To solve the given question using Fourier series, let's first understand the function given:
[tex]\[ f(x) = \left\{
\begin{array}{ll}
-k, & -\pi < x < 0 \\
k, & 0 < x < \pi
\end{array}
\right. \][/tex]
This piecewise function represents an odd periodic function that resembles a square wave, alternating between -k and k on the interval \(-\pi\) to \(\pi\).
When we proceed with finding the Fourier series for an odd periodic function like this, we notice that only the sine terms will be present because cosine terms vanish due to the odd symmetry of the function.
The Fourier series of such a function will look like:
[tex]\[ f(x) = \sum_{n=1,3,5,\ldots}^\infty b_n \sin(nx) \][/tex]
For the Fourier coefficients \( b_n \), we have:
[tex]\[ b_n = \frac{2k}{n\pi} \int_{0}^{\pi} \sin(nx) \, dx \][/tex]
The integral calculates as follows:
[tex]\[ \int_{0}^{\pi} \sin(nx) \, dx = \left[ -\frac{\cos(nx)}{n} \right]_{0}^{\pi} = \left( -\frac{\cos(n\pi)}{n} + \frac{\cos(0)}{n} \right) = \left( -\frac{(-1)^n}{n} + \frac{1}{n} \right) \][/tex]
For odd \( n \):
[tex]\[ \cos(n\pi) = (-1)^n \][/tex]
[tex]\[ \cos(n\pi) = (-1)^n = -1 \][/tex]
Now, substituting back:
[tex]\[ b_n = \frac{2k}{n\pi} \left(\frac{(-1)^n + 1}{n} \right) = \frac{2k}{n\pi} \left(\frac{(-1 + 1)}{n}\right) = \frac{2k}{n\pi} \left(\frac{0}{n} \right) \][/tex]
Thus, each specific Fourier coefficient can be calculated. However, the final Fourier series sum in its simplest form shows the series:
[tex]\[ k\left( 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots \right) \][/tex]
This series is recognized as the alternating series whose sum is known to involve the approximation \(\frac{\pi}{4}\).
Thus, the series sum:
[tex]\[ 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots \][/tex]
is equal to:
[tex]\[ \frac{\pi}{4} \][/tex]
Therefore, the correct answer is:
(c) [tex]\(\frac{\pi}{4}\)[/tex]
