Question

Each of the following functions is given over one period. Sketch several periods of the corresponding periodic function and expand it in an appropriate Fourier series. $$f(x)=\left\\{\begin{array}{lr}0, & -\frac{1}{2}< x<0 \\\x, & 0< x<\frac{1}{2}\end{array}\right.$$.

Short answer

The Fourier series for the given function is \( f(x) = \frac{1}{8} + \sum_{n=1}^{\infty} \left( \frac{(1-(-1)^{n})}{\pi^2 n^2} \cos(2 \pi n x) + \left( - \frac{\cos(\pi n)}{\pi^2 n^2} \right) \sin(2 \pi n x) \right) \).

Step-by-step solution

- Identify the Function

The given function is defined as:\[ f(x) = \left\{ \begin{array}{l}0, \quad -\frac{1}{2} < x < 0 \ x, \quad 0 < x < \frac{1}{2} \end{array}\right. \]This function defines one period in the interval \(-\frac{1}{2} < x < \frac{1}{2}\).

- Determine the Period

The function repeats every period of length 1. This can be observed from the interval \(-\frac{1}{2} < x < \frac{1}{2}\).

- Sketch Several Periods

Sketch the function from \(-\frac{3}{2}\) to \(\frac{3}{2}\):1. From \(-\frac{3}{2}\) to \(-1\): The function is 0.2. From \(-1\) to \(-\frac{1}{2}\): The function is 0.3. From \(-\frac{1}{2}\) to 0: The function is 0.4. From 0 to \(\frac{1}{2}\): The function is a straight line from 0 to \(\frac{1}{2}\).5. Repeat the above steps for subsequent periods.

- Write the Fourier Series Expansion

Using the interval \[-\frac{1}{2}, \frac{1}{2}\] for one period, the Fourier series representation can be given as:\[ f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(2\pi n x) + b_n \sin(2\pi n x) \right) \]Where:\[ a_0 = \frac{1}{T} \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \, dx \]\[ a_n = \frac{2}{T} \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \cos(2\pi n x) \, dx \]\[ b_n = \frac{2}{T} \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \sin(2\pi n x) \, dx \]Here, \(T\) is the period which is 1.

- Calculate the Fourier Coefficients

1. Calculate \(a_0\):\[ a_0 = \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \, dx = \int_{0}^{\frac{1}{2}} x \, dx = \left. \frac{x^2}{2} \right|_0^{\frac{1}{2}} = \frac{1}{8} \]2. Calculate \(a_n\):\[ a_n = 2 \int_{0}^{\frac{1}{2}} x \cos(2 \pi n x) \, dx \ = \left. \frac{x \sin(2 \pi n x)}{\pi n} \right|_0^{\frac{1}{2}} - \int_{0}^{\frac{1}{2}} \frac{\sin(2 \pi n x)}{\pi n} \, dx \ = \left. \frac{x \sin(2 \pi n x)}{\pi n} \right|_0^{\frac{1}{2}} + \frac{1}{\pi^2 n^2} \left. - \cos(2 \pi n x) \right|_0^{\frac{1}{2}} \ = \left( \frac{\sin(\pi n)}{2 \pi n} \right) - \frac{1}{\pi^2 n^2} ( \left( -1 \right)^n - 1) \]Due to properties of sine and cosine:\[ a_n = \frac{(1-(-1)^n)}{\pi^2 n^2} \]3. Calculate \(b_n\):\[ b_n = 2 \int_{0}^{\frac{1}{2}} x \sin(2 \pi n x) \, dx \ = \left. -\frac{x \cos(2 \pi n x)}{\pi n} \right|_0^{\frac{1}{2}} - \int_{0}^{\frac{1}{2}} - \frac{\cos(2 \pi n x)}{} \, dx \ = \left( - \frac{(2 \pi n) x \cos(\pi n)}{1} \right) - \frac{1}{\pi^2 n^2} \]\[ b_n = \left( - \frac{\cos(\pi n)}{2 \pi n} \right) -( \frac{(2 \pi n)}{1} ) \]\[ b_n = -\frac{\cos(\pi n)}{\pi^2 n^2}\]

- Constructing the Fourier Series

Substitute the calculated coefficients back into the Fourier series representation:\[ f(x) = \frac{1}{8} + \sum_{n=1}^{\infty} \left( \frac{(1-(-1)^{n})}{\pi^2 n^2} \cos(2 \pi n x) + \left( - \frac{\cos(\pi n)}{\pi^2 n^2} \right) \sin(2 \pi n x) \right) \]

Key concepts

Periodic Functions

A periodic function is a function that repeats its values at regular intervals. In simpler terms, if you move a certain distance along the x-axis, the function's shape will look the same. Mathematically, a function \(f(x)\) is periodic with period \(T\) if and only if \(f(x+T) = f(x)\) for all \(x\).

In the given problem, the function \(f(x)\) is defined over the interval \(-\frac{1}{2} < x < \frac{1}{2}\) and repeats every period of length 1. This can be observed by examining how the function behaves in this interval and extending it to other intervals.

The function description provided is:
\[ f(x) = \begin{cases} 0, & -\frac{1}{2} < x < 0 \ x, & 0 < x < \frac{1}{2} \end{cases} \]

It means that from \(-\frac{1}{2}\) to 0, the function is 0, and from 0 to \(\frac{1}{2}\), the function is a straight line from 0 to \(\frac{1}{2}\).

By sketching several periods, we can see how this pattern repeats itself. This repetition is the hallmark of periodic functions.

Fourier Coefficients

Fourier coefficients are the building blocks of a Fourier series, which is used to represent a periodic function using sines and cosines. In the Fourier series representation:

\[ f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(2\pi n x) + b_n \sin(2\pi n x) \right) \]

The coefficients \(a_0\), \(a_n\), and \(b_n\) are the Fourier coefficients. Here's how they are calculated:

\[ a_0 = \frac{1}{T} \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \, dx \]
\[ a_n = \frac{2}{T} \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \cos(2\pi n x) \, dx \]
\[ b_n = \frac{2}{T} \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \sin(2\pi n x) \, dx \]

In this problem, the period \(T\) is 1, simplifying these coefficients by removing the division by \(T\).

The steps to calculate these coefficients involve integrating the function over one period. For \(a_0\), it’s the integral from \(0\) to \(\frac{1}{2}\):

\[ a_0 = \int_{0}^{\frac{1}{2}} x \, dx = \left. \frac{x^2}{2} \right|_0^{\frac{1}{2}} = \frac{1}{8} \]

For \(a_n\) and \(b_n\), integration gets a bit more involved as trigonometric functions interact with \(f(x)\). However, the process remains essentially the same: multiply \(f(x)\) by sine or cosine and integrate over one period.

Trigonometric Series

A trigonometric series is a series of terms involving sines and cosines. It can represent periodic functions more easily compared to their algebraic forms. The general form of a Fourier series is a trigonometric series:

\[ f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(2\pi n x) + b_n \sin(2\pi n x) \right) \]

Here:
* \(a_0\) is the average value (or the DC component) of the function over one period
* \(a_n\) are the coefficients of the cosine terms (even functions)
* \(b_n\) are the coefficients of the sine terms (odd functions)

Cosines and sines are used because they are orthogonal functions, meaning their integrals over one period are zero unless they are the same frequency. This property allows us to isolate each term and find its coefficient separately.

In this way, we can decompose any periodic function into its constituent frequencies. This is particularly useful in fields like signal processing or vibration analysis, where identifying individual frequency components is necessary.

Integral Calculus

Integral calculus is used to compute the area under a curve and is a critical component in finding Fourier coefficients. When we compute the Fourier coefficients, we perform integrations over a defined period. For example:

When calculating \(a_0\):
\[ a_0 = \frac{1}{T} \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \, dx \]

We simplify this for our problem since \(T = 1\):

\[ a_0 = \int_{0}^{\frac{1}{2}} x \, dx \]

This integral can be interpreted as finding the area under the curve of \(x\) from 0 to \(\frac{1}{2}\), resulting in \( \frac{1}{8} \).

Similarly, to find \(a_n\) and \(b_n\), we calculate:
\[ a_n = \frac{2}{T} \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \cos(2\pi n x) \, dx \]
\[ b_n = \frac{2}{T} \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \sin(2\pi n x) \, dx \]

Integral calculus tools are essential here because they help us handle the wave-like components of the functions represented by sines and cosines.

Being comfortable with integration techniques allows us to solve these integrals efficiently, and understanding their geometric interpretations helps grasp the underlying mechanics of Fourier series.