- #1
jostpuur
- 2,116
- 19
Riemann says that the zeta function doesn't have zeros on the half plane [itex]\{z\in\mathbb{C}\;|\;\textrm{Re}(z)>1\}[/itex], because the sum
[tex]
\log(\zeta(z)) = \log\Big(\frac{1}{\underset{p\in\mathbb{P}}{\prod}\big(1 - \frac{1}{p^z}\big)}\Big) = -\sum_{p\in\mathbb{P}}\log\big(1 - \frac{1}{p^z}\big)
[/tex]
remains finite. Well, why does that sum remain finite? Doesn't look quite obvious to me.
hmhmh... what kind of set is the set
[tex]
\zeta(\{z\in\mathbb{C}\;|\;\textrm{Re}(z)>1\})\subset\mathbb{C}?
[/tex]
Does it fit in a domain of some logarithm, or does it wind around the origo?
[tex]
\log(\zeta(z)) = \log\Big(\frac{1}{\underset{p\in\mathbb{P}}{\prod}\big(1 - \frac{1}{p^z}\big)}\Big) = -\sum_{p\in\mathbb{P}}\log\big(1 - \frac{1}{p^z}\big)
[/tex]
remains finite. Well, why does that sum remain finite? Doesn't look quite obvious to me.
hmhmh... what kind of set is the set
[tex]
\zeta(\{z\in\mathbb{C}\;|\;\textrm{Re}(z)>1\})\subset\mathbb{C}?
[/tex]
Does it fit in a domain of some logarithm, or does it wind around the origo?
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