Wikipedia:Reference desk/Archives/Mathematics/2013 January 27

= January 27 =

Limit 2
$$\lim_{n\rightarrow \infty} \sum_{k=0}^n\frac{k^k(n-k)^{n-k}}{k!(n-k)!}n!n^{-n-1/2}$$

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 $$\lim_{n\rightarrow \infty} \sum_{k=0}^n\frac{k^k(n-k)^{n-k}}{k!(n-k)!}\sqrt{2\pi}e^{-n}$$. 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 $$n\to\infty$$ 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.

$$P(t)=\lim_{m\to\infty}P_0\left(1+\frac{r}{m}\right)^{mt}$$

Substitute $$n=\frac{m}{r}$$. No matter what r is as long as it's positive, as m→∞, n→∞, so you have

$$P(t)=\lim_{n\to\infty}P_0\left(1+\frac{1}{n}\right)^{(rn)t}$$

$$P(t)=P_0\left(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}\right)^{rt}$$

We can define $$e:=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}$$

so $$P(t)=P_0e^{rt}$$

72.128.82.131 (talk) 02:12, 28 January 2013 (UTC)