infinite products

hbghlyj

$$\prod_{n=1}^{+\infty}\left(1+x^{2 n-1}\right)^{8}-\prod_{n=1}^{+\infty}\left(1-x^{2 n-1}\right)^{8}=16 x \prod_{n=1}^{+\infty}\left(1+x^{2 n}\right)^{8}$$ 这其实是theta函数的四次方和性质,相当于1/lambda(z)+1/lambda(z+1)=1 For $|x|<1$, prove $\prod_{n \geq 1}\left(\frac{1}{1+x^{n}} \frac{1}{1+x^{2 n-1}}\right)^{8}+16 \prod_{n \geq 1}\left(\frac{1+x^{2 n}}{1+x^{2 n-1}}\right)^{8}=1$
$-\sum_{n \geq 1} \frac{n q^{n}}{1-q^{n}}-8 \sum_{n \geq 1} \frac{n q^{4 n}}{1-q^{4 n}}+6 \sum_{n \geq 1} \frac{n q^{2 n}}{1-q^{2 n}}=\sum_{n \geq 1} \frac{2 n q^{2 n}}{1+q^{2 n}}-\sum_{n \geq 1} \frac{(2 n-1) q^{2 n-1}}{1+q^{2 n-1}}$

kuing

将两层连成一层

kuing

回复 4# hbghlyj

其实我也好奇那堆空的表格到底起了什么作用？

青青子衿

\begin{align*}
\color{black}{\prod_{n=1}^{+\infty}\left[\left(1+q^{2 n-1}\right)^{8}\right]-\prod_{n=1}^{+\infty}\left[\left(1-q^{2 n-1}\right)^{8}\right]=16 q \prod_{n=1}^{+\infty}\left[\left(1+q^{2 n}\right)^{8}\right]}
\end{align*}
这个无穷乘积表达式好像和椭圆伊塔/西塔函数有关……

Czhang271828

回复 6# 青青子衿

找到一位玩这个的老哥...

www-elsa.physik.uni-bonn.de/~dieckman/InfProd … vingxThetaxFunctions

Czhang271828

hbghlyj 发表于 2022-1-5 13:04
For $|x|

An elementary proof without the knowlegde of $\eta$ -function, modular form, etc. Can be found in P155 in the book An Invitation to $q$ -Series. The book Eta Products and Theta Series Identities is one of my dictionaries in the study of modular form, where you can find the identity at page 152.

青青子衿

Czhang271828 发表于 2022-8-9 21:06
An elementary proof without the knowlegde of $\eta$ -function, modular form, etc. Can be found in  ...

\begin{align*}
\left.\partial_z\vartheta_1(z,q)\right|_{z=0}&=2 \sqrt[4]{q}\prod\limits_{i=1}^{+\infty\>\>}\left(1-q^{2 k}\right)^3\\
\vartheta_2(0,q)&=2 \sqrt[4]{q}\prod\limits_{i=1}^{+\infty\>\>}\left(1-q^{2 k}\right) \left(1+q^{2 k}\right)^2\\
\vartheta_3(0,q)&=\prod\limits_{i=1}^{+\infty\>\>}\left(1-q^{2 k}\right) \left(1+q^{2 k-1}\right)^2\\
\vartheta_4(0,q)&=\prod\limits_{i=1}^{+\infty\>\>}\left(1-q^{2 k}\right) \left(1-q^{2 k-1}\right)^2\\
\sqrt{k}&=\dfrac{\vartheta_2(0,q)}{\vartheta_3(0,q)}=2\sqrt[4]{q}\cdot\left[\dfrac{\prod\limits_{i=1}^{+\infty\>\>}(1+q^{2k})}{\prod\limits_{i=1}^{+\infty\>\>}(1+q^{2k-1})}\right]^2\\
\sqrt{k'}&=\dfrac{\vartheta_4(0,q)}{\vartheta_3(0,q)}=\left[\dfrac{\prod\limits_{i=1}^{+\infty\>\>}(1-q^{2k-1})}{\prod\limits_{i=1}^{+\infty\>\>}(1+q^{2k-1})}\right]^2\\
k^2+k'^2&=1
\end{align*}

1. 2 q^(1/4) Product[(1 - q^(2 k))^3, {k, 1, +Infinity}] /. q -> 0.3
2. EllipticThetaPrime[1, 0, 0.3]
3. 2 q^(1/4) Product[(1 - q^(2 k)) (1 + q^(2 k))^2, {k,
4.     1, +Infinity}] /. q -> 0.3
5. EllipticTheta[2, 0, 0.3]
6. Product[(1 - q^(2 k)) (1 + q^(2 k - 1))^2, {k, 1, +Infinity}] /.
7. q -> 0.3
8. EllipticTheta[3, 0, 0.3]
9. Product[(1 - q^(2 k)) (1 - q^(2 k - 1))^2, {k, 1, +Infinity}] /.
10. q -> 0.3
11. EllipticTheta[4, 0, 0.3]

Copy the Code

青青子衿

青青子衿 发表于 2022-1-7 10:35
\begin{align*}
\color{black}{\prod_{n=1}^{+\infty}\left[\left(1+q^{2 n-1}\right)^{8}\right]-\prod_{n=1}^{+\infty}\left[\left(1-q^{2 n-1}\right)^{8}\right]=16 q \prod_{n=1}^{+\infty}\left[\left(1+q^{2 n}\right)^{8}\right]}
\end{align*}
...

An Identity Relating a Theta Function to a Sum of Lambert Series
researchgate.net/publication/228887537_An_Ide … um_of_Lambert_Series