Partial Differential Equations: Exercises

10 Fourier series on different intervals

Exercise 1: Fourier cosine and sine series on \([0,\pi]\)

We again use the table for this exercise.

For each of the following functions, compute both the Fourier cosine and sine series on the interval \([0,\pi]\). (Attempt to use the table wherever possible!) To what values do these series converge when \(\theta=0\) and \(\theta=\pi\)?

  1. \(f(\theta) = 1\)

  2. \(f(\theta) = \pi - \theta\)

  3. \(f(\theta) = \theta^2\)

  1. The even extension of \(f\) is just the constant function \(f(\theta) = 1\) on \([-\pi,\pi]\), so the Fourier cosine series is just \(1\). The odd extension of \(f\) is the function in entry 6 of the table, so the Fourier sine series is \[ \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\sin{\left[(2n-1)\theta\right]}}{2n-1}. \] The latter converges to \(0\) at both \(\theta=0\) and \(\theta=\pi\).

  2. The odd extension of \(f\) is the function in entry 3 of the table, so the Fourier sine series is \[ 2 \sum_{n=1}^{\infty} \frac{\sin{n\theta}}{n}. \] This converges to \(\pi\) at \(\theta=0\) and to \(0\) at \(\theta=\pi\). The even extension of \(f\) is the function \(-g(\theta) + \pi\), where \(g(\theta)\) is function in entry 2 of the table. Thus the cosine series is \[ \frac{\pi}{2} + \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\cos{\left[(2n-1)\theta\right]}}{(2n-1)^2}. \] This converges to \(\pi\) at \(\theta=0\) and to \(0\) at \(\theta=\pi\).

  3. The even extension of \(f\) is the function in entry 16 of the table, so the cosine series is \[ \frac{\pi^2}{3}+ 4 \sum_{n=1}^\infty \frac{(-1)^n}{n^2} \cos{n\theta}. \] This converges to \(0\) at \(\theta=0\) and to \(\pi^2\) at \(\theta=\pi\). I don’t see a way to use the table for the odd extension of \(f\), so we compute the Fourier sine series from scratch: \[ b_n = \frac{2}{\pi} \int_0^\pi \theta^2 \sin{n\theta} \, d\theta = \frac{2}{\pi} \left( \frac{2-\pi^2n^2}{n^3}\cdot (-1)^n - \frac{2}{n^3} \right). \] Thus the sine series is \[ \frac{2}{\pi} \sum_{n=1}^\infty \left( \frac{2-\pi^2n^2}{n^3}\cdot (-1)^n - \frac{2}{n^3} \right)\sin{n\theta}. \] This converges to \(0\) at \(\theta=0\) and to \(\pi^2\) at \(\theta=\pi\).

Exercise 2: Fourier cosine and sine series on different intervals

You know the drill. Try to make use of the table rather than working from scratch.

Expand each of the following functions as a series of the indicated type.

  1. \(f(x)=1\) as a sine series on \((0,6\pi]\).

  2. The function \[ f(x) = \begin{cases} 1 & : 0 < x < 2, \\ -1 & : 2 \leq x \leq 4, \\ \end{cases} \] as a cosine series on \((0,4]\).

  1. We make the change of variables \(x = 6\theta\), so that the function \[ g(\theta) = f(6\theta) = 1 \] is defined on the interval \((0, \pi]\). Its odd extension is the function in entry 6 of the table, so the Fourier sine series is \[ \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\sin{\left[(2n-1)\theta\right]}}{2n-1}, \] which converges to \(0\) at \(\theta=0\) and \(\theta=\pi\). Substituting \(\theta = x/6\) gives the sine series for \(f\): \[ \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\sin{\left[\frac{(2n-1)x}{6}\right]}}{2n-1}. \]

  2. I still need to write up this solution. Check later!