I'm asked to find the second order taylor series expansion of $\cos(x+2y)e^{2x}$ around $(2, -1)$. It's from an old exam of a professor who doesn't give solutions to them, so I don't know if my results are correct. So I woud like that you tell me if my results are correct, and if not, where I made a mistake. Thank you.
As first partial derivatives, I get $$f_x=-\sin(x+2y)e^{2x}+2\cos(x+2y)e^{2x}$$ $$f_y=-2\sin(x+2y)e^{2x}$$ As second partial derivatives, I get
$$\begin{align}
f_{xx}&=-\cos(x+2y)e^{2x}-2\sin(x+2y)e^{2x}-2\sin(x+2y)e^{2x}+4\cos(x+2y)e^{2x}\\
&=3\cos(x+2y)e^{2x}-4\sin(x+2y)e^{2x}
\end{align}$$
$$f_{yy}=-4\cos(x+2y)e^{2x}$$ $$f_{xy}=f_{yx}=-2\cos(x+2y)e^{2x}-4\sin(x+2y)e^{2x}$$
So now, if I plug everything into the formula, I get
$$e^{4}+2e^{4}(x-2)+\frac{1}{2!}(3e^{4}(x-2)^2-4e^{4}(x-2)(y+1)-4e^{4}(y+1)^2)=e^{4}(1+2(x-2)+\frac{1}{2!}(3(x-2)^2-4(x-2)(y+1)-4(y+1)^2))$$
Are my results correct ? I just proceeded as usual, using the common two-variables taylor series expansion. Thanks for your help !
Best Answer
Yes, it is correct. An easy way to make sure is just to plot things: blue is the original function, red is your expansion
And this is the code to draw it