Binomial Coefficient is instance of Gaussian Binomial Coefficient

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\dbinom r m_q$ denote the Gaussian binomial coefficient:


Then:

$\ds \lim_{q \mathop \to 1^-} \dbinom r m_q = \dbinom r m$

where $\dbinom r m$ denotes the conventional binomial coefficient.


Proof

We have by definition of Gaussian binomial coefficient:

$\ds \dbinom r m_q = \prod_{k \mathop = 0}^{m - 1} \dfrac {1 - q^{r - k} } {1 - q^{k + 1} }$


Consider a typical factor of this continued product:

\(\ds \dfrac {1 - q^{r - k} } {1 - q^{k + 1} }\) \(=\) \(\ds \dfrac {\paren {1 - q^{r - k} } / \paren {1 - q} } {\paren {1 - q^{k + 1} } / \paren {1 - q} }\) multiplying top and bottom by $1 - q$
\(\ds \) \(=\) \(\ds \dfrac {\sum_{j \mathop = 0}^{r - k - 1} q^j} {\sum_{j \mathop = 0}^k q^j}\) Sum of Geometric Sequence
\(\ds \leadsto \ \ \) \(\ds \lim_{q \mathop \to 1^-} \dfrac {1 - q^{r - k} } {1 - q^{k + 1} }\) \(=\) \(\ds \lim_{q \mathop \to 1^-} \dfrac {\sum_{j \mathop = 0}^{r - k - 1} q^j} {\sum_{j \mathop = 0}^k q^j}\)
\(\ds \) \(=\) \(\ds \dfrac {\sum_{j \mathop = 0}^{r - k - 1} 1} {\sum_{j \mathop = 0}^k 1}\)
\(\ds \) \(=\) \(\ds \dfrac {r - k} {k + 1}\)


Thus:

\(\ds \lim_{q \mathop \to 1^-} \dbinom r k_q\) \(=\) \(\ds \prod_{k \mathop = 0}^{m - 1} \dfrac {r - k} {k + 1}\)
\(\ds \) \(=\) \(\ds \prod_{k \mathop = 1}^m \dfrac {r + 1 - k} k\) Translation of Index Variable of Product
\(\ds \) \(=\) \(\ds \dbinom r m\) Definition of Binomial Coefficient

$\blacksquare$


Sources