Closed form for $\sum_{m \geq 1} (-1)^m q^{m(m+1)/2 + m \Delta}$?

Question

Is there a useful closed form for the following series ($|\Delta|$ is a small integer)?

$$f(q,\Delta) =\sum_{m=1}^{\infty} (-1)^m q^{m(m+1)/2 + m \Delta}$$

It is a large-$n$ approximation of the polynomial $-[n+\Delta, n]_q$ discussed here.

EDIT: A more useful form, it turns out, is $ \tilde{f}(q,z) =\sum\limits_{m=1}^{\infty} (-1)^m q^{m(m+1)/2} z^m$. Its normal (non-$q$-analog) limit is trivial and appealing.

I guess it can be expressed in terms of theta functions (which see), but not in terms of, say, the functions of first-year calculus. — Gerry Myerson, Oct 18 '11 at 11:42
@GerryMyerson: thanks, Gerry! Theta-functions seems to be the key. And I don't care much about which year calculus it is, be it what it is. — Slava Kashcheyevs, Oct 18 '11 at 11:48
No link missing - "which see" just means I'm encouraging you to look up theta functions somewhere, as there is a lot of literature about them and I think they are what you want. — Gerry Myerson, Oct 18 '11 at 11:51
Of course, one only needs to replace $q^\Delta$ with $z$ in the $q$-hypergeometric expression given below. :) — J. M. ain't a mathematician, Oct 19 '11 at 18:01
:) Sure! Still there is a tiny hope that it is theta-something for which representations other that this defining series are workagle (like asymptotic expansion around $q \to 1$). — Slava Kashcheyevs, Oct 19 '11 at 18:07
@J.M. May be we can try expressing the sum via Ramanujan theta function and use http://dlmf.nist.gov/20.11#E3 Another lead that I'm investigating is Ramanujan's identity as quote in Eq. 3.4 of Bruce Berndt's introduction to q-series — Slava Kashcheyevs, Oct 19 '11 at 18:11
I'm rather unskilled in manipulating modular forms (sorry), but you do have a good reason to suspect that Ramanujan has considered series of this sort; it does look like something he'd have looked at... — J. M. ain't a mathematician, Oct 19 '11 at 18:15

score 3 · Accepted Answer · answered Oct 18 '11 at 12:17

3

A fair bit of massaging is needed here.

$$\begin{align*}\sum_{m=1}^{\infty} (-1)^m q^{m(m+1)/2 + m \Delta}&=\sum_{m=2}^{\infty} (-1)^{m-1} q^{m(m-1)/2}q^{(m-1)\Delta}\\&=-q^{-\Delta}\sum_{m=2}^{\infty} (-1)^m q^{m(m-1)/2}q^{m\Delta}\\&=q^{-\Delta}-1-q^{-\Delta}\sum_{m=0}^{\infty} (-1)^m q^{m(m-1)/2}q^{m\Delta}\\&=q^{-\Delta}-1-q^{-\Delta}\sum_{m=0}^{\infty}\frac{(q;q)_m}{(q;q)_m (0;q)_m} (-1)^m q^{m(m-1)/2}q^{m\Delta}\end{align*}$$

and finally we recognize the form of a basic hypergeometric function:

$$\sum_{m=1}^{\infty} (-1)^m q^{m(m+1)/2 + m \Delta}=q^{-\Delta}-1-q^{-\Delta}{}_1 \phi_1\left({q \atop 0};q,q^\Delta\right)$$

Probably there is an easier expression in terms of Jacobi theta functions, but I haven't tried that route...

answered Oct 18 '11 at 12:17

J. M. ain't a mathematician

76,540

I'm trying to adopt $\theta_1(z | \tau)$ to the case of $\Delta= 0, 1$. This would allow me to define the effective temperature that physicists would recognize and love... – Slava Kashcheyevs Oct 18 '11 at 12:20
@Slaviks: due to the $q^{m\Delta}$ bit, it won't be so straightforward... – J. M. ain't a mathematician Oct 18 '11 at 12:31
got a copy of Gasper's book on basic hypergeometric series. Fantastic! – Slava Kashcheyevs Oct 18 '11 at 12:47
yes, just realized that. Since your ${}_1 \phi_1$ is not limited to small $\Delta$, I hope to get the Fermi-Dirac function from there... – Slava Kashcheyevs Oct 18 '11 at 12:48
Gasper/Rahman is quite nice. :) If you can, get Fine's book too. – J. M. ain't a mathematician Oct 18 '11 at 12:50
May be you have a quick advice how to take $q \to 1$ limit while keeping $q^{\Delta} \not = 1, 0$? – Slava Kashcheyevs Oct 18 '11 at 12:51
Nothing obvious comes to mind. :( Sorry. – J. M. ain't a mathematician Oct 18 '11 at 12:56
I thnik I've got an idea - take $q=e^{-\epsilon}$ and convert the sum int he limit of $\epsilon \to 0 $ to an integral. I looks like an error integral, should be expandable then in $\epsilon$ to the leading order. – Slava Kashcheyevs Oct 18 '11 at 13:27

Closed form for $\sum_{m \geq 1} (-1)^m q^{m(m+1)/2 + m \Delta}$?

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