Mixing
Error analysis of approximating Fourier transforms
$begingroup$
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 ?
fourier-analysis fourier-transform approximation-theory approximate-integration
asked Jan 18 at 8:14
biryanibiryani
268413
$endgroup$
add a comment |
$begingroup$
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 ?
fourier-analysis fourier-transform approximation-theory approximate-integration
asked Jan 18 at 8:14
biryanibiryani
268413
$endgroup$
add a comment |
$begingroup$
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 ?
fourier-analysis fourier-transform approximation-theory approximate-integration
asked Jan 18 at 8:14
biryanibiryani
268413
$endgroup$
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
fourier-analysis fourier-transform approximation-theory approximate-integration
asked Jan 18 at 8:14
biryanibiryani
268413
asked Jan 18 at 8:14
biryanibiryani
268413
asked Jan 18 at 8:14
biryanibiryani
268413
asked Jan 18 at 8:14
biryanibiryani
268413
asked Jan 18 at 8:14
biryanibiryani
268413
268413
add a comment |
add a comment |
1 Answer
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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.
answered Jan 20 at 0:38
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Jan 20 at 0:38
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 20 at 0:38
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 20 at 0:38
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
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.
answered Jan 20 at 0:38
DisintegratingByPartsDisintegratingByParts
59.5k42580
answered Jan 20 at 0:38
DisintegratingByPartsDisintegratingByParts
59.5k42580
answered Jan 20 at 0:38
DisintegratingByPartsDisintegratingByParts
59.5k42580
answered Jan 20 at 0:38
DisintegratingByPartsDisintegratingByParts
59.5k42580
59.5k42580
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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