Question

In each of the following problems you are given a function on the interval \(-\pi < x < \pi\) Sketch several periods of the corresponding periodic function of period \(2 \pi\). Expand the periodic function in a sine-cosine Fourier series. $$f(x)=\left\\{\begin{array}{lr} \pi+x, & -\pi < x < 0, \\ \pi-x, & 0 < x < \pi. \end{array}\right.$$

Short answer

The periodic function is: \( f(x) = \pi + \sum_{n=1}^\infty \frac{8(-1)^{n+1}}{n} \sin(nx) \)

Step-by-step solution

- Sketch the function

First, let's sketch the given function for one period. The function is split into two parts: 1. For \(-\pi < x < 0\), the function is defined as \(\pi + x\)\.2. For \(-0 < x < \pi\), the function is defined as \(\pi - x\)\.Plot these two linear equations in the interval \(-\pi, \pi\), then repeat this pattern to sketch several periods.

- Define the Fourier Series formula

The Fourier series expansion of a periodic function with period \(2\pi \) can be written as: \[ f(x) = a_0 + \sum_{n=1}^\infty (a_n\cos(nx) + b_n\sin(nx)) \] where \( a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx \) \( a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) dx \) \( b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) dx \)

- Compute \(a_0\)

Calculate the 0th coefficient \(a_0\): \[ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx \]Split the integral into two parts: \[ a_0 = \frac{1}{2\pi} \left( \int_{-\pi}^{0} (\pi + x) dx + \int_{0}^{\pi} (\pi - x) dx \right) \]. Performing the integrals, we find: \[ a_0 = \pi \]

- Compute \(a_n\) coefficients

Calculate the cosine coefficients \(a_n\): \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) dx \]Split it again: \[ a_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} (\pi + x) \cos(nx) dx + \int_{0}^{\pi} (\pi - x) \cos(nx) dx \right) \] Evaluate these integrals to find: \[ a_n = 0 \text{ for all } n \]

- Compute \(b_n\) coefficients

Calculate the sine coefficients \(b_n\): \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) dx \]Again splitting into two parts: \[ b_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} (\pi + x) \sin(nx) dx + \int_{0}^{\pi} (\pi - x) \sin(nx) dx \right) \] Performing these calculations, we obtain: \[ b_n = \frac{4}{n} \sin(n\pi) - 4 \cdot \frac{(-1)^n}{n} = \frac{8(-1)^{n+1}}{n} \]

- Combine the results

Combine the coefficients to form the Fourier series: \[ f(x) = \pi + \sum_{n=1}^\infty \frac{8(-1)^{n+1}}{n} \sin(nx) \]

Key concepts

Periodic Functions

A periodic function repeats its values at regular intervals. For example, the sine and cosine functions repeat every \(2\pi\) and are common periodic functions studied in trigonometry and signal processing. The given problem involves a function defined in the interval \(-\pi < x < \pi\), which then repeats for other intervals of length \(2\pi\). This repeating nature is essential in Fourier series, which represent functions as sums of sine and cosine terms. It's crucial to understand that any function that continuously repeats its values over a fixed length is periodic.

Trigonometric Series

A trigonometric series expresses a function as a series of sine and cosine terms. The Fourier series is one such expansion, representing any periodic function as a sum:
\[ f(x) = a_0 + \sum_{n=1}^\infty (a_n \cos(nx) + b_n \sin(nx)) \]
where \(a_0\), \(a_n\), and \(b_n\) are coefficients. Trigonometric series are valuable because they break down complex periodic functions into simpler oscillating components. This decomposition helps in analyzing and understanding the behavior of periodic functions in various domains such as physics, engineering, and signal processing.

Coefficients Calculation

Calculating the coefficients \(a_0\), \(a_n\), and \(b_n\) is a key part of finding a Fourier series. Here are the integral formulas for these coefficients:

• Zero-term coefficient: \[ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx \]
• Cosine coefficients: \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) dx \]
• Sine coefficients: \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) dx \]

Using these integrals, we find the constants' values that multiply the sine and cosine terms. For example, in the given solution, \(a_0 = \pi\), \(a_n = 0\) for all \(n\), and \(b_n = \frac{8(-1)^{n+1}}{n}\).

Sine and Cosine Integrals

Integrals involving sine and cosine functions are at the heart of calculating Fourier series coefficients. These integrals exploit the orthogonality properties of sine and cosine functions over a complete period:

• For the cosine integral: \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) dx \]
• For the sine integral: \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) dx \]

In the given exercise, evaluating these integrals for the specific function \(f(x) = \pi + x\ \) for \(-\pi < x < 0\), and \(\pi - x\ \) for \(0 < x < \pi\), involves splitting the integrals into two parts and simplifying the expressions. These steps help derive the Fourier coefficients that combine to form the Fourier series, providing a full representation of the original periodic function.