Generatingfunctionology 1.6 Stirling Number

Question

I am reading Herbert Wilf's Generatingfunctionology these days. And on Page 21, Section 1.6 Another 2-Variable Case, I was confused by the operation he did on formula (1.36), which is:

...

The partial fraction expansion in question has the form:

$$\frac{1}{(1-x)(1-2x)...(1-kx)}=\sum_{j=1}^{k}\frac{a_j}{1-jx}$$

To find the $\alpha$'s, fix $r$, $1\le r \le k$, multiply both sides by $1-rx$, and let $x=\frac{1}{r}$.

As there is such a term $(1-rx)$ on the LHS as part of the divisor, I thought that that implies that $(1-rx)\ne0$, which means $x\ne\frac{1}{r}$. But later on, as mentioned above, the author just let $x=\frac{1}{r}$, which doesn't seems to make sense to me.

My question is why is this operation allowed? Is there any special theorem related to this?

Thanks in advance.

Answer 1

Generating functions are often in a class of objects called formal power series, where the notion of convergence need not apply (though it often does for small values of $x$). That is to say, you are not actually evaluating the series for any value of $x$; it is being treated as an algebraic object in its own right. All you care about are the coefficients for each power of $x$, and this substitution is merely an algebraic trick to find them.

Answer 2

The left-hand side of (1) \begin{align*} \frac{1}{(1-x)(1-2x)\cdots(1-kx)}=\sum_{j=1}^{k}\frac{a_j}{1-jx}\tag{1} \end{align*} is a rational function in the variable $x$ with poles $\frac{1}{r}, 1\leq r\leq k$. The denominator is zero at the poles and the function is not defined there. The right-hand side is the same rational function but written using a partial fraction decomposition.

If we now fix an $r$ with $1\leq r\leq k$ and multiply the rational function with $(1-rx)$. We get \begin{align*} \frac{1-rx}{(1-x)(1-2x)\cdots(1-kx)}=\sum_{j=1}^{k}\frac{a_j(1-rx)}{1-jx}\tag{2} \end{align*}

In (2) we still have the poles $\frac{1}{r}, 1\leq r\leq k$ as in (1). But we can now cancel the term $1-rx$ resulting in \begin{align*} &\frac{1}{(1-x)(1-2x)\cdots(1-(r-1)x)(1-(r+1)x)\cdots (1-kx)}\\ &\qquad\qquad=\sum_{j=1}^{r-1}\frac{a_j(1-rx)}{1-jx}+a_r+\sum_{j=r+1}^{k}\frac{a_j(1-rx)}{1-jx}\tag{3} \end{align*}

The cancellation of the factor $1-rx$ in (3) removes the pole $\frac{1}{r}$, which is a so-called removable singularity. Contrary to (1) and (2) there is no harm to evaluate the rational function in (3) at $x=\frac{1}{r}$.

We can now evaluate (3) at $x=\frac{1}{r}$ and get the values $a_r, 1\leq r\leq k$.