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   Author  Topic: asymptotic expansion  (Read 3089 times)
Moon
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asymptotic expansion  
« on: Apr 27th, 2004, 11:24pm »
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Please help me with this problem
 
Show that
 
 [int]0 [infty](1+t/k)kexp(-t)dt~1/2+1/(8k)-1/(32k2), k-> [infty]
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Re: asymptotic expansion  
« Reply #1 on: May 1st, 2004, 9:38am »
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Is this a "method of steepest descents problem"?
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Moon
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Re: asymptotic expansion  
« Reply #2 on: May 1st, 2004, 3:11pm »
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I think this is a Laplace's Method.
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Re: asymptotic expansion  
« Reply #3 on: May 1st, 2004, 8:40pm »
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If you mean the Laplace transform, that may be the origin of this problem, but I don't think it offers any help in how to solve it.
 
The problem as stated is not quite as bad as it seems. The limit of the right as k [to] [infty] is 1/2. Hence to show the desired asymptotic equivalence all you need to do is prove that the limit of the integral as k [to] [infty] is also 1/2. I.e., you don't need the 1/(8k) - 1/(32k2) to prove the asymptotic equivalence.
 
This is an interesting integral. If you could exchange integral and limit, taking the limit of the function as k [to] [infty] yields [int]0[supinfty] ete-t dt, which is infinite. But such exchange of the integral and limit is not justified in this situation.
 
Right now, the best I can suggest is to look at the method of steepest descents. It is used for producing such asymptotic formulas.
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Re: asymptotic expansion  
« Reply #4 on: May 6th, 2004, 5:21pm »
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You may or may not have noticed that I posted the integral here in the medium forum, where it would get more attention (I am not sure what other regulars visit this forum, but I seem to be the only one providing much in the way of answers in it.)
 
T&B posted that version on sci.math, where for some reason it got better response than what I assume is your own post there.
 
Several responders there and two on the other thread here demonstrated that the limit of the integral is [infty], not 1/2 as the other side of your asymptotic equivalence demands.
 
The most comprehensive result was posted by David C. Ullrich on sci.math:
limk k-1/2[int]0[supinfty](1 + t/k)ke-tdt = [surd]([pi]/2).

 
[surd]([pi]k/2) is definitely not ~ 1/2 + 1/(8k) - 1/(32k2).
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Alexey Sukhinin
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Re: asymptotic expansion  
« Reply #5 on: May 7th, 2004, 12:47pm »
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If you make a substitution 1+t/k = exp(u) and then integrate
the result by parts you'll get the needed result.
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Re: asymptotic expansion  
« Reply #6 on: May 7th, 2004, 6:56pm »
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I may get the "needed" result but I certainly don't get the desired result! Let f(n) = [int]0[supinfty] (1+t/k)ne-t dt. I prefer the substitution x = k + t. The bookkeeping is slightly easier. The substitution gives
f(n) = ekk-n[int]k[supinfty] xne-xdx.
Integrate by parts (u=xn, v=-e-x) to get
 
f(n)=ekk-n[ -xne-x]k[supinfty] +(n/k)ekk1-n [int]k[supinfty] xn-1e-x dx
 
f(n) = 1 + (n/k)f(n-1).
 
So, assuming k is an integer,  
 
f(k) = 1 + (k/k)(1 + ((k-1)/k)(1 + ((k-2)/k)(1 + ... + (1/k)f(0))...)
Clearly f(0)=1, and expanding the expression above yields:
 
f(k) = 1 + k/k + k(k-1)/k2 + k(k-1)(k-2)/k3 + ... + k!/kk = [sum]n=0k k!/(k-n)! k-n.
 
This not only does not look like 1/2+1/(8k)-1/(32k2), but is easily seen not to be asymptotically equivalent:
 
Each term in the sum = C(k,n)n!k-n > C(k,n)k-n. The sum of the latter terms is (1+1/k)k, which converges to e as k [to] [infty]. So whatever limit f(k) has as k [to] [infty], it has be [ge] e, but the limit of 1/2 + 1/(8k) - 1(32k2) is 1/2.
 
In fact, as has been demonstrated in the calculations cited, limk[to][subinfty] f(k) = [infty].
 
I suppose that a different approach might actually give you f(k) = 1/2 + 1/(8k) - 1/(32k2) + g(k) for some function g. But the limit of g(k) as k [to] [infty] will not be zero, as the supposed asymptotic equivalence requires.
« Last Edit: May 7th, 2004, 6:58pm by Icarus » IP Logged

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