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Riemann's Functional equation: $\zeta(-z)$=${-2*z!\over(2\pi)^{z+1}}$$sin({\pi z\over2})$$\zeta(z+1)$

This formulas expresses $\zeta(-z)$ in terms of $\zeta(z+1)$
Note: I read that the author said, suppose $z=x+iy$ is a complex number in the right half plane, meaning $x \ge 0$ then $-z$ is in the left half plane. Since $z+1$ is in the right half plane, Riemann's functional equation allows us to indirectly define $\zeta(-z)$. For example, if we take $z=2+3i$, the point $z+1=3+3i$ is located in the half plane where $x \ge 1$ but we already know how to compute $\zeta(z+1)$ using the zeta series and a computer and it will show that $\zeta(3+3i)\approx 0.94+0.008i$

My first question is how do I compute $\zeta(z+1)$ using the zeta series or a computer for fact checking?
My 2nd question is how do I compute Riemann's functional equation when $z$ is equal to a negative number because I can solve this when $z \ge 0$

For example, when $z=2$ in the Riemann functional equation we have:
$\zeta(-2)$=${-2*2!\over(2\pi)^3}$$sin({2\pi \over2})$$\zeta(3)$
Here we have $sin\pi=0$ which indicates that $\zeta(-2)=0$
What if we had $z=-2$, how do I solve this?
Can someone show me in detail when $z=-2$ because the equation would now look like:

$\zeta(2)$=${(-2)(-2!)\over(2\pi)}$$sin({-2\pi \over2})$$\zeta(-1)$

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    $\begingroup$ The functional equation is $\zeta(s) = \chi(s) \zeta(1-s)$ where $\chi(s) = \pi^{s-1} 2^s \Gamma(1-s) \sin(\pi s/2)$. This is an equality of meromorphic functions. From the zeros-poles locations of $\Gamma(1-s)$ you know those of $\chi(s)$. It is always true that $\zeta(s) = \lim_{z \to s}\chi(z) \zeta(1-z)$ and $\zeta(1-s) = \lim_{z \to s}\frac{1}{\chi(z)} \zeta(z)$ and depending if $\chi$ has no pole or zero at $s$ then you can remove the $\lim_{z \to s}$ and use direct evaluation $\zeta(s) = \chi(s) \zeta(1-s),\zeta(1-s) = \frac{1}{\chi(s)} \zeta(s)$ $\endgroup$ – reuns Jan 23 at 0:10
  • $\begingroup$ $\chi(s)$ has a zero at $s=-2$ so $\zeta(-2) = \chi(-2) \zeta(3) = 0$. And $\frac{1}{\chi(s)}$ has a pole at $s=-2$ so $\zeta(3) = \lim_{s \to -2}\zeta(1-s)=\lim_{s \to -2} \frac{\zeta(s)}{\chi(s)} $, as the pole is simple the latter $=\frac{\zeta'(-2)}{\chi'(-2)} $ whose only the denominator has a closed-form $\endgroup$ – reuns Jan 23 at 0:16
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Your problem boils down to finding $\zeta(2k)$. In fact, it can be shown that $$\zeta(2k)=(-1)^{k+1}\frac{B_{2k} 2^{2k-1}\pi^{2k}}{(2k)!}$$ Here are some proofs. No known values of the zeta function exist except for the even positive integers and negative integers.

EDIT: $\zeta(z)$ is equal to $0$ at the notorious non-trivial zeroes as well as negative even integers.

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  • $\begingroup$ What about for odd numbers? I looked at the proof section also $\endgroup$ – DDP Jan 22 at 22:42
  • $\begingroup$ @DDP no known values have been found for odd positive integers. $\endgroup$ – aleden Jan 22 at 22:43
  • $\begingroup$ okay. I was just fact checking for all real numbers(negative numbers included) in the functional equation for $z$ $\endgroup$ – DDP Jan 22 at 22:45

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