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Ejercicios resueltos de series de Fourier.
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Ejercicios resueltos de series de Fourier.
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Series de Fourier.El producto interno es un número real el cual se denota por < 𝑓, 𝑔 >= 𝑓1 𝑔1 + 𝑓2 𝑔2 + 𝑓3 𝑔3 + ⋯ + 𝑓𝑛 𝑔𝑛 .
Si 𝑓1 y 𝑓2 son dos funciones escalares, definidas en un intervalo [𝑎; 𝑏], entonces:
𝑏
< 𝑓1 , 𝑓2 >= ∫𝑎 𝑓1 (𝑥)𝑓2 (𝑥)𝑑𝑥, es la definición de producto interno.
𝑏
Una función ortogonal es definida como < 𝑓1 , 𝑓2 >= 0, esto es ∫𝑎 𝑓1 (𝑥)𝑓2 (𝑥)𝑑𝑥 = 0.
Ejemplo Si 𝑓1 (𝑥) = 𝑥 4 y 𝑓2 (𝑥) = 𝑥 5 determine si las funciones son ortogonales en el intervalo [-2;2].
2
2 2 𝑥 10 210 (−2)10
< 𝑓1 , 𝑓2 >= ∫−2 𝑥 4 𝑥 5 𝑑𝑥 ; < 𝑓1 , 𝑓2 >= ∫−2 𝑥 9 𝑑𝑥; < 𝑓1 , 𝑓2 >= [ ] ; < 𝑓1 , 𝑓2 >= [ − ] =0
10 −2 10 10
Ejemplo Si 𝑓1 (𝑥) = 𝑥 2 y 𝑓2 (𝑥) = 𝑥 4 determine si las funciones son ortogonales en el intervalo [-2;2].
2
2 2 𝑥7 27 (−2)7 28
< 𝑓1 , 𝑓2 >= ∫−2 𝑥 2 𝑥 4 𝑑𝑥 ; < 𝑓1 , 𝑓2 >= ∫−2 𝑥 6 𝑑𝑥; < 𝑓1 , 𝑓2 >= [ ] ; < 𝑓1 , 𝑓2 >= [ − ] =
7 −2 7 7 7
Conjuntos ortogonales. Si 𝑎 = {𝑓0 , 𝑓1 , 𝑓3 (𝑥), … , 𝑓𝑛 (𝑥)} es ortogonal en el intervalo [𝑎; 𝑏]
𝑏
si < 𝑓𝑚 , 𝑓𝑛 >= ∫𝑎 𝑓𝑚 (𝑥)𝑓𝑛 (𝑥)𝑑𝑥 = 0;
Ejemplo Dado el conjunto 𝐵 = {1, cos(𝑥) , cos(2𝑥) , cos(3𝑥) , … cos(𝑚𝑥) , … , cos(𝑛𝑥)},
pruebe que es ortogonal en el intervalo [−𝜋; 𝜋]
Respuesta:
𝜋
a) para 𝑓1 (𝑥) = 1 y 𝑓𝑛 (𝑥) = cos (𝑛𝑥) entonces < 𝑓1 , 𝑓𝑛 >= ∫−𝜋(1)(cos (𝑛𝑥))𝑑𝑥;
1 1
< 𝑓1 , 𝑓𝑛 >= [𝑠𝑒𝑛(𝑛𝑥)]𝜋−𝜋 ; < 𝑓1 , 𝑓𝑛 >= [𝑠𝑒𝑛(𝑛𝜋) − 𝑠𝑒𝑛(−𝑛𝜋)] ;
𝑛 𝑛
nota: 𝑠𝑒𝑛(−𝜃) = −𝑠𝑒𝑛(𝜃)
1
< 𝑓1 , 𝑓𝑛 >= [2𝑠𝑒𝑛(𝑛𝜋)] como 𝑠𝑒𝑛(𝑛𝜋) = 0 ∴ < 𝑓1 , 𝑓𝑛 >= 0 son ortogonales.
𝑛
𝜋
b) para 𝑓1 (𝑥) = 1 y 𝑓𝑛 (𝑥) = sen(𝑛𝑥) entonces < 𝑓1 , 𝑓𝑛 >= ∫−𝜋(1)(sen (𝑛𝑥))𝑑𝑥;
−1 −1
< 𝑓1 , 𝑓𝑛 >= [𝑐𝑜𝑠(𝑛𝑥)]𝜋−𝜋 ; < 𝑓1 , 𝑓𝑛 >= [𝑐𝑜𝑠(𝑛𝜋) − 𝑐𝑜𝑠(−𝑛𝜋)]
𝑛 𝑛
nota: cos(−𝜃) = cos (𝜃)
−1
< 𝑓1 , 𝑓𝑛 >= [0] ∴ < 𝑓1 , 𝑓𝑛 >= 0 son ortogonales.
𝑛
c) Para 𝑚 ≠ 𝑛 𝑓𝑚 (𝑥) = cos (𝑚𝑥) y 𝑓𝑛 (𝑥) = cos (𝑛𝑥) entonces
𝜋
< 𝑓𝑚 , 𝑓𝑛 >= ∫−𝜋(cos (𝑚𝑥))(cos (𝑛𝑥))𝑑𝑥; empleando identidades trigonométricas.
𝜋 1
< 𝑓𝑚 , 𝑓𝑛 >= ∫−𝜋 (cos ((𝑚 + 𝑛)𝑥)) + (cos ((𝑚 − 𝑛)𝑥))𝑑𝑥
2
1
Nota: cos(𝐴𝑥) cos(𝐵𝑥) = 2 [𝑐𝑜𝑠((𝐴 + 𝐵)𝑥) + cos ((𝐴 − 𝐵)𝑥)]
1 𝜋 1 𝜋
< 𝑓𝑚 , 𝑓𝑛 >= 2(𝑚+𝑛) ∫−𝜋(𝑚 + 𝑛)cos ((𝑚 + 𝑛)𝑥) 𝑑𝑥 +2(𝑚−𝑛) ∫−𝜋(𝑚 − 𝑛) cos((𝑚 − 𝑛)𝑥) 𝑑𝑥
1 1
< 𝑓𝑚 , 𝑓𝑛 >= 𝑠𝑒𝑛((𝑚 + 𝑛)𝑥) + 𝑠𝑒𝑛((𝑚 − 𝑛)𝑥) ∴ < 𝑓𝑚 , 𝑓𝑛 >= 0 son ortogonales
2(𝑚+𝑛) 2(𝑚−𝑛)
d) Para 𝑚 ≠ 𝑛 𝑓𝑚 (𝑥) = sen (𝑚𝑥) y 𝑓𝑛 (𝑥) = sen (𝑛𝑥) entonces
𝜋
< 𝑓𝑚 , 𝑓𝑛 >= ∫−𝜋(sen(𝑚𝑥))(sen (𝑛𝑥))𝑑𝑥; empleando identidades trigonométricas.
𝜋 1
< 𝑓𝑚 , 𝑓𝑛 >= ∫−𝜋 2 (cos ((𝑚 + 𝑛)𝑥)) + (cos ((𝑚 − 𝑛)𝑥))𝑑𝑥
1
Nota: sen(𝐴𝑥) sen(𝐵𝑥) = 2 [𝑐𝑜𝑠((𝐴 − 𝐵)𝑥) − cos ((𝐴 + 𝐵)𝑥)]
1 𝜋 1 𝜋
< 𝑓𝑚 , 𝑓𝑛 >= 2(𝑚−𝑛) ∫−𝜋(𝑚 − 𝑛)cos ((𝑚 − 𝑛)𝑥) 𝑑𝑥 − 2(𝑚+𝑛) ∫−𝜋(𝑚 + 𝑛) cos((𝑚 + 𝑛)𝑥) 𝑑𝑥
, 1 1
< 𝑓𝑚 , 𝑓𝑛 >= 2(𝑚−𝑛) 𝑐𝑜𝑠((𝑚 − 𝑛)𝑥) − 2(𝑚+𝑛) 𝑐𝑜𝑠((𝑚 + 𝑛)𝑥) ∴ < 𝑓𝑚 , 𝑓𝑛 >= 0 son ortogonales
e) Para 𝑚 ≠ 𝑛 𝑓𝑚 (𝑥) = sen (𝑚𝑥) y 𝑓𝑛 (𝑥) = cos (𝑛𝑥) entonces
𝜋
< 𝑓𝑚 , 𝑓𝑛 >= ∫−𝜋(sen(𝑚𝑥))(cos (𝑛𝑥))𝑑𝑥; empleando identidades trigonométricas.
𝜋 1
< 𝑓𝑚 , 𝑓𝑛 >= ∫−𝜋 (cos ((𝑚 + 𝑛)𝑥)) + (cos ((𝑚 − 𝑛)𝑥))𝑑𝑥
2
1
Nota: sen(𝐴𝑥) cos(𝐵𝑥) = 2 [𝑠𝑒𝑛((𝐴 − 𝐵)𝑥) − sen ((𝐴 + 𝐵)𝑥)]
1 𝜋 1 𝜋
< 𝑓𝑚 , 𝑓𝑛 >= 2(𝑚−𝑛) ∫−𝜋(𝑚 − 𝑛)sen ((𝑚 − 𝑛)𝑥) 𝑑𝑥 − 2(𝑚+𝑛) ∫−𝜋(𝑚 + 𝑛) sen((𝑚 + 𝑛)𝑥) 𝑑𝑥
−1 𝜋 1
< 𝑓𝑚 , 𝑓𝑛 >= 2(𝑚−𝑛) [𝑐𝑜𝑠((𝑚 − 𝑛)𝑥)]−𝜋 + 2(𝑚+𝑛) [𝑐𝑜𝑠((𝑚 + 𝑛)𝑥)]𝜋−𝜋
−1 1
< 𝑓𝑚 , 𝑓𝑛 >= 2(𝑚−𝑛) (𝑐𝑜𝑠((𝑚 − 𝑛)𝜋 − cos (𝑚 − 𝑛)(−𝜋)) + 2(𝑚+𝑛) (𝑐𝑜𝑠((𝑚 + 𝑛)𝜋 − cos(𝑚 + 𝑛) (−𝜋))
Nota: cos(−𝜃) = cos (𝜃) ∴ < 𝑓𝑚 , 𝑓𝑛 >= 0 son
f) Para 𝑚 = 𝑛 𝑓𝑛 (𝑥) = cos (𝑛𝑥) y 𝑓𝑛 (𝑥) = cos (𝑛𝑥) entonces :
𝜋 1 𝜋
< 𝑓𝑛 , 𝑓𝑛 ≥ ∫−𝜋 𝑐𝑜𝑠 2 (𝑛𝑥)𝑑𝑥; < 𝑓𝑛 , 𝑓𝑛 >= ∫−𝜋(1 + cos(2𝑛𝑥))𝑑𝑥
2
1 𝜋 1 𝜋 1 1
< 𝑓𝑛 , 𝑓𝑛 >= 2 ∫−𝜋 𝑑𝑥 + 4𝑛 ∫−𝜋 2ncos (2𝑛𝑥)𝑑𝑥 ; < 𝑓𝑛 , 𝑓𝑛 >= 2 [𝑥]𝜋−𝜋 + 4𝑛 [𝑠𝑒𝑛(2𝑛𝑥)]𝜋−𝜋
1 1
< 𝑓𝑛 , 𝑓𝑛 >= 2 (𝜋 − (−𝜋)) + 4𝑛 (𝑠𝑒𝑛(2𝑛𝜋) − 𝑠𝑒𝑛(2𝑛(−𝜋))
1
< 𝑓𝑛 , 𝑓𝑛 >= 𝜋 + 𝑠𝑒𝑛(2𝑛𝜋) = 𝜋.
2𝑛
g) Para 𝑚 = 𝑛 𝑓𝑛 (𝑥) = sen (𝑛𝑥) y 𝑓𝑛 (𝑥) = sen (𝑛𝑥) entonces :
𝜋 1 𝜋
< 𝑓𝑛 , 𝑓𝑛 ≥ ∫−𝜋 𝑠𝑒𝑛2 (𝑛𝑥)𝑑𝑥; < 𝑓𝑛 , 𝑓𝑛 >= 2 ∫−𝜋(1 − cos(2𝑛𝑥))𝑑𝑥
1 𝜋 −1 𝜋 −1 1
< 𝑓𝑛 , 𝑓𝑛 >= ∫−𝜋 𝑑𝑥 + ∫−𝜋 −2n ∙ sen (2𝑛𝑥)𝑑𝑥 ; < 𝑓𝑛 , 𝑓𝑛 >= [𝑥]𝜋−𝜋 + [𝑐𝑜𝑠(2𝑛𝑥)]𝜋−𝜋
2 4𝑛 2 4𝑛
1 1
< 𝑓𝑛 , 𝑓𝑛 >= (𝜋 − (−𝜋)) + (𝑐𝑜𝑠(2𝑛𝜋) − 𝑐𝑜𝑠(2𝑛(−𝜋))
2 4𝑛
Nota: cos(−𝜃) = cos (𝜃)
1
< 𝑓𝑛 , 𝑓𝑛 >= 𝜋 + 4𝑛 (0) = 𝜋.
Si consideramos la sucesión de funciones {𝑓0 (𝑥), 𝑓1 (𝑥), 𝑓2 (𝑥), … , 𝑓𝑛 (𝑥)} es un
conjunto ortogonal, tal que entonces se puede desarrollar formalmente la función 𝑓,
como como una serie ortogonal: 𝑓(𝑥) = 𝐶0 𝑓0 + 𝐶1 𝑓1 + 𝐶2 𝑓2 + ⋯ + 𝐶𝑛 𝑓𝑛 , donde
𝐶0 , 𝐶1 , 𝐶2 , … + 𝐶𝑛 se encuentran utilizando el concepto del producto interno.
Fourier propone inicialmente la sucesión de funciones siguiente:
𝜋𝑥 2𝜋𝑥 3𝜋𝑥 𝑛𝜋𝑥 𝜋𝑥 2𝜋𝑥 3𝜋𝑥 𝑛𝜋𝑥
{1, cos ( 𝑝 ) , cos ( 𝑝
) , cos ( 𝑝 ) , … cos ( 𝑝 ) , … 𝑠𝑒𝑛 ( 𝑝 ) , 𝑠𝑒𝑛 ( 𝑝 ) , 𝑠𝑒𝑛 ( 𝑝 ) , … , 𝑠𝑒𝑛 ( 𝑝 )}
Para conformar una serie tal que
𝑎0 𝑛𝜋 𝑛𝜋
𝑓(𝑥) = 2
+ ∑∞ ∞
𝑛=1 𝑎𝑛 cos ( 𝑝 𝑥) + ∑𝑛=1 𝑏𝑛 𝑠𝑒𝑛( 𝑝 𝑥) de esta serie es necesario conocer los coeficientes
𝑎0 , 𝑎1 , 𝑎2 , … , 𝑎𝑛 y 𝑏1 , 𝑏2 , 𝑏3, … , 𝑏𝑛
El procedimiento para conocer la primera constante 𝑎0 es:
𝑎0 𝑛𝜋 𝑛𝜋
𝑓(𝑥) = 2
+ ∑∞
𝑛=1(𝑎𝑛 cos ( 𝑝 𝑥) + 𝑏𝑛 𝑠𝑒𝑛( 𝑝 𝑥))
𝑝 𝑝 𝑎0 𝑝 𝑛𝜋 𝑝 𝑛𝜋
∫−𝑝 𝑓(𝑥)𝑑𝑥 = ∫−𝑝 2
𝑑𝑥 + ∑∞
𝑛=1 ∫−𝑝 𝑎𝑛 cos ( 𝑝 𝑥)𝑑𝑥 + ∫−𝑝 𝑏𝑛 𝑠𝑒𝑛 ( 𝑝 𝑥) 𝑑𝑥
De acuerdo con los incisos a) y b) de las funciones ortogonales las integrales
𝑝 𝑛𝜋 𝑝 𝑛𝜋
∫−𝑝 𝑎𝑛 cos ( 𝑝 𝑥)𝑑𝑥 = 0 y ∫−𝑝 𝑏𝑛 𝑠𝑒𝑛 ( 𝑝 𝑥) 𝑑𝑥 = 0,
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