Multivariate Taylor Series Derivation (2D) I am trying to understand the derivation here
kiwi.atmos.colostate.edu/group/dave/pdf/TaylorSeries.pdf
I understand how first, second total differentials are derived. I do not understand how they are plugged into a form that is compatible with the single variable Taylor Series, which is
$f(x) = f(a) + f'(a)(x-a) + \frac{1}{2!}f''(a) (x-a)^2 + ...$ 
The coefficients of multivariate expansion are exactly the same as the single variable version. Sure given the total differential, $(x-a)$ is substituted for $dx$, $(y-b)$ for $dy$, but it's not shown why, or how the rest of $f(x,y)$ expansion should mirror the single variable form based on the total differential. Expansion for $f(x,y)$ is below
$f(x,y) = f(a,b) + 
\bigg[(x-a)\frac{\partial}{\partial x} + (y-b)\frac{\partial}{\partial x}\bigg]f + 
\frac{1}{2!}
\bigg[(x-a)\frac{\partial}{\partial x} + (y-b)\frac{\partial}{\partial x}\bigg]^2 f + ...
$
If you know a better 2D Taylor Series derivation, that would be welcome as well. 
Thanks,
 A: So we want to express $f(a+u,b+v)$ in terms of $f(a,b)$, the increment $(u,v)$ and the partial derivatives of $f$ at $(a,b)$. To this end we look at the auxiliary function
$$\phi(t):=f(a+t u, b+t v)\qquad(0\leq t\leq 1)\ .$$
By the one-variable Taylor theorem we have
$$f(a+u,b+v)=\phi(1)=\sum_{r=0}^n{1\over r!}\phi^{(r)}(0)+{\rm error\ term}\ .\qquad(*)$$
Therefore we have to compute the successive derivatives of $\phi$ at $t=0$.
Obviously $$\phi^{(0)}(t)=\phi(t)=f(a+tu,b+tv)$$ and therefore $\phi^{(0)}(0)=f(a,b)$. To compute $\phi'(t)$ we need the chain rule:
$$\phi'(t)={d\over dt}\phi(t)={d\over dt}f(a+t u,b+ t v)=f_1(a+t u,b+ tv)u+f_2(a+ tu, b+ tv) v\ ,$$
whence $\phi'(0)=f_1(a,b)u+f_2(a,b)v$. Proceeding in this manner we get
$$\eqalign{\phi''(t)&={d\over dt}\bigl(f_1(a+t u,b+ tv)u+f_2(a+ tu, b+ tv) v\bigr) \cr 
&=f_{11}(a+tu,b+tv)u^2+2f_{12}(a+tu,b+tv) uv +f_{22}(a+tu,b+tv)v^2\cr}$$
and therefore
$\phi''(0)=f_{11}(a,b)u^2 + 2f_{12}(a,b)uv+f_{22}(a,b) v^2\ .$
And so on for the higher derivatives of $\phi$. The values $\phi(0)$, $\phi'(0)$, $\phi''(0)$, $\ldots$ so obtained have to be plugged into the equation $(*)$
Of course all of this could be written up more elegantly using vector notation.
A: If you are looking for a physical intuition behind multivariate Taylor series, remember that $ T_x = \partial_x$ is the generator of translations, as for suitable class of functions:
$$
   (T_x f)(x) = \lim_{\epsilon \to 0} \frac{ f(x+\epsilon) - f(x)}{\epsilon}
$$
Thus $\mathcal{T}_a = \mathrm{e}^{a T_x}$ is the operator of translation by $a$:
$$
    (\mathcal{T}_a f)(x) = f(x+a)
$$
Writing defining series of exponential give univariate Taylor expansion:
$$
    f(x+a) = (1 + a T_x + \frac{a^2}{2!} T_x \circ T_x + \ldots ) \circ f(x)
$$
Taking this to mulrivariate, the translation by vector $\vec{u}$ becomes 
$$
 \mathcal{T}_\vec{u} = \exp( \sum_{i=1}^n u_i T_{x_i} ) = \circ_{i=1}^n \mathcal{T}_{u_i}
$$ 
You can now develop each univariate translation operator into a series, giving the multivariate Taylor expansion formula or, alternatively, you could expand the combined exponential:
$$
   f(\vec{x} + \vec{u}) = \sum_{n=0}^\infty \frac{1}{n!} \left( \langle \vec{u}, \mathcal{T}_{\vec{x}} \rangle \right)^n \circ f(\vec{x}) 
$$
Hope this helps.
