Talk:Laurent series

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comments[edit]

I wonder if it is useful to allow for infinitely many negative-exponent terms, so that functions with essential singularities can be described? You can't talk formally about those of course. AxelBoldt, Thursday, July 11, 2002

The problem is that if you allow for infinitely many negative-exponent terms, Laurent series stop being unique. For instance:

1 + 1/x + 1/x2 + 1/x3 + ... = 1/(1 - 1/x) = -x/(1 - x) = -x(1 + x + x2 + x3 + ...) = -x - x2 - x3 - x4 - ...

Josh Grosse, Thursday, July 11, 2002

But the left hand series converges for |x| > 1, while the right hand side converges for |x| < 1, so the two Laurent series don't really overlap. AxelBoldt

One can certainly discuss such series, but I've never seen them referred to as "Laurent series". Of course, if you have infinitely many terms of negative degree but only finitely many terms of positive degree, (and if c = 0), then you have a "Laurent series in 1/z". But as far as I can tell, this is no more special than a Laurent series in any other expression. — Toby Bartels, Friday, July 12, 2002

In algebra, formal Laurent series are certainly restricted to finitely many negative indices, since that's the only thing that makes sense. In complex analysis however, things seem to be different:

I guess it's kind of nice if you can define the residual of a function simply by pointing to the -1 coefficient of its Laurent series; otherwise you need a case distinction for poles vs. essential singularities. AxelBoldt

Upon further consideration (also involving looking in some textbooks), you're absolutely right, Axel. — Toby Bartels, Friday, July 12, 2002


The article talks about formal Laurent series and appears to allow infinitely many negative degree terms. I don't think it is possible to define those in any meaningful way, if one wants a ring at least. AxelBoldt 14:46 Aug 13, 2002 (PDT)

Hey, you're the one that wanted these terms in the first place! Of course, I realise that you didn't mean that in the formal case. There may be a way to cure this since we need to identify certain infinitely negative series with certain infinitely positive ones (as per Josh Grosse's example above).

I'll have to think about it. Meanwhile, I've asked a colleague that apparently knows about this how it works. — Toby 13:47 Aug 14, 2002 (PDT)

The book Complex Variables and Applications by Ruel V. Churchill of the University of Michigan allows an infinite number of negative terms, and I see no reason why the definition should not allow for an infinite number of negative terms, as this would allow for expansions of √x and lnx around the singularities at 0 that converge for all x, as well as a infinitely convergent series for |x| around the singularity. Speaking of which, if anyone does know about these, have they ever heard of the ones mentioned above?Scythe33 01:16, 10 July 2005 (UTC)[reply]

The article states: "Similarly, the sum of two convergent Laurent series need not converge". I am having a hard time seeing how this can be. Are we sure this statement is correct? If this simply refers to the case where the annuli of convergence are disjoint, I would say that this is obvious anyway and the statement just confuses things by seeming to state something stronger. Danpovey (talk) 07:16, 10 January 2010 (UTC)[reply]

I agree with this and would take it a little further. If (formal) Laurent series don't converge in the same domain, I don't know what value can be placed in statements about the convergence properties of their sums or products. But if two Laurent series do converge, then they do so uniformly in the interior of their annuli of convergence. Consequently, if these regions overlap, their sums and products must also converge (also term-by-term derivatives, integrals, etc.), and I think the statements in the article about sums and products not converging is potentially confusing and/or misleading to students of complex analysis. Snowjeep (talk) 00:26, 19 April 2012 (UTC)[reply]

Merge with z transform[edit]

Should this article be merged with Z-transform? --Richard Clegg 14:39, 24 August 2006 (UTC)[reply]

No I think. The Z-transform is a very important topic in signal processing, and while it does use Laurent series, the primary focus of that article is not the series in itself, rather, how a function changes under the Z transform and how this can be applied to solving problems. In short, the Z-transform is much more than just an application of Laurent series. Oleg Alexandrov (talk) 16:08, 24 August 2006 (UTC)[reply]

correction[edit]

hi. i think i spotted an error. in section Multiplication, down the page shouldnt it say: (one cannot take the convolution of infinite sequences) instead of: (one cannot take the convolution of integer sequences) ? —Preceding unsigned comment added by Liquid.phynix (talkcontribs) 18:26, 11 June 2010 (UTC)[reply]

Unmentioned fact[edit]

The section Multiplication and sum begins with this paragraph:

"Laurent series cannot in general be multiplied. Algebraically, the expression for the terms of the product may involve infinite sums which need not converge (one cannot take the convolution of integer sequences). Geometrically, the two Laurent series may have non-overlapping annuli of convergence."

But this (and the rest of this section) do not mention the basic fact that if two functions f and g are both analytic on the same annulus A, then so is their product, which therefore possesses a Laurent series converging on A. And so in this case, the Laurent series of the product must be computable from the Laurent series of the functions f and g. Surely this problem has been solved long ago. Ideally someone knowledgeable on this subject will include this in the article.2601:200:C000:1A0:9D6A:3426:156B:13FB (talk) 01:56, 18 June 2022 (UTC)[reply]

There are many unmentioned facts here. Notice this talk page describes the article as "C class" on the Wikiproject Mathematics quality scale, and there are very few references in the article. Large books could be written on this topic and the article here just scratches the surface. If you know more about Laurent series and have good references, don’t hesitate to make improvements. –jacobolus (t) 03:29, 18 June 2022 (UTC)[reply]
There is more to this particular unmentioned fact. First, it is (if true) an utterly basic fact. It is mentioned (in the above quote) that the two functions' annuli of convergence may not overlap. Because of this, it would be entirely appopriate to state (if true) that two functions with Laurent series in the same annulus can in fact be multiplied to obtain the product of the two functions. 2601:200:C000:1A0:9D6A:3426:156B:13FB (talk) 19:43, 18 June 2022 (UTC)[reply]