4
$\begingroup$

By Cauchy identity, $${}_1\phi_0(a;—;q,z)=\sum_{n\geq0}\frac{(a;q)_n}{(q;q)_n}z^n=\frac{(az;q)_{\infty}}{(z;q)_\infty},\quad|z|<1,|q|<1,$$ we can obtain the $q-$analogue of $(1-z)^{-a}(1-z)^{-b}=(1-z)^{-a-b},$ $${}_1\phi_0(a;—;q,z){}_1\phi_0(b;—;q,az)={}_1\phi_0(ab;—;q,z),$$ which is $$\sum_{n\geq0}\frac{(a;q)_n}{(q;q)_n}z^n\sum_{m\geq0}\frac{(b;q)_m}{(q;q)_m}(az)^m=\sum_{n\geq0}\frac{(ab;q)_n}{(q;q)_n}z^n.$$

Comparing the coefficients of $z$ in the both side of equation above, the identity in the title can be established as $$\sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i}=\frac{(ab;q)_n}{(q;q)_n}.$$

However, what I really concern is that there is any possible for us to prove this identity directly without utilizing Cauchy identity. Maybe we can use mathematical induction on $n$ or some other combinatorial or algebraic proofs?

$\endgroup$
3
$\begingroup$

Here's a proof that doesn't involve Cauchy's identity. I will begin by stating two elementary lemmas.

Lemma 1: Let $f \in \mathbb{C}[x,y,z]$ be a polynomial. If $f$ vanishes on the set $$\{ (a,a^i,a^j) : a \in \mathbb{C} , i,j \in \mathbb{N}_{>0} \},$$ then $f$ is identically 0.

Lemma 2: Let $\binom{n}{k}_q :=\frac{(q;q)_n}{(q;q)_k (q;q)_{n-k}}$ denote the $q$-binomial coefficients. Then the following Vandermonde-like identity holds: $$\sum_{i} \binom{i+A-1}{i}_q \binom{n-i+B-1}{n-i}_q q^{A(n-i)} = \binom{n+A+B-1}{n}_q.$$


Given these two lemmas, we proceed as follows: Let $$S(q,a,b) = (q;q)_n \sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i} - (ab;q)_n.$$ Your identity is equivalent to $S$ vanishing identically. Since $S$ is polynomial in its variables, by Lemma 1 it suffices to show that $$(*) S(q,q^A, q^B) = 0$$ for all $q,A,B$. Since $(q^A;q)_i =\frac{(q;q)_{i+A-1}}{(q;q)_{A-1}}$, $(*)$ is equivalent to $$\sum_{i=0}^{n} \frac{(q;q)_{i+A-1}}{(q;q)_{A-1} (q;q)_i} \frac{(q;q)_{n-i+B-1}}{(q;q)_{B-1} (q;q)_{n-i}} q^{A(n-i)} = \frac{(q;q)_{n+A+B-1}}{(q;q)_{A+B-1} (q;q)_{n}},$$ or, in $q$-nomial notation, $$\sum_{i} \binom{i+A-1}{i}_q \binom{n-i+B-1}{n-i}_q q^{A(n-i)} = \binom{n+A+B-1}{n}_q,$$ which is true by Lemma 2.


Lemma 1 may be proved, for instance, using the Combinatorial Nullstellensatz. Lemma 2 is proved just like the q-Vandermonde identity, only that the proof uses the identity $\prod_{k=0}^{n-1} (1-q^k t)^{-1} = \sum_{k} \binom{n+k-1}{k}_q t^k$ instead of $\prod_{k=0}^{n-1} (1+q^k t) = \sum_{k} \binom{n}{k}_q q^{\binom{k}{2}} t^k$.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Below is a simplification of the argument by Ofir Gorodetsky.

At first, we denote $a=1/c$ and multiply both sides by $c^n (q;q)_n$, we get an equivalent identity $$ \sum_{i=0}^n \binom{n}i_q(c-1)(c-q)\dots (c-q^{i-1})(1-b)(1-bq)\dots (1-bq^{n-i-1})=\\ =(c-b)(c-qb)\dots (c-q^{n-1}b). $$ Consider both parts as polynomials in $b,c$ of degree at most $n$. It suffices to check that they take the same values on a "triangle" $\{c=q^k,b=q^{-m}:k,m\in \{0,1,2\dots,n\}, k+m\leqslant n\}$ (this is a useful lemma in Combinatorial Nullstellensatz spirit, it may be deduced from CN itself). But this is almost obvious: if $k+m<n$, both parts take zero values (any summand in LHS takes zero value), and if $k+m=n$, there is unique non-zero summand in LHS corresponding to $i=k$, and it is straightforward that it equals RHS.

Note: taking homogeneous (in $b,c$) parts of this identity we get $q$-binomial theorem. Specific summands in LHS are not homogeneous, but the total sum somehow is.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I see. By plugging negative powers of $q$, instead of positive ones as I did, one gets a much simpler proof. Very nice! $\endgroup$ – Ofir Gorodetsky Oct 26 '16 at 5:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.