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Let there be given a set of $n$ real numbers, $\{r_i\} \subset (0,1)$; is it possible to find some conditions satisfied by a real number $m \in \mathbb{R}$ to ensure that:

$\hspace{6cm}r_1^m+r_2^m+r_3^m+...+r_n^m=1$

For example, it might be possible to define such an $m$ by an infinite series.

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    @DJC: edited to make the interval open instead of closed. Thx. In response to your second comment, I am asking that question implicitly as well.2011-02-28

2 Answers 2

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A hint: For given $r_i\in ]0,1[$ consider the function $f(x):=\sum_{i=1}^n r_i^x$ and see what happens for $x\to 0$ resp. $x\to\infty$.

My response to the comments:

One has $f(0)=n>1$ and $\lim_{x\to\infty} f(x)=0$; furthermore $f$ is monotonically decreasing, so there is exactly one $x>0$ with $f(x)=1$. Unless all $r_i$ are equal (and maybe in some other special cases) it is not possible to solve the equation $f(x)=1$ explicitly. For a numerical solution assume $r_1\leq r_2\leq\ldots\leq r_n$ and put $a:={\log n\over \log(1/r_1)},\qquad b:={\log n\over \log(1/r_n)}.$ Then $f(a)\geq 1$ and $f(b)\leq 1$. Now use binary search.

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    @Raskolnikov: I actually don't have any particular set of $r_i$ in mind! @Christian: I'm afraid you might be correct that only numerical methods will work for this problem, but I don't see why in a clear way. Would you mind linking me or explaining a short proof of this fact? Thx!2011-03-01
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It is possible to tighten the range of possible values for $m$ to the interval

$-\frac{\ln n }{\ln \min_i r_i} \leq m \leq -\frac{\ln n }{\ln \max_i r_i}$

which is not so good if the maximum and the minimum of the $r_i$ are very close to 1 and 0 respecively.