6
$\begingroup$

There are lots of pages on MSE and other websites regarding few first zeros of The Riemann Zeta Function. On MSE [this] is by far the closest (or the only one I've found) explanation for finding the first zero, but again it doesn't explain how to calculate $t \approx 14.134725$ numerically in details. And, I don't think solving $\zeta(z)=\frac 1{1-2^{1-z}}\sum_{n=1}^\infty \frac {(-1)^{n-1}}{n^z}=0$ would be a plausible approach(?)

I have heard that Riemann computed the first three zeros himself and I suppose he didn't have an access a computer so what are the numerical methods/formulas/algorithms to arrive at $t \approx 14.134725$ through calculation by hands?

$\endgroup$

1 Answer 1

5
$\begingroup$

A formula Riemann derived, now called the Riemann Siegel Formula, gives a way to compute the zeta function on the critical line with extreme accuracy. He used the contour integral representation of the Zeta function that was slightly modified (more similar to the hurwitz zeta function but it doesn't really matter) and expanded the contour to give some easily computable terms, then redrew another contour to "concentrate" the value of the rest of the integral and was able to give some asymptotic remainder terms, all of which give a surprisingly accurate (and fairly simple) way to calculate the value of the zeta function on the critical strip. You can see the formula on Wikipedia (Specifically for the critical strip) and the generalized formula here Using Desmos I was able to insert the formula and I tested out the accuracy and was surprised-- the first few zeroes of the zeta function on the critical strip lie around: t=14.134725... t=21.02203... t=25.0108... And here is a picture of a graph of the formula: https://i.stack.imgur.com/3kQNz.jpg

And it's accuracy maintains for large values as well--two zeroes that lie incredibly close to each other and describe "Lehmer's Phenomenon" , both lying at around t=7005 are shown with the formula aswell: https://i.stack.imgur.com/yGxSg.jpg

If you're interested in a thourough derivation of the formula here is a pretty good explanation--there's several versions of the formula that are far more complex and have a much more complex derivation but this one is still suitable. I cant upload photos on here yet and I'm on mobile so I'm sorry if this is messy--hope this helps.

$\endgroup$
0

You must log in to answer this question.