Mixing
$p$ is adherent value of $left(sum_{i=1}^{p} z_{i}^nright)_{ninmathbb{N}}$ where $z_{i}$ are complex of...
Here is an exercise I'm trying to solve :
Let $z_{1}, ldots,z_{p}$ be some complex numbers of modulus 1 and, for $ninmathbb{N}$, $u_n = sumlimits_{i=1}^{p} z_{i}^n$.
Show that $p$ is adherent value of $(u_n)$
My attempts : Let $v = (theta_1, ldots,theta_p)$ where $theta_i$ is the argument of $z_i$. Since $mathbb{Z}v + 2pimathbb{Z}^p$ is a subgroup of $mathbb{R}^p$, some properties of subgroups of $mathbb{R}^p$ might help to conclude?
sequences-and-series diophantine-approximation
edited Nov 23 '18 at 0:24
rtybase
10.4k21433
asked Nov 20 '18 at 13:17
曾靖國
3868
add a comment |
Here is an exercise I'm trying to solve :
Let $z_{1}, ldots,z_{p}$ be some complex numbers of modulus 1 and, for $ninmathbb{N}$, $u_n = sumlimits_{i=1}^{p} z_{i}^n$.
Show that $p$ is adherent value of $(u_n)$
My attempts : Let $v = (theta_1, ldots,theta_p)$ where $theta_i$ is the argument of $z_i$. Since $mathbb{Z}v + 2pimathbb{Z}^p$ is a subgroup of $mathbb{R}^p$, some properties of subgroups of $mathbb{R}^p$ might help to conclude?
sequences-and-series diophantine-approximation
edited Nov 23 '18 at 0:24
rtybase
10.4k21433
asked Nov 20 '18 at 13:17
曾靖國
3868
You can tackle the problem by using simultaneous version of the Dirichlet's approximation theorem
– rtybase
Nov 22 '18 at 23:32
add a comment |
Here is an exercise I'm trying to solve :
Let $z_{1}, ldots,z_{p}$ be some complex numbers of modulus 1 and, for $ninmathbb{N}$, $u_n = sumlimits_{i=1}^{p} z_{i}^n$.
Show that $p$ is adherent value of $(u_n)$
My attempts : Let $v = (theta_1, ldots,theta_p)$ where $theta_i$ is the argument of $z_i$. Since $mathbb{Z}v + 2pimathbb{Z}^p$ is a subgroup of $mathbb{R}^p$, some properties of subgroups of $mathbb{R}^p$ might help to conclude?
sequences-and-series diophantine-approximation
edited Nov 23 '18 at 0:24
rtybase
10.4k21433
asked Nov 20 '18 at 13:17
曾靖國
3868
Here is an exercise I'm trying to solve :
Let $z_{1}, ldots,z_{p}$ be some complex numbers of modulus 1 and, for $ninmathbb{N}$, $u_n = sumlimits_{i=1}^{p} z_{i}^n$.
Show that $p$ is adherent value of $(u_n)$
My attempts : Let $v = (theta_1, ldots,theta_p)$ where $theta_i$ is the argument of $z_i$. Since $mathbb{Z}v + 2pimathbb{Z}^p$ is a subgroup of $mathbb{R}^p$, some properties of subgroups of $mathbb{R}^p$ might help to conclude?
sequences-and-series diophantine-approximation
sequences-and-series diophantine-approximation
edited Nov 23 '18 at 0:24
rtybase
10.4k21433
asked Nov 20 '18 at 13:17
曾靖國
3868
edited Nov 23 '18 at 0:24
rtybase
10.4k21433
asked Nov 20 '18 at 13:17
曾靖國
3868
edited Nov 23 '18 at 0:24
rtybase
10.4k21433
edited Nov 23 '18 at 0:24
rtybase
10.4k21433
edited Nov 23 '18 at 0:24
rtybase
10.4k21433
10.4k21433
asked Nov 20 '18 at 13:17
曾靖國
3868
asked Nov 20 '18 at 13:17
曾靖國
3868
asked Nov 20 '18 at 13:17
曾靖國
3868
3868
You can tackle the problem by using simultaneous version of the Dirichlet's approximation theorem
– rtybase
Nov 22 '18 at 23:32
add a comment |
You can tackle the problem by using simultaneous version of the Dirichlet's approximation theorem
– rtybase
Nov 22 '18 at 23:32
You can tackle the problem by using simultaneous version of the Dirichlet's approximation theorem
– rtybase
Nov 22 '18 at 23:32
You can tackle the problem by using simultaneous version of the Dirichlet's approximation theorem
– rtybase
Nov 22 '18 at 23:32
add a comment |
2 Answers
2
active
oldest
votes
As per the comments, we can use simultaneous version of the Dirichlet's approximation theorem (svDAT). But before, let's suppose $z_k=e^{ialpha_k}$ (I will use $k$ as the index to avoid confusions with complex $i$) where $|z_k|=1$. From svDAT applied to $frac{alpha_k}{2pi}$ there are integers $t_{1},ldots ,t_{p}$ and $nin mathbb {Z} ,1leq nleq N$ such that
$$left|frac{alpha_{k}}{2pi}-{frac {t_{k}}{n}}right|leq {frac {1}{nN^{1/p}}} Rightarrow
left|nalpha_{k}-2pi t_{k}right|leq {frac {2pi}{N^{1/p}}}$$
For large enough $N$ we have ${frac {2pi}{N^{1/p}}} < varepsilon$ or
$$nalpha_k approx 2pi t_k Rightarrow z_k^n=e^{i nalpha_k} approx e^{i 2pi t_k}=1 tag{1}$$
or
$$sumlimits_{k=1}^{p} z_{k}^n approx p$$
To motivate $(1)$, if we have $|alpha -beta| < varepsilon$ then
$$left| e^{ialpha} -e^{ibeta}right|=
left| cos{alpha}+isin{alpha} -cos{beta}-isin{beta}right|=
left| cos{alpha}-cos{beta}+i(sin{alpha}-sin{beta})right|=\
sqrt{(cos{alpha}-cos{beta})^2+(sin{alpha}-sin{beta})^2}=\
sqrt{left(-2sin{frac{alpha-beta}{2}}sin{frac{alpha+beta}{2}}right)^2+left(2sin{frac{alpha-beta}{2}}cos{frac{alpha+beta}{2}}right)^2}=\
2left|sin{frac{alpha-beta}{2}}right| <
2left|frac{alpha-beta}{2}right| < varepsilon$$
from here.
answered Nov 23 '18 at 0:20
rtybase
10.4k21433
add a comment |
Sorry to answer my own question. Here is my solution :
Denote by $theta_{i}$ the argument of $z_i$ for all $i$. We show that for all $epsilon>0$, there exists $ninmathbb{Z}$ nonzero such that $(ntheta_{1}, ldots,ntheta_{p})in[0, epsilon]^{p}$ (mod $2pi$) :
Fix $epsilon>0$. Denote $A_{m,k} = [frac{2pi(k-1)}{m}, frac{2pi k}{m}]$ for all $m, kinmathbb{N}, 1leq kleq m$. Then $[0, 1]^psubsetcup_{(k_1, ldots,k_p)in[|1, m|]^p}prod_{i=1}^{p}A_{m, k_{i}}$.
Let $Minmathbb{N}$ such that $M>1/epsilon$ and $Ninmathbb{N}$ such that $N>M^p$. By pigeonhole principle, there must exists $l, sin[|1, N+1|]$ distinct such that $(ltheta_{1}, ldots,ltheta_{p}), (stheta_{1}, ldots,stheta_{p})inprod_{i=1}^{p}A_{m, k_{i}}$ (mod $2pi$) for some $(k_1, ldots,k_p)in[|1, M|]^p$. Then $((l-s)theta_{1}, ldots,(l-s)theta_{p})in[0, epsilon]^{p}$ (mod $2pi$). Done.
answered Nov 22 '18 at 21:00
曾靖國
3868
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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oldest
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As per the comments, we can use simultaneous version of the Dirichlet's approximation theorem (svDAT). But before, let's suppose $z_k=e^{ialpha_k}$ (I will use $k$ as the index to avoid confusions with complex $i$) where $|z_k|=1$. From svDAT applied to $frac{alpha_k}{2pi}$ there are integers $t_{1},ldots ,t_{p}$ and $nin mathbb {Z} ,1leq nleq N$ such that
$$left|frac{alpha_{k}}{2pi}-{frac {t_{k}}{n}}right|leq {frac {1}{nN^{1/p}}} Rightarrow
left|nalpha_{k}-2pi t_{k}right|leq {frac {2pi}{N^{1/p}}}$$
For large enough $N$ we have ${frac {2pi}{N^{1/p}}} < varepsilon$ or
$$nalpha_k approx 2pi t_k Rightarrow z_k^n=e^{i nalpha_k} approx e^{i 2pi t_k}=1 tag{1}$$
or
$$sumlimits_{k=1}^{p} z_{k}^n approx p$$
To motivate $(1)$, if we have $|alpha -beta| < varepsilon$ then
$$left| e^{ialpha} -e^{ibeta}right|=
left| cos{alpha}+isin{alpha} -cos{beta}-isin{beta}right|=
left| cos{alpha}-cos{beta}+i(sin{alpha}-sin{beta})right|=\
sqrt{(cos{alpha}-cos{beta})^2+(sin{alpha}-sin{beta})^2}=\
sqrt{left(-2sin{frac{alpha-beta}{2}}sin{frac{alpha+beta}{2}}right)^2+left(2sin{frac{alpha-beta}{2}}cos{frac{alpha+beta}{2}}right)^2}=\
2left|sin{frac{alpha-beta}{2}}right| <
2left|frac{alpha-beta}{2}right| < varepsilon$$
from here.
answered Nov 23 '18 at 0:20
rtybase
10.4k21433
add a comment |
As per the comments, we can use simultaneous version of the Dirichlet's approximation theorem (svDAT). But before, let's suppose $z_k=e^{ialpha_k}$ (I will use $k$ as the index to avoid confusions with complex $i$) where $|z_k|=1$. From svDAT applied to $frac{alpha_k}{2pi}$ there are integers $t_{1},ldots ,t_{p}$ and $nin mathbb {Z} ,1leq nleq N$ such that
$$left|frac{alpha_{k}}{2pi}-{frac {t_{k}}{n}}right|leq {frac {1}{nN^{1/p}}} Rightarrow
left|nalpha_{k}-2pi t_{k}right|leq {frac {2pi}{N^{1/p}}}$$
For large enough $N$ we have ${frac {2pi}{N^{1/p}}} < varepsilon$ or
$$nalpha_k approx 2pi t_k Rightarrow z_k^n=e^{i nalpha_k} approx e^{i 2pi t_k}=1 tag{1}$$
or
$$sumlimits_{k=1}^{p} z_{k}^n approx p$$
To motivate $(1)$, if we have $|alpha -beta| < varepsilon$ then
$$left| e^{ialpha} -e^{ibeta}right|=
left| cos{alpha}+isin{alpha} -cos{beta}-isin{beta}right|=
left| cos{alpha}-cos{beta}+i(sin{alpha}-sin{beta})right|=\
sqrt{(cos{alpha}-cos{beta})^2+(sin{alpha}-sin{beta})^2}=\
sqrt{left(-2sin{frac{alpha-beta}{2}}sin{frac{alpha+beta}{2}}right)^2+left(2sin{frac{alpha-beta}{2}}cos{frac{alpha+beta}{2}}right)^2}=\
2left|sin{frac{alpha-beta}{2}}right| <
2left|frac{alpha-beta}{2}right| < varepsilon$$
from here.
answered Nov 23 '18 at 0:20
rtybase
10.4k21433
add a comment |
As per the comments, we can use simultaneous version of the Dirichlet's approximation theorem (svDAT). But before, let's suppose $z_k=e^{ialpha_k}$ (I will use $k$ as the index to avoid confusions with complex $i$) where $|z_k|=1$. From svDAT applied to $frac{alpha_k}{2pi}$ there are integers $t_{1},ldots ,t_{p}$ and $nin mathbb {Z} ,1leq nleq N$ such that
$$left|frac{alpha_{k}}{2pi}-{frac {t_{k}}{n}}right|leq {frac {1}{nN^{1/p}}} Rightarrow
left|nalpha_{k}-2pi t_{k}right|leq {frac {2pi}{N^{1/p}}}$$
For large enough $N$ we have ${frac {2pi}{N^{1/p}}} < varepsilon$ or
$$nalpha_k approx 2pi t_k Rightarrow z_k^n=e^{i nalpha_k} approx e^{i 2pi t_k}=1 tag{1}$$
or
$$sumlimits_{k=1}^{p} z_{k}^n approx p$$
To motivate $(1)$, if we have $|alpha -beta| < varepsilon$ then
$$left| e^{ialpha} -e^{ibeta}right|=
left| cos{alpha}+isin{alpha} -cos{beta}-isin{beta}right|=
left| cos{alpha}-cos{beta}+i(sin{alpha}-sin{beta})right|=\
sqrt{(cos{alpha}-cos{beta})^2+(sin{alpha}-sin{beta})^2}=\
sqrt{left(-2sin{frac{alpha-beta}{2}}sin{frac{alpha+beta}{2}}right)^2+left(2sin{frac{alpha-beta}{2}}cos{frac{alpha+beta}{2}}right)^2}=\
2left|sin{frac{alpha-beta}{2}}right| <
2left|frac{alpha-beta}{2}right| < varepsilon$$
from here.
answered Nov 23 '18 at 0:20
rtybase
10.4k21433
As per the comments, we can use simultaneous version of the Dirichlet's approximation theorem (svDAT). But before, let's suppose $z_k=e^{ialpha_k}$ (I will use $k$ as the index to avoid confusions with complex $i$) where $|z_k|=1$. From svDAT applied to $frac{alpha_k}{2pi}$ there are integers $t_{1},ldots ,t_{p}$ and $nin mathbb {Z} ,1leq nleq N$ such that
$$left|frac{alpha_{k}}{2pi}-{frac {t_{k}}{n}}right|leq {frac {1}{nN^{1/p}}} Rightarrow
left|nalpha_{k}-2pi t_{k}right|leq {frac {2pi}{N^{1/p}}}$$
For large enough $N$ we have ${frac {2pi}{N^{1/p}}} < varepsilon$ or
$$nalpha_k approx 2pi t_k Rightarrow z_k^n=e^{i nalpha_k} approx e^{i 2pi t_k}=1 tag{1}$$
or
$$sumlimits_{k=1}^{p} z_{k}^n approx p$$
To motivate $(1)$, if we have $|alpha -beta| < varepsilon$ then
$$left| e^{ialpha} -e^{ibeta}right|=
left| cos{alpha}+isin{alpha} -cos{beta}-isin{beta}right|=
left| cos{alpha}-cos{beta}+i(sin{alpha}-sin{beta})right|=\
sqrt{(cos{alpha}-cos{beta})^2+(sin{alpha}-sin{beta})^2}=\
sqrt{left(-2sin{frac{alpha-beta}{2}}sin{frac{alpha+beta}{2}}right)^2+left(2sin{frac{alpha-beta}{2}}cos{frac{alpha+beta}{2}}right)^2}=\
2left|sin{frac{alpha-beta}{2}}right| <
2left|frac{alpha-beta}{2}right| < varepsilon$$
from here.
answered Nov 23 '18 at 0:20
rtybase
10.4k21433
answered Nov 23 '18 at 0:20
rtybase
10.4k21433
answered Nov 23 '18 at 0:20
rtybase
10.4k21433
answered Nov 23 '18 at 0:20
rtybase
10.4k21433
10.4k21433
add a comment |
add a comment |
Sorry to answer my own question. Here is my solution :
Denote by $theta_{i}$ the argument of $z_i$ for all $i$. We show that for all $epsilon>0$, there exists $ninmathbb{Z}$ nonzero such that $(ntheta_{1}, ldots,ntheta_{p})in[0, epsilon]^{p}$ (mod $2pi$) :
Fix $epsilon>0$. Denote $A_{m,k} = [frac{2pi(k-1)}{m}, frac{2pi k}{m}]$ for all $m, kinmathbb{N}, 1leq kleq m$. Then $[0, 1]^psubsetcup_{(k_1, ldots,k_p)in[|1, m|]^p}prod_{i=1}^{p}A_{m, k_{i}}$.
Let $Minmathbb{N}$ such that $M>1/epsilon$ and $Ninmathbb{N}$ such that $N>M^p$. By pigeonhole principle, there must exists $l, sin[|1, N+1|]$ distinct such that $(ltheta_{1}, ldots,ltheta_{p}), (stheta_{1}, ldots,stheta_{p})inprod_{i=1}^{p}A_{m, k_{i}}$ (mod $2pi$) for some $(k_1, ldots,k_p)in[|1, M|]^p$. Then $((l-s)theta_{1}, ldots,(l-s)theta_{p})in[0, epsilon]^{p}$ (mod $2pi$). Done.
answered Nov 22 '18 at 21:00
曾靖國
3868
add a comment |
Sorry to answer my own question. Here is my solution :
Denote by $theta_{i}$ the argument of $z_i$ for all $i$. We show that for all $epsilon>0$, there exists $ninmathbb{Z}$ nonzero such that $(ntheta_{1}, ldots,ntheta_{p})in[0, epsilon]^{p}$ (mod $2pi$) :
Fix $epsilon>0$. Denote $A_{m,k} = [frac{2pi(k-1)}{m}, frac{2pi k}{m}]$ for all $m, kinmathbb{N}, 1leq kleq m$. Then $[0, 1]^psubsetcup_{(k_1, ldots,k_p)in[|1, m|]^p}prod_{i=1}^{p}A_{m, k_{i}}$.
Let $Minmathbb{N}$ such that $M>1/epsilon$ and $Ninmathbb{N}$ such that $N>M^p$. By pigeonhole principle, there must exists $l, sin[|1, N+1|]$ distinct such that $(ltheta_{1}, ldots,ltheta_{p}), (stheta_{1}, ldots,stheta_{p})inprod_{i=1}^{p}A_{m, k_{i}}$ (mod $2pi$) for some $(k_1, ldots,k_p)in[|1, M|]^p$. Then $((l-s)theta_{1}, ldots,(l-s)theta_{p})in[0, epsilon]^{p}$ (mod $2pi$). Done.
answered Nov 22 '18 at 21:00
曾靖國
3868
add a comment |
Sorry to answer my own question. Here is my solution :
Denote by $theta_{i}$ the argument of $z_i$ for all $i$. We show that for all $epsilon>0$, there exists $ninmathbb{Z}$ nonzero such that $(ntheta_{1}, ldots,ntheta_{p})in[0, epsilon]^{p}$ (mod $2pi$) :
Fix $epsilon>0$. Denote $A_{m,k} = [frac{2pi(k-1)}{m}, frac{2pi k}{m}]$ for all $m, kinmathbb{N}, 1leq kleq m$. Then $[0, 1]^psubsetcup_{(k_1, ldots,k_p)in[|1, m|]^p}prod_{i=1}^{p}A_{m, k_{i}}$.
Let $Minmathbb{N}$ such that $M>1/epsilon$ and $Ninmathbb{N}$ such that $N>M^p$. By pigeonhole principle, there must exists $l, sin[|1, N+1|]$ distinct such that $(ltheta_{1}, ldots,ltheta_{p}), (stheta_{1}, ldots,stheta_{p})inprod_{i=1}^{p}A_{m, k_{i}}$ (mod $2pi$) for some $(k_1, ldots,k_p)in[|1, M|]^p$. Then $((l-s)theta_{1}, ldots,(l-s)theta_{p})in[0, epsilon]^{p}$ (mod $2pi$). Done.
answered Nov 22 '18 at 21:00
曾靖國
3868
Sorry to answer my own question. Here is my solution :
Denote by $theta_{i}$ the argument of $z_i$ for all $i$. We show that for all $epsilon>0$, there exists $ninmathbb{Z}$ nonzero such that $(ntheta_{1}, ldots,ntheta_{p})in[0, epsilon]^{p}$ (mod $2pi$) :
Fix $epsilon>0$. Denote $A_{m,k} = [frac{2pi(k-1)}{m}, frac{2pi k}{m}]$ for all $m, kinmathbb{N}, 1leq kleq m$. Then $[0, 1]^psubsetcup_{(k_1, ldots,k_p)in[|1, m|]^p}prod_{i=1}^{p}A_{m, k_{i}}$.
Let $Minmathbb{N}$ such that $M>1/epsilon$ and $Ninmathbb{N}$ such that $N>M^p$. By pigeonhole principle, there must exists $l, sin[|1, N+1|]$ distinct such that $(ltheta_{1}, ldots,ltheta_{p}), (stheta_{1}, ldots,stheta_{p})inprod_{i=1}^{p}A_{m, k_{i}}$ (mod $2pi$) for some $(k_1, ldots,k_p)in[|1, M|]^p$. Then $((l-s)theta_{1}, ldots,(l-s)theta_{p})in[0, epsilon]^{p}$ (mod $2pi$). Done.
answered Nov 22 '18 at 21:00
曾靖國
3868
answered Nov 22 '18 at 21:00
曾靖國
3868
answered Nov 22 '18 at 21:00
曾靖國
3868
answered Nov 22 '18 at 21:00
曾靖國
3868
3868
add a comment |
add a comment |
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You can tackle the problem by using simultaneous version of the Dirichlet's approximation theorem
– rtybase
Nov 22 '18 at 23:32