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Magazine Chapter 12: Problem 2
In Problems, find the Fourier series of \(f\) on the given interval.
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f(x)=\left\\{\begin{array}{lr}
-1, & -\pi
Short Answer
Expert verified
The Fourier series is \( f(x) = \frac{1}{2} + \sum_{n=1, \, n\text{ odd}}^{\infty} \frac{6}{n\pi} \sin(nx) \).
Step by step solution
01
Define the Fourier Series
The Fourier series of a function \( f(x) \) on the interval \( [-L, L] \) is given by: \[ f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos \frac{n\pi x}{L} + b_n \sin \frac{n\pi x}{L} \right) \] where, \[ a_0 = \frac{1}{2L} \int_{-L}^{L} f(x) \, dx \] \[ a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos \frac{n\pi x}{L} \, dx \] \[ b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin \frac{n\pi x}{L} \, dx \] For the given problem, the interval is \( [-\pi, \pi] \), so \( L = \pi \).
02
Calculate \(a_0\)
Calculate \( a_0 \) using the formula:\[ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx \]Substitute \( f(x) \) into the integral:\[ a_0 = \frac{1}{2\pi} \left( \int_{-\pi}^{0} (-1) \, dx + \int_{0}^{\pi} 2 \, dx \right) \]This simplifies to:\[ a_0 = \frac{1}{2\pi} \left( [-x]_{-\pi}^{0} + [2x]_{0}^{\pi} \right) \]\[ a_0 = \frac{1}{2\pi} ((0 - \pi) + (2\pi - 0)) = \frac{1}{2\pi} \pi = \frac{1}{2} \]
03
Calculate \(a_n\)
Calculate \( a_n \) using the formula: \[ a_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} (-1) \cos \frac{n\pi x}{\pi} \, dx + \int_{0}^{\pi} 2 \cos \frac{n\pi x}{\pi} \, dx \right) \]For the first integral:\[ \int_{-\pi}^{0} (-1) \cos(n x) \, dx = -\left[ \frac{\sin(nx)}{n} \right]_{-\pi}^{0} = 0 \]For the second integral:\[ \int_{0}^{\pi} 2 \cos(nx) \, dx = 2\left[ \frac{\sin(nx)}{n} \right]_{0}^{\pi} = 0 \]Thus, \( a_n = 0 \) for all \( n \geq 1 \).
04
Calculate \(b_n\)
Calculate \( b_n \) using the formula: \[ b_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} (-1) \sin \frac{n\pi x}{\pi} \, dx + \int_{0}^{\pi} 2 \sin \frac{n\pi x}{\pi} \, dx \right) \]For the first integral:\[ \int_{-\pi}^{0} (-1) \sin(nx) \, dx = \left[ \frac{\cos(nx)}{n} \right]_{-\pi}^{0} = \frac{1 - (-1)^n}{n} \]For the second integral:\[ \int_{0}^{\pi} 2 \sin(nx) \, dx = 2\left[ -\frac{\cos(nx)}{n} \right]_{0}^{\pi} = \frac{2((-1)^{n+1} - 1)}{n} \]Thus,\[ b_n = \frac{1}{\pi} \left( \frac{1 - (-1)^n}{n} + \frac{2((-1)^{n+1} - 1)}{n} \right) = \begin{cases} 0, & \text{if } n \text{ is even} \ \frac{6}{n\pi}, & \text{if } n \text{ is odd} \end{cases} \]
05
Write the Fourier Series
Using the calculated values, the Fourier series becomes:\[ f(x) = \frac{1}{2} + \sum_{n=1, \, n\, \text{odd}}^{\infty} \frac{6}{n\pi} \sin(nx) \]This includes only sine terms for odd \( n \) because \( a_n = 0 \) and even \( n \) leads to \( b_n = 0 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Series
The Fourier series is an essential tool in mathematics that expresses a function as the sum of sine and cosine terms, known as a trigonometric series. These series transform complex functions into simpler, periodic forms, allowing for easier analysis and problem-solving.
- The general form of a Fourier series for a function \( f(x) \), defined on an interval \([-L, L]\), is:
- \(a_0\) is the average or mean value of the function over one period, calculated using constant integration.
- \(a_n\) pertains to the cosine terms, emphasizing how much cosine contributes to the function.
- \(b_n\) is associated with the sine terms, illustrating the function's sine component.
- Each term holds specific coefficients, \(a_0, a_n,\) and \(b_n\), calculated through integral formulas.
Orthogonality of Sine and Cosine
In the Fourier series, orthogonality refers to how sine and cosine functions work together matter-of-factly. This property significantly simplifies the calculations of the coefficients \(a_n\) and \(b_n\).
- Sine and cosine functions are orthogonal over an interval \([-L, L]\), meaning their integral products over this interval yield zero, whenever they aren't the same frequency.
- \( \int_{-L}^{L} \cos(nx) \cos(mx) \, dx = 0, \) for \( n eq m \)
- \( \int_{-L}^{L} \sin(nx) \sin(mx) \, dx = 0, \) for \( n eq m \)
- \( \int_{-L}^{L} \sin(nx) \cos(mx) \, dx = 0 \)
Piecewise Functions
Piecewise functions are defined by different expressions in distinct intervals, a common format for real-world applications. The function \(f(x)\) given in the original exercise is an example of this. It takes on different values based on the interval of \(x\):
- \(f(x) = -1\) for \(-\pi < x < 0\)
- \(f(x) = 2\) for \(0 \leq x < \pi\)
- The Fourier series of a piecewise function potentially converges differently at these discontinuities, often tending towards the average of the limits from either side due to Gibbs phenomenon.
