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Magazine Chapter 23: Problem 22
Obtain the Fourier series over the indicated interval for the given function.
Always sketch the function that is the sum of the series obtained.
Interval, \(-c
Short Answer
Expert verified
The Fourier series is just the constant term \(\frac{c^4}{5}\).
Step by step solution
01
Understand the Given Interval and Function
The problem states that we need to find the Fourier series for the function \(f(x) = x^4\) over the interval \(-c < x < c\). This means that the periodicity of the function will be from \(-c\) to \(c\).
02
Determine the Fourier Series Coefficients
The Fourier series representation of a function \(f(x)\) over the interval \(-c < x < c\) is given by \(f(x) \sim a_0 + \sum_{n=1}^{\infty} \left( a_n \cos\left(\frac{n \pi x}{c}\right) + b_n \sin\left(\frac{n \pi x}{c}\right) \right)\). Here, \(a_0, a_n,\) and \(b_n\) are the Fourier coefficients we need to calculate.
03
Calculate \(a_0\)
The coefficient \(a_0\) is given by \(a_0 = \frac{1}{2c} \int_{-c}^{c} x^4 \, dx\). Evaluating the integral, we find:\[ a_0 = \frac{1}{2c} \left[ \frac{x^5}{5} \right]_{-c}^{c} = \frac{1}{2c} \left( \frac{c^5}{5} - \frac{(-c)^5}{5} \right) = \frac{2c^5}{5 \times 2c} = \frac{c^4}{5}. \]
04
Calculate \(a_n\)
The coefficients \(a_n\) are given by \(a_n = \frac{1}{c} \int_{-c}^{c} x^4 \cos\left(\frac{n \pi x}{c}\right) \, dx\). Using integration by parts generally proves that for a polynomial times a cosine function, the coefficients \(a_n\) will be zero for even degree polynomials integrated over a symmetric interval like \(-c\) to \(c\), due to symmetry and cancellation of terms.
05
Calculate \(b_n\)
Similarly, \(b_n\) is given by \(b_n = \frac{1}{c} \int_{-c}^{c} x^4 \sin\left(\frac{n \pi x}{c}\right) \, dx\). Due to the even nature of \(x^4\) and the odd nature of the sine function, this integral results in \(b_n = 0\) for all \(n\).
06
Write the Resulting Fourier Series
Based on our calculations, the Fourier series for \(f(x) = x^4\) over the interval \(-c < x < c\) simplifies to:\[ f(x) \sim \frac{c^4}{5}. \]This shows that the function is approximated by its constant term.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Fourier coefficients
Fourier coefficients are the building blocks of a Fourier series. These coefficients allow us to convert a periodic function into a sum of sines and cosines, which can simplify complex calculations.
The Fourier series for a function \( f(x) \) over an interval like \(-c < x < c\)\, is given by \( f(x) \sim a_0 + \sum_{n=1}^{\infty} \( a_n \cos\left(\frac{n \pi x}{c}\right) + b_n \sin\left(\frac{n \pi x}{c}\right) \)\). Each term involves a Fourier coefficient, e.g., \( a_0\)\, \( a_n\), and \( b_n\).
To compute these coefficients:
The Fourier series for a function \( f(x) \) over an interval like \(-c < x < c\)\, is given by \( f(x) \sim a_0 + \sum_{n=1}^{\infty} \( a_n \cos\left(\frac{n \pi x}{c}\right) + b_n \sin\left(\frac{n \pi x}{c}\right) \)\). Each term involves a Fourier coefficient, e.g., \( a_0\)\, \( a_n\), and \( b_n\).
To compute these coefficients:
- \( a_0 \): Represents the average value of the function over the interval. It is calculated using the integral \( \frac{1}{2c} \int_{-c}^{c} f(x) \, dx \). For \( f(x) = x^4\), \( a_0 \) becomes \( \frac{c^4}{5}\).
- \( a_n \) and \( b_n \): Are derived using integrals involving \( \cos\left(\frac{n \pi x}{c}\right) \) and \( \sin\left(\frac{n \pi x}{c}\right) \), respectively.
Integration by parts
Integration by parts is a mathematical technique useful for integrating the product of two functions. It follows the formula:
To apply:
- \[ \int u \, dv = uv - \int v \, du \]
To apply:
- Choose \( u \) and \( dv \) from the integral. Typically, you set the polynomial part as \( u\).
- Use differentiation to get \( du \) and integration to find \( v \).
- Plug into the formula above to simplify the integral.
Symmetrical interval analysis
Symmetrical interval analysis is a powerful concept in Fourier series when evaluating functions over symmetric ranges such as \(-c < x < c\). These symmetrical intervals greatly simplify calculations because they exploit the properties of even and odd functions.
Here's how it works:
Here's how it works:
- Even functions, like \( x^4\), have a property that \( f(-x) = f(x) \), and odd functions have \( f(-x) = -f(x) \).
- When integrated over symmetric intervals, even functions result in non-zero values while odd functions cancel out, leading to zero.
- This specific symmetry makes calculations of Fourier series coefficients more straightforward, as seen with \( b_n = 0\) for our given function.
- The alternating properties often simplify series calculations and the understanding of how a Fourier series represents a function.
