serie de fourier
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=
1π
[ 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=
1π
[−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=
1π
¿
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=
1π
[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=
2π
[sinx⎥0π2 ]
Ao=2π [sin π2−sin 0 ]Ao=
2π
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=
1π
[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=
1π
[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.