Psi > Pi

by holomorph, Jun 17, 2006, 8:21 am

$\displaystyle \psi > \pi$

test no. 2

by holomorph, Jun 17, 2006, 8:21 am

[tex]\displaystyle \zeta(s)=\frac{\mathit{\Gamma}(1-s)}{2\pi i}\oint_{C}\frac{u^{s-1}}{e^{-u}-1}\,du [/tex]

[tex]
\displaystyle\frac{1}{n}\sum_{k=1}^{n}d(n)=\log n + 2\gamma - 1 + O(n^{-\tfrac{1}{2}})
[/tex]

test 1

by holomorph, Jun 17, 2006, 8:21 am

[tex]\displaystyle 1 - \int_1^{\infty}\frac{x-\lfloor x \rfloor}{x^2}\,dx=\gamma := \lim_{n\to\infty}\left(\, \sum_{k=1}^n\frac{1}{n}-\log n \right)[/tex]

[tex]\displaystyle\int\limits_{-\infty}^{+\infty}}e^{-x^2}\,dx=\sqrt{\pi}[/tex]

works :ninja:

Hmm

by holomorph, Jun 17, 2006, 8:21 am

$\displaystyle \psi(x) = x - \sum_{\rho} \frac{x^{\rho}}{\rho} - \log(2\pi) - \tfrac{1}{2}\log(1-x^{-2})$

$\displaystyle L(n) \geqslant n! \prod_{k=1}^{n-1} \left[ (n-k)^n n!/n^n \right] = (n!)^{2n}/n^{n^2}$

$\displaystyle L(n)^{1/n^2} \sim \frac{n}{e^2}$

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4 blog entries • Page 1 of 1

Cosmin Davidescu

holomorph

yay for smileys

By marisa, on Dec 10, 2005, 2:59 pm
is that all emoticons? woah!

By riddler, on Jul 27, 2005, 2:38 pm
By holomorph, on Jul 12, 2005, 4:17 pm
wow...

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By holomorph, on May 29, 2005, 4:57 am

Cosmin's Mathematics Page

Psi > Pi

by holomorph, Jun 17, 2006, 8:21 am

test no. 2

by holomorph, Jun 17, 2006, 8:21 am

test 1

by holomorph, Jun 17, 2006, 8:21 am

Hmm

by holomorph, Jun 17, 2006, 8:21 am