Approximate numbers by certains rationals












4












$begingroup$


Let $n in mathbb N$ and $k_n in left{0,..,n right}$ then we define the numbers



$$x_{n,k_n} = frac{k_n+n^2}{n^3+n^2}.$$



It is easy to see that these numbers satisfy



$$x_{n,0} = frac{1}{n+1} le x_{n,k_n} le x_{n,n} =frac{1}{n}.$$



I would like to know whether there exist three constants $C_1,C_2,C_3>0$ and an integer $i in mathbb N$ such that we can find for every $x_{n,k_n}$ a $reduced$ fraction $$frac{p_{n,k_n}}{q_{n,k_n}}$$
such that two conditions hold:



1.) The denominator can be controlled nicely:



$$ frac{C_1}{n^i} le frac{1}{q_{n,k_n}} le frac{C_2}{n^3}$$
and



2.) The approximation is sufficiently good:



$$leftvert x_{n,k_n}-frac{p_{n,k_n}}{q_{n,k_n}} rightvert le frac{C_3}{n^3}.$$



So to summarize: I am wondering whether one can approximate the $x_{n,k_n}$ by reduced fractions up to an error of order $1/n^3$ and whether those fractions can have a denominator that is always between two different powers of $1/n^k.$






real-analysis calculus number-theory analysis elementary-number-theory







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    Let $n in mathbb N$ and $k_n in left{0,..,n right}$ then we define the numbers



    $$x_{n,k_n} = frac{k_n+n^2}{n^3+n^2}.$$



    It is easy to see that these numbers satisfy



    $$x_{n,0} = frac{1}{n+1} le x_{n,k_n} le x_{n,n} =frac{1}{n}.$$



    I would like to know whether there exist three constants $C_1,C_2,C_3>0$ and an integer $i in mathbb N$ such that we can find for every $x_{n,k_n}$ a $reduced$ fraction $$frac{p_{n,k_n}}{q_{n,k_n}}$$
    such that two conditions hold:



    1.) The denominator can be controlled nicely:



    $$ frac{C_1}{n^i} le frac{1}{q_{n,k_n}} le frac{C_2}{n^3}$$
    and



    2.) The approximation is sufficiently good:



    $$leftvert x_{n,k_n}-frac{p_{n,k_n}}{q_{n,k_n}} rightvert le frac{C_3}{n^3}.$$



    So to summarize: I am wondering whether one can approximate the $x_{n,k_n}$ by reduced fractions up to an error of order $1/n^3$ and whether those fractions can have a denominator that is always between two different powers of $1/n^k.$






    real-analysis calculus number-theory analysis elementary-number-theory







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      3



      $begingroup$


      Let $n in mathbb N$ and $k_n in left{0,..,n right}$ then we define the numbers



      $$x_{n,k_n} = frac{k_n+n^2}{n^3+n^2}.$$



      It is easy to see that these numbers satisfy



      $$x_{n,0} = frac{1}{n+1} le x_{n,k_n} le x_{n,n} =frac{1}{n}.$$



      I would like to know whether there exist three constants $C_1,C_2,C_3>0$ and an integer $i in mathbb N$ such that we can find for every $x_{n,k_n}$ a $reduced$ fraction $$frac{p_{n,k_n}}{q_{n,k_n}}$$
      such that two conditions hold:



      1.) The denominator can be controlled nicely:



      $$ frac{C_1}{n^i} le frac{1}{q_{n,k_n}} le frac{C_2}{n^3}$$
      and



      2.) The approximation is sufficiently good:



      $$leftvert x_{n,k_n}-frac{p_{n,k_n}}{q_{n,k_n}} rightvert le frac{C_3}{n^3}.$$



      So to summarize: I am wondering whether one can approximate the $x_{n,k_n}$ by reduced fractions up to an error of order $1/n^3$ and whether those fractions can have a denominator that is always between two different powers of $1/n^k.$






      real-analysis calculus number-theory analysis elementary-number-theory







      share|cite|improve this question











      $endgroup$




      Let $n in mathbb N$ and $k_n in left{0,..,n right}$ then we define the numbers



      $$x_{n,k_n} = frac{k_n+n^2}{n^3+n^2}.$$



      It is easy to see that these numbers satisfy



      $$x_{n,0} = frac{1}{n+1} le x_{n,k_n} le x_{n,n} =frac{1}{n}.$$



      I would like to know whether there exist three constants $C_1,C_2,C_3>0$ and an integer $i in mathbb N$ such that we can find for every $x_{n,k_n}$ a $reduced$ fraction $$frac{p_{n,k_n}}{q_{n,k_n}}$$
      such that two conditions hold:



      1.) The denominator can be controlled nicely:



      $$ frac{C_1}{n^i} le frac{1}{q_{n,k_n}} le frac{C_2}{n^3}$$
      and



      2.) The approximation is sufficiently good:



      $$leftvert x_{n,k_n}-frac{p_{n,k_n}}{q_{n,k_n}} rightvert le frac{C_3}{n^3}.$$



      So to summarize: I am wondering whether one can approximate the $x_{n,k_n}$ by reduced fractions up to an error of order $1/n^3$ and whether those fractions can have a denominator that is always between two different powers of $1/n^k.$






      real-analysis calculus number-theory analysis elementary-number-theory




      real-analysis calculus number-theory analysis elementary-number-theory






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 29 '18 at 2:06







      Sascha

















      asked Apr 13 '18 at 10:18









      SaschaSascha

      88318




      88318






















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          +100







          $begingroup$

          I guess we can even keep $q_{n,k_n}=n^3+n^2$. Without this strict restriction let $q_{n,k_n}=q$ be the largest prime number less than $n^3$. R. C. Baker, G. Harman, and J. Pintz in a paper “The difference between consecutive primes II” proved that for all $x > x_0$ the interval $[x-x^{0.525},x]$ contains prime numbers (with enough effort, the value of $x_0$ could be determined effectively).
          Thus $q=n^3-O(n^{1.575})$. Also, if $nge 2$ then by Bertrand’s postulate, $qgefrac {n^3}2$. Since all fractions $frac rq$ with $1le r<q$ are reduced and a difference between any two consecutive of them is $frac 1q$, for each $k_n$ there exists $r$ such that $left|x_{n,k_n}-frac rqright|le frac 1{2q}le frac 1{n^{3}}$.






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            +100







            $begingroup$

            I guess we can even keep $q_{n,k_n}=n^3+n^2$. Without this strict restriction let $q_{n,k_n}=q$ be the largest prime number less than $n^3$. R. C. Baker, G. Harman, and J. Pintz in a paper “The difference between consecutive primes II” proved that for all $x > x_0$ the interval $[x-x^{0.525},x]$ contains prime numbers (with enough effort, the value of $x_0$ could be determined effectively).
            Thus $q=n^3-O(n^{1.575})$. Also, if $nge 2$ then by Bertrand’s postulate, $qgefrac {n^3}2$. Since all fractions $frac rq$ with $1le r<q$ are reduced and a difference between any two consecutive of them is $frac 1q$, for each $k_n$ there exists $r$ such that $left|x_{n,k_n}-frac rqright|le frac 1{2q}le frac 1{n^{3}}$.






            share|cite|improve this answer











            $endgroup$


















              2





              +100







              $begingroup$

              I guess we can even keep $q_{n,k_n}=n^3+n^2$. Without this strict restriction let $q_{n,k_n}=q$ be the largest prime number less than $n^3$. R. C. Baker, G. Harman, and J. Pintz in a paper “The difference between consecutive primes II” proved that for all $x > x_0$ the interval $[x-x^{0.525},x]$ contains prime numbers (with enough effort, the value of $x_0$ could be determined effectively).
              Thus $q=n^3-O(n^{1.575})$. Also, if $nge 2$ then by Bertrand’s postulate, $qgefrac {n^3}2$. Since all fractions $frac rq$ with $1le r<q$ are reduced and a difference between any two consecutive of them is $frac 1q$, for each $k_n$ there exists $r$ such that $left|x_{n,k_n}-frac rqright|le frac 1{2q}le frac 1{n^{3}}$.






              share|cite|improve this answer











              $endgroup$
















                2





                +100







                2





                +100



                2




                +100



                $begingroup$

                I guess we can even keep $q_{n,k_n}=n^3+n^2$. Without this strict restriction let $q_{n,k_n}=q$ be the largest prime number less than $n^3$. R. C. Baker, G. Harman, and J. Pintz in a paper “The difference between consecutive primes II” proved that for all $x > x_0$ the interval $[x-x^{0.525},x]$ contains prime numbers (with enough effort, the value of $x_0$ could be determined effectively).
                Thus $q=n^3-O(n^{1.575})$. Also, if $nge 2$ then by Bertrand’s postulate, $qgefrac {n^3}2$. Since all fractions $frac rq$ with $1le r<q$ are reduced and a difference between any two consecutive of them is $frac 1q$, for each $k_n$ there exists $r$ such that $left|x_{n,k_n}-frac rqright|le frac 1{2q}le frac 1{n^{3}}$.






                share|cite|improve this answer











                $endgroup$



                I guess we can even keep $q_{n,k_n}=n^3+n^2$. Without this strict restriction let $q_{n,k_n}=q$ be the largest prime number less than $n^3$. R. C. Baker, G. Harman, and J. Pintz in a paper “The difference between consecutive primes II” proved that for all $x > x_0$ the interval $[x-x^{0.525},x]$ contains prime numbers (with enough effort, the value of $x_0$ could be determined effectively).
                Thus $q=n^3-O(n^{1.575})$. Also, if $nge 2$ then by Bertrand’s postulate, $qgefrac {n^3}2$. Since all fractions $frac rq$ with $1le r<q$ are reduced and a difference between any two consecutive of them is $frac 1q$, for each $k_n$ there exists $r$ such that $left|x_{n,k_n}-frac rqright|le frac 1{2q}le frac 1{n^{3}}$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 30 '18 at 13:28

























                answered Dec 29 '18 at 7:02









                Alex RavskyAlex Ravsky

                42.6k32383




                42.6k32383






























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