Re: Re: Tough Limit
- To: mathgroup at smc.vnet.net
- Subject: [mg34283] Re: [mg34251] Re: [mg34235] Tough Limit
- From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
- Date: Mon, 13 May 2002 05:54:49 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Strictly speaking you may be right, but it seemed to me pretty clear that the questioner meant to take absolute value and from his reply I gather that I was right in supposing so. Andrzej Kozlowski Toyama International University JAPAN http://platon.c.u-tokyo.ac.jp/andrzej/ On Monday, May 13, 2002, at 02:04 AM, DrBob wrote: > Taking the absolute value changes the problem, not the answer. > > The following: > > FullSimplify[Binomial[-1/2, n] > /Binomial[2n,n],Element[n,Integers]]//InputForm > > evaluates to > > (-1/4)^n > > confirming the identity you mentioned below. > > So... I let > > a=Binomial[-1/2,n] > b=((-1/4)^n*Gamma[1 + 2*n])/Gamma[1 + n]^2 > c=b/.Gamma[1+n_]->n! > d=c/.k_!->(k^(1/2 + k)*Sqrt[2*Pi])/E^k > e=d Sqrt[n]*Sqrt[Pi] > > Skipping most of the outputs, the last one evaluates to: > > (-1)^n > > This has no limit as n tends to infinity (unless you change the problem > by taking the absolute value). > > An earlier answer from Vladimir Bondarenko that drew plots to show the > series is all over the interval {-1, 1} was misleading because the plots > aren't restricted to integer n. (MY earlier answer was even MORE bogus. > Mea culpa!!) > > Bobby Treat > > -----Original Message----- > From: Andrzej Kozlowski [mailto:andrzej at platon.c.u-tokyo.ac.jp] To: mathgroup at smc.vnet.net > Sent: Sunday, May 12, 2002 2:26 AM > Subject: [mg34283] [mg34251] Re: [mg34235] Tough Limit > > First of all, you need to take the absolute value here, since your > expression is negative for odd n, e.g. > > In[3]:= > ((Binomial [-1/2 ,n ] * Sqrt[ n* Pi ])/.n->99)//N > > Out[3]= > -0.998738 > > I don't think Mathematica can do solve this problem without a lot of > human human help. Here is one way that makes just a slight use of > Mathematica (for the sake of decency). > First we use the identity Binomial[-1/2, n] == (-1/4)^n Binomial[2n, n], > > which unfortunately Mathematica does not know but you can find it > somewhere in Graham, Knuth, Patashink, "Concrete Mathemaitcs" together > with about a million other formulas of this type. Hence Abs[Binomial > [-1/2, n]] == (2n)!/(4^n (n!)^2). Now we use the assymptotic Sterling > formula (also proved in the same reference) and let Mathematica do the > cancellation: > > In[4]:= > Sqrt[n*Pi]*((2*n)!/(4^n*n!^2)) /. k_! -> Sqrt[2*Pi*k]*(k/E)^k > > Out[4]= > 1 > > > On Saturday, May 11, 2002, at 05:04 PM, RJMilazzo wrote: > >> Can anyone suggest how I can use Mathematica to get the following > limit: >> >> limit ( Binomial [-1/2 ,n ] * Sqrt[ n* Pi ] ) as n-> Infinity >> >> I have tried both Calculus`Limit` and the standard Limit functions. I >> can >> verify with NLimit that this limit equals approximately 1. I don't >> think that >> this is rigorous enough for a proof. >> >> Thanks >> James >> rjmilazzo at aol.com >> >> >> > > > >