Re: Re: infinite product

• To: mathgroup at smc.vnet.net
• Subject: [mg65327] Re: [mg65309] Re: infinite product
• From: Devendra Kapadia <dkapadia at wolfram.com>
• Date: Mon, 27 Mar 2006 06:55:52 -0500 (EST)
• References: <6334718.1143190534402.JavaMail.jakarta@nitrogen.mathforum.org> <240320060230425202%bruck@math.usc.edu> <20060324081024.411$gC@newsreader.com> <200603251017.FAA02277@smc.vnet.net> • Sender: owner-wri-mathgroup at wolfram.com On Sat, 25 Mar 2006, Ronald Bruck wrote: > In article <20060324081024.411$gC at newsreader.com>, David W. Cantrell
> <DWCantrell at sigmaxi.org> wrote:
>
>> Ronald Bruck <bruck at math.usc.edu> wrote:
>>> In article
>>> <6334718.1143190534402.JavaMail.jakarta at nitrogen.mathforum.org>, eugene
>>> <jane1806 at mail.ru> wrote:
>>>
>>>> Could you please help me to calculate the following product
>>>> \prod_{n=2}^{\infty} (n^2-1)/(n^2+1).
>>>>
>>>> In the case \prod (n^3-1)/(n^3+1) we can easily cancel out the
>>>> multipliers in the numerator and denomonator and it can easily be
>>>> proved that the values in this case is 2/3. But i have no ideas to deal
>>>> with our case with squares.
>>>
>>> I wouldn't expect this to be anything simple, but I plugged it into
>>> Mathematica anyway.  The result may be very interesting to fans of
>>> Mathematica:
>>>
>>> {(E^(LogGamma[2 - I] + LogGamma[2 + I])*Pi*Csch[Pi]*   Gamma[1 +
>>> SystemSeriesDumpk]^2)/
>>> (20*InternalErdelyiBernoulliB[SystemSeriesDumpk,     3 - I, 2 -
>>> I]*InternalErdelyiBernoulliB[    SystemSeriesDumpk, 3 + I, 2 + I]*
>>> Gamma[(-2 - I) + SystemSeriesDumpk]*   Gamma[(-2 + I) +
>>> SystemSeriesDumpk]),  (E^((-2*I)*SystemSeriesDumpk*Pi + LogGamma[2
>>> - I] +      LogGamma[2 + I])*Gamma[-2 - I]*Gamma[-2 + I]*   Gamma[1 +
>>> SystemSeriesDumpk]^2)/
>>> (2*InternalErdelyiBernoulliB[SystemSeriesDumpk,     3 - I, 2 -
>>> I]*InternalErdelyiBernoulliB[    SystemSeriesDumpk, 3 + I, 2 + I]*
>>> Gamma[(-2 - I) + SystemSeriesDumpk]*   Gamma[(-2 + I) +
>>> SystemSeriesDumpk]),  (E^(LogGamma[2 - I] + LogGamma[2 +
>>> I])*Pi*Csch[Pi]*   Gamma[1 + SystemSeriesDumpk]^2)/
>>> (20*InternalErdelyiBernoulliB[SystemSeriesDumpk,     3 - I, 2 -
>>> I]*InternalErdelyiBernoulliB[    SystemSeriesDumpk, 3 + I, 2 + I]*
>>> Gamma[(-2 - I) + SystemSeriesDumpk]*   Gamma[(-2 + I) +
>>> SystemSeriesDumpk])}
>>>
>>> Question is, what are "SystemSeriesDumpk" and "ErdelyiBernoulliB"?
>>> The former is clearly some sort of abort, and the latter an internal
>>> routine.
>>>
>>> The numerical value is approximately 0.272029.
>>
>> I've gotten garbage like that from Mathematica before too. But I don't in
>> this case, using version 5.1:
>>
>> In[1]:= FullSimplify[Product[(n^2 - 1)/(n^2 + 1), {n, 2, Infinity}]]
>>
>> Out[1]= Pi*Csch[Pi]
>>
>> which agrees with the answer derived earlier by Boudewijn. I wonder what
>> version of Mathematica you're using.
>
> OK, I've now tried Version 5.2 on three machines:  two Mac OS X 10.4.5,
> and one Windows XP.  All give the same result.
>
> So 5.2 is a slight "downgrade".  I'll add a followup to
> comp.soft-sys.math.mathematica.
>
>
Hello,

Thank you for reporting the problem with the above infinite product.

For this example, we first calculate the corresponding finite product
which is easy since the input is a rational function. The bad result
containing internal variables arises while taking the limit at
Infinity using Series.

We expect to fix this issue in a future release. A possible
workaround for the problem is to use Limit as shown below to
arrive at the result given in Mathematica 5.1.

=========================================

In[1]:= \$Version

Out[1]= 5.2 for Linux (June 27, 2005)

In[2]:= p = Product[(n^2 - 1)/(n^2 + 1), {n, 2, m}]

Pi Csch[Pi] Gamma[m] Gamma[2 + m]
Out[2]= -------------------------------------
Gamma[(1 - I) + m] Gamma[(1 + I) + m]

In[3]:= p1 = Limit[p, m -> Infinity]

Out[3]= Pi Csch[Pi]

In[4]:= N[p1]

Out[4]= 0.272029

In[5]:= NProduct[(n^2 - 1)/(n^2 + 1), {n, 2, Infinity}]

Out[5]= 0.272029 + 0. I

========================================

I apologize for the inconvenience caused by this problem.

Sincerely,