Talk:Analytic continuation

Trim

Ughh. Someone should copy the bottom half of the article on Riemann surfaces into this article. The article on Riemann surfaces should be trimmed to remove the parts about analytic continuation, and just point to this article instead. Who is up for that? linas 19:43, 22 Jan 2005 (UTC)

There also should be a mentioning of the generalization to meromorphic continuation since the link points to this article currently. - Gauge 05:16, 20 May 2005 (UTC)[reply]

extending a relation by transitivity

Can someone explain what is meant by "extending by transitivity"? Does it mean transitive closure? Do we also have to do some kind of "symmetric closure" to make this relation symmetric, as it must if we're to make an equivalence relation? -Lethe | Talk 22:33, May 31, 2005 (UTC)

Hmmm - 'not symmetric' is true, as defined. But the transitive closure is symmetric. This may really need a picture. As defined, g ≥ h can happen and the radius of convergence of h can be much smaller than that of g: so the power series for h can't 'reach' far enough. But by taking enough small steps with other functions, one can 'reach' the point about which g is defined. Say for example g is defined at 0, h at 1, and they both represent a function that has a singularity at 1.01. Then the radius of convergence of h is (at most) 0.01. You can expand h about a point like 0.99, and get a slightly larger radius of convergence, 0.015. Expand that about 0.98, say, with radius of convergence 0.02. Continuing this way, one gets the relation h R g where R is the transitive closure of ≥. Charles Matthews 06:20, 1 Jun 2005 (UTC)

Hmm. Let me see if I have it. Transitivity fails because b can be in a's radius of convergence, and c can b in b's, but c need not be in a's. Our transitive closure will have a>c when there is an intermediate chain of germs that arrive at c. And I can sort of imagine how we can use a similar chain to see that symmetry holds too, once we have transitivity. A chain of germs getting bigger until they get a's center. Or something. I'm going to add some words to that sentence. -Lethe | Talk 11:09, Jun 1, 2005 (UTC)

Hmm. I would believe your assertion that I can go back from h to g by some finite number of steps, because the picture you give feels somehow right. However, this fact doesn't look easy to prove and so is not at all trivial, and I think the sentence should be changed accordingly. "If we extend the relation by transitivity, we obtain a symmetric relation" conveys the impression that just taking the transitive closure implies also symmetry, which isn't the case in general. Ezander (talk) 10:55, 17 March 2011 (UTC)[reply]

Examples of analytic continuation

The formula

L(z-1)=\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k}}(z-1)^{k}

is weird. Perhaps the author wanted to write

L(z)=\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k}}(z-1)^{k}

Bo Jacoby 13:47, 10 October 2005 (UTC)[reply]

OTOH perhaps the author meant log(z) = L(z-1) = ... and "L(z-1)" is only there to emphasize the fact that z0=1. 194.78.219.157 18:28, 10 October 2005 (UTC)[reply]

Perhaps, but that doesn't agree with the notation used in Analytic_continuation#Formal_definition. I dare to correct it. Bo Jacoby 06:39, 12 October 2005 (UTC)[reply]

And its not really an example of analytic continuation is it (its a power series with a statement of what its germ is)? Its a concept that I am just trying to understand and there are surprisingly few examples of it being done. Instead one reads lots of earnest accounts of how an analytic continutation (if it exists) is unique in certain circumstances and how easy it is to do (or is it). Nowhere is it said (as far as I can see) under what conditions such a thing exists or how to find one (although its implied that this is so easy it should be obvious). Fine for a typical maths textbook, but not (surely) an encylopedia. How about an *example*? Francis Davey 19:02, 11 May 2006 (UTC)[reply]

Global obstructions

The reference to "global obstructions" is not immediately clear to me. The article on "Sheaf Cohomology" doesn't help about. Better specify. 9:05, 20 May 2006

I've written this out in a longer explanation. Charles Matthews 11:17, 20 May 2006 (UTC)[reply]

this sentence

"That is because the difference is an analytic function vanishing on a non-empty open set."

This needs further comments, because i don't get it.

I gave it a try. How is it now? -lethe ^{talk +} 16:12, 11 June 2006 (UTC)[reply]

If by "difference" you mean the "difference of two analytic functions" and by "vanish" you mean "[the value of the difference] approach zero".

Not "approaches zero", but rather "is equal to zero at every point of the domain". -lethe ^{talk +} 06:14, 12 June 2006 (UTC)[reply]

Monodromy theorem

I added the monodromy theorem to this page. I tried to state it also within the language of sheaves, but I am not really familiar with those, so maybe some correction is needed. This unsigned comment was added by KennyDC at 00:42, 16 Sep 2006.

History of analytic continuation

I'm a little curious about something, and I'm wondering if anyone else agrees with me, or has a different perspective. Whichever – I'd like to start a discussion here.

I think that the idea of "analytic continuation" has its roots in real analysis. A good example is the gamma function, Γ(z). The factorials were well understood by guys like Pascal and Newton. Euler developed an integral formula

\Gamma (z)=\int _{0}^{\infty }e^{-t}t^{z-1}\mathrm {d} t\,

which, in the first instance, can be regarded as an extension of the domain of the factorial function from the non-negative integers into the real numbers > −1, since the improper integral exists if z > 0 is real. Euler's approach can be extended into the half-plane where ℜ(z) > 0, and Euler himself gave a product formula that defines Γ(z) everywhere in ℂ where the function is analytic (that is, everywhere except at the poles of Γ). I'm pretty sure I can come up with some other examples ... the sequence of perfect squares was almost certainly the motivation for thinking about the real-valued function y = x² a long time ago, for instance.

Another good example is the exponential function, and the natural logarithm ln(x). The natural log, in particular, was well understood as an exclusively real-valued function long before Euler extended the definition of e^x into the complex plane.

I've got a book (Edwards, on Riemann's Zeta Function) that talks about some of this, particularly the difference between "analytic continuation" as Riemann understood it, and the essentially Weierstrassian approach to analytic continuation that is the exclusive focus of this article. So anyway, here's my question. Would this article benefit by adding some of these ideas, either in a "Motivation" or "History" section? Or should this article restrict itself to the (quite modern, and very narrow) generally accepted definition of "analytic continuation"? DavidCBryant 17:13, 17 January 2007 (UTC)[reply]

Maybe that would be a nice touch. Sometimes, as with the zeta function and its functional equation, a function's domain can be extended in one fell swoop by a nice trick. But this is by far the exception, not the rule.

On the other hand, the Weierstrassian approach of using a sequence of power series centered at various points along a path, so that their regions of convergence successively overlap, works in all cases where analytic continuation is possible.Daqu (talk) 10:07, 23 May 2008 (UTC)[reply]

The example of log(z) seems wrong

The article states:

"Sum_{1 <= k < oo} (-1)^(k+1) (z-1)^k is a power series corresponding to the natural logarithm near z = 1. This power series can be turned into a germ g = (1, 0, 1, −1, 1, −1, 1, −1, ...)."

Shouldn't the germ be g = (1, 0, 1, -1/2, 1/3, -1/4, ...) (using the notation for germs described in the paragraph above this statement) ???Daqu (talk) 09:57, 23 May 2008 (UTC)[reply]

agreed. done. Bo Jacoby (talk) 12:27, 23 May 2008 (UTC).[reply]

Natural barrier

Whoever pre-empted the term "Natural barrier" for this mathematical usage got it hugely wrong. In normal usage the term is geographical and refers to obstacles to movement, especially of people and especially at modest technological levels. I'm removing the redirect for Natural barrier immediately. I suggest anyone who wants to see the mathematical usage represented should create a disambig page. Philcha (talk) 10:44, 24 May 2008 (UTC)[reply]

Philcha, the fact that a term has one established meaning does not suggest whether or not it also has another meaning, especially in a different field. As it happens, "natural barrier" is used in complex analysis to mean a set -- on the boundary of the domain of definition of an analytic function -- beyond which the function cannot be analytically continued.

For better or worse, mathematics has adopted a large number of common terms and assigned technical meanings to them within mathematics.

(It is also true that within math, the term "natural boundary" is considerably more common than "natural barrier", and is used to mean exactly the same thing.)Daqu (talk) 16:36, 10 June 2008 (UTC)[reply]

Significant omission: the connection between analytic continuation and the Riemann surface of a function

There is an intimate connection between analytic continuation of an analytic function, and the Riemann surface of the function. This is not discussed in the article, which strikes me as a significant omission.

The article does include the briefest mention of Riemann surfaces, in the following passage:

"The concept of a universal cover was first developed to define a natural domain for the analytic continuation of an analytic function. The idea of finding the maximal analytic continuation of a function in turn led to the development of the idea of Riemann surfaces."

But an article on analytic continuation requires more than merely the briefest mention of Riemann surfaces. In addition, the first quoted sentence is confusing, since, for example, the natural domain for the sqrt(z) function is the double cover of C* (the complex plane with the origin removed) -- and not the universal cover of C*.Daqu (talk) 02:36, 11 June 2008 (UTC)[reply]

Daqu, it's not improved yet - after almost 9 years ...! :-(. The para you quoted still has the same form, and is equally uninformative as it was then. This is yet another place where a bit of history would help readers immeasurably in understanding the motivations of, and the paths taken, by the mathematicians who pioneered these techniques. And your example about needing the double cover (rather than the universal cover) as the natural domain of the sqrt(z) function illustrates the need for caution and precision when analytically continuing any given function.

And another major omission is this: a clear example showing exactly how one does perform the analytic continuation. After all, an article that claims to describe a technique, yet fails to do so, nor even exemplify it, is nearly useless for those wanting to learn what must be done to use this technique! yoyo (talk) 11:16, 5 May 2017 (UTC)[reply]

Confusing sentence: Parts of a germ

This sentence: "The base g0 of g is z0, the stem of g is (α0, α1, α2, ...) and the top g1 of g is α0. The top of g is the value of f at z0, the bottom of g." I couldn't find this terminology by a Google search. Also, z0 is referred to as both the 'base' and the 'bottom' of g. Lateralrust (talk) 21:36, 9 March 2011 (UTC)[reply]

Seconded. This sentence in particular is confusing, as it doesn't even read like a complete sentence: "The top of g is the value of f at z0, the bottom of g." —Preceding unsigned comment added by 128.163.128.193 (talk) 04:53, 30 March 2011 (UTC)[reply]

Typo?

"This is because F1 − F2 is an analytic function which vanishes on the open, connected domain U of f and hence must vanish on its entire domain. This follows directly from the identity theorem for holomorphic functions."

Is this correct? F1 - F2? It seems that would be zero, which is only an analytic continuation of 0. Was this meant to imply something more along the lines of F1 "slash? F2? Using "and" seems like it would be better and avoid extremely likely confusion. 67.194.8.120 (talk) 00:54, 28 March 2011 (UTC)[reply]

Unhelpful

The following objection was added by User:Swanrt to the body of the page. I sympathise with it, but it clearly does not belong there:

Note: This discussion mixes a basic idea, in the title, analytic continuation, with a complex abstract idea, germs, their topology, and related subjects. The post would benefit if someone took the time to define analytic continuation, give its motivation, and give two examples, very concretely. A separate post should be written on the notion of a "germ." Warning: the current post will likely not be helpful to the student learning complex analysis for the first time or curious about this aspect of analytic functions.

Writing

Is there some chance that someone who has both an understanding of the mathematics and a reasonable ability to write comprehensibly might take a pass over this article? I have a B.A. in mathematics, an M.S. in computer science, plenty of theoretical background in mathematical analysis and other related fields, decades of experience as a software developer, etc., and I can't make head or tail of this. If I can't read it, what person who is not already solid on the mathematics in question is going to be able to do so? - Jmabel | Talk 06:47, 29 January 2014 (UTC)[reply]

Well said! And I second your request, Jmabel. I, too, have a similar background. Yet we're diving into abstract sheaves before we've even bothered to explain the general notion of what we want to achieve or how to go about it. Just because we may love pure mathematics doesn't mean we're not interested in the concrete details of solving specific problems ... yoyo (talk) 11:32, 5 May 2017 (UTC)[reply]

Misleading illustration

The picture purporting to illustrate analytic continuation, having the caption "Analytic continuation of natural logarithm (imaginary part)", is misleading.

In analytic continuation, the center of each disk in a sequence of function elements must lie inside the previous disk. In this picture, not even one successive disk's center lies within the previous disk.

This is not helpful for illustrating analytic continuation. Can someone please replace this picture with one that correctly depicts each successive disk as having its center inside the previous disk? (It would be a good idea to display a small dot at each disk's center.)Daqu (talk) 16:34, 12 January 2015 (UTC)[reply]

Good point, Daqu! I'm willing to try improving it, but need some guidance. I've examined the picture file (on Wikimedia Commons), but that doesn't give any indication of the tools used for creating it, so we'd have to start from scratch (or try to persuade the original creator to improve the graphic. That was 13 years ago, and the user mostly contributes in Japanese, so good luck if you want to try that way.) Any recommendations for suitable graphic tools for illustrating complex analysis? (I mostly use photographic and paint programs.) And presumably, a sequence of disks each of constant radius R, with centres separated by some large fraction of R (e.g. 99%), would meet the requirements? yoyo (talk) 11:47, 5 May 2017 (UTC)[reply]

Article becomes inappropriately abstract way too soon

The description of analytic continuation in terms of germs and sheaves is completely inappropriate as the first definition that the reader encounters. It would be entirely appropriate if these topics are placed toward the end of the article, as the "modern viewpoint" of analytic continuation. But they are unnecessarily abstract for the first definition that a reader encounters.

The first definition that a reader encounters ought to be simply about a sequence of function elements (power series converging on an open set) such that the center of each successive function element lies within the disk of convergence of the previous one, and such that each successive function element agrees with the previous one where their domains overlap. This may be thought of as continuation along the curve consisting of the polygon connecting successive function elements' centers with straight lines.

Then, a topology for the function elements should be defined by declaring that continuation along two polygons with the same endpoints is equivalent when the polygons are homotopic to each other, holding the endpoints fixed, within an open region of analyticity. This (almost) defines the topology of the function (its Riemann surface). For more information on this, the reader should be referred to the article on Riemann surfaces.

Also: All the stuff about natural boundaries should have its own article. The concept of "natural boundary" is entirely appropriate for this article, but the theorems about which power series have a natural boundary are too removed from the subject of analytic continuation.Daqu (talk) 19:45, 12 January 2015 (UTC)[reply]

Natural boundary

The current article states that "The circle is a natural boundary if all its points are singular." But isn't the correct definition "The circle is a natural boundary if the singular points are dense in the circle."? In my experience, the normal case is that there are an infinite but countable number of singular points that are dense on the circle. It is because they are dense, that one cannot analytically continue past the natural boundary. I also believe that its impossible to have an uncountable number of singular points on the natural boundary; if one did, the function inside the circle wouldn't be analytic. So an uncountable number of singular points on the boundary is "crazy talk", from what I can tell. Finally, I think that it is important to note that, if one removes the singular points from the boundary, what remains is a cantor set, viz, the function is regular on a cantor set. One still cannot analytically continue past this, since each "element" of the cantor set is "infinitely small"; the radius of convergence is zero at each such regular point. I am 99.9% certain that this is the case, but am not aware of which theorems or books convert this into a more precise and accurate statement. (I'm thinking that maybe books on Hardy spaces make such statements, make this more precise.) 67.198.37.16 (talk) 22:20, 10 May 2018 (UTC)[reply]

Small open set?

In the section titled "Formal definition of a germ" the following appears:

"... it would be equivalent to begin with an analytic function defined on some small open set."

What is meant by "small" in this context?

I suggest "small" be either removed or replaced by some specific quantifier.

Dratman (talk) 08:37, 18 March 2019 (UTC)[reply]

As far as I perceive the ubiquitous use of the epitheton small ornating open intervals, open balls, open neighborhoods, ..., it is always for allowing these domains to be arbitrarily small in some metric

(\varepsilon >0),

just for being able to keep undesired behavior outside the considered range, and since demanding the existence of something that may be arbitrarily small sounds a bit restrictive, the more laissez faire formulation of just small took over. As I perceive it, just removing it may go a bit too far. Purgy (talk) 11:28, 18 March 2019 (UTC)[reply]

Last step in 'Worked Example'

I can follow it up to the last step, but the last integral just seems to vanish. I assume it is some kind of trick, does anyone know how to do it?

Darcourse (talk) 14:05, 28 April 2022 (UTC)[reply]

Solved. The last three steps are:

1) $e^{2i(m-k)\pi }=1$ unless $k=m$

2) The integral is constant w.r.t. $\theta$

3) GF for ${\frac {1}{(1+(a-1))^{k-1}}}$ (more commonly ${\frac {1}{(1-x)^{k}}}$ )

Darcourse (talk) 15:23, 28 April 2022 (UTC)[reply]

worked example error?

After the shift to the new center, the result is: $f(z)={\frac {1}{a-z}}$

Shouldn't it be just $f(z)={\frac {1}{z}}$ Otherwise, the original function and the shifted function are not equal in the overlapping region. 2A01:C22:D416:8200:58E8:6611:CDBE:5215 (talk) 19:47, 12 April 2023 (UTC)[reply]

I agree, so does WolframAlpha: https://www.wolframalpha.com/input?i=1%2Fa*sum+from+k%3D0+to+infinity+%281-z%2Fa%29%5Ek 2001:4CA0:250D:0:5914:C38E:67DB:9480 (talk) 16:15, 13 June 2023 (UTC)[reply]

Worked example

Can we add a few explanations on this set of expressions in the worked example section, I cannot understand some of the simplifications:

${\begin{aligned}f(z)&=\sum _{k=0}^{\infty }a_{k}(z-a)^{k}\\&=\sum _{k=0}^{\infty }(-1)^{k}a^{-k-1}(z-a)^{k}\\&={\frac {1}{a}}\sum _{k=0}^{\infty }\left(1-{\frac {z}{a}}\right)^{k}\\&={\frac {1}{a}}{\frac {1}{1-\left(1-{\frac {z}{a}}\right)}}\\&={\frac {1}{z}}\\&={\frac {1}{(z+a)-a}}\end{aligned}}$

For instance, how do we get the second expression?

${\begin{aligned}&=\sum _{k=0}^{\infty }(-1)^{k}a^{-k-1}(z-a)^{k}\\&={\frac {1}{a}}\sum _{k=0}^{\infty }\left(1-{\frac {z}{a}}\right)^{k}\end{aligned}}$ Zeyn1 (talk) 11:44, 10 February 2024 (UTC)[reply]

In the chain of equations, the second equation

\sum _{k=0}^{\infty }a_{k}(z-a)^{k}=\sum _{k=0}^{\infty }(-1)^{k}a^{-k-1}(z-a)^{k}

holds because of the calculation immediately above this one in the article. The third equation

\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}

holds because of simpler algebra (multiplying the

(-1)^{k}a^{-k-1}

into the parentheses, with one factor of

a^{-1}

left over outside). Was your question about one of those equations? Mgnbar (talk) 14:44, 10 February 2024 (UTC)[reply]

Thanks. It's clear now. Zeyn1 (talk) 12:46, 22 February 2024 (UTC)[reply]

not understanding the end of the calculation

Hello, could we add few steps in one demonstration because I am not understanding two things in those three lines: ${\begin{aligned}a_{k}&={\textrm {...}}\\&={\frac {1}{2\pi }}\sum _{n=0}^{\infty }(-1)^{n}\int _{0}^{2\pi }{\binom {n}{k}}(a-1)^{n-k}d\theta \\&=\sum _{n=0}^{\infty }(-1)^{n}{\binom {n}{k}}(a-1)^{n-k}\\&=(-1)^{k}a^{-k-1}\end{aligned}}$ So apparently for all $m\neq k$ the integration is zero because of the loop provided by $e^{ik(m-\theta )}$ . Ok, but then summing over $n$ in ${\binom {n}{k}}$ creates terms that can't exist if $n<k$ . So should we not change $\sum _{n=0}^{\infty }$ to $\sum _{n=k}^{\infty }$ ?

In addition, $\sum _{n=0}^{\infty }(-1)^{n}{\binom {n}{k}}(a-1)^{n-k}=(-1)^{k}a^{-k-1}$ is certainly not the Newton binomial formula, it might be related, but I'm not seeing it.Klinfran (talk) 12:38, 16 February 2024 (UTC)[reply]

Traditionally, the binomial coefficient is zero if

n<k.

Nevertheless, an easy change of summation index avoids the use of this convention. The last relation does not results from the binomial coefficient, but from the derivation of the geometric series. I have edited the article for making the article clearer. D.Lazard (talk) 15:44, 16 February 2024 (UTC)[reply]

germs as power series vs. as equivalence classes

The article on germs themselves defines them as equivalence classes of functions that agree on some neighborhood, but here they're defined essentially as the data for a specific power series. Shouldn't there be some note explaining that usage differs in this context (or whatever the case is)? 24.56.238.67 (talk) 20:44, 29 May 2024 (UTC)[reply]