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IN:={Ø,{Ø},{Ø,{Ø}},{Ø,{Ø},{Ø,{Ø}}},…}

Differentiation June 28, 2007

Filed under: Analysis — abi007 @ 8:07 pm

\displaystyle\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}

\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{d}{dt}\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right)=\frac{\frac{d^{2}y}{dt^{2}}\frac{dx}{dt}-\frac{dy}{dt}\frac{d^{2}x}{dt^{2}}}{\left(\frac{dx}{dt}\right)^{2}}

 

Riemann vs. Lebesgue June 26, 2007

Filed under: Analysis,Integrale,Tagebuch — abi007 @ 2:42 pm

ein uneigentliches Riemannintegral ist genau dann ein Lebesgueintegral, wenn es absolut konvergiert.

\displaystyle f:\Omega\rightarrow\mathbb{R},\int_{\Omega}{f(x)}dx<\infty.\leftrightarrow f ist Lebesgueintegrabel.

Es seien \displaystyle f:[0,\infty[\rightarrow[0,\infty[ und g:[0,\infty[\rightarrow[0,\infty[ nicht-negative Funktionen, welche über [a, b] Riemann-integrierbar sind für alle 0\leq a<b[. Gilt f(x)\leq g(x) für alle genügend großen x und konvergiert das uneigentliche Riemann-Integral \displaystyle \int_0^{\infty}g(x)dx, so konvergiert auch \displaystyle \int_0^{\infty}f(x)dx

  • FourierTransformation:

\displaystyle \int_0^{\infty}\frac{sin(x)}{x}dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{sin(x)}{x}dx
\displaystyle =\frac{1}{2}(\sqrt{2\pi}\cdot\frac{1}{\sqrt{2\pi}})\cdot\int_{-\infty}^{\infty}\frac{sin(x)}{x}\cdot e^{-i\cdot x \cdot 0}dx
\displaystyle =\frac{1}{2}\sqrt{2\pi}\cdot\sqrt{\frac{\pi}{2}}\cdot 1_{([-1,1])}(0)
\displaystyle =\frac{1}{2}\sqrt{2\pi}\cdot\sqrt{\frac{\pi}{2}}\cdot 1
\displaystyle =\frac{\pi}{2}

(seite 74.)

 

Kategorien June 25, 2007

Filed under: CAT,Gruppen — abi007 @ 11:21 am
  • \forall f\in \text{Hom}(G,H)\exists^{=1} CD:

cd1

  • M,N\triangleright G, M\subseteq N \Longrightarrow \exists^{=1} CD:

cd2

  • N\triangleright G, U\leq G \Longrightarrow \exists^{=1} CD:

cd3

  • \forall f\in \text{R-Modul-Hom}(G,H)\exists^{=1} CD:

cd4

  • U\subseteq V; U,V\text{ Untermodule von }M\Longrightarrow \exists^{=1} CD:

cd5

  • U,V\text{ Untermodule von }M\Longrightarrow \exists^{=1} CD:

cd6

 

Adjungierte Matrix June 24, 2007

Filed under: Lineare Algebra — abi007 @ 12:03 pm

Adjungierte Matrix (conjugate matrix)

\begin{matrix} A*=A^H & = & \bar{A}^T \\ \bar{z} & = & a-b\cdot i \\ \langle Av, w\rangle & = & \langle v, A^*\,w\rangle, v, w \in\Bbb K^n \\ \Bbb K =\Bbb R & \rightarrow & A^*=A^T \\\end{matrix}

A=A* \Longleftrightarrow selbstadjungiert, symmetrisch; hermitesch

\begin{matrix}\left(A + B\right)^* &=& A^* + B^*\\rA)^* &=& \overline{r}A^*\\\left(AB\right)^* &=& B^*A^*\\\left(A^*\right)^* &=& A\\\left(A^{-1}\right)^* &=& \left(A^*\right)^{-1}\\det(A^*) &=&\overline{\det(A)}\\ trace(A^*) &=& \overline{trace(A)}\end{matrix}

adj(A) = \tilde A^T = \begin{pmatrix} \tilde a_{11} & \tilde a_{21} & \cdots & \tilde a_{n1}\\ \tilde a_{12} & \tilde a_{22} & & \tilde a_{n2}\\ \vdots & & \ddots & \vdots\\ \tilde a_{1n} & \tilde a_{2n} & \cdots & \tilde a_{nn}\end{pmatrix}

 

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)

  • \sum_{i=m}^n x = (n-m+1)x
  • \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}
  • \sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2
  • \sum_{i=1}^n i^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}
  • \sum_{i=0}^n x^i = \frac{x^{n+1}-1}{x-1} (special case of the above where m = 0)
  • \sum_{i=0}^{n-1} {i \choose k} = {n \choose k+1}
  • \left(\sum_i a_i\right)\left(\sum_j b_j\right) = \sum_i\sum_j a_ib_j
  • \left(\sum_i a_i\right)^2 = 2\sum_i\sum_{j<i} a_ia_j + \sum_i a_i^2
  • \displaystyle\sum_{k=0}^{n} {{n-k} \choose k} = \mathrm{F}(n+1).
 

Prime June 22, 2007

Filed under: Primzahlen — abi007 @ 1:15 pm
  • \displaystyle p\in\mathbb{P}\rightarrow p\equiv\pm 1 (mod 6)
  • \displaystyle p\in\mathbb{P},p\equiv 3 (mod 4)\rightarrow

\displaystyle \sum_{x=1}^{p-1}x(\frac{x}{p})<0

\displaystyle x^2+a\nmid p

\displaystyle (1-2(2|p))\sum_{r=1}^{\frac{p-1}{2}}r(r|p)=p\frac{1-(2|p)}{2}\sum_{r=1}^{p}(r|p)

\displaystyle ((2|p)-2)\sum_{r=1}^{p-1}r(r|p)=p\sum_{r=1}^{\frac{p-1}{2}}(r|p)

欧拉定理说:

\displaystyle a^\varphi (m)\equiv 1 (mod m)

如果 m 是素数，那么

\displaystyle \varphi (m)=m-1

立即得到:

\displaystyle a^{m-1}\equiv 1 (mod m)

如果n不是素数，也一样,比如说:

计算 \displaystyle 7^m (mod 15)

\displaystyle 7^{\varphi(15)}=7^8\equiv 1 (mod 15)

然后因为:

\displaystyle 7^1\equiv 7 (mod 15)

你也可以算出所有的

\displaystyle 7^m(mod 15)

至于它的证明，也就只不过是欧拉定理罢了。

 

Theoreme June 21, 2007

Filed under: Diophantine,Zahlentheorie — abi007 @ 9:09 pm
  • Catalan’sche Vermutung:

\displaystyle x^p-y^q=1, x,y,p,q>1 \rightarrow 3^2-2^3=1

  • Fermat’s Little Theorem:

\displaystyle a\in\mathbb{Z}, p\in\mathbb{P}\rightarrow
\displaystyle a^p \equiv a \pmod{p}\,\!

  • Prime number theorem:

\displaystyle \pi(x)\sim\frac{x}{\ln x}

\displaystyle Li(n)=\int_2^n\frac{dx}{ln(x)}\sim\pi(n)

  • Bertrand-Chebyshev theorem

\displaystyle \forall n>1\exists p\in\mathbb{P}:n<p<2n

  • Quadratic Reciprocity Law

\displaystyle (\frac{p}{q})(\frac{q}{p})=(-1)^{\frac{(p-1)(q-1)}{4}}

  • Sophie Germain:

\displaystyle a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)

 

Möbius June 21, 2007

Filed under: Zahlentheorie — abi007 @ 4:14 pm

\displaystyle\sum\limits_{d\left| n \right.} {\mu (d)\varphi (d) = \prod\limits_{p\left| n \right.} {(2 - p)} },n \in \mathbb{N}

\displaystyle\varphi_1(n)=n\sum\limits_{d\left| n \right.} \frac{|\mu(d)|}{d}\Longrightarrow \varphi_1(n)=\sum\limits_{d^2\left| n \right.}\mu(d)\sigma(\frac{n}{d^2})

\displaystyle S(x)=\sum\limits_{d\left.\leq x \right.}\sigma(k) \Longrightarrow \sum\limits_{n\left.\leq x \right.}\varphi_1(n)=\sum\limits_{d\left.\leq \sqrt{x} \right.} \mu(d)S(\frac{x}{d^2})

\displaystyle\mu(n):=\begin{cases}1 & \mbox{wenn } n=1 \\ (-1)^k & \mbox{wenn } n \mbox{ quadratfrei, } k \mbox{ ist die Anzahl der Primfaktoren} \\ 0 & \mbox{sonst} \end{cases}

 

 
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