6
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For the $q$-exponential $$e_q(u) = \sum_{n=0}^{\infty} \frac{u^n}{[n]_q!}$$ with $[k]_q=\frac{1-q^k}{1-q}$ and $[n]_q! = [n]_q [n-1]_q \cdots [1]_q$, we don't have the property $e_q(u) e_q(v) = e_q (u+v)$. I learned this the hard way when trying to compute the infinite product $$\prod_{i=0}^{\infty} e_q (x q^i)$$ where $x$ is an indeterminate.

  • Does anyone see a way to get a closed form expression for this product?

Any help would be greatly appreciated!

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    $\begingroup$ What do you consider closed form here? A MacMahon type product? A generating function for plane partitions? Something else?... $\endgroup$ – Gjergji Zaimi Mar 19 '13 at 3:25
  • $\begingroup$ Hopefully something else - but you're right that for a certain value of $x$ as a function of $q$ this will return the MacMahon generating function for plane partitions. $\endgroup$ – Alexander Moll Mar 19 '13 at 3:32
  • $\begingroup$ I'm not seeing how to write the generating function for plane partitions in terms of the q-exponential. Can you explain, or provide a reference please? Or is it really easy and I'm just being stupid? I guess I'm mainly asking out of personal interest, but it might be helpful. $\endgroup$ – Benjamin Young Mar 19 '13 at 4:01
  • $\begingroup$ Hey Ben - use the last comment in en.wikipedia.org/wiki/Q-exponential $\endgroup$ – Alexander Moll Mar 19 '13 at 15:28
  • $\begingroup$ Ah, as I expected, I was being stupid. :) $\endgroup$ – Benjamin Young Mar 19 '13 at 20:42

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