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I want to find a linear combination of $\cos nx$, $n\ge 0$ (so it's an even trigonometric polynomial) with integer coefficients having small absolute values in a given range. To be specific, I want to find $a_0,a_1,\cdots,a_m \in \mathbb{Z}$ (not all of them zero) such that $f(x)=a_0+a_1 \cos x + \cdots + a_m \cos mx$ satisfies $|f(x)|<1$ for all $x\in I$, where an interval (or a finite union of intervals) $I$ is given.

Because of the symmetry, WLOG we can assume that $I \subset [0,\pi]$. For example, $f(x)=1+2\cos x + 2\cos 2x - 2\cos 4x - 2\cos 5x$ satisfies $|f(x)|<1$ for $x \in [2,\pi]$. It will be helpful for me if one can find such $f$ for "longer" $I$.

(EDIT: I'd like to exclude the simplest cases, monomials $\cos mx$.)

Therefore..
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  • For the longest possible case, $I = [0,\pi]$, I believe it is known that one cannot do better than just the monomials $\cos(mx)$, meaning that $I$ must be strictly shorter. Possibly you are already aware of this since the question focuses on the strict inequality :). – Erick Wong Aug 24 '17 at 02:10
  • After setting the bounty, I'm not sure the question is clear enough. :/ Is the interval $I$ required to be of the form $[a,\pi]$? Otherwise, $f(x) = \cos x$ satisfies $|f|<1$ on $[\epsilon, \pi-\epsilon]$ which is as much as one can hope for, so why look for anything else? –  Aug 24 '17 at 02:19
  • @Michelle Well, ultimately I want to use this to solve another problem. Specifically, I'd like to use a root of polynomial corresponding to $f(x)$. However, for monomials, the corresponding root is a root of unity which is excluded in original problem. So $f(x)$ should be more 'complex' than just monomials, i.e., with more terms or higher coefficients. – Therefore.. Aug 24 '17 at 15:36
  • The shorter the sum, the better the chance to get a larger interval. Single $\cos(mx)$ are excluded, therefore it will be hard to get an interval which has at least the half size of $[0,\pi]$ .Examples:$(1)$ $1+\cos(x)$ for $x\in[\tfrac \pi 2,\pi]$ , $(2)$ $\cos(x)+ \cos(3x)+\cos(6x) +\cos(9x)+\cos(12x) $ for $x\in[0.21,1.66]$ . – user90369 Aug 25 '17 at 09:18

2 Answers2

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Standing the Parseval's Theorem you cannot get $|f(t)|<1$ over the whole period ($(-\pi,\pi]$).
So you shall look for a function which is quite high in a short interval, and falls within $\pm 1$ for the rest of the period.

One of the candidates looks to be partial expansion of the delta function, i.e. with all coefficients equal to $1$.

G Cab
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This is only a trivial elaboration of G Cab's good idea. The Dirichlet kernel $$ D_n(x)=1+2\sum_{k=1}^n \cos(kx)=\frac{\sin( (n+\frac12)x)}{\sin \frac{x}{2}} $$ is a linear combination of cosines with integer coefficients and it has the property that $$\lim_{n\to \infty} \sup_{|x|>\delta} |D_n(x)| =0.$$ In particular, for any fixed $\delta>0$, there exists a degree $N(\delta)$ such that $|D_n(x)|<1$ for $x\in [-\pi, -\delta] \cup [\delta, \pi]$ and $n\ge N(\delta)$.

Giuseppe Negro
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