1
$\begingroup$

Why do negative even numbers plugged into the Zeta function produce a zero? The Riemann Hypothesis implies that the non-trivial zeros are connected to the primes, so how does that fit with negative even numbers?

Improve this question
$\endgroup$
  • 1
    $\begingroup$ That's an analytic continuation of $\zeta$, which is not very related to its series definition on $\left]1,+\infty\right[$ so dont be surprised if negative even numbers are zeros. If you want a proof, it uses the Gamma function 's analytic extension. $\endgroup$ – Atmos Jan 14 '18 at 22:57
  • 1
    $\begingroup$ this answer is not so simple. Take a look here for a proof. $\endgroup$ – Masacroso Jan 14 '18 at 22:58
  • 1
    $\begingroup$ For your second question, the effect of a zero $\rho$ of the zeta function on the distribution of the primes is, roughly speaking, to put a term $\sim x^\rho$ into the explicit formula for the Chebyshev function (and thus an analogous term in the explicit formula for the prime counting function). The nontrivial zeros have $\Re(\rho)>0$ so they are important at large $x.$ Whereas the trivial zeros have $\Re(\rho)<0$ so they don't contribute anything at large $x.$ $\endgroup$ – spaceisdarkgreen Jan 14 '18 at 23:09
1
$\begingroup$

The functional equation of the zeta function, needed to extend anatically that function to $\;\Bbb C\setminus\{1\}\;$ and one of the most astonishingly beautiful equations in mathematics, is

$$\zeta(s)=2^s\pi^{s-1}\,\sin\frac{\pi s}2\,\Gamma(1-s)\,\zeta(1-s)$$

Well, now for $\;s=-2n\;,\;\;n\in\Bbb N\;$ , you get $\;\zeta(-2n)=0\;$ ...

Improve this answer
$\endgroup$
-2
$\begingroup$

For example, the zeta function at $-2$ is $1^2 + 2^2 + 3^2 + ...$ which is trivially equal to 0.

Similarly, the zeta function at $-4$ is $1^4 + 2^4 + 3^4 + ... $ which is also trivially equal to zero.

It is trivial to see that the same holds for the function at $-6$, $-8$, ... etc. For that reason, these are considered the trivial zeros of the Zeta function.

Improve this answer
$\endgroup$
  • $\begingroup$ Also, it is trivial to see that 1 + 2 + 3 + 4 +... (the sum of all positive integers) is equal to -1/12. $\endgroup$ – murvaja Apr 27 '20 at 20:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy