Question

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

Answer 1

`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}`