Question

(a) Sketch the graph of the given function for three periods. (b) Find the Fourier series for the given function. $$ f(x)=\left\\{\begin{array}{ll}{0,} & {-2 \leq x \leq-1} \\ {x,} & {-1 < x < 1,} \\ {0,} & {1 \leq x < 2}\end{array} \quad f(x+4)=f(x)\right. $$

Short answer

#Answer# (a) When sketching the graph for three periods, you should see a triangle wave repeating every period of 4 units. (b) The Fourier series of f(x) can be expressed as: $$ f(x) = \sum_{n=1}^∞ bₙ \sin \left( \frac{nπx}{2} \right) $$ Where the bₙ coefficients are obtained by solving the integral: $$ bₙ = \frac{1}{2} \int_{-1}^1 x \sin \left( \frac{nπx}{2} \right) dx $$

Step-by-step solution

(a) Sketch the graph for three periods

Since the function f(x) is defined in three different intervals, we need to sketch the graph accordingly. First, we identify the pattern in each interval: 1. When -2 ≤ x ≤ -1, f(x) = 0. 2. When -1 < x < 1, f(x) = x. 3. When 1 ≤ x < 2, f(x) = 0. This is a periodic function with a period of 4, so we can replicate this pattern in the next periods. After knowing the pattern, sketch the graph for three periods. You should see a triangle wave for every period.

(b) Find the Fourier series

To find the Fourier series, we need to calculate the Fourier coefficients, which are a₀, aₙ, and bₙ. Since f(x) is an odd function, all the aₙ coefficients will be zero. So, we only need to find a₀ and bₙ. First, let's find a₀. $$ a₀ = \frac{1}{L} \int_{-L}^L f(x)dx $$ Since L = 2, and f(x) is odd, the integral over symmetric limits will be zero. Therefore, a₀ = 0. Now, let's find bₙ. $$ bₙ = \frac{1}{L} \int_{-L}^L f(x) \sin \left( \frac{nπx}{L} \right)dx $$ Since the function is defined in three different intervals, we need to solve the integral for each interval: 1. -2 ≤ x ≤ -1. In this interval, f(x) = 0, so the integral will also be zero. 2. -1 < x < 1. In this interval, f(x) = x, so the integral will be: $$ bₙ = \frac{1}{2} \int_{-1}^1 x \sin \left( \frac{nπx}{2} \right) dx $$ Solve the integral to find the coefficient. 3. 1 ≤ x < 2. In this interval, f(x) = 0, so the integral will also be zero. Now, we can write the Fourier series in the following form: $$ f(x) = \sum_{n=1}^∞ bₙ \sin \left( \frac{nπx}{2} \right) $$ Replace the values of bₙ from the calculations in step 2 to obtain the Fourier series of f(x).

Key concepts

Periodic Function

A periodic function is a function that repeats its values in regular intervals or periods. The function described in the exercise, \(f(x)\), is periodic with a period of 4. This means that every 4 units along the x-axis, the function's pattern repeats itself. In the context of time, like hours of daylight, temperature over seasons, or sound waves, periodic functions are key to modeling repeating phenomena.

To illustrate this, think of a waveform, which is a perfect example of periodic behavior. If you draw the function for one complete cycle and then replicate it endlessly on both sides, you would generate a graph that looks the same every cycle. This is exactly what the function \(f(x)\) is doing after every 4 units along the x-axis.

A periodic function has certain intervals where its rule remains constant, or follows the same mathematical pattern until the next period begins. In this case, you would observe the function being zero where \(-2 \leq x \leq -1\) and \(1 \leq x < 2\), while it follows the line \(f(x) = x\) in the interval \(-1 < x < 1\). This repeating triangle waveform, characteristic of the described function, is easily predictable, thanks to its periodic nature.

Fourier Coefficients

Fourier coefficients play a critical role in expressing periodic functions as an infinite sum of sines and cosines, known as a Fourier series. They essentially "weight" each sine and cosine wave's contribution to the overall function. In this context, these coefficients break down a complex waveform into simpler wave components.

There are three types of Fourier coefficients: \(a_0\), \(a_n\), and \(b_n\). Each one serves a different purpose in the Fourier series, with \(a_n\) and \(b_n\) representing the amplitudes of cosine and sine terms, respectively. In this exercise, however, since \(f(x)\) is an odd function, all \(a_n\) coefficients are zero, leaving only the sine components, \(b_n\), to characterize the series. The constant term \(a_0\) also becomes zero because the integral over one period equals zero for odd functions.

To calculate \(b_n\), you'd integrate the product of \(f(x)\) and \(\sin\left( \frac{n\pi x}{2} \right)\) over the symmetric limits of your interval \([-L, L]\). These coefficients reflect how the function decomposes into harmonic components, which particularly helps in signal processing or any field involving wave analysis.

Odd Function

An odd function is a type of function that satisfies the property \(f(-x) = -f(x)\). This characteristic implies that the function is symmetrical around the origin, and flipping it over both axes would produce the same graph. Understanding odd functions is significant because it simplifies the computation of Fourier series.

In our exercise with function \(f(x)\), we already know it's odd because one half of its graph is the negative mirror image of the other. Therefore, when calculating the Fourier series, this property immediately tells us that the cosine terms, represented by \(a_n\), will be zero. Why? Because cosine terms are even functions and they "cancel out" when integrated over symmetric limits with an odd function.

This leads to a series representation that merely consists of sine terms, simplifying the Fourier series calculations. As a result, for the provided \(f(x)\), finding the Fourier series primarily involves computing the sine coefficients, \(b_n\). This efficient approach allows for a deeper understanding of the function's structure and its harmonic components.