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Prime products

  1. Jul 14, 2005 #1
    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..
     
    Last edited: Jul 14, 2005
  2. jcsd
  3. Jul 14, 2005 #2

    matt grime

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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.
     
  4. Jul 14, 2005 #3

    Hurkyl

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    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.
     
  5. Jul 15, 2005 #4
    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?..
     
    Last edited: Jul 15, 2005
  6. Jul 15, 2005 #5

    matt grime

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    did you not read the posts above?
     
  7. Jul 15, 2005 #6
    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]
     
    Last edited: Jul 15, 2005
  8. Jul 15, 2005 #7

    matt grime

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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.
     
  9. Jul 15, 2005 #8
    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]
     
    Last edited: Jul 15, 2005
  10. Jul 15, 2005 #9

    matt grime

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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?
     
    Last edited: Jul 15, 2005
  11. Jul 15, 2005 #10

    Zurtex

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    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.
     
  12. Jul 15, 2005 #11

    matt grime

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    well, that never exists since the second product never exists; the terms either converge to zero or diverge to (minus) infinity.
     
  13. Jul 15, 2005 #12
    can you help me i need to find a product so:

    [tex]1+1/\zeta(s)=\prod f(p) [/tex]
     
  14. Jul 15, 2005 #13

    matt grime

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    almost certainly not since there isn't even a product for zeta(s) that works for all s.
     
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