<s>[INST] Finding taylor series in Real Analysis In the following question I am trying to find the Taylor series for,
$$f(x)=x^3\sin(x^2)$$
centered at $x=0$.
My first thought was to find the Taylor series for $\sin x$ then derive it to get my series I wanted but I am not quite sure how to do that. 
I worked out the Taylor series for $f(x)=\sin(x)$ centered at $x=0$ and got,
$f(x)=\sin x, f(0)=0$
$f'(x)=\cos x, f'(0)=1$
$f"(x)=-\sin x, f''(0)=0$
$f'''(x)=-\cos x, f'''(0)=-1$
$f''''(x)=\sin x, f''''(0)=0$
So the Taylor series for $\sin x$ centered at $x=0$ is,
$\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...$
But how do I use this to find the Taylor series for $f(x)=x^3\sin(x^2)$
 [/INST] We have
$$\sin x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...\tag 1$$
This series converges for all $x$ (just use the Ratio Test to verify it). Thus, replacing $x$ by $x^2$ in $(1)$ we get
$$\sin (x^2)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2{(2n+1)}}}{(2n+1)!}=x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}...\tag 2$$ Then multiply $x^3$ to $(2)$ we get
$$x^3\sin (x^2)=\sum_{n=0}^{\infty}(-1)^nx^3\frac{x^{2{(2n+1)}}}{(2n+1)!}=x^5-\frac{x^9}{3!}+\frac{x^{13}}{5!}-\frac{x^{17}}{7!}...\tag 3$$ Hence,
$$x^3\sin (x^2)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n+5}}{(2n+1)!}=x^5-\frac{x^9}{3!}+\frac{x^{13}}{5!}-\frac{x^{17}}{7!}...\tag 4$$ 
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