Dtjytyk

Quoting Keynes in a lecture

Is there a RAID 0 Equivalent for RAM?

What's the difference between releasing hormones and tropic hormones?

Is there a way to get `mathscr' with lower case letters in pdfLaTeX?

Why is the "ls" command showing permissions of files in a FAT32 partition?

Limits and Infinite Integration by Parts

How to fade a semiplane defined by line?

What are some good ways to treat frozen vegetables such that they behave like fresh vegetables when stir frying them?

Does the Linux kernel need a file system to run?

How should I respond when I lied about my education and the company finds out through background check?

Creepy dinosaur pc game identification

Pre-mixing cryogenic fuels and using only one fuel tank

I'm the sea and the sun

How does the math work for Perception checks?

Why Shazam when there is already Superman?

Why is it that I can sometimes guess the next note?

Multiplicative persistence

Does an advisor owe his/her student anything? Will an advisor keep a PhD student only out of pity?

It grows, but water kills it

How do you respond to a colleague from another team when they're wrongly expecting that you'll help them?

Biological Blimps: Propulsion

How much character growth crosses the line into breaking the character

What are the advantages of simplicial model categories over non-simplicial ones?

Why does a simple loop result in ASYNC_NETWORK_IO waits?

Lattice points below a curve

Complex elliptic curve for the “conjugate” latticeNumber of lattice pointsFind lattice points on a planar curveLattice points in spheresHow to count lattice points on a line.Help in Understanding the Formula for The Lattice Point Counting in Triangles with Rational CoordinatesIs the area of a convex set approximated by the number of its lattice points, in the following sense?The number of lattice points under a rational functionFinding integer points in Lattice?Counting Lattice Points

1

$begingroup$

Assume a curve represented by a function
$f(b)=0.5(sqrt{N-b^2}-b+1)$ with $1leq b leq sqrt{frac{N}{2}}$. I want to count the lattice points below this curve, more specifically, I would like a closed form expression. I can assume this figure to be a parabola and try curve fitting and then use the method given in this paper to sort of get an approximation. But I am not convinced. Can someone please offer me a better solution?

Thanks

integer-lattices

share|cite|improve this question

edited Mar 13 at 17:10

RTn

asked Mar 13 at 16:20

RTnRTn

306

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Important question: are you counting lattice points with $f(b)<0$ as negative?
  $endgroup$
  – R. Burton
  Mar 13 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No. In the interval $1leq b leq sqrt{N/2}$, $f(b)$ is always positive. The choice of $N$ is thus made to ensure that.
  $endgroup$
  – RTn
  Mar 13 at 17:21
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Assume a curve represented by a function
$f(b)=0.5(sqrt{N-b^2}-b+1)$ with $1leq b leq sqrt{frac{N}{2}}$. I want to count the lattice points below this curve, more specifically, I would like a closed form expression. I can assume this figure to be a parabola and try curve fitting and then use the method given in this paper to sort of get an approximation. But I am not convinced. Can someone please offer me a better solution?

Thanks

integer-lattices

share|cite|improve this question

edited Mar 13 at 17:10

RTn

asked Mar 13 at 16:20

RTnRTn

306

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Important question: are you counting lattice points with $f(b)<0$ as negative?
  $endgroup$
  – R. Burton
  Mar 13 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No. In the interval $1leq b leq sqrt{N/2}$, $f(b)$ is always positive. The choice of $N$ is thus made to ensure that.
  $endgroup$
  – RTn
  Mar 13 at 17:21
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Assume a curve represented by a function
$f(b)=0.5(sqrt{N-b^2}-b+1)$ with $1leq b leq sqrt{frac{N}{2}}$. I want to count the lattice points below this curve, more specifically, I would like a closed form expression. I can assume this figure to be a parabola and try curve fitting and then use the method given in this paper to sort of get an approximation. But I am not convinced. Can someone please offer me a better solution?

Thanks

integer-lattices

share|cite|improve this question

edited Mar 13 at 17:10

RTn

asked Mar 13 at 16:20

RTnRTn

306

$endgroup$

share|cite|improve this question

edited Mar 13 at 17:10

RTn

asked Mar 13 at 16:20

RTnRTn

306

share|cite|improve this question

edited Mar 13 at 17:10

RTn

asked Mar 13 at 16:20

RTnRTn

306

asked Mar 13 at 16:20

RTnRTn

306

asked Mar 13 at 16:20

RTnRTn

306

1 Answer
1

active

oldest

votes

0

$begingroup$

If I understand what you are asking correctly, then the number of lattice points below the curve can be treated as a sort of "discrete integral".

Given that the number of lattice points below a point $(b,f(b))$ is, at most $f(b)$, the number of lattice points under the curve from $1$ to $leftlfloorsqrtfrac{N}{2}rightrfloor$ could be obtained from summing the number of lattice points below each $f(n)$ for $n=1,ldots,lfloor sqrt{frac{N}{2}}rfloor$

$$F(b)=sum_{n=1}^{leftlfloor sqrt{frac{N}{2}}rightrfloor} lfloor f(n)rfloor$$

Where $F(b)$ coincides with the number of lattice points and $lfloorcdotrfloor$ is the flooring function.

---

Edit: Okay, it seems that I messed up while entering this into Wolfram. After working it out by hand, the closest approximation to the graph of $F(b)$ is a parabola. However the closed form of the summation (if it exists) cannot be a polynomial of order less than 3. This is because the flooring function in the summation makes the graph asymetrical about the vertical line through the maximum for most $N$.

share|cite|improve this answer

edited Mar 13 at 17:35

answered Mar 13 at 16:37

R. BurtonR. Burton

670110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, indeed. But I don't get how you arrived at the Expression for $F(b)$. Assume $N=100$, $1leq b leq 6$, then the number of lattice points is $37$. For $b=6$, $f(b)=3$ and so according to your formula, $F(b)=21-lfloor f(1)rfloor$. This seems not right.. Am I missing something here?
  $endgroup$
  – RTn
  Mar 13 at 16:58
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, I forgot to add that the summation should run from your lower to upper bound, otherwise you have to take the difference, just like taking a definite integral.
  $endgroup$
  – R. Burton
  Mar 13 at 17:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was wondering if there is a closed form expression for this instead of a summation. Like that which exists for a hyperbola using the Dirichlet method. It would be nice to have a closed form expression.
  $endgroup$
  – RTn
  Mar 13 at 17:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If there is a closed form, it will be for the summation $sum_{n=1}^{lfloorsqrt{frac{N}{2}}rfloor}lfloor f(n)rfloor$. I'll see if I can fix it.
  $endgroup$
  – R. Burton
  Mar 13 at 17:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh wonderful. I will modify the original equation a bit to include a 0.5 as a multiplier. I tried taking a finite continuous integral of this function in the said interval and then subtracted the coefficient of $b^2$ that I obtained by curve fitting $f(b)$ to a parabola and it seems to work. The only thing is that it seems a bit arbitrary to do that.
  $endgroup$
  – RTn
  Mar 13 at 17:09
  
  
  

  
  
  
  

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
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3146809%2flattice-points-below-a-curve%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

If I understand what you are asking correctly, then the number of lattice points below the curve can be treated as a sort of "discrete integral".

Given that the number of lattice points below a point $(b,f(b))$ is, at most $f(b)$, the number of lattice points under the curve from $1$ to $leftlfloorsqrtfrac{N}{2}rightrfloor$ could be obtained from summing the number of lattice points below each $f(n)$ for $n=1,ldots,lfloor sqrt{frac{N}{2}}rfloor$

$$F(b)=sum_{n=1}^{leftlfloor sqrt{frac{N}{2}}rightrfloor} lfloor f(n)rfloor$$

Where $F(b)$ coincides with the number of lattice points and $lfloorcdotrfloor$ is the flooring function.

---

Edit: Okay, it seems that I messed up while entering this into Wolfram. After working it out by hand, the closest approximation to the graph of $F(b)$ is a parabola. However the closed form of the summation (if it exists) cannot be a polynomial of order less than 3. This is because the flooring function in the summation makes the graph asymetrical about the vertical line through the maximum for most $N$.

share|cite|improve this answer

edited Mar 13 at 17:35

answered Mar 13 at 16:37

R. BurtonR. Burton

670110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, indeed. But I don't get how you arrived at the Expression for $F(b)$. Assume $N=100$, $1leq b leq 6$, then the number of lattice points is $37$. For $b=6$, $f(b)=3$ and so according to your formula, $F(b)=21-lfloor f(1)rfloor$. This seems not right.. Am I missing something here?
  $endgroup$
  – RTn
  Mar 13 at 16:58
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, I forgot to add that the summation should run from your lower to upper bound, otherwise you have to take the difference, just like taking a definite integral.
  $endgroup$
  – R. Burton
  Mar 13 at 17:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was wondering if there is a closed form expression for this instead of a summation. Like that which exists for a hyperbola using the Dirichlet method. It would be nice to have a closed form expression.
  $endgroup$
  – RTn
  Mar 13 at 17:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If there is a closed form, it will be for the summation $sum_{n=1}^{lfloorsqrt{frac{N}{2}}rfloor}lfloor f(n)rfloor$. I'll see if I can fix it.
  $endgroup$
  – R. Burton
  Mar 13 at 17:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh wonderful. I will modify the original equation a bit to include a 0.5 as a multiplier. I tried taking a finite continuous integral of this function in the said interval and then subtracted the coefficient of $b^2$ that I obtained by curve fitting $f(b)$ to a parabola and it seems to work. The only thing is that it seems a bit arbitrary to do that.
  $endgroup$
  – RTn
  Mar 13 at 17:09
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

If I understand what you are asking correctly, then the number of lattice points below the curve can be treated as a sort of "discrete integral".

Given that the number of lattice points below a point $(b,f(b))$ is, at most $f(b)$, the number of lattice points under the curve from $1$ to $leftlfloorsqrtfrac{N}{2}rightrfloor$ could be obtained from summing the number of lattice points below each $f(n)$ for $n=1,ldots,lfloor sqrt{frac{N}{2}}rfloor$

$$F(b)=sum_{n=1}^{leftlfloor sqrt{frac{N}{2}}rightrfloor} lfloor f(n)rfloor$$

Where $F(b)$ coincides with the number of lattice points and $lfloorcdotrfloor$ is the flooring function.

---

Edit: Okay, it seems that I messed up while entering this into Wolfram. After working it out by hand, the closest approximation to the graph of $F(b)$ is a parabola. However the closed form of the summation (if it exists) cannot be a polynomial of order less than 3. This is because the flooring function in the summation makes the graph asymetrical about the vertical line through the maximum for most $N$.

share|cite|improve this answer

edited Mar 13 at 17:35

answered Mar 13 at 16:37

R. BurtonR. Burton

670110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, indeed. But I don't get how you arrived at the Expression for $F(b)$. Assume $N=100$, $1leq b leq 6$, then the number of lattice points is $37$. For $b=6$, $f(b)=3$ and so according to your formula, $F(b)=21-lfloor f(1)rfloor$. This seems not right.. Am I missing something here?
  $endgroup$
  – RTn
  Mar 13 at 16:58
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, I forgot to add that the summation should run from your lower to upper bound, otherwise you have to take the difference, just like taking a definite integral.
  $endgroup$
  – R. Burton
  Mar 13 at 17:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was wondering if there is a closed form expression for this instead of a summation. Like that which exists for a hyperbola using the Dirichlet method. It would be nice to have a closed form expression.
  $endgroup$
  – RTn
  Mar 13 at 17:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If there is a closed form, it will be for the summation $sum_{n=1}^{lfloorsqrt{frac{N}{2}}rfloor}lfloor f(n)rfloor$. I'll see if I can fix it.
  $endgroup$
  – R. Burton
  Mar 13 at 17:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh wonderful. I will modify the original equation a bit to include a 0.5 as a multiplier. I tried taking a finite continuous integral of this function in the said interval and then subtracted the coefficient of $b^2$ that I obtained by curve fitting $f(b)$ to a parabola and it seems to work. The only thing is that it seems a bit arbitrary to do that.
  $endgroup$
  – RTn
  Mar 13 at 17:09
  
  
  

  
  
  
  

add a comment | 

$begingroup$

If I understand what you are asking correctly, then the number of lattice points below the curve can be treated as a sort of "discrete integral".

Given that the number of lattice points below a point $(b,f(b))$ is, at most $f(b)$, the number of lattice points under the curve from $1$ to $leftlfloorsqrtfrac{N}{2}rightrfloor$ could be obtained from summing the number of lattice points below each $f(n)$ for $n=1,ldots,lfloor sqrt{frac{N}{2}}rfloor$

$$F(b)=sum_{n=1}^{leftlfloor sqrt{frac{N}{2}}rightrfloor} lfloor f(n)rfloor$$

Where $F(b)$ coincides with the number of lattice points and $lfloorcdotrfloor$ is the flooring function.

---

Edit: Okay, it seems that I messed up while entering this into Wolfram. After working it out by hand, the closest approximation to the graph of $F(b)$ is a parabola. However the closed form of the summation (if it exists) cannot be a polynomial of order less than 3. This is because the flooring function in the summation makes the graph asymetrical about the vertical line through the maximum for most $N$.

share|cite|improve this answer

edited Mar 13 at 17:35

answered Mar 13 at 16:37

R. BurtonR. Burton

670110

$endgroup$

share|cite|improve this answer

edited Mar 13 at 17:35

answered Mar 13 at 16:37

R. BurtonR. Burton

670110

answered Mar 13 at 16:37

R. BurtonR. Burton

670110

answered Mar 13 at 16:37

R. BurtonR. Burton

670110