Talk:Meromorphic function

must the set of poles be finite? -- Tarquin

Nope. 1/sin(z) is meromorphic. AxelBoldt

I don't think I completely understand the definition.

Can there be an infinite number of poles? And if so, do they have to be countable? --anon

Yes, they can be an infinite set. Hopefully my rewrite shows that.

They really do have to be isolated? --anon Yes, by definition. If they are not isolated, it is impossible to prove that a meromorphic function is a ratio of two holomorphic functions. Oleg Alexandrov 17:28, 11 August 2005 (UTC)Reply

http://mathworld.wolfram.com/MeromorphicFunction.html

here they speak of

if this a different definition or is it some non trivial theorem that if your definition is true, the number of poles is countable

thanks --anon

I don't see any contradiction between our page and mathworld. The poles of a meromorphic function must be isolated, by definition. Now, one can prove a theorem saying that a set of isolated points is finite or countable. So, is it the proof of this theorem that you are interested in? Oleg Alexandrov 20:44, 11 August 2005 (UTC)Reply

Outside of every neighbourhood of origin the function sin(1/z) is bounded in bounded sets. Thus origin cannot be it's accumulation point of poles. Perhaps here is a typo and it should be 1/sin(1/z)? —Preceding unsigned comment added by J Kataja (talk • contribs) 10:01, 9 January 2008 (UTC)Reply

You're right. The function ${\displaystyle \sin(1/z)}$  is actually not meromorphic in the origin, but for a different reason. -- EJ (talk) 12:00, 9 January 2008 (UTC)Reply

The statement concerning elliptic functions and elliptic curves sounds wrong. What "elliptic curves" did the author have in mind? An ellipse in the plane? Then it is simply wrong: elliptic functions are the inverse functions of the functions used to calculate the area of an ellipse.

Also elliptic functions are defined on a "period parallelogram", i.e. a fundamental region of a discrete lattice in the complex plane. That does not look like an "elliptic curve" to me. —Preceding unsigned comment added by 204.119.233.250 (talk) 21:16, 8 February 2008 (UTC)Reply

See elliptic function. Is is not about ellipses. Oleg Alexandrov (talk) 04:11, 9 February 2008 (UTC)Reply

This article says that "[meromorphic] functions are sometimes said to be regular functions or regular on D." However, I was unable to find a source which defines 'regular' synonymous to 'meromorphic'. Rather, most sources say that 'regular' = 'holomorphic'.[1][2][3] Especially, the last source is mathworld which User:Linas submitted as a reference when s/he wrote the sentence. If nobody opposes, I'll remove the sentence within a week, which may be added to holomorphic function article with a slightly changed wording. --Acepectif (talk) 19:15, 16 March 2008 (UTC)Reply

Heh. Neither planetmath nor mathworld state what you claim they state. Springer does call a holomorphic function a "regular analytic function"; however, it is generally understood that meromorphic functions are regular except at those points at which they're not. :-) This is just common(-sense) usage; the singular points are not regular, so a function is never regular at its singular points, and so a meromorphic function is regular everywhere where its not singular. If you can figure out a way of saying this simply, that would be good.

Please note that the phrase "regular function" enjoys broader usage than just in complex analysis; so, for example, in algebraic geometry, the points of a variety that are not singular are in fact regular, and so, by analogy the functions there are also called "regular functions". (Varieties have, of course the equivalent concept of meromorphic functions as wel, etc.) linas (talk) 21:38, 16 March 2008 (UTC)Reply

First of all, it is that you should prove with sources that 'regular' = 'meromorphic' (at least often enough to be mentioned in this article), rather than that I should prove that 'regular' = 'holomorphic', to prohibit me from deleting your sentence. Despite this, if I prove the latter, it will be obvious that your sentence must change its wording, if not deleted. So let's see whether planetmath and mathworld state what I claim they state.

For planetmath, see the bottom of this 'holomorphic' page. "Other names: holomorphic function, regular function, complex differentiable" - these are all synonyms. You can't find the word 'regular' from the 'meromorphic' page of this site. Also, mathworld clearly state that "A function is termed regular iff it is analytic and single-valued throughout a region R." If it saw 'regular' is 'meromorphic', it would have said "throughout a region R except a set of isolated points."

Even though I felt that it may be somewhat out of topic, I also wanted to mention the usage in algebraic geometry when I wrote the post above. In algebraic geometry, regular functions are everywhere-defined polynomials, i.e. they are analogues of holomorphic functions, not meromorphic ones. Rational functions or fractions of regular functions may correspond to meromorphic functions. --Acepectif (talk) 04:17, 17 March 2008 (UTC)Reply

Sounds good to me. Do it. linas (talk) 16:24, 17 March 2008 (UTC)Reply

The article states:

• Somewhat similarly, the function

${\displaystyle f(z)={\frac {z}{e^{z}-1}}}$ 

has singularities at all points in the form ${\displaystyle z=2n\pi i}$ , for ${\displaystyle n\in \mathbb {Z} }$ . However, it is not meromorphic on all of ${\displaystyle \mathbb {C} }$ , since the singularity at ${\displaystyle z=0}$  is removable: ${\displaystyle \lim _{z\rightarrow 0}f(z)=1}$ .

Why cancellation of a zero of denominator by the zero of the numerator breaks the meromorphicity? I would say, such ${\displaystyle f}$  is meromorphic. dima (talk) 00:54, 30 September 2008 (UTC)Reply

Well, the definition as given in the article indeed requires all singularities to be poles, disallowing removable singularities. (OTOH, the extension of f obtained by putting f(0) = 1 is meromorphic on ${\displaystyle \mathbb {C} }$ .) For a more striking example, if you take the function which is constant 0 on ${\displaystyle \mathbb {C} \setminus \{0\}}$ , and undefined in 0, it is not meromorphic on ${\displaystyle \mathbb {C} }$  according to the definition. I also find this convention counterintuitive, but I am not in a position to judge which definition is more standard. — Emil J. 11:30, 3 October 2008 (UTC)Reply

But a removable singularity is a pole, namely a pole of order 0. See e.g. the article on poles, which states: A pole of order 0 is a removable singularity. And there is no requirement for the poles of a meromorphic function to have order of at least 1. Furthermore, this restriction would contradict the statement, that all rational functions are meromorphic (x/x is certainly rational and has a removable singularity at 0). I think this example is wrong and should be removed. -- (ezander) 134.169.77.186 (talk) 07:58, 14 September 2009 (UTC)Reply

I agree that this example is not a good one, and I have removed it. The argument that "the definition as given in the article indeed requires all singularities to be poles, disallowing removable singularities" is not valid, as the definition specifies that we are dealing with a function "on an open subset D of the complex plane", which presumably is meant to mean that the function is defined on that open subset, in which case any removable singularities are by definition excluded. In fact I shall amend the definition to make this explicit.

Incidentally, the concept of a pole of order 0 is not a standard one: every text I have examined requires the order of a pole to be positive, and if there are any that don't then I think they must be a small minority. JamesBWatson (talk) 10:36, 17 September 2009 (UTC)Reply

Sorry, but your reformulation does not make any sense. If the function is defined in all of D, then it is holomorphic on D, not just meromorphic. The whole point of the concept is that a function meromorphic on D can be undefined in some points of D (which are isolated singularities). The only question is whether it is required to have poles (of positive order) in all these undefined points, or whether it can have removable singularities there. As for the poles of order 0 issue: removable singularities can also be zeros of positive order, and these are not poles of order 0, they would have to be "poles of negative order". Somehow I do not think that introducing those is a terribly good idea. — Emil J. 12:01, 17 September 2009 (UTC)Reply

Well, yes I see what you mean, and I did not express what I was trying to say accurately. Certainly it is possible to make what I said correct by taking the range of the function to be the extended complex plane, which personally I think is by far the best treatment, but I accept this is probably not the usual approach. Would it be acceptable to say "holomorphic on all D except a set of isolated points, which are poles or removable singularities"? Whatever the form of words it seems to me that removable singularities do not stop a function from being meromorphic. JamesBWatson (talk) 16:07, 17 September 2009 (UTC)Reply

Fine with me. — Emil J. 17:22, 17 September 2009 (UTC)Reply

I dimly remember that there is a theorem which states that you can expand any meromorphic function into a summation over its poles and the residues at those poles. Surely this article here should mention such an important theorem? --173.180.169.187 (talk) 16:06, 19 April 2013 (UTC)Reply

If someone knows about that or perhaps could provide a reference for it at the bottom of the pae, that would be great.2601:7:6580:5E3:7D6D:997D:F92D:729D (talk) 11:37, 21 January 2015 (UTC)Reply

I would clarify that the definition requires that the isolated singularities be poles rather than essential singularities. The fact that the phrase "the poles of the function" is in parentheses makes it ambiguous whether this is actually an additional nontrivial restriction on the definition, or simply a clarifying/explanatory comment for the reader which is being sloppy about distinguishing between poles and essential singularities.

"By using analytic continuation to eliminate removable singularities": Removable singularities are by definition isolated singularities that can be eliminated by extending the function holomorphically. Mentioning the general concept of analytic continuation makes this seem more complicated than it is. Maybe one should mention Removable singularity#Riemann's theorem instead. BSpringborn (talk) 09:19, 16 June 2017 (UTC)Reply

For example, I propose replacing the paragraph

> On a non-compact Riemann surface, every meromorphic function can be realized as a quotient of two (globally defined) holomorphic functions. In contrast, on a compact Riemann surface, every holomorphic function is constant, while there always exist non-constant meromorphic functions.

with

> On a non-compact Riemann surface, every meromorphic function into the complex plane can be realized as a quotient of two (globally defined) holomorphic functions. In contrast, on a compact Riemann surface, every holomorphic function into the complex plane is constant, while there always exist non-constant meromorphic functions.

I had to double check the original paragraph because it didn't pass my smell test.

I haven't made this change. I'm concerned my version might be too wordy. — Preceding unsigned comment added by Svennik (talk • contribs) 21:21, 25 August 2019 (UTC)Reply

What is the reference for this, what does it mean to have "functions on an elliptic curve" and is the statement even true? Kotika98 (talk) 12:53, 26 August 2021 (UTC)Reply