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Suppose we have three pairwise disjoint infinite sets $A,B,C$ which are subsets of $[0,1]$.

It can be assumed that $A,B,C$ are countable. Also, each sum converges absolutely.

Their elements are unknown.

Given three functions $f_A,f_B,f_C:[0,1]\to \mathbb R$, we know that for every integer $n\ge 0$, $$\sum_{p\in A} f_A(p)~p^{n} +\sum_{p\in B}f_B(p)~p^n+n\sum_{p\in C}f_{C}(p)p^{n-1}=C_n$$ where $C_1,C_2,\cdots$ are known.

Under what conditions can we obtain a unique solution of the elements in $A,B,C$?

In general, when does a system of infinite equations (and thus infinite unknowns) has a unique solution? Moreover, is there a branch of mathematics that specifically studies this topic?

Szeto
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  • I suppose $A,B,C$ are countable? – Jack M Sep 18 '18 at 07:48
  • @JackM Sure, you can assume that. – Szeto Sep 18 '18 at 08:10
  • Also, if $f$ is not positive, then $\sum_{p\in A} f(p)$ may not be well-defined. How do we get around that? – Jack M Sep 18 '18 at 08:14
  • @JackM The sum is $\sum_{p\in A}f(p)p^n$. I have implicitly assumed its convergence. Sorry for not mentioning that explicitly. – Szeto Sep 18 '18 at 08:18
  • No, the problem isn't that it might not converge, the problem is: in what order do you take the terms? You're summing over a set, not a sequence. If $f$ isn't positive, it may be that the value of the sum depends on how you order the elements of $A$. – Jack M Sep 18 '18 at 08:19
  • @JackM I again missed something. Please see my edit. Thank you for your patience. – Szeto Sep 18 '18 at 08:32
  • Okay, so we should assume that $\sum_{p\in P} |f_A(p)p^n|$ converges for every countable set $P\subset [0, 1]$? – Jack M Sep 18 '18 at 08:40
  • @JackM If this assumption can lead to some interesting results, you may assume that. – Szeto Sep 18 '18 at 08:56
  • Well, in the meantime, the problem of finding functions $f$ on $[0, 1]$ such that for any countable set $X$, the sum $\sum_{x\in X} x|f(x)|$ is convergent seems plenty difficult enough to be getting on with before attacking the full problem you describe in the question. – Jack M Sep 18 '18 at 12:38

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