91Ó°ÊÓ

First, identify the periodic extension of the function. The function is given as: \(f(x)= \begin{cases}0, & -\pi < x < 0 \ x, & 0 < x < \pi .\end{cases}\) We need to extend this function periodically for the interval \(-\infty < x < \infty\) with a period of \2\pi\. To visualize this, sketch the function from \(-\pi < x < \pi\), and then repeat this shape every \2\pi\ units.

Sketch several periods of the function. For the interval \[-2\pi, 2\pi\], repeat the given shapes as below:For \-2\pi < x < -\pi\: \f(x) = 0\ For \-\pi < x < 0\: \f(x) = 0\ For \0 < x < \pi\: \f(x) = x\ For \pi < x < 2\pi\: \repeat the shape of [0, \pi] interval

Next, calculate the Fourier coefficients \a_0, a_n,\ and \b_n\. The formulas are:\( a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx\)\( a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx\)\( b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx\)Apply these formulas to find the coefficients.

Compute \a_0\:\(a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx = \frac{1}{2\pi} \left( \int_{-\pi}^{0} 0 \, dx + \int_{0}^{\pi} x \, dx \right)\)\(a_0 = \frac{1}{2\pi} \left( 0 + \frac{x^2}{2} \bigg|_{0}^{\pi} \right) = \frac{1}{2\pi} \left( \frac{\pi^2}{2} \right) = \frac{\pi}{4}\)

Compute \a_n\ for \ \geq 1dx.\(a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx = \frac{1}{\pi} \left( \int_{-\pi}^{0} 0 \cos(nx) \, dx + \int_{0}^{\pi} x \cos(nx) \, dx \right)\)Since the first term is zero:\(a_n = \frac{1}{\pi} \int_{0}^{\pi} x \cos(nx) \, dx\)Using integration by parts:\( u = x, \, dv = \cos(nx)dx\)\( du = dx, \, v = \frac{\sin(nx)}{n} \)\(\int x \cos(nx)dx = x \frac{\sin(nx)}{n} \bigg|_{0}^{\pi} - \int \frac{\sin(nx)}{n} \, dx\)The definite integrals lead to:\(a_n = \frac{1}{\pi} \left[ \frac{\pi \sin(n\pi)}{n} - \frac{\cos(nx)}{n^2} \bigg|_{0}^{\pi} \right]\)Since \sin(n\pi) = 0\ and \cos(n\pi) = (-1)^n\, results in:\(a_n = \frac{1}{n^2 \pi} \left[ -\cos(n\pi) - (-1)^n \right] = 0\)

Compute \b_n\ for \:\(b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx = \frac{1}{\pi} \left( \int_{-\pi}^{0} 0 \sin(nx)dx + \int_{0}^{\pi} x \sin(nx)dx \right)\)Using integration by parts:\( u = x, \, dv = \sin(nx)dx\)\( du = dx, \, v = -\frac{\cos(nx)}{n} \)\(b_n = \frac{1}{\pi} \right[ x \frac{-\cos(nx)}{n} \bigg|_{0}^{\pi} + \int \frac{\cos(nx)}{n} dx\)The definite with results:\(b_n = \frac{(-1)^n -1}{n}\)

Combine the computed Fourier coefficients into the Fourier series:\(f(x) = \frac{\pi}{4} + \sum_{n=1}^{\infty} \left (\frac{2 (-1)^n-1}{n}\right)\sin(nx)\)