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Out of curiosity, I was taking a look at the R {pracma} package zeta() function, and specifically the example

## First zero on the critical line s = 0.5 + i t example in the documentation:

x <- seq(0, 20, len=1001)
z <- 0.5 + x*1i
fr <- Re(zeta(z))
fi <- Im(zeta(z))
fa <- abs(zeta(z))

enter image description here

the first root corresponds to

min(fa)  0.00418561897356751

which happens to be the absolute value of the zeta function at $0.5\,+\,14.14\,i$:

x[which.min(fa)] 14.14

However, $\small 0.00418561897356751$ is not too close an approximation to $\small 0$. For instance Wolfram alpha approximates the first non-trivial zero to $0.5 + 14.1347\,i$:

$$\tiny 0.5 + 14.13472514173469379045725198356247027078425711569924317568556746014996342980925676494901039317\,i$$

Questions:

[After the helpful comments received, it is clear that the roots of the Riemann zeta function are likely approximated numerically].

  1. What degree of precision is needed to claim one additional root in the trillions of roots already found in the critical line?

  2. Is there a way of using the zeta() function to look for roots of the zeta function? Are there other, perhaps more sanctioned (?) computer algorithms?

NOTE: I am aware of the note on the linked page: The zeros are accurate to an absolute precision of $\pm 2\times 10^{−102}$, but I don't know its source.

Antoni Parellada
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  • seq(0, 20, len=1001) says you look at values that are $0.02 i$ apart, so you only look at $\frac{1}{2} + 14.12 i,; \frac{1}{2} + 14.14 i,; \frac{1}{2} + 14.16 i$, and you don't come particularly close to the zero. If you feed the function a better approximation of the zero, I'm convinced it will compute a value much closer to $0$. – Daniel Fischer Jun 12 '17 at 14:17
  • @DanielFischer Thank you! When one reads that there are trillions of zeros already found along the $0.5$ line, are these points approximations, or is there a way of getting the exact values. – Antoni Parellada Jun 12 '17 at 14:21
  • Approximations. Unless one happens to know a closed form expression, that's pretty much all one can do. We can write exact expressions (e.g. integrals) for the zeros, but those can (so far at least) only be numerically evaluated [other than naming the zero $\rho_n$ and exactly evaluating the expression as $\rho_n$]. But one can approximate to every desired precision (of course higher precision requires more work). – Daniel Fischer Jun 12 '17 at 14:30
  • @DanielFischer Thanks, again. It's amazing then that trillions of values could have been calculated, isn't it? Just to follow up on your initial comment, by setting up x <- seq(0, 20, 0.0001), min(fa) 1.9941387992206e-05 and x[which.min(fa)] 14.1347 – Antoni Parellada Jun 12 '17 at 14:36
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    To prove a zero $\zeta(s_0)=0, s_0 \approx 1/2+iy$ is exactly on the critical line, it is enough to prove the real function $Z(t) = \zeta(1/2+it) e^{i \text{arg}(\Gamma(1/4+it/2)} \pi^{-it/2}$ changes of sign around $t= y$. We know $Z(t)$ is real for $t$ real thanks to the functional equation. @DanielFischer – reuns Jun 12 '17 at 19:21
  • @user1952009 Or we can calculate the number of zeros in the critical strip with imaginary part in $(t_1,t_2)$, where $t_1 < y < t_2$. If that number is $1$, it follows that the zero is on the critical line. As long as all zeros are simple, that's good. Both ways would have problems with zeros of multiplicity $> 1$, though. The real part is simple (unless RH turns out to be false). But to determine the imaginary part, we need to approximate. – Daniel Fischer Jun 12 '17 at 19:44
  • @DanielFischer Look at $Z'(t)$ if the zero is double. What I wrote provides a lower bound for the number of zeros on the critical line, what you wrote (the argument principle) an upper bound, and with the two we show the RH is true for $|\Im(s) | \le T$. – reuns Jun 12 '17 at 19:48
  • @user1952009 Please note I posted a question based on your comment above in case you happen to be interested, or have time for it. – Antoni Parellada Jun 13 '17 at 19:13

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