$p$ is adherent value of $left(sum_{i=1}^{p} z_{i}^nright)_{ninmathbb{N}}$ where $z_{i}$ are complex of...











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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?










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  • You can tackle the problem by using simultaneous version of the Dirichlet's approximation theorem
    – rtybase
    Nov 22 at 23:32















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down vote

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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?










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  • You can tackle the problem by using simultaneous version of the Dirichlet's approximation theorem
    – rtybase
    Nov 22 at 23:32













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





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?










share|cite|improve this question















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






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edited Nov 23 at 0:24









rtybase

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asked Nov 20 at 13:17









曾靖國

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  • You can tackle the problem by using simultaneous version of the Dirichlet's approximation theorem
    – rtybase
    Nov 22 at 23:32


















  • You can tackle the problem by using simultaneous version of the Dirichlet's approximation theorem
    – rtybase
    Nov 22 at 23:32
















You can tackle the problem by using simultaneous version of the Dirichlet's approximation theorem
– rtybase
Nov 22 at 23:32




You can tackle the problem by using simultaneous version of the Dirichlet's approximation theorem
– rtybase
Nov 22 at 23:32










2 Answers
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accepted










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.






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    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.






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      2 Answers
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      2 Answers
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      up vote
      1
      down vote



      accepted










      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.






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted










        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.






        share|cite|improve this answer























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          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.






          share|cite|improve this answer












          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.







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          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 23 at 0:20









          rtybase

          10.1k21433




          10.1k21433






















              up vote
              0
              down vote













              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.






              share|cite|improve this answer

























                up vote
                0
                down vote













                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.






                share|cite|improve this answer























                  up vote
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                  down vote










                  up vote
                  0
                  down vote









                  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.






                  share|cite|improve this answer












                  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.







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                  share|cite|improve this answer



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                  answered Nov 22 at 21:00









                  曾靖國

                  3217




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