[math-fun] Periodic functions & Fourier transforms ?
It is the smoothness. Any fourier series that converges everywhere can be regarded as a maclaurin series of an function that's analytic and bounded within the unit disk (fourier is when restrict to the disk boundary). Well anyhow for the right kind of fourier series. Analytic functions are described by a countable number of coefficients. It is a delusion to think there is an uncountable number of degrees of freedom. So to make that clear, what we want is some theorem saying: Specifying the value of the analytic function at a countable set of points within the disk, will suffice to specify the function. Is such a theorem known? I can't think of one off the top of my head. But here, voila, I proved it: use monte carlo integration to compute the maclaurin series coefficients. With probability=1, any particular series coefficient will get computed in this way with error that in the infinite-points limit goes to zero. That proves that a countable set of points must exist, that works to compute any finite-numbered coefficient exactly; and hence, theorem proven. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
Analytic functions are described by a countable number of coefficients. It is a delusion to think there is an uncountable number of degrees of freedom.
So to make that clear, what we want is some theorem saying: Specifying the value of the analytic function at a countable set of points within the disk, will suffice to specify the function.
Is such a theorem known? I can't think of one off the top of my head. But here, voila, I proved it:
We can make it much easier than that, while proving a stronger theorem. Theorem: a (merely) continuous function from R^n to R^n is determined by countably many numbers. Proof: specify its values at rational points (of which there are only countably many); its values elsewhere are determined by continuity. -- g
At 03:26 AM 12/21/2015, Gareth McCaughan wrote: Theorem: a (merely) continuous function from R^n to R^n is determined
by countably many numbers.
Proof: specify its values at rational points (of which there are only countably many); its values elsewhere are determined by continuity.
That's a cool theorem! Does it have a name?
Don't you need the set of countable points to be dense in R^n for this to work? Cheers Seb On Dec 21, 2015 12:27 PM, "Gareth McCaughan" <gareth.mccaughan@pobox.com> wrote:
Analytic functions are described by a countable number of coefficients.
It is a delusion to think there is an uncountable number of degrees of freedom.
So to make that clear, what we want is some theorem saying: Specifying the value of the analytic function at a countable set of points within the disk, will suffice to specify the function.
Is such a theorem known? I can't think of one off the top of my head. But here, voila, I proved it:
We can make it much easier than that, while proving a stronger theorem.
Theorem: a (merely) continuous function from R^n to R^n is determined by countably many numbers.
Proof: specify its values at rational points (of which there are only countably many); its values elsewhere are determined by continuity.
-- g
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An analytic function is completely determined by its values on a countable set of points that has a limit point, provided it is analytic at that limit point. -- Gene From: Gareth McCaughan <gareth.mccaughan@pobox.com> To: math-fun@mailman.xmission.com Sent: Monday, December 21, 2015 3:26 AM Subject: Re: [math-fun] Periodic functions & Fourier transforms ?
Analytic functions are described by a countable number of coefficients. It is a delusion to think there is an uncountable number of degrees of freedom.
So to make that clear, what we want is some theorem saying: Specifying the value of the analytic function at a countable set of points within the disk, will suffice to specify the function.
Is such a theorem known? I can't think of one off the top of my head. But here, voila, I proved it:
We can make it much easier than that, while proving a stronger theorem. Theorem: a (merely) continuous function from R^n to R^n is determined by countably many numbers. Proof: specify its values at rational points (of which there are only countably many); its values elsewhere are determined by continuity. -- g
A variant of Jim Propp's original question, as I understood it, is: "Let X=\R/\Z. How can a person give an intuitive explanation of the Hilbert space isomorphism L^2(X) \to \ell^2(X) when the frequency space \ell^2 seems so much smaller than the physical space L^2?" I don't think that there is an easy answer, but coming to understand the issues is rewarding. This is an isomorphism of Hilbert spaces, not of sets, so a person needs an intuitive feel for the topologies arising from L^2-norm and the \ell^2-norm, neither of which is the (continuous or discrete) C^0-norm or any C^k-norm. L^2-functions don't have pointwise values, and this is a very hard sell. L^2(X) is naturally the completion of C^\infty(X) under the L^2-norm, but this is also a hard sell because people seem to find the more ad hoc construction of L^2(X) as {square-integrable Lebesgues}/{equality a.e.} more tangible. Bessel/Parseval statements about Fourier series representation don't say what a person hopes because they assume a basis, and the existence of the basis (or the fact that the "obvious" basis really is one) is the real issue. The old-fashioned proof-of-basis via the "Fejer kernel" is (imho) unexplanatory. A clear argument interweaves the Banach spaces C^k(X) with Sobolev spaces, which are Hilbert even though the C^k-spaces aren't, and then argues that the spectral theorem basis of differentiation-eigenfunctions in L^2 (L^2 functions can be differentiated despite not having pointwise values, using a Friedrichs extension of differentiation on C^\infty, even though that operator is "unbounded" [discontinuous])... a clear argument interweaves and then argues that the spectral theorem basis of differentiation-eigenfunctions must lie in the littlest space C^\infty, and in that space we know what they are: the oscillations. I doubt that any of this is helpful to Jim. On the other hand, these ideas scale to many spaces X, such as Euclidean space (where the basis of Laplacian eigenfunctions isn't the oscillations, which aren't L^2, but rather comprises the Gaussian times harmonic polynomials), spheres, and quotients of complex upper half spaces, where we are talking about automorphic forms and nobody knows a basis of cuspforms but we know that there is one. Jerry Shurman
It would be helpful to the non-analysts amongst us to indicate where this argument might be explained in greater detail --- though I have a nasty feeling that considerable expansion may be required! WFL On 12/21/15, Jerry Shurman <jerry@reed.edu> wrote:
A variant of Jim Propp's original question, as I understood it, is:
"Let X=\R/\Z. How can a person give an intuitive explanation of the Hilbert space isomorphism L^2(X) \to \ell^2(X) when the frequency space \ell^2 seems so much smaller than the physical space L^2?"
I don't think that there is an easy answer, but coming to understand the issues is rewarding. This is an isomorphism of Hilbert spaces, not of sets, so a person needs an intuitive feel for the topologies arising from L^2-norm and the \ell^2-norm, neither of which is the (continuous or discrete) C^0-norm or any C^k-norm. L^2-functions don't have pointwise values, and this is a very hard sell. L^2(X) is naturally the completion of C^\infty(X) under the L^2-norm, but this is also a hard sell because people seem to find the more ad hoc construction of L^2(X) as {square-integrable Lebesgues}/{equality a.e.} more tangible. Bessel/Parseval statements about Fourier series representation don't say what a person hopes because they assume a basis, and the existence of the basis (or the fact that the "obvious" basis really is one) is the real issue. The old-fashioned proof-of-basis via the "Fejer kernel" is (imho) unexplanatory. A clear argument interweaves the Banach spaces C^k(X) with Sobolev spaces, which are Hilbert even though the C^k-spaces aren't, and then argues that the spectral theorem basis of differentiation-eigenfunctions in L^2 (L^2 functions can be differentiated despite not having pointwise values, using a Friedrichs extension of differentiation on C^\infty, even though that operator is "unbounded" [discontinuous])... a clear argument interweaves and then argues that the spectral theorem basis of differentiation-eigenfunctions must lie in the littlest space C^\infty, and in that space we know what they are: the oscillations.
I doubt that any of this is helpful to Jim. On the other hand, these ideas scale to many spaces X, such as Euclidean space (where the basis of Laplacian eigenfunctions isn't the oscillations, which aren't L^2, but rather comprises the Gaussian times harmonic polynomials), spheres, and quotients of complex upper half spaces, where we are talking about automorphic forms and nobody knows a basis of cuspforms but we know that there is one.
Jerry Shurman
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On Tue, 22 Dec 2015, Fred Lunnon wrote:
It would be helpful to the non-analysts amongst us to indicate where this argument might be explained in greater detail --- though I have a nasty feeling that considerable expansion may be required! WFL
For the modern part of it, Paul Garrett's web site, especially the writeups from his courses. The Friedrichs extension idea is incredible, I think. The Fejer kernel argument that the oscillations are a basis of L^2(\R/\Z) should be in any respectable undergraduate text on Fourier analysis. For the complex analytic arguments being made in this thread, a \Z-periodic function on the complex upper half plane is analytic on the punctured disk under the change of variables q=e^{2\pi iz}, and so it has a representation \sum_{n\in\Z}a_n q^n, though the representation needn't be a power series or even a Laurent series. But this is a much smaller setting than L^2.
So to make that clear, what we want is some theorem saying: Specifying the value of the analytic function at a countable set of points within the disk, will suffice to specify the function.
Is such a theorem known? I can't think of one off the top of my head. But here, voila, I proved it: use monte carlo integration to compute the maclaurin series coefficients. With probability=1, any particular series coefficient will get computed in this way with error that in the infinite-points limit goes to zero. That proves that a countable set of points must exist, that works to compute any finite-numbered coefficient exactly; and hence, theorem proven.
I don't follow your proof, but it must have a flaw, because the theorem is false. The exponential function and the constant function 1 agree at 2k * pi * i, a countable number of points. Choose some simply connected region containing all of these, and use the Riemann mapping theorem to get two analytic functions on the (open) disk that agree at countably many points. One of these functions will not extend to an analytic function on the closed disk, because the actual theorem is that if two complex analytic functions agree on a sequence of points that has a limit, and at that limit point, and are analytic at that limit point, they are identical. I don't remember the proof, but I think it was covered in my intro complex analysis course, so it doesn't use a lot of machinery. Andy Latto andy.latto@pobox.com
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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participants (8)
-
Andy Latto -
Eugene Salamin -
Fred Lunnon -
Gareth McCaughan -
Henry Baker -
Jerry Shurman -
Seb Perez-D -
Warren D Smith