Answer to Question #87391 in Calculus for Samuel

Question

Find the maclaurent series of arctan x

Answer 1

To obtain the maclaurent series of arctan x, use the standard integral

$$\arctan x=\int_{0}^{x}{\frac{dt}{1+{{t}^{2}}}}$$

Consider the sum of a geometric series

$$\frac{1}{1+{{t}^{2}}}=\frac{1}{1-\left( -{{t}^{2}} \right)}=1-{{t}^{2}}+{{t}^{4}}-{{t}^{6}}+...$$

Substituting this sum into the integral, we obtain

$$\int_{0}^{x}{\frac{dt}{1+{{t}^{2}}}}=\int_{0}^{x}{\left( 1-{{t}^{2}}+{{t}^{4}}-{{t}^{6}}+... \right)dt}$$

$$=\left( t-\frac{{{t}^{3}}}{3}+\frac{{{t}^{5}}}{5}-\frac{{{t}^{7}}}{7}+... \right)_{0}^{x}=x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}-\frac{{{x}^{7}}}{7}+...$$

Thus

$$\arctan x=x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}-\frac{{{x}^{7}}}{7}+...$$

or

$$\arctan x=\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}\frac{{{x}^{2n+1}}}{2n+1}}$$

for

$$\left| x \right|\le 1$$

This is the Maclaurent series of arctan x.