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        general >> complex analysis >> Exponential function
        
(Message started by: comehome1981 on Feb 22^nd, 2009, 12:38pm)

Title: Exponential function
Post by comehome1981 on Feb 22^nd, 2009, 12:38pm

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 (*).

Title: Re: Exponential function
Post by towr on Feb 22^nd, 2009, 1:26pm

~~Taking 1/n_i = i/(N*(N+1)/2) seems to bring the product swiftly to zero.~~

Of course, the main reason why 1/n_i = 1/N given a series that converges to exp(1) is because of the binomial theorem.
(1+1/N)^N =
sum_i=1..N choose(N,i) 1/Nⁱ
= sum_i=1..N N!/i!/(N-i)! 1/Nⁱ
= sum_i=1..N 1/i! N!/(N-i)! 1/Nⁱ
~= sum_i=1..N 1/i! = exp(1)

The most important steps here don't apply to different 1/n_i. 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.

Title: Re: Exponential function
Post by comehome1981 on Feb 22^nd, 2009, 1:56pm

on 02/22/09 at 13:26:10, towr wrote:

Taking 1/n_i = 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

Title: Re: Exponential function
Post by towr on Feb 22^nd, 2009, 2:11pm

on 02/22/09 at 13:56:15, 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.

Title: Re: Exponential function
Post by Eigenray on Feb 22^nd, 2009, 5:29pm

Yes it is still true, because
x - x²/2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif log(1+x) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif x
and
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif1/n_i² http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif max(1/n_i) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 0.
Assuming that's what you mean. If you only mean that for each i,
1/n_i http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 0 as N http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif,
then the claim is false. For example, take n₁= .... =n_N-1 = (N-1)/c, where 0 < c < 1, and let n_N = 1/(1-c). Then
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/prod.gif(1+1/n_i) = (1+c/(N-1))^N-1 (2-c) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif e^c(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, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/prod.gif(1+a_i) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 1 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifa_i = 2, so the limit is never less than 2. So the possible limits are exactly those numbers between 2 and e, inclusive.

Title: Re: Exponential function
Post by comehome1981 on Feb 22^nd, 2009, 6:24pm

on 02/22/09 at 17:29:34, Eigenray wrote:

Yes, this is what I want.

Thanks so much.