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Topic: Exponential function (Read 9107 times) |
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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 (*).
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| « Last Edit: Feb 22nd, 2009, 12:39pm by comehome1981 » |
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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.
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| « Last Edit: Feb 22nd, 2009, 2:57pm by towr » |
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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. |
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The product would be greater than 1.
And also, I run program in your example, it approaches to exp(1) as N-> infinity
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| « Last Edit: Feb 22nd, 2009, 2:04pm by comehome1981 » |
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towr
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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
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Yeah, I seem to have forgotten the "1+" when I ran with that. Whoops 
The fact that whatever you do the minimum is at the very least 1 ought to have tipped me off.
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Eigenray
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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.
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| « Last Edit: Feb 22nd, 2009, 5:44pm by Eigenray » |
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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. |
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Yes, this is what I want.
Thanks so much.
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