Finding the power series of $f(x) = (x^2 - \sin(x^2))/x^6$

Question

Suppose $f(x) = (x^2 - \sin(x^2))/x^6$ By using the Maclaurin series for $\sin x$ at $x = 0$, it is asked to find the power series for $f$ at $x = 0$. I understand how to obtain the Maclaurin series for $\sin(x)$ and how to find the Maclaurin series for $\sin(x^2)$ but I can't seem to progress any further than that. I can't think of any way to incorporate my other factors $(x^2,x^6)$ into the power series at $x = 0$.

$$\boxed{f(x)=\frac{x^2-\sin(x^2)}{x^6}}$$ $$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\dots(-1)^n\frac{x^{2n+1}}{(2n+1)!}$$ $$\sin(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}\implies$$ $$\sin(x^2)=\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n+1}}{(2n+1)!}\implies$$ $$\sin(x^2)=\sum_{n=0}^\infty(-1)^n\frac{x^{4n+2}}{(2n+1)!}$$

Answer 1

Using the well known series for sine:

$$\frac{x^2-\sin x^2}{x^6}=\frac{x^2-\sum\limits_{n=0}^\infty (-1)^n\frac{x^{2(2n+1)}}{(2n+1)!}}{x^6}=\frac1{x^6}\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2(2n+1)}}{(2n+1)!}=$$

$$=\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{4(n-1)}}{(2n+1)!}$$

Answer 2

Begin with determining the Maclaurin series for the numerator: by substitution, \begin{align} x^2-\sin x^2&=x^2-\biggl(x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}+\dotsm\biggr) \\ &= \frac{x^6}{3!}-\frac{x^{10}}{5!}+\frac{x^{14}}{7!}-\dotsm \end{align} There remains to find a general formula, and divide each term by $x^6$.