Finding taylor series in Real Analysis

Question

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 $f(x)=x^3\sin(x^2)$

Answer 1

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, $$x^3\sin (x^2)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n+5}}{(2n+1)!}=x^5-\frac{x^9}{3!}+\frac{x^{13}}{5!}-\frac{x^{17}}{7!}...\tag 4$$

Answer 2

You can find the Taylor series for $\sin(x^2)$ by plugging $x^2$ into the Taylor series for $\sin$. So $\sin(x^2) = x^2-x^6/3!+x^{10}/5!\ldots.$ Then multiply each term by $x^3$ to get the series for $x^3\sin(x^2).$

Answer 3

HINT:

$$\sin(x^2)=\sum_{k=0}^\infty(-1)^k\frac{x^{2(2k+1)}}{(2k+1)!}$$