Section 11.9

Fourier Series and Related Questions

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Problem 4857

Expand in the simplest Fourier series the function f(x)f(x), defined on the interval (−π,π)(-\pi ,\pi ) by the conditions:

f(x)={π4for 0<x<π;−π4for −π<x<0. f(x) = \begin{cases} \dfrac {\pi }{4} & \text{for } 0 < x < \pi ; \\ -\dfrac {\pi }{4} & \text{for } -\pi < x < 0 . \end{cases}
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Problem 4858

Expand in the simplest Fourier series the function f(x)f(x), defined on the interval (−π,π)(-\pi ,\pi ) by the conditions:

f(x)={0for −π<x<0;1for 0<x<π. f(x) = \begin{cases} 0 & \text{for } -\pi < x < 0; \\ 1 & \text{for } 0 < x < \pi . \end{cases}
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Problem 4859

Expand in the simplest Fourier series the function f(x)=∣x∣f(x) = \left|x\right| for −π<x<π-\pi < x < \pi.

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Problem 4860

Expand in the simplest Fourier series the function f(x)=xf(x) = x for −π<x<π-\pi < x < \pi.

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Problem 4861

Expand in the simplest Fourier series the function f(x)f(x), defined on the interval (−π,π)(-\pi ,\pi ) by the conditions:

f(x)={sin⁡12x,0<x<π;1+cos⁡12x,−π<x<0. f(x) = \begin{cases} \sin \dfrac {1}{2}x, & 0 < x < \pi ; \\ 1+\cos \dfrac {1}{2}x, & -\pi < x < 0 . \end{cases}
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Problem 4862

Expand in the simplest Fourier series the function f(x)f(x), defined on the interval (−π,π)(-\pi ,\pi ) by the conditions:

f(x)={π−x2x,0<x<π;−π+x2x,−π<x<0. f(x) = \begin{cases} \dfrac {\pi -x}{2}x, & 0 < x < \pi ; \\ -\dfrac {\pi +x}{2}x, & -\pi < x < 0 . \end{cases}
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Problem 4863

Expand in the simplest Fourier series the function f(x)=x2f(x) = x^{2} for −π<x<π-\pi < x < \pi.

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Problem 4864

Expand in the simplest Fourier series the function f(x)f(x), defined on the interval (−π,π)(-\pi ,\pi ) by the conditions:

f(x)={cos⁡x,0<x<π;−cos⁡x,−π<x<0. f(x) = \begin{cases} \cos x, & 0 < x < \pi ; \\ -\cos x, & -\pi < x < 0 . \end{cases}
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Problem 4865

Expand in the simplest Fourier series the function f(x)=∣sin⁡x∣f(x) = \left|\sin x\right| for −π<x<π-\pi < x < \pi.

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Problem 4866

Expand in the simplest Fourier series the function f(x)=cos⁡axf(x) = \cos ax for −π<x<π-\pi < x < \pi, aa not an integer.

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Problem 4867

Expand in the simplest Fourier series the function f(x)=sin⁡axf(x) = \sin ax for −π<x<π-\pi < x < \pi, aa not an integer.

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Problem 4868

Expand in the simplest Fourier series the function f(x)f(x), defined on the interval (−π,π)(-\pi ,\pi ) by the conditions: f(x)=0f(x) = 0 for −π<x<0-\pi < x < 0; f(x)=sin⁡2xf(x) = \sin^{2} x for 0<x<π0 < x < \pi.

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Problem 4869

Expand in a Fourier series the periodic function: f(x)={x}f(x) = \left\{ x\right\}, the fractional part of xx, i.e. the difference x−[x]x-[x], where [x][x] is the integer part of xx.

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Problem 4870

Expand in a Fourier series the periodic function: f(x)=(x)f(x) = (x), the distance from xx to the nearest integer.

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Problem 4871

Expand in a Fourier series the periodic function: f(x)=xf(x) = x, 11, or 3−x3-x, according as xx lies in the interval (0,1)(0,1), (1,2)(1,2), or (2,3)(2,3), assuming that the period of f(x)f(x) equals 33.

If the function is not given as periodic and one seeks an expansion valid only on a given interval, then it can be expanded in a Fourier series with period equal to the length of the given interval, or with any larger period, but in the latter case the function must be extended in some way (as long as the Dirichlet conditions are satisfied). This can be used to obtain an expansion in cosines alone or in sines alone, or to restrict oneself to only the sines or cosines of odd multiples of the angle.

In the problems that follow, it will be assumed that the period is to be chosen as the smallest one for which the requirements of the problem are satisfied.

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Problem 4872

On the interval (−h,h)(-h,h), expand the function exe^{x}.

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Problem 4873

Expand ch⁡ax\operatorname {ch}ax on the interval (−π,π)(-\pi ,\pi ).

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Problem 4874

Expand sh⁡ax\operatorname {sh}ax on the interval (−π,π)(-\pi ,\pi ).

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Problem 4875

On the interval (0,π)(0,\pi ), expand x3x^{3} in cosines.

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Problem 4876

On the interval (0,π)(0,\pi ), expand x3x^{3} in sines.

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Problem 4877

On the interval (0,π)(0,\pi ), expand sin⁡x\sin x in cosines.

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Problem 4878

The function f(x)=x2(a−x)f(x) = x^{2}(a-x) is given on the interval (0,a)(0,a). Expand it in a Fourier series with period aa.

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Problem 4879

The function f(x)=x2(a−x)f(x) = x^{2}(a-x) is given on the interval (0,a)(0,a). Expand it in a Fourier series as an even function.

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Problem 4880

The function f(x)=x2(a−x)f(x) = x^{2}(a-x) is given on the interval (0,a)(0,a). Expand it in a Fourier series as an odd function.

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Problem 4881

The function f(x)=xf(x) = x is given on the interval (0,a)(0,a). Expand it in a Fourier series with period aa.

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Problem 4882

The function f(x)=xf(x) = x is given on the interval (0,a)(0,a). Expand it in a Fourier series as an even function.

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Problem 4883

The function f(x)=xf(x) = x is given on the interval (0,a)(0,a). Expand it in a Fourier series as an odd function.

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Problem 4884

The function f(x)=xf(x) = x is given on the interval (0,a)(0,a). Expand it in a Fourier series as an odd function, in sines of odd multiples of the angle.

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Problem 4885

Functions rational in cos⁡x\cos x and sin⁡x\sin x can be expanded in a Fourier series without using the Fourier formulas. To do this we write:

cos⁡x=t2+12t,sin⁡x=t2−12ti, \cos x = \frac{t^{2}+1}{2t}, \qquad \sin x = \frac{t^{2}-1}{2ti},

where t=exit = e^{xi}. The problem then reduces to expanding the function in positive and negative powers of tt. Passing back from tt to xx, we obtain the desired Fourier series expansion.

By the method just described and similar ones, obtain the following expansions:

1−a21−2acos⁡x+a2=1+2∑n=1∞ancos⁡nx;∣a∣<1. \frac{1-a^{2}}{1-2a\cos x+a^{2}} = 1+2\sum _{n=1}^{\infty } a^{n}\cos nx; \quad \left|a\right| < 1 .
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Problem 4886

By the method just described, obtain the following expansion:

1−acos⁡x1−2acos⁡x+a2=∑n=0∞ancos⁡nx;∣a∣<1. \frac{1-a\cos x}{1-2a\cos x+a^{2}} = \sum _{n=0}^{\infty } a^{n}\cos nx; \quad \left|a\right| < 1 .
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Problem 4887

By the method just described, obtain the following expansion:

asin⁡x1−2acos⁡x+a2=∑n=0∞ansin⁡nx;∣a∣<1. \frac{a\sin x}{1-2a\cos x+a^{2}} = \sum _{n=0}^{\infty } a^{n}\sin nx; \quad \left|a\right| < 1 .
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Problem 4888

By the method just described, obtain the following expansion:

ln⁡(1−2acos⁡x+a2)=−2∑n=1∞anncos⁡nx;∣a∣<1. \ln (1-2a\cos x+a^{2}) = -2\sum _{n=1}^{\infty } \frac{a^{n}}{n}\cos nx; \quad \left|a\right| < 1 .
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Problem 4889

By the method just described, obtain the following expansion:

ln⁡∣2sin⁡x2∣−1=cos⁡x+cos⁡2x2+cos⁡3x3+… \ln \left|2\sin \frac{x}{2}\right|^{-1} = \cos x+\frac{\cos 2x}{2}+\frac{\cos 3x}{3}+\ldots
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Problem 4890

The function f(x)f(x), given on the interval (0,π2)\left(0,\dfrac {\pi }{2}\right), is continued onto the interval (π2,π)\left(\dfrac {\pi }{2},\pi \right), so that f(π2+x)=−f(π2−x)f\left(\dfrac {\pi }{2}+x\right) = -f\left(\dfrac {\pi }{2}-x\right); prove that for 0<x<π0 < x < \pi the equality holds

f(x)=∑ancos⁡(2n+1)x, f(x) = \sum a_{n}\cos (2n+1)x,

where

an=4π∫0π/2f(x)cos⁡(2n+1)x dx. a_{n} = \frac{4}{\pi } \int _{0}^{\pi /2} f(x)\cos (2n+1)x\, dx .
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Problem 4891

How should the function f(x)f(x), given on the interval (0,π2)\left(0,\dfrac {\pi }{2}\right), be continued onto the interval (π2,π)\left(\dfrac {\pi }{2},\pi \right) so that for 0<x<π0 < x < \pi

f(x)=∑bnsin⁡(2n+1)x f(x) = \sum b_{n}\sin (2n+1)x

and how can the coefficients be computed?

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Problem 4892

Prove, for 0<x<π0 < x < \pi, the following equalities:

cos⁡2x1⋅3+cos⁡4x3⋅5+cos⁡6x5⋅7+…=12−π4sin⁡x. \frac{\cos 2x}{1 \cdot 3}+\frac{\cos 4x}{3 \cdot 5}+\frac{\cos 6x}{5 \cdot 7}+\ldots = \frac{1}{2}-\frac{\pi }{4}\sin x .
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Problem 4893

Prove, for 0<x<π0 < x < \pi, the following equalities:

sin⁡x13+sin⁡3x33+sin⁡5x53+…=πx8(π−x). \frac{\sin x}{1^{3}}+\frac{\sin 3x}{3^{3}}+\frac{\sin 5x}{5^{3}}+\ldots = \frac{\pi x}{8}(\pi -x) .
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Problem 4894

Prove the equality:

∑n=1∞sin⁡(2n+2)θn(n+1)= \sum _{n=1}^{\infty } \frac{\sin (2n+2)\theta }{n(n+1)} = =sin⁡2θ−(π−2θ)sin⁡2θ−sin⁡θcos⁡θln⁡4sin⁡2θ. = \sin 2\theta -(\pi -2\theta )\sin ^{2}\theta -\sin \theta \cos \theta \ln 4\sin ^{2}\theta .
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Problem 4895

Prove the equality:

∑n=1∞cos⁡(2n+1)θn2(n+1)2= \sum _{n=1}^{\infty } \frac{\cos (2n+1)\theta }{n^{2}(n+1)^{2}} = =2(π−2θ)sin⁡θ+(π23−2πθ+2θ2−3)cos⁡θ. = 2(\pi -2\theta )\sin \theta +\left(\frac{\pi ^{2}}{3}-2\pi \theta +2\theta ^{2}-3\right)\cos \theta .
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Problem 4896

Prove the correctness of the expansion:

cos⁡3x1⋅3⋅5−cos⁡5x3⋅5⋅7+cos⁡7x5⋅7⋅9−…=π8cos⁡2x−13cos⁡x \frac{\cos 3x}{1 \cdot 3 \cdot 5}-\frac{\cos 5x}{3 \cdot 5 \cdot 7}+\frac{\cos 7x}{5 \cdot 7 \cdot 9}-\ldots = \frac{\pi }{8}\cos ^{2}x-\frac{1}{3}\cos x

on the interval 0<x<π20 < x < \dfrac {\pi }{2}.

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Problem 4897

Prove the equalities:

∑n=−∞∞e−πa(n+x)2=1a∑n=−∞∞e−πn2acos⁡2πnx,a>0. \sum _{n=-\infty }^{\infty } e^{-\pi a(n+x)^{2}} = \frac{1}{\sqrt{a}} \sum _{n=-\infty }^{\infty } e^{-\frac{\pi n^{2}}{a}} \cos 2\pi nx, \quad a > 0 .
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Problem 4898

Prove the equalities:

12+∑n=1∞e−πn2x=1x(12+∑n=1∞e−πn2x),x>0. \frac{1}{2}+\sum _{n=1}^{\infty } e^{-\pi n^{2} x} = \frac{1}{\sqrt{x}}\left(\frac{1}{2}+\sum _{n=1}^{\infty } e^{-\frac{\pi n^{2}}{x}}\right), \quad x > 0 .
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Problem 4899

Prove the equalities:

lim⁡x→1−01−x (x+x4+x9+…)=12π. \lim _{x \rightarrow 1-0} \sqrt{1-x}\, (x+x^{4}+x^{9}+\ldots ) = \frac{1}{2}\sqrt{\pi } .
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Problem 4900

Prove the equalities:

lim⁡x→1−0(x−x4+x9−x16+…)=12. \lim _{x \rightarrow 1-0} (x-x^{4}+x^{9}-x^{16}+\ldots ) = \frac{1}{2} .
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Problem 4901

Prove the equality (Poisson's formula):

∑n=abf(n)= \sum _{n=a}^{b} f(n) = =f(a)+f(b)2−∫abf(x) dx+2∑n=1∞∫abf(x)cos⁡2πnx dx. = \frac{f(a)+f(b)}{2}-\int _{a}^{b} f(x)\, dx+2\sum _{n=1}^{\infty } \int _{a}^{b} f(x)\cos 2\pi nx\, dx .
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Problem 4902

The modulus of continuity ω(h)\omega (h) is the exact upper bound of the oscillation of the function f(x)f(x) on intervals of length hh, i.e. of the numbers ∣f(x2)−f(x1)∣\left|f(x_{2})-f(x_{1})\right|, where ∣x2−x1∣<h\left|x_{2}-x_{1}\right| < h. Prove that for a continuous function f(x)f(x) the integrals

∫abf(x)cos⁡nx dxand∫abf(x)sin⁡nx dx \int _{a}^{b} f(x)\cos nx\, dx \quad \text{and} \quad \int _{a}^{b} f(x)\sin nx\, dx

are, in absolute value, no greater than Cω(πn)C\omega \left(\dfrac {\pi }{n}\right), where CC is a corresponding constant.

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Problem 4903

Prove the equality:

∫0π/2∣sin⁡mx∣sin⁡x dx=2πln⁡m+O(1), \int _{0}^{\pi /2} \frac{\left|\sin mx\right|}{\sin x}\, dx = \frac{2}{\pi }\ln m+O(1),

where O(1)O(1), as m→∞m \rightarrow \infty, is a bounded quantity.

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Problem 4904

Prove Lebesgue's inequality:

∣a02+∑n=1m(ancos⁡nx+bnsin⁡nx)∣<CMln⁡m. \left|\frac{a_{0}}{2}+\sum _{n=1}^{m} (a_{n}\cos nx+b_{n}\sin nx)\right| < CM\ln m .

Here ana_{n} and bnb_{n} are the Fourier coefficients of the function f(x)f(x), and MM is an upper bound for ∣f(x)∣\left|f(x)\right|.

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Problem 4905

Prove that for 0<x<π0 < x < \pi the equality holds:

∑ν=1nsin⁡νxν=−x2+∫0x/2sin⁡(2n+1)tt dt+ \sum _{\nu =1}^{n} \frac{\sin \nu x}{\nu } = -\frac{x}{2}+\int _{0}^{x/2} \frac{\sin (2n+1)t}{t}\, dt+ +∫0x/2(1sin⁡t−1t)sin⁡(2n+1)t dt. +\int _{0}^{x/2} \left(\frac{1}{\sin t}-\frac{1}{t}\right) \sin (2n+1)t\, dt .
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Problem 4906

Prove that for any real xx and integer n>0n > 0 the equality holds

∑ν=1nsin⁡νxν=2πθ;∣θ∣<1. \sum _{\nu =1}^{n} \frac{\sin \nu x}{\nu } = 2\pi \theta ; \quad \left|\theta \right| < 1 .
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Problem 4907

For 0<x<π0 < x < \pi, prove the equality:

∑ν=1nsin⁡(2ν−1)x2ν−1= \sum _{\nu =1}^{n} \frac{\sin (2\nu -1)x}{2\nu -1} = =12∫0xsin⁡2ntt dt+12∫0x(1sin⁡t−1t)sin⁡2nt dt. = \frac{1}{2}\int _{0}^{x} \frac{\sin 2nt}{t}\, dt+\frac{1}{2}\int _{0}^{x} \left(\frac{1}{\sin t}-\frac{1}{t}\right) \sin 2nt\, dt .
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Problem 4908

For any real xx and integer n>0n > 0, prove the equality:

∑ν=1nsin⁡(2ν−1)x2ν−1=32θ;∣θ∣<1. \sum _{\nu =1}^{n} \frac{\sin (2\nu -1)x}{2\nu -1} = \frac{3}{2}\theta ; \quad \left|\theta \right| < 1 .
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Problem 4909

Prove Fejér's inequality:

∣cos⁡xn+cos⁡2xn−1+…+cos⁡nx1−cos⁡(n+1)x1−cos⁡(n+2)x2−…−cos⁡2nxn∣<10. \left|\frac{\cos x}{n}+\frac{\cos 2x}{n-1}+\ldots +\frac{\cos nx}{1}-\frac{\cos (n+1)x}{1}-\frac{\cos (n+2)x}{2}-\ldots -\frac{\cos 2nx}{n}\right| < 10 .
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Problem 4910

The Fejér sum σn\sigma_{n} is the arithmetic mean of the first nn Fourier sums of the function f(x)f(x):

nσn=s0+s1+…+sn−1, n\sigma _{n} = s_{0}+s_{1}+\ldots +s_{n-1},

where for m⩾1m \geqslant 1

sm=a02+∑ν=1m(aνcos⁡νx+bνsin⁡νx);s0=a02, s_{m} = \frac{a_{0}}{2}+\sum _{\nu =1}^{m} (a_{\nu }\cos \nu x+b_{\nu }\sin \nu x); \quad s_{0} = \frac{a_{0}}{2},

aνa_{\nu }, bνb_{\nu } being the Fourier coefficients of the function f(x)f(x). Prove that ∣σn(x)∣<M\left|\sigma_{n}(x)\right| < M, if ∣f(x)∣<M\left|f(x)\right| < M throughout the period.

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Problem 4911

Prove that if f(x)f(x) is a continuous function on the interval [−π,π]\left[-\pi ,\pi \right], l⩽f(x)⩽Ll \leqslant f(x) \leqslant L, and for every mm, ∣mam∣<A\left|ma_{m}\right| < A, ∣mbm∣<B\left|mb_{m}\right| < B, then

l−(A+B)⩽sn(x)⩽L+(A+B); l-(A+B) \leqslant s_{n}(x) \leqslant L+(A+B);

where ama_{m}, bmb_{m} are the Fourier coefficients, sn(x)s_{n}(x) is the nn-th partial sum of the Fourier series of f(x)f(x).

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Problem 4912

Prove that for every continuous function f(x)f(x) on the interval [−π,π]\left[-\pi ,\pi \right], σn(x)→f(x)\sigma_{n}(x) \rightarrow f(x) as n→∞n \rightarrow \infty.

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Problem 4913

Prove that for a periodic function f(x)f(x) satisfying the Cauchy--Lipschitz condition

∣f(x2)−f(x1)∣<C∣x2−x1∣, \left|f(x_{2})-f(x_{1})\right| < C\left|x_{2}-x_{1}\right|,

the inequality ∣σn−f(x)∣<Aln⁡nn\left|\sigma_{n}-f(x)\right| < \dfrac {A\ln n}{n} holds, where AA is a constant.

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Problem 4914

Taking f(x)f(x), wherever it occurs, to be a continuous function with period 2π2\pi, prove the following formulas and assertions:

The integral 2π⋅(2n)!!(2n−1)!!∫−ππf(x+u)cos⁡2nu2 du\dfrac {2}{\pi } \cdot \dfrac {(2n)!!}{(2n-1)!!} \int_{-\pi }^{\pi } f(x+u) \cos^{2n}\dfrac {u}{2}\, du is a trigonometric polynomial that tends to f(x)f(x) as n→∞n \rightarrow \infty.

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Problem 4915

Prove that

∫0π/2Fm(u) du>π3m−111m2;2π∫0π/2uFm(u) du<cm2, \int _{0}^{\pi /2} F_{m}(u)\, du > \frac{\pi }{3m}-\frac{1}{11m^{2}}; \qquad 2\pi \int _{0}^{\pi /2} u F_{m}(u)\, du < \frac{c}{m^{2}},

where Fm(u)=(sin⁡mumsin⁡u)4F_{m}(u) = \left(\dfrac {\sin mu}{m\sin u}\right)^{4}, and cc is some constant.

∫0π/2Fm(u) du=π3m+π6m3=12Am. \int _{0}^{\pi /2} F_{m}(u)\, du = \frac{\pi }{3m}+\frac{\pi }{6m^{3}} = \frac{1}{2A_{m}} .
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Problem 4916

The quantity Tm(x)T_{m}(x), where

Tm(x)=Am∫−π/2π/2f(x+2u)Fm(u) du, T_{m}(x) = A_{m} \int _{-\pi /2}^{\pi /2} f(x+2u) F_{m}(u)\, du,

(Fm(u)F_{m}(u) and AmA_{m} are defined in problem 4915) is a trigonometric polynomial of order 2m−22m-2.

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Problem 4917

Prove the equality:

f(x)−Tm(x)=Am∫−π/2π/2(f(x)−f(x+2u))Fm(u) du. f(x)-T_{m}(x) = A_{m} \int _{-\pi /2}^{\pi /2} \left(f(x)-f(x+2u)\right) F_{m}(u)\, du .

For the value of AmA_{m}, see problem 4915.

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Problem 4918

If f(x)f(x) satisfies the condition

∣f(x1)−f(x2)∣<C∣x2−x1∣, \left|f(x_{1})-f(x_{2})\right| < C\left|x_{2}-x_{1}\right|,

then Jackson's inequality holds:

∣f(x)−Tm(x)∣<Mm, \left|f(x)-T_{m}(x)\right| < \frac{M}{m},

where MM is a corresponding constant.

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Problem 4919

Let f(x)f(x) be a function integrable on the interval (0,2π)(0,2\pi ), let ana_{n} and bnb_{n} be its Fourier coefficients, let Sm=a02+∑n=1m(ancos⁡nx+bnsin⁡nx)S_{m} = \dfrac {a_{0}}{2}+\sum_{n=1}^{m} (a_{n}\cos nx+b_{n}\sin nx) be its Fourier sum, and let Pm(x)P_{m}(x) be any trigonometric polynomial of order mm, different from Sm(x)S_{m}(x). Prove the inequality:

∫02π(f(x)−Sm(x))2 dx<∫02π(f(x)−Pm(x))2 dx. \int _{0}^{2\pi } \left(f(x)-S_{m}(x)\right)^{2}\, dx < \int _{0}^{2\pi } \left(f(x)-P_{m}(x)\right)^{2}\, dx .
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Problem 4920

Prove Bessel's inequality:

a022+∑ν=1∞(aν2+bν2)⩽1π∫02π∣f(x)∣2 dx. \frac{a_{0}^{2}}{2}+\sum _{\nu =1}^{\infty } (a_{\nu }^{2}+b_{\nu }^{2}) \leqslant \frac{1}{\pi } \int _{0}^{2\pi } \left|f(x)\right|^{2}\, dx .
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Problem 4921

Starting from Bessel's inequality and the inequality 2ab⩽a2+b22ab \leqslant a^{2}+b^{2}, prove the convergence of the series ∑ν=1∞(aν2+bν2)\sum_{\nu =1}^{\infty } (a_{\nu }^{2}+b_{\nu }^{2}) and ∑ν=1∞∣aν∣+∣bν∣να\sum_{\nu =1}^{\infty } \dfrac {\left|a_{\nu }\right|+\left|b_{\nu }\right|}{\nu^{\alpha }}, where α>12\alpha > \dfrac {1}{2}.

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Problem 4922

Prove the inequality:

∫ab∣f(x)−φ(x)∣2 dx<2∫ab∣f(x)−ψ(x)∣2 dx+2∫ab∣ψ(x)−φ(x)∣2 dx. \int _{a}^{b} \left|f(x)-\varphi (x)\right|^{2}\, dx < 2\int _{a}^{b} \left|f(x)-\psi (x)\right|^{2}\, dx+2\int _{a}^{b} \left|\psi (x)-\varphi (x)\right|^{2}\, dx .
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Problem 4923

Let f(x)f(x) be a function integrable on (0,2π)(0,2\pi ), let ε\varepsilon be an arbitrary positive constant, let φ(x)\varphi (x) be a continuous function such that ∫02π∣f(x)−φ(x)∣2 dx<ε\int_{0}^{2\pi } \left|f(x)-\varphi (x)\right|^{2}\, dx < \varepsilon, and let σn(x)\sigma_{n}(x) be the Fejér sum (see problem 4910) for φ(x)\varphi (x), differing from φ(x)\varphi (x) by less than ε2π\sqrt{\dfrac {\varepsilon }{2\pi }}. Prove the inequality:

∫02π∣f(x)−σn(x)∣2 dx<4ε. \int _{0}^{2\pi } \left|f(x)-\sigma _{n}(x)\right|^{2}\, dx < 4\varepsilon .
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Problem 4924

Using the preceding results, prove A.M. Lyapunov's theorem (the completeness theorem):

a022+∑ν=1∞(aν2+bν2)=1π∫02πf2(x) dx, \frac{a_{0}^{2}}{2}+\sum _{\nu =1}^{\infty } (a_{\nu }^{2}+b_{\nu }^{2}) = \frac{1}{\pi } \int _{0}^{2\pi } f^{2}(x)\, dx,

valid for any integrable function f(x)f(x).

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Problem 4925

Prove the equality:

1π∫02πφ(x)ψ(x) dx=a0α02+∑ν=1∞(aναν+bνβν), \frac{1}{\pi } \int _{0}^{2\pi } \varphi (x)\psi (x)\, dx = \frac{a_{0}\alpha _{0}}{2}+\sum _{\nu =1}^{\infty } (a_{\nu }\alpha _{\nu }+b_{\nu }\beta _{\nu }),

where φ(x)\varphi (x) and ψ(x)\psi (x) are integrable on the interval (0,2π)(0,2\pi ), and aνa_{\nu }, bνb_{\nu }, αν\alpha_{\nu }, βν\beta_{\nu } are the Fourier coefficients of these functions.

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Problem 4926

Prove V.A. Steklov's theorem:

12h∫x−hx+hf(u) du=a02+∑ν=1∞(aνcos⁡νx+bνsin⁡νx)sin⁡νhνh. \frac{1}{2h} \int _{x-h}^{x+h} f(u)\, du = \frac{a_{0}}{2}+\sum _{\nu =1}^{\infty } (a_{\nu }\cos \nu x+b_{\nu }\sin \nu x)\frac{\sin \nu h}{\nu h} .
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Problem 4927

Prove V.A. Steklov's inequalities:

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(a)

∫02π∣f′(x)∣2 dx⩾∫02πf2(x) dx\displaystyle \int_{0}^{2\pi } \left|f'(x)\right|^{2}\, dx \geqslant \int_{0}^{2\pi } f^{2}(x)\, dx, if ∫0πf(x) dx=0\displaystyle \int_{0}^{\pi } f(x)\, dx = 0;

(b)

∫0π∣f′(x)∣2 dx⩾∫0πf2(x) dx\displaystyle \int_{0}^{\pi } \left|f'(x)\right|^{2}\, dx \geqslant \int_{0}^{\pi } f^{2}(x)\, dx; f(0)=f(π)=0f(0) = f(\pi ) = 0.

Problem 4928

Prove the equality:

1π∫02π(f(x)−σn(x))2 dx=1n2∑ν=1n−1ν2(aν2+bν2)+∑ν=n∞(aν2+bν2), \frac{1}{\pi } \int _{0}^{2\pi } \left(f(x)-\sigma _{n}(x)\right)^{2}\, dx = \frac{1}{n^{2}} \sum _{\nu =1}^{n-1} \nu ^{2}(a_{\nu }^{2}+b_{\nu }^{2})+\sum _{\nu =n}^{\infty } (a_{\nu }^{2}+b_{\nu }^{2}),

where σn(x)\sigma_{n}(x) is the Fejér sum of the function f(x)f(x) (see 4910).

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Problem 4929

Prove that for a function f(x)f(x) satisfying the Cauchy--Lipschitz condition

∣f(x2)−f(x1)∣<A⋅∣x2−x1∣α, \left|f(x_{2})-f(x_{1})\right| < A \cdot \left|x_{2}-x_{1}\right|^{\alpha },

the inequality holds:

∑ν=n2n(aν2+bν2)<kn2α, \sum _{\nu =n}^{2n} (a_{\nu }^{2}+b_{\nu }^{2}) < \frac{k}{n^{2\alpha }},

where kk is a corresponding constant.

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Problem 4930

Prove S.N. Bernstein's theorem: the Fourier series of a function satisfying the Cauchy--Lipschitz condition with α>12\alpha > \dfrac {1}{2} converges absolutely.

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Problem 4931

The parametric equation of a closed curve can be put in the form

x=∑n=1∞(ancos⁡nt+bnsin⁡nt);y=∑n=1∞(αncos⁡nt+βnsin⁡nt). x = \sum _{n=1}^{\infty } (a_{n}\cos nt+b_{n}\sin nt); \qquad y = \sum _{n=1}^{\infty } (\alpha _{n}\cos nt+\beta _{n}\sin nt) .

Using the formulas ds2=dx2+dy2ds^{2} = dx^{2}+dy^{2} and

A=12∫Sx dy−y dx, A = \frac{1}{2} \int _{S} x\, dy-y\, dx,

where AA is the area bounded by the closed curve SS, prove the inequality 4πA⩽L24\pi A \leqslant L^{2} (LL being the length of the curve SS), which expresses that among all closed curves of length LL the circle has the greatest area (Hurwitz's proof).

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Problem 4932

Expand in Chebyshev polynomials Tn(x)T_{n}(x) the function defined on the interval (−1,1)(-1,1) by the condition f(x)=x3f(x) = x^{3}.

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Remark.
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The Chebyshev polynomials are defined by the formula Tn(x)=12n−1cos⁡(narccos⁡x)T_{n}(x) = \dfrac {1}{2^{n-1}}\cos (n\arccos x) and have the orthogonality property: for n≠mn \neq m, ∫−11Tn(x)Tm(x) dx1−x2=0\displaystyle \int_{-1}^{1} \frac{T_{n}(x)T_{m}(x)\, dx}{\sqrt{1-x^{2}}} = 0, while for n=mn = m the right-hand side equals π22n−1\dfrac {\pi }{2^{2n-1}}.

Problem 4933

Expand in Chebyshev polynomials Tn(x)T_{n}(x) the function defined on the interval (−1,1)(-1,1) by the conditions:

f(x)={0,−1<x<0;1,0<x<1. f(x) = \begin{cases} 0, & -1 < x < 0; \\ 1, & 0 < x < 1 . \end{cases}
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Remark.
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The Chebyshev polynomials are defined by the formula Tn(x)=12n−1cos⁡(narccos⁡x)T_{n}(x) = \dfrac {1}{2^{n-1}}\cos (n\arccos x) and have the orthogonality property: for n≠mn \neq m, ∫−11Tn(x)Tm(x) dx1−x2=0\displaystyle \int_{-1}^{1} \frac{T_{n}(x)T_{m}(x)\, dx}{\sqrt{1-x^{2}}} = 0, while for n=mn = m the right-hand side equals π22n−1\dfrac {\pi }{2^{2n-1}}.

Problem 4934

Expand in Chebyshev polynomials Tn(x)T_{n}(x) the function defined on the interval (−1,1)(-1,1) by the condition f(x)=∣x∣f(x) = \left|x\right|.

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Remark.
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The Chebyshev polynomials are defined by the formula Tn(x)=12n−1cos⁡(narccos⁡x)T_{n}(x) = \dfrac {1}{2^{n-1}}\cos (n\arccos x) and have the orthogonality property: for n≠mn \neq m, ∫−11Tn(x)Tm(x) dx1−x2=0\displaystyle \int_{-1}^{1} \frac{T_{n}(x)T_{m}(x)\, dx}{\sqrt{1-x^{2}}} = 0, while for n=mn = m the right-hand side equals π22n−1\dfrac {\pi }{2^{2n-1}}.

Problem 4935

Expand in Chebyshev polynomials Tn(x)T_{n}(x) the function defined on the interval (−1,1)(-1,1) by the condition f(x)=x2nf(x) = x^{2n}.

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Remark.
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The Chebyshev polynomials are defined by the formula Tn(x)=12n−1cos⁡(narccos⁡x)T_{n}(x) = \dfrac {1}{2^{n-1}}\cos (n\arccos x) and have the orthogonality property: for n≠mn \neq m, ∫−11Tn(x)Tm(x) dx1−x2=0\displaystyle \int_{-1}^{1} \frac{T_{n}(x)T_{m}(x)\, dx}{\sqrt{1-x^{2}}} = 0, while for n=mn = m the right-hand side equals π22n−1\dfrac {\pi }{2^{2n-1}}.

Problem 4936

Expand, for −1<x<1-1 < x < 1, in Legendre polynomials the function defined by the conditions:

f(x)={0,−1<x<0;1,0<x<1. f(x) = \begin{cases} 0, & -1 < x < 0; \\ 1, & 0 < x < 1 . \end{cases}
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Remark.
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The Legendre polynomials are defined by the formula Pn(x)=12n⋅n!dn(x2−1)ndxnP_{n}(x) = \dfrac {1}{2^{n} \cdot n!}\dfrac {d^{n}(x^{2}-1)^{n}}{dx^{n}} and have the orthogonality property: for n≠mn \neq m, ∫−11Pn(x)Pm(x) dx=0\displaystyle \int_{-1}^{1} P_{n}(x)P_{m}(x)\, dx = 0, while for n=mn = m the right-hand side equals 22n+1\dfrac {2}{2n+1}.

Problem 4937

Expand, for −1<x<1-1 < x < 1, in Legendre polynomials the function defined by the condition f(x)=∣x∣f(x) = \left|x\right|.

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Remark.
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The Legendre polynomials are defined by the formula Pn(x)=12n⋅n!dn(x2−1)ndxnP_{n}(x) = \dfrac {1}{2^{n} \cdot n!}\dfrac {d^{n}(x^{2}-1)^{n}}{dx^{n}} and have the orthogonality property: for n≠mn \neq m, ∫−11Pn(x)Pm(x) dx=0\displaystyle \int_{-1}^{1} P_{n}(x)P_{m}(x)\, dx = 0, while for n=mn = m the right-hand side equals 22n+1\dfrac {2}{2n+1}.

Problem 4938

Expand in Laguerre polynomials Ln(x)L_{n}(x), for x>0x > 0, the function f(x)=e−axf(x) = e^{-ax}.

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Remark.
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The Abel--Laguerre polynomials are defined by the formula Ln(x)=exn!dnxne−xdxnL_{n}(x) = \dfrac {e^{x}}{n!}\dfrac {d^{n}x^{n}e^{-x}}{dx^{n}} and have the orthogonality property: for n≠mn \neq m, ∫0∞e−xLn(x)Lm(x) dx=0\displaystyle \int_{0}^{\infty } e^{-x}L_{n}(x)L_{m}(x)\, dx = 0, while for n=mn = m the right-hand side equals 11.

Problem 4939

Expand in Laguerre polynomials Ln(x)L_{n}(x), for x>0x > 0, the function f(x)=xnf(x) = x^{n}.

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Remark.
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The Abel--Laguerre polynomials are defined by the formula Ln(x)=exn!dnxne−xdxnL_{n}(x) = \dfrac {e^{x}}{n!}\dfrac {d^{n}x^{n}e^{-x}}{dx^{n}} and have the orthogonality property: for n≠mn \neq m, ∫0∞e−xLn(x)Lm(x) dx=0\displaystyle \int_{0}^{\infty } e^{-x}L_{n}(x)L_{m}(x)\, dx = 0, while for n=mn = m the right-hand side equals 11.

Problem 4940

Expand in Laguerre polynomials Ln(x)L_{n}(x), for x>0x > 0, the function f(x)=sin⁡axf(x) = \sin ax.

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Remark.
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The Abel--Laguerre polynomials are defined by the formula Ln(x)=exn!dnxne−xdxnL_{n}(x) = \dfrac {e^{x}}{n!}\dfrac {d^{n}x^{n}e^{-x}}{dx^{n}} and have the orthogonality property: for n≠mn \neq m, ∫0∞e−xLn(x)Lm(x) dx=0\displaystyle \int_{0}^{\infty } e^{-x}L_{n}(x)L_{m}(x)\, dx = 0, while for n=mn = m the right-hand side equals 11.

Problem 4941

Expand in Laguerre polynomials Ln(x)L_{n}(x), for x>0x > 0, the function f(x)=cos⁡axf(x) = \cos ax.

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Remark.
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The Abel--Laguerre polynomials are defined by the formula Ln(x)=exn!dnxne−xdxnL_{n}(x) = \dfrac {e^{x}}{n!}\dfrac {d^{n}x^{n}e^{-x}}{dx^{n}} and have the orthogonality property: for n≠mn \neq m, ∫0∞e−xLn(x)Lm(x) dx=0\displaystyle \int_{0}^{\infty } e^{-x}L_{n}(x)L_{m}(x)\, dx = 0, while for n=mn = m the right-hand side equals 11.

Problem 4942

Expand in Chebyshev--Hermite polynomials, for −∞<x<∞-\infty < x < \infty, the function

sign⁡x={−1for x<0;1for x>0. \operatorname {sign} x = \begin{cases} -1 & \text{for } x < 0; \\ 1 & \text{for } x > 0 . \end{cases}
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Remark.
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The Chebyshev--Hermite polynomials are defined (in the present chapter) by the formula Hn(x)=ex2/2n!dne−x2/2dxnH_{n}(x) = \dfrac {e^{x^{2}/2}}{n!}\dfrac {d^{n}e^{-x^{2}/2}}{dx^{n}} and have the orthogonality property: for n≠mn \neq m, ∫−∞∞e−x2/2Hn(x)Hm(x) dx=0\displaystyle \int_{-\infty }^{\infty } e^{-x^{2}/2}H_{n}(x)H_{m}(x)\, dx = 0, while for n=mn = m the right-hand side equals 2πn!\dfrac {\sqrt{2\pi }}{n!}.

Problem 4943

Expand in Chebyshev--Hermite polynomials, for −∞<x<∞-\infty < x < \infty, the function f(x)=∣x∣f(x) = \left|x\right|.

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Remark.
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The Chebyshev--Hermite polynomials are defined (in the present chapter) by the formula Hn(x)=ex2/2n!dne−x2/2dxnH_{n}(x) = \dfrac {e^{x^{2}/2}}{n!}\dfrac {d^{n}e^{-x^{2}/2}}{dx^{n}} and have the orthogonality property: for n≠mn \neq m, ∫−∞∞e−x2/2Hn(x)Hm(x) dx=0\displaystyle \int_{-\infty }^{\infty } e^{-x^{2}/2}H_{n}(x)H_{m}(x)\, dx = 0, while for n=mn = m the right-hand side equals 2πn!\dfrac {\sqrt{2\pi }}{n!}.

Problem 4944

Expand in Chebyshev--Hermite polynomials, for −∞<x<∞-\infty < x < \infty, the function f(x)=e−axf(x) = e^{-ax}.

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Remark.
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The Chebyshev--Hermite polynomials are defined (in the present chapter) by the formula Hn(x)=ex2/2n!dne−x2/2dxnH_{n}(x) = \dfrac {e^{x^{2}/2}}{n!}\dfrac {d^{n}e^{-x^{2}/2}}{dx^{n}} and have the orthogonality property: for n≠mn \neq m, ∫−∞∞e−x2/2Hn(x)Hm(x) dx=0\displaystyle \int_{-\infty }^{\infty } e^{-x^{2}/2}H_{n}(x)H_{m}(x)\, dx = 0, while for n=mn = m the right-hand side equals 2πn!\dfrac {\sqrt{2\pi }}{n!}.

Problem 4945

Expand in Chebyshev--Hermite polynomials, for −∞<x<∞-\infty < x < \infty, the function f(x)=cos⁡axf(x) = \cos ax.

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Remark.
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The Chebyshev--Hermite polynomials are defined (in the present chapter) by the formula Hn(x)=ex2/2n!dne−x2/2dxnH_{n}(x) = \dfrac {e^{x^{2}/2}}{n!}\dfrac {d^{n}e^{-x^{2}/2}}{dx^{n}} and have the orthogonality property: for n≠mn \neq m, ∫−∞∞e−x2/2Hn(x)Hm(x) dx=0\displaystyle \int_{-\infty }^{\infty } e^{-x^{2}/2}H_{n}(x)H_{m}(x)\, dx = 0, while for n=mn = m the right-hand side equals 2πn!\dfrac {\sqrt{2\pi }}{n!}.

Problem 4946

Expand in Chebyshev--Hermite polynomials, for −∞<x<∞-\infty < x < \infty, the function f(x)=sin⁡axf(x) = \sin ax.

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Remark.
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The Chebyshev--Hermite polynomials are defined (in the present chapter) by the formula Hn(x)=ex2/2n!dne−x2/2dxnH_{n}(x) = \dfrac {e^{x^{2}/2}}{n!}\dfrac {d^{n}e^{-x^{2}/2}}{dx^{n}} and have the orthogonality property: for n≠mn \neq m, ∫−∞∞e−x2/2Hn(x)Hm(x) dx=0\displaystyle \int_{-\infty }^{\infty } e^{-x^{2}/2}H_{n}(x)H_{m}(x)\, dx = 0, while for n=mn = m the right-hand side equals 2πn!\dfrac {\sqrt{2\pi }}{n!}.

Problem 4947

Represent as a Fourier integral the function

f(x)={e−mx,x>0;emx,x<0;m>0. f(x) = \begin{cases} e^{-mx}, & x > 0; \\ e^{mx}, & x < 0; \quad m > 0 . \end{cases}
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Problem 4948

Represent as a Fourier integral the function

f(x)={e−mx,x>0;−emx,x<0;m>0. f(x) = \begin{cases} e^{-mx}, & x > 0; \\ -e^{mx}, & x < 0; \quad m > 0 . \end{cases}
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Problem 4949

Represent as a Fourier integral the function

f(x)={e−mx,x>0;0,x<0;m>0. f(x) = \begin{cases} e^{-mx}, & x > 0; \\ 0, & x < 0; \quad m > 0 . \end{cases}
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Problem 4950

Represent as a Fourier integral the function

f(x)={1,∣x∣<τ;0,∣x∣>τ. f(x) = \begin{cases} 1, & \left|x\right| < \tau ; \\ 0, & \left|x\right| > \tau . \end{cases}
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Problem 4951

Represent as a Fourier integral the function

f(x)={1,0<x<τ;−1,−τ<x<0;0,∣x∣>τ. f(x) = \begin{cases} 1, & 0 < x < \tau ; \\ -1, & -\tau < x < 0; \\ 0, & \left|x\right| > \tau . \end{cases}
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Problem 4952

Represent as a Fourier integral the function

f(x)={A(1−∣x−x0∣a),∣x−x0∣<a;0,∣x−x0∣>a. f(x) = \begin{cases} A\left(1-\dfrac {\left|x-x_{0}\right|}{a}\right), & \left|x-x_{0}\right| < a; \\ 0, & \left|x-x_{0}\right| > a . \end{cases}
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Problem 4953

Represent as a Fourier integral the function

f(x)={e−kxsin⁡ωx,x>0;0,x<0;k>0. f(x) = \begin{cases} e^{-kx}\sin \omega x, & x > 0; \\ 0, & x < 0; \quad k > 0 . \end{cases}
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Problem 4954

Represent as a Fourier integral the function

f(x)={sin⁡x,0<x<nπ;n an integer;0,x<0; x>nπ. f(x) = \begin{cases} \sin x, & 0 < x < n\pi ; \quad n \text{ an integer}; \\ 0, & x < 0; \, x > n\pi . \end{cases}
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