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Ejercicios de ecuaciones de series de Fourier de senos y cosenos
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Ejercicios de ecuaciones de series de Fourier de senos y cosenos
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Series de Fourier de senos y de cosenos.Si una función es par o impar, la serie de Fourier se simplifica al hacer las operaciones siguiendo
los conceptos del siguiente teorema.
a) El producto de dos funciones par es par.
b) El producto de dos funciones impar es par.
c) El producto de una función par y una impar es impar.
d) La suma o diferencia de dos funciones par es par.
e) La suma o diferencia de dos funciones impar es impar.
𝑝 𝑝
f) Si 𝑓(𝑥) es par, entonces ∫−𝑝 𝑓(𝑥)𝑑𝑥 = 2 ∫0 𝑓(𝑥)𝑑𝑥
𝑝
g) Si 𝑓(𝑥) es impar, entonces ∫−𝑝 𝑓(𝑥)𝑑𝑥 = 0.
Función par 𝑓(𝑥) = 𝑓(−𝑥) Función impar 𝑓(𝑥) = −𝑓(−𝑥)
𝑓(𝑥) = 𝑥 2 ; 𝑓(−𝑥) = (−𝑥)2 ; 𝑓(−𝑥) = 𝑥 2 𝑓(𝑥) = 𝑥 3 ; 𝑓(−𝑥) = (−𝑥)3 ; 𝑓(−𝑥) = −𝑥 3 ; −𝑓(−𝑥) = 𝑥 3
El eje de simetría "𝑦" El origen (0,0) es el punto de simetría
Función no par, no impar 𝑓(𝑥) ≠ 𝑓(−𝑥)
𝑥+1 −𝑥+1 𝑥−1
𝑓(𝑥) = ; 𝑓(−𝑥) = ; 𝑓(−𝑥) =
𝑥−1 −𝑥−1 𝑥+1
Serie de cosenos
1 𝑝 2 𝑝
𝑎0 = 𝑝 ∫−𝑝 𝑓(𝑥)𝑑𝑥 ; si 𝑓(𝑥) es par entonces 𝑎0 = 𝑝 ∫0 𝑓(𝑥)𝑑𝑥
1 𝑝 𝑛𝜋 2 𝑝 𝑛𝜋
𝑎𝑛 = ∫−𝑝 𝑓(𝑥) cos ( 𝑥) 𝑑𝑥 ; Si 𝑓(𝑥) es par entonces 𝑎𝑛 = ∫0 𝑓(𝑥) cos ( 𝑥) 𝑑𝑥
𝑝 𝑝 𝑝 𝑝
1 𝑝 𝑛𝜋
𝑏𝑛 = ∫−𝑝 𝑓(𝑥) sen ( 𝑥) 𝑑𝑥 Si 𝑓(𝑥) es par entonces 𝑏𝑛 = 0
𝑝 𝑝
Serie de senos
1 𝑝
𝑎0 = ∫−𝑝 𝑓(𝑥)𝑑𝑥 ; si 𝑓(𝑥) es impar entonces 𝑎0 = 0
𝑝
1 𝑝 𝑛𝜋
𝑎𝑛 = ∫−𝑝 𝑓(𝑥) cos ( 𝑥) 𝑑𝑥 ; Si 𝑓(𝑥) es impar entonces 𝑎𝑛 = 0
𝑝 𝑝
1 𝑝 𝑛𝜋 2 𝑝 𝑛𝜋
𝑏𝑛 = ∫−𝑝 𝑓(𝑥) sen ( 𝑥) 𝑑𝑥 Si 𝑓(𝑥) es impar entonces 𝑏𝑛 = ∫−𝑝 𝑓(𝑥) sen ( 𝑥) 𝑑𝑥
𝑝 𝑝 𝑝 𝑝
Ejemplo.
Desarrolle para la función dada, la serie de Fourier adecuada de senos o cosenos.
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