The function f(x)= 2(x^2)arctan(x^10) is represented as a power series. What is the lowest term with a nonzero coeddicient.
I cant seem to get the answer for this problem and any help is appreciated. Thanks.
The power series at 0 for arctan is
$\displaystyle \tan^{-1}(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)}$
so
$\displaystyle 2x^2\tan^{-1}(x^{10})=2x^2\sum_{n=0}^{\infty}\frac{(-1)^n(x^{10})^{2n+1}}{(2n+1)}=2\sum_{n=0}^{\infty}\ frac{(-1)^nx^{20n+12}}{(2n+1)}$
From here just plug in n=0 to get the first term