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Riemann's zeta function fractal because of Voronin?

  1. Dec 12, 2012 #1
    Riemann's zeta function "fractal" because of Voronin?

    I am not sure which rubric this belongs to, but since the zeta function is involved, I am putting it here.
    I noticed a comment (but was in too much of a hurry to remember the source) that, because of the "universality" of the Riemann zeta function as expressed in the Voronin Theorem, that the zeta function was in some form "fractal". I do not understand what aspect of fractals was being invoked in that comment, since the function does not fit the classic definition of a fractal as far as I can see. Any clues? Thanks.
     
  2. jcsd
  3. Feb 2, 2015 #2
  4. Feb 4, 2015 #3
    Many thanks, Churnhouse!
     
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