Editing Analytic continuation (section)

==Worked example==
[[File:AnalyticContinuationGraphic.pdf|316px|right|thumb|Analytic continuation from ''U'' (centered at 1) to ''V'' (centered at a=(3+i)/2)]]
Begin with a particular analytic function <math>f</math>. In this case, it is given by a [[power series]] centered at <math>z=1</math>:

<math display="block">f(z) = \sum_{k=0}^\infty (-1)^k (z-1)^k.</math>

By the [[Cauchy–Hadamard theorem]], its radius of convergence is 1. That is, <math>f</math> is defined and analytic on the open set <math>U = \{|z-1|<1\}</math> which has boundary <math>\partial U = \{|z-1|=1\}</math>. Indeed, the series diverges at <math>z=0 \in \partial U</math>.

Pretend we don't know that <math>f(z)=1/z</math>, and focus on recentering the power series at a different point <math>a \in U</math>:

<math display="block">f(z) = \sum_{k=0}^\infty a_k (z-a)^k.</math>

We'll calculate the <math>a_k</math>'s and determine whether this new power series converges in an open set <math>V</math> which is not contained in <math>U</math>. If so, we will have analytically continued <math>f</math> to the region <math>U \cup V</math> which is strictly larger than <math>U</math>.

The distance from <math>a</math> to <math>\partial U</math> is <math>\rho = 1 - |a-1| > 0</math>. Take <math>0 < r < \rho</math>; let <math>D</math> be the disk of radius <math>r</math> around <math>a</math>; and let <math>\partial D</math> be its boundary. 
Then <math>D \cup \partial D \subset U</math>. Using [[Cauchy's differentiation formula]] to calculate the new coefficients, one has
<math display="block">\begin{align}
a_k &= \frac{f^{(k)}(a)}{k!} \\
&=\frac{1}{2\pi i} \int_{\partial D} \frac{f(\zeta) d \zeta}{(\zeta -a)^{k+1}} \\
&=\frac{1}{2\pi i} \int_{\partial D} \frac{\sum_{n=0}^\infty (-1)^n (\zeta-1)^n d \zeta}{(\zeta -a)^{k+1}} \\
&=\frac{1}{2\pi i} \sum_{n=0}^\infty (-1)^n \int_{\partial D} \frac{(\zeta-1)^n d\zeta}{(\zeta -a)^{k+1}} \\
&=\frac{1}{2\pi i} \sum_{n=0}^\infty (-1)^n \int_0^{2\pi} \frac{(a+re^{i \theta}-1)^n rie^{i \theta}d\theta}{(re^{i \theta})^{k+1}} \\
&=\frac{1}{2\pi} \sum_{n=0}^\infty (-1)^n \int_0^{2\pi} \frac{(a-1+re^{i \theta})^n d\theta}{(re^{i \theta})^{k}}
\\
&=\frac{1}{2\pi} \sum_{n=0}^\infty (-1)^n \int_0^{2\pi} \frac{\sum_{m=0}^n \binom{n}{m} (a-1)^{n-m} (re^{i \theta})^m d\theta}{(re^{i \theta})^{k}} 
\\
&=\frac{1}{2\pi} \sum_{n=0}^\infty (-1)^n \sum_{m=0}^n \binom{n}{m} (a-1)^{n-m} r^{m-k} \int_0^{2\pi} e^{i (m-k)\theta} d\theta 
\\
&=\frac{1}{2\pi} \sum_{n=k}^\infty (-1)^n \binom{n}{k} (a-1)^{n-k}\int_0^{2\pi}  d\theta \\
&=\sum_{n=k}^\infty (-1)^n \binom{n}{k} (a-1)^{n-k} 
\\
&=(-1)^k \sum_{m=0}^\infty  \binom{m+k}{k} (1-a)^m \\
&=(-1)^k a^{-k-1}
\end{align}.</math>

The last summation results from the {{mvar|k}}th derivation of the [[geometric series]], which gives the formula 
<math display = "block">\frac 1{(1-x)^{k+1}} = \sum_{m=0}^\infty \binom{m+k}k x^m.</math>

Then,
<math display = "block"> \begin{align}
f(z) &= \sum_{k=0}^\infty a_k (z-a)^k \\
&= \sum_{k=0}^\infty (-1)^k a^{-k-1} (z-a)^k \\
&= \frac{1}{a} \sum_{k=0}^\infty \left ( 1 - \frac{z}{a} \right )^k \\
&= \frac{1}{a} \frac{1}{1 - \left(1 - \frac{z}{a}\right)} \\
&= \frac{1}{z} \\
&= \frac{1}{(z + a) - a}
\end{align}</math>

which has radius of convergence <math>|a|</math> around <math>0</math>. If we choose <math>a \in U</math> with <math>|a|>1</math>, then <math>V</math> is not a subset of <math>U</math> and is actually larger in area than <math>U</math>. The plot shows the result for <math>a = \tfrac{1}{2}(3+i).</math>

We can continue the process: select <math>b \in U \cup V</math>, recenter the power series at <math>b</math>, and determine where the new power series converges. If the region contains points not in <math>U \cup V</math>, then we will have analytically continued <math>f</math> even further. This particular <math>f</math> can be analytically continued to the whole punctured complex plane <math>\Complex \setminus \{0\}.</math>

In this particular case the obtained values of <math>f(-1)</math> are the same when the successive centers have a positive imaginary part or a negative imaginary part. This is not always the case; in particular this is not the case for the [[complex logarithm]], the [[antiderivative]] of the above function.