Question

Find a power series representation for f(x) = arctan(x).

Answer 1

Recall that

$\dfrac{\mathrm d(\arctan x)}{\mathrm dx}=\dfrac1{1+x^2}$

and that for $|x|$, we have

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

So we get

$\displaystyle\frac1{1+x^2}=\frac1{1-(-x^2)}=\sum_{n=0}^\infty(-x^2)^n=\sum_{n\ge0}(-1)^nx^{2n}$

Taking the integral then gives

$\displaystyle\int\frac{\mathrm dx}{1+x^2}=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}+C$

We know that $\arctan0=0$, so if we evaluate both sides of the equation above for $x=0$, we would get a series that converges to 0, so that

$0=0+C\implies C=0$

and so

$\arctan x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}$