28

I'm converting the comment above to an answer:
Shapiro made the following conjecture in the paper "The expansion of mean-periodic functions in series of exponentials" Comm. Pure and Appl. Math. 11 (1958) 1–22:
If two exponential polynomials have infinitely many zeros in common they are both multiples of some third (entirely transcendental) ...

17

Assuming the $b_i$ are all distinct (or at least non-zero for $i \neq 0$), this is not possible. (Otherwise there are trivial examples, e.g. $1 + 2 \cos(0 \theta)+ \cos(0 \theta) \geq 0$ or $1 + 4 \cos \theta + \cos(2\theta) + 2 \cos(0 \theta) \geq 0$.)
Suppose that $\sum_{i=0}^k a_i \cos b_i \theta \geq 0$. Since $a_{i_0} = \sum_{i \neq i_0} a_i$, this ...

17

Jurkat and van Horne showed that the $L^1$ norm is asymptotic to a constant times $\sqrt{N}$ (see Theorems 4 and 5 in their paper which compute all moments). For other related work see Jurkat and van Horne and Marklof. Finally Vaughan and Wooley considered Weyl sums for powers larger than $2$, and formulated some conjectures -- the case for squares behaves ...

17

The magic words are $\tan(\theta/2).$ That substitution reduces your question to asking which rational functions $\mathbb{R} \rightarrow \mathbb{R}$ are homeomorphisms. Those are precisely the functions whose derivative does not change sign, so differentiating our function we get a rational function which does not change sign. This is true if and only if ...

answered Jul 21 '13 at 0:14

Igor Rivin

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16

You are looking at the sum of $m$-th powers of the imaginary parts of $n$-th roots of $-1.$ First, note that these can be expressed as polynomials in elementary symmetric functions (of the imaginary parts), so you just need to find the polynomial of which they are roots. Writing
$$(x+i y)^n +1 = 0,$$ you get two equations, in $x$ and $y,$ eliminate $x$ and ...

answered Jan 23 '17 at 13:39

Igor Rivin

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13

I don't know if this is irreducible, but it gives an answer: Let $\zeta$ be a primitive $n/2$-root of unity. Multiply out
$$\prod_{a=1}^{n/2} \prod_{b=1}^{n/2} (\zeta^a x^{2/n} + \zeta^b y^{2/n} - 1 ).$$
The result is a polynomial in $x$ and $y$, and one of the factors is $x^{2/n} + y^{2/n}-1$, so it vanishes when $(x,y) = (\cos^n t, \sin^n t)$.

answered Sep 20 '19 at 13:55

David E Speyer

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11

Since $a,b$ are incommensurable, $(ax,bx)$ is asymptotically equidistributed
in the torus $({\bf R} / 2\pi{\bf Z})^2$.
[One proof is via a continuous version of Weyl's equidistribution criterion:
for any integers $r,s$ with $(r,s) \neq (0,0)$ we have
$$
\frac1m \int_0^m e^{i(rax+sbx)}\, dx = O_{r,s}(1/m) \to 0
$$
as $m \to \infty$.] Therefore $\{x > 0 \...

answered May 13 at 18:16

Noam D. Elkies

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10

A compactness argument shows that for sufficiently large $X$ one has the bound
$$ \sup_{x \in {\mathbb T} \backslash [-1/X,1/X]} |f(x)| \gg \sup_{x \in [-1/X,1/X]} |f(x)|$$
whenever $f$ is a trigonometric polynomial of degree at most $X$; this would imply that $B_X \gg X$. (Perhaps there is a normalising factor of $1/X$ missing in your question?)
Proof: ...

9

You can use polynomial elimination, which is implemented in Macaulay2.
In fact, the parametric equations of your affine curve are $$x=\frac{(2t)^n}{(1+t^2)^n}, \quad y=\frac{(1-t^2)^n}{(1+t^2)^n},$$
so that an implicit equation $f(x, \,y)=0$ for it is obtained by eliminating the variable $t$ among these, i.e. eliminating $t$ in the ideal $$I=(x(1+t^2)^n-(2t)...

answered Sep 20 '19 at 14:06

Francesco Polizzi

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8

Here is how to prove it with more standard methods. First of all, let me
restate your identity:
Definition. Let $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $. A
partition shall mean an integer
partition, i.e., a
weakly decreasing finite list of positive integers. If $\lambda$ is a partition
and $i$ is a positive integer, then $m_{i}\left( \lambda\...

answered Feb 15 '20 at 2:22

darij grinberg

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5

$$\prod_{k=1}^{n}\left(\cos kx-1\right)=2^{-n} e^{-\frac{1}{2} i n (n+1) x} \left(e^{i x};e^{i x}\right)_n^2,$$
with $(\cdot;\cdot)_n$ the q-Pochhammer symbol.
I guess this counts as a "closed form", but of course it's just a rewriting of the product in terms of some named quantity.
Steven Stadnicki has suggested to compute the Fourier ...

5

I think I can sketch a shorter proof.
Let $z_j = x_j+x_j^{-1}$, and let $p_m$ and $h_m$ denote the power-sum and complete homogeneous symmetric polynomial.
Then (see e.g p.3 in this preprint)
$$
2 T_m(z_j/2) = p_m(x_j,x_j^{-1})
\text{ and }
U_m(z_j/2) = h_m(x_j,x_j^{-1})
$$
Now, we can use the Newton identities, to express $h_m$
in terms of the power-sum ...

answered Feb 15 '20 at 14:36

Per Alexandersson

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5

this is an answer to the question as originally posed, without the additional conditions on $a_0$ and $b_0$
for example,
$$6+\cos \theta+2 \cos 2 \theta+3 \cos 3 \theta\geq 0,$$
or more generally
$$\tfrac{1}{2}k(k+1)+\sum_{n=1}^k n \cos n\theta\geq 0.$$

answered Feb 5 '20 at 19:00

Carlo Beenakker

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5

$f(x)=\frac 12(1+\sin nx)$ is, indeed, the optimal choice. To see it, let's normalize a bit differently by $0\le f\le 2$. Then $f=1+g$ where $g$ is a real trigonometric polynomial of degree $n$ bounded by $1$. We need the following classical
Lemma. $g^2+n^{-2}(g')^2\le 1$
(I learned it from Alexandre Eremenko when we were discussing another MO question).
...

5

Yes, something like that is proved in the paper by Simon Kokkendorff:
Kokkendorff, Simon L., Polar duality and the generalized law of sines, J. Geom. 86, No. 1-2, 140-149 (2006). ZBL1115.51010.
It would be too cumbersome to define everything here, but the paper is very nicely written.

answered Nov 18 '17 at 21:27

Igor Rivin

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5

Letting $\xi=e^{\frac{2\pi i}{2n}}$, a primitive root of unity $z^{2n}=1$. Then, $\sin\left(\frac{\pi k}n\right)=\frac{\xi^{k}-\xi^{-k}}{2i}$ and hence
\begin{align}
\sum_{k=0}^{n-1}\sin^m\left(\frac{\pi k}n\right)
&=\frac1{(2i)^m}\sum_{k=0}^{n-1}\sum_{j=0}^m\binom{m}j(-1)^j\xi^{-jk}\xi^{(m-j)k} \\
&=\frac1{(2i)^m}\sum_{j=0}^m(-1)^j\binom{m}j\sum_{k=...

answered Jan 23 '17 at 21:54

T. Amdeberhan

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4

So we start with a rational $r=\frac{p}{q}$.
1) Say $2\not|p,2\not|q$. Then taking $x=q\pi$, we get:
$$f_r(x)=\cos(x)+\cos(rx)=\cos(q\pi)+\cos(p\pi)=-2$$
so $r$ is not the argmax.
2) Say $2$ divides exactly one of $p,q$. Then there exists an odd integer $k$ that solves the congruence equation:
$$kp=q+1 \pmod{2q}$$
we take $x=k\pi$, and we have:
$$f_r(x)=-1+...

4

Appendix A4 of the book
P. Borwein, T. Erdelyi, Polynomials and Polynomial inequalities, Graduate Texts in Mathematics 161, Springer
should be a good source for your question. In particular, (A.4.22) gives
$$\|P'\|_p\leq cn^2\|P\|_p,$$
for every polynomial $P$ of degree $n$ and $0<p<\infty$. Apparently, finding the best possible constant $c$ is ...

4

For even integer exponents, say $p=2k$ and $p \geq2$, the quantity is just the $k$-order additive energy of the set $S \subset \mathbb{Z}$ of non-zero Fourier coefficients. It is easy to see that this is maximized by any arithmetic progression of the desired length (which coincides with the $n$-order Dirichlet kernel).
In the case of $p=4$ this is just the ...

3

Let $r$ be an irrational real number. For real $x>0$, let $U_x$ be a random variable (r.v.) uniformly distributed on the interval $[0,x]$, and then let
$$C_x:=\cos rU_x+2\cos U_x.$$ Then the problem can be restated as follows: Is it true that
\begin{equation*}
P(C_x>0)\to1/2\,\text{?} \tag{1}
\end{equation*}
Everywhere here, the limits are taken ...

2

Let $\gamma(x) = (\alpha(x), \beta(x)),$ where $\alpha, \beta$ are the real and imaginary parts. A self-intersection corresponds to a simultaneous zero of $(\alpha(x)-\alpha(y))/(x-y)$ and $(\beta(x)-\beta(y)/(x-y),$ If you use the rational parametrization for the circle (the $\tan t/2$ trick), both expressions become polynomials, and the number of self-...

answered Mar 13 '15 at 14:32

Igor Rivin

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2

Typically, one does "Prony method": considers an infinite (or just long enough) sequence $c=(c_1,c_2,\dots)$ and a system of equations of the form $V(x)Z=c$, with $Z$ a vector of non-0 unknowns, and $V(x)$ the Vandermonde matrix
$$
V(x)=\begin{pmatrix}
1&1&\dots&1\\
x_1& x_2&\dots &x_k\\
x_1^2& x_2^2&\dots &x_k^2\\
&\...

2

Let $\zeta=e^{2\pi i/N}$ and $\alpha=e^{2\pi i/(2N)}$. Your assumption was $\Re(\alpha(\zeta^{m_3}-2\zeta^{m_2}+\zeta^{m_1}))=0$. This is equivalent to $\alpha(\zeta^{m_3}-2\zeta^{m_2}+\zeta^{m_1}))+\bar\alpha(\zeta^{-m_3}-2\zeta^{-m_2}+\zeta^{-m_1})=0$. Multiplying by $\bar\alpha$ and using the fact that $\alpha^2=\zeta$, you get $\zeta^{m_3}-2\zeta^{m_2}+\...

1

Here are some small explicit solutions and general comments.
Since the polynomials are symmetric, one would expect them to be more nicely expressed using $$S=x+y \\ M=x-y\\ P=xy.$$
Here $M$ should only appear to even powers and $M^2=-(x-y)(y-x).$
It seems convenient to use all three although either $P$ or $M^2$ suffices with $S$ since
$4P=S^2-M^2.$
...

1

Estimation for almost periodic processes (2006), provides a general method to determine the lines of support of the spectra, with applications to bias and covariance estimation.

answered Mar 24 '19 at 9:40

Carlo Beenakker

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1

This is very broad, and largely depend on some details you have not provided. Usually the questions are -
How smooth is $f$?
Is $P$ a measure with any special properties, e.g., has an associated Sturm Liouville operator?
What is the choice of the basis $\phi _1, \ldots , \phi _N, \ldots$.
All these questions change the answer considerably. For example, ...

reference-request fourier-analysis na.numerical-analysis approximation-theory trigonometric-polynomials

1

Well I'm not sure if this'll be much help, but I'm doubtful that conditions like those you've described, involving inequalities, will get you what you want. Heuristically at least, I think the condition that $| f(t) | < 1$ for $t \neq t_{0}$ will only be satisfied for $(t_{0}, \ldots, t_{n})$ in an $n$-dimensional subset of $[0, 1]^{n + 1}$, whereas the ...

ca.classical-analysis-and-odes real-analysis fourier-analysis harmonic-analysis trigonometric-polynomials

1

This solution is based on the suggestion of Ian. First we need an extension of the Nazarov-Turán Lema in infinite dimentions. It can be found here.
The formulation of the Nazarov-Turán Lema in higher dimensions is the following:
(Nazarov-Turán Lemma) Let $p:\mathbb T^n\to\mathbb C$ a trigonometric polynomial definded by $$ p(\boldsymbol z)=\sum_k c_k\...

1

Grepstad has carried out the details of Beurling's paper in her Master thesis, see
https://core.ac.uk/download/pdf/52106196.pdf
For the question you ask, see formula (3.6). I do not know whether the
exponential type obtained is sharp...

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