Fouriertransformationseigensch < Transformationen < Analysis < Hochschule < Mathe < Vorhilfe
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| Aufgabe | Sei die Fouriertransformation definiert wie folgt:
[mm] $F[\Psi(x);k] \equiv \tilde{\Psi}(k) [/mm] = [mm] \int_{-\infty}^{\infty} e^{-ikx}\Psi [/mm] (x) dx$
[mm] $\Psi(x) [/mm] = [mm] \int_{-\infty}^{\infty} e^{ikx}\tilde{ \Psi} [/mm] (k) [mm] \frac{dk}{2\pi}$
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Man zeige die folgenden Eigenschaften:
$i) [mm] F[a\Psi_{1}(x) [/mm] + [mm] b\Psi_{2}(x);k] [/mm] = [mm] aF[\Psi_{1}(x);k]+bF[\Psi_{2}(x);k]$
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$ii) [mm] F[\Psi(x-a);k]=e^{-ika}F[\Psi(x);k]$
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$iii) [mm] F[\Psi [/mm] (ax);k] = [mm] \frac{1}{a} F[\Psi [/mm] (x);k/a] \ \ [mm] \forall [/mm] a > 0$
$iv) [mm] F(\Psi [/mm] (-x); k ] = [mm] F[\Psi(x);-k]$
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$v) [mm] F[\frac{d}{dx} \Psi(x); [/mm] k] = ik [mm] F[\Psi(x);k]$
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$vi) [mm] F[x\Psi(x);k] [/mm] = i [mm] \frac{d}{dk} F[\Psi(x);k]$ [/mm] |
Hallo,
i) [mm] $aF[\Psi_{1}(x);k]+ bF[\Psi_{2} [/mm] (x);k] [mm] =a\int_{-\infty}^{\infty}e^{-ikx}\Psi_{1}(x)dx [/mm] + [mm] b\int_{-\infty}^{\infty} e^{-ikx}\Psi_{2}(x) [/mm] dx = [mm] \int_{-\infty}^{\infty}ae^{-ikx}\Psi_{1}(x)dx [/mm] + [mm] \int_{-\infty}^{\infty}be^{-ikx}\Psi_{2}(x)dx [/mm] = [mm] \int_{-\infty}^{\infty}e^{-ikx}(a\Psi_{1}(x) [/mm] + [mm] b\Psi_{2}(x)) [/mm] dx = [mm] F[a\Psi_{1}(x)+b\Psi_{2}(x);k]$
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ii) $y:=x-a$ substituieren...
iii) $y:=ax$ substituieren
iv) [mm] $F[\Psi(x);-k] [/mm] = [mm] \int_{-\infty}^{\infty} e^{ikx} \Psi(x) [/mm] dx = [mm] \int_{-\infty}^{\infty} e^{ikx}(\int_{-\infty}^{\infty} e^{-ikx}\tilde{\Psi}(-k) \frac{dk}{2\pi}) [/mm] dx = [mm] \int_{-\infty}^{\infty} e^{-ikx} \Psi(-x) [/mm] dx = [mm] F[\Psi(-x); [/mm] k]$
v) partiell integrieren
vi) unter dem Integral differenzieren (nach k) ...
Stimmt das so?
Vielen Dank für jede Hilfe.
Gruss
kushkush
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(Mitteilung) Reaktion unnötig  | | Datum: | 16:20 Sa 24.09.2011 | | Autor: | matux |
$MATUXTEXT(ueberfaellige_frage)
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