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Disclaimer: The original version of this question focused on $2^n$ in lieu of $n^2$. It is in the hope that the question is easier with $n^2$ that I changed it.

I have an always-nonnegative (on the nonnegative integers) trigonometric polynomial¹ $P$: $$P(n) = \mathcal{R}\left(\sum_{j=1}^k a_j e^{i \theta_j n}\right),$$ with $\lim \limits_{n \to \infty} P(n^2) = 0$.

I want to show that $P(n^2) = 0$.

Here are a few basic facts on trigonometric polynomials that may help. A trigonometric polynomial is an almost-periodic function. This implies in particular that:

  • For $\epsilon > 0$, there is a number $L=L(\epsilon)$ such that any interval of length $L$ on the real line has an $\epsilon$-translation integer, that is, an integer $t$ such that $|P(n) - P(n +t)| < \epsilon$ for any $n$.

  • From any sequence $\{P(n + m_k)\}$, one can extract a subsequence which converges uniformly with respect to $n$. Moreover the function to which it converges is almost-periodic itself. ($\{m_k\}$ is an arbitrary sequence of numbers).

  • In particular, for any sequence $\{m_k\}$, for any $\epsilon$, there exist $i \neq j$ such that $m_i - m_j$ is an $\epsilon$-translation number.

An even simpler case is to consider that $\lim \limits_{n \to \infty} P(n) = 0$. A proof that it implies that $P(n) = 0$ for any $n$ is the following. Suppose there is an $m$ such that $P(m) > 0$, and set $\epsilon = P(m)/2$. Now for any $N$, there exists an $\epsilon$-translation number $t$ of $P$ with $t > N$ an integer. Then $|P(m) - P(m+t)| < \epsilon$, thus $P(m+t) > \epsilon$. Thus $\lim \limits_{n \to \infty} P(n)$ is not null.

¹: Trigonometric polynomial is taken in the sense of, say, Corduneanu (in Almost Periodic Functions). Wikipedia seems to have a different definition.

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Always positive at integer points $n$? How it couples with the conclusion $P(2^n) = 0$? –  Andrew Jul 10 '12 at 4:07
    
@Andrew: I meant "nonnegative," my bad, this comes from my French training :-) –  Michaël Cadilhac Jul 10 '12 at 16:14

1 Answer 1

[Edit: This answer concerns a previous version of the question.]

The following is a counter-example (assuming "positive" is meant to be read "non-negative"):

$$P(x)=\left(1+\cos\pi(x-1)+\cos(\sqrt2-1)\pi(x-1)+\cos\sqrt2\pi(x-1)\right)^2$$

It is trivially non-negative for all $x\in\Bbb R$, since it is a square. For all $k\in\Bbb N_0$ we have:

$$\begin{align}P(2k)&=0\tag{1}\\P(2k+1)&=16 \cos^4k \sqrt2\pi\tag{2}\end{align}$$

which we can show using the usual trigonometric formulas:

$$\begin{align}P(2k)&=\left(1+\cos\pi(2k-1)+\cos(\sqrt2-1)\pi(2k-1)+\cos\sqrt2\pi(2k-1)\right)^2\\&=\left(1+(-1)+\cos(\sqrt2\pi(2k-1)-\pi(2k-1))+\cos\sqrt2\pi(2k-1)\right)^2\\&=\left(\cos(\sqrt2\pi(2k-1)+\pi)+\cos\sqrt2\pi(2k-1)\right)^2\\&=\left(-\cos(\sqrt2\pi(2k-1))+\cos\sqrt2\pi(2k-1)\right)^2\\&=0\\ \\P(2k+1)&=\left(1+\cos\pi2k+\cos(\sqrt2-1)\pi2k+\cos\sqrt2\pi2k\right)^2\\&=\left(1+1+\cos(2\sqrt2k\pi-2k\pi)+\cos2\sqrt2k\pi\right)^2\\&=\left(2+2\cos2\sqrt2k\pi\right)^2\\&=\left(4\cos^2\sqrt2k\pi\right)^2\\&=16 \cos^4 \sqrt2k\pi\end{align}$$

Also, $P$ is not a periodic function, which I think we can most likely show by choosing an arbitrary $T\neq 0$ and showing that $P(x+T)-P(x)$ is non-zero for some $x\in\Bbb R$. This seems a bit tedious, so I shall supply the calculation later if I have some more time. [Edit: As user10676 notes, this calculation is not necessary. $P$ is not periodic because $P(x)=16$ only for $x=1$.]

It is quite clear though that the period cannot be an integer, which can be seen directly by examining $(1)$ and $(2)$ and using the irrationality of $\sqrt2$. Since this is what the question is asking about, we have indeed found a relevant counter-example.

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2  
$P$ is not periodic because it takes value 16 only for $x=1$. –  user10676 Jul 10 '12 at 14:48
    
@user10676: beautiful! Thanks, I'll edit accordingly. –  Dejan Govc Jul 10 '12 at 14:53
    
@DejanGovc: Great counter example to the perodicity of such polynomials! Alas, and this is very bad StackExchange practice from me, the question was not carefully worded... One of my main focus really is the title of the question, which you did not address. I slightly reworded the question so as to emphasize my main question; in particular, the periodicity I was referering to is that of zeros, and again, not all of them. Of course, I'll award the bounty to you if no other answer is given. Cheers! –  Michaël Cadilhac Jul 10 '12 at 15:51
    
@MichaëlCadilhac: Glad you like it! If I think of something useful, I'll post another answer. –  Dejan Govc Jul 10 '12 at 16:16
    
Thanks! Note that I added the possibility to take the $a_j$'s to be real numbers, thus getting rid of phases in the cosines --- I hope this helps. –  Michaël Cadilhac Jul 10 '12 at 17:18

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