Editing Analytic continuation (section)

==Natural boundary==
Suppose that a power series has radius of convergence ''r'' and defines an analytic function ''f'' inside that disc.  Consider  points on the circle of convergence.  A point for which there is a neighbourhood  on which ''f'' has an analytic extension is ''regular'', otherwise ''singular''.  The circle is a '''natural boundary''' if all its points are singular.

More generally, we may apply the definition to any open connected domain on which ''f'' is analytic, and classify the points of the boundary of the domain as regular or singular: the domain boundary is then a natural boundary if all points are singular, in which case the domain is a ''[[domain of holomorphy]]''.

===Example I: A function with a natural boundary at zero (the prime zeta function)===
For <math>\Re(s) > 1</math> we define the so-called [[prime zeta function]], <math>P(s)</math>, to be

:<math>P(s) := \sum_{p\ \text{ prime}} p^{-s}.</math>

This function is analogous to the summatory form of the [[Riemann zeta function]] when <math>\Re(s) > 1</math> in so much as it is the same summatory function as <math>\zeta(s)</math>, except with indices restricted only to the [[prime numbers]] instead of taking the sum over all positive [[natural numbers]]. The prime zeta function has an analytic continuation to all complex ''s'' such that <math>0 < \Re(s) < 1</math>, a fact which follows from the expression of <math>P(s)</math> by the logarithms of the [[Riemann zeta function]] as

:<math>P(s) = \sum_{n \geq 1} \mu(n)\frac{\log\zeta(ns)}{n}.</math>

Since <math>\zeta(s)</math> has a simple, non-removable pole at <math>s := 1</math>, it can then be seen that <math>P(s)</math> has a simple pole at <math>s := \tfrac{1}{k}, \forall k \in \Z^{+}</math>. Since the set of points

:<math>\operatorname{Sing}_P := \left\{k^{-1} : k \in \Z^+\right\} = \left \{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4},\ldots \right \}</math>

has accumulation point 0 (the limit of the sequence as <math>k\mapsto\infty</math>), we can see that zero forms a natural boundary for <math>P(s)</math>. This implies that <math>P(s)</math> has no analytic continuation for ''s'' left of (or at) zero, i.e., there is no continuation possible for <math>P(s)</math> when <math>0 \geq \Re(s)</math>. As a remark, this fact can be problematic if we are performing a complex contour integral over an interval whose real parts are symmetric about zero, say <math>I_F \subseteq \Complex \ \text{such that}\ \Re(s) \in (-C, C), \forall s \in I_F</math> for some <math>C > 0</math>, where the integrand is a function with denominator that depends on <math>P(s)</math> in an essential way.

===Example II: A typical lacunary series (natural boundary as subsets of the unit circle)===
For integers <math>c \geq 2</math>, we define the [[lacunary series]] of order ''c'' by the power series expansion

:<math>\mathcal{L}_c(z) := \sum_{n \geq 1} z^{c^n}, |z| < 1.</math>

Clearly, since <math>c^{n+1} = c \cdot c^{n}</math> there is a functional equation for <math>\mathcal{L}_c(z)</math> for any ''z'' satisfying <math>|z| < 1</math> given by <math>\mathcal{L}_c(z) = z^{c} + \mathcal{L}_c(z^c)</math>. It is also not difficult to see that for any integer <math>m \geq 1</math>, we have another functional equation for <math>\mathcal{L}_c(z)</math> given by

:<math>\mathcal{L}_c(z) = \sum_{i=0}^{m-1} z^{c^{i}} + \mathcal{L}_c(z^{c^m}), \forall |z| < 1.</math>

For any positive natural numbers ''c'', the lacunary series function diverges at <math>z = 1</math>. We consider the question of analytic continuation of <math>\mathcal{L}_c(z)</math> to other complex ''z'' such that <math>|z| > 1.</math> As we shall see, for any <math>n \geq 1</math>, the function <math>\mathcal{L}_c(z)</math> diverges at
the <math>c^{n}</math>-th roots of unity. Hence, since the set formed by all such roots is dense on the boundary of the unit circle, there is no analytic continuation of <math>\mathcal{L}_c(z)</math> to complex ''z'' whose modulus exceeds one.

The proof of this fact is generalized from a standard argument for the case where <math>c := 2.</math><ref>See the example given on the ''MathWorld'' page for [http://mathworld.wolfram.com/NaturalBoundary.html  natural boundary].</ref> Namely, for integers <math>n \geq 1</math>, let

:<math>\mathcal{R}_{c,n} := \left \{z \in \mathbb{D} \cup \partial{\mathbb{D}}: z^{c^n} = 1 \right \},</math>

where <math>\mathbb{D}</math> denotes the open [[unit disk]] in the complex plane and <math>|\mathcal{R}_{c,n} | = c^n</math>, i.e., there are <math>c^n</math> distinct [[complex number]]s ''z'' that lie on or inside the unit circle such that <math>z^{c^n} = 1</math>. Now the key part of the proof is to use the functional equation for <math>\mathcal{L}_c(z)</math> when <math>|z| < 1</math> to show that

:<math>\forall z \in \mathcal{R}_{c,n}, \qquad \mathcal{L}_c(z) = \sum_{i=0}^{c^n-1} z^{c^i} + \mathcal{L}_c(z^{c^n}) = \sum_{i=0}^{c^n-1} z^{c^i} + \mathcal{L}_c(1) = +\infty.</math>

Thus for any arc on the boundary of the unit circle, there are an infinite number of points ''z'' within this arc such that <math>\mathcal{L}_c(z) = \infty</math>. This condition is equivalent to saying that the circle <math>C_1 := \{z: |z| = 1\}</math> forms a natural boundary for the function <math>\mathcal{L}_c(z)</math> for any fixed choice of <math>c \in \Z \quad c > 1.</math> Hence, there is no analytic continuation for these functions beyond the interior of the unit circle.