Mixing
Proof that $1^alpha$ is dense on the unit circle
$begingroup$
We take an irrational, real $alpha$.
I have got as far to show that
$1^{alpha}=e^{log(1^{alpha})}=e^{alpha ln(1)+ialpha arg(1)}=e^{ialpha arg(1)}=e^{ialpha 2pi k},kinmathbb{Z}$.
After that I show that $e^{ialpha 2pi k_1+2pi k_2}neq e^{ialpha 2pi k_1k_3}$ for any $k_1,k_2,k_3inmathbb{Z}$ by noting that $alpha2pi k_1k_3equiv 0modalpha$ but $alpha 2pi k_1+2pi k_2notequiv 0modalpha$ because $alpha$ is irrational.
This entails that no element of $1^{alpha}=e^{ialpha 2pi k}$ is coterminal with any other.
Then I assume you use a pigeonhole argument of some kind, but I am entirely new to this and unsure how to do this or make it rigorous. Additionally, as per instructions, I would like to solve this problem by dividing the unit circle into $ninmathbb{N}$ segments and proving that for any segment, some $1^{alpha}$ is contained in it.
Please help. Thanks :)
complex-analysis number-theory analysis complex-numbers proof-explanation
asked Jan 23 at 3:50
Tejas RaoTejas Rao
33711
$endgroup$
add a comment |
$begingroup$
We take an irrational, real $alpha$.
I have got as far to show that
$1^{alpha}=e^{log(1^{alpha})}=e^{alpha ln(1)+ialpha arg(1)}=e^{ialpha arg(1)}=e^{ialpha 2pi k},kinmathbb{Z}$.
After that I show that $e^{ialpha 2pi k_1+2pi k_2}neq e^{ialpha 2pi k_1k_3}$ for any $k_1,k_2,k_3inmathbb{Z}$ by noting that $alpha2pi k_1k_3equiv 0modalpha$ but $alpha 2pi k_1+2pi k_2notequiv 0modalpha$ because $alpha$ is irrational.
This entails that no element of $1^{alpha}=e^{ialpha 2pi k}$ is coterminal with any other.
Then I assume you use a pigeonhole argument of some kind, but I am entirely new to this and unsure how to do this or make it rigorous. Additionally, as per instructions, I would like to solve this problem by dividing the unit circle into $ninmathbb{N}$ segments and proving that for any segment, some $1^{alpha}$ is contained in it.
Please help. Thanks :)
complex-analysis number-theory analysis complex-numbers proof-explanation
asked Jan 23 at 3:50
Tejas RaoTejas Rao
33711
$endgroup$
$begingroup$
Look at this answer to a very closely related question.
$endgroup$
– Greg Martin
Jan 23 at 4:42
add a comment |
$begingroup$
We take an irrational, real $alpha$.
I have got as far to show that
$1^{alpha}=e^{log(1^{alpha})}=e^{alpha ln(1)+ialpha arg(1)}=e^{ialpha arg(1)}=e^{ialpha 2pi k},kinmathbb{Z}$.
After that I show that $e^{ialpha 2pi k_1+2pi k_2}neq e^{ialpha 2pi k_1k_3}$ for any $k_1,k_2,k_3inmathbb{Z}$ by noting that $alpha2pi k_1k_3equiv 0modalpha$ but $alpha 2pi k_1+2pi k_2notequiv 0modalpha$ because $alpha$ is irrational.
This entails that no element of $1^{alpha}=e^{ialpha 2pi k}$ is coterminal with any other.
Then I assume you use a pigeonhole argument of some kind, but I am entirely new to this and unsure how to do this or make it rigorous. Additionally, as per instructions, I would like to solve this problem by dividing the unit circle into $ninmathbb{N}$ segments and proving that for any segment, some $1^{alpha}$ is contained in it.
Please help. Thanks :)
complex-analysis number-theory analysis complex-numbers proof-explanation
asked Jan 23 at 3:50
Tejas RaoTejas Rao
33711
$endgroup$
We take an irrational, real $alpha$.
I have got as far to show that
$1^{alpha}=e^{log(1^{alpha})}=e^{alpha ln(1)+ialpha arg(1)}=e^{ialpha arg(1)}=e^{ialpha 2pi k},kinmathbb{Z}$.
After that I show that $e^{ialpha 2pi k_1+2pi k_2}neq e^{ialpha 2pi k_1k_3}$ for any $k_1,k_2,k_3inmathbb{Z}$ by noting that $alpha2pi k_1k_3equiv 0modalpha$ but $alpha 2pi k_1+2pi k_2notequiv 0modalpha$ because $alpha$ is irrational.
This entails that no element of $1^{alpha}=e^{ialpha 2pi k}$ is coterminal with any other.
Then I assume you use a pigeonhole argument of some kind, but I am entirely new to this and unsure how to do this or make it rigorous. Additionally, as per instructions, I would like to solve this problem by dividing the unit circle into $ninmathbb{N}$ segments and proving that for any segment, some $1^{alpha}$ is contained in it.
Please help. Thanks :)
complex-analysis number-theory analysis complex-numbers proof-explanation
complex-analysis number-theory analysis complex-numbers proof-explanation
asked Jan 23 at 3:50
Tejas RaoTejas Rao
33711
asked Jan 23 at 3:50
Tejas RaoTejas Rao
33711
edited Jan 23 at 4:39
Tejas Rao
asked Jan 23 at 3:50
Tejas RaoTejas Rao
33711
asked Jan 23 at 3:50
Tejas RaoTejas Rao
33711
asked Jan 23 at 3:50
Tejas RaoTejas Rao
33711
33711
$begingroup$
Look at this answer to a very closely related question.
$endgroup$
– Greg Martin
Jan 23 at 4:42
add a comment |
$begingroup$
Look at this answer to a very closely related question.
$endgroup$
– Greg Martin
Jan 23 at 4:42
$begingroup$
Look at this answer to a very closely related question.
$endgroup$
– Greg Martin
Jan 23 at 4:42
$begingroup$
Look at this answer to a very closely related question.
$endgroup$
– Greg Martin
Jan 23 at 4:42
add a comment |
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$begingroup$
Look at this answer to a very closely related question.
$endgroup$
– Greg Martin
Jan 23 at 4:42