RE: Re: Re: Tough Limit
- To: mathgroup at smc.vnet.net
- Subject: [mg34301] RE: [mg34284] Re: [mg34251] Re: [mg34235] Tough Limit
- From: "DrBob" <majort at cox-internet.com>
- Date: Tue, 14 May 2002 04:10:38 -0400 (EDT)
- Reply-to: <drbob at bigfoot.com>
- Sender: owner-wri-mathgroup at wolfram.com
So long as we're beating this dead horse anyway...
Here's a more complete analysis. (Note I've used a better approximation
to n! An even MORE complete analysis would address precisely how the
approximation might affect our answer.)
gosper[n_]:=(n^n*Sqrt[1/3+2*n]*Sqrt[Pi])/E^n
a=Binomial[-1/2, n];
b=Gamma[1 + 2*n]/Gamma[1 + n]^2;
c=FullSimplify[a /b]
d=b c/.Gamma[1+n_]->n!
e=d/.k_!->gosper[k]
f=e Sqrt[n*Pi]
Limit[f,n->Infinity]
Limit[Abs[f],n->Infinity]
g=f/Cos[n*Pi];
Limit[g,n->Infinity]
The last four outputs (in InputForm) are:
(Sqrt[n]*Sqrt[1/3 + 4*n]*Cos[n*Pi])/(1/3 + 2*n)
Interval[{-1,1}]
Interval[{0,1}].
1
Note that Cos[n Pi] appears in the result of FullSimplify above. When n
isn't restricted to being integer, this clearly explains the cyclical
behavior Vladimir plotted earlier, and the Interval results above.
Bobby Treat
-----Original Message-----
From: Andrzej Kozlowski [mailto:andrzej at platon.c.u-tokyo.ac.jp]
To: mathgroup at smc.vnet.net
Subject: [mg34301] Re: [mg34284] Re: [mg34251] Re: [mg34235] Tough Limit
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: [mg34301] [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: [mg34301] [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: [mg34301] [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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