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A question of roots of riemann function

  1. Aug 9, 2005 #1
    let be the quotient:

    [tex] Lim_{x->c}\frac{\zeta(1-x)}{\zeta(x)} [/tex]
    where x=c is a root of riemann function... then my question is if that limit is equal to exp(ik) with k any real constant...thanks... the limit is wehn x tends to c bieng c a root of riemann constant
     
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  3. Aug 9, 2005 #2

    shmoe

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    If c is a trivial root, the answer is clearly no.

    If c is a non-trivial root, if you use the functional equation the answer should be clear.
     
  4. Aug 9, 2005 #3

    matt grime

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    no, that can't be true, k surely cannot be *any* real constant? it can at best be *some* real constant.
     
  5. Aug 9, 2005 #4
    let,s suppose c is a trivial root i mean 0<re(c)<1 i have checked with the functional equation the limit [tex]Limx->c\frac{\zeta(1-x)}{zeta(x)} [/tex] my question is if we can express this limit as exp(ik) with k some real constant...

    i have checked the functional equation for zeta if we call [tex]\zeta(1-s)=F(s)\zeta(s) [/tex] the only thinkg i have proved is that F^*(a+ib)=F(a-ib) (with * complex conjugate) but i don,t know if it will be true that the coefficient is equal to exp(ik).
     
  6. Aug 9, 2005 #5

    shmoe

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    if 0<Re(c)<1 it is a NON-trivial root.

    [tex]\frac{\zeta(1-s)}{\zeta(s)}=\chi(1-s)[/tex], and I'm sure you know what [tex]\chi[/tex] is despite your continued use of "F".

    So it depends on [tex]\chi(1-c)[/tex]. If c is on the critical line, this has modulus 1. If c is off the critical line, this may have modulus 1 as well, or it may not.
     
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