The qPochhammer Symbol
The Pochhammer symbol, also known as the rising factorial, is a notation that is used in combinatorics and special functions.
Definition of Pochhammer Symbol
It is defined as follows:
\[(a)_n = \begin{cases} 1 & \text{if } n = 0 \\ a(a+1)(a+2)\dotsm(a+n1) & \text{if } n > 0 \end{cases}\]The Pochhammer symbol has a number of applications in various areas of mathematics, including combinatorics, special functions, and algebraic geometry. It is named after German mathematician Ernst Pochhammer, who introduced it in a paper published in 1877.
Definition of qPochhammer Symbol
The qPochhammer symbol is a generalization of the Pochhammer symbol to the realm of qseries, which are series that involve powers of the variable q. The qPochhammer symbol was introduced by Leonard Carlitz in 1935. It has many of the same properties as the Pochhammer symbol, but is defined in terms of qseries rather than ordinary series.
The qPochhammer symbol is defined as follows:
\[(a;q)_n = \begin{cases} 1 & \text{if } n = 0 \\ \displaystyle\prod_{i=0}^{n1} (1aq^i) = (1a)(1aq)(1aq^2)\dotsm(1aq^{n1}) & \text{if } n > 0 \end{cases}\]Properties
Here are some properties of the qPochhammer symbol:

Recurrence relation: For all nonnegative integers $n$ and $m$, we have
\[(a;q)_{n+m} = (a;q)_n (aq^n;q)_m\] 
Infinity identity: For all complex numbers $a$ and $q$, we have
\[(a;q)_n = \frac{(a;q)_\infty}{(aq^n;q)_\infty}\] 
Binomial Equality: The Gaussian binomial coefficients are defined as follows:
\[\binom{n}{k}_q = \frac{(1q)(1q^2)\dotsm(1q^n)}{(1q)(1q^2)\dotsm(1q^k)(1q^{k+1})\dotsm(1q^n)}\]where $n$ and $k$ are nonnegative integers. If $k$ > $n$, this evaluates to $0$. For $k = 0$, the value is $1$ since both the numerator and denominator are empty products.
For all nonnegative integers $n$ and all complex numbers $a$ and $b$, we have
\[\sum_{k=0}^n \binom{n}{k}_q a^k b^{nk} = (a;q)_n (b;q)_n\]
Proofs
Here are the proofs for the above properties:
Recurrence relation
We can prove the recurrence relation by induction on $m$:

For the base case $m=0$, we have
\[\begin{aligned} (a;q)_n &= (a;q)_n (a;q)_0 \\ &= (a;q)_n \cdot 1 \end{aligned}\]This holds, so the base case is proven.

Inductive step: Assume that the formula holds for some arbitrary value of $\textcolor{purple}{m = k}$:
\[\textcolor{purple}{(a;q)_{n+m} = (a;q)_{n+k} = (a;q)_n (aq^n;q)_k}\]We will now prove that it holds for $m = k+1$. We will prove the recurrence formula for the qPochhammer symbol using induction on $m$.
\[\begin{aligned} (a;q)_{n+m} &= (a;q)_{n+k+1} \\ &= (1a)(1aq)(1aq^2)\dotsm(1aq^{n+k1})(1aq^{n+k}) \\ &= \underbrace{(a;q)_{n+k}}_{\textcolor{purple}{\text{true for }m=k}}(1aq^{n+k}) \\ &= \overbrace{(a;q)_n \textcolor{green}{(aq^n;q)_k}}(1aq^{n+k}) \\ &= (a;q)_n\textcolor{green}{(1aq^n)(1aq^{n+1})(1aq^{n+2})\dotsm(1aq^{n+k1})} (1aq^{n+k})\\ &= (a;q)_n (aq^n;q)_{k+1} \\ &= (a;q)_n (aq^n;q)_{m} \end{aligned}\]Thus, the formula holds for $m = k+1$. By induction, the formula holds for all values of $m$.
Infinity identity
We can prove the infinity identity as follows:
\[\begin{aligned} \frac{(a;q)_\infty}{(aq^n;q)_\infty} &= \frac{\displaystyle\lim_{n \to +\infty}(a;q)_n }{\displaystyle\lim_{n \to +\infty}(aq^n;q)_n} = \frac{\displaystyle\prod_{i=0}^\infty (1aq^{i})}{ \displaystyle \prod_{i=0}^\infty (1aq^{i+n})}= \frac{\displaystyle\prod_{i=0}^{n1} (1aq^{i}) \prod_{i=n}^{\infty} (1aq^{i})}{ \displaystyle\prod_{i=0}^\infty (1aq^{i+n})}\\ &= \frac{\displaystyle\prod_{i=0}^{n1} (1aq^{i}) \prod_{k=0}^{\infty} (1aq^{k+n})}{ \displaystyle\prod_{i=0}^\infty (1aq^{i+n})}= \prod_{i=0}^{n1} (1aq^{i}) \frac{\displaystyle\prod_{k=0}^{\infty} (1aq^{k+n})}{ \displaystyle\prod_{i=0}^\infty (1aq^{i+n})}\\ &= \prod_{i=0}^{n1} (1aq^{i}) \cdot 1 \\ &= (a;q)_n \\ \end{aligned}\]Binomial Equality
We can prove the binomial equality by induction on $n$:
 For the base case $n=0$, we have
 Inductive step: Assume that the formula holds for some arbitrary value of $\textcolor{purple}{n}$.We will now prove that it holds for $n+1$:
If you found this post or this website helpful and would like to support our work, please consider making a donation. Thank you!
Help Us