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January 27

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Limit 2

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Already simplified as much as I can. --AnalysisAlgebra (talk) 17:50, 27 January 2013 (UTC)[reply]

In this context, --AnalysisAlgebra (talk) 17:53, 27 January 2013 (UTC)[reply]

Have you tried using Stirling's approximation? Sławomir Biały (talk) 18:17, 27 January 2013 (UTC)[reply]
Yes, I have. That gives . I don't see how that makes it any easier. --AnalysisAlgebra (talk) 18:20, 27 January 2013 (UTC)[reply]
Would it be allowed to apply Stirling for k and (n-k) as well? If you substitute k^k and (n-k)^(n-k), you get a much simpler formula it seems. But I don't know if that substitution is valid ... Ssscienccce (talk) 01:48, 28 January 2013 (UTC)[reply]
The approximation is only valid for large or , but they are not all large (k is taken from 0). --AnalysisAlgebra (talk) 11:59, 28 January 2013 (UTC)[reply]
You can in fact use Stirling's formula here, for a simple reason - it's indeed inaccurate for small k and n-k, but as the summands where k or n-k are small become a vanishing part of the entire sum, so their accuracy has no effect on the result for the limit. -- Meni Rosenfeld (talk) 14:08, 29 January 2013 (UTC)[reply]
Have you tried applying the binomial theorem? Sławomir Biały (talk) 20:02, 27 January 2013 (UTC)[reply]
How could you use that? --AnalysisAlgebra (talk) 03:25, 28 January 2013 (UTC)[reply]

Observation, if it is helpful...     .     EdChem (talk) 02:59, 28 January 2013 (UTC)[reply]

It's not clear to me that the limit exists, but numerically it looks as though it does. Using R, the values for n=1e6,2e6 and 4e6 give 1.253981, 1.253786 and 1.253648 respectively. HTH, Robinh (talk) 08:20, 28 January 2013 (UTC)[reply]

continuous compounding formula derivation proof

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Is this how its done or is there another simpler way?

link — Preceding unsigned comment added by Ap-uk (talkcontribs) 23:43, 27 January 2013 (UTC)[reply]

That is OK but not really rigorous.


Substitute . No matter what r is as long as it's positive, as m→∞, n→∞, so you have


We can define
so
72.128.82.131 (talk) 02:12, 28 January 2013 (UTC)[reply]