How to find a certain coefficient from a series that is given by a generating function of two variables?

Question

How to find the coefficient of $x^1y^3$ from the series that is given by the $\frac{x y}{(y-1) (x+y-2)}$?

The coefficient is according to Wolfram Mathematica $\text{SeriesCoefficient}\left[\frac{x y}{(y-1) (x+y-2)},\{x,0,1\},\{y,0,3 \}\right]=\frac{7}{8}$.

The function is from the article dealing with the Problem of Points - page 61, left column in the middle.

I found a lot of articles about how to find out g.f. for a series. But nothing clear (to me) about how to find out the series from a g. f. of two variables. There seems to be a procedure for one variable that starts with dividing into partial fractions. I guess I understand that division, but probably not very well, because I can't use it with two variables.

I know how to enumerate/calculate result using the original method (which is listed first in the article (Method of Enumeration) and also many times on the Internet). But I am interested in the procedure to find out the coefficient from the g.f.

Answer 1

You can more or less use the tools from univariate case by extracting coefficients individually, i.e. $[x^my^n]f(x,y)=[y^n]([x^m]f(x,y))$. So for your function:

\begin{align} [x^my^n]\frac{x y}{(1-y) (2-y-x)} &=[y^n]\left([x^m]\frac{x y}{(1-y) (2-y-x)}\right) \\ &=[y^n]\left(\frac{y}{(1-y)(2-y)}[x^m]\frac{x}{(1-\frac{x}{2-y})}\right) \\ &=[y^n]\left(\frac{y}{(1-y)(2-y)}[x^m]\sum_{i=1}^{\infty} \frac{x^{i}}{(2-y)^{i-1}}\right) \tag{1}\\ &=[y^n]\frac{y}{(1-y)(2-y)^m} \\ &=\frac{1}{2^m}[y^n]\left(\frac{y}{1-y}\cdot\left(1-\frac{y}{2}\right)^{-m}\right) \\ &=\frac{1}{2^m}[y^n]\left(\sum_{i=1}^{\infty}y^i\cdot\sum_{j=0}^{\infty}\binom{-m}{j}\left(-\frac{1}{2}\right)^jy^j\right) \tag{2} \\ &=\frac{1}{2^m}\sum_{j=0}^{n-1}\binom{-m}{j}\left(-\frac{1}{2}\right)^j \tag{3} \\ &=\frac{1}{2^m}\sum_{j=0}^{n-1}\binom{m-j-1}{j}\left(\frac{1}{2}\right)^j \tag{4} \\ \end{align}

where:

1. Infinite geometric series
2. Binomial series
3. Extracting product coefficient using Cauchy product
4. $\binom{-m}{j}(-1)^j=\frac{-m(-m-1)\cdots(-m-j+1)}{j!}(-1)^j=\frac{(m+j-1)\cdots(m+1)m}{j!}=\binom{m+j-1}{j}$

So in your example $m=1,n=3$ we have $$ \frac{1}{2}\left(\binom{0}{0}+\binom{1}{1}\frac{1}{2}+\binom{2}{2}\frac{1}{4}\right)=\frac{7}{8}. $$