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Topic: Exponential function (Read 8853 times) 

comehome1981
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Exponential function
« on: Feb 22^{nd}, 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 22^{nd}, 2009, 12:39pm by comehome1981 » 
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towr
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Re: Exponential function
« Reply #1 on: Feb 22^{nd}, 2009, 1:26pm » 
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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/N^{i} = sum_{i=1..N} N!/i!/(Ni)! 1/N^{i} = sum_{i=1..N} 1/i! N!/(Ni)! 1/N^{i} ~= 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!/(Ni)! 1/N^{i} ~= 1 (for large N) might still not hold.

« Last Edit: Feb 22^{nd}, 2009, 2:57pm by towr » 
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comehome1981
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Re: Exponential function
« Reply #2 on: Feb 22^{nd}, 2009, 1:56pm » 
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on Feb 22^{nd}, 2009, 1:26pm, 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

« Last Edit: Feb 22^{nd}, 2009, 2:04pm by comehome1981 » 
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towr
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Re: Exponential function
« Reply #3 on: Feb 22^{nd}, 2009, 2:11pm » 
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on Feb 22^{nd}, 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.


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Eigenray
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Re: Exponential function
« Reply #4 on: Feb 22^{nd}, 2009, 5:29pm » 
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Yes it is still true, because x  x^{2}/2 log(1+x) x and 1/n_{i}^{2} max(1/n_{i}) 0. Assuming that's what you mean. If you only mean that for each i, 1/n_{i} 0 as N , then the claim is false. For example, take n_{1}= .... =n_{N1} = (N1)/c, where 0 < c < 1, and let n_{N} = 1/(1c). Then (1+1/n_{i}) = (1+c/(N1))^{N1} (2c) e^{c}(2c). 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+a_{i}) 1 + a_{i} = 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 22^{nd}, 2009, 5:44pm by Eigenray » 
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comehome1981
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Re: Exponential function
« Reply #5 on: Feb 22^{nd}, 2009, 6:24pm » 
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on Feb 22^{nd}, 2009, 5:29pm, Eigenray wrote:Yes it is still true, because x  x^{2}/2 log(1+x) x and 1/n_{i}^{2} max(1/n_{i}) 0. Assuming that's what you mean. 
 Yes, this is what I want. Thanks so much.


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