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Magazine Chapter 7: Problem 22
Find the third-order Fourier approximation to fon \([-\pi, \pi]\). $$f(x)=x^{2}$$
Short Answer
Expert verified
The third-order Fourier approximation is \( f(x) \approx \frac{\pi^2}{3} - 4\cos(x) - \frac{1}{2}\cos(2x) - \frac{4}{9}\cos(3x) \).
Step by step solution
01
Understand the Fourier Series
The Fourier series approximates a function as a sum of sine and cosine terms. For a function \( f(x) \) defined on \([-\pi, \pi]\), the Fourier series can be written as: \( a_0 + \sum_{{n=1}}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right) \), where \( a_n \) and \( b_n \) are the Fourier coefficients.
02
Calculate the Constant Term \(a_0\)
The constant term \( a_0 \) is calculated as \( a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} x^2 \, dx \). To solve this: 1. Calculate the definite integral: \( \int x^2 \, dx = \frac{x^3}{3} \). 2. Evaluate from \(-\pi\) to \(\pi\): \( \frac{\pi^3}{3} - \left(-\frac{\pi^3}{3}\right) = \frac{2\pi^3}{3} \). 3. Substitute in the formula for \( a_0 \): \( a_0 = \frac{1}{2\pi} \times \frac{2\pi^3}{3} = \frac{\pi^2}{3} \).
03
Calculate \(a_n\) Coefficients
The coefficients \( a_n \) are found using the formula: \( a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cos(nx) \, dx \). This integral evaluation becomes a bit complex as we need integration by parts. For our purpose, the even nature of \( f(x) \) and \( \cos(nx) \) simplifies to: For \( n \ge 1 \), \( a_n = (-1)^n \frac{4}{n^2} \).
04
Calculate \(b_n\) Coefficients
Since \( f(x) = x^2 \) is even, the sine terms \( b_n \) are zero for all \( n \). This is because the integral of the product of an even function and an odd function over a symmetric interval is zero: \( b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \sin(nx) \, dx = 0 \).
05
Assemble the Third-Order Fourier Approximation
For the third-order approximation, we use terms up to \( n = 3 \). This includes the constant term \( a_0 \) and the three cosine terms:\[ f(x) \approx \frac{\pi^2}{3} + \left(-\frac{4}{1^2}\right) \cos(x) + \left(-\frac{4}{2^2}\right) \cos(2x) + \left(-\frac{4}{3^2}\right) \cos(3x) \]This simplifies to:\[ f(x) \approx \frac{\pi^2}{3} - 4\cos(x) - \frac{1}{2}\cos(2x) - \frac{4}{9}\cos(3x) \]
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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 at the heart of decomposing a function into a series of sinusoidal components. When dealing with a Fourier series, we typically work with coefficients labeled as \( a_0 \), \( a_n \), and \( b_n \). Each of these coefficients plays a different role in the approximation of a function:
- **\( a_0 \)**: This is the constant term which accounts for the average value of the function over one period. It is computed using the integral \( a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx \).
- **\( a_n \) Coefficients**: These coefficients correspond to the cosine terms in the series. They help in capturing the even components of the function. For our specific function \( f(x) = x^2 \), we found that \( a_n = (-1)^n \frac{4}{n^2} \) using integral calculus techniques.
- **\( b_n \) Coefficients**: These coefficients correspond to the sine terms. For functions that are symmetric (even functions), like \( x^2 \), the sine terms contribute no value, i.e., \( b_n = 0 \).
integral calculus
Integral calculus is essential when determining Fourier coefficients. It involves finding the area under the curve of a function, which effectively lets us measure different components of the function we want to approximate. The Fourier series requires evaluating definite integrals in this context:
- For the constant term \( a_0 \), calculate \( \frac{1}{2\pi} \int_{-\pi}^{\pi} x^2 \, dx \). Here, integral calculus gives \( \frac{\pi^2}{3} \) as the coefficient, indicating the series' baseline.
- To solve for \( a_n \), we use the integral \( \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cos(nx) \, dx \). This task can become intricate, necessitating tools like integration by parts to handle the multiplication of polynomials and trigonometric functions efficiently.
- The evaluation of \( b_n \) through \( \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \sin(nx) \, dx \) simplifies significantly for even functions, due to symmetry properties, yielding zero.
trigonometric approximation
Trigonometric approximation via Fourier series transforms a complex periodic function into a series of simpler wave functions. This approach simplifies analysis, especially in engineering and physics applications. For our function \( f(x) = x^2 \), this process involved:
- Calculating the Fourier coefficients using integral calculus to get specific values for \( a_0 \), \( a_n \), and \( b_n \).
- Constructing the Fourier approximation by summing these coefficients with their sine and cosine counterparts. For a third-order approximation, only a limited number of terms (up to \( n = 3 \)) are included.
- Utilizing the resulting expression \( \frac{\pi^2}{3} - 4\cos(x) - \frac{1}{2}\cos(2x) - \frac{4}{9}\cos(3x) \) to approximate the original function \( x^2 \).
