Editing Analytic continuation (section)

===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.