Aproximación mediante una serie finita de Fourier. Primera parte. Fourier.

Sea

$\displaystyle S_k (t) = \frac{1}{2} a_0 + \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})}$

la suma de los primeros $(2k+1)$ términos de la serie de Fourier que representa $f(t)$ en el intervalo $\displaystyle -\frac{T}{2} < t < \frac{T}{2}$ .

Si $f(t)$ se aproxima por $S_k (t)$ , es decir,

$\displaystyle f(t) = \frac{1}{2} a_0 + \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} + \varepsilon_k(t)$

$\displaystyle f(t) = S_k (t) + \varepsilon_k(t)$

despejando $\varepsilon_k(t)$ ,

$\displaystyle \varepsilon_k(t) = f(t) - S_k (t)$

donde $\varepsilon_k (t)$ es la diferencia o el error entre $f(t)$ y su aproximación, entonces el error cuadrático medio $E_k$ está definido por

$\displaystyle E_k (t) = \frac{1}{T} \int_{-T/2}^{T/2}{[\varepsilon_k (t)]^2 \, dt} = \frac{1}{T} \int_{-T/2}^{T/2}{[f(t) - S_k (t)]^2 \, dt}$

Ahora se va a demostrar que si se aproxima una función $f(t)$ por una serie finita de Fourier $S_k (t)$ , entonces esta aproximación tiene la propiedad de ser el mínimo error cuadrático medio. Para ello, se toma la ecuación del error cuadrático medio

$\displaystyle E_k (t) = \frac{1}{T} \int_{-T/2}^{T/2}{[f(t) - S_k (t)]^2 \, dt}$

Y sustituyendo $\displaystyle S_k (t) = \frac{1}{2} a_0 + \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})}$ en esta ecuación, se tiene los siguiente

$\displaystyle E_k (t) = \frac{1}{T} \int_{-T/2}^{T/2}{\left\{f(t) - \left[\frac{1}{2} a_0 + \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})}\right] \right\}^2 \, dt}$

$\displaystyle E_k (t) = \frac{1}{T} \int_{-T/2}^{T/2}{\left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right]^2 \, dt}$

Y $E_k$ se transformó en una función de $a_0$ , $a_n$ y $b_n$ .

Para que el error cuadrático medio $E_k$ sea mínimo, sólo basta con que sus derivadas parciales (con respecto a $a_0$ , $a_n$ y $b_n$ ) deban ser iguales a cero. Es decir,

$\displaystyle \frac{\partial E_k (t)}{\partial a_0} = 0$ , $\displaystyle \frac{\partial E_k (t)}{\partial a_n} = 0$ , $\displaystyle \frac{\partial E_k (t)}{\partial b_n} = 0$

Para $n=1,2,3,\cdots$

Derivando $E_k$ parcialmente con respecto a $a_0$ , resulta

$\displaystyle E_k (t) = \frac{1}{T} \int_{-T/2}^{T/2}{\left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right]^2 \, dt}$

$\displaystyle \frac{\partial E_k (t)}{\partial a_0} = \frac{\partial}{\partial a_0} \left\{ \frac{1}{T} \int_{-T/2}^{T/2}{\left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right]^2 \, dt} \right\}$

$\displaystyle \frac{\partial E_k (t)}{\partial a_0} = \frac{1}{T} \int_{-T/2}^{T/2}{\frac{\partial}{\partial a_0} \left\{ \left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right]^2 \right\} \, dt}$

$\displaystyle \frac{\partial E_k (t)}{\partial a_0} = \frac{1}{T} \int_{-T/2}^{T/2}{\left\{ 2 \left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right] \right\} \cdot \left(- \frac{1}{2} \right)\, dt}$

$\displaystyle \frac{\partial E_k (t)}{\partial a_0} = - \frac{1}{T} \int_{-T/2}^{T/2}{\left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right] \, dt}$

$\displaystyle \frac{\partial E_k (t)}{\partial a_0} = - \frac{1}{T} \int_{-T/2}^{T/2}{f(t) \, dt} + \frac{1}{T} \int_{-T/2}^{T/2}{\frac{1}{2} a_0 \, dt} + \frac{1}{T} \int_{-T/2}^{T/2}{\sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \, dt}$

$\displaystyle \frac{\partial E_k (t)}{\partial a_0} = - \frac{1}{T} \int_{-T/2}^{T/2}{f(t) \, dt} + \frac{1}{T} \cdot \frac{1}{2} a_0 \int_{-T/2}^{T/2}{dt} + \frac{1}{T} \sum_{n=1}^{k}{\int_{-T/2}^{T/2}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t}) \, dt}}$

$\displaystyle \frac{\partial E_k (t)}{\partial a_0} = - \frac{1}{T} \int_{-T/2}^{T/2}{f(t) \, dt} + \frac{1}{2T} a_0 \int_{-T/2}^{T/2}{dt} + \frac{1}{T} \sum_{n=1}^{k}{\left\{a_n \int_{-T/2}^{T/2}{\cos{n \omega_0 t} \, dt} + b_n \int_{-T/2}^{T/2}{\sin{n \omega_0 t} \, dt} \right\} }$

En la tercera integral, si $n \ne 0$ , por propiedades de ortogonalidad

$\displaystyle = - \frac{1}{T} \int_{-T/2}^{T/2}{f(t) \, dt} + \frac{1}{2T} a_0 \int_{-T/2}^{T/2}{dt} + \frac{1}{T} \sum_{n=1}^{k}{\left(a_n \cdot 0 + b_n \cdot 0 \right) }$

$\displaystyle = - \frac{1}{T} \int_{-T/2}^{T/2}{f(t) \, dt} + \frac{1}{2T} a_0 \int_{-T/2}^{T/2}{dt} = - \frac{1}{T} \int_{-T/2}^{T/2}{f(t) \, dt} + \frac{1}{2T} a_0 \left[ t \right]_{-T/2}^{T/2}{dt}$

$\displaystyle = - \frac{1}{T} \int_{-T/2}^{T/2}{f(t) \, dt} + \frac{1}{2T} a_0 \left[ \frac{T}{2} - \left( -\frac{T}{2} \right) \right] = - \frac{1}{T} \int_{-T/2}^{T/2}{f(t) \, dt} + \frac{1}{2T} a_0 \left( \frac{T}{2} + \frac{T}{2} \right)$

$\displaystyle = - \frac{1}{T} \int_{-T/2}^{T/2}{f(t) \, dt} + \frac{1}{2T} a_0 \cdot (T)$

$\displaystyle \frac{\partial E_k (t)}{\partial a_0} = - \frac{1}{T} \int_{-T/2}^{T/2}{f(t) \, dt} + \frac{1}{2} a_0 = 0$

Lo cual, esto queda demostrado la primera parte.

Derivando $E_k$ parcialmente con respecto a $a_n$ , resulta

$\displaystyle E_k (t) = \frac{1}{T} \int_{-T/2}^{T/2}{\left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right]^2 \, dt}$

$\displaystyle \frac{\partial E_k (t)}{\partial a_n} = \frac{\partial}{\partial a_n} \left\{ \frac{1}{T} \int_{-T/2}^{T/2}{\left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right]^2 \, dt} \right\}$

$\displaystyle = \frac{1}{T} \int_{-T/2}^{T/2}{\frac{\partial}{\partial a_n} \left\{ \left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right]^2 \right\} \, dt}$

$\displaystyle = \frac{1}{T} \int_{-T/2}^{T/2}{\left\{ 2 \left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right] \right\} \cdot \frac{\partial}{\partial a_n} \left[- (a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})\right] \, dt}$

$\displaystyle = \frac{1}{T} \int_{-T/2}^{T/2}{\left\{ 2 \left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right] \right\} \cdot \left[- \left( \frac{\partial}{\partial a_n} (a_n \cos{n \omega_0 t}) + \frac{\partial}{\partial a_n}(b_n \sin{n \omega_0 t}) \right) \right] \, dt}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{\left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right] \cdot \left( \cos{n \omega_0 t} + 0 \right) \, dt}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{\left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right] \cdot \cos{n \omega_0 t} \, dt}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{\left[f(t) \cos{n \omega_0 t} - \frac{1}{2} a_0 \cos{n \omega_0 t} - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} \cos{n \omega_0 t} + b_n \sin{n \omega_0 t} \cos{n \omega_0 t})} \right] \, dt}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + \frac{2}{T} \int_{-T/2}^{T/2}{\frac{1}{2} a_0 \cos{n \omega_0 t} \, dt} + \frac{2}{T} \int_{-T/2}^{T/2}{\sum_{n=1}^{k}{(a_n \cos^2{n \omega_0 t} + b_n \sin{n \omega_0 t} \cos{n \omega_0 t})} \, dt}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + \frac{2}{T} \cdot \frac{1}{2} a_0 \int_{-T/2}^{T/2}{\cos{n \omega_0 t} \, dt} + \frac{2}{T} \sum_{n=1}^{k}{\int_{-T/2}^{T/2}{(a_n \cos^2{n \omega_0 t} + b_n \sin{n \omega_0 t} \cos{n \omega_0 t}) \, dt}}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + \frac{1}{T} a_0 \int_{-T/2}^{T/2}{\cos{n \omega_0 t} \, dt} + \frac{2}{T} \sum_{n=1}^{k}{\left[\int_{-T/2}^{T/2}{a_n \cos^2{n \omega_0 t} \, dt} + \int_{-T/2}^{T/2}{b_n \sin{n \omega_0 t} \cos{n \omega_0 t} \, dt} \right]}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + \frac{1}{T} a_0 \int_{-T/2}^{T/2}{\cos{n \omega_0 t} \, dt} + \frac{2}{T} \sum_{n=1}^{k}{\left[a_n \int_{-T/2}^{T/2}{\cos^2{n \omega_0 t} \, dt} + b_n \int_{-T/2}^{T/2}{\sin{n \omega_0 t} \cos{n \omega_0 t} \, dt} \right]}$

Si $m \ne 0$ , $n \ne 0$ y $m \ne n$ , por propiedades de ortogonalidad

$\displaystyle \frac{\partial E_k (t)}{\partial a_n} = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + \frac{1}{T} a_0 (0) + \frac{2}{T} \sum_{n=1}^{k}{\left[a_n \int_{-T/2}^{T/2}{\cos^2{n \omega_0 t} \, dt} + b_n (0) \right]}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + \frac{2}{T} \sum_{n=1}^{k}{\left[a_n \int_{-T/2}^{T/2}{\cos^2{n \omega_0 t} \, dt} \right]}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + \frac{2}{T} \sum_{n=1}^{k}{\left[a_n \int_{-T/2}^{T/2}{\left(\frac{1}{2} + \frac{1}{2} \cos{2n \omega_0 t} \right) \, dt} \right]}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + \frac{2}{T} \sum_{n=1}^{k}{\left\{a_n \left[\frac{1}{2}t + \frac{1}{2n\omega_0} \sin{n \omega_0 t} +C \right]_{-T/2}^{T/2} \right\}}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + \frac{2}{T} \sum_{n=1}^{k}{\left\{a_n \left[\frac{1}{2}(\frac{T}{2}) + \frac{1}{2n\omega_0} \sin{n \omega_0 (\frac{T}{2})} - \frac{1}{2}(-\frac{T}{2}) - \frac{1}{2n\omega_0} \sin{n \omega_0 (-\frac{T}{2})} \right] \right\}}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + \frac{2}{T} \sum_{n=1}^{k}{\left\{a_n \left[\frac{T}{4} + \frac{1}{2n (\frac{2\pi}{T})} \sin{n (\frac{2\pi}{T}) (\frac{T}{2})} + \frac{T}{4} - \frac{1}{2n (\frac{2\pi}{T})} \sin{n \frac{2\pi}{T} (-\frac{T}{2})} \right] \right\}}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + \frac{2}{T} \sum_{n=1}^{k}{\left\{a_n \left[\frac{T}{4} + \frac{T}{4n\pi} \sin{n \pi} + \frac{T}{4} + \frac{T}{4n \pi} \sin{n \pi} \right] \right\}}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + \frac{2}{T} \sum_{n=1}^{k}{a_n \left(\frac{T}{2} + \frac{T}{2n\pi} \sin{n \pi} \right)}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + \sum_{n=1}^{k}{a_n \left(1 + \frac{1}{n\pi} \sin{n \pi} \right)}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + \sum_{n=1}^{k}{a_n (1)} + \sum_{n=1}^{k}{a_n \cdot \frac{1}{n\pi} \sin{n \pi}}$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + \sum_{n=1}^{k}{a_n} + 0$

$\displaystyle = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + \sum_{n=1}^{k}{a_n}$

$\displaystyle \frac{\partial E_k}{\partial a_n} = - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} + a_n = 0$

Lo cual, esto queda demostrado la segunda parte.

Derivando $E_k$ parcialmente con respecto a $b_n$ , resulta

$\displaystyle E_k (t) = \frac{1}{T} \int_{-T/2}^{T/2}{\left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right]^2 \, dt}$

$\displaystyle \frac{\partial E_k (t)}{\partial b_n} = \frac{\partial}{\partial b_n} \left\{ \frac{1}{T} \int_{-T/2}^{T/2}{\left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right]^2 \, dt} \right\}$

$\displaystyle = \frac{1}{T} \int_{-T/2}^{T/2}{\frac{\partial}{\partial b_n} \left\{ \left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right]^2 \right\} \, dt}$

$\displaystyle = \frac{1}{T} \int_{-T/2}^{T/2}{2 \cdot \left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right] \cdot \frac{\partial}{\partial b_n} \left[ (a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t}) \right] \, dt}$

$\displaystyle = \frac{1}{T} \int_{-T/2}^{T/2}{2 \cdot \left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right] \cdot \left[ \frac{\partial}{\partial b_n} (a_n \cos{n \omega_0 t}) + \frac{\partial}{\partial b_n} (b_n \sin{n \omega_0 t}) \right] \, dt}$

$\displaystyle = \frac{1}{T} \int_{-T/2}^{T/2}{2 \cdot \left[f(t) - \frac{1}{2} a_0 - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right] \cdot \sin{n \omega_0 t} \, dt}$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{\left[f(t) \sin{n \omega_0 t} - \frac{1}{2} a_0 \sin{n \omega_0 t} - \sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} \sin{n \omega_0 t} + b_n \sin{n \omega_0 t} \sin{n \omega_0 t}} \right] \, dt}$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - \frac{2}{T} \int_{-T/2}^{T/2}{\frac{1}{2} a_0 \sin{n \omega_0 t} \, dt} - \frac{2}{T} \int_{-T/2}^{T/2}{\sum_{n=1}^{k}{(a_n \cos{n \omega_0 t} \sin{n \omega_0 t} + b_n \sin^2{n \omega_0 t}}) \, dt}$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - \frac{2}{T} \cdot \frac{1}{2} a_0 \int_{-T/2}^{T/2}{\sin{n \omega_0 t} \, dt} - \frac{2}{T} \sum_{n=1}^{k}{\int_{-T/2}^{T/2}{(a_n \cos{n \omega_0 t} \sin{n \omega_0 t} + b_n \sin{n \omega_0 t} \sin{n \omega_0 t}) \, dt}}$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - \frac{1}{T} a_0 \int_{-T/2}^{T/2}{\sin{n \omega_0 t} \, dt} - \frac{2}{T} \sum_{n=1}^{k}{\left[\int_{-T/2}^{T/2}{a_n \cos{n \omega_0 t} \sin{n \omega_0 t} \, dt} + \int_{-T/2}^{T/2}{b_n \sin^2{n \omega_0 t} \, dt}\right]}$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - \frac{1}{T} a_0 \int_{-T/2}^{T/2}{\sin{n \omega_0 t} \, dt} - \frac{2}{T} \sum_{n=1}^{k}{\left[a_n \int_{-T/2}^{T/2}{\cos{n \omega_0 t} \sin{n \omega_0 t} \, dt} + b_n \int_{-T/2}^{T/2}{\sin^2{n \omega_0 t} \, dt}\right]}$

Si $m \ne 0$ , $n \ne 0$ y $m \ne 0$ , por propiedades de ortogonalidad, resulta lo siguiente

$\displaystyle \frac{\partial E_k}{\partial b_n} = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - \frac{1}{T} a_0 (0) - \frac{2}{T} \sum_{n=1}^{k}{\left[a_n (0) + b_n \int_{-T/2}^{T/2}{\sin^2{n \omega_0 t} \, dt} \right]}$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - \frac{2}{T} \sum_{n=1}^{k}{\left[b_n \int_{-T/2}^{T/2}{\sin^2{n \omega_0 t} \, dt} \right]}$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - \frac{2}{T} \sum_{n=1}^{k}{\left[b_n \int_{-T/2}^{T/2}{ \left(\frac{1}{2} - \frac{1}{2} \cos{2n \omega_0 t} \right) \, dt} \right]}$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - \frac{2}{T} \sum_{n=1}^{k}{b_n \left[\frac{1}{2} t - \frac{1}{4n\omega_0} \sin{2n \omega_0 t} \right]_{-T/2}^{T/2}}$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - \frac{2}{T} \sum_{n=1}^{k}{b_n \left\{\left[\frac{1}{2} \left(\frac{T}{2}\right) - \frac{1}{4n\omega_0} \sin{2n \omega_0 \left(\frac{T}{2}\right)} \right] - \left[ \frac{1}{2} \left(\frac{T}{2}\right) - \frac{1}{4n\omega_0} \sin{2n \omega_0 \left(\frac{T}{2}\right)} \right] \right\}}$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - \frac{2}{T} \sum_{n=1}^{k}{b_n \left\{\left[\frac{1}{2} \left(\frac{T}{2}\right) - \frac{1}{4n \left(\frac{2\pi}{T}\right)} \sin{2n \left(\frac{2\pi}{T}\right) \left(\frac{T}{2}\right)} \right] - \left[ \frac{1}{2} \left(\frac{T}{2}\right) - \frac{1}{4n \left(\frac{2\pi}{T}\right)} \sin{2n \left(\frac{2\pi}{T}\right) \left(\frac{T}{2}\right)} \right] \right\}}$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - \frac{2}{T} \sum_{n=1}^{k}{b_n \left\{\left[\frac{T}{4} - \frac{T}{8n\pi} \sin{2n\pi} \right] - \left[-\frac{T}{4} - \frac{T}{8n\pi} \sin{2n\pi} \right] \right\}}$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - \frac{2}{T} \sum_{n=1}^{k}{b_n \left(\frac{T}{4} - \frac{T}{8n\pi} \sin{2n\pi} + \frac{T}{4} + \frac{T}{8n\pi} \sin{2n\pi} \right)}$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - \frac{2}{T} \sum_{n=1}^{k}{b_n \left(\frac{T}{2} - \frac{T}{4n\pi} \sin{2n\pi} \right)}$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - \sum_{n=1}^{k}{b_n \left(1 - \frac{1}{2n\pi} \sin{2n\pi} \right)}$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - \sum_{n=1}^{k}{\left(b_n - \frac{1}{2n\pi} b_n \sin{2n\pi} \right)}$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - \sum_{n=1}^{k}{b_n} - \sum_{n=1}^{k}{\frac{1}{2n\pi} b_n \sin{2n\pi} }$

$\displaystyle = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - b_m - 0 = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - b_n$

$\displaystyle \frac{\partial E_k}{\partial b_n} = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} - b_n = 0$

Lo cual, esto queda demostrado la última parte.

Se concluye que el error cuadrático medio $E_k$ es mínimo.

Aproximación mediante una serie finita de Fourier. Primera parte. Fourier.

Publicado por Cesar Reyes

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Aproximación mediante una serie finita de Fourier. Primera parte. Fourier.

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Publicado por Cesar Reyes

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