Talk:Radius of convergence

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[edit] L and C: root and ratio test

Why must 1/L and 1/C be used for the radius of convergence? Why can't we invert the equations as they stand? For example {1 \over L} = R = \mbox{Radius Of Convergence} = \lim_{n\rightarrow\infty} \left| \frac{c_{n}}{c_n+1} \right| . Why can't we put that up. Do C and L have special uses?? - Fresheneesz

Alright that it, I'm freaking changing it. If someone wants to change it back, discuss it *here*. Fresheneesz 08:25, 27 March 2006 (UTC)
A possible reason is consistency with root test and ratio test. But feel free to change it. -- Jitse Niesen (talk) 12:03, 27 March 2006 (UTC)
Ratio test doesn't have a letter assocaited with it on the page at least (anymore?). But I don't want to change it if people actually use the letters L or C to mean something. If its 100% arbitrarily used.... Well, yea i guess it would be a good parallel. I think i'll fix it up so it is less arbitrary. Fresheneesz 09:23, 28 March 2006 (UTC)

[edit] Radius of convergence of other series

I've looked it up, and I can't seem to find any reference to the radius of convergence for series that are not power series. Is radius of convergence only definined for power series? Also, the ratio and root tests given on this page are not the same as the ratio and root tests that this page links to (I think), because here, the root test is applied to only the coeffecients of the term of the power series, while on the root test page, the entire term is applied. Fresheneesz 01:21, 29 March 2006 (UTC)

Well, you have to do a bit of work. A power series is a series of the form
\sum_{n=0}^\infty c_n z^n.
I set the centre a to zero because it does not matter. As you say, the root test should be applied to the whole term, so the root test says that the series converges if
\limsup_{n\to\infty} \sqrt[k]{|c_k z^k|} < 1.
This condition is equivalent to
|z| < \frac{1}{\limsup_{n\to\infty} \sqrt[k]{|c_k|}},
which is the formulation on this page.
For a general series depending on a variable z, the region of convergence will most likely not be a circle, and thus it makes no sense to talk of a radius of convergence. -- Jitse Niesen (talk) 07:29, 29 March 2006 (UTC)
Oh, thats interesting, I should have seen that. That would be an insightful addition to this page, I think i'll add it. I noticed that "region of convergence" redirects here, while it has no explanation whatsoever. I don't know what I can about it, but something should be added about it. Fresheneesz 10:19, 29 March 2006 (UTC)

[edit] Merge with power series

Since it looks as if a "radius of convergence" is a term only used for power series, and that page already has something on radius of convergence, it seems only natural to merge the two. Any comments? Fresheneesz 10:29, 29 March 2006 (UTC)

I disagree. This is an important enough concept to warrant its own article. Besides, the radius of convergence is only for power series of one variable, while power series can be in many variables. Oleg Alexandrov (talk) 17:42, 29 March 2006 (UTC)
I bet in power series of many variables one has many radii to define a polydisc of convergence, or even polyballs("Hartogs stuff') or more general cartesian products of balls of different dimension.Rich 21:33, 31 October 2006 (UTC)

[edit] odd wording

"The root test is defined as: C=..." - That doesn't make sense. A test is not an equation, nor is it a real number. Also for "ratio test" later. --Zerotalk 08:59, 14 August 2006 (UTC)

I agree. I have no idea what is being said. Can someone fix this? A5 21:06, 31 October 2006 (UTC)

I also agree. I've done some cleanup accordingly. The article titled root test actually asserted that the root test is a number. I've done some editing on that one too. Michael Hardy 21:54, 31 October 2006 (UTC)

[edit] 3-sphere of convergence?

Suppose one had a power series w/radius of conv = 1, say, around z=0, say. If z were allowed to be quaternionic, then the boundary of convergence would be a 3 sphere of rad 1. I bet the subset of the boundary where the series converges could be quite beautiful and fascinating. Does anyone know of results/research in this area? It would be a great addition to put in this article.Rich 07:47, 25 September 2006 (UTC)

[edit] What about

What about |z-a| =r \!? --Abdull 17:38, 11 February 2007 (UTC)

That's a limiting case. Many things can happen depending on the function. You may have divergence, or convergence, or conditional convergence. Oleg Alexandrov (talk) 00:07, 12 February 2007 (UTC)
In fact, the series MUST be bad at at least one point on the circle |z-a|=r since the radius of convergence is equal to the distance from point "a" to the nearest singularity. --Rocketman768 18:52, 1 May 2007 (UTC)
I think you mean "to the nearest SINGULAR POINT", not "singularity". It sounds like you're saying by "series MUST be bad at at least one point" that the series must diverge somewhere on the circle of convergence, which I disagree with, see article. Also see Konrad Knopp, Theory of Functions Part I, Dover, page 100, and Exercise 2 on page 73. --Rich Peterson