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## Hot answers tagged trigonometric-polynomials

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 ...

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 ...

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 $(x,y) = (\cos^n t, \sin^n t)$.

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 $\{x > 0 \... 10 A compactness argument shows that for sufficiently large$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$tin the ideal $$I=(x(1+t^2)^n-(2t)... 8 Here is how to prove it with more standard methods. First of all, let me restate your identity: Definition. Let \mathbb{N}=\left\{ 0,1,2,\ldots\right\} . A partition shall mean an integer partition, i.e., a weakly decreasing finite list of positive integers. If \lambda is a partition and i is a positive integer, then m_{i}\left( \lambda\... 5$$\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}) $$Now, we can use the Newton identities, to express h_m in terms of the power-sum ... 5 this is an answer to the question as originally posed, without the additional conditions on a_0 and b_0 for example,$$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 0.5 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. 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=... 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 polynomialP$of degree$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-... 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.$$Here M should only appear to even powers and M^2=-(x-y)(y-x). It seems convenient to use all three although either P or M^2 suffices with S since 4P=S^2-M^2. ... 1 Estimation for almost periodic processes (2006), provides a general method to determine the lines of support of the spectra, with applications to bias and covariance estimation. 1 This is very broad, and largely depend on some details you have not provided. Usually the questions are - How smooth is f? Is P a measure with any special properties, e.g., has an associated Sturm Liouville operator? What is the choice of the basis \phi _1, \ldots , \phi _N, \ldots. All these questions change the answer considerably. For example, ... 1 Well I'm not sure if this'll be much help, but I'm doubtful that conditions like those you've described, involving inequalities, will get you what you want. Heuristically at least, I think the condition that | f(t) | < 1 for t \neq t_{0} will only be satisfied for (t_{0}, \ldots, t_{n}) in an n-dimensional subset of [0, 1]^{n + 1}, whereas the ... 1 This solution is based on the suggestion of Ian. First we need an extension of the Nazarov-Turán Lema in infinite dimentions. It can be found here. The formulation of the Nazarov-Turán Lema in higher dimensions is the following: (Nazarov-Turán Lemma) Let p:\mathbb T^n\to\mathbb C a trigonometric polynomial definded by$$ 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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