q-exponential functions

Question

We know that $$e_q(z)=\sum_{j\geq 0} \frac{z^j}{(q;q)_j}=\frac{1}{(z;q)_{\infty}}$$

where $(a;q)_{\infty}=\prod_{i=0}^{\infty}(1−aq^i)$ denotes the q-shifted factorial.

The limit between the $q$-exponential and the ordinary exponential is

$$\lim_{q\rightarrow1}e_q((1-q)z)=\lim_{q\rightarrow1}\frac{1}{((1-q)z;q)_{\infty}}=e^z.$$

My question is how to prove

$$\lim_{q\rightarrow1}\frac{1}{((1-q)\sqrt{1-q^2}x;q)_{\infty}}=e^{\sqrt{2}x}?$$

I'm glad for your help.

Somos · Accepted Answer · 2017-08-19T20:41:38.237

3

We know that $(q;q)_n=(1-q)^n[n]_q!\;$ and $\lim_{q\to1} [n]_q!=n!.\quad$ Therefore $$e_q((1-q)(1+q)x) = \sum_{n=0}^\infty \frac{(1-q)^n(1+q)^nx^n}{(q;q)_n}=\sum_{n=0}^\infty \frac{(1+q)^n x^n}{[n]_q!}.$$ The limit as $q\to1$ gives $e^{2x}$ which is justified because of absolute convergence.

edited Aug 19 '17 at 20:41

answered Aug 17 '17 at 12:21

Somos

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Mr @Somos Can we permute the limit and the sum ? – othmane Aug 18 '17 at 20:24
@othmane Yes, go ahead. See what you get and let us know. – Somos Aug 18 '17 at 20:42
Mr @somos it's my bad i forgot a number, in fact this is my question – othmane Aug 19 '17 at 20:35
Mr@somos it's my bad i forgot a number, in fact this is my question $$\lim_{q\rightarrow 1} e_q((1-q)\sqrt{1-q}(1+q)x) $$ – othmane Aug 19 '17 at 20:40
@othmane I didn't understand your comment at first. Please see this MSE question. – Somos Aug 19 '17 at 20:48
The problem is in the limit, i have edited my question above . – othmane Aug 20 '17 at 08:11
@othmane Are you sure you want $((1-q)\sqrt{1-q^2}x;q){\infty}$? I did some computation and I think you want $((1-q)\sqrt{1+q}x;q){\infty}$ to get the $e^{\sqrt{2}x}$ result. Please try a numerical example to check your limit. – Somos Aug 20 '17 at 12:11

q-exponential functions

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