upper bound for $L(1,chi^2)$
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Suppose $chi$ is a complex number. We know $chi bar{chi}=chi^2.$
Can anyone give me a reference for an upper bound for $L(1,chi^2)?$
I know there exists upper bounds for $L(1,chi),$ but I could not find any for $L(1,chi^2)$.
analytic-number-theory
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up vote
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Suppose $chi$ is a complex number. We know $chi bar{chi}=chi^2.$
Can anyone give me a reference for an upper bound for $L(1,chi^2)?$
I know there exists upper bounds for $L(1,chi),$ but I could not find any for $L(1,chi^2)$.
analytic-number-theory
1
For $chi$ primitive, $chi^m$ can be non-primitive or even the principal character, that's the only difference. If the order of $chi$ is coprime with $m$ then $chi^m$ is primitive.
– reuns
Nov 21 at 14:22
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up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose $chi$ is a complex number. We know $chi bar{chi}=chi^2.$
Can anyone give me a reference for an upper bound for $L(1,chi^2)?$
I know there exists upper bounds for $L(1,chi),$ but I could not find any for $L(1,chi^2)$.
analytic-number-theory
Suppose $chi$ is a complex number. We know $chi bar{chi}=chi^2.$
Can anyone give me a reference for an upper bound for $L(1,chi^2)?$
I know there exists upper bounds for $L(1,chi),$ but I could not find any for $L(1,chi^2)$.
analytic-number-theory
analytic-number-theory
asked Nov 21 at 13:40
usere5225321
612412
612412
1
For $chi$ primitive, $chi^m$ can be non-primitive or even the principal character, that's the only difference. If the order of $chi$ is coprime with $m$ then $chi^m$ is primitive.
– reuns
Nov 21 at 14:22
add a comment |
1
For $chi$ primitive, $chi^m$ can be non-primitive or even the principal character, that's the only difference. If the order of $chi$ is coprime with $m$ then $chi^m$ is primitive.
– reuns
Nov 21 at 14:22
1
1
For $chi$ primitive, $chi^m$ can be non-primitive or even the principal character, that's the only difference. If the order of $chi$ is coprime with $m$ then $chi^m$ is primitive.
– reuns
Nov 21 at 14:22
For $chi$ primitive, $chi^m$ can be non-primitive or even the principal character, that's the only difference. If the order of $chi$ is coprime with $m$ then $chi^m$ is primitive.
– reuns
Nov 21 at 14:22
add a comment |
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For $chi$ primitive, $chi^m$ can be non-primitive or even the principal character, that's the only difference. If the order of $chi$ is coprime with $m$ then $chi^m$ is primitive.
– reuns
Nov 21 at 14:22