Question

Find the Fourier series of \(f\) on \([-L, L]\) and determine its sum for \(-L \leq x \leq L .\) Where indicated by \(C\), graph \(f\) and $$ F_{m}(x)=a_{0}+\sum_{n=1}^{m}\left(a_{n} \cos \frac{n \pi x}{L}+b_{n} \sin \frac{n \pi x}{L}\right) $$ on the same axes for various values of \(m\). $$ L=1 ; \quad f(x)=\left\\{\begin{array}{cc} 0, & -1

Short answer

#teacher_short_answer# For the given function \(f(x)\), $$ a_0 = \frac{2}{\pi}, a_n = \frac{2}{n \pi}[1-(-1)^n], b_n = 0. $$ The sum of the Fourier series for \(f(x)\) is $$ f(x)\approx \frac{2}{\pi}+\sum_{n=1}^{\infty} \frac{2}{n \pi}[1-(-1)^n] \cos (n \pi x) $$ on the interval \(-1 \leq x \leq 1\).

Step-by-step solution

Find the Fourier coefficients \(a_n\) and \(b_n\)

To find the Fourier coefficients, we first need to calculate the \(a_n\) and \(b_n\) coefficients using the following formulas: $$ a_{n}=\frac{1}{L} \int_{-L}^{L} f(x) \cos \frac{n \pi x}{L} dx $$ and $$ b_{n}=\frac{1}{L} \int_{-L}^{L} f(x) \sin \frac{n \pi x}{L} dx $$ Here we know that \(L=1\) and the function \(f(x)\) is given by: $$ f(x)=\left\{ \begin{array}{cc} 0, & -1<x<\frac{1}{2}, \\ |\sin \pi x|, & -\frac{1}{2}<x<\frac{1}{2}, \\ 0, & \frac{1}{2}<x<1 \end{array} \right. $$ To find \(a_n\) and \(b_n\), we will integrate from \(-\frac{1}{2}\) to \(\frac{1}{2}\) since the function is zero elsewhere. For the \(a_n\) coefficients, we have: $$ a_{n}=\int_{-\frac{1}{2}}^{\frac{1}{2}} |\sin \pi x | \cos \frac{n \pi x}{1} dx $$ For \(a_0\), we have: $$ a_{0}=\int_{-\frac{1}{2}}^{\frac{1}{2}} |\sin \pi x| dx $$ For the \(b_n\) coefficients, we have: $$ b_{n}=\int_{-\frac{1}{2}}^{\frac{1}{2}} |\sin \pi x| \sin \frac{n \pi x}{1} dx $$

Calculate \(a_n\) and \(b_n\)

To calculate the \(a_n\) coefficients, split the integrals for \(a_n\) and \(a_0\) for when the sine function is positive and negative. Calculating \(a_0\): $$ a_{0}=\int_{-\frac{1}{2}}^{0} -\sin\pi x\, dx + \int_{0}^{\frac{1}{2}} \sin\pi x\, dx $$ $$ a_{0} = \frac{2}{\pi} $$ Calculating \(a_n\) for \(n>0\): $$ a_{n}=\int_{-\frac{1}{2}}^{0} -\sin \pi x \cos(n \pi x) dx + \int_{0}^{\frac{1}{2}} \sin \pi x \cos(n \pi x) dx $$ $$ a_{n}=\frac{2}{n \pi}[1-(-1)^n] $$ Calculating \(b_n\): $$ b_{n}=\int_{-\frac{1}{2}}^{0} -\sin \pi x \sin(n \pi x) dx + \int_{0}^{\frac{1}{2}} \sin\pi x \sin(n \pi x) dx $$ $$ b_{n}=0 $$ Notice that the \(b_n\) coefficients are zero.

Write the Fourier series and determine its sum

Now we can write the Fourier series using the coefficients calculated above: $$ f(x)\approx F_m(x)=\frac{2}{\pi}+\sum_{n=1}^{m} \frac{2}{n \pi}[1-(-1)^n] \cos (n \pi x) $$ The sum of the Fourier series for the given function \(f(x)\) is $$ f(x)\approx \frac{2}{\pi}+\sum_{n=1}^{\infty} \frac{2}{n \pi}[1-(-1)^n] \cos (n \pi x) $$ on the interval \(-1 \leq x \leq 1\).

Graph the function and its Fourier series for various values of \(m\)

To graph the function and its Fourier series for various values of \(m\), use the sum formula for \(F_m(x)\) above and the given function for \(f(x)\). For each value of \(m\), plot both the original function \(f(x)\) and the Fourier series approximation \(F_m(x)\) on the same axes. Observe the approximation improve with increasing values of \(m\).

Key concepts

Fourier Coefficients

Understanding Fourier coefficients is fundamental to grasping Fourier series. In the world of mathematics, especially in the study of signals or periodic functions, these coefficients perform the crucial task of breaking down complex waveforms into simpler trigonometric functions.

For a given periodic function, Fourier coefficients, denoted as a_n and b_n, represent the amplitude of cosine and sine components respectively. They're found by integrating the original function multiplied by corresponding trigonometric functions over one period of the function. In the case of even-symmetry, the sine terms (associated with b_n) tend to zero, simplifying the series.

In practice, we compute these coefficients to reconstruct or approximate the original function through a summation of its trigonometric components. This is exactly what was done in the step-by-step solution provided: after finding the formulae to calculate a_n and b_n, we integrate, taking the particular symmetry of f(x) into account, and thus, determine their values for n greater than zero.

Fourier Series Approximation

The objective of a Fourier series approximation is to model a more complex function using simpler, sinusoidal components. By doing so, one gains the ability to study and manipulate the original function in a more manageable form.

As shown in the exercise, after calculating the Fourier coefficients, constructing the Fourier series approximation becomes straightforward. It is essentially a trigonometric series, with the computed coefficients a_n and b_n determining the contribution of each harmonic component.

Convergence of the Series

The convergence of a Fourier series is intriguing. As more terms are included (increasing m), the series converges to the original function, recreating its shape with more accuracy. This aspect was demonstrated when graphing the function f(x) and its Fourier series F_m(x) for various values of m. With mathematics, we can navigate complex functions like f(x), which have different expressions over different intervals, offering a potent tool for analysis in engineering and physics.

Boundary Value Problems

Boundary Value Problems (BVPs) are pivotal within the territory of differential equations and mathematical physics. They revolve around finding a solution to a differential equation that must satisfy certain conditions, termed as 'boundary conditions,' at the extremes of the domain.

The exercise at hand did not delve into differential equations explicitly. However, understanding BVPs can enrich a student's interpretation of Fourier series. When we resolve a BVP using Fourier series, we're essentially turning these conditions into a conversational language for functions over specific domains. The problem becomes a question of decomposing the restricted behavior into an infinite sum of sinusoids which, like the pieces of a puzzle, fit within the boundaries of our conditions.

Fourier series are notably valuable in solving BVPs with periodic or segmentally defined functions, as it allows one to deal with each piece on its domain simplifying the overall process. This is where the connection between BVPs and the exercise becomes clear; Fourier series provide a pathway to answer these complicated problems.