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   Exponential function
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   Author  Topic: Exponential function  (Read 8741 times)
comehome1981
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Exponential function  
« on: Feb 22nd, 2009, 12:38pm »
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It is known that
 
(1+1/n)^n -> exp(1).
 
You can imagine that we partition the unit length into n pieces with equal length 1/n, then
               (1+1/n)^n -> exp(1).   .....(*)
 
But now we partition the the unit length into N unequal length, say, 1/n_i with i=1,2,...,N
        sum_i   n_i=1    and    n_i ->0 as N -> infinity
 
Is it still ture that
 
Prod^{N}_{i=1} (1+1/n_i)  -> exp(1)  as N->infinity  
 
 
It is kind of generalizing the (*).
« Last Edit: Feb 22nd, 2009, 12:39pm by comehome1981 » IP Logged
towr
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Re: Exponential function  
« Reply #1 on: Feb 22nd, 2009, 1:26pm »
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Taking 1/ni = i/(N*(N+1)/2) seems to bring the product swiftly to zero.
 
Of course, the main reason why 1/ni = 1/N given a series that converges to exp(1) is because of the binomial theorem.
(1+1/N)N =  
sumi=1..N choose(N,i) 1/Ni
= sumi=1..N N!/i!/(N-i)! 1/Ni
= sumi=1..N 1/i!  N!/(N-i)! 1/Ni
~= sumi=1..N 1/i! = exp(1)
 
The most important steps here don't apply to different 1/ni. They all need to be the same to apply the binomial theorem; and even then the approximation step N!/(N-i)! 1/Ni ~= 1 (for large N) might still not hold.
« Last Edit: Feb 22nd, 2009, 2:57pm by towr » IP Logged

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comehome1981
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Re: Exponential function  
« Reply #2 on: Feb 22nd, 2009, 1:56pm »
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on Feb 22nd, 2009, 1:26pm, towr wrote:
Taking 1/ni = i/(N*(N+1)/2) seems to bring the product swiftly to zero.

 
The product would be greater than 1.  
 
And also, I run program in your example, it approaches to exp(1) as N-> infinity
« Last Edit: Feb 22nd, 2009, 2:04pm by comehome1981 » IP Logged
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Re: Exponential function  
« Reply #3 on: Feb 22nd, 2009, 2:11pm »
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on Feb 22nd, 2009, 1:56pm, comehome1981 wrote:
The product would be greater than 1.  
 
And also, I run program in your example, it approaches to exp(1) as N-> infinity
Yeah, I seem to have forgotten the "1+" when I ran with that. Whoops Embarassed
The fact that whatever you do the minimum is at the very least 1 ought to have tipped me off.
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Re: Exponential function  
« Reply #4 on: Feb 22nd, 2009, 5:29pm »
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Yes it is still true, because
x - x2/2 log(1+x)   x
and
1/ni2   max(1/ni) 0.
Assuming that's what you mean.  If you only mean that for each i,
1/ni 0 as N ,
then the claim is false.  For example, take n1= .... =nN-1 = (N-1)/c, where 0 < c < 1, and let nN = 1/(1-c).  Then
(1+1/ni) = (1+c/(N-1))N-1 (2-c) ec(2-c).
For 0 < c < 1, this can be any number between 2 and e exclusive.  We can also make the limit 2 if we let c depend on N, say c = 1/N.
 
On the other hand, (1+ai) 1 + ai = 2, so the limit is never less than 2.  So the possible limits are exactly those numbers between 2 and e, inclusive.
« Last Edit: Feb 22nd, 2009, 5:44pm by Eigenray » IP Logged
comehome1981
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Re: Exponential function  
« Reply #5 on: Feb 22nd, 2009, 6:24pm »
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on Feb 22nd, 2009, 5:29pm, Eigenray wrote:
Yes it is still true, because
x - x2/2 log(1+x)   x
and
1/ni2   max(1/ni) 0.
Assuming that's what you mean.

 
Yes, this is what I want.  
 
Thanks so much.
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