Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For real $0<q<1$, integer $n >0 $ and integer $k\ge 0$, define $$[k, n]_q \equiv -\sum_{m=1}^{n} q^{m(k+1)} (q^{-n}; q)_m = -\sum_{m=1}^{n} q^{m(k+1)} \prod_{l=0}^{m-1} (1-q^{l-n})$$

where $(\cdot\; ; q)_n$ is a $q$-Pochhammer symbol.

These functions express exact occupation numbers of $k$-th energy level in an ideal Fermi gas with equidistant spectrum and exactly $n$ fermions. (For physicists, $q$ is just $e^{-\Delta \epsilon/ kT}$).

My numerical experiments with Mathematica show so far :

  1. All $[k, n]_q$ are polynomials in $q$.
  2. $0<[k, n]_q<1$ for $0<q<1$.
  3. $\lim\limits_{q\to 1} [k, n]_q = 0$.
  4. $\lim\limits_{q\to 0} [k, n]_q = \begin{cases} 1, & k < n \\ 0, & k \ge n \end{cases}$.

Points (2.) to (4.) can be proven from the physics starting point, but I'm totally puzzled by (1.). The product from $l=0$ to $m-1$ contains negative powers of $q$ because of $n >l$, nevertheless, they conspire to cancel in the final sum.

Why are these functions polynomials? What would be the optimal way to compute their coefficients? Are there deeper mathematical properties to them?

Combinatorial context. The original "combinatorial physics" definition of these numbers can be written as

$$[k, n]_q =\frac{\sum_{ \{ \nu_k \} } \nu_k q^{\sum_k k \nu_k} \delta_{n, \sum_k \nu_k}}{\sum_{ \{ \nu_k \} }q^{\sum_k k \nu_k} \delta_{n, \sum_k \nu_k}}$$

where the summation indices run as $\nu_k=0,1$ for $k=0, 1, 2 \ldots$. Properties (2.)-(4.) follow easily from this definition. More physics context for the problem is being prepared for publication, see also a related post at TheoreticalPhysics.SE.

Update: A physics paper describing these polynomials has been posted to arXiv, includes a reference to this question.

share|cite|improve this question
It might be helpful if you use the expression $$\sum_{m=1}^n \prod_{\ell=0}^{m-1} \left(q^{k+1}-q^{k+\ell-n+1}\right)$$; for $n \leq k+1$, it's clear why they're polynomials. Why they're still polynomials when $n > k+1$ does not look so straightforward... – J. M. Oct 17 '11 at 10:04
...and since you're in Mathematica anyway: QHypergeometricPFQ[{q^(-n), q}, {0}, q, q^(k + 1)] - 1 looks to be a more compact way of expressing your polynomials. – J. M. Oct 17 '11 at 10:19
@J.M. useful indeed, thanks! (you've flipped the overall minus sign) – Slaviks Oct 17 '11 at 10:31
Wait, doesn't point 4 imply that every negative power of $q$ in the function's expansion has coefficient zero? Though asking for a combinatorial or generatingfunctionological approach is intriguing. – anon Oct 17 '11 at 10:46
I have a suspicion that your polynomial is a special case of one of the $q$-Hahn class of orthogonal polynomials; I haven't fully browsed the usual database, but methinks it's in there somewhere... – J. M. Oct 17 '11 at 10:54
up vote 7 down vote accepted

Some digging through Koekoek and Swarttouw's The Askey-scheme of hypergeometric orthogonal polynomials and its $q$-analogue reveals that $[k,n]_q$ is related to the Al-Salam-Carlitz I polynomials (see page 115 of Koekoek and Swarttouw). More precisely,

$$1-[k,n]_q={}_2\phi_1\left({{q^{-n},q}\atop{0}}; q, q^{k+1}\right)=(-1)^n q^{\frac{n}2(2k-n+3)}U_n^{(q^{-k-1})}\left(\frac1{q};q\right)$$

In particular, letting $S(n,k;q)=1-[k,n]_q$, there is the three-term recurrence


with the initial conditions $S(0,k;q)=1,\quad S(1,k;q)=q^{k+1}-q^k+1$.

From the relation with the Al-Salam-Carlitz II polynomials (see page 116 of Koekoek and Swarttouw), we have another basic hypergeometric expression:

$$1-[k,n]_q={}_2\phi_0\left({{q^n,\frac1{q}}\atop{-}}; \frac1{q}, q^{k-n+1}\right)$$

(Mathematica note: unfortunately Mathematica doesn't have support for ${}_2\phi_0$... yet.)

One can also derive a "reversal" identity by reversing the summation order:

$$\begin{align*}1-[k,n]_q&=q^{n(k+1)}(q^{-n};q)_n\; {}_1 \phi_1\left({{q^{-n}}\atop{q^{-n}}};q,q^{-k}\right)\\&=(-1)^n q^{\frac{n}{2}(2k-n+1)}(q;q)_n\; {}_1 \phi_1\left({{q^{-n}}\atop{q^{-n}}};q,q^{-k}\right)\end{align*}$$

I'll update this post if I manage to dig up more information...

share|cite|improve this answer
Confirm you three-term recurrence wtih small exmaples. It is also consistent with @PeterTaylor result. – Slaviks Oct 18 '11 at 6:04
I have a problem with your last equation: using the definition of ${}_2\phi_0$ and a $q$-Pochhammer identity I recover the original definition in terms of $(q^{-n} ; q)_m$ but with an extra factor of $(q^{-1} ; q^{-1})_m$ (coming from the second upper argument of ${}_2\phi_0$). Something is wrong, apparently... – Slaviks Oct 18 '11 at 8:24
@Slaviks: Hmm, maybe I did not do the proper substitutions for Al-Salam-Carlitz II; let me check again, and I'll edit later... – J. M. Oct 18 '11 at 9:29
I retract my comment - just realized the extra $q$-Pochhammer is coming from the definition. Thanks! – Slaviks Oct 18 '11 at 13:05

Hope this does not goes against the rules here, but I wanted to post a more permanent summary of the brainstorming that took place in the comments to the question.

  • J.M. suggested writing

$$[k,n]_q = -\sum_{m=1}^n \prod_{\ell=0}^{m-1} \left(q^{k+1}-q^{k+\ell-n+1}\right)$$


$$[k,n]_q = 1-{}_2\phi_1\left({{q^{-n},q}\atop{0}}; q, q^{k+1}\right)$$

  • anon noticed that point (4.) implies point (1.)
share|cite|improve this answer
A summary is perfectly fine methinks. :) Did you already try browsing the database I gave you? (I'm also thinking you'll eventually need to read up on basic hypergeometric series anyway...) – J. M. Oct 17 '11 at 12:05
No, not yet - it appears quite dense to me. I'm presently struggling to transform your basic hypergeometric function form into somethings that allows me to compute the $q \to 1^{-}$ via this property. – Slaviks Oct 17 '11 at 12:13
Well, I'll try searching+transforming myself, but maybe later (there are other things I need to look at for the time being). I was about to refer you to the DLMF for the time being, but I see you've gotten there already... :) – J. M. Oct 17 '11 at 12:18
@J.M.: not without your help: :) – Slaviks Oct 17 '11 at 12:27
Another alternative expression, obtained by reversing summation/multiplication order, is $$\sum _{m=0}^n \prod _{\ell=m+1}^n \left(q^{k+1}-q^{k-\ell+1}\right)$$ – J. M. Oct 19 '11 at 7:52

Not a full answer, but there's a recurrence $$[k,1]_q = q^{k}-q^{k+1}$$ $$[k,n+1]_q = \left(q^{k+1}-q^{k-n}\right) \left([k,n]_q -1\right)$$

One derivation is $$[k,n+1]_q = -\sum_{m=1}^{n+1} \prod_{\ell=0}^{m-1} \left(q^{k+1}-q^{k+\ell-n}\right)$$ $$ = -\left(q^{k+1}-q^{k-n}\right) \sum_{m=1}^{n+1} \prod_{\ell=1}^{m-1} \left(q^{k+1}-q^{k+\ell-n}\right)$$ Subst. $\ell^\prime = \ell - 1, \; m^\prime = m-1$ $$ = -\left(q^{k+1}-q^{k-n}\right) \sum_{m^\prime=0}^{n} \prod_{\ell^\prime=0}^{m^\prime-1} \left(q^{k+1}-q^{k+\ell^\prime-n+1}\right)$$ $$ = \left(q^{k+1}-q^{k-n}\right) \left(-1-\sum_{m^\prime=1}^{n} \prod_{\ell^\prime=0}^{m^\prime-1} \left(q^{k+1}-q^{k+\ell^\prime-n+1}\right)\right)$$ $$ = \left(q^{k+1}-q^{k-n}\right) \left([k,n]_q -1\right) $$

share|cite|improve this answer
An alternative base case is $[k, 0]_q = 0$ – Peter Taylor Oct 17 '11 at 21:21

An alternative form, conjectured by my colleague:

$$[k,n]_q =1+\sum_{i=1}^{k+1}(-1)^i \frac{q^{(n-k)i+i(i-1)/2}}{1-q^{k+1}} (q^{k+1}\; ;q^{-1})_i= 1 - \sum_{i=1}^{k+1} q^{i(i+1)/2+n-k-1} \prod_{j=1}^{i-1}(q^{n-j}-q^{n-k-1}) $$

Checked for small $k$ and $n$, still need to prove. But this form explicitly proves that $[k,n]_q$ are polynomials for $k<n$. Also, the number of terms does not grow with $n$.

EDIT: Exploring the new form a bit more leads to a striking "reversal" relation:

$$[k,n]_q=1-q^{n-k} \sum_{m=0}^k q^{m(n+1)} (q^{-k} \;; q)_m=1-q^{n-k} \left (1 - [n,k]_q \right) $$

EDIT-2 Another empirical observation: $\frac{[k,n]_q}{(1-q^n)} q^{-k+(n-1)n/2}$ is a polynomial as well.

share|cite|improve this answer

Your Answer


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.