Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Let $a_1\ge a_2\geq\cdots\geq a_n$ be real numbers. And let $r=(r_1,\ldots,r_n)$ be sequence of random variables taking on values $1$ and $-1$ and such that $\sum_{i=1}^n r_i=0$.

I am wondering if one can estimate from above and from below $\cos\left(\sum_{i=1}^na_ir_i\right)$

share|cite|improve this question
Almost surely, and by something more precise than -1 and +1? – Did May 7 '12 at 18:55
something better then $-1$ and $1$. – David May 7 '12 at 18:57
Unless your $a_i$ satisfy more interesting constraints you can effectively make it anything you want. Take $a_2,\ldots,a_n=-1$, so that $\sum_{i=1}^na_ir_i=(a_1+1)r_1$ and now set $a_1$ to anything you desire above $-1$. $\cos$ is an even function so not much is lost with $a_1\geq -1$. – Alex R. May 7 '12 at 19:02
@Sam : But could it be that the question is whether you can give estimates that depend on $a_1,\ldots,a_n$? – Michael Hardy May 7 '12 at 19:11
Yes, I would like to get bound which would depend on $a_i, i=1, \ldots n$ – David May 7 '12 at 19:55
up vote 0 down vote accepted

This is just an observation, but doesn't fit neatly into a comment.

$n$ must be even, otherwise it is impossible to have the sum of $r_i$ be zero.

Let $\sigma$ be a permutation so that $|a_{\sigma_1}|\geq ... \geq |a_{\sigma_n}|$. Then we have $$| \sum_{i=1}^n a_i r_i | \leq |a_{\sigma_1}|+...+|a_{\sigma_{\frac{n}{2}}}|-(|a_{\sigma_{\frac{n}{2}+1}}| +...+ |a_{\sigma_n}|)$$ and the bounds are achieved with appropriate (legal) choice of $r_i$. I have no idea how this translates into bounds on $\cos(\sum_{i=1}^n a_i r_i)$ without more information about the $a_i$.

share|cite|improve this answer

Your Answer


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.