Solution

Function Analysis

The given function $ f(x) $ is periodic with period $ T = 4 $. This means the fundamental frequency is $ \omega_0 = \frac{2\pi}{T} = \frac{\pi}{2} $. The function is defined piecewise:

• $ f(x) = 0 $ if $ -2 < x < -1 $
• $ f(x) = 2k $ if $ -1 < x < 1 $
• $ f(x) = 0 $ if $ 1 < x < 2 $

Observing the definition, we see that $ f(-x) = f(x) $ for all $ x $ within the interval $ [-2, 2] $. This indicates that $ f(x) $ is an even function. Consequently, its Fourier series will consist only of a constant term ($ a_0 $) and cosine terms ($ a_n $); all sine coefficients ($ b_n $) will be zero.

Fourier Series Calculation

The general form of a Fourier series for a function with period $ T $ is $ f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(n\omega_0 x) + b_n \sin(n\omega_0 x) \right) $. Since $ f(x) $ is even, $ b_n = 0 $ for all $ n $, and the series simplifies to $ f(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos(n\omega_0 x) $.

Constant Term $a_0$

The constant term $ a_0 $ is calculated using the formula:

$ a_0 = \frac{1}{T} \int_{-T/2}^{T/2} f(x) dx $

Substituting the given values ($ T=4 $, $ L=T/2=2 $):

$ a_0 = \frac{1}{4} \int_{-2}^{2} f(x) dx $

Integrating over the defined intervals:

$ a_0 = \frac{1}{4} \left( \int_{-2}^{-1} 0 dx + \int_{-1}^{1} 2k dx + \int_{1}^{2} 0 dx \right) = \frac{1}{4} \int_{-1}^{1} 2k dx $

$ a_0 = \frac{1}{4} [2kx]_{-1}^{1} = \frac{1}{4} (2k(1) - 2k(-1)) = \frac{1}{4} (2k + 2k) = \frac{4k}{4} = k $

Cosine Coefficients $a_n$

The coefficients $ a_n $ are calculated as:

$ a_n = \frac{2}{T} \int_{-T/2}^{T/2} f(x) \cos(n\omega_0 x) dx $

$ a_n = \frac{2}{4} \int_{-2}^{2} f(x) \cos\left(\frac{n\pi}{2} x\right) dx $

Due to the even symmetry of $ f(x) $, we can simplify the integral range:

$ a_n = \frac{2}{2} \int_{0}^{2} f(x) \cos\left(\frac{n\pi}{2} x\right) dx = \int_{0}^{1} 2k \cos\left(\frac{n\pi}{2} x\right) dx $

Evaluating the integral:

$ a_n = 2k \left[ \frac{\sin(\frac{n\pi}{2} x)}{\frac{n\pi}{2}} \right]_0^1 = 2k \left[ \frac{2}{n\pi} \sin\left(\frac{n\pi}{2} x\right) \right]_0^1 $

$ a_n = \frac{4k}{n\pi} \left( \sin\left(\frac{n\pi}{2}\right) - \sin(0) \right) = \frac{4k}{n\pi} \sin\left(\frac{n\pi}{2}\right) $

Series Construction

The term $ \sin\left(\frac{n\pi}{2}\right) $ yields the sequence $ 1, 0, -1, 0, 1, \dots $ for $ n = 1, 2, 3, 4, 5, \dots $. Non-zero coefficients $ a_n $ occur only for odd values of $ n $.

The series expansion includes:

• For $ n=1 $: $ a_1 = \frac{4k}{\pi} \sin(\frac{\pi}{2}) = \frac{4k}{\pi} $
• For $ n=3 $: $ a_3 = \frac{4k}{3\pi} \sin(\frac{3\pi}{2}) = -\frac{4k}{3\pi} $
• For $ n=5 $: $ a_5 = \frac{4k}{5\pi} \sin(\frac{5\pi}{2}) = \frac{4k}{5\pi} $

Combining the constant term and the cosine series terms:

$ f(x) = k + \frac{4k}{\pi} \left( \cos \frac{\pi}{2} x - \frac{1}{3} \cos \frac{3\pi}{2} x + \frac{1}{5} \cos \frac{5\pi}{2} x -+ ... \right) $

This result matches option C.