Problem on computing the Fourier transform of the Gaussian
$begingroup$
The problem sounds like this.
Show that $stoint_mathbb{R}e^{-(x+is)^2}dx$ is constant wrt
$sinmathbb{R}.$ Then use this fact to shot that
$mathcal{F}(e^{-a|x|^2})=e^{-frac{|x|^2}{a}}$ for $a>0,$ where
$mathcal{F}$ is the Fourier transform on $mathbb{R}$
I know the traditional ODE way to show this fact about the Gaussian bell but this proof got me interested and I don't have any clue on solving this. I've tried to differentiate wrt $s$ but no luck.
fourier-analysis fourier-transform gaussian-integral
asked Dec 10 '18 at 23:58
Hurjui IonutHurjui Ionut
501412
$endgroup$
add a comment |
$begingroup$
The problem sounds like this.
Show that $stoint_mathbb{R}e^{-(x+is)^2}dx$ is constant wrt
$sinmathbb{R}.$ Then use this fact to shot that
$mathcal{F}(e^{-a|x|^2})=e^{-frac{|x|^2}{a}}$ for $a>0,$ where
$mathcal{F}$ is the Fourier transform on $mathbb{R}$
I know the traditional ODE way to show this fact about the Gaussian bell but this proof got me interested and I don't have any clue on solving this. I've tried to differentiate wrt $s$ but no luck.
fourier-analysis fourier-transform gaussian-integral
asked Dec 10 '18 at 23:58
Hurjui IonutHurjui Ionut
501412
$endgroup$
1
$begingroup$
Yes the claim is more than sloppy. For $s in imathbb{R}$ the constantness is obvious from a change of variable. But the function is analytic in $s in mathbb{C}$, thus being constant on some interval implies constant everywhere, in particular constant in $s in mathbb{R}$
$endgroup$
– reuns
Dec 11 '18 at 0:10
$begingroup$
Ok thanks, now the part about that constant function is clear. I still don't know how to show the last part of the problem.
$endgroup$
– Hurjui Ionut
Dec 11 '18 at 0:20
1
$begingroup$
Expand $(x+is)^2$ you'll see the Fourier transform of $e^{-x^2}$ and with a change of variable the FT of $e^{-a x^2}$
$endgroup$
– reuns
Dec 11 '18 at 0:21
$begingroup$
I still don't get it..
$endgroup$
– Hurjui Ionut
Dec 11 '18 at 0:33
add a comment |
$begingroup$
The problem sounds like this.
Show that $stoint_mathbb{R}e^{-(x+is)^2}dx$ is constant wrt
$sinmathbb{R}.$ Then use this fact to shot that
$mathcal{F}(e^{-a|x|^2})=e^{-frac{|x|^2}{a}}$ for $a>0,$ where
$mathcal{F}$ is the Fourier transform on $mathbb{R}$
I know the traditional ODE way to show this fact about the Gaussian bell but this proof got me interested and I don't have any clue on solving this. I've tried to differentiate wrt $s$ but no luck.
fourier-analysis fourier-transform gaussian-integral
asked Dec 10 '18 at 23:58
Hurjui IonutHurjui Ionut
501412
$endgroup$
The problem sounds like this.
Show that $stoint_mathbb{R}e^{-(x+is)^2}dx$ is constant wrt
$sinmathbb{R}.$ Then use this fact to shot that
$mathcal{F}(e^{-a|x|^2})=e^{-frac{|x|^2}{a}}$ for $a>0,$ where
$mathcal{F}$ is the Fourier transform on $mathbb{R}$
I know the traditional ODE way to show this fact about the Gaussian bell but this proof got me interested and I don't have any clue on solving this. I've tried to differentiate wrt $s$ but no luck.
fourier-analysis fourier-transform gaussian-integral
fourier-analysis fourier-transform gaussian-integral
asked Dec 10 '18 at 23:58
Hurjui IonutHurjui Ionut
501412
asked Dec 10 '18 at 23:58
Hurjui IonutHurjui Ionut
501412
asked Dec 10 '18 at 23:58
Hurjui IonutHurjui Ionut
501412
asked Dec 10 '18 at 23:58
Hurjui IonutHurjui Ionut
501412
asked Dec 10 '18 at 23:58
Hurjui IonutHurjui Ionut
501412
501412
1
$begingroup$
Yes the claim is more than sloppy. For $s in imathbb{R}$ the constantness is obvious from a change of variable. But the function is analytic in $s in mathbb{C}$, thus being constant on some interval implies constant everywhere, in particular constant in $s in mathbb{R}$
$endgroup$
– reuns
Dec 11 '18 at 0:10
$begingroup$
Ok thanks, now the part about that constant function is clear. I still don't know how to show the last part of the problem.
$endgroup$
– Hurjui Ionut
Dec 11 '18 at 0:20
1
$begingroup$
Expand $(x+is)^2$ you'll see the Fourier transform of $e^{-x^2}$ and with a change of variable the FT of $e^{-a x^2}$
$endgroup$
– reuns
Dec 11 '18 at 0:21
$begingroup$
I still don't get it..
$endgroup$
– Hurjui Ionut
Dec 11 '18 at 0:33
add a comment |
1
$begingroup$
Yes the claim is more than sloppy. For $s in imathbb{R}$ the constantness is obvious from a change of variable. But the function is analytic in $s in mathbb{C}$, thus being constant on some interval implies constant everywhere, in particular constant in $s in mathbb{R}$
$endgroup$
– reuns
Dec 11 '18 at 0:10
$begingroup$
Ok thanks, now the part about that constant function is clear. I still don't know how to show the last part of the problem.
$endgroup$
– Hurjui Ionut
Dec 11 '18 at 0:20
1
$begingroup$
Expand $(x+is)^2$ you'll see the Fourier transform of $e^{-x^2}$ and with a change of variable the FT of $e^{-a x^2}$
$endgroup$
– reuns
Dec 11 '18 at 0:21
$begingroup$
I still don't get it..
$endgroup$
– Hurjui Ionut
Dec 11 '18 at 0:33
1
1
$begingroup$
Yes the claim is more than sloppy. For $s in imathbb{R}$ the constantness is obvious from a change of variable. But the function is analytic in $s in mathbb{C}$, thus being constant on some interval implies constant everywhere, in particular constant in $s in mathbb{R}$
$endgroup$
– reuns
Dec 11 '18 at 0:10
$begingroup$
Yes the claim is more than sloppy. For $s in imathbb{R}$ the constantness is obvious from a change of variable. But the function is analytic in $s in mathbb{C}$, thus being constant on some interval implies constant everywhere, in particular constant in $s in mathbb{R}$
$endgroup$
– reuns
Dec 11 '18 at 0:10
$begingroup$
Ok thanks, now the part about that constant function is clear. I still don't know how to show the last part of the problem.
$endgroup$
– Hurjui Ionut
Dec 11 '18 at 0:20
$begingroup$
Ok thanks, now the part about that constant function is clear. I still don't know how to show the last part of the problem.
$endgroup$
– Hurjui Ionut
Dec 11 '18 at 0:20
1
1
$begingroup$
Expand $(x+is)^2$ you'll see the Fourier transform of $e^{-x^2}$ and with a change of variable the FT of $e^{-a x^2}$
$endgroup$
– reuns
Dec 11 '18 at 0:21
$begingroup$
Expand $(x+is)^2$ you'll see the Fourier transform of $e^{-x^2}$ and with a change of variable the FT of $e^{-a x^2}$
$endgroup$
– reuns
Dec 11 '18 at 0:21
$begingroup$
I still don't get it..
$endgroup$
– Hurjui Ionut
Dec 11 '18 at 0:33
$begingroup$
I still don't get it..
$endgroup$
– Hurjui Ionut
Dec 11 '18 at 0:33
add a comment |
0
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1
$begingroup$
Yes the claim is more than sloppy. For $s in imathbb{R}$ the constantness is obvious from a change of variable. But the function is analytic in $s in mathbb{C}$, thus being constant on some interval implies constant everywhere, in particular constant in $s in mathbb{R}$
$endgroup$
– reuns
Dec 11 '18 at 0:10
$begingroup$
Ok thanks, now the part about that constant function is clear. I still don't know how to show the last part of the problem.
$endgroup$
– Hurjui Ionut
Dec 11 '18 at 0:20
1
$begingroup$
Expand $(x+is)^2$ you'll see the Fourier transform of $e^{-x^2}$ and with a change of variable the FT of $e^{-a x^2}$
$endgroup$
– reuns
Dec 11 '18 at 0:21
$begingroup$
I still don't get it..
$endgroup$
– Hurjui Ionut
Dec 11 '18 at 0:33