Is there an analogue of the Mertens function for the generalised 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 generalized Riemann conjecture ? (say a generalized Mertens function defined somehow with characters, and an asymptotic behavior)
characters riemann-hypothesis
asked Nov 18 at 14:21
MikeTeX
1,234412
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show 1 more 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 generalized Riemann conjecture ? (say a generalized Mertens function defined somehow with characters, and an asymptotic behavior)
characters riemann-hypothesis
asked Nov 18 at 14:21
MikeTeX
1,234412
1
Sure $L(s,chi)$ has no zeros for $Re(s) > 1/2$ iff $sum_{n le x} mu(n) chi(n) = O(x^{1/2+epsilon})$. For other L-functions it works quite the same way (but this is special to them, see $eta(s-1)$ whose series only converges for $Re(s) > 1$)
– reuns
Nov 18 at 15:25
do you have a source for this ?
– MikeTeX
Nov 18 at 15:30
The proof is the same as for $zeta(s)$ and is quite similar to the PNT. Search about the PNT in arithmetic progressions.
– reuns
Nov 18 at 15:31
1
math.stackexchange.com/questions/1619762/…
– Peter Humphries
Nov 22 at 8:32
Yes, this completely answers my question. Thanks.
– MikeTeX
Nov 25 at 8:39
|
show 1 more 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 generalized Riemann conjecture ? (say a generalized Mertens function defined somehow with characters, and an asymptotic behavior)
characters riemann-hypothesis
asked Nov 18 at 14:21
MikeTeX
1,234412
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 generalized Riemann conjecture ? (say a generalized Mertens function defined somehow with characters, and an asymptotic behavior)
characters riemann-hypothesis
characters riemann-hypothesis
asked Nov 18 at 14:21
MikeTeX
1,234412
asked Nov 18 at 14:21
MikeTeX
1,234412
asked Nov 18 at 14:21
MikeTeX
1,234412
asked Nov 18 at 14:21
MikeTeX
1,234412
asked Nov 18 at 14:21
MikeTeX
1,234412
1,234412
1
Sure $L(s,chi)$ has no zeros for $Re(s) > 1/2$ iff $sum_{n le x} mu(n) chi(n) = O(x^{1/2+epsilon})$. For other L-functions it works quite the same way (but this is special to them, see $eta(s-1)$ whose series only converges for $Re(s) > 1$)
– reuns
Nov 18 at 15:25
do you have a source for this ?
– MikeTeX
Nov 18 at 15:30
The proof is the same as for $zeta(s)$ and is quite similar to the PNT. Search about the PNT in arithmetic progressions.
– reuns
Nov 18 at 15:31
1
math.stackexchange.com/questions/1619762/…
– Peter Humphries
Nov 22 at 8:32
Yes, this completely answers my question. Thanks.
– MikeTeX
Nov 25 at 8:39
|
show 1 more comment
1
Sure $L(s,chi)$ has no zeros for $Re(s) > 1/2$ iff $sum_{n le x} mu(n) chi(n) = O(x^{1/2+epsilon})$. For other L-functions it works quite the same way (but this is special to them, see $eta(s-1)$ whose series only converges for $Re(s) > 1$)
– reuns
Nov 18 at 15:25
do you have a source for this ?
– MikeTeX
Nov 18 at 15:30
The proof is the same as for $zeta(s)$ and is quite similar to the PNT. Search about the PNT in arithmetic progressions.
– reuns
Nov 18 at 15:31
1
math.stackexchange.com/questions/1619762/…
– Peter Humphries
Nov 22 at 8:32
Yes, this completely answers my question. Thanks.
– MikeTeX
Nov 25 at 8:39
1
1
Sure $L(s,chi)$ has no zeros for $Re(s) > 1/2$ iff $sum_{n le x} mu(n) chi(n) = O(x^{1/2+epsilon})$. For other L-functions it works quite the same way (but this is special to them, see $eta(s-1)$ whose series only converges for $Re(s) > 1$)
– reuns
Nov 18 at 15:25
Sure $L(s,chi)$ has no zeros for $Re(s) > 1/2$ iff $sum_{n le x} mu(n) chi(n) = O(x^{1/2+epsilon})$. For other L-functions it works quite the same way (but this is special to them, see $eta(s-1)$ whose series only converges for $Re(s) > 1$)
– reuns
Nov 18 at 15:25
do you have a source for this ?
– MikeTeX
Nov 18 at 15:30
do you have a source for this ?
– MikeTeX
Nov 18 at 15:30
The proof is the same as for $zeta(s)$ and is quite similar to the PNT. Search about the PNT in arithmetic progressions.
– reuns
Nov 18 at 15:31
The proof is the same as for $zeta(s)$ and is quite similar to the PNT. Search about the PNT in arithmetic progressions.
– reuns
Nov 18 at 15:31
1
1
math.stackexchange.com/questions/1619762/…
– Peter Humphries
Nov 22 at 8:32
math.stackexchange.com/questions/1619762/…
– Peter Humphries
Nov 22 at 8:32
Yes, this completely answers my question. Thanks.
– MikeTeX
Nov 25 at 8:39
Yes, this completely answers my question. Thanks.
– MikeTeX
Nov 25 at 8:39
|
show 1 more comment
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1
Sure $L(s,chi)$ has no zeros for $Re(s) > 1/2$ iff $sum_{n le x} mu(n) chi(n) = O(x^{1/2+epsilon})$. For other L-functions it works quite the same way (but this is special to them, see $eta(s-1)$ whose series only converges for $Re(s) > 1$)
– reuns
Nov 18 at 15:25
do you have a source for this ?
– MikeTeX
Nov 18 at 15:30
The proof is the same as for $zeta(s)$ and is quite similar to the PNT. Search about the PNT in arithmetic progressions.
– reuns
Nov 18 at 15:31
1
math.stackexchange.com/questions/1619762/…
– Peter Humphries
Nov 22 at 8:32
Yes, this completely answers my question. Thanks.
– MikeTeX
Nov 25 at 8:39