Partial Differential Equations: Exercises

06 A first look at Fourier series, part 2

Exercise 1: More practice computing Fourier series

This exercise refers to the Fourier series listed in rows 1-20 of the file here, the same file as in Exercise 1 of the previous section.

  1. Verify, by hand, that the Fourier series for the function in row 7 of the table in the file above is the one shown.

  2. Verify, by hand, that the Fourier series for the function in row 16 of the table in the file above is the one shown.

For extra fun, open up an instance of Desmos, and see if you can manage to plot a few partial sums of the Fourier series.

The solutions are given in the PDF itself! Make sure your calculations match the Fourier series shown in the table.

Exercise 2: Bessel’s inequality (\(ab\)-version)

We saw the following “\(ab\)-version” of Bessel’s inequality in the slides: If \(f:\mathbb{R} \to \mathbb{C}\) is a \(2\pi\)-periodic function, integrable on \([-\pi, \pi]\), and if \(a_n\) and \(b_n\) are its Fourier coefficients, then \[ \frac{|a_0|^2}{4} + \frac{1}{2}\sum_{n=1}^\infty \left( |a_n|^2 + |b_n|^2 \right) \leq \frac{1}{2\pi} \int_{-\pi}^\pi |f(\theta)|^2 \, d\theta. \]

In this exercise, you will derive the \(ab\)-version of Bessel’s inequality from the \(c\)-version of Bessel’s inequality that we discussed and proved in class.

  1. First, show that \[ |a_n|^2 + |b_n|^2 = 2\left(|c_n|^2 + |c_{-n}|^2\right), \quad n\geq 1. \] You’ll need to use the conversion formulas between \(a_n\), \(b_n\) and \(c_n\), that we discussed in class.

  2. Then, derive the \(ab\)-version from the \(c\)-version.

  1. We know that \(a_n = c_n + c_{-n}\) and \(b_n = i(c_n - c_{-n})\). So \[ \begin{align*} |a_n|^2 + |b_n|^2 &= a_n \overline{a_n} + b_n \overline{b_n} \\ &= (c_n + c_{-n})(\overline{c_n} + \overline{c_{-n}}) + i(c_n - c_{-n})(-i)(\overline{c_n} - \overline{c_{-n}}) \\ &= 2c_n \overline{c_n} + 2c_{-n} \overline{c_{-n}} \\ &= 2\left(|c_n|^2 + |c_{-n}|^2\right). \end{align*} \]

  2. We have that \[ \sum_{n=-\infty}^\infty |c_n|^2 = |c_0|^2 + \sum_{n=1}^\infty \left( |c_n|^2 + |c_{-n}|^2 \right) = \frac{|a_0|^2}{4} + \frac{1}{2}\sum_{n=1}^\infty \left( |a_n|^2 + |b_n|^2 \right), \] where we used (a) and the fact that \(a_0 = 2c_0\). Now apply the \(c\)-version of Bessel’s inequality to get the \(ab\)-version: \[ \frac{|a_0|^2}{4} + \frac{1}{2}\sum_{n=1}^\infty \left( |a_n|^2 + |b_n|^2 \right) = \sum_{n=-\infty}^\infty |c_n|^2 \leq \frac{1}{2\pi} \int_{-\pi}^\pi |f(\theta)|^2 \, d\theta. \]