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11.3: Serie Fourier-Legendre

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    113741
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    Dado que la ecuación de Legendre es autoadjoint, podemos mostrar que\(P_n(x)\) forma un conjunto ortogonal de funciones. Para descomponer funciones como series en polinomios de Legendre necesitaremos las integrales

    \[\int_{-1}^1 P_n^2(x) dx = \frac{2n+1}{2}, \nonumber \]

    que se puede determinar usando la relación 5. dos veces para obtener una relación de recurrencia

    \[\begin{aligned} \int_{-1}^1 P_n^2(x) dx &= \int_{-1}^1 P_n(x) \frac{(2n-1)x P_{n-1}(x)-(n-1)P_{n-2}(x)}{n} dx \nonumber\\ &=\frac{(2n-1)}{n}\int_{-1}^1 x P_n(x) P_{n-1}(x) dx \nonumber\\ &=\frac{(2n-1)}{n}\int_{-1}^1 \frac{(n+1)P_{n+1}(x) + n P_{n-1}(x)}{2n+1} P_{n-1}(x) dx \nonumber\\ &=\frac{(2n-1)}{2n+1}\int_{-1}^1 P_{n-1}^2(x) dx,\end{aligned} \nonumber \]

    y el uso de una integral muy simple para fijar este número para\(n=0\),

    \[\int_{-1}^1 P_0^2(x) dx = 2. \nonumber \]

    Así que ahora podemos desarrollar cualquier función\([-1,1]\) en una serie de Fourier-Legendre

    \[\begin{aligned} f(x) & = \sum_n A_n P_n(x) \nonumber\\ A_n & = \frac{2n+1}{2} \int_{-1}^1 f(x) P_n(x) dx\end{aligned} \nonumber \]

    Ejercicio\(\PageIndex{1}\): Fourier-Legendre series

    Encuentra la serie Fourier-Legendre para

    \[f(x) = \left\{ \begin{array}{ll} 0, & -1 < x < 0\\ 1, & 0 < x <1 \end{array} \right. . \nonumber \]

    Contestar

    ENCONTRAMOS\[\begin{aligned} A_0 &= \dfrac{1}{2} \int_0^1 P_0(x) dx = \dfrac{1}{2},\\ A_1 &= \frac{3}{2} \int_0^1 P_1(x) dx= \frac{1}{4}, \\ A_2 &= \frac{5}{2} \int_0^1 P_2(x) dx= 0, \\ A_3 &= \frac{7}{2} \int_0^1 P_3(x) dx= -\frac{7}{16} .\end{aligned} \nonumber \]

    Todos los demás coeficientes para par\(n\) son cero, para impares\(n\) pueden evaluarse explícitamente.


    This page titled 11.3: Serie Fourier-Legendre is shared under a CC BY-NC-SA 2.0 license and was authored, remixed, and/or curated by Niels Walet via source content that was edited to the style and standards of the LibreTexts platform.