Question

In each of the following problems you are given a function on the interval \(-\pi

Short answer

The Fourier series is \[f(x) = \frac{1}{4} + \sum_{n=1}^{\infty} -\frac{1}{n\pi} \left[(-1)^n - \cos \left(\frac{n\pi}{2} \right)\right] \sin(nx)\]

Step-by-step solution

- Understand the given function

The function is defined in two parts over the interval \footnotesize(-\pi < x < \pi):1. \(0\) for \(-\pi < x < \frac{\pi}{2}\)2. \(1\) for \(\frac{\pi}{2} < x < \pi\)

- Sketch several periods of the function

Since the function is periodic with period \(2\pi\), repeat the given function over multiple intervals of length \(2\pi\). This involves extending the definition of the function to all x by periodically repeating it.

- Determine the Fourier coefficients

Use the formulas for the sine-cosine Fourier series coefficients:\[a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx\]\[a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx\]\[b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx\]

- Calculate \(a_0\)

Compute \(a_0\): \[a_0 = \frac{1}{2\pi} \left( \int_{-\pi}^{\frac{\pi}{2}} 0 \, dx + \int_{\frac{\pi}{2}}^{\pi} 1 \, dx \right) = \frac{1}{2\pi} \left( 0 + \int_{\frac{\pi}{2}}^{\pi} 1 \, dx \right) = \frac{1}{2\pi} \left( \pi - \frac{\pi}{2} \right) = \frac{1}{2\pi} \cdot \frac{\pi}{2} = \frac{1}{4}\]

- Calculate \(a_n\)

Compute \(a_n\): \[a_n = \frac{1}{\pi} \left( \int_{-\pi}^{\frac{\pi}{2}} 0 \cos(nx) \, dx + \int_{\frac{\pi}{2}}^{\pi} 1 \cos(nx) \, dx \right) = \frac{1}{\pi} \int_{\frac{\pi}{2}}^{\pi} \cos(nx) \, dx\]\[a_n = \frac{1}{\pi} \left[\frac{\sin(nx)}{n} \right]_{\frac{\pi}{2}}^{\pi} = \frac{1}{\pi} \cdot \frac{\sin(n\pi) - \sin\left(\frac{n\pi}{2}\right)}{n}\]\(a_n = 0 \) for integer n.

- Calculate \(b_n\)

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

- Form the Fourier series

The Fourier series expansion of the function is:\[f(x) = \frac{1}{4} + \sum_{n=1}^{\infty} \left[0 \cdot \cos(nx) + b_n \cdot \sin(nx)\right]\]Since \(a_n = 0\), the Fourier series simplifies to:\[f(x) = \frac{1}{4} + \sum_{n=1}^{\infty} b_n \sin(nx)\]With \( b_n = -\frac{1}{n\pi} \left[ (-1)^n - \cos \left(\frac{n\pi}{2} \right) \right] \)

Key concepts

Periodic Function

A periodic function repeats its values at regular intervals over time. The given function of the exercise is a simple example of a periodic function with a period of \( 2\pi\). This means the function repeats its pattern every \( 2\pi\) units. To fully capture the behavior of this function, imagine extending it infinitely in both directions. This is done by replicating the provided interval \(-\pi < x < \pi\) repeatedly.
When sketching multiple periods of the function, ensure each segment between \(-\pi < x < \pi\) is identical. This repeated pattern highlights the essence of a periodic function. For instance, for the given function:

• \( f(x) = 0 \) when \(-\pi < x < \frac{\pi}{2}\)
• \( f(x) = 1 \) when \( \frac{\pi}{2} < x < \pi\)

The pattern will continue endlessly in both directions along the x-axis.

Fourier Coefficients

Fourier coefficients are essential in breaking down a periodic function into its sine and cosine components. These coefficients help us understand which sine and cosine waves make up our original function. They are represented as \( a_0, a_n, \) and \( b_n \) and are calculated using integrals over one period of the function.
The coefficients are:

• \( a_0 \) (the average value or DC component):\[ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx \]
• \( a_n \) (coefficients for cosine terms): \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx \]
• \( b_n \) (coefficients for sine terms): \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \]

In our exercise's context, we calculated \( a_0 \), \( a_n \), and \( b_n \) for the given piecewise function:
After some integration and simplification, we found:

• \( a_0 = \frac{1}{4} \)
• \( a_n = 0 \) for integer \( n \)
• \( b_n = -\frac{1}{n\pi} \left[ (-1)^n - \cos\left(\frac{n\pi}{2}\right) \right] \)

These coefficients are crucial for constructing the Fourier series of the function.

Sine-Cosine Series

The Fourier series is a way to represent a periodic function as a sum of sine and cosine terms. Each term in the series corresponds to a frequency component of the original function. The general form of the Fourier series expansion is:
\[ f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right) \]
In this exercise, given the calculations of the Fourier coefficients, the Fourier series for the function \( f(x)\) becomes:
\[ f(x) = \frac{1}{4} + \sum_{n=1}^{\infty} \left[ 0 \cdot \cos(nx) + b_n \cdot \sin(nx) \right] \]
Since \( a_n = 0 \), the series simplifies to:
\[ f(x) = \frac{1}{4} + \sum_{n=1}^{\infty} b_n \sin(nx) \]
Substituting \( b_n \):
\[ f(x) = \frac{1}{4} + \sum_{n=1}^\infty \left( -\frac{1}{n\pi} \left[ (-1)^n - \cos \left( \frac{n\pi}{2} \right) \right] \sin (nx) \right) \]
By analyzing the terms of this series, we can see how sine and cosine functions with different frequencies combine to form the original periodic function. This breakdown into sine and cosine components allows better understanding and analysis of the function’s behavior, especially in applications such as signal processing and solving differential equations.