Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 7: Problem 22
You are given \(f(x)\) on an interval, say \(0
Short Answer
Expert verified
Identify the even, odd, and periodic extension functions for the given interval and then find the Fourier expansion for each using the respective series formulas.
Step by step solution
01
- Define the Even Function
The even function, denoted as \( f_{c}(x) \), has period \( 2b \) and mirrors the values of \( f(x) \) for \( x > b \) back to \( -b < x < 0 \). For \( f(x) \) defined as \[ f(x)= \begin{cases} 10, & 0<x<10 \ 20, & 10<x<20 \end{cases} \]and period \( b = 30 \), the values for the even function are: \( f_{c}(x) = f(|x|) \), so on the interval \( [0, 60] \): \[ f_c(x) = \begin{cases} 10, & 0 < x < 10 \ 20, & 10 < x < 20 \ 20, & 20 < x < 30 \ 10, & 30 < x < 40 \ 10, & 40 < x < 50 \ 20, & 50 < x < 60 \end{cases} \]
02
- Define the Odd Function
The odd function, denoted as \( f_{s}(x) \), has period \( 2b \) and reflects \( f(x) \) around the origin to give: \( f_{s}(x) = -f(-x) \) for \( x < 0 \). For \( f(x) \) defined above: \( b = 30 \): the values on the interval \( [-30, 30] \): \[ f_s(x) = \begin{cases} 10, & 0 < x < 10 \ 20, & 10 < x < 20 \ 20, & 20 < x < 30 \ -10, & -10 < x < 0 \ -20, & -20 < x < -10 \ -20, & -30 < x < -20 \end{cases} \]
03
- Define the periodic extension with period \( b \)
The periodic function \( f_{p}(x) \), with period \( b \), simply extends \( f(x) \) to the entire real line with period \( b \). For \( f(x) \) defined above: \( b = 20 \) \[ f_p(x) = f(x \, \text{mod} \, 20) = \begin{cases} 10, & x \, \text{mod} \, 20 < 10 \ 20, & x \, \text{mod} \, 20 > 10 \end{cases} \]
04
- Fourier Series for Even Function
To find the Fourier series of \( f_c(x) \): Expand in cosine terms only since it's an even function. \[ f_c(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos\left( \frac{n\pi x}{b} \right) \] Compute \( a_0 \) and \( a_n \): \[ a_0 = \frac{2}{2b} \int_{0}^{b} f_c(x) \, dx \] \[ a_n = \frac{2}{2b} \int_{0}^{b} f_c(x) \cos\left( \frac{n\pi x}{b} \right) \, dx \]
05
- Fourier Series for Odd Function
To find the Fourier series of \( f_s(x) \): Expand in sine terms only since it's an odd function. \[ f_s(x) = \sum_{n=1}^{\infty} b_n \sin\left( \frac{n\pi x}{b} \right) \] Compute \( b_n \): \[ b_n = \frac{2}{2b} \int_{0}^{b} f_s(x) \sin\left( \frac{n\pi x}{b} \right) \, dx \]
06
- Fourier Series for Periodic Extension \( f_p(x) \)
To find the Fourier series of \( f_p(x) \): Expand in combination of sine and cosine terms. \[ f_p(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos\left( \frac{2n\pi x}{b} \right) + b_n \sin\left( \frac{2n\pi x}{b} \right) \right) \] Compute \( a_0 \), \( a_n \), and \( b_n \): \[ a_0 = \frac{1}{b} \int_{0}^{b} f_p(x) \, dx \] \[ a_n = \frac{2}{b} \int_{0}^{b} f_p(x) \cos\left( \frac{2n\pi x}{b} \right) \, dx \] \[ b_n = \frac{2}{b} \int_{0}^{b} f_p(x) \sin\left( \frac{2n\pi x}{b} \right) \, dx \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
even function
An even function is one that remains unchanged when its argument is negated. Mathematically, this means that for any even function \(f(x)\), it satisfies \(f(-x) = f(x)\). These functions are symmetric about the y-axis. In the context of Fourier series, even functions are expanded using only cosine terms because the cosine function itself is even. For the given problem, the even extension would be defined by reflecting the function about the y-axis and repeating over the interval since it has a periodicity of \(2b\).
odd function
An odd function changes sign when its argument is negated, meaning \(f(-x) = -f(x)\). These functions are symmetric about the origin. When dealing with Fourier series, odd functions are expanded using only sine terms, as sine functions are inherently odd. For the specified interval, the odd extension involves reflecting the function across the y-axis, flipping its values to negative in the corresponding interval. This reflection and flipping create a repeating pattern of period \(2b\).
periodic function
A periodic function repeats its values in regular intervals or periods. The mathematical definition is \(f(x + P) = f(x)\) for a period \(P\). In our scenario, the periodic extensions are based on the given piecewise function, either repeating every interval \(b\) or \(2b\). Thus forming the foundation for constructing the Fourier series by defining the pieces across repeated intervals.
Fourier coefficients
Fourier coefficients are constants that multiply the sine and cosine terms in the Fourier series expansion. These coefficients, \(a_n\) and \(b_n\), are determined using integrals of the function over one period. For an even function, only cosine terms with coefficients \(a_n\) are needed. For an odd function, only sine terms with coefficients \(b_n\) are necessary. For arbitrary periodic functions, both sine and cosine terms are employed in the series.
cosine series
A cosine series is a series expansion using only cosine terms. This is typically employed for expanding even functions, which are symmetrical about the y-axis. The general form of a cosine Fourier series is:
\( f(x) = a_0 + \sum_{n=1}^{\tiny \frac{\tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny 1}{2}}a_n \cos\frac{n\pi x}{b}\)
\( f(x) = a_0 + \sum_{n=1}^{\tiny \frac{\tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny \tiny 1}{2}}a_n \cos\frac{n\pi x}{b}\)
sine series
A sine series uses only sine terms and is ideal for representing odd functions, which are symmetrical about the origin. The form of the sine Fourier series is:
\( f(x) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{b}\right)\).
These expansions do not include a constant term but solely the sine terms weighted by their coefficients \(b_n\).
\( f(x) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{b}\right)\).
These expansions do not include a constant term but solely the sine terms weighted by their coefficients \(b_n\).
