09 Differentiation and integration of Fourier series

Differentiation, part 1

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 \(a_0' = c_0' = 0\), and for \(n \geq 1\) we have \[ a_n' = n b_n, \quad b_n' = -n a_n, \quad c_n' = i n c_n. \]

• 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, part 2

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 \(f'\).

• Why is the theorem true?
  • 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. Q.E.D.
• Notice that the Fourier series of \(f'\) is obtained by term-by-term differentiation of the Fourier series of \(f\): \[ f(\theta) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left(a_n \cos{n\theta} + b_n \sin{n\theta}\right) = \sum_{n=-\infty}^{\infty} c_n e^{i n \theta}. \]

Exercise 1: Term-by-term differentiation

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 \[ g(\theta) = \begin{cases} 1 & : 0 < \theta < \pi, \\ -1 & : -\pi < \theta < 0. \end{cases} \]

Exercise 2: Term-by-term differentiation doesn’t always work

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

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 \[ F(\theta) - c_0\theta = \frac{A_0}{2} + \sum_{n=1}^{\infty} \left(\frac{a_n}{n} \sin{n\theta} - \frac{b_n}{n} \cos{n\theta}\right) = C_0 + \sum_{n\neq 0} \frac{c_n}{i n} e^{i n \theta}, \tag{1} \] where \[ C_0 = \frac{A_0}{2} = \frac{1}{2\pi} \int_{-\pi}^{\pi} F(\theta) \, d\theta. \]

• Why is the theorem true?
  • Define \(g(\theta) = f(\theta) - c_0\), and set \[ G(\theta) = \int_0^{\theta} g(\phi) \, d\phi = F(\theta) - c_0\theta. \] Then \(G\) is continuous and piecewise smooth, since \(g\) is piecewise continuous.
  • But \(G\) is also \(2\pi\)-periodic, because: \[ G(\theta+ 2\pi) - G(\theta) = \int_\theta^{\theta + 2\pi} g(\phi) \, d\phi = \int_{-\pi}^{\pi} f(\phi) \, d\phi - \int_{\theta}^{\theta+2\pi}c_0 \, d\phi = 2\pi c_0 - 2\pi c_0 = 0. \] for all \(\theta\).
  • Thus, our Fourier Convergence Theorem applies and tells us that \(G\) converges to its Fourier series everywhere. Suppose its Fourier coefficients are \(A_n\), \(B_n\), and \(C_n\).
  • The Fourier coefficients of \(g\) are the same as those of \(f\), except that \(c_0 = 0\) for \(g\).
  • Since \(G'(\theta) = g(\theta)\) wherever \(g\) is continuous, we know that \[ a_n = nB_n, \quad b_n = -nA_n, \quad c_n = i n C_n. \]
  • Substituting these into the Fourier series of \(G\) gives the desired result. Note that \(C_0\) and \(A_0\) are just the usual Fourier coefficients of \(G\). Q.E.D.
• A few additional things to notice:
  • Note \(f\) 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!
  • Provided that \(c_0=a_0=0\), note that the series in (1) are obtained by term-by-term integration of the Fourier series of \(f\) from \(0\) to \(\theta\) (after first replacing \(\theta\) with \(\phi\)).

Exercise 1: Term-by-term integration

Consider the \(2\pi\)-periodic function \(f\) with

\[ f(\theta) = \begin{cases} -1 & : -\pi < \theta < 0, \\ 1 & : 0 < \theta < \pi. \end{cases} \]

Then we know that

\[ f(\theta) = \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\sin{[(2n-1)\theta]}}{2n-1} \]

at all points of continuity. Identify the function \(F(\theta) = \int_0^{\theta} f(\phi) \, d\phi\), and use the integration theorem to obtain its Fourier series.

Uniform and absolute convergence of Fourier series

• Our Fourier Convergence Theorem only guarnatees pointwise convergence of the Fourier series of a piecewise smooth function \(f\). That is, for each fixed \(\theta\) at which \(f\) is continuous, we have \[ f(\theta) = \lim_{N\to \infty} S_N^f(\theta), \] where \(S_N^f\) is the \(N\)-th partial sum of the Fourier series of \(f\).
• But the manner in which this limit approaches \(f(\theta)\) can be vary from point to point. This can make it difficult to work with Fourier series.
• Instead, we seek a stronger notion of convergence, called uniform and absolute convergence, which says intuitively that the Fourier series converges to \(f\) “in the same way” for all \(\theta\). (Hence the term “uniform”.)
  • This buys us things like:
    • Continuity of \(f\) is guaranteed. The uniform limit of continuous functions is known to be continuous.
    • An integral sign \(\int_a^b\) can be pulled past the \(\sum_{n=1}^\infty\) sign. The uniform limit of integrable functions is known to be integrable, and the integral of the limit is the limit of the integrals.
    • If the Fourier series of \(f'\) converges uniformly, then the Fourier series of \(f\) can be differentiated term-by-term to obtain the Fourier series of \(f'\).
    • We may rearrange the terms of the Fourier series of \(f\) without changing its sum.
• When can we guarantee uniform and absolute convergence of the Fourier series of a function \(f\)? By what we said above, if this type of convergece held, then \(f\) must be continuous. It turns out that, along with piecewise smoothness, this is also sufficient:

Absolute and Uniform Convergence Theorem

Suppose \(f: \mathbb{R} \to \mathbb{R}\) is \(2\pi\)-periodic, continuous, and piecewise smooth. Then the Fourier series of \(f\) converges uniformly and absolutely to \(f\) on all of \(\mathbb{R}\).