Question

How to prove that $\sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i}=\frac{(ab;q)_n}{(q;q)_n}$?

score 102 · Answer 1 · 2026-02-28 03:01:28

102

Views

How to prove that $\sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i}=\frac{(ab;q)_n}{(q;q)_n}$?

Published on 28 Feb 2026 - 3:01

score 136 · Answer 2 · 2015-11-05 14:15:59

136

Views

a conjecture of two equivalent q-continued fractions related to the reciprocal of the Göllnitz-Gordon continued fraction A111374-OEIS

Published on 05 Nov 2015 - 14:15

score 726 · Answer 3 · 2015-11-22 19:29:08

726

Views

Closed form of the integral ${\large\int}_0^\infty e^{-x}\prod_{n=1}^\infty\left(1-e^{-24\!\;n\!\;x}\right)dx$

Published on 22 Nov 2015 - 19:29

score 1.7k · Answer 4 · 2016-01-03 18:33:04

1.7k

Views

Why $e^{\pi}-\pi \approx 20$, and $e^{2\pi}-24 \approx 2^9$?

Published on 03 Jan 2016 - 18:33

score 104 · Answer 5 · 2016-03-13 01:01:15

104

Views

Weighted Q-binomial Coefficients

Published on 13 Mar 2016 - 1:01

score 55 · Answer 6 · 2016-03-29 11:18:43

55

Views

show $\sum_{j=0}^n (-1)^j {n \brack j}_q =0$ for n odd

Published on 29 Mar 2016 - 11:18

score 77 · Answer 7 · 2013-06-28 23:34:29

77

Views

A q-series related to adjoint representation of lie group

Published on 28 Jun 2013 - 23:34

score 4k · Answer 8 · 2013-07-03 01:07:31

4k

Views

Motivation for/history of Jacobi's triple product identity

Published on 03 Jul 2013 - 1:07

score 375 · Answer 9 · 2026-03-24 21:48:06

375

Views

q-series identities

Published on 24 Mar 2026 - 21:48

score 145 · Answer 10 · 2026-03-24 18:47:06

145

Views

Q-series identities #2

Published on 24 Mar 2026 - 18:47

score 625 · Answer 11 · 2013-12-11 14:11:15

625

Views

Combinatorial Identity $ \sum_{k=1}^n (-1)^{k-1} \cdot q^{\frac{k(k-1)}{2}} \cdot \frac{\prod_{i=n-k+1}^n(1-q^i)}{\prod_{i=1}^k(1-q^i)} = 1 $

Published on 11 Dec 2013 - 14:11

score 3.8k · Answer 12 · 2013-12-12 23:11:40

3.8k

Views

A problem with the geometric series and matrices?

Published on 12 Dec 2013 - 23:11

score 180 · Answer 13 · 2014-01-01 23:19:58

180

Views

Ramanujan Notebook Part 1 (1.16): $\sum q^{n^2} = (-q;q^2)_\infty^2(q^2;q^2)_\infty=\frac{(-q;-q)_\infty}{(q;-q)_\infty}$

Published on 01 Jan 2014 - 23:19

score 348 · Answer 14 · 2026-03-24 19:52:33

348

Views

$q$-series identity

Published on 24 Mar 2026 - 19:52

score 301 · Answer 15 · 2017-04-16 18:12:01

301

Views

Differentiating q-Series Expression

Published on 16 Apr 2017 - 18:12

List Question

How to prove that $\sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i}=\frac{(ab;q)_n}{(q;q)_n}$?

a conjecture of two equivalent q-continued fractions related to the reciprocal of the Göllnitz-Gordon continued fraction A111374-OEIS

Closed form of the integral ${\large\int}_0^\infty e^{-x}\prod_{n=1}^\infty\left(1-e^{-24\!\;n\!\;x}\right)dx$

Why $e^{\pi}-\pi \approx 20$, and $e^{2\pi}-24 \approx 2^9$?

Weighted Q-binomial Coefficients

show $\sum_{j=0}^n (-1)^j {n \brack j}_q =0$ for n odd

A q-series related to adjoint representation of lie group

Motivation for/history of Jacobi's triple product identity

q-series identities

Q-series identities #2

Combinatorial Identity $ \sum_{k=1}^n (-1)^{k-1} \cdot q^{\frac{k(k-1)}{2}} \cdot \frac{\prod_{i=n-k+1}^n(1-q^i)}{\prod_{i=1}^k(1-q^i)} = 1 $

A problem with the geometric series and matrices?

Ramanujan Notebook Part 1 (1.16): $\sum q^{n^2} = (-q;q^2)_\infty^2(q^2;q^2)_\infty=\frac{(-q;-q)_\infty}{(q;-q)_\infty}$

$q$-series identity

Differentiating q-Series Expression

Trending Questions

Popular # Hahtags

Popular Questions