MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer
    
Well, that is the best uniform bound, but what about specific $m$ like $m=N+1$? That is still interesting, imo. – Brendan McKay Dec 3 '12 at 0:22
    
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 $m$.) – Pietro Majer Dec 3 '12 at 15:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.