       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, 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:= Series[Exp[(-y)*(Sqrt[1 + m^2*u^2]/m)], {u, Infinity, 10}]
>> Out= E^(-((Sqrt[1 + m^2*u^2]*y)/m))
>>
>> What do I miss here?
>>
>> Thanks
>> Dimitris
>>
>>
>

```

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