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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
>>>
>>>
>>>
>>
>>
>>
>>
>
>
>
>






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