Problem: Given three positive integers $0 < n_1 < n_2 < n_3$ such that $$n_1 + n_2 \ne n_3, \quad n_2 \ne 2n_1, \quad n_3 \ne 2n_1, \quad n_3 \ne 2n_2,$$ is there always a real number $x$ such that $$\cos n_1 x + \cos n_2 x + \cos n_3 x < -2?$$

This is a follow-up of the question in MO: Is there always a real $x$ such that $\cos n_1 x + \cos n_2 x + \cos n_3 x < -2$?

I checked that (brute force), the statement is true for all positive integers $1\le n_1 < n_2 < n_3 \le 1000$ with $n_1 + n_2 \ne n_3, \, n_2 \ne 2n_1, \, n_3 \ne 2n_1, \, n_3 \ne 2n_2$. For $n_3 > 1000$, I randomly generate some $n_1, n_2, n_3$ without finding a counterexample.

I also found a related problem in MO: The maximum of a real trigonometric polynomial.
Perhaps the problem here is easier to deal with.

My idea is to consider the case $\cos n_1 x = -1$ or $\cos n_2 x = -1$ or $\cos n_3 x = -1$.
For example, $\cos n_3 x = -1$ leads to $x = \frac{(2k_3 + 1)\pi}{n_3}$ where $0\le k_3 \le n_3; k_3\in \mathbb{Z}$;
We consider $f(k_3) = \cos \frac{(2k_3 + 1)n_1\pi}{n_3} + \cos\frac{(2k_3 + 1)n_2\pi}{n_3} - 1$. However, I failed to go proceed.

I also tried non-negative trigonometric polynomials and found this article: "Extremal Positive Trigonometric Polynomials", https://www.dcce.ibilce.unesp.br/~dimitrov/papers/main.pdf
Theorem 4 gives the necessary and sufficient condition for $f(\theta) = a_0/2 + \sum_{k=1}^n a_k \cos k \theta$ to be non-negative on $\mathbb{R}$.

  • 4
    $\begingroup$ Excluding the examples from the previous answer does not mean the inequality is true for the remaining cases. It is possible that you need to exclude other possibilites of the form $a_1 n_1+a_2n_2+a_3n_3=0$ with $a_i$ integers. $\endgroup$ Dec 7, 2021 at 12:02
  • 1
    $\begingroup$ @BeniBogosel Thanks. Yes, we only show it for small $n_3$, far from the truth. We need a counterexample. $\endgroup$
    – River Li
    Dec 7, 2021 at 12:47
  • 2
    $\begingroup$ An alternative formalism is whether there exist $a,b\in\Bbb Q^+$ such that $\cos x+2(\cos^2ax+\cos^2bx)$ is always non-negative where $a\ne1$, $b\not\in\{1,a+1/2,2a\}$ and $a<b$. $\endgroup$ Dec 7, 2021 at 13:40
  • 2
    $\begingroup$ Experimentation appears to show that $\cos x^*+2(\cos^2ax^*+\cos^2bx^*)<0$ for some $x^*$ that solves $a\sin2ax^*+b\sin2bx^*=0$ (setting derivative of second term to zero). $\endgroup$ Dec 7, 2021 at 14:15
  • 3
    $\begingroup$ Just wanted to mention Chowla's cosine problem: for any set $A \subseteq \mathbb{Z}$ of $n$ integers, the minimum of $\sum_{a \in A} \cos(ax)$ is less than $-c\sqrt{n}$ (for some absolute $c>0$). $\endgroup$ Dec 8, 2021 at 10:00

1 Answer 1


In principle this problem can be resolved numerically in finite time, by exploiting the dichotomy between structure (small linear relations between the frequencies $n_1,n_2,n_3$) and randomness (equidistribution), though I do not know if the approach below can actually be implemented in a feasible amount of time. (One could in fact use quantitative equidistribution theorems on tori for this, such as Proposition 1.1.17 of this book of mine, but I will take a more explicit approach here as it will likely give better numerical constants.)

We first claim that if $\cos(n_1 x) + \cos(n_2 x) + \cos(n_3 x) \geq -2$ for all $x$ then there must be a non-trivial linear relation $a_1 n_1 + a_2 n_2 + a_3 n_3 = 0$ between the $n_1,n_2,n_3$ with integers $a_1,a_2,a_3$ with $|a_1| + |a_2| + |a_3| \leq C$ for some effectively computable absolute constant $C$. Indeed, use the Weierstrass approximation theorem to find a polynomial $P: {\bf R} \to {\bf R}$ such that $$ \int_0^1 \int_0^1 \int_0^1 P( \cos(2\pi \theta_1) + \cos(2\pi \theta_2) + \cos(2\pi \theta_3) )\ d\theta_1 d\theta_2 d\theta_3 > \sup_{-2 \leq x \leq 3} P(x)$$ (this can be done by choosing $P$ to be large and positive on $[-3,-2]$ and small on $[-2,3]$). Then we have $$ \int_0^1 \int_0^1 \int_0^1 P( \cos(2\pi \theta_1) + \cos(2\pi \theta_2) + \cos(2\pi \theta_3) )\ d\theta_1 d\theta_2 d\theta_3 \neq \int_0^1 P( \cos(2\pi n_1 \theta) + \cos(2\pi n_2 \theta) + \cos(2\pi n_3 \theta) )\ d\theta.$$ But expanding out the polynomial and extracting the Fourier coefficients we see that the LHS and RHS in fact agree unless there is a non-trivial collision $a_1 n_1 + a_2 n_2 + a_3 n_3 = 0$ with $|a_1| + |a_2| + |a_3| \leq \mathrm{deg} P$. In order to make this algorithm run in as feasible a time as possible it is desirable to get $\mathrm{deg} P$ as small as one can; presumably this can be done numerically since for each fixed choice of degree, one has to solve a linear program in the coefficients of $P$.

Next, once one has restricted to the case $a_1 n_1 + a_2 n_2 + a_3 n_3=0$ for some fixed $a_1,a_2,a_3$, one can perform a similar argument to see that either one has $\cos(2\pi \theta_1) + \cos(2\pi \theta_2) + \cos(2\pi \theta_3) \geq -2$ on the entire hyperplane $\{ (\theta_1,\theta_2,\theta_3): a_1 \theta_1 + a_2 \theta_2 + a_3 \theta_3 =0\}$ (in which the answer to your question is negative), or else there must be a second constraint $b_1 n_1 + b_2 n_2 + b_3 n_3 = 0$ with $(b_1,b_2,b_3)$ independent of $(a_1,a_2,a_3)$ and with $|b_1|+|b_2|+|b_3|$ also bounded. After reducing $n_1,n_2,n_3$ to lowest terms, this leaves one with an explicit finite list of candidate triples $(n_1,n_2,n_3)$ for which the question can be decided by case-by-case check. (To reduce the number of cases, one can normalise $0 \leq a_1 \leq a_2 \leq a_3$ (with $n_1,n_2,n_3$ now arbitrary integers) rather than $0 < n_1 < n_2 < n_3$.)

  • 7
    $\begingroup$ To start the bidding, I get that $P(t) = (t+2)(t+1/3)^2(1-4/3)^2(t-3)$ obeys Terry's condition (the integral is 3.965 and the sup is 0). $\endgroup$ Dec 7, 2021 at 16:33
  • 6
    $\begingroup$ Degree 6, not bad! By my count this means we can restrict $(a_1,a_2,a_3)$ (after some easy normalisations and removal of degenerate cases) to $(0,1,2)$, $(0,1,3)$, $(0,1,4)$, $(0,1,5)$, $(0,2,3)$, $(1,1,1)$, $(1,1,2)$, $(1,1,3)$, $(1,1,4)$, $(1,2,2)$, $(1,2,3)$, which is a manageable number of cases. $\endgroup$
    – Terry Tao
    Dec 7, 2021 at 17:06
  • 1
    $\begingroup$ @DavidESpeyer: In fact the integral is $571/144$ (according to Mathematica 10.1). $\endgroup$
    – GH from MO
    Dec 7, 2021 at 18:36
  • $\begingroup$ @TerryTao Could you explain a little bit where does the motivation for the first part of proof (i.e. proving $|a_1|+|a_2|+|a_3|\ll 1$) come from? I am having hard time understanding how one could think of introducing the polynomial $P$ into the picture. Is this motivated by some more general approach to these sorts of problems? $\endgroup$
    – user0820
    Aug 8, 2022 at 16:35
  • 2
    $\begingroup$ It comes from making Quantitative the classical equidistribution theory of Weyl that is based on exponential sums (or exponential integrals in this case since x is not restricted to be discrete). My book linked to in my answer has further discussion. $\endgroup$
    – Terry Tao
    Aug 8, 2022 at 19:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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