text:
# Is it really so difficult to draw smooth lines in Unity?

I been trying for a while to draw smooth lines in Unity but with Line Renderer I obtained only jagged lines with the corners not rounded, in particular when the angle of curvature is really small . I incresed the value of antialiasing in quality settings and tried different materials but nothing changed. I also tried to instantiate a sphere every time the mouse move but it creates some gaps between the various spheres, in particular when the mouse go fast. I know there is a plugin called Vectrosity for this but there is a way to achieve this whitout it?

• You probably want a custom material for the line renderer, one that puts a small edge of alpha along the edges. The hard corners is tougher to fix, as it requires backtracking and inserting additional segments. Also, your question doesn't follow the Good Question format of SO – Draco18s Apr 21 '17 at 17:12
• Thank you! Indeed I thought to instantiate a segment where the line renderer curves but I don't know how to find the point where the curves is. And about the material for adding alpha do you mean to add a transparent edges around the material ? I also noticed that when I draw the lines fast the width of the line seems to decrease, is that normal? – rickyxd Apr 21 '17 at 17:27
• It's been a while since I poked at the Line Renderer, and it has some weird behaviors by default. So it might be getting smaller or it might be that the quad is twisting from "vertical" to "horizontal" and camera angle only makes it look like its getting narrower. And yes, I meant a transparent edge around the material. – Draco18s Apr 21 '17 at 17:29
• I'm a beginner in Unity and seems strange that it doesn't have a good class for drawing paths with mouse/fingers, anyway thank you for your tips I will try them! – rickyxd Apr 21 '17 at 17:35
• As a beginner I'd look at a problem like this and think "man, in like six months I could solve this" and just live with it for now. – Draco18s Apr 21 '17 at 17:36

I obtained only jagged lines with the corners not rounded, in particular when the angle of curvature is really small .

This was a problem in Unity 5.4 and below. This problem has been fixed in Unity 5.5 and above after LineRenderer was completely re-designed.

All you have to do is update to Unity 5.5 or version above and this problem should go away.

There is a new variable called LineRenderer.numCornerVertices. You can use that to set how smooth you want the line to be. The value of 5 seems fine for this.

There is also another new variable called LineRenderer.numCapVertices that can be used to set how smooth the end of the line should be.

This is a screenshot that demonstrate between 5.4 and 5.5 the changes:

• The lines are still jaggy if zoom in your left(5.5) image, why is it like that? Even in Unity 2017, same jaggy. Not fine and smooth as the original Graphic API of iOS, any way to approve this in Unity? – Suge Oct 31 '17 at 3:36
• Sorry for making you angry, hope I can cancel the downvote. – Suge Oct 31 '17 at 5:45
• Not angry. If you want help from another user then downvote their answer before their response, how do you expect them to help? I mean that's fine if the answer doesn't make sense at-all or is poor. This is your decision and you shouldn't have to remove the downvote. I actually don't want you to remove it. You have the right to downvote any question/answer. Happy coding! – Programmer Oct 31 '17 at 5:47
• This answer not working for me, situation remain as it in - Unity 2017 : stackoverflow.com/questions/47922491/… – Siddharth Dec 21 '17 at 12:02
• There are 3 answers in this. You can try the other ones(jpgm's answer). If not working then make your own. See Iggy's answer for how to do so. If you can't make your own from that answer then buy a plugin from the assetstore. – Programmer Dec 21 '17 at 12:05

You can get some good results by generating a mesh from a set of points.

The algorithm for it is as follows:

1. You have a set of points, could be generated with bezier curve.

1. For each point, take a directional vector to the next point v = (p2 - p1) (marked in blue). Then rotate that vector by 90 degrees normal = v.y, -v.x marked in red.

1. This illustrates that we will use each normal from the point position. You can now multiply this vector in both directions by the desired width of the line.

1. Create the vertices at those positions.

1. Add indices to form triangles. It will be something like [i, w/2 + i, w/2 + i + 1] where i is the current index, and w is the total number of vertices.

1. Create the other triangles. Again something like [i, w/2 * i + 1, i + 1]

1. And the final result. You can add more points to make the line smoother.

Thanks to @Iggy 's inspiration and tutorials on catlikecoding.com (where the spline code I'm using comes from), I created a component that will create a mesh based on a spline given a width and sample frequency. Higher sample frequency = smoother curve of course.

using UnityEngine;

[ExecuteInEditMode]
[RequireComponent(typeof(MeshFilter), typeof(MeshRenderer), typeof(BezierSpline))]
public class SplineMesh : MonoBehaviour {

[Range(1, 20)]
public int sampleFrequency = 5;

[Range(0, 5f)]
public float lineWidth = 0.3f;

BezierSpline spline;
Mesh mesh;

private void Awake () {
spline = GetComponent<BezierSpline>();
mesh = GetComponent<Mesh>();
}

void Update()
{
/*
for(int i = 0; i <= sampleFrequency; i++){
float interval = i / (float)sampleFrequency;

var point = spline.GetPoint(interval);
var direction = spline.GetDirection(interval);

var perpendicularLeftVec = PerpendicularLeft(direction) * lineWidth;
var perpendicularRightVec = PerpendicularRight(direction) * lineWidth;

Debug.DrawLine(point, point + (Vector3)perpendicularLeftVec, Color.magenta, 0.5f, false);
Debug.DrawLine(point, point + (Vector3)perpendicularRightVec, Color.cyan, 0.5f, false);
}
*/
}

Vector2 PerpendicularRight(Vector2 orig){
var vec = new Vector2(orig.y, -orig.x);
vec.Normalize();
return vec;
}
Vector2 PerpendicularLeft(Vector2 orig){
var vec = new Vector2(orig.y, -orig.x);
vec.Normalize();
return vec * -1;
}

private Vector3[] vertices;

public void GenerateMesh(){
vertices = new Vector3[(sampleFrequency + 1) * 2];

//iterate over our samples adding two vertices for each one
for(int s = 0, i = 0; s <= sampleFrequency; s++, i += 2){
float interval = s / (float)sampleFrequency;

//get point along spline, and translate to local coords from world
var point = transform.InverseTransformPoint(spline.GetPoint(interval));
var direction = spline.GetDirection(interval);

var perpendicularLeftVec = PerpendicularLeft(direction) * lineWidth;
var perpendicularRightVec = PerpendicularRight(direction) * lineWidth;
// var perpendicularVec = turnLeft ? PerpendicularLeft(diffVector) : PerpendicularRight(diffVector);

vertices[i] = point + (Vector3)perpendicularLeftVec;
vertices[i + 1] = point + (Vector3)perpendicularRightVec;
}

GetComponent<MeshFilter>().mesh = mesh = new Mesh();
mesh.name = "Spline Mesh";

mesh.vertices = vertices;

//now figure out our triangles
int [] triangles = new int[sampleFrequency * 6];
for(int s = 0, ti = 0, vi = 0; s < sampleFrequency; s++, ti += 6, vi += 2){
//first tri
triangles[ti] = vi;
triangles[ti + 1] = vi + 3;
triangles[ti + 2] = vi + 1;
//second matching tri
triangles[ti + 3] = vi;
triangles[ti + 4] = vi + 2;
triangles[ti + 5] = vi + 3;
}

mesh.triangles = triangles;
mesh.RecalculateNormals();

Debug.Log("Generated Spline Mesh");
}

}

I had this issue with Unity 2017.2. I tried changing my AA settings to max to get rid of line render jaggies. Didn't work and it was frustrating.

My solution was resolving the problem that MSAA was turned off on the camera because rendering was deferred. The camera has a setting for "use graphics settings" which should have never messed up in the first place, but I'm a beginner - I don't know much. I changed the setting to "forward" and my jaggies disappeared into the mist.

If I was more industrious, I would post before and after images.

I finally got this working thanks to @Iggy's excellent answer.

Create a new Image on your canvas, delete the Image script and replace it with UICubicBezier.

To give:

using UnityEngine;
using UnityEngine.UI;

[ExecuteInEditMode]
{
public float thickness = 2;

public int anchors = 20;

protected override void OnPopulateMesh(VertexHelper vh)
{
// draws a cubic bezier curve from the lower left hand corner (start)
// to the upper right hand corner (end).
vh.Clear();

var rt = this.rectTransform;
var rect = rt.rect;

var start = new Vector2(-rect.width / 2, -rect.height / 2);
var cp1 = new Vector2(-rect.width / 6, -rect.height / 2);
var cp2 = new Vector2(rect.width / 6, rect.height / 2);
var end = new Vector2(rect.width / 2, rect.height / 2);
var data = new BezierData(start, cp1, cp2, end);

// all you need to know is that data.GetPoint generates a sequence of points
// between the start and end points.
var points = new Vector2[this.anchors];
for (var anchor = 0; anchor < points.Length; anchor++)
{
var t = (float)anchor / this.anchors;
points[anchor] = data.GetPoint(t);
}

// because the normals are at the mid-points between vertexes the start and end
// points don't touch the bounding box. to fix this some vertexes are added to
// the start and end that touch the bounding box.
this.DrawStartVertexes(vh, start);

for (var anchor = 0; anchor < points.Length - 1; anchor++)
{
this.DrawVertexes(vh, points[anchor], points[anchor + 1]);
}

this.DrawEndVertexes(vh, end);

for (var v = 0; v + 2 < vh.currentVertCount; v += 2)
{
vh.AddTriangle(v, v + 1, v + 2);
}

for (var v = 0; v + 3 < vh.currentVertCount; v += 2)
{
vh.AddTriangle(v + 1, v + 2, v + 3);
}
}

private void DrawStartVertexes(VertexHelper vh, Vector2 start)
{
// d = thickness * \sqrt{2}, so the distance between the vertexes
// is equal to the thickness (https://en.wikipedia.org/wiki/Triangle#Right_triangles)
var d = this.thickness * 0.70710678118f;

var vertex = UIVertex.simpleVert;
vertex.color = this.color;

vertex.position = new Vector2(start.x, start.y + d);

vertex.position = new Vector2(start.x + d, start.y);
}

private void DrawEndVertexes(VertexHelper vh, Vector2 end)
{
// d = thickness * \sqrt{2}, so the distance between the vertexes
// is equal to the thickness (https://en.wikipedia.org/wiki/Triangle#Right_triangles)
var d = this.thickness * 0.70710678118f;

var vertex = UIVertex.simpleVert;
vertex.color = this.color;

vertex.position = new Vector2(end.x - d, end.y);

vertex.position = new Vector2(end.x, end.y - d);
}

private void DrawVertexes(VertexHelper vh, Vector2 start, Vector2 end)
{
var v = end - start;
var mid = start + v / 2; // the mid-point between start and end.
var perp = Vector2.Perpendicular(v.normalized); // vector of length 1 perpendicular to v.

var vertex = UIVertex.simpleVert;
vertex.color = this.color;

// move half the thickness away from the mid-point.
vertex.position = mid + (perp * this.thickness / 2);

// move half the thickness away from the mid-point in the opposite direction.
vertex.position = mid - (perp * this.thickness / 2);
}

private struct BezierData
{

public BezierData(Vector2 start, Vector2 cp1, Vector2 cp2, Vector2 end)
{

this.start = start;
this.cx = 3 * (cp1.x - start.x);
this.bx = 3 * (cp2.x - cp1.x) - this.cx;
this.ax = end.x - start.x - this.cx - this.bx;

this.cy = 3 * (cp1.y - start.y);
this.by = 3 * (cp2.y - cp1.y) - this.cy;
this.ay = end.y - start.y - this.cy - this.by;
}

public Vector2 GetPoint(float t)
{
var tSquared = t * t;
var tCubed = tSquared * t;

return new Vector2(
(this.ax * tCubed) + (this.bx * tSquared) + (this.cx * t) + this.start.x,
(this.ay * tCubed) + (this.by * tSquared) + (this.cy * t) + this.start.y);
}
}
}


text:
# Europhysics Conference on High-Energy Physics 2011

Jul 20 – 27, 2011
Alpes Congrès - Alpexpo
Europe/Paris timezone

## Production of lepton, quark and meson pairs in peripheral ulrarelativistic heavy ion collisions

Jul 21, 2011, 6:15 PM
15m
Lesdiguières (Alpes Congrès - Alpexpo)

### Lesdiguières

#### Alpes Congrès - Alpexpo

Parallel session talk

### Speaker

Prof. Antoni Szczurek (Rzeszow University and Institute of Nuclear Physics PAN)

### Description

We discuss exclusive production of lepton-antilepton, quark-antiquark, $\pi \pi$ and $\rho^0 \rho^0$ and $D \bar D$ meson pairs in ultraperipheral, ultrarelativistic heavy-ion collisions i.e. processes initiated by photon-photon fusion which could be studied at RHIC and LHC. The cross sections for exclusive muon pair production in nucleus - nucleus collisions was calculated and several differential distributions are shown. Realistic (Fourier transform of charge density) charge form factors of nuclei are used and the corresponding results are compared with the cross sections calculated with monopole charged form factor of a nucleus often used in the literature. Absorption effects are discussed and quantified. The cross sections obtained with realistic form factors are significantly smaller than those obtained with the monopole form factor. The effect is bigger for large muon rapidities and/or large muon transverse momenta. We show predictions for STAR and PHENIX collaborations at RHIC as well as ALICE and CMS collaborations at LHC. The cross section for exclusive heavy quark and heavy antiquark pair ($Q \bar Q$) production in peripheral ultrarelativistic heavy ion collisions was calculated for the LHC energy $\sqrt{s_{NN}}$ = 5.5 TeV. We present results in the impact parameter equivalent photon approximation (EPA) and compare some of them with results obtained by exact calculations of the Feynman diagrams in the momentum space. We include both $Q \bar Q$, $Q \bar Q g$ and $Q \bar Q q \bar q$ final states as well as photon single-resolved components. Realistic charge densities in nuclei were taken in the calculation. The different components give contributions of the same order of magnitude to the nuclear cross section. We present for the first time cross section for exclusive production of $\pi^+ \pi^-$ and $\pi^0 \pi^0$ pairs. The elementary process $\gamma \gamma \to \pi \pi$ is discussed in detail. We concentrate on high-$p_t$ processes. We consider pQCD processes ala Brodsky-Lepage or alternatively hand-bag mechanism. The nuclear cross section is calculated within b-space EPA for RHIC and LHC. Similar analysis is performed for $\rho^0 \rho^0$ production, where the elementary cross section is less known. Our analysis includes a low-energy phenomenon (close-to-threshold enhancement of the cross section). The cross section for the low-energy phenomenon is parametrized and the high-energy cross section is calculated in a simple Regge model. Predictions for heavy ion collisions are presented. The cross sections for exclusive $D^+ D^-$ and $D^0 \bar {D^0}$ meson pair production in peripheral nucleus - nucleus collisions are calculated and several differential distributions are presented. The calculation of the elementary $\gamma \gamma \to D \bar D$ cross section is done within the heavy-quark approximation and in the Brodsky-Lapage formalism with distribution amplitudes describing recent CLEO data on leptonic $D^+$ decay. The cross sections of a few nb are predicted for RHIC and of a few hundreds of nb for LHC with details depending on the approximation made in calculating elementary $\gamma \gamma \to D \bar D$ cross sections. The speaker will present these different results to domonstrate how reach is the option of ultrarelativistic ultraperipheral collisions at RHIC and LHC. Literature: \item M. K{\l}usek, W. Sch\"afer and A. Szczurek, "Exlusive production of $\rho^0 \rho^0$ pairs in $\gamma \gamma$ collisions at RHIC", Phys. Lett. {\bf B674} (2009) 92. \item K{\l}usek-Gawenda and A. Szczurek, Exclusive muon-pair production in utrarelativistic heavy-ion collisions - realistic nucleus charge form factor and differential distributions'', Phys. Rev. {\bf C82} (2010) 014904. \item M. K{\l}usek-Gawenda, A. Szczurek, M. Machado and V. Serbo, Double -- photon exclusive processes with heavy quark -- heavy antiquark pairs in high-energy Pb-Pb collisions at LHC'', Phys. Rev. {\bf C83} (2011) 024903. \item M. {\L}uszczak and A. Szczurek, Exclusive $D \bar D$ meson pair production\\ in peripheral ultrarelativistic heavy ion collisions'', arXiv:1103.4268, in print in Phys. Lett.B \item M. K{\l}usek-Gawenda and A, Szczurek, arXiv:1104.0571, in print in Phys. Lett.B

### Primary author

Prof. Antoni Szczurek (Rzeszow University and Institute of Nuclear Physics PAN)

### Presentation materials

 Transparents

format: plain


text:
### abitnav's blog

By abitnav, history, 12 days ago, ,

There is a group on telegram where people copy and ask solutions openly during live contests

I request all of you reading this post to report this group.

The following Screenshots are for today's contest Educational Codeforces Round 92 (Rated for Div. 2)

Openly discussing A.

I am thankful for my friend jhanurag for providing the Screenshots

• +315

 » 12 days ago, # |   +91 Oh I have yet another suggestion for a case like this. If Codeforces could send a few coordinators on telegram in search of groups like that who would behave like cheaters and when the time comes they would circulate a specific solution in the group which can be caught during plag check or just somehow collect the cf id's of those cheaters and ban them all..Ik it's a bit filmy but it can work.
•  » » 12 days ago, # ^ |   +55 That's FBI move
•  » » 12 days ago, # ^ |   +5 I don't think that will work. They aren't fools to simply submit what you share. They will make necessary changes and then submit, clearing the threshold for Plag check. Also, they might just take the idea of the solution and implement it themselves. Chances of this working are quite low.
•  » » » 12 days ago, # ^ | ← Rev. 2 →   +10 Yeah, I agree with your point that they are not simply fools and it will eventually catch just a few of them. But, They are not fools but they are greedy and we can take it to our advantage. The right time to circulate the solution would be 5-10 minutes before the contest ends. In that small window, they can only manipulate variable names or add unnecessary loops or do that 'tutiya' thing. It can easily be detected using a checker that can filter solution based on function calls and removing unnecessary variables. Even I can catch a few of them by simply reading them off. We do alot of it during hacking phase. Circulating an apt code which is neither too easy to understand & implement nor too fishy to ignore and not submit is ideal. We just have to induce a sense of fear among them so that they think twice before asking for codes or help during the contest. Mercilessly banning a few of the cheaters and sending a message in that group for the same is enough. It's harder to catch the people who work privately in groups but it all starts from a public group so.. X
•  » » » » 12 days ago, # ^ |   +4 Okay, that seems more likely to work.
 » 12 days ago, # |   +69 For the life of me, i cannot understand what these people gain by cheating in contests. There is no direct benefit, and cheating devoids one completely of the satisfaction and sense of achievement that we yearn for. These people will never actually be able to learn, grow and become competent and self-sufficient coders. They harm themselves the most.While there is absolutely no place for cheating and plagiarism in this community, and it must be treated severely, I realise that there is no solution potent enough to completely stop it, and those of us who are honest must not allow this to dishearten us even one bit.
•  » » 12 days ago, # ^ |   +12 Because people think they may get shortlisted for an interview based on ratings, which is kind of true because people like me don't get referrals that easily. The system is corrupted and don't value the hardwork.
•  » » » 12 days ago, # ^ |   +6 If you want to use your cf rating as a way of getting an interview, I think the rating should be high enough. I don't see how it is possible to become at least red with the techniques like that.
•  » » » » 12 days ago, # ^ |   0 Candidate master is more than enough to proove a person . And you can get referal for interview. For lower companies even expert works .Industry don't have such high standard that they would only consider red (even for faang referals).
•  » » » » » 11 days ago, # ^ |   +10 Then let's put it that way:I'm an international master. How do I get a referral?
•  » » » » » » 11 days ago, # ^ |   +13 Just ask any employee of faang on LinkedIn, and show them your rating . You will definitely get referal .Also do mention this rating in your resume , it will help you while interview too. At this rating you already won the half game even before interview.
•  » » » » » » 11 days ago, # ^ |   -27 Yes, In India it is correct. Expert is considered good coder in India.
 » 12 days ago, # | ← Rev. 3 →   -11 Sad to see India tag on the blog.Edit: It is being removed.
•  » » 12 days ago, # ^ |   +8 This same thing I felt too, really sad to look at it
 » 12 days ago, # |   -96 There are already many comments on this so i think this is just a Contibution increasing Post
•  » » 12 days ago, # ^ | ← Rev. 2 →   +89 Or maybe a cheaters recruitment post by group members. SpoilerOtherwise, I don't see any reason advantage of publicly posting group invite link.
•  » » » 12 days ago, # ^ | ← Rev. 2 →   0 Seems , like they are inviting other who don't know about it.
•  » » » 12 days ago, # ^ |   0 And among all of this, innocent jhanurag becomes the prison snitch
•  » » » 12 days ago, # ^ |   0 is there no system of banning such groups straightaway?
 » 12 days ago, # |   +95 I can hear their accents
•  » » 11 days ago, # ^ |   +4
 » 12 days ago, # |   +42 Why do you share the link? More cheaters will join.
 » 12 days ago, # |   +8 All we can do is not to pay attention. If someone wants to 'play unfairly', we can do nothing. But actually with this your post you just share the link and probably that someone will use it...
 » 12 days ago, # |   0 "Its beyond science" to explain what they gain with useless copy! XD
 » 12 days ago, # |   0 ONe day They Will suffer ...
•  » » 12 days ago, # ^ |   +1 Before or after the death?
 » 12 days ago, # |   -74
•  » » 12 days ago, # ^ |   +127 Why? He did a good thing by deviating them that you need to use FFT to solve B.
•  » » » 12 days ago, # ^ |   -18 Is this allowed in the contest time?
•  » » » » 12 days ago, # ^ | ← Rev. 2 →   +22 Please google what is FFT. FFT is Fast Fourier Transformation. It is a method for multiplying two polynomials in $O(nlog(n))$ instead of $O(n^2)$. It is usually used in signal processing. In CP, it is usually for Div 1 D or even higher. Some grandmasters I know don't know how FFT works as it is advanced. So basically, FFT in Div 2 B is impossible. It is a troll.
•  » » » » » 12 days ago, # ^ |   -88 FFT is Fast Fourier transform, not transformation. You should have googled that instead.
•  » » » » » » 12 days ago, # ^ |   +29 That explains your contribution.
•  » » » » » » » 12 days ago, # ^ |   -63 Someday I will get a good contribution and will call you on your mobile phone, if you mind.
•  » » » » » » 12 days ago, # ^ |   +9 What's the difference? Noun and Verb?
•  » » » » » » » 12 days ago, # ^ |   -28 Ok, don't deeply dive into my comment.
•  » » 12 days ago, # ^ |   +19 Ah yes, FFT, the best solution available for B.
•  » » 10 days ago, # ^ | ← Rev. 3 →   +11 You are joking, right? I found this group in a recent codeforces blog. Then I joined this group and was messing with people who wanted hints for problems during contest time. And funny thing is that most of them even took me seriously! and No way FFT could be a solution for a div2B problem!
 » 12 days ago, # |   +245 asking for a hint for A, how pathetic
 » 12 days ago, # |   -37 Whatever they do, copy, share, cheat, or might even get to Red by this. But, in the end, it's all about our DS, Algo, and problem-solving skills that'll be tested on the real grounds. They'll feel to have cheated themselves when they won't be able to solve problems in the real world.
 » 12 days ago, # |   +3 What were you guys doing in the group?
•  » » 12 days ago, # ^ |   0 I would say they joined for discussions of problem or asking for help but not in live contest and now they are just exposer.
 » 12 days ago, # |   +2 The HELL is wrong with people. WHY THE FUCK do they cheat????? As if by becoming purple or even master they will get direct interview call from FAANG. Special appeal to INDIANS... The world does not give a FUCK's ASS to your rating on CODEFORCES or CODECHEF or ANY OJ.ITS A SPORT.TREAT IT LIKE ONE. DONT SPOIL THE NAME OF INDIA.
 » 12 days ago, # | ← Rev. 2 →   0 Please guys I don't want to rude but how about you stop this fuc**ing behavior, because of you guys everyone is calling Indian cheaters and I don't like that. How about you feel some shame, you won't achieve anything doing this.Man, this makes me sad and angry at the same time.
•  » » 12 days ago, # ^ |   0 There are many Indian cheaters, because Indians are so many.
 » 12 days ago, # |   +3 Telegram is great when it comes to getting useful and great study material, but it sucks when you see this kind of unfair activities going on during live contests. It need to stop, maybe we need a mole in such groups who could leak the cf ids of people plag. the solutions.
 » 11 days ago, # |   0 hahaha
 » 10 days ago, # |   +1 Sorry for asking,but why is that most of cheaters found are always Indians?
•  » » 10 days ago, # ^ |   -6 This is because of the mindset that marks are more important than knowledge which is developed by our education system. That is why some students just want good ratings (analogous to marks).Its just my prescriptive. Sorry if it hurts someone. BTW i am also Indian.
 » 10 days ago, # |   +7 Is is sooo funny that they are saying "hint for ..." bro you are burning problems why you want hints!? why you just dont say give me the fu**ing solution
 » 10 days ago, # |   0 As far as I can see, he kept asking for a hint for Problem C and someone gave him GFG link. LOl!
 » 9 days ago, # |   +2 They are cheating themselves !!


text:
Plane Intersecting a Sphere

Aristarchan

New member
I was thinking this morning of X,Y and Z axes, visualizing a plane intersecting a sphere at varying depths from it's surface. Would the movement of that plane in a direction perpendicular to it's two dimensions, so that a circle indicating where the sphere intersects with the plane changes diameter but stays centered on the same point in that plane, be always referred to as movement along a Z axis?

lev888

Full Member
Not sure I understand the question. How exactly is the coordinate system defined? In any case, a plane moving in the direction of its normal will intersect any sphere in the way you describe.

pka

Elite Member
I was thinking this morning of X,Y and Z axes, visualizing a plane intersecting a sphere at varying depths from it's surface. Would the movement of that plane in a direction perpendicular to it's two dimensions, so that a circle indicating where the sphere intersects with the plane changes diameter but stays centered on the same point in that plane, be always referred to as movement along a Z axis?
Consider the sphere $$x^2+y^2+z^2=25$$ see here.
Now the points $$P_1: (5,0,0),~P_2,~(0,5,0)~\&~P_3,~(0,0,5)$$ are on that sphere.
Thus the plane determined by those points is $$x+y+z=5$$ SEE HERE

LCKurtz

Junior Member
I had a little time to kill, so here's an animation showing a plane passing through a sphere, intersecting in various size of circles:

Not sure what the OP's question is, but the plane can cut the sphere at any angle.


text:
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 On the minimum order of 4-lazy cops-win graphs Bull. Korean Math. Soc. 2018 Vol. 55, No. 6, 1667-1690 https://doi.org/10.4134/BKMS.b170948Published online November 1, 2018 Kai An Sim, Ta Sheng Tan, Kok Bin Wong University of Reading Malaysia, University of Malaya, University of Malaya Abstract : We consider the minimum order of a graph $G$ with a given lazy cop number $c_L(G)$. Sullivan, Townsend and Werzanski \cite{k=3} showed that the minimum order of a connected graph with lazy cop number 3 is 9 and $K_3$ \Box $K_3$ is the unique graph on nine vertices which requires three lazy cops. They conjectured that for a graph $G$ on $n$ vertices with $\Delta (G) \geq n-k^2$, $c_L(G) \leq k$. We proved that the conjecture is true for $k=4$. Furthermore, we showed that the Petersen graph is the unique connected graph $G$ on 10 vertices with $\Delta(G) \leq 3$ having lazy cop number 3 and the minimum order of a connected graph with lazy cop number 4 is 16. Keywords : Cops and Robbers, vertex-pursuit games, minimum order MSC numbers : 05C57, 05C80 Downloads: Full-text PDF


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# 28.2 Simultaneity and time dilation  (Page 5/8)

 Page 5 / 8

The situation would seem different to the astronaut. Because motion is relative, the spaceship would seem to be stationary and the Earth would appear to move. (This is the sensation you have when flying in a jet.) If the astronaut looks out the window of the spaceship, she will see time slow down on the Earth by a factor of $\gamma =\text{30}\text{.}0$ . To her, the Earth-bound sister will have aged only 2/30 (1/15) of a year, while she aged 2.00 years. The two sisters cannot both be correct.

As with all paradoxes, the premise is faulty and leads to contradictory conclusions. In fact, the astronaut’s motion is significantly different from that of the Earth-bound twin. The astronaut accelerates to a high velocity and then decelerates to view the star system. To return to the Earth, she again accelerates and decelerates. The Earth-bound twin does not experience these accelerations. So the situation is not symmetric, and it is not correct to claim that the astronaut will observe the same effects as her Earth-bound twin. If you use special relativity to examine the twin paradox, you must keep in mind that the theory is expressly based on inertial frames, which by definition are not accelerated or rotating. Einstein developed general relativity to deal with accelerated frames and with gravity, a prime source of acceleration. You can also use general relativity to address the twin paradox and, according to general relativity, the astronaut will age less. Some important conceptual aspects of general relativity are discussed in General Relativity and Quantum Gravity of this course.

In 1971, American physicists Joseph Hafele and Richard Keating verified time dilation at low relative velocities by flying extremely accurate atomic clocks around the Earth on commercial aircraft. They measured elapsed time to an accuracy of a few nanoseconds and compared it with the time measured by clocks left behind. Hafele and Keating’s results were within experimental uncertainties of the predictions of relativity. Both special and general relativity had to be taken into account, since gravity and accelerations were involved as well as relative motion.

1. What is $\gamma$ if $v=0\text{.650}c$ ?

## Solution

$\gamma =\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}=\frac{1}{\sqrt{1-\frac{\left(0\text{.}\text{650}c{\right)}^{2}}{{c}^{2}}}}=1\text{.}\text{32}$

2. A particle travels at $1\text{.}\text{90}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ and lives $2\text{.}\text{10}×{\text{10}}^{-8}\phantom{\rule{0.25em}{0ex}}s$ when at rest relative to an observer. How long does the particle live as viewed in the laboratory?

$\Delta t=\frac{{\Delta }_{t}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}=\frac{2\text{.}\text{10}×{\text{10}}^{-8}\phantom{\rule{0.25em}{0ex}}\text{s}}{\sqrt{1-\frac{\left(1\text{.}\text{90}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}{\right)}^{2}}{\left(3\text{.}\text{00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}{\right)}^{2}}}}=2\text{.}\text{71}×{\text{10}}^{-8}\phantom{\rule{0.25em}{0ex}}\text{s}$

## Section summary

• Two events are defined to be simultaneous if an observer measures them as occurring at the same time. They are not necessarily simultaneous to all observers—simultaneity is not absolute.
• Time dilation is the phenomenon of time passing slower for an observer who is moving relative to another observer.
• Observers moving at a relative velocity $v$ do not measure the same elapsed time for an event. Proper time $\Delta {t}_{0}$ is the time measured by an observer at rest relative to the event being observed. Proper time is related to the time $\Delta t$ measured by an Earth-bound observer by the equation
$\Delta t=\frac{{\Delta t}_{0}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}={\gamma \Delta t}_{0},$

where

$\gamma =\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}.$
• The equation relating proper time and time measured by an Earth-bound observer implies that relative velocity cannot exceed the speed of light.
• The twin paradox asks why a twin traveling at a relativistic speed away and then back towards the Earth ages less than the Earth-bound twin. The premise to the paradox is faulty because the traveling twin is accelerating. Special relativity does not apply to accelerating frames of reference.
• Time dilation is usually negligible at low relative velocities, but it does occur, and it has been verified by experiment.

## Conceptual questions

Does motion affect the rate of a clock as measured by an observer moving with it? Does motion affect how an observer moving relative to a clock measures its rate?

To whom does the elapsed time for a process seem to be longer, an observer moving relative to the process or an observer moving with the process? Which observer measures proper time?

How could you travel far into the future without aging significantly? Could this method also allow you to travel into the past?

## Problems&Exercises

(a) What is $\gamma$ if $v=0\text{.}\text{250}c$ ? (b) If $v=0\text{.}\text{500}c$ ?

(a) 1.0328

(b) 1.15

(a) What is $\gamma$ if $v=0\text{.}\text{100}c$ ? (b) If $v=0\text{.}\text{900}c$ ?

Particles called $\pi$ -mesons are produced by accelerator beams. If these particles travel at $2\text{.}\text{70}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ and live $2\text{.}\text{60}×{\text{10}}^{-8}\phantom{\rule{0.25em}{0ex}}\text{s}$ when at rest relative to an observer, how long do they live as viewed in the laboratory?

$5\text{.}\text{96}×{\text{10}}^{-8}\phantom{\rule{0.25em}{0ex}}\text{s}$

Suppose a particle called a kaon is created by cosmic radiation striking the atmosphere. It moves by you at $0\text{.}\text{980}c$ , and it lives $1\text{.}\text{24}×{\text{10}}^{-8}\phantom{\rule{0.25em}{0ex}}\text{s}$ when at rest relative to an observer. How long does it live as you observe it?

A neutral $\pi$ -meson is a particle that can be created by accelerator beams. If one such particle lives $1\text{.}\text{40}×{\text{10}}^{-\text{16}}\phantom{\rule{0.25em}{0ex}}\text{s}$ as measured in the laboratory, and $0\text{.}\text{840}×{\text{10}}^{-\text{16}}\phantom{\rule{0.25em}{0ex}}\text{s}$ when at rest relative to an observer, what is its velocity relative to the laboratory?

$0.800c$

A neutron lives 900 s when at rest relative to an observer. How fast is the neutron moving relative to an observer who measures its life span to be 2065 s?

If relativistic effects are to be less than 1%, then $\gamma$ must be less than 1.01. At what relative velocity is $\gamma =1\text{.}\text{01}$ ?

$0\text{.}\text{140}c$

If relativistic effects are to be less than 3%, then $\gamma$ must be less than 1.03. At what relative velocity is $\gamma =1\text{.}\text{03}$ ?

(a) At what relative velocity is $\gamma =1\text{.}\text{50}$ ? (b) At what relative velocity is $\gamma =\text{100}$ ?

(a) $0\text{.}\text{745}c$

(b) $0\text{.}\text{99995}c$ (to five digits to show effect)

(a) At what relative velocity is $\gamma =2\text{.}\text{00}$ ? (b) At what relative velocity is $\gamma =\text{10}\text{.}0$ ?

Unreasonable Results

(a) Find the value of $\gamma$ for the following situation. An Earth-bound observer measures 23.9 h to have passed while signals from a high-velocity space probe indicate that $\text{24.0 h}$ have passed on board. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

(a) 0.996

(b) $\gamma$ cannot be less than 1.

(c) Assumption that time is longer in moving ship is unreasonable.

I really need lots of questions on frictional force
Shii
I can help answering what I can
Shii
does friction also need some force to perform?
Mohit
no friction is a force just like the gravitational force
clifford
yeah but u can't apply friction anywhere else like other forces
Mohit
I don't understand that question. friction does work alongside other forces based on the situation.
clifford
eg. when walking there are two forces acting on us gravitational and frictional force. friction helps us move forward and gravity keeps us on the ground
clifford
friction is a contact force. Two surfaces are necessary for the force to work.
clifford
hope this helped
clifford
the friction force which oppose while it contact with surrounding. there are two kind of friction. slidding and rolling friction.
Neyaz
What is physics?
physics is a branch of science in which we are dealing with the knowledge of our physical things. macroscopic as well as microscopic. we are going look inside the univers with the help of physics. you can learn nature with the help of physics. so many branches of physics you have to learn physics.
vijay
What are quarks?
6 type of quarks
Neyaz
what is candela
Candela is the unit for the measurement of light intensity.
Osei
any one can prove that 1hrpower= 746 watt
Newton second is the unit of ...............?
Neyaz
Impulse and momentum
Fauzia
force×time and mass× velocity
vijay
Good
Neyaz
What is the simple harmonic motion?
oscillatory motion under a retarding force proportional to the amount of displacement from an equilibrium position
Yuri
Straight out of google, you could do that to, I suppose.
Yuri
*too
Yuri
ok
Fauzia
Oscillatory motion under a regarding force proportional to the amount of displacement from an equilibrium position
Neyaz
examples of work done by load of gravity
What is ehrenfest theorem?
You can look it up, faster and more reliable answer.
Yuri
That isn't a question to ask on a forum and I also have no idea what that is.
Yuri
what is the work done by gravity on the load 87kj,11.684m,mass xkg[g=19m/s
Maureen
What is law of mass action?
rate of chemical reactions is proportional to concentration of reactants ...
ok thanks
Fauzia
what is lenses
lenses are two types
Fauzia
concave and convex
right
Fauzia
speed of light in space
in vacuum speed of light is 3×10^8 m/s
vijay
ok
Vikash
2.99×10^8m/s
Umair
2.8820^8m/s
Muhammed
Vikash
he is correct but we can round up in simple terms
vijay
3×10^8m/s
vijay
is it correct
Fauzia
I mean 3*10^8 m/s ok
vijay
299792458 meter per second
babar
3*10^8m/s
Neyaz
how many Maxwell relations in thermodynamics
vijay
how we can do prove them?
vijay
What is second law of thermodynamics?
Neyaz
please who has a detailed solution to the first two professional application questions under conservation of momentum
I want to know more about pressure
Osei
I can help
Emeh
okay go on
True
I mean on pressure
Emeh
definition of Pressure
John
it is the force per unit area of a substance.S.I unit is Pascal 1pascal is defined as 1N acting on 1m² area i.e 1pa=1N/m²
Emeh
pls explain Doppler effect
Emmex
solve this an inverted differential manometer containing oil specific gravity 0.9 and manometer reading is 400mm find the difference of pressure
Einstine claim that nothing can go with the speed of light even its half (50%) but in to make antimatter they they hit the sub atomic particals 99.9%the speed of light how is it possible
nothing with physical properties. this doesn't include things like particles and gravitational waves
Mustafa
that particles are of very small mass.... near equals to massless
Aritra
but they exist
vijay
yes they exist but mass is too less
Aritra
ok
vijay
greet all
Abayomi
the unit of radioactivity is .....?
Neyaz
Great Sharukh ! Do you have question in physics?
book says that when wave enter from one medium to another its wavelenght changes but frequency not how ? and f is inversely related to wavelenth
Sharukh
yes but how comes
Sani
what is normal force?
the force that pushes upward on us. the force that opposes gravity
clifford


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# Permittivity of free space

The permittivity of free space, ε0, is a physical constant used often in electromagnetism. It represents the capability of a vacuum to permit electric fields. It is also connected to the energy stored within an electric field and capacitance. Perhaps more surprisingly, it's fundamentally related to the speed of light. The permittivity of free space, ε0, is defined as:[1]

where

An electric field, , in a region of space has field energy associated with it, that energy density is:[2]

The energy stored in a capacitor (with no dielectric) is:

Energy

where

• is the area of the plates,
• is the distance between the plates
• is the voltage between the plates.

The permittivity of free space can also be used to find the Coulomb force. The constant gives how strong the force is between two charges separated by a distance:[1]

where

• is the Coulomb force,
• and are two charges, and
• is the separation between the charges.


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# Crazy Calculus 2

Calculus Level pending

If $$\displaystyle a= \lim _{x \to 0^+} \ x^ m(\ln x)^n$$, where $$m$$ and $$n$$ are natural numbers, then which of the following is true?

×


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# Bijection between reals and irrationals

1. Homework Statement

Prove that |R \ |Q ~ |R ... the irrationals are equinumerous to the reals

2. Homework Equations

3. The Attempt at a Solution

I can prove it using the Cantor Schroeder Bernstein theorem, but i was wondering if there
is a clever way of constructing the bijection explicitely ...
thanks,
g.
1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution


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Current Page
Cardioid

# Cardioid

Object type: Plane curve

### Definition

A cardioid is a curve traced out by a point (the 'drawing point') on the perimeter of a circle as the circle is rolling without slipping along the perimeter of another, stationary, circle of the same radius. Below, the common radius is 1, the stationary circle is centred at the origin, and the point of contact between the drawing point and the stationary circle is chosen to be $(1,0)$.

The specific cardioid shown above is the image $\mathbf{r}(\left[0,2\pi\right[)$ where $$\mathbf{r}(t) = \basis\begin{pmatrix}2\cos t-\cos{2t}\\2\sin t-\sin{2t}\end {pmatrix}.$$


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Hello Jetters,

Welcome to another/first edition of JetDAO Community Call Notes. This month’s call happened on the 3rd of August, with James as the participant and Dante as the host. We’ll cover the important points of the call in this article but if you want to listen to the whole call, please head over here.

The call consists of multiple sections in which we heard information about the new ‘’Jet DAO Asset Onboarding/Application’’ processes. In the first section, James told us about the previously onboarded assets, including WBTC, SOL, and USDC, and the mechanism behind their onboarding. Previously, these three asset applications were posted on the Jet Governance Forum by team members. By doing so, the on-chain governance mechanism is tested & the feedback from Jet users was collected. The first onboarded asset types are blue chips and stablecoins. In Jet Protocol V2, James told us that other types of crypto assets might be onboarded via governance as well. This will help Jet users to unlock the features of Solana DeFi.

After some basic explanation about the asset onboarding, James started another section that consists mostly of the details about how a new asset is to be onboarded to the Jet V2. Everything starts with the application form which can be found in the JetDAO GitBook. There are no restrictions neither for applied assets nor for the appliers. Even if you are not from the team of an asset that you’re applying for, it’s possible for you to submit the form. Besides that, there is no rule to hold any amount of $JET tokens to apply for. Anyone with bare hands with empty pockets can apply for the onboarding process. By doing so, the Jet team wishes to reveal the hidden feature of many assets created on the Solana chain. If you wish to see assets on Jet V2, you must clarify the exact benefits & features, first. So, it happens without monetary support. 在对资产上线进行了一些基本解释之后，James 开始讲解另一部分，主要包含如何将新资产加入/上线到 Jet V2 的详细信息。一切从申请表开始，表可以在 JetDAO GitBook 找到。申请的资产和申请人都没有限制。即使您不是来自申请的资产的团队，也可以提交表格。除此之外，没有规定必须持有任何数量的$JET 代币才可以申请。任何”赤手空拳“的人都可以进入上线申请流程。通过这样做，Jet 团队希望揭示Solana 链上创建的许多资产的隐藏特征。如果您希望看到资产在 Jet V2 （上线），您必须首先明确其优势和功能。因此，上述事件的发生并不以金融支持为前提。

From now on, James reminded us that anyone can request help from Discord or Governance Forum before and/or during the application process. Every Jet DAO contributor champ to see new assets so they’re keen to help. During the call, Norman Ives posted into chat that anyone who is interested in risk management may reach out to him, too. Join the Discord server here to start your Jet adventure.

Just before the call ended, James talked about the future improvements in the Jet roadmap. He said that one of the very first changes in the roadmaps will be about the onboarding process. When the DAO gets bigger and stronger, the team wishes to create ‘’working groups’’ for asset onboarding. These are passionate community members that are dedicated to finding better assets for the protocol. By doing so, member reputation will rocket within the community and it helps them for future interventions. And when we think about the processes cumulatively, the more they engage with different communities the easier the process will be. In the end, it results in a totally improved ecosystem! This is best for every Solana lover.

The call was all about the onboarding process! But within the sentences of James, you understand that the V2 launch is so so so close! He said ‘’it’s a matter of days!’’ Please expect to see the Jet V2 launch soon and stay with our community.


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# NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4

These NCERT Solutions for Class 10 Maths Chapter Ex 4.4 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.4

Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x² – 3x + 5 = 0
(ii) 3x2 – 4√3x + 4 = 0
(iii) 2x2-6x + 3 = 0
Solution:
(i) We have,
2x² – 3x + 5 = 0
Here, a = 2, b = – 3 and c = 5
∴ Discriminant (D) = b² – 4ac = (- 3)² – 4 x 2 x 5 = 9 – 40 = – 31
∴ D= – 31
Here, discriminant (D) < 0 Therefore, the given quadratic equation has no real roots.

(ii) We have, 3x² – 4$$\sqrt{3x}$$ + 4 = 0
Here, a = 3, b= – 4$$\sqrt{3x}$$ and c = 4
∴ Discriminant (D) = b² – 4ac = $$(-4 \sqrt{3 x})^{2}$$ x 4 = 48 – 48
∴ Discriminant (D) = 0
Therefore, the given quadratic equation has two equal real roots.

(iii) We have, 2x² – 6x + 3 = 0
Here, a = 2,b = – 6, c = 3
∴ Discriminant (D) = b² – 4ac = (- 6)2 – 4 x 2 x 3 = 36 – 24 = 12
∴ D= 12 Discriminant (D) > 0
Therefore, the given quadratic equation has two distinct real roots. Distinct roots $$\sqrt{3x}$$

Question 2.
Find the value of k for each of the following quadratic equations, so that they have two real equal roots.
(i) 2x² + kx + 3 = 0
(ii) kx(x – 2) + 6 = 0
Solution:
(i) We have,
2x² + kx + 3 = 0
Here, a = 2, b = k and c = 3
∴ Discriminant (D) = b² – 4ac = k² – 4 x 2 x 3
D = k² – 24
But we have given that equation has two real and equal roots.
∴ D = 0
k² – 24 = 0
k² = 24
∴ k = $$\sqrt{24}$$
k = ±2$$\sqrt{6}$$

(ii) We have,
kx (x – 2) + 6 = 0
or, kx² – 2kx + 6 = 0
Here, a = k, b = – 2k and c = 6
∴ Discriminant (D) = b²- 4ac
= (- 2k)² – 4 x k x 6
D= 4k² – 24k
But we have given that equation has two real equal roots.
∴ D = 0
4k² – 24k = 0
or, 4k(k – 6) = 0
So, 4k = 0 or k – 6 = 0
Therefore, k = 0 or k = 6

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800m2 ? If so, find its length and breadth.
Solution:
Let breadth of rectangular mango grove = x
∴ length of rectangular mago grove = 2x
∴ Area of rectangular mango grove = l x b
or 800 = x × 2x [∴ Area of rectangle = Length x Breadth]
or 800 = 2x²
or x = $$\frac { 800 }{ 2 }$$
∴ x = $$\sqrt{400}$$ = ± 20
Since, dimensions cannot be in negative, so we reject the value of x = – 20.
∴ Breadth of rectangular mango grove = x = 20 m
and length of rectangular mango grove = 2x – 2 x 20 = 40m

Question 4.
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Let the present age of one friend be x
∴ the present age of other friend be 20 – x
Four years ago age of first friend = x – 4
and four years ago age of second friend = (20 – x) – 4 = 16 – x
∴ According to question,
(x – 4) (16 – x) = 48
or 16x – x² – 64 + 4x = 48
or – x² + 20x – 64 = 48
or x² – 20x + 64 + 48 = 0
or x² – 20x + 112 = 0
Here, a 1,b = -20 and c = 112
∴ Discriminant (D) b² – 4ac
= (-20)2 – 4 x 1 x 112
= 440 – 448 = -48
∴ D<0
So, the equation has no any real roots.
Therefore, the given situation is not possible.

Question 5.
It is possible to design a rectangulat park of perimeter 80 m and area 400 m²? If so find its length and breadth.
Solution:
Let the length of the rectangular park = x
∴ the breadth of the rectangular park = (40 – x)
∴ Area of rectangular park = Length x Breadth
or 400 = x(40 – x)
or 400 = 40x – x²
or x² – 40x + 400 = 0
Here, a = 1, b = -40 and c = 400
∴ Discriminant (D) = b² – 4ac
= (- 40)² – 4 x 1 x 400
= 1600 – 1600
D = 0
Therefore, equation has two real and equal roots.
\begin{aligned} x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ &=\frac{-(-40) \pm \sqrt{0}}{2 \times 1}=\frac{40}{2}=20 \end{aligned}
So, the length of rectangular park = x = 20 m
and the breadth of rectangular park = (40 -x)
= 40 – 20 = 20 m.

error: Content is protected !!


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In the original paper, Eq. (9) (p. 041701-3) and the expression for parameters λ (p. 041701-4, after Eq. (32)) and ρ (p. 041701-5, after Eq. (40)) are not correct.

$Qx=∂Mx∂x=−2M0g2(ξ)le+Q0g4(ξ)$
$λ=νE(1t1−1t2)2Dk2le22$
$ρ=(1−2β)δ+βωle1$


text:
Biased reward learning
post by Stuart Armstrong 372 days ago | Ryan Carey likes this | discuss

A putative new idea for AI control; index here.

What are the biggest failure modes of reward learning agents?

The first failure mode is when the agent directly (or indirectly) chooses its reward function.

For instance, imagine a domestic robot that can be motivated to tidy (reward $$R_0$$) or cook (reward $$R_1$$). It has a switch that allows the human to choose the correct reward function. However, cooking gives a higher expected reward than tidying, and the agent may choose to set the switch directly (or manipulate the human’s choice). In that case, it will set it to cook’.

In that case, the agent biases its reward learning process.

A second failure mode (this version due to Jessica, original idea here) is when the agent influences its reward function without biasing it.

For example, the domestic robot might be waiting for the human to arrive in an hour’s time. It expected the human will be 50% likely to choose $$R_0$$ (tidying) versus 50% likely to choose $$R_1$$ (cooking). If instead the robot can randomise its reward switch now (with equal odds on $$R_0$$ and $$R_1$$), it can know its reward function early, and get in a full extra hour of tidying/cooking.

A subsequent post will formalise influence, here let’s look at bias.

# Formalising bias

First of all, for a given policy $$\pi$$, we can say that $$\vdh$$ is unbiased for $$\pi$$, if $$\pi$$ preserves the expectation of $$\vdh$$. That is:

• For all histories $$h_t$$ with $$t < m$$, $$\vdh(\cdot\mid h_t)=\mathbb{E}_{\mu}^{\pi}[\vdh(\cdot\mid h_{t+1})\mid h_t]$$.

If the expectation of $$\vdh$$ is preserved by any policy, then we can say that $$\vdh$$ itself is unbiased:

• The prior $$\vdh$$ is unbiased is $$\vdh$$ is unbiased for $$\pi$$ for all policies $$\pi$$.

Recall that $$\vdh=\vd$$ on histories of length $$m$$. So $$\vdh$$ being unbiased implies restrictions on $$\vd$$:

• If $$\vdh$$ is unbiased, then for all $$h_t$$ with $$t<m$$ and for all policies $$\pi$$, $$\vdh(\cdot \mid h_t)=\mathbb{E}_{\mu}^{\pi}[\vd(\cdot\mid h_m)\mid h_t]$$.

Since $$\vdh$$ being unbiased imposes restrictions on $$\vd$$, we can directly define:

• The posterior $$\vd$$ is unbiased if there exists a possible prior $$\vdh'$$ with $$\vdh'=\vd$$ on histories of length $$m$$, and $$\vdh'$$ is unbiased.

So what does unbiased mean in practice? It simply means that whatever actions or policies the agent follows, they cannot change the expectation of their values.

# Bias and learning incentives

This is an opportunity to put the learning and biasing graph:

The x-axis represents the probability $$\vdh(R_1\mid h_t)$$ of $$R_1$$ being the correct’ reward function. The current value is $$\vdh(R_1\mid h_t)=p$$.

The orange curve (which is always convex, though not necessarily strictly so) represents, for a given probability $$q$$ of correctness of $$R_1$$, the expected value the agent could get if it knew it would never learn anything more about the correct value.

If it learnt immediately and costlessly about the correct values, it would go to $$q=0$$ or $$q=1$$ with probability $$1-p$$ and $$p$$, respectively. Thus its expected reward is the point on the blue curve at the x-coordinate $$p$$.

Thus the green arrow represents the incentive to learn. But, if it can’t learn easily, it may try and randomise its reward function, so the green arrow also represents the incentive to (unbiased) influence.

The shape of the orange curve itself represents the incentive to bias.

If the orange curve is flat, it is equal to the blue one, so there is no incentive to learn. If the orange curve is flat and horizontal, there is no incentive to bias, either.

### NEW DISCUSSION POSTS

This is exactly the sort of
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text:
# How could I optimize the following high-dimensional table with a lot of duplicates?

Union[FullSimplify[Flatten[
Table[Sum[Exp[2 \[Pi] I Subscript[n, i]/5], {i, 0, 6}],
{Subscript[n, 0], 0, 4}, {Subscript[n, 1], 0, 4},
{Subscript[n, 2], 0, 4}, {Subscript[n, 3], 0, 4},
{Subscript[n, 4], 0, 4}, {Subscript[n, 5], 0, 4},
{Subscript[n, 6], 0, 4}]]]]

This Table goes through a lot of redundant calculations, blows up unnecessarily and then throws them away. Is there a more efficient way to just generate unique values of this kind?

All I'm trying to do here is to generate a list of complex points that can be expressed as arbitrary sums of, in this case, six fifth roots of unity.

That job still finished in a reasonable amount of time, but I'd like to do the same thing for a higher number of higher-order roots of unity.
Ideally, I'd like to be able to generate the generic limiting pattern, if it exists, for an infinite number of n-order roots. - for 2nd, 3rd, 4th and 6th order roots that generates, respectively, all integers, a triangular grid, a square grid and another triangular grid. I wonder how cases look like that can't have such regular grids - so any other Integer, essentially.

-
Please post working code - this entails getting rid of all the _ and ensuring correct syntax in general. –  Yves Klett Feb 21 '13 at 14:09
@YvesKlett This now is straight from Mathematica. I just wanted to make it more readible. –  kram1032 Feb 21 '13 at 14:15
well, fancy syntax FullForm obfuscation is not optimal (you could format it for easy reading yet), but incorrect code is much worse still. BTW, your code is still not working. Copy and paste back into your session to make sure it works. –  Yves Klett Feb 21 '13 at 14:21
@YvesKlett as said, it should work now. Unless I somehow can't directly copy Mathematica code from Mathematica 8 and post it as-is in a codeblock –  kram1032 Feb 21 '13 at 14:23
I just realized there might be a much simpler way to solve this particular kind of problem but even so, if there is a generic way to skip redundant steps in Tables like this, it should be a valuable question non-the-less. –  kram1032 Feb 21 '13 at 14:52

Much faster:

t = Table[Exp[2 I Pi n/5], {n, 0, 4}]
u = Union[Total /@ Tuples[t, 7]]

ListPlot[{Re@#, Im@#} & /@  u, AspectRatio -> 1]

Edit

You may look at the points added each time you add a term to the sum:

ListPlot[Transpose /@ ({Re@#, Im@#} & /@ (Union[Total /@ Tuples[t, #]] & /@
Reverse@Range@8)), AspectRatio -> 1, PlotStyle -> Table[Hue[1/x], {x, 8}]]

-
perfect, thank you! –  kram1032 Feb 21 '13 at 15:43
For easier generalization, is it the same to do Tuples[t,7] compared to Tuples[{t,t,t,t,t,t,t}]? - at least the resulting image seems to be the same. That way you'd in general get: t = Table[Exp[2 I Pi i/k], {i, 0, k-1}]; u = Union[Total /@ Tuples[t,n]]; ListPlot[{Re@#, Im@#} & /@ u, AspectRatio -> 1, Axes->None] for sums over n order-k roots of unity. –  kram1032 Feb 21 '13 at 16:02
@kram1032 Yup, the same thing –  belisarius Feb 21 '13 at 16:19


text:
# Unbiased Estimation

Printer-friendly version

On the previous page, we showed that if Xi are Bernoulli random variables with parameter p, then:

$\hat{p}=\dfrac{1}{n}\sum\limits_{i=1}^n X_i$

is the maximum likelihood estimator of p. And, if Xi are normally distributed random variables with mean μ and variance σ2, then:

$\hat{\mu}=\dfrac{\sum X_i}{n}=\bar{X}$  and   $\hat{\sigma}^2=\dfrac{\sum(X_i-\bar{X})^2}{n}$

are the maximum likelihood estimators of μ and σ2, respectively. A natural question then is whether or not these estimators are "good" in any sense. One measure of "good" is "unbiasedness."

 Definition. If the following holds: $E[u(X_1,X_2,\ldots,X_n)]=\theta$ then the statistic $u(X_1,X_2,\ldots,X_n)$ is an unbiased estimator of the parameter θ. Otherwise,  $u(X_1,X_2,\ldots,X_n)$ is a biased estimator of θ.

### Example

If Xi is a Bernoulli random variable with parameter p, then:

$\hat{p}=\dfrac{1}{n}\sum\limits_{i=1}^nX_i$

is the maximum likelihood estimator (MLE) of p. Is the MLE of p an unbiased estimator of p?

Solution. Recall that if Xi is a Bernoulli random variable with parameter p, then E(Xi) = p. Therefore:

$E(\hat{p})=E\left(\dfrac{1}{n}\sum\limits_{i=1}^nX_i\right)=\dfrac{1}{n}\sum\limits_{i=1}^nE(X_i)=\dfrac{1}{n}\sum\limits_{i=1}^np=\dfrac{1}{n}(np)=p$

The first equality holds because we've merely replaced p-hat with its definition. The second equality holds by the rules of expectation for a linear combination. The third equality holds because E(Xi) = p. The fourth equality holds because when you add the value p up n times, you get np. And, of course, the last equality is simple algebra.

In summary, we have shown that:

$E(\hat{p})=p$

Therefore, the maximum likelihood estimator is an unbiased estimator of p.

### Example

If Xi are normally distributed random variables with mean μ and variance σ2, then:

$\hat{\mu}=\dfrac{\sum X_i}{n}=\bar{X}$  and   $\hat{\sigma}^2=\dfrac{\sum(X_i-\bar{X})^2}{n}$

are the maximum likelihood estimators of μ and σ2, respectively. Are the MLEs unbiased for their respective parameters?

Solution. Recall that if Xi is a normally distributed random variable with mean μ and variance σ2, then E(Xi) = μ and Var(Xi) = σ2 . Therefore:

$E(\bar{X})=E\left(\dfrac{1}{n}\sum\limits_{i=1}^nX_i\right)=\dfrac{1}{n}\sum\limits_{i=1}^nE(X_i)=\dfrac{1}{n}\sum\limits_{i=1}\mu=\dfrac{1}{n}(n\mu)=\mu$

The first equality holds because we've merely replaced X-bar with its definition. Again, the second equality holds by the rules of expectation for a linear combination. The third equality holds because E(Xi) = μ. The fourth equality holds because when you add the value μ up n times, you get nμ. And, of course, the last equality is simple algebra.

In summary, we have shown that:

$E(\bar{X})=\mu$

Therefore, the maximum likelihood estimator of μ is unbiased. Now, let's check the maximum likelihood estimator of σ2. First, note that we can rewrite the formula for the MLE as:

$\hat{\sigma}^2=\left(\dfrac{1}{n}\sum\limits_{i=1}^nX_i^2\right)-\bar{X}^2$

because:

Then, taking the expectation of the MLE, we get:

$E(\hat{\sigma}^2)=\dfrac{(n-1)\sigma^2}{n}$

as illustrated here:

\begin{align}
E(\hat{\sigma}^2) &= E\left[\dfrac{1}{n}\sum\limits_{i=1}^nX_i^2-\bar{X}^2\right]=\left[\dfrac{1}{n}\sum\limits_{i=1}^nE(X_i^2)\right]-E(\bar{X}^2)\\
&= \dfrac{1}{n}\sum\limits_{i=1}^n(\sigma^2+\mu^2)-\left(\dfrac{\sigma^2}{n}+\mu^2\right)\\
&= \dfrac{1}{n}(n\sigma^2+n\mu^2)-\dfrac{\sigma^2}{n}-\mu^2\\
&= \sigma^2-\dfrac{\sigma^2}{n}=\dfrac{n\sigma^2-\sigma^2}{n}=\dfrac{(n-1)\sigma^2}{n}\\
\end{align}

The first equality holds from the rewritten form of the MLE. The second equality holds from the properties of expectation. The third equality holds from manipulating the alternative formulas for the variance, namely:

$Var(X)=\sigma^2=E(X^2)-\mu^2$ and   $Var(\bar{X})=\dfrac{\sigma^2}{n}=E(\bar{X}^2)-\mu^2$

The remaining equalities hold from simple algebraic manipulation. Now, because we have shown:

$E(\hat{\sigma}^2) \neq \sigma^2$

the maximum likelihood estimator of σis a biased estimator.

### Example

If Xi are normally distributed random variables with mean μ and variance σ2, what is an unbiased estimator of σ2? Is S2 unbiased?

Solution. Recall that if Xi is a normally distributed random variable with mean μ and variance σ2, then:

$\dfrac{(n-1)S^2}{\sigma^2}\sim \chi^2_{n-1}$

Also, recall that the expected value of a chi-square random variable is its degrees of freedom. That is, if:

$X \sim \chi^2_{(r)}$

then E(X) = r. Therefore:

$E(S^2)=E\left[\dfrac{\sigma^2}{n-1}\cdot \dfrac{(n-1)S^2}{\sigma^2}\right]=\dfrac{\sigma^2}{n-1} E\left[\dfrac{(n-1)S^2}{\sigma^2}\right]=\dfrac{\sigma^2}{n-1}\cdot (n-1)=\sigma^2$

The first equality holds because we effectively multiplied the sample variance by 1. The second equality holds by the law of expectation that tells us we can pull a constant through the expectation. The third equality holds because of the two facts we recalled above. That is:

$E\left[\dfrac{(n-1)S^2}{\sigma^2}\right]=n-1$

And, the last equality is again simple algebra.

In summary, we have shown that, if Xi is a normally distributed random variable with mean μ and variance σ2, then S2 is an unbiased estimator of σ2. It turns out, however, that S2 is always an unbiased estimator of σ2, that is, for any model, not just the normal model. (You'll be asked to show this in the homework.) And, although S2 is always an unbiased estimator of σ2, S is not an unbiased estimator of σ. (You'll be asked to show this in the homework, too.)

Sometimes it is impossible to find maximum likelihood estimators in a convenient closed form. Instead, numerical methods must be used to maximize the likelihood function. In such cases, we might consider using an alternative method of finding estimators, such as the "method of moments." Let's go take a look at that method now.


text:
# Expanding Gamma function around poles

1. ### DMESONS

27
Can someone help me to expand the following gamma functions around the pole ε, at fisrt order in ε

$\Gamma[(1/2) \pm (ε/2)]$

where ε= d-4

2. ### Bill_K

4,159
Γ(½ ± ε/2) ≈ Γ(½) ± ε/2 Γ'(½)

No, seriously..

Well, you also need to use the digamma function, ψ(x) = Γ'(x)/Γ(x). And the values Γ(½) = √π and ψ(½) = - γ - 2 ln 2 where γ is Euler's constant.

3. ### Hepth

508
$$\Gamma(\frac{1}{2} - \frac{\epsilon}{2}) = \sqrt{\pi }+\frac{1}{2} \sqrt{\pi } \epsilon (\gamma_E +\log (4))+O\left(\epsilon ^2\right)$$

$$\Gamma(\frac{1}{2} + \frac{\epsilon}{2}) = \sqrt{\pi }+\frac{\sqrt{\pi } \epsilon \psi ^{(0)}\left(\frac{1}{2}\right)}{2}+O\left(\epsilon ^2\right)$$

4. ### DMESONS

27
Bill_K and Hepth, I am so grateful for your help

I am new in this subject


text:
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

by Siddarth Ramesh, Data Scientist, Microsoft

I’m an R programmer. To me, R has been great for data exploration, transformation, statistical modeling, and visualizations. However, there is a huge community of Data Scientists and Analysts who turn to Python for these tasks. Moreover, both R and Python experts exist in most analytics organizations, and it is important for both languages to coexist.

Many times, this means that R coders will develop a workflow in R but then must redesign and recode it in Python for their production systems. If the coder is lucky, this is easy, and the R model can be exported as a serialized object and read into Python. There are packages that do this, such as pmml. Unfortunately, many times, this is more challenging because the production system might demand that the entire end to end workflow is built exclusively in Python. That’s sometimes tough because there are aspects of statistical model building in R which are more intuitive than Python.

Python has many strengths, such as its robust data structures such as Dictionaries, compatibility with Deep Learning and Spark, and its ability to be a multipurpose language. However, many scenarios in enterprise analytics require people to go back to basic statistics and Machine Learning, which the classic Data Science packages in Python are not as intuitive as R for. The key difference is that many statistical methods are built into R natively. As a result, there is a gap for when R users must build workflows in Python. To try to bridge this gap, this post will discuss a relatively new package developed by Microsoft, revoscalepy

### Why revoscalepy?

Revoscalepy is the Python implementation of the R package RevoScaleR included with Microsoft Machine Learning Server.

The methods in ‘revoscalepy’ are the same, and more importantly, the way the R user can view data is the same. The reason this is so important is that for an R programmer, being able to understand the data shape and structure is one of the challenges with getting used to Python. In Python, data types are different, preprocessing the data is different, and the criteria to feed the processed dataset into a model is different.

To understand how revoscalepy eases the transition from R to Python, the following section will compare building a decision tree using revoscalepy with building a decision tree using sklearn. The Titanic dataset from Kaggle will be used for this example. To be clear, this post is written from an R user’s perspective, as many of the challenges this post will outline are standard practices for native Python users.

### revoscalepy versus sklearn

revoscalepy works on Python 3.5, and can be downloaded as a part of Microsoft Machine Learning Server. Once downloaded, set the Python environment path to the python executable in the MML directory, and then import the packages.

The first chunk of code imports the revoscalepy, numpy, pandas, and sklearn packages, and imports the Titatic data. Pandas has some R roots in that it has its own implementation of DataFrames as well as methods that resemble R’s exploratory methods.

import revoscalepy as rp;
import numpy as np;
import pandas as pd;
import sklearn as sk;

### Preprocessing with sklearn

One of the challenges as an R user with using sklearn is that the decision tree model for sklearn can only handle the numeric datatype. Pandas has a categorical type that looks like factors in R, but sklearn’s Decision Tree does not integrate with this. As a result, numerically encoding the categorical data becomes a mandatory step. This example will use a one-hot encoder to shape the categories in a way that sklearn’s decision tree understands.

The side effect of having to one-hot encode the variable is that if the dataset contains high cardinality features, it can be memory intensive and computationally expensive because each category becomes its own binary column. While implementing one-hot encoding itself is not a difficult transformation in Python and provides good results, it is still an extra step for an R programmer to have to manually implement. The following chunk of code detaches the categorical columns, label and one-hot encodes them, and then reattaches the encoded columns to the rest of the dataset.

from sklearn import tree
le = sk.preprocessing.LabelEncoder()
x = titanic_data.select_dtypes(include=[object])
x = x.drop(['Name', 'Ticket', 'Cabin'], 1)
x = pd.concat([x, titanic_data['Pclass']], axis = 1)
x['Pclass'] = x['Pclass'].astype('object')
x = pd.DataFrame(x)
x = x.fillna('Missing')
x_cats = x.apply(le.fit_transform)
enc = sk.preprocessing.OneHotEncoder()
enc.fit(x_cats)
onehotlabels = enc.transform(x_cats).toarray()
encoded_titanic_data =
pd.concat([pd.DataFrame(titanic_data.select_dtypes(include=[np.number])),
pd.DataFrame(onehotlabels)], axis = 1)

At this point, there are more columns than before, and the columns no longer have semantic names (they have been enumerated). This means that if a decision tree is visualized, it will be difficult to understand without going through the extra step of renaming these columns. There are techniques in Python to help with this, but it is still an extra step that must be considered.

### Preprocessing with revoscalepy

Unlike sklearn, revoscalepy reads pandas’ ‘category’ type like factors in R. This section of code iterates through the DataFrame, finds the string types, and converts those types to ‘category’. In pandas, there is an argument to set the order to False, to prevent ordered factors.

titanic_data_object_types = titanic_data.select_dtypes(include = ['object'])
titanic_data_object_types_columns = np.array(titanic_data_object_types.columns)
for column in titanic_data_object_types_columns:
titanic_data[column] = titanic_data[column].astype('category', ordered = False)
titanic_data['Pclass'] = titanic_data['Pclass'].astype('category', ordered = False)

This dataset is already ready to be fed into the revoscalepy model.

### Training models with sklearn

One difference between implementing a model in R and in sklearn in Python is that sklearn does not use formulas.

Formulas are important and useful for modeling because they provide a consistent framework to develop models with varying degrees of complexity. With formulas, users can easily apply different types of variable cases, such as ‘+’ for separate independent variables, ‘:’ for interaction terms, and ‘*’ to include both the variable and its interaction terms, along with many other convenient calculations. Within a formula, users can do mathematical calculations, create factors, and include more complex entities third order interactions. Furthermore, formulas allow for building highly complex models such as mixed effect models, which are next to impossible build without them. In Python, there are packages such as ‘statsmodels’ which have more intuitive ways to build certain statistical models. However, statsmodels has a limited selection of models, and does not include tree based models.

With sklearn, model.fit expects the independent and dependent terms to be columns from the DataFrame. Interactions must be created manually as a preprocessing step for more complex examples. The code below trains the decision tree:

model = tree.DecisionTreeClassifier(max_depth = 50)
x = encoded_titanic_data.drop(['Survived'], 1)
x = x.fillna(-1)
y = encoded_titanic_data['Survived']
model = model.fit(x,y)

### Training models with revoscalepy

revoscalepy brings back formulas. Granted, users cannot view the formula the same way as they can in R, because formulas are strings in Python. However, importing code from R to Python is an easy transition because formulas are read the same way in the revoscalepy functions as the model fit functions in R. The below code fits the Decision Tree in revoscalepy.

#rx_dtree works with formulas, just like models in R
form = 'Survived ~ Pclass + Sex + Age  + Parch  + Fare + Embarked'
titanic_data_tree = rp.rx_dtree(form, titanic_data, max_depth = 50)

The resulting object, titanic_data_tree, is the same structural object that RxDTree() would create in R. Because the individual elements that make up the rx_dtree() object are the same as RxDTree(), it allows R users to easily understand the decision tree without having to translate between two object structures.

### Conclusion

From the workflow, it should be clear how revoscalepy can help with transliteration between R and Python. Sklearn has different preprocessing considerations because the data must be fed into the model differently. The advantage to revoscalepy is that R programmers can easily convert their R code to Python without thinking too much about the ‘Pythonic way’ of implementing their R code. Categories replace factors, rx_dtree() reads the R-like formula, and the arguments are similar to the R equivalent. Looking at the big picture, revoscalepy is one way to ease Python for the R user and future posts will cover other ways to transition between R and Python.

Microsoft Docs: Introducing revoscalepy


text:
# Solving linear inequalities over rings

The concrete problem: for any given $N\ge 1$ I have a system of $2^N-1$ linear inequalities over $\mathbb{Z}_6^N$ which looks like this: for every nonempty $S\subseteq[N]$ there is some $b_S\in\mathbb{Z}_6$ and the inequality $\sum_{i\in S}x_i\ne b_S$. I want to find a solution to all the inequalities at once, of course.

Is there an efficient way to find solution to such a system? To count the number of solutions? To check whether a solution exists? Also, what about more general cases (any number of inequalities, any type of ring, no restriction on the coefficients of the variables in the inequalities)?

-
What order are you using in the inequalities? – user23211 May 13 '12 at 9:54
I'm afraid I don't understand the question. – Gadi A May 13 '12 at 10:02
In an inequality you use the symbol $\leq$ or a similar one. What does the symbol mean in this case? – user23211 May 13 '12 at 10:07
Maybe "inequality" is the wrong word (but I don't know what's the right word). There's no order here, only expressions of the form $x\ne y$ (not equal). – Gadi A May 13 '12 at 11:41
Oh, I see. Sorry, I misunderstood your question. – user23211 May 13 '12 at 11:48


text:
Home OALib Journal OALib PrePrints Submit Ranking News My Lib FAQ About Us Follow Us+
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Search Results: 1 - 10 of 100 matches for " "
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 Physics , 2013, DOI: 10.1007/s11207-013-0360-2 Abstract: Some studies have shown that our knowledge on solar activity in the years 1640, 1652, and 1741 can be improved. In this contribution, we revise the annual group sunspot numbers for these years from original observations. For the years 1640, 1652 and 1741, we have obtained the corrected values 15.2, 1.8, and 27.3, respectively (instead of the original values 15.0, 4.0, and 57.7).
 Fernanda Fioravante Diálogos , 2010, DOI: 10.4025/dialogos.v13i3.436 Abstract: O presente artigo tem por objetivo discutir a receita e as despesas da camara de S o Jo o del Rei entre os anos de 1719 e 1750. Para tanto, procurei realizar uma análise comparativa com outras camaras em diferentes partes do Império Português, focando principalmente a camara de Vila Rica entre os anos de 1721 e 1734. Assim sendo, foram pesquisadas as “rela es de receita e despesa” das respectivas vilas. The present article aims to discuss the revenue and expenses of the S o Jo o del Rei municipal council between the years 1719 and 1750. To that end, I sought to make a comparative analysis with other municipal councils in different parts of the Portuguese Empire, focusing primarily on the Vila Rica council between the years 1721 and 1734. The “revenue and expense relationships” of the respective villages were analyzed. Este artículo tiene por objetivo discutir los ingresos y egresos de la Cámara de S o Jo o del Rei, entre 1719 y 1750. Para ello, busqué realizar un análisis comparativo con otras cámaras municipales de diferentes partes del Imperio Portugués, especialmente, con la Cámara municipal de Vila Rica entre los a os 1721 y 1734. Por lo tanto, fueron investigadas las “relaciones de rendimientos y gastos” de ambas localidades.
 Climate of the Past Discussions , 2013, DOI: 10.5194/cpd-9-1013-2013 Abstract: This paper advances the current debate on causes and effects of famines. Since Sen's food entitlement decline theory emerged in the 1980's, climate and environmental factors are widely excluded in famine analysis. Studying the causation and the processes of famines as well as the adaptations to it before the 20th century will enhance modern famine theories and lead to a rethinking of the role of climate/environmental aspects in current research. In our case study, the "Famine Vulnerability Analysis Model" (FVAM) serves as an explanatory model and will open up new perspectives on famines. Special emphasis will be put on the Europe-wide crises of 1740–1741, with a focus on the famine of the "great frost" in Ireland. The interaction of demographic, political, economic and environmental aspects is characteristic in this famine.
 Physics , 2015, Abstract: We present the results of two Chandra X-ray Observatory ( CXO) observations of TeV gamma-ray source HESS J1741-302A/B. Our analysis also includes GeV gamma-ray and radio data. We investigate whether there is any connection between HESS~J1741-302A/B and the sources seen at lower energies. One of the brightest X-ray sources in the HESS J1741-302B field, CXOU~J174112.1-302908, appears to be associated with a low-mass star (possibly representing a quiescent LMXB or CV), hence, it is unlikely to be a source of TeV gamma-rays. In the same field we detected X-rays from WR 98a, which is likely to be a colliding wind binary with massive stars, however, no TeV emission has been reported so far from such systems although a predictions have been made. Finally, Suzaku source J1740.5-3014 (which is not covered by the CXO observations) appears to be a hard X-ray source detected by INTERGAL ISGRI which supports the magnetized CV classification and also makes its association with the TeV emission unlikely. The young pulsar, undetected in X-rays and located near the CV, may be the contributor of relativistic particles responsible for the TeV emission. Alternatively, HESS J1741-302 could be a new type of accelerator, which is dark at lower energies.
 J.P. van Dijk Tijdschrift voor de Geschiedenis der Geneeskunde, Natuurwetenschappen, Wiskunde en Techniek , 1982, Abstract: Classifications of death causes from 1750 till 1900 The systematic registration of the causes of death takes place in the Netherlands and other European countries in the middle of the nineteenth century. This is preceded by a period in which the registration of causes of death is a local matter, in other words the lists of causes of death differ per town. So the registration in the Netherlands begins in The Hague in the year 1755, and does not become national until 1865 when the Medical Inspectorate of Public Health comes into being. Complete centralization takes place after the Central Bureau of Statistics has taken over the registration in 1900.
 Physics , 1999, DOI: 10.1086/308779 Abstract: (Abridged) We report multi-epoch VLBI observations of the source PKS 1741-038 as it underwent an extreme scattering event. Observations at four epochs were obtained, and images were produced at three of these. During the event the source consisted of a dominant, compact component, essentially identical to the structure seen outside the event. However, the source's diameter increased slightly at 13 cm during the ESE. An increase in the source's diameter is inconsistent with a simple refractive model. We also see no evidence for ESE-induced substructure within the source or the formation of multiple images, as would occur in a strongly refractive lens. However, a model in which the decrease in flux density during the ESE occurs solely because of stochastic broadening within the lens requires a larger broadening diameter during the ESE than is observed. Thus, the ESE toward 1741-038 involved both stochastic broadening and refractive defocussing within the lens. If the structure responsible for the ESE has a size of order 1 AU, the level of scattering within an ESE lens may be a factor of 10^7 larger than that in the ambient medium. A filamentary structure could reduce the difference between the strength of scattering in the lens and ambient medium, but we conclude that, if ESEs arise from filamentary structures, they occur when the filamentary structures are seen lengthwise. We predict the amount of pulse broadening that would result from a comparable lens passing in front of a pulsar. The pulse broadening would be no more than 1.1 microseconds, consistent with the lack of pulse broadening detected during ESEs toward the pulsars PSR B1937+21 and PSR J1643-1224.
 Alexandros M. Goulielmos Modern Economy (ME) , 2017, DOI: 10.4236/me.2017.82022 Abstract: The theory of “long waves”, due to Russian economist Kondratieff, which appeared in the 1920s, is presented, and a search whether a similar pattern exists in shipping economy since 1741 is made. Kondratieff believed in the existence —inherent in capitalistic system—of at least 21/2 long cycles of a non-random duration of 54 years on average since 1780—with 27 years up and 27 years down. As we showed—using mainly a diagrammatic analysis—economic history verified—till end-2008—Kondratieff’s theories, though it rejected his claimed harmony in their duration. Given that history may be repeated, and only for this is useful, we tried a nonlinear dynamic forecasting model for dry cargo shipping market for 20 years ahead using “Rescaled Range Analysis” and “Kernel Density estimation method” and found that cycles are here to stay. Presented also is the merger-waves in capitalistic economies since 1893— a factor ignored by Joan Robinson in her macroeconomic analysis for the trade cycle in the 1960s. Further we confirm/reject previous analysts, including Schumpeter, who claimed: 1) that technology has created the long shipping cycles (since 1741); 2) that the impact of major wars since 1740 on freight markets was another cyclical cause; 3) that economies of scale play the significant role and 4) that the impact of risk emerged at a certain time though this outcome was too ignored by macroeconomists. The paper “discovered” the secular shipping cycles of a total duration of 139 years using the proper extra-long time data. This means that the discovery of unknown cycles is a matter of data’s longevity. We have also verified Kondratieff’s cycles—when data were divided in 3 separate technological periods: 1741-1871; 1872-1947 and 1947-today, but lasting longer i.e. 58 years on average. We saw that the great gift of capitalism, which is the “free will”, is to be accompanied by “rationality”, and by a deep knowledge of how capitalistic system functions—a warning placed by Kondratieff a century ago—so that to avoid crises like that at end-2008; also globalization has to be used for the benefit of all nations. This paper indicated new tools: the “Joseph effect”—for cycles and the “Hurst exponent” for
 Physics , 2000, DOI: 10.1086/318262 Abstract: We report multi-epoch VLA H I absorption observations of the source 1741-038 (OT-068) before and during an extreme scattering event (ESE). Observations at four epochs, three during the ESE, were obtained. We find no changes in the equivalent width, maximum optical depth, or velocity of maximum optical depth during the ESE, but we do find a secular trend of decreasing maximum optical depth between our observations and ones by other observers a decade prior. The resulting limit on the H I column density change during the ESE for a structure with a spin temperature T_s is 6.4 x 10^{17} cm^{-2} (T_s/10 K). Tiny-scale atomic structures (TSAS), with a column density N_H ~ 3 x 10^{18} cm^{-2}, are ruled out marginally by this limit, though geometric arguments may allow this limit to be relaxed. Galactic halo molecular clouds, that are opaque in the H I line, cannot be excluded because the observed velocity range covers only 25% of their allowed velocity range.
 Physics , 2003, DOI: 10.1086/378851 Abstract: We report Chandra and XMM-Newton observations of the transient neutron star low-mass X-ray binary GRS 1741.9-2853. Chandra detected the source in outburst on 2000 October 26 at an X-ray luminosity of ~10^{36} erg/s (2--8 keV; 8 kpc), and in quiescence on 2001 July 18 at ~10^{32} erg/s. The latter observation is the first detection of GRS 1741.9-2853 in quiescence. We obtain an accurate position for the source of 17h 45m 2.33s, -28o 54' 49.7" (J2000), with an uncertainty of 0.7". GRS 1741.9-2853 was not detected significantly in three other Chandra observations, nor in three XMM-Newton observations, indicating that the luminosity of the source in quiescence varies by at least a factor of 5 between (< 0.9 - 5.0) \times 10^{32} erg/s (2--8 keV). A weak X-ray burst with a peak luminosity of 5 \times 10^{36} erg/s above the persistent level was observed with Chandra during the outburst on 2000 October 26. The energy of this burst, 10^{38} erg, is unexpectedly low, and may suggest that the accreted material is confined to the polar caps of the neutron star. A search of the literature reveals that GRS 1741.9-2853 was observed in outburst with ASCA in Fall 1996 as well, when the BeppoSAX WFC detected the three previous X-ray bursts from this source. The lack of X-ray bursts from GRS 1741.9-2853 at other epochs suggests that it produces bursts only during transient outbursts when the accretion rate onto the surface of the neutron star is about 10^{-10} M_sun/yr. A similar situation may hold for other low-luminosity bursters recently identified from WFC data.
 Physics , 2002, DOI: 10.1086/341333 Abstract: Hickson Compact Group (HCG) 31, consisting of the Wolf-Rayet galaxy NGC 1741 and its irregular dwarf companions, was observed using the Infrared Space Observatory. The deconvolved ISOCAM maps of the galaxies using the 7.7 micron and 14.3 micron (LW6 and LW3) filters are presented, along with ISOPHOT spectrometry of the central starburst region of NGC 1741 and the nucleus of galaxy HCG 31A. Strong mid-IR emission was detected from the central burst in NGC 1741, along with strong PAH features and a blend of features including [S IV] at 10.5 micron. The 14.3/6.75 micron flux ratio, where the 6.75 micron flux was synthesized from the PHT-S spectrum, and 14.3/7.7 micron flux ratios suggest that the central burst within NGC 1741 may be moving towards the post-starburst phase. Diagnostic tools including the ratio of the integrated PAH luminosity to the 40 to 120 micron infrared luminosity and the far-infrared colours reveal that despite the high surface brightness of the nucleus, the properties of NGC 1741 can be explained in terms of a starburst and do not require the presence of an AGN. The Tycho catalogue star TYC 04758-466-1, with m$_{V}$ = 11.3 and spectral type F6, was detected at 7.7 and 14.3 microns.
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Defective 2-colorings of sparse graphs. (English) Zbl 1282.05041
Summary: A graph $$G$$ is $$(j,k)$$-colorable if its vertices can be partitioned into subsets $$V_1$$ and $$V_2$$ such that every vertex in $$G[V_1]$$ has degree at most $$j$$ and every vertex in $$G[V_2]$$ has degree at most $$k$$. We prove that if $$k\geqslant 2j+2$$, then every graph with maximum average degree at most $$2\left(2-\frac{k+2}{(j+2)(k+1)}\right)$$ is $$(j,k)$$-colorable. On the other hand, we construct graphs with the maximum average degree arbitrarily close to $$2\left(2-\frac{k+2}{(j+2)(k+1)}\right)$$ (from above) that are not $$(j,k)$$-colorable.In fact, we prove a stronger result by establishing the best possible sufficient condition for the $$(j,k)$$-colorability of a graph $$G$$ in terms of the minimum, $$\phi_{j,k}(G)$$, of the difference $$\phi_{j,k}(W,G)=\left(2-\frac{k+2}{(j+2)(k+1)}\right)| W| -| E(G[W])|$$ over all subsets $$W$$ of $$V(G)$$. Namely, every graph $$G$$ with $$\phi_{j,k}(G)>\frac{-1}{k+1}$$ is $$(j,k)$$-colorable. On the other hand, we construct infinitely many non-$$(j,k)$$-colorable graphs $$G$$ with $$\phi_{j,k}(G)=\frac{-1}{k+1}$$.

MSC:
 05C15 Coloring of graphs and hypergraphs 05C42 Density (toughness, etc.) 05C35 Extremal problems in graph theory 05C07 Vertex degrees
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References:
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