[Date Index]
[Thread Index]
[Author Index]
Re: asymptotics
*To*: mathgroup at smc.vnet.net
*Subject*: [mg76828] Re: asymptotics
*From*: dimitris <dimmechan at yahoo.com>
*Date*: Mon, 28 May 2007 00:59:41 -0400 (EDT)
*References*: <200705241023.GAA21917@smc.vnet.net><f36dl8$772$1@smc.vnet.net>
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
Prev by Date:
**Re: Re: Stopping Automatic Animation in v6**
Next by Date:
** Re: ListIntegrate3D ?**
Previous by thread:
**Re: asymptotics**
Next by thread:
**Re: Re: asymptotics**
| |