How should i approximate this? Find the taylor series for the given function about the given point.
$$f(x,y)= e^{x^2+y^2}$$
My solution I thought was $e^{x^2}\cdot e^{y^2}$.
How should I make a summation about this?. 
 A: There are several ways to proceed.  It depends on the representation that is fit for purpose.
In general, the product of convergent series is given by the Cauchy Product
$$\sum_{n=0}^\infty a_n\sum_{m=0}^\infty b_m=\sum_{n=0}^\infty\sum_{p=0}^na_pb_{n-p}$$
For $a_n=\frac{x^{2n}}{n!}$ and $b_n=\frac{y^{2n}}{n!}$ we find that 
$$\begin{align}
e^{x^2}e^{y^2}&=\sum_{n=0}^\infty\sum_{p=0}^n \frac{x^{2p}}{p!}\frac{y^{2n-2p}}{(n-p)!}\\\\
&=\sum_{n=0}^\infty \frac{y^{2n}}{n!}\underbrace{\sum_{p=0}^n\binom{n}{p}\left(\frac{x^2}{y^2}\right)^p}_{\text{Binomial Expansion of}\,\,(1+x^/y^2)^n}\\\\
&=\sum_{n=0}^\infty \frac{y^{2n}}{n!}\left(1+\frac{x^2}{y^2}\right)^n\\\\
&=\sum_{n=0}^\infty \frac{(x^2+y^2)^n}{n!}\\\\
&=e^{x^2+y^2}
\end{align}$$
from which we have a variety of series forms. 
A: $e^{x^2} = 1 + x^2 + \frac 12 x^4 + \frac 16 x^6\cdots\\
(e^{x^2})(e^{y^2} = (1 + x^2 + \frac 12 x^4 + \frac 16 x^6\cdots)(1 + y^2 + \frac 12 y^4 + \frac 16 y^6\cdots)\\
(e^{x^2})(e^{y^2} = 1 + x^2 + y^2 + \frac 12 (x^4 + y^4 + 2x^2y^2)+\frac 16 (x^6 + y^6 + 3x^4y^2 + 3x^2y^4) \cdots$
I am just multiplying the terms from one expansion by the terms in the other expansion.
but more generally
$f(x,y) = f(x_0,y_0) + f_x (x_0,y_0)(x-x_0) + f_y (x_0,y_0)(x-x_0) + \frac 12 (f_{xx}(x_0,y_0)(x=x_0)^2 + 2f_{xy}(x_0,y_0)(x-x_0)(y-y_0)+f_{yy}(x_0,y_0)(y-y_0)^2)$ 
