Theorem
Suppose \(f: \mathbb{R} \to \mathbb{R}\) is \(2\pi\)-periodic, continuous, and piecewise smooth. Let \(a_n\), \(b_n\), and \(c_n\) be its Fourier coefficients, and let \(a_n'\), \(b_n'\), and \(c_n'\) be the Fourier coefficients of its derivative \(f'\). Then \[ a_n' = n b_n, \quad b_n' = -n a_n, \quad c_n' = i n c_n. \]
Let’s examine \(c_n'\), to see why this is true, which is given by the standard formula: \[ c_n' = \frac{1}{2\pi} \int_{-\pi}^{\pi} f'(\theta) e^{-i n \theta} \, d\theta. \]
Then, using integration by parts, we get that \[ c_n' = \frac{1}{2\pi} f(\theta) e^{-i n \theta} \Big|_{-\pi}^{\pi} + \frac{in}{2\pi} \int_{-\pi}^{\pi} f(\theta) e^{-i n \theta} \, d\theta = i n c_n, \] since the first term vanishes because \(f\) is periodic and \(e^{-in\pi} - e^{in\pi} = 2i \sin{n\pi} = 0\).
Differentiation Theorem
Suppose \(f: \mathbb{R} \to \mathbb{R}\) is \(2\pi\)-periodic, continuous, and piecewise smooth, and suppose that its derivative \(f'\) is also piecewise smooth. If \(a_n\), \(b_n\), and \(c_n\) are the Fourier coefficients of \(f\), then
\[ f'(\theta) = \sum_{n=1}^{\infty} \left(n b_n \cos{n\theta} - n a_n \sin{n\theta}\right) = \sum_{n=-\infty}^{\infty} i n c_n e^{i n \theta}. \]
for all \(\theta\) at which \(f'\) is continuous. At discontinuities of \(f'\), the Fourier series on the right converge to the average of the left- and right-hand limits of \(f'\).
Since \(f'\) is piecewise smooth, our main convergence theorem applies and tells us that \(f'\) converges to its Fourier series at all points of continuity, and to the average of the left- and right-hand limits at discontinuities.
But the Fourier coefficients of \(f'\) were computed on the previous slide. The result follows.
Notice that the Fourier series of \(f'\) is obtained by term-by-term differentiation of the Fourier series of \(f\): \[ f(\theta) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left(a_n \cos{n\theta} + b_n \sin{n\theta}\right) = \sum_{n=-\infty}^{\infty} c_n e^{i n \theta}. \]
Consider the \(2\pi\)-periodic function \(f\) with \(f(\theta) = |\theta|\) for \(\theta \in (-\pi, \pi)\). We know its Fourier series is given by \[ f(\theta) = \frac{\pi}{2} - \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\cos{\left[(2n-1)\theta\right]}}{(2n-1)^2} . \]
Using this, obtain the Fourier series of the function \[ g(\theta) = \begin{cases} 1 & : 0 < \theta < \pi, \\ -1 & : -\pi < \theta < 0. \end{cases} \]
Consider the \(2\pi\)-periodic function \(f\) with \(f(\theta) = \theta\) for \(\theta \in (-\pi, \pi)\). We know its Fourier series is given by \[ f(\theta) = 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin{n\theta}. \]
Then the term-by-term differentiated series is \[ 2 \sum_{n=1}^{\infty} (-1)^{n+1} \cos{n\theta}. \]
Does this converge to \(f'(\theta) = 1\)? Discuss.
Integration Theorem
Suppose \(f\) is \(2\pi\)-periodic and piecewise continuous, and let \(a_n\), \(b_n\), and \(c_n\) be its Fourier coefficients. Define \[ F(\theta) = \int_0^{\theta} f(\phi) \, d\phi. \]
Then for all \(\theta\) we have \[ F(\theta) = \frac{a_0}{2}\theta + \sum_{n=1}^{\infty} \left(\frac{a_n}{n} \sin{n\theta} - \frac{b_n}{n} \cos{n\theta}\right) = c_0\theta + \sum_{n\neq 0} \frac{c_n}{i n} e^{i n \theta} . \tag{1} \]
Notice that \(f\) is not assumed piecewise smooth, so we cannot apply our convergence theorem to guarantee that it converges to its Fourier series. However, the theorem is still true!
Notice also that the series on the right-hand side is obtained by term-by-term integration of the Fourier series of \(f\), \[ \frac{a_0}{2} + \sum_{n=1}^{\infty} \left(a_n \cos{n\theta} + b_n \sin{n\theta}\right) = \sum_{n=-\infty}^{\infty} c_n e^{i n \theta}, \] by first replacing \(\theta\) with \(\phi\) and then integrating from \(0\) to \(\theta\).
Finally, notice that the series on the right-hand side of (1) is not a Fourier series, because of the \(a_0 \theta / 2\) and \(c_0 \theta\) terms. However, it is a Fourier series if \(f\) has mean value zero, i.e., if \(a_0 = 0\).