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Problem with Infinite products

  • To: mathgroup at
  • Subject: [mg65251] Problem with Infinite products
  • From: Roger Bagula <rlbagulatftn at>
  • Date: Wed, 22 Mar 2006 06:13:40 -0500 (EST)
  • Sender: owner-wri-mathgroup at

I've gotten the Sum versions of these functions to work
but had to use finite limits there:

Clear[f, zta]
f[n_, 1] := If[Mod[Prime[n], 12] - 1 == 0, Prime[n], 1/2]
f[n_, 2] := If[Mod[Prime[n], 12] - 5 == 0, Prime[n], 1/2]
f[n_, 3] := If[Mod[Prime[n], 12] - 7 == 0, Prime[n], 1/2]
f[n_, 4] := If[Mod[Prime[n], 12] - 11 == 0, Prime[n], 1/2]
zta[x_, m_] := Product[f[n, m]^(x)/(-1 + f[n, m]^(x)), {n, 1, Infinity}]
zta[2, 1]
zta[2, 2]
zta[2, 3]
zta[2, 4]
N[Product[zta[2, n], {n, 1, 4}]/Zeta[2]]

These won't even work right in finite limits.

Factoring the Zeta function using modulo 12:

  I realized that one could use the {1,5,7,11} prime endings modulo 12
  to factor the zeta function.
I solved the problem using the fact that the a0 and bo terms are:
By plotting w[n[ I found that:
With that out of the way the resulting other factor terms become:
And the Zeta function is factored  as:
This factoriozation uses the fact
that in the Infinite Limit the populations of the {1,5,7,11} states are 
with the solution:
So to find the factors one has to only find the sums
and compute the factors from them.
So we actually have five factors.
An important thing to notice is that the b(0) factor is always rational
for integer s. That is the  division
Zeta[s]/(1/(1 - 1/2^s))*(1/(1 - 1/3^s))
gives a term in Pi.->Pi2/9 for s=2
Looking at the b(i) that I got solving for Zeta[2]
one sees this:
That's as far as I got last night when I realized that the genus[n]
of the primes had to be a complex number and started working on that.
So that:
with the result that:

The functions so far don't give any different answers fron the regular 
Zeta function.
But if we change the sign of one or more terms ( 4 ways for one sign, 6 
ways for two signs)
then the result is a new set of functions which are all still founded in 
the original zeta function.
The two sign functions should be always very near zero.

Anyway I did the forms of one minus : the important thing seem to be 
that none are now infinite at s=1
like the Zeta function is. At Zeta[1] they vary from near 4 to just 
above five.
This is just a limited numerical approximation but it works:
f[n_, 1] := If[Mod[Prime[n], 12] - 1 == 0, 1/n, 0]
f[n_, 2] := If[Mod[Prime[n], 12] - 5 == 0, 1/n, 0]
f[n_, 3] := If[Mod[Prime[n], 12] - 7 == 0, 1/n, 0]
f[n_, 4] := If[Mod[Prime[n], 12] - 11 == 0, 1/n, 0]
zta[x_, m_] := Sum[f[n, m]^(x), {n, 1, 1000}]
r[x_] := 1 + 1/2^x
zeta1[x_, n_] := r[x] + Sum[zta[x, m], {m, 1, 4}] - 2*zta[x, n]

Table[N[zeta1[1, n]], {n, 1, 4}]
{5.01617, 4.06003, 4.31921, 4.57554}

In any case it would be nice to have direct product versions of the 
functions working.

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