User:Van bui 1

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Derived formulae
Total : 5

$$\frac{J_2\left(0,e^{-\pi\sqrt{n}}\right){\cdot}J_3^7\left(0,e^{-\pi\sqrt{n}}\right)} {J_4^2\left(0,e^{-\pi\sqrt{n}}\right)}=\frac{2^4{\cdot}G_n^{12}{\cdot\psi^6\left(e^{-\pi\sqrt{n}}\right)}} {e^{\frac{3\pi\sqrt{n}}{4}}}\cdots(1)$$

$$J_3\left(0,q^{-1}\right)=\prod_{n=1}^{\infty}\left(\frac{q^n+1}{q^n-1}\right)^{(-1)^{n+1}} \cdots(2)$$

$$J_4^2\left(0,q^{-1}\right)=2\prod_{n=1}^{\infty}\left(\frac{q^{2n}+1}{q^{2n}-1}\right)^{2(-1)^{n+1}} -\prod_{n=1}^{\infty}\left(\frac{q^n+1}{q^n-1}\right)^{2(-1)^{n+1}}\cdots(3)$$

$$J_3\left(0,q^{-1}\right){\cdot}J_4\left(0,q^{-1}\right)= \prod_{n=1}^{\infty}\left(\frac{q^{2n}+1}{q^{2n}-1}\right)^2\cdots(4)$$

$$\prod_{n=1}^{\infty}\left(\frac{q^{2n-1}-1}{q^{2n-1}+1}\right)^2\left(\frac{q^{2n}-1}{q^{2n}+1} \right)^2=2\prod_{n=1}^{\infty}\left(\frac{q^{2n}+1}{q^{2n}-1}\right)^{2(-1)^{n+1}} -\prod_{n=1}^{\infty}\left(\frac{q^n+1}{q^n-1}\right)^{2(-1)^{n+1}}\cdots(5)$$

$$\prod_{k=1}^{\infty}\left(1+\frac{1}{q^{2k-1}}\right)=\frac{q}{q-1}{\cdot}\frac{q^2}{q^2+1}{\cdot}\frac{q^3}{q^3-1} {\cdot}\frac{q^4}{q^4+1}{\cdot}=\prod_{k=1}^{\infty}\left(\frac{q^k}{q^k+(-1)^k}\right)\cdots(6)$$

Jacobi theta function
Total : 5

$$J_3\left(0,e^{-4\pi}\right)=\frac{\pi^{\frac{1}{4}}}{\Gamma\left(\frac{3}{4}\right)} \left[\frac{1}{2^{\frac{5}{4}}}+\frac{1}{2^{\frac{9}{4}}}\left(2+\frac{3}{\sqrt{2}} \right)^{\frac{1}{2}}\right]^{\frac{1}{2}}\cdots(1)$$

$$J_3\left(0,e^{\frac{-\pi}{2}}\right){\cdot}J_4\left(0,e^{-\frac{\pi}{2}}\right)= \frac{\pi^{\frac{1}{4}}}{2^{\frac{1}{2}}{\cdot}\Gamma^2\left(\frac{3}{4}\right)}\cdots(2)$$

$$\psi\left(e^{-\frac{\pi}{2}}\right)=\frac{\pi^{\frac{1}{4}}{\cdot}e^{\frac{\pi}{16}}}{\Gamma\left(\frac{3}{4}\right)} \left(\frac{1}{2^{\frac{7}{4}}}+\frac{1}{2^{\frac{5}{4}}}\right)^{\frac{1}{4}}\cdots(3)$$

$$\psi\left(e^{-\frac{\pi}{4}}\right)=\frac{\pi^{\frac{1}{4}}{\cdot}e^{\frac{\pi}{32}}}{\Gamma\left(\frac{3}{4}\right)} \left(1+\frac{1}{2^{\frac{1}{4}}}\right)^{\frac{1}{2}}\left(\frac{1}{2^{\frac{7}{4}}}+\frac{1}{2^{\frac{5}{4}}}\right)^{\frac{1}{8}}\cdots(4)$$

$$J_2\left(0,e^{-\frac{\pi}{4}}\right)=\frac{\pi^{\frac{1}{4}}}{\Gamma\left(\frac{3}{4}\right)} \left(2^{\frac{11}{4}}+2^{\frac{9}{4}}\right)^{\frac{1}{4}}\cdots(5)$$

Infinite product
Total : 1

$$\frac{\sqrt{2}}{e^{\frac{\pi}{6}}{\cdot}\left(4+3\sqrt{2}-8\right)^{\frac{1}{8}}}= \prod_{n=1}^{\infty}\left[1-\frac{1}{e^{4\pi(2n-1)}}\right]\cdots(1)$$

Infinite sum
Total : 13

$$256=1+240\sum_{n=1}^{\infty}\frac{n^3}{e^{\frac{n\pi}{2}}-1}\cdots(1)$$

$$\frac{33\pi^2}{4\Gamma^8\left(\frac{3}{4}\right)}= 1+240\sum_{n=1}^{\infty}\frac{n^3}{e^{n\pi}-1}\cdots(2)$$

$$\frac{1089\pi^4}{16\Gamma^{16}\left(\frac{3}{4}\right)}= 1+480\sum_{n=1}^{\infty}\frac{n^7}{e^{n\pi}-1}\cdots(3)$$

$$\frac{1}{24}=\sum_{n=1}^{\infty}\frac{n^3}{e^{n\pi}-1}- 11\sum_{n=1}^{\infty}\frac{n^3}{e^{2n\pi}-1}\cdots(4)$$

$$\frac{1}{16}=\sum_{n=1}^{\infty}\frac{n^3}{e^{n\pi}-1}- 16\sum_{n=1}^{\infty}\frac{n^3}{e^{4n\pi}-1}\cdots(5)$$

$$\frac{1}{4}=\sum_{n=1}^{\infty}\frac{n^7}{e^{n\pi}-1}- 11^2\sum_{n=1}^{\infty}\frac{n^7}{e^{2n\pi}-1}\cdots(6)$$

$$\frac{17}{32}=\sum_{n=1}^{\infty}\frac{n^7}{e^{n\pi}-1}- 16^2\sum_{n=1}^{\infty}\frac{n^7}{e^{4n\pi}-1}\cdots(7)$$

$$\frac{\pi^2}{2^5\cdot\Gamma^8\left(\frac{3}{4}\right)}=\sum_{n=1}^{\infty}\frac{n^3}{e^{n\pi}-1}- \sum_{n=1}^{\infty}\frac{n^3}{e^{2n\pi}-1}\cdots(8)$$

$$\frac{33\cdot\pi^2}{2^{10}\cdot\Gamma^8\left(\frac{3}{4}\right)}=\sum_{n=1}^{\infty}\frac{n^3}{e^{n\pi}-1}- \sum_{n=1}^{\infty}\frac{n^3}{e^{4n\pi}-1}\cdots(9)$$

$$\frac{\pi^2}{2^{10}\cdot\Gamma^8\left(\frac{3}{4}\right)}=\sum_{n=1}^{\infty}\frac{n^3}{e^{2n\pi}-1}- \sum_{n=1}^{\infty}\frac{n^3}{e^{4n\pi}-1}\cdots(10)$$

$$\frac{3^2\cdot\pi^4}{2^{6}\cdot\Gamma^{16}\left(\frac{3}{4}\right)}= \sum_{n=1}^{\infty}\frac{n^7}{e^{n\pi}-1}- \sum_{n=1}^{\infty}\frac{n^7}{e^{2n\pi}-1}\cdots(11)$$

$$\frac{17\cdot{33^2}\cdot\pi^4}{2^{17}\cdot\Gamma^{16}\left(\frac{3}{4}\right)}= \sum_{n=1}^{\infty}\frac{n^7}{e^{n\pi}-1}- \sum_{n=1}^{\infty}\frac{n^7}{e^{4n\pi}-1}\cdots(12)$$

$$\frac{3^4\cdot\pi^4}{2^{17}\cdot\Gamma^{16}\left(\frac{3}{4}\right)}= \sum_{n=1}^{\infty}\frac{n^7}{e^{2n\pi}-1}- \sum_{n=1}^{\infty}\frac{n^7}{e^{4n\pi}-1}\cdots(13)$$