Taylor expansion of $(1-x)(1-y)$. I am trying to compute the Taylor expansion to second order of the following function about point $(x,y)=(a,a)$, with $a<1$, 
$$f(x,y)=(1-x)(1-y).$$
If I am not mistaken, this expansion can be given by 
$$(1-a) (1-a)-(x-a)(1-a)- (y-a)(1-a),$$
where it can be seen that e.g. there are not terms in $(x-a)^2$, which I think it results from the fact that $\frac{\partial^2 f}{\partial x^2}=0$.
Is my analysis correct ? and if so, can I claim that the Taylor expansion to order $n \ge 2$ will be the same as for $n=1$?
 A: Recall that the Taylor expansion of order $k$ around a point $p \in \mathbf R^d$ of a $C^k$-function $f \colon U \subseteq \mathbf R^d \to \mathbf R^{d'}$ is given by 
$$ T^p_kf(x)= \sum_{|\alpha|\le k} \frac{1}{\alpha!}\partial^\alpha f(p)(x-p)^\alpha 
$$
for $k = 2$ we have 
$$ 
T^{(a,a)}_2 f(x) = f(a,a) + \partial_x f(a,a)(x-a) + \partial_y f(a,a)(y-a) + \frac 12 \partial_x^2 f(a,a)(x-a)^2 + \partial_{x}\partial_{y}f(a,a)(x-a)(y-a) + \frac 12 \partial_y^2f(a,a)(y-a)^2 $$
We have
\begin{align*}
 \partial_x f(x,y) &= -(1-y)\\
 \partial_y f(x,y) &= -(1-x)\\
 \partial_x\partial_y f(x,y) = 1\\
 \partial_x^2 f = \partial_y^2 f &= 0
\end{align*}
This gives 
$$
  T^{(a,a)}_2f(x,y) = (1-a)(1-a) - (1-a)(x-a) - (1-a)(y-a) + (x-a)(y-a)
$$
You are missing the $x$-$y$-term, note that $\partial_x\partial_y f\ne 0$. You are correct in saying, that there is no $(x-a)^2$ term due to $\partial_x^2 f = 0$. As $\partial^\alpha f = 0$ for indices $\alpha$ of order $\ge 3$, we have that all expansions of order $\ge 3$ are equal to the one for $k = 2$.
A: The function $f(x,y)$ is a polynomial in two variables and following identity holds:
$$f(x,y)=(1-x)(1-y)=(1-a)(1-a)-(1-a)(x-a)-(1-a)(y-a)+(x-a)(y-a).$$
Therefore your expansion for $n=1$ is correct, but for $n=2$ something is missing. 
Note that, since the degree of $f(x,y)$ is two, the expansions for $n\geq 2$ will be all the same.
