Equality and Convergence of Prime Products

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Discussion Overview

The discussion centers around the equality and convergence of infinite products involving prime numbers, specifically examining the conditions under which certain products converge and whether specific equalities hold. Participants explore theoretical implications, mathematical definitions, and the relationships between various products and the Riemann zeta function.

Discussion Character

  • Debate/contested
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant questions the exactness of the equality \(\Pi(a_p+b_p)= \Pi(a_p)+\Pi(b_p)\) under the assumption that both products converge.
  • Another participant argues that the conjecture is vacuous and suggests that there is no situation where all three products can converge based on standard definitions of convergent products.
  • A different participant asserts that the equality is not exact and claims it is impossible for all three infinite products to converge.
  • One participant proposes specific values for \(a_p\) and \(b_p\) and inquires whether the equality \(1-\zeta(s)=\Pi\frac{p^{-s}}{1-p^{-s}}\) is exact, suggesting a connection to the Riemann zeta function.
  • Another participant challenges the validity of the inquiry, emphasizing that if certain products converge, it implies that others cannot.
  • Several participants express skepticism about the convergence of proposed products and question the definitions and indices used in the formulations.
  • One participant mentions the conditions under which a product converges, suggesting it occurs approximately when \(s < -1\), which is outside the convergence region for the zeta function.
  • A later reply indicates that the second product in a proposed formulation never exists, as the terms either converge to zero or diverge.
  • Another participant expresses doubt about finding a product that satisfies the equation \(1+1/\zeta(s)=\prod f(p)\), noting that there isn't a product for \(\zeta(s)\) that works for all \(s\).

Areas of Agreement / Disagreement

Participants do not reach consensus on the validity of the proposed equalities or the convergence of the products. Multiple competing views remain, with ongoing debate about the conditions necessary for convergence and the implications of the conjectures presented.

Contextual Notes

Participants highlight limitations in definitions and the conditions under which products converge, as well as unresolved mathematical steps regarding the proposed equalities. The discussion reflects a complex interplay of mathematical reasoning without definitive resolutions.

eljose
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Is this equaltiy exact?:

[tex]\Pi(a_p+b_p)= \Pi(a_p)+\Pi(b_p)[/tex]

where both products a_p,b_p and (a_p+b_p) converge

another qeustion [tex]\Pi 1=1[/tex] ?

all products are made respect to all primes..
 
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your first conjecture is vacuous i think, though i haven't checked properly: there is no situation where all three products can exist for the usual definitions of convergent products, namely, prod(1+x_n) converges iff sum(x_n) converges. and besides, does it even seem reasonable? if the index set were finite it is false.

and yes for the second one.
 
The equality is not exact -- it's not in the ballpark. It's impossible for all three of those infinite products to converge.

Yes, the infinite product of the constant 1 is, in fact, 1, as can be easily seen by taking the limit of the partial products.
 
how about [tex]a_p=1 , b_p=\frac{1}{1-p^{-s}}[/tex]they both converge, if the equality is exact the sum:

[tex]a_p-b_p=\frac{p^{-s}}{1-p^{-s}}[/tex] the product would converge to

[tex]1-\zeta(s)[/tex] being R(s) riemann,s zeta function..is that true?..
 
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did you not read the posts above?
 
yes i have read the post, i have given two converging products and i would like to know if the equality [tex]1-\zeta(s)=\Pi\frac{p^{-s}}{1-p^{-s}}[/tex] is exact, because you can see that all the products

a(p)=1 b(p)=1/1-p^{-s} and a(p)+b(p) converge for certain values of s and how about?...

[tex]1+1/\zeta(s)=\Pi(2+p^{-s}[/tex]
 
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so, you'vwe been told that if prod(1+x_n) is convergent, and so is prod(1+y_n) then the prod(1+x_n+1+y_n) cannot be convergent. so why are you still asking if it is possible?

let p+n be the n'th prime.

let 1+y_n = a_{p_n}-b_{p_n} and 1+x_n= b_{p_n}. you are claiming these both converge, therefore, for whatever value of s this correpsonds to, to a_{p_n} cannot converge. but a_p_n=1 so actually (at least one of) the other two products cannot exist.
 
could you help me with these product and say if they converge and what would be their value?..

[tex]\Pi\frac{2+p^{-s}}{1+p^{-s}} \Pi2-p^{-s}[/tex]
 
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it converges iff the sum of 1/(1+p^{-s}) work out when this can happen, eg by the sandwich principle say. i think you'll find it's approximately when s<-1 (assuming s real) which as we know is out of the range of the region of convergence for the zeta function.


edit: now you've changed it. and [itex]\prod 2[/itex] doesn't make sense, really. what's the insdex? why is there a stray p^-s involved? why am i bothering to help you since you never listen to me anyway?
 
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  • #10
matt grime said:
it converges iff the sum of 1/(1+p^{-s}) work out when this can happen, eg by the sandwich principle say. i think you'll find it's approximately when s<-1 (assuming s real) which as we know is out of the range of the region of convergence for the zeta function.


edit: now you've changed it. and [itex]\prod 2[/itex] doesn't make sense, really. what's the insdex? why is there a stray p^-s involved? why am i bothering to help you since you never listen to me anyway?
Reading his code, not that it really matters anyway, it's meant to be displayed as:

[tex]\prod \left(\frac{2+p^{-s}}{1+p^{-s}}\right) \prod \left(2 - p^{-s}\right)[/tex]

Although to me it still seems quite ill-defined, but never mind.
 
  • #11
well, that never exists since the second product never exists; the terms either converge to zero or diverge to (minus) infinity.
 
  • #12
can you help me i need to find a product so:

[tex]1+1/\zeta(s)=\prod f(p)[/tex]
 
  • #13
almost certainly not since there isn't even a product for zeta(s) that works for all s.
 

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