What is meant by "expand f[x,y]=xy in powers of x-1 and y-1"? This question refers to C.H. Edwards, Jr.'s Advanced Calculus of Several Variables, Chapter II-7, Example 2.

Suppose we want to expand $f[x,y]=xy$ in powers of $x-1$ and $y-1$.  Of course the result will be
$xy=1+(x-1)+(y-1)+(x-1)(y-1)$,
but let us obtain this result by calculating the second degree Taylor polynomial $P_2[\mathbf{h}]$ of $f[x,y]$ at $\mathbf{a}=\{1,1\}$ with $\mathbf{h}=\{h_1,h_2\}=\{x-1,y-1\}$. ...

Establishing the equivalence is a matter of simple algebra, but "expand $f[x,y]=xy$ in powers of $x-1$ and $y-1$" has no clear meaning to me.  Edwards obviously does not mean to use a Taylor polynomial, because he wants to show the above result is equivalent to the Taylor polynomial.
What exactly does it mean to expand a function in powers of some given pattern?  I'm assuming this is something I should have learned in high school.
 A: $\newcommand{\dd}{\partial}$To expand a polynomial $p(x, y)$ in powers of $x - a$ and $y - b$ is to find coefficients $c_{jk}$ (which turn out to exist uniquely, and with all but finitely many equal to $0$) such that
$$
p(x, y) = \sum_{j,k=0}^{\infty} c_{j,k} (x - a)^{j} (y - b)^{k}.
$$
(If the total degree of $p$ is $N$, the sum is limited to indices with $j + k \leq N$.)
One way to achieve this expansion is to write, using your example $p(x, y) = xy$,
\begin{align*}
  xy &= \bigl(a + (x - a)\bigr) \bigl(b + (y - b)\bigr) \\
  &= ab + b(x - a) + a(y - b) + (x - a)(y - b).
  \tag{1}
\end{align*}
(Analogous use of the binomial theorem handles arbitrary monomials $x^{\ell} y^{m}$, and every polynomial in two variables is a finite sum of monomials.)
Another way approach is to invoke Taylor's theorem:
$$
p(x, y) = \sum_{n=0}^{\infty} \sum_{j=0}^{n} \frac{\dd^{n} p}{\dd^{j} x\, \dd^{n-j} y}(a, b)\, \frac{(x - a)^{j}}{j!}\, \frac{(y - b)^{n-j}}{(n - j)!}.
\tag{2}
$$
Again taking $p(x, y) = xy$, one finds
$$
c_{0,0} = p(a, b) = ab,\quad
c_{1,0} = \frac{\dd p}{\dd x}(a, b) = b,\quad
c_{0,1} = \frac{\dd p}{\dd y}(a, b) = a,\quad
c_{1,1} = \frac{\dd^{2} p}{\dd x\, \dd y}(a, b) = 1,
$$
with all other partial derivatives vanishing, so that (2) agrees with (1). This is presumably the content of Edwards' example.
Generally, (2) yields the coefficients
$$
c_{j,k} = \frac{\dd^{j+k} p}{\dd^{j} x\, \dd^{k} y}(a, b)\, \frac{1}{j!\, k!}.
$$
There are similar formulas for sufficiently smooth (non-polynomial) functions, perhaps with infinitely many non-zero coefficients, and with "error" or "remainder" terms.
A: Rewrite $x$ as $(x-1)+1$ and similarly for $y=(y-1)+1$. Now let $X=x-1$ and $Y=y-1$. So
\begin{eqnarray*}
f(x,y)=xy&=&((x-1)+1)((y-1)+1)=(X+1)(Y+1) \\
\end{eqnarray*}
\begin{eqnarray*}
&=&XY&+&X&+&Y&+&1 \\
&=&(x-1)(y-1)&+&(x-1)&+&(y-1)&+&1
\end{eqnarray*}
So the solution you give in the question is spot on.
