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

1 Answer
Sep 20, 2014

#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}#