I'm having trouble getting the power series representation of this function.
(My work and the question are attached)
Any help would be greatly appreciated!
Thanks in advance!
$\displaystyle \arctan x = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + ... \ \ |x|<1 $
$\displaystyle \arctan x^{5} = x^{5} - \frac{x^{15}}{3} + \frac{x^{25}}{5} - \frac{x^{35}}{7} + ... $
$\displaystyle 2x^{2} \arctan x^{5} = 2x^{7} - \frac{2x^{17}}{3} + \frac{2x^{27}}{5} - \frac{2x^{37}}{7} + ... $
Not necessarily memorized. The important point is that to find the Taylor's series for $\displaystyle 2x^2 arctan(x^5)$ you don't have to work directly the entire thing. Start by finding the Taylor's series for arctan(x):
arctan(0)= 0.
$\displaystyle (arctan(x))'= \frac{1}{x^2+ 1}$ (That's a very important formula- you should have it memorized!) so (arctan(0))'= 1.
$\displaystyle (arctan(x))"= \left(\frac{1}{x^2+1}\right)'= \frac{-2x}{(x^2+1)^2}$ which is 0 at x= 0
etc.
$\displaystyle \frac{d}{dx} \arctan x = \frac{1}{1+x^{2}} = \sum_{n = 0}^{\infty} (-1)^n x^{2n}, \ |x|<1 $
so $\displaystyle \arctan x = \int \sum^{\infty}_{n=0} (-1)^{n} x^{2n} \ dx = \sum^{\infty}_{n=0} (-1)^{n} \int x^{2n} \ dx = \sum^{\infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}$
and $\displaystyle \arctan x^5 = \sum^{\infty}_{n=0} (-1)^{n}\frac{(x^5)^{2n+1}}{2n+1}= \sum^{\infty}_{n=0} (-1)^{n}\frac{x^{10n+5}}{2n+1}$
and finally $\displaystyle 2x^{2} \arctan x^5 = 2 \sum^{\infty}_{n=0} (-1)^{n}\frac{x^{10n+7}}{2n+1}$