Question

Let $$ f(x)= \begin{cases}x, & 0 \leq x \leq \frac{\pi}{2}, \\ \pi-x, & \frac{\pi}{2} \leq x \leq \pi,\end{cases} $$ and let \(\tilde{f}\) be the odd extension of \(f\) to \([-\pi, \pi]\). Find the Fourier series of \(\tilde{f}\) on \([-\pi, \pi]\).

Short answer

The Fourier series of \(\tilde{f}\) contains only sine terms, based on its odd extension over \([-\pi, \pi]\).

Step-by-step solution

Analyze the Piecewise Function

The function \( f(x) \) is defined as \( f(x) = x \) when \( 0 \leq x \leq \frac{\pi}{2} \) and \( f(x) = \pi - x \) when \( \frac{\pi}{2} \leq x \leq \pi \). This is a piecewise function that is defined on the interval \([0, \pi]\).

Define the Odd Extension

The odd extension \( \tilde{f}(x) \) of a function \( f(x) \) is defined as \( \tilde{f}(x) = -f(-x) \). Thus, for the function \( f(x) \), the odd extension \( \tilde{f} \) on \([-\pi, \pi]\) will be \( \tilde{f}(x) = f(x) \) for \( 0 \leq x \leq \pi \) and \( \tilde{f}(x) = -f(-x) \) for \( -\pi \leq x < 0 \).

Determine the Fourier Series Components

Since \( \tilde{f} \) is an odd function, the Fourier series will only contain sine terms (\( b_n \)). The coefficient for each term is given by:\[b_n = \frac{2}{\pi} \int_0^{\pi} \tilde{f}(x) \sin\left(\frac{n\pi x}{\pi}\right) \,dx\]We need to calculate this integral for each \( n \).

Calculate the Fourier Coefficients

Due to symmetry and periodicity, evaluate:\[b_n = \frac{2}{\pi} \left( \int_0^{\frac{\pi}{2}} x \sin(nx) \,dx + \int_{\frac{\pi}{2}}^{\pi} (\pi-x) \sin(nx) \,dx \right)\]Compute these integrals separately for each section using integration by parts.

Integrate by Parts for Each Segment

Use integration by parts for both segments:1. \( \int_0^{\frac{\pi}{2}} x \sin(nx) \,dx \): Let \( u = x \), \( dv = \sin(nx) \, dx \).2. \( \int_{\frac{\pi}{2}}^{\pi} (\pi-x) \sin(nx) \,dx \): Let \( u = \pi - x \), \( dv = \sin(nx) \, dx \).Solve separately, simplifying their resulting expressions.

Construct the Complete Fourier Series

After calculating the coefficients \( b_n \), write the Fourier series as:\[\tilde{f}(x) = \sum_{n=1}^{\infty} b_n \sin(nx)\]Substitute the computed \( b_n \) values to get the final series expression.

Key concepts

Understanding Piecewise Functions

Piecewise functions are defined by different expressions over different parts of their domain. They can appear complicated at first, but breaking them into pieces makes them simpler to analyze.
For the function \( f(x) = \begin{cases}x, & 0 \leq x \leq \frac{\pi}{2}, \ \pi-x, & \frac{\pi}{2} \leq x \leq \pi,\end{cases} \), this means:

• When \( x \) is between 0 and \( \frac{\pi}{2} \), the function behaves like the line \( y = x \).
• When \( x \) is between \( \frac{\pi}{2} \) and \( \pi \), the function follows the line \( y = \pi - x \), which is a reflection of the previous segment over the line \( x = \frac{\pi}{2} \).

By understanding each piece, we can later perform helpful operations like extensions and integrations more easily.

Creating an Odd Function Extension

An odd function is symmetric about the origin, meaning that \( f(-x) = -f(x) \). To extend a piecewise function like \( f(x) \) to be odd over \([-\pi, \pi]\), we define:

• \( \tilde{f}(x) = f(x) \) for \( 0 \leq x \leq \pi \)
• \( \tilde{f}(x) = -f(-x) \) for \( -\pi \leq x \leq 0 \)

This method allows us to create a function that naturally fits into the symmetry rules of Fourier series analysis.
Odd functions only have sinus components in their Fourier series, making them perfect candidates for sine-only Fourier expansions.

Mastering Integration by Parts

Integration by Parts is a technique often used to solve the integral of a product of functions. It is based on the rule: \[ \int u \, dv = uv - \int v \, du \]
In the context of finding Fourier coefficients, this technique is invaluable. Taking the integral \( \int_0^{\frac{\pi}{2}} x \sin(nx) \,dx \) as an example:

• Choose \( u = x \) so that \( du = dx \)
• Choose \( dv = \sin(nx) \, dx \), leading to \( v = -\frac{1}{n} \cos(nx) \)

Substitute into the integration by parts formula to solve this integral efficiently. Repeat similar steps for other segments and functions as needed.
This technique simplifies what could otherwise be complex calculations, providing manageable solutions.

Extracting Fourier Coefficients

Fourier coefficients are key to building a Fourier series, allowing a function to be expressed as a sum of sinusoidal functions. For odd functions, only the \( b_n \) coefficients (linked to sine terms) are non-zero.

• The coefficients are calculated using:\[ b_n = \frac{2}{\pi} \int_0^{\pi} \tilde{f}(x) \sin\left(\frac{n\pi x}{\pi}\right) dx \]

In practice, compute them by evaluating separate integrals for each piece of your function.
Using our function, the work involves evaluating two integrals through integration by parts. Each results in specific \( b_n \) values that form a series.
Assembling these allows a description of \( \tilde{f}(x) \) as a convergent series:\[ \tilde{f}(x) = \sum_{n=1}^{\infty} b_n \sin(nx) \]This series captures the essence of \( \tilde{f}(x) \) over \([-\pi, \pi]\), crafted entirely from its sine wave components.