Error analysis of approximating Fourier transforms












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$begingroup$


Consider the problem of computing the Fourier transform of a function, $f(x).$ $$ hat{f}(k) = int_{-infty}^{infty} dx~ f(x)~ e^{i k x} .$$



Suppose I want to approximate this transform by a discrete, truncated version,
$$ hat{F}_{Delta, ~L}(k) = sum_{n = -L}^L Delta~ f(Delta n)~ e^{i k Delta n} .$$



I want the approximation to work with error $epsilon$ in some interval $[0, k_{max}]$,



$$sup_{k in [0, k_{max}]} ~| hat{f}(k) - hat{F}_{Delta, ~L}(k) | leq epsilon $$



What values of $Delta$, $L$ should I choose to achieve this error. The answer will obviously depend on the properties of $hat{f}(k)$, like how fast it decays. I have seen this done for specific functions but haven't seen any general rigorous result. Since this seems to be a problem with many practical applications it seems unlikely that no one has worked it out. Are there any rigorous results known for this general case ?










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    2












    $begingroup$


    Consider the problem of computing the Fourier transform of a function, $f(x).$ $$ hat{f}(k) = int_{-infty}^{infty} dx~ f(x)~ e^{i k x} .$$



    Suppose I want to approximate this transform by a discrete, truncated version,
    $$ hat{F}_{Delta, ~L}(k) = sum_{n = -L}^L Delta~ f(Delta n)~ e^{i k Delta n} .$$



    I want the approximation to work with error $epsilon$ in some interval $[0, k_{max}]$,



    $$sup_{k in [0, k_{max}]} ~| hat{f}(k) - hat{F}_{Delta, ~L}(k) | leq epsilon $$



    What values of $Delta$, $L$ should I choose to achieve this error. The answer will obviously depend on the properties of $hat{f}(k)$, like how fast it decays. I have seen this done for specific functions but haven't seen any general rigorous result. Since this seems to be a problem with many practical applications it seems unlikely that no one has worked it out. Are there any rigorous results known for this general case ?










    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Consider the problem of computing the Fourier transform of a function, $f(x).$ $$ hat{f}(k) = int_{-infty}^{infty} dx~ f(x)~ e^{i k x} .$$



      Suppose I want to approximate this transform by a discrete, truncated version,
      $$ hat{F}_{Delta, ~L}(k) = sum_{n = -L}^L Delta~ f(Delta n)~ e^{i k Delta n} .$$



      I want the approximation to work with error $epsilon$ in some interval $[0, k_{max}]$,



      $$sup_{k in [0, k_{max}]} ~| hat{f}(k) - hat{F}_{Delta, ~L}(k) | leq epsilon $$



      What values of $Delta$, $L$ should I choose to achieve this error. The answer will obviously depend on the properties of $hat{f}(k)$, like how fast it decays. I have seen this done for specific functions but haven't seen any general rigorous result. Since this seems to be a problem with many practical applications it seems unlikely that no one has worked it out. Are there any rigorous results known for this general case ?










      share|cite|improve this question









      $endgroup$




      Consider the problem of computing the Fourier transform of a function, $f(x).$ $$ hat{f}(k) = int_{-infty}^{infty} dx~ f(x)~ e^{i k x} .$$



      Suppose I want to approximate this transform by a discrete, truncated version,
      $$ hat{F}_{Delta, ~L}(k) = sum_{n = -L}^L Delta~ f(Delta n)~ e^{i k Delta n} .$$



      I want the approximation to work with error $epsilon$ in some interval $[0, k_{max}]$,



      $$sup_{k in [0, k_{max}]} ~| hat{f}(k) - hat{F}_{Delta, ~L}(k) | leq epsilon $$



      What values of $Delta$, $L$ should I choose to achieve this error. The answer will obviously depend on the properties of $hat{f}(k)$, like how fast it decays. I have seen this done for specific functions but haven't seen any general rigorous result. Since this seems to be a problem with many practical applications it seems unlikely that no one has worked it out. Are there any rigorous results known for this general case ?







      fourier-analysis fourier-transform approximation-theory approximate-integration






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      asked Jan 18 at 8:14









      biryanibiryani

      268413




      268413






















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          $begingroup$

          If you assume that $f in L^1$, then approximations are not so terrible because $hat{f}$ is uniformly continuous, with $|hat{f}|_{infty}le |f|_{1}$. So, for $epsilon > 0$, there exists $R > 0$ large enough such that
          $$
          left|hat{f}(k)-frac{1}{sqrt{2pi}}int_{-R}^{R}f(x)e^{-ikx}dxright| \
          le frac{1}{sqrt{2pi}}int_{|u|ge R}|f(u)|du < frac{epsilon}{2},;;; kinmathbb{R}.
          $$

          So the Fourier transform is uniformly approximated by the truncated Fourier series integral on $[-R,R]$. Then you can approximate the Fourier integral over $[-R,R]$ by a discrete sum
          $$
          hat{f}(k)approxsum_{n=-N}^{N-1}frac{1}{sqrt{2pi}}int_{Rn/N}^{R(n+1)/N}f(x)dxe^{-ikRn/N}.
          $$

          I'm not sure if that's the type of approximation you want or not.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So you assume $f$ is continuous ($f(x)=e^{-x^2}(1 - 1_{xin mathbb{Q}})$)
            $endgroup$
            – reuns
            Jan 20 at 2:24












          • $begingroup$
            This is the kind of approximation I use. The question is how does $epsilon$ depend on $R$ and $N$.
            $endgroup$
            – biryani
            Jan 21 at 10:39












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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          If you assume that $f in L^1$, then approximations are not so terrible because $hat{f}$ is uniformly continuous, with $|hat{f}|_{infty}le |f|_{1}$. So, for $epsilon > 0$, there exists $R > 0$ large enough such that
          $$
          left|hat{f}(k)-frac{1}{sqrt{2pi}}int_{-R}^{R}f(x)e^{-ikx}dxright| \
          le frac{1}{sqrt{2pi}}int_{|u|ge R}|f(u)|du < frac{epsilon}{2},;;; kinmathbb{R}.
          $$

          So the Fourier transform is uniformly approximated by the truncated Fourier series integral on $[-R,R]$. Then you can approximate the Fourier integral over $[-R,R]$ by a discrete sum
          $$
          hat{f}(k)approxsum_{n=-N}^{N-1}frac{1}{sqrt{2pi}}int_{Rn/N}^{R(n+1)/N}f(x)dxe^{-ikRn/N}.
          $$

          I'm not sure if that's the type of approximation you want or not.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So you assume $f$ is continuous ($f(x)=e^{-x^2}(1 - 1_{xin mathbb{Q}})$)
            $endgroup$
            – reuns
            Jan 20 at 2:24












          • $begingroup$
            This is the kind of approximation I use. The question is how does $epsilon$ depend on $R$ and $N$.
            $endgroup$
            – biryani
            Jan 21 at 10:39
















          2












          $begingroup$

          If you assume that $f in L^1$, then approximations are not so terrible because $hat{f}$ is uniformly continuous, with $|hat{f}|_{infty}le |f|_{1}$. So, for $epsilon > 0$, there exists $R > 0$ large enough such that
          $$
          left|hat{f}(k)-frac{1}{sqrt{2pi}}int_{-R}^{R}f(x)e^{-ikx}dxright| \
          le frac{1}{sqrt{2pi}}int_{|u|ge R}|f(u)|du < frac{epsilon}{2},;;; kinmathbb{R}.
          $$

          So the Fourier transform is uniformly approximated by the truncated Fourier series integral on $[-R,R]$. Then you can approximate the Fourier integral over $[-R,R]$ by a discrete sum
          $$
          hat{f}(k)approxsum_{n=-N}^{N-1}frac{1}{sqrt{2pi}}int_{Rn/N}^{R(n+1)/N}f(x)dxe^{-ikRn/N}.
          $$

          I'm not sure if that's the type of approximation you want or not.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So you assume $f$ is continuous ($f(x)=e^{-x^2}(1 - 1_{xin mathbb{Q}})$)
            $endgroup$
            – reuns
            Jan 20 at 2:24












          • $begingroup$
            This is the kind of approximation I use. The question is how does $epsilon$ depend on $R$ and $N$.
            $endgroup$
            – biryani
            Jan 21 at 10:39














          2












          2








          2





          $begingroup$

          If you assume that $f in L^1$, then approximations are not so terrible because $hat{f}$ is uniformly continuous, with $|hat{f}|_{infty}le |f|_{1}$. So, for $epsilon > 0$, there exists $R > 0$ large enough such that
          $$
          left|hat{f}(k)-frac{1}{sqrt{2pi}}int_{-R}^{R}f(x)e^{-ikx}dxright| \
          le frac{1}{sqrt{2pi}}int_{|u|ge R}|f(u)|du < frac{epsilon}{2},;;; kinmathbb{R}.
          $$

          So the Fourier transform is uniformly approximated by the truncated Fourier series integral on $[-R,R]$. Then you can approximate the Fourier integral over $[-R,R]$ by a discrete sum
          $$
          hat{f}(k)approxsum_{n=-N}^{N-1}frac{1}{sqrt{2pi}}int_{Rn/N}^{R(n+1)/N}f(x)dxe^{-ikRn/N}.
          $$

          I'm not sure if that's the type of approximation you want or not.






          share|cite|improve this answer









          $endgroup$



          If you assume that $f in L^1$, then approximations are not so terrible because $hat{f}$ is uniformly continuous, with $|hat{f}|_{infty}le |f|_{1}$. So, for $epsilon > 0$, there exists $R > 0$ large enough such that
          $$
          left|hat{f}(k)-frac{1}{sqrt{2pi}}int_{-R}^{R}f(x)e^{-ikx}dxright| \
          le frac{1}{sqrt{2pi}}int_{|u|ge R}|f(u)|du < frac{epsilon}{2},;;; kinmathbb{R}.
          $$

          So the Fourier transform is uniformly approximated by the truncated Fourier series integral on $[-R,R]$. Then you can approximate the Fourier integral over $[-R,R]$ by a discrete sum
          $$
          hat{f}(k)approxsum_{n=-N}^{N-1}frac{1}{sqrt{2pi}}int_{Rn/N}^{R(n+1)/N}f(x)dxe^{-ikRn/N}.
          $$

          I'm not sure if that's the type of approximation you want or not.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 20 at 0:38









          DisintegratingByPartsDisintegratingByParts

          60.5k42681




          60.5k42681












          • $begingroup$
            So you assume $f$ is continuous ($f(x)=e^{-x^2}(1 - 1_{xin mathbb{Q}})$)
            $endgroup$
            – reuns
            Jan 20 at 2:24












          • $begingroup$
            This is the kind of approximation I use. The question is how does $epsilon$ depend on $R$ and $N$.
            $endgroup$
            – biryani
            Jan 21 at 10:39


















          • $begingroup$
            So you assume $f$ is continuous ($f(x)=e^{-x^2}(1 - 1_{xin mathbb{Q}})$)
            $endgroup$
            – reuns
            Jan 20 at 2:24












          • $begingroup$
            This is the kind of approximation I use. The question is how does $epsilon$ depend on $R$ and $N$.
            $endgroup$
            – biryani
            Jan 21 at 10:39
















          $begingroup$
          So you assume $f$ is continuous ($f(x)=e^{-x^2}(1 - 1_{xin mathbb{Q}})$)
          $endgroup$
          – reuns
          Jan 20 at 2:24






          $begingroup$
          So you assume $f$ is continuous ($f(x)=e^{-x^2}(1 - 1_{xin mathbb{Q}})$)
          $endgroup$
          – reuns
          Jan 20 at 2:24














          $begingroup$
          This is the kind of approximation I use. The question is how does $epsilon$ depend on $R$ and $N$.
          $endgroup$
          – biryani
          Jan 21 at 10:39




          $begingroup$
          This is the kind of approximation I use. The question is how does $epsilon$ depend on $R$ and $N$.
          $endgroup$
          – biryani
          Jan 21 at 10:39


















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