Exercícios Série de Fourier

Prévia do material em texto

1) Considere a extensão periódica de f (x )={ 1 , se 0< x<π0 , se −π≤x≤0} , (L=π)
Calcule a série de Fourier de f (x )
a0=
1
π∫
−π
π
f ( x)dx =
1
π∫
−π
0
0dx+
1
π∫
0
π
1dx =
1
π∫
0
π
1dx =
1
π ( x )0
π
= ππ = 1
an=
1
π∫
−π
π
f ( x)cos (n x )dx =
1
π∫
−π
0
0 cos (n x )dx+
1
π∫
0
π
1 cos(n x )dx =
1
π∫
0
π
cos (n x)dx
=
1
π(sen (n x)n )0
π
=
1
π(sen (nπ)n )−1π(
sen (0)
n ) = 0
bn=
1
π∫
−π
π
f (x )sen(n x)dx =
1
π∫
−π
0
0sen (n x)dx+
1
π∫
0
π
1sen(n x)dx =
1
π∫
0
π
sen (n x)dx
=
1
π(−cos (n x)n )0
π
=
1
π(−cos (nπ)n )−1π(
−cos(0)
n ) =
1
π(−(−1)
n
n )−1π(−1n )
=
1
π((−1)
n+1
n )+1π(1n) = ((−1)
n+1
πn )+( 1πn) = (−1)
n+1
+1
πn
f (x )=
1
2
+
1
π∑
n=1
+∞
((−1)
n+1
+1
n )sen (nx)
f (x )=
1
2
+
2
π(sen ( x)+sen(3 x)3 +
sen (5 x)
5
+
sen (7 x)
7
...)
2) Considere a extensão periódica de f (x )=x , x∈(−π ,π) , (L=π)
Calcule a série de Fourier de f (x )
Função ímpar
a0=0 , an=0
bn=
2
π∫
0
π
x sen (n x)dx
f (x )=∑
n=1
+∞
bnsen (n x )
∫ x sen(n x)dx integração por partes
u= x
du=dx
v=−
cos(n x)
n
dv=sen(n x )dx
, ∫u dv=uv−∫v du = −
x cos (n x )
n
−∫(− cos(n x )n )dx
= −
x cos (n x )
n
+
1
n
∫cos (n x)dx = −
x cos (n x )
n
+
1
n
sen(n x)
n
= −
x cos (n x )
n
+
sen (n x )
n2
bn=
2
π∫
0
π
x sen (n x)dx =
2
π(− x cos (n x )n +sen (n x)n2 )0
π
=
2
π(−π cos (nπ)n +sen (nπ)n2 )−
2
π(−0cos (0)n +sen(0)n2 ) =
2
π(−π cos(nπ)n )
= 2(−(−1)
n
n ) = 2(−1)
n+1
n
f (x )=2∑
n=1
+∞
((−1)
n+1
n )sen (n x )
f (x )=2(sen(x )−sen(2 x)2 +
sen (3 x )
3
−
sen (4 x)
4
+
sen(5 x )
5
...)
3) Considere a extensão periódica de f (x )=x2 , x∈(−π ,π) , (L=π)
Calcule a série de Fourier de f (x ) usando integração de série de Fourier
g ( x)=2 x , ∫ g ( x)dx= x2+C
série de Fourier de
g ( x)=2 x = 4∑
n=1
+∞
((−1)
n+1
n )sen (n x ) , onde a0=0
∫
0
x
2t dt=4∫
0
x
∑
n=1
+∞
((−1)
n+1
n )sen (n t )dt = 4(∑n=1
+∞
((−1)
n+1
n )(− cos(n t)n ))0
x
= 4(∑
n=1
+∞
((−1)
n+1
n )(− cos(n x )n ))−4(∑n=1
+∞
((−1)
n+1
n )(−cos (0)n ))
= 4∑
n=1
+∞ (−1)n cos(n x)
n2
−4∑
n=1
+∞ (−1)n
n2
f (x )=
a0
2
+∑
n=1
+∞
an cos(n x )+bnsen (n x) = ∑
n=1
+∞ 4 (−1)n cos (n x )
n2
−4∑
n=1
+∞ (−1)n
n2
an=
4(−1)n
n2
a0
2
=−4∑
n=1
+∞ (−1)n
n2
, a0=−8∑
n=1
+∞ (−1)n
n2
a0=
1
π∫
−π
π
x2dx = 1π( x
3
3 )−π
π
=
1
π(π
3
3 )−
1
π((−π)
3
3 ) = π
3
3π
+ π
3
3π
= 2 π
2
3
f (x )=π
2
3
+∑
n=1
+∞ 4 (−1)n cos(n x )
n2
f (x )=π
2
3
+4(−cos (x )+ cos(2 x)4 −
cos (3 x)
9
+
cos(4 x )
16
−
cos(5 x )
25
...)
OBS: lim
x→π
f ( x)=π2=π
2
3
+4(−cos(π)+cos (2π)4 −
cos (3π)
9
+
cos (4π)
16
−
cos (5π)
25
...)
π
2
=π
2
3
+4(1+14+
1
9
+
1
16
+
1
25
...) , π2−π
2
3
=4(1+ 14+
1
9
+
1
16
+
1
25
...)
2 π
2
3
=4(1+14+
1
9
+
1
16
+
1
25
...) , π
2
6
=1+
1
4
+
1
9
+
1
16
+
1
25
...
4) Considere a extensão periódica de f (x )=x2 , x∈(−π ,π) , (L=π)
Use a identidade de Parseval para mostrar que π
4
90
=1+
1
16
+
1
81
+
1
256
+
1
625
...
1
L
∫
−L
L
( f (x ))
2
dx=
a0
2
2
+∑
n=1
+∞
an
2
+bn
2
1
π∫
−π
π
( x2)
2
dx=
a0
2
2
+∑
n=1
+∞
an
2
+bn
2 , onde a0=2
π
2
3
e an=
4(−1)n
n2
e bn=0
1
π∫
−π
π
x4dx=
(2 π
2
3 )
2
2
+∑
n=1
+∞
(4(−1)
n
n2 )
2
,
1
π∫
−π
π
x4dx=
(4 π
4
9 )
2
+∑
n=1
+∞
(16n4 )
1
π∫
−π
π
x4dx=
(4 π
4
9 )
2
+16∑
n=1
+∞
( 1n4) ,
1
π( x
5
5 )−π
π
=2 π
4
9
+16∑
n=1
+∞
( 1n4)
1
π(π
5
5 )−
1
π((−π)
5
5 )=2 π
4
9
+16∑
n=1
+∞
( 1n4) ,
1
π(π
5
5 )+
1
π (π
5
5 )=2 π
4
9
+16∑
n=1
+∞
( 1n4)
2
π(π
5
5 )=2 π
4
9
+16∑
n=1
+∞
( 1n4) , 2 π
4
5
=2 π
4
9
+16∑
n=1
+∞
( 1n4) , π
4
5
=π
4
9
+8∑
n=1
+∞
( 1n4)
π
4
5
−π
4
9
=8∑
n=1
+∞
( 1n4) , 4 π
4
45
=8∑
n=1
+∞
( 1n4) então π
4
90
=∑
n=1
+∞
( 1n4)