How can I show that this isn't a counterexample to the Riemann Hypothesis?

I was joking with my friend about disproving the Riemann Hypothesis and I randomly gave him this number: $$z_? = \frac13 + 5466322356764788987534453212467843688257746873357895.6356798776664333i$$

He plugged it into Mathematica and got $\zeta(z_?)=0.000000000445763 + 0.000000011155i$. This was remarkably close to zero for a completely random choice. To investigate, we ran a small program that calculated $\min_{k\in S}\zeta(z_?+ik\epsilon)$ where $\epsilon=10^{-8}$ and $S$ was some "large" subset of $\mathbb{Z}$ (namely $[-10^{10},\ 10^{10}]\cap\mathbb{Z}$). After about 10 minutes of computations, Mathematica spit out this:

$$\zeta(z_!) = 0.\times10^{-33}+0.\times10^{-33}i \\ z_! = \frac13 + 5466322356764788987534453212467843688237746873357395. 635356798779776664433i$$

Now I know $z_!$ wouldn't be an exact value for the root, but that's damn close if it is.

I REFUSE to believe I randomly guessed a counterexample to RH. I simply REFUSE. How can I prove that this isn't a counterexample? Can I use the asymptotics of the zeta function?

Edit: Something that made me suspicious is that the computation only took 10 minutes. I don't know what the implementation of the zeta fuction is in Mathematica, but doing 10 billion computations of the zeta function for an argument that large with high enough accuracy seems like it should take more than 10 minutes, which made me think this is purely a floating-point error, or a rounding error, or something non-analytic.

• +1 for critical thinking. This could so easily have been an "I have disproved RH, where should I publish?" question. – Arthur Nov 30 '16 at 8:15
• $10$ billion computations in $10$ minutes yields a rate of $16.66$MHz. This doesn't sound like it should be that much of a problem if the guys at Wolfram put a reasonable server at the back end (assuming that you've used Mathematica online). – barak manos Nov 30 '16 at 8:29
• My Mathematica (v.9) documentation tells that evaluating the imaginary parts requires extra precision. Without adjusting the value of $MaxExtraPrecision the calculation of$\zeta(1/3+10^ki)$begins to lose accuracy at$k=12$, and it simply refuses to go up to$k\approx50$what redoing this would amount to. I am running a small test with$MaxExtraPrecision=200 to see what (if anything) it spews out with $k=13,14,\ldots$. IOW, I suspect that the algorithm in use just doesn't work well (or at all) with large imaginary parts. – Jyrki Lahtonen Nov 30 '16 at 8:54
• More precisely: Mathematica gives $\zeta(1/3+10^{10}i)$ with what it thinks is 20 digit precision almost instantly, but only gives 7 (resp. 6) digit precision for $\zeta(1/3+10^ki)$ with $k=11$ (resp. $k=12$). After that it slows down to a crawl. If I ask for $\zeta(1/3+10^{20}i)$, then it instantly gives an error message infinite expression in a way that looks like it attempted to divide by zero. – Jyrki Lahtonen Nov 30 '16 at 9:03
• You could prove RH. – user60589 Nov 30 '16 at 13:48

Most of computing algorithms depend on series expansions which could be represented by conditional loops statements and bit shifting operations. You can write your own program to calculate Riemann-Zeta function inside the critical strip depending on Dirichlet-Eta function. \begin{align} & \eta(s)=\left(1-2^{1-s}\right)\zeta(s)=-\sum_{n=1}^{\infty}\frac{(-1)^n}{n^s} \\[4mm] & s=a-i\,b \Rightarrow \frac{1}{n^s}=\frac{n^{i\,b}}{n^a}=\frac{e^{i\,b\log n}}{e^{a\log n}}=\frac{\cos(b\log n)+i\,\sin(b\log n)}{\exp(a\log n)} \Rightarrow \\[4mm] & \color{blue}{-\eta(a-i\,b)=\sum_{n=1}^{\infty}(-1)^n\frac{\cos(b\log n)}{\exp(a\log n)}+i\,\sum_{n=1}^{\infty}(-1)^n\frac{\sin(b\log n)}{\exp(a\log n)}} \\[4mm] & \text{Where}\colon \\ & \qquad \log n=-\log(1/n)=-\log\left(1-(1-1/n)\right)=\sum_{m=1}^{\infty}\frac{(1-1/n)^m}{m}=\sum_{m=1}^{\infty}\frac{(n-1)^m}{m\,n^m} \\ & \qquad \exp(-a\log n)=\sum_{m=0}^{\infty}(-1)^m\frac{(a\log n)^m}{m!} \\ & \qquad \cos(b\log n)=\sum_{m=0}^{\infty}(-1)^m\frac{(b\log n)^{2m}}{(2m)!} \\ & \qquad \sin(b\log n)=\sum_{m=0}^{\infty}(-1)^m\frac{(b\log n)^{2m+1}}{(2m+1)!} \\ & \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \\ \end{align} Separate functions could be written to calculate the functions {Log(x), Exp(x), Cos(x), Sin(x)} using conditional loops that break at a desired error tolerance.
As well as, if you wish to reduce the floating point error and increase the accuracy, you could re-define the basic operations {Add($x + y$), Sub($x - y$), Mul($x \times y$), Div($x \div y$)} with new functions of your own, using the bit shifting operations over concatenated variables. This will give you the ability to do calculation with much more than 64 bit.
• I downvoted because : no you can't use $\zeta(s) = \frac{1}{1-2^{1-s}}\sum_{n=1}^\infty (-1)^{n+1} n^{-s}$ for computing $\zeta(1/3+i 10^{30})$ in a reasonable time. Take a look instead at en.wikipedia.org/wiki/… . And proving that $\eta(s_0) \ne 0$ doesn't mean there isn't a zero very close to $s_0$, see instead the en.wikipedia.org/wiki/Argument_principle – reuns Dec 3 '16 at 17:26
• I did not say there are no zeros around $s_0$. I share with you my suspicions about the long execution time for large values with/without using series acceleration techniques/algorithms. Nevertheless, I do appreciate your time reviewing the answer and commenting. Thank you. – Hazem Orabi Dec 3 '16 at 22:17