Fourier Series Derivation for Periodic Function f(x)

The given function is:

$f(x) = \begin{cases} -1, & \text{when } -\pi < x < 0 \\ 1, & \text{when } 0 < x < \pi \end{cases} ; f(x + 2\pi) = f(x)$

This function is defined over the interval $(-\pi, \pi)$ and is periodic with period $2\pi$. We observe that $f(-x) = -f(x)$, indicating that $f(x)$ is an odd function.

Function Properties Analysis

• Since $f(x)$ is an odd function, its Fourier series expansion will only contain sine terms. The constant term ($a_0$) and cosine terms ($a_n$) are zero.
• The Fourier series has the form: $f(x) = \sum_{n=1}^{\infty} b_n \sin(nx)$.

Fourier Coefficient Calculation

The formula for the sine coefficient $b_n$ is:

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

Because $f(x)$ is an odd function and $\sin(nx)$ is also an odd function, their product $f(x)\sin(nx)$ is an even function. Therefore, the integral can be simplified:

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

In the interval $(0, \pi)$, $f(x) = 1$. Substituting this value:

$b_n = \frac{2}{\pi} \int_{0}^{\pi} (1) \sin(nx) dx$ $b_n = \frac{2}{\pi} \left[ -\frac{\cos(nx)}{n} \right]_{0}^{\pi}$ $b_n = -\frac{2}{n\pi} [\cos(n\pi) - \cos(0)]$

Using $\cos(n\pi) = (-1)^n$ and $\cos(0) = 1$:

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

Determining Values of $b_n$

• For even $n$: Let $n = 2k$ where $k$ is an integer. Then $(-1)^n = 1$. $b_n = -\frac{2}{n\pi} [1 - 1] = 0$
• For odd $n$: Let $n = 2k-1$ where $k$ is an integer. Then $(-1)^n = -1$. $b_n = -\frac{2}{n\pi} [-1 - 1] = -\frac{2}{n\pi} [-2] = \frac{4}{n\pi}$

So, the non-zero coefficients occur only for odd values of $n$. The Fourier series becomes:

$f(x) = \sum_{n \text{ odd}, n\ge 1}^{\infty} \frac{4}{n\pi} \sin(nx)$ $f(x) = \frac{4}{\pi} \sum_{n \text{ odd}, n\ge 1}^{\infty} \frac{\sin(nx)}{n}$ $f(x) = \frac{4}{\pi} \left[ \sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \dots \right]$

Final Fourier Series Expansion

Comparing this result with the given options, the correct Fourier series expansion for $f(x)$ is:

$f(x) = \frac{4}{\pi} \left[ \sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \dots \right]$