How do you find the power series representation for the function $f(x)=tan^(-1)(x)$ ?

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Answer 1 · 2014-09-20T11:57:56

$f(x)=tan^{-1}x=sum_{n=0}^infty(-1)^n x^{2n+1}/{2n+1}$

Let us look at some details.

Let us find the power series for $f'(x)$

By taking the derivative,

$f'(x)=1/{1+x^2}=1/{1-(-x^2)}$

We know the power series

$1/{1-x}=sum_{n=0}^infty x^n$

by replacing $x$ by $-x^2$ ,

$Rightarrow1/{1-(-x^2)}=sum_{n=0}^infty(-x^2)^n$

So, we have

$f'(x)=sum_{n=0}^infty(-1)^nx^{2n}$

By integrating $f'(x)$ ,

$f(x)=int sum_{n=0}^infty (-1)^nx^{2n}dx=sum_{n=0}^infty(-1)^n x^{2n+1}/{2n+1}+C$

Since $f(0)=tan^{-1}(0)=0$ , $C=0$ .

Hence,

$tan^{-1}x=sum_{n=0}^infty(-1)^n x^{2n+1}/{2n+1}$

How do you find the power series representation for the function f(x)=tan^(-1)(x)f(x)=tan^(-1)(x) ?