Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Define asymptotic as a class of sequences {$ x_i$},$_{i\in\mathbb N}$ modulo equivalence {$x_i$}={$y_i$} if $\lim_{i\to\infty} (x_i/y_i)=c\in\mathbb R,c\ne 0$.

More, we define $X= \{x_i\} \lt Y= \{ y_i \}$ if $\lim_{i\to\infty} (x_i/y_i)=0$.

We can interpret $x_i$ as $f(1/i)$ and say that these asymptotics are precisely aymptotics of functions at 0.

There is a classic object - Puiseux series $\{\sum\limits_{\alpha\in I}a_\alpha t^{\alpha}\}$ (where $I$ is well-ordered set).

Asymptotics and Puiseux series look quite similar, isn't it?

My question is a bit vague.

Suppose one has a finite set of Puiseux series $X_1\leq X_2\leq\dots X_n$ , linear dependencies between them(so, one can build matroids etc). One proves something about this set - more or less using only valuation map $val(\sum\limits_{\alpha\in I}) = -\min\limits_{\alpha\in I}\alpha$, which is just "order" for an asymptotics.

How to argue that all these arguments which work for Puiseux series work for asymptotics as well?

In fact one can rewrite all proofs but the problem is that it is not clear what is "order" of an asymptotic (and order of a Puiseux series is just a real number).

share|improve this question
    
I'm afraid I cannot see the similarity between asymptotics and Puiseaux series that you claim (but that could be my own problem). I think a more fundamental error lies in what you say about asymptotics of functions at zero: the function $x \mapsto \sin \pi/x$ vanishes on reciprocals of integers, but the function does not asymptotically vanish near zero. –  S. Carnahan May 2 '13 at 3:15
    
That is true, but in a question I have a finite set of pairwise comparable asymptotics, so, I evoke for properties of "good" asymptotics –  Nikita Kalinin May 2 '13 at 10:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.