Is the inverse Fourier transform of a radial function radial?
Let $f: Bbb{R}^d to Bbb{R}$ be a rapidly decreasing function in $Bbb{R}^d$. Concerning Fourier transforms in $Bbb{R}^d$, we define $hat{f}(xi) = int_{Bbb{R}^d} f(x) e^{-2 pi i xi cdot x}dx $, for $xi in Bbb{R}^d$. And the inverse of $g(xi)$ is $int_{Bbb{R}^d}g(xi)e^{2 pi i xi cdot x}dxi$. A function $f: Bbb{R}^d to Bbb{R}$ is said to be radial iff there is $F: Bbb{R} to Bbb{R}$ such that $f(x) = F(|x|)$ for $x in Bbb{R}^d$. It is easy to prove that the transform of a radial function is radial. Does the same hold for inverse transforms?
fourier-transform
asked Nov 28 at 11:24
Nuntractatuses Amável
58712
add a comment |
Let $f: Bbb{R}^d to Bbb{R}$ be a rapidly decreasing function in $Bbb{R}^d$. Concerning Fourier transforms in $Bbb{R}^d$, we define $hat{f}(xi) = int_{Bbb{R}^d} f(x) e^{-2 pi i xi cdot x}dx $, for $xi in Bbb{R}^d$. And the inverse of $g(xi)$ is $int_{Bbb{R}^d}g(xi)e^{2 pi i xi cdot x}dxi$. A function $f: Bbb{R}^d to Bbb{R}$ is said to be radial iff there is $F: Bbb{R} to Bbb{R}$ such that $f(x) = F(|x|)$ for $x in Bbb{R}^d$. It is easy to prove that the transform of a radial function is radial. Does the same hold for inverse transforms?
fourier-transform
asked Nov 28 at 11:24
Nuntractatuses Amável
58712
2
the inverse transform of a function $f$ is the fourier trnasform of $xmapsto f(-x)$
– Surb
Nov 28 at 11:28
Yes, convert the cartesian n-dimensional Fourier Transform to n-spherical (?) coordinates and integrate over the angular variables. You will be left with a transform that is a single integral in terms of the radial variable. You can do the same for the inverse transform. For n = 1, you will get the Fourier cosine transform. For n = 2, you will get the zero order Hankel Transform. For higher n, see math.stackexchange.com/a/3007081/441161
– Andy Walls
Nov 28 at 19:36
add a comment |
Let $f: Bbb{R}^d to Bbb{R}$ be a rapidly decreasing function in $Bbb{R}^d$. Concerning Fourier transforms in $Bbb{R}^d$, we define $hat{f}(xi) = int_{Bbb{R}^d} f(x) e^{-2 pi i xi cdot x}dx $, for $xi in Bbb{R}^d$. And the inverse of $g(xi)$ is $int_{Bbb{R}^d}g(xi)e^{2 pi i xi cdot x}dxi$. A function $f: Bbb{R}^d to Bbb{R}$ is said to be radial iff there is $F: Bbb{R} to Bbb{R}$ such that $f(x) = F(|x|)$ for $x in Bbb{R}^d$. It is easy to prove that the transform of a radial function is radial. Does the same hold for inverse transforms?
fourier-transform
asked Nov 28 at 11:24
Nuntractatuses Amável
58712
Let $f: Bbb{R}^d to Bbb{R}$ be a rapidly decreasing function in $Bbb{R}^d$. Concerning Fourier transforms in $Bbb{R}^d$, we define $hat{f}(xi) = int_{Bbb{R}^d} f(x) e^{-2 pi i xi cdot x}dx $, for $xi in Bbb{R}^d$. And the inverse of $g(xi)$ is $int_{Bbb{R}^d}g(xi)e^{2 pi i xi cdot x}dxi$. A function $f: Bbb{R}^d to Bbb{R}$ is said to be radial iff there is $F: Bbb{R} to Bbb{R}$ such that $f(x) = F(|x|)$ for $x in Bbb{R}^d$. It is easy to prove that the transform of a radial function is radial. Does the same hold for inverse transforms?
fourier-transform
fourier-transform
asked Nov 28 at 11:24
Nuntractatuses Amável
58712
asked Nov 28 at 11:24
Nuntractatuses Amável
58712
asked Nov 28 at 11:24
Nuntractatuses Amável
58712
asked Nov 28 at 11:24
Nuntractatuses Amável
58712
asked Nov 28 at 11:24
Nuntractatuses Amável
58712
58712
2
the inverse transform of a function $f$ is the fourier trnasform of $xmapsto f(-x)$
– Surb
Nov 28 at 11:28
Yes, convert the cartesian n-dimensional Fourier Transform to n-spherical (?) coordinates and integrate over the angular variables. You will be left with a transform that is a single integral in terms of the radial variable. You can do the same for the inverse transform. For n = 1, you will get the Fourier cosine transform. For n = 2, you will get the zero order Hankel Transform. For higher n, see math.stackexchange.com/a/3007081/441161
– Andy Walls
Nov 28 at 19:36
add a comment |
2
the inverse transform of a function $f$ is the fourier trnasform of $xmapsto f(-x)$
– Surb
Nov 28 at 11:28
Yes, convert the cartesian n-dimensional Fourier Transform to n-spherical (?) coordinates and integrate over the angular variables. You will be left with a transform that is a single integral in terms of the radial variable. You can do the same for the inverse transform. For n = 1, you will get the Fourier cosine transform. For n = 2, you will get the zero order Hankel Transform. For higher n, see math.stackexchange.com/a/3007081/441161
– Andy Walls
Nov 28 at 19:36
2
2
the inverse transform of a function $f$ is the fourier trnasform of $xmapsto f(-x)$
– Surb
Nov 28 at 11:28
the inverse transform of a function $f$ is the fourier trnasform of $xmapsto f(-x)$
– Surb
Nov 28 at 11:28
Yes, convert the cartesian n-dimensional Fourier Transform to n-spherical (?) coordinates and integrate over the angular variables. You will be left with a transform that is a single integral in terms of the radial variable. You can do the same for the inverse transform. For n = 1, you will get the Fourier cosine transform. For n = 2, you will get the zero order Hankel Transform. For higher n, see math.stackexchange.com/a/3007081/441161
– Andy Walls
Nov 28 at 19:36
Yes, convert the cartesian n-dimensional Fourier Transform to n-spherical (?) coordinates and integrate over the angular variables. You will be left with a transform that is a single integral in terms of the radial variable. You can do the same for the inverse transform. For n = 1, you will get the Fourier cosine transform. For n = 2, you will get the zero order Hankel Transform. For higher n, see math.stackexchange.com/a/3007081/441161
– Andy Walls
Nov 28 at 19:36
add a comment |
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2
the inverse transform of a function $f$ is the fourier trnasform of $xmapsto f(-x)$
– Surb
Nov 28 at 11:28
Yes, convert the cartesian n-dimensional Fourier Transform to n-spherical (?) coordinates and integrate over the angular variables. You will be left with a transform that is a single integral in terms of the radial variable. You can do the same for the inverse transform. For n = 1, you will get the Fourier cosine transform. For n = 2, you will get the zero order Hankel Transform. For higher n, see math.stackexchange.com/a/3007081/441161
– Andy Walls
Nov 28 at 19:36