Half-range Fourier expansions

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Expand the function f(x)=x2f(x)=x^{2} into a Fourier series:

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(a)

as a cosine expansion;

(b)

as a sine expansion;

(c)

as an expansion on the interval (0,2π)(0,2\pi ).

Plot the graph of the function and the graphs of the sums of the Fourier series in cases (1), (2), and (3). Using these expansions, find the sums of the series:

∑n=1∞1n2,∑n=1∞(−1)n+1n2and∑n=1∞1(2n−1)2. \sum _{n=1}^{\infty } \frac{1}{n^{2}}, \quad \sum _{n=1}^{\infty } \frac{(-1)^{n+1}}{n^{2}} \quad \text{and} \quad \sum _{n=1}^{\infty } \frac{1}{(2n-1)^{2}}.

How should the integrable function f(x)f(x), given on the interval (0,π2)\left(0,\dfrac {\pi }{2}\right), be extended to the interval (−π,π)(-\pi ,\pi ) so that its expansion into a Fourier series has the form

f(x)=∑n=1∞ancos⁡(2n−1)x(−π<x<π)? f(x)=\sum _{n=1}^{\infty } a_{n}\cos (2n-1)x \quad (-\pi <x<\pi )?
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How should the integrable function f(x)f(x), given on the interval (0,π2)\left(0,\dfrac {\pi }{2}\right), be extended to the interval (−π,π)(-\pi ,\pi ) so that its expansion into a Fourier series has the form

f(x)=∑n=1∞bnsin⁡(2n−1)x(−π<x<π)? f(x)=\sum _{n=1}^{\infty } b_{n}\sin (2n-1)x \quad (-\pi <x<\pi )?
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On the interval (0,π2)\left(0,\dfrac {\pi }{2}\right), expand the function

f(x)=x(π2−x) f(x)=x\left(\frac{\pi }{2}-x\right)

as follows:

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(a)

in cosines of odd multiples of the angle;

(b)

in sines of odd multiples of the angle.

Sketch the graphs of the sums of the Fourier series in cases (1) and (2).

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

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

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

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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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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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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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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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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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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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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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Expand into a Fourier series the function f(x)=π−2x,0<x⩽πf(x) = \pi -2 x, 0 < x \leqslant \pi, extending it to the segment [−π;0][-\pi ; 0]:

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(1)

evenly;

(2)

oddly.

In both cases, draw the graph of the sum of the series.

Expand into a Fourier series the function f(x)f(x), given on the segment [0;π][0 ; \pi ] and extended to the segment [−π;0][-\pi ; 0] evenly. Draw the graph of the sum of the series:

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(1)

f(x)={x,0⩽x⩽π/2π/2,π/2<x⩽πf(x) = \begin{cases} x, & 0 \leqslant x \leqslant \pi / 2 \\ \pi / 2, & \pi / 2 < x \leqslant \pi \end{cases}

(2)

f(x)=2π(π2−x),0⩽x⩽πf(x) = \frac{2}{\pi }\left(\frac{\pi }{2}-x\right), 0 \leqslant x \leqslant \pi.

Expand the function f(x)=x,0⩽x⩽πf(x) = x, 0 \leqslant x \leqslant \pi, into a Fourier series in cosines.

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Expand the function f(x)=cos⁡2x,0⩽x⩽πf(x) = \cos 2 x, 0 \leqslant x \leqslant \pi, into a Fourier series in sines.

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Expand into a Fourier series on (0;π)(0 ; \pi ) in cosines the function

f(x)={π/2−x,0<x<π/20,π/2⩽x<π f(x) = \begin{cases} \pi / 2-x, & 0 < x < \pi / 2 \\ 0, & \pi / 2 \leqslant x < \pi \end{cases}
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Expand into a Fourier series on the interval (0;π)(0 ; \pi ) in sines the function

f(x)={sin⁡x,0⩽x⩽π/20,π/2<x<π f(x) = \begin{cases} \sin x, & 0 \leqslant x \leqslant \pi / 2 \\ 0, & \pi / 2 < x < \pi \end{cases}
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Expand the function

f(x)={x,0⩽x⩽12−x,1<x⩽2 f(x) = \begin{cases} x, & 0 \leqslant x \leqslant 1 \\ 2-x, & 1 < x \leqslant 2 \end{cases}

into a Fourier series in cosines on the segment [0;2][0 ; 2].

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Expand the function f(x)=x2f(x) = x^{2} into a Fourier series:

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(1)

on the segment [−π;π][-\pi ; \pi ] in cosines;

(2)

on the interval (0;π)(0 ; \pi ) in sines;

(3)

on the interval (0;2π)(0 ; 2 \pi ) in sines and cosines.

Using these expansions, find the sums of the series

S1=∑n=1∞1n2,S2=∑n=1∞(−1)n+1n2,S3=∑n=1∞1(2n−1)2 S_{1} = \sum _{n = 1}^{\infty } \frac{1}{n^{2}}, \quad S_{2} = \sum _{n = 1}^{\infty } \frac{(-1)^{n+1}}{n^{2}}, \quad S_{3} = \sum _{n = 1}^{\infty } \frac{1}{(2 n-1)^{2}}

Expand into a Fourier series the function f(x)=x−x2/2,0⩽x⩽1f(x) = x-x^{2} / 2,0 \leqslant x \leqslant 1:

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(1)

in cosines;

(2)

in sines.

Expand into a Fourier series in sines the function

f(x)=xsin⁡x,0⩽x⩽π. f(x) = x \sin x, \quad 0 \leqslant x \leqslant \pi .
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Prove that 3x2−6πx+2π212=∑n=1∞cos⁡nxn2,0⩽x⩽π\frac{3 x^{2}-6 \pi x+2 \pi^{2}}{12} = \sum_{n = 1}^{\infty } \frac{\cos n x}{n^{2}}, 0 \leqslant x \leqslant \pi.

Hint. Use the results of problems 40 and 45.

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Expand into a Fourier series in cosines the function f(x)=exf(x) = e^{x} on the interval (0;ln⁡2)(0 ; \ln 2).

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Expand into a Fourier series in sines the function

f(x)={1,0<x<π/20,π/2<x<π f(x) = \begin{cases} 1, & 0 < x < \pi / 2 \\ 0, & \pi / 2 < x < \pi \end{cases}
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Expand into a Fourier series in sines the function

f(x)={x,0⩽x<12−x,1⩽x<2 f(x) = \begin{cases} x, & 0 \leqslant x < 1 \\ 2-x, & 1 \leqslant x < 2 \end{cases}
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Expand the function f(x)=eax,0<x<πf(x) = e^{a x}, 0 < x < \pi:

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(1)

into a Fourier series in cosines;

(2)

into a Fourier series in sines.

Expand the function f(x)=sin⁡ax,0⩽x⩽πf(x) = \sin a x, 0 \leqslant x \leqslant \pi, into a Fourier series in cosines.

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Expand into a Fourier series in cosines the function

f(x)={1,0⩽x⩽a0,a<x⩽π f(x) = \begin{cases} 1, & 0 \leqslant x \leqslant a \\ 0, & a < x \leqslant \pi \end{cases}
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Expand into a Fourier series in cosines the function

f(x)={1−x/2a,0⩽x⩽2a0,2a<x⩽π f(x) = \begin{cases} 1-x / 2 a, & 0 \leqslant x \leqslant 2 a \\ 0, & 2 a < x \leqslant \pi \end{cases}
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Prove that

∑n=1∞cos⁡(2n−1)x(2n−1)4=196π(π−2x)(π2+2πx−2x2),0⩽x⩽π \sum _{n = 1}^{\infty } \frac{\cos (2 n-1) x}{(2 n-1)^{4}} = \frac{1}{96} \pi (\pi -2 x)\left(\pi ^{2}+2 \pi x-2 x^{2}\right), \quad 0 \leqslant x \leqslant \pi
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Prove that if a function ff, absolutely integrable on the segment [0;π][0 ; \pi ], satisfies the condition f(π−x)=f(x)f(\pi -x) = f(x), then its Fourier coefficients have the following properties:

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(1)

when ff is expanded into a Fourier series in cosines, a2n−1=0,n∈Na_{2 n-1} = 0, n \in N;

(2)

when ff is expanded into a Fourier series in sines, b2n=0,n∈Nb_{2 n} = 0, n \in N.

How should a function, absolutely integrable on the segment [0;π/2][0 ; \pi / 2], be extended to the segment [−π;π][-\pi ; \pi ] so that its Fourier series has the form

∑n=1∞ancos⁡(2n−1)x? \sum _{n = 1}^{\infty } a_{n} \cos (2 n-1) x?
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How should a function, absolutely integrable on the segment [0;π/2][0 ; \pi / 2], be extended to the segment [−π;π][-\pi ; \pi ] so that its Fourier series has the form

∑n=1∞bnsin⁡(2n−1)x? \sum _{n = 1}^{\infty } b_{n} \sin (2 n-1) x?
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Expand the function f(x)=x(π/2−x)f(x) = x(\pi / 2-x) into a Fourier series on the segment [0;π/2][0 ; \pi / 2]:

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(1)

with respect to the system {cos⁡(2n−1)x},n∈N\left\{ \cos (2 n-1) x\right\} , n \in N;

(2)

with respect to the system {sin⁡(2n−1)x},n∈N\left\{ \sin (2 n-1) x\right\} , n \in N.

Prove that a piecewise smooth function on the segment [0;l][0 ; l] can be expanded into a series of the form

∑n=1∞b2n−1sin⁡(2n−1)πxl \sum _{n = 1}^{\infty } b_{2 n-1} \sin \frac{(2 n-1) \pi x}{l}
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Prove that a piecewise smooth function on the segment [0;l][0 ; l] can be expanded into a series of the form

∑n=1∞a2ncos⁡2nπxl \sum _{n = 1}^{\infty } a_{2 n} \cos \frac{2 n \pi x}{l}
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Prove that if the function ff is continuously differentiable on the segment [x;x+π][x ; x+\pi ], then

f(x+π)−f(x)=−4π∑n=0∞∫xx+πf(t)cos⁡((2n+1)(t−x))dt f(x+\pi )-f(x) = -\frac{4}{\pi } \sum _{n = 0}^{\infty } \int _{x}^{x+\pi } f(t) \cos ((2 n+1)(t-x)) d t

(Steklov).

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References

[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.
[ККЧ+03] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Интегралы. Ряды. 2-е изд., ФИЗМАТЛИТ, 2003.