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MathGroup Archive 2005

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Re: Re: Product {for p=2 to infinity} (p^2+1)/(p^2-1)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg53997] Re: [mg53968] Re: [mg53937] Product {for p=2 to infinity} (p^2+1)/(p^2-1)
  • From: DrBob <drbob at bigfoot.com>
  • Date: Sun, 6 Feb 2005 00:45:18 -0500 (EST)
  • References: <200502040912.EAA01094@smc.vnet.net> <200502050816.DAA22265@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

Daniel,

Your Out[9] differs from the result I'm getting in version 5.1:

Clear[f]
f[n_] = Sum[Log[p^2 + 1] - Log[p^2 - 1], {p, 2, n}]
ff = Limit[f[n], n -> Infinity]

Log[2] - Log[Gamma[n]] +
   Log[Gamma[(1 - I) + n]] +
   Log[Gamma[(1 + I) + n]] -
   Log[Gamma[2 + n]] -
   LogGamma[2 - I] -
   LogGamma[2 + I]
(1/3)*(Log[8] -
    3*Log[Gamma[2 - I]] -
    3*Log[Gamma[2 + I]])

Bobby

On Sat, 5 Feb 2005 03:16:29 -0500 (EST), Daniel Lichtblau <danl at wolfram.com> wrote:

> Zak Seidov wrote:
>
>> Dear Math gurus,
>> I try to copy Math session:
>> \!\(FullSimplify[\[Product]\+\(p = 2\)\%\[Infinity]\((\(p\^2 +
>> 1\)\/\(p\^2 - \1\))\)]
>> n  Sinh[\[Pi]]\/\[Pi]\),
>> that is,
>> Product {for p=2 to infinity} (p^2+1)/(p^2-1)=>
>> sinh(pi)/pi=3.67608 -
>> is it OK,
>> or this should be 3/2?
>> Please email me your help:
>> seidovzf at yahoo.com
>>
>> Many thanks,
>> Zak
>>
>>
>
> The result you get is consistent with a numeric approximation e.g. by
> taking the product to 10^3.
>
> Alternatively you can try to verify that the sum of logs is correct by
> doing an indefinite sum. First we do the sum to infinity and recover the
> log of your result.
>
> In[5]:= Sum[Log[(p^2+1)/(p^2-1)], {p,2,Infinity}]
> Out[5]= Log[2] - LogGamma[2 - I] - LogGamma[2 + I]
>
> Fine so far. Now we do the indefinite summation, that is, sum to
> arbitrary 'n'.
> In[8]:= f[n_] = Sum[Log[p^2+1]-Log[p^2-1], {p,2,n}]
> Out[8]= Log[2] - Log[Gamma[n]] + Log[Gamma[(1 - I) + n]] +
>   Log[Gamma[(1 + I) + n]] - Log[Gamma[2 + n]] - LogGamma[2 - I] -
>   LogGamma[2 + I]
>
> We check that this agrees in the limit as n goes to infinity.
>
> In[9]:= InputForm[ff = Limit[f[n], n->Infinity]]
> Out[9]//InputForm=
> ((3*I)*Pi + Log[8] - 3*Log[-Gamma[2 - I]] - 3*Log[Gamma[2 + I]])/3
>
> In[10]:= InputForm[FullSimplify[ff]]
> Out[10]//InputForm= Log[Sinh[Pi]/Pi]
>
> Finally one can ask whether the indefinite sum is correct. To show this
> you would start with:
>
> In[31]:= InputForm[g[n_] = FullSimplify[f[n]-f[n-1],
> Assumptions->{n>2,Element[n,Integers]}]]
> Out[31]//InputForm=
> -Log[Gamma[-I + n]] - Log[Gamma[I + n]] +
>  Log[Gamma[(1 - I) + n]/(-1 + n^2)] + Log[Gamma[(1 + I) + n]]
>
> You can now try various methods to confirm that this is your summand.
> I'll leave the symbolic proof to others and just show a quick numerical
> verification.
>
> In[34]:= FullSimplify[g[5] - (Log[5^2+1] - Log[5^2-1])]
> Out[34]= 0
>
>
> Daniel Lichtblau
> Wolfram Research
>
>
>
>
>
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


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