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Anybody with q-series experience ?

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- Jan 17, 2013

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Anybody with q-series experience ?

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- Feb 7, 2012

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I had to deal with them when working with compact quantum groups. What sort of experience were you looking for?Anybody with q-series experience ?

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I am looking for a proof for the following

\(\displaystyle \frac{(-b)_{\infty}}{(a)_{\infty}} = \sum_{k \geq 0} \frac{\left(-b/a\right)_k} {(q)_k} a^k \)

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That looks like a variation on the q-binomial theorem. Have you looked for a proof in Gasper and Rahman?

I am looking for a proof for the following

\(\displaystyle \frac{(-b)_{\infty}}{(a)_{\infty}} = \sum_{k \geq 0} \frac{\left(-b/a\right)_k} {(q)_k} a^k \)

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Please don't expect to understand everything that Ramanujan could do! That way madness lies.

There are results that Ramanujan somehow apprehended by intuition, that even today cannot be proved although they appear to be true, and nobody knows how he arrived at them. He must have had some quite unique insight that probably even he could not have explained.

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\(\displaystyle (a)_k = a(a+1)(a+2) \cdots (a+k-1)\)

\(\displaystyle (a)_k =(a;q)_k = \prod_{n=0}^{k-1} (1-aq^n)\)

The latter defines a base $q$ .

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I was assuming that $(a)_k$ was an abbreviation for $(a;q)_k$ (as here, for example), with the $q$ not explicitly mentioned. But I could well be wrong.

\(\displaystyle (a)_k = a(a+1)(a+2) \cdots (a+k-1)\)

\(\displaystyle (a)_k =(a;q)_k = \prod_{n=0}^{k-1} (1-aq^n)\)

The latter defines a base $q$ .

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Look at Pochhammer symbol.

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\(\displaystyle \lim_{q \to 1}\frac{(a;q)_{\infty}}{(aq^x;q)_{\infty}}= (1-a)^x \)

It is kind of easy to prove it for $x\in \mathbb{Z}^+$

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\(\displaystyle (a;q)_{\infty}= \prod_{k=0}(1-aq^k)\)

Similarilry we have

\(\displaystyle (aq^x;q)_{\infty}= \prod_{k=0}(1-aq^{k+x})\)

So we have to prove that

\(\displaystyle \lim_{q \to 1}\prod_{k=0} \frac{1-aq^k}{1-aq^{k+x}} = (1-a)^ x\)

Any clue ?

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Ok , I think I got it , this is a simple consequence of the q-binomial theorem

\(\displaystyle \lim_{q \to 1}\frac{(a;q)_{\infty}}{(aq^x;q)_{\infty}}= (1-a)^x \)

It is kind of easy to prove it for $x\in \mathbb{Z}^+$

Consider the following

\(\displaystyle {}_1\phi_0 (a;- ;q,z) = \sum_{k\geq 0}\frac{(a;q)_k}{(q;q)_k}z^k=\frac{(az;q)_{\infty}}{(z;q)_{\infty}}\) (1)

In (1) let $a = q^{x}$ and $z = a$

\(\displaystyle {}_1\phi_0 (q^x;- ;q,a) = \sum_{k\geq 0}\frac{(q^x;q)_k}{(q;q)_k}a^k=\frac{(aq^{x};q)_ {\infty} }{(a;q)_{\infty}}\)

Hence we have

\(\displaystyle \frac{(aq^{x};q)_{\infty}}{(a;q)_{\infty}}= \sum_{k\geq 0}\frac{(q^x;q)_k}{(q;q)_k}a^k\)

Now consider the limit

\(\displaystyle \lim_{q \to 1}\frac{(aq^{x};q)_{\infty}}{(a;q)_{\infty}}= \lim_{q \to 1} \sum_{k\geq 0}\frac{(q^x;q)_k}{(q;q)_k}a^k\)(2)

Suppose that \(\displaystyle |a|<1\) and \(\displaystyle |q|<1\) so the sum is uniformly convergent on any sub-disk . So we have to approach $1$ from the left to stay in the disk !

The idea is use the L'Hospitale rule

\(\displaystyle \lim_{q \to 1^-}\frac{(q^x;q)_k}{(q;q)_k} = \lim_{q \to 1^-} \frac{(1-q^x)\cdot (1-q^{x+1}) \cdot(1-q^{x+2}) \cdots (1-q^{x+k-1}) }{(1-q)\cdot(1-q^2)\cdot(1-q^3) \cdots(1-q^k)}\)

which can be written as

\(\displaystyle \lim_{q \to 1^-}\frac{(q^x;q)_k}{(q;q)_k} = \lim_{q \to 1^-} \frac{(1-q^x)}{1-q}\cdot \lim_{q \to 1^-}\frac{(1-q^{x+1})}{1-q^2} \cdot \lim_{q \to 1} \frac{(1-q^{x+2})}{1-q^3} \cdots \lim_{q \to 1^-} \frac{(1-q^{x+k-1}) }{(1-q^k)}\)

\(\displaystyle \lim_{q \to 1^-}\frac{(q^x;q)_k}{(q;q)_k} = \frac{x (x+1)(x+2)\cdots (x+k-1)}{1\cdot 2 \cdot 3 \cdots k} = \frac{(x)_k}{k!} \)

Substitute in (2)

\(\displaystyle \lim_{q \to 1^-}\frac{(aq^{x};q)_{\infty}}{(a;q)_{\infty}}= \sum_{k\geq 0}\frac{(x)_k}{k!}a^k\)

The sum on the right is well-know $(1-x)^{-a}$

\(\displaystyle \lim_{q \to 1^-}\frac{(aq^{x};q)_{\infty}}{(a;q)_{\infty}}= (1-x)^{-a} \) (3)

From (3) we conclude that

\(\displaystyle \lim_{q \to 1^-}\frac{(a;q)_{\infty}}{ (aq^{x};q)_{\infty}}= (1-x)^{a} \)

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