asymptotic expansion
Leslaw Bieniasz
Hello,
I need an asymptotic expansion for the expression:
1/( 1 + a*exp(b/x) )
where a >> 1, b >= 0 and x > 0,
in the limit of x -> 0+
Ideally, I'd like to have a simple expression dependent
on a power of x or 1/x.
Can anybody help me to derive such an expansion?
Is this possible at all?
Leslaw
Alois Steindl
Hello,
since I don't know what type of answer you expect, you could start
with the simpler function
a*exp(-b/x)
From there it is quite simple to answer your question.
Alois
--
Alois Steindl, Tel.: +43 (1) 58801 / 32558
Inst. for Mechanics and Mechatronics Fax.: +43 (1) 58801 / 32598
Vienna University of Technology, A-1040 Wiedner Hauptstr. 8-10
David C. Ullrich
Well, it's asyptotic to 1/( a*exp(b/x) ) as x -> 0+,
so it tends to 0 faster than any power of x. So I
guess no, what you want is not possible.
>Leslaw
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
Alois Steindl
>
> Well, it's asyptotic to 1/( a*exp(b/x) ) as x -> 0+,
> so it tends to 0 faster than any power of x. So I
> guess no, what you want is not possible.
>
>
> David C. Ullrich
Hello,
I would like to object: It seems to me that the result is extremely
simple. It might be so simple, that it isn't of any use, but it is
possible.
Best wishes
Alois
David C. Ullrich
Really? Then what power of x _is_ 1/( 1 + a*exp(b/x) )
asymptotic to as x -> 0?
>Best wishes
>Alois
Herman Rubin
Leslaw Bieniasz <nbbi...@cyf-kr.edu.pl> wrote:
>Hello,
If b > 0, there is no such expression, as the rate of
approach to 0 is too great. The "1+" in the denominator
is irrelevant in this case. Of course, if b=0, the
expression is a constant 1/(1+a).
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Alois Steindl
>
> Really? Then what power of x _is_ 1/( 1 + a*exp(b/x) )
> asymptotic to as x -> 0?
>
The power series expansion of f(x)=1/( 1 + a*exp(b/x) ) (for a > 0, b>0,
x>0) is simply 0, like the series expansion of exp(-1/x^2) = 0.
Unfortunately the radius of convergence is also 0, but one has
f(x) = o(x^k) for all k > 1.
Best wishes
Alois
David C. Ullrich
Of course f(x) = o(x^k) for all k > 0.
The question was what power of x is asymptotic
to f(x). You just explained why the answer is
"there is no such power of x".
Hence when the OP said he wanted to know how to
find what power of x was asymptotic to f(x) the
correct answer is that that can't be done. So
exactly what was your "objection" to my
statement "what you want is not possible"?
Jack Schmidt
>> f(x)=1/( 1 + a*exp(b/x) ) (for a > 0, b>0, x>0)
>> is simply 0, like the series expansion of exp(-1/x^2)
>> = 0. Unfortunately the radius of convergence is also 0,
>> but one has f(x) = o(x^k) for all k > 1.
>
> Of course f(x) = o(x^k) for all k > 0.
>
> The question was what power of x is asymptotic
> to f(x). You just explained why the answer is
> "there is no such power of x".
The confusion here is probably the difference between
the subject of the thread and the question actually
asked. The asymptotic expansion of 1/(1+a*exp(b/x))
is just 0, as was claimed, and the proof is precisely
that it is o(x^k) for all k.
However, 1/(1+a*exp(b/x)) is not asymptotic to 0 (even
if the definition made sense for the zero function).
This discrepancy is apparently quite important in
applications, at least according to this enjoyable
survey:
John P. Boyd
The Devil's Invention: Asymptotic, Superasymptotic
and Hyperasymptotic Series
Acta Appl. Math. 56 (1999), no. 1, 1--98.
http://www.ams.org/mathscinet-getitem?mr=1698036
http://dx.doi.org/10.1023/A:100614590362
http://www-personal.umich.edu/~jpboyd/boydactaapplicreview.pdf
David C. Ullrich
<Jack.Schmi...@gmail.com> wrote:
>>> The power series expansion of
>>> f(x)=1/( 1 + a*exp(b/x) ) (for a > 0, b>0, x>0)
>>> is simply 0, like the series expansion of exp(-1/x^2)
>>> = 0. Unfortunately the radius of convergence is also 0,
>>> but one has f(x) = o(x^k) for all k > 1.
>>
>> Of course f(x) = o(x^k) for all k > 0.
>>
>> The question was what power of x is asymptotic
>> to f(x). You just explained why the answer is
>> "there is no such power of x".
>
>The confusion here is probably the difference between
>the subject of the thread and the question actually
>asked. The asymptotic expansion of 1/(1+a*exp(b/x))
>is just 0, as was claimed,
I don't see anywhere in the thread where anyone's claimed
that. But never mind.
>and the proof is precisely
>that it is o(x^k) for all k.
>
>However, 1/(1+a*exp(b/x)) is not asymptotic to 0 (even
>if the definition made sense for the zero function).
Exactly what is the definition of "the asyptotic expansion"?
>This discrepancy is apparently quite important in
>applications, at least according to this enjoyable
>survey:
>
>John P. Boyd
>The Devil's Invention: Asymptotic, Superasymptotic
>and Hyperasymptotic Series
>Acta Appl. Math. 56 (1999), no. 1, 1--98.
>
>http://www.ams.org/mathscinet-getitem?mr=1698036
>http://dx.doi.org/10.1023/A:100614590362
>http://www-personal.umich.edu/~jpboyd/boydactaapplicreview.pdf
David C. Ullrich