# 2 variables Taylor Series expansion at center other than (0,0)

When finding the Taylor Series expansion for a function of 2 variables which can be written as a product of two single variable functions, one can multiply their respective Taylor Series expansions to obtain a result.

For instance, considering the Taylor Series expansions for $$e^{-(x^2+y^2)}$$ , centered at $$(0,0)$$, one can find that it is simply

$$e^{-(x^2+y^2)}=\Bigg(\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{n!}\Bigg)\Bigg(\sum_{n=0}^{\infty}(-1)^n\frac{y^{2n}}{n!}\Bigg)$$

Because $$e^{-(x^2+y^2)}$$ can be written as $$e^{-x^2}e^{-y^2}$$

Now my question is this. Suppose we wanted to find the Taylor Series expansion for this same function, ($$e^{-(x^2+y^2)}$$), however centered at $$(x,y)=(1,2)$$. Could we proceed by finding the Taylor Series expansion for $$e^{-x^2}$$ centred at x=1, and likewise for $$e^{-y^2}$$, centred at y=2, and then multiplying them together as above? Why or why not?

Yes, the product of those series equals $$e^{-(x^2+y^2)}.$$ But that product is not quite the Taylor series of this function centered at $$(0,0);$$ that would be
$$\sum_{m=0}^{\infty}\sum_{n=0}^{\infty} (-1)^{m+n}\frac{x^{2m}}{m!}\frac{y^{2n}}{n!}.$$
The same ideas apply at $$(1,2).$$