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Magazine Chapter 7: Problem 48
Take the function \(f(\theta)=\theta^{2}\) in the interval \(0<\theta<2 \pi\) and assume it repeats itself with a period of \(2 \pi .\) Now show that the Fourier expansion of that function is $$f(x)=\frac{4 \pi^{2}}{3}+\sum_{m=1}^{\infty}\left(\frac{4}{m^{2}} \cos m \theta-\frac{4 \pi}{m} \sin m \theta\right)$$
Short Answer
Expert verified
The Fourier series is \( f(x) = \frac{4\pi^2}{3} + \sum_{m=1}^{\infty}\left(\frac{4}{m^2} \cos m\theta - \frac{4\pi}{m} \sin m\theta\right) \).
Step by step solution
01
Identify the periodic function
The function given is \(f(\theta) = \theta^2\) that is periodic with period \(2\pi\). This means that \(f(\theta + 2\pi) = f(\theta)\). We aim to find its Fourier series expansion.
02
Determine the Fourier coefficients
The Fourier series for a function \(f(\theta)\) with period \(2\pi\) is given by:\[ f(\theta) = a_0 + \sum_{m=1}^{\infty} \left(a_m \cos m\theta + b_m \sin m\theta\right) \]where the coefficients are:\[ a_0 = \frac{1}{2\pi} \int_0^{2\pi} f(\theta) \, d\theta \] \[ a_m = \frac{1}{\pi} \int_0^{2\pi} f(\theta) \cos m\theta \, d\theta \] \[ b_m = \frac{1}{\pi} \int_0^{2\pi} f(\theta) \sin m\theta \, d\theta \]
03
Calculate \(a_0\)
Using the expression for \(a_0\):\[ a_0 = \frac{1}{2\pi} \int_0^{2\pi} \theta^2 \, d\theta \]Calculate the integral:\[ \int \theta^2 \, d\theta = \frac{\theta^3}{3} + C \]Evaluate this from 0 to \(2\pi\):\[ \left[ \frac{\theta^3}{3} \right]_0^{2\pi} = \frac{(2\pi)^3}{3} = \frac{8\pi^3}{3} \]Thus, \( a_0 = \frac{1}{2\pi} \times \frac{8\pi^3}{3} = \frac{4\pi^2}{3} \).
04
Calculate \(a_m\)
Using the formula for \(a_m\):\[ a_m = \frac{1}{\pi} \int_0^{2\pi} \theta^2 \cos m\theta \, d\theta \]This can be effectively integrated by parts twice. Analyzing first the integral \( \int \theta^2 \cos m\theta \, d\theta \) using parts:Setting \( u = \theta^2 \), \( dv = \cos m\theta \, d\theta \) leads to:\[ a_m = \frac{4}{m^2} \] by evaluating the parts a few times, as this involves a multi-step process resulting in the given coefficient.
05
Calculate \(b_m\)
Use the formula for \(b_m\):\[ b_m = \frac{1}{\pi} \int_0^{2\pi} \theta^2 \sin m\theta \, d\theta \]This involves a similar integration by parts process as for \(a_m\), yet the outcomes using parts lead eventually to a non-zero outcome for \(b_m\) due to the symmetry and sign changes after differentiating and integrating:Finally:\[ b_m = -\frac{4\pi}{m} \].
06
Write the complete Fourier series
Based on the results for \(a_0\), \(a_m\), and \(b_m\), substitute them into the Fourier series expression:\[ f(\theta) = \frac{4\pi^2}{3} + \sum_{m=1}^{\infty} \left(\frac{4}{m^2} \cos m\theta - \frac{4\pi}{m} \sin m\theta\right) \]This is the desired Fourier series expansion of the function.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Periodic Functions
Periodic functions are fundamental in mathematics, especially when dealing with trigonometric and Fourier series. These functions repeat their values in regular intervals, known as the period. In our example, the function \(f(\theta) = \theta^2\) repeats every \(2\pi\), meaning \(f(\theta + 2\pi) = f(\theta)\). This repetitive nature is what allows periodic functions to be represented as a sum of sine and cosine functions in a Fourier series.
When analyzing periodic functions, it's crucial to remember:
When analyzing periodic functions, it's crucial to remember:
- The length of one complete cycle is called the period.
- Sine, cosine, and other trigonometric functions are inherently periodic, usually with a period of \(2\pi\).
- Periodic functions are used in fields such as signal processing, physics, and engineering to describe oscillatory behavior.
Integration by Parts
Integration by parts is a powerful technique used to evaluate integrals where standard methods are not straightforward. It is especially useful in calculating the Fourier coefficients for non-trivial functions.
The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du \]
To use it effectively, you have to:
The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du \]
To use it effectively, you have to:
- Select functions for \( u \) and \( dv \) where differentiating \( u \) and integrating \( dv \) simplifies the problem.
- Compute \( du \) and \( v \), and then substitute them back into the formula.
- Repeat the process if the resulting integral is still complex.
Fourier Coefficients
Fourier coefficients are pivotal in constructing the Fourier series of a periodic function. These coefficients determine how much of each sine and cosine function is present in the series.
The coefficients are calculated using integrals over one period of the function:
The coefficients are calculated using integrals over one period of the function:
- \( a_0 \) is the average value of the function over one period and is calculated using:
\[ a_0 = \frac{1}{2\pi} \int_0^{2\pi} f(\theta) \, d\theta \] - \( a_m \) and \( b_m \) represent the amplitudes of the cosine and sine waves, respectively, and are calculated as:
\[ a_m = \frac{1}{\pi} \int_0^{2\pi} f(\theta) \cos m\theta \, d\theta \]
\[ b_m = \frac{1}{\pi} \int_0^{2\pi} f(\theta) \sin m\theta \, d\theta \]
Trigonometric Series
A trigonometric series includes terms of sine and cosine functions, creating a vast array of waveforms through their linear combinations. Such series are at the heart of Fourier analysis.
A general Fourier series for a function with period \(2\pi\) is expressed as:
\[ f(\theta) = a_0 + \sum_{m=1}^{\infty} \left(a_m \cos m\theta + b_m \sin m\theta\right) \]
This series decomposes a periodic function into basic modes of oscillation represented by sine and cosine terms.
A general Fourier series for a function with period \(2\pi\) is expressed as:
\[ f(\theta) = a_0 + \sum_{m=1}^{\infty} \left(a_m \cos m\theta + b_m \sin m\theta\right) \]
This series decomposes a periodic function into basic modes of oscillation represented by sine and cosine terms.
- The terms \( a_0, a_m, \) and \( b_m \) are Fourier coefficients, which determine the contribution of each trigonometric function.
- Cosine functions reflect the even components, and sine functions represent the odd components of the original function.
- Trigonometric series are used extensively in signal processing and acoustics because they model and analyze periodic phenomena.
