Question

In each case, find the Fourier approximation \(f_{5}\) of the given function in \(\mathbf{C}[-\pi, \pi]\) a. \(f(x)=\pi-x\) b. \(f(x)=|x|=\left\\{\begin{aligned} x & \text { if } 0 \leq x \leq \pi \\\\-x & \text { if }-\pi \leq x<0 \end{aligned}\right.\) c. \(f(x)=x^{2}\) d. \(f(x)=\left\\{\begin{array}{ll}0 & \text { if }-\pi \leq x<0 \\ x & \text { if } 0 \leq x \leq \pi\end{array}\right.\)

Short answer

Compute the Fourier coefficients for terms 0 to 5 and sum them to form the approximation \( f_5(x) \).

Step-by-step solution

Understand the Fourier Series

The Fourier series of a function on the interval [-\pi, \pi] can be expressed as:\[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right) \]To find the Fourier approximation, we'll compute the coefficients \( a_0, a_n, \) and \( b_n \) for \( n = 1, 2, \ldots, 5 \) and sum the series up to these terms.

Compute Coefficients for Part (a)

Given \( f(x) = \pi - x \), we compute the coefficients:\[ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} (\pi - x) \, dx \]\[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} (\pi - x) \cos(nx) \, dx \]\[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} (\pi - x) \sin(nx) \, dx \]Solving these integrals will give us the specific coefficients needed for the Fourier terms up to \( n = 5 \).

Compute Coefficients for Part (b)

The function is piecewise:\[ f(x) = \begin{cases} x, & 0 \leq x \leq \pi \ -x, & -\pi \leq x < 0 \end{cases} \]Find Fourier coefficients:\[ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} |x| \, dx \] \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} |x| \cos(nx) \, dx \]\[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} |x| \sin(nx) \, dx \]Evaluate these integrals for the first 5 terms.

Compute Coefficients for Part (c)

For \( f(x) = x^2 \), calculate:\[ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \, dx \]\[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cos(nx) \, dx \]\[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \sin(nx) \, dx \]Solving these integrals provides the Fourier coefficients for terms up to \( n = 5 \).

Compute Coefficients for Part (d)

For \( f(x) \) being piecewise:\[ f(x) = \begin{cases} 0, & -\pi \leq x < 0 \ x, & 0 \leq x \leq \pi \end{cases} \]Determine the Fourier coefficients:\[ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx \]\[ a_n = \frac{1}{\pi} \int_{0}^{\pi} x \cos(nx) \, dx \]\[ b_n = \frac{1}{\pi} \int_{0}^{\pi} x \sin(nx) \, dx \]Evaluate these integrals for the first 5 terms.

Construct Fourier Approximations

Combine the coefficients computed in previous steps into the Fourier series formula for each part. Sum the series terms up to \( n = 5 \) using:\[ f_5(x) = \frac{a_0}{2} + \sum_{n=1}^{5} (a_n \cos(nx) + b_n \sin(nx)) \]This gives you the desired approximations \( f_5(x) \) for each function \( f(x) \).

Key concepts

Fourier coefficients

Fourier coefficients are essential components of a Fourier series. They help express a periodic function on the interval \([-\pi, \pi]\) as a combination of sine and cosine functions. These coefficients determine the amplitude and the phase of each sine and cosine component in the Fourier series expansion. Let's break it down further:

• \(a_0\) coefficient: Represents the average value of the function over the given interval. It's found using the integral: \[ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx \]
• \(a_n\) coefficients: These coefficients indicate how much of each cosine wave is needed in the series: \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx \]
• \(b_n\) coefficients: These determine the sine components of the function: \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \]

To compute these coefficients, you'll integrate over the function's specified interval. Each term in the series relies on these coefficients, enabling us to accurately create approximations, like the given \(f_5(x)\). Understanding these coefficients' roles helps you appreciate the Fourier series' capability to approximate any periodic function.

trigonometric series

A trigonometric series refers to an infinite sum of sines and cosines.The Fourier series is a type of trigonometric series, critical for representing periodic functions. Here's why it matters:

• Nature of sine and cosine functions: These functions are periodic and oscillatory, making them ideal for constructing series that mimic other periodic functions over specific intervals.
• Fourier series as special trigonometric series: It takes the form: \[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right) \] In this series, each term contributes to the overall shape of the function.

By summing these terms to a specified \(n\), such as \(n=5\) as in the exercise, we can build accurate "snapshots" of more complex periodic functions. Through the trigonometric series, students gain valuable insights into the synthesis of complex waveforms, which is integral to physics and engineering.

piecewise functions

Piecewise functions define different expressions over distinct parts of their domains. These functions can change behavior abruptly, which makes them challenging yet essential to understand when finding Fourier series:

• Segmented Definition: A piecewise function might look like: \[ f(x) = \begin{cases} x, & 0 \leq x \leq \pi \ -x, & -\pi \leq x < 0 \end{cases} \] Here, \(f(x)\) is defined by two expressions over two intervals.
• Handling in Fourier Series: Integrals need to be computed separately for each segment. For example, when computing the Fourier coefficients, we treat each piece independently in its respective range: \[ \int |x| \cos(nx) \, dx = \int_0^{\pi} x \cos(nx) \, dx + \int_{-\pi}^{0} -x \cos(nx) \, dx \]

Due to their segmented nature, piecewise functions offer a practical view into how Fourier series can be employed to accommodate and approximate, challenging real-world problems, including signal processing and mechanical modeling. Working with piecewise functions equips students with the tools to handle discontinuities efficiently.