The problem asks us to construct the Fourier series of the function $f(x) = x$ on the interval $-\pi < x < \pi$.

1. Problem Description

The problem asks us to construct the Fourier series of the function $f(x) = x$ on the interval $-\pi < x < \pi$.

2. Solution Steps

The Fourier series of a function $f(x)$ defined on the interval $(-L, L)$ is given by

$$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos\left(\frac{n\pi x}{L}\right) + \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{L}\right),$$

where the coefficients are given by

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

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

$$b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) \, dx.$$

In our case, $f(x) = x$ and $L = \pi$.

First, let's calculate $a_0$:

$$$$

a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x \, dx = \frac{1}{\pi} \left[\frac{x^2}{2}\right]_{-\pi}^{\pi} = \frac{1}{\pi} \left(\frac{\pi^2}{2} - \frac{(-\pi)^2}{2}\right) = \frac{1}{\pi} \left(\frac{\pi^2}{2} - \frac{\pi^2}{2}\right) =

0. $$

Next, let's calculate $a_n$:

$$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \cos(nx) \, dx.$$

Since $x \cos(nx)$ is an odd function and the integral is taken over a symmetric interval, the integral is zero. Therefore, $a_n = 0$ for all $n$.

Now, let's calculate $b_n$:

$$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin(nx) \, dx.$$

Since $x \sin(nx)$ is an even function, we can write

$$b_n = \frac{2}{\pi} \int_{0}^{\pi} x \sin(nx) \, dx.$$

We integrate by parts. Let $u = x$ and $dv = \sin(nx) \, dx$. Then $du = dx$ and $v = -\frac{1}{n}\cos(nx)$.

$$\int x \sin(nx) \, dx = -\frac{x}{n} \cos(nx) + \int \frac{1}{n} \cos(nx) \, dx = -\frac{x}{n} \cos(nx) + \frac{1}{n^2} \sin(nx).$$

Thus,

$$b_n = \frac{2}{\pi} \left[-\frac{x}{n} \cos(nx) + \frac{1}{n^2} \sin(nx)\right]_0^{\pi} = \frac{2}{\pi} \left[-\frac{\pi}{n} \cos(n\pi) + \frac{1}{n^2} \sin(n\pi) - (0 + 0)\right] = \frac{2}{\pi} \left[-\frac{\pi}{n} (-1)^n\right] = -\frac{2}{n}(-1)^n = \frac{2}{n}(-1)^{n+1}.$$

Therefore, the Fourier series is

$$f(x) = \sum_{n=1}^{\infty} \frac{2}{n}(-1)^{n+1} \sin(nx) = 2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin(nx).$$

3. Final Answer

The Fourier series of $f(x) = x$ on $-\pi < x < \pi$ is

$f(x) = 2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin(nx)$.