Forgot password?
 Create new account
View 104|Reply 0

q级数的相关的四次恒等式

[Copy link]

462

Threads

969

Posts

9934

Credits

Credits
9934

Show all posts

青青子衿 Posted at 2025-2-19 18:13:34 |Read mode

\begin{align*}
x&= -q^{25/56}\prod_{n=1}^\infty(1-q^{7n-1})(1-q^{7n-6})(1-q^{7n})\\
&=\sum_{n\in{\mathbb{Z}}}(-1)^nq^{(14n^2+5)^2/56}\\
y&=\; q^{9/56}\;\prod_{n=1}^\infty(1-q^{7n-2})(1-q^{7n-5})(1-q^{7n})\\
&=\sum_{n\in{\mathbb{Z}}}(-1)^nq^{(14n^2+3)^2/56}\\
z&=  q^{1/56}\,\prod_{n=1}^\infty (1-q^{7n-3})(1-q^{7n-4})(1-q^{7n})\\
&=\sum_{n\in{\mathbb{Z}}}(-1)^nq^{(14n^2+1)^2/56}\\
\end{align*}

\begin{align*}
x^3y+y^3z+z^3x = 0
\end{align*}


Ramanujan Modular Forms and the Klein Quartic
researchgate.net/publication/239931633
Modular relations for the nonic analogues of
the Rogers–Ramanujan functions with
applications to partitions
core.ac.uk/download/pdf/82473664.pdf

手机版Mobile version|Leisure Math Forum

2025-4-20 12:09 GMT+8

Powered by Discuz!

× Quick Reply To Top Return to the list