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• ... Hmm.... And I thought I had proven that the product converged to zero. David, what is wrong with my Proof below which I had already sent to Tim? Kermit
Message 1 of 8 , May 5, 2010
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> Messages in this topic (5)
> ________________________________________________________________________
> 1b. Re: product convergence
> Date: Tue May 4, 2010 1:17 pm ((PDT))
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>> prod(2<p<x, (p-2)/p) = O(1/log(x)^2)
>>
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> So let's work out the constant, say K, in
>
> prod(2<p<x, (p-2)/p) ~ K/log(x)^2
>
> We should use the twin-prime constant
>
> C2 = prod (2<p, p*(p-2)/(p-1)^2) = 0.6601618158...
>
> and then use the square of Mertens' formula,
> remembering that the latter includes p = 2.
>
> K = C2*(exp(-Euler)*2)^2 =
> 0.832429065661945278030805943531465575045445318077417053240894...
>
> Sanity check:
>
> default(primelimit,10^8);
> \p5
> P=1.;x=10^8;forprime(p=3,x,P*=1-2/p);print(P*log(x)^2);
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> 0.83242
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> Looks OK to me...
>
> David
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>

Hmm.... And I thought I had proven that the product converged to zero.

David, what is wrong with my "Proof" below which I had already sent to Tim?

Kermit

> *****************************************************
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> Hello Tim.
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> http://en.wikipedia.org/wiki/Infinite_product
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> The infinite product 3/5 5/7 9/11 ...
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> converges to zero.
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> would converge to a positive number between 0 and 1 only if
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> the infinite product
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> 5/3 7/5 11/9 ..... converged to a positive number > 1.
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> 5/3 7/5 11/9 ..... = (1 + 2/3) (1 + 2/5) (1 + 2/9) (1 + 2/11) (1 +
> 2/15) (1 + 2/17) ...
>
>
> 1 + 2/3 + 2/5 + 2/9 + 2 / 11 + .... =< (1 + 2/3) (1 + 2/5) (1 + 2/9)
> (1 + 2/11) (1 + 2/15) (1 + 2/17) ... =< exp( 2/3 + 2/5 + 2/9 + 2/11
>
• ... Now let x tend to infinity and I think that you will see that K/log(x)^2 vanishes, not so :-? David
Message 1 of 8 , May 5, 2010
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Kermit Rose <kermit@...> wrote:

> I thought I had proven that the product converged to zero.

It does. I wrote:

> prod(2<p<x, (p-2)/p) ~ K/log(x)^2
> K = 0.832429065661945278030805943531465575045445318077417053240894...

Now let x tend to infinity and I think that you
will see that K/log(x)^2 vanishes, not so :-?

David
• ... It began OK and then fizzled out. Here is an elementary proof. Proposition: The product prod(p 2, (p-2)/p) over primes p 2 vanishes. Proof: It suffices to
Message 1 of 8 , May 8, 2010
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Kermit Rose <kermit@...> wrote:

> what is wrong with my "Proof"

It began OK and then fizzled out.
Here is an elementary proof.

Proposition: The product prod(p>2, (p-2)/p) over primes p>2 vanishes.

Proof: It suffices to show that the contrary proposition is absurd.
Suppose that prod(p>2, (p-2)/p) did not vanish.
Then, by taking logs, we would conclude that
sum(p>2, log(p) - log(p-2)) is finite. But
log(p) - log(p-2) > 2/p. Thus
sum(p>2, 1/p) would also be finite.
Yet that is easily proven to be absurd:
http://primes.utm.edu/infinity.shtml#punchline
Hence the product prod(p>2, (p-2)/p) does indeed vanish.

Comment: By the same argument, it follows that
the product prod(p>x, (p-x)/p) over primes p > x
vanishes for all real x > 0.

David
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