Re: asymptotics

*To*: mathgroup at smc.vnet.net*Subject*: [mg76629] Re: [mg76613] asymptotics*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Fri, 25 May 2007 06:23:10 -0400 (EDT)*References*: <200705241023.GAA21917@smc.vnet.net>

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

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