Taylor series for $x\sin(x^2)$, looking for $a_n$

Question

I had to find the coefficient $a_n \in \mathbb{R}$ so that $\sum\limits_{n=0}^\infty a_nx^n=x\sin(x^2)$,$\forall x \in \mathbb{R}$

Hmm I thought since:

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

$x\sin(x^2)=x\sum\limits_{n=0}^\infty(-1)^n\frac{(x)^{2(2n+1)}}{(2n+1)!}=x\sum\limits_{n=0}^\infty(-1)^n\frac{(x)^{4n+2}}{(2n+1)!}=\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}(x)^{4n+3}$

But can I get this in term of only $x^n$? Or is this the solution?

I took some derivatives before and figured that indeed for $x_0=0$ as our center the first non-zero derivative is the 3rd.

So I dont see how we could express our function as a series with $x^n$ instead of $x^{4n+3}$

Could someone tell me if the solution for $a_n$ is indeed $\frac{(-1)^n}{(2n+1)!}$

Answer 1

As you found, for every $x \in \Bbb{R}$, \begin{align} x \sin(x^2) &= \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)!}x^{4n+3} \end{align} This can of course be written as $\sum_{k=0}^{\infty}a_k x^k$ for specifically chosen coefficients $a_k$, but I doubt doing so explicitly is very illuminating. Anyway, here it is: define for each integer $k \geq 0$, \begin{align} a_k := \begin{cases} 0 & \text{if $k \not\equiv 3 \mod 4$} \\\\ \dfrac{(-1)^{\frac{k-3}{4}}}{\left(\frac{k-3}{2} + 1 \right)!}&\text{if $k \equiv 3 \mod 4$} \end{cases} \end{align} Then, $x \sin(x^2) = \sum\limits_{k=0}^{\infty}a_k x^k$. All I did was to do a case by case definition of what $a_k$ is, and in the second part, I simply set $k = 4n+3$, and solved for $n$, then plugged it into $\dfrac{(-1)^n}{(2n+1)!}$. So, again, while this is a complete expression for the various coefficients, I'm not sure how helpful you'll find it.

Answer 2

$$ x\sin(x^2)=x\sum\limits_{n=0}^\infty(-1)^n\frac{(x)^{2(2n+1)}}{(2n+1)!}=x^{3}\sum\limits_{n=0}^\infty(-1)^n\frac{(x)^{2n+1}}{(2n+1)!}=x^3 \sum\limits_{j=1 ,\text{odd}}^\infty\frac{(-1)^{\frac{j-1}{2}}}{j!}x^{j}$$

Answer 3

By using the definition of Taylor Series, we write$$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n,$$where $f(x)=x\sin(x^2).$ So$$x\sin(x^2)=\sum_{n=0}^{\infty}\frac{(x\sin(x^2))^{(n)}(x_0)}{n!}(x-x_0)^n$$and if we let $x_0=0$, we get $$x\sin(x^2)=\sum_{n=0}^{\infty}\frac{(x\sin(x^2))^{(n)}(0)}{n!}(x)^n.$$ By using the general Leibniz rule, we get $$(x\sin(x^2))^{(n)}=\sum_{k=0}^n\binom{n}{k}x^{(n-k)}\sin^{(k)}(x^2),$$but if $n-k>1$, then $x^{(n-k)}=0,$ so$$(x\sin(x^2))^{(n)}=\binom{n}{n-1}\sin^{(n-1)}(x^2)+\binom{n}{n}x\sin^{(n)}(x^2)=n\sin^{(n-1)}(x^2)+x\sin^{(n)}(x^2).$$ If we plug this into series, we have$$x\sin(x^2)=\sum_{n=0}^{\infty}\frac{n\sin^{(n-1)}(x^2)|_{x=0}}{n!}x^n.$$