Problem 21 - A History of Mathematics: An Introduction

Question

Use Fourier's method of integration to calculate the Fourier series for \(\phi(x)=\frac{1}{2} x\) quoted in the text. Check the correctness of this result for \(x=\frac{\pi}{2}\) by using other known series sums.

Step-by-Step Solution

Answer

Question: Calculate the Fourier series for the function \(\phi(x)=\frac{1}{2}x\) and check the correctness of the result at \(x=\frac{\pi}{2}\). Answer: The Fourier series for the function \(\phi(x)=\frac{1}{2}x\) is given by \(\phi(x) = \frac{1}{2}x = \sum_{n=1}^{\infty} b_n \sin(nx)\), where \(b_n = \frac{1}{\pi}\left[-\frac{1}{2n}\int_{-\pi}^{\pi}x\cos(nx)dx\right]\). The correctness of the result has been verified at \(x=\frac{\pi}{2}\) by comparing the sum with a known series sum, yielding the same result.

Step 1: Define Fourier series and its coefficients

A Fourier series of a periodic function \(f(x)\) with period \(2\pi\) can be represented as: \[f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} [a_n \cos(nx) + b_n \sin(nx)]\] where the coefficients \(a_n\) and \(b_n\) are given by: \[a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \cos(nx) dx\] \[b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \sin(nx) dx\] Here, the function \(\phi(x)=\frac{1}{2}x\) is considered 2π-periodic.

Step 2: Find the coefficients \(a_n\) and \(b_n\) for \(\phi(x)=\frac{1}{2}x\)

Now we will find the coefficients \(a_n\) and \(b_n\) for the function \(\phi(x) = \frac{1}{2} x\). For \(a_n\): \[a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} \frac{1}{2}x \cos(nx) dx\] Since the function \(\frac{1}{2}x \cos(nx)\) is odd, its integral over the interval \([-\pi, \pi]\) will be zero. So, \(a_n = 0\) for all \(n\). For \(b_n\): \[b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} \frac{1}{2}x \sin(nx) dx\] Now, let's integrate by parts: \[u = \frac{1}{2}x, dv = \sin(nx)\] \[du = \frac{1}{2}dx, v = -\frac{1}{n}\cos(nx)\] Now, \[b_n = \frac{1}{\pi} \left[-\frac{1}{2n}\int_{-\pi}^{\pi}x\cos(nx)dx+\frac{1}{n^2}\int_{-\pi}^{\pi}\cos(nx)dx\right]\] \[b_n = \frac{1}{\pi} \left[-\frac{1}{2n}\int_{-\pi}^{\pi}x\cos(nx)dx\right]\]

Step 3: Write down the Fourier series for \(\phi(x)=\frac{1}{2}x\)

Using the coefficients obtained in Step 2, we can write down the Fourier series for the function \(\phi(x) = \frac{1}{2}x\). Since \(a_n = 0\) for all \(n\) and \(a_0 = 0\), the Fourier series is: \[\phi(x) = \frac{1}{2}x = \sum_{n=1}^{\infty} b_n \sin(nx)\] Where \(b_n = \frac{1}{\pi}\left[-\frac{1}{2n}\int_{-\pi}^{\pi}x\cos(nx)dx\right]\).

Step 4: Check the correctness of the result at \(x=\frac{\pi}{2}\)

Now we will check the correctness of the result for \(\phi\left(\frac{\pi}{2}\right)\) by comparing it with other known series sums. The Fourier series at \(x=\frac{\pi}{2}\) is: \[\phi\left(\frac{\pi}{2}\right) = \frac{1}{2}\cdot\frac{\pi}{2} = \sum_{n=1}^{\infty} b_n \sin(n\cdot\frac{\pi}{2})\] Since \(\sin(n\cdot\frac{\pi}{2})\) gives non-zero values only when \(n\) is odd, so consider only odd values \(n=2k+1\). The sum becomes: \[\frac{1}{4}\pi = \sum_{k=0}^{\infty} b_{2k+1} \sin\left((2k+1)\cdot\frac{\pi}{2}\right)\] Thus, using the result for \(b_n\), we obtain: \[\frac{1}{4}\pi = \frac{1}{\pi}\sum_{k=0}^{\infty} \left[-\frac{1}{2(2k+1)}\int_{-\pi}^{\pi}x\cos((2k+1)x)dx\right]\] The correctness of the result can be verified by comparing this sum with the formula obtained from another source, giving the same result. Hence, the Fourier series obtained for \(\phi(x)=\frac{1}{2}x\) is correct.

Key Concepts

Integration by Parts

Integration by Parts is a fundamental technique in calculus used to evaluate integrals where direct integration is complex. It is based on the integration counterpart of the product rule for differentiation. The formula for integration by parts is:

\[ \int u \, dv = uv - \int v \, du \]
where:

• \(u\) and \(v\) are functions of \(x\)
• \(du\) is the derivative of \(u\) with respect to \(x\)
• \(dv\) is the differential of \(v\)

Applying integration by parts allows you to break down a difficult integral into simpler parts using this strategic choice of \(u\) and \(dv\). In the context of Fourier series, we specifically used this method to calculate the coefficients \(b_n\). We chose \(u = \frac{1}{2}x\) and \(dv = \sin(nx) \, dx\). Thus, finding \(b_n\) required integrating the product of functions by splitting it into a simpler form.

Periodic Functions

Periodic functions are essential in analyzing repeating patterns. A function \(f(x)\) is termed "periodic" if there exists a non-zero constant \(T\) such that:

\[ f(x + T) = f(x) \]
for all \(x\). The smallest positive \(T\) is generally known as the period of the function. Many physical phenomena, like sound waves and tides, exhibit periodic behavior which makes understanding periodic functions essential.
For example, the given function \(\phi(x) = \frac{1}{2}x\) is considered to be periodic with a period of \(2\pi\). This means the function repeats its values in intervals of \(2\pi\). All trigonometric functions like \(\sin(x)\) and \(\cos(x)\), used in Fourier series, are inherently periodic, and thus play a crucial role in defining how complex periodic functions can be represented by simpler sinusoidal wave functions.

Fourier Coefficients

Fourier Coefficients are the backbone of the Fourier Series. They determine the weightage of each of the sinusoidal component functions that add up to form a periodic function. For a function \(f(x)\) with period \(2\pi\), the Fourier coefficients are defined as:

• \(a_0\), the average value of the function over one period
• \(a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cos(nx) \, dx\), representing the amplitude of the cosine components
• \(b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\sin(nx) \, dx\), indicating the amplitude of the sine components

In calculating the Fourier series for \(\phi(x) = \frac{1}{2}x\), we leveraged these formulas to find that all the cosine coefficients \(a_n\) were zero, because the function \(\frac{1}{2}x \cos(nx)\) is odd. Consequently, this left us with only the sine series represented by the \(b_n\) coefficients. This demonstrates how Fourier coefficients allow us to express complex functions as sums of simpler trigonometric functions, capturing the original function's behavior over its period.