0
$\begingroup$

I need to find $f(x)$,

If $$ \frac{1/x_m}{\sum_u 1/x_u} = \frac{f(x_m)}{\sum_u f(x_u)} $$ $x_u > 0, \forall u$, $u = 0,1,2,3,...,U$, and $x_m = x_u$ if $m = u$

Could $f(x)$ be any thing other than $f(x) = 1/x$,

Is there any proof?

Or there is no way to know the function $f(x_m)$.

  • 0
    What range $u$ are the sums taken over, and how are $x_k$ chosen? – 2017-01-12
  • 0
    $u = 0, 1, 2, ..., U$ and $x_k $ can be any value $> 0$ – 2017-01-12
  • 0
    Is $U$ related to $m\,$? Does the equality hold for one fixed $U\,$, or for *any* $U\,$? – 2017-01-12
  • 0
    yes $x_m = x_u$ if $m = u$ – 2017-01-12
  • 0
    Sorry, but that's obvious, and it doesn't answer the question. You said that `u=0,1,2,...,U` which means that $\sum_u \equiv \sum_{u=0}^U$. Did you mean $\sum_{u=0}^m$ maybe? – 2017-01-12
  • 0
    No, just there is a limited number of terms in the summation. There is no relation between U and m. The equation holds for any U – 2017-01-12
  • 0
    @dxiv I think $U$ is a constant, $u$ is the index of the sum, and $m$ varies to give $U+1$ equations. – 2017-01-12
  • 0
    @BobJones The OP just replied that `There is no relation between U and m`. – 2017-01-12
  • 0
    @dxiv That is what I believed too, but I think my description is a bit more clear. – 2017-01-12

0 Answers 0