<s>[INST] How do you find the Taylor Series of $f(x) = x^{2}*\sin(\frac{x^{2}}{3})$ about $a = 0$? Please without getting too technical, this is calculus 2. I know that the general form of a Taylor Series about 0 (Maclaurin) is: $$\sum_{n=0}^{\infty} \frac{f^{n}(0)}{n!} * x^{n}$$ and so, $$f(x) = f(0)/0! + f (0)/1!*x + f (0)/2!*x^2 + f (0)/3!*x^3 + .... $$
Just very unsure how this one would be carried out and simplified. Is it simply taking derivatives? Can it be put into general summation notation? 
Please help
Thank you
 [/INST] Use Taylor Series for $\sin x$ 
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!} -\frac{x^7}{7!}\ldots$$
$$\sin {\left(\frac{x^2}{3} \right)}=x-\frac{{\left(\frac{x^2}{3} \right)}^3}{3!}+\frac{{\left(\frac{x^2}{3} \right)}^5}{5!} -\frac{{\left(\frac{x^2}{3} \right)}^7}{7!}\ldots$$
You want $x^2 \sin {\left(\frac{x^2}{3} \right)}$, so multiply the above series by $x^2$.
$$x^2 \sin {\left(\frac{x^2}{3} \right)} = x^3- \frac{x^8}{162} +\frac{x^{12}} {29160} \ldots$$
In general :
$$x^2 \sin {\left(\frac{x^2}{3} \right)} = \sum _{n=0}^{\infty}\frac{(-1)^n x^{4n+6}}{3^{2n+1} (2n+1)!}$$
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