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| − | Instead of doing the substitution it should also be possible to write:
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| − | $$ \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(x-y) g(y) e^{-i<k,x>} dy \: dx $$
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| − | as
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| − | $$ \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(x-y) g(y) e^{-i<k,x-y+y>} dy \: dx $$
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| − | $$ = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(x-y) e^{-i<k,x-y>} g(y) e^{-i<k,y>} dy \: dx $$
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| − | and then use Fubini
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| − | [[User:Un vieil homme|Un vieil homme]] ([[User talk:Un vieil homme|talk]]) 20:43, 31 December 2014 (CET)
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| − | In 1c)the derivation got exchanged with the integral without proofing that this is allowed:
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| − | $$ \frac{d}{dk} \widehat{F_c}(k) = \frac{d}{dk} (\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F_c(x)e^{-ikx} dx) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F_c(x) \frac{d}{dk} e^{-ikx} dx = \frac{-i}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-cx^{2}}xe^{-ikx} dx $$
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| − | [[User:Alex|Alex]] ([[User talk:Alex|talk]]) 14:55, 4 January 2015 (CET)
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| − | In part d) of the problem used \( e^{-ax^{2}}\sqrt{2\pi} \widehat{F_l}(s) \) with \( s = 2axi \). But a Fourier-transform can't take complex arguments - I didn't know that either so I checked here: [http://en.wikipedia.org/wiki/Fourier_transform#Definition] So you might rather use the trick we've used in problem 3 with completing the square. Actually you can copy that from the guys of group Y on the Dropbox, they got a pretty solid solution. - A.
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