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