Expansion of function in polar coordinates












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I'd like to expand a function in polar coordinates to something that splits radius and angle



$f(r,theta)=sum_i A_i(r)B_i(theta)$



I've found some hints on the internet by the name of polar Fourier transform, but I didn't find a Wikipedia page or a good explanation. What is the name for such a decomposition and how to find a basic description?



Is using Bessel function the only way to find such a factor decomposition?



Is it possible (or useful) to write this operation in complex number representation for the 2D coordinates?










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  • $begingroup$
    I like how this is explained in the book on harmonic analysis by Stein and Weiss, chapter on spherical harmonics.
    $endgroup$
    – Giuseppe Negro
    Oct 30 '18 at 10:02
















3












$begingroup$


I'd like to expand a function in polar coordinates to something that splits radius and angle



$f(r,theta)=sum_i A_i(r)B_i(theta)$



I've found some hints on the internet by the name of polar Fourier transform, but I didn't find a Wikipedia page or a good explanation. What is the name for such a decomposition and how to find a basic description?



Is using Bessel function the only way to find such a factor decomposition?



Is it possible (or useful) to write this operation in complex number representation for the 2D coordinates?










share|cite|improve this question











$endgroup$












  • $begingroup$
    I like how this is explained in the book on harmonic analysis by Stein and Weiss, chapter on spherical harmonics.
    $endgroup$
    – Giuseppe Negro
    Oct 30 '18 at 10:02














3












3








3


3



$begingroup$


I'd like to expand a function in polar coordinates to something that splits radius and angle



$f(r,theta)=sum_i A_i(r)B_i(theta)$



I've found some hints on the internet by the name of polar Fourier transform, but I didn't find a Wikipedia page or a good explanation. What is the name for such a decomposition and how to find a basic description?



Is using Bessel function the only way to find such a factor decomposition?



Is it possible (or useful) to write this operation in complex number representation for the 2D coordinates?










share|cite|improve this question











$endgroup$




I'd like to expand a function in polar coordinates to something that splits radius and angle



$f(r,theta)=sum_i A_i(r)B_i(theta)$



I've found some hints on the internet by the name of polar Fourier transform, but I didn't find a Wikipedia page or a good explanation. What is the name for such a decomposition and how to find a basic description?



Is using Bessel function the only way to find such a factor decomposition?



Is it possible (or useful) to write this operation in complex number representation for the 2D coordinates?







fourier-series






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edited Jan 27 '15 at 14:59









Arjang

5,57362363




5,57362363










asked Jan 27 '15 at 14:43









GerenukGerenuk

1,0481920




1,0481920












  • $begingroup$
    I like how this is explained in the book on harmonic analysis by Stein and Weiss, chapter on spherical harmonics.
    $endgroup$
    – Giuseppe Negro
    Oct 30 '18 at 10:02


















  • $begingroup$
    I like how this is explained in the book on harmonic analysis by Stein and Weiss, chapter on spherical harmonics.
    $endgroup$
    – Giuseppe Negro
    Oct 30 '18 at 10:02
















$begingroup$
I like how this is explained in the book on harmonic analysis by Stein and Weiss, chapter on spherical harmonics.
$endgroup$
– Giuseppe Negro
Oct 30 '18 at 10:02




$begingroup$
I like how this is explained in the book on harmonic analysis by Stein and Weiss, chapter on spherical harmonics.
$endgroup$
– Giuseppe Negro
Oct 30 '18 at 10:02










1 Answer
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1












$begingroup$

Consider the 2D Fourier transform in rectangular coordinates,



$$
F(u,v) = int_{-infty}^{infty} left[ int_{-infty}^{infty} f(x,y)e^{2ipi ux}dx right]e^{2ipi vy}dy
$$
One can take 1D FT in X direction then in Y direction. But for Polar coordinates the expression changes,
Let , $ux+vy = rho r cos (phi-theta)$, then
$$
F(u,v) = int_{0}^{2pi} int_{0}^{infty} f(r,theta)e^{2ipirho r cos(phi- theta)}rdr dtheta
$$
Now the function $f(r,theta) $ has seperability in polar coordinates if it can be written in the form ,
$$f(r,theta)= f_r(r)f_{theta}(theta)$$
Then the function $f(r,theta)$ is a circularly symmetric with $f_{theta}(theta)=1$ , examples include like that of a cylinder etc. Read more about Hankel transform, 2D Polar Fourier transform and functions of this type.






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    protected by user26857 Nov 4 '15 at 22:03



    Thank you for your interest in this question.
    Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).



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    1 Answer
    1






    active

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Consider the 2D Fourier transform in rectangular coordinates,



    $$
    F(u,v) = int_{-infty}^{infty} left[ int_{-infty}^{infty} f(x,y)e^{2ipi ux}dx right]e^{2ipi vy}dy
    $$
    One can take 1D FT in X direction then in Y direction. But for Polar coordinates the expression changes,
    Let , $ux+vy = rho r cos (phi-theta)$, then
    $$
    F(u,v) = int_{0}^{2pi} int_{0}^{infty} f(r,theta)e^{2ipirho r cos(phi- theta)}rdr dtheta
    $$
    Now the function $f(r,theta) $ has seperability in polar coordinates if it can be written in the form ,
    $$f(r,theta)= f_r(r)f_{theta}(theta)$$
    Then the function $f(r,theta)$ is a circularly symmetric with $f_{theta}(theta)=1$ , examples include like that of a cylinder etc. Read more about Hankel transform, 2D Polar Fourier transform and functions of this type.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Consider the 2D Fourier transform in rectangular coordinates,



      $$
      F(u,v) = int_{-infty}^{infty} left[ int_{-infty}^{infty} f(x,y)e^{2ipi ux}dx right]e^{2ipi vy}dy
      $$
      One can take 1D FT in X direction then in Y direction. But for Polar coordinates the expression changes,
      Let , $ux+vy = rho r cos (phi-theta)$, then
      $$
      F(u,v) = int_{0}^{2pi} int_{0}^{infty} f(r,theta)e^{2ipirho r cos(phi- theta)}rdr dtheta
      $$
      Now the function $f(r,theta) $ has seperability in polar coordinates if it can be written in the form ,
      $$f(r,theta)= f_r(r)f_{theta}(theta)$$
      Then the function $f(r,theta)$ is a circularly symmetric with $f_{theta}(theta)=1$ , examples include like that of a cylinder etc. Read more about Hankel transform, 2D Polar Fourier transform and functions of this type.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Consider the 2D Fourier transform in rectangular coordinates,



        $$
        F(u,v) = int_{-infty}^{infty} left[ int_{-infty}^{infty} f(x,y)e^{2ipi ux}dx right]e^{2ipi vy}dy
        $$
        One can take 1D FT in X direction then in Y direction. But for Polar coordinates the expression changes,
        Let , $ux+vy = rho r cos (phi-theta)$, then
        $$
        F(u,v) = int_{0}^{2pi} int_{0}^{infty} f(r,theta)e^{2ipirho r cos(phi- theta)}rdr dtheta
        $$
        Now the function $f(r,theta) $ has seperability in polar coordinates if it can be written in the form ,
        $$f(r,theta)= f_r(r)f_{theta}(theta)$$
        Then the function $f(r,theta)$ is a circularly symmetric with $f_{theta}(theta)=1$ , examples include like that of a cylinder etc. Read more about Hankel transform, 2D Polar Fourier transform and functions of this type.






        share|cite|improve this answer











        $endgroup$



        Consider the 2D Fourier transform in rectangular coordinates,



        $$
        F(u,v) = int_{-infty}^{infty} left[ int_{-infty}^{infty} f(x,y)e^{2ipi ux}dx right]e^{2ipi vy}dy
        $$
        One can take 1D FT in X direction then in Y direction. But for Polar coordinates the expression changes,
        Let , $ux+vy = rho r cos (phi-theta)$, then
        $$
        F(u,v) = int_{0}^{2pi} int_{0}^{infty} f(r,theta)e^{2ipirho r cos(phi- theta)}rdr dtheta
        $$
        Now the function $f(r,theta) $ has seperability in polar coordinates if it can be written in the form ,
        $$f(r,theta)= f_r(r)f_{theta}(theta)$$
        Then the function $f(r,theta)$ is a circularly symmetric with $f_{theta}(theta)=1$ , examples include like that of a cylinder etc. Read more about Hankel transform, 2D Polar Fourier transform and functions of this type.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 6 '17 at 9:36









        Faraad Armwood

        7,5342619




        7,5342619










        answered Aug 2 '15 at 18:37









        Syed Alam AbbasSyed Alam Abbas

        333410




        333410

















            protected by user26857 Nov 4 '15 at 22:03



            Thank you for your interest in this question.
            Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).



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