= Analytic continuation
{wiki}
An analytic continuation extends a <holomorphic function> beyond its original domain without changing its existing values. Formally, if $U\subset V$ are domains and $f$ is <holomorphic> on $U$, a <holomorphic function> $F$ on $V$ with $F|_U=f$ is an analytic continuation. More locally, <holomorphic functions> on overlapping <open sets> continue one another when they agree on the overlap, allowing a chain of patches along a path. The <identity theorem for holomorphic functions> forces agreement throughout a <connected> overlap once they agree on a nonempty <open set> in it. It also makes an extension to a fixed <connected> domain unique. https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/af8c9c78c3915aaf3fd1695f8d32a3b3_MIT18_04S18_topic13.pdf[Orloff, MIT 18.04 Topic 13, section 13.2 and Theorem 13.2]{ref}
The <geometric series> illustrates why a formula can reach farther than the <power series> that first represents it. For $z\ne1$, its <partial sum> is
$$
S_N(z)=\sum_{n=0}^Nz^n=\frac{1-z^{N+1}}{1-z}.
$$
When $|z|<1$, $z^{N+1}\to0$, so the <series> converges to $f(z)=1/(1-z)$. The <rational function> $F(z)=1/(1-z)$ is <holomorphic> on $\mathbb C\setminus\{1\}$ and agrees there with the sum on the <unit disc>. It therefore continues $f$ to every point except its <simple pole> at $1$. In particular,
$$
f(1/2)=2,\qquad F(2)=-1,\qquad S_N(2)=2^{N+1}-1\longrightarrow+\infty.
$$
Thus <analytic continuation> gives a value at $2$ without making the original <geometric series> converge there. Its <radius of convergence> remains $1$: on $|z|\ge1$ the terms $z^n$ do not even tend to zero.
\Image[/analytic-continuation-geometric-series.png]
{title=The original geometric series converges only inside the unit circle, while its rational continuation exists everywhere except the pole at 1}
The <pole> at $1$ limits the original <Taylor series>, although most points of its convergence circle are regular for $F$. Recentring at $-1$ gives another genuinely convergent <power series>,
$$
F(z)=\frac1{2-(z+1)}=\frac12\sum_{n=0}^{\infty}\left(\frac{z+1}{2}\right)^n,\qquad |z+1|<2.
$$
At $z=-3/2$ this yields $\tfrac12\sum_{n\ge0}(-1/4)^n=2/5$, even though the original powers of $-3/2$ grow. This is the useful idea: overlapping local descriptions can represent one <holomorphic function> on a larger region.
Uniqueness on a <connected> overlap does not imply that every path leads to the same terminal <germ of a holomorphic function>. Continuing a <complex logarithm> around $0$ changes its value by $2\pi i$, giving <monodromy>; the return patch can carry a different <branch of a multivalued function>. If two patches have a disconnected overlap, agreement on one component imposes no agreement on the others. https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/af8c9c78c3915aaf3fd1695f8d32a3b3_MIT18_04S18_topic13.pdf[Orloff, section 13.2.1, Example 13.4]{ref} Continuation using the <rational function> $1/(1-z)$ is single-valued: any path avoiding $1$ gives the same terminal <germ of a holomorphic function>, even if it winds around the <pole>.
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