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?
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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
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1$\begingroup$ this answer is not so simple. Take a look here for a proof. $\endgroup$ – Masacroso Jan 14 '18 at 22:58
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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
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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 $\;\zeta(-2n)=0\;$ ...
answered Jan 14 '18 at 22:57
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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.
