) d t \rightarrow \text { (2) }$

Similarly in the third integral, butting t=u+2 T, we have $\int_{2 T}^{3 T} e^{-p t} F(t) d t=\int_0^T e^{-p(u+2 T)} F(u+2 T) d u$

= $\int_0^T e^{-p u} \cdot e^{-2 p T} F(u) d u$

= $e^{-2 p T} \int_0^T e^{-p t} F(t) d t \rightarrow$(3)

Repeating the same substitution and replacing the integrals by equations (2). (3) and so on, we get $L[F(t)]=\int_0^T e^{-p} F(t) d t+e^{-p T} \int_0^T e^{-p t} F(t) d t+e^{-2 p T} \int_0^T e^{-p t} F(t) d t$

+ $e^{-3 p T} \int_0^T e^{-p t} F(t) d t+\cdots \text { to } \infty$

= $\left(1+e^{-p T}+e^{-2 p T}+e^{-3 p T}+\cdots \infty\right) \int_0^T e^{-p t} F(t) d t=\frac{1}{1-e^{-p T}} \int_0^T e^{-p t} \cdot F(t) d t$

114. Find the Laplace transform of the triangular wave of period ‘ 2a ‘ given by $F(t)=\left\{\begin{array}{cl}
t, & 0 2 a-t, & a \end{array}\right.$

Solution:

In this case, the period is T=2 a.

L$[F(t)]=\frac{1}{1-e^{-p T}} \int_0^T e^{-p t} F(t) d t=\frac{1}{1-e^{-2 a p}}\left[\int_0^a e^{-p t} F(t) d t+\int_a^{2 a} e^{-p t} F(t) d t\right]$

= $\frac{1}{1-e^{-2 a p}}\left[\int_0^a e^{-p t} \cdot t d t+\int_a^{2 a} e^{-p t}(2 a-t) d t\right]$

= $\frac{1}{1-e^{-2 a p}}\left[t \cdot\left(\frac{e^{-p t}}{-p}\right)-1\left(\frac{e^{-p t}}{p^2}\right)_0^a+\left\{(2 a-t)\left(\frac{e^{-p t}}{-p^p}\right)-(-1)\left(\frac{e^{-p t}}{-p^2}\right)\right\}_a^{2 a}\right]$

= $\frac{1}{1-e^{-2 a p}}\left[\frac{-a e^{-p a}}{p}-\frac{e^{-p a}}{p^2}+\frac{1}{p^2}+\frac{e^{-2 a p}}{p^2}+\frac{a \cdot e^{-a p}}{p}-\frac{e^{-a p}}{p^2}\right]$

= $\frac{1}{1-e^{-2 a p}}\left[\frac{1-2 e^{-a p}+e^{-2 a p}}{p^2}\right]=\frac{\left(1-e^{-a p}\right)^2}{p^2\left(1+e^{-a p}\right)\left(1-e^{-a p}\right)}$

= $\left[\frac{1-e^{-a p}}{p^2\left(1+e^{-a p}\right)}\right]=\frac{1}{p^2} \cdot \frac{\left(1-e^{-a p}\right) e^{+\frac{a p}{2}}}{\left(1+e^{-a p}\right) e^{+\frac{a p}{2}}}=\frac{1}{p^2} \cdot \frac{e^{a p / 2}-e^{-a p / 2}}{e^{a p / 2}+e^{-a p / 2}}=\frac{1}{p^2} \tanh \left(\frac{a p}{2}\right)$.

115. Find the Laplace transform of the square wave function of the period ‘ a ‘ defined as \(F(t)= \begin{cases}1, & 0

Solution:

In this case, the period T=a

L$[F(t)]=\frac{1}{1-e^{-a p}} \int_0^a e^{-p t} F(t) d t=\frac{1}{1-e^{-a p}}\left[\int_0^{a / 2} e^{-p t} F(t) d t+\int_{a / 2}^a e^{-p t} F(t) d t\right]$

= $\frac{1}{1-e^{-a p}}\left[\int_0^{a / 2} e^{-p t} \cdot 1 \cdot d t+\int_{a / 2}^a e^{-p t} \cdot(-1) d t\right]=\frac{1}{1-e^{-a p}}\left[\frac{e^{-p t}}{-p}\right]_0^{a / 2}-\left[\frac{e^{-p t}}{-p}\right]_{a / 2}^a$

= $\frac{1}{1-e^{-a p}}\left[\frac{e^{-p a / 2}}{-p}+\frac{1}{p}+\frac{e^{-a p}}{p}-\frac{e^{-p a / 2}}{p}\right]=\frac{1}{1-e^{-a p}} \cdot \frac{1-2 e^{-\frac{a p}{2}}+e^{-a p}}{p}=\frac{1}{p}\left[\frac{1+2 e^{-\frac{a p}{2}}+e^{-a p}}{1-e^{-a p}}\right]$

116. Find the Laplace transform of the output of a full sine wave rectifier given as : $F(t)=\left\{\begin{array}{cc}
\sin w t, & 0 0, & t>\pi / w
\end{array}\right.$

Solution:

The period T = $\frac{\pi}{w}$

L$[F(t)]=\frac{1}{1-e^{-p \pi / w}} \int_0^{\pi / w} e^{-p t} F(t) d t=\frac{1}{1-e^{-p \pi / w}} \int_0^{\pi / w} e^{-p t} \cdot E \sin w t \cdot d t$

= $\frac{E}{1-e^{-p \pi / w}}\left[\frac{e^{-p t}}{p^2+w^2}(-p \sin w t-w \cos w t)\right]_0^{\pi / w}$

= $\frac{E}{\left(1-e^{-p \pi / w}\right)\left(p^2+w^2\right)}\left[e^{-p \pi / w} \cdot w+w\right]=\frac{E w}{\left(1-e^{-p \pi / w}\right)\left(p^2+w^2\right)}\left(1+e^{-p \pi / w}\right)$

= $\frac{E w}{\left(p^2+w^2\right)} \cdot\left[\frac{e^{p \pi / 2 w}+e^{-p \pi / 2 w}}{e^{p \pi / 2 w}-e^{-p \pi / 2 w}}\right]=\frac{E w}{p^2+w^2} \cosh (p \pi / 2 w)$

117. Find the Laplace transform of $F(t)=e^{-t}, 0

Solution:

In this case, the period T=2

∴ \(L[F(t)]=\frac{1}{1-e^{-2 p}} \int_0^2 e^{-p t} F(t) d t=\frac{1}{1-e^{-2 p}} \int_0^2 e^{-p t} e^{-t} d t=\frac{1}{1-e^{-2 p}} \int_0^2 e^{-(p+1) t} d t$

= $\frac{1}{1-e^{-2 p}}\left[\frac{e^{-(p+1) t}}{-(p+1)}\right]_0^2=\frac{1}{1-e^{-2 p}}\left[\frac{e^{-2(p+1)}}{-(p+1)}+\frac{1}{p+1}\right]=\frac{1}{1-e^{-2 p}}\left[1-\frac{e^{-2(p+1)}}{p+1}\right]$

118. Find the Laplace transform of the function F(t)=1,06.

Solution:

F(t) = $1[u(t-0)-u(t-2)]+2[u(t-2)-u(t-4)]+3[u[t-4]-u(t-6)]$

= $u(t-0)+u(t-2)+u(t-4)-3 u(t-6)$

L$\{F(t)\}=L\{u(t-0)\}+L\{u(t-2)\}+L\{u(t-4)\}-3 L\{u(t-6)\}$

= $\frac{1}{p}+\frac{e^{-2 p}}{p}+\frac{e^{-4 p}}{p}-\frac{3 e^{-6 p}}{p}$

119. Find the Laplace Transform of the square-wave function of period 2a defined as $F(t)=\left\{\begin{array}{c}
k \text { when } 0 -k \text { when } a \end{array}\right.$

Solution:

Since F(t) is a periodic function with period T=2 a

Lf(t)$=\frac{1}{1-e^{-2 a p}} \int_0^{2 a} e^{-p t} F(t) d t=\frac{1}{1-e^{-2 a p}}\left[\int_0^a k e^{-p t} d t+\int_a^{2 a}(-k) e^{-p t} d t\right]$

= $\frac{1}{1-e^{-2 a p}}\left[k\left(\frac{e^{-p t}}{-p}\right)_0^a+k\left(\frac{e^{-p}}{p}\right)_0^{2 a}\right]=\frac{1}{1-e^{-2 a p}} \frac{k}{p}\left(1-e^{-a p}\right)^2=\frac{1}{1-\left(e^{-a p}\right)^2} \frac{k}{p}\left(1-e^{-a p}\right)^2$

= $\frac{1}{\left(1+e^{-a p}\right)\left(1-e^{-a p}\right)} \frac{k}{p}\left(1-e^{-a p}\right)^2=\frac{k\left(1-e^{-a p}\right)}{p\left(1+e^{-a p}\right)}$

120. Find L{F(t)}, where F(t) is a periodic function of period 2π and it is given by $F(t)=\left\{\begin{array}{cc}
\sin t, & 0 0, & \pi \end{array}\right.$

Solution:

Since F(t) s a periodic function with period $2 \pi \text {, }$

∴ $L\{F(t)\}=\frac{1}{1-e^{-2 \pi p}} \int_0^2 \pi e^{-p t} F(t) d t=\frac{1}{1-e^{-2 \pi p}}\left[\int_0^\pi e^{-p t} F(t) d t+\int_\pi^{2 \pi} e^{-p t} F(t) d t\right]$

= $\frac{1}{1-e^{-2 \pi p}}\left[\int_0^\pi e^{-p t} \sin t d t+\int_\pi^{2 \pi} e^{-p t} \cdot 0 d t\right]=\frac{1}{1-e^{-2 \pi p}} \int_0^\pi e^{-p t} \sin t d t$

= $\frac{1}{1-e^{-2 \pi p}}\left\{\frac{e^{-p t}}{p^2+1}(-p \sin t-\cos t)\right\}_0^\pi$ $\left[\int e^{a x} \sin b x=\frac{e^{a x}}{a^2+b^2}(a \sin b x-b \cos b x)\right]$

= $\frac{1}{1-e^{-2 \pi \dot{p}}}\left\{\frac{e^{-\pi p}}{p^2+1}(1)-\frac{1}{p^2+1}(-1)\right\}=\frac{e^{-\pi p}+1}{\left(p^2+1\right)\left(1-e^{-2 \pi p}\right)}=\frac{1+e^{-\pi p}}{\left(p^2+1\right)\left(1+e^{-\pi p}\right)\left(1-e^{-\pi p}\right)}$

Hence, $L\left\{F^{-}(t)\right\}=\frac{1}{\left(p^2+1\right)\left(1-e^{-\pi p}\right)} \text {. }$

The post Laplace Transform – II Exercise Solved Problems appeared first on Answer Key for Math.

Laplace Transform – 1 Exercise Solved Problems

Kumar — Wed, 23 Aug 2023 11:15:57 +0000

Laplace Transform – 1 Exercise 1

Laplace Transform Solved Examples Step-By-Step

1. Define Integral transform and Laplace transform.

Solution:

Integral Transform: Let K(p, t) be a function of two variables p and t, where p is a parameter real or complex independent of t. The function f(p) defined by the integral $\int_{-\infty}^{\infty} K(p, t) F(t) d t$ is called the integral transform of the function F(t) and is denoted by T[F(t)].

The function K(p, t) is called the kernel of the transformation.

Laplace transform: Let a function F(t) be continuous and defined for all positive values of t. The Laplace transformation or Laplace transform of F(t) is defined by $L[F(t)]=\int_0^{\infty} e^{-p t}. F(t) d t$ where ‘ p’ is a parameter, and the transform is denoted by f(p) or $\bar{F}(p)$.

2. Define piecewise continuous function.

Solution:

Piecewise continuous function

A function F(t) is said to be piecewise continuous or sectionally continuous in a closed interval [a, b] if it is defined in that interval and is such that the interval can be divided into a finite number of sub-intervals in each of which F(t) is continuous and has finite left and right-hand limits.

3. Define a function of exponential order a.

Solution:

A function of exponential order a

A function F(t) is said to be of exponential order a as t→∞ if there exists a positive real number M, a number a, and a finite number $t_0$ such that $\left|e^{-a t} F(t)\right|
Exercise Problems On Laplace Transform With Solutions

4. Define a function of class A.

Solution:

A function of class A

A function F(t) is said to be of class A if F(t) is piecewise continuous on every finite interval in the range t≥0 and is of exponential order as t→∞

5. If a function F(t) is piecewise continuous on every finite interval in the range t≥0 and satisfies \(|F(t)|a.

Solution:

Given

If a function F(t) is piecewise continuous on every finite interval in the range t≥0 and satisfies \(|F(t)|
L[F(t)] = \(\int_0^{\infty} e^{-p t} F(t) d t=\int_0^{t_0} e^{-p t} F(t) d t+\int_{t_0}^{\infty} e^{-p t} F(t) d t \rightarrow \text { (1) }$

Since F(t) is piecewise continuous on $\left[0, t_0\right]$, the integral $\int_0^{t_0} e^{-p t} F(t) d t$ exists.

∴ $\int_{t_0}^{\infty} e^{-p t} F(t) d t|\leq \int_{t_0}^{\infty}| e^{-p t} F(t) d t$l,< $\int_{t_0}^{\infty} e^{-p t} M e^{a t} d t$

= $\int_{t_0}^{\infty} M e^{-(p-a) t} d t=M [\frac{e^{-(p-a) t}}{-(p-a)}]_{t_0}^{\infty}$

= $\frac{M e^{-(p-a) t_0}}{p-a}$ for p>a .

But $\frac{M e^{-(p-a) t_0}}{p-a}$ can be made as small as we require by taking $t_0$ sufficiently large. Thus L[F(t)] exists for all p>a.

6. Show that $L\left(\frac{1}{\sqrt{t}}\right)$ exists for p≥0; even if $\frac{1}{\sqrt{t}}$t is not piecewise continuous in the range t≥0. Does this function belong to a function of class A? Justify your answer.

Solution:

The function $F(t)=\frac{1}{\sqrt{t}}$ is not piecewise continuous on every finite interval in the range t≥0. But

L[F(t)] = $\int_0^{\infty} e^{-p t} F(t) d t=\int_0^{\infty} e^{-p t} \frac{1}{\sqrt{t}} d t=\int_0^{\infty} e^{-x^2} d x$ where $\sqrt{p t}=x$

= $\frac{2}{\sqrt{p}} \frac{\sqrt{\pi}}{2}=\frac{\sqrt{\pi}}{\sqrt{p}}, p>0$ ∴L[F(t)] exists for p>0.

7. If the Laplace transforms of two functions $F_1(t), F_2(t)$) exists then prove that $L\left[F_1(t)+F_2(t)\right]=L\left[F_1(t)\right]+L\left[F_2(t)\right]$.

Solution:

L$\left[F_1(t)+F_2(t)\right]=\int_0^{\infty}\left[F_1(t)+F_2(t)\right] e^{-p t} \cdot d t$

= $\int_0^{\infty} F_1(t) e^{-p t} d t+\int_0^{\infty} F_2(t) e^{-p t} d t$

= $L\left[F_1(t)\right]+L\left[F_2(t)\right]$

8. If $k_1, k_2$ are two constants and the Laplace transforms of the functions $F_1(t), F_2(t)$ exists then prove that $L\left[k_1 F_1(t)+k_2 F_2(t)\right]=k_1 L\left[F_1(t)\right]+k_2 L\left[F_2(t)\right]$.

Solution:

Given

If $k_1, k_2$ are two constants and the Laplace transforms of the functions $F_1(t), F_2(t)$

L$\left[k_1 F_1(t)+k_2 F_2(t)\right]=\int_0^{\infty}\left[k_1 F_1(t)+k_2 F_2(t)\right] e^{-p t} \cdot d t$

= $\int_0^{\infty} k_1 F_1(t) e^{-p t} d t+\int_0^{\infty} k_2 F_2(t) e^{-p t} d t$

= $k_1 \int_0^{\infty} F_1(t) e^{-p t} d t+k_2 \int_0^{\infty} F_2(t) e^{-p t} d t$

= $k_{\mathrm{i}} L\left[F_1(t)\right]+k_2 L\left[F_2(t)\right]$

9. Show that the Laplace transform of the unit function F(t)=1 is 1/p.

Solution:

L[F(t)] = $\int_0^{\infty} 1 \cdot e^{-p t} d t=\left[\frac{e^{-p t}}{-p}\right]_0^{\infty}=\frac{1}{p}, \text { provided } p>0$

10. Show that Laplace transforms of $F(t)=t^n,-1

Solution:

Here F(t) \(\rightarrow \infty$ as $t \rightarrow 0$ for $t \geq 0$, i.e. the function is not piecewise continuous on every finite interval in the range $t \geq 0$.

We have $\underset{t \rightarrow \infty}{\text{Lt}}\left\{e^{a t} F(t)\right\}=\underset{t \rightarrow \infty}{\text{Lt}}\left(\frac{t^n}{e^{a t}}\right)=\underset{t \rightarrow \infty}{\text{Lt}} \frac{1}{t^n e^{a t}}=\underset{t \rightarrow \infty}{L t} \frac{1}{t^m e^{a t}}$ where 00.

∴ F(t) = $t^n$ is of exponential order.

Since $F(t)=t^n$ is not piecewise continuous over every finite interval in the range $t \geq 0$, hence it is not a function of class A. But $t^n$ is integrable from o to any positive number $t_0$.

Now $L\{F(t)\}=\int_0^{\infty} e^{-p t} F(t) d t=\int_0^{\infty} e^{-p t} t^n d t=\int_0^{\infty} e^{-x}\left(\frac{x}{p}\right)^n \frac{1}{p} d x$

putting pt=x so that $d t=\frac{1}{p} d x$ and taking p>0

= $\frac{1}{p^{n+1}} \int_0^{\infty} e^{-x} x^{(n+1)-1} d x=\frac{\Gamma(n+1)}{p^{n+1}}$, if p>0 and n+1>0 i.e. n>-1.

Hence the Laplace transform of $t^n, 0>n>-1$ exists, although it is not a function of class A.

Laplace Transform Practice Exercises With Answers

11. If n is a positive integer, then show that $L\left(t^n\right)=\frac{n !}{p^{n+1}}, p>0$

Solution:

Given

n is a positive integer

L$\left(t^n\right)=\int_0^{\infty} t^n e^{-p t} d t=\left[\frac{t^n \cdot e^{-p t}}{-p}\right]_0^{\infty}-\int_0^{\infty} \frac{e^{-p t}}{-p} \cdot n t^{n-1} d t=\frac{n}{p} \int_0^{\infty} e^{-p t} t^{n-1} d t=\frac{n}{p} L\left(t^{n-1}\right)$

Similarly $L\left(t^{n-1}\right)=\frac{n-1}{p} \cdot L\left(t^{n-2}\right)$ and so on.

Hence $L\left(t^n\right)=\frac{n}{p} \cdot \frac{n-1}{p} \cdot \frac{n-2}{p} \cdots \frac{2}{p} \cdot \frac{1}{p} \cdot L\left(t^{n-n}\right)=\frac{n !}{p^n} L\left(t^{\circ}\right)=\frac{n !}{p^n} L(1)=\frac{n !}{p^n} \cdot \frac{1}{p}=\frac{n !}{p^{n+1}}$

12. If ‘ a ‘ is a constant, then show that $L\left(e^{a t}\right)=\frac{1}{p-a}$

Solution:

L$\left(e^{a t}\right)=\int_0^{\infty} e^{-p t} e^{a t} d t=\int_0^{\infty} e^{-(p-a) t} d t=-\left[\left.\frac{e^{-(p-a) t}}{(p-a)}\right|_0 ^{\infty}=\frac{1}{p-a}, \text { provided } p>a\right.$

13. Find the Laplace transform of $e^{-a t}$ by using the basic definition.

Solution:

L$\left(e^{-a t}\right)=\int_0^{\infty} e^{-p t} e^{-a t} d t=\int_0^{\infty} e^{-(p+a) t} d t=-\left[\left.\frac{e^{-(p+a) t}}{(p+a)}\right|_0 ^{\infty}=\frac{1}{p+a}, \text { provided } p>a\right.$

14. If ‘ a ‘ is a constant, then show that $L(\sin a t)=\frac{a}{p^2+a^2}$.

Solution:

L$(\sin a t)=\int_0^{\infty} e^{-p t} \sin a t d t=\left[\frac{e^{-p t}}{p^2+a^2}(-p \sin a t-a \cos a t)\right]_0^{\infty}=\frac{a}{p^2+a^2}$

15. If ‘ a ‘ is a constant, then show that $L(\cos a t)=\frac{p}{p^2+a^2}$.

Solution:

L$(\cos a t)=\int_0^{\infty} e^{-p x} \cos a t d t=\left[\frac{e^{-p t}}{p^2+a^2}(-p \cos a t+a \sin a t)\right]_0^{\infty}=\frac{p}{p^2+a^2}$

16. If ‘ a ‘ is a constant, then show that $L(\sinh a t)=\frac{a}{p^2-a^2}$.

Solution:

L$(\sinh a t)=\int_0^{\infty} e^{-\not t} \sinh a t d t=\int_0^{\infty} e^{-p t}\left[\frac{e^{a t}-e^{-a t}}{2}\right] d t=\frac{1}{2}\left[\int_0^{\infty} e^{-p t} e^{a t} d t-\int_0^{\infty} e^{-p t} e^{-a t} d t\right]$

= $\frac{1}{2}\left[\frac{1}{p-a}-\frac{1}{p+a}\right]=\frac{a}{p^2-a^2}$

Applications Of Laplace Transform With Solved Examples

17. If ‘ a ‘ is a constant, then show that $L(\cosh a t)=\frac{p}{p^2-a^2}$.

Solution:

L$(\cosh a t)=\int_0^{\infty} e^{-p t} \cosh a t d t=\int_0^{\infty} e^{-p t}\left[\frac{e^{a t}+e^{-a t}}{2}\right] d t=\frac{1}{2}\left[\int_0^{\infty} e^{-p t} e^{a t} d t+\int_0^{\infty} e^{-p t} e^{-a t} d t\right]$

= $\frac{1}{2}\left[\frac{1}{p-a}+\frac{1}{p+a}\right]=\frac{p}{p^2-a^2}$

18. Define a gamma function.

Solution:

Gamma function

If n>0 then the Gamma function Γ(n) is defined as $\Gamma(n)=\int_0^{\infty} e^{-x} x^{n-1} d x$.

19. Define Heaviside’s unit step function.

Solution:

Heaviside’s unit step function

The Heavisides unit step function is denoted by U(t-a) or u(t-a) or H(t-a) and is defined as H(t-a)=0 if ta.

20. Show that $L[H(t-a)]=\frac{e^{-a p}}{p}$.

Solution:

L$\left[H(t-a)=\int_0^{\infty} e^{-p t} H(t-a) d t=\int_0^a e^{-p t} H(t-a) d t+\int_a^{\infty} e^{-p t} H(t-a) d t\right.$

= $\int_0^a e^{-p t}(0) d t+\int_a^{\infty} e^{-p t} 1 d t=\int_a^{\infty} e^{-p t} d t=\left[\frac{e^{-p t}}{-p}\right]_a^{\infty}=\frac{e^{-a p}}{p} .$

21. Define Unit Impulse Function.

Solution:

Unit Impulse Function

The Unit Impulse Function or Dirac Delta Function at a is denoted by δ(t-a) and is defined as $\delta(t-a)=\underset{h \rightarrow 0}{L t} I(h, t-a)$, where $I(h, t-a)=\frac{1}{h}$ if aa+h.

22. Show that $L[\delta(t-a)]=e^{-a p}$.

Solution:

We know that I(h, t-a) = $\frac{1}{h}[H(t-a)-H(t-a-h)]$

L$[I(h, t-a)]=\frac{1}{h}\{L[H(t-a)]-L[H(t-a-h)]\}=\frac{1}{h}\left[\frac{e^{-a p}}{p}-\frac{e^{-(a+h) p}}{p}\right]=\frac{e^{-a p}}{p}\left[1-e^{-h p}\right]$

L$\left[\delta(t-a)=\underset{h \rightarrow 0}{L t} L[I(h, t-a)]={ }_{h \rightarrow 0}^{L t} \frac{e^{-a p}}{p h}\left[1-e^{-h p}\right]=e^{-a p}\right.$

23. Prove that the function $F(t)=t^2$ is of exponential order 3 .

Solution:

⇒ $\ Lim_{t \rightarrow \infty} e^{-3 t} F(t)=\ Lim_{t \rightarrow \infty} e^{-3 t} \cdot t^2=\ Lim_{t \rightarrow \infty} \frac{t^2}{3 t}\left(\text { form } \frac{\infty}{\infty}\right)=\ Lim_{t \rightarrow \infty} \frac{2 t}{3 e^{3 t}}$

= $\ Lim_{t \rightarrow \infty} \frac{2}{9 e^{3 t}}=\frac{2}{\infty}=0$

Hence $t^2$ is exponential order 3.

Laplace Transform Examples And Detailed Solutions

24. Prove that the Laplace transform of $e^{t^2}$ does not exist.

Solution:

Let $F(t)=e^{t^2} \text {. Then } \ Lim e^{-a t} F(t)=\ Lim_{t \rightarrow \infty} e^{-a t} e^{t^2}=\ Lim_{t \rightarrow \infty} e^{\left(t^{t^2}-a t\right)}$

= $\infty \text { i.e. } f(t)=e^{t^2}$ is not of exponential order. Hence its Laplace transform does not exist.

25. Find the Laplace transform of $e^{2 t}+4 t^3-2 \sin 3 t+3 \cos 3 t$

Solution:

L$\left\{e^{2 t}+4 t^3-2 \sin 3 t+3 \cos 3 t\right\}$

= $\frac{1}{p-2}+4 \cdot \frac{3 !}{p^4}-2 \cdot \frac{3}{p^2+9}+3 \cdot \frac{p}{p^2+9}=\frac{1}{p-2}+\frac{24}{p^4}-\frac{6}{p^2+9}+\frac{3 p}{p^2+9} .$

Step-By-Step Guide To Solving Laplace Transform Exercises

26. Find $L\left(t^3+2 t^2-4 t+6\right)$.

Solution:

L$\left(t^3+2 t^2-4 t+6\right)=L\left(t^3\right)+2 L\left(t^2\right)-4 L(t)+6 L(1)$

= $\frac{3 !}{p^4}+2 \cdot \frac{2 !}{p^3}-4 \cdot \frac{1}{p^2}+6 \cdot \frac{1}{p}=\frac{6}{p^4}+\frac{4}{p^3}-\frac{4}{p^2}+\frac{6}{p}$

27. Find the Laplace transforms of $(t-2)^3$.

Solution:

L$\left\{(t-2)^3\right\}=L\left\{t^3-6 t^2+12 t-8\right\}=L\left\{t^3\right\}-6 L\left\{t^2\right\}+12 L\{t\}-8 L\{1\}$

= $\frac{3 !}{p^4}-6 \frac{2 !}{p^3}+12 \frac{1 !}{p^2}-8 \frac{1}{p}=\frac{6}{p^4}-\frac{12}{p^3}+\frac{12}{p^2}-\frac{8}{p}$

28. Find the Laplace transform of $\left(t^2+1\right)^2$.

Solution:

L$\left\{\left(t^2+1\right)^2\right\}=L\left\{t^4+2 t^2+1\right\}=L\left\{t^4\right\}+2 L\left\{t^2\right\}+L\{1\}$

= $\frac{4 !}{p^5}+2 \cdot \frac{2 !}{p^3}+\frac{1}{p}=\frac{24}{p^5}+\frac{4}{p^3}+\frac{1}{p}=\frac{1}{p^5}\left(p^4+4 p^2+24\right), p>0 .$

29. Find $L\left\{\left(\sqrt{t}+\frac{1}{\sqrt{t}}\right)^3\right\}$.

Solution:

L$\left\{\left(\sqrt{t}+\frac{1}{\sqrt{t}}\right)^3\right\}=L\left\{t^{3 / 2}+3 t^{1 / 2}+3 t^{-1 / 2}+t^{-3 / 2}\right\}$

= $\frac{\Gamma(5 / 2)}{p^{5 / 2}}+3 \frac{\Gamma(3 / 2)}{p^{3 / 2}}+3 \frac{\Gamma(1 / 2)}{p^{1 / 2}}+\frac{\Gamma(-1 / 2)}{p^{-1 / 2}}$

= $\left(\frac{3}{2}\right)\left(\frac{1}{2}\right) \frac{\sqrt{\pi}}{p^{5 / 2}}+(3)\left(\frac{1}{2}\right) \frac{\sqrt{\pi}}{p^{3 / 2}}+(3) \frac{\sqrt{\pi}}{p^{1 / 2}}-\frac{2 \sqrt{\pi}}{p^{-1 / 2}}=\frac{\sqrt{\pi}}{4}\left[\frac{3}{p^{5 / 2}}+\frac{6}{p^{3 / 2}}+\frac{12}{p^{1 / 2}}-8 \sqrt{p}\right] .$

30. Find the Laplace transform of $\frac{e^{-a t}-1}{a}$

Solution:

L$\left\{\frac{e^{-a t}-1}{a}\right\}=\frac{1}{a} L\left\{e^{-a t}-1\right\}=\frac{1}{a}\left[L\left\{e^{-a t}\right\}-L\{1\}\right]=\frac{1}{a}\left[\frac{1}{p+a}-\frac{1}{p}\right]=-\frac{1}{p(p+a)}$

31. Find the Laplace transform of $e^{-t}-t^4+3 \sin (\sqrt{3} t)+4 \cos t$.

Solution:

L$\left\{e^{-t}-t^4+3 \sin \sqrt{3} t+4 \cos t\right\}=\frac{1}{p+1}-\frac{4 !}{p^5}+3 \cdot \frac{\sqrt{3}}{p^2+3}+4 \cdot \frac{p}{p^2+1} .$

32. Find $L\left\{7 e^{2 t}+9 e^{-2 t}+5 \cos t+7 t^3+5 \sin 3 t+2\right\}$.

Solution:

L$\left\{7 e^{2 t}+9 e^{-2 t}+5 \cos t+7 t^3+5 \sin 3 t+2\right\}$

= $7 L\left\{e^{2 t}\right\}+9 L\left\{e^{-2 t}\right\}+5 L\{\cos t\}+7 L\left\{t^3\right\}+5 L\{\sin 3 t\}+2 L\{1\}$

= $\frac{7}{p-2}+\frac{9}{p+2}+\frac{5 p}{p^2+1}+\frac{21}{p^4}+\frac{15}{p^2+9}+\frac{2}{p}$

Solved Problems On Inverse Laplace Transform”

33. Find $L\left\{e^{-t}+t^4+3 \sin 4 t-3 \sinh 2 t\right\}$.

Solution:

L$\left\{e^{-t}+t^4+3 \sin 4 t-3 \sinh 2 t\right\}=L\left\{e^{-t}\right\}+L\left\{t^4\right\}+3 L\{\sin 4 t\}$

–$3 L\{\sinh 2 t\}$ $\frac{1}{p+1}+\frac{4 !}{p^5}+3\left(\frac{4}{p^2+16}\right)-3\left(\frac{2}{p^2-4}\right)=\frac{1}{p+1}+\frac{24}{p^5}+\frac{12}{p^2+16}-\frac{6}{p^2-4}$

34. Find $L\left\{3 t^4-2 t^3+4 e^{-3 t}-2 \sin 5 t+3 \cos 2 t\right\}$.

Solution:

L$\left\{3 t^4-2 t^3+4 e^{-3 t}-2 \sin 5 t+3 \cos 2 t\right\}$

3$L\left\{t^4\right\}-2 L\left\{t^3\right\}+4 L\left\{e^{-3 t}\right\}-2 L\{\sin 5 t\}+3 L\{\cos 2 t\}$

= $\frac{3(4 !)}{p^5}-\frac{2(3 !)}{p^4}+\frac{4}{p+3}-2 \frac{5}{p^2+5^2}+\frac{3 p}{p^2+2^2}$

= $\frac{72}{p^5}-\frac{12}{p^4}+\frac{4}{p+3}-\frac{10}{p^2+25}+\frac{3 p}{p^2+4}$

35. Find $L\left\{\left(5 e^{2 t}-3\right)^2\right\}$

Solution:

L$\left\{\left(5 e^{2 t}-3\right)^2\right\}=L\left\{25 e^{4 t}+9-30 e^{2 t}\right\}=25 L\left\{e^{4 t}\right\}-30 L\left\{e^{2 t}\right\}+9 L\{1\}$

= $\frac{25}{p-4}-\frac{30}{p-2}+\frac{9}{p}$.

36. Evaluate L{3 sin⁡ 2t – 5 cos⁡ 2t}.

Solution:

L$\{3 \sin 2 t-5 \cos 2 t\}=3 L\{\sin 2 t\}-5 L\{\cos 2 t\}$

= $3\left(\frac{2}{p^2+2^2}\right)-5\left(\frac{p}{p^2+2^2}\right)=\frac{6-5 p}{p^2+4}$

37. Find the Laplace transform of sin ⁡2t cos⁡ t.

Solution:

L$\{\sin 2 t \cos t\}=L\left\{\frac{1}{2}(\sin 3 t+\sin t)\right\}=\frac{1}{2}[L\{\sin 3 t\}+L\{\sin t\}]$

= $\frac{1}{2}\left[\frac{3}{p^2+9}+\frac{1}{p^2+1}\right]=\frac{2\left(p^2+3\right)}{\left(p^2+1\right)\left(p^2+9\right)}$

38. Find L(sin⁡ 2t cos⁡ 3t).

Solution:

L$(\sin 2 t \cos 3 t)=L\left(\frac{\sin 5 t-\sin t}{2}\right)=\frac{1}{2}[L(\sin 5 t)-L(\sin t)]$

= $\frac{1}{2}\left[\frac{5}{p^2+25}-\frac{1}{p^2+1}\right]$

= $\frac{1}{2}\left[\frac{5 p^2+5-p^2-25}{\left(p^2+25\right)\left(p^2+1\right)}\right]=\frac{1}{2}\left[\frac{4 p^2-20}{\left(p^2+25\right)\left(p^2+1\right)}\right]=\frac{2 p^2-10}{\left(p^2+25\right)\left(p^2+1\right)}$

Laplace Transform In Differential Equations Solved Problems

39. Find L[sin ⁡3t cos⁡ 2t].

Solution:

⇒ $\sin 3 t \cos 2 t=\frac{1}{2}(\sin 5 t+\sin t)$

L$\{\sin 3 t \cos 2 t\}=\frac{1}{2} L\{\sin 5 t+\sin t\}=\frac{1}{2}[L\{\sin 5 t\}+L\{\sin t\}]$

= $\frac{1}{2}\left[\frac{5}{p^2+25}-\frac{1}{p^2+1}\right]=\frac{1}{2}\left[\frac{5 p^2+5-p^2-25}{\left(p^2+25\right)\left(p^2+1\right)}\right]=\frac{1}{2}\left[\frac{4 p^2-20}{\left(p^2+25\right)\left(p^2+1\right)}\right]$

= $\frac{2\left(p^2-5\right)}{\left(p^2+25\right)\left(p^2+1\right)}$

40. Find the Laplace transform of cos⁡ 3t sin⁡ 5t.

Solution:

L$\{\sin 5 t \cos 3 t\}=L\left\{\frac{1}{2}(\sin 8 t+\sin 2 t)\right\}=\frac{1}{2}[L\{\sin 8 t\}+L\{\sin 2 t\}]$

= $\frac{1}{2}[\dot{L}\{\sin 5 t\}+L\{\sin t\}]=\frac{1}{2}\left[\frac{8}{p^2+64}+\frac{2}{p^2+4}\right]$

= $\frac{1}{2}\left[\frac{8 p^2+32+2 p^2+128}{\left(p^2+64\right)\left(p^2+4\right)}\right]$

= $\frac{1}{2}\left[\frac{10 p^2+160}{\left(p^2+64\right)\left(p^2+4\right)}\right]=\frac{5 p^2+80}{\left(p^2+64\right)\left(p^2+4\right)}$

41. Find the Laplace transform of cos⁡ t cos⁡ 2t cos⁡ 3t.

Solution:

L$\{\cos t \cos 2 t \cos 3 t\}=L\left\{\frac{\cos t}{2}(\cos 5 t+\cos t)\right\}$

= $L\left\{\frac{1}{4}\left[2 \cos 5 t \cos t+2 \cos ^2 t\right]\right\}$

= $\frac{1}{4} L\{\cos 6 t+\cos 4 t+\cos 2 t+1\}=\frac{1}{4}[L\{\cos 6 t\}$ $+L\{\cos 4 t\}+L\{\cos 2 t\}+L\{1\}]$

= $\frac{1}{4}\left[\frac{p}{p^2+6^2}+\frac{p}{p^2+4^2}+\frac{p}{p^2+2^2}+\frac{1}{p}\right]=\frac{1}{4}\left[\frac{1}{p}+\frac{p}{p^2+4}+\frac{p}{p^2+16}+\frac{p}{p^2+36}\right]$

42. Find the Laplace transform of $(\sin t+\cos t)^2$.

Solution:

⇒ $(\sin t+\cos t)^2=\sin ^2 t+\cos ^2 t+2 \sin t \cos t=1+\sin 2 t$

∴ $L\left\{(\sin t+\cos t)^2\right\}=L\{1+\sin 2 t\}=\frac{1}{p}+\frac{2}{p^2+4}=\frac{p^2+2 p+4}{p\left(p^2+4\right)}$

43. Find $L\left[(\sin t-\cos t)^2\right]$.

Solution:

⇒ $(\sin t-\cos t)^2=\sin ^2 t+\cos ^2 t-2 \sin t \cos t=1-\sin 2 t$

∴ $L\left\{(\sin t-\cos t)^2\right\}=L\{1-\sin 2 t\}=\frac{1}{p}-\frac{2}{p^2+4}=\frac{p^2-2 p+4}{p\left(p^2+4\right)}$

Worked Examples Of Laplace Transform In Control Systems

44. Find $L\left(\sin ^2 4 t\right)$.

Solution:

L$\left(\sin ^2 4 t\right)=L\left[\frac{1-\cos 8 t}{2}\right]=\frac{1}{2} L(1)-\frac{1}{2} L(\cos 8 t)$

= $\frac{1}{2} \cdot \frac{1}{p}-\frac{1}{2} \cdot \frac{p}{p^2+64}=\frac{1}{2}\left[\frac{1}{p}-\frac{p}{p^2+64}\right]=\frac{1}{2}\left[\frac{p^2+64-p^2}{p\left(p^2+64\right)}\right]=\frac{32}{p\left(p^2+64\right)}$

45. Find $L\left\{\sin ^2 a t\right\}$.

Solution:

L$\left\{\sin ^2 a t\right\}=L\left\{\frac{1-\cos 2 a t}{2}\right\} \doteq \frac{1}{2} L\{1\}-\frac{1}{2} L\{\cos 2 a t\}$

= $\frac{1}{2} \frac{1}{p}-\frac{1}{2} \frac{p}{p^2+4 a^2}=\frac{1}{2}\left[\frac{1}{p}-\frac{p}{p^2+4 a^2}\right]=\frac{1}{2}\left[\frac{4 a^2}{p\left(p^2+4 a^2\right)}\right]=\frac{2 a^2}{p\left(p^2+4 a^2\right)}$

46. Find $L\left\{\sin ^2 a t+\left(t^2+1\right)^2\right\}$.

Solution:

L$\left\{\sin ^2 a t+\left(t^2+1\right)^2\right\}=L\left\{\frac{1-\cos 2 a t}{2}+t^4+2 t^2+1\right\}=L\left\{\frac{3}{2}-\frac{1}{2} \cos 2 a t+t^4+2 t^2\right\}$

= $\frac{3}{2} L\{1\}-\frac{1}{2} L\{\cos 2 a t\}+L\left\{t^4\right\}+2 L\left\{t^2\right\}$

= $\frac{3}{2}\left(\frac{1}{p}\right)-\frac{1}{2}\left(\frac{p}{p^2+4 a^2}\right)+\frac{4 !}{p^5}+2\left(\frac{2 !}{p^3}\right)=\frac{3}{2 p}-\frac{p}{2\left(p^2+4 a^2\right)}+\frac{24}{p^5}+\frac{4}{p^3}$

47. Find the Laplace transform of $\cos ^3 2 t$.

Solution:

cos 6 t = $\cos 3(2 t)=4 \cos ^3 2 t-3 \cos 2 t$

∴ $\cos ^3 2 t=\frac{1}{4}(3 \cos 2 t+\cos 6 t)$

Hence $L\left\{\cos ^3 2 t\right\}=L\left\{\frac{1}{4}(3 \cos 2 t+\cos 6 t)\right\}=\frac{3}{4} L\{\cos 2 t\}+\frac{1}{4} L\{\cos 6 t\}$

= $\frac{3}{4} \cdot \frac{p}{p^2+4}+\frac{1}{4} \cdot \frac{p}{p^2+36}=\frac{p}{4}\left(\frac{3}{p^2+4}+\frac{1}{p^2+36}\right)=\frac{p}{4}\left[\frac{4 p^2+112}{\left(p^2+4\right)\left(p^2+36\right)}\right]$

= $\frac{p\left(p^2+28\right)}{\left(p^2+4\right)\left(p^2+36\right)}$

48. Find Laplace Transform of $\sin ^3 2 t$.

Solution:

L$\left\{\sin ^3 2 t\right\}=L\left\{\frac{3 \sin 2 t-\sin 6 t}{4}\right\}=\frac{3}{4} L\{\sin 2 t\}-\frac{1}{4} L\{\sin 6 t\}=\frac{3}{4} \frac{2}{p^2+4}-\frac{1}{4} \frac{6}{p^2+36}$

= $\frac{3}{2}\left[\frac{1}{p^2+4}-\frac{1}{p^2+36}\right]=\frac{3}{2}\left[\frac{p^2+36-p^2-4}{\left(p^2+4\right)\left(p^2+36\right)}\right]=\frac{48}{\left(p^2+4\right)\left(p^2+36\right)}$

49. Find $L\left(\cos ^3 3 t\right)$.

Solution:

We know that $cos 9 t=\cos 3(3 t)=4 \cos ^3 3 t-3 \cos 3 t$

Hence $\cos ^3 3 t=\frac{1}{4} \cos 9 t+\frac{3}{4} \cos 3 t$

L$\left(\cos ^3 3 t\right)=\frac{1}{4} L(\cos 9 t)+\frac{3}{4} L(\cos 3 t)=\frac{1}{4} \frac{p}{p^2+81}+\frac{3}{4} \cdot \frac{p}{p^2+9}$

= $\frac{p\left(p^2+9+3 p^2+243\right)}{4\left(p^2+81\right)\left(p^2+9\right)}=\frac{p\left(4 p^2+252\right)}{4\left(p^2+81\right)\left(p^2+9\right)}=\frac{p\left(p^2+63\right)}{\left(p^2+81\right)\left(p^2+9\right)}$

50. Find the Laplace transform of $(\sin t-\cos t)^3$.

Solution:

($sin t-\cos t)^3=\sin ^3 t-3 \sin ^2 t \cos t+3 \sin t \cos ^2 t-\cos ^3 t$

= $\sin ^3 t-3 \cos t\left(1-\cos ^2 t\right)+3 \sin t\left(1-\sin ^2 t\right)-\cos ^3 t$

= $\sin ^3 t-3 \cos t+3 \cos ^3 t+3 \sin t-3 \sin ^3 t-\cos ^3 $

= $2 \cos ^3 t-3 \cos t+3 \sin t-2 \sin ^3 t$

= $\frac{3 \cos t+\cos 3 t}{2}-3 \cos t+3 \sin t-\frac{3 \sin t-\sin 3 t}{2}$

= $\frac{1}{2} \cos 3 t-\frac{3}{2} \cos t+\frac{3}{2} \sin t+\frac{1}{2} \sin 3 t$

L$\left\{(\sin t-\cos t)^3\right\}=\frac{1}{2} L\{3 \sin t-3 \cos t+\sin 3 t+\cos 3 t\}$

= $\frac{1}{2}\left[\frac{3}{p^2+1}-\frac{3 p}{p^2+1}+\frac{3}{p^2+9}+\frac{p}{p^2+9}\right]$

51. Find the Laplace transform of $\cosh ^2 2 t$.

Solution:

L$\left\{\cosh ^2 2 t\right\}=L\left\{\frac{1}{2}(1+\cosh 4 t)\right\}=\frac{1}{2}[L\{1\}+L\{\cosh 4 t\}]$

= $\frac{1}{2}\left[\frac{1}{p}+\frac{p}{p^2-16}\right]=\frac{p^2-8}{p\left(p^2-16\right)}$

52. Find the Laplace transform of $L\left\{\sinh ^2 3 t\right\}$.

Solution:

⇒ $\sinh ^2(3 t)=\left(\frac{e^{3 t}-e^{-3 t}}{2}\right)^2=\frac{e^{9 t}+e^{-9 t}-2}{4}$

Hence $L\left\{\sinh ^2 3 t\right\}=\frac{1}{4}\left[L\left\{e^{9 t}\right\}+L\left\{e^{-9 t}\right\}-2 L\{1\}\right]=\frac{1}{4}\left[\frac{1}{p-9}+\frac{1}{p+9}-\frac{2}{p}\right] \text {. }$

53. Find the Laplace transform of $\sin h^3 2 t$.

Solution:

L$\left\{\sin h^3 2 t\right\}=L\left\{\left(\frac{e^{2 t}-e^{-2 t}}{2}\right)^3\right\}=L\left\{\frac{1}{8}\left(e^{6 t}-3 e^{2 t}+3 e^{-2 t}-e^{-6}\right)\right\}$

= $=\frac{1}{8} \cdot \frac{1}{p-6}-\frac{3}{8} \cdot \frac{1}{p-2}+\frac{3}{8} \cdot \frac{1}{p+2}-\frac{1}{8} \cdot \frac{1}{p+6}$

= $\frac{1}{8}\left(\frac{1}{p-6}-\frac{1}{p+6}\right)-\frac{3}{8}\left(\frac{1}{p-2}-\frac{1}{p+2}\right)$

= $\frac{1}{8}\left(\frac{12}{p^2-36}-\frac{12}{p^2-4}\right)=\frac{48}{\left(p^2-4\right)\left(p^2-36\right)}$

54. Find the Laplace transform of $\cosh ^3 2 t$.

Solution:

⇒ $\cosh 6 t=4 \cosh ^3 2 t-3 \cosh 2 t \Rightarrow \cosh ^3 2 t=\frac{\cosh 6 t+3 \cosh 2 t}{4}$

∴ $L\left(\cosh ^3 2 t\right)=\frac{1}{4}\left[L\{\cosh 6 t\}+3 L\{\cosh 2 t\}=\frac{1}{4}\left[\frac{p}{p^2-36}+\frac{3 p}{p^2-4}\right]\right.$

= $\frac{p}{4}\left[\frac{1}{p^2-36}+\frac{3}{p^2-4}\right]=\frac{p\left(p^2-28\right)}{\left(p^2-4\right)\left(p^2-36\right)}$

55. Prove that $L\left\{e^{a t} \sinh b t\right\}=\frac{b}{(p-a)^2-b^2}$.

Solution:

L$\left\{e^{a t} \sinh b t\right\}=\int_0^{\infty} e^{-p t} e^{a t} \sinh b t d t=\int_0^{\infty} e^{-(p-a) t} \sinh b t d t$

= $\int_0^{\infty} e^{-(p-a) t}\left(\frac{e^{b t}-e^{-b t}}{2}\right) d t=\frac{1}{2} \int_0^{\infty}\left(e^{-(p-a-b) t}-e^{-(p-a+b) t}\right) d t$

= $\frac{1}{2}\left[\frac{e^{-(p-a-b) t}}{-(p-a-b)}-\frac{e^{-(p-a+b) t}}{-(p-a+b)}\right]_b^p=\frac{1}{2}\left[\frac{1}{p-a-b}-\frac{1}{p-a+b}\right]$

= $\frac{1}{2}\left[\frac{2 b}{(p-a)^2-b^2}\right]=\frac{b}{(p-a)^2-b^2}$

56. Prove that $L\left\{e^{a t} \cosh b t\right\}=\frac{p-a}{(p-a)^2-b^2}$.

Solution:

L$\left\{e^{a t} \cosh b t\right\}=\int_0 e^{-p t} e^{a t} \cosh b t d t=\int_0^{\infty} e^{-(p-a) t} \frac{e^{b t}+e^{-b t}}{2} d t$

= $\frac{1}{2} \int_0^{\infty}\left[e^{-(p-a-b) t}+e^{-(p-a+b) t}\right] d t=\frac{1}{2}\left[\frac{e^{-(p-a-b) t}}{-(p-a-b)}+\frac{e^{-(p-a+b) t}}{-(p-a+b)}\right]_0^{\infty}$

= $\frac{1}{2}\left[\frac{1}{p-a-b}+\frac{1}{p-a+b}\right]=\frac{p-a}{(p-a)^2-b^2}$

57. Find the Laplace transform of f(t)=|t-1|+|t+1|,t≥0.

Solution:

0
∴ $F(t)=(1-t)+(t+1)=2, \text { when } 0
When \(t>1 \Rightarrow|t-1|=t-1 \text {. }$

∴ $F(t)=(t-1)+(t+1)=2 t \text {, when } t>1$

Thus $L\{F(t)\}=\int_0^{\infty} F(t) e^{-p t} d t=\int_0^1 2 e^{-p t} d t+\int_1^{\infty} 2 t e^{-p t} d t=\left(\frac{e^{-p t}}{-p}\right)_0^1+2\left[t\left(\frac{e^{-p t}}{-p}\right)-\left(\frac{e^{-p t}}{p^2}\right)\right]_1^{\infty}$

= $2\left(\frac{e^{-p}}{-p}+\frac{1}{p}\right)+2\left(\frac{e^{-p}}{p}+\frac{e^{-p}}{p^2}\right)=\frac{2}{p}\left(1+\frac{e^{-p}}{p}\right) .$

58. Find the Laplace transform of $f(t)=\left\{\begin{array}{c}
e^t \text { when } 0 0 \text { when } t>1
\end{array}\right.$.

Solution:

L[f(t)]=$\int_0^{\infty} e^{-p t} f(t) d t=\int_0^1 e^{-p t} f(t) d t+\int_1^{\infty} e^{-p t} f(t) d t$ $\int_0^1 e^{-p t} e^t d t+\int_1^{\infty} e^{-p t}(0) d t$

= $\int_0^1 e^{(1-p) t} d t=\left[\frac{e^{(1-p) t}}{1-p}\right]_0^1=\frac{1}{1-p}\left(e^{1-p}-1\right)$

59. Obtain the Laplace transform of the function $F(t)=\left\{\begin{array}{cl}
(t-1)^2, & t>1 \\
0, & 0 \end{array}\right.$.

Solution:

By defination, $L\{F(t)\}=\int_0^{\infty} e^{-p t} F(t) d t=\int_0^1 e^{-p t} F(t) d t+\int_1^{\infty} e^{-p t} F(t) d t$

= $\int_0^1 e^{-p t} \cdot 0 d t+\int_1^{\infty} e^{-p t}(t-1)^2 d t=\int_1^{\infty} e^{-p t}(t-1)^2 d t$

= $\left[(t-1)^2\left(\frac{e^{-p t}}{-p}\right)-2(t-1)\left(\frac{e^{-p t}}{p^2}\right)-2\left(\frac{e^{-p t}}{p^3}\right)\right]_1^{\infty}$

= $\left[(0-0-0)-\left\{0-0-\frac{2}{p^3} e^{-p}\right\}\right]=\frac{2}{p^3} e^{-p}$

60. Find the Laplace transform of $F(t)=\left\{\begin{array}{cl}
\sin t, & 0 0 & t>\pi
\end{array}\right.$

Solution:

L$\{F(t)\}=\int_0^{\infty} e^{-p t} F(t) d t=\int_0^\pi e^{-p t} F(t) d t+\int_\pi^{\infty} e^{-p t} F(t) d t$

= $\int_0^\pi e^{-p t} \sin t d t+\int_\pi^{\infty} e^{-p t}(0) d t$

= $\int_0^\pi e^{-p t} \sin t d t=\left[\frac{e^{-p t}}{p^2+1}(-p \sin t-\cos t)\right]_0^\pi=\frac{1+e^{-\pi p}}{p^2+1}$

61. Find the Laplace transform of F(t) defined as $F(t)= \begin{cases}e^t & \text { when } 05\end{cases}$

Solution:

L$\{F(t)\}=\int_0^{\infty} e^{-p t} F(t) d t=\int_0^5 e^{-p t} F(t) d t+\int_5^{\infty} e^{-p t} F(t) d t$

= $\int_0^5 e^{-p t} e^t d t+\int_5^{\infty} e^{-p t} 3 d t=\int_0^5 e^{-(p-1) t} d t+3 \int_5^{\infty} e^{-p t} d t=\left[\frac{e^{-(p-1) t}}{-(p-1)}\right]_0^5+3\left(\frac{e^{-p}}{-p}\right)_5^{\infty}$

= $\left[\frac{-e^{-5(p-1)}}{p-1}+\frac{1}{p-1}\right]+3 \frac{e^{-5 p}}{p}=\frac{1-e^{-5(p-1)}}{p-1}+\frac{3}{p} e^{-5 p}$

62. Find the Laplace transform of $F(t)= \begin{cases}2 t, & 05\end{cases}$.

Solution:

L$[f(t)]=\int_0^{\infty} e^{-p t} f(t) d t=\int_0^5 e^{-p t} f(t) d t+\int_5^{\infty} e^{-p t} f(t) d t=\int_0^5 e^{-p t} 2 t d t+\int_5^{\infty} e^{-p t}(1) d t$

= $\left(\frac{2 t e^{-p t}}{-p}\right)_0^5-\int_0^5 2 \frac{e^{-p t}}{-p} d t+\left(\frac{e^{-p t}}{-p}\right)_5^{\infty}=\frac{-10}{p} e^{-5 p}-\left(\frac{2 e^{-p t}}{p^2}\right)_0^5+\frac{e^{-5 p}}{p}$

= $-\frac{9}{p} e^{-5 p}-\frac{2}{p^2} e^{-5 p}+\frac{2}{p^2}=\frac{2}{p^2}\left(1-e^{-5 p}\right)-\frac{9}{p} e^{-5 p}$

63. Find L{F(t)} where $F(t)= \begin{cases}0, & 02\end{cases}$

Solution:

L$\{\hat{F}(t)\}=\int_0^{\infty} e^{-p t} F(t) d t=\int_2^{\infty} 4 e^{-p t} d t=\left[\frac{4 e^{-p t}}{-p}\right]_2^{\infty}=\frac{4}{p} e^{-2 p}$

64. Find the Laplace Transform of F(t) defined as $F(t)= \begin{cases}1, & 06\end{cases}$

Solution:

L$\{f(t)\}=\int_0^2 e^{-p t}(1) d t+\int_2^4 e^{-p t} 2 d t+\int_4^6 e^{-p t} 3 d t+\int_6^{\infty} e^{-p t}(0) d t$

=$\left(\frac{e^{-p t}}{-p}\right)_0^2+2\left(\frac{e^{-p t}}{-p}\right)_2^4+3\left(\frac{e^{-p t}}{-p}\right)_4^6=-\frac{1}{p}\left(e^{-2 p}-1\right)-\frac{2}{p}\left(e^{-4 p}-e^{-2 p}\right)-\frac{3}{p}\left(e^{-6 p}-e^{-4 p}\right)$

= $\frac{1}{p}\left(1+e^{-2 p}+e^{-4 p}-3 e^{-6 p}\right)$

65. Find L{F(t)} if $F(t)= \begin{cases}0, & 02\end{cases}$

Solution:

L$\{F(t)\}=\int_0^{\infty} e^{-p t} F(t) d t=\int_0^1 e^{-p t} F(t) d t+\int_1^2 e^{-p t} \cdot F(t) d t+\int_2^{\infty} e^{-p t} F(t) d t$

= $\int_0^1 e^{-p t}(0) d t+\int_1^2 e^{-p t}(t) d t+\int_2^{\infty} e^{-p t}(0) F(t) d t=\int_1^2 t e^{-p t} d t$

= $\left[t\left(\frac{e^{-p t}}{-p}\right)-1\left(\frac{e^{-p t}}{p^2}\right)\right]_1^2=-\left[\frac{t}{p} e^{-p t}+\frac{1}{p^2} e^{-p t}\right]_1^2=-\left[\frac{2}{p} e^{-2 p}+\frac{1}{p^2} e^{-2 p}\right]+\left[\frac{1}{p} e^{-p}+\frac{1}{p^2} e^{-p}\right]$

= $-e^{-2 p}\left(\frac{2}{p}+\frac{1}{p^2}\right)+e^{-p}\left(\frac{1}{p}+\frac{1}{p^2}\right)$

66. Find L{f(t)} where $F(t)=\left\{\begin{array}{cc}
e^{t-a}, & t>a \\
0, & t \end{array}\right.$

Solution:

L$\{F(t)\}=\int_0^{\infty} e^{-p t} F(t) d t=\int_0^a e^{-p t} F(t) d t+\int_a^{\infty} e^{-p t} F(t) d t=\int_0^a e^{-p t}(0) d t+\int_a^{\infty} e^{-p t} e^{t-a} d t$

= $\int_a^{\infty} e^{-a} e^{-(p-1) t} d t=e^{-a}\left[\frac{e^{-(p-1) t}}{-(p-1)}\right]_a^{\infty}=e^{-a}\left[\frac{-e^{-(p-1) a}}{-(p-1)}\right]=\frac{e^{-p a}}{p-1},(p>1)$

67. Find the Laplace transform of the function $F(t)= \begin{cases}4, & 01\end{cases}$

Solution:

L$\{F(t)\}=\int_0^{\infty} e^{-p t} F(t) d t=\int_0^1 F(t) e^{-p t} d t+\int_1^{\infty} F(t) e^{-p t} d t=\int_0^1 4 e^{-p t} d t+\int_1^{\infty} 3 e^{-p t} d t$

= $4\left[\frac{e^{-p t}}{-p}\right]_0^1+3\left[\frac{e^{-p t}}{-p}\right]_1^{\infty}=4\left[\frac{e^{-p}-e^0}{-p}\right]+3\left[\frac{e^{-\infty}-e^{-p}}{-p}\right]=4\left[\frac{e^{-p}-1}{-p}\right]+3\left[\frac{-e^{-p}}{-p}\right]$

= $\frac{4 e^{-p}-4-3 e^{-p}}{-p}=\frac{-e^{-p}+4}{p}=\frac{4-e^{-p}}{p}$

68. Show that $L\left\{\frac{1}{\sqrt{\pi t}}\right\}=\frac{1}{\sqrt{p}}$.

Solution:

L$\left\{\frac{1}{\sqrt{\pi} t}\right\}=\frac{1}{\sqrt{\pi}} L\left\{\frac{1}{\sqrt{t}}\right\}=\frac{1}{\sqrt{\pi}} L\left\{t^{-1 / 2}\right\}=\frac{1}{\sqrt{\pi}} \frac{\Gamma(-1 / 2+1)}{p^{-1 / 2+1}}=\frac{1}{\sqrt{\pi}} \frac{\Gamma(1 / 2)}{p^{1 / 2}}=\frac{1}{\sqrt{\pi}} \cdot \frac{\sqrt{\pi}}{\sqrt{p}}=\frac{1}{\sqrt{p}}$

69. Using expansion $\sin x=x-\frac{x^3}{3 !}+\frac{x^5}{5 !}-\frac{x^7}{7 !}+\ldots$, show that $L\{\sin \sqrt{t}\}=\frac{\sqrt{\pi}}{2 p^{3 / 2}} e^{-1 / 4 p}$.

Solution:

sin $\sqrt{t}=\sqrt{t}-\frac{(\sqrt{t})^3}{3!}+\frac{(\sqrt{t})^5}{5!}-\frac{(\sqrt{t})^7}{7!}+\cdots=t^{1 / 2}-\frac{t^{3 / 2}}{3!}+\frac{t^{5 / 2}}{5!}-\frac{t^{7 / 2}}{7!}+\cdots$

∴ $L\{\sin \sqrt{t}\}=L\left\{t^{1 / 2}\right\}-\frac{1}{3!} L\left\{t^{3 / 2}\right\}+\frac{1}{5!} L\left\{t^{5 / 2}\right\}-\frac{1}{7!} L\left\{t^{7 / 2}\right\}+\cdots$

= $\frac{\Gamma\left(\frac{3}{2}\right)}{p^{3 / 2}}-\frac{1}{3!} \frac{\Gamma\left(\frac{5}{2}\right)}{p^{5 / 2}}+\frac{1}{5!} \frac{\Gamma\left(\frac{7}{2}\right)}{p^{7 / 2}}-\frac{1}{7!} \frac{\Gamma\left(\frac{9}{2}\right)}{p^{9 / 2}}+\cdots$

therefore $L\left\{t^n\right\}=\frac{\Gamma(n+1)}{p^{n+1}}$

= $\frac{\frac{1}{2} \Gamma\left(\frac{1}{2}\right)}{p^{3 / 2}}-\frac{1}{3!} \frac{\frac{3}{2} \cdot \frac{1}{2} \Gamma\left(\frac{1}{2}\right)}{p^{5 / 2}}+\frac{1}{5!} \frac{\frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \Gamma\left(\frac{1}{2}\right)}{p^{7 / 2}}-\frac{1}{7!} \frac{\frac{7}{2} \cdot \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \Gamma\left(\frac{1}{2}\right)}{p^{9 / 2}}+\cdots$

= $\frac{\sqrt{\pi}}{2 p^{3 / 2}}\left[1-\frac{3}{2 \cdot 3!p}+\frac{5 \cdot 3}{2 \cdot 2 \cdot 5!p^2}-\frac{7 \cdot 5 \cdot 3}{2 \cdot 2 \cdot 2 \cdot 7!p^3}+\cdots\right]$ (because $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$

= $\frac{\sqrt{\pi}}{2 p^{3 / 2}}\left[1-\frac{1}{2^2} p+\frac{1}{\left(2^2 p\right)^2 2!}-\frac{1}{\left(2^2 p\right)^3 3!}+\cdots\right]=\frac{\sqrt{\pi}}{2 p^{3 / 2}}\left(e^{-1 /\left(2^2 p\right)}\right)=\frac{\sqrt{\pi}}{2 p^{3 / 2}} e^{-1 / 4 p}$

70. State and prove first translation or shifting theorem

Solution:

Statement: If L[F(t)]=f(p), then

1) $L\left[e^{-a t} F(t)\right]=f(p+a)$

2) $L\left[e^{a t} F(t)\right]=f(p-a)$

Proof: By definition, $L[F(t)]=\int_0^{\infty} e^{-p t} \cdot F(t) d t$

1) $L\left[e^{-a t} \cdot \dot{F}(t)\right]=\int_0^{\infty} e^{-p t} e^{-a t} F(t) d t=\int_0^{\infty} e^{-(p+a) t} F(t) d t=\int_0^{\infty} e^{-u t} F(t) d t$

where u=p+a. Hence $L\left[e^{-a t} F(t)\right]=f(p+a)$

2) $L\left[e^{a t} F(t)\right]=\int_0^{\infty} e^{- p t} e^{a t} F(t) d t=\int_0^{\infty} e^{-(p-a) t} F(t) d t=\int_0^{\infty} e^{-u t} F(t) d t$

where u=p-a =f(u)=f(p-a). Hence $L\left[e^{a t} F(t)\right]=f(p-a)$.

71. Show that

1) $L\left[e^{-a t} \cos b t\right]=\frac{p+a}{(p+a)^2+b^2}$

2) $L\left[e^{a t} \cos b t\right]=\frac{p-a}{(p-a)^2+b^2}$

Solution:

1) By first shifting theorem, $L[\cos b t]=\left[\frac{p}{p^2+b^2}\right] \Rightarrow L\left[e^{-a t} \cos b t\right]=\left[\frac{p}{p^2+b^2}\right]_{p \rightarrow p+a}=\frac{p+a}{(p+a)^2+b^2}$.

2) By first shifting theorem, $L[\cos b t]=\left[\frac{p}{p^2+b^2}\right] \Rightarrow L\left[e^{a t} \cos b t\right]=\left[\frac{p}{p^2+b^2}\right]_{p \rightarrow p-a}=\frac{p-a}{(p-a)^2+b^2}$.

72. Show that

1) $L\left[e^{-a t} \sin b t\right]=\frac{b}{(p+a)^2+b^2}$

2) $L\left[e^{a t} \sin b t\right]=\frac{b}{(p-a)^2+b^2}$

Solution:

1) By first shifting theorem, $L[\sin b t]=\left[\frac{b}{p^2+b^2}\right] \Rightarrow L\left[e^{-a t} \sin b t\right]=\left[\frac{b}{p^2+b^2}\right]_{p \rightarrow p+a}=\frac{b}{(p+a)^2+b^2} .$.

2) By first shifting theorem, $L[\sin b t]=\left[\frac{b}{p^2+b^2}\right] \Rightarrow L\left[e^{a t} \sin b t\right]=\left[\frac{b}{p^2+b^2}\right]_{p \rightarrow p-a}=\frac{b}{(p-a)^2+b^2}$.

73. Show that

1) $L\left[e^{-a t} t^n\right]=\frac{n !}{(p+a)^{n+1}}$

2) $L\left[e^{a t} t^n\right]=\frac{n !}{(p-a)^{n+1}}$

Solution:

1) By first shifting theorem, $L\left[t^n\right]=\frac{n !}{p^{n+1}} \Rightarrow L\left[e^{-a t} t^n\right]=\left[\frac{n !}{p^{n+1}}\right]_{p \rightarrow p+a}=\frac{n !}{(p+a)^{n+1}}$.

2) By first shifting theorem, $L\left[t^n\right]=\frac{n !}{p^{n+1}} \Rightarrow L\left[e^{a t} t^n\right]=\left[\frac{n !}{p^{n+1}}\right]_{p \rightarrow p-a}=\frac{n !}{(p-a)^{n+1}}$.

74. Show that

1) $L\left[e^{-a t} \cosh b t\right]=\frac{p+a}{(p+a)^2-b^2}$

2) $L\left[e^{a t} \cosh b t\right]=\frac{p-a}{(p-a)^2-b^2}$

Solution:

1) By first shifting theorem, L$L[\cosh b t]=\frac{p}{p^2-b^2} \Rightarrow L\left[e^{-a t} \cosh b t\right]=\left[\frac{p}{p^2-b^2}\right]_{p \rightarrow p+a}=\frac{p+a}{(p+a)^2-b^2}$.

2) By first shifting theorem, $L[\cosh b t]=\frac{p}{p^2-b^2} \Rightarrow L\left[e^{a t} \cos b t\right]=\left[\frac{p}{p^2-b^2}\right]_{p \rightarrow p-a}=\frac{p-a}{(p-a)^2-b^2}$.

75. Show that

1) $L\left[e^{-a t} \sinh b t\right]=\frac{b}{(p+a)^2-b^2}$

2) $L\left[e^{a t} \sinh b t\right]=\frac{b}{(p-a)^2-b^2}$

Solution:

1) By first shifting theorem, $L\left[[\sinh b t]=\frac{b}{p^2-b^2} \Rightarrow L\left[e^{-a t} \sinh \ b t\right]=\left[\frac{b}{p^2-b^2}\right]_{p \rightarrow p+a}=\frac{b}{(p+a)^2-b^2}\right.$.

2) By first shifting theorem, $L[\sinh b t]=\frac{b}{p^2-b^2} \Rightarrow L\left[e^{a t} \sinh b t\right]=\left[\frac{b}{p^2-b^2}\right]_{p \rightarrow p-a}=\frac{b}{(p-a)^2+b^2}$.

76. Find the Laplace transform of $e^{-t} \cos 2 t$.

Solution:

L$\{\cos 2 t\}=\frac{p}{p^2+4}\left[\ L(\cos a t)=\frac{p}{p^2+a^2}\right]$

Now Applying the First shifting theorem, we get

L$\left\{e^{-t} \cos 2 t\right\}=[L\{\cos 2 t\}]_{p \rightarrow p+1}=\left[\frac{p}{p^2+4}\right]_{p \rightarrow p+1}=\frac{p+1}{(p+1)^2+4}=\frac{p+1}{p^2+2 p+5}$

77. Find the Laplace transform of $e^{-3 t} \sin t$.

Solution:

L$\{\sin t\}=\frac{1}{p^2+1} \text {. So } L\left\{e^{-3 t} \sin t\right\}=\left(\frac{1}{p^2+1}\right)_{p \rightarrow p+3}=\frac{1}{(p+3)^2+1}=\frac{1}{p^2+6 p+10}$

78. Find $L\left[e^{-2 t} \sin 3 t\right]$.

Solution:

We know that, $L(\sin 3 t)=\frac{3}{p^2+9}$

L$\left(e^{-2 t} \sin 3 t\right)=\left(\frac{3}{p^2+9}\right)_{p \rightarrow p+2}=\frac{3}{(p+2)^2+9}=\frac{3}{p^2+4 p+13}$

79. Find $L\left\{e^{3 t} \sin 4 t\right\}$.

Solution:

We have, $L\{\sin 4 t\}=\frac{4}{p^2+16}=f(p)$

Using First Shifting Theorem, $L\left\{e^{3 t} \sin 4 t\right\}=6 f(p-3)=\frac{4}{(p-3)^2+16}=\frac{4}{p^2-6 p+25}$

80. Find $L\left\{e^{4 t} \cos 5 t\right\}$.

Solution:

We know $L[\cos 5 t]=\left[\frac{p}{p^2+5^2}\right] \text {. By first shifting theorem, }$

L$\left[e^{4 t} \cos 5 t\right]=\left[\frac{p}{p^2+5^2}\right]_{p \rightarrow p-4}=\frac{p-4}{(p-4)^2+5^2}=\frac{p-4}{p^2-8 p+41} .$

81. Find the Laplace transform of $e^{-3 t}(2 \cos 5 t-3 \sin 5 t)$.

Solution:

We have $L(2 \cos 5 t-3 \sin 5 t)=2 \cdot \frac{p}{p^2+35}-3 \cdot \frac{5}{p^2+25}=\frac{2 p-15}{p^2+25}$

Now Applying the First Shifting Theorem, we have

L$\left\{e^{-3 t}(2 \cos 5 t-3 \sin 5 t)\right\}=\left(\frac{2 p-15}{p^2+25}\right)_{p \rightarrow p+3}=\frac{2(p+3)-15}{(p+3)^2+25}=\frac{2 p-9}{p^2+6 p+34} .$.

82. Find $L\left\{e^{-2 t}(3 \cos 6 t-5 \sin 6 t)\right\}$.

Solution:

By First Shifting Theorem,

L$\left\{e^{-2 t}(3 \cos 6 t-5 \sin 6 t)\right\}=[L\{3 \cos 6 t-5 \sin 6 t\}]_{p \rightarrow p+2}$

= $\left[\frac{3 p}{p^2+36}-\frac{30}{p^2+36}\right]_{p \rightarrow p+2}=\frac{3(p+2)-30}{(p+2)^2+36}=\frac{3 p-24}{p^2+4 p+40}$

83. Find $L\left\{e^{-t} \cos ^2 t\right\}$.

Solution:

We have $L\left\{\cos ^2 t\right\}=L\left\{\frac{1+\cos 2 t}{2}\right\}=\frac{1}{2}[L\{1\}+L\{\cos 2 t\}]=\frac{2\left(p^2+2\right)}{p\left(p^2+4\right)}=f(p)$

By First Shifting Theorem,

L$\left\{e^{-t} \cos ^2 t\right\}=f(p+1)=\left[\frac{2\left(p^2+2\right)}{p\left(p^2+4\right)}\right]_{p \rightarrow p+1}=\frac{2\left[(p+1)^2+2\right]}{(p+1)\left[(p+1)^2+4\right]}$

= $\frac{2\left(p^2+2 p+3\right)}{(p+1)\left(p^2+2 p+5\right)}$

84. Find $L\left(e^t \cos ^2 t\right)$.

Solution:

L$\left\{\cos ^2 t\right\}=L\left\{\frac{1+\cos 2 t}{2}\right\}=\frac{1}{2}\left[\frac{1}{p}+\frac{p}{p^2+4}\right]=\frac{2(p^2+2)}{p\left(p^2+4\right)}=f(p)$

By First Shifting Theorem,$L\left\{e^t \cos ^2 t\right\}=f(p-1)$

= $\left[\frac{2\left(p^2+2\right)}{p\left(p^2+4\right)}\right]_{p \rightarrow p-1}=\frac{2\left[(p-1)^2+2\right]}{(p-1)\left[(p-1)^2+4\right]}=\frac{2\left(p^2-2 p+3\right)}{(p1)\left(p^2-2 p+5\right)} \text {. }$

85. Find $L\left\{e^{3 t} \sin ^2 t\right\}$.

Solution:

We have $ L\left\{\sin ^2 t\right\}=L\left\{\frac{1-\cos 2 t}{2}\right\}=\frac{1}{2}[L\{1\}-L\{\cos 2 t\}]$

= $\frac{1}{2}\left[\frac{1}{p}-\frac{p}{p^2+4}\right]=\frac{2}{p\left(p^2+4\right)}=f(p), \text { say }$

By First Shifting Theorem, $L\left\{e^{3 t} \sin ^2 t\right\}=f(p-3)$

= $\left[\frac{2}{p\left(p^2+4\right)}\right]_{p \rightarrow p-3}=\frac{2}{(p-3)\left[(p-3)^2+4\right]}=\frac{2}{(p-3)\left(p^2-6 p+13\right)}$

86. Find $L\left(e^t \cos t \sin t\right)$.

Solution:

L$(\sin t \cos t)=L\left(\frac{\sin 2 t}{2}\right)=\frac{1}{2} L(\sin 2 t)=\frac{1}{2} \cdot \frac{2}{p^2+4}=\frac{1}{p^2+4}$

L$\left(e^t \cdot \sin t \cos t\right)=\left(\frac{1}{p^2+4}\right)_{p \rightarrow p-1}=\frac{1}{(p-1)^2+4}=\frac{1}{p^2+2 p+5} .$

87. Find $L\left\{e^t \sin 2 t \cos t\right\}$.

Solution:

L$\{\sin 2 t \cos t\}=L\left\{\frac{\sin 3 t+\sin t}{2}\right\}=\frac{1}{2}[L\{\sin 3 t\}+L\{\sin t\}]=\frac{1}{2}\left[\frac{3}{p^2+9}+\frac{1}{p^2+1}\right]$

By First Shifting Theorem, $L\left\{e^t \sin 2 t \cos t\right\}$

= $\frac{1}{2}\left[\frac{3}{p^2+9}+\frac{1}{p^2+1}\right]_{p \rightarrow p-1}=\frac{1}{2}\left[\frac{3}{(p-1)^2+9}+\frac{1}{(p-1)^2+1}\right]$

= $\frac{1}{2}\left[\frac{3}{p^2-2 p+10}+\frac{1}{p^2-2 p+2}\right] .$

88. Find $L\left\{e^{4 t} \sin 2 t \cos t\right\}$.

Solution:

We have $L\{\sin 2 t \cos t\}=L\left\{\frac{1}{2}(2 \sin 2 t \cos t)\right\}=L\left\{\frac{1}{2}(\sin 3 t+\sin t)\right\}$

= $\frac{1}{2}[L\{\sin 3 t\}+L\{\sin t\}]=\frac{1}{2}\left[\frac{3}{p^2+3^2}+\frac{1}{p^2+1^2}\right]$

By First Shifting Theorem, $L\left\{e^{4 t} \cdot \sin 2 t \cos t\right\}=\frac{1}{2}\left[\frac{3}{p^2+9}+\frac{1}{p^2+1}\right]_{p \rightarrow p-4}$

= $\frac{1}{2}\left[\frac{3}{(p-4)^2+9}+\frac{1}{(p-4)^2+1}\right]=\frac{1}{2}\left[\frac{3}{p^2-8 p+25}+\frac{1}{p^2-8 p+17}\right]$

89. Find $L\left[e^{-2 t} \sin 2 t \sin 3 t\right]$.

Solution:

L$[\sin 2 t \sin 3 t]=L\left[\frac{\cos t-\cos 5 t}{2}\right]=\frac{1}{2}[L(\cos t)-L(\cos 5 t)]$

= $\frac{1}{2}\left[\frac{p}{p^2+1}-\frac{p}{p^2+25}\right]=\frac{12 p}{\left(p^2+1\right)\left(p^2+25\right)}$

L$\left(e^{-2 t} \sin 2 t \sin 3 t\right)=\left[\frac{12 p}{\left(p^2+1\right)\left(p^2+25\right)}\right]_{p \rightarrow p+2}=\frac{12(p+2)}{\left\{(p+2)^2+1\right\}\left\{(p+2)^2+25\right\}}$

= $\frac{12 p+24}{\left(p^2+4 p+5\right)\left(p^2+4 p+29\right)} .$

90. Find $L\left\{e^{-t}(3 \sin 2 t-5 \cosh 2 t)\right\}$.

Solution:

L$\langle 3 \sin 2 t-5 \cosh 2 t\}=3 L\{\sin 2 t\}-5 \dot{L}[\cosh 2 t]$

= $3 \frac{2}{p^2+2^2}-5 \frac{p}{p^2-2^2}=\frac{6}{p^2+4}-\frac{5 p}{p^2-4}$

L$\left\{e^{-t}(3 \sin 2 t-5 \cosh 2 t)\right\}=\left[\frac{6}{p^2+4}-\frac{5 p}{p^2-4}\right]_{p \rightarrow p+2}=\frac{6}{(p+1)^2+4}-\frac{5(p+1)}{(p+1)^2-4}$

= $\frac{6}{p^2+2 p+5}-\frac{5 p+5}{p^2+2 p-3}$

91. Evaluate $L\left\{e^t\left(\cos 2 t+\frac{1}{2} \sinh 2 t\right)\right\}$.

Solution:

L$\left\{\cos 2 t+\frac{1}{2} \sinh 2 t\right\}=L\{\cos 2 t\}+\frac{1}{2} L[\sinh 2 t]=\frac{p}{p^2+2^2}+\frac{1}{2} \cdot \frac{2}{p^2-2^2}$

= $\frac{p}{p^2+4}+\frac{1}{p^2-4}=f(p), \text { say. }$ ∴ By First Shifting Theorem,

L$\left\{e^t\left(\cos 2 t+\frac{1}{2} \sinh 2 t\right)\right\}=f(p-1)=\left(\frac{p}{p^2+4}+\frac{1}{p^2-4}\right)_{p \rightarrow p-1}$

= $\frac{p-1}{(p-1)^2+4}+\frac{1}{(p-1)^2-4}=\frac{p-1}{p^2-2 p+5}+\frac{1}{p^2-2 p-3} .$

92. Evaluate $L\left\{e^{-t}\left(\cos 2 t+\frac{1}{2} \sinh 2 t\right)\right\}$.

Solution:

L$\left\{\cos 2 t+\frac{1}{2} \sinh 2 t\right\}=L\{\cos 2 t\}+\frac{1}{2} L[\sinh 2 t]=\frac{p}{p^2+2^2}+\frac{1}{2} \cdot \frac{2}{p^2-2^2}$

+ $\frac{p}{p^2+4}+\frac{1}{p^2-4}=f(p) \text {, say. }$ ∴ By First Shifting Theorem,

L$\left\{e^{-t}\left(\cos 2 t+\frac{1}{2} \sinh 2 t\right)\right\}=f(p+1)=\left(\frac{p}{p^2+4}+\frac{1}{p^2-4}\right)_{p \rightarrow p+1}$

= $\frac{p+1}{(p+1)^2+4}+\frac{1}{(p+1)^2-4}=\frac{p+1}{p^2+2 p+5}+\frac{1}{p^2+2 p-3} .$

93. Find Laplace transform of $L\left\{e^{2 t}(3 \sinh 2 t-5 \cosh 2 t)\right\}$.

Solution:

By First Shifting Theorem,

L$\left\{e^{2 t}(3 \sinh 2 t-5 \cosh 2 t)\right\}=[L\{3 \sinh 2 t-5 \cosh 2 t\}]_{p \rightarrow p-2}$

= $\left[\frac{6}{p^2-4}-\frac{5 p}{p^2-4}\right]_{p \rightarrow p-2}=\frac{6-5(p-2)}{(p-2)^2-4}=\frac{16-5 p}{p^2-4 p} .$

94. Find L(sinh ⁡at cos ⁡at).

Solution:

L$(\sinh a t \cos a t)=L\left[\left(\frac{e^{a t}+e^{-a t}}{2}\right) \cos a t\right]=\frac{1}{2} L\left(e^{a t} \cos a t\right)+\frac{1}{2} L\left(e^{-a t} \cos a t\right)$

L$(\cos a t)=\frac{p}{p^2+a^2} \Rightarrow L\left(e^{a t} \cos a t\right)=\left(\frac{p}{p^2+a^2}\right)_{p \rightarrow p-a}=\frac{p-a}{(p-a)^2+a^2}=\frac{p-a}{p^2-2 a p+2 a^2}$

Similarly $L\left(e^{-a t} \cos a t\right)=\frac{p+a}{p^2+2 a p+2 a^2}$

∴ $L(\sinh a t \cos a t)=\frac{1}{2}\left[\frac{p-a}{p^2-2 a p+2 a^2}\right]+\frac{1}{2}\left[\frac{p+a}{p^2+2 a p+2 a^2}\right]$

95. Find L{sinh⁡ at sin ⁡at}.

Solution:

Sice $\sinh a t \sin a t=\frac{1}{2}\left(e^{a t}-e^{-a t}\right) \sin a t \text {, we have }$

L$\{\sinh a t \sin a t\}=\frac{1}{2}\left[L\left\{e^{a t} \sin a t\right\}-L\left\{e^{-a t} \sin a t\right\}\right]$

= $\frac{1}{2}\left[\frac{a}{(p-a)^2+a^2}-\frac{a}{(p+a)^2+a^2}\right]=\frac{a}{2}\left[\frac{(p+a)^2+a^2-(p-a)^2-a^2}{\left\{(p-a)^2+a^2\right\}\left\{(p+a)^2+a^2\right\}}\right]$

= $\frac{a}{2} \frac{4 a p}{\left(p^2+2 a^2\right)^2-(2 a p)^2}=\frac{2 a^2 p}{p^4+4 a^4}$

96. Find L{cosh ⁡at cos⁡ at}

Solution:

L$\{\cosh a t \cos a t\}=\frac{1}{2} L\left\{\left(e^{a t}+e^{-a t}\right) \cos a t\right\}=\frac{1}{2}\left[L\left\{e^{a t} \cos a t\right\}+L\left\{e^{-a t} \cos a t\right\}\right]$

= $\frac{1}{2}\left[\frac{(p-a)}{(p-a)^2+a^2}+\frac{p+a}{(p+a)^2+a^2}\right]$, Using First Shifting Theorem

= $\frac{1}{2}\left[\frac{(p-a)\left(p^2+2 a^2+2 a p\right)+(p+a)\left(p^2+2 a^2-2 a p\right)}{p^4+4 a^4}\right]=\frac{1}{2}\left[\frac{2 p\left(p^2+2 a^2\right)-4 p a^2}{p^4+4 a^4}\right]$

= $\frac{p\left(p^2+2 a^2-2 a^2\right)}{p^4+4 a^4}=\frac{p^3}{p^4+4 a^4}$

97. Find L{cosh ⁡at sin⁡ bt}.

Solution:

L$\{\cosh a t \sin b t\}=L\left\{\left(\frac{e^{a t}+e^{-a t}}{2}\right) \sin b t\right\}=\frac{1}{2}\left[L\left\{e^{a t} \sin b t\right\}+L\left\{e^{-a t} \sin b t\right\}\right]$

By First Shifting Theorem,

L$\{\cosh a t \sin b t\}=\frac{1}{2}\left[\{L(\sin b t)\}_{p \rightarrow p-a}+\{L(\sin b t)\}_{p \rightarrow p+a}\right]$

= $\frac{1}{2}\left[\left(\frac{b}{p^2+b^2}\right)_{p \rightarrow p-a}+\left(\frac{b}{p^2+b^2}\right)_{p \rightarrow p+a}\right]=\frac{1}{2}\left[\frac{b}{(p-a)^2+b^2}+\frac{b}{(p+a)^2+b^2}\right]$

= $\frac{b}{2}\left[\frac{1}{p^2-2 a p+a^2+b^2}+\frac{1}{p^2+2 a p+a^2+b^2}\right]=\frac{b}{2}\left[\frac{\dot{2}\left(p^2+a^2+b^2\right)}{\left[(p-a)^2+b^2\right]\left[(p+a)^2+b^2\right]}\right]$

= $\frac{b\left(p^2+a^2+b^2\right)}{\left[(p-a)^2+b^2\right]\left[(p+a)^2+b^2\right]}$

98. Evaluate $L\left\{t^2 e^{-2 t}\right\}$.

Solution:

We know that $L\left(t^2\right)=\frac{2 !}{t^3}=\frac{2}{p^3} \text { then } L\left(e^{-2 t} t^2\right)=\left(\frac{2}{p^3}\right)_{p \rightarrow p+2}=\frac{2}{(p+2)^3} \text {. }$

99. Find $L\left(e^{-4 t} t^2\right)$.

Solution:

We know that $L\left(t^2\right)=\frac{2 !}{t^3}=\frac{2}{p^3} \text { then } L\left(e^{-4 t} t^2\right)=\left(\frac{2}{p^3}\right)_{p \rightarrow p+4}=\frac{2}{(p+4)^3}$

100. Find $\dot{L}\left\{e^{-3 t}\left(t^2+1\right)\right\}$.

Solution:

L$\left\{\left(t^2+1\right)\right\}=L\left\{t^2\right\}+L\{1\}=\frac{2 !}{p^3}+\frac{1}{p}$

L$\left\{e^{-3 t}\left(t^2+1\right)\right\}=\left(\frac{2}{p^3}+\frac{1}{p}\right)_{p \rightarrow p+3}=\frac{2}{(p+3)^3}+\frac{1}{p+3}=\frac{2+(p+3)^2}{(p+3)^3}=\frac{p^2+6 p+11}{(p+3)^3}$

101. Find $L\left\{(t+3)^2 e^t\right\}$.

Solution:

We have $L\left\{(t+3)^2\right\}=L\left\{t^2+6 t+9\right\}=\frac{2}{p^3}+\frac{6}{p^2}+\frac{9}{p}=f(p)$

By First Shifting Theorem, $L\left\{e^t(t+3)^2\right\}=f(p-1)=\left[\frac{2}{p^3}+\frac{6}{p^2}+\frac{9}{p}\right]_{p \rightarrow p-1}$

= $\frac{2}{(p-1)^3}+\frac{6}{(p-1)^2}+\frac{9}{(p-1)}=\frac{9 p^2-12 p+5}{(p-1)^3}$

102. Find $L\left\{(t+2)^2 e^t\right\}$.

Solution:

By First Shifting Theorem,

L$\left\{(t+2)^2 e^t\right\}=\left[L\left\{(t+2)^2\right\}\right]_{p \rightarrow p-1}=\left[L\left\{t^2+4 t+4\right\}\right]_{p \rightarrow p-1}$

= $\left[\frac{2}{p^3}+\frac{4}{p^2}+\frac{4}{p}\right]_{p \rightarrow p-1}=\left[\frac{2+4 p+4 p^2}{p^3}\right]_{p \rightarrow p-1}=\frac{2+4(p-1)+4(p-1)^2}{(p-1)^3}$

= $\frac{2\left[1+2 p-2+2 p^2-4 p+2\right]}{(p-1)^3}=\frac{2\left(2 p^2-2 p+1\right)}{(p-1)^3}$

103. Find $L\left\{\left(1+t e^{-t}\right)^3\right\}$.

Solution:

We have $\left(1+t e^{-t}\right)^3=1+3 t e^{-t}+3 t^2 e^{-2 t}+t^3 e^{-3 t} \text {. }$

∴ $L\left\{\left(1+t e^{-t}\right)^3\right\}=L\left\{1+3 t e^{-t}+3 t^2 e^{-2 t}+t^3 e^{-3 t}\right\}$

= $L\{1\}+3 L\left\{t e^{-t}\right\}+3 L\left\{t^2 e^{-2 t}\right\}+L\left\{t^3 e^{-3 t}\right\}, \text { using Linearity Property }$

= $\frac{1}{p}+3 \cdot \frac{1}{(p+1)^2}+3 \cdot \frac{2 !}{(p+2)^2}+\frac{3 !}{(p+3)^4}=\frac{1}{p}+\frac{3}{(p+1)^2}+\frac{6}{(p+2)^3}+\frac{6}{(p+3)^4}$

104. Show that

1) $L\{t \sin a t\}=\frac{2 a p}{\left(p^2+a^2\right)^2}$

2) $L\{t \cos a t\}=\frac{p^2-a^2}{\left(p^2+a^2\right)^2}$

Solution:

Since, $L\{t\}=\frac{1}{p^2} \text {, we have } L\left\{t e^{i a t}\right\}=\frac{1}{(p-i a)^2}=\frac{(p+i a)^2}{(p-i a)^2(p+i a)^2}$

Equating the real and imaginary parts on both sides, we obtain

⇒ $L\{t(\cos a t+i \sin a t)\}=\frac{\left(p^2-\dot{a}^2\right)+i 2 a p}{\left(p^2+a^2\right)^2}$

1) $L\{t \sin a t\}=\frac{2 a p}{\left(p^2+a^2\right)^2}$

2) $L\{t \cos a t\}=\frac{p^2-a^2}{\left(p^2+a^2\right)^2}$

105. Find the Laplace transform of $t e^{2 t}$ sin⁡3t.

Solution:

Since $L\{t\}=\frac{1}{p^2}$, we have $L\left\{t e^{i 3 t}\right\}=\frac{1}{(p-3 i)^2}=\frac{(p+3 i)^2}{(p-3 i)^2(p+3 i)^2}$

⇒ $L\{t \cos 3 t+i t \sin 3 t\}=\frac{\left(p^2-9\right)+i 6 p}{\left(p^2+9\right)^2}$

Equating imaginary parts on both sides, we have $L\{t \sin 3 t\}=\frac{6 p}{\left(p^2+9\right)^2}$

Now applying the First Shifting theorem, we have $L\left\{e^{2 t} t \sin 3 t\right\}=[L\{t \sin 3 t\}]_{p \rightarrow p-2}=\left[\frac{6 p}{\left(p^2+9\right)^2}\right]_{p \rightarrow p-2}=\frac{6(p-2)}{\left(p^2-4 p+13\right)^2}$

106. State and prove the second translation or shifting theorem.

Solution:

Statement: If L[F(t)]=f(p) and $G(t)=\left\{\begin{array}{l}F(t-a), t>a \text {. } \\ 0, t
Proof: \(L(G(t)]=\int_0^{\infty} e^{-s t} G(t) d t=\int_0^a e^{-p t} G(t) d t+\int_a^{\infty} e^{-p t} G(t) d t$

= $\int_0^a e^{-p t} 0 d t+\int_a^{\infty} e^{-p t} F(t-a) d t=\int_a^{\infty} e^{-p t} F(t-a) d t$

Put u=t-a. Then, t=u+a and dt=du when t=a, u=0 and $t \rightarrow \infty, u \rightarrow \infty$

∴ $L[G(t)]=\int_0^{\infty} e^{-p(u+a)} F(u) d u=e^{-a p} \int_0^{\infty} e^{-p u} F(u) d u=e^{-a p} f(p)$

107. If L[F(t)]=f(p) and a>0, then prove that $L\{F(t-a) H(t-a)\}=e^{-a p} f(p)$, where $H(t)= \begin{cases}1, & \text { if } t>0 \\ 0, & \text { if } t<0\end{cases}$

Solution:

By definition, $L\{F(t-a) H(t-a)\}=\int_0^{\infty} e^{-p t} F(t-a) H(t-a) d t \rightarrow(1)$

Put t-a=u so that dt=du.

Also when t=0, u=-a when $t \rightarrow \infty, u \rightarrow \infty$.

Then $L\{F(t-a) H(t-a)\}=\int_0^{\infty} e^{-p(u+a)} F(u) H(u) d u$ by (1)

⇒ $\int_{-a}^0 e^{-p(u+a)} F(u) H(u) d u+\int_0^{\infty} e^{-p(u+a)} F(u) H(u) d u$

= $\int_{-a}^0 e^i-p(u+a) F(u) \cdot 0 d u+\int_0^{\infty} e^{-p(u+a)} F(u) \cdot 1 d u$, by Definition of H(t)

= $\int_{-a}^0 e^{-p(u+a)} F(u) d u=e^{-p a} \int_0^{\infty} e^{-p u} F(u) d u=e^{-p a} \int_0^{\infty} e^{-p t} F(t) d t,$

= $e^{-p a} L\{F(t)\}=e^{-a p} f(p)$

108. Find L[G(t)], where $G(t)= \begin{cases}e^{t-a}, & t>a \\ 0, & t

Solution:

Let \(F(t)=e^t \text {. Then, } L[F(t)]=f(p)=\frac{1}{p-1}$

∴ $G(t)=\left\{\begin{array}{l}
F(t-a), t>a. \\
0, \quad t \end{array}\right.$

Then $L[G(t)]=e^{-a p} f(p)=e^{-a p} \frac{1}{p-1}=\frac{e^{-a p}}{p-1}, p>1$

109. Find L{G(t)} where $G(t)=\left\{\begin{array}{cc}
\sin (t-2 \pi / 3), & t>2 \pi / 3 \\
0, & t<2 \pi / 3
\end{array}\right.$

Solution:

Let $F(t)=\sin t \text { so that } G(t)=\left\{\begin{array}{cl}
F(t-2 \pi / 3), & t>2 \pi / 3 . \\
0, & t<2 \pi / 3
\end{array}\right.$

We know that, $L(F(t)]=L(\sin t)=\frac{1}{p^2+1}=f(p)$

∴ $L[G(t)]=e^{\frac{-2 \pi}{3} p} \frac{1}{p^2+1}=\frac{e^{-2 \pi p / 3}}{p^2+1}, p>0$

110. Find L{F(t)} where $F(t)=\left\{\begin{array}{rr}
\sin \left(t-\frac{\pi}{3}\right), & t>\frac{\pi}{3} \\
0, & t<\frac{\pi}{3}
\end{array}\right\}$.

Solution:

Let $G(t)=\sin t \text {. }$

∴ $L\{G(t)\}=L\{\sin t\}=\frac{1}{p^2+1}=f(p)$

Now, $F(t)=\left\{\begin{aligned}
\sin (t-\pi / 3), & t>\pi / 3 \\
0, & t>\pi / 3
\end{aligned}\right.$

Applying second shifting theorem $L\{G(t)\}=e^{-\pi p / 3} f(p)=\frac{e^{-\pi p / 3}}{p^2+1}$

111. Find L[G(t)], where $G(t)=\left\{\begin{array}{cc}
\sin 2 t, & 0 0, & t>\pi
\end{array}\right.$.

Solution:

L$[G(t)]=\int_0^{\infty} e^{-p t} G(t) d t 2=\int_0^\pi e^{-p t} \sin 2 t d t+\int_\pi^{\infty} e^{-p t}(0) d t$

= $\left[e^{-p t} \frac{-(p \sin 2 t+2 \cos 2 t)}{p^2+4}\right]_0^\pi=0$

∴ $\left[ \int e^{a x} \sin b x d x=\frac{e^{a x}}{a^2+b^2}(a \sin b x-b \cos b x)\right]$

112. Find L{G(t)} where $G(t)=\left\{\begin{array}{cl}
\cos (-t-\pi / 3), & \text { if } t>\pi / 3 \\
0, & \text { if } t<\pi / 3
\end{array}\right.$

Solution:

Let $F(t)=\cos t \text {. }$

∴ $L\{F(t)\}=L\{\cos t\}=\frac{p}{p^2+1}=f(p)$

Now $G(t)=\left\{\begin{array}{cc}
F(t-\pi / 3)=\cos (t-\pi / 3), & t>\pi / 3 \\
0, & t<\pi / 3
\end{array}\right.$

Applying the second Shifting theorem, we get $L\{(t)\}=e^{-\pi p / 3}\left(\frac{p}{p^2+1}\right)=\frac{p e^{-\pi p / 3}}{p^2+1}$

Proceeding as above, we get L$L[G(t)]=e^{-2 \pi p / p}\left(\frac{1}{p^2+1}\right)=\frac{p e^{-2 \pi p / 3}}{p^2+1}$

113. Find L{F(t)} where $F(t)=\left\{\begin{array}{cl}
\cos (t-2 \pi / 3), & \text { if } t>2 \pi / 3 \\
0, & \text { if } t<2 \pi / 3
\end{array}\right.$

Solution:

Let $G(t)=\cos t \text {. }$

∴ $L\{G(t)\}=L\{\cos t\}=\frac{p}{p^2+1}=f(p)$

Now, $F(t)=\left\{\begin{array}{cc}
G(t-2 \pi / 3)=\cos (t-2 \pi / 3), & t>2 \pi / 3 \\
0, & t<2 \pi / 3
\end{array}\right.$

Applying second shifting theorem$L\{F(t)\}=e^{-2 \pi p / 3}\left(\frac{p}{p^2+1}\right)=\frac{p e^{-2 \pi p / 3}}{p^2+1}$

114. Find the Laplace transform of $(t-2)^3 u(t-2)$.

Solution:

Comparing the given function with $f(t-a) u(t-a) \text {, we have } a=2 \text { and } f(t)=t^3$

∴ $L\{F(t)\}=L\left\{t^3\right\}=\frac{3 !}{p^4}=\frac{6}{p^4}=f(p)$

Applying second shifting theorem $L\left\{(t-2)^3 u(t-2)\right\}=e^{-2 p} \frac{6}{p^4}=\frac{6 e^{-2 p}}{p^4}$

115. Find the Laplace transform of $e^{-3 t} u(t-2)$.

Solution:

L$\left\{e^{-3 t} u(t-2)\right\}=L\left\{e^{-3(t-2)} e^{-6} u(t-2)\right\}=e^{-6} L\left\{e^{-3(t-2)} u(t-2)\right\}$

Taking,$F(t)=e^{-3 t}, f(p)=\frac{1}{p+3}$and using second shifting theorem, we have

L$\left\{e^{-3 t} u(t-2)\right\} = e^{-6} e^{-2 p} \frac{1}{p+3}=\frac{e^{-2(p+3)}}{p+3}$

116. State and prove a change of scale property.

Solution:

Statement: If $L[F(t)]=f(p), \text { then } L[F(a t)]=\frac{1}{a} f\left(\frac{p}{a}\right)$

Proof: We know that $L[F(a t)]=\int_0^{\infty} e^{-p t} F(a t) d t$

Put u = at Then, t = $\frac{u}{a}$ and $d t=\frac{d u}{a}$

∴ $L[F(a t)]=\int_0^{\infty} e^{-p u / a} F(u) \frac{d u}{a}=\frac{1}{a} \int_0^{\infty} e^{-p W a} F(u) d u=\frac{1}{a} \int_0^{\infty} e^{-(b / a) u} F(u) d u=\frac{1}{a} f\left(\frac{p}{a}\right)$

117. Find L(cos⁡ 4t) by using the change of scale property.

Solution:

We have that $L(\cos t)=\frac{p}{p^2+1}=f(p)$

Then $L(\cos 4 t)=\frac{1}{4} f\left(\frac{p}{4}\right)=\frac{1}{4} \cdot \frac{p / 4}{(p / 4)^2+1}=\frac{p}{p^2+16}, p>0 \text {. }$

118. Find L{cos⁡ 5t} by using the change of scale property.Solution:

We have that $L(\cos t)=\frac{p}{p^2+1}=f(p)$

Then $L(\cos 5 t)=\frac{1}{5} f\left(\frac{p}{5}\right)=\frac{1}{5} \cdot \frac{p / 5}{(p / 5)^2+1}=\frac{p}{p^2+25}, p>0 \text {. }$

119. Find L[sinh ⁡2t] by using the change of scale property.

Solution:

We have $L[\sinh t]=\frac{1}{p^2-1}=f(p)$

∴ $L[\sinh 2 t]=\frac{1}{2} \cdot f\left(\frac{p}{2}\right)=\frac{1}{2} \cdot \frac{1}{(p / 2)^2-1}=\frac{2}{p^2-4}$

120. Find $L\left\{\sin ^2(a t)\right\}$ by using change of scale property.

Solution:

Since $\sin ^2(a t)=\frac{1-\cos 2 a t}{2} \text {, we have }$

L$\left\{\sin ^2(a t)\right\}=\frac{1}{2} L\{(1-\cos 2 a t)\}=\frac{1}{2}[L(1)-L\{\cos 2 a t\}]=\frac{1}{2} \cdot \frac{1}{p}-\frac{1}{2} L\{\cos 2 a t\}$

Taking $F(t)=\cos t, f(p)=\frac{p}{p^2+1}$

∴ $L\{\cos (2 a t)\}=\frac{1}{2 a} \frac{p / 2 a}{(P / 2 a)^2+1}=\frac{p}{p^2+4 a^2}$

Hence $L\left\{\sin ^2 a t\right\}=\frac{1}{2 p}-\frac{1}{2} \frac{p}{p^2+4 a^2}=\frac{1}{2}\left[\frac{1}{p}-\frac{p}{p^2+4 a^2}\right]=\frac{2 a^2}{p\left(p^2+4 a^2\right)} \text {. }$

121. If $L\{F(t)\}=\frac{1}{p} e^{-1 / p}$, prove that $L\left\{e^{-t} F(3 t)\right\}=\frac{e^{-3 /(p+1)}}{p+1}$

Solution:

L$\{F(t)\}=\frac{1}{p} e^{-1 / p}=f(p)$

∴ By the Change of Scale property, we have $F(3 t)=\frac{1}{3} f\left(\frac{p}{3}\right)=\frac{1}{3} \cdot \frac{3}{p} e^{-3 / p}=\frac{1}{p} e^{-3 / p}$

Now applying the First Shifting theorem, we get $L\left\{e^{-t} F(3 t)\right\}=\frac{e^{-3 /(p+1)}}{p+1}$.

122. If $L[F(t)]=\frac{9 p^2-12 p+15}{(p-1)^3}$ using change of scale property

Solution:

Given $L[F(t)]=\frac{9 p^2-12 p+15}{(p-1)^3}=f(p)$. By Change of Scale Property,

L$[F(3 t)]=\frac{1}{3} f\left(\frac{p}{3}\right)=\frac{1}{3} \cdot \frac{9(p / 3)^2-12(p / 3)+15}{(p / 3-1)^3}=\frac{9\left(p^2-4 p+15\right)}{(p-3)^3}$

123. If $L\{F(t)\}=\frac{p^2-p+1}{(2 p+1)^2(p-1)}$ then show that $L\{F(2 t)\}=\frac{p^2-2 p+4}{4(p+1)^2(p-2)}$ by applying change of scale property,

Solution:

Given $L\{F(t)\}=\frac{p^2-p+1}{(2 p+1)^2(p-1)}=f(p)$

By Change of scale property,

L$\{F(2 t)\}=\frac{1}{2} f\left(\frac{p}{2}\right)=\frac{1}{2}\left[\frac{\left(\frac{p}{2}\right)^2-\frac{p}{2}+1}{\left(2 \cdot \frac{p}{2}+1\right)\left(\frac{p}{2}-1\right)}\right]=\frac{1}{2}\left[\frac{-\frac{p^2-2 p+4}{4}}{\frac{(p+1)^2(p-2)}{4}}\right]=\frac{1}{4} \frac{p^2-2 p+4}{(p+1)^2(p-2)}$

The post Laplace Transform – 1 Exercise Solved Problems appeared first on Answer Key for Math.

Bessel’s Equations Solved Exercise Problems

Kumar — Mon, 21 Aug 2023 09:47:30 +0000

Bessel’s Equations Exercise 5

Solved Exercise Problems On Bessel’S Equations

1. Define Bessel’s differential equation.

Solution:

The differential equation of the form $\frac{d^2 y}{d x^2}+\frac{1}{x} \frac{d y}{d x}+\left(1-\frac{n^2}{x^2}\right) y=0$ is called Bessel’s differential equation or Bessel’s equation.

Bessel’s Equations step-by-step solutions

2. Prove that $a_0 x^n\left[1+(-1) \frac{x^2}{2^2 \cdot 1 !(n+1)}+(-1)^2 \frac{x^4}{2^4 2 !(n+1)(n+2)}+\ldots\right]$and $a_0 x^{-n}\left[1+(-1) \frac{x^2}{2^2 \cdot 1 !(-n+1)}+(-1)^2 \frac{x^4}{2^4 2 !(-n+1)(-n+2)}+\ldots .\right]$ are solutions of Bessel’s differential equation:

Solution:

Given

The Bessel’s differential equation is $\frac{d^2 y}{d x^2}+\frac{1}{x} \frac{d y}{d x}+\left(1-\frac{n^2}{x^2}\right) y=0 \rightarrow \text { (1) }$

Let us assume that its series solution is $y=\sum_{r=0}^{\infty} a_r x^{k+r}$.

∴ $\frac{d y}{d x}=\sum_{r=0}^{\infty} a_r(k+r) x^{k+r-1}$ and $\frac{d^2 y}{d x^2}=\sum_{r=0}^{\infty} a_r(k+r)(k+r-1) x^{k+r-2}$

Substituting these values in (1) we have $\sum_{r=0}^{\infty} a_r\left[(k+r)(k+r-1) x^{k+r-2}+\frac{1}{x}(k+r) x^{k+r-1}+\left(1-\frac{n^2}{x^2}\right) x^{k+r}\right]=0$

⇒ $\sum_{r=0}^{\infty}\left[a_r\left\{(k+r)^2-n^2\right\} x^{k+r-2}+x^{k+r}\right]=0 \rightarrow \text { (2) }$

Since (2) is an identity, the coefficients of various powers of x must be zero.

∴ Equating to zero the coefficient of lowest power of x, i.e., of $x^{k-2}$ in (2), we have $a_0\left(k^2-n^2\right)=0 \text {. }$

Now $a_0 \neq 0$ as it is the coefficient of the first term with which we begin the series.

∴ $k^2-n^2=0 \Rightarrow k= \pm n \rightarrow \text { (3) }$

Now equating to zero the coefficient of $x^{k-1}$ in (2), we have $a_1\left\{(k+1)^2-n^2\right\}=0$

Now $k= \pm n \Rightarrow(k+1)^2-n^2 \neq 0 \Rightarrow a_1=0$.

Again equating to zero the coefficient of the general term i.e. of $x^{k+r}$ in (2), we have

⇒ $a_{r+2}\left\{(k+r+2)^2-n^2\right\}+a_r=0 \Rightarrow a_{r+2}(k+r+n+2)(k+r-n+2)=-a_r$

⇒ $a_{r+2}=-\frac{a_r}{(k+r+n+2)(k+r-n+2)} \rightarrow \text { (4) }$

Putting r=1 in (4), we have $a_3=\frac{a_1}{(k+n+3)(k-n+3)}=0$, since $a_1=0$.

Similarly putting $r=3,5, \ldots$ etc. in (4), we have $a_1=a_3=a_5 \ldots=0$ (each).

Now two cases arise.

Case 1: When k=n, from (4), we have $a_{r+2}=-\frac{a_r}{(2 n+r+2)(r+2)}$

Putting r=0,2,4 etc, we have $a_2=-\frac{a_0}{(2 n+2)(2)}=-\frac{a_0}{2^2 \cdot 1(n+1)}$

⇒ $a_4=-\frac{a_2}{(2 n+4)(4)}=-\frac{a_2}{2^2 \cdot 2!(n+2)}=\frac{a_0}{2^4 \cdot 2!(n+1)(n+2)} \text { etc. }$

∴ y = $a_0\left[x^n-\frac{x^{n+2}}{2^2 \cdot 1!(n+1)}+\frac{x^{n+4}}{2^4 2!(n+1)(n+2) \ldots}\right]$

= $a_0 x^n\left[1+(-1) \frac{x^2}{2^2 \cdot 1!(n+1)}+(-1)^2 \frac{x^4}{2^4 2!(n+1)(n+2)}+\cdots\right]$

Case 2: When k=-i. The series solution is obtained by replacing n by -n as $a_0 x^{-n}\left[1+(-1) \frac{x^2}{2^2 \cdot 1!(-n+1)}+(-1)^2 \frac{x^4}{2^4 2!(-n+1)(-n+2)}+\cdots\right]$

Examples Of Bessel’s Equations Solved Problems

3. Define Bessel’s function of the first kind of order n.

Solution:

If $a_0=\frac{1}{2^n \Gamma(n+1)}$, then the solution of the Bessel’s equation, $y=a_0 x^n\left[1+(-1) \frac{x^2}{2^2 \cdot 1!(n+1)}+(-1)^2 \frac{x^4}{2^4 2!(n+1)(n+2)}+\cdots\right]$ is called

Bessel’s function of the first kind of order n and it is denoted by $J_n(x)$.

∴ $J_n(x)=\frac{x^n}{2^n \Gamma(n+1)}\left[1+(-1) \frac{x^2}{2^2 \cdot 1!(n+1)}+(-1)^2 \frac{x^4}{2^4 \cdot 2!(n+1)(n+2)}+.\right]$

= $\frac{x^n}{2^n \Gamma(n+1)} \sum_{r=0}^{\infty}(-1)^r \frac{x^{2 r}}{2^{2 r} r!(n+1)(n+2) \ldots(n+r)}=\sum_{r=0}^{\infty}(-1)^r\left(\frac{x}{2}\right)^{n+2 r} \frac{1}{r!\Gamma(n+r+1)}$

Bessel’s Differential Equation Solved Problems

4. Show that

1) $J_{-n}(x)=(-1)^n J_n(x)$ when n is a positive integer and

2) $J_n(-x)=(-1)^n J_n(x)$ for + ve or – ve integers.

Solution:

1) We have $\int_{-n}(x)=\sum_{r=0}^{\infty}(-1)^r\left(\frac{x}{2}\right)^{-n+2 r} \frac{1}{r!\Gamma(-n+r+1)^r}$

Since if p is an integer, then $\Gamma(-p)$ is infinity for p>0,

∴ we get terms in $J_{-n}$ equal to zero till -n+r+1<1, i.e., r
Hence $J_{-n}(x)=\sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(-n+r+1)}\left(\frac{x}{2}\right)^{-n+2 r}=\sum_{s=0}^{\infty} \frac{(-1)^{n+s}}{(n+s)!\Gamma(s+1)}\left(\frac{x}{2}\right)^{n+2 s}$

= $(-1)^n \sum_{s=0}^{\infty} \frac{(-1)^s}{\Gamma(n+s+1) s!}\left(\frac{x}{2}\right)^{n+2 s}=(-1)^n J_n(x) \text {. }$

2) We know that $J_n(x)=\sum_{r=0}^{\infty}(-1)^r \frac{1}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r}$

Case 1. Let n be a+v e integer. Replacing x bY -x, we have

⇒ $J_n(-x)=\sum_{r=0}^{\infty}(-1)^r \frac{1}{r!\Gamma(n+r+1)}\left(\frac{-x}{2}\right)^{n+2 r}$

= $(-1)^n \sum_{r=0}^{\infty}(-1)^r \frac{1}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r}$

= $(-1)^n J_n(x)$

Case 2. If n is – ve integer, say n=-m where m is a + ve integer:

∴ $J_n(x)=J_{-m}(x)=(-1)^m J_m(x)$

Replacing x by -x

⇒ $J_n(-x)=(-1)^m J_m(-x)=(-1)^m(-1)^m J_m(x)=(-1)^m J_{-m}(x)=(-1)^{2 m}(-1)^{-m} J_{-m}(x)$

= $(-1)^{-m} J_{-m}(x)=(-1)^n J_n(x)$

Hence $J_n(-x)=(-1)^n J_n(x)$ for + ve or – ve integers.

Step-By-Step Guide To Solving Bessel’s Equations

5. If $Y_n(x)$ is defined by $Y_n(x)=\frac{\cos n \pi J_n(x)-J_{-n}(x)}{\sin n \pi}$, when n is not an integer and $Y_n(x)=\lim _{r \rightarrow n} \frac{\cos r \pi J_r(x)-J_{-r}(x)}{\sin r \pi}$, when n is an integer then show that $J_n(x)$ and $Y_n(x)$are two independent solutions of Bessel’s equation.

solution:

Case 1. Suppose n is not an integer.

Since n is not an integer, $\sin n \pi \neq 0$.

Hence $Y_n(x)$ is a linear combination of $J_n(x)$ and $J_{-n}(x)$. But $J_n(x)$ and $J_{-n}(x)$ are two independent solutions of Bessel’s equation.

Thus $J_n(x)$ and $Y_n(x)$ are two independent solutions of Bessel’s equation.

Case 2. Suppose n is an integer.

Since n is an integer, $\cos n \pi=(-1)^n, \cos n \pi=0$ and $J_{-n}(x)=(-1)^n J_n$

Hence $Y_n(x)$ is undefined. To make it meaningful, we define $Y_n(x)=\lim _{r \rightarrow n} Y_r(x)=\lim _{r \rightarrow n} \frac{\cos r \pi J_r(x)-J_{-r}(x)}{\sin r \pi}=\lim _{r \rightarrow n} \frac{\frac{\partial}{\partial r}\left[\cos r \pi J_r(x)-J_{-r}(x)\right]}{\frac{\partial}{\partial r}[\sin r \pi]}$

= $\left[\frac{-\pi \sin r \pi J_r(x)+\cos r \pi \frac{\partial}{\partial r} J_r(x)-\frac{\partial}{\partial r} J_{-r}(x)}{\pi \cos r \pi}\right]_{r=n}$

= $\frac{\cos n \pi\left[\frac{\partial}{\partial r} J_r(x)\right]_{r=n}-\left[\frac{\partial}{\partial r} J_{-r}(x)\right]_{r=n}}{\pi \cos r \pi}$

= $\frac{(-1)^n\left[\frac{\partial}{\partial r} J_r(x)\right]_{r=n}-(-1)^{2 n}\left[\frac{\partial}{\partial r} J_{-r}(x)\right]_{r=n}}{\pi(-1)^n}$

= $\frac{1}{\pi}\left[\frac{\partial}{\partial r} J_r(x)-(-1)^n \frac{\partial}{\partial r} J_{-r}(x)\right]_{r=n}$

Since $J_r(x)$ and $J_{-r}(x)$ are two solutions of Bessel’s equation, it can be proved that $Y_n(x)$ is a solution of Bessel’s equation and it is independent with $J_n(x)$.

Bessel’s Equations Solutions With Applications

6. Define Bessel’s function of the second kind of order n.

Solution:

The function $Y_n(x)$ defined by $Y_n(x)=\frac{\cos n \pi J_n(x)-J_{-n}(x)}{\sin n \pi}$, when n is not an integer and $Y_n(x)=\lim _{r \rightarrow n} \frac{\cos r \pi J_r(x)-J_{-r}(x)}{\sin r \pi}$, when n is an integer is called Bessel’s function of the second kind of order n.

7. Write the general solution of the following equations

1) $x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+\left(x^2-64\right) y=0$

2) $x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+\left(x^2-\frac{9}{16}\right) y=0$.

Solution:

1) Given equation is $x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+\left(x^2-64\right) y=0$

⇒ $x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+\left(x^2-8^2\right) y=0$, which is a Bessel’s equation of order 8 , an integer

∴ It’s solution is $y=a J_8(x)+b Y_8(x)$, where a and b are arbitrary constants.

2) Given equation is $x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+\left(x^2-\frac{9}{16}\right) y=0$

⇒ $x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+\left[x^2-\left(\frac{3}{4}\right)^2\right] y=0$, which is a Bessel’s equation of order 3/4, not an integer.

∴ It’s solution is $y=a J_8(x)+b J_{-3 / 4}(x)$, where a and b are arbitrary constants.

8. $y=a_0\left(1-\frac{x^2}{2^2}+\frac{x^4}{2^2 \cdot 4^2}-\frac{x^6}{2^2 \cdot 4^2 \cdot 6^2} \cdots\right)$ is a solution of Bessel’s differential equation for n=0.

Solution:

The Bessel’s differential equation for n=0 is $\frac{d^2 y}{d x^2}+\frac{1}{x} \frac{d y}{d x}+y=0 \rightarrow$ (1).

Let us assume that its solution is $y^{\prime}=\sum_{r=0}^{\infty} a_r x^{k+r}$

∴ $\frac{d y}{d x}=\sum_{n=0}^{\infty} a_r(k+r) x^{k+r-1} \text { and } \frac{d^2 y}{d x^2}=\sum_{r=0}^{\infty} a_r(k+r)(k+r-1) x^{r+r-2}$

Substituting these values in (1), we get $\sum_{r=0}^{\infty} a_r\left[(k+r)(k+r-1) x^{k+r-2}+\frac{1}{x}(k+r) x^{k+r-1}+x^{k+r}\right]=0$

⇒ $\sum_{r=0}^{\infty} a_r\left[(k+r)^2 x^{k+r-2}+x^{k+r}\right]=0 \rightarrow \text { (2) }$

which is an identity.

Equating to zero, the coefficient of lowest power of x i.e. $x^{k-2}$, we have $a_0 k^2=0$.

Since $a_0 \neq 0$ we get $k^2=0 \Rightarrow k=0 \rightarrow(3)$

Now equating to zero the coefficient of next power of x, i.e., $x^{k-1}$, we have $a_1(k+1)^2=0 \text {. }$

Since $k+1 \neq 0$ by virtue of (3), we have $a_1=0$.

Again equating to zero the coefficient of the general term, i.e. $x^{k-r}$, we have $a_{r+2}(k+r+2)^2+a_r=0 \Rightarrow a_{r+2}=-\frac{a_r}{(k+r+2)^2}$.

When $k=0, a_{r+2}=-\frac{a_r}{(r+2)^2}$.

Putting $r=1,3,5, \ldots$ etc., we have $a_1=a_3=a_5=\cdots=0$ (each).

Again putting r=0,2,4, etc., we have $a_2=-\frac{a_0}{2^2}, a_4=-\frac{a_2}{4^2}=\frac{a_0}{2^2 \cdot 4^2}$ etc.

Since $y=\sum_{r=0}^{\infty} a_r x^r$, when k=0, we have $y=a_0\left(1-\frac{x^2}{2^2}+\frac{x^4}{2^2 \cdot 4^2}-\frac{x^6}{2^2 \cdot 4^2 \cdot 6^2} \cdots\right)$

Advanced Solved Examples On Bessel’s Equations

9. Prove that $\ Lt_{x \rightarrow 0} \frac{J_n(x)}{x^n}=\frac{1}{2^n \Gamma(n+1)}$.

Solution:

⇒ $J_n(x)=\frac{x^n}{2^n \Gamma(n+1)}\left[1-\frac{x^2}{4(n+1)}+\frac{x^4}{4 \cdot 8(n+1)(n+2)}-\cdots\right]$

⇒ $\ Lt_{x \rightarrow 0} \frac{J_n(x)}{x^n}=L_{x \rightarrow 0} \frac{1}{2^n \Gamma(n+1)}\left[1-\frac{x^2}{4(n+1)}+\frac{x^4}{4 \cdot 8(n+1)(n+2)}-\ldots\right]=\frac{1}{2^n \Gamma(n+1)} .$

10. Show that

1) $J_{-1 / 2}(x)=\sqrt{\left(\frac{2}{\pi x}\right)} \cos x$,

2) $J_{1 / 2}(x)=\sqrt{\left(\frac{2}{\pi x}\right)} \sin x$ and

3) $\left[J_{1 / 2}(x)\right]^2+\left[J_{-1 / 2}(x)\right]^2=\frac{2}{\pi x}$

Solution:

We know that $J_n(x)=\frac{x^n}{2^n \Gamma(n+1)}\left[1-\frac{x^2}{2(2 n+2)}+\frac{x^4}{2 \cdot 4(2 n+2)(2 n+4)}+\cdots\right]$

1) Putting $n=-\frac{1}{2}$ in $J_n(x)$, we get $J_{-1 / 2}(x)=\frac{x^{-1 / 2}}{2^{-1 / 2} \Gamma(1 / 2)}\left[1-\frac{x^2}{2(-1+2)}+\frac{x^4}{2 \cdot 4(-1+2)(-1+4)}-\cdots\right]$

= $\sqrt{\left(\frac{2}{\pi x}\right)}\left[1-\frac{x^2}{2!}+\frac{x^4}{4!} \cdots\right]=\sqrt{\left(\frac{2}{\pi x}\right)} \cos x$

2) Putting $n=\frac{1}{2}$, in $J_n(x)$, we have

⇒ $J_{1 / 2}(x)=\frac{x^{1 / 2}}{2^{1 / 2} \Gamma(3 / 2)}\left[1-\frac{x^2}{2(1+2)}+\frac{x^4}{2 \cdot 4 \cdot(1+2)(1+4)} \cdots\right]$

= $\sqrt{\left(\frac{x}{2}\right)} \frac{1}{\frac{1}{2} \sqrt{\pi}} \frac{1}{x}\left[x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right]=\sqrt{\left(\frac{2}{\pi x}\right)} \sin x$

3) $\left[J_{1 / 2}(x)\right]^2+\left[J_{-1 / 2}(x)\right]^2=\left(\frac{2}{\pi x}\right) \cos ^2 x+\left(\frac{2}{\pi x}\right) \sin ^2 x=\frac{2}{\pi x}$

11. Prove that $\sqrt{\left(\frac{\pi x}{2}\right)} J_{3 / 2}(x)=\frac{1}{x} \sin x-\cos x$.

Solution:

We know that $J_n(x)=\frac{x^n}{2^n \Gamma(n+1)}\left[1-\frac{x^2}{2(2 n+2)}+\frac{x^4}{2 \cdot 4(2 n+2)(2 n+4)} \cdots\right]$

Putting n=3 / 2, we have $J_{3 / 2}(x)=\frac{x^{3 / 2}}{2^{3 / 2} \Gamma(5 / 2)}[1-\frac{x^2}{2 \cdot 5}+\frac{x^4}{2 \cdot 4 \cdot 5 \cdot 7}$

– $\frac{x^6}{2 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 9} \cdots]$

= $\frac{x \sqrt{x}}{2 \sqrt{2}(3 / 2)(1 / 2) \sqrt{\pi}}\left[1-\frac{x^2}{2 \cdot 5}+\frac{x^4}{2 \cdot 4 \cdot 5 \cdot 7}-\frac{x^6}{2 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 9}+\cdots\right]$

∴ $\sqrt{\left(\frac{\pi x}{2}\right)} J_{\frac{3}{2}}(x)=\frac{1}{3}\left[x^2-\frac{x^4}{2 \cdot 5}+\frac{x^6}{2 \cdot 4 \cdot 5 \cdot 7}-\frac{x^8}{2 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 9}\right]$

= $\frac{2 x^2}{3!}-\frac{4 x^4}{5!}+\frac{6 x^6}{7!}-\frac{8 x^8}{9!}+\cdots$

= $\left(\frac{1}{2!}-\frac{1}{3!}\right) x^2-\left(\frac{1}{4!}=\frac{1}{5!}\right) x^4+\left(\frac{1}{6!}-\frac{1}{7!}\right) x^6-\left(\frac{1}{8!}-\frac{1}{9!}\right) x^8 \cdots$

= $\left(\frac{x^2}{2!}-\frac{x^4}{4!}+\frac{x^6}{6!} \cdots\right)+\left(-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!} \cdots\right)$

= $-\left(1-\frac{x^2}{2!}+\frac{x^4}{4}-\frac{x^6}{6!} \cdots\right)+\frac{1}{x}\left[x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} \cdots\right]=-\cos x+\frac{1}{x} \sin x .$

Hence $\sqrt{\left(\frac{\pi x}{2}\right)} J_{3 / 2}(x)=\frac{1}{x} \sin x-\cos x$

Worked examples of Bessel’s equations problems

12. Show that

1) $J_{-3 / 2}(x)=-\sqrt{\left(\frac{2}{\pi x}\right)}\left[\sin x+\frac{1}{x} \cos x\right]$.

2) $J_{5 / 2}(x)=\left[\frac{2}{\pi x}\right]^{1 / 2}\left[\frac{3-x^2}{x^2} \sin x-\frac{3}{x} \cos x\right]$.

Solution

We know that $J_n(x)=\frac{x^{\prime \prime}}{2^n \Gamma(n+1)}\left[1-\frac{x^2}{2(2 n+2)}+\frac{x^4}{2 \cdot 4(2 n+2)(2 n+4)}-\cdots\right]$ → (1)

1) Put n=-3 / 2 in (1). Then

Put n=-3 / 2 in (1).

⇒ $J_{-3 / 2}(x)=\frac{x^{-3 / 2}}{2^{-3 / 2} \Gamma\left(-\frac{3}{2}+1\right)}\left[1-\frac{x^2}{2(-3+2)}+\frac{x^4}{2 \cdot 4 \cdot(-3+2)(-3+4)}-\cdots\right]$

= $\frac{2 \sqrt{2}}{x^{3 / 2} \Gamma \Gamma\left(\frac{-1}{2}\right)^2}\left[1+\frac{x^2}{2}-\frac{x^4}{8}+\frac{x^0}{144}-\cdots\right]$

= $\frac{2 \sqrt{2}}{x^{3 / 2}(-2 \sqrt{\pi})}\left[1+\frac{x^2}{2}-\frac{x^4}{8}+\frac{x^6}{144}-\cdots\right]$

= $-\sqrt{\frac{2}{\pi x}}\left[\frac{1}{x}+\frac{x}{2}-\frac{x^3}{8}+\frac{x^5}{144}-\cdots\right]$

= $-\sqrt{\frac{2}{\pi r}}\left[\frac{1}{x}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots\right)+\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)\right]$

= $-\sqrt{\frac{2}{\pi x}}\left[\frac{1}{x} \cos x+\sin x\right]$

2) Put n=5 / 2 in (1).

Then $J_{5 / 2}(x)=\frac{x^{5 / 2}}{2^{5 / 2} \Gamma\left(\frac{5}{2}+1\right)}[1-\frac{x^2}{2 \cdot 7}$

+ $\frac{x^4}{2 \cdot 4 \cdot 7 \cdot 9}-\frac{x^6}{2 \cdot 4 \cdot 6 \cdot 7 \cdot 9 \cdot 11}+\cdots]$

= $\frac{x^{5 / 2}}{4 \sqrt{2}\left(\frac{5}{2}\right)\left(\frac{3}{2}\right)\left(\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)}\left[1-\frac{x^2}{14}+\frac{x^4}{504}-\frac{x^6}{33264}+\cdots\right]$

= $\frac{x^2 \sqrt{2 x}}{15 \sqrt{\pi}}\left[1-\frac{x^2}{14}+\frac{x^4}{504}-\frac{x^6}{33264}+\cdots\right]$

= $\left[\frac{2}{\pi x}\right]^{1 / 2} \frac{1}{15}\left[x^3-\frac{x^5}{14}+\frac{x^7}{504}-\cdots\right]$

= $\left[\frac{2}{\pi x}\right]^{1 / 2}\left[\frac{x^3}{15}-\frac{x^5}{210}+\frac{x^7}{7560}-\cdots\right]$

= $\left[\frac{2}{\pi x}\right]^{1 / 2}\left[\frac{3}{x^2}\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)-\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)-\frac{3}{x}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots\right)\right]$

= $\left[\frac{2}{\pi x}\right]^{1 / 2}\left[\frac{3}{x^2} \sin x-\sin x-\frac{3}{x} \cos x\right]=\left[\frac{2}{\pi x}\right]^{1 / 2}\left[\frac{3-x^2}{x^2} \sin x-\frac{3}{x} \cos x\right]$

13. Show that $\int_0^1 \frac{u J_0(x u)}{\left(1-u^2\right)^{1 / 2}}=\frac{\sin x}{x}$.

Solution:

We have $J_0(x)=1-\frac{x^2}{2^2}+\frac{x^4}{2^2 \cdot 4^2}-\frac{x^6}{2^2 \cdot 4^2 \cdot 6^2}+\cdots \rightarrow(1)$

∴ $\int_0^1 \frac{u J_0(x u)}{\left(1-u^2\right)^{1 / 2}} d u$

= $\int_0^1 \frac{u}{\left(1-u^2\right)^{1 / 2}}\left(1-\frac{x^2}{4} u^2+\frac{x^4}{4 \cdot 16} u^4 \ldots\right) d u^{\pi / 2}$

= $\int_0^\pi \frac{\sin \theta}{\cos \theta}\left(1-\frac{x^2}{4} \sin ^2 \theta+\frac{x^4}{64} \sin ^4 \theta-\cdots\right) \cos \theta d \theta$

(On putting $u=\sin \theta$ and $u=\cos \theta d \theta$)

= $\int_0^{\pi / 2} \sin \theta d \theta-\frac{x^2}{4} \int_0^{\pi / 2} \sin ^3 \theta d \theta+\frac{x^4}{64} \int_0^{\pi / 2} \sin ^5 \theta d \theta \ldots$

= $[-\cos \theta]_0^{\pi / 2}-\frac{x^2}{4} \times \frac{2}{3}+\frac{x^4}{64} \times \frac{4 \cdot 2}{5 \cdot 3}-\cdots$

= $1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots=\frac{1}{x}\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots\right)=\frac{\sin x}{x}$

Solutions for Bessel’s differential equation practice problems

14. Show that $\int_0^{\pi / 2} J_1(z \cos \theta) d \theta=\frac{1-\cos z}{z}$.

Solution:

We have $J_n(x)=\sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{2 r+n} \rightarrow(1)$

(1) ⇒ $J_1(z \cos \theta)=\sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(1+r+1)}\left(\frac{z \cos \theta}{2}\right)^{2 r+1}$

∴ $\int_0^{\pi / 2} J_1(z \cos \theta) d \theta=\sum_{r=0}^{\infty} \frac{(-1)^r z^{2 r+1}}{r!(r+1)!2^{2 r+1}} \int_0^{\pi / 2} \cos ^{2 r+1} \theta d \theta$

= $\sum_{r=0}^{\infty} \frac{(-1)^r z^{2 r+1}}{r!(r+1)!2^{2 r+1}} \cdot \frac{2 r(2 r-2) \ldots 4 \cdot 2}{(2 r+1)(2 r-1) \ldots 5 \cdot 3}$

= $\sum_{r=0}^{\infty} \frac{(-1)^r z^{2 r+1}}{r!(r+1)!2^{2 r+1}} \cdot \frac{[2 r(2 r-2) \ldots 4 \cdot 2]^2}{(2 r+1) 2 r(2 r-1)(2 r-2) \ldots 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$

= $\sum_{r=0}^{\infty} \frac{(-1)^r z^{2 r+1}}{r!(r+1)!2^{2 r+1}} \cdot \frac{2^2(r!)^2}{(2 r+1)!}=\sum_{r=0}^{\infty} \frac{(-1)^r z^{2 r+1}}{2(r+1)(2 r+1)!}$

= $\frac{1}{z}-\frac{1}{z}+\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots=\frac{1}{z}-\frac{1}{z}\left(1+\frac{z^2}{2!}-\frac{z^4}{4!}+\frac{z^6}{6!}-\cdots\right)=\frac{1-\cos z}{z}$

15. Prove that Recurrence Formula I : $x J_n^{\prime}(x)=n J_n(x)-x J_{n+1}(x)$.

Solution:

We have $J_n(x)=\sum_{r=0}^{\infty}(=1)^r\left(\frac{x}{2}\right)^{n+2 r} \frac{1}{r!\Gamma(n+r+1)}$ where n is a positive integer.

Differentiating w.r.t. x, we have $J_n^{\prime}(x)=\sum_{r=0}^{\infty}(-1)^r \frac{(n+2 r)}{r!\Gamma(n+r+1)} \frac{1}{2}\left(\frac{x}{2}\right)^{n+2 r-1}$

∴ $x J_n^{\prime}(x)=x \sum_{r=0}^{\infty}(-1)^r \frac{(n+2 r)}{r!\Gamma(n+r+1)} \frac{1}{2}\left(\frac{x}{2}\right)^{n+2 r-1}$

= $\sum_{r=0}^{\infty}(-1)^r \frac{(n+2 r)}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r}$

= $\sum_{r=0}^{\infty}(-1)^r \frac{n}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r}$

+ $\sum_{r=0}^{\infty}(-1)^r \frac{2 r}{r!\Gamma(n+r+1)} \frac{x}{2}\left(\frac{x}{2}\right)^{n+2 r-1}$

= $n J_n(x)+x \sum_{r=1}^{\infty}(-1)^r \frac{1}{(r-1)!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r-1}$ Putting r-1=s

= $n J_n(x)-x \sum_{s=0}^{\infty}(-1)^s \frac{1}{s!\Gamma(n+1+s+1}\left(\frac{x}{2}\right)^{n+1+2 s}=n J_n(x)-x J_{n+1}(x) .$

Hence $x J_n^{\prime}=n J_n(x)-x J_{n+1}(x)$ which may also be written as $\frac{d}{d x}\left(x^{-n} J_n\right)=-x^{-n} J_{n+1}$.

16. Prove that Recurrence Formula II : $x J_n^{\prime}(x)=-n J_n(x)+x J_{n-1}(x)$.

Solution:

As in Recurrence Formula I, we have

⇒ $x J_n^{\prime}(x)=\sum_{r=0}^{\infty}(-1)^r \frac{(n+2 r)}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r}$

= $\sum_{r=0}^{\infty}(-1)^r \frac{(2 n+2 r-n)}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2}$

= $-n \sum_{r=0}^{\infty}(-1)^r \frac{1}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r}$

+ $\sum_{r=0}^{\infty}(-1)^r \frac{(2 n+2 r)}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r}$

= $-n J_n(x)+\sum_{r=0}^{\infty}(-1)^r \frac{\dot{2}(n+r)}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r-1} \cdot\left(\frac{x}{2}\right)$

= $-n J_n(x)+x \sum_{r=0}^{\infty}(-1)^r \frac{1}{r!\Gamma(n+r)}\left(\frac{x}{2}\right)^{n+2 r-1}$

= $-n J_n(x)+x \sum_{r=0}^{\infty}(-1)^r \frac{1}{r!\Gamma(n-1+r+1)}\left(\frac{x}{2}\right)^{n-1+2 r}=-n J_n(x)+x J_{n-1}(x)$

Hence $x J_n^{\prime}(x)=-n J_n(x)+x J_{n-1}(x)$ which may also be written as $\frac{d}{d x}\left(x^n J_n\right)=x^n J_{n-1}$

17. Prove that Recurrence Formula 3: $2 J_n^{\prime}(x)=J_{n-1}(x)-J_{n+1}(x)$.

Solution:

Recurrence formulas 1 and 2 are $x J_n^{\prime}(x)=n J_n(x)-x J_{n+1}(x)$ and $x J_n^{\prime}(x)=-n J_n(x)+x J_{n-1}(x) \text {. }$

Adding we get $2 x J_n^{\prime}(x)=x\left[J_{n-1}(x)-J_{n+1}(x)\right]$.

Hence $2 J_n^{\prime}(x)=J_{n-1}(x)-J_{n+1}(x)$.

Aliter: $J_n(x)=\sum_{r=0}^{\infty}(-1)^r \frac{1}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r}$

⇒ $2 J_n^{\prime}(x)=\sum_{r=0}^{\infty}(-1)^r \frac{2}{r!\Gamma(n+r+1)}(n+2 r) \frac{1}{2}\left(\frac{x}{2}\right)^{n+2 r-1}$

= $\sum_{r=0}^{\infty}(-1)^r \frac{(n+r+r)}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r-1}$

= $\sum_{r=0}^{\infty}(-1)^r \frac{(n+r)}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r-1}+\sum_{r=0}^{\infty}(-1)^r \frac{r}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r-1}$

= $\sum_{r=0}^{\infty}(-1)^r \frac{1}{r!\Gamma(n+r)}\left(\frac{x}{2}\right)^{n+2 r-1}-\sum_{r=0}^{\infty}(-1)^{r-1} \frac{1}{(r-1)!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r-1}$

= $\sum_{r=0}^{\infty}(-1)^r \frac{1}{r!\Gamma(n-1+r+1)}\left(\frac{x}{2}\right)^{n-1+2 r}-\sum_{s=0}^{\infty}(-1)^s \frac{1}{s!\Gamma(1+n+s+1)}\left(\frac{x}{2}\right)^{n+1+2 s}$

by putting r-1=s.

= $J_{n-1}(x)-J_{n+1}(x) \text {. Hence } 2 J_n^{\prime}(x)=J_{n-1}(x)-J_{n+1}(x) \text {. }$

18. Prove that Recurrence Formula 4: $2 n J_n(x)=x\left[J_{n-1}(x)+J_{n+1}(x)\right]$.

Solution:

Writing Recurrence formula 1 and 2, we have $x J_n^{\prime}(x)=n J_n(x)-x J_{n+1}(x)$ and $x J_n^{\prime}(x)=-n J_n(x)+x J_{n-1}(x) \text {. }$

Subtracting, we have $0=2 n J_n(x)-x\left[J_{n+1}(x)+J_{n-1}(x)\right]$.

Hence $2 n J_n(x)=x\left[J_{n+1}(x)+J_{n-1}(x)\right]$

Aliter: $J_n(x)=\sum_{r=0}^{\infty}(-1)^r \frac{1}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r}$

∴ $2 n J_n(x)=\sum_{r=0}^{\infty}(-1)^r \frac{(2 n+2 r-2 r)}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r}$

= $\sum_{r=0}^{\infty}(-1)^r \frac{2(n+r)}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r}-\sum_{r=0}^{\infty}(-1)^r-\frac{2 r}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r}$

= $x \sum_{r=0}^x(-1)^r \frac{1}{r!\Gamma(n-1+r-1)}\left(\frac{x}{2}\right)^{n-1+2}+x \sum_{r=0}^{\infty}(-1)^{r-1} \frac{1}{(r-1)!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r-1}$

= $x J_{n-1}(x)+x \sum_{s=0}(-1)^s \frac{1}{s!\Gamma(n+1+s+1)} \times\left(\frac{x}{2}\right)^{n+1+2 s}$, by putting r-1=s

= $x J_{n-1}(x)+x J_{n+1}(x)$

Hence $2 n J_n(x)=x\left[J_{n-1}(x)+J_{n+1}(x)\right]$.

19. Prove that Recurrence Formula 5: $\frac{d}{d x}\left[x^{-n} J_n(x)\right]=x^{-n} J_{n+1}(x)$.

Solution:

⇒ $\frac{d}{d x}\left[x^{-n} J_n(x)\right]=-n x^{-n-1} J_n(x)+x^{-n} J_n^{\prime}(x)=x^{-n-1}\left[-n J_n(x)+x J_n^{\prime}(x)\right]$

= $x^{-n-1}\left[-n J_n(x)+\left\{n J_n(x)-x J_{n+1}(x)\right\}\right]=x^{-n-1}\left[-x J_{n+1}(x)\right]=-x^{-n} J_{n+1}(x)$

Hence $\frac{d}{d x}\left[x^{-n} J_n(x)\right]=-x^{-n} J_{n+1}(x)$

20. Prove that Recurrence Formula VI : $\frac{d}{d x}\left[x^n J_n(x)\right]=x^n J_{n-1}(x)$.

Solution:

⇒ $\frac{d}{d x}\left[x^n J_n(x)\right]=n x^{n-1} J_n(x)+x^n J_n^{\prime}(x)$

= $x^{n-1}\left[n J_n(x)+x J_n^{\prime}(x)\right]=x^{n-1}\left[n J_n(x)+\left\{-n J_n(x)+x J_{n-1}(x)\right\}\right]$

= $x^{n-1}\left[x J_{n-1}(x)\right]=x^n J_{n-1}(x)$

Hence $\frac{d}{d x}\left[x^n J_n(x)\right]=x^n J_{n-1}(x)$

21. Prove that $e^{x\left(z-\frac{1}{z}\right)}=\sum^{\infty} z^n J_n(x)$.

Solution:

⇒ $e^{r(z-1 / z) / 2}=e^{n / 2} e^{-x / 2 z} $

= $\left[1+\frac{x z}{2}+\frac{1}{2!}\left(\frac{x z}{2}\right)^2+\cdots+\frac{1}{n!}\left(\frac{x z}{2}\right)^n+\frac{1}{(n+1)!}\left(\frac{x z}{2}\right)^{n+1}+\frac{1}{(n+2)!}\left(\frac{x z}{2}\right)^{n+2}+\cdots\right]$

⇒ ${\left[1-\frac{x}{2 z}+\frac{1}{2!}\left(\frac{x}{2 z}\right)^2+\cdots+\frac{(-1)^n}{n!}\left(\frac{x}{2 z}\right)^n+\frac{(-1)^{n+1}}{(n+1)!}\left(\frac{x}{2 z}\right)^{n+1}+\frac{(-1)^{n+2}}{(n+2)!}\left(\frac{x}{2 z}\right)^{n+2}+\cdots\right]}$

Coefficient of $z^n$ in this product

= $\frac{1}{n!}\left(\frac{x}{2}\right)^n-\frac{1}{(n+1)!}\left(\frac{x}{2}\right)\left(\frac{x}{2}\right)^{n+1}+\frac{1}{(n+2)!} \frac{1}{2!}\left(\frac{x}{2}\right)^2\left(\frac{x}{2}\right)^{n+2}+\cdots$

= $\frac{1}{n!}\left(\frac{x}{2}\right)^n-\frac{1}{(n+1)!}\left(\frac{x}{2}\right)^{n+2}+\frac{1}{2!(n+2)!}\left(\frac{x}{2}\right)^{n+4}+\cdots$

= $\frac{(-1)^0}{\Gamma(n+1)}\left(\frac{x}{2}\right)^n+\frac{(-1)}{1!\Gamma(n+2)}\left(\frac{x}{2}\right)^{n+2}+\frac{(-1)^2}{2!\Gamma(n+3)}\left(\frac{x}{2}\right)^{n+4}+\cdots$

= $\sum_{r=0}^{\infty}(-1)^r \frac{1}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r}=J_n(x)$

Similarly, the coefficient of $z^{-n}$ in the product

= $\frac{(-1)^n}{n!}\left(\frac{x}{2}\right)^n+\frac{(-1)^{n+1}}{(n+1)!} \frac{x}{2}\left(\frac{x}{2}\right)^{n+1}+\frac{(-1)^{n+2}}{(n+2)!} \frac{1}{2!}\left(\frac{x}{2}\right)^2\left(\frac{x}{2}\right)^{n+2}+\cdots$

= $(-1)^n\left[\frac{1}{n!}\left(\frac{x}{2}\right)^n+\frac{(-1)}{\Gamma(n+2)}\left(\frac{x}{2}\right)^{n+2}+\frac{(-1)^2}{2!\Gamma(n+3)}\left(\frac{x}{2}\right)^{n+4}+. .\right]=(-1)^n J_n(x) .$

Hence $e^{x\left(z-\frac{1}{2}\right)^2}=J_0+\left(z-\frac{1}{z}\right) J_1+\left(z^2+\frac{1}{z^2}\right)_2+\cdots\left[z^n+(-1)^n \frac{1}{z^n}\right] J_n+\cdots$

= $\sum_{-\infty}^{\infty} J_n(x), \text { since } J_{-n}(x)=(-1)^n J_n(x)$

22. Prove that

1) $\cos (x \sin \theta)=J_0+2 J_2 \cos 2 \theta+2 J_4 \cos 4 \theta+\ldots$

2) $\sin (x \sin \theta)=2 \sin \theta J_1+2 \sin 3 \theta J_3+\ldots+2 J_{2 m+1} \sin (2 m+1) \theta+\ldots$

3) $\cos x=J_0-2 J_2+2 J_4 \ldots$

4) $\sin x=2 J_1-2 J_3+2 J_5 \ldots$

Solution:

We know that $e^{(z-1 / z) / 2}=J_0+\left(z-\frac{1}{z}\right) J_1+\left(z^2+\frac{1}{z^2}\right) J_2+\left(z^3-\frac{1}{z^3}\right) J_3+\cdots$

Putting $z=e^{i \theta}$, we have

⇒ $e^{x\left(\frac{e^{i \theta}-e^{-i \theta}}{2}\right)}=J_0+\left(e^{i \theta}-e^{-i \theta}\right) J_1+\left(e^{2 i \theta}+e^{-2 i \theta}\right) J_2+\left(e^{3 i \theta}-e^{-3 i \theta}\right) J_3+\cdots$

⇒ $e^{r i \sin \theta}=J_0+(2 i \sin \theta) J_1+(2 \cos 2 \theta) J_2+(2 i \sin 3 \theta) J_3+\cdots$

⇒ $\cos (x \sin \theta)+i \sin (x \sin \theta)$

= $\left(J_0+2 \cos 2 \theta J_2+2 \cos 4 \theta J_4+\cdots\right)+i\left(2 \sin \theta J_1+2 \sin 3 \theta J_3+\cdots\right)$

Equating real and imaginary parts, we have

1) $\cos (x \sin \theta)=J_0+2 J_2 \cos 2 \theta+2 J_4 \cos 4 \theta+\cdots+2 J_{2 m} \cos 2 m \theta+\cdots \rightarrow$ (1)

2) $\sin (x \sin \theta)=2 \sin \theta J_1+2 \sin 3 \theta J_3+\cdots+2 J_{2 m+1} \sin (2 m+1) \theta+\cdots \rightarrow$ (2)

Putting $\theta=\frac{\pi}{2}$ in (1) and (2) we have

3) $\cos x=J_0-2 J_2+2 J_4+\cdots$ and

4) $\sin x=2 J_1-2 J_3+2 J_5-\cdot$

23. Express $J_4(x)$ in terms of $J_0 \text { and } J_1$.

Solution:

From recurrence relation 4, we have $J_{n+1}(x)=\left(\frac{2 n}{x}\right) J_n(x)-J_{n-1}(x) \rightarrow$ (1)

Putting n=3 in (1), we get $J_4(x)=\left(\frac{6}{x}\right) J_3(x)-J_2(x) \rightarrow(2)$

Putting n=2 in (1), we get $J_3(x)=\left(\frac{4}{x}\right) J_2(x)-J_1(x) \Rightarrow$ (3)

Substituting (3) in (2), we get $J_4(x)=\frac{6}{x}\left[\frac{4}{x} J_2(x)-J_1(x)\right]-J_2(x)$

⇒ $J_4(x)=\left(\frac{24}{x^2}-1\right) J_2(x)-\frac{6}{x} J_1(x) \rightarrow(4)$

Next, putting n=1 in (1) we get $J_2(x)=\left(\frac{2}{x}\right) J_1(x)-J_0(x) \rightarrow$ (5)

Substituting (5) in (4), we get $J_4(x)=\left(\frac{24}{x^2}-1\right)\left[\frac{2}{x} J_1(x)-J_0(x)\right]-\frac{6}{x} J_1(x)$

⇒ $J_4(x)=\left(\frac{48}{x^3}-\frac{8}{x}\right) J_1(x)-\left(\frac{24}{x^2}-1\right) J_0(x)$

24. Prove that

1) $J_0^{\prime}=-J_1$

2) $J_2=J_0^{\prime \prime}-x^{-1} J_0^{\prime}$ and

3) $J_2-J_0=2 J_0^{\prime \prime}$.

Solution:

1) From Recurrence formula $\mathrm{I}$, we have $x J_n^{\prime}=n J_n-x J_{n+1}$.

Putting n=0, we have $x J_0^{\prime}=-x J_1$

∴ $J_0^{\prime}=-J_1$

2) From Recurrence formula {1}, we have $x J_n^{\prime}=n J_n-x J_{n+1}$.

Putting n=1, we have $x J_1^{\prime}=J_1-x J_2 \rightarrow$ (1).

But from (1): $J_0^{\prime}=-J_1$.

∴ Differentiating, wehave $J_0^{\prime \prime}=-J_1^{\prime}$.

Substituting these results in (1), we have $-x J_0^{\prime \prime}=-J_0^{\prime}-x J_2 \Rightarrow x J_2=x J_0^{\prime \prime}-J_0^{\prime}$.

‍‍∴ $J_2=J_0^{\prime \prime}-x^{-1} J_0^{\prime} \text {. }$

3) From Recurrence formula 3, we have $2 J_n^{\prime}=J_{n-1}-J_{n+1} \rightarrow(A)$

Differentiating both sides w.r.t. x and multiplying by 2 we have $2^2 J_n^{\prime \prime}=2 J^{\prime}{ }_{n-1}-2 J^{\prime}{ }_{n+1} \rightarrow(B)$

From (A), replacing n by (n-1) and (n+1) respectively, we have $2 J_{n-1}^{\prime}=J_{n-2}-J_n \text { and } 2 J_{n+1}^{\prime}=J_n-J_{n+2}$

Substituting these values in (B), we have $2^2 J_n^{\prime \prime}=\left(J_{n-2}-J_n\right)-\left(J_n-J_{n+2}\right)=J_{n-2}-2 J_n+J_{n+2}$

Putting n=0, we get $2^2 J_0^{\prime \prime}=J_{-2}-2 J_0+J_2=(-1)^2 J_2-2 J_0+J_2=2 J_2-2 J_0$ since $J_{-n}=(-1)^n J_n$

Hence $2 J_0^{\prime \prime}=J_2-J_0$

25. Prove that $4 \frac{d^2}{d x^2}\left[J_n(x)\right]=J_{n-2}(x)-2 J_n(x)+J_{n+2}(x)$.

Solution:

Recurrence formula 3 is $2 J_n^{\prime}(x)=J_{n-1}(x)-J_{n+1}(x) \rightarrow(1)$

Replacing n by n-1 in (1), we get $2 J_{n-1}^{\prime}(x)=J_{n-2}(x)-J_n(x) \rightarrow$ (2)

Replacing n by n+1 in (1), we get $2 J_{n+1}^{\prime}(x)=J_n(x)-J_{n+2}(x) \rightarrow$ (3)

From (1), (2) and (3); $4 \frac{d^2}{d x^2}\left[J_n(x)\right]=2 \frac{d}{d x}\left\{2 J_n^{\prime}(x)\right\}=2 \frac{d}{d x}\left\{J_{n-1}(x)-J_{n+1}(x)\right\}=2 J_{n-1}^{\prime}(x)-2 J_{n+1}^{\prime}(x)$

= $J_{n-2}(x)-J_n(x)-J_n(x)+J_{n+2}(x)=J_{n-2}(x)-2 J_n(x)+J_{n+2}(x)$

26. Establish the differential formula $x^2 J_n^{\prime \prime}(x)=\left(n^2-n-x^2\right) J_n(x)+x J_{n+1}(x) \text { for } n=0,1,2, \ldots$

Solution:

Writing recurrence formula I, we have $x J_n^{\prime}(x)=n J_n(x)-x J_{n+1}(x) \rightarrow(1)$

Differentiating both sides w.r.t. x, we have

⇒ $x J_n^{\prime \prime}(x)+J_n^{\prime}(x)=n J_n^{\prime}(x)-x J_{n+1}^{\prime}(x)-J_{n+1}(x)$

⇒ $x^2 J_n^{\prime \prime}(x)=(n-1) x J_n^{\prime}(x)-x \cdot x J_{n+1}^{\prime}(x)-x J_{n+1}(x) \rightarrow(2)$

From recurrence formula 2, we have $x J_n^{\prime}(x)=-n J_n(x)+x J_{n-1}(x)$.

Writing (n+1) for n; we have $x J_{n+1}^{\prime}(x)=-(n+1) J_{n+1}(x)+x J_n(x) \rightarrow$ (3)

Substituting for $x J_n^{\prime}(x)$ from (1) and for $x J_{n+1}^{\prime}(x)$ from (3) in (2),

we have $x^2 J_n^{\prime \prime}(x)=(n-1)[n J_n(x)-x J_{n+1}(x)-x\left[-(n+1) J_{n+1}(x)+x J_n(x)\right]-x J_{n+1}(x)]$

= $\left(n^2-n-x^2\right) J_n(x)+x J_{n+1}(x) .$

27. Prove that

1) $J_{n-1}=\frac{2}{x}\left[n J_n-(n+2) J_{n+2}+(n+4) J_{n+4} \cdots\right]$

2) $\frac{1}{2} x J_n=(n+1) J_{n+1}-(n+3) J_{n+3}+(n+5) J_{n+5}+\ldots$

3) $J_n^{\prime}=\frac{2}{x}\left[\frac{n}{2} J_n-(n+2) J_{n+2}+(n+4) J_{n+4} \cdots \cdots\right]$

Solution:

1) From recurrence formula 4, we have $x\left(J_{n-1}+J_{n+1}\right)=2 n J_n \Rightarrow J_{n-1}+J_{n+1}=\frac{2}{x} n J_n \rightarrow \text { (1). }$

Replacing n by (n+2), and changing sign, we get $-J_{n+1}-J_{n+3}=\frac{-2}{x}(n+2) J_{n+2} \rightarrow(2)$

Again replacing n by (n+4) in (1), we have $J_{n+3}+J_{n+5}=\frac{2}{x}(n+4) J_{n+4} \rightarrow$ (3)

Again replacing n by (n+6) in (1) and changing the sign, we have $-J_{n+5}-J_{n+7}=\frac{2}{x}(n+6) J_{n+6} \rightarrow(4)$

……..

……..

Adding (1), (2), (3), (4), etc., we have $J_{n-1}=\frac{2}{x}\left[n J_n-(n+2) J_{n+2}+(n+4) J_{n+4}-(n+6) J_{n+6} \cdots\right]$

2) Replacing n by (n+1), we have $\frac{x}{2} J_n=(n+1) J_{n+1}-(n+3) J_{n+3}+(n+5) J_{n+5} \cdots$

3) From recurrence formula II, we have $J_n^{\prime}=-\frac{n}{x} J_n+J_{n-1}$.

Substituting the value of $J_{n-1}$ from (1), we have $J_n^{\prime}=-\frac{n}{x} J_n+\frac{2}{x}\left[n J_n-(n+2) J_{n+2}+(n+4) J_{n+4} \cdots\right]$

= $\frac{2}{x}\left[\frac{n}{2} J_n-(n+2) J_{n+2}+(n+4) J_{n+4} \cdots\right]$

28. Prove that

1) $\frac{d}{d x}\left[J_n^2+J_{n+1}^2\right]=2\left(\frac{n}{x} J_n^2-\frac{n+1}{x} J_{n+1}^2\right)$

2) $J_0^2+2\left(J_1^2+J_2^2+J_3^2+\ldots\right)=1$

3) $\left|J_0(x)\right| \leq 1,\left|J_n(x)\right| \leq 2^{-1 / 2},(n \geq 1)$

Solution:

From recurrence formula 1, we have $J_n^{\prime}=\frac{n}{x} J_n-J_{n+1} \rightarrow$ (1)

Also from recurrence formula 2, we have $J_n^{\prime}=-\frac{n}{x} J_n+J_{n-1}$

Replacing n by (n+1), we have $J_{n+1}^{\prime}=-\frac{n+1}{x} J_{n+1}+J_n \rightarrow$ (3)

Substituting the values of $J_n^{\prime}$ and $J_{n+1}^{\prime}$ from (2) and (3), we have

⇒ $\frac{d}{d x}\left(J_n^2+J_{n+1}^2\right)=2 J_n J_n^{\prime}+2 J_{n+1} J_{n+1}^{\prime}$

= $2 J_n\left(\frac{n}{\chi} J_n-J_{n+1}\right)+2 J_{n+1}\left(-\frac{n+1}{x} J_{n+1}+J_n\right)=2\left(\frac{n}{x} J_n^2-\frac{n+1}{x} J_{n+1}^2\right)$

2) From (1), we have $\frac{d}{d x}\left(J_n^2+J_{n+1}^2\right)=2\left(\frac{n}{x} J_n^2-\frac{n+1}{x} J_{n+1}^2\right)$

Putting n=0,1,2,3, ….. etc., we have

⇒ $\frac{d}{d x}\left(J_0^2+J_1^2\right)=2\left(0-\frac{1}{x} J_1^2\right)$

⇒ $\frac{d}{d x}\left(J_1^2+J_2^2\right)=2\left(\frac{1}{x} J_1^2-\frac{2}{x} J_2^2\right)$

⇒ $\frac{d}{d x}\left(J_2^2+J_3^2\right)=2\left(\frac{2}{x} J_2^2-\frac{3}{x} J_3^2\right)$

………

………

Adding all these, we have $\frac{d}{d x}\left[J_0^2+2\left(J_1^2+J_2^2+J_3^2+..\right)\right]=0$

Integrating, we have $J_0^2+2\left(J_1^2+J_2^2+J_3^2+\cdots\right)=c$ (constant), Putting x=0, we get $\left(J_0^2\right)_{x=0}=c$, since

⇒ $\left(J_1\right)_{x=0}=0,\left(J_2\right)_{x=0}=0$ etc.

⇒ $\left(J_0^2\right)_{x=0}=c \Rightarrow 1=c$.

Hence $J_0^2+2\left(J_1^2+J_2^2+J_3^2+\cdots\right)=1$.

3) We have proved that $J_0^2+2\left(J_1^2+J_2^2+J_3^2 \cdots+J_n^2+\cdots\right)=1$

Since $J_1^2, J_2^2, J_3^2 \ldots$ are all positive or zero.

∴ $J_0^2 \leq 1 \Rightarrow\left|J_0\right| \leq 1$, i.e.. $\left|J_0(x)\right| \leq 1$.

Also $|2 J_n^2| \leq 1 \Rightarrow W_n |\leq \frac{1}{\sqrt{2}}$, i.e., $\left|J_n(x)\right| \leq 2^{-1 / 2}$ (for $n \geq 1$).

29. Prove that $J_n(x)=0$ has no repeated roots except at x=0.

Solution:

If possible let α be a repeated root of $J_n(x)=0$.

∴ $J_n(\alpha)=0 \text { and } J_n^{\prime}(\alpha)=0 \rightarrow(1)$

From Recurrence formulas 1 and 2, we have $J_{n+1}(x)=\frac{n}{x} J_n-J_n^{\prime}(x)$ and $J_{n-1}(x)=\frac{n}{x} J_n+J_n^{\prime}(x)$

∴ $J_{n+1}(\alpha)=0$ and $J_{n-1}(\alpha)=0$ with the help of (1), i.e., for the same value of $x, J_n(x), J_{n+1}(x)$ and $J_{n-1}(x)$ are all zero, which is absurd as we cannot have two power series having the same sum function. Thus $J_n(x)=0$ cannot have repeated roots except at x=0.

30. Show that $\int_0^x x^n J_{n-1}(x) d x=x^n J_n(x)$.

Solution:

Recurrence formula VI is $\frac{d}{d x}\left[x^n J_n(x)\right]=x^n J_{n-1}(x)$

Integrate this between the limits 0 to x, we get

∴ $\left[x^n J_n(x)\right]_0^x=\int_0^x x^n J_{n-1}(x) d x \Rightarrow \int_0^x x^n J_{n-1}(x) d x=x^n J_n(x) .$.

31. Show that $\int_0^x x^{n+1} J_n(x) d x=x^{n+1} J_{n+1}(x)$.

Solution:

Recurrence formula VI is $\frac{d}{d x}\left[x^n J_n(x)\right]=x^n J_{n-1}(x)$

Replace n by n+1, we get $\frac{d}{d x}\left[x^{n+1} J_{n+1}(x)\right]=x^{n+1} J_n(x)$

Integrate this between the limits 0 to x we get

$\left[x^{n+1} J_{n+1}(x)\right]=\int_0^x x^{n+1} J_n(x) d x \Rightarrow \int_0^x x^{n+1} J_n(x) d x=x^{n+1} J_{n+1}(x) .$.

32. Prove that $\int_a^b J_0(x) J_1(x) d x=\frac{1}{2}\left[J_0^2(a)-J_0^2(b)\right]$.

Solution:

Recurrence relation VII is $\frac{d}{d x}\left[x^{-n} J_n(x)\right]=-x^{-n} J_{n+1}(x) \Rightarrow \frac{d}{d x} J_0(x)=-J_1(x)$

∴$\int_a^b J_0(x) J_1(x) d x=-\int_a^b J_0(x) J_0^{\prime}(x) d x=-\left[\frac{\left[J_0(x)\right]^2}{2}\right]_a^b=\frac{1}{2}\left[J_0^2(a)-J_0^2(b)\right]$.

33. Prove that

1) $\frac{d}{d x}\left(x J_n J_{n+1}\right)=x\left(J_n^2-J_{n+1}^2\right)$

2) $x=2 J_0 J_1+6 J_1 J_2+\ldots+2(2 n+1) J_n J_{n+1}+\ldots$

Solution:

1) $\frac{d}{d x}\left(x J_n J_{n+1}\right)=J_n J_{n+1}+x\left(J_n^{\prime} J_{n+1}+J_n J_{n+1}^{\prime}\right)$

= $J_n J_{n+1}+\left(x J_n^{\prime}\right) J_{n+1}+J_n\left(x J_{n+1}^{\prime}\right) \rightarrow \text { (1) }$

From Recurrence formulas 1 and 2, we have $x J_n^{\prime}=n J_n-x J_{n+1} \rightarrow$ (2)

and $x J_n^{\prime}=-n J_n+x J_{n-1} \rightarrow \text { (3). }$

Replacing n by (n+1) in (3), we have $x J_{n+1}^{\prime}=-(n+1) J_{n+1}+x J_n \rightarrow$ (4)

Substituting the values of $x J_n^{\prime}$ and $x J_{n+1}^{\prime}$ from (2) and (4), in (1), we have

∴ $\frac{d}{d x}\left(x J_n J_{n+1}\right)=J_n J_{n+1}+\left(n J_n-x J_{n+1}\right) J_{n+1}+J_n\left\{-(n+1) J_{n+1}+x J_n\right\}=x\left(J_n^2-J_{n+1}^2\right) \text {. }$

2) From (1), we have $\frac{d}{d x}\left(x J_n J_{n+1}\right)=x\left(J_n^2-J_{n+1}^2\right)$

Putting n=0,1,2,3, ….. we have

⇒ $\frac{d}{d x}\left(x J_0 J_1\right)=x\left(J_0^2-J_1^2\right) \rightarrow(1)$

⇒ $\frac{d}{d x}\left(x J_1 J_2\right)=x\left(J_1^2-J_2^2\right) \rightarrow(2)$

⇒ $\frac{d}{d x}\left(x J_2 J_3\right)=x\left(J_2^2-J_3^2\right) \rightarrow(3)$

⇒ $\frac{d}{d x}\left(x J_3 J_4\right)=x\left(J_3^2-J_4^2\right) \rightarrow(4)$

………..

…….

Multiplying (1), (2), (3)… by 1,3,5, ………. respectively and adding we have

⇒ $\frac{d}{d x} {\left[x\left(J_0 J_1+3 J_1 J_2+5 J_2 J_3+\cdots\right)\right]=x\left[\left(J_0^2-J_1^2\right)+3\left(J_1^2-J_2^2\right)+5\left(J_2^2-J_3^2\right)+\cdots\right] }$

= $x\left[J_0^2+2\left(J_1^2+J_2^2+\cdots\right)\right]=x \cdot 1=x$

Integrating both sides, we have $x\left(J_0 J_1+3 J_1 J_2+5 J_2 J_2+\cdots\right)=\frac{x^2}{2}+c$ (constant)

Putting x=0, we get 0=0+c ∴ c=0

Hence $x\left(J_0 J_1+3 J_1 J_2+5 J_2 J_3+\cdots\right)=\frac{x^2}{2}$

⇒ $2 J_0 J_1+6 J_1 J_2+10 J_2 J_3+\cdots+2(2 n+1) J_n J_{n+1}+\cdots=x$

34. Show that $2^r J_n^r=J_{n-r}-r J_{n-r+2}+\frac{r(r-1)}{2 !} J_{n-r+4}-\cdots+(-1)^r J_{n+r}$

Solution:

Recurrence formula 3 is $2 J_n{ }^{\prime}=J_{n-1}-J_{n+1} \rightarrow$ (1)

Replacing n by (n-1) and (n+1) respectively in (1), we get $2 J_{n-1}^{\prime}=J_{n-2}-J_n \rightarrow$ (2)

and $2 J_{n+1}^{\prime}=J_n-J_{n+2} \rightarrow \text { (3) }$

Differentiating (1), w.r.t. ‘ x ‘, we have $2 J_n^{\prime \prime}=J_{n-1}^{\prime}-J^{\prime}{ }_{n+1} \Rightarrow 2^2 J_n^{\prime \prime}=2 J_{n-1}^{\prime}-2 J_{n+1}^{\prime} \rightarrow \text { (4) }$

Substituting the values of $2 J_{n-1}^{\prime}$ and $2 J_{n+1}^{\prime}$ from (2) and (3) obtained in (4), we get $2^2 J_n^{\prime \prime}=\left(J_{n-2}^{\prime}-J_n\right)-\left(J_n-J_{n+2}\right)=J_{n-2}-2 J_n+J_{n+2}^{\prime \prime} \text {. }$

Differentiating (4) again, and multiplying by 2, we have $2^3 J_n^{\prime \prime \prime}=2 J_{n-2}^{\prime}-2^2 J_n^{\prime}+2{ }_{n+2}^{\prime} \rightarrow \text { (5). }$

Substituting in (5), the values of $2 J_n^{\prime}, 2 J_{n-2}^{\prime}, 2 J_{n+2}^{\prime}$ from (1), the last two obtained by replacing n by (n-2) and (n+2) respectively in (1),

we get $^3 J_n^{\prime \prime \prime}=\left(J_{n-3}-J_{n-1}\right)-2\left(J_{n-1}-J_{n+1}\right)+\left(J_{n+1}-J_{n+2}\right)$

= $J_{n-2}-3 J_{n-1}+3 J_{n+1}-J_{n+2}=J_{n-2}-{ }^3 C_1 J_{n-1}+{ }^2 C_2 J_{n+1}-{ }^3 C_3 J_{n+3}$

Applying the same process again, and again, we have $2^r J_n^r=J_{n-r}-{ }^r C_1 J_{n-r+2}+{ }^r C_2 J_{n-r+4}+\cdots+(-1)^r \cdot{ }^r C_r J_{n+r}$

= $J_{n-r}-r J_{n-r+2}+\frac{r(r-1)}{2!} J_{n-r+4}+\cdots+(-1)^r J_{n+r}$

35. Prove that $J_n J_{-n}^{\prime}-J_n^{\prime} J_{-n}=-\frac{2 \sin n \pi}{x \pi}$.

Solution:

We know that $J_n$ and $J_{-n}$ are solutions of Bessel’s equation $y^{\prime \prime}+\frac{1}{x} y^{\prime}+\left(1-\frac{n^2}{x^2}\right) y=0$

∴ $J_n^{\prime \prime}+\frac{1}{x} J_n^{\prime}+\left(1-\frac{n^2}{x^2}\right) J_n=0 \rightarrow$ (1)

and $J_{-n}^{\prime \prime}+\frac{1}{x} J_{-n}^{\prime}+\left(1-\frac{n^2}{x^2}\right) J_{-n}=0 \rightarrow \text { (2) }$

Multiplying (1) by $J_{-n}$ and (2) by $J_n$ and then substracting. we get $J_n^{\prime \prime} J_{-n}-J_{-n}^{\prime \prime} J_n+\frac{1}{x}\left(J_n^{\prime} J_{-n}-J_{-n}^{\prime} J_n\right)=0 \rightarrow (3)$

Let $J_n{ }^{\prime} J_{-n}-J_{-n}^{\prime} J_n=v \rightarrow(4)$

Differentiating (4) w.r.t. x we get $\left(J_n^{\prime \prime} J_{-n}+J_n^{\prime} J_{-n}^{\prime}\right)-\left(J_{-n}^{\prime \prime} J_n+J_{-n}^{\prime} J_n^{\prime}\right)=v^{\prime}$

⇒ $J_n^{\prime \prime} J_{-n}-J_{-n}^{\prime \prime} J_n=v^{\prime} \rightarrow \text { (5) }$

Using (4) and (5), (3) becomes $v^{\prime}+\left(\frac{1}{x}\right) v=0 \Rightarrow \frac{d v}{d x}+\frac{1}{x} v=0 \Rightarrow \frac{1}{v} d v+\frac{1}{x} d x=0$

Integrating, we get $\log v+\log x=\log c \Rightarrow v=\frac{c}{x} \Rightarrow J_n^{\prime} J_{-n}-J_{-n}^{\prime} J_n=\frac{c}{x} \rightarrow$ (6) by (4)

Now, $J_n(x)=\frac{1}{2^n \Gamma(n+1)}\left[x^n-\frac{x^{n+2}}{4(n+1)}+\frac{x^{n+4}}{4 \cdot 8 \cdot(n+1)(n+2)}-\cdots\right]$

and $J_{-n}(x)=\frac{1}{2^{-n} \Gamma(-n+1)}\left[x^{-n}-\frac{x^{2-n}}{4(1-n)}+\frac{x^{4-n}}{4 \cdot 8 \cdot(1-n)(2-n)}-\cdots\right]$

∴ Using the above values of $J_n$ and $J_{-n}$, and from (6) we have

⇒ $\frac{1}{2^n \Gamma(n+1)}\left[n x^{n-1}-\frac{(n+2) x^{n+1}}{4(n+1)}+(n+4) \frac{x^{n+3}}{4 \cdot 8 \cdot(1-n)(2-n)}-\cdots\right]$

x $\frac{1}{2^{-n} \Gamma(-n+1)}\left[x^{-n}-\frac{x^{2-n}}{4(1-n)}+\frac{x^{4-n}}{4 \cdot 8 \cdot(1-n)(2-n)}-\cdots\right]$

– $\frac{1}{2^{-n} \Gamma(-n+1)}\left[-n x^{-n-1}-(2-n) \frac{x^{1-n}}{4(1-n)}+(4-n) \frac{x^{3-n}}{4 \cdot 8 \cdot(1-n)(2-n)}-\cdots\right]$

x $\frac{1}{2^n \Gamma(n+1)}\left[x^n-\frac{x^{n+2}}{4(n+1)}+\frac{x^{n+4}}{4 \cdot 8 \cdot(n+1)(n+2)}-\cdots\right]=\frac{c}{x}$

Now comparing the coefficients of $\frac{1}{x}$ from both sides, we get

⇒ $\frac{n}{2^n \Gamma(n+1) \cdot 2^{-n} \Gamma(-n+1)}+\frac{n}{2^{-n} \cdot \Gamma(-n+1) 2^n \Gamma(n+1)}=c$

⇒ $c=\frac{2 n}{n \Gamma(n) \Gamma(1-n)} \Rightarrow c=\frac{2}{(\pi / \sin n \pi)}=\frac{2 \sin n \pi}{\pi}$

(because $\Gamma(n) \Gamma(1-n)=\frac{\pi}{\sin n \pi}$)

Putting this value of c in (6) and multiplying both sides by (-1), we get $J_n J_{-n}^{\prime}-J_n^{\prime} J_{-n}=-\frac{2 \sin n \pi}{\pi x}$

36. Prove that $\frac{J_{n+1}}{J_n}=\frac{(x / 2)}{(n+1)-\frac{(x / 2)^2}{(n+2)-\frac{(x / 2)^2}{(n+3)} \cdots}}$

Solution:

Since $J_n$ and $J_{-n}$ are the solutions of the Bessel’s differential equation.

⇒ $\frac{d^2 y}{d x^2}+\frac{1}{x}-\frac{d y}{d x}+\left(1-\frac{n^2}{x^2}\right) y=0$, we have $\frac{d^2 J_n}{d x^2}+\frac{1}{x} \frac{d J_n}{d x}+\left(1-\frac{n^2}{x^2}\right) J_n=0 \rightarrow(1)$ and $\frac{d^2 J_{-n}}{d x^2}+\frac{1}{x} \frac{d J_{-n}}{d x}+\left(1-\frac{n^2}{x^2}\right) J_{-n}=0 \rightarrow (2)$

Multiplying (1) by $J_{-n}$ and (2) by $J_n$ and subtracting, we have

⇒ $\left(J_{-n} \frac{d^2 J_n}{d x^2}-J_n \frac{d^2 J_{-n}}{d x^2}\right)+\frac{1}{x}\left(J_{-n} \frac{d J_n}{d x}-J_n \frac{d J_{-n}}{d x}\right)=0$

⇒ $\frac{J_{-n} \frac{d^2 J_n}{d x^2}-J_n}{J_{-n} \frac{d J_n}{d x}-J_n}$ $\frac{\frac{d^2 J_{-n}}{d x^2}}{\frac{d J_{-n}}{d x}}=-\frac{1}{x}$

Integrating both the sides w.r.t x, we have $\log \left(J_{-n} \frac{d J_n}{d x}-J_n \frac{d J_n}{x}\right)=-\log x+\log A$

where A is an arbitrary constant. ∴ ${-n} \frac{d J_n}{d x}-J_n \frac{d J_{-n}}{d x}=\frac{A}{x} \rightarrow(3)$

∴ $\frac{1}{2^{-n} \Gamma(-n+1)}\left[x^{-n}-\frac{x^{-n+2}}{2(-2 n+2)}+\frac{x^{-n+4}}{2 \cdot 4(-2 n+2)(-2 n+4)} \cdots\right]$

x $\frac{1}{2^n \Gamma(n+1)}\left[n x^{n-1}-\frac{(n+2) x^{n+1}}{2 \cdot(2 n+2)}+\frac{(n+4) x^{n+3}}{2 \cdot 4(2 n+2)(2 n+4)} \cdots\right]$

– $\frac{1}{2^n \Gamma(n+1)} \cdot\left[x^n-\frac{x^{n+2}}{2(2 n+2)}+\frac{x^{n+4}}{2 \cdot 4(2 n+2)(2 n+4)}\right]$

x $\frac{1}{2^{-n} \Gamma(-n+1)}\left[-n x^{-n-1}-\frac{(-n+2) x^{-n+1}}{2 \cdot(-2 n+2)}+\frac{(-n+4) x^{-n+3}}{2 \cdot 4(-2 n+2)(-2 n+4)}\right]=\frac{A}{x}$.

Comparing coefficient of $\frac{1}{x}$ on both the sides, we have $\frac{1}{\Gamma(-n+1) \Gamma(n+1)}[n-(-n)]=A \Rightarrow A=\frac{2 n}{\Gamma(-n+1) \Gamma(n+1)}=\frac{2}{\Gamma(1-n) \Gamma(n)}$

= $\frac{2}{\frac{\pi}{\sin n \pi}}=\frac{2 \sin n \pi}{\pi}$, since $\Gamma(n) \Gamma(1-n)=\frac{\pi}{\sin n \pi}$

Hence from (3), we have $J_{-n} \frac{d J_n}{d x}-J_n \frac{d J_{-n}}{d x}=\frac{2 \sin n \pi}{\pi x}$

Dividing both sides by $-J_n^2$, we get $\frac{\frac{d J_{-n}}{d x} J_n-J_{-n} \frac{d J_n}{d x}}{J_n^2}=-\frac{2 \sin n \pi}{\pi x J_n^2} \Rightarrow \frac{d}{d x}\left(\frac{J_{-n}}{J_n}\right)=-\frac{2 \sin n \pi}{\pi x J_n^2}$.

37. Prove that $\frac{d}{d x}\left(\frac{J_{-n}}{J_n}\right)=-\frac{2 \sin n \pi}{\pi x J_n^2}$.

Solution:

1) Recurrence relation 6 is $\left.\frac{d}{d x}\left(x^n J_n\right)=x^n J_{n-1} \Rightarrow\left(\frac{1}{x} \frac{d}{d x}\right)\left(x^n J_n\right)=x^{n-1} J_{n-1}\right)$

Now, $\left(\frac{1}{x} \frac{d}{d x}\right)^m\left(x^n J_n\right)$

= $\left(\frac{1}{x} \frac{d}{d x}\right)^{m-1}\left(\frac{1}{x} \frac{d}{d x}\right)\left(x^n J_n\right)=\left(\frac{1}{x} \frac{d}{d x}\right)^{m-1}\left(x^{n-1} J_{n-1}\right) \cdots$

= $x^{n-m} J_{n-m}$, on preceeding as before m times more.

2) Recurrence relation 7 is $\frac{d}{d x}\left(x^{-n} J_n\right)=-x^{-n} J_{n+1} \Rightarrow\left(\frac{1}{x} \frac{d}{d x}\right)\left(x^{-n} J_{n+1}\right)=(-1)^1 x^{-n-1} J_{n+1}$

Now $\left(\frac{1}{x} \frac{d}{d x}\right)^m\left(x^{-n} J_n\right)=\left(\frac{1}{x} \frac{d}{d x}\right)^{m-1}\left(\frac{1}{x} \frac{d}{d x}\right)\left(x^{-n} J_n\right)=\left(\frac{1}{x} \frac{d}{d x}\right)^{m-1}(-1)^1 x^{n-1} J_{n+1}$

……. = $(-1)^m x^{-n-m} J_{n+m} \rightarrow(3)$, on proceeding as before m times more.

3) Replacing n by o and m by n in part $\left(\frac{1}{x} \frac{d}{d x}\right)^m\left(x^{-n} J_n\right)=(-1)^m x^{-n-m} J_{n+m}$, we get

⇒ $\left(\frac{1}{x} \frac{d}{d x}\right)^n J_0=(-1)^n x^{-n} J_n \Rightarrow J_n(x)=(-1)^n x^n\left(\frac{1}{x} \frac{d}{d x}\right)^n J_0(x)$
38. Prove that
1) $\left(\frac{1}{x} \frac{d}{d x}\right)^m\left(x^n J_n\right)=x^{n-m} J_{n-m}$, where m is positive integer and m
2) $\left(\frac{1}{x} \frac{d}{d x}\right)^m\left(x^{-n} J_n\right)=(-1)^m x^{-n-m} J_{n+m}$

3) $J_n(x)=(-1)^n x^n\left(\frac{1}{x} \frac{d}{d x}\right)^n J_0(x)$,n being positive integer

Solution:

From recurrence formula 4, we have $J_{n-1}+J_{n+1}=\frac{2 n}{x} J_n \Rightarrow J_{n-1}=\frac{2 n}{x} J_n-J_{n+1}$.

Replacing n by (n+1), we have $J_n=\frac{2(n+1)}{x} J_{n+1}-J_{n+2} \Rightarrow \frac{J_n}{J_{n+1}}=\frac{2(n+1)}{x}-\frac{J_{n+2}}{J_{n+1}} \rightarrow \text { (1). }$

Replacing n by (n+1) in (1), we have $\frac{J_{n+1}}{J_{n+2}}=\frac{2(n+2)}{x}-\frac{J_{n+3}}{J_{n+2}} \rightarrow$ (2) and etc.

⇒ $\frac{J_{n+1}}{J_n}=\frac{1}{\frac{J_n}{J_{n+1}}}=\frac{1}{\frac{2(n+1)}{x}-\frac{J_{n+2}}{J_{n+1}}}$

= $\frac{1}{\frac{2(n+1)}{x}-\frac{1}{\frac{J n+1}{J n+2}}}=\frac{1}{\frac{2(n+1)}{x}-\frac{1}{\frac{2(n+2)}{x}-\frac{J_{n+3}}{J_{n+2}}}}$

= $\frac{1}{\frac{2(n+1)}{x}-\frac{1}{\frac{2(n+2)}{x}-\frac{1}{J n+2}}}=\frac{1}{\frac{2(n+1)}{x}-\frac{1}{\frac{2(n+2)}{x}-\frac{1}{\frac{2(n+3)}{x} \ldots}}}$

= $\frac{x / 2}{(n+1)-\frac{x / 2}{\frac{2(n+2)}{x}-\frac{1}{\frac{2(n+3)}{x}} \cdots}}=\frac{x / 2}{(n+1)-\frac{x}{(x / 2)^2}}$

= $\frac{(x / 2)}{(n+1)-\frac{(x / 2)^2}{(n+2)-\frac{(x / 2)^2}{(n+3)} \cdots}}$

39. Show that $x^n J_n(x)$ is a solution of $x \frac{d^2 y}{d x^2}+(1-2 n) \frac{d y}{d x}+x y=0$.

Solution:

Given equation is $x \frac{d^2 y}{d x^2}+(1-2 n) \frac{d y}{d x}+x y=0 \rightarrow$ (1)

Let $y=x^n J_n(x) \rightarrow(2)$

Recurrence relation 6 is $\frac{d}{d x}\left[x^n J_n(x)\right]=x^n J_{n-1}(x) \Rightarrow \frac{d y}{d x}=x^n J_{n-1}$

and hence $\frac{d^2 y}{d x^2}=x^n J_{n-1}^{\prime}+n x^{n-1} J_{n-1}$

From recurrence relation 1 we have $J_n^{\prime}(x)=\frac{n}{x} J_n-J_{n+1}(x) \Rightarrow J_{n-1}^{\prime}-\frac{n-1}{x} J_{n-1}=-J_n(x) \rightarrow \text { (5) }$

∴ $x\left(x^n J_{n-1}^{\prime}+n x^{n-1} J_{n-1}\right)+(1-2 n) x^n J_{n-1}+x^{n+1} J_n$

=$x^{n+1} J_{n-1}^{\prime}-(n-1) x^n J_{n-1}+x^{n+1} J_n$

= $x^{n+1}\left[J_{n-1}^{\prime}-\frac{(n-1)}{x} J_{n-1}\right]+x^{n+1} J_n=x^{n+1} J_n-x^{n+1} J_n=0$

Hence $x^n J_n$ is a solution of $x \frac{d^2 y}{d x^2}+(1-2 n) \frac{d y}{d x}+x y=0$.

40. Show that $x^{-n} J_n(x)$ is a solution of $x \frac{d^2 y}{d x^2}+(1+2 n) \frac{d y}{d x}+x y=0$.

Solution:

Given equation is $x \frac{d^2 y}{d x^2}+(1+2 n) \frac{d y}{d x}+x y=0 \rightarrow$ (1)

Let $y=x^{-n} J_n(x) \rightarrow(2)$

Recurrence relation V is $\frac{d}{d x}\left\{-x^n J_n(x)\right\}=-x^{-n} J_{n+1}(x) \Rightarrow \frac{d y}{d x}=-x^{-n} J_{n+1}$

and hence $\frac{d^2 y}{d x^2}=-x^{-n} J_{n+1}^{\prime}+n x^{-n-1} J_{n+1} \rightarrow$ (3)

From recurrence relation 2, we have $J_n^{\prime}(x)=J_{n-1}(x)-\frac{n}{x} J_n \Rightarrow J_{n+1}^{\prime}+\frac{n+1}{x} J_{n+1}=J_n(x)$

∴ $x\left(-x^{-n} J_{n+1}^{\prime}+n x^{-n-1} J_{n+1}\right)+(1+2 n)\left(-x^{-n} J_{n+1}\right)+x^{-n+1} J_n$

= $-x^{-n+1} J_{n+1}^{\prime}-(n+1) x^{-n} J_{n+1}+x^{-n+1} J_n$

=$-x^{-n+1}\left[J_{n+1}^{\prime}+\frac{n+1}{x} J_{n+1}\right]+x^{-n+1} J_n=-x^{-n+1} J_n+x^{-n+1} J_n=0$

Hence $x^{-n} J_n$ is a solution of $x \frac{d^2 y}{d x^2}+(1+2 n) \frac{d y}{d x}+x y=0$

41. Show that $y=x^{-n / 2} J_n(2 \sqrt{x})$ satisfies the differential equation $x \frac{d^2 y}{d x^2}+(n+1) \frac{d y}{d x}+y=0$.

Solution:

Put $2 \sqrt{x}=t$. Then $x=\frac{t^2}{4} \Rightarrow \frac{d x}{d t}=\frac{t}{2}$

Now $y=x^{-n / 2} J_n(2 \sqrt{x})=\left(\frac{t^2}{4}\right)^{-n / 2} J_n(t)=\left(\frac{t}{2}\right)^{-n} J_n(t)$

⇒ $ \frac{d y}{d x}=\frac{d y}{d t} \frac{d t}{d x}=\frac{2}{t} \frac{d}{d t}\left\{\left(\frac{t}{2}\right)^{-n} J_n(t)\right\}=\frac{2^{n+1}}{t} \frac{d}{d t}\left\{t^{-n} J_n(t)\right\}$

= $-\frac{2^{n+1}}{t} t^{-n} J_{n+1}(t)=-2^{n+1} t^{-n-1} J_{n+1}(t)$

⇒ $\frac{d^2 y}{d x^2}=\frac{d}{d x}\left\{\frac{d y}{d x}\right\}=\frac{d}{d t}\left\{\frac{d y}{d x}\right\} \frac{d t}{d x}=\frac{2}{t} \frac{d}{d t}\left\{-2^{n+1} t^{-n-1} J_{n+1}(t)\right\} $

= $-\frac{2^{n+2}}{t} \frac{d}{d t}\left\{t^{-(n+1)} J_{n+1}(t)\right\}=\frac{2^{n+2}}{t} t^{-(n+1)} J_{n+2}(t)=2^{n+2} t^{-n-2} J_{n+2}(t)$

x $\frac{d^2 y}{d x^2}+(n+1) \frac{d y}{d x}+y$

= $\frac{t^2}{4} 2^{n+2} t^{-n-2} J_{n+2}(t)+(n+1)\left(-2^{n+1}\right) t^{-n-1} J_{n+1}(t)+2^n t^{-n} J_n(t)$

= $2^n t^{-n-1}\left[t J_{n+2}(t)-2 J_{n+1}(t)(n+1)+t J_n(t)\right]=0$

y = $x^{-n / 2} J_n(2 \sqrt{x}) \text { satisfies } x \frac{d^2 y}{d x^2}+(n+1) \frac{d y}{d x}+y=0$ .

42. Show that $y=\frac{1}{\pi} \int_0^\pi \cos (x \sin \phi) d \phi$ satisfies Bessel’s equation of zeroeth order.

Solution:

Let y = $J_0(x)=\frac{1}{\pi} \int_0^\pi \cos (x \sin \phi) d \phi \rightarrow \text { (1) }$

∴ $\frac{d y}{d x}=-\frac{1}{\pi} \int_0^\pi \sin (x \sin \phi) \sin \phi d \phi \rightarrow(2)$

and $\frac{d^2 y}{d x^2}=-\frac{1}{\pi} \int_0^\pi \cos (x \sin \phi) \sin ^2 \phi d \phi \rightarrow \text { (3) }$

Evaluating the R.H.S. of (2) by the method of integration by parts, we have $\frac{d y}{d x}=-\frac{1}{\pi}\left[\{-\sin (x \sin \phi) \cos \phi\}_0^\pi+\int_0^\pi \cos (x \sin \phi) x \cos ^2 \phi d \phi\right]$

= $-\frac{x}{\pi} \int_0^\pi \cos (x \sin \phi) \cos ^2 \phi d \phi=-\frac{x}{\pi} \int_0^\pi \cos (x \sin \phi) \cdot\left(1-\sin ^2 \phi\right) d \phi$

= $-\frac{x}{\pi} \int_0^\pi \cos (x \sin \phi) d \phi+\frac{x}{\pi} \int_0^\pi \cos (x \sin \phi) \sin ^2 \phi d \phi=-x y-x \frac{d^2 y}{d x^2}$

from (1) and (3)

∴ $\frac{d^2 y}{d x^2}+\frac{1}{x} \frac{d y}{d x}+y=0$ which is Bessel’s equation for n=0

Hence $y=\frac{1}{x} \int_0^\pi \cos (x \sin \phi) d \phi$ satisfies the Bessel’s equation of the zeroeth order.

43. Show that $y=A J_n(x) \int_0^x \frac{d x}{x J_n^2(x)}+B J_n(x)$ is the complete solution of Bessel’s equation.

Solution:

The Bessel’s equation is $y^{\prime \prime}+\frac{1}{x} y+\left(1-\frac{n^2}{x^2}\right) y=0 \rightarrow$ (1)

We know that a solution (1) is $u=J_n(x) \rightarrow$ (2)

Let the complete solution of (1) be $y=u v \rightarrow$(3)

Comparing (1) with $y^{\prime \prime}+P y^{\prime}+Q y=R$, we get $P=\frac{1}{x}$ and Q = $1-\frac{n^2}{x^2}, R=0$

Then, we know that v is given by $\frac{d^2 v}{d x^2}+\left(P+\frac{2}{u} \frac{d u}{d x}\right) \frac{d v}{d x}=\frac{R}{u}$

⇒ $\frac{d^2 v}{d x^2}+\left(\frac{1}{x}+\frac{2 J_n^{\prime}(x)}{J_n(x)}\right) \frac{d v}{d x}=0 \rightarrow(4)$

Let $\frac{d v}{d x}=q$ so that $\frac{d^2 v}{d x^2}=\frac{d q}{d x} \rightarrow (5)$

Then (4) ⇒ $\frac{d q}{d x}+\left(\frac{1}{x}+\frac{2 J_n^{\prime}}{J_n}\right) q=0 \Rightarrow \frac{d q}{q}+\left(\frac{1}{x}+\frac{2 J_n^{\prime}}{J_n}\right) d x=0$.

Integrating, we get $\log q+\log x+2 \log J_n=\log A \Rightarrow q x J_n^2=A$

⇒ q = $\frac{d v}{d x}=\frac{A}{\left(x J_n^2\right)} \Rightarrow d v=\left\{\frac{A}{\left(x J_n^2\right)}\right\} d x$

Integrating, we get $v=A \int_0^x \frac{d x}{x J_n^2}+B \rightarrow$ (6) where A and B are arbitrary constants.

From (2), (3) and (6) the required complete solution is

y = $J_n(x)\left(A \int_0^x \frac{d x}{x J_n^2}+B\right) \Rightarrow y=A J_n(x) \int_0^x \frac{d x}{x J_n^2(x)}+B J_n(x)$

44. If n>-1, show that $\int_0^x x^{-n} J_{n+1}(x) d x=\frac{1}{2^n \Gamma(n+1)}-x^{-n} J_n(x)$.

Solution:

From recurrence formula V, we have $x^{-n} J_{n+1}(x)=-\frac{d}{d x}\left\{x^{-n} J_n(x)\right\}$

Integrating between the limits o and x, we have $\int_0^x x^{-n} J_{n+1}(x) d x=-\left[x^{-n} J_n(x)\right]=-x^{-n} J_n(x)+L t \frac{J_n(x)}{x \rightarrow 0}$

(Diffe. Nr. and Dr. n times)

= $-x^{-n} J_n(x)+\underset{x \rightarrow 0}{ } \frac{J_n^n(x)}{n!}=-x^{-n} J_n(x)+\frac{n!}{2^n \Gamma(n+1) n!}=-x^{-n} J_n(x)+\frac{1}{2^n \Gamma(n+1)}$

45. Show that when n is integral, 1) $\pi J_n=\int_0^\pi \cos (n \theta-x \sin \theta) d \theta$

2) $\pi J_0=\int_0^\pi \cos (x \cos \phi) d \phi=\int_0^\pi \cos (x \sin \phi) d \phi$ and hence deduce that $J_0(x)=1-\frac{x^2}{2^2}+\frac{x^4}{2^2 \cdot 4^2}-\frac{x^6}{2^2 \cdot 4^2 \cdot 6^2}+\ldots=\sum_{r=0}^{\infty} \frac{(-1)^r x^{2 r}}{\left(2^r r !\right)^2}$.

Solution:

We have $\cos (x \sin \theta)=J_0+2 J_2 \cos 2 \theta+2 J_4 \cos 4 \theta+\cdots+2 J_{2 m} \cos 2 m \theta+\cdots \rightarrow (1)$

and $\sin (x \sin \theta)$ = $2 \sin \theta J_1+2 \sin 3 \theta J_3+\cdots+2 J_{2 m+1} \sin (2 m+1) \theta+\cdots \rightarrow (2)$

1. Multiplying both sides of (1) by $\cos 2 m \theta$ and then integrating between the limits o to π, $\int_0^\pi \cos (x \sin \theta) \cos 2 m \theta d \theta$

= $J_0 \int_0^\pi \cos 2 m \theta d \theta+2 J_2 \int_0^\pi \cos 2 \theta \cos 2 m \theta d \theta+\cdots$

+ $2 J_{2 m} \int_0^\pi \cos ^2 2 m \theta d \theta+\cdots=0+0+\cdots+J_{2 m} \int_0^\pi(1+\cos 4 m \theta) d \theta+\cdots=\pi J_{2 m}$.

Similarly we can prove that $\int_0^\pi \cos (x \sin \theta) \cos (2 m+1) \theta d \theta=0$.

Again multiplying both sides of (2) by $\sin (2 m+1) \theta$ and then integrating between the limits o to π,

⇒ $\int_0^\pi \sin (x \sin \theta) \sin (2 m+1) \theta d \theta=2 J_1 \int_0^\pi \sin \theta \sin (2 m+1) \theta d \theta$

+ $2 J_3 \int_0^\pi \sin 3 \theta \sin (2 m+1) \theta d \theta+\cdots+2 J_{2 m+1} \int_0^\pi \sin ^2(2 m+1) \theta d \theta+\cdots$

= $0+0+\cdots+J_{2 m+1} \int_0^\pi\{1-\cos 2(2 m+1) \theta\} d \theta+\cdots=J_{2 m+1}(\theta)=\pi J_{2 m+1}$

Similarly, $\int_0^\pi \sin (x \sin \theta) \sin 2 m \theta d \theta=0$

Therefore, $\int_0^\pi \cos (2 \min \theta-x \sin \theta) d \theta$

= $\int_0^\pi \cos 2 m \theta \cos (x \sin \theta) d \theta+\int_0^\pi \sin 2 m \theta \sin (x \sin \theta) d \theta=\pi J_{2 m}$

Also $\int_0^\pi \cos [(2 m+1) \theta-x \sin \theta] d \theta$

= $\int_0^\pi \cos (2 m+1) \theta \cos (x \sin \theta) d \theta+\int_0^\pi \sin (2 m+1) \theta \sin (x \sin \theta) d \theta=\pi J_{2 m+1} \text {. }$

Hence for all positive integral n, we have $int_0^\pi \cos (n \theta-x \sin \theta) d \theta=\pi J_n$

If n is negative, say n=-m,where m is +ve then $\int_0^\pi \cos (n \theta-x \sin \theta) d \theta$

= $\int_0^\pi \cos \{-m(\pi-\phi)-x \sin (\pi-\phi)\} d \phi, \text { putting } \theta=\pi-\phi$

= $\int_0^\pi \cos \{-m \pi+(m \phi-x \sin \phi)\} d \phi$

= $\int_0^\pi\{\cos m \pi \cdot \cos (m \phi-x \sin \phi)+\sin m \pi \sin (m \phi-x \sin \phi)\} d \phi$

= $(-1)^m \int \pi \cos (m \phi-x \sin \phi) d \phi=(-1)^m \pi J_m(x)=\pi J_{-m}(x)=\pi J_n(x)$

Hence for all integral values of $n, \int_0^\pi \cos (n \theta-x \sin \theta) d \theta=\pi J_n$.

2. Putting $\theta=\frac{\pi}{2}+\phi$ in the value of $\cos (x \sin \theta)$ from (1), we have

⇒ $\cos (x \cos \phi)=J_0-2 J_2 \cos 2 \phi+2 J_4 \cos 4 \phi-\cdots$

∴ $\int_0^\pi \cos (x \cos \phi) d \phi=J_0 \int_0^\pi d \theta-2 J_2 \int_0^\pi \cos 2 \phi d \phi+\cdots$

From (1) we have $\cos (x \sin \phi)=J_0+2 J_2 \cos 2 \phi+2 J_4 \cos 4 \phi+\cdots$

∴ $\int_0^\pi \cos (x \sin \phi) d \phi=J_0 \int_0^\pi d \phi+2 J_2 \int_0^\pi \cos 2 \phi d \phi+\cdots=\pi J_0$

Deduction: We have proved that $J_0(x)=\frac{1}{\pi} \int_0^\pi \cos (x \cos \phi) d \phi$

= $\frac{1}{\pi} \int_0^\pi\left(1-\frac{x^2 \cos ^2 \phi}{2!}+\frac{x^4 \cos ^4 \phi}{4!}-\frac{x^6 \cos ^6 \phi}{6!}+\cdots\right) d \phi \rightarrow \text { (1) }$

Since $\int_0^\pi \cos ^{2 r} \phi d \phi=\frac{1 \cdot 3 \cdot 5 \cdot(2 r-1)}{2 \cdot 4 \cdot 6 \cdots(2 r)} \pi$ from definite integrals.

From (1) we have $J_0(x)=\frac{1}{\pi}\left[\pi-\frac{x^2}{2!} \cdot \frac{1}{2} \pi+\frac{x^4}{4!} \cdot \frac{1 \cdot 3}{2 \cdot 4} \pi-\frac{x^6}{6!} \cdot \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \pi+\cdots\right]$

= $1-\frac{x^2}{2^2}+\frac{x^4}{2^2 \cdot 4^2}-\frac{x^6}{2^2 \cdot 4^2 \cdot 6^2}+\cdots=1-\frac{x^2}{2^2}+\frac{x^4}{2^4(2!)^2}-\frac{x^6}{2^6(3!)^2}+\cdots$

= $\sum_{r=0}^{\infty} \frac{(-1)^r x^{2 r}}{\left(2^r r!\right)^2}$

46. Show that $J_{2 n}(x)=(-1)^n \frac{2}{\pi} \int_0^{\pi / 2} \cos 2 n \phi \cos (x \sin \phi) d \phi$.

Solution:

We have $\cos (x \sin \theta)=J_0+2 J_2 \cos 2 \theta+2 J_4 \cos 4 \theta+\cdots+2 J_{2 m} \cos 2 m \theta+\cdots \rightarrow$ (1)

Multiplying both sides of (1) by $\cos 2 m \theta$ and then integrating between the limits o to $\pi$, $\int_0^\pi \cos (x \sin \theta) \cos 2 m \theta d \theta=J_0^\pi \int_0^\pi \cos 2 m \theta d \theta+2 J_2 \int_0^\pi \cos 2 \theta \cos 2 m \theta d \theta+\cdots$

+ $2 J_{2 m} \int_0^\pi \cos ^2 2 m \theta d \theta+\cdots=0+0+\cdots+J_{2 m} \int_0^\pi(1+\cos 4 m \theta) d \theta+\cdots=\pi J_{2 m}$

⇒ $J_{2 m}=\frac{1}{\pi} \int_0^\pi \cos (x \sin \theta) \cos 2 m \theta d \theta \Rightarrow J_{2 n}=\frac{1}{\pi} \int_0^\pi \cos (x \sin \theta) \cos 2 n \theta d \theta$

Put $\theta=\phi+\frac{\pi}{2}$. Then $d \theta=d \phi ; \theta=0, \pi \Rightarrow \phi=-\frac{\pi}{2}, \frac{\pi}{2}$

∴ $J_{2 n}=\frac{1}{\pi} \int_{-\pi / 2}^{\pi / 2} \cos (x \cos \phi) \cos (n \pi+2 n \phi) d \phi=\frac{(-1)^n}{\pi} \int_{-\pi / 2}^{\pi / 2} \cos (x \cos \phi) \cos 2 n \phi d \phi$

= $(-1)^n \frac{2}{\pi} \int_0^{\pi / 2} \cos (x \cos \phi) \cos 2 n \phi d \phi$

47. Show that $J_{2 n-1}(x)=-(-1)^n \frac{2}{\pi} \int_0^{\pi / 2} \cos (2 n+1) \theta \sin (x \cos \theta) d \theta$.

Solution:

We have $\sin (x \sin \theta)=2 \sin \theta J_1+2 \sin 3 \theta J_3+\cdots+2 J_{2 m+1} \sin (2 m+1) \theta+\cdots \rightarrow(1)$

Multiplying both sides of (1) by sin (2 m+1)θ and then integrating between the limits o to π

⇒ $\int_0^\pi \sin (x \sin \theta) \sin (2 m+1) \theta d \theta=2 J_1 \int_0^\pi \sin \theta \sin (2 m+1) \theta d \theta$

+ $J_3 \int_0^\pi \sin 3 \theta \sin (2 m+1) \theta d \theta+\cdots+2 J_{2 m+1} \int_0^\pi \sin ^2(2 m+1) \theta d \theta+\cdots$

= $0+0+\cdots+J_{2 m+1} \int_0^\pi\{1-\cos 2(2 m+1) \theta\} d \theta+\cdots=J_{2 m+1}(\theta)=\pi J_{2 m+1}$

⇒ $J_{2 m+1}=\frac{1}{\pi} \int_0^\pi \sin (x \sin \theta) \sin (2 m+1) \theta d \theta$

⇒ $J_{2 n+1}=\frac{1}{\pi} \int_0^\pi \sin (x \sin \theta) \sin (2 n+1) \theta d \theta$

Put $\theta=\phi+\frac{\pi}{2}$. Then $d \theta=d \phi ; \theta=0, \pi \Rightarrow \phi=-\frac{\pi}{2}, \frac{\pi}{2}$

∴ $J_{2 n+1}=\frac{1}{\pi} \int_{-\pi / 2}^{\pi / 2} \sin (x \cos \phi) \sin \left[(2 n+1)\left(\phi+\frac{\pi}{2}\right)\right] d \phi$

= $\frac{1}{\pi} \int_{-\pi / 2}^{\pi / 2} \sin (x \cos \phi)(-1)^n \cos (2 n+1) \phi d \phi$

= $(-1)^n \frac{2}{\pi} \int_{-\pi / 2}^{\pi / 2} \sin (x \cos \phi)(-1)^n \cos (2 n+1) \phi d \phi$

48. Prove that $J_n(x)=\frac{(x / 2)^n}{\sqrt{\pi} \Gamma(n+1 / 2)} \int_{-1}^{+1}\left(1-t^2\right)^{n-1 / 2} e^{i x t} d t,\left(n>-\frac{1}{2}\right)$. And deduce that $\int_0^1\left(1-t^2\right)^{n-1 / 2} \cos (x t) d t=\frac{2^{n-1} \Gamma(1 / 2) \Gamma(n+1 / 2) J_n(x)}{x^n}$.

Solution:

Let I = $\int_{-1}^{+1}\left(1-t^2\right)^{n-1 / 2} e^{i x t} d t=\int_{-1}^{+1}\left(1-t^2\right)^{n-1 / 2}\left(\sum_{r=0}^{\infty} \frac{\left(i x t^r\right.}{r!}\right) d t$

= $\sum_{r=0}^{\infty} \frac{(i x)^r}{r!} \int_{-1}^{+1}\left(1-t^2\right)^{n-1 / 2} r d t$

Now, if r is odd, the integrand in $I$ is an odd function of t. ∴ I=0.

And if r is even, the integrand in I is an even function of t.

∴ I = $\sum_{r=0}^{\infty} \frac{(i x)^{2 s}}{(2 s)!} 2 \int_0^1\left(1-t^2\right)^{n-1 / 2} t^{2 s} d t$

= $\sum_{t=0}^{\infty} \frac{(i x)^{2 s}}{(2 s)!} \int_0^1(1-v)^{n-1 / 2} \cdot v^{s-1 / 2} d v$ ; Putting $t^2=v$ so that 2 t d t=d v

= $\sum_{t=0}^{\infty} \frac{(i x)^{2 s}}{(2 s)!} B\left(n+\frac{1}{2}, s+\frac{1}{2}\right)$, since $B(m, n)=\int_0^1(1-t)^{m-1} t^{n-1}$

= $\sum_{r=0}^{\infty} \frac{(i x)^2}{(2 s)!} \frac{\Gamma\left(-n+\frac{1}{2}\right) \Gamma\left(s+\frac{1}{2}\right)}{\Gamma\left(n+\frac{1}{2}+s+\frac{1}{2}\right)},\left(n>-\frac{1}{2}\right)$ since $B(m, n)=\frac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)}$

= $\Gamma\left(n+\frac{1}{2}\right)_{s=0}^{\infty} \frac{\left(i^2\right)^s x^{2 s}\{(2 s+1) / 2\}}{(2 s)!\Gamma(n+s+1)}$

= $\Gamma\left(n+\frac{1}{2}\right) \sum_{s=0}^{\infty} \frac{(-1)^s x^{2 s}}{(2 s)!\Gamma(n+s+1)} \frac{(2 s)!}{2^{2 s} s!} \sqrt{\pi}, \text { since } \Gamma \cdot\left(\frac{2 s+1}{2}\right)=\frac{(2 s)!\sqrt{\pi}}{2^{2 s} s!}$

= $\Gamma\left(n+\frac{1}{2}\right)\left(\frac{x}{2}\right)^{-n} \sum_{s=0}^{\infty} \frac{(-1)^s}{s!\Gamma(n+s+1)}\left(\frac{x}{2}\right)^{n+2 s} \sqrt{\pi}=\Gamma\left(n+\frac{1}{2}\right)\left(\frac{x}{2}\right)^{-n} \sqrt{\pi} J_n(x)$

∴ $J_n(x)=\frac{(x / 2)^n}{\sqrt{\pi} \Gamma\left(n+\frac{1}{2}\right)} \int_{-1}^{+1}\left(1-t^2\right)^{n-1 / 2} e^{i x t} d t \rightarrow(1)$

Deduction: From (1), we have $J_n(x)=\frac{x^n}{2^n \Gamma\left(\frac{1}{2}\right) \Gamma\left(n+\frac{1}{2}\right)} \int_{-1}^1\left(1-t^2\right)^{n-1 / 2}(\cos x t+i \sin x t) d t$

Equating real parts from both sides, we get $J_n(x)=\frac{x^n}{2^n \Gamma\left(\frac{1}{2}\right) \Gamma\left(n+\frac{1}{2}\right)} \int_{-1}^1\left(1-t^2\right)^{n-1 / 2} \cos (x t) d t$

= $\frac{x^n}{2^n \Gamma\left(\frac{1}{2}\right) \Gamma\left(n+\frac{1}{2}\right)} 2 \int_0\left(1-t^2\right)^{n-1 / 2} \cos (x t) d t$

⇒ $\int_0\left(1-t^2\right)^{n-1 / 2} \cos (x t) d t=\frac{2^{n-1} \Gamma\left(\frac{1}{2}\right) \Gamma\left(n+\frac{1}{2}\right) J_n(x)}{x^n}$

49. Prove that $J_n(x)=\frac{2(x / 2)^{n-m}}{\Gamma(n-m)} \int_0^1(1-t)^{n-m-1} t^{m+1} J_m(x t) d t$.

Solution:

Let I = $\int_0^1\left(1-t^2\right)^{n-m-1} t^{m+1} J_m(x t) d t$

= $\int_0^1\left(1-t^2\right)^{n-m-1} t^{m+1} \sum_{r=0}^{\infty}(-1)^r \frac{1}{r!\Gamma(m+r+1)}\left(\frac{x t}{2}\right)^{2 r+m} d t$

= $\sum_{r=0}^{\infty} \frac{(-1)^r(x / 2)^{2 r+m}}{r!\Gamma(m+r+1)} \int_0^1\left(1-t^2\right)^{n-m-1} t^{2 r+2 m+1} d t$

= $\sum_{r=0}^{\infty} \frac{(-1)^r(x / 2)^{2 r+m}}{r!\Gamma(m+r+1)} \int_0^1\left(1-t^2\right)^{n-m-1}\left(t^2\right)^{m+r} t d t$

= $\sum_{r=0}^{\infty} \frac{(-1)^r(x / 2)^{2 r+m}}{r!\Gamma(m+r+1)} \frac{1}{2} \int_0^1(1-z)^{n-m-1} Z^{(m+r+1)-1} d z \text { on putting } t^2=z$

= $\sum_{r=0}^{\infty} \frac{(-1)^r(x / 2)^{2 r+m}}{r!\Gamma(m+r+1)} \frac{1}{2} B(n-m, m+r+1), \text { if } n-m>0, m+r+1>0$

= $\sum_{r=0}^{\infty} \frac{(-1)^r(x / 2)^{2 r+m}}{r!\Gamma(m+r+1)} \frac{1}{2} \frac{\Gamma(n-m) \Gamma(m+r+1)}{\Gamma(n+r+1)}$

= $\frac{\Gamma(n-m)}{2}\left(\frac{x}{2}\right)^{m-n} \sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2 r}=\frac{\Gamma(n-m)}{2}\left(\frac{x}{2}\right)^{m-n} J_n(x)$

∴ $ J_n(x)=\frac{2(x / 2)^{n-m}}{\Gamma(n-m)} \int_0^1\left(1-t^2\right)^{n-m-1} t^{m+1} J_m(x t) d t$

50. Prove that $J_n(x)=\frac{-x^n}{2^{n-1} \Gamma(n)} \int_0^{\pi / 2} \sin \theta \cos ^{2 n-1} \theta J_0(x \sin \theta) d \theta \text {, where } n>-\frac{1}{2}$.

Solution:

We have $J_n(x)=\sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{2 r+n}$

(1) ⇒ $J_0(x \sin \theta)=\sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(r+1)}\left(\frac{x \sin \theta}{2}\right)^{2 r} \text {. }$…(2)

Let I = $\int_0^{\pi / 2} \sin \theta \cos ^{2 n-1} \theta J_0(x \sin \theta) d \dot{\theta} \rightarrow(3)$

∴ I = $\int_0^{\pi / 2} \sin \theta \cos ^{2 n-1} \theta\left(\sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(r+1)} \frac{x^{2 r}}{2^{2 r}} \sin ^{2 r} \theta\right) d \theta$

= $\sum_{r=0}^{\infty} \frac{(-1)^r x^{2 r}}{(r!)^2 \cdot 2^{2 r}} \int_0^{\pi / 2} \cos ^{2 n-1} \theta \sin ^{2 r+1} \theta d \theta$

= $\sum_{r=0}^{\infty} \frac{(-1)^r x^{2 r}}{(r!)^2 2^{2 r}} \frac{\Gamma(n) \Gamma(r+1)}{\Gamma(n+r+1)}=\frac{\Gamma(n)}{2} \sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{2 r}$

⇒ $\frac{x^n}{2^{n-1} \Gamma(n)} I=\left(\frac{x}{2}\right)^n \sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{2 r}=\sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{2 r+n}$

⇒ $\frac{x^n}{2^{n-1} \Gamma(n)} \int_0^{\pi / 2} \sin \theta \cos ^{2 n-1} \theta J_0(x \sin \theta) d \theta=J_n(x)$, by (1) and (3).

51. Prove that $J_n(x)=(-2)^n \cdot x^n \frac{d^n}{d\left(x^2\right)^n} J_0(x)$.

Solution:

Bessel’s equation for zeroeth order is $y_2+\frac{1}{x} y_1+y=0 \rightarrow$ (1) whose solution is $J_0(x)$

Changing the independent variable from x to X, by the relation $x^2=X$, so that

⇒ $y_1=\frac{d y}{d x}=\frac{d y}{d X} \frac{d x}{d x}=2 x \frac{d y}{d x}=2 \sqrt{X} \frac{d y}{d X}$ and

⇒ $y_2=\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(2 \sqrt{X} \frac{d y}{d X}\right)$

= $\frac{d}{d x}\left(2 \sqrt{X} \frac{d y}{d X}\right) \frac{d x}{d x}=\left(2 \sqrt{X} \frac{d^2 y}{d X^2}+\frac{1}{\sqrt{X}} \frac{d y}{d X}\right) 2 \sqrt{X}$

= $4 X \frac{d^2 y}{d X^2}+2 \frac{d y}{d X}$

Substituting in (1), we have $\left(4 X \frac{d^2 y}{d X^2}+2 \frac{d y}{d X}\right)+\frac{1}{\sqrt{X}} 2 \sqrt{X} \frac{d y}{d X}+y=0$

∴ $4 X \frac{d^2 y}{d X^2}+4 \frac{d y}{d X}+y=0 \rightarrow \text { (4) }$…..(4)

Differentiating (2), n times w.r.t. X by using Leibnitz’s theorem we have

4$\left[X \frac{d^{n+2} y}{d X^{n+2}}+n 1 \frac{d^{n+1} y}{d X^{n+1}}\right]+4 \frac{d^{n+1} y}{d X^{n+1}}+\frac{d^n y}{d X^n}$=0

⇒ $4 X \frac{d^{n+2} y}{d X^{n+2}}+4(n+1) \frac{d^{n+1} y}{d X^{n+1}}+\frac{d^n y}{d X^n}=0$

Putting $y=\frac{d^n y}{d X^n}=\frac{d^n J_0(x)}{d X^n}$, it becomes $4 X \frac{d^2 Y}{d X^2}+4(n+1) \frac{d Y}{d X}+Y=0$

Again, $J_n(x)$ is the solution of the Bessel’s equation $\frac{d^2 y}{d x^2}+\frac{1}{x} \frac{d y}{d x}+\left(1-\frac{n^2}{x^2}\right) y=0 \rightarrow$ (4)

Putting $y=x^n z$, so that $\frac{d y}{d x}=x^n \frac{d z}{d x}+n x^{n-1} Z$

and $\frac{d^2 y}{d x^2}=x^n \frac{d^2 z}{d x^2}+2 n x^{n-1} \frac{d z}{d x}+n(n-1) x^{n-2} z$=0

Substituting in (4), we have $x^n \frac{d^2 z}{d x^2}+2 n x^{n-1} \frac{d z}{d x}+n(n-1) x^{n-2} z+\frac{1}{x}\left[x^n \frac{d z}{d x}+n x^{n-1} z\right]+\left[1-\frac{n^2}{x^2}\right] x^n z=0$

⇒ $x^n \frac{d^2 z}{d x^2}+(2 n+1) x^{n-1} \frac{d z}{d x}+x^n z=0 \Rightarrow \frac{d^2 z}{d x^2}+(2 n+1) \frac{1}{x} \frac{d z}{d x}+z=0 \rightarrow \text { (5). }$

Comparing (3) and (5), we get $z=Y=\frac{d^n J_0(x)}{d X^n}=\frac{d^n J_0(x)}{d\left(x^2\right)^n}$

But y = $x^n Z$

Hence $J_n(x)=c x^n \frac{d^n J_0(x)}{d\left(x^2\right)^n} \rightarrow(6)$ where c is a constant to be determined.

We know that $J_0(x)=\sum_{r=0}^{\infty} \frac{(-1)^r x^{2 r}}{\left(2^r r!\right)^2}$

∴ $\frac{d^2 J_0(x)}{d\left(x^2\right)^n}=\frac{d^n}{d\left(x^2\right)^n} \sum_{r=0}^{\infty} \frac{(-1)^r x^{2 r}}{\left(2^r r!\right)^2}=\frac{d^n}{d\left(x^2\right)^n} \sum_{r=0}^{\infty} \frac{(-1)^{n+r}\left(x^2\right)^{n+r}}{\left[2^{n+r}(n+r)!\right]^2}$

(Since all those terms in which the index of x is less than 2n will vanish on differentiation n times w.r.t. $x^2$)

= $\sum_{r=0}^{\infty} \frac{(-1)^{n+r}(n+r)(n+r-1) \ldots(r+1)\left(x^2\right)^r}{\left[2^{n+r} \cdot(n+r)!\right]^2}=(-1)^n \sum_{r=0}^{\infty}(-1)^r \frac{(n+r)!}{r!2^{3(n+r)} \cdot[(n+r)!]^2} x^{2 r}$

∴ $J_n(x)=c x^n(-1)^n \sum_{r=0}^{\infty}(-1)^r \frac{1}{r!2^{2 n+2 r}(n+r)!} x^{2 r}$

= $\frac{c(-1)^r}{2^n} \sum_{r=0}^{\infty}(-r)^r \frac{1}{r!\Gamma(n+r+1)}\left[\frac{x}{2}\right]^{n+2 r}=\frac{c}{(-2)^n} J_n(x)$ . ∴ $c=(-2)^n$

Hence $J_n(x)=(-2)^n x^n \frac{d^n J_n(x)}{d\left(x^2\right)^n}$

52. Prove that $\int_0^{\infty} \frac{J_n(x)}{x} d x=\frac{1}{n}$.

Solution:

If n is a non-negative integer then we have $\int_0^{\infty} J_n(x) d x=1$

From recurrence relation 4, $\frac{2 n}{x} J_n(x)=J_{n-1}^0(x)+J_{n+1}(x) \rightarrow$ (1)

Integrating (1) both sides with respect to x, from o to $\infty$, we get

2n $\int_0^{\infty} \frac{J_n(x)}{x}=\int_0^{\infty} J_{n-1}(x)+\int_0^{\infty} J_{n+1}(x) \Rightarrow 2 n \int_0^{\infty} \frac{J_n(x)}{x}$=1+1

⇒ $\int_0^{\infty} \frac{J_n(x)}{x}=1 \text {. }$

53. Show that $\int_0^{\pi / 2} \sqrt{(\pi x)} \cdot J_{1 / 2}(2 x) d x=1$.

Solution:

⇒ $\int_0^{\pi / 2} \sqrt{\pi x} J_{1 / 2}(2 x) d x=\int_0^{\pi / 2} \sqrt{\pi x} \sqrt{\frac{2}{\pi(2 x)}} \sin 2 x d x=\int_0^{\pi / 2} \sin 2 x d x=\left[-\frac{1}{2} \cos 2 x\right]^{\pi / 2}$

= $-\frac{1}{2}[-1-1]=1$

54. Show that $\int_0^1 J_0[\sqrt{x(t-x)}] d x=2 \sin \frac{1}{2} t$.

Solution:

We know that $J_n=\sum_{r=0}^{\infty}(-1)^r\left(\frac{x}{2}\right)^{n+2 r} \frac{1}{r!\Gamma(n+r+1)} \rightarrow(1)$

Putting n=0 and $x=\{x(t-x)\}^{1 / 2}$ in (1), we get

⇒ $J_0(\{x(t-x)\}^{1 / 2}=\sum_{r=0}^{\infty}(-1)^r \frac{1}{r!\Gamma(r+1)}\left[\frac{x\{x(t-x)\}^{1 / 2}}{2}\right]^{2 r}=\sum_{r=0}^{\infty} \frac{(-1)^r x^r(t-x)^r}{2^{2 r} r!\Gamma(r+1)}$

∴ $\int_0^1 J_0\left(\{x(t-x)\}^{1 / 2}\right) d x=\int_0^1 \sum_{r=0}^{\infty} \frac{(-1)^r x^r(t-x)^r}{2^{2 r} r!\Gamma(r+1)} d x=\sum_{r=0}^{\infty} \frac{(-1)^r}{2^{2 r} r!\Gamma(r+1)} \int_0^1 x^r(t-x)^r d x$

= $\sum_{r=0}^{\infty} \frac{(-1)^r t^{2 r+1}}{2^{2 r} r!\Gamma(r+1)} \int_0^1 y^r(1-y)^r d y$, putting x = ty and d x=t dy

= $\sum_{r=0}^{\infty} \frac{(-1)^r t^{2 r+1}}{2^{2 r} r!\Gamma(r+1)} B(r+1, r+1)=\sum_{r=0}^{\infty} \frac{(-1)^r t^{2 r+1}}{2^{2 r} r!\Gamma(r+1)} \frac{\Gamma(r+1) \Gamma(r+1)}{\Gamma(2 r+1)}$

= $\sum_{r=0}^{\infty} \frac{(-1)^r t^{2 r+1}}{2^{2 r} r!r!} \frac{r!r!}{\Gamma(2 r+1)}=\sum_{r=0}^{\infty} \frac{(-1)^r t^{2 r+1} r!}{2^{2 r} r!\Gamma(2 r+1)}=2 \sum_{r=0}^{\infty}(-1)^r \frac{(t / 2)^{2 r+1}}{(2 r+1)!}$

= $2\left\{\frac{t}{2}-\frac{(t / 2)^3}{3!}+\frac{(t / 5)^5}{5!}-. .\right\}=2 \sin \frac{t}{2}$

55. Prove that $\int_0^b x J_0(a x) d x=\frac{b}{a} J_1(a b)$.

Solution:

Put ax = t Then a dx=dt and x=0, b ⇒ t=0, a b

⇒ $\int_0^b x J_0(a x) d x=\int_0^{a b} \frac{t}{a} J_0(t) \frac{d t}{a}=\frac{1}{a^2} \int_0^{a b} t J_0(t) d t=\frac{1}{a^2}\left[t J_1(t)\right]=\frac{a b}{a} J_1(a b) .$

56. Evaluate $\int J_3(x) d x$ and express the result in terms of $J_0 \text { and } J_1$.

Solution:

Given
$\int J_3(x) d x$
From recurrence 7, we have $x^{-n} J_{n+1}=\frac{d}{d x}\left\{x^{-n} J_n(x)\right\}$.

Integrating it, $\int x^{-n} J_{n+1}(x) d x=-x^{-n} J_n(x) \rightarrow$ (1)

Now $\int J_3(x) d x=\int x^2\left[x^{-2} J_3(x)\right] d x=x^2\left[-x^{-2} J_2(x)\right]-\int 2 x\left[-x^{-2} J_2(x)\right] d x$

= $-J_2(x)+2 \int x^{-1} J_2(x) d x=-J_2(x)+2\left[-x^{-2} J_1(x)\right]+c$

∴ $\int J_3(x) d x=-J_2(x)-2 x^{-1} J_1(x)+c \rightarrow \text { (2) }$

From recurrence relation 4, we have $\left(\frac{2 n}{x}\right) J_n(x)=J_{n-1}(x)+J_{n+1}(x) \rightarrow(3)$

Putting n=1 in (3), we get $J_2(x)=\frac{2 J_1(x)}{x}-J_0(x) \rightarrow$ (4)

Substituting (4) in (2), we get $\int J_3(x) d x=-\left(\frac{2 J_1(x)}{x}-J_0(x)\right)-2 x^{-1} J_2(x)+c$

⇒ $\int J_3(x) d x=J_0(x)-\frac{4 J_1(x)}{x}+c$, where c is an arbitrary constant.

57. Evaluate $\int x^3 J_3(x) d x$.

Solution:

Given

$\int x^3 J_3(x) d x$.

Since, $\frac{d}{d x}\left\{x^{-n} J_n\right\}=-x^n J_{n+1}$, we have $\int x^{-n} J_{n+1} d x=-x^{-n} J_n \rightarrow(1)$

Now, $\int x^3 J_3(x) d x=\int x^5\left(x^{-2} J_3\right) d x=x^5\left(-x^{-2} J_2\right)-\int 5 x^4\left(-x^{-2} J_2\right) d x$

= $-x^3 J_2+\int 5 x^2 J_2 d x=-x^3 J_2+5 \int x^3\left(x^{-1} J_2\right) d x$

= $-x^3 J_2+5\left[x^3\left(-x^{-1} J_1\right)-\int 3 x^2\left(-x^{-1} J_1\right) d x\right]$

= $-x^3 J_2-5 x^2 J_1+15 \int x J_1 d x=-x^3 J_2-5 x^2 J_1+15 \int x\left(-J_0^{\prime}\right) d x$

= $-x^3 J_2-5 x^2 J_1-15 \int x J_0^{\prime} d x=-x^3 J_2-5 x^2 J_1=15\left[x J_0-\int 1 J_0 d x\right]$

= $-x^3 J_2-5 x^3 J_1-15 x J_0+15 x \int J_0 d x$

58. Evaluate $\int x^4 J_1(x) d x$.

Solution:

Given

$\int x^4 J_1(x) d x$.

Since, $\frac{d}{d x}\left\{x^n J_n\right\}=x^n J_{n-1}$, we have $\int x^n J_{n-1} d x=x^n J_n \rightarrow$ (1)

Now, $\int x^4 J_1 d x=\int x^2\left(x^2 J_1\right) d x=x^2\left(x^2 J_2\right)-\int 2 x\left(x^2 J_2\right) d x$

= $x^4 J_2-2 \int x^3 J_2 d x=x^4 J_2-2 x^3 J_3+c$

59. Express $\int x^{-3} J_4(x) d x \text { in terms of } J_0 \text { and } J_1$.

Solution:

Putting n=3 in recurrence relation 7, we have $\frac{d}{d x}\left\{x^{-n} J_n\right\}=-x^{-n} J_{n+1} \Rightarrow \frac{d}{d x}\left\{x^{-3} J_3\right\}=-x^{-3} J_4$

Integrating, we get $\int x^{-3} J_4(x) d x=-x^3 J_3+c \rightarrow$ (1), c being an arbitrary constant

From recurrence relation 6, we have $J_{n+1}=\left(\frac{2 n}{x}\right) J_n-J_{n-1} \rightarrow$ (2)

Putting n=2 and n=1 successively in (2), we get $J_3=\left(\frac{4}{x}\right) J_2-J_1 \rightarrow(3)$ and $J_2=\left(\frac{2}{x}\right) J_1-J_0 \rightarrow \text { (4) }$

Substituting (4) in (3), we get $J_3=\left(\frac{4}{x}\right)\left[\left(\frac{2}{x}\right) J_1-J_0\right]-J_1=\left(\frac{8}{x^2}-1\right) J_1-\left(\frac{4}{x}\right) J_0 \rightarrow$ (5)

Substituting (5) in (1), we get $\int x^{-3} J_4(x) d x=-x^{-3}\left[\left(\frac{8}{x^2}-1\right) J_1-\frac{4}{x} J_0\right]+c=\left(-\frac{1}{x^3}-\frac{8}{x^5}\right) J_1+\frac{4}{x^4} J_0 \text {. }$

60. Prove that $\int x^{-1} J_4(x) d x=-x^{-1} J_3(x)-2 x^{-2} J_2(x)+c$.

Solution:

Recurrence relation 7 is $\frac{d}{d x}\left\{x^{-n} J_n(x)=-x^{-n} J_{n+i}(x)\right.$.

Integrating it, we get $\int x^{-n} J_{n+1}(x) d x=-x^{-n} J_n(x) \rightarrow$ (1)

∴ $x^{-1} J_4(x) d x=\int x^2\left[x^{-3} J_4(x)\right] d x=x^2\left[-x^{-3} J_3(x)\right]-\int 2 x \times\left[-x^{-3} J_3(x)\right] d x$

= $-x^{-1} J_3(x)+2 \int x^{-2} J_3(x) d x=-x^{-1} J_3(x)+2\left[-x^{-2} J_2(x)\right]+c$

= $-x^{-1} J_3(x)-2 x^{-2} J_2(x)+c$ .

61. Prove that $\int J_3(x) d x=-J_2(x)-\frac{2}{x} J_0(x)+c$.

Solution:

⇒ $\frac{d}{d x}\left\{x^{-n} J_n(x)\right\}=-x^{-n} J_{n+1}(x) \Rightarrow \int x^{-n} J_{n+1}(x) d x=-x^{-n} J_n(x) \rightarrow(1)$

Now, $\int J_3(x) d x=\int x^2\left\{x^{-2} J_3(x)\right\} d x=x^2 \int x^{-2} J_3(x) d x-\int 2 x\left(\int x^{-2} J_3(x) d x\right) d x$

= $x^2\left[-x^{-2} J_2(x)\right]-\int 2 x\left(-x^{-2} J_2(x)\right) d x$, using (1) for n=2

= $-J_2(x)+2 \int x^{-1} J_2(x) d x=-J_2(x)-2 x^{-1} J_1(x)+c$, using (1) for n=1 .

62. Show that
1) $\int_0^x x^3 J_0(x) d x=x^3 J_1(x)-2 x^2 J_2(x)$

2) $\int_0^1 x^3 J_0(x) d x=2 J_0(1)-3 J_1(1)$.

Solution:

1) $\frac{d}{d x}\left\{x^n J_n(x)\right]=x^n J_{n-1}(x) \Rightarrow \int x^n J_{n-1}(x) d x=x^n J_n(x) \rightarrow \text { (1) }$

∴ $\int_0^x x^3 J_0(x) d x=\int_0^x x^2\left[x J_0(x)\right] d x=\left[x^2\left\{x J_1(x)\right\}\right]-\int_0^x 2 x\left\{x J_1(x)\right\} d x$ using (1) for n=1

= $x^3 J_1(x)-2 \int_0^x x^2 J_1(x) d x=x^3 J_1(x)-2\left[x^2 J_2(x)\right]$, using (1) for n=2

= $x^3 J_1(x)-2 x^2 J_2(x)$ as $J_2(0)=0$.

2) From (1), we have $\int_0^x x^3 J_0(x) d x=x^3 J_1(x)-2 x^2 J_2(x) \rightarrow$ (2)

Putting x=1 in (2), $\int_0^1 x^3 J_0(x) d x=J_1(1)-2 J_2(1) \rightarrow$ (3)

Recurrence relation IV is $J_{n-1}(x)+J_{n+1}(x)=\frac{2 n}{x} J_n(x)$.

Putting n=1 and x=1, we have $J_2(1)=2 J_1(1)-J_0(1) \rightarrow(4)$

Substituting the above value of $J_2(1)$ in (3), we have $\int_0^1 x^3 J_0(x) d x=J_1(1)-2\left[2 J_1(1)-J_0(1)\right]=2 J_0(1)-3 J_1(1) \text {. }$

63. Prove that $\int_0^{\infty} e^{-a x} J_0(b x) d x=\frac{1}{\sqrt{\left(a^2+b^2\right)}}, a>0$

Solution:

We have $J_0(x)=\frac{1}{\pi} \int_0^\pi \cos (x \sin \phi) d x$

∴ $\int_0^{\infty} e^{-a x} J_0(b x) d x=\int_0^{\infty} e^{-a x}\left\{\frac{1}{\pi} \int_0^\pi \cos (b x \sin \phi) d \phi\right\} d x$

= $\frac{1}{\pi} \int_0^\pi\left[\int_0^{\infty} e^{-a x} \cos (b x \sin \phi) d x\right] d \phi$

= $\frac{1}{\pi} \int_0^\pi\left[\int_0^{\infty} e^{-a x} \cdot \frac{e^{i(b x \sin \phi)}+e^{-i(b x \sin \phi)}}{2} d x\right] d \phi$

= $\frac{1}{2 \pi} \int_0^\pi\left[\int_0^{\infty}\left\{e^{-(a-i b \sin \phi) x}+e^{-(a+i b \sin \phi) x}\right\} d x\right] d \phi$

= $\left.\frac{1}{2 \pi} \int_0^\pi\left[\frac{e^{-(a-i b \sin \phi) x}}{-(a-i b \sin \phi)}-\frac{e^{-(a+i b \sin \phi) x}}{a+i b \sin \phi}\right] d \phi\right]$

= $\frac{1}{2 \pi} \int_0^\pi\left[\frac{1}{a-i b \sin \phi}+\frac{1}{a+i b \sin \phi}\right] d \phi$

= $\frac{1}{2 \pi} \int_0^\pi \frac{2 a d \phi}{a^2+b^2 \sin ^2 \phi}=2 \cdot \frac{a}{\pi} \int_0^{\pi / 2} \frac{\text{cosec}^2 \phi d \phi}{b^2+a^2 \text{cosec}^2 \phi}$

= $2 \frac{a}{\pi} \int_0^{\pi / 2} \frac{\text{cosec}^2 \phi d \phi}{\left(a^2+b^2\right)+a^2 \cot ^2 \phi}$

= $2 \frac{a}{\pi}\left(\frac{1}{a \sqrt{\left(a^2+b^2\right)}} \cot ^{-1} \frac{a \cot \phi}{\sqrt{\left(a^2+b^2\right)}}\right)_0^{\pi / 2}$

= $\frac{2}{\pi \sqrt{\left(a^2+b^2\right)}} \cdot\left(\cot ^{-1} 0-\cot ^{-1} \infty\right)=\frac{1}{\sqrt{\left(a^2+b^2\right)}}$

64. Prove that $\int_0^x t\left\{J_n(t)\right\}^2 d t=\frac{x^2}{2}\left\{J_n^2(x)-J_{n-1}(x) J_{n+1}(x)\right\}$.

solution:

We have $\frac{d}{d t}\left[\frac{t^2}{2}\left\{J_n^2(t)-J_{n=1}(t) J_{n+1}(t)\right\}\right]$

= $t\left(J_n^2-J_{n-1}^{\prime} J_{n+1}+\frac{t^2}{2}\left[2 J_n J_n^{\prime}-J_{n-1}^{\prime} J_{n+1}-J_{n-1} J_{n+1}^{\prime}\right]\right.$

= $t\left(J_n^2-J_{n-1} J_{n+1}\right)+\frac{t^2}{2}\left[J_n\left(J_{n-1}-J_{n+1}\right)-\left(\frac{n-1}{t} J_{n-1}-J_n\right) J_{n+1}-J_{n-1}\left(-\frac{n+1}{t} J_{n+1}+J_n\right)\right]$

(Substituting the value of $2 J_n^{\prime}$ from recurrence formula 3 and the value of $2 J_{n-1}^{\prime}$ and $J^{\prime} n+1$ obtained by replacing n by n-1 and n+1 in recurrence formula 1 and 2 respectively.

∴ $\frac{d}{d t}\left[\frac{t^2}{2}\left\{J_n^2(t)-J_{n=1}(t) J_{n+1}(t)\right\}\right]=t J_n^2(t)$

Integrating both sides from o to x, we have $\int_0^x t\left\{J_n(t)\right\}^2 d t=\left[\frac{t^2}{2}\left\{J_n^2(t)-J_{n-1}(t) J_{n+1}(t)\right]_0^x=\frac{x^2}{2}\left\{J_n^2(x)-J_{n-1}(x) J_{n+1}(x)\right\} .\right.$

The post Bessel’s Equations Solved Exercise Problems appeared first on Answer Key for Math.

Ring Theory & Vector Calculus Notes

Alekhya — Thu, 10 Aug 2023 12:27:19 +0000

Ring Theory And Vector Calculus

Ring Theory Notes With Solved Examples

Definition of Homomorphism – Homomorphic Image – Elementary Properties of Homomorphism -Kernel of a Homomorphism – Fundamental theorem of Homomorphism – Maximal Ideals – Prime Ideals.

2 Homomorphism of Rings, Maximal and Prime Ideals
3 Homomorphism of Rings, Maximal and Prime Ideals

In groups, we have learned that one way of knowing more information about a group is to examine its interaction with other groups by using homomorphism.

The concept of homomorphism in rings is analogous to that of homomorphism in groups. The homomorphism in rings is a mapping that preserves the relations a + b = c and ab = d, the addition and multiplication operations.

Definition. (Homomorphism). Let R, R₀ be two rings. A mapping f: R → R₀ is – said to be a homomorphism if (a)f(a + b) = f(a) + f(b) and (b)f(ab) = f(a)f(b) for all a, b ∈ R.

Note 1. The operations +, · on the left-hand side of the properties (a), (b) are that of the ring R, while the operations +, · on the right-hand side of the properties (a), (b) are that of the ring R₀.

Note 2. Since R, R₀ are commutative groups under addition clearly property (a) shows that a ring homomorphism is a group homomorphism from (R, +) to (R₀, +).

Definition. If f: R → R₀ is a homomorphism of a ring R into R₀ then the image set f(R) = R¯ = {f(x) | x ∈ R} is called the f – homomorphic image of R.

Definition. Let R, R₀ be two rings. A homomorphism f: R → R₀ is called an epimorphism or onto homomorphism if f is onto mapping.

A homomorphism f: R → R₀ is called a monomorphism if f is one-one mapping

A homomorphism f: R → R₀ is called an isomorphism if f is both one-one and onto mapping.

A homomorphism f: R → R of a ring R into itself is called an endomorphism.

A homomorphism f: R → R which is both one-one and onto is called an automorphism.

Notation. 1. If f: R → R₀ is an onto homomorphism or epimorphism then R 0 is the homomorphic image of R and we write R ‘ R₀.

If f: R → R₀ is an isomorphism then we say that R is isomorphic to R 0 or R, R 0 are isomorphic and we write R ∼= R₀.

Note:

1. If f: R → R₀ is an onto homomorphism then f(R) = R₀.

2. If U is an ideal of the ring R, then R/U = {x + U | x ∈ R} is also a ring w.r.t. addition and multiplication of cosets. Then the mapping f: R → R/U defined by f(x) = x + U for all x ∈ R is called the natural homomorphism from R onto R/U.

3. A homomorphism is used to simplify a ring while retaining certain of its features. An isomorphism is used to show that two rings are algebraically identical.

Ring Theory And Vector Calculus Detailed Study Guide

Example 1. Let R, R₀ be two rings, and f: R → R₀ be defined by f(x) = 00∀x ∈ R, where 00 ∈ R₀ is the zero element.

Let a, b ∈ R. Then f(a) = 00, f(b) = 00 and hence f(a + b) = 00, f(ab) = 00.

Then a + b, ab ∈ R.

f(a + b) = 00 = 00 + 00 = f(a) + f(b) and f(ab) = 00 = 00.00 = f(a) · f(b).

∴ f is a homomorphism from R into R0. This is called Zero homomorphism.

Example 2. Let R be a ring and f: R → R be defined by f(x) = x∀x ∈ R.

Let a, b ∈ R so that a + b, ab ∈ R.

By definition, f(a + b) = a + b = f(a) + f(b) and f(ab) = ab = f(a)f(b).

Also for each y ∈ R (codomain), there exists y ∈ R (domain) so that f(y) = y

⇒ f is onto mapping.

Further for a, b ∈ R, f(a) = f(b) ⇒ a = b ⇒ f is one-one mapping.

Hence f is an automorphism. This is called Identity homomorphism.

Example 3. Let Z be the ring of integers and f: Z → 2Z be defined by f(n) = 2n∀n ∈ Z.

Let m, n ∈ Z. Then m+n, mn ∈ Z. Then f(m+n) = 2(m+n) = 2m+2n = f(m) + f(n). But f(mn) = 2(mn) 6= (2m)(2n) = f(m)f(n).

∴ f is not a ring homomorphism.

Although, the group (Z, +) is isomorphic to the group (2Z, +), the ring

(Z, +, ·) is not isomorphic to the ring (2Z, +, ·).

Theorem 1. Let f: R → R 0 be a homomorphism of a ring R into the ring R 0 and 0 ∈ R, 00 ∈ R0 be the zero elements. Then (1) f(0) = 0 f(−a) = −f(a)∀a ∈ R and (3) f(a − b) = f(a) − f(b) for all a,b ∈ R.

Proof. (1) For 0 ∈ R we have 0 + 0 = 0.

∴ f(0 + 0) = f(0) ⇒ f(0) + f(0) = f(0) + 00 ( ∵ f is homomorphism )

∴ f(0) = 00 ( by left cancellation law in R0 ) (2) For a ∈ R there exists −a ∈ R so that a + (−a) = 0.

∴ f(a + (−a)) = f(0) ⇒ f(a) + f(−a) = f(0) = 00. ( ∵ f is homomorphism)

∴ f(−a) = −f(a).

( ∵ f(a),f(−a) ∈ R₀ , ring )

(3) For a,b ∈ R; f(a − b) = f(a + (−b)) = f(a) + f(−b) = f(a) − f(b) (f(−b) = −f(b)By(2))

Note 1. A homomorphism maps the zero element of R into the zero element of R₀.

2. A homomorphism maps the negative ’ −a ’ of each element a ∈ R into the negative in R₀ of the corresponding element a’.

Remark. If the rings R, R’ have unity elements 1, and 1₀ respectively then it does not necessarily follow that f(1) = 1₀ is true.

However, if R₀ is an integral domain then f(1) = 1₀ is true.

Properties And Theorems Of Ring Theory With Notes

Theorem 2. The homomorphic image of a ring is a ring.

Proof. Let R, R’ be two rings and f: R → R₀ be a homomorphism.

By definition, homomorphic image of R = R¯ = f(R) = {f(x) ∈ R₀ | x ∈ R}.

To prove that f(R) is a ring, we show that f(R) = R¯ is a subring of R₀.

For 0 ∈ R, f(0) = 00 ∈ R₀.

∴ f(0) = 00 ∈ R¯ and hence R¯ ⊂ R₀.

Let a₀, b₀ ∈ R¯.

∴ There exists a,b ∈ R so that f(a) = a’,f(b) = b’.

Since a, b ∈ R we have a − b, ab ∈ R and hence f(a − b), f(ab) ∈ R¯.

Now a’ − b’ = f(a) − f(b) = f(a − b) ∈ R¯ (Theorem (1))

a₀b₀ = f(a)f(b) = f(ab) ∈ R.¯ (Homomorphism property (2))

Thus a’,b; ∈ R¯ ⇒ a’ − b’,a’b’ ∈ R¯.

∴ R¯ is a subring of R0 and hence R¯ is a ring.

Corollary. The homomorphic image of a commutative ring is a commutative ring.

Proof. Let R be a commutative ring and R¯ = f(R) be its holomorphic image.

a₀,b₀ ∈ R¯ ⇒ a₀b₀ = f(a)f(b) where a,b ∈ R

= f(ab) = f(ba) = f(b)f(a) = b₀a₀.

( ∵ R is commutative )

∴ f(R) = R¯ is commutative.

Theorem 3. If f: R → R₀ be an isomorphism from the ring R to the ring R₀ then

f(0) = 00 where 0, 00 are the zero elements of R, R’.

For each a ∈ R, f(−a) = −f(a).

R₀ is a commutative ring if R is a commutative ring,

R₀ is an integral domain if R is an integral domain. and

R₀ is a field if R is a field.

Proof. For (1), (2), and (3) see the proof of Theorem (1) of Art 3.1 and its corollary.

(4) Since f(0) = 0′ and f is one-one we have that 0 ∈ R is the only element whose image is 0′ ∈ R’.

∴ Let a’, b’ ∈ R’ and a’ ≠ 0′, b’ ≠0′.

Then there exists a, b ∈ R and a≠0, b 6= 0 so that f(a) = a’, f(b) = b₀.

a, b ∈ R, a ≠ 0, b ≠ 0 and R has no zero divisors ⇒ ab ≠ 0.

⇒ f(ab) ≠ f(0) ( ∵ f is one-one )

⇒ f(a)f(b) ≠ 00 ⇒ a0b0 6= 00.

∴ R₀ is without zero divisors.

Let 1 ∈ R be the unity element.

Then f(1) ∈ R₀ and say f(1) = 1₀.

For a₀ ∈ R₀ there exists unique a ∈ R so that f(a) = a₀.

For each a₀ ∈ R₀, a₀1₀ = f(a)f(1) = f(a1) = f(a) = a₀.

∴ a₀1₀ = 1₀a₀ = a₀ ⇒ f(1) = 1₀ is the unity element of R₀.

Hence R₀ is an integral domain.

(5) If R is a field then

R is commutative,

R has a unity element and

Every non-zero element of R has a multiplicative inverse.

By (3) and (5) R₀ is commutative and has unity element 1₀ = f(1) for 1 ∈ R.

Let a’ ∈ R’ and a’ ≠ 0′. There exists a ∈ R so that f(a) = a₀.

a = 0 ⇒ f(a) = f(0) ⇒ a₀ = 00 and hence a ≠ 0.

Since R is a field, there exists a− 1 ∈ R so that aa− 1 = 1 = a− 1a.

∴ f a’− 1 = f(1) ⇒ f(a)f a− 1 = 10 = f a− 1 f(a).

Hence f a− 1 = f(a)− 1 is the multiplicative inverse of f(a) = a0.

∴ R’ is a field.

Vector Calculus Solved Examples And Exercises With Notes

Theorem 4. Let R, R₀ be two rings and f: R → R₀ be a homomorphism. For every ideal U’ in the ring R’, f⁻¹ (U’) is an ideal in R.

Proof. Let U = f⁻¹ (U’) = {x ∈ R | f(x) ∈ U0}.

f(0) = 0′ ∈ U’ ⇒ 0 ∈ f⁻¹(U’) = U.

∴ U ≠ φ and U ⊂ R.

Let a, b ∈ U. By the def. of U = f− 1 (U’); f(a), f(b) ∈ U’.

U’ is an ideal, f(a), f(b) ∈ U₀ ⇒ f(a) − f(b) ∈ U’

⇒ f(a − b) ∈ U’ ⇒ a − b ∈ f− 1 (U’) = U

∴ a, b ∈ U ⇒ a − b ∈ U

Let a ∈ U, r ∈ R. Then f(a) ∈ U’ and f(r) ∈ R’.

Since U₀ is an ideal in R’; f(a)f(r), f(r)f(a) ∈ U’

⇒ f(ar), f(ra) ∈ U ⇒ ar, ra ∈ U, ∴ a ∈ U, r ∈ R ⇒ ar, ra ∈ U

Hence U = f⁻¹(U) is an ideal in R.

Note. If S₀ is a subring of R₀ then f⁻¹ (S’) is a subring of R.

Theorem. 5. Let R, R₀ be two rings and f: R → R’ be a homomorphism. For every ideal U in R,f(U) is an ideal in R¯ = f(R)

Proof. f(U) = {f(x) | x ∈ U}.

0 ∈ U ⇒ f(0) = 0′ ∈ f(U) ⇒ f(U) 6= φ and f(U) ⊂ f(R).

Let a’,b’ ∈ f(U). ’There exist a,b ∈ U such that f(a) = a₀,f(b) = b₀.

∴ a’ − b’ = f(a) − f(b) = f(a − b) ∈ f(U) ( ∵ a,b ∈ U and U is an ideal )

Let a’ ∈ f(U) and r’ ∈ f(R) = R¯.

There exist a ∈ U,r ∈ R such that f(a) = a’,f(r) = r’

a ∈ U,r ∈ R and U is an ideal ⇒ ar,ra ∈ U ⇒ f(ar),f(ra) ∈ f(U)

⇒ f(a) · f(r),f(r) · f(a) ∈ f(U) ( ∵ f is homomorphism )

⇒ a’r’,r ‘a’ ∈ f(U)

From (1) and (2): f(U) is an ideal in f(R) = R¯.

Note 1. If f: R → R’ is onto homomorphism then for every ideal U in R,f(U) is an ideal in R’.

2. The above theorem is true for a subring.

Ring Theory And Vector Calculus Kernel Of A Homomorphism

Definition. (Kernel). Let R, R’ be two rings and f: R → R’ be a homomorphism. The set {x ∈ R | f(x) = 0′} where 00 ∈ R’ is the zero element, is defined as the Kernel of the homomorphism f. The kernel of the homomorphism f: R → R’ is denoted by Ker for I(f).

Note 1. If f: R → R’ is a homomorphism then Ker f = f− 1 {00} ⊂ R.

2. For 0 ∈ R we have f(0) = 0′. Therefore 0 ∈ Ker f and hence Ker f 6= φ.

Example 1. Consider the Zero homomorphism f: R → R0 defined by f(x) = 0’∀x ∈ R.

Ker f = {x ∈ R | f(x) = 0′} = {x ∈ R | ∀x ∈ R} = R.

Example 2. Consider the identity homomorphism f: R → R defined by f(x) = x∀x ∈ R.

Ker f = {x ∈ R | f(x) = 0} = {x ∈ R | x = 0 only }( ∵ f(0) = 0) = {0}.

Step-By-Step Solutions For Vector Calculus Problems With Notes

Theorem 1. If f is a homomorphism of a ring R into a ring R’ then Ker f is an ideal of R.

Proof. If 0 ∈ R is the zero element of R then f(0) = 0′ the zero element of R’.

∴ 0 ∈ Ker f and hence Ker f 6= φ, Ker f ⊂ R.

Let a,b ∈ Ker f and r ∈ R. Then f(a) = 0’f(b) = 0′.

f(a − b) = f(a) − f(b) = 0′ − 0′ = 0′ ⇒ a − b ∈ Ker f

f(ar) = f(a)f(r) = 0’f(r) = 0′ and f(ra) = f(r)f(a) = f(r)00 = 00 ⇒ ar,ra ∈ Ker f.

∴ a,b ∈ Ker f,r ∈ R ⇒ a − b ∈ Ker f and ar,ra ∈ Ker f.

Hence Ker f is an ideal of R.

Theorem 2. If f is a homomorphism of a ring R into the ring R 0 then f is an isomorphism if and only if Ker f = {0}.

Proof. Let f be an into isomorphism. That is, f is one-one homomorphism.

We prove that Ker f = {0}.

a ∈ R, f(a) = 0′ ⇒ f(a) = f(0) ⇒ a = 0 ( ∵ f is one-one )

∴ 0 ∈ R is the only element in R so that f(0) = 0′.

∴ By definition, Ker f = {0}.

Conversely, let Ker f = {0}. We now prove that f is one-one.

a, b ∈ R and f(a) = f(b) ⇒ f(a) − .f(b) = 0′ ⇒ f(a − b) = 00

⇒ a − b ∈ Ker f = {0} ⇒ a − b = 0 ⇒ a = b. ∴ f is one-one.

Note. Ker f = {0} ⇔ f is one-one.

Theorem 3. If U is an ideal of a ring R then the quotient ring R/U is a homomorphic image of R. or Every quotient ring of a ring is a homomorphic image of the ring.
Proof.

We know that R/U = {x + U | x ∈ R} is a ring with respect to the addition and multiplication of cosets defined as (a + U) + (b + U) = (a + b) + U and (a + U) : (b + U) = ab + U where a + U, b + U ∈ R/U.

Let f: R → R/U be a mapping defined by f(a) = a + U for all a ∈ R.

For a, b ∈ R, a = b ⇒ a + U = b + U ⇒ f(a) = f(b).

∴ the mapping f is well-defined.

For a, b ∈ R; f(a + b) = (a + b) + U = (a + U) + (b + U) = f(a) + f(b)

and f(ab) = ab + U = (a + U) · (b + U) = f(a) · f(b)

∴ f is a homomorphism.

Let x + U ∈ R/U. Then x ∈ R and for this x ∈ R we have f(x) = x + U.

∴ for each x + U ∈ R/U there exists x ∈ R so that f(x) = x + U.

∴ f is onto mapping.

Hence f: R → R/U is an onto homomorphism.

Note. Ker f = {x ∈ R | f(x) = 0 + U} = {x ∈ R | x + U = 0 + U} = {x ∈ R | x ∈ U} = U.

In view of this result, the above theorem can also be stated as follows: “Every ideal in a ring R is the Kernel of some homomorphism defined on R”.

f: R → R/U is called Canonical homomorphism.

Example Z₆ = {0, 1, 2, 3, 4, 5} under addition and multiplication modulo −6 is a ring.

U = {0, 3} is an ideal of Z₆ and Z₆/U = {0 + U, 1 + U, 2 + U} = set of 3 elements.

Take Z₃ = {0, 1, 2}.

By the correspondence f(0) = 0 + U, f(1) = 1 + U and f(2) = 2 + U; Z ₃ and Z₆/U are isomorphic.

Ring Theory And Its Properties Explained With Examples

4 Theorem 4. (Fundamental theorem of homomorphism)

Let R, R’ be two rings and f: R → R ‘ be homomorphism with Kernel U. Then R is isomorphic to R/U.

Proof. a ∈ Ker f = U ⇒ f(a) = 00 where 00 is the zero element of R’.

Since U is an ideal of R, R/U = {x + U | x ∈ R} is the quotient ring of cosets under the addition and multiplication of cosets.

Since f: R → R’ is a homomorphism, f(R) = R¯ is a ring.

That is, for each f(x) ∈ R¯ we have x ∈ R.

Define φ = R/U → R¯ by φ(x + U) = f(x)∀x + U ∈ R/U a + U,b + U ∈ R/U and a + U = b + U ⇔ a − b ∈ U

⇔ f(a − b) = 00 ⇔ f(a) − f(b) = 00 = f(0) ⇔ f(a) = f(b) ⇔ φ(a + U) = φ(b + U).

∴ φ is well-defined and one-one mapping.

Let y ∈ R¯.

Since f: R → R¯ is onto, there exists x ∈ R so that f(x) = y. For this x ∈ R

we have x + U ∈ R/U.

∴ for each y ∈ R¯ there exists x + U ∈ R/U so that φ(x + U) = f(x) = y.

∴ φ is onto mapping.

Let a + U,b + U ∈ R/U. Then a,b ∈ R

φ[(a + U) + (b + U)] = φ[(a + b) + U] = f˙(a + b) (Definition of φ )

= f(a) + f(b) = φ(a + U) + φ(b + U) ( ∵ f is homomorphism )

φ[(a + U)(b + U)] = φ[a.b + U] = f(ab)

= f(a)f(b) = φ(a + U)φ(b + U) ( ∵ f is homomorphism )

∴ φ is a homomorphism.

Hence φ is an isomorphism from R/U to R¯ = f(R).

Note. 1. Every homomorphic image of a ring R is isomorphic to some quotient ring thereof.

2. If f: R → R₀ is onto homomorphism from a ring R to the ring R’ and U is an ideal of R then R/U is isomorphic to R₀. Then f(R) = R¯ = R0. In the above proof replace R¯ by R₀.

3. R¯ = f(R) = the homomorphic image of R.

Let Ψ: R → R/U be the Canonical homomorphism. By fundamental theorem φ: R/U → R¯ is isomorphism.

For x ∈ R we have Ψ(x) = x + U ∈ R/U.

For this x + U ∈ R/U we have φ(x + U) = f(x) ∈ f(R) = R¯

Also, for x ∈ R we have f(x) ∈ f(R) = R¯. Therefore, φ · Ψ = f.

Ring Theory And Vector Calculus Solved Problems

Example. 1. Is the ring 2Z isomorphic to ring 3Z?

Solution.

2Z = {2n | r ∈ Z} and 3Z = {3n | n ∈ Z}.

Define f : 2Z → 3Z by f(2x) = 3x∀2x ∈ 2Z. Let 2m, 2n ∈ 2Z.

f(2m + 2n) = f(2(m + n)) = 3(m + n) = 3m + 3n = f(2m) + f(2n)

f(2m.2n) = f(2(2mn)) = 3(2mn) 6= 3m.3n = f(2m)f(2n)

∴ The correspondence f does not preserve multiplication.

∴ Ring 2Z is not isomorphic to ring 3Z.

Vector Calculus Problems In Physics With Detailed Solutions

Example. 2. Let Z₄, Z₁₀ be modulo-4 and modulo – 10 rings. If f: Z₄ → Z₁₀ is defined by f(x) = 5x∀x ∈ Z₄ then prove that f is a homomorphism.

Solution.

Given

Let Z₄, Z₁₀ be modulo-4 and modulo – 10 rings. If f: Z₄ → Z₁₀ is defined by f(x) = 5x∀x ∈ Z₄

Let a, b ∈ Z₄.

Let a+b = 4q₁+r₁ and a·b = 4q₂+r₂ where 0 ≤ r₁ , r₂ < 4.

f(a + b) =f (r₁) = 5r₁ = 5 (a + b − 4q₁) =5a + 5b − 20q₁ = 5a + 5b( modulo 10) = f(a) + f(b) in Z₁₀.

f(a : b) =f (r₂) = 5r₂ = 5 (ab − 4q₂) = 5ab − 20q₂ =5ab(mod10) = 25ab(mod10) = 5a · 5b = f(a) · f(b) in Z₁₀.

∴ Z₄ is homomorphic to Z₁₀.

Example. 3. Prove or disprove that f: M₂(Z) → Z defined by

⇒ $f\left(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\right)=a \forall\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \in \mathrm{M}_2(\mathrm{Z}) \text { is a ringhomomorphism. } $

⇒ $\text { Sol. Let } \mathrm{A}_1=\left[\begin{array}{ll}
a_1 & b_1 \\
c_1 & d_1
\end{array}\right], \mathrm{A}_2=\left[\begin{array}{ll}
a_2 & b_2 \\
c_2 & d_2
\end{array}\right] \in \mathrm{M}_2(\mathrm{Z})$

where a₁,b₁,c₁,d₁,a₂,b₂,c₂,d₂ ∈

Z. By definition of f,f ( A₁) = a1,f ( A₂) = a2.

f ( A1 + A2) = f $ \left(\begin{array}{ll}
a_1+a_2 & b_1+b_2 \\
c_1+c_2 & d_1+d_2
\end{array}\right)$ = a₁ + a₂ = f ( A₁) + f ( A₂)

∴ f preserves addition.

f ( A₁ · A₂) = f $\left(\begin{array}{cc}
a_1 a_2+b_1 c_2 & a_1 b_2+b_1 d_2 \\
c_1 a_2+d_1 c_2 & c_1 b_2+d_1 d_2
\end{array}\right) $ = a1a2 + b1c2 6= a1a2 =

f ( A₁) · f ( A₂) f does not preserve multiplication.

∴ f is not a homomorphism.

Example. 4. Let Z(√2) = {m + n√2 | m,n ∈ Z} be a ring under addition and multiplication of numbers. Prove that f : Z(√2) → Z(√2) defined by f(m + n√2) = m − n√2∀m + n√2 ∈ Z(√2) is an automorphism. Also, find Ker f.

Solution.

Given

Let Z(√2) = {m + n√2 | m,n ∈ Z} be a ring under addition and multiplication of numbers.

Let a,b ∈ Z(√2) so that a = m₁ + n₁√2,b = m₂ + n2√2 where m₁,n₁,m₂,n₂ ∈ Z.

Then we have a+b = (m₁ + m₂)+(n₁ + n₂) √2 and ab = (m₁m₂ + 2n₁n₂)+ (m₁n₂ + m₂n₁) √2. Clearly, f is well-defined.

By definition of f;

f(a + b) = (m₁ + m₂) − (n₁ + n₂) √2 = m₁ − n₁√2 + m₂ − n₂√2 = f(a) + f(b) and f(ab) = (m₁m₂ + 2n₁n₂)−(m₁n₂ + m₂n₁) √2 =(m₁ − n₁√2 m₂ − n₂√2 )= f(a)f(b)

∴ f is an endomorphism.

a,b ∈ Z(√2); f(a) = f(b) ⇒ m₁ − n₁√2 = m₂ − n₂√2

⇒ m₁ = m₂ and n₁ = n₂ ⇒ m₁ + n₁√2 = m₂ + n₂√2 ⇒ a = b

∴ f is one-one.

Let y = m + n√2 ∈ Z(√2), the co-domain of f.

Then x = m − n√2 ∈ Z(√2), the domain of f, exists so that

f˙(x) = f(m − n√2) = m − (−n√2) = m + n√2 = y.

∴ For each y ∈ Z(√2) there exists x ∈ Z(√2) so that f(x) = y.

∴ f is onto. Hence f is an automorphism.

f(m+n√2) = m−n√2 = 0, zero element of Z(√2) ⇒ m = 0,n = 0 ⇒ m+n√2 = 0, zero element.

∴ ker f = {0}.

Ring Theory Theorems And Proofs Explained In Study Notes

Example. 5. Let Zb e is the ring of integers and Zn is the ring of residue classes modulo n. If a mapping f: Z → Zn is defined by f(x) = ¯r∀x ∈ Z where x ≡ r(modn) proves that f is a homomorphism. Also, find Ker f.

Solution.

Given

Let Zb e is the ring of integers and Zn is the ring of residue classes modulo n. If a mapping f: Z → Zn is defined by f(x) = ¯r∀x ∈ Z where x ≡ r(modn)

Let x, y ∈ Z.

By the definition of f; f(x) = ¯r, f(y) = ¯s where x ∼= r(modn) and y ≡ s(modn).

Clearly, f is well-defined.

We know that (1)x ≡ r(modn),

y ≡ s(modn) ⇒ x + y ≡ r + s(modn) and xy ≡ rs(modn) and (2)r + s = r¯ + ¯s, rs = ¯rs¯.

Now f(x+y) = r + s = ¯r+¯s = f(x)+f(y) and f(xy) = rs = ¯rs¯ = f(x)f(y).

∴ f is a homomorphism.

Hence Zn is a homomorphic image of Z.

This is called the natural homomorphism from Z to Z_n.

Example. 6. Let C be the ring of Complex numbers and M₂(R) be the ring of 2 × 2 matrices. If f : C → M₂(R) is defined by f(a + ib) =$\left(\begin{array}{cc}a & b \\ -b & a\end{array}\right)$ prove that f is an into isomorphism or monomorphism. Also, find Ker f

Solution.

Given

Let C be the ring of Complex numbers and M₂(R) be the ring of 2 × 2 matrices. If f : C → M₂(R) is defined by f(a + ib) =$\left(\begin{array}{cc}a & b \\ -b & a\end{array}\right)$

Let Z₁, Z₂ ∈ C and Z₁ = x₁ + iy 1, Z₂ = x₂ + iy₂ where x₁, y₁, x₂, y₂ ∈ R.

Then f (Z₁) = f (x₁ + iy₁) = $\left(\begin{array}{cc}
x_1 & y_1 \\
-y_1 & x_1
\end{array}\right) $ and f (Z₂) = f (x₂ + iy₂) = $ \left(\begin{array}{cc}
x_2 & y_2 \\
-y_2 & x_2
\end{array}\right)$

f (Z₁ + Z₂) = f ((x₁ + x₂) + i (y₁ + y₂)) = $ \left(\begin{array}{cc}
x_1+x_2 & y_1+y_2 \\
-\left(y_1+y_2\right) & x_1+x_2
\end{array}\right) $=

⇒ $\left(\begin{array}{cc}
x_1 & y_1 \\
-y_1 & x_1
\end{array}\right)+\left(\begin{array}{cc}
x_2 & y_2 \\
-y_2 & x_2
\end{array}\right) $ = f (Z₁) + f (Z₂)

f (Z₁ · Z₂) = f ((x₁x₂ − y₁y₂) + i (x₁y₂ + x₂y₁)) =$ \left(\begin{array}{cc}
x_1 x_2-y_1 y_2 & x_1 y_2+x_2 y_1 \\
-\left(x_1 y_2+x_2 y_1\right) & x_1 x_2-y_1 y_2
\end{array}\right)$

⇒ $\left(\begin{array}{cc}
x_1 & y_1 \\
-y_1 & x_1
\end{array}\right) \cdot\left(\begin{array}{cc}
x_2 & y_2 \\
-y_2 & x_2
\end{array}\right)$ = f (Z₁) · f (Z₂)

∴ f is a homomorphism from C to M2(R).

f (Z₁) = f (Z₂) ⇒ $\left(\begin{array}{cc}
x_1 & y_1 \\
-y_1 & x_1
\end{array}\right)=\left(\begin{array}{cc}
x_2 & y_2 \\
-y_2 & x_2
\end{array}\right)$ ⇒ x₁ = x_2, y₁ = y₂

⇒ x₁ + iy₁ = x₂+ iy₂⇒ Z₁ = Z₂.

∴ f is one-one.

⇒ $\text { For }\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right) \in \mathrm{M}_2(\mathrm{Z}) \text { and }\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right) \neq\left(\begin{array}{cc}
a & b \\
-b & a
\end{array}\right)$

there is no complex number a + ib ∈ C satisfying the correspondence.

∴ f is not onto. Hence f is a monomorphism or an isomorphism.

Note. Instead of M₂(R) if we take ring of 2×2 matrices = S =$\left\{\left(\begin{array}{cc}a & b \\ -b & a\end{array}\right) \mid a, b \in \mathrm{R}\right\}$

then f: C → S will be isomorphism onto. 1 = 1 + 0i ∈ C is the unity in C, and

f(1) = f(1 + 0i) = $\left(\begin{array}{cc}1 & 0 \\ -0 & 1\end{array}\right)=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$ = unit matrix in M_{2 (R)}

f(a + ib) =$\left(\begin{array}{cc}a & b \\ -b & a\end{array}\right)=\left(\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right)$ = zero element in M₂(R) ⇒ a = 0, b = 0

⇒ a + ib = 0 + i0 = 0 = Zero element in C.

∴ Ker f = {0}

Example. 7. Let R be the ring of integers and R 0 be the set of even integers in which addition is the same as that of integers and multiplication (∗) is defined by a ∗ b = ab/2∀a, b ∈ R 0. Prove that R is isomorphic to R’.

Solution.

Given

Let R be the ring of integers and R 0 be the set of even integers in which addition is the same as that of integers and multiplication (∗) is defined by a ∗ b = ab/2∀a, b ∈ R 0

We know that R₀ = {2n | n ∈ Z} is a commutative group under addition.

Let a, b, c ∈ R₀ so that a = 2m, b = 2n, c = 2p where m, n, p ∈ Z.

a, b ∈ R₀ ⇒ a ∗ b = (ab/2) = (2m)(2n)/2 = 2(mn) = 2q where q ∈ Z.

∴ ∗ is a binary operation in R₀.

∴ ∗ is a binary operation in R₀. a, b, c ∈ R₀ ⇒ (a ∗ b) ∗ c = ab² ∗ c = (ab/22)c = ABC 4 = a(bc/2)² = a ∗ bc² = a ∗ (b ∗ c)

∴ ∗ is associative in R’.

a, b, c ∈ R’ ⇒ a ∗ (b + c) = a(b²+c) = ab² + ac² = a ∗ b + a ∗ c.

Similarly, (b + c) ∗ a = b ∗ a + c ∗ a. ∴ ∗ is distributive over addition.

a, b ∈ R 0 and a ∗ b = ab² = ba² = b ∗ a ⇒ ∗ is commutative in R 0.

Hence (R₀, +, ∗) is a commutative ring.

Define f: R → R₀ by f(x) = 2x∀x ∈ R.. Obviously f is well defined.

Let x, y ∈ R so that x + y, xy ∈ R. Then f(x) = 2x, f(y) = 2y.

Now f(x + y) = 2(x + y) = 2x + 2y = f(x) + f(y) and

f(xy) = 2(xy) = (2x)(22 y) = 2x ∗ 2y = f(x) ∗ f(y).

∴ f is a homomorphism from R into R 0.

x, y ∈ R, f(x) = f(y) ⇒ 2x = 2y ⇒ x = y ⇒ f is one-one.

Let b ∈ R₀. Then b = 2a where a ∈ R and for a ∈ R we have f(a) = 2a = b.

∴ for each b ∈ R 0 there exists a ∈ R so that b = f(a) ⇒ f is onto .

Hence f is an isomorphism from R to R 0.

Example. 8. Prove that any homomorphism defined on a field is either isomorphism or zero homomorphism. Or, prove that a field has no proper homomorphic image.

Solution. Let F be a field R be a ring and f: F → R be a homomorphism.

Then we know that Ker f is an ideal of the field F.

Since a field has no proper ideals either Ker f = F or Ker f = {0} where ’ 0 ’ is the zero element of F.

Let Ker f = F.

By definition of Ker f., we have f(x) = 00∀x ∈ F where 00 ∈ R is the zero element.

∴ homomorphic image of F = f(F) = {0′}.

Hence, in this case, f is a zero homomorphism. Let Ker f = {0}.

∴ By theorem (2) Art. 3.2, f is an isomorphism from F to R.

Hence, in this case, a homomorphic image of F = f(F) is also a field.

Example. 9. Prove that Z/hnior Z/nZ is isomorphic to Z_n:

Solution. Define f: Z → Z_n as f(x) = ¯r∀x ∈ Z.

Then f is a homomorphism

∀r¯ ∈ Zn we have r ∈ Z and for this r ∈ Z, f(r) = ¯r ( ∵ r ≡ r mod n).

∴ f: Z → Zn is onto homomorphism. But Ker f = nZ = hni.

By fundamental theorem; Z/ Ker f ∼= Z_n(i.e.,) Z/hni ∼= Z_n.

Further, if n is a prime = p then Zp is a field.

Z/hpi ∼= Z_p ⇒ that a quotient ring of an integral domain is isomorphic to the field Z_p.

Hence a quotient ring of an integral domain may be a field.

Note. Z₆= {0, 1, 2, 3, 4, 5} is a ring and U = {0, 3} is an ideal of Z_6.
Z₆/U = {0 + U, 1 + U, 2 + U} contains only 3 elements.

If we define φ : Z₃ → Z₆/N as φ(0) = 0 + U, φ(1) = 1 + U, φ(2) = 2 + U, then φ is an isomorphism.

Ring Theory And Vector Calculus Maximal Ideals

The concept of a maximal ideal of a ring is analogous to the idea of the maximum normal subgroup in Group Theory.

Definition. (Maximal Ideal). A maximal ideal M of a ring R is an ideal different from R such that there is no proper ideal U of R properly containing M.

(or)

Let R be a ring and M be an ideal of R so that M 6= R. M is said to be a maximal ideal of R if whenever U is an ideal of R such that M ⊂ U ⊂ R then either R = U or U = M.

Note 1. M is a maximal ideal of R if any ideal U of R containing M is either R or M.

2. An ideal M of a ring R is called a maximal ideal if M is not included in any other ideal of R except R itself. That is, the only ideal that properly contains a maximal ideal is the entire ring.

3. If M is a maximal ideal of the ring R then there exists no ideal U of R so that M ⊂ U ⊂ R

Theorem. 1. In the ring Z of integers the ideal generated by a prime integer is a maximal ideal.

Proof. Let p be the prime integer and M = HPI = pZ = {pn | n ∈ Z} be the ideal generated by p.

Let U be any ideal so that M ⊂ U ⊂ Z

Since every ideal of Z is a principal ideal, U is a principal ideal so U = hqi where q is an integer.

M ⊂ U ⊂ Z ⇒ hpi ⊂ hqi ⊂ Z ⇒ p ∈ hqi ⇒ p = qm, m ∈ Z.

Since p is a prime, either q = 1 or m = 1

m = 1 ⇒ p = q ⇒ hpi = hqi ⇒ M = U; q = 1 ⇒ hqi = Z ⇒ U = Z

∴ M is a maximal ideal.

Note. An ideal generated by a composite integer is not a maximal ideal.

Consider M = h6i = {. . . , −12, −6, 0, 6, 12, . . .}, the ideal generated by composite integer 6.

There exists ideal U = h3i = {. . . , −12, −9, −6, −3, 0, 3, 6, . . .} so that M ⊂ U ⊂ Z

Theorem. 2. If M is a maximal ideal of the ring of integers Z then M is generated by prime integer.

Proof. Let M = hni where n ∈ Z be a maximal ideal of Z.

We prove that n is a prime integer.

If possible, suppose that n = ab where a, and b are prime integers.

Then U = hai is an ideal of Z and U ⊃ M so that M ⊂ U ⊂ Z

Since M is the maximal ideal of Z, by the definition either U = Z or M = U.

Case (1). Let U = Z. Then U = hai = h1i so that a = 1

∴ n = ab = b ⇒ n is a prime integer.

Case (2). Let M = U

Then U = hai = M ⇒ a ∈ M ⇒ a ∈ hni ⇒ a = rn for some r ∈ Z.

∴ n = ab = (rn)b = n(rb) ⇒ 1 = rb ⇒ r = 1, b = 1.

∴ n = a(1) = a ⇒ n is a prime integer.

From cases (1) and (2) we have that n is a prime integer.

Note. 1. For the ring of integers Z, any ideal generated by a prime integer is a maximal ideal.

2. A ring may have more than one maximal ideal. For example, the ring Z has h2i, h3i, h5i, . . . as maximal ideals.

Theorem. 3. An ideal in Z is a maximal ideal if and only if it is generated by a prime integer.

Proof. Write the proofs of Theorems (1) and (2).

Theorem. 4. An ideal U of a commutative ring R with unity is maximal if and only if the quotient ring RU is a field.

Proof. R is a commutative ring with unity and U is an ideal ⇒ the quotient ring R/U = {x + U | x ∈ R} is commutative and has a unity element.

Zero element of R/U = 0 + U = U where 0 ∈ R is the zero element in R.

Unity element of R/U = 1 + U where 1 ∈ R is the unity element in R. It is to be noted that a + U = U zero element of R/U ⇔ a ∈ U

(1) Suppose that U is a maximal ideal of R. We prove that R/U is a field.

To prove that R/U is a field we have to show that every non-zero element of R/U has a multiplicative inverse.

Let x + U ∈ R/U and x + U be non-zero elements. Then x /∈ U

If hxi is the principal ideal of R then hxi + U is also an ideal of R.

( ∵ The sum of two ideals is also an ideal. Ex. 4)

x /∈ U ⇒ U ⊂ hxi + U

Now we have, U ⊂ hxi + U ⊆ R and U is maximal ideal ⇒ hxi + U = R = h1i

⇒ there exists a ∈ U and α ∈ R such that a + xα = 1

∴ 1 + U = (a + xα) + U = (a + U) + (xα + U) (sum of cosets)

= U + (xα + U) = (0 + U) + (xα + U)

( ∵ a ∈ U ⇒ a + U = U)

= xα + U = (x + U)(α + U) (Product of cosets)

∴ for non-zero element x + U ∈ R/U there exists α + U ∈ R/U such that (x + U)(α + U) = 1 + U.

Hence every non-zero element of R/U is invertible.

∴ R/U is a field.

(2) suppose that R/U is a field. We prove that U is the maximal ideal.

Let U₀ be an ideal of R so that U 0 ⊃ U and U 0 6= U

Now we show that U₀ = R

Since U₀ ⊃ U and U₀ 6= U, there exists α ∈ U₀ such that α /∈ U

α /∈ U ⇒ α + U is non-zero element of R/U

R/U is a field of α + U is a non-zero element of R/U

⇒ α + U has multiplicative inverse, say x + U.

∴ (α + U)(x + U) = 1 + U

⇒ αx + U = 1 + U ⇒ 1 − αx ∈ U ⊂ U 0 ( ∵ a + U = b + U ⇒ a − b ∈ U)

x ∈ R, α ∈ U₀ and U₀ is an ideal ⇒ αx ∈ U₀.

αx ∈ U₀, 1 − αx ∈ U₀ ⇒ αx + (1 − αx) = 1 ∈ U₀

∴ 1 ∈ U₀ and U 0 is an ideal ⇒ U₀ = R.

Hence U is a maximal ideal.

Ring Theory And Vector Calculus Prime Ideals

Definition. (Prime Ideal) An ideal U₆= R of a commutative ring R is said to be prime ideal if for all a, b ∈ R and a, b ∈ U ⇒ a ∈ U or b ∈ U.

Example. For an integral domain R, the null ideal is a prime ideal.

∵ a, b ∈ R, ab ∈ h0i ⇒ ab = 0 ⇒ a = 0 or b = 0

Theorem. An ideal U≠ R of a commutative ring R, is a prime idea if and only if R/U is an integral domain.

Proof. Let R/U be an integral domain.

We now prove that U is a prime ideal of R.

∀a, b ∈ R and a, b ∈ U ⇒ ab + U = U ⇒ (a + U).(b + U) = 0 + U

⇒ a + U = 0 + U or b + U = 0 + U (0 + U is the zero element of R/U)

⇒ a ∈ U or b ∈ U.

∴ U is a prime ideal of R.

Conversely, let U be a prime ideal of R.

We now prove that R/U is an integral domain.

a + U, b + U ∈ R/U and (a + U) · (b + U) = 0 + U

⇒ ab + U = 0 + U ⇒ ab ∈ U ⇒ a ∈ U or b ∈ U ( ∵ U is prime ideal)

⇒ a + U = 0 +. U or b + U = 0 + U.

∴ R/U has no zero divisors and hence is an integral domain.

Corollary. Every maximal ideal of a commutative ring R with unity is a prime ideal.

Proof. Let U be a maximal ideal of a ring R.

By Theorem (3) of Art. 3.3; R/U is a field. ⇒ R/U is an integral domain.

By Theorem (4) of Art. 3.3; U is a prime ideal.

Thus every maximal ideal of R is a prime ideal.

Note. 1. The converse of the above corollary is not true.

That is, a prime ideal of a commutative ring with unity need not be a maximal ideal.

Consider the integral domain Z of integers.

The null ideal = h0i of Z is a prime ideal. But the h0i ideal is not the maximal ideal.

There exists ideal = h2i of Z such that h0i ⊂ h2i ⊂ Z and h2i 6= h0i, h2i 6= Z.

Example. If R = {0, 2, 4, 6} is a ring with respect to addition and multiplication modulo 8, then show that M = {0, 4} is a maximal ideal of R but not a prime ideal.

Solution.

Given

R = {0, 2, 4, 6} is a ring with respect to addition and multiplication modulo 8

For 0, 0 ∈ M, 0 − 0 = 0 ∈ M, For 0,4 or 4, 0 ∈ M,

0 − 4 = 8 − 4 = 4 ∈ M, 4 − 0 = 4 ∈ M. For 4, 4 ∈ M, 4 − 4 = 0 ∈ M.

For 0 ∈ M, 0, 2, 4, 6 ∈ R we have 0.0 = 0, 0.2 = 0, 0.4 = 0, 0.6 = 0 ∈ M For

4 ∈ M, 0, 2, 4, 6 ∈ R we have 4.0 = 0 ∈ M, 4.2 = 8 = 0 ∈ M, 4.4 = 16 = 0 ∈ M,

4.6 = 24 = 0 ∈ M.

∴ M is an ideal of R.

U₁ = {0, 2, 4} is not an ideal, for, 2, 4 ∈ U1 we have 2 − 4 = −2 = 6 ∈/ U₁

U₂ = {0, 4, 6} is not ideal, for, 6, 4 ∈ U 2 we have 6 − 4 = 2 ∈/ U₂.

∴ There is no ideal U of R such that M ⊂ U ⊂ R.

∴ M = {0, 4} is a maximal ideal.

For 2, 6 ∈ R and 2.6 = 12 = 4 ∈ M does not imply either 2 ∈ M or 6 ∈ M

∴ M is not a prime ideal.

Ring Theory And Vector Calculus Field Of Quotients Of An Integral Domain

If an integral domain is such that every non-zero element of it has a multiplicative inverse the n it is a field. However many integral domains do not form fields.

Though the integral domain of integers is not a field, it is such that it can be embedded in the field of rational numbers. In this section, we show that every integral domain can be regarded as being contained in a certain field. The minimal field containing an integral domain is called a field of quotients of an integral domain.

Theorem. Every integral domain can be embedded in a field. (or) An integral domain D can be embedded in a field F such that every element of F can be regarded a quotient of two elements of D.

Proof. Let D be an integral domain with at least two elements.

Consider S = {(a, b) | a, b ∈ D, b 6= 0}. Then S 6= φ and S ⊂ D × D.

For all (a, b), (c, d) ∈ S define a relation ∼ on S as (a, b) ∼ (c, d) if and only if ad = bc. We now prove that ∼ is an equivalence relation on S.

(1) For each (a, b) ∈ S we have ab = ba which implies that (a, b) ∼ (a, b).

(2) (a, b), (c, d) ∈ S and (a, b) ∼ (c, d) ⇒ ad = bc ⇒ cb = da ⇒ (c, d) ∼ (a, b).

(3) (a, b), (c, d), (e, f) ∈ S and (a, b) ∼ (c, d), (c, d) ∼ (e, f) ⇒ ad = bc, cf = de.

⇒ (ad)f = (bc)f; cf = de ⇒ (af)d = b(de) = d(be) ⇒ af = be ( ∵ d 6= 0)

⇒ (a, b) ∼ (e, f)

∴ ’ 0 ’ is an equivalence relation on S.

The equivalence relation partitions the set S into equivalence classes which are either identical or disjoint.

For (a, b) ∈ S let a/b denote the equivalence class of (a, b). Then a/b = {(x, y) ∈ S | (x, y) ∼ (a, b)}. If a/b, c/d are the equivalence classes of (a, b), (c, d) ∈ S then either a/b = c/d or a/b ∩ c/d = φ. It is evident that a/b = c/d if and only if ad = bc.

Let F denote the set of all the equivalence classes or the set of quotients.

Then F = {n a/b | (a, b)} ∈ S˙ o.

Since D has at least two elements, say, 0, a ∈ D

we have quotients $\frac{0}{a}, \frac{a}{a} \in F and \frac{0}{a} \neq \frac{a}{a}$

∴ the set F has at least two elements.

For a/b, c/d ∈ F define addition (+) and multiplication ( · ) as

⇒ $\frac{a}{b}+\frac{c}{d}=\frac{a d+b c}{b d} and \frac{a}{b}, \frac{c}{d}=\frac{a c}{b d}$

Since D is without zero divisors, b ≠ 0, d ≠ 0 ∈ D ⇒ bd≠ 0

So $\frac{a d+b c}{b d}, \frac{a c}{b d} \in F$

Now we prove that the addition and multiplication defined above are well defined.

Let $\frac{a}{b}=\frac{a^{\prime}}{b^{\prime}} and \frac{c}{d}=\frac{c^{\prime}}{d^{\prime}}. Then a b^{\prime}=a^{\prime} b$ and c d^{\prime}=c^{\prime} d$

⇒ $Now (I) \Rightarrow a b^{\prime} d d^{\prime}=a^{\prime} b d d^{\prime} and b b^{\prime} c d^{\prime}=b b^{\prime} c^{\prime} d$

⇒ $\text { (2) For } \frac{a}{b}, \frac{c}{d} \in F ; \frac{a}{b}+\frac{c}{d}=\frac{a d+b c}{b d}=\frac{b c+a d}{d b}=\frac{c}{d}+\frac{a}{b} \Rightarrow$

⇒ $(3) For $u \neq 0 \in D$ we have $\frac{0}{u} \in F$ such that $\frac{0}{u}+\frac{a}{b}=\frac{0 b+u a}{u b}=\frac{u a}{u b}=\frac{a}{b} \forall \frac{a}{b} \in F. \therefore \frac{0}{u} \$in F is the zero element.

⇒ $ \frac{a d+b c}{b d}=\frac{a^d d+b c}{b^{\prime} d^{\prime}} $

Also (I) ⇒$ a b^{\prime} c d^{\prime}=a^{\prime} b c^{\prime} d \Rightarrow(a c)\left(b^{\prime} d^{\prime}\right)=\left(a^{\prime} c^{\prime}\right)(b d) \Rightarrow \frac{a c}{b d}=\frac{a^{\prime} c^{\prime}}{b^{\prime} d^{\prime}} $

∴ The addition and multiplication of quotients are well-defined binary operations on F.

We now prove that (F, +, ·) is a field.

⇒ $\text { (1) For } \frac{a}{b}, \frac{c}{d}, \frac{c}{f} \in F ;\left(\frac{a}{b}+\frac{c}{d}\right)+\frac{c}{f}=\frac{a d+b c}{b d}+\frac{c}{b}=\frac{(a d+b c) f+(b d) e}{(b d) f}=\frac{a(d f)+(c f+d e) b}{b(d f)}= $

⇒ $\frac{a}{b}+\frac{c f+d c}{d f}=\frac{a}{b}+\left(\frac{c}{d}+\frac{c}{f}\right) \Rightarrow $ addition is assocative.

⇒ $\text { (2) For } \frac{a}{b}, \frac{c}{d} \in F ; \frac{a}{b}+\frac{c}{d}=\frac{a d+b c}{b d}=\frac{b c+a d}{d h}=\frac{c}{d}+\frac{a}{b} \Rightarrow \text { addition is commutative. } $

(3) For $u \neq 0 \in D we have \frac{0}{u} \in F such that \frac{0}{u}+\frac{a}{b}=\frac{0 b+u a}{u b}=\frac{u a}{u b}=\frac{a}{b} \forall \frac{a}{b} \in F. therefore \frac{0}{u} \in F$is the zero element.

(4) Let$\frac{a}{b} \in F. Then \frac{-a}{b} \in F such that \frac{a}{b}+\frac{-a}{b}=\frac{a b+(-a) b}{b^2}=\frac{0}{b^2}=\frac{0}{u}\left(because 0 u=0 b^2\right)$ therefore every element in F has additive inverse.

(5) For $\frac{a}{b}, \frac{c}{d}, \frac{c}{f} \in F ;\left(\frac{a}{b}, \frac{c}{d}\right) \cdot \frac{c}{f}=$

⇒ $\frac{a c}{b d} \cdot \frac{c}{f}=\frac{(a c) c}{(b d) f}=\frac{a(c)}{b(d f)}=\frac{a}{b} \cdot \frac{c c}{d f}=\frac{a}{b} \cdot\left(\frac{c}{d} \cdot \frac{c}{f}\right)$

∴ multiplication is associative.

⇒ $\text { (6) For } \frac{a}{b}, \frac{c}{d} \in F ; \frac{a}{b} \cdot \frac{c}{d}=\frac{a c}{b d}=\frac{c a}{d b}=\frac{c}{d} \cdot \frac{a}{b}$

∴ multiplication is commutative

(7) for $u \neq 0 \in D we have frac{u}{u} \in Fsuch that \frac{a}{b} \cdot \frac{u}{u}=\frac{a u}{b u}=\frac{a}{b} \forall \frac{a}{b} \in Ftherefore frac{u}{u} \in F$ is the unity element.

(8) Let$\frac{a}{b} \in F and \frac{a}{b} \neq \frac{\underline{0}}{u}.$

⇒ $Then a u \neq 0 which implies that a \neq 0 asu \neq 0. therefore b \neq 0 and a \neq 0 \Rightarrow \frac{b}{a} \in F$

⇒ $therefore for \frac{a}{b}\left(\neq \frac{a}{u}\right) \in F there exists\frac{b}{a} \in F such that$

⇒ $ \frac{a}{b} \cdot \frac{b}{a}=\frac{a b}{b a}=\frac{u}{u} \quad(because(a b) u=(b a) u)$

∴ every non-zero element in F has multiplicative inverse

⇒ $\text { (9) For } \frac{a}{b}, \frac{c}{d}, \frac{c}{f} \in F ; \frac{a}{b} \cdot\left(\frac{c}{d}+\frac{c}{f}\right)=\frac{a}{b} \cdot \frac{c f+d c}{d f}=\frac{a(c f+d c)}{b(d f)}=\frac{(a c f+a d c)(b d f)}{(b d f) b d f)}$

⇒ $\frac{a c f \text { bdf+adebdf }}{(b a d) b d f)}=\frac{a c f}{b d f}+\frac{a d e}{b d f}=\frac{a c}{b d}+\frac{a c}{b f}=\frac{a}{b} \cdot \frac{c}{d}+\frac{a}{b} \cdot \frac{c}{f}$

Similarly we can prove that $ \left(\frac{c}{d}+\frac{c}{f}\right) \cdot \frac{a}{b}=\frac{c}{d} \cdot \frac{a}{b}+\frac{e}{f} \cdot \frac{a}{b}$

∴ Multiplication is distributive over addition.

In view of (1), (2), (3), (4), (5), (6), (7), (8) and (9) (F, +, ·) is a field.

Now we have to prove that D is embedded in the field F, that is, we have to show that there exists an isomorphism of D into F.

Define the mapping φ : D → F by φ(a) = ax/x ∀a ∈ D and x(6= 0) ∈ D.

⇒ $a, b \in D \text { and } \phi(a)=\phi(b) \Rightarrow \frac{a x}{x}=\frac{b x}{\tau} \Rightarrow(a x)^x x=(b x) x$

⇒ $\Rightarrow(a-b) x^2=0 \Rightarrow a-b=0 \text { since } x^2 \neq 0 \Rightarrow a=b . \quad therefore \phi \text { is one – one. }$

⇒ $\text { For } a, b \in D ; \phi(a+b)=\frac{(a+b) x}{x}=\frac{(a+b) x x}{x x}=\frac{a x x+b x x}{x x}=\frac{a x}{x}+\frac{b x}{x}=\phi(a)+\phi(b)$

⇒ $\phi(a b)=\frac{(a b) x}{x}=\frac{(a b) x x}{x x}=\frac{a x}{x}, \frac{b x}{x}=\phi(a) \phi(b)$

∴ φ is a homomorphism. Hence φ is an isomorphism of D into F.

∴ the integral domain D is embedded in the field F.

Note. 1. Every element in the field F is in the form of a quotient of two elements in D. So, the field F is called the “field of quotients of D ”

2. The equivalence class of (a, b) ∈ S is also denoted as [(a, b)] or [a, b] or (a, b) Then [(a, b)] = [(c, d)] ⇔ ad = bc, [(a, b)] + [(c, d)] = [(ad + bc, bd)], [(a, b)] · [(c, d)] = [(ac, bd)], the zero element of F = [(0, 1)] and the unity element of F = [(1, 1)]

3. If D is the ring of integers then the field F, constructed in the above theorem, would be the field Q of rational numbers.

Ring Theory And Vector Calculus Prime Fields

Definition. A field is said to be prime if it has no subfield other than itself.

The field Zp = {0, 1, 2, . . . , p − 1} where p is a prime and the field Q, of the set of all rational numbers, are prime fields. The field of real numbers R is not a prime field.
We now establish that any field F contains a subfield isomorphic to Zp or contains a subfield isomorphic to Q.

Theorem. 1. If R is a ring with unity element ’ 1 ’ then f: Z → R defined by f(x) = n · 1∀n ∈ Z is a homomorphism.

Proof. Let m, n ∈ Z. Then f(m) = m.1, f(n) = n.1.

f(m + n) = (m + n) · 1 = m · 1 + n · 1 = f(m) + f(n)

Let m > 0, n > 0.

(mn) · 1 = 1 + 1 + . . . + 1(mn times ) = {(1 + 1 + . . . + 1)m times } {(1 +1 + . . . + 1)n times } = (m.1)(n.1) (using Distributivity in the ring R )

Similarly, ∀m, n ∈ Z we can prove that (mn).1 = (m.1)(n.1) using Distributivity.

∴ f(mn) = (mn) · 1 = (m · 1)(n · 1) = f(m)f(n).

Hence f: Z → R is a homomorphism.

Theorem. 2. If R is a ring with unity element ’ I ’ and characteristic of R = n > 0 Then R contains a subring isomorphic to Zn.

Proof. Consider the homomorphism f: Z → R defined by f(m) = m.1∀m ∈ Z.

∴ Ker f is an ideal of Z.

But every ideal in Z is the form hsi = sZ where s ∈ Z.

Characteristic of ring R = n > 0 ⇒ n is the least positive integer such that n.1 = 0.

∴ Ker f = nZ = hni By fundamental Theorem, f(Z) ⊆ R is isomorphic to Z/nz = Z/hni.

But Z/nz = Z/hni is isomorphic to Zn.

∴ f(Z) ⊆ R is isomorphic to Zn.

Theorem.3. If R is a ring with unity element ’ 1 ’ and characteristic of R = 0 then R contains a subring isomorphic to Z.

Proof. Consider the homomorphism f: Z → R defined by f(m) = m: 1∀m ∈ Z.

Characteristic of R = 0 ⇒ m.1 ≠ 0∀m ∈ Z and m≠ 0.

Ker f = {m ∈ Z | f(m) = m.1 = 0} = {0}.

∴ f(Z) ⊆ R is isomorphic to Z.

Corollary. A field F of prime characteristic = p contains a subfield isomorphic to Zp and a field F of characteristic zero contains a subfield isomorphic to Q, the field of rational numbers.

Proof. Let F be the field of characteristic = p, a prime.

Then p is the least positive integer such that p.1 = 0.

∴ Ker f = pZ = hpi,

Hence, by the above theorem; F contains a subfield isomorphic to Zp.

Let F be the field of characteristic = 0.

By the above Theorem; F contains a subring isomorphic to Z.

But the field F contains a field of quotients of Z which is the field Q of rational numbers. Thus we have established that apart from isomorphism the only prime fields are Q and Zp.

Ring Theory And Vector Calculus Multiple Choice Questions:

1. For any two elements a,b in a ring, a(−b) =

(1) −(ab)

(2) ab

(3) −(ba)

(4) None

Answer: 1. −(ab)

2. If (R, +, ·) is a ring then (R, +) is

(1) A group

(2) An abelian group

(3) A finite group

(4) Semigroup

Answer: (2) An abelian group

3. The residue classes modulo 11 with respect to addition and multiplication modulo 11 is

(1) Commutative ring

(2) An integral domain

(3) A field

(4) Skew field

Answer: (3) A field

4. The characteristic of the residue classes mod 8 is

(1) 0

(2) 2

(3) 8

(4) none

Answer: (3) 8

5. If F is a field then the number of ideals in F is

(1) 0

(2) 1

(3) 2

(4) Infinite

Answer: (3) 2

6. If a, b are nilpotent elements in a commutative ring then ab is

(1) Nilpotent

(2) Not nilpotent

(3) Idempotent

(4) Zero

Answer: (1) nilpotent

7. The characteristic of the field of rational numbers is.

(1) 0

(2) ∞

(3) A prime

(4) None

Answer: (1) 0

8. With the usual addition and multiplication, the set of all even integers is

(1) A ring

(2) A field

(3) An integral domain

(4) None

Answer: (1) a ring

9. The number of proper ideals of a field is

(1) 0

(2) 1

(3) 2

(4) None of these

Answer: (3) 2

10. The set a + bi | a, b ∈ Z, i 2 = −1 of Gaussian integers is

(1) Ring

(2) Integral do ain

(3) Field

(4) None

Answer: (1) Ring

11. A commutative ring satisfying cancellation laws is a

(1) Field

(2) Skew field

(3) Integral domain

(4) None

Answer: (3) integral domain

12. In the ring Z of integers the ideal generated by 7 is

(1) Prime ideal

(2) Maximal ideal

(3) Not maximal

(4) None

Answer: (4) none

13. For the homomorphism f : R → R defined by f(x) = x∀x ∈ R Ker f =

(1) {0}

(2) R

(3){0, 1}

(4) none

Answer: (1) {0}

14. In the ring of integers Z, the units are (1) 0,1 (2).1 only (3) 1, −1 only (4) none 17. If a, b are two non-zero elements of a Euclidean ring R and b is a unit in R, then

(1) d(ab) = d(1)

(2) d(ab) > d(1)

(3) d(ab) < d(1)

(4) None

Answer: (3) d(ab) < d(1)

15. If F is a field and f: F → R is a homomorphism so that Ker f = {0} then f is

(1) Isomorphism

(2) Monomorphism

(3) Zero homomorphism

(4) None

Answer: (3) zero homomorphism

16. In the ring Z[i] of Gaussian integers 1 + i is

(1) Unit

(2) Unity element

(3) Prime element

(4) None

Answer: (3) prime element

17. If U is an ideal of ring R with unity 1 such that 1 ∈ U then U is

(1) U

(2) R

(3) ⊆ R

(4) none

Answer: (1) U

18. Let R be a commutative ring with unity and a ∈ R, then U = {ra | r ∈ R} is

(1) left ideal only

(2) ideal only

(3) prime ideal

(4) smallest ideal containing ’ a ’

Answer: (3) prime ideal

19. Every ring of numbers with unity is

(1) integral domain

(2) division ring

(3) field

(4) none

Answer: (1) integral domain

20. The ring R = {a + b√2 | a, b ∈ Q} is

(1) Integral domain

(2) Skew field

(3) Field

(4) None

Answer: (2) Skew field

21. If S₁, S₂ are two subrings of a ring R then S₁ + S₂ is

(1) Subring

(2) Ideal

(3) Need not be a subring

Answer: (3) need not be a subring

22. A subring S of a ring R is called ideal if

(1) α ∈ S, a ∈ R ⇒ αa ∈ S

(2) α ∈ S, a ∈ R ⇒ aα ∈ S

(3) α ∈ S, a ∈ R ⇒ αa, aα ∈ S

(4) none

Answer: (1) α ∈ S, a ∈ R ⇒ αa ∈ S

23. The set Q of rational numbers is

(1) subring

(2) ideal

(3) not subring

(4) not ideal, for the ring of real numbers

Answer: (1) subring

24. A homomorphic image of an integral domain is

(1) a ring

(2) integral domain

(3) need not be an integral domain

(4) none

Answer: (1) a ring

25. If R is a non-zero ring so that a₂ = a∀∈ R then the characteristic of R =

(1) 0

(2) 1

(3) 2

(4) Prime

Answer: (3) 2

26. If the characteristic of a ring R is 2 and a, b ∈ R then (a + b) 2 =

(1) a² + 2ab + b²

(2) a² + ab + ba + b²

(3) a² + b²

(4) none

Answer: (1) a² + 2ab + b²

27. If p is a prime, the ring of integers modulo p is.

(1) Field

(2) Integral domain

(3) Skew field

(4) None

Answer: (1) Field

Ring Theory & Vector Calculus Fill In The Blanks:

1. If R is a ring without zero divisors then hold in R.

Answer: Cancellation laws

2. A ring R has no zero divisors if

Answer: There exist a, b ∈ R and ab = 0 ⇒ a = 0 or b = 0

3. A division ring has divisors.

Answer: No zero-divisors

4. A finite integral domain is a

Answer: Field

5. In a ring R if a₂ = a for a ∈ R₀a₀ is called w. r. t. multiplication.

Answer: Idempotent element

6. a ∈ 0 ∈ R, R is a ring, is called nilpotent element if there exists

Answer: There exist a, b ∈ R and ab = 0 ⇒ a = 0 or b = 0

7. If characteristic of a ring R = 2 and a, b ∈ R commute then (a − b)² =

Answer: n ∈ N so that an = 0 46. a² + b²

8. A subring of (Z₆, +6, × 6) is

Answer: {0, 3}

9. A field has ideals.

Answer: Need not be an ideal

10. The union of two ideals of a ring R, of R.

Answer: No proper

11. For a field every ideal is

Answer: Need not be an ideal

12. A subring of (R, +, ·) which is not an ideal is

Answer: A principal ideal

13. In the quotient ring R/U the zero element is and the unity element is

Answer: (Q, +, ·) 52. U, 1 + U

14. An ideal U of a ring R is a prime ideal if.

Answer: U ⊂ U 0 ⊂ R

15. For the ring of integers any ideal generated by a prime integer is a

Answer: an ideal of R

16. For a commutative ring R, with unity if U is a maximal ideal then R/U is a

Answer: Maximal ideal

17. If f: R → R’ is a ring isomorphism and R is an integral domain then R’ is

Answer: associates

The post Ring Theory & Vector Calculus Notes appeared first on Answer Key for Math.

The Plane Definition Theorems Proofs Solved Problems Exercise Orthogonal Projection On Planes

Tanuja — Thu, 10 Aug 2023 03:52:37 +0000

Chapter 3 The Plane

Definition Of A Plane In Geometry With Examples And Proofs

Definition: A Plane is a surface such that if any two points are taken on it, the line joining them lies wholly on the surface.

Theorem.1 Every equation of the first degree in x, y, z represents a plane.

Proof. Let ax + bx + cz + d = 0, a₂ + b₂ + c₂ ≠ 0 …..(1)

be the first-degree equation in x, y, z.

If we have to show that (1) represents the equation to the plane, we prove that every point on the line joining any two points on (1) also lies on the locus (1).

Let p(x, y, z) and Q(x₂, y₂, z₂) be any two points on the locus (1).

Then we have ax₁ + bx₁ + cz₁ + d = 0 …..(2)

and ax₂ + by₂ + cz₂ + d = 0 …..(3)

Let R be any point on the line segment joining the points P and Q. Suppose R divides PQ in the ratio K : 1.

then $\mathrm{R}=\left(\frac{\mathrm{K} x_2+x_1}{\mathrm{~K}+1}, \frac{\mathrm{K} y_2+y_1}{\mathrm{~K}+1}, \frac{\mathrm{K} z_2+z_1}{\mathrm{~K}+1}\right), \mathrm{K}+1 \neq 0$

We have to show that R lies on the locus (1) for all values of K(≠ -1).

Substituting the coordinates of R in the LHS of (1), we get

⇒ $\frac{a\left(\mathrm{~K} x_2+x_1\right)}{\mathrm{K}+1}+\frac{b\left(\mathrm{~K} y_2+y_1\right)}{\mathrm{K}+1}+\frac{c\left(\mathrm{~K} z_2+z_1\right)}{\mathrm{K}+1}+d$

= a(Kx₂ + x₁) + b(Ky₂ + y₁) + c(Kz₂ + z₁) + d(K + 1)

= K(ax₂ + by₂ + cz₂ + d) + (ax₁ + by₁ + cz₁ + d) = K(0) + = 0 which shows that R lies on the locus (1).

Since R is an arbitrary point on the line joining P and Q, every point on PQ lies on (1)

∴ The equation ax + by + cz + d = 0, a₂ + b₂ + c₂ ≠ 0 always represents a plane.

Chapter 3 The Plane Converse Of The Above Theorem

Theorem.2 The equation to every plane is of the first degree in x, y, z.

Proof. Let π be the plane and O be the origin.

Case(1). Let O ∉ π, and let M be the foot of the perpendicular from O in π.

Let OM = p (>0). Let p(x, y, z) be any point in the plane.

Let [l, m, n] be Dc’s of the perpendicular OM.

p ≠ M. jOIN OP. OM = Projection of OP along OM

⇒ p = l(x-0)+m(y-0)+n(z-0) = lx+my+nz

Case.(2). Let O ∈ π, then p = 0.

p ∈ π <=> lx + my + nz = 0 <=> lx + my + nz = p where p = 0

Since any point P on π satisfies the equation lx + my + nz = p

it represents the equation of the plane.

The equation to π is lx + my + nz = p where p ≥ 0.

Hence the equation to the plane lx + my + nz = p is a first-degree equation in x, y, z.

Note.

1. If O ∈ π, equation to the plane is lx + my nz = 0.

2. lx + my + nz = p is called the normal form of the equation to the plane. Coefficients of x, y, z in the equation are l, m, n, and [l, m,n] are Dc’s of the normal OM to the plane, where p(≥ 0) is the distance of the origin to the plane.

3. An equation to the plane, in general, is taken as ax + by + cz + d = 0.

Theorems Related To Planes With Solved Problems Step-By-Step

Theorem: The Plane Transformation Of The Equation To The Plane Into The Normal Form

Let the equation to the plane be ax + by + cz + d = 0, a2 + b2 + c2 ≠ 0 …..(1)

We can take d ≥ 0 or d ≤ 0.

ax + by + cz + d = 0 <=> ax + by + cz = -d. Dividing by $\sqrt{a^2+b^2+c^2}$, we get

⇒ $\frac{-a}{\sqrt{a^2+b^2+c^2}}+\frac{-b}{\sqrt{a^2+b^2+c^2}}+\frac{-c}{\sqrt{a^2+b^2+c^2}}=\frac{d}{\sqrt{a^2+b^2+c^2}}$ …..(2)

or $\frac{a}{\sqrt{a^2+b^2+c^2}} x+\frac{b}{\sqrt{a^2+b^2+c^2}} y+\frac{c}{\sqrt{a^2+b^2+c^2}} z=-\frac{d}{\sqrt{a^2+b^2+c^2}}$ …..(3)

This ((3) or (2)) is of the form lx + my + nz = p [p ≥ 0]

Where $[l=\pm \frac{a}{\sqrt{\sum a^2}}, m=\pm \frac{b}{\sqrt{\sum a^2}}, n=\pm \frac{c}{\sqrt{\sum a^2}}, p=\mp \frac{d}{\sqrt{\sum a^2}}]$

∴ The normal form of the equation to the plane (1) is

⇒ $\pm \frac{a x}{\sqrt{\sum a^2}} \pm \frac{b y}{\sqrt{\sum a^2}} \pm \frac{c z}{\sqrt{\sum a^2}}=\mp \frac{d}{\sqrt{\sum a^2}}(d \leq 0 \text { or } d \geq 0)$

Note.

1. Direction ratios of a normal to the plane ax + by + cz + d = 0 are (a, b, c).

i.e., the coefficients of x, y, z in the equation.

2. Distance of the origin from the plane ax + by + cz + d = 0 is $\frac{|d|}{\sqrt{\left(a^2+b^2+c^2\right)}}$

3. First-degree equation in x, y, z without constant term <=> plane is passing through the origin.

4. abc ≠ 0 and $\left(\frac{1}{a}\right) x+\left(\frac{1}{b}\right) y+\left(\frac{1}{c}\right) z+(-1)=0$. This equation represents a plane intersecting the x-axis in the point (a, 0, 0), intersecting the y-axis in the point (0, b, 0), and intersecting the z-axis in the point (0, 0, c).

Chapter 3 The Plane (1). Consider the equation lx + my = p (l≠0, m≠0) of a plane (π), d.cs. of a normal to it being l,m,0. Since 0, 0, 1 are dc.s. of the z-axis and l.0+m.0+0.1 = 0, the normal to π is perpendicular to the z-axis i.e., π is parallel to the z-axis.

Hence lx + my = p is the equation to a plane parallel to the z-axis.

Similarly, lx + my = p is the equation to a plane parallel to the y-axis and my + nz = p is the equation to a plane parallel to the x-axis.

(2) Consider the equation lx = p(l≠0) of a plane (π), d.cs. of a normal to it being l, 0, 0. Since o, 1, 0 are d.cs. of the y-axis i.e., π is parallel to the y-axis. Similarly, π is also parallel to the z-axis. Hence π is a plane parallel to the yz plane (x=0).

Similarly, my = p is a plane parallel to the zx plane (y=0) and nz = p is a plane parallel to the xy plane (z=0).

Orthogonal Projection Of Points And Lines On Planes Examples

Theorem.3 If the equation a₁x + b₁y + c₁z + d₁ = 0, a₂x + b₂y + c₂z + d₂ = 0 represents the same plane, then a₁:b₁:c₁:d₁ = a₂:b₂:c₂:d₂.

Proof. Given equations are a₁x + b₁y + c₁z + d₁ = 0 …..(1)

a₂x + b₂y + c₂z + d₂ = 0 …..(2)

∴ (a_1,b_1,c₁), (a_2,b_2,c₂) are d.rs. of normals to the same plane.

Since the normals are either equal (coincident) or parallel,

we have a₁:a₂ = b₁:b₂ = c₁:c₂ = λ(say) (λ≠0) or (a_1,b_1,c₁)=λ(a_2,b_2,c₂)

Let (x₁, y₁, z₁) be any point in the plane represented by (1) and (2).

∴ d₁ = -(a₁x₁ + b₁y₁ + c₁z₁)

= -(a_1,b_1,c₁).(x₁, y₁, z₁) = -λ(a_2,b_2,c₂).(x₁, y₁, z₁)

= -λ(a₂x₁ + b₂y₁ + c₂z₁) = -λd₂

∴ a₁:a₂ = b₁:b₂ = c₁:c₂ = d₁:d₂

Chapter 3 Angles Between Two Planes

Definition. Angles between two planes are equal to the angles between their normals.

Angles between the planes a₁x + b₁y + c₁z = d₁ , a₂x + b₂y + c₂z = d₂

Let the equation to the planes be a₁x + b₁y + c₁z + d₁ = 0 …..(1)

a₂x + b₂y + c₂z + d₂ = 0 …..(2)

Dc’s of the normal to (1) =

⇒ $m_1=\left(\frac{a_1}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)}}, \frac{b_1}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)}}, \frac{c_1}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)}}\right)$ and

Dc’s of the normal to (2) =

⇒ $m_2=\left(\frac{a_2}{\sqrt{\left(a_2{ }^2+b_2{ }^2+c_2{ }^2\right)}}, \frac{b_2}{\sqrt{\left(a_2{ }^2+b_2{ }^2+c_2{ }^2\right)}}, \frac{c_2}{\sqrt{\left(a_2{ }^2+b_2{ }^2+c_2{ }^2\right)}}\right)$

Let θ be one of the angles between the planes.

∴ θ = one of the angles between the normals m1, m2

=$\cos ^{-1}\left(\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)} \sqrt{\left(a_2^2+b_2^2+c_2^2\right)}}\right)$

The other angle between the planes is 180° – θ.

Cor.1. Condition of parallelism.

Planes are parallel ⇒ θ = 0° or 180° ⇒ $\pm 1=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)} \sqrt{\left(a_2^2+b_2^2+c_2^2\right)}}$

⇒ $\left(a_1^2+b_1^2+c_1^2\right)\left(a_2^2+b_2^2+c_2^2\right)=\left(a_1 a_2+b_1 b_2+c_1 c_2\right)^2$

⇒ a₁²b₂² + a₁²c₂² + b₁²a₂² + b₁²c₂² + c₁²a₂² + c₁²b₂² – 2a₁a₂b₁b₂ – 2b₁b₂c₁c₂ – 2c₁c₂a₁a₂ = 0

⇒ $\left(a_1 b_2-a_2 b_1\right)^2+\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2=0$

⇒ $a_1 b_2-a_2 b_1=b_1 c_2-b_2 c_1=c_1 a_2-c_2 a_1=0 \Rightarrow a_1: a_2=b_1: b_2=c_1: c_2$

OR: Planes are parallel ⇒ their normals are parallel

⇒ d.rs of normals are proportional ⇒ a1:a2 = b1:b2 = c1:c2.

Cor. 2. Condition of perpendicularity.

Planes are perpendicular ⇒ θ = 90° ⇒ a1a2 + b1b2 + c1c2 = 0.

example. The plane x + 2y – 3z + 4 = 0 is perpendicular to the plane 2x + 5y + 4z + 1 = 0

since (1)(2) + (2)(5) + (-3)(4) = 0

OR: Planes are perpendicular ⇒ their normals are perpendicular

⇒ (a_1,b_1,c₁).(a_2,b_2,c₂) = 0 ⇒ a₁a₂ + b₁b₂ + c₁:c₂ = 0

Note.

1. The equation a₁x + b₁y + c₁z + d₁ = 0, a₁x + b₁y + c₁z + d₂ = 0 represent a pair of parallel planes.

2. A plane parallel to ax + by + cz + d = 0 is ax + by + cz + d = k, where k is an unknown real number.

example 1. The equation of the plane through the point (x₁, y₁, z₁) and parallel to the plane ax + by + cz + d = 0 is ax + by + cz = a₁x + b₁y + c₁z .

example 2. The normals to the plane as x – y + z – 1 = 0, 3x + 2y – z + 2 = 0 are perpendicular since (1)(3) + (-1)(2) + 1(-1) = 0.

Chapter 3 The Plane Determination Of A Plane Under Given Conditions

Consider the equation ax + by + cz + d = 0 of a plane. Since (a, b, c) ≠ (0, 0, 0), without loss of generally, we can take a≠0.

∴ Equation of the plane is $x+\frac{b}{a} y+\frac{c}{a} z+\frac{d}{a}=0$

∴ To know uniquely $\frac{b}{a}, \frac{c}{a}, \frac{d}{a}$ we require three conditions.

For example, we can find the equation to a plane, if (1) three non-collinear points in the plane are given. (2) if two points in the plane and a plane perpendicular to the required plane (3) if one point in the plane and two planes perpendicular to the required plane are given.

Definition. If a plane π intersects the coordinate axes at (a, 0, 0), (0, b, 0), (0, 0, c) then a, b, c are respectively called the x-intercept, the y-intercept, the z-intercept of the plane π.

If the plane lx + my + nz = p intersects the x-axis at (a, 0, 0) then its x-intercept = a = $\frac{p}{l}$.

Similarly its y-intercept = b = $\frac{p}{m}$, its z-intercept = c = $\frac{p}{n}$

Note. If abc ≠ 0 and ax + by + cz + d = 0 …..(1) is a plane, then its x-intercept = $-\frac{d}{a}$, (by putting y = 0, z = 0 in (1))

y-intercept = $-\frac{d}{b}$, z-intercept = $-\frac{d}{c}$.

example. The intercepts made by the plane x – 12y – 2z = 9 with the axes are $\frac{9}{1},-\frac{9}{12},-\frac{9}{2}, \text { i.e., } 9,-\frac{3}{4},-\frac{9}{2}$.

Step-By-Step Solutions For Orthogonal Projection Problems On Planes

Theorem.4 Equation to the plane making intercepts a, b, c on the coordinate axes is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$.

Proof. Let π be the plane making intercepts,

a, b, c on the coordinate axes

Let A = (a, 0, 0), B = (0, b, 0), and C = (0, 0, c).

∴ abc ≠ 0.

Clearly, A, B, C are non-collinear. Let the equation to the plane π in the normal form be

lx + my + nz = p …..(1)

Let M be the foot of the perpendicular

from O to π and let [l, m, n] be the Dc’s of OM. Let OM = p.

p = OM = Projection of OA on OM = al.

Similarly p = bm, and p = cn. ∴ From(1)

equation to the plane π is $\frac{p}{a} x+\frac{p}{b} y+\frac{p}{c} z=p \Rightarrow \frac{x}{a}+\frac{x}{b}+\frac{z}{c}=1$ (∵ p ≠ 0)

Note. Equation to the plane ABC is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$. This is called the intercept form of the equation to the plane and this plane does not pass through the origin.

OR: Let A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c). Let $\mathrm{P}=\bar{r}=(x, y, z)$.

Now A, B, C are non-collinear. $P \in \overleftrightarrow{A B C}(P \neq A \text { or } P=A)$ (∵ abc ≠ 0)

<=> $\overline{\mathrm{AP}}, \overline{\mathrm{AB}}, \overline{\mathrm{AC}}$ are coplanar or $\overline{\mathrm{AP}}(=\overline{0}), \overline{\mathrm{AB}}, \overline{\mathrm{AC}}$ are three vectors.

<=> $\bar{r}-\bar{a}(\neq \overline{0}), \bar{b}-\bar{a}, \bar{c}-\bar{a}$ are coplanar or $\bar{r}-\bar{a}(=\overline{0}), \bar{b}-\bar{a}, \bar{c}-\bar{a}$ are three vectors.

<=> [(x-a, y, z),(-a, b, 0), (-a, 0, c)] = 0

<=> $\left|\begin{array}{ccc}
x-a & y & z \\
-a & b & 0 \\
-a & 0 & c
\end{array}\right|=0 \Leftrightarrow(x-a) b c+y a c+z a b=0$

<=> $\frac{x}{a}-1+\frac{y}{b}+\frac{z}{c}=0 $

<=> $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$.

∴ Equation to $\overleftrightarrow{\mathrm{ABC}} \text { is } \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$.

Note. $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ is called the intercept form of the equation of the plane and this plane does not pass through the origin.

Properties Of Planes And Their Orthogonal Projections With Examples

Theorem. Equation to the plane determined by three non-collinear points A(x₁, y₁, z₁), B(x₂, y₂, z₂), C(x₃, y₃, z₃) is

⇒ $\left|\begin{array}{llll}
x & y & z_1 & 1 \\
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1
\end{array}\right|=0$

Proof. Let the equation of the required plane be ax + by + cz + d = 0 …..(1)

This passes through the given points if a₁x + b₁y + c₁z + d = 0 …..(2)

a₂x + b₂y + c₂z + d = 0 …..(3) ax₃ + by₃ + cz₃ + d = 0 …..(4)

Eliminating a, b, c, d from the above equations (1), (2), (3), (4), we have

⇒ $\left|\begin{array}{llll}
x & y & z & 1 \\
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1
\end{array}\right|=0$ …..(5)

This is the equation to the required plane.

OR :

Proof. Let $\mathrm{A}=\bar{a}=\left(x_1, y_1, z_1\right), \quad \mathrm{B}=\bar{b}=\left(x_2, y_2, z_2\right), \mathrm{C}=\bar{c}=\left(x_3, y_3, z_3\right)$

Let $\bar{r}=(x, y, z) . \quad \bar{r}(\neq \bar{a}) \in \pi, \bar{r}(=\bar{a}) \in \pi$

<=> $\bar{r}-\bar{a}, \bar{b}-\bar{a}, \bar{c}-\bar{a}$ are coplanar or $\bar{r}-\bar{a}(=\overline{0}), \bar{b}-\bar{a}, \bar{c}-\bar{a}$ are three vectors

<=> $ [\bar{r}-\bar{a}, \bar{b}-\bar{a}, \bar{c}-\bar{a}]=0$

<=> $\left[\left(x-x_1, y-y_1, z-z_1\right),\left(x_2-x_1, y_2-y_2, z_2-z_1\right),\left(x_3-x_1, y_3-y_1, z_3-z_1\right)\right]=0$

<=> $\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
x_3-x_1 & y_3-y_1 & z_3-z_1
\end{array}\right|=0$ …..(1)

<=> $\left|\begin{array}{cccc}
x-x_1 & y-y_1 & z-z_1 & 0 \\
x_1 & y_1 & z_1 & 1 \\
x_2-x_1 & y_2-y_1 & z_2-z_1 & 0 \\
x_3-x_1 & y_3-y_1 & z_3-z_1 & 0
\end{array}\right|=0$

<=> $\left|\begin{array}{cccc}
x & y & z & 1 \\
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1
\end{array}\right|=0 \mid \begin{aligned}
& \mathrm{R}_2+\mathrm{R}_1 \\
& \mathrm{R}_3+\mathrm{R}_1 \\
& \mathrm{R}_4+\mathrm{R}_1
\end{aligned}$

Note. equation (1) may also be taken as the equation of the required plane.

Note. If the points (x₁, y₁, z₁), (x₂, y₂, z₂), (x₃, y₃, z₃), (x₄, y₄, z₄) are such that

⇒ $\left|\begin{array}{llll}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
x_4 & y_4 & z_4 & 1
\end{array}\right|=0$, then the points are coplanar.

⇒ If $\left|\begin{array}{llll}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
x_4 & y_4 & z_4 & 1
\end{array}\right| \neq 0$, then the points are non-coplanar.

Applications Of Orthogonal Projection On Planes In Mathematics

Theorem. Equation to the plane (π) through the point A(x₁, y₁, z₁) and perpendicular to the line (L) with d.rs. (a, b, c) is a(x-x₁)+b(y-y₁)+c(z-z₁) = 0.

Proof: Let p ∈ π and p = (x, y, z). A = (x₁, y₁, z₁)

and d.rs. of L are (a, b, c)

Now d.rs. of AP = (x – x1, y – y1, z – z1).

⇒ $\overleftrightarrow{A P} \in \pi \Leftrightarrow \overleftrightarrow{A P} \perp L$

<=> a(x-x1) + b(y-y1) + c(z-z1) = 0

∴ Equation to π is a(x – x1) + b(y – y1) + c(z – z1) = 0.

Chapter 3 The Plane Parametric Equation Of A Plane

Theorem. An equation to the plane passing through three non-collinear points A(a), B(b), C(c) is r = (1 – t – s) a + sb + tc, where s, t are any scalars (real numbers)

Proof: $\bar{r} \in \overleftrightarrow{\mathrm{ABC}} \Rightarrow \bar{r}-\bar{a}, \bar{b}-\bar{a}, \bar{c}-\bar{a}$ are coplanar;

or $\bar{r}-\bar{a}(\not 0), \bar{b}-\bar{a}, \bar{c}-\bar{a}$ are three vectors

<=> $\bar{r}-\bar{a}=s(\bar{b}-\bar{a})+t(\bar{c}-\bar{a})$ where s, t are any scalars

<=> $\bar{r}=(1-s-t) \bar{a}+s \bar{b}+t \bar{c}$ is the equation of the plane through $\bar{a}, \bar{b}, \bar{c}$.

Note. Let $\bar{r}=(x, y, z), \bar{a}=\left(x_1, y_1, z_1\right), \bar{b}=\left(x_2, y_2, z_2\right), \bar{c}=\left(x_3, y_3, z_3\right)$

Then parametric equation to the plane $\overleftrightarrow{\mathrm{ABC}}$ is

(x, y, z) = (1 – s – t)( x₁, y₁, z₁) + s(x₂, y₂, z₂) + t(x₃, y₃, z₃)

i.e. x = x₁ + s(x₂ – x₁) + t(x₃ – x₁),

y = y₁ + s(y₂ – y₁) + t(y₃ – y₁)

z = z₁ + s(z₂ – z₁) + t(z₃ – z₁)

Cor. Points $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ are coplanar

<=> $\overline{\bar{d}}=(1-s-t) \bar{a}+s \bar{b}+t \bar{c}$

<=> $(1-s-t) \bar{a}+s \bar{b}+t \bar{c}+(-1) \bar{d}=0$

<=> $\lambda_1 \bar{a}+\lambda_2 \bar{b}+\lambda_3 \bar{c}+\lambda_4 \bar{d}=0$,

λ₁ + λ₂ + λ₃ + λ₄= 1 – s – t + s + t – 1 = 0 and (λ₁ + λ₂ + λ₃ + λ₄) ≠ (0, 0, 0, 0)

Chapter 3 The Plane Two Sides Of A Plane

A Plane π divides the space into two half-spaces. Let a line $\overleftrightarrow{\mathrm{AB}}$ intersect π in C. If A, B are in the same half-space, then (C; A, B) is negative, and A, B lie on the same side of C. If A, B are in different half spaces, then (C; A, B) is positive, and A, B lie on either side of C.

Theorem. If the line through A(x₁, y₁, z₁) and B(x₂, y₂, z₂) intersect the plane ax + by + cz + d = 0 in C, then (C; A, B) = -(a₁x + b₁y + c₁z + d):(a₂x + b₂y + c₂z + d).

Proof. Let A = (x₁, y₁, z₁) and B = (x₂, y₂, z₂) and C divide AB in the ratio λ1 : λ2 (λ1 + λ2 ≠ 0).

∴ (C; A, B) = λ₁ : λ₂ (λ₁ + λ₂ ≠ 0) and

⇒ $\mathrm{C}=\left(\frac{\lambda_2 x_1+\lambda_1 x_2}{\lambda_1+\lambda_2}, \frac{\lambda_2 y_1+\lambda_1 y_2}{\lambda_1+\lambda_2}, \frac{\lambda_2 z_1+\lambda_1 z_2}{\lambda_1+\lambda_2}\right)$

⇒ $\mathrm{C} \in \pi \Leftrightarrow a\left(\frac{\lambda_2 x_1+\lambda_1 x_2}{\lambda_1+\lambda_2}\right)+b\left(\frac{\lambda_2 y_1+\lambda_1 y_2}{\lambda_1+\lambda_2}\right)+c\left(\frac{\lambda_2 z_1+\lambda_1 z_2}{\lambda_1+\lambda_2}\right)+d=0$

<=> $\lambda_2\left(a x_1+b y_1+c z_1+d\right)+\lambda_1\left(a x_2+b y_2+c z_2+d\right)=0$

<=> $\lambda_1\left(a x_2+b y_2+c z_2+d\right)=-\lambda_2\left(a x_1+b y_1+c z_1+d\right)$

<=> $\lambda_1: \lambda_2=-\left(a x_1+b y_1+c z_1+d\right):\left(a x_2+b y_2+c z_2+d\right)$

<=> $(\mathrm{C} ; \mathrm{A}, \mathrm{B})=-\left(a x_1+b y_1+c z_1+d\right):\left(a x_2+b y_2+c z_2+d\right)$

OR :
Proof: Let $\mathrm{A}=\bar{a}=\left(x_1, y_1, z_1\right) \text { and } \mathrm{B}=\bar{b}=\left(x_2, y_2, z_2\right)$

Let $\mathrm{C}=\bar{c}$ divide AB in the ratio λ₁ : λ₂.

∴ (C; A, B) = λ₁ : λ₂ (λ₁ + λ₂ ≠ 0)

Then $\bar{c}=\frac{\lambda_2 \bar{a}+\lambda_1 \bar{b}}{\lambda_1+\lambda_2}$

Let the plane represented by ax + by + cz + d = 0 be $\bar{r} \cdot \bar{m}=q$ where

⇒ $\bar{r}=(x, y, z), \bar{m}=(a, b, c) \text { and } q=-d$

∴ $\mathrm{C} \in \pi \Leftrightarrow \bar{c} \cdot \bar{m}-q=0$

<=> $\left(\frac{\lambda_2 \bar{a}+\lambda_1 \bar{b}}{\lambda_1+\lambda_2}\right) \cdot \bar{m}-q=0$

<=> $\lambda_2(\bar{a} \cdot \bar{m})+\lambda_1(\bar{b} \cdot \bar{m})=q \lambda_1+q \lambda_2 \Leftrightarrow \lambda_1(\bar{b} \cdot \bar{m}-q)=-\lambda_2(\bar{a} \cdot \bar{m}-q)$

<=> $\lambda_1: \lambda_2=-(\bar{a} \cdot \bar{m}-q):(\bar{b} \cdot \bar{m}-q)$

<=> $\lambda_1: \lambda_2=-\left\{\left(x_1, y_1, z_1\right) \cdot(a, b, c)+d\right\}:\left\{\left(x_2, y_2, z_2\right) \cdot(a, b, c)+d\right\}$

<=> $\lambda_1: \lambda_2=-\left(a x_1+b y_1+c z_1+d\right):\left(a x_2+b y_2+c z_2+d\right)$

<=> $(\mathrm{C} ; \mathrm{A}, \mathrm{B})=-\left(a x_1+b y_1+c z_1+d\right):\left(a x_2+b y_2+c z_2+d\right)$

Note 1. A, B lie in the same half-space.

<=> a₁x + b₁y + c₁z + d, a₂x + b₂y + c₂z + d are of the same sign and A, B lie in the different half-spaces.

<=> a₁x + b₁y + c₁z + d, a₂x + b₂y + c₂z + d are of the different signs.

example

1. The points (1, 2, -5), (0, 4, -7) lie in the different half space (on the opposite sides) of the plane x + 2y + 2z – 9 = 0 since 2 + 2(3) + 2(5) – 9 > 0 and 0 + 2(4) + 2(-7) – 9 <0.

2. The points (1, 2, -5),(0, 4, -7) lie in the same half-space (on the same side) of the plane x + 2y + 2z – 9 = 0 since (1) + 2(2) + 2(-5) – 9 < 0 and 0 + 2(4) + 2(-7) – 9 < 0.

3. The points (1, -1, 3) and (3, 3, 3) lie on different sides of the plane x + 2y – 7z + 9 = 0 since 5(1) + 2(-) – 7(3) + 9 = -9 < 0 and 5(3) + 2(3) – 7(3) + 9 = 9 > 0.

Worked Examples Of Orthogonal Projection Onto Planes In Geometry

Chapter 3 The Plane Perpendicular Distance Of A Point From A Plane

Theorem. The distance of A(x₁, y₁, z₁) from the plane ax + by + cz + d = 0 i.e. length of the perpendicular from the point A(x₁, y₁, z₁) to the plane ax + by + cz + d = 0 is

⇒ $\frac{\left|a x_1+b y_1+c z_1+d\right|}{\sqrt{\left(a^2+b^2+c^2\right)}}$.

Proof: Let π be the plane ax + by + cz + d = 0 …..(1)

Let A = (x₁, y₁, z₁) be the point (A ∉ π) from which the perpendicular drawn to the plane π meets it in C.

Let the normal form of π be lx + my + nz = p …..(2)

the equation to the plane parallel to (2) and passing through the point A be

lx + my + nz = p₁ …..(3) where lx₁+ my₁ + nz₁ = p₁ …..(4)

Let ODE be perpendicular to (2) and (3) as shown.

⇒ AC = p1 – p

⊥r distance of A to the plane π

= AC = OE – OD = lx₁+ my₁ + nz₁ – p

= $+\frac{a}{\sqrt{\sum a^2}} x_1+\frac{b}{\sqrt{\sum a^2}} y_1+\frac{c}{\sqrt{\sum a^2}} z_1 \pm \frac{d}{\sqrt{\sum a^2}}$

i.e., $\pm \frac{\left(a x_1+b y_1+c z_1+d\right)}{\sqrt{a^2+b^2+c^2}} \text { or } \frac{\left|a x_1+b y_1+c z_1+d\right|}{\sqrt{a^2+b^2+c^2}}$

OR : Proof: Let $\mathrm{A}=\bar{a}=\left(x_1, y_1, z_1\right)$.

Let π be the plane ax + by + cz + d = 0.

Equation to π can be taken as $\bar{r} \cdot \bar{m}=q \text { where } \bar{r}=(x, y, z)$.

⇒ $\bar{m}=(a, b, c) \text { and } q=-d$.

Now, $|\bar{m}|=\sqrt{\left(a^2+b^2+c^2\right)}$.

Let C be the foot of the perpendicular from A to π.

Let B(≠C) be $\bar{b}$ in π.

∴ $\bar{b} \cdot \bar{m}=q$ …..(1)

∴ $\mathrm{AC}=\frac{|\overline{\mathrm{AB}} \cdot \overline{\mathrm{AC}}|}{|\overline{\mathrm{AC}}|}=\frac{|(\bar{b}-\bar{a}) \cdot \bar{m}|}{|\bar{m}|}=\frac{|\bar{b} \cdot \bar{m}-\bar{a} \cdot \bar{m}|}{|\bar{m}|}=\frac{|\bar{a} \cdot \bar{m}-\bar{b} \cdot \bar{m}|}{|\bar{m}|}=\frac{|\bar{a} \cdot \bar{m}-\bar{q}|}{|\bar{m}|}$

= $\frac{\left|\left(x_1, y_1, z_1\right) \cdot(a, b, c)-q\right|}{\sqrt{a^2+b^2+c^2}}=\frac{\left|a x_1+b y_1+c z_1+d\right|}{\sqrt{a^2+b^2+c^2}}$

example. The distance of the points (2, 3, 4) and (1, 1, 4) from the plane 3x – 6y + 2z + 11 = 0

=$\left|\frac{3(2)-6(3)+2(4)+11}{\sqrt{(9+36+4)}}\right|=1 \text { and }\left|\frac{3-6+8+11}{\sqrt{9+36+4}}\right|=\frac{16}{7}$

Chapter 3 The Plane Distance Between Parallel Planes

Theorem. Distance between parallel planes ax + by + cz + d₁ = 0, ax + by + cz + d₂ = 0 is $\frac{\left|d_1-d_2\right|}{\sqrt{a^2+b^2+c^2}}$, d₁ < 0, d₂ > 0.

Proof. The equations to the planes are

ax + by + cz + d₁ = 0 …..(1) ax + by + cz + d₂ = 0 …..(2)

the Dc’s of the normal to the planes are

⇒ $\left(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}\right)$

Now p₁ and p₂ be the perpendicular distances to the planes from the origin

⇒ $p_1=\frac{-d_1}{\sqrt{a^2+b^2+c^2}}, p_2=\frac{-d_2}{\sqrt{a^2+b^2+c^2}}$

⇒ Distance between the parallel planes = $\left|p_1-p_2\right|=\frac{\left|d_2-d_1\right|}{\sqrt{a^2+b^2+c^2}}$.

Note. Equation to the plane parallel to (1) and (2) and midway between (1) and (2)

(1) When the origin lies on the same side of both (1) and (2) is

i.e., $a x+b y+c z=\frac{-\left(d_1+d_2\right)}{2}$

(2) When origin lies in between (1) and (2) is $ax+by+c z=\frac{\left|d_1-d_2\right|}{2}$

example. The distance between the planes 2x – y + 3z = 6 and -6x + 3y – 9z = 5

= The distance between the planes 2x – y + 3z = 6 and $2 x-y+3 z=-\frac{5}{3}$

=$\frac{\left|6-\left(-\frac{5}{3}\right)\right|}{\sqrt{(4+1+9)}}=\frac{23}{3 \sqrt{14}}$

Chapter 3 The Plane Solved Problems

Example. 1. Find the point P equidistant from A(4, -3, 7) and B(2, -1, 1) and lying on y-axis. Hence find the equation to the plane through P and perpendicular to $\overleftrightarrow{\mathrm{AB}}$.

Solution:

Given

A(4, -3, 7) and B(2, -1, 1)

Let P = (0, y, 0). Since PA = PB, PA² = PB²

⇒ (0 – 4)2 + (y + 3)2 + (0 – 7)2 = 4 + (y + 1)2 + 1 ⇒ 4y = -68 ⇒ y = -17

∴ P = (0, -17, 0) and d.rs of $\overleftrightarrow{\mathrm{AB}}$ are 2, -2, 6.

∴ The equation to the plane through P, and perpendicular $\overleftrightarrow{\mathrm{AB}}$ is

2(x – 0) – 2(y + 17) + 6(z – 0) = 0 i.e., 2x – 2y + 6z – 34 = 0,

i.e., x – y + 3z – 17 = 0

Geometric Interpretation Of Planes And Orthogonal Projections

Example.2. Show that the line joining the points (6, -4, 4), (0, 0, -4) intersects the line joining the points (-1, -2, -3), (1, 2, -5).

Solution.

Given

(6, -4, 4), (0, 0, -4)

And (-1, -2, -3), (1, 2, -5)

Let A = (6, -4, 4), B = (0, 0, 4), C = (-1, -2, -3), D = (1, 2, -5).

∴ $\overline{\mathrm{AB}}=(-6,4,-8) \text { and } \overline{\mathrm{CD}}=(2,4,-2)$

⇒ $\overline{\mathrm{AB}}$ is neither perpendicular not parallel to $\overline{\mathrm{CD}}$.

∴ If we prove that A, B, C, D are coplanar, then $\overleftrightarrow{\mathrm{AB}}$ intersects $\overleftrightarrow{\mathrm{CD}}$.

Now equation to the plane $\overleftrightarrow{\mathrm{ABC}}$ is $\left|\begin{array}{ccc}
x-6 & y+4 & z-4 \\
-6 & 4 & -8 \\
-7 & 2 & -7
\end{array}\right|=0$

⇒ (-28 + 16)(x – 6) – (42 – 56)(y + 4) + (-12 + 28)(z – 4) = 0

⇒ 6(x – 6) – 7(y + 4) – 8(z – 4) = 0 ⇒ 6x – 7y – 8z – 32 = 0 …..(1)

Substituting D in the L.H.S of (1), we get 6 – 14 + 40 – 32 = 0 = R.H.S.

∴ $D \in \overleftrightarrow{\mathrm{ABC}}$

∴ $\overleftrightarrow{\mathrm{AB}} \text { and } \overleftrightarrow{\mathrm{CD}}$ intersect.

Example.3. Obtain the equation to the plane containing (0, 4, 3) and the line through the points (-1, -5, -3), (-2, -2, 1). Hence show that (0, 4, 3), (-1, -5, -3), (-2, -2, 1) and (1, 1, -1) are coplanar.

Solution.

Given

(0, 4, 3) (-1, -5, -3), (-2, -2, 1)

Let π be the required plane. Let a, b, c be d.rs. of a normal to it.

Let A = (0, 4, 3), B = (-1, -5, -3), C = (-2, -2, 1)

D.rs. of $\overleftrightarrow{\mathrm{AB}}$ are -1, -9, -6 and d.rs. of $\overleftrightarrow{\mathrm{AC}}$ are -2, -6, -2.

Since $\overleftrightarrow{\mathrm{AB}}, \overleftrightarrow{\mathrm{AC}}$ are in π,

⇒ $\left.\begin{array}{r}
-a-9 b-6 c=0 \\
-2 a-6 b-2 c=0
\end{array}\right\} \text {. Solving, } \frac{a}{-18}=\frac{b}{10}=\frac{c}{-12} \text { i.e., } \quad \frac{a}{9}=\frac{b}{-5}=\frac{c}{6}$

∴ Equation to π is 9(x – 0) – 5(y – 4) + 6(z – 3) = 0 i.e., 9x – 5y + 6z + 2 = 0 …..(1)

Clearly (1, 1, -1) lies on (1). Hence the points are coplanar.

∴ The equation to the plane containing the points is (1).

Example.4. Find the equation of the plane through (4, 4, 0) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 3z – 8 = 0.

Solution.

Given

(4, 4, 0)

x + 2y + 2z = 5 and 3x + 3y + 3z – 8 = 0

Let π be the required plane. Let a, b, c be d.rs of normal to π.

Since π passes through (4, 4, 0) equation to π is a(x – 4) + b(y – 4) + c(z – 0) = 0

But π is perpendicular to x + 2y + 2z = 5 and 3x + 3y + 2z – 8 = 0

∴ $\left.\begin{array}{r}
a+2 b+2 c=0 \\
3 a+3 b+2 c=0
\end{array}\right\}$.

∴ $\frac{a}{-2}=\frac{b}{4}=\frac{c}{-3}$.

∴ Equation to π is -2(x – 4) + 4(y – 4)-3z = 0 [using(1)]

i.e., 2x – 4y + 3z + 8 = 0.

Solved Problems On Equations Of Planes And Orthogonal Projection

Example.5. Find the equation of the plane passing through (1, 0, -2) and perpendicular to the planes 2x + y – 2 = z; x – y – z = 3

Solution.

Given

(1, 0, -2)

2x + y – 2 = z; x – y – z = 3

Let π be the required plane. Let a, b, c be the drs. of the above plane.

The equation of the plane passing through (1, 0, -2) and having a, b, c as drs. is a(x-1) + b(y-0) + c(z+2) = 0 ⇒ a(x-1) + by + c(z+2) = 0 …..(1)

But the π plane is perpendicular to the planes 2x + y – z = 2 and x – y – z = 3.

∴ 2a + b – c = 0 …..(2), a – b – c = o …..(3)

Solving (2) and (3) $\frac{a}{2}=\frac{b}{-1}=\frac{c}{3}$.

Equation of the π plane is 2(x-1)-y+3(z+2)=0 i.e., 2x – y + 3z + 4 = 0.

Example.6. Find the angles between the planes 2x – 3y – 6z = 6 and 6x + 3y – 2z = 18.

Solution.

Given Planes

2x – 3y – 6z = 6 and 6x + 3y – 2z = 18.

Let θ be one of the angles between the given planes.

∴ $\theta=\cos ^{-1} \frac{2(6)-3(3)-6(-2)}{\sqrt{(4+9+36)} \sqrt{(36+9+4)}}=\cos ^{-1}\left(\frac{15}{49}\right)$

The other angle between the planes is 180° – θ i.e., $180^{\circ}-\cos ^{-1}\left(\frac{15}{49}\right)$

Example.7. Find the locus of the point whose distance from the origin is three times its distance from the plane 2x – y + 2z = 3.

Solution.

Given

2x – y + 2z = 3

Let O be the origin and P be the point (x₁, y₁, z₁) such that OP is equal to 3 times its distance from the plane 2x – y + 2z = 3

∴ $\mathrm{OP}^2=9 \cdot \frac{\left(2 x_1-y_1+2 z_1-3\right)^2}{4+1+4}$

⇒ x₁² + y₁² + z₁² = 4x₁² + y₁² + 4z₁² + 9 – 4x₁y₁ – 4y₁z₁ + 8x₁z₁ – 12x₁ + 6y₁ – 12z₁

⇒ 3x₁²+ 3z₁² – 4x₁y₁ – 4y₁z₁ + 8x₁z₁– 12x₁ + 6y₁ – 12z₁ + 9 = 0

∴ Locus of P is 3x² + 3z² – 4xy – 4yz + 8xy – 12x + 6y – 12z + 9 = 0

Chapter 3 The Plane Systems Of Planes

Consider the equation ax + by + cz + d = 0, (a, b, c) ≠ (0, 0, 0),$\lambda_1=\frac{b}{a}, \lambda_2=\frac{c}{a}, \lambda_3=\frac{d}{a}$ of a plane.

When three conditions satisfying the equation are given, λ₁, λ₂, λ₃ can be uniquely determined and hence a plane can be uniquely determined.

When two conditions satisfying the equation are given, one of λ₁, λ₂, λ₃ say, λ₁ cannot be found uniquely and λ₁ is called a parameter. Since λ₁ can be assigned any real vaule, an infinite number of planes arise, and these planes are called a system of planes.

When one condition satisfying the equation is given we have two parameters, say, λ₁, λ₂ giving rise to a system of planes for different values of λ₁, λ₂.

We give below a few systems of planes involving one or two parameters.

(1) The equation ax + by + cz + λ = 0 represents the system of planes parallel to a given plane ax + by + cz + d = 0, λ being the parameter.

(2) The equation ax + by + cz + λ = 0 represents the system of planes perpendicular to lines with d.rs. a, b, c; λ being the parameter.

(3) The equation a(x – x₁)+b(y – y₁)+c(z – z₁) = 0, (a, b, c) ≠ (0, 0, 0)

i.e., (x – x₁) + λ₁ (y – y₁) + λ₂ (z – z₁)= 0 where a ≠ 0(say), λ₁ = (b/a), λ₂= (c/a) represents the system of planes passing through the point (x₁, y₁, z₁); λ₁, λ₂ being the parameters.

(4) The equation λ₁(a₁x + b₁y + c₁z + d₁ ) + λ₂ (a₂x + b₂y + c₂z + d₂) = 0 represents the system of planes through the line of intersection of the planes a₁x + b₁y + c₁z + d₁ = 0, a₂x + b₂y + c₂z + d₂ = 0; λ₁, λ₂ being parameters and (λ₁, λ₂) ≠ (0, 0).

The truth of the statement can be seen from the theorem proved in the ensuing article.

Theorem. π₁ ≡ a₁x + b₁y + c₁z + d₁ = 0, π₂≡ a₂x + b₂y + c₂z + d₂ = 0 represent two intersecting planes.

(A) If (λ₁, λ₂) ≠ (0, 0), then λ₁π₁ + λ₂π₂ = 0 represents a plane passing through the line L of the intersection of π₁ and π₂.

(B) Any plane passing through the line L of the intersection of π₁ and π₂ is given by λ₁π₁ + λ₂π₂ = 0, (λ₁, λ₂) ≠ (0, 0)

Proof: (A) Let S ≡ λ₁π₁ + λ₂π₂ = 0, (λ₁, λ₂) ≠ (0, 0)

S is a first-degree equation in x, y, z and hence represents a plane.

Now λ₁ = 0 ⇒ S = π₂, λ₂= 0 ⇒ S = π₁. Let P(x₁, y₁, z₁) ∈ L

∴ a₁x₁+ b₁y₁ + c₁z₁ + d₁ = 0 …..(1) a₂x₁ + b₂y₁ + c₂z₁ + d₂ = 0 …..(2)

Also from (1) and (2), P ∈ S, when (λ₁, λ₂) ≠ (0, 0)

∴ S represents a plane through the line L of the intersection of π₁ and π₂.

If λ₁ ≠ 0, λ₂ ≠ 0, for different values of λ₁, λ₂; S represents any plane through the line L of intersection π₁ and π₂ and different from π₁ and π₂.

(B) Let P(x₁, y₁, z₁), Q(x₂, y₂, z₂) be different points on L such that x₁ ≠ x₂ (say).

Let S ≡ αx + βy + γz + δ = 0 be a plane through L and hence

αx₁ + βy₁ + γz₁ + δ = 0 …..(3) αx₂ + βy₂ + γz₂ + δ = 0 …..(4)

Let l, m, n be d.rs. of L, the line of intersection of the planes π₁ and π₂.

∴ $\left.\begin{array}{c}
a_1 l+b_1 m+c_1 n=0 \\
a_2 l+b_2 m+c_2 n=0
\end{array}\right\} \frac{l}{b_1 c_2-b_2 c_1}=\frac{m}{c_1 a_2-c_2 a_1}=\frac{n}{a_1 b_2-a_2 b_1}$

Since (b₁c₂ – b₂c₁, c₁a₂– c₂a₁, a₁b₂ – a₂b₁) ≠ (0, 0, 0) without loss of generality we can take b₁c₂ – b₂c₁ ≠ 0.

For λ₁, λ₂ and (λ₁, λ₂) ≠ (0,0) there exist equations λ₁b₁ + λ₂b₂ = β, λ₁c₁ + λ₂c₂ = γ

such that they have a unique solution λ₁ and λ₂.

∴ αx + βy + γz + δ ≡ αx + (λ₁b₁ + λ₂b₂)y (λ₁c₁ + λ₂c₂)z + δ

≡ λ₁(a₁x + b₁y + c₁z + d₁) + λ₂(a₂x + b₂y + c₂z + d₂) + αx – λ₁a₁x – λ₂a₂x – λ₁d₁ – λ₂d₂ + δ

≡ λ₁(a₁x + b₁y + c₁z + d₁) + λ₂(a₂x + b₂y + c₂z + d₂) + (α – λ₁a₁– λ₂a₂) x + (δ – λ₁d₁– λ₂d₂)

≡ λ₁(a₁x + b₁y + c₁z + d₁) + λ₂(a₂x + b₂y + c₂z + d₂) + λ₃x + λ₄ …..(5)

Where λ3 = α – λ₁a₁– λ₂a₂, λ₄ = δ – λ₁d₁– λ₂d₂

∴ P ∈ L ⇒ P ∈ S ⇒ αx₁ + βy₁ + γz₁ + δ = 0 ⇒ λ₃x₁ + λ₄= 0 …..(6)

using (3) and (5), and Q ∈ L ⇒ Q ∈ S ⇒ λ₃x₂ + λ₄ = 0 …..(7)

using (4) and (5),

∴ (6) – (7) ⇒ λ₃(x₁ – x₂) = 0 ⇒ λ₃ = 0 (∵ x₁ ≠ x₂)

∴ From (6), λ₄ = 0.

∴ S ≡ λ₁π₁ + λ₂π₂ = 0 is the plane passing through the line of intersection π₁and π₂.

Note. Let λ₁ ≠ 0 (say). Now equation to the plane (distinct from π₁, π₂) passing through the line of intersection of planes π₁ and π₂ can be taken as $\pi_1+\left(\frac{\lambda_2}{\lambda_1}\right) \pi_2=0$ i.e. π₁ + λπ₂ = 0 where $\lambda=\frac{\lambda_2}{\lambda_1}$. This form of equation might be taken while doing problems.

Chapter 3 The Plane Planes Bisecting The Angles Between Two Planes.

Theorem. π₁ ≡ a₁x + b₁y + c₁z + d₁= 0, π₂ ≡ a₂x + b₂y + c₂z + d₂ = 0 and d_1d2 > 0. Equation to the plane bisecting the angle containing the origin between the planes

⇒ $\pi_1, \pi_2 \text { is } \frac{a_1 x+b_1 y+c_1 z+d_1}{\sqrt{a_1^2+b_1^2+c_1^2}}=+\frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{a_2^2+b_2^2+c_2^2}}$

and to the plane bisecting the other angle between the planes

⇒ $\pi_1, \pi_2 \text { is } \frac{a_1 x+b_1 y+c_1 z+d_1}{\sqrt{a_1^2+b_1^2+c_1^2}}=-\frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{a_2^2+b_2^2+c_2^2}}$

Proof: Equations to π₁, π₂ are

a₁x + b₁y + c₁z + d₁ = 0 …..(1) a₂x + b₂y + c₂z + d₂ = 0 …..(2) and d₁ d₂>0

we know that if p(x, y, z) is any point on one of the planes bisecting the angle between π₁, π₂ then the perpendicular distances of P from π₁, π₂ are equal (in magnitude) so that

⇒ $\frac{a_1 x+b_1 y+c_1 z+d_1}{\sqrt{a_1^2+b_1^2+c_1^2}}=\pm \frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{a_2^2+b_2^2+c_2^2}}$ are the equations to the bisecting planes.

∴ The equation to the plane bisecting the angle containing the origin is

⇒ $\frac{a_1 x+b_1 y+c_1 z+d_1}{\sqrt{a_1^2+b_1^2+c_1^2}}=+\frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{a_2^2+b_2^2+c_2^2}} \text { if } d_1>0, d_2>0$ …..(7)

This plane bisects the angle containing the origin also bisects the vertically opposite angle.

∴ The equation to the plane bisecting the other angle and its vertically opposite angle is

⇒ $\frac{a_1 x+b_1 y+c_1 z+d_1}{\sqrt{a_1^2+b_1^2+c_1^2}}=-\frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{a_2^2+b_2^2+c_2^2}} \text { if } d_1>0, d_2>0$ …..(8)

Note 1. In bisecting planes (7) and (8), one bisects the acute and the other bisects the obtuse angle between the given planes π₁, π_2.

The bisecting plane of the acute angle makes with either of the planes π₁, π₂ an angle less than 45° and the bisecting plane of the obtuse angle makes with either of the planes π₁, π₂ an angle greater than 45° (of course < 90°). This gives a test for determining which angle (acute or obtuse) each bisecting plane bisects.

2. Even if d₁ < 0, d₂ < 0, the theorem holds.

But if d₁ > 0, d₂< 0 or d₁ < 0, d₂ > 0, equation to the plane bisecting the angle containing the origin is $\frac{a_1 x+b_1 y+c_1 z+d_2}{\sqrt{a_1^2+b_1^2+c_1^2}}=-\frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{a_2^2+b_2^2+c_2^2}}$ and the other equation gives the other bisecting plane.

3. l₁x + m₁y + n₁z = q₁, l₂x + m₂y + n₂z = q₂ are two intersecting planes such that (l₁, m₁, n₁), (l₂, m₂, n₂) are unit points and q₁q₂ > 0 (q_{1 ,}q₂ are of the same sign).

∴ The equation to the plane bisecting the angle containing the origin is (l₁ – l₂)x + (m₁ – m₂)y + (n₁ – n₂)z = q₁ – q₂ and the equation to the plane bisecting the other angle is (l₁ + l₂)x + (m₁ + m₂)y + (n₁ + n₂)z = q₁ + q₂

Chapter 3 The Plane Solved Problems

Example. 1. Find the equation to the plane through the point (x₁, y₁, z₁) and parallel to the plane ax + by + cz + d = 0

Solution.

Given point (x₁, y₁, z₁) and plane ax + by + cz + d = 0

Let ax + by + cz + λ = 0 …..(1)

be the plane parallel to ax + by + cz + d = 0 …..(2)

for all values of λ.

If (1) passes through (x₁, y₁, z₁) then ax₁ + by₁ + cz₁ + λ = 0 i.e., λ = – ax₁ – by₁ – cz₁

∴ Required plane is ax + by + cz – ax₁ – by₁ – cz₁ = 0

i.e., a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

Example. 2. Find the equations of the planes through the intersection of the planes x + 3y + 6 = 0 and 3x – y – 4z = 0 such that the perpendicular distances of each from the origin are unity.

Solution.

Given planes x + 3y + 6 = 0 and 3x – y – 4z = 0

Let the plane passing through the intersection of the planes x + 3y + 6 = 0, 3x – y – 4z = 0 be (x + 3y + 6) + λ(3x – y – 4z) = 0

⇒ (1 + 3λ)x + (3 – λ)y – 4λz + 6 = 0 …..(1)

Perpendicular distance of origin from (1) = 1

⇒ $\frac{6}{\sqrt{(1+3 \lambda)^2+(3-\lambda)^2+16 \lambda^2}}=1 \Rightarrow 26 \lambda^2=26 \Rightarrow \lambda=\pm 1$

∴ Required planes are 4x + 2y – 4z + 6 = 0, -2x + 4y + 4z + 6 = 0

i.e., 2x + y – 2z + 3 = 0, x – 2y – 2z – 3 = 0

Example. 3. Find the equation to the plane through the intersection of the planes x + 2y + 3z + 4 = 0 and 4x + 3y + 3z + 1 = 0 and perpendicular to the plane x + y + z + 9 = 0

Solution.

Given planes x + 2y + 3z + 4 = 0 and 4x + 3y + 3z + 1 = 0 and perpendicular to the plane x + y + z + 9 = 0

Let the plane through the intersection of the planes

x + 2y + 3z + 4 = 0, 4x + 3y + 3z + 1 = 0 be (x + 2y + 3z + 4) + λ(4x + 3y + 3z + 1) = 0

⇒ (1 + 4λ)x + (2 + 3λ)y + (3 + 3λ)z + (4 + λ) = 0 …..(1)

If (1) is perpendicular to x + y + z + 9 = 0, then (1 + 4λ).1 + (2 + 3λ).1 + (3 + 3λ).1 = 0

i.e., 10λ = -6 i.e., λ = -3/5.

∴ Required plane is 7x – y – 6z – 17 = 0

Example. 4. Find the equation to the plane through the line of intersection of x – y + 3z + 5 = 0 and 2x + y – 2z + 6 = 0 and passing through (-3, 1,1)

Solution.

Given Planes x – y + 3z + 5 = 0 and 2x + y – 2z + 6 = 0 and passing through (-3, 1,1)

Let the equation to the plane through the intersection of the planes x – y + 3z + 5 = 0, be (x – y + 3z + 5) + λ(2x + y – 2z + 6) = 0 …..(1), for and λ.

Let (1) pass through the point (-3, 1,1).

∴ -3 – 1 + 3 + 5 + λ(-6 + 1 -2 + 6) = 0 i.e., -λ + 4 = 0 i.e., λ = 4

∴ Equation to the required plane is 9x + 3y – 5z + 29 = 0

Example. 5. Find the equation to the plane through (2, -3, ) and is normal to the line joining (3, 4, -1) and (2, -1, 5).

Solution.

Given

(2, -3, ) (3, 4, -1) and (2, -1, 5)

Let the plane through (2, -3, 1) and perpendicular to the join of p(3, 4, -1) and Q(2, -1, 5) be a(x – 2) + b(y + 3) + c(z – 1) = 0.

Since d.rs of $\overleftrightarrow{\mathrm{PQ}}$ are (3 – 2, 4 + 1, -1 -5) i.e., (1, 5, -6)

We have $\frac{a}{1}=\frac{b}{5}=\frac{c}{-6}=\lambda, \text { say }$.

∴ Required plane is λ(x – 2) + 5λ(y + 3) – 6λ(z – 1) = 0

i.e., x – 2 + 5(y + 3) – 6(z – 1) = 0 i.e., x + 5y – 6z = -19

Example. 6. A variable plane passes through a fixed point (a, b, c). It meets the axes of reference in A, B, and C. Show that the locus of the point of intersection of the planes through A, B, C, and parallel to the coordinate planes is ax^-1 + by^-1 + cz^-1 = 1

Solution.

Given

A variable plane passes through a fixed point (a, b, c). It meets the axes of reference in A, B, and C.

Let the variable plane meeting the coordinates axes in A, B,C be $\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=1$ …..(1)

∴ A = (α, 0, 0), B = (0, β, 0), C = (0, 0, γ)

Also (1) passes through the fixed point (a, b, c)

∴ $\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}=1$

But equations to the planes through A, B, C and parallel to the coordinate planes are x = α, y = β, z = γ.

Clearly they intersect at P = (α, β, γ)

∴ Locus of P from (2) is $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=1 \text { i.e., } ax^{-1}+b y^{-1}+c z^{-1}=1$

Example. 7. Find the bisecting plane of the acute angle between the planes 3x – 2y – 6z + 2 = 0, -2x + y – 2z – 2 = 0

Solution.

Equations to the given planes are taken as 3x – 2y – 6z + 2 = 0 …..(1)

2x – y + 2x + 2 = 0 …..(2)

(constant terms are taken as +ve)

∴ Equations to the bisecting planes between the given planes are

⇒ $\frac{3 x-2 y+6 z+2}{\sqrt{(9+4+36)}}=\pm \frac{2 x-y+2 z+2}{\sqrt{(4+1+4)}}$

i.e., 5x – y – 4z + 8 = 0 …..(3)

23x – 13y + 32z + 20 = 0 …..(4)

Let θ be the acute angle between (2) and (3)

∴ $\cos \theta=\left|\frac{10+1-8}{\sqrt{9} \cdot \sqrt{(25+1+16)}}\right|=\frac{1}{(\sqrt{42})}$

∴ $\tan \theta=\sqrt{(41)}>1$

Hence 2θ, the angle between the planes

(1) and (2) are greater than 90° i.e., obtuse.

∴ (3) is the equation to the plane bisecting the obtuse angle between (1) and (2).

∴ (4) is the equation to the plane bisecting the acute angle between (1) and (2).

Note. 5x – y – 4z + 8 = 0 is the plane bisecting the angle containing the origin between (1) and (2).

Chapter 3 The Plane Joint Equation Of A Pair Of Planes

Consider the pair of planes π₁, π₂ whose respective equations are

l₁x + m₁y + n₁z + d₁= 0 …..(1) l₂x + m₂y + n₂z + d₂= 0 …..(2)

Consider the equation (l₁x + m₁y + n₁z + d₁)(l₂x + m₂y + n₂z + d₂) = 0 …..(3)

Let P = (x₁, y₁, z₁).

P ∈ π₁ ⇒ l₁x₁ + m₁y₁ + n₁z₁ + d₁ = θ ⇒ (l₁x₁ + m₁y₁ + n₁z₁ + d₁)(l₂x₁ + m₂y₁ + n₂z₁ + d₂) = 0

P ∈ π₂ ⇒ l₂x₁ + m₂y₁ + n₂z₁ + d₂ = θ ⇒ (l₁x₁ + m₁y₁ + n₁z₁ + d₁)(l₂x₁ + m₂y₁ + n₂z₁ + d₂) = 0

P lies on (3) ⇒ (l₁x₁ + m₁y₁ + n₁z₁ + d₁)(l₂x₁ + m₂y₁ + n₂z₁ + d₂) = 0

⇒ l₁x₁ + m₁y₁ + n₁z₁ + d₁ = 0 or ⇒ l₂x₁ + m₂y₁ + n₂z₁ + d₂ = 0 ⇒ P ∈ π₁ or ⇒ P ∈ π₂

∴ We have P ∈ π₁ or P ∈ π₂ : P lies on (3)

i.e., an equation is satisfied if and only if a point lies on the one plane or the other plane or both.

∴ (3) represents the joint or combined equation to the plane π₁ and π₂.

Note 1. a₁x + b₁y + c₁z + d₁= 0, a₂x + b₂y + c₂z + d₂ = 0 are two intersecting planes.

The combined equation to the pair of planes bisecting the angles between them is

⇒ $\left[\frac{a_1 x+b_1 y+c_1 z+d_1}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)}}-\frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{\left(a_2^2+b_2^2+c_2^2\right)}}\right]$

⇒ $\left[\frac{a_1 x+b_1 y+c_1 z+d_1}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)}}+\frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{\left(a_2^2+b_2^2+c_2^2\right)}}\right]=0$

i.e., $\frac{\left(a_1 x+b_1 y+c_1 z+d_1\right)^2}{a_1^2+b_1^2+c_1^2}-\frac{\left(a_2 x+b_2 y+c_2 z+d_2\right)^2}{a_2^2+b_2^2+c_2^2}=0$

Theorem. S ≡ ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0 represents a pair of planes π₁, π₂.

Then H ≡ ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 represents a pair of planes through the origin and parallel to the planes π₁, π₂.

Proof. Let the planes π₁, π₂ represented by S = 0 be respectively l₁x + m₁y + n₁z + d₁= 0, l₂x + m₂y + n₂z + d₂ = 0

∴ ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d

≡ (l₁x + m₁y + n₁z + d₁)(l₂x + m₂y + n₂z + d₂)

⇒ l₁l₂ = a,m₁m₂ = b,n₁n₂ = c, l₁m₂ + l₂m₁ = 2h, m₁n₂ + m₂n₁ = 2f, n₁l₂ + n₂l₁= 2g.

Joint equation of the planes passing through the origin and parallel to π₁, π₂ is

(l₁x + m₁y + n₁z)(l₂x + m₂y + n₂z) = 0

i.e., l₁l₂x² + m₁m₂y² + n₁n₂z² + (l₁m₂+ l₂m₁)xy + (m₁n₂ + m₂n₁)yz + (n₁l₂ + n₂l₁)zx = 0

i.e., ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 i.e., H = 0.

Definition. If H ≡ ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 and D = $\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|$, then D is called the determinant of H.

Theorem. H ≡ ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 represents the equation of a pair of planes or a plane if D = 0, f² ≥ bc, g² ≥ac, h²≥ ab.

Proof. H = 0 represents the equation to a pair of planes or a plane ⇒ H can be expressed as a product of two linear factors in x, y, z.

Let the factors be l₁x + m₁y + n₁z + d₁, l₂x + m₂y + n₂z + d₂

where (l₁, m₁, n₁) ≠ (0, 0, 0)

∴ ax² + by² + cz² + 2fyz + 2gzx + 2hxy ≡ (l₁x + m₁y + n₁z + d₁)(l₂x + m₂y + n₂z + d₂)

⇒ l₁l₂ = a,m₁m₂ = b,n₁n₂ = c, l₁m₂ + l₂m₁ = 2h, m₁n₂ + m₂n₁ = 2f, n₁l₂ + n₂l₁= 2g.

l₁d₂ + l₂d₁ = 0, m₁d₂ + m₂d₁ = 0, n₁d₂ + n₂d₁ = 0, d₁d₂ = 0

Now d₁d₂ = 0 ⇒ d₁ = 0 or d₂ = 0.

d₂ = 0 ⇒ l₂d₁ = 0, m₂d₁ = 0, n₂d₁ = 0

⇒ d₁= 0 (∵ at least one of l₂, m₂, n₂ is not equal to zero)

Similarly d₁ = 0 ⇒ d₂ = 0. ∴ d₁ = d₂ = 0

We know that $\left[\begin{array}{ccc}
l_1 & l_2 & 0 \\
m_1 & m_2 & 0 \\
n_1 & n_2 & 0
\end{array}\right]\left[\begin{array}{ccc}
l_2 & m_2 & n_2 \\
l_1 & m_1 & n_1 \\
0 & 0 & 0
\end{array}\right]$

= $\left[\begin{array}{ccc}
2 l_1 l_2 & l_1 m_2+l_2 m_1 & l_1 n_2+l_2 n_1 \\
l_2 m_1+l_1 m_2 & 2 m_1 m_2 & m_1 n_2+m_2 n_1 \\
n_1 l_2+l_1 n_2 & n_1 m_2+n_2 m_1 & 2 n_1 n_2
\end{array}\right]=\left[\begin{array}{ccc}
2 a & 2 h & 2 g \\
2 h & 2 b & 2 f \\
2 g & 2 f & 2 c
\end{array}\right]$

∴ $\operatorname{det}\left\{\left[\begin{array}{ccc}
l_1 & l_2 & 0 \\
m_1 & m_2 & 0 \\
n_1 & n_2 & 0
\end{array}\right]\left[\begin{array}{ccc}
l_2 & m_2 & n_2 \\
l_1 & m_1 & n_1 \\
0 & 0 & 0
\end{array}\right]\right\}=\operatorname{det}\left[\begin{array}{ccc}
2 a & 2 h & 2 g \\
2 h & 2 b & 2 f \\
2 g & 2 f & 2 c
\end{array}\right]$

⇒ $\left|\begin{array}{ccc}
l_1 & l_2 & 0 \\
m_1 & m_2 & 0 \\
n_1 & n_2 & 0
\end{array}\right|\left|\begin{array}{ccc}
l_1 & m_2 & n_2 \\
l_1 & m_1 & n_1 \\
0 & 0 & 0
\end{array}\right|=8\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & e
\end{array}\right| \Rightarrow 0 \times 0=8 \mathrm{D} \Rightarrow \mathrm{D}=0$.

Also, 4f² – 4bc = (m₁n₂ + m₂n₁)² – 4m₁n₂m₂n₁ = (m₁n₂ – m₂n₁)² ≥ 0 ⇒ f² ≥ bc.

Similarly, we can prove that g² ≥ ac, h²≥ ab.

We give below two theorems, for which proofs may be supplied by the readers if needed.

1. H = 0 represents a pair of planes if (1) D = 0 and (2) at least one of h² – ab, f² – bc, g² – ac is + ve and the remaining two are non-negative.

2. H = 0 represents a plane if D = 0, h² – ab, f² – bc, g² – ac. In this case, H takes the form (a₁x + b₁y + c₁z)².

Theorem. If θ(≤ π/2) is the angle between the pair of planes H = 0, then $\cos \theta=\left|\frac{a+b+c}{\sqrt{\left\{(a+b+c)^2+4\left(f^2+g^2+h^2-a b-b c-c a\right)\right\}}}\right|$

Proof. Let the pair of planes represented by H = 0 be

l₁x + m₁y + n₁z = 0, l₂x + m₂y + n₂z = 0.

∴ ax² + by² + cz² + 2fyz + 2gzx + 2hxy ≡ (l₁x + m₁y + n₁z)(l₂x + m₂y + n₂z)

⇒ l₁l₂ = a, m₁m₂ = b, n₁n₂ = c

l₁m₂ + l₂m₁ – 2h, m₁n₂ + m₂n₁ = 2f, n₁l₂ + n₂l₁ = 2g

Since θ(≤ (π/2)) is the angle between the planes,

⇒ $\cos \theta=\mid \frac{l_1 l_2+m_1}{\sqrt{\left(l_1^2+m_1^2+n_1^2\right)}}\frac{m_2+n_1 n_2}{\sqrt{\left(l_2^2+m_2^2+n_2^2\right)}}|$

=$\left|\frac{a+b+c}{\sqrt{\left\{\left(a^2+b^2+c^2+4 h^2-2 a b+4 f^2-2 b c+4 g^2-2 a c\right)\right\}}}\right|$

=$\left|\frac{a+b+c}{\sqrt{\left\{(a+b+c)^2+4\left(f^2+g^2+h^2-b c-c a-a b\right)\right\}}}\right|$

Cor. 1. Planes are perpendicular <=> θ = 90°

<=> cosθ = 0 <=> a + b + c = 0

<=> Coefficient of x2 + Coefficient of y2 + Coefficient of z2 = 0

2. Planes are identical (coincident) <=> θ = 0°

<=> $\cos \theta=1 \Leftrightarrow\left|\frac{a+b+c}{\sqrt{\left\{(a+b+c)^2+4\left(f^2+g^2+h^2-a b-b c-c a\right)\right\}}}\right|=1$

<=> (a + b + c)² = (a + b + c)² + 4(f² +g²+ h² – ab – bc – ca)

<=> f² + g² + h² – bc – ca – ab = 0 <=> (f² – bc) + (g² – ac) + (h² – ab) = 0

<=> f² = bc, g² = ac, h² = ab (∵ f² ≥ bc, g² ≥ ac, h² ≥ ab)

Note. 1. l₁x + m₁y + n₁z = 0, l₂x + m₂y + n₂z = 0 are two planes intersecting in a line with d.rs., l, m, n.

⇒ l₁l + m₁m + n₁n = 0, l₂l + m₂m + n₂n = 0 ⇒ $\frac{l}{m_1 n_2-m_2 n_1}=\frac{m}{n_1 l_2-n_2 l_1}=\frac{n}{l_1 m_2-l_2 m_1}$

⇒ $\frac{l}{\left[\left(m_1 n_2-m_2 n_1\right)^2-4 m_1 n_2 n_1 m_2\right]}=\frac{m}{\sqrt{\left[\left(n_1 l_2-n_2 l_1\right)^2-4 n_1 l_2 n_2 l_1\right]}}=\frac{n}{\sqrt{\left[\left(l_1 m_2-l_2 m_1\right)^2-4 l_1 m_2 l_2 m_1\right]}}$

⇒$\frac{l}{\sqrt{\left(4 f^2-4 b c\right)}}=\frac{m}{\left(\sqrt{\left.4 g^2-4 a c\right)}\right.}=\frac{n}{\sqrt{\left(4 h^2-4 a b\right)}}\frac{l}{\overline{\left(f^2-b c\right)}}=\frac{m}{\left(\sqrt{\left.g^2-a c\right)}\right.}=\frac{n}{\sqrt{\left(h^2-a b\right)}}$

⇒ d.rs. of the common line are $\sqrt{\left(f^2-b c\right)}, \sqrt{\left(g^2-a c\right)}, \sqrt{\left(h^2-a b\right)}$

2. If θ(≤ π/2) is the angle between the pair of intersecting planes given by ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0, then

⇒ $\cos \theta=\left|\frac{a+b+c}{\sqrt{\left\{(a+b+c)^2+4\left(f^2+g^2+h^2-a b-b c-c a\right)\right\}}}\right|$.

d.rs. of the line of intersection are $\sqrt{\left(f^2-b c\right)}, \sqrt{\left(g^2-a c\right)}, \sqrt{\left(h^2-a b\right)}$

Chapter 3 The Plane Solved Problems

Example. 1. Prove that the equation 2x² – 6y² – 12z² + 18yz + 2zx + xy = 0 represents a pair of planes, and find the angle between them.

Solution.

Let the given equation be ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 …..(1)

comparing the given equation to (1), a =2, b = -6, c = -12, f = 9, g = 1, h = 1/2

∴ $\mathrm{D}=\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{ccc}
2 & 1 / 2 & 1 \\
1 / 2 & -6 & 9 \\
1 & 9 & -12
\end{array}\right|$

= $2(72-81)-\frac{1}{2}(-6-9)+1\left(\frac{9}{2}+6\right)=-18+\frac{15}{2}+\frac{21}{2}=0$

f² = 81, bc = 72 ⇒ f² > bc, ac = -24 ⇒ g² > ac, h² = 1/4, ab = -12 ⇒ h² > ab.

∴ The given equation represents a pair of planes through the origin.

Let θ be the acute angle between the planes.

∴ $\cos \theta=\left|\frac{a+b+c}{\sqrt{\left[(a+b+c)^2+4\left(f^2+g^2+h^2-a b-b c-c a\right)\right]}}\right|$

OR : 2x² – 6y² – 12z² + 18yz + 2zx + xy = 0

⇒ 2x² + x(y + 2z) – (6y² + 12z²– 18yz) = 0

⇒ $x=\frac{-(y+2 z) \pm \sqrt{\left[(y+2 z)^2+8\left(6 y^2+12 z^2-18 y z\right)\right]}}{4}$

⇒ $4 x=-(y+2 z) \pm \sqrt{\left(49 y^2-140 y z+100 z^2\right)} \Rightarrow 4 x=-y-2 z \pm(7 y-10 z)$

⇒ 4x – 6y + 12z = 0, 4x + 8y – 8z = 0 ⇒ 2x – 3y + 6z = 0, x + 2y – 2z = 0

∴ The given equation represents a pair of planes through the origin.

⇒ $\cos \theta=\frac{2(1)-3(2)+6(-2)}{\sqrt{(4+9+36)} \cdot \sqrt{(1+4+4)}}=\frac{+16}{21}$ ∴ $\theta=\cos ^{-1}\left(\frac{16}{21}\right)$

Example. 2. If a² + b² + c² > 2ab + 2bc + 2ca, show that the equation ax² + by² + cz² -(a + b – c)xy – (b + c – a)yz – (c + a – b)zx = 0 represents a pair of planes. Also show that the line of intersection of the planes makes equal angles with the coordinate axes.

Solution.

Given

a² + b² + c² > 2ab + 2bc + 2ca,

Let H ≡ ax² + by² + cz² -(a + b – c)xy – (b + c – a)yz – (c + a – b)zx = 0 ……(1)

∴ Determinant of H = D

=$\left|\begin{array}{ccc}
a & \frac{-(a+b-c)}{2} & \frac{-(c+a-b)}{2} \\
\frac{-(a+b-c)}{2} & b & \frac{-(b+c-a)}{2} \\
\frac{-(c+a-b)}{2} & \frac{-(b+c-a)}{2} & c
\end{array}\right|$

= $-\frac{1}{8}\left|\begin{array}{ccc}
2 a & (a+b-c) & (c+a-b) \\
(a+b-c) & -2 b & (b+c-a) \\
(c+a-b) & (b+c-a) & -2 c
\end{array}\right| R_1=R_1+R_2+R_3$.

= $-\frac{1}{8}\left|\begin{array}{ccc}
0 & 0 & 0 \\
a+b-c & -2 b & b+c-a \\
c+a-b & b+c-a & -2 c
\end{array}\right|=0$

[f² > bc be condition]:

⇒ $\left(\frac{b+c-a}{2}\right)^2-b c=\frac{a^2+b^2+c^2-(2 a b+2 b c+2 c a)}{4}>0 \text { (by hyp.) }$

Similarly [g2 > ac, h2 ≥ ab conditions] $\left(\frac{c+a-b}{2}\right)^2-a c>0,\left(\frac{a+b-c}{2}\right)^2-a b>0$

∴ (1) represents a pair of intersecting planes.

For the common line $\overrightarrow{\mathrm{PQ}}$ of intersection of the planes, d.rs. are

⇒ $\sqrt{\left[\left(\frac{b+c-a}{2}\right)^2-b c\right]}, \sqrt{\left[\left(\frac{c+a-b}{2}\right)^2-c a\right]}, \sqrt{\left[\left(\frac{a+b-c}{2}\right)^2-a b\right]}$

i.e. $\frac{\sum a^2-2 a b}{4}, \frac{\sum a^2-2 a b}{4}, \frac{\sum a^2-a b}{4}$

But d.cs. of $\overrightarrow{\mathrm{OX}}, \overrightarrow{\mathrm{OY}}, \overrightarrow{\mathrm{OZ}} \text { are } 1,0,0 ; 0,1,0 ; 0,0,1$.

Let $\theta=\left(\begin{array}{ll}
\overrightarrow{P Q} & \overrightarrow{O X}
\end{array}\right)$

∴ $\cos \theta=\frac{1.1+1.0+1.0}{\sqrt{3} \sqrt{1}}=\frac{1}{\sqrt{3}} \text { i.e. } \theta=\cos ^{-1} \frac{1}{\sqrt{3}}$

Similarly we can observe that $\overrightarrow{\mathrm{PO}} \overrightarrow{\mathrm{OY}}=\operatorname{Cos}^{-1}(1 / \sqrt{3}),(\overrightarrow{\mathrm{PQ}} \overrightarrow{\mathrm{OZ}})=(1 / \sqrt{3})$

∴ The common line $\overleftrightarrow{P Q}$ makes equal with the axes.

Example.3. Show that the equation x² + 4y² + 9z²– 12yz – 6zx + 4xy + 5x + 10y – 15z + 6 = 0 represents a pair of parallel planes and find the distance between them.

Solution.

Given

x² + 4y² + 9z² – 12yz – 6zx + 4xy = (x + 2y – 3z)2

∴ x² + 4y² + 9z² – 12yz – 6zx + 4xy + 5x + 10y – 15z + 6

≡ (x + 2y – 3z + k)(x + 2y – 3z – l) where

k + l = 5, 2k + 2l = 10, – 3k – 3l = -15, kl = 6 i.e., k = 3, l = 2

∴ The given equation represents the planes

x + 2y – 3z + 3 = 0, x + 2y – 3z + 2 = 0 which are parallel.

∴ Distance between the parallel planes =$\frac{|3-2|}{\sqrt{(1+4+9)}}=\frac{1}{\sqrt{(14)}}$

The post The Plane Definition Theorems Proofs Solved Problems Exercise Orthogonal Projection On Planes appeared first on Answer Key for Math.

Cosets and Lagrange’s Theorem And Examples

Kumar — Wed, 09 Aug 2023 06:20:50 +0000

Cosets and Lagrange’s Theorem And Examples Coset

Definition. Let (H,.) be a subgroup of the group (G,.).

Let a ∈ G. Then the set aH = {ah | h ∈ H} is called a left coset of H in G generated by a and the set Ha = {ha | h ∈ H} is called a right coset of H in G generated by a.

Here the operation in G is denoted multiplicatively. Also aH, and Ha are called cosets of H generated by a in G.

Since every element of aH or Ha is in G, aH and Ha are complexes of G.

If e is the identity in G, then eH = {eh \ h ∈ H} = {h | h e H} = H and He = {he | he H} = H. Hence the subgroup of G is itself a left and a right coset of H in G.

If e is the identity in G, it is also the identity in H.

Therefore, for a ∈ G,e ∈ H we have ea ∈ Ha and ae ∈ aH.

Hence the left coset or the right coset of H generated by a is non-empty. Further a ∈ Ha, a ∈ aH and $\mathbf{H} a \cap a \mathbf{H} \neq \phi$.

If group G is abelian, then for every h ∈ H, we shall have ah = ha. Hence aH = Ha. However, even if G is not abelian, also we may have aH = Ha or aH ≠ Ha.

Note

1. The left or right coset of any subgroup in a group is called residue class modulo the subgroup of the group.

2. If the operation in G is denoted additively, then the left subset of H in G is generated by a, denoted by a + H = {a + h | h ∈ H}

i.e. a + H =.{a + h | h ∈ H}.

Similarly, the right coset of H is G generated by a.

= H + a = {h + a | h ∈ H}.

3. Let H be a subgroup of the group G and a, b ∈ G.

Then (1) a (bH) = (ah) H and (Hb)a = H(ba).

(2) x ∈ aH => yx ∈ y(aH) for y ∈ G => yx e (ya) H.

4. The clement a is called the coset representative of aH (Ha).

| aH |, | Ha | denotes the number of elements in aH, Ha respectively,

Example 1. Consider the group of symmetries of the square

i.e. $\left(\mathbf{D}_4, o\right)$ where $\mathbf{D}_4=\left\{r_{90}, r_{180}, r_{360}, x, y, d_1, d_2\right\}$

(H,o) where $\mathbf{H}=\left\{r_{180}, r_{360}, x, y\right\}$ is a subgroup of $\left(\mathbf{D}_4, o\right)$.

Cosets Definition And Examples In Group Theory

Then all the left cosets of H in G are

$r_{90} \mathbf{H}=\left\{r_{90} \circ r_{180}, r_{90} \circ r_{360}, r_{90} \circ x, r_{90} \circ y\right\}$ $=\left\{r_{270}, r_{90}, d_2, d_1\right\}$

$r_{180} \mathbf{H}=\left\{r_{180} o r_{180}, r_{180} o r_{360}, r_{180} o x, r_{180} o y\right\}$ $=\left\{r_{360}, r_{180}, x, y\right\}=\mathbf{H}$

$r_{270} \mathbf{H}=\left\{r_{270} o r_{180}, \ldots \ldots\right\}$ $=\left\{r_{90} ; r_{270}, d_1, d_2\right\}$

$r_{360} \mathbf{H}=\left\{r_{360} o r_{180}, \ldots \ldots\right\}$ $=\left\{r_{180}, r_{360}, x, y\right\}=\mathbf{H}$

$x \mathbf{H}=\left\{x \circ r_{180}, \ldots \ldots\right\}$ $=\left\{x, y, r_{360}, r_{180}\right\}=\mathbf{H}$

$y \mathbf{H}=\left\{y o r_{180}, \ldots \ldots\right\}$ $=\left\{x, y, r_{180}, r_{360}\right\}=\mathbf{H}$

$d_1 \mathbf{H}=\left\{d_1 o r_{180}, \ldots \ldots\right\}$ $=\left\{d_2, d_1, r_{270}, r_{90}\right\}$

$d_2 \mathbf{H}=\left\{d_2 o r_{180}, \ldots \ldots\right\}$ $=\left\{d_1, d_2, r_{90}, r_{270}\right\}$

We have two distinct left cosets, namely,

∴ $\left\{r_{90}, r_{270}, d_1, d_2\right\} \text { and }\left\{r_{180}, r_{360}, x, y\right\}=\mathbf{H}$

These two may be taken as $r_{90} \mathbf{H} \text { and } \mathbf{H}$

Obviously $r_{90} \mathbf{H} \cap \mathbf{H}=\phi \text { and } r_{90} \mathbf{H} \cup \mathbf{H}=\mathbf{D}_4$

Similarly, we can have all the right cosets of $\mathrm{H} \text { in } \mathbf{D}_4$.

Note. If a ∈ H, then aH = H = Ha.

Lagrange’s Theorem In Group Theory Explained With Examples

Example 2. Let G be the additive group of integers.

Now G = {…… -3, -2, -1, 0, 2, 3,…….. } and 0 is the identity in G. Also G is abelian.

Let H be a subset of G where elements of H are obtained by multiplying each element of G by 3 (say) i.e., H = { ……..,-9, -6,-3, 0,-3, 6, 9……}

Clearly, H is a subgroup of (G,+). (∵ $n_1, n_2 \in \mathbf{H} \Rightarrow n_1-n_2 \in \mathbf{H}$)

Since G is abelian, the left coset of H of an element in G = right coset of H in G.

∴ 0 + H = H = {………, -9,-6,-3, 0, 3, 6,……}.

Since 1 ∈ G, 1 + H = { ………..,-8, -5, -2, 1, 4, 7,……}

Since 2 ∈ G, 2 + H = {………..,-7, -4, -1, 2, 5, 8,……}

Observe that (1) 3 + H = 6 + H = …… = 0 + H ,

4 + H = 7 + H = ……… = 1 + H,

5 + H = 8 + H = …….. = 2 + H.

(2) 0 + H, 1 + H, 2 + H are disjoint.

(3) 0 + H ∪ 1 + H ∪ 2 + H = G

Cosets and Lagrange’s Theorem And Examples Properties of Cosets

Theorem 1. H is any subgroup of a group (G,) and h ∈ G. Then h ∈ H iff hH = H = Hh.

Proof. (1) h ∈ H to prove that hH = H = Hh.

Let ‘h’ be an arbitrary element of H. Then is an arbitrary element of hH.

Since H is a subgroup of G, h, h’ ∈ H => hh’∈ H

Thus every element of hH is also an element of H.

∴ hH ⊆ H …………………(1)

Again $h^{\prime}=e h^{\prime}=\left(h h^{-1}\right) h^{\prime}=h\left(h^{-1} h^{\prime}\right) \in h \mathbf{H}$

(∴ e is the identity in H, $h \in \mathbf{H} \Rightarrow h^{-1} \in \mathbf{H} \text { and } h^{-1} \in \mathbf{H}, h^{\prime} \in \mathbf{H} \Rightarrow h^{-1} h^{\prime} \in \mathbf{H}$)

∴ H ⊆ hH

From (1) and (2), hH = H.

Similarly, we can prove H = Hh

∴ h ∈ H

∴ hH = H = Hh.

(2) Let hH = H = Hh . To prove that he H.

Now h ∈ G. Since h = he, h ∈ hH.

But hH = H

∴ h ∈ H

Similarly Hh = H => h ∈ H

∴ hH = H= Hh => h ∈ H

Step-By-Step Guide To Understanding Cosets And Lagrange’s Theorem

Theorem 2. If a,b are any two elements of a group (G,.) and H any subgroup of G, then $\mathbf{H} a=\mathbf{H} b \Leftrightarrow ab^{-1} \in \mathbf{H}$ and aH = bH ⇔ $a^{-1} b \in \mathbf{H}$.

Proof. $a \in \mathbf{H} a, \mathbf{H} a=\mathbf{H} b \Rightarrow a \in \mathbf{H} b \Rightarrow a b^{-1} \in(\mathbf{H} b) b^{-1}$
$\Rightarrow a b^{-1} \in \mathbf{H}\left(b b^{-1}\right) \Rightarrow a b^{-1} \in \mathbf{H} e \Rightarrow a b^{-1} \in \mathbf{H}$
Now $a b^{-1} \in \mathbf{H} a b^{-1}=\mathbf{H} \Rightarrow \mathbf{H} a b^{-1} b=\mathbf{H} b$
$\Rightarrow \mathbf{H} a\left(b^{-1} b\right)=\mathbf{H} b \Rightarrow \mathbf{H} a e=\mathbf{H} b \quad \Rightarrow \mathbf{H} a=\mathbf{H} b$
Similarly we can Prove that $a \mathbf{H}=b \mathbf{H} \Leftrightarrow a^{-1} b \in \mathbf{H}$

Note. $\text { If } a b^{-1} \in \mathbf{H} \text { then }\left(a b^{-1}\right)^{-1} \in \mathbf{H} \Rightarrow\left(b^{-1}\right)^{-1} a^{-1} \in \mathbf{H} \Rightarrow b a^{-1} \in \mathbf{H}$

Similarly $a^{-1} b \in \mathbf{H} \Rightarrow b^{-1} a \in \mathbf{H}$

Solved Examples Of Cosets In Mathematics

Theorem 3. If a,b are any two elements of a group G and H any subgroup of G, then $a \in b \mathbf{H} \Leftrightarrow a \mathbf{H}=b \mathbf{H} \text { and } a \in \mathbf{H} b \Leftrightarrow \mathbf{H} a=\mathbf{H} b$

Proof. $a \in b \mathbf{H} \Rightarrow b^{-1} a \in b^{-1} b \mathbf{H}$
$\Rightarrow b^{-1} a \in e \mathbf{H} \Rightarrow b^{-1} a \in \mathbf{H}$ $\Rightarrow b^{-1} a \mathbf{H}=\mathbf{H} \Rightarrow b b^{-1} a \mathbf{H}=b \mathbf{H} \Rightarrow a \mathbf{H}=b \mathbf{H}$
Converse: Let aH = bH

∴ $a \in a \mathbf{H} \Rightarrow a \in b \mathbf{H}$

Similarly, other results can be proved.

Proof Of Lagrange’s Theorem With Examples

Theorem 4. Any two left (right) cosets of a subgroup are either disjoint or identical.

Proof. Let H be a subgroup of group G. Let aH and bH be two left cosets of H in G. If aH and bH are disjoint, there is nothing to prove.

If $a \mathbf{H} \cap b \mathbf{H} \neq \phi$, then there exists at least one element c such that c ∈ aH and c ∈ bH.

Let $c=a h_1 \text { and } c=b h_2 \text { where } h_1, h_2 \in \mathbf{H}$

∴ $a h_1=b h_2 \Rightarrow a h_1 h_1^{-1}=b h_2 h_1^{-1} \Rightarrow a e=b\left(h_2 h_1^{-1}\right) \Rightarrow a=b\left(h_2 h_1^{-1}\right)$

Since H is a subgroup, $h_2 h_1^{-1} \in \mathbf{H} \cdot \text { Let } h_3=h_2 h_1^{-1}$

∴ $h_3 \in \mathbf{H}$

Now $a=b h_3$

∴ $a \mathbf{H}=b h_3 \mathbf{H}=b \mathbf{H}$ (∵ $h_3 \in \mathbf{H} \Rightarrow h_3 \mathbf{H}=\mathbf{H}$ )

Two left cosets are identical if they are not disjoint.

∴ $a \mathbf{H} \cap b \mathbf{H}=\phi \quad \text { or } \quad a \mathbf{H}=b \mathbf{H}$

Similarly, we can prove that $\mathbf{H} a \cap \mathbf{H} b=\phi \text { or } \mathbf{H} a=\mathbf{H} b$.

Cor. H is any subgroup of a group G. If the cosets aH, bH, cH, … are all disjoint, then $\mathbf{G}=\mathbf{H} \cup a \mathbf{H} \cup b \mathbf{H} \cup c \mathbf{H} \ldots$ where H is the cost corresponding to the identity element in G.

Also $\mathbf{G}=\mathbf{H} \cup \mathbf{H} a \cup \mathbf{H} b \cup \mathbf{H} c \cup \ldots$

Congruence Modulo H

Definition. Let (G,.) be a group and (H, .) be a subgroup of G. For a, b ∈ G, if $b^{-1} a \in \mathbf{H}$ we say that a ≡ b (mod H)

Theorem 5. H is a subgroup of group G, for a,b ∈ G the relation a ≡ b (mod H) is an equivalence relation.

Proof. (1) Reflexive: Let e be the identity in (G,.).

Since H is a subgroup of G, e is the identity in H.

Let a ∈ G. Since $a^{-1} a=e \text {, we have } a^{-1} a \in \mathbf{H}$

∴ a ≡ a (mod H) => relation is reflexive.

(2) Symmetric: Let a ≡ b (mod H) for a, b ∈ G.

∴ $b^{-1} a \in \mathbf{H} \Rightarrow\left(b^{-1} a\right)^{-1}, \in \mathbf{H} \Rightarrow a^{-1} b \in \mathbf{H}$

b ≡ a (mod H )=> relation is symmetric.

(3) Transitive: Let a ≡ b (mod H) and b ≡ c (mod H) for a,b,c ∈ G.

∴ $b^{-1} a \in \mathbf{H} \text { and } c^{-1} b \in \mathbf{H} \Rightarrow\left(c^{-1} b\right)\left(b^{-1} a\right) \in \mathbf{H}$
$\Rightarrow c^{-1}\left(b b^{-1}\right) a \in \mathbf{H} \Rightarrow c^{-1}(e a) \in \mathbf{H}$
$\Rightarrow c^{-1} a \in \mathbf{H} \Rightarrow a \equiv c(\bmod \mathbf{H}) \Rightarrow$ relation is transitive.

Since the congruence modulo H is reflexive, symmetric, and transitive, it is an equivalence relation.

Note: Let H be a subgroup of group G and a ∈ G, then the equivalence class containing a w.r.t. the equivalence relation ( ≡ mod H ) is denoted by $\bar{a}$.

Cosets And Lagrange’s Theorem In Abstract Algebra

Theorem 6. Let (H, .) be a subgroup of a group (G,.). For a ∈ G, let the equivalence class $\bar{a}=\{x \in \mathbf{G} / x≡a(\bmod \mathbf{H})\}$. Then $\bar{a}=a \mathbf{H}$

Proof. To prove that $\bar{a}=a \mathbf{H}$

Let e be the identity in G.

∴ e is also the identity in H.
$x \in \bar{a} \Leftrightarrow x \equiv a(\bmod \mathbf{H})$ $\Leftrightarrow a^{-1} x \in \mathbf{H}$
$\Leftrightarrow a^{-1} x=h \in \mathbf{H} \text { for some } h \in \mathbf{H}$$\Leftrightarrow a\left(a^{-1} x\right)=a h \in a \mathbf{H} \text { for some } h \in \mathbf{H}$
$\Leftrightarrow\left(a a^{-1}\right) x=a h \in a \mathbf{H} \text { for some } h \in \mathbf{H}$ $\Leftrightarrow e x=a h \in a \mathbf{H} \text { for some } h \in \mathbf{H}$ $\Leftrightarrow x=a h \in a \mathbf{H} \text { for some } h \in \mathbf{H}$ $\Leftrightarrow x \in a \mathbf{H}$
∴ $\bar{a}=a \mathbf{H}$

Note 1. The equivalence relation a ≡ b (mod H) induces a partition in G which is nothing but the left coset decomposition of G. w.r.t. H. No left coset of H in G will be empty. Every element of G belongs to one and only one left coset of G.

2. The relation in G, defined by a ≡ b (mod H) if $a b^{-1} \in \mathbf{H}$, is an equivalence relation. This relation induces a partition in G which is nothing but the right coset decomposition of G.

Theorem 7. Let (H,) be a subgroup of a group (G.). Then there exists a bijection between any two left cosets of H in G.

Proof. Let aH,bH be two left cosets of H for a, b ∈ G.

Define $f: a \mathbf{H} \rightarrow b \mathbf{H} \text { such that } f(a h)=b h \text { for } h \in \mathbf{H}$

For $h_1, h_2 \in \mathbf{H}, a h_1, a h_2 \in a \mathbf{H} \text { and } b h_1, b h_2 \in b \mathbf{H}$

Now $f\left(a h_1\right)=f\left(a h_2\right) \Rightarrow b h_1=b h_2 \Rightarrow h_1=h_2 \Rightarrow a h_1=a h_2$

∴ $f \text { is } 1-1$

Now $b h \in b \mathbf{H} \Rightarrow \exists h \in \mathbf{H} \text { such that } b h \in b \mathbf{H}$

⇒ $\exists h \in \mathbf{H} \text { such that } a h \in a \mathbf{H}$

∴ For ah ∈ aH , f(ah) = bh

∴ f is onto.

∴ f is a bijection and there exists 1 — 1 correspondence between any two left cosets of H in G.

Note 1. Let H be a subgroup of a finite group G. Since there is 1—1 correspondence between any two left cosets of H, every left coset has the same number of elements including H (v H is also a left coset).

2. The above theorem can be proved between two right cosets. Also, every right coset of H of a finite group G has the same number of elements including H. (∵ H is also a right coset).

Properties Of Cosets Explained With Solved Problems

Theorem 8. If w is a subgroup of a group G then there is a one-to-one correspondence between the set of all distinct left cosets of H in G and the set of all distinct right cosets of H in G.

Proof. In G, let $\mathrm{G}_1$ = the set of all distinct left cosets

and $\mathrm{G}_2$ = the set of all distinct right cosets.

Define a mapping $f: \mathbf{G}_1 \rightarrow \mathbf{G}_2 \text { such that } f(a \mathbf{H})=\mathbf{H} a^{-1} \forall a \in \mathbf{G}$

For: $\text { Let } a \mathbf{H}, b \mathbf{H} \in \mathbf{G}_1$

Now $a \mathbf{H}=b \mathbf{H} \Rightarrow b^{-1} a \in \mathbf{H} \Rightarrow\left(b^{-1} a\right)^{-1} \in \mathbf{H}$

⇒ $a^{-1}\left(b^{-1}\right)^{-1} \in \mathbf{H} \quad \Rightarrow \mathbf{H} a^{-1}=\mathbf{H} b^{-1}$

⇒ $f(a \mathbf{H})=f(b \mathbf{H})$

⇒ $f \text { is one-one : Let } a \mathbf{H}, b \mathbf{H} \in \mathbf{G}_1$

∴ $f(a \mathbf{H})=f(b \mathbf{H}) \Rightarrow \mathbf{H} a^{-1}=\mathbf{H} b^{-1}$

⇒ $a^{-1}\left(b^{-1}\right)^{-1} \in \mathbf{H} \Rightarrow a^{-1} b \in \mathbf{H} \Rightarrow\left(a^{-1}b\right)^{-1} \in \mathbf{H}$

⇒ $b^{-1} a \in \mathbf{H} \Rightarrow a \mathbf{H}=b \mathbf{H}$

∴ $f \text { is } 1-1$.

f is onto Let $\mathbf{H} a \in \mathbf{G}_2$.

Since $a \in \mathbf{G}, a^{-1} \in \mathbf{G}$

∴ $a^{-1} \mathbf{H} \in \mathbf{G}_1 \text { and } f\left(a^{-1} \mathbf{H}\right)=\mathbf{H}\left(a^{-1}\right)^{-1}=\mathbf{H} a$

∴ $f \text { is onto }$

There is a one-to-one correspondence between $\mathbf{G}_1 \text { and } \mathbf{G}_2$.

Note 1. If H is a subgroup of a finite group G, then the number of distinct left cosets of H in G is the same as the number of distinct right cosets of H in G.

2. Since H is common to both the set of left cosets of H of a finite group G and the set of right cosets of H of the finite group G, the number of elements in a left coset of H is equal to the number of elements in a right coset of H.

Cosets and Lagrange’s Theorem And Examples Index of a subgroup of a finite group.

Definition. If H is a subgroup of a finite group G, then the number of distinct left (right) cosets of H in G is called the index of H in G. It is denoted by (G: H) or 

Lagrange’s Theorem

Theorem .9. The order of a subgroup of a finite group divides the order of the group.

Proof. Since H is a subgroup of a finite group G, H is finite.

(1) If H = G, then O(H) / O(G) .

(2) If H ≠ G, let O(G) = n and O(H) = m

We know that every right coset of H in G has the same number of elements and the number of right cosets of H in G is finite.

Also since H = He, H is a right coset of H in G.

∴ If Ha, Hb, Hc, are right cosets of H in G, then

O (Ha) = O(Hb) = O(Hc) = …….. = O (H) = m

Let the number of distinct right cosets of H of G be k

All these right cosets are disjoint and induce a partition of G.

∴ $\mathbf{O}(\mathbf{G})=\mathbf{O}(\mathbf{H} a)+\mathbf{O}(\mathbf{H} b)+\mathbf{O}(\mathbf{H} c)+\ldots \ldots+\mathbf{O}(\mathbf{H})(k \text { terms })$

= $m+m+m+\ldots+m(k \text { times }) \Rightarrow n=k m \Rightarrow k=\frac{n}{m}$

∴ $\mathbf{O}(\mathbf{H}) \text { divides } \mathbf{O}(\mathbf{G}) \text { i.e. } \mathbf{O}(\mathbf{H}) / \mathbf{O}(\mathbf{G})$

Note 1. Lagrange’s theorem can also be proved by taking the right cosets of H in G.

2. Lagrange’s theorem deals with finite groups only.

Let O(G) = n. If m is not a divisor of n, then there can be no subgroup of G of order m.

3. Since $k=\frac{n}{m}$ number of the distinct left (right) cosets of H in $\mathbf{G}=\frac{|\mathbf{G}|}{|\mathbf{H}|}$.

= $\frac{\text { order of the group } \mathbf{G}}{\text { order of the subgroup } \mathbf{H} \text { of } \mathbf{G}}=\text { Index of } \mathbf{H} \text { in } \mathbf{G}=(\mathbf{G}: \mathbf{H})$

4. The converse of Lagrange’s theorem is not true.

(1) Consider G = {1, – 1,i, – i} . Clearly. G is a group of order 4 w.r.t. multiplication. Since 2 is a divisor of 4 i.e. the order of the group G, let us examine whether a complex H (of order 2) of G, which is a subgroup of G, exists.

Consider a complex $\mathbf{H}_1=\{i, \quad-i\}$

Since – i. i = 1 and since $1 \notin \mathbf{H}_1 . \mathbf{H}_1$ is not a subgroup of G.

Again consider a complex $\mathbf{H}_2=\{1,-1\}$. Clearly, $\mathbf{H}_2$ is a subgroup of G.

∴ In conclusion, even if m is a divisor of n, a subgroup of order m in G need not exist.

(2) G is a finite group of order 6. Since 3 is a divisor of 6 i.e. the order of the group G, let us examine whether a complex H (of order 3) of G, which is a subgroup of G, exists.

Consider a complex $\mathbf{H}_1=\left\{r_0, f_1, f_2\right\} \text { of } \mathbf{G}$

Since $f_1 \circ f_2=r_1$ and since $r_1 \notin \mathbf{H}_1, \mathbf{H}_1$ is not a subgroup of G.

Again consider a complex $\mathbf{H}_2=\left\{r_0, r_1, r_2\right\}$.

Clearly $\mathbf{H}_2$ is a subgroup of G with identity $r_0$ and with $r_0^{-1}=r_0, r_1^{-1}=r_2$ and

∴ $r_2^{-1}=r_1$.

∴ In conclusion, even if m is a divisor of n, a subgroup of order m in G need not exist.

Thus the converse of Lagrange’s Theorem does not hold.

Cor. : The order of an element of a finite group divides the order of the group.

Theorem 10. Suppose H and K are subgroups of a group G such that K ≤ H ≤ G and suppose (H: K) and (G: H) are both finite. Then (G: K) is finite, and (G: K) = (G:H)(H: K)

Proof: H and K are Subgroups of a group G such that K ≤ H ≤ G and suppose (H: K) and (G: H) are both finite.

(G: H) = the index of subgroup H in G is the number of distinct left cosets of H in G and (H: K) = the index of subgroup K in H is the number of distinct left cosets of K in H.

Thus by Lagrange’s Theorem : $(\mathbf{G}: \mathbf{H})=\frac{|\mathbf{G}|}{|\mathbf{H}|} \text { and }(\mathbf{H}: \mathbf{K})=\frac{|\mathbf{H}|}{|\mathbf{K}|}$

∴ $(\mathbf{G}: \mathbf{H})(\mathbf{H}: \mathbf{K})=\frac{|\mathbf{G}|}{|\mathbf{H}|} \cdot \frac{|\mathbf{H}|}{|\mathbf{K}|}=\frac{|\mathbf{G}|}{|\mathbf{K}|}=(\mathbf{G}: \mathbf{K})$ implying that (G: K) is finite and (G: K) = (G: H) (H: K)

OR :

Suppose that the collection of distinct left cosets of H in $\mathbf{G}=\left\{a_i \mathbf{H}: i=1,2, \ldots \ldots, r\right\}$ and the collection of distinct left cosets of K in $\mathbf{H}=\left\{b_j \mathbf{K}: j=1,2, \ldots \ldots, s\right\}$.

Now we show that $\left\{a_i b_j \mathbf{K}: i=1 ; 2, \ldots . . r, j=1,2, \ldots . s\right\}$ is the collection of distinct left cosets of K in G.

G $=\bigcup_{i=1 \text { to r }} a_i \mathbf{H}, a_i \in \mathbf{G} \text { and } \mathbf{H}=\bigcup_{i=1 \text { to s }} b_j \mathbf{K}, b_j \in \mathbf{G}$

Now $x \in \mathbf{G} \Rightarrow x \in \bigcup_i a_i \mathbf{H} \Rightarrow x=a_1 h, h \in \mathbf{H} \text { and }$

h$\in \mathbf{H} \Rightarrow h \in \bigcup_j b_j \mathbf{K} \Rightarrow h=b_j \mathbf{K}, k \in \mathbf{K}$

∴ $x=a_i h=a_i b_j \mathbf{K}, k \in \mathbf{K} \Rightarrow x \in \bigcup_{i, j} a_i b_j \mathbf{K} \Rightarrow \mathbf{G}=\bigcup_{i, j} a_i b_j \mathbf{K}$

Now we show $a_i b_j \mathbf{K}=a_{i^{\prime}} b_{j^{\prime}} \mathbf{K} \Leftrightarrow i=i^{\prime}, j=j^{\prime}$

If $i=i^{\prime}, j=j^{\prime}, \text { then } a_i b_j \mathbf{K}=a_{i^{\prime}} b_{j^{\prime}} \mathbf{K}$

If possible $xa_i b_j \mathbf{K}=a_{i^{\prime}} b_{j^{\prime}} \mathbf{K} \text { when } i=i^{\prime}, j \neq j^{\prime}$

Then $b_j \mathbf{K} \cap b_{j^{\prime}} \mathbf{K}=\phi \Rightarrow a_i b_j \mathbf{K} \cap a_{i^{\prime}} b_{j^{\prime}} \mathbf{K}=\phi$

=> it is a contradiction.

∴ $j=j^{\prime}$

If possible$a_i b_j \mathbf{K}=a_{i^{\prime}} b_{j^{\prime}} \mathbf{K} \text { when } i \neq i^{\prime}, j=j^{\prime}$

Then $b_j \mathbf{K}=b_{j^{\prime}} \mathbf{K} \text { and } a_i \mathbf{H} \cap a_{i^{\prime}} \mathbf{H}=\phi \Rightarrow a_i b_j \mathbf{K} \cap a_{i^{\prime}} b_{j^{\prime}} \mathbf{K}=\phi$

=> it is a contradiction.

∴ $i=i^{\prime}$

When $i \neq i^{\prime}, j=j^{\prime}, a_i \mathbf{H} \cap a_{i^{\prime}} \mathbf{H}=\phi \text { and }$

∴ $b_i \mathbf{K} \cap b_i, \mathbf{K}=\phi \Rightarrow a_i b_j \mathbf{K} \cap a_{i^{\prime}} b_{j^{\prime}} \mathbf{K}=\phi$

∴ $a_i b_j \mathbf{K}=a_{i^{\prime}} b_{j^{\prime}} \mathbf{K} \Leftrightarrow i=i^{\prime}, j=j^{\prime}$

Thus G is the collection of distinct left cosets of K in G.

Hence (G: K) is finite and (G: K) = (G: H) (H: K)

($\text { Euler’s } \phi \text {-function }$ It is the function $\phi: Z^{+} \rightarrow Z^{+}$ defined as (1) $\text { For } 1 \in Z^{+}, \phi(1)=1$ and (2) for $n(>1) \in Z^{+}, \phi(n)=$ the number of positive integers less than n and relatively prime to n.)

Theorem 11. If n is a positive integer and a is an integer relatively prime to n then $a^{(x)}≡1(\bmod n) \text { where } \phi \text { is the Euler’s } \phi \text {-function. }$

Proof: Let x be any integer. Let [x] denote the residue class of the set of integers mod n.

G = {[a]/a is an integer relatively prime to n}.

Then G is a group of order ∅(n) with respect to the multiplication of residue classes. The identity in G is [1].

⇒ $[a] \in \mathbf{G} \Rightarrow[a]^{0(\mathbf{G})}=[1] \Rightarrow[a]^{\phi(n)}=[1]$

a a a a } . . . to $\phi((n) \text { times }]$ =[1]

⇒ $\left[a^{\phi(n)}\right]=[1] \Rightarrow a^{\phi(n)} \equiv 1(\bmod n)$

This theorem is known as Euler’s theorem.

Normalizer Of An Element Of A Group

Definition. If a is an element of a group G, then the normalizer on a in G is the set of all those elements of G which commute with a. The normalizer of an in G is denoted, by N (a) where N (a) = {x ∈ G / ax = xa).

The normalizer N (a) is a subgroup of G.

Note. If e is the identity in group G, $e x=x e=x \forall x \in \mathbf{G} \Rightarrow \mathbf{N}(e)=\mathbf{G}$

Cosets and Lagrange’s Theorem And Examples Solved Problems

Example. 1. Use Lagrange’s Theorem to prove that a finite group cannot be expressed as the union of two of its proper subgroups.

Solution: Let G be a finite group of order n. Assume that $\mathbf{H} \cup \mathbf{K}=\mathbf{G}$ where H, K are two proper subgroups of G.

Since e ∈ H and e ∈ K at least one of H, K (say H) must contain more than half the number of elements of G.

Let O(H) = p

∴ $\frac{n}{2}H is a proper subgroup of G )

∴ n is not divisible by p which contradicts Lagrange’s theorem.

Hence our assumption that \(\mathbf{H} \cup \mathbf{K}=\mathbf{G}$ is wrong.

∴ A finite group cannot be expressed as the union of two of its proper subgroups.

Example. 2. Show that two right, cosets Ha, Hb of a group G only if the two left cosets $a^{-1} \mathbf{H}, b^{-1} \mathbf{H}$ of G are distinct.

Solution: Suppose that (Ha) = (Hb).

H$a=\mathbf{H} b \Leftrightarrow a b^{-1} \in \mathbf{H} \Leftrightarrow a b^{-1} \mathbf{H}=\mathbf{H}$

⇒ $\Leftrightarrow a^{-1} a b^{-1} \cdot \mathbf{H}=a^{-1} \mathbf{H}$$\Leftrightarrow b^{-1} \mathbf{H}=a^{-1} \mathbf{H} \Leftrightarrow a^{-1} \mathbf{H}=b^{-1} \mathbf{H}$

∴ Ha, Hb are distinct iff $a^{-1} \mathbf{H} \text { and } b^{-1} \mathbf{H}$ are distinct.

Example. 3. Show that every finite group of prime order does not have any proper subgroup.

Solution: Let G be a finite group of order n where n is prime.

If possible, let H be a subgroup of order m, say

Then m ≤ n. But by Lagrange’s theorem m is a divisor of n.

Also since n is prime, either m = 1 or m = n.

∴ H = {e} or H = G. But these two are improper subgroups of G.

∴ Any group of prime order does not have any proper subgroup.

Note. Thus the total number of subgroups of a group of prime order is 2.

Example. 4. $\mathbf{P}_3=\left\{f_1, f_2, f_3, f_4, f_5, f_6\right\}$ is a non-abelian group over S.

⇒ $\mathbf{H}=\left\{f_1, f_2\right\}$ is a subgroup of $\mathbf{P}_3$.

Let us form the left cosets of H in $\mathbf{P}_3$.

⇒ $f_1 \mathbf{H}=\mathbf{H}, f_2 \mathbf{H}=\mathbf{H}, f_3 \mathbf{H}=\mathbf{H}=\left\{f_3, f_6\right\}, f_4 \mathbf{H}=\left\{f_4, f_5\right\}$

⇒ $f_5 \mathbf{H}=\mathbf{H}=\left\{f_5, f_4\right\}, f_6 \mathbf{H}=\left\{f_6, f_3\right\}$

Thus we get only three distinct left cosets.i.e. H, $f_3 \mathbf{H}, f_4 \mathbf{H} \text { of } \mathbf{H} \text { in } \mathbf{P}_3$

Thus $\mathbf{P}_3=\mathbf{H} \cup f_3 \mathbf{H} \cup f_4 \mathbf{H}$ and index of subgroup H in $\mathbf{P}_{\mathbf{3}} \text { is } 3$.

Observe that the number of elements in each left coset is the same as in H.

Further $f_3 \mathbf{H} \neq \mathbf{H} f_3 \text { since } \mathbf{H} f_3=\left\{f_3, f_5\right\}$.

We can observe similar results by taking all the right cosets of H in G.

Note. $\mathbf{A}_3=\left\{f_1, f_5, f_6\right\}$ is a commutative subgroup of $\mathbf{P}_3$.

Two distinct left cosets of $\mathbf{A}_3$ are $\mathbf{A}_3, f_2 \mathbf{A}_3$ where $f_2 \mathbf{A}_3=\left\{f_2, f_3, f_4\right\}$.

Also $\mathbf{P}_2=\mathbf{A}_3 \cup f_2 \mathbf{A}_3$ and index of a subgroup of $\mathbf{A}_3 \text { in } \mathbf{P}_3 \text { is } 2$.

Example. 5. Let H be a subgroup of group G and let $\mathbf{T}=\{x \in \mathbf{G} / x \mathbf{H}=\mathbf{H} x\}$. Show that T is a subgroup of G.

Solution: H is a subgroup of group G.

Let $x_1, x_2 \in \mathbf{T}$

∴ $x_1 \mathbf{H}=\mathbf{H} x_1, x_2 \mathbf{H}=\mathbf{H} x_2$

Now $x_2 \mathbf{H}=\mathbf{H} x_2 \Rightarrow x_2^{-1}\left(x_2 \mathbf{H}\right) x_2^{-1}=x_2^{-1}\left(\mathbf{H} x_2\right) x_2^{-1}$

⇒ $\mathbf{H} x_2^{-1}=x_2^{-1} \mathbf{H} \Rightarrow x_2^{-1} \in \mathbf{T}$

Also $\left(x_1 x_2^{-1}\right) \mathbf{H}=x_1\left(x_2^{-1} \mathbf{H}\right)=x_1\left(\mathbf{H} x_2^{-1}\right)=\left(x_1 \mathbf{H}\right) x_2^{-1}$

= $\left(\mathbf{H} x_1\right) x_2^{-1}=\mathbf{H}\left(x_1 x_2^{-1}\right) \Rightarrow x_1 x_2^{-1} \in \mathbf{T}$
Thus $x_1, x_2 \in \mathbf{T} \Rightarrow x_1, x_2^{-1} \in \mathbf{T}$

∴ T is a subgroup of G.

Conjugate Element Of A Group Self

Definition. (G,.) is a group and a ∈ G such that $a=x^{-1} a x \forall x \in \mathbf{G}$. Then a is called a self-conjugate element of G. A self-conjugate element is sometimes called an invariant element.

Here $a=x^{-1} a x \Rightarrow x a=a x \forall x \in \mathbf{G}$

Cosets and Lagrange’s Theorem And Examples The centre of a group.

Definition. The set Z of all self-conjugated elements of group G is called the centre of group G.

Thus $\mathbf{Z}=\{z \in \mathbf{G} / z x=x z \forall x \in \mathbf{G}\}$

If G is abelian, then the centre of G is G. (Vide Theorem 18 Chapter 5)

Cosets and Lagrange’s Theorem And Examples Exercise 4

1. If H = {1, – 1} and G = {1, – 1, i, – i] then prove that (H, .) is a subgroup of the group (G,.). Find all the right cosets of H in G.

Solution:

1 $\mathbf{H}=\mathbf{H},(-1) \mathbf{H}=\{-1,1\}, i \mathbf{H}=\{i,-i\},(-i) \mathbf{H}=\{-i, i\}$1

2. Prove that  is a subgroup of . Find the left cosets of the above subgroup in . Find the index of the subgroup in G.

Solution:

Given

If H = {1, – 1} and G = {1, – 1, i, – i]

0 + ${ }_{15} \mathbf{H}=\{0,3,6,9,12\}=\mathbf{H}, 1+{ }_{15} \mathbf{H}=\{1,4,7,10,13\}, 2+_{15} \mathbf{H}=\{2,5,8,11,14\}$

3+ ${ }_{15} \mathbf{H}=\mathbf{H}, 4+_{15} \mathbf{H}=1+_{15} \mathbf{H}=1+_{15} \mathbf{H}, 5+_{15} \mathbf{H}=2+_{15} \mathbf{H} \text {, etc. }$

3. (1) Determine the coset decomposition of the additive group of integers relative to a subgroup of all integral multiples of 4 = 4Z.

(2) Find all co-sets and index of the subgroup < 4 > of $\mathrm{z}_{12}$.

Solution:

(1) $\mathbf{Z}=(0+\mathbf{H}) \cup(1+\mathbf{H}) \cup(2+\mathbf{H}) \cup(3+\mathbf{H})$

(2) $\mathbf{Z}_{12}=0+{ }_{12} \mathbf{H} \cup 1+{ }_{12} \mathbf{H} \cup 2+_{12} \mathbf{H} \cup 3+_{12} \mathbf{H}$ where H = {0, 4, 8} = <4>. The index of the subgroup H of$\mathrm{Z}_{12}$ is 4.

The post Cosets and Lagrange’s Theorem And Examples appeared first on Answer Key for Math.

Rings, Integral Domains & Fields Examples

Alekhya — Tue, 08 Aug 2023 06:49:44 +0000

Rings, Integral Domains And Fields

Examples Of Rings In Mathematics With Solutions

The ring is the second algebraic system of the subject of Modern Algebra. The abstract concept of rings has its origin from the set of integers. Even though integers, real numbers, integers modulo − n, and Matrices are endowed with two binary operations, when dealt them in Groups we have considered only one binary operation ignoring the other.

The concept of Ring will take into account both addition and multiplication. The algebra of rings will follow the pattern already laid out for groups.

Definition. (Ring.) Let R be a non-empty set and +, · be two binary operations in R. (R, +, ·) is said to be a ring if, for a,b,c ∈ R;

R₁ · a + b = b + a

R₂· (a + b) + c = a + (b + c)

R₃. there exists 0 ∈ R such that a + 0 = a for a ∈ R.

R₄. there exists −a ∈ R such that a + (−a) = 0 for a ∈ R.

R₅· (a · b) · c = a · (b · c) and

R₆· a · (b + c) = a · b + a · c and (b + c) · a = b · a + c · a.

Note

1. The operation ’+’ is called the addition and the operation ’ · ’ is called the multiplication in the ring (R, +, •).

2. The ring (R, +, ·) is also called the ring R.

3. We write a.b as ab.

4. The properties R₁,R₂,R₃,R₄ merely state that (R, +) is a commutative group. Thus (R, +) is called the additive group of the ring R.

5. The identity element ’ 0 ’ in (R, +) is called the zero of the ring R. The zero of a ring should not be confused with the zero of the numbers.

6. By R₃, 0 + 0 = 0 for 0 ∈ R.

7. The properties R₅, and R₆ may be respectively called the Associative and Distributive laws.

In view of Note (4) and Note (7) a ring may also be defined as follows:

Definition. (Ring.) Le t.R be a non-empty set and +, · b e two binary operations in R.(R, +, ·) is said to be a ring, if

(1) (R, +) is a commutative group,

(2) (R, ·) is a semigroup and

(3) Distributive laws hold.

Definition. (Unity Element.) In a ring (R, +, ·) if there exists 1 ∈ R such that a.1 = 1. a = a for every a ∈ R then we say that R is a ring with unity element or identity element.

Note 1. If R is a ring with identity element, by R₄, we have −1 ∈ R so that

1 + (−1) = 0.

2. A ring with unity element contains at least two elements 0 and 1 if R 6= {0}.

Definition. In a ring (R, +, ·) if a.b = b.a for a, b ∈ R then we say that R is a commutative ring.

Imp. A ring R (1) need not be commutative under multiplication and (2) need not have an identity (unity) element under multiplication, unless or otherwise stated.

Example. 1. Let R = {0} and +, · be the operations defined by 0 + 0 = 0 and 0 · 0 = 0. Then (R, +, ·) is clearly a ring called the Null ring or Zero ring.

Integral Domains Definition And Examples Step-By-Step

Example 2. The set Z of integers w.r.t. usual addition and multiplication is a commutative ring with unity element.

For (1) (Z, +) is a commutative group (2) Multiplication is associative in Z and (3) Multiplication is distributive over addition.

Example. 3. The set N of natural numbers is not a ring w.r.t. usual addition and multiplication, because, (N, +) is not a group.

Example. 4. The sets Q, R, and C are rings under the usual addition and multiplication of numbers.

Example 5. The set of integers mod under the addition and multiplication mode is a ring.

Example. 6. The set of irrational numbers under addition and multiplication is not a ring as there is no zero element.

Let (R, +, ·) be a commutative ring with unity element. Then (R, +) is a commutative group and (R, ·) is a semi-group with identity. element 1. So we have the following results which are obvious from the theory of groups.

1. The zero element of R is unique and a + 0 = a for every element ’ a ’ in R.

2. For a ∈ R the additive inverse −a ∈ R is unique and a + (−a) = 0.

3. The identity element 1 ∈ R is unique and a · 1 = 1.a = a for every a ∈ R.

4. For a ∈ R, −(−a) = a. 5. For 0 ∈ R, −0 = 0. 6. For a, b ∈ R, −(a + b) = −a − b.

5. For a, b, c ∈ R, a + b = a + c ⇒ b = c and b + a = c + a ⇒ b = c

6. For a, b, x ∈ R, the equations a+x = b and x+a = b have unique solutions.

7. For 1 ∈ R, the identity element, −(−1) = 1.

8. For a, b₁, b₂, . . . . . . .b_n ∈ R, from R₆, we have a (b₁ + b₂ + . . . .. + b_n) = ab₁ + ab₂ + . . . + ab_n and (b₁ + b₂ + . . . .. + b_n) a = b₁a + b₂a + . . . . + b_na.

Notation. 1. If R is a ring and a, b ∈ R then a + (−b) ∈ R, a + (−b) is written as a − b.

9. If R is a ring and a ∈ R then a + a ∈ R and a + a is written as 2a.

10. If R is a ring and a ∈ R then a · a ∈ R and a · a is written as a2.

Rings, Integral Domains, And Fields Some Basic Properties Of Rings

Theorem 1. If R is a ring and 0, a, b ∈ R, then

0a = a0 = 0,

a(−b) = (−a)b = −(ab)

(−a)(−b) = ab

a(b − c) = ab − ac. )

Proof. (1) 0a = (0 + 0)a ⇒ 0 + 0a = 0a + 0a (By R₃, R₆)

∴ 0 = 0a (By right cancellation law of (R, +) )

Similarly, we can prove that a0 = 0. Hence 0a = a0 = 0

(2) To prove that a(−b) = −(ab) we have to show that a(−b) + (ab) = 0.

a(−b) + ab = a{(−b) + b} = a0 = 0 ( By R6, R4) ⇒ a(−b) = −(ab)

Similarly, we can prove that (−a)b = −(ab).

Hence, a(−b) = (−a)b =−(ab).

(3) (−a)(−b) = −{(−a)b} = −{−(ab)} = ab ( by (2 ))[ ∵ (R, +) is a group ]

(4) a(b − c) = a[b + (−c)] = ab + a(−c) = ab − ac (By R6 ) [By theorem (1), (2)]

Similarly, we can prove that (b − c)a = ba − ca.

Solved Problems On Rings, Integral Domains, And Fields

Theorem. 2. If (R, +, ·) is a ring with unity then this unity 1 is the only multiplicative identity.

Proof. Suppose that there exist 1, 10 ∈ R such that 1.x = x.1 = x and 10 · x = x · 10 = x∀x ∈ R

Regarding 1 as identity, 1.10 = 10. Regarding 10 as identity, 1.10 = 1

Thus 10 = 1.10 = 1. ∴ 1 is the only multiplicative identity.

Theorem, 3. If R is a ring with unity element 1 and a ∈ R then

(−1)a = −a

(−1)(−1) = 1

Proof. (1) (−1)a+a = (−1)a+1a = {(−1)+1}a = 0a = 0 ( ∵ a = 1a, R6, R4)

∴ (−1)a = −a

(2) For a ∈ R we have (−1)a = −a. . Taking a = −1, (−1)(−1) = −(−1) =1

Rings, Integral Domains, And Fields Boolean Ring

Definition. In a ring R if a₂ = a∀a ∈ R then R is called a Boolean ring.

Theorem 1. If R is a Boolean ring then (1) a + a = 0∀a ∈ R (2) a + b = 0 ⇒ a = b and (3) R is commutative under multiplication. Or, Every Boolean ring is abelian.

Proof. (1) a ∈ R ⇒ a + a ∈ R.

Since a₂ = a∀a ∈ R, we have (a + a)₂ = a + a ⇒ (a + a)(a + a) = a + a ⇒ a(a + a) + a(a + a) = a + a ⇒ a₂ + a₂ + a₂ + a₂ = a + a (By R₆) ⇒

(a+a)+(a+a) = a+a ( ∵ R is Boolean ) ⇒ (a+a)+(a+a) = (a+a)+0 (By R₃ ) ⇒ a + a = 0 [By left cancellation law of group (R, +)]

(2) For a, b ∈ R, a + b = 0 ⇒ a + b = a + a ⇒ b = a [By (1)]

(3) a, b ∈ R ⇒ a + b˙ ∈ R ⇒ (a + b)₂ = a + b

(∵ R is Boolean) ⇒ (a + b)(a + b) = a + b ⇒ a(a + b) + b(a + b) = a + b (By R₆) ⇒ a₂ + ab + ba + b₂ = a + b (By R₆ ) ⇒ (a + ab) + (ba + b) =

a + b. (∵ R is Boolean) ⇒ (a + b) + (ab + ba) = a + b [ ∵ (R, +) is a group ] ⇒ (a + b) + (ab + ba) = (a + b) + 0 ⇒ ab + ba = 0 ⇒ ab = ba (By (2))

Rings, Integral Domains And Fields Solved Problems

Example. 1. If R is a ring with identity element 1 and 1 = 0 then R = {0}.

Solution. x ∈ R ⇒ x = 1x ⇒ x = 0x ⇒ x = 0

[By Theorem 1(1)]

∴ R = {0}

Thus a ring R with unity has at least two elements if R₆= {0}.

Rings And Their Properties With Detailed Examples

Example. 2. Prove that the set of even integers is a ring, commutative without unity under the usual addition and multiplication of integers.

Solution. Let R = the set of even integers. Then R = {2x | x ∈ Z}.

a, b, c ∈ R ⇒ a = 2m, b = 2n, c = 2p where m, n, p ∈ Z.

(R, +) is a commutative group. (see ex. in groups)

a · b = (2m)(2n) = 2l where l = 2mn ∈ Z

∴ Multiplication ( · of integers is a binary operation in R.

(a · b) · c = (2m · 2n) · 2p = 8mnp and a · (b · c) = 2m · (−2n · 2p) = 8mnp

∴ (a.b).c = a.(b.c) ⇒ Multiplication (•) is associative in R.

a. (b + c) = 2m(2n + 2p) = 2m · 2n + 2m · 2p = a · b + a · c

Similarly, (b + c) · a = b · a + c · a

∴ Distributive laws hold in R. Hence (R, +, ·) is a ring.

Since ’ 1 ’ is not an even integer; 1 ∈/ R and hence R has no unity element.

Example. 3. (R, +) is an abelian group. Show that (R, +, ·) is a ring if multiplication (·) is defined as a.b=0 ∀a, b ∈ R.

Solution. To prove that (R, +, ·) is a ring we have to show that (R, ·) is semigroup and distributive laws hold.

∀a, b ∈ R, a.b = 0 where 0 ∈ R is the zero element in the group.

∴ multiplication ‘0 ’ is a binary operation in R.

Let a, b, c ∈ R. Then (a · b) · c = 0.c = 0; a.(b · c) = a · 0 = 0 (By Def.)

∴ (a · b) · c = a · (b, c)∀a, b, c ∈ R ∴ (R, ·) is a semi group.

Let a, b, c ∈ R. a ∈ R, b + c ∈ R ⇒ a · (b + c) = 0

a ∈ R, b ∈ R ⇒ ab = 0; a ∈ R, c ∈ R ⇒ ac = 0 ⇒ ab + ac = 0 + 0 = 0.

Hence a · (b + c) = a · b + a · c

Similarly, we can prove that (b + c) · a = b · a + c· a.

∴ Distributive laws hold.

Example. 4. Prove that Z _m = {0, 1, 2, . . . ., m− 1} is a ring with respect to addition and multiplication modulo m.

Solution. We denote addition modulo m by +_mand multiplication modulo m by × _m

We also know that a+_m b = a + b(modm) = r where r is the remainder when a + b is divided by m. a × m b = ab(modm) = s where s is the remainder when ab is divided by m.

Let a, b, c ∈ Z_m. a+_m b = a + b( mod m) ∈ Z_m⇒ +_m is a binary operation in Z_m. a +_m b = a + b(modm) = b + a(modm) = b +m a ⇒ +_m is commutative in Z_m

(a+_m b) + mc = (a + b) + c(modm) = a + (b + c)(modm) = a +m (b +_m c)

∴ +m is associative in Z_m

There exists 0 ∈ Z_m

such that 0 +m a = 0 + a(modm) = a(modm) = a.

⇒ 0 is the zero element.

For 0 ∈ Z_m, we have 0 +m 0 = 0(modm) ⇒ additive inverse of 0 = 0.

For a 6= 0 ∈ Z_mwe have 0 < a < m ⇒ 0 < m − a < m ⇒ m − a ∈ Z_m

a +_m (m − a) = a + (m − a)(modm) = m(modm) = 0(modm)

∴ inverse of a 6= 0 ∈ Z_mis m − a ∈ Z_m

Hence ( Z_m, +) is an abelian group.

Let a, b, c ∈ Z_m. a× _m b = ab(modm) ∈ Z_m⇒ × _mis a binary operation in Z_m

(a × _mb) × m c = (ab)c(modm) = a(bc)(modm) = a × _m(b × _mc)

∴ × _m is associative in Z_m.

a× m (b+_m c) = a(b+ c)( mod m) = ab+ ac( mod m) = (a × _mb)+m (a × _mc) and (b +_m c) × m a = (b × _ma) +m (c × _ma) so that distributive laws hold.

∴ ( Z_m,+_m, × _m) is a ring.

Note. Put m = 6 in the above proof to prove that Z₆ is a ring.

Applications Of Rings, Integral Domains, And Fields In Mathematics

Example. 5. Prove that the set R = {a, b} with addition ( +) and multiplication (·) defined as follows is a ring

+ a b

a a b

b b a

and

. a b

a a a

b a b

Solution. From the above tables, clearly +,· are binary operations in R.

1. (a + a) + b = a + b = b; a + (a + b) = a + b = b ⇒ (a + a) + b = a + (a + b) (a + b) + a = b + a = b; a + (b + a) = a + b = b ⇒ (a + b) + a = a + (b + a), etc,

∴ Associativity is true.

2. a ∈ R is the zero element because a + a = a, b + a = b

3. a + b = b = b + a ⇒ commutativity is true.

4. a + a = a ⇒ additive inverse of a = a and b + b = a ⇒ additive inverse of b = b.

5. a · (a · b) = a · a = a; (a · a) · b = a · b = a ⇒ a · (a · b) = (a · a) · b, etc

∴ Associativity is true.

6. a · (b + a) = a · b = a; a · b + a · a = a + a = a ⇒ a · (b + a) = a · b + a · a (b + a) · a = b · a = a; b · a + a · a = a + a = a ⇒ (b + a) · a = b · a + a · a, etc.

∴ Distributive laws are true.

Hence (R, +, ·) is a ring.

Example. 6. If R is a ring and a, b, c, d ∈ R then prove that

(a + b)(c + d) = ac + ad + bc + bd, and

a + b = c + d ⇔ a − c = d − b

Solution.

(1) (a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd ( By R₆)

(2) a + b = c + d ⇔ (a + b) + (−b) = (c + d) + (−b)

⇔ a + (b + (−b)) = (c + d) + (−b)

⇔ a + 0 = (c + d) + (−b)

⇔ a + (−c) = (−c) + {(c + d) + (−b)}

⇔ a − c = ((−c) + c) + {d + (−b)}

⇔ a − c = 0 + (d − b) ⇔ a − c = d − b (By R4 )

Rings, Integral Domains And Fields Exercise 1

1( a )
1. If R is a ring and a, b, c ∈ R prove that (a − b) − c = (a-c) − b

2. If R is a ring and a, b ∈ R then prove that the equation a + x = b has unique solution in R.

3. In a ring R if ’ a ’ commutes with ’ b ’ prove that ’ a ’ commutes with ’ −b0 ’ where a, b ∈ R.

4. If R is a ring with unity element ’ 1 ’ and R₆= {0} prove that 1 6= 0 where 0 ∈ R is the zero element.

5. R is a Boolean ring and for a ∈ R, 2a = 0 ⇒ a = 0 then prove that R = {0}. 6. If R is a commutative ring prove that (a + b)² = a² + 2ab + b²∀a, b ∈ R.

6. If R is a ring and a, b, c, d ∈ R evaluate (a − b)(c − d).

7. If R = {a√2 | a ∈ Q} is (R, +, ·) under ordinary addition and multiplication, a ring ?

8. Is the set of all pure imaginary numbers = {iy | y ∈ R} a ring with respect to addition and multiplication of complex numbers?

9. If Z = the set of all integers. and ’ n ’ is a fixed integer prove that the set nZ = {nx | x ∈ Z} is a ring under ordinary addition and multiplication of integers.

Answers

2. b − a 7. ac + bd − ad − bc 8. Not a ring 9. Not a ring

Rings, Integral Domains, And Fields Zero Divisors Of A Ring

Though rings are a generalisation of the number system some algebraic properties of the number system need not hold in general rings.

The product of two numbers can only be zero if at least one of them is zero, whereas in any ring it may not be true.

For example, in the ring (Z₆, +, ·) of modulo −6, we have 2.3 = 0 with neither 2 = 0 nor 3 = 0.

Definition. (Zero Divisors). Two non-zero elements a, and b of a ring R are said to be zero divisors (divisors of zero) if ab = 0, where 0 ∈ R is the zero element.

In particular,’ a ’ is the left zero divisor, and ’ b ’ is the right zero divisor.

Definition. (Zero Divisor). a 6= 0 ∈ R is a zero divisor if there exists b 6= 0 ∈ R such that ab = 0.

Note. 1. In a commutative ring there is no distinction between left and right zero divisors.

2. A ring R has no zero divisors ⇔ a, b ∈ R and ab = 0 ⇒ a = 0 or b = 0

Example.1. The ring of integers Z has no zero divisors.

Step-By-Step Guide To Understanding Rings, Integral Domains, Fields

Example. 2. In the ring (Z₁₂, +, ·), the elements 2, 3, 4, 6, 8, 9, 10 are zero divisors.

For 2 · 6 = 0, 3 : 4 = 0, 3 · 8 = 0, 4 · 6 = 0, 4 · 9 = 0, 6 · 10 = 0

Observe that the G. C. D of any of {2, 3, 4, 6, 8, 9, 10} and 12 6= 1.

Example3. The ring (M₂, +, ·) of 2 × 2 matrices whose elements are in Z, has zero divisors.

For, we have A = 10006= O, B = 00106= O, where O is a zero matrix, such that AB = O.

Example 4. The ring (Z₃, +, ·) of modulo −3 has no zero divisors.

Example 5. The ring Z × Z = {(a, b) | a, b ∈ Z} has zero divisors.

For, (0, 1), (1, 0) ∈ Z × Z ⇒ (0, 1) · (1, 0) = (0, 0) = zero element in Z × Z.

Imp. In the ring of integers Z, all the solutions of x² − 4x + 3 = 0 are obtained by factoring as x² − 4x + 3 = (x − 1)(x − 3) and equating each factor to zero. While doing so, we are using the fact that Z is an Integral Domain, so that it has no zero divisors.

But if we want to find all solutions of an equation in a ring R which has zero divisors, we can do so, by trying every element in the Ring by substitution in the product (x − 1)(x − 3) for zero.

Example 6. In the ring Z of integers, the equation x² +2x+4 = 0 i.e. (x+1) 2 +3 = 0 has no solution as (x + 1)² + 3 ≥ 3∀x ∈ Z.

But, in the ring Z₆ = {0, 1, 2, 3, 4, 5}, (x+1)² +3 takes respectively the values 4, 1, 0, 1, 4, 3 for x = 0, 1, 2, 3, 4, 5 ∈ Z₆.

∴ x² + 2x + 4 = 0 has 2 ∈ Z₆ as solution.

Rings, Integral Domains, And Fields Cancellation Laws In A Ring

If (R, +, ·) is a ring, then (R, +) is an abelian group. So, cancellation laws with respect to addition are true in R. Now, we are concerned about the cancellation laws in R, namely ab = ac ⇒ b = c, ba = ca ⇒ b = c for a, b, c ∈ R with respect to multiplication.

Definition. (Cancellation laws). In a ring R, for a, b, c ∈ R if a₆= 0, ab = ac ⇒ b = c and a₆ = 0, ba = ca ⇒ b = c then we say that cancellation laws hold in R.

Theorem. A ring R has no zero divisors if and only if the cancellation laws hold in R.

Proof. Let the ring have zero divisors. We prove that cancellation laws hold in R.

a, b, c ∈ R and a₆= 0, ab = ac ⇒ ab − ac = 0

⇒ a(b − c) = 0 ⇒ b − c = 0 ( ∵ a₆= 0) ⇒ b = c

Similarly we can prove a₆= 0, ba = ca ⇒ b = c

Conversely, let the cancellation laws hold in R. We prove that R has no zero divisors. If possible, suppose that there exist a, b ∈ R such that a₆= 0, b₆= 0, and ab = 0.

ab = 0 ⇒ ab = a0 ⇒ b = 0 (By cancellation law)

This is a contradiction.

∴ a₆= 0, b₆= 0 and ab = 0 is not true in R.

∴ R has no zero divisors.

Note. The importance of having no zero divisors in a ring R, is, that an equation ax = b where a₆ = 0, b ∈ R can have at most one solution in R.

For x₁, x₂ ∈ R if ax₁ = b and ax₂ = b then ax₁ = ax₂ ⇒ x₁ = x₂ (By cancellation law)

If a₆ = 0 ∈ R has a multiplicative inverse, say, a− 1 ∈ R then the solution is a− 1b ∈ R.

Rings, Integral Domains, And Fields Solved Problems

Example. 1. Find the zero-divisors of Z₁₂, the ring of residue classes modulo – 12.

Solution. Z₁₂ = {0, 1, . . . . . . .11}.

For a¯ 6= 0 ∈ Z₁₂ there should exist ¯b ∈ Z₁₂ such that a¯ × ¯b ≡ 0(mod12)

We have 2 × 6 = 0, 3 × 4 = 0, 4 × 3 = 0, 6 × 2 = 0, 8 × 3 = 0, 8 × 6 = 0, 8 × 9 = 0, 10 × 6 = 0

∴ 2, 3, 4, 6, 8, 9, 10 are zero divisors.

Worked Examples Of Integral Domains In Abstract Algebra

Example. 2. Solve the equation x₂ − 5x + 6 = 0 in the ring Z₁₂.

Solution.

Given

x₂ − 5x + 6 = 0

In the ring of integers Z, which has no zero divisors, x₂ − 5x + 6 = 0 ≡ (x − 2)(x − 3) = 0 has two solutions 2, 3 ∈ Z.

But in Z₁₂; for x = 6, (x − 2)(x − 3) = (4)(3) = 12 = 0 and for x = 11, (x − 2)(x − 3) = (9)(8) = 72 = 0.

∴ the given equation has 4 solutions, namely, 2,3,6, and 11 in the ring Z₁₂.

Example. 3. In the ring Z_n, show that the zero divisors are precisely those elements that are not relatively prime to n. (or) show that every non-zero element of Zn is a unit or zero divisor:

Solution. Let m ∈ Z_n = {0, 1, 2, . . . , n − 1} and m 6= 0. Let m be not relatively prime to n.

Then G. C.D of m, n = (m, n) 6= 1.

Let (m, n) = d.

We have (m, n) = d ⇒ md , nd = 1 ⇒ md , nd ∈ Zn and md 6= 0, nd 6= 0

∴ m nd = md n = 0(modn). Thus m 6= 0, nd 6= 0 ⇒ m nd = 0

⇒ m is a zero divisor.

∴ Every m ∈ Zn which is not relatively prime to n is a zero divisor.

Let m ∈ Zn be relatively prime to n.

Then (m, n) = 1

∴ Let mr = 0 for some r ∈ Zn.

We have mr = 0(modn) ⇒ n| mr ⇒ n| r ( ∵ (m, n) = 1) ⇒ r = 0 (0 ≤ r < n − 1)

∴ If m ∈ Zn is relatively prime to n then m is not a zero divisor.

Note. If p is a prime, then Z_p ring has no zero divisors.

Theorems And Properties Of Rings And Fields With Examples

1. (g) Some Special Types Of Rings

Definition. (Integral Domain)

A commutative ring with unity containing no zero divisors is an Integral Domain.

Note. 1. Some authors define integral domain without a unity element.

2. For “Integral Domain” we simply use the word “Domain” and denote by the symbol D. Imp. D is an integral domain ⇔

D is a ring,

D is commutative,

D has a unity element and

D has no zero divisors.

Example 1. The ring of integers Z is naturally an integral domain.

1 ∈ Z is the unity element and ∀a, b ∈ Z we have ab = ba (commutativity) and ab = 0 ⇒ a = 0 or b = 0 (no zero divisors).

Example 2. (Z₆, +, ·) where Z₆ = {0, 1, 2, 3, 4, 5}, the set of integers under the modulo −6 system, is a ring. 1 ∈ Z₆ is the unity element and ∀a, b ∈ Z₆ we have ab(mod6) = ba(mod6)( commutative ) But, for 2 6= 0, 3 6= 0(mod6), 2.3 = 6(mod6) = 0 and hence Z₆ has zero divisors. Therefore, Z₆ is not an integral domain.

Example. 3. If Q = the set of all rational numbers and R = the set of all real numbers then (Q, +, ·) and (R, +, ·) are integral domains.

Example 4. -7 = {0, 1, 2, 3, 4, 5, 6}, the set of all integers under modulo −7 is an integral domain with respect to addition and multiplication modulo −7.

Example 5. The ring (M₂, +, ·) of 2× 2 matrices is not an integral domain because it is not commutative and has zero divisors.

Example 6. Z × Z = {(a, b) | a, b ∈ Z} is not an Integral Domain under the addition and multiplication of components.

Theorem. 1. In an integral domain, cancellation laws hold.(Write the proof of (1) part of a theorem in Art. 1.6)

Theorem. 2. A commutative ring with unity is an integral domain if and only if the cancellation laws hold. (Write the proof of a theorem in Art. 1.6)

Definition. (Multiplicative Inverse). Let R be a ring with the unity element ’

1 ’. A non-zero element a ∈ R˙ is said to be invertible under multiplication, if there exists b ∈ R such that ab = ba = 1, b ∈ R is called the multiplicative inverse of a ∈ R.

From the theory of groups, the multiplicative inverse of a 6= 0 ∈ R, if exists, is unique. It is denoted by a − 1. Also aa− 1 = a− 1a = 1.

Definition. (Unit of a Ring). Let R be a ring with unity. An element u ∈ R is said to be a unit of R if it has a multiplicative inverse in R.

Note.

1. The zero element of a ring is not a unit.

2. The unity element of a ring and the unit of a ring R is different. The unity element is the multiplicative identity while a unit of a ring is an element of the ring having a multiplicative inverse in the ring. Of course, the unity element is a unit.

Integral Domains And Fields Classification With Solved Examples

Theorem. 3. In a ring R with unity, if a (6= 0) ∈ R has a multiplicative inverse, then it is unique.

Proof. Suppose that there exist b, b₀ ∈ R such that ab = ba = 1 and ab₀ = b₀a = 1.

Then ab = ab₀ = 1.

By the definition of cancellation law, b = b₀.

Example 1. It is a ring with a unity element = 1. We have 1.1 = 1 and (−1)(−1) = 1 for −1, 1 ∈ Z.

If a 6= ±1 ∈ Z then there exists no b ∈ Z such that ab = ba = 1.

Therefore, −1;1 are the only units in the ring Z.

Observe that any element is also a unit.

Example 2. Consider the ring Z7 = {0, 1, 2, 3, 4, 5, 6} under addition and multiplication modulo −7.

It has a unity element = 1 which is also a unit.

Further 24 = 4.2 = 1(mod7), 3.5 = 5.3 = 1(mod7) and 6.6 = 1(mod7). Thus every non-zero element is a unit.

Example 3. Consider the ring Z × Z = {(m, n); m, n ∈ Z}.

The unity element = (1, 1) which is also a unit: Also, (1, −1)(1, −1) = (1.1, (−1)(−1)) = (1, 1); (−1, 1)(−1, 1) = (1, 1) and (−1, −1)(−1, −1) = (1, 1)

Thus (1, 1), (1, −1), (−1, 1) and (−1, −1) are units in Z × Z.

Definition. (Division Ring or Skew Field) Let R be a ring with a unity element.

If every non-zero element of R is a unit then R is a Division Ring. (R, +,)isaDivisionring ⇔ (1)R is a ring, (2) R has a unity element and (3) every non-zero element in R is invertible under multiplication.

Example 1. ( Z, +,) is not a division ring, for, 2 6= 0 ∈ Z has no multiplicative inverse in Z.

Example 2. (Q, +, ·) and (R, +, ·) are division rings.

Example 3. The ring (M₂, +, ·) of non-singular 2 × 2 matrices is a division ring.

Definition. (Field) Let R be a commutative ring with a unity element. If every nonzero element of R is invertible under multiplication then R is a field.

Another Definition. A commutative ring with unity is called a field if every nonzero element is a unit.

(R, +, ·) is a field ⇔ (1)R is a ring, (2) R is commutative (3) R has a unity element and (4) every non-zero element of R is a unit. Usually, a field is denoted by the symbol F.

Note. 1 . A division ring which is also commutative is a field.

2. In a field, the zero element and the unity element are different. Therefore, a field has at least two elements.

Example 1. We know that (Q, +) where Q = the set of all rational numbers is an additive group and (Q − {0}, ·) is a multiplicative group. Further distributive laws hold. Therefore (Q, +, ·) is a field.

Example 2. (Z, +, ·) where Z = the set of all integers is not a field, because all non-zero elements of Z are not units.

Example 3. (Z₇, +, ·) where Z₇ = the set of integers under modulo −7 is a field.

Theorem. 4. A field has no zero – divisors.

Proof. Let (F, +, ·) be a field.

Let a, b ∈ F and a₆= 0.

a₆= 0 ∈ F, F is a field ⇒ there exists a−1 ∈ F such that aa−1 = a−1a = 1.

ab = 0. ⇒ a−1(ab) = a−10 ⇒ a−1a b = 0 ⇒ 1b = 0 ⇒ b = 0

Thus a, b ∈ R, a₆= 0 and ab = 0 ⇒ b = 0.

Similarly, we can prove that, a, b ∈ R, b₆= 0 and ab = 0 ⇒ a = 0.

∴ F has no zero divisors.

Note. A division ring has no zero divisors. (Write the proof of the theorem (3))

Theorem. 5. Every field is an integral domain.

Proof. Let (F, +, ·) be a field. Then the ring F is a commutative ring with unity and having every non-zero element as a unit.

But an integral domain is a commutative ring with unity and having no zero divisors. So, we have to prove that F has no zero divisors. (Write the proof of the above Theorem (3))

Note. The converse of the above theorem need not be true. However, an integral domain with a finite number of elements can become a field.

Theorem 6. Every finite integral domain is a field.

Proof. Let 0, 1, a₁, a₂, . . . . . . , a_n be all the elements of the integral domain D.

Then D has n + 2 elements which is finite. Integral domain D is a commutative ring with unity and having no zero divisors.

So, we have to prove that every non-zero element of D has a multiplicative inverse in D.

Let a ∈ D and a₆= 0

Now consider the n + 1 products a₁, aa₁, aa₂, . . . ., aan.

If possible, suppose that aai = aaj for i 6= j.

Since a₆= 0, by cancellation law, we have ai = aj.

This is a contradiction since i 6= j.

Therefore a₁, aa₁, aa₂, . . . . . . . . . , aan are (n + 1) distinct elements in D.

Since D has no zero divisors, none of these (n +1) elements is zero element.

Hence, by counting ;

a₁, aa₁, aa₂, . . . . . . .., aan are the (n+1) elements 1, a₁, a₂, . . . . . . , a_n in some order.

∴ a₁ = 1 or a = 1 or aai = 1 for some i.

For a₆= 0 ∈ D there exists b = ai ∈ D such that ab. = 1

⇒ a₆= 0 ∈ D has a multiplicative inverse in D.

∴ D is a field.

Theorem 7. If p is a prime then Z_p, the ring of integers modulo p, is a field.

Proof. In Example 4 on page 191, we proved that (Z_p, +, ·) is a ring.

Since Z

p = {0, 1, 2, . . . . . . , p − 1} has p distinct elements, and Z_p is a finite ring.

We prove now that Z_p is an integral domain.

Clearly, 1 ∈ Z_pis the unity element.

For a, b ∈ Z_p, ab(modp) ≡ ba(modp) ⇒ ab = ba and hence Zp is commutative.

For a, b ∈ Z_p and ab = 0 ⇒ ab ≡ 0(modp) ⇒ p|ab ⇒ p|a or p | b ( ∵ p is prime)

⇒ a ≡ 0(modp) or b ≡ 0(modp) ⇒ a = 0 or b = 0.

∴ Z_p has no zero divisors. Thus (Z_p, +, ·) is a finite integral domain .

∴ Z_p is a field.

Theorem. 8. If (Zn, +, ·) is a field then n is a prime number.

Proof. If possible let m be a divisor of n.

∴ there exists q ∈ Z such that n = mq. Clearly 1 ≤ m, q ≤ n.

mq = n ⇒ mq ≡ 0(modn). Since Zn is a field, Zn has no zero divisors.

∴ mq ≡ 0(modn) ⇒ m = 0(modn) or q = 0(modn)

⇒ m = n or q = n ⇒ m = n or m = 1( ∵ mq = n). ∴ n is a prime number.

Theorem. 9. Z p = {0, 1, 2, . . . , p − 1} is a field if and only if p is a prime number.

Proof. Write the proofs of Theorem 7 and Theorem 8.

Note. In the field Z_p= {0, 1, 2, . . . . . . , p − 1} where p is a prime, 1 and p − 1

are the only elements that are their own multiplicative inverses.

Rings, Integral Domains, And Fields Solved Problems

Example. 4. Find all solutions of x₂ − x + 2 = 0 over Z₃[1].

Solutions.

Given

x₂ − x + 2 = 0

We have Z₃ = {0, 1, 2} under modulo – 3 system. Z₃[1] = {a + ib | a, b ∈ Z3 and i₂ = −1 = {0, 1, 2, i, 1 + i, 2 + i, 2i, 1 + 2i, 2 + 2i}, containing 9 elements.

Let P(x) = x₂ − x+2. Then P(0) 6= 0, P(1) 6= 0, P(2) 6= 0, P(i) = −1− i+2 6= 0

P(1 + i) = (1 − 1 + 2i) − 1 − i + 2 6= 0, P(2 + i) = (4 − 1 + 4i) − (2 + i) + 2 6= 0, P(2i) = −4 6= 0

P(1+2i) = (1−4+4i)−(1+2i)+2 6= 0, P(2+2i) = (4−4+8i)−(2+2i)+2 6= 0

∴ x₂ − x + 2 = 0 has no solution over Z3[1].

Example. 5. Show that 1, p − 1 are the only elements of the field Z_p, p is prime, that are their own multiplicative inverses.

Solution. Observe that, in the Z_p field, x2 − 1 = 0 has only two solutions.

x₂ − 1 = 0 ⇒ x₂ = 1 ⇒ x = 1 ⇒ Multiplicate inverse of x = x.

So, we have to prove that 1, p − 1 are solutions of x₂ − 1 = 0 in Z_p.

1 ∈ Z

p ⇒ 12 − 1 = 1 − 1 = 0

p − 1 ∈ Zp ⇒ (p − 1) 2 − 1 = p2 − 2p + 1 − 1 = p2 − 2p

= p(p − 2) = 0(p − 2) = 0 ( ∵ p = 0(modp))

Example. 6. In a ring R with unity if a ∈ R has multiplicative inverse then a ∈ R is not a zero divisor.

Solution. a ∈ R has multiplicative inverse

⇒ There exists a−1 ∈ R, such that aa−1 = a−1a = 1, where 1 ∈ R is the unity element, To prove that a ∈ R is not a zero divisor we have to prove that

for b ∈ R so that ab = 0 or ba = 0 ⇒ b = 0 only.

ab = 0 ⇒ a−1(ab) = a−10 ⇒ 1b = 0 ⇒ b = 0; ba = 0 ⇒ (ba)a−1 = 0a−1 ⇒ b1 = 0 ⇒ b = 0

∴ a ∈ R is not a zero divisor.

Example. 7. Construct a field of two elements.

Solution. Let F = {0, 1} and addition ( +), multiplication ( ) in F be defined as follows :

+ 0 1

0 0 1

1 1 0

. 0 1

0 0 0

1 0 1

Clearly, + and – are binary operations in F.

We have 0 + 1 = 1 + 0 and 0.1 = 1.0 and hence +, · are commutative.

The two operations are associative.

0 ∈ F is the zero elements and 1 ∈ F is the unity element.

Clearly, distributivity is also true.

Additive inverse of 0 = 0, additive inverse of 1 = 1.

The multiplicative inverse of 16= 0 ∈ F is 1 .

Hence ({0, 1}, +, ·) is a field.

Example. 8. Show that the set R of all real-valued continuous functions defined on [0, 1] is a commutative ring with unity, with respect to addition (+) and multiplication ( •) of functions defined as (f + g)(x) = f(x) + g(x) and (f · g)(x) = f(x) · g(x)∀x ∈ [0, 1] and f, g ∈ R.

Solution. f, g are real-valued continuous functions on [0, 1] ⇒ (1)f + g and f.g are real-valued continuous functions on [0, 1] and (2) f(x), g(x) are real numbers for x ∈ [0, 1].

∴ The addition and multiplication of functions are binary operations in R.

Let f, g, h ∈ R ∀x ∈ [0, 1], ((f + g) + h)(x) = (f + g)(x) + h(x) = (f(x) + g(x)) + h(x)

= f(x) + (g(x) + h(x)) = f(x) + (g + h)(x) = (f + (g + h))(x)

∴ (f + g) + h = f + (g + h)∀f, g, h ∈ R

If O(x) = 0∀x ∈ [0, 1] then O is a real-valued continuous function.

Therefore there exists O ∈ R so that (f + O)(x) = f(x) + O(x) = f(x)∀x ∈ [0, 1] and f ∈ R.

If f is a real-valued continuous function on [0, 1] then ’ −f0 ’ is also a real-valued continuous function so that (−f)(x) = −f(x)∀x ∈ [0, 1].

Therefore for f ∈ R there exists −f ∈ R so that (f + (−f))(x) = f(x) − f(x) = 0 = 0(x)∀x ∈ [0, 1]

That is, additive inverse exists ∀f ∈ R. ∴ (R, +) is a commutative group.

∀x ∈ [0, 1]; ((fg)h)(x) = (fg)(x)h(x) = (f(x)g(x))h(x) = f(x)(g(x)h(x)) = f(x)(gh)(x) = (f(gh))(x)

∴ (fg)h˙ = f(gh)∀f, g, h ∈ R

∀x ∈ [0, 1]; (f(g + h))(x) = f(x)(g + h)(x) = f(x)(g(x) + h(x)) = f(x)g(x) + f(x)h(x) = (fg)(x) + (fh)(x) = (fg + fh)(x)

∴ f(g + h) = fg + fh∀f, g, h ∈ R

Similarly (g + h)f = gf + hf ∀f, g, h ∈ R. Hence (R, +, ·) is a ring.

∀x ∈ [0, 1], (fg)(x) = f(x)g(x) = g(x)f(x) = (gf)(x)

∴ fg = gf∀f, g ∈ R.

∴ R is a commutative ring.

The constant function e(x) = 1∀x ∈ [0, 1] is real-valued and continuous.

Also e ∈ R is such that (ef)(x) = e(x)f(x) = f(x)∀x ∈ [0, 1]

∴ e ∈ R defined as above is the unity element.

Example 9. Prove that the set Z[1] = a + bi | a, b ∈ Z, i ²= −1 of Gaussian integers is an integral domain with respect to the addition and multiplication of numbers. Is it a field?

Solution. Let Z(1) = {a + bi | a,b ∈ Z}.

Let x, y ∈ Z (i) so that x = a + bi,y = c + di where a,b,c,d ∈ Z

x + y = (a + c) + (b + d)i = a₁ + b₁ i where a₁ = a + c, b₁ = b + d ∈ Z

x,y = (ac − bd) + (ad + bc)i = a₂ + b₂ i where a₂ = ac − bd,b₂ = ad + bc ∈ Z

∴ +, · are binary operations in Z(1).

Since the elements of Z (1) are complex numbers we have that

Addition and multiplication are commutative in Z(1),

Addition and multiplication are associative in Z (1) and

Multiplication is distributive over addition in Z(1).

Clearly zero element = 0 + 0i = 0 and unity element = 1 + 0i = 1.

Further, for every x = a + ib ∈ Z(i) we have −x = (−a) + i(−b) ∈ Z(i) so that x + (−x) = {a + (−a)} + i{b + (−b)} = 0 + i0 = 0 ⇒ Additive inverse exists.

∴ Z(i) is a commutative ring with a unity element.

For x,y ∈ Z(i),x.y = 0 ⇒ x = 0 or y = 0 since x,y are complex numbers.

Hence Z(i) is an integral domain with a unity element.

For α = 3 + 4i 6= 0 ∈ Z(i) we have β = 253 − i 254 so that α · β˙ = 259 + 1625 + i −2512 + 1225 = 1 + i0 = 1. But β /∈ Z(i) as 253 , − 254 ∈/ Z.

So, every non-zero element of Z(i) is not invertible. ∴ Z(i) is not a field.

Example. 10. Prove that Q[√2] = {a + b√2 | a,b ∈ Q} is a field with respect to ordinary addition and multiplication of numbers.

Solution. Let x,y,z ∈ Q[√2] so that

x = (a₁ + b₁)√2,y = (a₂ + b₂)√2,z = (a₃ + b₃)√2 where a₁,b₁,a₂,b₂,a₃,b₃ ∈ Q

x+y = (a₁ + a₂)+(b₁ + b₂) √2 = a+b√2 where a₁ +a₂ = a,b₁ +b₂ = b ∈ Q

x · y = (a₁a₂ + 2b₁b₂)+(a₁b₂ + a₂b₁) √2 = c+d√2 where c = a₁a₂ +2b₁b₂ ∈ Q and d = a₁b₂ + a₂b₁ ∈ Q

∴ Addition (+) and multiplication ( ) of numbers are binary operations in Q[√2].

x + y = (a₁ + a₂) + (b₁ + b₂) √2 = (a₂ + a₁) + (b₂ + b₁) √2 = (a₂ + b₂)√2+ (a₁ + b₁)√2 = y + x ⇒ Addition is commutative.

(x + y) + z = (a₁ + a₂ + a₃) + (b₁ + b₂ + b₃) √2 and x + (y + z) = (a₁ + a₂ + a₃) + (b₁ + b₂ + b₃) √2 ⇒ (x + y) + z = x + (y + z) ⇒ Addition is associative.

For 0 ∈ Q we have 0 + 0√2 = 0 ∈ Q[√2] so that x + 0 = x for x ∈ Q[√2] ⇒ 0 ∈ Q[√2] is the zero element.

For x = (a₁ + b₁)√2 ∈ Q[√2] we have −x = (−a₁) + (−b₁) √2 ∈ Q[√2] so that x + (−x) = 0 ⇒ Additive inverse exists.

∴ (Q[√2], +) is a commutative group.

x · y = (a₁ + b₁)√2 · (a₂ + b₂)√2 = (a₁a₂ + 2b₁b₂) + (a₁b₂ + a₂b₁) √2 = (a₂a₁ + 2b₂b₁)+(a₂b₁ + b₂a₁) √2 = y·x ⇒ Multiplication is commutative.

(x · y) · z = (a₁a₂ + 2b₁b₂ + a₁b₂ + a₂b₁)√2 · (a₃ + b₃)√2 = (a₁a₂a₃ + 2b₁b₂a₃ + 2a₁b₂b₃ + 2a₃b₁b₃)+(a₁a₂b₃ + 2b₁b₂b₃ + a₁a₃b₂ + a₂a₃b₁) √2 and x · (y · z)

= (a₁ + b₁)√2 (a₂a₃ + 2b₂b₃ + a₂b₃ + a₃b₂)√2 = (a₁a₂a₃ + 2a₁b₂b₃ + 2a₂b₁b₃ + 2a 3b₁b₂)+(a₁a₂b₃ + a₁a₃b₂ + a₂a₃b₁ + 2b₁b₂b₃) √2

∴ (x, y), z = x.(y, z) ⇒ Multiplication is associative.

x · (y + z) = (a₁ + b₂)√2 (a₂ + a₃ + b₂ + b₃)√2 = (a₁a₂ + a₁a₃ + 2b₁b₂ + 2b₁b₃) + (a₁b₂ + a₁b₃ + a₂b₁ + a₃b₁) √2 and x·y+x, z = (a₁a₂ + 2b₁b₂ + a₁b₂ + a₂b₁)√2 +(a₁a₃ + 2b₁b₃ + a₁b₃ + a₃b₁)√2 = (a₁a₂ + 2b₁b₂ + a₁a₃ + 2b₁b₃) + (a₁b₂ + a₂b₁ + a₁b 3 + a₃b₁) √2

∴ x · (y + z) = x · y + x · z ⇒ Distributivity is true.

Hence (Q[√2], +, ·) is a ring.

1 = 1 + 0√2 ∈ Q[√2] so that x₁ = (a₁ + b₁)√2 (1 + 0√2) = x∀x ∈ Q[√2].

∴ Q[√2] is a commutative ring with a unity element.

To show that Q[√2] is a field we have to prove further every non-zero element in Q[√2] has a multiplicative inverse.

Let a + b√2 ∈ Q[√2] and a 6= 0 or b 6= 0

⇒ Then $\frac{1}{a+b \sqrt{2}}=\frac{a-b \sqrt{2}}{a^2-2 b^2}=\left(\frac{a}{a^2-2 b^2}\right)+\left(\frac{-b}{a^2-2 b^2}\right) \sqrt{2}$

⇒ since $a^2-2 b^2 \neq 0 for a \neq 0 or b \neq 0. a, b \in Q \Rightarrow \frac{a}{a^2-2 b^2}, \frac{-b}{a^2-2 b^2} \in Q$

⇒ For $a+b \sqrt{2} \neq 0 \in Q[\sqrt{2}]$

there exists $\left(\frac{a}{a^2-2 b^2}\right)+\left(\frac{-b}{a^2-2 b^2}\right) \sqrt{2} \in Q[\sqrt{2}]$

⇒ such $t(a+b \sqrt{2})\left[\left(\frac{a}{a^2-2 b^2}\right)+\left(\frac{-b}{a^2-2 b^2}\right) \sqrt{2}\right]=1=1+0 \sqrt{2}$

∴ Every non-zero element of Q[√2] is invertible.

Hence Q[√2] is a field. Ex.

Example11. If Z = the set of integers then prove that the set z × z = {(m, n) | m, n ∈ z} with respect to addition (+) and multiplication (•) defined as (m₁, n₁) + (m₂, n₂) = (m₁ + m₂, n₁ + n₂) and (m₁, n₁) · (m₂, n₂) = (m₁m₂, n₁n₂) ∀ (m₁, n₁) , (m₂, n₂) ∈ z × z is a ring and not an integral domain.
Solution.

Let x = (m₁, n₁) , y = (m₂, n₂) , z = (m₃, n₃) ∈ z×z so that m₁, n₁, m₂, n₂, m₃, n₃ ∈ Z

(1) x + y = (m₁, n₁) + (m₂ + n₂) = (m₁ + m₂, n₁ + n₂) ∈ z × z x · y = (m₁, n₁) · (m₂, n₂) = (m₁m₂, n₁n₂) ∈ z × z as m₁ + m₂, n₁ + n₂, m₁m₂, n₁n₂ ∈ z

∴ + and · are binary operations in z × z

(2) x + y = (m₁ + m₂, n₁ + n₂) = (m₂ + m₁, n₂ + n₁) = y + x and x.y = (m₁m₂, n₁n₂) = (m₂m₁, n₂n₁) = yx ⇒ + and · are commutative.

(3) (x + y) + z = (m₁ + m₂, n₁ + n₂) + (m₃, n₃) = (m₁ + m₂ + m₃, n₁ + n₂ + n₃) = (m₁ + m₂ + m₃, n₁ + n₂ + n₃) = x + (y + z) (x · y) · z = (m₁m₂, n₁n₂) · (m₃, n₃) = ((m₁m₂) m₃, (n₁, n₂) n₃) = (m₁ (m₂m₂) , n₁ (n₂n₃)) = x.(y.z). ⇒ + and – are associative.

(4) x · (y + z) = (m₁, n₁) · (m₂ + m₃, n₂ + n₃) = (m₁ (m₂ + m₃) , n₁ (n₂ + n₃)) = (m₁m₂ + m₁m₃, n₁n₂ + n₁n₃) = (m₁m₂, n₁n₂) + (m₁m₃, n₁n₃) = x · y + x · z

Since multiplication is commutative, (y + z) · x = y · x + z · x

∴ Distributivity is true.

(5) For 0 ∈ z we have (0, 0) ∈ z×z and (m, n)+(0, 0) = (m+0, n+0) = (m, n)

∴ (0, 0) ∈ z × z is the zero elements.

(6) For 1 ∈ z we have (1, 1) ∈ z × z and (m, n).(1, 1) = (m · 1, n · 1) = (m, n)

∴ (1, 1) ∈ z × z is the unity element. Hence z × z is a commutative ring with unity.

But we have, (0, 1), (1, 0) ∈ z × z and (0, 1) 6= (0, 0), (1, 0) 6= (0, 0) such that (0, 1) · (1, 0) = (0 · 1, 1 · 0) = (0, 0)

∴ (0, 1), (1, 0) are zero divisors in z× z. Hence z× z is not an integral domain.

Rings, Integral Domains, And Fields Exercise 2

1. List all zero divisors in the ring Z₂₀. Also, find the units in Z₂₀. Is there any relationship between zero divisors and units?

2. Solve the equation 3x = 2 in (a) Z₇ (b)Z₂₃

3.

Find all solutions of x³ − 2x² − 3x = 0 in Z_12.

Find all solutions of x² + x − 6 = 0 in Z₁₄.

4. Describe all units in (a) Z₄(b) Z₅. Prove that Z₂× Z₂ = {(0, 0), (0, 1), (1, 0), (1, 1)} under componentwise addition and multiplication is a Boolean ring.

5. Find all solutions of a₂ + b₂ = 0 inZ₇.

6. Write the multiplication table for Z₃[1] = {0, 1, 2,i, 1+i, 2+i, 2i, 1+2i, 2+ 2i}.

7. R is a set of real numbers. Show that R × R forms a field under addition and multiplication defined by (a,b) + (c,d) = (a + c,b + d) and (a,b) : (c,d) = (ac − bd,ad + bc) is a field. (Hint. R × R = C = a + ib | a,b ∈ R,i2 = −1 )

9. If Z is the set of all integers and addition ⊕, multiplication (x) is defined in Z as a ⊕ b = a + b − 1 and a × b = a + b − ab∀a,b ∈ Z then prove that (Z, ⊕, ×) is a commutative ring.

10. Let (R, +) be an abelian group. If multiplication (·) in R is defined as a · b = 0, 0000 is the zero element in R, ∀a,b ∈ R then prove that (R, +, ·) is a ring.

11. If R = {0, 1, 2, 3, 4} prove that (R, +5, x5) under addition and multiplication modulo – 5 is a field.

12. Give examples of (1) a commutative ring with unity (2) an integral domain and (3) a Division ring.

13. If R = the set of all even integers and (+) is ordinary addition and multiplication (∗) is defined as a ∗ b = ab₂ ∀a,b ∈ R then prove that (R, +, ∗) is
a commutative ring.

14. S is a non-empty set containing n elements. Prove that P(S) forms finite Boolean ring w.r.t ’+’ and ’ ’ ’defined as A + B = (A ∩ B) − (A ∪ B) and
A · B = A ∩ B∀A, B ∈ P(S). Find addition and multiplication tables when S = {a,b}.

15. If R₁, R₂,……, R_n are rings, then prove that R₁ × R₂ × ……. R_n = {(r₁,r₂,……, r_n) | ri ∈ Ri} forms a ring under componentwise addition and multiplication, that is, (a₁,a₂,……, a_n)+(b₁,b₂,……, b_n) = (a₁ + b₁,a₂ + b₂,…, an + bn) and (a₁,a₂,……, a_n) · (b₁,b₂,……, b_n) = (a₁b₁,a₂b₂,…., a_nb_n)

Rings, Integral Domains & Fields Integral Multiples And Integral Powers Of An Element

Integral multiples: Let (R, +, ·) be a ring and a ∈ R. We define 0a = O where ’ 0 ’ is the integer and O is the zero element of the ring

If n ∈ N we define na = a + a + . . . + a, (n terms ).

If n is negative integer, na = (−a) + (−a) + . . . + (−a), (−n terms ).

(−n)a = (−a) + (−a) + . . . + (−a), n terms = n(−a) = −(na) where n ∈ N.

The set {na | n ∈ Z, a ∈ R} is called the set of integral multiples of an element ’ a ’. It may be noted that for n ∈ Z, a ∈ R we have na ∈ R. Theorem.

If m, n ∈ Z and a, b ∈ R, a ring, then (1)(m + n)a = ma + an, (2) m(na) = (mn)a, (3) m(a + b) = ma + mb and (iv)m(ab) = (ma)b. (Proof is left as an exercise )

Note 1. If the ring R has unity element then for n ∈ Z and a ∈ R we have na = (n1)a = (n1)a

2. If m, n ∈ Z and a, b ∈ R, a ring then we have (ma)(nb) = m{a(nb)} = m{(na)b} = m{n(ab)} = (mn)(ab).

Integral powers: Let (R, +, ) be a ring and a ∈ R.

For n ∈ N we write an = a.a . . . a(n times ).

It may be noted that an = an − 1.a.

Theorem. If m, n ∈ N and a, b ∈ R, a ring then

am · an = am +n and

(am)n = amn (Proof is left as an exercise )

Rings, Integral Domains, And Fields Idempotent Element And Nilpotent Element Of A Ring

Definition. In a ring R, if a 2 = a for a ∈ R then ’ a ’ is called an idempotent element of R with respect to multiplication.

Theorem. 1. If a 6= 0 is an idempotent element of an integral domain with unity then a = 1.
Proof. Let (R, +, ·) be an integral domain.

a₆= 0 ∈ R is an idempotent element ⇒ a 2 = a

∵ a₂ = a₁ ( ∵ al = a)

⇒ a₂ − a₁ = 0 ⇒ a(a − 1) = 0 (’ 0 ’ is the zero element)

⇒ a − 1 = 0 since R has no zero divisors ⇒ a = 1.

Note. 1. An integral domain with unity contains only two idempotent elements ’ 0 ’ and ’1’.

2. A division ring contains exactly two idempotent elements.

3. The product of two idempotent elements in a commutative ring R is idempotent.

For, (ab)² = (ab)(ab) = a(ba)b = a(ab)b = (aa)(bb) = a²b² = ab for a, b ∈ R which are idempotent:

Definition. Let R be a ring and a 6= 0 ∈ R. If there exists n ∈ N such that an = 0 then ’ a ’ is called a nilpotent element of R.

Theorem. 2. An integral domain has no nilpotent element other than zero.

Proof. Let R be an integral domain and a₆= 0 ∈ R.

we have a 1 = a₆= 0, a₂ = a · a₆= 0 since R has no zero divisors.

Let an ≠ 0 for n ∈ N.

Then an +1 = an. a₆= 0, since R has no zero divisors.

∴ By induction, a₆ = 0 for every n ∈ N.

Hence a₆= 0 ∈ R is not a nilpotent element.

Example1. In the ring (Z₆, +, ·), 3 and 4 and idempotent elements, for 32 = 3 and 42 = 4.

Example 2. In the ring (Z₈, +, ·), there are no idempotent elements.

Example 3. In the ring (Z₈, +, ·), 2 and 4 are nilpotent elements, for 23 = 0 and 42 = 0.

Example 4. In the ring Z₆, +˙, there are no nilpotent elements.

Example. 1. If a,b are nilpotent elements in a commutative ring R then prove that a + b, a · b are also nilpotent elements.

Solution.

Given

a,b are nilpotent elements in a commutative ring R

a,b ∈ R are nilpotent elements

⇒ there exists m,n ∈ N such that am = 0,bn = 0. We have

⇒ (a + b)m+n = am+n + (m + n)C1 · am+n−1 · b + … + (m + n)Cn · am · bn + … + … + bm+n = am an + (m + n)C1an−1 · b + (m + n)C2 · an−2 · b2 + … + (m + n)Cn · bn + (m + n)Cn+1 · am−1 · b + (m + n)Cn+2 · am−2 · b2 + … + bm bn = 0 ( ∵ am = 0 = bn.)

Also, (ab)mn = amn · bmn = (am)n (bn)m = 0 ( ∵ R is commutative ).

∴ a + b and a · b are nilpotent elements.

Rings, Integral Domains And Fields Characteristic Of A Ring

Definition. The characteristic of a ring R is defined as the least positive integer p such that pa = 0 for all a ∈ R. In case such a positive integer p does not exist then we say that the characteristic of R is zero or infinite.

Note.

1. If R is a ring and Z = {n ∈ N | na = 0∀a ∈ R} 6= φ then the least element in Z is the characteristic of R.

2. If the ring R has characteristic zero then ma = 0 ’ where a 6= 0 can hold only if m = 0.

3. If the characteristic of a ring R is not zero then we say that the characteristic of R is finite.

4. As the integral domain, division ring and field are also ring characteristics that have meaning for these structures. Imp. If for some a ∈ R, pa 6= 0 then characteristic of R 6= p. Characteristic of a ring R = p ⇒ pa = 0∀a ∈ R.

Example 1 . R = {0, 1, 2, 3, 4, 5, 6} = Z7 is a ring under addition and multiplication modulo 7. Zero element of R = 0.

∀a ∈ R we have 7a ≡ 0(mod7) ⇒ 7a = 0∀a ∈ R. Further for 1 ∈ R, p(1) = p 6= 0 where p 6= 0 and 0 < p < 7.

∴ 7 is the least positive integer so that 7a = 0∀a ∈ R ⇒ Characteristic of R = 7.

Example. 2. The characteristic of the ring (Z, +, ·) is zero. For, there is no positive integer n so na = 0 for all a ∈ Z.

Example3. If R 6= {0} and the characteristic of R is not zero then the characteristic of R > 1.

Characteristic of R = 1 ⇒ 1a = 0∀a ∈ R ⇒ a = 0∀a ∈ R ⇒ R = {0}.

Example 4. For any element x ∈ Z3[1] ring, we have 3x = 0∀x ∈ Z3[1] ⇒ characteristic of Z3[i] = 3.

Example. 5. In the ring, R = {0, 3, 6, 9} ⊂ Z12, 4x = 0∀x ∈ R, and ’ 4 ’ is the least positive integer.

∴ Characteristic of R = {0, 3, 6, 9} = 0

Theorem 1. If R is a ring with a unity element, then R has characteristic p > 0 if and only if p is the least positive integer such that p₁ = 0.
Proof.

Let characteristic of R = p(> 0)

By definition, pa = 0∀a ∈ R. In particular p₁ = 0.

Conversely, let p be the least positive integer such that p₁ = 0.

∴ q < p and q ∈ N ⇒ q₁ 6= 0. Then for any a ∈ R we have

p · a = a + a + . . . + a(p terms ) = a(1 + 1 + . . . + 1) = a(p₁) = a0 = 0.

∴ p is the least positive integer so that p.a = 0∀a ∈ R. ∴ Characteristic of R = p.

Theorem 2. The characteristic of a ring with a unity element is the order of the unity element regarded as a member of the additive group.

Proof. Let (R, +, ·) be a ring so that (R, +) is its additive group.

Case 1. Let O(1) = 0 when the unity, element 1 is regarded as an element of (R, +). By the definition of the order of an element in a group, there exists no positive integer n so that n₁ = 0.

∴ Characteristic of R = 0.

Case 2. Let O(1) = p(6= 0).

By the definition of the order of elements in a group, p is the least positive integer,

so p₁ = 0. For any a ∈ R, pa = p(1a) = (pl)a = 0a = 0

∴ Characteristic of R = p.

For example., For the commutative ring Z× Z, the zero element = (0, 0) and the unity element = (1, 1). By the definition of the order of an element in the additive group Z × Z, there exists no positive integer m such that m(1, 1) = (m, m) = (0, 0).

Therefore, the characteristic of Z × Z is zero.

Theorem 3. The characteristic of an integral domain is either a prime or zero.

Proof. Let (R, +, ·) be an integral domain. Let the characteristic of R = p(6= 0).

If possible, suppose that p is not a prime.

Then p = mn where 1 < m, n < p. a 6= 0 ∈ R ⇒ a · a = a2 ∈ R and a2 6= 0 ( ∵ R is integral domain )

pa 2 = 0 ⇒ (mn)a 2 = 0 ⇒ (ma)(na) = 0 ⇒ ma = 0 or na = 0 ( ∵ R is integral domain )

Let ma = 0. For any x ∈ R, (ma)x = 0 ⇒ a(mx) = 0 ⇒ mx = 0 ( ∵ a 6= 0)

This is absurd, as 1 < m < p and characteristic of R = p.

∴ ma 6= 0. Similarly, we can prove that na 6= 0.

This is a contradiction and hence p is a prime.

Theorem 4. The characteristic of a field is either a prime or zero.

Proof. Since every field is an integral domain, by the above theorem the characteristic of a field is either a prime or zero.

Note.

1. The characteristic of a division ring is either a prime or zero.
2. The characteristic of Z p, where p is a prime, is p.

Rings, Integral Domains & Fields Solved Problems

Example. 2. The characteristic of an integral domain ( R, +, ·) is zero or a positive integer according as the order of any non-zero element of R regarded as a member of the group ( R, +).

Solution. Let a ∈ R and a 6= 0.

Case (1). Let O(a) = 0 when ’ a ’ is regarded as a member of (R, +).

By the definition of order, there exists no positive integer n so that na = 0.

∴ Characteristic of R = 0.

Case (2). Let O(a) = p.

By the definition of order, p is the least positive integer, so that pa = 0.

For any x ∈ R, pa = 0 ⇒ (pa)x = 0x ⇒ a(px) = 0 ⇒ px = 0 since a 6= 0.

∴ p is the least positive integer so that px = 0∀x ∈ R.

Hence characteristic of R = p.

Example. 3. If R is a non-zero ring so that a 2 = a∀a ∈ R proves that the characteristic of R = 2 or proves that the characteristic of a Boolean ring is 2.

Solution. Since a₂ = a∀a ∈ R, we have (a + a)₂ = a + a

⇒ (a + a)(a + a) = a + a ⇒ a(a + a) + a(a + a) = a + a

⇒ a₂ + a₂ + a₂ + a₂ = a + a ⇒ (a + a) + (a + a) = (a + a) + 0 ⇒ a + a = 0 ⇒ 2a = 0.

∴ for every a ∈ R, we have 2a = 0. Further for a 6= 0, 1a = a 6= 0.

∴ 2 is the least positive integer so 2a = 0∀a ∈ R.

Hence characteristic of R = 2.

Example. 4. Find the characteristic of the ring Z₃ × Z₄.
Solution.

The characteristic of the ring Z₃ × Z₄

We have Z₃ = {0, 1, 2}, Z₄ = {0, 1, 2, 3} Z₃ × Z₄= {(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3), (2, 0), (2, 1), (2, 2), (2, 3)} contains 12 ordered pairs as elements. Zero element (0, 0) and unity element = (1, 1)

We have, 1(1, 1) = (1, 1) ≠ (0, 0); 2(1, 1) = (2, 2) ≠ (0, 0);

3(1, 1) = (3, 3) = (0, 3)≠  (0, 0); 4(1, 1) = (4, 4) = (1, 0) ≠  (0, 0);

5(1, 1) = (5, 5) = (2, 1)≠ (0, 0); 6(1, 1) = (6, 6) = (0, 2) ≠  (0, 0);

7(1, 1) = (7, 7) = (1, 3) ≠  (0, 0); 8(1, 1) = (8, 8) = (2, 0)≠  (0, 0);

9(1, 1) = (9, 9) = (0, 1)≠  (0, 0); 10(1, 1) = (10, 10) = (1, 2)≠  (0, 0);

11(1, 1) = (11, 11) = (2, 3) ≠  (0, 0); 12(1, 1) = (12, 12) = (0, 0);

L Least positive integer = 12. Hence characteristic of Z₃ × Z₄ = 12.

Also, G. C. D of 3, 4 = (3, 4) = 1 ⇒ the additive group Z₃ × Z₄ is isomorphic with Z₁₂

∴ Characteristic of Z₃ × Z₄ = Characteristic of Z₁₂ = 12.

Example. 5. If the characteristic of a ring is 2 and the elements a, b of the ring commute prove that (a + b)₂ = a₂ + b₂ = (a − b)₂.

Solution. Since characteristic of the ring R = 2 ⇒ 2x = 0∀x ∈ R.

a, b ∈ R commute ⇒ ab = ba.

(a+ b)₂ = (a+ b)(a+ b) = a(a+ b)+ b(a+ b) = a₂ + ab+ ba+ b₂ = a₂ +2ab+ b₂

a, b ∈ R ⇒ ab ∈ R and 2(ab) = 0. ( ∵ characteristic of R = 2)

(a + b)₂ = a₂ + 0 + b₂ = a₂ + b₂.

Similarly, we can prove that (a − b)₂ = a₂+ b₂.

Example. 6. If R is a commutative ring with unity of characteristic = 3 then prove that (a + b)₃ = a₃ + b3∀a, b ∈ R

Solution. R is a ring with characteristic = 3 ⇒ 3x = 0, zero elements of R∀x ∈ R.

Since R is a commutative ring, by Binomial Theorem, (a + b)₃ = a₃ +3a₂b + 3ab₂ + b₃

a, b ∈ R ⇒ a2b, ab₂ ∈ R ⇒ 3a₂b = 0, 3ab₂ = 0. ∴ (a + b)₃ = a₃ + b₃.

Rings, Integral Domains, And Fields Exercise 3

1. Prove that the characteristic of the ring Zn = {0, 1, 2, . . . , n − 1} under addition and multiplication modulo n, is n.

2. Prove that the characteristic of a field is either prime or zero.

3. Prove that the characteristic of a finite integral domain is finite.

4. Prove that any two non-zero elements of an integral domain regarded as the members of its additive group are of the same order.

5. Give examples of a field with zero characteristics and a field with characteristics.

6. Find the characteristics of the rings (1) 2Z (2) Z × Z

7. If R is a commutative ring with unity of characteristic = 4 then simplify (a + b) 4 for all a, b ∈ R.

8. If R is a commutative ring with unity of characteristic = 3 compute and simplify (1)(x + y) 6(2)(x + y) 9∀x, y ∈ R

Answers

5. Zn ring, Z5 ring
6. (1) 0 (2) 0 7. a4 + a3b + ab3 + b4
7. (1)x6 + 2x3y3 + y6 (2) x9 + y9

Rings, Integral Domains & Fields Divisibility Units, Associates And Primes In A Ring.

Definition. (Divisor or Factor) Let R be a commutative ring and a 6= 0, b ∈ R. If there exists q ∈ R such that b = aq then a’ is said to divide ’ b ’.

Notation. ’ a ’ divides ’ b ’ is denoted by a | b and a ’ does not divide ’ b ’ is denoted by aχb.

Note.

1. If ’ a ’ divides ’ b ’ then we say that a ’ is a divisor or factor of ’ b ’.

2. For a₆= 0, 0 ∈ R we have a · 0 = 0 and hence every non-zero element of a ring R is a divisor of 0 ’ 0 ’ Zero element of R.

3. a₆= 0, b ∈ R and a | b ⇔ b = aq for some q ∈ R.

For example., 1. In the ring Z of integers, 3115 and 317.

For example., 2. In the ring Q of rational numbers, 3/7 because there exists (7/3) ∈ Q such that 7 = 3.(7/3),

For example., 3. In a field F, two non-zero elements are divisors to each other.

For example., 4. In the ring Z₆, 4 | 2, In the ring Z₈, 3 | 7 and in the ring Z₁₅, 9 | 12.

For example., 5. The unit of a ring R divides every element of the ring if a ∈ R is a unit then aa −1 = a−1a = 1 where a−1 ∈ R.

For any b ∈ R we have b = 1b = aa−1b = a a−1b ⇒ a | b.

Theorem. 1. If R is a commutative ring with unity and a b, c ∈ R then

a | a (2) a | b and b|c → a|c

a|b → a|bx∀x ∈ R

a | b and a|c ⇒ a|bx + cy∀x, y ∈ R.

Proof.

(1) If 1 ∈ R is the unity element in R then a = a.1 which implies that a | a.

(2) a | b ⇒ b = aq₁ for some q₁ ∈ R; b | c ⇒ c = bq₂ for some q₂ ∈ R.

Now c = bq₂ = (aq₁) q₂ = a (q₁ · q₂) = aq where q = q₁q₂ ∈ R ⇒ a | c.

(3) a | b ⇒ b = aq for some q ∈ R.

Now bx = (aq)x = a(qx) = aq₁ where q₁ = qx ∈ R ⇒ a | bx.

(4) a|b ⇒ a|bx∀x ∈ R; a|c ⇒ a|cy∀y ∈ R

a | bx ⇒ bx = aq₁ for some q₁ ∈ R; a | cy ⇒ cy = aq₂ for some q₂ ∈ R.

∴ bx+cy = aq₁+aq₂ = a (q₁ + q₂) = aq where q = q₁+q₂ ∈ R ⇒ a | (bx+cy).

Note. a | b and a|c ⇒ a|b± c. Definition. (Greatest Common Divisor G.C.D)

Let R be a commutative ring and a, b ∈ R, d ∈ R is said to be the greatest common divisor of ’ a ’ and ’ b ’ if

d | a and d | b and

whenever c | a and c | b

where c ∈ R then c | d. Notation. If ’ d ’ is a greatest common divisor (G. C. D) of ’ a ’ and ’ b ’ then we write d = (a, b).

Definition. (Unit) Let R be a commutative ring with unity. An element a ∈ R is said to be a unit in R if there exists an element b ∈ R such that ab = 1 in R. However, the unity element ’ 1 ’ is also a unit because 1, 1 = 1.

2. In a ring, the unity element is unique, while, units may be more than one.

3. If ab = 1 then a− 1 = b. So, a unit in a ring R is an element of the ring so its multiplicative inverse is also in the ring. That is a ∈ R is a unit of R means that the element d a is invertible.

4. Units of a ring are in fact the two divisors of the unity element in the ring.

5. ’ a ’ is a unit in R ⇒ ab = 1 for some b ∈ R ⇒ ’ b ’ is also a unit of R. example. In a field F, every non-zero element has a multiplicative inverse. So, every non-zero element in a field is a unit.

Theorem 2. Let D be an integral domain. For a, b ∈ D, if both a | b and b | a are true then a = ub where u is a unit in D.

Proof. a | b ⇒ b = aq₁ for some q₁ ∈ D; b | a ⇒ a = bq₂ for some q₂ ∈ D.

b = aq₁ = (bq₂) q₁ = b (q₂q₁) ⇒ 1 = q₂q₁, by using cancellation property in integral domain.

∴ q₂q₁ = 1 ⇒ q₂ is a unit in D.

Hence a = bq₂ where q₂ is a unit in D.

Definition. (Associates) Let R be a commutative ring with unity. Two elements ’ a ’ and ’ b ’ in R are said to be associates if b = ua for some unit u in R.

Note. The relation of being associated in a ring R is an equivalence relation in R.

For example., 1. If ’ 1 ’ is the unity element in the ring R then ’ 1 ’ is a unit in R. For a(6= 0) ∈ R we have a = 1. a and a, a are associates in R.

For example., 2. In the ring Z of integers, the units are 1 and −1 only. For a₆= 0 ∈ Z, we have a = a : 1 and a = (−a)(−1) only. Therefore, a ∈ Z has only two associates, namely, a, −a.

For example., 3. In the ring Z₆ = {0, 1, 2, 3, 4, 5} of integers modulo −6, the units are 1,5 only.

For 2 ∈ Z₆; 2 ≡ 2.1(mod6) and 2 ≡ 4.5(mod6)

∴ 2 has two associates 2,4.

Theorem. 3. In an integral domain D, two non-zero elements a, b ∈ D are associates iff a | b and b | a.

Proof. From Theorem (2) we see that a | b and b | a

⇒ there exists unit u ∈ D such that a = ub ⇒ a, b are associates. a, b are associates in D ⇒ there exists unit u in D such that a = ub ⇒ b | a. u is unit in D ⇒ there exists unit v ∈ D such that uv = 1.

Now a = ub ⇒ va = v(ub) ⇒ va = (vu)b ⇒ va = (1) b ⇒ b = va ⇒ a | b.

Hence a | b and b | a.

Definition. (Trivial Divisors and Proper Divisors)

Let a 6= 0 be an element in the integral domain D. The units in D and the associates of ’ a ’ are divisors of ’ a ’. These divisors of ’ a ’ is called Trivial divisors of ’ a ’. The remaining divisors of ’ a ’ are called the proper divisors of ’ a ’.

For example., Consider the integral domain (Z, +, ·). The units in Z are 1 and −1 only.

For a₆= 0 ∈ Z, the trivial divisors are 1, −1, a, −a only. The remaining divisors of ’ a ’ are proper divisors.

3 ∈ Z has only trivial divisors ±1, ±3 and no proper divisors.

6 ∈ Z has trivial divisors ±1, ±6 and also proper divisors ±2, ±3.

Definition. (Prime and Composite elements)

Let ’ a ’ be a non-zero and non-unit element in an integral domain D. If ’ a ’ has no proper divisors in D then ’ a ’ is called a prime element in D. If ’ a ’ has proper divisors in D then ’ a ’ is called Composite element in D.

Note. a ∈ D is a prime element and a = bc then one of b or c is a unit in D.

For example., 1. In the integral domain (Z, +, ·);

5 ∈ Z is a prime element and 6 ∈ Z is a composite element.

For example., 2. In the integral domain (Q, +, ·); 6 ∈ Q is the prime element since all its divisors are units.

Rings, Integral Domains, And Fields Solved Problems

Example.1. Find all the units of Z₁₂ the ring of residue classes modulo 12.
Solutions.

We have Z₁₂ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}.

Clearly, the unity element = 1 is a unit. a¯ ∈ Z₁₂ is a unit if there exists ¯b ∈ Z₁₂ such that a¯¯b = 1.

For ~a = even there is no ¯b so that a¯¯b ≡ 1(mod12), as a¯¯b, is even. So we have to verify for a¯ = odd.

For 5 ∈ Z₁₂ we have 5 × 5 = 1; 7 ∈ Z₁₂we have 7 × 7 = 1 and 11 ∈ Z₁₂. we have 11 × 11 = 1. ∴ 1, 5, 7, and 11 are the units in Z₁₂

Example.2. Prove that ±1, ±i are the only four units in the domain of Gaussian integers.

Solution. Z[1] =a + ib | a,b ∈ Z, i²= −1 is the integral domain of Gaussian integers. 1 + 0i = 1 is the unity element. Let x + iy ∈ Z[i] be a unit. By the definition, there exists u + iv ∈ Z[1] such that (x + iy)(u + iv) = 1 ⇒ |(x + iy)(u + iv)| = 1 ⇒ x² + y² u² + v²= 1

⇒ x² = 1,y² = 0 or x² = 0 or y² = 1 ⇒ x = ±1,y = 0 or x = 0,y = ±1.

∴ ±1 + 0i, 0 ± 1 i i.e., ±1; ±i are the possible units.

Example.3. Find all the associates of (2 − i) in the ring of Gaussian integers. (N. U. 97).

Solution. We have 2 − i = (2 − i) · 1; 2 − i = (−2 + i) · (−1); (2 − i) = (−2i − 1) · I and 2 − i = (2i + 1) · (−i).

∴ 2 − i, −2 + i, −2i − 1 and 2i + 1 are the associates.

Example 4. In the domain of Gaussian integers, prove that the associates of a + ib are a + ib, −a − ib, ia − b, −i a + b.
Solution.

Since ±1 and ±i are the four units of Z[1], a + ib = (a + ib).1; a + ib = (−a − ib) · (−1); a + ib = (ia − b) · (−i) and a + ib = (−i a + b) · i

∴ a + ib, −a − ib, i a − b, and −i a + b are the associates of a + ib.

Example. 5. If D is an integral domain and U is a collection of units in D, Prove that (U, ·) is a group.
Solution. (Left to the reader)

Example. 6. Find all units of Z₁₄.
Solution.

Z14 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14} Unity element = 1 is a unit. Since 14 is even, the number in Z14 cannot be a unit.

For 3 ∈ Z₁₄. we have 3.5 = 15 = 1 ⇒ 3, 5 are units.

For 9 ∈ Z_14, we have 9.11 = 99 = 1 ⇒ 9, 11 are units.

For 13 ∈ Z₁₄ we have 13.13 = 169 = 1 ⇒ 13 is a unit.

Example. 7. Find all the units in the matrix ring M₂ (Z₂)
Solution.

We have Z₂ = {0, 1} so that 0 + 0 = 0, 0 + 1 = 1 + 0 = 1 and 1 + 1 = 0.

⇒ $\mathrm{M}_2\left(\mathrm{Z}_2\right)=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right) where a, b, c, d \in\{0,1\}$

⇒ Number of elements in $M_2\left(Z_2\right)=2^4=16$

⇒ Clearly $\mathrm{I}_2=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$ is the unity element and hence an unit.

A ∈ M₂ (Z₂) is a unit in M₂ if there exists a B ∈ M₂ such that AB = I2 the unity element. AB = I2 happens when A is non-singular and B = A−1.

Hence the · · · its of M₂ (Z₂) are all the non-singular matrices. Matrices having only one ’ 0 ’ and three ’ 1 ’s are :

⇒ $\left(\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right),$

⇒ $\left(\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right),$

⇒ $\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right),$

⇒ $\left(\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right)$

which are non-singular Hence the above 4 matrices are units. Matrices having two ‘ 0 ‘s and two 1 “s are

⇒ $\left(\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right),$

⇒ $\left(\begin{array}{ll}
0 & 0 \\
1 & 1
\end{array}\right),$

⇒ $\left(\begin{array}{ll}
1 & 0 \\
1 & 0
\end{array}\right),$

⇒ $\left(\begin{array}{ll}
0 & 1 \\
0 & 1
\end{array}\right),$

⇒ $\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right),$

⇒ $\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right)$

Among the above six matrices of

⇒ $\mathrm{M}_2 ;\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)$

⇒ $\left(\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right)$

are only nonsingular. Hence these two matrices are units.

Matrices having three ‘0’s and one ‘1 are:

⇒ $\left(\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right),$

⇒ $\left(\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right),$

⇒ $\left(\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right),$

⇒ $\left(\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right)$

which are all singular.

The zero matrix

⇒ $\mathrm{O}=\left(\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right)$

and the matrix having all ‘l’s

⇒ $\left(\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right)$

are both singular. Hence the units of

⇒ $\mathrm{M}_2\left(\mathrm{Z}_2\right) $are

⇒ $\left(\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right),$

⇒ $\left(\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right),$

⇒ $\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right),$

⇒ $\left(\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right),$

⇒ $\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$

⇒ $\text { and }\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right) \text { which are six in number. }$

Rings, Integral Domains, And Fields Some Noncommutative Examples.

There are many rings that are not Commutative under multiplication. We study three non-commutative rings, namely, the ring of square matrices over a field, the ring of endomorphisms of an abelian group, and the Quaternions.

Rings, Integral Domains And Fields Solve Problems

Example 1. Prove that the set of all 2 × 2 matrices over the field of Complex numbers is a ring with unity under addition and multiplication of matrices.

Solution. Let $R=\left\{\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]: a, b, c, d \in \mathrm{C}\right\} be the set of 2 \times 2 matrices over \mathrm{C}$
Let A=

⇒ $\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]=\left[a_{i j}\right]_{2 \times 2}, B=\left[\begin{array}{ll}b_{11} & b_{12} \\ b_{21} & b_{22}\end{array}\right]=\left[b_{i j}\right]_{2 \times 2} and C=$

⇒ $[\left[\begin{array}{ll}
c_{11} & c_{12} \\
c_{21} & c_{22}
\end{array}\right]=\left[c_{i j}\right]_{2 \times 2} $

be three elements in \mathrm{R}.

(1) A + B = [aij] 2× 2 + [bij] 2× 2 = [aij + bij] 2× 2 and

A + B = [bij + aij] 2× 2 = B + A ( ∵ aij,bij ∈ C)

∴ Addition is a binary operation and also Commutative. (2) (A + B) + C = [aij + bij] 2× 2 + [cij] 2× 2 = [(aij + bij) + cij] 2× 2 = [aij + (bij + cij)] 2× 2 = A + (B + C) ( ∵ aij,bij,cij ∈ C)

∴ Addition is associative.

⇒ (3) We have O = $\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=[0]_{2 \times 2} \in \mathrm{R} such that A+\mathrm{O}=$

⇒ $\left[a_{i j}+0\right]_{2 \times 2}= \left.a_{i j}\right]_{2 \times 2} = A$

= Therefore O = $\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$ is the Zero element.

(4) For A − [aij] 2× 2 ,aij ∈ C ⇒ −aij ∈ C so that aij + (−aij) = o ∈ C.

∴ there exists −A = [−aij] 2× 2 ∈ R such that A+(−A) = [aij + (−aij)] 2× 2 = [0] 2× 2 = O.

∴ (R, +) is an abelian group.

(5) Let A = [aij] 2× 2 ,B = [bjk] 2× 2 ,C = [ckl] 2× 2 ∈ R.

From the definition of multiplication; AB = [aij] [bjk] = [uik] 2× 2 where uik = P2j=1 aij bjk = ai1b1k + ai2b2k ∈ C

∴ Multiplication is a binary operation.

(6) $(A B) C=\left[\sum_{j=1}^2 a_{i j} b_{j k}\right]_{2 \times 2}\left[c_{k l}\right]_{2 \times 2}$

=$\left[\sum_{k=1}^2\left(\sum_{j=1}^2 a_{i j} b_{j k}\right) c_{k l}\right]$

=$\left[\sum_{j=1}^2 a_{i j}\left(\sum_{k=1}^2 b_{j k} c_{k l}\right)\right]$

=A(B C) therefore Multiplication is associative.

(7) $A(B+C)=\left[a_{i j}\right]_{2 \times 2}\left[b_{j k}+c_{j k}\right]_{2 \times 2}$

=$\left[\sum_{j=1}^2 a_{i j}\left(b_{j k}+c_{j k}\right)\right]$

= $\left[\sum_{j=1}^2\left(a_{i j} b_{j k}+a_{i j} c_{j k}\right)\right]$

= $\left[\sum_{j=1}^2 a_{i j} b_{j k}\right]+\left[\sum_{j=1}^2 a_{i j} c_{j k}\right]=A B+A C$

Similarly, we can prove that (B + C)A = BA + CA. ∴ Distributive laws hold.

Hence (R, +, ·) is a ring

⇒ $\text { Since } 1 \in \mathrm{C}, I=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \in \mathbb{R} \text {. For } A$

⇒ $=\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]_{2 \times 2}$

⇒ $A I=\left[\begin{array}{ll}
a_{11}+0 & 0+a_{12} \\
a_{21}+0 & 0+a_{22}
\end{array}\right]_{2 \times 2}=\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right] $ = A. Also IA = A.

⇒ $\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] $

⇒ $
\text { Let } A=\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right] \text { and } B=\left[\begin{array}{ll}
2 & 0 \\
0 & 1
\end{array}\right] \text {. }$ Then AB =

⇒ $
\left[\begin{array}{ll}
2+0 & 0+2 \\
2+0 & 0+2
\end{array}\right]=$

⇒ $
\left[\begin{array}{ll}
2 & 2 \\
6 & 4
\end{array}\right]
$

and BA =$
\left[\begin{array}{ll}
2+0 & 4+0 \\
0+4 & 0+4
\end{array}\right]=\left[\begin{array}{ll}
2 & 4 \\
4 & 4
\end{array}\right] $

so that AB 6= BA

Hence (R, +, ·) is not a commutative ring.

Notation. The ring of all 2 × 2 matrices over the field of complex numbers C is denoted by M₂(C). If F is a field the ring of all n × n matrices over F is denoted M n( F). The zero element in M_n( F) is denoted by O·n× n and the unity element by I

Zero divisors in M₂(C) : A =

⇒ $
\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right] \neq \mathrm{O} \text { and } B=\left[\begin{array}{ll}
0 & 0 \\
0 & 1
\end{array}\right] \neq \mathrm{O}$

Then AB = $
\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right] \neq \mathrm{O} \text { and } B A=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=\mathrm{O}
$

We observe that AB 6= BA and A₆ = O, B₆= O ⇒ BA = O. Therefore there exist Zero divisors in M₂(F) where F is a field.

Nilpotent element in M₂(C) :

For A = $
\left[\begin{array}{ll}
0 & 2 \\
0 & 0
\end{array}\right]
$

we have A₂ = $
\left[\begin{array}{ll}
0 & 2 \\
0 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 2 \\
0 & 0
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=\mathrm{O} $

Therefore A is a nilpotent matrix in M₂(C).

⇒ $ B=\left[\begin{array}{cc}
1 & 1 \\
-1 & -1
\end{array}\right]$ is also a nilpotent matrix element in M₂(C).

Example.2. The set of 2 × 2 matrices of the form

⇒ $
\left[\begin{array}{cc}
x & y \\
-\bar{y} & \bar{x}
\end{array}\right]
$

where x, y are complex numbers and x,¯ y¯ denote the complex conjugates of xy; is a skew field for compositions of matrix addition and multiplication.

Solution. Let M =$
\left\{\left[\begin{array}{cc}
\frac{x}{-y} & \bar{x}
\end{array}\right]: x=a+i b, y=c+i d ; a, b, c, d \in \mathrm{R}\right\} $ be the set
of 2 × 2 matrices.

⇒ $
\text { Let } A=\left[\begin{array}{cc}
x_1 & y_1 \\
-y_1 & \bar{x}_1
\end{array}\right], B=\left[\begin{array}{cc}
x_2 & y_2 \\
-y_2 & \bar{x}_2
\end{array}\right], C=\left[\begin{array}{cc}
x_3 & y_3 \\
-y_3 & \overline{x_3}
\end{array}\right] \in \mathrm{M} $

⇒ $
\text { (1) } A+B=\left[\begin{array}{cc}
x_1+x_2 & y_1+y_2 \\
-\bar{y}_1-\bar{y}_2 & \overline{x_1}+\bar{x}_2
\end{array}\right]=\left[\begin{array}{cc}
x_1+x_2 & y_1+y_2 \\
-\overline{y_1+y_2} & \underline{x_1+x_2}
\end{array}\right] \in \mathrm{M} \text {, since }
$

¯Z₁ ± Z₂ = Z¯₁ ± Z¯₂ for Z¯₁, Z¯₂ ∈ C

⇒ $
\text { A. } B=\left[\begin{array}{ll}
x_1 & y_1 \\
-y_1 & \bar{x}_1
\end{array}\right]\left[\begin{array}{cc}
x_2 & y_2 \\
-y_2 & -x_2
\end{array}\right]=\left[\begin{array}{cc}
x_1 x_2-y_1 y_2 & x_1 y_2+y_1 x_2 \\
-y_1 x_2-x_1 y_2 & -y_1 y_2+x_1 x_2
\end{array}\right]$

If u = x₁x₂ − y₁y¯₂ and ν = x₁y₂ + y₁x¯₂ then u = ¯x₁x¯₂ − y¯₁y₂ and v¯ = x¯₁y¯₂ + ¯y₁x₂

A,B =$
\left[\begin{array}{cc}
u & v \\
-\bar{v} & \bar{u}
\end{array}\right] \in \mathrm{M}
$

Hence addition ( + ) and multiplication

(1) are binary operations.

(2) Clearly A + B = B + A for any A,B ∈ M.

(3) Clearly (A + B) + C = A + (B + C) and (A,B) · C = A(B,C) for any A,B,C ∈ M because addition and multiplication of matrices are associative.

(4) There exists O =

⇒ $
\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=\left[\begin{array}{cc}
0+i 0 & 0+i 0 \\
-0+i 0 & 0+i 0
\end{array}\right] \in
$

M so that A +O = A for any A ∈ M.

(5) For A =$
\left[\begin{array}{ll}
x & y \\
-y & \bar{x}
\end{array}\right] \text { there exists }-A=\left[\begin{array}{ll}
-x & -y \\
\bar{y} & -x
\end{array}\right] \text { so that } A+(-A)=
$

⇒ $
\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=0
$

= 0 (Zero matrix)

(6) For any A,B,C ∈ M, distributive laws, namely, A· (B+C) = A·B+A·C

and (B + C) · A = B · A + C · A are clearly true. Hence (M, +, −) is a ring

⇒ $
\text { (7) We have } I=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1+i 0 & 0+i 0 \\
-0+i 0 & \frac{1+i \cdot 0}{1+}
\end{array}\right] \in \text { so that } A \cdot I=I \cdot A=A \text { for any } A \in \mathrm{M} \text {. }
$

∴ the ring M has unity element I.

(8) Let A 6= 0 ∈ M so that A =$
\left[\begin{array}{cc}
x & y \\
-y & \bar{x}
\end{array}\right]=\left[\begin{array}{cc}
a+i b & c+i d \\
-c+i d & a-i b
\end{array}\right]
$

where a,b,c,d are not all zero.

Det A = (a + ib)(a − ib) − (c + id)(−c + id) = a₂ + b₂ + c₂ + d₂ 6= 0

Since det A 6= 0, A 6= O is invertible. Hence (M, +, ·) is a skew field.

Note. The matrix $
\left[\begin{array}{cc}
x & y \\
-y & \bar{x}
\end{array}\right] \text { is also given as }\left[\begin{array}{cc}
a+i b & c+i d \\
-c+i d & a-i b
\end{array}\right]
$ in the problem.

Rings, Integral Domains & Fields Ring Of Quaternions

Example 3. Prove that the set of Quaternions is a skew field.

Solution. Let Q = R × R × R × R = {α₀ + α₁i + α₂j + α₃k | α₀,α₁,α₂,α₃ ∈ R} where i,j,k are quaternion units satisfying the relations :

i ² = j² = k ²  = i · jk = −1, ij = −ji = k, jk = −kj = i, ki = −ik = j. Let

X, Y, Z ∈ Q so that X = α₀ + α₁i + α₂j + α₃k, Y = β₀ + β₁i + β₂j + β₃k and Z = γ₀+ γ₁i + γ₂j + γ₃k where α_t, β_t, γ_tfor t = 0, 1, 2, 3 are real numbers.

We define X = Y ⇔ α_t= βt for t = 0, 1, 2, 3.

We define addition (+) as X + Y = (α₀ + β₀) + (α₁ + β1) i + (α2 + β2) j +

(α₃ + β₃) k and multiplication (·) as X.Y = ( α₀β₀− α₁β₁− α₂β₂ − α₃β₃) + (α₀β₁+ α₁β₀ + α₂β₃ − α₃β₂) i

+ (α₀β₂ + α₂β₀ + α₃β₁ − α₁β₃) j + (α₀β₃ + α₃β₀ + α₁β₂ − α₂β₁) k.

(1) ∀X, Y ∈ Q; X + Y = (α₀+ β₀) + (α₁+ β₁) i + (α₂ + β₂) j + (α₃ + β₃) k

As αt + βt for t = 0, 1, 2, 3 ∈ R, X + Y ∈ Q. ∴ addition (+) is a binary operation.

(2) ∀X, Y ∈ Q; X + Y = (α₀+ β₀) + (α₁+ β₁) i + (α₂+ β₂) j + (α₃ + β₃) k

= (β0 + α₀) + (β₁ + α₁) i + (β₂ + α₂) j + (β₃ + α₃) k˙ = Y + X

∴ addition is commutative:

( ∵ α_t + β_t= β_t+ α_t for t = 0, 1, 2, 3) .

(3) ∀X, Y, Z ∈ Q

(X + Y) + Z = {(α₀ + β₀) + (α₁ + β₁) i + (α₂ + β₂) j + (α₃+ β₃) k} + (γ₀ + γ₁i + γ₂j + γ₃k)

= [(α₀ + β₀) + γ₀] + [(α₁ + β₁) + γ₁] i + [(α₂ + β₂) + γ₂] j + [(α₃ + β₃) + γ₃] k

= [α₀ + (β₀+ γ₀)] + [α₁ + (β₁ + γ₁)] i + [α₂ + (β₂ + γ₂)] j + [α₃+ (β₃ + γ₃)] k

= X + (Y + Z). ( ∵ (α_t + βt) + γ_t = α_t + (β_t + γ_t) for t = 0, 1, 2, 3) .

∴ addition is associative.

(4) For O = 0 + 0i + 0j + 0k ∈ Q and X = α₀ + α₁i + α₂j + α₃k we have

O + X = (0 + α₀) + (0 + α₁) i + (0 + α₂) j + (0 + α₃) k = α₀ + α₁i + α₂j + α₃k =

X = X + O

∴ O = 0 + 0i + 0j + 0k is the additive identity.

(5) For X = α₀+ α₁i + α₂j + α₃k there exists −X = (−α₀) + (−α₁) i +

(−α₂) j + (−α₃) k ∈ Q such that X + (−X) = [α₀+ (−α₀)] + [α₁ + (−α₁)] i +

[α₂+ (−α₂)] j + [α₃ + (−α₃)] k = 0 + 0i + 0j + 0k = 0, the additive identity.

∴ every element has an additive inverse.

Hence, from (1), (2), (3), (4) and (5) : (Q, +) is abelian group.

(6) X.Y = b₀ + b₁i + b₂j + b₃k where b₀ = α₀β₀− α₁β₁ − α₂β₂ − α₃β₃
b₁ = α₀β₁+ α₁β₀ + α₂β₃− α₃β₂, b₂ = α₀β₂ + α₂β₀ + α₃β₁ − α₁β₃
and b3 = α0β3 + α3β0 + α1β2 − α2β1 are real numbers.

∴ multiplication (·) is a binary operation.

(7) (X : Y) · Z = (b₀ + b₁i + b₂j + b₃k) · (γ₀ + γ₁i + γ₂j + γ₃k)

= (b₀γ₀ − b₁γ₁ − b₂γ₂ − b₃γ₃) + (b₀γ₁ + b₁γ₀ + b₂γ₃ − b₃γ₂) i + (b₀γ₂ + b₂γ₀ + b₃γ₁ − b₁γ₃) j

+ (b₀γ₃ + b₃γ₀ + b₁γ₂ − b₂γ₁) k

Y.Z = c₀+ c₁i + c₂j + c₃k where c₀ = β₀γ₀ − β₁γ₁− β₂γ₂ − β₃γ₃,

c₁ = β₀γ₁ + β₁γ₀ + β₂γ₃ − β₃γ₂,

c₂ = β₀γ₂ + β₂γ₀+ β₃γ₁ − β₁γ₃,

c₃ = β₀γ₃ + β₃γ₀ + β₁γ2 − β2γ₁.

X.(Y.Z) = (α₀ + α₁i + α₂j + α₃k) . (c₀ + c₁i + c₂j + c₃k)

= (α₀c₀ − α₁c₁ − α₂c₂ − α₃c₃) + (α₀c₁ + α₁c₀ + α₂c₃− α₃c₂) i

+ (α₀c₂ + α₂c₀ + α₃c₁ − α₁c₃) j + (α₀c₃ + α₃c₀ + α₁c₂− α₂c₁) k

Since the corresponding terms of (X.Y). Z and X.(Y, Z) are equal we have

∴ multiplication is associative.

(8) Both the distributive laws, namely, X.(Y + Z) = X.Y + X.Z and (Y + Z) · X = Y · X + Z · X can be proved to be true.

From the truth of the above 8 properties, we establish that (Q+, ·) is a ring.

(9) There exists 1 = 1 + 0i + 0j + 0k ∈ Q such that

∀X ∈ Q we have 1.X = (1 + 0i + 0j + 0k) · α0˙ + α₁i + α₂j + α₃k

= (1.α₀− 0.α₁ − 0.α₂− 0.α₃) + (1.α₁ + α₀· 0 + 0.α₃ − 0.α₂) i

+ (1.α₂ + 0.α₀ + 0.α₁ − α₁· 0) j + (1 · α₃+ 0.α₀ + 0 · α₂ − 0.α₁) k

= α₀ + α₁i + α₂j + α₃k = X. Also X.1= X.

∴ 1 = 1 + 0i + 0j + 0k ∈ Q is the unity element.

(10) Let X 6= 0, the zero element. Then not all α0, α1, α2, α3 are zero ∈ R.

∴ α²₀ + α² ₁+ α²₂+ α²₃ = ∆ ≠0 ∈ R.

For the real numbers α ₀

⇒ $
\frac{\alpha_0}{\Delta}, \frac{-\alpha_1}{\Delta}, \frac{-\alpha_2}{\Delta}, \frac{-\alpha_3}{\Delta}
$ there exists

⇒ $
\mathrm{X}^1=\frac{\alpha_0}{\Delta}-\frac{\alpha_1}{\Delta} i-\frac{\alpha_2}{\Delta} j-\frac{\alpha_3}{\Delta} k \in \mathrm{Q} . \quad \text { Further } \mathrm{X} \cdot \mathrm{X}^1=\left(\frac{\alpha_0^2}{\Delta}+\frac{\alpha_1^2}{\Delta}+\frac{\alpha_2^2}{\Delta}+\frac{\alpha_3^2}{\Delta}\right)
$

⇒ $
+\left(\frac{-\alpha_0 \alpha_1}{\Delta}+\frac{\alpha_0 \alpha_1}{\Delta}-\frac{\alpha_2 \alpha_3}{\Delta}+\frac{\alpha_3 \alpha_2}{\Delta}\right) i+\left(\frac{-\alpha_0 \alpha_2}{\Delta}+\frac{\alpha_2 \alpha_0}{\Delta}-\frac{\alpha_3 \alpha_1}{\Delta}+\frac{\alpha_1 \alpha_3}{\Delta}\right) j
$

⇒ $
+\left(\frac{-\alpha_0 \alpha_3}{\Delta}+\frac{\alpha_3 \alpha_0}{\Delta}-\frac{\alpha_1 \alpha_2}{\Delta}+\frac{\alpha_2 \alpha_1}{\Delta}\right)
$

k = 1 + 0i + 0j + 0k = 1, the unity element.

Similarly, we can prove that X1 · X = 1.

We have X.Y = (α₀β₀− α₁β₁ − α₂β₂ − α₃β₃)+(α₀β₀ + α₁β₀+ α₂β₃− α₃β₂) i

+ (α₀β₂ + α₂β₀+ α₃β₁ − α₁β₃) j + (α₀β₃ + α₃β₀ + α₁β₂ − α₂β₁) k

= b₀+ b₁i + b₂j + b₃k and

∴ every non-zero element of Q has a multiplicative inverse.

Y.X = (β₀α₀− β₁α₁ − β₂α₂− β₃α₃) + (β₀α₁ + β₁α₀ + β₂α₃ − β₃α₂) i

+ (β₀α₂ + β₂α₀ + β₃α₁ − β₁α₃) j + (β₀α₃ + β₃α₀ + β₁α₂− β₂α₁) k = a₀ + a₁i + a₂j + a₃k.

we observe that b ₀ = a ₀,b ₁ ≠ a ₁,b ₂ ≠a ₂,_b₃ ≠a ₃.

∴ multiplication is not commutative.

Hence (Q+, ·) is a Division ring or Skew field.

Note. We can take 1 = (1, 0, 0, 0),i = (0, 1, 0, 0),j = (0, 0, 1, 0) and k = (0, 0, 0, 1)

2. The set G = {±1, ±i, ±j, ±k} form a nonabelian group of order 8 under multiplication (·) defined as follows:

i ₂ = j ₂ = k ₂ = i,k = −1,ij = −ji = k; jk = −kj = i and ki = −ik = j

Rings, Integral Domains, And Fields Ring Of Endomorphisms Of An Abelian Group.

Let G be an abelian group. A homomorphism of G into itself is an endomorphism of G. The set of all endomorphisms of G is denoted by Hom (G, G) or

Hom (G).

For f,g ∈ Hom(G, G) if we define addition (+) and multiplication (·) of two

endomorphisms as (f + g)(x) = f(x) + g(x) and (f,g)(x) = f(g(x))∀x ∈ G then < Hom(G, G), +, > is a ring.

Note. Hom (G, G) is not commutative as the composition of functions is not commutative.

The post Rings, Integral Domains & Fields Examples appeared first on Answer Key for Math.

Binary Operations (Definition, Types, and Examples)

Teja — Sun, 06 Aug 2023 12:59:23 +0000

The use of numbers was existed for many centuries and we are familiar with the types of numbers – integers, rational numbers, real numbers, and complex numbers together with certain operations, such as addition and multiplication, defined on them.

The addition is basically just such a rule that people learn, enabling them to associate, with two numbers in a given order, some number as the answer. Multiplication is also such a rule, but a different rule.

But with the use of arbitrary quantities a,b,c,…, for numbers the subject, Algebra which is the generalization of Arithmetic, came into being. For many years mathematicians concentrated on improving the methods to use numbers, and not on the structure of the number system.

In the nineteenth century, mathematicians came to know that the methods to use numbers are not limited to only sets of numbers but also to other types of sets. A set with a method of combination of the elements of it is called an algebraic structure and we can have many algebraic structures.

The study of algebraic structures which have been subjected to axiomatic development in terms of Abstract Algebra. In what follows we study Group Theory i.e. the study of the algebraic structure. Group, which is rightly termed the basis of Abstract Algebra.

In Group Theory the basic ingredients are sets, relations, and mappings. It is expected that the student is very much familiar with them. However, we introduce and discuss some of the aspects connected with them which will be useful to us in our future study.

Binary Operations Equality Of Sets A And B

A ⊆ B And B ⊆ A ⇔ A = B

Binary Operations Union And Intersection Of Sets A₁, A₂, . . . , A_n

A₁ ∪ A₂ ∪ A₃ . . . ∪ A_n = S n_i =1 A_iand A₁ ∩ A₂ ∩ A₃ . . . ∩ A_n = Tn_i =1 A_i

Binary Operations f Is A Relation From A Set A TO A Set B

⇔ f ⊆ A × B. ⇔ f ⊆ {(a, b) : a ∈ A, b ∈ B}

We write (a, b) ∈ f as afb and we say that a is f related to b.

Sometimes we write ∼ for f. In such a case we write a ∼ b.

If A = B, then we say that f is a relation in A.

If f ⊆ A × B, we write f⁻¹ = {(b, a)/(a, b) ∈ f} ⊆ B × A and f⁻¹ is called

inverse relation of f and it is from B to A.

The domain of f is equal to the range of f⁻¹and the range of f is equal to the domain of f⁻¹. Further (f⁻¹)^-1= f.

f is a relation in A ⇔ f ⊆ A × A ⇔ f{(a, b)/a, b ∈ A} ⊆ A ∈ A.

Binary Operations Types Of Relations

f is a relation in a set A.

(1) If ∀x ∈ A, (x, x) ∈ f then f is said to be reflexive in A.

(2) If ∀(x, y) ∈ f ⇒ (y, x) ∈ f, then f is said to be symmetric in A.

(3) If (x, y) ∈ f and (y, z) ∈ f ⇒ (x, z) ∈ f, then f is said to be transitive in A.

(4) If f is reflexive, symmetric and transitive, then f is said to be an equivalence relation.

Example: 1. In the set of triangles in a plane, the relation of similarity is an equivalence relation.

2. A = {1, 2, 3} the relation f = {(1, 1), (2, 2), (3, 3)} is an equivalence relation in A.

Partition of a set

A partition of a set S is a set of non-empty subsets S_i, with i in some index set ∆, such that:

(1). s = ∪_i∈∆s_i and

(2) s_i ∩ s_j = φ for i 6= j.

That is, a partition of a set S is a collection of disjoint subsets of S whose union is the whole set S.

Binary Operations Partition Of A Set

f is an equivalence relation in a non-empty set S and a is an element of S.

The subset of elements which are f related to a constitutes an equivalence class of a.

The equivalence class of a is denoted by a¯ or [a] or {a}. Thus a¯ = {x ∈ S afx} and a¯ ⊆ S.

Further

(1) a ∈ a¯,
(2) b ∈ a¯ ⇒ ¯b = ¯a.

For, b ∈ a¯ ⇒ afb and x is any element of ¯b ⇒ bfx.

Now a fb, bfx ⇒ afx ⇒ x ∈ a¯ ⇒ ¯b ⊆ a¯

Again y is any element of a¯ ⇒ afy.

Since f symmetric a f b ⇒ bfa.

Now bfa, afy ⇒ bfy ⇒ y ∈ ¯b ⇒ a¯ ⊆ ¯b and hence ¯b = ¯a using (1).

(3) a¯ = ¯b ⇒ afb

For a¯ = ¯b ⇒ a ∈ ¯b ⇒ bfa ⇒ afb.

(4) afb˙ ⇒ a¯ = ¯b

For x ∈ a, afb¯ ⇒ afx, bfa ⇒ bfa, afx ⇒ bfx ⇒ x ∈ ¯b ⇒ a¯ ⊆ ¯b . . …

Again y is any element of ¯b ⇒ bfy.

Now a fb ⇒ afy ⇒ y ∈ a¯ ⇒ ¯b ⊆ a¯ and hence a¯ = ¯b using (2).

Theorem 1. If f is an equivalence relation in a non-empty set S and a, b are two arbitrary elements of S, then

(1) a¯ = ¯b or a¯ ∩ ¯b = φ

(2) a¯ ∪ ¯b ∪ c¯ ∪ .. = S

Proof.

Given

If f is an equivalence relation in a non-empty set S and a, b are two arbitrary elements of S,

(1) If a¯ ∩ ¯b = φ, there is nothing to prove. Let a¯ ∩ ¯b = φ.

Then there exists some element x such that x ∈ a¯ and x ∈ ¯b.

∴ afx and bfx ⇒ afx and xfb ⇒ af : b ⇒ a¯ = ¯b

Hence we must have either a¯ = ¯b if a¯ ∩ ¯b 6= φ or a¯ 6= ¯b if a¯ ∩ ¯b = φ.

(2) Let c be any element of S.

If afc, then a¯ = ¯c, and if bfc then ¯b = ¯c.

If a¯ 6= ¯c or ¯b 6= ¯c, then a¯ ∩ ¯b ∩ c¯ = φ.

∴ Every element of S must belong to some equivalence class of S

i.e. all the elements of S must belong to the disjoint equivalence classes of S

i.e. a¯ ∪ ¯b ∪ c¯ ∪ . . . = S.

Note. If f is an equivalence relation defined in a non-empty set S, the set of equivalence classes related to f is a partition of S.

That is, two equivalence classes related to f are

(1) either identical or disjoint and

(2) the union of all the disjoint equivalence classes of f is the set S.

Theorem 2. For any given partition of a set S, there exists an equivalence relation f in S such that the set of equivalence classes related to f is the given partition.

Proof. Let P = {S_a, S_b, S_c, . . . ..} be any partition of S.
Let p, q ∈ S.

Let us define a relation f in S by pfq if there is a S_i in the partition such that p, q ∈ S_i.

(1) Since S = S a ∪ S b ∪ S c ∪ . . . . . . , ∀x˙ ∈ S, there exists S_i ∈ P such that x ∈ S_i.

Hence x, x ∈ S_i ⇒ xfx.

∴ f is reflexive in S.

(2) If xfy, then there exists S_i ∈ P such that x, y ∈ S_i.

But x, y ∈ S_i ⇒ y, x ∈ S_i ⇒ yfx. Hence xfy ⇒ yfx.

∴ f is symmetric in S.

(3) Let xfy and yfz then by the definition of f, there exist subsets S_j and S_k (not necessarily distinct) of S such that x, y ∈ S_j and y, z ∈ S_k.

Since y ∈ S_j and also y ∈ S_k, we have S_j ∩ S_k 6= φ.

But S_j, S_k belong to the partition of S.

∴ S_j ∩ S_k 6= φ ⇒ S_j = S_k

Then x, z ∈ S_jand hence xfz.

Hence f is transitive in S.

∴ f is an equivalence relation in S.

Binary Operations Functions Or Mappings

Definition. A, B are non-empty sets. If f ⊆ A × B such that the following conditions are true, then f is called a function from A, to B.

(1) ∀x ∈ A∃y ∈ B

such that (x, y) ∈ f.

(2) (x, y), (x, z) ∈ f ⇒ y = z.

If f is a function from A to B then we say that f is a mapping from A to B and we write f: A → B.

Domain of f is A and range of f is f(A) and f(A) ⊆ B.

Alternatively, if f is a relation that associates every element of A to an element of B, and if x = y ⇒ f(x) = f(y) for x, y ∈ A, then f is a function from A to B.

In this context, we say that the function is well-defined.

Transformation. If f: A → A then the function f is called an operator or transformation on A.

Equality of Functions. If f : A → B and g : A → B and if f(x) = g(x) for every x ∈ A then f = g. If ∃x ∈ A such that f(x) 6= g(x) then we say that f ≠ g.

Binary Operations 1.8 Types Of Functions Or Mappings

(1) If f; A → B is such that there is at least one element in B which is not the f image of any element in A, then we say that f is a mapping from A into B

i.e. f maps A into B.

(2) If f : A → B is such that f(A) = B, then we say that f is a mapping from A into B · f is also called a surjection or a surjective mapping.

If ∃ some element b ∈ B such that f(a) 6= b for some a ∈ A, then f is not onto.

(3) If f : A → B is such that for x, y ∈ A, f(x) = f(y) ⇒ x = y, then f is said to be a one-one or one-to-one function or an injection or an injective mapping. We write f as 1 − 1.

If x, y ∈ A and x 6= y ⇒ f(x) 6= f(y), then f is 1 − 1.

This is equivalent to the above condition.

If f(x) = f(y) does not imply x = y then we say that f is not 1 − 1.

(4) If f : A → B is 1 − 1 and onto, then f is called a bijection.

In other words we say that f is a 1 − 1 function from A onto B. Here f is called a one-one correspondence between A and B.

If A, B are finite and if f : A → B is a bijection then the number of elements in A, B are equal.

(5) If f : A → B is such that every element of A is mapped into one and only one element of B, then f is called a constant function. Here f (A) is a singleton set.

(6) If f : A → A is such that f(x) = x for every x ∈ A, then f is called the identity function on A. It is denoted by I A or simply I. I is always 1 − 1 and onto.

(7) If f : A → B is a bijection then f − 1 : B → A is unique and is also a bijection. If f : A → B, is one-one and onto, then f − 1 : B → A where f − 1 = {(b, a)/(a, b) ∈ f) is called the inverse mapping of f.

Here f(a) = b ⇔ f − 1(b) = a.

Only bijections possess inverse mappings.

Binary Operations Product Or Composite Of Mappings And Some Of Their Properties

1. Let f : A → B and g : B → C.

Then the composite function of f and g, denoted by gof is a mapping from A to C.

i.e. gof : A → C such that (g ◦ f)(x) = g[f(x)], ∀x ∈ A.

Here fog cannot be defined. Even if it is possible to define fog and gof, then we may have fog 6= gof. Thus composition of mappings is not commutative:

2. If f : A → B and g : B → C are one-one, then gof : A → C is one-one.

If f, g are onto, then gof is onto.

If f, g are functions such that gof, is one-one, then f is one-one.

If f, g are functions such that gof is onto, then g is onto.

3. If f: A → B is a bijection, thenf⁻¹: B → A. Also f⁻¹of = I A and fof⁻¹=I_B_.

In particular, if f: A → A is a bijection, then f⁻¹ : A → A.
Also  of = f◦ f⁻¹ = I.

4. If f : A → B then I Bof = f and foI A = f.

In particular, if f: A → A then Iof = foI = f.

5. If f : A → B and g : B → C are bijections, then gof : A → C is also a bijection. If f, g are functions such that gof is a bijection then f is one-one and g is onto. In particular, if f, and g are bijections on A, then gof is also a bijection on A. Also (g ◦ f)⁻¹ is a bijection and (g ◦ f)− 1 = f⁻¹og⁻¹

6. If f : A → B, g : B → C and h : C → D, then (h ◦ g) of = ho( gof ) i.e. composition of mappings is associative.

Definition. f : A → A is a function. f_n: A → A where n ∈ Z is defined as follows.

If n = 0, f 0 = I the identity mapping on A.

If n = 1, f 1 = f

If n ≥ 2 and n ∈ N, fn = f◦f_n − 1 = f_n − 1 of .

If n is a negative integer and f is a bijection then f_n = f⁻¹ −n.

7. f₂ = f ◦ f, f₃ = f ◦ f 2 = fofof . . . . . . f_n = fo f⁻¹= fofof ….. to n times where n ∈ N.

8. If n is a negative integer and n = −m so that m is a positive integer, then fn = f−m = f⁻¹ m = f⁻¹o f⁻¹o f⁻¹o . . . . . . to m times
= ( fofof . . . . . . to m times ) ⁻¹ = (fm)⁻¹

9. If f : A → A is a bijection, then fn : A → A for n ∈ Z is also a bijection.

10. If m, n ∈ Z then fmofn = fm +n = fn +m = fnofm.

Binary Operations Binary Operations

Let R be the set of real numbers and addition (+), multiplication (x), and subtraction (−) be the operations in R. For every pair of numbers a, b ∈ R, we have unique elements a + b, a × b, a − b ∈ R.

Thus we can look upon addition, multiplication, and subtraction as three mappings of R × R into R, which for each element (a, b) of R × R determine the elements a + b, a × b, a − b respectively of R, Also one can define many mappings from R× R into R.

All these mappings are examples of binary operations on R. The idea of binary operation is not limited only to the sets of numbers. For example, the operations of union (∪), intersection (∩), and difference (−) are binary operations in P(A), the power set of A.

Binary operation

Definition. Let S be a non-empty set. If f : S × S → S is a mapping, then f. is called binary operation or binary composition in S (or on S ).
Thus
1. If a relation in S is such that every pair (distinct or equal) of elements of S taken in definite order is associated with a unique element of S then it is called a binary operation is S. Otherwise the relation is not a binary operation in S and the relation is simply an operation in S.

2. (a, b˙ ) ∈ S × S, ∃ a unique image f(a, b) ∈ S.

We observe that addition (+), multiplication (× or :), and subtraction (−) are binary operations in R and division (÷) is not a binary operation in R.
( ∵ division by 0 is not defined.)

Symbolism. It is customary to denote the binary operation in S by o (read as a circle) or (read as star) or . or + and to take a, b, c ∈ S as arbitrary elements a, b, c of S.

1. For a ∈ S, b ∈ S ⇒ a + b ∈ S ⇒ + is a binary operation in S. Also + is called addition, + is called usual addition if S ⊆ C and a + b is called the sum of a and b. Addition (+) is to be understood depending upon the set over which the operation is to be taken.

2. For a ∈ S, b ∈ S ⇒ a · b ∈ S ⇒. is a binary operation in S. Also . is called multiplication, . is called usual multiplication if S ⊆ C and a.b is called product of a, b. Multiplication (.) is to be understood depending upon the set over which the operation is to be taken.

3. For a ∈ S, b ∈ S ⇒ aob ∈ S ⇒ o is a binary operation in S.

4. For a ∈ S, b ∈ S ⇒ a ∗b ∈ S ⇒ ∗ is a binary operation in S.

This is called closure law.

Sometimes we write products a · b or a ∗b of a and b as ab.

If a, b ∈ S such that aob /∈ S then o is not a binary operation in S. In this case we say that S is not closed under; o.

+ is a binary operation in the set of natural numbers N as (a, b) ∈ N ⇒ a + b ∈ N.

• is not a binary operation in N as (a, b) ∈ N does not imply a − b ∈ N.

o is a binary operation in S. The image elements under the mapping o are in S. If a and b are elements of a subset H of S, it may or may not happen that ab ∈ H.

But if ab ∈ H for arbitrary elements a, b ∈ H the subset H is said to be closed under the operation o.It may be observed that if o is a binary operation in S, it is implied that S is closed under the operation o :

o is a binary operation in S. If a, b ∈ S and a 6= b, we know that (a, b) 6= (b, a), and hence, in general, it is not necessary that the images in S of (a, b) and (b,a) under the binary operation o must be equal. In other words, if o is a binary composition in S it is not necessary that a, b ∈ S must hold aob = boa.

+ is a binary operation in R. If a, b, c ∈ R, then a + b ∈ R, b + c ∈ R, (a + b) + c ∈ R and a + (b + c) ∈ R. We observe that (a + b) + c = a + (b + c).

Again – is a binary operation in R. If a, b, c ∈ R as above we observe that (a − b) − c ∈ R and a − (b − c) ∈ R. But (a − b) − c 6= a − (b − c).

Definition. o is a binary operation in a set S. If for a, b ∈ S, aob = boa, then o is said to be commutative in S. This is called commutative law. Otherwise, o

is said to be not-commutative, in S.

Definition. o is a binary operation in a set S. If for a, b, c ∈ S, (aob)oc = ao(boc) then o is said to be associative in S.

This is called associative law. Otherwise, o is said to be not associative in S.

Note. If o is associative in S, then we write (aob)oc = ao(boc) = aoboc e.g. is a commutative binary operation on N ⇒ a ∗ (b ∗ c) = (b ∗ c) ∗ a = (c ∗ b) ∗ a

Definition. o, ∗ are binary operations in a set S.

If a, b, c ∈ S,

(1) ao(b ∗ c) = (a ◦ b) ∗ (a ◦ c),

(2) (b ∗ c)oa = (b ◦ a) ∗ (c ◦ a),

then o is said to be distributive w.r.t. the operation *.

(1) is called the left distributive law and

(2) is called the right distributive law.

(1) and (2) are called distributive laws.

It is customary in mathematics to omit the words and only if from a definition. Definitions are always understood to be if and only if statements. Theorems are not always if and only if statements and no such convention is ever used for theorems.

Note. To prove that a binary operation in S obeys (follows) a law (commutative law, associative law, etc., ) we must prove that elements of every ordered pair obey the law i.e., the law must be proved by taking arbitrary elements.

But to prove that a binary operation in S does not obey a particular law, it is sufficient if we give a counter-example. This method of proving the result is called the proof by counter-example.

Example: 1. Set of Natural numbers N

(1) +,. are binary’operations in N, since for a, b ∈ N ⇒ a + b ∈ N and ab ∈ N. In other words, N is said to be closed w.r.t. the operation + and.

(2) +,. are commutative in N since for a, b ∈ N, a + b = b + a and ab = ba.

(3) +, are associative in N since for a, b, c ∈ N. (a + b) + c = a + (b + c) and a(bc) = (ab)c.

(4) – is distributive w.r.t. the operation + in N since for a, b, c ∈ N. a · (b + c) = a · b + a · c and (b + c) · a = b · a + c.a.

(5) The operations subtraction (−) and division (÷) are not binary operations in N for 3, 5 ∈ N does not imply 3 − 5 ∈ N and 3/5∈ N.

(6) Operations +, −, · are binary operations on R but ÷ is not. However, ÷ is a binary operation on R ∗ = R − {0}.

(7) On Z+, ∀a, b ∈ Z+if ∗ is defined as a ∗ b = ab or a ∗ b = |a − b|, then ∗ is a binary operation.

(8)In N, o is a binary operation defined as aob = L. C. M. for every a, b ∈ N. Then 7o5 = 35 and 16o20 = 80.

Example: 2. A is the set of even integers.

(1) +,. are binary operations in A since for a, b ∈ A, a + b ∈ A and ab ∈ A.

(2) +,. are commutative in A since for a, b ∈ A, a + b = b + a and ab = ba.

(3) +, . . . are associative in A since for a, b, c ∈ A,(a + b) + c = a + (b + c) and a(bc) = (ab)c.

(4) is distributive w.r.t. the operation + in A since for a, b, c ∈ A, a.(b + c) = a · b + a · c and (b + c) · a = b · a + c · a.

Example: 3. A is the set of odd integers.

(1) – is a binary operation in A. Also: is associative and commutative in A.

(2) + is not a binary operation in A since 3, 5 ∈ A does not imply 3 + 5 = 8 ∈ A.

Example: 4. S is the set of all m × n matrices such that each element of any matrix is a complex number.

The addition of matrices, denoted by +, is a binary operation in S. Also +)is commutative and associative in S.

Example: 5. S is the set of all vectors.

(1) Addition of vectors, denoted by + is a binary operation in S. Also + is commutative and associative in s.

(2) Dot product of vectors, denoted ., is not a binary operation in S since for a,¯ ¯b ∈ S, a¯ · ¯b /∈ S.

(3) Cross product of vectors denoted by × is a binary operation in S since for a,¯ ¯b ∈ S, a¯ × ¯b = ¯c ∈ S.

Example: 6. In N the operation o defined by aob = aab+b is not a binary operation.

Example: 7. ’ o ’ is a composition in R such that aob = a + 3b for a, b ∈ R.

(1) Since a, b ∈ R, a + 3b is a real number and hence a + 3b ∈ R i.e. aob ∈ R. Therefore o is a binary operation in R.

(2) a ◦ b = a + 3b and boa = b + 3a Since abb 6= boa for a, b ∈ R, 0o ’ is not commutative in R.

(3) (a ◦ b)oc = (aob) + 3c = (a + 3b) + 3c and ao(boc) = a + 3(boc) = a + 3(b + c) = a + 3b + 9c.

Since (aob)oc 6= ao(boc) for a, b, c ∈ R, o is not associative in R.

Example: 8. On Q define ∗ such that a ∗ b = ab + 1 for every a, b ∈ Q.

(1) Since ab +1 ∈ Q for every a, b ∈ Q then ∗ is a binary operation.

(2) Since a ∗ b = ab + 1 = ba + 1 = b ∗ a, then ∗ is commutative.

(3) ∀a, b, c ∈ Q, (a ∗ b) ∗ c = (ab + 1) ∗ c = (ab + 1)c + 1 = abc + c + 1and a ∗ (b ∗ c) = a ∗ (bc + 1) = a(bc + 1) + 1 = abc + a + 1

⇒ (a ∗ b) ∗ c 6= a ∗ (b ∗ c) ⇒ ∗ is not associative in Q.

Example: 9. On R − {−1} define o such that aob = $\frac{a}{b+1} $for every a, b ∈ R − {−1}.

(1) Since $\frac{a}{b+1} $∈ R − {−1} for every a, b ∈ R − {−1}, then o is a binaryoperation.

(2) Since aob = $\frac{a}{b+1} $and boa = $\frac{b}{a+1} $then aob ≠ boa and hence o is not commutative.

(3) ∀a, b, c ∈ R − {−1}, (aob)oc = $\frac{a}{b+1} $oc = =$\frac{\frac{a}{b+1}}{c+1}$and ao(boc) = ao $\frac{b}{a+1} $

⇒ 0 is not associative R − {−1}.

Composition table for an operation on finite sets (Cayley’s composition table) Sometimes an operation o on a finite set can conveniently be specified by a table called the composition table. The construction of the table is explained below.

Let S = {a₁, a₂, . . . a_i, a_j, . . . a_n} be a finite set with n elements. Let a table with (n + 1) rows and (n + 1) columns be taken. Let the squares in the first row be filled in with a, a₁, a₂, . . . a_n, and the squares in the first column be filled in with a, a₁, a₂, . . . a_n. Let a_i(1 ≤ i ≤ n) and a_j(1 ≤ j ≤ n) be any two elements of S.

Let the product a_ioa_j. obtained by operating a_i with a_j be placed in the square which is at the intersection of the row headed by ai and the column headed by a_j. Thus the following table be got.

From the composition table, we can infer the following laws.

(1) Closure law. If all the products formed in the table are the elements of s, the ’ o ’ is said to be a binary operation in S, and S is said to be closed under the composition ’ o ’.

Otherwise, o is not a binary operation in S and the set S is not closed under the operation o.

(2) Commutative law. If the elements in every row are identical with the corresponding elements in the corresponding column, then the composition o is said to be commutative in S. Otherwise, the binary operation o is not commutative in S.

(3) Associative law. Also, we can know from the table whether the binary operation follows associative law or not.

Note: The diagonal through a 1oa 1 and anoan is called the leading diagonal in the table. If the elements in the table are symmetric about the leading diagonal, then we infer that o is commutative in S.

Binary Operations Identity Element.

Definition. Let o be a binary operation on a non-empty set S. If there exists an element e ∈ S such that aoe = a = eoa ∀a ∈ A, then e is called Identity of S w.r.t. the operation o. If e is an identity of S w.r.t. o, then it can be proved to be unique.

Example: 1 In Z, 0 is the identity w.r.t. ’ + ’ since a + 0 = a = 0 + a, ∀a ∈ Z. But in N, 0 is not the idntity w.r.t. + since 0 ∈/ N and 1 is the identity w.r.t. ’ ’ as a · 1 = a = 1 · a∀a ∈ N.

Example: 2 In R, 0 is the identity w.r.t. + since a + 0 = a = 0 + a, ∀a ∈ R. In R, 1 is the identity w.r.t. ’!’ since a · 1 = a = 1 · a, ∀a ∈ R.

Note: Operations (−), (÷) are not binary operations in N. But +, −, · are binary operations in R and ÷ is a binary operation in R ∗ (non-zero real number set). Also o is the identity in R w.r.t. +, 1 is the identity in R (w.r.t. ’ · ’ where as ’ − ’ and 0 ÷ 0 ’ do not have identity element in R.)

Invertible element. Definition.

Let e be the identity element in S w.r.t. the binary operation o. An element a ∈ S is said to be invertible w.r.t. o, if there is an element b in S such that aob = e = boa and b is called inverse of a. If o is associative in S, then inverse of a is unique in S and is denoted by a− 1 or sometimes as 1/a if the operation is · and by −a if the operation is +.

Note
1. a ◦ a− 1 = a− 1oa = e and e− 1oe = eoe− 1 = e. Also a− 1− 1 = a.

2. In R, −a is the inverse w.r.t. ’ + ’ and 1/a(a 6= 0) is the inverse w.r.t. ’.’of a. For : a + (−a) = 0 = (−a) + a and a · a1 = 1 = a1 · a(a 6= 0)

3. −a is not the inverse of a in N w.r.t. + and a− 1 is not the inverse of a in N w.r.t. •.

Also a− 1 is the inverse of a in R. w.r.t. ’ ’ ’ and −a is the inverse of a. w.r.t. ’+.’ in R.

Example: 1. S = {1, −1, i, −i} and usual multiplication is the operation in S.
Then we have the following composition table. We can clearly see that is a binary operation in N following commutative law and associative law.

Example: 2. Consider the binary operation on the set {1, 2, 3, 4, 5} defined by aob = min{a, b}. Composition table is :

Example: 3. Define a binary operation ∗ on the set A = {0, 1, 2, 3, 4, 5} as a ∗ b = a + b, −if6a, +if ab <+ b6 ≥ 6 .0 is the identity w.r.t. ∗ and each element (a 6= 0) of the set is invertible with 6 − a being the inverse of a.

For: Composition table is :

(1) ∗ is binary since every entry belongs to A.

(2) Every row is same as the corresponding column ⇒ ∗ is commutative.

(3) Since every element of the first row = every corresponding element of the top row, the identity element exists and it is 0 .

since 0 ∗ 0 = 0, 0 ∗ 1 = 1, . . . ., 0 ∗ 5 = 5 and 0 ∗ 0, 1 ∗ 0 := 1, 2 ∗ 0 =2, . . . . . . . . . , 5 ∗ 0 = 5.

(4) Since 1 ∗ 5 = 0 = 5 ∗ 1, 1− 1 = 5 and 5− 1 = 1;

Since 2 ∗ 4 = 0 = 4 ∗ 2, 2− 1 = 4 and 4− 1 = 2;

Since 3 ∗ 3 = 0, 3− 1 = 3. Also 0− 1 = 0.

Solved Problems

Example: 1. (1) Show that the operation o given by aob = ab. is a binary operation. on the set of natural numbers N. Is this operation associative and commutative in N?
(2) a ∗ b = smaller of a and b (O.U.A12)

Solution. (1) N is the set of natural numbers and o is the operation defined in N such that aob = ab for a, b ∈ N. When a, b ∈ N, ab = a × a × . . . b times is also a natural number and hence ab ∈ N.

∴ o is binary operation in N. Let a, b, c ∈ N.

∴ (aob)oc = (aob)c = abc = abc and ao(boc) = aobc = abc

∴ (aob)oc 6= ao(boc) and o is not associative in N.

Since ab 6= ba i.e. ’ o0 is not commutative in N.

(2) a, b, c ∈ Z +. Now a ∗ b = smaller of a and b.

Suppose a < b. Then a ∗ b = a, and b ∗ a = a

⇒ a ∗ b = b ∗ a ⇒ ∗ is commutative.

Now suppose (a ∗ b) ∗ c = a ∗ c = a and

a ∗ (b ∗ c) = a ∗ b = a ⇒ (a ∗ b) ∗ c = a ∗ (b ∗ c)

⇒ ∗ is associative.

Example: 2. Let S be a non-empty set and o be an operation on S defined by aob = a for a, b ∈ S. Determine whether o is commutative and associative in S.

Solution. Since aob = a for a, b ∈ S and boa = b for a, b ∈ S, aob 6= boa.

∴ o is not commutative in S. Since (aob)oc = aoc = a

ao(boc) = aob = a for a, b, c ∈ S

∴ o is associative in S.

Example: 3. o is an operation defined on Z such that aob = a + b − ab for a, b ∈ Z. Is the operation, o a binary operation in Z ? If so, is it associative and commutative in Z?

Solution.

Given

o is an operation defined on Z such that aob = a + b − ab for a, b ∈ Z. Is the operation, o a binary operation in Z

If a, b ∈ Z we have a + b ∈ Z, ab ∈ Z

a + b − ab ∈ Z.

∴ aob = a + b − ab ∈ Z

∴ o is a binary operation in Z.

Since aob = a + b − ab = b + a − ba = boa, ’ o ’ is commutative in Z.

Now. (aob)ac = (aab) + c − (aob)c

= a + b − ab + c − (a + b − ab)c

= a + b − ab + c − ac − bc + abc and ao(boc) = a + (boc) − a(boc)

= a + b + c − bc − a(b + c − bc)

= a + b + c − bc − ab − ac + abc

= a + b − ab + c − ac − bc + abc

∴ (aabˆ )oc = ao(boc) and hence o is associative in Z.

Example: 4. S = {a, b, c} and o is an operation on S for which the following composition table is formed. Is the operation o a binary operation in S ? Is the operation o in S commutative and associative?

Solution. All the products formed are the elements of S.

∴ o is a binary operation in S and hence S is closed under the operation o.

Since the elements in every row are identical with corresponding elements in the corresponding column, o is commutative in S.

Since (aob)oc = boc = a = aoa = ao(boc), (boa) oc = boc = bo(aoc) etc., o is associative in S.

Example: 5. Fill in the blanks in the following composition table so that o is associative in S˙ = {a, b, c, d}

Solution.

Using associative law and by trial and error :

Since do(aoa) = doa and (doa)oa = do(aoa) = doa, doa must be equal to d only.

Since do(boa) = dob and (dob)oa = do(boa), dob must be equal to d only.

Since do(coa) = doc and (doc)oa = do(coa), doc must be equal to a only.

Since do(doa) = dod and do(doa) = (dod)oa, dod must be equal to a.

Thus d, d, a, a are respectively the four products.

Example: 6. Let P (S) be the power set of a non-empty set S. Let ’ ∩ ’ be an operation in P( S). Prove that associative law and commutative law are true for the operation ∩ in P(S) :

Solution.

Given

Let P (S) be the power set of a non-empty set S. Let ’ ∩ ’ be an operation in P( S).

P (S) = Set of all possible subsets of S.

Let A, B, C ∈ P(S). Since A ⊆ S, B ⊆ S ⇒ A ∩ B ⊆ S ⇒ A ∩ B ∈ P(S)

Also B ∩ A ⊆ S ⇒ B ∩ A ∈ P(S).

∴ ∩ is a binary operation in P(S).

Also A ∩ B = B ∩ A.

∴ ∩ is commutative in P(S).

Again A ∩ B, B ∩ C, (A ∩ B) ∩ C, and A ∩ (B ∩ C) are subsets of S.

∴ (A ∩ B) ∩ C, A ∩ (B ∩ C) ∈ P(S).

Since (A ∩ B) ∩ C = A ∩ (B ∩ C), ∩ is associative in P(S).

Example: 7. A = {a, b}. Consider the set S of all mappings from A → A. Is the composition of mappings denoted by o is a binary composition in S.

Solution.

Given

A = {a, b}. Consider the set S of all mappings from A → A.

Total number of possible mappings from A → A is 4.

Let them be I : A → A = {(a, a), (b, b)}

f₁ : A → A = {(a, b), (b, a)}

f₂ : A → A = {(a, a), (b, a)}

f₃ : A → A = {(a, a), (b, b)}

∴ S = {I, f₁, f₂, f₃}. Let the composition of mappings be denoted by o.

Composition table is :

Clearly (1) o is a binary operation in S,

(2) o is not commutative in S and

(3) o is not associative in S. sin a(f₂of₃) of 16= f₂o (f₃of₁)

The post Binary Operations (Definition, Types, and Examples) appeared first on Answer Key for Math.

Abstract Algebra Subgroups Notes

Kumar — Sun, 06 Aug 2023 12:33:39 +0000

COMPLEX DEFINITION

Any subset of a group G is called a complex of G.

Example. 1. The set of integers is a complex of the group (R,+).

Example. 2. The set of even integers is a complex of the group (Z,+).

Example. 3. The set of odd integers is a complex of the group. (R, +).

Example. 4 The set (1, – 1) is a complex of the multiplicative group G = (1, -1, i,-i)

Multiplication Of Two Complexes.

Definition: If M and N are any two complexes of group G then

MN = (mn ∈ G / m ∈ M, n ∈ N)

Clearly, MN ⊆ G and MN is called the product of the complexes M, N of G.

Theorem 1: The multiplication of complexes of a group G is associative.

Proof: Let M, N, and P be any three complexes in a group G.

Let m ∈ M, n ∈ N, p ∈ P so that m, n, p ∈ G.

We have MN = (mn ∈ G / m ∈ M, n ∈ N} so that

(MN) P = ((mn)p ∈ G / mn ∈ MN, p ∈ P) = (m (n p) ∈ G / m ∈ M, np ∈ NP)

= M (NP) (∵ associativity is true in G )

Note. If HK = KH then we cannot imply that hk = kh for all h ∈ H and for all k ∈ K. What we imply is HK ⊆ KH and KH ⊆ HK.

Definition: If M is a complex in a group G, then we define $\mathbf{M}^{-1}=\left\{m^{-1} \in \mathbf{G} / m \in \mathbf{M}\right\} \text { i.e. } \mathbf{M}^{-1}$ is the set of all inverses of the elements of M.

Clearly $\mathbf{M}^{-1} \subseteq \mathbf{G}$.

Theorem 2: If M, N are any two complexes in group G then $(\mathbf{M N})^{-1}=\mathbf{N}^{-1} \mathbf{M}^{-1}$

Proof. We have MN = {mn ∈ G / m ∈ M, n ∈ N)

Now $(\mathbf{M N})^{-1}$ = $\left\{(m n)^{-1} \in \mathbf{G} / m \in \mathbf{M}, n \in \mathbf{N}\right\}$

= $\left\{n^{-1} m^{-1} \in \mathbf{G} / m \in \mathbf{M}, n \in \mathbf{N}\right\}=\mathbf{N}^{-1} \mathbf{M}^{-1}$

Subgroups

Definition: Let (G,.) be a group. Let H be a non-empty subset of G such that (H,.) be a group. Then H is called a subgroup of G.
It is denoted by H ≤ G or G ≥ H. And H < G or G > H we mean H ≤ G, but H ≠ G.

Note: A complex of group G. is only a subset of G but a subgroup of group G is a group. The binary operations in a group and its subgroup are the same.

Example. 1. (Z,.) is a subgroup of (Q,.). Also (Q+,.) ‘is a subgroup of (R+,.)

Example. 2. The additive group of even integers is a subgroup of the additive group of all integers. •

Example. 3. The multiplicative group {1,-1} is a subgroup of the multiplicative group (1,-1, i,-i}.

For: G = {1,-1,i,-i} is a group under usual multiplication.
The composition table is:

Here 1 is the identity and $(i)^{-1}=-i,(-i)^{-1}=i,(-1)^{-1}=-1$

Consider H = {1,-1} which is.a subset of a group (G,.).

Clearly (H,.) is a group.

Here 1 is the identity, $(-1)^{-1}=-1$

∴ H is a subgroup of G.

Similarly ({1},.),({1,-1,i,-i},.) are subgroups of (G,.).

Example. 4. (N,+) is not a subgroup of the group (Z,+) since identity does not exist in N under +.

Note 1: Every group having at least two elements has at least two subgroups. Suppose e is the identity element in a group G. Then $\{e\} \subseteq \mathbf{G}$ and we have $e e=e, e^{-1}=e$, etc.

So {e} is a subgroup of G. Also $\mathbf{G} \subseteq \mathbf{G}$. So G is also a subgroup of G. These two subgroups {e}, G of G are called trivial or improper subgroups of G. All other subgroups, if they exist, are called non-trivial or proper subgroups of G.

Note 2. A complex of a group need not be a subgroup of the group. But a subgroup of a group is always complex of the group.

Note 3. A complex of a group (G,.) need not be a subgroup w.r.t. the binary operation, but it can be a group w.r.t. another binary operation. For example, the complex $\left\{3^n, n \in z\right\}$ of the group (Z, +) is not a subgroup of (Z, +) w.r.t. binary operation + whereas the same subset is a group under multiplication.

It is clear that every subgroup of an abelian group is abelian. But for the non-abelian group, it may not be true.

⇒ $\mathbf{P}_3=\left\{f_1, f_2, f_3, f_4, f_5, f_6\right\}$ set of all bisections on three symbols is a non-abelian group.

But $\mathbf{A}_3=\left\{f_1, f_5, f_6\right\}$ and H = $\left\{f_1, f_2\right\}$ an abelian subgroup of P3.

Lattice Diagram. Often it is useful to show the subgroups of a group by a Lattice diagram. In this diagram, we show the larger group near the top of the diagram followed by a line running toward a subgroup of the group.

We give below the Lattice Diagram for the multiplicative group $\{1,-1, i,-i\}$.

Abstract Algebra Subgroups Notes The Identity And Inverse Of An Element Of A Subgroup H Of A Group G.

Theorem 3: The identity of a subgroup H of a group is the same as the G

Proof. Let a ∈ H and e’ be the identity of H.

Since H is a group, ae’ = a ……………(l)

Let e be the identity in G.

Again a ∈ H => a ∈ G

∴ ae = a ………….(2)

Also e’ ∈ H => e’ ∈ G

From (1) and (2), ae’ = ae => e’= e (using left cancellation law).

Theorem 4: The inverse of any element of a subgroup H of group G is the same as the inverse of that element regarded as an element of group G.

Proof. Let e be the identity in -G.

Since H is a subgroup of G, e is also the identity in H.

Let a ∈ H.

∴ a ∈ G.

Let b be the inverse of a in H and c be the inverse of an in G. Then ab = e and ac = e.

=> ab = ac => b – c (using left cancellation law)

Theorem 5. If H is any subgroup of a group G . then $\mathbf{H}^{-1}=\mathbf{H}$.

Proof. Let H be a subgroup of group G. Let $h^{-1} \in \mathbf{H}^{-1}$

By def. of $\mathbf{H}^{-1}, h \in \mathbf{H}$

Since H is a subgroup of a group G, $h^{-1} \in \mathbf{H}$

∴ $h^{-1} \in \mathbf{H}^{-1} \Rightarrow h^{-1} \in \mathbf{H}$

∴ $\mathbf{H}^{-1} \subseteq \mathbf{H}$

Again $h \in \mathbf{H} \Rightarrow h^{-1} \in \mathbf{H} \Rightarrow\left(h^{-1}\right)^{-1} \in \mathbf{H}^{-1} \Rightarrow h \in \mathbf{H}^{-1}$

∴ $\mathbf{H} \subseteq \mathbf{H}^{-1} \text {. Hence } \mathbf{H}^{-1}=\mathbf{H}$

Note. The converse of the above theorem is not true i.e. if H is any complex of a group G such that $\mathbf{H}^{-1}=\mathbf{H}$, then H need not be a subgroup of G.

e.g. H = {-1} is a complex of the multiplicative group G = (1, – 1}.

Since the inverse of -1 is -1, then $\mathbf{H}^{-1}=\{-1\}$.

But H = {-1} is not a group under multiplication since (-1)(-1) = 1 ∉ H (Closure is not true) i.e. H is not a subgroup of G.

Hence even if $\mathbf{H}^{-1}=\mathbf{H}$ , H is not a subgroup of G.

Theorem 6. If H is any subgroup of group G, then HH = H.

Proof. Let x ∈ HH so that x = $h_1 \cdot h_2$ where $h_1 \in \mathbf{H}$ and $h_2 \in \mathbf{H}$. Since H is a subgroup, $h_1 h_2 \in \mathbf{H}$

∴ x ∈ H

∴ HH ⊆ H.

Let $h_3 \in \mathbf{H}$ and e be the identity in H.

Then $h_3=h_3 e \in \mathbf{H H}$

∴ H ⊆ HH

∴ HH = H

Abstract Algebra Subgroups Notes Criterion For A Complex To Be A Subgroup

Theorem 7. A non-empty complex H of a group G is a subgroup of G if and only if

(1) $a \in \mathbf{H}, b \in \mathbf{H} \Rightarrow a b \in \mathbf{H}$

(2) $a \in \boldsymbol{H}, a^{-1} \in \mathbf{H}$

Proof. The conditions are necessary.

Let H be a subgroup of the group G.

• To prove that (1), (2) are true

∴ H is a group.

∴ By closure axiom a, b ∈ H => ab ∈ H and by inverse axiom $a \in \mathbf{H} \Rightarrow a^{-1} \in \mathbf{H}$

The conditions are sufficient.

Let (1) and (2) be true.

To prove that H is a subgroup of G.

1. By (1) closure axiom is true in H.

2. The elements of H are also elements of G. Since G is a group, the composition ‘ in G is associative and hence the composition in H is associative.

3. Since H is non-empty, let a ∈ H.

∴ By (2) $a^{-1} \in \mathbf{H}$

∴ $a \in \mathbf{H}, a^{-1} \in \mathbf{H} \Rightarrow a a^{-1} \in \mathbf{H}$.

=> e ∈ H (∵ $a a^{-1} \in \mathbf{H} \Rightarrow a a^{-1} \in \mathbf{G} \Rightarrow a a^{-1}=e$ where e is the identity in G ).

=> e is the identity in H.

4. Since we have $a \in \mathbf{H} \Rightarrow a^{-1} \in \mathbf{H}$ and $a a^{-1}=e$ evety element of H possess inverse in H. Hence H itself is a group for the composition in G. So H is a subgroup of G.

Note 1. If the operation in G is +, then the conditions in the above theorem can be stated as follows :

a, $ b \in \mathrm{H} \Rightarrow a+b \in \mathrm{H}$ , (ii) $a \in \mathbf{H} \Rightarrow-a \in \mathbf{H}$

2. It is called a Two-step subgroup Test.

Theorem 8. H is a non-empty complex of a group G. The necessary and sufficient condition for H to be a subgroup of G is $a, b \in \mathbf{H} \Rightarrow a b^{-1} \in \mathbf{H}$ where $b^{-1}$ is the inverse of. b in G.

Proof.

Given

H is a non-empty complex of a group G. The necessary and sufficient condition for H to be a subgroup of G is $a, b \in \mathbf{H} \Rightarrow a b^{-1} \in \mathbf{H}$ where $b^{-1}$ is the inverse of. b in G.

The condition is necessary.

Since H is a group by itself, $b \in \mathbf{H} \Rightarrow b^{-1} \in \mathbf{H}$

∴ $a \in \mathbf{H}, b \in \mathbf{H} \Rightarrow a \in \mathbf{H}, b^{-1} \in \mathbf{H} \Rightarrow a b^{-1} \in \mathbf{H}$ (by closure axiom).

The condition is sufficient.

1. Since $\mathbf{H} \neq \phi, \text { let } a \in \mathbf{H}$

By hyp. $a \in \mathbf{H}, b \in \mathbf{H} \Rightarrow a b^{-1} \in \mathbf{H}$

a $\in \mathbf{H}, a^{-1} \in \mathbf{H} \Rightarrow a a^{-1} \in \mathbf{H} \Rightarrow a a^{-1} \in \mathbf{G}$

But in G, $a \in \mathbf{G} \Rightarrow a a^{-1}=e$, e is the identity in G.

∴ e ∈ H

2. We have $e \in \mathbf{H}, a \in \mathbf{H} \Rightarrow e a^{-1} \in \mathbf{H} \Rightarrow a^{-1} \in \mathbf{H}$

∴ $a \in \mathbf{H} \Rightarrow a^{-1} \in \mathbf{H}$

3. $b \in \mathbf{H} \Rightarrow b^{-1} \in \mathbf{H}$

∴ $a \in \mathbf{H}, b \in \mathbf{H} \Rightarrow a, b^{-1} \in \mathbf{H} \Rightarrow a\left(b^{-1}\right)^{-1} \in \mathbf{H}$

=> $a b \in \mathbf{H}$

4. Since all the elements of H are in G and since the composition is associative in

G, the composition is associative in H.

∴ H is a group for the composition in G and hence H is a subgroup of G.

Note 1. If the operation in G is +, then the condition in the above theorem can be stated as follows :

a$\in \mathbf{H}, b \in \mathbf{H} \Rightarrow a-b \in \mathbf{H}$

2. The above theorem can be used to prove that a certain non-empty subset of a given group is a subgroup of the group. It is called the One-step subgroup Test.

Theorem 9. A necessary and sufficient condition for a non-empty complex H of a group G to be a subgroup of G is that .

Proof. The condition is necessary.

Let H be a subgroup of G.

Let $a b^{-1} \in \mathbf{H H}^{-1}$ so that a ∈ H, b ∈ H

Since H is a group we have $b^{-1} \in \mathbf{H}$.

∴ $a \in \mathbf{H} \Rightarrow b^{-1} \in \mathbf{H} \Rightarrow a b^{-1} \in \mathbf{H}$ , (By closure axiom)

∴ $\mathbf{H H}^{-1} \subseteq \mathbf{H}$

The condition is sufficient.

Let $\mathrm{HH}^{-1} \subseteq \mathbf{H}$. Let a, b ∈H.

∴ $a b^{-1} \in \mathbf{H H}^{-1}$

Since $\mathbf{H H}^{-1} \subseteq \mathbf{H}, a b^{-1} \in \mathbf{H}$

∴ H is a subgroup of G.

Theorem 10. A necessary and sufficient condition for a non-empty subset H of a group G to be a subgroup of G is that $\mathbf{H H}^{-1}=\mathbf{H}$.

Proof. The condition is necessary.

Let H be a subgroup of G. Then we have $\mathrm{HH}^{-1} \subseteq \mathrm{H}$.

Let e be the identity in G.

∴ e is the identity in H.
Let h ∈ H

∴ $h=h e=h e^{-1} \in \mathbf{H H}^{-1}$

∴ $\mathbf{H} \subseteq \mathbf{H H}^{-1}$

∴ $\mathbf{H H}^{-1} \subseteq \mathbf{H}$

The condition is sufficient.

Let $\mathbf{H} \mathbf{H}^{-1}=\mathbf{H}$.

∴ $\mathbf{H H}^{-1} \subseteq \mathbf{H}$

∴ H is a subgroup of G.

Theorem 11. The necessary and sufficient condition for a finite complex H of a group G is $a, b \in \mathbf{H} \Rightarrow a b \in \mathbf{H}$

OR

Prove that a non-empty finite subset of a group which is closed under multiplication is a subgroup of G.

Proof. The condition is necessary.

H be a subgroup of G. By closure axiom, .

The condition is sufficient.

Let $a, b \in \mathbf{H} \Rightarrow a b \in \mathbf{H}$ (∵ H ≠ ∅)

Let a ∈ H.

By hyp. we have $a \in \mathbf{H}, a \in \mathbf{H} \Rightarrow a \cdot a \in \mathbf{H} \Rightarrow a^2 \in \mathbf{H}$

Again $a^2 \in \mathbf{H}, a \in \mathbf{H} \Rightarrow a^2, a \in \mathbf{H} \Rightarrow a^3 \in \mathbf{H}$

By induction, we can prove that $a^n \in \mathbf{H}$ where n is any positive integer.

Thus all the elements $a, a^2, a^3, \ldots, a^n, \ldots$ belong to H and they are infinite in number.

But H is a finite subset of G.

But H is a finite subset of G.

Therefore, there must be repetitions in the collection of elements.

Let $a^r=a^s$ for some positive integers r and s such that r > s

∴ $a^r \cdot a^{-s}=a^s \cdot a^{-s}=a^{s-s}=a^0=e$ where e is the identity in G.

Since r – s is a positive integer, $a^{r-s} \in \mathbf{H}^{\prime} \Rightarrow e \in \mathbf{H}$

∴ $e=a^0 \in \mathbf{H}$.

Now r > s => r – s ≥ 1

∴ r – s – 1 ≥ 0 and hence $a^{r-s-1} \in \mathbf{H}$

Also $a^{r-s-1} a=a^{r-s}=e=a \cdot a^{r-s-1}$

∴ We have for $a \in \mathbf{H}, a^{r-s-1} \in \mathbf{H}$ as the inverse of a.

Thus each element of H is invertible.

Since all the elements of H are elements of G, associativity is satisfied.

H is a group for the composition in G and hence H is a subgroup of G.

Cor. A finite non-empty subset H of a group G is also a subgroup of G if and only if HH = H.

e.g. $\left(\mathbf{Z}_6,+_6\right)$ is a group and $\left(\mathbf{H}=\{0,2,4\},+_6\right)$ is a subgroup of it.

For: $\mathbf{H} \subset \mathbf{Z}_6$. Also $0+_6 0=0,0+_6 2=2,0+_6 4=4,2+_6 4=0,4+_6 4=2$ etc.

So $a, b \in \mathbf{H} \Rightarrow a+_6 b \in \mathbf{H}$

Note. The criterion given in the above theorem is valid only for finite subsets of a group G. It is not valid for an infinite subset of an infinite group G.

e.g. (Z, +) is a group. Let N be the set of all positive integers.

N ⊂ Z. Also (N, +) is not a group even though a, b ∈ N => a + b ∈ N.

∴ (N,+) is not a subgroup of (Z,+). Hence the above theorem is not satisfied.

Theorem 12. A non-empty subset H of a finite group G is a subgroup if $a \in \mathbf{H}, b \in \mathbf{H} \Rightarrow a b \in \mathbf{H}$

OR

A necessary and sufficient condition for a complex H of a finite group G to be a subgroup is that $a \in \mathbf{H}, b \in \mathbf{H} \Rightarrow a b \in \mathbf{H} .$.

Proof. The condition is necessary.

Let H be a subgroup of a finite group G.

Then H is closed w.r.t. the composition in G.

∴ $a \in \mathbf{H}, b \in \mathbf{H} \Rightarrow a b \in \mathbf{H}$

The condition is sufficient.

H is a non-empty subset (complex of G ) of a finite group G such that

a $\in \mathbf{H}, b \in \mathbf{H} \Rightarrow a b \in \mathbf{H}$

Now we have to prove that H is a subgroup of G.

Associativity. Since H is a subset of G, all the elements of H are the elements of G and hence associativity is true in H w.r.t. the composition in G.

Existence of identity. Let a ∈ H.

a ∈ G. Since G is finite and since every element of a finite group is of finite order, it follows that the order of a is finite.
Let 0 (a) = n

∴ $a^n=e$ where e is the identity in G.

By closure law in H, we have $a^2, a^3, \ldots, a^n, \ldots \in \mathbf{H}$. …………..(1)

Since $a^n=e=a^0$, we have $a^0=e \in \mathbf{H}$ i.e. identity exists in H.

Existence of inverse. Let a ∈ H Here $e=a^n=a^0$.

∴ a ∈ G and 0(a) = n => n is the least positive integer such that

=> (n – 1) > 0. $a^n=e$

By (1), $a^{n-1} \in \mathbf{H}$.

Now in G, $a^{n-1} a=a^n=a a^{n-1}$.

=> $a^{n-1} a=a a^{n-1}=e$ true in H .

=> $a^{-1}=a^{n-1}$

=> Every element of H is invertible.

=> H is a group and hence a subgroup of G.

Abstract Algebra Subgroups Notes Criterion For The Product Of Two Subgroups To Be A Subgroup

Theorem 13. If H and K are two subgroups of a group G, then HK is a subgroup of G iff HK = KH.

Proof. Let H, K be any two subgroups of G.

1st part. Let HK = KH. To prove that HK is a subgroup of G.

So it is sufficient to prove that $(\mathbf{H K})(\mathbf{H K})^{-1}=\mathbf{H K}$

⇒ $(\mathbf{H K})(\mathbf{H K})^{-1}$

= $(\mathbf{H K})\left(\mathbf{K}^{-1} \mathbf{H}^{-1}\right)$(Theorem 2.)

= $\mathbf{H}\left(\mathbf{K K}^{-1}\right) \mathbf{H}^{-1}$ (∵ Complex multiplication is associative).

= $\mathbf{H}(\mathbf{K}) \mathbf{H}^{-1}$ (Theorem 10) = $(\mathbf{H K}) \mathbf{H}^{-1}$ (Theorem 1) .

= $\text { (KH) } \mathbf{H}^{-1}$ (Hyp.) = $\mathbf{K}\left(\mathbf{H H}^{-1}\right)=\mathbf{K H}=\mathbf{H K}$.

HK = KH => HK is a subgroup of G.

2nd part. Let HK be a subgroup of group G.

∴ $(\mathrm{HK})^{-1}=\mathrm{HK} \Rightarrow \mathrm{K}^{-1} \mathbf{H}^{-1}=\mathrm{HK} \Rightarrow \mathrm{KH}=\mathrm{HK}$

(∵ K is a subgroup, $\mathbf{K}^{-1}=\mathbf{K}$ and H is a subgroup, $\mathbf{H}^{-1}=\mathbf{H}$)

Cor. If H, K are subgroups of an abelian group G, then HK is a subgroup of G.

For: Since G is abelian, HK = KH. By the above theorem, HK is a subgroup of G.

Abstract Algebra Subgroups Notes Solved Problems

Example.1: If Z is the additive group of integers, then prove that the set of all multiples of integers by a fixed integer m is a subgroup of Z.
Solution:

Given

Z is the additive group of integers

We have Z = { ….,3,-2,-1, 0,1,2,3,…}

Let H = {…,- 3m, – 2m, – m, 0, m, 2m, 3m,…} = mz

where m is a fixed integer.

Let a = rm, b = sm be any two elements of H where r, s are integers.

Then a – b = rm – sm = (r – s) m – pm where p is an integer

=> a – b ∈ H.

∴ a, b ∈ H => a – b ∈ H.

∴ H is a subgroup of Z.

Example. 2: Prove that in the dihedral group of order 8, denoted by $\mathbf{D}_4$, the subset $\mathbf{H}=\left\{r_{360}, r_{180}, x, y\right\}$ is subgroup of $\mathbf{D}_4$.

Solution: We can observe from the composition table of $\mathbf{D}_4$.

(1) Closure is obvious.

(2) Associativity is evident since the composition of maps is associative

(3) The identity element of H is $r_{360}$.

(4) Each element of H is inverse of itself. I

∴ H is a group. Here H is a subgroup of $\mathbf{D}_4$.

Example. 3 : $\mathbf{P}_3$ is a non-abelian group of order 6. $\mathbf{A}_3$ is a subgroup of $\mathbf{P}_3$. Also $\mathbf{A}_3$ is an abelian subgroup of $\mathbf{P}_3$.

Example. 4: We have $\mathbf{G}=\left\{r_0, r_1, r_2, f_1, f_2, f_3\right\}$ the set of all symmetries of cm Unilateral triangle, as a non-abelian group.

Consider $\mathrm{H}=\left\{r_0, r_1, r_2\right\}$ . We can see from the composition table that H is a subgroup of G. Also H is abelian. Hence a non-abelian group can have an abelian subgroup.

Example. 5: S, the set of all ordered pairs (a,b) of real numbers for which a ≠ 0 w.r.t the operation x defined by (a, b) x (c, d) = (ac, be + d) is a non-abelian group.

Let H = {(1, b) | b ∈ R} be a subset of S. Show that H is a subgroup of the group (S, x).

Solution:

Given

S the set of all ordered pairs (a,b) of real numbers for which a ≠ 0 w.r.t the operation x defined by (a, b) x (c, d) = (ac, be + d) is a non-abelian group.

Let H = {(1, b) | b ∈ R} be a subset of S.

Identity in S is (1, 0). Clearly’ (1, 0) ∈ H.

The inverse of (a,b) in S is $\left(\frac{1}{a},-\frac{b}{a}\right)$ (∵ a ≠ 0)

The inverse of (1, c) in S is $\left(\frac{1}{1},-\frac{c}{1}\right)$ i.e (1, -c)

Clearly (1 ,c) ∈ H. Let (1,b) ∈ H.

∴ $(1, b)(1, c)^{-1}=(1, b) \times(1,-c)=(1,1, b .1-c)=(1, b-c) \in \mathbf{H} \text { since } b-c \in \mathbf{R}$

∴ $(1, b),(1, c) \in \mathrm{H} \Rightarrow(1, b) \times(1, c)^{-1} \in \mathrm{H}$

∴ H is a subgroup of (S, x).

Note. 

H is an abelian subgroup of the non-abelian group (S, x).

Hence a non-abelian group can have an abelian subgroup.

UNION AND INTERSECTION OF SUBGROUPS

Theorem 14. If $\mathrm{H}_1 \text { and } \mathrm{H}_2$ are two subgroups of a group G then $\mathrm{H}_1 \cap \mathrm{H}_2$, is also a subgroup of G.

(OR)

If H and K are subgroups of a group G show that $\mathbf{H} \cap \mathbf{K}$ is also a subgroup of G. )

Proof. Let $\mathbf{H}_1 \text { and } \mathbf{H}_2$ be two subgroups of G.

Let e be the identity in G.

∴ $e \in \mathbf{H}_1 \text { and } e \in \mathbf{H}_2 \Rightarrow e \in \mathbf{H}_1 \cap \mathbf{H}_2$

∴ $\mathbf{H}_1 \cap \mathbf{H}_2 \neq \phi$

Let $a \in \mathbf{H}_1 \cap \mathbf{H}_2, b \in \mathbf{H}_1 \cap \mathbf{H}_2$

∴ $a \in \mathbf{H}_1, a \in \mathbf{H}_2 \text { and } b \in \mathbf{H}_1, b \in \mathbf{H}_2$

Since $\mathbf{H}_1$ is a subgroup, $a \in \mathbf{H}_1 \text { and } b \in \mathbf{H}_1 \Rightarrow a b^{-1} \in \mathbf{H}$

Similarly $a b^{-1} \in \mathbf{H}_2$.

∴ $a b^{-1} \in \mathbf{H}_1 \cap \mathbf{H}_2$

Thus we have $a \in \mathbf{H}_1 \cap \mathbf{H}_2, b \in \mathbf{H}_1 \cap \mathbf{H}_2 \Rightarrow a b^{-1} \in \mathbf{H}_1 \cap \mathbf{H}_2$.

∴ $\mathbf{H}_1 \cap \mathbf{H}_2$ is a subgroup of G.

Note 1. The intersection of an arbitrary family of subgroups of a group is a subgroup of the group i.e. if $\left\{\mathbf{H}_i / i \in \Delta\right\}$ is any set of subgroups of a group G, then $\bigcap_{i \in \Delta} \mathbf{H}_i$ is a subgroup of G.

2. $\mathbf{H}_1 \cap \mathbf{H}_2$ is the largest subgroup of G contained in $\mathbf{H}_1 \cap \mathbf{H}_2$ is the subgroup contained in $\mathbf{H}_1 \text { and } \mathbf{H}_2$ and is the subgroup that contains every subgroup of G contained in both $\mathbf{H}_1 \text { and } \mathbf{H}_2$.

3. The union of two subgroups of a group need not be a subgroup of the group.

e.g. Let (Z,+) be the group of all integers.

Let $\mathbf{H}_1=\{\ldots-6,-4,-2,0,2,4 \ldots\}=2 \mathbf{Z}$ and

⇒ $\mathbf{H}_2=\{\ldots-12,-9,-6,-3,0,3,6,9 \ldots\}=3 \mathbf{Z}$ be two subgroups of Z.

We have $\mathbf{H}_1 \cup \mathbf{H}_2=\{\ldots,-12,-9,-6,-4,-3,-2,0,2,3,4,6,9 \ldots\}$.

Since $4 \in \mathbf{H}_1 \cup \mathbf{H}_2, 3 \in \mathbf{H}_1 \cup \mathbf{H}_2$ does not imply $4+3 \in
\mathbf{H}_1 \cup \mathbf{H}_2, \mathbf{H}_1 \cup \mathbf{H}_2$ is not closed under +.
∴ $\mathbf{H}_1 \cup \mathbf{H}_2$ is not a subgroup of (Z,+).

So the intersection of two subgroups of a group is a subgroup of the group whereas the union of two subgroups of a group need not be a subgroup of the group.

Thus we conclude: An arbitrary intersection of subgroups of a group G is a subgroup but the union of subgroups need not be a subgroup.

Theorem 15. The union of two subgroups of a group is a subgroup if one is contained in the other.

(OR)

If H and K are subgroups of a group Q then show that H ∪ K is a subgroup if either H ⊆ K or K ⊆ H

Proof. Let $\mathbf{H}_1 \text { and } \mathbf{H}_2$ be two subgroups of a group (G,.)

To prove that $\mathbf{H}_1 \cup \mathbf{H}_2$ is a subgroup ⇔ $\mathbf{H}_1 \subseteq \mathbf{H}_2 \text { or } \mathbf{H}_2 \subseteq \mathbf{H}_1$.

The condition is necessary.

Let $\mathbf{H}_1 \subseteq \mathbf{H}_2$

∴ $\mathbf{H}_1 \cup \mathbf{H}_2=\mathbf{H}_2$

Since $\mathbf{H}_2$ is a subgroup of G, $\mathbf{H}_1 \cup \mathbf{H}_2$ is a subgroup of G.

Similarly, $\mathbf{H}_2 \subseteq \mathbf{H}_1$ => $\mathbf{H}_1 \cup \mathbf{H}_2$ is a subgroup of G.

The condition is sufficient.

Let $\mathbf{H}_1 \cup \mathbf{H}_2$ be a subgroup of G.

We prove that $\mathbf{H}_1 \subseteq \mathbf{H}_2 \text { or } \mathbf{H}_2 \subseteq \mathbf{H}_1$

Suppose that $\mathbf{H}_1 \not \subset \mathbf{H}_2 \text { and } \mathbf{H}_2 \not \subset \mathbf{H}_1$

Since $\mathbf{H}_1 \not \subset \mathbf{H}_2, \exists a \in \mathbf{H}_1 \text { and } a \notin \mathbf{H}_2$.

Again $\mathrm{H}_2 \subset \mathrm{H}_1 \Rightarrow 3 b \in \mathrm{H}_2 \text { and } b \notin \mathrm{H}_1$.

From (1) and (2) we have that $a \in \mathbf{H}_1 \cup \mathbf{H}_2 \text { and } b \in \mathbf{H}_1 \cup \mathbf{H}_2$.

Since $\mathbf{H}_1 \cup \mathbf{H}_2$ is a subgroup, we have $a b \in \mathrm{H}_1 \cup \mathrm{H}_2$.

∴ $a b \in \mathbf{H}_1 \text { or } a b \in \mathbf{H}_2 \text { or } a b \in \mathbf{H}_1 \cap \mathbf{H}_2$.

Suppose $a b \in \mathrm{H}_1$.

Since $\mathrm{H}_1$ is a subgroup, $a \in \mathrm{H}_1 \text { and } a b \in \mathrm{H}_1$

=> $a^{-1} \in \mathbf{H}_1 \text { and } a b \in \mathbf{H}_1 \Rightarrow a^{-1}(a b) \in \mathbf{H}_1$

=> $\left(a^{-1} a\right) b \in \mathbf{H}_1 \Rightarrow c b \in \mathbf{H}_1 \Rightarrow b \in \mathbf{H}_1$ which is absurd by (2).

∴ ab ∉ $\mathbf{H}_1$

Similarly, we can show that ab ∉ $\mathrm{H}_2$.

∴ ab∉ $\mathbf{H}_1 \cap \mathbf{H}_2$

∴ ab ∉ $\mathrm{H}_1 \cup \mathrm{H}_2$ which is a contradiction that $\mathrm{H}_1 \cup \mathrm{H}_2$ is a group.

∴ we must have $\mathrm{H}_1 \subseteq \mathrm{H}_2 \text { or } \mathrm{H}_2 \subseteq \mathrm{H}_1$.

Note. $\left(Z_{16,}+16\right)$ is a group. $\mathrm{S}=\{0,8\}, \mathrm{T}=\{0,4,8,12\}$ under $+_{16}$ are two groups. Clearly, they are subgroups of $Z_{16}$. Since S ∪ T = {0,4,8,12} = T, we have $\left(S \cup T_{,}+{ }_{16}\right)$

as a subgroup of $Z_{16}$. Observe that $S \subset T$ i.e. S is contained in T.

Example.6. Prove that a set of all multiples of 3 is a subgroup of the group of integers under addition.

Solution: Consider $3 Z=\{3 n / n \in Z\}$

3$Z \neq \phi$ and 3Z is a subset of Z.

Let $3 m, 3 n \in 3 Z \Rightarrow m, n \in Z$

3$m-3 n=3(m-n) \in 3 Z$

∴ (3Z,+) is a subgroup of (Z, +) (using Th.8)

Example. 7. G & a group of non-zero real numbers under multiplication. Prove that

(1) $\mathbf{H}=\{x \in \mathbf{G} / x=1 \text { or } x \text { is irrational }\}$

(2) $\mathbf{K}=\{x \in \mathbf{G} / x \geq 1\}$ are not subgroups of G

Solution: (1) $\sqrt{2}, \sqrt{2} \in H$ but $\sqrt{2} \cdot \sqrt{2}=2 z \mathrm{H}$.

So H is not a subgroup even though H ⊂ G.

(2) 1 is the identity in G and K ⊂ G.

2 ∈ K but $2^{-1}=(1 / 2) \notin K$ . So K is not a subgroup.

Example. 8. $\left(Z_6=\{0,1,2,3,4,5\}_{,}+6\right)$ is a group. Prove that S = [0,2,4}, T = {0,3} are subgroups of $Z_6$ and S ∪ T, not a subgroup of $\mathrm{z}_6$.

Solution: S = {0,2,4}, T = {0,3} are subsets of $\mathrm{z}_6$. and From the tables, 0 is the identity

(1) $0^{-1}=0,2^{-1}=4,4^{-1}=2$

(2)$0^{-1}=0,3^{-1}=3$

Clearly $\left(\mathbf{S},+_6\right),\left(\mathbf{T},+_6\right)$ are subgroups of $\mathrm{z}_6$.

Now S ∪ T = {0,2,3,4} is not a subgroup of $\mathrm{z}_6$ as $1,5 \notin \mathbf{S} \cup \mathbf{T}$

The post Abstract Algebra Subgroups Notes appeared first on Answer Key for Math.

Orthogonal Trajectories

Kumar — Sat, 05 Aug 2023 04:43:59 +0000

Orthogonal Trajectories Of A Family Of Curves

Trajectory

Definition. If a carve C cuts every member of a given family of curves T according to some specified law, then the curve C is called the trajectory of the given family of curves T.

Orthogonal Trajectory

Definition. If a curve C cuts every member of a given family of curves T at a right angle, then the curve C is called an orthogonal trajectory of the family T.

example. Consider a family of concentric circles $x^2+y^2=a^2$ with centre at the origin, where ‘a’ is a parameter. We know that the line y = x passing through the origin cuts each member of the above family of circles at a right angle. Therefore y = x is an orthogonal trajectory of the above family.

Infact y = mx where m is any real number is also an orthogonal trajectory.

Thus the family of straight lines y = mx where m is a parameter form orthogonal trajectories for the family of concentric circles.

Conversely, for the family of straight lines y = mx passing through the origin, the family of concentric circles $x^2+y^2=a^2$ form orthogonal trajectories.

Orthogonal Trajectories Cartesian Coordinates

Theorem 1. The family of curves represented by the differential equation $F\left(x, y,-1 / y^{\prime}\right)=0$ to form the orthogonal trajectories of the family of given curves represented by $F\left(x, y, y^{\prime}\right)=0$.

Proof.

Let/(x,y,c) = 0 ….(1) be the equation of the family of curves, where c is the parameter.

Differentiating (1) w. r. t. x. and eliminating c from (1), we get the differential equation of the family (1).

Let the differential equation of the family (1) be F (x, y, y’) = 0.

Let C be a member of the family of curves (1) and T be a member of the family of trajectories of (1).

Let $\left(x_0, y_0\right)$ be any point on T.

Let two curves C and T intersect at P(x,y) at a right angle.

For each point on C, we associate an ordered triad (x,y,y’). where x,y are the coordinates of the point and y’ is the slope of the tangent line at that point to C.

Similarly, for each point on T we associate an ordered triad $\left(x_0, y_0, y_0^{\prime}\right)$.

If $\left(x, y, y^{\prime}\right) \text { and }\left(x_0, y_0, y_0^{\prime}\right)$ are ordered triads corresponding to P, we have $x=x_0, y=y_0$ and $y^{\prime}=-1 / y_0^{\prime}$since the two curves C and T cut each other at a right angle at P.

Since the ordered triad (x,y,y’) = 0 satisfies F (x,y,y’) = 0, it implies that the ordered triad $\left(x_0, y_0,-1 / y_0^{\prime}\right)$ satisfies the equation $\mathrm{F}\left(x_0, y_0,-1 / y_0^{\prime}\right)=0$ …(2)

But the ordered triad $\left(x_0, y_0-1 / y_0^{\prime}\right)$ represents any point on T. (2) is true for any point on T. Thus T is the locus of the point $\left(x_0, y_0\right)$.

Hence the differential equation satisfying T is F (x,y,-1/ y’) = 0

∴ The family of curves of F (x,y,-1/y’) = 0 are the orthogonal trajectories of the family of F (x,y,y’) = 0.

Working Rule To Find The Orthogonal Trajectories Of A Given Family Of Curves

Let f(x,y,c) = 0 be the equation of the given family of curves. ………………..(1)

(1)Differentiate (1) w.r.t. x and obtain the differential equation F (x,y,y’) = 0 by eliminating the parameter c. …………………..(2)

Replace $\frac{d y}{d x} \text { by }-\frac{1}{d y / d x}$ in (2)

Then the differential equation of the orthogonal trajectories if $\mathrm{F}\left(x, y,-\frac{d x}{d y}\right)=0$ …………………..(3)

(3) Solve the equation (3) to get the equation of the orthogonal trajectories of (1).

Self Orthogonal Family Of Curves

Definition. If each member of a given family of curves cuts every other member of the family at right angle, then the given family of curves is said to be self-orthogonal.

Note. For self orthogonal family of curves, the differential equation of the family is the same as the differential equation of the family of orthogonal trajectories.

Orthogonal Trajectories Solved Problems

Example 1: Find the orthogonal trajectories of the family of curves $y=\frac{1}{\log c_1 x}$ where $c_1$ is the parameter.
Solution.

Given equation of the family of curves is $y=\frac{1}{\log c_1 x} \Rightarrow \log c_1 x=\frac{1}{y}$ …………………..(1)

Differentiating (1) w. r. tx : $x: \frac{1}{c_1 x} \cdot c_1=-\frac{1}{y^2} \frac{d y}{d x}$

⇒ $\Rightarrow \frac{1}{x}=-\frac{1}{y^2} \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=-\frac{y^2}{x}$ …………………(2)

(2) is free from the parameter $c_1$.

Hence (2) is the differential equation of the given family (1).

Replacing $\frac{d y}{d x}$ by $\frac{d x}{d y}$ in (2), the differential equation of the orthogonal trajectories is

⇒ $-\frac{d x}{d y}=-\frac{y^2}{x} \Rightarrow y^2 d y=x d x \Rightarrow \int y^2 d y=\int x d x+c_2^{\prime}$

⇒ $\frac{y^3}{3}=\frac{x^2}{2}+c_2^{\prime} \Rightarrow 2 y^3=3 x^2+6 c_2^{\prime}=3 x^2+c_2$

∴ The equation of the orthogonal trajectories of the given family is $2 y^3=3 x^2+c_2$.

Orthogonal Trajectories Definition And Examples

Example. 2: Find the orthogonal trajectories of the family of curves $y=\frac{x}{1+c_1 x}$where $c_1$ is the parameter.
Solution.

Given, the equation of the family of curves is $y=\frac{x}{1+c_1 x}$ …………………(1)

Differentiating (1) i.e. $1+c_1 x=x / y \text { w.r.t. } x: c_1=\frac{y \cdot 1-x y^{\prime}}{y^2}$ ……………………(2)

Eliminating  from (1) and (2): $1+x\left(\frac{y-x y^{\prime}}{y^2}\right)=\frac{x}{y} \Rightarrow 1+\frac{x}{y}-\frac{x^2}{y^2} \frac{d y}{d x}=\frac{x}{y}$

⇒ $ 1=\frac{x^2}{y^2} \frac{d y}{d x}$ …………………….(3)

Now (3) is the differential equation of the given family (1).

Replacing $\frac{d y}{d x}$ by $-\frac{d x}{d y}$ in (3):

The Differential equation of the family of Orthogonal trajectories is

⇒ $1=\frac{x^2}{y^2}\left(-\frac{d x}{d y}\right) \Rightarrow y^2 \frac{d y}{d x}=-x^2$

⇒ $\int y^2 d y=-\int x^2 d x+c_2^{\prime} \Rightarrow \frac{y^3}{3}=-\frac{x^3}{3}+c_2^{\prime} \Rightarrow x^3+y^3=3 c_2^{\prime}=c_2$

∴ The equation of the orthogonal trajectories of the given family is $x^3+y^3=c_2$.

Example. 3. Find the orthogonal trajectories of the family of curves $\left(\frac{d y}{d x}\right)^2=\frac{a}{x}$
Solution.

Given the differential equation of the family of curves is

⇒ $\left(\frac{d y}{d x}\right)^2=\frac{a}{x}$ where a is any real number. ………………………(1)

Replacing $\frac{d y}{d x}$ by $-\frac{d x}{d y}$ in (1), the differential equation (D.E.) of the orthogonal trajectories

is $\left(-\frac{d x}{d y}\right)^2=\frac{a}{x} \Rightarrow\left(\frac{d x}{d y}\right)^2=\frac{a}{x} \Rightarrow \frac{d x}{d y}=\pm \frac{\sqrt{a}}{\sqrt{x}} \Rightarrow d y=\pm \frac{\sqrt{x}}{\sqrt{a}} d x \Rightarrow \int d y=\pm \frac{1}{\sqrt{a}} \int \sqrt{x} d x-c$ ………………………..(2)

⇒ $\Rightarrow y+c=\pm \frac{1}{\sqrt{a}} \frac{x^{3 / 2}}{(3 / 2)} \Rightarrow 3 \sqrt{a}(y+c)=\pm 2 x^{3 / 2} \Rightarrow 9 a(y+c)^2=4 x^3$ ……………………….(3)

∴ The equation of orthogonal trajectories is $9 a(y+c)^2=4 x^3$ where c is the parameter.

Example. 4. Find the orthogonal trajectories of the family of carves $x^{2 / 3}+y^{2 / 3}=a^{2 / 3}$, where ‘a’ is the parameter.
Solution.

Given equation of family of curves is x$x^{2 / 3}+y^{2 / 3}=a^{2 / 3}$ …………………….(1)

Differentiating (1) w. r. t. $x: \frac{2}{3} x^{-1 / 3}+\frac{2}{3} y^{-1 / 3} \frac{d y}{d x}=0$

⇒ $\Rightarrow x^{-1 / 3}+y^{-1 / 3} \frac{d y}{d x}=0 \Rightarrow x^{1 / 3} \frac{d y}{d x}+y^{1 / 3}=0$ ……………………….(2)

(2) is the differential equation of the given family (1).

Replacing $\frac{d y}{d x}$ by $\frac{d x}{d y}$ in (2), we get the D.E. of the orthogonal trajectories of (1) as $y^{1 / 3}-x^{1 / 3} \frac{d x}{d y}=0 \Rightarrow y^{1 / 3} d y-x^{1 / 3} d x=0$.

Integrating: $\int y^{1 / 3} d y-\int x^{1 / 3} d x=\frac{3}{4} c \Rightarrow \frac{y^{4 / 3}}{4 / 3}-\frac{x^{4 / 3}}{4 / 3}=\frac{c}{4 / 3}$

⇒ $y^{4 / 3}-x^{4 / 3}=c$ is the required equation of the orthogonal trajectories.

Solved Problems On Orthogonal Trajectories Step-By-Step

Example.5. Find the orthogonal trajectories of the family of parabolas through the origin and foci on the y-axis.
Solution.

The equation of the family of parabolas through the origin and foci on the y-axis is $x^2=4 a y$ …………………(1) where a is the parameter.

Differentiating (1) w. r. t. $x: 2 x=4 a \frac{d y}{d x}$ ………………….(2)

Eliminating the parameter a from (1) and (2): $\frac{d y}{d x}=\frac{2 y}{x}$ ………………………(3)

(3) is the differential equation of the family (1).

Replacing $\frac{d y}{d x}$ by $\left(-\frac{d x}{d y}\right)$ in (3), the D.E. of the orthogonal trajectories to (1) is

⇒ $-\frac{d x}{d y}=\frac{2 y}{x} \Rightarrow \frac{d y}{d x}=\frac{-x}{2 y} \Rightarrow 2 y d y=-x d x$

Integrating : $\int 2 y d y=\int-x d x+c \Rightarrow y^2=-\left(\frac{x^2}{2}\right)+c$

∴ The equation of the orthogonal trajectories of the family of curves (1) is

⇒ $y^2=-\frac{x^2}{2}+c \Rightarrow \frac{y^2}{c}+\frac{x^2}{2 c}=1$ ……………………….(4)

Note. The orthogonal trajectories of the family of parabolas (1) are ellipses (4) with centres at the origin and with centres on x-axis

Example. 6. Find the orthogonal trajectories of the family of rectangular hyperbolas $x y=a^2$ where a is the parameter.
Solution.

Given the equation of the family of curves is $x y=a^2$ where a is the parameter. (1)

Differentiating (1) w.r. t. x : $x \frac{d y}{d x}+y=0$ ……………………..(2) is free from ‘a’ the parameter.

Hence (2) is the differential equation of the given family of curves (1).

Replacing $\frac{d y}{d x}$ by $\left(-\frac{d x}{d y}\right)$ in (2), the differential equation of the orthogonal trajectories is: $x\left(-\frac{d x}{d y}\right)+y=0 \Rightarrow y d y=x d x$.

Integrating: $\int y d y=\int x d x+\frac{c}{2} \Rightarrow \frac{y^2}{2}=\frac{x^2}{2}+\frac{c}{2}$

⇒ $y^2-x^2=c \Rightarrow \frac{y^2}{c}-\frac{x^2}{c}=1, c \neq 0$

∴ The equation of the orthogonal trajectories of the given family of rectangular hyperbolas is $\frac{y^2}{c}-\frac{x^2}{c}=1(c \neq 0)$ which again represents a family of hyperbolas.

Step-By-Step Guide To Solving Orthogonal Trajectories Problems

Example. 7: Find the orthogonal trajectories of the family of curves $4 y+x^2+1+c_1 e^{2 y}=0$ where $c_1$ is the parameter.
Solution.

Given equation of family of curves is $4 y+x^2+1+c_1 e^{2 y}=0$ ………………..(1)

Differentiating (1) w. r. t. $x: 4 \frac{d y}{d x}+2 x+2 c_1 e^{2 y} \frac{d y}{d x}=0$ ……………………….(2)

Eliminating $c_1$ from (1) and (2): $4 \frac{d y}{d x}+2 x+2\left(-4 y-x^2-1\right) \frac{d y}{d x}=0$

⇒ $\left(4-8 y-2 x^2-2\right) \frac{d y}{d x}+2 x=0 \Rightarrow\left(1-4 y-x^2\right) \frac{d y}{d x}+x=0$ …………………………(3)

Now (3) is the differential equation of the given family (1).

Replacing $\frac{d y}{d x}$ by $-\frac{d x}{d y}$ in (3).

The Differential equation of the family of Orthogonal trajectories is

⇒ $\left(1-4 y-x^2\right)\left(-\frac{d x}{d y}\right)+x=0 \Rightarrow x^2+4 y-1+x \frac{d y}{d x}=0 \Rightarrow x \frac{d y}{d x}+4 y=1-x^2$

⇒ $\frac{d y}{d x}+\left(\frac{4}{x}\right) y=\frac{1-x^2}{x}$

This is a linear equation in y.

⇒ $\text { I.F. }=e^{\int(4 / x) d x}=e^{4 \log x}=e^{\log x^4}=x^4$

∴ The G.S is $y x^4=\int\left(\frac{1-x^2}{x}\right) x^4 d x+c_2$

⇒ $y x^4=\int x^3 d x-\int x^5 d x+c_2=\frac{x^4}{4}-\frac{x^6}{6}+c_2 \Rightarrow y=\frac{1}{4}-\frac{x^2}{6}+c_2 x^{-4}$

∴ The equation of the orthogonal trajectories of the given family is

⇒ $y=(1 / 4)-(1 / 6) x^2+c_2 x^{-4}$

Example. 8: Find the members of the orthogonal trajectories for the curve $x+y=c e^y$ which passes through (0,5).
Solution.

Given family of curves is $x+y=c e^y$ ……………………(1)

Differentiating (1) w. r. t. $1+\frac{d y}{d x}=c e^y \frac{d y}{d x}$ ……………………..(2)

Eliminatingcfrom(1)and(2): $1+\frac{d y}{d x}=(x+y) \frac{d y}{d x} \Rightarrow \frac{d y}{d x}(x+y-1)=1$ ………………………(3)

(3)is the differential equation of the given family of curves.

Replacing $\frac{d y}{d x}$ by $-\frac{d x}{d y}$ in (3):

The D. E. of the family of O. T. is $-\frac{d x}{d y}(x+y-1)=1 \Rightarrow \frac{d y}{d x}=-x-y+1$

⇒ $\frac{d y}{d x}+y=1-x$ …………………..(4)

This is a linear equation in y.

I F = $e^x$

Solution is $y e^x=e^x-\left(x e^x-e^x\right)+c_2 \Rightarrow y e^x=2 e^x-x e^x+c_2$

∴ The equation of the O. T. of the given family of curves is $y e^x=2 e^x-x e^x+c_2$.

Given this passes through (0,5) $\Rightarrow 5=2-0+c_2 \Rightarrow c_2=3$

∴ The equation of the orthogonal trajectories of the given family of curves is $y e^x=2 e^x-x e^x+3 \Rightarrow y=2-x+3 e^{-x}$

Example. 9. Find the orthogonal trajectories of the family of coaxial circles $x^2+y^2+2 g x+c=0$, where g is a parameter.
Solution.

Given family of coaxial circles is $x^2+y^2+2 g x+c=0$ …………………….(1)

Differentiating (1) w.r.t. $x: 2 x+2 y \frac{d y}{d x}+2 g=0 \Rightarrow g=-x-y \frac{d y}{d x}$ …………………….(2)

Eliminating g from (1) and (2) :

⇒ $x^2+y^2-2 x^2-2 x y \frac{d y}{d x}+c=0 \Rightarrow y^2-x^2-2 x y \frac{d y}{d x}+c=0$ ……………………..(3)

(3) is the differential equation of the given family of curves. Replacing $\frac{d y}{d x}$ by $-\frac{d x}{d y}$ in (3)

⇒ the D.E. of the family of orthogonal trajectories is $y^2-x^2+(2 x y) \frac{d x}{d y}+c=0$

⇒ $(2 x y) \frac{d x}{d y}-x^2=-y^2-c \Rightarrow 2 x \frac{d x}{d y}-\frac{1}{y} x^2=-y-\frac{c}{y}$ …………………..(4)

(4)is a linear equation.in x.

Putting $x^2=v \Rightarrow 2 x \frac{d x}{d y}=\frac{d v}{d y}$ ……………………(5)

(4) and (5) $frac{d v}{d y}-\frac{1}{y} v=-y-\frac{c}{y} \text { where } P=-\frac{1}{y}, Q=-y-\frac{c}{y}$ ……………………(6)

⇒ $\text { I.F. }=\exp \left(\int-\frac{1}{y} d y\right)=e^{-\log y}=e^{\log y^{-1}}=\frac{1}{y}$

General solution of (6) is v(I.F) = $\int Q(\mathrm{I} . \mathrm{F}) d y+c \Rightarrow \frac{v}{y}=\int\left(-y-\frac{c}{y}\right) \frac{1}{y} d y+c_1$

⇒ $v\left(\frac{1}{y}\right)=-\int d y-\int \frac{c}{y^2} d y+c_1=-y+\frac{c}{y}+c_1$

⇒ $x^2+y^2-c_1 y-c=0$ is the equation of the orthogonal trajectories of the given family of circles where $c_1$ is the parameter.

Aliter to solve : $y^2-x^2+2 x y \frac{d x}{d y}+c=0 \Rightarrow y(2 x) d x-x^2 d y+\left(y^2+c\right) d y=0$

⇒ $\frac{y(2 x) d x-x^2 d y}{y^2}+d y+\frac{c}{y^2} d y=0 \Rightarrow\left(x^2 / y\right)+y-(c / y)+2 k=0 \Rightarrow x^2+y^2+2 k y-c=0$

Worked Examples Of Orthogonal Trajectories In Mathematics

Example. 10. Find the orthogonal trajectories of the family of curves $\frac{x^2}{a^2}+\frac{y^2}{a^2+\lambda}=1$, where λ is the parameter
Solution.

Given family of curves is $\frac{x^2}{a^2}+\frac{y^2}{a^2+\lambda}=1$ ……………………(1) where λ is the parameter.

Differentiating (1) w.r.t. $x: \frac{2 x}{a^2}+\frac{2 y}{a^2+\lambda} \cdot \frac{d y}{d x}=0$

⇒ $-\frac{x}{a^2}=\frac{y}{a^2+\lambda} \frac{d y}{d x} \Rightarrow \frac{1}{a^2+\lambda}=-\frac{x}{a^2 y} \frac{d x}{d y}$ ………………………(2)

Eliminating λ from (1) and (2): $\frac{x^2}{a^2}+y^2\left(-\frac{x}{a^2 y} \frac{d x}{d y}\right)=1$

⇒ $\frac{x^2}{a^2}-\frac{x y}{a^2} \frac{1}{d y / d x}=1$ ……………………..(3)

(3) is the differential equation of the given family of curves (1).

Replacing $\frac{d y}{d x}$ by $\left(-\frac{d x}{d y}\right)$ in (3), the DE. of the orthogonal trajectories is:

⇒ $\frac{x^2}{a^2}-\frac{x y}{a^2}\left(\frac{-1}{d x / d y}\right)=1 \Rightarrow \frac{x^2}{a^2}+\frac{x y}{a^2} \cdot \frac{d y}{d x}=1$

⇒ $\frac{x y}{a^2} \frac{d y}{d x}=1-\frac{x^2}{a^2}=\frac{a^2-x^2}{a^2} \Rightarrow y d y=\left(\frac{a^2}{x}-x\right) d x$

Integrating: $\int y d y=\int \frac{a^2}{x} d x-\int x d x+\frac{c}{2} \Rightarrow \frac{y^2}{2}=a^2 \log |x|-\frac{x^2}{2}+\frac{c}{2}$

⇒ $x^2+y^2-2 a^2 \log x=c$ where c is the parameter, is the required equation of the orthogonal trajectories.

Example. 11. Show that the family of confocal conics $\frac{x^2}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}=1$ is self-orthogonal, where λ is the parameter
Solution.

Given family of confocal conics is $\frac{x^2}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}=1$ ………………………..(1)

where λ is the parameter

Differentiating (1) w. r. t. $x: \frac{2 x}{a^2+\lambda}+\frac{2 y}{b^2+\lambda} \cdot \frac{d y}{d x}=0$

For convenience write $\frac{d y}{d x}=p$.

Then $\frac{x}{a^2+\lambda}+\frac{y}{b^2+\lambda} \cdot p=0$ …………………………(2)

⇒ $x\left(b^2+\lambda\right)+y\left(a^2+\lambda\right) p=0$

⇒ $\lambda(x+y p)=-\left(a^2 y p+b^2 x\right) \Rightarrow \lambda=\frac{-\left(b^2 x+a^2 y p\right)}{x+y p}$

⇒ $a^2+\lambda=a^2-\frac{\left(b^2 x+a^2 y p\right)}{x+y p}=\frac{a^2 x+a^2 y p-b^2 x-a^2 y p}{x+y p} \Rightarrow a^2+\lambda=\frac{\left(a^2-b^2\right) x}{x+y p}$

Also $b^2+\lambda=\frac{-\left(a^2-b^2\right) y p}{x+y p}$ ………………………(3)

Eliminating λ from (1) and (3):

⇒ $\frac{x(x+y p)}{a^2-b^2}-\frac{y(x+y p)}{\left(a^2-b^2\right) p}=1 \Rightarrow \frac{x+y p}{a^2-b^2}\left(x-\frac{y}{p}\right)=1$

⇒ $(x+y p)\left(x-\frac{y}{p}\right)=a^2-b^2$ …………………….(4)

(4) is the differential equation of the given family of curves (1).

Replacing p by -1/ p in (4), the differential equation of the orthogonal trajectories is $\left(x-\frac{y}{p}\right)(x+y p)=a^2-b^2$ ………………………..(5)

(5) is the same as the differential equation (4) of the given family of curves (1).

Hence the given family of confocal conics (1) is self-orthogonal. That is each member of the given family of curves intersects its own members orthogonally.

Orthogonal Trajectories Exercise 3(a)

1. Find the orthogonal trajectories of the following family of curves :

(1) y = cx where ‘c’ is the parameter.

Solution: $x^2+y^2=k^2$

(2) $y=a x^n$ where ‘a’ is the parameter.

Solution: $n y^2+x^2=c$

(3) $\sinh y=c_1 x$ where $c_1$ is the parameter.

Solution: $2 \log (\cosh y)+x^2=c_2$

(4) $c_1 x^2+y^2=1$ where $c_1$ is the parameter.

Solution: $2 \log |y|=x^2+y^2+c_2$

(5) $x^{1 / 3}+y^{1 / 3}=c_1$ where $c_1$ is the parameter.

Solution: $y^{5 / 3}-x^{5 / 3}=c_2$

(6) $y^3+3 x^2 y=c_1$ where $c_1$ is the parameter.

Solution: $y^2-x^2=c_2 x$

(7) $3 x y=x^3-a^3$ where ‘a’ is the parameter.

Solution: $2 x^2=(2 y-1)+c e^{-2 y}$

2. Find the orthogonal trajectories of

(1) the family of circles $x^2+y^2=a^2$ where ‘a’ is the parameter.

Solution: $x=c y$

(2)the family of circles $x^2+y^2=c x$, c being the parameter.

Solution: $x^2+y^2=c_1 y$

(3)the family of circles $x^2+y^2+2 f y+1=0$, f being the parameter.

Solution: $x^2+y^2+c x-1=0$

(4)the family of semicubical parabolas $a y^2=x^3$, ‘a’ being the parameter.

Solution: $2 x^2+3 y^2=c$

(5)the family of parabolas $y^2=4 a x$ where ‘a’ is the parameter.

Solution: $2 x^2+y^2=c$

3. (1) Find the orthogonal trajectories of the family of rectangular hyperbolas $y=c_1 / x$, $c_1$ being the parameter.

Solution: $y^2-x^2=c_2$

(2) Find the orthogonal trajectories of the family of curves whose differential equation is $\frac{d y}{d x}=\frac{y^2-x^2}{2 x y}$

Solution: $x^2+y^2=c y$

4. (1) Find the orthogonal trajectories of the family of straight lines in a plane and passing through the origin.

Solution: $x^2+y^2=c^2$

(2) Find the orthogonal trajectories of the family of circles through the origin and with centres on x – axis.

Solution: $x^2+y^2=c y$

(3) Find the orthogonal trajectories of the family of parabolas opening in the y – direction with the vertex at (1,2).

Solution: $2(y-2)^2+(x-1)^2=c_2^2$

5. (1) Find the orthogonal trajectories of the family $y=x+c e^{-x}$ and determine that particular member of each family that passes through (0,3).

Solution: $y=x+3 e^{-x}, x-y+2+e^{3-y}$

(2) Find the family orthogonal to the family $y=c e^{-x}$ of exponential curves. Determine the member of each family passing through (0,4).

Solution: $y=4 e^{-x}, y^2=2(x+8)$

Orthogonal Trajectories: Polar Coordinates

Theorem 2. If f(r,θ,c) = 0, c being the parameter, is the polar equation of the family of curves, then the differential equation of the family of its orthogonal trajectories is $\mathrm{F}\left(r, \theta,-r^2 \frac{d \theta}{d r}\right)=0$

Proof.

Given equation of the family of curves is f(r,θ,c) = 0 …………………..(1)

where c is the parameter.

Differentiating (1) w. r. t. θ and eliminating c from (1), we get the differential equation of the family (1).

Let the differential equation of the family (1) be

⇒ $\mathrm{F}\left(r, \theta, \frac{d r}{d \theta}\right)=0$ ……………………(2)

Let C be a member of the family of the curves (1) and T be a member of the family of the trajectories of (1).

Let $\left(r_0, \theta_0\right)$ be any point on T. Let the two curves C and T intersect at P (r,θ). At P, let $\phi$ and $\phi_0$ be the angles which the tangets to the two curves C and T make with the common radius vector OP.

We have tan $\tan \phi=r \frac{d \theta}{d r}, \tan \phi_0=r_0 \frac{d \theta_0}{d r_0}$

Also $\phi_0-\phi=\pi / 2 \Rightarrow \phi_0=(\pi / 2)+\phi$

⇒ $\tan \phi_0=\tan (\pi / 2+\phi)=-\cot \phi=-1 / \tan \phi \Rightarrow \tan \phi \tan \phi_0=-1 \Rightarrow\left(r \frac{d \theta}{d r}\right)\left(r_0 \frac{d \theta_0}{d r_0}\right)=-1$

⇒ $\frac{d r}{d \theta}=-r r_0 \frac{d \theta_0}{d r_0}$ ………………………(3)

Also at $r=r_0 \Rightarrow \theta=\theta_0$ ………………………(4)

Eliminating $r, \theta, \frac{d r}{d \theta}$, from (2), (3), (4) we get :

⇒ $\mathrm{F}\left(r_0, \theta_0,-r_0^2 \frac{d \theta_0}{d r_0}\right)=0$ …………………….(5)

(5) is true for any point on T.

Thus T is the locus of the point $\left(r_0, \theta_0\right)$.

Hence the differential equation satisfying T is $\mathrm{F}\left(r, \theta,-r^2 \frac{d \theta}{d r}\right)=0$

Working Rule To Find The Orthogonal Trajectories Of A Given Family Of Curves In Polar Coordinates

Let f(r.θ,c)=0 be the equation of the given family of curves …………………….(1)

(1) Differentiate (1) w. r. t. θ and obtain the differential equation $\mathrm{F}\left(r, \theta, \frac{d r}{d \theta}\right)=0$ by eliminating the parameter c. ……………………………(2)

(2) Replace $\frac{d r}{d \theta}$ by $-r^2 \frac{d \theta}{d r}$ in (2)

Then the differential equation of the orthogonal trajectories is $\mathrm{F}\left(r, \theta,-r^2 \frac{d \theta}{d r}\right)=0$ ……………………(3)

(3) Solve the equation (3) to get the equation of the orthogonal trajectories of (1).

Orthogonal Trajectories Solved Problems

Example. 1. Find the orthogonal trajectories of the family of curves r=aθ where a is the parameter.
Solution.

Given equation of family of curves is r = aθ ………………………..(1)

Differentiating (1) w. r. t. $\theta \Rightarrow \frac{d r}{d \theta}=a$ …………………….(2)

Eliminating ‘a’ from (1) and (2) ⇒ $r=\theta \frac{d r}{d \theta} \Rightarrow \frac{d r}{d \theta}=\frac{r}{\theta}$ ………………………(3)

(3) is the differential equation of the given family (1).

Changing $\frac{d r}{d \theta}$ by $\left(-r^2 \frac{d \theta}{d r}\right)$ in (3), the differential equation of required orthogonal trajectory is

⇒ $-r^2 \frac{d \theta}{d r}=\frac{r}{\theta} \Rightarrow \frac{d r}{r}=-\theta d \theta \Rightarrow \int \frac{d r}{r}=-\int \theta d \theta+\log c \Rightarrow \log r=-\frac{\theta^2}{2}+\log c$

⇒ $\log r-\log c=-\frac{\theta^2}{2} \Rightarrow \log (r / c)=-\frac{\theta^2}{2} \Rightarrow r / c=\exp \left(-\theta^2 / 2\right)$

∴ The required family of the orthogonal trajectories of the family (1) is .

Example. 2. Find an equation of the orthogonal trajectory of the family of circles having a polar equation r = 2acosθ where ‘a’ is the parameter.
Solution.

Given equation of the family of circles is r = 2a cosθ ……………….(1)

where ‘a’ is the parameter

Differentiating (1) w. r. t. $\theta \Rightarrow \frac{d r}{d \theta}=-2 a \sin \theta$ ………………….(2)

Eliminating ‘a’ from (1) and (2): $r=-\frac{1}{\sin \theta} \cdot \frac{d r}{d \theta} \cdot \cos \theta \Rightarrow r=-(\cot \theta) \frac{d r}{d \theta}$ …………………..(3)

(3) is the differential equation of the given family (1).

Replace $\frac{d r}{d \theta}$ by $\left(-r^2 \frac{d \theta}{d r}\right)$ in (3), the differential equation of the required orthogonal trajectories is

⇒ $r=(-\cot \theta)\left(-r^2 \frac{d \theta}{d r}\right) \Rightarrow \frac{d r}{d \theta}=r \cot \theta$

⇒ $\frac{d r}{r}=\cot \theta d \theta \Rightarrow \int \frac{d r}{r}=\int \cot \theta d \theta \Rightarrow \log r=\log \sin \theta+\log 2 c$

⇒ $\log r=\log 2 c \sin \theta \Rightarrow r=2 c \sin \theta$

∴ The equation of the orthogonal trajectories of the given family is r = 2c sin θ.

Example. 3. Find the orthogonal trajectories of the family of curves $r^2=a^2 \cos 2 \theta$ where ‘a’ is the parameter.
Solution.

Given equation of the family of curves $r^2=a^2 \cos 2 \theta$ ……………………(1)

Differentiating (1) w.r.t. $\theta \Rightarrow 2 r \frac{d r}{d \theta}=-2 a^2 \sin 2 \theta \Rightarrow \frac{d r}{d \theta}=-\frac{a^2}{r} \sin 2 \theta$ …………………….(2)

Eliminating  from (1) and (2) $\Rightarrow \frac{d r}{d \theta}=-\left(\frac{r^2}{\cos 2 \theta}\right) \frac{1}{r} \sin 2 \theta \Rightarrow \frac{d r}{d \theta}=-r \tan 2 \theta$ ……………………..(3)

(3) is the differential equation of the given family (1).

Replacing $\frac{d r}{d \theta}$ by $\left(-r^2 \frac{d \theta}{d r}\right)$ in (3), the differential equation of the required orthogonal trajectories is

⇒ $-r^2 \frac{d \theta}{d r}=-r \tan 2 \theta \Rightarrow r \frac{d \theta}{d r}=\tan 2 \theta \Rightarrow \frac{d r}{r}=\cot 2 \theta d \theta$.

Integrating : $\int \frac{d r}{r}=\int \cot 2 \theta d \theta+\log c \Rightarrow \log r=\frac{\log \sin 2 \theta}{2}+\log c$

⇒ $2 \log r=\log \sin 2 \theta+2 \log c \Rightarrow \log r^2=\log c^2 \sin 2 \theta$

⇒ $r^2=c^2$ sin 2θ is the equation of the orthogonal trajectories of the family (1).

Example. 4. Find the orthogonal trajectories of the family of curves r = a(1 – cos θ) where ‘a ‘is the parameter
Solution.

Given equation of the family of curves is

r = a(1 – cos θ) where a is the parameter. …………………..(1)

Differentiating (1) w.r.t. $\theta: \frac{d r}{d \theta}=a \sin \theta \Rightarrow a=\frac{1}{\sin \theta} \frac{d r}{d \theta}$ ……………………….(2)

Eliminating a from (1) and (2)

⇒ $r=\frac{1}{\sin \theta} \frac{d r}{d \theta}(1-\cos \theta) \Rightarrow r=\tan \frac{\theta}{2} \frac{d r}{d \theta} \Rightarrow \frac{d r}{d \theta}=r \cot \frac{\theta}{2}$ ………………………(3)

(3) is the differential equation of the given family (1).

Replacing $\frac{d r}{d \theta}$ by $\left(-r^2 \frac{d \theta}{d r}\right)$ in (3), the differential equation of the required orthogonal trajectories is

⇒ $-r^2 \frac{d \theta}{d r}=r \cot \frac{\theta}{2} \Rightarrow \frac{d r}{r}=\left(-\tan \frac{\theta}{2}\right) d \theta$

⇒ $\int \frac{d r}{r}=-\int \tan \frac{\theta}{2} d \theta+\log c \Rightarrow \log r=2 \log \cos \left(\frac{\theta}{2}\right)+\log c$

⇒ $\log r=\log c \cos ^2\left(\frac{\theta}{2}\right) \Rightarrow r=c \cos ^2\left(\frac{\theta}{2}\right)=\left(\frac{c}{2}\right)(1+\cos \theta)$

∴ The equation of the. orthogonal trajectories of the given family is r = k(1 + cos θ) where k = c/2.

Example. 5. Find the orthogonal trajectories of the families of curves $r=\frac{2 a}{1+\cos \theta}$ a is a parameter.
Solution.

Given equation of the family of curves is $r=\frac{2 a}{1+\cos \theta}$ where a is a parameter

⇒ $ r=\frac{2 a}{2 \cos ^2(\theta / 2)}=a \sec ^2 \frac{\theta}{2}$ …………………….(1)

Differentiating (1) w. r. t. $\theta: \frac{d r}{d \theta}=2 a \sec \frac{\theta}{2} \cdot \sec \frac{\theta}{2} \tan \frac{\theta}{2} \cdot \frac{1}{2} \Rightarrow \frac{d r}{d \theta}=a \sec ^2 \frac{\theta}{2} \tan \frac{\theta}{2}$ …………….(2).

Eliminating a from (1) and (2) : $\frac{d r}{d \theta}=r \tan \frac{\theta}{2}$ ………………….(3)

which is the differential equation of the family (1).

Replacing $\frac{d r}{d \theta}$ by$\left(-r^2 \frac{d \theta}{d r}\right)$ in (3), the differential equation of the required orthogonal trajectories is

⇒ $-r^2 \frac{d \theta}{d r}=r \tan \frac{\theta}{2} \Rightarrow-r \frac{d \theta}{d r}=\tan \frac{\theta}{2} \Rightarrow \frac{d r}{r}=\left(-\cot \frac{\theta}{2}\right) d \theta$

⇒ $\int \frac{d r}{r}=-\int \cot \frac{\theta}{2} d \theta+\log c \Rightarrow \log r=-2 \log \sin (\theta / 2)+\log c \Rightarrow \log r \sin ^2(\theta / 2)=\log c$

⇒ $r \sin ^2(\theta / 2)=c \Rightarrow r=\frac{2 c}{1-\cos \theta}$ is the required family of orthogonal trajectories of the given family.

Example. 6. Find the orthogonal trajectories of the family of curves $r^n \sin (n \theta)=a^n$ where a is the parameter.
Solution.

The given equation of the family of curves is $r^n \sin (n \theta)=a^n$ where a is the parameter. …………………(1)

Taking log on both sides of (1) $\Rightarrow n \log r+\log \sin (n \theta)=n \log a$ ……………….. (2)

Differentiating (2) w.r.t. $\theta: \frac{n}{r} \frac{d r}{d \theta}+\frac{n \cos n \theta}{\sin n \theta}=0 \Rightarrow \frac{1}{r} \frac{d r}{d \theta}=-\cot n \theta \Rightarrow \frac{d r}{d \theta}=-r \cot n \theta$ ……………………..(3)

(3) is the differential equation of the given family of curves(1).

Replacing $\frac{d r}{d \theta}$ by $-r^2 \frac{d \theta}{d r}$ in (3), the differential equation of the required orthogonal trajectories is

⇒ $-r^2 \frac{d \theta}{d r}=-r \cot n \theta \Rightarrow \frac{d r}{r}=\tan (n \theta) d \theta$

Integrating: $\int \frac{d r}{r}=\int \tan (n \theta) d \theta+\log c \Rightarrow \log r=-\frac{1}{n} \log \cos (n \theta)+\log c \Rightarrow n \log r$$+\log \cos (n \theta)=n \log c$

⇒ $r^n \cos n \theta=c^n$ is the required family of the orthogonal trajectories of (1).

Orthogonal Trajectories Exercise 3(b)

1. Find the orthogonal trajectories of the families of curves in polar coordinates :
(1) rθ = a, a being the parameter.

Solution: $r^2=c e^{\theta^2}$

(2) r = a( 1 + cos θ) where a is the parameter.

Solution: $r=c(1-\cos \theta)$

(3) r sin 2θ = λ, λ being the parameter.

Solution: $r^4 \cos 2 \theta=c^2$

(4) $r^n=a^n \cos n \theta$ where a is the parameter.

Solution: $r^n=c^n \sin (n \theta)$

(5) $r^n \cos n \theta=a^n$ where a is the parameter.

Solution: $r^n \sin (n \theta)=c^n$

2. Find the orthogonal trajectory of $r=c_1(1-\sin \theta), c_1$ being the parameter.

Solution: $r=c_2(1+\sin \theta)$

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