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comehome1981
Newbie  Posts: 20 Exponential function   « on: Feb 22nd, 2009, 12:38pm » Quote Modify

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
wu::riddles Moderator
Uberpuzzler      Some people are average, some are just mean.

Gender: Posts: 13730 Re: Exponential function   « Reply #1 on: Feb 22nd, 2009, 1:26pm » Quote Modify

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
Newbie  Posts: 20 Re: Exponential function   « Reply #2 on: Feb 22nd, 2009, 1:56pm » Quote Modify

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
towr
wu::riddles Moderator
Uberpuzzler      Some people are average, some are just mean.

Gender: Posts: 13730 Re: Exponential function   « Reply #3 on: Feb 22nd, 2009, 2:11pm » Quote Modify

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 The fact that whatever you do the minimum is at the very least 1 ought to have tipped me off. IP Logged

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Eigenray
wu::riddles Moderator
Uberpuzzler      Gender: Posts: 1948 Re: Exponential function   « Reply #4 on: Feb 22nd, 2009, 5:29pm » Quote Modify

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
Newbie  Posts: 20 Re: Exponential function   « Reply #5 on: Feb 22nd, 2009, 6:24pm » Quote Modify

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. IP Logged

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