Section 6.3

Some Special Trigonometric Polynomials

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Problem 6.16
12+cos⁡ϑ+cos⁡2ϑ+⋯+cos⁡nϑ=sin⁡2n+12ϑ2sin⁡ϑ2, \frac{1}{2} + \cos \vartheta + \cos 2\vartheta + \cdots + \cos n\vartheta = \frac{\sin \dfrac {2n+1}{2}\vartheta }{2 \sin \dfrac {\vartheta }{2}}, cos⁡ϑ+cos⁡2ϑ+cos⁡3ϑ+⋯+cos⁡nϑ=sin⁡n2ϑcos⁡n+12ϑsin⁡ϑ2, \cos \vartheta + \cos 2\vartheta + \cos 3\vartheta + \cdots + \cos n\vartheta = \frac{\sin \dfrac {n}{2}\vartheta \cos \dfrac {n+1}{2}\vartheta }{\sin \dfrac {\vartheta }{2}}, cos⁡ϑ+cos⁡3ϑ+cos⁡5ϑ+⋯+cos⁡(2n−1)ϑ=sin⁡2nϑ2sin⁡ϑ, \cos \vartheta + \cos 3\vartheta + \cos 5\vartheta + \cdots + \cos (2n-1)\vartheta = \frac{\sin 2n\vartheta }{2 \sin \vartheta }, sin⁡ϑ+sin⁡2ϑ+sin⁡3ϑ+⋯+sin⁡nϑ=sin⁡n2ϑsin⁡n+12ϑsin⁡ϑ2. \sin \vartheta + \sin 2\vartheta + \sin 3\vartheta + \cdots + \sin n\vartheta = \frac{\sin \dfrac {n}{2}\vartheta \sin \dfrac {n+1}{2}\vartheta }{\sin \dfrac {\vartheta }{2}}.
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Problem 6.17
sin⁡ϑsin⁡ϑ+sin⁡3ϑsin⁡ϑ+sin⁡5ϑsin⁡ϑ+⋯+sin⁡(2n−1)ϑsin⁡ϑ=(sin⁡nϑsin⁡ϑ)2. \frac{\sin \vartheta }{\sin \vartheta } + \frac{\sin 3\vartheta }{\sin \vartheta } + \frac{\sin 5\vartheta }{\sin \vartheta } + \cdots + \frac{\sin (2n-1)\vartheta }{\sin \vartheta } = \left(\frac{\sin n\vartheta }{\sin \vartheta }\right)^2.

What may be deduced from this for ϑ=0\vartheta = 0?

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Problem 6.18
n+12+ncos⁡ϑ+(n−1)cos⁡2ϑ+⋯+cos⁡nϑ=12(sin⁡(n+1)ϑ2sin⁡ϑ2)2. \frac{n+1}{2} + n \cos \vartheta + (n-1) \cos 2\vartheta + \cdots + \cos n\vartheta = \frac{1}{2} \left( \frac{\sin (n+1)\dfrac {\vartheta }{2}}{\sin \dfrac {\vartheta }{2}} \right)^2.
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Problem 6.19

Where do the zeros of the following trigonometric polynomials lie:

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(1)

12+cos⁡ϑ+cos⁡2ϑ+⋯+cos⁡nϑ\dfrac {1}{2} + \cos \vartheta + \cos 2\vartheta + \cdots + \cos n\vartheta;

(2)

cos⁡ϑ+cos⁡2ϑ+⋯+cos⁡nϑ\cos \vartheta + \cos 2\vartheta + \cdots + \cos n\vartheta;

(3)

cos⁡ϑ+cos⁡3ϑ+cos⁡5ϑ+⋯+cos⁡(2n−1)ϑ\cos \vartheta + \cos 3\vartheta + \cos 5\vartheta + \cdots + \cos (2n-1)\vartheta;

(4)

sin⁡ϑ+sin⁡2ϑ+⋯+sin⁡nϑ\sin \vartheta + \sin 2\vartheta + \cdots + \sin n\vartheta;

(5)

sin⁡ϑ+sin⁡3ϑ+⋯+sin⁡(2n−1)ϑ\sin \vartheta + \sin 3\vartheta + \cdots + \sin (2n-1)\vartheta;

(6)

n+12+ncos⁡ϑ+(n−1)cos⁡2ϑ+⋯+cos⁡nϑ\dfrac {n+1}{2} + n \cos \vartheta + (n-1) \cos 2\vartheta + \cdots + \cos n\vartheta?

Problem 6.20

Derive the identity:

cos⁡ϑ+cos⁡2ϑ+⋯+cos⁡nϑ=12sin⁡(n+1)ϑcot⁡ϑ2−cos⁡2n+12ϑ. \cos \vartheta + \cos 2\vartheta + \cdots + \cos n\vartheta = \frac{1}{2} \sin (n+1)\vartheta \cot \frac{\vartheta }{2} - \cos ^2 \frac{n+1}{2}\vartheta .
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Problem 6.21

Show that

sin⁡ϑ+sin⁡2ϑ+⋯+sin⁡nϑ+sin⁡(n+1)ϑ2 \sin \vartheta + \sin 2\vartheta + \cdots + \sin n\vartheta + \frac{\sin (n+1)\vartheta }{2}

is non-negative for 0≦ϑ≦π0 \leqq \vartheta \leqq \pi.

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Problem 6.22

The arithmetic means of the partial sums of the series

12+cos⁡ϑ+cos⁡2ϑ+cos⁡3ϑ+⋯+cos⁡nϑ+⋯ \frac{1}{2} + \cos \vartheta + \cos 2\vartheta + \cos 3\vartheta + \cdots + \cos n\vartheta + \cdots

are non-negative for all ϑ\vartheta; they converge as nn tends to infinity uniformly to 00 if ε≦ϑ≦2π−ε\varepsilon \leqq \vartheta \leqq 2\pi - \varepsilon, ε>0\varepsilon > 0.

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Problem 6.23

The trigonometric polynomial

A(n,ϑ)=sin⁡ϑ+sin⁡2ϑ2+sin⁡3ϑ3+⋯+sin⁡nϑn A(n, \vartheta ) = \sin \vartheta + \frac{\sin 2\vartheta }{2} + \frac{\sin 3\vartheta }{3} + \cdots + \frac{\sin n\vartheta }{n}

has in the interval 0≦ϑ≦π0 \leqq \vartheta \leqq \pi at each of the points

πn+1,3πn+1,5πn+1,…,(2q−1)πn+1 \frac{\pi }{n+1}, \quad 3\frac{\pi }{n+1}, \quad 5\frac{\pi }{n+1}, \quad \ldots , \quad (2q-1)\frac{\pi }{n+1}

(and only at these points) a relative maximum, and at each of the points

2πn,22πn,32πn,…,(q−1)2πn \frac{2\pi }{n}, \quad 2\frac{2\pi }{n}, \quad 3\frac{2\pi }{n}, \quad \ldots , \quad (q-1)\frac{2\pi }{n}

(and only at these points) a relative minimum, where q=[(n+1)/2]q = [(n+1)/2].

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Problem 6.24

(Continuation of problem 23.) The maxima of A(n,ϑ)A(n, \vartheta ) in the interval 0≦ϑ≦π0 \leqq \vartheta \leqq \pi decrease monotonically, so that the absolute maximum of A(n,ϑ)A(n, \vartheta ) in the whole interval 0≦ϑ≦π0 \leqq \vartheta \leqq \pi is equal to A(n,π/(n+1))A(n, \pi /(n+1)).

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Remark.
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See problem 20.

Problem 6.25

(Continuation of problem 23.) The maxima A(n,π/(n+1))A(n, \pi /(n+1)) increase monotonically with nn, and we have

lim⁡n→∞A(n,πn+1)=∫0πsin⁡ϑϑ dϑ=1.8519…. \lim _{n \to \infty } A\left(n, \frac{\pi }{n+1}\right) = \int _0^{\pi } \frac{\sin \vartheta }{\vartheta } \, d\vartheta = 1.8519\ldots .
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Problem 6.26

The trigonometric polynomial

B(n,ϑ)=cos⁡ϑ+cos⁡2ϑ2+cos⁡3ϑ3+⋯+cos⁡nϑn B(n, \vartheta ) = \cos \vartheta + \frac{\cos 2\vartheta }{2} + \frac{\cos 3\vartheta }{3} + \cdots + \frac{\cos n\vartheta }{n}

has in the interval 0≦ϑ≦π0 \leqq \vartheta \leqq \pi at each of the points

0,2πn,22πn,…,p2πn 0, \quad \frac{2\pi }{n}, \quad 2\frac{2\pi }{n}, \quad \ldots , \quad p\frac{2\pi }{n}

(and only at these points) a relative maximum, and at each of the points

2πn+1,22πn+1,32πn+1,…,q2πn+1 \frac{2\pi }{n+1}, \quad 2\frac{2\pi }{n+1}, \quad 3\frac{2\pi }{n+1}, \quad \ldots , \quad q\frac{2\pi }{n+1}

(and only at these points) a relative minimum, where p=[n/2]p = [n/2], q=[(n+1)/2]q = [(n+1)/2].

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Problem 6.27

(Continuation of problem 26.) The trigonometric polynomial B(n,ϑ)B(n, \vartheta ) assumes its least value for ϑ=[(n+1)/2] 2π/(n+1)\vartheta = [(n+1)/2] \, 2\pi /(n+1).

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Remark.
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See problem 21.

Problem 6.28

For 0≦ϑ≦2π0 \leqq \vartheta \leqq 2\pi and n=1,2,3,…n = 1, 2, 3, \ldots

B(n,ϑ)=cos⁡ϑ+cos⁡2ϑ2+cos⁡3ϑ3+⋯+cos⁡nϑn≧−1. B(n, \vartheta ) = \cos \vartheta + \frac{\cos 2\vartheta }{2} + \frac{\cos 3\vartheta }{3} + \cdots + \frac{\cos n\vartheta }{n} \geqq -1.
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