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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:
  • \( \int \sin \theta \, d\theta = -\cos \theta \), 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.
By understanding these integration techniques, calculating coefficients for Fourier series becomes more straightforward.
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 integration of an even function times an odd function over a symmetric interval results in zero.
  • Thus, \( b_n = \frac{1}{\pi} \int_{0}^{2\pi} |\sin \theta| \sin(n\theta) \, d\theta = 0 \).
This knowledge reduces the complexity of the Fourier series and eliminates unnecessary calculations.
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:
  • \( \int_0^{2\pi} \cos(nx) \sin(mx) \, dx = 0 \) for any integer \( n eq m \).
  • Similarly, \( \int_0^{2\pi} \cos(nx) \cos(mx) \, dx = 0 \) for any distinct integers \( n eq m \).
The orthogonality conditions lead to simple evaluations of integrals where cross terms vanish, allowing us to solve for coefficients like \( a_n \) efficiently. For \( |\sin \theta| \), this principle allows only specific terms to survive, as seen in this exercise where only every second \( a_n \) is non-zero due to symmetry and periodicity conditions. This results in a Fourier series that exclusively contains cosine terms with certain values of \( n \).

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