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analytic_function [2014/02/21 11:46] nikolaj |
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| ===== Analytic function ===== | ===== Analytic function ===== | ||
| ==== Set ==== | ==== Set ==== | ||
| - | | @#88DDEE: $\mathcal O\subset \mathbb C$ | | + | | @#55CCEE: context | @#55CCEE: $\mathcal O\subset \mathbb C$ | |
| - | | @#FFBB00: $f\in \mathrm{it}$ | | + | | @#FFBB00: definiendum | @#FFBB00: $f\in \mathrm{it}$ | |
| - | | @#AAFFAA: $f:\mathcal O\to\mathbb C$ | | + | | @#AAFFAA: inclusion | @#AAFFAA: $f:\mathcal O\to\mathbb C$ | |
| - | | @#FFFDDD: $c$ ... series in $\mathbb C$ | | + | | @#FFFDDD: for all | @#FFFDDD: $c$ ... series in $\mathbb C$ | |
| >todo, roughly | >todo, roughly | ||
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| {{fence_height.png}} | {{fence_height.png}} | ||
| - | A complex function $f(z)$ consists of two real functions, so it's fence height is just given by the sum of the fence heigh of \mathrm{Re}\,f(z)$ and $i\,\mathrm{Im}\,f(z)$. Let's consider the function $f(z):=z=r\,\cos\theta+i\,r\,\sin\theta$. We have | + | A complex function $f(z)$ consists of two real functions, so it's fence height is just given by the sum of the fence heigh of $\mathrm{Re}\,f(z)$ and $i\,\mathrm{Im}\,f(z)$. Let's consider the function $f(z):=z=r\,\cos\theta+i\,r\,\sin\theta$. We have |
| $z^n=r^n\,\mathrm{e}^{i\,n\,\theta}=r\,\cos(n\,\theta)+i\,r\,\sin(n\,\theta)$ | $z^n=r^n\,\mathrm{e}^{i\,n\,\theta}=r\,\cos(n\,\theta)+i\,r\,\sin(n\,\theta)$ | ||
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| == Cauchy's integral formula == | == Cauchy's integral formula == | ||
| $\frac{1}{n!}f^{(n)}(p) = \frac{1}{2\pi\, i} \oint_\gamma \frac{f(z)}{(z-p)^{n+1}}\, \mathrm dz$ | $\frac{1}{n!}f^{(n)}(p) = \frac{1}{2\pi\, i} \oint_\gamma \frac{f(z)}{(z-p)^{n+1}}\, \mathrm dz$ | ||
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| + | Roughly, the Laplace transform uses this for a re-encoding of a functions $f:\mathbb R^+\to\mathbb R$ with Taylor expansion $f(t)=\sum_{n=0}^\infty a_n t^n$, namely by mapping $t^n$ to $s^{-n}\cdot \frac{1}{s}$. | ||
| === Reference === | === Reference === | ||
