Reihen « 0,1,2,3,…

Wolstenholme December 28, 2008

Filed under: Primzahlen,Reihen,Zahlentheorie — abi007 @ 12:34 pm

Wilson: $\displaystyle p\in\mathbb{P} \Leftrightarrow (p-1)!\equiv -1\pmod{p}\Rightarrow\forall p\in\mathbb{P},n\in\mathbb{N}:$
$\displaystyle {{np-1}\choose{p-1}} \equiv 1 \pmod{p}$

Charles Babbage bewies 1819, dass für jede Primzahl $p>2$ diese Kongruenz gilt:

$\displaystyle{{2p-1}\choose{p-1}} \equiv 1 \pmod{p^2}$

Der Mathematiker Joseph Wolstenholme (1829-1891) bewies dann 1862, dass für jede Primzahl $p>3$ die folgende Kongruenz gilt:

$\displaystyle{{2p-1}\choose{p-1}} \equiv 1 \pmod{p^3}$

Satz von Wolstenhome.
Für $5\leq p\in\mathbb{P}$ gilt
$\displaystyle \frac{a}{b}=1+\frac12+\frac13+\frac14+...+\frac{1}{p-1} \Rightarrow a|p^2$

Der Satz ist äquivalent dazu:
$\displaystyle \frac{a}{b}= 1+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\ldots+\frac1{(p-1)^2} \Rightarrow a|p$

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sin, cos, tan, sec, csc, ct …… March 22, 2008

Filed under: Analysis,Identität,Reihen — abi007 @ 4:46 am

$\displaystyle\sin\,(\pi\,x)\;=\;(\pi\,x)\prod_{k=1}^\infty\;\left(1-\dfrac{x^2}{k^2}\right)$

http://www.mathlinks.ro/weblog.php?w=993

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Zeta September 8, 2007

Filed under: Analysis,Funktion,Primzahlen,Reihen,Zahlentheorie — abi007 @ 7:23 am

$\displaystyle\zeta(n)=\sum_{k=1}^{\infty}\frac1{k^n}$
$\displaystyle\zeta(s) = \prod_{p\ \mathrm{prim}}\frac1{1-1/p^s}=\frac1{(1-1/2^s)(1-1/3^s)(1-1/5^s)\cdots}.$
$\displaystyle \zeta_K(s)=\sum_{I\subset\mathcal{O}_K} (N^K_{\mathbb{Q}}(I))^{-s}= \prod_{\mathfrak{p}}\frac{1}{1-(N^K_{\mathbb{Q}}(\mathfrak{p}))^{-s}}$

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

$\displaystyle \zeta(2)=\frac{\pi^2}6$

http://www.mathlinks.ro/viewtopic.php?t=159424&start=20

http://www.secamlocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf

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$\displaystyle\gamma:=\lim_{n\to\infty} \left(H_n-\ln n\right)= \lim_{n\rightarrow \infty}\; \sum_{i=1}^n \left[\frac{1}{i} - \ln \left( 1 + \frac{1}{i} \right) \right]$

$\displaystyle\gamma= -\int\limits_0^1 \ln(-\ln x)\, \mathrm{d}x$

$\displaystyle\gamma = -\int\limits_0^\infty e^{-x}\ln x\, \mathrm{d}x$

$\displaystyle\gamma = -\int\limits_0^\infty (\frac1{1-e^{-x}}-\frac1x)e^{-x} \mathrm{d}x$

$\displaystyle\gamma = -\int\limits_0^\infty \frac1x(\frac1{1+x}-e^{-x}) \mathrm{d}x$

$\displaystyle \gamma\;\;=\;\;\int_{0}^{1}\;\;\frac{1}{x+1}\cdot\left(\sum_{k=1}^{\infty}\;x^{2^{k}-1}\right)\;\textbf dx$

$\displaystyle \gamma\;\;=\;\; \int_{0}^{1}\;\;\frac{x+2}{x^{2}+x+1}\cdot\left(\sum_{k=1}^{\infty}\;x^{3^{k}-1}\right)\;\textbf dx$

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$\displaystyle \gamma = - \Gamma'(1)$

$\displaystyle \Gamma(z) =\lim_{n \to \infty} \frac{n! \; n^z}{z \; (z+1)\cdots(z+n)}$

$\displaystyle\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{n=1}^{\infty}(1 + \frac{z}{n})^{-1}e^{\frac{z}{n}}$

$\displaystyle \Gamma(x)=\int_0^\infty t^{x-1}e^{-t}dt$

$\displaystyle =[-t^{x-1}e^{-t}]_0^\infty+\int_0^\infty(x-1)t^{x-2}e^{-t}dt$

$\displaystyle =(x-1)\int_0^\infty t^{x-2}e^{-t}dt$

$\displaystyle =(x-1)\Gamma(x-1).$

$\displaystyle\Gamma \left( x \right)\Gamma \left( {1 - x} \right) = \pi \csc \pi x$

$\displaystyle\int_{0}^{\infty} \frac{x^n}{e^x-1}=\zeta(n+1)\Gamma(n+1)$

http://en.wikipedia.org/wiki/Planck_law#Appendix

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$\displaystyle\int_0^{\frac{1}{2}} \; \;\frac{{\tan ^{ - 1} \,x}}{x}\;\;{\mathbf{d}}x = \sum\limits_{n = 0}^{ + \infty } {\frac{{( - 1)^n }}{{(2n + 1)^2 \cdot2^{2n + 1} }}}$

$\displaystyle{\rm K} = \beta (2) = \sum\limits_{n = 0}^\infty {\frac{{( - 1)^n }}{{(2n + 1)^2 }}} = \frac{1}{{1^2 }} - \frac{1}{{3^2 }} + \frac{1}{{5^2 }} - \frac{1}{{7^2 }} + \cdots$

$\displaystyle = - \int_0^1 {\frac{{\ln (t)}}{{1 + t^2 }}} {\text{ d}}t$

$\displaystyle = \int_0^1 {\int_0^1 {\frac{1}{{1 + x^2 y^2 }}} } dxdy = \frac{1}{2}\int_0^1 {\text{K}} (x)\,dx = \int_0^1 {\frac{{\tan ^{ - 1} x}}{x}} dx$

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series June 22, 2007

Filed under: Analysis,Reihen — abi007 @ 1:45 pm

$\displaystyle (x^n - y^n ) = (x - y) \cdot \sum\limits_{k = 1}^n {(x^{k - 1} \cdot y^{n - k} )}$
Binomialsatz:

$\displaystyle (x+y)^n = \sum_{k=0}^{n}{n \choose k} x^{n-k}y^{k}$

$\displaystyle (x_1+\ldots+x_r)^n=\sum_{k_1+\ldots+k_r=n}{n\choose k_1,\ldots,k_r}\cdot x_1^{k_1}\cdots x_r^{k_r}.$
$\displaystyle {n \choose k_1, \dots , k_r} := \frac{n!}{k_1!\cdot \dots \cdot k_r!}$

$\displaystyle \frac{1}{(1-x)^r}=\sum_{k=0}^\infty {r+k-1 \choose k} x^k \equiv \sum_{k=0}^\infty {r+k-1 \choose r-1} x^k.$

$\displaystyle\sum_{k=0}^{\infty}\frac{(-1)^{k}}{5k+1}= \int^{1}_{0}\frac{1}{1+x^{5}}\,dx$

$\displaystyle\sum_{k=1}^\infty\;\frac{1}{k^{3}\;(k+1)^{3}}\;=\;\boxed{10-6\;\zeta{(2)}}$

$\displaystyle\frac{1}{k^{3}(k+1)^{3}}$

$\displaystyle =\left(\frac{1}{k}-\frac{1}{k+1}\right)^{3}$

$\displaystyle = \left(\frac{1}{k^{3}}-\frac{1}{(k+1)^{3}}\right)+6 \left(\frac{1}{k}-\frac{1}{k+1}\right)-3 \frac{1}{(k+1)^{2}}-3 \frac{1}{k^{2}}$

$\displaystyle\sum_{k=1}^{ \infty}\left(\frac{1}{k^{3}}-\frac{1}{(k+1)^{3}}\right) = 1$
$\displaystyle\sum_{k=1}^{ \infty}\left( \frac{1}{k}-\frac{1}{k+1}\right) = 1$

$\displaystyle\sum_{k=1}^{ \infty}\frac{1}{(k+1)^{2}}= \zeta(2)-1$

$\displaystyle\sum_{k=1}^{ \infty}\frac{1}{k^{2}}= \zeta(2)$

$\displaystyle\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\;\;\;\cdots \;\;\;+\;\;\; -\;\;\;\cdots\right)^{3}\;$

$\displaystyle=\; \frac{1}{2}\;\left(1-\frac{1}{3^{3}}+\frac{1}{5^{3}}-\frac{1}{7^{3}}+\frac{1}{9^{3}}-\frac{1}{11^{3}}+\;\;\;\cdots \;\;\;+\;\;\; -\;\;\;\cdots\right)$