Questions tagged [q-series]
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9
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Fitting a product into the quintuple or Jacobi triple product
The Rogers-Ramanujan functions fit nicely into the QPI or JTP. In fact we have that $$(q^{5};q^{5})_{\infty}(q,q^{4};q^{5})_{\infty}=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{\frac{(5n^{2}-3n)}{2}}$$ and we ...
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Reference for modularity of the Andrews–Gordon–Rogers–Ramanujan identities?
The right-hand side of the identity https://mathworld.wolfram.com/Andrews-GordonIdentity.html is a $q$-series $\frac{(q^i,q^{2k+1-i},q^{2k+1};q^{2k+1})_\infty}{(q;q)_\infty}$; is there a reference of ...
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A curious $q$-series identity on a truncated Euler function
Recall that a $q$-Pochhammer symbol is defined as
$$
(x)_n = (x;q)_n := \prod_{l=0}^{n-1}(1-q^l x).
$$
I found the following curious $q$-series identity that seems to hold for any $n\geq 0$:
$$
(-1)^{...
- 1,410
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q-binomial-like series with exponentials defining probability distribution
Recently I encountered the series
$$f(d) = \frac{1}{(t;t)_\infty} \sum_{k=0}^\infty \frac{(-1)^k t^{\binom{k}{2}}}{(t;t)_k} e^{-t^{d-k}}$$
where $(t;t)_n = \prod_{i=1}^n (1-t^i)$, and $0 < t < 1$...
- 538
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Addition formulas for q-analogs of trigonometric functions
Sine and Cosine functions possess notable formulas for addition of angles
$$
\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b) \qquad \text{or} \qquad \cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b).
$$
One can ...
- 106
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Analytic function with q- difference equation involving theta
Consider the analytic space $\mathbb{C}^{*}$ with coordinate $z$. Let $q$ be some parameter with $|q|<1$ and define the analytic function $$\theta(z;q):=\sum_{n\in\mathbb{Z}}q^{\binom{n}{2}}(-z)^{n}...
user108998
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Concerning the coefficient $[q^n]\sum_{n\ge1}\frac{(aq)^n}{(1-bq^n)^2}$
I posted this on MSE, and not even @ParamanandSingh could answer, so I thought I should post it here
Background:
While trying to answer this question, I came up with a question of my own.
Let $|a|,|b|,...
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15
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q-series identity related to Jackson-Slater, proof required
The question:
I have been trying to prove the following $q$-series identity for quite some time now:
$$
\sum_{k \geq 0} \frac{q^{2k^2}}{(q)_{2k}} = \sum_{m,k \geq 0} \frac{q^{m^2 + 3k m + 4k^2}}{(q)...
- 3,202
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Finite version special case Jacobi triple product formula
In this paper, Shanks uses the following formula:
$$ \sum_{s=0}^{n-1}q^{s(2n+1)} \times \left( \prod_{k=s+1}^{n} \dfrac{1-q^{2k}}{1-q^{2k-1}}\right) = \sum_{s=1}^{2n} q^{\frac{s(s-1)}{2}}$$
to get a ...
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