ラマヌジャン

http://www.nn.iij4u.or.jp/~hsat/misc/math/ramanujan.html

<文字等式>
$\left(6a^2 - 4ab + 4b^2\right)^3 + \left(3b^2 + 5ab - 5a^2\right)^3=\left(6b^2 - 4ab + 4a^2\right)^3 + \left(3a^2 + 5ab - 5b^2\right)^3$

<数等式>
$\left(2^\frac{1}{3} -1 \right)^{\frac{1}{3}} = \left({\frac{1}{9}}\right)^{\frac{1}{3}} -\left({\frac{2}{9}}\right)^{\frac{1}{3}} +\left({\frac{4}{9}}\right)^{\frac{1}{3}}$

$\left(4^{\frac{1}{5}} +1\right)^{\frac{1}{2}} = {{16^{\frac{1}{5}}+8^{\frac{1}{5}}+2^{\frac{1}{5}}-1} \over {5^{\frac{1}{2}}}$

$\left( \frac{3+2\times 5^{\frac{1}{4}}}{3-2\times 5^{\frac{1}{4}}} \right)^{\frac{1}{4}} = \frac{5^{\frac{1}{4}}+1}{5^{\frac{1}{4}}-1}$

$\left(5^{\frac{1}{3}}-4^{\frac{1}{3}}\right)^{\frac{1}{2}} = \frac{2^{\frac{1}{3}}+20^{\frac{1}{3}}-25^{\frac{1}{3}}}{3}$

$12^3 + 1^3 = 10^3 + 9^3 =1729$

$174^3 + 133^3 = 45^3 + 14^6=7620661$

<級数>
$\sum_{i=0}^{\infty}{i \over {e^{2i\pi}-1}} = {\frac{1}{24}} - {\frac{1}{8\pi}}$

$\sum_{i=0}^{\infty} {{i^{1 \over 3}} \over {e^{2i\pi}-1}} = \frac{1}{24}$

$\sum_{n=o}^{\infty}(-1)^n = \frac{1^3+1^2}{0!}-\frac{3^3+3^2}{1!}+\frac{5^3+5^2}{2!}-...=0$

<積分>
$\phi (t) = \int_0^{\infty}{{\cos tx} \over {\exp (2\pi x^{\frac{1}{2}})-1}}dx \left,(but \exp (x) = e^x\right)$ と置くとき
$\phi \left(\pi \right) = {2-2^{\frac{1}{2}} \over 8}, \phi \left( {{2\pi} \over 3} \right) = \frac{1}{3}-3^{\frac{1}{2}} \left( \frac{3}{16}-\frac{1}{8\pi} \right), \phi \left( {\pi \over 5} \right) = \frac{6+5^{\frac{1}{2}}}{4}-\frac{5\times 10^{\frac{1}{2}}}{8}$

<無限積>
$\prod_{i=0}^{\infty} \left\{1+\exp \left( -\left(2i+1\right)\pi\times{1353}^{\frac{1}{2}} \right) \right\}$

$= 2^{1 \over 12}\exp\left(-\pi\times \frac{1353^{1 \over 2}}{24}\right)\left(\left( \frac{569+99\times 33^{1 \over 2}}{8}\right)^{1 \over 2} + \left(\frac{561+99\times 33^{1 \over 2}}{8}\right)^{1 \over 2}\right)^{1 \over 2}$
$\times \left( \left( \frac{25+3\times 33^{1 \over 2}}{8} \right)^{1 \over 2} + \left(\frac{17+3\times 33^{1 \over 2}}{8}\right)^{1 \over 2}\right)^{1 \over 2} \left(123+11^{1\over 2}\right)^{1 \over 4} \left(10+3\times 11^{1 \over 2}\right)^{1 \over 8}\left(26+15\times 3^{1 \over 2}\right)^{1 \over 8}$

Rogers-Ramanujan の恒等式[Rogers 1894, Ramanujan 1913]
1 +∑n= 1∞(xn*n / Πi = 1n(1 - xi)) = 1/Πn = 0∞*1,
1 +∑n= 1∞(xn(n+1) / Πi = 1n(1 - xi)) = 1/Πn = 0∞*2.

<円周率>
$\frac{1}{\pi}=\frac{2\sqrt{2}}{99^2} \sum_{n=0}^\infty \frac{(4n)!(1103+26390n)}{(4^n99^nn!)^4}$