TITLE: Derivation of multivariable Taylor series
QUESTION [13 upvotes]: I am having trouble grokking why it is, assuming that the function is analytic everywhere (and many other assumptions that I am, no doubt, naively assuming), that this is true:
$f(x,y)=f(x_0,y_0)+[f'_x(x_0,y_0)(x-x_0)+f'_y(x_0,y_0)(y-y_0)]+\frac{1}{2!}[f''_{xx}(x_0,y_0)(x-x_0)+2f''_{yx}(x_0,y_0)(x-x_0)(y-y_0)+f''_{yy}(x_0,y_0)(y-y_0)^2]+...$
I am familiar with the one-variabled Taylor series, and intuitively feel why the 'linear' multivariable terms should be as they are.
In short, I ask for a proof of this equality. If possible, it would be nice to have an answer free of unnecessary compaction of notation (such as table of partial derivatives).
As a auxiliary question, I see a direct analogy with the first 2 terms  $f(x,y)=f(x_0,y_0)+[f'_x(x_0,y_0)(x-x_0)+f'_y(x_0,y_0)(y-y_0)]$ and the total differential $f(x,y)-f(x_0,y_0)=\Delta f(x,y)=f'_x(x_0,y_0)\Delta x+f'_y(x_0,y_0)\Delta y$.
When $\Delta x $ and  $\Delta y $ are not infinitesimally small, can I use the third term in the Taylor multivariable series to get closer to the real total differential?

REPLY [6 votes]: I think the easiest way to understand this is coming from the place of operators and linear transformations. A Taylor series in one dimension can be understood by exponentiating the derivative operator:
$$
f(x+a) = e^{a\frac{d}{dx}}f(x) = f(x) + af^\prime(x) + \frac{1}{2!}a^2f^{\prime\prime}(x)+...
$$
You can see this in one way as follows. The infinitesimal (linear order) transformation $f(x+dx) = f(x) + dx f^\prime(x)$ is known, and we can build up the finite transformation by an infinite succession of infinitesimal transformations:
$$
f(x+a) = \lim_{N\rightarrow\infty} \left(1+\frac{a}{N}\frac{d}{dx}\right)^N f(x) = e^{a\frac{d}{dx}} f(x).
$$
It is straightforward to extend this to multiple variables if we know the infinitesimal transformation (sometimes referred to as the generator), which you intuitively know as, $f(x+dx, y+dy) = f(x,y) + dx\frac{\partial}{\partial x}f(x,y) + dy\frac{\partial}{\partial y}f(x,y)$.
The finite transformation is then,
$$
f(x+a,y+b) = e^{a\frac{\partial}{\partial x}+b\frac{\partial}{\partial y}} f(x,y)\\
= \left[1+a\frac{\partial}{\partial x} + b\frac{\partial}{\partial y} + \frac{1}{2!}\left(a^2\frac{\partial^2}{\partial x^2} + 2a\frac{\partial}{\partial x}b\frac{\partial}{\partial y}+ b^2\frac{\partial^2}{\partial y^2}\right) + ...\right]f(x,y).
$$