User:Mathsformulae314

$$\left[2J_1{'}(0,q^{-\frac{1}{2}}){\cdot}{J_1{'}(0,q^{-1})}\right]^{\frac{1}{6}}=\left[J_2(0,q^{-\frac{1}{2}}) {\cdot}{J_4(0,q^{-1})}\right]^{\frac{1}{2}}$$

$$2^5J_1^{'}(0,q^{-1})=q^{\frac{1}{2}}J_4^3(0,q^{-1})J_2^6(0,q^{-\frac{1}{2}})?$$

$$\psi(q^{-n})=\prod_{k=1}^{\infty}\left(\frac{q^{kn}}{q^{kn}-1}\right)^{(-1)^{k+1}}$$

$$ln\left[\psi(q^{-1})\right]=\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{1}{q^k+1}\right)$$

$$ln\left[\frac{R(q)}{q^{\frac{1}{5}}}\right]=\sum_{n=1}^{\infty}\frac{q^n-q^{2n}-q^{3n}+q^{4n}}{k(q^{5n}-1)}$$

$$arctan\left(\frac{2a^3+2a^2-4}{a^4-2a^2-4a}\right)=2arctan\left(\frac{1}{a-1}\right)$$

$$arctan\left(\frac{2a^3-2a^2+4}{a^4-2a^2+4a}\right)=2arctan\left(\frac{1}{a+1}\right)$$

$$arctan\left(\frac{2a}{a^2-2}\right)=\sum_{n=0}^{\infty}\frac{(-1)^n2^{2n+1}}{a^{4n+3}}\left(\frac{a^2}{4n+1}+\frac{2}{4n+3}\right)$$

$$\arctan\left(\frac{1}{a}\right)=\sum_{n=0}^{\infty}\frac{(-1)^n}{a^{2n+1}(2n+1)}$$

$$\frac{4}{\phi+2}=\prod_{n=1}^{\infty}(1+\frac{1}{25n^2-1})^{(-1)^{n+1}}\cdot{\sum_{n=0}^{\infty}\frac{(n!)^2}{\phi^{2n}(2n+1)!}}$$

Alternating symmetric formula
$$\frac{2^3}{\pi}=\frac{\prod_{n=1}^{\infty}\left(1+\frac{1}{4n^2-1}\right)^{(-1)^{n+1}}} {\sum_{n=1}^{\infty}\frac{1}{4n^2-1}}$$

$$ln\left[\frac{\pi}{asin\left(\frac{\pi}{a}\right)}\right]=\sum_{n=1}^{\infty}\frac{\zeta(2n)}{a^{2n}\cdot{n}}$$

$$ln\left[\frac{2sin\left(\frac{\pi}{2a}\right)}{sin\left(\frac{\pi}{a}\right)}\right]=\sum_{n=1}^{\infty}\frac{\lambda(2n)}{a^{2n}\cdot{n}}$$

$$ln\left[\frac{4asin^2\left(\frac{\pi}{2a}\right)}{{\pi}sin\left(\frac{\pi}{a}\right)}\right]=\sum_{n=1}^{\infty}\frac{\eta(2n)}{a^{2n}\cdot{n}}$$

$$\frac{2sin\left(\frac{\pi}{2a}\right)}{sin\left(\frac{\pi}{a}\right)}=\prod_{n=0}^{\infty}\left[1+\frac{1}{\left[a(2n+1)\right]^2-1}\right]$$

$$\frac{4asin^2\left(\frac{\pi}{2a}\right)}{{\pi}sin\left(\frac{\pi}{a}\right)}=\prod_{n=1}^{\infty}\left[1-\frac{1}{(an)^2-1}\right]^{(-1)^{n+1}}$$

$$\frac{2sin\left(\frac{\pi}{2a}\right)}{tan\left(\frac{\pi}{a}\right)}=\frac{a^2-2^2}{a^2-1}\cdot\frac{(3a)^2-2^2}{(3a)^2-1}\cdot\frac{(5a)^2-2^2}{(5a)^2-1}\cdots$$

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

$$\frac{\pi}{\Gamma^4\left(\frac{3}{4}\right)}=1+8 \sum_{n=1}^{\infty}\frac{n}{e^{n\pi}+(-1)^n}\cdots(12)$$

$$ln2=24\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(e^{n\pi}-1)}-8\sum_{n=1}^{\infty}\frac{1}{n(e^{n\pi}-1)}$$

$$2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(e^{kn\pi}-1)}=\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{e^{kn\pi}+1}+\frac{1}{e^{kn\pi}-1}\right)$$

$$\pi=60\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(e^{n\pi}-1)}+24\sum_{n=1}^{\infty} \frac{(-1)^n}{n(e^{2n\pi}-1)}+6\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{1} {e^{n\pi}-1}+\frac{1}{e^{n\pi}+1}\right)\cdots(1)$$

$$\pi+ln\left(\frac{1}{2^3}\right)=24\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(e^{n\pi}-1)}\cdots(1)$$

$$2\pi+ln\left(\frac{1}{2^9}\right)=24\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(e^{2n\pi}-1)}\cdots(3)$$

$$ln36+\pi=8\sum_{n=0}^{\infty}\frac{1}{4n+1}\left(\frac{1} {2^{4n+1}}+\frac{1}{3^{4n+1}}\right)\cdots(3)$$

$$\frac{\pi}{2}=arctan(2)+\sum_{k=0}^{\infty} \frac{(-1)^k\cdot2^{2k+1}\left[\lambda(4k+2)-1\right]}{2k+1}\cdots(7)$$

$$\frac{\pi}{4}=\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1} \sum_{r=1}^{\infty}\frac{1}{(r^2+r+1)^{2k+1}}\cdots(6)$$

ln 0f 3
$$ln3=1+\frac{1}{4^2}+\frac{1}{7^2}+\frac{1}{10^2}\cdots(1)$$