Diferenciación de la serie de Fourier. Fourier.

Se observa que la diferenciación término por término de una serie trigonométrica

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

multiplica los coeficientes $a_n$ y $b_n$ por $\pm n omega_0$ . Donde la diferenciación tiende a disminuir la convergencia y puede resultar en divergencia.

Demostración del teorema de diferenciación de la serie de Fourier

Teorema. Si $f(t)$ es continua cuando $-T/2 < t < T/2$ con $f(-T/2) = f(T/2)$ , y si la derivada $f'(t)$ es continua por tramos, y diferenciable, entonces la serie de Fourier

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

se puede diferenciar término por término para obtener

$\displaystyle f'(t) = \sum_{n=1}^{k}{n\omega_0 (-a_n \sin{n\omega_0 t} + b_n \cos{n \omega t})}$

Demostración.

Partiendo de la serie de Fourier, se deriva la función con respecto a « $t$ «

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

$\displaystyle \frac{d}{dt} f(t) = \frac{1}{2} \frac{d}{dt}(a_0) + \frac{d}{dt} \left[\sum_{n=1}^{\infty}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})} \right]$

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

Atendiendo sólo el segundo término de la derivada del miembro derecho

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

Se deriva lo siguiente

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

Derivando el primer término de la suma

$\displaystyle \frac{d}{dt} \left[ \frac{2}{T} \left( \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} \right) \cdot \cos{n \omega_0 t} \right]$

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

$\displaystyle = \left(\frac{2}{T} \int_{-T/2}^{T/2}{\left(f'(t) \cos{n \omega_0 t} - n \omega_0 f(t) \sin{n \omega_0 t} \right) \, dt} \right) \cos{n \omega_0 t} - n \omega_0 \left(\frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} \right) \sin{n \omega_0 t}$

$\displaystyle = \left( \frac{2}{T}\int_{-T/2}^{T/2}{f'(t) \cos{n \omega_0 t} \, dt} - n \omega_0 \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} \right) \cos{n \omega_0 t} - n \omega_0 \left( \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} \right) \sin{n \omega_0 t}$

$\displaystyle = \left(\frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \cos{n \omega_0 t} \, dt} \right) \cos{n \omega_0 t} - n \omega_0 \left(\frac{2}{T}\int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} \right) \cos{n \omega_0 t} - n \omega_0 \left( \frac{2}{T}\int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} \right) \sin{n \omega_0 t}$

Derivando el segundo término de la suma

$\displaystyle \frac{d}{dt} \left[\left(\frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n\omega_0 t} \, dt} \right) \cdot \sin{n \omega_0 t} \right]$

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

$\displaystyle = \left(\frac{2}{T} \int_{-T/2}^{T/2}{\left(f'(t) \sin{n\omega_0 t} + n\omega_0 \cdot f(t) \cos{n\omega_0 t} \right) \, dt} \right) \sin{n \omega_0 t} + \left(\frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n\omega_0 t} \, dt} \right) (n \omega_0 \cos{n \omega_0 t})$

$\displaystyle = \left(\frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \sin{n\omega_0 t} \, dt}+ n\omega_0 \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n\omega_0 t} \, dt} \right) \sin{n \omega_0 t} + \left(\frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n\omega_0 t} \, dt} \right) (n \omega_0 \cos{n \omega_0 t})$

$\displaystyle = \left(\frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \sin{n\omega_0 t} \, dt} \right) \sin{n \omega_0 t} + n\omega_0 \left(\frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n\omega_0 t} \, dt} \right) \sin{n \omega_0 t} + n\omega_0 \left(\frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n\omega_0 t} \, dt} \right) \cos{n \omega_0 t}$

Regresado a la suma

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

$\displaystyle = \sum_{n=1}^{\infty}{\left[ \left(\frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \cos{n \omega_0 t} \, dt} \right) \cos{n \omega_0 t} - n \omega_0 \left(\frac{2}{T}\int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} \right) \cos{n \omega_0 t} - n \omega_0 \left(\frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt} \right) \sin{n \omega_0 t} + \right.}$

$\displaystyle \left. + \left(\frac{2}{T}\int_{-T/2}^{T/2}{f'(t) \sin{n\omega_0 t} \, dt} \right) \sin{n \omega_0 t} + n\omega_0 \left(\frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n\omega_0 t} \, dt} \right) \sin{n \omega_0 t} + n\omega_0 \left(\frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n\omega_0 t} \, dt} \right) \cos{n \omega_0 t} \right]$

$\displaystyle = \sum_{n=1}^{\infty}{\left[ \left(\frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \cos{n \omega_0 t} \, dt} \right) \cos{n \omega_0 t} + \left(\frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \sin{n\omega_0 t} \, dt} \right) \sin{n \omega_0 t} \right]}$

Regresando a la primera derivada de la serie de Fourier y sustituyendo los resultados obtenidos

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

$\displaystyle f'(t) = \frac{1}{2} \frac{d}{dt}\left[\frac{2}{T} \int_{-T/2}^{T/2}{f(t) \, dt} \right] + \sum_{n=1}^{\infty}{\left[ \left(\frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \cos{n \omega_0 t} \, dt} \right) \cos{n \omega_0 t} + \left(\frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \sin{n\omega_0 t} \, dt} \right) \sin{n \omega_0 t} \right]}$

$\displaystyle f'(t) = \frac{1}{2} \cdot \frac{2}{T} \int_{-T/2}^{T/2}{\frac{d}{dt}[f(t)] \, dt} + \sum_{n=1}^{\infty}{\left[ \left(\frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \cos{n \omega_0 t} \, dt} \right) \cos{n \omega_0 t} + \left(\frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \sin{n\omega_0 t} \, dt} \right) \sin{n \omega_0 t} \right]}$

$\displaystyle f'(t) = \frac{1}{2} \cdot \frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \, dt} + \sum_{n=1}^{\infty}{\left[ \left(\frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \cos{n \omega_0 t} \, dt} \right) \cos{n \omega_0 t} + \left(\frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \sin{n\omega_0 t} \, dt} \right) \sin{n \omega_0 t} \right]}$

$\displaystyle f'(t) = \frac{1}{2} \alpha_0 + \sum_{n=1}^{\infty}{(\alpha_n \cos{n \omega_0 t} + \beta_n \sin{n \omega_0 t})}$

Donde $\displaystyle \alpha_0 = \frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \, dt}$ , $\displaystyle \alpha_n = \frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \cos{n \omega_0 t} \, dt}$ y $\displaystyle \beta_n = \frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \sin{n\omega_0 t} \, dt}$ .

De $\alpha_0$ , tiene una siguiente expresión (utilizando integración directa)

$\displaystyle \alpha_0 = \frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \, dt}$

$\displaystyle \alpha_0 = \frac{2}{T} \left[f(t) + C \right]_{-T/2}^{T/2}$

$\displaystyle \alpha_0 = \frac{2}{T} \left[f \left(\frac{T}{2}\right) - f \left(-\frac{T}{2} \right) \right]$

Si $\displaystyle f \left( \frac{T}{2} \right) = f \left( - \frac{T}{2} \right)$ , entonces

$\displaystyle \alpha_0 = 0$

De $\alpha_n$ , tiene una siguiente expresión (utilizando integración por partes)

$\displaystyle \alpha_n = \frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \cos{n \omega_0 t} \, dt}$

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

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

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

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

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

$\displaystyle \alpha_n = n \omega_0 b_n$

Recordando que $\displaystyle f \left( \frac{T}{2} \right) = f \left( - \frac{T}{2} \right)$ .

De $\beta_n$ , tiene una siguiente expresión (utilizando integración por partes)

$\displaystyle \beta_n = \frac{2}{T} \int_{-T/2}^{T/2}{f'(t) \sin{n\omega_0 t} \, dt}$

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

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

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

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

$\displaystyle \beta_n = \frac{4}{T} f \left( \frac{T}{2} \right) \sin{n \omega_0 (\frac{T}{2})} - n\omega_0 \cdot \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n \omega_0 t} \, dt}$

$\displaystyle \beta_n = \frac{4}{T} f \left( \frac{T}{2} \right) \sin{n \omega_0 (\frac{T}{2})} - n\omega_0 a_n$

Recordando que $\displaystyle f \left( \frac{T}{2} \right) = f \left( - \frac{T}{2} \right)$ .

Regresando con la continuación de la derivada de $f'(t)$ y sustituyendo

$\displaystyle f'(t) = \frac{1}{2} \alpha_0 + \sum_{n=1}^{\infty}{(\alpha_n \cos{n \omega_0 t} + \beta_n \sin{n \omega_0 t})}$

$\displaystyle f'(t) = \frac{1}{2} (0) + \sum_{n=1}^{\infty}{\left[n \omega_0 b_n \cos{n \omega_0 t} + \left[ \frac{4}{T} f \left( \frac{T}{2} \right) \sin{n \omega_0 (\frac{T}{2})} - n\omega_0 a_n \right] \sin{n \omega_0 t}\right]}$

$\displaystyle f'(t) = \sum_{n=1}^{\infty}{n \omega_0 b_n \cos{n \omega_0 t}} + \sum_{n=1}^{\infty}{ \frac{4}{T}f \left( \frac{T}{2} \right) \sin{n \omega_0 (\frac{T}{2})} \sin{n\omega_0 t}} - \sum_{n=1}^{\infty}{n\omega_0 a_n \sin{n \omega_0 t}}$

$\displaystyle f'(t) = \sum_{n=1}^{\infty}{n \omega_0 b_n \cos{n \omega_0 t}} + \frac{4}{T}f \left( \frac{T}{2} \right) \sum_{n=1}^{\infty}{\sin{n \omega_0 (\frac{T}{2})} \sin{n\omega_0 t}} - \sum_{n=1}^{\infty}{n\omega_0 a_n \sin{n \omega_0 t}}$

$\displaystyle f'(t) = \sum_{n=1}^{\infty}{n \omega_0 b_n \cos{n \omega_0 t}} + \frac{4}{T}f \left( \frac{T}{2} \right) \sum_{n=1}^{\infty}{\sin{n \pi} \sin{n\omega_0 t}} - \sum_{n=1}^{\infty}{n\omega_0 a_n \sin{n \omega_0 t}}$

$\displaystyle f'(t) = \sum_{n=1}^{\infty}{n \omega_0 b_n \cos{n \omega_0 t}} - \sum_{n=1}^{\infty}{n\omega_0 a_n \sin{n \omega_0 t}}$

$\displaystyle f'(t) = \sum_{n=1}^{\infty}{(n \omega_0 b_n \cos{n \omega_0 t} - n\omega_0 a_n \sin{n \omega_0 t})}$

$\displaystyle f'(t) = \sum_{n=1}^{\infty}{(n \omega_0)(b_n \cos{n \omega_0 t} - a_n \sin{n \omega_0 t})}$

$\displaystyle \therefore f'(t) = \sum_{n=1}^{\infty}{n \omega_0(-a_n \sin{n \omega_0 t} + b_n \cos{n \omega_0 t})}$

Y el teorema queda demostrado.

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Demostración del teorema de diferenciación de la serie de Fourier

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