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A real number definition involving Bruijn-Newmann constant

  1. Nov 8, 2006 #1
    A "real" number definition involving Bruijn-Newmann constant..

    Ok, ths isn't anything new, but i would like to discuss this possibility, taking into account the function:

    [tex] \xi(1/2+iz)=A(\lambda)^{-1/2}\int_{-\infty}^{\infty}dxe^{(-\lambda)^{-1} B (x-z)^{2}}H(\lambda, x) [/tex]

    then with the expression above we could study all the values of "lambda" so the Wiener-Hopf integral above involving a symmetryc tranlational Kernel has only real roots, or use it to prove that for `[tex] \lambda >0 [/tex] has always real roots so RH would be proved and Bruijn constant would be [tex] 6.10^{-9}<\Lambda <0 [/tex]:grumpy: :frown:
     
    Last edited: Nov 8, 2006
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  3. Nov 8, 2006 #2

    arildno

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    Nice try, eljose, but I'm sure this doesn't work either.
     
  4. Nov 10, 2006 #3
    It would be nice if someone could explain me the trick Newmann use to prove that [tex] \lambda >1/2 [/tex] to take a look at it...:frown: :frown: i'm not a mathematician but i believe that a real function should always have real roots :confused: am i wrong??
     
  5. Nov 10, 2006 #4

    Office_Shredder

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    x2+1=f(x) is a real function with no real roots
     
  6. Nov 10, 2006 #5
    oh..sorry you are rigth and also the functions (non polynomials) have complex roots, such us:

    [tex] exp(x)+x+1=0 [/tex] [tex] e^{x^2}+1=0 [/tex]

    amazingly the complex function [tex] exp(2i \pi x)-1=0 [/tex] has only real roots.

    the definition of a function f(x) is more than 300 years old, i don't know why mathematicians don't have some criteria to decide wether a real function has only real roots or complex appart from knowing that if f(x) and f(x*) (complex conjugate) are equal then there are pairs or complex roots changing only their imaginary part b to -b
     
  7. Nov 10, 2006 #6

    matt grime

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    We do have a simple criterion: f has real zeroes if and only if f(x)=0 implies x is in R. You're once more confusing several different issues, Jose. Testing when something satisfies some property is (always?) a hard problem except in toy examples as anyone can tell just by thinking about it for a few seconds, instead of offering yet another damning indictment of the stupidity of mathematicians.
     
  8. Nov 11, 2006 #7
    I didn't want to offend mathematician :frown: i only questioned that such a "easy" (in appearance) question was not answered and that you could find some theorems for much more difficult questions,.. that's all, in fact Newmann could prove that for [tex] \lambda > 1/2 [/tex] H(z,\lambda) had real roots the question is why this can't be applied for the other positive values of Lambda :confused:
     
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