Difference between revisions of "MediaWiki:Sitenotice id"
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[[User:Un vieil homme|Un vieil homme]] ([[User talk:Un vieil homme|talk]]) 20:43, 31 December 2014 (CET) | [[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) |
Revision as of 13:55, 4 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:
$$ \frac{d}{dk} \widehat{F_c}(k) = \frac{d}{dk} (\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F_c(x)e^{-ikx} dx) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F_c(x) \frac{d}{dk} e^{-ikx} dx = \frac{-i}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-cx^{2}}xe^{-ikx} dx $$