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

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

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