Section 6.2

General Problems on Trigonometric Polynomials

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Problem 6.8

A cosine polynomial of order nn may always be written in the form P(cos⁡ϑ)P(\cos \vartheta ), where P(x)P(x) is a polynomial of degree nn. The converse is also true.

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Problem 6.9

A sine polynomial of order nn may always be written in the form sin⁡ϑ P(cos⁡ϑ)\sin \vartheta \, P(\cos \vartheta ), where P(x)P(x) is a polynomial of degree n−1n-1. The converse is also true.

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Problem 6.10

The product of two trigonometric polynomials of order mm and nn respectively is a trigonometric polynomial of order m+nm+n.

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Problem 6.11

A trigonometric polynomial of order nn with real coefficients λr,μs\lambda_r, \mu_s, r=0,1,2,…r = 0, 1, 2, \ldots, s=1,2,…s = 1, 2, \ldots,

g(ϑ)=λ0+λ1cos⁡ϑ+μ1sin⁡ϑ+λ2cos⁡2ϑ+μ2sin⁡2ϑ+⋯+λncos⁡nϑ+μnsin⁡nϑ(ϑ real) g(\vartheta ) = \lambda _0 + \lambda _1 \cos \vartheta + \mu _1 \sin \vartheta + \lambda _2 \cos 2\vartheta + \mu _2 \sin 2\vartheta + \cdots + \lambda _n \cos n\vartheta + \mu _n \sin n\vartheta \qquad (\vartheta \text{ real})

may be written in the form

g(ϑ)=e−inϑG(eiϑ), g(\vartheta ) = e^{-in\vartheta } G(e^{i\vartheta }),

where G(z)=u0+u1z+u2z2+⋯+u2nz2nG(z) = u_0 + u_1 z + u_2 z^2 + \cdots + u_{2n} z^{2n} denotes a polynomial of degree 2n2n which remains unchanged if one forms the reciprocal polynomial and simultaneously replaces the coefficients by their complex conjugates:

G(z)=u‾2n+u‾2n−1z+u‾2n−2z2+⋯+u‾0z2n=z2nG‾(z−1). G(z) = \overline{u}_{2n} + \overline{u}_{2n-1} z + \overline{u}_{2n-2} z^2 + \cdots + \overline{u}_0 z^{2n} = z^{2n} \overline{G}(z^{-1}).

Calculate the coefficients u0,u1,u2,…,u2nu_0, u_1, u_2, \ldots , u_{2n}.

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Problem 6.12

Denote by G(z)G(z) a polynomial of degree 2n2n satisfying the identity

z2nG‾(z−1)=G(z). z^{2n} \overline{G}(z^{-1}) = G(z).

Then for ϑ\vartheta real

e−inϑG(eiϑ)=g(ϑ) e^{-in\vartheta } G(e^{i\vartheta }) = g(\vartheta )

is a trigonometric polynomial in ϑ\vartheta of order nn with real coefficients.

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Problem 6.13

Let G(z)G(z) be a polynomial of degree 2n2n and

z2nG‾(z−1)=G(z). z^{2n} \overline{G}(z^{-1}) = G(z).

How are the zeros of G(z)G(z) distributed in the complex plane?

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Problem 6.14

A trigonometric polynomial of order nn with real coefficients

g(ϑ)=λ0+λ1cos⁡ϑ+μ1sin⁡ϑ+λ2cos⁡2ϑ+μ2sin⁡2ϑ+⋯+λncos⁡nϑ+μnsin⁡nϑ, g(\vartheta ) = \lambda _0 + \lambda _1 \cos \vartheta + \mu _1 \sin \vartheta + \lambda _2 \cos 2\vartheta + \mu _2 \sin 2\vartheta + \cdots + \lambda _n \cos n\vartheta + \mu _n \sin n\vartheta ,

in which λn\lambda_n and μn\mu_n are not both zero, has exactly 2n2n zeros, if both real and complex values are admitted and multiple zeros are counted according to their multiplicity.

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Remark.
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ϑ\vartheta and ϑ+2π\vartheta + 2\pi are regarded as not distinct.

Problem 6.15

Determine all trigonometric polynomials of order nn

g(ϑ)=λ0+λ1cos⁡ϑ+μ1sin⁡ϑ+λ2cos⁡2ϑ+μ2sin⁡2ϑ+⋯+λncos⁡nϑ+μnsin⁡nϑ, g(\vartheta ) = \lambda _0 + \lambda _1 \cos \vartheta + \mu _1 \sin \vartheta + \lambda _2 \cos 2\vartheta + \mu _2 \sin 2\vartheta + \cdots + \lambda _n \cos n\vartheta + \mu _n \sin n\vartheta ,

which have real coefficients and satisfy identically in α\alpha and β\beta the relation

∑ν=0ng(α−νπn+1)g(νπn+1−β)=g(α−β). \sum _{\nu =0}^{n} g\left(\alpha - \frac{\nu \pi }{n+1}\right) g\left(\frac{\nu \pi }{n+1} - \beta \right) = g(\alpha - \beta ).
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