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Recall: Term-by-Term Differentiation for Power Series

If $f(z) = \sum_{n=0}^{\infty} a_nz^n$ for |z| < R (R > 0), then f is analytic on B(0; R) and $f'(z) = \sum_{n=1}^{\infty} na_nz^{n-1}$ for |z| < R.

Power series as a limit.. A power series $f(z) = \sum_{n=0}^{\infty} a_nz^n$ defines a function by the limit of its partial sums: $f(z)= \lim_{N \to \infty} S_N(z), \text{ where } S_N(z)=\sum_{n=0}^{N}a_nz^n$.

The domain of f is the set of points where this limit exists.
Radius of convergence. By the root or ratio test, there is a unique $R \in [0, \infty]$ (radius of convergence) such that (i) the series converges for all |z| < R; (ii) diverges for all |z| > R; (iii) on the boundary |z| = R, it must be examined or checked case‑by‑case.
At z = 0, for any $N \ge 0, S_N(0)=a_0 \cdot 0^0 + a_1 \cdot 0^1+\dots +a_N \cdot0^N$. In power-series contexts we adopt the convention $z^0 = 1$ for all z, including z = 0. This convention eliminates the usual $0^0$ indeterminacy and makes the constant term well defined. Hence, $S_N(0) = a_0, \forall N$, so the limit as $N \to \infty$ is also $a_0, f(0)=\lim_{N \to \infty} S_N(0) = a_0$. A verbose restatement of the fact is as follows $f(z) = \begin{cases} a_0, &z = 0 \\\\ \sum_{n=0}^{\infty} a_nz^n, &z \ne 0 \end{cases}$

Example: the exponential series

We begin by defining a function $f(z)$ as a power series, $f(z) = \sum_{n=0}^{\infty} \frac{z^n}{n!}$. Our first step is to find out where this function is well-defined, i.e., for which complex numbers z the series converges.

We use the Ratio Test for absolute convergence. A series converges absolutely if the limit of the ratio of its terms is less than 1.

$\left| \frac{a_{n+1}(z)}{a_n(z)} \right| = \left| \frac{\frac{z^{n+1}}{(n+1)!}}{\frac{z^n}{n!}} \right| = \left| \frac{z^{n+1}}{(n+1)!} \cdot \frac{n!}{z^n} \right| = \left| \frac{z}{n+1} \right|$

Now, we take the limit as $n \to \infty, L = \lim_{n \to \infty} \left| \frac{z}{n+1} \right| =[\text{For any fixed complex number z, |z| is just a constant}] |z| \cdot \lim_{n \to \infty} \frac{1}{n+1} = |z| \cdot 0 = 0$. Since L = 0 < 1 for any complex number z, the condition for convergence, L < 1, is always satisfied, the series converges for every $z \in \mathbb{C}$ and its radius of convergence is $R = \infty$. We call such functions entire.

Differentiate term by term

A fundamental theorem for power series states that we can differentiate them term by term inside their radius of convergence. Since $R = \infty$, we can do this for all of ℂ.

f’(z) = $\sum_{n=1}^{\infty} \frac{nz^{n-1}}{n!} = \sum_{n=1}^{\infty} \frac{z^{n-1}}{(n-1)!} =[\text{Readjusting the index}] \sum_{n=0}^{\infty} a_nz^n$ = f(z).

n = 0: The first term is $\frac{d}{dz}(\frac{z^0}{0!}) = \frac{d}{dz}(1) = 0$. That’s why the derivative sum starts at n = 1. By convention, 0! = 1 and in power-series $0^0 = 1$.

Besides, f(0) = $\sum_{n=0}^{\infty} \frac{0^n}{n!} = \frac{0^0}{0!} + \frac{0^1}{1!} + \frac{0^2}{2!} + \dots = 1$ and the given power series is a solution to the initial value problem defined by f’(z) = f(z) with f(0) = 1.

The Existence and Uniqueness Theorem for ODEs guarantees that this IVP has a unique solution. Since the function $e^z$ is known from real analysis to satisfy this exact same IVP, the power series f(z) must be the complex exponential $e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$.

Local vs global power-series representations

A key question arises: is every analytic function (a function f is analytic at a point a if it is differentiable at a and in a small disk around a) equal to a power series or representable as a power series within its domain?

The answer is nuanced. $\frac{1}{1-z}$ is well-defined and differentiable everywhere in the complex plane except at the single point z = 1 where it blows up, so its domain of analyticity is ℂ \ {1}. It can be expressed as a power series in a disk of convergence: $\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n, \forall |z| \lt 1$.

The function f(z) is perfectly analytic at z = 2 (since $2 \neq 1$), but its power series $\sum z^n$ diverges there. This is not a contradiction! It just means that a single power series centered at a = 0 is not a global representation of the function (this is because the power series cannot go past the problem point z = 1).

The big question is: Is every analytic function equal to some power series locally?, i.e. in some B(a; r) for r > 0 where a is a point of analyticity for f. The answer is yes, and Taylor’s Theorem is the theorem that proves it.

Not all analytic functions can be represented by a single power series over their entire domain. However, every analytic function can be approximated by a power series in a small neighborhood around each point of analyticity.

Taylor’s Theorem. If f is analytic on an open disk B(a; r) (a disk of radius R centered at a), then f(z) can be represented exactly by a unique power series within that disk: $f(z) = \sum_{n=0}^{\infty}a_n (z - a)^n, \forall z \in B(a; r)$

It states that if a function is differentiable in a disk, it must also be infinitely differentiable and representable by a power series.

Furthermore, there exist unique constants $a_n = \frac{f^{(n)}(a)}{n!} = \frac{1}{2\pi i}\int_{C_r} \frac{f(w)}{(w-a)^{n+1}}$ where $C_r$ is a circle of radius r < R centered at a and oriented in the counterclockwise direction.

As we have demonstrated earlier, if a function $f$ is analytic, it is automatically infinitely differentiable, i.e., the expression $f^{(n)}(a)$ (the $n$-th derivative at $a$) makes sense. The Cauchy Integral Formula for Derivatives gives us a way to calculate $f^{(n)}(a)$ using an integral: $f^{(n)}(a) = \frac{n!}{2\pi i} \oint_{C_r} \frac{f(w)}{(w-a)^{n+1}} dw$