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Amount of singular points

by jostpuur
Tags: points, singular
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jostpuur
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Dec30-07, 12:15 AM
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Is it possible for a complex analytic function to have an uncountable set of singular points?
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HallsofIvy
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Dec30-07, 06:00 AM
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Analytic where? Obviously, a function that is analytic everywhere has NO singular points. Am I correct that by "singular point" you mean a point at which the function is not analytic? Certainly it would be possible to define a function that would be analytic everywhere except at certain points and I see no reason why one could not do that for and uncountable set of points. The only requirement would be that the set of points on which the function is not analytic would be a closed set.
mathwonk
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Dec30-07, 12:41 PM
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it depends what kind of functions you want to allow. you want of course a function which is analytic on some open, presumably connected set in C, and which cannot be analytically continued outside that set to another strictly larger such open connected set, right?

well it is a theorem that every open connected set in C has such functions. A maximal domain for a complex holomorphic function, i.e. one for which there exists a function with no larger holomorphic extensions, is called a "domain of holomorphy", as i think i recall from my brief stint as a complex analyst in 1974-5.

this term is coined in the theory of several complex variables, since there it is no longer true that all open connected sets are such domains. but in one complex variable, all open connecetd sets are domains of holomorphy.

so i agree with halls that essentially the only requirement for the set of non holomorphic points is to be closed, in particular you could even have the open domain be dense in C. for references you can see gunning and rossi, or hormander.

there are other equivalent conditions in more variables as i recall, such as having a strictly pseudo convex boundary in C^n, or for a manifold having vanishing analytic sheaf cohomology ???, or ... hmmm memory fades after 30 years of disuse.

but

http://en.wikipedia.org/wiki/Domain_of_holomorphy

seems to confirm most of what i said.

jostpuur
#4
Dec30-07, 04:22 PM
P: 2,023
Amount of singular points

Quote Quote by HallsofIvy View Post
Analytic where? Obviously, a function that is analytic everywhere has NO singular points. Am I correct that by "singular point" you mean a point at which the function is not analytic?
I was thinking about some kind of divergence. Log(z) is not analytic on the negative real line, but that is not a kind of example I was thinking.

Certainly it would be possible to define a function that would be analytic everywhere except at certain points and I see no reason why one could not do that for and uncountable set of points. The only requirement would be that the set of points on which the function is not analytic would be a closed set.
[tex]
\sum_{n=1}^{\infty} \frac{1}{(z-n)^2}
[/tex]

is an example of a function that has countably infinite set of singular points. But how could one get a concrete example of uncountable set on singular points? You cannot define it with sum like this at least.

Quote Quote by mathwonk
it depends what kind of functions you want to allow. you want of course a function which is analytic on some open, presumably connected set in C, and which cannot be analytically continued outside that set to another strictly larger such open connected set, right?
Yes the analytic continuation is what I've been wondering, although I didn't make it clear yet.

For example a function

[tex]
f:\;]-\infty,1[\;\to\mathbb{R},\quad\quad f(x) = \frac{1}{1-x}
[/tex]

cannot be extended to values x>1 using analytic continuation on the real line. However it is possible through analytic continuation into to the complex plane, because there (1,0) is a point that can be gone around.

So I was thinking, that could similar obstacles come on the way of continuation also on the complex plane. For example a line [itex]l=\{x_0+iy\in\mathbb{C}\;|\;y\in\mathbb{R}\}[/itex], where x0 is a constant, so that [itex]f(z)\to\infty[/itex] when [itex]z\to l[/itex].

This seems to answer my original problem.


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