Summer 2015 – Q.4

4. (a) If $latex u=2xy,\;y=x^2-y^2$ and $latex x=r\;\cos\theta,\;y=r\;\sin\theta$ then find $latex \frac{\partial\left(u_1\;v\right)}{\partial\left(\partial_1\;\theta\right)}$ [6M]

(b) If $latex i^{i^{i^{.^{.^{.^\infty}}}}}=A+iB,$ prove that $latex \left(\frac{\mathrm{πA}}2\right)=\frac BA\;and\;A^2+B^2=e^{-\pi B}$ [6M]

$latex \begin{array}{l}3x+2y+7z=-4\\\\2x+3y+z=5\\\\3x+4y+z=7\end{array}$ [8M]

5. (a) By using De Moiverse theorem Express $latex \frac{\sin7\theta}{\sin\theta}$ in powers of sinθ only. [6M]

Using De Meiver’s theorem,

expand $latex \frac{\sin\;7\theta}{\sin\;\theta}$ in terms of sinθ

$latex \cos7\theta+i\sin7\theta=(\cos\theta+i\sin\theta)^7$

$latex =\cos^7\theta+^7c_1\cos^6\theta(i\sin\theta)+^7c_2\cos^5\theta(i\sin\theta)^2+^7c_3\cos^4\theta(i\sin\theta)^3+^7c_4\cos^3\theta(i\sin\theta)^4+^7c_5\cos^2\theta(i\sin\theta)^5+^7c_6\cos\theta(\sin\theta)^6+^7c_7(i\sin\theta)^7$

Equating imaginary parts,we get

$latex \sin7\theta=7\cos^6\theta\sin\theta-35\cos^4\theta\sin^3\theta+21\cos^2\theta\sin^5\theta-\sin^7\theta$

$latex =7(1-\sin^2\theta)^3\sin\theta-35(1-\sin^2\theta)^2\sin^3\theta+21(1-\sin^2\theta)\sin^5\theta-\sin^7\theta$

$latex =7(1-3\sin^2\theta+3\sin^4\theta-\sin^6\theta)\sin\theta-35(1-2\sin^2\theta+\sin^4\theta)\sin^3\theta+21\sin^5\theta-21\sin^7\theta-\sin^7\theta$

$latex =7\sin\theta-21\sin^3\theta+21\sin^5\theta-7\sin^7\theta-35\sin^3\theta+70\sin^5\theta-35\sin^7\theta+21\sin^5\theta-22\sin^7\theta$

$latex =7\sin\theta-56\sin^3\theta+112\sin^5\theta-64\sin^7\theta$

$latex =\frac{\sin7\theta}{\sin\theta}=7-56\sin^2\theta+112\sin^4\theta-64\sin^6\theta$

(b) By using Taylor’s series expand tan^-1 x in positive powers of (x – 1) upto first four non-zero terms. [6M]

Using Tayler series Expand $latex \tan^{-1}x$ in power of $latex (x-1)$

We know,$latex f(a+h)=f(a)+hf'(a)+\frac{h^2}{2!}f”(a)+\frac{h^3}{3!}f”(a)+…..$

Here h = x-1

$latex f(x-1+1)=f(1)+\frac{x-1}{1!}f'(1)+\frac{(x-1)^2}{2!}f”(1)+\frac{(x-1)^3}{3!}f”'(1)$

Here a = 1

$latex f(x)=\tan^{-1}x$

$latex f'(x)=\frac1{1+x^2}=(1+x^2)^{-1}$

$latex f”(x)=-1(1+x^2)^{-2}(2x)$

$latex =2\left[x(1+x^2)^{-2}\right]$

$latex f”'(x)=-2\left[x(-2)(1+x^2)^{-3}2x+(1+x^2)^{-2}\right]$

$latex =-2\left[-4x^2(1+x^2)^{-3}+(1+x^2)^{-2}\right]$

$latex f(1)=\tan^{-1}(1)=\frac x4$

$latex f'(1)=(1+1)^{-1}=\frac12$

$latex f”(1)=-2(1+1)^{-2}$

$latex f”(1)=\frac{-2}4=\frac{-1}2$

$latex f”'(1)=-2\left[-4(1+1)^{-3}+(1+1)^{-2}\right]$

$latex -2\left[\frac{-4}8+\frac14\right]$

$latex -2\times\frac{-4+2}8=\frac12$

Substituting in (1), we get

$latex \tan^{-1}x=\frac\pi4+\frac{x-1}{1!}\left(\frac12\right)+\frac{(x-1)^2}{2!}\times\left(\frac{-1}2\right)+\frac{(x-1)^3}{3!}\left(\frac12\right)+…….$

$latex \frac\pi4+\frac{x-1}2-\frac{(x-1)^2}4+\frac{(x-1)^3}{12}+……$

$latex \tan\left(\frac\pi4+\frac x2\right)=\frac{\tan{\displaystyle\frac\pi4}+\tan{\displaystyle\frac x2}}{1-\tan{\displaystyle\frac\pi4}\tan{\displaystyle\frac x2}}$

$latex \frac{1+\tan{\displaystyle\frac x2}}{1-\tan{\displaystyle\frac x2}}….(2)$

$latex u=\log\left[\tan\left(\frac\pi4+\frac x2\right)\right]$

Hence Proved

(c) If $latex y=\sin\left[\log\left(x^2+2x+1\right)\right]$ prove that $latex \left(x+1\right)^2y_{n+2}+\left(2n+1\right)\left(x+1\right)y_{n+1}+\left(n^2+4\right)y_n=0$ [8M]

T.P.T $latex \left(x+1\right)^2y_{n+2}+\left(2n+1\right)\left(x+1\right)y_{n+1}+\left(n^2+4\right)y_n=0$

$latex \begin{array}{l}y=\sin\left[\log\left(x+1\right)^2\right]\\\\\;\;\;=\sin\left[2\log\left(x+1\right)\right]\\\\y_1=\frac{2\;\cos\left[2\log\left(x+1\right)\right]}{x+1}\\\\y_1\left(x+1\right)=2\;\cos\left[2\log\left(x+1\right)\right]\end{array}$

Taking second derivative

$latex \begin{array}{l}\left(x+1\right)y_1+\left(x+1\right)^2y_2=\frac{-4\sin\left(2\log\left(x+1\right)\right)}{x+1}\\\\=-4y\\\\\left(x^2+1+2x\right)y_2+\left(x+1\right)y_1+4y=0\end{array}$

Differentiating n times by Leibnitz’s theorem, we get

$latex \begin{array}{l}\left(1+x^2+2x\right)y_{n+2}+n\left(2x+2\right)y_{n+1}+\frac{n\left(n-1\right)}22y_n+\left(x+1\right)y_{n+1}+ny_n+4y_n=0\\\\\left(x+1\right)^2y_{n+2}+y_{n+1}\left(2nx+2n+x+1\right)+y_n\left(n^2-n+n+4\right)=0\\\\\left(x+1\right)^2y_{n+2}+\left(2n+1\right)\left(x+1\right)y_{n+1}+y_n\left(n^2+4\right)=0\end{array}$

Hence Proved.