Verifying very high Riemann zeros. Using some newly derived formulas for the n-th Riemann zero on the critical line,
I calculated the 10^(10^6)'th zero to 1 million decimal places rather easily.
Can anyone suggest an alternative way to verify this?
I'm confident it is correct,  but it cannot be simply verified with Mathematica for instance since it cannot compute the Riemann zeta function for such a large ordinate.  
The result is here:  http://www.lepp.cornell.edu/~leclair/10106zero.pdf
 A: I'm not an expert by any means, but I think one way is to use the argument principle: numerically evaluate a contour integral of the logarithmic derivative around your zero.  The answer is going to be an integer, so if you can provably approximate it you can prove numerically that it is one.  This would require high-precision approximations to the logarithmic derivative.
[Update:]
I just realized that you are simply trying to find a single zero, not to count the zeroes in an interval.  This is done by finding sign changes. Try the discussions of Turing's work by Hejhal--Odlyzko and Booker: set $\Gamma_\mathbb{R}(s) = \pi^{-s/2}\Gamma(\frac{s}{2})$, and $\xi(s) = \Gamma_\mathbb{R}(s)\zeta(s)$.  By the functional equation, $\xi(s)$ is real on the critical line, so $\Xi(t) = \xi(\frac{1}{2}+it)$ is a real-valued function of $t$.  To show your zero is accurate you need to find a sign change (conjecturally your zero, like all other zeroes, is simple): if $\gamma$ is the height of your zero and you'd to show it is accurate to within $\epsilon$ you need to show that $\Xi(\gamma\pm\epsilon)$ have distinct signs.  Now the $\Gamma$ factor is very small, but dividing by its absolute value won't affect signs.  The Riemann-Siegel formula (see Booker's paper) allows you to accurately evaluate the zeta function at those points.
