User:Nadifa's journey to mathematics

http://home.earthlink.net/~jsondow/

http://numbers.computation.free.fr/Constants/constants.html

Picture of Adam (me) http://en.wikipedia.org/wiki/File:Syndrome.jpg

http://www.google.co.uk/search?hl=en&gs_rn=9&gs_ri=psy-ab&pq=chines+baby&cp=12&gs_id=1c&xhr=t&q=blasian+baby&bav=on.2,or.r_qf.&biw=1280&bih=653&um=1&ie=UTF-8&tbm=isch&source=og&sa=N&tab=wi&ei=0R5rUbf7BtOz0QXMvoG4Dg#imgrc=k5iw4ueIOjj_yM%3A%3BcE_fgt6R2lyqAM%3Bhttp%253A%252F%252F2.bp.blogspot.com%252F_ySjExcU0kyU%252FSQz2MJ-zCnI%252FAAAAAAAAIxs%252F1Dmxio9DVDc%252FLOVE%252BMATH%252BBABY.jpg%3Bhttp%253A%252F%252Fajilbab.com%252Fmixed%252Fmixed-race-baby-boy-names.htm%3B485%3B323

This is me relaxing after teaching

http://www.nowloss.com/ways-to-reduce-stress.htm

http://www.youtube.com/user/bullcleo1?feature=watch

Proof the following,
Definition

$$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$$

$$\eta(s)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}$$

$$\lambda(s)=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^s}$$

$$E_{2n}$$ see

http://mathworld.wolfram.com/EisensteinSeries.html

(33)

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where m ≥ 2

$$\frac{(2\pi)^{\frac{3m-3}{2}}\Gamma(m)}{m^{\frac{2m+1}{2}}\Gamma{\left(\frac{1}{m}\right)} \Gamma{\left(\frac{2}{m}\right)}\cdots\Gamma{\left(\frac{m-1}{m}\right)}}= \prod_{k=1}^{\infty}\frac{mk}{mk-1}\cdot\frac{mk}{mk+1}\times\cdots\times{\frac{mk}{mk-(m-1)}\cdot\frac{mk}{mk+(m-1)}}$$

Let m = 2 and we have,

Wallis's formula

$$\frac{\pi}{2}=\prod_{k=1}^{\infty}\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}$$

More work on the above

$$\Gamma\left(\frac{1}{m}\right)\Gamma\left(\frac{2}{m}\right)\cdots\Gamma\left(\frac{m-1}{m}\right)= \frac{(2\pi)^{\frac{m-1}{2}}}{\sqrt{m}}$$

$$\frac{(2\pi)^{m-1}\Gamma(m)}{m^m}= \prod_{k=1}^{\infty}\frac{mk}{mk-1}\cdot\frac{mk}{mk+1}\times\cdots\times{\frac{mk}{mk-(m-1)}\cdot\frac{mk}{mk+(m-1)}}$$

$$\frac{(2\pi)^{m-1}\Gamma(m)}{m^m}=\Gamma\left(1-\frac{1}{m}\right)\Gamma\left(1+\frac{1}{m}\right) \Gamma\left(1-\frac{2}{m}\right)\Gamma\left(1+\frac{2}{m}\right)\cdots \Gamma\left(1-\frac{m-1}{m}\right)\Gamma\left(1+\frac{m-1}{m}\right)$$

(32)

$$2\pi^{2k-1}=\sum_{n=1}^{\infty}\frac{1}{n^{2k-1}}\left[A\coth\left(\frac{n\pi}{2}\right)- B\coth(n\pi)+C\coth(2n\pi)\right]$$

$$2\pi=\sum_{n=1}^{\infty}\frac{1}{n}\left[72\coth\left(\frac{n\pi}{2}\right)- 96\coth(n\pi)+24\coth(2n\pi)\right]$$

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$$\left(1+\frac{D-E+F}{2}\right)\zeta(2k+1)=\sum_{n=1}^{\infty}\frac{1}{n^{2k+1}}\left(D\coth\left(\frac{n\pi}{2}\right)- E\coth(n\pi)+F\coth(2n\pi)\right)$$

-- $$E\sum_{n=1}^{\infty}\frac{\coth(n\pi)}{n^{4k-1}}=F\sum_{n=1}^{\infty}\frac{\coth(2n\pi)}{n^{4k-1}}+ D\sum_{n=1}^{\infty}\frac{\coth(\frac{n\pi}{2})}{n^{4k-1}}$$

$$37\sum_{n=1}^{\infty}\frac{\coth(n\pi)}{n^3}=7\sum_{n=1}^{\infty}\frac{\coth(2n\pi)}{n^3}+ 28\sum_{n=1}^{\infty}\frac{\coth(\frac{n\pi}{2})}{n^3}$$

see http://www.plouffe.fr/simon/inspired22.html

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(31)

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$$\sum_{n=1}^{\infty}\left(\frac{x}{n}-\ln{\left(\frac{n+x}{n}\right)}\right)=x\gamma+\ln{x!}$$

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(30)

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$$-\frac{1}{2}(2\pi)^{4m-1}\sum_{j=0}^{2m}(-1)^j\frac{B_{2j}}{(2j)!}\frac{B_{4m-2j}}{(4m-2j)!}= \sum_{n=1}^{\infty}\frac{coth(n\pi)}{n^{4m-1}}$$

m =1, yield Ramanujan's equation

$$\frac{7\pi^3}{180}=\sum_{n=1}^{\infty}\frac{coth(n\pi)}{n^3}$$

$$2\pi+3\ln2=12\sum_{n=1}^{\infty}(-1)^{n-1}\frac{coth(n\pi)}{n}$$

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(29)

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$$5\ln2-\pi=8\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n\left(e^{n\pi}+1\right)}$$

(28)

Nice pattern!

Where $$A=2n-1$$

$$\frac{2A}{1+A^2}=\frac{2n-1}{(n-1)^2+n^2}$$

$$\frac{1+3A^2}{3A+A^3}=\frac{3n^2-3n+1}{(n-1)^3+n^3}$$

$$\frac{4A+4A^3}{1+6A^2+A^4}=\frac{4n^3-6n^2+4n-1}{(n-1)^4+n^4}$$

(27)

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$$P_n(Not-Primes)=\frac{(3n-1)(3n+2)-1}{3}=3n^2+n-1$$

for all $$n=3k+1$$, and where k = 1, 2, 3, 4, ...

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(26)

$$2artanh\left(\frac{x}{2a+x}\right)=\ln\left(\frac{a+x}{a}\right)$$

$$2artanh\left(\frac{x}{2a-x}\right)=\ln\left(\frac{a}{a-x}\right)$$

$$2artanh\left(\frac{x}{2a+x}\right)+2artanh\left(\frac{x}{2a-x}\right)=\ln\left(\frac{a+x}{a-x}\right)$$

$$2artanh\left(\frac{x}{2a+x}\right)-2artanh\left(\frac{x}{2a-x}\right)=\ln\left(\frac{a^2-x^2}{a^2}\right)$$

$$\ln2=2artanh\left(\frac{1}{3}\right)$$

$$\ln3=2artanh\left(\frac{1}{2}\right)$$

$$\frac{artanh\left(\frac{1}{\phi}\right)}{artanh\left(\frac{1}{\phi^3}\right)}=3$$

$$\ln3=\sum_{n=0}^{\infty}\frac{1}{(2n+1)2^{2n}}$$

$$2artanh\left(\frac{1}{2}\right)-2artanh\left(\frac{1}{5}\right)=\ln2$$

$$artanh\left(\frac{1}{2}\right)+artanh\left(\frac{1}{7}\right)=\ln2$$

(25)

$$\lim_{n\to \infty}n\left(a^{\frac{1}{n}}-b^{\frac{1}{n}}\right)=\ln\left(\frac{a}{b}\right)$$

$$\lim_{n\to \infty}n\left(a^{\frac{1}{n}}+b^{\frac{1}{n}}-2\right)=\ln{ab}$$

Trapezium rule (sort of)

$$\lim_{k,n \to \infty}n\left[1^{\frac{1}{n}}-2\left(2^{\frac{1}{n}}-3^{\frac{1}{n}}+4^{\frac{1}{n}}-5^{\frac{1}{n}}+\cdots+(2k)^{\frac{1}{n}}\right)+(2k+1)^{\frac{1}{n}}\right]=\ln\left(\frac{2}{\pi}\right)$$

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(24)

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where x ≥ 0 and k > 1

$$\prod_{n=0}^{\infty}\Gamma^{\frac{k-1}{k^{n+1}}}\left(n+x\right)=\Gamma(x)\prod_{n=1}^{\infty}\left(n+x\right)^{\frac{1}{k^n}}$$

and let $$x=\frac{1}{y}$$ and we have

y ≠ 0 and k > 1

$$\prod_{n=0}^{\infty}\Gamma^{\frac{k-1}{k^{n+1}}}\left(n+\frac{1}{y}\right)=\frac{1}{y^{\frac{1}{k-1}}}\Gamma\left(\frac{1}{y}\right)\prod_{n=1}^{\infty}(ny+1)^{\frac{1}{k^n}}$$

see http://mathworld.wolfram.com/GammaFunction.html

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(23)

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$$\prod_{k=2}^{\infty}\left(\prod_{n=0}^{\infty}(n!)^{\frac{k-1}{k^{n+1}}}\right)=\prod_{n=1}^{\infty}n^{\zeta(n)-1}=2.47795...$$

$$\prod_{k=2}^{\infty}\left(\prod_{n=0}^{\infty}(n!)^{\frac{k-1}{k^{n+1}}}\right)^{(-1)^{k-1}} =\prod_{n=1}^{\infty}n^{\eta(n)-1}$$

$$\prod_{k=2,4...}^{\infty}\left(\prod_{n=0}^{\infty}(n!)^{\frac{k-1}{k^{n+1}}}\right)=\prod_{n=1}^{\infty}n^{\frac{\zeta(n)}{2^n}}$$

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(22) --

$$E_{12}\left(e^{-\pi}\right)=\frac{3969}{15202}\left[J_{2}^{24}\left(e^{-\pi}\right)+ J_{3}^{24}\left(e^{-\pi}\right)+J_{4}^{24}\left(e^{-\pi}\right) \right]$$

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(21)

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

$$\left(\frac{3}{4}\cdot\frac{\pi^2}{\Gamma^8\left(\frac{3}{4}\right)}\right)^2=1+480\sum_{n=1}^{\infty}\frac{n^7}{e^{2n\pi}-1}$$

$$\frac{441}{691}\cdot\left(\frac{3}{4}\cdot\frac{\pi^2}{\Gamma^8\left(\frac{3}{4}\right)}\right)^3=1+\frac{65520}{691}\sum_{n=1}^{\infty}\frac{n^{11}}{e^{2n\pi}-1}$$

$$\frac{1617}{3617}\cdot\left(\frac{3}{4}\cdot\frac{\pi^2}{\Gamma^8\left(\frac{3}{4}\right)}\right)^4=1+\frac{16320}{3617}\sum_{n=1}^{\infty}\frac{n^{15}}{e^{2n\pi}-1}$$

see http://en.wikipedia.org/wiki/Eisenstein_series Identities involving Eisenstein series

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(20)

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$$\sum_{n=1}^{\infty}\frac{n^3}{e^{\frac{n\pi}{2^{2k+1}}}-1}= 2\sum_{n=1}^{\infty}\frac{n^7}{e^{\frac{n\pi}{2^{k}}}-1}$$

$$\sum_{n=1}^{\infty}\frac{n^3}{e^{2^{-k}n\pi}-1}=\sum_{n=0}^{k}16^{n-1}$$

$$\sum_{n=1}^{\infty}\frac{n^7}{e^{2^{-k}n\pi}-1}=\frac{1}{2}\sum_{n=0}^{2k+1}16^{n-1}$$

(19)

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

$$\left(\frac{33}{4}\cdot\frac{\pi^2}{\Gamma^8\left(\frac{3}{4}\right)}\right)^2=1+480\sum_{n=1}^{\infty}\frac{n^7}{e^{n\pi}-1}$$

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

$$\left(\frac{3}{4}\cdot\frac{\pi^2}{\Gamma^8\left(\frac{3}{4}\right)}\right)^2=1+480\sum_{n=1}^{\infty}\frac{n^7}{e^{2n\pi}-1}$$

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

$$\left(\frac{33}{4^3}\cdot\frac{\pi^2}{\Gamma^8\left(\frac{3}{4}\right)}\right)^2=1+480\sum_{n=1}^{\infty}\frac{n^7}{e^{4n\pi}-1}$$

(18)

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Where $$T_n$$ is triangular numbers, 1,3,6,10,... and $$Fn$$ is the Fibonacci numbers, 1,1,2,3,5,...

$$\sum_{k=1}^{2n-1}T_{n+1-k}F^2_k=F_{2n}F_{2n+1}-n$$

$$\sum_{k=1}^{2n}T_{n+1-k}F^2_k=F_{2n+1}F_{2n+2}-n-1$$

$$\sum_{k=1}^{n}(n+1-k)^2F^2_k=F_{n+1}(F_n+F_{n+2})-n-1$$

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(17)

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$$\ln\left(\frac{4}{\pi}\right)^k+k\gamma+\ln{k!}=\sum_{n=1}^{\infty}\left(\frac{k\lambda(2n)}{n}-\ln\frac{n+k}{n}\right)$$

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(16) -

Glaisher-Kinkelin constant A = 1.28242712...

$$\eta(s)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^s}$$

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

$$\ln\left(\frac{2^{\frac{1}{6}}\sqrt{\pi}e}{A^6}\right)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n+1}\eta(n)$$

$$\ln\left(\frac{A^6\sqrt{\pi}}{2^{\frac{7}{6}}e}\right)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\eta(n)$$

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(15)

$$\phi=\frac{1+\sqrt{5}}{2}$$

$$\sum_{i=1}^{n}\phi^{2n+1-i}F_i-\sum_{i=1}^{n}\phi^{n+1-i}F_{n+i}=F_n^2$$

$$\sum_{n=1}^{2m+1}(-1)^{n+1}\phi^{(-1)^{n+1}n}F_n={\phi}F_{2m+1}F_{2m+2}$$

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(14)

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$$\frac{1}{12}\approx\lim_{n \to \infty}\frac{n!-\sqrt{2n\pi}\left(\frac{n}{e}\right)^n}{(n-1)!}$$

(13)

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Where $$\gamma$$ is Euler's constant

$$\gamma=\lim_{n \to \infty}-\frac{n}{\alpha}\tanh^{-1}\left[\frac{1-\frac{n}{\Gamma\left(\frac{1}{n}\right)}} {1-\frac{n}{\Gamma\left(\frac{1}{n}\right)}+\frac{1}{\alpha}}\right] $$

$$\gamma=\lim_{n \to \infty}\frac{\sum_{i=1}^{k}\left[n^i-\Gamma^i\left(\frac{1}{n}\right)\right]} {\sum_{i=1}^{k}in^{i-1}}$$

set $$i=1$$

see http://mathworld.wolfram.com/Euler-MascheroniConstant.html on (51)

$$\gamma=\lim_{n \to \infty}\frac{n^ke^{\frac{\alpha\gamma}{n^k}}-e^{\frac{\beta\gamma}{n^k}}\Gamma^{k}\left(\frac{1}{n}\right)}{\alpha-\beta+kn^{k-1}}$$

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(12)

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Lucas sequence

$$L_1=1, L_2=3, L_n+L_{n+1}=L_{n+2}$$

$$\phi=\frac{\sqrt{5}+1}{2}$$

$$\frac{1}{L_{2n}-2}=\sum_{m=1}^{\infty}m\phi^{-2nm}$$

(1)

$$3-\zeta(2)=\sum_{m=3}^{\infty}m\left[\zeta(m)-1\right]$$

(2)

$$3-\zeta(2)-\frac{7}{4}=\sum_{m=3}^{\infty}(-1)^{m-1}m\left[\zeta(m)-1\right]$$

(11)

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$$\sqrt{x}J_3\left(0,e^{-x\pi}\right)=1+2\sum_{n=1}^{\infty}e^{\frac{-n^2\pi}{x}}$$

(10)

Where $$J_n\left(0,q\right)$$ is Jacobi theta function

$$J_2\left(e^{\frac{-\pi}{n}}\right)=f(n)\frac{\pi^{\frac{1}{4}}}{\Gamma\left(\frac{3}{4}\right)}\approx\sqrt{n}$$

$$J_3\left(e^{\frac{-\pi}{n}}\right)=g(n)\frac{\pi^{\frac{1}{4}}}{\Gamma\left(\frac{3}{4}\right)}\approx\sqrt{n}$$ ;

$$J_4\left(e^{\frac{-\pi}{n}}\right)=h(n)\frac{\pi^{\frac{1}{4}}}{\Gamma\left(\frac{3}{4}\right)}\approx{0}$$

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

$$\sqrt{\pi}=\lim_{k \to \infty}\frac{2}{e^{\frac{1}{4k}}\sqrt{k}}\sum_{n=0}^{\infty}e^{\frac{-n(n+1)}{k}}$$

$$\sqrt{\pi}=\lim_{k \to \infty}\frac{12}{\sqrt{k}}\sum_{n=1}^{\infty}\frac{n}{e^{\frac{n}{k}}+1}$$

Comments: k decimal digits of $$\pi$$, e.g

when k = 8, the sum will gives 8 decimal digits of pi

$$\sqrt{\pi}=\lim_{k \to \infty}\frac{4}{\sqrt{k}}\sum_{n=1,3,5,...}^{\infty}e^{\frac{-n^2}{k}}$$

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(9)

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Where $$\sigma=1.66168794\cdots$$, is Somos's constant

By definition $$(-1)!!=1$$

$$\sigma=\prod_{n=0}^{\infty}(n!)^{\frac{1}{2^{n+1}}}$$

$$2\sigma=\prod_{n=0}^{\infty}\left(\frac{(2n)!}{2n+1}\right)^{\frac{1}{2^{n+1}}}= \prod_{n=0}^{\infty}\left(\frac{(2n)!}{(2n-1)!!}\right)^{\frac{1}{2^{n+1}}}$$

(8)

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$$\sum_{k=1}^{n}\sin^{2p}\left(\frac{k\pi}{2n+1}\right)=\frac{{2p \choose p}(2n+1)}{2^{2p+1}}$$

$$\sum_{k=1}^{n}\cos^{2p}\left(\frac{k\pi}{2n+1}\right)=\frac{{2p \choose p}(2n-1)}{2^{2p+1}}$$

(7)

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Where m and k are integers >=1 and $$B_m$$ are Bernoulli numbers

Here are a few Bernoulli numbers

$$B_2=-\frac{1}{30}, B_3=\frac{1}{42}, B_4=-\frac{1}{30}$$

$$\frac{B_{m+1}}{4m+4}\left[1+(-1)^m2^{2(m+1)(k+1)}\right]=\sum_{n=1}^{\infty}\frac{n^{2m+1}}{e^{2^{-k}n\pi}-1}$$

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(6)

$$\sum_{n=1}^{\infty}\frac{1}{F_n}=\frac{\sqrt{5}}{1\sqrt{5}+\frac{1^2\cdot5}{-2\sqrt{5}+\frac{1^2\cdot5}{3\sqrt{5}+\frac{2^2\cdot5}{-5\sqrt{5}+\frac{3^2\cdot5}{8\sqrt{5}+\frac{5^2\cdot5}{-13\sqrt{5}+\cdots}}}}}}$$

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(5)

Where $$\phi=\frac{1+\sqrt{5}}{2}$$

$$1+\sqrt{\phi}=\frac{1}{\sqrt{\phi^3}-\phi}$$

$$\frac{1}{\sqrt{2\phi^5}-\phi^3}=1+\sqrt{\sqrt{5}-1}$$

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(4)

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These two formulae are from Sondow website (Guillera and Sondow)

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

and

$$e^{\gamma}=\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n}\left(\frac{k+2}{k+1}\right)^{(-1)^k}\right)^{\frac{1}{n+2}}$$

I slightly rewrite it as Wallis format. From there I noticed the following pattern and I can't proof it(don't have the knowledge to proof it)

This is very similar to the formulas on the connection between Wallis and Vieta by Thomas Olser paper

$$\frac{\pi}{2}=\prod_{k=1}^{m}\left(\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}\right)

\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n}\left(\frac{k+2m+2}{k+2m+1}\right)^{(-1)^k}\right)^{\frac{1}{2^{n+1}}}$$

$$m=0$$

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

$$m=1$$

$$\frac{\pi}{2}=\frac{2}{1}\cdot\frac{2}{3}\cdot\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n}\left(\frac{k+4}{k+3}\right)^{(-1)^k}\right)^{\frac{1}{2^{n+1}}}$$

$$\frac{\pi}{2}=\left(\frac{2}{1}\cdot\frac{2}{3}\right)\left(\frac{4}{3}\right)^{\frac{1}{2}}\left(\frac{4}{3}\cdot\frac{4}{5}\right)^{\frac{1}{4}}\left(\frac{4}{3}\cdot\frac{4^2}{5^2}\cdot\frac{6}{5}\right)^{\frac{1}{8}}\left(\frac{4}{3}\cdot\frac{4^3}{5^3}\cdot\frac{6^3}{5^3}\cdot\frac{6}{7}\right)^{\frac{1}{16}}\cdots $$

$$m=2$$

$$\frac{\pi}{2}=\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n}\left(\frac{k+6}{k+5}\right)^{(-1)^k}\right)^{\frac{1}{2^{n+1}}}$$

$$\frac{\pi}{2}=\left(\frac{2}{1}\cdot\frac{2}{3}\right)\cdot\left(\frac{4}{3}\cdot\frac{4}{5}\right) \left(\frac{6}{5}\right)^{\frac{1}{2}}\left(\frac{6}{5}\cdot\frac{6}{7}\right)^{\frac{1}{4}}\left(\frac{6}{5}\cdot\frac{6^2}{7^2}\cdot\frac{8}{7}\right)^{\frac{1}{8}}\left(\frac{6}{5}\cdot\frac{6^3}{7^3}\cdot\frac{8^3}{7^3}\cdot\frac{8}{9}\right)^{\frac{1}{16}}\cdots $$

Other infinite products
that exhibit this kind of connection From sondow and Guillera (papers 2005)

$$\left[\frac{\Gamma\left(\frac{1}{4}\right)}{2\Gamma\left(\frac{3}{4}\right)}\right]^2=\left(\frac{3}{1}\right)^{\frac{1}{2}}\left(\frac{3}{1}\cdot\frac{3}{5}\right)^{\frac{1}{4}}\left(\frac{3}{1}\cdot\frac{3^2}{5^2}\cdot\frac{7}{5}\right)^{\frac{1}{8}}\left(\frac{3}{1}\cdot\frac{3^3}{5^3}\cdot\frac{7^3}{5^3}\cdot\frac{7}{9}\right)^{\frac{1}{16}}\cdots $$

Examples of relation

$$\left[\frac{\Gamma\left(\frac{1}{4}\right)}{2\Gamma\left(\frac{3}{4}\right)}\right]^2=\left(\frac{3}{1}\cdot\frac{3}{5}\right)\left(\frac{7}{5}\right)^{\frac{1}{2}}\left(\frac{7}{5}\cdot\frac{7}{9}\right)^{\frac{1}{4}}\left(\frac{7}{5}\cdot\frac{7^2}{9^2}\cdot\frac{11}{9}\right)^{\frac{1}{8}}\left(\frac{7}{5}\cdot\frac{7^3}{9^3}\cdot\frac{11^3}{9^3}\cdot\frac{11}{13}\right)^{\frac{1}{16}}\cdots $$

$$\left[\frac{\Gamma\left(\frac{1}{4}\right)}{2\Gamma\left(\frac{3}{4}\right)}\right]^2=\left(\frac{3}{1}\cdot\frac{3}{5}\right)\cdot\left(\frac{7}{5}\cdot\frac{7}{9}\right) \left(\frac{11}{9}\right)^{\frac{1}{2}}\left(\frac{11}{9}\cdot\frac{11}{13}\right)^{\frac{1}{4}}\left(\frac{11}{9}\cdot\frac{11^2}{13^2}\cdot\frac{15}{13}\right)^{\frac{1}{8}}\left(\frac{11}{9}\cdot\frac{11^3}{13^3}\cdot\frac{15^3}{13^3}\cdot\frac{15}{17}\right)^{\frac{1}{16}}\cdots $$

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where $$H_{x}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{x}$$

$$H_0=0$$

$$e^{\gamma}=\frac{e^{H_{x-1}}}{x}

\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n}\left(\frac{k+x+1}{k+x}\right)^{(-1)^k}\right)^{\frac{1}{n+2}}$$

I rewrite it as $$\frac{x}{e^{\psi_0(x)}} =

\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n}\left(\frac{k+x+1}{k+x}\right)^{(-1)^k}\right)^{\frac{1}{n+2}}$$

and where $$\psi_0$$ is Digamma function (see wolfram site)

(3)

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$$\lim_{n,m \to \infty}\left(\frac{\pi}{2}\right)^2\left[(1+n)^2+(1+m)^2-(n-m)(n-m+1)\right]=Z$$ $$Z=(m+1)^2\prod_{k=1}^{n}\left(\frac{2k}{2k-1}\right)^2\prod_{k=1}^{m}\left(\frac{2k}{2k+1}\right)^2+ (n+1)^2\prod_{k=1}^{n}\left(\frac{2k}{2k+1}\right)^2\prod_{k=1}^{m}\left(\frac{2k}{2k-1}\right)^2 $$

let n = m

and we have Wallis formula

$$\frac{\pi}{2}=\prod_{k=1}^{\infty}\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}$$

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(2)

$$\sqrt{\frac{1}{4}+\frac{\phi}{2}}=\sqrt{\frac{2+\sqrt{5}}{2+\sqrt{2+\sqrt{5}}}}\cdot \sqrt{\frac{2+\sqrt{2+\sqrt{5}}}{2+\sqrt{2+\sqrt{2+\sqrt{5}}}}}\cdots$$

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(1)

Let $$\alpha=\frac{2\pi}{7}$$

$$\left(cos2\alpha\right)^{2n-1}\left(\frac{cos\alpha}{cos4\alpha}\right)^{\frac{1}{3}}+ \left(cos4\alpha\right)^{2n-1}\left(\frac{cos2\alpha}{cos\alpha}\right)^{\frac{1}{3}}+ \left(cos\alpha\right)^{2n-1}\left(\frac{cos4\alpha}{cos2\alpha}\right)^{\frac{1}{3}}= -\frac{{2n \choose n}\sqrt[3]{49}}{(n+1)2^{2n-1}}$$