What is the Taylor series of #f(x)=arctan(x)#?

1 Answer
Sep 25, 2014

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

Let us look at some details.

#f(x)=arctanx#

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

Remember that the geometric power series

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

by replacing #x# by #-x^2#,

#Rightarrow 1/{1-(-x^2)}=sum_{n=0}^infty(-x^2)^n=sum_{n=0}^infty(-1)^nx^{2n}#

So,

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

By integrating,

#f(x)=int sum_{n=0}^infty(-1)^nx^{2n}dx#

by putting the integral sign inside the summation,

#=sum_{n=0}^infty int (-1)^n x^{2n}dx#

by Power Rule,

#=sum_{n=1}^infty(-1)^n{x^{2n+1}}/{2n+1}+C#

Since #f(0)=arctan(0)=0#,

#f(0)=sum_{n=1}^infty(-1)^n{(0)^{2n+1}}/{2n+1}+C=C Rightarrow C=0#

Hence,

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