Problem 11 In each of the following problem... [FREE SOLUTION]

Chapter 7: Problem 11

In each of the following problems you are given a function on the interval \(-\pi

Short Answer

Expert verified

The Fourier series of the periodic function is \( f(x) = \sum_{n=1}^{\infty} \frac{1 - \cos(n\pi)}{2}\sin(nx) \).

Step by step solution

- Understanding the Function

Given the function is defined as: \[ f(x)=\begin{cases} 0, & -\pi<x<0, \ \sin x, & 0<x<\pi \end{cases} \] This function is defined piecewise on the interval \( -\pi < x < \pi \). One part is zero and the other part is sin(x).

- Sketching the Periodic Function

Since the function is periodic with a period of \( 2\pi \), we need to sketch it for multiple periods. One period is from \( -\pi \) to \( \pi \). Then repeating this pattern, another period would be from \( \pi \) to \( 3\pi \), and so on. The sketch will look like a series of segments of sin(x) followed by zeros.

- Expressing the Function Mathematically

To find the Fourier series, we first express \( f(x) \) as a periodic function with period \( 2\pi \): \[ f(x) = f(x + 2n\pi) \] where \( n \) is an integer.

- Fourier Coefficient Calculation

We calculate the Fourier coefficients: \[ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx \] Since the integral of \( f(x) \) over \( -\pi \) to \( 0 \) is zero and over \( 0 \) to \( \pi \) is \( \int_0^\pi \sin(x) \, dx \), we can solve: \[ a_0 = \frac{1}{2\pi} \left[ \int_0^\pi \sin(x) \, dx \right] = 0 \] since \( \int_0^\pi \sin(x) \, dx \) is zero.

- Sine and Cosine Coefficients

Now we find \( a_n \) and \( b_n \): \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cos(nx) \, dx \]For \( f(x) \), the integral splits into two intervals: \( -\pi \) to \( 0 \) and \( 0 \) to \( \pi \), leading to: \[ a_n = \frac{1}{\pi} \left[ \int_0^\pi \sin(x) \cos(nx) \, dx \right] \]Using integration by parts, \( a_n = 0 \).

- Finding \( b_n \)

We find \( b_n \) as follows: \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\sin(nx) \, dx \]This splits into two intervals: \( -\pi \) to \( 0 \) and \( 0 \) to \( \pi \), simplifying to: \[ b_n = \frac{1}{\pi} \left[ \int_0^\pi \sin(x) \sin(nx) \, dx \right] \] Using the orthogonality of sines, \( b_n = \frac{1}{2}(1 - \cos(n\pi)) \).

- Constructing the Fourier Series

Combining \( a_0 \), \( a_n \), and \( b_n \), the Fourier series is: \[ f(x) = \sum_{n=1}^{\infty} b_n \sin(nx) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Periodic Function

A periodic function repeats its values at regular intervals, known as the period. In this exercise, the function given is defined within the interval \(-\pi < x < \pi\). To understand this function over all real values of x, we extend it by repeating the given interval. This extension creates a periodic function with period \(2\pi\). This means that \(f(x) = f(x + 2n\pi)\) for any integer \(n\). By sketching multiple periods, we can visually observe the repetitive nature of the function.

Sine-Cosine Expansion

The sine-cosine expansion is a way to represent a periodic function as a sum of sines and cosines. This method is fundamental in Fourier series analysis. The general form of the Fourier series for a function \(f(x)\) is written as: \[ f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n\cos(nx) + b_n\sin(nx) \right) \] In our example, the function is odd, and it features sine only, leading to a simplified series with no cosine terms. Thus, the series becomes \[ f(x) = \sum_{n=1}^{\infty} b_n \sin(nx) \].

Fourier Coefficients

Fourier coefficients \(a_n\) and \(b_n\) are essential to constructing the Fourier series. They measure how much each sine and cosine function contributes to the overall shape of the periodic function. For our function, \(a_n \) coefficients turn out to be zero due to the specific nature of the function when integrated over the symmetrical interval: \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cos(nx) \, dx = 0 \]. The sine coefficients, \(b_n \), are calculated using the formula: \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\sin(nx) \, dx \]. Substituting values and solving through integration, we find \[ b_n = \frac{1}{2}(1 - \cos(n\pi)) \].

Piecewise Functions

Piecewise functions are functions defined by different expressions depending on the interval in which the input falls. For \(f(x)\): \[ f(x) = \left\{ \begin{array}{cr} \ 0, & -\pi < x < 0, \ \sin x, & 0 < x < \pi \ \end{array} \right. \], There are two pieces: one where the function equals zero and another where it equals \(sin x\). Such piecewise definitions are common in real-world scenarios where behavior changes under different conditions. It is crucial in analysis to handle each piece separately before combining results, especially in integral calculations.

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