Question

You are given \(f(x)\) on an interval, say \(0< x< b\). Sketch several periods of the even function \(f_{c}\) of period \(2 b,\) the odd function \(f_{s}\) of period \(2 b,\) and the function \(f_{p}\) of period \(b\), each of which equals \(f(x)\) on \(0< x< b\). Expand each of the three functions in an appropriate Fourier series. $$f(x)=\left\\{\begin{aligned} 1, & 0< x< \frac{1}{2} \\ -1, & \frac{1}{2}< x< 1 \end{aligned}\right.$$

Short answer

Plot the even function, the odd function, and the periodic function. Expand each using their respective Fourier series formulas.

Step-by-step solution

- Function Definitions

Given the function \[f(x)=\begin{aligned} 1, & 0< x< \frac{1}{2} \ -1, & \frac{1}{2}< x< 1 \end{aligned}\]Let's construct the even function \(f_{c}\), the odd function \(f_{s}\), and the function \(f_{p}\) of period \(b\), each of which equals \(f(x)\) on the interval \(0< x< b\).

- Even Function \(f_{c}\)

An even function \(f_{c}\) satisfies \(f_{c}(x) = f_{c}(-x)\). For \(f(x)\) on \(0 < x < 1\):\(f_{c}(x) = \begin{aligned} 1, & 0< x < \frac{1}{2} \ -1, & \frac{1}{2} < x < 1 \ -1, & -1 < x < -\frac{1}{2} \ 1, & -\frac{1}{2} < x < 0 \end{aligned}\)

- Odd Function \(f_{s}\)

An odd function \(f_{s}\) satisfies \(f_{s}(x) = -f_{s}(-x)\). For \(f(x)\) on \(0 < x < 1\):\[f_{s}(x) = \begin{aligned} 1, & 0< x < \frac{1}{2} \ -1, & \frac{1}{2} < x < 1 \ 1, & -1 < x < -\frac{1}{2} \ -1, & -\frac{1}{2}< x < 0 \end{aligned}\]

- Function \(f_{p}\)

The function \(f_{p}\) of period \(b\) repeats \(f(x)\) over each period of \(b\):\[f_{p}(x) = f(x \,\bmod \,b)\]

- Fourier Series Expansion: Even Function \(f_{c}\)

The Fourier series expansion for an even function is given by:\[f_{c}(x) = a_{0} + \sum_{n=1}^{\infty}a_{n}\cos\left(\frac{n\pi x}{L}\right)\]where \[ a_{0} = \frac{1}{L} \int_{0}^{L} f_{c}(x) dx \]and \[ a_{n} = \frac{2}{L} \int_{0}^{L} f_{c}(x)\cos\left(\frac{n\pi x}{L}\right) dx\]

- Fourier Series Expansion: Odd Function \(f_{s}\)

The Fourier series expansion for an odd function is given by:\[f_{s}(x) = \sum_{n=1}^{\infty}b_{n}\sin\left(\frac{n\pi x}{L}\right)\]where \[ b_{n} = \frac{2}{L} \int_{0}^{L} f_{s}(x) \sin\left(\frac{n\pi x}{L}\right) dx \]

- Fourier Series Expansion: Function \(f_{p}\)

The Fourier series expansion for \(f_{p}\) is:\[f_{p}(x) = a_{0} + \sum_{n=1}^{\infty}\left(a_{n}\cos\left(\frac{2n\pi x}{L}\right) + b_{n}\sin\left(\frac{2n\pi x}{L}\right)\right)\]where \[a_{0} = \frac{1}{L} \int_{0}^{L} f_{p}(x) dx\]\[ a_{n} = \frac{2}{L} \int_{0}^{L} f_{p}(x)\cos\left(\frac{2n\pi x}{L}\right) dx \]\[ b_{n} = \frac{2}{L} \int_{0}^{L} f_{p}(x) \sin\left(\frac{2n\pi x}{L}\right) dx \]

Key concepts

even and odd functions

Even and odd functions play an important role in Fourier series. They help simplify problems and reduce the complexity of the calculations needed for expansions.

An even function, denoted by \(f_{c}(x)\), satisfies the property \(f_{c}(x) = f_{c}(-x)\). This means the function is symmetric with respect to the y-axis.

For example, if \(f(x) = 1\) for \(0 < x < \frac{1}{2}\) and \(f(x) = -1\) for \(\frac{1}{2} < x < 1\), the even extension for \(-1 < x < 1\) would be:
\[ f_{c}(x) = \begin{aligned} 1, & \, 0< x < \frac{1}{2} \ -1, & \, \frac{1}{2} < x < 1 \ -1, & \, -1 < x < -\frac{1}{2} \ 1, & \, -\frac{1}{2} < x < 0 \end{aligned} \]

Odd functions, on the other hand, satisfy the property \(f_{s}(x) = -f_{s}(-x)\). So, these functions are symmetric about the origin.

Using the previous example, the odd extension for \(-1 < x < 1\) would be:
\[ f_{s}(x) = \begin{aligned} 1, & \, 0 < x < \frac{1}{2} \ -1, & \, \frac{1}{2} < x < 1 \ 1, & \, -1 < x < -\frac{1}{2} \ -1, & \, -\frac{1}{2} < x < 0 \end{aligned} \]

Understanding these properties helps in constructing the correct Fourier series expansions.

periodic functions

A periodic function is one that repeats its values in regular intervals or periods. The period of a function \(T\) is the smallest positive number such that \(f(x + T) = f(x)\).

Periodic functions are fundamental in Fourier series because they allow us to expand functions over a single period.

If a function has period \(b\), it replicates its pattern every \(b\) units. For example, the function \(f(x)\) given in the exercise is defined on the interval \(0 < x < 1\):
\[ f(x) = \begin{aligned} 1, & \, 0< x < \frac{1}{2} \ -1, & \, \frac{1}{2} < x < 1 \end{aligned} \]

If we extend this function to be periodic with period \(b\), the function over any interval of length \(b\) will look exactly the same:
\( f_{p}(x) = f(x \, \text{mod} \, b) \). This periodic nature is crucial for the Fourier expansion, which decomposes the function into sinusoidal components repeating over this interval.

Fourier coefficients

The Fourier series represents a periodic function as a sum of sines and cosines. The coefficients in this series are called Fourier coefficients. They are crucial because they determine the amplitude of each sinusoidal component.

For an even function, the Fourier series expansion is:
\[ f_{c}(x) = a_{0} + \sum_{n=1}^{\infty}a_{n}\cos\left(\frac{n\pi x}{L}\right) \]
The coefficients \(a_{0}\) and \(a_{n}\):
\[ a_{0} = \frac{1}{L} \int_{0}^{L} f_{c}(x) dx \]
\[ a_{n} = \frac{2}{L} \int_{0}^{L} f_{c}(x)\cos\left(\frac{n\pi x}{L}\right) dx \]

For an odd function, the Fourier series is:
\[ f_{s}(x) = \sum_{n=1}^{\infty}b_{n}\sin\left(\frac{n\pi x}{L}\right) \]
The coefficients \(b_{n}\):
\[ b_{n} = \frac{2}{L} \int_{0}^{L} f_{s}(x) \sin\left(\frac{n\pi x}{L}\right) dx \]

For a general periodic function, we use both sines and cosines:
\[ f_{p}(x) = a_{0} + \sum_{n=1}^{\infty} \left( a_{n}\cos\left(\frac{2n\pi x}{L}\right) + b_{n} \sin\left(\frac{2n\pi x}{L}\right) \right) \]

Each term's coefficient tells us how much of that term is in the original function. Calculating these coefficients involves integrating the product of the function and the sine or cosine terms over a period.