Problem 12 - Advanced Engineering Mathematics

Question

Expand the given function in an appropriate cosine or sine series. $$ f(x)=\left\\{\begin{array}{lr} 1, & -2

Step-by-Step Solution

Answer

The function is expanded into a cosine series as: \( f(x) = 1 + \frac{4}{\pi} \cos\left(\frac{\pi x}{2}\right) + \frac{4}{3\pi} \cos\left(\frac{3\pi x}{2}\right) + \dots \).

Step 1: Identify the Function's Interval and Symmetry

The function \( f(x) \) is defined on the interval \( -2 < x < 2 \). Upon close examination, we see that the function exhibits even symmetry, as \( f(x) = f(-x) \) holds true throughout the interval. Thus, expanding \( f(x) \) into a Fourier cosine series is appropriate.

Step 2: Write the General Fourier Cosine Series Form

The Fourier cosine series for an even function \( f(x) \) defined on \( -L < x < L \) is given by:\[ f(x) = A_0 + \sum_{n=1}^{\infty} A_n \cos\left(\frac{n\pi x}{L}\right) \]where \( L = 2 \) in our problem.

Step 3: Calculate the Constant Term \( A_0 \)

The term \( A_0 \) is calculated by:\[ A_0 = \frac{1}{L} \int_{-L}^{L} f(x) \, dx = \frac{1}{2} \int_{-2}^{2} f(x) \, dx \]Split the integral into regions:\[ A_0 = \frac{1}{2} \left( \int_{-2}^{-1} 1 \, dx + \int_{-1}^{1} 0 \, dx + \int_{1}^{2} 1 \, dx \right) = \frac{1}{2} \times 2 = 1 \]Thus, \( A_0 = 1 \).

Step 4: Compute the Coefficients \( A_n \)

For each \( n \geq 1 \), \[ A_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) \, dx = \frac{1}{2} \int_{-2}^{2} f(x) \cos\left(\frac{n\pi x}{2}\right) \, dx \]Split as before:\[ A_n = \frac{1}{2} \left( \int_{-2}^{-1} \cos\left(\frac{n\pi x}{2}\right) dx + \int_{1}^{2} \cos\left(\frac{n\pi x}{2}\right) dx \right) \]Evaluate these integrals to find each \( A_n \). Notably, due to symmetry and periodic properties of cosine, it results in:\[ A_n = \frac{4}{n\pi} \sin\left(\frac{n\pi}{2}\right) \text{ for odd } n \]Due to the integral of cosine over its period, \( A_n = 0 \) for even \( n \).

Step 5: Write the Full Cosine Series Expansion

Substitute the calculated coefficients into the general form:\[ f(x) = 1 + \sum_{\text{odd } n} \frac{4}{n\pi} \sin\left(\frac{n\pi}{2}\right) \cos\left(\frac{n\pi x}{2}\right) \]Thus, the cosine series expansion of \( f(x) \) is:\[ f(x) = 1 + \left(\frac{4}{\pi} \cos\left(\frac{\pi x}{2}\right) + \frac{4}{3\pi} \cos\left(\frac{3\pi x}{2}\right) + \dots \right) \]

Key Concepts

Even Symmetry

Understanding symmetry plays a crucial role in simplifying problems in mathematics. The function given in this exercise exhibits even symmetry, which means that it is symmetric about the y-axis. In simpler terms, this means that for any value of x, the function's value at x is the same as at -x. Mathematically, this is expressed as \( f(x) = f(-x) \).
This particular symmetry makes the Fourier cosine series the perfect tool for expansion because it inherently uses cosine functions, which are also symmetric around the y-axis. As a result, the Fourier cosine series will accurately represent an even function over a given interval without needing sine terms, which represent odd symmetry.

Fourier Series Coefficients

Fourier series coefficients are vital in constructing a Fourier series, as they determine how much of each function (sine and cosine) is needed to approximate the original function. In a Fourier cosine series for an even function, we focus only on the cosine coefficients \( A_n \). These coefficients are calculated by integrating the product of the original function and cosine functions over the function's interval of definition.
The constant term, \( A_0 \), represents the average value of the function over the interval, given by \( A_0 = \frac{1}{L} \int_{-L}^{L} f(x) \, dx \).
Subsequent coefficients \( A_n \) are calculated as:

• \( A_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) \, dx \)

These coefficients capture the amount of each cosine function needed to match the shape of \( f(x) \).

Cosine Function

The cosine function, represented as \( \cos(x) \), is a fundamental trigonometric function with a periodic and even nature. This means it repeats its pattern over regular intervals and is symmetric with respect to the y-axis. In the Fourier cosine series, these properties of the cosine function make it an ideal match for even functions.
When using cosine functions in Fourier series, they help reconstitute the waveform of the original function by adding waves atop each other in specific amounts defined by the coefficients. The combination of these cosine waves through different scales and shifts can accurately represent any well-behaved function over a given interval.
Cosine is crucial here as they ensure all portions of the function, whether large or small, across the interval of definition are well-represented, benefiting analysis and solving periodic function problems.

Piecewise Function

A piecewise function is a function defined by multiple sub-functions, each applying to a specific interval in the domain. The given function in the problem is piecewise, meaning it has different values depending on the segment of x it applies to:

• Value 1 for \(-2 < x < -1\) and \(1 < x < 2\).
• Value 0 for \(-1 < x < 1\).

This kind of representation allows complex functions to be expressed easily when they exhibit different behaviors over their domains. In terms of Fourier series, the challenge is to approximate each portion of the piecewise function with harmonics that smoothly transition between these segments, achieving seamless continuity and minimum error throughout.

Interval of Definition

The interval of definition for a function is the range of x-values over which the function is defined. In our exercise, the function \( f(x) \) is defined on the interval \(-2 < x < 2\). This interval serves as the basis for computation of the Fourier series and is critical for understanding where the function behaves in a particular manner.
The interval reveals where changes occur and helps in splitting the integral during the calculation of Fourier coefficients. For even functions in Fourier cosine series, this interval is mirrored around zero. Therefore, when integrating, particularly for computing coefficients, understanding this interval ensures that all parts of the function are integrated accurately for precise series terms.