       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

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 + a 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 + a 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
> 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:= Series[Exp[(-y)*(Sqrt[1 + m^2*u^2]/m)], {u, Infinity, 10}]
> >>> Out= 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 // N
> >>   5.318723018905205*10^-2317
> >>
> >> g // N
> >>   5.318723018905206*10^-2317
> >>
> >> which are pretty close. For larger values they are even closer:
> >>
> >>   f // N
> >>   1.811911734460420*10^-23163
> >>
> >>   g // 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
> >
> >

```

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