Problem 9 - Advanced Engineering Mathematics

Question

In Problems, find the Fourier series of \(f\) on the given interval. $$ f(x)=\left\\{\begin{array}{lr} 0, & -\pi

Step-by-Step Solution

Answer

The Fourier series is \( f(x) = \sin(x) \) for \( 0 \leq x < \pi \).

Step 1: Identify the Function Properties

First, notice that the given function \( f(x) \) is piecewise with periods from \( -\pi \) to \( \pi \). It is important to note the two parts of the function: \( f(x) = 0 \) for \( -\pi < x < 0 \) and \( f(x) = \sin(x) \) for \( 0 \leq x < \pi \). This is a half-range expansion problem focusing on the sine function.

Step 2: Establish the Fourier Sine Series Formula

For a function defined as \( f(x) \) on \( 0 \leq x < L \), the Fourier sine series is given by:\[ f(x) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{L}\right) \]where the coefficients \( b_n \) are calculated as:\[ b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) \, dx \]In this case, the interval is \( [0, \pi] \), and \( L = \pi \).

Step 3: Calculate the Fourier Coefficients

Using the formula for \( b_n \), we find:\[ b_n = \frac{2}{\pi} \int_{0}^{\pi} \sin(x) \sin\left(\frac{n\pi x}{\pi}\right) \, dx \]This simplifies to:\[ b_n = \frac{2}{\pi} \int_{0}^{\pi} \sin(x) \sin(nx) \, dx \]

Step 4: Solve the Integral

The integral \( \int_{0}^{\pi} \sin(x) \sin(nx) \, dx \) uses the product-to-sum identities:\[ \sin(x) \sin(nx) = \frac{1}{2} \left[\cos((n-1)x) - \cos((n+1)x)\right] \]So, the integral becomes:\[ \int_{0}^{\pi} \frac{1}{2} \left[ \cos((n-1)x) - \cos((n+1)x) \right] \, dx \]We evaluate each component separately. The integrals of \( \cos((n-1)x) \) and \( \cos((n+1)x) \) over the interval \( [0, \pi] \) will both be zero for all \( n eq 1 \) due to the properties of cosine integrals over \( [0, \pi] \) which oscillate to zero.

Step 5: Coefficient Evaluation

For \( n = 1 \), the integral \[ \int_{0}^{\pi} \cos((n-1)x) \, dx = \int_{0}^{\pi} 1 \, dx = \pi \]Hence:\[ b_1 = \frac{2}{\pi} \cdot \frac{\pi}{2} = 1 \]For \( n eq 1 \), \( b_n = 0 \). Thus, the series only contains the term for \( n = 1 \).

Step 6: Write the Fourier Series

Since all \( b_n \) are zero except for \( b_1 \), the Fourier series is:\[ f(x) = \sin(x) \]This matches the original piecewise definition in the interval \( 0 \le x < \pi\).

Key Concepts

Piecewise Functions

A piecewise function is a type of function that has different expressions based on different intervals of the input variable, denoted as "pieces".It allows a function to possess distinct behaviors on different sections of its domain. A piecewise function is usually defined using inequalities that specify the variable's range for each piece.
In this exercise, we deal with the function \( f(x) \) defined on the interval \(-\pi\) to \(\pi\):

• \( f(x) = 0 \) when \(-\pi < x < 0\)
• \( f(x) = \sin(x) \) when \(0 \leq x < \pi\)

This kind of function provides a clear illustration of how complex behaviors can be simplified by splitting them into easier-to-handle segments. Understanding piecewise functions is crucial when working with the Fourier series as it helps in accurately defining the behavior of a function across its entire domain.

Sine Function

The sine function is a fundamental component in trigonometry and periodic functions. It is defined as the y-coordinate of a point on the unit circle, as it moves through an angle \( x \) from the positive x-axis. The sine function is known for its smooth oscillating pattern, repeating every \( 2\pi \).

• The form \( \sin(x) \) yields values between -1 and 1.
• It is an odd function, meaning \( \sin(-x) = -\sin(x) \).

In the context of Fourier series, the sine function is crucial in expressing waveforms as sums of sine and cosine waves, allowing the analysis of periodic functions. In our problem, the piece of \( f(x) = \sin(x) \) on the interval \(0 \leq x < \pi\) is used to construct a Fourier sine series, highlighting its relevance in decomposing complex functions into simpler trigonometric components.

Integral Calculation

Integrals are used to calculate areas under curves and are essential in determining Fourier coefficients. In this exercise, we perform an integral calculation to find these coefficients.

• The coefficients \( b_n \) are found using the formula \( b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) \, dx \).
• Specifically, these coefficients determine the weights of the sine components in the Fourier sine series.

For the function \( f(x) = \sin(x) \), the integral \( \int_{0}^{\pi} \sin(x) \sin(nx) \, dx \) employs integration techniques linking products of sine waves via product-to-sum identities. This resolves into simpler expressions that are easier to evaluate, essential for deriving the function's Fourier series representation. Understanding the integral calculation process is crucial to understanding how individual sine waves combine to approximate the function.

Product-to-Sum Identities

Product-to-sum identities are mathematical formulas used to transform products of trigonometric functions into sums or differences, making integration and differentiation manageable. These identities are invaluable when working with problems like calculating Fourier series coefficients. For sine functions,

• \( \sin(x) \sin(y) = \frac{1}{2} [\cos(x-y) - \cos(x+y)] \).

By transforming products into sums, evaluating integrals becomes straightforward since trigonometric identities simplify complex oscillations into simpler components that are easy to integrate over.
In the exercise, these identities reduce the integral \( \int_{0}^{\pi} \sin(x) \sin(nx) \, dx \) to an addition of cosine terms. This simplification is particularly helpful because the resulting cosine terms often integrate to zero over intervals like \([0, \pi]\), simplifying calculations. Such use illuminates the power of product-to-sum identities in mathematical problem-solving.