Difference between revisions of "MediaWiki:Sitenotice id"

Revision as of 16:11, 14 January 2015

Instead of doing the substitution it should also be possible to write:

$$ \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(x-y) g(y) e^{-i<k,x>} dy \: dx $$

as

$$ \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(x-y) g(y) e^{-i<k,x-y+y>} dy \: dx $$

$$ = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(x-y) e^{-i<k,x-y>} g(y) e^{-i<k,y>} dy \: dx $$

and then use Fubini

Un vieil homme (talk) 20:43, 31 December 2014 (CET)

In 1c)the derivation got exchanged with the integral without proofing that this is allowed:

$$ \frac{d}{dk} \widehat{F_c}(k) = \frac{d}{dk} (\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F_c(x)e^{-ikx} dx) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F_c(x) \frac{d}{dk} e^{-ikx} dx = \frac{-i}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-cx^{2}}xe^{-ikx} dx $$

Alex (talk) 14:55, 4 January 2015 (CET)

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: [1] 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.

@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.

Proposal for alternative solution in (1d)

In (1b), we showed that for \(f, g \in L^1(\mathbb{R}^n)\), the Fourier transform of the convolution is: $$\widehat{f*g}(k) = \sqrt{2\pi}^n \hat f(k) \hat g(k)$$

Here, we have \(n=1\) and can thus write: $$\widehat{F_a*F_b}(k) = \sqrt{2\pi} \ \hat F_a(k) \hat F_b(k)$$

We insert the Fourier transform of \(F_a\) and \(F_b\), which we found in (1c): $$\begin{align} \widehat{F_a*F_b}(k) &= \sqrt{2\pi} \ \left(\frac{1}{\sqrt{2a}} e^{-\frac{k^2}{4a}}\; \frac{1}{\sqrt{2b}} e^{-\frac{k^2}{4b}}\right) \\ &= \sqrt{\frac{\pi}{2ab}}\exp\left(-\frac{k^2}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\right) \\ &= \sqrt{\frac{\pi}{a+b}} \sqrt{\frac{1}{2\frac{ab}{a+b}}} \exp\left(-\frac{k^2}{4}\frac{a+b}{ab}\right) \\ &= \alpha \hat F_c(k) \end{align}$$ for \(c=\frac{ab}{a+b}\) and \(\alpha = \sqrt{\frac{\pi}{a+b}}\)

Because of the linearity of the Fourier transform, we can conclude: $$(F_a*F_b)(x) = \alpha F_c(x)$$

Nik (talk) 14:38, 14 January 2015 (CET)

Proposal for alternate solution in (1c)

Proposition: $$ \frac{d}{dk}\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-cx^2}e^{-ikx} dx \right) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{d}{dk} \left(e^{-cx^2}e^{-ikx}\right) dx$$

Proof: The lecture notes tell us (\(\rightarrow\) look here, p.95) that $$\widehat{x\psi}(k) = i\frac{d}{dk}\hat \psi(k)$$ under some conditions which I'll assume to hold in this case because I'm a lazy physicist.

I'm a not so lazy mathematician, so I checked this for you. Everything is alright since \( F \) is a Schwartz-function.

Now, taking a close look at the proposition yields $$\text{LHS} = \frac{d}{dk}\ \widehat{e^{-cx^2}}\,(k)$$ $$\text{RHS} = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty -ix e^{-cx^2}e^{-ikx} dx = -i\widehat{\;xe^{-cx^2}\,}(k)$$ Aaaand... we're done.

Nik (talk) 17:00, 14 January 2015 (CET)