4
$\begingroup$

Let $k_1,\ldots,k_n$ be distinct integers. Let $s_n(t)=\cos (k_1t)+\cdots+\cos (k_nt)$ be a trigonometric sum. Consider any interval $I\subset [-\pi,\pi)$ of length $\delta=\delta(n)$. Let $\,U$ be a uniform distribution in the interval $I$. I am interested in the quantity $\mathbb{P}(s_n(U)>0)$.

Questions: 1) (strong form) How large should $\delta(n)$ be so that we would have $\mathbb{P}(s_n(U)>0)\approx 1/2$?

2) (weak form) How large should $\delta$ be so that we would have $\mathbb{P}(s_n(U)>0)$ is bounded away from zero and one independently of $n$?

Comment: If $s_n$ is the Dirichlet kernel, that is, $k_i=i$, it is easy to see that we must have $\delta_n>>n^{-1}$. I would be content if one of the latter statements was true with $\delta(n)=1/\log (n)$.

Improve this question
$\endgroup$
6
  • $\begingroup$ Need to explain what is $f$. $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 21, 2014 at 9:28
  • $\begingroup$ @Liviu Nicolaescu Sorry, $f=s_n$, I corrected the typo. $\endgroup$
    – TOM
    Commented Oct 21, 2014 at 9:42
  • 1
    $\begingroup$ Do you have any particular reason to believe that even, say, $\delta_n=\pi/2$ works for large $n$? $\endgroup$
    – fedja
    Commented Oct 23, 2014 at 0:06
  • $\begingroup$ Take a very fast growing sequence of $k_j>0$. Then $f=2s_n/\sqrt{n}$ has essentially standard normal distribution and $f(x)^2-2+n^{-1/2}f(2x)$ has all coefficients equal. However the probability that the standard normal random variable is between $-\sqrt 2$ and $\sqrt 2$ is not $\frac 12$. This shows that we can skew the sign distribution somewhat at any scale. The question is how much... $\endgroup$
    – fedja
    Commented Oct 23, 2014 at 11:45
  • $\begingroup$ @fedja Could you elaborate the statement about similarity with the normal distributions? $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 23, 2014 at 16:41

1 Answer 1

Reset to default
3
$\begingroup$

enter image description here

Above you see a $12$-frame animation of the family of trig polynomials

$$ P_k(t) =\cos t+\cos 2t+\cos 3t+\cos kt,\;\;\; k=4,\dotsc, 15. $$

I have included it to illustrate the fact that there seems to be another quantity relevant to your question, besides $n$, namely the degree $d=\max\{k_i;\;\;i=1.\dotsc, n\}$. In the above example $n=4$, but the degree varies from $4$ to $15$.

The next animation may be more suggestive because you can see large intervals where the trig polynomials are negative. More precisely, below is a $10$-frame animation of the trig polynomials

$$P_k(t) =\cos 2t+\cos 3t+ \cos(4k+1)t, \;\;k=4,\dotsc, 13. $$

enter image description here

Improve this answer
$\endgroup$
4
  • $\begingroup$ I can't prove it, but I expect "highly structured" examples to be the worst- the ones with lengthy arithmetic progressions ${k_1,\ldots k_n}$. And then degree plays no part. It may be that I am wrong, but as I said, I expect good behavior in sufficiently long intervals. Say $\delta=1/\log(n)$. For $n$ large, say $n=100$ and $\delta=1/\log(100)\approx 0.2$ I would expect many roots in the interval and thus many intervals many sing changes. For $n=4$ one cannot see this effect as the length $\delta$ is comparable to the interval length where the sum of cosines changes sign. $\endgroup$
    – TOM
    Commented Oct 23, 2014 at 2:29
  • $\begingroup$ After looking at many examples, your conjecture seems more and more plausible (and difficult). Fedja.s $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 23, 2014 at 9:06
  • $\begingroup$ Fedja's question is an excellent place to start. It suggests that maybe the $1/2$ in your question should be replaced by some universal number $c\in (0,1)$. $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 23, 2014 at 9:13
  • $\begingroup$ the latter is exactly the weaker form I asked (question 2). $\endgroup$
    – TOM
    Commented Oct 23, 2014 at 10:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .