Re: asymptotics

• To: mathgroup at smc.vnet.net
• Subject: [mg76798] Re: asymptotics
• From: m.r at inbox.ru
• Date: Sun, 27 May 2007 05:06:05 -0400 (EDT)
• References: <f33qsc\$mc6\$1@smc.vnet.net>

```On May 24, 5:54 am, dimitris <dimmec... at yahoo.com> 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*=
> 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

You can divide out the part that contributes the essential
singularity:

In[1]:= E^(-y u Sqrt[m^2]/m) Series[
E^(-y Sqrt[1 + m^2 u^2]/m + y u Sqrt[m^2]/m),
{u, Infinity, 4}] // InputForm

Out[1]//InputForm= SeriesData[u, Infinity, {1, -(Sqrt[m^2]*y)/(2*m^3),
y^2/(8*m^4),
(6*(m^2)^(3/2)*y - Sqrt[m^2]*y^3)/(48*m^7),
(-24*m^2*y^2 + y^4)/(384*m^8)}, 0, 5, 1]/E^((Sqrt[m^2]*u*y)/m)

Maxim Rytin
m=2Er at inbox.ru

```

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