algebraic conjugate and sum of roots of unity
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In above lemma, why $|a'| leq 1$ still holds? I didn't see how it relates to "algebraic conjugate of a root of unity is also a root of unity", since $a$ is the sum of unity.
(definition of algebraic conjugate: http://planetmath.org/algebraicconjugates)
abstract-algebra number-theory algebraic-number-theory
asked Jan 16 '15 at 22:43
CYC
940711
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up vote
1
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In above lemma, why $|a'| leq 1$ still holds? I didn't see how it relates to "algebraic conjugate of a root of unity is also a root of unity", since $a$ is the sum of unity.
(definition of algebraic conjugate: http://planetmath.org/algebraicconjugates)
abstract-algebra number-theory algebraic-number-theory
asked Jan 16 '15 at 22:43
CYC
940711
Each $|epsilon_i|$. Their average is inside their convex hull, their convex hull is inside the unit disc.
– Pp..
Jan 16 '15 at 22:48
@Pp.. How is that related to $a'$?
– CYC
Jan 16 '15 at 22:50
1
When you conjugate $a$ you get an average of some $epsilon_i$. You know, the $epsilon_i$ just get moved around. $sigma(fracsum epsilon_in)=fracsumsigma(epsilon_i)n$
– Pp..
Jan 16 '15 at 22:57
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In above lemma, why $|a'| leq 1$ still holds? I didn't see how it relates to "algebraic conjugate of a root of unity is also a root of unity", since $a$ is the sum of unity.
(definition of algebraic conjugate: http://planetmath.org/algebraicconjugates)
abstract-algebra number-theory algebraic-number-theory
asked Jan 16 '15 at 22:43
CYC
940711
In above lemma, why $|a'| leq 1$ still holds? I didn't see how it relates to "algebraic conjugate of a root of unity is also a root of unity", since $a$ is the sum of unity.
(definition of algebraic conjugate: http://planetmath.org/algebraicconjugates)
abstract-algebra number-theory algebraic-number-theory
abstract-algebra number-theory algebraic-number-theory
asked Jan 16 '15 at 22:43
CYC
940711
asked Jan 16 '15 at 22:43
CYC
940711
edited Jan 16 '15 at 22:56
asked Jan 16 '15 at 22:43
CYC
940711
asked Jan 16 '15 at 22:43
CYC
940711
asked Jan 16 '15 at 22:43
CYC
940711
940711
Each $|epsilon_i|$. Their average is inside their convex hull, their convex hull is inside the unit disc.
– Pp..
Jan 16 '15 at 22:48
@Pp.. How is that related to $a'$?
– CYC
Jan 16 '15 at 22:50
1
When you conjugate $a$ you get an average of some $epsilon_i$. You know, the $epsilon_i$ just get moved around. $sigma(fracsum epsilon_in)=fracsumsigma(epsilon_i)n$
– Pp..
Jan 16 '15 at 22:57
add a comment |Â
Each $|epsilon_i|$. Their average is inside their convex hull, their convex hull is inside the unit disc.
– Pp..
Jan 16 '15 at 22:48
@Pp.. How is that related to $a'$?
– CYC
Jan 16 '15 at 22:50
1
When you conjugate $a$ you get an average of some $epsilon_i$. You know, the $epsilon_i$ just get moved around. $sigma(fracsum epsilon_in)=fracsumsigma(epsilon_i)n$
– Pp..
Jan 16 '15 at 22:57
Each $|epsilon_i|$. Their average is inside their convex hull, their convex hull is inside the unit disc.
– Pp..
Jan 16 '15 at 22:48
Each $|epsilon_i|$. Their average is inside their convex hull, their convex hull is inside the unit disc.
– Pp..
Jan 16 '15 at 22:48
@Pp.. How is that related to $a'$?
– CYC
Jan 16 '15 at 22:50
@Pp.. How is that related to $a'$?
– CYC
Jan 16 '15 at 22:50
1
1
When you conjugate $a$ you get an average of some $epsilon_i$. You know, the $epsilon_i$ just get moved around. $sigma(fracsum epsilon_in)=fracsumsigma(epsilon_i)n$
– Pp..
Jan 16 '15 at 22:57
When you conjugate $a$ you get an average of some $epsilon_i$. You know, the $epsilon_i$ just get moved around. $sigma(fracsum epsilon_in)=fracsumsigma(epsilon_i)n$
– Pp..
Jan 16 '15 at 22:57
add a comment |Â
2 Answers
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0
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Just use the triangle inequality.
(Each conjugate of each $varepsilon_i$ is a root of unity, and so has absolute value $1$.)
answered Jan 17 '15 at 2:23
tracing
4,337815
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0
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We have $sigma (sum epsilon_i/n) = sumsigma(epsilon_i)/n$, since each $|sigma(epsilon_i)|=1$, $|a'|=|sigma(a)|=1$.
answered Sep 9 at 11:47
CYC
940711
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Just use the triangle inequality.
(Each conjugate of each $varepsilon_i$ is a root of unity, and so has absolute value $1$.)
answered Jan 17 '15 at 2:23
tracing
4,337815
add a comment |Â
up vote
0
down vote
Just use the triangle inequality.
(Each conjugate of each $varepsilon_i$ is a root of unity, and so has absolute value $1$.)
answered Jan 17 '15 at 2:23
tracing
4,337815
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Just use the triangle inequality.
(Each conjugate of each $varepsilon_i$ is a root of unity, and so has absolute value $1$.)
answered Jan 17 '15 at 2:23
tracing
4,337815
Just use the triangle inequality.
(Each conjugate of each $varepsilon_i$ is a root of unity, and so has absolute value $1$.)
answered Jan 17 '15 at 2:23
tracing
4,337815
answered Jan 17 '15 at 2:23
tracing
4,337815
answered Jan 17 '15 at 2:23
tracing
4,337815
answered Jan 17 '15 at 2:23
tracing
4,337815
4,337815
add a comment |Â
add a comment |Â
up vote
0
down vote
accepted
We have $sigma (sum epsilon_i/n) = sumsigma(epsilon_i)/n$, since each $|sigma(epsilon_i)|=1$, $|a'|=|sigma(a)|=1$.
answered Sep 9 at 11:47
CYC
940711
add a comment |Â
up vote
0
down vote
accepted
We have $sigma (sum epsilon_i/n) = sumsigma(epsilon_i)/n$, since each $|sigma(epsilon_i)|=1$, $|a'|=|sigma(a)|=1$.
answered Sep 9 at 11:47
CYC
940711
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
We have $sigma (sum epsilon_i/n) = sumsigma(epsilon_i)/n$, since each $|sigma(epsilon_i)|=1$, $|a'|=|sigma(a)|=1$.
answered Sep 9 at 11:47
CYC
940711
We have $sigma (sum epsilon_i/n) = sumsigma(epsilon_i)/n$, since each $|sigma(epsilon_i)|=1$, $|a'|=|sigma(a)|=1$.
answered Sep 9 at 11:47
CYC
940711
answered Sep 9 at 11:47
CYC
940711
answered Sep 9 at 11:47
CYC
940711
answered Sep 9 at 11:47
CYC
940711
940711
add a comment |Â
add a comment |Â
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Each $|epsilon_i|$. Their average is inside their convex hull, their convex hull is inside the unit disc.
– Pp..
Jan 16 '15 at 22:48
@Pp.. How is that related to $a'$?
– CYC
Jan 16 '15 at 22:50
1
When you conjugate $a$ you get an average of some $epsilon_i$. You know, the $epsilon_i$ just get moved around. $sigma(fracsum epsilon_in)=fracsumsigma(epsilon_i)n$
– Pp..
Jan 16 '15 at 22:57