Chapter 7: Problem 36
Show that the Fourier series representation of the function \(f(\theta)=|\sin \theta|\) is $$ f(\theta)=\frac{2}{\pi}-\frac{4}{\pi} \sum_{m=1}^{\infty} \frac{\cos 2 m \theta}{4 m^{2}-1} $$
Short Answer
Expert verified
The Fourier series is \( \frac{2}{\pi} - \frac{4}{\pi} \sum_{m=1}^{\infty} \frac{\cos(2m\theta)}{4m^2 - 1} \).
Step by step solution
01
Understand the problem
We need to find the Fourier series representation of the function \( f(\theta) = |\sin \theta| \). This function is periodic with period \( 2\pi \).
02
Write out the definition of Fourier series
The Fourier series for a function \( f(x) \) over a period \( 2\pi \) is given by \( a_0 + \sum_{n=1}^{\infty} \left(a_n \cos(nx) + b_n \sin(nx)\right) \), where the coefficients \( a_n \) and \( b_n \) are determined by specific integrals.
03
Determine \( a_0 \)
The term \( a_0 \) is calculated as follows:\[ a_0 = \frac{1}{\pi} \int_{0}^{2\pi} |\sin \theta| \, d\theta \]Since \(|\sin \theta|\) is symmetric about \(\theta = \pi/2\), integrate from 0 to \(\pi\) and double the result:\[ a_0 = \frac{2}{\pi} \int_{0}^{\pi} \sin \theta \, d\theta = \frac{2}{\pi} \left[-\cos \theta\right]_{0}^{\pi} = \frac{2}{\pi} \times 2 = \frac{4}{\pi} \]
04
Determine \( a_n \)
For \( a_n \) for \( n \geq 1 \), use:\[ a_n = \frac{1}{\pi} \int_{0}^{2\pi} |\sin \theta| \cos(n\theta) \, d\theta \]Again due to symmetry, restrict from 0 to \(\pi\):\[ a_n = \frac{2}{\pi} \int_{0}^{\pi} \sin \theta \cos(n\theta) \, d\theta \]Using the product-to-sum identities, solve the integral:\[ \sin \theta \cos(n\theta) = \frac{1}{2} \left(\sin((n+1)\theta) + \sin((n-1)\theta)\right) \]
05
Evaluate the integral for \( a_{2m} \)
For \( a_{2m} \) (since the series involves only even \( n = 2m \)), only cosine terms remain:\[ a_{2m} = \frac{2}{\pi} \int_{0}^{\pi} \sin \theta \cos(2m\theta) \, d\theta \] Integrate by parts or use orthogonality conditions of sine and cosine functions over \([0,\pi]\). This gives:\[ a_{2m} = \frac{-4}{\pi} \cdot \frac{1}{4m^2 - 1} \]
06
Determine \( b_n \)
Since \( |\sin \theta| \) is an even function, all \( b_n = 0 \). \[ b_n = \frac{1}{\pi} \int_{0}^{2\pi} |\sin \theta| \sin(n\theta) \, d\theta = 0 \] This is because the integral of an even function times an odd function over a symmetric interval is zero.
07
Compile the Fourier series
Combine all the coefficients obtained:- \( a_0 = \frac{4}{\pi} \)- Only even terms \( a_{2m} \):\[ f(\theta) = \frac{2}{\pi} + \sum_{m=1}^{\infty} a_{2m} \cos(2m\theta) = \frac{2}{\pi} - \frac{4}{\pi} \sum_{m=1}^{\infty} \frac{\cos(2m\theta)}{4m^2 - 1} \]
08
Conclusion
The Fourier series representation for \( f(\theta) = |\sin \theta| \) is verified as:\[ f(\theta) = \frac{2}{\pi} - \frac{4}{\pi} \sum_{m=1}^{\infty} \frac{\cos(2m\theta)}{4m^2 - 1} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration of trigonometric functions
Integrating trigonometric functions involves finding the area under a curve described by a trigonometric function. This is an essential skill in calculus, useful for solving many problems, like computing coefficients in a Fourier series. When dealing with functions like \( |\sin \theta| \), the absolute value affects the integration process. The absolute function \( |\sin \theta| \) is symmetric over its period, simplifying integration by allowing us to focus on half the interval and then double the result.
To integrate \( \sin \theta \) over the interval \([0, \pi]\), we calculate:
To integrate \( \sin \theta \) over the interval \([0, \pi]\), we calculate:
- \( \int \sin \theta \, d\theta = -\cos \theta \), which gives us the antiderivative of \( \sin \theta \).
- We evaluate this antiderivative at the bounds of integration: \([0, \pi]\).
- The result is \( [-\cos \theta]_0^\pi = [-\cos(\pi) + \cos(0)] = 2 \), meaning the integral over \([0, \pi]\) is 2.
Even and odd functions
Understanding even and odd functions is crucial in Fourier analysis because it simplifies the process of determining Fourier coefficients. An even function is symmetric about the y-axis, meaning that \( f(x) = f(-x) \). For example, \( \cos(x) \) is an even function. On the other hand, an odd function is symmetric about the origin, meaning that \( f(x) = -f(-x) \). A common example is \( \sin(x) \).
The function \( |\sin \theta| \) is even. This is important because it affects the values of the Fourier coefficients. For even functions, all \( b_n \) coefficients, which multiply \( \sin(n\theta) \) in the series, are zero:
The function \( |\sin \theta| \) is even. This is important because it affects the values of the Fourier coefficients. For even functions, all \( b_n \) coefficients, which multiply \( \sin(n\theta) \) in the series, are zero:
- The integration of an even function times an odd function over a symmetric interval results in zero.
- Thus, \( b_n = \frac{1}{\pi} \int_{0}^{2\pi} |\sin \theta| \sin(n\theta) \, d\theta = 0 \).
Orthogonality conditions in Fourier series
Orthogonality is a fundamental concept when working with Fourier series. It refers to the perpendicularity of sine and cosine functions in the context of inner products over a defined interval. This property greatly simplifies the calculation of coefficients in a Fourier series.
In a Fourier series, any two functions \( \cos(nx) \) and \( \sin(mx) \) are orthogonal over an interval \([0, 2\pi]\). This means:
In a Fourier series, any two functions \( \cos(nx) \) and \( \sin(mx) \) are orthogonal over an interval \([0, 2\pi]\). This means:
- \( \int_0^{2\pi} \cos(nx) \sin(mx) \, dx = 0 \) for any integer \( n eq m \).
- Similarly, \( \int_0^{2\pi} \cos(nx) \cos(mx) \, dx = 0 \) for any distinct integers \( n eq m \).