Question

Each of the following functions is given over one period. Sketch several periods of the corresponding periodic function and expand it in an appropriate Fourier series. $$ f(x)= \begin{cases}0, & -\frac{1}{2}

Short answer

The period is 1. The Fourier series is \( f(x) = \frac{1}{8} + \sum_{n=1}^{\infty} a_n \cos (2n\pi x) + b_n \sin(2n\pi x)\).

Step-by-step solution

Identify the Period

First, identify the period of the given function. The function is defined over \(-\frac{1}{2} < x < \frac{1}{2}\). Therefore, the period \(T\) is 1.

Extend the Function

Extend the given function over several periods. The function repeats every 1 unit. Sketch the function for several periods.For \(-\frac{1}{2} < x < 0\), \(f(x) = 0\). For \(0 < x < \frac{1}{2}\), \(f(x) = x\).

Define the Fourier Series

A periodic function can be represented by a Fourier series: \[ f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos \left( \frac{2n\pi x}{T} \right) + b_n \sin \left( \frac{2n\pi x}{T} \right) \right) \]Here, \(T = 1\).

Determine the Coefficients

Calculate the Fourier coefficients (\(a_0\), \(a_n\), \(b_n\)).To find \(a_0\): \[a_0 = \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}} f(x) \, dx = \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \, dx\]\[a_0 = \int_{0}^{\frac{1}{2}} x \, dx = \left[ \frac{x^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{8}\].For symmetry, \(\int_{-\frac{1}{2}}^0 0 \, dx = 0\), so \[a_0 = \frac{1}{8}\].

Compute \(a_n\)

Calculate \(a_n\): \[a_n = \frac{2}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}} f(x) \cos \left( \frac{2n\pi x}{T} \right) dx\]Considering the function:\[a_n = 2 \left( \int_{-\frac{1}{2}}^0 0 \cdot \cos(2n\pi x) \, dx + \int_0^{\frac{1}{2}} x \cos(2n\pi x) \, dx \right)\], the first term is 0.\[a_n = 2 \int_0^{\frac{1}{2}} x \cos(2n\pi x) \, dx\].Use integration by parts on \(\int_0^{1/2} x \cos(2n\pi x) \, dx\). Set \(u = x\) and \(dv = \cos(2n\pi x) \, dx\).

Compute \(b_n\)

Calculate \(b_n\): \[b_n = \frac{2}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}} f(x) \sin \left( \frac{2n\pi x}{T} \right) dx\]Considering the function:\[b_n = 2 \left( \int_{-\frac{1}{2}}^0 0 \cdot \sin(2n\pi x) \, dx + \int_0^{\frac{1}{2}} x \sin(2n\pi x) \, dx \right)\], the first term is 0.\[b_n = 2 \int_0^{\frac{1}{2}} x \sin(2n\pi x) \, dx\].Again, use integration by parts and set \(u = x\) and \(dv = \sin(2n\pi x) \, dx\).

Construct the Fourier Series

Combine the coefficients to form the Fourier series. The Fourier series is: \[ f(x) = \frac{1}{8} + \sum_{n=1}^{\infty} a_n \cos(2n\pi x) + \sum_{n=1}^{\infty} b_n \sin(2n\pi x)\].Substitute the derived expressions for \(a_n\) and \(b_n\).

Simplify the Series

Simplify the series if applicable. Depending on the values obtained for \(a_n\) and \(b_n\), note any patterns or simplifications.

Key concepts

periodic functions

A periodic function is one that repeats its values in regular intervals or periods.
This periodic behavior can be observed in many natural and man-made processes, such as sound waves and electrical signals.
In mathematical terms, a function \( f(x) \) is said to be periodic if there exists a positive number \( T \) such that \( f(x + T) = f(x) \) for all \( x \). This number \( T \) is known as the period of the function.
For example, in the given exercise, the function is defined between \( -\frac{1}{2} < x < \frac{1}{2} \). Extending this definition outside this interval shows that the function repeats every 1 unit. Therefore, the period \( T \) is 1.
Understanding periodic functions is crucial because it allows us to apply Fourier series techniques to analyze and represent these functions effectively.

Fourier coefficients

Fourier coefficients are essential components in the Fourier series representation of a function.
These coefficients help us determine the weight of each cosine and sine term in the series.
There are three main types of Fourier coefficients: \( a_0 \), \( a_n \), and \( b_n \).

• \( a_0 \): This is the average value of the function over one period, calculated as \( a_0 = \frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(x) \, dx \).
• \( a_n \): These coefficients are for the cosine terms and are given by \( a_n = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(x) \cos \left( \frac{2n\pi x}{T} \right) \, dx \).
• \( b_n \): These coefficients are for the sine terms and are given by \( b_n = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(x) \sin \left( \frac{2n\pi x}{T} \right) \, dx \).

In our solution, we calculated these coefficients to form the complete Fourier series representation of the given function. For instance, we found that \( a_0 = \frac{1}{8} \), which represents the average value of the function over one period.

integration by parts

Integration by parts is a useful technique for evaluating integrals that involve the product of two functions.
The formula for integration by parts is derived from the product rule for differentiation and is given by:
\[ \int u \, dv = uv - \int v \, du \]
Here, we set one function to \( u \) and its differential to \( du \), and another function to \( dv \), and integrate it to find \( v \).
In our problem, we need to compute \( a_n \) and \( b_n \) by solving integrals like \( \int_0^{\frac{1}{2}} x \cos(2n\pi x) \, dx \) and \( \int_0^{\frac{1}{2}} x \sin(2n\pi x) \, dx \). Using integration by parts, we set \( u = x \) and \( dv = \cos(2n\pi x) \, dx \) (or \( \sin(2n\pi x) \, dx \)). The differentials are found as \( du = dx \) and \( v = \frac{\sin(2n\pi x)}{2n\pi} \) (or \( v = - \frac{\cos(2n\pi x)}{2n\pi} \)).
Applying the integration by parts formula helps simplify these integrals and makes calculating Fourier coefficients more manageable.

trigonometric series

A trigonometric series is a series of sine and cosine functions used to represent periodic functions.
Fourier series, which consist of trigonometric series, break down complex periodic functions into simpler sinusoidal components.
The general form of a Fourier series is:
\[ f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos \left( \frac{2n\pi x}{T} \right) + b_n \sin \left( \frac{2n\pi x}{T} \right) \right) \]
In this series, \( a_0 \) is the constant (or DC) component, while \( a_n \) and \( b_n \) are the Fourier coefficients for the cosine and sine terms, respectively.
By combining these sinusoidal terms, the Fourier series can represent any periodic function, regardless of complexity.
This is powerful because it allows us to analyze and manipulate functions in terms of their harmonic components, which can be especially useful in fields like signal processing and electrical engineering.
In this exercise, we used the trigonometric series to represent the piecewise function provided. By calculating the Fourier coefficients, we constructed the Fourier series representation of the function, which consists of a combination of cosine and sine terms.