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RH proof ... last look

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  • Bill Bouris
    Reimann did the all of the hard work in his 1859 paper... he found the following two remarkable formulas... The Eta(t) function and the meromorphic
    Message 1 of 1 , Aug 20, 2005
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      Reimann did the all of the hard work in his 1859 paper... he found the following two remarkable formulas... The Eta(t) function and the meromorphic condition... (both can be located on the first page of an 11-page abstract written by T. Bombieri... on the Millennium Prize website).

      so...

      if F1(s) = [Eta(t)] / [(-1/2)*s] and F3(s) = [Eta(t)] / [(-1/2)(s-1)], then either...

      F1(s) <= F3(s) or F1(s) => F3(s) is true for a given number, where G(x) is Euler's gamma fctn and Z(x) is Reimann's zeta fctn.

      so we have... (1-s)*pi^(-s/2)*G(s/2)*Z(s) <= or => s*pi^(-s/2)*G(s/2)*Z(s) but not both at the same time(or does equality hold?)...

      so choose the <= case(the => case has the same argument)...

      and by meromorphic substitution... F2(s) = the [Eta(t) mero-equivalent] / [(-1/2)*s]...

      (1-s)*pi^(-s/2)*G(s/2)*Z(s) <= F2(s) <= s*pi^(-s/2)*G(s/2)*Z(s)

      (1-s)*pi^(-s/2)*G(s/2)*Z(s) <= (1-s)*pi^(-(1-s)/2)*G((1-s)/2)*Z(1-s) <= s*pi^(-s/2)*G(s/2)*Z(s)

      counterexamples can be found to refute inequality... (also in the => argument)... as follows...

      let s = 2
      F1(2) = -0.481218
      F2(2) = 175.9293
      F3(2) = 0.962435

      let s = 1/3
      F1(1/3) = 14.64149
      F2(1/3) = 4.177407
      F3(1/3) = 7.320746

      let s = 2/3
      F1(2/3) = 2.088703
      F2(2/3) = 7.320746
      F3(2/3) = 4.177407

      let s = -1
      F1(-1) = 35.18586
      F2(-1) = 0.030635
      F3(-1) = 175.9293

      so these points exhibit counter examples to both the <=,<= and the =>,=> cases...
      and finally if F1(s) = F2(s) = F3(s)... then from F1 and F3 the coefficients of s and 1-s have to be equal (the rest of their formulas are the same... thus s = (1-s) implies that s =1/2 (trivial).

      convincing enough?






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