Negated Upper Index of Gaussian Binomial Coefficient

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Theorem

Let $q \in \R_{\ne 1}, r \in \R, k \in \Z$.

Then:

$\dbinom r k_q = \paren {-1}^k \dbinom {k - r - 1} k_q q^{k r - k \paren {k - 1} / 2}$

where $\dbinom r k_q$ is a binomial coefficient.


Proof

First note that:

\(\ds 1 - q^t\) \(=\) \(\ds q^{-t} \dfrac {1 - q^t} {q^{-t} }\)
\(\ds \) \(=\) \(\ds \dfrac {q^{-t} - 1} {q^{-t} }\)
\(\ds \) \(=\) \(\ds q^t \paren {q^{-t} - 1}\)
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds -q^t \paren {1 - q^{-t} }\)


Then:

\(\ds \binom r k_q\) \(=\) \(\ds \prod_{j \mathop = 0}^{k - 1} \dfrac {1 - q^{r - j} } {1 - q^{j + 1} }\) Definition of Gaussian Binomial Coefficient
\(\ds \) \(=\) \(\ds \prod_{j \mathop = 0}^{k - 1} \dfrac {-q^{r - j} \paren {1 - q^{-\paren {r - j} } } } {1 - q^{j + 1} }\) from $(1)$
\(\ds \) \(=\) \(\ds \paren {-1}^k \prod_{j \mathop = 0}^{k - 1} \dfrac {q^{r - j} \paren {1 - q^{-\paren {r - j} } } } {1 - q^{j + 1} }\)
\(\ds \) \(=\) \(\ds \paren {-1}^k \prod_{j \mathop = 0}^{k - 1} \dfrac {q^{-\paren {j - r} } \paren {1 - q^{j - r} } } {1 - q^{j + 1} }\)
\(\ds \) \(=\) \(\ds \paren {-1}^k \prod_{j \mathop = 0}^{k - 1} \dfrac {q^{-\paren {\paren {k - 1} - j - r} } \paren {1 - q^{\paren {k - 1} - j - r} } } {1 - q^{j + 1} }\) Permutation of Indices of Product
\(\ds \) \(=\) \(\ds \paren {-1}^k \prod_{j \mathop = 0}^{k - 1} \dfrac {1 - q^{\paren {k - r - 1} - j} } {1 - q^{j + 1} } \prod_{j \mathop = 0}^{k - 1} q^{-\paren {\paren {k - 1} - j - r} }\)
\(\ds \) \(=\) \(\ds \paren {-1}^k \prod_{j \mathop = 0}^{k - 1} \dfrac {1 - q^{\paren {k - r - 1} - j} } {1 - q^{j + 1} } \prod_{j \mathop = 1}^k q^j \prod_{j \mathop = 0}^{k - 1} q^{r - k}\)
\(\ds \) \(=\) \(\ds \paren {\paren {-1}^k \prod_{j \mathop = 0}^{k - 1} \dfrac {1 - q^{k - r - 1 - j} } {1 - q^{j + 1} } } q^{k r - k \paren {k - 1} / 2}\) Closed Form for Triangular Numbers and algebra
\(\ds \) \(=\) \(\ds \paren {-1}^k \dbinom {k - r - 1} k_q q^{k r - k \paren {k - 1} / 2}\) Definition of Gaussian Binomial Coefficient

$\blacksquare$


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