Mixing
Fourier Analysis - Functions on a Circle
$begingroup$
In general, if a continuous function $g(x)$ is defined on the interval $[-pi,pi]$, can I say it is definitely possible to extend this function to be a $2pi$-period function? I think we need to make sure the endpoints match? Functions are $2pi$-period on the real line are also called functions on a circle in Fourier analysis. But what about $$g(x) = x \on [-pi,pi]$$
This is clearly continuous, but $g(-pi) ne g(pi)$. How can I still extend this function to a continuous $2pi$-period function? Or do I have some conceptual misunderstandings?
Thanks so much for your help.
fourier-analysis fourier-series periodic-functions
asked Feb 2 at 21:39
xf16xf16
937
$endgroup$
add a comment |
$begingroup$
In general, if a continuous function $g(x)$ is defined on the interval $[-pi,pi]$, can I say it is definitely possible to extend this function to be a $2pi$-period function? I think we need to make sure the endpoints match? Functions are $2pi$-period on the real line are also called functions on a circle in Fourier analysis. But what about $$g(x) = x \on [-pi,pi]$$
This is clearly continuous, but $g(-pi) ne g(pi)$. How can I still extend this function to a continuous $2pi$-period function? Or do I have some conceptual misunderstandings?
Thanks so much for your help.
fourier-analysis fourier-series periodic-functions
asked Feb 2 at 21:39
xf16xf16
937
$endgroup$
add a comment |
$begingroup$
In general, if a continuous function $g(x)$ is defined on the interval $[-pi,pi]$, can I say it is definitely possible to extend this function to be a $2pi$-period function? I think we need to make sure the endpoints match? Functions are $2pi$-period on the real line are also called functions on a circle in Fourier analysis. But what about $$g(x) = x \on [-pi,pi]$$
This is clearly continuous, but $g(-pi) ne g(pi)$. How can I still extend this function to a continuous $2pi$-period function? Or do I have some conceptual misunderstandings?
Thanks so much for your help.
fourier-analysis fourier-series periodic-functions
asked Feb 2 at 21:39
xf16xf16
937
$endgroup$
In general, if a continuous function $g(x)$ is defined on the interval $[-pi,pi]$, can I say it is definitely possible to extend this function to be a $2pi$-period function? I think we need to make sure the endpoints match? Functions are $2pi$-period on the real line are also called functions on a circle in Fourier analysis. But what about $$g(x) = x \on [-pi,pi]$$
This is clearly continuous, but $g(-pi) ne g(pi)$. How can I still extend this function to a continuous $2pi$-period function? Or do I have some conceptual misunderstandings?
Thanks so much for your help.
fourier-analysis fourier-series periodic-functions
fourier-analysis fourier-series periodic-functions
asked Feb 2 at 21:39
xf16xf16
937
asked Feb 2 at 21:39
xf16xf16
937
asked Feb 2 at 21:39
xf16xf16
937
asked Feb 2 at 21:39
xf16xf16
937
asked Feb 2 at 21:39
xf16xf16
937
937
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If the values at the endpoints don't match, you can't extend to a continuous function. But you can perform Fourier analysis even with functions admitting some discontinuities. Pretty much any periodic function which is not completely insane admits a Fourier expansion (of course the type and rate of convergence depends on regularity of the function; if you insist on some particular type of convergence, e.g. uniform convergence, then my statement is strictly speaking false).
answered Feb 2 at 21:46
BlazejBlazej
1,632620
$endgroup$
$begingroup$
Hi Blazej, is that possible we can use change of variables to change the domain from $[-pi, pi]$ to $[-pi/2, pi/2]$ and then say we can extend this function to a new function $F(x)$ such that in $[-pi/2, pi/2]$ $F(x)$ is just equal to the same values and we require $F(x)$ to be continuous and $F(pi) = F(-pi)$. Then from there, we extend $F(x)$ to be $2pi$-periodic?
$endgroup$
– xf16
Feb 2 at 21:58
$begingroup$
Dear xf16, why do you want to insist on continuity?Given your input on $[-pi,pi]$ and assumption of periodicity you can define your function ambigously everywhere, except for $pi, 3 pi$ etc. Whatever value you choose at this point, Fourier series will be unchanged. Actually the natural choice would be the arithmetic average of the two values, which is $0$ in your case. Then Fourier series converges pointwise.
$endgroup$
– Blazej
Feb 2 at 22:20
$begingroup$
Hi Blazej, the reason I hope for continuity is that I kind of want to extend to a function which is continuous on the circle so that I can use Fejer's theorem on this function to say that there exists a trigonometric polynomial to uniformly approximate this function.
$endgroup$
– xf16
Feb 2 at 22:25
$begingroup$
This is impossible if you want to approximate the function on the whole domain. If you extend to a discontinuous function you will get a uniform approximation on compact subsets not including the points of discontinuity. At the point of discontinuity Fourier series converges to the arithmetic average of the two values. Moreover this convergence is only pointwise, not uniform. Look up Gibbs phenomenon.
$endgroup$
– Blazej
Feb 2 at 23:10
add a comment |
$begingroup$
If $g(-pi) ne g(pi)$, then you cannot extend to a $2pi$ periodic function because a $2pi$ periodic function $h$ on $mathbb{R}$ is required to satisfy $h(x)=h(x+2pi)$ for all $x$, and your $g$ does not satisfy that for $x=-pi$. You can extend any function defined on $[-pi,pi)$ or on $(-pi,pi]$ to a $2pi$ periodic function on $mathbb{R}$, but not if it is defined on $[-pi,pi]$ with $h(-pi)ne h(pi)$.
answered Feb 3 at 4:15
DisintegratingByPartsDisintegratingByParts
60.4k42681
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
If the values at the endpoints don't match, you can't extend to a continuous function. But you can perform Fourier analysis even with functions admitting some discontinuities. Pretty much any periodic function which is not completely insane admits a Fourier expansion (of course the type and rate of convergence depends on regularity of the function; if you insist on some particular type of convergence, e.g. uniform convergence, then my statement is strictly speaking false).
answered Feb 2 at 21:46
BlazejBlazej
1,632620
$endgroup$
$begingroup$
Hi Blazej, is that possible we can use change of variables to change the domain from $[-pi, pi]$ to $[-pi/2, pi/2]$ and then say we can extend this function to a new function $F(x)$ such that in $[-pi/2, pi/2]$ $F(x)$ is just equal to the same values and we require $F(x)$ to be continuous and $F(pi) = F(-pi)$. Then from there, we extend $F(x)$ to be $2pi$-periodic?
$endgroup$
– xf16
Feb 2 at 21:58
$begingroup$
Dear xf16, why do you want to insist on continuity?Given your input on $[-pi,pi]$ and assumption of periodicity you can define your function ambigously everywhere, except for $pi, 3 pi$ etc. Whatever value you choose at this point, Fourier series will be unchanged. Actually the natural choice would be the arithmetic average of the two values, which is $0$ in your case. Then Fourier series converges pointwise.
$endgroup$
– Blazej
Feb 2 at 22:20
$begingroup$
Hi Blazej, the reason I hope for continuity is that I kind of want to extend to a function which is continuous on the circle so that I can use Fejer's theorem on this function to say that there exists a trigonometric polynomial to uniformly approximate this function.
$endgroup$
– xf16
Feb 2 at 22:25
$begingroup$
This is impossible if you want to approximate the function on the whole domain. If you extend to a discontinuous function you will get a uniform approximation on compact subsets not including the points of discontinuity. At the point of discontinuity Fourier series converges to the arithmetic average of the two values. Moreover this convergence is only pointwise, not uniform. Look up Gibbs phenomenon.
$endgroup$
– Blazej
Feb 2 at 23:10
add a comment |
$begingroup$
If the values at the endpoints don't match, you can't extend to a continuous function. But you can perform Fourier analysis even with functions admitting some discontinuities. Pretty much any periodic function which is not completely insane admits a Fourier expansion (of course the type and rate of convergence depends on regularity of the function; if you insist on some particular type of convergence, e.g. uniform convergence, then my statement is strictly speaking false).
answered Feb 2 at 21:46
BlazejBlazej
1,632620
$endgroup$
$begingroup$
Hi Blazej, is that possible we can use change of variables to change the domain from $[-pi, pi]$ to $[-pi/2, pi/2]$ and then say we can extend this function to a new function $F(x)$ such that in $[-pi/2, pi/2]$ $F(x)$ is just equal to the same values and we require $F(x)$ to be continuous and $F(pi) = F(-pi)$. Then from there, we extend $F(x)$ to be $2pi$-periodic?
$endgroup$
– xf16
Feb 2 at 21:58
$begingroup$
Dear xf16, why do you want to insist on continuity?Given your input on $[-pi,pi]$ and assumption of periodicity you can define your function ambigously everywhere, except for $pi, 3 pi$ etc. Whatever value you choose at this point, Fourier series will be unchanged. Actually the natural choice would be the arithmetic average of the two values, which is $0$ in your case. Then Fourier series converges pointwise.
$endgroup$
– Blazej
Feb 2 at 22:20
$begingroup$
Hi Blazej, the reason I hope for continuity is that I kind of want to extend to a function which is continuous on the circle so that I can use Fejer's theorem on this function to say that there exists a trigonometric polynomial to uniformly approximate this function.
$endgroup$
– xf16
Feb 2 at 22:25
$begingroup$
This is impossible if you want to approximate the function on the whole domain. If you extend to a discontinuous function you will get a uniform approximation on compact subsets not including the points of discontinuity. At the point of discontinuity Fourier series converges to the arithmetic average of the two values. Moreover this convergence is only pointwise, not uniform. Look up Gibbs phenomenon.
$endgroup$
– Blazej
Feb 2 at 23:10
add a comment |
$begingroup$
If the values at the endpoints don't match, you can't extend to a continuous function. But you can perform Fourier analysis even with functions admitting some discontinuities. Pretty much any periodic function which is not completely insane admits a Fourier expansion (of course the type and rate of convergence depends on regularity of the function; if you insist on some particular type of convergence, e.g. uniform convergence, then my statement is strictly speaking false).
answered Feb 2 at 21:46
BlazejBlazej
1,632620
$endgroup$
If the values at the endpoints don't match, you can't extend to a continuous function. But you can perform Fourier analysis even with functions admitting some discontinuities. Pretty much any periodic function which is not completely insane admits a Fourier expansion (of course the type and rate of convergence depends on regularity of the function; if you insist on some particular type of convergence, e.g. uniform convergence, then my statement is strictly speaking false).
answered Feb 2 at 21:46
BlazejBlazej
1,632620
answered Feb 2 at 21:46
BlazejBlazej
1,632620
answered Feb 2 at 21:46
BlazejBlazej
1,632620
answered Feb 2 at 21:46
BlazejBlazej
1,632620
1,632620
$begingroup$
Hi Blazej, is that possible we can use change of variables to change the domain from $[-pi, pi]$ to $[-pi/2, pi/2]$ and then say we can extend this function to a new function $F(x)$ such that in $[-pi/2, pi/2]$ $F(x)$ is just equal to the same values and we require $F(x)$ to be continuous and $F(pi) = F(-pi)$. Then from there, we extend $F(x)$ to be $2pi$-periodic?
$endgroup$
– xf16
Feb 2 at 21:58
$begingroup$
Dear xf16, why do you want to insist on continuity?Given your input on $[-pi,pi]$ and assumption of periodicity you can define your function ambigously everywhere, except for $pi, 3 pi$ etc. Whatever value you choose at this point, Fourier series will be unchanged. Actually the natural choice would be the arithmetic average of the two values, which is $0$ in your case. Then Fourier series converges pointwise.
$endgroup$
– Blazej
Feb 2 at 22:20
$begingroup$
Hi Blazej, the reason I hope for continuity is that I kind of want to extend to a function which is continuous on the circle so that I can use Fejer's theorem on this function to say that there exists a trigonometric polynomial to uniformly approximate this function.
$endgroup$
– xf16
Feb 2 at 22:25
$begingroup$
This is impossible if you want to approximate the function on the whole domain. If you extend to a discontinuous function you will get a uniform approximation on compact subsets not including the points of discontinuity. At the point of discontinuity Fourier series converges to the arithmetic average of the two values. Moreover this convergence is only pointwise, not uniform. Look up Gibbs phenomenon.
$endgroup$
– Blazej
Feb 2 at 23:10
add a comment |
$begingroup$
Hi Blazej, is that possible we can use change of variables to change the domain from $[-pi, pi]$ to $[-pi/2, pi/2]$ and then say we can extend this function to a new function $F(x)$ such that in $[-pi/2, pi/2]$ $F(x)$ is just equal to the same values and we require $F(x)$ to be continuous and $F(pi) = F(-pi)$. Then from there, we extend $F(x)$ to be $2pi$-periodic?
$endgroup$
– xf16
Feb 2 at 21:58
$begingroup$
Dear xf16, why do you want to insist on continuity?Given your input on $[-pi,pi]$ and assumption of periodicity you can define your function ambigously everywhere, except for $pi, 3 pi$ etc. Whatever value you choose at this point, Fourier series will be unchanged. Actually the natural choice would be the arithmetic average of the two values, which is $0$ in your case. Then Fourier series converges pointwise.
$endgroup$
– Blazej
Feb 2 at 22:20
$begingroup$
Hi Blazej, the reason I hope for continuity is that I kind of want to extend to a function which is continuous on the circle so that I can use Fejer's theorem on this function to say that there exists a trigonometric polynomial to uniformly approximate this function.
$endgroup$
– xf16
Feb 2 at 22:25
$begingroup$
This is impossible if you want to approximate the function on the whole domain. If you extend to a discontinuous function you will get a uniform approximation on compact subsets not including the points of discontinuity. At the point of discontinuity Fourier series converges to the arithmetic average of the two values. Moreover this convergence is only pointwise, not uniform. Look up Gibbs phenomenon.
$endgroup$
– Blazej
Feb 2 at 23:10
$begingroup$
Hi Blazej, is that possible we can use change of variables to change the domain from $[-pi, pi]$ to $[-pi/2, pi/2]$ and then say we can extend this function to a new function $F(x)$ such that in $[-pi/2, pi/2]$ $F(x)$ is just equal to the same values and we require $F(x)$ to be continuous and $F(pi) = F(-pi)$. Then from there, we extend $F(x)$ to be $2pi$-periodic?
$endgroup$
– xf16
Feb 2 at 21:58
$begingroup$
Hi Blazej, is that possible we can use change of variables to change the domain from $[-pi, pi]$ to $[-pi/2, pi/2]$ and then say we can extend this function to a new function $F(x)$ such that in $[-pi/2, pi/2]$ $F(x)$ is just equal to the same values and we require $F(x)$ to be continuous and $F(pi) = F(-pi)$. Then from there, we extend $F(x)$ to be $2pi$-periodic?
$endgroup$
– xf16
Feb 2 at 21:58
$begingroup$
Dear xf16, why do you want to insist on continuity?Given your input on $[-pi,pi]$ and assumption of periodicity you can define your function ambigously everywhere, except for $pi, 3 pi$ etc. Whatever value you choose at this point, Fourier series will be unchanged. Actually the natural choice would be the arithmetic average of the two values, which is $0$ in your case. Then Fourier series converges pointwise.
$endgroup$
– Blazej
Feb 2 at 22:20
$begingroup$
Dear xf16, why do you want to insist on continuity?Given your input on $[-pi,pi]$ and assumption of periodicity you can define your function ambigously everywhere, except for $pi, 3 pi$ etc. Whatever value you choose at this point, Fourier series will be unchanged. Actually the natural choice would be the arithmetic average of the two values, which is $0$ in your case. Then Fourier series converges pointwise.
$endgroup$
– Blazej
Feb 2 at 22:20
$begingroup$
Hi Blazej, the reason I hope for continuity is that I kind of want to extend to a function which is continuous on the circle so that I can use Fejer's theorem on this function to say that there exists a trigonometric polynomial to uniformly approximate this function.
$endgroup$
– xf16
Feb 2 at 22:25
$begingroup$
Hi Blazej, the reason I hope for continuity is that I kind of want to extend to a function which is continuous on the circle so that I can use Fejer's theorem on this function to say that there exists a trigonometric polynomial to uniformly approximate this function.
$endgroup$
– xf16
Feb 2 at 22:25
$begingroup$
This is impossible if you want to approximate the function on the whole domain. If you extend to a discontinuous function you will get a uniform approximation on compact subsets not including the points of discontinuity. At the point of discontinuity Fourier series converges to the arithmetic average of the two values. Moreover this convergence is only pointwise, not uniform. Look up Gibbs phenomenon.
$endgroup$
– Blazej
Feb 2 at 23:10
$begingroup$
This is impossible if you want to approximate the function on the whole domain. If you extend to a discontinuous function you will get a uniform approximation on compact subsets not including the points of discontinuity. At the point of discontinuity Fourier series converges to the arithmetic average of the two values. Moreover this convergence is only pointwise, not uniform. Look up Gibbs phenomenon.
$endgroup$
– Blazej
Feb 2 at 23:10
add a comment |
$begingroup$
If $g(-pi) ne g(pi)$, then you cannot extend to a $2pi$ periodic function because a $2pi$ periodic function $h$ on $mathbb{R}$ is required to satisfy $h(x)=h(x+2pi)$ for all $x$, and your $g$ does not satisfy that for $x=-pi$. You can extend any function defined on $[-pi,pi)$ or on $(-pi,pi]$ to a $2pi$ periodic function on $mathbb{R}$, but not if it is defined on $[-pi,pi]$ with $h(-pi)ne h(pi)$.
answered Feb 3 at 4:15
DisintegratingByPartsDisintegratingByParts
60.4k42681
$endgroup$
add a comment |
$begingroup$
If $g(-pi) ne g(pi)$, then you cannot extend to a $2pi$ periodic function because a $2pi$ periodic function $h$ on $mathbb{R}$ is required to satisfy $h(x)=h(x+2pi)$ for all $x$, and your $g$ does not satisfy that for $x=-pi$. You can extend any function defined on $[-pi,pi)$ or on $(-pi,pi]$ to a $2pi$ periodic function on $mathbb{R}$, but not if it is defined on $[-pi,pi]$ with $h(-pi)ne h(pi)$.
answered Feb 3 at 4:15
DisintegratingByPartsDisintegratingByParts
60.4k42681
$endgroup$
add a comment |
$begingroup$
If $g(-pi) ne g(pi)$, then you cannot extend to a $2pi$ periodic function because a $2pi$ periodic function $h$ on $mathbb{R}$ is required to satisfy $h(x)=h(x+2pi)$ for all $x$, and your $g$ does not satisfy that for $x=-pi$. You can extend any function defined on $[-pi,pi)$ or on $(-pi,pi]$ to a $2pi$ periodic function on $mathbb{R}$, but not if it is defined on $[-pi,pi]$ with $h(-pi)ne h(pi)$.
answered Feb 3 at 4:15
DisintegratingByPartsDisintegratingByParts
60.4k42681
$endgroup$
If $g(-pi) ne g(pi)$, then you cannot extend to a $2pi$ periodic function because a $2pi$ periodic function $h$ on $mathbb{R}$ is required to satisfy $h(x)=h(x+2pi)$ for all $x$, and your $g$ does not satisfy that for $x=-pi$. You can extend any function defined on $[-pi,pi)$ or on $(-pi,pi]$ to a $2pi$ periodic function on $mathbb{R}$, but not if it is defined on $[-pi,pi]$ with $h(-pi)ne h(pi)$.
answered Feb 3 at 4:15
DisintegratingByPartsDisintegratingByParts
60.4k42681
answered Feb 3 at 4:15
DisintegratingByPartsDisintegratingByParts
60.4k42681
answered Feb 3 at 4:15
DisintegratingByPartsDisintegratingByParts
60.4k42681
answered Feb 3 at 4:15
DisintegratingByPartsDisintegratingByParts
60.4k42681
60.4k42681
add a comment |
add a comment |
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