Quantifying how crowded points are in $d$ dimensional cube
Let $x_1, cdots, x_n$ be distinct points in the $d$ dimensional cube $[-1,1]^d$. What is a lower bound on the quantity:
$$ sum_{1 le j < k le n} frac{1}{||x_k-x_j||}$$
where $|| cdot ||$ is the Euclidean norm.
An asymptotic answer for large $n$ is also fine and bounds on any quantity that looks similar, such as replacing the norm with norm squared, is also welcomed.
My motivation is a problem that appeared in the book The Cauchy-Schwarz Master Class which proved the following:
If $-1 le x_1 < x_2 < cdots < x_n le 1$ then $$ sum_{1 le j < k le n} frac{1}{x_k-x_j} ge frac{n^2 log n}8.$$
geometry inequality upper-lower-bounds
add a comment |
Let $x_1, cdots, x_n$ be distinct points in the $d$ dimensional cube $[-1,1]^d$. What is a lower bound on the quantity:
$$ sum_{1 le j < k le n} frac{1}{||x_k-x_j||}$$
where $|| cdot ||$ is the Euclidean norm.
An asymptotic answer for large $n$ is also fine and bounds on any quantity that looks similar, such as replacing the norm with norm squared, is also welcomed.
My motivation is a problem that appeared in the book The Cauchy-Schwarz Master Class which proved the following:
If $-1 le x_1 < x_2 < cdots < x_n le 1$ then $$ sum_{1 le j < k le n} frac{1}{x_k-x_j} ge frac{n^2 log n}8.$$
geometry inequality upper-lower-bounds
As it stands your question is a bit too broad, because it is simultaneously asking for the best bound, an asymptotic bound, and "bounds on any quantity that looks similar". Could you please edit your question to make it more focused?
– Lord_Farin
Dec 24 '18 at 9:13
Ok, I edited a little. Since I dont know the difficulty of this question its bound to be a little vague but the whole point is to simulate some discussion and get any useful bounds.
– Sandeep Silwal
Dec 24 '18 at 16:25
These kind of "simple" questions very quickly get out of hand...
– Fabian
Jan 1 at 4:35
add a comment |
Let $x_1, cdots, x_n$ be distinct points in the $d$ dimensional cube $[-1,1]^d$. What is a lower bound on the quantity:
$$ sum_{1 le j < k le n} frac{1}{||x_k-x_j||}$$
where $|| cdot ||$ is the Euclidean norm.
An asymptotic answer for large $n$ is also fine and bounds on any quantity that looks similar, such as replacing the norm with norm squared, is also welcomed.
My motivation is a problem that appeared in the book The Cauchy-Schwarz Master Class which proved the following:
If $-1 le x_1 < x_2 < cdots < x_n le 1$ then $$ sum_{1 le j < k le n} frac{1}{x_k-x_j} ge frac{n^2 log n}8.$$
geometry inequality upper-lower-bounds
Let $x_1, cdots, x_n$ be distinct points in the $d$ dimensional cube $[-1,1]^d$. What is a lower bound on the quantity:
$$ sum_{1 le j < k le n} frac{1}{||x_k-x_j||}$$
where $|| cdot ||$ is the Euclidean norm.
An asymptotic answer for large $n$ is also fine and bounds on any quantity that looks similar, such as replacing the norm with norm squared, is also welcomed.
My motivation is a problem that appeared in the book The Cauchy-Schwarz Master Class which proved the following:
If $-1 le x_1 < x_2 < cdots < x_n le 1$ then $$ sum_{1 le j < k le n} frac{1}{x_k-x_j} ge frac{n^2 log n}8.$$
geometry inequality upper-lower-bounds
geometry inequality upper-lower-bounds
edited Dec 24 '18 at 16:25
asked Dec 24 '18 at 1:47
Sandeep Silwal
5,83811236
5,83811236
As it stands your question is a bit too broad, because it is simultaneously asking for the best bound, an asymptotic bound, and "bounds on any quantity that looks similar". Could you please edit your question to make it more focused?
– Lord_Farin
Dec 24 '18 at 9:13
Ok, I edited a little. Since I dont know the difficulty of this question its bound to be a little vague but the whole point is to simulate some discussion and get any useful bounds.
– Sandeep Silwal
Dec 24 '18 at 16:25
These kind of "simple" questions very quickly get out of hand...
– Fabian
Jan 1 at 4:35
add a comment |
As it stands your question is a bit too broad, because it is simultaneously asking for the best bound, an asymptotic bound, and "bounds on any quantity that looks similar". Could you please edit your question to make it more focused?
– Lord_Farin
Dec 24 '18 at 9:13
Ok, I edited a little. Since I dont know the difficulty of this question its bound to be a little vague but the whole point is to simulate some discussion and get any useful bounds.
– Sandeep Silwal
Dec 24 '18 at 16:25
These kind of "simple" questions very quickly get out of hand...
– Fabian
Jan 1 at 4:35
As it stands your question is a bit too broad, because it is simultaneously asking for the best bound, an asymptotic bound, and "bounds on any quantity that looks similar". Could you please edit your question to make it more focused?
– Lord_Farin
Dec 24 '18 at 9:13
As it stands your question is a bit too broad, because it is simultaneously asking for the best bound, an asymptotic bound, and "bounds on any quantity that looks similar". Could you please edit your question to make it more focused?
– Lord_Farin
Dec 24 '18 at 9:13
Ok, I edited a little. Since I dont know the difficulty of this question its bound to be a little vague but the whole point is to simulate some discussion and get any useful bounds.
– Sandeep Silwal
Dec 24 '18 at 16:25
Ok, I edited a little. Since I dont know the difficulty of this question its bound to be a little vague but the whole point is to simulate some discussion and get any useful bounds.
– Sandeep Silwal
Dec 24 '18 at 16:25
These kind of "simple" questions very quickly get out of hand...
– Fabian
Jan 1 at 4:35
These kind of "simple" questions very quickly get out of hand...
– Fabian
Jan 1 at 4:35
add a comment |
2 Answers
2
active
oldest
votes
This is an answer only up to $n = exp(ad)$ for some $a in theta(1)$, but may be enough to get started: There are as many as $n = exp(ad)$ points $y_1,ldots, y_n$ in ${-1,1}^d$ that satisfy the following:
$||y_i-y_j||_1 in theta(d)$ for each $i not = j$, where $|| cdot ||_1$ denotes the Manhattan metric.
[Google error-correcting codes]
Then for each $i not = j$, the following holds:
$$frac{1}{||y_i-y_j||_2} leq frac{1}{csqrt{d}}$$
for some $c in Omega(1)$, which implies the following:
Inequality 1: $$sum_{1 le i < j le n} frac{1}{||y_i-y_j||_2} leq frac{n^2}{2csqrt{d}}$$
However. not that for any distinct $x,y$, the following holds:
Inequality 2: $$frac{1}{||x-y||_2} ge frac{1}{sqrt{d}} $$
So Inequalities 1 and 2 together imply that for $y_1,ldots, y_n$ as above,
$sum_{1 le i < j le n} frac{1}{||y_i-y_j||_2}$ is no more than a constant multiple larger than $sum_{1 le i < j le n} frac{1}{||x_i-x_j||_2}$ for any other $x_1,ldots, x_n in [-1,1]^d$.
add a comment |
The situation seems to be less interesting asymptotically for fixed $d geq 2$. We have
$$frac{1}{||x-y||_2} geq frac{1}{2sqrt{d}}$$ for any $x, y in [-1, 1]^d$, so there's an easy lower bound of
$$sum_{1 leq j < k leq n} frac{1}{||x_k - x_j||} geq frac{1}{2sqrt{d}} binom{n}{2} = Omega(n^2).$$ This lower bound is actually achieved (up to a constant) by letting the $x_j$ lie on a grid, say $Lambda^+ = {n^{-1/d}, 2n^{-1/d}, dots, 1}^d$.
First note that if we set
$$Lambda = {-1, dots, -2n^{-1/d}, -n^{-1/d}, 0, n^{-1/d}, 2n^{-1/d}, dots, 1}^d$$
we have
$$sum_{j neq k} frac{1}{||x_k - x_j||} < sum_{v in Lambda, v neq 0} frac{1}{||v||}$$
for each $k$.
Also, since $d geq 2$, the integral
$$int_{[-1, 1]^d} frac{dx}{||x||}$$
actually converges, say to some $I_d < infty$, meaning in particular that
$$frac{1}{n} sum_{v in Lambda, v neq 0} frac{1}{||v||} to I_d$$
(this is just a standard grid approximation).
Putting these together, we get
$$sum_{1 leq j < k leq n} frac{1}{||x_k - x_j||} = frac{1}{2} sum_k sum_{j neq k} frac{1}{||x_k - x_j||} < frac{n}{2} sum_{v in Lambda, v neq 0} frac{1}{||v||} sim frac{n^2}{2} I_d$$
so our sum is $O(n^2)$, matching the lower bound.
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%2f3050869%2fquantifying-how-crowded-points-are-in-d-dimensional-cube%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is an answer only up to $n = exp(ad)$ for some $a in theta(1)$, but may be enough to get started: There are as many as $n = exp(ad)$ points $y_1,ldots, y_n$ in ${-1,1}^d$ that satisfy the following:
$||y_i-y_j||_1 in theta(d)$ for each $i not = j$, where $|| cdot ||_1$ denotes the Manhattan metric.
[Google error-correcting codes]
Then for each $i not = j$, the following holds:
$$frac{1}{||y_i-y_j||_2} leq frac{1}{csqrt{d}}$$
for some $c in Omega(1)$, which implies the following:
Inequality 1: $$sum_{1 le i < j le n} frac{1}{||y_i-y_j||_2} leq frac{n^2}{2csqrt{d}}$$
However. not that for any distinct $x,y$, the following holds:
Inequality 2: $$frac{1}{||x-y||_2} ge frac{1}{sqrt{d}} $$
So Inequalities 1 and 2 together imply that for $y_1,ldots, y_n$ as above,
$sum_{1 le i < j le n} frac{1}{||y_i-y_j||_2}$ is no more than a constant multiple larger than $sum_{1 le i < j le n} frac{1}{||x_i-x_j||_2}$ for any other $x_1,ldots, x_n in [-1,1]^d$.
add a comment |
This is an answer only up to $n = exp(ad)$ for some $a in theta(1)$, but may be enough to get started: There are as many as $n = exp(ad)$ points $y_1,ldots, y_n$ in ${-1,1}^d$ that satisfy the following:
$||y_i-y_j||_1 in theta(d)$ for each $i not = j$, where $|| cdot ||_1$ denotes the Manhattan metric.
[Google error-correcting codes]
Then for each $i not = j$, the following holds:
$$frac{1}{||y_i-y_j||_2} leq frac{1}{csqrt{d}}$$
for some $c in Omega(1)$, which implies the following:
Inequality 1: $$sum_{1 le i < j le n} frac{1}{||y_i-y_j||_2} leq frac{n^2}{2csqrt{d}}$$
However. not that for any distinct $x,y$, the following holds:
Inequality 2: $$frac{1}{||x-y||_2} ge frac{1}{sqrt{d}} $$
So Inequalities 1 and 2 together imply that for $y_1,ldots, y_n$ as above,
$sum_{1 le i < j le n} frac{1}{||y_i-y_j||_2}$ is no more than a constant multiple larger than $sum_{1 le i < j le n} frac{1}{||x_i-x_j||_2}$ for any other $x_1,ldots, x_n in [-1,1]^d$.
add a comment |
This is an answer only up to $n = exp(ad)$ for some $a in theta(1)$, but may be enough to get started: There are as many as $n = exp(ad)$ points $y_1,ldots, y_n$ in ${-1,1}^d$ that satisfy the following:
$||y_i-y_j||_1 in theta(d)$ for each $i not = j$, where $|| cdot ||_1$ denotes the Manhattan metric.
[Google error-correcting codes]
Then for each $i not = j$, the following holds:
$$frac{1}{||y_i-y_j||_2} leq frac{1}{csqrt{d}}$$
for some $c in Omega(1)$, which implies the following:
Inequality 1: $$sum_{1 le i < j le n} frac{1}{||y_i-y_j||_2} leq frac{n^2}{2csqrt{d}}$$
However. not that for any distinct $x,y$, the following holds:
Inequality 2: $$frac{1}{||x-y||_2} ge frac{1}{sqrt{d}} $$
So Inequalities 1 and 2 together imply that for $y_1,ldots, y_n$ as above,
$sum_{1 le i < j le n} frac{1}{||y_i-y_j||_2}$ is no more than a constant multiple larger than $sum_{1 le i < j le n} frac{1}{||x_i-x_j||_2}$ for any other $x_1,ldots, x_n in [-1,1]^d$.
This is an answer only up to $n = exp(ad)$ for some $a in theta(1)$, but may be enough to get started: There are as many as $n = exp(ad)$ points $y_1,ldots, y_n$ in ${-1,1}^d$ that satisfy the following:
$||y_i-y_j||_1 in theta(d)$ for each $i not = j$, where $|| cdot ||_1$ denotes the Manhattan metric.
[Google error-correcting codes]
Then for each $i not = j$, the following holds:
$$frac{1}{||y_i-y_j||_2} leq frac{1}{csqrt{d}}$$
for some $c in Omega(1)$, which implies the following:
Inequality 1: $$sum_{1 le i < j le n} frac{1}{||y_i-y_j||_2} leq frac{n^2}{2csqrt{d}}$$
However. not that for any distinct $x,y$, the following holds:
Inequality 2: $$frac{1}{||x-y||_2} ge frac{1}{sqrt{d}} $$
So Inequalities 1 and 2 together imply that for $y_1,ldots, y_n$ as above,
$sum_{1 le i < j le n} frac{1}{||y_i-y_j||_2}$ is no more than a constant multiple larger than $sum_{1 le i < j le n} frac{1}{||x_i-x_j||_2}$ for any other $x_1,ldots, x_n in [-1,1]^d$.
answered Dec 27 '18 at 19:16
Mike
3,089211
3,089211
add a comment |
add a comment |
The situation seems to be less interesting asymptotically for fixed $d geq 2$. We have
$$frac{1}{||x-y||_2} geq frac{1}{2sqrt{d}}$$ for any $x, y in [-1, 1]^d$, so there's an easy lower bound of
$$sum_{1 leq j < k leq n} frac{1}{||x_k - x_j||} geq frac{1}{2sqrt{d}} binom{n}{2} = Omega(n^2).$$ This lower bound is actually achieved (up to a constant) by letting the $x_j$ lie on a grid, say $Lambda^+ = {n^{-1/d}, 2n^{-1/d}, dots, 1}^d$.
First note that if we set
$$Lambda = {-1, dots, -2n^{-1/d}, -n^{-1/d}, 0, n^{-1/d}, 2n^{-1/d}, dots, 1}^d$$
we have
$$sum_{j neq k} frac{1}{||x_k - x_j||} < sum_{v in Lambda, v neq 0} frac{1}{||v||}$$
for each $k$.
Also, since $d geq 2$, the integral
$$int_{[-1, 1]^d} frac{dx}{||x||}$$
actually converges, say to some $I_d < infty$, meaning in particular that
$$frac{1}{n} sum_{v in Lambda, v neq 0} frac{1}{||v||} to I_d$$
(this is just a standard grid approximation).
Putting these together, we get
$$sum_{1 leq j < k leq n} frac{1}{||x_k - x_j||} = frac{1}{2} sum_k sum_{j neq k} frac{1}{||x_k - x_j||} < frac{n}{2} sum_{v in Lambda, v neq 0} frac{1}{||v||} sim frac{n^2}{2} I_d$$
so our sum is $O(n^2)$, matching the lower bound.
add a comment |
The situation seems to be less interesting asymptotically for fixed $d geq 2$. We have
$$frac{1}{||x-y||_2} geq frac{1}{2sqrt{d}}$$ for any $x, y in [-1, 1]^d$, so there's an easy lower bound of
$$sum_{1 leq j < k leq n} frac{1}{||x_k - x_j||} geq frac{1}{2sqrt{d}} binom{n}{2} = Omega(n^2).$$ This lower bound is actually achieved (up to a constant) by letting the $x_j$ lie on a grid, say $Lambda^+ = {n^{-1/d}, 2n^{-1/d}, dots, 1}^d$.
First note that if we set
$$Lambda = {-1, dots, -2n^{-1/d}, -n^{-1/d}, 0, n^{-1/d}, 2n^{-1/d}, dots, 1}^d$$
we have
$$sum_{j neq k} frac{1}{||x_k - x_j||} < sum_{v in Lambda, v neq 0} frac{1}{||v||}$$
for each $k$.
Also, since $d geq 2$, the integral
$$int_{[-1, 1]^d} frac{dx}{||x||}$$
actually converges, say to some $I_d < infty$, meaning in particular that
$$frac{1}{n} sum_{v in Lambda, v neq 0} frac{1}{||v||} to I_d$$
(this is just a standard grid approximation).
Putting these together, we get
$$sum_{1 leq j < k leq n} frac{1}{||x_k - x_j||} = frac{1}{2} sum_k sum_{j neq k} frac{1}{||x_k - x_j||} < frac{n}{2} sum_{v in Lambda, v neq 0} frac{1}{||v||} sim frac{n^2}{2} I_d$$
so our sum is $O(n^2)$, matching the lower bound.
add a comment |
The situation seems to be less interesting asymptotically for fixed $d geq 2$. We have
$$frac{1}{||x-y||_2} geq frac{1}{2sqrt{d}}$$ for any $x, y in [-1, 1]^d$, so there's an easy lower bound of
$$sum_{1 leq j < k leq n} frac{1}{||x_k - x_j||} geq frac{1}{2sqrt{d}} binom{n}{2} = Omega(n^2).$$ This lower bound is actually achieved (up to a constant) by letting the $x_j$ lie on a grid, say $Lambda^+ = {n^{-1/d}, 2n^{-1/d}, dots, 1}^d$.
First note that if we set
$$Lambda = {-1, dots, -2n^{-1/d}, -n^{-1/d}, 0, n^{-1/d}, 2n^{-1/d}, dots, 1}^d$$
we have
$$sum_{j neq k} frac{1}{||x_k - x_j||} < sum_{v in Lambda, v neq 0} frac{1}{||v||}$$
for each $k$.
Also, since $d geq 2$, the integral
$$int_{[-1, 1]^d} frac{dx}{||x||}$$
actually converges, say to some $I_d < infty$, meaning in particular that
$$frac{1}{n} sum_{v in Lambda, v neq 0} frac{1}{||v||} to I_d$$
(this is just a standard grid approximation).
Putting these together, we get
$$sum_{1 leq j < k leq n} frac{1}{||x_k - x_j||} = frac{1}{2} sum_k sum_{j neq k} frac{1}{||x_k - x_j||} < frac{n}{2} sum_{v in Lambda, v neq 0} frac{1}{||v||} sim frac{n^2}{2} I_d$$
so our sum is $O(n^2)$, matching the lower bound.
The situation seems to be less interesting asymptotically for fixed $d geq 2$. We have
$$frac{1}{||x-y||_2} geq frac{1}{2sqrt{d}}$$ for any $x, y in [-1, 1]^d$, so there's an easy lower bound of
$$sum_{1 leq j < k leq n} frac{1}{||x_k - x_j||} geq frac{1}{2sqrt{d}} binom{n}{2} = Omega(n^2).$$ This lower bound is actually achieved (up to a constant) by letting the $x_j$ lie on a grid, say $Lambda^+ = {n^{-1/d}, 2n^{-1/d}, dots, 1}^d$.
First note that if we set
$$Lambda = {-1, dots, -2n^{-1/d}, -n^{-1/d}, 0, n^{-1/d}, 2n^{-1/d}, dots, 1}^d$$
we have
$$sum_{j neq k} frac{1}{||x_k - x_j||} < sum_{v in Lambda, v neq 0} frac{1}{||v||}$$
for each $k$.
Also, since $d geq 2$, the integral
$$int_{[-1, 1]^d} frac{dx}{||x||}$$
actually converges, say to some $I_d < infty$, meaning in particular that
$$frac{1}{n} sum_{v in Lambda, v neq 0} frac{1}{||v||} to I_d$$
(this is just a standard grid approximation).
Putting these together, we get
$$sum_{1 leq j < k leq n} frac{1}{||x_k - x_j||} = frac{1}{2} sum_k sum_{j neq k} frac{1}{||x_k - x_j||} < frac{n}{2} sum_{v in Lambda, v neq 0} frac{1}{||v||} sim frac{n^2}{2} I_d$$
so our sum is $O(n^2)$, matching the lower bound.
answered yesterday
user125932
58529
58529
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%2f3050869%2fquantifying-how-crowded-points-are-in-d-dimensional-cube%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
As it stands your question is a bit too broad, because it is simultaneously asking for the best bound, an asymptotic bound, and "bounds on any quantity that looks similar". Could you please edit your question to make it more focused?
– Lord_Farin
Dec 24 '18 at 9:13
Ok, I edited a little. Since I dont know the difficulty of this question its bound to be a little vague but the whole point is to simulate some discussion and get any useful bounds.
– Sandeep Silwal
Dec 24 '18 at 16:25
These kind of "simple" questions very quickly get out of hand...
– Fabian
Jan 1 at 4:35