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MathGroup Archive 2006

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Re: Problem with Infinite products

  • To: mathgroup at smc.vnet.net
  • Subject: [mg65299] Re: Problem with Infinite products
  • From: Maxim <m.r at inbox.ru>
  • Date: Sat, 25 Mar 2006 05:17:34 -0500 (EST)
  • References: <dvrbsp$a3a$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On Wed, 22 Mar 2006 11:19:53 +0000 (UTC), Roger Bagula  
<rlbagulatftn at yahoo.com> wrote:

> 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.
> Zeta[s]=Sum[a(i),{i,0,4}]=Product[b(i),{i,0,4}]
> I solved the problem using the fact that the a0 and bo terms are:
> (1+1/2^s)=((1/(1-1/2^s))*(1/(1-1/3^s))^w[s]=a0=b0^w[s]
> w[s_]=Log[(1+1/2^s)]/Log[((1/(1-1/2^s))*(1/(1-1/3^s))]
> By plotting w[n[ I found that:
> Limit[w[s],s->Infinity]=1
> w[2]=0.55034
> With that out of the way the resulting other factor terms become:
> b(i)=a(i)^s1(s)
> And the Zeta function is factored  as:
> Zeta[s]=a(0)^(1/w[s])*Product[a(i)^si(s),{i,1,4}]
> This factoriozation uses the fact
> that in the Infinite Limit the populations of the {1,5,7,11} states are
> equal.
> (Zeta[s]/a(0))^s1(s)=Zeta[s]/b(o)
> with the solution:
> s1(s)=Log[Zeta[s]/b(o)]/Log[Zeta[s]/a(0)]
> 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:
> c(i)*(Pi/3)^(1/2)=b(i)=a(i)^s1(s)
> c(i)=Sqrt[3/Pi]*a(i)^s1(s)
> 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.
> Limit[b[0][s],s->Infinity]=1
> So that:
> Limit[Product[a(i)^si(s),{i,1,4}],s->Infinity]=1
> with the result that:
> Limit[c(i),s->Infinity]=Sqrt[3/Pi]
>
> 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.
>

This is because of the way Mathematica evaluates symbolic sums/products:

In[1]:= Sum[Sum[(-1/2)^(k^2 + 2*k), {k, Infinity}], {k, 1}]

Out[1]= -Infinity

Because the summation range is finite, the outer Sum evaluates the summand  
for each value of k; this works effectively like Block, so we get

Block[{k = 1}, Sum[(-1/2)^(k^2 + 2*k), {k, Infinity}]]

The inner sum has a 'symbolic' summation range, so first the summand is  
evaluated symbolically within the context set up by Block, and (-1/2)^(k^2  
+ 2*k) evaluates to -1/8, which is summed from k = 1 to Infinity to give  
-Infinity. This is more or less explained in the documentation, but this  
isn't the whole story:

In[2]:= Sum[Sum[(-1/2)^(k^2 + k), {k, Infinity}], {k, 1}]

Out[2]= 1/2*(-2 - (-1)^(3/4)*2^(1/4)*EllipticTheta[2, 0, -1/2])

Why does it work differently here? If we evaluate

Block[{k = 1}, Sum[Print[k]; (-1/2)^(k^2 + k), {k, Infinity}]]

we'll see that the summand is first evaluated with k renamed to K$n, and  
the outcome depends on whether Mathematica is able to find the value of  
the sum at this step or whether the evaluation proceeds further. In fact,  
the result in such cases can be said to be undefined, because we don't  
know whether Mathematica will find a closed form answer or not in each  
particular case (the inner sum can be expressed in terms of EllipticTheta  
in both cases, it's just that Mathematica -- currently -- cannot do it for  
the sum from In[1]).

A simple recommendation is to always wrap the iterator variables of Sum,  
Product, Table, FindRoot, etc. in Module:

zta[x_, m_] := Module[{n},
   Product[f[n, m]^(x)/(-1 + f[n, m]^(x)), {n, 1, Infinity}]]

Then Product[zta[2, n], {n, 1, 4}] won't evaluate to zero.

Maxim Rytin
m.r at inbox.ru


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