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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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| − | ''@Un vieil homme:'' I think the Fubini idea doesn't work. We've proven Fubini at Kowalski's for Riemann-integrable functions so I guess that's the theorem we can use. Fubini only works for compact intervals which is not the case above. - A.
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| − | ----
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| − | '''Proposal for alternative solution in (1d)'''
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| − | In (1b), we showed that for \(f, g \in L^1(\mathbb{R}^n)\), the Fourier transform of the convolution is:
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| − | $$\widehat{f*g}(k) = \sqrt{2\pi}^n \hat f(k) \hat g(k)$$
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| − | Here, we have \(n=1\) and can thus write:
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| − | $$\widehat{F_a*F_b}(k) = \sqrt{2\pi} \ \hat F_a(k) \hat F_b(k)$$
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| − | We insert the Fourier transform of \(F_a\) and \(F_b\), which we found in (1c):
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| − | $$\begin{align}
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| − | \widehat{F_a*F_b}(k) &= \sqrt{2\pi} \ \left(\frac{1}{\sqrt{2a}} e^{-\frac{k^2}{4a}}\; \frac{1}{\sqrt{2b}} e^{-\frac{k^2}{4b}}\right) \\
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| − | &= \sqrt{\frac{\pi}{2ab}}\exp\left(-\frac{k^2}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\right) \\
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| − | &= \sqrt{\frac{\pi}{a+b}} \sqrt{\frac{1}{2\frac{ab}{a+b}}} \exp\left(-\frac{k^2}{4}\frac{a+b}{ab}\right) \\
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| − | &= \alpha \hat F_c(k)
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| − | \end{align}$$
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| − | for \(c=\frac{ab}{a+b}\) and \(\alpha = \sqrt{\frac{\pi}{a+b}}\)
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| − | Because of the linearity of the Fourier transform, we can conclude:
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| − | $$(F_a*F_b)(x) = \alpha F_c(x)$$
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| − | [[User:Nik|Nik]] ([[User talk:Nik|talk]]) 14:38, 14 January 2015 (CET)
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| − | ----
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| − | '''Proposal for alternate solution in (1c)'''
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| − | '''Proposition:'''
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| − | $$ \frac{d}{dk}\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-cx^2}e^{-ikx} dx \right) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{d}{dk} \left(e^{-cx^2}e^{-ikx}\right) dx$$
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| − | '''Proof:'''
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| − | The lecture notes tell us (\(\rightarrow\) look [http://www.math.ethz.ch/~gruppe5/group5/lectures/mmp/hs14/Files/Fourier-Heat-etc..pdf here], p.95) that
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| − | $$\widehat{x\psi}(k) = i\frac{d}{dk}\hat \psi(k)$$
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| − | under some conditions which I'll assume to hold in this case because I'm a lazy physicist.
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| − | Now, taking a close look at the proposition yields
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| − | $$\text{LHS} = \frac{d}{dk}\ \widehat{e^{-cx^2}}\,(k)$$
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| − | $$\text{RHS} = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty -ix e^{-cx^2}e^{-ikx} dx = -i\widehat{\;xe^{-cx^2}\,}(k)$$
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| − | Aaaand... we're done.
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| − | <p style="text-align:center"><img src="http://mmp.vsos.ethz.ch/100problems.jpg" style="height:250px"></p>
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