32

Let $f$ be an integrable real valued function defined on $[0,\infty)$. Let $$m_n=\int_0^\infty f(x)x^n \mathrm dx$$ be the $n^{th}$ moment, and suppose that all of these integrals converge absolutely. Are there conditions that can we impose on $f$ that would allow us to write $f$ explicitly in terms of its moments and certain simple functions.

This idea is similar to the fact that we can reconstruct sufficiently nice functions on $[0,1]$ from a sum of their Fourier coefficients and $e^{inx}$. Also, on $(-\infty,\infty)$ we can reconstruct $f$ from an integral over the real line.

The analogy for $[0,\infty)$ and $x^s$ is of course the Mellin transform, which has an inversion formula as a line integral in the complex plane.

My question is then: Can we impose nice enough (non-trivial) conditions on a class of functions so that we can invert the Mellin transform on the real line based only on its values at the positive integers?

Thanks!

Note: I am not asking about the moment problem and the requirements for uniqueness. (Such as Carleman's Condition etc..)

J. M. ain't a mathematician
  • 75,051
Eric Naslund
  • 72,099
  • 4
    Surely you are aware of Post's formula http://en.wikipedia.org/wiki/Post%27s_inversion_formula to inverse a Laplace transform. To fit into your question, one can restrict the limit to a sequence of integer values. – Did Apr 13 '11 at 05:50
  • If derivatives of Laplace transforms are not allowed, a solution when $f$ is continuous might be to approach it on $[0,T]$ by a Bernstein polynomial (the ones which yield a proof of Stone-Weierstrass approximation theorem based on the law of large numbers) of degree $N$ large enough and to use the explicit bounds of the error to let $T$ and $N$ go to infinity. But your functions $f$ are not continuous, are they? – Did Apr 13 '11 at 06:10
  • @Didier: Thanks, this is very interesting. At the moment, I don't see how to make it fit my question exactly, or just Mellin Transforms, but I think I might find a way by reading the derivation. (Wikipedia gives a link to http://www.rose-hulman.edu/~bryan/invlap.pdf) – Eric Naslund Apr 13 '11 at 06:21
  • @Didier: I am interested in the most general case, but continuous should be nice enough. The function $f$ which motivated this question was actually analytic, and monotonic on $\mathbb{R}^+$. (But that is a bit too narrow a criterion!) – Eric Naslund Apr 13 '11 at 06:25
  • Eric: OK. Rereading myself, I fail to see how the "Stone-Weierstrass" comment applies, so you can probably forget it. Sorry about the noise. – Did Apr 13 '11 at 07:05
  • I haven't gotten to the thing, but this article might have something you can use... – J. M. ain't a mathematician May 01 '11 at 21:26
  • Here is another article you might want to look at. – J. M. ain't a mathematician May 09 '11 at 17:05

2 Answers2

2

I don't know why you restrict your expression to positive x, because if you let your integral range from negative to positive infinity, the sum of the moments seems to have a pretty simple interpretation. Just take the Fourier Transform of your expression with f(x) and you get the definition of the Taylor series for F(w). (Remember that the Fourier Transforms of the powers of x are just the derivatives of the delta function.) In other words, the functions that are expressible in terms of the sum of their moments are just the functions whose Fourier Transforms are analytical.

Marty Green
  • 1,917
  • 1
    Marty: I also have tried in such a style, but seems OP is searching for something else. Likely, http://en.wikipedia.org/wiki/Stieltjes_moment_problem is also not appropriate. – Alex 'qubeat' Jun 03 '11 at 12:13
1

(Partial answer) You might try looking at Newton interpolation and then an inverse Mellin transform on the result:

With $\bigtriangledown^{s-1}_n=\sum_{n=0}^{\infty}(-1)^n \binom{s-1}{n}[...]$, formally

$g(s)=\bigtriangledown^{s-1}_{n} \bigtriangledown^{n}_{j} \frac{m_j}{j!}$

$=\int_0^\infty f(x)\bigtriangledown^{s-1}_{n} \bigtriangledown^{n}_{j} \frac{x^j}{j!} dx$

$=\int_0^\infty f(x) \frac{x^{s-1}}{(s-1)!}dx$= a modified Mellin transform, and consequently,

$m_j=j!g(j+1)=\int_0^\infty f(x) x^jdx$.

Then performing the modified inverse Mellin transform

$f(x)=\displaystyle\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+ i\infty} \frac{\pi}{sin(\pi s)} g(s) \frac{x^{-s}}{(-s)!} ds$.

You could then investigate what classes of functions allow these maneuvers.

Trivial example:

With $m_j=a^j$, then $g(s)=\frac{a^{s-1}}{(s-1)!}$ analytically continued from the Newton interpolation to all $s$, and

$f(x)=\displaystyle\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+ i\infty} a^{s-1} x^{-s} ds=\delta(x/a-1)/a= \delta(x-a)$.

Another example:

Let $U(\alpha,\beta,z)$ be the confluent hypergeometric function of the second kind.

With $m_j=j!U(j+1,j+1+b,a)$, then $g(s)=U(s,s+b,a)$ analytically continued from the Newton interpolation to all $s$, and

$f(x)=\displaystyle\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+ i\infty} \frac{\pi}{sin(\pi s)} U(s,s+b,a) \frac{x^{-s}}{(-s)!} ds=e^{-ax}(1+x)^{b-1}$.

Edit

(Perhaps more to the point of your question.) The mathoverflow link in the comment below illustrates Ramanujan's Master Theorem / Formula, which gives f(x) as a Taylor or power series in terms of the moments for a certain class of functions. See also Chaudry and Quadir "Extension of Hardy's Class for Ramanujan's Interpolation Formula and Master Theorem with Applications" and Amdeberhan, Espinosa, Gonzalez, Harrison, Moll, and Straub "Ramanujan's Master Theorem."

Tom Copeland
  • 2,416