[Resolução] III.6.16c
Posted: 07 Dec 2022 20:02
\( f(x)=x^2, x \in [0,\pi] \)
(I) Série de Fourier em senos:
\( S_f(x) = \sum^{\infty}_{n=1} b_n sin(\frac{n\pi x}{L}) \)
Calculamos \( b_n \):
\( b_n = \frac{2}{L}\int_{0}^{L} x^2 sin(\frac{n\pi x}{L}) dx = \frac{2}{\pi}\int_{0}^{\pi} x^2 sin(nx) dx \)
\( = \frac{2}{\pi} \left [- \frac{x^2cos(nx)}{n} + \frac{2}{n}\int x cos(nx) dx \right ] \)
\( = \frac{2}{\pi} \left [- \frac{x^2cos(nx)}{n} + \frac{2xsin(nx)}{n^2} - \frac{2}{n^2}\int sin(nx) dx \right ] \)
\( = \frac{2}{\pi} \left [- \frac{x^2cos(nx)}{n} + \frac{2xsin(nx)}{n^2} + \frac{2cos(nx)}{n^3} \right ]^\pi_0 \)
\( = \frac{2}{\pi} \left [- \frac{x^2cos(nx)}{n} + \frac{2xsin(nx)}{n^2} + \frac{2cos(nx)}{n^3} \right ]^\pi_0 = \frac{2}{\pi} \left [- \frac{\pi^2cos(n\pi)}{n} + \frac{2cos(n\pi)}{n^3} \right ] - \left [\frac{2}{n^3} \right ] \)
\( = \frac{2}{\pi} \left [- \frac{n^2\pi^2cos(n\pi)+2cos(n\pi) - 2}{n^3} \right ] = \frac{2}{\pi} \left [- \frac{n^2\pi^2(-1)^n+2(-1)^n - 2}{n^3} \right ] = \frac{2}{\pi} \left [\frac{(-1)^n(2-n^2\pi^2) - 2}{n^3} \right ] \)
Após isso, inserimos \( b_n \) em \( S_f(x) \):
\( S_f(x) = \sum^{\infty}_{n=1}\frac{2}{\pi} \left [\frac{(-1)^n(2-n^2\pi^2) - 2}{n^3} \right ]sin(nx) \)
(II) Série de Fourier em cossenos:
\( S_f(x) = \frac{a_0}{2} + \sum^{\infty}_{n=1} a_n cos(\frac{n\pi x}{L}) \)
Primeiro, calculamos \( a_0 \):
\( a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2dx = \frac{1}{\pi}\left [ \frac{x^3}{3} \right ]^\pi_{-\pi} =\frac{1}{\pi}\frac{2\pi^3}{3}=\frac{2\pi^2}{3} \)
Após isso, calculamos \( a_n \):
\( a_n = \frac{2}{L}\int_{0}^{L} x^2 cos(\frac{n\pi x}{L}) dx = \frac{2}{\pi}\int_{0}^{\pi} x^2 cos(nx) dx \)
\( = \frac{2}{\pi} \left [\frac{x^2sin(nx)}{n} - \frac{2}{n}\int x sin(nx) dx \right ] \)
\( = \frac{2}{\pi} \left [\frac{x^2sin(nx)}{n} + \frac{2xcos(nx)}{n^2} - \frac{2}{n^2}\int cos(nx) dx \right ] \)
\( = \frac{2}{\pi} \left [\frac{x^2sin(nx)}{n} + \frac{2xcos(nx)}{n^2} - \frac{2sin(nx)}{n^3} \right ]^\pi_0 \)
\( = \frac{2}{\pi} \left [\frac{2\pi cos(n\pi)}{n^2} \right ] = \frac{4 cos(n\pi)}{n^2} = \frac{4 (-1)^n}{n^2} \)
Após isso, inserimos \( a_0 \) e \( a_n \) em \( S_f(x) \):
\( S_f(x) = \frac{\pi^2}{3} + \sum^{\infty}_{n=1} \left [ \frac{4 (-1)^n}{n^2} \right ]cos(nx) \)
(I) Série de Fourier em senos:
\( S_f(x) = \sum^{\infty}_{n=1} b_n sin(\frac{n\pi x}{L}) \)
Calculamos \( b_n \):
\( b_n = \frac{2}{L}\int_{0}^{L} x^2 sin(\frac{n\pi x}{L}) dx = \frac{2}{\pi}\int_{0}^{\pi} x^2 sin(nx) dx \)
\( = \frac{2}{\pi} \left [- \frac{x^2cos(nx)}{n} + \frac{2}{n}\int x cos(nx) dx \right ] \)
\( = \frac{2}{\pi} \left [- \frac{x^2cos(nx)}{n} + \frac{2xsin(nx)}{n^2} - \frac{2}{n^2}\int sin(nx) dx \right ] \)
\( = \frac{2}{\pi} \left [- \frac{x^2cos(nx)}{n} + \frac{2xsin(nx)}{n^2} + \frac{2cos(nx)}{n^3} \right ]^\pi_0 \)
\( = \frac{2}{\pi} \left [- \frac{x^2cos(nx)}{n} + \frac{2xsin(nx)}{n^2} + \frac{2cos(nx)}{n^3} \right ]^\pi_0 = \frac{2}{\pi} \left [- \frac{\pi^2cos(n\pi)}{n} + \frac{2cos(n\pi)}{n^3} \right ] - \left [\frac{2}{n^3} \right ] \)
\( = \frac{2}{\pi} \left [- \frac{n^2\pi^2cos(n\pi)+2cos(n\pi) - 2}{n^3} \right ] = \frac{2}{\pi} \left [- \frac{n^2\pi^2(-1)^n+2(-1)^n - 2}{n^3} \right ] = \frac{2}{\pi} \left [\frac{(-1)^n(2-n^2\pi^2) - 2}{n^3} \right ] \)
Após isso, inserimos \( b_n \) em \( S_f(x) \):
\( S_f(x) = \sum^{\infty}_{n=1}\frac{2}{\pi} \left [\frac{(-1)^n(2-n^2\pi^2) - 2}{n^3} \right ]sin(nx) \)
(II) Série de Fourier em cossenos:
\( S_f(x) = \frac{a_0}{2} + \sum^{\infty}_{n=1} a_n cos(\frac{n\pi x}{L}) \)
Primeiro, calculamos \( a_0 \):
\( a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2dx = \frac{1}{\pi}\left [ \frac{x^3}{3} \right ]^\pi_{-\pi} =\frac{1}{\pi}\frac{2\pi^3}{3}=\frac{2\pi^2}{3} \)
Após isso, calculamos \( a_n \):
\( a_n = \frac{2}{L}\int_{0}^{L} x^2 cos(\frac{n\pi x}{L}) dx = \frac{2}{\pi}\int_{0}^{\pi} x^2 cos(nx) dx \)
\( = \frac{2}{\pi} \left [\frac{x^2sin(nx)}{n} - \frac{2}{n}\int x sin(nx) dx \right ] \)
\( = \frac{2}{\pi} \left [\frac{x^2sin(nx)}{n} + \frac{2xcos(nx)}{n^2} - \frac{2}{n^2}\int cos(nx) dx \right ] \)
\( = \frac{2}{\pi} \left [\frac{x^2sin(nx)}{n} + \frac{2xcos(nx)}{n^2} - \frac{2sin(nx)}{n^3} \right ]^\pi_0 \)
\( = \frac{2}{\pi} \left [\frac{2\pi cos(n\pi)}{n^2} \right ] = \frac{4 cos(n\pi)}{n^2} = \frac{4 (-1)^n}{n^2} \)
Após isso, inserimos \( a_0 \) e \( a_n \) em \( S_f(x) \):
\( S_f(x) = \frac{\pi^2}{3} + \sum^{\infty}_{n=1} \left [ \frac{4 (-1)^n}{n^2} \right ]cos(nx) \)