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MathGroup Archive 2007

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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




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