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Higher Mathematics. Part 3

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cos nπ

sin nx

−π

−

∫ sin nxdx

π2

n+1

2π

(−1)

n+1

(−1)

n (−1)

cos nx

πn3

Therefore

∞

2π

f (x) = ∑

(−1)n+1 sin nx + ∑

(−1)n −1 sin nx

n=1

πn3

− 1 .

If п is even number then (−1)n −1 = 0 .

Therefore

sin x

sin 2x

sin 3x

8 sin x

sin 3x

sin 5x

f (x) = 2π

−

− ... −

+ ... .

This equality is fulfilled for

x [0;

π] ,

except point

x = π,

in which the

sum of series equals 0, but

f (π) = π2 .

Example 5. Expand function

f (x) =

−1, if

x (−1; 0],

( f (x + 2)

= f (x)

(Fig. 3.6)) in Fourier series.

x (0; 1),

2x, if

π2

–2π

–π

2π

–1

– π2

–1

Fig. 3.5

Fig. 3.6

Solution. Function f (x) is piecewise monotone and periodic with period

2l = 2 . Therefore we can expand it in Fourier series. We find Fourier coefficients by formulas (3.13):

0	1	0	1
a0 = ∫ (−1)dx + ∫ 2xdx = − x		+ x2	= −1+ 1 = 0;
−1	0	−1	0

sin nπx

= ∫ (−1) cos nπxdx + ∫ 2x cos nπxdx = −

−1

nπ

−1

sin nπx

+2x

−

∫ sin nπxdx =

cos nπx

nπ

((−1)

− 1)

n = 2k,

(cos nπ − 1) =

, if n = 2k

π2

−

(2k − 1)

cos nπx

= ∫ (−1) sin nπxdx + ∫ 2x sin nπxdx =

−1

nπ

−1

− cos nπx

− cos nπx dx =

(1− (−1)n )−

+2x

−

∫ 2

cos nπ +

nπ

−

n = 2k,

sin nπx

−

3(−1)

) =

n2π2

nπ

(2k −1)π , if n = 2k − 1.

Therefore Fourier series looks like

∞		−4			∞	1
f (x) = ∑					cos(2k −1)πx −∑		sin 2kπx +
	(	)	2
k =1				π2	k =1 kπ
		2k − 1

∞4

+k∑=1 (2k − 1)π sin(2k − 1) πx.

Sum of Fourier series S(x) is

f (x), if x (−1+ 2k; 2k) (2k; 1		+ 2k),
	− 0,5, if x = 2k,	k Z.
S(x) =
	0,5, if x = −1+ 2k,

− 1;

0 +

Example 6. Expand function (Fig. 3.7, а) in Fourier series on segment [1; 4] . Solution. Let’s write the given function analytically:

	1, if	x [1;3),
f (x) =	4 − x, if	x [3;4].
	4 − x, if	x [3;4].

Let’s extend this function periodically with a period T = 2l = 3 . For calculation of Fourier coefficients we’ll use formulas (3.14). To simplify a calculation let’s consider a new function f1 (x) :

f1		1, if	x [0;3),	(Fig.3.7, b).
	(x) =	4 − x, if	x [3;4]

On the segment [1; 4]

the function

f1 (x)

coincides with the function

f (x) .

Let’s extend

f1 (x)

on the interval [−4; 0) in an odd way (Fig. 3.7, c).

y = f(x)

y = f1(x)

О 1

3 4

О 1

3 4

–4

3 4

–1

Fig. 3.7

On the segment [−4; 4]

this new function is odd and its Fourier series does

not contain cosines: a0

= an

∞

= 0 . That is f (x) = ∑ bn sin πnx , l = 4 .

Let’s find bn :

n=1

2 4

πnx

∫

f1 (x) sin

dx =

∫

sin

dx +

∫

(4 − x) sin

4 0

dx =

2 0

4 cos

πnx

−4 cos

πnx

−

+ (4 − x)

−

∫3

cos

πn

cos

3πn

− 1

3πn

sin πnx

−

cos

−

πn

3πn

−

cos

− 1

cos

sin

πn

Therefore Fourier series is

∞

3πn

πnx

f (x) = ∑

sin

πn

n 1

This equality is fulfilled for all

x [−4; 0) (0; 4] .

Micromodule 3

CLASS AND HOME ASSIGNMENTS

Expand the given

2π -periodic functions defined on interval

(−π; π) in

Fourier series.

3, if

x (−π;0),

1. f (x) = x.

2. f (x) = 1+

3. f (x) =

x [0;

π).

−1, if

Expand

the

given 2π -periodic

function

Fourier

series

f (x) = x ,

x (0; 2 π],

f (x + 2π) = f (x) .

Expand the given functions defined on interval (0; π) in cosine series:

5. f (x) = x

1, if

x (0;π / 2),

6. f (x) =

0, if x [π / 2; π).

7. Expand the function

f (x) = π

−

defined on interval (0; π) in sine series.

Expand the functions in Fourier series.

1, if

x [−1; 0),

9. f (x) =

0, if

x [−3;1),

f (x) =

x [0; 1).

x [1;3].

−1, if

x, if

10. Expand the function

f (x) = 2x defined on interval (0; 2)

in cosine

series.

Answers

∞

sin kx

sin 2x

sin 3x

sin 4x

∞

sin(2k − 1)x

1. 2∑(−1)k

2.1 + sin x −

−

+ ….

1 −

∑

k =1

2k − 1

∞

sin kx

π2

∞

coskx

∞

cos(2k − 1)x

∞

sin 2kx

4. π − 2∑

+ 4∑(−1)k

∑(−1)k +1

. 7.

∑

2k − 1

k 1

k =1

π k =1

k 1

∞

sin(2k − 1)π x

∞

πn

πnx

1 ∞

−

∑

sin

(−1)

− cos

cos

∑×

2k − 1

πn

π k =1

3 n=1

πn

πnx

∞

cos

π(2k − 1)x

n+1

3(−1)

+ cos

sin

. 10.

4 −

∑

(2k − 1)

πn

k =1

Micromodule 3

SELF-TEST ASSIGNMENTS

3.1. Expand 2π -periodic function f (x) defined on interval (−π; π)

in Fourier

series. Sketch the graph of Fourier series sum.

	0,				if x (−π; 0),
3.1.1. f (x) =						x [0; π).
	x −1, if					x [0; π).
	0,				if	x (−π; 0),
3.1.3. f (x) =							x [0; π).
	x + 2, if						x [0; π).
3.1.5.	0,				if	x (−π; 0),
3.1.5.	f (x) =					x [0; π).
	x / 2, if					x [0; π).
3.1.7.	3 − x, if						x (−π; 0),
3.1.7.	f (x) =		1,		if		x [0;π).
			1,		if		x [0;π).
			0,		if		x (−π; 0),
3.1.9. f (x) =							x [0;π).
	4x − 3, if						x [0;π).
3.1.11.	f (x) =			1,		if		x (−π; 0),
3.1.11.	f (x) =							x [0;π).
		2x − 5, if						x [0;π).
3.1.13.	f (x) =			1,		if		x (−π; 0),
3.1.13.	f (x) =
		π − 2x, if x [0; π).
3.1.15.	f (x) =				0,	if		x (−π; 0),
3.1.15.	f (x) =							x [0;π).
		1− 2x, if						x [0;π).
3.1.17.	f (x) =				0,	if		x (−π; 0),
3.1.17.	f (x) =							x [0;π).
		4 − 2x, if						x [0;π).
3.1.19.	f (x) =				0,	if		x (−π; 0),
3.1.19.	f (x) =							x [0;π).
		6x − 5, if						x [0;π).
					0,	if		x (−π; 0),
3.1.21.	f (x) =		π
3.1.21.	f (x) =		π	− x, if				x [0; π).
			4	− x, if				x [0; π).
			4
3.1.23.	f (x) =				0,	if		x (−π; 0),
3.1.23.	f (x) =							x [0; π).
		2x − 3, if						x [0; π).
					0,		if	x (−π; 0),
3.1.25.	f (x) =		π
3.1.25.	f (x) =		π	+ 2x, if x [0; π).
			4	+ 2x, if x [0; π).
			4

3.1.2.	2x, if						x (−π; 0),
	f (x) =			if			x [0;π).
	1,
3.1.4.	1− x, if							x (−π; 0),
	f (x) =					if		x [0; π).
	−1,
			0,			if		x (−π; 0),
3.1.6. f (x) =								x [0;π).
	2x + 3, if
3.1.8.			0,			if		x (−π; 0),
	f (x) =							x [0; π).
	x − 2, if
3.1.10.	f (x) =	4 − x, if						x (−π; 0),
			− 1,				if	x [0; π).

3.1.12.	f (x) =	3 − 2x, if							x (−π; 0),
				0,			if		x [0;π).

3.1.14.	f (x) =	5x + 1, if							x (−π; 0),
				0,			if		x [0;π).

3.1.16.	f (x) =	3x + 2, if							x (−π; 0),
				0,			if		x [0;π).

3.1.18.	f (x) =	x + π, if							x (−π; 0),
				0,			if		x [0;π).

3.1.20.	f (x) =	π − 2x, if x (−π; 0),
				0,			if		x [0;π).

3.1.22.	f (x) =	π + x, if							x (−π; 0),
				0,			if		x [0;π).

		π		− 2x, if					x (−π; 0),
3.1.24.	f (x) =		2

				0,			if		x [0;π).

					x				x (−π; 0),
3.1.26.	f (x) =	1		−			, if
					2

				0,			if		x [0;π).

									85

			0,	if	x (−π; 0),		1+ 2x, if			x (−π; 0),
3.1.27.						3.1.28.
3.1.27.	f (x) = x		− 1, if		x [0;π).	3.1.28.	f (x) =	0,	if	x [0;π).

		3
		3
3.1.29.			0,	if	x (−π; 0),	3.1.30.	1− 3x, if			x (−π; 0),
3.1.29.	f (x) =					3.1.30.	f (x) =	0,	if	x [0;π).
	3x − 5, if x [0; π).							0,	if	x [0;π).

3.2.Expand the function f (x) defined on segment [0;l] : а) in sine series;

b)in cosine series.

	x		for x [0;1),
3.2.1. f (x) =			for x [1;2].
	2
	x − 1 for x [0;2),
3.2.3. f (x) =		1		for x [2;4].

3.2.5.	1− x				for x [0;1),
	f (x) =		−1		for x [1;2].
	x
3.2.7.	2 − x				for x [0;2),
	f (x) =		0		for x [2;3].

3.2.9.	x − 1				for x [0;1),
	f (x) =		0		for x [1;4].

		x		for x [0;1),
3.2.11.	f (x) =			for x [1;2],
		1
				for x (2;3].
		2
3.2.13. f (x) =		2 − x			for x [0; 4].
			x		for x [0;1),
3.2.15.	f (x) =		1		for x [1;2],

		3 − x			for x (2;3].

3.2.17.	f (x) =	2 − x for x [0;1),
			1		for x [1;4].

			0		for x [0;1),
3.2.19.	f (x) =
		x − 2 for x [1;2],
		3 − x			for x (2;3].

		1			for x [0;2),
3.2.2. f (x) =					for x [2;3].
	3 − x
3.2.4.	−1					for x [0;1),
	f (x) =		− x			for x [1;3].
	2
3.2.6.	2				for x [0;2),
	f (x) =				for x [2;5].
	−1
3.2.8.			x			for x [0;1),
	f (x) =		− x			for x [1;2].
	2
3.2.10.	f (x) =	0				for x [0;2),
				− x		for x [2;4].
		2
		− x			for x [0;1),
3.2.12.	f (x) =		1		for x [1; 2],


		−1 for x (2;3].
3.2.14.	f (x) = x			for x [0;6].
		1		− x		for x [0;1),
3.2.16.	f (x) =			0		for x [1; 2],

				− 2		for x (2;3].
		x
3.2.18.	f (x) =	− x				for x [0;1),
				− 2		for x [1;2].
		x
				1		for x [0;1),
3.2.20.	f (x) =
		2 − x for x [1; 2],
				0		for x (2;3].

3.2.21. f (x)

3.2.23. f (x)

3.2.25. f (x)

3.2.27. f (x)

3.2.29. f (x)

−1 for x [0;1),

=x − 2 for x [1;2].

2 for x [0;1),

=1 for x [1;2],0 for x (2;3].

− 1		for x [0;1),
	x	for x [1;2],
=
	2	for x (2;3].

= x − 3		for x [0;6].
= − 2		for x [0; 2),
3		for x [2; 3].

3.2.22.	− 2 for x [0; 2),
	f (x) =	1	for x [2;4].

	−1		for x [0;1),
3.2.24.		0	for x [1;3],
	f (x) =
		2	for x (3; 4].

	− x		for x [0;1),
3.2.26.		0	for x [1;3],
	f (x) =
			for x (3;4].
	x
3.2.28.	f (x) = − x		for x [0;4].
3.2.30.	f (x) = − 3		for x [0; 3),
	1		for x [3; 5].

Micromodule 4

BASIC THEORETICAL INFORMATION. FOURIER INTEGRAL

Fourier integral. Fourier integral for odd and even functions. Complex form of Fourier integral. Fourier transformation.

Key words: amplitude spectrum — амплітудний спектр, phase spectrum —

фазовий спектр, Fourier cosine-transform — косинус-перетворення Фур’є, Fourier sine-transform — синус-перетворення Фур’є, wave numbers — хви-

льові числа, spectrum density — спектральна щільність.

Literature: [9, chapter 9, §4], [14, chapter 3, §5], [15, chapter 13, part 13.5], [16, chapter 17, §12–14], [17, chapter 6, §22].

4.1. Fourier Integral. Fourier Transformation

Any piecewise monotone function f (x) defined on interval (a; b) may be represented as a Fourier series. Fourier integral is the analog of Fourier series for functions defined on (−∞; ∞) .

Suppose a non-periodic function f (x) defined on (−∞; ∞)			obeys all con-
ditions of Dirichlet’s theorem on any finite segment [−l; l] . Let			f (x) be abso-
lutely integrable. That is,
∞
∫	f (x)	dx < ∞ .

−∞
			87

Then we can expand this function in Fourier series (3.12) on a segment [−l; l] . If we substitute values of coefficients obtained in (3.13), we receive

∞

f (x) =

∫ f (t)dt + 1

∑

∫ f (t)(cosωnt cosωn x + sin ωnt sin ωn x)dt =

−l

n=1

−l

∞

∫ f (t)dt +

∑

∫ f (t)cosωn (t − x)dt.

(4.1)

−l

n=1

−l

Here ω = πn ( n = 1, 2,... ) are the wave numbers of a function

f (x) .

Let’s denote

= π

− ω

= Δω

( n = 1, 2,... ).

n+1

Then formula (4.1) will be

∞

f (x) =

∫ f (t)dt +

∑

∫ f (t)cosωn (t − x)dt π ,

−l

π n=1

−l

∞

f (x) =

∫

f (t)dt +

∑

∫

(t)cosω (t − x)dt Δω .

−l

π n=1

−l

Let’s consider a limit as l → ∞ . The first term tends to zero because

lim

∫ f (t)dt ≤ lim

∫

f (t)

dt < lim

= 0.

l→∞

−l

→∞

2l −l

l→∞

Let’s denote

Then

ϕ(ωn ) = ∫

−l

f (

f (t)cosωn (t − x)dt.

	1 lim	∞
x) =	1 lim	∑ϕ(ωn )Δωn .	(4.2)
	π l→∞ n=1

The right side of formula (4.2) is the integral sum of the function

ϕ(ω) = ∫ f (t)cosω(t − x)dt, ω (0; ∞).

−l

Suppose l → ∞ Δωn → 0, that is, wave numbers ωn get all possible values from 0 to +∞ . Then point spectrum is continuous. Formula (4.2) looks like

f (x) 1 ∞ ϕ( )d

= π ∫0 ω ω

	1	∞ ∞
f (x) =		∫ ∫	f (t) cos ω(t − x)dt dω.	(4.3)

	π 0 −∞

Definition. Formula (4.3) is called a Fourier formula. The right side of it is

called Fourier integral of a function f (x) .

This formula is true for all points of continuity of a function f (x) . If

x0 is a

point of discontinuity then

1 ∞ ∞

f (x + 0) + f (x − 0)

∫

f (t) cos ω(t − x)dt dω =

−∞

Let’s write Fourier integral as

∞

∫

f (x) =

f (t) cos(ωt − ωx)dt dω =

0 −∞

∞

∫

f (t)(cos ωt cos ωx

+ sin ωt sin ωx)dt dω =

0 −∞

∞

∫

(t) cos ωtdt cos ωxdω +

f (t) sin ωtdt sin

ωxdω.

0 −∞

Let’s denote

∞

a(ω) =

∫ f (t) cos ωtdt, b(ω) =

∫

f (t) sin ωtdt,

(4.4)

−∞

then

∞

f (x) = ∫ (a(ω) cos ωx + b(ω) sin ωx)dω.

(4.5)

4.2. Fourier Integral for Odd and Even Functions

Suppose f (x)

is an even function then

f (t) cos ωt

is an even function too and

f (t) sin ωt

is an odd function.

Suppose f (x)

is an odd function then

f (t) cos ωt is an odd function too

and f (t) sin ωt

is an even function.

In these cases formulas (4.4), (4.5) look like those in Table 4.1.

								Table 4.1

Property of	f (x) is an even function				f (x)			is an odd function
a function f (x)	f (x) is an even function				f (x)			is an odd function
a function f (x)

		∞				∞
Fourier integral	f (x) = ∫ a(ω) cos ωxdω				f (x) = ∫ b(ω) sin ωxdω
	0					0
		2	∞		a(ω) = 0,
Fourier	a(ω) =	2		f (t) cos ωtdt ,			∞
Fourier	a(ω) =			f (t) cos ωtdt ,		2	∞
coefficients		π		∫0	b(ω) =	2	∫	f (t) sin ωtdt
	B(ω) = 0.				b(ω) =	π	∫	f (t) sin ωtdt
	B(ω) = 0.					π	0

4.3. Complex Form of Fourier Integral.

Fourier Transformation

Suppose a function f (x) is presented by Fourier integral (4.5). We use formulas

cos ωx =

eiωx

+ e−iωx

, sin ωx =

eiωx − e−iωx

Then

∞

f (x) =

∫

F(ω)eiωx dω,

(4.6)

2π

−∞

where

∞

F(ω) = ∫

f (t)e−iωt dt.

(4.7)

−∞

It follows from formulas (4.6) and (4.7) that

∞

−iωx

iωx

f (x) =

∫

f (t)e

dω.

(4.8)

2π

dt e

−∞ −∞

Definition. The right side of a formula (4.8) is called a Fourier integral in complex form for the function f (x) .

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