Wikipedia:Reference desk/Archives/Mathematics/2013 January 29

= January 29 =

Proof of existence would be a useful start
The limit $$\lim_{n\rightarrow \infty} \sum_{k=0}^n\frac{k^k(n-k)^{n-k}}{k!(n-k)!}n!n^{-n-1/2}$$ is apparently difficult to find (see above); a proof that it simply exists would still be useful though. --AnalysisAlgebra (talk) 11:36, 29 January 2013 (UTC)


 * A trick with this sort of thing is to turn it into a function between 0 an 1 and integrate that, so do do the substitution $$k=nx$$ and also you have to multiply the expression by $n$ to compensate for it now only going from 0 to 1 instead of 0 to $n$. I had a quick go at that then substituting using Stirlings approximation but unfortunately I seem to have come to the result of it should be the integral of $$\frac{1}{\sqrt {2\pi(1-x)}}$$ which comes to $$\sqrt {\frac {2} {\pi}}$$ which is .7 something whereas your result looks more like the inverse of that i.e. $$\sqrt {\frac {\pi} {2}}$$! Dmcq (talk) 12:51, 29 January 2013 (UTC)
 * Fixed it, never was good at math ;-) I'd missed out a factor of sqrt x in the bottom so it should have been the integral of $$\frac{1}{\sqrt {2\pi x(1-x)}}$$ and the answer does come to $$\sqrt {\frac {\pi} {2}}$$ which is what you got near when approximating. Dmcq (talk) 13:25, 29 January 2013 (UTC)
 * I'm afraid I don't understand; can you explain what you're doing in detail? --AnalysisAlgebra (talk) 13:28, 29 January 2013 (UTC)
 * Using Stirling's approximation (see my comment to the previous question for justification), we have
 * $$\lim_{n\rightarrow \infty} \sum_{k=0}^n\frac{k^k(n-k)^{n-k}}{k!(n-k)!}n!n^{-n-1/2} = \lim_{n\rightarrow \infty} \sum_{k=1}^{n-1}\frac{k^k(n-k)^{n-k}}{k!(n-k)!}n!n^{-n-1/2} = $$
 * $$\lim_{n\to\infty} \sum_{k=1}^{n-1}\frac{1}{\sqrt{2\pi k(n-k)}} = \lim_{n\to\infty} \sum_{k=1}^{n-1}\frac{1}{n\sqrt{2\pi (k/n)(1-k/n)}} = \int_0^1\frac{dx}{\sqrt{2\pi x(1-x)}}=\sqrt{\frac{\pi}{2}}.$$
 * -- Meni Rosenfeld (talk) 14:15, 29 January 2013 (UTC)
 * We are summing $$n$$ terms which individually go to zero, so what I was doing was treating the sum as a numerical integration of a function from 0 to 1 where each little bit had width $$\frac 1 n$$, thus such a function would have to be $$n$$ times larger at each point to get the same small bit. Except for a smaller an smaller bit at the ends the factorials will be approximated well by Stirlings approximation so we get an accurate result by substituting using it. There's singularities at the ends of the range but if you do the substitution $$x = \sin^2 \theta$$ you can see they aren't bad and that gives you the result of the integration. Dmcq (talk) 14:55, 29 January 2013 (UTC)

That's great! Thanks. --AnalysisAlgebra (talk) 15:51, 29 January 2013 (UTC)

Generalizing integral
We know $$\int_0^1\frac{dx}{\sqrt{x(1-x)}}=\pi.$$ The indefinite integral is quite hard though.

I would like to know if this integral can be generalized in a certain way. According to Wolfram Alpha, these are the values for the following integrals:

$$\int_0^1\int_0^{1-x}\frac{dydx}{\sqrt{xy(1-x-y)}} = 2\pi$$

$$\int_0^1\int_0^{1-x}\int_0^{1-x-y}\frac{dzdydx}{\sqrt{xyz(1-x-y-z)}} = \pi^2$$

Is there a general solution for an integral of this type of arbitrary dimension? --AnalysisAlgebra (talk) 16:02, 29 January 2013 (UTC)\theta
 * Try $$x=\sin^2 \theta, y=\cos^2\theta \sin^2 \phi$$ etc. p.s. if anybody has trouble about the change of variables see jacobian determinant. Dmcq (talk) 17:48, 29 January 2013 (UTC)
 * I mean, is there a closed form formula for the result based on the dimension (number of variables being integrated)? --AnalysisAlgebra (talk) 19:10, 29 January 2013 (UTC)
 * I haven't the foggiest if there is a simple formula. Have you tried finding out for yourself? A few more values might trigger an idea or you might be able to see a pattern by just doing the first few integrations or a general one by hand. How about this for a closed form - the value for n is AAGI(n) where AAGI is the AnalysisAlgebra Generalised Integral function ;-) Dmcq (talk) 00:39, 30 January 2013 (UTC)


 * By converting the first few at least to polar integrals they appear to be of the form of surface integrals over n-dimensional hemispheres. In which case they'd be expected to follow a pattern π, 2π, π2, 4/3π2, 1/2π3 ... as outlined here. — Preceding unsigned comment added by 123.136.64.14 (talk) 02:50, 30 January 2013 (UTC)
 * The first few terms are $$\pi, 2 \pi, \pi^2, (4 \pi^2)/3, \pi^3/2, (8 \pi^3)/15, \pi^4/6, (16 \pi^4)/105, \pi^5/24$$. Then Mathematica chokes. So we definitely have the pattern alluded to by anon.
 * Code:

Table[Integrate[1/Sqrt[ Product[a[k], {k, 1, n}] (1 - Sum[a[k], {k, 1, n}])], Apply[Sequence, Table[{a[k], 0, 1 - Sum[a[j], {j, 1, k - 1}]}, {k, 1, n}]]], {n, 1, 10}]
 * -- Meni Rosenfeld (talk) 08:08, 30 January 2013 (UTC)

trial and error method
What is this trial and error method used to solve the the rate and the year of payment when it comes to annuities? Is there a website to show an example? Thanks.--Donmust90 (talk) 16:03, 29 January 2013 (UTC)Donmust90


 * Let say your goal is to distribute $100,000 over 10 years. So, just pay out $10,000 at the end of each year, right ?  Nope, because it's also earning interest as it sits, let's say at 10%.  So, we have to pay out a bit more to account for the interest.  Let's say we pay out $12,000 a year and run the numbers:

Start                End Year  Amount   Interest    Amount   Distribution --     ---      1    100,000   +10,000   110,000        -12,000     2     98,000    +9,800   107,800        -12,000   3     95,800    +9,580   105,380        -12,000   4     93,380    +9,338   102,718        -12,000                          .                          ..


 * So, we'd continue for the full 10 years. If money was left over, then the distribution is too low.  If the money runs out too fast, then the distribution is too high.  Then we pick another number and try again, until we get the result we want.  This might seem painful, but a computer program can do all this quite quickly.


 * There are also formulas that could be used to figure it out without trial and error, but some get fairly complex, at least for the average banker to use, when we add in continuous compounding, changing tax rates, etc. Again, a program can be used to do the math.  StuRat (talk) 06:32, 31 January 2013 (UTC)

Value of pi
Who proclaimed that π[py] = 22/7. it is mathematical question. please answer — Preceding unsigned comment added by SIVA SUBRAMANIAN.S (talk • contribs) 16:29, 29 January 2013 (UTC)


 * Question header added. AndrewWTaylor (talk) 16:36, 29 January 2013 (UTC)


 * According to Chronology of computation of π the approximation π = 22/7 was known to the Egyptians in the 26th Century BC. AndrewWTaylor (talk) 16:39, 29 January 2013 (UTC)


 * supposedly the legislature of a US state legislated that pi = 22/7. I forgot the details and I'm not sure it is true. Bubba73 You talkin' to me? 04:39, 31 January 2013 (UTC)


 * See this. Bubba73 You talkin' to me? 04:42, 31 January 2013 (UTC)


 * We also have an article on the Indiana pi bill. --Trovatore (talk) 04:45, 31 January 2013 (UTC)


 * Yes, legislators have their dirty little fingers in everybody's pi. StuRat (talk) 05:28, 31 January 2013 (UTC)


 * At least the Indiana bill would have made the pi higher. —Tamfang (talk) 01:17, 3 July 2013 (UTC)