Re: Tough Limit
- To: mathgroup at smc.vnet.net
- Subject: [mg34322] Re: [mg34235] Tough Limit
- From: Andrzej Kozlowski <andrzej at platon.c.u-tokyo.ac.jp>
- Date: Wed, 15 May 2002 03:35:09 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
If you do not make these assumptions the answer is of course no. 1) If n is an odd integer Binomial[-1/2,n] is negative. 2) If n is an even integer Binomial[-1/2,n] is positive. 2) If n is not required to be an integer than Binomial[-1/2,n] = 0 whenever n=m/2 for an odd integer m. Proof: trivial exercise. Andrzej On Monday, May 13, 2002, at 10:01 PM, DrBob wrote: > 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 > Sent: Monday, May 13, 2002 4:55 AM > Subject: [mg34322] [mg34296] [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: [mg34322] [mg34296] [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: [mg34322] [mg34296] [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 >>>> >>>> >>>> >>> >>> >>> >>> >> >> >> >> > > > > >