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Hi,

I am trying to put a bound on a sum. Given $\omega=\exp(2\pi i/3)$ and $n$ positive real numbers

$ 0=\tau_0 < \tau_1 < \tau_2 < ...\tau_{n-1}< \tau_n=1 $

such that

$\sum_{i=1}^n\omega^i (\tau_i^m-\tau_{i-1}^m)=0$ for $m=0,1,2,..,N $, what is the upper bound on

$|\sum_{i=1}^n\omega^i (\tau_i^m-\tau_{i-1}^m)|$

for $m >N$?

Thanks.

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1 Answer 1

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The only relevant facts, $0=\tau_0 < \tau_1 < \dots < \tau_n= 1$ and $|\omega|=1$ imply

$$\Big|\sum_{i=1}^n\omega^i (\tau_i^m-\tau_{i-1}^m)\Big| \le \sum_{i=1}^n|\tau_i^m-\tau_{i-1}^m|=1\, . $$ On the other hand, $\Big|\sum_{i=1}^n\omega^i (\tau_i^m-\tau_{i-1}^m)\Big|\to1$ as $m\to+\infty $.

So $1$ is the best possible bound, and the other information is not needed.

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  • $\begingroup$ Well, that is the best uniform bound, but what about specific $m$ like $m=N+1$? That is still interesting, imo. $\endgroup$
    – Brendan McKay
    Commented Dec 3, 2012 at 0:22
  • $\begingroup$ I agree; but a constant $C(m)$ uniform in $n$ , or a constant $C(n,m)$ for specific $m$ and $n$? (the determinative article in the upper bound made me think the question referred to a universal bound, uniform both in $n$ and $m$.) $\endgroup$
    – Pietro Majer
    Commented Dec 3, 2012 at 15:36

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