advanced28 min studyLesson 364 in Mathematics

Analytic continuation and identity theorems

Use interior accumulation and overlapping analytic descriptions without confusing one Taylor disk with an entire function's domain.

Before you begin

On this page

The question for today

What you’ll understand

State the identity theorem with connectedness and an interior accumulation point.
Continue an analytic function using overlapping Taylor disks.
Distinguish uniqueness of a continuation from its existence and path independence.

A Taylor series is one local description of a holomorphic function, not necessarily its largest possible domain. Overlapping disks can carry the same analytic information farther than the original radius allows. The identity theorem makes the overlap rigid: two holomorphic descriptions that agree on enough nearby points must agree throughout their connected common domain. This gives uniqueness where continuation is possible, while leaving separate questions about singularities, existence along a chosen path, and whether different paths return the same analytic branch.

Zeros cannot accumulate in the interior unless everything vanishes

Suppose a holomorphic function on a connected open domain is not identically zero. At any zero, its Taylor series has a first nonzero coefficient of some finite order. Factoring out the corresponding power leaves a holomorphic factor nonzero at that point and therefore nonzero in a smaller neighborhood. The original zero is isolated. If all Taylor coefficients vanished at one point, the function would vanish on a disk; connectedness then propagates that local vanishing through the domain.

Apply this argument to the difference of two functions. If they agree at distinct points accumulating at an interior point of their connected domain, they agree everywhere on that domain. Agreement on a real interval inside a complex domain is enough, even though the interval is not open in the plane. Agreement at finitely many points is not enough. An accumulation point on the boundary is also insufficient because the local Taylor argument needs a disk around the accumulation point contained in the domain.

f(a)=0,\ f\not\equiv0\quad\Longrightarrow\quad f(z)=(z-a)^m h(z),\quad m\ge1,\quad h(a)\ne0

For a holomorphic nonzero function on a connected domain, the local factorization shows that each zero is isolated and has a finite multiplicity.↔ Scroll wide equations horizontally.

Continue the function rather than summing outside a convergence disk

An analytic continuation extends a holomorphic description to a larger domain while agreeing on a nonempty open overlap. A chain of overlapping disks can implement this extension by repeatedly re-expanding at a new center. Each series is used only inside its own convergence disk. The original coefficients do not license substitution beyond their radius; instead, the compatible new series provides values there. A singularity can stop one route while another route remains available.

For a rational function, poles visibly constrain Taylor disks. A series centered at zero may stop at a pole on one side even though the function is perfectly regular far away on the other side. Re-expanding toward the regular side can reach those points. More generally, the identity theorem ensures that two putative extensions agree on a connected overlap when they match there, but it does not guarantee that an extension exists through a singular boundary point. Uniqueness is not a substitute for constructing or proving existence.

\frac1{1-z}=\sum_{n=0}^{\infty}z^n\quad(|z|<1),\qquad \frac1{1-z}=\frac23\sum_{n=0}^{\infty}\left(\frac23(z+\tfrac12)\right)^n\quad(|z+\tfrac12|<\tfrac32)

The second Taylor series is centered at −1/2. The two disks overlap and both describe the same rational function, but their convergence regions differ.↔ Scroll wide equations horizontally.

Boundary behavior and path dependence need their own analysis

Boundary accumulation can coexist with a nonzero holomorphic function. On the unit disk, $\sin(\pi/(1-z))$ has zeros at $1-1/n$ for integers n≥2, accumulating only at the excluded boundary point 1. There is no contradiction with isolated interior zeros. Likewise, a Taylor series failing to converge at a point does not by itself prove that the underlying function has a singularity there; the function might be represented by another continuation through a neighborhood of that point.

Continuation around holes can return a different local branch, as with logarithms. The identity theorem compares single-valued functions on a specified common domain, whereas a sequence of local continuations records a path. On suitable simply connected domains, a monodromy theorem can remove this path dependence when continuation exists along every path. Those existence and topology hypotheses will matter later. For now, keep track of the domain of each local description and justify agreement on every overlap.

Work it through

Reach a regular point beyond the first Taylor disk

Use a Taylor expansion centered at −1/2 to find the analytic continuation of $\sum_{n\ge0}z^n$ at z=−1. Explain why directly substituting into the original series fails.

On its original unit disk the geometric series equals $1/(1-z)$ . At center a=−1/2, rewrite the denominator as $(3/2)[1-(2/3)(z+1/2)]$ .
The new geometric expansion is $(2/3)\sum_{n\ge0}[(2/3)(z+1/2)]^n$ . It converges when $|z+1/2|<3/2$ , so it contains z=−1 and overlaps the original disk.
At z=−1 the new ratio is −1/3. Summing gives $(2/3)/(1+1/3)=1/2$ . The original series would become $1-1+1-1+\cdots$ , whose ordinary partial sums do not converge.

The result

The continued analytic value is 1/2, supplied by a compatible convergent series. This is not ordinary convergence of the original series at its boundary point.

The living notebook

Explore the model

Interactive lab

Interactive lab · spectral

See why finitely many matching values are insufficient

Pause and Reset; set Mode count N=1 and Decay p=1, keeping time zero. Compare with a fresh Reset at N=2,p=1. Predict that both finite sine sums vanish at x=0,π,2π but differ between those locations. These are real traces of entire finite sine sums at the reset time. The graph illustrates finite agreement only; it does not test an infinite interior accumulation set or perform analytic continuation.

Finite sumHighest included mode

Sum at x = π/2: 0.867
Smallest wavelength: 2π / 5
Highest-mode coefficient: 0.2

Mode count N5Coefficient decay p1

t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe identity theorem needs an interior accumulation of distinct matching points, rather than a finite list of coincidences.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Interior accumulation makes the difference have a nonisolated zero, forcing it to vanish identically. Connectedness propagates that conclusion through the domain.

Your turn

Try it yourself

Functions f and g are holomorphic on the unit disk and satisfy $f(1/n)=g(1/n)$ for every integer n≥2. Must they agree everywhere? Would the same argument work with matching points $1-1/n$ instead?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Locate the two accumulation points and check whether each belongs to the open domain.

Compare with the solution

The first points accumulate at 0, an interior point, so the identity theorem gives f=g throughout the disk. The second points accumulate at the boundary point 1, so that argument does not apply. Indeed f(z)=0 and $g(z)=\sin(\pi/(1-z))$ agree at all $1-1/n$ yet are not identical; for example g(−1/2)=sin(2π/3) is nonzero.

Keep the idea

Take it with you

Interior accumulation determines a holomorphic function on a connected domain.
Overlapping Taylor disks can extend a function without extending one series' convergence radius.
Uniqueness, existence, and path independence of continuation are distinct claims.

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