Fourier Series Expansion of

We need to find the Fourier series for $f(x) = |\cos(x)|$ on $(-\pi, \pi)$. The function $f(x) = |\cos(x)|$ is an even function.

The general Fourier series for an even function is:

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

The provided series form is $\alpha + \beta \left[ \frac{1}{3}\cos(2x) - \frac{1}{15}\cos(4x) + \dots \right]$. Comparing this with the standard form, we identify $\alpha = \frac{a_0}{2}$ (the constant term) and $\beta$ is a factor related to the subsequent cosine coefficients.

Calculating Constant Term $\alpha$

First, compute $a_0$, which is the average value of $f(x)$ over the interval multiplied by $\pi$. Specifically, $a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} |\cos(x)| dx$. Since the function is even, we can write:

$a_0 = \frac{2}{\pi} \int_{0}^{\pi} |\cos(x)| dx$

The integral needs to be split because $\cos(x)$ changes sign at $x = \pi/2$:

$a_0 = \frac{2}{\pi} \left[ \int_{0}^{\pi/2} \cos(x) dx + \int_{\pi/2}^{\pi} (-\cos(x)) dx \right]$

Evaluate the integrals:

$a_0 = \frac{2}{\pi} \left[ [\sin(x)]_{0}^{\pi/2} - [\sin(x)]_{\pi/2}^{\pi} \right]$

$a_0 = \frac{2}{\pi} \left[ (\sin(\pi/2) - \sin(0)) - (\sin(\pi) - \sin(\pi/2)) \right]$

$a_0 = \frac{2}{\pi} \left[ (1 - 0) - (0 - 1) \right] = \frac{2}{\pi} (1 + 1) = \frac{4}{\pi}$

The constant term $\alpha$ is half of $a_0$:

$\alpha = \frac{a_0}{2} = \frac{1}{2} \left( \frac{4}{\pi} \right) = \frac{2}{\pi}$

Calculating Coefficient $\beta$

Next, consider the cosine coefficients $a_n$. For $n \ge 1$, $a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} |\cos(x)| \cos(nx) dx$. Due to symmetry, this simplifies to $a_n = \frac{2}{\pi} \int_{0}^{\pi} |\cos(x)| \cos(nx) dx$.

It can be shown that $a_n = 0$ for all odd integers $n$. For even integers $n=2k$ where $k \ge 1$, the coefficients are:

$a_{2k} = \frac{4(-1)^k}{\pi(1-4k^2)}$

Calculate the specific coefficients needed for the series structure:

• For $k=1$ ($n=2$): $a_2 = \frac{4(-1)^1}{\pi(1-4(1)^2)} = \frac{-4}{\pi(1-4)} = \frac{-4}{-3\pi} = \frac{4}{3\pi}$
• For $k=2$ ($n=4$): $a_4 = \frac{4(-1)^2}{\pi(1-4(2)^2)} = \frac{4}{\pi(1-16)} = \frac{4}{-15\pi} = -\frac{4}{15\pi}$

The Fourier series begins:

$f(x) = \frac{a_0}{2} + a_2 \cos(2x) + a_4 \cos(4x) + \dots$

$f(x) = \frac{2}{\pi} + \frac{4}{3\pi} \cos(2x) - \frac{4}{15\pi} \cos(4x) + \dots$

Rewrite this by factoring out $\frac{4}{\pi}$ from the cosine terms:

$f(x) = \frac{2}{\pi} + \frac{4}{\pi} \left[ \frac{1}{3}\cos(2x) - \frac{1}{15}\cos(4x) + \dots \right]$

Comparing this calculated series $ \frac{2}{\pi} + \frac{4}{\pi} \left[ \dots \right] $ with the given form $ \alpha + \beta \left[ \dots \right] $, we identify:

$\alpha = \frac{2}{\pi}$

$\beta = \frac{4}{\pi}$

Final Values

Thus, the values are $\alpha = \frac{2}{\pi}$ and $\beta = \frac{4}{\pi}$.