Analogue of the Mertens function for the extended Riemann conjecture
It is known that the Riemann conjecture is equivalent to
$$M(x) = O(x^{frac12+epsilon}),$$
where M(x) is the Mertens function.
Does there exist an analogue to this equivalence for the extended Riemann conjecture that is, the Riemann conjecture for the Dedekind Zeta function
$$zeta_K (s) = sum_{I subseteq mathcal{O}_K} frac{1}{(N_{K/mathbf{Q}} (I))^{s}} ?$$
Edit: this is not a duplicate of the same question for the generalized Riemann hypothesis, as the extended Riemann Zeta function is not the same as the generalized Dirichlet L-function.
riemann-zeta zeta-functions
add a comment |
It is known that the Riemann conjecture is equivalent to
$$M(x) = O(x^{frac12+epsilon}),$$
where M(x) is the Mertens function.
Does there exist an analogue to this equivalence for the extended Riemann conjecture that is, the Riemann conjecture for the Dedekind Zeta function
$$zeta_K (s) = sum_{I subseteq mathcal{O}_K} frac{1}{(N_{K/mathbf{Q}} (I))^{s}} ?$$
Edit: this is not a duplicate of the same question for the generalized Riemann hypothesis, as the extended Riemann Zeta function is not the same as the generalized Dirichlet L-function.
riemann-zeta zeta-functions
2
Possible duplicate of Is there an analogue of the Mertens function for the generalised Riemann conjecture
– Winther
Nov 29 at 14:00
1
No, this is not a duplicate. The generalized Zeta function is not the same as the extended Zeta function.
– MikeTeX
Nov 29 at 14:20
add a comment |
It is known that the Riemann conjecture is equivalent to
$$M(x) = O(x^{frac12+epsilon}),$$
where M(x) is the Mertens function.
Does there exist an analogue to this equivalence for the extended Riemann conjecture that is, the Riemann conjecture for the Dedekind Zeta function
$$zeta_K (s) = sum_{I subseteq mathcal{O}_K} frac{1}{(N_{K/mathbf{Q}} (I))^{s}} ?$$
Edit: this is not a duplicate of the same question for the generalized Riemann hypothesis, as the extended Riemann Zeta function is not the same as the generalized Dirichlet L-function.
riemann-zeta zeta-functions
It is known that the Riemann conjecture is equivalent to
$$M(x) = O(x^{frac12+epsilon}),$$
where M(x) is the Mertens function.
Does there exist an analogue to this equivalence for the extended Riemann conjecture that is, the Riemann conjecture for the Dedekind Zeta function
$$zeta_K (s) = sum_{I subseteq mathcal{O}_K} frac{1}{(N_{K/mathbf{Q}} (I))^{s}} ?$$
Edit: this is not a duplicate of the same question for the generalized Riemann hypothesis, as the extended Riemann Zeta function is not the same as the generalized Dirichlet L-function.
riemann-zeta zeta-functions
riemann-zeta zeta-functions
edited Nov 29 at 14:21
asked Nov 29 at 13:59
MikeTeX
1,227412
1,227412
2
Possible duplicate of Is there an analogue of the Mertens function for the generalised Riemann conjecture
– Winther
Nov 29 at 14:00
1
No, this is not a duplicate. The generalized Zeta function is not the same as the extended Zeta function.
– MikeTeX
Nov 29 at 14:20
add a comment |
2
Possible duplicate of Is there an analogue of the Mertens function for the generalised Riemann conjecture
– Winther
Nov 29 at 14:00
1
No, this is not a duplicate. The generalized Zeta function is not the same as the extended Zeta function.
– MikeTeX
Nov 29 at 14:20
2
2
Possible duplicate of Is there an analogue of the Mertens function for the generalised Riemann conjecture
– Winther
Nov 29 at 14:00
Possible duplicate of Is there an analogue of the Mertens function for the generalised Riemann conjecture
– Winther
Nov 29 at 14:00
1
1
No, this is not a duplicate. The generalized Zeta function is not the same as the extended Zeta function.
– MikeTeX
Nov 29 at 14:20
No, this is not a duplicate. The generalized Zeta function is not the same as the extended Zeta function.
– MikeTeX
Nov 29 at 14:20
add a comment |
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2
Possible duplicate of Is there an analogue of the Mertens function for the generalised Riemann conjecture
– Winther
Nov 29 at 14:00
1
No, this is not a duplicate. The generalized Zeta function is not the same as the extended Zeta function.
– MikeTeX
Nov 29 at 14:20