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

Sep 25, 2014

$f \left(x\right) = {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} \frac{{x}^{2 n + 1}}{2 n + 1}$

Let us look at some details.

$f \left(x\right) = \arctan x$

$f ' \left(x\right) = \frac{1}{1 + {x}^{2}} = \frac{1}{1 - \left(- {x}^{2}\right)}$

Remember that the geometric power series

$\frac{1}{1 - x} = {\sum}_{n = 0}^{\infty} {x}^{n}$

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

$R i g h t a r r o w \frac{1}{1 - \left(- {x}^{2}\right)} = {\sum}_{n = 0}^{\infty} {\left(- {x}^{2}\right)}^{n} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n}$

So,

$f ' \left(x\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n}$

By integrating,

$f \left(x\right) = \int {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n} \mathrm{dx}$

$= {\sum}_{n = 0}^{\infty} \int {\left(- 1\right)}^{n} {x}^{2 n} \mathrm{dx}$

by Power Rule,

$= {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} \frac{{x}^{2 n + 1}}{2 n + 1} + C$

Since $f \left(0\right) = \arctan \left(0\right) = 0$,

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

Hence,

$f \left(x\right) = {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} \frac{{x}^{2 n + 1}}{2 n + 1}$