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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:


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

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Firstly, you should probably require $r_i \in (0,1)$ instead of $[0,1]$. If $r_i = 0$ then you throw it out, and it does not change the answer. If $r_i = 1$, then all other terms $r_j$ with $i \neq j$ must necessarily be $0$, and then it matters not what power of $m$ you have. Secondly, I am not sure of what you are truly asking. Are you asking, "Given a set of $n$ integers $r_1 , \dots r_n$, how does one find a real number $m$ such that $r_1^m + \dots + r_n^m = 1$"? Or are you asking, "For what values of $m$ can one find $r_1, \dots , r_n$ with $r_1^m + \dots + r_n^m = 1$"? – JavaMan Feb 28 '11 at 16:09
After re-reading your entry, you may be asking a third question. Namely, "Is there any number $m$ such that for any sequence $\{r_i\}_{i=1}^n$, we have $r_1^m + \dots + r_n^m = 1$? – JavaMan Feb 28 '11 at 16:20
@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. – Matt Calhoun Feb 28 '11 at 20:40
up vote 6 down vote accepted

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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so it has at least one solution... how does that answer the question? – yoyo Feb 28 '11 at 16:57
@yoyo: It actually has a unique solution, since $f(x)$ is monotone. – Yuval Filmus Feb 28 '11 at 19:19
I'm wondering since you phrased this as a hint, do you know the solution to this problem, or were you trying to show that the problem would or wouldn't have a solution? – Matt Calhoun Mar 1 '11 at 1:03
@Matt, he meant to show that there is a solution, not to construct it. I don't think there is a general rule to construct a "closed formula" for the solution. But you could use some numerical method like Newton-Raphson once you establish there is a unique solution. If we know more about the $r_i$ though, we can make more informed guesses as to what the solution is. – Raskolnikov Mar 1 '11 at 9:15
@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! – Matt Calhoun Mar 1 '11 at 14:41

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.

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