Chapter 23: Problem 21
Obtain the Fourier series over the indicated interval for the given function. Always sketch the function that is the sum of the series obtained. \(\begin{aligned} \text { Interval, }-c< x< c ; \text { function, } f(x) &=c+x, &-c< x< 0 \\ &=0, & 0< x< c \end{aligned}\)
Short Answer
Expert verified
The Fourier series is comprised of constant \(a_0 = \frac{c}{4}\) and cosine terms for odd \(n\). The series oscillates with more terms.
Step by step solution
01
Define the function and interval
The function is defined as follows:For \(-c < x < 0, \) the function is \(f(x) = c + x.\) For \(0 < x < c, \) the function is \(f(x) = 0. \)The interval is \(-c < x < c.\)
02
Set up the Fourier series formula
The Fourier series is given by:\[f(x) = a_0 + \sum_{n=1}^{\infty} \left(a_n \cos\left(\frac{n\pi x}{c}\right) + b_n \sin\left(\frac{n\pi x}{c}\right)\right)\]where \(a_0, a_n,\) and \(b_n\) are the Fourier coefficients we need to determine.
03
Calculate the constant term \(a_0\)
The constant term is calculated using:\[a_0 = \frac{1}{2c} \int_{-c}^{c} f(x) \, dx\]Considering the piecewise nature of the function:\[a_0 = \frac{1}{2c} \left(\int_{-c}^{0} (c + x) \, dx + \int_{0}^{c} 0 \, dx\right) = \frac{1}{2c} \int_{-c}^{0} (c + x) \, dx\]Solving this integral:\[a_0 = \frac{1}{2c} \left[cx + \frac{x^2}{2}\right]_{-c}^{0} = \frac{1}{2c} \left((0) - (-\frac{c^2}{2})\right) = \frac{c}{4}\]
04
Calculate the coefficients \(a_n\)
The cosine coefficients \(a_n\) are calculated using:\[a_n = \frac{1}{c} \int_{-c}^{c} f(x) \cos\left(\frac{n \pi x}{c}\right) \, dx\]This becomes:\[a_n = \frac{1}{c} \left(\int_{-c}^{0} (c + x) \cos\left(\frac{n \pi x}{c}\right) \, dx + \int_{0}^{c} 0 \cdot \cos\left(\frac{n \pi x}{c}\right) \, dx\right) \]\[= \frac{1}{c} \int_{-c}^{0} (c + x) \cos\left(\frac{n \pi x}{c}\right) \, dx\]Solving this involves integrating by parts and using trigonometric identities, leading to:\(a_n = \frac{2c}{n^2\pi^2}(1 - (-1)^n) \)This coefficient is zero for even \(n\), and simplifies to:\[a_n = \frac{4c}{n^2\pi^2}\text{ for odd } n\] after integrating.
05
Calculate the coefficients \(b_n\)
The sine coefficients \(b_n\) are calculated using:\[b_n = \frac{1}{c} \int_{-c}^{c} f(x) \sin\left(\frac{n \pi x}{c}\right) \, dx\]This separation also simplifies to:\[b_n = \frac{1}{c} \int_{-c}^{0} (c + x) \sin\left(\frac{n \pi x}{c}\right) \, dx\]Given \(f(x)\) is an even function over the symmetrical interval around zero, \(b_n\) comes out to zero after the calculations, indicating no sine component:\(b_n = 0\text{ for all }n\).
06
Sketch the function from the Fourier Series
The Fourier series representation of \(f(x)\) is:\[ f(x) = \frac{c}{4} + \sum_{n=1, \text{odd}}^{\infty} \frac{4c}{n^2\pi^2} \cos\left(\frac{n\pi x}{c}\right)\] Sketching this series involves plotting terms to see how the sum approaches the step-like function defined by \(f(x)\). This plotting typically results in oscillations (Gibbs phenomenon) near the discontinuities at \(x = 0, \pm c\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Fourier Coefficients
The Fourier coefficients are essential in constructing the Fourier series of a function. In the given exercise, the function is defined piecewise, and finding the Fourier coefficients will help convert this function into a sum of sine and cosine terms. Let's break down what these coefficients mean and how they are calculated:
- **Constant Term \(a_0\):** This term represents the average value of the function over the interval. It's computed using the integral \(a_0 = \frac{1}{2c} \int_{-c}^{c} f(x) \, dx\). Here, the result was \(\frac{c}{4}\), indicating that the average height of the function is \(\frac{c}{4}\).
- **Cosine Coefficients \(a_n\):** These coefficients reflect the function's even symmetry and are found using \(a_n = \frac{1}{c} \int_{-c}^{c} f(x) \cos\left(\frac{n \pi x}{c}\right) \, dx\). For our piecewise function, these coefficients turn out to be zero for even \(n\) and calculated as \(\frac{4c}{n^2\pi^2}\) for odd \(n\).
- **Sine Coefficients \(b_n\):** Since the function is even, all \(b_n\) coefficients are zero, implying no sine components are necessary to represent these functions.
Piecewise Functions
A piecewise function is defined by different expressions at different intervals for the domain. These functions are useful when a single formula cannot describe the whole function across its interval.
The function in this exercise is an example of a piecewise function where:
Piecewise functions represent scenarios like electrical signal changes, stock prices over specific periods, or other real-world systems where conditions change within the observed timeframe.
The function in this exercise is an example of a piecewise function where:
- For \(-c < x < 0\), the function is \(f(x) = c + x\).
- For \(0 < x < c\), the function is simply zero: \(f(x) = 0\).
Piecewise functions represent scenarios like electrical signal changes, stock prices over specific periods, or other real-world systems where conditions change within the observed timeframe.
Gibbs Phenomenon
Gibbs phenomenon describes the peculiar behavior that occurs when approximating a discontinuous function, such as a piecewise function, using a Fourier series. Particularly, it refers to the ripples or overshoots near the discontinuity points of a function.
When plotting the Fourier series of the given function \(f(x)\) (which includes discontinuities at \(x = 0\) and \(x = \pm c\)), you'll notice oscillations that do not vanish as more terms are taken. Instead, these oscillations tend to a specific height (~9% of the jump) as more terms are added.
Key aspects of Gibbs Phenomenon include:
When plotting the Fourier series of the given function \(f(x)\) (which includes discontinuities at \(x = 0\) and \(x = \pm c\)), you'll notice oscillations that do not vanish as more terms are taken. Instead, these oscillations tend to a specific height (~9% of the jump) as more terms are added.
Key aspects of Gibbs Phenomenon include:
- It does not disappear as more terms are added to the Fourier series, though the effects become narrower.
- The overshoot remains approximately the same magnitude, showing a fundamental limitation in representing discontinuous functions perfectly with a Fourier series.
- The presence of this effect highlights how Fourier series can converge to a function pointwise but not always uniformly at discontinuities.