Question

Each of the functions is given over one period. For each function, sketch several periods and decide whether it is even or odd. Then use (9.4) or (9.5) to expand it in an appropriate Fourier series. $$ f(x)=\left\\{\begin{array}{rr} -1, & -l

Short answer

\( f(x) \) is odd; its Fourier series is \( f(x) = \sum_{n=1, 3, 5}^{\text{∞}} \frac{4}{nπ}\text {sin}\left( \frac{nπx}{l} \right) \)

Step-by-step solution

Sketch Several Periods

First, sketch the given piecewise function over a single period $$ f(x)=\left\{\begin{array}{rr} -1 & -l<x<0 \ 1 & 0<x<l \ \end{array}\right. $$ Now extend this to several periods to observe its overall behavior.

Determine if the Function is Even or Odd

A function is even if \( f(-x) = f(x) \) and odd if \( f(-x) = -f(x) \). In this case, \( f(-x) = -f(x) \), so the function is odd.

Choose Appropriate Fourier Series Expression

Since the function is odd, use the sine series expansion (9.4): \[ f(x) = \sum_{n=1}^{\text{∞}}b_n\text{sin}\frac{nπx}{l} \]

Find Fourier Coefficients

To find the coefficient \( b_n \) use the formula: \[ b_n = \frac{2}{l}\int_{0}^{l}f(x)sin\frac{nπx}{l}dx \] Here, we must compute the integrals for the two pieces: \[ b_n = \frac{2}{l}\int_{0}^{l} f(x)\text{sin}\left(\frac{nπx}{l}\right) dx = \frac{2}{l}\left[ \int_{0}^{l} (1) \text{sin}\left(\frac{nπx}{l}\right) dx + \int_{-l}^{0} (-1) \text{sin}\left(\frac{nπx}{l}\right) dx\right] \] By symmetry and periodicity:

Evaluate the Integrals

Compute the integrals: \[ b_n = \frac{2}{l} \int_{0}^{l} \text{sin}\left(\frac{nπx}{l}\right) dx=\frac{2}{l} \left[ -\frac{l}{nπ} cos\left( \frac{nπx}{l}\right) \right]_{0}^{l} \] \[ = \frac{2}{l} \left( -\frac{l}{nπ} (cos(nπ) - cos(0)) \right) = \frac{2}{l}\left( -\frac{l}{nπ} (-1-1) \right)= \frac{4}{nπ} \]

Write the Complete Fourier Series

Combine the coefficients to form the final solution: \( f(x) = \sum_{n=1, 3, 5}^{\text{∞}} \frac{4}{nπ} \text{sin}\left( \frac{nπx}{l} \right) \). This is the Fourier sine series expansion for the given function.

Key concepts

Fourier Coefficients

To understand Fourier series, it's crucial to learn about Fourier coefficients. These coefficients, denoted as \(a_n\) and \(b_n\), represent the weights for each term in the series. They determine how much of each sine and cosine function is needed to approximate the original function. For our piecewise function, since it is an odd function, we only need to calculate the sine coefficients \(b_n\). The formula for calculating \(b_n\) is given by:

\[ b_n = \frac{2}{l} \int_{0}^{l} f(x)\text{sin} \left(\frac{n\pi x}{l} \right) dx\]

This integral essentially breaks down the function into its sine components by integrating over one period of the function.

Piecewise Functions

Piecewise functions are those defined by different expressions depending on the interval of the input variable. In our problem, the function \(f(x)\) is defined differently on two segments:

\(f(x) =\)\[\begin{array}{rr}-1 & -lUnderstanding piecewise functions is important as it allows us to break down complex functions into simpler segments, making them easier to analyze.

Even and Odd Functions

A function is even if it remains unchanged when we replace \(x\) with \(-x\), mathematically: \[ f(-x) = f(x) \]. Conversely, a function is odd if flipping the sign of \(x\) flips the sign of the function, i.e., \[ f(-x) = -f(x) \]. For our example, the function is odd because: \( f(-x) = -f(x) \). Recognizing this property can simplify the computation of Fourier series. For odd functions, we use the Fourier sine series, and for even ones, we use the cosine series.

Sine Series Expansion

When a function is odd, its Fourier series representation only involves sine terms. This is because sine functions are themselves odd functions. For the given problem, we use the sine series expansion formula:

\[ f(x) = \sum_{n=1}^{\text{∞}} b_n \text{sin}\left( \frac{n\pi x}{l} \right) \].

In this series, the coefficients \(b_n = \frac{4}{n\pi}\) represent the contributions of the sine functions to the shape of the original function. By summing these terms, we can represent the original function as a series of sines.