Lecture 16 — Examples of Derivatives and Taylor Series

Mid-semester exam: Wednesday 22/04/2020 9am.

Remember from real analysis that we have functions differentiable once but not twice.

Continuing with derivatives, consider \frac{d}{dz} \log z where |z| > 0. Recall that in \mathbb C, \log z = \ln |z| + i \arg z=\ln r + i\theta. Looking at the second expression in its components, u = \ln r and v = \theta so u_r = 1/r, u_\theta = 0, v_r = 0 and v_\theta = 1. Checking C/R in polar coordinates, we need ru_r = v_\theta \quad \text{and}\quad u_\theta = -rv_r which we do have. We need to make \log a function so it can be continuous; we need to choose a branch. Pick a subset of \mathbb C_* such that \alpha < \theta < \alpha + 2\pi then \log is differentiable. From Lecture 15, \frac d{dz} \log z = e^{-i\theta}(u_r + iv_r) = e^{-i\theta}/r = 1/z. For example, \frac d{dz} \operatorname{Log} z = 1/z for -\pi < \operatorname{Arg}z < \pi and |z| > 0.

For f(z) = z^c where c \in \mathbb C_* is fixed, we have f(z) = \exp (c \log z) and f'(z) = c\exp (c \log z)/z by the chain rule and using the derivative of \log. We can also write this as z^c c/z = cz^{c-1} which is valid on any domain of the form \{z : |z| > 0, \alpha < \arg z < \alpha + 2\pi\}, due to the branch cut of \log.

Remark: Try this for g(z) = c^z.

Notation from real analysis

Given \Omega \subseteq \mathbb R^n,

• C(\Omega) = C^0(\Omega) is the set of continuous functions \Omega \to \mathbb R.
• C^k(\Omega) is the set of functions whose derivatives/partial derivatives of order 0, 1, \ldots, k exist and are continuous. (0-th order derivative is f itself.)
• C^\infty(\Omega) is the set of functions whose derivatives/partial derivatives exist and are continuous. That is, smooth functions.
• C^\omega(\Omega) is the set of real analytic functions. That is, for all x_0 \in \Omega,
  • 1. f has a power series expansion about x_0 (namely, its Taylor series), and
  • 2. f is given by its power series expansion (i.e. the power series converges to f on some neighbourhood of x_0.)

Note that (i) implies f is smooth, and in \mathbb R^n, (i) does not imply (ii).

Example: Consider an example to illustrate this past point. f(x) = \begin{cases} e^{-1/x^2} & x >0 \\ 0 & x \le 0 \end{cases} Then, f^{(n)}(x) exists for all x \ne 0 trivially and f^{(n)}(0) = 0 for all n. Also, f^{(n)} is continuous on \mathbb R. However, the Taylor series of f about 0 is \sum_{n=0}^\infty \frac{f^{(n)}(0) x^n}{n!} \equiv 0 so f is not equal to its Taylor series in a neighbourhood of 0. Therefore, f \in C^\infty(\mathbb R) but f \notin C^\omega(\mathbb R).

In real analysis, we have C^\omega \subsetneq C^\infty \subsetneq \cdots \subsetneq C^{1000} \subsetneq \cdots \subsetneq C^1 \subsetneq C^0. Next, we will be moving onto integration but there are some problems. There was the intuition of ‘area’ but how does this translate to \mathbb C? We could look at something like a two-dimensional volume under a hypersurface but that doesn’t really work. Instead, we can revert to a complex valued function of real parameters. Next lecture, we will see why this makes sense and how it leads to the familiar integration.