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5.2: Serie Trigonométrica de Fourier
https://espanol.libretexts.org/Matematicas/Ecuaciones_diferenciales/Un_Segundo_Curso_en_Ecuaciones_Diferenciales_Ordinarias%3A_Sistemas_Din%C3%A1micos_y_Problemas_de_Valor_L%C3%ADmite_(Herman)/05%3A_Serie_de_Fourier/5.02%3A_Serie_Trigonom%C3%A9trica_de_Fourier
$\dfrac{a_{0}}{2} \int_{0}^{2 \pi} \cos m x d x+\sum_{n=1}^{\infty}\left[a_{n} \int_{0}^{2 \pi} \cos n x \cos m x d x+b_{n} \int_{0}^{2 \pi} \sin n x \cos m x d x\right] \text {. } \label{5.6}$ \ i... $\dfrac{a_{0}}{2} \int_{0}^{2 \pi} \cos m x d x+\sum_{n=1}^{\infty}\left[a_{n} \int_{0}^{2 \pi} \cos n x \cos m x d x+b_{n} \int_{0}^{2 \pi} \sin n x \cos m x d x\right] \text {. } \label{5.6}$ \ int_ {0} ^ {2\ pi}\ cos n x\ cos m x d x &=\ dfrac {1} {2}\ int_ {0} ^ {2\ pi} [\ cos (m+n) x+\ cos (m-n) x] d x\\ $\int_{0}^{2 \pi} \sin m x \cos m x d x=\dfrac{1}{2} \int_{0}^{2 \pi} \sin 2 m x d x=\dfrac{1}{2}\left[\dfrac{-\cos 2 m x}{2 m}\right]_{0}^{2 \pi}=0. \nonumber$

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