url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://au.mathworks.com/matlabcentral/cody/problems/768-are-all-the-three-given-point-in-the-same-line/solutions/2783779 | 1,600,917,112,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400213006.47/warc/CC-MAIN-20200924002749-20200924032749-00605.warc.gz | 283,746,195 | 16,718 | Cody
# Problem 768. Are all the three given point in the same line?
Solution 2783779
Submitted on 5 Aug 2020 by Jerry Wright
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
X = [-1 1 0]; Y=[-1 1 0]; y_correct = 1; assert(isequal(your_fcn_name(X,Y),y_correct))
xy = -1 -1 1 1 0 0 ans = logical 1
2 Pass
X = [0 3 6]; Y=[5 7 9]; y_correct = 1; assert(isequal(your_fcn_name(X,Y),y_correct))
xy = 0 5 3 7 6 9 ans = logical 1
3 Pass
X = [0 7 14]; Y=[-7 -5 -3]; y_correct = 1; assert(isequal(your_fcn_name(X,Y),y_correct))
xy = 0 -7 7 -5 14 -3 ans = logical 1
4 Pass
X = [10 12 -10]; Y=[-7 -3 47]; y_correct = 0; assert(isequal(your_fcn_name(X,Y),y_correct))
xy = 10 -7 12 -3 -10 47 ans = logical 0
5 Pass
X = [0 0.8 1]; Y=[0.2 1 1]; y_correct = 0; assert(isequal(your_fcn_name(X,Y),y_correct))
xy = 0 0.2000 0.8000 1.0000 1.0000 1.0000 ans = logical 0 | 403 | 985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2020-40 | latest | en | 0.586701 |
https://math.stackexchange.com/questions/939710/a-collection-of-pairwise-disjoint-open-intervals-must-be-countable | 1,632,010,001,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056578.5/warc/CC-MAIN-20210918214805-20210919004805-00069.warc.gz | 439,929,314 | 39,093 | # A collection of pairwise disjoint open intervals must be countable
Let $U$ be a collection of pairwise disjoint open intervals. That is, members of $U$ are open intervals in $\mathbb{R}$ and any two distinct members of $U$ are disjoint. Show that $U$ is countable.
My attempted proof:
Let $U \subseteq \mathbb{R}$ be a collection of pairwise disjoint opern intervals. Let $(a,b) \subseteq U$ be one such interval. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, there exists a rational number $q_{a,b} \in (a,b)$. Define $f: U \to \mathbb{Q}$ by $f((a,b))=q_{a,b}$. As $f(a_1,b_1)=f(a_2,b_2)$, or $q_{a_1,b_1}=q_{a_2,b_2}$, implies $(a_1,b_1)=(a_2,b_2)$, we deduce that $f$ is injective. It was proved already in my lecture class that $\mathbb{Q}$ is countable, and that union of countable sets is countable. Thus, we deduce that $U$ is countable. (End of attempted proof)
I feel confident about my proof except the injectivity part. I was looking for help there, assuming the rest of my proof is cogent. If not, please let me know how I can improve on it.
• Jul 25 '16 at 16:11
• Where did you use disjointness? I think the fact that the intervals are disjoint must be important, yet nowhere in the proof did you say "since the intervals are disjoint . . ."
– bof
Jul 26 '16 at 1:01
• It's not any union of countable sets that is countable, it's countable union of countable sets that's countable. But I don't understand why you bring this up. How do you use this fact in your argument? What "union of countable sets" are you talking about??
– bof
Jul 26 '16 at 1:05
Once you have set up a map $f:U \to \Bbb Q$ such that $f(I) \in \Bbb Q$ and $f(I) \in I$ for each $I \in U$ (and I think you might need the Axiom of Choice to do this; but I'm no expert on set theory so if someone more knowledgeable wants to chime in, I for one would welcome it . . .), then injectivity follows from the fact that members of $U$ are disjoint, for if $I_1, I_2 \in U$ with $f(I_1) = f(I_2) = q \in \Bbb Q$, we have $q \in I_1$ and $q \in I_2$, contradicting $I_1 \cap I_2 = \emptyset$. Since $\text{Image}(f) \subset \Bbb Q$, it is countable by virtue of being a subset of a countable set. It seems to me that that the OP's argument for injectivity is basically the same as mine, but I don't quite see where one needs the fact that a (countable) union of countable sets is countable.
• The Axiom of Choice is not needed for this. You can set $f(I) =$ the first rational in Cantor's enumeration of $\Bbb Q$ which happens to belongs to $I.$ (By the way, just as you mentioned, the OP uses the fact that a subset of a countable set is countable, not the fact that a countable union of countable sets is countable.) Jul 25 '16 at 16:19 | 828 | 2,720 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-39 | latest | en | 0.897382 |
https://www.gauthmath.com/video-detail/1697449158835206 | 1,685,945,559,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224651325.38/warc/CC-MAIN-20230605053432-20230605083432-00636.warc.gz | 862,101,972 | 26,039 | Gauthmath
# Proportional Function
## Proportional Function
#### Context of this video
"We are all in the gutter but some of us are looking at the stars". Oscar Wilde .
Human beings have been exploring the universe for a long time, trying to prove whether we are the one and only in the void. Or are we just a dust in the universe?
To find the answer, humans utilized their wisdom and invented spaceships, traversing to the depths of the universe.
In 1977, United States launched Voyager 1. Its speed could reach to 61452km/h.
Given the travel distance y, travel time t, and travel speed, we can express the relationship between travel distance and time as y equals sixty-one thousand four hundred and fifty two x.
This equation is in the form of Y equals kx, where k is a nonzero constant.
This is proportional function, and k is the constant of proportionality.
Let's take a look at this example, we are going to freeze a coke which is zero degree Celsius at the moment. We make the temperature decrease by 2 degrees Celsius every minute.So how does the coke's temperature change with time x?
Apparently, y equals negative two times x. Is this proportional function?
Usually, function that meets the following two conditions is proportional function: first, k is not equal to 0; second, the degree of x is 1.
Let's take a closer look at y equals negative two x. Even though this function has a negative constant, k, this function also belongs to the proportional function category.
So, what effect does the sign of k have on the function?
Let's take a look at the graphs of proportional functions.
We can directly find out the effect of k on function: as k is a nonzero constant, the graph of y equals kx is a straight line which passes through the origin.
When k is positive, the line y equals kx passes through the first and third Quadrants and rises from left to right. That means, y increases as x increases.
If k is negative, the line y equals kx passes through the second and fourth Quadrants and declines from left to right. That means, y decreases as x increases.
As we all know, any two distinct points in a plane determine a line. We usually only use the origin,（zero, zero）and point (one,k) to graph the function.
We can also obtain the function expression from any point other than the origin.
Have proportional function questions? Come and find an expert to help you on Gauthmath. Homework solver, step by step. See you on Gauthmath. | 542 | 2,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-23 | latest | en | 0.932172 |
community.quant-aq.com | 1,701,812,620,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100568.68/warc/CC-MAIN-20231205204654-20231205234654-00795.warc.gz | 212,426,358 | 5,431 | # Hygroscopicity and density
Hi QuantAQ team,
I am analyzing data from ModulairPM. I was wanting to know the range of Hygroscopicity/kappa values used in the model to post process the data. I want to know this so that I can play with different Hygroscopic values from other instruments like AMS. I checked the documentation of the device and I found the sigmoid function graph for both kappa and density functions against diameter. But to replicate those graphs I would need some values like highest y axis value, x0 etc (basically more details on numbers required to re create those graphs). Can you please help me with this?
Hi Shivang,
You can get the values for your device by clicking on the device model page. To get there, navigate to the page for the sensor of interest and click on whichever model you’re interested in, in the lower right-hand side of the page. Let us know if you need more info and we can post a video or screenshot of where to find these.
Hi David,
Hey David, thank you. it helped. So I was able to capture the sigmoid functions for both kappa and density given in the technical notes. I am trying to post-process PM2.5 and PM10. I am taking PM1 from your final post-processed results and adding my results of PM2.5-1 and PM10-1 post-processed values. For some reason I am not able to capture PM2.5 and PM10 when RH is above 80%. I am underestimating them by a lot (when compared with your post-processed PM). Can you help me understand why is that the case? I mean do I need to add something to get the right values? This is important piece because there are lots of values with high RH days.
In this figure, x axis is your post-processed PM concentrations and y axis is mine. Red scatter is when RH>80%. R2 is 0.92 (for PM10) if I calculate it amongst PM10 values for when RH<80%.
`new_bins[:, 1] = old_bins[:, 1] / (1 + kappa_mid * (rh / (100. - rh)))**(1./3.)`
I am doing this for all RH values, and treating RH = 100% as 99.9%. I am wondering if something needs to be tweaked here for high RH values.
If you cannot disclose this because of your intellectual property rights, I will completely understand and would love if you could point me to right literature.
Hmm. Are you able to post the code you’re using? Are you applying the other corrections before or after adjusting the bin diameters?
Hi David,
I have sent you an email with complete python code I used to make our post-processing model. Thank you for helping, appreciate it
Hey @shivang I didn’t receive an email. Can you please send again to david.hagan@qaunt-aq.com. Thanks!
Hey David,
Just sent it to you now. Thank you for the follow up | 628 | 2,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-50 | longest | en | 0.915156 |
https://the-algorithms.com/algorithm/vector-cross-product | 1,721,937,256,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763861452.88/warc/CC-MAIN-20240725175545-20240725205545-00732.warc.gz | 485,767,071 | 22,803 | #### Vector Cross Product
A
```/**
* @file
*
* @brief Calculates the [Cross Product](https://en.wikipedia.org/wiki/Cross_product) and the magnitude of two mathematical 3D vectors.
*
*
* @details Cross Product of two vectors gives a vector.
* Direction Ratios of a vector are the numeric parts of the given vector. They are the tree parts of the
* vector which determine the magnitude (value) of the vector.
* The method of finding a cross product is the same as finding the determinant of an order 3 matrix consisting
* of the first row with unit vectors of magnitude 1, the second row with the direction ratios of the
* first vector and the third row with the direction ratios of the second vector.
* The magnitude of a vector is it's value expressed as a number.
* Let the direction ratios of the first vector, P be: a, b, c
* Let the direction ratios of the second vector, Q be: x, y, z
* Therefore the calculation for the cross product can be arranged as:
*
* ```
* P x Q:
* 1 1 1
* a b c
* x y z
* ```
*
* The direction ratios (DR) are calculated as follows:
* 1st DR, J: (b * z) - (c * y)
* 2nd DR, A: -((a * z) - (c * x))
* 3rd DR, N: (a * y) - (b * x)
*
* Therefore, the direction ratios of the cross product are: J, A, N
* The following C++ Program calculates the direction ratios of the cross products of two vector.
* The program uses a function, cross() for doing so.
* The direction ratios for the first and the second vector has to be passed one by one seperated by a space character.
*
* Magnitude of a vector is the square root of the sum of the squares of the direction ratios.
*
* ### Example:
* An example of a running instance of the executable program:
*
* Pass the first Vector: 1 2 3
* Pass the second Vector: 4 5 6
* The cross product is: -3 6 -3
* Magnitude: 7.34847
*
* @author [Shreyas Sable](https://github.com/Shreyas-OwO)
*/
#include <iostream>
#include <array>
#include <cmath>
#include <cassert>
/**
* @namespace math
* @brief Math algorithms
*/
namespace math {
/**
* @namespace vector_cross
* @brief Functions for Vector Cross Product algorithms
*/
namespace vector_cross {
/**
* @brief Function to calculate the cross product of the passed arrays containing the direction ratios of the two mathematical vectors.
* @param A contains the direction ratios of the first mathematical vector.
* @param B contains the direction ration of the second mathematical vector.
* @returns the direction ratios of the cross product.
*/
std::array<double, 3> cross(const std::array<double, 3> &A, const std::array<double, 3> &B) {
std::array<double, 3> product;
/// Performs the cross product as shown in @algorithm.
product[0] = (A[1] * B[2]) - (A[2] * B[1]);
product[1] = -((A[0] * B[2]) - (A[2] * B[0]));
product[2] = (A[0] * B[1]) - (A[1] * B[0]);
return product;
}
/**
* @brief Calculates the magnitude of the mathematical vector from it's direction ratios.
* @param vec an array containing the direction ratios of a mathematical vector.
* @returns type: double description: the magnitude of the mathematical vector from the given direction ratios.
*/
double mag(const std::array<double, 3> &vec) {
double magnitude = sqrt((vec[0] * vec[0]) + (vec[1] * vec[1]) + (vec[2] * vec[2]));
return magnitude;
}
} /// namespace vector_cross
} /// namespace math
/**
* @brief test function.
* @details test the cross() and the mag() functions.
*/
static void test() {
/// Tests the cross() function.
std::array<double, 3> t_vec = math::vector_cross::cross({1, 2, 3}, {4, 5, 6});
assert(t_vec[0] == -3 && t_vec[1] == 6 && t_vec[2] == -3);
/// Tests the mag() function.
double t_mag = math::vector_cross::mag({6, 8, 0});
assert(t_mag == 10);
}
/**
* @brief Main Function
* @details Asks the user to enter the direction ratios for each of the two mathematical vectors using std::cin
* @returns 0 on exit
*/
int main() {
/// Tests the functions with sample input before asking for user input.
test();
std::array<double, 3> vec1;
std::array<double, 3> vec2;
/// Gets the values for the first vector.
std::cout << "\nPass the first Vector: ";
std::cin >> vec1[0] >> vec1[1] >> vec1[2];
/// Gets the values for the second vector.
std::cout << "\nPass the second Vector: ";
std::cin >> vec2[0] >> vec2[1] >> vec2[2];
/// Displays the output out.
std::array<double, 3> product = math::vector_cross::cross(vec1, vec2);
std::cout << "\nThe cross product is: " << product[0] << " " << product[1] << " " << product[2] << std::endl;
/// Displays the magnitude of the cross product.
std::cout << "Magnitude: " << math::vector_cross::mag(product) << "\n" << std::endl;
return 0;
}
``` | 1,297 | 4,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-30 | latest | en | 0.823725 |
http://www.cfd-online.com/W/index.php?title=Meshing&diff=prev&oldid=13666 | 1,480,751,807,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540909.75/warc/CC-MAIN-20161202170900-00152-ip-10-31-129-80.ec2.internal.warc.gz | 384,248,843 | 11,205 | # Meshing
(Difference between revisions)
Revision as of 02:36, 27 July 2010 (view source)Peter (Talk | contribs)m (Reverted edits by Jheena789 (Talk) to last version by Dingxiwang)← Older edit Revision as of 00:06, 3 January 2012 (view source)Newer edit → Line 12: Line 12: ---- ---- || '''Introduction''' | [[Mesh classification|Mesh Classification >>]] || '''Introduction''' | [[Mesh classification|Mesh Classification >>]] + [http://www.prlog.org/11289974-phone-number-lookup-verizon-phone-number-reverse-lookup-to-get-information-you-need-quickly.html reverse lookup] + + [http://thetvtopc.com/Reverse_Cell_Phone_Lookup_Number reverse phone lookup cell]
## Revision as of 00:06, 3 January 2012
Introduction Mesh classification Structured mesh generation Unstructured mesh generation Special topics
## Introduction
The partial differential equations that govern fluid flow and heat transfer are not usually amenable to analytical solutions, except for very simple cases. Therefore, in order to analyze fluid flows, flow domains are split into smaller subdomains (made up of geometric primitives like hexahedra and tetrahedra in 3D and quadrilaterals and triangles in 2D). The governing equations are then discretized and solved inside each of these subdomains. Typically, one of three methods is used to solve the approximate version of the system of equations: finite volumes, finite elements, or finite differences. Care must be taken to ensure proper continuity of solution across the common interfaces between two subdomains, so that the approximate solutions inside various portions can be put together to give a complete picture of fluid flow in the entire domain. The subdomains are often called elements or cells, and the collection of all elements or cells is called a mesh or grid. The origin of the term mesh (or grid) goes back to early days of CFD when most analyses were 2D in nature. For 2D analyses, a domain split into elements resembles a wire mesh, hence the name.
An example of a 2D analysis domain (flow over a backward facing step) and its mesh are shown in pictures below.
The process of obtaining an appropriate mesh (or grid) is termed mesh generation (or grid generation), and has long been considered a bottleneck in the analysis process due to the lack of a fully automatic mesh generation procedure. Specialized software progams have been developed for the purpose of mesh and grid generation, and access to a good software package and expertise in using this software are vital to the success of a modeling effort.
|| Introduction | Mesh Classification >> reverse lookup | 564 | 2,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2016-50 | latest | en | 0.901725 |
https://www.scribd.com/document/279707176/Fluent-Multiphase-16-0-L04-Gas-Liquid-Flows | 1,569,179,865,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575627.91/warc/CC-MAIN-20190922180536-20190922202536-00079.warc.gz | 1,001,072,195 | 75,458 | You are on page 1of 62
# Lecture 4:
Gas-Liquid Flows
16.0 Release
ANSYS Fluent
1
## April 24, 2015
ANSYS Confidential
Outline
Introduction
Conservation equations
Modelling strategies : Euler-Lagrangian and Eulerian
Interfacial Forces
Drag
Non-Drag Forces
Turbulence Interaction
Mixture Model
Validation example
2
## April 24, 2015
ANSYS Confidential
Introduction
Gasliquid flows occur in many
applications. The motion of bubbles in a
liquid as well as droplets in a conveying
gas stream are examples of gasliquid
flows
Rain/Hail
Stones
Combustion
Spray Drying
Bubble Column
## Bubble columns are commonly used in
several process industries
## Atomization to generate small droplets
for combustion is important in power
generation systems
## April 24, 2015
Distillation
Process
Boiling
Process
Absorption
Process
ANSYS Confidential
## Why Study Gas-Liquid Flows
The main interests in studying gas-liquid flows, in devices like bubble columns or
stirred tank reactors, are:
Design and scale-up
Fluid dynamics and regime analysis
Hydrodynamic parameters
## April 24, 2015
ANSYS Confidential
Bubble Columns
To design bubble column reactors, the following hydrodynamic parameters are required:
Specic gasliquid interfacial area ( )
Sauter mean bubble diameter, ( )
## Axial and radial dispersion coefcients of the gas and liquid, ( )
Heat and mass transfer coefcients, (, )
Gas holdup, ( )
Physicochemical properties of the liquid medium, (, )
## April 24, 2015
ANSYS Confidential
Regime Analysis
Two types of ow regimes are commonly observed in
bubble columns:
## The bubbly flow regime,
Gas velocity < 5cm/s
Bubbles are of relatively uniform small sizes (db =
2 to 6 mm)
Rise velocity does exceed 0.025m/s
Holdup shows linear dependence with the flow
## April 24, 2015
ANSYS Confidential
< .
< . /
Regime Analysis
> .
> . /
## The churn turbulent flow regime
Gas velocity > 5cm/s
Bubble are Large bubbles ( > ) and show wide
size distribution
Rise velocity is in the range of 1-2m/s
Most frequently
observed flow
regime in industrialsize, large diameter
columns
7
## April 24, 2015
ANSYS Confidential
## Photographic Representation of Bubbly and
Churn-Turbulent Flow Regimes
## Churn Turbulent Flow Regime
ANSYS Confidential
## Design and Scale-up of Bubble Column Reactors
Bubble have significant effect on hydrodynamics well as heat and mass transfer
## coefcients in a bubble columns
The average bubble size and rise velocity in a bubble column is found to be affected by:
## April 24, 2015
ANSYS Confidential
Euler-Lagrangian Method
In this approach, a single set of
conservation equations is solved for a
continuous phase
Eulerian Cell
Gravity
## The dispersed phase is explicitly tracked
by solving an appropriate equation of
motion in the lagrangian frame of
reference through the continuous phase
flow field
## The interaction between the continuous
and the dispersed phase is taken into
account with separate models for drag,
and non-drag forces
10
## April 24, 2015
Buoyancy
Liquid Flow
ANSYS Confidential
Eulerian Approach
In the Eulerian approach, both the continuous and dispersed phases are
considered to be interpenetrating continua
The Eulerian model describes the motion for each phase in a macroscopic sense
The flow description therefore consists of differential equations describing the
conservation of mass, momentum and energy for each phase separately
11
## April 24, 2015
ANSYS Confidential
Conservation Equations
Continuity equation:
q q q q v q m pq mqp Sq
t
p 1
source
mass transfer
Momentum equation:
Forces
Drag
q q v q q q v q q p
q q q g K pq v p v q m pq v pq mqp v qp
p 1
t
Interfacial Force
Friction
Pr essure
Bouyancy
mass transfer
Drag Forces
Non
Fq F
F wl,q
F vm,q
F td,q
lift,q
external Lift
Wall Lubrication
Virtual Mass
Turbulent Dispersion
Force
Force
Force
Force
Force
12
## April 24, 2015
ANSYS Confidential
## Interphase Momentum Exchange
A key question is how to model the inter-phase
momentum exchange
Drag
This is the force that acts on the bubble and takes into
Virtual
Mass
account:
Interphase
Momentum
Exchange
## Effect of multi-bubble interaction
Gas holdup
Turbulent modulation
Turbulent Dispersion
Turbulent Interaction
13
## April 24, 2015
Lift
Turbulent
Interaction
ANSYS Confidential
Turbulent
Dispersion
Drag Force
We can think of drag as a hydrodynamic
friction between the liquid phase and the
dispersed phase
## We can also think of drag as a
hydrodynamic resistance to the motion of
the particle through the water. The
source of this drag is shape of particle
14
## April 24, 2015
ANSYS Confidential
Drag Force
For a single spherical bubble, rising at steady state, the drag force is given by:
FD
C
D
Ap
drag coefficient
vp
vq v p vq
2 slip
velocity
For a swarm of bubbles the drag, in absence of bubble-bubble interaction, is given by:
6 p
q
FD , swarm NFD
C A
v p vq v p vq
d 3 D p 2
p
q
3 p
v p vq v p vq
CD
4
dp
15
## April 24, 2015
ANSYS Confidential
Drag Force
In order to ensure that the interfacial force vanishes in absence any dispersed phase, the
drag force needs to multiplied by as shown:
FD , swarm
3 p q
4
In Fluent
q
CD
d
p
v p vq v p vq
18 q d p
FD , swarm K pq v p v q p p
Ai f v p v q
d 2
p p 6
18 q d p C D Re
p p
Ai
v p vq
d 2
p p 6 24
16
## April 24, 2015
ANSYS Confidential
Drag Force
To estimate the drag force bubble diameter, ,is needed
The is often taken as the mean bubble size
For bubble columns operating at low gas superficial velocities (< 5 cm/s) works
reasonably well
For bubble columns operating at higher gas superficial velocities (> 5 cm/s), bubble
breakup and coalesce dominate and bubble size is no longer uniform and mean bubble
size approach may not be adequate
17
## April 24, 2015
ANSYS Confidential
Glycerol
Drag Coefficient
The drag coefficient is likely to be different for a
## single bubble and a bubble swarm. This is because the
shape and size of a bubble in a swarm is different than
that of an isolated bubble
## When the bubble size is small ( < 1mm in water):
bubble is approximately spherical
## When the bubble size is large ( > 18mm in water):
bubble is approximately a spherical cap
complex shapes
18
## April 24, 2015
ANSYS Confidential
Water /
Bubble Shape
We can use the Eotvos number () together with
the Morton number () to characterize the
shape of bubbles or drops moving in a surrounding
fluid or continuous phase
Eo
Number
gd p2
## Ratio of bouncy force and surface tension force and
essentially gives a measure of the volume of the
bubble
Number
Mo
## April 24, 2015
Lorond Eotvos
gq4
q
Ratio of physical properties
Constant for a given incompressible two-phase system.
Water has a Morton number of .
19
## 3mm air bubble rising in tap water
ANSYS Confidential
## Drag Laws for Small and Constant Bubble Sizes
At low flow rates bubbles assume an approximately spherical shape while they rise in a
rectilinear path
Schiller and Naumann (1978)
CD
24
1 0.15 Re 0.687
Re
0.44
CD
for : Re 1000
for : Re 1000
C D a1
a
a2
32
Re Re
Re
q v p vq d p
q
## When Reynolds number is
small ( < 1) these
correlations essentially
reduce to the well known
Stokes drag law = 24
The density and the viscosity are calculated from volume averaged properties and is given by
Schiller Naumann model
20
## April 24, 2015
ANSYS Confidential
## Drag Laws for Variable Bubble Sizes
For all other flow rate, bubble size and shapes varies
## Consequently, different drag correlations are needed
Several drag correlation are found in literature
Grace drag law
Tomiyama drag law
Universal drag law
## As bubble size increases,
spherical caps may be formed
21
## April 24, 2015
ANSYS Confidential
## Terminal Rise Velocity for Bubbles
The drag correlations for
large bubbles are very
different from those for
spherical particles
Spherical
Bubble
Correlation
Grace Correlation
22
## April 24, 2015
ANSYS Confidential
Bubble Regimes
Viscous
Regime
## Viscous and inertial forces are
important
the function is given by an
empirical correlation e.g. SN
Distorted
Bubble Regime
is proportional to the size
of bubble
is independent of viscosity
Cap Regime
23
## 2015 ANSYS, Inc.
24
CD max
1 0.15 Re0 .687 ,0.44
Re
## The drag coefficient on the Reynolds number decreases with
increasing values of the Reynolds number
C D ,
constant value
## April 24, 2015
ANSYS Confidential
CD
2
dp
3
8
3
## Automatic Regime Detection
Flow regime automatically determined from
continuity of drag coefficient
3cm/s
C D ,viscous C D , distorted C D ,viscous
## C D ,viscous C D , distorted C D min C D ,viscous , C D , distorted
The determined by
choosing minimum of
vicious regime and capped
regime
24
## April 24, 2015
35cm/s
ANSYS Confidential
## Drag Laws for Variable Bubble Sizes
Universal Drag Law (for Bubbly Flow)
Viscous regime
CD
q v q v p d p
q
24
1 0.1 Re 0 .75 ; Re
;e
Re
e
1 p
Distorted regime
CD
2
dp
3
g 1 17.67 f
18.67 f
6/7
f (1 p )1.5
Capped regime
As the bubble size increases the bubble become spherical caped shaped
CD
25
## April 24, 2015
8
1 - p 2
3
ANSYS Confidential
## Drag Laws for Variable Bubble Sizes
Grace Drag Law
The flow regime transitions between the viscous and distorted particle flow and can
expressed as follows.
Viscous regime
Distorted regime
CD
24
1 0.15 Re 0.687
Re
CD
4
g
dp
2
3
vt
q
vt
d
q p
0.757
,
0.94 H
J
0.441
,
3.42 H
26
## 2015 ANSYS, Inc.
CD
April 24, 2015
Mo 0.149 ( J 0.857)
2 H 59.3
H 59.3
q
4
H EoMo 0.149
3
ref
Capped regime
- 0.14
, ref 9 x10 4 kg / ms
8
3
ANSYS Confidential
## Drag Laws for Variable Bubble Sizes
Tomiyama Model (1998)
24
72 8 Eo
C D max min
(1 0.15 Re 0.687 ),
,
Re
Re p
3 Eo 4
Viscous
Regime
Distorted
Regime
Cap
Regime
Like the Grace et al model and universal drag model the Tomiyama model is well suited to gasliquid flows in which the bubbles can have a range of shapes
27
## April 24, 2015
ANSYS Confidential
Non-Drag Forces
For gasliquid flows, non-drag forces have a profound influence on the flow characteristics,
## especially in dispersed flows
Bubbles rising in a liquid can be subject to a additional forces including:
Lift Force
Wall Lubrication Force
## Virtual Mass Force
Turbulence Dispersion Force
28
## April 24, 2015
ANSYS Confidential
Lift Force
When the liquid flow is non-uniform or rotational, bubbles experience a lift force
This lift force depends on the bubble diameter, the relative velocity between the phases, and the vorticity
and is given by the following form
Flift CL p q v q v p v q
## The lift coefficient, , often is approximately constant
in inertial flow regime ( < < ),
set to 0.5
## Lift forces are primarily responsible for inhomogeneous
radial distribution of the dispersed phase holdup and
could be important to include their effects in CFD
simulations
29
## April 24, 2015
ANSYS Confidential
## Lift Coefficients: Saffman Mei Model
Saffman and Mei developed an expression for lift force
constant by combining the two lift forces:
Classical aerodynamics lift force resulting from interaction
between bubble and liquid shear
Lateral force resulting from interaction between bubbles and
vortices shed by bubble wake
Shear Lift Force
q d p2
3
'
CL
CL ; Re
q
q
2 Re
1 - 0.3314 1 Re
2 Re
'
C L 6.46
Re
0.0524
;
30
( 0.1Re)
1 Re
e
0.3314
;
2
Re
## April 24, 2015
for : 40 Re 100
for : Re 40
ANSYS Confidential
Vorticity induced
Lift Force
Suitability
## Mainly spherical rigid particles
Could be applied to small liquid
drops
## Lift Coefficients: Moraga et al Model
Moraga et al. (1999) proposed an al alternative expression for the lift coefficient that correlated with
the product of bubble and shear Reynolds numbers
0.0767
Re Re
Re Re
7
C L 0.12 0.2e 36000 e 3e
-0.6353
31
## April 24, 2015
for Re Re 6000
for 6000
Re Re 5 107
for Re Re 5 107
ANSYS Confidential
Suitability
## Mainly spherical rigid particles
Could be applied to small liquid
drops
## Lift Coefficients: Legendre and Magnaudet Model
Legendre and Magnaudet proposed an expression for the lift coefficient that is a
## function of bubble Reynolds number and dimensionless shear rate
This model accounts of induced circulation inside bubbles
C L C L2,low Re C L2,high Re ,
6
0.5 '
Re
Sr
J
2
1 1 16 Re 1
C L ,high Re
2 1 29 Re 1
C L ,low Re
J
'
2.255
1 0.1
2
32
q d p2
2
1 Re
,
,
, Re
q
3
Re
2 Re
q
2
## April 24, 2015
ANSYS Confidential
Suitability
and liquid drops
## Lift Coefficients Tomiyama Model
Tomiyama et al correlated the lift coefficient for larger
bubbles with a modified Etvs number and accounts
for bubble deformation
## min 0.288 tanh 0.121 Re, f Eo '
C L f Eo '
0.27
'
f Eo 0.00105 Eo '3 - 0.0159 Eo ' 2 0.474
for Eo ' 4
for Eo ' 10
10 Eo '
Eo '
g q p d H2
d H d p 1 0.163Eo 0.757
1
3
Eo
g q p d p2
Suitability
## All shape and size of bubble
and drops
Dependence of lift coefficient on bubble diameter
33
## April 24, 2015
ANSYS Confidential
## Wall Lubrication Force
This is a force that prevents the bubbles from
touching
The main effect of this force is to ensure zero void
fraction (found experimentally) near vertical walls
Wall lubrication force is normally correlated with slip
velocity and can be expressed as force is defined as:
gas void fraction
FWL CWL p q v p v q n w
||
34
## April 24, 2015
ANSYS Confidential
## Wall Lubrication Coefficient: Antal et al Model
Antal et al. (1991) proposed a wall lubrication force coefficient according to:
C
C
CWL max 0, W 1 W 2
d
yw
p
CW 1 0.01
CW 2 0.05
yw distance to nearest wall
CW 2
yw
C
W1
d b 5d b
Suitability
## Mainly small bubbles
Requires Fine Mesh
As a result, the Antal model will only be active on a sufficiently fine mesh
35
## April 24, 2015
ANSYS Confidential
## Wall Lubrication Coefficient: Tomiyama Model
Modified the Antal model for special case of pipe flow and accordingly:
CWL CW
dp 1
1
2
2 yw D yw 2
0.47
0 .933 Eo 0 .179
e
CW
0.00599 Eo 0.0187
0.179
D Pipe Diameter
for Eo 1
for 1 Eo 5
for 5 Eo 33
for 33 Eo
Coefficients were developed on a single air bubble in a glycerol solution but results have
## been extrapolated to air-water system
Depends on Eotvos number, hence accounts for dependence of wall lubrication force on
bubble shape
Suitability
36
## Viscous Fluids and all bubble size and shapes
Could be used for low air-water system
ANSYS Confidential
## Wall Lubrication Coefficient: Frank Model
Generalised Tomiyama model to be geometry independent
Model constants calibrated and validated for bubbly flow in vertical pipes
yw
CWC d b
1
CWL CW max 0,
m 1
yw
CWD
yw
C
d
WC b
for Eo 1
0.47
0 .933Eo 0 .179
for 1 Eo 5
e
CW
for 5 Eo 33
0.00599 Eo 0.0187
0.179
for 33 Eo
Suitability
m 1.7
37
## April 24, 2015
ANSYS Confidential
## Viscous Fluids and all bubble size and
shapes in vertical pipe flows
Could be used for low air-water system
## Wall Lubrication Coefficient: Hosokawa Model
Hosokawa et al. (2002) investigated the influence of the Morton number and developed a
new correlation for the coefficient:
## CWL max 1.9 ,0.0217 Eo
Re
Includes the effects of Eotvos number and bubble relative Reynolds number on the lift
coefficient
Suitability
38
## April 24, 2015
ANSYS Confidential
## All bubble size and shapes
Could be used for low air-water system
## Turbulent Dispersion forces
The turbulent dispersion force accounts for an
## interaction between turbulent eddies and particles
Results in a turbulent dispersion and homogenization
of the dispersed phase distribution
The simplest way to model turbulent dispersion is to
turb.
dispersion
force
gas void fraction
FTD CTD q k q p
fluid vel.
39
## April 24, 2015
ANSYS Confidential
## Turbulent Dispersion Models
Lopez de Bertodano Model, Default CTD = 1
CTD = 0.1 to 0.5 good for medium sized bubbles in ellipsoidal flow regime. However, C TD up to 500
required for small bubbles
## Burns et al. Model Default CTD = 1
The defaults value of CTD are appropriate for bubbly flows
## Simonin Model Default CTD = 1
Same as Burns et al. Model
## Diffusion in VOF Model
Instead of modelling the turbulent dispersion as an interfacial momentum force in the phase
momentum equations, we can model it as a turbulent diffusion term in the phasic continuity
equation
40
## April 24, 2015
ANSYS Confidential
Turbulent Interaction
Turbulence in bubbly flows are very complex due:
Bubble-induced turbulence
Interaction between bubble-induced and shearinduced
turbulences
Direct interaction between bubbles and turbulence eddies
and
## Turbulence Dispersion Models in Fluent
Sato
Simonin
Only available when dispersed and per phase turbulence
models are enabled
Troshko and Hassan
Alternative to Simonin Model
41
## April 24, 2015
ANSYS Confidential
## Virtual Mass Force
The virtual mass force represents the force due to
inertia of the dispersed phase due to relative
acceleration
f vm
Dv q
Dv p
; CVM 0.5
CVM p q
Dt
Dt
## Large continuous-dispersed phase density ratios, e.g.
bubbly flows
Transient Flows can affect period of oscillating bubble
plume.
Strongly Accelerating Flows e.g. bubbly flow through
narrow constriction.
42
## April 24, 2015
ANSYS Confidential
## Dip your palms into the water and slowly
bring them together. Such a movement will
require small effort. Now try to clap your
hands frequently. The speed of hands now is
low and will require considerable effort
43
## April 24, 2015
ANSYS Confidential
Introduction
The mixture model, like the Eulerian model, allows the phases to be interpenetrating. It
differs from the Eulerian model in three main respects:
Solves one set of momentum equations for the mass averaged velocity and tracks volume fraction
of each fluid throughout domain
Particle relaxation times < 0.001 - 0.01 s
## Local equilibrium assumption to model algebraically the relative velocity
This approach works well for flow fields where both phases generally flow in the same
direction and in the absence of sedimentation
44
## April 24, 2015
ANSYS Confidential
## Underlying Equations of the Mixture Model
Solves one equation for continuity of mixture
m
mum 0
t
Solves one equation for the momentum of the mixture
T
r r
um
mumum p eff um um m g F k k uk uk
t
k 1
## Solves for the transport of volume fraction of each secondary phase
( p p ) .( p p um ) .( p p u p )
t
45
## April 24, 2015
ANSYS Confidential
Constitutive Equations
n
Average density
k k
k 1
n
k 1 k k uk
um
m
Drift velocity
ukr uk um
Slip Velocity
u pq u p uq
46
## April 24, 2015
ANSYS Confidential
n
r
uk u pq k k uqk
k 1 m
Relative Velocity
If we assume the particles follows the mixture flow path, then, the slip velocity between
the phases is
um
a g um um
a v p m
u pq
f drag p
In turbulent flows, the relative velocity should contain a diffusion term in the momentum
equation for the disperse phase. FLUENT adds this dispersion to the relative velocity as
follows:
u pq
47
## April 24, 2015
a p p m
fdrag
ANSYS Confidential
m
q
p D
## Validation of the Multiphase Flow in
Rectangular Bubble Column
16.0 Release
48
## April 24, 2015
ANSYS Confidential
Objectives
Investigate air-water bubbly flow in a rectangular
bubble column as investigated at HZDR by Krepper et
al., Experimental and numerical studies of void
fraction distribution in rectangular bubble columns,
Nuclear Engineering and Design Vol. 237, pp. 399-408,
2007
## Validation of Momentum Exchange Models for
disperse bubbly flows accounting:
49
Drag force
Lift force
Turbulent dispersion
Turbulence Interaction
Turbulence models
2015 ANSYS, Inc.
## April 24, 2015
ANSYS Confidential
Computational Geometry
Duct Dimensions:
Outlet:
Degassing or Pressure
Outlet
Height: 1.0 m
Width: 0.1 m
Depth: 0.01 m
LW 0.020.01 m
Inlet:
Velocity or mass
inlet
50
## April 24, 2015
ANSYS Confidential
Materials Setups
Gas Bubble
air
Water
## FLUENT Fluid Materials:
water-liquid (h2o<l>)
Phases Setup
51
Phase Specification
Primary Phase:
Secondary Phase:
## water (Material: water)
gas bubble (diameter: 3mm with Material: air)
Phase Interaction
Drag:
Lift:
Wall Lubrication:
Turbulent Dispersion
Turbulent Interaction
Surface Tension Coeff.:
## Grace Drag Force
Tomiyama lift force
Antal et al (default coeff.)
Burns et al. (cd=0.8)
Sato Model (default coeff.)
0.072
## April 24, 2015
ANSYS Confidential
Boundary Conditions
Boundary Patch
Inlet
Properties
Type:
Gas Bubble:
Water:
52
Outlet
Type:
Degassing outlet:
Walls
No Slip
## Mass flow inlet
2.37E-05 kg/s
Gas Volume Fraction (VF): 1.0
Turbulence Intensity 10%
Viscosity Ratio 10
mass flow rate: 0 kg/s
Water VF: 0.0
Degassing
Symmetry for water
Sink for air
ANSYS Confidential
## Solution Methods and Control
Solution Methods
Pres.-Vel. Coupling
Coupled Scheme
Spatial Discretization
Momentum:
Volume Fraction:
TKE:
## Least Squared Cell Based
QUICK
QUICK
1st Order Upwind
Bounded 2nd Order Implicit
Transient Formulation
Solution Controls
Courant No.
53
200
## Explicit Relax. Factors
Momentum:
Under-Relax. Factors
Density:
1
Body Forces:
Volume Fraction:
0.5 TKE:
Specific. Diss. Rate: 0.8 Turb. Viscosity:
## April 24, 2015
0.75 Pressure:
ANSYS Confidential
0.75
0.5
0.8
0.5
## Instantaneous Gas Volume Fraction
k- Troshko-Hassan
k-SST-Sato
54
## April 24, 2015
ANSYS Confidential
## Mean gas volume fraction
distribution at plane y=0.63m
55
## Mean gas volume fraction
distribution at plane y=0.08m
ANSYS Confidential
## Mean gas volume fraction
distribution at plane y=0.63m
56
## Mean gas volume fraction
distribution at plane y=0.08m
ANSYS Confidential
## Summary and Conclusions
It was found that the most appropriate drag which is in good accordance with the
57
## measurements is the Grace Drag law
The k- turbulence model combined with the Sato Model reproduced well the
experiments with no fundamental differences to the k- SST plus the Sato Model. This
may indicate that the bubble induced turbulence is quite significant in this bubble column
The Troshko-Hassan k- turbulence model performed well, particularly near the injection
point, a region of interest as it seemed to be problematic when the validations were
carried out with ANSYS CFX using k- SST plus the Sato Model
## April 24, 2015
ANSYS Confidential
58
## April 24, 2015
ANSYS Confidential
## Numerical schemes for multiphase flows
Three algorithms available for solving the pressure-velocity coupling
Phase coupled SIMPLE (PC-SIMPLE)
Pressure Coupled (Volume Fraction solved in a segregated manner)
Full multiphase coupled (Volume Fraction solved along with pressure and momentum)
A possibility of solving all primary and secondary phase volume fractions directly rather
than solving only the secondary phases directly
59
## April 24, 2015
ANSYS Confidential
## Multiphase coupled solver
Simultaneous solution of the equations of a multiphase system offers a more robust
alternative to the segregated approach
## Can be extended to volume fraction correction (Full multiphase coupled)
For steady state problems the coupled based methodology is more efficient than
segregated methodology
For transient problems the efficiency is not as good as for steady, particularly for small
time steps. Solver efficiency increases with increase in time steps used for discretization
of the transient terms.
60
## April 24, 2015
ANSYS Confidential
Solution Strategies
Solution controls for PC-SIMPLE
Conservative solution control settings are shown
If convergence is slow, try reducing URFs for volume
fraction and turbulence
## Tighten the multi-grid settings for pressure (lower it by
two orders of magnitude). Default is 0.1
61
## April 24, 2015
ANSYS Confidential
Solution Strategies
For steady state problems using coupled multiphase
solver is effective
Use lower courant numbers for steady state and higher
URFs for momentum and pressure
Recommended values
Courant number = 20
URF pressure and momentum = 0.5 - 0.7
URF volume fraction = 0.2 - 0.5
62
## good as for steady, particularly for small time steps.
Use larger time steps and high courant numbers (1E7)
for coupled solvers and high URFs (> 0.7)
## April 24, 2015
ANSYS Confidential | 6,832 | 24,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-39 | latest | en | 0.813457 |
http://ps-2.kev009.com/CATIA-B18/lo1ug_C2/lo1ugut1202.htm | 1,652,882,167,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522270.37/warc/CC-MAIN-20220518115411-20220518145411-00707.warc.gz | 48,518,316 | 3,310 | # Dimensions and Tolerances
As dimension commands work as in the Interactive Drafting workbench, this section of the documentation provides links to the Interactive Drafting User's Guide. As such, the information detailed in this section is presented in an Interactive Drafting context. You should note that the Interactive Drafting User's Guide contains images that correspond to the Interactive Drafting workbench and therefore illustrate dimensions in an environment that is different from the 2D Layout for 3D Design environment (symbols and background color, for example).Refer to Dimensioning in a 2D Layout for 3D Design Context which explains what is specific to creating dimensions in the 2D Layout for 3D Design workbench as opposed to doing so in the Interactive Drafting workbench.
Before you begin: You should be familiar with basic concepts. Create dimensions: Create dimensions by clicking elements. Create half-dimensions: Create half-dimensions on distance, angle, diameter, cylinders, diameter edges and diameter tangents but not on cumulated dimensions. Create explicit dimensions: Create dimensions by selecting the desired icon and the required geometrical elements. Create/modify angle dimensions: Create an angle dimension and perform the following kinds of modification: new angle sector or turn an angle sector into a supplementary sector. Creating fillet radius dimensions: Create a fillet radius dimension in a projected view. Create chamfer dimensions: Create a chamfer dimension using selection. Create thread dimensions: Create associative thread dimensions. Create/modify coordinate dimensions: Automatically create coordinate dimensions on elements. Create/modify curvature radius dimensions: Create and modify a curvature radius dimension. This lets you know the curvature radius at a given point on a curve (spline, ellipse and so on). Create overall curve dimensions: You can create dimensions on the overall size of any kind of curve, whether it is canonical or not (line, circle, ellipse, spline and so on). You can also create dimensions on the overall size between two curves, or between a curve and a line, for example. Create curvilinear length dimensions: You can create dimensions for the curvilinear length of a curve, that is measure the overall length of a curve. Create partial curvilinear length dimensions: You can create dimensions for the curvilinear length of a curve portion, that is measure the partial length of a curve. Create dimensions along a reference direction: You can create dimensions along a direction of measure. In other words, you can measure the projection of a segment/distance onto a direction. Create dimensions between intersection points: You can create dimensions between an intersection point and an element or between two intersection points. Create dimensions between an element and a view axis: Create dimensions between an element and a view axis (one of the two axes or the origin). Create driving dimensions: Create dimensions that will drive associated constrained geometry. Modify the dimension type: Modify the dimension type as you create a dimension. In other words, you modify the dimension attributes. Re-route dimensions: Recalculate dimensions taking into account new geometrical elements. Interrupt one or more extension lines: Interrupt manually one or more extension lines of one or more dimensions, either using the contextual menu or the Insert menu bar command. Modify the dimension line location: Use the mouse to modify dimension line location either before or after creating dimensions. Modify the dimension value text position: Use the cursor to modify dimension value text position. Specify the dimension value position: Automatically or explicitly position the dimension value inside or outside the area between extremity symbols. Add text before/after the dimension value: Insert text before or after the dimension value. Modify the dimension overrun/blanking: Modify the dimension extension line overrun and/or blanking either together or separately. Scaling a dimension: Apply (or not) a scale to the dimension text when a scale is applied to the object containing the dimension. Line up dimensions (free space): Line up dimensions relative to a point in the free space. Line up dimensions (reference): Line up dimensions according to a given reference. Create a datum feature: Use the Datum Feature Creation dialog box to create a datum feature. Modify a datum feature: Modify a datum feature by editing it. Create a geometrical tolerance: Use the Geometric Dimensioning And Tolerancing Parameters dialog box to create geometrical tolerances. Modify a geometrical tolerance: Use the Geometric Dimensioning And Tolerancing Parameters dialog box to modify geometrical tolerances. Copy a geometrical tolerance: Copy an existing geometrical tolerance and then edit the content for creating a new one. | 920 | 4,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-21 | latest | en | 0.818951 |
https://www.physicsforums.com/threads/regarding-fermats-principle.877747/ | 1,586,454,790,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371861991.79/warc/CC-MAIN-20200409154025-20200409184525-00458.warc.gz | 1,057,360,514 | 17,339 | # Regarding fermats principle
• I
## Main Question or Discussion Point
In fermats principle what if there is more than one path for which the variation in optical path length is stationary.Then which path would light take to reach from one point to another.
## Answers and Replies
Related Classical Physics News on Phys.org
vanhees71
Gold Member
2019 Award
Do you have an example where this problem occurs? Of course, there's always a problem at the "shadow boundary" concerning the eikonal approximation, which leads to ray optics as an approximation for wave optics and which (in lowest order) can be formulated in terms of Fermat's principle.
Sorry vanhees,I don't have an example to give.This is just something that came to my mind while I was reading the topic.But is it actually possible to have 2 paths for which variation in optical path length is stationary?And if you can could you explain it to me classically if possible without going to quantum mech,its just that I am just a sophomore.
Nugatory
Mentor
In fermats principle what if there is more than one path for which the variation in optical path length is stationary.Then which path would light take to reach from one point to another.
Would you consider light emitted from a point source and being focused by a lens on another point to be an example? The optical path length is the same along all paths through points equidistant from the central axis from source to center of lens to focal point.
Yeah good point,I hadn't considered it that way.One more thing(this might be a stupid question,bare with me) what happens if there is just one photon and we send it one at a time,would we see that the photon is taking all of the paths at once or each path with equal probability of having the photon in it.
"if there is just one photon and we send it one at a time "
Say again?????
:),what I meant was we are sending one photon at a time
Nugatory
Mentor
Yeah good point,I hadn't considered it that way.One more thing(this might be a stupid question,bare with me) what happens if there is just one photon and we send it one at a time,would we see that the photon is taking all of the paths at once or each path with equal probability of having the photon in it.
Now you're proposing a variant of the quantum mechanical double slit experiment. If we have detectors along the possible paths, then only one detector will trigger for each photon; if we do not have detectors along the possible paths then it is meaningless to ask about which path each photon follows.
I very very strongly recommend that you nail down your understanding of this problem using classical optics before you jump into the world of quantum mechanics.
Thank you for that. | 577 | 2,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-16 | latest | en | 0.958356 |
http://mymathforum.com/pre-calculus/345412-removable-discontinuities-rational-functions.html | 1,571,086,240,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655310.17/warc/CC-MAIN-20191014200522-20191014224022-00201.warc.gz | 134,331,014 | 9,937 | My Math Forum removable discontinuities in rational functions
User Name Remember Me? Password
Pre-Calculus Pre-Calculus Math Forum
December 4th, 2018, 08:19 PM #1 Senior Member Joined: Sep 2015 From: USA Posts: 2,574 Thanks: 1422 removable discontinuities in rational functions So I was taught that $f(x) = \dfrac{(x+1)(x-2)}{x+1} = x-2$ is continuous. I'm reading online now that $x=-1$ is considered a removable discontinuity, i.e. a "hole" If this is the case what stops us from creating infinite holes in any continuous function by multiplying by 1 in the fashion $\tilde{f}(x) = f(x)\prod \limits_{k=1}^\infty \dfrac{x-a_k}{x-a_k}$ where $a_k$ is any sequence of real numbers What's the current convention on removable singularities?
December 5th, 2018, 12:09 AM #2 Global Moderator Joined: Dec 2006 Posts: 21,019 Thanks: 2254 Nothing stops you from doing that. It's equivalent to not defining the function at infinitely many values of $x$.
December 5th, 2018, 05:08 AM #3
Senior Member
Joined: Sep 2016
From: USA
Posts: 666
Thanks: 437
Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
Originally Posted by romsek So I was taught that $f(x) = \dfrac{(x+1)(x-2)}{x+1} = x-2$ is continuous.
Strictly speaking, $\dfrac{(x+1)(x-2)}{x+1} = x-2$ is not true which is essentially the point. The expressions are equal except when $x=-1$ in which case the expression on the left is not defined while the expression on the right is $-3$. In this case, we would define the function $f_1(x) = \dfrac{(x+1)(x-2)}{x+1}$ and the function $f_2(x) = x-2$, and then $f_2$ is referred to as the continuous extension of $f_1$.
This difference is often hammered during a Calc 1 course since students at this level don't understand limits, and they don't understand that functions aren't just formulas.
At the level of a first real analysis course or higher, it is very typical to make no distinction between a function with removable singularities and its continuous extension. For example, a statement like:
"$f(x) = \frac{\sin x}{x}$ is continuous on $\mathbb{R}$" would not be atypical to see and should be considered as a true statement.
What is meant by this, is that $\lim_{x \to 0} f(x) = 1$ so the function being described is assigned the value $f(0) = 1$ which makes it continuous. This is cumbersome to say over and over so when the audience understands limits and continuity a long explanation of this is typically omitted.
### questioning answering forum of probility
Click on a term to search for related topics.
Thread Tools Display Modes Linear Mode
Similar Threads Thread Thread Starter Forum Replies Last Post Zery Pre-Calculus 3 August 14th, 2015 05:21 PM Zery Pre-Calculus 3 August 8th, 2015 10:38 AM edwinandrew Algebra 1 November 27th, 2013 04:31 PM mathkid Algebra 5 September 22nd, 2012 06:44 PM Paper_Scars Algebra 1 March 23rd, 2012 09:02 AM
Contact - Home - Forums - Cryptocurrency Forum - Top | 833 | 2,956 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2019-43 | latest | en | 0.935152 |
https://ilnumerics.net/apidoc/html/Overload_ILNumerics_ILMath_max.htm | 1,726,576,338,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651773.64/warc/CC-MAIN-20240917104423-20240917134423-00770.warc.gz | 270,944,190 | 3,563 | ILMath.max Method
ILNumerics Ultimate VS Documentation
ILNumerics - Technical Application Development
NameDescription
max(BaseArrayByte, BaseArrayByte)
[ILNumerics Computing Engine]
max(BaseArrayInt16, BaseArrayInt16)
[ILNumerics Computing Engine]
max(BaseArrayInt32, BaseArrayInt32)
[ILNumerics Computing Engine]
max(BaseArrayInt64, BaseArrayInt64)
[ILNumerics Computing Engine]
max(BaseArraySByte, BaseArraySByte)
[ILNumerics Computing Engine]
max(BaseArrayUInt16, BaseArrayUInt16)
[ILNumerics Computing Engine]
max(BaseArrayUInt32, BaseArrayUInt32)
[ILNumerics Computing Engine]
max(BaseArrayUInt64, BaseArrayUInt64)
[ILNumerics Computing Engine]
max(BaseArraycomplex, BaseArraycomplex, Boolean)
Gives the maximum of corresponding elements from A and B. Recognizes NaN values.
[ILNumerics Computing Engine]
max(BaseArrayfcomplex, BaseArrayfcomplex, Boolean)
Gives the maximum of corresponding elements from A and B. Recognizes NaN values.
[ILNumerics Computing Engine]
max(BaseArrayDouble, BaseArrayDouble, Boolean)
Gives the maximum of corresponding elements from A and B. Recognizes NaN values.
[ILNumerics Computing Engine]
max(BaseArraySingle, BaseArraySingle, Boolean)
Gives the maximum of corresponding elements from A and B. Recognizes NaN values.
[ILNumerics Computing Engine]
max(BaseArraycomplex, OutArrayInt64, Int32, Boolean)
Computes the maximum of elements of A along the specified dimension.
[ILNumerics Computing Engine]
max(BaseArrayfcomplex, OutArrayInt64, Int32, Boolean)
Computes the maximum of elements of A along the specified dimension.
[ILNumerics Computing Engine]
max(BaseArrayByte, OutArrayInt64, Int32, Boolean)
Computes the maximum of elements of A along the specified dimension.
[ILNumerics Computing Engine]
max(BaseArrayDouble, OutArrayInt64, Int32, Boolean)
Computes the maximum of elements of A along the specified dimension.
[ILNumerics Computing Engine]
max(BaseArrayInt16, OutArrayInt64, Int32, Boolean)
Computes the maximum of elements of A along the specified dimension.
[ILNumerics Computing Engine]
max(BaseArrayInt32, OutArrayInt64, Int32, Boolean)
Computes the maximum of elements of A along the specified dimension.
[ILNumerics Computing Engine]
max(BaseArrayInt64, OutArrayInt64, Int32, Boolean)
Computes the maximum of elements of A along the specified dimension.
[ILNumerics Computing Engine]
max(BaseArraySByte, OutArrayInt64, Int32, Boolean)
Computes the maximum of elements of A along the specified dimension.
[ILNumerics Computing Engine]
max(BaseArraySingle, OutArrayInt64, Int32, Boolean)
Computes the maximum of elements of A along the specified dimension.
[ILNumerics Computing Engine]
max(BaseArrayUInt16, OutArrayInt64, Int32, Boolean)
Computes the maximum of elements of A along the specified dimension.
[ILNumerics Computing Engine]
max(BaseArrayUInt32, OutArrayInt64, Int32, Boolean)
Computes the maximum of elements of A along the specified dimension.
[ILNumerics Computing Engine]
max(BaseArrayUInt64, OutArrayInt64, Int32, Boolean)
Computes the maximum of elements of A along the specified dimension.
[ILNumerics Computing Engine]
Top | 738 | 3,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-38 | latest | en | 0.478136 |
http://www.netlib.org/scalapack/explore-html/dc/d2a/pcunmtr_8f_source.html | 1,656,671,528,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103940327.51/warc/CC-MAIN-20220701095156-20220701125156-00040.warc.gz | 101,519,110 | 11,453 | ScaLAPACK 2.1 2.1 ScaLAPACK: Scalable Linear Algebra PACKage
pcunmtr.f
Go to the documentation of this file.
1 SUBROUTINE pcunmtr( SIDE, UPLO, TRANS, M, N, A, IA, JA, DESCA,
2 \$ TAU, C, IC, JC, DESCC, WORK, LWORK, INFO )
3 *
4 * -- ScaLAPACK routine (version 1.7) --
5 * University of Tennessee, Knoxville, Oak Ridge National Laboratory,
6 * and University of California, Berkeley.
7 * May 1, 1997
8 *
9 * .. Scalar Arguments ..
10 CHARACTER SIDE, TRANS, UPLO
11 INTEGER IA, IC, INFO, JA, JC, LWORK, M, N
12 * ..
13 * .. Array Arguments ..
14 INTEGER DESCA( * ), DESCC( * )
15 COMPLEX A( * ), C( * ), TAU( * ), WORK( * )
16 * ..
17 *
18 * Purpose
19 * =======
20 *
21 * PCUNMTR overwrites the general complex M-by-N distributed matrix
22 * sub( C ) = C(IC:IC+M-1,JC:JC+N-1) with
23 *
24 * SIDE = 'L' SIDE = 'R'
25 * TRANS = 'N': Q * sub( C ) sub( C ) * Q
26 * TRANS = 'C': Q**H * sub( C ) sub( C ) * Q**H
27 *
28 * where Q is a complex unitary distributed matrix of order nq, with
29 * nq = m if SIDE = 'L' and nq = n if SIDE = 'R'. Q is defined as the
30 * product of nq-1 elementary reflectors, as returned by PCHETRD:
31 *
32 * if UPLO = 'U', Q = H(nq-1) . . . H(2) H(1);
33 *
34 * if UPLO = 'L', Q = H(1) H(2) . . . H(nq-1).
35 *
36 * Notes
37 * =====
38 *
39 * Each global data object is described by an associated description
40 * vector. This vector stores the information required to establish
41 * the mapping between an object element and its corresponding process
42 * and memory location.
43 *
44 * Let A be a generic term for any 2D block cyclicly distributed array.
45 * Such a global array has an associated description vector DESCA.
46 * In the following comments, the character _ should be read as
47 * "of the global array".
48 *
49 * NOTATION STORED IN EXPLANATION
50 * --------------- -------------- --------------------------------------
51 * DTYPE_A(global) DESCA( DTYPE_ )The descriptor type. In this case,
52 * DTYPE_A = 1.
53 * CTXT_A (global) DESCA( CTXT_ ) The BLACS context handle, indicating
54 * the BLACS process grid A is distribu-
55 * ted over. The context itself is glo-
56 * bal, but the handle (the integer
57 * value) may vary.
58 * M_A (global) DESCA( M_ ) The number of rows in the global
59 * array A.
60 * N_A (global) DESCA( N_ ) The number of columns in the global
61 * array A.
62 * MB_A (global) DESCA( MB_ ) The blocking factor used to distribute
63 * the rows of the array.
64 * NB_A (global) DESCA( NB_ ) The blocking factor used to distribute
65 * the columns of the array.
66 * RSRC_A (global) DESCA( RSRC_ ) The process row over which the first
67 * row of the array A is distributed.
68 * CSRC_A (global) DESCA( CSRC_ ) The process column over which the
69 * first column of the array A is
70 * distributed.
71 * LLD_A (local) DESCA( LLD_ ) The leading dimension of the local
72 * array. LLD_A >= MAX(1,LOCr(M_A)).
73 *
74 * Let K be the number of rows or columns of a distributed matrix,
75 * and assume that its process grid has dimension p x q.
76 * LOCr( K ) denotes the number of elements of K that a process
77 * would receive if K were distributed over the p processes of its
78 * process column.
79 * Similarly, LOCc( K ) denotes the number of elements of K that a
80 * process would receive if K were distributed over the q processes of
81 * its process row.
82 * The values of LOCr() and LOCc() may be determined via a call to the
83 * ScaLAPACK tool function, NUMROC:
84 * LOCr( M ) = NUMROC( M, MB_A, MYROW, RSRC_A, NPROW ),
85 * LOCc( N ) = NUMROC( N, NB_A, MYCOL, CSRC_A, NPCOL ).
86 * An upper bound for these quantities may be computed by:
87 * LOCr( M ) <= ceil( ceil(M/MB_A)/NPROW )*MB_A
88 * LOCc( N ) <= ceil( ceil(N/NB_A)/NPCOL )*NB_A
89 *
90 * Arguments
91 * =========
92 *
93 * SIDE (global input) CHARACTER
94 * = 'L': apply Q or Q**H from the Left;
95 * = 'R': apply Q or Q**H from the Right.
96 *
97 * UPLO (global input) CHARACTER
98 * = 'U': Upper triangle of A(IA:*,JA:*) contains elementary
99 * reflectors from PCHETRD;
100 * = 'L': Lower triangle of A(IA:*,JA:*) contains elementary
101 * reflectors from PCHETRD.
102 *
103 * TRANS (global input) CHARACTER
104 * = 'N': No transpose, apply Q;
105 * = 'C': Conjugate transpose, apply Q**H.
106 *
107 * M (global input) INTEGER
108 * The number of rows to be operated on i.e the number of rows
109 * of the distributed submatrix sub( C ). M >= 0.
110 *
111 * N (global input) INTEGER
112 * The number of columns to be operated on i.e the number of
113 * columns of the distributed submatrix sub( C ). N >= 0.
114 *
115 * A (local input) COMPLEX pointer into the local memory
116 * to an array of dimension (LLD_A,LOCc(JA+M-1)) if SIDE='L',
117 * or (LLD_A,LOCc(JA+N-1)) if SIDE = 'R'. The vectors which
118 * define the elementary reflectors, as returned by PCHETRD.
119 * If SIDE = 'L', LLD_A >= max(1,LOCr(IA+M-1));
120 * if SIDE = 'R', LLD_A >= max(1,LOCr(IA+N-1)).
121 *
122 * IA (global input) INTEGER
123 * The row index in the global array A indicating the first
124 * row of sub( A ).
125 *
126 * JA (global input) INTEGER
127 * The column index in the global array A indicating the
128 * first column of sub( A ).
129 *
130 * DESCA (global and local input) INTEGER array of dimension DLEN_.
131 * The array descriptor for the distributed matrix A.
132 *
133 * TAU (local input) COMPLEX array, dimension LTAU, where
134 * if SIDE = 'L' and UPLO = 'U', LTAU = LOCc(M_A),
135 * if SIDE = 'L' and UPLO = 'L', LTAU = LOCc(JA+M-2),
136 * if SIDE = 'R' and UPLO = 'U', LTAU = LOCc(N_A),
137 * if SIDE = 'R' and UPLO = 'L', LTAU = LOCc(JA+N-2).
138 * TAU(i) must contain the scalar factor of the elementary
139 * reflector H(i), as returned by PCHETRD. TAU is tied to the
140 * distributed matrix A.
141 *
142 * C (local input/local output) COMPLEX pointer into the
143 * local memory to an array of dimension (LLD_C,LOCc(JC+N-1)).
144 * On entry, the local pieces of the distributed matrix sub(C).
145 * On exit, sub( C ) is overwritten by Q*sub( C ) or Q'*sub( C )
146 * or sub( C )*Q' or sub( C )*Q.
147 *
148 * IC (global input) INTEGER
149 * The row index in the global array C indicating the first
150 * row of sub( C ).
151 *
152 * JC (global input) INTEGER
153 * The column index in the global array C indicating the
154 * first column of sub( C ).
155 *
156 * DESCC (global and local input) INTEGER array of dimension DLEN_.
157 * The array descriptor for the distributed matrix C.
158 *
159 * WORK (local workspace/local output) COMPLEX array,
160 * dimension (LWORK)
161 * On exit, WORK(1) returns the minimal and optimal LWORK.
162 *
163 * LWORK (local or global input) INTEGER
164 * The dimension of the array WORK.
165 * LWORK is local input and must be at least
166 *
167 * If UPLO = 'U',
168 * IAA = IA, JAA = JA+1, ICC = IC, JCC = JC;
169 * else UPLO = 'L',
170 * IAA = IA+1, JAA = JA;
171 * if SIDE = 'L',
172 * ICC = IC+1; JCC = JC;
173 * else
174 * ICC = IC; JCC = JC+1;
175 * end if
176 * end if
177 *
178 * If SIDE = 'L',
179 * MI = M-1; NI = N;
180 * LWORK >= MAX( (NB_A*(NB_A-1))/2, (NqC0 + MpC0)*NB_A ) +
181 * NB_A * NB_A
182 * else if SIDE = 'R',
183 * MI = M; MI = N-1;
184 * LWORK >= MAX( (NB_A*(NB_A-1))/2, ( NqC0 + MAX( NpA0 +
185 * NUMROC( NUMROC( NI+ICOFFC, NB_A, 0, 0, NPCOL ),
186 * NB_A, 0, 0, LCMQ ), MpC0 ) )*NB_A ) +
187 * NB_A * NB_A
188 * end if
189 *
190 * where LCMQ = LCM / NPCOL with LCM = ICLM( NPROW, NPCOL ),
191 *
192 * IROFFA = MOD( IAA-1, MB_A ), ICOFFA = MOD( JAA-1, NB_A ),
193 * IAROW = INDXG2P( IAA, MB_A, MYROW, RSRC_A, NPROW ),
194 * NpA0 = NUMROC( NI+IROFFA, MB_A, MYROW, IAROW, NPROW ),
195 *
196 * IROFFC = MOD( ICC-1, MB_C ), ICOFFC = MOD( JCC-1, NB_C ),
197 * ICROW = INDXG2P( ICC, MB_C, MYROW, RSRC_C, NPROW ),
198 * ICCOL = INDXG2P( JCC, NB_C, MYCOL, CSRC_C, NPCOL ),
199 * MpC0 = NUMROC( MI+IROFFC, MB_C, MYROW, ICROW, NPROW ),
200 * NqC0 = NUMROC( NI+ICOFFC, NB_C, MYCOL, ICCOL, NPCOL ),
201 *
202 * ILCM, INDXG2P and NUMROC are ScaLAPACK tool functions;
203 * MYROW, MYCOL, NPROW and NPCOL can be determined by calling
204 * the subroutine BLACS_GRIDINFO.
205 *
206 * If LWORK = -1, then LWORK is global input and a workspace
207 * query is assumed; the routine only calculates the minimum
208 * and optimal size for all work arrays. Each of these
209 * values is returned in the first entry of the corresponding
210 * work array, and no error message is issued by PXERBLA.
211 *
212 *
213 * INFO (global output) INTEGER
214 * = 0: successful exit
215 * < 0: If the i-th argument is an array and the j-entry had
216 * an illegal value, then INFO = -(i*100+j), if the i-th
217 * argument is a scalar and had an illegal value, then
218 * INFO = -i.
219 *
220 * Alignment requirements
221 * ======================
222 *
223 * The distributed submatrices A(IA:*, JA:*) and C(IC:IC+M-1,JC:JC+N-1)
224 * must verify some alignment properties, namely the following
225 * expressions should be true:
226 *
227 * If SIDE = 'L',
228 * ( MB_A.EQ.MB_C .AND. IROFFA.EQ.IROFFC .AND. IAROW.EQ.ICROW )
229 * If SIDE = 'R',
230 * ( MB_A.EQ.NB_C .AND. IROFFA.EQ.ICOFFC )
231 *
232 * =====================================================================
233 *
234 * .. Parameters ..
235 INTEGER BLOCK_CYCLIC_2D, CSRC_, CTXT_, DLEN_, DTYPE_,
236 \$ lld_, mb_, m_, nb_, n_, rsrc_
237 parameter( block_cyclic_2d = 1, dlen_ = 9, dtype_ = 1,
238 \$ ctxt_ = 2, m_ = 3, n_ = 4, mb_ = 5, nb_ = 6,
239 \$ rsrc_ = 7, csrc_ = 8, lld_ = 9 )
240 * ..
241 * .. Local Scalars ..
242 LOGICAL LEFT, LQUERY, NOTRAN, UPPER
243 INTEGER IAA, IAROW, ICC, ICCOL, ICOFFC, ICROW, ICTXT,
244 \$ iinfo, iroffa, iroffc, jaa, jcc, lcm, lcmq,
245 \$ lwmin, mi, mpc0, mycol, myrow, ni, npa0, npcol,
246 \$ nprow, nq, nqc0
247 * ..
248 * .. Local Arrays ..
249 INTEGER IDUM1( 4 ), IDUM2( 4 )
250 * ..
251 * .. External Subroutines ..
252 EXTERNAL blacs_gridinfo, chk1mat, pchk2mat, pcunmql,
253 \$ pcunmqr, pxerbla
254 * ..
255 * .. External Functions ..
256 LOGICAL LSAME
257 INTEGER ILCM, INDXG2P, NUMROC
258 EXTERNAL ilcm, indxg2p, lsame, numroc
259 * ..
260 * .. Intrinsic Functions ..
261 INTRINSIC cmplx, ichar, max, mod, real
262 * ..
263 * .. Executable Statements ..
264 *
265 * Get grid parameters
266 *
267 ictxt = desca( ctxt_ )
268 CALL blacs_gridinfo( ictxt, nprow, npcol, myrow, mycol )
269 *
270 * Test the input parameters
271 *
272 info = 0
273 IF( nprow.EQ.-1 ) THEN
274 info = -(900+ctxt_)
275 ELSE
276 left = lsame( side, 'L' )
277 notran = lsame( trans, 'N' )
278 upper = lsame( uplo, 'U' )
279 *
280 IF( upper ) THEN
281 iaa = ia
282 jaa = ja+1
283 icc = ic
284 jcc = jc
285 ELSE
286 iaa = ia+1
287 jaa = ja
288 IF( left ) THEN
289 icc = ic + 1
290 jcc = jc
291 ELSE
292 icc = ic
293 jcc = jc + 1
294 END IF
295 END IF
296 *
297 * NQ is the order of Q
298 *
299 IF( left ) THEN
300 nq = m
301 mi = m - 1
302 ni = n
303 CALL chk1mat( mi, 4, nq-1, 4, iaa, jaa, desca, 9, info )
304 ELSE
305 nq = n
306 mi = m
307 ni = n - 1
308 CALL chk1mat( ni, 5, nq-1, 5, iaa, jaa, desca, 9, info )
309 END IF
310 CALL chk1mat( mi, 4, ni, 5, icc, jcc, descc, 14, info )
311 IF( info.EQ.0 ) THEN
312 iroffa = mod( iaa-1, desca( mb_ ) )
313 iroffc = mod( icc-1, descc( mb_ ) )
314 icoffc = mod( jcc-1, descc( nb_ ) )
315 iarow = indxg2p( iaa, desca( mb_ ), myrow, desca( rsrc_ ),
316 \$ nprow )
317 icrow = indxg2p( icc, descc( mb_ ), myrow, descc( rsrc_ ),
318 \$ nprow )
319 iccol = indxg2p( jcc, descc( nb_ ), mycol, descc( csrc_ ),
320 \$ npcol )
321 mpc0 = numroc( mi+iroffc, descc( mb_ ), myrow, icrow,
322 \$ nprow )
323 nqc0 = numroc( ni+icoffc, descc( nb_ ), mycol, iccol,
324 \$ npcol )
325 *
326 IF( left ) THEN
327 lwmin = max( ( desca( nb_ ) * ( desca( nb_ ) - 1 ) ) / 2,
328 \$ ( mpc0 + nqc0 ) * desca( nb_ ) ) +
329 \$ desca( nb_ ) * desca( nb_ )
330 ELSE
331 npa0 = numroc( ni+iroffa, desca( mb_ ), myrow, iarow,
332 \$ nprow )
333 lcm = ilcm( nprow, npcol )
334 lcmq = lcm / npcol
335 lwmin = max( ( desca( nb_ ) * ( desca( nb_ ) - 1 ) )
336 \$ / 2, ( nqc0 + max( npa0 + numroc( numroc(
337 \$ ni+icoffc, desca( nb_ ), 0, 0, npcol ),
338 \$ desca( nb_ ), 0, 0, lcmq ), mpc0 ) ) *
339 \$ desca( nb_ ) ) + desca( nb_ ) * desca( nb_ )
340 END IF
341 *
342 work( 1 ) = cmplx( real( lwmin ) )
343 lquery = ( lwork.EQ.-1 )
344 IF( .NOT.left .AND. .NOT.lsame( side, 'R' ) ) THEN
345 info = -1
346 ELSE IF( .NOT.upper .AND. .NOT.lsame( uplo, 'L' ) ) THEN
347 info = -2
348 ELSE IF( .NOT.lsame( trans, 'N' ) .AND.
349 \$ .NOT.lsame( trans, 'C' ) ) THEN
350 info = -3
351 ELSE IF( .NOT.left .AND. desca( mb_ ).NE.descc( nb_ ) ) THEN
352 info = -(900+nb_)
353 ELSE IF( left .AND. iroffa.NE.iroffc ) THEN
354 info = -12
355 ELSE IF( left .AND. iarow.NE.icrow ) THEN
356 info = -12
357 ELSE IF( .NOT.left .AND. iroffa.NE.icoffc ) THEN
358 info = -13
359 ELSE IF( left .AND. desca( mb_ ).NE.descc( mb_ ) ) THEN
360 info = -(1400+mb_)
361 ELSE IF( ictxt.NE.descc( ctxt_ ) ) THEN
362 info = -(1400+ctxt_)
363 ELSE IF( lwork.LT.lwmin .AND. .NOT.lquery ) THEN
364 info = -16
365 END IF
366 END IF
367 *
368 IF( left ) THEN
369 idum1( 1 ) = ichar( 'L' )
370 ELSE
371 idum1( 1 ) = ichar( 'R' )
372 END IF
373 idum2( 1 ) = 1
374 IF( upper ) THEN
375 idum1( 2 ) = ichar( 'U' )
376 ELSE
377 idum1( 2 ) = ichar( 'L' )
378 END IF
379 idum2( 2 ) = 2
380 IF( notran ) THEN
381 idum1( 3 ) = ichar( 'N' )
382 ELSE
383 idum1( 3 ) = ichar( 'C' )
384 END IF
385 idum2( 3 ) = 3
386 IF( lwork.EQ.-1 ) THEN
387 idum1( 4 ) = -1
388 ELSE
389 idum1( 4 ) = 1
390 END IF
391 idum2( 4 ) = 16
392 IF( left ) THEN
393 CALL pchk2mat( mi, 4, nq-1, 4, iaa, jaa, desca, 9, mi, 4,
394 \$ ni, 5, icc, jcc, descc, 14, 4, idum1, idum2,
395 \$ info )
396 ELSE
397 CALL pchk2mat( ni, 5, nq-1, 5, iaa, jaa, desca, 9, mi, 4,
398 \$ ni, 5, icc, jcc, descc, 14, 4, idum1, idum2,
399 \$ info )
400 END IF
401 END IF
402 *
403 IF( info.NE.0 ) THEN
404 CALL pxerbla( ictxt, 'PCUNMTR', -info )
405 RETURN
406 ELSE IF( lquery ) THEN
407 RETURN
408 END IF
409 *
410 * Quick return if possible
411 *
412 IF( m.EQ.0 .OR. n.EQ.0 .OR. nq.EQ.1 )
413 \$ RETURN
414 *
415 IF( upper ) THEN
416 *
417 * Q was determined by a call to PCHETRD with UPLO = 'U'
418 *
419 CALL pcunmql( side, trans, mi, ni, nq-1, a, ia, jaa, desca,
420 \$ tau, c, icc, jcc, descc, work, lwork, iinfo )
421 *
422 ELSE
423 *
424 * Q was determined by a call to PCHETRD with UPLO = 'L'
425 *
426 CALL pcunmqr( side, trans, mi, ni, nq-1, a, iaa, jaa, desca,
427 \$ tau, c, icc, jcc, descc, work, lwork, iinfo )
428 *
429 END IF
430 *
431 work( 1 ) = cmplx( real( lwmin ) )
432 *
433 RETURN
434 *
435 * End of PCUNMTR
436 *
437 END
cmplx
float cmplx[2]
Definition: pblas.h:132
max
#define max(A, B)
Definition: pcgemr.c:180
pchk2mat
subroutine pchk2mat(MA, MAPOS0, NA, NAPOS0, IA, JA, DESCA, DESCAPOS0, MB, MBPOS0, NB, NBPOS0, IB, JB, DESCB, DESCBPOS0, NEXTRA, EX, EXPOS, INFO)
Definition: pchkxmat.f:175
pcunmqr
subroutine pcunmqr(SIDE, TRANS, M, N, K, A, IA, JA, DESCA, TAU, C, IC, JC, DESCC, WORK, LWORK, INFO)
Definition: pcunmqr.f:3
pcunmql
subroutine pcunmql(SIDE, TRANS, M, N, K, A, IA, JA, DESCA, TAU, C, IC, JC, DESCC, WORK, LWORK, INFO)
Definition: pcunmql.f:3
pcunmtr
subroutine pcunmtr(SIDE, UPLO, TRANS, M, N, A, IA, JA, DESCA, TAU, C, IC, JC, DESCC, WORK, LWORK, INFO)
Definition: pcunmtr.f:3
chk1mat
subroutine chk1mat(MA, MAPOS0, NA, NAPOS0, IA, JA, DESCA, DESCAPOS0, INFO)
Definition: chk1mat.f:3
pxerbla
subroutine pxerbla(ICTXT, SRNAME, INFO)
Definition: pxerbla.f:2 | 6,074 | 15,526 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-27 | latest | en | 0.518097 |
https://hackage-origin.haskell.org/package/elynx-tree-0.6.0.0/candidate/docs/ELynx-Tree-Bipartition.html | 1,642,713,276,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302622.39/warc/CC-MAIN-20220120190514-20220120220514-00262.warc.gz | 320,554,729 | 4,482 | elynx-tree-0.6.0.0: Handle phylogenetic trees
ELynx.Tree.Bipartition
Description
Creation date: Fri Aug 30 15:28:17 2019.
Synopsis
# Documentation
groups :: Tree e a -> Tree e [a] Source #
Each node of a tree is root of an induced subtree. Set the node labels to the leaves of the induced subtrees.
# Data type
data Bipartition a Source #
A bipartition of a tree is a grouping of the leaves of the tree into two non-overlapping, non-empty sub sets.
For unrooted trees:
• Each branch partitions the leaves of the tree into two subsets, or a bipartition.
For rooted trees:
• A bifurcating root node induces a bipartition; see bipartition.
• Each inner node induces a bipartition by taking the leaves of the sub tree and the complement leaf set of the full tree.
The order of the two subsets of a Bipartition is meaningless. That is, Bipartitions are weird in that
Bipartition x y == Bipartition y x
is True. Also,
Bipartition x y > Bipartition y x
is False, even when x > y. That's why we have to make sure that for
Bipartition x y
we always have x >= y. We ensure by construction that the larger subset comes first, and so that equality checks are meaningful; see bp and bpUnsafe.
#### Instances
Instances details
Eq a => Eq (Bipartition a) Source # Instance detailsDefined in ELynx.Tree.Bipartition Methods(==) :: Bipartition a -> Bipartition a -> Bool #(/=) :: Bipartition a -> Bipartition a -> Bool # Ord a => Ord (Bipartition a) Source # Instance detailsDefined in ELynx.Tree.Bipartition Methodscompare :: Bipartition a -> Bipartition a -> Ordering #(<) :: Bipartition a -> Bipartition a -> Bool #(<=) :: Bipartition a -> Bipartition a -> Bool #(>) :: Bipartition a -> Bipartition a -> Bool #(>=) :: Bipartition a -> Bipartition a -> Bool #max :: Bipartition a -> Bipartition a -> Bipartition a #min :: Bipartition a -> Bipartition a -> Bipartition a # (Read a, Ord a) => Read (Bipartition a) Source # Instance detailsDefined in ELynx.Tree.Bipartition Methods Show a => Show (Bipartition a) Source # Instance detailsDefined in ELynx.Tree.Bipartition MethodsshowsPrec :: Int -> Bipartition a -> ShowS #show :: Bipartition a -> String #showList :: [Bipartition a] -> ShowS # NFData a => NFData (Bipartition a) Source # Instance detailsDefined in ELynx.Tree.Bipartition Methodsrnf :: Bipartition a -> () #
bp :: Ord a => Set a -> Set a -> Either String (Bipartition a) Source #
Create a bipartition from two sets.
Ensure that the larger set comes first.
Return Left if one set is empty.
bpUnsafe :: Ord a => Set a -> Set a -> Bipartition a Source #
Create a bipartition from two sets.
Ensure that the larger set comes first.
toSet :: Ord a => Bipartition a -> Set a Source #
Conversion to a set containing both partitions.
bpHuman :: Show a => Bipartition a -> String Source #
Show a bipartition in a human readable format. Use a provided function to extract information of interest.
# Work with Bipartitions
bipartition :: Ord a => Tree e a -> Either String (Bipartition a) Source #
For a bifurcating root, get the bipartition induced by the root node.
Return Left if - the root node is not bifurcating; - a leave set is empty.
bipartitions :: Ord a => Tree e a -> Either String (Set (Bipartition a)) Source #
Get all bipartitions of the tree.
Return Left if the tree contains duplicate leaves.
getComplementaryLeaves :: Ord a => Set a -> Tree e (Set a) -> [Set a] Source #
Report the complementary leaves for each child.
bipartitionToBranch :: (Semigroup e, Ord a) => Tree e a -> Either String (Map (Bipartition a) e) Source #
Convert a tree into a Map from each Bipartition to the branch inducing the respective Bipartition.
Since the induced bipartitions of the daughter branches of a bifurcating root node are equal, the branches leading to the root have to be combined in this case. See http://evolution.genetics.washington.edu/phylip/doc/treedist.html and how unrooted trees are handled.
Further, branches connected to degree two nodes also induce the same bipartitions and have to be combined.
For combining branches, a binary function is required. This requirement is encoded in the Semigroup type class constraint (see prune).
Return Left if the tree contains duplicate leaves. | 1,106 | 4,237 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-05 | latest | en | 0.837054 |
https://lists.boost.org/boost-users/2007/06/28948.php | 1,723,194,660,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640762343.50/warc/CC-MAIN-20240809075530-20240809105530-00198.warc.gz | 288,973,404 | 3,724 | # Boost Users :
From: Aaron Windsor (aaron.windsor_at_[hidden])
Date: 2007-06-26 07:40:57
On 6/22/07, gast128 <gast128_at_[hidden]> wrote:
> Dear all,
>
> we have a simple problem: we want to know all vertices between a source and
> sink vertex in adirected graph. So my thoughts were to see if a vertex is
> reachable from both source and sink (by doing a breadth_first_search on the
> graph from source and a breadth_first_search from the sink on the reverse
> graph) and if reachable they lay in between.
>
> However this counts too much vertices. Consider graph So -> Si -> C. C is
> reachable from both So as reachable from Si using in edges. This is not good.
> The breadth_first_search should not perform an out_edge discover from Si.
>
> Do I have to implement my own breadth_first_search_with_stop or are there
> alternatives?
>
> Wkr,
> me
I'm assuming you mean simple paths? (a simple path doesn't allow
repeated vertices). There's a simple path between vertices s and t
that passes through x exactly when the following three conditions
hold: (1) there's a path from s to x AND (2) there's a path from x to
t AND (3) if you remove any single vertex from the graph besides x,
there's still either a path from s to x or a path from x to t.
You can use the BGL's depth-first or breadth-first search to determine
if there's a path between two vertices, and you can use the BGL's
filtered_graph to remove one vertex at a time from the graph and test
for reachability. If you have a graph with n vertices, you can do the
test on the entire graph in time O(n^3).
Regards,
Aaron | 407 | 1,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-33 | latest | en | 0.936799 |
http://capitalistimperialistpig.blogspot.com/2005/05/this-weeks-physics-puzzler.html | 1,726,065,188,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651387.63/warc/CC-MAIN-20240911120037-20240911150037-00646.warc.gz | 6,028,288 | 37,359 | ### This week's physics puzzler.
This question is prompted by a passage from Brian Greene's Fabric of the Cosmos. Greene is discussing the fact that in our usual experience entropy is proportional to volume (other things being equal) whereas for black holes the entropy is proportional to the area of the horizon. On page 479 of chapter 16 he says
Were you to double the radius of a black hole, its volume would increase by a factor of 8 while its surface area would only increase by a factor of 4.
This is misleading. His radius, area, volume relations are only valid for flat space, in curved space the ratios are different. The radius circumference relations for a circle on the surface of a two sphere illustrate the point.
So here's the question, in two parts:
a) What are the radius, area, volume relations (for a surface r=constant) in a Schwarzschild space-time?
b) Do the notions of radius and volume even make sense for a black hole?
1. Anonymous5:41 PM
I don't remember enough GR to calculate the 'volume' relations for Schwarzschild spacetime, but for (b) I think it's fair to say that familiar notions of radius and volume are pretty much kaput for a black hole.
We can obviously see the event horizon, and thus a familiar surface area, but when we go inside there our radial coordinate becomes timelike and conversely.
So for a BH radius, I think people just talk about the pseudo-radius that would correspond to the observed surface area were it embedded in a flat spacetime.
BH "volume" is more confusing. To give a familiar measurement of volume, you need to convert the within-horizon radial measurement into a timelike measurement for outside guys. Googling this a neat answer comes up:
BH_volume = surface area * lifetime
2. Anonymous6:56 PM
Hartle calculates length, area, volume and four-volume on ppgs. 146-148 of his book Gravity for general diagonal metrics (including Schwarzschild spacetime), so that solves (a). The calculation should break down at the horizon for a BH, on account of the coordinate singularity. Inside the BH, all calculations collide with the singularity at the "center." | 468 | 2,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-38 | latest | en | 0.932686 |
https://www.biostars.org/p/451378/ | 1,603,821,951,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894426.63/warc/CC-MAIN-20201027170516-20201027200516-00387.warc.gz | 637,091,362 | 7,693 | Question: How to compute Spearman correlation between numeric and non-numeric data variables ?
0
3 months ago by
prithvi.mastermind0 wrote:
I am working on correlation analysis and stuck on How to calculate the correlation between numerical data and non-numerical data? From the sample data given below, as usual, correlation is computed between two numerical data For e.g. between mRNA expression of myc and expression of P53. But how to calculate the correlation between a numeric and a non-numeric data For e.g between mRNA expression and mutation status of TP53. IS there any method or package in R that would compute the value of correlation 'r' with P values. Any help would be highly appreciated.
Sample MYC exp TP53 exp TP53 status 1 11.68583 6.338739 WT 2 9.668901 6.192507 MUT 3 8.080415 6.404516 WT 4 8.296929 6.869241 WT 5 9.446335 6.337951 MUT 6 7.958141 5.419711 MUT 7 7.423971 5.992706 WT 8 7.394608 6.16542 WT 9 10.97504 6.220372 MUT 10 5.756091 6.411477 WT
gene R genome • 290 views
modified 3 months ago by dsull1.6k • written 3 months ago by prithvi.mastermind0
0
3 months ago by
dsull1.6k
UCLA
dsull1.6k wrote:
When you have numerical variables, use `cor.test` to get the pearson r correlation and the p-values.
When you have your WT/MUT categories, all you have to do is set WT = 0 and MUT = 1, and then calculate the pearson r correlation (some people call this point-biserial correlation) and get p-values like you'd normally do. The p-values are derived from the t-statistic (you're technically performing a t-test).
Personally, I don't find this correlation metric that informative. It's much more informative to look at the mean of WT group and the mean of MUT group. (From there, you can see how different the WT and MUT means are, or report a fold change between WT and MUT, etc.)
EDIT: Misread the title -- if you want to do spearman's, not possible.
Thanks dsull for your valuable suggestion.
I'm using graph pad prism for correlation analysis. So is it necessary to calculate the p-values by applying t-test? Or the p-values are automatically generated through Pearson correlation are good to go? | 570 | 2,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-45 | longest | en | 0.892363 |
http://www.lmfdb.org/Character/Dirichlet/101/ | 1,571,489,916,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986693979.65/warc/CC-MAIN-20191019114429-20191019141929-00022.warc.gz | 279,197,771 | 6,785 | # Properties
Modulus 101 Structure $$C_{100}$$ Order 100
Show commands for: SageMath / Pari/GP
sage: from dirichlet_conrey import DirichletGroup_conrey # requires nonstandard Sage package to be installed
sage: H = DirichletGroup_conrey(101)
pari: g = idealstar(,101,2)
## Character group
sage: G.order() pari: g.no Order = 100 sage: H.invariants() pari: g.cyc Structure = $$C_{100}$$ sage: H.gens() pari: g.gen Generators = $\chi_{101}(2,\cdot)$
## First 32 of 100 characters
Each row describes a character. When available, the columns show the orbit label, order of the character, whether the character is primitive, and several values of the character.
orbit label order primitive -1 1 2 3 4 5 6 7 8 9 10 11
$$\chi_{101}(1,\cdot)$$ 101.a 1 No $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$
$$\chi_{101}(2,\cdot)$$ 101.i 100 Yes $$-1$$ $$1$$ $$e\left(\frac{1}{100}\right)$$ $$e\left(\frac{69}{100}\right)$$ $$e\left(\frac{1}{50}\right)$$ $$e\left(\frac{6}{25}\right)$$ $$e\left(\frac{7}{10}\right)$$ $$e\left(\frac{9}{100}\right)$$ $$e\left(\frac{3}{100}\right)$$ $$e\left(\frac{19}{50}\right)$$ $$i$$ $$e\left(\frac{13}{100}\right)$$
$$\chi_{101}(3,\cdot)$$ 101.i 100 Yes $$-1$$ $$1$$ $$e\left(\frac{69}{100}\right)$$ $$e\left(\frac{61}{100}\right)$$ $$e\left(\frac{19}{50}\right)$$ $$e\left(\frac{14}{25}\right)$$ $$e\left(\frac{3}{10}\right)$$ $$e\left(\frac{21}{100}\right)$$ $$e\left(\frac{7}{100}\right)$$ $$e\left(\frac{11}{50}\right)$$ $$i$$ $$e\left(\frac{97}{100}\right)$$
$$\chi_{101}(4,\cdot)$$ 101.h 50 Yes $$1$$ $$1$$ $$e\left(\frac{1}{50}\right)$$ $$e\left(\frac{19}{50}\right)$$ $$e\left(\frac{1}{25}\right)$$ $$e\left(\frac{12}{25}\right)$$ $$e\left(\frac{2}{5}\right)$$ $$e\left(\frac{9}{50}\right)$$ $$e\left(\frac{3}{50}\right)$$ $$e\left(\frac{19}{25}\right)$$ $$-1$$ $$e\left(\frac{13}{50}\right)$$
$$\chi_{101}(5,\cdot)$$ 101.g 25 Yes $$1$$ $$1$$ $$e\left(\frac{6}{25}\right)$$ $$e\left(\frac{14}{25}\right)$$ $$e\left(\frac{12}{25}\right)$$ $$e\left(\frac{19}{25}\right)$$ $$e\left(\frac{4}{5}\right)$$ $$e\left(\frac{4}{25}\right)$$ $$e\left(\frac{18}{25}\right)$$ $$e\left(\frac{3}{25}\right)$$ $$1$$ $$e\left(\frac{3}{25}\right)$$
$$\chi_{101}(6,\cdot)$$ 101.e 10 Yes $$1$$ $$1$$ $$e\left(\frac{7}{10}\right)$$ $$e\left(\frac{3}{10}\right)$$ $$e\left(\frac{2}{5}\right)$$ $$e\left(\frac{4}{5}\right)$$ $$1$$ $$e\left(\frac{3}{10}\right)$$ $$e\left(\frac{1}{10}\right)$$ $$e\left(\frac{3}{5}\right)$$ $$-1$$ $$e\left(\frac{1}{10}\right)$$
$$\chi_{101}(7,\cdot)$$ 101.i 100 Yes $$-1$$ $$1$$ $$e\left(\frac{9}{100}\right)$$ $$e\left(\frac{21}{100}\right)$$ $$e\left(\frac{9}{50}\right)$$ $$e\left(\frac{4}{25}\right)$$ $$e\left(\frac{3}{10}\right)$$ $$e\left(\frac{81}{100}\right)$$ $$e\left(\frac{27}{100}\right)$$ $$e\left(\frac{21}{50}\right)$$ $$i$$ $$e\left(\frac{17}{100}\right)$$
$$\chi_{101}(8,\cdot)$$ 101.i 100 Yes $$-1$$ $$1$$ $$e\left(\frac{3}{100}\right)$$ $$e\left(\frac{7}{100}\right)$$ $$e\left(\frac{3}{50}\right)$$ $$e\left(\frac{18}{25}\right)$$ $$e\left(\frac{1}{10}\right)$$ $$e\left(\frac{27}{100}\right)$$ $$e\left(\frac{9}{100}\right)$$ $$e\left(\frac{7}{50}\right)$$ $$-i$$ $$e\left(\frac{39}{100}\right)$$
$$\chi_{101}(9,\cdot)$$ 101.h 50 Yes $$1$$ $$1$$ $$e\left(\frac{19}{50}\right)$$ $$e\left(\frac{11}{50}\right)$$ $$e\left(\frac{19}{25}\right)$$ $$e\left(\frac{3}{25}\right)$$ $$e\left(\frac{3}{5}\right)$$ $$e\left(\frac{21}{50}\right)$$ $$e\left(\frac{7}{50}\right)$$ $$e\left(\frac{11}{25}\right)$$ $$-1$$ $$e\left(\frac{47}{50}\right)$$
$$\chi_{101}(10,\cdot)$$ 101.c 4 Yes $$-1$$ $$1$$ $$i$$ $$i$$ $$-1$$ $$1$$ $$-1$$ $$i$$ $$-i$$ $$-1$$ $$i$$ $$i$$
$$\chi_{101}(11,\cdot)$$ 101.i 100 Yes $$-1$$ $$1$$ $$e\left(\frac{13}{100}\right)$$ $$e\left(\frac{97}{100}\right)$$ $$e\left(\frac{13}{50}\right)$$ $$e\left(\frac{3}{25}\right)$$ $$e\left(\frac{1}{10}\right)$$ $$e\left(\frac{17}{100}\right)$$ $$e\left(\frac{39}{100}\right)$$ $$e\left(\frac{47}{50}\right)$$ $$i$$ $$e\left(\frac{69}{100}\right)$$
$$\chi_{101}(12,\cdot)$$ 101.i 100 Yes $$-1$$ $$1$$ $$e\left(\frac{71}{100}\right)$$ $$e\left(\frac{99}{100}\right)$$ $$e\left(\frac{21}{50}\right)$$ $$e\left(\frac{1}{25}\right)$$ $$e\left(\frac{7}{10}\right)$$ $$e\left(\frac{39}{100}\right)$$ $$e\left(\frac{13}{100}\right)$$ $$e\left(\frac{49}{50}\right)$$ $$-i$$ $$e\left(\frac{23}{100}\right)$$
$$\chi_{101}(13,\cdot)$$ 101.h 50 Yes $$1$$ $$1$$ $$e\left(\frac{33}{50}\right)$$ $$e\left(\frac{27}{50}\right)$$ $$e\left(\frac{8}{25}\right)$$ $$e\left(\frac{21}{25}\right)$$ $$e\left(\frac{1}{5}\right)$$ $$e\left(\frac{47}{50}\right)$$ $$e\left(\frac{49}{50}\right)$$ $$e\left(\frac{2}{25}\right)$$ $$-1$$ $$e\left(\frac{29}{50}\right)$$
$$\chi_{101}(14,\cdot)$$ 101.e 10 Yes $$1$$ $$1$$ $$e\left(\frac{1}{10}\right)$$ $$e\left(\frac{9}{10}\right)$$ $$e\left(\frac{1}{5}\right)$$ $$e\left(\frac{2}{5}\right)$$ $$1$$ $$e\left(\frac{9}{10}\right)$$ $$e\left(\frac{3}{10}\right)$$ $$e\left(\frac{4}{5}\right)$$ $$-1$$ $$e\left(\frac{3}{10}\right)$$
$$\chi_{101}(15,\cdot)$$ 101.i 100 Yes $$-1$$ $$1$$ $$e\left(\frac{93}{100}\right)$$ $$e\left(\frac{17}{100}\right)$$ $$e\left(\frac{43}{50}\right)$$ $$e\left(\frac{8}{25}\right)$$ $$e\left(\frac{1}{10}\right)$$ $$e\left(\frac{37}{100}\right)$$ $$e\left(\frac{79}{100}\right)$$ $$e\left(\frac{17}{50}\right)$$ $$i$$ $$e\left(\frac{9}{100}\right)$$
$$\chi_{101}(16,\cdot)$$ 101.g 25 Yes $$1$$ $$1$$ $$e\left(\frac{1}{25}\right)$$ $$e\left(\frac{19}{25}\right)$$ $$e\left(\frac{2}{25}\right)$$ $$e\left(\frac{24}{25}\right)$$ $$e\left(\frac{4}{5}\right)$$ $$e\left(\frac{9}{25}\right)$$ $$e\left(\frac{3}{25}\right)$$ $$e\left(\frac{13}{25}\right)$$ $$1$$ $$e\left(\frac{13}{25}\right)$$
$$\chi_{101}(17,\cdot)$$ 101.e 10 Yes $$1$$ $$1$$ $$e\left(\frac{3}{10}\right)$$ $$e\left(\frac{7}{10}\right)$$ $$e\left(\frac{3}{5}\right)$$ $$e\left(\frac{1}{5}\right)$$ $$1$$ $$e\left(\frac{7}{10}\right)$$ $$e\left(\frac{9}{10}\right)$$ $$e\left(\frac{2}{5}\right)$$ $$-1$$ $$e\left(\frac{9}{10}\right)$$
$$\chi_{101}(18,\cdot)$$ 101.i 100 Yes $$-1$$ $$1$$ $$e\left(\frac{39}{100}\right)$$ $$e\left(\frac{91}{100}\right)$$ $$e\left(\frac{39}{50}\right)$$ $$e\left(\frac{9}{25}\right)$$ $$e\left(\frac{3}{10}\right)$$ $$e\left(\frac{51}{100}\right)$$ $$e\left(\frac{17}{100}\right)$$ $$e\left(\frac{41}{50}\right)$$ $$-i$$ $$e\left(\frac{7}{100}\right)$$
$$\chi_{101}(19,\cdot)$$ 101.g 25 Yes $$1$$ $$1$$ $$e\left(\frac{24}{25}\right)$$ $$e\left(\frac{6}{25}\right)$$ $$e\left(\frac{23}{25}\right)$$ $$e\left(\frac{1}{25}\right)$$ $$e\left(\frac{1}{5}\right)$$ $$e\left(\frac{16}{25}\right)$$ $$e\left(\frac{22}{25}\right)$$ $$e\left(\frac{12}{25}\right)$$ $$1$$ $$e\left(\frac{12}{25}\right)$$
$$\chi_{101}(20,\cdot)$$ 101.h 50 Yes $$1$$ $$1$$ $$e\left(\frac{13}{50}\right)$$ $$e\left(\frac{47}{50}\right)$$ $$e\left(\frac{13}{25}\right)$$ $$e\left(\frac{6}{25}\right)$$ $$e\left(\frac{1}{5}\right)$$ $$e\left(\frac{17}{50}\right)$$ $$e\left(\frac{39}{50}\right)$$ $$e\left(\frac{22}{25}\right)$$ $$-1$$ $$e\left(\frac{19}{50}\right)$$
$$\chi_{101}(21,\cdot)$$ 101.h 50 Yes $$1$$ $$1$$ $$e\left(\frac{39}{50}\right)$$ $$e\left(\frac{41}{50}\right)$$ $$e\left(\frac{14}{25}\right)$$ $$e\left(\frac{18}{25}\right)$$ $$e\left(\frac{3}{5}\right)$$ $$e\left(\frac{1}{50}\right)$$ $$e\left(\frac{17}{50}\right)$$ $$e\left(\frac{16}{25}\right)$$ $$-1$$ $$e\left(\frac{7}{50}\right)$$
$$\chi_{101}(22,\cdot)$$ 101.h 50 Yes $$1$$ $$1$$ $$e\left(\frac{7}{50}\right)$$ $$e\left(\frac{33}{50}\right)$$ $$e\left(\frac{7}{25}\right)$$ $$e\left(\frac{9}{25}\right)$$ $$e\left(\frac{4}{5}\right)$$ $$e\left(\frac{13}{50}\right)$$ $$e\left(\frac{21}{50}\right)$$ $$e\left(\frac{8}{25}\right)$$ $$-1$$ $$e\left(\frac{41}{50}\right)$$
$$\chi_{101}(23,\cdot)$$ 101.h 50 Yes $$1$$ $$1$$ $$e\left(\frac{43}{50}\right)$$ $$e\left(\frac{17}{50}\right)$$ $$e\left(\frac{18}{25}\right)$$ $$e\left(\frac{16}{25}\right)$$ $$e\left(\frac{1}{5}\right)$$ $$e\left(\frac{37}{50}\right)$$ $$e\left(\frac{29}{50}\right)$$ $$e\left(\frac{17}{25}\right)$$ $$-1$$ $$e\left(\frac{9}{50}\right)$$
$$\chi_{101}(24,\cdot)$$ 101.g 25 Yes $$1$$ $$1$$ $$e\left(\frac{18}{25}\right)$$ $$e\left(\frac{17}{25}\right)$$ $$e\left(\frac{11}{25}\right)$$ $$e\left(\frac{7}{25}\right)$$ $$e\left(\frac{2}{5}\right)$$ $$e\left(\frac{12}{25}\right)$$ $$e\left(\frac{4}{25}\right)$$ $$e\left(\frac{9}{25}\right)$$ $$1$$ $$e\left(\frac{9}{25}\right)$$
$$\chi_{101}(25,\cdot)$$ 101.g 25 Yes $$1$$ $$1$$ $$e\left(\frac{12}{25}\right)$$ $$e\left(\frac{3}{25}\right)$$ $$e\left(\frac{24}{25}\right)$$ $$e\left(\frac{13}{25}\right)$$ $$e\left(\frac{3}{5}\right)$$ $$e\left(\frac{8}{25}\right)$$ $$e\left(\frac{11}{25}\right)$$ $$e\left(\frac{6}{25}\right)$$ $$1$$ $$e\left(\frac{6}{25}\right)$$
$$\chi_{101}(26,\cdot)$$ 101.i 100 Yes $$-1$$ $$1$$ $$e\left(\frac{67}{100}\right)$$ $$e\left(\frac{23}{100}\right)$$ $$e\left(\frac{17}{50}\right)$$ $$e\left(\frac{2}{25}\right)$$ $$e\left(\frac{9}{10}\right)$$ $$e\left(\frac{3}{100}\right)$$ $$e\left(\frac{1}{100}\right)$$ $$e\left(\frac{23}{50}\right)$$ $$-i$$ $$e\left(\frac{71}{100}\right)$$
$$\chi_{101}(27,\cdot)$$ 101.i 100 Yes $$-1$$ $$1$$ $$e\left(\frac{7}{100}\right)$$ $$e\left(\frac{83}{100}\right)$$ $$e\left(\frac{7}{50}\right)$$ $$e\left(\frac{17}{25}\right)$$ $$e\left(\frac{9}{10}\right)$$ $$e\left(\frac{63}{100}\right)$$ $$e\left(\frac{21}{100}\right)$$ $$e\left(\frac{33}{50}\right)$$ $$-i$$ $$e\left(\frac{91}{100}\right)$$
$$\chi_{101}(28,\cdot)$$ 101.i 100 Yes $$-1$$ $$1$$ $$e\left(\frac{11}{100}\right)$$ $$e\left(\frac{59}{100}\right)$$ $$e\left(\frac{11}{50}\right)$$ $$e\left(\frac{16}{25}\right)$$ $$e\left(\frac{7}{10}\right)$$ $$e\left(\frac{99}{100}\right)$$ $$e\left(\frac{33}{100}\right)$$ $$e\left(\frac{9}{50}\right)$$ $$-i$$ $$e\left(\frac{43}{100}\right)$$
$$\chi_{101}(29,\cdot)$$ 101.i 100 Yes $$-1$$ $$1$$ $$e\left(\frac{91}{100}\right)$$ $$e\left(\frac{79}{100}\right)$$ $$e\left(\frac{41}{50}\right)$$ $$e\left(\frac{21}{25}\right)$$ $$e\left(\frac{7}{10}\right)$$ $$e\left(\frac{19}{100}\right)$$ $$e\left(\frac{73}{100}\right)$$ $$e\left(\frac{29}{50}\right)$$ $$-i$$ $$e\left(\frac{83}{100}\right)$$
$$\chi_{101}(30,\cdot)$$ 101.h 50 Yes $$1$$ $$1$$ $$e\left(\frac{47}{50}\right)$$ $$e\left(\frac{43}{50}\right)$$ $$e\left(\frac{22}{25}\right)$$ $$e\left(\frac{14}{25}\right)$$ $$e\left(\frac{4}{5}\right)$$ $$e\left(\frac{23}{50}\right)$$ $$e\left(\frac{41}{50}\right)$$ $$e\left(\frac{18}{25}\right)$$ $$-1$$ $$e\left(\frac{11}{50}\right)$$
$$\chi_{101}(31,\cdot)$$ 101.g 25 Yes $$1$$ $$1$$ $$e\left(\frac{21}{25}\right)$$ $$e\left(\frac{24}{25}\right)$$ $$e\left(\frac{17}{25}\right)$$ $$e\left(\frac{4}{25}\right)$$ $$e\left(\frac{4}{5}\right)$$ $$e\left(\frac{14}{25}\right)$$ $$e\left(\frac{13}{25}\right)$$ $$e\left(\frac{23}{25}\right)$$ $$1$$ $$e\left(\frac{23}{25}\right)$$
$$\chi_{101}(32,\cdot)$$ 101.f 20 Yes $$-1$$ $$1$$ $$e\left(\frac{1}{20}\right)$$ $$e\left(\frac{9}{20}\right)$$ $$e\left(\frac{1}{10}\right)$$ $$e\left(\frac{1}{5}\right)$$ $$-1$$ $$e\left(\frac{9}{20}\right)$$ $$e\left(\frac{3}{20}\right)$$ $$e\left(\frac{9}{10}\right)$$ $$i$$ $$e\left(\frac{13}{20}\right)$$ | 4,908 | 11,110 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2019-43 | longest | en | 0.418484 |
https://www.jiskha.com/questions/1071740/suppose-a-coin-is-tossed-100-times-57-heads-and-43-tails-what-is-the-probability-the-next | 1,585,482,063,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370494331.42/warc/CC-MAIN-20200329105248-20200329135248-00067.warc.gz | 1,010,395,878 | 5,140 | # finite math
Suppose a coin is tossed 100 times 57 heads and 43 tails what is the probability the next one will be heads.
1. 👍 0
2. 👎 0
3. 👁 178
1. 1/2
1. 👍 0
2. 👎 0
## Similar Questions
1. ### Math
Luis has a coin that is weighted so that the probability that Heads appears when it is tossed is 0.55. Suppose that the coin is tossed 3 times. What is the probability that all 3 tosses are Heads? please help ,e to solve this
asked by sajid on February 17, 2012
2. ### Probability
Suppose that an unfair coin comes up heads 54.1% of the time. The coin is flipped a total of 13 times. a) What is the probability that you get exactly 6 heads? b) What is the probability that you get exactly 6 tails? c) What is
asked by Vikki on February 24, 2012
3. ### math
In a game, a player tosses a coin 4 times. If the player gets 3 or 4 heads, he/she wins. What is the theoretical probability of winning this game? I just need to know the outcomes. I don't know how to get them. Please and Thank
asked by Angela on May 30, 2007
4. ### stats
Suppose you toss a coin 100 times and get 82 heads and 18 tails. Based on these results, what is the probability the next flip will be a tails?
asked by john on September 7, 2015
5. ### Math
10. A coin is loaded so that the probability of heads is 0.55 and the probability of tails is 0.45, Suppose the coin is tossed twice and the results of tosses are independent. a.What is the probability of obtaining exactly two
asked by Renee on March 2, 2014
1. ### common fraction
Sara tossed a fair coin five times, and Kaleb tossed a fair coin three times. There were five heads and three tails in the eight tosses. What is the probability that either Sara or Kaleb tossed exactly three heads? Express your
asked by Anonymous on June 27, 2012
2. ### Statistics
A coin is tossed five times. Find the probability of getting four tails. I realize that getting four tails is the exact same thing as getting only one heads, but how would you figure out this problem using probability rules?
asked by Emily on January 2, 2011
3. ### Prealgebra/probability
A coin is tossed three times. Use a tree diagram to find the number of possible outcomes that could produce exactly two heads. I don't know how to write out a tree diagram on here, but I think this one is heads -> heads, tails ->
asked by Fawn on July 3, 2016
4. ### math
A student claims that if a fair coin is tossed and comes up heads 5 times in a row, then, according to the law of averages, the probability of tails on the next toss is greater than the probability of heads. What is your reply?
asked by crystal on April 20, 2010
5. ### math-probability
A student claims that if a fair coin is tossed and comes up heads 5 times in a row, then, according to the law of averages, the probability of tails on the next toss is greater than the probability of heads. What is your reply?
asked by jennifer on April 19, 2010
More Similar Questions | 790 | 2,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2020-16 | latest | en | 0.974324 |
https://community.qlik.com/thread/36653 | 1,531,952,157,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.62/warc/CC-MAIN-20180718213135-20180718233135-00470.warc.gz | 629,691,370 | 22,851 | 4 Replies Latest reply: Oct 15, 2011 7:02 AM by Mohammed Siraj
# How to show current month percentage
hi,
As u see the data in attached text file, for Prodct A planned its till december and Prduct A actual its till July which is my current month and every month the data will be updated, i want to have a input box where i can enter the current month name, simultaneously i want a table wch only show current month in it and both the Prduct A planned and Product A Actual value, and a separate text box whc must calculate the percentage (i.e (a/b)-1), for clear understanding i have attached the pic how i want it to show, Requred urgent help,Thanks
• ###### Re: How to show current month percentage
Hi, Would something like this work for you?
Thank
Steve
• ###### How to show current month percentage
hi Steve,
i am working on Personal Edition verions of Qlik View so can able to open the file wch you have send it to me, it will be great help if you can write it in notepad or reply me back
• ###### Re: How to show current month percentage
Hi well the graph is just the sum of your two measures against month, then a table with the two measures as :
=sum({<Months= {\$(=[Enter Month Name])} >} [Product A Actual Sales])
Sum ({<Months= {\$(=[Enter Month Name])} >} [Product A Planned])
then a text field of :
= 'Global Product A is ' & ( num( (sum({<Months= {\$(=[Enter Month Name])} >} [Product A Planned] )
/
sum({<Months= {\$(=[Enter Month Name])} >} [Product A Actual Sales] ))-1,'0.0%') ) &' Below Budget for ' & [Enter Month Name] &' 2011'
where [Enter Month Name] is a variable that you set with the month name.
Thanks
Steve
Hi if you really wanted to use set analysis you could do something like this but the question would be why not just use your If statement??
= sum( {<prd_failures={\$(=if(vfact = 1,'*') )} > } prd_failures)/
sum( {<measurement_failures={\$(=if(vfact = 1,'*') )} > } measurement_failures)
Thanks
Steve
=sum({<Months= {\$(=[Enter Month Name])} >} [Product A Actual Sales])
Sum ({<Months= {\$(=[Enter Month Name])} >} [Product A Planned])
= 'Global Product A is ' & ( num( (sum({<Months= {\$(=[Enter Month Name])} >} [Product A Planned] )
/
sum({<Months= {\$(=[Enter Month Name])} >} [Product A Actual Sales] ))-1,'0.0%') ) &' Below Budget for ' & [Enter Month Name] &' 2011'
• ###### How to show current month percentage
Hi,
Sorry for the late response, ur script has worked, but i still have one doubt as u seen the text in the blue box, here i have one more condition, if the percentage vlaue is negative dn it mst show "Below" and if Positive then it must show Above, if u can help on these dn it be great | 732 | 2,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-30 | latest | en | 0.811628 |
https://www.uopassignments.com/res-342-uop | 1,628,020,533,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154471.78/warc/CC-MAIN-20210803191307-20210803221307-00712.warc.gz | 1,093,703,919 | 6,338 | # RES 342
#### RES 342 Entire Course Recent
RES 342 Entire Course Recent RES 342 Week 1 Individual Assignment Hypothesis Identification Artic..
\$25.00 \$45.00
This Tutorial Purchased: 9 Times Rating: A+
#### RES 342 Week 1 Individual Assignment E-Text Recent
RES 342 Week 1 Individual Assignment E-Text Recent Resource: Ch. 8 of Applied Statistics in Busin..
\$7.00 \$14.00
This Tutorial Purchased: 5 Times Rating: A+
#### RES 342 Week 1 Individual Assignment Hypothesis Identification Article Analysis Recent
RES 342 Week 1 Individual Assignment Hypothesis Identification Article Analysis Recent Use the Un..
\$7.00 \$14.00
This Tutorial Purchased: 5 Times Rating: A+
#### RES 342 Week 2 Individual Assignment E-Text Recent
RES 342 Week 2 Individual Assignment E-Text Recent Resource: Ch. 9 of Applied Statistics in Busin..
\$7.00 \$14.00
This Tutorial Purchased: 6 Times Rating: A+
#### RES 342 Week 2 Individual Quiz Recent
RES 342 Week 2 Individual Quiz Recent Prepare to take a quiz on topics covered in previous weeks...
\$7.00 \$14.00
This Tutorial Purchased: 5 Times Rating: A+
#### RES 342 Week 2 Team Assignment One Sample Hypothesis Testing Paper Recent
RES 342 Week 2 Team Assignment One Sample Hypothesis Testing Paper Recent Use the ratio or interv..
\$7.00 \$14.00
This Tutorial Purchased: 5 Times Rating: A+
#### RES 342 Week 3 Individual Assignment Applying Analysis of Variance (ANOVA) and Nonparametric Tests Simulation Recent
RES 342 Week 3 Individual Assignment Applying Analysis of Variance (ANOVA) and Nonparametric Tests S..
\$7.00 \$13.00
This Tutorial Purchased: 5 Times Rating: A+
#### RES 342 Week 3 Individual Assignment E-Text Recent
RES 342 Week 3 Individual Assignment E-Text Recent Resource: Ch. 10 & 11 of Applied Statistic..
\$7.00 \$14.00
This Tutorial Purchased: 5 Times Rating: A+
#### RES 342 Week 3 Team Assignment Two or More Sample Hypothesis Testing Paper Recent
RES 342 Week 3 Team Assignment Two or More Sample Hypothesis Testing Paper Recent Use the numeric..
\$7.00 \$14.00
This Tutorial Purchased: 6 Times Rating: A+
#### RES 342 Week 4 Individual Assignment E Text Recent
RES 342 Week 4 Individual Assignment E Text Recent Resource: Ch. 15 of Applied Statistics in Busi..
\$7.00 \$13.00
This Tutorial Purchased: 5 Times Rating: A+
#### RES 342 Week 4 Individual Assignment Parametric and Nonparametric Data Identification Assignment Recent
RES 342 Week 4 Individual Assignment Parametric and Nonparametric Data Identification Assignment Rec..
\$7.00 \$14.00
This Tutorial Purchased: 6 Times Rating: A+
#### RES 342 Week 4 Individual Quiz Recent
RES 342 Week 4 Individual Quiz Recent Prepare to take a quiz on topics covered in previous weeks...
\$7.00 \$14.00
This Tutorial Purchased: 6 Times Rating: A+
#### RES 342 Week 4 Team Assignment Nonparametric Hypothesis Testing Paper Recent
RES 342 Week 4 Team Assignment Nonparametric Hypothesis Testing Paper Recent Use the same r..
\$7.00 \$14.00
This Tutorial Purchased: 6 Times Rating: A+
#### RES 342 Week 5 Individual Assignment Applying Time Series Methodologies Simulation Recent
RES 342 Week 5 Individual Assignment Applying Time Series Methodologies Simulation Recent Resourc..
\$7.00 \$12.00
This Tutorial Purchased: 5 Times Rating: A+
#### RES 342 Week 5 Individual E Text Recent
RES 342 Week 5 Individual E Text Recent Resource: Ch. 12–14 of Applied Statistics in Business and..
\$7.00 \$13.00
This Tutorial Purchased: 5 Times Rating: A+
Showing 1 to 15 of 16 (2 Pages) | 904 | 3,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-31 | latest | en | 0.654677 |
https://smartconversion.com/perimeter-calculator-ellipse-naive | 1,708,840,742,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474581.68/warc/CC-MAIN-20240225035809-20240225065809-00716.warc.gz | 538,470,110 | 7,595 | Report a Problem
Suggestions
Calculate The Perimeter Of An Ellipse
Last updated: Saturday, June 24, 2023
Select an approximation formula below
Naïve Formula
Euler's Formula
Kepler's Formula
Peano's Formula
Ramanujan Formula 1
Ramanujan Formula 2
The perimeter of an ellipse is a bit more complex than that of a circle. It cannot be expressed using a simple formula like that of the circle's circumference. However, there are several approximate formulas that can be used to estimate the perimeter of an ellipse, depending on the desired level of accuracy.
This calculator uses the Naive Formula, while this formula is relatively simple, it is only an approximation and may not be accurate for highly eccentric ellipses. In those cases, more advanced methods may be required to calculate the perimeter. Nonetheless, the naive formula can be a useful tool for estimating the perimeter of an ellipse in many practical applications.
The Naive Formula is one of the many approximation formulas used for determining the perimeter of an ellipse, it has a relative error of approximately -21.5%.
$$P$$ $$=$$ $$\pi$$ $$\cdot$$ $$(a$$ $$+$$ $$b)$$
$$P$$: the perimeter of the ellipse
$$a$$: the length of the major axis
$$b$$: the length of the minor axis
$$\pi$$: A mathematical constant with an infinite decimal tail
The SI unit of perimeter is $$meter\text{ }(m)$$
Find $$P$$
Use this calculator to determine the perimeter of an ellipse when both lengths of the minor and major axis are given.
Hold & Drag
CLOSE
the length of the major axis
$$a$$
$$meter$$
the length of the minor axis
$$b$$
$$meter$$
$$\pi$$ : A mathematical constant with an infinite decimal tail
Bookmark this page or risk going on a digital treasure hunt again
1 (3)
6 (20)
6 (6)
1 (2)
1 (2)
1 (2) | 434 | 1,769 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-10 | latest | en | 0.843634 |
www.ojasonline.in | 1,627,315,822,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.81/warc/CC-MAIN-20210726152107-20210726182107-00006.warc.gz | 951,615,667 | 21,456 | Developed By sarkar
Mathematical symbols are used to perform various functions. Symbols make it easy to refer to math quantities. It is interesting to note that mathematics is entirely based on numbers and symbols. The symbols of mathematics not only refer to different quantities but also represent the relationship between two quantities. Mathematical symbols are mainly used to conduct mathematical operations under various concepts. As we know, the concept of mathematics relies purely on numbers and symbols.
Maths has many symbols that have some predefined value. To simplify expressions, we can use those types of values instead of symbols. Some examples are symbols of pi (π), which have a value of 22/7 or 3.17, and an e-symbol in maths that holds the value e = 2.718281828…. This symbol is known as the E-constant or Euler's constant. The table below lists all the common math symbols with meanings and examples.
There are many mathematical symbols which are very important for students. To make it easier to understand, here is the list of mathematical symbols with definitions and examples. There are many signs and symbols from simple integration concept to complex integration concept hint. Here, the list of mathematical symbols is provided in a tabular form, and those assumptions are categorized conceptually.
List of mathematical symbols
Basic Math Symbol
Argument mark
Calculus and analysis symbol
Valence symbol
Greek letters
General number sign
Importance | 290 | 1,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-31 | latest | en | 0.940676 |
https://math.stackexchange.com/questions/2149225/bayesian-inference-about-a-proportion-nfl-playoffs-and-coin-flips | 1,726,277,102,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.77/warc/CC-MAIN-20240913233654-20240914023654-00349.warc.gz | 358,846,104 | 39,325 | # Bayesian inference about a proportion (NFL playoffs and coin flips)
Question: In the NFL, a professional American football league, there are 32 teams, of which 12 make the playoffs. In a typical season, 20 teams (the ones that don’t make the playoffs) play 16 games, 4 teams play 17 games, 6 teams play 18 games, and 2 teams play 19 games. At the beginning of each game, a coin is flipped to determine who gets the football first. You are told that an unknown team won ten of its coin flips last season. Given this information, what is the posterior probability that the team did not make the playoffs (i.e. played 16 games)?
Hint: Conduct both a Bayesian and Frequentist analysis of data to make inferences about a proportion
Why might a Bayesian and Frequentist analysis be necessary? This strikes me as a Bayesian problem.
Is there a relationship between whether a football team wins a coin flip and whether they make the playoffs? The two probabilities seem independent.
Below are the probabilities that I'm deducing. Please help me find out the problems with my logic.
• Posterior = P(did not make playoffs | won 10 out of 16 coin flips)
• Prior = P(did not make playoffs) = $\frac{20}{32}$ = 0.625
• Likelihood = P(won 10 out of 16 coin flips | did not make playoffs) = $\binom{16}{10}(1-0.5)^{10}(1-0.5)^{6}$ = 0.122. There are two teams per game, and I assume that each has an equal probability of winning since we have no further information.I am assuming that the coin is fair, so p = 0.5 and q = 0.5.
• Data = P(won 10 coin out of 16 coin flips) = 0.122
Posterior = $\frac{Likelihood * Prior}{Data}$ = $\frac{0.122*0.625}{0.122}$ = 0.625
Am I calculating the likelihood improperly because there is a relationship between the coin flip and the probability of winning? Any help is appreciated.
• Why are you dividing by $0.122$? as P(data)? That's $P(data|did not make playoffs).$ Commented Feb 17, 2017 at 20:13
• @spaceisdarkgreen: I was assuming that the data is won 10 out of 16 coin flips. If winning coin flips is independent of making the playoffs, shouldn't be be dividing by 0.122? Commented Feb 17, 2017 at 20:15
• No, you divide by total probability of the data, not the conditional probability of the data (which you did calculate correctly). See answer. Commented Feb 17, 2017 at 20:19
• @spaceisdarkgreen: Thanks for all your help today. I understand the logic now. :) Commented Feb 17, 2017 at 20:23
Let $p_n$ be the probability that our team played exactly $n$ games.
Our prior: $$p_{16}=\frac {20}{32},\,p_{17}=\frac 4{32},\,p_{18}=\frac 6{32},\,p_{19}=\frac 2{32}$$
Now, if there were $n$ coin tosses, the probability that our guys won exactly $10$ of them is $P(W=10\,|\,n)=\binom n{10}2^{-n}$. We can work this out case by case to see that $$n= \{16,17,18,19\}\implies P(W=10\,|\,n)=\{0.122192383,0.148376465,0.166923523,0.176197052\}$$
Of course our revised estimate that our team played only $16$ games is then $$\hat {p_{16}}=\frac {P(W=10\,|\,16)\times p_{16}}{\sum_{n=16}^{19}P(W=10\,|\,n)p_n}\approx 0.556521739$$
• Thank you so much. I didn't understand that the prior should contain information for all the teams, and not just the team for which we are trying to estimate the parameter. Commented Feb 17, 2017 at 20:19
• Yes....your prior is just the Frequentist's prior. After you add new information, that changes your estimates.
– lulu
Commented Feb 17, 2017 at 20:22
Your problem is in dividing by $P(data).$ $P(data)$ is the total probability of the data but you plugged in the probability conditional on not making the playoffs. (Would it really make sense if that canceled out every single time... why wouldn't the formula just be posterior = prior in that case?)
To get the total probability of the data, you need to sum over all the possible $n$'s: $$p(data) = p(data|n=16) P(n=16) + p(data|n=17)P(n=17)+p(data|n=18)P(n=18)+p(data|n=19)P(n=19)$$
and compute all those just like you computed $p(data) = p(data|n=16) P(n=16)=0.122*0.625$ | 1,183 | 4,000 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-38 | latest | en | 0.952871 |
https://www.tumblr.com/search/double%20width | 1,510,950,714,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934803906.12/warc/CC-MAIN-20171117185611-20171117205611-00546.warc.gz | 915,917,023 | 86,042 | # double width
flickr
How do you determine the different levels in step sequences? Everything is so fast, I don't get how people can count the edges and turns?! Love ur blog btwn ❤️❤️❤️
(Huge thanks to @drag0ness for this reply!)
Counting step sequences is difficult and requires a lot of practice to be able to discern the steps and turns the skater is doing in order to achieve the highest level. Most of this is from the ISU Handbook (for singles skating; step sequences in ice dance are a whole other beast) but I think watching a step sequence with all the steps/turns written out helps. There’s a video listing all the turns and steps in Yuzuru Hanyu’s 2017 ACI SP here; you can watch it in slow-mo on Youtube.
The basic things that skaters do with their feet (gifs of ice dancers doing some of these turns/steps can be seen here and here):
Turns: twizzles, brackets, loops, counters, rockers, three turns
Steps: toe steps, chasses, mohawks, choctaws, change of edge, cross rolls
Turns/steps that are considered “difficult” (by the ISU):
• Twizzles (turn)
• Brackets (turn)
• Loops (turn)
• Counters (turn)
• Rockers (turn)
• Choctaws (step)
All of these are turns on one foot. The choctaw uses 2 feet but you are stepping to change to the opposite edge on the other foot (eg. from a backward outside edge to a forward inside edge).
To get a level 4, you need to do a minimum of 11 difficult turns and steps and you have to do the steps in both clockwise and counterclockwise directions. Each step cannot be counted more than twice. In terms of ice coverage, it must either use the entire length of the ice or double the width of the ice.
Other level features:
1) Rotations in both directions with full body rotation covering at least 1/3 of the pattern in total for each rotational direction.
2) Use of upper body movements for at least 1/3 of the pattern. Easy to understand - is the skater using varied arm movement? Arms moving around affect your balance, can help you steer your body around the turns. Is their torso moving too or does it only stay upright over their center of gravity?
3) Two different combinations of 3 difficult turns executed with a clear rhythm. By this they mean that within the 3 different, difficult (of the 5 turns listed above) turns you do, the ending edge of the 1st turn must be the starting edge of the 2nd turn and the ending edge of the 2nd turn must be starting edge for the 3rd turn. By 2 different combinations, it means 1 on right foot, 1 on left. (For example, in the Yuzuru video above, he does the first combination on his left foot at 00:13 and the second combination on his right foot at 00:23.)
Some mistakes that can affect levels:
1) A skater may not be able to hold an edge during a turn and as a result do a hop; this may make them lose 1 of the 11 difficult turns/steps they have planned and lose a level.
2) The edges in the turns are considered too shallow to be counted in the step sequence; counted as not enough steps/turns.
3) Not enough ice coverage; this can invalidate the entire step sequence.
TL;DR Did the skater do at least 11 hard thingies with their feet? Did they do the hard thingies in both directions and used both their feet? Is their upper body moving too? Did they cover a lot of ice and show deep edges? If so, it’s probably a level 4.
“Just write. It doesn’t matter what you write. Just sit at your desk and write.”
Listen to Emma Thompson’s discuss her writing process in depth.
Find out more about her lecture here.
Aaron Hotchner / Choices | An Age Gap
As per anon’s request for 500 followers extravaganza:
## Imagine having an age gap while dating Aaron Hotchner
Okay, okay sorry for the delay on the 500 Follower fics, but I’ve been busy studying, which I need to start doing now unfortunately… But! This fic ended up being waaaaay longer than I intended, but it was worth the effort! I really enjoyed writing this one. This is a spiritual sequel to Afraid, but you don’t technically have to read it. But I hope you guys enjoy!!
Audio 11: Teasing The Fatty
A long teasing audio :) I didn’t go too mean since that’s not everyone’s cup of tea, but I did include some name calling and a bit of a mocking tone.
Here’s something happy that happens a bit after Jeremy and Michael eventually make up after their huge fight. God bless these boys. Thank you, @potterheadbbc <3
The Old Meets The New
Author: AvengeSuperWhoLock
Word Count: 2714
Summary: Loki is up to his usual antics. And by antics I mean trying to poach you from the Avengers and turn you Loki-evil (outside of a coffee shop no less). But when your ex-husband approaches, he finds out that he’s not the scariest thing in your life. When he finds out why you’re more scared of a Midgardian than you are of him, things get messy.
Part 2: A Question Of Morals
Part 3: Old Habits Die Hard
Part 4: Friends And Enemies
Can’t deny I’ve been looking forward to this one ♥ Double-width dimensions this time because Kel’s such a flamboyant sod. Dat waistcoat.
Previous portraits can be found here: Allrianne | Breeze | Elend
Phoenix Rising - Chapter 8 (Eric X OC)
Rating: M (violence/swearing/smut :p)
Genre: General/Drama/Angst
Thanks everyone for the re-blogs and support!!! IT IS SO AWESOME!!!
************************************************************************************************************************************************************************************************************************************************************************************************************
Finally, after a hellishly long time that to Fox felt like forever, she heard the doctor speak.
“Okay! Scan’s done, prep for sedation!”
Fox kept her eyes locked on Eric’s, crooning softly to him. The panic in his eyes had faded, but had been replaced with misery and anguish instead. He stared desperately at Fox, pleading with her to stay, please don’t leave me alone like this.
I’m not leaving you baby, ever.
I’m so scared.
I’ve done a double-height one of these before, but this is the first double-width I’ve needed. I think Deerling is just adorable, even if it’s oddly forgettable for something that has both a special gimmick and a unique type combination.
I got a job interview! It’s for a place that would have hired me two years ago except they had to give the position to a transfer from another store. Let’s hope that doesn’t happen again!
Zaibatsu Cards! Collect em’ all!
Astrid’s Skirt
I’m going to start the Astrid tutorials with her skirt! This was the most time consuming part of the project due to several reason. 1) I didn’t have the appropriate needles for my sewing machine so I had to sew by hand, 2) I did a lot of trials in order to achieve the right look, 3) it was repetitive work (not difficult, just repetitive). I’d also like to note that I used pleather for her skirt because I like to avoid using leather if possible. So this tutorial may/may not be helpful if you decide to use leather instead because the two materials are used differently.
Materials
• About ½ yard fake brown leather
• Three sheets of 12x18″ foam (the thin stuff aka craft foam), preferably brown or black
• Brown/black thread and a LARGE sewing needle
• Bobbie pins or other clips
• Measuring tape
• Paper, pens, ruler, markers, etc. (for template making)
Two colored Scarf Tutorial
Hey so if you want to cosplay characters like Roxy Lalonde, Eridan Ampora, Trickster!Roxy, Erisolsprite or any other one with a two colored striped scarf this could be useful for you.
I’m going to show you how I made my Roxy Lalonde scarf but this will work for all of the above characters, just choose fabrics and their colors appropriateley.
So if this applies to you or you just like diy’s or are curious how I made this then read on :D
This is how mine looks:
Three Days in the Highlands Chapter 32
DISCLAIMER:
This is a complete work of fiction and as such is an entirely fabricated tale created in my imagination. I appreciate you taking the time to read my story . Many thanks xox
SYNOPSIS:
Sam joins Caitriona in the bath and she is overwhelmed by how he makes her feel and vice versa. Highly Suggestive.
Special thanks to the lovely @sassylover-stuff for her bubble bath manip picture used below. xox
Harvest Goddess Magical Melody Wig Tutorial
Okay so, keep in mind, this tutorial is kind of long, the entirety of this wig is very expensive (overall, it cost me a little under 250\$ CAD but if you’re NOT dealing with the horrific USD/CAD exchange rate, it probably won’t be as expensive). Also, disclaimer that I’m not a professional, most of this is pure bullshit, and if I ever did this wig again I’d want to change a LOT on how I went about it.
Also, if this wasn’t clear before, I don’t recommend this as a beginner project if you don’t already have experience with some intermediate cosplay stuff.
Some tutorials I based this off was echoing-artemis’s (DA) Rapunzel wig tutorial and quite a few scattered wig dyeing tutorials.
The stuff I got for this were:
2x Epic-cosplay Asteria Wigs in Fusion Vanilla Purple
3x Matching long wefts (also from Epic Cosplay)
Wig clips
Those little hair combs you sometimes see added to Arda wigs. I didn’t get these because I had no idea where to get them from and I ran out of time but it probably would’ve made a huge difference.
1/2m of thick batting (idk how much I used really but 1/2m is probably a safe bet)
FW ink in purple for the gradient
2x bottles of rubbing alcohol
Empty spray bottle
Gloves for dyeing
Long necked wig head. This will subsequently be destroyed
2x Wig stands. Okay, you’ll need at least one, but mine BROKE halfway through so…keep a spare one handy. Or you’ll be duct taping your wig head to the table and also plying friends to come over to hold your wig head for you.
Wefting needles. Whatever kind you like working with.
Tacky glue
Caulking
Hairspray (I used got2b)
Spray adhesive. Everything is temporary, except for Kryolan Easy-Tack. Kryolan Easy-Tack is forever….(seriously, protect your surfaces when working with this shit)
Wig combs/brushes and spray detangler
Probably some kind of container your wig can drip into while it dries from the gradient dye.
(^2x wigs + 3x wefts)
First, take your first wig, and harvest it’s wefts. I basically took off all the wefts and left the skin top. So yeah, very carefully get all your wefts out. You now have an entire wig’s worth of wefts. Sew this into your second wig, making sure to brush out the wig roughly every couple wefts you sew in, because if you don’t it WILL get tangled just because of how friking long this thing is.
Next, add that first pack of wefts to it. Make sure you also add the wefts for the bangs, don’t skimp on the bangs.
(^ still wefting…)
Next, you want to sew in your wig clips. I just put two at the very front, because I’m a butt and forgot to put two at the back. If you do have those hair combs, definitely insert those as well. The more support the better. I didn’t put it on a scale to weigh it, but I did hold up my 10lb dog and also held up the wig, and they weighed pretty much the same so…
Okay now dye it. Go search up some tutorials on wig dyeing. I just mixed 12 drops of FW ink into my spray bottle, and stuck in about half a cup of alcohol. That’s supposed to be overdoing it, but my wig did not end up crunchy or felt like it had product in it, so I just went with it because I didn’t want to dye it 4 times since the alcohol smell gave me a headache. You might want a face mask in this step it gets pretty stinky. Make sure you protect your surfaces somehow because this mixture will stain. Spray dye your wig up to the gradient. Let it dry. Repeat. I did mine twice until I was happy with the colour.
(^ dyed and brushed out)
Okay so this is where the fun ended and I started hating my life. Brush out your wig. Make it nice and smooth. Brush it out again, if you have any tangles here, GET RID OF THEM, or there is no going back once you proceed. First, I divided the wig into two large sections, plus two smaller sections for the braids. I braided the two smaller sections and just put some elastic on it and pinned them at the top so they’re out of the way.
Now, with your two large sections, divide those again. The Harvest Goddess’s hair in Magical Melody are these two big twists. They’re not really braids, they’re just like…twisted pieces?? So there’s your twisted pieces, you have pieces 1, 2, 3, 4.
Cut a length of batting matched roughly to the length of the wig. My batting wasn’t thick enough by itself, so I cut double the width, folded it lengthwise and just quickly basted down along the edge using my sewing machine so it stayed put.
(^ cut batting)
(^ folded in half and basted. You can see the one I already attached to the wig in the background)
Cut the batting to the length of your wig. You don’t want it exactly as long as your wig because you do want it slightly shorter so you can get that point at the end of her twists.
Get your glues (I used tacky glue and Kryolan Easy Tack for this step). I would recommend actually just taking that batting, spraying easy tack all over one side. I originally tried to do it with tacky glue and tbh wasn’t worth it, the hairs ended up tangled and everywhere.
So now you have that batting nice and soaked on ONE SIDE, attach that one side carefully to one of your sections of hair. Make sure you smooth everything down so that at least the majority of your hair pieces are stuck to the batting. Don’t worry if not all of them are. You just want to cover the batting in as much hair as possible so the batting doesn’t show.
Now spray the other side. You’ll want to place some kind of protective plastic sheet or something behind so you don’t get easy tack on anything other than that batting. Attach the other side. Okay great, now repeat for all four sections. Now you should have four wig covered batting sections. This bulks out your wig.
(^ you can see where I did the braids and pinned them on top of the wig just to get them out of the way. I used some large hair clips to keep each section of hair separated. The big thick chunk in front is the fully attached wig-covered batting piece. What I’m holding up is the massacred remains of the first wig that I took apart for wefts)
Take the remnants from the skin top of the first massacred wig and start caulking the ends and filling any gaps. The less gaps you have now the easier it’ll be in the next step.
Once everything is dry, you can start twisting the hair together to form her giant twists. This is the part where my wig stand finally broke, and my wig came crashing to the floor which made all the fibres mess up so…that was a lot of clean up and fixing to do. Hopefully this doesn’t happen to you…
(^Wig laying on the ground. One side is halfway done being twisted, and the other is the two separate sections ready to be twisted)
Twist those giant sections together, and then secure the end with an elastic. I used caulking to style the ends of her twists into a point.
Then, take your second pack of wefts, cut off the bulky top part so you have pure fibre, caulk the fibres together, and start patching up the wig again. Patch. Keep patching. Don’t stop patching until everything is nice and filled and you cannot see batting. Blend into the original wig fibres. You can do this on the stand or on the floor. I really don’t know if most wig stands can take this weight right now tbh so the floor might be safer.
Also, don’t forget to cover up that elastic at the end with a lock of wig hair. It looks a lot better that way trust me.
Take your hairspray and kind of spray that wig in place. I didn’t glue all the fibres down pat because I didn’t want it to look like it was stiff, but you’ll want to spray that then so those stray wig fibres have less of a chance of getting loose. Style your braids accordingly. Good thing you pinned them up beforehand huh.
At this point, I actually redyed my wig a THIRD time because those wefts I put in to patch up the wig no longer matched the gradient colour.
Style the bangs. I just honestly pushed them back and sprayed/glued them in place. I wanted a lace front at first for her hairline, but there was just no way I could find a lace front wig in that length, nor that colour. If you have the time to turn your lace into a lace front, look into that I guess? For me, just pushing back the bangs and spraying them worked fine.
This probably is letting some cosplay secrets out of the bag, but I got a bunch of the fake flowers you see on the final cosplay and glued them on my wig in parts where I could see either patches of batting showering, or where my patched fibres weren’t blending in as nicely with the original wig fibres. Shh….no one will know in the final piece…
Plus they matched the ones on my dress so…
I don’t know if you want to do this too, but I attached fishing line from my wrist to the wig so that I could pose my hair. It didn’t actually relieve that much of the weight, but it did make it so it didn’t just drop down and so it could hold that curved shape that you see in the games. To do that, I looped fishing wire through a small button and just hot glued it straight to the ends of each twist/braid where I had secured the elastic. I made little like…fabric bracelets, put some vines and tiny flowers on it to make it look pretty and just stuck the end of the fishing wire straight through the fabric, knotted it, and hot glued in place again. The fabric bracelets had a snap closure to go around my wrist. This method worked for me, I didn’t have any costume malfunctions and the hot glue was strong enough to hold it. I just stuck a flower over where the fishing line attached to the wig to cover it up.
I think that’s it? Overall, it took me about a month and a half and I never want to do this again. If I could do it again, I would probably not make that wig one big piece and do the braids separately and just attach them to the base wig.
Again, this is just how I did it. I can pretty confidently say this was the wrong way to do it, but it worked out for me in the end so take what you will from this tutorial. I ended up staying 6.5 hours in the cosplay, and most of the time was in horrific pain from how heavy that wig was. It’s not going to be a comfortable cosplay to be in so I’m not going to sugarcoat it but it was a good learning experience and it was pretty fun being a beautiful fairy thing and also being too large to fit in the washroom. Anyway, remember, if it looks bad, stick a flower on it.
I had a few questions for your wonderful Sonia costume. First, did you buy it or make it, and if you made it, what materials and the like did you use? And did you buy the blouse like it was or did you make it? Also where'd you get that pretty wig? And finally, the one thing that keeps frustrating me, where did you get your blue bow pin? I'm having trouble finding anything to use omg sorry for all the questions, I keep asking around for advice but no ones really giving a simple answer omg
Sonia Tutorial
First of all, I made/editted the entire cosplay. The dress, the shirt and both bows were all made from scratch. The Thigh highs were made from stockings, the brooch made editted from a brooch purchased, and the shoes painted.
With The World At Stake… Who’s Side Are You On?…
4
quickie tutorial time! most of the cosplay bow tutorials i see out there result in a pillowy, unrealistic-looking bow since they usually involve no loops. this is how i make bows that result in one that looks slightly more “tied” in place with no visible seams.
1. start with a long rectangle that’s double the width of your bow and at least twice the length. if your fabric is too flimsy to hold its shape, you may want to add interfacing. double this over and sew to make a long tube, which you will want to press with the seam centered in the tube.
2. with the seam on the inside, pin the two ends where you want it to meet in the middle so it is the width you want. cut off any excess because we don’t want extra bulk.
3. sew that pinned bit together to make your tube into a ring. if you want your bow to have tails, now is the time to make them. you can do so using the same method as the first tube, but then closing off the ends.
4. tie it together in the middle, being careful to “crimp” the middle in a way that will be visually-pleasing. make a small second tube that’s long enough to fit around the middle and then some. if you have tails, you may wish to tie them onto your bow now.
5. hand sew the middle tube around the bow, and ta-da! you’re done!
there’s tons of different ways to attach this to your hair or wherever. you can sew/glue on clips or snaps, or if you’re super lazy like me, you can just loop a hairband through that middle part on the back. it’s more likely to sort of rotate that way, but meh. good enough for me!
DIY Steampunk Goggles
So I was inspired to do my own Goggles by Mo'Brien‘s video. The only problem was that we don’t have a Dollar Store like that in Austria. So I made many mistakes, and I didn’t really like my first version.
A friend helped me out and showed me another way to make my googles, and that’s what I did. :)
All in all I needed a transparent cover, fabric, corrugating medium, a glue gun, some wire, nail polish, painting colours, pringels, some sticky tape and thumb tacks. :D
That’s what my friend did for me, I only had to do the other one, and find out how to put them together.
I’ve alredy had the the brown form from my first try. I’ve cut them out of the pringels box and then glued the ends together, so that the goggle had a smaller radius. Don’t forget to paint them! ;)
After that I had to take the corrugating medium and form it like one of the googles. The corrugating medium has to fit inside the main form. Close the ends of the corrugating medium with some sticky tape. After that I had to glue the transparent cover over the corrugating medium form and push it into the main form. It fitted, so I glued it inside of the main form with the glue gun.
When that was done I was able to decorate my goggles with thumb tacks and corrugating medium.
I’m not good at sewing, but I’ve managed to do it. :D My fabric was to thin for my opinion. That’s why I doubled the width.
I’ve just sewd with no length in mind. The longer the better in this case.
Then I had to turn around the fabric. I didn’t like that part. So, yeah … here’s a heart. <3
I’m sorry to say that but I had to iron the haert out.
Repeat the sewing steps. And repeat it again with a piece of fabric that will sit perfectly over your nose!
First I thought it was a good idea to stick the fabric to lenses with my glue gun. It didn’t look that good, but I’ve let the smaller piece for the nose stay in it’s space.
I took some of my wire and formed it in a way so it would be the perfect holder. I’ve painted the wire golden to fit the colours and glued it to the sides of the main form with the glue gun.
I thought It wouldn’t hurt either to paint the cold glue golden, too.
After that I cut the other fabric strips, so that they would fit my head perfectly. Then I had to sew one end together and put the other end around the holder and the sew it together. | 5,514 | 23,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-47 | latest | en | 0.922887 |
https://root.cern/doc/v624/ErrorIntegral_8C_source.html | 1,709,046,011,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474676.26/warc/CC-MAIN-20240227121318-20240227151318-00634.warc.gz | 504,145,219 | 6,212 | ROOT Reference Guide
Searching...
No Matches
ErrorIntegral.C
Go to the documentation of this file.
1/// \file
2/// \ingroup tutorial_fit
3/// \notebook -js
4/// Estimate the error in the integral of a fitted function
5/// taking into account the errors in the parameters resulting from the fit.
6/// The error is estimated also using the correlations values obtained from
7/// the fit
8///
9/// run the macro doing:
10///
11/// ~~~{.cpp}
12/// .x ErrorIntegral.C
13/// ~~~
14///
15/// \macro_image
16/// \macro_output
17/// \macro_code
18///
19/// \author Lorenzo Moneta
20
21#include "TF1.h"
22#include "TH1D.h"
23#include "TFitResult.h"
24#include "TMath.h"
25#include <assert.h>
26#include <iostream>
27#include <cmath>
28
29TF1 * fitFunc; // fit function pointer
30
31const int NPAR = 2; // number of function parameters;
32
33//____________________________________________________________________
34double f(double * x, double * p) {
35 // function used to fit the data
36 return p[1]*TMath::Sin( p[0] * x[0] );
37}
38
39//____________________________________________________________________
40void ErrorIntegral() {
41 fitFunc = new TF1("f",f,0,1,NPAR);
42 TH1D * h1 = new TH1D("h1","h1",50,0,1);
43
44 double par[NPAR] = { 3.14, 1.};
45 fitFunc->SetParameters(par);
46
47 h1->FillRandom("f",1000); // fill histogram sampling fitFunc
48 fitFunc->SetParameter(0,3.); // vary a little the parameters
49 auto fitResult = h1->Fit(fitFunc,"S"); // fit the histogram and get fit result pointer
50
51 h1->Draw();
52
53 /* calculate the integral*/
54 double integral = fitFunc->Integral(0,1);
55
56 auto covMatrix = fitResult->GetCovarianceMatrix();
57 std::cout << "Covariance matrix from the fit ";
58 covMatrix.Print();
59
60 // need to pass covariance matrix to fit result.
61 // Parameters values are are stored inside the function but we can also retrieve from TFitResult
62 double sigma_integral = fitFunc->IntegralError(0,1, fitResult->GetParams() , covMatrix.GetMatrixArray());
63
64 std::cout << "Integral = " << integral << " +/- " << sigma_integral
65 << std::endl;
66
67 // estimated integral and error analytically
68
69 double * p = fitFunc->GetParameters();
70 double ic = p[1]* (1-std::cos(p[0]) )/p[0];
71 double c0c = p[1] * (std::cos(p[0]) + p[0]*std::sin(p[0]) -1.)/p[0]/p[0];
72 double c1c = (1-std::cos(p[0]) )/p[0];
73
74 // estimated error with correlations
75 double sic = std::sqrt( c0c*c0c * covMatrix(0,0) + c1c*c1c * covMatrix(1,1)
76 + 2.* c0c*c1c * covMatrix(0,1));
77
78 if ( std::fabs(sigma_integral-sic) > 1.E-6*sic )
79 std::cout << " ERROR: test failed : different analytical integral : "
80 << ic << " +/- " << sic << std::endl;
81}
#define f(i)
Definition RSha256.hxx:104
1-Dim function class
Definition TF1.h:213
virtual Double_t Integral(Double_t a, Double_t b, Double_t epsrel=1.e-12)
IntegralOneDim or analytical integral.
Definition TF1.cxx:2515
virtual Double_t * GetParameters() const
Definition TF1.h:520
virtual void SetParameters(const Double_t *params)
Definition TF1.h:644
virtual void SetParameter(Int_t param, Double_t value)
Definition TF1.h:634
virtual Double_t IntegralError(Double_t a, Double_t b, const Double_t *params=0, const Double_t *covmat=0, Double_t epsilon=1.E-2)
Return Error on Integral of a parametric function between a and b due to the parameter uncertainties ...
Definition TF1.cxx:2692
1-D histogram with a double per channel (see TH1 documentation)}
Definition TH1.h:618
virtual void FillRandom(const char *fname, Int_t ntimes=5000, TRandom *rng=nullptr)
Fill histogram following distribution in function fname.
Definition TH1.cxx:3525
virtual TFitResultPtr Fit(const char *formula, Option_t *option="", Option_t *goption="", Double_t xmin=0, Double_t xmax=0)
Fit histogram with function fname.
Definition TH1.cxx:3892
virtual void Draw(Option_t *option="")
Draw this histogram with options.
Definition TH1.cxx:3073
Double_t x[n]
Definition legend1.C:17
TH1F * h1
Definition legend1.C:5
Double_t Sin(Double_t)
Definition TMath.h:639 | 1,170 | 4,007 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-10 | latest | en | 0.348636 |
https://socratic.org/questions/a-charge-of-15-c-passes-through-a-circuit-every-12-s-if-the-circuit-can-generate-2#262367 | 1,653,039,824,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531779.10/warc/CC-MAIN-20220520093441-20220520123441-00327.warc.gz | 610,043,376 | 5,999 | # A charge of 15 C passes through a circuit every 12 s. If the circuit can generate 6 W of power, what is the circuit's resistance?
May 5, 2016
I found: $3.84 \Omega$
#### Explanation:
The current will be:
$I = \frac{q}{t} = \frac{15}{12} = 1.25 A$
The power is:
$P = V I = {I}^{2} R$ (using Ohm's Law $V = R I$)
so:
$R = \frac{P}{I} ^ 2 = \frac{6}{1.25} ^ 2 = 3.84 \Omega$ | 152 | 381 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-21 | latest | en | 0.756706 |
https://www.crazy-numbers.com/en/6715 | 1,709,497,154,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476397.24/warc/CC-MAIN-20240303174631-20240303204631-00870.warc.gz | 714,958,774 | 6,326 | # Everything about number 6715
Discover a lot of information on the number 6715: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 6715
Is 6715 a prime number? No
Is 6715 a perfect number? No
Number of divisors 8
List of dividers 1, 5, 17, 79, 85, 395, 1343, 6715
Sum of divisors 8640
Prime factorization 5 x 17 x 79
Prime factors 5, 17, 79
## How to write / spell 6715 in letters?
In letters, the number 6715 is written as: Six thousand seven hundred and fifteen. And in other languages? how does it spell?
6715 in other languages
Write 6715 in english Six thousand seven hundred and fifteen
Write 6715 in french Six mille sept cent quinze
Write 6715 in spanish Seis mil setecientos quince
Write 6715 in portuguese Seis mil setecentos quinze
## Decomposition of the number 6715
The number 6715 is composed of:
1 iteration of the number 6 : The number 6 (six) is the symbol of harmony. It represents balance, understanding, happiness.... Find out more about the number 6
1 iteration of the number 7 : The number 7 (seven) represents faith, teaching. It symbolizes reflection, the spiritual life.... Find out more about the number 7
1 iteration of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1
1 iteration of the number 5 : The number 5 (five) is the symbol of freedom. It represents change, evolution, mobility.... Find out more about the number 5
## Mathematical representations and links
Other ways to write 6715
In letter Six thousand seven hundred and fifteen
In roman numeral MMMMMMDCCXV
In binary 1101000111011
In octal 15073
In US dollars USD 6,715.00 (\$)
In euros 6 715,00 EUR (€)
Some related numbers
Previous number 6714
Next number 6716
Next prime number 6719
## Mathematical operations
Operations and solutions
6715*2 = 13430 The double of 6715 is 13430
6715*3 = 20145 The triple of 6715 is 20145
6715/2 = 3357.5 The half of 6715 is 3357.500000
6715/3 = 2238.3333333333 The third of 6715 is 2238.333333
67152 = 45091225 The square of 6715 is 45091225.000000
67153 = 302787575875 The cube of 6715 is 302787575875.000000
√6715 = 81.945103575504 The square root of 6715 is 81.945104
log(6715) = 8.8120991089573 The natural (Neperian) logarithm of 6715 is 8.812099
log10(6715) = 3.8270460170047 The decimal logarithm (base 10) of 6715 is 3.827046
sin(6715) = -0.98811980842798 The sine of 6715 is -0.988120
cos(6715) = -0.15368553670483 The cosine of 6715 is -0.153686
tan(6715) = 6.4294912170284 The tangent of 6715 is 6.429491 | 833 | 2,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-10 | latest | en | 0.739665 |
http://puzzlefry.com/puzzles/gold-coins-puzzle/ | 1,563,788,705,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527907.70/warc/CC-MAIN-20190722092824-20190722114824-00113.warc.gz | 127,016,126 | 40,925 | Five greedy pirates and gold coin distribution Puzzle
Five puzzleFry ship’s pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treacherous and selfish (especially the captain).
The captain always proposes a distribution of the loot. All pirates vote on the proposal, and if half the crew or more go “Aye”, the loot is divided as proposed, as no pirate would be willing to take on the captain without superior force on their side.
If the captain fails to obtain support of at least half his crew (which includes himself), he faces a mutiny, and all pirates will turn against him and make him walk the plank. The pirates start over again with the next senior pirate as captain.
What is the maximum number of coins the captain can keep without risking his life?
ravi Expert Asked on 19th July 2015 in
• 6 Answer(s)
Answer – 98 Coins
The captain says he will take 98 coins, and will give one coin to the third most senior pirate and another coin to the most junior pirate. He then explains his decision in a manner like this…
If there were 2 pirates, pirate 2 being the most senior, he would just vote for himself and that would be 50% of the vote, so he’s obviously going to keep all the money for himself.
If there were 3 pirates, pirate 3 has to convince at least one other person to join in his plan. Pirate 3 would take 99 gold coins and give 1 coin to pirate 1. Pirate 1 knows if he does not vote for pirate 3, then he gets nothing, so obviously is going to vote for this plan.
If there were 4 pirates, pirate 4 would give 1 coin to pirate 2, and pirate 2 knows if he does not vote for pirate 4, then he gets nothing, so obviously is going to vote for this plan.
As there are 5 pirates, pirates 1 & 3 had obviously better vote for the captain, or they face choosing nothing or risking death.
John123 Expert Answered on 19th July 2015.
I didn’t get the case right from 3 pirates. If the captain offers 1 coin to pirate1, pirate1 won’t go with him knowing that the other pirate won’t go with the decision if he himself denies the proposal otherwise he won’t get anything. And if he denies, captain walks down the plank, he will be the senior most and he can then take the whole share as he owns at least half the vote now.
Now take the case for 5 pirates. If he takes 98 coins and give 1 coins to pirate 1 and 3. Now both pirate 1 and pirate 3 knows that if either of them deny the support, they will get the majority from the vote of 2 other pirates who are not getting any coin and thereby getting chance for more coins. So I think your answer is incorrect.
on 3rd February 2016.
Solution is great.
Try calculate pirat numbers from other order.
pirat5 is captan, pirat1 is lowest status pirat
case1) pirat2, pirat1,
case2) pirat3, pirat3, pirat1
….
on 5th February 2016.
Dear john123, could you please check my answer to the puzzle:
I think it was be 50 (considering that its a rule that the captain would get the highest number of gold coin)
Let the pirates in order of seniority be A, B, C, D, E
The first captain wld be A. First he would propose that he takes 98 and give 1 each to C and E (as mentioned by you)
In this case, B would tell E to reject the proposal and he would give him more than 1(so, at least 2). E knows that he would get nothing if he gets A,B and C killed, as D would take 100 for himself and vote for the proposal himself. That would gain him half of the vote and the proposal would be accepted
So, if it comes down to C, he would get a max of 1 coin, as he would choose 1 over 0.
B offers him more than 1. E knows that B is sure to keep his promise as if B offers him 0 coins, there is no way E would accept it and if B offers him 1 coin, he can reject B and move to C’s proposal, where he surely will get 1 coin, and there still a very, very small chance that he gets more than 1.
This can be sure by an example: if a greedy man is given 2 options of boxes, one sure to contain \$10 and the other sure to have a minimum of 10, the man would choose the second one as he would not face a loss in comparison to the first one, and would also have a chance to gain a bonus.
So, this shows that to survive, B has to give at least 2 coin to E and would surely do it. C will think all this and come to a conclusion that if B distributes, he would get 0, so it is better to except 1. There is no way E would reject B so B is bound to get accepted and thus he can’t gain more than 1. This secures 1 vote for A
Now thinking all this, A would think that to get E to accept his proposal, A had to give E the max number he can get. In worse case, as captain had to keep the highest number, if B were to secure 1 vote, B would only divide it among him and E, which would be 51 for B and 49 for E. Also, E knows that after A gets killed, B won’t give more than 2 coins as proved above( in C’s proposal).
So, if A offers E 49, E would sure to get 49 and will not reject the proposal for B.
Another example for this, if the greedy man is offered 2 boxes, 1 sure to contain 10 rupees and the other might have less than 10 but a max of 10, he would choose the first box, as he the second box may indulge in a loss in comparison to the first box.
Thus, A would propose 50 coins for himself, 1 for C and 49 for E
on 9th September 2016.
Either the 50 I mentioned above or 96, 1 for C and 3 for E. E cant trust B, as if A gets killed be will give only 2 coins to E and E will have to accept this. So, if A would offer E, E would gladly accept
on 9th September 2016.
Add Comment
I think the answer should be 25 coins. Divide 100 coins among 4 people and leave 1 with no coin. Now 4 pirates are satisfied. To create a mutiny for anyone junior in rank, he has to satisfy one more pirate other than the one not getting anything as if he goes against the captain he may have only 1 support in the person who is not getting any money.
Chi Starter Answered on 3rd February 2016.
The captain divides 100 coins 3 ways by giving himself 33 and two other pirates get 33 each and throws the remaining one coin in the sea.. this way he gets maximum number of coins and enough number of supporters to get his decision approved.
eidalrashidi Starter Answered on 26th April 2016.
060 can also be the solution……he (captain) divides the share into 5 equal parts and would give 2 pirates their respective shares. so he will keep the share of 3 people.
Pramod C.M Starter Answered on 17th June 2016.
I think it was be 50 (considering that its a rule that the captain would get the highest number of gold coin)
Let the pirates in order of seniority be A, B, C, D, E
The first captain wld be A. First he would propose that he takes 98 and give 1 each to C and E (as mentioned by john123)
In this case, B would tell E to reject the proposal and he would give him more than 1(so, at least 2). E knows that he would get nothing if he gets A,B and C killed, as D would take 100 for himself and vote for the proposal himself. That would gain him half of the vote and the proposal would be accepted
So, if it comes down to C, he would get a max of 1 coin, as he would choose 1 over 0.
B offers him more than 1. E knows that B is sure to keep his promise as if B offers him 0 coins, there is no way E would accept it and if B offers him 1 coin, he can reject B and move to C’s proposal, where he surely will get 1 coin, and there still a very, very small chance that he gets more than 1.
This can be sure by an example: if a greedy man is given 2 options of boxes, one sure to contain \$10 and the other sure to have a minimum of 10, the man would choose the second one as he would not face a loss in comparison to the first one, and would also have a chance to gain a bonus.
So, this shows that to survive, B has to give at least 2 coin to E and would surely do it. C will think all this and come to a conclusion that if B distributes, he would get 0, so it is better to except 1. There is no way E would reject B so B is bound to get accepted and thus he can’t gain more than 1. This secures 1 vote for A
Now thinking all this, A would think that to get E to accept his proposal, A had to give E the max number he can get. In worse case, as captain had to keep the highest number, if B were to secure 1 vote, B would only divide it among him and E, which would be 51 for B and 49 for E. Also, E knows that after A gets killed, B won’t give more than 2 coins as proved above( in C’s proposal).
So, if A offers E 49, E would sure to get 49 and will not reject the proposal for B.
Another example for this, if the greedy man is offered 2 boxes, 1 sure to contain 10 rupees and the other might have less than 10 but a max of 10, he would choose the first box, as he the second box may indulge in a loss in comparison to the first box.
Thus, A would propose 50 coins for himself, 1 for C and 49 for E
Bhavuk Starter Answered on 9th September 2016.
Either this or A will keep 96 for himself and offer 3 to E and 1 to C. This is because E cant trust B as if A gets killed, to gain the max profit B would only give 2 to E. E can be certain that he wouldn’t gain more than 2, so even 3 should satisfy him.
on 9th September 2016.
Add Comment
96 coins
The problem has a recursive framework, such that the pirates could continue to rebel against the current captain. The ending condition is when only two pirates are left because one of their votes will meet the 50% required vote and no more mutiny can occur. Due to the leadership specification (where each successive group of pirates will lose their highest ranked pirate), these are the possible subsets of pirates throughout the recursive process: {1,2,3,4,5}, {2,3,4,5}, {3,4,5}, {4,5}
As you can see, the ending condition is when only pirates 4 and 5 are left, so lets work backwards through the logic from this point!
{4,5}
Distribution: 4=100, 5=0
Reasoning: If only pirates 4 and 5 are left, pirate 4 can give himself all the coins because his vote will meet the 50% requirement.
{3,4,5}
Distribution: 3=99, 5=1
Reasoning: Pirate 4 will certainly vote against pirate 3 because he always stands to gain far more through mutiny, so pirate 3 needs pirate 5’s vote to avoid mutiny. All he has to do is offer a single coin because it’s better than the zero coins pirate 5 will earn if he rebelled.
{2,3,4,5}
Distribution: 2=98, 5=2
Reasoning: Pirate 2 needs only a single vote to help him avoid mutiny, so he needs only offer pirate 5 a coin more than he’d earn in a mutinous situation.
{1,2,3,4,5}
Distribution: 1=96, 3=1, 5=3
Reasoning: The captain, pirate 1, needs two additional votes to avoid mutiny. He can buy pirate 5’s vote by offering him a coin more than he’d earn in his next best situation. Since pirate 3 will earn zero coins in the next situation, he needs only offer a single coin to earn his vote.
So the captain should give himself 96, the 3rd pirate 1, and the 5th pirate 3.
I had initially guessed 34 based on my intuition, but you can see how wrong that really is!
RenaissanceV2 Starter Answered on 17th October 2016.
In the {2,3,4,5} scenario, the distribution will be 2=99, 4=1 ( not 2=99, 5=2) and 4 will accept the one coin as if it goes to the next scenario, 4 will be left with zero coins…
So if it comes to this scenario, 3 and 5 will not get any coins so they would rather each accept 1 coin given to them by 5
Therefore answer is 98 coins for captain 1 and one coin each for pirates 3 and 5 ( as mentioned by john 123 above)
on 9th June 2019.
Add Comment
• Your Answer
By posting your answer, you agree to the privacy policy and terms of service.
• More puzzles to try-
• What is the logic behind these ?
3 + 3 = 3 5 + 4 = 4 1 + 0 = 3 2 + 3 = 4 ...Read More »
• Which letter replaces the question mark
Which letter replaces the question markRead More »
• Defective stack of coins puzzle
There are 10 stacks of 10 coins each. Each coin weights 10 gms. However, one stack of coins is defective ...Read More »
• Dr. Willam and surgical gloves puzzle
Dr.Willam wants to operate for three different persons who were wounded. But he had only two surgical gloves. There is ...Read More »
• Which clock works best?
Which clock works best? The one that loses a minute a day or the one that doesn’t work at all?Read More »
• To which one will the needle touch 1 OR 2
In the gear arrangement, To which one will the needle touch, 1 OR 2Read More »
• How did he know?
A man leaves his house in the morning to go to office and kisses his wife. In the evening on ...Read More »
• Which room is safest puzzle
A murderer is condemned to death. He has to choose between three rooms. The first is full of raging fires, ...Read More »
• Crossing the Bridge Puzzle (Bridge and torch problem)
A bridge will collapse in 17 minutes. 4 people want to cross it before it will collapse. It is a ...Read More »
• Probability of having boy
In a country where everyone wants a boy, each family continues having babies till they have a boy. After some ...Read More »
• Magical flowers!!
A devotee goes to three temples, temple1, temple2 and temple3 one after the other. In front of each temple, there ...Read More »
• How many times a day do all three hands of analog watch overlaps?
There is a perfectly accurate analog watch with hour, minute, and second hands. How many times a day do all three ...Read More »
• Tuesday, Thursday what are other two days staring with T?
Four days are there which start with the letter ‘T‘. I can remember only two of them as “Tuesday , Thursday”. ...Read More »
• How could only 3 apples left
Two fathers took their sons to a fruit stall. Each man and son bought an apple, But when they returned ...Read More »
• How Many Eggs ?
A farmer is taking her eggs to the market in a cart, but she hits a pothole, which knocks over ...Read More »
• Minimum number of persons needed to cross a Desert
In the middle of the confounded desert, there is the lost city of “Ash”. To reach it, you will have ...Read More »
• Lateral thinking sequence Puzzle
Solve this logic sequence puzzle by the correct digit- 8080 = 6 1357 = 0 2022 = 1 1999 = ...Read More »
• Pizza Cost Math Brain Teaser
Jasmine, Thibault, and Noah were having a night out and decided to order a pizza for \$10. It turned out ...Read More »
• Overthrow your math teacher!
Some numbers are very mysterious. They are all integers. They have more than one digit. If you multiply them with ...Read More »
• Richie’s Number System
Richie established a very strange number system. According to her claim for different combination of 0 and 2 you will ...Read More » | 3,792 | 14,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-30 | latest | en | 0.976453 |
https://mathematica.stackexchange.com/questions/201489/how-to-test-whether-an-improper-integral-converges-or-diverges | 1,596,595,094,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735906.77/warc/CC-MAIN-20200805010001-20200805040001-00382.warc.gz | 345,014,196 | 33,053 | # How to test whether an improper integral converges or diverges?
To verify the improper integral $$\int_{1}^{\infty}x\cdot\left | \sin (x^4)\cdot\sin x \right |dx$$ converges or diverges,I code
Integrate[x*Abs[Sin[x^4]*Sin[x]], {x, 1, Infinity}]
then feedback is the expression of $$\int_{1}^{\infty}x\cdot\left | \sin (x^4)\cdot\sin x \right |dx$$.Is that means the improper integral $$\int_{1}^{\infty}x\cdot\left | \sin (x^4)\cdot\sin x \right |dx$$ converges?
• Obviously, the integral diverges. But it is surprising that the system does not give a message, and estimates the numerical value as 1.5408*10^147 – Alex Trounev Jul 3 '19 at 15:33
• @AlexTrounev: Yes,I proved it diverges in the past .But after using mathematica to verify my conclusion, I puzzled ! Thanks for your confirmation ! What's worng with it? – AplehKevin Jul 3 '19 at 15:43
• Not all cases have been studied and not all are entered into the database. – Alex Trounev Jul 3 '19 at 16:27
• @AlexTrounev：Is there any better way to deal with this kind of problem？ – AplehKevin Jul 3 '19 at 16:36
• This is rather a math question than a Mathematica question: see math.stackexchange.com/questions/1018348/… . – user64494 Jul 4 '19 at 17:04 | 383 | 1,215 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-34 | latest | en | 0.72267 |
http://research.stlouisfed.org/fred2/series/DDOE01ARA086NWDB | 1,429,923,136,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246645538.5/warc/CC-MAIN-20150417045725-00023-ip-10-235-10-82.ec2.internal.warc.gz | 219,528,832 | 19,704 | Consumer Price Index for Argentina
2011: 175.95990 Index 2005=100 (+ see more)
Annual, End of Year, Not Seasonally Adjusted, DDOE01ARA086NWDB, Updated: 2013-06-06 7:00 PM CDT
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
Consumer price index reflects changes in the cost to the average consumer of acquiring a basket of goods and services that may be fixed or changed at specified intervals, such as yearly. The Laspeyres formula is generally used.
Source Code: GFDD.OE.01
Source: World Bank
Release: Global Financial Development
Restore defaults | Save settings | Apply saved settings
w h
Graph Background: Plot Background: Text:
Color:
(a) Consumer Price Index for Argentina, Index 2005=100, Not Seasonally Adjusted (DDOE01ARA086NWDB)
Consumer price index reflects changes in the cost to the average consumer of acquiring a basket of goods and services that may be fixed or changed at specified intervals, such as yearly. The Laspeyres formula is generally used.
Source Code: GFDD.OE.01
Consumer Price Index for Argentina
Integer Period Range: to copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
Graph Data
Graph Image
Suggested Citation
``` World Bank, Consumer Price Index for Argentina [DDOE01ARA086NWDB], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/DDOE01ARA086NWDB/, April 24, 2015. ```
Retrieving data.
Graph updated.
Recently Viewed Series
Subscribe to our newsletter for updates on published research, data news, and latest econ information.
Name: Email: | 575 | 2,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2015-18 | latest | en | 0.814702 |
https://www.stchristophersprimary.com/monday-56/ | 1,656,575,362,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103669266.42/warc/CC-MAIN-20220630062154-20220630092154-00527.warc.gz | 1,074,269,764 | 20,595 | # Monday
English
Physical Activity
'Just Dance' Dance 1 Dance Monkey - & Dance 2 - Savage Love. Create a warm up, complete the 2 dances and create a cool down. Enjoy!
Maths
Today we are going to recap on the place value of numbers between 11 and 20.
Here are some ideas to support your child:
• Pick a number between 11 and 20, write the numeral and number word for it.
Can they show it in tens and ones?
What is one more or one less than that number?
• Pick two numbers (between 11 and 20). Compare the numbers, which is greater than, less than or equal to.
• Order groups of objects between 11 and 20. Can you order them from least to greatest? What about from the least to greatest? | 181 | 698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-27 | latest | en | 0.902086 |
https://gateoverflow.in/247106/probability-gravner-74 | 1,670,248,364,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711017.45/warc/CC-MAIN-20221205132617-20221205162617-00299.warc.gz | 297,115,792 | 21,774 | 534 views
Let X be a $N(\mu , \sigma^2)$ random variable and let $Y = \alpha X+\beta$, with $\alpha$ > $0$. How is $Y$ distributed?
$Y = \alpha X + \beta$
Mean = $\mu$
Standard Deviation = $\sigma$
Variance = $\sigma ^2$
Let Z be a random variable such that
$Z = \alpha X$
So Z is scaled by some $\alpha > 0$
$\mu _ Z = \alpha \mu$
$\sigma _Z = \alpha \sigma$
$Z \sim$ $N(\mu _ Z , \sigma _Z ^2 ) \sim N(\alpha \mu, (\alpha \sigma)^2)$
Now,
$Y = Z + \beta$
So the distribution will be shifted depending upon the magnitude and sign of $\beta$,
But it doesn't change the standard deviation but it does change the mean.
$\mu_y = \mu_z + \beta = \alpha \mu + \beta$
Hence,
$Y \sim N(\alpha \mu + \beta, (\alpha \sigma)^2)$
### 1 comment
what about mode , median here ?? they also change ...
$E(Y)=E(\alpha x+\beta )=E(\alpha x)+E(\beta )=\alpha E(x)+\beta =\alpha \mu +\beta$
Since E(ax)=aE(x) and E(constant)=constant
$Var(Y)=Var(\alpha x+\beta )=Var(\alpha x)+Var(\beta )=\alpha^{2} Var(x)+\beta =\alpha^{2} \sigma ^{2}$
Since Var(ax)=$a^{2}$Var(x) and Var(constant)=0
Hence, Y is a N($\alpha \mu +\beta$, $\alpha^{2} \sigma ^{2}$) random variable
by
1
696 views | 413 | 1,188 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2022-49 | latest | en | 0.650423 |
https://converter.ninja/volume/imperial-quarts-to-us-dry-barrels/174-imperialquart-to-drybarrel/ | 1,656,814,643,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104209449.64/warc/CC-MAIN-20220703013155-20220703043155-00334.warc.gz | 220,676,274 | 5,397 | # 174 imperial quarts in US dry barrels
## Conversion
174 imperial quarts is equivalent to 1.71028136712522 US dry barrels.[1]
## Conversion formula How to convert 174 imperial quarts to US dry barrels?
We know (by definition) that: $1\mathrm{imperialquart}\approx 0.00982920325934033\mathrm{drybarrel}$
We can set up a proportion to solve for the number of US dry barrels.
$1 imperialquart 174 imperialquart ≈ 0.00982920325934033 drybarrel x drybarrel$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{drybarrel}\approx \frac{174\mathrm{imperialquart}}{1\mathrm{imperialquart}}*0.00982920325934033\mathrm{drybarrel}\to x\mathrm{drybarrel}\approx 1.7102813671252175\mathrm{drybarrel}$
Conclusion: $174 imperialquart ≈ 1.7102813671252175 drybarrel$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 US dry barrel is equal to 0.584699113971453 times 174 imperial quarts.
It can also be expressed as: 174 imperial quarts is equal to $\frac{1}{\mathrm{0.584699113971453}}$ US dry barrels.
## Approximation
An approximate numerical result would be: one hundred and seventy-four imperial quarts is about one point seven one US dry barrels, or alternatively, a US dry barrel is about zero point five seven times one hundred and seventy-four imperial quarts.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic.
Was it helpful? Share it! | 426 | 1,544 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2022-27 | latest | en | 0.741844 |
https://calcitfast.com/quarts-to-gallons | 1,716,814,046,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00161.warc.gz | 127,442,225 | 9,589 | # Quarts to Gallons
Quarts (qt) to gallons (gal) conversion calculator
## How many quarts in a gallon
1 quart = 0.25 gallons
1 gallon = 4 quarts
## How to convert quarts to gallons
To find gallons from quarts, you need to devide quarts value by 4. See example below:
For example, let's convert 80 quarts to gallons:
gallons = 80 / 4 = 20 gallons
In result we'll receive that 80 quarts are equal to 20 gallons.
Quarts (qt)
Gallons (gal)
• 0.25 gal
• 0.50 gal
• 0.75 gal
• 1.00 gal
• 1.25 gal
• 1.50 gal
• 1.75 gal
• 2.00 gal
• 2.25 gal
• 2.50 gal
• 2.75 gal
• 3.00 gal
• 3.25 gal
• 3.50 gal
• 3.75 gal
• 4.00 gal
• 4.25 gal
• 4.50 gal
• 4.75 gal
• 5.00 gal
• 5.25 gal
• 5.50 gal
• 5.75 gal
• 6.00 gal
• 6.25 gal
• 6.50 gal
• 6.75 gal
• 7.00 gal
• 7.25 gal
• 7.50 gal
• 7.75 gal
• 8.00 gal
• 8.25 gal
• 8.50 gal
• 8.75 gal
• 9.00 gal
• 9.25 gal
• 9.50 gal
• 9.75 gal
• 10.00 gal
• 10.25 gal
• 10.50 gal
• 10.75 gal
• 11.00 gal
• 11.25 gal
• 11.50 gal
• 11.75 gal
• 12.00 gal
• 12.25 gal
• 12.50 gal
• 12.75 gal
• 13.00 gal
• 13.25 gal
• 13.50 gal
• 13.75 gal
• 14.00 gal
• 14.25 gal
• 14.50 gal
• 14.75 gal
• 15.00 gal
• 15.25 gal
• 15.50 gal
• 15.75 gal
• 16.00 gal
• 16.25 gal
• 16.50 gal
• 16.75 gal
• 17.00 gal
• 17.25 gal
• 17.50 gal
• 17.75 gal
• 18.00 gal
• 18.25 gal
• 18.50 gal
• 18.75 gal
• 19.00 gal
• 19.25 gal
• 19.50 gal
• 19.75 gal
• 20.00 gal
• 20.25 gal
• 20.50 gal
• 20.75 gal
• 21.00 gal
• 21.25 gal
• 21.50 gal
• 21.75 gal
• 22.00 gal
• 22.25 gal
• 22.50 gal
• 22.75 gal
• 23.00 gal
• 23.25 gal
• 23.50 gal
• 23.75 gal
• 24.00 gal
• 24.25 gal
• 24.50 gal
• 24.75 gal
• 25.00 gal | 833 | 1,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-22 | latest | en | 0.267271 |
https://www.cram.com/essay/comparing-the-enthalpy-changes-of-combustion-of/PKJT7FK35C | 1,586,455,694,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371861991.79/warc/CC-MAIN-20200409154025-20200409184525-00011.warc.gz | 843,099,849 | 9,692 | # Comparing the Enthalpy Changes of Combustion of Different Alcohols
4538 Words 19 Pages
Comparing the enthalpy changes of combustion of different alcohols
Planning
What are alcohols?
==================
Alcohol is the common family name for the hydrocarbon group alkanols.
They are part of a homologous series. At least one of the hydrogen groups in the molecule is replaced by an OH group.
GRAPH
Propane
Propan-2-ol
They are all organic compounds.
The general formulas for the alcohols are:
CnH(2n+1)OH
Where n represents a number.
The first and simplest member of the alkanols family is methanol. Its molecular formula is CH3OH.
You can now see that each member of the alcohol family has a different number of carbons in its structure. They increase by one carbon atom and two
…show more content…
So as the number of carbon atoms increases the enthalpy of combustion will become more negative. I think this because as you add a H-C-H bond you increase the energy required to break the bonds by 826KJ/Mol.
This is worked out by the average bond enthalpy of the C-H bond, which is 413kJmol-1 being multiplied by two because there are two extra C-H bonds in the molecule.
Using this I will predict the enthalpy change of combustion value for methanol. To do this I will use the equation
CH3OH(l) + 1½O2 O(g) † CO2(g) + 2H2O(l)
In order to make this easier to work out and for people to understand
I will multiply everything by 2 so that I don’t have any values in the equation. This gives the new equation of;
2CH3OH(l) + 3O2 O(g) † 2CO2(g) + 4H2O(l)
On the left hand side of this equation you have
2 x (3 x C-H(413))
2 x (1 x C-O(358))
2 x (1 x O-H(464))
3 x (1 x O=O(498))
In total the total energy required to break the bonds is
(6 x 413 kJmol-1) + (2 x 358 kJmol-1) + (2 x 464 kJmol-1) + (3 x 498 kJmol-1)
=
2478 kJmol-1 + 716 kJmol-1 + 928 kJmol-1 + 1494 kJmol-1 = 5616kJmol-1
On the right hand side of the equation
2 x (2 x C=O(-805 kJmol-1))
4 x (2 x O-H(-464 kJmol-1))
In total the total energy required to form the bonds is
(4 x -805 kJmol-1) + (8 x -464 kJmol-1)
-3220 kJmol-1 + -3712
## Related Documents
• ###### Investigating the Dfference in Enthalpy of Combustion for a Number of Alcohols
Investigate the difference in enthalpy of combustion for a number of alcohols Enthalpy I am going to investigate the difference in enthalpy of combustion for a number of alcohols, the enthalpy of combustion being the 'enthalpy change when one mole of any substance is completely burnt in oxygen under the stated conditions'. I will be attempting to find how the number of carbon atoms the alcohol contains effects the enthalpy change that occurs during the combustion of the alcohol. Method I plan to measure…
Words: 3891 - Pages: 16
• ###### The Burning of Primary and Secondary Alcohols Extended Experimental Investigation
The Burning of Primary and Secondary Alcohols Year 11A Chemistry By Jarrod Ahern Abstract The aim of the investigation was to determine whether primary alcohols use less energy than secondary alcohols for fuel. The hypothesis is if primary alcohols are heated and results are taken, they produce a lesser heat of reaction then secondary alcohols. The method used was to find the average bond energies of three relating primary and secondary alcohols and compare it with the average theoretical…
Words: 2730 - Pages: 11
• ###### Chem Syllabus Essay examples
communicate scientific information and appreciate the contribution that a study of chemistry makes to their understanding of the world. The syllabus places greater emphasis on the understanding and application of chemical concepts and principles and different learning styles and needs, so that students will develop skills that will be of long term value in an increasingly technological world, rather than focusing on large quantities of factual information. In addition, it encourages the use of various…
Words: 24323 - Pages: 98 | 1,026 | 3,955 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2020-16 | latest | en | 0.866632 |
http://www.idealessaywriters.com/intermediate-algebra/ | 1,511,482,630,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807044.45/warc/CC-MAIN-20171123233821-20171124013821-00527.warc.gz | 433,271,675 | 13,782 | ## Intermediate Algebra
Assignment 5: Exercise on Algebraic Expressions
solve the questions, showing your work, and submit it to the W1: Assignment 5 Dropbox. Your facilitator will discuss questions, which seem troublesome, in this Discussion Area.
1. Simplify: 4[a – b(2 + 3a)] – 3a[2+b(4 – a)]
2. Simplify: (7x2 – 2y + z) + (3y – z) + (4x2 – 6y)
3. Simplify: (8x2 – xy + 4y2) – (12x2 – y2 + 7xy)
4. Simplify: (12x2y2 – 15xy2)/(3xy)
5. Simplify: 5x[(3x2 + x – 4) + (9x2 – 6x)/(3x)]
6. Simplify: 8 – {3x + 9y – (2x – y) + 8x + 5y}
7. If P = 2y – 6xy – 3x and Q = 4 – 5y, find the value of 2P – 3Q.
8. A woman buys three board games for \$107.10, including tax. If each board game cost \$34, what was the sales tax rate?
9. Sandy has \$3.25 in change. She has five more dimes than quarters but twice as many nickels as dimes. Express the given information as an algebraic expression and simplify. [You are not required to solve the problem.]
10. Jim’s nephew owns a triangular plot of land. The perimeter fence that surrounds the land measures 121 feet. The shortest side is 15 feet longer than one-half of the longest side. The second longest side is 4 feet shorter than the longest side. What is the length of each side? | 415 | 1,228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-47 | longest | en | 0.892123 |
https://www.physicsforums.com/threads/conservation-of-energy-please-help-me.269711/ | 1,508,581,889,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824733.32/warc/CC-MAIN-20171021095939-20171021115939-00563.warc.gz | 973,817,359 | 14,281 | # Conservation of Energy-please help me
1. Nov 5, 2008
### RAKINMAZID
Conservation of Energy--please help me!!
1. The problem statement, all variables and given/known data
A block is pushed against the spring with spring constant 5.7 kN/m (located on the left-hand side of the track) and compresses the spring a distance 4.6 cm from its equilibrium position. The block starts at rest, is accelerated by the compressed spring, and slides across a frictionless track except for a small rough area on a horizontal section of the track. It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9.8 m/s^2.
1) What is the speed v of the block when it leaves the track?
2) What is the horizontal distance x the block travels in the air?
3) What is the total speed of the block when it hits the ground?
2. Relevant equations
KE = (1/2)mv^2
Ue = (1/2)kx^2
3. The attempt at a solution
To find velocity I set KE equal to Ue (def. of conservation of energy)
(1/2)mv^2 = (1/2)kx^2
mv^2 = kx^2
v = sqrt [(kx^2)/m]
2. Nov 6, 2008
### StatusX
Re: Conservation of Energy--please help me!!
That sounds right for (1). For (2) and (3) you need the height it falls, so I hope you've been given that. And I don't know what the deal with this rough area is. Have they given you any more information about that? Because if it slows the block down by friction, conservation of energy will be useless.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 417 | 1,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-43 | longest | en | 0.903667 |
https://schoollearningcommons.info/question/90-2-prove-that-the-alternate-angles-formed-by-a-transversal-of-two-parallel-lines-are-of-equal-24885459-60/ | 1,632,354,950,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057403.84/warc/CC-MAIN-20210922223752-20210923013752-00314.warc.gz | 512,509,537 | 13,694 | ## 90° 2. Prove that, the alternate angles formed by a transversal of two parallel lines are of equal measures. Given: line l ll
Question
90°
2. Prove that, the alternate angles formed by a transversal of two parallel lines are of
equal measures.
Given: line l ll line m
line n is a transversal.
To Prove: Zd= 2b
Proof: 2d + 20 =
(1) (Angles in linear pair)
Lct
… (2) (property of interior angles of parallel
lines)
cd +
+ Zb… from (1) and (2)
d/o
16
17
= Lb
Q3. B) Solve the following questions (3 Marks each)
figure 2
1.
Prove that opposite angles formed by two intersecting lines are of equal measures.
in progress 0
2 weeks 2021-09-07T07:27:12+00:00 2 Answers 0 views 0 | 220 | 676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-39 | latest | en | 0.874755 |
https://www.jiskha.com/display.cgi?id=1331864105 | 1,501,262,803,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550975184.95/warc/CC-MAIN-20170728163715-20170728183715-00647.warc.gz | 800,917,226 | 3,830 | # College Physics
posted by .
A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 28.2 ° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 46.7 kg, and the coefficient of kinetic friction between the skis and the snow is 0.226. Find the magnitude of the force that the tow bar exerts on the skier.
• College Physics -
constant velocity= zero acceleration, so force pulling= forcefriction+forcegravitydownslope
force pulling= mg*mu*cosTheta+mgSinTheta
## Respond to this Question
First Name School Subject Your Answer
## Similar Questions
1. ### DrBob222 physics
A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 27.2 ° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 53.7 kg, …
2. ### Physics
A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 27.2 ° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 53.7 kg, …
3. ### Physics
A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 24.0° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 60.0 kg, …
4. ### AP physics
A water skier, moving at a speed of 8.61 m/s, is being pulled by a tow rope that makes an angle of 34.9 ° with respect to the velocity of the boat. The tow rope is parallel to the water. The skier is moving in the same direction as …
5. ### physics
A skier is pulled by a tow rope up a frictionless ski slope that makes an angle of 12 degrees with the horizontal. The rope moves parallel to the slope with a constant speed of 1.0 m/s. The force of the rope does 880 J of work on the …
6. ### physics
A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 25.9° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 55.3 kg, …
7. ### physics
A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 25.9° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 55.3 kg, …
8. ### physics
A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 25.9° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 55.3 kg, …
9. ### physics
A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 22.2° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 56.0 kg, …
10. ### Physics
A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 23.0° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 55.3 kg, …
More Similar Questions
Post a New Question | 848 | 3,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-30 | longest | en | 0.92581 |
https://web2.0calc.com/questions/geometry_4215 | 1,660,390,635,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571950.76/warc/CC-MAIN-20220813111851-20220813141851-00342.warc.gz | 544,716,472 | 5,508 | +0
geometry
0
192
2
AD and BE are altitudes of triangle ABC. If AC = 10, BC = 18, and AD = 5, then find BE.
Jun 16, 2021
#1
+124526
+1
A
5 10
B D C
18
Area of triangle = (1/2) ( AD) ( BC) = (1/2) ( 5) ( 18) = 45
Also ....
Area of triangle = (1/2) ( AC) (BE
45 = (1/2) (10) (BE)
90 = 10 BE
BE = 90 / 10 = 9
Jun 16, 2021
#2
0
AD and BE are altitudes of triangle ABC. If AC = 10, BC = 18, and AD = 5, then find BE.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Angle A = sin-1(5/10)
Altitude BE = sin(A) * BC
Jun 16, 2021 | 268 | 641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2022-33 | latest | en | 0.168744 |
https://www.doorsteptutor.com/Exams/NEET/Physics/Questions/Part-219.html | 1,501,178,419,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549428325.70/warc/CC-MAIN-20170727162531-20170727182531-00168.warc.gz | 773,817,677 | 13,032 | # NEET Physics: Questions 983 - 987 of 2142
Get 1 year subscription: Access detailed explanations (illustrated with images and videos) to 2142 questions. Access all new questions we will add tracking exam-pattern and syllabus changes. View Sample Explanation or View Features.
Rs. 550.00 or
## Question number: 983
MCQ▾
### Question
The refractive indices of material of a prism for blue and red colors are 1.532 and 1.514 respectively. Calculate angular dispersion produced by the prism if angle of prism is 8°.
### Choices
Choice (4) Response
a.
0.133°
b.
0.111°
c.
0.122°
d.
0.144°
## Question number: 984
MCQ▾
### Question
In an EM wave the amplitude of electric field is 10 V/m. The frequency of wave is 5 x 10 4 Hz. The wave is propagating along Z-axis. Then the average energy density of magnetic field is-
### Choices
Choice (4) Response
a.
2 x 10 -8 J/m 3
b.
2.21 x 10 -8 J/m 3
c.
2.21 x 10 -10 J/m 3
d.
2 x 10 -10 J/m 3
## Question number: 985
MCQ▾
### Question
A ray of light is incident at an angle of 60° on one face of a prism which has an angle of 30°. The ray emerging out of the prism makes an angle of 30° with the incident ray. Calculate the refractive index of the material of the prism,
### Choices
Choice (4) Response
a.
2
b.
1
c.
d.
## Question number: 986
MCQ▾
### Question
Which of the following pairs of space and time varying E and B fields would generate a plane electromagnetic wave travelling in (-Z) direction?
### Choices
Choice (4) Response
a.
E X, B Z
b.
E Y, B X
c.
E Z, B Y
d.
E X, B Y
## Question number: 987
» Ray Optics » Reflection and Refraction of Light » Spherical Surface
MCQ▾
### Question
Two mirrors are inclined at an angle of 50°. Then what is the number of images formed for an object placed in between the mirrors?
### Choices
Choice (4) Response
a.
7
b.
1
c.
3
d.
6
f Page | 557 | 1,895 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-30 | longest | en | 0.743437 |
https://www.picotech.com/support/topic27711.html?p=140233 | 1,568,559,642,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514571506.61/warc/CC-MAIN-20190915134729-20190915160729-00053.warc.gz | 982,903,836 | 10,299 | ## Integral, time, complex formula issues.
Forum for discussing PicoScope version 6 (non-automotive version)
lab!fyi
Newbie
Posts: 0
Joined: Mon Dec 12, 2016 6:58 pm
Contact:
### Integral, time, complex formula issues.
Hello!
Scope fails integral(derivative(A)) math test. Started investigating, in the process did find out:
- derivative goes off the scale at the start of capture
- derivative autoranging does not work
- tried to solve with time skew, discovered that:
-- A[0.00000005] works
-- A[50n] does not give error but also no output
- time skew cannot be used in complex formulas
Tried on 2 different scopes, official and beta software.
Attachments
Martyn
Posts: 3655
Joined: Fri Jun 10, 2011 8:15 am
Location: St. Neots
### Re: Integral, time, complex formula issues.
One thing to note is that you only have 100 true data samples across the screen, the rest of the sample points are created from interpolation.
What is the purpose of these calculations ? What are you trying to determine from the data ?
With the equations you are creating you are performing maths, on maths, on maths, so a better understanding of the required results would help.
Martyn
Technical Support Manager
lab!fyi
Newbie
Posts: 0
Joined: Mon Dec 12, 2016 6:58 pm
Contact:
### Re: Integral, time, complex formula issues.
Martyn wrote:What is the purpose of these calculations ? What are you trying to determine from the data ?
Purpose is functional testing of the scope. Reversing the operation is very good for that. Before processing real data and drawing (sometimes very important) conclusions these kinds of tests are essential. I have stuck with Pico because do not know any other scope that can support page wide formulas in real time without time-consuming use of external software.
Specific test was requested from me by scientist to see if Pico performs better than scope from other company (that he is currently using) with differentials/integrals. It would, if not described bugs.
Picture of similar bug attached. Problem is not the sample points count. Probably error lies in incorrect internal boundary condition handling at the start of the trace.
Bug 1:
integral(A[0.0001]) works (integral on time advanced channel)
integral(A[-0.0001]) does not work (integral on time delayed channel)
Bug 2:
Time skew does not work in complex formula,
will display "Invalid formula".
Example: A-B phase calculation:
acos(integral(A[0.0001]*B)/(sqrt(integral(A[0.0001]*A[0.0001]))*sqrt(integral(B*B))))/pi*180
Suggestion: Channel time skew compensation is basic feature of DSO. It must work with full complexity math. However it is cumbersome to have it only in math channels. Would be very helpful to also adjust global time skew for channel in channel menu (where probe X is set etc) and in math independently. VHF freq range is problematic with multiple channels and BNC cables w/o time skew compensation (non-equal length cables may be a problem etc).
Attachments
lab!fyi
Newbie
Posts: 0
Joined: Mon Dec 12, 2016 6:58 pm
Contact:
### Re: Integral, time, complex formula issues.
Hi again, just discovered that last stuff I posted was already gotten a ticket #TS00079098 by e-mail discussion. Had not noticed reply e-mail somehow... I was asked several questions in e-mail and just sent reply with explanations why is it so I want time-skew math feature with at least 1 picosecond resolution
Martyn
Posts: 3655
Joined: Fri Jun 10, 2011 8:15 am
Location: St. Neots
### Re: Integral, time, complex formula issues.
I have taken a look at the email, and the other two you have sent, and will follow up through the ticket system. Our offices are closed until the 3rd of January, and I don't have access to any scopes to test with.
Martyn
Technical Support Manager
lab!fyi
Newbie
Posts: 0
Joined: Mon Dec 12, 2016 6:58 pm
Contact:
### Re: Integral, time, complex formula issues.
Hi,
magenta trace - derivative(A)
lime trace - (A[0.000000002]-A)/0.000000002
Latter is 1 sample point time advanced (2ns) due to nature of function.
Cant integrate none of them. Only way to make ref wfm and integrate then.
Result perhaps sheds some light why it is not working in real time.
purple trace - integral({derivative(A)})
green trace - integral({(A[0.000000002]-A)/0.000000002})
None of them lands on where it should, purple on blue, green on (blue - 2ns).
Attachments
PS6_derivative_integral.psdata | 1,093 | 4,400 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-39 | latest | en | 0.932674 |
http://openwetware.org/wiki/User:Carl_Boettiger/Notebook/Stochastic_Population_Dynamics/2010/04/11 | 1,481,280,764,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542693.41/warc/CC-MAIN-20161202170902-00150-ip-10-31-129-80.ec2.internal.warc.gz | 206,115,385 | 5,865 | # User:Carl Boettiger/Notebook/Stochastic Population Dynamics/2010/04/11
Stochastic Population Dynamics Main project page
Previous entry Next entry
## Effective warning signals
• Predicting variance of variance by direct calculation -- still need to crunch some math for the expected convergence.
• Still, the approach should be able to do more than describe single points as unexpected deviates.
• Essentially the same as the phyolgenentic problem -- one rate vs two rates. Model selection approaches?
• So far theory is essentially built on a model selection between linear models.
• Calculate the eigenvalue directly rather than ratio of eigenvalue to noise:
• Estimate the eigenvalue from the correlation function and from power spectrum, rather than the lag-1 autocorrelation, or variance.
• Proper signal processing techniques for detecting bifurcations?
### Coding Progress
• Added a proper autocorrelation function calculation, log transform and linear regression gives the eigenvalue and the variance.
• Tested using the Langevin model
$x_{t+1} = x_t \left(1- \frac{\kappa}{\gamma} \Delta t \right) + \Delta t \sqrt{\frac{2 K_B T}{\gamma} } \xi_t$
Whose correlation function is given by
$\frac{K_BT}{\gamma} e^{-\kappa t / \gamma}$ | 288 | 1,253 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2016-50 | latest | en | 0.761252 |
https://it.mathworks.com/matlabcentral/cody/problems/624-get-the-length-of-a-given-vector/solutions/1174032 | 1,601,263,707,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401583556.73/warc/CC-MAIN-20200928010415-20200928040415-00717.warc.gz | 438,546,583 | 16,362 | Cody
Problem 624. Get the length of a given vector
Solution 1174032
Submitted on 28 Apr 2017 by Ernest Woei
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
x = [1 2 3]; y_correct = 3; assert(isequal(VectorLength(x),y_correct))
2 Pass
x = 1:10; y_correct = 10; assert(isequal(VectorLength(x),y_correct))
3 Pass
x = rand(1,928); y_correct = 928; assert(isequal(VectorLength(x),y_correct)) | 156 | 511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-40 | latest | en | 0.652752 |
https://www.ilearnengineering.com/aerospace/flight-mechanics-of-a-typical-aircraft-flight | 1,702,186,625,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101195.85/warc/CC-MAIN-20231210025335-20231210055335-00694.warc.gz | 653,138,421 | 80,375 | # Flight Mechanics of a Typical Aircraft Flight
Updated: Nov 14, 2022
This article will give a brief overview of a typical flight of an aircraft, discussing the different stages within a typical flight and link this to some of the topics in flight mechanics.
Figure 1 – Phases within a typical flight
Figure 1 has split a typical flight into various phases. The phases that w shall discuss are the Take-off, Climb, Cruise, Descent, and Landing phases.
## Take Off
Take-off is the phase of flight in which an aircraft goes through a transition from moving along the ground to flying in the air. Usually, the engines are operated at full power during take-off. Following the taxi motion, the aircraft stops at the starting line of the runway. Before take-off, the engines, particularly piston engines, are routinely run up at high power to check for engine related problems. When the pilot releases the brakes, the aircraft starts accelerating rapidly until the necessary speed for take-off is achieved. The increase in velocity dramatically increases the lift force generated from the wings.
The take-off speed required varies with air density (and hence the altitude of the runway relative to sea-level), aircraft weight, and aircraft configuration. During take-off an aircraft will typically deploy and extend the flaps by 10 to 15 degrees, increasing the wing area and camber and hence considerably increasing the lift generated. Air density is affected by factors such as elevation and air temperature, and more lift is required when density is lower at higher altitude runways for instance.
## Climb
Following take-off, the aircraft must climb to a certain altitude (typically 30,000 ft or 10 km for commercial jet airliners) before it can cruise at this altitude in a safe and economical way. A climb is carried out by increasing the lift from the wings supporting the aircraft until their lifting force exceeds the weight of the aircraft. Once this occurs, the aircraft will climb to a higher altitude until the lifting force and weight are again in balance. The increase in lift may be accomplished by increasing the angle of attack of the wings, by increasing the thrust of the engines to increase speed or by increasing the surface area or shape of the wing. In most cases, engine thrust, and angle of attack are simultaneously increased to produce a climb.
Because lift diminishes with decreasing air density, a climb, once initiated, will end by itself when the diminishing lift with increasing altitude drops to a point that equals the weight of the aircraft. At that point, the aircraft will return to level flight at a constant altitude. At this point, or just after take-off, aircraft will often begin a turn using a combination of ailerons to bank or roll and the rudder to combat adverse yaw. Aircraft may begin turning as early as possible to adhere to noise restrictions above urban areas or to avoid regions with high amounts of air traffic.
## Cruise
Cruise is the level portion of aircraft travel where flight is most fuel efficient. It occurs between ascent and descent phases and is usually most of the journey. Technically, cruising consists of heading changes only at a constant airspeed and altitude. It ends as the aircraft approaches the destination where the descent phase of flight commences in preparation for landing.
Commercial or passenger aircraft are usually designed for optimum performance at their cruise speed. There is also an optimum cruise altitude for a particular aircraft type and conditions including payload weight, centre of gravity, air temperature, humidity, and airspeed. This altitude is usually where the drag is minimum, and the lift is maximum (often referred to as the maximum lift to drag ratio). The aircraft will be in straight and level flight at this stage.
## Descent
A descent during air travel is any portion where an aircraft decreases altitude. Descents are an essential component of an approach to landing. Other partial descents might be to avoid traffic, poor flight conditions, clouds, to see something lower, to enter warmer air, or to take advantage of wind direction.
Normal descents take place at a constant airspeed and constant angle of descent. The pilot controls the angle of descent by varying engine power and pitch angle to keep the airspeed constant. At the beginning of and during the descent phase, the engine noise diminishes further as the engines are operated at low power settings. This reduces the thrust of the aircraft and decreases its speed, and therefore lift. However, towards the end of the descent phase, the passenger can feel further accelerations and an increase in the noise.
## Landing
Landing is the last part of a flight, where the aircraft returns to the ground. Aircraft usually land at an airport on a firm runway, generally constructed of asphalt concrete, concrete, gravel or grass. To land, the airspeed and the rate of descent are reduced to where the object descends at a slow enough rate to allow for a gentle touch down. Landing is accomplished by slowing down and descending to the runway. This speed reduction is achieved by reducing thrust and inducing a greater amount of drag using flaps, landing gear or speed brakes.
As the plane approaches the ground, the pilot will execute a flare (to induce a gentle landing). Although pilots are trained to perform the landing operation, many aircraft rely on so-called “Instrument Landing Systems (ILS)” in most airports to help land the aircraft safely and accurately. ILS provides precision guidance to an aircraft approaching and landing on a runway, using a combination of radio signals and, in many cases, high-intensity lighting arrays to enable a safe landing during instrument meteorological conditions (IMC), such as low ceilings or reduced visibility due to fog, rain, or blowing snow.
## Accredited Aerospace Engineering Courses
This is a small excerpt from iLearn Engineering®‘s suite of accredited Aerospace Engineering courses all of which are available for enrolment 365 days a year.
Why not check out the online engineering short courses specifically in aerospace engineering:
Diploma in Aircraft Design
Diploma in Aerospace Structures
Diploma in Principles of Flight
Diploma in Aerodynamics, Propulsion and Space
Alternatively, you can view all our online engineering courses here.
## Recent Posts
### Best practice standards used in the construction industry
Best practice standards used in the construction industry. If you’ve read our article on the civil engineering code of conduct, you’ll be familiar with some best practices in construction projects. In this article we’re going to focus on the best practice standards used in the construction industry. What are best practices? In civil engineering and […]
### Civil Engineering Code of Conduct
Civil Engineering Code of Conduct If you’ve read our article on sustainable design in a construction project you’ll already be familiar with some ethics around construction. However, there’s a lot more to it than that, and we’ll find out some of the considerations an engineer must think about in this article. Business ethics for engineers […] | 1,395 | 7,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-50 | longest | en | 0.951781 |
https://codebreaker.xyz/problem/decayedbridges | 1,723,624,005,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641104812.87/warc/CC-MAIN-20240814061604-20240814091604-00390.warc.gz | 134,634,146 | 4,363 | Problem Statement
There are N islands and M bridges.
The i-th bridge connects the Ai-th and Bi-th islands bidirectionally.
Initially, we can travel between any two islands using some of these bridges.
However, the results of a survey show that these bridges will all collapse because of aging, in the order from the first bridge to the M-th bridge.
Let the inconvenience be the number of pairs of islands (a, b) (a < b) such that we are no longer able to travel between the a-th and b-th islands using some of the bridges remaining.
For each i (1 ≤ i ≤ M), find the inconvenience just after the i-th bridge collapses.
Constraints
• All values in input are integers.
• 2 ≤ N ≤ 105
• 1 ≤ M ≤ 105
• 1 ≤ Ai < Bi ≤ N
• All pairs (Ai, Bi) are distinct.
• The inconvenience is initially 0.
Input
Input is given from Standard Input in the following format:
```N M
A1 B1
A2 B2
:
AM BM
```
Output
In the order i = 1, 2, ..., M, print the inconvenience just after the i-th bridge collapses. Note that the answer may not fit into a 32-bit integer type.
```4 5
1 2
3 4
1 3
2 3
1 4
```
Sample Output 1
```0
0
4
5
6
```
For example, when the first to third bridges have collapsed, the inconvenience is 4 since we can no longer travel between the pairs (1, 2), (1, 3), (2, 4) and (3, 4).
```6 5
2 3
1 2
5 6
3 4
4 5
```
```8
9
12
14
15
```
```2 1
1 2
```
```1
```
Submitting .cpp to 'decayedbridges'
Time Limit: 1 Seconds
Memory Limit: 1024MB
Source: ABC 120D | 454 | 1,465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-33 | latest | en | 0.860146 |
https://www.sefsd.org/builds-reviews/scaling-and-comparing-performances-of-aircraft-models-2d-3d-wing-loading/ | 1,675,652,525,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500303.56/warc/CC-MAIN-20230206015710-20230206045710-00708.warc.gz | 999,937,442 | 22,682 | [This article was borrowed from the April issue of the Ampeer. Thanks to their Editor: Ken Myers]
By Andrej Marinsek
1. Introduction
Many years ago (Model Airplane News, Dec. 1997) an article was published in this magazine titled “3D Wing Loadings” (Three dimensional wing loadings) by Larry Renger; it was recently published again on the internet in a slightly cleaned up version. Its different approach to a specific modeling subject is interesting but, as it will be shown later, has some problems. The concept of the 3DWL, though correct in one respect, has otherwise rather limited reach and leads to some vague interpretations and questionable conclusions. The 3DWL persists around in different forms and publications and seems to be, nowadays, the most advertised and supposedly even the only appropriate approach for estimation and comparison of some model performances. This is somehow surprising, so it needs to be addressed in some way.
2. General remarks
Coherent units from the International System of Units (SI) are used in calculations as they are clearer. In most cases only one unit is attributed to a certain physical property and numerical transformations are simpler or not needed at all.
Instead of the term weight (W), which is strictly speaking, a kind of force, the expression mass is used (designated by the letter m), which is the proper name for the physical property measured in kg (lb., oz., etc), and is employed in all calculations here.
3. Agility of models
The motion of models in the air can be on one side described by the words like “agile” or “hot” or “docile” or “flyable” or whatever expression is used to appreciate the performance of models in flight. However this can be pretty undetermined and subjective.
On the other hand, some objective (given by numbers) performance parameters exist. With regard to the lateral axis of models, some performances directly depend on the lift force. These are the minimal speed in horizontal flight vm (stall speed), the minimal absolute turning (or circling) radius Rm and the minimal relative turning radius (RTm), which will be defined and discussed a bit later.
Also, some settings (such as the center of gravity) and a number of model properties, for instance wing profile, low/high wing, aspect ratio, tail (distance from the wing, area, position), the size of rudders, propulsion, thrust vectoring, etc. considerably affect certain performance parameters. | 518 | 2,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-06 | longest | en | 0.944395 |
https://nl.mathworks.com/matlabcentral/cody/problems/350-back-to-basics-7-equal-nans/solutions/289205 | 1,606,630,627,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141196324.38/warc/CC-MAIN-20201129034021-20201129064021-00250.warc.gz | 418,115,854 | 16,952 | Cody
# Problem 350. Back to basics 7 - Equal NaNs
Solution 289205
Submitted on 25 Jul 2013 by MKN
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = [1:10]; y = x; z = true; assert(isequal(are_equal(x,y),z));
2 Pass
%% x = [1:10 NaN]; y = x; z = true; assert(isequal(are_equal(x,y),z));
3 Pass
%% x = [2 4 6 NaN 8]; y = [2 4 NaN 6 8]; z = false; assert(isequal(are_equal(x,y),z));
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 203 | 641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-50 | latest | en | 0.655317 |
https://iitutor.com/courses/qce-mathematical-methods-algebra-statistics-and-functions/ | 1,709,067,122,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474686.54/warc/CC-MAIN-20240227184934-20240227214934-00450.warc.gz | 313,736,707 | 51,487 | # QCE Mathematical Methods – Algebra, Statistics and Functions
1.1 Arithmetic Sequences and Series
1.2 Geometric Sequences and Series
1.3 Functions
1.6 Polynomials
1.7 Set Theory
1.8 Probability
1.9 Binomial Theorem
1.10 Exponents
In this course, students will develop the mathematical understanding and skills to solve problems relating to:
• Arithmetic and Geometric Sequences and Series 1
• Functions and Graphs
• Counting and Probability
• Exponential Functions 1
• Arithmetic and Geometric Sequences and Series 2
Arithmetic and geometric sequences are introduced, and their applications are studied. Simple relationships between variable quantities are reviewed, which are used to introduce the key concepts of a function and its graph. Quadratic functions and index rules are revised. The study of inferential statistics begins in this unit with a review of the fundamentals of probability and the introduction of conditional probability and independence concepts. The algebraic expansion of powers of a binomial is found using the binomial theorem.
Source – QCAA General Senior Syllabus 2025
QCE Mathematical Methods Courses
QCE Mathematical Methods Syllabus
## Course Content
### Arithmetic Sequences and Series
Lesson Content
0% Complete 0/15 Steps
Lesson Content
0% Complete 0/10 Steps
Lesson Content
0% Complete 0/6 Steps
Lesson Content
0% Complete 0/9 Steps
Lesson Content
0% Complete 0/9 Steps
Geometric Sequences and Series
Lesson Content
0% Complete 0/10 Steps
Lesson Content
0% Complete 0/11 Steps
Lesson Content
0% Complete 0/6 Steps
Lesson Content
0% Complete 0/5 Steps
Lesson Content
0% Complete 0/8 Steps
Lesson Content
0% Complete 0/6 Steps
Lesson Content
0% Complete 0/1 Steps
Lesson Content
0% Complete 0/7 Steps
Lesson Content
0% Complete 0/8 Steps
Lesson Content
0% Complete 0/10 Steps
Functions
Lesson Content
0% Complete 0/8 Steps
Lesson Content
0% Complete 0/3 Steps
Lesson Content
0% Complete 0/5 Steps
Lesson Content
0% Complete 0/8 Steps
Lesson Content
0% Complete 0/3 Steps
Lesson Content
0% Complete 0/4 Steps
Lesson Content
0% Complete 0/4 Steps
Lesson Content
0% Complete 0/2 Steps
Lesson Content
0% Complete 0/4 Steps
Lesson Content
0% Complete 0/6 Steps
Lesson Content
0% Complete 0/4 Steps
Lesson Content
0% Complete 0/4 Steps
Lesson Content
0% Complete 0/7 Steps
Lesson Content
0% Complete 0/6 Steps
Lesson Content
0% Complete 0/2 Steps
Lesson Content
0% Complete 0/3 Steps
Lesson Content
0% Complete 0/6 Steps
Lesson Content
0% Complete 0/4 Steps
Lesson Content
0% Complete 0/4 Steps
Lesson Content
0% Complete 0/4 Steps
Lesson Content
0% Complete 0/5 Steps
Lesson Content
0% Complete 0/6 Steps
Lesson Content
0% Complete 0/4 Steps
Lesson Content
0% Complete 0/8 Steps
Lesson Content
0% Complete 0/9 Steps
Lesson Content
0% Complete 0/4 Steps
Lesson Content
0% Complete 0/2 Steps
Lesson Content
0% Complete 0/6 Steps
Lesson Content
0% Complete 0/2 Steps
Lesson Content
0% Complete 0/6 Steps
Lesson Content
0% Complete 0/15 Steps
Lesson Content
0% Complete 0/9 Steps
Lesson Content
0% Complete 0/10 Steps
Lesson Content
0% Complete 0/10 Steps
Lesson Content
0% Complete 0/3 Steps
Lesson Content
0% Complete 0/13 Steps
Lesson Content
0% Complete 0/12 Steps
Lesson Content
0% Complete 0/8 Steps
Lesson Content
0% Complete 0/7 Steps
Lesson Content
0% Complete 0/7 Steps
Lesson Content
0% Complete 0/9 Steps
Lesson Content
0% Complete 0/10 Steps
Polynomials
Lesson Content
0% Complete 0/5 Steps
Lesson Content
0% Complete 0/5 Steps
Lesson Content
0% Complete 0/3 Steps
Lesson Content
0% Complete 0/10 Steps
Lesson Content
0% Complete 0/2 Steps
Lesson Content
0% Complete 0/4 Steps
Lesson Content
0% Complete 0/7 Steps
Lesson Content
0% Complete 0/13 Steps
Lesson Content
0% Complete 0/18 Steps
Lesson Content
0% Complete 0/9 Steps
Lesson Content
0% Complete 0/12 Steps
Set Theory
Lesson Content
0% Complete 0/4 Steps
Lesson Content
0% Complete 0/4 Steps
Lesson Content
0% Complete 0/8 Steps
Lesson Content
0% Complete 0/3 Steps
Lesson Content
0% Complete 0/4 Steps
Lesson Content
0% Complete 0/1 Steps
Lesson Content
0% Complete 0/2 Steps
Lesson Content
0% Complete 0/6 Steps
Lesson Content
0% Complete 0/6 Steps
Probability
Lesson Content
0% Complete 0/7 Steps
Lesson Content
0% Complete 0/4 Steps
Lesson Content
0% Complete 0/11 Steps
Lesson Content
0% Complete 0/2 Steps
Lesson Content
0% Complete 0/6 Steps
Lesson Content
0% Complete 0/4 Steps
Lesson Content
0% Complete 0/8 Steps
Lesson Content
0% Complete 0/13 Steps
Lesson Content
0% Complete 0/7 Steps
Lesson Content
0% Complete 0/6 Steps
Lesson Content
0% Complete 0/13 Steps
Binomial Theorem
Lesson Content
0% Complete 0/2 Steps
Lesson Content
0% Complete 0/10 Steps
Lesson Content
0% Complete 0/7 Steps
Exponents
Lesson Content
0% Complete 0/6 Steps
Lesson Content
0% Complete 0/9 Steps
Lesson Content
0% Complete 0/10 Steps
Lesson Content
0% Complete 0/5 Steps
Lesson Content
0% Complete 0/9 Steps
Lesson Content
0% Complete 0/12 Steps
Lesson Content
0% Complete 0/8 Steps
Lesson Content
0% Complete 0/3 Steps
Not Enrolled
• 99 Lessons
• 665 Topics
• 160 Quizzes | 1,491 | 5,070 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-10 | latest | en | 0.780093 |
https://www.statsmedic.com/apstats-day1 | 1,591,189,321,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347434137.87/warc/CC-MAIN-20200603112831-20200603142831-00456.warc.gz | 935,155,023 | 105,789 | ## Day 1
Here we gooooooo! (Does anyone else hearplaying in the background when they read that? No, just me? Cool.) Anyways, here we go! Day 1 of AP Stats. We’ve got 150 days to get these students ready for the AP test. This year we’re doing things a little different than we have in the past. This revolves around our focus on having students “Experience First, Formalize Later.” Our goal for the year is that for every lesson students will begin experiencing statistics through an activity that will require little to no teacher input. After working through this, we will go back through and formalize their learning with vocabulary and formulas, etc. We did this two years ago when we piloted an It went so well we felt we had to do it for AP Stats also. We know this is a huge undertaking, but we will be posting the daily lessons in real time. Follow along with us and feel free to use and adapt all of our activities. Knowing that we are helping AP Stats teachers and students is what keeps us motivated!
##### Activity:
This activity comes to us from Doug Tyson. We attended his session at the last year and were blown away. We immediately did the activity with students in Intro Stats when we got back and the engagement and understanding of students was incredible. Check out his site for the full version of the activity, as well as a video explaining the lesson more thoroughly.
Today is the first day of school. While in the past we spent this day doing introductions and such, this year we really want to get students thinking about statistics right away. We want to hook students, to show them the relevance and importance of understanding statistics. For some students, the thought of AP Stats has been stressing them out all summer. We’re going to show them today that they CAN do statistics through this activity’s accessibility.
You will need to make a set of 12 'T-shirts' for each group of 4. We used card stock so that you couldn't see through them.
Begin by giving students some information about Parkinson's disease so everyone is on the same page. Then, show the video below:
After the video, give a brief recap and ask the students to discuss whether or not they think Joy can smell Parkinson’s disease and to complete the front page of the activity. Walk around the room and ask groups what they think. Can Joy really smell Parkinson’s disease? If they say no, follow up by asking what it would take to convince them. If they say yes, ask why. Is it possible she’s just a really good guesser? If she got 10 out of 12 correct would you still believe her?
Share out answers and discuss with the class that in statistics we don’t always know for sure if a claim is true. It is possible that Joy is just a good guesser. Even if she got 12 out of 12 correct, it would still be possible that she had just guessed. But in statistics we figure out the likelihood of something happening purely by chance and if it is unlikely to occur by chance have strong evidence to make a conclusion. This is really the main goal for this course.
That being said, we want to investigate, what is the likelihood of Joy guessing 11 or more correct if she really can’t smell Parkinson’s disease (just by guessing)? Pick a student from the class to smell a shirt and guess whether it has been worn by someone with Parkinson’s or not. Show the student the shirt and have them guess “yes” or “no”. Reveal the answer. Try a couple more.
Have students complete the simulation in their groups. When they’re done they should add a sticker to the dotplot at the front of the room for the number of correct guesses.
You can also show students how to simulate using technology. The variable name is “Guesses”, the categories are “Correct” and “Incorrect”. Start with Joy’s 11 correct and 1 incorrect. Scroll to the bottom pull down menu and choose “Simulate sample count”. Set the hypothesized proportion at 0.5 and start taking samples. The applet will also allow you to find the percent of values greater than or equal to 11 (the P-value!).
After students have completed the rest of the activity, you should still have some students who think Joy cannot smell the disease. Show the second video, pause after 7 seconds, then mic drop. Your students are now hooked on statistics. | 930 | 4,285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2020-24 | latest | en | 0.967782 |
https://www.nag.com/numeric/cl/nagdoc_cl26/html/g01/g01fcc.html | 1,627,301,789,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152129.33/warc/CC-MAIN-20210726120442-20210726150442-00652.warc.gz | 912,966,689 | 4,294 | # NAG Library Function Document
## 1Purpose
nag_deviates_chi_sq (g01fcc) returns the deviate associated with the given lower tail probability of the ${\chi }^{2}$-distribution with real degrees of freedom.
## 2Specification
#include #include
double nag_deviates_chi_sq (double p, double df, NagError *fail)
## 3Description
The deviate, ${x}_{p}$, associated with the lower tail probability $p$ of the ${\chi }^{2}$-distribution with $\nu$ degrees of freedom is defined as the solution to
$PX≤xp:ν=p=12ν/2Γν/2 ∫0xpe-X/2Xv/2-1dX, 0≤xp<∞;ν>0.$
The required ${x}_{p}$ is found by using the relationship between a ${\chi }^{2}$-distribution and a gamma distribution, i.e., a ${\chi }^{2}$-distribution with $\nu$ degrees of freedom is equal to a gamma distribution with scale parameter $2$ and shape parameter $\nu /2$.
For very large values of $\nu$, greater than ${10}^{5}$, Wilson and Hilferty's normal approximation to the ${\chi }^{2}$ is used; see Kendall and Stuart (1969).
## 4References
Best D J and Roberts D E (1975) Algorithm AS 91. The percentage points of the ${\chi }^{2}$ distribution Appl. Statist. 24 385–388
Hastings N A J and Peacock J B (1975) Statistical Distributions Butterworth
Kendall M G and Stuart A (1969) The Advanced Theory of Statistics (Volume 1) (3rd Edition) Griffin
## 5Arguments
1: $\mathbf{p}$doubleInput
On entry: $p$, the lower tail probability from the required ${\chi }^{2}$-distribution.
Constraint: $0.0\le {\mathbf{p}}<1.0$.
2: $\mathbf{df}$doubleInput
On entry: $\nu$, the degrees of freedom of the ${\chi }^{2}$-distribution.
Constraint: ${\mathbf{df}}>0.0$.
3: $\mathbf{fail}$NagError *Input/Output
The NAG error argument (see Section 3.7 in How to Use the NAG Library and its Documentation).
## 6Error Indicators and Warnings
On any of the error conditions listed below except ${\mathbf{fail}}\mathbf{.}\mathbf{code}=$ NE_ALG_NOT_CONV nag_deviates_chi_sq (g01fcc) returns $0.0$.
NE_ALG_NOT_CONV
The algorithm has failed to converge in $〈\mathit{\text{value}}〉$ iterations. The result should be a reasonable approximation.
NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 2.3.1.2 in How to Use the NAG Library and its Documentation for further information.
NE_GAM_NOT_CONV
The series used to calculate the gamma function has failed to converge. This is an unlikely error exit.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
See Section 2.7.6 in How to Use the NAG Library and its Documentation for further information.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 2.7.5 in How to Use the NAG Library and its Documentation for further information.
NE_PROBAB_CLOSE_TO_TAIL
The probability is too close to $0.0$ or $1.0$.
NE_REAL_ARG_GE
On entry, ${\mathbf{p}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{p}}<1.0$.
NE_REAL_ARG_LE
On entry, ${\mathbf{df}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{df}}>0.0$.
NE_REAL_ARG_LT
On entry, ${\mathbf{p}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{p}}\ge 0.0$.
## 7Accuracy
The results should be accurate to five significant digits for most argument values. Some accuracy is lost for $p$ close to $0.0$.
## 8Parallelism and Performance
nag_deviates_chi_sq (g01fcc) is not threaded in any implementation.
For higher accuracy the relationship described in Section 3 may be used and a direct call to nag_deviates_gamma_dist (g01ffc) made.
## 10Example
This example reads lower tail probabilities for several ${\chi }^{2}$-distributions, and calculates and prints the corresponding deviates until the end of data is reached.
### 10.1Program Text
Program Text (g01fcce.c)
### 10.2Program Data
Program Data (g01fcce.d)
### 10.3Program Results
Program Results (g01fcce.r)
© The Numerical Algorithms Group Ltd, Oxford, UK. 2017 | 1,185 | 3,950 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 39, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-31 | latest | en | 0.669551 |
https://www.scribd.com/document/26497999/Section-1-2 | 1,571,218,088,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986666959.47/warc/CC-MAIN-20191016090425-20191016113925-00290.warc.gz | 1,093,852,010 | 72,223 | You are on page 1of 25
# Section 1.
## 2 29 Conditional and Biconditional
Section 1.2:
1.2: Conditional and Biconditional Connectives
## Purpose of Lesson: To introduce the conditional and biconditional connectives
and various equivalent forms.
The Conditional
Conditional Sentence
The most important way to combine sentences in mathematics is the
implication which has the form “if P then Q.” In a
conditional sentence (or implication),
purely logical sense, conditional sentences do not necessarily imply a cause
and effect between the components P and Q, although in mathematics and in
general discourse they do. From a logical point of view the sentence
## is a legitimate implication, although there is no relationship between the
component parts. On the other hand, when we write
## If N is an integer, then 2N is an even integer
there is a definite cause and effect between the components. The reader has
seen conditional sentences in Euclidean geometry where much of the subject
is explained through implications. The sentence “If a polygon has three sides,
then it is a triangle,” is a conditional sentence.
## Conditional Sentence : If P and Q are sentences, then the conditional sentence
“if P then Q” is denoted symbolically by
P⇒Q
and the truth values of the sentence are defined by the truth table:
P Q P⇒Q
T T T ⇒ is false if
T F F T ⇒ F and
F T T otherwise true
F F T
## The sentence P is called the assumption
assumption (or premise or antecedent
antecedent)
ntecedent of the
1
implication and Q is called the conclusion (or consequent ).
1
In pure logical systems P and Q are generally called the antecedent and consequent. In mathematics they
are more likely to be called the assumption and conclusion.
Section 1.2 30 Conditional and Biconditional
## The conditional statement P ⇒ Q can be visualized by the Euler (or Venn)
Venn
diagram as shown in Figure 1.
Figure 1
## i) If f is a real-valued differentiable function on ( −∞, ∞ ) , then f is continuous on
( −∞, ∞ ) .
ii) If N is an even number greater than 2, then N is the sum of two primes. (This
is true and you get an A for the course if you can prove this. Just slide your
solution under my door.)
iii) If a and b are the lengths of the legs of a right triangle, and c is the length of
the hypotenuse, then c 2 = a 2 + b 2 .
## Understanding the Conditional Sentence: The conditional sentence “if P then
Q“ is best understood as a promise, where if the promise is kept, the
Section 1.2 31 Conditional and Biconditional
## conditional sentence is true, otherwise the sentence is false. As an illustration
suppose your professor makes you the promise:
## If pigs fly, then you will receive an A for the course.
you an A, your professor has kept his or her promise and the conditional
sentence “if …. then” is true. But, suppose pigs fly but your professor
reneges and you do not get an A. Then your professor has broken the bond
and the sentence “if … then” is false.
Now (here is where it gets confusing) suppose pigs don’t fly, then what
should your professor do? In this case the professor can do anything he or
she so desires and the promise is still kept, the argument being that the
sentence “if … then” is true since the professor only promised an A if in fact
pigs fly2. This line of reasoning jives with the truth table for the conditional
sentence.
## The conditional sentence P ⇒ Q is sometimes called an inference,
inference and we say
that P implies Q. Another way of stating P ⇒ Q in English is to say that P is
a sufficient condition for Q, which means the truth of P guarantees the truth of
Q. We can also say Q is a necessary condition for P, meaning that Q
necessarily follows from P.
## Example 2: Necessary Conditions and Sufficient Conditions.
P Q Condition
Q is necessary for
being pregnant being female
P
2N is an
N is an integer P is sufficient for Q
integer
Q is necessary for
life on earth air
P
Getting run over by a steam
squashed P is sufficient for Q
roller
Necessary Conditions and Sufficient Conditions
Table 1
## The implication P ⇒ Q gives rise to three related implications
2
Some people might argue that if pigs don’t fly and the professor gives the student an A, then the
sentence “if … then” should be considered false.
Section 1.2 32 Conditional and Biconditional
## Implication Converse Inverse Contrapositive
P⇒Q Q⇒P ∼ P ⇒∼ Q ∼ Q ⇒∼ P
## It is not difficult to show by truth tables that
converse: P⇒Q ≡ Q⇒ P
inverse: P ⇒ Q ≡ ∼ P ⇒∼ Q
contrapositive: P ⇒ Q ≡ ∼ Q ⇒ ∼ P
## Martin Note: Ad agencies often use the technique of equating an implication
with its (non equivalent) converse for the purpose of selling products. For
example, the slogan “Buy U.S. Olympic accredited sports gear, the designated
gear of the U.S. Olympic team, is an attempt to get you to buy into the
converse argument that if you wear accredited Olympic gear, then you have
some status as an Olympic athlete. You can think of a thousand other
## Martin Note: In mathematics when one writes the implication P ⇒ Q one
generally assumes that the sentence is true since we have assumed a true
assumption P and have proven that the conclusion Q is true. In formal logic
however, one allows for the possibility that P may be either true or false.
Martin Note: True can’t imply false, but false can imply anything.
## which is equivalent to proving that
( P ⇒ Q ) ∧ ( Q ⇒ R ) ⇒ P ⇒ R
is a tautology. We can verify this by noting all the T’s in column (5) of the truth table:
(1) ( 2) ( 3) ( 4) (5)
P Q R P⇒Q Q⇒R P⇒ R ( P ⇒ Q ) ∧ ( Q ⇒ R ) ⇒ P ⇒ R
( P ⇒ Q ) ∧ (Q ⇒ R )
Section 1.2 33 Conditional and Biconditional
T T T T T T T T
T T F T F F F T
T F T F T T F T
T F F F T F F T
F T T T T T T T
F T F T F T F T
F F T T T T T T
F F F T T T T T
Table 2
Biconditional
Theorems of the form “P if and only if Q” are highly valued in
mathematics, giving equivalent and interesting new ways to say exactly the same
thing. For example, here is biconditional from number theory you may not be aware
of: A positive integer n is divisible by 3 if and only if the sum of the digits of n is
divisible by 3.
## Definition: If P and Q are sentences, then the biconditional sentence “ P if and
only if Q” is denoted by
P⇔Q
whose truth values are given by the truth table
P Q P⇔Q
⇔ is true if
T T T P and Q are
T F F
the same, otherwise
F T F
false
F F T
## P ⇔ Q is often read as " P if and only if Q " or P iff Q for shorthand.
Another phrasing of P ⇔ Q is P is a necessary and sufficient condition for Q.
Note: Possibly the most famous conditional statement of all time is due
to the French philosopher/mathematician Rene Decartes (1591-1650)
with his famous quote in Latin “Cogito ergo sum”, which means “I think
therefore I am”, which in more traditional conditional, but less elegant
form, would be “If I think then I am.”
Section 1.2 34 Conditional and Biconditional
## Example 3: Biconditional Equivalent to two Implications
Show that the biconditional P ⇔ Q is equivalent to ( P ⇒ Q ) ∧ ( Q ⇒ P ) .
Solution The truth values in the truth table under ( 5 ) and ( 6 ) are the same
as seen in Table 3.
(1) ( 2) ( 3) ( 4)
P Q P⇒Q Q⇒P ( P ⇒ Q ) ∧ (Q ⇒ P ) P⇔Q
T T T T T T
T F F T F F
F T T F F F
F F T T T T
Equivalence of ( P ⇒ Q ) ∧ (Q ⇐ P ) ≡ P ⇔ Q
Table 3
## Warning Be careful not to confuse the biconditional connective P ⇔ Q with
Warning:
P ≡ Q which says P and Q are logical equivalent compound sentences. The
biconditional P ⇔ Q does not necessarily mean P and Q have the same truth
values (since the biconditional can be false), whereas P ≡ Q does.
Margin Note:
Note British mathematicians George Boole (1815-1664) and Augustus
DeMorgan (1806-1871) started a renaissance of logic in the mid-1800s. De
Morgan’s major contributions to logic include De Morgan’s laws. George
Boole approached logic in a new way, reducing it to simple algebra, replacing
the truth value T by 1 and the truth value F by 0. Logical “and” becomes
multiplication and logical “or” becomes addition in this new system. The
result was Boolean algebra, the mathematical foundation for much of digital
computers.
## Biconditional Truth Value
1 + 3 = 5 if and only if 3 + 1 = 5 True
The moon is round iff the earth is round True
d t
e = tet
1 + 2 = 3 if and only if False
dt
A necessary and sufficient condition for 1 = 0 is 6 / 3 = 2 False
## Equivalent Forms of the Biconditional
Section 1.2 35 Conditional and Biconditional
## Equivalent Biconditional Forms
Forms Example
P⇔Q x − 1 = 0 iff x = 1
P if and only if Q x − 1 = 0 if and only if x = 1
P iff Q x − 1 = 0 iff x = 1
If P then Q and conversely if x − 1 = 0 then x = 1 and conversely
If Q then P and conversely If x = 1 then x − 1 = 0 and conversely
P is a necessary and sufficient condition for Q x − 1 = 0 is a necessary and sufficient condition for x = 1
Q is a necessary and sufficient condition for P x = 1 is a necessary and sufficient condition for x − 1 = 0
## Although the two conditionals P ⇒ Q and Q ⇒ P we do have the following
equivalent conditionals
P⇒Q ≡ ∼Q⇒ ∼ P
which can be verified by constructing a truth table for each conditional. An example
of this equivalence would be
If it rains then the grass gets wet is equivalent to If grass does not get wet then it does
not rain.
Note: Greek philosophers called Modus Ponens the valid argument that if P is
true and if P ⇒ Q is true, then Q is true, which in sentential logic notation is
P ∧ ( P ⇒ Q ) ⇒ Q .
## Biconditional sentences allow one to replace one mathematical fact with
another. For example, we solve the first-order linear differential equation by
using a series of “if and only if” statements. We apologize to those students
who have yet to take calculus, but nevertheless the demonstration shows how
problems can be solved by a sequence of “if and only if” statements.
Section 1.2 36 Conditional and Biconditional
dy dy
+ ay = 0 ⇔ e at + ay = 0
dt dt
d ( e at y (t ) )
⇔ =0
dt
at
⇔ e y (t ) = c
⇔ y (t ) = c e− at
Note: The origin of the “iff” notation first appeared in print in 1955 in the text
General Topology by John Kelly although its invention is generally credited to
the Hungarian/American Paul Halmos.
Section 1.2 37 Conditional and Biconditional
## Working Definitions: The following definitions are needed in some problems
in this and if following sections.
▪ An integer n divides an integer m (and we write n m ) if there exists
an integer q such that m = n × q .
## 1. Identify the assumption and conclusion in the following conditional
sentences and tell if the implication is true or false.
## a) If pigs fly then I am richer than Bill Gates.
Ans: Assumption: pigs fly, Conclusion: I am richer than Bill Gates, True
b) If a person got the plague in the 17th century they were in trouble.
## Ans: Assumption: a person got the plague in the 17th century
Conclusion: they were in trouble, True
c) If you miss class over 75% of the time you are in trouble.
Ans: Assumption: you miss class over 75% of the time, Conclusion: you are in
trouble, True
## Ans: Assumption: If the determinant of a matrix is nonzero, Conclusion: the
matrix has an inverse, True
Section 1.2 38 Conditional and Biconditional
Ans:
## a) If I am not richer than Bill Gates, then pigs do not fly.
b) If people in the 17th century were not in trouble, then they never got the
plague.
c) If you are not in trouble, then you did not miss class 75% of the time.
## e) If x + y is not prime, then either x is not prime or y is not prime.
f) If a matrix does not have an inverse, then the determinant of the matrix is
zero.
## g) If f does not have an inverse, then f is not a 1-1 function.
3. Let P be the sentence "4 > 6" , Q the sentence "1 + 1 = 2" , and R the
sentence "1 + 1 = 3" . What is the truth value of the following sentences?
a) P∧∼ Q Ans: F
b) ∼ ( P ∧ Q) Ans: T
c) ∼ ( P ∨ Q) Ans: F
d) ∼ P ∧ ∼ Q Ans: T
e) P ∧ Q Ans: F
f) P ⇒ Q Ans: T
g) Q ⇔ R Ans: F
h) P ⇒ (Q ⇒ R ) Ans: T
i) ( P ⇒ Q) ⇒ R Ans: F
j) ( R ∨ Q ∨ R) ⇔ ( P ∧ Q ∧ R) Ans: F
4. Let P be the sentence “Jerry is richer than Mary”, Q is the sentence “Jerry
is taller than Mary”, and R is the sentence “Mary is taller than Jerry.” For the
Section 1.2 39 Conditional and Biconditional
following sentences what can you conclude about Jerry and Mary if the
sentences are true. Express the information in a convenient form.
a) P∨Q
b) P ∧Q
c) ∼ P∨Q
d) Q∧R
e) ∼ Q∧ ∼ R
f) P ∧ ( P ⇒ Q)
## Ans: Both P and Q are true.
g) P ⇔ (Q ∨ R )
Ans: Either all P,Q,R are true or P is false and one (or both) Q and R are
false.
h) Q ∧ (P ⇒ R)
Ans: Either all P,Q,R are true or Q is false and P is true and R false.
i) P∨Q∨ R
## Ans: Either one of P,Q,R are true.
j) P ∨ ( Q ∧ R )
Section 1.2 40 Conditional and Biconditional
## Ans: Either P is true or both Q and R are true.
5. Construct truth tables to show the following sentences mean the same
thing.
## a) P iff Q means the same as ∼ P iff ∼ Q
Ans:
P Q P ⇔Q ∼P ∼Q ∼ P ⇔∼ Q
T T T F F T
T F F F T F
F T F T F F
F F T T T T
## b) ∼ ( P ⇔ Q ) means the same as ( P ∧ ∼ Q ) ∨ ( ∼ P ∧ Q )
Ans:
P Q P ⇔Q ∼ ( P ⇔ Q) ∼P ∼Q P∧ ∼ Q ∼ P ∧Q ( P∧ ∼ Q ) ∨ ( ∼ P ∧ Q )
T T T F F F F F F
T F F T F T T F T
F T F T T F F T T
F F T F T T F F F
## c) P ⇒ Q means the same as ∼ P ∨ Q
Ans:
P Q P ⇒Q ∼P ∼ P∨Q
T T T F T
T F F F F
F T T T T
F F T T T
## 6. Translate the given sentences in English to conditional form.
Section 1.2 41 Conditional and Biconditional
## a) Unless you study you won’t get a good grade.
Ans: If you do not study, then you will not get a good grade, or its
Ans:
contrapositive, which states: if you did get a good grade then you
studied.
## d) Get out or I’ll call the cops.
Ans: If you do not get out, then I will call the cops. (Normally one drops the
“then” in normal conversation.)
## f) Criticize her and she will slap you.
Ans: If you criticize her, (then) she will slap you. (Often people drop the
“then.”)
## 7. (In Plain English)
English Without making a truth table, say why the following are
true.
a) ( P ∨ Q ) ∧ ∼ P ⇒ Q
Ans:
Ans If we know that P or Q but that P is not true, then we have no choice
but to assume Q is true.
b) P ∧ Q ∧ ∼ Q ⇒ ∼ P
( )
Section 1.2 42 Conditional and Biconditional
Ans:
Ans If by assuming P and from this you deduce a contradiction, then you
must assume your assumption P is false.
c) ( P ∨ Q) ⇒ (∼ P ⇒ Q)
Ans: P ∨ Q says that P or Q is true. Hence, if P is not true then Q must be
true.
8. (Distributive Laws for AND and OR) For the sentences P, Q and R verify
the distributive laws
a) P ∧ (Q ∨ R) ≡ ( P ∧ Q) ∨ ( P ∧ R)
Ans: Note that the columns ( 2 ) and ( 5 ) are the same.
(1) ( 2) ( 3) ( 4) (5)
P Q R Q∨ R P ∧ (Q ∨ R) P ∧Q P∧R ( P ∧ Q) ∨ ( P ∧ R)
T T T T T T T T
T T F T T T F T
T F T T T F T T
T F F F F F F F
F T T T F F F F
F T F T F F F F
F F T T F F F F
F F F F F F F F
b) P ∨ (Q ∧ R) ≡ ( P ∨ Q) ∧ ( P ∨ R)
Ans: Note that the columns ( 2 ) and ( 5 ) are the same.
(1) ( 2) ( 3) ( 4) (5)
P Q R Q∧R P ∨ (Q ∧ R) P∨Q P∨ R ( P ∨ Q) ∧ ( P ∨ R)
T T T T T T T T
T T F F T T T T
T F T F T T T T
T F F F T T T T
F T T T T T T T
F T F F F T F F
F F T F F F T F
F F F F F F F F
Section 1.2 43 Conditional and Biconditional
## 9. (Inverse, Converse, and Contrapositive) One of the following sentences has
the same meaning as P ⇒ Q . Which one is it?
inverse : ∼ P ⇒ ∼ Q
converse : Q ⇒ P
contrapositive : ∼ Q ⇒ ∼ P
For the two sentences which are not always true, give examples where they
are true.
10. (True or False?) Is the following statement ever true? Is it ever false?
( P ⇒ Q ) ∧ Q ⇒ P
## Ans: From the truth table
(1) ( 2) ( 3)
P Q P ⇒Q ( P ⇒ Q) ∧ Q ( P ⇒ Q ) ∧ Q ⇒ P
T T T T T
T F F F T
F T T T F
F F T F F
the statement is TRUE when P is true, otherwise false. Hence, the statement
is not a tautology or a contradiction.
## a) P ⇒ Q (direct form of an implication)
b) ∼ Q ⇒ ∼ P (contrapositive form)
c) ( P ∧ ∼ Q ) ⇒ ∼ P (proof by contradiction)
d) ( P ∧ ∼ Q ) ⇒ Q (proof by contradiction)
e) ( P ∧ ∼ Q ) ⇒ R ∧ ∼ R (proof by reduction ad absurdum)
Ans:
Ans Draw truth tables
(
12. (Hmmmmmmmmmm) Is the statement ( P ∨ Q ) ⇔ P ∨ ∼ Q true for all )
truth values of P and Q , or is it false for all values, or is it sometimes true and
sometimes false?
Section 1.2 44 Conditional and Biconditional
Ans:
Ans It is true then P is true and Q is false, otherwise the statement is false
as can be seen from the truth table.
(1) ( 2) ( 3) ( 4)
P Q P∨Q ∼Q P ∨∼Q ( P ∨ Q) ⇔ ( P ∨ ∼ Q)
T T T F T T
T F T T T T
F T T F F F
F F F T T F
13. (
(Interesting Biconditional) Is the statement ( P ∨ Q ) ⇔ ∼ P ∨ ∼ Q true )
for all truth values of P and Q , or is it false for all values, or is it sometimes
true and sometimes false?
Ans It is true when exactly one of P and Q is true, and false when both are
Ans:
true or both are false as can be seen from the truth table.
(1) ( 2) ( 3) ( 4) (5)
P Q P∨Q ∼P ∼Q ∼P ∨∼Q ( P ∨ Q) ⇔ (∼ P ∨ ∼ Q)
T T T F F F F
T F T F T T T
F T T T F T T
F F F T T T F
## 14. Find the negation of the following sentences.
a) ( P ∨ Q) ∧ R
Ans: ∼ ( P ∨ Q ) ∧ R ≡ ∼ ( P ∨ Q ) ∨ ∼ R ≡ ( ∼ P ∧ ∼ Q ) ∨ R
b) ( P ∨ Q) ∧ ( R ∨ S )
Ans: ∼ [∼ Q ⇒ ∼ P] ≡ ∼ (Q ∨ ∼ P ) ≡ ∼ Q ∧ P
c) (∼ P ∨ Q) ∧ R
Ans: ∼ ( ∼ P ∨ Q ) ∧ R ≡ ∼ ( ∼ P ∨ Q ) ∨ ∼ R ≡ ( P ∧ ∼ Q ) ∨ ∼ R
Section 1.2 45 Conditional and Biconditional
## a) the contrapositive is true
Ans:
Ans If a function f is differentiable then f is continuous.
be.
## c) the converse is true
Ans: If today is Monday, then it is the second day of the week. The
converse of this sentence states that if this is the second day of the week,
then it is Monday, which is also true. When both a conditional and its
converse are true the hypothesis and conclusion are logically equivalent.
## d) the converse is false
Ans:
If a function f is differentiable then f is continuous. Here the converse
states that if a function is continuous, then it is differentiable, which is
false. The function f ( x ) = x is a counterexample of a continuous
function that is not differentiable.
## a) Prove or disprove that an implication and its inverse are equivalent.
Ans: Note that columns (1) and ( 4 ) are not the same. Hence the inverse
of an implication is not equivalent to the implication.
(1) ( 2) ( 3) ( 4)
P Q P ⇒Q ∼P ∼Q ∼ P ⇒∼ Q
T T T F F T
T F F F T T
F T T T F F
F F T T T T
b) What are the truth values of P and Q for which an implication and its
inverse are both true?
Section 1.2 46 Conditional and Biconditional
Ans When both P and Q are true or when they are both false.
Ans:
c) What are the truth values of P and Q for which the implication and its
inverse are both false?
## “If N is an integer, then 2N is an even integer” write the converse,
contrapositive, and inverse sentences.
## Ans Converse: If 2N is an even integer, then N is an integer.
Ans:
Contrapositive: If 2N is an odd integer, then N is not an integer.
Inverse: If N is not an integer, then 2N is not an even integer.
## 18. Let P,Q, and R be sentences. Show
a) P ⇒ ( Q ⇔ R ) requires paranthesis
Ans: One simply verifies that the truth tables for P ⇒ ( Q ⇔ R ) and
(Q ⇒ P ) ⇔ R are not identical.
b) ( P ∧ Q) ∨ R requires paranthesis
Ans One simply verifies that the truth tables for ( P ∧ Q ) ∨ R and
Ans:
P ∧ ( Q ∨ R ) are not identical.
c) (∼ P ∨ Q) ⇒ R may be written ∼ P ∨ Q ⇒ R
## Ans: One simply verifies the truth tables for ( ∼ P ∨ Q ) ⇒ R and ∼ P ∨ Q ⇒ R
are the same.
19. (Challenge)
(Challenge) Rewrite the sentence
P ⇒ (Q ⇒ R )
## in an equivalent form in which the symbol " ⇒ " does no occur.
Section 1.2 47 Conditional and Biconditional
## Ans: Since (Q ⇒ R ) ≡ ( ∼ P ∨ Q ) we can write
(
P ⇒ ( Q ⇒ R ) ≡ ∼ P ∨ ( Q ⇒ R ) ≡ ∼ P ∨ ∼ Q ∨ R )
20. (Non Obvious Statement)
Statement) The statement
P ⇒ (Q ⇒ P )
can be read “If P is true, then P follows from any Q ” Is this a tautology,
contradiction, or does its truth value depend on the truth or falsity of
P and Q ?
Ans:
Ans Making a truth table for the sentence, we find
P Q Q⇒P P ⇒ (Q ⇒ P )
T T T T
T F T T
F T F T
F F T T
## 21. (Another Non Obvious Statement)
Statement) The statement
(Q ⇒ P ) ∨ ( P ⇒ Q )
can be read “For any two sentences P and Q , it is always true that
P implies Q or Q implies P ” Is this a tautology, contradiction, or does its true
value depend on the truth or falsity of P and Q ?
Ans:
Ans Making a truth table for the sentence, we find
P Q P⇒Q Q⇒P ( P ⇒ Q ) ∨ (Q ⇒ P )
T T T T T
T F F T T
F T F F T
F F T T T
## Hence, the statement is a tautology.
Section 1.2 48 Conditional and Biconditional
22. (Three-
(Three-Valued Logic) Two-valued (T and F) truth tables were basic in
logic until 1921 when the Polish logician Jan Lukasiewicz (1878-1956) and
American logician Emil Post (1897-1954) introduced n -valued logical
systems where n is any integer greater than 1. For example, sentences in a
three-valued logic might have values True, False, and Unknown. Three-value
logic is useful in computer science in database work. The truth tables for the
AND, OR, and NOT connectives are as follows:
A B A OR B A AND B NOT A
True True True True False
True Unknown True Unknown False
True False True False False
Unknown True True Unknown Unknown
Unknown Unknown Unknown Unknown Unknown
Unknown False Unknown False Unknown
False True True False True
False Unknown Unknown False True
False False False False True
From these connectives, derive the connectives for the conditional and
biconditional connectives.
(1) ( 2) ( 3)
P Q ∼P ∼ P∨Q P⇒Q
T T F T T
T U F U U
T F F F F
U T U T T
U U U U U
U F U U U
F T T T T
F U T T T
F F T T T
## Using the fact that p P ⇔ Q ≡ ( P ⇒ Q ) ∧ ( Q ⇒ P ) we have the biconditional in
column ( 6 ) .
Section 1.2 49 Conditional and Biconditional
(1) ( 2) ( 3) ( 4) (5) ( 6)
P Q ∼P ∼ P∨Q P⇒Q ∼Q ∼ Q∨ P Q⇒P ( P ⇒ Q ) ∧ (Q ⇒ P )
T T F T T F T T T
T U F U U U T T U
T F F F F T T T F
U T U T T F U U U
U U U U U U U U U
U F U U U T T T U
F T T T T F F F F
F U T T T U U U U
F F T T T T T T T
23. (Modus Ponens and Modus Tollens) Modus Ponens3 and Modus Tollens4
are systematic ways of making logical arguments that takes the form
If P then Q If P then Q
P ∼Q
Therefore Q Therefore ∼ P
Modus Ponens Modus Tollens
Show that Modus Ponens and Modus Tollens are both tautologies,.
## Ans: Modus Ponens is a tautology which can be seen by constructiong the
truth table for ( P ⇒ Q ) ∧ P ⇒ Q and observing there are all T’s in column
(4) below:
(1) ( 2) ( 3) ( 4)
P Q P ⇒Q ( P ⇒ Q) ∧ P ( P ⇒ Q ) ∧ P ⇒ Q
T T T T T
T F F F T
F T T F T
F F T F T
Modus Tollens is a tautology which can be seen by constructing the truth table
for ( P ⇒ Q ) ∧ ∼ Q ⇒ ∼ P and observing there are T’s in column (4) below:
3
Latin: mode that affirms.
4
Latin mode that denies.
Section 1.2 50 Conditional and Biconditional
(1) ( 2) ( 3) ( 4)
P Q P ⇒Q ∼Q ( P ⇒ Q) ∧ ∼ Q ( P ⇒ Q ) ∧ ∼ Q ⇒ ∼ Q
T T T F F T
T F F T F T
F T T F F T
F F T T T T
## 24. (Interesting) Are the following two statements equivalent?
P ∧ (Q ⇒ R)
( P ∧ Q) ⇒ R
Ans: Yes, as can be seen from the following truth table.
(1) ( 2) ( 3) ( 4)
P Q R Q⇒R P ∧Q P ∧ (Q ⇒ R) ( P ∧ Q) ⇒ R
T T T T T T T
T T F F T F F
T F T T F T T
T F F T F T T
F T T T F F T
F T F F F F T
F F T T F F T
F F F T F F T
25. Determine which of the following statements is true for the following
diagram.
Section 1.2 51 Conditional and Biconditional
a) All the squares are red if and only if all the squares are green.
b) Either there is a green circle or there are no yellow circles.
c) If there is red circle, then there is not a green circle.
Ans:
a) Both statements “all the squares are red” and “all squares are green” are
false,so the biconditional is true.
## c) The premise is false to the conditional is true.
26. (Logical Puzzle) On an isolated island the inhabitants either always tell the
truth or always lie. You make a visit to the island and ask the first person you
meet about their favorite baseball team. The person says two things:
## ▪ I like the Red Sox.
▪ If I like the Red Sox, then I like the Yankees.
Does the islander like the Red Sox? Does the islander like the Yankees?
Section 1.2 52 Conditional and Biconditional
## R : I LIKE THE RED SOX
R ⇒ Y : IF I LIKE THE RED SOX I LIKE THE YANKEES
## R : I LIKE THE RED SOX
R ⇒ Y : IF I LIKE THE RED SOX I LIKE THE YANKEES
can not both be false since if R is false, then the conditional R ⇒ Y will
always be true. Hence we conclude both statements R ⇒ Y and R are true.
But if both these statements are true, so is Y . In other words, the islander
likes both the Red Sox and the Yankees.
27. (Professor
Professor Snarf’s Birthday)
Birthday)
Professor Snarf tells Mary the month of his birthday and tells Dave the
day of his birthday, and tells them not to pass this information to the other.
Professor Snarf then tells them if they can deduce his birthday, they will each
receive an A in their logic course, knowing it is impossible. However, Professor
Snarf does not realize how smart Mary and Dave are in logic.
Day
1 2 3 4 5 6
Jan Jan 3 Jan 4 Jan 6
Feb Feb 3 Feb 5
Montth
March March 1 March 4
April April 1 April 2 April 6
## After a moment’s thought:
Statement 1: Mary says, “If I don’t know the answer, then Dave doesn’t know
either.
Statement 2:
2 Dave says “I didn’t know the answer before, but now I know.”
Statement 3:
3 Mary says: “I now know the answer too.”
whereupon they correctly tell Professor Snarf his birthday and he grudgenly
gives them their As. What was the correct answer that Mary and Dave gave
Professor Snarf?
Section 1.2 53 Conditional and Biconditional
Ans: Let us rename Mary by MONTH and Dave by DAY to so we don’t forget
that MONTH knows the month and DAY knows the day.
## In a nutshell, MONTH is going to pass some valuable information to DAY (enough
so DAY knows the birthday), and then DAY passes back some information to
MONTH (enough so that MONTH knows the birthday), and then MONTH passes
you the reader information so YOU know the birthday.
At the start MONTH (who knows the month but not the day) does not know the
birthday and so Statement 1, which is an implication of the form P ⇒ Q implies
that neither does DAY This is MONTH’s clever day of telling DAY that the
month is not February or April, since if the birthday were in one of those two
months, DAY would know inasmuch as day 2 only occurs in April and day 5 only
in Feb. So now MONTH and DAY reduced the number of possible birthdays from
10 to 5 as illustrated below.
Day
1 2 3 4 5 6
Jan Jan 3 Jan 4 Jan 6
Montth
March March 1 March 4
It is now DAY’s turn to deduce the answer and pass enough information back to
MONTH so she can deduce the answer. DAY says he knows the birthday, which
tells us MONTH it is not on the 4th of the month. (It is easy for him to find the
answer, he just looks in the column of the day Professor Snarf gave him to find
the month. So now we are down to three possibilities March 1, Jan 3, or Jan 6.
Day
1 2 3 4 5 6
Jan Jan 3 Jan 6
Montth
March March 1
Now, it is MONTH’s turn to deduce the birthday and tell you ! She says she
knows the birthday, which means the birthday must by March 1, otherwise if her
month was Jan she would be unable to decide between Jan 3 and Jan 6. Hence,
Professor Snarf’s birthday is March 1. | 8,634 | 27,304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2019-43 | latest | en | 0.928487 |
https://oysterstreetpottery.com/qa/what-does-the-r-with-two-lines-mean-in-math.html | 1,606,858,474,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141681524.75/warc/CC-MAIN-20201201200611-20201201230611-00082.warc.gz | 428,736,884 | 8,149 | # What Does The R With Two Lines Mean In Math?
## What do double vertical lines mean in math?
The double bar symbol is used to denote certain kinds of norms in mathematics (e.g., or ).
It is also used to denote parallel lines, as in , and in an older notation for the covariant derivative.
In this work, the notation or.
is reserved for matrix and function norms, respectively..
## What does ∼ mean?
approximately equal”∼” is one of many symbols, listed in the Wikipedia article on approximation, used to indicate that one number is approximately equal to another. Note that “approximately equal” is reflexive and symmetric but not transitive. “∼” is one of many symbols used in logic to indicate negation.
## What does Z+ mean in math?
set of integersThe following letters describe what set each letter represents: N is the set of natural numbers (ℕ) Z is the set of integers (ℤ) Q is the set of rational numbers (ℚ) R is the set of real numbers (ℝ) C is the set of complex numbers (ℂ) ∅ is the empty set In this tutorial we also cover how to write each of the following …
## What does ø mean in math?
Slashed zeroThe letter “Ø” is sometimes used in mathematics as a replacement for the symbol “∅” (Unicode character U+2205), referring to the empty set as established by Bourbaki, and sometimes in linguistics as a replacement for same symbol used to represent a zero. … Slashed zero is an alternate glyph for the zero character.
## What is this symbol called?
The at sign, @, is normally read aloud as “at”; it is also commonly called the at symbol or commercial at.
## What do double lines mean?
double-bar-line. Noun. (plural double bar lines) A symbol in musical notation used to separate two sections or phrases of music, and to start and end a musical piece.
## What is the horizontal line symbol called?
VinculumA vinculum is a horizontal line used in mathematical notation for a specific purpose. It may be placed as an overline (or underline) over (or under) a mathematical expression to indicate that the expression is to be considered grouped together.
## What does the R in math mean?
real numbersList of Mathematical Symbols • R = real numbers, Z = integers, N=natural numbers, Q = rational numbers, P = irrational numbers. Page 1. List of Mathematical Symbols. • R = real numbers, Z = integers, N=natural numbers, Q = rational numbers, P = irrational numbers.
## What does ≡ mean in math?
In computer applications (like Excel) the symbols <> mean not equal. ≡ means identical to. … ≈ means approximately equal to, or almost equal to. The two sides of a relationship indicated by this symbol will not be accurate enough to manipulate mathematically.
## What is this symbol name?
Know Names of Keyboard Symbols in Computer KeyboardSymbolName&ersand or and’apostrophe or single quote*asterisk@at32 more rows•Apr 9, 2020
## What is the vertical line symbol called?
The vertical bar, | , is a glyph with various uses in mathematics, computing, and typography. It has many names, often related to particular meanings: Sheffer stroke (in logic), verti-bar, vbar, stick, vertical line, vertical slash, bar, pike, or pipe, and several variants on these names. | 729 | 3,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-50 | latest | en | 0.934439 |
http://stackoverflow.com/questions/3472311/what-is-a-portable-method-to-find-the-maximum-value-of-size-t | 1,462,247,808,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860118369.35/warc/CC-MAIN-20160428161518-00134-ip-10-239-7-51.ec2.internal.warc.gz | 270,082,600 | 21,671 | # What is a portable method to find the maximum value of size_t?
I'd like to know the maximum value of size_t on the system my program is running. My first instinct was to use negative 1, like so:
``````size_t max_size = (size_t)-1;
``````
But I'm guessing there's a better way, or a constant defined somewhere.
-
That's one very smart trick you've got there. +1! – Mhmmd Aug 12 '10 at 21:49
Yep, what you have is fine (you don't need the cast, by the way). – Stephen Canon Aug 12 '10 at 21:56
@Craig: One possible reason could be to set that as an invalid value for a `size_t` type variable. For instance, `std::string::npos` is set to `(size_t)-1` (at least in the MSVC implementation). – Praetorian Aug 13 '10 at 3:09
Can someone explain, what `size_t max_size = (size_t)-1;` actually does and how? Thank you. – Kolyunya Jul 1 '13 at 12:13
`size_t` is an unsigned type according to the standard. So say it's defined as a 32-bit value. A -1 is represented as 0xffffffff for a signed value using two's complement. However if we cast this to size_t which is an unsigned type, it is the max value instead. `(size_t)(-1)` is the same as `(size_t)(0xffffffff)` on a 32-bit system. It is better to use the -1 since that will work if it's 16-bit (0xffff) or 64-bit as well. – Joakim Feb 15 '14 at 12:24
A manifest constant (a macro) exists in C99 and it is called `SIZE_MAX`. There's no such constant in C89/90 though.
However, what you have in your original post is a perfectly portable method of finding the maximum value of `size_t`. It is guaranteed to work with any unsigned type.
-
+1 Excellent answer. – Stephen Canon Aug 12 '10 at 21:56
@jamesdlin: The signed-to-unsigned conversion is always well-defined in C. It is required to follow the rules of typical unsigned modulo arithmetics with modulo equal to the largest value of the target unsigned type plus 1. So, in the above case you will get `-1 mod (<max-value> + 1)`, which is always just `<max-value>`. – AnT Aug 12 '10 at 22:24
@jamesdlin: §5.2.4.2.1 guarantees that `-1` is representable as an `int`. §6.3.1.3 guarantees that it converts to a valid `size_t` value. – Stephen Canon Aug 12 '10 at 22:35
@jamesdlin: another way to see that is that `size_t` is an unsigned type, so all values are valid. This can't be a trap representation since there is no such trap. – Jens Gustedt Aug 13 '10 at 6:36
@jamesdlin: yes it does, unsigned types are really simple minded ;-) from "`6.2.6.2 Integer types`": `If there are N value bits, each bit shall represent a different power of 2 between 1 and 2^N−1..` So for unsigned integer types there are really no surprises possible. – Jens Gustedt Aug 13 '10 at 12:45
``````#define MAZ_SZ (~(size_t)0)
``````
or `SIZE_MAX`
-
The `size_t max_size = (size_t)-1;` solution suggested by the OP is definitely the best so far, but I did figure out another, more convoluted, way to do this. I'm posting it just for academic curiosity.
``````#include <limits.h>
size_t max_size = ((((size_t)1 << (CHAR_BIT * sizeof(size_t) - 1)) - 1) << 1) + 1;
``````
-
If I should ever find that in some code (especially without a comment), it would make me very grumpy. :) – Craig McQueen Jun 20 '14 at 2:55
As an alternative to bit-operations suggested in the other answers, you could do this in C++
``````#include <limits>
size_t maxvalue = std::numeric_limits<size_t>::max()
``````
-
`std::numeric_limits<size_t>::max()` is not a `constexpr`, and it is not optimized well by some compilers, like Clang. GCC, ICC and MSC handle it fine. Its often better to stick with the `#define`. – jww Nov 18 '15 at 10:07 | 1,059 | 3,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2016-18 | longest | en | 0.932718 |
https://openi.nlm.nih.gov/detailedresult.php?img=PMC4587369_pone.0138355.g004&req=4 | 1,521,756,913,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648003.58/warc/CC-MAIN-20180322205902-20180322225902-00104.warc.gz | 683,866,163 | 12,064 | The Effects of Thermal Radiation on an Unsteady MHD Axisymmetric Stagnation-Point Flow over a Shrinking Sheet in Presence of Temperature Dependent Thermal Conductivity with Navier Slip. Mondal S, Haroun NA, Sibanda P - PLoS ONE (2015) Bottom Line: The flow is due to a shrinking surface that is shrunk axisymmetrically in its own plane with a linear velocity.The magnetic field is imposed normally to the sheet.The model equations that describe this fluid flow are solved by using the spectral relaxation method. View Article: PubMed Central - PubMed Affiliation: University of KwaZulu-Natal, School of Mathematics, Statistics and Computer Science, Private Bag X01, Scottsville, Pietermaritzburg 3209, South Africa. ABSTRACTIn this paper, the magnetohydrodynamic (MHD) axisymmetric stagnation-point flow of an unsteady and electrically conducting incompressible viscous fluid in with temperature dependent thermal conductivity, thermal radiation and Navier slip is investigated. The flow is due to a shrinking surface that is shrunk axisymmetrically in its own plane with a linear velocity. The magnetic field is imposed normally to the sheet. The model equations that describe this fluid flow are solved by using the spectral relaxation method. Here, heat transfer processes are discussed for two different types of wall heating; (a) a prescribed surface temperature and (b) a prescribed surface heat flux. We discuss and evaluate how the various parameters affect the fluid flow, heat transfer and the temperature field with the aid of different graphical presentations and tabulated results. No MeSH data available. Related in: MedlinePlus © Copyright Policy Related In: Results - Collection License getmorefigures.php?uid=PMC4587369&req=5 .flowplayer { width: px; height: px; } pone.0138355.g004: Effect of α on velocity profiles h(η) for M = 0.1, δ = 0.2, β = −0.25, Pr = 0.72, Nr = 0.2, Ec = 1.0, R = 1.0 and ϵ = 0.5. Mentions: We note that for a stretching sheet α is positive and for a shrinking sheet α is negative while α = 0 represents Hiemenz flow. Figs 3 and 4 show the effect of α on the vertical velocity components f′(η) and h(η). We observe that f′(η) increases with increases in α while the value of h(η) decreases with increases in the values of α.
The Effects of Thermal Radiation on an Unsteady MHD Axisymmetric Stagnation-Point Flow over a Shrinking Sheet in Presence of Temperature Dependent Thermal Conductivity with Navier Slip.
Mondal S, Haroun NA, Sibanda P - PLoS ONE (2015)
Related In: Results - Collection
Show All Figures
getmorefigures.php?uid=PMC4587369&req=5
pone.0138355.g004: Effect of α on velocity profiles h(η) for M = 0.1, δ = 0.2, β = −0.25, Pr = 0.72, Nr = 0.2, Ec = 1.0, R = 1.0 and ϵ = 0.5.
Mentions: We note that for a stretching sheet α is positive and for a shrinking sheet α is negative while α = 0 represents Hiemenz flow. Figs 3 and 4 show the effect of α on the vertical velocity components f′(η) and h(η). We observe that f′(η) increases with increases in α while the value of h(η) decreases with increases in the values of α.
Bottom Line: The flow is due to a shrinking surface that is shrunk axisymmetrically in its own plane with a linear velocity.The magnetic field is imposed normally to the sheet.The model equations that describe this fluid flow are solved by using the spectral relaxation method.
View Article: PubMed Central - PubMed
Affiliation: University of KwaZulu-Natal, School of Mathematics, Statistics and Computer Science, Private Bag X01, Scottsville, Pietermaritzburg 3209, South Africa.
ABSTRACT
In this paper, the magnetohydrodynamic (MHD) axisymmetric stagnation-point flow of an unsteady and electrically conducting incompressible viscous fluid in with temperature dependent thermal conductivity, thermal radiation and Navier slip is investigated. The flow is due to a shrinking surface that is shrunk axisymmetrically in its own plane with a linear velocity. The magnetic field is imposed normally to the sheet. The model equations that describe this fluid flow are solved by using the spectral relaxation method. Here, heat transfer processes are discussed for two different types of wall heating; (a) a prescribed surface temperature and (b) a prescribed surface heat flux. We discuss and evaluate how the various parameters affect the fluid flow, heat transfer and the temperature field with the aid of different graphical presentations and tabulated results.
No MeSH data available.
Related in: MedlinePlus | 1,078 | 4,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-13 | longest | en | 0.858438 |
https://www.jiskha.com/display.cgi?id=1288236787 | 1,503,305,077,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107744.5/warc/CC-MAIN-20170821080132-20170821100132-00217.warc.gz | 923,904,375 | 3,650 | # algebra
posted by .
how do I simplify -2xyz^3/5x^2z^2 times 10xy/x^3
• algebra -
-2xyz^3 / 5x^2z^2 * 10xy / x^3,
-2x^-1yz / 5 * 10x^-2y / 1,
-20x^-3y^2z / 5,
-4y^2z / x^3.
## Similar Questions
1. ### algerba
simplify the expression -10xy - 15y / 5y -10xy -3y^2 is my answer rigth or wrong
2. ### algebra
Factor the Following: 3x^2 + 10xy + 7y^2
3. ### Algebra
what is the greatest common monomial factor of 6x^2y^3z^4-4x^3y^3z^3+10xy^3z^4
4. ### algebra
want to know if this right simplify and show all work -3x^2+4xy+7y^2+2(-3x^2+5xy+11y^2) -3x^2+4xy+7y^2-6y^2+10xy+22y^2 -3x^2+14xy+45y^2 and -6a-4a-2(100^2+3a) -6a-4a^2+200+3a 3a-4a^2+200
5. ### Math
Here are the questions I need help with, please help. [(1/4)^-3 divide sign (1/4)^2]^-2 I got (1/4)^10 but I got it wrong (3^1/2)^3 _______ I got 1/216 9 (-56r^2)divide sign (-7r^2)= I got 8r x(y)(x)(2y) (12) (10xy^3) (3x^2y^2) (5x) …
6. ### Math Ms. Sue
Here are the questions I need help with, please help. [(1/4)^-3 divide sign (1/4)^2]^-2 I got (1/4)^10 but I got it wrong (3^1/2)^3 _______ I got 1/216 9 (-56r^2)divide sign (-7r^2)= I got 8r x(y)(x)(2y) (12) (10xy^3) (3x^2y^2) (5x) …
7. ### Math Please Urgent
Here are the questions I need help with, please help. [(1/4)^-3 divide sign (1/4)^2]^-2 I got (1/4)^10 but I got it wrong (3^1/2)^3 _______ I got 1/216 9 (-56r^2)divide sign (-7r^2)= I got 8r x(y)(x)(2y) (12) (10xy^3) (3x^2y^2) (5x) …
8. ### Math To: Reiny
This is one whole question: (12) (10xy^3) (3x^2y^2) (5x) (2y) I don't know how to do any of these: simplify sqaure root of 160000 simplify sqaure root of 0.04 simplify 3 * square root 81 simplify square root of 900 simplify square …
9. ### maths
Simplify; (x - y)^3 divide x^2z - 2xyz + y^2z
10. ### Math
For each of the following , simplify by combining like terms : 4x^3y+5x(6xy-7y^2) - 8y(9x^2-10xy) - 11xy^2
More Similar Questions | 868 | 1,879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-34 | latest | en | 0.719156 |
http://msdn.microsoft.com/en-us/library/bb206365(d=printer,v=vs.85).aspx | 1,411,491,465,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657139314.4/warc/CC-MAIN-20140914011219-00001-ip-10-234-18-248.ec2.internal.warc.gz | 172,532,533 | 11,186 | World Transform (Direct3D 9)
The discussion of the world transformation introduces basic concepts and provides details on how to set up a world transform.
What Is a World Transform?
A world transform changes coordinates from model space, where vertices are defined relative to a model's local origin, to World Space, where vertices are defined relative to an origin common to all the objects in a scene. In essence, the world transform places a model into the world; hence its name. The following diagram shows the relationship between the world coordinate system and a model's local coordinate system.
The world transform can include any combination of translations, rotations, and scalings.
Setting Up a World Matrix
As with any other transform, create the world transform by concatenating a series of matrices into a single matrix that contains the sum total of their effects. In the simplest case, when a model is at the world origin and its local coordinate axes are oriented the same as world space, the world matrix is the identity matrix. More commonly, the world matrix is a combination of a translation into world space and possibly one or more rotations to turn the model as needed.
The following example, from a fictitious 3D model class written in C++, uses the helper functions included in the D3DX utility library to create a world matrix that includes three rotations to orient a model and a translation to relocate it relative to its position in world space.
```
/*
* For the purposes of this example, the following variables
* are assumed to be valid and initialized.
*
* The m_xPos, m_yPos, m_zPos variables contain the model's
* location in world coordinates.
*
* The m_fPitch, m_fYaw, and m_fRoll variables are floats that
* contain the model's orientation in terms of pitch, yaw, and roll
*/
void C3DModel::MakeWorldMatrix( D3DXMATRIX* pMatWorld )
{
D3DXMATRIX MatTemp; // Temp matrix for rotations.
D3DXMATRIX MatRot; // Final rotation matrix, applied to
// pMatWorld.
// Using the left-to-right order of matrix concatenation,
// apply the translation to the object's world position
// before applying the rotations.
D3DXMatrixTranslation(pMatWorld, m_xPos, m_yPos, m_zPos);
D3DXMatrixIdentity(&MatRot);
// Now, apply the orientation variables to the world matrix
if(m_fPitch || m_fYaw || m_fRoll) {
// Produce and combine the rotation matrices.
D3DXMatrixRotationX(&MatTemp, m_fPitch); // Pitch
D3DXMatrixMultiply(&MatRot, &MatRot, &MatTemp);
D3DXMatrixRotationY(&MatTemp, m_fYaw); // Yaw
D3DXMatrixMultiply(&MatRot, &MatRot, &MatTemp);
D3DXMatrixRotationZ(&MatTemp, m_fRoll); // Roll
D3DXMatrixMultiply(&MatRot, &MatRot, &MatTemp);
// Apply the rotation matrices to complete the world matrix.
D3DXMatrixMultiply(pMatWorld, &MatRot, pMatWorld);
}
}
```
After you prepare the world matrix, call the IDirect3DDevice9::SetTransform method to set it, specifying the D3DTS_WORLD macro for the first parameter.
Note Direct3D uses the world and view matrices that you set to configure several internal data structures. Each time you set a new world or view matrix, the system recalculates the associated internal structures. Setting these matrices frequently-for example, thousands of times per frame-is computationally time-consuming. You can minimize the number of required calculations by concatenating your world and view matrices into a world-view matrix that you set as the world matrix, and then setting the view matrix to the identity. Keep cached copies of individual world and view matrices so that you can modify, concatenate, and reset the world matrix as needed. For clarity, in this documentation Direct3D samples rarely employ this optimization. | 832 | 3,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2014-41 | latest | en | 0.855729 |
morgan431.omaralashi.com | 1,632,007,677,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056578.5/warc/CC-MAIN-20210918214805-20210919004805-00532.warc.gz | 469,730,201 | 5,997 | # Forms of Bets on the Roulette Table
The table where a roulette game is playing is called the “rolette table.” The numbers with this roulette table range from game to game. The chance of winning on a single spin of a wheel is known as the “roll.” The size of each bet in a game is also variable. In the event that you place a bet and win, that bet will undoubtedly be doubled if you choose the same number from another players in a casino game that you play.
A Roulette table layout is used to help place bets. You can find two forms of Roulette table layouts, even and odd. When playing Roulette the game is a lot easier to take care of whenever there are even numbers up for grabs. Odd numbers include even numbers and multiples of even numbers. When placing bets, odd numbers like an odd number four always have the highest probability of being won by you.
The most typical type of roulette table used in most casinos in the us and Europe may be the push-tabard roulette table. This is a smaller version of a standard roulette table that makes it easier to place and win bets. This table layout is used more often in European and American casinos than it is in home casinos in the us and Europe. Many home casinos in America and Europe have adopted the push toward layout for roulette tables due to its smaller size.
Another type of Roulette table may be the non-push layout. 로투스 바카라 In this sort of table all of the symbols are shown on the left hand side. The numbers are printed on the proper and all of the outcomes of spins of the wheels are read from top to bottom. For instance, in a 4 wheel spin the dealer will first show the ace symbol then the numbers, and the third wheel symbol showing that the 3rd spin occurred.
In the end of this information is shown the odds of each result is then given on the left hand side. The odds on a Roulette table payouts table can be extremely complicated if you do not have some basic knowledge about probability. The easiest method to find out what the chances are on the Roulette table payouts is to place your bets on a live game and take time to study the odds. The Internet also has an abundance of information about the many odds for different game variations.
Before placing your bets on a Roulette table payout you should learn about the various kinds of spins that are found in the game. All of the spins except for the straight flush require that the winning number is known beforehand. Once the wheels are spinning at full speed you have two options. You can either win a number it doesn’t represent the odds on the Roulette table, or you can end up with chips which are worth less than the number of chips you have in your bank roll.
The “ringing bet” is the first bet you make once you enter the game. This bet is placed on the proper or the outer corner of your seat. It represents the initial non-zero value seen on the Roulette wheel although it spins. If the consequence of this bet is really a straight flush or an Ace, you have lost all of your previous chip burn.
Just about the most popular types of bets may be the outside bet or “one number bet”. That is a risky bet, as it could put you at an increased risk for losing more chips than you have in your bankroll. The exterior bet, also called the blind or spread bet, is made when the dealer reveals only one number that is on the Roulette wheel. In virtually all Roulette games, the dealer will place this number within an exact location on the Roulette table (externally or the middle of the circle). You must place your bet before the dealer reveals this one number. | 748 | 3,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-39 | latest | en | 0.934072 |
http://metamath.tirix.org/mpests/spanval.html | 1,716,081,329,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057631.79/warc/CC-MAIN-20240519005014-20240519035014-00268.warc.gz | 17,551,928 | 2,724 | # Metamath Proof Explorer
## Theorem spanval
Description: Value of the linear span of a subset of Hilbert space. The span is the intersection of all subspaces constraining the subset. Definition of span in Schechter p. 276. (Contributed by NM, 2-Jun-2004) (Revised by Mario Carneiro, 23-Dec-2013) (New usage is discouraged.)
Ref Expression
Assertion spanval ${⊢}{A}\subseteq ℋ\to \mathrm{span}\left({A}\right)=\bigcap \left\{{x}\in {\mathbf{S}}_{ℋ}|{A}\subseteq {x}\right\}$
### Proof
Step Hyp Ref Expression
1 df-span ${⊢}\mathrm{span}=\left({y}\in 𝒫ℋ⟼\bigcap \left\{{x}\in {\mathbf{S}}_{ℋ}|{y}\subseteq {x}\right\}\right)$
2 sseq1 ${⊢}{y}={A}\to \left({y}\subseteq {x}↔{A}\subseteq {x}\right)$
3 2 rabbidv ${⊢}{y}={A}\to \left\{{x}\in {\mathbf{S}}_{ℋ}|{y}\subseteq {x}\right\}=\left\{{x}\in {\mathbf{S}}_{ℋ}|{A}\subseteq {x}\right\}$
4 3 inteqd ${⊢}{y}={A}\to \bigcap \left\{{x}\in {\mathbf{S}}_{ℋ}|{y}\subseteq {x}\right\}=\bigcap \left\{{x}\in {\mathbf{S}}_{ℋ}|{A}\subseteq {x}\right\}$
5 ax-hilex ${⊢}ℋ\in \mathrm{V}$
6 5 elpw2 ${⊢}{A}\in 𝒫ℋ↔{A}\subseteq ℋ$
7 6 biimpri ${⊢}{A}\subseteq ℋ\to {A}\in 𝒫ℋ$
8 helsh ${⊢}ℋ\in {\mathbf{S}}_{ℋ}$
9 sseq2 ${⊢}{x}=ℋ\to \left({A}\subseteq {x}↔{A}\subseteq ℋ\right)$
10 9 rspcev ${⊢}\left(ℋ\in {\mathbf{S}}_{ℋ}\wedge {A}\subseteq ℋ\right)\to \exists {x}\in {\mathbf{S}}_{ℋ}\phantom{\rule{.4em}{0ex}}{A}\subseteq {x}$
11 8 10 mpan ${⊢}{A}\subseteq ℋ\to \exists {x}\in {\mathbf{S}}_{ℋ}\phantom{\rule{.4em}{0ex}}{A}\subseteq {x}$
12 intexrab ${⊢}\exists {x}\in {\mathbf{S}}_{ℋ}\phantom{\rule{.4em}{0ex}}{A}\subseteq {x}↔\bigcap \left\{{x}\in {\mathbf{S}}_{ℋ}|{A}\subseteq {x}\right\}\in \mathrm{V}$
13 11 12 sylib ${⊢}{A}\subseteq ℋ\to \bigcap \left\{{x}\in {\mathbf{S}}_{ℋ}|{A}\subseteq {x}\right\}\in \mathrm{V}$
14 1 4 7 13 fvmptd3 ${⊢}{A}\subseteq ℋ\to \mathrm{span}\left({A}\right)=\bigcap \left\{{x}\in {\mathbf{S}}_{ℋ}|{A}\subseteq {x}\right\}$ | 857 | 1,896 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 15, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-22 | latest | en | 0.358294 |
https://gmatclub.com/forum/subject-verb-agreement-1-noting-that-the-price-of-oil-and-145727.html | 1,722,689,512,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640368581.4/warc/CC-MAIN-20240803121937-20240803151937-00521.warc.gz | 227,818,196 | 138,415 | Last visit was: 03 Aug 2024, 05:51 It is currently 03 Aug 2024, 05:51
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
# Subject Verb Agreement 1. Noting that the price of oil and
SORT BY:
Tags:
Show Tags
Hide Tags
Alum
Joined: 19 Mar 2012
Posts: 4330
Own Kudos [?]: 51748 [71]
Given Kudos: 2326
Location: United States (WA)
Concentration: Leadership, General Management
Schools: Ross '20 (M)
GMAT 1: 760 Q50 V42
GMAT 2: 740 Q49 V42 (Online)
GMAT 3: 760 Q50 V42 (Online)
GPA: 3.8
WE:Marketing (Non-Profit and Government)
Retired Moderator
Joined: 23 Jul 2010
Posts: 402
Own Kudos [?]: 1853 [2]
Given Kudos: 370
GPA: 3.5
WE:Business Development (Health Care)
Intern
Joined: 22 Jan 2012
Posts: 8
Own Kudos [?]: 8 [0]
Given Kudos: 2
Intern
Joined: 29 Dec 2012
Posts: 41
Own Kudos [?]: 153 [0]
Given Kudos: 6
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
1 C
2 B
3 A
4 C
5 E
6 A
7 E
8 A
9 B
10 C
11 C
12 E
13 C
14 D
15 D
16 C
17 D
18 A
19 A
Manager
Joined: 19 Sep 2008
Status:Please do not forget to give kudos if you like my post
Posts: 69
Own Kudos [?]: 206 [0]
Given Kudos: 257
Location: United States (CA)
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
My responses:
souvik101990 can you post the solution?
Intern
Joined: 04 Jan 2012
Posts: 15
Own Kudos [?]: 9 [0]
Given Kudos: 11
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
We need the official solution to this. Please post an official solution.
Intern
Joined: 04 Jan 2012
Posts: 15
Own Kudos [?]: 9 [1]
Given Kudos: 11
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
1
Bookmarks
Should 13 be C or D ?
It refers to "number" of https://gmatclub.com/chat. So shouldn't it be "has" ? I went for D on this one.
Intern
Joined: 04 Jan 2012
Posts: 15
Own Kudos [?]: 9 [1]
Given Kudos: 11
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
1
Kudos
Also, for 9th, I went for E. Since it is singular "The banana", so it should be followed by "contains". If you disagree, please clarify.
Senior Manager
Joined: 05 Nov 2012
Posts: 338
Own Kudos [?]: 4652 [0]
Given Kudos: 606
Concentration: Technology, Other
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
Q19: Every workday at dawn, the patriarch of one of the city’s five richest families leaves his
mansion and walks to city hall.
Every workday at dawn,
the patriarch[/color] of one of the city’s five richest families
leaves his mansion and
walks to city hall.
• richest families leaves his mansion and walks -Correct
• richest families leave his mansion and walk
• richest families leaves his mansion and walk
richer families leave his mansion and walks
richer families leaves his mansion and walks
Senior Manager
Joined: 05 Nov 2012
Posts: 338
Own Kudos [?]: 4652 [1]
Given Kudos: 606
Concentration: Technology, Other
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
1
Kudos
17. Since the last election,
the lobbying effort
initiated by environmental organizations, homeowners, and small business owners
have increased awareness of pending environmental legislation.
have increased awareness of pending environmental legislation
have increased awareness about pending legislation dealing with the environment
• has increased awareness about pending environmental legislation
• has increased awareness of pending environmental legislation
• has increased awareness of environmental legislation that is still pending [wordy]
awareness of Vs awareness about:
https://www.manhattangmat.com/forums/awa ... t6770.html
Senior Manager
Joined: 05 Nov 2012
Posts: 338
Own Kudos [?]: 4652 [1]
Given Kudos: 606
Concentration: Technology, Other
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
1
Bookmarks
16. Since 1989,
after the Berlin Wall had been demolished,
-- one of the most problematic ethnic groups in the reunified Germany, in cultural and economic assimilation terms,
-- were the former East Germans,
--- who have had to acclimate to an entirely different political system.
• after the Berlin Wall had been demolished, one of the most problematic ethnic groups in the reunified German, in cultural and economic assimilation terms, were the former East Germans
[ in cultural and economic assimilation terms: I believe this is also wrong here? It shall be in X and Y]
• after the Berlin Wall was demolished, one of the most problematic ethnic groups in the reunified Germany, in cultural terms as well as those of economic assimilation, were the former East Germans
• when the Berlin Wall was demolished, one of the reunified Germany's most problematic ethnic groups, in terms of cultural and economic assimilation,was the former East Germans
• when the Berlin Wall was demolished, one of the most problematic ethnic groups in the reunified Germany, in terms of cultural and economic assimilation, has been the former East Germans
• after the Berlin Wall had been demolished, one of the most problematic ethnic groups in the reunified Germany, in both terms of cultural and economic assimilation, have been the former East Germans
C Vs D:
The present perfect is used with for and since, when the actions started in the past have not finished or whose consequences still goes on in the present:
I have lived in Victoria for five years.(I still live in Victoria.)
The simple past is used with for and since, when the actions have already finished: I lived in Victoria for five years.(I don't live in Victoria now.)
Senior Manager
Joined: 05 Nov 2012
Posts: 338
Own Kudos [?]: 4652 [0]
Given Kudos: 606
Concentration: Technology, Other
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
15. According to a recent study conducted at over 100 American universities,
the number of college graduates
--interested in pursuing a career in financial services
is likely to double by 2010.
• is likely to double
Are likely to increase by twice
Are likely to double
Will double
Will increase by twice
1: "number" is singular => "is"
2. A Vs D: A bec. study can't conclude the something will happen. [Am i correct in my understanding?]
3. double vs increase by twice: Which one is better?
Senior Manager
Joined: 05 Nov 2012
Posts: 338
Own Kudos [?]: 4652 [0]
Given Kudos: 606
Concentration: Technology, Other
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
13. Since 1929,
-when the global telegraph business peaked,
--the number of https://gmatclub.com/chat
---delivered annually
--have decreased from 200 million to only 21,000 last year.
have decreased from 200 million
have been reduced from 200 million
• has decreased from 200 million
has been reduced from 200 million down
• has decreased from 200 million down
14. Mathematical analysis of humpback whale sounds
provide evidence
-- that animals other than humans
-- use a hierarchical structure of communication.
provide evidence that animals other than humans
• provides evidence that animals other than humans
provide evidence that an animal other than humans
• provides evidence that an animal other than a human
provide evidence that animals, like humans,
Senior Manager
Joined: 05 Nov 2012
Posts: 338
Own Kudos [?]: 4652 [0]
Given Kudos: 606
Concentration: Technology, Other
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
11. According to industry analysts,
the recent growth
--in the number of hybrid motor vehicles in major metropolitan areas
are
likely to accelerate in the future.
• the number of hybrid motor vehicles in major metropolitan areas are
• the numbers of hybrid motor vehicles in major metropolitan areas are
• the number of hybrid motor vehicles in major metropolitan areas is
the numbers of hybrid motor vehicles in major metropolitan areas is [the numbers is a correct usage but here it's usage is incorrect.]
This chart lists the numbers of various crimes committed in the city last year.
--> There's a different number for each kind of crime, so singular "number" would make no sense here.
• hybrid motor vehicles’ numbers in major metropolitan areas are
12. Before the Civil War,
Harriet Tubman, along with other former slaves and white abolitionists,
helped create what had become known as the Underground Railroad,
and were
responsible for leading hundreds, if not thousands, of slaves to freedom.
had become known as the Underground Railroad, and were
• would become known as the Underground Railroad, and were
had become known as the Underground Railroad, and was
has been becoming known as the Underground Railroad,and was
• would become known as the Underground Railroad, and was
Manager
Joined: 23 Jan 2013
Posts: 99
Own Kudos [?]: 172 [1]
Given Kudos: 41
Concentration: Technology, Other
Schools: Berkeley Haas
GMAT Date: 01-14-2015
WE:Information Technology (Computer Software)
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
1
Kudos
1. Noting that the price of oil and other fuel components, a major factor in the cost structure of
an airline, have risen and will continue to rise, the company management was pessimistic
about their outlook for the upcoming quarter.
• have risen and will continue to rise, the company management was pessimistic about their : Management is singular has to be its , Eliminate A
• have risen and will continue to rise, the company management was pessimistic about the : The price is being spoken about it is singular .. 'have' is incorrect .
• will continue to rise, the company management was pessimistic about the : Correct answer
• has risen and will continue to rise, the company management was pessimistic about their : Management is singular has to be its , Eliminate D
• will continue to rise, the company management was pessimistic about their Management is singular , Their is incorrect , Eliminate E .
Manager
Joined: 23 Jan 2013
Posts: 99
Own Kudos [?]: 172 [0]
Given Kudos: 41
Concentration: Technology, Other
Schools: Berkeley Haas
GMAT Date: 01-14-2015
WE:Information Technology (Computer Software)
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
The Daughters of the American Revolution, a volunteer service organization, admits as members only women who can prove lineal descent from a patriot of the American Revolution.
' The Daughters of the American Revolution ' subject is an organization it is singular requires singular verb B , D are out .
E : Changes the meaning of the sentence.
A,C : Both are close A stresses on the fact that only women who can prove lineal descent can be admitted .
A wins
A. admits as members only women who can prove lineal descent
B. admit as members only women who can prove lineal descent
C. admits as members women who can prove lineal descent
D. only admit as members women who can prove lineal descent
E. admits as members women who can prove lineal descent only
Manager
Joined: 06 Nov 2012
Status:Manager
Affiliations: Manager
Posts: 100
Own Kudos [?]: 420 [0]
Given Kudos: 111
Location: India
Concentration: Entrepreneurship, Sustainability
GMAT 1: 650 Q49 V29
GMAT 2: 680 Q49 V33
GPA: 3
WE:Supply Chain Management (Energy and Utilities)
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
1.C
2.B
3.A
4.C
5.E
6.A
7.E
8.A
9.B
10.C
11.C
12.E
13.C
14.B
15.A
16.D
17.C
18.A
19.A
OAs and OEs please...
Manager
Joined: 23 Jan 2013
Posts: 99
Own Kudos [?]: 172 [0]
Given Kudos: 41
Concentration: Technology, Other
Schools: Berkeley Haas
GMAT Date: 01-14-2015
WE:Information Technology (Computer Software)
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
6. The governor’s team of advisors, including her education and political strategists, has not been
available for comment since the governor released her controversial education reform
proposal.
Solution : Governor's team : Subject , requires singular verb has , B , C are out .
Eliminated D , E since the pronoun she cannot refer to governor's possesive noun requires her .
A is a clear winner
• has not been available for comment since the governor released her controversial education
reform proposal
• have not been available for comment since the governor released her controversial education
reform proposal
• have not been available for comment since she released her proposal on controversial
education reform
• has not been available for comment since she released her controversial education reform
proposal
• has not been available to make comments since she released her proposal on controversial
reform in education
Manager
Joined: 23 Jan 2013
Posts: 99
Own Kudos [?]: 172 [0]
Given Kudos: 41
Concentration: Technology, Other
Schools: Berkeley Haas
GMAT Date: 01-14-2015
WE:Information Technology (Computer Software)
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
7. The amount of car accidents caused by faulty brakes, like accidents caused by faulty wiring,
has increased significantly since regulations on manufacturing have been relaxed.
Solution : 'The number of ' should be used since car accidents are countable So A , B are out .
C - uses have , but 'the number of ' requires singular verb
D - uses incorrect comparison , E is a clear winner..
• The amount of car accidents caused by faulty brakes,like accidents caused by faulty wiring,
has increased significantly since regulations on manufacturing have been relaxed.
• The amount of car accidents caused by faulty brakes,like those caused by faulty wiring, have
increased significantly since regulations on manufacturing were relaxed.
• The number of car accidents caused by faulty brakes,like the number caused by faulty wiring,
have increased significantly since regulations on manufacturing were relaxed.
• The number of car accidents caused by faulty brakes,like accidents caused by faulty wiring,
has increased significantly since regulations on manufacturing had been relaxed.
• The number of car accidents caused by faulty brakes,like the number caused by faulty wiring,
has increased significantly since regulations on manufacturing were relaxed.
Manager
Joined: 07 Dec 2009
Posts: 69
Own Kudos [?]: 102 [0]
Given Kudos: 375
GMAT Date: 12-03-2014
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
Hi Souvik,
Request you to Provide the OAs
Many Thanks
Re: Subject Verb Agreement 1. Noting that the price of oil and [#permalink]
1 2 3
Moderators:
GMAT Club Verbal Expert
6999 posts
GMAT Club Verbal Expert
236 posts
GRE Forum Moderator
13959 posts | 3,755 | 15,095 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-33 | latest | en | 0.827303 |
https://forensicreader.com/ridge-tracing-in-whorl-fingerprint/ | 1,713,821,890,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00785.warc.gz | 225,371,996 | 36,972 | # Ridge Tracing in Whorl Fingerprint: Rules, Classification, Calculation With Images
Ridge tracing in fingerprint is used to obtain the classification value for Whorl by drawing an imaginary line from the extreme left to right delta and then counting the intermediate ridges.
The techniques depend upon the establishment of the focal points of the deltas and then tracing an imaginary line between them.
## What is the Need for Ridge Tracing in Fingerprint?
Ridge tracing values are used to classify the fingerprints in simpler terms. As fingerprints are unique, it is nearly impossible to capture each whorl pattern and then match them one by one. This not only takes a lot of effort and time which is not an efficient way.
So, to classify whorl patterns in fingerprint systems such as Henry classification, these whorl patterns must be valued either by numbers or alphabets that are unique to a specific set of parameters in a pattern. And that’s what ridge tracing is needed for.
Specifically, in the Henry Classification system, ridge tracing values are used in the Major division and subsecondary division of fingerprints.
## Why Ridge Tracing is Only for Whorls?
Ridge Tracing is only for whorls, as it required two deltas to work on. In other words, ridge tracing is only made to calculate the values of Whorls.
Other major patterns such as arches and loops have their own marking parameters such as:
• For Arches: capital and small letters are used.
• For Loops: ridge counting is employed.
• Similarly, for Whorls: Ridge tracing is used in some classification systems.
## Rules of Ridge Tracing of Whorls Fingerprint Patterns
### Rule 1: Choosing Deltas
Ridge tracing always starts from the extreme left delta and ends at the extreme right delta or nearest point.
For patterns such as simple whorls, they only have two deltas, so the left one is considered as the starting point and the right one as the endpoint.
The following image represents some of the various whorl patterns and their respective left and right deltas.
However, for accidental/composite whorl, there might be more than two deltas, in that case, you have to choose the extreme left and right deltas. This is an example of accidental whorls with three deltas.
### Rule 2: When Ridges Abruptly End
While tracing ridges, when a ridge ends abruptly, tracing drops to the next lower ridge (immediately beneath the point where it ends) and continues.
Here are given image, Ridge (1) abruptly ends, so as per the rule ridge (2) is selected which ends after small curving, which then continued to ridge (3) then ridge (4), and finally continued to ridge (5) that is the nearest point to the right delta.
### Rule 3: When Delta Has a Dot Pattern
It follows the same rules of abruptly ending. That means the very next lower ridge is used for tracing and continues in the same fashion.
In the given image, you can see a dot pattern is present at the left delta, which passes the tracing to the next lower ridge which continues to the right delta.
### Rule 4: When Ridge Bifurcation Appears
A bifurcation can be represented as branched ‘Y’. A ridge divided into two and then continued.
So, if the fingerprint whorl pattern has a bifurcation, ridge tracing is continued through the lower branch of ridge bifurcation.
### Rule 5: Choosing Intervening Ridges
The intervening ridges should be counted in numbers between the traced imaginary line and the right delta. Here is an example of choosing the correct ridges.
## Classification of Ridge Tracing Patterns for Whorls
There are three main categories on which ridge tracing can be classified. These are:
1. Inner Whorl
2. Outer Whorl
3. Meeting Whorl
### 1. Inner Whorl
If the traced ridge goes above or reverse back inner before reaching to right delta and there are three or more interviewing ridges between the traced ridge and the right delta. It is considered as INNER (I).
An imaginary line recurving above the right delta doesn’t always call to be Inner. It must fulfill the second criterion i.e. 3 or more intervening ridges between deltas and traced ridges.
### 2. Outer Whorl
If the ridge tracing followed near the right delta’s relative point above or below with three or more intervening ridges between such delta and traced ridge. It is called an OUTER (O).
Similarly to Inner, not all whorls that end below the right delta are considered outer. It must have 3 or more intervening ridges between deltas and traced ridges.
### 3. Meeting Whorl
If the tracing of the ridge starts from the left delta and is followed to the right delta or nearest to it with one or two intervening ridges between the delta and traced ridge.
In other words, all other tracings that end, above or below the right delta and have less than 3 intermediate ridges, are always classified as Meet.
## How to Calculate the Ridge Tracing Values for Whorl Fingerprint Patterns?
### Step 1: Choosing Deltas
Choose the left and right delta points in a given whorl pattern.
### Step 2: Start Ridge Tracing from Left Delta
Ridge tracing starts from the left delta and proceeds toward the right delta.
### Step 3: Continue the Abrupt Ends
When the ridge ends abruptly, tracing drops to the next lower ridge and continues.
### Step 4: Extends the Abrupts Ends to Right Delta
When the ridge tracing reached up to the right delta, draw a perpendicular line from the tracing line to the right delta.
### Step 5: Count Intervening Ridges
From the perpendicular line, count intervening ridges between the right delta and the respected traced ridge line.
Based on the following table, assign the values.
As for this example, the ridge tracing value comes to be 4 outside (below) of the right delta. So the corresponding designation will be “Outer” and the value will be ‘O’
## Practice Worksheet For Ridge Tracing in Fingerprints
1. Choose the left delta and right delta points.
1. Ridge tracing starts from the left delta and continues towards the right delta.
1. Ridge tracing continues towards the right delta; There is ridge bifurcation so as per the rule ridge tracing continues to the lower branch of bifurcation.
1. After completion of ridge tracing, draw a perpendicular line from the right delta to the tracing line.
1. Count the intervening ridges which fall on a perpendicular line between the right delta and the respected traced ridges.
1. From counting, the values come to ‘1’ so “Meeting”.
## References:
• Fundamentals of Fingerprint Analysis, Second Edition By Hillary Moses Daluz [Book]
• Manual in Basic Finger Printing’ 2008 Ed. By P. Tubid [Book]
• Intelligent Biometric Techniques in Fingerprint and Face Recognition by I. Hayashi, et.al [Book]
• FBI Advanced Latent Fingerprint School By United States. Federal Bureau of Investigation [Books]
• FBI Law Enforcement Bulletin by FBI [Books]
Alpana Verma holds a master’s degree in forensics from Banaras Hindu University. She has also earned a forensic fingerprinting certification from the Central Forensic Science Laboratory, CBI. Currently, Alpana is a research intern in the entomology field, working within the Zoology department at BHU.
Alpana Verma
Co-Author, Fingerprint Expert | 1,580 | 7,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-18 | longest | en | 0.913637 |
https://www.termpaperwarehouse.com/essay-on/Micro-Computers-Quiz-1/365800 | 1,611,433,483,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538431.77/warc/CC-MAIN-20210123191721-20210123221721-00304.warc.gz | 982,812,593 | 16,333 | # Micro Computers Quiz 1
Submitted By mscott656
Words 318
Pages 2
Part 1 1. What are the Sales Amount by Channel?
SELECT DimChannel.ChannelName, Sum(FactSales.SalesAmount) AS SumOfSalesAmount
FROM DimChannel INNER JOIN FactSales ON DimChannel.ChannelKey = FactSales.channelKey
GROUP BY DimChannel.ChannelName; a. By Channel by Year
SELECT DimChannel.ChannelName, DimDate.FiscalYear, Sum(FactSales.SalesAmount) AS SumOfSalesAmount
FROM DimDate INNER JOIN (DimChannel INNER JOIN FactSales ON DimChannel.ChannelKey = FactSales.channelKey) ON DimDate.Datekey = FactSales.DateKey
GROUP BY DimChannel.ChannelName, DimDate.FiscalYear; b. Show the results with the best year at the top
SELECT DimChannel.ChannelName, DimDate.FiscalYear, Sum(FactSales.SalesAmount) AS SumOfSalesAmount
FROM DimDate INNER JOIN (DimChannel INNER JOIN FactSales ON DimChannel.ChannelKey = FactSales.channelKey) ON DimDate.Datekey = FactSales.DateKey
GROUP BY DimChannel.ChannelName, DimDate.FiscalYear
ORDER BY Sum(FactSales.SalesAmount) DESC; 2. What are the different types of Promotions?
SELECT DimPromotion.PromotionName
FROM DimPromotion; c. What are the start and end dates for each promotion. Put the oldest start dates first
SELECT DimPromotion.PromotionName, DimPromotion.StartDate, DimPromotion.EndDate
FROM DimPromotion
ORDER BY DimPromotion.StartDate; 3. What are the quantity and the amounts of items returned grouped by Channel then by Manufacturer?
SELECT DimChannel.ChannelName, DimProduct.Manufacturer, Sum(FactSales.ReturnQuantity) AS SumOfReturnQuantity, Sum(FactSales.ReturnAmount) AS SumOfReturnAmount
FROM DimProduct INNER JOIN (DimChannel INNER JOIN FactSales ON DimChannel.ChannelKey = FactSales.channelKey) ON DimProduct.ProductKey = FactSales.ProductKey
GROUP BY DimChannel.ChannelName, DimProduct.Manufacturer; 4. What is the average Weight of products sold grouped by Product Subcategory?
SELECT...
### Similar Documents
#### Economic
...OBJECTIVES VISION In pursuit of perfection, the University of Baguio is committed to provide balanced quality education by nurturing academic excellence, relevant social skills and ethical values in a fun-learning environment. MISSION The University of Baguio educates individuals to be empowered professionals in a global community. OBJECTIVES The University of Baguio aims to produce a graduate who: exemplifies a higher standard of learning; manifests the mastery of relevant skills; upholds a conduct that is rightful and just; undertakes scientific and significant researches; advocates sustainable programs for the community and the environment; and leads and demonstrates exemplary performance in the field of specialization. IV. V. VI. 1. 2. 3. 4. 5. 6. VII. SCHOOL MISSION AND OBJECTIVES SCHOOL OF DENTISTRY MISSION The University of Baguio educates individuals to be empowered professionals in a global community. The School of Dentistry is committed in providing excellent dental education to produce competitive and socially responsible dentists. OBJECTIVE The UB School of Dentistry provides a conducive and responsive learning environment which aims to produce an epitome of a...
Words: 1484 - Pages: 6
#### Essay
...available. Topics include contemporary issues such as inflation, unemployment, economic growth, international dependencies, and how public policy deals with them. | Instructor Contact Information: | Campus Location: | Prestonsburg | Building & Room: | Pike 215D | Office Hours: | N/A | Office Phone Number: | 1-888-641-4132 | Alternate Number: | 886-3863, Ext. 64781 | Best Times to Call: | The best place to communicate with me is in the message area of the blackboard course shell. | KCTCS Email: | Daniel.bell@kctcs.edu | Special Instructions: | N/A | Supervisor Contact Information: | Name: | Kelli Hall | Campus Location: | Pikeville | Building & Room: | N 102 | Office Phone Number: | 606-218-1275 or Ext 81275 | KCTCS Email: | Kelli.Hall@kctcs.edu | Text and Supplies: Macroeconomics, McConnell & Brue, Flynn 19th Edition – Copyright 2012 by McGraw-Hill-Irwin. | | Approved Course Competencies General Education: (KCTCS General Education Competency Statements and General Education Requirements) | I. Communicate Effectively | 1. Read and listen with comprehension. 2. Speak and write clearly using Standard English. 3. Interact cooperatively with others using both verbal and non-verbal means....
Words: 2128 - Pages: 9
Free Essay
#### Paper
...Department Head: Kim Breivogel 919-866-5203 or kbbreivogel@waketech.edu Course Delivery/Location: Delivery of this online course will take place through Blackboard. Students must log in and check Blackboard regularly for class policies, resources, assignments, and essential communications from your instructor. Blackboard can be accessed at https://dist-ed.waketech.edu/webapps/ login/. Wake Tech email (mywaketech.edu) will also be used and must be checked regularly. **This syllabus is subject to change. IMPORTANT DATES : All online students are expected to enter this class on the first day of class : Wednesday, September 14, 2011. Week 1 Assignments must be completed by midnight on Tuesday, September 20, 2011. 10% date : Wednesday, September 21, 2011 If a student fails to complete the assignments for Week 1 or does not communicate with the instructor by September 21, 2011, he or she will be dropped from the course by the instructor. Withdrawing with a W grade : Tuesday, November 8, 2011 The last day to drop this course with a non-punitive grade of W is Tuesday, November 8, 2011. If you wish to drop this course, you may do so yourself, or I can submit an online withdrawal form for you. Course Description: This course introduces the scientific study of human society, culture and social interactions. Topics include...
Words: 5763 - Pages: 24
Free Essay
#### Jkgkjk
... |Office Phone: (305) 348-3466 | |12:30-1:30 p.m. and 3:30-4:00 p.m. | | |Office: DM 281A |E-mail: Use Moodle course mail ONLY! | | |FIU e-mail: shpurikm@fiu.edu | |Teaching Assistant (TA): Jesse Slappey |TA Email: Use Moodle course mail ONLY! | |PSY 2012 Training and Development Center: Room DM 284, phone: 305-348-6892 | |Office hours: Mondays 10 a.m. - 2 p.m. (TA Ben) | |Tuesdays 3.30 - 6.30 p.m. (TA Ben) | |Wednesdays 10.30 a.m. - 1.30 p.m. (TA Brittany) | |Thursdays 8.30 a.m. - 3.30 p.m. (TA Jesse) | |Fridays 10 a.m. - 1 p.m. (TA Ben)...
Words: 4292 - Pages: 18
#### Economics
...Intermediate Macroeconomics Semester 1, 2015 Unit of Study Outline Unit Coordinator: Dr. Stella Huangfu Location: Room 339, Merewether Building Email address: stella.huangfu@sydney.edu.au Phone: +61-2-9036 9311 Consultation Hours: Thursdays 2-4 pm or other times by appointment. Tutors: Email address: Consultation Hours: Class times: Thursday 11am-1pm Venue: Bosch Lecture Theatre 2 This Unit of Study Outline MUST be read in conjunction with the Faculty of Arts and Social Sciences Student Administration Manual (http://sydney.edu.au/arts/current_students/student_admin_manual.shtml) and all applicable University policies. In determining applications and appeals, it will be assumed that every student has taken the time to familiarise themselves with these key policies and procedures. [ECOS2002 Intermediate Macroeconomics] UNIT DESCRIPTION This unit of study develops models of the goods, money and labour markets, examines issues in macroeconomic policy. Macroeconomic relationships, covering consumption, investment, money and employment, are explored in detail. Macro-dynamic relationships, especially those linking inflation and unemployment, are also considered. Exchange rates and open economy macroeconomics are also addressed. In the last part of the unit, topics include the determinants and theories of economic growth, productivity and technology, the dynamics of the business cycle, counter-cyclical policy and the relationship between micro and macro policy in......
Words: 3541 - Pages: 15
Free Essay
#### Technology
...VOCATIONAL HIGHER SECONDARY TEACHER’S SOURCE BOOK COMPUTER SCIENCE FIRST YEAR Government of Kerala Department of Education SCERT - 2005-06 STATE COUNCIL OF EDUCATIONAL RESEARCH AND TRAINING Vidyabhavan, Poojappura, Thiruvananthapuram-12 Prepared by: State Council of Educational Research & Training (SCERT) Vidyabhavan, Poojappura, Thiruvananthapuram -12 Kerala E-mail:scertkerala@asianetindia.com Type setting by: SCERT Computer Lab. © Government of Kerala Education Department 2005 P REFACE Dear Teachers, Computer Science is concerned with the gathering, manipulation, classification, storage and retrival of knowledge. Understanding Computer Science, is necessary because of its power and influence in modern society. Learning of Computer Science should be activity based, process oriented, student-centred, environmental based and life oriented. The approach to learning is based on five domains of science: Knowledge domain, process domain, application and connection domain, domain of attitudes and values and creativity domain. But we know that most of the present vocational higher secondary teachers are not familiar with this paradigm. Hence for the first time we are introducing sourcebooks for all the subjects in the vocational higher secondary curriculum. This source book for computer science aims to provide guidelines to the teachers of our state to change their pedagogy from the conventional content-based approach to the process...
Words: 27034 - Pages: 109
Free Essay
#### Abcd
...2015-17 Semester II Course Outline 1 Shanti Business School, Ahmedabad PGDM 2015-17: Program Structure Semester-1 Semester-2 Semester-3 Semester-4 Course credit Course credit Course credit Language Skills @ Written Analysis & Communication @ Soft skills II @ Employability Skills @ IT & MIS 2 Soft skills I @ Computing skills 2 Social Media Marketing @ 2 Legal Aspects of Business 2 Business Strategy 3 Management Control Systems 3 Micro Economics 3 Macro Economics 3 Business Environment 3 Business Ethics & Corporate Governance 2 Quantitative Methods-1 3 Business Research Methods 3 Quantitative Methods-2 3 Core Elective-1 3 Core Elective1 3 Core Elective-2 3 Core Elective2 3 Elective-1 3 Elective-1 3 Elective-2 3 Elective-2 3 Grand Project-1 3 Grand Project-2 3 Principles of Management Basic Building Blocks Autumn Break Executive Skills Organisational Behavior Human Resources Management 3 Marketing Management 1 3 Marketing Management -2 3 Understanding Financial Statements 3 Financial Mgt 3 Operation Management Management Domain 3 3 Basics of Business Planning 2 Electives Credits Autumn Break credit SUMMER INTERNSHIP Course S 1 22 S......
Words: 7010 - Pages: 29
Free Essay
#### Innovation and Erp Systems
...Scheme and Syllabus of B.E. (Computer Science and Engineering) 3 rd th TO 8 Semester 2014-2015 University Institute of Engineering and Technology, Panjab University, Chandigarh DEPARTMENT: COMPUTER SCIENCE AND ENGINEERING VISION: To be recognized as an international leader in Computer Science and Engineering education and research to benefit society globally. MISSION: · · · · To move forward as frontiers of human knowledge to enrich the citizen, the nation, and the world. To excel in research and innovation that discovers new knowledge and enables new technologies and systems. To develop technocrats, entrepreneurs, and business leaders of future who will strive to improve the quality of human life. To create world class computing infrastructure for the enhancement of technical knowledge in field of Computer Science and Engineering. PROGRAMME: B.E. CSE (UG PROGRAMME) PROGRAMME EDUCATIONAL OBJECTIVES: I. Graduates will work as software professional in industry of repute. II. Graduates will pursue higher studies and research in engineering and management disciplines. III. Graduates will work as entrepreneurs by establishing startups to take up projects for societal and environmental cause. PROGRAMME OUTCOMES: A. Ability to effectively apply knowledge of computing, applied sciences and mathematics to computer science & engineering problems. B. Identify, formulate, research literature, and analyze complex computer science & engineering problems reaching substantiated......
Words: 23989 - Pages: 96
Free Essay
#### Ink Out of Tea Bags
...CHAPTER 2 This chapter includes the review of related literature, review of related studies and review of other related Computer Assisted Instruction (CAI) which helped the researchers in designing and developing their own CAI that was used in this research. REVIEW OF RELATED LITERATURE The review includes discussion of technical terms like Computer-Assisted Instruction (CAI), multimedia, and review of software packages in mathematics, studies on development, evaluation and validation of CAI materials. Instruction and Learning Instruction is the arrangement of information and environment to facilitate learning. By environment, we mean not only where instruction takes place but also the methods, media and equipment needed to convey information and guide the learner’s study. Information and environment can be arranged on a continuum from very formal to very informal. The relationship between information and environment can change depending on the instructional goal. The arrangement of information and the environment is normally the responsibility of the instructor and the designers of media. The choice of the strategy of instruction determines the environment (the methods, media, equipment and facilities) and how the information is assembled and used. The method can range from teacher control to learner control. Learning is the development of new knowledge, skills, or attitudes when the individual interacts with information and environment. Learning takes......
Words: 15415 - Pages: 62
Free Essay
#### B.Tech (Eee) Syllabus for 2012-13 Admitted
...GITAM UNIVERSTIY (Declared as Deemed to be University U/S 3 of UGC Act, 1956) [pic] REGULATIONS & SYLLABUS Of B.Tech. (Electrical & Electronics & Engineering) (W.e.f 2012-13 admitted batch) Gandhi Nagar Campus, Rushikonda VISAKHAPATNAM – 530 045 Website: www.gitam.edu REGULATIONS (W.e.f. 2012-013 admitted batch) 1. ADMISSIONS 1. Admissions into B.Tech (Electrical & Electronics Engineering) programme of GITAM University are governed by GITAM University admission regulations. 2. ELIGIBILTY CRITERIA 1. A pass in 10+2 or equivalent examination approved by GITAM University with Physics, Chemistry and Mathematics. 2. Admissions into B.Tech will be based on an All India Entrance Test (GAT) conducted by GITAM University and the rule of reservation, wherever applicable. 3. STRUCTURE OF THE B.Tech. PROGRAMME 3.1 The Programme of instruction consists of: i) A general core programme comprising Basic Sciences, Basic Engineering, Humanities & Social Sciences and Mathematics. ii) An engineering core programme imparting to the student the fundamentals of engineering in the branch concerned. iii) An elective programme enabling the students to take up a group of departmental / interdepartmental courses of interest to him/her. In addition, a student has to i)...
Words: 3379 - Pages: 14
#### Celluar
...Western International University IT-330-6940-Emerging Technologies Instructor Parikh, Raj By Shanon Ellis June 18, 2012 In recent years, with the demand for communication and information, the number of people using mobile phones has developed rapidly. People are more and more dependable on these small devices. It is clear that cell phone have benefited to every user in many sides of their lives. First of all, mobile phones, as a communication tool, connect people around the world. Many years ago, it was very difficult to contact with your friends, your partners immediately. Business man had to use their telephone in their office. When they went out, it was impossible to keep in touch with their business partners. That has led to the necessary to have mobile phone. Therefore, the mobility of owing a mobile phone has encouraged using this device. The number of people having mobile phone made a boom in last ten years. Another benefit of mobile phones is to help people in entertainment and supporting material for learning. These mobile phones have a lot of functions. Furthermore, smart phones have its own operating system such as Symbian, window mobile, android, or IOS. Users can set up applications to help us study such as dictionary. It is also an assistant for business people. These people schedule their works on this smart phones, it will remind them for that works. In some cases, with the GPS function, phone will guide you if you are in the new...
Words: 2223 - Pages: 9
Free Essay
#### Communications Training Module
...Personal Safety: What are they thinking? Training Session By: Edward J. Hamilton Business Communication PRCM325J Ohio University 2010 Table of Contents Overview 3 Behavioral Objectives 4 Outline of Training Session 5 Literature Review: What are they thinking? 7 Bibliography- Literature Review 10 Attachment 1: Study Abroad Handbook: Risk Factors and Strategies to Reduce Risk 12 Attachment 2: Proxemic circle 15 Attachment 3: Multi-Cultural meanings of Paralanguage Gestures 16 Personal Safety: What are they thinking? Overview Criminals are bolder and have less regard for their victims than ever before. Places that once seemed safe are no longer immune from danger. The death spiral of the economy plays a major factor in the increase of crime. There are two parties involved in the commission of crimes, to both persons and property; those two parties are the potential victim and perpetrator. Within each of those parties at the time of a commission of a crime, they possess an innate sense of non-verbal and verbal communication that engage, takes over and governs each party’s behavior throughout the event. For each respective party in a personal contact situation, that sense comprises of a series of cues which could take the form of voice intonation, body language, strength in numbers, and the surprise of direct violence. In some cases, the exchange can be analogous to a chess game in hyper drive. In other cases, the exchange may not be......
Words: 7335 - Pages: 30
#### Gujarat Technological University Syllabus for New Mba Program Effective from Academic Year 2011-12 Mba I Semester I
...Gujarat Technological University Syllabus for New MBA Program effective from Academic Year 2011-12 MBA I Semester I Accounting for Managers (AFM) 1. Course Objective: The objective of the course is to acquaint the students with the language of Accounting and to develop in them the ability to evaluate and use accounting data as an aid to decision making. The main purpose is to assist the students in developing skills in problem solving and decision making in the financial area. Emphasis is laid on analysis and utilization of financial and accounting data for planning and control. 2. Course Duration: The course duration is of 36 sessions of 75 minutes each i.e. 45 hours. 3. Course Contents: Module No: Module Content No. of Sessions 70 Marks (External Evaluation) 17 I II Fundamentals of Accounting Basic understanding of accounting, Accounting Concepts, Conceptual framework of financial statements, Accounting Policies, Journal Entries and preparation of accounts - Trial Balance to Balance sheet and profit and loss Account, Recognition of Income and Expenses, Provisions, Contingent Liabilities Accounting Standards and Applicability: Disclosure of Accounting Policies (AS-1), Valuation of Inventories (AS-2), Depreciation Accounting (AS-6), Income Recognition & Accrual Income (AS-9), Accounting of Fixed Assets (AS-10), Accounting for Intangible Assets (AS-26), Accounting for Investments (AS-13), 7 7 17 III IV V Preparing and Understanding Financial......
Words: 5698 - Pages: 23
#### Test
...Post Graduate Program in Management (PGPM) Faculty Name | Prof. S.K. Shanthi | Year/Term | 2014-15/Term-1 | Course Name | Micro-economics | No. of Credits | 3 | No of Contact Hours | 20 | Session Duration | 90 Min. | About the Instructor: Prof. S.K. Shanthi shanthi.sk@greatlakes.edu.in Course Objectives and Key Take Aways This course is designed to introduce students to basic microeconomic theory at a relatively rapid pace. The focus will be on fundamental economic principles that can be used by managers to think about business problems, including those inside the firm (relating to the problem of administering the firm’s resources in the most cost-effective way) and those outside the firm (relating to broader market and government forces that can affect the behavior of firms). In particular, it addresses how businesses make choices concerning the use of scarce resources, how prices and incomes are determined in competitive markets, and how market power affects the prices and quantities of goods available to society. We will also examine the impact of government intervention in the economy. At the end of the course the student should be able to: 1. Explain price and output changes in markets using supply and demand. 2. Analyse the usefulness of the concept of elasticity of demand in pricing, market segmentation decisions. 3. Find the profit maximizing level of various firm activities using marginal analysis. 4.......
Words: 2234 - Pages: 9 | 4,756 | 21,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-04 | latest | en | 0.485086 |
https://grandcoder.in/staircase-in-java/ | 1,674,779,703,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494852.95/warc/CC-MAIN-20230127001911-20230127031911-00414.warc.gz | 326,987,091 | 55,788 | ## Staircase in java
In this question you need to write a program that prints a staircase of size ‘n’.
This is a staircase of size n=4:
``` #
##
###
####
```
Its base and height are both equal to n. It is drawn using # symbols and spaces. The last line is not preceded by any spaces.
Write a program that prints a staircase of size n.
#### Input
A single integer, n, denoting the size of the staircase.
#### Output
Print a staircase of size n using # symbols and spaces.
Note: The last line must not have spaces in it.
6
#### Sample Output
``` #
##
###
####
#####
######
```
#### Explanation
The staircase is right-aligned, composed of # symbols and spaces, and has a height and width of n=6. You might notice, first line has 5 spaces (n-1) followed by 1 # and the last line has 0 spaces (n-n) and 6 # (n)
## Solution:–
```import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner scanner = new Scanner(System.in);
int x= scanner.nextInt();
int row=x;
for(int i=1;i<=row;i++)
{
for(int j=1;j<=row-i;j++)
{
System.out.print(" ");
}
for(int k=1;k<=i;k++)
{
System.out.print("#");
}
System.out.println();
}
}
}``` | 319 | 1,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2023-06 | latest | en | 0.710035 |
https://www.jobilize.com/online/course/6-2-line-integrals-vector-calculus-by-openstax?qcr=www.quizover.com&page=5 | 1,569,011,226,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574077.39/warc/CC-MAIN-20190920200607-20190920222607-00297.warc.gz | 895,749,546 | 23,574 | 6.2 Line integrals (Page 6/20)
Page 6 / 20
Reversing orientation
Find the value of integral ${\int }_{C}\text{F}·d\text{r},$ where $C$ is the semicircle parameterized by $\text{r}\left(t\right)=⟨\text{cos}\phantom{\rule{0.2em}{0ex}}t+\pi ,\text{sin}\phantom{\rule{0.2em}{0ex}}t⟩,0\le t\le \pi$ and $\text{F}=⟨\text{−}y,x⟩.$
Notice that this is the same problem as [link] , except the orientation of the curve has been traversed. In this example, the parameterization starts at $\text{r}\left(0\right)=⟨\pi ,0⟩$ and ends at $\text{r}\left(\pi \right)=⟨0,0⟩.$ By [link] ,
$\begin{array}{cc}\hfill {\int }_{C}\text{F}·d\text{r}& ={\int }_{0}^{\pi }⟨\text{−}\text{sin}\phantom{\rule{0.2em}{0ex}}t,\text{cos}\phantom{\rule{0.2em}{0ex}}t+\pi ⟩·⟨\text{−}\text{sin}\phantom{\rule{0.2em}{0ex}}t+\pi ,\text{cos}\phantom{\rule{0.2em}{0ex}}t⟩dt\hfill \\ & ={\int }_{0}^{\pi }⟨\text{−}\text{sin}\phantom{\rule{0.2em}{0ex}}t,\text{−}\text{cos}\phantom{\rule{0.2em}{0ex}}t⟩·⟨\text{sin}\phantom{\rule{0.2em}{0ex}}t,\text{cos}\phantom{\rule{0.2em}{0ex}}t⟩dt\hfill \\ & ={\int }_{0}^{\pi }\left(\text{−}{\text{sin}}^{2}t-{\text{cos}}^{2}t\right)dt\hfill \\ & ={\int }_{0}^{\pi }-1dt\hfill \\ & =\text{−}\pi .\hfill \end{array}$
Notice that this is the negative of the answer in [link] . It makes sense that this answer is negative because the orientation of the curve goes against the “flow” of the vector field.
Let C be an oriented curve and let − C denote the same curve but with the orientation reversed. Then, the previous two examples illustrate the following fact:
${\int }_{C}\text{F}·d\text{r}=\text{−}{\int }_{C}\text{F}·d\text{r}.$
That is, reversing the orientation of a curve changes the sign of a line integral.
Let $\text{F}=x\text{i}+y\text{j}$ be a vector field and let C be the curve with parameterization $⟨t,{t}^{2}⟩$ for $0\le t\le 2.$ Which is greater: ${\int }_{C}\text{F}·\text{T}ds$ or ${\int }_{\text{−}C}\text{F}·\text{T}ds?$
${\int }_{C}\text{F}·\text{T}ds$
Another standard notation for integral ${\int }_{C}\text{F}·d\text{r}$ is ${\int }_{C}Pdx+Qdy+Rdz.$ In this notation, P , Q , and R are functions, and we think of d r as vector $⟨dx,dy,dz⟩.$ To justify this convention, recall that $d\text{r}=\text{T}ds={r}^{\prime }\left(t\right)dt=⟨\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}⟩dt.$ Therefore,
$\text{F}·d\text{r}=⟨P,Q,R⟩·⟨dx,dy,dz⟩=Pdx+Qdy+Rdz.$
If $d\text{r}=⟨dx,dy,dz⟩,$ then $\frac{d\text{r}}{dt}=⟨\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}⟩,$ which implies that $\frac{d\text{r}}{dt}=⟨\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}⟩dt.$ Therefore
$\begin{array}{cc}\hfill {\int }_{C}\text{F}·d\text{r}& ={\int }_{C}Pdx+Qdy+Rdz\hfill \\ & =\left(P\left(\text{r}\left(t\right)\right)\frac{dx}{dt}+Q\left(\text{r}\left(t\right)\right)\frac{dy}{dt}+R\left(\text{r}\left(t\right)\right)\frac{dz}{dt}\right)dt.\hfill \end{array}$
Finding the value of an integral of the form ${\int }_{C}Pdx+Qdy+Rdz$
Find the value of integral ${\int }_{C}zdx+xdy+ydz,$ where C is the curve parameterized by $\text{r}\left(t\right)=⟨{t}^{2},\sqrt{t},t⟩,1\le t\le 4.$
As with our previous examples, to compute this line integral we should perform a change of variables to write everything in terms of t . In this case, [link] allows us to make this change:
$\begin{array}{cc}\hfill {\int }_{C}zdx+xdy+ydz& ={\int }_{1}^{4}\left(t\left(2t\right)+{t}^{2}\left(\frac{1}{2\sqrt{t}}\right)+\sqrt{t}\right)dt\hfill \\ & ={\int }_{1}^{4}\left(2{t}^{2}+\frac{{t}^{3\text{/}2}}{2}+\sqrt{t}\right)dt\hfill \\ & ={\left[\frac{2{t}^{3}}{3}+\frac{{t}^{5\text{/}2}}{5}+\frac{2{t}^{3\text{/}2}}{3}\right]}_{t=1}^{t=4}\hfill \\ & =\frac{793}{15}.\hfill \end{array}$
Find the value of ${\int }_{C}4xdx+zdy+4{y}^{2}dz,$ where $C$ is the curve parameterized by $\text{r}\left(t\right)=⟨4\phantom{\rule{0.2em}{0ex}}\text{cos}\left(2t\right),2\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2t\right),3⟩,0\le t\le \frac{\pi }{4}.$
$-26$
We have learned how to integrate smooth oriented curves. Now, suppose that C is an oriented curve that is not smooth, but can be written as the union of finitely many smooth curves. In this case, we say that C is a piecewise smooth curve . To be precise, curve C is piecewise smooth if C can be written as a union of n smooth curves ${C}_{1},{C}_{2}\text{,…},{C}_{n}$ such that the endpoint of ${C}_{i}$ is the starting point of ${C}_{i+1}$ ( [link] ). When curves ${C}_{i}$ satisfy the condition that the endpoint of ${C}_{i}$ is the starting point of ${C}_{i+1},$ we write their union as ${C}_{1}+{C}_{2}+\cdots +{C}_{n}.$
The next theorem summarizes several key properties of vector line integrals.
Properties of vector line integrals
Let F and G be continuous vector fields with domains that include the oriented smooth curve C . Then
1. ${\int }_{C}\left(\text{F}+\text{G}\right)·d\text{r}={\int }_{C}\text{F}·d\text{r}+{\int }_{C}\text{G}·d\text{r}$
2. ${\int }_{C}k\text{F}·d\text{r}=k{\int }_{C}\text{F}·d\text{r},$ where k is a constant
3. ${\int }_{C}\text{F}·d\text{r}={\int }_{\text{−}C}\text{F}·d\text{r}$
4. Suppose instead that C is a piecewise smooth curve in the domains of F and G , where $C={C}_{1}+{C}_{2}+\cdots +{C}_{n}$ and ${C}_{1},{C}_{2}\text{,…},{C}_{n}$ are smooth curves such that the endpoint of ${C}_{i}$ is the starting point of ${C}_{i+1}.$ Then
${\int }_{C}\text{F}·d\text{s}={\int }_{{C}_{1}}\text{F}·d\text{s}+{\int }_{{C}_{2}}\text{F}·d\text{s}+\cdots +{\int }_{{C}_{n}}\text{F}·d\text{s}.$
Notice the similarities between these items and the properties of single-variable integrals. Properties i. and ii. say that line integrals are linear, which is true of single-variable integrals as well. Property iii. says that reversing the orientation of a curve changes the sign of the integral. If we think of the integral as computing the work done on a particle traveling along C , then this makes sense. If the particle moves backward rather than forward, then the value of the work done has the opposite sign. This is analogous to the equation ${\int }_{a}^{b}f\left(x\right)dx=\text{−}{\int }_{a}^{b}f\left(x\right)dx.$ Finally, if $\left[{a}_{1},{a}_{2}\right],\left[{a}_{2},{a}_{3}\right]\text{,…},\left[{a}_{n-1},{a}_{n}\right]$ are intervals, then
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers! | 3,135 | 9,416 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 49, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-39 | longest | en | 0.654953 |
http://www.jiskha.com/display.cgi?id=1272412997 | 1,495,751,117,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608617.6/warc/CC-MAIN-20170525214603-20170525234603-00313.warc.gz | 524,602,313 | 4,035 | # Chemistry II
posted by on .
What am I doing wrong? I am trying to balance the equation:
NO3^-(aq)+Sn^2+(aq)-->Sn^4+(aq)+NO(g)
here are the coefficients I have. what is wrong with them?
2NO3^-(aq)+3Sn^2+(aq)+8OH-(aq)-->3Sn^4+(aq)+2NO(g)+4H2O(l)
• Chemistry II - ,
What's wrong or what are the correct coefficients?
1. O doesn't balance--14 O on left; 6 on right.
2. Charge doesn't balance. 4- on left and 12+ on right.
I would balance it this way. I assume it is an acid solution.
8H^+ + 2NO3^- + 3Sn^+2 ==> 3Sn^+4 + 2NO + 4H2O
atoms:
1. 8 H on left and right.
2. 2 N on left and right.
3. 6 O on left and right.
4. 3 Sn on left and right.
charge:
12+ on left and right.
electron change:
N = +10 - +4 = +6
Sn = +6 - 12 = -6 | 271 | 733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-22 | latest | en | 0.91034 |
https://ez.analog.com/ez-blogs/b/engineerzone-spotlight/posts/garbled-circuits-part-2-a-general-solution | 1,722,940,803,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640484318.27/warc/CC-MAIN-20240806095414-20240806125414-00113.warc.gz | 190,034,283 | 35,789 | # Garbled Circuits Part 2: A General Solution
In Part 1 of our Garbled Circuits series, we found a problem-specific solution for rescuing our shipwrecked friends. However, the solution didn't provide a general approach for computing functions privately (without revealing their inputs). One way to construct a general solution for evaluating a particular function f(x) is to design a circuit that maps the possible inputs x to the possible outputs f(x). For example, consider a NAND gate:
The gate maps all possible values of its input wires ({0,1}, {0,1}) to a value of its output wire {0,1}. However, in order to find the value of the output wire, the evaluator must know the values of the input wires. Our target problems require that the input wires remain private, so we need to modify this approach.
Garbled Circuits
The general solution was given in 1986 by Turing Award laureate Andrew Yao [1]. Incredibly, Yao proved that any polynomial-time function can be computed securely (without disclosing a player’s inputs) in polynomial time by “garbling” regular circuits. In this introduction, we will consider the simplest case where there are only two participants, Alice and Bob. Each has a private input bit that should not be disclosed to the other, and each would like to learn the result of NAND(Alice Input, Bob Input). Since any function can be constructed from NAND gates, it is sufficient to show how to garble only this gate. We will have Alice generate (construct) the garbled circuit, and Bob will evaluate the garbled circuit to recover the result.
Generator Steps (Alice)
The first step for the generator is to replace the wire inputs {0,1} with an independent and identically distributed (i.i.d.) random value K. These random values will be used as encryption keys for a symmetric cipher, such as AES. In our notation, the binary value {0,1} that K maps to is the superscript, while the input wire {1,2} that K corresponds to is the subscript. In our example, Alice will provide the input to Wire 1, and Bob will provide the input to Wire 2.
Next, the values of the output wire are encrypted using the key K corresponding to the appropriate input values:
Applying this to the NAND gate in our running example, the Generator produces this garbled circuit:
As Alice (Wire 1) knows her input bit b, she simply removes her other key corresponding to 1-b. However, how will Alice send Bob the key corresponding to his input bit?
There are issues with the obvious solutions:
• If Bob asks Alice for the key corresponding to his bit b, then he has revealed his private input.
• If Alice sends Bob both keys for b and 1-b, then Bob can evaluate f(x) on two inputs, rather than just one. This reveals additional information, including possibly Alice’s private input.
To understand why sending both keys reveals additional information, consider an example where Alice’s input bit is 0 and Bob’s input bit is 0. The output of NAND(0,0) is 1. If Bob only knows that his input bit is 0 and the result is 0, then Alice’s input bit could be either 0 or 1. However, if Bob was given the ability to evaluate the gate on both 0 and 1, he would discover NAND(A,0)=1 and NAND(A,1)=1, which reveals that Alice’s input bit must be 0. This is an unnecessary disclosure of Alice’s private input bit.
Since Bob can’t ask for his input key, and Alice can’t give him both possible keys, we need a solution where Bob receives only the key for his input bit, and Alice doesn’t know which one she sent to Bob. Impossible? We'll find out in Part 3 of our #garbledcircuits blog series!
1. Yao, A. Protocols for Secure Computations. In 27th Annual Symposium on Foundations of Computer Science, 1986, pages 162-167. IEEE, 1986. | 857 | 3,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-33 | latest | en | 0.888263 |
http://forums.devx.com/showthread.php?171461-Graph-Traversal-Breadth-First-Seach-and-Depth-First-Search&p=518116 | 1,524,319,482,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945222.55/warc/CC-MAIN-20180421125711-20180421145711-00378.warc.gz | 121,278,332 | 21,344 | Graph Traversal: Breadth First Seach and Depth First Search
DevX Home Today's Headlines Articles Archive Tip Bank Forums
# Thread: Graph Traversal: Breadth First Seach and Depth First Search
1. Registered User
Join Date
Mar 2009
Posts
24
## Graph Traversal: Breadth First Seach and Depth First Search
I don't understand BFS and DFS. How do I visit the vertex and mark it as visited? Not sure how to do DFS with an arraylist of iterators. Need to some how iterate through the adjacency list for each vertex.
Code:
```public class Graph
{
// translate integer vertex to string name
private final List<String> vertexName = new ArrayList<String>();
// translate name to integer form of vertex
private final Map<String, Integer> nameToVertexMap = new HashMap<String, Integer>();
// represent adjacency list of the graph
private final List<List<Integer>> adjacencyList = new ArrayList<List<Integer>>();
// Reads data from a scanner and builds the graph.
public void buildGraph(Scanner sc)
{
while (sc.hasNext())
{
// get data for the adjacency list of one vertex
String initVertexName = sc.next();
int initVertex = this.getVertexForName(initVertexName);
if (initVertex == -1)
List<Integer> adjList = this.getNeighbors(initVertex);
// get the outdegree
int outDegree = sc.nextInt();
// collect list of neighbors for this vertex
for (int k = 0; k < outDegree; k++)
{
// get the string name of the terminal vertex for an edge
String finalVertexName = sc.next();
//get the vertex number for this vertex name
int finalVertex = this.getVertexForName(finalVertexName);
// if the vertex name has no number, get it a number
if (finalVertex == -1)
}
}
}
// string representation of the adjacency list of the graph
@Override
public String toString()
{
StringBuilder sb = new StringBuilder();
for (int v = 0; v < this.adjacencyList.size(); v++)
{
sb.append(this.vertexName.get(v) + " : ");
for (Integer i : adjL)
sb.append(vertexName.get(i) + " ");
sb.append("\n");
}
return sb.toString();
}
/**
* @param vertex: integer representation of a vertex.
* @return the adjacency list for a specified vertex
*/
public List<Integer> getNeighbors(int vertex)
{
}
// returns the number of vertices in this graph
public int getNumberOfVertices()
{
}
// returns string name of the vertex to use for output.
public String getVertexName(int vertex)
{
return vertexName.get(vertex);
}
/**
* @param name: name of a vertex, read from an input stream or scanner
* @return the integer (vertex) assigned to that name, or -1 if no integer
* vertex has been assigned to that name.
*/
public int getVertexForName(String name)
{
return vertexName.indexOf(name);
}
/**
* Adds a new vertex with a given name to the graph and associates
* the string name of the vertex with an integer used to identify the vertex.
* @param name: string name for a vertex
* @return an integer value that will be used to identify the vertex.
*/
private int addNewAssociation(String name)
{
int vertex = getVertexForName(name);
if (vertex != -1) throw new IllegalStateException("There is already a " +
"vector with name " + name);
// The integer associated with the name is its index in the list vertexName
vertex = vertexName.size() - 1;
nameToVertexMap.put(name, vertex);
return vertex;
}
}```
Code:
```public class GraphTraversal
{
/**
* This version of BFS will stop as soon as it finds a path from the source
* to the destination.
* @param source: the vertex where the search should begin.
* @param dest: the vertex where the search should end.
* @param graph: the graph to be searched.
* @param visited: a boolean array used to mark vertices when they are
* visited.
* @param pred: an integer array that assigns to each vertex its predecessor
* in the path that is found.
*/
static void breadthFirstSearch(int source, int dest, Graph graph,
boolean [] visited, int [] pred)
{
visited = new boolean [graph.getNumberOfVertices()];
pred = new int [graph.getNumberOfVertices()];
Queue<Integer> q = new LinkedList<Integer>();
// visit source
// mark source
q.offer(source);
while(!q.isEmpty())
{
int f = q.peek();
List<Integer> w = graph.getNeighbors(f);
for(int i : w)
{
// visit w
// mark w
}
}
q.remove();
}
// This version of DFS will stop as soon as it finds a path from the source to the destination.
static void depthFirstSearch(int source, int dest, Graph graph,
boolean [] visited, int [] pred)
{
ArrayList<Iterator> i = new ArrayList<Iterator>();
}
/**
* breadthFirst performs a BFS of a graph.
* @param source: the initial vertex for the search
* @param graph: This is the graph to be traversed
* @param visited: this array indicates which vertices have been visited.
* BFS will set visited[v] = for only those vertices that were
* visited. Here v is the index that identifies a vertex in
* the graph.
* @param pred: for each vertex v that is visited, pred[v] will be set to its
* predecessor, that is, the vertex that led to v in the BFS.
*/
static void breadthFirstSearch(int source, Graph graph, boolean [] visited,
int [] pred)
{
breadthFirstSearch(source, graph.getNumberOfVertices(), graph, visited,
pred);
}
static void breadthFirstSearch(String s_source, Graph graph,
boolean [] visited, int [] pred)
{
int source = graph.getVertexForName(s_source);
breadthFirstSearch(source, graph.getNumberOfVertices(), graph, visited,
pred);
}
static void breadthFirstSearch(String s_source, String s_dest, Graph graph,
boolean [] visited, int [] pred)
{
int source = graph.getVertexForName(s_source);
int dest = graph.getVertexForName(s_dest);
breadthFirstSearch(source, dest, graph, visited, pred);
}
static void depthFirstSearch(String source, Graph graph, boolean[] visited,
int [] pred)
{
int sourceVertex = graph.getVertexForName(source);
depthFirstSearch(sourceVertex, graph.getNumberOfVertices(), graph,
visited, pred);
}
static void depthFirstSearch(int source, Graph graph, boolean [] visited,
int [] pred)
{
depthFirstSearch(source, graph.getNumberOfVertices(), graph, visited,
pred);
}
static void depthFirstSearch(String source, String dest, Graph graph,
boolean [] visited, int [] pred)
{
int sourceVertex = graph.getVertexForName(source);
int destVertex = graph.getVertexForName(dest);
depthFirstSearch(sourceVertex, destVertex, graph, visited, pred);
}
/**
* getPath finds the path that goes from a given start vertex to a
* given destination vertex
* @param source: the start vertex for a path.
* @param dest: the end vertex for a path.
* @param pred: an array that specifies the predecessor of each vertex in
* a path. If a vertex has no predecessor in the path, its
* predecessor is null.
* @return a list of vertices that form a path from a source vertex to
* a destination vertex.
*/
public static List<Integer> getPath(int source, int dest, int [] pred)
{
List<Integer> path = new LinkedList<Integer>();
int current = dest;
while (current != source)
{
current = pred[current];
if (current == -1)
{
path.clear();
return path;
}
}
return path;
}
/**
* getPath finds a path from a given start vertex to a given destination vertex.
* It allows the vertices to be specified by name instead of by numeric values.
* @return a list of vertex names for the vertices on the path.
*/
static List<String> getPath(String source, String dest, int [] pred, Graph graph)
{
int sourceVertex = graph.getVertexForName(source);
int destVertex = graph.getVertexForName(dest);
List<Integer> path = getPath(sourceVertex, destVertex, pred);
List<String> sPath = new ArrayList<String>();
for (int i : path)
return sPath;
}
}```
2. Senior Member
Join Date
Dec 2004
Location
San Bernardino County, California
Posts
1,468
have you thought of the vertices as objects of a class? if this were so, what could you do to mark an object as "visited"?
3. Senior Member
Join Date
Dec 2004
Location
San Bernardino County, California
Posts
1,468
To keep track of visited vertices: have a data structure which is easily searched so you can look-up that vertex.
DFS: An easy implementation is using a stack. As you you visit a vertex, store all of the adjacent edges in the stack. When you've finished adding edges, now pop off the edge at the top [or if you are using a list, add to the back of the list and then remove from the back of the list]. Is the terminus of the edge that has been popped been visited? If so, discard that edge and pop another. If it has not been visited, mark it as visited and then visit it.
4. Registered User
Join Date
Mar 2009
Posts
24
This is what I had written for BFS. I'm supposed to add the source to the queue. Then visit source and mark it as visited by adding it to an array of booleans. After I visit the source I'm supposed to check it's neighbors and add them to the queue. I keep checking the neighbors of each vertex in the queue. What I don't get is how do I visit the vertex and it's neighbors and how do I add it to an array of booleans. Also how do I use pred? I'm not sure where to put that in the code.
Code:
```/**
* This version of BFS will stop as soon as it finds a path from the source
* to the destination.
* @param source: the vertex where the search should begin.
* @param dest: the vertex where the search should end.
* @param graph: the graph to be searched.
* @param visited: a boolean array used to mark vertices when they are
* visited.
* @param pred: an integer array that assigns to each vertex its predecessor
* in the path that is found.
*/
static void breadthFirstSearch(int source, int dest, Graph graph,
boolean [] visited, int [] pred)
{
visited = new boolean [graph.getNumberOfVertices()];
pred = new int [graph.getNumberOfVertices()];
Queue<Integer> q = new LinkedList<Integer>();
// visit source
// mark source
q.offer(source);
while(!q.isEmpty())
{
int f = q.peek();
List<Integer> w = graph.getNeighbors(f);
for(int i : w)
{
// visit w
// mark w
}
}
q.remove();
}```
5. Senior Member
Join Date
Dec 2004
Location
San Bernardino County, California
Posts
1,468
You read the first edge in the queue. You "visit" the terminus of the destination of that edge by:
1. Look in your boolean array to see if this vertex has been visited (is the value at visited[new vertex] TRUE or FALSE). If true, then it has been visited and you do nothing more with this vertex, remove the first edge in the queue, and start over.
2. If it has not been visited, store the value of the start vertex for this edge in pred[new vertex]; test to see if this is your destination [if so, you're done]; read each of the vertices in the adjacency list for the vertex you're visiting, adding each edge to the end of the queue. Set the value of visited[new vertex] as TRUE. You're done, so now remove this edge from the queue and start over.
Last edited by nspils; 03-16-2009 at 08:13 AM.
6. Registered User
Join Date
Mar 2009
Posts
24
I'm still confused.
So is this what I'm supposed to do:
- Create an empty visited and pred arrays
- Create an empty queue
- Add source (vertex) to the queue
- Visit the source (vertex) - Not sure how to do this. What does it mean to visit a vertex? Does visiting the vertex mean you peek at it in the queue?
- Mark it as visited - If the visited arrays are empty how do I know if it has been visited. Can I say that if it has been visited set it to true in the visited array?
- While the queue is not empty get the neighbors of the first element in the queue.
- Visit/Mark the neighbor
- Keep visiting/marking neighbors of the first element in the queue until the first element is the destination
Last edited by justinsbabe5000; 03-16-2009 at 04:03 PM.
7. Senior Member
Join Date
Dec 2004
Location
San Bernardino County, California
Posts
1,468
the arrays and the queue exist before you do your traversal. Don't create new ones.
Visit means to do work "at" that vertex. So when you "visit" the source, you do all your processes for that vertex. Then you "discard" that vertex and move to the next you are going to visit - if it has not been visited yet.
When you visit, check if the destination; check if visited; record pred for this node; add all "neighbors" to the queue ... you don't move on to any, yet. Mark this vertex as having been visited.
think of this as your graph (written as adjacency list form)
0 - - 1, 2, 3, 4
1 - - 5
2 - - 6, 7, 8
3 - - null
4 - - 9, 10
5 -- 11, 12
6 -- null
7 -- 13
8 -- 14, 15, 16
9 -- null
10 -- null
11 -- 17, 18
12 -- null
13 -- null
14 -- null
15 -- null
16 -- null
17 -- null
18 -- null
0 is the source, 9 is your destination
Now, sketch out your processes as you visit - what will the values of the arrays and the queue before you visit 0? How do the arrays and the queue change as you visit 0? Now, move to the first neighbor. What does the queue look like? What are the values of the arrays?
Now, ask and answer your questions. Plus: how best do we store the value of both pred and destination of each "edge" as an Integer? How do you mark the value for pred? How do you initialize the arrays so that the values you want as default values are there?
8. Registered User
Join Date
Mar 2009
Posts
24
Thanks. I kinda got it to work. I did some search online and found an example of how to do both BFS and DFS and I took what you said and I figured it out.
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•
FAQ Latest Articles Java .NET XML Database Enterprise
Questions? Contact us. C++ Web Development Wireless Latest Tips Open Source | 3,291 | 13,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-17 | latest | en | 0.503917 |
https://www.physicsforums.com/threads/feynman-path-integral-and-events-beyond-the-speed-of-light.1006969/ | 1,708,923,357,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474650.85/warc/CC-MAIN-20240226030734-20240226060734-00512.warc.gz | 963,084,644 | 35,190 | # Feynman path integral and events beyond the speed of light
• I
• kurt101
In summary, Feynman discusses how to calculate the probability that light from a certain source will be reflected and detected by a mirror. He explains how to subtract the probabilities of reflections around the mirror to get the final probability. He also explains how to include paths that are not forbidden by the problem in the calculation. If you turn on a light source for a short amount of time and only detect the light that could have reflected from the mirror in the time it takes for light from the source to reflect and arrive at the detector, you would eliminate the contributions from the reflections around the mirror because they could not have possibly happened in that time due to the speed of light. However, if you ran enough trials, the uncertainty in when a photon
#### kurt101
TL;DR Summary
Do events that can't actually happen because of the speed of light limitations contribute to probabilities that are calculated using the Feynman's path integral method?
In Richard Feynman's book "The Strange Theory of Light and Matter", in chapter 2, he explains how to calculate the probability that light from some source will be reflected by a mirror and be detected at some location. He explains how you sum up all of the probability amplitudes (represented as vectors) for each way light can reflect off of a mirror and end up at the detector and then square this result to get the final probability. Here is the illustration of this:
Feynman then explains how the probability amplitudes of reflections around A and B tend to cancel out and thus don't contribute very much to the final probability. Here is the illustration of this:
Feynman then explains that if you remove sections of the mirror at A and B that you can change the contributions of the probability amplitudes around A and B so that they do contribute significantly to the final probability. Here is the illustration of this:
My Question:
What happens if you turn on the light source for a small amount of time and only detect the light that could have reflected from the middle of the mirror (around say E through I) in the amount of time that it takes light from the source traveling at the speed of light to reflect and arrive at the detector. In this case would you eliminate the contributions to the final probability from the reflections around A and B because they could not have possibly contributed because of the speed of light?
There's no certainty in when a photon is emitted, so you can't have QED on the one hand and classical precise events in spacetime on the other.
In fact, the arguments in that book generally rely on this uncertainty in order to have the interference in the first place.
Remember also that interference is generally a photon interfering with itself and not different photons interfering with each other. Although the full picture is more complicated than that.
vanhees71
PeroK said:
There's no certainty in when a photon is emitted, so you can't have QED on the one hand and classical precise events in spacetime on the other.
In fact, the arguments in that book generally rely on this uncertainty in order to have the interference in the first place.
But you can know that the amount of time the light source was turned on is well between the times it took for reflection in the middle of the mirror (say source -> G -> detector) and the time it took for reflection at the end of the mirror (say source -> A -> detector). So you can run the experiment with these constraints and see if having the missing sections of mirror at A and B has any change on the probability. And so does having or not having the sections with this constraint change the probability?
And maybe there would be some variation in light detection based on the uncertainty, but if you ran enough trials the uncertainty should wash out.
If you manipulate the light in time as you describe it will necessarilly not be monochromatic and you will wash out potential interference patterns. In rereading this I am uncertain as to what the actual question is.
kurt101 said:
But you can know that the amount of time the light source was turned on ...
The emission of a photon, as with practically everything in quantum theory, is a probabilistic event. If you try to have a photon at spacetime coordinates ##(t, x, y, z)## moving with speed ##c## in some direction ##\vec n##, then that is not quantum theory in the first place. That's a classical view of the photon as a massless particle with a classical trajectory.
QED by Feynman, if it does anything, is trying to get you NOT to think of light like that!
sysprog and vanhees71
kurt101 said:
Summary:: Do events that can't actually happen because of the speed of light limitations contribute to probabilities that are calculated using the Feynman's path integral method?
In the path integral method photons do not have a well-defined "speed"; paths in which photons travel at different speeds all contribute to the path integral. In more technical language, worldlines which are timelike and spacelike, as well as worldlines which are null, make nonzero contributions to the path integral.
In many cases (such as light traveling over long distances in free space), the contributions to the path integral from any worldlines other than the classical null worldline are negligible; in these cases, we can view the light as simply traveling at the speed of light through free space. But there are also many cases where this is not true. The general rule is that you must include in the path integral all paths that are not forbidden by some condition of the problem (such as the presence of an opaque screen), even if those paths would make no sense classically.
sysprog and vanhees71
One should take the "null world lines" as what they really are, namely the characteristics of the eikonal equation, and this is just an approximate solution of the wave equation of the electromagnetic field. The phasor approach in Feynman's popular book is nothing else than the classical refraction theory in Fraunhofer approximation, i.e., it solves the wave equation for the electromagnetic field, and thus it is consistent with the propagation of these waves with the speed of light.
sysprog
vanhees71 said:
One should take the "null world lines" as what they really are, namely the characteristics of the eikonal equation, and this is just an approximate solution of the wave equation of the electromagnetic field. The phasor approach in Feynman's popular book is nothing else than the classical refraction theory in Fraunhofer approximation, i.e., it solves the wave equation for the electromagnetic field, and thus it is consistent with the propagation of these waves with the speed of light.
For the specific examples Feynman discusses, which all involve light propagation (reflection off a mirror, diffraction gratings, bending when passing from one medium to another), yes, he is only considering null paths through spacetime (different paths through space, but which are all traversed at the speed of light and therefore have different travel times from source to receiver). But the path integral formulation is much more general; for example, when modeling a static force between charged objects, the path integral will have significant contributions from timelike and spacelike paths through spacetime.
vanhees71 and sysprog
hutchphd said:
If you manipulate the light in time as you describe it will necessarilly not be monochromatic and you will wash out potential interference patterns.
Why does turning on and off a monochromatic light source make the light from that source NOT monochromatic?
hutchphd said:
In rereading this I am uncertain as to what the actual question is.
I will try phrasing it again.
To make my question as clear as possible I am using the same context as Feynman used in chapter 2 of his book "The Strange Theory of Light and Matter" where he describes how to calculate the probability of light from a monochromatic source reflecting off a mirror to a detector. After he explains the basic procedure he goes on to explain how removing sections at the end of the mirror can alter the probability that light reaches the detector. Ok with this part?
My question involves taking the Feynman scenario and putting the following constraints on the light source and the detector. For clarity sake I will call this my experiment and to make the sequence of events clear, I will put in some time values.
Lets say the following:
It takes roughly 5 seconds for light to travel from the light source to the middle of the mirror (point G in the diagram) to the detector.
It takes roughly15 second for light to travel from the light source to the end of the mirror (point A in the diagram) to the detector.
In my experiment:
At 0 seconds, I turn on the light source and start recording at the detector.
At 10 seconds, I turn off the light source and stop recording at the detector.
In the first version of my experiment, I make no changes to the mirror. The mirror at point A and B look like:
In the second version of my experiment, I remove sections at the end of the mirror. The mirror at point A and B now looks like:
I run many trials of both versions of my experiment. Over all my trials I compare the average amount of light I measured in the first version of my experiment to the average amount of light measured in the second version of my experiment. Is the amount of light measured between the first version of my experiment the the same or different than the second version of my experiment?
kurt101 said:
Lets say the following:
It takes roughly 5 seconds for light to travel from the light source to the middle of the mirror (point G in the diagram) to the detector.
It takes roughly15 second for light to travel from the light source to the end of the mirror (point A in the diagram) to the detector.
Light can travel from Earth four times further than the Moon in 5 seconds.
sysprog and kurt101
kurt101 said:
Why does turning on and off a monochromatic light source make the light from that source NOT monochromatic?
Because one of the requirements for a source to be truly monochromatic is that it emits light continuously forever, infinitely into the past and infinitely into the future. Which means, of course, that no real light source is ever truly monochromatic.
hutchphd
PeroK said:
The emission of a photon, as with practically everything in quantum theory, is a probabilistic event. If you try to have a photon at spacetime coordinates ##(t, x, y, z)## moving with speed ##c## in some direction ##\vec n##, then that is not quantum theory in the first place. That's a classical view of the photon as a massless particle with a classical trajectory.
QED by Feynman, if it does anything, is trying to get you NOT to think of light like that!
I am trying to understand how QED applies to the real world. I think that was one of the major reasons Feynman wrote the book. I think that is the major purpose of quantum mechanics. In the real world, when I turn on the light, I expect to detect photons in the detector at some later time that is roughly the distance of the shortest path of light divided by 3 x 10^8 m/s. I don't know how the light actually traveled from the source to the detector and I have tried to avoid any kind of interpretation in the presentation of my question.
PeroK said:
Light can travel from Earth four times further than the Moon in 5 seconds.
Ha, ha, I guess the mirror is very large.
PeterDonis said:
In the path integral method photons do not have a well-defined "speed"; paths in which photons travel at different speeds all contribute to the path integral. In more technical language, worldlines which are timelike and spacelike, as well as worldlines which are null, make nonzero contributions to the path integral.
In many cases (such as light traveling over long distances in free space), the contributions to the path integral from any worldlines other than the classical null worldline are negligible; in these cases, we can view the light as simply traveling at the speed of light through free space. But there are also many cases where this is not true. The general rule is that you must include in the path integral all paths that are not forbidden by some condition of the problem (such as the presence of an opaque screen), even if those paths would make no sense classically.
I have done a lot of searching for answers related to my question prior to posting it. I get the sense that there is a lot of uncertainty on what actually contributes to the probability in the path integral formulation. While I believe what you told me is correct, you left it sufficiently vague (i.e. "The general rule is that you must include in the path integral all paths that are not forbidden by some condition of the problem") and so I not able to answer my question with this rule. In other words I don't know what those conditions are.
kurt101 said:
I get the sense that there is a lot of uncertainty on what actually contributes to the probability in the path integral formulation.
No, there is no uncertainty at all. The math is perfectly clear.
For you, the problem is that you aren't looking at the math. You're looking at various people's attempts to describe what the math is saying in ordinary language. And there is no way to do that with 100% accuracy because our ordinary language simply doesn't have the words or the concepts required. That's why physicists don't use ordinary language to do physics; they use math.
kurt101 said:
I don't know what those conditions are.
There will be some paths through spacetime that are forbidden because of the presence of something (such as an opaque screen, which was the example I gave) that absorbs light. (You could, if you really wanted to, include the details of the absorption of the light by that something in your analysis, but all that would do is complicate the math even more to get the same answer--that none of that light makes it to your detector.)
PeterDonis said:
For the specific examples Feynman discusses, which all involve light propagation (reflection off a mirror, diffraction gratings, bending when passing from one medium to another), yes, he is only considering null paths through spacetime (different paths through space, but which are all traversed at the speed of light and therefore have different travel times from source to receiver). But the path integral formulation is much more general; for example, when modeling a static force between charged objects, the path integral will have significant contributions from timelike and spacelike paths through spacetime.
But of course the path integral for relativistic QT is not a path integral over paths in phase space of point particles (or after integrating out the momenta, if they appear only quadratically in the action, over paths in configuration space) in non-relativistic QT ("1st quantization") but a path integral over field configurations.
PeterDonis said:
No, there is no uncertainty at all. The math is perfectly clear.
If it is so clear to you then help me understand how to apply the algorithm Feynman describes to answer the question I am asking.
@vanhees71 I have read many of your discussions around microcausality in the context of QFT. For example you wrote:
vanhees71 said:
This together with the fact that a local relativistic QFT cannot describe any faster-than-light signal propagation (due to the microcausality built in this kind of relativistic QFTs) one must conclude that the correlation is not caused by the local measurements on each photon at far distant places but it is due to the preparation in the entangled state.
Does the microcausality condition require local (in a spatial sense) preparation?
In the scenario I am discussing in this thread where removing sections of a mirror (i.e. the grating) affect the probability of the light being detected; what is the (local?) preparation between the light source and the grating?
Ultimately in this thread, I am trying to understand how something distant affects where the light ends up. I am trying to understand this in the context of the QED algorithm Feynman describes.
For someone who truly understands QFT, my question about whether having the grating or not has an effect on the detection probability with the time window constraint I have added, should be trivial to answer.
vanhees71
kurt101 said:
If it is so clear to you then help me understand how to apply the algorithm Feynman describes to answer the question I am asking.
The problem here is that we can’t do anything with a description of an algorithm. To do something we need the real thing, and “QED: The strange theory….” is not that; it’s an extended multi-page analogy that offers non-specialists some intuition for how local interference can yield many of the macroscopic behaviors of light. You should not be surprised to find that its math-free toy model becomes vague and mushy when you push it.
One particularly important limitation of the model is that it does not work for light that is switched on and off (more precisely, is not kept on for a time that is long compared with the time it takes for light to reach all parts of the setup). Thus it‘s not a good place to start when you’re trying to understand which paths to include and why some seem to be arbitrarily excluded. The real theory, expressed mathematically, does not have this limitation.
MikeWhitfield, mattt, PeterDonis and 1 other person
kurt101 said:
If it is so clear to you then help me understand how to apply the algorithm Feynman describes to answer the question I am asking.
hutchphd anwered your intial question: "If you manipulate the light in time as you describe it will necessarilly not be monochromatic and you will wash out potential interference patterns." You slightly modified your question in response to his answer, but basically his answer still applies. If you don't understand his answer in the context of your initial question, then you will understand it even less in the context of your modified question.
In QED, Feynman describes how coherent superposition works. But what happens in your setup is partial coherent superposition, i.e. an incoherent average over coherent superpositions (what hutchphd described as "... will wash out ..."). Perhaps trying to understand the picture and explanations on the last slide of this extract will help you a bit.
To make everything even more confusing, partial coherence can be "reduced" to full coherence by going to a (church of the) larger Hilbert space. But if you don't already understand partial coherence, this construction won't help you either.
vanhees71 and PeroK
kurt101 said:
If it is so clear to you then help me understand how to apply the algorithm Feynman describes to answer the question I am asking.
You can't apply the specific algorithm Feynman describes to the specific case given in your OP, because, as has already been pointed out, his algorithm doesn't work for light sources that are switched on and off. More precisely, it doesn't work for light sources that are switched on and off on a time scale comparable to the time scale of the experiment you are doing.
You can use the math of QED directly to answer a question like the one you ask in your OP, but, as has already been pointed out, the math of QED is not the same as the heuristic ordinary language description that Feynman gives. If you really want to learn how to apply QED to general cases, not just the specific cases Feynman describes in his book, you will need to learn how to use the math of QED. There are no shortcuts that let you use vague ordinary language to actually do physics.
mattt, Vanadium 50, gentzen and 1 other person
PeterDonis said:
You can use the math of QED directly to answer a question like the one you ask in your OP, but, as has already been pointed out, the math of QED is not the same as the heuristic ordinary language description that Feynman gives.
Note that the last paragraph of my post #14 was an attempt to at least convey a more general sense of what the math of QED is doing, beyond what Feynman says in his book. There are other possible reasons besides the presence of an opaque screen for particular paths through spacetime being forbidden: one such reason is that the light source is turned off, so paths through spacetime that originate from the source when it is turned off will not have any nonzero amplitude associated with them. But you can't analyze such cases using the heuristic ordinary language picture Feynman gives in his book.
vanhees71
kurt101 said:
Is the amount of light measured between the first version of my experiment the the same or different than the second version of my experiment?
1. You already know that if a diffraction grating is placed at appropriate spots on the mirror, there will be no contribution to the final result from that section. Any way that contribution is alternatively excluded gives the same results as if a diffraction grating were present. There is no faster-than-light effect present in any version of this scenario.2. As a practical matter, using some classical reasoning to consider your version of the experiment: In a femtosecond (10^-15 seconds), light travels about about 333 nanometers (nm) - little shorter than a wavelength of visible light. The difference X in distance light travels when it is reflected from the middle (your B) versus when it is further away by a single wavelength (i.e. ever so slightly closer to your A or C) is extremely small. Using some "reasonable" parameters for visible light, approximately 45 degree angle of incidence, 1000 wavelength distance to mirror, back of a napkin calculations... that X works out to be on the order of magnitude of 1 nanometer (nm). That translates to a time window of roughly 0.003 femtoseconds.
In other words, we would need to nail down the precise time of emission AND detection to less than approximately .003 femtoseconds to be able to perform this experiment. The best detectors today have a comparable resolution of no better than 1 picosecond (1000 femtoseconds), or nearly a million times longer that what you'd need to even think about running your experiment. And that doesn't begin to consider a variety of quantum effects and uncertainty you'd encounter.3. Think of the Feynman paths as follows: Because of inherent uncertainty in photon emission times: a detection at experimentally precise (if there was such a thing) time T3 could result from a slightly shorter path traversed by a photon B emitted at time T2, or a slightly longer path traversed by a photon A emitted slightly earlier at time T1. Both of those possibilities contribute to the result. You could perform this experiment a photon at a time, but it is not possible to force (or restrict) a single photon to be emitted at a specific time with sufficient resolution to distinguish between scenarios A and B. To the extent that you could perform this impossible task, you would only succeed in demonstrating what we already know would occur. And that would not show any FTL effect at all.
vanhees71
It's best not to try to imagine the Feynman paths as real processes or worry about FTL paths and so on. Most of the "paths" contributing to the path integral aren't even functions but distributions and so in no sense describe actual paths, i.e. assign a position for each time.
MikeWhitfield and vanhees71
gentzen said:
hutchphd anwered your intial question: "If you manipulate the light in time as you describe it will necessarilly not be monochromatic and you will wash out potential interference patterns." You slightly modified your question in response to his answer, but basically his answer still applies. If you don't understand his answer in the context of your initial question, then you will understand it even less in the context of your modified question.
I added more detail to my question, but did not change it. In Feynman's book, chapter 2 that I used as the basis for my description, the light source is described as monochromatic and I left that detail out. Here is the actual quote from the book:
"At S we have a source that emits light of one color at very low intensity (let’s use red light again). The source emits one photon at a time." - Feynman, Richard P.. QED: The Strange Theory of Light and Matter (Princeton Science Library) (p. 38). Princeton University Press. Kindle Edition.
Also I do not say in my question that the light source emits one photon at a time, but I do not think that is relevant to my question. Is it important to the result at the detector that the light source emits 1 photon at a time?
gentzen said:
In QED, Feynman describes how coherent superposition works. But what happens in your setup is partial coherent superposition, i.e. an incoherent average over coherent superpositions (what hutchphd described as "... will wash out ..."). Perhaps trying to understand the picture and explanations on the last slide of this extract will help you a bit.
I understand the light source in Feynman's example is not coherent; at least in the sense that he does not describe it as a laser where light is in phase. That said, I don't understand how turning on the light makes the light any different (i.e. more or less coherent) than how Feynman already describes it. Unless I am mistaken about Feynman's example and he does mean to imply that is light source is fully coherent even though he never states it. I would be surprised if this is the case.
kurt101 said:
I added more detail to my question, but did not change it. In Feynman's book, chapter 2 that I used as the basis for my description, the light source is described as monochromatic and I left that detail out. Here is the actual quote from the book:
"At S we have a source that emits light of one color at very low intensity (let’s use red light again). The source emits one photon at a time." - Feynman, Richard P.. QED: The Strange Theory of Light and Matter (Princeton Science Library) (p. 38). Princeton University Press. Kindle Edition.
Also I do not say in my question that the light source emits one photon at a time, but I do not think that is relevant to my question. Is it important to the result at the detector that the light source emits 1 photon at a time?
I understand the light source in Feynman's example is not coherent; at least in the sense that he does not describe it as a laser where light is in phase. That said, I don't understand how turning on the light makes the light any different (i.e. more or less coherent) than how Feynman already describes it. Unless I am mistaken about Feynman's example and he does mean to imply that is light source is fully coherent even though he never states it. I would be surprised if this was the case.
So, why don't we just stick with one photon? The whole point of that section is to describe how the probability of the photon being detected depends on the diffraction grating.
Why have you brought all these added compexities into the equation?
PeterDonis said:
There are other possible reasons besides the presence of an opaque screen for particular paths through spacetime being forbidden: one such reason is that the light source is turned off, so paths through spacetime that originate from the source when it is turned off will not have any nonzero amplitude associated with them.
This sounds promising to helping me answer my question, but in the context of everything else told to me, I am far from certain.
What about when the the light source is turned on? Do paths that don't yet exist through spacetime (outside of the light source's light cone), have a nonzero amplitude?
PeroK
kurt101 said:
What about when the the light source is turned on? Do paths that don't yet exist through spacetime (outside of the light source's light cone), have a nonzero amplitude?
You may have completely lost the point of this treatment of QED. I think you should ask yourself whether you are intent on studying the material in Feynman's book, or whether you'd rather let your mind wander onto vague and superfluous questions.
Perhaps I'm too harsh. If so I apologise. But, if you were my student I'd tell you to knuckle down and concentrate.
kurt101 said:
I understand the light source in Feynman's example is not coherent; at least in the sense that he does not describe it as a laser where light is in phase.
He describes the light source as monochromatic. This is what makes it coherent. To get quasi-monochromatic light from a non-laser source, you have to filter out most frequencies, and only keep a small portion of the frequency spectrum. That is typically still wider than for light from a laser, but the coherence length can be made longer than the distance-differences in your optical instruments. And that is enough for treating the light as quasi-monochromatic.
kurt101 said:
That said, I don't understand how turning on the light makes the light any different (i.e. more or less coherent) than how Feynman already describes it.
Let me first descibe it in Feynman's photon picture. Assume that each photon would have exactly one frequency, and nicely form a coherent superposition with itself. If different photons have slightly different frequencies, then their interference patterns are slightly different. And the sum of those interference patterns no longer shows sharp interference peaks, but appears washed out. But to see an interference pattern at all, you must look at the image produced by many photons, so all you can see is the washed out pattern.
If your light has the frequency f, and you switch it on for t seconds, then you end up with an frequency distribution in the range f +- 1/t. If f is big compared to 1/t, then it makes sense to talk of quasi-monochromatic light. And if c*t is big compared to the distance-differences in your optical instruments, then treating the light as monochromatic should work well.
vanhees71
DrChinese said:
1. You already know that if a diffraction grating is placed at appropriate spots on the mirror, there will be no contribution to the final result from that section.
It is actually the opposite in the example, the diffraction grating placed at the end of the mirror adds a contribution, but I could see where one placed at the middle of the mirror counteracts the contribution, and so I get what you are saying.
DrChinese said:
Any way that contribution is alternatively excluded gives the same results as if a diffraction grating were present.
I am not sure if I follow what you are saying. Are you saying that given the constraints of my experiment, you will see the same results with or without the diffraction grating? (i.e. were you giving a direct answer to my question?).
DrChinese said:
There is no faster-than-light effect present in any version of this scenario.
Ultimately this is what I want to understand. I have had trouble finding clear statements in any of the descriptions of the Fenman path integral (both with math and without) that clearly say contributions prohibited by the speed of light don't count.
That said, I still don't understand if contributions that are prohibited by the speed of light in my example are from the light sources perspective or say the mirror's perspective. Hopefully you understand what I am saying here and this would be a part 2 question for this thread and so I am reluctant to get into it too much before I am comfortable that I understand my current question.
DrChinese said:
2. As a practical matter, using some classical reasoning to consider your version of the experiment: In a femtosecond (10^-15 seconds), light travels about about 333 nanometers (nm) - little shorter than a wavelength of visible light. The difference X in distance light travels when it is reflected from the middle (your B) versus when it is further away by a single wavelength (i.e. ever so slightly closer to your A or C) is extremely small. Using some "reasonable" parameters for visible light, approximately 45 degree angle of incidence, 1000 wavelength distance to mirror, back of a napkin calculations... that X works out to be on the order of magnitude of 1 nanometer (nm). That translates to a time window of roughly 0.003 femtoseconds.
In other words, we would need to nail down the precise time of emission AND detection to less than approximately .003 femtoseconds to be able to perform this experiment. The best detectors today have a comparable resolution of no better than 1 picosecond (1000 femtoseconds), or nearly a million times longer that what you'd need to even think about running your experiment. And that doesn't begin to consider a variety of quantum effects and uncertainty you'd encounter.
I know my example is not a practical one, at least the numbers I gave, but it is the principle of the example that is important and I still have the understanding that in principle you can run my experiment. If that is not the case, and I am still missing something, please make that clear to me.
DrChinese said:
3. Think of the Feynman paths as follows: Because of inherent uncertainty in photon emission times: a detection at experimentally precise (if there was such a thing) time T3 could result from a slightly shorter path traversed by a photon B emitted at time T2, or a slightly longer path traversed by a photon A emitted slightly earlier at time T1. Both of those possibilities contribute to the result. You could perform this experiment a photon at a time, but it is not possible to force (or restrict) a single photon to be emitted at a specific time with sufficient resolution to distinguish between scenarios A and B. To the extent that you could perform this impossible task, you would only succeed in demonstrating what we already know would occur. And that would not show any FTL effect at all.
If you changed the time constraint by adding additional detection time to be sure scenarios A and B (with and without the grating) should give a different result; then ran scenarios A and B many times; then took the averages; I would expect on average you would see a consistent difference between A and B, where scenario B with the grating that add contributions would on average detect more photons than A.
PeroK said:
You may have completely lost the point of this treatment of QED. I think you should ask yourself whether you are intent on studying the material in Feynman's book, or whether you'd rather let your mind wander onto vague and superfluous questions.
Perhaps I'm too harsh. If so I apologise. But, if you were my student I'd tell you to knuckle down and concentrate.
That is an interesting perspective. I don't think my questions are vague or superfluous. I think understanding the constraints for any algorithm, formula, etc. is extremely important. So I don't know where you are coming from.
My goal is one of understanding out of curiosity and not one of learning QM so I can earn a living from it (at least not right now (no jokes here)) and so maybe that plays a part in the types of questions I ask. I am biased towards a causal deterministic perspective and tend to think the universe is not fundamentally statistical and so maybe that bias shows (certainly obvious in my earlier threads) and plays a part in the types of questions I ask. That said, I enjoy reading the different perspectives on this forum and I am open to all well supported positions.
You also said "knuckle down and concentrate". I do read carefully most everyone's responses and try very hard to understand what is being said to me. So I hope you are not saying this because you don't think I am not reading or not trying to understand what is being said to me. If I ask more questions or try to clarify where I am coming from that does not mean I am rejecting what others are telling me. It might mean I still don't understand and/or I think I have been misunderstood.
weirdoguy
PeroK said:
So, why don't we just stick with one photon? The whole point of that section is to describe how the probability of the photon being detected depends on the diffraction grating.
I am ok with 1 photon at a time versus many. I did not specify either way. I still don't think it matters for my question.
PeroK said:
Why have you brought all these added compexities into the equation?
Because I am trying to understand how distant objects (e.g. the grating) affect the result. Its kind of like what others are telling me, the ordinary language does not matter, you have to understand the math. However in my case I would say it is the experiment that ultimately tells me the truth and while I don't doubt the math is correct from some perspective, math obviously has its limitations on where and how it can be used, and so it is helpful for me to understand what the actual experiment says.
kurt101 said:
I do not say in my question that the light source emits one photon at a time, but I do not think that is relevant to my question. Is it important to the result at the detector that the light source emits 1 photon at a time?
It is important to the result at the detector to know what state the light emitted from the source is in. Saying "the source emits one photon at a time" does not tell you that. That is one of the key limitations of trying to learn physics from a pop science book written in ordinary language--even one by Feynman, who was one of the best physicists I know of at minimizing the distortion inherent in any such presentation.
The issue here, which is not necessarily important to the exposition Feynman was trying to make, but is very important if you are actually trying to understand the underlying physics in detail, is that "monochromatic" light is not in an eigenstate of photon number. "Monochromatic" light (I put the term in quotes because of the issue already raised, that truly "monochromatic" light can only be emitted by a source that emits light forever), such as light emitted by a laser, so called because (at least to whatever level of approximation is appropriate for the specific problem) it has a well-defined frequency, is in a coherent state. A coherent state does not have a well-defined photon number, since it is not in an eigenstate of photon number. So it is really not correct to describe such a source as emitting "one photon at a time". A strictly correct statement would be something like "the intensity of the source is such that the expected number of clicks at the photon detector over some standard interval of time is one". (Note that it is the detector that produces discrete clicks, or some other discrete phenomenon, that we associate with the term "photon".)
An eigenstate of photon number is called a Fock state, and is a very different state that requires a very different type of source (one which is much more difficult to make than a laser; experiments have been done using such sources, but they are much less common). That is not the kind of source Feynman was talking about; he was talking about something like a laser.
vanhees71
PeroK said:
why don't we just stick with one photon?
Because, as I noted in my previous post just now, "one photon" is a misleading description of the actual state of the light emitted by the kind of source (something like a laser) that Feynman was talking about.
vanhees71
kurt101 said:
What about when the the light source is turned on? Do paths that don't yet exist through spacetime (outside of the light source's light cone), have a nonzero amplitude?
You are not thinking correctly about spacetime. Spacetime includes time. A spacetime model already includes all the information about when (i.e., at what events in spacetime) the light source is turned on. Only paths originating from those events in spacetime will have a nonzero amplitude. There is no such thing as "paths that don't yet exist through spacetime"; spacetime is not something that "changes" as things happen. A spacetime model already includes all the information about everything that happens.
MikeWhitfield and vanhees71
kurt101 said:
Unless I am mistaken about Feynman's example and he does mean to imply that is light source is fully coherent even though he never states it.
If the light is monochromatic, it is in a coherent state, as I have already said. A source like an ordinary light bulb does not produce monochromatic light. You need something like a laser.
vanhees71 | 8,599 | 40,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-10 | latest | en | 0.93452 |
https://iris.uniroma3.it/handle/11590/423870 | 1,716,318,246,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058512.80/warc/CC-MAIN-20240521183800-20240521213800-00495.warc.gz | 265,159,510 | 13,024 | First-year mathematics courses in architecture aim to provide students with scientific language, increase spatial, creative, thinking together with the ability to recognize and create forms, and enable informed use of design software. However, future Italian architects often consider these courses marginal in their education (Pagano, & Tedeschini Lalli, 2005). In order to overcome this criticality, we developed an experimental 4-hour workshop following the DBR methodology (Brown ,1992; Barab & Squire, 2004). The didactic content is the parabola, familiar to first-year students, rediscovered with activities aimed at revealing the connection between geometric form and analytical description. Students fold (on paper) the envelope of a parabola, verify the reflection property, thus "discovering" the algebraic description of the curve and, finally, apply it to a problem of architectural luminosity. After the design phase, the lab was experimented with two groups of 75 first-year students from Politecnico di Torino and Università Roma Tre in a.y. 2021-22. Comparison of our field notes with comparative analysis of the responses of a final questionnaire provided us with encouraging results on conceptual learning and engagement, with an impact on mathematics beyond the specific example considered in the workshop.
I corsi di matematica del primo anno in architettura mirano a fornire agli studenti il linguaggio scientifico, aumentare il pensiero spaziale, creativo, insieme alla capacità di riconoscere e creare forme e consentire un uso consapevole dei software di progettazione. Tuttavia, i futuri architetti italiani considerano spesso questi corsi marginali nella loro formazione (Pagano & Tedeschini Lalli, 2005). Per superare questa criticità, abbiamo sviluppato un’officina sperimentale di 4 ore seguendo la metodologia DBR (Brown ,1992; Barab & Squire, 2004). Il contenuto didattico è la parabola, familiare agli studenti del primo anno, riscoperta con attività volte a svelare il legame tra forma geometrica e descrizione analitica. Gli studenti piegano (su carta) l'inviluppo di una parabola, ne verificano la proprietà di riflessione, "scoprendo" così la descrizione algebrica della curva e, infine, la applicano ad un problema di luminosità in architettura. Dopo la fase di progettazione, il laboratorio è stato sperimentato con due gruppi di 75 studenti del primo anno del Politecnico di Torino e dell'Università Roma Tre nell'a.a. 2021-22. Il confronto delle nostre note sul campo con l'analisi comparativa delle risposte di un questionario finale ci ha fornito risultati incoraggianti sull'apprendimento concettuale e sul coinvolgimento, con un impatto sulla matematica al di là dell'esempio specifico considerato nell’officina.
Magrone, P., Spreafico, M.L. (2022). Nuovi approcci nei corsi di Matematica per l’Architettura: connettere forme e formule in geometria attraverso esperienze laboratoriali. ANNALI ONLINE... DELLA DIDATTICA E DELLA FORMAZIONE DOCENTE, 14(24), 99-122.
### Nuovi approcci nei corsi di Matematica per l’Architettura: connettere forme e formule in geometria attraverso esperienze laboratoriali
#### Abstract
First-year mathematics courses in architecture aim to provide students with scientific language, increase spatial, creative, thinking together with the ability to recognize and create forms, and enable informed use of design software. However, future Italian architects often consider these courses marginal in their education (Pagano, & Tedeschini Lalli, 2005). In order to overcome this criticality, we developed an experimental 4-hour workshop following the DBR methodology (Brown ,1992; Barab & Squire, 2004). The didactic content is the parabola, familiar to first-year students, rediscovered with activities aimed at revealing the connection between geometric form and analytical description. Students fold (on paper) the envelope of a parabola, verify the reflection property, thus "discovering" the algebraic description of the curve and, finally, apply it to a problem of architectural luminosity. After the design phase, the lab was experimented with two groups of 75 first-year students from Politecnico di Torino and Università Roma Tre in a.y. 2021-22. Comparison of our field notes with comparative analysis of the responses of a final questionnaire provided us with encouraging results on conceptual learning and engagement, with an impact on mathematics beyond the specific example considered in the workshop.
##### Scheda breve Scheda completa Scheda completa (DC)
2022
I corsi di matematica del primo anno in architettura mirano a fornire agli studenti il linguaggio scientifico, aumentare il pensiero spaziale, creativo, insieme alla capacità di riconoscere e creare forme e consentire un uso consapevole dei software di progettazione. Tuttavia, i futuri architetti italiani considerano spesso questi corsi marginali nella loro formazione (Pagano & Tedeschini Lalli, 2005). Per superare questa criticità, abbiamo sviluppato un’officina sperimentale di 4 ore seguendo la metodologia DBR (Brown ,1992; Barab & Squire, 2004). Il contenuto didattico è la parabola, familiare agli studenti del primo anno, riscoperta con attività volte a svelare il legame tra forma geometrica e descrizione analitica. Gli studenti piegano (su carta) l'inviluppo di una parabola, ne verificano la proprietà di riflessione, "scoprendo" così la descrizione algebrica della curva e, infine, la applicano ad un problema di luminosità in architettura. Dopo la fase di progettazione, il laboratorio è stato sperimentato con due gruppi di 75 studenti del primo anno del Politecnico di Torino e dell'Università Roma Tre nell'a.a. 2021-22. Il confronto delle nostre note sul campo con l'analisi comparativa delle risposte di un questionario finale ci ha fornito risultati incoraggianti sull'apprendimento concettuale e sul coinvolgimento, con un impatto sulla matematica al di là dell'esempio specifico considerato nell’officina.
Magrone, P., Spreafico, M.L. (2022). Nuovi approcci nei corsi di Matematica per l’Architettura: connettere forme e formule in geometria attraverso esperienze laboratoriali. ANNALI ONLINE... DELLA DIDATTICA E DELLA FORMAZIONE DOCENTE, 14(24), 99-122.
File in questo prodotto:
Non ci sono file associati a questo prodotto.
I documenti in IRIS sono protetti da copyright e tutti i diritti sono riservati, salvo diversa indicazione.
Utilizza questo identificativo per citare o creare un link a questo documento: `https://hdl.handle.net/11590/423870`
• ND
• ND
• ND | 1,650 | 6,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-22 | latest | en | 0.497865 |
https://flatearthsoc.com/like-flat-earth-news-blog-2017-flat-earth-news-nick-davies.html | 1,571,236,229,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986668994.39/warc/CC-MAIN-20191016135759-20191016163259-00021.warc.gz | 497,097,702 | 8,252 | The circumstances which attend bodies which are caused merely to fall from a great height prove nothing as to the motion or stability of the Earth, since the object, if it be on a thing that is in motion, will participate in that motion; but, if an object be thrown, upwards from a body at rest, and, again, from a body in motion, the circumstances attending its descent will be very different. In the former case, it will fall, if thrown vertically upwards, at the place from whence it was projected; in the latter case, it will fall behind the moving body from which it is thrown will leave it in the rear. Now, fix a gun, muzzle upwards, accurately, in the ground; fire off a projectile; and it will fall by the gun. If the Earth traveled eleven hundred miles a minute, the projectile would fall behind the gun, in the opposite direction to that of the supposed motion. Since, then, this is NOT the case, in fact, the Earth's fancied motion is negatived and we have a proof that the Earth is not a, globe.
As we have seen that there is, really no south point (or pole) but an infinity of points forming, together, a vast circumference -- the boundary of the known world, with its battlements of icebergs which bid defiance to man's onward course, in a southerly direction - so there can be no east or west "points,' just as there is no "yesterday," and no "tomorrow." In fact, as there is one point that is fixed (the North), it is impossible for any other point to be fixed likewise. East and west are, therefore, merely directions at right angles with a north and south line: and as the south point of the compass shifts round to all parts of the circular boundary, (as it may be carried round the central North) so the directions east and west, crossing this line, continued to form a circle at any latitude. A westerly circumnavigation, is going around with the North Star continually on the right hand, and an easterly circumnavigation is performed only when the reverse condition of things is maintained, the North Star being on the left hand as the journey is made. These facts, taken together, form a beautiful proof that the Earth is not a globe.
If the Earth were a globe, people - except those on the top - would, certainly, have to be "fastened" to its surface by some means or other, whether by the "attraction" of astronomers or by some other undiscovered and undiscoverable process! But, as we know that we simply walk on its surface without any other aid than that which is necessary for locomotion on a plane, it follows that we have, herein, a conclusive proof that Earth is not a globe.
43.) The circumstances which attend bodies which are caused merely to fall from a great height prove nothing as to the motion or stability of the Earth, since the object, if it be on a thing that is in motion, will participate in that motion; but, if an object be thrown, upwards from a body at rest, and, again, from a body in motion, the circumstances attending its descent will be very different. In the former case, it will fall, if thrown vertically upwards, at the place from whence it was projected; in the latter case, it will fall behind the moving body from which it is thrown will leave it in the rear. Now, fix a gun, muzzle upwards, accurately, in the ground; fire off a projectile; and it will fall by the gun. If the Earth traveled eleven hundred miles a minute, the projectile would fall behind the gun, in the opposite direction to that of the supposed motion. Since, then, this is NOT the case, in fact, the Earth's fancied motion is negatived and we have a proof that the Earth is not a, globe.
It is in evidence that, if a projectile be fired from a rapidly moving body in an opposite direction to that in which the body is going, it will fall short of the distance at which it would reach the ground if fired in the direction of motion. Now, since the Earth is said to move at the rate of nineteen miles in, a second of time, "from west to east," it would make all the difference imaginable if the gun were fired in an opposite direction. But, as, in practice, there is not the slightest difference, whichever way the thing may be done, we have a forcible overthrow of all fancies relative to the motion of the Earth, and a striking proof that the Earth is not a globe.
37.) If the Earth were a globe, there would, very likely, be (for nobody knows) six months day and six months night at the arctic and antarctic regions, as astronomers dare to assert there is: – for their theory demands it! But, as this fact – the six months day and six months night – is; nowhere found but in the arctic regions, it agrees perfectly with everything else that we know about the Earth as a plane, and, whilst it overthrows the "accepted theory," it furnishes a striking proof that Earth is not a globe.
29. If the Earth were a globe, it would, unquestionably, have the same general characteristics - no matter its size - as a small globe that may be stood upon the table. As the small globe has top, bottom, and sides, so must also the large one - no matter how large it be. But, as the Earth, which is "supposed" to be a large globe, bas no sides or bottom as the small globe has, the conclusion is irresistible that it is a proof that the Earth is not a globe.
```66) Dr. Rowbotham conducted several other experiments using telescopes, spirit levels, sextants and “theodolites,” special precision instruments used for measuring angles in horizontal or vertical planes. By positioning them at equal heights aimed at each other successively he proved over and over the Earth to be perfectly flat for miles without a single inch of curvature. His findings caused quite a stir in the scientific community and thanks to 30 years of his efforts, the shape of the Earth became a hot topic of debate around the turn of the nineteenth century.
```
37.) If the Earth were a globe, there would, very likely, be (for nobody knows) six months day and six months night at the arctic and antarctic regions, as astronomers dare to assert there is: – for their theory demands it! But, as this fact – the six months day and six months night – is; nowhere found but in the arctic regions, it agrees perfectly with everything else that we know about the Earth as a plane, and, whilst it overthrows the "accepted theory," it furnishes a striking proof that Earth is not a globe.
##### If the Earth were a globe, there certainly would be - if we could imagine the thing to be peopled all round - "antipodes:" "people who," says the dictionary, "living exactly on the opposite side of the globe to ourselves, have their feet opposite to ours: - people who are hanging heads downwards whilst we are standing heads up! But, since the theory allows us to travel to those parts of the Earth where the people are said to be heads downwards, and still to fancy ourselves to be heads upwards and our friends whom we have left behind - us to be heads downwards, it follows that the whole thing is a myth - a dream - a delusion - and a snare; and, instead of there being any evidence at all in this direction to substantiate the popular theory, it is a plain proof that the Earth is not a globe.
22.) God's Truth never – no, never – requires a falsehood to help it along. Mr. Proctor, in his " Lessons," says: Men " have been able to go round and round the Earth in several directions." Now, in this case, the word " several will imply more than two, unquestionably: whereas, it is utterly impossible to circumnavigate the Earth in any other than an easterly or a westerly direction; and the fact is perfectly consistent and clear in its relation to Earth as a Plane.. Now, since astronomers would not be so foolish as to damage a good cause by misrepresentation, it is presumptive evidence that their cause is a bad one, and – a proof that Earth is not a globe.
Consider a sphere. Since a sphere has a consistent shape, no matter where on it you stand, you have exactly the same amount of sphere under you. (Imagine an ant walking around on a crystal ball. From the insect's point of view, the only indication of movement would be the fact the ant is moving its feet—the shape of the surface would not change at all.) A sphere's center of mass is in the center of the sphere, which means gravity will pull anything on the surface of the sphere straight down toward the center of the sphere. This will occur no matter where on the surface the object is located.
I started reading this e-book today and finished it today. Great Work ! I have to say, even if approached flat-earth model fistly 3 month ago, this Book really opened my eyes completely and still i know how strong the Brainwash is and even if you - by ratio - see something clear, it will still need a time to be completely internalized. Thx alot, greetingz from Germany also i was able to watch the video below, which oc is blocked in Germany, but this led me to HotSpotShield, also very important for a native German, greetingz from Frankfurt/Germany, Benjamin
It is supposed," in the regular course of the Newtonian theory, that the Earth is, in June, about 190 millions of miles (190,000,000) away from its position in December. Now, since we can, (in middle north latitudes), see the North Star, on looking out of a window that faces it - and out of the very same corner of the very same pane of glass in the very same window - all the year round, it is proof enough for any man in his senses that we have made no motion at all. It is a proof that the Earth is not a globe.
41.) When astronomers assert that it is "necessary" to make "allowance for curvature" in canal construction, it is, of course, in order that, in their idea, a level cutting may be had, for the water. How flagrantly, then, do they contradict themselves when the curved surface of the Earth is a "true level!" What more can they want for a canal than a true level? Since they contradict themselves in such an elementary point as this, it is an evidence that the whole thing is a delusion, and we have a proof that the Earth is not a globe.
58.) Astronomers have never agreed amongst themselves about a rotating Moon revolving round a rotating and revolving Earth – this Earth, Moon, planets and their satellites all, at the same time dashing through space, around the rotating and revolving Sun, towards the constellation Hercules, at the rate of four millions of miles a day! And they never will: agreement is impossible! With the a Earth a plane and without motion, the whole thing is clear. And if a straw will show which way the wind blows, this may be taken as a pretty strong proof that the Earth is not a globe.
157) If “gravity” magically dragged the atmosphere along with the spinning ball Earth, that would mean the atmosphere near the equator would be spinning around at over 1000mph, the atmosphere over the mid-latitudes would be spinning around 500mph, and gradually slower down to the poles where the atmosphere would be unaffected at 0mph. In reality, however, the atmosphere at every point on Earth is equally unaffected by this alleged force, as it has never been measured or calculated and proven non-existent by the ability of airplanes to fly unabated in any direction without experiencing any such atmospheric changes.
If the Earth were a globe, it would, if we take Valentia to be the place of departure, curvate downwards, in the 1665 miles across the Atlantic to Newfoundland, according to the astronomers' own tables, more than three hundred miles; but, as the surface of the Atlantic does not do so - the fact of its levelness having been clearly demonstrated by Telegraph Cable surveyors, - it follows that we have a grand proof that Earth is not a globe.
His first musical stirrings were at the age of eight, when his parents gave him a secondhand radiogram which included a few records left by the previous owner. Among them were Drummin' Man by drumming legend Gene Krupa, and, in Davies's own words, "it hit like a thunderbolt". "I must have played it 2,000 times," he said. "That was it."[5] A friend of the family made Rick a makeshift drum kit out of a biscuit tin, and at the age of 12 he joined the British Railways Staff Association Brass and Silver Jubilee Band as a snare drummer.[6] In an interview in 2002 he said: "As a kid, I used to hear the drums marching along the street in England, in my home town, when there was some kind of parade, and it was the most fantastic sound to me. Then, eventually, I got some drums and I took lessons. I was serious about it... I figured if I could do that – I mean a real drummer, read music and play with big bands, rock bands, classical, Latin, and know what I was going to do – I would be in demand and my life was set... Eventually, I started fiddling with the keyboards, and that seemed to go over better than my drumming, for some reason. So you've gotta go with what people react to." He never had lessons for keyboards, but, according to Betty Davies, "taught himself most of what he knows about music".[5]
## Light travels at a finite speed, with the speed depending upon the medium. With mirages, the medium is air. In air, the speed of light is only slightly less than it is in a vacuum, and the speed of light in air depends upon the temperature of the air. Simply put, the speed of light is slightly greater in warmer air than it is in cooler air. In physics, we usually express this behavior reciprocally as the index of refraction, n:
It is well known that the law, regulating the apparent decrease in the size of objects as we leave them in the distance (or as they leave us) is very different with luminous bodies from what it is in the case of those which are non-luminous. Sail past the light of a small lamp in a row-boat on a dark night, and it will seem to be no smaller when a mile off than it was when close to it. Proctor says, in speaking of the Sun: "his apparent size does not change!" - far off or near. And then he forgets the fact! Mr. Proctor tells us, subsequently, that, if the traveler goes so far south that the North Star appears on the horizon, "the Sun should therefore look much larger" - if the Earth were a plane! Therefore, he argues, "the path followed cannot have been the straight course," - but a curved one. Now, since it is nothing but common scientific trickery to bring forward, as an objection to stand in the way of a plane Earth, the non-appearance of a thing which has never been known to appear at all, it follows that, unless that which appears to be trickery were an accident, it was the only course open to the objector - to trick. (Mr. Proctor, in a letter to the "English Mechanic" for Oct. 20,1871, boasts of having turned a recent convert to the Zetetic Philosophy by telling him that his arguments were all very good, but that "it seems as though [Mark the language!] the sun ought to look nine times larger in summer." And Mr. Proctor concludes thus: "He saw, indeed, that, in his faith in "Parallax," he had "written himself down an ass.") Well, then: trickery or no trickery on the part of the objector, the objection is a counterfeit - a fraud - no valid objection at all; and it follows that the system which does not purge itself of these things is a rotten system, and the system which advocates, with Mr. Proctor at their head, a weapon to use - the Zetetic philosophy of "Parallax" - is destined to live! This is a proof that the Earth is not a globe.
The most commonly accepted explanation of this is that the space agencies of the world are involved in a conspiracy faking space travel and exploration. This likely began during the Cold War's 'Space Race', in which the USSR and USA were obsessed with beating each other into space to the point that each faked their accomplishments in an attempt to keep pace with the other's supposed achievements. Since the end of the Cold War, however, the conspiracy is most likely motivated by greed rather than political gains, and using only some of their funding to continue to fake space travel saves a lot of money to embezzle for themselves. | 3,594 | 16,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-43 | latest | en | 0.969105 |
https://www.jiskha.com/questions/1636775/Betty-decides-to-support-local-artists-by-hanging-their-artwork-in-her-shop-When | 1,537,891,591,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161661.80/warc/CC-MAIN-20180925143000-20180925163400-00159.warc.gz | 782,104,269 | 5,080 | # Math
Betty decides to support local artists by hanging their artwork in her shop. When decorating the coffee shop, Betty comes across the following formula for placing pictures on her walls: 1/3 |a-12|+ 2/3|a|where a equals the number of pictures that she would like to hang.
What is the value of this expression when a space equals space 9?
1. "when a space equals space 9"
you mean: a = 9 ??
Why not just plug in a = 9 and evaluate ?
btw, I have no clue what that formula is supposed to represent?
posted by Reiny
## Similar Questions
1. ### Math
Betty decides to support local artists by hanging their artwork in her shop. When decorating the coffee shop, Betty comes across the following formula for placing pictures on her walls: 1 third open vertical bar A-12 close
2. ### Sign Language
De'VIA artwork was started by hearing people to support Deaf artists focuses on being Deaf and growing up Deaf*** can only be done by Deaf artists is always 3D artwork
3. ### Art
when criticizing an artists techincal skills the critic should consider 1)his personal opinion of the artwork 2)the style of the artwork 3) the creativity of the artwork 4) the beauty of the artwork I think it's 1)
4. ### Art
The design for Cubist artwork was dramatically different from what had come before in the history of painting. Why did the artists involved in the creation of Cubism choose this radical and innovative approach to the design of
5. ### Adding and subtracting positive and negative numb
Betty is in the process of opening her own business – a coffee shop in downtown Fargo, North Dakota. With the Grand Opening a month away, she runs the numbers involved in her upfront expenses: \$5,400 for local advertising,
6. ### business law
On March 5, 2009, Andy, a local MP3 supplier, calls Betty, a local electronics chain retailer, offering to supply her with 1000 red IPODS, 1000 black IPODS, 1000 white IPODS and 1000 green IPODS at \$ 200 per IPOD which Betty
7. ### algebra
Betty is in the process of opening her own business – a coffee shop in downtown Fargo, North Dakota. With the Grand Opening a month away, she runs the numbers involved in her upfront expenses: \$5,400 for local advertising,
8. ### Math
Betty is in the process of opening her own business – a coffee shop in downtown Fargo, North Dakota. With the Grand Opening a month away, she runs the numbers involved in her upfront expenses: \$5,400 for local advertising,
9. ### decorating today
The answers are a. interior designers b. Upper class c. Architects d. artists I think it is b
10. ### Math
Betty has worked out the financing of her coffee shop. She plans to use the money in a savings account for her start-up costs. The savings account was opened by her grandmother five years ago. To figure out the annual interest
More Similar Questions | 665 | 2,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-39 | latest | en | 0.954234 |
https://statmodeling.stat.columbia.edu/category/miscellaneous-statistics/page/4/ | 1,582,647,262,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146123.78/warc/CC-MAIN-20200225141345-20200225171345-00307.warc.gz | 552,008,980 | 13,314 | Archive of posts filed under the Miscellaneous Statistics category.
## The methods playroom: Mondays 11-12:30
Each Monday 11-12:30 in the Lindsay Rogers room (707 International Affairs Bldg, Columbia University): The Methods Playroom is a place for us to work and discuss research problems in social science methods and statistics. Students and others can feel free to come to the playroom and work on their own projects, with the understanding that […]
## What’s the origin of the term “chasing noise” as applying to overinterpreting noisy patterns in data?
Roy Mendelssohn writes: In an internal discussion at work I used the term “chasing noise”, which really grabbed a number of people involved in the discussion. Now my memory is I first saw the term (or something similar) in your blog. But it made me interested in who may have first used the term? Did […]
## Beyond Power Calculations: Some questions, some answers
Brian Bucher (who describes himself as “just an engineer, not a statistician”) writes: I’ve read your paper with John Carlin, Beyond Power Calculations. Would you happen to know of instances in the published or unpublished literature that implement this type of design analysis, especially using your retrodesign() function [here’s an updated version from Andy Timm], […]
## More on the piranha problem, the butterfly effect, unintended consequences, and the push-a-button, take-a-pill model of science
The other day we had some interesting discussion that I’d like to share. I started by contrasting the butterfly effect—the idea that a small, seemingly trivial, intervention at place A can potentially have a large, unpredictable effect at place B—with the “PNAS” or “Psychological Science” view of the world, in which small, seemingly trivial, intervention […]
## You should (usually) log transform your positive data
The reason for log transforming your data is not to deal with skewness or to get closer to a normal distribution; that’s rarely what we care about. Validity, additivity, and linearity are typically much more important. The reason for log transformation is in many settings it should make additive and linear models make more sense. […]
## “The issue of how to report the statistics is one that we thought about deeply, and I am quite sure we reported them correctly.”
Ricardo Vieira writes: I recently came upon this study from Princeton published in PNAS: Implicit model of other people’s visual attention as an invisible, force-carrying beam projecting from the eyes In which the authors asked people to demonstrate how much you have to tilt an object before it falls. They show that when a human […]
## “I feel like the really solid information therein comes from non or negative correlations”
Steve Roth writes: I’d love to hear your thoughts on this approach (heavily inspired by Arindrajit Dube’s work, linked therein): This relates to our discussion from 2014: My biggest takeaway from this latest: I feel like the really solid information therein comes from non or negative correlations: • It comes before • But it doesn’t […]
## What can be learned from this study?
James Coyne writes: A recent article co-authored by a leading mindfulness researcher claims to address the problems that plague meditation research, namely, underpowered studies; lack of or meaningful control groups; and an exclusive reliance on subjective self-report measures, rather than measures of the biological substrate that could establish possible mechanisms. The article claims adequate sample […]
## Amending Conquest’s Law to account for selection bias
Robert Conquest was a historian who published critical studies of the Soviet Union and whose famous “First Law” is, “Everybody is reactionary on subjects he knows about.” I did some searching on the internet, and the most authoritative source seems to be this quote from Conquest’s friend Kingsley Amis: Further search led to this elaboration […]
## Here are some examples of real-world statistical analyses that don’t use p-values and significance testing.
Joe Nadeau writes: I’ve followed the issues about p-values, signif. testing et al. both on blogs and in the literature. I appreciate the points raised, and the pointers to alternative approaches. All very interesting, provocative. My question is whether you and your colleagues can point to real world examples of these alternative approaches. It’s somewhat […]
## You are invited to join Replication Markets
Anna Dreber writes: Replication Markets (RM) invites you to help us predict outcomes of 3,000 social and behavioral science experiments over the next year. We actively seek scholars with different voices and perspectives to create a wise and diverse crowd, and hope you will join us. We invite you – your students, and any other […]
## Are supercentenarians mostly superfrauds?
Ethan Steinberg points to a new article by Saul Justin Newman with the wonderfully descriptive title, “Supercentenarians and the oldest-old are concentrated into regions with no birth certificates and short lifespans,” which begins: The observation of individuals attaining remarkable ages, and their concentration into geographic sub-regions or ‘blue zones’, has generated considerable scientific interest. Proposed […]
## Causal Inference and Generalizing from Your Data to the Real World (my talk tomorrow, Sat., 6pm in Berlin)
For the Berlin Bayesians meetup, organized by Eren Elçi: Causal Inference and Generalizing from Your Data to the Real World Andrew Gelman, Department of Statistics and Department of Political Science, Columbia University Learning from data involves three stages of extrapolation: from sample to population, from treatment group to control group, and from measurement to the […]
## I don’t have a clever title but this is an interesting paper
Why do we, as a discipline, have so little understanding of the methods we have created and promote? Our primary tool for gaining understanding is mathematics, which has obvious appeal: most of us trained in math and there is no better form of information than a theorem that establishes a useful fact about a method. […]
## Just forget the Type 1 error thing.
John Christie writes: I was reading this paper by Habibnezhad, Lawrence, & Klein (2018) and came across the following footnote: In a research program seeking to apply null-hypothesis testing to achieve one-off decisions with regard to the presence/absence of an effect, a flexible stopping-rule would induce inflation of the Type I error rate. Although our […]
## Swimming upstream? Monitoring escaped statistical inferences in wild populations.
Anders Lamberg writes: In my mails to you [a few years ago], I told you about the Norwegian practice of monitoring proportion of escaped farmed salmon in wild populations. This practice results in a yearly updated list of the situation in each Norwegian salmon river (we have a total of 450 salmon rivers, but not […]
## What’s published in the journal isn’t what the researchers actually did.
David Allison points us to these two letters: Alternating Assignment was Incorrectly Labeled as Randomization, by Bridget Hannon, J. Michael Oakes, and David Allison, in the Journal of Alzheimer’s Disease. Change in study randomization allocation needs to be included in statistical analysis: comment on ‘Randomized controlled trial of weight loss versus usual care on telomere […]
## Calibrating patterns in structured data: No easy answers here.
“No easy answers” . . . Hey, that’s a title that’s pure anti-clickbait, a veritable kryptonite for social media . . . Anyway, here’s the story. Adam Przedniczek writes: I am trying to devise new or tune up already existing statistical tests assessing rate of occurrences of some bigger compound structures, but the most tricky […]
## The garden of 603,979,752 forking paths
Amy Orben and Andrew Przybylski write: The widespread use of digital technologies by young people has spurred speculation that their regular use negatively impacts psychological well-being. Current empirical evidence supporting this idea is largely based on secondary analyses of large-scale social datasets. Though these datasets provide a valuable resource for highly powered investigations, their many […]
## Harvard dude calls us “online trolls”
Story here. Background here (“How post-hoc power calculation is like a shit sandwich”) and here (“Post-Hoc Power PubPeer Dumpster Fire”). OK, to be fair, “shit sandwich” could be considered kind of a trollish thing for me to have said. But the potty language in this context was not gratuitous; it furthered the larger point I […] | 1,745 | 8,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-10 | latest | en | 0.926193 |
https://www.singaporemath.com/collections/distributor-us-canada/products/dimensions-math-tests-3b | 1,709,301,835,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475311.93/warc/CC-MAIN-20240301125520-20240301155520-00310.warc.gz | 996,688,198 | 48,342 | # Dimensions Math Tests 3B
\$14.00
Test books are essential for evaluating student progress in an organized way. They are closely aligned with Textbook content.
Tests for Grades 1–5 have differentiated assessments. Each chapter has both a Test A and Test B, with Test A focusing on key concepts and fundamental problem-solving skills, and Test B focusing on the application of analytical skills and heuristics. Tests consist of multiple-choice questions that assess comprehension of key concepts, and free response questions for students to demonstrate their problem-solving skills.
SKU: DMTST3B
ISBN: 9781947226524
Pagecount: 166
Dimensions: 8.5 x 11 x 0.75 in
Binding: Standard
Color: Grayscale
Cover: Soft
Perforated: No
Sample Pages
Chapter 1: Numbers to 10
Test A
Test B
Chapter 2: Number Bonds
Test A
Test B
Test A
Test B
Chapter 4: Subtraction
Test A
Test B
Continual Assessment 1
Test A
Test B
Chapter 5: Numbers to 20
Test A
Test B
Test A
Test B
Chapter 7: Subtraction Within 20
Test A
Test B
Chapter 8: Shapes
Test A
Test B
Chapter 9: Ordinal Numbers
Test A
Test B
Continual Assessment 2
Test A
Test B
A & B Books: Our programs divide the school year into two semesters. “A” level books are for the first half of the school year. “B” level books are for the second half of the school year. You need both “A” and “B” material for a complete school year.
Required Components: Textbooks, Workbooks, and Guides (either Home Instructor’s Guides or Teacher’s Guides) are all necessary components. These three elements each serve a unique function and work together to build math mastery.
Printouts: Dimensions Math PK-5 Resources
Recommended Manipulatives: Dimensions Math Grades PK-5 Recommended Manipulatives
Answer Key: Included in the back of the book
Dimensions Math At Home™ Videos
Invite a professional Singapore math teacher into your home classroom. This subscription of pre-recorded lessons covers all Textbook and Workbook material for an entire school year and can be done at your own pace.
Manipulatives
Manipulatives help students visualize and represent math concepts. Stock up on suggested items to deepen engagement during math activities and lessons.
Supplementary Math
Supplementary math helps students of all abilities: struggling, advanced, or on level. Find just the right addition to round out your core program. | 558 | 2,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-10 | latest | en | 0.868409 |
https://www.jiskha.com/display.cgi?id=1320538468 | 1,501,168,341,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549428300.23/warc/CC-MAIN-20170727142514-20170727162514-00579.warc.gz | 812,964,334 | 5,192 | # math - functions
posted by .
I compute (109/3)^1/4 and store that value in A.
Store 3 in P.
Now compute f(7)= Pa^7 by typing:
P x A x^y 7
***My problem is that I don't know WHERE the x^y key is on a TI -83 Plus.
Then I have to subtract this answer from the answer using a rounded value of a (which I found to be 1831.)The difference will show the error used by rounding a and will not be zero.
• math - functions -
• math - functions -
So I punched it in and the calculator reads:
P * A ^ 7
1613.078231
Did I do it correctly?
• math - functions -
Yes, I got the same answer.
The best way to check is to use numbers that you know, such as:
2^6=64,
2^20=1048576
3^4=81,
etc.
## Similar Questions
1. ### Computers
1. There is a requirement to store 750 records. The number of buckets is 10,000. The number of slots per bucket is 3. What is the PPD?
2. ### Computers
1. There is a requirement to store 750 records. The number of buckets is 10,000. The number of slots per bucket is 3. What is the PPD?
3. ### stastics
johns runs a computer software store.he counted 128 people who walkrd by his store in a day , 60 of whom came into the store, of the 60 only 21 brought something in the store.estimate the prob that that a person who walks by the storeenter …
4. ### calculus
consider the initial value problem dy/dx=y^2+1, y(0)=0. estimate y(2) 1) use h=1 to compute your estimate I got x1=1 x2=2 y=1=1 y2=3 2)use h=1/2 to compute your estimate I got x1=1/2 x2=1 x3=1.5 x4=2 y1=1/2 y2=9/8 y3=289/128 y4=5.306671143 …
5. ### Statistics
John runs a computer software store. Yesterday he counted 127 people whwo walked by his store, 58 of whom came into the store. OF the 58, only 25 bought something in the store. A. What is probability that a person who walks by the …
6. ### 7th grade math Ms. Sue last two questions thanks
15. In Store A, a book that regularly sells for \$24.99 is on sale at 15% off. In Store B, the same book regularly sells for \$27.99 and is on sale at 25% off. Which store sells the book for the lower sale price?
7. ### STAT
An independent-measures research study compares three treatment conditions with a sample of n = 10 in each condition. The sample means are M1 = 2, M2 = 3, and M3 = 7 a. Compute SS for the set of 3 treatment means. (Use the three means …
8. ### math
A store is located at (-1, -2). the owner of the store plans to build another store that is 5 blocks east and 4 blocks north of the original store. plot both stores on the coordinate plane. Give directions from the original store to …
9. ### statistics
John runs a computer software store. Yesterday he counted 133 people who walked by the store, 66 of whom came into the store. Of the 66, only 24 bought something in the store.
10. ### Math
Frank bought supplies for school. In the first store, he spent half of his money plus \$10. In the second store, he spent half of what he had left, plus \$10. In the third store, he spent 80% of what he had left. He came home with \$5. …
More Similar Questions
Post a New Question | 870 | 3,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2017-30 | longest | en | 0.911067 |
https://bgr.com/2015/09/08/iphone-6-battery-life-2/ | 1,542,209,277,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742117.38/warc/CC-MAIN-20181114150002-20181114171323-00030.warc.gz | 553,953,455 | 26,346 | While Apple remains obsessed with making its devices as thin as possible, most iPhone owners truly couldn’t care less. In every poll we’ve seen on the topic, iPhone owners have unanimously indicated that they’d rather have a new device with greater battery life than one that’s even thinner than previous models. In short, the iPhone, for most people, is already thin enough.
DON’T MISS: Canon has an insane 250 megapixel sensor that can read lettering on a plane flying 11 miles away
Over the past few years, successive iPhone models have gotten thinner and thinner while battery life has more or less remained the same. That being the case, an interesting hypothetical question posed on Reddit wondered how long battery life on the iPhone 6 (7.67mm thick with the protruded lens) would be if the device was as thick as the original iPhone (11.6mm).
Well, no need to wonder as someone actually up and did the math. Suffice it to say, if Apple really wanted to go to town on battery life, it would have no problem creating a monster of a device.
The framework for the calculation reads as follows:
[The iPhone 6 battery] is 3.75 in x 1.5 in x 0.13 in for a total volume of 0.73125 in^3. According to wikipedia, the battery capacity is 1810 mA.h (milliamp hours). This gives us a energy density of 2475 [mA.h/in^3].
Also according to wikipedia, the iPhone 6 is 0.185 inches thinner than the OG iPhone. This would increase the volume of the iPhone 6 by 2.657 in^3. If this were totally filled with batteries of the same energy density as the iPhone 6 has now, it would give you 6576 mA.h of additional battery capacity.
And just how would that translate into real world usage? The chart below is rather telling.
The iPhone 6 would deliver more than two full days! of Wi-Fi browsing while the 6 Plus would deliver a whopping 55 hours of video playback. Impressive figures, to be sure, but the obvious trade-off is that these battery oriented iPhones would be rather heavy, with the hypothetical iPhone 6 and iPhone 6 Plus weighing in at 8.1 and 10.5 oz respectively. By way of contrast, the real iPhone 6 weighs in at 4.55 ounces. | 509 | 2,135 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-47 | longest | en | 0.960323 |
http://www.devhardware.com/forums/ti-83-84-102/programming-factors-to-a-list-470642.html | 1,386,547,098,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163835370/warc/CC-MAIN-20131204133035-00075-ip-10-33-133-15.ec2.internal.warc.gz | 327,875,863 | 17,157 | Forums: » Register « | User CP | Games | Calendar | Members | FAQs | Sitemap | Support |
January 21st, 2013, 03:27 PM
Seaspider1
n00b DevH'er
Join Date: Jan 2013
Posts: 1
Time spent in forums: 10 m 27 sec
Reputation Power: 0
Programming Factors...to a list?
I wrote a program for my TI-84 Plus Silver Edition to find all factors of a number inputed. For instance, the home screen would read when executed for the number 102:
prgmFACTOR
(Input)?102
1
102
2
51
3
34
6
17
(Disp) "NUMBER OF FACTORS"
8
Done
I have the program check numbers between 1 and the sqrt(inputed number) with each lower factor (1, 2, 3, 6) as a variable and the upper factor (102, 51, 34, 17) as another variable. Both of these variables go through a loop of adding one to the lower factor, checking, and displaying both. When a large number is factored, I can't scroll up to view the factors that have been calculated because all history of the execution disappears.
My question is whether I can, still using the loop, put the lower factors in List 1 and the upper factors in List 2, lining up respectively so that it can be more easily viewed. When I tried to write it out, it would only put one factor in each list and then as it looped it would replace the factor with the next, leaving both List 1 and List 2 with no more than one number each.
If it helps at all, I have "A" as the original number and "E" as the lower factor being increased incrementally. When A/E = a number without a decimal:
Disp A, A/E
So to further clarify, I need help with getting "A" in List 1 and "A/E" in List 2.
Any help would be greatly appreciated!
January 21st, 2013, 06:34 PM
Weregoose
Contributing User
Join Date: Dec 2011
Posts: 40
Time spent in forums: 16 h 2 m 21 sec
Reputation Power: 2
Using matrices:
:Input A
:[[A][1
:For(E,2,√(A
:If not(fPart(A/E
:augment(Ans,[[A/E][E
:End
:Ans
Viewing: Dev Hardware Forums > SOFTWARE > TI 83/84 > Programming Factors...to a list? | 585 | 1,955 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2013-48 | longest | en | 0.914726 |
http://www.jiskha.com/display.cgi?id=1364349185 | 1,493,417,628,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123097.48/warc/CC-MAIN-20170423031203-00577-ip-10-145-167-34.ec2.internal.warc.gz | 574,301,503 | 4,001 | Math
posted by on .
i need help on a couple of questions. I did all my other questions but i need a lot of help with the word problems...
1. The side length of a small square is s. A larger square has a perimeter of 124.8 cm. Its sides are 3.2 cm longer than those of the small square.
a) represent the situation with an equation of the form a(x + b) = c. Then determine the side length of the smaller square.
b.) verify your solution by using a model.
2. Valerie bought five packages of golf balls on sale for \$29.50. Each package had a discount of \$2.75. Write and solve an equation to determine the regular price of each package.
thanks :) and im sorry i reposted this
• Math - ,
a. 4(x+3.2) = 124.8
Solve for x.
b. I cannot model.
c. 5(x-2.75) = 29.50
Solve for x. | 218 | 782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2017-17 | latest | en | 0.910565 |
https://freewallpaperspot.com/1000924/free-4th-grade-math-worksheets-on-angles/ | 1,618,568,821,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038056325.1/warc/CC-MAIN-20210416100222-20210416130222-00441.warc.gz | 376,800,272 | 12,145 | # Free 4th Grade Math Worksheets On Angles
Also angles on a straight line concept are emphasized. Draw the Line of Symmetry.
Basic Geometry Terms Worksheet Worksheets For All Download And Share Worksheets Free On Bonlacfoods Co Geometry Worksheets Fourth Grade Math 4th Grade Math
### In these printable measuring angles worksheets we will learn to measure angles using a protractor.
Free 4th grade math worksheets on angles. The worksheets start out with the base leg of the angle always laying horizontal which is the easiest way to visualize whether the angle is acute or obtuse. Use this fun math challenge worksheet to review all kinds. 4th Grade Math Worksheets.
Vividly illustrated to bring math to life these fourth grade geometry worksheets make complex concepts enjoyable even for reluctant students. Grade 4 Measuring Angles – Displaying top 8 worksheets found for this concept. Expert-level skills arent built in a day to acquire superior skills in estimating angles 4th grade and 5th grade children need to bolster practice with our printable estimating angles worksheets.
These Angles Worksheets are great for teaching the different classification of angles. You will solve different equations to find the value of missing angles. This page features many worksheets and a set of task cards.
Measuring angles is much easy after we started to use a mathematical tool the protractor. Learn some basic geometry with this worksheet all about the angle. Naming Angles Worksheets These Angles Worksheets are great for teaching the correct nomenclature to identify angles and sides of angles.
Earlier worksheets also avoid angles that arent quite right angles unless the angle really IS a right angle to make the acuteobtuse determination more obvious. Count on these worksheets to generously support your efforts while drawing angles. With each page containing six printed protractors these pdf worksheets offer students of 3rd grade and 4th grade tremendous preparation material for drawing angles with 1-degree increment.
Drawing Angles with a Protractor 1-Degree Increment. Work your way through this compilation of worksheets and examine the angles on a straight line that add up to 180. Practice identifying 2D and 3D shapes calculating area and perimeter mapping on a coordinate grid measuring angles and more.
Gain a deep understanding of classifying angles with these printable worksheets diligently prepared for students of grade 4 grade 5 and grade 6. Practice identifying 2D and 3D shapes calculating area and perimeter mapping on a coordinate grid measuring angles and more. They will be asked to label the vertex and sides of angles and name all angles with a given vetex.
Complementary and Supplementary Angles. Help your students learn to about complementary and supplementary angles with these printable teaching resources. An acute angle is an angle between 0 and 90.
Some of the worksheets for this concept are Grade 4 geometry work Reading protractor level 1 s1 Abc def acute obtuse a d c Measuring angles Grade 4 supplement Math mammoth grade 4 a Math mammoth grade 4 b complete curriculum Introduction. In these worksheets students classify angles as straight right acute or obtuse. Draw the Line of Symmetry.
Worksheets for grade 6 through grade 8 students in this section have problems exclusively on adjacent vertically opposite and linear pairs of angles. Our grade 4 math worksheets help build mastery in computations with the 4 basic operations delve deeper into the use of fractions and decimals and introduce the concept of factors. Geometry Free geometry worksheets Our grade 4 geometry worksheets cover topics such as classifying angles triangles and quadrilaterals areas and perimeters and coordinate grids.
Grade 3 Geometry Free Printable Worksheets. Grade 4 math worksheets on classifying acute obtuse and right angles. You can use at school and at home the angles worksheets pdf.
Free pdf worksheets from K5 Learnings online reading and math program. With adequate exercises in multi-digit multiplication and division equivalent fractions addition and subtraction of fractions with like denominators and multiplication of fractions by whole numbers analyzing and classifying geometric figures based on their. Endorsing learning and practice our printable 4th grade math worksheets with answer keys amazingly fit into your curriculum.
Grade 5 math worksheets on classifying and measuring angles. A right angle is an angle of exactly 90. This page has worksheets for teaching students to measure and draw angles with a protractor.
We have already learnt that two rays originating from a common end point form an angle. An obtuse angle is more than 90 and less than 180. Measuring Drawing Angles.
Please click the link for different angles worksheets. These angles worksheets will produce 20 problems for the student to identify whether the angle is acute obtuse right or straight. Grade 4 math worksheets from K5 Learning.
Upon regular practice and revision with this set of pdf exercises children can distinguish themselves in identifying-types-of-angles tasks. Free pdf worksheets from K5 Learnings online reading and math program. In this second grade geometry worksheet your child will practice finding the line or lines of symmetry on a variety of different shapes.
Vividly illustrated to bring math to life these fourth grade geometry worksheets make complex concepts enjoyable even for reluctant students. Angles on a Straight Line. Angles worksheets pdf is a good resource for children in Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade and 5th Grade.
Right Angles Geometry Worksheet Education Com Geometry Worksheets Math Geometry Third Grade Math Worksheets
Worksheets Word Lists And Activities Greatschools Math Division Worksheets 4th Grade Math 4th Grade Math Worksheets
8 4th Grade Geometry Worksheets Printable Geometry Worksheets Fourth Grade Math 4th Grade Math
Geometry Worksheets Geometry Worksheets For Practice And Study Geometry Worksheets Angles Worksheet Algebra Worksheets
4th Grade Math Worksheets Angle Classification 3ans Gif 1000 1294 4th Grade Math Worksheets Math Worksheets Mathematics Worksheets
4th Grade Worksheets Best Coloring Pages For Kids Geometry Worksheets Fourth Grade Math 4th Grade Math
Drawing And Measuring Angles Maths Worksheet And Answers 9 1 Gcse Foundation Grade 2 Year 7 Angles Worksheet Measuring Angles Worksheet Year 7 Maths Worksheets
Grade 5 Geometry Worksheet Classify And Measure Angles Geometry Worksheets Angles Worksheet 4th Grade Math Worksheets
Pinterest In 2021 Geometry Worksheets Fourth Grade Math 4th Grade Math
Angles Geometry Worksheets Fourth Grade Math 4th Grade Math
An Exercise In Deducing Angles Free Math Worksheets Geometry Worksheets Math Worksheet
Angles Worksheet Angles Worksheet Math Math Geometry
Printable Geometry Worksheets Find The Missing Angle 1 Geometry Worksheets Angles Worksheet 8th Grade Math Worksheets
Finding Missing Angles Worksheet Angles Worksheet Finding Missing Angles Worksheet Geometry Worksheets
Image Result For Reflex Angles Worksheets Grade 4 Maths Angles Worksheet 4th Grade Math Worksheets
4th Grade Geometry 4th Grade Math Worksheets Geometry Worksheets Geometry Vocabulary
Pin By Vicky S On 4th Grade Classroom Angles Worksheet Math Worksheets Free Math Worksheets
4th Grade Geometry Geometry Worksheets 4th Grade Math Worksheets Math Geometry
Finding Supplementary Angles Worksheet Angles Worksheet Math Worksheets Geometry Angles
#### Author: Bertha Natasha
My name is Bertha Natasha, also known as Bertha, and I am USA, although I currently live in Seattle, WA. Many of my recipes are inspired by healthy food, by my mom's American-Mexican cuisine, who grew up in New Mexico, and also by my granny's kitchen in Texas. | 1,510 | 7,819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-17 | latest | en | 0.877756 |
https://studylib.net/doc/14335495/one-way-anova | 1,611,426,181,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538226.66/warc/CC-MAIN-20210123160717-20210123190717-00290.warc.gz | 577,933,464 | 13,608 | # One-Way ANOVA
### One-Way ANOVA
. A design with a single categorical predictor variable is called a oneway ANOVA design. For example, a study of 4 different fertilizers used on different individual plants could be analyzed via one-way ANOVA, with four levels for the factor
Fertilizer.
In genera, consider a single categorical predictor variable
A
with 1 case in each of its 3 categories. Using the sigma-restricted coding of A into 2 quantitative contrast variables, the matrix
X
defining the between design is
That is, cases in groups
A
1
,
A
2
, and
A
3
are all assigned values of 1 on
X
0
(the intercept), the case in group
A
1
is assigned a value of 1 on
X
1
and a value 0 on
X
2
, the case in group
A
2
is assigned a value of 0 on
X
1
and a value 1 on
X
2
, and the case in group
A
3
is assigned a value of -1 on
X
1
and a value -1 on
X
2
. Of course, any additional cases in any of the 3 groups would be coded similarly. If there were 1 case in group
A
1
, 2 cases in group
A
2
, and 1 case in group
A
3
, the
X
matrix would be where the first subscript for
A
gives the replicate number for the cases in each group. For brevity, replicates usually are not shown when describing ANOVA design matrices . | 367 | 1,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-04 | latest | en | 0.895991 |
http://www.chegg.com/homework-help/questions-and-answers/input-signal-10-sin-2-pi-1592-t-0degrees-applies-series-rc-circuit-resistor-r-1000-ohms-ca-q2272227 | 1,472,586,145,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471983001995.81/warc/CC-MAIN-20160823201001-00274-ip-10-153-172-175.ec2.internal.warc.gz | 372,747,698 | 13,449 | an input signal of 10 sin (2 pi 1592 t + 0degrees) is applies to a series RC circuit with resistor R=1000 ohms and capacitor C =0.1 uF. The magnitude of the total circuit impedence is...
a. 1000 ohms
b. 1414 ohms
c 1592 ohms
d 2000 ohms
e. none of the above | 91 | 257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2016-36 | latest | en | 0.647379 |
https://vidsbook.com/characteristics-of-a-function-in-math-latest-2023-4/ | 1,674,785,468,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494852.95/warc/CC-MAIN-20230127001911-20230127031911-00409.warc.gz | 638,905,008 | 18,321 | # Characteristics Of A Function In Math latest 2023
You are searching about Characteristics Of A Function In Math, today we will share with you article about Characteristics Of A Function In Math was compiled and edited by our team from many sources on the internet. Hope this article on the topic Characteristics Of A Function In Math is useful to you.
Page Contents
## Data Analysis – March Madness Predictions
I didn’t really pay attention to college basketball this year, so I decided to take a different approach to filling out my bracket.
I started by downloading the full Division 1 men’s basketball schedule (taken from rivals.yahoo.com), along with each game’s score, date and home team.
In the model, I assume that each team has two (unknown) vectors of real numbers describing the quality of their offense and defense in several attributes, respectively. For example, we might want to represent the quality of each team’s guards, the quality of forwards, and the quality of crosses, both in attack and defense. We could do this using an offensive and defensive vector:
Offense: [5, 10, 4]
Defense: [2, 3, 10]
This means the guards are a 5 in attack and a 2 in defense, etc. In my model, it will be easier if we assume that high numbers are better for attacks and low numbers are better for defenses.
The score of a match between team i and team j can then be generated as the dot product of team i’s offensive vector with team j’s defensive vector, and vice versa. In our race example, if our forward team was playing against a team with vectors:
Offense: [3, 2, 4]
Defense: [2, 5, 5]
Then the score of the first team should be 5 * 2 + 10 * 5 + 4 * 5 = 80
and the score of the second team should be 3 * 2 + 2 * 3 + 4 * 10 = 52
What an eruption!
Now the only problem is that we don’t actually know the vectors describing each team’s offense and defense. It’s OK – we’ll learn them from the data.
Formally, the goal is to find latent matrices O and D that minimize the sum of the squared error between the predicted scores and the observed scores. In mathematics,
sum_g (score_gi – O_i: * D_j:)^2 + (score_gj – O_j: * D_i:)^2
where I use the notation that team i played against team j in game g (i and j depend on g, but I drop this dependency in notation to keep things simple)*.
I won’t go into detail, but we can take the derivative of the error function with respect to each latent vector in order to find changes in the vectors that will make them closer to the results of all games earlier in the game. season. I repeat this until there are no more changes that will improve the error (batch gradient descent, for the detail minded).
Results In the case where I choose 1-dimensional latent vectors, I get an offensive and defensive rating for each team as output. Remember that to predict the first team’s score against another team, multiply the first team’s offensive rating (higher is better) by the second team’s defensive rating (lower is better) .
Here are the top 10 attacks and defenses, as learned by the 1D version of my model:
Offenses
North Carolina (9.79462281797)
Pittsburgh (9.77375501699)
Connecticut (9.74628326851)
Memphis (9.71693872544)
Louisville (9.69785532917)
Duke (9.65866585522)
UCLA (9.59945808934)
West Virginia (9.56811566735)
Arizona Street (9.56282860536)
Missouri (9.55043151623)
tusks
North Carolina (7.02359489844)
Pittsburgh (7.0416251036)
Memphis (7.05499448413)
Connecticut (7.07696194481)
Louisville (7.14778041166)
Duke (7.18950625894)
UCLA (7.21883856723)
Gonzaga (7.22607569868)
Kansas (7.2289767174)
Missouri (7.2395184452)
And here are the results of the full tournament simulation with a 5-dimensional model. For each game I report the predicted score, but for the range I just choose the predicted winner.
==================== ROUND 1 ====================
Ohio St. 74.9907105909, Siena 69.6702059811
Utah 69.7205426091, Arizona 69.2592708246
Wake Forest 72.3264784371 Cleveland Street 64.3143396939
West Virginia 66.7025939102, Dayton 57.550404701
Kansas 84.0565034675 North Dakota Street 71.281863854
Coll. from Boston. 65.0669174572, USC 68.7027018576
Michigan Street 77.3858437718, Robert Morris 59.6407479
Connecticut 91.9763662649, Chattanooga 63.9941388666
BYU 74.7464520646, Texas A&M 70.5677646712
Purdue 69.8634461612, Northern Iowa 59.4892887466
Washington 81.8475059935 Mississippi St. 74.6374151171
Marquette 73.4307446299 Utah St. 69.1796188404
Missouri 83.8888903275, Cornell 68.1053984941
California 74.9638076999, Maryland 71.2565877894
Memphis 78.3145709447, CSU Northridge 59.0206289492
Pittsburgh 85.5983991252 E. Tennessee Street 64.8099546261
Oklahoma St. 81.6131739754, Tennessee 81.8021658489
Florida Street 59.994769086, Wisconsin 60.9139371828
Xavier 77.3537694 Portland Street 63.8161558802
UCLA 76.790261041, VCU 65.2726887151
Villanova 72.9957948506, American 58.6863439306
Texas 64.5805075558, Minnesota 62.3595994418
Duke 85.084666484, Binghamton 61.1984347353
LSU 65.0807263343, Butler 64.9895028812
Illinois 70.6250577544, West. Kentucky 57.6646396014
Gonzaga 75.0447785407, Akron 61.0678281691
Arizona St. 64.7151394863, Temple 58.0578420156
Syracuse 74.7825424779, Stephen F. Austin 60.5056731732
Clemson 74.4054903161, Michigan 70.8395522274
Oklahoma 78.5992492855 Morgan St. 59.7587888038
==================== ROUND 2 ====================
Louisville 67.3059313968 Ohio St. 60.5835683909
Utah 71.3007847464, Wake Forest 73.2895225467
West Virginia 67.9574088476, Kansas 67.4869037187
USC 62.1192840465 Michigan St. 64.56295945
Connecticut 76.8719158147, BYU 71.8412099454
Purdue 74.245343296, Washington 73.6100911982
Marquette 76.4607554812, Missouri 80.5497967091
California 64.7143532135, Memphis 70.9373235427
Pittsburgh 79.1278381289, Tennessee 70.6786108051
Wisconsin 63.0943233452, Xavier 63.5379857382
UCLA 74.1282015782, Villanova 71.4919550735
Texas 66.3817261194, Duke 70.9875941571
North Carolina 86.2296333847, LSU 73.8695973309
Illinois 62.6218220536, Gonzaga 65.6078661776
Arizona St. 74.0588194422, Syracuse 71.254787147
Clemson 76.9943827197, Oklahoma 78.9108038697
==================== SOFT 16 ====================
Louisville 72.8097088102, Wake Forest 68.2411945982
West Virginia 66.1905929215 Michigan St. 65.2198396254
Connecticut 70.4975234274, Purdue 67.014115714
Missouri 66.6046145365, Memphis 69.9964130636
Pittsburgh 72.8975484716, Xavier 64.848615134
UCLA 72.3676109557, Duke 73.1522519556
North Carolina 84.6606149747, Gonzaga 80.3910425893
Arizona St. 67.8668018941, Oklahoma 67.0441371239
==================== ELITE EIGHT ====================
Louisville 64.0822047092, West Virginia 61.7652102534
Connecticut 64.875382557, Memphis 65.9485921907
Pittsburgh 72.8027424093, Duke 70.5222034022
North Carolina 76.2640153058, Arizona St. 72.3363504426
==================== FINAL FOUR ====================
Louisville 60.7832463768, Memphis 61.4830569498
Pittsburgh 80.3421788636, North Carolina 81.0056716364
==================== FINAL GAME ====================
Memphis 73.8935857273, North Carolina 74.259537592
In the end, these predictions were enough to win my slice. Obviously, everything must be taken with a grain of salt, but being a PhD student in Machine Learning [http://www.machinelearningphdstudent.com/]it was fun to put my money where my mouth was and have a little fun.
Oh, and let me know if you want the data I gathered or the code I wrote to make this work – I’m happy to share it.
* I also regularize the latent vectors by adding independent zero-mean Gaussian priors (or equivalently, a linear penalty on the L2 norm squared of the latent vectors). This is known to improve these matrix factorization type models by encouraging them to be simpler and less prone to picking up spurious features in the data.
## Question about Characteristics Of A Function In Math
If you have any questions about Characteristics Of A Function In Math, please let us know, all your questions or suggestions will help us improve in the following articles!
The article Characteristics Of A Function In Math was compiled by me and my team from many sources. If you find the article Characteristics Of A Function In Math helpful to you, please support the team Like or Share!
Rate: 4-5 stars
Ratings: 5729
Views: 68437946
## Search keywords Characteristics Of A Function In Math
Characteristics Of A Function In Math
way Characteristics Of A Function In Math
tutorial Characteristics Of A Function In Math
Characteristics Of A Function In Math free
## Do You Need Math For Real Estate latest 2023
You are searching about Do You Need Math For Real Estate, today we will share with you article about Do You Need Math For Real Estate was…
## Cool Math Tricks For Elementary Students latest 2023
You are searching about Cool Math Tricks For Elementary Students, today we will share with you article about Cool Math Tricks For Elementary Students was compiled and…
## Cool Math Me And The Key latest 2023
You are searching about Cool Math Me And The Key, today we will share with you article about Cool Math Me And The Key was compiled and…
## Cool Math Learn To Fly 1 latest 2023
You are searching about Cool Math Learn To Fly 1, today we will share with you article about Cool Math Learn To Fly 1 was compiled and… | 2,681 | 9,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-06 | latest | en | 0.946476 |
https://www.unitconverters.net/flow/acre-foot-day-to-kilobarrel-us-day.htm | 1,660,854,277,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573399.40/warc/CC-MAIN-20220818185216-20220818215216-00491.warc.gz | 899,392,055 | 3,373 | Home / Flow Conversion / Convert Acre-foot/day to Kilobarrel (US)/day
# Convert Acre-foot/day to Kilobarrel (US)/day
Please provide values below to convert acre-foot/day [ac*ft/d] to kilobarrel (US)/day, or vice versa.
From: acre-foot/day To: kilobarrel (US)/day
### Acre-foot/day to Kilobarrel (US)/day Conversion Table
Acre-foot/day [ac*ft/d]Kilobarrel (US)/day
0.01 ac*ft/d0.0775839838 kilobarrel (US)/day
0.1 ac*ft/d0.7758398381 kilobarrel (US)/day
1 ac*ft/d7.7583983805 kilobarrel (US)/day
2 ac*ft/d15.5167967611 kilobarrel (US)/day
3 ac*ft/d23.2751951416 kilobarrel (US)/day
5 ac*ft/d38.7919919027 kilobarrel (US)/day
10 ac*ft/d77.5839838053 kilobarrel (US)/day
20 ac*ft/d155.1679676107 kilobarrel (US)/day
50 ac*ft/d387.9199190266 kilobarrel (US)/day
100 ac*ft/d775.8398380533 kilobarrel (US)/day
1000 ac*ft/d7758.3983805327 kilobarrel (US)/day
### How to Convert Acre-foot/day to Kilobarrel (US)/day
1 ac*ft/d = 7.7583983805 kilobarrel (US)/day
1 kilobarrel (US)/day = 0.1288925821 ac*ft/d
Example: convert 15 ac*ft/d to kilobarrel (US)/day:
15 ac*ft/d = 15 × 7.7583983805 kilobarrel (US)/day = 116.375975708 kilobarrel (US)/day | 425 | 1,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-33 | longest | en | 0.520601 |
http://forums.theregister.co.uk/forum/1/2009/11/18/surtees/ | 1,438,490,361,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042988962.66/warc/CC-MAIN-20150728002308-00329-ip-10-236-191-2.ec2.internal.warc.gz | 101,831,975 | 14,772 | # E-car driven from Dover to Calais
Veteran racing driver John Surtees has become the first man to drive from Britain to France in an electric car. John_Surtees_at_midpoint John Surtees at the Channel Tunnel's halfway point Credit: David Barzilay Surtees – the only man to ever win world championships on both four and two wheels – yesterday drove a prototype …
This topic is closed for new posts.
#### New campaign:
It's high time one of the service tunnels was opened to cyclists. Then we can stop all this John O'Groats to Land's End nonsense and do John O'Groats to Gibraltar!!!!!
Why is the limit 31 mph?
#### Pity.
Proof that 'leccy cars are down the tubes.
#### re neil 21
because 31mph is roughly equal to 50kph
#### Not the halfway point...
"The racing veteran also stopped at the tunnel’s halfway point, where French and British diggers first met in 1990 for a photo call as they signed the tunnel’s wall."
The French and British tunnelers did not meet at the halfway point of the tunnel (well, not as commonly defined). The actual meeting point was 13.9 miles from the English Coast and 9.7 miles from the French. The reason being that the geographical conditions on the French side are less favourable to tunneling.
31mph == 50kph
#### @The Indomitable Gall
I'm in Gibraltar now, I'll start cycling up and by that the time I get there they should have opened up that tunnel!
#### Speed Limit
The speed limit in the service passageways is actually 30.87 MPH, but they rounded up for print.
The reason is that if they went any faster, the drag caused by the car passing through the air would generate too much heat, which when added to the friction of the trains going through the tunnel, would throw the tunnel's environmental systems out of equilibrium.
Of course there's an alternative explanation that the service passageways are just REALLY narrow, but that's just crazy talk...
#### @Neil 21
its probably 50 km/h, which converted into mph is 'about' 31mph. Although why Europe gets to dictate the speed limit is beyond me.
#### 31mph
Eurotunnel safety manager Mr N Tuffnel explained that it was "1 faster" than 30mph
#### Not the first electric car from UK to France
First electric car from UK to France.. rubbish!
Eurotunnel has a fleet of electric Peugeot 106s that are used in the service tunnels that lock onto a guidance rail in the floor so that they can pass easily in the narrow tunnel.
They also have a buzzer that goes off on the dashboard that you have to press a button to de-activate after a number of crashes caused by dozing drivers! | 594 | 2,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2015-32 | longest | en | 0.958119 |
https://www.coursehero.com/file/6524223/EnergyinNutLab/ | 1,524,717,481,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948064.80/warc/CC-MAIN-20180426031603-20180426051603-00050.warc.gz | 746,590,566 | 61,004 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
EnergyinNutLab
# EnergyinNutLab - LAB HOW MUCH ENERGY IN A NUT INTRODUCTION...
This preview shows pages 1–2. Sign up to view the full content.
L AB : HOW MUCH ENERGY IN A NUT I NTRODUCTION In this activity you will determine the amount of energy stored in a nut. Water will act as the heat sink and therefore the calorie as a measure of heat will be used. MATERIALS balance an empty soft drink can (aluminum) graduated cylinder a thermometer a paper clip matches a pecan half a large clamp and ring stand 1. Mass a pecan half on the balance to the nearest 0.01 g and record the mass in your data table. 2. Fashion a stand from the paper clip and impale the nut on one end of the paper clip. 3. Measure 100 mL of water in a graduated cylinder and pour it into the soft drink can. 4. Clamp the soft drink can to a ring stand and suspend the can about an inch above the nut. Place a thermometer in the soft drink can and take the initial temperature of the water in the can and record it in your data table. 5. Now light the nut. It may take a little trying to start burning, but it will ignite. Stir the water in the can slowly as long as the nut burns.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]} | 337 | 1,412 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-17 | latest | en | 0.867387 |
http://openstudy.com/updates/5145d8dbe4b04cdfc58275fb | 1,444,361,797,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737913406.61/warc/CC-MAIN-20151001221833-00154-ip-10-137-6-227.ec2.internal.warc.gz | 230,234,540 | 9,559 | ## BrennaM 2 years ago Which expression is the factored equivalent of 4x2 + 23x + 15
1. DHASHNI
4x2 + 23x + 15 =4x2 +20x +3x +15 => 4x(x+5)+3(x+5) =>(x+5)(4x+3)
2. BrennaM
Thanks :)
Find more explanations on OpenStudy | 95 | 222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2015-40 | longest | en | 0.643512 |
http://iim-cat-questions-answers.2iim.com/DI_word_problems/wordprob1_4.shtml | 1,521,311,294,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645280.4/warc/CC-MAIN-20180317174935-20180317194935-00535.warc.gz | 140,676,720 | 9,568 | # DI Word Problems: The Cryptic Analyst
You are here: Home CAT Questionbank Di Word Problems The Cryptic Analyst Q.4
## DI Word Problems: The Cryptic Analyst
Five Companies A, B, C, D and E saw growth rates ranging from 10% to 50% in the year 2015. The company A with the least revenues of Rs. 600 crores in 2015 saw the maximum growth rate of 50% and the Company D with the highest revenue saw the least growth rate of 10%. Company B’s revenues in 2016 was equal to that of Company D in 2015, while Company C’s 2016 revenue was equal to that of Company B’s in 2015, Company A’s 2016 revenue was equal to that of Company E in 2015.
John, an accountant observes that one of the companies has twice the growth rate of another. Mathew, his colleague corrects him and says that this is the case in two different instances.
Company E has a revenue equal to the average seen in Company A and D, and growth rate equal to the average growth rate of A and D.
Ram, known for his cryptic-speak mentioned that if company C had grown at the rate seen by company A in 2015 would have reached the revenues seen by Company B in 2016.
In absolute terms, which company added the maximum revenue in 2016?
1. A
2. B
3. D
4. E
Choice A
## Detailed Solution
General Solution
We are going to construct and fill a table with details that we can infer from the statements. Company A has Revenues of Rs. 600 crores and sees a growth rate of 50%. Let Company D’s revenue in 2015 be X. The revenue of Company B in 2016 should also be X. Let revenue of Company B in 2015 be Y, this should be equal to revenue of Company C in 2016. Let us capture this in a table:
Company A’s 2016 revenue was equal to that of Company E in 2015.
The next statement that is very useful is “Company E has a revenue equal to the average seen in Company A and D, and growth rate equal to the average growth rate of A and D.”. So, Company E should have seen a growth rate of 30%. Revenue of Company D should be Rs. 1200 crores:
Or, if Company C had grown by 50%, it would have reached a revenue of X. Or, Company C had a revenue of 2/3 X in 2015.
Now, we move to these two statements – “John, an accountant observes that one of the companies has twice the growth rate of another. Mathew, his colleague corrects him and says that this is the case in two different instances.” Among the three % growths we already have, none is twice any other. So, the other two growth rates seen in B and C should somehow incorporate this.
So, the growth rates seen in B and C should be from either 2 * growth rates of A, D or E or ½ of growth rates seen in A, D or E. But since the growth rates should be between 10% and 50%, the only possibilities we have are 15%, 20% and 25%.
From the numbers, we can see that Company C should have grown by 25% and Company B by 20%
Solution to Question 2
Company A added Rs. 300 crores, the maximum addition in absolute terms.
## Our Online Course, Now on Google Playstore!
### Fully Functional Course on Mobile
All features of the online course, including the classes, discussion board, quizes and more, on a mobile platform.
### Cache Content for Offline Viewing
Download videos onto your mobile so you can learn on the fly, even when the network gets choppy! | 820 | 3,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-13 | latest | en | 0.968 |
https://mathoverflow.net/questions/157820/extreme-points-and-centroid | 1,607,044,421,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141733120.84/warc/CC-MAIN-20201204010410-20201204040410-00331.warc.gz | 377,841,415 | 28,799 | # Extreme points and centroid
It is well-known that the centroid of a triangle is the intersection point of its three medians. The medians happen to be area bisectors, but it seems that most (all?) other lines through the centroid are not area bisectors. With other polygons there are area bisectors which pass through the centroid but not through a vertex (for instance any line which passes through the centroid of a square), but based on the examples that I am able to compute it seems that any line which passes through a vertex and the centroid is an area bisector. This leads me to pose the following general question:
Let $C$ be a convex body in the plane and let $L$ be a line which passes through the centroid of $C$ and an extreme point of $C$. Is $L$ necessarily an area bisector?
I couldn't find any relevant tools in my usual convex geometry references, but I apologize if I missed something obvious.
• Take a line which goes through the centroid and is far from being an area bisector. You can deform C infinitesimally so that the intersection of L with the boundary is now a vertex. Now the line connecting the new vertex with the centroid is still far from being an area bisector. – Yoav Kallus Feb 17 '14 at 14:08
• Is it clear that you can deform $C$ as you describe while preserving convexity? – Paul Siegel Feb 17 '14 at 14:17
• Yes, I guess it is: e.g. replace the edge of a triangle with the arc of a circle with a very large radius. – Paul Siegel Feb 17 '14 at 14:18
• Please use top-level tags. – user9072 Feb 17 '14 at 14:25
The statement is obviously wrong. Take a square with side $a$ and triangle of hight $a\sqrt3$ on one of its sides. This side passes through the centroid but is not an area bisector.
• Sorry, what do you mean by the "triangle of height $a \sqrt{3}$ on one of its sides"? – Paul Siegel Feb 17 '14 at 14:16
• ??? Triangle whose base is the side and height equals $a\sqrt3$. Actually, Yoav's comment answers this even better. – Alex Degtyarev Feb 17 '14 at 14:21 | 511 | 2,013 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-50 | latest | en | 0.939629 |
https://community.nanog.org/t/satellite-latency-fwd/20534 | 1,718,349,420,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861521.12/warc/CC-MAIN-20240614043851-20240614073851-00165.warc.gz | 152,424,994 | 4,886 | # Satellite latency (fwd)
>The radius of the earth is about 6400km. Geostationary orbit is, as
>you note, 36000km above the equator. The path from the satellite to
>the North Pole is the hypotenuse of a right triangle with legs of
>6400km and (6400+36000)km. That gives a distance from the North Pole
>to the satellite of 43000km. It's reasonable to conclude that the
>distance from either New York or San Diego is less than that.
Is that 36000 km above the equator measures from the surface of the earth,
The surface of the Earth. (Or, more precisely, the surface of the
Earth that is directly below the satellite. It's further to other
parts of the surface of the earth. So, geostationary orbit is 36000km
above the equator. Or (36000+6400)km from the center of the earth.)
Looking at a spam I just received from satcast.com, it looks like they are
considering the distance from the dish to the satellite to be roughly 44000
miles. That's roughly 70000 km. Assuming that they are stating that from
the dish to the bird, that would account for the larger RTT times that
people see.
I'm not sure why spam should be considered a reliable source of
information, but, anyway, there's nowhere on earth where the distance
from the dish to a geosynch satellite is 44000 miles. I would guess
that they are referring to the round-trip distance -- 22K miles up,
22K miles down. That's not surprising, since geostationary orbit is
22K miles above the equator. (They are presumably ignoring the
slightly increased distance to the satellite from anywhere not directly
under the satellite.)
-- Brett | 402 | 1,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.942877 |
http://www.chegg.com/homework-help/grob-s-basic-electronics-student-cd-11th-edition-chapter-15.4-solutions-9780077427108 | 1,448,703,040,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398451744.67/warc/CC-MAIN-20151124205411-00288-ip-10-71-132-137.ec2.internal.warc.gz | 346,130,701 | 16,159 | View more editions
TEXTBOOK SOLUTIONS FOR Grob s Basic Electronics Student CD 11th Edition
• 1343 step-by-step solutions
• Solved by publishers, professors & experts
• iOS, Android, & web
Over 90% of students who use Chegg Study report better grades.
May 2015 Chegg Study Survey
PROBLEM
Chapter: Problem:
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 11
a. i = 2.5mA
• Step 2 of 11
b. i = 4.33mA
• Step 3 of 11
c. i = 5mA
• Step 4 of 11
d. i = 4.33mA
• Step 5 of 11
e. i = 2.5mA
• Step 6 of 11
f. i = 0mA
• Step 7 of 11
g. i = -2.5mA
• Step 8 of 11
h. i = -4.33mA
• Step 9 of 11
i. i = -5mA
• Step 10 of 11
j. i = -4.33mA
• Step 11 of 11
k. i = -2.5mA
Corresponding Textbook
Grob s Basic Electronics Student CD | 11th Edition
9780077427108ISBN-13: 0077427106ISBN: Mitchel SchultzAuthors: | 330 | 843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2015-48 | latest | en | 0.435151 |
https://www.physicsforums.com/threads/ether-and-how-it-was-disproved.304704/ | 1,529,281,640,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859904.56/warc/CC-MAIN-20180617232711-20180618012711-00567.warc.gz | 899,519,690 | 22,035 | # Ether and how it was disproved.
1. Apr 3, 2009
### RelativityKid
Hello, i am relatively new to Physics and find the subject fascinating, really seems to give insights into the unvierse like religion/spirituality does.
I was reading "A brief history of time" by Stephen Hawking and in the chapter "Space and time" he discusses Ether and how it was disproved.
For any who has read that book, could you explain to me how it was disproved?
It talks about relative speeds and how at right angles to this ether it SHOULD go at the speed of light, i have no idea what this means and whether it means the earths orbit of the sun or the earth spinning.
I would really appreciate a baby step explanation of this as i would really like to get my head around the topic.
2. Apr 3, 2009
3. Apr 3, 2009
### RelativityKid
Yes thankyou, this helped me.
But i am still unsure why light travelling at the same speed somehow disproves the existance of ether?
Please elaborate, as the article cleared up some of my doubts but not fully.
4. Apr 3, 2009
### Dmitry67
Before Tam came here :)
In fact, the story is not so straightforward. Special Realitvity technically had not disaproved the existance of ether: Lorentz had developed the theory called LET - it gives exactly the same predictions as SR.
In modern terms, LET is an interpretation of SR with a hidden variable (the absolute rest frame of the ether). So LET is worse then SR for the following reasons:
* ether is undetectable, remember about the Ocamms razor
* ether has too many weird and isconsistent properties
* hidden variables
Finally even Lorentz had abandoned that idea.
What is more important, SR is just a first step into GR, and there are no scientifically accepted analogs of LET which are compatible with GR.
5. Apr 3, 2009
### Bob S
Suppose the solar system were filled with ether. Would our planet be moving in it? How fast? Would the speed of light always be c when an observer were at rest in the ether, and not when the observer were in a another moving inertial frame? Would the Earth ever be at rest in the ether? Would this be summer, winter, spring or fall? Is our planet dragging the ether along with it? Or is the ether always at rest in every inertial system? How can this be?
6. Apr 3, 2009
### RelativityKid
Thankyou; but:
Why does the experiment disprove/nearly disprove/whatever ether?
7. Apr 3, 2009
### neopolitan
The Michelson Morley experiment failed to provide support for aether.
The scientific method goes a little like this (although it was not formalised as such at the time of Michelson and Morley):
Think up a theoretical model for how things work.
Create a falsifiable hypothesis from the model, or a number of hypotheses.
Create, and then conduct, a repeatable experiment which tests those hypotheses.
Report findings to your peers who can then repeat the experiment.
So long as your model and hypotheses continue to pass repeated tests, it becomes a Theory.
Michelson and Morely were testing for aether wind, which was a hypothetical consequence of the aether theory. They didn't find it and repeated experiments failed to find it.
If for there to be aether, there must be aether wind, then the absence of aether wind disproves aether.
As for LET (Lorentz Ether Theory, I prefer the older spelling version, since http://en.wikipedia.org/wiki/Ether" [Broken] refers to a chemical compound), LET is a model in which there is an aether, but it is undetectable because of Relativity. Some LET theorists could argue that there is no difference between aether and the "spacetime continuum" or the curved space that you see bent in many diagrams depicting the effects of gravity.
The problem with aether, is that by making it undetectable, you make a step in the scientific method very very difficult, if not impossible. You can't falsify aether if it is undetectable.
Einstein did make a comment at some stage to the effect that he hadn't necessarily destroyed aether (or the theoretical framework around it), but that he had reduced it to such an extent that it was as good as non-existent.
So, really, aether is not disproved - it's not disprovable! - but aether wind was disproved. And as DaleSpam says on occasion, there are benefits in thinking about some things in terms of LET, just as there are benefits in thinking about other things in other terms.
neopolitan
Last edited by a moderator: May 4, 2017
8. Apr 3, 2009
### yogi
What gets disproved is the notion of space being filled with a conventional like medium - one that acts like a gas or a fluid - when you try to measure the velocity of light with respect to this funny substance, the velocity always turns out to be the same in all directions - the null result was unexpected by M and M - and many interesting theories have been advanced to save the idea of an ether - you can read Einstein's own statments about a different kind of ether in some of his later writings - he attributes some of the properties of space as being ether-like ...the problem is, whatever space is, it can't be modeled like a conventional medium - the properties of space are sui generis - it has a characteristic impedance of 377 ohms as all antenna designers know - it has a capacity per unit length and an inductance per unit length - and it appears in some theories as having a characteristic that imparts inertia to particles - this would seem to be Einsteins's view - but others may interpret his comments differently
9. Apr 4, 2009
### clem
These numbers are just due to a mismatch of units in SI and have nothing to do with properties of space.
10. Apr 4, 2009
### Aether
LET and the standard formulation of SR are not different physical theories. They are the SAME physical theory cast in different coordinate systems. LET does not have any physically "weird and inconsistent properties" that are not also shared by the standard formulation of SR. LET is just as physically compatible with GR as is the standard formulation of SR.
11. Apr 4, 2009
### yogi
Re Post #9 - Einstein would disagree with you - the properties of space are more than a units mismatch - You would also find strong disagreement from from Maxwell and a host of others. If space isn't something how is it curved - and how does it contain energy? To relegate space to nothing discredits much of what Einstein attempted to convey. The units ratio is historic - it does not detract from the reality of spacetime.
12. Apr 5, 2009
### Dmitry67
Lets say LET and SR are different interpretations of the same theory, I mentioned it before
So LET=SR
Whats about GR? You say there is GLET=GR? Tam argued in the famous thread that GLET exists, but do you have any links?
13. Apr 5, 2009
### clem
Please read my post again. I wrote "These numbers are just due to a mismatch of units in SI and have nothing to do with properties of space." I read that to mean that space has properties that those numbers have nothing to do with.
14. Apr 5, 2009
### Aether
In GR, physics is modeled with tensor equations. We don't need a different "GLET" because the tensor equations of LET are known to be valid (e.g., they are the same as those given by the standard formulation of SR in at least one inertial frame).
I don't know what thread you are referring to here. There is no "GLET" afaik, but there is a metric tensor for LET (aka, GGT) that can be used in tensor equations.
Last edited: Apr 5, 2009
15. Apr 5, 2009
### Histspec
Specific models like Fresnel's (nearly) stationary aether were disproved by the Michelson-Morley-Experiment. Also models in which the Aether is fully dragged by matter were disproved by the Sagnac effect etc.
However, as it was shown by Lorentz and Poincaré, one can create an aether model (LET) in which the aether is totally undetectable. For example, see Lorentz's famous 1904-paper:
http://en.wikisource.org/wiki/Electromagnetic_phenomena
(Electromagnetic phenomena in a system moving with any velocity smaller than that of light)
However, as it was told above, LET was superseded by SR because an undetectable aether is considered an unnecessary artifact within a modern scientific theory.
Regards,
16. Apr 7, 2009
### yogi
True -the units for the electrical correlation are derived from basic units a la the SI - if these numbers have nothing to do with the characteristics of space - would you say the same with regard to a coax transmission line?
17. Jul 4, 2009
### matheinste
I believe there has been some discussion as to exactly what Einstein meant as regards an aether.
But anyway, leaving aside whatever Einstein meant, if as you believe, this new aether and spacetime have identical features, presumably we cannot tell them apart. So why not stick with spacetime.
Matheinste
Last edited by a moderator: Sep 25, 2014
18. Jul 4, 2009
### Bob S
Suppose we redefined ε0 and μ0 = 1. Then the speed of light in vacuum would be 1/sqrt(ε0 μ0) = 1, and the impedance of free space would be sqrt(μ00) = 1. But we would still have D = εE and B = μH (dielectric and magnetic materials). Maxwell's equations would not be fundamentally changed. We would always have inductance L and capacitance C. The velocity and impedance of a coaxial transmission line would still be 1/sqrt(LC) and sqrt(L/C) respectively. But in these units, the speed limit on our highways might be 9.5 x 10-8 c. We chose our units to be convenient in ordinary life. Cubits, stones, and days were all everyday units only a few hundred years ago, but none were very precise.
α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω
19. Jul 6, 2009
### Nickelodeon
Not sure what happened to Latin of Lite's post that you quoted - I couldn't find any evidence of it.
That aside, I think the reason for adopting a 'new aether' philosophy is that GR doesn't explain the mechanics of why space is curved in the vicinity of matter but is happy to regard it as a natural property of matter and stops there. The 'new aether', which seems to have an uncanny resemblance to dark energy, may take things a bit further.
20. Jul 9, 2009
### yogi
With respect to magnitudes - I would agree - but the ratio of units, irrespective of the magnitude(s) such as meters/sec squared conveys a physical thing ...farads/meter conveys a physical property
21. Jul 10, 2009
### pellman
Isn't the aether the same thing as the electromagnetic field? What defining property does the aether have which is not shared with the electromagnetic field?
I think the only difference between them is that when we say "aether" we mean to communicate we are referring to an old view of EM which supposes there must be a preferred reference frame. However, this is not a defining property of the aether. By definition the aether is simply "the thing that waves to produce the wave phenomenon 'light'". Which is, again by definition, the EM field.
It is often pointed out the aether had troublesome properties. For instance,
Nevertheless, by this point the mechanical qualities of the aether had become more and more magical: it had to be a fluid in order to fill space, but one that was millions of times more rigid than steel in order to support the high frequencies of light waves. It also had to be massless and without viscosity, otherwise it would visibly affect the orbits of planets. Additionally it appeared it had to be completely transparent, non-dispersive, incompressible, and continuous at a very small scale.
http://en.wikipedia.org/wiki/Luminiferous_aether
Well, does not the EM field have these very same properties?
22. Jul 10, 2009
### ZapperZ
Staff Emeritus
Not it's not. The classical ether was the "medium" for EM wave propagation.
Zz.
23. Jul 10, 2009
### pellman
Why don't we consider the EM field the medium for EM wave propagation?
Ok. Wait. I think I got it. With, say, a sound wave the wave corresponds to the variation in time and space of the quantity pressure. The answer to the question "Pressure of what?" is "Air," the medium. With a surface water wave, the wave corresponds to the variation in space of time of the quantity height. The answer to "The height of what?" is "The water," the medium.
With an EM wave, the wave corresponds to the variation in time and space of the quantity we call the EM tensor, or, if you prefer, the force per unit charge on an imaginary charge. But there is no "of what" underneath. There is only the EM tensor at each point. It is as if a wave could just be the variation of bare pressure with no associated fluid which it is the pressure of. Or a wave of varying height but not the height of anything.
In these latter cases the absence of the medium makes it nonsensical. But when you think about, you see that the presence of a medium for EM, while not non-sensical, is just unnecessary.
24. Jul 10, 2009
### [Nicolette]
I too am reading this book and had a few questions about this chapter. all the replies so far seem pretty deep rather than "baby step" so i thought I would throw in what I thought of it because my "understanding" is very simple.
so i thought the idea was that light waves would travel through the ether in the same manner that sound waves travel through air. so think about the doppler effect in the sense that the sound wave will appear to move at different speeds if you are moving toward the source as opposed to if you were moving perpendicular to the previous path, or not moving at all. but this was not the case for light, it travelled at the same speed during both instances.
please correct me if i have made a mistake, as i too am trying to better understand it. | 3,201 | 13,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-26 | latest | en | 0.968122 |
http://forum.roulette30.com/index.php?topic=426.0 | 1,519,353,598,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814311.76/warc/CC-MAIN-20180223015726-20180223035726-00700.warc.gz | 116,378,695 | 14,696 | ### Author Topic: Genius Bank System By Ahcoln! (Read 5979 times)
0 Members and 1 Guest are viewing this topic.
#### Reyth
• Global Moderator
• Hero Member
• Posts: 4189
• Thanked: 1446 times
##### Genius Bank System By Ahcoln!
« on: June 26, 2015, 05:47:23 PM »
Ahcoln's Super Recoup System
Hello everyone,
This is a very special newsletter. It will provide you with the full explanation of Ahcoln's Super Recoup system.
Ahcoln won the second place after my FRS in the GG roulette contest with his New Recoup system. Super Recoup is even a more enhanced system.
Ahcoln's Super Recoup is a sophisticated system. So, you can really make the most out of it.
Super Recoup is the kind of system, which lets you control your losses, bets flat during normal circumstances and escalates your winnings during a winning streak. So, it's an ideal system.
It bets on even bets, so you can use it for all even bets in Roulette, Craps and Baccarat.
The bet selection can be the player's choice. If you have a good bet selection, go ahead and use it. The one Ahcoln has chosen and the one, which will be given in this newsletter is simply follow the last decision. In Roulette, that would mean to bet Red if the last decision was Red and to bet Black if the last decision was Black. On a Green zero, you skip the next bet. The examples that will be given here will also be for Roulette for bets placed on Odd and Even.
The Super Recoup uses three betting stages.
- The "N"ormal stage
- The "R"ecoup stage
- The "P"arlay stage
The "N"ormal stage bets 1 unit flat on each decision.
The "R"ecoup stage tries to recover losses and bets the contents of the negative bankroll of various banks the system uses. If say the balance of a bank is -7 units, one will bet 7 units when one reaches the R stage.
If the bet is won, the negative balance of that bank is cleared and one is back to the "N"ormal stage. If the bet is lost, one freezes that balance in that bank and opens a new bank with that negative balance to be cleared at a later stage.
This technique avoids the betting of large amounts and also the reaching of the maximum bet limits.
We will elaborate on the use of various banks a bit later.
During the "P"arlay stage bets are increased during a winning streak, that is during consecutive wins.
During the "P" stage bets are increased according to the Fibonacci sequence: 1,1,2,3,5,8,13,21, etc., where your next bet is the sum of the previous two.
Do not get confused on the term Parlay, which usually means to let the bet ride on a win. It simply bets according to the numbers in the Fibonacci series, as per Ahcoln's terminology.
Since you are in "P"arlay stage after 2 consecutive wins in the "N"ormal stage, you start betting 2 units, then 3, 5, 8, as you keep winning, until your first loss. When you finally lose your bet, that loss eliminates the profits of the last 2 wins. However, if you have been winning a few bets consecutively, you will be able to store some profits (up until your last win minus the last 2 wins) into the profit bank (PB).
It's important to know when to switch between those stages.
Obviously we start betting in the "N"ormal stage, placing 1 unit flat on each spin.
2 consecutive wins will trigger either a "R"ecoup or a "P"arlay stage.
If you have no debts to the banks you're using, then 2 wins will trigger a "P"arlay. If you owe money to at least one of the banks, then 2 wins will trigger a "R"ecoup stage.
Now, it's a good time to talk about those banks. You will have one profit bank and a few recovery or recoup banks. Should you make profits up front, all your profits go to the profit bank and they are stored there.
For example, you start with the "N"ormal stage and bet 1 unit. If you win that bet you store your profit in the profit bank. Then you reset. Should you lose your second bet, your loss goes to the first recoup bank (RB_0). If you lose another bet, now you owe two units to the first recoup bank. Now your subsequent wins and losses are taken from the first recoup bank until you no longer owe that bank.
So if your balance is say -3 units in the RB_0 or recoup bank 0 (the first recoup bank), your next win of 1 unit in the "N"ormal stage is deducted from RB_0, brining your balance to -2 units.
As you keep betting in the "N"ormal stage with flat bets of 1 unit, you update your bank balance in the first recoup bank, should your balance be negative. Should your balance become positive, then you store your wins in the profit bank, since you would owe nothing to the recoup bank.
Now, if you have 2 wins in a row, and you owe some money to the recoup bank, this triggers a "R"ecoup stage, during which you bet all you owe in the first recoup bank. So if your balance is -4 units in the RB_0 (recoup bank 0), you bet 4 units to clear that debt. If you win that bet, you owe nothing to the first recoup bank . Then you're back to betting 1 unit flat in the "N"ormal stage. After that if you win twice in a row again, you are in the "P"arlay stage.
If you lose that 4 unit bet, you freeze the -4 units in the first recoup bank and open a new balance in a new recoup bank (we'll call it RB_1) and transfer those lost 4 units into that bank.
Your subsequent bets in the "N"ormal stage will now update the balance of RB_1.
If you have 2 wins in a row again, you're in "R"ecoup stage and you bet the entire balance of RB_1. If you win that bet, you are still in "R"ecoup stage and this time you bet the entire balance of RB_0. So debts in the recoup banks are cleared one at a time by consecutive wins.
Now, let's take some more complex stages, where there is a special recoup bank management method. If you have several recoup banks open, with balances of say -4, -5, -3, -6, you can eliminate one of the recoup banks by using the so called "add-2, adjust-1" method. You simply add the balances of those 4 banks: -4-5-3-6 = -18 and divide it by 3 = -6. So now you have 3 recoup banks instead of 4 with a balance of -6 on each.
2-adjust-1 is used when there are 2,4,6,8 or 10 recoup banks (multiples of 2 or even number of recoup banks) in debt. If there is only one bank, there is no adjustment, you just add 1 bank on a loss. If the current number of recoup banks is 2, you adjust before you add to 3 banks.
If you cancel a bank, reduce from 4-3, next you will add a bank if lose on "R" due current bank=3.
if you cancel a bank, reduce from 3-2, next you will adjust a bank if lose on "R" due to current bank=2.
All the above will be soon very clear when we illustrate all rules with a detailed example.
Just to summarize the rules before that:
The bet selection is follow the last decision. On a zero you skip the next bet.
Your profits go to the profit bank (PB).
Your losses go to the recoup bank(s), RB_0, RB_1, etc.
A new recoup bank is opened only if one loses in the "R"ecoup stage.
If your debt is clear to the recoup banks, 2 consecutive wins trigger a parlay, where you bet according to Fibonacci series.
If there are debts to the recoup banks, then 2 consecutive wins trigger the "R"ecoup stage, where you bet the entire balance of the most recent recoup bank opened. If the recoup wins, you apply another recoup if there are still debts in the recoup banks, betting from the last towards the first recoup bank. If not you reset to normal stage betting one unit.
Total number of spins 244
Total bets 584
Maximum recoup bet: 24
Maximum parlay bet: 8
Most -ve: -128
Ttotal winning sessions: 12
Total units in RB -4
Final net win: 8 units
Edge: 1.37%
The above was the way Ahcoln has played his Super Recoup. You are free to set your own rules when it comes to switching stages between "N", "R", and "P". You might trigger a parlay after 3 or 4 wins for instance. You may want to use a different progression than Fibonacci. You may simply let your chips ride after 4 wins and remove 1 chip for your profit bank. Or you may want to use a March progression after a few number of wins, where you bet 1 unit more every time you win.
Quoting from Ahcoln's explanation: "Switching states is the most important part and most original idea of this system. You will need to decide the condition on when to switch between the states of “N” to “R” or “P”. For example, in blackjack, you’ll play “N” until reaching the state where the shoe is rich with 10’s and Aces, here you will switch to “R” state to recoup all your losses, and when all Recoup banks are cancelled, you’ll switch to “P”."
I hope you liked Ahcoln's Super Recoup.
Thanking Ahcoln for his great contribution and for granting his permission to publish this strategy.
« Last Edit: June 26, 2015, 05:57:07 PM by Reyth »
#### Reyth
• Global Moderator
• Hero Member
• Posts: 4189
• Thanked: 1446 times
##### Re: Genius Bank System By Ahcoln!
« Reply #1 on: June 26, 2015, 05:59:37 PM »
Even though this system is way too much of a grind for far too little profit, this genius banking system can be used with any betting system with success. I think it is probably the best money/risk management tool I have ever seen.
I wanted to make sure I posted it so I don't forget about it.
#### GameNeverOver
• Mature Member
• Posts: 208
• Thanked: 12 times
##### Re: Genius Bank System By Ahcoln!
« Reply #2 on: June 26, 2015, 08:28:38 PM »
@Reyth
do some testing and write here the results from your test. There are lots of systems across the web that are stating the same (that they gave the player an EDGE) but when a little deeper testing is done, the systems fails..
#### Reyth
• Global Moderator
• Hero Member
• Posts: 4189
• Thanked: 1446 times
##### Re: Genius Bank System By Ahcoln!
« Reply #3 on: June 26, 2015, 08:32:26 PM »
I don't plan on using this system because in 244 spins it only generated 8 units of profit. What I possibly plan on doing is using the BANK SYSTEM in my Kav play so I can possibly reduce the bankroll requirements.
« Last Edit: June 26, 2015, 08:39:37 PM by Reyth »
#### GameNeverOver
• Mature Member
• Posts: 208
• Thanked: 12 times
##### Re: Genius Bank System By Ahcoln!
« Reply #4 on: June 26, 2015, 08:38:43 PM »
8 units per 244 spins? Thats the average profit? With what bankroll?
« Last Edit: June 26, 2015, 08:39:45 PM by Reyth »
#### Reyth
• Global Moderator
• Hero Member
• Posts: 4189
• Thanked: 1446 times
##### Re: Genius Bank System By Ahcoln!
« Reply #5 on: June 26, 2015, 08:40:39 PM »
I think 200 units would of covered everything.
#### GameNeverOver
• Mature Member
• Posts: 208
• Thanked: 12 times
##### Re: Genius Bank System By Ahcoln!
« Reply #6 on: June 26, 2015, 08:45:38 PM »
8 units avg profit per 244 spins with that low bankroll is actually very good. 8 units profit per sessions means doubling your bankroll every 25 sessions..
« Last Edit: June 26, 2015, 09:09:47 PM by Reyth »
#### GameNeverOver
• Mature Member
• Posts: 208
• Thanked: 12 times
##### Re: Genius Bank System By Ahcoln!
« Reply #7 on: June 26, 2015, 09:55:58 PM »
8 units avg profit per 244 spins with that low bankroll is actually very good. 8 units profit per sessions means doubling your bankroll every 25 sessions..
@Reyth
You wrote 24 spins, and 8 units profit per 24 spin is not good, it is EXCELLENT. But in 244 spins (around 6 to 8 hours constant playing) this is poor average profit.
But other important question here is:
is this profit consistent?
Test it when you'll have more free time, I'm curious to see the results.
#### Reyth
• Global Moderator
• Hero Member
• Posts: 4189
• Thanked: 1446 times
##### Re: Genius Bank System By Ahcoln!
« Reply #8 on: June 26, 2015, 09:59:49 PM »
I don't think I will be testing it because I don't need to spin 250 times to gain 10 units of profit. All I know is that it won an award in a contest of system submissions (it came in 2nd place). You or anyone else is free to test it however and I will be happy to try and help with any questions.
« Last Edit: June 26, 2015, 10:03:57 PM by Reyth »
#### roulettefan
• New
• Posts: 15
##### Re: Genius Bank System By Ahcoln!
« Reply #9 on: June 27, 2015, 07:19:38 AM »
hello reth congratulation for your work
where you find this system competition ?
thank
#### Reyth
• Global Moderator
• Hero Member
• Posts: 4189
• Thanked: 1446 times
##### Re: Genius Bank System By Ahcoln!
« Reply #10 on: June 27, 2015, 11:13:01 AM »
It took place on the Gambler's Glen forum:
t's been a busy month but here's the results. Some small errors in w[/size]eek one have been corrected.We used all 9 tables from Dec 1-14 at Hamburg, Germany, which was theequivalent of sitting at one table for about 3.5 months! So it was a good test of each method. The 'en prison' rule was active, which gave back 1/2 of even bets when the zero came up.CONGRATULATIONS TO THE WINNER: Izak and his "FRS II, method 2", whichbarely beat out Ahcoln's "New Recoup" for 1st place. FRS II only had 2 losing days, and New Recoup remarkably had none. (I am rechecking the New Recoup spreadsheet which Ahcoln graciously provided me, to see if there are any errors in it...)Third place was JBoyd's "Even Parlay 21." Also coming out positive for the two week, 28,429 spin contest were Tony's "Double Win & Out" method, and JBoyd's "Even Parlay 13." Here are the top five:Place___Name____Method__________________________Total___Wk_1____Wk_21_______Izak____FRS II - 2 Dozens, method 2___________7177____3961____32162_______Ahcoln__New Recoup______________________6580____3247____33333_______JBoyd___Even Parlay 21____________________2794____-398____31924_______Tony____Double Win and Out_________________715_____-62_____7775_______JBoyd___Even Parlay13_____________________424_______2_____422The following did not make a net profit, but at least they did not lose their 3000 unit bankroll!6_______Rob_________54_________________________-200______525_____-7257_______Izak________FRS II - 2 Dozens, method 1_______-469_____-630______1618_______Izak________Five Step Marty_________________-560_____-795______2359_______Kelly_______Classic Two Step_______________-1765_____-717____-104810______Sir Ivor______Hi / Low (hit and run)____________-1900_____-840____-106011______Puppy Chow__Investment Roulette____________-2296______597____-289312______Alltime_______20/20 sans progression_________-2732_____-505____-2227The following players lost all of their 3000 unit bankroll the secondweek:13______Spino___The Spino______________-3000___-1764___-123614______Rev_____Profit Maker____________-3000___-2128____-87215______Docker__Docker's System_________-3000___-2771____-22916______Elt_____Against All Odds__________-3000___-2049____-951And these players lost their bankroll the first week:17______Buffster___Luck Is Everything_________-3000_____-3000___018______Belle______Repeating Numbers II______-3000_____-3000___019______Tony_______Pretend System__________-3000_____-3000___020______Hernan_____Perdedor II_____________-3000_____-3000___021______Bill B_____Hi / Low Dozen___________-3000_____-3000___022______Grenoble Scientist____The Winner_____-3000_____-3000___023______Alltime____20/20___________________-3000_____-3000___024______Donali_____Donali's Secret___________-3000_____-3000___025______Carsch_____18 Number System_______-3000_____-3000___026______Free Roulette__Cover Method_________-3000_____-3000___027______Hernan_____Perdedor I______________-3000_____-3000___028______EuroRoul___D'Alembert______________-3000_____-3000___0Finally, my apologies to Turbo (Dozen Eliminator), Serena (14 # group), Ahcoln (Quarter Hit), Tater (Street Sleeper), OHPC (International Method), Roulette France (Mr. Opps Tribute), JapJap (Tokyo Streets) andJoe Daw (Joe Daw's Method), whose methods were either too complex forme to program, or I didn't get them finished yet. Perhaps next contest? I am working on Turbo's and Serena's, but may just put them in the next contest.-------------------------------------------------------------------I'll have more stats later, including a more full description of the methods that came out positive. I am also working on the "consolation prize" for the method that gained the "most per unit bet." It will be one of the top five.Thanks to all who entered. If you think I calculated your method wrongly, send me your corrections or suggestions or questions. It was fun, let's do it again next year.JB[Edited by JBoyd333 on 23-Dec-04 01:25][/color]
#### Reyth
• Global Moderator
• Hero Member
• Posts: 4189
• Thanked: 1446 times
##### Re: Genius Bank System By Ahcoln!
« Reply #11 on: June 27, 2015, 11:14:54 AM »
That's alot of editing. If you read carefully you can decipher the profitable ones at the top.
#### BlueAngel
• I always express my opinion
• Hero Member
• Posts: 1574
• Thanked: 244 times
• Gender:
• Do you want truth? You cannot handle the truth!
##### Re: Genius Bank System By Ahcoln!
« Reply #12 on: June 27, 2015, 01:00:06 PM »
That's alot of editing. If you read carefully you can decipher the profitable ones at the top.
Why not you to explain us the FRS II by Izak which was the winner instead of the runner??
Btw, Izak is Izak Mataya??
#### Reyth
• Global Moderator
• Hero Member
• Posts: 4189
• Thanked: 1446 times
##### Re: Genius Bank System By Ahcoln!
« Reply #13 on: June 27, 2015, 01:48:06 PM »
I don't think Izak Mataya openly reveals anything specific concerning his methods.
Also, the only reason I mentioned it is for the BANK system which I find ingenious. The actual system I find quite inefficient.
In fact, I am currently testing this bank system with my Kav system to see how it works... : )
« Last Edit: June 27, 2015, 01:50:24 PM by Reyth »
#### BlueAngel
• I always express my opinion
• Hero Member
• Posts: 1574
• Thanked: 244 times
• Gender:
• Do you want truth? You cannot handle the truth!
##### Re: Genius Bank System By Ahcoln!
« Reply #14 on: June 27, 2015, 01:52:24 PM »
I don't think Izak Mataya openly reveals anything specific concerning his methods.
Also, the only reason I mentioned it is for the BANK system which I find ingenious. The actual system I find quite inefficient.
But you are not giving straight answers, so I'm going to repeat: Why not the FRS II which was number 1 in the competition? | 5,002 | 18,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-09 | latest | en | 0.913128 |
https://www.graphpad.com/guides/prism/latest/curve-fitting/reg_analysis_checklist_simple_logistic.htm | 1,669,645,585,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710533.96/warc/CC-MAIN-20221128135348-20221128165348-00831.warc.gz | 842,832,110 | 14,011 | Please enable JavaScript to view this site.
Analysis checklist: Simple logistic regression
To check that simple logistic regression is an appropriate analysis for your these data, ask yourself these questions:
Is the outcome (Y) variable binary (dichotomous)?
The independent (Y) variable may only take on two values and in Prism, these must be coded as a 0 and a 1.
Are the observations (rows) independent?
One of the fundamental assumptions of logistic regression is that each row of data is a unique, independent observation. An example of independent observations is a study on 100 randomly selected people where a 1 indicates a positive outcome and a 0 a negative outcome, and each person is recorded on a single row. If each person was measured more than once (say at various time points in the study), then the observations are not independent and logistic regression isn't appropriate. Independence would also be questionable if some of the participants were in the same family as almost any outcome is likely to be more similar between two individuals from one family than from two unrelated individuals.
Does the model fit and predict the data well?
All models are wrong, but some are useful…
Prism offers a variety of metrics to evaluate how well the simple logistic model fits to the entered data. However, you should keep in mind that fitting models to data and interpretation of model fits is - to some extent - subjective. Some possibilities to consider when evaluating a given model include:
Does the model classify data well? In other words, does the model correctly predict the observed 0s and 1s? You can evaluate this in Prism a number of different ways such as with Tjur’s R squared, an ROC plot (with area under the ROC curve), and the row classification table.
Does the logistic model out-perform an intercept-only model? Prism tests this concept in two related, but slightly different ways: using the Wald test to examine if β1 is significantly non-zero, and using the likelihood ratio test to directly compare the given model with an intercept-only model.
Do you have sufficient data to trust your results?
As with all stats modeling, the more data (generally) the better. At the bottom of the tabular results sheet of the analysis results, Prism reports how many observations were included in the model (Rows analyzed). For simple logistic regression, a general rule of thumb is to have at least ten observations with an outcome of zero and ten observations with an outcome of 1.
Are you underfitting?
In the case of simple logistic regression, it’s possible that your predictor (X) variable is only one of multiple variables that affect if an outcome is a success or not. If the model prediction performance isn’t as good as desired, perhaps you’re missing some key variable(s) that you either didn’t measure or chose not to model. If you simply chose not to model them, you should definitely investigate their impact using multiple logistic regression - which is a natural extension of simple logistic regression. Read more about multiple logistic regression here. On the other hand, if the key variable is one that you didn’t measure, you’re out of luck. Go back and repeat the experiment with a focus on collecting more information. THEN come back and perform multiple logistic regression! | 658 | 3,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-49 | latest | en | 0.947508 |
http://mathhelpforum.com/calculus/78102-integral-question-print.html | 1,527,052,532,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865450.42/warc/CC-MAIN-20180523043959-20180523063959-00637.warc.gz | 178,129,422 | 2,650 | # integral question..
• Mar 11th 2009, 12:29 AM
transgalactic
integral question..
$\displaystyle \int \frac{x}{\sin ^2x}dx=-arcotx -\int -arccot x dx$
u=x u'=1
$\displaystyle v'=\frac{1}{\sin ^2x}dx$
how to solve integral of arc cot??
• Mar 11th 2009, 05:32 AM
mr fantastic
Quote:
Originally Posted by transgalactic
$\displaystyle \int \frac{x}{\sin ^2x}dx=-arcotx -\int -arccot x dx$
u=x u'=1
$\displaystyle v'=\frac{1}{\sin ^2x}dx$
how to solve integral of arc cot??
$\displaystyle \int u \, dv = uv - \int v \, du$.
$\displaystyle u = x \Rightarrow du = dx$
$\displaystyle dv = \frac{1}{\sin^2 x} \, dx \Rightarrow v = - \cot x$.
So your integral becomes $\displaystyle -x \cot x + \int \cot x \, dx$, which is simple to find.
• Mar 11th 2009, 05:38 AM
transgalactic
thanks :) | 289 | 789 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2018-22 | latest | en | 0.520037 |
http://thisquantumworld.com/wp/the-technique-of-quantum-mechanics/quantum-states-output/ | 1,474,983,934,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661087.58/warc/CC-MAIN-20160924173741-00024-ip-10-143-35-109.ec2.internal.warc.gz | 258,530,812 | 9,685 | # 11 Quantum states (output)
Let’s take stock. We learned that measurement outcomes are represented by the subspaces of a vector space. Because subspaces correspond one-to-one to projectors, this is the same as saying that
• Measurement outcomes are represented by the projectors of a vector space.
We also learned that
• The outcomes of compatible elementary tests correspond to commuting projectors.
Finally, we decided that if the subspaces A and B represent two possible outcomes of a measurement with three possible outcomes, then p(AUB) = p(A) + p(B). If we further take into account that
1. two subspaces corresponding to different outcomes of the same measurement are orthogonal (operationally this means that the probability of obtaining different outcomes in one and the same measurement is zero, formally this means that every vector in one subspace is orthogonal to every vector in the other),
2. projectors are said to be orthogonal if the corresponding subspaces are,
3. the sum of two orthogonal projectors is another orthogonal projector,
we arrive at the following:
• If PA and PB are orthogonal projectors, then the probability of the outcome represented by the sum of projectors PA + PB is the sum of the probabilities of the outcomes represented by PA and PB, respectively.
These three items (or “postulates”) allow us[1] to prove Gleason’s theorem[2], which holds for vector spaces with at least three dimensions. (More recently the validity of Gleason’s theorem has been extended to include 2-dimensional vector spaces.) The theorem states that the probability of obtaining the outcome represented by the projector P is given by
(Trace Rule) p(P) = Tr(WP),
where W is a unique operator, known as density operator, whose properties will be listed presently. To obtain the trace of an operator X, we apply X to the basis vectors a1, a2, a3,…, take the inner product of the resulting vectors with the same basis vectors, and sum over the basis vectors:
Tr(X) = <a1|Xa1> + <a2|Xa2> + <a3|Xa3> + ···
If P projects into a 1-dimensional subspace containing the unit vector v, the trace rule reduces to
p(P) = <v|Wv>.
The properties of the density operator (and the reasons why it has them) are as follows:
• W is linear. This ensures that the third postulate is satisfied: p(P1+P2) = p(P1) + p(P2), where P1 and P2 are orthogonal.
• W is self-adjoint. This ensures that the probability <v|Wv> is a real number. (What could be the meaning of a complex probability?)
• W is positive. This ensures that <v|Wv> does not come out negative. (What could be the meaning of a negative probability?)
• The trace of W equals 1. This ensures that the probabilities of the possible outcomes of a measurement add up to 1. Together with the positivity of the density operator, it ensures that no probability comes out greater than 1. (What could be the meaning of a probability greater than 1?)
• W2 = W or W2 < W.
The equality of an operator with its own square is characteristic of a projector. Since the trace of W equals 1, and since the trace of a projector equals the dimension of the subspace into which it projects, the equality W2 = W tells us that W projects into a 1-dimensional subspace. Since we began by upgrading from a point in a phase space to a line (or 1-dimensional subspace) in a vector space, we are not surprised by this result. If that subspace contains the unit vector w, the trace rule reduces to
p(P) = <w|Pw>,
and if P projects into a 1-dimensional subspace containing the unit vector v, it further reduces to
(Born’s Rule) p(P) = |<v|w>|2.
If the density operator satisfies the equality W2 = W, it is known as (or said to describe) a pure state, and the unit vector w is the so-called state vector. (Although W is uniquely determined by w, the converse is not true. If w1 and w2 only differ by their phases, they determine the same density operator and, hence, yield the same probabilities. They are therefore physically equivalent.)
If the density operator satisfies the inequality W2 < W, it is known as (or said to describe) a mixture or mixed state. A pure state defines probabilities distributions. It is a machine with inputs and outputs: insert the possible outcomes of the measurement you are going to make, insert the time of the measurement, and get the probabilities with which those outcomes are obtained. A mixed state defines a probability distribution over probability distributions. It adds a second layer of uncertainty to the uncertainty inherent in a pure state.
There are situations in which this additional uncertainty is subjective in the same sense in which probability distributions over a classical phase space are subjective: the uncertainty arises from a lack of knowledge of relevant facts. But there are also situations in which the additional uncertainty is due to a lack of relevant facts. In these situations it represents an additional objective fuzziness, over and above that associated with the individual algorithms. Later we shall come across examples of this kind of uncertainty.
Next
1. [↑] Peres, A. (1995). Quantum Theory: Concepts and Methods, Kluwer, p. 190.
2. [↑] Gleason, A.M. (1957). Measures on the closed subspaces of a Hilbert space, Journal of Mathematics and Mechanics 6, 885–894. | 1,195 | 5,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2016-40 | longest | en | 0.944759 |
https://math.stackexchange.com/questions/3830204/what-is-the-trick-needed-to-compute-the-integral-int-frac1x2x1-dx/3830247 | 1,719,003,205,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862157.88/warc/CC-MAIN-20240621191840-20240621221840-00010.warc.gz | 334,066,633 | 39,439 | # What is the trick needed to compute the integral $\int \frac{1}{x^{2}+x+1} dx$?
Working through Spivak's Calculus and using old assignments from the course offered at my school I'm working on the following problem, asking me to find the integral $$\int \frac{1}{x^{2}+x+1} dx$$
Looking through Spivak and previous exercises I worked on, I thought using a partial fraction decomposition would be the technique, but even in Spivak the only exercises I've seen which are similar involve:
$$\int \frac{1}{(x^{2}+x+1)^{n}} dx\ ,\text{where}\ n> 1$$
In which case it is pretty straightforward to solve. So there must be a reason why the exercise isn't presented unless it is so straightforward.
Integration by parts and substitution (at least for now) have proven fruitless as well. So I come here to ask if I'm missing any special trick to compute this integral ?
• Using $x^2+x+1 = (x+\frac12)^2+\frac34$ and use trigonometric substitution. Commented Sep 17, 2020 at 18:43
• Yoo could split the fraction into partial fractions, although I agree that the constants involved will be very messy. Commented Sep 17, 2020 at 19:11
It's actually very simple to identify this integral as this:
$$\int \frac{1}{x^2+x+1} dx = \int \frac{1}{\left(x+\frac{1}{2} \right)^2+ \frac{3}{4}} dx.$$
Now you can see that you can use the following rule for integration: $$\int \frac{1}{x^2+a^2}dx=\frac{1}{a} \arctan \left(\frac{x}{a} \right)+ C.$$
And now you get $$\int \frac{1}{\left(x+\frac{1}{2} \right)^2+ \frac{3}{4}} dx = \frac{2}{\sqrt{3}} \arctan \left(\frac{2x+1}{\sqrt{3}} \right) + C.$$
• Post in same timing Commented Sep 17, 2020 at 18:53
I'll take a different approach from the answers so far. For this approach, you'll have to be at least a bit comfortable working with complex numbers.
We can factor $$x^2 + x + 1$$ as $$\left(x - \frac{1 + \sqrt{3}i}{2}\right)\left(x - \frac{1 - \sqrt{3}i}{2}\right)$$.
That means we can rewrite the integral using partial fractions: $$\int \frac{dx}{x^2+x+1} = \int \frac{A dx}{x - \frac{1 +\sqrt{3}i}{2}} + \int \frac{B dx}{x - \frac{1 -\sqrt{3}i}{2}}$$ for some $$A, B \in \mathbb{C}$$. We can find $$A$$ and $$B$$ easily: $$Ax - A \frac{1 +\sqrt{3}i}{2} + Bx - B \frac{1 -\sqrt{3}i}{2} = 1$$. This gives us that $$A = -B$$, and $$A = -\frac{\sqrt{3}i}{3}$$.
These integrals are of the form $$\int \frac{Kdx}{x-L}$$, and are easy enough to solve: just use the substitution $$u = x - \frac{1 +\sqrt{3}i}{2}$$ for the first one, and $$u =x - \frac{1 -\sqrt{3}i}{2}$$ for the second one.
Though the efficient way has been given several times, decomposition in simple fractions remains your good friend.
The polynomial $$x^2+x+1$$ has complex roots, let $$\omega$$ and $$\omega^*$$, which, incidentally, are cube roots of unity. Now,
$$\frac1{x^2+x+1}=\frac1{(x-\omega)(x-\omega^*)}=\frac1{2i\Im(\omega)}\left(\frac1{x-\omega}-\frac1{x-\omega^*}\right)$$ and after integration,
$$\frac1{2i\Im(\omega)}\log\frac{x-\omega}{x-\omega^*}.$$
As $$|x-\omega|=|x-\omega^*|$$, the logarithm reduces to the difference of the arguments,
$$-\frac1{\Im(\omega)}\arctan\frac{\Im(\omega)}{x-\Re(\omega)}=-\frac2{\sqrt3}\arctan\frac{\sqrt3}{2x+1}.$$
There is even a faster way, noticing that
$$\frac1{x-\omega}=\frac{x-\omega^*}{(x-\omega)(x-\omega^*)}=\frac{x-\Re(\omega)}{x^2+x+1}+i\frac{\Im(\omega)}{x^2+x+1}.$$
Hence, taking the imaginary part,
$$\int\frac{dx}{x^2+x+1}=\frac{\Im(\log(x-\omega))}{\Im(\omega)}=-\frac1{\Im(\omega)}\arctan\frac{\Im(\omega)}{x-\Re(\omega)}.$$
As a byproduct,
$$\int\frac{{x-\Re(\omega)}}{x^2+x+1}dx=\frac{\Re(\log(x-\omega))}{\Im(\omega)}=\frac1{\Im(\omega)}\log\sqrt{(x-\Re(\omega))^2+\Im^2(\omega)}.$$
Hint : $$x^2+x+1 = (x+\frac{1}{2})^{2}+\frac{3}{4}$$
• going to give this a try. Will report back with progress....Thanks. Commented Sep 17, 2020 at 18:44
$$\int \frac{1}{(x+1/2)^2 + 3/4} dx= \frac{2}{\sqrt3}\tan^{-1}\frac{2x+1}{\sqrt3} +c$$ | 1,407 | 3,923 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 29, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-26 | latest | en | 0.878389 |
https://www.physicsforums.com/threads/electric-field-of-extended-mass.578189/ | 1,542,491,002,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743854.48/warc/CC-MAIN-20181117205946-20181117231946-00250.warc.gz | 946,677,594 | 15,596 | # Electric Field of extended mass
1. Feb 16, 2012
### Redoctober
So here is the scenario (see attachment) - I have a semicircle wire (radius R=15.9cm) which is made of insulator material , the semicircle consist of two combined quartercircle wires parts where one has equally distributed charge +Q and the other has -Q . Required is find the Electric field in direction of x at the origin . Q=5.33nC
My approach was as follows
Let E = 1/(4*pi*e)∫1/(R^2).dQ r
dQ=λ*ds and ds=R*dθ and i also know that unit vector r = cosθ*i+sinθ*j
therefore for the E in x direction i get this expression
E = 1/(4*pi*e)*1/(R^2)*λ*R∫cosθ.dθ
Integrating from 0 to pi ( thus taking only half of the semicircle ) and using λ as 2/(pi*r)
I get Q/(2*pi^2*e*R^2) .
Because the other half has opposite charge i can say that the Etot = Eneg +Epos
Therefore i multiply the equation by two to finaly get
Q/(pi^2*e*R^2)
If i put the values given i get as absolute value 2413 N/C for Electric field at origin of circel in the direction of x
Unfortunately it is a wrong solution :( !! What is the mistake i hv done ?? Can anyone spot it ? Thanks in advance
#### Attached Files:
• ###### Picture.png
File size:
1.1 KB
Views:
113
2. Feb 16, 2012
### tiny-tim
Hi Redoctober!
noooo
0 to π/2
3. Feb 16, 2012
### Redoctober
Oh !! srrry typo error :S :S , i integrated from 0 to pi/2 .
The answer 2413 N/C is wrong :/ !
4. Feb 16, 2012
### Redoctober
Anyone has a solution ?? :/ !
5. Feb 17, 2012
### JHamm
The E field for the right part of your semi circle is given by
$$E = \int_0^{\frac{\pi}{2}} \frac{-Q}{4\pi\epsilon_0r^2} d\theta$$
While the E field for the left part is the same but with +Q charge and integrated from $\frac{\pi}{2}$ to $\pi$. Just add these two together to get the total field.
6. Feb 17, 2012
### vanhees71
Are you sure? An electric field is a vector, and the correct formula is
$$\vec{E}(\vec{x})=\frac{1}{4 \pi \epsilon_0} \int_{\mathbb{R}^3} \rho(\vec{x}') \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}.$$
Of course, in the here considered case, you have to integrate along the semicircle and use the charge per length instead of the bulk-charge density. However, you should check, whether you have all the geometrical factors for you vector component right.
7. Feb 17, 2012
### Redoctober
Vanhees is right , Electric field is a vector .
So regarding the E will be in radial direction .
For the Y axis we need to consider the j component .
Vanhees , your mathematical expression is a bit high beyond my math skill xD !
My question is why the final expression for the problem is wrong :/
i considered the +Q and -Q E vector addition
I reached to the conclusion of
|E| = Q/(4*pi^2*e*r^2) (In the direction of J
Btw : how do u write the math expressions in that style xD . my way of typing the math is silly :/ !!
8. Feb 19, 2012
### Redoctober
The final equation i got was actually correct
i made an error with the calculation thats why my answer was wrong :)
Last edited: Feb 19, 2012
9. Feb 19, 2012
### vanhees71
Yes, I've checked your result too. It's correct. I parametrized the line charge with help of the charge per angle:
$$\lambda(\varphi)=\begin{cases} -\frac{2 Q}{\pi} & \text{for} \quad 0 \leq \varphi \leq \pi/2 \\ +\frac{2Q}{\pi} & \text{for} \quad \pi/2<\varphi\leq \pi \\ 0 & \text{elsewhere}. \end{cases}$$
Then you can use my formula for $\vec{x}=0$ to get
$$\vec{E}=\frac{1}{4 \pi \epsilon_0 r^2} \int_0^{2 \pi} \mathrm{d} \varphi' \lambda(\varphi') \begin{pmatrix} -\cos \varphi' \\ -\sin \varphi' \\0 \end{pmatrix}.$$
Integrating over the two regions gives finally
$$\vec{E}=\frac{Q}{\pi^2 \epsilon_0 r^2} \vec{e}_x.$$ | 1,188 | 3,682 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2018-47 | latest | en | 0.868878 |
http://studylib.net/doc/5430716/ppt | 1,529,451,107,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863259.12/warc/CC-MAIN-20180619232009-20180620012009-00027.warc.gz | 312,533,220 | 15,734 | ```L20: Putting it together:
Tree Search (Ch. 6)
November 29, 2011
• Next homework, CUDA, MPI (Ch. 3) and Apps (Ch. 6)
- Goal is to prepare you for final
- We’ll discuss it in class on Thursday
- Solutions due on Monday, Dec. 5 (should be straightforward
if you are in class)
• Poster dry run on Dec. 6, final presentations on Dec.
8
• Optional final report (4-6 pages) due on Dec. 14 can
that
Outline
• Next homework
• Poster information
• SC Followup: Student Cluster Competition
• Chapter 6 shows two algorithms (N-body and Tree
Search) written in the three programming models
- How to approach parallelization of an entire algorithm
(Foster’s methodology is used in the book)
- What do you have to worry about that is different for each
programming model?
Homework 4, due Monday, December 5 at
11:59PM
Instructions: We’ll go over these in class on December 1. Handin
“handin cs4961 hw4 <probfile>”
1.
Given the following sequential code, sketch out two CUDA
implementations for just the kernel computation (not the host
code). Determine the memory access patterns and whether
you will need synchronization. Your two versions should (a)
use only global memory; and, (b) use both global and shared
memory. Keep in mind capacity limits for shared memory and
limits on the number of threads per block.
…
int a[1024][1024], b[1024];
for (i=0; i<1020; i++) {
for (j=0; j<1024-i; j++) {
b[i+j] += a[j][i] + a[j][i+1] + a[j][i+2] + a[j][i+3];
}
}
Homework 4, cont.
2. Programming Assignment 3.8, p. 148. Parallel merge sort
starts with n/comm_sz keys assigned to each process. It
ends with all the keys stored on process 0 in sorted order. …
when a process receives another process’ keys, it merges the
new keys into its already sorted list of keys. … parallel
mergesort … Then the processes should use tree-structured
communication to merge the global list onto process 0 , which
prints the result.
3. Exercise 6.27, p. 350. If there are many processes and many
redistributions of work in the dynamic MPI implementation of
the TSP solver, process 0 could become a bottleneck for
energy returns. Explain how one could use a spanning tree of
processes in which a child sends energy to its parent rather
than process 0.
4. Exercise 6.30, p. 350 Determine which of the APIs is
preferable for the n-body solvers and solving TSP.
a.
How much memory is required… will data fit into the memory …
b.
How much communication is required by each of the parallel algorithms
(consider remote memory accesses and coherence as communication)
c.
Can the serial programs be easily parallelized by the use of OpenMP
directives?Doe they need synchronization constructs such as condition
Overview
• Last time we talked about solving n-body problems.
- “Regular” computation on a grid
- Parallelization relatively straightforward through standard
data and computation partitioning
• This lecture on Tree Search (specifically, singlesource shortest path)
- “Irregular”
- amount of work associated with a node or path varies
- Graph may be represented by dense or sparse adjacency
matrix (sparse adjacency matrix is also irregular)
- Impact on implementation
- Termination
- Collecting result plus path, synchronization
Tree search problem (TSP)
• An NP-complete problem.
• No known solution to TSP that is better in all cases
than exhaustive search.
• Ex., the travelling salesperson problem, finding a
minimum cost tour.
• (Also called single-source shortest path)
A Four-City TSP
0 is root or single source. What is the shortest path?
Pseudo-code for a recursive solution to TSP
using depth-first search
We’ll need to eliminate recursion
• Non-recursive implementation is more flexible
• Shared data structures
• Freedom to schedule threads/processes flexibly
• How to eliminate recursion
• Explicit management of “stack” data structure
• Loop instead of recursive calls
Pseudo-code for a second solution to TSP that
doesn’t use recursion (two solutions in text)
Termination and find best solution
on path?
new to path?
shorter than existing paths?
Global variable – stores “best_tour”
• How to know whether global “best tour” value is
better?
• Guarantee that any writes will be performed
atomically
• Textbook suggests reading it without locking
• Value may be slightly out of date, but just
increases work slightly since value is
monotonically decreasing
• Cost of synchronizing read not worth it
• If the process’s value is not as good, not
updated
Making sure we have the “best tour”
• In the case where a thread tests and decides
it has a better global solution, we need to
ensure two things:
1) That the process locks the value with a mutex,
preventing a race condition.
2) In the possible event that the first check was
against an old value while another process was
updating, we do not put a worse value than the
new one that was being written.
• We handle this by locking, then testing again.
First scenario
process x
local
tour value
global
tour value
30
22
27
22
process y
local
tour value
27
3. test
1. test
6. lock
2. lock
7. test again
4. update
8. update
5. unlock
9. unlock
Second scenario
process x
local
tour value
global
tour value
30
27
29
process y
local
tour value
27
3. test
1. test
6. lock
2. lock
7. test again
4. update
8. unlock
5. unlock
Pseudo-code for a Pthreads implementation of a
statically parallelized solution to TSP
Thread operates on its private stack
Needs mutex (next slides)
A Careful Look at Load Imbalance
• The master thread is partitioning the work
• Some expensive paths will be pruned early
• What to do?
• Schedule some work statically and some
dynamically for under-utilized processors
• Minimize cost of moving work
• This could be done a lot smarter than it is
• Challenges:
• Which processes are available to take on more
work?
• When does execution terminate?
Dynamic Parallelization of Tree Search
• Termination issues.
• Code executed by a thread before it splits:
- It checks that it has at least two tours in its stack.
- It checks that there are threads waiting.
- It checks whether the new_stack variable is NULL.
Do I have extra work, and are there idle threads?
Termination test
Pseudo-Code for Pthreads Terminated Function, cont.
Grouping the termination variables
Parallelizing the Tree Search Programs
Using OpenMP
• Same basic issues implementing the static and dynamic
parallel tree search programs as Pthreads.
• A few small changes can be noted.
Other OpenMP changes
• Use a critical section instead of a mutex to guard
selection of “best tour”, but similar “optimistic”
synchronization
• Need a different construct for conditional wait (not
clean)
- Next slide, but more in textbook to select particular idle
- Requires auxiliary stack of idle threads to select
OpenMP emulated condition wait
MPI is similar
• Static and dynamic partitioning schemes
• Maintaining “best_tour” requires global
synchronization, could be costly
- May be relaxed a little to improve efficiency
- Alternatively, some different communication constructs can
be used to make this more asynchronous and less costly
- MPI_Iprobe checks for available message rather than
actually receiving
- MPI_Bsend and other forms of send allow aggregating
results of communication asynchronously
Sending a different number of objects to
each process in the communicator to
distribute initial stack
Gathering a different number of objects from
each process in the communicator
Checking to see if a message is available
Terminated Function for
a Dynamically
Partitioned TSP solver
that Uses MPI.
Printing the best tour
Terminated Function for a Dynamically
Partitioned TSP solver with MPI (1)
Terminated Function for a Dynamically
Partitioned TSP solver with MPI (2)
Packing data into a buffer of contiguous
memory
Unpacking data from a buffer of contiguous
memory
Summary of Lecture
• This “tree search” is the hardest parallel
programming challenge we have studied
which in turn leads to termination challenges requiring elaborate
synchronization and communication
• This may be too hard for an efficient OpenMP implementation
and is also challenging for MPI
• We did not even talk about locality!
• This chapter is dense, but getting through it is worth
it
• It shows REAL parallel programming and the need
for elaborate language constructs
• Other “irregular” algorithms are even more difficult
``` | 1,986 | 8,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-26 | latest | en | 0.869285 |
https://metanumbers.com/326685823 | 1,632,202,604,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057158.19/warc/CC-MAIN-20210921041059-20210921071059-00502.warc.gz | 436,267,234 | 7,511 | # 326685823 (number)
326,685,823 (three hundred twenty-six million six hundred eighty-five thousand eight hundred twenty-three) is an odd nine-digits composite number following 326685822 and preceding 326685824. In scientific notation, it is written as 3.26685823 × 108. The sum of its digits is 43. It has a total of 2 prime factors and 4 positive divisors. There are 325,522,960 positive integers (up to 326685823) that are relatively prime to 326685823.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 9
• Sum of Digits 43
• Digital Root 7
## Name
Short name 326 million 685 thousand 823 three hundred twenty-six million six hundred eighty-five thousand eight hundred twenty-three
## Notation
Scientific notation 3.26685823 × 108 326.685823 × 106
## Prime Factorization of 326685823
Prime Factorization 281 × 1162583
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 326685823 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 326,685,823 is 281 × 1162583. Since it has a total of 2 prime factors, 326,685,823 is a composite number.
## Divisors of 326685823
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 3.27849e+08 Sum of all the positive divisors of n s(n) 1.16286e+06 Sum of the proper positive divisors of n A(n) 8.19622e+07 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 18074.5 Returns the nth root of the product of n divisors H(n) 3.98581 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 326,685,823 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 326,685,823) is 327,848,688, the average is 81,962,172.
## Other Arithmetic Functions (n = 326685823)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 325522960 Total number of positive integers not greater than n that are coprime to n λ(n) 162761480 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 17597031 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 325,522,960 positive integers (less than 326,685,823) that are coprime with 326,685,823. And there are approximately 17,597,031 prime numbers less than or equal to 326,685,823.
## Divisibility of 326685823
m n mod m 2 3 4 5 6 7 8 9 1 1 3 3 1 2 7 7
326,685,823 is not divisible by any number less than or equal to 9.
## Classification of 326685823
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (326685823)
Base System Value
2 Binary 10011011110001101010001111111
3 Ternary 211202201100200221
4 Quaternary 103132031101333
5 Quinary 1132112421243
6 Senary 52230002211
8 Octal 2336152177
10 Decimal 326685823
12 Duodecimal 914a6367
20 Vigesimal 521feb3
36 Base36 5ei0e7
## Basic calculations (n = 326685823)
### Multiplication
n×y
n×2 653371646 980057469 1306743292 1633429115
### Division
n÷y
n÷2 1.63343e+08 1.08895e+08 8.16715e+07 6.53372e+07
### Exponentiation
ny
n2 106723626949187329 34865095903440241755536767 11389932549189303909226493534154241 3720929488746395730382774333609356830025343
### Nth Root
y√n
2√n 18074.5 688.721 134.441 50.4459
## 326685823 as geometric shapes
### Circle
Diameter 6.53372e+08 2.05263e+09 3.35282e+17
### Sphere
Volume 1.46043e+26 1.34113e+18 2.05263e+09
### Square
Length = n
Perimeter 1.30674e+09 1.06724e+17 4.62004e+08
### Cube
Length = n
Surface area 6.40342e+17 3.48651e+25 5.65836e+08
### Equilateral Triangle
Length = n
Perimeter 9.80057e+08 4.62127e+16 2.82918e+08
### Triangular Pyramid
Length = n
Surface area 1.84851e+17 4.10889e+24 2.66738e+08
## Cryptographic Hash Functions
md5 09a856dac41ad679bc7b0e95f849edc0 e82810b9e792fb320da7f07115bd6c011e0f4674 8a1609acb3ab517e195a7387b95cc195199e5ce091f1cc8beb66f55b9d593ffa ec81bbc65138c1f70208a6a6bca4a19bbb37d696ef9fb614afd45709862c7aa717cbf71fa161c3539a973220346f60ce2913c048953ec24ac9ce13b51ee9712b a8ab0cac225a67f0b7449c948963e3773e1f04d8 | 1,615 | 4,624 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-39 | latest | en | 0.788873 |
https://acm.timus.ru/forum/thread.aspx?id=24241&upd=634130627407968750&locale=ru | 1,709,208,401,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474808.39/warc/CC-MAIN-20240229103115-20240229133115-00685.warc.gz | 79,486,811 | 3,268 | ENG RUS Timus Online Judge
Online Judge
Задачи
Авторы
Соревнования
О системе
Часто задаваемые вопросы
Новости сайта
Форум
Ссылки
Архив задач
Отправить на проверку
Состояние проверки
Руководство
Регистрация
Исправить данные
Рейтинг авторов
Текущее соревнование
Расписание
Прошедшие соревнования
Правила
вернуться в форум
## Обсуждение задачи 1359. Стройка
How to solve this one?
Послано Douglas Cardoso 25 июн 2010 11:39
I really couldn't figure out how to solve it. The way I tried even gives the wrong output for the sample! I modelled like this:
Each integer coordinates point represents a node of a graph. There is a edge form point (a, b) to (c, d) iff a >= c and b > d. The edge's weight is equal to the time to go from one point to another one adjacent to the first, which is calculated like this:
if theta is the angle between the segment (a, b)-(c, d) and the x-axis, then due to the gravity there is a 10*cos(theta) acceleration in the x-axis direction. So the time needed to go from (a, b) to (c, d) is sqrt(2*(c-a)/(10*cos(theta))).
Finally, what I do is to run a shortest-path algorithm in this graph. But for the sample my program answers 0.752121.
Does any one knows if ignored any detail? Or maybe if my entire model is wrong?
Thanks! | 388 | 1,257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-10 | latest | en | 0.428954 |
https://gmatclub.com/forum/a-retailer-sells-only-radios-and-clocks-160076.html?fl=similar | 1,487,965,998,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171629.92/warc/CC-MAIN-20170219104611-00646-ip-10-171-10-108.ec2.internal.warc.gz | 719,879,747 | 62,141 | A retailer sells only radios and clocks : GMAT Data Sufficiency (DS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
It is currently 24 Feb 2017, 11:53
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# A retailer sells only radios and clocks
Author Message
TAGS:
### Hide Tags
Current Student
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 64
Kudos [?]: 604 [1] , given: 355
### Show Tags
18 Sep 2013, 14:50
1
KUDOS
1
This post was
BOOKMARKED
00:00
Difficulty:
55% (hard)
Question Stats:
63% (02:28) correct 37% (01:58) wrong based on 104 sessions
### HideShow timer Statistics
A retailer sells only radios and clocks. If she currently has 44 total items in inventory, how many of them are radios?
(1) The retailer has more than 28 radios in inventory
(2) The retailer has less than twice as many radios as clocks in inventory.
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 37108
Followers: 7252
Kudos [?]: 96513 [0], given: 10753
### Show Tags
18 Sep 2013, 14:59
Expert's post
3
This post was
BOOKMARKED
A retailer sells only radios and clocks. If she currently has 44 total items in inventory, how many of them are radios?
R + C = 44.
(1) The retailer has more than 28 radios in inventory --> R > 28. Not sufficient.
(2) The retailer has less than twice as many radios as clocks in inventory --> R < 2C --> . Not sufficient.
(1)+(2) Since from the stem we know that C = 44 - R, then from (2) we get that R < 2(44 - R) --> R < 88/3=~29.something. So, we have that 28 < R < 29.something, which means that R = 29. Sufficient.
_________________
Intern
Joined: 17 May 2013
Posts: 2
Followers: 0
Kudos [?]: 0 [0], given: 1
### Show Tags
15 Oct 2013, 12:18
Hi Bunuel,
Please correct me if I am wrong.
As per my understanding, the second statement equation becomes C<2R
Math Expert
Joined: 02 Sep 2009
Posts: 37108
Followers: 7252
Kudos [?]: 96513 [1] , given: 10753
### Show Tags
15 Oct 2013, 12:39
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
aniketrathi wrote:
Hi Bunuel,
Please correct me if I am wrong.
As per my understanding, the second statement equation becomes C<2R
No.
Consider this: say we are told that "the retailer has twice as many radios as clocks in inventory", then we would have R=2C (twice as many radios as clocks).
But we are told that "the retailer has less than twice as many radios as clocks in inventory", so it should be R<2C.
Hope it's clear.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13955
Followers: 590
Kudos [?]: 167 [0], given: 0
### Show Tags
23 Jun 2015, 23:01
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
SVP
Joined: 17 Jul 2014
Posts: 2328
Location: United States (IL)
Concentration: Finance, Economics
Schools: Stanford '19 (S)
GMAT 1: 560 Q42 V26
GMAT 2: 550 Q39 V27
GMAT 3: 560 Q43 V24
GMAT 4: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Followers: 23
Kudos [?]: 272 [0], given: 145
### Show Tags
05 Apr 2016, 17:25
1. r>28 - not sufficient.
2. r<2C - not sufficient.
1+2
r=29 - yes
r=30 - no
r=29
sufficient. C
Re: A retailer sells only radios and clocks [#permalink] 05 Apr 2016, 17:25
Similar topics Replies Last post
Similar
Topics:
A certain company sells only cars and trucks, and the average selling 0 21 Feb 2017, 17:23
5 A certain company that sells only cars and trucks reported 5 13 Oct 2013, 00:08
A certain bakery only sells bread. It sells bread at a 3 27 May 2012, 08:55
16 Company Z only sells chairs and tables. what percent of its 5 29 Oct 2010, 03:23
Company z sells only chairs and tables. What percent of 5 16 Jul 2010, 15:55
Display posts from previous: Sort by | 1,396 | 4,638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-09 | longest | en | 0.907905 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.