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Are there exact roots to any of the Zeta zeroes? For example the first one 1/2 +14.134725I, is there a nice looking polynomial that has an exact solution? I would assume if there is an exact value, than there would also be a conjugate. The conjugate would be 1/2 -14.134725I. I imagine it would also include negatives. I've seen polynomials with exact zeta conjugates. Here is a polynomial with zeta conjugates in it. Sorry it's so big.

$${\frac {256\,{x}^{20}-640\,{x}^{18}+560\,{x}^{16}-200\,{x}^{14}+25\,{x }^{12}-{x}^{10}+25\,{x}^{8}-200\,{x}^{6}+560\,{x}^{4}-640\,{x}^{2}+256 }{{x}^{10}}}$$

output of nth root as input for Zeta function

  1. 0.77196759e-1-40.855557*I
  2. 0.77196759e-1+40.855557*I
  3. 0.75855086e-1-4.7847003*I
  4. 0.75855086e-1+4.7847003*I
  5. 0.50739489e-1-.95183871*I
  6. 0.50739489e-1+.95183871*I
  7. 0.21253471e-1-.54700119*I
  8. 0.21253471e-1+.54700119*I
  9. -0.83307942e-1+0.40490429e-2*I
  10. -0.83307942e-1-0.40490429e-2*I
  11. -0.81509478e-1+0.34644023e-1*I
  12. -0.81509478e-1-0.34644023e-1*I
  13. -0.49293923e-1+.17966248*I
  14. -0.49293923e-1-.17966248*I
  15. -0.15358441e-1+.31869497*I
  16. -0.15358441e-1-.31869497*I
  17. 0.68554652e-1+1.8289288*I
  18. 0.68554652e-1-1.8289288*I
  19. -0.71850985e-1+0.91582220e-1*I
  20. -0.71850985e-1-0.91582220e-1*I
Thomas Olson
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  • What is your question? Are you asking if any of the roots of the Riemann Zeta function are roots of polynomials with some restricted class of coefficients? – Daniel McLaury Oct 16 '14 at 05:25
  • What are "exact zeta conjugates" and where do you claim them to be in your list of zeros to the polynomial? – Daniel R Oct 16 '14 at 07:03
  • I didn't include the list of zeros. Those are the zeta values of the zeros. I guess I didn't really need to include that. Not that it matters, but those zeros are exactly divisable by pi if you use arcsin or one of those functions. 1/20, 3/20, 5/20, 7/20, 9/20, they are some kind of trigonometric constant involving inverses. – Thomas Olson Oct 16 '14 at 07:23
  • Well, I'm clueless to what you are asking here. And the polynomial, if divided by $x^{10}$, is not a polynomial. – Daniel R Oct 16 '14 at 15:53
  • Ok then, rational functions, whatever. Do what the computer does and chuck out the denominator. Lucian answered the question, don't worry about it. – Thomas Olson Oct 16 '14 at 18:13
  • Are you honestly trying to understand the question or are you one of those guys that want a 10 page abstract with 50 pages of definitions? – Thomas Olson Oct 16 '14 at 18:50
  • Why wouldn't I honestly be trying to understand your question? – Daniel R Oct 16 '14 at 20:14

1 Answers1

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In other words, you are asking us if the non-trivial zeroes of the Riemann $\zeta$ function are algebraic. The official answer to this question is that we don't know yet, i.e., it hasn't been proven either one way or the other. However, as far as educated guesses are concerned, no mathematician expects this to be the case; more to the point, they are expected to be transcendental. Indeed, the higher the numerical precision with which we compute the root in question, the larger the polynomial degree, and the greater its coefficients.

Lucian
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  • Yeah, that's what I'm asking. The Riemann Zeta function appears to have algebraic and trigonometric qualities according to that polynomial I just posted, that's why I asked. The conjugates are roots of inverses. The negatives are slightly off though. I guess if it were completely algebraic the negatives would mirror the positives. Instead there is a small offset between them. I took a peak at the actual conjugates of the zeros of the zeta and they approach 0 on the other side of the imaginary value. I thought that was cool. – Thomas Olson Oct 16 '14 at 08:03