матмод (4сем) / 2кр ряды фурье / примеры fourier4

1

Za

b

(x) sin

nπx

1

Z0

2

bn =

f

dx

=

f (x) sin(nπx) dx

L

L

1

=

Z0

1

0 · sin(nπx) dx + Z1

2 x · sin(nπx) dx

integrate by parts: u = x,

dv = sin(nπx) dx

1

2

1

2

=

0 −

x cos(nπx) 1

+

Z1

cos(nπx) dx

nπ

nπ

1

1

2

=

−

[2 cos(2nπ) − cos(nπ)] +

sin(nπx) 1

nπ

n2π2

1

1

2

− (−1)

n

n

=

−

[2

− (−1)

] +

[sin(2nπ) − sin(nπ)] =

−

+ 0

nπ

n2π2

nπ

bn = −

2 − (−1)n

n = 1, 2, . . .

nπ

a0

∞

nπx

nπx

f (x) =

F (x) =

+ n=1 han cos

+ bn sin

i

for 0 < x < 2

2

L

L

X

3

∞

1

− (−1)n

2 − (−1)n

sin(nπx)

=

+ n=1

4

n2π2

cos(nπx)

− nπ

X

f (x) =

4

+ π n=1 n "

1

nπ

cos(nπx) − [ 2 − (−1)n ] sin(nπx)#

0 < x < 2

3

1

∞

1

− (−1)n

X

2

F(x)

1

0

-3

-2

-1

0

1

2

3

x

2

f0(x) =

x,

−2 < x < 1

fo(x)1

0, −1 ≤ x ≤ 1

0

x,

1 < x < 2

-1

-2

-2

-1

0

1

2

x

∞

nπx

f (x) = FS (x) = n=1 bn sin

,

with L = 2.

L

X

bn

=

L Z0

L

f (x) sin

L

dx

=

2

Z0

2

f (x) sin

2

dx

2

nπx

2

nπx

=

Z0

1

0 · sin

2

2

x · sin

2

dx

dx + Z1

nπx

nπx

nπx

2

nπx

2

2

2

nπx

2

0 − nπ x cos

1

integrate by parts:

u = x,

dv = sin

dx

=

2

+ nπ Z1

cos

2

dx

=

−nπ

2

+

n2π2

sin

2

2

2 cos(nπ) − cos

1

2

h

nπ

i

4

nπx

2

n

nπ

4

nπ

=

−

h2(−1)

− cos

i +

hsin(nπ) − sin

i

nπ

2

n2π2

2

=

−

2

h2(−1)n − cos

nπ

i +

4

h0 − sin

nπ

i

nπ

2

n2π2

2

bn

=

2

2(−1)n − cos

nπ

+

2

sin

nπ

n = 1, 2, . . .

−

nπ

2

nπ

2

∞

nπx

f (x) = FS (x) =

n=1 bn sin

for 0 < x < 2

L

X

f (x) =

2 ∞ 1

2(−1)n − cos

nπ

2

nπ

sin

nπx

0 < x < 2

−π n=1 n

2

+ nπ sin

2

2

X

2

1

FS (x)

0

-1

-2

-5

-4

-3

-2

-1

0

1

2

3

4

5

x

2

−x,

−2 < x < 1

fe(x)

fe(x) =

0, −1 ≤ x ≤ 1

1

x,

1 < x < 2

0

-2

-1

0

1

2

x

Note that f (x) is defined on the interval [0,2), whereas fe(x) (the even extension of f (x)) is defined on the

interval (-2,2). The Fourier Cosine series of f (x) on the interval [0,2) is equivalent to the Fourier

series of fe(x) on the interval (-2,2). In this case, we have L = 2 and bn = 0 for all n = 1, 2, . . ..

For 0 < x < 2, set

a0

∞

nπx

f (x) = FC (x) =

+ n=1 an cos

with L = 2.

2

L ,

X

2

L

2

2

1

2

3

a0 =

Z0

f (x) dx

=

Z0

f (x) dx = Z0

0 dx + Z1

x dx = 0 +

L

2

2

a0

=

3

2

an

=

L

f (x) cos

L

dx

=

2

2

f (x) cos

2

dx

L Z0

Z0

2

nπx

2

nπx

=

1

0 · cos

2

x · cos

2

dx

Z0

2 dx + Z1

nπx

nπx

nπx

2

nπx

2

2

2

nπx

=

= cos

2

0 + nπ x sin

1

integrate by parts:

u

x,

dv

dx

=

2

− nπ Z1

sin

2

dx

2

nπ

4

nπx

2

=

+

cos

1

nπ h2 sin(nπ) − sin

2

i

n2π2

2

2

nπ

4

nπ

=

h0 − sin

i +

hcos(nπ) − cos

i

nπ

2

n2π2

2

=

2

sin

nπ

+

4

h(−1)n − cos

nπ

i

−

nπ

2

n2π2

2

2

2

nπ

nπ

an

=

(−1)n − cos

− sin

n = 1, 2, . . .

nπ

nπ

2

2

a0

∞

nπx

f (x) = FC (x) =

+ n=1 an cos

for 0 < x < 2

2

L

X

f (x) =

3

+

2 ∞ 1

2

(−1)n − cos

nπ

− sin

nπ

cos

nπx

0 < x < 2

4

π n=1 n

nπ

2

2

2

X