Question

assume that the given function is periodically extended outside the original interval. (a) Find the Fourier series for the extended function. (b) Sketch the graph of the function to which the series converge for three periods. $$ f(x)=\left\\{\begin{array}{cc}{-1,} & {-1 \leq x<0} \\ {1,} & {0 \leq x<1}\end{array}\right. $$

Short answer

Based on the steps provided, the Fourier series of the given piecewise function is: $$ f(x) = \sum_{n=1}^\infty \left[\frac{2}{n\pi}(-1)^n \cos(n\pi x)\right] $$This Fourier series converges to the average of the left and right endpoint values of the original function at the jump discontinuity, resembling a square wave with rounded edges at the transition points between -1 and 1, which is due to the Gibbs phenomenon. When sketching the graph for three periods in the range \([-5,7]\), the function shows a periodic pattern with period 2, as expected. The more terms included in the Fourier series, the closer the graph gets to the exact extended function.

Step-by-step solution

Determine the period and frequency of the given function

The function is defined on the interval \(-1 \leq x < 1\), so it has a period of 2 units. This means the function will be extended periodically with a period of 2 units. Therefore, the frequency of the function is \(1/2\).

Calculate the Fourier coefficients

We are given a piecewise function, so we will compute the Fourier coefficients using the integral formula for a piecewise function. Let's calculate the Fourier coefficients \(a_n\) and \(b_n\). $$ a_0 = \frac{1}{2} \int_{-1}^1 f(x) dx $$Since \(f(x)=-1\) for \(-1\leq x<0\) and \(f(x)=1\) for \(0\leq x<1\), the integral can be separated into two parts:$$ a_0 = \frac{1}{2} \left[\int_{-1}^0 (-1) dx + \int_{0}^1 (1) dx\right] $$Solving the integrals and adding them, we get:$$ a_0 = \frac{1}{2} (1-1) = 0 $$For \(n \geq 1\), we compute the coefficients \(a_n\) and \(b_n\) as follows:$$ a_n = \frac{1}{1}\int_{-1}^1 f(x) \cos(\frac{n\pi x}{1}) dx = \left[\int_{-1}^0 (-1) \cos(n\pi x) dx + \int_{0}^1 (1) \cos(n\pi x) dx\right] $$$$ b_n = \frac{1}{1}\int_{-1}^1 f(x) \sin(\frac{n\pi x}{1}) dx = \left[\int_{-1}^0 (-1) \sin(n\pi x) dx + \int_{0}^1 (1) \sin(n\pi x) dx\right] $$Now, we can compute \(a_n\) and \(b_n\) for each piece of the function: For \(a_n\):$$ \int_{-1}^0 (-1) \cos(n\pi x) dx = -\frac{1}{n\pi}\left[\sin(n\pi x)\right]_{-1}^0 = \frac{2}{n\pi}(-1)^n $$$$ \int_{0}^1 (1) \cos(n\pi x) dx = \frac{1}{n\pi}\left[\sin(n\pi x)\right]_{0}^1 = 0 $$For \(b_n\):$$ \int_{-1}^0 (-1) \sin(n\pi x) dx = \frac{1}{n\pi}\left[\cos(n\pi x)\right]_{-1}^0 = \frac{1}{n\pi}((-1)^n - 1) $$$$ \int_{0}^1 (1) \sin(n\pi x) dx = -\frac{1}{n\pi}\left[\cos(n\pi x)\right]_{0}^1 = -\frac{1}{n\pi}(1 - (-1)^n) $$Adding these integrals for each piece, we have:$$ a_n = \frac{2}{n\pi}(-1)^n, \quad\quad b_n = 0 $$

Construct the Fourier series

Using the Fourier coefficients we calculated in step 2, we can now construct the Fourier series for the given piecewise function. The Fourier series is given by:$$ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left[a_n \cos\left(\frac{n\pi x}{1}\right) + b_n \sin\left(\frac{n\pi x}{1}\right)\right] $$As we calculated earlier, \(a_0=0\) and \(b_n=0\). Therefore, the Fourier series becomes:$$ f(x) = \sum_{n=1}^\infty \left[\frac{2}{n\pi}(-1)^n \cos(n\pi x)\right] $$This is the Fourier series for the periodically extended function.

Sketch the graph of the function for three periods

Now we will sketch the graph of the function to which the Fourier series converges for three periods. Since the period of the function is 2 units, this means that we will sketch the graph for the extended function in the range \([-5,7]\). To do this, consider a finite number of terms in the Fourier series and plot the resulting function. For illustration purposes, we will consider the first five terms. The more terms we include, the closer we get to the exact extended function. Note that the Fourier series converges to the average of the left and right endpoint values of the original function at the jump discontinuity. The graph will resemble a square wave with rounded edges at the transition points between -1 and 1, which is due to the Gibbs phenomenon.

Key concepts

Piecewise Functions

Piecewise functions are mathematical expressions that have different definitions over different intervals. They can be visualized as a graph that consists of distinct sections, each with its own rule.

For example, in the exercise provided, the function \( f(x) = \left \{\begin {array}{cc}{-1,} & {-1 \leq x<0} \ {1,} & {0 \leq x<1}\end{array}\right. \) is defined in two pieces: it takes the value -1 when \( x \) is between -1 and 0, and 1 when \( x \) is between 0 and 1. When we extend this function periodically, each interval of length 2 will have the same piecewise definition.

Understanding how a piecewise function behaves is crucial because when we calculate the Fourier coefficients, we need to pay attention to how the function is defined over each interval. This directly influences the integration process used to find the Fourier coefficients.

Fourier Coefficients

Fourier coefficients are the parameters that, when used in a Fourier series, allow us to represent a periodic function as a sum of sines and cosines. The coefficients \( a_n \) and \( b_n \) essentially capture the function's behavior over its domain.

To calculate these coefficients for a piecewise function like the one given, we integrate the function multiplied by sine and cosine functions corresponding to different harmonics or frequencies. For the given exercise, \( a_n \) is found to be related to the cosine terms and turns out to be \( a_n = \frac{2}{n\pi}(-1)^n \), while \( b_n \) is zero because the integral of an odd function over a symmetric interval around the origin is zero.

These coefficients are particularly important as they determine the amplitude of each corresponding sine or cosine wave in the series. Without knowing the Fourier coefficients, one cannot reconstruct the original function from its Fourier series.

Periodic Functions

A periodic function is one that repeats its values in regular intervals or periods. The fundamental period of a function is the positive length \( P \) such that \( f(x + P) = f(x) \) for all \( x \) in the function's domain.

In the context of our piecewise function, the period is 2 units, meaning after every 2 units the function's values repeat. Recognizing the period of a function is essential for determining the correct form of the Fourier series, as the frequency of the sines and cosines in the series depends on the period. The periodic extension involves replicating the given function's interval \( [-1,1] \) across the entire real number line in such a way that the patterns recur every 2 units.

Convergence of Fourier Series

The convergence of a Fourier series refers to its ability to represent a function accurately within its domain. When the Fourier series converges, the sum of sine and cosine terms approaches the value of the function at most points.

In many cases, including the example of the piecewise function provided, the Fourier series will converge to the function itself at points of continuity and to the average of the left and right limits at points of discontinuity. This is a manifestation of the Gibbs phenomenon, where the series overshoots at discontinuities, creating a characteristic 'ringing' effect near the edges.

For this piecewise function, the Fourier series represents the original function well, except at the jump discontinuities, where it converges to the midpoint of the jump. The more terms we include in the series, the better the approximation, especially in the regions away from discontinuities.