Section 6.7

Maximum-Minimum Problems on Trigonometric Polynomials

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Problem 6.50

Let

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

(ϑ\vartheta real) be a real non-negative trigonometric polynomial of nnth order, and let the constant term in g(ϑ)g(\vartheta ) be equal to 11, i.e.

λ0=12π∫02πg(ϑ) dϑ=1. \lambda _0 = \frac{1}{2\pi } \int _0^{2\pi } g(\vartheta ) \, d\vartheta = 1.

Then

g(ϑ)≦n+1. g(\vartheta ) \leqq n+1.
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Problem 6.51

With the notation of problem 50, we have

λn2+μn2≦1. \sqrt{\lambda _n^2 + \mu _n^2} \leqq 1.
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Problem 6.52

With the notation of problem 50, we have

λ12+μ12≦2cos⁡πn+2. \sqrt{\lambda _1^2 + \mu _1^2} \leqq 2\cos \frac{\pi }{n+2}.
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Problem 6.53

Let h(z)=c(1+z1z)(1+z2z)⋯(1+znz)h(z) = c(1+z_1 z)(1+z_2 z) \cdots (1+z_n z), ∣zν∣≦1\left|z_\nu \right| \leqq 1, ν=1,2,…,n\nu = 1, 2, \ldots , n, cc real, c>0c > 0. Then, setting g(ϑ)=∣h(eiϑ)∣2g(\vartheta ) = \left|h(e^{i\vartheta })\right|^2, we have

12π∫02πlog⁡g(ϑ) dϑ=log⁡c2. \frac{1}{2\pi } \int _0^{2\pi } \log g(\vartheta ) \, d\vartheta = \log c^2.
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Problem 6.54

Let g(ϑ)=∣x0+x1z+x2z2+⋯+xnzn∣2g(\vartheta ) = \left|x_0 + x_1 z + x_2 z^2 + \cdots + x_n z^n\right|^2, z=eiϑz = e^{i\vartheta }, where x0,x1,…,xnx_0, x_1, \ldots , x_n range over all complex numbers such that

12π∫02πlog⁡g(ϑ) dϑ=0 \frac{1}{2\pi }\int _0^{2\pi } \log g(\vartheta ) \, d\vartheta = 0

(that is, with the notation of problem 53, c=1c = 1). Determine the maximum of g(ϑ0)g(\vartheta_0), ϑ0\vartheta_0 fixed.

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Problem 6.55

With the notation of problems 53, 54, determine the maximum of λ0\lambda_0.

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Problem 6.56

With the notation of problems 53, 54, determine the maximum of λν2+μν2\sqrt{\lambda_\nu^2 + \mu_\nu^2}, ν=1,2,…,n\nu = 1, 2, \ldots , n.

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Problem 6.57

A real trigonometric polynomial without a constant term

g(ϑ)=λ1cos⁡ϑ+μ1sin⁡ϑ+λ2cos⁡2ϑ+μ2sin⁡2ϑ+⋯+λncos⁡nϑ+μnsin⁡nϑ g(\vartheta ) = \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

cannot have the same sign for all real ϑ\vartheta unless it is identically zero. This theorem is to be proved without the use of integral calculus.

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Problem 6.58

Let −m-m and MM be the minimum and maximum, respectively, of a real trigonometric polynomial of nnth order of the form

g(ϑ)=λ1cos⁡ϑ+μ1sin⁡ϑ+λ2cos⁡2ϑ+μ2sin⁡2ϑ+⋯+λncos⁡nϑ+μnsin⁡nϑ, g(\vartheta ) = \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 ,

(ϑ\vartheta real) m≧0m \geqq 0, M≧0M \geqq 0 (problem 57). Then we have

M≦nm,m≦nM. M \leqq nm, \qquad m \leqq nM.
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Problem 6.59

The first mean-value theorem of integral calculus may be strengthened for real trigonometric polynomials as follows: Let g(ϑ)g(\vartheta ) be a real trigonometric polynomial of nnth order and let mm be the minimum and MM the maximum of g(ϑ)g(\vartheta ) (ϑ\vartheta real). Then we have

m+M−mn+1≦12π∫02πg(ϑ) dϑ≦M−M−mn+1. m + \frac{M-m}{n+1} \leqq \frac{1}{2\pi } \int _0^{2\pi } g(\vartheta ) \, d\vartheta \leqq M - \frac{M-m}{n+1}.
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Problem 6.60

Let −m-m be the minimum and MM the maximum of a real trigonometric polynomial of nnth order

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 ,

(ϑ\vartheta real). Then either mm or MM is greater than λn2+μn2\sqrt{\lambda_n^2 + \mu_n^2}. We have m=M=λn2+μn2m = M = \sqrt{\lambda_n^2 + \mu_n^2} only if g(ϑ)=ccos⁡n(ϑ−ϑ0)g(\vartheta ) = c\cos n(\vartheta - \vartheta_0).

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Problem 6.61

Compound nn simple harmonic motions with periods in the ratio

1:12:13:⋯:1n 1 : \frac{1}{2} : \frac{1}{3} : \cdots : \frac{1}{n}

and with arbitrary phases. The greatest displacement of the resulting motion is then equal to at least the arithmetic mean of the amplitudes of the separate simple harmonic motions.

With the notation of problem 58, we are dealing with the inequality:

M≧λ12+μ12+λ22+μ22+⋯+λn2+μn2n. M \geqq \frac{\sqrt{\lambda _1^2+\mu _1^2} + \sqrt{\lambda _2^2+\mu _2^2} + \cdots + \sqrt{\lambda _n^2+\mu _n^2}}{n}.
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Remark.
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Strengthening of problem 50.