1

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.

ockin
  • 13
  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Apr 21 '23 at 07:19
  • Any ideas from your side? – Cesareo Apr 21 '23 at 08:26
  • In this specific case $[x^1y^3]\frac{x y}{(y-1) (x+y-2)}=[x^0y^2]\frac{1}{(y-1) (x+y-2)}=[y^2]\frac{1}{(y-1) (y-2)}$ – Sil Apr 21 '23 at 20:54
  • Thanks @Sil . It seems that your simplification can be applied to this as well: $[x^1y^2z^2]\frac{x y z}{(1-x) (1-z) (3-x-y-z)}=[y^1z^1]\frac{1}{(1-z) (3-y-z)}$ – ockin Apr 22 '23 at 20:27
  • The power series of an analytic function is its Taylor series at the origin; you can always extract one particular coefficient by taking partial derivatives etc. – JBL Apr 23 '23 at 14:06

1 Answers1

0

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 $$ \frac{1}{2}\left(\binom{0}{0}+\binom{1}{1}\frac{1}{2}+\binom{2}{2}\frac{1}{4}\right)=\frac{7}{8}. $$

Sil
  • 16,612