Aproximación mediante la serie finita de Fourier. Segunda parte. Fourier.

Reducción del error cuadrático medio (otra manera de expresarlo)

Ahora se va a demostrar que el error cuadrático medio $E_k$ en una aproximación a $f(t)$ por $S_k (t)$ , definida por la ecuación

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

se puede reducir a

$\displaystyle E_k (t) = \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} - \frac{1}{4} a_0^2 - \frac{1}{2} \sum_{n=1}^{k}{(a_n^2 + b_n^2)}$

Partiendo de la primera ecuación del error cuadrático medio, se desarrolla el binomio al cuadrado

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

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

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

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

La primera integral no se altera. En la segunda, tendrá el siguiente resultado

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

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

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

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

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

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

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

$\displaystyle = \frac{1}{2} a_0 \cdot a_0 + \sum_{n=1}^{k}{\left(a_n \cdot a_n + b_n \cdot b_n \right)} = \frac{1}{2} a_0^2 + \sum_{n=1}^{k}{\left(a_n^2 + b_n^2 \right)}$

Para la segunda integral el resultado es

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

En la tercera integral, resulta lo siguiente

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

$\displaystyle = \frac{1}{T} \int_{-T/2}^{T/2}{\left\{\frac{1}{4} a_0^2 + 2 \cdot \frac{1}{2} a_0 \sum_{n=1}^k{(a_n \cos{n\omega_0 t} + b_n \sin{n \omega_0 t})} + \left[ \sum_{n=1}^k{(a_n \cos{n\omega_0 t} + b_n \sin{n \omega_0 t})} \right] \right\}^2 \, dt}$

$\displaystyle = \frac{1}{T} \int_{-T/2}^{T/2}{\left\{\frac{1}{4} a_0^2 + a_0 \sum_{n=1}^k{(a_n \cos{n\omega_0 t} + b_n \sin{n \omega_0 t})} + \left[ \sum_{n=1}^k{(a_n^2 \cos^2{n\omega_0 t} + 2a_n b_n \sin{n \omega_0 t} \cos{n \omega_0 t} + b_n^2 \sin^2{n \omega_0 t})} \right] \right\} \, dt}$

$\displaystyle = \frac{1}{T} \int_{-T/2}^{T/2}{\frac{1}{4} a_0^2 \, dt} + a_0 \cdot \frac{1}{T} \int_{-T/2}^{T/2}{\sum_{n=1}^k{(a_n \cos{n\omega_0 t} + b_n \sin{n \omega_0 t})} \, dt} + \frac{1}{T} \int_{-T/2}^{T/2}{\left[ \sum_{n=1}^k{(a_n^2 \cos^2{n\omega_0 t} + 2a_n b_n \sin{n \omega_0 t} \cos{n \omega_0 t} + b_n^2 \sin^2{n \omega_0 t})} \right] \, dt}$

$\displaystyle = \frac{1}{T} \cdot \frac{1}{4} a_0^2 \int_{-T/2}^{T/2}{dt} + a_0 \cdot \frac{1}{T} \sum_{n=1}^k{\int_{-T/2}^{T/2}{(a_n \cos{n\omega_0 t} + b_n \sin{n \omega_0 t}) \, dt}} + \frac{1}{T} \sum_{n=1}^k{\int_{-T/2}^{T/2}{(a_n^2 \cos^2{n\omega_0 t} + 2a_n b_n \sin{n \omega_0 t} \cos{n \omega_0 t} + b_n^2 \sin^2{n \omega_0 t}) \, dt}}$

$\displaystyle = \frac{1}{4T} a_0^2 \int_{-T/2}^{T/2}{dt} + a_0 \cdot \frac{1}{T} \sum_{n=1}^k{\left[\int_{-T/2}^{T/2}{a_n \cos{n\omega_0 t} \, dt} + \int_{-T/2}^{T/2}{b_n \sin{n \omega_0 t} \, dt} \right]} + \frac{1}{T} \sum_{n=1}^k{\left[\int_{-T/2}^{T/2}{a_n^2 \cos^2{n\omega_0 t} \, dt} + \int_{-T/2}^{T/2}{2 a_n b_n \sin{n \omega_0 t} \cos{n \omega_0 t} \, dt} + \int_{-T/2}^{T/2}{b_n^2 \sin^2{n \omega_0 t} \, dt} \right]}$

$\displaystyle = \frac{1}{4T} a_0^2 \int_{-T/2}^{T/2}{dt} + a_0 \cdot \frac{1}{T} \sum_{n=1}^k{\left[a_n \int_{-T/2}^{T/2}{\cos{n\omega_0 t} \, dt} + b_n \int_{-T/2}^{T/2}{\sin{n \omega_0 t} \, dt} \right]} + \frac{1}{T} \sum_{n=1}^k{\left[a_n^2 \int_{-T/2}^{T/2}{\cos^2{n\omega_0 t} \, dt} + 2 a_n b_n \int_{-T/2}^{T/2}{\sin{n \omega_0 t} \cos{n \omega_0 t} \, dt} + b_n^2 \int_{-T/2}^{T/2}{\sin^2{n \omega_0 t} \, dt} \right]}$

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

$\displaystyle \frac{1}{T} \int_{-T/2}^{T/2}{[S_k(t)]^2 \, dt} = \frac{1}{4T} a_0^2 \int_{-T/2}^{T/2}{dt} + a_0 \cdot \frac{1}{T} \sum_{n=1}^k{\left[a_n \cdot (0) + b_n \cdot (0) \right]} + \frac{1}{T} \sum_{n=1}^k{\left[a_n^2 \int_{-T/2}^{T/2}{\left(\frac{1}{2} + \frac{1}{2} \cos{n\omega_0 t} \right) \, dt} + 2 a_n b_n (0) + b_n^2 \int_{-T/2}^{T/2}{\left(\frac{1}{2} - \frac{1}{2} cos{n \omega_0 t} \right) \, dt} \right]}$

$\displaystyle = \frac{1}{4T} a_0^2 \left[t + C \right]_{-T/2}^{T/2} + a_0 \cdot \frac{1}{T} (0) + \frac{1}{T} \sum_{n=1}^k{\left\{a_n^2 \left[\frac{1}{2} t + \frac{1}{2n\omega_0} \sin{n\omega_0 t} +C \right]_{-T/2}^{T/2} + b_n^2 \left[\frac{1}{2} t - \frac{1}{2n\omega_0} \sin{n \omega_0 t} +C \right]_{-T/2}^{T/2} \right\}}$

$\displaystyle = \frac{1}{4T} a_0^2 \left[\frac{T}{2} - (-\frac{T}{2})\right] + \frac{1}{T} \sum_{n=1}^k{\left\{a_n^2 \left\{ \left[ \frac{1}{2} \left(\frac{T}{2} \right) + \frac{1}{2n\omega_0} \sin{n\omega_0 \left(\frac{T}{2} \right)} \right] - \left[ \frac{1}{2} \left(-\frac{T}{2} \right) + \frac{1}{2n\omega_0} \sin{n\omega_0 \left(-\frac{T}{2} \right)} \right] \right\} + b_n^2 \left\{ \left[\frac{1}{2} \left(\frac{T}{2} \right) - \frac{1}{2n\omega_0} \sin{n \omega_0 \left(\frac{T}{2} \right)} \right] - \left[\frac{1}{2} \left(-\frac{T}{2} \right) - \frac{1}{2n\omega_0} \sin{n \omega_0 \left(-\frac{T}{2} \right)} \right] \right\} \right\}}$

$\displaystyle = \frac{1}{4T} a_0^2 \left(\frac{T}{2} + \frac{T}{2}\right) + \frac{1}{T} \sum_{n=1}^k{\left\{a_n^2 \left\{ \left[ \frac{T}{4} + \frac{1}{2n\omega_0} \sin{n\omega_0 \left(\frac{T}{2} \right)} \right] - \left[- \frac{T}{4} + \frac{1}{2n\omega_0} \sin{n\omega_0 \left(-\frac{T}{2} \right)} \right] \right\} + b_n^2 \left\{ \left[\frac{T}{4} - \frac{1}{2n\omega_0} \sin{n \omega_0 \left(\frac{T}{2} \right)} \right] - \left[-\frac{T}{4} - \frac{1}{2n\omega_0} \sin{n \omega_0 \left(-\frac{T}{2} \right)} \right] \right\} \right\}}$

$\displaystyle = \frac{1}{4T} a_0^2 \left(T \right) + \frac{1}{T} \sum_{n=1}^k{\left\{a_n^2 \left[\frac{T}{4} + \frac{1}{2n\omega_0} \sin{n\omega_0 \left(\frac{T}{2} \right)} + \frac{T}{4} - \frac{1}{2n\omega_0} \sin{n\omega_0 \left(-\frac{T}{2} \right)} \right] + b_n^2 \left[\frac{T}{4} - \frac{1}{2n\omega_0} \sin{n \omega_0 \left(\frac{T}{2} \right)} + \frac{T}{4} + \frac{1}{2n\omega_0} \sin{n \omega_0 \left(-\frac{T}{2} \right)} \right] \right\}}$

$\displaystyle = \frac{1}{4} a_0^2 + \frac{1}{T} \sum_{n=1}^k{\left\{a_n^2 \left[\frac{T}{4} + \frac{1}{2n\omega_0} \sin{n\omega_0 \left(\frac{T}{2} \right)} + \frac{T}{4} + \frac{1}{2n\omega_0} \sin{n\omega_0 \left(\frac{T}{2} \right)} \right] + b_n^2 \left[\frac{T}{4} - \frac{1}{2n\omega_0} \sin{n \omega_0 \left(\frac{T}{2} \right)} + \frac{T}{4} - \frac{1}{2n\omega_0} \sin{n \omega_0 \left(\frac{T}{2} \right)} \right] \right\}}$

$\displaystyle = \frac{1}{4} a_0^2 + \frac{1}{T} \sum_{n=1}^k{\left\{a_n^2 \left[\frac{T}{2} + \frac{1}{n\omega_0} \sin{n\omega_0 \left(\frac{T}{2} \right)} \right] + b_n^2 \left[\frac{T}{2} - \frac{1}{n\omega_0} \sin{n \omega_0 \left(\frac{T}{2} \right)} \right] \right\}}$

$\displaystyle = \frac{1}{4} a_0^2 + \frac{1}{T} \sum_{n=1}^k{\left[a_n^2 \left( \frac{T}{2} \right) + b_n^2 \left(\frac{T}{2} \right) \right]} = \frac{1}{4} a_0^2 + \frac{1}{T} \cdot \frac{T}{2} \sum_{n=1}^k{\left(a_n^2 + b_n^2 \right)}$

$\displaystyle = \frac{1}{4} a_0^2 + \frac{1}{2} \sum_{n=1}^k{\left(a_n^2 + b_n^2 \right)}$

Para la tercera integral, el resultado es

$\displaystyle \frac{1}{T} \int_{-T/2}^{T/2}{[S_k(t)]^2 \, dt} = \frac{1}{4} a_0^2 + \frac{1}{2} \sum_{n=1}^k{\left(a_n^2 + b_n^2 \right)}$

Regresando y sustituyendo

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

$\displaystyle = \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} - \left[\frac{1}{2} a_0^2 + \sum_{n=1}^{k}{\left(a_n^2 + b_n^2 \right)} \right]+ \left[ \frac{1}{4} a_0^2 + \frac{1}{2} \sum_{n=1}^k{\left(a_n^2 + b_n^2 \right)} \right]$

$\displaystyle = \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} - \frac{1}{2} a_0^2 - \sum_{n=1}^{k}{\left(a_n^2 + b_n^2 \right)} + \frac{1}{4} a_0^2 + \frac{1}{2} \sum_{n=1}^k{\left(a_n^2 + b_n^2 \right)}$

$\displaystyle = \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} - \frac{1}{4} a_0^2 - \frac{1}{2} \sum_{n=1}^k{\left(a_n^2 + b_n^2 \right)}$

Finalmente, el error cuadrático medio se reduce a

$\displaystyle \therefore E_k (t) = \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} - \frac{1}{4} a_0^2 - \frac{1}{2} \sum_{n=1}^k{\left(a_n^2 + b_n^2 \right)}$

Comprobando la siguiente desigualdad

Se demostrará la siguiente desigualdad

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

Partiendo de la ecuación del error cuadrático medio

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

Se dice que

$\displaystyle E_k (t) \ge 0$

$\displaystyle \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)-S_k (t)]^2 \, dt} \ge 0$

$\displaystyle \frac{1}{T} \int_{-T/2}^{T/2}{ \left\{ [f(t)]^2 - 2 f(t) S_k (t) + [S_k (t)]^2 \right\} \, dt} \ge 0$

$\displaystyle \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) S_k (t) \, dt} + \frac{1}{T} \int_{-T/2}^{T/2}{[S_k (t)]^2 \, dt} \ge 0$

Despejando la primera integral

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

Recordando que la primera integral del segundo miembro es equivalente a

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

Y la segunda integral de ese mismo miembro es equivalente a

$\displaystyle \frac{1}{T} \int_{-T/2}^{T/2}{[S_k(t)]^2 \, dt} = \frac{1}{4} a_0^2 + \frac{1}{2} \sum_{n=1}^k{\left(a_n^2 + b_n^2 \right)}$

Entonces, sustituyendo en la desigualdad, resulta

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

$\displaystyle \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} \ge \left[ \frac{1}{2} a_0^2 + \sum_{n=1}^{k}{\left(a_n^2 + b_n^2 \right)} \right] - \left[ \frac{1}{4} a_0^2 + \frac{1}{2} \sum_{n=1}^k{\left(a_n^2 + b_n^2 \right)} \right]$

$\displaystyle \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} \ge \frac{1}{2} a_0^2 + \sum_{n=1}^{k}{\left(a_n^2 + b_n^2 \right)} - \frac{1}{4} a_0^2 - \frac{1}{2} \sum_{n=1}^k{\left(a_n^2 + b_n^2 \right)}$

$\displaystyle \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} \ge \frac{1}{4} a_0^2 + \frac{1}{2} \sum_{n=1}^k{\left(a_n^2 + b_n^2 \right)}$

Y con esto queda establecida la desigualdad.

El teorema de Parseval

El teorema de Parseval afirma que si $a_0$ , $a_n$ y $b_n$ para $n=1, 2, 3, \cdots$ son los coeficientes en la expansión de Fourier de una función periódica $f(t)$ con período $T$ , entonces

$\displaystyle \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} = \frac{1}{4} a_0^2 + \sum_{n=1}^{\infty}{(a_n^2 + b_n^2)}$

Para demostrar este teorema, se parte con la aproximación del error cuadrático medio $E_k$ de $f(t)$ por $S_k (t)$

$\displaystyle E_k (t) = \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} - \frac{1}{4} a_0^2 - \frac{1}{2} \sum_{n=1}^k{\left(a_n^2 + b_n^2 \right)}$ y $\displaystyle E_k(t) = \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)-S_k (t)]^2 \, dt} \ge 0$

$\displaystyle E_{k+1} = E_k - \frac{1}{2} (a_{k+1}^2 + b_{k+1}^2)$

Las desigualdades $\displaystyle E_k (t) \ge 0$ y $\displaystyle E_{k+1} = E_k - \frac{1}{2} (a_{k+1}^2 + b_{k+1}^2)$ presenta una sucesión en donde $|E_k|$ solo tiene términos no negativos y no es creciente, por lo que, la sucesión converge. Recordando la ecuación de $\varepsilon_k (t)$

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

Tomando el límite cuando $k \rightarrow \infty$

$\displaystyle \lim_{k \rightarrow \infty}{\varepsilon_k (t)} = \lim_{k \rightarrow \infty}{f(t)} - \lim_{k \rightarrow \infty}{S_k (t)}$

Ahora, recordando la desigualdad $\displaystyle E_k (t) \ge 0$ , se hará que $E_k (t) = 0$ . Entonces, tomando el límite en ambos miembros cuando $k \rightarrow \infty$ , se brinda lo siguiente

$E_k (t) = 0$

$\displaystyle \lim_{k \rightarrow \infty}{E_k (t)} = \lim_{k \rightarrow \infty}{0}$

$\displaystyle \lim_{k \rightarrow \infty}{E_k (t)} = 0$

$\displaystyle \lim_{k \rightarrow \infty}{\frac{1}{T} \int_{-T/2}^{T/2}{[f(t)-S_k (t)]^2 \, dt}} = 0$

$\displaystyle \lim_{k \rightarrow \infty}{\frac{1}{T} \int_{-T/2}^{T/2}{ \left\{ [f(t)]^2 - 2 f(t) S_k (t) + [S_k (t)]^2 \right\} \, dt}} = 0$

$\displaystyle \lim_{k \rightarrow \infty}{ \left\{ \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} - \frac{2}{T} \int_{-T/2}^{T/2}{f(t) S_k (t) \, dt} + \frac{1}{T} \int_{-T/2}^{T/2}{[S_k (t)]^2 \, dt} \right\}} = 0$

$\displaystyle \lim_{k \rightarrow \infty}{ \left\{ \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} \right\}} - \lim_{k \rightarrow \infty}{ \left\{ \frac{2}{T} \int_{-T/2}^{T/2}{f(t) S_k (t) \, dt}\right\}} + \lim_{k \rightarrow \infty}{ \left\{\frac{1}{T} \int_{-T/2}^{T/2}{[S_k (t)]^2 \, dt} \right\}} = 0$

$\displaystyle \lim_{k \rightarrow \infty}{ \left\{ \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} \right\}} = \lim_{k \rightarrow \infty}{ \left\{ \frac{2}{T} \int_{-T/2}^{T/2}{f(t) S_k (t) \, dt}\right\}} - \lim_{k \rightarrow \infty}{ \left\{\frac{1}{T} \int_{-T/2}^{T/2}{[S_k (t)]^2 \, dt} \right\}}$

$\displaystyle \lim_{k \rightarrow \infty}{ \left\{ \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} \right\}} = \lim_{k \rightarrow \infty}{ \left\{ \frac{1}{2} a_0^2 + \sum_{n=1}^{k}{\left(a_n^2 + b_n^2 \right)} \right\}} - \lim_{k \rightarrow \infty}{ \left\{ \frac{1}{4} a_0^2 + \frac{1}{2} \sum_{n=1}^{k}{\left(a_n^2 + b_n^2 \right)} \right\}}$

$\displaystyle \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} = \frac{1}{2} a_0^2 + \sum_{n=1}^{\infty}{\left(a_n^2 + b_n^2 \right)} - \frac{1}{4} a_0^2 - \frac{1}{2} \sum_{n=1}^{\infty}{\left(a_n^2 + b_n^2 \right)}$

$\displaystyle \frac{1}{T} \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} = \frac{1}{4} a_0^2 + \frac{1}{2} \sum_{n=1}^{\infty}{\left(a_n^2 + b_n^2 \right)}$

Y el teorema de Parseval queda demostrado.

4 comentarios sobre “Aproximación mediante la serie finita de Fourier. Segunda parte. Fourier.”

Llento Soto Oliveira dice:

24 abril, 2020 a las 1:00 pm

En las sucesiones de Ek me perdí, dices que no hay términos negativos pero…y ¿ese menos de Ek+1 = Ek – (1/2)((ak+1)^2 + (bk+1)^2)?

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1. Cesar Augusto dice:
  
  24 abril, 2020 a las 4:30 pm
  
  La desigualdad menciona E_k (t) >=0, eso significa que jamás habrá valores negativos, por eso tiene expresado |E_k| en valor absoluto (un equivalente). Del valor que dé E_(k+1) siempre será positivo, por muy pequeño que sea, ya que sólo depende de E_k, a_(k+1) y b_(k+1); de la expresión que me dices referente al signo «-» no tiene porque interpretarse de que E_(k+1) dará en algún momento valores negativos.
  
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  1. LLENTO SOTO OLIVEIRA dice:
    
    24 abril, 2020 a las 5:57 pm
    
    Muchas gracias por la aclaración, malinterpreté esa parte del texto…e igualmente gracias por compartir la demostración de estos teoremas, me sirvieron mucho para entender los fundamentos de la serie de fourier.
    
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  2. Cesar Augusto dice:
    
    24 abril, 2020 a las 9:11 pm
    
    Claro, un gusto por atenderte…
    
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