Editing Gaussian binomial coefficient (section)

==q identities==

===Analogs of Pascal's identity===

The analogs of [[Pascal's identity]] for the Gaussian binomial coefficients are:<ref>Mukhin, Eugene, chapter 3</ref>

:<math>{m \choose r}_q = q^r {m-1 \choose r}_q + {m-1 \choose r-1}_q</math>

and

:<math>{m \choose r}_q = {m-1 \choose r}_q + q^{m-r}{m-1 \choose r-1}_q.</math>

When <math>q=1</math>, these both give the usual binomial identity. We can see that as <math>m\to\infty</math>, both equations remain valid.

The first Pascal analog allows computation of the Gaussian binomial coefficients recursively (with respect to ''m'' ) using the initial values

:<math>{m \choose m}_q ={m \choose 0}_q=1 </math>

and also shows that the Gaussian binomial coefficients are indeed polynomials (in ''q'').

The second Pascal analog follows from the first using the substitution <math> r \rightarrow m-r </math> and the invariance of the Gaussian binomial coefficients under the reflection <math> r \rightarrow m-r </math>.

These identities have natural interpretations in terms of linear algebra. Recall that <math>\tbinom{m}{r}_q</math> counts ''r''-dimensional subspaces <math>V\subset \mathbb{F}_q^m</math>, and let <math>\pi:\mathbb{F}_q^m \to \mathbb{F}_q^{m-1} </math> be a projection with one-dimensional nullspace <math>E_1 </math>. The first identity comes from the bijection which takes <math>V\subset \mathbb{F}_q^m </math> to the subspace <math>V' = \pi(V)\subset \mathbb{F}_q^{m-1}</math>; in case <math>E_1\not\subset V</math>, the space <math>V'</math> is ''r''-dimensional, and we must also keep track of the linear function <math>\phi:V'\to E_1</math> whose graph is <math>V</math>; but in case <math>E_1\subset V</math>, the space <math>V'</math> is (''r''−1)-dimensional, and we can reconstruct <math>V=\pi^{-1}(V')</math> without any extra information. The second identity has a similar interpretation, taking <math>V</math> to <math>V' = V\cap E_{n-1}</math> for an (''m''−1)-dimensional space <math>E_{m-1}</math>, again splitting into two cases.

===Proofs of the analogs===

Both analogs can be proved by first noting that from the definition of <math>\tbinom{m}{r}_q</math>, we have:

{{NumBlk|:|<math>\binom{m}{r}_q = \frac{1-q^m}{1-q^{m-r}} \binom{m-1}{r}_q</math>|{{EquationRef|1}}}}

{{NumBlk|:|<math>\binom{m}{r}_q = \frac{1-q^m}{1-q^r} \binom{m-1}{r-1}_q</math>|{{EquationRef|2}}}}

{{NumBlk|:|<math>\frac{1-q^r}{1-q^{m-r}}\binom{m-1}{r}_q = \binom{m-1}{r-1}_q</math>|{{EquationRef|3}}}}

As

:<math>\frac{1-q^m}{1-q^{m-r}}=\frac{1-q^r+q^r-q^m}{1-q^{m-r}}=q^r+\frac{1-q^r}{1-q^{m-r}}</math>

Equation ({{EquationNote|1}}) becomes:

:<math>\binom{m}{r}_q = q^r\binom{m-1}{r}_q + \frac{1-q^r}{1-q^{m-r}}\binom{m-1}{r}_q</math>

and substituting equation ({{EquationNote|3}}) gives the first analog.

A similar process, using

:<math>\frac{1-q^m}{1-q^r}=q^{m-r}+\frac{1-q^{m-r}}{1-q^r}</math>

instead, gives the second analog.

===''q''-binomial theorem===

There is an analog of the [[binomial theorem]] for ''q''-binomial coefficients, known as the Cauchy binomial theorem:

:<math>\prod_{k=0}^{n-1} (1+q^kt)=\sum_{k=0}^n q^{k(k-1)/2} 
{n \choose k}_q t^k .</math>

Like the usual binomial theorem, this formula has numerous generalizations and extensions; one such, corresponding to Newton's generalized binomial theorem for negative powers, is

:<math>\prod_{k=0}^{n-1} \frac{1}{1-q^kt}=\sum_{k=0}^\infty  
{n+k-1 \choose k}_q t^k. </math>

In the limit <math>n\rightarrow\infty</math>, these formulas yield

:<math>\prod_{k=0}^{\infty} (1+q^kt)=\sum_{k=0}^\infty \frac{q^{k(k-1)/2}t^k}{[k]_q!\,(1-q)^k}</math>

and

:<math>\prod_{k=0}^\infty \frac{1}{1-q^kt}=\sum_{k=0}^\infty  
\frac{t^k}{[k]_q!\,(1-q)^k}</math>.

Setting <math>t=q</math> gives the generating functions for distinct and any parts respectively. (See also [[Basic hypergeometric series]].)

===Central ''q''-binomial identity===

With the ordinary binomial coefficients, we have:

:<math>\sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}</math>

With ''q''-binomial coefficients, the analog is:

:<math>\sum_{k=0}^n q^{k^2}\binom{n}{k}_q^2 = \binom{2n}{n}_q</math>