Hola Joaquín, antes que nada deberás recordar la forma que tiene la serie de Fourier:
\( f(t)= \displaystyle\frac{a_0}{2}\displaystyle\sum_{n=1}^\infty \left [ a_n cos(\displaystyle\frac{nπt}{L}) + b_n sen(\displaystyle\frac{nπt}{L}) \right ]
[quote]Donde [tex]L \) es el extremo del intervalo, \( a_o , a_n , b_n \) son los coeficientes de Fourier que toman los valores:[/quote]
\( a_0=\displaystyle\frac{1}{L}\displaystyle\int_{-L}^{L} f(t)\, dt , a_n=\displaystyle\frac{1}{L}\displaystyle\int_{-L}^{L} f(t)cos(\displaystyle\frac{nπt}{L})\,dt
, b_n=\displaystyle\frac{1}{L}\displaystyle\int_{-L}^{L} f(t)sen(\displaystyle\frac{nπt}{L})\,dt \)
Ahora tenemos que comenzar a calcular estos coeficientes.
Como tenemos una función por dos partes, debemos calcular dos integrales en cada formula y unirlas mediante una suma:
\( a_0=\displaystyle\frac{1}{L}\displaystyle\int_{-L}^{L} f(t)\, dt \) = \( \displaystyle\frac{1}{2}\displaystyle\int_{-2}^{0} -1\, dt \) + \( \displaystyle\frac{1}{2}\displaystyle\int_{0}^{2} 1\, dt \) = \( \displaystyle\frac{-1}{2} t |_{-2}^0 \) + \( \displaystyle\frac{1}{2} t |_0^2 \) =\( -1+1 \)=\( 0 \)
Luego \( a_0 = 0 \)
\( a_n=\displaystyle\frac{1}{2}\displaystyle\int_{-L}^{L} f(t)cos(\displaystyle\frac{nπt}{L})\,dt = \displaystyle\frac{1}{L}\displaystyle\int_{-2}^{0} {-1}cos(\displaystyle\frac{nπt}{2})\,dt + \displaystyle\frac{1}{2}\displaystyle\int_{0}^{2} 1cos(\displaystyle\frac{nπt}{2})\,dt= \)
Utilizamos lo siguiente: \( \displaystyle\int cos(at)dt= \displaystyle\frac{1}{a}sen (at) \)
\( =\displaystyle\frac{-1}{2}\displaystyle\frac{1}{\displaystyle\frac{nπ}{2}}sen(\displaystyle\frac{nπt}{2})|_{-2}^0 + \displaystyle\frac{1}{2}\displaystyle\frac{1}{\displaystyle\frac{nπ}{2}}sen(\displaystyle\frac{nπt}{2})|_{0}^{-2} = \displaystyle\frac{-1}{nπ}[sen(0) - sen({-nπ})] + \displaystyle\frac{1}{nπ}[sen(nπ) - sen({0})] \)
Como resulta que por trigonometría sen(nπ)=0 y sen(0)=0
Se concluye que \( a_n =0 \)
\( b_n=\displaystyle\frac{1}{L}\displaystyle\int_{-L}^{L} f(t)sen(\displaystyle\frac{nπt}{L})\,dt= \displaystyle\frac{1}{2}\displaystyle\int_{-2}^{0} -1sen(\displaystyle\frac{nπt}{2})\,dt + \displaystyle\frac{1}{2}\displaystyle\int_{0}^{2} 1sen(\displaystyle\frac{nπt}{2})\,dt= \)
Utilizamos lo siguiente: \( \displaystyle\int sen(at)dt= \displaystyle\frac{-1}{a}cos (at) \)
\( = \displaystyle\frac{-1}{2}(\displaystyle\frac{-1}{\displaystyle\frac{nπ}{2}})cos(\displaystyle\frac{nπt}{2})|_{-2}^0 + \displaystyle\frac{1}{2}(\displaystyle\frac{-1}{\displaystyle\frac{nπ}{2}})cos(\displaystyle\frac{nπt}{2})|_{0}^2= \displaystyle\frac{1}{nπ}[cos(0)-cos(-π)]-\displaystyle\frac{-1}{nπ}[cos(nπ)-cos(0)] \)
Por identidades trigonométricas \( cos(nπ)=cos(-nπ)=-1^n ; {cos(0)=1}, reemplazamos y aplicamos distributividad \)
\( =\displaystyle\frac{1}{nπ} - \displaystyle\frac{1}{nπ}(-1)^n - \displaystyle\frac{1}{nπ}(-1)^n + \displaystyle\frac{1}{nπ}= \displaystyle\frac{2}{nπ}[1-(-1)^n] \)
Hemos encontrado los valores de los coeficientes, ahora solo resta reemplazar para tener nuestra serie de Fourier
