0
$\begingroup$

Consider a set of integrable functions on the interval $(0,1)$.

Let's introduce an operation $\operatorname{eval}f=\int_0^1 f(x)\,dx$ (which is the mean value of the function).

In such system the function $f(x)=x$ will play the role of an umbra with moments $1,1/2,1/3,1/4,...$ because $\int_0^1 x^n \, dx=\frac 1{n+1}$.

My question is: is it always possible to find a function $f(x)$ such that $\int_0^1 f(x)^n \, dx = a_n$ where $a_n$ is an arbitrary sequence, for instance, Bernoulli numbers? If so, what function would be a representation of Bernoulli umbra?

Improve this question
$\endgroup$
11
  • $\begingroup$ Consider the simplest cast $a_0 =1$, $a_1 = 1$, and $a_k =0$ otherwise. $a_1 =1$ determines $f(x) = 1$, but then this determines $a_k=1$ for $k >1$. $\endgroup$
    – Tom Copeland
    Commented Feb 12 at 23:54
  • $\begingroup$ @TomCopeland hmm, no. $a_1=1$ does not determine $f(x)=1$. $\endgroup$
    – Anixx
    Commented Feb 12 at 23:58
  • $\begingroup$ @TomCopeland but I agree that a sequence with zeros at even positions cannot be represented this way. $\endgroup$
    – Anixx
    Commented Feb 13 at 0:02
  • 2
    $\begingroup$ There are many restrictions, for example $a_1^2\le a_2$, by Cauchy-Schwarz. $\endgroup$
    – Christian Remling
    Commented Feb 13 at 0:04
  • 1
    $\begingroup$ My bad, you're right. $\int_{0}^x f(t) dt = b_0 x + b_1 x^2/2! + ...$ if $f(x) = b_0 + b_1 x+b_2x^2/2! + \cdots$. $\endgroup$
    – Tom Copeland
    Commented Feb 13 at 0:04

0

Reset to default

You must log in to answer this question.