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January 27

Limit 2

$\underset{n\to \mathrm{\infty }}{\lim }\sum _{k=0}^n\frac{k^k(n-k{)}^{n-k}}{k!(n-k)!}n!n^{-n-1/2}$

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

In this context, $0^0:=1$ --AnalysisAlgebra (talk) 17:53, 27 January 2013 (UTC)

Have you tried using Stirling's approximation? Sławomir Biały (talk) 18:17, 27 January 2013 (UTC)

Yes, I have. That gives $\underset{n\to \mathrm{\infty }}{\lim }\sum _{k=0}^n\frac{k^k(n-k{)}^{n-k}}{k!(n-k)!}\sqrt{2\pi }{e}^{-n}$. I don't see how that makes it any easier. --AnalysisAlgebra (talk) 18:20, 27 January 2013 (UTC)

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)

The approximation is only valid for large $k$ or $(n-k)$, but they are not all large (k is taken from 0). --AnalysisAlgebra (talk) 11:59, 28 January 2013 (UTC)

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 $n\to \mathrm{\infty }$ 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)

Have you tried applying the binomial theorem? Sławomir Biały (talk) 20:02, 27 January 2013 (UTC)

How could you use that? --AnalysisAlgebra (talk) 03:25, 28 January 2013 (UTC)

Observation, if it is helpful...     $\frac{k^k(n-k{)}^{n-k}}{k!(n-k)!}n!n^{-n-1/2}=\frac{n!}{k!(n-k)!}.\frac{k^k(n-k{)}^n}{(n-k{)}^k}.\frac{1}{n^n\sqrt{n}}=\frac{n!}{k!(n-k)!}.\frac{(1-\frac{k}{n}{)}^n}{(\frac{n}{k}-1{)}^k}.\frac{1}{\sqrt{n}}$.     EdChem (talk) 02:59, 28 January 2013 (UTC)

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)

continuous compounding formula derivation proof

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

link — Preceding unsigned comment added by Ap-uk (talk • contribs) 23:43, 27 January 2013 (UTC)

That is OK but not really rigorous.

$P(t)=\underset{m\to \mathrm{\infty }}{\lim }{P}_0{(1+\frac{r}{m})}^{mt}$
Substitute $n=\frac{m}{r}$. No matter what r is as long as it's positive, as m→∞, n→∞, so you have
$P(t)=\underset{n\to \mathrm{\infty }}{\lim }{P}_0{(1+\frac{1}{n})}^{(rn)t}$
$P(t)=P_0{\left(\underset{n\to \mathrm{\infty }}{\lim }{(1+\frac{1}{n})}^n\right)}^{rt}$
We can define $e:=\underset{n\to \mathrm{\infty }}{\lim }{(1+\frac{1}{n})}^n$
so $P(t)=P_0{e}^{rt}$
72.128.82.131 (talk) 02:12, 28 January 2013 (UTC)