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:

\[(a)_n = \begin{cases} 1 & \text{if } n = 0 \\ a(a+1)(a+2)\dotsm(a+n-1) & \text{if } n > 0 \end{cases}\]

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 q-Pochhammer Symbol

The q-Pochhammer symbol is a generalization of the Pochhammer symbol to the realm of q-series, which are series that involve powers of the variable q. The q-Pochhammer 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 q-series rather than ordinary series.

The q-Pochhammer symbol is defined as follows:

\[(a;q)_n = \begin{cases} 1 & \text{if } n = 0 \\ \displaystyle\prod_{i=0}^{n-1} (1-aq^i) = (1-a)(1-aq)(1-aq^2)\dotsm(1-aq^{n-1}) & \text{if } n > 0 \end{cases}\]

Properties

Here are some properties of the q-Pochhammer 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:

    \[\binom{n}{k}_q = \frac{(1-q)(1-q^2)\dotsm(1-q^n)}{(1-q)(1-q^2)\dotsm(1-q^k)(1-q^{k+1})\dotsm(1-q^n)}\]

    where $n$ and $k$ are non-negative 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

    \[\sum_{k=0}^n \binom{n}{k}_q a^k b^{n-k} = (a;q)_n (b;q)_n\]

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 q-Pochhammer symbol using induction on $m$.

    \[\begin{aligned} (a;q)_{n+m} &= (a;q)_{n+k+1} \\ &= (1-a)(1-aq)(1-aq^2)\dotsm(1-aq^{n+k-1})(1-aq^{n+k}) \\ &= \underbrace{(a;q)_{n+k}}_{\textcolor{purple}{\text{true for }m=k}}(1-aq^{n+k}) \\ &= \overbrace{(a;q)_n \textcolor{green}{(aq^n;q)_k}}(1-aq^{n+k}) \\ &= (a;q)_n\textcolor{green}{(1-aq^n)(1-aq^{n+1})(1-aq^{n+2})\dotsm(1-aq^{n+k-1})} (1-aq^{n+k})\\ &= (a;q)_n (aq^n;q)_{k+1} \\ &= (a;q)_n (aq^n;q)_{m} \end{aligned}\]

    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:

\[\begin{aligned} \frac{(a;q)_\infty}{(aq^n;q)_\infty} &= \frac{\displaystyle\lim_{n \to +\infty}(a;q)_n }{\displaystyle\lim_{n \to +\infty}(aq^n;q)_n} = \frac{\displaystyle\prod_{i=0}^\infty (1-aq^{i})}{ \displaystyle \prod_{i=0}^\infty (1-aq^{i+n})}= \frac{\displaystyle\prod_{i=0}^{n-1} (1-aq^{i}) \prod_{i=n}^{\infty} (1-aq^{i})}{ \displaystyle\prod_{i=0}^\infty (1-aq^{i+n})}\\ &= \frac{\displaystyle\prod_{i=0}^{n-1} (1-aq^{i}) \prod_{k=0}^{\infty} (1-aq^{k+n})}{ \displaystyle\prod_{i=0}^\infty (1-aq^{i+n})}= \prod_{i=0}^{n-1} (1-aq^{i}) \frac{\displaystyle\prod_{k=0}^{\infty} (1-aq^{k+n})}{ \displaystyle\prod_{i=0}^\infty (1-aq^{i+n})}\\ &= \prod_{i=0}^{n-1} (1-aq^{i}) \cdot 1 \\ &= (a;q)_n \\ \end{aligned}\]

Binomial Equality

We can prove the binomial equality by induction on $n$:

  • For the base case $n=0$, we have
\[\sum_{k=0}^0 \binom{0}{k}_q a^k b^{0-k} = \binom{0}{0}_q a^0 b^{0-0} = 1 = (a;q)_0 (b;q)_0\]
  • Inductive step: Assume that the formula holds for some arbitrary value of $\textcolor{purple}{n}$.We will now prove that it holds for $n+1$:
\[\begin{align*} \sum_{k=0}^{n+1} \binom{n+1}{k}_q a^k b^{n+1-k} &= \sum_{k=0}^n \binom{n+1}{k}_q a^k b^{n+1-k} + \binom{n+1}{n+1}_q a^{n+1} b^{n+1-(n+1)} \\ &= \textcolor{purple}{\sum_{k=0}^n \binom{n}{k}_q a^k b^{n-k}} + \frac{(1-q^{n+1})}{(1-q)} \cdot a^{n+1} b^0 \\ &= \textcolor{purple}{(a;q)_n (b;q)_n} + \frac{(1-q^{n+1})}{(1-q)} \cdot a^{n+1} b^0 \\ &= (a;q)_{n+1} (b;q)_{n+1} \end{align*}\]