# Is there a $q$-L'Hospital's Rule?

Let $\binom{n}{j}_q$ be a $q$-binomial coefficient and $(x;q)_n = (1-x)(1-qx)\cdots(1-q^{n-1}x).$ Consider the sum $$f(n,m,r,k)= \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}q^{mj^2+rj} \binom{2n}{j}_{q^k}$$ for integers $m,r$ and positive integers $k$.

Note that a famous result of Gauss says that $f(n,0,0,1)=(q;q^2)_n$.

For $n=1$ we get $f(1,m,r,k)=1-q^{m+r}-q^{k+m+r}+q^{4m+2r}$

and therefore

$$\frac{f(1,m,r,k) } {1-q} ={\frac{1-q^{k+m+r}}{1-q}}-q^{m+r}{\frac{1-q^{3m+r}}{1-q}}$$

which for ${q\to1}$ converges to ${(k+m+r)-(3m+r)}={k-2m}.$

Computer experiments suggest that for each positive integer $n$

$$\lim_ {q\to1}\frac{f(n,m,r,k) } {{(q;q^2)_n} }=(k-2m)^n.$$

Till now I did not find a method which leads to a proof of this result. I asked the special case $\lim_ {q\to1}\frac{f(n,m,m,1)}{f(n,1,1,1)}$ of this Problem already in Mathematics Stack Exchange 1359886.

• This limit appears to diverge, and it doesn't match the limit in the Stack Exchange question. Could you please clarify? Aug 3, 2015 at 19:00
• The Stack Exchange question is in fact a special case of this question. I have edited my question in order to clarify the Situation. Aug 4, 2015 at 8:22

I think you can achieve this using the actual $q$-L'Hospital's Rule. Consider

$$\frac{ \partial^i f(n,m,r,k)}{\partial q^i} (1)= \sum_{i=0}^a \binom{a}{i} \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}\frac{\partial^a q^{(m-k/2)j^2+rj}}{ \partial q^a}(1) \frac{ \partial ^{i-a} q ^{ (k/2) j^2} \binom{2n}{j}_{q^k}}{ \partial q^{i-a}}(1)$$

Now $\frac{\partial^a q^{(m-k/2) j^2+rj}}{ \partial q^a}(1)$ is a polynomial of degree $2a$ in $j$. So we can write this in terms of

$$F(b,c) = \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}j^b \frac{ \partial ^{c} q^{(k/2) j^2} \binom{2n}{j}_{q^k}}{ \partial q^{c}}(1)$$

where $b+2c \leq 2i$.

We want to show that this vanishes for $i < n$ and evaluate it for $i=n$. For the first, it suffices to show that for all $b$ and $c$ with $b+2c < 2n$ we have:

$$F(b,c)=0$$

But these terms all show up in computing the partial derivatives for the sum

$$\sum\limits_{j = 0}^{2n} {( - 1)}^{ j} q^{(k/2) j^2 + rj} \binom{2n}{j}_{q^k}$$

$$\frac{ \partial ^i \sum\limits_{j = 0}^{2n} {( - 1)}^{ j} q^{(k/2) j^2 + rj} \binom{2n}{j}_{q^k}}{ \partial q^i} (1) = \sum_{i=0}^a \binom{a}{i} \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}\frac{\partial^a q^{rj}}{ \partial q^a}(1) \frac{ \partial ^{i-a} q ^{ (k/2) j^2} \binom{2n}{j}_{q^k}}{ \partial q^{i-a}}(1)$$

$$= \sum_{i=0}^a \binom{a}{i} \sum\limits_{j = 0}^{2n} {( - 1)}^{ j} (rj)(rj-1) \dots (rj+1-a) \frac{ \partial ^{i-a} q ^{ (k/2) j^2} \binom{2n}{j}_{q^k}}{ \partial q^{i-a}}(1)$$

$$= \sum_{i=0}^a \binom{a}{i} \sum_{b=0}^a[j^b] (rj)(rj-1) \dots (rj+1-a) F(b,i-a)$$

Now suppose by induction that $F(b,c)=0$ for $b+c < i$ for some fixed $i \leq 2n$. Then this sum simplifies to only include the terms where $b=a$. Here

$$[j^a] (rj)(rj-1) \dots (rj+1-a) = r^a$$

so we obtain

$$= \sum_{i=0}^a \binom{a}{i} r^a F(a,i-a)$$

If $i<2n$ then this polynomial is identically $0$ by Max's computation, so all the coefficients vanish, so $F(a,i-a)=0$. This verifies the induction step.

Hence $F(b,c)=0$ for $b+c<2n$, which includes $b+2c<2n$, so the first $n-1$ partial derivatives vanish.

What about the $n$th partial derivative? Well the only $F(b,c)$ with $b + 2c \leq 2n$ that we don't know is $0$ is $F(2n,0)$. So dropping all the other terms we get

$$\frac{ \partial^n f(n,m,r,k)}{\partial q^n} (1)= F(2n,0) [ j^{2n}]\frac{\partial^n q^{(m-k/2)j^2+rj}}{ \partial q^n}(1)$$

Clearly

$$[ j^{2n}]\frac{\partial^n q^{(m-k/2)j^2+rj}}{ \partial q^n}(1) = [j^{2n}] ( (m-k/2)j^2+rj) ((m-k/2)j^2+rj-1) \dots ((m-k/2)j^2+rj+1-n) = (m-k/2)^{n}$$

So we get a constant depending only on $n$ times $(2k-m)^n$. To get the correct constant, you can compute it directly, or just note that it is sufficient to handle one special case for each $n$ and note that you did the $f(n,0,1,2)$ case in the comments to Max's answer.

• Very nice! We just happen to be lucky that $F(b,c)=0$ for $b+c<2n$, which was not obvious at all. Aug 12, 2015 at 2:30
• This is a beautiful proof and deserves the bounty. Aug 12, 2015 at 10:54
• @MaxAlekseyev Well it's not purely luck, because once you find evidence for a nice formula, it gets a lot more likely that there is some nice reason that a nice formula holds. Aug 12, 2015 at 12:59
• This answer is cited in this paper of Johann Cigler's arxiv.org/abs/1602.07850
– j.c.
Sep 27, 2017 at 15:54

A couple of remarks.

First, it easy to see that if $q=1-t$, then $$(q;q^2)_n = \prod_{j=0}^{n-1} (1-q^{2j+1}) = \prod_{j=0}^{n-1} ((2j+1)t+O(t^2)) = (2n-1)!!\cdot t^n + O(t^{n+1}).$$ The limit in question can therefore be restated as $$\lim_{q\to 1} \frac{f(n,m,r,k)}{(1-q)^n} = (k-2m)^n\cdot (2n-1)!!.$$

Second, the limit can be easily proved for $k=2m$. In fact, we can even get a closed form for $f(n,m,r,2m)$. Using generating function for $q$-binomial coefficients, we have $$f(n,m,r,2m) = \sum_{j=0}^{2n} (-1)^j q^{mj^2+rj} \binom{2n}{j}_{q^{2m}}$$ $$=\sum_{j=0}^{2n} (-1)^j q^{(r+m)j} q^{2mj(j-1)/2} \binom{2n}{j}_{q^{2m}} =\sum_{j=0}^{2n} (-1)^j q^{(r+m)j} [z^j]\ (-z,q^{2m})_{2n}$$ $$=\sum_{j=0}^{2n} (-q^{r+m})^j [z^j]\ (-z;q^{2m})_{2n} = (q^{r+m};q^{2m})_{2n}.$$ Here $[z^j]$ is the operator of extracting the coefficient of $z^j$.

Again letting $q=1-t$, we trivially get $(q^{r+m};q^{2m})_{2n} = O(t^{2n})$, which implies the zero limit for any $n>0$.

• Yes, this is one of the special cases, where I had a proof. The other one is $f(n,0,r,1)$. Aug 5, 2015 at 11:03
• Another special case with a closed formula is $f(n,0,1,2)=(q;q^2)_n(-q^2;q^2)_n.$ This implies $\lim_ {q\to1}\frac{f(n,0,2r+1,2)}{{(q;q^2)_n} }=2^n.$ Aug 11, 2015 at 7:13