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Mobile App Chapter 7: Problem 19
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
Expert verified
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
01
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.
02
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 \(0 < x < \frac{1}{2}\), \(f(x) = x\).
03
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, \(T = 1\).
04
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 \[a_0 = \frac{1}{8}\].
05
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 \(dv = \cos(2n\pi x) \, dx\).
06
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 \(dv = \sin(2n\pi x) \, dx\).
07
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 \(b_n\).
08
Simplify the Series
Simplify the series if applicable. Depending on the values obtained for \(a_n\) and \(b_n\), note any patterns or simplifications.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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 \( b_n \).
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 \( b_n = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(x) \sin \left( \frac{2n\pi x}{T} \right) \, dx \).
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.
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.
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.
