text:
# Meep C-plus-plus Tutorial

(Difference between revisions)
 Revision as of 06:22, 22 November 2013 (edit)Ardavan (Talk | contribs)← Previous diff Current revision (15:05, 18 April 2017) (edit)Ardavan (Talk | contribs) Line 1: Line 1: {{Meep}} {{Meep}} - Instead of using the libctl/Scheme interface described in the [[Meep tutorial]], one can also access Meep as a C++ library by writing a C++ program that links to it. The C++ interface provides the greatest level of flexibility in designing FDTD simulations. + Instead of using the libctl/Scheme interface described in the [[Meep tutorial]], Meep is also callable as a C++ library by writing a C++ program that links to it. The C++ interface provides the greatest level of flexibility in setting up FDTD simulations. There are 19 examples of such C++ programs in the tests/ directory of the Meep codebase which cover a wide range of Meep's functionality. These are a good reference. Finally, we should also note that, while Meep is nominally in C++, it is perhaps better described as "C+". That is, most of the coding style is C-like with a few C++ features. - + - (This is, in fact, the original way in which we used Meep, and the way in which all of the Meep test programs, in its tests/ directory, are written.) + - + - We should also note that, while Meep is nominally in C++, it is perhaps better described as "C+". That is, most of the coding style is C-like with a few C++ features. + == Differences from libctl == == Differences from libctl == Line 158: Line 154: == Compiling == == Compiling == - In order to compile your code and link against the Meep library, you must link in several additional libraries that Meep requires. There are two ways to do this: + In order to compile your code and link against the Meep library, you must link in several additional libraries that Meep requires. There are three possible ways to do this: - 1. After compiling the MEEP package following the instructions elsewhere in this wiki, place myfile.cpp (where myfile.cpp is a MEEP C++ file of your choosing) in the tests/ subdirectory, cd to the same directory, and type: + 1. After compiling the Meep package following the instructions elsewhere in this wiki, place myfile.cpp (where myfile.cpp is a MEEP C++ file of your choosing) in the tests/ subdirectory, cd to the same directory, and invoke: make myfile.dac make myfile.dac - The resulting executable may be invoked from the tests/ directory via: + Run the resulting executable via: ./myfile.dac ./myfile.dac

## Current revision

Instead of using the libctl/Scheme interface described in the Meep tutorial, Meep is also callable as a C++ library by writing a C++ program that links to it. The C++ interface provides the greatest level of flexibility in setting up FDTD simulations. There are 19 examples of such C++ programs in the tests/ directory of the Meep codebase which cover a wide range of Meep's functionality. These are a good reference. Finally, we should also note that, while Meep is nominally in C++, it is perhaps better described as "C+". That is, most of the coding style is C-like with a few C++ features.

## Differences from libctl

The C++ interface has several differences from the libctl interface besides the obvious difference in syntax.

The most notable difference is that, while the libctl interface puts the origin (0,0,0) at the center of the computational cell, the C++ interface by default puts the origin at the corner of the computational cell. That is, an $L\times L\times L$ cell goes from ( − L / 2, − L / 2, − L / 2) to (L / 2,L / 2,L / 2) in libctl, but from (0,0,0) to (L,L,L) in C++. (You can change this by calling grid_volume::shift_origin.)

## Overview

We begin with a very brief outline of a Meep C++ program, with minimal explanations, leaving more details for the examples below and the C++ reference page. Your C++ program should begin with:

#include <meep.hpp>
using namespace meep;

to include the definitions of the Meep routines. Later, when you compile (see below), you must also link to the Meep libraries.

Your main program should then initialize Meep, and will generally then define a computational grid_volume and the associated structure (describing the geometry and materials), initialize the fields, add sources, and then time-step. In short:

int main(int argc, char **argv) {
initialize mpi(argc, argv); // do this even for non-MPI Meep
double resolution = 20; // pixels per distance
grid_volume v = vol2d(5,10, resolution); // 5x10 2d cell
structure s(v, eps, pml(1.0));
fields f(&s);

f.output_hdf5(Dielectric, v.surroundings());

double freq = 0.3, fwidth = 0.1;
gaussian_src_time src(freq, fwidth);
while (f.time() < f.last_source_time()) {
f.step();
}

f.output_hdf5(Hz, v.surroundings());

return 0;
}

This example, doesn't do much — it just runs a Gaussian source and outputs the $\mathbf{H}_z$ field at the end. The dielectric structure is determined by the user-defined function eps, which has the form:

double eps(const vec &p) {
if (p.x() < 2 && p.y() < 3)
return 12.0;
return 1.0;
}

which returns the dielectric function $\varepsilon(\mathbf{x})$ (here, just a 2×3 rectangle of ε=12 in the upper-left corner). Unlike in the libctl front-end, by default the origin 0 of the coordinate system is at the corner of the computational cell.

Now that you have the basic flavor, we can proceed to some more specific examples.

## Computing the Quality Factor of a Resonator

In this first tutorial, we will write the script to compute the quality factor of a 1D Fabry-Perot cavity. For a 1D system, Meep considers a computational cell along the z coordinate. The control file will be a C++ file (having extension *.cpp). In order to use all the classes and subroutines available in Meep, the first two lines of any control file must be the following:

#include <meep.hpp>
using namespace meep;

The particular Fabry-Perot cavity we will investigate consists of an air region bounded by two distributed Bragg reflectors (quarter-wave stack) of epsilon 12 and 1. We choose the size of the defect to be twice as large as the air thickness in the quarter-wave stack, so that a defect mode is found near midgap. This structure will include N = 5 periods in the Bragg reflector on either side of the defect. We can use a larger N but the quality factor may then be too large to compute. The parameters are set up as follows:

const double eps1 = 12.0; // epsilon of layer 1
const double eps2 = 1.0; // epsilon of layer 2
const double grating_periodicity = 1.0;
const double d1 = sqrt(eps2) / (sqrt(eps1)+sqrt(eps2)); // quarter wave stack dimensions
const double d2 = sqrt(eps1) / (sqrt(eps1)+sqrt(eps2));
const double half_cavity_width = d2;
const int N = 5;

Meep supports perfectly matching layers (PML) as absorbing boundary conditions. The PML begins at the edge of the computational volume and works inwards. Hence, we specify the size of the computational cell as follows (note that z_center is half the cell length):

const double pml_thickness = 1.0;
const double z_center = half_cavity_width + N*grating_periodicity + pml_thickness;

To specify a dielectric structure in Meep, we define a function that takes as input one parameter, a position vector, and returns the value of the dielectric at that position.

double eps(const vec &p) {
vec r = p - vec(z_center);
if (abs(r.z()) < half_cavity_width)
return 1.0;
else {
double dz = abs(r.z()) - half_cavity_width;
while (dz > grating_periodicity)
dz -= grating_periodicity;
return (dz < d1) ? eps1 : eps2;
}
}

We are now ready to set up the computational cell, excite the sources, time step the fields, and compute the resulting quality factor. Here we set up the main part of the control file incorporating some of Meep's classes and sub routines. We will excite the mode at the midgap of the bandgap, which we expect to have the largest quality factor since it will have the largest exponential decay.

int main(int argc, char **argv){
initialize mpi(argc,argv);
const double amicron = 10.0;
const grid_volume vol = vol1d(2*z_center,amicron);
structure s(vol,eps,pml(pml_thickness));
fields f(&s);

Note the constructor for the grid_volume class in 1D which takes as parameters the size of the cell and the resolution. The sources in Meep are excitations of the polarization vector in $\mathbf{D}=\epsilon\mathbf{E}+\mathbf{P}$. The polarization can be any one of the six cartesian or cylindrical fields. There are a variety of sources including dipole and current sources, gaussian pulses and a continuous wave sources. For now, we use a gaussian source centered at the midgap frequency with a narrow width, along with the start time and end time of the source.

 const double w_midgap = (sqrt(eps1)+sqrt(eps2))/(4*sqrt(eps1)*sqrt(eps2));
gaussian_src_time src(w_midgap, 1.0, 0.0, 5.0);

Here we make use of the built in function, vol.center(), that automatically computes the geometric center of the of the computational volume. To see what the unit cell of the dielectric function looks, output as an HDF5 file, we invoke the built in command:

f.output_hdf5(Dielectric, v.surroundings())

This versatile command can be used to visualize all of the field components, field energy when called at a particular time step. The resulting output file will be eps-000000.00.h5 and to view it, we first need to convert it to the portable network graphics (PNG) format with the h5topng tool. We are now ready to begin the time stepping of the fields, and do so in a simple loop:

while (f.time() < f.last_source_time()) f.step();

To calculate the quality factor, we need to monitor a field beyond the time at which our point source has been turned off. We can do this in Meep by establishing an array where all fields at a particular point can be monitored. We set up the area as well as the output arrays we will need later as such:

int maxbands = 5;
int ttot = int(400/f.dt)+1;
complex<double> *p = new complex<double>[ttot];
complex<double> *amps = new complex<double>[maxbands];
double *freq_re = new double[maxbands];
double *freq_im = new double[maxbands];

The variable maxbands is the number of mode frequencies we will be looking for while ttot represents the total number of timesteps for which we will monitor the field component at. For this simulation given the large quality factor that we expect, we will time step for a long period of time which in this case is 400 unit seconds. Now we are ready to time step and collect the field information:

 int i=0;
while (f.time() < f.last_source_time() + 400) {
f.step();
p[i++] = f.get_field(Ex,vol.center());
}

The get_field function does exactly that, obtains the value of the field at a given position within the computational cell using linear interpolation where necessary. Now we have all the information we need and must obtain the frequencies of the modes which we do by invoking a special tool for harmonic inversion, harminv, to extract the real and imaginary frequencies:

 int num =  do_harminv(p, ttot, f.dt, 0.8*w_midgap, 1.2*w_midgap, maxbands, amps, freq_re, freq_im);

The integer returned by harminv is the number of mode frequencies obtained from the input data p. The particular call to harminv included passing the arrays by value and telling harminv to look for frequencies within 20% of the mid gap frequency up to a maximum of 5 bands. At this point, the necessary information to compute the quality has been stored in the freq_re and freq_im arrays and we compute the quality factor using the formula, Q = − ωr / 2ωi. All that is left to do, is to print these results with the quality factor:

master_printf("frequency,amplitude,quality factor\n");
for (int i=0; i<num; ++i)
master_printf("%0.6g%+0.6gi,%0.5g%+0.5gi,%g\n",freq_re[i],freq_im[i],real(amps[i]),imag(amps[i]),-0.5*freq_re[i]/freq_im[i]);
return 0;
}

That's it, we are done. The output for this program is 3.26087e+06. The large quality factor is to be expected since our system includes a relatively high number of bragg reflectors (N=5) in each direction and we are exciting the mode at mid gap. Suppose we want to see what the resonant mode looks like, say over one period. The following block of code time steps the field for exactly one period while outputting the Ex field for the computational cell at each time step:

double curr_time = f.time();
while (f.time() < curr_time + 1/w_midgap) {
f.output_hdf5(Ex, vol.surroundings());
f.step();
}

After we convert the hdf5 files to PNG format and superimpose the images on the dielectric background to produce an animated, we obtain the following:

## Compiling

In order to compile your code and link against the Meep library, you must link in several additional libraries that Meep requires. There are three possible ways to do this:

1. After compiling the Meep package following the instructions elsewhere in this wiki, place myfile.cpp (where myfile.cpp is a MEEP C++ file of your choosing) in the tests/ subdirectory, cd to the same directory, and invoke:

make myfile.dac

Run the resulting executable via:

./myfile.dac

2. Use the w:pkg-config program, which is installed by default on most GNU/Linux systems. Then, to compile a file foo.cpp using Meep, do:

g++ pkg-config --cflags meep foo.cpp -o foo pkg-config --libs meep

(Naturally, replace g++ with the name of your C++ compiler if you are not using the GNU compilers.)

3. Compile with g++, this time invoking each library separately:

g++ -malign-double foo.cpp -o foo -lmeep -lhdf5 -lz -lgsl -lharminv -llapack -lcblas -latlas -lfftw3 -lm


text:
# What pressure (in atmospheres) is required to compress 1.00 L of gas at 760. mm Hg pressure to a volume of 50.0 mL?

$\text{20 atmospheres........}$
${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$, now ${P}_{1} = 1 \cdot a t m$, and so we solve for ${P}_{2}$:
P_2=(P_1V_1)/V_2=(1.00*cancelLxx1*atm)/(0.050*cancelL)=??atm

format: plain


text: Given a set A and a relation R in A, R is reflexive iff all the ordered pairs of the form are in R for every x in A. A relation which fails to be reflexive is called Some important results concerning Rosenberg partial hypergroupoids, induced by relations, are generalized to the case of Degree of Relationship Degree of relationship: describes the number of entities involved in a relationship Unary (one entity) Binary (two entities) Ternary (three entities) N’ary(more than 3) Binary (two entities) relationship is most common 20 The binary operations associate any two elements of a set. Binary relation for sets This video is about: Introduction to Binary Relation. If a is an element of a set A, then we write a A∈ and say a belongs to A or a is in A or a is a member of A.If a does not belongs to A, we write Set alert. ��I7���v7]��҈jt�ۮ]���}��|qYonc��3at[�P�*ct���M�!ǣ��" ���=䑍F���4~G�͐Ii]���מS�=96���G����_J���c0�dD�_�|>��)��|V�MTpPn� -����x�Լ�7z�ǋ�'ESF��(��R9�c�bS� ㉇�ڟio�����XO��^Fߑ��&�*�"�;�0 Jyv��&��2��Y,��E��ǫ�DҀ�y�dX2 �)I�k ��nj]��gw�e����"φ�0)��?]�]��O!���C�s�D�Y}?�? A binary relation A is a poset iff A does not admit an embedding of the following finite relations: The binary relation … Binary Relations A binary relation over a set A is some relation R where, for every x, y ∈ A, the statement xRy is either true or false. Properties of Binary Relations 87 (8) R is transitive in X iﬀ forx,y,z st x ∈ X & y ∈ X & z ∈ X & hx,yi ∈ R & hy,zi ∈ R holds hx,zi ∈ R. We now deﬁne several new predicates. %äüöß The wife-husband relation R can be deﬂned from X to Y. 3 0 obj Binary Relations 6 Exercise: Given set A = {r, o, t, p, c} and set B = {discrete, math, proof, proposition}, and corresponding relation R ⊆ A × B such that the tuple (letter, word) is in the relation if that letter occurs somewhere in the word. :��&i�c�*��ANJ#2�W !jZ�� eT�{}���t�;���]�N��?��ͭ�kM[�xOӷ. Properties Properties of a binary relation R on a set X: a. reflexive: if for every x X, xRx holds, i.e. Download Binary Relation In Mathematics With Example doc. In Studies in Logic and the Foundations of Mathematics, 2000. A binary relation A is a poset iff A does not admit an embedding of the following finite relations: The binary relation … Binary relations. We consider here certain properties of binary relations. >> stream Let R is a relation on a set A, that is, R is a relation from a set A to itself. For instance, let X denote the set of all females and Y the set of all males. 1. M���LZ��l�G?v�P:�9Y\��W���c|_�y�֤#����)>|��o�ޣ�f{}d�H�9�vnoﺹ��k�I��0Kq)ө�[��C�O;��)�� &�K��ea��*Y���IG}��t�)�m�Ú6�R�5g |1� ܞb�W���������9�o�D�He夵�fݸ���-�R�2G�\{�W� �)Ԏ Binary relations establish a relationship between elements of two sets Definition: Let A and B be two sets.A binary relation from A to B is a subset of A ×B. x��T˪�0��+�X�����&�����tצ���f���. View Relation.pdf from COMPUTERSC CS 60-231 at University of Windsor. 1.1.2 Preorders A preorder or ordered set is a pair (X,≤) where Xis a set and ≤ is a reﬂexive transitive binary relation on X. 2.1: Binary Relations - Mathematics LibreTexts Skip to main content We can also represent relations graphicallyor using a table endstream 1.1.2 Preorders A preorder or ordered set is a pair (X,≤) where Xis a set and ≤ is a reﬂexive transitive binary relation on X. We express a particular ordered pair, (x, y) R, where R is a binary relation, as xRy. About this page. Reflexivity. @*�d)���7�t�a���M�Y�F�6'{���n endobj Math 461 Relations and Orders 8 Linear orders Deﬂnition 8.1. Such classes are typically speci ed in terms of the properties required for membership. Example 1.6. 9�����D���-��XE��^8� Sets are usually denoted by capital letters A B C, , ,K and elements are usually denoted by small letters a b c, , ,... . Given a set A and a relation R in A, R is reflexive iff all the ordered pairs of the form are in R for every x in A. We express a particular ordered pair, (x, y) R, where R is a binary relation, as xRy. + : R × R → R e is called identity of * if a * e = e * a = a i.e. | Find, read and cite all the research you need on ResearchGate Finally, Jason Joan Yihui Binary relation Definition: Let A and B be two sets. Binary relations generalize further to n-ary relations as a set of n-tuples indexed from 1 to n, and yet further to I-ary relations where Iis an arbitrary index set. Formally, De nition 1.1 A binary relation in a set A is a subset RˆA A. Also, R R is sometimes denoted by R 2. 5.2.1 Characterization of posets, chains, trees. The wife-husband relation R can be thought as a relation from X to Y.For a lady The following de nitions for these properties are not completely standard, in that they mention only those ordered pairs Formally, a binary relation R over a set X is symmetric if: ∀, ∈ (⇔). A function f WA !B is a special case of binary relation in which A binary relation from A to B is a subset of a Cartesian product A x B. R t•Le A x B means R is a set of ordered pairs of the form (a,b) where a A and b B. 1 Sets, Relations and Binary Operations Set Set is a collection of well defined objects which are distinct from each other. Download Binary Relation In Mathematics With Example pdf. 5 0 obj Then R R, the composition of R with itself, is always represented. Similarly, the subset relation relates a set, A, to another set, B, precisely when A B. The symmetric component Iof a binary relation Ris de ned by xIyif and only if xRyand yRx. The logical operations treat a binary relation purely as a set, ignoring the nature of its ele-ments. The dual R0of a binary relation Ris de ned by xR0yif and only if yRx. Introduction to Relations 1. Examples: < can be a binary relation over ℕ, ℤ, ℝ, etc. Binary operations on a set are calculations that combine two elements of the set (called operands) to produce another element of the same set. View 5 - Binary Relations.pdf from CS 2212 at Vanderbilt University. <> 5 Binary Relation Wavelet Trees (BRWT) We propose now a special wavelet tree structure to represent binary relations. We can define binary relations by giving a rule, like this: a~b if some property of a and b holds This is the general template for defining a relation. Subscribe to our YouTube channel to watch more Math lectures. ≡ₖ is a binary relation over ℤ for any integer k. A binary relation R on X is apreorderif R is re exive and transitive. 5.2.1 Characterization of posets, chains, trees. Chapter 4: Binary Operations and Relations 4.1: Binary Operations DEFINITION 1. Just as we get a number when two numbers are either added or subtracted or multiplied or are divided. Draw the following: 1. Preference Relations, Social Decision Rules, Single-Peakedness, and Social Welfare Functions 1 Preference Relations 1.1 Binary Relations A preference relation is a special type of binary relation. Binary Relations and Preference Modeling 51 (a,b) 6∈Tor a¬Tb. This wavelet tree contains two bitmaps per level at each node v, Bvl and Bvr . <> Binary Relations De nition: A binary relation between two sets X and Y (or between the elements of X and Y) is a subset of X Y | i.e., is a set of ordered pairs (x;y) 2X Y. Theory of Relations. • We use the notation a R b to denote (a,b) R and a R b to denote (a,b) R. If a R b, we say a is related to b by R. 2 0 obj 0 denotes the empty relation while 1 denoted (prior to the 1950’s)1 the complete relation … In Studies in Logic and the Foundations of Mathematics, 2000. De nition of a Relation. 0 denotes the empty relation while 1 denoted (prior to the 1950’s)1 the complete relation … A relation which fails to be reflexive is called Relations 1.1. binary relations for Ais a set C 2 Aof binary relations on A. De nition: A binary relation from a set A to a set Bis a subset R A B: If (a;b) 2Rwe say ais related to bby R. Ais the domain of R, and Bis the codomain of R. If A= B, Ris called a binary relation … 2. 4.4 Binary Relations Binary relations deﬁne relations between two objects. Therefore, such a relationship can be viewed as a restricted set of ordered pairs. Binary Relations Intuitively speaking: a binary relation over a set A is some relation R where, for every x, y ∈ A, the statement xRy is either true or false. 3 0 obj Some relations, such as being the same size as and being in the same column as, are reflexive. Binary Relations A binary relationRfrom a set Ato a set Bis a subset of A X B Example: •Let A = {0,1,2}andB = {a,b} •{(0, a), (0, b), (1,a) , (2, b)} is a relation from Ato B. Request PDF | On Jan 1, 2008, Violeta Leoreanu Fotea and others published n-hypergroups and binary relations. The predicate Ris … Week 4-5: Binary Relations 1 Binary Relations The concept of relation is common in daily life and seems intuitively clear. Since binary relations are sets, we can apply the classical operations of set theory to them. Properties Properties of a binary relation R on a set X: a. reflexive: if for every x X, xRx holds, i.e. stream Binary Relations Prof. Susan Older 29 September 2016 (CIS 375) Relations 29 September 2016 1 / 7 Others, such as being in front of or x��[[���~ϯ�("�t� '��-�@�}�w�^&�������9$wF��rҼ�#��̹~��ן��{�.G�Kz����r�8��2�������Y�-���Sb�\mUow����� #�{zE�A����������|� �V����11|LjD�����oRo&n��-�A��EJ��PD��Z��Z��~�?e��EI���jbW�a���^H���{�ԜD LzJ��U�=�]J���|CJtw��׍��.C�e��2nJ;�r]n�$\�e�K�u�����G墲t�����{"��4�0�z;f ���Ř&Y��s�����-LN�$��n�P��/���=���W�m5�,�ð�*����p[T���V$��R�aFG�H�R!�xwS��� ryX�q�� ��p�p�/���L�#��L�H��N@�:���7�_ҧ�f�qM�[G4:��砈+2��m�T�#!���բJ�U!&'l�( ��ɢi��x�&���Eb��*���zAz��md�K&Y�ej6 �g���\��Q���SlwmY\uS�cά�u��p�f��5;¬_����z�5r#���G�D��?��:�r���Q$��Q Relations and Their Properties 1.1. +|!���T �MP�o)�K �[��N?��xr_|����e���t�J���CX����L\�!��H�2ű���b����H=��_n�K+�����[���:� �mS�׮x�n���R���x�o�5,��W�>^��-t*v5VkX�>$�4�˴�B��jp_6\�fw�ˈ�R�-��u'#2��}�d�4���Υx� �t&[�� A binary relation R on X is aweak orderor acomplete preorderif R is complete and transitive. We ask that binary relation mathematics example of strict weak orders is related to be restricted to be restricted to the only if a reflexive relation Every set and binary in most Remark 2.1. Except when explicitly mentioned otherwise, we will suppose in all what follows that the set Ais ﬁnite . A binary relation is a set of pairs of elements assumed to be drawn from an indeterminate but ﬁxed set X. Namely, we provide a technique that enables us to prove an upper bound on the number of mistakes made by our polynomial-time algorithm for learning non-pure binary relations. 7 Binary Relations • Let A, B be any two sets. Types of Relations • Let R be a binary relation on A: – R is reflexive if xRx for every x in A – R is irreflexive if xRx for every x in A – R is symmetric if xRy implies yRx for every x,y in A – R is antisymmetric if xRy and yRx together imply x=y for every x,y in A – R is transitive if xRy and yRz imply xRz for every x,y,z in A /Filter /FlateDecode Binary relations A (binary) relation R between the sets S and T is a subset of the cartesian product S ×T. Except when explicitly mentioned otherwise, we will suppose in all what follows that the set Ais ﬁnite . For example, if a relation R is such that everything stands in the relation R to itself, R is said to be reflexive . The logical operations treat a binary relation purely as a set, ignoring the nature of its ele-ments. Rsatisﬂes the trichotomy property iﬁ … If R is a relation between X and Y (i.e., if R X Y), we often write xRy instead of (x;y) 2R. In other words, a binary relation R … It often happens that a binary relation R defined on a topological space (Formula presented.) Download as PDF. The resultant of the two are in the same set. Examples: < can be a binary relation over ℕ, ℤ, ℝ, etc. 5 Binary Relation Wavelet Trees (BRWT) We propose now a special wavelet tree structure to represent binary relations. A binary relation over a set $$A$$ is some relation $$R$$ where, for every $$x, y \in A,$$ the statement $$xRy$$ is either true or false. 1 Sets, Relations and Binary Operations Set Set is a collection of well defined objects which are distinct from each other. If a is an element of a set A, then we write a A∈ and say a belongs to A or a is in A or a is a member of A.If a does not belongs to A, we write Dynamic binary relations k 2 -tree a b s t r a c t introduce ofa binarydynamic relationsdata ⊆structure × .the compact representation R A B The data structure is a dynamic variant of the k2-tree, a static compact representation that takes advantage of clustering in the binary relation to achieve compression. /Length 3527 Remark 2.1. The matrix representation of the relation. The wife-husband relation R can be thought as a relation from X to Y.For a lady Binary Relations A binary relationRfrom a set Ato a set Bis a subset of A X B Example: •Let A = {0,1,2}andB = {a,b} •{(0, a), (0, b), (1,a) , (2, b)} is a relation from Ato B. A binary relation is a set of pairs of elements assumed to be drawn from an indeterminate but ﬁxed set X. << Sets are usually denoted by capital letters A B C, , ,K and elements are usually denoted by small letters a b c, , ,... . Binary Relations November 4, 2003 1 Binary Relations The notion of a relation between two sets of objects is quite common and intuitively clear. The arrow diagram representation of the relation. 2. %PDF-1.5 All these properties apply only to relations in (on) a (single) set, i.e., in A ¥ A for example. (x, x) R. b. ↔ can be a binary relation over V for any undirected graph G = (V, E). About this page. Properties of binary relations Binary relations may themselves have properties. x��TMk�0��W�9�ꌾ,�1�G�-����[��� $���̌d��˶��P�H���4���h,�ٙ!�_~7�/�!���O�2��$�o�ĺv;�HPf7�����1WC�2#xY8��/WL�a���᡹��� � ��h\A��~uh���c�(A� 3.y�skt��7� %���� �6"����f�#�����h���uL��\$�,ٺ4����h�4 ߑ+�a�z%��і��)�[��WNY��4/y!���U?�Ʌ�w�-� Definition (binary relation): A binary relation from a set A to a set B is a set of ordered pairs where a is an element of A and b is an element of B. Let Rbe a binary relation on A. domain, with elements of another set called the codomain . %PDF-1.4 Reflexivity. M.�G�ٔ�e��!���"ix61����i�ţ��}S\pX%_�hr���u�a�s���X��v�iI�ZWT�� A binary relation from A to B is a subset of a Cartesian product A x B. R t•Le A x B means R is a set of ordered pairs of the form (a,b) where a A and b B. Similarly, R 3 = R 2 R = R R R, and so on. stream We implement the above idea in CASREL, an end-to-end cascade binary tagging framework. Then the complement of R can be deﬁned by R = f(a;b)j(a;b) 62Rg= (A B) R Inverse Relation A binary relation associates elements of one set called the . Binary Relations and Preference Modeling 51 (a,b) 6∈Tor a¬Tb. This wavelet tree contains two bitmaps per level at each node v, Bvl and Bvr . Brice Mayag (LAMSADE) Preferences as binary relations Chapter 1 7 / 16 relation to Paul. When an ordered pair is in a relation R, we write a R b, or R. It means that element a is related to element b in relation R. We can also represent relations graphicallyor using a table Let us consider R. The predicate Ris reﬂexive is deﬁned by R is reﬂexive in ﬁeldR. Let's see how to prove it. Let Aand Bbe sets and deﬁne their Cartesian product to be the set of all pairwise Consider the binary relation ~defined over the set ℤ: a~b if a+bis even Some examples: 0~4 1~9 2~6 5~5 Turns out, this is an equivalence relation! learning non-pure binary relations, and demonstrate how the robust nature of WMG can be exploited to handle such noise. It is an operation of two elements of the set whose … Since binary relations are sets, we can apply the classical operations of set theory to them. Knowledge Hypergraphs: Prediction Beyond Binary Relations Bahare Fatemi1; 2y, Perouz Taslakian , David Vazquez2 and David Poole1 1University of British Columbia 2Element AI fbfatemi, pooleg@cs.ubc.ca, fperouz,dvazquezg@elementai.com, Abstract Knowledge graphs store facts using relations … Relations and Their Properties 1.1. Abinary relation from A to B is a subset of A B . De nition 1.5. • We use the notation a R b to denote (a,b) R and a R b to denote (a,b) R. If a R b, we say a is related to b by R. A binary relation R from A to B, written R : A B, is a subset of the set A B. Complementary Relation Deﬁnition: Let R be the binary relation from A to B. 6.042 6.003 6.012 6.004 . In this paper, we introduce and study the notion of a partial n-hypergroupoid, associated with a binary relation. De nition: A binary relation from a set A to a set Bis a subset R A B: If (a;b) 2Rwe say ais related to bby R. Ais the domain of R, and Bis the codomain of R. If A= B, Ris called a binary relation … relation to Paul. https://www.tutorialspoint.com/.../discrete_mathematics_relations.htm All these properties apply only to relations in (on) a (single) set, i.e., in A ¥ A for example. Binary Relations Any set of ordered pairs defines a binary relation. A binary relation is essentially just any set of ordered pairs. Theory of Relations. endobj A partial order is an antisymmetric preorder. a + e = e + a = a This is only possible if e = 0 Since a + 0 = 0 + a = a ∀ a ∈ R 0 is the identity element for addition on R Others, such as being in front of or VG�%�4��슁� Some relations, such as being the same size as and being in the same column as, are reflexive. lec 3T.3 . If (a,b) ∈ R, we say a is in relation R to be b. A binary relation R on X is atotal orderor alinear orderif R is complete, antisymmetric and transitive. De nition of a Relation. Knowledge Hypergraphs: Prediction Beyond Binary Relations Bahare Fatemi1; 2y, Perouz Taslakian , David Vazquez2 and David Poole1 1University of British Columbia 2Element AI fbfatemi, pooleg@cs.ubc.ca, fperouz,dvazquezg@elementai.com, Abstract Knowledge graphs store facts using relations … A symmetric relation is a type of binary relation.An example is the relation "is equal to", because if a = b is true then b = a is also true. Albert R Meyer . The binary operation, *: A × A → A. Download as PDF. The relation R S is known the composition of R and S; it is sometimes denoted simply by RS. Interpretation. CS 2212 Discrete Structures 5. (x, x) R. b. Ris irre°exive iﬁ ha;ai2=Rfor all a2A. For example, “less-than” on the real numbers relates every real number, a, to a real number, b, precisely when a


text:
# What is the parametric equation of ellipse ?

## Solution :

The equation x = acos$$\theta$$ & y = bsin$$\theta$$ together represent the parametric equation of ellipse $${x_1}^2\over a^2$$ + $${y_1}^2\over b^2$$ = 1, where $$\theta$$ is a parameter.

Note that if P($$\theta$$) = (acos$$\theta$$, bsin$$\theta$$) is on the ellipse then ; Q($$\theta$$) = (acos$$\theta$$, bsin$$\theta$$) is on auxilliary circle.

A circle described on major axis as diameter is called auxilliary circle.


text:
# B Quantum Darwinism question

1. Oct 18, 2016

### cube137

Quantum Darwinism is about fragments and how different observers should observe the same thing, may I know where the born rule is applied. does the proper mixed state (or one outcome) occurs after the fragments leave the objects, halfway or when the observers received the fragments?

2. Oct 18, 2016

### Staff: Mentor

3. Oct 18, 2016

### cube137

4. Oct 19, 2016

### Staff: Mentor

As far as I can tell, it isn't; the Zurek paper derives all of its key conclusions without ever using the Born rule. The paper also stresses that it is "interpretation free"; there is no claim that collapse occurs or that it doesn't occur. That question is simply not addressed.

5. Oct 19, 2016

### cube137

The paper quoted below (I'd like to understand what it means exactly by "Quantum states acquire objective existence when reproduced in many copies", I hope seasoned quantumists like Stevendaryl or Simon Phoenix can comment. Peterdonis is a relativist. And also what it means "Consensus between records deposited in fragments of E looks like collapse"):

Last edited by a moderator: Oct 19, 2016
6. Oct 19, 2016

### Staff: Mentor

First you have to understand what he means by "objective existence". Basically, it looks to me like he means "repeatability": repeated measurements of a system must give consistent results. The "many copies" is what allows that to happen: each measurement basically takes its information from a different copy. And each measurement transfers the information in the copy to the measuring device, so each copy can only be used once to obtain information about the measured system--that's why you need many copies to have repeated measurements give consistent results.

I think the key here is "looks like collapse". What he is trying to explain is why, regardless of whether there actually is collapse or not, our observations make it look like there is collapse--where here "look like collapse" basically means the same thing as "objective existence" above, i.e., repeated measurements giving consistent results. And, as above, what allows that to happen is that there are many "records" (copies), each one in a different fragment of E (the environment), and all of them containing the same information about the measured system.

Again, this is all "interpretation free", as the paper says. It doesn't say that collapse "really" happens; it leaves that question open. It just explains why it looks that way to us.

Just to clarify, I do post most often in the relativity forum, but I'm reasonably conversant with QM and QFT. The main reason I don't post more often in this forum is that it seems far more prone to threads that are more about philosophy than physics (QM is funny that way). But this thread topic does not fall into that category.

7. Oct 20, 2016

### cube137

Ok. To understand what Zurek was driving at. Supposed there was not "many copies".. but only one copy, what would happen?

This is how I understood Zurek. In the second paragraph he wrote:

"Fragility of states is the second problem with quantumclassical
correspondence: Upon measurement, a general
preexisting quantum state is erased - it "collapses" into
an eigenstate of the measured observable. How is it then
possible that objects we deal with can be safely observed,
even though their basic building blocks are quantum?"

Can we take as example a macroscopic object like a piece of wood. Is this what Zurek refer to as objects where their basic building blocks are quantum? Or was Zurek referring to object as a small isolated quantum system like a buckyball or a particle? Without the "many copies".. did Zurek mean if different people interact with the same wood, they can re-prepare it and the wood can change shape? Or was he referring only to atoms or tiny isolated quantum systems where people can re-prepare it by measuring it? But I don't think he was referring only to electrons, atoms, or other microscopic systems but big objects like wood too. Is that right?

8. Oct 20, 2016

### secur

"Preferred states of S become objective" when "many observers find out the state of S independently". So the answer to your question is the third option: "when the observers received the fragments". The outcome occurs when multiple observers receive consensus information about the state. However, it can occur without humans: i.e. the environment itself can "witness" those multiple fragments.

The fragments concept is not very important in this work, neither is the "Darwinian" ansatz. The key concepts are pointer states, einselection, envariance, equiprobability (symmetry, swap), finegraining (using ancilla C) and a couple others. Some of it's pretty clever but it's not "interpretation-free". Quantum Darwinism is a flavor of MWI.

More or less nothing. Only one copy would not be enough to allow objective outcome.

Yes.

No. "Re-prepare" applies only to quantum objects.

9. Oct 20, 2016

### cube137

You agreed that a piece of wood is what Zurek refers to as objects where their basic building blocks are quantum. Hence wood can be considered as quantum objects in quantum Darwinism... hence you can re-prepare a piece of wood in the world of Zurek. But then, maybe let's refer to "Re-prepare" for electrons, photons, and small quantum system, while reserved the word "Transmute" for changing the shape of the wood. If you won't agree. But Zurek just said macroscopic object can considered as quantum object too like buckyball. Remember in the world of quantum decoherence, everything is quantum. Our classical bias only came because of Copenhagen and because of Everett where each branch is a solid classical world hence trapping one to the mindset that there is a classical world when everything is quantum at the core.

10. Oct 20, 2016

### Staff: Mentor

If there were only one copy, you would not be observing a macroscopic object and you would not be getting measurement results of the sort we usually associate with such objects. You would be looking at, say, a quantum computing experiment where a single qubit is being manipulated. And the results you get would not look like something with "objective existence" in the sense Zurek is using that term.

All objects have basic building blocks that are quantum.

No. Zurek is saying that it is impossible for a piece of wood not to have "many copies" of the information about it in the environment, because it is composed of such a huge number of quantum building blocks and it is not at all isolated--each building block is continually interacting with other building blocks and with the environment. To be able to "re-prepare" a piece of wood, you would have to first isolate it from its environment, and then control all the interactions between its quantum building blocks with sufficient accuracy to allow quantum interference effects to be observable. In principle that could be done, but in practice it is impossible.

The reason we can "re-prepare" atoms or tiny isolated quantum systems (like qubits in a quantum computer) by measuring them is that they are tiny isolated quantum systems; we can prevent them from interacting with their environment, and they contain a small enough number of quantum building blocks (just one, in the ideal case) that we can control all their interactions.

11. Oct 20, 2016

### Staff: Mentor

I think this is misstated. A correct statement would be that re-preparing is only possible in practical terms for objects containing a small number of quantum building blocks which can be isolated sufficiently from the environment that all of their interactions can be controlled. But there is no magic boundary at which objects stop being quantum. All objects are composed of quantum building blocks. It's just that not all objects meet the practical conditions for us being able to re-prepare them.

12. Oct 20, 2016

### Staff: Mentor

Not in practical terms. See my previous posts.

13. Oct 20, 2016

### secur

To back up a little, QD attempts to answer the question why large-scale quantum systems - for instance a block of wood - present one stable state. I.e., why we cannot, in practice, "transmute" or re-prepare them. The answer: decoherence via interaction with the environment causes einselection of pointer states, wherein all phase information is lost. Once that happens we obtain "objective reality". All subsequent measurements will give the same state. You can no longer re-prepare that block of wood, as you can with small quantum objects such as an electron.

Obviously this is a practical program. QD tries to explain why, in practice, a block of wood can't be re-prepared, even though in theory it can. Here's a relevant quote:

Note, he never defines the term "re-prepare" - anywhere on the net, that I could find. We're all supposing it means a second change, after the initial "collapse". QD usually uses it in that sense. Also he doesn't define "quantum object" but it seems to mean the wavefunction itself. I was using it to mean "small objects"; that might not be exactly consistent with Zurek. Fortunately you both understood what I meant.

Note, I'm only answering the question, "What does QD say?" I don't necessarily agree with it.

This is, indeed, typical "quantum philosophy", contrary to your statement above. It can easily lead to picking terminological nits, attempting to mind-read Zurek, and - in a word - the endless philosophical debate of QM interpretation. So, if you still disagree with my take on "re-prepare", I'll happily concede the point.

OTOH QD is pretty interesting, and I'm up to speed on it now. So that topic is worth pursuing.

14. Oct 20, 2016

### Staff: Mentor

Yes, small quantum objects. (Which, if unpacked, basically means what I said before, objects made of a small enough number of quantum building blocks that they can be isolated and their interactions controlled.) But your previous post that I objected to didn't include the qualifier "small", which is why I objected and why it appears to have confused the OP.

Or, in terms that the OP seems more comfortable with, it tries to explain why, in practice, only small quantum objects can be re-prepared, even though all objects are quantum objects and in theory any quantum object can be re-prepared.

15. Oct 20, 2016

### cube137

Supposed for sake of discussion there were really only one copy or not even a single copy. So what would the piece of wood look like? (please don't use example of single qubit but a big object like wood). Would the wood become invisible or became a blob in superposition? Or what should the piece of wood look like (just a rough description if you can't describe completely)?

16. Oct 20, 2016

### Staff: Mentor

In order to do this, you would have to completely isolate the piece of wood from its environment. That is, you would have to somehow prevent all interactions between every single quantum building block in the piece of wood and anything else. In practice that is impossible, by many, many orders of magnitude.

But suppose you could, in fact, do the above. You would still have to deal with the fact that the piece of wood is composed of a huge number of quantum building blocks, something like $10^{25}$ of them, and they are interacting with each other, and you can't control the interactions, which means you can't run controlled experiments on the piece of wood to test its quantum state, the way you can for small quantum systems like qubits. Nor can we possibly model the detailed interactions between all the building blocks mathematically.

So I don't think anyone knows what such a piece of wood would look like, nor does it matter since we can't make one anyway.

17. Oct 20, 2016

### cube137

If that's true. Then the "objects" he meant in the following are only quantum objects like electrons, photons, and small quantum system and not a block of wood or larger object where it can't be isolated. Correct? But the way he writes it.. it sounds like any objects like cars.

"Fragility of states is the second problem with quantumclassical
correspondence: Upon measurement, a general
preexisting quantum state is erased - it "collapses" into
an eigenstate of the measured observable. How is it then
possible that objects we deal with can be safely observed,
even though their basic building blocks are quantum?"

In the following article. http://www.nature.com/news/2004/041223/full/news041220-12.html They emphasized the "objects" meant macroscopic object like the Buckingham palace:

"If it wasn't for quantum darwinism, the researchers suggest in Physical Review Letters1, the world would be very unpredictable: different people might see very different versions of it. Life itself would then be hard to conduct, because we would not be able to obtain reliable information about our surroundings... it would typically conflict with what others were experiencing.

The difficulty arises because directly finding out something about a quantum system by making a measurement inevitably disturbs it. "After a measurement," say Wojciech Zurek at Los Alamos National Laboratory in New Mexico and his colleagues, "the state will be what the observer finds out it is, but not, in general, what it was before."

Because, as Zurek says, "the Universe is quantum to the core," this property seems to undermine the notion of an objective reality. In this type of situation, every tourist who gazed at Buckingham Palace would change the arrangement of the building's windows, say, merely by the act of looking, so that subsequent tourists would see something slightly different.

So this is very bad example? They should not use example of macroscopic object like Buckingham Palace! But then.. for other mentors.. in case there is a chance PeterDonis may have misunderstood the issue. Please let us know the case. Thanks.

18. Oct 20, 2016

### Staff: Mentor

No. All objects are quantum objects, in the sense that they are composed of quantum building blocks. But the number of building blocks in the object makes a difference. An object with only 1 building block, like an electron, is very different from an object with $10^{25}$ building blocks, like a piece of wood. Just because everything is a quantum object doesn't mean everything has to behave exactly the same.

First of all, this isn't strictly true. If the system is already in an eigenstate of the observable being measured, then the measurement doesn't change its state. But that's not really a practical issue, because if we already know the system is in an eigenstate, we don't need to measure it anyway because we already know its state.

However, once again, the size of the disturbance relative to the size of the object matters. If you are measuring an object that has only one quantum building block, like an electron, any measurement you make is going to disturb it significantly--heuristically, because the measurement itself has a minimum size which is basically one quantum building block. (For example, if we try to measure the electron by bouncing photons off of it, the minimum measurement we can make is to use one photon.) But if you are measuring an object with $10^{25}$ building blocks, like a piece of wood, there are lots of ways to measure it without significantly affecting its state, simply because of the huge number of building blocks. In fact, measuring an object of that size is really no different from what its environment is continually doing to it anyway--which is part of Zurek's point. The reason macroscopic objects like pieces of wood or Buckingham Palace look the same to everybody is that none of us have to do anything special to measure them; they're already being measured, all the time, just by being embedded in their environment. All Zurek is doing is giving more details about how that works and why it privileges particular states, the ones we think of as "classical" states and are intuitively familiar with.

No. See above.

19. Oct 20, 2016

### cube137

Yes. That's what standard decoherence does.. constantly interacting with objects and environment where they are entangled. And the improper mixed state becomes proper mixed state. We accept the proper mixed state which could be Bohmian, branches in Many Worlds, etc.

What Zurek was trying to do (please confirm my understanding) was like trying to derive how improper mixed state becomes proper mixed state. In the objects, they are in improper mixed state with environment. Then decoherence chose the pointer states and there are many fragments in the environment to make up classical world. In Copenhagen, Bohmian, Many worlds.. Classical world is a priori, observation is the primitive so you don't need quantum Darwinism.. but in Zurek quantum Darwinism, quantum state is the primitive. So far my analysis is correct (let's now use the natural language of Density Matrix)? So by this context if there is a way to block the many fragments (just theoretically), then the object would become invisible (because you can't perceive superposition and the eigenstate information flow is blocked). This is just for sake of illustration although I know there is no way to block the fragments.

20. Oct 20, 2016

### Staff: Mentor

Not quite. See below.

I don't think this is correct. Decoherence explains why we can neglect interference terms between different "classical" alternatives, but that means going from a superposition to a mixed state, not from an improper mixed state to a proper mixed state.

No. What Zurek was trying to do is to explain why, when decoherence happens, it always ends up putting systems in mixed states with "classical" alternatives--live cats or dead cats, blocks of wood over here or over there, Buckingham Palace built vs. not built, etc.--instead of mixed states that don't look anything like the classical states we observe--for example, a mixture of ("block of wood here" plus "block of wood there") and ("block of wood here" minus "block of wood there"), which is a perfectly valid mixed state, but not one that describes any block of wood we've ever observed.

In other words, according to Zurek, it's not enough to explain, as standard decoherence theory does, why macroscopic systems end up in mixed states, with interference terms negligible. You have to also explain why they end up in the particular mixed states we actually observe, and never all the other possible mixed states that are possible mathematically. As I understand it, that is what Quantum Darwinism tries to do.


text:
# Best way to solve $X^3-X^2-X-1=0$

can anyone help me for this cubic equation ? can be solved without delta method?

$X^3-X^2-X-1=0$

(answer is $\sim 1.8393$)

• Welcome to MSE! Please note, for further reference, the editing I did to your post. $\LaTeX$ is the language of choice for writing math here. Enjoy! – Andreas Caranti Jun 3 '14 at 16:48
• You can rewrite the equation as $x^3 = x^2 + x + 1$, so $x = \left(x^2 + x + 1\right)^{1/3}$. Write this formula in your calculator, start with $x = 1$, and repeatedly substitute the answer as $x$, and you'll get closer and closer to that answer. See en.wikipedia.org/wiki/Fixed-point_iteration – M. Vinay Jun 3 '14 at 16:55
• @ReneSchipperus The other two roots are complex. – M. Vinay Jun 3 '14 at 16:56
• Some better methods are bisection method, false position (regula-falsi) method, and Newton-Raphson method. But fixed-point iteration is easy and works in this case. – M. Vinay Jun 3 '14 at 16:59
• see mathworld.wolfram.com/TribonacciConstant.html by Cardano's method – Will Jagy Jun 3 '14 at 17:37

Here's a solution that finds all three roots of the function

$$x^3-x^2-x-1=0$$ Substitute $y=x-\frac{1}{3}$: $$-\frac{4}{3}-y-(y+\frac{1}{3})^2+(y+\frac{1}{3})^3=0$$ Expanding this gives: $$y^3-\frac{4}{3}y-\frac{38}{27}=0$$ If $y=\frac{\lambda}{z}+z$ then $z=\frac{1}{2}\left(y+\sqrt{y^2-4\lambda}\right)$ $$-\frac{38}{27}-\frac{4}{3}\left(z+\frac{\lambda}{z}\right)+\left(z+\frac{\lambda}{z}\right)^3=0$$ Multiply both sides by $z^3$ $$z^6+z^4\left(3\lambda-\frac{4}{3}\right)-\frac{38z^3}{27}+z^2\left(3\lambda^2-\frac{4\lambda}{3}\right)+\lambda^3=0$$ Substitute $\lambda=\frac{4}{9}$ and $u=z^3$ $$u^2-\frac{38}{27}u+\frac{64}{729}=0$$ Choose one of the solutions $$u=\frac{1}{27}(19+3\sqrt{33})$$ Substitute back for $u=z^3$ $$z^3=\frac{1}{27}(19+3\sqrt{33})$$ Solving for $z$ $$z=\frac{1}{3}\sqrt[3]{19+3\sqrt{33}}$$ $$z=-\frac{1}{3}\sqrt[3]{-19+3\sqrt{33}}$$ $$z=\frac{1}{3}(-1)^{2/3}\sqrt[3]{19+3\sqrt{33}}$$ Starting with the first solution, we substitute back for $z=\frac{y}{2}+\frac{1}{2}\sqrt{y^2-\frac{16}{9}}$ Solving this for $y$ gives $$y=\frac{4}{3\sqrt[3]{19+3\sqrt{33}}}+\frac{1}{3}\sqrt[3]{19+3\sqrt{33}}$$ Substitute back for $y=x-\frac{1}{3}$ $$x=\frac{1}{3}+\frac{4}{3\sqrt[3]{19+3\sqrt{33}}}+\frac{1}{3}\sqrt[3]{19+3\sqrt{33}}$$ This is the first solution for $x$, going on to the second solution $$z=-\frac{1}{3}\sqrt[3]{-19-3\sqrt{33}}$$ Substitute back for $z=\frac{y}{2}+\frac{1}{2}\sqrt{y^2-\frac{16}{9}}$ $$\frac{y}{2}+\frac{1}{2}\sqrt{y^2-\frac{16}{9}}=-\frac{1}{3}\sqrt[3]{-19-3\sqrt{33}}$$ Solve for $y$ $$y=\frac{4(-1)^{2/3}}{3\sqrt[3]{19+3\sqrt{33}}}-\frac{1}{3}\sqrt[3]{-19-3\sqrt{33}}$$ Substitute back for $y=x-\frac{1}{3}$ $$x=\frac{1}{3}+\frac{4(-1)^{2/3}}{3\sqrt[3]{19+3\sqrt{33}}}-\frac{1}{3}\sqrt[3]{-19-3\sqrt{33}}$$ Here's the second solution $$z=\frac{1}{3}(-1)^{2/3}\sqrt[3]{19+3\sqrt{33}}$$ For the last solution we substitute back for $z=\frac{y}{2}+\frac{1}{2}\sqrt{y^2-\frac{16}{9}}$ $$\frac{y}{2}+\frac{1}{2}\sqrt{y^2-\frac{16}{9}}=\frac{1}{3}(-1)^{2/3}\sqrt[3]{19+3\sqrt{33}}$$ Solve the equation for $y$ $$y=\frac{(-1)^{2/3}}{3}\sqrt[3]{19+3\sqrt{33}}-\frac{4}{3}\sqrt[3]{\frac{-1}{19+3\sqrt{33}}}$$ Substitute back for $x=y-\frac{1}{3}$ $$x=\frac{1}{3}+\frac{(-1)^{2/3}}{3}\sqrt[3]{19+3\sqrt{33}}-\frac{4}{3}\sqrt[3]{\frac{-1}{19+3\sqrt{33}}}$$ I have collected the 3 solutions that I found above here. $$x=\frac{1}{3}+\frac{4}{3\sqrt[3]{19+3\sqrt{33}}}+\frac{1}{3}\sqrt[3]{19+3\sqrt{33}}\approx1.83928675521416113255$$ $$x=\frac{1}{3}+\frac{4(-1)^{2/3}}{3\sqrt[3]{19+3\sqrt{33}}}-\frac{1}{3}\sqrt[3]{-19-3\sqrt{33}}\approx-0.41964-0.60629i$$ $$x=\frac{1}{3}+\frac{(-1)^{2/3}}{3}\sqrt[3]{19+3\sqrt{33}}-\frac{4}{3}\sqrt[3]{\frac{-1}{19+3\sqrt{33}}}\approx-0.41964+0.60629i$$

@Darksonn 's answer certainly works, but if you only want the positive real solution, there is a "cubic formula" that is reasonably useful in this case.

Theorem: If the cubic equation $$X^3 + pX + q$$ ($p, q$ real) satisfies $$\frac{p^3}{27} + \frac{q^2}{4} \geq 0,$$ then a solution to the cubic equation is $$X = \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{p^3}{27} + \frac{q^2}{4}}} + \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{p^3}{27} + \frac{q^2}{4}}}.$$

This is the Cardano-Tartaglia Formula.

In our case, we take @Darksonn 's $y$-substitution $\displaystyle y = x - \frac{1}{3}$ and get

$$y^3 - \frac{4}{3}y - \frac{38}{27} =0.$$

So $p = \displaystyle - \frac{4}{3}$, $q = \displaystyle - \frac{38}{27}$; one may check that in this case, $$\frac{p^3}{27} + \frac{q^2}{4} = \frac{- 64 \cdot 4 + 38^2}{27^2 \cdot 4} = \frac{297}{27^2}, \quad -\frac{q}{2} = \frac{19}{27},$$ so that the answer becomes \begin{align} y &= \sqrt[3]{\frac{19}{27} + \sqrt{\frac{297}{27^2}}} + \sqrt[3]{\frac{19}{27} - \sqrt{\frac{297}{27^2}}}\\ & = \sqrt[3]{\frac{19 + \sqrt{297}}{27}} + \sqrt[3]{\frac{19 - \sqrt{297}}{27}} \\ &= \frac{1}{3} \left(\sqrt[3]{19 + 3\sqrt{33}} + \sqrt[3]{19 - 3\sqrt{33}} \right). \end{align}

Since $\displaystyle y = x - \frac{1}{3}$, $\displaystyle x = y + \frac{1}{3}$, and we get $$x = \frac{1}{3} + \frac{1}{3} \left(\sqrt[3]{19 + 3\sqrt{33}} + \sqrt[3]{19 - 3\sqrt{33}} \right).$$

To see that our answer matches @Darksonn 's, note that the terms not already matching are $\sqrt[3]{19 - 3\sqrt{33}}$ on the one side and $\displaystyle \frac{4}{\sqrt[3]{19 + 3 \sqrt{33}}}$ on the other side. To see this equality, note that for all positive $a, b$, $$a - b = \frac{a^2 - b^2}{a + b}$$ and letting $a = 19$, $b = 3 \sqrt{33}$, we get $$19 - 3 \sqrt{33} = \frac{64}{19 + 3 \sqrt{33}}$$ Taking the cube root of both sides, we're done.

If you wish to handle the case where the discriminant $$\frac{p^3}{27} + \frac{q^2}{4}$$ is negative, or to understand the shortcomings of the above formula, I strongly recommend looking at Lecture 4, "Equations of Degree Three and Four," in Fuchs and Tabachnikov, Mathematical Omnibus [Amer. Math. Soc., Providence, 2007]. It is a good "elementary" explanation of what is going on.

Yet other good methods are mentioned in the answers to the similar problem https://math.stackexchange.com/questions/612765/find-roots-of-the-cubic-equation-x3-x2-3-0-without-using-calculator?rq=1 .


text:
# Expectation value of the expectation value

## Homework Statement

The expectation value of the position observable x is <x> = ∫ψ*xψdx. The expectation value of the expectation value, <<x>>=<x>, is still the expectation value...why?

## The Attempt at a Solution

All I can think of is that the expectation value is essentially an average, and the average of an average is the original average.

i.e.) A={2,8} The average is (2+8)/2 = {5}
The average of {5} = {5}.


text:
# networkx.algorithms.bipartite.cluster.robins_alexander_clustering¶

robins_alexander_clustering(G)[source]

Compute the bipartite clustering of G.

Robins and Alexander [1] defined bipartite clustering coefficient as four times the number of four cycles C_4 divided by the number of three paths L_3 in a bipartite graph:

$CC_4 = \frac{4 * C_4}{L_3}$
Parameters
Ggraph

a bipartite graph

Returns
clusteringfloat

The Robins and Alexander bipartite clustering for the input graph.

References

1

Robins, G. and M. Alexander (2004). Small worlds among interlocking directors: Network structure and distance in bipartite graphs. Computational & Mathematical Organization Theory 10(1), 69–94.

Examples

>>> from networkx.algorithms import bipartite
>>> G = nx.davis_southern_women_graph()
>>> print(round(bipartite.robins_alexander_clustering(G), 3))
0.468


text: Can you please let me know how can i input biomass in my case. ): O. Kubaschewski, C.B. Data from NIST Standard Reference Database 69: The National Institute of Standards and Technology (NIST) ΔHfo(CO(g)) = -110.5 Find the enthalpy of the reaction of Fe2O3 with CO. Fe(s) + 3 H2O() -> Fe(OH)3(s) + 3/2 H2(g) DrH° = +160.9 kJ/mol-rxn given the enthalpies of reaction below. I have found my experience with Elsevier to be increasingly frustrating over the years, especially with the copy-editing. I am wondering what is your experience with other publishers, e.g. How to calculate FeO from Fe2O3 (Whole rock analysis?). The requested data can be retrieved from a variety of printed secondary references (e.g. S° = A*ln(t) + B*t + C*t2/2 + D*t3/3 − How do i calculate for percentage porosity in a ceramic material? NIST-JANAF Themochemical Tables, Fourth Edition,     S° = standard entropy (J/mol*K) H° − H°298.15= A*t + B*t2/2 + Requires a JavaScript / HTML 5 canvas capable browser. Is there a desciption of the reaction types of aspen plus? Thanks Any one please tell suitable solvent for PLA polymers? E/(2*t2) + G there is a formula but asks for hydrogen and water content in the fuel, which is hard to find hydrogen contents in that fuel thank you. i realised the archimedes method has many different formula. Fe 2 O 3 (s) + 3 H 2 O(l) → 2 Fe(OH) 3 (s) Δ r H° = +288.6 kJ/mol-rxn Determine the standard enthalpy of formation of calcium carbonate from the thermochemical equations given below. Whether, this FeO can be used for altered metavolcanics? a. Have questions or comments? Alcock, P.J. All rights reserved. National Institute of Standards and How to calculate low heat value from high heating value? Once a paper gets accepted (assuming it is in TeX format with perfect formatting), a month can easily pass by until the corrected proofs are actually available. Data, Monograph 9, 1998, 1-1951. The heat of formation of Fe2O3(s) is 826.0 kJ/mol. But for biomass I have only data of heating value, proximate analysis, Ulitimate analysis. However, NIST makes no warranties to that effect, and NIST in my recent Corrosion Science paper I had to submit a long list of comments when they simply messed up the paper layout completely and inserted incomprehensible symbols ("-->") everywhere in the text.     H° = standard enthalpy (kJ/mol) For more information contact us at [email protected] or check out our status page at https://status.libretexts.org. I dont have data for sulfate analysis. we have considered a biochar combustor to Supply an auxiliary heat for gasification reactor and now the problem with this biochar combustor is that we need its enthalpy as a function of temperature to determine the biochar combustion temperature. Now I am deliberately trying to choose suitable journals that are not published by Elsevier for my next publication, and so far I've found one belonging to AIP and another to IOP. Is there any handbook or any other data available regarding the thermal conductivity values of metal oxides? Legal. Answer to: Find the heat of reaction for 2Fe3O4 (s) + O2 (g) \rightarrow 3Fe2O3 (s). J. Phys. 1.784 × 104 kJ c. 638.9 kJ … Calculate the enthalpy of reaction for the following reaction: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ΔHfo(Fe2O3(s)) = -824.2 kJ/mol. Go To: Top, Solid Phase Heat Capacity (Shomate Equation), Notes, Chase, 1998 E/t2 The publisher seems to be careless and irresponsible about the typesetting and copy-editing (which is the only job they actually do in addition to selling our work! I want to know the thermophysical properties of Fe2O3? N.B. Find the enthalpy for : 4Fe + 3O2 = 2Fe2O3 I got the following informations: Fe + 3H2O = Fe(OH)3 + 3/2H2 - Enthalpy is 160.9 kj H2 + 1/2O2 = H2O - Enthalpy is -285.8 kj Fe2O3 + 3H2O = 2Fe(OH)3 - Enthalpy is 288.6 I try using Hess Law but cannot solve it. Copyright for NIST Standard Reference Data is governed by Enthalpy of Reaction from Standard Enthalpies of Formation. ), e.g. What is ΔrH° for the following phase change? I am looking for a document that describes the different reaction types. © 2008-2020 ResearchGate GmbH. H 2 (g) + 1/2 O 2 (g) → H 2 O(l) Δ r H° = –285.8 kJ/mol-rxn. Standard Reference Data Act. Can anyone help me find the enthalpy? Exercise $$\PageIndex{3}$$ Determine the standard enthalpy of formation of Fe 2 O 3 (s) given the thermochemical equations below.. Fe(s) + 3 H 2 O(l) → Fe(OH) 3 (s) + 3/2 H 2 (g) Δ r H° = +160.9 kJ/mol-rxn. C(s) + 2 S(s) → CS2(l)                         ΔrH° = +89.4 kJ/mol-rxn, C(s) + 2 S(s) → CS2(g)                        ΔrH° = +116.7 kJ/mol-rxn, Reverse (1) and add to (2)                    ΔHrxn = -ΔH1 + ΔH2, (1) CS2(l) → C(s) + 2 S(s)                               = -89.4 kJ/mol + 116.7 kJ/mol = 27.3 kJ/mol. C*t3/3 + D*t4/4 − E/t + F − H Fe(s) + 3 H2O(l) → Fe(OH)3(s) + 3/2 H2(g)                    ΔrH° = +160.9 kJ/mol-rxn, H2(g) + 1/2 O2(g) → H2O(l)                                          ΔrH° = –285.8 kJ/mol-rxn, Fe2O3(s) + 3 H2O(l) → 2 Fe(OH)3(s)                              ΔrH° = +288.6 kJ/mol-rxn, 2 Fe(s) + 6 H2O(l) → 2 Fe(OH)3(s) + 3 H2(g)                    ΔrH° = +321.8 kJ/mol-rxn, 3 H2(g) + 3/2 O2(g) → 3 H2O(l)                                     ΔrH° = –857.4kJ/mol-rxn, 2 Fe(OH)3(s) → Fe2O3(s) + 3 H2O(l)                               ΔrH° = -288.6 kJ/mol-rxn, 3 H2(g) + 3/2 O2(g) → 3 H2O(l)                                       ΔrH° = –857.4kJ/mol-rxn, 2 Fe(OH)3(s) → Fe2O3(s) + 3 H2O(l)                                 ΔrH° = -288.6 kJ/mol-rxn, 2 Fe(s) + 3/2 O2(g) → Fe2O3(s)                                       ΔHf = -824.2 kJ/mol. © 2018 by the U.S. Secretary of Commerce Data compilation copyright The requested data can be retrieved from a variety of printed secondary references (e.g. Fe2O3 (s) + 3CO (s) → 2Fe (s) + 3CO2 (g) ΔH = -28.0 kJ 3Fe (s) + 4CO2 (s) → 4CO (g) + Fe3O4 (s) ΔH = +12.5 kJ the enthalpy of the reaction of Fe2O3 with CO Missed the LibreFest? Technology, Office of Data Chase, M.W., Jr., Ref. ): O. Kubaschewski, C.B. Cp° = A + B*t + C*t2 + D*t3 + been selected on the basis of sound scientific judgment. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. How to calculate FeO from Fe2O3 (Whole rock analysis?). these are my fuels 2-Butoxyethanol, Diethyl ether and Ethylene glycol. shall not be liable for any damage that may result from What are the thermal conductivity values of various metal oxide particles such as Aluminium oxide and Copper oxide? Calculate ΔrH° for the combustion of ammonia, using standard molar enthalpies of formation: ΔH°rxn = Σ ΔH°f(products) - Σ ΔH°f(reactants), = [(4 * ΔH°f(NO2)) + (6 * ΔH°f(H2O))] - [(4 * ΔH°f(NH3)) + (7 * ΔH°f(O2))], = [(4 * 33.1) + (6 * -285.8)] - [(4 * -45.9) + (7 * 0)]. All rights reserved. View plot The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.The standard pressure value p ⦵ = 10 5 Pa (= 100 kPa = 1 bar) is recommended by IUPAC, although prior to 1982 the value 1.00 atm (101.325 kPa) was used. Expert Answer 100% (7 ratings) Previous question Next question Get more help from Chegg. B. which one is suitable say for rhombohedral crystal structure? the Database and to verify that the data contained therein have Alcock, P.J. we are doing a research on simulation and modeling of a novel energy system. What is the overall chemical equation that results from the sum of the given steps? by the U.S. Secretary of Commerce on behalf of the U.S.A. Spencer, "Materials Thermochemistry", 6. If sulfate is compulsory then what is the procedure for sulfate analysis for my biomass. [all data], Go To: Top, Solid Phase Heat Capacity (Shomate Equation), References. i need to dissolve the PLA polymer for electrospinnig.i dont know which solvent i can use except chloroform and DMF. and Informatics, Solid Phase Heat Capacity (Shomate Equation), hematite phase; Data last reviewed in June, 1965. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. do you have any idea how to determine it? -109 kJ. Fe2O3(s) + 3CO(g) yields 2Fe(s) + 3CO2(s) (Change in enthalpy= -28 kJ) 3Fe(s) + 4CO2(s) yields 4CO(g) + FE3O4(s) (Change in enthalpy= +12.5 kJ) the enthalpy of the reaction of Fe 2 O 3 with CO (3Fe2O3(s) + CO(g) yields CO2(g) + 2Fe3O4(s)) is.. A. errors or omissions in the Database. : There's almost a monopoly of the above-mentioned publisher in certain scientific fields, which is alarming to me! Ca(OH)2(s) → CaO(s) + H2O(l)                                ΔrH° = 65.2 kJ/mol-rxn, Ca(OH)2(s) + CO2(g) → CaCO3(s) + H2O(l)              ΔrH° = −113.8 kJ/mol-rxn, C(s) + O2(g) → CO2(g)                                           ΔrH° = −393.5 kJ/mol-rxn, 2 Ca(s) + O2(g) → 2 CaO(s)                                    ΔrH° = −1270.2 kJ/mol-rxn.


text:
# prove or disprove that every positive integer is the sum of a perfect square and a prime number or 1 [duplicate]

I am absolutely clueless on how to prove this statement and what makes it more difficult is not knowing if it is true or false to try and find a way to prove it. Tried numbers up to 20 and it was true for all but that is not a proof that it is true for all integers. How do i go about answering this question?

## marked as duplicate by Ross Millikan, JonMark Perry, Claude Leibovici, Daniel W. Farlow, Daniel Robert-NicoudMar 19 '16 at 14:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• hmmm. how did you come by this question? – Will Jagy Dec 19 '13 at 19:31
• I am going back through a course i took a few years back and solving every question in the book but i am stuck on this 1 – george Dec 19 '13 at 19:32
• What about 25?? – Mike Dec 19 '13 at 19:35
• does bunyakovsky imply this? – Jorge Fernández Hidalgo Dec 19 '13 at 19:57
• What was being discussed in the book in the chapter that includes this problem? What was the book? – Will Jagy Dec 19 '13 at 20:33

## 1 Answer

Primes are positive.

So, 25 is not so expressible. The sequence of non-expressible integers begins 25,34,58,64,85,91,121,130,169,196,214,289,324,... and is A014090 at the OEIS.

There, Dean Hickerson gives a proof that this sequence is infinite. Note that the allowance of the number 1 as an honorary prime does not cause difficulty, since a square cannot differ by 1 from a square.

Added: Dean's proof goes like this. Let $n$ be a positive integer such that $2n-1$ is composite. Then if we can write $n^2$ as the sum of a square and a prime, we'll have $$n^2 = m^2 + p$$ for some integer $m$ and some prime $p>0$. Then, $$p = n^2-m^2 = (n-m)(n+m).$$ Since $p$ is prime, we must have $$n-m=1$$ and so $p=n+m=2n-1$, which contradicts the primality of $p$. So $n^2$ is not so representable.

Now, if we allow negative primes, we simply modify this and consider positive $n$ such that $2n-1$ and $2n+1$ are composite. Then if $n^2=m^2+p$ for some integer $m$ and some prime $p$, possibly negative, we have $$p =n^2-m^2 = (n-m)(n+m)$$ and so $n-m=\pm 1$ and $n+m=2n+1$ or $n+m=2n-1$. Since both values are composite, this contradicts $p$'s primality. So $n^2$ is not so representable.

Now, there are infinitely many such $n$. Consider, for example, $n$ of the form $$n = 13+15r$$ for $r$ a positive integer. Then $$2n-1 = 25+30r = 5(5+6r)$$ and $$2n+1 = 27+30r = 3(9+10r)$$ are both composite, and so there are infinitely many integers not representable as the sum of a square and a prime (positive or negative).

• not according to the text i am using hungerford, he is defining a prime number to be an integer other than 1 or -1 whose only devisors are p, -p, 1, -1 – george Dec 20 '13 at 1:48
• How odd. This breaks unique factorization, among other things. – Matthew Conroy Dec 20 '13 at 2:20
• I did find this problem in a set of lecture notes by Ulrich Meierfrankenfeld, at Michigan State. But I do not see it in the original book by Hungerford. So it is possible that Prof. Meierfrankenfeld put in this question after saying that primes were always positive for his course. – Will Jagy Dec 20 '13 at 3:56
• i dont think that this breaks unique factorisation because he does say that every number can be written uniquely as a product of positive primes – george Dec 20 '13 at 4:03
• it is on page 18 question 24 in hungerford so supposedly a beginning problem – george Dec 20 '13 at 4:07


text:
Free Version
Easy

# Solving Proportions: Getting a Hit at the Plate

SATSTM-YTLWO5

Danny is a baseball player who gets a hit $6$ times out of every $20$ times at bat.

How many hits will Danny expect to get if he is at bat $130$ times?

A

$\text{120 hits}$

B

$\text{65 hits}$

C

$\text{45 hits}$

D

$\text{39 hits}$

E

$\text{18 hits}$


text:
## Engage NY Eureka Math 8th Grade Module 6 Lesson 9 Answer Key

### Eureka Math Grade 8 Module 6 Lesson 9 Exercise Answer Key

Example 1: Crocodiles and Alligators
Scientists are interested in finding out how different species adapt to finding food sources. One group studied crocodilians to find out how their bite force was related to body mass and diet. The table below displays the information they collected on body mass (in pounds) and bite force (in pounds).

Data Source: http://journals.plos.org/plosone/article?id = 10.1371/journal.pone.0031781#pone – 0031781 – t001
(Note: Body mass and bite force have been converted to pounds from kilograms and newtons, respectively.)
As you learned in the previous lesson, it is a good idea to begin by looking at what a scatter plot tells you about the data. The scatter plot below displays the data on body mass and bite force for the crocodilians in the study.

Exercises 1–6

Exercise 1.
Describe the relationship between body mass and bite force for the crocodilians shown in the scatter plot.
As the body mass increases, the bite force tends to also increase.

Exercise 2.
Draw a line to represent the trend in the data. Comment on what you considered in drawing your line.
The line should be as close as possible to the points in the scatter plot. Students explored this idea in Lesson 8.

Exercise 3.
Based on your line, predict the bite force for a crocodilian that weighs 220 pounds. How does this prediction compare to the actual bite force of the 220 – pound crocodilian in the data set?
Answers will vary. A reasonable prediction is around 650 to 700 pounds. The actual bite force was 1,000 pounds, so the prediction based on the line was not very close for this crocodilian.

Exercise 4.
Several students decided to draw lines to represent the trend in the data. Consider the lines drawn by Sol, Patti, Marrisa, and Taylor, which are shown below.

For each student, indicate whether or not you think the line would be a good line to use to make predictions.
a. Sol’s line
In general, it looks like Sol’s line overestimates the bite force for heavier crocodilians and underestimates the bite force for crocodilians that do not weigh as much.

b. Patti’s line
Patti’s line looks like it fits the data well, so it would probably produce good predictions. The line goes through the middle of the points in the scatter plot, and the points are fairly close to the line.

c. Marrisa’s line
It looks like Marrisa’s line overestimates the bite force because almost all of the points are below the line.

d. Taylor’s line
It looks like Taylor’s line tends to underestimate the bite force. There are many points above the line.

Exercise 5.
What is the equation of your line? Show the steps you used to determine your line. Based on your equation, what is your prediction for the bite force of a crocodilian weighing 200 pounds?
Answers will vary. Students have learned from previous modules how to find the equation of a line. Anticipate students to first determine the slope based on two points on their lines. Students then use a point on the line to obtain an equation in the form y = mx + b (or y = a + bx). Students use their lines to predict a bite force for a crocodilian that weighs 200 pounds. A reasonable answer would be around 800 pounds.

Exercise 6.
Patti drew vertical line segments from two points to the line in her scatter plot. The first point she selected was for a dwarf crocodile. The second point she selected was for an Indian gharial crocodile.

a. Would Patti’s line have resulted in a predicted bite force that was closer to the actual bite force for the dwarf crocodile or for the Indian gharial crocodile? What aspect of the scatter plot supports your answer?
The prediction would be closer to the actual bite force for the dwarf crocodile. That point is closer to the line (the vertical line segment connecting it to the line is shorter) than the point for the Indian gharial crocodile.

b. Would it be preferable to describe the trend in a scatter plot using a line that makes the differences in the actual and predicted values large or small? Explain your answer.
It would be better for the differences to be as small as possible. Small differences are closer to the line.

Exercise 7: Used Cars

Exercise 7.
Suppose the plot below shows the age (in years) and price (in dollars) of used compact cars that were advertised in a local newspaper.

a. Based on the scatter plot above, describe the relationship between the age and price of the used cars.
The older the car, the lower the price tends to be.

b. Nora drew a line she thought was close to many of the points and found the equation of the line. She used the points (13,6000) and (7,12000) on her line to find the equation. Explain why those points made finding the equation easy.

The points are at the intersection of the grid lines in the graph, so it is easy to determine the coordinates of these points.

c. Find the equation of Nora’s line for predicting the price of a used car given its age. Summarize the trend described by this equation.
Using the points, the equation is y = – 1000x + 19000, or Price = – 1000(age) + 19000. The slope of the line is negative, so the line indicates that the price of used cars decreases as cars get older.

d. Based on the line, for which car in the data set would the predicted value be farthest from the actual value? How can you tell?
It would be farthest for the car that is 10 years old. It is the point in the scatter plot that is farthest from the line.

e. What does the equation predict for the cost of a 10 – year – old car? How close was the prediction using the line to the actual cost of the 10 – year – old car in the data set? Given the context of the data set, do you think the difference between the predicted price and the actual price is large or small?
The line predicts a 10 – year – old car would cost about $9,000. – 1000(10) + 19000 = 9000. Compared to$4,040 for the 10 – year – old car in the data set, the difference would be $4,960. The prediction is off by about$5,000, which seems like a lot of money, given the prices of the cars in the data set.

f. Is $5,000 typical of the differences between predicted prices and actual prices for the cars in this data set? Justify your answer. Answer: No, most of the differences would be much smaller than$5,000. Most of the points are much closer to the line, and most predictions would be within about $1,000 of the actual value. ### Eureka Math Grade 8 Module 6 Lesson 9 Problem Set Answer Key Question 1. The Monopoly board game is popular in many countries. The scatter plot below shows the distance from “Go” to a property (in number of spaces moving from “Go” in a clockwise direction) and the price of the properties on the Monopoly board. The equation of the line is P = 8x + 40, where P represents the price (in Monopoly dollars) and x represents the distance (in number of spaces). Price of Property Versus Distance from “Go” in Monopoly a. Use the equation to find the difference (observed value – predicted value) for the most expensive property and for the property that is 35 spaces from “Go.” Answer: The most expensive property is 39 spaces from “Go” and costs$400. The price predicted by the line would be 8(39) + 40, or $352. Observed price – predicted price would be$400 – $352 =$48. The price predicted for 35 spaces from “Go” would be 8(35) + 40, or $320. Observed price – predicted price would be$200 – $320 = –$120.

b. Five of the points seem to lie in a horizontal line. What do these points have in common? What is the equation of the line containing those five points?
These points all have the same price. The equation of the horizontal line through those points would be
P = 200.

c. Four of the five points described in part (b) are the railroads. If you were fitting a line to predict price with distance from “Go,” would you use those four points? Why or why not?
Answers will vary. Because the four points are not part of the overall trend in the price of the properties, I would not use them to determine a line that describes the relationship. I can show this by finding the total error to measure the fit of the line.

Question 2.
The table below gives the coordinates of the five points shown in the scatter plots that follow. The scatter plots show two different lines.

a. Find the predicted response values for each of the two lines.

b. For which data points is the prediction based on Line 1 closer to the actual value than the prediction based on Line 2?
It is only for data point A. For data point C, both lines are off by the same amount.

c. Which line (Line 1 or Line 2) would you select as a better fit? Explain.
Line 2 is a better fit because it is closer to more of the data points.

Question 3.
The scatter plots below show different lines that students used to model the relationship between body mass (in pounds) and bite force (in pounds) for crocodilians.
Match each graph to one of the equations below, and explain your reasoning. Let B represent bite force (in pounds) and W represent body mass (in pounds).
Equation 1
B = 3.28W + 126

Equation 2
B = 3.04W + 351

Equation 3
B = 2.16W + 267

Equation:

Equation: 3
The intercept of 267 appears to match the graph, which has the second largest intercept.

Equation:

Equation: 2
The intercept of Equation 2 is larger, so it matches Line 2, which has a y – intercept closer to 400.

Equation:

Equation: 1
The intercept of Equation 1 is the smallest, which seems to match the graph.

b. Which of the lines would best fit the trend in the data? Explain your thinking.
Answers will vary. Line 3 would be better than the other two lines. Line 1 is not a good fit for larger weights, and Line 2 is above nearly all of the points and pretty far away from most of them. It looks like Line 3 would be closer to most of the points.

Question 4.
Comment on the following statements:
a. A line modeling a trend in a scatter plot always goes through the origin.
Some trend lines go through the origin, but others may not. Often, the value (0, 0) does not make sense for the data.

b. If the response variable increases as the independent variable decreases, the slope of a line modeling the trend is negative.
If the trend is from the upper left to the lower right, the slope for the line is negative because for each unit increase in the independent variable, the response decreases.

### Eureka Math Grade 8 Module 6 Lesson 9 Exit Ticket Answer Key

Question 1.
The scatter plot below shows the height and speed of some of the world’s fastest roller coasters. Draw a line that you think is a good fit for the data.

Students would draw a line based on the goal of a best fit for the given scatter plot. A possible line is drawn below.

Question 2.
m = $$\frac{115 – 85}{500 – 225}$$ ≈ 0.11


text:
# Project 3: Intersect (Algo Answers)

## Cameras

When implementing ray tracing, you will often need to transform points between different coordinate spaces, e.g. between the world and camera spaces.

Remember that a camera's view matrix describes a transformation which takes points from world space to camera space.

Suppose you are given the position, look vector, and up vector of a virtual camera. How would you use this information to compute the camera's view matrix? If you prefer, you may describe what should be computed instead of writing mathematical equations.

The basis vectors of the camera's coordinate space, , , and , can be computed based on the and vectors:

From these, we can construct the rotation matrix :

Since we're given the camera's position, , we can also construct the translation matrix :

Finally, we can compose and to obtain :

Explanation:

Refer to our lecture slides on camera models & viewing transforms or your lab 4 code.

## Generating Rays

### Pixels In Space

Recall that the view plane describes an imaginary 2D plane, perpendicular to the camera's look vector, which is some distance away from the camera. Here's a diagram:

Now, suppose you are given the following information about a specific pixel in an image:

• : image width, in pixels
• : image height, in pixels
• : our pixel's row index
• : our pixel's column index
• : depth of the view plane from the camera position
• : height view angle
• : width view angle

Please calculate the 3D point on the view plane, (x, y, z), in camera space, which corresponds with this pixel.

For the purpose of this algo, we use to express the depth of the view plane from camera position. However, this value is arbitrary—in your implementation, feel free to set it to . Why do you think that is the case? (Not an algo question)

As always, please assume that the pixel (0, 0) corresponds to the top left pixel of the image.

Explanation:

and give the relative coordinates of x and y on the unit view plane. Next, we just need to figure out the actual size of the view plane, which can be computed via height angle and the depth of the view plane.

Hence, the , are:

### Rays

We would like to shoot a ray from our camera, located at in world space, through a particular pixel's point on the view plane, in camera space. See Figure 1 for an example.

Please write an expression for this ray in the form , where is the origin of the ray, and is a vector in the direction it's being shot at. Both and must be in world space, so you're going to also need , the camera's view matrix.

Explanation:

We compute our pixel's point on the view plane in world space as . We can then obtain the ray's direction, , by subtracting the camera's position in world space.

Meanwhile, the ray's origin, , is simply the camera's position in world space. With this information, we can easily express our ray as above.

Note: you can choose not to normalize . We chose to do so as we like unit-length direction vectors.

## Implicit Shape Primitives

In lecture, we derived implicit equations for spheres and cylinders to perform intersection tests. In this question, you will attempt to do the same with a cone.

Begin by writing out the implicit equations for a cone with the following parameters:

• Radius = 0.5 at the flat base.
• Height = 1
• Origin is halfway along the height of the cone.

You will have to write two implicit equations, one for the conical top and one for the flat base. Each should be of the form , and each should have zero or more boundary conditions (e.g. ).

Next, you will solve these implicit equations for their intersection points with some arbitrary ray . You will have to find values of such that satisfies our implicit equations.

Please find the values of at the points along the ray which intersect with the cone. Give in terms of and , then state for what conditions this solution is valid.

Example of Boundary Checking:

• solves for the flat base, but it is not a valid solution.

Your final answer might be pretty long. Feel free to use sub-expressions!

Conical top

Solution is valid if .

Flat base

Solution is valid if .

Explanation:

Conical top

The radius of the cone at some is given by this expression:

Where and is the slope, which can easily be found to be in our case.

Hence, the equation of the conical top is:

We can express the intersection point in terms of :

By solving this equation, we can obtain the answer above.

Flat base

The equation for the flat base is:

This follows from the equation for a circle.

If you got this question correct, you got an extra point on the algo assignment!

## Transforming Object Normals

In the first part of the ray project, you will need to compute the lighting equation using the world space normal at the intersection point. From the previous question you may notice we have only computed the object space postion and normal. Thus we are going to have to transform the intersection data from object space to world space.

For positions, you already know how to do this (think back to lab 4!). However, transforming normals is not as simple as multiplying by the model matrix.

Write down the equation for the world space normal vector, , in terms of the following...

• : the normal vector in object-space.
• : the object's model matrix

, where is the upper-left 3x3 portion of the object's modeling transformation matrix, and is the inverse of its transpose.


text:
Free Version
Moderate

# Money Expansion Shortfalls

APMACR-PVFJUE

Suppose a bank received a demand deposit of \$1000 and the Required Reserve Ratio was 10%. If, in the final analysis, the actual amount of money creation was only \$800, then which of the following is a likely explanation for the money creation shortfall?

I. Currency Drain

II. Excess Reserves

III. Liability trap

A

I only

B

II only

C

III only

D

I and II

E

I, II and III