serie de fourier

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DESCRIPTION

algunos problemas de serie de fourier resueltos.

TRANSCRIPT

Tema #2

Serie de Fourier

Las series de Fourier tienen la forma:

Donde   y   se denominan coeficientes de Fourier de la serie de Fourier de la

función 

Práctica

1. f ( x )={0 ,−π <x<01 ,0<x<π }

Ao=1π [∫

−π

0

0dx+∫0

π

1dx ]Ao=

[ x⎥0π ]

Ao=1π

[π ]

Ao=1

An=1π [∫

−π

0

0cos nxdx+∫0

π

cos nxdx ]An=

1π [ sinnxn ⎥0

π]An=

1π [ sinnπn −

sinn (0 )n ]

An=0

bn=1π [∫

−π

0

0sinnx dx+∫0

π

sinnx dx ]bn=

−1π [ cosnxn ⎥0

π]bn=

1π [−cosnπn

+cosn (0 )n ]

bn=1nπ

[1−(−1)n ]

f ( x ) 12+∑n=1

{ 1nπ [1−(−1)n ]sinnx }

2. f ( x )={−1 ,−π<x<02 ,0<x<π }Ao=

1π [∫

−π

0

−1dx+∫0

π

2dx ]Ao=

[−x⎥−π0 +2 x⎥0

π ]

Ao=1π

[ (0−π )+(2π−0)]

Ao=1π

(π )

Ao=1

An=1π [−∫

−π

0

cos nxdx+2∫0

π

cosnx dx ]An=

1π [−sinnxn

⎥−π0 +2 sinnx

n⎥0π]

An=1π [−sinn (0)n

−sinnπn

+ 2sinnπn

−2sinn (0 )

n ]An=0

bn=1π [−∫

−π

0

sinnx dx+2∫0

π

sinnx dx]bn=

1π [ cosnxn ⎥−π

0 −2cosnxn

⎥0π]

bn=1π [ cosn(0)n

− cosnπn

−2cosnπn

+2cosn (0 )

n ]bn=

1π [ 1n−1n (−1)n−2n (−1)n+ 2

n ]bn=

1π [ 3n+ 3n (−1)n+1]

bn=3nπ

[1+(−1)n+1 ]

f ( x ) 12+∑n=1

{ 3nπ [1+(−1)n+1 ]sinnx }

3. f ( x )={1 ,−1≤ x<0x ,0<x<1 } Ao=∫

−1

0

1dx+∫0

1

xdx

Ao=x⎥−10 + x

2

2⎥01

Ao=0−(−1 )+ 12

2−0

2

2

Ao=1+12

Ao=32

An=∫−1

0

cosnπx dx+∫0

1

xcos nπxdx

An=sinnπxnπ

⎥−10 + xsin nπx

nπ⎥01−∫

0

1sinnπxnπ

dx

An=sinnπxnπ

⎥−10 + xsin nπx

nπ⎥01+ cosnπx

(nπ )2⎥01

An=0+0+cos nπ (1 )

(nπ )2−cosnπ (0 )

(nπ )2

An=1

(nπ )2[ (−1 )n−1 ]

bn=∫−1

0

sinnπx dx+∫0

1

xsin nπxdx

bn=−cosnπxnπ

⎥−10 − xcos nπx

nπ⎥01+∫

0

1cos nπxnπ

dx

bn=−cosnπxnπ

⎥−10 − xcos nπx

nπ⎥01+ sennπx

(nπ )2⎥01

bn=−cosnπ (0 )

nπ+cosnπ (−1 )

nπ−

(1 )cos nπ (1 )nπ

+(0)cosnπ (0 )

nπ+0

bn=1nπ

[−1+(−1 )n−(−1 )n ]

bn=−1nπ

f ( x ) 34+∑n=1

{ 1(nπ )2

[(−1)n−1 ]cosnπx− 1nπsinnπx }

4. f ( x )={0 ,−1≤ x<0x ,0<x<1 } Ao=∫

−1

0

0dx+∫0

1

xdx

Ao=x2

2⎥01

Ao=12

2−0

2

2

Ao=12

An=∫−1

0

0cosnπx dx+∫0

1

xcosnπx dx

An=xsin nπxnπ

⎥01−∫

0

1sinnπxnπ

dx

An=xsin nπxnπ

⎥01+ cosnπx

(nπ )2⎥01

An=0+cosnπ (1 )

(nπ )2−0−

cos nπ (0 )(nπ )2

An=1

(nπ )2[(−1)n−1 ]

bn=∫−1

0

0sinnπx dx+∫0

1

xsinnπx dx

bn=−xcosnπx

nπ⎥01+∫

0

1cosnπxnπ

dx

bn=−xcosnπx

nπ⎥01+ sen nπx

(nπ )2⎥01

bn=−cosnπ (1 )

nπ+0+

(0 ) cosnπ (0 )nπ

−0

bn=1nπ

(−1 )n+1

f ( x ) 14+∑n=1

{ 1(nπ )2

[(−1)n−1 ]cosnπx+ 1nπ

(−1 )n+1 sinnπx }

5. f ( x )={0 ,−π <x<0x2 ,0≤ x<π }

Ao=1π [∫

−π

0

0dx+∫0

π

x2dx ]

Ao=1π [ x33 ⎥0π]

Ao=1π [ π33 −0

3

3 ]Ao=

π2

3

An=1π [∫

−π

0

0cos nxdx+∫0

π

x2cos nx dx]An=

1π [ x2 sinnxn

⎥0π−2n∫0

π

xsin nxdx ]An=

1π [ x2 sinnxn

⎥0π−2n (−xcos nxn

⎥0π+ 1n∫0

π

cos nxdx )]An=

1π [ x2 sinnxn

⎥0π+ 2 xcosnx

n2⎥0π−2sinnx

n3⎥0π]

An=1π [ π2 sinnπn

−0+ 2πcos nπn2

−0−2sinnπn3

+0]An=

1π [ 2 πn2 (−1)n]

An=2

n2(−1)n

bn=1π [∫

−π

0

0sinnx dx+∫0

π

x2sinnx dx ]bn=

1π [−x2 cosnxn

⎥0π+ 2n∫0

π

xcos nx dx]bn=

1π [−x2 cosnxn

⎥0π+ 2n ( xsinnxn ⎥0

π−1n∫0

π

sinnx dx)]bn=

1π [−x2 cosnxn

⎥0π+ 2xsin nx

n2⎥0π+ 2cosnx

n3⎥0π ]

bn=1π [−π2 cosnπn

+0+ 2 πsinnπn2

−0+ 2cos nπn3

−2cosn (0)

n3 ]bn=

1π [−π2n (−1 )n+ 2

n3(−1 )n− 2

n3 ]bn=

πn

(−1 )n+1+ 2

n3π(−1 )n− 2

n3π

bn=πn

(−1 )n+1+ 2

n3π[ (−1 )n−1 ]

f ( x ) π2

6+∑n=1

{ 2n2 (−1)ncosnx+( πn (−1 )n+1+ 2n3π

[ (−1 )n−1 ])sinnx }

6. f ( x )={ π2 ,−π<x<0π2−x2 ,0≤ x<π } Ao=

¿

Ao=1π [ x π2⎥−π

0 +x π2⎥0π− x

3

3⎥0π]

Ao=1π [ (0 ) π2−(−π )π 2+(π ) π2−(0 ) π2−π

3

3+ 0

3

3 ]Ao=

1π [ π3+π 3−π33 ]Ao=

1π [ 5 π33 ]

Ao=5 π2

3

An=1π

¿

An=1π [∫

−π

0

π2 cosnxdx+∫0

π

π 2cosnxdx−∫0

π

x2 cosnxdx ]An=

1π [ π2 sinnxn

⎥−π0 + π

2 sinnxn

⎥0π−( x2 sinnxn

⎥0π−2n∫0

π

xsinnxdx)]

An=1π [0+0−(0−2n∫0

π

xsinnxdx)]An=

1π [ 2n (−xcosnxn

⎥0π+ 1n∫0

π

cosnxdx)]An=

1π [−2 xcosnxn2

⎥0π+ 2 sinnx

n3⎥0π ]

An=1π [−2 πcosnπn2

+0+0−0]An=

1π [ 2 πn2 (−1)n+1]

An=2

n2(−1)n+1

bn=1π

¿

bn=1π [∫

−π

0

π2 sinnxdx+∫0

π

π 2 sinnxdx−∫0

π

x2 sinnxdx ]bn=

1π [−π2 cosnxn

⎥−π0 −π

2cosnxn

⎥0π−(−x2 cosnxn

⎥0π+ 2n∫0

π

xcosnxdx)]bn=

1π [−π2 cosnxn

⎥−π0 −π

2cosnxn

⎥0π+ x

2 cosnxn

⎥0π−2n ( xsinnxn ⎥0

π−1n∫0

π

sinnxdx )]bn=

1π [−π2 cosnxn

⎥−π0 −π

2cosnxn

⎥0π+ x

2 cosnxn

⎥0π−2 xsinnx

n2⎥0π−2cosnx

n3⎥0π]

bn=1π [−π2 cosn (0 )

n+π2cosn (−π )

n− π

2 cosnπn

+π2 cosn (0 )

n+ π

2cosnπn

−0cosn (0 )n

−2πsinnπn2

+0 sinn (0 )n2

−2cosnπn3

+2cosn (0 )n3 ]

bn=1π [ π2 cosn (π )

n−2cosnπ

n3+2cosn (0)n3 ]

bn=1π [ π2n (−1 )n− 2

n3(−1 )n+ 2

n3 ]bn=

πn

(−1 )n+ 2

n3π[1−(−1 )n ]

f ( x ) 5 π2

6+∑n=1

{ 2n2 (−1)n+1 cosnx+( πn (−1 )n+ 2n3π

[1−(−1 )n ])sinnx}

7. f ( x )= {x+π ,−π<x<π }

Ao=1π [∫

−π

π

( x+π )dx ]Ao=

1π [∫

−π

π

xdx+∫−π

π

πdx ]Ao=

1π [ x22 ⎥−π

π +πx ⎥−ππ ]

Ao=1π [ π22 −

(−π )2

2+π (π )−π (−π )]

Ao=1π

[2π2 ]Ao=2π

An=1π [∫

−π

π

( x+π ) cosnxdx ]An=

1π [∫

−π

π

xcosnxdx+∫−π

π

πcosnxdx ]An=

1π [ xsinnxn ⎥−π

π −1n∫−π

π

sinnxdx+ πsinnxn

⎥−ππ ]

An=1π [ xsinnxn ⎥−π

π + cosnxn2

⎥−ππ + πsinnx

n⎥−ππ ]

An=1π [0−0+ cosnπn2 − cosnπ

n2+0−0]

An=0

bn=1π [∫

−π

π

( x+π ) sinnxdx ]bn=

1π [∫

−π

π

xsinnxdx+∫−π

π

πsinnxdx]bn=

1π [−xcosnxn

⎥−ππ + 1

n∫−π

π

cosnxdx− πcosnxn

⎥−ππ ]

bn=1π [−xcosnxn

⎥−ππ + sinnx

n2⎥−ππ − πcosnx

n⎥−ππ ]

bn=1π [−πcosnπn

−πcosnπn

+0+0−πcosnπn

+ πcosnπn ]

bn=1π [−πcosnπn

−πcosnπn ]

bn=1π [ (−1 ) 2 π

n(−1 )n]

bn=2n

(−1 )n+1

f ( x ) π+∑n=1

{2n (−1 )n+1 sinnx}

8. f ( x )= {3−2 x ,−π< x<π }

Ao=1π [∫−π

π

(3−2x )dx ]Ao=

1π [∫

−π

π

3dx−∫−π

π

2xdx ]Ao=

[3 x⎥−ππ −x2⎥−π

π ]

Ao=1π

[3 (π )−3 (−π )−π2+(−π2 ) ]

Ao=1π

[6 π ]

Ao=6

An=1π [∫

−π

π

(3−2x ) cosnxdx ]An=

1π [∫

−π

π

3cosnxdx−∫−π

π

2 xcosnxdx]An=

1π [ 3 sinnxn ⎥−π

π −2( xsinnxn ⎥−ππ −1

n∫−π

π

sinnxdx)]An=

1π [0−0−2cosnxn2

⎥−ππ ]

An=1π [−2cosnπn2

+ 2cosnπn2 ]

An=1π [−2n2 (−1 )n+ 2

n2(−1 )n]

An=0

bn=1π [∫

−π

π

(3−2x ) sinnxdx ]

bn=1π [∫

−π

π

3 sinnxdx−∫−π

π

2 xsinnxdx]bn=

1π [−3cosnxn

⎥−ππ −2(−xcosnxn

⎥−ππ + 1

n∫−π

π

cosnxdx)]bn=

1π [−3cosnxn

⎥−ππ + 2xcosnx

n⎥−ππ −2 sinnx

n2⎥−ππ ]

bn=1π [−3cosnπn

+3cosnπn

+ 2πcosnπn

+ 2 πcosnπn

−0+0]bn=

1π [−3n (−1 )n+ 3

n(−1 )n+ 2π

n(−1 )n+ 2 π

n(−1 )n]

bn=4n(−1)n

f ( x ) 3+∑n=1

{4n (−1)n sinnx}

9. f ( x )={ 0 ,−π<x<0sinx ,0≤x<π }Ao=

1π [∫

0

π

sinxdx ]Ao=

−1π

[cosx ⎥0π ]

Ao=−1π

[cosπ−cos0 ]

Ao=2π

An=1π [∫

0

π

sinxcosnxdx ]An=

12 π

∫0

π

[sin (1+n ) x+sin (1−n)x ] dx

An=12 π [−cos (1+n) x1+n

⎥0π−cos (1−n)x1−n

⎥0π]

An=12 π [−cos (1+n )π

1+n+cos (1+n ) (0 )1+n

−cos (1−n )π1−n

+cos (1−n ) (0 )1−n ]

An=12 π [(−1)2+n1+n

+(−1)2−n

1−n+ 21−n2 ]

bn=1π [∫

0

π

sinxsinnx dx]bn=

12π

∫0

π

[cos (1−n ) x−cos (1+n)x ]dx

bn=12π [−sin (1+n ) x

1+n⎥0π+sin (1−n)x1−n

⎥0π]

bn=12π [−sin (1+n )π

1+n+sin (1+n ) (0 )1+n

+sin (1−n )π1−n

−sin (1−n)(0)

1−n ]bn=0

f ( x ) 1π

+∑n=1

∞ { 12π [(−1)2+n1+n+(−1)2−n

1−n+

2

1−n2 ]cosnx}

10. f ( x )={ 0 ,−π2 <x<0

cosx ,0≤ x<π2}

Ao=2π [∫0

π2

cosxdx ]Ao=

[sinx⎥0π2 ]

Ao=2π [sin π2−sin 0 ]Ao=

An=2π [∫

0

π2

cosxcos2nxdx ]An=

1π∫0

π2

[cos (1−2n ) x+cos (1+2n ) x ]dx

An=1π [ sin (1+2n ) x

1+2n⎥0π2+sin (1−2n ) x1−2n

⎥0π2 ]

An=1π [ sin (1+2n )( π

2)

1+2n−sin (1+2n ) (0 )1+2n

+sin (1−2n )( π2 )

1−2n−sin (1−2n)(0)

1−2n ]An=

1π [ 11+2n

+ 11−2n ]

An=1

π (1−4 n2)

bn=2π [∫

0

π2

cosxsin2nxdx ]bn=

1π∫0

π2

[sin (1+2n ) x−sin (1−2n ) x ] dx

bn=1π [−cos (1+2n ) x

1+2n⎥0π2+cos (1−2n ) x1−2n

⎥0π2 ]

bn=1π [−cos (1+2n )( π2 )

1+2n+cos (1+2n ) (0 )

1+2n+cos (1−2n )( π2 )

1−2n−cos (1−2n ) (0 )

1−2n ]bn=

1π [ 11+2n

− 11−2n ]

bn=4 n

π (4n2−1)

f ( x ) 1π

+∑n=1

{ 1π (1−4n2)

cos 2nx+ 4 n

π (4n2−1 )sin 2nx }

11. f ( x )={ 0 ,−2< x←1−2 ,−1≤ x<01 ,0≤x<1 }

Ao=12 [∫−10 −2dx+∫0

1

1dx ]Ao=

12

[−2 x⎥−10 +x⎥0

1 ]

Ao=12

[−2+1 ]

Ao=−12

An=12 [∫

−1

0

−2cos nπ2xdx+∫

0

1

cosnπ2xdx ]

An=12 [−4nπ sin nπ2 x⎥−1

0 + 2nπsinnπ2x⎥0

1]An=

12 [−4nπ sin nπ2 (0 )− 4

nπsinnπ2

+ 2nπsinnπ2

− 2nπsinnπ2

(0)]An=

12 [0− 2

nπsinnπ2

−0]An=

−1nπsinnπ2

bn=12 [∫−1

0

−2sin nπ2xdx+∫

0

1

sinnπ2xdx ]

bn=12 [ 4nπ cos nπ2 x⎥−1

0 − 2nπcos

nπ2x ⎥0

1]bn=

12 [ 4nπ cos nπ2 (0 )− 4

nπcos

nπ2

− 2nπcos

nπ2

+ 2nπcos

nπ2

(0)]bn=

12 [ 6nπ− 6

nπcos

nπ2 ]

bn=3nπ (1−cos nπ2 )

f ( x ) −14+∑n=1

{−1nπ sin nπ2 cos nπ2 x+ 3nπ (1−cos nπ2 )sin nπ2 x }

12. f ( x )={0 ,−2<x<0x ,0≤ x<11 ,1≤x<2 }

Ao=12 [∫−20 0dx+∫01 xdx+∫12 1dx ]Ao=

12 [ x

2

2⎥01+x ⎥1

2]Ao=

12 [ 12−0+2−1]Ao=

12 [ 32 ]

Ao=34

An=12 [∫

−2

0

0cosnπ2xdx+∫

0

1

xcosnπ2xdx+∫

1

2

cosnπ2xdx ]

An=12 [ 2 xnπ sin nπ2 x ⎥01− 2

nπ∫0

1

sinnπ2xdx+ 2

nπsinnπ2x ⎥1

2]An=

12 [ 2 xnπ sin nπ2 x ⎥01+ 4

(nπ )2cos

nπ2x ⎥0

1+2nπsinnπ2x⎥1

2]An=

12 [ 2nπ sin nπ2 −0+

4

(nπ )2cos

nπ2

−4

(nπ )2cos

nπ2

(0 )+ 2nπsinnπ2

(2 )− 2nπsinnπ2 ]

An=12 [ 4

(nπ )2cos

nπ2

−4

(nπ )2 ]An=

2

(nπ )2 [cos nπ2 −1]

bn=12 [∫−2

0

0sinnπ2xdx+∫

0

1

xsinnπ2xdx+∫

1

2

sinnπ2xdx ]

bn=12 [−2xnπ cos

nπ2x⎥0

1+ 2nπ

∫0

1

cosnπ2xdx− 2

nπcos

nπ2x ⎥1

2]bn=

12 [−2xnπ cos

nπ2x⎥0

1+4

(nπ )2sinnπ2x ⎥0

1−2nπcos

nπ2x⎥1

2]bn=

12 [−2nπ cos nπ2 +0+

4

(nπ )2sinnπ2

−4

(nπ )2sinnπ2

(0 )− 2nπcos

nπ2

(2 )+ 2nπcos

nπ2 ]

bn=12 [ 4

(nπ )2sinnπ2

−2nπcosnπ ]

bn=2

(nπ )2sinnπ2

+ 1nπ

(−1)n+1

f ( x ) 34+∑n=1

{ 2(nπ )2 [cos nπ2 −1]cos nπ2 x+[ 2

(nπ )2sinnπ2

+ 1nπ

(−1 )n+1]sin nπ2 x}

13. f ( x )={ 1 ,−5<x<01+x ,0≤ x<5}Ao=

15 [∫

−5

0

dx+∫0

5

dx+∫0

5

xdx ]Ao=

15 [ x⎥−5

0 + x⎥05+ x

2

2⎥05]

Ao=15 [5+5+ 252 ]Ao=

92

An=15 [∫

−5

0

cosnπ5xdx+∫

0

5

cosnπ5xdx+∫

0

5

xcosnπ5xdx ]

An=15 [ 5nπ sin nπ5 x ⎥−5

0 + 5nπsinnπ5x⎥0

5+ 5 xnπsinnπ5x⎥0

5− 5nπ

∫0

5

sinnπ5xdx ]

An=15 [0+ 25

(nπ )2cos

nπ5x⎥0

5]

An=15 [ 25(nπ )2

cosnπ5

(5 )− 25

(nπ )2cos

nπ5

(0 )]An=

15 [ 25(nπ )2

(−1 )n− 25

(nπ )2 ]An=

5

(nπ )2[(−1)n−1 ]

bn=15 [∫

−5

0

sinnπ5xdx+∫

0

5

sinnπ5xdx+∫

0

5

xsinnπ5xdx ]

bn=15 [−5nπ cos nπ5 x⎥−5

0 − 5nπcos

nπ5x ⎥0

5−5xnπcos

nπ5x⎥0

5+ 5nπ

∫0

5

cosnπ5xdx ]

bn=15 [−5nπ cos nπ5 x⎥−5

0 −5nπcos

nπ5x ⎥0

5−5xnπcos

nπ5x⎥0

5+25

(nπ )2sinnπ5x ⎥0

5]bn=

15 [−5nπ cos nπ5 (0 )+ 5

nπcos

nπ5

(5 )− 5nπcos

nπ5

(5 )+ 5nπcos

nπ5

(0 )−5 (5 )nπ

cosnπ55+0+

25

(nπ )2sinnπ5

(5 )− 25

(nπ )2sinnπ5

(0)]bn=

15 [−5nπ + 5

nπ(−1 )n− 5

nπ(−1 )n+ 5

nπ+ 25nπ

(−1 )n+1]bn=

15 [ 25nπ (−1 )n+1]

bn=15 [ 25nπ (−1 )n+1]

bn=5nπ

(−1 )n+1

f ( x ) 94+∑n=1

{ 5(nπ )2

[(−1)n−1 ]cos nπ5x+ 5nπ

(−1 )n+1 sin nπ5x}

14. f ( x )={2+x ,−2<x<02 ,0≤x<2 } Ao=12 [∫−20 2dx+∫−20 xdx+∫02 2dx ]Ao=

12 [2 x⎥−2

0 + x2

2⎥−20 +2x ⎥0

2]

Ao=12 [0−2 (−2 )+0−4

2+2 (2 )−0]

Ao=3

An=12 [∫

−2

0

2cosnπ2xdx+∫

−2

0

xcosnπ2xdx+∫

0

2

2cosnπ2xdx ]

An=12 [ 4nπ sin nπ2 x⎥−2

0 + 4nπsinnπ2x⎥0

2+ 2 xnπsinnπ2x⎥−2

0 − 2nπ

∫−2

0

sinnπ2xdx ]

An=12

¿

An=12

¿

An=12

¿

An=2¿¿

bn=12 [∫−2

0

2sinnπ2xdx+∫

0

2

2sinnπ2xdx+∫

−2

0

xsinnπ2xdx ]

bn=12 [−4nπ cos nπ2 x ⎥−2

0 − 4nπcos

nπ2x⎥0

2−2 xnπcos

nπ2x⎥−2

0 + 2nπ

∫−2

0

cosnπ2xdx ]

bn=12 [−4nπ cos nπ2 x ⎥−2

0 −4nπcos

nπ2x⎥0

2−2 xnπcos

nπ2x⎥−2

0 +4

(nπ )2sinnπ2x ⎥−2

0 ]bn=

12 [−4nπ cos (0 )+ 4

nπcosnπ− 4

nπcosnπ+ 4

nπcos (0 )−2 (0 )

nπcos 0+

2 (−2 )nπ

cosnπ ]bn=

12 [−4nπ + 4

nπ(−1 )n− 4

nπ(−1 )n+ 4

nπ− 4nπ

(−1 )n]bn=

12 [ 4nπ (−1 )n+1]

bn=2nπ

(−1 )n+1

f ( x ) 32+∑n=1

¿¿

15. f (x)={ex ,−π<x<π }

Ao=1π [∫−π

π

e xdx ]Ao=

[ex⎥−ππ ]

Ao=1π

(eπ−e−π )

An=1π [∫

−π

π

ex cosnxdx ]An=

1π [ e x sinnxn

⎥−ππ −1

n∫−π

π

ex sinnxdx]An=

1π [0−1n (−ex cosnxn

⎥−ππ + 1

n∫−π

π

ex cosnxdx)]An=

1π [ e xcosnxn2

⎥−ππ − 1

n2∫−π

π

excosnxdx ]An=

1π [ n2 ex

n2 (n2+1 )cosnx ⎥−π

π ]An=

1π [ ex

(n2+1 )cosnx ⎥−π

π ]An=

1π [ eπ

(n2+1 )cosnπ− e−π

(n2+1 )cosnπ ]

An=1π [ eπ

(n2+1 )(−1)n+ e−π

(n2+1 )(−1)n+1]

bn=1π [∫

−π

π

e x sinnxdx]bn=

1π [−ex cosnxn

⎥−ππ + 1

n∫−π

π

ex cosnxdx ]

bn=1π [−ex cosnxn

⎥−ππ + 1

n ( ex sinnxn⎥−ππ −1

n∫−π

π

ex sinnxdx)]bn=

1π [−ex cosnxn

⎥−ππ +0− 1

n2∫−π

π

e xsinnxdx ]bn=

1π [ −n2 ex

n (n2+1 )cosnx ⎥−π

π ]bn=

1π [ −ne x

(n2+1 )cosnx ⎥−π

π ]bn=

1π [ −neπ

(n2+1 )cosnπ+ ne−π

(n2+1 )cosnπ ]

bn=1π [ neπ(n2+1 )

(−1)n+1+ ne−π

(n2+1 )(−1)n]

f ( x ) 12 π

(eπ−e−π )+∑n=1

∞ {1π [ eπ

(n2+1 )(−1)n+ e−π

(n2+1 )(−1)n+1]cosnx+ 1π [ neπ(n2+1 )

(−1)n+1+ ne−π

(n2+1 )(−1)n] sinnx}

16. f (x)={ 0 ,−π<x<0ex−1 ,0< x<π } Ao=1π [∫

0

π

(ex−1)dx ]Ao=

[ex⎥0π−x⎥0π ]

Ao=1π

[eπ−e0−π ]

Ao=1π

[eπ−1−π ]

An=1π [∫

0

π

ex cosnxdx−∫0

π

cosnxdx ]An=

1π [ e xsinnxn

⎥0π−1n∫0

π

ex sinnxdx−sin nxn

⎥0π]

An=1π [0−1n (−ex cosnxn

⎥0π+ 1n∫0

π

e xcosnxdx)−0]An=

1π [ e xcosnxn2

⎥0π− 1n2∫0

π

ex cosnxdx ]An=

1π [ n2 ex

n2 (n2+1 )cosnx ⎥0

π ]An=

1π [ ex

(n2+1 )cosnx ⎥0

π ]An=

1π [ eπ

(n2+1 )cosnπ− e0

(n2+1 )cos0]

An=1

π (n2+1)[ eπ(−1)n−1 ]

bn=1π [∫

0

π

ex sinnxdx−∫0

π

sinnxdx]bn=

1π [−excos nxn

⎥0π+1n∫0

π

ex cosnxdx+ cosnxn

⎥0π]

bn=1π [−excos nxn

⎥0π+1n ( e

xsinnxn

⎥0π−1n∫0

π

ex sinnxdx)+ cosnxn ⎥0π ]

bn=1π [−excos nxn

⎥0π+ e

xsinnxn2

⎥0π− 1n2∫0

π

ex sinnxdx+ cosnxn

⎥0π]

bn=1π [−excos nxn

⎥0π+0− 1

n2∫0

π

ex sinnxdx+ cosnxn

⎥0π]

bn=1π

¿

bn=1π

¿

bn=1π

¿

bn=n

π (n¿¿2+1)[(−1)n−1−eπ (−1 )n+1 ]¿

bn=n

π (n¿¿2+1)[(−1)n−eπ (−1 )n ] ¿

f ( x ) 12 π

[eπ−1−π ]+∑n=1

¿¿

Conclusión

Las series de Fourier constituyen la herramienta matemática básica del análisis de Fourier empleado para analizar funciones periódicas a través de la descomposición. El nombre se debe al matemático francés Jean-Baptiste Joseph Fourier que desarrolló la teoría cuando estudiaba la ecuación del calor.

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.

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