UNIVERSIDAD NACIONAL “PEDRO RUIZ GALLO”
Facultad de Ingeniería Mecánica y Eléctrica.
CURSO: MATEMÁTICA 4.
ESCUELA: Ingeniería Mecánica y Eléctrica.
SEMESTRE ACADÉMICO: 2019 II.
CICLO ACADÉMICO: IV CICLO.
DOCENTE: Msc.Mardo Gonzales Herrera
INTEGRANTES:
Bravo Samamé Rodrigo
Flores Carpio Eduardo
Gómez Pita Raúl
Gutiérrez Rubio Sergio.
Millones Cava Manuel
LAMBAYEQUE,23 DE ENERO DEL 2020
Ejercicios: 1,2,5,7,8 de Matemática avanzada para ingeniería de Peter o Neill
Encuentre la serie de Fourier de las funciones en los intervalos:
1. 𝑓(𝑥) = 4 ; −3 ≤ 𝑥 ≤ 3
Hallamos los coeficientes de la ecuación general:
1 3
𝑎0 = ∫ 4 𝑑𝑥
3 −3
𝑎0 =
1
3
4𝑥|
3
−3
𝑎0 = 8
1 3
𝑛𝜋𝑥
𝑎𝑛 = ∫ 4 cos(
) 𝑑𝑥
3 −3
3
𝑎𝑛 =
𝑎𝑛 =
4 3
𝑛𝜋𝑥
3
)] |
[𝑠𝑒𝑛(
3 𝑛𝜋
3
−3
4
[𝑠𝑒𝑛(𝑛𝜋) − 𝑠𝑒𝑛(−𝑛𝜋)]
𝑛𝜋
𝑎𝑛 = 0
El resultado del coeficiente an ya era conocido por ser una multiplicación de función impar por
una par, dando 0 el resultado.
1 3
𝑛𝜋𝑥
𝑏𝑛 = ∫ 4 sen(
) 𝑑𝑥
3 −3
3
𝑏𝑛 =
𝑏𝑛 =
4 3
𝑛𝜋𝑥
3
)] |
[𝑐𝑜𝑠(
3 𝑛𝜋
3
−3
4
[𝑐𝑜𝑠(𝑛𝜋) − 𝑐𝑜𝑠(−𝑛𝜋)]
𝑛𝜋
𝑏𝑛 =
4
[2𝑐𝑜𝑠(𝑛𝜋)]
𝑛𝜋
𝑏𝑛 =
8
(−1)𝑛
𝑛𝜋
Reemplazando los valores obtenidos en la fórmula:
∞
1
𝑛𝜋𝑥
𝑛𝜋𝑥
𝑓(𝑥) = 𝑎0 + ∑ [𝑎𝑛 cos (
) + 𝑏𝑛 𝑠𝑒𝑛(
)]
2
𝐿
𝐿
𝑛=1
Obtenemos:
∞
8
8
𝑛𝜋𝑥
𝑓(𝑥) = + ∑ [
(−1)𝑛 𝑠𝑒𝑛(
)]
2
𝑛𝜋
3
𝑛=1
𝑓(𝑥) = 4
2.𝑓(𝑥) = −𝑥 ; −1 ≤ 𝑥 ≤ 1
Hallamos los coeficientes de la ecuación general:
1
𝑎0 = 1 ∫ −𝑥 𝑑𝑥
−1
𝑎0 = −
𝑥2 1
|
2 −1
𝑎0 = 0
1
𝑛𝜋𝑥
𝑎𝑛 = 1 ∫ −𝑥 cos(
) 𝑑𝑥
3
−1
Aplicamos integrales de chebyshev, obteniendo:
𝑎𝑛 = − [
𝑥
1
1
𝑠𝑒𝑛(𝑛𝜋𝑥) + 2 2 cos(𝑛𝜋𝑥)] |
𝑛𝜋
𝑛 𝜋
−1
𝑎𝑛 =
1
𝑛2 𝜋 2
[−𝑐𝑜𝑠(𝑛𝜋) + 𝑐𝑜𝑠(−𝑛𝜋)]
𝑎𝑛 = 0
1 1
𝑛𝜋𝑥
𝑏𝑛 = ∫ −𝑥 sen(
) 𝑑𝑥
3 −1
3
𝑏𝑛 = − [−
𝑥
1
1
𝑐𝑜𝑠(𝑛𝜋𝑥) + 2 2 sen(𝑛𝜋𝑥)] |
𝑛𝜋
𝑛 𝜋
−1
𝑏𝑛 =
1
[𝑐𝑜𝑠(𝑛𝜋) − 𝑐𝑜𝑠(−𝑛𝜋)]
𝑛𝜋
𝑏𝑛 =
1
[2𝑐𝑜𝑠(𝑛𝜋)]
𝑛𝜋
𝑏𝑛 =
2
(−1)𝑛
𝑛𝜋
Reemplazando los valores obtenidos en la fórmula:
∞
1
𝑛𝜋𝑥
𝑛𝜋𝑥
𝑓(𝑥) = 𝑎0 + ∑ [𝑎𝑛 cos (
) + 𝑏𝑛 𝑠𝑒𝑛(
)]
2
𝐿
𝐿
𝑛=1
Obtenemos:
∞
0
2
𝑛𝜋𝑥
𝑓(𝑥) = + ∑ [
(−1)𝑛 𝑠𝑒𝑛(
)]
2
𝑛𝜋
1
𝑛=1
∞
𝑓(𝑥) = ∑ [
𝑛=1
5.𝑓(𝑥) = {
2
𝑛𝜋𝑥
(−1)𝑛 𝑠𝑒𝑛(
)]
𝑛𝜋
1
−4, −𝜋 ≤ 𝑥 ≤ 0
4, 0 ≤ 𝑥 ≤ 𝜋
Hallamos los coeficientes de la ecuación general:
𝑎0 =
1 0
1 𝜋
∫ −4 𝑑𝑥 + ∫ 4 𝑑𝑥
𝜋 −𝜋
𝜋 0
1
0
1
𝜋
𝑎0 = − 4𝑥|
+ 4𝑥|
𝜋
−𝜋 𝜋
0
𝑎0 = 0
𝑎𝑛 =
1 0
𝑛𝜋𝑥
1 𝜋
𝑛𝜋𝑥
∫ −4 cos(
)𝑑𝑥 + ∫ 4 cos(
)𝑑𝑥
𝜋 −𝜋
𝜋
𝜋 0
𝜋
Aplicamos integrales de chebyshev ,obteniendo:
𝑎𝑛 = [
−4
0
4
𝜋
𝑠𝑒𝑛(𝑛𝑥)|
+
sen(𝑛𝑥)| ]
𝑛𝜋
−𝜋 𝑛𝜋
0
𝑎𝑛 = 0
𝑏𝑛 =
1 0
𝑛𝜋𝑥
1 𝜋
𝑛𝜋𝑥
∫ −4 sen(
) 𝑑𝑥 + ∫ 4 sen(
) 𝑑𝑥
𝜋 −𝜋
𝜋
𝜋 0
𝜋
𝑏𝑛 = [
4
0
4
𝜋
𝑐𝑜𝑠(𝑛𝑥)|
−
cos(𝑛𝑥)| ]
𝑛𝜋
−𝜋 𝑛𝜋
0
𝑏𝑛 =
4
4(−1)𝑛 4(−1)𝑛
4
−
−
+
𝑛𝜋
𝑛𝜋
𝑛𝜋
𝑛𝜋
−8(−1)𝑛
8
𝑏𝑛 =
+
𝑛𝜋
𝑛𝜋
Reemplazando los valores obtenidos en la fórmula:
∞
1
𝑛𝜋𝑥
𝑛𝜋𝑥
𝑓(𝑥) = 𝑎0 + ∑ [𝑎𝑛 cos (
) + 𝑏𝑛 𝑠𝑒𝑛(
)]
2
𝐿
𝐿
𝑛=1
∞
0
8
8
(−1)𝑛 + )𝑠𝑒𝑛(𝑛𝑥)]
𝑓(𝑥) = + ∑ [(
2
𝑛𝜋
𝑛𝜋
𝑛=1
Resolviendo esta sumatoria obtenemos:
𝑓(𝑥) =
16
𝑠𝑒𝑛(2𝑛 − 1)
(2𝑛 − 1)
7.𝑓(𝑥) = 𝑥 2 − 𝑥 + 3 ; −2 ≤ 𝑥 ≤ 2
Hallamos los coeficientes de la ecuación general:
1 2
𝑎0 = ∫ 𝑥 2 − 𝑥 + 3 𝑑𝑥
2 −2
1 𝑥3 𝑥2
2
𝑎0 = ( − + 3𝑥)|
2 3
2
−2
𝑎0 =
26
3
1 2
𝑛𝜋𝑥
𝑎𝑛 = ∫ (𝑥 2 − 𝑥 + 3) cos(
) 𝑑𝑥
3 −2
3
Aplicamos integrales de chebyshev, obteniendo:
1
2
𝑛𝜋𝑥
4
𝑛𝜋𝑥
8
𝑛𝜋𝑥
2
𝑎𝑛 = [(𝑥 2 − 𝑥 + 3)
𝑠𝑒𝑛(
) + (2𝑥 − 1) 2 2 𝑐𝑜𝑠(
) − (2) 3 3 𝑠𝑒𝑛(
)] |
2
𝑛𝜋
2
𝑛 𝜋
2
𝑛 𝜋
2
−2
1 12
20
𝑎𝑛 = [ 2 2 𝑐𝑜𝑠(𝑛𝜋) + 2 2 𝑐𝑜𝑠(−𝑛𝜋)]
2 𝑛 𝜋
𝑛 𝜋
1 32
𝑎𝑛 = [ 2 2 𝑐𝑜𝑠(𝑛𝜋)]
2 𝑛 𝜋
1 32
𝑎𝑛 = [ 2 2 (−1)𝑛 ]
2 𝑛 𝜋
𝑎𝑛 =
16
(−1)𝑛
𝑛2 𝜋 2
1 2
𝑛𝜋𝑥
𝑏𝑛 = ∫ (𝑥 2 − 𝑥 + 3) sen(
) 𝑑𝑥
3 −2
2
Aplicamos integrales de chebyshev, obteniendo:
1
2
𝑛𝜋𝑥
4
𝑛𝜋𝑥
8
𝑛𝜋𝑥
2
𝑏𝑛 = [−(𝑥 2 − 𝑥 + 3)
𝑐𝑜𝑠(
) + (2𝑥 − 1) 2 2 𝑠𝑒𝑛(
) + (2) 3 3 𝑐𝑜𝑠(
)] |
2
𝑛𝜋
2
𝑛 𝜋
2
𝑛 𝜋
2
−2
1 −10
18
16
16
𝑏𝑛 = [ 1 1 𝑐𝑜𝑠(𝑛𝜋) + 1 1 𝑐𝑜𝑠(−𝑛𝜋) + 3 3 𝑐𝑜𝑠(𝑛𝜋) − 3 3 𝑐𝑜𝑠(−𝑛𝜋)]
2 𝑛 𝜋
𝑛 𝜋
𝑛 𝜋
𝑛 𝜋
1 8
𝑏𝑛 = [ 1 1 𝑐𝑜𝑠(𝑛𝜋)]
2 𝑛 𝜋
𝑏𝑛 =
4
(−1)𝑛
𝑛𝜋
Reemplazando los valores obtenidos en la fórmula:
∞
1
𝑛𝜋𝑥
𝑛𝜋𝑥
𝑓(𝑥) = 𝑎0 + ∑ [𝑎𝑛 cos (
) + 𝑏𝑛 𝑠𝑒𝑛(
)]
2
𝐿
𝐿
𝑛=1
Obtenemos:
∞
26
16
𝑛𝜋𝑥
4
𝑛𝜋𝑥
𝑓(𝑥) =
+ ∑ [ 2 2 (−1)𝑛 cos (
)+
(−1)𝑛 𝑠𝑒𝑛(
)]
2.3
𝑛 𝜋
𝐿
𝑛𝜋
𝐿
𝑛=1
∞
13
16
𝑛𝜋𝑥
4
𝑛𝜋𝑥
𝑓(𝑥) =
+ ∑ [ 2 2 (−1)𝑛 cos (
)+
(−1)𝑛 𝑠𝑒𝑛(
)]
3
𝑛 𝜋
𝐿
𝑛𝜋
𝐿
𝑛=1
8. 𝑓(𝑥) = {
−𝑥, −5 ≤ 𝑥 ≤ 0
, 0≤𝑥≤5
Hallamos los coeficientes de la ecuación general:
1 0
1 𝜋
𝑎0 = ∫ −𝑥 𝑑𝑥 + ∫ 1 + 𝑥 2 𝑑𝑥
5 −5
5 0
𝑎0 = −
1 𝑥2 0
1
𝑥3 5
|
+ (𝑥 + )|
5 2 −5 5
3 0
𝑎0 =
5
15
+1+
2
3
𝑎0 =
𝑎𝑛 =
71
3
1 0
𝑛𝜋𝑥
1 5
𝑛𝜋𝑥
∫ −𝑥 cos(
)𝑑𝑥 + ∫ 1 + 𝑥 2 cos(
)𝑑𝑥
𝜋 −5
𝜋
𝜋 0
𝜋
Aplicamos integrales de chebyshev ,obteniendo:
1 5𝑥
𝑛𝜋𝑥
25
𝑛𝜋𝑥
0
𝑎𝑛 = − [ 𝑠𝑒𝑛 (
) + 2 2 cos(
)] |
5 𝑛𝜋
5
𝑛 𝜋
5
−5
1
5
𝑛𝜋𝑥
25
𝑛𝜋𝑥
125
𝑛𝜋𝑥 5
+ [(𝑥 2 + 1)
𝑠𝑒𝑛(
) + (2𝑥) 2 2 𝑐𝑜𝑠(
) − (2) 3 3 𝑠𝑒𝑛(
)] |
5
𝑛𝜋
5
𝑛 𝜋
5
𝑛 𝜋
5
0
1 25
25
50
𝑎𝑛 = − ( 2 2 − 2 2 (−1)𝑛 ) + ( 2 2 (−1)𝑛 )
5 𝑛 𝜋
𝑛 𝜋
𝑛 𝜋
𝑎𝑛 = −
5
𝑛2 𝜋 2
𝑎𝑛 =
𝑏𝑛 =
+
5
𝑛2 𝜋 2
5
𝑛2 𝜋 2
(−1)𝑛 +
50
(−1)𝑛
𝑛2 𝜋 2
[11(−1)𝑛 − 1]
1 0
𝑛𝜋𝑥
1 5
𝑛𝜋𝑥
∫ −𝑥 sen(
) 𝑑𝑥 + ∫ 1 + 𝑥 2 sen(
) 𝑑𝑥
𝜋 −5
𝜋
𝜋 0
𝜋
Aplicamos integrales de chebyshev ,obteniendo:
1 −5𝑥
𝑛𝜋𝑥
25
𝑛𝜋𝑥
0
𝑏𝑛 = − [
𝑐𝑜𝑠 (
) + 2 2 sen(
)] |
5 𝑛𝜋
5
𝑛 𝜋
5
−5
1
5
𝑛𝜋𝑥
25
𝑛𝜋𝑥
+ [−(𝑥 2 + 1)
𝑐𝑜𝑠(
) + (2𝑥) 2 2 𝑠𝑒𝑛(
)
5
𝑛𝜋
5
𝑛 𝜋
5
125
𝑛𝜋𝑥 5
+ (2) 3 3 𝑐𝑜𝑠(
)] |
𝑛 𝜋
5
0
𝑏𝑛 =
5
50
[1 − 21(−1)𝑛 ] + 3 3 [(−1)𝑛 − 1]
𝜋
𝑛 𝜋
Reemplazando los valores obtenidos en la fórmula:
∞
1
𝑛𝜋𝑥
𝑛𝜋𝑥
𝑓(𝑥) = 𝑎0 + ∑ [𝑎𝑛 cos (
) + 𝑏𝑛 𝑠𝑒𝑛(
)]
2
𝐿
𝐿
𝑛=1
∞
71
5
𝑛𝜋𝑥
5
𝑓(𝑥) =
+ ∑ [ 2 2 [11(−1)𝑛 − 1]cos (
) + [1 − 21(−1)𝑛 ]
2.3
𝑛 𝜋
𝐿
𝜋
𝑛=1
+
50
𝑛𝜋𝑥
[(−1)𝑛 − 1]𝑠𝑒𝑛(
)]
3
3
𝑛 𝜋
𝐿
SECCION 2.4
EN CADA PROBLEMA DE 1 AL 10 ESCRIBA LA SERIE DE FOURIER EN COSENOS Y LA SERIE DE
FOURIER EN SENOS DE LA FUNCION EN EL INTERVALO
DETERMINE LA SUMA DE CADA SERIE
1. 𝒇(𝒙)= 𝟒 ; 𝟎 ≤ 𝒙 ≤ 𝟑
SERIE FOURIER COSENO
∞
𝑎0
𝑛𝜋𝑥
𝑓(𝑥)= + ∑ 𝑎𝑛 cos(
)
2
𝐿
𝑛=1
DONDE
2 𝐿
𝑎0 = ∫ 𝑓(𝑥) 𝑑𝑥
𝐿 0
𝐿
2
𝑛𝜋𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos(
) 𝑑𝑥
𝐿 0
𝐿
REEMPLAZANDO PARA 𝑎0
2 𝐿
𝑎0 = ∫ 𝑓(𝑥) 𝑑𝑥
𝐿 0
2 3
𝑎0 = ∫ 4 𝑑𝑥
3 0
8 3
𝑎0 = 𝑥 {
3 0
8
𝑎0 = (3) = 8
3
REEMPLAZANDO PARA 𝑎𝑛
2 𝐿
𝑛𝜋𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos(
) 𝑑𝑥
𝐿 0
𝐿
2 3
𝑛𝜋𝑥
𝑎𝑛 = ∫ 4 cos(
) 𝑑𝑥
3 0
3
8 3
𝑛𝜋𝑥
𝑎𝑛 = ∫ cos(
) 𝑑𝑥
3 0
3
8 3
𝑛𝜋𝑥 3
𝑎𝑛 =
sen(
){
0
3 𝑛𝜋
3
𝑎𝑛 =
8
𝑛𝜋(3)
𝑛𝜋(0)
) − sen (
)]
[sen (
𝑛𝜋
3
3
8
[sen(𝑛𝜋) − sen(0)]; 𝑑𝑜𝑛𝑑𝑒 sen(𝑛𝜋) = 0
𝑛𝜋
𝑎𝑛 = 0
REMPLANZADO TODO EN LA FUNCION
𝑎𝑛 =
∞
𝑎0
𝑛𝜋𝑥
𝑓(𝑥)= + ∑ 𝑎𝑛 cos(
)
2
𝐿
𝑛=1
∞
8
𝑛𝜋𝑥
𝑓(𝑥)= + ∑(0) cos(
)
2
𝐿
𝑛=1
𝑓(𝑥)= 4
SERIE FOURIER SENO
∞
𝑛𝜋𝑥
𝑓(𝑥)= ∑ 𝑏𝑛 sen(
)
𝐿
𝑛=1
DONDE
2 𝐿
𝑛𝜋𝑥
𝑏𝑛 = ∫ 𝑓(𝑥) sen(
) 𝑑𝑥
𝐿 0
𝐿
REEMPLAZANDO PARA 𝑏𝑛
2 3
𝑛𝜋𝑥
𝑏𝑛 = ∫ 4 sen(
) 𝑑𝑥
3 0
3
8 3
𝑛𝜋𝑥
𝑏𝑛 = ∫ sen(
) 𝑑𝑥
3 0
3
8
3
𝑛𝜋𝑥 3
𝑏𝑛 = (−
cos(
){ )
0
3
𝑛𝜋
3
8
𝑛𝜋𝑥 3
𝑎𝑛 = −
cos(
){
0
𝑛𝜋
3
8
𝑛𝜋(3)
𝑛𝜋(0)
𝑏𝑛 = −
) − cos (
)]
[cos (
𝑛𝜋
3
3
8
[cos(𝑛𝜋) − cos(0)] ; 𝑑𝑜𝑛𝑑𝑒 cos(𝑛𝜋) = (−1)𝑛
𝑏𝑛 = −
𝑛𝜋
𝑏𝑛 = −
8
[(−1)𝑛 − 1]
𝑛𝜋
REMPLANZADO TODO EN LA FUNCION
∞
𝑛𝜋𝑥
𝑓(𝑥)= ∑ 𝑏𝑛 sen(
)
𝐿
𝑛=1
∞
𝑓(𝑥)= ∑(−
𝑛=1
8
𝑛𝜋𝑥
[(−1)𝑛 − 1]) sen(
)
𝑛𝜋
3
SUMA TOTAL
∞
𝑓(𝑥)= 4 + ∑(−
𝑛=1
8
𝑛𝜋𝑥
[(−1)𝑛 − 1]) sen(
)
𝑛𝜋
3
2. 𝒇(𝒙)= {
𝟏, 𝒑𝒂𝒓𝒂 𝟎 ≤ 𝒙 < 𝟏
−𝟏, 𝒑𝒂𝒓𝒂 𝟏 ≤ 𝒙 ≤ 𝟐
SERIE FOURIER COSENO
∞
𝑎0
𝑛𝜋𝑥
𝑓(𝑥)= + ∑ 𝑎𝑛 cos(
)
2
𝐿
𝑛=1
DONDE
2 𝐿
𝑎0 = ∫ 𝑓(𝑥) 𝑑𝑥
𝐿 0
𝐿
2
𝑛𝜋𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos(
) 𝑑𝑥
𝐿 0
𝐿
REEMPLAZANDO PARA 𝑎0
2 𝐿
𝑎0 = ∫ 𝑓(𝑥) 𝑑𝑥
𝐿 0
1
2
2 2
𝑎0 = ∫ 1 𝑑𝑥 + ∫ −1 𝑑𝑥
2 0
2 1
1
2
𝑎0 = 1 ∫ 𝑑𝑥 − 1 ∫ 𝑑𝑥
0
1
1
2
𝑎0 = 1𝑥 { − 1𝑥 {
0
1
𝑎0 = (1 − 0) − (2 − 1)
𝑎0 = 0
REEMPLAZANDO PARA 𝑎𝑛
2 𝐿
𝑛𝜋𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos(
) 𝑑𝑥
𝐿 0
𝐿
2 1
𝑛𝜋𝑥
2 2
𝑛𝜋𝑥
𝑎𝑛 = ∫ 1 cos(
) 𝑑𝑥 + ∫ (−1) cos(
) 𝑑𝑥
2 0
2
2 1
2
1
2
𝑛𝜋𝑥
𝑛𝜋𝑥
𝑎𝑛 = ∫ cos(
) 𝑑𝑥 + ∫ cos(
) 𝑑𝑥
2
2
0
1
𝑎𝑛 =
2
𝑛𝜋𝑥 1
2
𝑛𝜋𝑥 2
sen(
){ +
sen(
){
0
1
𝑛𝜋
2
𝑛𝜋
2
𝑎𝑛 =
𝑎𝑛 =
2
𝑛𝜋(1)
𝑛𝜋(0)
2
𝑛𝜋(2)
𝑛𝜋(1)
) − sen(
)] +
) − sen(
)]
[ sen(
[ sen(
𝑛𝜋
2
2
𝑛𝜋
2
2
2
𝑛𝜋
2
𝑛𝜋
[ sen( ) − sen(0)] +
[ sen(𝑛𝜋) − sen( )] ; 𝑑𝑜𝑛𝑑𝑒 sen(𝑛𝜋) = 0
𝑛𝜋
2
𝑛𝜋
2
𝑎𝑛 =
2
𝑛𝜋
2
𝑛𝜋
[ sen( ) − 0] +
[0 − sen( )]
𝑛𝜋
2
𝑛𝜋
2
𝑎𝑛 = 0
SERIE FOURIER SENO
∞
𝑛𝜋𝑥
𝑓(𝑥)= ∑ 𝑏𝑛 sen(
)
𝐿
𝑛=1
DONDE
2 𝐿
𝑛𝜋𝑥
𝑏𝑛 = ∫ 𝑓(𝑥) sen(
) 𝑑𝑥
𝐿 0
𝐿
REEMPLAZANDO PARA 𝑏𝑛
2 1
𝑛𝜋𝑥
2 2
𝑛𝜋𝑥
𝑏𝑛 = ∫ (1) sen(
) 𝑑𝑥 + ∫ (−1) sen(
) 𝑑𝑥
2 0
2
2 1
2
1
2
𝑛𝜋𝑥
𝑛𝜋𝑥
𝑏𝑛 = ∫ sen(
) 𝑑𝑥 − ∫ sen(
) 𝑑𝑥
2
2
0
1
𝑏𝑛 = −
2
𝑛𝜋𝑥 1 2
𝑛𝜋𝑥 2
cos (
){ +
cos (
){
0 nπ
1
nπ
2
2
2
𝑛𝜋(1)
𝑛𝜋(0)
2
𝑛𝜋(2)
𝑛𝜋(1)
[cos (
[cos (
) − cos (
) − cos (
)] +
)]
nπ
2
2
nπ
2
2
2
𝑛𝜋
2
𝑛𝜋
𝑏𝑛 = − [cos ( ) − cos(0)] +
[cos(𝑛𝜋) − cos ( )] ; 𝑑𝑜𝑛𝑑𝑒 cos(𝑛𝜋) = (−1)𝑛
nπ
2
nπ
2
𝑛𝜋
cos ( ) = 0
2
2
2
[(−1)𝑛 −0]
𝑏𝑛 = − [0 −1] +
nπ
nπ
2
[1 + (−1)𝑛 ]
𝑏𝑛 =
nπ
𝑏𝑛 = −
SUMA TOTAL
∞
2
𝑛𝜋𝑥
𝑓(𝑥)= ∑( [1 + (−1)𝑛 ]) sen(
)
nπ
2
𝑛=1
3. 𝒇(𝒙)= {
𝟎, 𝒑𝒂𝒓𝒂 𝟎 ≤ 𝒙 < 𝝅
𝒄𝒐𝒔(𝒙) , 𝒑𝒂𝒓𝒂 𝝅 ≤ 𝒙 ≤ 𝟐𝝅
SERIE FOURIER COSENO
∞
𝑎0
𝑛𝜋𝑥
𝑓(𝑥)= + ∑ 𝑎𝑛 cos(
)
2
𝐿
𝑛=1
DONDE
2 𝐿
𝑎0 = ∫ 𝑓(𝑥) 𝑑𝑥
𝐿 0
𝐿
2
𝑛𝜋𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos(
) 𝑑𝑥
𝐿 0
𝐿
REEMPLAZANDO PARA 𝑎0
2 𝐿
𝑎0 = ∫ 𝑓(𝑥) 𝑑𝑥
𝐿 0
𝜋
2
2 2𝜋
𝑎0 =
∫ (0) 𝑑𝑥 +
∫ (cos 𝑥) 𝑑𝑥
2𝜋 0
2𝜋 𝜋
1 2𝜋
∫ (cos 𝑥) 𝑑𝑥
𝜋 𝜋
1 2𝜋
𝑎0 = ∫ (cos 𝑥) 𝑑𝑥
𝜋 𝜋
1
2𝜋
𝑎0 = sin(𝑥) {
𝜋
𝜋
1
𝑎0 = [sin 2𝜋 − sin 𝜋]
𝜋
𝑎0 =
𝑎0 = 0
REEMPLAZANDO PARA 𝑎𝑛
2 𝐿
𝑛𝜋𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos(
) 𝑑𝑥
𝐿 0
𝐿
𝑎𝑛 =
2 𝜋
𝑛𝜋𝑥
2 2𝜋
𝑛𝜋𝑥
∫ (0) cos(
) 𝑑𝑥 +
∫ cos(𝑥) cos(
) 𝑑𝑥
2𝜋 0
2𝜋
2𝜋 𝜋
2𝜋
𝑎𝑛 =
𝑎𝑛 =
1 2𝜋
𝑛𝑥
∫ cos(𝑥) cos( ) 𝑑𝑥
𝜋 𝜋
2
2𝜋
2𝜋
1
𝑛
𝑛
[∫ cos (𝑥 − 𝑥) 𝑑𝑥 + ∫ cos (𝑥 + 𝑥) 𝑑𝑥 ]
2𝜋 𝜋
2
2
𝜋
𝑎𝑛 =
𝑎𝑛 =
𝑎𝑛 =
2𝜋
2𝜋
1
2−𝑛
2+𝑛
[∫ cos((
)𝑥) 𝑑𝑥 + ∫ cos ((
) 𝑥) 𝑑𝑥 ]
2𝜋 𝜋
2
2
𝜋
1
1
2−𝑛
1
2
2+𝑛
2𝜋
2𝜋
(sin(
)𝑥) { ] +
(sin(
)𝑥) { ]
[
[
𝜋
𝜋
2𝜋 1 − 𝑛
2
2𝜋 2 + 𝑛
2
1
2
2−𝑛
2−𝑛
1
2
2+𝑛
2+𝑛
(sin (
) 2𝜋 − sin (
) 𝜋)] +
(sin (
) 2𝜋 − sin (
) 𝜋)]
[
[
2𝜋 2 − 𝑛
2
2
2𝜋 2 + 𝑛
2
2
𝑎𝑛 =
1
2
2−𝑛 𝜋
1
2
2+𝑛 𝜋
(sin(2 − 𝑛)𝜋 − sin (
) )] +
(sin(2 + 𝑛)𝜋 − sin (
) )]
[
[
2𝜋 2 − 𝑛
2
2
2𝜋 2 + 𝑛
2
2
𝑎𝑛 =
1
2
2−𝑛 𝜋
1
2
2+𝑛 𝜋
−sin (
) ]+
− sin (
) ]
[
[
2𝜋 2 − 𝑛
2
2
2𝜋 2 + 𝑛
2
2
SERIE FOURIER SENO
∞
𝑛𝜋𝑥
𝑓(𝑥)= ∑ 𝑏𝑛 sen(
)
𝐿
𝑛=1
DONDE
2 𝐿
𝑛𝜋𝑥
𝑏𝑛 = ∫ 𝑓(𝑥) sen(
) 𝑑𝑥
𝐿 0
𝐿
REEMPLAZANDO PARA 𝑏𝑛
𝑏𝑛 =
2 𝜋
𝑛𝜋𝑥
2 2𝜋
𝑛𝜋𝑥
∫ (0) sen(
) 𝑑𝑥 +
∫ cos 𝑥 sen(
) 𝑑𝑥
2𝜋 0
2𝜋
2𝜋 𝜋
2𝜋
𝑏𝑛 =
2 2𝜋
𝑛𝜋𝑥
∫ cos 𝑥 sen(
) 𝑑𝑥
2𝜋 𝜋
2𝜋
𝑏𝑛 =
1 2𝜋
𝑛𝑥
∫ cos 𝑥 sen( ) 𝑑𝑥
𝜋 𝜋
2
𝑏𝑛 =
1 2𝜋
𝑛
1 2𝜋
𝑛
∫ sin( 𝑥 − 𝑥) 𝑑𝑥 +
∫ sin( 𝑥 + 𝑥)𝑑𝑥
2𝜋 𝜋
2
2𝜋 𝜋
2
𝑏𝑛 =
1 2𝜋
𝑛
1 2𝜋
𝑛
∫ sin( − 1)𝑥 𝑑𝑥 +
∫ sin( + 1)𝑥𝑑𝑥
2𝜋 𝜋
2
2𝜋 𝜋
2
𝑏𝑛 =
1 2𝜋
𝑛−2
1 2𝜋
𝑛+2
∫ sin(
)𝑥 𝑑𝑥 +
∫ sin(
)𝑥𝑑𝑥
2𝜋 𝜋
2
2𝜋 𝜋
2
𝑏𝑛 =
𝑏𝑛 = −
1 2𝜋
𝑛−2
1 2𝜋
𝑛+2
∫ sin(
)𝑥 𝑑𝑥 +
∫ sin(
)𝑥𝑑𝑥
2𝜋 𝜋
2
2𝜋 𝜋
2
1
2
𝑛−2
1
2
𝑛+2
2𝜋
2𝜋
(
) cos (
)𝑥{ −
(
) cos (
)𝑥{
𝜋 2𝜋 𝑛 + 2
𝜋
2𝜋 𝑛 − 2
2
2
𝑏𝑛 = −
1
2
𝑛−2
𝑛−2
(
) [cos (
) 2𝜋 − cos (
) 𝜋]
2𝜋 𝑛 − 2
2
2
1
2
𝑛+2
𝑛+2
−
(
) [cos (
) 2𝜋 − cos(
) 𝜋]
2𝜋 𝑛 + 2
2
2
SUMA TOTAL
∞
1
2
2−𝑛 𝜋
1
2
2+𝑛 𝜋
𝑛𝜋𝑥
𝑓(𝑥)= ∑( [
−sin (
) ]+
− sin (
) ]) cos(
)
[
2𝜋 2 − 𝑛
2
2
2𝜋 2 + 𝑛
2
2
2𝜋
𝑛=1
∞
+∑−
𝑛=1
−
1
2
𝑛−2
𝑛−2
(
) [cos (
) 2𝜋 − cos (
) 𝜋]
2𝜋 𝑛 − 2
2
2
1
2
𝑛+2
𝑛+2
𝑛𝜋𝑥
(
) [cos (
) 2𝜋 − cos(
) 𝜋] sen(
)
2𝜋 𝑛 + 2
2
2
2𝜋
4. 𝒇(𝒙)= 𝟐𝒙 ; 𝒑𝒂𝒓𝒂 𝟎 ≤ 𝒙 ≤ 𝟏
SERIE FOURIER COSENO
∞
𝑎0
𝑛𝜋𝑥
𝑓(𝑥)= + ∑ 𝑎𝑛 cos(
)
2
𝐿
𝑛=1
DONDE
2 𝐿
𝑎0 = ∫ 𝑓(𝑥) 𝑑𝑥
𝐿 0
𝐿
2
𝑛𝜋𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos(
) 𝑑𝑥
𝐿 0
𝐿
REEMPLAZANDO PARA 𝑎0
2 1
𝑎0 = ∫ 2𝑥 𝑑𝑥
1 0
1
𝑎0 = 4 ∫ 𝑥 𝑑𝑥
0
𝑥2 1
){
2 0
12
𝑎0 = 2 ( )
2
𝑎0 = 1
𝑎0 = 4 (
REEMPLAZANDO PARA 𝑎𝑛
2 𝐿
𝑛𝜋𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos(
) 𝑑𝑥
𝐿 0
𝐿
2 1
𝑎𝑛 = ∫ 2𝑥 cos(𝑛𝜋𝑥) 𝑑𝑥
1 0
1
𝑎𝑛 = 4 ∫ 𝑥 cos(𝑛𝜋𝑥) 𝑑𝑥
0
𝑥 sin(𝑛𝜋𝑥) cos(𝑛𝜋𝑥) 1
𝑎𝑛 = 4 [
−
]{
0
𝑛𝜋
𝑛2 𝜋 2
(1) sin(𝑛𝜋(1)) cos(𝑛𝜋(1))
(0) sin(𝑛𝜋(0)) cos(𝑛𝜋(0))
𝑎𝑛 = 4 [(
−
)−(
−
)]
𝑛𝜋
𝑛2 𝜋 2
𝑛𝜋
𝑛2 𝜋 2
cos(𝑛𝜋(1))
cos(𝑛𝜋(0))
𝑎𝑛 = 4 [(−
)+(
)]
2
2
𝑛 𝜋
𝑛2 𝜋 2
(−1)𝑛
1
𝑎𝑛 = 4 [(− 2 2 ) + ( 2 2 )]
𝑛 𝜋
𝑛 𝜋
SERIE FOURIER SENO
∞
𝑛𝜋𝑥
𝑓(𝑥)= ∑ 𝑏𝑛 sen(
)
𝐿
𝑛=1
DONDE
2 𝐿
𝑛𝜋𝑥
𝑏𝑛 = ∫ 𝑓(𝑥) sen(
) 𝑑𝑥
𝐿 0
𝐿
REEMPLAZANDO PARA 𝑏𝑛
2 1
𝑏𝑛 = ∫ 2𝑥 sen(𝑛𝜋𝑥) 𝑑𝑥
1 0
1
𝑏𝑛 = 4 ∫ 𝑥 sen(𝑛𝜋𝑥) 𝑑𝑥
0
𝑏𝑛 = 4 [−
𝑥 cos(𝑛𝜋𝑥) sin(𝑛𝜋𝑥) 1
+
]{
0
𝑛𝜋
𝑛2 𝜋 2
𝑏𝑛 = 4 [(−
(1) cos(𝑛𝜋(1)) sin(𝑛𝜋(1))
(0) cos(𝑛𝜋(0)) sin(𝑛𝜋(0))
+
+
) − (−
)]
2
2
𝑛𝜋
𝑛 𝜋
𝑛𝜋
𝑛2 𝜋 2
(1) cos(𝑛𝜋(1))
𝑏𝑛 = 4 [(−
)]
𝑛𝜋
(−1)𝑛
𝑏𝑛 = −4 [(
)]
𝑛𝜋
SUMA TOTAL
∞
∞
𝑛=1
𝑛=1
1
(−1)𝑛
1
(−1)𝑛
𝑓(𝑥)= + ∑(4 [(− 2 2 ) + ( 2 2 )]) cos(𝑛𝜋𝑥) + ∑(= −4 [(
)]) sen(𝑛𝜋𝑥)
2
𝑛 𝜋
𝑛 𝜋
𝑛𝜋
5. 𝒇(𝒙)= 𝒙𝟐 ; 𝟎 ≤ 𝒙 ≤ 𝟐
SERIE FOURIER COSENO
∞
𝑎0
𝑛𝜋𝑥
𝑓(𝑥)= + ∑ 𝑎𝑛 cos(
)
2
𝐿
𝑛=1
DONDE
2 𝐿
𝑎0 = ∫ 𝑓(𝑥) 𝑑𝑥
𝐿 0
𝐿
2
𝑛𝜋𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos(
) 𝑑𝑥
𝐿 0
𝐿
REEMPLAZANDO PARA 𝑎0
2 2
𝑎0 = ∫ 𝑥 2 𝑑𝑥
2 0
2
𝑎0 = ∫ 𝑥 2 𝑑𝑥
0
𝑥3 2
{
3 0
8
𝑎0 =
3
𝑎0 =
REEMPLAZANDO PARA 𝑎𝑛
2 𝐿
𝑛𝜋𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos(
) 𝑑𝑥
𝐿 0
𝐿
2 2
𝑛𝜋𝑥
𝑎𝑛 = ∫ 𝑥 2 cos(
) 𝑑𝑥
2 0
2
𝑛𝜋𝑥
𝑛𝜋𝑥
𝑛𝜋𝑥
2𝑥 2 sin( 2 ) 8𝑥 cos ( 2 ) 16 sin( 2 ) 2
𝑎𝑛 = (
+
−
){
0
𝑛𝜋
𝑛2 𝜋 2
𝑛3 𝜋 3
𝑎𝑛 =
16(−1)𝑛
𝑛2 𝜋 2
SERIE FOURIER SENO
∞
𝑛𝜋𝑥
𝑓(𝑥)= ∑ 𝑏𝑛 sen(
)
𝐿
𝑛=1
DONDE
2 𝐿
𝑛𝜋𝑥
𝑏𝑛 = ∫ 𝑓(𝑥) sen(
) 𝑑𝑥
𝐿 0
𝐿
REEMPLAZANDO PARA 𝑏𝑛
2 2
𝑛𝜋𝑥
𝑏𝑛 = ∫ 𝑥 2 sen(
) 𝑑𝑥
2 0
2
2
𝑛𝜋𝑥
𝑏𝑛 = ∫ 𝑥 2 sen(
) 𝑑𝑥
2
0
𝑛𝜋𝑥
𝑛𝜋𝑥
𝑛𝜋𝑥
−2𝑥 2 cos( 2 ) 8𝑥 sin( 2 ) 16 cos( 2 ) 2
𝑏𝑛 = [
+
+
]{
0
𝑛𝜋
𝑛2 𝜋 2
𝑛3 𝜋 3
𝑏𝑛 =
−8(−1)𝑛 16(−1)𝑛
16
+
+ 3 3
3
3
𝑛𝜋
𝑛 𝜋
𝑛 𝜋
SUMA TOTAL
∞
∞
𝑛=1
𝑛=1
4
16(−1)𝑛
𝑛𝜋𝑥
−8(−1)𝑛 16(−1)𝑛
16
𝑛𝜋𝑥
𝑓(𝑥)= + ∑( 2 2 ) cos (
) + ∑(
+
+ 3 3 ) sen(
)
3
3
3
𝑛 𝜋
2
𝑛𝜋
𝑛 𝜋
𝑛 𝜋
2
6. 𝑓(𝑥) = 𝑒 −𝑥 𝑝𝑎𝑟𝑎 0 ≤ 𝑥 ≤ 1
∞
𝐶(𝑥) = 1 − 𝑒
−1
+2∑
𝑛=1
1
[1 − (−1)𝑛 𝑒 −1 ] cos(𝑛𝜋𝑥) = 𝑒 −𝑥
1 + 𝑛2 𝜋 2
Para 0 ≤ 𝑥 ≤ 1
∞
𝑆(𝑥) = 2𝜋 ∑
𝑛=1
7. 𝑓(𝑥) = {
−𝑥
1
𝑠𝑖 0 < 𝑥 < 1
[1 − (−1)𝑛 𝑒 −1 ] sen(𝑛𝜋𝑥) = { 𝑒
2
2
0 𝑠𝑖 𝑥 = 0 𝑜 𝑥 = 1
1+𝑛 𝜋
𝑥 𝑝𝑎𝑟𝑎 0 ≤ 𝑥 ≤ 2
2 − 𝑥 𝑝𝑎𝑟𝑎 2 < 𝑥 ≤ 3
∞
1
4
2𝑛𝜋
12
2𝑛𝜋
6
𝑛𝜋𝑥
𝐶𝑥 = + ∑ [ sin(
) + 2 2 cos (
) − 2 2 [1 + (−1)−𝑛 ]] cos (
)
2
𝑛𝜋
3
𝑛 𝜋
3
𝑛 𝜋
3
𝑛=1
𝑥 𝑠𝑖 0 ≤ 𝑥 < 2
= { 1 𝑠𝑖 𝑥 = 2
2 − 𝑥 𝑠𝑖 2 < 𝑥 ≤ 3
𝑥 𝑠𝑖 0 ≤ 𝑥 < 2
12
2𝑛𝜋
4
2𝑛𝜋
2
𝑛𝜋𝑥
1 𝑠𝑖 𝑥 = 2
(−1)−𝑛 ] sen (
𝑆𝑥 = ∑ [ 2 2 sin(
)−
cos (
)−
)={
2 − 𝑥 𝑠𝑖 2 < 𝑥 ≤ 3
𝑛 𝜋
3
𝑛𝜋
3
𝑛𝜋
3
𝑛=1
0 𝑠𝑖 𝑥 = 3
∞
1, 𝑝𝑎𝑟𝑎 0 ≤ 𝑥 < 1
8. 𝑓(𝑥) = { 0, 𝑝𝑎𝑟𝑎 1 ≤ 𝑥 ≤ 3
−1, 𝑝𝑎𝑟𝑎 3 < 𝑥 ≤ 5
1 𝑠𝑖 0 ≤ 𝑥 < 1
1
𝑠𝑖 𝑥 = 1
∞
2
1 4
1
𝑛𝜋
2𝑛𝜋
𝑛𝜋𝑥
𝐶𝑥 = − + ∑ cos ( ) sin (
) cos
= 0 𝑠𝑖 1 < 𝑥 < 3
5 𝜋
𝑛
5
5
5
1
𝑛=1
− 𝑠𝑖 𝑥 = 3
2
{−1 𝑠𝑖 3 < 𝑥 ≤ 5
1 𝑠𝑖 0 ≤ 𝑥 < 1
1
𝑠𝑖 𝑥 = 1
∞
2
4
1
𝑛𝜋
2𝑛𝜋
𝑛𝜋𝑥
𝑆𝑥 = ∑
[1 + (−1)𝑛 − 2 cos
cos
] sin
= 0 𝑠𝑖 1 < 𝑥 < 3
𝜋
2𝑛
5
5
5
1
𝑛=1
− 𝑠𝑖 𝑥 = 3
2
{−1 𝑠𝑖 3 < 𝑥 ≤ 5
9. 𝑓(𝑥) = {
𝑥 2 , 𝑝𝑎𝑟𝑎 0 ≤ 𝑥 < 1
1, 𝑝𝑎𝑟𝑎 1 ≤ 𝑥 ≤ 4
∞
5 16
1
𝑛𝜋
4
𝑛𝜋
𝑛𝜋𝑥
𝑥2,
𝑝𝑎𝑟𝑎 0 ≤ 𝑥 < 1
𝐶𝑥 = + 2 ∑[ 2 cos ( ) − 3 sin ( )] cos
={
1, 𝑝𝑎𝑟𝑎 1 ≤ 𝑥 ≤ 4
6 𝜋
𝑛
4
𝑛 𝜋
4
4
𝑛=1
𝑥2,
𝑝𝑎𝑟𝑎 0 ≤ 𝑥 < 1
16
𝑛𝜋
64
𝑛𝜋
2(−1)𝑛
𝑛𝜋𝑥
𝑆𝑥 = ∑[ 2 2 sin
+ 3 3 [cos
− 1] −
] sin
= { 1, 𝑝𝑎𝑟𝑎 1 ≤ 𝑥 ≤ 4
𝑛 𝜋
4
𝑛 𝜋
4
𝑛𝜋
4
0, 𝑠𝑖 𝑥 = 4
𝑛=1
∞
10. 𝑓𝑥 = 1 − 𝑥 3 𝑝𝑎𝑟𝑎 0 ≤ 𝑥 ≤ 2
∞
24
1
4
𝑛𝜋𝑥
𝐶𝑥 = −1 − 2 ∑ 2 [2(−1)𝑛 + 2 2 [1 − (−1)𝑛 ]] cos
= 1 − 𝑥 3 𝑠𝑖 0 ≤ 𝑥 ≤ 2
𝜋
𝑛
𝑛 𝜋
2
𝑛=1
∞
𝑆𝑥 =
3
2
1
48
𝑛𝜋𝑥
∑ [1 + 7(−1)𝑛 − 2 2 (−1)𝑛 ] sin
= {1 − 𝑥 𝑠𝑖 0 < 𝑥 < 2
𝜋
𝑛
𝑛 𝜋
2
0 𝑠𝑖 𝑥 = 0 𝑜 𝑥 = 2
𝑛=1