parcial 3 práctica # 3
DESCRIPTION
practicaTRANSCRIPT
Tema #3
Series de Senos y Series de Cosenos
Propiedades
El producto de dos funciones pares es par. El producto de dos funciones impares es par. El producto de una función par por una impar es impar.
Conclusión
La serie de Fourier de una Función Par −p<x< p es la Serie Coseno
f ( x )a02
+∑n=1
∞
ancosnπxp
Donde a0=2p∫0
p
f ( x )dx an=2p∫0
p
f ( x ) cos nπxpdx
La serie de Fourier de una Función Impar −p<x< p es la Serie Seno
f ( x ) ∑n=1
∞
bnSennπxp
Donde bn=2p∫0
p
f ( x )Sen nπxpdx
Práctica
23. f (x)
bn=2p∫0
p
f ( x )sin nπpx dx
bn=2π∫0
p
1sin nππx dx
bn=2π∫0π
sinnxdx
bn=2π
¿
bn=2π
[−cosnπn
+cos n (0 )n
]
bn=2π
[−(−1)n
n+ 1n]
bn=2π
[1−(−1)n
n]
f (x)∑n=1
∞ 2[1−(−1 )n]πn
sin nx→f (x) 2π∑n=1
∞ 1−(−1 )n
nsin nx
- 1, - x 0
1, 0 x
24. f (x)
A0=2p∫0
p
f ( x )dx
A0=22∫0
2
1dx
A0=22∫0
2
dx
A0=22
¿
A0=22[2−0 ]
A0=2
f (x)∑n=1
∞ 1nπ
[1+(−1)n]sin nπ2x
bn=2p∫0
p
f ( x )sin nπpx dx
bn=22∫02
1sin nπ2x dx
bn=22∫02
sin nπ2xdx
bn=22¿
bn=22[−2cos nπ
2(2 )
nπ+2cos nπ
2(0)
nπ]
bn=22[−(−1)n
nπ+ 1nπ
]
bn=1nπ
[1+(−1)n]
1, -2 x -1
0, -1 x 1
1, 1 x 2
25. f(x)= |x|, - x
A0=2p∫0
p
f ( x )dx
A0=2π∫0
π
xdx
A0=2π
¿
A0=2π
[ (π )2
2−(0)2
2]
A0=2π
[ π2
2]
A0=π
f (x) π2∑n=1
∞ 2[ (−1 )n−1]π n2
cosnx
f ( x ) π2
+ 2π∑n=1
∞ (−1 )n−1n2
cosnx
An=2p∫0p
f ( x )cos nπpx dx
An=2π∫0
π
x cos nππx dx
An=2π∫0
π
x cosnx dx
An=2π
¿
An=2π
[ x sin nxn
+ cosnxn2
]0
π
An=2π
¿]
An=2π
[(−1 )n
n2− 1n2
]
An=1n2π
[(−1 )n−1]
26. f(x) = x, - x
bn=2p∫0
p
f ( x )sin nπpx dx
bn=2π∫0π
x sin nππx dx
bn=2π
¿
bn=2π
¿
bn=2π
[−x cosnxn
+ sin nxn2
]0
π
bn=2π
[−π cosnπn
+ sin nπn2
+ (0 )cos n (0 )n
− sin n (0 )n2
]
bn=2π
[−π (−1 )n
n]
bn=2π
[π (−1 )n+1
n]
bn=2 (−1 )n+1
nf ( x)2∑
n=1
∞ (−1 )n+1
nsin nx
27. f(x)= x2, -1 x 1
A0=2p∫0
p
f ( x )dx
A0=21∫01
x2dx
A0=2[x3
3]0
1
A0=2[(1 )3
3−
(0 )3
3]
A0=2[(1 )3
3]
A0=23
An=2p∫0
p
f ( x )cos nπpx dx
An=21∫01
x2 cos nπ1x dx
An=21∫01
x2cosnπ x dx
An=2¿
An=2¿
An=2[x2sin nπxnπ
+2 x cosnπxn2π2
−2sin nπxn3π3
]0
1
An=2¿]
An=2¿]
An=2[2cosnπn2π2
]
An=4cosnπn2π2
An=4(−1)n
n2π 2
f ( x ) 22 (3 ) ∑n=1
∞ 4 (−1 )n
n2π2cosn π x
f ( x ) 13+ 4π 2∑n=1
∞ (−1 )n
n2cosnπ x
28. f(x) = x |x|, -1 x 1
bn=2p∫0
p
f ( x )sin nπpx dx
bn=21∫0
1
( x )( x)sin nπ1xdx
bn=2∫0
1
x2 sin nπx dx
bn=2 ¿
bn=2 ¿
bn=2¿
bn=2[−x2 cosnπx
nπ+ 2 x sin nπx
n2π2−2(−cosnπxn3π3 )]
0
1
bn=2[−x2 cosnπx
nπ+ 2 x sin nπx
n2π2+ 2cosnπx
n3π3]0
1
bn=2 ¿ ]
bn=2 ¿]
bn=2[− (−1 )n
nπ+2 (−1 )n
n3π3− 2n3 π3
]
bn=2π
[(−1 )n+1
n+2 (−1 )n
n2π2− 2n2π2
]
f (x)∑n=1
∞ 2π [ (−1 )n+1
n+2 (−1 )n
n2π2− 2n2π2 ]sin n π x
f (x) 2π∑n=1
∞ [ (−1 )n+1
n+ 2 (−1 )n
n2π2− 2n2π2 ]sin n π x
29. f(x) = 2 - x2, - x
A0=2p∫0
p
f ( x )dx
A0=2π∫0
π
(π2−x2 )dx
A0=2π
¿ ]
A0=2π
¿
A0=2π
¿¿]
A0=2π
[π¿¿3−π3
3]¿
A0=2π
[ 3 π3−π3
3]
A0=2π
[ 2π3
3]
A0=4 π2
3
An=2p∫0p
f ( x )cos nπpx dx
An=2π∫0
π
(π¿¿2−x2)cos nππx dx¿
An=2π∫0
π
(π¿¿2−x2)cosnx dx¿
An=2π
[π2∫0
π
cosnx dx−∫0
π
x2cos nxdx ]
An=2π
{(π2 )( sin nxn )−[(x2 )( sinnxn )−2∫0
π
x sin nxn
dx]}
An=2π
[ π2 sin nxn
− x2sin nxn
+2∫0
π
x sinnxn
dx ]
An=2π
¿]
An=2π
¿]
An=2π
[ π2 sinnxn
− x2sinnxn
−2xcos nxn2
+ 2sin nxn3
]0
π
An=2π
[ π2 sinnπn
−(π )2 sin nπ
n−2π cosnπ
n2+ 2sin n π
n3−π2 sin n (0 )
n+
(0 )2sin nπn
+2 (0 )cos n (0 )
n2−2sin n (0 )n3
]
An=2π
[−2 π cosnπn2
]
An=2π
[−2 π (−1)n
n2]
An=2π
[2π (−1)n+1
n2]
An=4(−1)n+1
n2
f ( x ) 4π2
2(3)∑n=1∞ 4 (−1 )n+1
n2cosnx
f ( x ) 2 π2
3+4∑
n=1
∞ (−1 )n+1
n2cos nx
30. f(x) = x3, - x
bn=2p∫0
p
f ( x )sin nπpx dx
bn=2π∫0
π
x3 sin n xdx
bn=2π
¿
bn=2π
¿
bn=2π
¿
bn=2π
¿
bn=2π
¿
bn=2π
¿
bn=2π
¿
bn=2π
[−x3cosnxn
+ 3 x2sin n xn2
−6 x cosnxn3
−6sin nxn4
]0
π
bn=2π
[−(π )3 cosnπ
n+3(π )2sin n π
n2−6 (π)cosnπ
n3−6sin n π
n4]0
π
bn=2π
[− (π )3 cosnπ
n+3 (π )2 sin nπ
n2−6 (π ) cosnπ
n3−6sin n π
n4+
(0 )3 cosn (0 )n
−3 (0 )2 sin n (0 )
n2+6 (0 )cosn (0 )
n3+6sin n (0 )n4
]
bn=2π
[−π3 cosnπn
+ 3π2 sin nπn2
−6 πcosnπn3
−6sin n πn4
]
bn=2π
[−π3 cosnπn
−6 πcosnπn3
]
bn=2π
[−π3(−1)n
n−6π (−1)n
n3]
f (x)∑n=1
∞ 2π [−π3 (−1 )n
n+6 π (−1 )n
n3 ]sin nx
f ( x )2∑n=1
∞ [−π2 (−1 )n
n+6 (−1 )n
n3 ]sin nx
31. f(x)
bn=2p∫0
p
f ( x )sin nπpx dx
bn=2π∫0
π
(x+1)sin nππx dx
x – 1, - x 0
x + 1, 0 x
bn=2π∫0
π
(x+1)sin n xdx
bn=2π
¿]
bn=2π
[ ( x )(−cosnxn )−∫0
π−cosnxn
dx+(−cosnxn )]
bn=2π
[− x cosnxn 0
π
+∫0
π cos nxn
dx− cosnxn 0
π
]
bn=2π
[−x cosnxn
+ sinnxn2
− cosnxn
]0
π
bn=2π
[−πcos nπn
+ sin nπn2
− cosnπn
+ (0 ) cosn (0 )n
− sin n (0 )n2
+ cos n (0 )n
]
bn=2π
[−πcos nπn
−cos nπn
+ 1n]
bn=2π
[−π (−1 )n
n− (−1 )n
n+ 1n]
bn=2π
[−π (−1 )n− (−1 )n+1
n]
f (x)∑n=1
∞ 2π [1−π (−1)n−(−1)n
n ]sin nx
f (x) 2π∑n=1
∞ [1−(−1 )n(1+π)n ]sin nx
32. f(x)
bn=2p∫0
p
f ( x )sin nπpx dx
bn=21∫0
1
(x−1)sin nπ1xdx
bn=2π∫0
1
(x−1)sin nπ x dx
x + 1, -1 x 0
x – 1, 0 x 1
bn=2π
¿]
bn=2π
[ ( x )(−cosnxn )−∫0
1−cosnxn
dx−(−cos nxn )]
bn=2π
[− x cosnxn 0
1
+∫0
1 cos nxn
dx+ cos nxn 0
1
]
bn=2π
[−x cosnxn
+ sin nxn2
+cos nxn
]0
1
bn=2π
¿]
bn=2π
[−πcos nπnπ
− cosn πnπ
− 1nπ
]
bn=2π
[−π (−1 )n
nπ− (−1 )n
nπ− 1nπ
]
bn=2π
[−π (−1 )n− (−1 )n−1
nππ ]
f (x)∑n=1
∞ 2π [−π (−1 )n− (−1 )n−1
nπ ]sin nπx
f (x) 2π∑n=1
∞ [−π (−1 )n− (−1 )n−1nπ ]sin nπx
33. f(x)
A0=2p∫0
p
f ( x )dx
A0=22∫0
2
(x+1 ) dx
A0=1¿
A0=1¿
1, -2 x -1
-x, -1 x 0
x, 0 x 1
1, 1 x 2
A0=1[x2
2 0
1
+x12 ]
A0=1[(1 )2
2−
(0 )2
2+(2 )−(1 )]
A0=1[12+1]
A0=1[1+22
]
A0=1[ 32 ]
A0=32
An=2p∫0
p
f ( x )cos nπpx dx
An=22∫02
(x+1 ) cos nπ2x dx
An=1¿
An=1¿
An=1¿
An=1¿
An=1[2 x sin nπ
2x
nπ 0
1
+4 cos nπ
2x
n2π2 0
1
+2sin nπ2x
nπ 1
2
]
An=1[2 (1 )sin nπ
2(1 )
nπ+4cos nπ
2(1 )
n2π2−2 (0 ) sin nπ
2(0 )
nπ−4 cos nπ
2(0 )
n2π2+2sin nπ
2(2 )
nπ−2sin nπ
2(1 )
nπ]
An=1[4 cos nπ
2(1 )
n2π2−4cos nπ
2(0 )
n2 π2]
An=1[4 cos nπ
2n2π2
−4(1)n2π2
]
An=4cos nπ
2n2π2
−4(1)n2π2
An=4cos nπ
2−4
n2π 2
f ( x ) 32(2)∑n=1
∞
[4cos nπ
2−4
n2π 2¿]cos nπ
2x¿
f ( x ) 34+ 4π2
∑n=1
∞
[cos nπ
2−1
n2¿]cos nπ
2x ¿
34. f(x)
A0=2p∫0
p
f ( x )d x
-, -2 x -
x, - x
, x 2
A0=22 π∫0
2π
π dx
A0=22 π
(π )∫0
2π
dx
A0=1 [ x ]02π
A0=1[2π−0]
A0=1[2π ]
A0=2π
bn=2p∫0
p
f ( x )sin nπpx dx
bn=22π∫0
2π
π sin nπ2πxdx
bn=22π
(π )∫0
2π
sin n2xdx
bn=1π
(π )∫0
2π
sin n2x dx
bn=1∫0
2 π
sin n2x dx
bn=1[−2cos n
2x
n]0
2π
35. f(x) = |sen x|, - x
A0=2p∫0p
f ( x )dx
A0=2π∫0
π
sin x dx
A0=2π
¿¿
A0=2π
¿
A0=2π
[1−(−1 )]
A0=2π
[1+1]
A0=2π
[2]
A0=4π
An=2p∫0
p
f ( x )cos nπpx dx
An=2π∫0
π
(sin x ) cos nππx dx
An=2π∫0
π
(sin x ) cosnx dx
An=2π∫0
π 12¿¿
An=2π
¿
An=2π
¿¿
An=2π
¿
An=2π
[−(−1)n
2 (1+n )−
(−1)n
2 (1−n )+ 12 (1+n )
+ 12 (1−n )
]
An=2π
[(−1)n
(1−n2 )+ 1
(1−n2 )]
An=2π
[1+(−1 )n
(1−n2)]
f (x) 4π (2)∑n=2
∞ 2π [ 1+(−1 )n
(1−n2 ) ]cos nx
f ( x ) 2π +2π∑n=2
∞ [ 1+(−1 )n
(1−n2 ) ]cosnx
36. f(x) = cos x, -/2 x /2
A0=2p∫0
p
f ( x )dx
A0=2π2
∫0
π2
cos xdx
A0=4π
¿¿
A0=4π
¿
A0=4π
¿
A0=4π
[1]
A0=4π
An=2p∫0
p
f ( x )cos nπpx dx
An=2π2
∫0
π2
(cos x )cos nππ2
xdx
An=4π∫0
π2
(cos x ) cos2nx dx
An=4π∫0
π212¿¿
An=2π
¿
An=4π
¿
An=4π
¿
An=4π
[ (−1 )n
2 (1+2n )+ (−1)n
2 (1−2n )]
An=4π
[ (2+4n ) (−1 )n+(2−4 n)(−1)n
2 (1+2n )2 (1−2n )]
An=4π
[ (2+4n ) (−1 )n+(2−4 n)(−1)n
2 (1+2n )2 (1−2n )]
An=4π
[ (2+4n ) (−1 )n+(2−4 n)(−1)n
(2+4n ) (2−4 n )]
An=4π
[ (2+4n ) (−1 )n+(2−4 n)(−1)n
4−8n+8n−16n2]
An=4π
[ 2 (−1 )n+2(−1)n
4−16n2]
An=4π
[4 (−1 )n
4(1−4 n2)]
An=4π
[ (−1 )n
(1−4 n2)]
f ( x ) 4π (2)∑n=2
∞ 4π [ (−1 )n
(1−4 n2) ]cos nππ2
x
f ( x ) 2π
+ 4π∑n=2
∞ [ (−1 )n
(1−4n2) ]cos2n x37. f(x)
A0=1p∫− p
p
f ( x )dx
A0=112
∫0
1
(1+0)dx
A0=21
¿]
A0=2[∫0
12
dx ]A0=2[x ]0
12
1, 0 x 1/2
0, 1/2 x 1
A0=2[12−0 ]
A0=2[12]
A0=1
An=1p∫− p
p
f ( x ) cos nπpx dx
An=112
∫0
12
(1+0 ) cos nπ1x dx
An=21
¿
An=2¿
An=2[sin nπ ( 12 )nπ
−sin nπ (0 )nπ
]
An=2[sin nπ
2nπ
]
An=2π
[sin nπ
2n
]
f ( x ) 12∑n=1
∞ 2π [ sin nπ2n ]cos nπ1 x
f ( x ) 12+ 2π∑n=1
∞ [ sin nπ2n ]cosnπ xbn=
1p∫−p
p
f (x ) sin nπpx dx
bn=112
∫0
12
(1+0 ) sin nπ1xdx
bn=21
¿
bn=2¿]
bn=2[−cosnπ ( 12 )
nπ+cos nπ (0 )nπ
]
bn=2[−cos nπ
2nπ
+ 1nπ
]
bn=2[1−cos nπ
2nπ
]
bn=2π
[1−cos nπ
2n
]
f ( x )∑n=1
∞ 2π [1−cos nπ2n ]sin nπ1 x
f ( x ) 2π∑n=1
∞ [ 1−cos nπ2n ]sin nπ x38. f(x)
A0=1p∫− p
p
f ( x )dx
A0=11∫01
(0+1)dx
A0=1¿]
A0=1[∫12
1
dx ]A0=1[x ]1
2
1
A0=1[1−12]
A0=1[12]
0, 0 x 1/2
1, 1/2 x 1
A0=12
f ( x ) 12(2)∑n=1
∞
[−1nπ ]cos nπ1 xf ( x ) 1
4+ 1π∑n=1
∞ [−1n ]cos nπ x
An=1p∫− p
p
f ( x ) cos nπpx dx
An=11∫0
12
(0+1 )cos nπ1xdx
An=11
¿
An=1¿
An=1[sin nπ (1 )nπ
−sin nπ ( 12 )nπ
]
An=1[−sin nπ
2nπ
]
An=−1π
[sin nπ
2n
]
An=−1π
[sin nπ
2n
]
An=−1nπ
bn=1p∫−p
p
f (x ) sin nπpx dx
bn=11∫0
12
(0+1 ) sin nπ1xdx
bn=11
¿
bn=1¿]
bn=1[−cosnπ (1 )
nπ+cosnπ ( 12 )nπ
]
bn=1[−cosnπnπ
]
bn=1[−(−1)n
nπ]
bn=1[−(−1)n
nπ]
bn=(−1)n+1
nπ
f ( x )∑n=1
∞ [ (−1)n+1nπ ]sin nπ1 x
f ( x ) 1π∑n=1
∞ [ (−1)n+1n ]sinnπ x39. f(x) = cos x , 0 x /2
A0=2p∫0p
f ( x )dx
A0=2π2
∫0
π2
cos xdx
A0=( 21 )( 2π )∫0
π2
cos x dx
A0=( 4π )¿¿
A0=4π
¿
A0=4π
[sin(π2 )]
A0=4π
[1]
A0=4π
An=2p∫0
p
f ( x )cos nπpx dx
An=2π2
∫0
π2
cos x¿¿¿
An=( 21 )( 2π )∫0
π2
cos x¿¿¿
An=4π∫0
π2
cos x¿¿¿
An=4π∫0
π212¿¿
An=4π
¿
An=4π
¿
An=4π
[(−1 )n
2 (1+2n )+
(−1)n
2 (1−2n )]
An=4π
[ (2+4n ) (−1 )n+(2−4 n)(−1)n
2 (1+2n )2 (1−2n )]
An=4π
[ (2+4n ) (−1 )n+(2−4 n)(−1)n
(2+4n ) (2−4 n )]
An=4π
[(2+4n ) (−1 )n+(2−4 n)(−1)n
4−8n+8n−16n2]
An=4π
[2 (−1 )n+2(−1)n
4−16n2]
An=4π
[ 4 (−1 )n
4(1−4 n2)]
An=4π
[ (−1 )n
(1−4 n2)]
f ( x ) 4π (2)∑n=1
∞ 4π [ (−1 )n
(1−4 n2) ]cos nππ2
x
f ( x ) 2π
+ 4π∑n=1
∞ [ (−1 )n
(1−4n2) ]cos2n x
bn=2p∫0
p
f ( x )sin nπpx dx
bn=2π2
∫0
π2
cos x sin nππ2
x dx
bn=( 21 )( 2π )∫0
π2
cos x ¿¿¿
bn=4π∫0
π2
cos x¿¿¿
bn=4π∫0
π212¿¿
bn=4π
¿
bn=4π
[ 12 (1+2n )
− 12 (1−2n )
]
bn=4π [
(2−4 n ) (1 )−(2+4 n )(1)2 (1+2n )2 (1−2n )
]
bn=4π
[(2−4 n )−(2+4n )
(2+4n ) (2−4n )]
bn=4π
[ 2−4 n−2−4 n4−8n+8n−16n2
]
bn=4π
[−4n−4 n4−16n2
]
bn=4π
[ −8n4(1−4 n2)
]
bn=4π
[ −2n(1−4n2)
]
bn=4π
[ 2n(4 n2−1)
]
bn=8π
[ n(4 n2−1)
]
f ( x )∑n=1
∞ 8π [ n
(4 n2−1) ]sin nππ2
x
f ( x ) 8π∑n=1
∞
[ n(4 n2−1) ]sin 2n x
40. f(x) = sen x, 0 x
A0=2p∫0p
f ( x )dx
A0=2π∫0
π
sin x dx
A0=2π
¿¿
A0=2π
¿
A0=2π
¿
A0=2π
[2]
A0=4π
An=2p∫0p
f ( x )cos nπpx dx
An=2π∫0π
sin x ¿¿
An=2π∫0π
sin x ¿¿¿
An=2π∫0
π 12¿¿
An=2π
¿
An=2π
[−(−1 )n
2 (1+n )−
(−1 )n
2 (1−n )+ 12 (1+n )
+ 12 (1−n )
]
An=2π
[−2 (1−n ) (−1 )n−2 (1+n ) (−1 )n+2 (1−n ) (1 )+2 (1+n )(1)2 (1+n )2 (1−n )
]
An=2π
[(−2+2n ) (−1 )n (−2−2n ) (−1 )n+2−2n+2+2n
2 (1+n )2 (1−n )]
An=2π
[−2 (−1 )n+2n (−1 )n−2 (−1 )n−2n (−1 )n+2+2(2+2n ) (2−2n )
]
An=2π
[−2 (−1 )n−2 (−1 )n+44−4 n+4 n−4n2
]
An=2π
[−4 (−1 )n+44−4n2
]
An=2π
[ 4 {−(−1 )n+1 }4 (1−n2)
]
An=2π
[1− (−1 )n
(1−n2)]
f ( x ) 4π (2)∑n=1
∞ 2π [ 1−(−1 )n
(1−n2) ]cos nππ x
f ( x ) 2π
+ 2π∑n=1
∞ [ 1−(−1 )n
(1−n2) ]cos nx
41. f(x)
A0=2p∫0p
f ( x )dx
x, 0 x /2
- x, /2 x
A0=2π∫0
π
[ (x )+(π−x )]dx
A0=2π
¿
A0=2π
[ x2
2 0
π2+(πx− x
2
2)π2
π
]
A0=2π
[( π2 )
2
2−
(0 )2
2+π (π )−
(π )2
2−π ( π2 )+ ( π2 )
2
2]
A0=2π
[
π 2
42
+π 2−π2
2−π
2
2+
π2
42
]
A0=2π
[ π2
8+π2−π2+ π
2
8]
A0=2π
[ 2π2
8]
A0=2π
[ π2
4]
A0=π2
An=2p∫0p
f ( x )cos nπpx dx
An=2π∫0
π
[ (x )+(π−x ) ]cos nππx dx
An=2π
¿
An=2π
¿
An=2π
[ (x )( sinnxn )−∫0
π2sin nxn
dx+(π )( sin nxn )−∫π2
π
x cosnx dx ]
An=2π
[ x sin nxn 0
π2−∫
0
π2sin nxn
dx+ π sin nxn π
2
π
− x sin nxn π
2
π
+∫π2
πsin nxn
dx ]
An=2π
[ x sin nxn 0
π2+ cosnx
n2 0
π2+ π sin nx
n π2
π
− x sin nxn π
2
π
− cos nxn2 π
2
π
]
An=2π
[
π2sinn( π2 )n
+cos n( π2 )n2
−(0 )sin n (0 )
n−cos n (0 )n2
+π sin n (π )
n−
(π )sin n (π )n
−cosn (π )n2
−π sin n( π2 )
n+
(π2 )sin n( π2 )n
+cosn( π2 )n2
]
An=2π
[
π2n
+cos n( π2 )n2
− 1n2
−(−1)n
n2− πn+
π2n
+cos n( π2 )n2
]
An=2π
[(2) π2n
+2cosn( π2 )
n2− 1n2
−(−1)n
n2−πn
]
An=2π
[ πn+2cosn( π2 )
n2− 1n2
−(−1)n
n2−πn
]
An=2π
[2cosn( π2 )
n2−
(−1 )n
n2− 1n2
]
An=2π
[2cosn( π2 )−(−1 )n−1
n2]
f ( x ) π2(2)∑n=1
∞ 2π [ 2cosn( π2 )− (−1 )n−1
n2 ]cos nππ x
f ( x ) π4
+ 2π∑n=1
∞ 2π [ 2cosn( π2 )−(−1 )n−1
n2 ]cosn x
CONCLUSIÓN
Es una aplicación usada en muchas ramas de la ingeniería, además de ser una herramienta sumamente útil en la teoría matemática abstracta. Áreas de aplicación incluyen análisis vibratorio, acústica, óptica, procesamiento de imágenes y señales, y compresión de datos. En ingeniería, para el caso de los sistemas de telecomunicaciones, y a través del uso de los componentes espectrales de frecuencia de una señal dada, se puede optimizar el diseño de un sistema para la señal portadora del mismo. Refiérase al uso de un analizador de espectros.
Las series de Fourier tienen la forma:
Donde y se denominan coeficientes de Fourier de la serie de Fourier de la función