Ejercicios resueltos de coeficientes de Fourier

$f(x) = |x|$ es una función par, por lo que $b_n = 0$ para todo $n$.

$a_0 = \frac{1}{\pi}\int_{-\pi}^\pi |x| dx = \frac{2}{\pi}\int_0^\pi x dx = \frac{2}{\pi}\left[\frac{x^2}{2}\right]_0^\pi = \frac{2}{\pi} \cdot \frac{\pi^2}{2} = \pi$

$a_n = \frac{1}{\pi}\int_{-\pi}^\pi |x|\cos(nx) dx = \frac{2}{\pi}\int_0^\pi x\cos(nx) dx$

Sea $u = x$, $dv = \cos(nx)dx$, entonces $du = dx$, $v = \frac{\sin(nx)}{n}$:

$a_n = \frac{2}{\pi}\left[x\frac{\sin(nx)}{n}\right]_0^\pi - \frac{2}{\pi}\int_0^\pi \frac{\sin(nx)}{n} dx$

$= \frac{2}{\pi} \cdot 0 - \frac{2}{n\pi}\left[-\frac{\cos(nx)}{n}\right]_0^\pi$

$= \frac{2}{n^2\pi}[\cos(n\pi) - 1]$

Para $n$ par: $a_n = 0$

Para $n$ impar: $a_n = -\frac{4}{n^2\pi}$

$|x| = \frac{\pi}{2} - \frac{4}{\pi}\sum_{n \text{ impar}} \frac{\cos(nx)}{n^2} = \frac{\pi}{2} - \frac{4}{\pi}\left[\cos(x) + \frac{\cos(3x)}{9} + \frac{\cos(5x)}{25} + \cdots\right]$

Identificar el período $2L = 4$, por lo tanto $L = 2$

$f(x) = x^2$ es par, entonces $b_n = 0$ para todo $n$.

$a_0 = \frac{1}{L}\int_{-L}^L f(x) dx = \frac{1}{2}\int_{-2}^2 x^2 dx = \frac{1}{2} \cdot 2\int_0^2 x^2 dx$

$= \frac{1}{2} \cdot 2 \cdot \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}$

$a_n = \frac{1}{L}\int_{-L}^L f(x)\cos\left(\frac{n\pi x}{L}\right) dx = \frac{1}{2}\int_{-2}^2 x^2\cos\left(\frac{n\pi x}{2}\right) dx$

$= \int_0^2 x^2\cos\left(\frac{n\pi x}{2}\right) dx$

Primera integración: $u = x^2$, $dv = \cos\left(\frac{n\pi x}{2}\right)dx$

$a_n = \left[x^2 \cdot \frac{2\sin(n\pi x/2)}{n\pi}\right]_0^2 - \int_0^2 2x \cdot \frac{2\sin(n\pi x/2)}{n\pi} dx$

$= 0 - \frac{4}{n\pi}\int_0^2 x\sin\left(\frac{n\pi x}{2}\right) dx$

Segunda integración: $u = x$, $dv = \sin\left(\frac{n\pi x}{2}\right)dx$

$a_n = -\frac{4}{n\pi}\left[-x \cdot \frac{2\cos(n\pi x/2)}{n\pi} + \frac{2}{n\pi}\int_0^2 \cos\left(\frac{n\pi x}{2}\right) dx\right]$

$= \frac{8}{n^2\pi^2}[2\cos(n\pi) - 0] = \frac{16(-1)^n}{n^2\pi^2}$

$x^2 = \frac{4}{3} + \frac{16}{\pi^2}\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}\cos\left(\frac{n\pi x}{2}\right)$

$f(x)$ es par, por lo tanto $b_n = 0$ para todo $n$.

$a_0 = \frac{1}{\pi}\int_{-\pi}^\pi \left(1 - \frac{2|x|}{\pi}\right) dx = \frac{2}{\pi}\int_0^\pi \left(1 - \frac{2x}{\pi}\right) dx$

$= \frac{2}{\pi}\left[x - \frac{x^2}{\pi}\right]_0^\pi = \frac{2}{\pi}[\pi - \pi] = 0$

$a_n = \frac{2}{\pi}\int_0^\pi \left(1 - \frac{2x}{\pi}\right)\cos(nx) dx$

$= \frac{2}{\pi}\int_0^\pi \cos(nx) dx - \frac{4}{\pi^2}\int_0^\pi x\cos(nx) dx$

Primera integral: $\int_0^\pi \cos(nx) dx = \frac{\sin(n\pi)}{n} = 0$ para $n \geq 1$

Segunda integral (por partes):

$\int_0^\pi x\cos(nx) dx = \left[\frac{x\sin(nx)}{n}\right]_0^\pi - \int_0^\pi \frac{\sin(nx)}{n} dx$

$= 0 + \frac{1}{n^2}[\cos(nx)]_0^\pi = \frac{\cos(n\pi) - 1}{n^2}$

$a_n = -\frac{4}{\pi^2} \cdot \frac{\cos(n\pi) - 1}{n^2}$

Para $n$ par: $a_n = 0$

Para $n$ impar: $a_n = \frac{8}{n^2\pi^2}$

$a_1 = \frac{8}{\pi^2}$, $a_3 = \frac{8}{9\pi^2}$, $a_5 = \frac{8}{25\pi^2}$, $a_7 = \frac{8}{49\pi^2}$, $a_9 = \frac{8}{81\pi^2}$

La serie $\sum_{n \text{ impar}} \frac{1}{n^2}$ converge (serie p con $p = 2 > 1$), garantizando convergencia uniforme de la serie de Fourier.

De ejercicios anteriores: $x = 2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\sin(nx)$

$\frac{1}{\pi}\int_{-\pi}^\pi f^2(x) dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty} (a_n^2 + b_n^2)$

$\frac{1}{\pi}\int_{-\pi}^\pi x^2 dx = \frac{1}{\pi} \cdot \frac{2\pi^3}{3} = \frac{2\pi^2}{3}$

Para $f(x) = x$: $a_0 = 0$, $a_n = 0$, $b_n = \frac{2(-1)^{n+1}}{n}$

$\sum_{n=1}^{\infty} b_n^2 = \sum_{n=1}^{\infty} \frac{4}{n^2} = 4\sum_{n=1}^{\infty} \frac{1}{n^2}$

$\frac{2\pi^2}{3} = 4\sum_{n=1}^{\infty} \frac{1}{n^2}$

$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{2\pi^2}{3} \cdot \frac{1}{4} = \frac{\pi^2}{6}$

La identidad de Parseval se verifica y obtenemos el famoso resultado de Euler: $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$.