The Fourier series of the function $f(x) = |\cos(x)|$ on the interval $-\pi < x < \pi$ can be derived by expressing the function as a sum of sine and cosine terms. Since $|\cos(x)|$ is an even function, the Fourier series will consist only of cosine terms.

Step 1: Write the Fourier series expression for $f(x)$

The Fourier series for a function $f(x)$ defined on $-\pi < x < \pi$ is given by:

$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx)$

where $a_0$ and $a_n$ are the Fourier coefficients, given by:

$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx$ $a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx, \quad n = 1, 2, 3, \dots$

Step 2: Compute $a_0$

Since $f(x) = |\cos(x)|$, the integral for $a_0$ becomes:

$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} |\cos(x)| \, dx$

Due to the even symmetry, we can evaluate the integral over half the interval and multiply by 2:

$a_0 = \frac{2}{\pi} \int_0^{\pi} \cos(x) \, dx$

The integral of $\cos(x)$ over $[0, \pi]$ is zero because the positive and negative parts cancel out. Hence:

$a_0 = 0$

Step 3: Compute $a_n$

Next, we compute the Fourier coefficients $a_n$ for $n \geq 1$:

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

Using the product-to-sum identities, the integral can be evaluated. After performing the calculations, the result is:

\frac{4}{\pi(1 - n^2)} & \text{if } n \text{ is even} \\ 0 & \text{if } n \text{ is odd} \end{cases}$$ ### Step 4: Write the Fourier series Thus, the Fourier series for $$|\cos(x)|$$ is: \[ |\cos(x)| = \frac{4}{\pi} \left( \frac{1}{1^2 - 1^2} \right) + \sum_{n \text{ even}} \frac{4}{\pi