If I have $f(x,y) = \sin(2x)\sin(y)$ and I want to find the Taylor series around $(0,0)$ of $f(x, y)$, how should I do this?
What I've tried is: $\sin(2x)$ around $x = 0$ is $$ \sum_{n=0}^{\infty}(-1)^n\frac{(2x)^{2n+1}}{(2n+1)!} $$ And $\sin(y)$ around $y = 0$ is $$ \sum_{n=0}^{\infty}(-1)^n\frac{y^{2n+1}}{(2n+1)!} $$ And so the Taylor series of $f(x,y)$ would simply be $$ (\sum_{n=0}^{\infty}(-1)^n\frac{(2x)^{2n+1}}{(2n+1)!})\cdot(\sum_{n=0}^{\infty}(-1)^n\frac{y^{2n+1}}{(2n+1)!}) $$ But this seems like too simple of a solution and I wouldn't know how to multiply summations, as we haven't covered that yet. I can't find any similar problems in my book either.