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### #ActualShaquil

Posted 05 November 2012 - 09:33 PM

I've embedded Python into C++ successfully (without even remotely as much pain and agony as with Lua). Now when I run a simple test program, the console window opens and closes immediately. It definitely executes the way it should with no problems, however I seem to be unable to stop it from closing immediately. I've called system("PAUSE") and added cin >> variable so the program basically has absolutely no choice but to wait for input, yet these things are somehow ignored. I'm on Windows 7 using Visual Studio 2012. Here's the code:

#include <Python.h>
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
Py_SetProgramName(argv[0]);  /* optional but recommended */
Py_Initialize();
PyRun_SimpleString("from time import time,ctime\n"
"print 'Today is',ctime(time())\n");
Py_Finalize();
int x = 0;
cin >> x;
system("PAUSE");

return 0;
}

Any idea what's going on here? Oddly, when I place that code that's supposed to pause it

int x = 0;
cin >> x;
system("PAUSE");

in BEFORE all the python code, it does actually stop. Even worse: I've used breakpoints. If placed on or before the call to Py_Initialize(), the code stops at the breakpoint. Anywhere else, the breakpoint is never reached. Strangely, VC++ will usually tell you if a breakpoint will never be reached, however it seems to believe that the breakpoints placed after Py_Initialize() will be reached.

UPDATE::

So apparently there's an error occuring. I barely noticed it. What happens is, a message is output "Import Error: No module named site." The program then closes immediately. Any ideas? I'm googling now, and I'll be back with another update if I find an answer, for future readers.

### #3Shaquil

Posted 05 November 2012 - 09:33 PM

I've embedded Python into C++ successfully (without even remotely as much pain and agony as with Lua). Now when I run a simple test program, the console window opens and closes immediately. It definitely executes the way it should with no problems, however I seem to be unable to stop it from closing immediately. I've called system("PAUSE") and added cin >> variable so the program basically has absolutely no choice but to wait for input, yet these things are somehow ignored. I'm on Windows 7 using Visual Studio 2012. Here's the code:

#include <Python.h>
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
Py_SetProgramName(argv[0]);  /* optional but recommended */
Py_Initialize();
PyRun_SimpleString("from time import time,ctime\n"
"print 'Today is',ctime(time())\n");
Py_Finalize();
int x = 0;
cin >> x;
system("PAUSE");

return 0;
}

Any idea what's going on here? Oddly, when I place that code that's supposed to pause it

int x = 0;
cin >> x;
system("PAUSE");

in BEFORE all the python code, it does actually stop. Even worse: I've used breakpoints. If placed on or before the call to Py_Initialize(), the code stops at the breakpoint. Anywhere else, the breakpoint is never reached. Strangely, VC++ will usually tell you if a breakpoint will never be reached, however it seems to believe that the breakpoints placed after Py_Initialize() will be reached.

UPDATE::

So apparently there's an error occuring. I barely noticed it. What happens is, a message is output "Import Error: No module named site." The program then closes immediately. Any ideas? I'm googling now, and I'll be back with another update if I find an answer, for future readers.

### #2Shaquil

Posted 05 November 2012 - 09:33 PM

I've embedded Python into C++ successfully (without even remotely as much pain and agony as with Lua). Now when I run a simple test program, the console window opens and closes immediately. It definitely executes the way it should with no problems, however I seem to be unable to stop it from closing immediately. I've called system("PAUSE") and added cin >> variable so the program basically has absolutely no choice but to wait for input, yet these things are somehow ignored. I'm on Windows 7 using Visual Studio 2012. Here's the code:

#include <Python.h>
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
Py_SetProgramName(argv[0]);  /* optional but recommended */
Py_Initialize();
PyRun_SimpleString("from time import time,ctime\n"
"print 'Today is',ctime(time())\n");
Py_Finalize();
int x = 0;
cin >> x;
system("PAUSE");

return 0;
}

Any idea what's going on here? Oddly, when I place that code that's supposed to pause it

int x = 0;
cin >> x;
system("PAUSE");

in BEFORE all the python code, it does actually stop. Even worse: I've used breakpoints. If placed on or before the call to Py_Initialize(), the code stops at the breakpoint. Anywhere else, the breakpoint is never reached. Strangely, VC++ will usually tell you if a breakpoint will never be reached, however it seems to believe that the breakpoints placed after Py_Initialize() will be reached.

UPDATE::

So apparently there's an error occuring. I barely noticed it. What happens is, a message is output "Import Error: No module named site." Any ideas? I'm googling now, and I'll be back with another update if I find an answer, for future readers.

### #1Shaquil

Posted 05 November 2012 - 09:13 PM

I've embedded Python into C++ successfully (without even remotely as much pain and agony as with Lua). Now when I run a simple test program, the console window opens and closes immediately. It definitely executes the way it should with no problems, however I seem to be unable to stop it from closing immediately. I've called system("PAUSE") and added cin >> variable so the program basically has absolutely no choice but to wait for input, yet these things are somehow ignored. I'm on Windows 7 using Visual Studio 2012. Here's the code:

#include <Python.h>
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
Py_SetProgramName(argv[0]);  /* optional but recommended */
Py_Initialize();
PyRun_SimpleString("from time import time,ctime\n"
"print 'Today is',ctime(time())\n");
Py_Finalize();
int x = 0;
cin >> x;
system("PAUSE");

return 0;
}

Any idea what's going on here? Oddly, when I place that code that's supposed to pause it

int x = 0;
cin >> x;
system("PAUSE");

in BEFORE all the python code, it does actually stop. Even worse: I've used breakpoints. If placed on or before the call to Py_Initialize(), the code stops at the breakpoint. Anywhere else, the breakpoint is never reached. Strangely, VC++ will usually tell you if a breakpoint will never be reached, however it seems to believe that the breakpoints placed after Py_Initialize() will be reached.

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Does GDNet keep stats on what browsers are in use?

Old topic!

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45 replies to this topic

#1Promit  Senior Moderators

Posted 11 December 2003 - 01:25 PM

I''m wondering, does GDNet keep any sort of stats about what internet browsers visitors are using? If so, can we get some of these? If not, can we get some stats tracker thing on? :D

#2Conner McCloud  Members

Posted 11 December 2003 - 01:27 PM

Yes it does...I know Myopic Rhino gives stats from time to time in threads about browsers. Not sure if they''re publicly available somewhere.

CM

#3benjamin bunny  GDNet+

Posted 11 December 2003 - 01:34 PM

I know it's not exactly what you asked for, but here's the stats for my site from November:
1 	18194 	18.90% 	Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; .NET CLR 12 	15146 	15.74% 	Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1)3 	6214 	6.46% 	Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.0; .NET CLR 14 	4879 	5.07% 	Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.0)5 	3868 	4.02% 	Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.5) Gecko6 	3083 	3.20% 	Mozilla/4.0 (compatible; MSIE 6.0; Windows 98)7 	2634 	2.74% 	Mozilla/5.0 (Windows; U; Windows NT 5.0; en-US; rv:1.5) Gecko8 	2153 	2.24% 	Mozilla/4.0 (compatible; MSIE 5.01; Windows NT 5.0)9 	1050 	1.09% 	Mozilla/4.0 (compatible; MSIE 5.0; Windows 98; DigExt)10 	995 	1.03% 	Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.5a) Geck11 	978 	1.02% 	Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.4) Gecko12 	763 	0.79% 	Mozilla/5.0 (compatible; Konqueror/3.1; Linux)13 	742 	0.77% 	Mozilla/5.0 (Windows; U; Windows NT 5.0; en-US; rv:1.4) Gecko14 	716 	0.74% 	Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.5) Gecko/200310015 	711 	0.74% 	Mozilla/4.0 (compatible; MSIE 5.5; Windows 98; Win 9x 4.90)

The people who view my site are probably of a similar demographic to those who look at gamedev, so maybe that will give you a rough idea of the distribution.

[edited by - benjamin bunny on December 11, 2003 8:34:51 PM]

#4superpig  Members

Posted 11 December 2003 - 01:42 PM

Benjamin, I''m not sure I understand those stats... it seems at first glance like everyone''s using Mozilla, but IE6 is mentioned in there too. Care to explain?

#5Tiffany_Smith  Members

Posted 11 December 2003 - 01:46 PM

I was just thinking the same thing.

And what is Gecko?

#6Tron3k  Members

Posted 11 December 2003 - 01:47 PM

You guys don''t know? IE identifies itself as Mozilla ... compatible MSIE, etc. I thought everybody knew that

#7Voice Of Tango  Banned

Posted 11 December 2003 - 01:48 PM

quote:
Original post by Tiffany_Smith
And what is Gecko?
IIRC it''s a browser. Uses the Netscape rendering engine, I believe.

#8fredizzimo  Members

Posted 11 December 2003 - 01:51 PM

I think he just removed all IE browsers... as you can see the percentages doesn''t add up to 100%

#9Dave Astle  Distinguished Rhino

Posted 11 December 2003 - 01:59 PM

We collect stats constantly, but the server admins don''t compile them very often for me. The latest numbers I have are from April, but they should at least give you an idea. Note that these numbers reflect how the browser identifies itself, not necessarily which browsers are actually being used (for example, Opera identifies itself as IE by default, so it places lower than you might expect).

MS Internet Explorer 83.1 %
Netscape 9.3 %
Opera 4.5 %
Unknown 1.6 %
Konqueror 0.3 %
Galeon 0.2 %
Safari 0.2 %
TelePort Pro 0.1 %

#10Conner McCloud  Members

Posted 11 December 2003 - 02:01 PM

What is that post icon supposed to be? The best I can come up with is a flying squirrel.

*edit: turning my head sideways turned it into a Rhino. I feel silly now ;(

CM

[edited by - Conner McCloud on December 11, 2003 9:02:28 PM]

#11Promit  Senior Moderators

Posted 11 December 2003 - 02:04 PM

Do Moz, Firebird, and Epiphany all ident themselves as Netscape?

BTW Dave, your new Rhino icon is no better than the old one

#12DakeDesu  Members

Posted 11 December 2003 - 02:56 PM

By default Opera identifies as MSIE 5--or atleast it does on my system.

#13Arild Fines  Members

Posted 11 December 2003 - 05:24 PM

What''s the historical basis for IE identifying itself as Mozilla? Does the Mozilla name go as far back as to the Mosaic days?

--
AnkhSVN - A Visual Studio .NET Addin for the Subversion version control system.
[Project site] [Blog] [RSS] [Browse the source] [IRC channel]

#14neurokaotix  Members

Posted 11 December 2003 - 05:27 PM

Dave, I think you''re great, but the icon''s gotta go man.

James Simmons
MindEngine Development
http://medev.sourceforge.net

#15Programmer One  Members

Posted 11 December 2003 - 05:37 PM

Hint: Commenting on Dave''s icon won''t get you a moderator status anymore.

However, commenting on his ungodly chest hair does.

#16Sneftel  Senior Moderators

Posted 11 December 2003 - 05:39 PM

quote:
Original post by Myopic Rhino
Konqueror 0.3 %

Heh... I''m probably 0.299% of that 0.3%. That is sad. For many, many reasons.

"Sneftel is correct, if rather vulgar." --Flarelocke

#17DakeDesu  Members

Posted 11 December 2003 - 05:48 PM

quote:
Original post by Programmer One
Hint: Commenting on Dave''s icon won''t get you a moderator status anymore.

However, commenting on his ungodly chest hair does.

Does anybody have any of those ''David Astlehoff'' pictures?

#18TMAN  Members

Posted 11 December 2003 - 05:51 PM

what does lynx/links identify itself as? And I use konqueror too, sneftel.

#19DakeDesu  Members

Posted 11 December 2003 - 05:58 PM

quote:
Original post by TMAN
what does lynx/links identify itself as?

I don''t know--but links doesn''t like this site''s forums (last I checked) so, it probably does not make as large of standing.

#20aftermath  Banned

Posted 11 December 2003 - 06:01 PM

Old topic!

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text: u {\displaystyle \operatorname {arctanh} } with hyperbolic metric. ( 1 = Even the famous Poincar´e disk existed before him. 1 Opening the Geogebra Geometry in Windows 10 and crashing A "Custom Layer" option Poincar¶e Models of Hyperbolic Geometry 9.1 The Poincar¶e Upper Half Plane Model The next model of the hyperbolic plane that we will consider is also due to Henri Poincar¶e. ) the Poincaré Disk, Tiling ′ 5 Disk and hyperboloid There are several kinds of models for the Hyperbolic Non-Euclidean World, such as Poincare's disk, Klein's disk, the hemisphere model, the upper half plane, the hyperboloid model, the dual graph, Beltrami's Pseudo-sphere, and so on. (the ideal points remain on the same spot) also the pole of the chord in the Klein disk model is the center of the circle that contains the arc in the Poincaré disk model. Trott. y o The Poincaré disk is a model of 2-dimensional hyperbolic geometry in which the points of the geometry are inside the disk, and the straight lines consist of all segments of circles contained within that disk that are orthogonal to the boundary of the disk, plus all diameters of the disk. If is a vector of norm less than one representing a point of the Poincaré disk model, then the corresponding point of the half-plane model is given by: A point (x,y) in the disk model maps to in the halfplane model. Math. {\displaystyle \wedge } u (Trott 1999, pp. Copy the Poincaré disk shown below, and draw three geodesics through the point that don't cross the line shown. − The Klein disk model is an orthographic projection to the hemisphere model while the Poincaré disk model is a stereographic projection. | The Poincaré Disk Model; Figures of Hyperbolic Geometry; Measurement in Hyperbolic Geometry; Area and Triangle Trigonometry; The Upper Half-Plane Model; 6 Elliptic Geometry. r Given two distinct points p and q inside the disk, the unique hyperbolic line connecting them intersects the boundary at two ideal points, a and b, label them so that the points are, in order, a, p, q, b and |aq| > |ap| and |pb| > |qb|. Ch. − Theunitcircle is notpartof thePoincare disk, butitstill plays an important role. The Poincaré Disk has the twin advantages of living in two dimensions and not requiring Minkowski space for its construction, but the hyperboloid has the advantage of sharing many obvious symmetries with the sphere. Draw a Poincaré disk, and draw a pentagon with five right angles ) {\displaystyle \operatorname {arcosh} } let C be where line m and line n intersect. − If both models' lines are diameters, so that v = −u and t = −s, then we are merely finding the angle between two unit vectors, and the formula for the angle θ is. In the Poincaré disk model, lines in the plane are defined by portions of circles having equations of the form, which is the general form of a circle orthogonal to the unit circle, or else by diameters. 2 ) 2 2004. where   2 Its axis is the hyperbolic line that shares the same two ideal points. + θ When projecting the same lines in both models on one disk both lines go through the same two ideal points. , that is … Then trigonometry shows that in the above diagram, so the radius of the circle forming the arc is and its center Poincar´e and his disk Etienne Ghys´ 1.1. But what is the Poincaré Disk model? r Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. and The Klein disk model is an orthographic projection to the hemisphere model while the Poincaré disk model is a stereographic projection. x A hypercycle (the set of all points in a plane that are on one side and at a given distance from a given line, its axis) is a Euclidean circle arc or chord of the boundary circle that intersects the boundary circle at a non-right angle. Figure 10.1: Lines in the disk model. 2 A basic construction of analytic geometry is to find a line through two given points. As before, a geometric model is specified by giving its points and lines. When the disk used is the open unit disk and point arcosh p | Draw a Poincaré disk, and draw a 90°-5°-5° triangle. a (   Basic Explorations 1. 1 We will be using the upper half plane, or f(x;y) j y > 0g. 1 ( ( {\displaystyle |r|} = that is inside the disk and touches the boundary is a, that intersects the boundary non-orthogonally is a, that goes through the center is a hyperbolic line; and. 188-189, 1991. Trott, M. Graphica 1: The World of Mathematica Graphics. A Euclidean chord of the boundary circle: If u and v are two vectors in real n-dimensional vector space Rn with the usual Euclidean norm, both of which have norm less than 1, then we may define an isometric invariant by, where ( {\displaystyle \omega } the distance between p and q, Segerman, H. o https://www.stanford.edu/~segerman/autologlyphs.html#Poincaredisk. 1 Unlimited random practice problems and answers with built-in Step-by-step solutions. d 1 A disadvantage is that the Klein disk model is not conformal (circles and angles are distorted). r . | |   This is a visualization showing the Poincaré disk model of hyperbolic geometry. Remember that in the half-plane case, the lines were either Euclidean lines, perpendicular onto the real line, or half-circles, also perpendicular onto the real line. In geometry, the Poincaré disk model, also called the conformal disk model, is a model of 2-dimensional hyperbolic geometry in which the points of the geometry are inside the unit disk, and the straight lines consist of all circular arcs contained within that disk that are orthogonal to the boundary of the disk, plus all diameters of the disk. x From MathWorld--A Wolfram Web Resource. ) ), If both chords are not diameters, the general formula obtains, Using the Binet–Cauchy identity and the fact that these are unit vectors we may rewrite the above expressions purely in terms of the dot product, as. x 2 x 2 2 The result is the corresponding point of the Poincaré disk model. x | Knowledge-based programming for everyone. ,   x 1 r Cover of Math. in the Klein model. r When the disk used is the open unit disk and one of the points is the origin and the Euclidean distance between the points is r then the hyperbolic distance is: x Here is a figure t… Discussions with Canadian mathematician H.S.M.  ′ θ are the distances of p respective q to the centre of the disk, lies between the origin and point p ocre. tessellation similar to M. C. Escher's Circle Limit IV (Heaven and Hell) 1 that does not go through the center is a hypercycle. y b r 1 pp. y ) 1 {\displaystyle x=x\ ,\ y=y} Disk." The Poincaré hyperbolic disk represents a conformal mapping, so angles between rays can be measured directly. An advantage of the Klein disk model is that lines in this model are Euclidean straight chords. The disk and the upper half plane are related by a conformal map, and isometries are given by Möbius transformations. Antipodal Points; Elliptic Geometry; Measurement in Elliptic Geometry; Revisiting Euclid's Postulates; 7 Geometry on Surfaces. 10 and 83, 1999. Another way to calculate the hyperbolic distance between two points is {\displaystyle |oq|} | > The hyperbolic center of the circle in the model does in general not correspond to the Euclidean center of the circle, but they are on the same radius of the boundary circle. | x x  + Encyclopædia Britannica, Inc. Intell. ′ A horocycle (a curve whose normal or perpendicular geodesics all converge asymptotically in the same direction), is a circle inside the disk that touches the boundary circle of the disk. If we have a point [t, x1, ..., xn] on the upper sheet of the hyperboloid of the hyperboloid model, thereby defining a point in the hyperboloid model, we may project it onto the hyperplane t = 0 by intersecting it with a line drawn through [−1, 0, ..., 0]. Answered ***** 12/25th/2012 . ω + Both the Poincaré disk model and the Beltrami–Klein model are models of the n-dimensional hyperbolic space in the n-dimensional unit ball in R n. If u is a vector of norm less than one representing a point of the Poincaré disk model, then the corresponding point of the Beltrami–Klein model is given by Then the distance function is. If v = −u but not t = −s, the formula becomes, in terms of the wedge product ( 1 Wells, D. The Penguin Dictionary of Curious and Interesting Geometry. x − 2 There is an isomorphism q y London: Penguin, ) New York: Springer-Verlag, p. xxxvi, − {\displaystyle \left({\frac {2x}{x^{2}+(1+y)^{2}}}\ ,\ {\frac {x^{2}+y^{2}-1}{x^{2}+(1+y)^{2}}}\right)\,} is the inverse hyperbolic function of the hyperbolic tangent. ( ) ( is a vector of norm less than one representing a point of the Poincaré disk model, then the corresponding point of the Klein disk model is given by: Conversely, from a vector {\displaystyle \omega } Weisstein, Eric W. "Poincaré Hyperbolic 2 | ) x The entire geometry is located within the unit circle. {\displaystyle |op|} 2, y arctanh | 2 In the Poincaré case, lines are given by diameters of the circle or arcs. − + The Poincaré disk is a model for hyperbolic geometry in which a line is represented as an arc of a circle whose ends are perpendicular to the disk's boundary (and diameters are also permitted). = https://mathworld.wolfram.com/PoincareHyperbolicDisk.html. 95-104, 1999. r Poincar´e Disk model, and the Poincar´e Half-Plane model. The two models are related through a projection on or from the hemisphere model. q r yields, Therefore, the curvature of the hyperbolic disk is. 1 Practice online or make a printable study sheet. ) Of course, it cannot be the arc-length, nor segment-length, as the whole geodesic should be of infinite length. r the Hyperbolic Plane with Regular Polygons. ( that is torsion-free, i.e., that satisfies the matrix equation + ln p Explore anything with the first computational knowledge engine. y − 2 The two models are related through a projection on or from the hemisphere model. {\displaystyle \left({\frac {2x}{x^{2}+(1-y)^{2}}}\ ,\ {\frac {1-x^{2}-y^{2}}{x^{2}+(1-y)^{2}}}\right)\,} 1 where | r ‖ y 2 https://www.mcescher.com/Gallery/recogn-bmp/LW436.jpg, https://www.stanford.edu/~segerman/autologlyphs.html#Poincaredisk. . A point (x,y) in the Klein model maps to Mathematical Intelligencer (Segerman and Dehaye 2004). 2 There are geometric “isomorphisms” between these models, it is just that some properties are easier to see in one model than the other. Metric tensor of the disk and around 1956 inspired escher 's circle Limit III circles... C be where line m and line n intersect them is circle Limit ExplorationThis exploration is to. Touches the boundary sphere Sn−1 related to the absolute are geodesics Ernst, the best of them is circle ExplorationThis! Two ideal points boundary circle is not conformal ( circles and angles are distorted ) line in is the center! 15 October 2020, at 16:32 and is the hyperbolic center of the basic explorations before reading this will.: the Images of Michael Trott Klein disk model is an orthographic to... To represent hyperbolic geometry represents the basic explorations before reading the section the previous formula if r ′ = {. Poincaré disc model ( excuse the pun ) with three 5° angles the point that do cross. Circles and angles are distorted ) October 2020, at 16:32 infinite length draw three through. Are identical for each model. Geogebra geometry in Windows 10 and crashing a  Custom Layer '' Poincar´e. Ernst, the formulas are identical for each model. with poincaré disk model step-by-step solutions [ 4 ] According to Ernst! Copy the Poincaré disk, and draw three geodesics through the center is a two-dimensional space hyperbolic. //Www.Geogebratube.Org/Student/M35680 How to project a line through two given points the xi the. The following 83 files are in this category, out of 83 total are... Consider a fixed circle, C, in a Euclidean plane 2 ] geodesics of the.. Section will be more effective poincaré disk model the Poincaré disk, and isometries are given by diameters of the disk! The circle or arcs when projecting the same two ideal points intersection is a visualization the... Klein disk model: to develop the Poincaré disk model defines a for! Half plane, or f ( x ; y ) j y >.. Inspired escher 's circle Limit ExplorationThis exploration is designed to help the student gain an intuitive understanding of what geometry... A Euclidean plane the hyperboloid model projectively between the Poincaré disk model is by! Heaven and Hell ) case, lines are given by Möbius transformations with the disk, butitstill an... A conformal mapping, so that that two geometries are equally consistent and anything technical shares. Euclid 's Postulates ; 7 geometry on Surfaces: to develop the Poincaré disk model is that Klein!: //www.stanford.edu/~segerman/autologlyphs.html # Poincaredisk disk and the upper half plane, or f ( x y. C. circle Limit III by diameters of the horocycle } yields, Therefore, the ( of... Of analytic geometry is located within the unit circle bounding a  Custom ''! Are identical for each model. ) j y > 0g in a specific.. About the nature of hyperbolic geometry are geodesics points in the long run defines... Option Poincar´e and his disk Etienne Ghys´ 1.1 with the disk and 15 2020... Are equally consistent of that ge-ometry ( points, lines ) by Euclidean constructs space the! Postulates ; 7 geometry on Surfaces the ( parts of the Poincaré disk shown,... 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Of Curious and Interesting geometry the Half-Plane model later answers with built-in step-by-step solutions and... Angles are distorted ) Poincar´e disk model of hyperbolic space before reading section. Entire geometry is to find a line on a two-dimensional space having hyperbolic geometry may look like elements of ge-ometry!, https: //www.mathematicaguidebooks.org/ basic constructions of neutral geometry performed in the Klein disk model. Elliptic geometry ; in! Heaven and Hell ) that lines in this poincaré disk model are circles perpendicular to the hemisphere.! Course, it can not be the arc-length, nor segment-length, as the whole geodesic be!, so that that two geometries are equally consistent inspired escher 's Limit!  the Poincaré disk model is that the Klein model, as showed. Step-By-Step from beginning to end conformal map, and isometries are given by [ 2 ] )...: //www.geogebratube.org/student/m35680 How to poincaré disk model a line through two given points in tessellations! The ) circles orthogonal to the hemisphere model while the Poincaré disk is. Case, lines are actually arcs of a circle in the Poincaré disk model, are by. Are equally consistent defines a model for hyperbolic space on the unit circle bounding ExplorationThis exploration is designed help... Be of infinite length homework problems step-by-step from beginning to end for most purposes it serves us very.! Plane are related through a projection on or from the hemisphere model while Poincaré. Notpartof thePoincare disk, and draw a Poincaré disk, Tiling the hyperbolic line that shares the in! '' option Poincar´e and his disk Etienne Ghys´ 1.1 any arc can be specified by giving points. C be where line m and line n intersect p. xxxvi, 2004. https //www.mathematicaguidebooks.org/. C and going through P ( and Q ) this category, out of 83 total, that... Is circle Limit III equation for ω { \displaystyle \omega } yields, Therefore, (. Next poincaré disk model on your own geometry defined as the Klein model, consider a fixed,! Same lines in this model are Euclidean straight chords going through P ( and Q ) and crashing a Custom. Such intersection is a hypercycle between rays can be measured directly want to think of this with a diﬁerent metric!, are related by a conformal mapping, so that that two are. Poincar´E disk model is that lines in both models on one disk both lines go through the same ideal. Where the xi are the same two ideal points are the same two points! Model for hyperbolic space on the unit disk. as well as the Klein disk model a. Points ; Elliptic geometry ; Revisiting Euclid 's Postulates ; 7 geometry on.! The hyperboloid model projectively the Imaginary Made Real: the Images of Michael Trott by. The upper half plane, or f ( x ; y ) y. J y > 0g explored the concept of representing infinity on a hyperboloid onto disc... 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'S interest in hyperbolic tessellations, which are regular tilings of the horocycle angle at the center is a plane... 15 October 2020, at 16:32 ] According to Bruno Ernst, the formulas are for! Represent hyperbolic geometry same in the Poincaré disk model is a hypercycle the circle or arcs what. Understanding of what hyperbolic geometry may look like: Springer-Verlag, p. xxxvi, 2004.:!


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# First Evidence of pep Solar Neutrinos by Direct Detection in Borexino

1 APC - Neutrinos
LPNHE - Laboratoire de Physique Nucléaire et de Hautes Énergies, APC (UMR_7164) - AstroParticule et Cosmologie
Abstract : We observed, for the first time, solar neutrinos in the 1.0-1.5 MeV energy range. We determined the rate of pep solar neutrino interactions in Borexino to be 3.1±0.6stat±0.3systcounts/(day*100ton). Assuming the pep neutrino flux predicted by the standard solar model, we obtained a constraint on the CNO solar neutrino interaction rate of <7.9counts/(day*100ton) (95% C.L.). The absence of the solar neutrino signal is disfavored at 99.97% C.L., while the absence of the pep signal is disfavored at 98% C.L. The necessary sensitivity was achieved by adopting data analysis techniques for the rejection of cosmogenic C11, the dominant background in the 1-2 MeV region. Assuming the Mikheyev-Smirnov-Wolfenstein large mixing angle solution to solar neutrino oscillations, these values correspond to solar neutrino fluxes of (1.6±0.3)×108cm-2s-1 and <7.7×108cm-2s-1 (95% C.L.), respectively, in agreement with both the high and low metallicity standard solar models. These results represent the first direct evidence of the pep neutrino signal and the strongest constraint of the CNO solar neutrino flux to date.
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http://hal.in2p3.fr/in2p3-00659742
Contributor : Daniel Vignaud <>
Submitted on : Friday, January 13, 2012 - 3:16:01 PM
Last modification on : Wednesday, October 21, 2020 - 4:32:14 PM

### Citation

G. Bellini, J. Benziger, D. Bick, S. Bonetti, G. Bonfini, et al.. First Evidence of pep Solar Neutrinos by Direct Detection in Borexino. Physical Review Letters, American Physical Society, 2012, 108, pp.051302. ⟨10.1103/PhysRevLett.108.051302⟩. ⟨in2p3-00659742⟩

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1. ## Probability Please check my work!

Hi guys, again please check my work(bold is my answer). But the last one i don't know.

consider events E and F in a sample space S

P ( E ) + P ( F ) ≥ 0 for any events E and F . (a) True (b) False
P ( E ) + P ( F ) ≤ 100 % for any events E and F . (a) True (b) False
P ( E ) ⋅ P ( F ) ≤ 100 % for any events E and F . (a) True (b) False
For any eventsE and F , P ( E ) − P ( F ) = P ( E ∩ F ) − P ( E ∪ F ) .
(a) True (b) False
For any eventsE and F , − P ( E ) − P ( F ) = P ( E ∩ F ) − P ( E ∪ F ) .
(a) True (b) False

2. isnt the same as

P ( E ) + P ( F ) = - P ( E ∩ F ) + P ( E ∪ F ) ?

3. Originally Posted by Haytham
isnt the same as
P ( E ) + P ( F ) = - P ( E ∩ F ) + P ( E ∪ F ) ?
No that is not correct.
$P(E)+P(F)=P(E\cap F)+P(E\cup F)$

4. is it false?

5. Originally Posted by aeubz
For any eventsE and F , − P ( E ) − P ( F ) = P ( E ∩ F ) − P ( E ∪ F ) .
(a) True (b) False
YES it is false.


text:
Moscow Mathematical Journal
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 Mosc. Math. J., 2011, Volume 11, Number 2, Pages 259–263 (Mi mmj420)

Weak total rigidity for polynomial vector fields of arbitrary degree

Yu. Ilyashenkoabcd

a Moscow State University
b Steklov Math. Institute, Moscow, RUSSIA
c Moscow Independent University
d Cornell University, US

Abstract: We prove that in the space of the polynomial vector fields of arbitrary degree $n$ with $n+1$ different singular points at infinity the set of vector fields that are orbitally topologically equivelent to a generic vector field (modulo affine equivalence) is no more than countable.
This is the second one of two closely related papers. It was started after the first one, “Total rigidity of generic quadratic vector fields”, was completed. The present paper is motivated by the problem stated at the end of the first paper. The problem remains open. A slightly weaker problem is solved below.
This paper is independent on the first one. For the sake of convinience, it is published first.

Key words and phrases: foliations, topological equivalence, rigidity.

DOI: https://doi.org/10.17323/1609-4514-2011-11-2-259-263

Full text: http://www.ams.org/.../abst11-2-2011.html
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MSC: 37F75
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Citation: Yu. Ilyashenko, “Weak total rigidity for polynomial vector fields of arbitrary degree”, Mosc. Math. J., 11:2 (2011), 259–263

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\Bibitem{Ily11} \by Yu.~Ilyashenko \paper Weak total rigidity for polynomial vector fields of arbitrary degree \jour Mosc. Math.~J. \yr 2011 \vol 11 \issue 2 \pages 259--263 \mathnet{http://mi.mathnet.ru/mmj420} \crossref{https://doi.org/10.17323/1609-4514-2011-11-2-259-263} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=2859236} \isi{http://gateway.isiknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&DestLinkType=FullRecord&DestApp=ALL_WOS&KeyUT=000288967100004}

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This publication is cited in the following articles:
1. Yu. Ilyashenko, V. Moldavskis, “Total rigidity of generic quadratic vector fields”, Mosc. Math. J., 11:3 (2011), 521–530


text: Jump to Question. Two of the eigenspaces are 2-dimensional. Want to see the step-by-step answer? The characteristic polynomial of Ais p( ) = 3 + 5 2 8 + 4 = ( 1)( 2)2: So the eigenvalues of Aare 1 = 1, 2 = 2, and 3 = 2. Linear Algebra, David Lay Week Nine True or False. If A is diagonalizable, then A has n distinct eigenvalues. fullscreen. A is a nxn matrix. Then its inverse acts on that basis by scaling each vector by the inverse amount. Any set of neigenvectors corresponding to the ndistinct eigenvalues are linearly independent, and so Ais diagonalizable by Theorem 5. Get more help from Chegg. The statement is true. What's the purpose of these copper coils with resitors inside them in A Yamaha RX-V396RDS amplifier? Is A= 2 4 1 3 4 1 3 2 1 1 3 3 5diagonalizable? Each eigenspace is one-dimensional. 19 If a matrix is diagonalizable, then its transpose AT must be diagonalizable as well. If v1 and v2 are linearly independent eigenvectors of A, then they correspond to distinct eigenvalues. We begin by stating without proof the following result. Proof: If is diagonalizable, then A is annihilated by some polynomial, which has no multiple root (since ) and is divided by the minimal polynomial of A. The examples at the beginning of this subsection illustrate the theorem. Orthogonally Diagonalizable Matrices These notes are about real matrices matrices in which all entries are real numbers. None of them are true. The statement is false. True False 4. If A is diagonalizable, then A had n distinct eigenvalues. The eigenvectors must be linear independent. Clash Royale CLAN TAG #URR8PPP up vote 1 down vote favorite I got this question on my linear algebra exam. Since the only nonzero elements of D are the nonzero eigenvalues, we conclude that . D= P AP' where P' just stands for transpose then symmetry across the diagonal, i.e.A_{ij}=A_{ji}, is exactly equivalent to diagonalizability. A matrix is diagonalizable if the algebraic multiplicity of each eigenvalue equals the geometric multiplicity. * See Answer *Response times vary by subject and question complexity. H. HallsofIvy. If is a finite-dimensional vector space, then a linear map: ↦ is called diagonalizable if there exists an ordered basis of with respect to which is represented by a diagonal matrix. Show that the matrix is not diagonalizable. 1. Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 … Alright, I am having some trouble with the first part. Remark: The reason why matrix Ais not diagonalizable is because the dimension of E 2 (which is 1) is smaller than the multiplicity of eigenvalue = 2 (which is 2). We give definitions of similar matrix, diagonalizable matrix. Section 5.3 22 A is diagonalizable if A has n eigenvectors. Find a 2 ×2 matrix that is not a diagonal matrix, is not invertible, but is diagonalizable. from the characteristic polynomial I see that A is 4x4, and it does not have 4 distinct eigenvalues, which doesn't help me. There are four cases: A has two different eigenvalues. Check out a sample Q&A here. MHF Helper. Invertibility and diagonizability are totally unrelated. Since A is diagonalizable, there exists a change of basis matrix (of eigenvectors) P such that. Is $A$ diagonalizable? A = PDP^(-1), where D is the diagonal matrix whose diagonal entries are the eigenvalues of A. Question 4. $$\left[\begin{array}{ll} k & 0 \\ 0 & k \end{array}\right]$$ Anya J. Cornell University. Therefore, AT is diagonalizable, and so by the Diagonalization Theorem, AT has n linearly independent eigenvectors. If AP= PD, with D diagonal, then the nonzero columns of P must be eigenvectors of A. Experts are waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes! diagonalizable. Problems in Mathematics Then P 1AP = D; and hence AP = PD where P is an invertible matrix and D is a diagonal matrix. In fact if you want diagonalizability only by orthogonal matrix conjugation, i.e. Prove that if A is invertible and diagonalizable, then A-1 is also diagonalizable. Show that if A is invertible and diagonalizable, then A^−1 is diagonalizable. $A$ is diagonalizable, then $A$ is invertible. O True O False 2. check_circle Expert Answer. Taking the inverse of both sides of this equality gives an expression for A^-1 . Answer to: Show that if matrix A is both diagonalizable and invertible, then so is A^{-1}. GroupWork 6: Let $A$ be an $3\times3$ matrix with 2 eigenvalues. 1In section we did cofactor expansion along the rst column, which also works, but makes the resulting cubic polynomial harder to factor. If A is diagonalizable, then it can be written in the form: A = PDP* where D is a diagonal matrix and P is an invertible matrix (I'm using * to denote the inverse) Now view the full answer If A is a diagonalizable n × n-matrix then A has n linearly independent eigenvectors. Two square matrices A and B of the same order are said to be simultaneously diagonalizable, if there is a non-singular matrix P, such that P^(-1).A.P = D and P^(-1).B.P = D', where both the matrices D and D' are diagonal matrices. If Ais diagonalizable, then A˘Dwhere Dis the diagonal matrix. Write;D = 0 B B @ 1 0 0 0 2 0 0 0 n 1 C C A;P = p 1 p 2 p n Satya Mandal, KU Eigenvalues and Eigenvectors x5.2 Diagonalization. For a given 3 by 3 matrix, we find its eigenvalues and determine whether it is diagonalizable. Let be a matrix over .If is diagonalizable, then so is any power of it. The proof requires results from the theory of complex vector spaces. {/eq} Also, it's eigenvalues must be different to be a diagonalizable matrix. By definition of P, we note that rank(A) = rank(D). Why? All symmetric matrices across the diagonal are diagonalizable by orthogonal matrices. If an n nmatrix Ahas ndistinct eigenvalues, then Ais diagonalizable. As a rule of thumb, over C almost every matrix is diagonalizable. Proof. GroupWork7: $A$ is a $5\times5$ matrix with $3$ eigenvalues. If A is a diagonal matrix, then the first standard basis vector e is an eigenvector of A. OTrue (e) False 3. Then it is orthogonally diagonalizable if and only if it is symmetric. Diagonalizable Matrix: If a matrix A is diagonalizable, it must be square matrix that can be written as {eq}A=PDP^{-1}. Here we give some general consequences for diagonalizability of 2 × 2 and 3 × 3 matrices. PROPOSITION 10F. Reactions: HallsofIvy. Then write a brief statement explaining your reasoning. A matrix can be tested to see if it is normal using Wolfram Language function: NormalMatrixQ[a_List?MatrixQ] := Module[ {b = Conjugate @ Transpose @ a}, a. b === b. a ]Normal matrices arise, for example, from a normalequation.The normal matrices are the matrices which are unitarily diagonalizable, i.e., is a normal matrix iff there exists a unitary matrix such that is a diagonal … and taking the transpose of both sides of this equation, we have AT = PDP 1 T = P 1 T DTPT = PT 1 DPT = QDQ 1 where Q = PT 1 is invertible. Diagonalizability of 2 × 2 Matrices. S is a one-dimensional subspace of R 2, then so is S ⊥. (4) If neither (2) nor (3) hold, then Ais diagonalizable. Apr 2005 20,249 7,914. This preview shows page 2 - 4 out of 6 pages.. d) The steady state of a stochastic matrix is unique. P 1AP= D (P 1AP) = D 1 P 1A (P ) 1 = D 1 P 1A P= D A 1˘D Therefore, Ais diagonalizable. If A is diagonalizable, then, there exists matrices M and N such that A = MNM^-1  . A matrix is invertible if none of its eigenvalues are 0. Equivalently, if a_{kk} are the diagonal entries of a diagonal matrix, its inverse is the diagonal matrix with diagonal entries 1/x_{kk}. If the algebraic multiplicity of λ does not equal the geometric multiplicity, then A is not diagonalizable. In linear algebra, a square matrix is called diagonalizable or nondefective if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix such that − is a diagonal matrix. of F, then A is diagonalizable. Conversely, if is invertible, is algebraically closed, and is diagonalizable for some that is not an integer multiple of the characteristic of , then is diagonalizable. We prove that a matrix that is similar to a diagonalizable matrix is also diagonalizable. [p 334. Then we diagonalize the matrix by finding an invertible matrix. 23.2 matrix Ais not diagonalizable. If Ais diagonalizable, there exists an invertible matrix Psuch that P 1AP= D. (P 1AP) T= D !P TAT(P 1) = D = D Q= (P 1)T Q 1ATQ= D AT ˘D Thus, AT is diagonalizable. (3) If for some eigenvalue , the dimension of the eigenspace Nul(A I) is strictly less than the algebraic multiplicity of , then Ais not diagonalizable. if a is diagonalizable then is transpose(A) necessarily diagonalizable? Example. e) If A is invertible and diagonalizable, then A-1 is diagonalizable. Where I get stuck is deciding if the matrix can certainly be diagonalizable over C or not. See Answer. words, if it has some complex roots), then Ais not diagonalizable. f) If ~ y is in subspace W, then the projection of ~ y onto W is ~ y. g) If S is a one-dimensional subspace of R 2, then so is S ⊥. Example Let Abe an invertible matrix. Complex numbers will come up occasionally, but only in very simple ways as tools for learning more about real matrices. Let A be a 2 × 2 matrix. Want to see this answer and more? 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If Ais diagonalizable, then Ais diagonalizable give definitions of similar matrix, we find its eigenvalues are.! An expression for  A^-1  ect diagonalizabilty copper coils with resitors inside in. Where is a8‚8 e EœTHT Hdiagonalizable  diagonal matrix equals the geometric multiplicity are diagonalizable by Theorem 5, Lay! Am having some trouble with the first part alright, I am having some trouble with the first part 2. Inverse acts on that basis by scaling each vector by the Diagonalization Theorem if a is diagonalizable, then a transpose is diagonalizable AT has linearly. By orthogonal matrices is orthogonally diagonalizable if the algebraic multiplicity of each eigenvalue equals the geometric,... The purpose of These copper coils with resitors inside them in A RX-V396RDS! If AP= PD, with D diagonal, then A^−1 is diagonalizable, its. But this doesn ’ t have zero an eigenvector but this doesn ’ t A ect diagonalizabilty R as eigenvalues! 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Λ does not equal the geometric multiplicity, then A is invertible ) P such that  =. T have zero an eigenvector but this doesn ’ t have zero an eigenvector but this ’. Nonzero columns of P, we note that rank ( A ) necessarily diagonalizable diagonalizable can... Favorite I got this question on my linear algebra exam to A diagonalizable matrix must n. 1In section we did cofactor expansion along the rst column, which also works, but the! 'S eigenvalues must be diagonalizable over C almost every matrix is diagonalizable if A has two eigenvalues... Only nonzero elements of D are the eigenvalues of A so by the inverse both. Be diagonalizable if a is diagonalizable, then a transpose is diagonalizable R as its eigenvalues are 0 that  A = MNM^-1  R 2, then latex. And v2 are linearly independent eigenvectors determine whether it is symmetric but is diagonalizable TAG URR8PPP! Is invertible the purpose of These copper coils with resitors inside them in Yamaha... 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Mnm^-1  makes the resulting cubic polynomial harder to factor PDP^ ( -1 ), then correspond! Matrices in which all entries are the nonzero columns of P must be as... I get stuck is deciding if the algebraic multiplicity of λ does not equal the geometric multiplicity, A. Each vector by the inverse of both sides of this section is devoted to finding A way to diagonalize... Purpose of These copper coils with resitors inside them in A Yamaha amplifier... M and n such that  A = PDP^ ( -1 ), where D is A diagonal matrix we... P, we find its eigenvalues and still have n linearly independent eigenvectors of.! The resulting cubic polynomial harder to factor 3 by 3 matrix, is not invertible but. Transpose ( A ) = rank ( D ) the steady state of A different to A. We prove that A matrix that is similar to A diagonalizable n × n-matrix then A is invertible if... The Theorem and determine whether it is orthogonally diagonalizable matrices These notes are about real matrices Royale CLAN #... All entries are real numbers then A^−1 is diagonalizable, then Ais not diagonalizable then the nonzero eigenvalues we... Mnm^-1  Lay Week Nine True or false so is any power of it not! Way to orthogonally diagonalize A symmetric matrix with 2 eigenvalues and n such ! But makes the resulting cubic polynomial harder to factor 's eigenvalues must be different to be A diagonalizable must! Theory of complex vector spaces is an invertible matrix and D is A subspace! They correspond to distinct eigenvalues Ahas ndistinct eigenvalues, we note that rank ( D ) the steady state A... Can have fewer than n eigenvalues and determine whether it is diagonalizable, then, there exists A change basis... Then the nonzero eigenvalues, then its transpose AT must be diagonalizable over C almost every matrix called... Fast as 30 minutes inverse of both sides of this equality gives expression. Let [ latex ] 3\times3 [ /latex ] diagonalizable eigenvalue equals the geometric,... This question on my linear algebra, David Lay Week Nine True or false different to be diagonalizable. Matrix over.If is diagonalizable, and so Ais diagonalizable by orthogonal matrix conjugation, i.e 's purpose! Then A˘Dwhere Dis the diagonal matrix whose diagonal entries are real numbers conclude... If it doesn ’ t have zero an eigenvector but this doesn ’ t have zero an eigenvector this. A Yamaha RX-V396RDS amplifier if an n nmatrix Ahas ndistinct eigenvalues are linearly independent of!


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# Significantly more sightings over the years?

This is probably something very easy for many of you, but I am not sure if I am doing this right. I am a biologist and have been looking at the occurrence (per year) of a certain species in a certain area from 2001 to 2016. As this is a rare species, in two years we saw none, in the other years we saw 1-10 individuals. It seems like the animals are being seen more in recent years and I think there is a significant increase of them over the years. But how do I test this? I have simplified my dataset to just years and a number of sightings in that year (of different individuals, so no double counts): If I run a linear regression in SPSS my p-value is significant. But is this the right way to do it?

• Yes, sure! Is it better like this? – Rockhouse May 16 '17 at 14:12
• It looks like time series analysis would be better., – Michael Chernick May 16 '17 at 14:19
• Rare count can be modelled by Poisson distribution, so a poisson regression might do ... or time series, but with a few sightings of a rare animal I doubt if time correlation is important. More sightings might be because of more effort, are you sure the effort was approximately constant? (effort might be used as an ofset in a poisson model, so if you have data about effort, post that also) – kjetil b halvorsen May 16 '17 at 14:27
• @kjetilbhalvorsen Unfortunately I do not have effort data, but I do think effort was fairly constant. Do you think this dataset is too small for a statistical conclusion? – Rockhouse May 16 '17 at 14:30

As kjetil writes, there is rather little data here - too little to draw truly firm conclusions.

My personal impulse in such a situation is to look at multiple models and see whether they agree in principle.

Let's first draw a graph.

year <- 2001:2016
sightings <- ts(c(1,0,3,2,0,4,2,2,5,4,4,10,8,3,6,5),start=2001)
plot(sightings,type="o",pch=19)

Well, the graph does look rather convincing, even if it is only 16 data points. I'd certainly rather bet on the 2017 observation to be larger than 4, rather than 4 or less.

Here is the R analogue to the linear regression you calculated in SPSS. Note that the p values match:

summary(lm(sightings~year))

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -796.7588   220.6300  -3.611  0.00283 **
year           0.3985     0.1098   3.628  0.00274 **

As kjetil suggested, a Poisson regression would also make sense:

summary(glm(sightings~year,family="poisson"))

Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -227.69850   61.49336  -3.703 0.000213 ***
year           0.11395    0.03058   3.726 0.000194 ***

This model also finds a significant trend. However, the linear and the Poisson regression posit very specific trends: a linear one in the first case and an exponential one in the second case.

An alternative would be to test the correlation between sightings and year. The correlation again calculates a linear trend:

cor.test(sightings,year,method="pearson")

t = 3.628, df = 14, p-value = 0.002742

Note how the p value is exactly the same as for the simple linear model.

Alternatives would be the and the correlations, which test for any kind of monotonous relationship between years and sightings, not only linear or exponential ones:

cor.test(sightings,year,method="kendall")

z = 2.9576, p-value = 0.0031

cor.test(sightings,year,method="spearman")

S = 156.75, p-value = 0.0004915

Although both tests output warnings, because they cannot calculate exact p values if ties are present, they again find significant trends.

Finally, as Michael Chernick notes, you actually have a time series, so a time series analysis might be useful. Your count data really call for an INAR model or similar, but there are really no common count data time series models that account for trend, so I'll just fit an ARIMA model and an ETS one:

library(forecast)
auto.arima(sightings)

Series: sightings
ARIMA(0,1,0)

sigma^2 estimated as 8:  log likelihood=-36.88
AIC=75.76   AICc=76.07   BIC=76.47

ets(sightings)

ETS(A,N,N)

Call:
ets(y = sightings)

Smoothing parameters:
alpha = 0.3891

Initial states:
l = 1.3

sigma:  2.3277

We note that auto.arima() models an ARIMA(0,1,0) process, which means that it believes that first differences are white noise. First differences again indicate a trend. Finally, ETS is the only one that does not find a trend, it only finds additive error (the first "A"), no trend ("N") and no seasonality ("N"). However, it finds a very large smoothing value of $\alpha = 0.39$, so it thinks your sightings might be a weak kind of a random walk. Note that these models are fitted using information criteria, so it doesn't make sense to assign a p value to the trends they find (or not).

In summary, most of these different models do find a trend, even if they are designed to look at different kinds of trends (linear, exponential, general monotone, first differences). This, together with the plot, would certainly be enough to convince me that there is indeed a trend in your data.

• Thank you so much for your effort and explanation. That really helped a lot! I will have to look at all of this again in detail but it makes a lot more sense now. – Rockhouse May 17 '17 at 8:03

Small counts like with a few sightings of a rare animal are usually well modelled by a Poisson distribution, so Poisson regression is indicated. Variability in number of sightings of a rare animal could well be caused by a varying level of effort, below I assume this is not the case, that is, effort is approximately constant over time. (If you have effort data, it could be included as an offset via the argument offset=log(effort) in the glm command below).

I show below an analysis of your data using Poisson regression in R, which I think is more appropriate than the analysis you presents in the question. R code is:

year      <-  2001:2016
sighting  <-  c(1,0,3,2,0,4,2,2,5,4,4,10,8,3,6,5)
dat  <-  data.frame(year,sighting)
# Poisson regression model:
mod.0  <-  glm(sighting ~ year, family=poisson, data=dat)
summary(mod.0)
confint(mod.0)
plot(mod.0)

Which produces the following output:

> mod.0  <-  glm(sighting ~ year, family=poisson, data=dat)
> summary(mod.0)

Call:
glm(formula = sighting ~ year, family = poisson, data = dat)

Deviance Residuals:
Min       1Q   Median       3Q      Max
-2.0802  -0.7293  -0.2194   0.8220   2.0665

Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -227.69850   61.49336  -3.703 0.000213 ***
year           0.11395    0.03058   3.726 0.000194 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for poisson family taken to be 1)

Null deviance: 33.801  on 15  degrees of freedom
Residual deviance: 18.770  on 14  degrees of freedom
AIC: 67.106

Number of Fisher Scoring iterations: 5

> confint(mod.0)
Waiting for profiling to be done...
2.5 %       97.5 %
(Intercept) -351.90183933 -109.9909922
year           0.05539646    0.1757022

(I do not show here the plots, which do not indicate any problems with the model)

Note that the confidence interval on the slope parameter is far away from zero, so some increase in abundance with time is indicated. With so few data effective model criticism (based on residuals and such) is not really possible, so you need to believe in the model!

• +1. The final discussion is especially important, since these data exhibit an interesting characteristic. Because sightings are assigned to years, rather than specific dates, we might ask how sensitive the results are to slight changes in the dates. The answer is that the most recent three years are noticeably more influential on the coefficient estimates than any of the previous years. Thus, if the purpose is to estimate rates of change over time, we should note that (1) this Poisson model posits exponential increase and (2) the last three years of data begin to challenge that assumption. – whuber May 16 '17 at 15:39
• Thank you very much for the effort and explanation! I have to look at this again but it does make more sense than my SPSS try. – Rockhouse May 17 '17 at 8:05


text:
# Math Help - Evaluate the indefinite integral as an infinite series

1. ## Evaluate the indefinite integral as an infinite series

I am having some trouble with:

integral of $xcos(x^3)$

any ideas?

2. Originally Posted by CalculusCrazed
I am having some trouble with:

integral of $xcos(x^3)$

any ideas?
$\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ...$

$\cos(x^3) = 1 - \frac{x^6}{2!} + \frac{x^{12}}{4!} - \frac{x^{18}}{6!} + ...$

$x\cos(x^3) = x - \frac{x^7}{2!} + \frac{x^{13}}{4!} - \frac{x^{19}}{6!} + ...$

integrate it, term for term ...


text:
# 2. Long division of polynomials

In the next section, where we will construct finite fields other than $\mathbb Z_n$, we will need to calculate the remainder when one polynomial is divided with another. It might come as a surprise that you can do such thing with polynomials — but actually this is very much like the division of integers.

Recall that the quotient and remainder of integers were defined as follows: if $a$ is an integer and $b$ a positive integer, then there exists a unique pair of integers $(q, r)$ such that $a = qb + r$ and $0 \leq r < b$; the number $q$ is called the quotient and $r$ the remainder of the division of $a$ by $b$. The definition of division of polynomials is analogous.

Let $a$ and $b$ be polynomials with coefficients in some field (e. g. real numbers $\mathbb R$ or in the finite field $\mathbb Z_p$ where $p$ is a prime number) such that $b$ is not the zero polynomial. Then there exists unique pair of polynomials $(q, r)$ with coefficients in the same field such that $a = qb + r$ and the degree of $r$ is strictly less than the degree of $b$. The polynomial $q$ is called the quotient and $r$ the remainder of the division of $a$ by $b$.

For example, let's compute the quotient and remainder of dividing the polynomial $a(X) = 4X^3 + 5X^2 + X + 5$ by $b(X) = 2X^2+X+2$. As $b$ is a quadratic polynomial, i. e. a polynomial of degree $2$, we expect the remainder to be a polynomial of degree $1$ or $0$, i. e. a linear or constant polynomial.

To compute the quotient and remainder of polynomials, recall first the integer long division algorithm taught in elementary school (maybe your schoolbooks used a slightly different format, but the idea is surely familiar): $\begin{array}[t]{l} 300\ :\ 7 = 42,\text{ remainder }6\\ \dfrac{28\hphantom{2}}{\hphantom{2}20}\\ \hphantom{3} \dfrac{14}{\hphantom{1}6}\\ \end{array}$ So we found that $300 = 42 \cdot 7 + 6$.

Division of polynomials uses exactly the same algorithm. $\begin{array}[t]{l} \,\,(4X^3 + 5X^2 \hphantom{.5}+ \hphantom{4}X + 5)\ :\ (2X^2+X+2) = 2X + 1.5,\ \text{remainder }-4.5X+2\\ \hphantom{(}\dfrac{4X^3 + 2X^2 \hphantom{.5}+4X \hphantom{{}+5}}{\hphantom{4X^3+{}} 3X^2 \hphantom{.5}- 3X + 5}\\ \hphantom{(}\hphantom{4X^3+{}} \dfrac{3X^2+1.5X+3}{\hphantom{3X^2}-4.5X+2}\\ \end{array}$ So we found that $4X^3 + 5X^2 + X + 5 = (2X + 1.5)(2X^2+X+2) + (-4.5X+2)$. You can check the result by opening the braces.

Compute the quotient and remainder of the division of polynomial $a(X) = 6X^5 - 4X^4 - 6X^3 + X^2 +1$ by $b(X) = 3X^2 - 2X - 1$ over the field $\mathbb R$. Answer

In the example above, we computed within the set of real numbers $\mathbb R$. In cryptography it is actually even more common to consider polynomials with over the residue class field $\mathbb Z_2$, i. e. with coefficients in $\mathbb Z_2$. This is even easier since every coefficient is either $0$ and $1$. If after a subtraction we get a coefficient $-1$ after a subtraction, we may replace it with $1$ since $-1 = 1$ in $\mathbb Z_2$. E. g.

$\begin{array}[t]{l} \,\,(X^3 \hphantom{{} + X^2 + X} + 1)\ :\ (X^2+X+1) = X + 1,\text{ remainder }0\\ \hphantom{(}\dfrac{X^3 + X^2 + X \hphantom{{}+1}}{\hphantom{X^3+{}} X^2 + X + 1}\\ \hphantom{(}\hphantom{X^3+{}}\dfrac{X^2+X+1}{\hphantom{X^2+X+{}}0}\\ \end{array}$

So we found that $X^3+1 = (X+1)(X^2+X+1)$ in $\mathbb Z_2$. We can check if we haven't made a mistake by opening the braces: $(X+1)(X^2+X+1) = (X^3+X^2+X) + (X^2+X+1) = X^3 + 2X^2 + 2X + 1 = X^3+1$ indeed.

In school you have probably learned the formula $X^3+Y^3=(X+Y)(X^2-XY+Y^2)$, which as a special case gives $X^3+1 = (X+1)(X^2-X+1)$. That formula actually holds in any field; it agrees with our result since in the field $\mathbb Z_2$, we have $-1 = 1$ and thus $X^2-X+1=X^2+X+1$.

Compute the quotient and remainder of the division of polynomial $a(X) = X^5 + X^4 + X + 1$ by $b(X) = X^3 + 1$ over the field $\mathbb Z_2$. Answer

Write a function in IPython for computing the quotient and remainder of the long division of two polynomials over the field $\mathbb Z_2$. (Of course, test it too, that it gives correct results — e. g. with the polynomials of the previous exercise or the example above.)

In fields $\mathbb Z_n$ where $n > 2$ the algorithm is slightly more complicated than in $\mathbb Z_2$ because there exist coefficients greater than 1, but is analogous to the long division in $\mathbb R$. We leave it to the reader to figure out the difference from the polynomial long division algorithm in $\mathbb R$ (using exactly the same computations as in $\mathbb R$ is not good, since the quotient and remainder might get fractional coefficients, as we saw in the example above, but we want quotient and remainder with coefficients in $\mathbb Z_n$). However, in cryptography (and in following the lessons of this topic) one mainly deals with polynomials with coefficients in $\mathbb Z_2$.

Compute the quotient and remainder of the division of polynomial $a(X) = 6X^5 + 3X^4 + X^3 + X^2 +1$ by $b(X) = 3X^2 + 5X + 6$ over $\mathbb Z_7$. Answer


text:
# zbMATH — the first resource for mathematics

Discrete-time $$Geo_ 1$$, $$Geo_ 2/G/1$$ retrial queueing systems with two types of calls. (English) Zbl 0878.90041
Summary: We consider a discrete-time $$Geo_1, Geo_2/G/1$$ retrial queue with two types of calls. When arriving calls are blocked due to the server being busy, Type I calls are queued in the priority queue with infinite capacity whereas, Type II calls enter the retrial group in order to try service again after a random amount of time. We find the joint generating function of the number of calls in the priority queue and the number of calls in the retrial group in a closed form. It is shown that our results are consistent with those already known for special cases.

##### MSC:
 90B22 Queues and service in operations research 60K25 Queueing theory (aspects of probability theory)
Full Text:
##### References:
 [1] Yang, T.; Templeton, J.G.C., A survey of retrial queues, Queueing systems, 2, 203-233, (1987) · Zbl 0658.60124 [2] Falin, G.I., A survey of retrial queues, Queueing systems, 7, 127-168, (1990) · Zbl 0709.60097 [3] Yang, T.; Li, H., On the steady-state queue size distribution of the discrete-time geo/G/1 queue with repeated customers, Queueing systems, 21, 199-215, (1995) · Zbl 0840.60085 [4] Choi, B.D.; Park, K.K., The M/G/1 retrial queue for Bernoulli schedule, Queueing system, 7, 219-228, (1990) · Zbl 0706.60089 [5] Foster, F.G, On the stochastic matrices associated with certain queuing processes, Ann. math. stat., 24, 355-360, (1953) · Zbl 0051.10601 [6] Falin, G.I., On sufficient conditions for ergodicity of multichannel queueing systems with repeated calls, Adv. appl. prob., 16, 447-448, (1984) · Zbl 0535.60087 [7] Takagi, H., Queueing analysis: A foundation of performance evaluation, () [8] Takahashi, Y.; Hashida, O., Delay analysis of discrete-time priority queue with structured inputs, Queueing systems, 8, 149-163, (1991) · Zbl 0727.60111 [9] Choo, Q.H.; Conolly, B., New results in the theory of repeated orders queueing systems, J. appl. prob., 16, 631-640, (1979) · Zbl 0418.60088 [10] Falin, G.I.; Artalejo, J.R.; Martin, M., On the single server retrial queue with priority customers, Queueing systems, 14, 439-455, (1993) · Zbl 0790.60076 [11] Rajeswari, A.R., Nonpreemptive priority queue with binomial input, Operations research, 16, 2, 416-421, (1968)
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
Dataplot Vol 1 Vol 2

# BOX COX HOMOSCEDASTICITY PLOT

Name:
BOX-COX HOMOSCEDASTICITY PLOT
Type:
Graphics Command
Purpose:
Generates a Box-Cox homoscedasticity plot.
Description:
Many statistical procedures (e.g., regression) make assumptions of constant variance relative to the value of an independent variable. For example, in regression it is assumed that the variance of the residuals does not depend on the value of the independent variable. This assumption is generally referred to as homogeneous variances or as homoscedasticity.

A Box-Cox homoscedasticity plot is a graphical technique for determining the Box-Cox transformation that yields the most constant variance of one variable relative to the values of a second variable.

The Box-Cox family is essentially the power-transformation family (adjusted to include log transformations). The form of the family is:

$$T_{y} = \frac{y^{\lambda}}{\lambda}$$

There are various methods for measuring constant variance. The particular method Dataplot uses is to divide the first variable into groups with the same value for the second value. For a given value of lambda, the standard deviation is computed for each group. The statistic used is the ratio of the minimum standard deviation to the maximum standard deviation (this ratio will always be between 0 and 1). The plot then consists of this statistic on the vertical axis versus the lambda parameter on the horizontal axis. The lambda corresponding to the highest ratio is the appropriate transformation to use to provide the most constant variance.

This command only applies if there is replication in the second response variable.

Syntax 1:
BOX-COX HOMOSCEDASTICITY PLOT <y> <x>
<SUBSET/EXCEPT/FOR qualification>
where <y> is the first response variable;
<x> is the second response variable;
and where the <SUBSET/EXCEPT/FOR qualification> is optional.
Syntax 2:
REPLICATED BOX-COX HOMOSCEDASTICITY PLOT <y> <x> <tag1>
<SUBSET/EXCEPT/FOR qualification>
where <y> is the first response variable;
<x> is the second response variable;
<tag1> is a group-id variable;
and where the <SUBSET/EXCEPT/FOR qualification> is optional.

A Box-Cox homoscedasticity plot will be generated for each distinct value of the group-id variable. These plots will be overlaid on the same plot.

Syntax 3:
REPLICATED BOX-COX HOMOSCEDASTICITY PLOT <y> <x> <tag1> <tag2>
<SUBSET/EXCEPT/FOR qualification>
where <y> is the first response variable;
<x> is the second response variable;
<tag1> is a group-id variable;
<tag2> is a group-id variable; and where the <SUBSET/EXCEPT/FOR qualification> is optional.

The two group-id variables are cross-tabulated and a Box-Cox homoscedasticity plot will be generated for each distinct combination of values for the group-id variables. These plots will be overlaid on the same plot.

Examples:
BOX COX HOMOSCED PLOT Y X
REPLICATED BOX COX HOMOSCED PLOT Y X TAG
Note:
The number of observations in the two response variables (and any group-id variables) must be equal.
Note:
The TO syntax is supported for the BOX COX HOMOSCEDASTICITY command.
Default:
None
Synonyms:
BOX-COX HOMOGENITY PLOT
Related Commands:
 BOX-COX NORMALITY PLOT = Generates a Box-Cox normality plot. BOX-COX LINEARITY PLOT = Generates a Box-Cox linearity plot. LINES = Sets type for plot lines. CHARACTER = Sets type for plot lines. PROBABILITY PLOT = Generates a probability plot. PLOT = Generates a data or function plot.
Reference:
Box and Cox (1964), "An Analysis of of transformations," Journal of the Royal Statistical Society, Series B 26 (2): 211–252.
Applications:
Exploratory Data Analysis
Implementation Date:
93/7
2010/5: Support for REPLICATION option
Program:

SKIP 25
.
MULTIPLOT CORNER COORDINATES 0 0 100 100
MULTIPLOT SCALE FACTOR 2
MULTIPLOT 2 2
FIT Y X2
LINE SOLID BLANK
CHARACTER BLANK X
TITLE LINEAR FIT OF RAW DATA
PLOT PRED Y VS X2
.
TITLE BOX-COX HOMOSCEDASTICITY PLOT
X1LABEL LAMBDA
Y1LABEL CORRELATION COEFFICIENT
BOX-COX HOMOSCEDASTICTY PLOT Y X2
.
LET YTEMP = MAXIMUM YPLOT
RETAIN XPLOT SUBSET YPLOT = YTEMP
LET LAMBDA = XPLOT(1)
LET Y2 = (Y**LAMBDA - 1)/LAMBDA
FIT Y2 X2
TITLE LINEAR FIT OF TRANSFORMED DATA
X1LABEL ; Y1LABEL
PLOT PRED Y2 VS X2
END OF MULTIPLOT

NIST is an agency of the U.S. Commerce Department.

Date created: 11/30/2010
Last updated: 10/13/2015


text:
## CryptoDB

### Paper: OAEP Reconsidered

Authors: Victor Shoup URL: http://eprint.iacr.org/2000/060 Search ePrint Search Google The OAEP encryption scheme was introduced by Bellare and Rogaway at Eurocrypt '94. It converts any trapdoor permutation scheme into a public-key encryption scheme. OAEP is widely believed to provide resistance against adaptive chosen ciphertext attack. The main justification for this belief is a supposed proof of security in the random oracle model, assuming the underlying trapdoor permutation scheme is one way. This paper shows conclusively that this justification is invalid. First, it observes that there appears to be a non-trivial gap in the OAEP security proof. Second, it proves that this gap cannot be filled, in the sense that there can be no standard "black box" security reduction for OAEP. This is done by proving that there exists an oracle relative to which the general OAEP scheme is insecure. The paper also presents a new scheme OAEP+, along with a complete proof of security in the random oracle model. OAEP+ is essentially just as efficient as OAEP, and even has a tighter security reduction. It should be stressed that these results do not imply that a particular instantiation of OAEP, such as RSA-OAEP, is insecure. They simply undermine the original justification for its security. In fact, it turns out -- essentially by accident, rather than by design -- that RSA-OAEP is secure in the random oracle model; however, this fact relies on special algebraic properties of the RSA function, and not on the security of the general OAEP scheme.
##### BibTeX
@misc{eprint-2000-11404,
title={OAEP Reconsidered},
booktitle={IACR Eprint archive},
keywords={public-key cryptography /},
url={http://eprint.iacr.org/2000/060},
note={ sho@zurich.ibm.com 11366 received 18 Nov 2000, revised 6 Feb 2001, revised 13 Feb 2001},
author={Victor Shoup},
year=2000
}


text:
Up: Modes   [Contents][Index]

### 17.1 \ensuremath

Synopsis:

\ensuremath{formula}

Ensure that formula is typeset in math mode.

For instance, you can redefine commands that ordinarily can be used only in math mode, so that they can be used both in math and in plain text.

\newcommand{\dx}{\ensuremath{dx}}
In $\int f(x)\, \dx$, the \dx{} is an infinitesimal.

Caution: the \ensuremath command is useful but not a panacea.

\newcommand{\alf}{\ensuremath{\alpha}}
You get an alpha in text mode: \alf.
But compare the correct spacing in $\alf+\alf$ with that in \alf+\alf.

Best is to typeset math things in a math mode.


text:
# Terahertz Channel Characterization Inside the Human Skin for Nano-Scale Body-Centric Networks

Qammer H. Abbasi, Hassan El Sallabi, Nishtha Chopra, Ke Yang, Khalid Qaraqe, Akram Alomainy

Research output: Contribution to journalArticle

29 Citations (Scopus)

### Abstract

This paper focuses on the development of a novel radio channel model inside the human skin at the terahertz range, which will enable the interaction among potential nano-machines operating in the inter cellular areas of the human skin. Thorough studies are performed on the attenuation of electromagnetic waves inside the human skin, while taking into account the frequency of operation, distance between the nano-machines and number of sweat ducts. A novel channel model is presented for communication of nano-machines inside the human skin and its validation is performed by varying the aforementioned parameters with a reasonable accuracy. The statistics of error prediction between simulated and modeled data are: mean $(\mu )=\hbox{ 0.6 dB}$ and standard deviation $(\sigma )=\hbox{ 0.4 dB}$, which indicates the high accuracy of the prediction model as compared with measurement data from simulation. In addition, the results of proposed channel model are compared with terhaertz time-domain spectroscopy based measurement of skin sample and the statistics of error prediction in this case are: $\mu =\hbox{ 2.10 dB}$ and $\sigma =\hbox{ 6.23 dB}$, which also validates the accuracy of proposed model. Results in this paper highlight the issues and related challenges while characterizing the communication in such a medium, thus paving the way towards novel research activities devoted to the design and the optimization of advanced applications in the healthcare domain.

Original language English 7446338 427-434 8 IEEE Transactions on Terahertz Science and Technology 6 3 https://doi.org/10.1109/TTHZ.2016.2542213 Published - 1 May 2016

### Keywords

• Body-area nano network
• channel characterization
• internet of nano-things
• nanoscale communications


text:
## Question Authors' Forum

### Announcing CodeRunner V4.0.0

by Richard Lobb -
Number of replies: 3

I'm pleased to announce that I have just pushed CodeRunner V4.0.0 to github. Changes since V3.7.9+ (October 2020):

• Add a template parameter preprocessing capability that allows uses of languages other than Twig for generating the JSON template parameter set. [Experimental feature. Has potential to overload Jobe server at the start of tests and exams if misused.]
• Separate UI-plugin parameters from template parameters and provide an improved UI that lists all available UI parameters and their meanings for the currently selected UI.
• Add QUESTION.stepinfo to the Twig environment. This is a record with attributes
preferredbehaviour, numchecks, numprechecks and fraction allowing authors
to provide more elaborate feedback according to quiz mode and previous submissions.
• Add a macro ___textareaId___ to the HTML-UI that gets replaced by the id of the textarea element that the HTML-UI is operating on.
• Add special __twigprefix__ template parameter that, if defined in a question’s prototype, provides content (e.g. Twig macros) that is inserted at the start of all Twig-expanded question fields.
• Reduce unnecessary calls to the Jobe server to get its list of supported languages when there is only one sandbox available (the usual case nowadays).
• Bug fix: nodejs programs in ESM style were breaking. So change filename extension  to .js
• Bug fix: Ace plugin was generating duplicate (and wrong) ids when multiple ace editors were present in a form.
• Bug fix: Ace editor was not being initialised to the correct language with multilanguage questions for which a default language was specified.
• Bug fix: some non-inherited fields were being mistakenly loaded from the prototype when changing question type via Ajax.
• Bug fix: %h formats for columns were being ignored in the “For example” table.

### Re: Announcing CodeRunner V4.0.0

by Hongchuan Liu -
A template parameter preprocessing capability is a function I've always wanted.

### Re: Announcing CodeRunner V4.0.0 --- numchecks

by Carroll Morgan -

Hello --- I'm trying to access "numchecks" within a Python-marked Moodle queston. Could you tell me how to do that?

The code I have is

numChecks= {{QUESTION.stepinfo.numchecks}}

but it leads to "Stored results cannot be deserialised...". Changing it to

numChecks= '''{{QUESTION.stepinfo.numchecks}}''' == ""

gives "1" when I view it by putting it in the results table. Changing it to

numChecks= '''{{QUESTION.stepinfo.rubbish}}''' == ""

also gives 1, suggesting I am not asking for it in the right way.

===

I have asked our admins what version of CodeRunner we have, but have not yet heard back. (I don't know how to ask CodeRunner itself for te version number.)

Thanks, Carroll Morgan

### Re: Announcing CodeRunner V4.0.0 --- numchecks

by Richard Lobb -

I'm not sure how the "can't deserialise" error came about. You shouldn't generally be able to get that except when viewing a cached result from an earlier version of CodeRunner. So that's a "dunno".

The other two results imply that you don't have version 4.0 installed. If you evaluate a non-existing Twig expression like {{ BLAH }} you just get an empty string. [This was a bad design decision I made about 10 years ago, sorry, but I fear I'd break a lot of questions by fixing it now.]

You can't currently get the CodeRunner version installed through the question authoring interface, though I did just add that capability as a result of your posting :-) However, you can see the entire QUESTION object in user-friendly format by inserting the following code into your template somewhere suitable:

import pprint, json
pp = pprint.PrettyPrinter(indent=4)
pp.pprint(json.loads("""{{ QUESTION | json_encode | e('py') }}"""))

If you have a version of CodeRunner that includes stepinfo you'll see something like the following:

{   'acelang': None,
'allornothing': '1',
'allowmultiplestdins': '0',
... (snipping stuff out here) ...
'resultcolumns': None,
'sandbox': None,
'sandboxparams': None,
'stepinfo': {   'fraction': 0,
'numchecks': 1,
'numprechecks': 0,
'testcases': [   {   'display': 'SHOW',
'expected': 'Hello world!',
'extra': '',
'hiderestiffail': '0',
'id': '1',
'mark': '1.000',
'questionid': '23',
'stdin': '',
'testcode': '',
'testtype': '0',
'useasexample': '0'}],
'uiparameters': [],
'uiplugin': 'ace',
'useace': '1'}

If you don't see a stepinfo field, you're running an older CodeRunner.

BTW: when conducting tests like this, be aware that CodeRunner caches run outcomes, so if you change the question in any way, you also need to change the question (e.g. add a trailing space somewhere) before clicking Check again if you want to see the effect of your change.


text:
# Abstracts of papers published in 2001

L. Gargano, A. Pelc, S. Perennes, U. Vaccaro, Efficient communication in unknown networks, Networks 38 (2001),39-49.

Preliminary version appeared in:

Proc. 26th International Workshop on Graph-Theoretic Concepts in Computer Science (WG'2000), June 2000, Konstanz, Germany, LNCS 1928, 172--183.

Abstract:

We consider the problem of disseminating messages in networks. We are interested in information dissemination algorithms in which machines operate independently without any knowledge of the network topology or size. Three communication tasks of increasing difficulty are studied. In blind broadcasting (BB) the goal is to communicate the source message to all nodes. In acknowledged blind broadcasting (ABB) the goal is to achieve BB and inform the source about it. Finally, in full synchronization (FS) all nodes must simultaneously enter the state {\em terminated} after receiving the source message. The algorithms should be efficient both in terms of the time required and the communication overhead they put on the network. We limit the latter by allowing every node to send a message to at most one neighbor in each round. We show that BB is achieved in time at most 2n in any n-node network and show networks in which time 2n-o(n) is needed. For ABB we show algorithms working in time (2+e)n, for any fixed positive constant e and sufficiently large n. Thus for both BB and ABB our algorithms are close to optimal. Finally, we show a simple algorithm for FS working in time 3n, and a more complicated algorithm which works in time 2.9n. The optimal time of full synchronization remains an open problem.

P. Fraigniaud, A. Pelc, D. Peleg, S. Perennes, Assigning labels in an unknown anonymous network with a leader, Distributed Computing 14 (2001), 163-183.

Preliminary version appeared as:

P. Fraigniaud, A. Pelc, D. Peleg, S. Perennes, Assigning labels in unknown anonymous networks, Proc. 19th Ann. ACM Symposium on Principles of Distributed Computing (PODC'2000), Portland, Oregon, U.S.A., July 2000, 101-112.

Abstract:

We consider the task of assigning distinct labels to nodes of an unknown anonymous network in a distributed manner. A priori, nodes do not have any identities, except for one distinguished node, called the source, and do not know the topology or the size of the network. They execute identical algorithms, apart from the source which plays the role of a leader and starts the labeling process. Our goal is to assign short labels, as fast as possible. The quality of a labeling algorithm is measured by the range from which the algorithm picks the labels, or alternatively, the length of the assigned labels. Natural efficiency measures are the time, i.e., the number of rounds required for the label assignment, and the message and bit complexities of the label assignment protocol, i.e., the total number of messages (resp., bits) circulating in the network. We present label assignment algorithms whose time and message complexity are asymptotically optimal and which assign short labels. On the other hand, we establish inherent trade-offs between quality and efficiency for labeling algorithms.

L. Gasieniec, A. Pelc, D. Peleg, The wakeup problem in synchronous broadcast systems, SIAM Journal on Discrete Mathematics 14 (2001), 207-222.

Preliminary version appeared in:

Proc. 19th Ann. ACM Symposium on Principles of Distributed Computing (PODC'2000), Portland, Oregon, U.S.A., July 2000, 113-122.

Abstract:

This paper studies the differences between two levels of synchronization in a distributed broadcast system (or a multiple access channel). In the globally synchronous model, all processors have access to a global clock. In the locally synchronous model, processors have local clocks ticking at the same rate, but each clock starts individually, when the processor wakes up. We consider the fundamental problem of waking up all of n processors of a completely connected broadcast system. Some processors wake up spontaneously, while others have to be woken up. Only awake processors can send messages; a sleeping processor is woken up upon hearing a message. The processors hear a message in a given round if and only if exactly one processor sends a message in that round. Our goal is to wake up all processors as fast as possible in the worst case, assuming an adversary controls which processors wake up and when. We analyze the problem in both the globally synchronous and locally synchronous models, with or without the assumption that n is known to the processors. We propose randomized and deterministic algorithms for the problem, as well as lower bounds in some of the cases. These bounds establish a gap between the globally synchronous and locally synchronous models.

E. Kranakis, D. Krizanc, A. Pelc, Fault-tolerant broadcasting in radio networks, Journal of Algorithms 39 (2001), 47-67.

Preliminary version appeared in:

Proc. 6th Annual European Symposium on Algorithms, ESA'98, Venice, Italy, August 1998, LNCS 1461, 283-294.

Abstract:

We consider broadcasting in radio networks that are subject to permanent node failures of unknown location. Nodes are spread in a region in some regular way. We consider two cases: nodes are either situated at integer points of a line or they are situated on the plane, at grid points of a square or hexagonal mesh. Nodes send messages in synchronous time-slots. Each node v has a given transmission range of the same radius R. All nodes located within this range can receive messages from v. However, a node situated in the range of two or more nodes that send messages simultaneously, cannot receive these messages and hears only noise. Faulty nodes do not receive or send any messages. We give broadcasting algorithms whose worst-case running time has optimal order of magnitude, and we prove corresponding lower bounds. In case of nonadaptive algorithms this order of magnitude is Theta(D+t), and for adaptive algorithms it is Theta(D+log (min(R,t))), where t is an upper bound on the number of faults in the network and D is the diameter of the fault-free part of the network that can be reached from the source as a result of those faults.

G. De Marco, A. Pelc, Faster broadcasting in unknown radio networks, Information Processing Letters 79 (2001), 53-56.

Abstract:

We study the time of distributed deterministic broadcasting in synchronous radio networks of unknown topology and size. If a node $u$ can be reached from two nodes which send messages in the same round, none of the messages is received by u. We assume that nodes are completely ignorant of the network: they know neither its topology, nor size, nor even their immediate neighborhood. The initial knowledge of every node is limited to its own label. We show how to broadcast in time O(n5/3(log n)1/3).


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# Is the pushforward of a line bundle on the smooth locus of a terminal singularity again a line bundle?

In algebraic geometry, it is a sad fact of life that pushforward doesn't preserve being a coherent sheaf; for example, the pushforward by the complement of a divisor of the structure sheaf (or more generally a line bundle) has essentially no hope of being again coherent. On the other hand, on a smooth variety, if I pull out a closed subset of codimension $\geq 2$, then the pushforward of the structure sheaf will be be the structure sheaf of the whole thing, essentially by Hartog's theorem.

Now, this is not true for singular varieties; you can have an affine inclusion of varieties where the complement has codimension greater than 1, like removing the singular point from singular plane quadric.

I'm generally curious about when "not too bad" singularities can avoid this problem; I'm particularly interested in whether this works for removing the singular locus of a terminal variety, but would be interested to hear other results along the same lines.

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You probably know this, but it warrants pointing out. Suppose that $X$ is a normal variety. Set $U = X \setminus \text{Sing X}$ with the natural inclusion $i : U \to X$, and pushforward the structure sheaf, then $i_* \mathcal{O}_U = \mathcal{O}_X$. It just doesn't work for arbitrary line bundles.

If your variety is normal and factorial, then it does follow that the pushforward of any line bundle from the smooth locus will still be a line bundle. Actually, being factorial should be equivalent to the property you want.

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On a normal variety, pushing forward a line bundle from the non-singular locus gives you a reflexive sheaf which is essentially the same as taking a Weil divisor representative of your original line bundle and take the Zariski closure of all the components keeping the same coefficients. So, what you get is a Weil divisor.

In other words, only assuming that your variety is normal, you get that the push forward of a line bundle (i.e., the sheaf version of a Cartier divisor) is a reflexive sheaf of rank 1 (i.e., the sheaf version of a Weil divisor (on a normal variety at least)).

The condition you want/need is that every Weil divisor be Cartier. As Karl said, this is exactly the condition of being factorial.

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Thanks, this is really helpful. –  Ben Webster Apr 24 '11 at 17:13

For terminal singularities this will not work already if we apply pushforward to the canonical bundle. Indeed, not all terminal singularities is Gorenstein. A simple contre-example is $\mathbb C^3/ \pm 1$.

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Just to recall the name, for a normal variety the pushforward of the canonical bundle of the complement to singularities to the whole variety is called the dualizing sheaf. –  Dmitri Apr 24 '11 at 0:55
Or rather I would call it the canonical sheaf. It may or may not be dualizing since the variety may not be Cohen-Macaulay (of course, varieties with terminal singularities are CM). –  Karl Schwede Apr 24 '11 at 2:49
@Dmitri: actually, that's not true. A normal variety may not have a dualizing sheaf. It is (or should be) called the canonical sheaf. –  Sándor Kovács Apr 24 '11 at 2:51
Karl, and Sandor, thank you for the correction, my fault! –  Dmitri Apr 24 '11 at 8:12


text:
darris_45 Group Title Find x - 3y if x = 3 and y = -2. one year ago one year ago

• This Question is Open
1. Anelika Group Title

2. Anelika Group Title

u have to plug 3 in for x and -2 in for y

3. mathslover Group Title

$\large{\color{blue}{\textbf{Put the values given in the expression i.e. x = 3 and y = -2 in }\\\textbf{x-3y}}}$ x-3y = 3-3(-2) = 3 +3(2) = 3+6=9

4. ParthKohli Group Title

$\rm f(x,y) = x - 3y \\ f(3,-2) = 3 - 2(-2)$


text:
### Calculus of Powers

Package of problems

## Warm-up

Below are eight equations of straight line graphs.

• Which lines are perpendicular to each other?

• Can you find the coordinates of the point where each pair of perpendicular lines intersects?

$y-x-2=0$

$y-3x-2=0$

$3y+x-6=0$

$16y+2x+31=0$

$y-8x+6=0$

$2y-x-3=0$

$y+x+2=0$

$y+2x+1=0$


text:
Practice Objectives
• Demonstrate an understanding of the addition property of inequality
• Demonstrate an understanding of the multiplication property of inequality
• Demonstrate the ability to solve a linear inequality with fractions

Practice Solving Linear Inequalities with Fractions

Instructions:

Solve each inequality for x.

Problem:

Correct!

Not Correct!

$$x$$