# q-series

MHB Math Helper

#### Opalg

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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Great !

I am looking for a proof for the following

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

#### Opalg

##### MHB Oldtimer
Staff member
Great !

I am looking for a proof for the following

$$\displaystyle \frac{(-b)_{\infty}}{(a)_{\infty}} = \sum_{k \geq 0} \frac{\left(-b/a\right)_k} {(q)_k} a^k$$
That looks like a variation on the q-binomial theorem. Have you looked for a proof in Gasper and Rahman?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Thanks for the link , I really really appreciate it. I thought that q-series are interesting it reminds me of the famous relation between prime numbers and zeta function due to Euler . I was reading the NoteBook by Ramnujan and I was amazed by the vast results related to Hypergeometric functions which are a special case of q-series . I was hoping to read about the Jacobi theta function but I thought it would be better to start by q-series.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
By the way Opalg , do you think it is a hard thing to deal with because I was depressed to see how Ramanujan worked with these stuff and I was like what is that !

#### Opalg

##### MHB Oldtimer
Staff member
By the way Opalg , do you think it is a hard thing to deal with because I was depressed to see how Ramanujan worked with these stuff and I was like what is that !
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.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Hey , I am confused about the notations !

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

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

The latter defines a base $q$ .

#### Opalg

##### MHB Oldtimer
Staff member
Hey , I am confused about the notations !

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

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

The latter defines a base $q$ .
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.

MHB Math Helper

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
How to prove the following

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

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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
To simplify for those how don't understand the notations

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

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 ?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
How to prove the following

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

It is kind of easy to prove it for $x\in \mathbb{Z}^+$
Ok , I think I got it , this is a simple consequence of the q-binomial theorem

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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