Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I read in some papers that the Stirling numbers (of the second kind) have a natural $q$-analog $S_q(n,k)$, which satisfy the recurrence $$ S_q(n,k)=(k)_qS_q(n-1,k)+q^{k-1}S_q(n-1,k-1) $$ with the conditions that $S_q(0,k)=\delta_{0,k}$ and $S_q(n,0)=\delta_{n,0}$.

How is this recurrence arrived at? Even if this recurrence is taken as definition, there must be some motivation for it. Thank you.

share|improve this question
    
@anon: hmm, I don't want to make a broad discussion about that, but I think to remember there are different approaches to define q-analogues for the Stirling-numbers. So I think it should be not only legitime but also interesting to ask: "how did they arrive at that form of definition" which implies a small hint to the focus at "why did they prefer this specific definition", and not just "they are defined" –  Gottfried Helms Feb 11 '12 at 8:58

2 Answers 2

up vote 3 down vote accepted

The following justification is based on material in Johann Cigler’s lecture notes, Chapters 1 and 3.

The ordinary Stirling numbers of the second kind are characterized by the identity $$(xD)^n=\sum_k\left\{\matrix{n\\k}\right\}x^kD^k\;,\tag{1}$$ where $D$ is the ordinary differentiation operator. Thus, one approach to defining a $q$-analogue $S_q(n,k)$ is to require that it satisfy the analogue of $(1)$ with $D$ replaced by $D_q$, defined by $$D_qf(x)=\frac{f(qx)-f(x)}{qx-x}$$ and satisfying $$D_qx^n=\frac{(qx)^n-x^n}{(q-1)x}=\frac{q^n-1}{q-1}x^{n-1}=(n)_qx^{n-1}$$ and $$D_qx=qxD_q\;.$$

In other words, one might reasonably define $S_q(n,k)$ to satisfy

$$(xD_q)^n=\sum_{k=0}^nS_q(n,k)x^kD_q^k\;.\tag{2}$$

Assume that $(2)$ holds for some $n$; then

$$\begin{align*} xD_q(xD_q)^n&=\sum_kS_q(n,k)xD_qx^kD_q^k\\ &=\sum_kS_q(n,k)x\Big(q^kx^kD_q+(k)_qx^{k-1}\Big)D_q^k\\ &=\sum_kS_q(n,k)q^kx^{k+1}D_q^{k+1}+\sum_k(k)_qS_q(n,k)x^kD_q^k\\ &=\sum_k\left(S_q(n,k-1)q^{k-1}+(k)_qS_q(n,k)\right)x^kD_q^k\;, \end{align*}$$

so if we want $(2)$ to hold for $n+1$, we must set

$$S_q(n+1,k)=S_q(n,k-1)q^{k-1}+(k)_qS_q(n,k)\;.\tag{3}$$

$(2)$ clearly requires that $S_q(n,0)=\delta_{n,0}$; it imposes no constraint on $S_q(0,k)$ for $k>0$, but setting it to $0$ is the natural thing to do and is compatible with $(3)$.

share|improve this answer
    
Dear Brian, thanks for writing this up! I saw it just now. –  Dani Hobbes Mar 10 '12 at 22:21

If you can read German you find an answer in my lecture notes at University of Vienna, Elementare q-Identitäten, Chapter 3.

Your $S_q(n,k)$ is the same as my $q^{\binom{k}{2}} S[n,k]$

share|improve this answer
    
Thanks for these notes, Johann. –  Dani Hobbes Mar 10 '12 at 22:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.