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MathGroup Archive 2006

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Re: another problem with Infinite Products

  • To: mathgroup at smc.vnet.net
  • Subject: [mg65551] Re: another problem with Infinite Products
  • From: Roger Bagula <rlbagulatftn at yahoo.com>
  • Date: Fri, 7 Apr 2006 06:14:27 -0400 (EDT)
  • References: <200604051055.GAA21639@smc.vnet.net> <e12stv$jbq$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

First I want to thank Dr. Lichtenbau for taking time out to look at 
these problems. He is good at resolving doubts.
The Problem of a product form for the Adamson ultrazeta (1.291286)
is unresolved still.
Daniel Lichtblau wrote:
> I should mention, in case it is not already clear, that these products 
> do not have known closed forms in terms of defined functions. Hence they 
> return unevaluated. 

Neither of these candidates seem to fill the bill,
but as Dr. Lichtenbau points out, all three of these numbers are 
probably transcendental. And they seem to be new and unknown as well.
The may well not be important.
A=Product[1/(1 - 1/Prime[n]^f[n]), {n, 1, Infinity}]
exists such that
A=Sum[1/n^n,{n,1,Infinity}]=1.291..
and it appear this new function is such that everywhere:
f[n] >= Prime[n]

The idea presents itself that maybe:
Product[1/(1 - 1/Prime[n]^PrimePi[n]), {n, 1, Infinity}]
N[%]

Mathematica completely refuses to resolve that product.

Dr. Lichenbau's method:
pnpn[n_] := 1/Prime[n]^PrimePi[n]
logppreduced[m_] := Sum[Sum[pnpn[n]^j/j, {n, 1, m}],
    {j, 1, Infinity}]
ppreduced[m_] := Exp[logppreduced[m]]
InputForm[ppapproxtable10 =
    Table[ppreduced[m], {m, 1, 10}]]
N[%]
8.796649532382020782453191915* 10^10402
which is either a new prime or is singular ( humor).

Here is one based on a Bob Hanlon Binet solver that I don't expect to 
give an answer:
RSolve[{Product[1/(1 - 1/Prime[n]^f[n]), {n, 1, Infinity}] - Sum[1/n^n, 
{n, 1, Infinity}] == 0}, f[n], n]

I make the hypothesis that:
f[n]=k*Prime[n]

Here is a method that gives a much closer answer:
h[n_Integer?Positive] = 1/(1 
-1/Prime[n]^(1.3850602852044895*Prime[n]/1.2912859970626636))
Product[h[n], {n, 1, Infinity}]
N[%]
1.3311226406531202

Some semi-emprical work gives:k-->1.1359
h[n_Integer?Positive] = 1/(1 - 1/Prime[n]^(1.1359*Prime[n]))
1.291855277548342`
Still a little large but at least we have an idea of what Product give 
that sum. I realize that this isn't as satisfying as a closed form, but 
it is one step closer to undersatanding the dynamics involved.
Sqrt[1.2912859970626636]=1.1363476567770374

h[n_Integer?Positive] = 1/(1 - 1/Prime[n]^(1.1363476567770374*Prime[n]))
Product[h[n], {n, 1, Infinity}]
N[%]
1.29159935712232

1.29159935712232 - 1.2912859970626636=0.00031336
Which seems about as close as Mathematica products are to their sums 
unless the answer is known.
> Roger Bagula wrote:
> 
>>Gary Adamson a long time ago came up with a sort of ultra-zeta sum:
>>A=Sum[1/n^n,{n,1,Infinity}]=1.291..
>>That is very close to  2- Sqrt[2]...
>>One wonders if the Product:
>>Product[1/(1-1/Prime[n]^Prime[n]),{n,1,Infinity}]-->A
>>or is it?
>>Product[1/(1-1/Prime[n]^n),{n,1,Infinity}]-->A
>>It is clear that both these products should converge
>>and I think that the second should be A on reconsideration.
>>But Mathematica disagrees:
>>
>>A = Sum[1/n^n, {n, 1, Infinity}]
>>N[%]
>>
`
>>Product[1/(1 - 1/Prime[n]^Prime[n]), {n, 1, Infinity}]
>>N[%]
>>1.3850602852044895`
>>Product[1/(1-1/Prime[n]^n),{n,1,Infinity}]
>>N[%]
>>2.2691047868959395`
> 

> Daniel Lichtblau
> Wolfram Research
> 


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