0
$\begingroup$

Perturbative expansions of a function $f(x)$ around say $x=0$ cannot determine contributions from a function such as $e^{-1/x}$ since its Taylor series vanishes to all orders. This kind of contributions are usually called non-perturbative effects.

Asymptotic series approximate a function until some order but do not converge to the exact value.

I want to understand the relationship between non-perturbative effects and asymptotic series. Does every asymptotic series have contribution from functions with vanishing Taylor coefficients? Are contributions from functions with vanishing Taylor coefficients the (only) reason asymptotic series do not converge?

This seems unlikely to me because if we consider an exact function $p(x)$ describing the non-perturbative part of a function $f(x)$ and consider a Taylor series of $f(x)$ up to order $n$, let's call it $f_n(x)$ then I think we can have $f_n(x)$ running of to a random terrible function for large $n$ such that it is unlikely that $p(x)$ can correct it to the exact constant value at large $n$.

Can the contribution from functions such as $e^{-1/x}$ actually ever be reason that a series is asymptotic? It seems not. For example I can construct a function $f(x)= \sum_n^5 a_n x^n + e^{-1/x}$ which clearly converges in the sense that the value no longer changes after $n>5$. Thus, it is not a asymptotic series. Yet I would miss a non-perturbative contribution in the Taylor series.


It might be that there is no relation at all between asymptotic series and $e^{-1/x}$ type effects. These two things might be orthogonal in that both asymptotic and convergent series can have these effects. A good answer showing this would of course be appreciated. Yet one hint that there is a connection is for example that the Borel re-summation of a asymptotic series can contain information on $e^{-1/x}$ type effects.

$\endgroup$
0
$\begingroup$

Asymptotic series don't have vanishing Taylor series coefficients. In fact they tend to have the opposite problem of exploding Taylor series coefficients. The best example I know of an asymptotic expansion of a series is for the function $$ f(g) = \int_{-\infty}^{\infty} dx e^{-x^2-g x^4} $$ which of course converges for any $g \geq 0$ but its Taylor expansion has zero radius of convergence (because for $g<0$ it does not converge). However, for small values of $g$ the low-order series expansion in $g$ give a good approximation to its true value.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Of course I agree with the information above. And maybe that is the complete answer that there is no real relationship between asymptotic series and non-perturbative contributions (non-perturbative in the sense of vanishing Taylor coefficients). It might just happen to be that asymptotic series often also have non-perturbative contributions. But if that is so why can re-summations (Borel etc), capture non-perturbative effects for asymptotic series? So there must be some hidden connection, right? $\endgroup$ – Kvothe Oct 18 '19 at 12:41
  • $\begingroup$ Or am I making wrong assumptions and does Borel only retrieve non-perturbative effects in the other sense of the word, i.e. contributions with non-zero expansion coefficients? I do not believe that is the case though. For instance from wikipedia "Some of the singularities of the Borel transform are related to instantons and renormalons in quantum field theory". So the asymptotic expansion truly contains information on exp^{-1/x} effects, hinting at a link between asymptotic expansions and non-perturbative effects of this kind. $\endgroup$ – Kvothe Oct 18 '19 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.