The centre is a choice

Every series so far in this ladder has been written about zero, and zero has done nothing to earn the position. The construction asks for a point at which to measure derivatives, and any point where the function has them will serve.

That the interval moves is unremarkable. That its width changes, in a way that has no explanation anywhere on the line being drawn, is the whole subject of this rung.

The same function, a different polynomial #

Expanding about a point $a$ means measuring the derivatives there and building

$$f(x) \approx f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots$$

The powers are of $x - a$ rather than of $x$, which is the only change, and every argument from the rung below survives it verbatim — the coefficients still do not interfere, the remainder still has an unnamed point in it, and the point still lies between $a$ and $x$.

What changes is where the polynomial is good. A partial sum about $a$ is exact at $a$ and degrades away from it in both directions, so the centre is the one free parameter that decides where the accuracy goes. Nothing else about the construction is adjustable.

For $\ln(1+x)$ the arithmetic is transparent. The function fails at $x = -1$ and nowhere else, the series about $0$ reaches $1$, and the series about $1$ reaches $2$. The interval is always the largest one about the centre that avoids the point where the function stops existing, and it is symmetric because a power series’ region of convergence is always symmetric about its centre.

That last fact is worth pausing on. The series about $1$ converges out to $x = 3$, where nothing whatever is wrong with $\ln(1+x)$, purely because it also has to reach back to $-1$, where something is. A power series cannot converge on a lopsided interval, so a singularity on one side limits the reach on the other, and the excess is not usable.

Where the width comes from when there is nothing to see #

For $1/(1+x^2)$ that account fails, because the function is perfectly well behaved everywhere on the line and the interval is finite anyway.

The explanation is not on the line. Extend the function to complex inputs and $1 + z^2$ vanishes at $z = i$ and $z = -i$, where $f$ blows up. Those points are at distance exactly $1$ from the origin.

A Taylor series converges inside the largest disc about its centre containing no singularity, and diverges outside it. The radius is a distance in the complex plane, whether or not the function was ever meant to leave the real line, and the real interval a reader sees is the diameter of that disc lying along the axis.

Measured from $1.2$, the nearest of $\pm i$ is at distance $\sqrt{1.2^2 + 1} = 1.562$, which is the half-width in the hero figure. Measured from $0$ it is $1$. The number changes because the distance changes, and the distance is to a point with no position on the axis being drawn.

That is a genuinely startling arrangement. The behaviour of a real function of a real variable, on the real axis, is governed by two points that are not on the real axis, and the governing quantity is a distance measured in a plane the whole question never mentioned. It is the sharpest instance in this collection of a real problem whose answer is complex, and the reason multiplying is turning is a prerequisite to understanding it. The plane the argument happens in is the one a single extra point makes into a sphere, and on that sphere the two singularities of $1/(1+z^2)$ are two ordinary points like any other — which is the right way to see that nothing about them is special except their distance from the centre.

The coefficients already knew #

There is a second route to the radius that never mentions the complex plane, and putting the two side by side says something about where the information is kept.

The series $\sum c_k (x-a)^k$ converges exactly where $|x - a|$ is smaller than

$$R = \frac{1}{\limsup_k |c_k|^{1/k}},$$

which is a formula in the coefficients alone. For $1/(1+x^2)$ about zero the coefficients are $1, 0, -1, 0, 1, \ldots$, so $|c_k|^{1/k}$ is $1$ for every even $k$, the limsup is $1$, and $R = 1$. No singularity was mentioned, no plane was entered, and the answer came out.

So the coefficients know where the nearest singularity is, and they are computed from derivatives at a single real point. That is the fact worth carrying away from this rung. The information that $1/(1+z^2)$ misbehaves at $\pm i$ is present in the derivatives of $1/(1+x^2)$ at $x = 0$ — it has to be, since the derivatives determine the coefficients and the coefficients determine the radius — and the geometry of the disc is a translation of that arithmetic rather than an additional fact.

Which of the two accounts is the explanation depends on what is being asked. The coefficient formula computes the radius and says nothing about why it is that number; the disc says why and needs the singularities located first. In practice both directions are used: for a function given by a formula the singularities are found and the radius read off, and for a function known only through its coefficients — a solution of a differential equation, say, generated term by term — the growth of the coefficients is the only available evidence about where the trouble is.

The ratio test gives the same answer more cheaply when the coefficients are well behaved. For a geometric-looking series the ratio of consecutive terms settles on a constant, and the reciprocal of that constant is the radius; the limsup version exists because some series have coefficients that vanish periodically, as this one’s do, and a ratio of consecutive terms is then undefined half the time.

Walking #

Once the centre is a choice, a series can be moved. And moving it repeatedly does something the original series could not.

Start at $0$. The series there converges on $|z| < 1$ and knows nothing beyond. But it knows everything inside, including the value and all the derivatives at $0.8$ — so a new series can be built about $0.8$ from information the first series supplied, with no reference to the original formula at all. That new series converges out to $\sqrt{0.8^2+1} = 1.28$ from its own centre, which reaches to $2.08$: past the first disc entirely.

Repeat. Each step re-expands at a point the previous series covered, and each new disc reaches further because the centre has moved away from the singularities. Four steps take the function out to $x = 5.8$, where the first series diverges violently.

This is analytic continuation, and the surprising part is that it is well defined. A function is being extended using only its local data, and there is no obvious reason two different routes should agree at a point they both reach — but they do, whenever the region swept out has no hole in it, because two analytic functions agreeing on any disc agree everywhere they are both defined.

The agreement is not obvious and it is not a convention; it is a theorem, and the theorem is the reason the word the can be used about a continuation at all. Two chains reaching the same point through a region with no singularity in it produce the same series there, coefficient by coefficient, however different the routes look.

The proviso about holes is not a technicality. Walk a chain of discs around a singularity rather than away from it and the function that arrives back at the starting point need not be the one that left. For $\ln(1+z)$, a loop around $-1$ returns with $2\pi i$ added, which is why the logarithm has infinitely many values and why the plane has to be cut before it can have one.

What continuation costs and what it is worth #

Continuation is not a computational method. Each step needs every derivative at the new centre, computed from the old series, and the arithmetic degrades as the new centre approaches the edge of the old disc — the coefficients being summed grow like the reciprocal of the distance to the boundary, so a step that is nearly the full radius is a sum of enormous nearly-cancelling terms.

Taking small steps avoids the cancellation and needs many steps. Taking large ones needs few steps and loses digits in each. There is no setting at which the procedure is cheap, and nothing serious computes a function this way.

What it is worth is a definition. The zeta function is defined by a sum of reciprocal powers that converges only where the exponent is large enough, and every statement about it outside that region is a statement about its continuation; the same is true of the gamma function past the integral that defines it, and of a great deal else. The value of continuation is that it makes “the function” mean something at a point where the formula does not, and that the meaning is unique.

Uniqueness is the whole content. If a function had two legitimate continuations to the same point there would be nothing to talk about, and the reason it does not is that an analytic function is rigid: its values on any disc, however small, determine it everywhere it can be reached. That is the property the flat function of the rung below fails to have, and the contrast is exact — a smooth function is soft enough to be zero on an interval and positive elsewhere, and an analytic one is not.

Choosing a centre on purpose #

The practical version of all this is much duller and is worth having, because it is what actually gets used.

If a function needs to be evaluated accurately near $x = 3$, expanding about $3$ is better than expanding about $0$ and taking more terms. The improvement is not marginal. The error after $n$ terms goes like $|x - a|^{n+1}$, so halving the distance from the centre to the point of interest divides the error by $2^{n+1}$ — an improvement that compounds with the degree rather than adding to it.

That is why tables of special functions are built as a patchwork rather than as one series: the interval is cut into pieces, a different expansion is used on each, and each is centred in its own piece. The cost is a table of coefficients and a branch; the saving is orders of magnitude of accuracy for the same degree.

It is also why the choice of centre and the choice of node placement, which look like different questions, are the same question. A Taylor series concentrates all its accuracy at one point; an interpolating polynomial spreads it over several; and where those several are placed is the next rung, where putting them evenly turns out to be the worst available choice.

Where the account needs care #

The radius is about the nearest singularity, and a singularity is not only a place a function blows up. A branch point of $\sqrt{z}$ or $\ln z$ limits the radius exactly as a pole does, and so does an essential singularity, and none of them looks alike. What they share is that the function is not analytic there, which is the only property the theorem uses.

A function analytic everywhere has infinite radius from every centre. $e^z$, $\sin z$ and every polynomial qualify, and their series converge on the whole plane — which is why the rung below’s sine converges everywhere and why nothing in this essay applies to it.

Two singularities at the same distance both bind. The radius is the distance to the nearest, and when several tie the disc touches all of them; $1/(1+z^2)$ about zero is exactly that case, with the disc touching $i$ and $-i$ at once.

And the disc is open. On the boundary circle itself the series may converge at every point, at no point, or at some and not others, and which of the three happens is a genuinely delicate question that the radius says nothing about.

Weierstrass, and the definition that ran the other way #

The account above treats the function as given and the series as derived. Weierstrass built the whole of complex analysis the other way round in the 1860s and 1870s: a function element is a power series with a positive radius, and a function is a maximal collection of elements reachable from one another by continuation.

On that definition the walk in the figures is not a technique for extending a function; it is what the function is. A statement about $1/(1+z^2)$ at $x = 5$ is a statement about an element four steps along a chain, and there is no other kind of statement available.

The rival approach, Riemann’s, defines an analytic function by a differential condition and gets continuation as a consequence. The two give the same objects and read completely differently, and it is worth knowing that both exist because the vocabulary of each survives in the other’s textbooks. Analytic is Weierstrass’s word and means expandable in a series; holomorphic is closer to Riemann’s and means complex-differentiable; the theorem that they coincide is the central fact of the subject, and the reason a reader meets both words for one idea.

What the pictures cannot show #

The disc figures are drawn in the complex plane and every curve elsewhere on this page is drawn on the real line, and no figure shows the connection. What a reader would need is a graph of a complex function, which needs four dimensions, and the whole difficulty of the subject is that this picture does not exist.

The chain of discs shows the discs and not the functions. Each step re-derives a whole series from the previous one, and what is drawn is only where each one is valid — the arithmetic that carries the coefficients from one centre to the next, and the cancellation that makes it expensive, has no picture at all.

And the walk is drawn along the axis, which makes it look as though continuation is a one-dimensional business. It is not: the chain can go anywhere in the plane, and the interesting cases are precisely the ones where it goes around something and comes back changed. That is the situation the figures cannot draw, because the two arrivals differ in a value rather than in a position.

The ladder from here #

Rungs above: the points that ruin the fit, where a polynomial through samples fails on the same $1/(1+x^2)$ for a related but different reason. Complex differentiability, and why differentiable once implies analytic — the theorem that makes the disc argument work at all. Laurent series, which allow negative powers and so describe a function on an annulus around a singularity rather than a disc avoiding it. The monodromy theorem, which says exactly when a continuation around a loop comes back unchanged. And Padé approximants, which reach past the radius by allowing a denominator, and which converge where the series they are built from does not.

A parameter nobody was using #

The habit is worth naming because it is embarrassingly general: when a construction has a free choice in it that everybody makes the same way, ask what the other choices do.

The centre of a Taylor series is such a choice. It is set to zero in every textbook presentation, for the excellent reason that the algebra is tidier, and the consequence is that a reader can finish a course believing the radius of convergence is a property of the function. It is a property of the function and the centre, and the whole of the geometry above is invisible until the second one is allowed to move.

The same shape appears elsewhere in this collection whenever a construction has a base point. The fundamental group is defined at a point and the interesting question is what changes when the point moves; a coordinate system has an origin and the useful theorems are the ones that do not depend on it. In both cases the fixed choice is a convenience, and the mathematics that the convenience conceals is not small.