I am not very familiar with how generating functions are used but for something I was needing I ran into the following constructions,
- Let $c_{nk}$ be the number of solutions in $\{1,2,3,4...,\}$ for the equation $x_1+x_2+x_3+...+x_k = n$. Then one can easily show that $c_{nk} = ^{n-1}C_{k-1}$ But the following is not clear to me..one defines a "generating function" $c_k(x)$ as, $c_k(x) = \sum_{n=k}^{\infty} c_{nk} x^n$. Then one claims that the following is true,
$c_k(x) = (x+x^2+x^3+...)^k = x^k(1-x)^k$
I can't see from where the above came!
From the above it follows that, $\sum_{k=1} ^\infty c_k(x) = \frac{x}{1-2x}$
The idea seems to be that from the above one can read off that $\sum {k=1} ^n c_{nk} = 2^{n-1}$. (This was obvious from the initial formula given for $c_{nk}$) But this way of getting that result is not clear to me.
- Let $p(n)$ be the number of unordered partitions of $n$ (a positive integer). Then clearly $p(n)$ is the number of solutions to the equation, $x_1+x_2+x_3+...+x_n = n$ with $x_1\geq x_2 \geq x_3 ...x_n > 0$ After appropriate variable redefinition one can see that $p(n)$ is the number of solutions of $y_1 + 2y_2 + 3y_3 + ... + ny_n = n$ with $y_i \geq 0$ for all $i$. Now apparently $p(n)$ can be given through a generating function $P(x)$ defined as,
$P(x) = \sum _{n=0} ^\infty p(n)x^n$
Now I can't see how it follows that,
$P(x) = \prod _{k=1} ^\infty \frac{1}{(1-x^k)}$
(apparently the above can somehow be mapped to the familiar question of number of ways in which $n$ balls can be arranged in a line where $r_i$ of them are of the colour $i$ ($\sum r_i = n$). This is equal to the coefficient of $\prod _i x_{i}^{r_i}$ in $\prod _i (1+x_i + x_i^2 + x_i ^3 + ...)$. But the connection between this question and the generating function for unordered partitions of an integer is not cleat to me.)