Re: asymptotics

*To*: mathgroup at smc.vnet.net*Subject*: [mg76877] Re: asymptotics*From*: dimitris <dimmechan at yahoo.com>*Date*: Tue, 29 May 2007 04:57:23 -0400 (EDT)*References*: <200705241023.GAA21917@smc.vnet.net>

Thanks a lot for your new response. I think it should be another function (no Series) that do this kind of asymptotic expansions. There are some commands that the other CAS has and I wish Mathematica had (as probably there are various things of Mathematica that a user from the other CAS would want to his system!) but that's life. For example the function identify that appears in a recent question of Daniel Huber I think or the function convert that I really wish Mathematica had something similar. And if, you and Maxim pointed out an asymptotic series like this in the current thread is indeed a simple task for someone to do by the Series command, for something like the following what you would say convert(log(x)*BesselJ(0,x)/sin(x^2),MeijerG); (output is omitted) Dimitris / Andrzej Kozlowski : > This gives me an opportunity to write a few more words on asymptotic > expansions, including soem things that I used to know but forgot. > > There are in fact several concepts of defintions of "asymptotic > expansions". The most common defintion is that of an asymptotic-power > series, of the form > > f(x) = a[0] + a[1] x^(-1) + ... + a[n-1] x^(-n+1) + (a[n]+epsion[n] > [x]) x^-n > > where epsilon[n][x]->0 as x->0. > > This says that the difference between the sum of the first n-terms > of the expansion and the function f[x] can be made as small as > possible in comparison with 1/x^n when x is large enough. > > In the case of this kind of expansions we have a certain degree of > uniqueness: a given functions can have only one expansion for a given > range of values of x, but two different functions can have the same > asymptotic expansion. For example, the asymptotic expansion of > Exp[-x] > is > 0+0+..... so the asymptotic expansion of f[x] and f[x]+ Exp[-x] will > be the same (however, Mathematica, rather naturally will not give you > any expansion of Exp[-x] at Infinity - if it did it would have to > return 0). > > On the other hand, Exp[x], has not asymptotic expansion at all. > > Another common kind (about which I forgot but Maxim Rytin remembered) > is the case f[x]/g[x]. If f[x]/g[x] has an expansion of the above > kind then one says that > > f(x) = g[x] (a[0] + a[1] x + ... + a[n-1]/x^(n-1) + ...) > > is an asymptotic expansion of f. This is the kind of expansion that > the other CAS gave you and it has, in general, better properties than > the ones I considered. > > There are, however, much more general asymptotic expansions, which > is what i had in mind in my first replies to you. In fact, one can > expand functions in terms of essentially any given functions. I > think the most general "theory" of this kind was invented by Hoene- > Wronski but it has never been given a formal mathematical basis. The > subject is very complicated and obscure with many special techniques > and only very limited general theory. It does not seem to be suitable > for CAS. > > Now finally for your question. To tell the truth I do not know the > answer. When asked to expand a Bessel function at Infinity > Mathematica first produces messages about essential singularities > and then returns an expansion that looks like one of the expansions > you can see here > > http://functions.wolfram.com/BesselAiryStruveFunctions/BesselJ/06/02/02/ > > Probably the explanation is that in the case of well known special > functions like the Bessel functions someone has programmed > Mathematica to return these expansions while it makes no attempt to > find them in more general cases. But actually, it seems, the all that > the other CAS does is simply to also consider the case: > g[x]*(something that has a power series at infinity), where g[x] has > an essential singularity at Infinity. This kind of thing is simple > enough to do by oneself. > > Andrzej Kozlowski > > > > > > > > On 28 May 2007, at 13:59, dimitris wrote: > > > Hello. > > > > Hmmm! > > > >> Well, of course, Mathematica correctly does not expand your function > >> .as a power series about infinity because such an expansion does not > >> exist (or, if you prefer, is identically 0). Note that the > >> "assymtotic expansion" the other CAS gives you is into a power series > >> expansion: since after truncation it contains a factor Exp[-y u]. > > > > Because of applied mathematics background I may miss something > > fundamentally. > > So I apologize for it. > > > > Why for example Mathematica does expand BesselJ at infinity? > > There is an essential singularity there. > > > > Dimitris > > > > / Andrzej Kozlowski : > >> On 24 May 2007, at 19:23, dimitris wrote: > >> > >>> Sorry fellas if I ask something trivial > >>> but currently I can't find anything! > >>> > >>> In another CAS I took > >>> > >>> f:=asympt(exp(-y*sqrt(1+m^2*u^2)/m),u,5); > >>> > >>> / 2 1/2 2 2 1/2 2 2 > >>> | y (m ) y y (m ) (-6 m + y ) > >>> f := |1 - --------- + ------- - ---------------------- > >>> | 3 4 2 7 3 > >>> \ 2 m u 8 m u 48 m u > >>> > >>> 2 2 2 \ 2 1/2 > >>> y (-24 m + y ) 1 | / y (m ) u > >>> + ---------------- + O(----)| / exp(-----------) > >>> 8 4 5 | / m > >>> 384 m u u / > >>> > >>> ff:=simplify(convert(f,polynom)) assuming m>0; > >>> ff := 1/384*exp(- > >>> y*u)*(384*m^8*u^4-192*y*m^6*u^3+48*y^2*m^4*u^2 > >>> +48*y*m^4*u-8*y^3*m^2*u-24*y^2*m^2+y^4)/ > >>> m^8/u^4 > >>> > >>> In Mathematica I can't get the expansion in infinity > >>> > >>> In[113]:= Series[Exp[(-y)*(Sqrt[1 + m^2*u^2]/m)], {u, Infinity, 10}] > >>> Out[113]= E^(-((Sqrt[1 + m^2*u^2]*y)/m)) > >>> > >>> What do I miss here? > >>> > >>> Thanks > >>> Dimitris > >>> > >>> > >> > >> > >> Well, of course, Mathematica correctly does not expand your function > >> as a power series about infinity because such an expansion does not > >> exist (or, if you prefer, is identically 0). Note that the > >> "assymtotic expansion" the other CAS gives you is into a power series > >> expansion: since after truncation it contains a factor Exp[-y u]. > >> There are many such assymtotic expansions. I do not know how to use > >> Mathematica to get this particular one, but it is easy to get similar > >> ones. For example, here is one way to get an assymptotic expansion > >> pretty close to the one given by the other CAS: > >> > >> f[u_] = Simplify[ > >> Normal[Series[a^(Sqrt[m^2 + 1/u^2]/m), {u, Infinity, 10}]] /. > >> a -> Exp[(-y)*u], {m > 0, u > 0, y > 0}] > >> > >> (3840*u^9*m^10 - 1920*u^8*y*m^8 + 480*u^6*y*(u*y + 1)*m^6 - > >> 80*u^4*y*(u^2*y^2 + 3*u*y + 3)*m^4 + > >> 10*u^2*y*(u^3*y^3 + 6*u^2*y^2 + 15*u*y + 15)*m^2 - > >> y*(u^4*y^4 + 10*u^3*y^3 + 45*u^2*y^2 + 105*u*y + 105))/(E^ > >> (u*y) > >> *(3840* > >> m^10*u^9)) > >> > >> This looks a little more complicated than the expression given by the > >> other CAS (and is, of course, not equal to it) but they both give > >> good approximations of the original function at Infinity. You can > >> check it numerically as follows: > >> > >> g[u_] = 1/384* > >> Exp[-y* > >> u]*(384*m^8*u^4 - 192*y*m^6*u^3 + 48*y^2*m^4*u^2 + 48*y*m^4*u - > >> 8*y^3*m^2*u - 24*y^2*m^2 + y^4)/m^8/u^4; > >> > >> Let's choose some random values for m and y: > >> > >> m = Random[]; y = Random[]; > >> > >> then > >> > >> f[20000] // N > >> 5.318723018905205*10^-2317 > >> > >> g[20000] // N > >> 5.318723018905206*10^-2317 > >> > >> which are pretty close. For larger values they are even closer: > >> > >> f[200000] // N > >> 1.811911734460420*10^-23163 > >> > >> g[200000] // N > >> 1.811911734460420*10^-23163 > >> > >> Of course, as expected, the values are very close to zero. For the > >> function itself we get: > >> > >> N[Exp[(-y)*(Sqrt[1 + m^2*u^2]/m)] /. u -> 200000] > >> 1.81191173447088963591082`11.14761897993578*^-23163 > >> > >> Andrzej Kozlowski > > > >

**Follow-Ups**:**Re: Re: asymptotics***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

**References**:**asymptotics***From:*dimitris <dimmechan@yahoo.com>

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