Question

Prove by induction $\forall nϵNify=\frac{x}{x^2+a^2}$
then $y_n=\frac{(-1{)}^{n}n!}{a^{n+1}}si{n}^{n+1}\theta cos(n+1)\theta$
where $\theta =co{t}^{-1}\frac{x}{a}$

Answer 1

x = a cot $\theta$
$\therefore \frac{dx}{d\theta }=-acose{c}^2\theta$
$y=\frac{x}{x^2+a^2}=\frac{acot\theta }{a^{2}cose{c}^2\theta }=\frac{1}{a}sin\theta cos\theta$
$\therefore y_1=\frac{dy}{dx}=\frac{dy}{d\theta }.\frac{d\theta }{dx}=\frac{1}{a}(co{s}^2\theta -si{n}^2\theta )$
$(-\frac{1}{acose{c}^2\theta })$
$=-\frac{1}{a^2}si{n}^2\theta cos2\theta$
Thus p (n) holds for n = 1.
Assume p (n) holds good.
$\therefore y_n=\frac{(-1{)}^{n}n!}{a^{n+1}}si{n}^{n+1}\theta cos(n+1)\theta$
Differentiate both sides w.e.t. x.
$\therefore y_{n+1}=\frac{(-1{)}^{n}n!}{a^{n+1}}$
[(n +1) $si{n}^n\theta cos\theta cos(n+1)\theta -si{n}^{n+1}\theta .(n+1)sin(n+1)\theta \left]\right]\frac{d\theta }{dx}$
$=\frac{(-1{)}^{n}n!(n+1)}{a^{n+1}}si{n}^n\theta cos$ ( n + 1 + 1 )$\theta (-1\frac{1}{acose{c}^2\theta })$
$=\frac{(-1{)}^n(n+1)!}{a^{n+2}}si{n}^{n+2}\theta cos(n+2)\theta$
Hence p (n) $\Rightarrow$ p (n + 1).