Question

Assume that \(f\) has a Fourier sine series $$ f(x)=\sum_{n=1}^{\infty} b_{n} \sin (n \pi x / L), \quad 0 \leq x \leq L $$ (a) Show formally that $$ \frac{2}{L} \int_{0}^{L}[f(x)]^{2} d x=\sum_{n=1}^{\infty} b_{n}^{2} $$ This relation was discovered by Euler about \(1735 .\)

Short answer

Question: Prove that the sum of the square of the coefficients of the Fourier sine series representation of a function \(f(x)\) is equal to the integral of the square of the function multiplied by a constant. Answer: We showed that for a function \(f(x)\) with Fourier sine series representation, the following relationship holds: $$ \frac{2}{L} \int_{0}^{L} [f(x)]^2 dx = \sum_{n=1}^{\infty} b_n^2 $$ where \(b_n\) are the coefficients in the Fourier sine series representation of \(f(x)\) and \(L\) is the length of the interval on which the function is defined.

Step-by-step solution

1. Review the definition of Fourier sine series

A Fourier sine series of a function \(f(x)\) defined on the interval \(0 \leq x \leq L\) 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 found using the following formula: $$ b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx $$

2. Multiply both sides of the equation by \(f(x)\) and integrate with respect to \(x\) over the interval \([0, L]\)

We now want to investigate the integral of the square of \(f(x)\) over the interval \([0, L]\). We start by multiplying both sides of the Fourier sine series representation of \(f(x)\) by \(f(x)\) and integrating with respect to \(x\) over the interval \([0, L]\): $$ \int_{0}^{L} [f(x)]^2 dx = \int_{0}^{L}\left(f(x)\sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{L}\right)\right) dx $$

3. Use the orthogonality property of sine functions on the right-hand side of the equation

The orthogonality property states that for sine functions with different integers \(m\) and \(n\): $$ \int_{0}^{L}\sin\left(\frac{m\pi x}{L}\right)\sin\left(\frac{n\pi x}{L}\right)dx = \left\{\begin{matrix} 0, & m \neq n\\ \frac{L}{2}, & m = n \end{matrix}\right. $$ Utilizing the property we can rewrite the right-hand side of the equation as follows: $$ \int_{0}^{L} [f(x)]^2 dx = \sum_{n=1}^{\infty} b_n \left(\frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx \right) $$

4. Simplify the left-hand side of the equation and compare both sides

Now we recognize that the expression inside the summation on the right-hand side is simply the definition for \(b_n\): $$ \int_{0}^{L} [f(x)]^2 dx = \sum_{n=1}^{\infty} b_n^2 \left(\frac{L}{2}\right) $$ Now we can isolate the sum of the square of the coefficients on the right-hand side by multiplying both sides of the equation by \(\frac{2}{L}\): $$ \frac{2}{L} \int_{0}^{L} [f(x)]^2 dx = \sum_{n=1}^{\infty} b_n^2 $$ We have shown that the integral of the square of the function is equal to the sum of the square of its coefficients multiplied by a constant.

Key concepts

Orthogonality of Sine Functions

The orthogonality of sine functions is a key concept in understanding Fourier series. When we talk about orthogonal functions, we're referring to functions that are "independent" of each other in a specific sense. For sine functions within a Fourier series, this means they do not overlap when integrated over a specific interval.Consider two sine functions, \( \sin\left(\frac{m\pi x}{L}\right) \) and \( \sin\left(\frac{n\pi x}{L}\right) \). When you integrate the product of these functions over the interval from 0 to \( L \), something interesting happens:

• If \( m eq n \), the integral equals zero.
• If \( m = n \), the integral equals \( \frac{L}{2} \).

This orthogonality property is fundamental because it simplifies the Fourier series. It allows us to isolate each term in the series when finding the coefficients. This is useful in many applications, such as signal processing and solving differential equations.

Coefficients in Fourier Series

In a Fourier series, the coefficients play a crucial role as they determine the weight of each sine or cosine wave in the series. In the Fourier sine series described, the coefficients are denoted as \( b_n \).To find these coefficients, we use the formula:\[b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx\]This formula arises directly from the orthogonality property of sine functions. Because the sine functions are orthogonal, we can multiply the original function \( f(x) \) by \( \sin\left(\frac{n\pi x}{L}\right) \) and integrate over the defined interval, essentially "filtering out" all terms except those where the functions coincide.The value of each \( b_n \) tells us how much of each sine wave is needed to reconstruct \( f(x) \). When all terms are summed, they form an approximate representation of the original function. This is the essence of how Fourier series can represent periodic functions.

Integral of Function Squares

The integral of the square of a function is a concept that appears in many mathematical contexts, especially when dealing with Fourier series. It provides a measure of the "energy" or "power" of the function over a given interval.In the context of the problem you're solving, you're asked to demonstrate that:\[\frac{2}{L} \int_{0}^{L} [f(x)]^2 dx = \sum_{n=1}^{\infty} b_n^2\]This equation signifies that the integral of the square of the function \( f(x) \), scaled by \( \frac{2}{L} \), is exactly the sum of the squares of the Fourier coefficients. It connects the continuous domain, via the integral, with the discrete domain, via the Fourier coefficients.This result, identified by Euler, is powerful because it allows for the distribution of the function's energy across its frequency components, represented by these coefficients. Such insights are foundational in physical applications like signal processing and quantum mechanics, where understanding the distribution of energy is crucial.