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MathGroup Archive 2007

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Re: Re: asymptotics

  • To: mathgroup at smc.vnet.net
  • Subject: [mg76753] Re: [mg76654] Re: asymptotics
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Sat, 26 May 2007 04:53:35 -0400 (EDT)
  • References: <f33qsc$mc6$1@smc.vnet.net> <200705251036.GAA08078@smc.vnet.net>

This is all true, of course, except the last sentence. Obviously  
there cannot be a power series expansion of the such a  function at  
infinity since it has an essential singularity there (the function  
and all its "derivatives at infinity" are 0 so if a power series  
existed the function would have to be 0), but such functions can have  
various "asymptotic expansions", which are not power series (and  
hence not unique) but can be useful nevertheless. A simple example is  
the function Sin[1/Exp[x]]. It has no power series expansion at  
Infinity but Sin[1/x] does so we can get an "asymptotic expansion" in  
terms of Exp[x] as follows:

f[x_] = Normal[Series[Sin[1/x], {x, Infinity, 10}]] /. x -> Exp[x]

  1/(E^(9*x)*362880) - 1/(E^(7*x)*5040) + 1/(E^(5*x)*120) - 1/(E^(3*x) 
*6) + E^(-x)

For large x the values of f and of Sin[1/Exp[x]] are very close:

N[Sin[E^(-x)] /. x -> 10000, 20]
1.135483865314736098540938875066`19.999999999999833*^-4343

N[f[10000], 20]
1.135483865314736098540938875066`19.999999999999833*^-4343

The principle of this is pretty obvious: "expand what can be expanded  
in terms of what cannot be". This can sometimes be very useful, but  
to be useful, this kind of expansion has to be chosen for the  
specific purpose in mind. Because of this I doubt that having a CAS  
perform such expansions automatically is such a good idea. It is not  
clear to me exactly what the other CAS did to get that particular  
expansion and how it treats such cases in general. In any case I do  
not think we are much worse of for not having this feature in  
Mathematica.

Andrzej Kozlowski


On 25 May 2007, at 19:36, Jens-Peer Kuska wrote:

> Hi,
>
> the function has an essential singularity at u->Infinity and
> remember the definition of an essential singularity
> "A point a is an essential singularity if and only if the
> limit Lim[f,z->a] does not exist as a complex number nor equals  
> infinity"
>
> and this will make it impossible to get the zero order
> term of the series expansion.
>
> An asymtotic power series and x->Infinity exist only if
> for y[x] - Sum[a[n]*x^(-alpha*n),{n,0,N}] << x^(-alpha*N)
> for x->Infinity and every N. And the exponential function
> that you use has no powerseries expansion of this form.
>
> So the CAS may give you a result but this is nonsense.
>
> Regards
>    Jens
>
> 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
>>
>>
>



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