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Chapter 12: Problem 8

$$\begin{array}{l} a_{0}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) d x=\frac{1}{\pi} \int_{-\pi}^{\pi}(3-2 x) d x=6 \\ a_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x=\frac{1}{\pi} \int_{-\pi}^{\pi}(3-2 x) \cos n x d x=0 \\ b_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi}(3-2 x) \sin n x d x=\frac{4}{n}(-1)^{n} \\\ f(x)=3+4 \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \sin n x \end{array}$$

Short Answer

Expert verified
The Fourier series is: \( f(x) = 3 + 4 \sum_{n=1}^\infty \frac{(-1)^n}{n} \sin(nx) \).

Step by step solution

01

Calculate a鈧�

To find the Fourier coefficient \( a_0 \), we need to evaluate the integral: \[a_{0} = \frac{1}{\pi} \int_{-\pi}^{\pi} (3-2x) \, dx .\]The integral splits into two parts:\[\frac{1}{\pi} \int_{-\pi}^{\pi} 3\, dx - \frac{2}{\pi} \int_{-\pi}^{\pi} x\, dx.\]The first part is simple: \(\frac{1}{\pi}[3x]_{-\pi}^{\pi} = \frac{1}{\pi} [6\pi] = 6.\) The second part evaluates to zero because the antiderivative of \(x\) over a symmetric interval around zero results in zero:\[\frac{2}{\pi}\left[\frac{x^2}{2}\right]_{-\pi}^{\pi} = 0.\] Thus, \( a_0 = 6. \)
02

Calculate an for n 鈮� 1

For each \( n \geq 1 \), the Fourier cosine coefficient \( a_n \) is given by:\[a_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} (3 - 2x) \cos(nx) \, dx.\]Due to the fact that cosines of different frequencies are orthogonal, this integral evaluates to zero. This simplification could involve checking that the integral of either a constant times a cosine or a linear term times a cosine over a full period is zero:\[a_{n} = 0.\] This is confirmed as zero for all \( n \geq 1. \)
03

Calculate bn for n 鈮� 1

Now, let's find the Fourier sine coefficients \( b_n \) using:\[b_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} (3 - 2x) \sin(nx) \, dx.\]Calculate separately:\[b_{n} = \frac{3}{\pi} \int_{-\pi}^{\pi} \sin(nx) \, dx - \frac{2}{\pi} \int_{-\pi}^{\pi} x \sin(nx) \, dx.\]The first integral evaluates to zero as the integral of sine over a complete period is zero. The second integral simplifies to a known result wherein integration by parts shows:\[b_n = \frac{4}{n} (-1)^{n}.\]
04

Write the Fourier Series Representation

Now that we have all coefficients, the Fourier series representation of the function \( f(x) = 3 - 2x \) over \([ -\pi, \pi ]\) is:\[f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right).\]Substitute in the values we have: \[a_0=6, \quad a_n=0 \, \forall \, n \geq 1, \quad b_n = \frac{4}{n}(-1)^{n}. \rightarrow f(x) = 3 + 4 \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \sin(nx).\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fourier coefficients
When dealing with periodic functions, the Fourier series helps break down a complex function into a sum of simpler trigonometric functions. The coefficients in a Fourier series are crucial as they determine the amplitude of each trigonometric component. There are three main Fourier coefficients:
  • The constant term \(a_0\) represents the average value of the function over one period. It's calculated using the integral of the function without any sine or cosine involved.
  • The cosine coefficients \(a_n\), which describe how much cosine content (at different frequencies) exists in the function.
  • The sine coefficients \(b_n\), which determine the amount of sine content at varying frequencies.
For our function \(f(x) = 3 - 2x\), we determine these coefficients by integrating the function multiplied by sine and cosine terms over one period, from \(-\pi\) to \(\pi\). The results give us the representation in terms of simpler, periodic sine and cosine waves.
Orthogonal functions
In the realm of Fourier series, orthogonal functions hold significant value. But what does it mean for functions to be orthogonal in this context? Imagine orthogonality as a way for functions to not "interfere" with each other over a specific interval. For the Fourier series, the sine and cosine functions over \([-\pi, \pi]\) are orthogonal. This means that integrating their product over this interval results in zero unless they are of the same frequency. This property simplifies calculations greatly, allowing us to isolate and calculate Fourier coefficients efficiently. For example, when calculating \(a_n\) as shown in the exercise, the integral of the function times \(\cos(nx)\) results in zero due to this orthogonality, confirming no cosine content from \(n\geq1\) in our function.
Integration by parts
Integration by parts is a fundamental technique in calculus used when dealing with products of functions. It is particularly beneficial in solving integrals that are otherwise challenging. The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]This process requires choosing parts of the integral to set as \(u\) and \(dv\), differentiating and integrating accordingly. In the context of finding sine coefficients \(b_n\), integration by parts simplifies integrals involving products of polynomials and sine functions. This technique reveals key insights into the trigonometric components of a periodic function. It nicely underscores how even seemingly complex integrals can be tackled adequately by breaking them into more manageable parts.
Trigonometric series
A trigonometric series is quite literally adding up sine and cosine functions in various proportions and frequencies to approximate more complex functions. It's the backbone of Fourier series. Each sine and cosine in the series is not randomly chosen. Instead, their inclusion is guided by how well they match the function over an interval, which is determined through the Fourier coefficients as seen above.Ultimately, trigonometric series allow us to express non-sine/cosine functions as sums of these simple waves. The magic lies in their periodic nature and the symmetric properties of sine and cosine functions, making trigonometric series uniquely suited to model periodic behaviors, which is done effectively as seen in the provided Fourier series representation:\[f(x) = 3 + 4 \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \sin(nx)\]This beautifully expresses how a simple linear function \(3-2x\) can transform into a sum of sine waves.

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Most popular questions from this chapter

Using the facts that \(\phi_{0}\) and \(\phi_{1}\) are orthogonal to \(\phi_{n}\) for \(n>1\), we have $$\begin{aligned} \int_{a}^{b}(\alpha x+\beta) \phi_{n}(x) d x &=\alpha \int_{a}^{b} x \phi_{n}(x) d x+\beta \int_{a}^{b} 1 \cdot \phi_{n}(x) d x \\ &=\alpha \int_{a}^{b} \phi_{1}(x) \phi_{n}(x) d x+\beta \int_{a}^{b} \phi_{0}(x) \phi_{n}(x) d x \\ &=\alpha \cdot 0+\beta \cdot 0=0 \end{aligned}$$ for \(n=2,3,4, \dots\)

The eigenfunctions are \(\sin \alpha_{n} x\) where \(\tan \alpha_{n}=-\alpha_{n} .\) Thus $$\begin{aligned} \left\|\sin \alpha_{n} x\right\|^{2} &=\int_{0}^{1} \sin ^{2} \alpha_{n} x d x=\frac{1}{2} \int_{0}^{1}\left(1-\cos 2 \alpha_{n} x\right) d x \\ &=\left.\frac{1}{2}\left(x-\frac{1}{2 \alpha_{n}} \sin 2 \alpha_{n} x\right)\right|_{0} ^{1}=\frac{1}{2}\left(1-\frac{1}{2 \alpha_{n}} \sin 2 \alpha_{n}\right) \\ &=\frac{1}{2}\left[1-\frac{1}{2 \alpha_{n}}\left(2 \sin \alpha_{n} \cos \alpha_{n}\right)\right] \\ &=\frac{1}{2}\left[1-\frac{1}{\alpha_{n}} \tan \alpha_{n} \cos \alpha_{n} \cos \alpha_{n}\right] \\ &=\frac{1}{2}\left[1-\frac{1}{\alpha_{n}}\left(-\alpha_{n} \cos ^{2} \alpha_{n}\right)\right]=\frac{1}{2}\left(1+\cos ^{2} \alpha_{n}\right). \end{aligned}$$

For \(m \neq n\) $$\begin{array}{rl} \int_{0}^{\pi / 2} \cos (2 n+1) x & \cos (2 m+1) x d x \\ & =\frac{1}{2} \int_{0}^{\pi / 2}(\cos 2(n-m) x+\cos 2(n+m+1) x) d x \\ & =\left.\frac{1}{4(n-m)} \sin 2(n-m) x\right|_{0} ^{\pi / 2}+\left.\frac{1}{4(n+m+1)} \sin 2(n+m+1) x\right|_{0} ^{\pi / 2}=0. \end{array}$$ For \(m=n\) $$\begin{aligned} \int_{0}^{\pi / 2} \cos ^{2}(2 n+1) x d x &=\int_{0}^{\pi / 2}\left(\frac{1}{2}+\frac{1}{2} \cos 2(2 n+1) x\right) d x \\ &=\left.\frac{1}{2} x\right|_{0} ^{\pi / 2}+\left.\frac{1}{4(2 n+1)} \sin 2(2 n+1) x\right|_{0} ^{\pi / 2}=\frac{\pi}{4} \end{aligned}$$ so that $$\|\cos (2 n+1) x\|=\frac{1}{2} \sqrt{\pi}.$$

Since \(f(-x)=\sin (-3 x)=-\sin 3 x=-f(x), f(x)\) is an odd function.

Since \(f(x)\) is not defined for \(x<0\), it is neither even nor odd.
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