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}