Approximate numbers by certains rationals
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
asked Apr 13 '18 at 10:18
SaschaSascha
85318
add a comment |
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
asked Apr 13 '18 at 10:18
SaschaSascha
85318
add a comment |
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
asked Apr 13 '18 at 10:18
SaschaSascha
85318
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
asked Apr 13 '18 at 10:18
SaschaSascha
85318
asked Apr 13 '18 at 10:18
SaschaSascha
85318
edited Dec 29 '18 at 2:06
Sascha
asked Apr 13 '18 at 10:18
SaschaSascha
85318
asked Apr 13 '18 at 10:18
SaschaSascha
85318
asked Apr 13 '18 at 10:18
SaschaSascha
85318
85318
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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 $left|x_{n,k_n}-frac rqright|le frac 1{2q}le frac 1{n^{3}}$.
answered Dec 29 '18 at 7:02
Alex RavskyAlex Ravsky
39.4k32181
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2735235%2fapproximate-numbers-by-certains-rationals%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $left|x_{n,k_n}-frac rqright|le frac 1{2q}le frac 1{n^{3}}$.
answered Dec 29 '18 at 7:02
Alex RavskyAlex Ravsky
39.4k32181
add a comment |
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 $left|x_{n,k_n}-frac rqright|le frac 1{2q}le frac 1{n^{3}}$.
answered Dec 29 '18 at 7:02
Alex RavskyAlex Ravsky
39.4k32181
add a comment |
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 $left|x_{n,k_n}-frac rqright|le frac 1{2q}le frac 1{n^{3}}$.
answered Dec 29 '18 at 7:02
Alex RavskyAlex Ravsky
39.4k32181
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 $left|x_{n,k_n}-frac rqright|le frac 1{2q}le frac 1{n^{3}}$.
answered Dec 29 '18 at 7:02
Alex RavskyAlex Ravsky
39.4k32181
edited Dec 30 '18 at 13:28
answered Dec 29 '18 at 7:02
Alex RavskyAlex Ravsky
39.4k32181
answered Dec 29 '18 at 7:02
Alex RavskyAlex Ravsky
39.4k32181
answered Dec 29 '18 at 7:02
Alex RavskyAlex Ravsky
39.4k32181
39.4k32181
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2735235%2fapproximate-numbers-by-certains-rationals%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
