RE: Re: Re: Tough Limit
- To: mathgroup at smc.vnet.net
- Subject: [mg34293] RE: [mg34284] Re: [mg34251] Re: [mg34235] Tough Limit
- From: "DrBob" <majort at cox-internet.com>
- Date: Tue, 14 May 2002 04:10:03 -0400 (EDT)
- Reply-to: <drbob at bigfoot.com>
- Sender: owner-wri-mathgroup at wolfram.com
You may be right about what the questioner meant, but you also may NOT. Mathematica didn't assume what we assumed, so it didn't (couldn't) find a limit. In this, Mathematica had the right of it. I shouldn't have assumed integer n without being told, and you shouldn't have assumed the absolute value. Interpreting the question for someone can only postpone their learning to state questions properly. Bobby -----Original Message----- From: Andrzej Kozlowski [mailto:andrzej at tuins.ac.jp] To: mathgroup at smc.vnet.net Subject: [mg34293] [mg34284] Re: [mg34251] Re: [mg34235] Tough Limit As a general principle, it seems to me that the right interpretation of any question is the one which is most likely to have been what the questioner meant. It this case n has to be assumed to be an integer (the notation Binomial[n,k] is almost always used when k is an integer, since otherwise the term "binomial coefficient" does not make sense -- although Mathematica does indeed allow k to be any real) and one has to take absolute value. If one does not, the answer becomes trivially negative and it is unlikely that anyone would have referred to such a problem as a "tough limit". In any case Mathematica's Limit does not work with sequences. On Monday, May 13, 2002, at 02:48 PM, DrBob wrote: > OK. I may have been wrong to limit n to Integers, as well. > > Unless we do both (limit n to Integers AND take the absolute value), > there is no limit. Mathematica would do neither without being told, of > course. > > Bobby > > -----Original Message----- > From: Andrzej Kozlowski [mailto:andrzej at tuins.ac.jp] To: mathgroup at smc.vnet.net > Sent: Monday, May 13, 2002 12:45 AM > Cc: mathgroup at smc.vnet.net > Subject: [mg34293] [mg34284] Re: [mg34251] Re: [mg34235] Tough Limit > > 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: [mg34293] [mg34284] [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 >>> >>> >>> >> >> >> >> > > > >