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.