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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?

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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 $(1,2).$

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The expression is a product of two functions. Each function can be represented any way possible, such as a Taylor series around any point. The product of the two representations is perfectly valid.

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