Third order taylor polynomial of $f(x,y)=\sin(x) \sin(y)$?

Question

well, I have to find the Taylor polynomial of $f(x,y)=\sin(x)\sin(y)$ at $(0,\pi/4)$. I found:

Is $T_3(x,y)=-\frac{1}{12}\sqrt{2}x(16x^2+48y^2-24\pi+3\pi^2)$ correct?

Answer 1

Your answer doesn't look right.

The third order Taylor polynomial of $\sin x$ at $0$ is

$$\tag 1 x-x^3/6.$$

The third order Taylor polynomial of $\sin y$ at $\pi/4$ is

$$\tag 2 (1/\sqrt 2)[1+(y-\pi/4) - (y-\pi/4)^2/2 - (y-\pi/4)^3/6].$$

Multiply $(1),(2)$ two and discard any terms of total power $>3.$

Answer 2

No. The product of first order in each variable Taylor polynomials of $\sin(x) \sin(y)$ centered at $(0,\pi/4)$ is $$ \frac{4 xy +(4-\pi) x}{4 \sqrt{2}} \text{.} $$

The product of up to first order in each variable Taylor polynomials of your $T_3(x,y)$ centered at $(0,\pi/4)$ is $$ -2\sqrt{2} \pi xy + 2\sqrt{2} \pi x \text{.} $$

These would agree if your $T_3$ were correct. So there are (at least) defects in the first derivatives.