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Topic: Prime power moments (Read 1035 times) |
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Aryabhatta
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Prime power moments
« on: Jun 18th, 2008, 7:26pm » |
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if f:[0,1] -> R is a continuous function such that
Integral {0 to 1} of f(x) dx = 0
and
Integral {0 to 1} of xp f(x) dx = 0
for all primes p, show that f(x) = 0 for all x in [0,1].
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| « Last Edit: Jun 20th, 2008, 1:51am by Aryabhatta » |
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Aryabhatta
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Re: Prime power moments
« Reply #1 on: Jun 20th, 2008, 1:24am » |
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Sorry, this one might be unfairly hard. The only proof I know uses a theorem from functional analysis which I somehow stumbled upon recently.
I do not know of any elementary proof and I am not sure how well known that theorem would be among undergraduates.
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towr
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Re: Prime power moments
« Reply #2 on: Jun 20th, 2008, 2:28am » |
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If by any chance any function g(x) can be generated by a series a0+ ai xP(i) for some sequence ai, then I could see how to prove it. Because then we can exclude there is any range [a..b]  [0..1] where f(x) is non-zero. Which excludes that f(x) a continuous non-zero function.
It'd be akin to a Fourier transform; and I seem to recall it holds if you include all polynomials, so perhaps limiting oneself to prime polynomials works as well.
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| « Last Edit: Jun 20th, 2008, 2:41am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Aryabhatta
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Re: Prime power moments
« Reply #3 on: Jun 20th, 2008, 9:18am » |
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Good thinking! You are right... restricting oneself to use only {1, x^2, x^3, x^5, ..., x^p,...} works.
Perhaps an elementary proof would be to show that t^m for any positive integer m, can be written in that form...
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| « Last Edit: Jun 20th, 2008, 9:22am by Aryabhatta » |
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Obob
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Re: Prime power moments
« Reply #4 on: Jun 20th, 2008, 10:59am » |
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But this cannot be done. First of all, if a continuous function has a power series expansion, it is necessarily analytic. Secondly, if an analytic function has a power series expansion, that expansion is unique. So there are lots and lots of continuous functions which cannot be written in the form proposed by towr, and moreover any function t^m with m not prime cannot be written in that form, since the power series expansion of t^m around 0 is t^m=t^m.
What needs to be shown is that the span of {x^p:p prime or 0} forms a dense subspace of C([0,1]) with respect to the L^2 norm. For then the annihilator of this subspace is trivial, and there can be no such f. Conversely, if there were an f orthogonal to x^p for all p, then the span of {x^p:p prime or 0} would not be dense. So this is an equivalent statement.
The main tool we all know of for trying to prove this kind of statement is the Stone-Weierstrass theorem. The set {x^p:p prime or 0} does in fact seperate points and contains a nonzero constant. But it is not a subalgebra, so that theorem cannot be applied.
However, all we need to prove is L^2 convergence, not uniform convergence, which is weaker than what Stone-Weierstrass provides. So perhaps some other argument can be supplied. Certainly it would be enough to show that t^m for any m can be written as an infinite series sum a_i t^(p_i) in the L^2 norm, for then we could apply Stone-Weierstrass. But I have no reason to believe this can be done. Note that it is not possible to write t^m as such a series in the sup-norm, since this violates the uniqueness of Taylor series. So this sum will fail to converge to t^m pointwise.
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| « Last Edit: Jun 20th, 2008, 12:15pm by Obob » |
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Aryabhatta
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Re: Prime power moments
« Reply #5 on: Jun 20th, 2008, 1:40pm » |
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Yes, If I recall correctly, in fact, even in the "normal" fourier series, uniform (or even pointwise) convergence need not hold.
If we consider pointwise convergence, I think you are right that t^4 etc cannot be represented as such a series.
Also, even if we are talking about L2 convergence, we might need to ortho-normalize {1} U {x^p} first (though I am not sure about it).
Also, I do need to read up on these concepts, so please pardon me If I am speaking nonsense.
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Obob
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Re: Prime power moments
« Reply #6 on: Jun 20th, 2008, 1:47pm » |
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If the function is piecewise C1, then the normal Fourier series converges pointwise to the function anywhere the function is continuous. At discontinuities, the value of the Fourier series is the average of the left and right limits of the function at that point.
I doubt that orthonormalizing {1} U {x^p} will help much.
What is the theorem from functional analysis that you want to apply, Aryabhatta?
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Aryabhatta
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Re: Prime power moments
« Reply #7 on: Jun 20th, 2008, 1:50pm » |
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That theorem is called Muntz's Theorem
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Obob
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Re: Prime power moments
« Reply #8 on: Jun 20th, 2008, 1:57pm » |
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That is quite an interesting result!
I doubt though that there would be a simpler proof for the case of prime exponents than the general case, though.
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Aryabhatta
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Re: Prime power moments
« Reply #9 on: Jun 20th, 2008, 2:21pm » |
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on Jun 20th, 2008, 1:57pm, Obob wrote:That is quite an interesting result!
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Yes, I was surprised when I first came across it!
on Jun 20th, 2008, 1:47pm, Obob wrote:| If the function is piecewise C1, then the normal Fourier series converges pointwise to the function anywhere the function is continuous. |
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I was talking about C (and not C1), but I guess you had t^m in mind and for that, it is definitely true. In fact if I recall correctly, for C1, the convergence is uniform.
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Obob
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Re: Prime power moments
« Reply #10 on: Jun 20th, 2008, 2:38pm » |
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Fourier series of general continuous functions are terribly behaved. If you fix a point x in the unit interval, there is a dense subset of the space of continuous functions whose Fourier series fail to converge to the function at x. See Rudin, Real and Complex Analysis, p. 102.
Piecewise C1 is the hypothesis that gives you a nice theory.
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| « Last Edit: Jun 20th, 2008, 2:39pm by Obob » |
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