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Re: asymptotics
*To*: mathgroup at smc.vnet.net
*Subject*: [mg76634] Re: [mg76613] asymptotics
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Fri, 25 May 2007 06:25:46 -0400 (EDT)
*References*: <200705241023.GAA21917@smc.vnet.net> <5FACEAED-CA9A-4D1D-8093-09018D633612@mimuw.edu.pl>
On 24 May 2007, at 22:37, Andrzej Kozlowski wrote:
>
> 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
>
>
In fact, it seems the other CAS uses a somewhat different convention
from Mathematica, so that:
Simplify[Normal[Series[a^(Sqrt[m^2 + 1/u^2]/m), {u, Infinity, 9}]] /.
a -> Exp[(-y)*u], {m > 0, u > 0, y > 0}]
(1/(384*m^8*u^7))*((384*u^7*m^8 - 192*u^6*y*m^6 +
48*u^4*y*(u*y + 1)*m^4 -
8*u^2*y*(u^2*y^2 + 3*u*y + 3)*m^2 +
y*(u^3*y^3 + 6*u^2*y^2 + 15*u*y + 15))/E^(u*y))
gives an answer closer to the one the other CAS gives. It is still
not the same, but unlike power series, this kind of assymptotic
expansions are not unique.
Andrzej Kozlowski
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