Parcial 3 Práctica # 3

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