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 |
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https://www.calculatoratoz.com/en/rectangle-diagonal-in-terms-of-cosine-of-the-angle-that-adjacent-to-the-diagonal-and-the-adjacent-side-of-the-angle-calculator/Calc-595 | 1,610,774,907,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703500028.5/warc/CC-MAIN-20210116044418-20210116074418-00431.warc.gz | 724,538,505 | 33,999 | ## < ⎙ 11 Other formulas that you can solve using the same Inputs
Diagonal of a Rectangle when breadth and perimeter are given
Surface Area of Cuboid
Magnetic Flux
Magnetic Flux=Magnetic Field*Length*Breadth*cos(θ) GO
Diagonal of a Rectangle when breadth and area are given
Area of a Rectangle when breadth and diagonal are given
Area of a Rectangle when breadth and perimeter are given
Diagonal of a Rectangle when length and breadth are given
Length of rectangle when diagonal and breadth are given
Length of rectangle when perimeter and breadth are given
Area of a Rectangle when length and breadth are given
Length of rectangle when area and breadth are given
## < ⎙ 11 Other formulas that calculate the same Output
Diagonal of a Rectangle when breadth and perimeter are given
Diagonal of a Rectangle when length and perimeter are given
Diagonal=sqrt((2*(Length)^2)-(Perimeter*Length)+((Perimeter)^2/4)) GO
Diagonal of a Rectangle when breadth and area are given
Diagonal of a Rectangle when length and area are given
Diagonal=sqrt(((Area)^2/(Length)^2)+(Length)^2) GO
Diagonal of the rectangle when the radius of the circumscribed circle is given
Diagonal=2*Radius Of Circumscribed Circle GO
Diagonal of a Rectangle when length and breadth are given
Diagonal of a Square when perimeter is given
Diagonal=(Perimeter/4)*sqrt(2) GO
The maximum face diagonal length for cubes with a side length S
Diagonal=Side*(sqrt(2)) GO
Diagonal of a Square when side is given
Diagonal=Side*sqrt(2) GO
Diagonal of a Square when area is given
Diagonal=sqrt(2*Area) GO
Diagonal of a Cube
Diagonal=sqrt(3)*Side GO
### Rectangle diagonal in terms of cosine of the angle that adjacent to the diagonal and the adjacent side of the angle Formula
More formulas
Perimeter of a rectangle when length and width are given GO
Diagonal of a Rectangle when length and breadth are given GO
Perimeter of a rectangle when diagonal and length are given GO
Perimeter of rectangle when diagonal and width are given GO
Diagonal of a Rectangle when breadth and area are given GO
Diagonal of a Rectangle when length and area are given GO
Diagonal of a Rectangle when length and perimeter are given GO
Diagonal of a Rectangle when breadth and perimeter are given GO
Diagonal of the rectangle when the radius of the circumscribed circle is given GO
Rectangle diagonal in terms of sine of the angle GO
Perimeter of rectangle when area and rectangle length are given GO
The perimeter of the rectangle when the length and radius of the circumscribed circle are given GO
Perimeter of rectangle when breadth and radius of circumscribed circle are given GO
The perimeter of a rectangle when the diameter of circumscribed circle and length are given GO
Perimeter of rectangle when breadth and diameter of circumscribed circle GO
Rectangle circumscribed radius in terms of sine of the angle that adjacent to the diagonal and the opposite side of the angle GO
## What is Diagonal of a Rectangle?
A diagonal of a rectangle cut the rectangle into 2 right triangles with sides equal to the sides of the rectangle and with a hypotenuse that is diagonal. To find the diagonal of the rectangle divide breadth by the cosine of the angle formed between the diagonal and adjacent side.
## How to Calculate Rectangle diagonal in terms of cosine of the angle that adjacent to the diagonal and the adjacent side of the angle?
Rectangle diagonal in terms of cosine of the angle that adjacent to the diagonal and the adjacent side of the angle calculator uses Diagonal=Breadth/cos(Theta) to calculate the Diagonal, Rectangle diagonal in terms of cosine of the angle that adjacent to the diagonal and the adjacent side of the angle is a straight line joining two opposite corners of a rectangle. Diagonal and is denoted by d symbol.
How to calculate Rectangle diagonal in terms of cosine of the angle that adjacent to the diagonal and the adjacent side of the angle using this online calculator? To use this online calculator for Rectangle diagonal in terms of cosine of the angle that adjacent to the diagonal and the adjacent side of the angle, enter Breadth (b) and Theta (ϑ) and hit the calculate button. Here is how the Rectangle diagonal in terms of cosine of the angle that adjacent to the diagonal and the adjacent side of the angle calculation can be explained with given input values -> 2.309401 = 2/cos(30).
### FAQ
What is Rectangle diagonal in terms of cosine of the angle that adjacent to the diagonal and the adjacent side of the angle?
Rectangle diagonal in terms of cosine of the angle that adjacent to the diagonal and the adjacent side of the angle is a straight line joining two opposite corners of a rectangle and is represented as d=b/cos(ϑ) or Diagonal=Breadth/cos(Theta). Breadth is the measurement or extent of something from side to side and Theta is an angle that can be defined as the figure formed by two rays meeting at a common endpoint.
How to calculate Rectangle diagonal in terms of cosine of the angle that adjacent to the diagonal and the adjacent side of the angle?
Rectangle diagonal in terms of cosine of the angle that adjacent to the diagonal and the adjacent side of the angle is a straight line joining two opposite corners of a rectangle is calculated using Diagonal=Breadth/cos(Theta). To calculate Rectangle diagonal in terms of cosine of the angle that adjacent to the diagonal and the adjacent side of the angle, you need Breadth (b) and Theta (ϑ). With our tool, you need to enter the respective value for Breadth and Theta and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Diagonal?
In this formula, Diagonal uses Breadth and Theta. We can use 11 other way(s) to calculate the same, which is/are as follows -
• Diagonal=Side*sqrt(2)
• Diagonal=sqrt(3)*Side | 1,292 | 5,849 | {"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-2021-04 | latest | en | 0.79409 |
https://as.axahealthkeeper.com/98jkght/d891ri.php?35d6c0=vector-formulas-pdf | 1,618,245,694,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038067870.12/warc/CC-MAIN-20210412144351-20210412174351-00486.warc.gz | 186,966,763 | 29,033 | # vector formulas pdf
Password? Signup now to download free sample papers and notes. stream The length of the line shows its magnitude and the arrowhead points in the direction. All Rights Reserved, Your account is your portal to all things, By Login, you agree to our Terms and Privacy Policies. Account. a 2D vector f(t) for each t ∈ I, which may be regarded as the position vector of some point on the plane. r rx ry r r r = + Vector Addition by Components You can add two vectors by adding the components of the vector along each direction. Vector Formulas In these notes we use notation like F for vector valued functions and we use either F(t) = hf 1(t);f 2(t);f 3(t)i= f 1(t)i+ f 2(t)j + f 3(t)k for vector valued functions in R3 or F(t) = hf 1(t);f 2(t)i= f 1(t)i + f 2(t)j for vector valued functions in R2.In what follows we will usually give the formulas for R3.If a formula … 3 0 obj << For example, recall the Section Formula from Level 1. Conditions. The length of the line shows its magnitude and the arrowhead points in the direction. The fixed point is called the origin. The graph of a function of two variables, say, z=f(x,y), lies in Euclidean space, which in the Cartesian coordinate system consists of all ordered triples of real numbers (a,b,c). 2, 11 A vector of a of constant length (but varying direction) is a function of time, Show that da/dt is perpendicular to a. Please below, Didn't receive OTP ?Resend Chapter wise Sessions. 10.6.3 Vector (or cross) product of two vectors. Any vector can be broken down into components along the x and y axes. Back. Know More about these in Vector Algebra Class 12 Formulas PDF with Notes List. %���� Already have an account? you signed up Revision And MCQ Practice Of Physical Chemistry Topics For MHT-CET & NEET. for your Vidyakul account.We will send you an OTP (One Time Password) to reset your We can add two vectors by joining them head-to-tail: vector add a+b. Let PQ be any vector. // Figure 5 Theorem. Note that you can only add components which lie along the same direction. )������p�sغ} AkQW�pU/ ���N��]. A vector has magnitude (size) and direction: vector magnitude and direction. Please VECTOR CALCULUS: USEFUL STUFF Revision of Basic Vectors A scalar is a physical quantity with magnitude only A vector is a physical quantity with magnitude and direction A unit vector has magnitude one. vector perpendicular to the plane defined b y the tw o original v ectors when translated to a common origin, and of magnitude equal to the product of the absolute values of the original vectors multiplied by the sine of the angle b et w een them. Enter new password for your Vidyakul Please provide the Phone Number you used when IIT - Super 30 Team Live Class. We can add two vectors by joining them head-to-tail: vector add a+b. Vector Algebra x 13.1. All Chapters of Physics, chemistry & Maths are covered. The topics and sub-topics covered in Vector Algebra Class 12 Formulas PDF with Notes are: 10.5 Multiplication of a Vector by a Scalar, 10.6.1 Scalar (or dot) product of two vectors. We note in particular that -a is a vector with the same magnitUde as a but p0inting in the direction opposite to a. Figure 5 shows this vector, and as further examples of the multiplication of a vector by a scalar, the vectors .2a and -2a. A vector has magnitude (size) and direction: vector magnitude and direction. Two arrows represent the same vector if they have the same length and are parallel (see figure ... by the addition formula for the cosine. | 858 | 3,516 | {"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} | 3.625 | 4 | CC-MAIN-2021-17 | latest | en | 0.889778 |
https://sportsbizusa.com/20-compare-rational-numbers-worksheet/ | 1,675,512,814,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500126.0/warc/CC-MAIN-20230204110651-20230204140651-00281.warc.gz | 541,833,146 | 14,176 | # 20 Compare Rational Numbers Worksheet
pare and Order Rational Numbers Practice 22 2 Worksheet pare and order rational numbers worksheet pdf, ordering positive rational numbers worksheet, ordering rational and irrational numbers worksheet pdf, paring rational numbers worksheet, ordering rational numbers worksheet with answers, via: lessonplanet.com
Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their math education. The main reason behind this is that learning math can be done with the worksheets. With an organized worksheet, kids will be able to describe and explain the correct answer to any mathematical problem. But before we talk about how to create a math worksheet for kids, let’s have a look at how children learn math.
In elementary school, children are exposed to a number of different ways of teaching them how to do a number of different subjects. Learning these subjects is important because it would help them develop logical reasoning skills. It is also an advantage for them to understand the concept behind all mathematical concepts.
To make the learning process easy for children, the educational methods used in their learning should be easy. For example, if the method is to simply count, it is not advisable to use only numbers for the students. Instead, the learning process should also be based on counting and dividing numbers in a meaningful way.
The main purpose of using a worksheet for kids is to provide a systematic way of teaching them how to count and multiply. Children would love to learn in a systematic manner. In addition, there are a few benefits associated with creating a worksheet. Here are some of them:
Children have a clear idea about the number of objects that they are going to add up. A good worksheet is one which shows the addition of different objects. This helps to give children a clear picture about the actual process. This helps children to easily identify the objects and the quantities that are associated with it.
This worksheet helps the child’s learning. It also provides children a platform to learn about the subject matter. They can easily compare and contrast the values of various objects. They can easily identify the objects and compare it with each other. By comparing and contrasting, children will be able to come out with a clearer idea.
He or she will also be able to solve a number of problems by simply using a few cells. He or she will learn to organize a worksheet and manipulate the cells. to arrive at the right answer to any question.
This worksheet is a vital part of a child’s development. When he or she comes across an incorrect answer, he or she can easily find the right solution by using the help of the worksheets. He or she will also be able to work on a problem without having to refer to the teacher. And most importantly, he or she will be taught the proper way of doing the mathematical problem.
Math skills are the most important part of learning and developing. Using the worksheet for kids will improve his or her math skills.
Many teachers are not very impressed when they see the number of worksheets that are being used by their children. This is actually very much true in the case of elementary schools. In this age group, the teachers often feel that the child’s performance is not good enough and they cannot just give out worksheets.
However, what most parents and educators do not realize is that there are several ways through which you can improve the child’s performance. You just need to make use of a worksheet for kids. elementary schools.
As a matter of fact, there is a very good option for your children to improve their performance in math. You just need to look into it.
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Quiz & Worksheet Graph Rational Numbers on a Number Line via : study.com | 1,243 | 6,275 | {"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} | 4.0625 | 4 | CC-MAIN-2023-06 | latest | en | 0.955101 |
https://blog.csdn.net/u013345179/article/details/99693458 | 1,568,673,426,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572964.47/warc/CC-MAIN-20190916220318-20190917002318-00390.warc.gz | 401,248,023 | 20,665 | # 积累程度 poj3585
N≤2∗105N≤2∗105
1
5
1 2 11
1 4 13
3 4 5
4 5 10
26
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
typedef long long ll;
using namespace std;
const int N=700100;
int n;
bool v[N];
inline void adds(int x,int y,int z)
{
tot++;
point[tot]=y;
edge[tot]=z;
}
void dfsd(int x)
{
v[x]=1;
d[x]=0;
{
int y=point[i];
if(v[y])
continue;
dfsd(y);
if(du[y]==1)
d[x]+=edge[i];
else
d[x]+=min(edge[i],d[y]);
}
}
void dfs(int x)
{
v[x]=1;
{
int y=point[i];
if(v[y])
continue;
if(du[x]==1)
dp[y]=edge[i]+d[y];
else
{
dp[y]=d[y]+min(dp[x]-min(d[y],edge[i]),edge[i]);
}
dfs(y);
}
}
int main()
{
int jb;
scanf("%d",&jb);
while(jb--)
{
tot=0;
memset(Next,0,sizeof(Next));
memset(du,0,sizeof(du));
cin>>n;
for(int i=1; i<n; i++)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
du[x]++,du[y]++;
}
int root=1;
memset(v,0,sizeof(v));
dfsd(root);
dp[root]=d[root];
memset(v,0,sizeof(v));
dfs(root);
ll ans=-1;
for(int i=1; i<=n; i++)
{
ans=max(ans,dp[i]);
//cout<<dp[i]<<' '<<d[i]<<endl;
}
printf("%lld\n",ans);
}
} | 410 | 1,038 | {"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} | 2.71875 | 3 | CC-MAIN-2019-39 | latest | en | 0.122275 |
http://denysetiyadi.co.uk/pre-calculus/composite-inverse-functions/ | 1,558,610,285,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257243.19/warc/CC-MAIN-20190523103802-20190523125802-00462.warc.gz | 53,631,386 | 6,821 | # Composite and inverse functions
### 1) Composite functions
Two functions can be composed to form a new composite function. To write the composition of functions $f(x)$ and $g(x)$, write $f(x) \circ g(x)$ or $f(g(x))$. A composite function is sometimes called a compound function.
Question 1
Find the composition of the two functions:
$f(x)=x^{2}-2x+3$ and $g(x)=2x-1$.
$(f \circ g)(x)=(2x-1)^{2}-2(2x-1)+3=4x^{2}-8x+6$
$(g \circ f)(x)=2(x^{2}-2x+3)-1=2x^{2}-4x+5$.
Notice that $f\circ g \neq g\circ f$.
If you’re feeling confident, try this more difficult example. Don’t be worried if it looks hard, the process you go through is just the same as before. Replace with the function and then simplify.
Consider the functions $p(x)=\sqrt{x-1}$ and $q(x)=x^{2}$. Find $(q \circ p)(x)$.
Question 2
$f(x)=\frac{x-1}{x+1}$, find $h(x)=f(f(x))$
$f(f(x))=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}$
You now need to simplify the numerator and denominator separately.
Numerator $\frac{x-1}{x+1}-1=\frac{x-1}{x+1}-\frac{x+1}{x+1}=\frac{x-1-(x+1)}{x+1}=\frac{-2}{x+1}$
Denominator $\frac{x-1}{x+1}+1=\frac{x-1}{x+1}+\frac{x+1}{x+1}=\frac{x-1+(x+1)}{x+1}=\frac{2x}{x+1}$
$f(f(x))=\frac{\frac{-2}{x+1}}{\frac{2x}{x+1}}=\frac{-2}{x+1}\times\frac{x+1}{2x}=\frac{-2}{2x}=\frac{-1}{x}$
Therefore, $h(x)=\frac{-1}{x}$
< Previous | 1 | 2 | 3 | Next > | 541 | 1,348 | {"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": 19, "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-2019-22 | latest | en | 0.538787 |
https://testbook.com/question-answer/a-students-marks-were-wrongly-entered-as-96--5d1b14eefdb8bb6c1b7d7cdc | 1,642,532,349,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300997.67/warc/CC-MAIN-20220118182855-20220118212855-00357.warc.gz | 625,891,104 | 31,282 | # A student’s marks were wrongly entered as 96 instead of 69. Due to this, the average marks of the class went up by 1/3 marks. What is the number of students in the class?
1. 75
2. 81
3. 78
4. 63
5. 84
Option 2 : 81
Free
IBPS RRB Officer Scale- I Prelims Full Mock Test
124049
80 Questions 80 Marks 45 Mins
## Detailed Solution
⇒ Let the number of students in the class be X
⇒ When marks as entered as 96 instead of 69, the sum of marks of all students is increased by 96 - 69 = 27 marks.
∴ The average marks are increased by = 27/X
∴ 27/X = 1/3
⇒ X = 27 × 3 = 81 | 194 | 572 | {"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} | 4.125 | 4 | CC-MAIN-2022-05 | latest | en | 0.932703 |
https://studentshare.org/physics/1694716-dynamics | 1,540,281,850,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516117.80/warc/CC-MAIN-20181023065305-20181023090805-00418.warc.gz | 789,345,575 | 23,199 | Search
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# Dynamics - Assignment Example
Summary
The distance between the pulleys centres is 500mm. The larger pulley rotates at 600 R.P.M and maximum permissible belt tension is 900N. Calculate:
A multi- plate clutch has four contact surfaces (n = 4),…
## Extract of sample"Dynamics"
Download file to see previous pages Calculate the horizontal and vertical components of the mass. If the shaft rotating at 200RPM, determine the force excreted on the shaft.
Ashaft rotating at 500RPM carries masses of A = 3kg, B = 4kg and C = 2kg at distances of 0.9m, 0.7m, and 1.7m from the shaft respectively. The angles of A, B, and C respect to horizontal are 40o, 130o, and 290o respectively.
In an Epicyclic gearbox, the Annulus gear “A” is fixed and the input shaft is connected to the Sun gear “S” which rotates at 1800RPM. The Sun gear has 60 teeth and there are three planet gears “P”. The gear ratio between the Sun gear and Planet gear is 2:1.The input power is 45kW and the efficiency is 85%. Sketch a simple arrangement and calculate:
Pass: All questions in section A (including all parts) must be attempted and answered. The assessment criteria and the learning outcomes for Pass grade must be met and fundamental understanding must be demonstrated on this assignment.
Merit: All questions in section A (including all parts) and question B7 in section B must be answered with reasonable accuracy and with no major errors. A neat presentation and good communication of assignment is essential. The assessment criteria for Pass and Merit grade must be met and knowledge and understanding must be demonstrated on this assignment.
Distinction: All questionsin section A and section B (B7 & B8) must be answered fully correct with no errors. All the criteria for Pass, Merit and Distinction must be met. A neat presentation of assignment including any references/ bibliography is essential. All the answers in section B must be synthesised and clearly justified. All the SI units must be clearlyshown in front of calculations, showing full understanding on this ...Download file to see next pagesRead More
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https://admhelp.microfocus.com/ppm/en/9.50-9.51/Help/Content/UG/FMUG/106100_FMUG_5Tracking.htm | 1,544,412,235,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823236.2/warc/CC-MAIN-20181210013115-20181210034615-00561.warc.gz | 532,161,667 | 9,754 | # Cost Data Calculations and Formulas
Financial Management allows users to capture and track planned and actual cost information for their projects, giving visibility into project performance from a financial standpoint. Basic cost information can be captured on projects and tasks in the areas defined by Table 5-1. Cost data items and associated formulas.
Some cost data is derived from a project's active baseline work plan. For information about baselines, see the Project Management User’s Guide.
Table 5-1. Cost data items and associated formulas
Item
Definition
Formula
Planned Labor Cost
Cost of a work item (typically a task), which is a measure of the amount of scheduled effort on a task. This amount is task-specific and is rolled up to the project level.
Planned Labor Cost = Sum of (Scheduled Effort * Rate determined by cost rule for each task)
Planned Non-Labor Cost
Cost of non-labor items needed to complete a work item. This is not a direct measure of the effort to be spent on a work item. This figure is task-specific and is rolled up to the project level.
Manually entered
Planned Cost
Total planned cost represented by a work item.
Planned Cost = Planned Labor Cost + Planned Non-Labor Cost
Baseline Labor Cost
Labor cost for a work item in the active baseline taken of a project work plan.
Baseline Labor Cost = Planned Labor Cost at time of Baseline
Baseline Non-Labor Cost
Non-labor cost for a work item in the active baseline taken of a project work plan.
Baseline Non-Labor Cost = Planned Non-Labor Cost at time of Baseline
Baseline Cost
Total cost represented by the active baseline taken of a work item.
Baseline Cost = Baseline Labor Cost + Baseline Non-Labor Cost
Actual Labor Cost
Cost of the work performed on a work item.
Actual Labor Cost = Sum of (Actual Effort * Rate determined by cost rule for each work item)
Actual Non-Labor Cost
Total of all non-labor costs accrued in completing a work item.
Manually entered
Actual Cost
Total cost incurred in completing a work item.
Actual Cost = Actual Labor Cost + Actual Non-Labor Cost
Planned Value (PV)
Planned Value can be calculated one of two ways, depending on how the
PV_USE_ACTIVE_BASELINE_DATES `server.conf` parameter is set. If necessary, contact your PPM system administrator to verify this setting.
By default, the PV_USE_ACTIVE_BASELINE_DATES parameter is set to `false`. Therefore, by default, Planned Value is calculated by the portion of the Baseline Cost planned to be spent between the project's start date and the current date.
PV = Baseline Cost * [(Today's Date – Start Date) / (Finish Date – Start Date)]
If the PV_USE_ACTIVE_BASELINE_DATES parameter is set to `true`, Planned Value is calculated using a project's active baseline dates instead of its scheduled dates.
Note: When the Project Planned Value Update service runs for the first time after PV_USE_ACTIVE_BASELINE_DATES is enabled, projects with scheduled dates and active baseline dates that are both completely in the past are not calculated.
PV = Baseline Cost * [MIN (Today's Date, Baseline Finish Date) –Baseline Start Date] / (Baseline Finish Date – Baseline Start Date)
Earned Value (EV)
Earned Value can be calculated one of two ways, depending on how the
EV_ALLOW_PRORATING `server.conf` parameter is set. If necessary, contact your PPM system administrator to verify this setting.
By default, the EV_ALLOW_PRORATING `server.conf` parameter is set to `true.` Therefore, by default, Earned Value is calculated by the portion of the Baseline Cost for the entire project that has theoretically been spent by the current date, measured as a function of the amount of work performed thus far.
EV = Baseline Cost * % Complete
If the EV_ALLOW_PRORATING `server.conf `parameter is set to `false,` Earned Value is only acknowledged when a task or project is 100% complete. That is, if the task or project is less than 100% complete, the EV calculation is 0. When the task or project is 100% complete, the EV calculation is equal to the Baseline Cost of the task or project.
If % Complete < 100, EV = 0
If % Complete = 100, EV = Baseline Cost
Cost Performance Index (CPI)
Cost efficiency ratio of Earned Value to Actual Cost. CPI is used to calculate Projected Actual Cost for a project and predict the size of possible cost overrun.
CPI = EV / Actual Cost
Schedule Performance Index (SPI)
Schedule efficiency ratio of Earned Value to Planned Value. SPI describes what portion of the work plan or planned schedule has been accomplished in terms of its cost.
SPI = EV / PV
Cost Variance
Difference between the earned value and the actual cost for the project or task. Earned value compared with the actual cost incurred for the work performed provides an objective measure of planned and actual cost. Any difference is called a cost variance.
CV = EV – AC
Schedule Variance
Difference between the earned value and the planned value of the project or task. Planned value compared with earned value measures the dollar volume of work planned against the equivalent dollar volume of work accomplished. Any difference is called a schedule variance.
SV = EV – PV
Projected Actual Cost
Ratio of total cost represented by the latest baseline taken of a work item and the cost performance index (CPI).
Projected Actual Cost = Baseline Cost/CPI
Estimation At Completion (EAC)
The estimated cost of the project at the end of the project.
There are three methods to calculate EAC:
Variances are Typical - This method is used when the variances at the current stage are typical and are not expected to occur in the future.
Past Estimating Assumptions are not valid - This method is used when the past estimating assumptions are not valid and fresh estimates are applied to the project.
Variances will be present in the future - This method is used when the assumption is that the current variances will continue to be present in the future.
The calculation formulas for the three methods are as given below. You can select a desired formula in Project Settings to determine how the project calculates the EAC.
EAC = AC + ( BAC - EV ) / (SPI * CPI )
EAC = AC + ( BAC - EV )
EAC = BAC / CPI
To Complete Performance Index (TCPI)
To Complete Performance Index is an index showing the efficiency at which the resources on the project should be utilized for the remainder of the project. This can be calculated using the following formula:
TCPI = (BAC - EV) / (BAC - AC)
Note: Note the following:
• All cost information utilizing a formula is calculated automatically by Project Management.
• Calculations for SPI use the expected baseline cost of a project and do not involve Actual Cost. | 1,444 | 6,712 | {"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-2018-51 | latest | en | 0.865973 |
https://community.tableau.com/thread/281472 | 1,553,297,111,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202698.22/warc/CC-MAIN-20190322220357-20190323002357-00506.warc.gz | 445,067,272 | 30,207 | 10 Replies Latest reply on Sep 18, 2018 5:10 PM by Shinichiro Murakami
# Distance Calculation Using LOD Expressions
I am stumped and would appreciate some help. Please refer to the attached excel table. I have a large number of Locations (ID Column) with Values and Lat/Longs associated with each. For each Location I need to determine the average Value of all Locations that fall within a parameter established radius of the subject Location (yellow column). So for instance, in the case of Location A this would require that distances to the other 25 locations be determined. Let's assume that half of the 25 locations were less than a parameter established distance of 2 miles from Location A and the average Value associated with those locations would be the value I am trying to calculate (and would go in the yellow column). The process would then have to be repeated for each of the other 25 Locations.
I believe this can be done with LOD expressions but i haven't gotten past that. I would have attached a packaged workbook but it seemed easier to explain with the excel file. Thanks in advance for help.
• ###### 1. Re: Distance Calculation Using LOD Expressions
HI David,
This is very rough calc based on a certain latitude point (not far from your location)
So you need minor modification a little anyways,.
minor (?) Delta
If this helped, please marl my answer as correct / helped to close the thread, from original post, not inbox,
Thanks,
Shin
• ###### 2. Re: Distance Calculation Using LOD Expressions
Would showing the points on a map help too?
• ###### 3. Re: Distance Calculation Using LOD Expressions
Just following up.
If you have further questions, please reply to this.
If you think problem is solved, please mark my answer as correct / helped to close the thread not from inbox but from originals post.
Thanks,
Shin
• ###### 4. Re: Distance Calculation Using LOD Expressions
Shin
First let me thank you for helping me.
I have incorporated your input but now have run into another roadblock which I hope to explain with the attached Tableau workbook and accompanying excel data file.
I should also point out that I am using a “join” to accomplish what you had done by, I believe, creating a duplicate of the file. I tried that but the actual data files are so big I experienced significant computing delays while has not been the case with my creation of a separate file that only contains the Lat/Long data and join fields. I hope this data preparation approach is not causing the problem described below.
My objective is to create a “type curve” which represents the oil production that a new well (the “Valuation Selection”) could be expected to produce based upon an average of what similar wells in the immediate area (“Analogs”) have produced. The “Distance” calculation you helped me with determines which of the offset wells shall be considered an “Analog”. The problem then becomes determining the average oil production for the “Analogs”, including the “Valuation Selection”, on a “Month of Production” time sequence.
Please refer to attached Workbook, where worksheet “Distance” shows the Distance to offset wells that fall within 2.5 miles of each Valuation Selection and are therefore considered “Analogs”. You will notice that only two of the Valuation Selections have been chosen. On the “Map” worksheet, the locations of the two Valuation Selections are shown, along with the Analogs that fall within the Distance from each Valuation Selection. So for so good.
On the “Oil Prod” worksheet I want to show the average oil production for the Analogs on a “Month of Prod” basis. The top graph correctly reflects oil production for the two Valuation Selections without any regard to the Analogs. But the bottom chart does reflect the average oil production for all wells falling within X miles of the Valuation Selections, as I wanted it.
The problem seems to be with the “Average Oil Prod Within X” calculation. I have tried, unsuccessfully, various LOD Expressions to solve this problem and would again appreciate your help. And, anybody else that wants to join in the conversation is welcome.
Thanks Again,
David
• ###### 5. Re: Distance Calculation Using LOD Expressions
See attached excel file i neglected to attach to previous message
• ###### 6. Re: Distance Calculation Using LOD Expressions
Not exactly sure, but you mean this ?
Both top/bottom gray is same line. Top chart is only for the purpose of easier visual validation.
Thanks,
Shin
• ###### 7. Re: Distance Calculation Using LOD Expressions
No, i am sorry for not making it clearer. The bottom chart (Oil Prod worksheet) should reflect the average of the Analogs for each of the Valuation Selections. So, if the Apollo Unit is one of the Valuation Selections ( the orange line in this worksheet) the bottom chart should be the window average of the Apollo Unit Analogs which can be found on the Distance worksheet and are shown in orange--these Anlaogs to the Apollo Unit all are within 2.5 miles of the Apollo Unit. Likewise, the green line would reflect the window average of the Atlas Unit Analogs which are in green on the Distance worksheet. You can also look at the Map worksheet to see the two clusters of Analogs, one around the Apollo Unit (in orange) and the other around the Atlas Unit (in green).
And of course if i were to pick another Valuation Selection, say the Bean Hedtke Unit, this would create another cluster of Analogs within 2.5 miles and there would then be a third line on the bottom of the Oil Prod worksheet representing the average oil prod/Month of Prod for that Unit.
I hope this helps,
David
• ###### 8. Re: Distance Calculation Using LOD Expressions
Bit confused as you expect.
From below table on period "1"
APOLLO UNIT's average is average of red square, or you add some calculation beforehand with using distance or something?.
.
Thanks,
Shin
• ###### 9. Re: Distance Calculation Using LOD Expressions
Shin-We're getting there. I am out of town and will review tommorow and
then circle back with any questions. Thanks so much for your help,
David
On Fri, Sep 14, 2018 at 10:21 PM Shinichiro Murakami <
• ###### 10. Re: Distance Calculation Using LOD Expressions
Following up.
If you have further questions, please reply here.
If the post helped to solve your issue, please mark my answer as CORRECT / HELPED to close the thread, not from inbox but from original post.
Thanks,
Shin
1 of 1 people found this helpful | 1,424 | 6,514 | {"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} | 3.328125 | 3 | CC-MAIN-2019-13 | latest | en | 0.943164 |
https://brilliant.org/problems/finding-triangles-in-a-square/ | 1,529,894,147,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867364.94/warc/CC-MAIN-20180625014226-20180625034226-00397.warc.gz | 577,362,758 | 11,703 | # Finding Triangles in a Square
All horizontal and vertical adjacent points are at a distance $$1\text{ cm}$$ each. If the number of triangles of area $$1\text{ cm}^2$$ formed by using these points is $$\mu$$, evaluate $$\left\lfloor \dfrac { \mu }{ 8 } \right\rfloor$$.
Notation: $$\lfloor \cdot \rfloor$$ denotes the floor function.
× | 101 | 339 | {"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} | 2.703125 | 3 | CC-MAIN-2018-26 | latest | en | 0.755489 |
https://gmatclub.com/forum/in-the-new-building-city-residents-will-be-able-to-file-a-142814.html?kudos=1 | 1,508,438,374,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823360.1/warc/CC-MAIN-20171019175016-20171019195016-00258.warc.gz | 690,591,527 | 51,514 | It is currently 19 Oct 2017, 11:39
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# In the new building, city residents will be able to file a
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In the new building, city residents will be able to file a [#permalink]
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21 Nov 2012, 03:00
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In the new building, city residents will be able to file a complaint or fill out a police report and then
meet with a lawyer, get help with crisis and safety planning, housing and emergency grants
or meet
with advocates from the Asian American and Latino communities.
A. complaint or fill out a police report and then meet with a lawyer, get help with crisis and safety planning,
housing and emergency grants
B. complaint, fill out a police report, meet with a lawyer, or get help with crisis and safety planning as well as
housing and emergency grants
C. complaint, or fill out a police report and then meet with a lawyer to get help with crisis and safety planning,
housing, emergency grants
D. complaint; fill out a police report and then meet with a lawyer, get help with crisis, and safety planning,
housing and emergency grants,
E. complaint or fill out a police report and then meet with a lawyer; get help with crisis and safety planning,
housing, and emergency grants;
_________________
KUDOS me if you feel my contribution has helped you.
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Re: In the new building, city residents will be able to file [#permalink]
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21 Nov 2012, 07:48
1
KUDOS
In the new building, city residents will be able to file a complaint or fill out a police report and then
meet with a lawyer, get help with crisis and safety planning, housing and emergency grants
or meet
with advocates from the Asian American and Latino communities.
A. complaint or fill out a police report and then meet with a lawyer, get help with crisis and safety planning,
housing and emergency grantshousing and emergency grants modifies crisis and safety planning incorrectly
B. complaint, fill out a police report, meet with a lawyer, or get help with crisis and safety planning as well as
housing and emergency grantscomma, comma, comma, or is wrong - it suggests linking multiple and clauses finally with an or. The sequence then meet lawyer is lost.
C. complaint, or fill out a police report and then meet with a lawyer to get help with crisis and safety planning,
housing, emergency grantsmeeting with lawyer to get help with housing. emergency grants - incorrect
D. complaint; fill out a police report and then meet with a lawyer, get help with crisis, and safety planning,
housing and emergency grants,semicolon requires that following clauses should be independent except subordinate.
E. complaint or fill out a police report and then meet with a lawyer; get help with crisis and safety planning,
housing, and emergency grants;correct - get help with lists all items
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Re: In the new building, city residents will be able to file [#permalink]
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21 Nov 2012, 09:54
I agree with E. Semi-colons can be used as commas as well.
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Re: In the new building, city residents will be able to file a [#permalink]
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23 Jan 2013, 11:16
@Nanishora,
Could you explain as to how A is wrong. I could not understand your reason.....
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Re: In the new building, city residents will be able to file a [#permalink]
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24 Jan 2013, 04:59
Here, 'Complaint or' is the clue. or fill out a police report... So we can eliminate B,C and D. Is there any need to use comma after housing and before and?. No. Since it is wrong. Ans is 'A'
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Re: In the new building, city residents will be able to file a [#permalink]
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28 Jan 2013, 10:29
Can sum1 explain y E is tha ans???
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Re: In the new building, city residents will be able to file a [#permalink]
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03 Feb 2013, 16:26
IMO A
D & E are out because ';' can connect only main clause(complete sentence) which is absent in both the cases.
so left with A,B,C.
here structure seems to be:-
residents will be able to ***** or ***** or******
which is best stated in 'A'
br//suryav
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Re: In the new building, city residents will be able to file a [#permalink]
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06 Feb 2013, 04:06
Quote:
I agree with E. Semi-colons can be used as commas as well.
I agree with E, but I am doubtful about the semicolon in the end.
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Re: In the new building, city residents will be able to file a [#permalink]
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06 Feb 2013, 05:02
Did not catch it up until 4 min then got to find Answer is "A"
A. complaint or fill out a police report and then meet with a lawyer, get help with crisis and safety planning,
housing and emergency grants---"city residents meet with a lawyer or meet with meet with advocates" reveals a clear and concise sense also on right track of II'sm.
Do correct me if am wrong..
Kudos [?]: 21 [0], given: 56
Re: In the new building, city residents will be able to file a [#permalink] 06 Feb 2013, 05:02
Display posts from previous: Sort by | 1,852 | 6,824 | {"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-2017-43 | latest | en | 0.930838 |
http://www.chegg.com/homework-help/questions-and-answers/large-capacitor-charged-electrode--time-capacitor-connected-series-ammeters-anda-bulb-amme-q654621 | 1,369,153,689,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700212265/warc/CC-MAIN-20130516103012-00095-ip-10-60-113-184.ec2.internal.warc.gz | 372,530,434 | 8,930 | ## Capacitor Supplies Current to Bulb
A large capacitor is charged with on one electrode and on the other. At time , the capacitor is connected in series to two ammeters anda bulb. The ammeter connected to the positive side of the capacitorreads , and the ammeter connected to the negative side of thecapacitor reads .
Both ammeters will read positive if current flows in aclockwise sense through the circuit (from the + to the - terminalof the meter).
Immediately after time , what happens to the charge on the capacitor plates?
A.Electrons flow through the circuit from the positive to thenegative side of the capacitor.
B.Electrons flow through the circuit from the negative to thepositive side of the capacitor.
C.The positive and negative charges attract each other, so theystay in the capacitor.
D.Current flows clockwise through thecircuit.
E.Current flows counterclockwise through thecircuit.
List the letters corresponding to thecorrect statements in alphabetical order.
At any given instant after , what is the relationship between the currentflowing through the two ammeters, and , and the current through the bulb, ?
What is the sign of the quantity ? (Positive or Negative)
• Given :
.
(a) what happens to the charge on the capacitorplates?
.
B.Electrons flow through the circuit from the negativeto the positive side of the capacitor.
.
(b) Relation between IP IN andIB is :
.
.
(c) sign of thequantity = Positive
.
Hope this helps u! | 327 | 1,475 | {"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} | 3.046875 | 3 | CC-MAIN-2013-20 | latest | en | 0.862079 |
https://online.flippingbook.com/view/457957909/67/?sharedOn= | 1,719,129,269,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00837.warc.gz | 369,294,184 | 24,730 | # Algebra 1 Companion Book, Vol 1 – Summer Edition
1.2.3 Simplifying Expressions − Practice
1. Simplify the expression. 3 5 (12)(5)
2. Simplify the expression. + + 3 4 4
4 7
2 9
3 7
4. Simplify the expression 205 a − a by combining like terms.
3. Using the Distributive Property, find the product of the expression 5(17) and then simplify.
6. Simplify.
5. Simplify the expression 11 a 3 − 3 a by combining like terms.
2 3
5 3
+ x x 2
2
7. Simplify the expression 12 x + y 2 + 6 x + 3 y 2 by combining like terms.
8. Simplify by combining like terms. 7 a 3 + 5 t + 4 a 3 + 4 t + 3 a 2
9. Simplify.
5 + 2(3 x − 4) − x
57
Made with FlippingBook Digital Publishing Software | 261 | 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} | 2.9375 | 3 | CC-MAIN-2024-26 | latest | en | 0.704482 |
https://merjek.com/running-pearson-correlation/ | 1,695,871,413,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510358.68/warc/CC-MAIN-20230928031105-20230928061105-00730.warc.gz | 429,724,308 | 18,477 | #### A correlation measures the relationship between two variables — for example, is a person’s IQ related to their income?
For a Pearson correlation, we need two variables. Typically, both variables need to be continuous, normally distributed, and unbounded, like height or age. If a variable is categorical, like profession, or if there are a lot of bounded scores, like a lot of 0s or 100s on a test, it won’t work.
The test score for a Pearson correlation is r, which has a range from -1 to +1. The r score tells you two things about the relationship between the two variables: the strength and the direction of the relationship. The larger the absolute value of the r score, the stronger the relationship. If the number is positive, then the two variables are directly related: as one goes up, the other goes up. If the value is negative, then they are inversely related: as one goes up, the other goes down, and vice-versa.
It’s important to remember that although a Pearson correlation can identify a relationship between two variables, it cannot (by itself) determine whether there is a causal relationship, let alone which variable is causing the other. Some relationships are clearly the product of a third variable. For example, ice cream sales are positively correlated with drownings. Now, does buying ice cream cause people to drown? Of course not. In reality, a third variable (temperature) is responsible for the relationship between ice cream and drowning: as it gets hotter, people are more likely to eat ice cream and more likely to go swimming.
The r score is also associated with a p value, which tests for statistical significance. The p value assesses how likely we would obtain this dataset by chance, if the null hypothesis were true. So, the lower the p value, the less likely it is that the null hypothesis is true. Typically, our alpha level, the threshold for statistical significance, is set at .05. That is, if our p value is below .05, then we reject the null hypothesis.
The p value for a Pearson correlation is governed by two things: the strength of the relationship, and the degrees of freedom. The stronger the relationship (either negative or positive), the lower the p value. The degrees of freedom for a Pearson correlation is N minus 2, so the larger your sample size, the more degrees of freedom, and the lower your p value.
So now that we know what a correlation is, let’s look at an example. Let’s say that we want to know whether a person’s IQ is related to their income. We have the following dataset.
Our hypothesis is that smarter people are more skilled and in higher demand, and therefore make more money. However, the relationship between IQ and income isn’t perfect, is it? There’s a lot more that goes into a person’s income than just their IQ: what field they work in, how much experience they have, even where they live. So, it won’t be a perfect relationship between IQ and income, and it probably won’t even be a particularly strong relationship. So, we’ll hypothesize a moderate, positive relationship between IQ and income. In general, we want to have hypotheses that are backed by theory. That way we can avoid “fishing expeditions” which throw variables together randomly. Performing a test without a hypothesis grounded in theory increases the likelihood that any relationship you might find is just due to chance. In the last column, we also have the foot size of each individual. Obviously, we would not hypothesize any difference between foot size and either IQ or income.
So now that we have our hypothesis, let’s see how to perform a correlation on MagicStat (version 1.1.3).
1-) Select a data file
Select your own dataset by clicking the “Choose a data file” button. If you would like to use a sample data file, click “Sample datasets” on the toolbar, save it to your hard drive, then click “Choose a data file” and navigate to where you saved it.
2-) Explore the dataset
After you select your dataset, click the “Explore” button.
After you select your dataset, click the “Explore” button. On the right side of the window is information-at-a-glance about your dataset, including variable information, bar graphs, and histograms.
4- Choose the “Pearson Correlation” model
Click “Select a model to analyze your data”, and select “Pearson Correlation” on the dropdown.
5- Choose variables
Click the “Select variables” button, and pick which variables you want to include in the model. Here, we’re selecting “IQ”, “Income” and “Foot_Size”.
6- Analyze the dataset
Finally, click the “Analyze” button.
Interpreting results
Now it is time to interpret the results we obtained in the previous steps.
In a Pearson correlation, the degrees of freedom is purely a function of sample size, N minus 2. So, it is 52.
Next is our correlation table. We have a moderate correlation between IQ and Income at .41, as we hypothesized, and no correlation between Foot_Size and IQ or Income.
Below that is the p value for each relationship, and we see that moderate correlation between IQ and Income has a p value of 0.02, which means that if there were no relationship between IQ and Income, we’d expect to get this dataset about two times out of a thousand — not very likely!. And the p values for Foot_Size-IQ and Foot_Size-Income are close to 1, which means it’s not very likely that there is a relationship between them.
After the correlation table, MagicStat gives us some graphs. First is a correlation heatmap, to show where the strongest relationships are. We can see the moderate relationship between IQ and Income in purple, and the lack of relationship with Foot_Size in blue.
Then, we can select a scatterplot to visualize the relationship and check for outliers. If we look at the IQ-Income scatterplot, there do not seem to be any obvious outliers. With this scatterplot and a theoretical link between IQ and Income, we can feel confident in the relationship we found in our dataset.
Written by the MagicStat Team | 1,304 | 6,001 | {"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} | 3.875 | 4 | CC-MAIN-2023-40 | latest | en | 0.947265 |
https://discourse.processing.org/t/diffusion-algorithm-cc-13/27976 | 1,680,071,518,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948951.4/warc/CC-MAIN-20230329054547-20230329084547-00673.warc.gz | 256,558,443 | 4,708 | # Diffusion Algorithm (CC #13)
I wanted to port Dan’s p5.js code to processing using vectors as my objects.
I can get it to run using a component-wise function for the laplace of each A and B like he does in his YouTube Video.
My problem is getting it to work with a single vector function… it doesn’t look right? it diffuses from the corner in an odd way after the center block dissapears…
PVector grid [][]; //current state
PVector next [][]; //next state
float dA = 1.0; //diffusion rate of A
float dB = 0.5; //diffusion rate of B
float f = 0.055; //feed rate of A
float k = 0.062; //kill rate of B
void setup() {
frameRate(60);
size(600, 600, P2D);
pixelDensity(1);
grid = new PVector[width][height];
next = new PVector[width][height];
//fill grid with "A"
for (int x = 0; x < width; x++) {
for (int y = 0; y < height; y++) {
grid[x][y] = new PVector(1, 0);
next[x][y] = new PVector(1, 0);
}
}
for (int x = (width/2)-10; x < (width/2)+10; x++) {
for (int y = (height/2)-10; y < (height/2)+10; y++) {
grid[x][y].y = 1;
}
}
}
void draw() {
//colour in the pixels based on ammount of A and B
for (int x = 0; x < width; x++) {
for (int y = 0; y < height; y++) {
int pix = x + y * width;
float a = grid[x][y].x;
float b = grid[x][y].y;
pixels[pix] = color((a-b)*255);
}
}
updatePixels();
//determine next frame
for (int x = 1; x < width-1; x++) {
for (int y = 1; y < height-1; y++) {
PVector Del = laplace(x,y);
next[x][y].x = grid[x][y].x +
dA * Del.x -
grid[x][y].x * grid[x][y].y * grid[x][y].y +
f * (1 - grid[x][y].x);
next[x][y].y = grid[x][y].y +
dB * Del.y +
grid[x][y].x * grid[x][y].y * grid[x][y].y -
(k + f) * grid[x][y].y;
}
}
swap();
}
void swap() {
PVector temp [][] = grid;
grid = next;
next = temp;
}
PVector laplace(int x, int y) {
PVector sum = new PVector (0,0);
sum.add(grid[x][y].mult(-1)); // | 0.05 0.20 0.05 |
sum.add(grid[x-1][y-1].mult(0.05)); // | 0.20 -1.00 0.20 |
sum.add(grid[x-1][y+1].mult(0.05)); // | 0.05 0.20 0.05 | | 739 | 1,993 | {"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-2023-14 | latest | en | 0.531782 |
https://stats.stackexchange.com/questions/179937/transforming-cohens-d-effect-sizes-into-unit-changes | 1,627,133,492,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150266.65/warc/CC-MAIN-20210724125655-20210724155655-00148.warc.gz | 535,977,815 | 37,580 | # Transforming Cohen's d effect sizes into unit changes
I want to calculate unit changes (or percentile changes) from an intervention for a continuous outcome variable where I have the pre and post test means and standard deviations. From that I can calculate a Cohen's d, but I want to use the result for a CBA and therefore have to transform Cohen's d to unit changes. I am looking at symptoms for different mental disorders, and not specifically diagnosis/non-diagnosis.
An idea is to use a cut-off on the normal distribution, for instance 90th percentile, which then determines diagnosis/non-diagnosis. Is it a sound approach to (from a z-score table) start at the 90th percentile, see where I end up in the distribution after the given effect size (Cohen's d), and thereby find the percentile change from the intervention for the 90th percentile of the sample population?
If yes, can I then turn that percentile into percentages by dividing the percentile change by .5 and multiply it by 100? | 210 | 1,000 | {"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-2021-31 | latest | en | 0.922012 |
https://goprep.co/q26-the-mean-of-20-numbers-is-zero-of-them-at-the-most-how-i-1njypz | 1,624,464,385,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488539480.67/warc/CC-MAIN-20210623134306-20210623164306-00062.warc.gz | 265,552,969 | 38,389 | Q. 265.0( 1 Vote )
The mean of 20 nu
It’s given that mean of 20 numbers is 0, which implies that average of 20 numbers is 0.
This means that Sum of 20 numbers is 0.
If Sum of 19 numbers out of 20 is x(say), then the 20th number will be absolutely 0 for the average to be 0.
At the most, 19 numbers will be positive.
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http://landsurveyorsunited.com/group/kickassformulas/forum/topics/traverse-traverseline-and-how-does-gps-fit-into-all-of-this-with?page=2&commentId=544331%3AComment%3A325858&x=1 | 1,540,209,830,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515041.68/warc/CC-MAIN-20181022113301-20181022134801-00405.warc.gz | 202,472,445 | 30,667 | Next Generation Land Surveyor Community
# Traverse, Traverse Line and How does GPS fit into all of this with GPS Traverse
If you look the word traverse up it is used as a noun, a verb, a adverb, a adjective and the list goes on.It also has many definitions other than the survey term . In surveying it is in reference to a system of point who's beginning was also it's end.all the sums could be adjusted for a better fit. This is a Closed Traverse. A Traverse Line does not work like this. It's corrections or only averages of each part if this is applied to the survey.There maybe cases were only one angle and one distance is taken at each leg of this Line.This is wrong. Multiples must be taken in this system . This is also to make sure these were correct. Many times i have had to explain to someone your traverse line is not a traverse. Traverse Lines has stations ,a Traverse does not, a traverse can be adjusted, a traverse Line can not. A compound traverse is two traverse lines that have been tied together . This does not make them a Traverse. IN a old survey book of mine it says a traverse is a series of points that are connected by angles and distance and can be seen from one point to the next and to the next until they connect back to the first point. This allows the surveyor to make polar or rectangular adjustments to this system .
Can be seen? That's the only way you would turn the the angles and measure them fiscally . Why he said this i do not know, but that is the only way. So how does GPS " Traverse" fit into all of this . and is it really a traverse or is it only a system of Control points. Voice your conception of what you think and how this terminology is applied. so what makes a Traverse and a Traverse Line no matter what you use.
billy
This Content Originally Published by a land surveyor to Land Surveyors United Network
Views: 1187
### Replies to This Discussion
Salisu, Thank you for reply. the ? is not what is a traverse , but what is the meaning of a GPS Traverse,or is this possible . Can it have one, you know it's not the same on how the data is acquired . Do not make this connection or compare it to a conventional survey. But in GPS can one point be dependent on the one before it, just like in any traverse ,closed,open or a traverse line with stations or a compound traverse they all connect one point to the next, and are dependent on the one before it.this is what makes them a like. Make this connection. Or is GPS just a system of control points were this will never be the case. One answer is right before you and there are others.
billy
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\$2,599.99End Date: Monday Oct-29-2018 13:03:44 PDTBuy It Now for only: \$2,599.99Buy It Now | Add to watch list | 1,021 | 4,191 | {"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-2018-43 | latest | en | 0.961257 |
http://docplayer.net/21100834-St-math-jiji-answers-for-challenge-help.html | 1,713,814,418,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818337.62/warc/CC-MAIN-20240422175900-20240422205900-00230.warc.gz | 7,696,903 | 27,048 | # St Math Jiji Answers For Challenge Help
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12 views (last 30 days)
Heather Worth on 2 Feb 2022
Answered: Robert U on 2 Feb 2022
I need to plot the 500 iterations versus the iteration numbers, but with the code I'm using, it just gives me a blank graph.
lambda=6.544;
x0=0.65;
N=500;
for i=1:N
x1=lambda*(x0^2)*(1-x0);
x0=x1;
end
plot(N,x1)
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Answers (1)
Robert U on 2 Feb 2022
Hi Heather Worth,
you can store x1 each iteration in a vector and plot the vector at the end of the calculation.
lambda=6.544;
x0=0.65;
N=500;
for ik=1:N
x1(ik) = lambda*(x0^2)*(1-x0);
x0 = x1(ik);
end
plot(1:N,x1,'.','MarkerSize',12)
Or you can plot each iteration on the fly.
lambda=6.544;
x0=0.65;
N=500;
fh = figure;
ah = axes;
hold(ah,'on');
for ik=1:N
x1 = lambda*(x0^2)*(1-x0);
x0 = x1;
plot(ah,ik,x1,'bl.','MarkerSize',12)
end
Kind regards,
Robert
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47. Trigonometry
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48. Trigonometry
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49. Math trigonometry
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50. Trigonometry
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51. TRIGONOMETRY
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52. trigonometry
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53. trigonometry
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54. Trigonometry
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55. Trigonometry
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56. Trigonometry
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58. Trigonometry
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59. Trigonometry
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60. trigonometry
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61. Trigonometry
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62. Trigonometry
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63. math-geometry-soh-cah-toa???
Thank you for using the Jiskha Homework Help Forum. Hopefully the following sites will help you: 1l http://www.mathwords.com/s/sohcahtoa.htm 2. http://en.wikipedia.org/wiki/Trigonometry I get how to use soh-cah-toa and everything but could someone help me with how to plug it ...
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65. Trigonometry
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66. trigonometry
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67. Trigonometry
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68. trigonometry-math
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69. Trigonometry
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70. Trigonometry
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71. Trigonometry
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72. Trigonometry
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73. Trigonometry
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74. Trigonometry
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75. Trigonometry
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76. Trigonometry
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77. Math Trigonometry
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78. Trigonometry
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79. Trigonometry
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80. Trigonometry
Write sin(125°) in terms of its cofunction. Make sure your answer is a function of a positive angle.
81. Trigonometry
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82. Trigonometry
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83. Pre-Calculus/Trigonometry
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84. Pre-Calculus/Trigonometry
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85. Trigonometry
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86. Math: Trigonometry
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87. Trigonometry and Right Triangles!
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88. Trigonometry
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90. Algebra
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91. Trigonometry
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92. Trigonometry
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93. Trigonometry
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94. Trigonometry
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95. Trigonometry
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96. Trigonometry
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97. Trigonometry
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98. trig
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100. trigonometry
refer to the graph of y=sin(x) or cos(x) to find the exact values of (x) in the interval [0,4pi]that satisfy the equation. 4sin(x)=4 | 3,954 | 13,242 | {"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-2018-26 | latest | en | 0.827683 |
https://markxu.com/solomonoff | 1,642,933,925,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304217.55/warc/CC-MAIN-20220123081226-20220123111226-00612.warc.gz | 426,029,302 | 13,351 | # Artificially Intelligent
Any mimicry distinguishable from the original is insufficiently advanced.
# The Solomonoff Prior is Malign
| 5098 words
Crossposted from the AI Alignment Forum. May contain more technical jargon than usual.
This argument came to my attention from this post by Paul Christiano. I also found this clarification helpful. I found these counter-arguments stimulating and have included some discussion of them.
Very little of this content is original. My contributions consist of fleshing out arguments and constructing examples.
Thank you to Beth Barnes and Thomas Kwa for helpful discussion and comments.
# What is the Solomonoff prior?
The Solomonoff prior is intended to answer the question “what is the probability of X?” for any X, where X is a finite string over some finite alphabet. The Solomonoff prior is defined by taking the set of all Turing machines (TMs) which output strings when run with no input and weighting them proportional to $$2^{-K}$$, where $$K$$ is the description length of the TM (informally its size in bits).
The Solomonoff prior says the probability of a string is the sum over all the weights of all TMs that print that string.
One reason to care about the Solomonoff prior is that we can use it to do a form of idealized induction. If you have seen 0101 and want to predict the next bit, you can use the Solomonoff prior to get the probability of 01010 and 01011. Normalizing gives you the chances of seeing 1 versus 0, conditioned on seeing 0101. In general, any process that assigns probabilities to all strings in a consistent way can be used to do induction in this way.
# Why is it malign?
Imagine that you wrote a programming language called python^10 that works as follows: First, it takes all alpha-numeric chars that are not in literals and checks if they’re repeated 10 times sequentially. If they’re not, they get deleted. If they are, they get replaced by a single copy. Second, it runs this new program through a python interpreter.
Hello world in python^10:
ppppppppprrrrrrrrrriiiiiiiiiinnnnnnnnnntttttttttt('Hello, world!')
Luckily, python has an exec function that executes literals as code. This lets us write a shorter hello world:
eeeeeeeeexxxxxxxxxxeeeeeeeeeecccccccccc("print('Hello, world!')")
It’s probably easy to see that for nearly every program, the shortest way to write it in python^10 is to write it in python and run it with exec. If we didn’t have exec, for sufficiently complicated programs, the shortest way to write them would be to specify an interpreter for a different language in python^10 and write it in that language instead.
As this example shows, the answer to “what’s the shortest program that does X?” might involve using some roundabout method (in this case we used exec). If python^10 has some security properties that python didn’t have, then the shortest program in python^10 that accomplished any given task would not have these security properties because they would all pass through exec. In general, if you can access alternative ‘modes’ (in this case python), the shortest programs that output any given string might go through one of those modes, possibly introducing malign behavior.
Let’s say that I’m trying to predict what a human types next using the Solomonoff prior. Many programs predict the human:
1. Simulate the human and their local surroundings. Run the simulation forward and check what gets typed.
2. Simulate the entire Earth. Run the simulation forward and check what that particular human types.
3. Simulate the entire universe from the beginning of time. Run the simulation forward and check what that particular human types.
4. Simulate an entirely different universe that has reason to simulate this universe. Output what the human types in the simulation of our universe.
Which one is the simplest? One property of the Solmonoff prior is that it doesn’t care about how long the TMs take to run, only how large they are. This results in an unintuitive notion of “simplicity”; a program that does something $$2^{10}$$ times might be simpler than a program that does the same thing $$2^9 - 1$$ times because the number $$2^{10}$$ is easier to specify than $$2^9 - 1$$.
In our example, it seems likely that “simulate the entire universe” is simpler than “simulate Earth” or “simulate part of Earth” because the initial conditions of the universe are simpler than the initial conditions of Earth. There is some additional complexity in picking out the specific human you care about. Since the local simulation is built around that human this will be easier in the local simulation than the universe simulation. However, in aggregate, it seems possible that “simulate the universe, pick out the typing” is the shortest program that predicts what your human will do next. Even so, “pick out the typing” is likely to be a very complicated procedure, making your total complexity quite high.
Whether simulating a different universe that simulates our universe is simpler depends a lot on the properties of that other universe. If that other universe is simpler than our universe, then we might run into an exec situation, where it’s simpler to run that other universe and specify the human in their simulation of our universe.
This is troubling because that other universe might contain beings with different values than our own. If it’s true that simulating that universe is the simplest way to predict our human, then some non-trivial fraction of our prediction might be controlled by a simulation in another universe. If these beings want us to act in certain ways, they have an incentive to alter their simulation to change our predictions.
At its core, this is the main argument why the Solomonoff prior is malign: a lot of the programs will contain agents with preferences, these agents will seek to influence the Solomonoff prior, and they will be able to do so effectively.
## How many other universes?
The Solomonoff prior is running all possible Turing machines. How many of them are going to simulate universes? The answer is probably “quite a lot”.
It seems like specifying a lawful universe can be done with very few bits. Conway’s Game of Life is very simple and can lead to very rich outcomes. Additionally, it seems quite likely that agents with preferences (consequentialists) will appear somewhere inside this universe. One reason to think this is that evolution is a relatively simple mathematical regularity that seems likely to appear in many universes.
If the universe has a hospitable structure, due to instrumental convergence these agents with preferences will expand their influence. As the universe runs for longer and longer, the agents will gradually control more and more.
In addition to specifying how to simulate the universe, the TM must specify an output channel. In the case of Game of Life, this might be a particular cell sampled at a particular frequency. Other examples include whether or not a particular pattern is present in a particular region, or the parity of the total number of cells.
In summary, specifying lawful universes that give rise to consequentialists requires a very simple program. Therefore, the predictions generated by the Solomonoff prior will have some influential components comprised of simulated consequentialists.
## How would they influence the Solomonoff prior?
Consequentialists that find themselves in universes can reason about the fundamental laws that govern their universe. If they find that their universe has relatively simple physics, they will know that their behavior contributes to the Solomonoff prior. To gain access to more resources in other universes, these consequentialists might seek to act in ways that influence the Solomonoff prior.
A contrived example of a decision other beings would want to manipulate is “what program should be written and executed next?” Beings in other universes would have an incentive to get us to write programs that were aligned with their values. A particularly interesting scenario is one in which they write themselves into existence, allowing them to effectively “break into” our universe.
For example, somewhere in the Solomonoff prior there is a program that goes something like: “Simulate this universe. Starting from the year 2100, every hour output ‘1’ if there’s a cubic meter of iron on the Moon, else output ‘0’.” By controlling the presence/absence of a cubic meter of iron on the Moon, we would be able to influence the output of this particular facet of the Solomonoff prior.
This example is a very complicated program and thus will not have much weight in the Solomonoff prior. However, by reasoning over the complexity of possible output channels for their universe, consequentialists would be able to identify output channels that weigh heavily in the Solomonoff prior. For example, if I was in the Game of Life, I might reason that sampling cells that were living in the initial conditions of the universe is simpler than sampling other cells. Additionally, sampling cells and reporting their outputs directly is simpler than sampling cells and reversing their values. Therefore, I might choose to control regions close to the initial live cells of the universe for purposes of exerting influence over the Solomonoff prior.
Additionally, consequentialists can send across many plausible channels at once, focus efforts on channels that are easy to control, send through channels that would not unduly decrease their values for other reasons, etc. Since the proportional weight in the Solomonoff prior drops off exponentially with respect to complexity, it’s possible only very small regions of space would need to be used to achieve a large fraction of the maximum possible influence.
## How much influence will they have?
A simple application of the no free lunch theorem shows that there is no way of making predictions that is better than the Solomonoff prior across all possible distributions over all possible strings. Thus, agents that are influencing the Solomonoff prior cannot be good at predicting, and thus gain influence, in all possible worlds.
In practice, this is irrelevant because consequentialists do not want to achieve equal predictive accuracy in all worlds; they only care about worlds in which being predictive results in increased influence. If I’m exerting control over the Solomonoff prior, I only care about influencing it in ways that might affect material consequences in other universes. For example, I do not care about gaining influence in universes inhospitable to life. Thus I will be able to trade off predictive power in universes I don’t care about for predictive power in universes I do care about. This means that I should restrict my attention to all universes that have resources and the Solomonoff prior is being used to make important decisions, weighting appropriately.
Another interesting thing about the Solomonoff prior is it is actually a collection of priors. The “length” of a TM is defined relative to some universal TM. For any particular TM, it’s length with respect to different universal TMs will vary; thus, there are versions of the Solomonoff prior that give more or less weight to any given TM. (Note that you can simulate universal TMs with universal TMs, so the difference across all programs will be bounded by a constant factor. However, this constant factor can be large, so the difference in relative weight between different Solomonoff priors can also be large).
In particular, this suggests a good strategy for consequentialists: find a universe that is using a version of the Solomonoff prior that has a very short description of the particular universe the consequentialists find themselves in.
The combined strategy is thus to take a distribution over all decisions informed by the Solomonoff prior, weight them by how much influence can be gained and the version of the prior being used, and read off a sequence of bits that will cause some of these decisions to result in a preferred outcome.
The question of how much influence any given universe of consequentialists will have is difficult to answer. One way of quantifying this is to think about how many “universes they don’t care about” they’re trading off for “universes they do care about” (really we should be thinking in terms of sequences, but I find reasoning about universes to be easier).
Since the consequentialists care about exerting maximum influence, we can approximate them as not caring about universes that don’t use a version of the Solomonoff prior that gives them a large weight. This can be operationalized as only caring about universes that use universal TM in a particular set for their Solomonoff prior. What is the probability that a particular universe uses a universal TM from that set? I am not sure, but 1/million to 1/billion seems reasonable. This suggests a universe of consequentialists will only care about 1/million to 1/billion universes, which means they can devote a million/billion times the predictive power to universes they care about. This is sometimes called the “anthropic update”. (This post contains more discussion about this particular argument.)
Additionally, we might think about which decisions the consequentialists would care about. If a particular decision using the Solomonoff prior is important, consequentialists are going to care more about that decision than other decisions. Conservatively, perhaps 1/1000 decisions are “important” in this sense, giving another 1000x relative weighting.
After you condition on a decision being important and using a particular version of the Solomonoff prior, it thus seems quite likely that a non-trivial fraction of your prior is being controlled by consequentialists.
An intuition pump is that this argument is closer to an existence claim than a for-all claim. The Solomonoff prior is malign if there exists a simple universe of consequentialists that wants to influence our universe. This universe need not be simple in an absolute sense, only simple relative to the other TMs that could equal it in predictive power. Even if most consequentialists are too complicated or not interested, it seems likely that there is at least one universe that is.
## Example
### Complexity of Consequentialists
How many bits does it take to specify a universe that can give rise to consequentialists? I do not know, but it seems like Conway’s Game of Life might provide a reasonable lower bound.
Luckily, the code golf community has spent some amount of effort optimizing for program size. How many bytes would you guess it takes to specify Game of Life? Well, it depends on the universal TM. Possible answers include 6, 32, 39, or 96.
Since universes of consequentialists can “cheat” by concentrating their predictive efforts onto universal TMs in which they are particularly simple, we’ll take the minimum. Additionally, my friend who’s into code golf (he wrote the 96-byte solution!) says that the 6-byte answer actually contains closer to 4 bytes of information.
To specify an initial configuration that can give rise to consequentialists we will need to provide more information. The smallest infinite growth pattern in Game of Life has been shown to need 10 cells. Another reference point is that a self-replicator with 12 cells exists in HighLife, a Game of Life variant. I’m not an expert, but I think an initial configuration that gives rise to intelligent life can be specified in an 8x8 bounding box, giving a total of 8 bytes.
Finally, we need to specify a sampling procedure that consequentialists can gain control of. Something like “read <cell> every <large number> time ticks” suffices. By assumption, the cell being sampled takes almost no information to specify. We can also choose whatever large number is easiest to specify (the busy beaver numbers come to mind). In total, I don’t think this will take more than 2 bytes.
Summing up, Game of Life + initial configuration + sampling method takes maybe 16 bytes, so a reasonable range for the complexity of a universe of consequentialists might be 10-1000 bytes. That doesn’t seem like very many, especially relative to the amount of information we’ll be conditioning the Solomonoff prior on if we ever use it to make an important decision.
### Complexity of Conditioning
When we’re using the Solomonoff prior to make an important decision, the observations we’ll condition on include information that:
1. We’re using the Solomonoff prior
2. We’re making an important decision
3. We’re using some particular universal TM
How much information will this include? Many programs will not simulate universes. Many universes exist that do not have observers. Among universes with observers, some will not develop the Solomonoff prior. These observers will make many decisions. Very few of these decisions will be important. Even fewer of these decisions are made with the Solomonoff prior. Even fewer will use the particular version of the Solomonoff prior that gets used.
It seems reasonable to say that this is at least a megabyte of raw information, or about a million bytes. (I acknowledge some cart-horse issues here.)
This means that after you condition your Solomonoff prior, you’ll be left with programs that are at least a million bytes. As our Game of Life example shows, it only takes maybe 10-1000 of these bytes to specify a universe that gives rise to consequentialists. You have approximately a million bytes left to specify more properties of the universe that will make it more likely the consequentialists will want to exert influence over the Solomonoff prior for the purpose of influencing this particular decision.
# Why might this argument be wrong?
## Inaccessible Channels
### Argument
Most of the universe is outside of humanity’s light-cone. This might suggest that most “simple” ways to sample from our universe are currently outside our influence, meaning that the only portions of the Solomonoff prior we can control are going to have an extremely low weight.
In general, it might be the case that for any universe, consequentialists inside that universe are going to have difficulty controlling simple output channels. For example, in Game of Life, a simple way to read information might sample a cell particular cell starting at t=0. However, consequentialists in Game of Life will not appear until a much later time and will be unable to control a large initial chunk of that output channel.
### Counter-argument
Paul Christiano points out that the general form of this argument also applies to other TMs that compose of your Solomonoff prior. For example, when predicting what I’ll type next, you would “want” to simulate me and predict what I would type starting at some time T. However, this is a pretty complicated way of sampling. The fact that simple sampling procedures are less predictive doesn’t asymmetrically penalize consequentialists. The consequentialists universe and sampling method only have to be simple relative to other programs that are equally good at predicting.
One might also note that large numbers can be produced with relatively few bits, so “sample starting at <large number>” is not much more complicated than “sample starting at 0”.
## Speedy Channels
### Argument
There are many simple ways of sampling from universes very quickly. For example, in Game of Life, one can sample a cell every time-tick. It seems feasible for consequentialists to simulate Earth in the Game of Life, but not feasible to simulate Earth such that they can alter a specific cell every time tick per the simulation.
### Counter-argument
Consequentialists in the Game of Life could simply simulate Earth, compute the predictions, then later broadcast them along very fast sampling channels. However, it might be the case that building a machine that alters a cell arbitrarily every time tick is impossible. In our universe, there might be sample procedures that physics does not permit us to exert arbitrary control over, e.g. due to speed of light limitations. If this is the case, consequentialists will direct efforts towards the simplest channel they can control.
## Computational Burden
### Argument
Determining how to properly influence the Solomonoff prior requires massive computation resources devoted to simulating other universes and how they’re going to use the Solomonoff prior. While the Solomonoff prior does not penalize extremely long run-times, from the perspective of the consequentialists doing the simulating, run-times will matter. In particular, consequentialists will likely be able to use compute to achieve things they value (like we are capable of doing). Therefore, it would be extremely costly to exert influence over the Solomonoff prior, potentially to the point where consequentialists will choose not to do so.
### Counter-argument
The computational burden of predicting the use of the Solomonoff in other universes is an empirical question. Since it’s a relatively fixed cost and there are many other universes, consequentialists might reason that the marginal influence over these other universes is worth the compute. Issues might arise if the use of the Solomonoff prior in other universes is very sensitive to precise historical data, which would require a very precise simulation to influence, increasing the computational burden.
Additionally, some universes will find themselves with more computing power than other universes. Universes with a lot of computing power might find it relatively easy to predict the use of the Solomonoff prior in simpler universes and subsequently exert influence over them.
## Malign implies complex
### Argument
A predictor that correctly predicts the first N bits of a sequence then switches to being malign will be strictly more complicated than a predictor that doesn’t switch to being malign. Therefore, while consequentialists in other universes might have some influence over the Solomonoff prior, they will be dominated by non-malign predictors.
### Counter-argument
This argument makes a mistaken assumption that the malign influence on the Solomonoff prior is in the form of programs that have their “malignness” as part of the program. The argument given suggests that simulated consequentialists will have an instrumental reason to be powerful predictors. These simulated consequentialists have reasoned about the Solomonoff prior and are executing the strategy of “be good at predicting, then exert malign influence”, but this strategy is not hardcoded so exerting malign influence does not add complexity.
## Canceling Influence
### Argument
If it’s true that many consequentialists are trying to influence the Solomonoff prior, then one might expect the influence to cancel out. It’s improbable that all the consequentialists have the same preferences; on average, there should be an equal number of consequentialists trying to influence any given decision in any given direction. Since the consequentialists themselves can reason thus, they will realize that the expected amount of influence is extremely low, so they will not attempt to exert influence at all. Even if some of the consequentialists try to exert influence anyway, we should expect the influence of these consequentialists to cancel out also.
### Counter-argument
Since the weight of a civilization of consequentialists in the Solomonoff prior is penalized exponentially with respect to complexity, it might be the case that for any given version of the Solomonoff prior, most of the influence is dominated by one simple universe. Different values of consequentialists imply that they care about different decisions, so for any given decision, it might be that very few universes of consequentialists are both simple enough that they have enough influence and care about that decision.
Even if for any given decision, there are always 100 universes with equal influence and differing preferences, there are strategies that they might use to exert influence anyway. One simple strategy is for each universe to exert influence with a 1% chance, giving every universe 1/100 of the resources in expectation. If the resources accessible are vast enough, then this might be a good deal for the consequentialists. Consequentialists would not defect against each other for the reasons that motivate functional decision theory.
More exotic solutions to this coordination problem include acausal trade amongst universes of different consequentialists to form collectives that exert influence in a particular direction.
Be warned that this leads to much weirdness.
# Conclusion
The Solomonoff prior is very strange. Agents that make decisions using the Solomonoff prior are likely to be subject to influence from consequentialists in simulated universes. Since it is difficult to compute the Solomonoff prior, this fact might not be relevant in the real world.
However, Paul Christiano applies roughly the same argument to claim that the implicit prior used in neural networks is also likely to generalize catastrophically. (See Learning the prior for a potential way to tackle this problem).
Warning: highly experimental interesting speculation.
## Unimportant Decisions
Consequentialists have a clear motive to exert influence over important decisions. What about unimportant decisions?
The general form of the above argument says: “for any given prediction task, the programs that do best are disproportionately likely to be consequentialists that want to do well at the task”. For important decisions, many consequentialists would instrumentally want to do well at the task. However, for unimportant decisions, there might be consequentialists that want to make good predictions. These consequentialists would still be able to concentrate efforts on versions of the Solomonoff prior that weighted them especially high, so they might outperform other programs in the long run.
It’s unclear to me whether or not this behavior would be malign. One reason why it might be malign is that these consequentialists that care about predictions would want to make our universe more predictable. However, while I am relatively confident that arguments about instrumental convergence should hold, speculating about possible preferences of simulated consequentialists seems likely to produce errors in reasoning.
## Hail mary
Paul Christiano suggests that humanity was desperate enough to want to throw a “hail mary”, one way to do this is to use the Solomonoff prior to construct a utility function that will control the entire future. Since this is a very important decision, we expect consequentialists in the Solomonoff prior to care about influencing this decision. Therefore, the resulting utility function is likely to represent some simulated universe.
If arguments about acausal trade and value handshakes hold, then the resulting utility function might contain some fraction of human values. Again, this leads to much weirdness in many ways.
## Speed prior
One reason that the Solomonoff prior contains simulated consequentialists is that its notion of complexity does not penalize runtime complexity, so very simple programs are allowed to perform massive amounts of computation. The speed prior attempts to resolve this issue by penalizing programs by an additional logarithm of the amount of time for which it’s run.
The speed prior might reduce the relative weighting of universes with consequentialists because such programs have to be run for a very long time before they start producing reasonable predictions. The consequentialists have to gain control of their universe, understand their fundamental laws of physics, simulate other universes, then manipulate the speed prior. This might all take a very long time, causing consequentialists to be dominated by other programs.
In general, penalizing slowness might cause programs to “waste” less time on simulating consequentialists, devoting more computation towards performing the prediction task. However, it still might be the case that a universe that has very fast consequentialists might still end up dominating.
Evan Hubinger also provides an argument that even very fast programs are possibly malign. The key insight is that even though your program needs to be fast, it can be running programs that are themselves less fast. For example, one possible fast way to solve a classification problem is to search to find a neural network, then use that network for inference. However, if you wanted your search to find a fast neural network, then the search itself might take longer, resulting in a net increase in speed. Thus, time “waste” can manifest in programs that were explicitly optimized to not “waste” time. This “wasted” time could potentially be used for malign optimization, in this case for gaining influence over the speed prior.
## Randomized prior
A potential way to reduce the influence consequentialists have on a decision made by the Solomonoff prior is to randomize the particular version of the prior that gets used. For example, we might make the particular universal TM we use dependent on very precise historical data. Thus, it would be very costly for consequentialists to simulate this history, and thus costly to predict which form of the Solomonoff prior we used.
If consequentialists can’t predict which Solomonoff prior we are going to use, no particular civilization of consequentialists will have the large advantage conferred by the anthropic update. Therefore, one might hope that all civilizations of consequentialists will not care about that particular decision.
This argument makes a couple of assumptions. First, it assumes that simulating very precise histories is difficult; it might not be difficult for all universes. Second, it assumes that the universes through which influence is spread cannot coordinate, which might be possible for through acausal means.
## Symmetry considerations
The way that humanity reasons is evidence for the way that consequentialists in other universes will reason. If humanity reasons that the Solomonoff prior is malign and therefore is unwilling to use it to make decisions, then consequentialists in other universes might do likewise. These universes would not use the Solomonoff prior to make decisions.
The resulting state is that everyone is worried about the Solomonoff prior being malign, so no one uses it. This means that no universe will want to use resources trying to influence the Solomonoff prior; they aren’t influencing anything.
This symmetry obviously breaks if there are universes that do not realize that the Solomonoff prior is malign or cannot coordinate to avoid its use. One possible way this might happen is if a universe had access to extremely large amounts of compute (from the subjective experience of the consequentialists). In this universe, the moment someone discovered the Solomonoff prior, it might be feasible to start making decisions based on a close approximation.
## Recursion
Universes that use the Solomonoff prior to make important decisions might be taken over by consequentialists in other universes. A natural thing for these consequentialists to do is to use their position in this new universe to also exert influence on the Solomonoff prior. As consequentialists take over more universes, they have more universes through which to influence the Solomonoff prior, allowing them to take over more universes.
In the limit, it might be that for any fixed version of the Solomonoff prior, most of the influence is wielded by the simplest consequentialists according to that prior. However, since complexity is penalized exponentially, gaining control of additional universes does not increase your relative influence over the prior by that much. I think this cumulative recursive effect might be quite strong, or might amount to nothing. | 6,222 | 31,959 | {"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} | 3.140625 | 3 | CC-MAIN-2022-05 | latest | en | 0.933295 |
http://www.planningplanet.com/blog/how-use-earned-value-management-primavera-p6 | 1,540,024,076,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512679.76/warc/CC-MAIN-20181020080138-20181020101638-00419.warc.gz | 537,026,224 | 10,663 | # How to use Earned Value Management in Primavera P6
Primavera support you to control project performance by Earned Value Management technique.
I will show you how to use it in Primavera.
We have a simple project to finish casting 10 column.
• 1 column / day.
• 100 $/ column Each activity is assigned resource A. First we have to create baseline for this project. Go to Project -> Maintain Baselines. Click on Add and OK to create a baseline. We will assign this baseline for our project. Go to Project -> Assign Baselines. Click on Project Baseline and select our baseline. Then click OK. We can show columns as display in below picture to analyze Earned Value Management. Right click on Activity Table -> Column. Select columns in Earned Value group. At the end of day 5 of the project: How many column should have been built? (This is the Planned Value) The answer is 5 column (This value is automatically calculated by using Baseline we assign previously. Baseline said that Project’s duration is 10 days, today is 5th day, so 5 column should be completed) PV = 5 x 100 = 500$
How many column have actually been built? (This is the Earned Value)
We receive report from construction site and it said only 3 columns finish.
(This is value based on the Performance % Complete which is equal to Activity % Complete by default)
EV = 3 x 100 = 300 $How much did it cost to build those three column? (This is Actual Cost) We receive report from Accountant department and it said 200$ / column.
(This value is based on Actual Units)
AC = 3 x 200 = 600 $Then we can show “Schedule Variance” and “Cost Variance”. Schedule Variance SV = EV – PV = 300 – 500 = -200$
A negative number indicates that the project is behind schedule.
1 day for 1 column for 100 $. So it means we are late 2 days. Cost Variance CV = EV – AC = 300 – 600 = -300$
A negative number indicates that the project is over budget.
It means we are currently over budget 300 $. Then we can show “Estimate to Complete” and “Estimate at Completion” ETC = BAC – EV = 1000 – 300 = 700$
EAC = ETC + AC = 700 + 600 = 1300 $According to PMI standard, we should take CPI into account to calculate these 2 values. So we can change the way Primavera calculate by : Go to WBS window -> Earned Value tab. Check on PF = 1 / Cost Performance Index Now ETC = $\frac{BAC - EV}{CPI}$ = $\frac{700}{0.5}$ = 1400$
( CPI = $\frac{EV}{AC}$ = $\frac{300}{600}$ = 0.5 )
EAC = ETC + AC = 1400 + 600 = 2000 \$
However you have to configure this option at the beginning of project (before update anything). Otherwise it doesn’t affect.
We can also make diagram report
Click on Activity Usage Profile button.
Right click on the diagram and select Activity Usage Profile Option
Check in option as display in below picture
Now you can see the Earned Value diagram. You will the cost axis on the right side (not the left one).
I also attach a theory diagram of Earned Value Management for your reference.
With Earned Value Management in Primavera we can easily see our project performance and forecast the total cost at the end of project.
You can also find other post at my personal blog : https://doduykhuong.wordpress.com/ | 807 | 3,182 | {"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": 4, "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} | 3.0625 | 3 | CC-MAIN-2018-43 | longest | en | 0.897186 |
https://learnmetrics.com/200-amp-wire-size/ | 1,708,730,239,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474470.37/warc/CC-MAIN-20240223221041-20240224011041-00443.warc.gz | 379,249,394 | 83,775 | # Wire Size For 200 Amp Service: What Size For 100, 150 Feet Away? (NEC)
200 amp wires are quite big. They can be used as 200 amp ground wires, wires for 200 amp breaker, they can be copper or aluminum 200 amp wires. The common question for all these applications here is this:
What size wire do I need for 200 amp service?
If you look at the AWG gauge wire chart here, the answer seems to be quite clear (it’s not that simple, however):
As you can see, the obvious – but wrong – the choice is using a #000 AWG wire for 200 amp service (also known as 3/0 AWG) because it has a 200A ampacity. That, however, will likely end with you frying the wire.
For safety measures (in order to avoid frying a 200 amp circuit), the National Electric Code (NEC) dictates the following requirement:
Maximum loading for any branch circuit is 80% of the rating of the circuit for ampacity of wire for any load. (NEC 220-2)
This is known as NEC 80% requirement for wire sizing. In the case of 200 amps breakers, this means you will need a wire with at least 250A ampacity to connect a 200 amp service.
Looking at the AWG wire chart, you will need an even bigger than #0000 AWG wire for 200 amps. You have to go beyond the AWG chart at look at the KCMIL wire chart.
KCMIL wires are bigger and can handle currents of 200 amps and higher. Here is a shortlist of KCMIL wires that can be used for 200 amps:
• 250 KCMIL wire with 255A amperage.
• 300 KCMIL wire with 285A amperage.
• 350 KCMIL wire with 310A amperage.
• 400 KCMIL wire with 355A amperage.
• 500 KCMIL wire with 380A amperage.
The 250 KCMIL wire is the perfect size wire for 200 amp service because it has 255A ampacity (a minimum of 250A ampacity requirement is satisfied).
We will show you how this 250A ampacity is calculated. On top of that, you will also need a bigger wire gauge if you want to send electricity 100 feet or 150 feet away from sub panel (we will show you how to calculate what size wire you need in this case as well).
## 200 Amp Wire Gauge Calculation: What Size Wire You Need For 200 Amp Circuit?
To adequately determine the 200 amp wire gauge, you first need to calculate the minimum ampacity requirement as per the 80% NEC rule.
In the case of the 200 amp circuit, here is how the calculation looks like (you have to leave 20% extra amps capacity for safety reasons):
200 Amp Wire = 200A × 100% / 80% = 250A Ampacity
That means that you can to use a wire that has an ampacity of 250A or higher in order to set up a 200 amp circuit.
As you can see from the list of KMCIL wires above, the 250 KCMIL wire is perfect for a 200 amp service since it has an ampacity of 255A.
Now, if the sub panel is some distance away (50 feet, 100 feet, 150 feet, 200 feet, and so on), you will probably have to use an even bigger wire to counted the voltage drop.
The NEC 310-16 requirements cover this nicely:
### How To Account For Sub Panel Being 100 Feet Away?
When sending electricity a long way from the sub panel, the voltage will drop with the distance. To counter this drop, you have to boost the initial amps. That means that the circuit has to handle more amps; in turn, the wire you’re using has to handle more amps.
How much more exactly?
Roughly speaking, the NEC 310-16 requirement states that for a 200 amp copper or aluminum wire, you have to add 20% to the total amps for every 100 feet of distance.
Example: Let’s say you want to use 200 amp service 100 feet away from the sub panel. We already know that such a circuit will require wires with at least 250A ampacity. On top of this, we have to increase the amps by 20% to get the true size wire for 200 amps at such a distance:
200 Amp Wire (100 ft away) = 250A × 1.2 = 300A Ampacity
Now we have to find a wire with 300A ampacity or more. If you consult the KCMIL list above, you can see that the 350 KCMIL wire is perfect for 200 amp service 100 feet away. That’s because it has a 310A ampacity.
You can make a similar calculation for 50 feet (10% addition), 150 feet (30% addition), and 200 feet (40% addition), and choose the corresponding KCMIL wire.
Hopefully, now the 200 amp wire sizing is a bit clearer.
You should also look into different amp wires and similar calculations for:
If anything is unclear, you can use the comments to pose a question and we’ll try to help you out.
### 15 thoughts on “Wire Size For 200 Amp Service: What Size For 100, 150 Feet Away? (NEC)”
1. When sizing wire for a sub panel, do I use the total amperage of the sub panel or the expected maximum load? In my case, I want to install a 200 amp sub panel in an out building that will be used for a home auto / wood working shop. I can’t visualize ever drawing more than 60 amps but for safety margin let’s assume it is 100 amps. Do I size my wire for the full 200 amps or the 100 amps? This panel is 200′ from my main panel.
• Hello George, you will need to be aware of the voltage drop here. You are trying to think of the expected maximum load. For example, if you need 100 amps at the end of the wire, you might want a wire that can handle at least 200 amps.
2. Wire size for 200’ with 200 amp at end of wire coming from meter to house?
• Hello Chace, if we presume a linear voltage drop, and 100′ incurs a 20% drop, we are dealing with a 40% drop if the wire length is 200′. You need a minimum ampacity of 250A (accounting for the NEC 80% rule). Here’s how you calculate that: 250A*1.4 = 350A. So, you need a wire with 350A ampacity. That’s a massive amount of amps, by the way.
So massive that even if you consult this AWG wire size chart, you will see that no single wire has a 350A of higher ampacity (the biggest 4/0 AWG wire has 230A ampacity). That means you will have to either use several wires in one cable or a bigger kcmil wires. The best solution here is to get that big 400 kcmil copper wire; it has a 355A ampacity at 75°C. Hope this helps a bit.
3. Hello,
I need to run wire from my main breaker panel to my shop. I have 200amp service at the main panel in the house and plan on using a 100amp (prefer 125amp) breaker in that panel to service the shop. The distance is 220′ (probably actually just a few feet less) and will be run in underground conduit. It’s confusing but would 2/0 work for that amperage and distance? Thanks.
• Hello Roger, the 2/0 AWG copper wire has an ampacity of 175A at 75°C. With a 125A breaker, the 2/0 AWG is the right size wire in this situation.
4. At 125 feet I need to use 4/0 Al for a 100 amp sub panel. Will that panel take the 4/0 or do I need a higher amp sub panel?
• Hello Chuck, for a 125 feet 200A service you would need a wire with at least 360A ampacity. 4/0 aluminum wire has a median ampacity of 180A. So that would be too small. You can use two 4/0 wires for a combined ampacity of 360A. You can use a single 700 KCMIL wire with 375A ampacity as well.
5. I have a 200 amp metered panel. I have a garage 240′ away with a 200 amp panel.
I want to send 125 amps to garage. For a run that long, what would be the appropriate wire size? I plan to use aluminum.
Note: I have an electrician that I am working with and that will do the work. Building a house. PGE is installing power soon. Trying to decide on placement of customer power pole. It’s about 500 feet from house to garage. I am considering “splitting the difference”, placing pole roughly 250 spot from each location.
• Hello Micheal, 240 feet is quite a long way. The linear relationship of ‘20% drop per 100 ft’ might not apply at such long distances. You should consult with your electrician about the size of a 200 amp wire for 200 feet away or more.
Nonetheless, we can still apply this rule. That means that the drop would be about 48% if the garage is 240 feet away. Now, you want to send 125 amps, and we have to account for the 80% NEC rule as well. Here is a bit of how the calculation would go: 125A/0.52 = 240A. Now you multiply by 1.25 factor to account for 80% NEC rule, resulting in 300A. So, you are looking for an aluminum wire with 300A ampacity at a median 75°C temperature. You can consult this kcmil wire size chart here to find the wire; specifically, you would need a 500 kcmil aluminum wire with 310A ampacity. Hope this helps.
6. would a 2-0 2-0 2-0 aluminum direct burial wire be able to run a 80 amp load at 260 foot run? Thanks | 2,233 | 8,332 | {"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} | 2.65625 | 3 | CC-MAIN-2024-10 | latest | en | 0.918286 |
https://grabcad.com/questions/noob-question-how-is-the-thinking-logic-of-cut-planes-in-inventor-plz-see-examples | 1,529,704,861,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864795.68/warc/CC-MAIN-20180622201448-20180622221448-00027.warc.gz | 619,682,339 | 16,184 | # Noob question: how is the thinking logic of "cut planes" in Inventor? Plz see examples...
Hi to all here ,especially to those ones working in Inventor. Ok straight to the point on how to do (in Inventor) the sketches no. 3 , 5, 6, and 18 for example of the enclosed attachment.
These are VERY BANAL sketchs : i've tryed using working planes,working axes and point and I'm not able to achieve the given examples.My need is to understand how to cut a solid with a plane "passing through it" in a given point ,edge or surface. Hope you understand the BASIC concept I'm trying to learn...
Thank you very much for some clues on how I can achieve that and sorry if it's too simple,but for one new to a CAD system it could be a very daunting task.
Claudio
Step 1: Create a rectangle and extrude cut the following triangular shape from the side.
Step 2: Create a sketch on the front face of the solid. Project geometry as needed to generate the large diagonal line. It should connect to the corners shown.
Step 3: Create a workplane by using the workplane command at the top out of sketch mode, by selecting the large diagonal line from the sketch, and the face on the opposite end of the solid.
Step 4: Start a sketch on that workplane and project geometry needed to create the rectangle shown.
Step 5: Extrude cut the rectangle made away from the solid to create the next cut.
Step 6: Create a workplane by selecting the 3 highlighted corners in the image.
Step 7: Start a sketch on the generated workplane. Then project the same 3 points onto the sketch. Draw lines from each point to the other to create a triangle.
Step 8: Extrude cut the profile away from the solid and that will generate the finished shape.
Let me know if this works for you.
Are you making these simply as 2D sketches or as a 3D model and then creating views in a drawing?
Ok I think I understand. Its a matter of using 2D sketches to create a workplane, and then a sketch for the needed cut on the created workplane. Do these parts have dimensions? If not I can approximate the size and show you an example of how this is done.
Follow the steps below. I have comments following each image. THis should give you an idea on how to do the others.
Thank you very much Jonathan for your feedback! Very appreciated!
Hi Jonathan! I've tryed both but essentially I start with a 2D sketch making the consequent extrusion and build the next solid and so on...But where I struggle most is to understand HOW I cut a solid with a "plane cut" in a given angle as if I were a "Ninja with a katana"...lol
I'm going to try and show you using #3 | 613 | 2,606 | {"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} | 3.3125 | 3 | CC-MAIN-2018-26 | latest | en | 0.931963 |
https://thegardendc.com/site/49b241-1/49b241-650-joules-to-watts | 1,624,607,804,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487622113.11/warc/CC-MAIN-20210625054501-20210625084501-00571.warc.gz | 520,718,186 | 7,825 | The power P in watts (W) is equal to the energy E in joules (J), divided by the time period t in seconds (s): The unit symbol for joule per second is j/s. You can calculate watts from joules and seconds, but you can't convert joules to watts, since joule and watt units represent different quantities. Watts to joules calculator Joules to watts calculation. For example, if your input is in MegaJoules, you need to divide by 1,000,000 in order to get the result in Watts. The power P in watts (W) is equal to the energy E in joules (J), divided by the time period t in seconds (s): P (W) = E (J) / t (s) So. 1000 Joules per second = 1000 Watts 1000000 Joules per second = 1000000 Watts Embed this unit converter in your page or blog, by copying the following HTML code: Watt : The watt is a unit for power, is one of SI Units. Joules to Watts conversion example. This calculator has two text fields, where the first requires you to enter the energy in joules (J) while the next allows you to fill in the time in seconds denoted by (s). Joules to watts calculation formula. Sample task: convert 100 Joules, expended over an hour to Watts. Joules and watts are different quantities and therefore cannot be translated directly. One joule per second is equal to 0.001341022 horsepower. Joule/Second : The joule per second is a unit of power, equals to one watt. Its name is in honor of the Scottish engineer James Watt, the eighteenth-century developer of the steam engine. That is why watts are converted from joules and seconds. | 379 | 1,526 | {"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} | 2.8125 | 3 | CC-MAIN-2021-25 | latest | en | 0.91889 |
https://www.geeksforgeeks.org/product-of-elements-in-an-array-having-prime-frequency/?ref=rp | 1,590,515,845,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391277.13/warc/CC-MAIN-20200526160400-20200526190400-00195.warc.gz | 733,249,821 | 30,133 | # Product of elements in an array having prime frequency
Given an array arr[] of N elements, the task is to find the product of the elements which have prime frequencies in the array. Since, the product can be large so print the product modulo 109 + 7. Note that 1 is neither prime nor composite.
Examples:
Input: arr[] = {5, 4, 6, 5, 4, 6}
Output: 120
All the elements appear 2 times which is a prime
So, 5 * 4 * 6 = 120
Input: arr[] = {1, 2, 3, 3, 2, 3, 2, 3, 3}
Output: 6
Only 2 and 3 appears prime number of times i.e. 3 and 5 respectively.
So, 2 * 3 = 6
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Approach:
• Traverse the array and store the frequencies of all the elements in a map.
• Build Sieve of Eratosthenes which will be used to test the primality of a number in O(1) time.
• Calculate the product of elements having prime frequency using the Sieve array calculated in the previous step.
Below is the implementation of the above approach:
## C++
`// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` ` `#define MOD 1000000007 ` ` ` `// Function to create Sieve to check primes ` `void` `SieveOfEratosthenes(``bool` `prime[], ``int` `p_size) ` `{ ` ` ``// False here indicates ` ` ``// that it is not prime ` ` ``prime[0] = ``false``; ` ` ``prime[1] = ``false``; ` ` ` ` ``for` `(``int` `p = 2; p * p <= p_size; p++) { ` ` ` ` ``// If prime[p] is not changed, ` ` ``// then it is a prime ` ` ``if` `(prime[p]) { ` ` ` ` ``// Update all multiples of p, ` ` ``// set them to non-prime ` ` ``for` `(``int` `i = p * 2; i <= p_size; i += p) ` ` ``prime[i] = ``false``; ` ` ``} ` ` ``} ` `} ` ` ` `// Function to return the product of elements ` `// in an array having prime frequency ` `int` `productPrimeFreq(``int` `arr[], ``int` `n) ` `{ ` ` ``bool` `prime[n + 1]; ` ` ``memset``(prime, ``true``, ``sizeof``(prime)); ` ` ` ` ``SieveOfEratosthenes(prime, n + 1); ` ` ` ` ``int` `i, j; ` ` ` ` ``// Map is used to store ` ` ``// element frequencies ` ` ``unordered_map<``int``, ``int``> m; ` ` ``for` `(i = 0; i < n; i++) ` ` ``m[arr[i]]++; ` ` ` ` ``long` `product = 1; ` ` ` ` ``// Traverse the map using iterators ` ` ``for` `(``auto` `it = m.begin(); it != m.end(); it++) { ` ` ` ` ``// Count the number of elements ` ` ``// having prime frequencies ` ` ``if` `(prime[it->second]) { ` ` ``product *= (it->first % MOD); ` ` ``product %= MOD; ` ` ``} ` ` ``} ` ` ` ` ``return` `(``int``)(product); ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``int` `arr[] = { 5, 4, 6, 5, 4, 6 }; ` ` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` ` ` ``cout << productPrimeFreq(arr, n); ` ` ` ` ``return` `0; ` `} `
## Java
`// Java implementation of the approach ` `import` `java.util.*; ` ` ` `class` `GFG ` `{ ` `static` `int` `MOD = ``1000000007``; ` ` ` `// Function to create Sieve to check primes ` `static` `void` `SieveOfEratosthenes(``boolean` `prime[], ` ` ``int` `p_size) ` `{ ` ` ``// False here indicates ` ` ``// that it is not prime ` ` ``prime[``0``] = ``false``; ` ` ``prime[``1``] = ``false``; ` ` ` ` ``for` `(``int` `p = ``2``; p * p <= p_size; p++) ` ` ``{ ` ` ` ` ``// If prime[p] is not changed, ` ` ``// then it is a prime ` ` ``if` `(prime[p]) ` ` ``{ ` ` ` ` ``// Update all multiples of p, ` ` ``// set them to non-prime ` ` ``for` `(``int` `i = p * ``2``; ` ` ``i <= p_size; i += p) ` ` ``prime[i] = ``false``; ` ` ``} ` ` ``} ` `} ` ` ` `// Function to return the product of elements ` `// in an array having prime frequency ` `static` `int` `productPrimeFreq(``int` `arr[], ``int` `n) ` `{ ` ` ``boolean` `[]prime = ``new` `boolean``[n + ``1``]; ` ` ``for` `(``int` `i = ``0``; i < n; i++) ` ` ``prime[i] = ``true``; ` ` ` ` ``SieveOfEratosthenes(prime, n + ``1``); ` ` ` ` ``int` `i, j; ` ` ` ` ``// Map is used to store ` ` ``// element frequencies ` ` ``HashMap mp = ``new` `HashMap(); ` ` ` ` ``for` `(i = ``0` `; i < n; i++) ` ` ``{ ` ` ``if``(mp.containsKey(arr[i])) ` ` ``{ ` ` ``mp.put(arr[i], mp.get(arr[i]) + ``1``); ` ` ``} ` ` ``else` ` ``{ ` ` ``mp.put(arr[i], ``1``); ` ` ``} ` ` ``} ` ` ``long` `product = ``1``; ` ` ` ` ``// Traverse the map using iterators ` ` ``for` `(Map.Entry it : mp.entrySet()) ` ` ``{ ` ` ` ` ``// Count the number of elements ` ` ``// having prime frequencies ` ` ``if` `(prime[it.getValue()]) ` ` ``{ ` ` ``product *= (it.getKey() % MOD); ` ` ``product %= MOD; ` ` ``} ` ` ``} ` ` ``return` `(``int``)(product); ` `} ` ` ` `// Driver code ` `static` `public` `void` `main (String []arg) ` `{ ` ` ``int` `arr[] = { ``5``, ``4``, ``6``, ``5``, ``4``, ``6` `}; ` ` ``int` `n = arr.length; ` ` ` ` ``System.out.println(productPrimeFreq(arr, n)); ` `} ` `} ` ` ` `// This code is contributed by Rajput-Ji `
## Python3
`# Python3 implementation of the approach ` `MOD ``=` `1000000007` ` ` `# Function to create Sieve to check primes ` `def` `SieveOfEratosthenes(prime, p_size): ` ` ` ` ``# False here indicates ` ` ``# that it is not prime ` ` ``prime[``0``] ``=` `False` ` ``prime[``1``] ``=` `False` ` ` ` ``for` `p ``in` `range``(``2``, p_size): ` ` ` ` ``# If prime[p] is not changed, ` ` ``# then it is a prime ` ` ``if` `(prime[p]): ` ` ` ` ``# Update all multiples of p, ` ` ``# set them to non-prime ` ` ``for` `i ``in` `range``(``2` `*` `p, p_size, p): ` ` ``prime[i] ``=` `False` ` ` `# Function to return the product of elements ` `# in an array having prime frequency ` `def` `productPrimeFreq(arr, n): ` ` ``prime ``=` `[``True` `for` `i ``in` `range``(n ``+` `1``)] ` ` ` ` ``SieveOfEratosthenes(prime, n ``+` `1``) ` ` ` ` ``i, j ``=` `0``, ``0` ` ` ` ``# Map is used to store ` ` ``# element frequencies ` ` ``m ``=` `dict``() ` ` ``for` `i ``in` `range``(n): ` ` ``m[arr[i]] ``=` `m.get(arr[i], ``0``) ``+` `1` ` ` ` ``product ``=` `1` ` ` ` ``# Traverse the map using iterators ` ` ``for` `it ``in` `m: ` ` ` ` ``# Count the number of elements ` ` ``# having prime frequencies ` ` ``if` `(prime[m[it]]): ` ` ``product ``*``=` `it ``%` `MOD ` ` ``product ``%``=` `MOD ` ` ` ` ``return` `product ` ` ` `# Driver code ` `arr ``=` `[``5``, ``4``, ``6``, ``5``, ``4``, ``6``] ` `n ``=` `len``(arr) ` ` ` `print``(productPrimeFreq(arr, n)) ` ` ` `# This code is contributed by Mohit Kumar `
## C#
`// C# implementation for above approach ` `using` `System; ` `using` `System.Collections.Generic; ` ` ` `class` `GFG ` `{ ` `static` `int` `MOD = 1000000007; ` ` ` `// Function to create Sieve to check primes ` `static` `void` `SieveOfEratosthenes(``bool` `[]prime, ` ` ``int` `p_size) ` `{ ` ` ``// False here indicates ` ` ``// that it is not prime ` ` ``prime[0] = ``false``; ` ` ``prime[1] = ``false``; ` ` ` ` ``for` `(``int` `p = 2; p * p <= p_size; p++) ` ` ``{ ` ` ` ` ``// If prime[p] is not changed, ` ` ``// then it is a prime ` ` ``if` `(prime[p]) ` ` ``{ ` ` ` ` ``// Update all multiples of p, ` ` ``// set them to non-prime ` ` ``for` `(``int` `i = p * 2; ` ` ``i <= p_size; i += p) ` ` ``prime[i] = ``false``; ` ` ``} ` ` ``} ` `} ` ` ` `// Function to return the product of elements ` `// in an array having prime frequency ` `static` `int` `productPrimeFreq(``int` `[]arr, ``int` `n) ` `{ ` ` ``bool` `[]prime = ``new` `bool``[n + 1]; ` ` ``int` `i; ` ` ``for` `(i = 0; i < n; i++) ` ` ``prime[i] = ``true``; ` ` ` ` ``SieveOfEratosthenes(prime, n + 1); ` ` ` ` ``// Map is used to store ` ` ``// element frequencies ` ` ``Dictionary<``int``, ` ` ``int``> mp = ``new` `Dictionary<``int``, ` ` ``int``>(); ` ` ``for` `(i = 0 ; i < n; i++) ` ` ``{ ` ` ``if``(mp.ContainsKey(arr[i])) ` ` ``{ ` ` ``var` `val = mp[arr[i]]; ` ` ``mp.Remove(arr[i]); ` ` ``mp.Add(arr[i], val + 1); ` ` ``} ` ` ``else` ` ``{ ` ` ``mp.Add(arr[i], 1); ` ` ``} ` ` ``} ` ` ``long` `product = 1; ` ` ` ` ``// Traverse the map using iterators ` ` ``foreach``(KeyValuePair<``int``, ``int``> it ``in` `mp) ` ` ``{ ` ` ` ` ``// Count the number of elements ` ` ``// having prime frequencies ` ` ``if` `(prime[it.Value]) ` ` ``{ ` ` ``product *= (it.Key % MOD); ` ` ``product %= MOD; ` ` ``} ` ` ``} ` ` ``return` `(``int``)(product); ` `} ` ` ` `// Driver code ` `static` `public` `void` `Main (String []arg) ` `{ ` ` ``int` `[]arr = { 5, 4, 6, 5, 4, 6 }; ` ` ``int` `n = arr.Length; ` ` ` ` ``Console.WriteLine(productPrimeFreq(arr, n)); ` `} ` `} ` ` ` `// This code is contributed by Princi Singh `
Output:
```120
```
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• #### Verbal Reasoning - Mental Ability
Q:
Every rational number is an integer.
A) TRUE B) FALSE
Explanation:
Any integer 'i' can be written as i = i/1, which is a rational number. Clearly, 7/2,-5/4, etc... are rational numbers but they are not integers.
Hence, from the above analysis, every integer is a rational number but a rational number need not be an integer.
Therefore the given statement that 'Every rational number is an integer' is FALSE.
6 514
Q:
Two angles whose sum is 180 degrees are
A) Complimentary angles B) Supplementary angles C) Alternate angles D) None of the above
Explanation:
Two angles whose sum is 180 degrees are called supplementary angles.
3 514
Q:
Primary care physicians are responsible for diagnosis and treatment.
A) TRUE B) FALSE
Explanation:
Primary care physicians or General practitioners (GP) are the one you go first to get a check up, and if they deem that your problem more severe than they can help you, they will send you to a secondary care physicians like specialists in Cardiologists, Neurologists, Dentists, ...
Hence, the given statement, Primary care physicians are responsible for diagnosis and treatment is TRUE.
2 501
Q:
Which letter has at least one line of symmetry?
A) A B) C C) O D) All of the above
Explanation:
The letter A has a vertical axis of symmetry.
The letter C has a horizontal axis of symmetry.
The letter O has more than one axis of symmetry, including vertical axis and horizontal axis.
2 499
Q:
Which of the following activities are examples of data gathering?
A) Counting the votes in an election to determine the next governor B) Making a pie chart showing the kinds of television sets owned by people in a sample of 1,000 C) Finding the mean weight of cancer patients in a study D) Using the results of a mathematical model to predict the next day's weather
Explanation:
Option A) "Counting the votes in an election to determine the next governor" is the only one where data gathering is done.
In all other options, data is already collected and now analyzing of data is going on.
2 497
Q:
3x3x3+3+3/3 = ?
A) 11 B) 3.33 C) 31 D) None of the above
Explanation:
We all know about BODMAS rule, by using that
3x3x3+3+3/3
= 3x3x3+3+1
= 27+3+1
= 27 + 4
= 31
Hence, 3x3x3+3+3/3 = 31.
3 482
Q:
How many joins require for 5 tables to join straight?
A) 4 B) 5 C) 6 D) 10
Explanation:
To join 5 tables we require 4 joins.
5 477
Q:
A chord that passes through the center of a circle is
A) secant B) tangent C) diameter D) perigram | 677 | 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} | 3.609375 | 4 | CC-MAIN-2019-35 | latest | en | 0.925237 |
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Guess my Number (Posted on 2004-12-20)
The Mad Hatter poured Alice another cup of tea and set the teapot down.
"May I please have one of those biscuits?" asked Alice rather timedly, rather taken aback by the sight of her present company.
"A biscuit?? Oh, I don't know. Those are the March Hare's special un-birthday biscuits, don't you know. Can she have a biscuit, March ?" asked the Hatter.
"A biscuit?? One of mine? Hurmph. Ok, I'm thinking of an integer from one to three (inclusive). If you get it, you can have a biscuit. If not, I'm afraid you'll just have to leave" said the March Hare, leaning back.
"Oh goodness, I don't know" said Alice worriedly. "Couldn't I just get a hint?"
"Oh, all right..." the March Hare sighed. "You get to ask me one yes-or-no question, that's all"
Alice thought for a moment, then smiled, knowing exactly what to ask.
See The Solution Submitted by Sam Rating: 3.7647 (17 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): Independent solution | Comment 11 of 32 |
(In reply to re(2): Independent solution by Tristan)
If you want to be pedantic:-
"Alice puts her right hand behind her back. She asks, "Is the number four less than the number of fingers up on my right hand, or is it a two?" (This question is asked as a yes-or-no question, not a which-is-it question.)"
Alice has five fingers on her right hand (if you include a thumb as a finger) regardless of how many she is "holding".
Posted by Fletch on 2004-12-21 09:19:07
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# Got 50 on quantitative ->94% percentile?
Author Message
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Joined: 03 Jul 2012
Posts: 5
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Got 50 on quantitative ->94% percentile? [#permalink]
### Show Tags
04 Mar 2013, 08:36
Hey guys,
Got 50 on the quantitative section and I am in the 94% percentile? Could that be true?
btw a friend of mine got 51 on quantitative and is in the 98%?
Why is 51 not 99% percentile and why is 50 so "low"?
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Re: Got 50 on quantitative ->94% percentile? [#permalink]
### Show Tags
04 Mar 2013, 09:07
This post should help: Post
Senior Manager
Status: Making every effort to create original content for you!!
Joined: 23 Dec 2010
Posts: 490
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Concentration: Healthcare, Social Entrepreneurship
GMAT 1: 660 Q48 V34
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Re: Got 50 on quantitative ->94% percentile? [#permalink]
### Show Tags
04 Mar 2013, 09:16
pisco wrote:
Hey guys,
Got 50 on the quantitative section and I am in the 94% percentile? Could that be true?
Hi Pisco,
Yes that could be true, as the GMAT is getting older the percentiles corresponding to a particular score is getting lower. For example, 5 - 6 years back a 750 score corresponded to 99% percentile, but now it corresponds to 98% percentile. A possible reason could be that now a days a larger number of people are getting higher scores. 94%, anytime, is an awesome percentile, and congrats for that. All the best for your apps !!!
Vercules
_________________
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Re: Got 50 on quantitative ->94% percentile? [#permalink]
### Show Tags
04 Mar 2013, 09:21
Don't get too hung up on the percentiles. They are what they are:
http://www.mba.com/the-gmat/gmat-scores-and-score-reports/what-your-percentile-ranking-means.aspx
_________________
Re: Got 50 on quantitative ->94% percentile? [#permalink] 04 Mar 2013, 09:21
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Strategy for 50'th percentile 0 20 Aug 2013, 03:21
quantitative 1 23 Sep 2012, 01:14
1 In the quantitative section I got a scaled score of a 49 3 03 Aug 2010, 22:34
80 percentile 6 25 Apr 2010, 05:09
Percentile changes 4 27 Jun 2009, 06:40
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# Got 50 on quantitative ->94% percentile?
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# Combinations with duplicates! watch
1. Sean and Hana are going to bring 8 books with them. Sean has 16 books to choose from, and Hana has 24 books to choose from.
If Hana and Sean EACH pick 4 out of 40 books, and some of them were duplicates, how would you calculate the probability of this? (By duplicate, Sean and Hana both have a copy of the same book).
----------------------
My friend is saying that this can be done by
40!/8!*32!*2!, where 2! will eliminate the duplicate. But I don't think this is true???
2. (Original post by FuryBall)
Sean and Hana are going to bring 8 books with them. Sean has 16 books to choose from, and Hana has 24 books to choose from.
If Hana and Sean each pick 4 out of 40 books, and some of them were duplicates, how would you calculate the probability of this? (By duplicate, Sean and Hana both have a copy of the same book).
----------------------
My friend is saying that this can be done by
40!/8!*32!*2!, where 2! will eliminate the duplicate. But I don't think this is true???
There seems to be some confusion, even before you get to the duplicates.
From the first lines I get the impression Sean chooses 4 from his 16, and Hana chooses 4 from her 24. BUT then you say they choose 4 each from 40, which is a different situation.
3. (Original post by ghostwalker)
There seems to be some confusion, even before you get to the duplicates.
From the first lines I get the impression Sean chooses 4 from his 16, and Hana chooses 4 from her 24. BUT then you say they choose 4 each from 40, which is a different situation.
I mean that they EACH choose 4 from 40, so together it is still 8 books picked from 40.
4. (Original post by FuryBall)
I mean that they EACH choose 4 from 40, so together it is still 8 books picked from 40.
And what are you trying to calculate? The probability that they choose a duplicate book? Or is it the number of choices they can make in the 8, which seems to be what your friend is working on? The value is going to depend on the number of books that are duplicated as well - is it just one, or more than one?
5. (Original post by ghostwalker)
And what are you trying to calculate? The probability that they choose a duplicate book? Or is it the number of choices they can make in the 8, which seems to be what your friend is working on? The value is going to depend on the number of books that are duplicated as well - is it just one, or more than one?
The question from my project doesn't specify the number of duplicates as it just says "some duplicates". But I could specify it as 1 duplicate and then find a way of applying my method to x number of duplicates.
I think I have to find the way they can choose the books while eliminating/ignoring the probability of a duplicate.
I don't understand what my friend's calculation is exactly doing.
6. (Original post by FuryBall)
The question from my project doesn't specify the number of duplicates as it just says "some duplicates". But I could specify it as 1 duplicate and then find a way of applying my method to x number of duplicates.
I think I have to find the way they can choose the books while eliminating/ignoring the probability of a duplicate.
I don't understand what my friend's calculation is exactly doing.
Well your friends calculation is incorrect. You can safely divide by 2!, if each choice contained the two duplicates, but it doesn't; some choices will have one of the duplicate pair and some will have none.
This gives us the clue on how to break this down.
Either both duplicates are chosen, so we're left with how many ways of choosing the remaining 6 from the remaining 38; which I presume you're fine with.
Or one of the duplicates is chosen, so we then have to choose 7 from 38 (since we can't choose the 2nd duplicate).
Or neither duplicate is chosen, and we just choose 8 from 38.
7. (Original post by ghostwalker)
Well your friends calculation is incorrect. You can safely divide by 2!, if each choice contained the two duplicates, but it doesn't; some choices will have one of the duplicate pair and some will have none.
This gives us the clue on how to break this down.
Either both duplicates are chosen, so we're left with how many ways of choosing the remaining 6 from the remaining 38; which I presume you're fine with.
Or one of the duplicates is chosen, so we then have to choose 7 from 38 (since we can't choose the 2nd duplicate).
Or neither duplicate is chosen, and we just choose 8 from 38.
Interesting...
But if we have A A B C, the internal ordering of the duplicate A is 2!, and therefore 4!/2!*2!*2! should be correct? (2! in the case for each duplicate set. Divide by 2! * 2! for 2 titles of duplicates (I imagine something like A A B B C), and so on.
So could we say
40!(8!)(32!)(2!)^d with d being the set of duplicates.
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Top tips from students who have already aced their exams | 1,306 | 5,461 | {"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} | 4.0625 | 4 | CC-MAIN-2018-22 | latest | en | 0.974656 |
http://lists.tunes.org/archives/tunes/1998-October/001522.html | 1,701,199,618,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099942.90/warc/CC-MAIN-20231128183116-20231128213116-00046.warc.gz | 29,923,574 | 2,156 | # logical symbols
Francois-Rene Rideau fare@tunes.org
Sun, 25 Oct 1998 17:12:06 +0100
```> I was wondering if anybody can give me a
> sensible explanation of the reason for basing most logics on several
> certain operators.
Simplicity? See for instance how these operators arise simply
in a category-theory reconstruction of intuitionnistic logic/lambda-calculus.
For a more thorough discussion of the notion of simplicity/complexity,
see "An Introduction to Kolmogorov Complexity and its Applications"
by Li & Vitanyi.
> It seems that this operator-class constitutes some sort of dirty
> reflectivity with a limited (unary) scope.
Unarity is usually chosen because it leads
to a simple finitary presentation of the logic,
which allows for easier reflection.
Now, there are *lots* of variants of various calculi
with varying arity of operators.
For instance, people involved in Pi-calculi usually prefer polyadic variants.
> "For all variables x, A(...x...) is true" is equivalent to "There does
> not exist a variable x such that A(...x...) is not true"
This only holds in classical logic, but not in most intuitionnistic logic.
> these operators are declarations about symbols, not intensions.
As soon as you've got internal notions (be it in a meta-language),
you're dealing with extensions. Intension is an ever fleeing concept.
Whenever you closely look at intensions, they become extensions.
It only becomes back an intension when you don't look at it.
## Faré | VN: Уng-Vû Bân | Join the TUNES project! http://www.tunes.org/ ##
## FR: François-René Rideau | TUNES is a Useful, Not Expedient System ##
## Reflection&Cybernethics | Project for a Free Reflective Computing System ##
Mathematicians are like Frenchmen: whatever you say to them they
translate into their own language, and forthwith it is something
entirely different.
-- Johann Wolfgang von Goethe
``` | 453 | 1,889 | {"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-2023-50 | latest | en | 0.896829 |
https://www.hackmath.net/en/math-problem/59113 | 1,701,221,345,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100047.66/warc/CC-MAIN-20231129010302-20231129040302-00012.warc.gz | 889,487,531 | 8,865 | # Potatoes
In stock, they pack potatoes in bags weighing 10 1/2 kg. How many bags do they need if they have 378 kilograms of potatoes in stock?
n = 36
### Step-by-step explanation:
Did you find an error or inaccuracy? Feel free to write us. Thank you!
Tips for related online calculators
Need help with mixed numbers? Try our mixed-number calculator. | 88 | 356 | {"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-2023-50 | latest | en | 0.895898 |
https://math.stackexchange.com/questions/2840935/critical-point-confusion | 1,563,213,668,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195523840.34/warc/CC-MAIN-20190715175205-20190715201205-00506.warc.gz | 480,759,941 | 35,257 | # Critical point confusion
So I learned that in order to find the minimum/ maximum point of a function, I would take it's derivative and find critical points, points where the derivative is zero or undefined, and then put them on a number line and do strawberry field. If it's negative to the left of the point and positive to the right of the point, the point is a minimum, etc. What I'm confused about is when the derivative is undefined, or when the denominator of your fraction is equal to zero, isn't that where you have a vertical tangent line and not a candidate for a min/max point?
• What does the term "strawberry field" refer to? The derivative might be undefined where the function goes to infinity, in which case find out whether positive or negative from each side. – Joffan Jul 4 '18 at 17:31
• I think it's best to do these kinds of problems by sketching the graph and thinking about the picture rather than by trying to apply rules. Then you'll see the vertical tangents and corners and cusps. Corners and cusps might be maximum or minumum values. – Ethan Bolker Jul 7 '18 at 0:40
isn't that where you have a vertical tangent line and not a candidate for a min/max point?
Not necessarily.
• when the derivative is undefined
$f(x)=|x|$ is not differentiable at $x=0$, yet $x=0$ is a global minimum.
• or when the denominator of your fraction is equal to zero
$f(x)=\sqrt{|x|}$ has a vertical tangent line at $x=0$, yet $x=0$ is a global minimum.
You can apply this (first derivative test) only on intervals where the function is differentiable. | 377 | 1,569 | {"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} | 3.734375 | 4 | CC-MAIN-2019-30 | longest | en | 0.929396 |
http://www.jiskha.com/display.cgi?id=1303230486 | 1,495,764,788,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608622.82/warc/CC-MAIN-20170526013116-20170526033116-00133.warc.gz | 527,447,906 | 3,578 | # Physics
posted by on .
A ray of light of frequency is 5.09 x 10^14 Hz is traveling from water to medium X. The angle of incidence in water is 45 degrees and the angle of refraction in Medium X is 29 degrees. The index of refraction of Medium X is 1.94. Calculate the wavelength for each medium. What material is medium X? | 84 | 325 | {"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-2017-22 | latest | en | 0.922649 |
http://elliscomtech.blogspot.com/2010/03/introducingprobability.html | 1,529,473,105,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863463.3/warc/CC-MAIN-20180620050428-20180620070428-00419.warc.gz | 110,028,778 | 10,909 | ## Wednesday, March 24, 2010
### Introducing...PROBABILITY!
Hey Math Genius'!
Here is a webquest to help you understand PROBABILITY! Make sure to read through ALL of the website. You will have to have the following written in your diary:
1. A TITLE!
2. Definitions of EXPERIMENT, OUTCOME, EVENT, and PROBABILITY (Create a chart like the one on the website.)
3. Show the "formula" for the probability of an event (in the box with the BLUE headline)
4. Choose ONE of the examples to be copied into your diary.
5. Draw a picture to go with the example you choose. (example: Draw the spinner and include the different colors.)
You must also complete the 5 exercise problems at the bottom of the page. On a separate piece of paper write the question as well as the correct answer. This paper will be your ticket out. Work HARD! Be PROUD! | 200 | 841 | {"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-2018-26 | latest | en | 0.909769 |
https://slides.com/neeldhara/33-miniatures-4595eb | 1,713,497,271,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817253.5/warc/CC-MAIN-20240419013002-20240419043002-00216.warc.gz | 478,557,042 | 92,705 | Slides based on the book:
$$F_0 = 0, F_1 = 1$$
$$F_{n+2} = F_{n+1} + F_n$$ for $$n = 0,1,2,\ldots$$
$$F_0, F_1, F_2, \ldots$$
#1. Fibonacci Numbers, Quickly
The Fibonacci numbers were first described in Indian mathematics, as early as 200 BC in work by Pingala on enumerating possible patterns of Sanskrit poetry formed from syllables of two lengths.
Hemachandra (c. 1150) is credited with knowledge of the sequence as well, writing that "the sum of the last and the one before the last is the number ... of the next mātrā-vṛtta."
$$\underbrace{\left(\begin{array}{cc}\star & \star \\\star & \star\end{array}\right)}_{M}\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right) = \left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$$
#1. Fibonacci Numbers, Quickly
$$\underbrace{\left(\begin{array}{cc}{\color{IndianRed}\star} & {\color{IndianRed}\star} \\\star & \star\end{array}\right)}_{M}\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right) = \left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$$
#1. Fibonacci Numbers, Quickly
$$\underbrace{\left(\begin{array}{cc}{\color{IndianRed}1} & {\color{IndianRed}1} \\\star & \star\end{array}\right)}_{M}\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right) = \left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$$
#1. Fibonacci Numbers, Quickly
$$\underbrace{\left(\begin{array}{cc}{\color{IndianRed}1} & {\color{IndianRed}1} \\{\color{DodgerBlue}\star} & {\color{DodgerBlue}\star}\end{array}\right)}_{M}\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right) = \left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$$
#1. Fibonacci Numbers, Quickly
$$\underbrace{\left(\begin{array}{cc}{\color{IndianRed}1} & {\color{IndianRed}1} \\{\color{DodgerBlue}1} & {\color{DodgerBlue}0}\end{array}\right)}_{M}\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right) = \left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$$
#1. Fibonacci Numbers, Quickly
$$\underbrace{\left(\begin{array}{cc}{\color{IndianRed}1} & {\color{IndianRed}1} \\{\color{DodgerBlue}1} & {\color{DodgerBlue}0}\end{array}\right)}_{M}\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right) = \left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$$
$$M\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right)=\left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$$
#1. Fibonacci Numbers, Quickly
$$\underbrace{\left(\begin{array}{cc}{\color{IndianRed}1} & {\color{IndianRed}1} \\{\color{DodgerBlue}1} & {\color{DodgerBlue}0}\end{array}\right)}_{M}\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right) = \left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$$
$$M\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{2} \\F_{1}\end{array}\right)$$
$$M\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right)=\left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$$
#1. Fibonacci Numbers, Quickly
$$\underbrace{\left(\begin{array}{cc}{\color{IndianRed}1} & {\color{IndianRed}1} \\{\color{DodgerBlue}1} & {\color{DodgerBlue}0}\end{array}\right)}_{M}\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right) = \left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$$
$$M\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{2} \\F_{1}\end{array}\right)$$
#1. Fibonacci Numbers, Quickly
$$\underbrace{\left(\begin{array}{cc}{\color{IndianRed}1} & {\color{IndianRed}1} \\{\color{DodgerBlue}1} & {\color{DodgerBlue}0}\end{array}\right)}_{M}\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right) = \left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$$
$$M\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{2} \\F_{1}\end{array}\right)$$
$$M\left(\begin{array}{c}F_2 \\F_1\end{array}\right)=\left(\begin{array}{c}F_{3} \\F_{2}\end{array}\right)$$
#1. Fibonacci Numbers, Quickly
$$\underbrace{\left(\begin{array}{cc}{\color{IndianRed}1} & {\color{IndianRed}1} \\{\color{DodgerBlue}1} & {\color{DodgerBlue}0}\end{array}\right)}_{M}\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right) = \left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$$
$$M\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{2} \\F_{1}\end{array}\right)$$
$$M{\color{Orange}\left(\begin{array}{c}F_2 \\F_1\end{array}\right)}=\left(\begin{array}{c}F_{3} \\F_{2}\end{array}\right)$$
#1. Fibonacci Numbers, Quickly
$$\underbrace{\left(\begin{array}{cc}{\color{IndianRed}1} & {\color{IndianRed}1} \\{\color{DodgerBlue}1} & {\color{DodgerBlue}0}\end{array}\right)}_{M}\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right) = \left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$$
$$M\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{2} \\F_{1}\end{array}\right)$$
$$M \cdot {\color{Orange}M \left(\begin{array}{c}1 \\0\end{array}\right)}=\left(\begin{array}{c}F_{3} \\F_{2}\end{array}\right)$$
#1. Fibonacci Numbers, Quickly
$$\underbrace{\left(\begin{array}{cc}{\color{IndianRed}1} & {\color{IndianRed}1} \\{\color{DodgerBlue}1} & {\color{DodgerBlue}0}\end{array}\right)}_{M}\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right) = \left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$$
$$M\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{2} \\F_{1}\end{array}\right)$$
$$M^2\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{3} \\F_{2}\end{array}\right)$$
#1. Fibonacci Numbers, Quickly
$$\underbrace{\left(\begin{array}{cc}{\color{IndianRed}1} & {\color{IndianRed}1} \\{\color{DodgerBlue}1} & {\color{DodgerBlue}0}\end{array}\right)}_{M}\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right) = \left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$$
$$M\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{2} \\F_{1}\end{array}\right)$$
$$M^n\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{n+1} \\F_{n}\end{array}\right)$$
#1. Fibonacci Numbers, Quickly
$$M^n\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{n+1} \\F_{n}\end{array}\right)$$
#1. Fibonacci Numbers, Quickly
$$\underbrace{M \longrightarrow M^{2} \longrightarrow M^{4} \longrightarrow M^{8} \longrightarrow M^{16} \longrightarrow M^{32} \longrightarrow \cdots}_{}$$
repeated squaring
$$\mathcal{O}(\log_2 n)$$ multiplications of $$2 \times 2$$ matrices.
#1. Fibonacci Numbers, Quickly
$$\underbrace{M \longrightarrow M^{2} \longrightarrow M^{4} \longrightarrow M^{8} \longrightarrow M^{16} \longrightarrow M^{32} \longrightarrow \cdots}_{}$$
repeated squaring
$$\mathcal{O}(\log_2 n)$$ multiplications of $$2 \times 2$$ matrices.
$$M^n = M^{(2^{k_1} + 2^{k_2} + \cdots + 2^{k_\ell})} = M^{2^{k_1}} M^{2^{k_2}} \cdots M^{2^{k_\ell}}$$
$$\mathcal{O}(\log_2 n)$$ multiplications of $$2 \times 2$$ matrices.
#1. Fibonacci Numbers, Quickly
If we want to compute the Fibonacci numbers by this method, we have to be careful, since the $$F_n$$ grow very fast.
#1. Fibonacci Numbers, Quickly
As we will see later, the number of decimal digits of $$F_n$$ is of order $$n$$.
Thus we must use multiple precision arithmetic, and so the arithmetic operations will be relatively slow.
#1. Fibonacci Numbers, Quickly
#2. Fibonacci Numbers, The Formula
Consider the vector space of all infinite sequences:
$$(u_0, u_1, u_2, \ldots )$$
...with coordinate-wise addition and multiplication by real numbers.
In this space we define a subspace $$\mathcal{W}$$ of all sequences satisfying the equation:
$$u_{n+2} = u_{n+1}+u_n$$ for all $$n = 0, 1, ...$$
Easy to verify: $$\mathcal{W}$$ is a subspace.
Claim: $$dim(\mathcal{W}) = 2$$.
Observe that $$(0,1,1,2,3,\ldots)$$ and $$(1,0,1,1,2,\ldots)$$ is a basis.
Goal. Find a closed-form formula for $$F_n$$.
$$= (0,1,1,2,3,5,8,13,21,\ldots)$$
$$= (\ldots,\alpha^n,\ldots)$$
$$= (\ldots,\beta^n,\ldots)$$
=
+
$$\mathcal{W}$$
#2. Fibonacci Numbers, The Formula
$$(\tau^0,\tau^1,\tau^2,\tau^3,\cdots) \in \mathcal{W}$$
$$\tau^k = \tau^{k-1}+\tau^{k-2}$$
$$\tau^2 = \tau+1$$
$$\tau_{1,2}=(1 \pm \sqrt{5}) / 2$$
$$\alpha=(1 + \sqrt{5}) / 2$$
$$\beta=(1 - \sqrt{5}) / 2$$
#2. Fibonacci Numbers, The Formula
=
+
$$F = c \cdot (\alpha^0,\alpha^1,\alpha^2,\alpha^3,\cdots) + d \cdot (\beta^0,\beta^1,\beta^2,\beta^3,\cdots)$$
$$0 = c + d$$
$$1 = c\alpha + d\beta$$
$$d = 1/(\beta -\alpha)$$
$$c = 1/(\alpha -\beta)$$
#2. Fibonacci Numbers, The Formula
#2. Fibonacci Numbers, The Formula
$$F_n = c \cdot \alpha^n + d \cdot \beta^n$$
$$d = 1/(\beta -\alpha)$$
$$c = 1/(\alpha -\beta)$$
$$\alpha=(1 + \sqrt{5}) / 2$$
$$\beta=(1 - \sqrt{5}) / 2$$
$$F_n = \frac{1}{\sqrt{5}} \cdot \left[ \left( \frac{1+\sqrt{5}}{2} \right)^n - \left( \frac{1-\sqrt{5}}{2} \right)^n \right]$$
#2. Fibonacci Numbers, The Formula
$$F_n = \left\lfloor \frac{1}{\sqrt{5}} \cdot \left( \frac{1+\sqrt{5}}{2} \right)^n \right\rfloor$$
Exercise 1. Show that:
Exercise 2. Use this method to work out a closed form for:
$$y_{n+2}=2 y_{n+1}-y_n$$
(Source: generatingfunctionology, Wilf;
h/t Matthew Drescher and John Azariah for a fun Twitter discussion on this.)
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
Their main occupation was forming various clubs,
which at some point started threatening the very survival of the city.
⚠️ There could be as many as $$2^n$$ distinct clubs!
(Well, $$2^n - 1$$ if you would prefer to exclude the empty club.)
In order to limit the number of clubs, the city council decreed the following innocent-looking rules:
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
Their main occupation was forming various clubs,
which at some point started threatening the very survival of the city.
⚠️ There could be as many as $$2^n$$ distinct clubs!
(Well, $$2^n - 1$$ if you would prefer to exclude the empty club.)
In order to limit the number of clubs, the city council decreed the following innocent-looking rules:
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
Their main occupation was forming various clubs,
which at some point started threatening the very survival of the city.
⚠️ There could be as many as $$2^n$$ distinct clubs!
(Well, $$2^n - 1$$ if you would prefer to exclude the empty club.)
In order to limit the number of clubs, the city council decreed the following innocent-looking rules:
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
Their main occupation was forming various clubs,
which at some point started threatening the very survival of the city.
⚠️ There could be as many as $$2^n$$ distinct clubs!
(Well, $$2^n - 1$$ if you would prefer to exclude the empty club.)
In order to limit the number of clubs, the city council decreed the following innocent-looking rules:
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
Example: $$\{\{1\},\{2\}, \ldots, \{n\}\}$$.
(The Singletons Clubs.)
Example: $$\{\{1,2,3\},\{1,2,4\}, \ldots, \{1,2,n\}\}$$.
(Where 1 and 2 are popular.)
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
subsets of $$[n]$$
vectors in $$n$$-dimensional space
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
sets that satisfy (1) & (2)
linearly independent vectors
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
1
0
0
0
0
1
1
1
$$\in \mathbb{F}^n_2$$
1
0
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
1
0
0
0
0
1
1
1
1
0
$$\{1,3,5,6,7\} \subseteq [10] \longrightarrow (1,0,1,0,1,1,1,0,0,0)$$
Another Example:
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
1
0
0
0
0
1
1
1
1
0
$$\{{\color{IndianRed}1},{\color{DodgerBlue}3},{\color{SeaGreen}5},{\color{Tomato}6},{\color{Purple}7}\} \subseteq [10] \longrightarrow ({\color{IndianRed}1},0,{\color{DodgerBlue}1},0,{\color{SeaGreen}1},{\color{Tomato}1},{\color{Purple}1},0,0,0)$$
Another Example:
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
Claim. If $$\{S_1, \ldots, S_t\}$$ forms a valid set of clubs over $$[n]$$,
then $$\{v_1, \ldots v_t\}$$ is a linearly independent collection of vectors in $$\mathbb{F}_2^n$$, where $$v_i := f(S_i)$$.
$$\alpha_1 v_1 + \cdots + \alpha_i v_i + \cdots + \alpha_j v_j + \cdots + \alpha_t v_t = 0$$
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
Claim. If $$\{S_1, \ldots, S_t\}$$ forms a valid set of clubs over $$[n]$$,
then $$\{v_1, \ldots v_t\}$$ is a linearly independent collection of vectors in $$\mathbb{F}_2^n$$, where $$v_i := f(S_i)$$.
$$\alpha_1 (v_1 \cdot v_i) + \cdots + \alpha_i (v_i \cdot v_i) + \cdots + \alpha_j (v_j \cdot v_i) + \cdots + \alpha_t (v_t \cdot v_i) = 0$$
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
Claim. If $$\{S_1, \ldots, S_t\}$$ forms a valid set of clubs over $$[n]$$,
then $$\{v_1, \ldots v_t\}$$ is a linearly independent collection of vectors in $$\mathbb{F}_2^n$$, where $$v_i := f(S_i)$$.
$$\alpha_1 (v_1 \cdot v_i) + \cdots + \alpha_i ({\color{IndianRed}v_i \cdot v_i}) + \cdots + \alpha_j (v_j \cdot v_i) + \cdots + \alpha_t (v_t \cdot v_i) = 0$$
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
$$\{{\color{IndianRed}1},{\color{DodgerBlue}3},{\color{SeaGreen}5},{\color{Tomato}6},{\color{Purple}7}\} \subseteq [10] \longrightarrow ({\color{IndianRed}1},0,{\color{DodgerBlue}1},0,{\color{SeaGreen}1},{\color{Tomato}1},{\color{Purple}1},0,0,0)$$
$$({\color{IndianRed}1},0,{\color{DodgerBlue}1},0,{\color{SeaGreen}1},{\color{Tomato}1},{\color{Purple}1},0,0,0)$$
$$({\color{IndianRed}1},0,{\color{DodgerBlue}1},0,{\color{SeaGreen}1},{\color{Tomato}1},{\color{Purple}1},0,0,0)$$
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
Claim. If $$\{S_1, \ldots, S_t\}$$ forms a valid set of clubs over $$[n]$$,
then $$\{v_1, \ldots v_t\}$$ is a linearly independent collection of vectors in $$\mathbb{F}_2^n$$, where $$v_i := f(S_i)$$.
$$\alpha_1 (v_1 \cdot v_i) + \cdots + \alpha_i ({\color{IndianRed}v_i \cdot v_i}) + \cdots + \alpha_j (v_j \cdot v_i) + \cdots + \alpha_t (v_t \cdot v_i) = 0$$
$${\color{IndianRed}|S_i| \mod 2}$$
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
Claim. If $$\{S_1, \ldots, S_t\}$$ forms a valid set of clubs over $$[n]$$,
then $$\{v_1, \ldots v_t\}$$ is a linearly independent collection of vectors in $$\mathbb{F}_2^n$$, where $$v_i := f(S_i)$$.
$$\alpha_1 (v_1 \cdot v_i) + \cdots + \alpha_i ({\color{IndianRed}v_i \cdot v_i}) + \cdots + \alpha_j ({\color{DodgerBlue}v_j \cdot v_i}) + \cdots + \alpha_t (v_t \cdot v_i) = 0$$
$${\color{IndianRed}|S_i| \mod 2}$$
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
$$\{{\color{IndianRed}1},{\color{DodgerBlue}3},{\color{Tomato}5},{\color{Tomato}6},{\color{Tomato}7}\} \subseteq [10] \longrightarrow ({\color{IndianRed}1},0,{\color{DodgerBlue}1},0,{\color{Tomato}1},{\color{Tomato}1},{\color{Tomato}1},0,0,0)$$
$$\{{\color{IndianRed}1},{\color{DodgerBlue}3},{\color{SeaGreen}4},{\color{SeaGreen}8},{\color{SeaGreen}9}\} \subseteq [10] \longrightarrow ({\color{IndianRed}1},0,{\color{DodgerBlue}1},{\color{SeaGreen}1},0,0,0,{\color{SeaGreen}1},{\color{SeaGreen}1},0)$$
$$({\color{IndianRed}1},0,{\color{DodgerBlue}1},0,0,0,0,0,0,0)$$
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
Claim. If $$\{S_1, \ldots, S_t\}$$ forms a valid set of clubs over $$[n]$$,
then $$\{v_1, \ldots v_t\}$$ is a linearly independent collection of vectors in $$\mathbb{F}_2^n$$, where $$v_i := f(S_i)$$.
$$\alpha_1 (v_1 \cdot v_i) + \cdots + \alpha_i ({\color{IndianRed}v_i \cdot v_i}) + \cdots + \alpha_j ({\color{DodgerBlue}v_j \cdot v_i}) + \cdots + \alpha_t (v_t \cdot v_i) = 0$$
$${\color{IndianRed}|S_i| \mod 2}$$
$${\color{DodgerBlue}|S_i \cap S_j| \mod 2}$$
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
Claim. If $$\{S_1, \ldots, S_t\}$$ forms a valid set of clubs over $$[n]$$,
then $$\{v_1, \ldots v_t\}$$ is a linearly independent collection of vectors in $$\mathbb{F}_2^n$$, where $$v_i := f(S_i)$$.
$$\alpha_1 (v_1 \cdot v_i) + \cdots + \alpha_i ({\color{IndianRed}v_i \cdot v_i}) + \cdots + \alpha_j ({\color{DodgerBlue}v_j \cdot v_i}) + \cdots + \alpha_t (v_t \cdot v_i) = 0$$
$${\color{IndianRed}1}$$
$${\color{DodgerBlue}0}$$
#3. The Clubs of OddTown
There are $$n$$ citizens living in Oddtown.
(1) Each club has to have an odd number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$n$$ clubs.
Claim. If $$\{S_1, \ldots, S_t\}$$ forms a valid set of clubs over $$[n]$$,
then $$\{v_1, \ldots v_t\}$$ is a linearly independent collection of vectors in $$\mathbb{F}_2^n$$, where $$v_i := f(S_i)$$.
$${\color{Silver}\alpha_i (v_1 \cdot v_i) + \cdots +} \alpha_i ({\color{IndianRed}v_i \cdot v_i}){\color{Silver} + \cdots + \alpha_j (v_j \cdot v_i) + \cdots + \alpha_t (v_t \cdot v_i)} = 0$$
$$\implies \alpha_i = 0$$, $$\forall i \in [n]$$
#3. The Clubs of OddTown
#3. The Clubs of OddTown
There are $$n$$ citizens living in Eventown.
(1) Each club has to have an even number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$2^{\lfloor \frac{n}{2} \rfloor}$$ clubs.
Claim. If $$\{S_1, \ldots, S_t\}$$ is a maximal and valid set of clubs over $$[n]$$,
then $$\{v_1, \ldots v_t\}$$ is a totally isotropic subspace of dimension at most $$\lfloor \frac{n}{2} \rfloor$$.
Note that $$v_i \cdot v_j = 0$$ for all $$1 \leqslant i,j \leqslant t$$
Let $$X := \{v_1, \ldots, v_t\}$$.
In other words, $$X \perp X$$, implying that $$X \subseteq X^{\perp}$$.
#3. The Clubs of OddTown
There are $$n$$ citizens living in Eventown.
(1) Each club has to have an even number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$2^{\lfloor \frac{n}{2} \rfloor}$$ clubs.
Claim. If $$\{S_1, \ldots, S_t\}$$ is a maximal and valid set of clubs over $$[n]$$,
then $$\{v_1, \ldots v_t\}$$ is a totally isotropic subspace of dimension at most $$\lfloor \frac{n}{2} \rfloor$$.
Let $$X := \{v_1, \ldots, v_t\}$$.
In other words, $$X \perp X$$, implying that $$X \subseteq X^{\perp}$$.
If $$v$$ is in span$$(X)$$, then $$v \perp X$$: therefore $$X$$ is closed
(since $$\{S_1, \ldots, S_t\}$$ is maximal).
#3. The Clubs of OddTown
There are $$n$$ citizens living in Eventown.
(1) Each club has to have an even number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$2^{\lfloor \frac{n}{2} \rfloor}$$ clubs.
Claim. If $$\{S_1, \ldots, S_t\}$$ is a maximal and valid set of clubs over $$[n]$$,
then $$\{v_1, \ldots v_t\}$$ is a totally isotropic subspace of dimension at most $$\lfloor \frac{n}{2} \rfloor$$.
Let $$X := \{v_1, \ldots, v_t\}$$.
In other words, $$X \perp X$$, implying that $$X \subseteq X^{\perp}$$.
Therefore, $$X$$ is a subspace.
#3. The Clubs of OddTown
There are $$n$$ citizens living in Eventown.
(1) Each club has to have an even number of members.
(2) Every two clubs must have an even number of members in common.
Under these rules, it is impossible to form more than $$2^{\lfloor \frac{n}{2} \rfloor}$$ clubs.
Claim. If $$\{S_1, \ldots, S_t\}$$ is a maximal and valid set of clubs over $$[n]$$,
then $$\{v_1, \ldots v_t\}$$ is a totally isotropic subspace of dimension at most $$\lfloor \frac{n}{2} \rfloor$$.
Let $$X := \{v_1, \ldots, v_t\}$$.
In other words, $$X \perp X$$, implying that $$X \subseteq X^{\perp}$$.
$$\dim(X) + \dim(X^\perp) = n \implies \dim(X) \leqslant \lfloor \frac{n}{2} \rfloor$$.
#4. Same-Size Intersections
Generalized Fisher inequality.
If $$C_1, C_2, \ldots, C_m$$ are
distinct and nonempty subsets
of an $$n$$-element set
such that
all the intersections $$C_i \cap C_j, i \neq j$$,
have the same size (say $$t$$),
then
$$m \leqslant n$$.
#4. Same-Size Intersections
Given: $$C_1, C_2, \ldots, C_m \in 2^{[n]}$$
where $$|C_i \cap C_j| = t$$ for all $$1\leqslant i < j \leqslant m$$
To Prove: $$m \leqslant n$$.
Case 1. $$|C_i| = t$$ for some $$i \in [m]$$.
At most $$n-t$$ other sets.
Total # of sets $$\leqslant n-t + 1 \leqslant n$$.
#4. Same-Size Intersections
Given: $$C_1, C_2, \ldots, C_m \in 2^{[n]}$$
where $$|C_i \cap C_j| = t$$ for all $$1\leqslant i < j \leqslant m$$
To Prove: $$m \leqslant n$$.
Case 2. $$|C_i| > t$$ for all $$i \in [m]$$.
$$a_{i j}= \begin{cases}1 & \text { if } j \in C_i, \text { and } \\ 0 & \text { otherwise }\end{cases}$$
Let $$A$$ be the $$m \times n$$ matrix with entries:
#4. Same-Size Intersections
Given: $$C_1, C_2, \ldots, C_m \in 2^{[n]}$$
where $$|C_i \cap C_j| = t$$ for all $$1\leqslant i < j \leqslant m$$
To Prove: $$m \leqslant n$$.
Case 2. $$|C_i| > t$$ for all $$i \in [m]$$.
$$\{{\color{IndianRed}1,2,5}\},\{{\color{DodgerBlue}2,3}\},\{{\color{DarkSeaGreen}3,4,5}\}$$
$$\left(\begin{array}{ccccc}{\color{IndianRed}1} & {\color{IndianRed}1} & 0 & 0 & {\color{IndianRed}1} \\0 & {\color{DodgerBlue}1} & {\color{DodgerBlue}1} & 0 & 0 \\ 0 & 0 & {\color{DarkSeaGreen}1} & {\color{DarkSeaGreen}1} & {\color{DarkSeaGreen}1} \end{array}\right)$$
#4. Same-Size Intersections
Given: $$C_1, C_2, \ldots, C_m \in 2^{[n]}$$
where $$|C_i \cap C_j| = t$$ for all $$1\leqslant i < j \leqslant m$$
To Prove: $$m \leqslant n$$.
Case 2. $$|C_i| > t$$ for all $$i \in [m]$$.
$$B := AA^T$$
$$=$$
$$A$$
$$A^T$$
$$B$$
$$C_i$$
$$C_j$$
#4. Same-Size Intersections
$$=$$
$$A$$
$$A^T$$
$$B$$
$$C_i$$
$$C_j$$
$$t$$
#4. Same-Size Intersections
$$=$$
$$A$$
$$A^T$$
$$B$$
$$C_i$$
$$C_i$$
#4. Same-Size Intersections
$$=$$
$$A$$
$$A^T$$
$$B$$
$$C_i$$
$$C_i$$
$$d_i$$
#4. Same-Size Intersections
#4. Same-Size Intersections
Given: $$C_1, C_2, \ldots, C_m \in 2^{[n]}$$
where $$|C_i \cap C_j| = t$$ for all $$1\leqslant i < j \leqslant m$$
To Prove: $$m \leqslant n$$.
Case 2. $$|C_i| > t$$ for all $$i \in [m]$$.
$$B := AA^T$$
$$B=\left(\begin{array}{ccccc}d_1 & t & t & \ldots & t \\t & d_2 & t & \ldots & t \\\vdots & \vdots & \vdots & \vdots & \vdots \\t & t & t & \ldots & d_m\end{array}\right)$$
Recall that $$A$$ is a $$m \times n$$ matrix, so: rank$$(A) \leqslant n$$
#4. Same-Size Intersections
Given: $$C_1, C_2, \ldots, C_m \in 2^{[n]}$$
where $$|C_i \cap C_j| = t$$ for all $$1\leqslant i < j \leqslant m$$
To Prove: $$m \leqslant n$$.
Case 2. $$|C_i| > t$$ for all $$i \in [m]$$.
$$B := AA^T$$
$$B=\left(\begin{array}{ccccc}d_1 & t & t & \ldots & t \\t & d_2 & t & \ldots & t \\\vdots & \vdots & \vdots & \vdots & \vdots \\t & t & t & \ldots & d_m\end{array}\right)$$
$${\color{White}m = }$$rank$$(B) \leqslant$$ rank$$(A) \leqslant n$$
#4. Same-Size Intersections
Given: $$C_1, C_2, \ldots, C_m \in 2^{[n]}$$
where $$|C_i \cap C_j| = t$$ for all $$1\leqslant i < j \leqslant m$$
To Prove: $$m \leqslant n$$.
Case 2. $$|C_i| > t$$ for all $$i \in [m]$$.
$$B := AA^T$$
$$B=\left(\begin{array}{ccccc}d_1 & t & t & \ldots & t \\t & d_2 & t & \ldots & t \\\vdots & \vdots & \vdots & \vdots & \vdots \\t & t & t & \ldots & d_m\end{array}\right)$$
$$m =$$ rank$$(B) \leqslant$$ rank$$(A) \leqslant n$$
Given: $$C_1, C_2, \ldots, C_m \in 2^{[n]}$$
where $$|C_i \cap C_j| = t$$ for all $$1\leqslant i < j \leqslant m$$
To Prove: $$m \leqslant n$$.
Case 2. $$|C_i| > t$$ for all $$i \in [m]$$.
$$B := AA^T$$
$$B=\left(\begin{array}{ccccc}d_1 & t & t & \ldots & t \\t & d_2 & t & \ldots & t \\\vdots & \vdots & \vdots & \vdots & \vdots \\t & t & t & \ldots & d_m\end{array}\right)$$
$${\color{red}m = }$$ rank$${\color{red}(B)} \leqslant$$ rank$$(A) \leqslant n$$
#4. Same-Size Intersections
#4. Same-Size Intersections
$$B=\left(\begin{array}{ccccc}d_1 & t & t & \ldots & t \\t & d_2 & t & \ldots & t \\\vdots & \vdots & \vdots & \vdots & \vdots \\t & t & t & \ldots & d_m\end{array}\right)$$
Suffices to show: $$\mathbf{x}^T B \mathbf{x}>0 \text { for all nonzero } \mathbf{x} \in \mathbb{R}^m \text {. }$$
because once we have the above,
if $$B\mathbf{x} = \mathbf{0}$$, then $$\mathbf{x}^TB\mathbf{x} = \mathbf{x}^T\mathbf{0} = 0$$, hence $$\mathbf{x} = 0$$.
#4. Same-Size Intersections
$$B=\left(\begin{array}{ccccc}d_1 & t & t & \ldots & t \\t & d_2 & t & \ldots & t \\\vdots & \vdots & \vdots & \vdots & \vdots \\t & t & t & \ldots & d_m\end{array}\right)$$
Suffices to show: $$\mathbf{x}^T B \mathbf{x}>0 \text { for all nonzero } \mathbf{x} \in \mathbb{R}^m \text {. }$$
We can write $$B=t J_n+D$$, where $$J_n$$ is the all 1's matrix and
$$D$$ is the diagonal matrix with $$d_1-t, d_2-t, \ldots, d_n-t$$ on the diagonal.
#4. Same-Size Intersections
$$B=\left(\begin{array}{ccccc}d_1 & t & t & \ldots & t \\t & d_2 & t & \ldots & t \\\vdots & \vdots & \vdots & \vdots & \vdots \\t & t & t & \ldots & d_m\end{array}\right)$$
$$\left(\begin{array}{ccccc}d_1 & t & t & t \\t & d_2 & t & t \\t & t & d_3 & t \\t & t & t & d_4\end{array}\right) =$$
$$\left(\begin{array}{ccccc}0 & ~t~ & ~t~ & ~t~ \\t & ~0~ & ~t~ & ~t~ \\t & ~t~ & ~0~ & ~t~ \\t & ~t~ & ~t~ & ~0~ \end{array}\right)$$
+ $$\left(\begin{array}{ccccc}d_1 & 0 & 0 & 0 \\0 & d_2 & 0 & 0 \\0 & 0 & d_3 & 0 \\0 & 0 & 0 & d_4\end{array}\right)$$
#4. Same-Size Intersections
$$B=\left(\begin{array}{ccccc}d_1 & t & t & \ldots & t \\t & d_2 & t & \ldots & t \\\vdots & \vdots & \vdots & \vdots & \vdots \\t & t & t & \ldots & d_m\end{array}\right)$$
$$\left(\begin{array}{ccccc}~~~~~~t~~~~~~ & ~~~~~~t~~~~~~ & ~~~~~~t~~~~~~ & ~~~~~~t~~~~~~ \\~~~~~~t~~~~~~ & ~~~~~~t~~~~~~ & ~~~~~~t~~~~~~ & ~~~~~~t~~~~~~ \\~~~~~~t~~~~~~ & ~~~~~~t~~~~~~ & ~~~~~~t~~~~~~ & ~~~~~~t~~~~~~ \\~~~~~~t~~~~~~ & ~~~~~~t~~~~~~ & ~~~~~t~~~~~ & ~~~~~t~~~~~ \end{array}\right)$$
+ $$\left(\begin{array}{ccccc}(d_1-t) & 0 & 0 & 0 \\0 & (d_2-t) & 0 & 0 \\0 & 0 & (d_3-t) & 0 \\0 & 0 & 0 & (d_4-t)\end{array}\right)$$
$$\left(\begin{array}{ccccc}d_1 & t & t & t \\t & d_2 & t & t \\t & t & d_3 & t \\t & t & t & d_4\end{array}\right) =$$
#4. Same-Size Intersections
$$B=\left(\begin{array}{ccccc}d_1 & t & t & \ldots & t \\t & d_2 & t & \ldots & t \\\vdots & \vdots & \vdots & \vdots & \vdots \\t & t & t & \ldots & d_m\end{array}\right)$$
Suffices to show: $$\mathbf{x}^T B \mathbf{x}>0 \text { for all nonzero } \mathbf{x} \in \mathbb{R}^m \text {. }$$
We can write $$B=t J_n+D$$, where $$J_n$$ is the all 1's matrix and
$$D$$ is the diagonal matrix with $$d_1-t, d_2-t, \ldots, d_n-t$$ on the diagonal.
$$\mathbf{x}^T B \mathbf{x}=\mathbf{x}^T\left(t J_n+D\right) \mathbf{x}=t \mathbf{x}^T J_n \mathbf{x}+\mathbf{x}^T D \mathbf{x}$$
#4. Same-Size Intersections
$$B=\left(\begin{array}{ccccc}d_1 & t & t & \ldots & t \\t & d_2 & t & \ldots & t \\\vdots & \vdots & \vdots & \vdots & \vdots \\t & t & t & \ldots & d_m\end{array}\right)$$
Suffices to show: $$\mathbf{x}^T B \mathbf{x}>0 \text { for all nonzero } \mathbf{x} \in \mathbb{R}^m \text {. }$$
We can write $$B=t J_n+D$$, where $$J_n$$ is the all 1's matrix and
$$D$$ is the diagonal matrix with $$d_1-t, d_2-t, \ldots, d_n-t$$ on the diagonal.
$$\mathbf{x}^T B \mathbf{x}=\mathbf{x}^T\left(t J_n+D\right) \mathbf{x}=t {\color{IndianRed}\mathbf{x}^T J_n \mathbf{x}}+\mathbf{x}^T D \mathbf{x}$$
$${\color{IndianRed}\sum_{i, j=1}^n x_i x_j=\left(\sum_{i=1}^n x_i\right)^2 \geqslant 0}$$
#4. Same-Size Intersections
$$B=\left(\begin{array}{ccccc}d_1 & t & t & \ldots & t \\t & d_2 & t & \ldots & t \\\vdots & \vdots & \vdots & \vdots & \vdots \\t & t & t & \ldots & d_m\end{array}\right)$$
Suffices to show: $$\mathbf{x}^T B \mathbf{x}>0 \text { for all nonzero } \mathbf{x} \in \mathbb{R}^m \text {. }$$
We can write $$B=t J_n+D$$, where $$J_n$$ is the all 1's matrix and
$$D$$ is the diagonal matrix with $$d_1-t, d_2-t, \ldots, d_n-t$$ on the diagonal.
$$\mathbf{x}^T B \mathbf{x}=\mathbf{x}^T\left(t J_n+D\right) \mathbf{x}=t \mathbf{x}^T J_n \mathbf{x}+{\color{Olive}\mathbf{x}^T D \mathbf{x}}$$
$${\color{Olive}\mathbf{x}^T D \mathbf{x}=\sum_{i=1}^n\left(d_i-t\right) x_i^2>0}$$
#4. Same-Size Intersections
$$B=\left(\begin{array}{ccccc}d_1 & t & t & \ldots & t \\t & d_2 & t & \ldots & t \\\vdots & \vdots & \vdots & \vdots & \vdots \\t & t & t & \ldots & d_m\end{array}\right)$$
Suffices to show: $$\mathbf{x}^T B \mathbf{x}>0 \text { for all nonzero } \mathbf{x} \in \mathbb{R}^m \text {. }$$
We can write $$B=t J_n+D$$, where $$J_n$$ is the all 1's matrix and
$$D$$ is the diagonal matrix with $$d_1-t, d_2-t, \ldots, d_n-t$$ on the diagonal.
$$\mathbf{x}^T B \mathbf{x}=\mathbf{x}^T\left(t J_n+D\right) \mathbf{x}=t \mathbf{x}^T J_n \mathbf{x}+\mathbf{x}^T D \mathbf{x}$$
$$> 0$$
#4. Same-Size Intersections
Let $$L$$ be a set of $$s$$ nonnegative integers and
$$\mathscr{F}$$ a family of subsets of an $$n$$-element set $$X$$.
Suppose that for any two distinct members $$A, B \in \mathscr{F}$$ we have $$|A \cap B| \in L$$.
Assuming in addition that $$\mathscr{F}$$ is uniform,
i.e. each member of $$\mathscr{F}$$ has the same cardinality,
a celebrated theorem of D. K. Ray-Chaudhuri and R. M. Wilson asserts that:
$$|\mathscr{F}| \leqq\left(\begin{array}{l}n \\ s\end{array}\right)$$.
#4. Same-Size Intersections
Let $$L$$ be a set of $$s$$ nonnegative integers and
$$\mathscr{F}$$ a family of subsets of an $$n$$-element set $$X$$.
Suppose that for any two distinct members $$A, B \in \mathscr{F}$$ we have $$|A \cap B| \in L$$.
P. Frankl and R. M. Wilson proved that
without the uniformity assumption, we have:
$$|\mathscr{F}| \leqq\left(\begin{array}{l}n \\s\end{array}\right)+\left(\begin{array}{c}n \\s-1\end{array}\right)+\ldots+\left(\begin{array}{l}n \\0\end{array}\right)$$
#4. Same-Size Intersections
#5. Error Correcting Codes
1 0 1 1 0 0 1 0 0 1 1 0 1 1 1 0
1 0 1 1 0 0 1 0 0 1 1 0 1 1 0 1
Noisy
Channel
#5. Error Correcting Codes
1 0 1 1 0 0 1 0 0 1 1 0 1 1 1 0
1 0 1 1 0 0 1 0 0 1 1 0 1 1 0 1
Noisy
Channel
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
1 0 1 1 0 0 1 0 0 1 1 0 1 1 1 0
1 0 1 1 0 0 1 0 0 1 1 0 1 1 1 0
1 0 1 1 0 0 1 0 0 1 1 0 1 1 1 0
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
0.33... = message | 0.66... = redundancy
#5. Error Correcting Codes
$$\mathbb{F}^k_2$$
$$\mathbb{F}^n_2$$
#5. Error Correcting Codes
$$\mathbb{F}^k_2$$
$$\mathbb{F}^n_2$$
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
1 0 1 1
0 0 1 0
0 1 0 0
1 1 1 0
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
1
1
0
1
0 0 0 1
1
0.64 = message | 0.35 = redundancy
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X A B 1
C 0 1 1
D 0 0 1
0 0 1 1
1 0 1 1 0 0 1 0 0 1 1
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X A B 1
C 0 1 1
D 0 0 1
0 0 1 1
1 0 1 1 0 0 1 0 0 1 1
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 0 B 1
C 0 1 1
D 0 0 1
0 0 1 1
1 0 1 1 0 0 1 0 0 1 1
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 0 B 1
C 0 1 1
D 0 0 1
0 0 1 1
1 0 1 1 0 0 1 0 0 1 1
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 0 B 1
C 0 1 1
D 0 0 1
0 0 1 1
1 0 1 1 0 0 1 0 0 1 1
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 0 0 1
C 0 1 1
D 0 0 1
0 0 1 1
1 0 1 1 0 0 1 0 0 1 1
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 0 0 1
C 0 1 1
D 0 0 1
0 0 1 1
1 0 1 1 0 0 1 0 0 1 1
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 0 0 1
C 0 1 1
D 0 0 1
0 0 1 1
1 0 1 1 0 0 1 0 0 1 1
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 0 0 1
0 0 1 1
D 0 0 1
0 0 1 1
1 0 1 1 0 0 1 0 0 1 1
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 0 0 1
0 0 1 1
D 0 0 1
0 0 1 1
1 0 1 1 0 0 1 0 0 1 1
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 0 0 1
0 0 1 1
D 0 0 1
0 0 1 1
1 0 1 1 0 0 1 0 0 1 1
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 0 0 1
0 0 1 1
1 0 0 1
0 0 1 1
1 0 1 1 0 0 1 0 0 1 1
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 0 0 1
0 0 1 1
1 0 0 1
0 0 1 1
1 0 1 1 0 0 1 0 0 1 1
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
1 0 1 1 0 0 1 0 0 1 1
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 0 0 1
0 0 1 1
1 0 0 1
0 0 1 1
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 1 1 1
0 1 0 0
1 1 1 1
1 1 1 0
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 1 1 1
0 1 0 0
1 1 1 1
1 1 1 0
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 1 1 1
0 1 0 0
1 1 1 1
1 1 1 0
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 1 1 1
0 1 0 0
1 1 1 1
1 1 1 0
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 1 1 1
0 1 0 0
1 1 1 1
1 1 1 0
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 1 1 1
0 1 0 0
1 1 1 1
1 1 1 0
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 1 1 1
0 1 0 0
1 1 1 1
1 1 1 0
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 1 1 1
0 1 0 0
1 1 1 1
1 1 1 0
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 1 1 1
0 1 0 0
1 1 1 1
1 1 1 0
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 1 1 1
0 1 0 0
1 1 1 1
1 1 1 0
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 1 1 1
0 1 0 0
1 1 1 1
1 1 1 0
1 1 0 0 1 0 1 1 1 1 0
#5. Error Correcting Codes
Goal: Detect and correct as many errors as possible.
Hope: minimize tool usage: it is expensive!
A baby step: correct any one bit flip.
X 1 1 1
0 1 0 0
1 1 1 1
1 1 1 0
1 1 0 0 1 0 1 1 1 1 0
Exercise: Use the X-bit to detect if there is more than one error.
#5. Error Correcting Codes
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
#5. Error Correcting Codes
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
#5. Error Correcting Codes
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
#5. Error Correcting Codes
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
#5. Error Correcting Codes
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
#5. Error Correcting Codes
$$M_{16 \times 11} \cdot [1,1,0,0,1,0,1,1,1,1,0]^T = w$$
$$\begin{bmatrix} \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \end{bmatrix}$$
#5. Error Correcting Codes
$$M_{16 \times 11} \cdot [{\color{IndianRed}1},{\color{IndianRed}1},0,{\color{IndianRed}0},{\color{IndianRed}1},0,{\color{IndianRed}1},1,{\color{IndianRed}1},1,{\color{IndianRed}0}]^T = w$$
$$\begin{bmatrix} \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \end{bmatrix}$$
#5. Error Correcting Codes
$$M_{16 \times 11} \cdot [{\color{IndianRed}1},{\color{IndianRed}1},0,{\color{IndianRed}0},{\color{IndianRed}1},0,{\color{IndianRed}1},1,{\color{IndianRed}1},1,{\color{IndianRed}0}]^T = w$$
$$\begin{bmatrix} \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \end{bmatrix}$$
#5. Error Correcting Codes
$$M_{16 \times 11} \cdot [{\color{IndianRed}1},1,{\color{IndianRed}0},{\color{IndianRed}0},1,{\color{IndianRed}0},{\color{IndianRed}1},1,1,{\color{IndianRed}1},{\color{IndianRed}0}]^T = w$$
$$\begin{bmatrix} \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1\\\star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star \\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \end{bmatrix}$$
#5. Error Correcting Codes
$$M_{16 \times 11} \cdot [{\color{IndianRed}1},1,{\color{IndianRed}0},{\color{IndianRed}0},1,{\color{IndianRed}0},{\color{IndianRed}1},1,1,{\color{IndianRed}1},{\color{IndianRed}0}]^T = w$$
$$\begin{bmatrix} \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \end{bmatrix}$$
#5. Error Correcting Codes
$$M_{16 \times 11} \cdot [1,1,0,0,1,0,1,1,1,1,0]^T = w$$
$$\begin{bmatrix} \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \end{bmatrix}$$
#5. Error Correcting Codes
$$M_{16 \times 11} \cdot [1,1,0,0,1,0,1,1,1,1,0]^T = w$$
$$\begin{bmatrix} \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \end{bmatrix}$$
#5. Error Correcting Codes
$$M_{16 \times 11} \cdot [1,{\color{IndianRed}1},{\color{IndianRed}0},{\color{IndianRed}0},1,0,1,{\color{IndianRed}1},{\color{IndianRed}1},{\color{IndianRed}1},{\color{IndianRed}0}]^T = w$$
$$\begin{bmatrix} \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \end{bmatrix}$$
#5. Error Correcting Codes
$$M_{16 \times 11} \cdot [1,{\color{IndianRed}1},{\color{IndianRed}0},{\color{IndianRed}0},1,0,1,{\color{IndianRed}1},{\color{IndianRed}1},{\color{IndianRed}1},{\color{IndianRed}0}]^T = w$$
$$\begin{bmatrix} \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \end{bmatrix}$$
#5. Error Correcting Codes
$$M_{16 \times 11} \cdot [1,1,0,0,1,0,1,1,1,1,0]^T = w$$
$$\begin{bmatrix} \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1\\ 0 & {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \end{bmatrix}$$
#5. Error Correcting Codes
$$M_{16 \times 11} \cdot [1,1,0,0,{\color{IndianRed}1},{\color{IndianRed}0},{\color{IndianRed}1},{\color{IndianRed}1},{\color{IndianRed}1},{\color{IndianRed}1},{\color{IndianRed}0}]^T = w$$
$$\begin{bmatrix} \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1\\ 0 & {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \end{bmatrix}$$
#5. Error Correcting Codes
$$M_{16 \times 11} \cdot [1,1,0,0,{\color{IndianRed}1},{\color{IndianRed}0},{\color{IndianRed}1},{\color{IndianRed}1},{\color{IndianRed}1},{\color{IndianRed}1},{\color{IndianRed}0}]^T = w$$
$$\begin{bmatrix} \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1\\ 0 & {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ \end{bmatrix}$$
#5. Error Correcting Codes
$$M_{16 \times 11} \cdot [1,1,0,0,1,0,1,1,1,1,0]^T = w$$
$$\begin{bmatrix} \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\ 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1\\ 0 & {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\0 & 0 & 0 & 0 & {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & {\color{DodgerBlue}1} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & {\color{DodgerBlue}1} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & {\color{DodgerBlue}1} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & {\color{DodgerBlue}1} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & {\color{DodgerBlue}1} \\ \end{bmatrix}$$
#5. Error Correcting Codes
$$M_{16 \times 11} \cdot [1,1,0,0,1,0,1,1,1,1,0]^T = w$$
$$\begin{bmatrix} \star & \star & \star & \star & \star & \star & \star & \star & \star & \star & \star\\{\color{Blue}1} & {\color{Blue}1} & 0 & {\color{Blue}1} & {\color{Blue}1} & 0 & {\color{Blue}1} & 0 & {\color{Blue}1} & 0 & {\color{Blue}1}\\ {\color{Orange}1} & 0 & {\color{Orange}1} & {\color{Orange}1} & 0 & {\color{Orange}1} & {\color{Orange}1} & 0 & 0 & {\color{Orange}1} & {\color{Orange}1} \\ {\color{Thistle}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & {\color{SeaGreen}1} & {\color{SeaGreen}1} & {\color{SeaGreen}1} & 0 & 0 & 0 & {\color{SeaGreen}1} & {\color{SeaGreen}1} & {\color{SeaGreen}1} & {\color{SeaGreen}1}\\ 0 & {\color{Thistle}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & {\color{Thistle}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & {\color{Thistle}1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \ 0 & 0 & 0 & 0 & {\color{IndianRed}1} & {\color{IndianRed}1} & {\color{IndianRed}1} & {\color{IndianRed}1} & {\color{IndianRed}1} & {\color{IndianRed}1} & {\color{IndianRed}1}\\0 & 0 & 0 & 0 & {\color{Thistle}1} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & {\color{Thistle}1} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & {\color{Thistle}1} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & {\color{Thistle}1} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & {\color{Thistle}1} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & {\color{Thistle}1} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & {\color{Thistle}1} \\ \end{bmatrix}$$
#5. Error Correcting Codes
$$x_{1} \oplus x_{3} \oplus x_{5} \oplus x_{7} \oplus x_{9} \oplus x_{11} \oplus x_{13} \oplus x_{15} = 0 \\ x_{2} \oplus x_{3} \oplus x_{6} \oplus x_{7} \oplus x_{10} \oplus x_{11} \oplus x_{14} \oplus x_{15} = 0 \\ x_{4} \oplus x_{5} \oplus x_{6} \oplus x_{7} \oplus x_{12} \oplus x_{13} \oplus x_{14} \oplus x_{15} = 0 \\ x_{8} \oplus x_{9} \oplus x_{10} \oplus x_{11} \oplus x_{12} \oplus x_{13} \oplus x_{14} \oplus x_{15} = 0 \\$$
#5. Error Correcting Codes
$$\underbrace{\begin{bmatrix} 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\end{bmatrix}}_{\text{Parity Check Matrix: } P}$$
What is the null space of P?
The set of all valid code words.
#5. Error Correcting Codes
$$\underbrace{\begin{bmatrix} 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\end{bmatrix}}_{\text{Parity Check Matrix: } P}$$
$$Pw = P(v + e)$$
$$P(v + e) = Pv + Pe$$
$$Pv + Pe = 0 + Pe = Pe$$
#5. Error Correcting Codes
$$\underbrace{\begin{bmatrix} 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\end{bmatrix}}_{\text{Parity Check Matrix: } P}$$
$$\begin{bmatrix} 0 \\ 0 \\ {\color{IndianRed}1} \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$
#5. Error Correcting Codes
$$\underbrace{\begin{bmatrix} 0 & 1 & {\color{IndianRed}0} & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & {\color{IndianRed}1} & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & {\color{IndianRed}0} & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & {\color{IndianRed}0} & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\end{bmatrix}}_{\text{Parity Check Matrix: } P}$$
$$\begin{bmatrix} 0 \\ 0 \\ {\color{IndianRed}1} \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$
#5. Error Correcting Codes
$$\underbrace{\begin{bmatrix} 0 & 1 & {\color{IndianRed}0} & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & {\color{IndianRed}1} & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & {\color{IndianRed}0} & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & {\color{IndianRed}0} & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\end{bmatrix}}_{\text{Parity Check Matrix: } P}$$
$$\begin{bmatrix} 0 \\ 0 \\ {\color{IndianRed}1} \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$
$$\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$$
=
#5. Error Correcting Codes
The (4 → 7) Hamming Code
001
010
011
100
101
110
111
001
010
011
100
101
110
111
001
010
011
100
101
110
111
001
010
011
100
101
110
111
001 010 011 100 101 110 111
x
y
z
a
b
c
d
x = a + b + d
y = a + c + d
z = b + c + d
#5. Error Correcting Codes
I have a function $$f$$ that takes a 16-bit string as input & produces a number between 0 and 15 as output.
You give me a string $$s \in \{0,1\}^{16}$$ and a number $$n \in \{0,...,15\}$$.
It will turn out that $$f(s) = n$$.
Well, this is a bit too much :)
#5. Error Correcting Codes
I have a function $$f$$ that takes a 16-bit string as input & produces a number between 0 and 15 as output.
You give me a string $$s \in \{0,1\}^{16}$$ and a number $$n \in \{0,...,15\}$$.
It will turn out that $$f(t) = n$$.
I will flip one bit in $$s$$ to get $$t$$.
#5. Error Correcting Codes
I have a function $$f$$ that takes a 16-bit string as input & produces a number between 0 and 15 as output.
You give me a string $$s \in \{0,1\}^{16}$$ and a number $$n \in \{0,...,15\}$$.
It will turn out that $$f(t) = n$$.
I will flip one bit in $$s$$ to get $$t$$.
#6. Odd Distances
There are no four points in the plane such that
the distance between each pair is an odd integer.
#6. Odd Distances
Are there four points in the plane such that
the distance between each pair is an even integer?
#6. Odd Distances
How many points can we have on a plane
so that their pairwise distances are integers?
#6. Odd Distances
There are no four points in the plane such that
the distance between each pair is an odd integer.
Let us suppose for contradiction that
there exist 4 points with all the distances odd.
We can assume that one of them is $$\mathbf{0}$$,
and we call the three remaining ones $$\mathbf{a}, \mathbf{b}, \mathbf{c}$$.
Then $$\|\mathbf{a}\|,\|\mathbf{b}\|,\|\mathbf{c}\|,\|\mathbf{a}-\mathbf{b}\|$$, $$\|\mathbf{b}-\mathbf{c}\|$$, and $$\|\mathbf{c}-\mathbf{a}\|$$ are odd integers.
And also $$\|\mathbf{a}\|^2,\|\mathbf{b}\|^2,\|\mathbf{c}\|^2,\|\mathbf{a}-\mathbf{b}\|^2$$, $$\|\mathbf{b}-\mathbf{c}\|^2$$, and $$\|\mathbf{c}-\mathbf{a}\|^2$$ are $$\equiv 1 \bmod 8$$.
#6. Odd Distances
There are no four points in the plane such that
the distance between each pair is an odd integer.
$$m$$ odd $$\implies m^2 \equiv 1$$ mod $$8$$.
$$m^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4(k^2 + k) + 1 = 4k(k+1) + 1$$
#6. Odd Distances
There are no four points in the plane such that
the distance between each pair is an odd integer.
$$m$$ odd $$\implies m^2 \equiv 1$$ mod $$8$$.
$$m^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4(k^2 + k) + 1 = 4{\color{DodgerBlue}k}(k+1) + 1$$
#6. Odd Distances
There are no four points in the plane such that
the distance between each pair is an odd integer.
$$m$$ odd $$\implies m^2 \equiv 1$$ mod $$8$$.
$$m^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4(k^2 + k) + 1 = 4k({\color{DodgerBlue}k+1}) + 1$$
#6. Odd Distances
There are no four points in the plane such that
the distance between each pair is an odd integer.
Let us suppose for contradiction that
there exist 4 points with all the distances odd.
We can assume that one of them is $$\mathbf{0}$$,
and we call the three remaining ones $$\mathbf{a}, \mathbf{b}, \mathbf{c}$$.
Then $$\|\mathbf{a}\|,\|\mathbf{b}\|,\|\mathbf{c}\|,\|\mathbf{a}-\mathbf{b}\|$$, $$\|\mathbf{b}-\mathbf{c}\|$$, and $$\|\mathbf{c}-\mathbf{a}\|$$ are odd integers.
And also $$\|\mathbf{a}\|^2,\|\mathbf{b}\|^2,\|\mathbf{c}\|^2,\|\mathbf{a}-\mathbf{b}\|^2$$, $$\|\mathbf{b}-\mathbf{c}\|^2$$, and $$\|\mathbf{c}-\mathbf{a}\|^2$$ are $$\equiv 1 \bmod 8$$.
#6. Odd Distances
There are no four points in the plane such that
the distance between each pair is an odd integer.
$$2\langle\mathbf{a}, \mathbf{b}\rangle = \|\mathbf{a}\|^2+\|\mathbf{b}\|^2-\|\mathbf{a}-\mathbf{b}\|^2 \equiv 1(\bmod 8)$$
$$\|\mathbf{a}\|,\|\mathbf{b}\|,\|\mathbf{c}\|,\|\mathbf{a}-\mathbf{b}\|$$, $$\|\mathbf{b}-\mathbf{c}\|$$, and $$\|\mathbf{c}-\mathbf{a}\|$$ are odd integers.
$$\|\mathbf{a}\|^2,\|\mathbf{b}\|^2,\|\mathbf{c}\|^2,\|\mathbf{a}-\mathbf{b}\|^2$$, $$\|\mathbf{b}-\mathbf{c}\|^2$$, and $$\|\mathbf{c}-\mathbf{a}\|^2$$ are $$\equiv 1 \bmod 8$$.
$$2\langle\mathbf{a}, \mathbf{c}\rangle = \|\mathbf{a}\|^2+\|\mathbf{c}\|^2-\|\mathbf{a}-\mathbf{c}\|^2 \equiv 1(\bmod 8)$$
$$2\langle\mathbf{b}, \mathbf{c}\rangle = \|\mathbf{b}\|^2+\|\mathbf{c}\|^2-\|\mathbf{b}-\mathbf{c}\|^2 \equiv 1(\bmod 8)$$
#6. Odd Distances
There are no four points in the plane such that
the distance between each pair is an odd integer.
$$2\langle\mathbf{a}, \mathbf{b}\rangle = \|\mathbf{a}\|^2+\|\mathbf{b}\|^2-\|\mathbf{a}-\mathbf{b}\|^2 \equiv 1(\bmod 8)$$
$$2\langle\mathbf{a}, \mathbf{c}\rangle = \|\mathbf{a}\|^2+\|\mathbf{c}\|^2-\|\mathbf{a}-\mathbf{c}\|^2 \equiv 1(\bmod 8)$$
$$2\langle\mathbf{b}, \mathbf{c}\rangle = \|\mathbf{b}\|^2+\|\mathbf{c}\|^2-\|\mathbf{b}-\mathbf{c}\|^2 \equiv 1(\bmod 8)$$
$$2 \cdot \underbrace{\left(\begin{array}{ccc}\langle\mathbf{a}, \mathbf{a}\rangle & \langle\mathbf{a}, \mathbf{b}\rangle & \langle\mathbf{a}, \mathbf{c}\rangle \\\langle\mathbf{b}, \mathbf{a}\rangle & \langle\mathbf{b}, \mathbf{b}\rangle & \langle\mathbf{b}, \mathbf{c}\rangle \\\langle\mathbf{c}, \mathbf{a}\rangle & \langle\mathbf{c}, \mathbf{b}\rangle & \langle\mathbf{c}, \mathbf{c}\rangle\end{array}\right)}_{\text{rank} = 3} \equiv \left(\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\1 & 1 & 2 \end{array}\right) \bmod 8$$
#6. Odd Distances
There are no four points in the plane such that
the distance between each pair is an odd integer.
$$2 \cdot \underbrace{\left(\begin{array}{ccc}\langle\mathbf{a}, \mathbf{a}\rangle & \langle\mathbf{a}, \mathbf{b}\rangle & \langle\mathbf{a}, \mathbf{c}\rangle \\\langle\mathbf{b}, \mathbf{a}\rangle & \langle\mathbf{b}, \mathbf{b}\rangle & \langle\mathbf{b}, \mathbf{c}\rangle \\\langle\mathbf{c}, \mathbf{a}\rangle & \langle\mathbf{c}, \mathbf{b}\rangle & \langle\mathbf{c}, \mathbf{c}\rangle\end{array}\right)}_{\text{rank} = 3} \equiv \underbrace{\left(\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\1 & 1 & 2 \end{array}\right)}_{\text{det} = 4} \bmod 8$$
$$\overbrace{\left(\begin{array}{cc} a_1 & a_2 \\ b_1 & b_2 \\ c_1 & c_2 \end{array}\right) \cdot \left(\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right)}^{\text{rank} = 2}$$
$$=$$
$$\det(2B) \equiv 4 \bmod 8 \neq 0 \implies \det(B) \neq 0.$$
#7. Are These Distances Euclidean?
Can we find three points $$p, q, r$$ in the plane whose mutual Euclidean distances are all one?
*picture not to scale 😀
#7. Are These Distances Euclidean?
Can we find $$\mathbf{p}, \mathbf{q}, \mathbf{r}$$ with
$$\|\mathbf{p}-\mathbf{q}\|=\|\mathbf{q}-\mathbf{r}\|=1 \text { and }\|\mathbf{p}-\mathbf{r}\|=3$$?
$$\underbrace{\|\mathbf{p}-\mathbf{r}\| \leqslant{\color{DodgerBlue}\|\mathbf{p}-\mathbf{q}\|}+{\color{SeaGreen}\|\mathbf{q}-\mathbf{r}\|}}_{\text{{\color{IndianRed}Triangle Inequality}}}$$
$$\mathbf{p}$$
$$\mathbf{q}$$
$$\mathbf{r}$$
#7. Are These Distances Euclidean?
It turns out that the triangle inequality is the only obstacle for three points.
Whenever nonnegative real numbers $$x, y, z$$ satisfy $$x \leqslant y+z, y \leqslant x+z$$, and $$z \leqslant x+y$$,
then there are $$\mathbf{p}, \mathbf{q}, \mathbf{r} \in \mathbb{R}^2$$ such that:
$$\|\mathbf{p}-\mathbf{q}\|=x,\|\mathbf{q}-\mathbf{r}\|=y$$, and $$\|\mathbf{p}-\mathbf{r}\|=z$$.
These are well known conditions for the existence of a triangle with given side lengths.
#7. Are These Distances Euclidean?
Whenever nonnegative real numbers $$x, y, z$$ satisfy $$x \leqslant y+z, y \leqslant x+z$$, and $$z \leqslant x+y$$,
then there are $$\mathbf{p}, \mathbf{q}, \mathbf{r} \in \mathbb{R}^2$$ such that:
$${\color{DodgerBlue}\|\mathbf{p}-\mathbf{q}\|=x},{\color{SeaGreen}\|\mathbf{q}-\mathbf{r}\|=y}$$, and $${\color{Orange}\|\mathbf{p}-\mathbf{r}\|=z}$$.
$$\mathbf{p}$$
$$\mathbf{q}$$
$$\mathbf{r}$$
#7. Are These Distances Euclidean?
#7. Are These Distances Euclidean?
What about four points in $$\mathbb{R}^3$$?
What about four points in $$\mathbb{R}^2$$?
#7. Are These Distances Euclidean?
Theorem. Let $$m_{i j}, i, j=0,1, \ldots, n$$, be nonnegative real numbers with $$m_{i j}=m_{j i}$$ for all $$i, j$$ and $$m_{i i}=0$$ for all $$i$$.
Then points $$\mathbf{p}_0, \mathbf{p}_1, \ldots, \mathbf{p}_n \in \mathbb{R}^n$$ with $$\left\|\mathbf{p}_i-\mathbf{p}_j\right\|=m_{i j}$$ for all $$i, j$$ exist if and only if the $$n \times n$$ matrix $$G$$ with
$$g_{i j}=\frac{1}{2}\left(m_{0 i}^2+m_{0 j}^2-m_{i j}^2\right)$$
is positive semidefinite.
#7. Are These Distances Euclidean?
Fact. An real symmetric $$n \times n$$ matrix $$A$$ is positive semidefinite
if and only if
there exists an $$n \times n$$ real matrix $$X$$ such that $$A=X^T X$$.
#7. Are These Distances Euclidean?
$$\begin{bmatrix}\frac{1}{2}(m_{0 1}^2+m_{0 1}^2-m_{1 1}^2) & \frac{1}{2}(m_{0 1}^2+m_{0 2}^2-m_{1 2}^2) & \frac{1}{2}(m_{0 1}^2+m_{0 3}^2-m_{1 3}^2) \\ & & \\ \frac{1}{2}(m_{0 2}^2+m_{0 1}^2-m_{2 1}^2) & \frac{1}{2}(m_{0 2}^2+m_{0 2}^2-m_{2 2}^2) & \frac{1}{2}(m_{0 2}^2+m_{0 3}^2-m_{2 3}^2) \\ & & \\ \frac{1}{2}(m_{0 3}^2+m_{0 1}^2-m_{3 1}^2) & \frac{1}{2}(m_{0 3}^2+m_{0 2}^2-m_{3 2}^2) & \frac{1}{2}(m_{0 3}^2+m_{0 3}^2-m_{3 3}^2) \end{bmatrix}$$
#7. Are These Distances Euclidean?
$$2\langle\mathbf{x}, \mathbf{y}\rangle = \|\mathbf{x}\|^2+\|\mathbf{y}\|^2-\|\mathbf{x}-\mathbf{y}\|^2$$
$$\|\mathbf{x}\|^2+\|\mathbf{y}\|^2-\|\mathbf{x}-\mathbf{y}\|^2 =$$
$$- ((x_1 - y_1)^2 + (x_2-y_2)^2 + (x_3-y_3)^2)$$
$$2 \langle \mathbf{x}, \mathbf{y} \rangle = 2 (x_1 \cdot y_1 + x_2 \cdot y_2 + x_3 \cdot y_3)$$
$$(x_1^2 + x_2^2 + x_3^2) + (y_1^2 + y_2^2 + y_3^2)$$
#7. Are These Distances Euclidean?
$$\mathbf{x}_i:=\mathbf{p}_i-\mathbf{p}_0, i=1,2, \ldots, n$$
$$\langle \mathbf{x_i}, \mathbf{x_j} \rangle = \frac{1}{2} \left(\|\mathbf{x_i}\|^2+\|\mathbf{x_j}\|^2-\|\mathbf{x_i}-\mathbf{x_j}\|^2\right)$$
$$= \frac{1}{2} \left(\|\mathbf{x_i}\|^2+\|\mathbf{x_j}\|^2-\|(\mathbf{p_i}-\mathbf{p_0})- (\mathbf{p_j}-\mathbf{p_0})\|^2\right)$$
$$= \frac{1}{2} \left(\|\mathbf{x_i}\|^2+\|\mathbf{x_j}\|^2-\|(\mathbf{p_i} - \mathbf{p_j})\|^2\right)$$
$$= \frac{1}{2} \left(\|\mathbf{x_i}\|^2+\|\mathbf{x_j}\|^2-m_{ij}^2\right)$$
$$= \frac{1}{2} \left(\|\mathbf{p_i}-\mathbf{p_0}\|^2+\|\mathbf{p_j}-\mathbf{p_0}\|^2-m_{ij}^2\right)$$
$$= \frac{1}{2} \left( m_{0i}^2+m_{0j}^2-m_{ij}^2\right) = g_{ij}$$
#7. Are These Distances Euclidean?
$$\mathbf{x}_i:=\mathbf{p}_i-\mathbf{p}_0, i=1,2, \ldots, n$$
$$\begin{bmatrix} \langle x_1, x_1 \rangle & \langle x_1, x_2 \rangle & \cdots & \langle x_1, x_n \rangle \\ & & & \\ \langle x_2, x_1 \rangle & \langle x_2, x_2 \rangle & \cdots & \langle x_2, x_n \rangle \\ & & & \\ \vdots & \vdots & \ddots & \vdots \\ & & & \\ \langle x_n, x_1 \rangle & \langle x_n, x_2 \rangle & \cdots & \langle x_n, x_n \rangle \end{bmatrix}$$
#7. Are These Distances Euclidean?
$$\mathbf{x}_i:=\mathbf{p}_i-\mathbf{p}_0, i=1,2, \ldots, n$$
$$\begin{bmatrix}\rule{1.6cm}{0.4pt} & x_1 & \rule{1.6cm}{0.4pt} \\\rule{1.6cm}{0.4pt} & x_2 & \rule{1.6cm}{0.4pt} \\& \vdots & \\\rule{1.6cm}{0.4pt} & x_i & \rule{1.6cm}{0.4pt} \\& \vdots & \\\rule{1.6cm}{0.4pt} & x_j & \rule{1.6cm}{0.4pt} \\& \vdots & \\\rule{1.6cm}{0.4pt} & x_n & \rule{1.6cm}{0.4pt} \\\end{bmatrix}$$
$$\begin{bmatrix}\rule{1.6cm}{0.4pt} & x_1 & \rule{1.6cm}{0.4pt} \\\rule{1.6cm}{0.4pt} & x_2 & \rule{1.6cm}{0.4pt} \\& \vdots & \\\rule{1.6cm}{0.4pt} & x_i & \rule{1.6cm}{0.4pt} \\& \vdots & \\\rule{1.6cm}{0.4pt} & x_j & \rule{1.6cm}{0.4pt} \\& \vdots & \\\rule{1.6cm}{0.4pt} & x_n & \rule{1.6cm}{0.4pt} \\\end{bmatrix}^T$$
#7. Are These Distances Euclidean?
$$\mathbf{x}_i:=\mathbf{p}_i-\mathbf{p}_0, i=1,2, \ldots, n$$
$$\begin{bmatrix}\rule{1.6cm}{0.4pt} & x_1 & \rule{1.6cm}{0.4pt} \\\rule{1.6cm}{0.4pt} & x_2 & \rule{1.6cm}{0.4pt} \\& \vdots & \\{\color{IndianRed}\rule{1.6cm}{0.4pt}} & {\color{IndianRed}x_i} & {\color{IndianRed}\rule{1.6cm}{0.4pt}}\\& \vdots & \\\rule{1.6cm}{0.4pt} & x_j & \rule{1.6cm}{0.4pt} \\& \vdots & \\\rule{1.6cm}{0.4pt} & x_n & \rule{1.6cm}{0.4pt} \\\end{bmatrix}$$
$$\begin{bmatrix}\rule{1.6cm}{0.4pt} & x_1 & \rule{1.6cm}{0.4pt} \\\rule{1.6cm}{0.4pt} & x_2 & \rule{1.6cm}{0.4pt} \\& \vdots & \\\rule{1.6cm}{0.4pt} & x_i & \rule{1.6cm}{0.4pt} \\& \vdots & \\{\color{IndianRed}\rule{1.6cm}{0.4pt}} & {\color{IndianRed}x_j} & {\color{IndianRed}\rule{1.6cm}{0.4pt}} \\& \vdots & \\\rule{1.6cm}{0.4pt} & x_n & \rule{1.6cm}{0.4pt} \\\end{bmatrix}^T$$
#7. Are These Distances Euclidean?
$$\mathbf{x}_i:=\mathbf{p}_i-\mathbf{p}_0, i=1,2, \ldots, n$$
#7. Are These Distances Euclidean?
What we showed. Let $$m_{i j}, i, j=0,1, \ldots, n$$, be nonnegative real numbers with $$m_{i j}=m_{j i}$$ for all $$i, j$$ and $$m_{i i}=0$$ for all $$i$$.
If there are points $$\mathbf{p}_0, \mathbf{p}_1, \ldots, \mathbf{p}_n \in \mathbb{R}^n$$ with $$\left\|\mathbf{p}_i-\mathbf{p}_j\right\|=m_{i j}$$ for all $$i, j$$, then the $$n \times n$$ matrix $$G$$ with
$$g_{i j}=\frac{1}{2}\left(m_{0 i}^2+m_{0 j}^2-m_{i j}^2\right)$$
is positive semidefinite.
#7. Are These Distances Euclidean?
Up next. Let $$m_{i j}, i, j=0,1, \ldots, n$$, be nonnegative real numbers with $$m_{i j}=m_{j i}$$ for all $$i, j$$ and $$m_{i i}=0$$ for all $$i$$.
Consider the $$n \times n$$ matrix $$G$$ with
$$g_{i j}=\frac{1}{2}\left(m_{0 i}^2+m_{0 j}^2-m_{i j}^2\right).$$
If $$G$$ is positive semidefinite, there exist points $$\mathbf{p}_0, \mathbf{p}_1, \ldots, \mathbf{p}_n \in \mathbb{R}^n$$ with $$\left\|\mathbf{p}_i-\mathbf{p}_j\right\|=m_{i j}$$ for all $$i, j$$.
#7. Are These Distances Euclidean?
Up next. Let $$m_{i j}, i, j=0,1, \ldots, n$$, be nonnegative real numbers with $$m_{i j}=m_{j i}$$ for all $$i, j$$ and $$m_{i i}=0$$ for all $$i$$.
Consider the $$n \times n$$ matrix $$G$$ with
$$g_{i j}=\frac{1}{2}\left(m_{0 i}^2+m_{0 j}^2-m_{i j}^2\right).$$
If $${\color{Red}G}$$ is positive semidefinite, there exist points $$\mathbf{p}_0, \mathbf{p}_1, \ldots, \mathbf{p}_n \in \mathbb{R}^n$$ with $$\left\|\mathbf{p}_i-\mathbf{p}_j\right\|=m_{i j}$$ for all $$i, j$$.
#7. Are These Distances Euclidean?
Up next. Let $$m_{i j}, i, j=0,1, \ldots, n$$, be nonnegative real numbers with $$m_{i j}=m_{j i}$$ for all $$i, j$$ and $$m_{i i}=0$$ for all $$i$$.
Consider the $$n \times n$$ matrix $$G$$ with
$$g_{i j}=\frac{1}{2}\left(m_{0 i}^2+m_{0 j}^2-m_{i j}^2\right).$$
If $${\color{Red}G = XX^T}$$, there exist points $$\mathbf{p}_0, \mathbf{p}_1, \ldots, \mathbf{p}_n \in \mathbb{R}^n$$ with $$\left\|\mathbf{p}_i-\mathbf{p}_j\right\|=m_{i j}$$ for all $$i, j$$.
Set $$\mathbf{p_0} = 0$$ and let $$\mathbf{p_i}$$ be the $$i$$-th column of $$X$$.
#7. Are These Distances Euclidean?
$$\langle x_i, x_j \rangle = \frac{1}{2} \left( m_{0i}^2 + m_{0j}^2 - m_{ij}^2 \right)$$
$$\langle p_i, p_j \rangle = \frac{1}{2} \left( m_{0i}^2 + m_{0j}^2 - m_{ij}^2 \right)$$
$$\frac{1}{2} \left(\|\mathbf{p_i}\|^2+\|\mathbf{p_j}\|^2-\|\mathbf{p_i}-\mathbf{p_j}\|^2\right) = \frac{1}{2} \left( m_{0i}^2 + m_{0j}^2 - m_{ij}^2 \right)$$
$$\frac{1}{2} \left(m_{0i}^2+m_{0j}^2-\|\mathbf{p_i}-\mathbf{p_j}\|^2\right) = \frac{1}{2} \left( m_{0i}^2 + m_{0j}^2 - m_{ij}^2 \right)$$
$$\|\mathbf{p_i}-\mathbf{p_j}\| = m_{ij}$$
#7. Are These Distances Euclidean?
Theorem (ext). Let $$m_{i j}, i, j=0,1, \ldots, n$$, be nonnegative reals with $$m_{i j}=m_{j i}$$ for all $$i, j$$ and $$m_{i i}=0$$ for all $$i$$.
For any $${\color{IndianRed}1 \leqslant d \leqslant n}$$, points $$\mathbf{p}_0, \mathbf{p}_1, \ldots, \mathbf{p}_n \in \mathbb{R}^{{\color{IndianRed}d}}$$ with $$\left\|\mathbf{p}_i-\mathbf{p}_j\right\|=m_{i j}$$ for all $$i, j$$ exist if and only if the $$n \times n$$ matrix $$G$$ with
$$g_{i j}=\frac{1}{2}\left(m_{0 i}^2+m_{0 j}^2-m_{i j}^2\right)$$
is such that $$G = XX^T$$ for some matrix $$X$$ of rank at most $$d$$.
#8. Packing Complete Bipartite Graphs
#8. Packing Complete Bipartite Graphs
How many battles need to be fought
in order for every pair of nations to have fought exactly once?
#8. Packing Complete Bipartite Graphs
#8. Packing Complete Bipartite Graphs
#8. Packing Complete Bipartite Graphs
#8. Packing Complete Bipartite Graphs
#8. Packing Complete Bipartite Graphs
#8. Packing Complete Bipartite Graphs
#8. Packing Complete Bipartite Graphs
#8. Packing Complete Bipartite Graphs
#8. Packing Complete Bipartite Graphs
#8. Packing Complete Bipartite Graphs
#8. Packing Complete Bipartite Graphs | 33,805 | 80,827 | {"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": 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} | 4.21875 | 4 | CC-MAIN-2024-18 | latest | en | 0.628809 |
http://www.jiskha.com/display.cgi?id=1409790244 | 1,498,468,355,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320695.49/warc/CC-MAIN-20170626083037-20170626103037-00264.warc.gz | 559,360,608 | 3,751 | Math
posted by .
What are these fractions listed from least to greatest: 1/7,1/9,2/7,1/12?
• Math -
Change them all to either decimals or equivalent fractions with a common denominator.
The easiest way is to use a calculator and divide the numerator by the denominator to get the decimal equivalent.
1/7 = 0.143
etc. | 82 | 321 | {"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} | 3.03125 | 3 | CC-MAIN-2017-26 | latest | en | 0.879977 |
https://javatutoring.com/author/kanakesh-marri/page/3/ | 1,716,240,922,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00359.warc.gz | 282,219,995 | 16,856 | ## Call for Code Challenge – Commit To The Cause
Natural Disasters can be considered as one of the most devastating events as far as life on Planet Earth is concerned. It is a seething inevitability that is extremely tough to avoid. Certain changes in our planet forces life, particularly humankind in general, to suffer from the deadly consequences. It is our utmost responsibility to tackle these situations and get ...
## Java Program to Find Additive Inverse | Java Programs
Finding Additive Inverse through a Java Program – This specific article expresses the code for finding the additive inverse Java Program of a number. The problem here is to find the additive inverse of a user-defined number. The additive inverse of a number a is the number that, when added to a, yields 0 (example: Additive Inverse of 2 is ...
## C Program Find Circumference Of A Circle | 3 Ways
C Program to find the Circumference of a Circle – In this specific article, we will brief in on the methods of finding the circumference of a circle in C Programming. With the help of this piece, we will explain all the types of ways the circumference of a circle can be calculated. The various ways by which the circumference ...
## C Program Area Of Rectangle | C Programs
C Program to find the area of a rectangle – In this particular article, we will discuss how to calculate the area of a rectangle in C Programming. The methods used in this specific article are as follows: Using Standard Method Using Function Using Pointers Using Macros As we all know, rectangles are extremely commonly used quadrilaterals. In a rectangle, ...
## C Program Area Of Rhombus – 4 Ways | C Programs
C Program to find the area of a rhombus – In this specific article, we will detail in on how to calculate the area of a rhombus in C Programming. The methods used in this particular piece are as follows: Using Standard Method Using Function Using Pointers Using Macros A Rhombus is a quadrilateral whose opposite sides are parallel and ...
## Java Farm Management Project Source Code | Java Programs
Java Program for Farm Management – Here is an in-depth article that expresses the Farm Management Project or the Farm Fest Java Program along with suitable examples as well as the source code with output. In this challenge, a farmer is asking you to tell him how many legs can be counted among all his animals. The farmer breeds three ...
## Java Program Compare Strings by Characters | Java Programs
Compare Strings by Count of Character through Java Program – In this particular article, we will be talking about the code to compare whether two user-defined strings are equal in length using Java language. Suitable examples and sample output will be provided accordingly. The problem here is to check whether two given strings are equal in length. The input here ...
## Simple Java Program To Count Syllables | Java Programs
Count syllables in large string in Java – This specific article will be dealing with the Java program to count syllables. Suitable examples, as well as sample output, will be provided. This code is for counting the number of words in a user input string using Java language. The problem here is to count the number of words in a ... | 656 | 3,241 | {"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} | 3.390625 | 3 | CC-MAIN-2024-22 | latest | en | 0.8766 |
http://mathhelpforum.com/differential-geometry/91881-tricky-advanced-calc-problem-print.html | 1,529,392,371,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861980.33/warc/CC-MAIN-20180619060647-20180619080647-00410.warc.gz | 211,772,263 | 2,742 | # Tricky Advanced Calc Problem
• Jun 5th 2009, 07:22 AM
joeyjoejoe
Tricky Advanced Calc Problem
Suppose that f is continuous and differentiable on [a,b], f(a) = f(b) = 0. Show that for every real number $\displaystyle \lambda$ there is a real number c $\displaystyle \in$ [a,b] such that f'(c) = $\displaystyle \lambda$f(c).
Tried to use MVT (and Rolle's) to no avail. Any suggestions?
• Jun 5th 2009, 08:27 AM
HallsofIvy
Quote:
Originally Posted by joeyjoejoe
Suppose that f is continuous and differentiable on [a,b], f(a) = f(b) = 0. Show that for every real number $\displaystyle \lambda$ there is a real number c $\displaystyle \in$ [a,b] such that f'(c) = $\displaystyle \lambda$f(c).
Tried to use MVT (and Rolle's) to no avail. Any suggestions?
Since F is continuous on [a, b] it is integrable there. Let $\displaystyle F(x)= \int_a^x f(t)dt$ and apply Rolle's theorem to $\displaystyle f(x)- \lambda F(x)$.
• Jun 11th 2009, 09:54 AM
pankaj
Let $\displaystyle g(x)=e^{-\lambda x}f(x)$
g(a)=g(b)=0.Now use Rolle's Theorem | 345 | 1,032 | {"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-2018-26 | latest | en | 0.843044 |
https://vectorlinux.com/how-many-mondays-between-two-dates-sql/ | 1,719,285,531,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865545.19/warc/CC-MAIN-20240625005529-20240625035529-00199.warc.gz | 520,550,740 | 18,683 | How Many Mondays Between Two Dates Sql
SQL Programming
Have you ever wondered how many Mondays there are between two specific dates in SQL? Well, I must say that I found this question quite intriguing. As someone who spends a significant amount of time working with databases and SQL queries, I decided to dive deep into this topic.
First, let’s start by understanding the problem at hand. Given two dates, we want to determine the number of Mondays that fall between them. This may seem like a simple task, but trust me, there is more to it than meets the eye.
To solve this problem, we need to utilize the power of SQL’s date functions. One useful function is the `DATEPART` function, which allows us to extract specific parts of a date, such as the day of the week.
Let’s assume we have two dates: `start_date` and `end_date`. We can use the following SQL query to calculate the number of Mondays between them:
```SELECT COUNT(*) AS num_mondays FROM your_table WHERE DATEPART(dw, date_column) = 2 AND date_column >= start_date AND date_column <= end_date;```
Let me break down this query for you. The `DATEPART(dw, date_column)` part extracts the day of the week (represented as an integer) from the `date_column`. In SQL, Monday is typically represented by the number 2 (where Sunday is 1, Monday is 2, and so on).
The `WHERE` clause filters the results by only selecting rows where the day of the week is equal to 2 (Monday). Additionally, we include conditions to ensure that the date falls within the specified range, using the `start_date` and `end_date` variables.
Executing this query will give you the desired result - the number of Mondays between the two dates.
Now, let's add a personal touch to this topic. As someone who loves planning and organizing my week, knowing the number of Mondays between two dates can be quite handy. It allows me to set specific goals or milestones that I want to achieve by each Monday. It also helps me in creating a schedule and allocating time for different tasks.
Moreover, this SQL query can be extended to calculate the number of any other day of the week between two dates. Simply change the number in the `DATEPART(dw, date_column) = X` condition, where X represents the desired day (1 for Sunday, 2 for Monday, and so on).
In conclusion, determining the number of Mondays between two dates in SQL may seem like a simple task, but it requires a bit of SQL knowledge and the clever use of date functions. By leveraging the `DATEPART` function and a few additional conditions, we can easily calculate the desired result. So, the next time you need to plan your week or track specific events, give this SQL query a try! | 585 | 2,681 | {"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-2024-26 | latest | en | 0.927207 |
https://www.teacherspayteachers.com/Store/Common-Core-And-More-Va-And-Md | 1,529,584,029,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864148.93/warc/CC-MAIN-20180621114153-20180621134153-00034.warc.gz | 924,256,659 | 29,001 | # Common Core and More VA and MD
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United States - Maryland - FREDERICK
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TEACHING EXPERIENCE
ESOL teacher Alternative Education third, fourth, and fifth grade teacher middle and high school Group Home Counselor Kindergarten First Grade Second Grade
MY TEACHING STYLE
SIOP Model
HONORS/AWARDS/SHINING TEACHER MOMENT
2 Time Teacher of the Year nominee
MY OWN EDUCATIONAL HISTORY
BA in Spanish, Fisk University MA in Elementary Education
Taught VA SOLs Currently Teach in MD Common Core | 1,115 | 4,323 | {"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-2018-26 | latest | en | 0.8605 |
http://mathhelpforum.com/differential-equations/100519-transfer-function-differential-equation.html | 1,506,400,172,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818694719.89/warc/CC-MAIN-20170926032806-20170926052806-00090.warc.gz | 216,066,846 | 10,741 | # Thread: Transfer function by differential equation
1. ## Transfer function by differential equation
How can I obtain the transfer function of system Y(s)/ X(s), which has the differential equation below ?
d^3/dt^3 + 3(d^2*y/dt^2) +5(dy/dt) + y = d^3*x/dt^3 + 4(d^2* x/dt^2) + 6(dx/dt) + 8x
If possible give me a tip to insert special characters here, as pi and division sign with a number over the other
2. Originally Posted by moisesbr
How can I obtain the transfer function of system Y(s)/ X(s), which has the differential equation below ?
d^3/dt^3 + 3(d^2*y/dt^2) +5(dy/dt) + y = d^3*x/dt^3 + 4(d^2* x/dt^2) + 6(dx/dt) + 8x
If possible give me a tip to insert special characters here, as pi and division sign with a number over the other
For mathematical type setting we use LaTeX, the tutorial is here
CB
3. Originally Posted by moisesbr
How can I obtain the transfer function of system Y(s)/ X(s), which has the differential equation below ?
d^3/dt^3 + 3(d^2*y/dt^2) +5(dy/dt) + y = d^3*x/dt^3 + 4(d^2* x/dt^2) + 6(dx/dt) + 8x
If possible give me a tip to insert special characters here, as pi and division sign with a number over the other
Write this as:
$D^3y+3D^2y+5Dy+y=D^3x+4D^2x+6Dx+8x$
Then if we assume that at time $t=0$ that the input and out put and all relevant derivatives are zero, and we take the Laplace transform of the above equation we get:
$
s^3Y(s)+3s^2Y(s)+5sY(s)+Y(s)=s^3X(s)+4s^2X(s)+6sX( s)+8X(s)
$
then the transfer function is:
$
T(s)=\frac{Y(s)}{X(s)}=\frac{s^3+4s^2+6s+8}{s^3+3s ^2+5s+1}
$
CB
4. How did you pass from first to second line below ?
Sorry for the silly question but I've been a long time out of maths
and I having a hard time to remind all in a new course I am starting
d^3/dt^3 + 3(d^2*y/dt^2) +5(dy/dt) + y = d^3*x/dt^3 + 4(d^2* x/dt^2) + 6(dx/dt) + 8x
5. Originally Posted by moisesbr
How did you pass from first to second line below ?
Sorry for the silly question but I've been a long time out of maths
and I having a hard time to remind all in a new course I am starting
$D^n \equiv \frac{d^n}{dt^n}$ | 723 | 2,080 | {"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": 0, "codecogs_latex": 5, "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} | 3.765625 | 4 | CC-MAIN-2017-39 | longest | en | 0.857218 |
http://www.docstoc.com/docs/14715284/Physics-Final | 1,433,272,109,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1433195036112.6/warc/CC-MAIN-20150601214356-00020-ip-10-180-206-219.ec2.internal.warc.gz | 337,906,202 | 43,054 | # Physics Final by DynamiteKegs
VIEWS: 324 PAGES: 10
• pg 1
``` PHYSICS 123 FINAL EXAM December 10, 2001
1
GIVEN: x = xo + ax t2 + vox t; vx = vox + ax t 2 2
vx = vox + 2ax (x − xo )
2
1
y = yo + ay t2 + voy t; vy = voy + ay t 2 2
vy = voy + 2ay (y − yo )
2 2
Fnet = ma; g = 9.8m/s2 = 32 ft/s2 ; Fcentripetal = mv
r
GM1 M2 1
Fgravity = P Egravity = mgh; P Espring = kx2 ;
r2 2
W = F · s; W = F · ds = area under plot of force vs displacement. ∆KE = Wnet
dU (x) dp mv 2 mi xi
p = mv; F (x) = − ; F = ; Fcentripetal = ; xCM = i
dx dt r Mtotal
1 1
KEtrans = mv 2 ; KErot = Iω 2 L = Iω
2 2
1
θ = αt2 + ω0 t; ω = ω0 + αt 2
ω 2 = ω0 + 2α(θ − θ0 )
2
1 2 1
Irod−about−center = M l2 Isphere = M R2 Ihoop = M R2 Idisk = Isolid−cylinder = M R2
12 5 2
1 d
Irod−about−end = M l2 τ =r×F =r×p τ=
3 dt
dp
τ = I α; |τ | = |F ||r| sin θ p = mv; F = ; 2 π radians = 360 degrees
dt
Parallel Axis Theorem: IP = ICM + M L2 where L is distance from CM to pivot point P .
PART ONE: MULTIPLE CHOICE ( 4 points each)
Circle or check the correct response. Use GIVEN data on above if needed. .
1. A baseball is thrown vertically upwards. The acceleration of the ball at its highest point is:
(a) zero (b) 9.8 m/s2 upward (c) 9.8 m/s2 downward
(d) changing suddenly from 9.8 m/s2 upward to 9.8 m/s2 downward
2. A freely falling body (neglect air resistance) has an acceleration of 9.8 m/s2 downward. This
means that:
(a) the body falls 9.8 meters during each second
(b) the body falls 9.8 meters during the first second only
(c) the speed of the body increases by 9.8 m/s during each second
(d) the acceleration of the body increases by 9.8 m/s2 in each second
(e) the acceleration of the body decreases by 9.8 m/s2 in each second
3. Which of the following is NOT an example of accelerated motion?
(a) vertical component of projectile motion
(b) circular motion at constant speed
(c) the motion of the Earth around the sun
(d) a swinging pendulum
(e) horizontal component of projectile motion
1
4. An object is moving at constant speed in a circular path. The instantaneous velocity and
instantaneous acceleration vectors are:
(a) both tangent to the circular path (b) perpendicular to each other
(c) both perpendicular to the circular path (d) none of the above
5. A boat is traveling upstream at 14 mph with respect to the water in a river. The current in
the river moves at a speed of 6 mph with respect to the ground. A man on the boat runs across
the boat, perpendicular to the river, from one side to the other at 6 mph. The speed of the man
with respect to the ground is:
(a) 10 mph (b) 14 mph (c) 18.5 mph (d) 21 mph (e) 26 mph
6. A 1000 kg elevator is moving downward with an downward acceleration of 5 m/s2 . The tension
in the elevator cable is: (It might be helpful if you draw a FBD for the elevator):
(a) 6800 N (b) 1000 N (c) 3000 N (d) 9800 N (e) 4800 N
7. A 50 kg woman is ice skating towards the east on a frozen lake when she collides with a
2000 sport utility vehicle (SUV) which is driving on the ice towards the west.The maximum force
exerted on the woman by the SUV is 800 N, westward. What is the force that the woman exerts
on the truck?
(a) 800 N eastward (b) 50 N westward (c) 50 N eastward (d) 2000 N eastward
8. A crate rests on a horizontal surface and a woman pulls on it with a 10 N force. No matter
what the orientation of the force, the crate does not move. Rank the situations shown below
according to the magnitude of the friction force exerted by the surface on the crate, smallest first:
(a) 1, 2, 3 (b) 2,1,3 (c) 2,3,1 (d) 1,2,3 (e) 3,2,1
9. A physics textbook is on a spring scale in an elevator. Of the following, the scale shows the
(a) moves downward with decreasing speed
(b) moves downward with increasing speed
(c) moves downward with constant speed
(d) moves upward with constant speed
10. A car of mass m travels around a flat circular track of radius r, with speed v. Which of the
following is true?
(a) The net force on the car is zero.
2
(b) the gravity force on the car equals mv
r
2
(c) The friction force on the car equals mv
r
2
(d) The normal force on the car equals mv r
2
11. A pile driver is pounding a steel beam into the ground. The hammer (the falling object) has
mass 500 kg and falls from a height of 2 m onto the beam, starting at rest. The kinetic energy
of the hammer just before it hits the beam is:
(a) 9800 Joules (b) 4900 Joules (c) 1000 Joules (d) 500 Joules
12. In the situation described in Question 11, the beam then penetrates 10 cm into the ground.
What is the average force the ground exerts on the beam as it is forced downward?
(a) 5000 N (b) 98000 N (c) 4900 N (d) 49000 N
13. A roller coaster is pulled to the top of its track and allowed to roll down. For a bigger thrill
you want to re-design the track so the car is going twice as fast at the bottom of the run. How
much higher should you make the track?
(a) the same height (b) four times as high (c) half as high (d) twice as high
3
14. A person attempts to knock down a large wooden bowling pin by throwing a ball at it. The
person has one ball made of putty and a ball of the same mass made of rubber. The rubber ball
bounces back while the ball made of putty sticks to the pin. Which ball is more likely to topple
the pin?
(a) the rubber ball (b) the putty ball (c) both are equally likely
15. A compact car and a large truck collide and stick together. Which undergoes the larger
momentum change?
(a) the car (b) the truck (c) both have the same momentum change
(d) impossible to predict
16. A 60 kg hunter gets a rope around a 300 kg polar bear. They are at rest on frictionless ice
30 meters apart. The hunter pulls the polar bear towards him. When they meet the polar bear
will move:
(a) 30 m (b) 25 m (c) 15 m (d) 10 m (e) 5 m
17. A rider in a “barrel of fun” at the amusement park finds herself stuck with her back to the
wall. As the floor is lowered and the ride rotates she remains stuck. Which diagram correctly
shows the forces acting on her?
18. The fan shown below has been turned on and is speeding up as it rotates clockwise. The
direction of the acceleration of the point X on the fan tip is best represented by which arrow
below?
4
19. A meter stick shown below rotates about an axis through the dot at the 20 cm point, Five
forces act on the meter stick as shown. Rank the forces according to the magnitudes of the torque
they produce about the pivot point, greatest to least.
(a) F1 > F2 > F3 > F4 > F5
(b) F1 = F3 > F2 > F5 > F4
(c)F3 = F1 > F4 > F5 = F2
(d) F1 = F2 > F4 > F2 = F5
20. The object below consists of four identical particles of very small radius and mass m which
are connected by very light sticks to form a rigid body. The rotational inertia of the object about
the y axis shown below is:
(a) 3ma2 (b) 4ma2 (c) 2ma2 (d) ma2
21. Two disks are mounted on low friction bearings on a common shaft. THe first disc has
rotational inertia I and is spinning with angular velocity ω. The second disk has rotational
inertia 2I and is spinning in the same direction as the first disk with angular velocity 2ω as
shown. The disks are slowly forced together along the shaft until they couple and have a final
angular velocity ωf . What is ωf ?
(a) 5ω/3 (b) ω/3 (c) ω/ (3) (d) 5ω/2
5
22. An astronaut in an orbiting spacecraft feels “weightless” because she:
(a) has no acceleration.
(b) is beyond the gravity pull of the Earth.
(c) is pulled outwards by centrifugal force.
(d) is in the same free-fall orbit as the spacecraft.
23. An artificial Earth satellite is moved from a circular orbit of radius R to a circular orbit of
(a) the gravitational force does positive work, the kinetic energy of the satellite increases,
and the potential energy of the Earth-satellite system increases.
(b) the gravitational force does positive work, the kinetic energy of the satellite increases,
and the potential energy of the Earth-satellite system decreases.
(c) the gravitational force does negative work, the kinetic energy of the satellite decreases,
and the potential energy of the Earth-satellite system increases.
(d) the gravitational force does negative work, the kinetic energy of the satellite decreases,
and the potential energy of the Earth-satellite system decreases.
24. A particle is placed, in turn, a distance d from four objects, each of mass m: a uniform solid
sphere, a large uniform spherical shell, a small solid sphere, and a small spherical shell. Which
object exerts the largest gravitational force on the particle?
(a) the large solid sphere (b) the small solid sphere (c) the large spherical shell
(d) the small spherical shell (e) all exert the same force
25. The figure below shows a uniform beam of mass m which is acted on by a force F and is held
vertical by a cable. Draw a free body diagram for the beam, clearly labeling all of the forces that
act on it.
6
PART TWO: PROBLEMS (25 points each)
Work each problem on the exam page. Be sure all answers are clearly labeled, with units. Show
all work clearly. Answers with no clear reasoning will receive little or no credit.
1. In the figure below, a stone is projected at a cliff of height h with an initial speed or 42.0 m/s,
directed 60◦ above the horizontal. The stone hits point A 4.5 s after it is launched.
(a) Find the height h of the cliff.
(b) Find the speed of the stone just before impact.
(c) Find the maximum height H of the stone.
7
2. A car moves around a circular banked track as shown below.
(a) Draw a free body diagram for the car, showing all forces.
(b) If there is no friction between the car and the track, what is the speed of the car? Express
your answer algebraically in terms of the banking angle θ, the mass M of the car, the radius R
of the track, and g.
8
3. A 40 kg child and her father simultaneously dive from a 100 kg boat that is initially motionless.
Ths child dives horizontally with a speed of 2.0 m/s toward the east. Her father dives toward the
south with a velocity of 1.5 m/s at an angle of 37◦ above the horizontal. Find the magnitude and
direction of the velocity of the boat immediately after the dives. Assume the vertical component
of the father’s dive velocity does not affect the motion of the boat along the surface of the water.
9
4. In the figure below a wheel of radius 0.20 m is mounted on frictionless bearings on a horizontal
axis. The rotational inertia of the wheel about the axis is 0.40 kg-m2 . A massless cord is wrapped
around the wheel and is attached to a 6.0 kg box. A short time after being released from rest,
the box has a kinetic energy of 6.0 J. The string does not slip.
(a) What is the kinetic energy of the pulley at the time the box has kinetic energy 6.0 J?
(b) What is the distance the box has fallen at that instant?
10
```
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http://www.wtharvey.com/melk.html | 1,550,716,516,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247497858.46/warc/CC-MAIN-20190221010932-20190221032932-00028.warc.gz | 489,555,233 | 3,066 | Ongoing series from wtharvey.com
Here are some chess puzzles from the games of Hrant Melkumyan. The color disk on the diagram indicates who moves first.
Solutions are between the brackets under each puzzle.
Drag your cursor from one bracket to the other.
a)
White mates in 4.
Hrant Melkumyan vs Artyom Benza, Belgorod, 2008
r3q1n1/pppb4/3p2Nk/3P3p/2PQB3/6PP/PP3R2/6K1 w - - 1 0
[ Qh8+ Kg5 Qg7 ]
b)
Hrant Melkumyan vs Alexey Dreev, Warsaw, 2011
5rk1/p1qn1ppp/QpN5/2b5/8/PP3BP1/5PKP/2R5 w - - 1 0
[ b4 Bd4 Ne7+ ]
c)
Hrant Melkumyan vs Zoltan Zambo, Warsaw, 2011
r3k2r/pp3pp1/2p1bb1p/q3N3/2B1Q3/P3P3/1P3PPP/1K1R3R w kq - 1 0
[ Nxf7 Kxf7 Qxe6+ if Kg6 Bd3+ mates ]
d)
Hrant Melkumyan vs Csaba Balogh, Legnica, 2013
3r1rk1/1p1n1p1p/p1n3p1/3N4/qP2BP2/2Q5/5RPP/2R3K1 w - - 1 0
[ Ra1 Qb5 Ra5 ]
e)
Hrant Melkumyan vs Gregory Canfell, Sydney, 2014
5bk1/3r1p2/3P2r1/2Q1p3/2N1q1p1/4N3/PP3P2/2KR4 w - - 1 0
[ Qc8 if Qb7 Qxb7 Rxb7 d7 ]
Hrant Melkumyan Puzzles, Part II.
The color disk on the diagram indicates who moves first.
a)
Black mates in 4.
Avetik Grigoryan vs Hrant Melkumyan, Martuni, 2007
3R4/p4pkp/1p2bp2/4q3/P2R1Q1P/5BP1/2r2PK1/4r3 b - - 0 1
[ ...Rg1+ Kxg1 Qe1+ ]
b)
Homayoon Toufighi vs Hrant Melkumyan, Kemer, 2007
3k4/pp6/7p/2p2p2/1P1pn1q1/3Q4/P1P2P2/4RK2 b - - 0 1
[ ...Qh4 if Qf3 Nd2+ or if Ke2 Qxf2+ mates ]
c)
Michal Matuszewski vs Hrant Melkumyan, Warsaw, 2009
r5k1/1p1nppbp/5np1/qN6/p2P4/P3PQ1P/1Br1BPP1/1R3RK1 b - - 0 1
[ ...Qd2 ]
d)
Black mates in 3.
Alexei Fedorov vs Hrant Melkumyan, Warsaw, 2011
8/1b4pk/6qp/p1p1p3/1pPp1r2/3P1r1P/PP4P1/R3RBQK b - - 0 1
[ ...Rxh3+ if Qh2 Rxf1+ Rxf1 Qxg2# ]
e)
Sargis Manukyan vs Hrant Melkumyan, Yerevan, 2014
3r2k1/2R2pbp/1n2p1p1/8/2b1PB2/2N2P2/1P3KPP/r4B1R b - - 0 1
[ ...Bd4+ if Be3 Bxe3+ Kxe3 Rxf1 ]
Hrant Melkumyan Puzzles, Part III.
The color disk on the diagram indicates who moves first.
a)
Sargis Manukyan vs Hrant Melkumyan, Yerevan, 2014
3r2k1/2R2pbp/1n2p1p1/8/2b1PB2/2N2P2/1P3KPP/r4B1R b - - 0 1
[ ...Bd4+ ]
b)
Nikita Matinian vs Hrant Melkumyan, Sitges, 2017
r3rnk1/ppq1bppp/2p1n3/3pP3/4b3/3BBNNP/PPPQ1PP1/3RR1K1 b - - 0 1
[ 1...Bxf3 2.gxf3 d4 if 3.Bf4 g5 ]
c)
White mates in 2.
Hrant Melkumyan vs Aaron Do, Melbourne, 2018
4R3/p3r1p1/P1pk1b1p/2p1pB2/1PK5/2P2PP1/2P4P/8 w - - 1 0
[ 1.bxc5+ Kc7 2.Rc8# ]
d)
Zhen Low vs Hrant Melkumyan, Cha-Am, 2018
6k1/ppb3pp/b3qp2/8/3B4/1PPN1P1P/2Q3P1/6K1 b - - 0 1
[ 1...Bxd3 if 2.Qxd3 Qe1+ 3.Qf1 Bh2+ ]
Back to the Index | 1,331 | 2,430 | {"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-2019-09 | latest | en | 0.606861 |
https://www.askiitians.com/forums/7-grade-maths/find-ten-rational-numbers-between-3-5-and-find_272437.htm | 1,708,633,257,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.45/warc/CC-MAIN-20240222193722-20240222223722-00653.warc.gz | 659,984,633 | 42,303 | # Find ten rational numbers between 3/5 and ¾, Find ten rational numbers between 3/5 and ¾,
Harshit Singh
askIITians Faculty 5963 Points
3 years ago
Dear Student
Let us make the denominators same, say 80.
3/5 = (3 × 16)/(5× 16) = 48/80
3/4=(3×20)/(4×20)=60/80
Ten rational numbers between 3/5 and 3⁄4 = ten rational numbers between 48/80 and 60/80 Therefore, ten rational numbers between 48/80 and 60/80 = 49/80, 50/80, 51/80, 52/80, 54/80, 55/80, 56/80, 57/80, 58/80, 59/80
Thanks | 194 | 484 | {"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} | 3.859375 | 4 | CC-MAIN-2024-10 | latest | en | 0.810034 |
http://teaching.statistics-is-awesome.org/category/data/ | 1,582,980,942,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875149238.97/warc/CC-MAIN-20200229114448-20200229144448-00163.warc.gz | 133,750,866 | 44,756 | ## Age is just a number
Today I demonstrated some in-class interactive activities that I had developed for my super large intro statistics lectures at a teaching and learning symposium. I’ve shared a summary of the activities and the data below.
## Quick summary of the activity
### Some other ideas
If you haven’t already, check out learning.statistics-is-awesome.org/different_strokes/, where you can sample some cat (and other) drawings and learn more about how people draw in the Google game Quick, Draw!
I also get students to draw things in class and use their drawings as data. Below are all the drawings of cats made from the demonstration today, and also from the awesome teachers who helped me out last night. If you click/touch and hold a drawing you will be able to drag it around. How many different ways can you sort the drawings into groups?
## Different strokes?
Recently I’ve been developing and trialling learning tasks where the learner is working with a provided data set but has to do something “human” that motivates using a random sample as part of the strategy to learn something from the data.
Since I already had a tool that creates data cards from the Quick, Draw! data set, I’ve created a prototype for the kind of tool that would support this approach using the same data set.
For this new tool, called different strokes, users sort drawings into two or more groups based on something visible in the drawing itself. Since you have the drag the drawings around to manually “classify” them, the larger the sample you take, the longer it will take you.
There’s also the novelty and creativity of being able to create your own rules for classifying drawings. I’ll use cats for the example below, but from a teaching and assessment perspective there are SO many drawings of so many things and so many variables with so many opportunities to compare and contrast what can be learned about how people draw in the Quick, Draw!
Here’s a precis of the kinds of questions I might ask myself to explore the general question What can we learn from the data about how people draw cats in the Quick, Draw! game?
• Are drawings of cats more likely to be heads only or the whole body? [I can take a sample of cat drawings, and then sort the drawings into heads vs bodies. From here, I could bootstrap a confidence interval for the population proportion].
• Is how someone draws a cat linked to the game time? [I can use the same data as above, but compare game times by the two groups I’ve created – head vs bodies. I could bootstrap a confidence interval for the difference of two population means/medians]
• Is there a relationship between the number of strokes and the pause time for cat drawings? [And what do these two variables actually measure – I’ll need some contextual knowledge!]
• Do people draw dogs similarly to cats in the Quick, Draw! game? [I could grab new samples of cat and dog drawings, sort all drawings into “heads” or “bodies”, and then bootstrap a confidence interval for the difference of two population proportions]
Check out the tool and explore for yourself here: http://learning.statistics-is-awesome.org/different_strokes/
A little demo of the tool in action!
## You say data, I say data cards …
This long weekend (in Auckland anyway!), I spent some time updating the Quick! Draw! sampling tool (read more about it here Cat and whisker plots: sampling from the Quick, Draw! dataset). You may need to clear your browser cache/data to see the most recent version of the sampling tool.
One of the motivations for doing so was a visit to my favourite kind of store – a stationery store – where I saw (and bought!) this lovely gadget:
It’s a circle punch with a 2″/5 cm diameter. When I saw it, my first thought was “oh cool I can make dot-shaped data cards”, like a normal person right?
Using data cards to make physical plots is not a new idea – see censusatschool.org.nz/resource/growing-scatterplots/ by Pip Arnold for one example:
But I haven’t seen dot-shaped ones yet, so this led me to re-develop the Quick! Draw! sampling tool to be able to create some 🙂
I was also motivated to work some more on the tool after the fantastic Wendy Gibbs asked me at the NZAMT (New Zealand Association of Mathematics Teachers) writing camp if I could include variables related to the times involved with each drawing. I suspect she has read this super cool post by Jim Vallandingham (while you’re at his site, check out some of his other cool posts and visualisations) which came out after I first released the sampling tool and compares strokes and drawing/pause times for different words/concepts – including cats and dogs!
So, with Quick! Draw! sampling tool you can now get the following variables for each drawing in the sample:
The drawing and pause times are in seconds. The drawing time captures the time taken for each stroke from beginning to end and the pause time captures all the time between strokes. If you add these two times together, you will get the total time the person spent drawing the word/concept before either the 20 seconds was up, or Google tried to identify the word/concept. Below the word/concept drawn is whether the drawing was correctly recognised (true) or not (false).
I also added three ways to use the data cards once they have been generated using the sampling tool (scroll down to below the data cards). You can now:
3. show the sample data as a HTML table (which makes it easy to copy and paste into a Google sheet for example)
In terms of options (2) and (3) above, I had resisted making the data this accessible in the previous version of the sampling tool. One of the reasons for this is because I wanted the drawings themselves to be considered as data, and as human would be involved in developed this variable, there was a need to work with just a sample of all the millions of drawings. I still feel this way, so I encourage you to get students to develop at least one new variable for their sample data that is based on a feature of the drawing 🙂 For example, whether the drawing of a cat is the face only, or includes the body too.
There are other cool things possible to expand the variables provided. Students could create a new variable by adding drawing_time and pause_time together. They could also create a variable which compares the number_strokes to the drawing_time e.g. average time per stroke. Students could also use the day_sketched variable to classify sketches as weekday or weekend drawings. Students should soon find the hemisphere is not that useful for comparisons, so could explore another country-related classification like continent. More advanced manipulations could involve working with the time stamps, which are given for all drawings using UTC time. This has consequences for the variable day_sketched as many countries (and places within countries) will be behind or ahead of the UTC time.
If you’ve made it this far in the post…. why not play with a little R 🙂
I wonder which common household pet Quick! drawers tend to use the most strokes to draw? Cats, dogs, or fish?
Have a go at modifying the R code below, using the iNZightPlots package by Tom Elliott and my [very-much-in-its-initial-stages-of-development] iNZightR package, to see what we can learn from the data 🙂 If you’re feeling extra adventurous, why not try modifying the code to explore the relationship between number of strokes and drawing time!
## Game of data
This post is second in a series of posts where I’m going to share some strategies for getting real data to use for statistical investigations that require sample to population inference. As I write them, you will be able to find them all on this page.
I read an article posted on fivethirtyeight about the worst board games ever invented and it got me thinking about the board games I like to play. The Game of life has a low average rating on the online database of games referred to in this article but I remember kind of enjoying playing it as a kid. boardgamegeek.com features user-submitted information about hundreds of thousands of games (not just board games) and is constantly being updated. While there are some data sets out there that already feature data from this website (e.g. from kaggle datasets), I am purposely demonstrating a non-programming approach to getting this data that maximises the participation of teachers and students in the data collection process.
To end up with data that can be used as part of a sample to population inference task:
1. You need a clearly defined and nameable population (in this case, all board games listed on boardgamegeek.com)
2. You need a sampling frame that is a very close match to your population.
3. You need to select from your sampling frame using a random sampling method to obtain the members of your sample.
4. You need to define and measure variables from each member of the sample/population so the resulting data is multivariate.
boardgamegeek.com actually provide a link that you can use to select one of the games on their site at random (https://boardgamegeek.com/boardgame/random), so using this “random” link (hopefully) takes care of (2) and (3). For (4), there are so many potential variables that could be defined and measured. To decide on what variables to measure, I spent some time exploring the content of the webpages for a few different games to get a feel for what might make for good variables. I decided to stick to variables that are measured directly for each game, rather than ones that were based on user polls, and went with these variables:
• Millennium the game was released (1000, 2000, all others)
• Number of words in game title
• Minimum number of players
• Maximum number of players
• Playing time in minutes (if a range was provided, the average of the limits was used)
• Minimum age in years
• Game type (strategy or war, family or children’s, other)
• Game available in multiple languages (yes or no)
Time to play!
I’ve set up a Google form with instructions of how you can help create a random sample of games from boardgamegeek.com at this link: https://goo.gl/forms/8yBqryGTzrZGhEVx2. As people play along, the sample data will be added here: https://docs.google.com/spreadsheets/d/e/2PACX-1vSzR_VSVzaaeWpCvYbAQCUewaM3Tad2zfTBO7AWuDgFFTj5Jaq2TBo6N-gQGCe5e5t_qKW7Knuq6-pr/pub?gid=552938859&single=true&output=csv . The URL to the game is included so that the data can be checked. Feel free to copy and adapt however you want, but do keep in mind that nature of the variables you use. In particular, be very careful about using any of the aggregate ratings measures (and another great article by fivethirtyeight about movie ratings explains some of the reasons why.)
Bonus round
I wrote a post recently – Just Google it – which featured real data distributions. boardgamegeek.com also provides simple graphs of the ratings for each game, so we can play a similar matching game. You could also try estimating the mean and standard deviation of the ratings from the graph, with the added game feature of reverse ordering!
Which games do you think match which ratings graphs?
1. Monopoly
2. The Lord of the Rings: The Card Game
3. Risk
4. Tic-tac-toe
I couldn’t find a game that had a clear bi-modal distribution for its ratings but I reckon there must be games out there that people either love or hate 🙂 Let me know if you find one! To get students familiar with boardgamegeek.com, you could ask them to first search for their favourite game and then explore what information and ratings have been provided for this on the site. Let the games begin 🙂
## Finding real data for real data stories
This post is first in a series of posts where I’m going to share some strategies for getting real data for real data stories, specifically to use for statistical investigations that require sample to population inference. As I write them, you will be able to find them all on this page.
Key considerations for finding real data for sample to population inference tasks
It’s really important that I stress that the approaches I’ll discuss are not necessarily what I would typically use when finding data to explore. Generally, I’d let the data drive the analysis not the analysis drive the data I try to find. These are specific examples so that the data that is obtained can be used sensibly to perform sample to population inference. It’s also really important to talk about why I’m stressing the above 🙂 In NZ we have specific standards that are designed to assess understanding of sample to population inference, using informal and formal methods that have developed by exploring the behaviour of random samples from populations (AS91035, AS91264, AS91582). So, for the students’ learning about rules of thumbs and confidence intervals to make sense, we need to provide students with clearly defined named populations with data that are (or are able to be) randomly sampled from these populations. At high school level at least, these strict conditions are in place so that students can focus on one central question: What can and can’t I say about the population(s) based on the random sample data?
For all the examples I’ll cover in this series of posts, there are four key considerations/requirements:
1. You need a clearly defined and nameable population. Ideally this should be as simple and clear as possible to help students out but to ensure (2) the “name” can end up being quite specific.
2. You need a sampling frame that is a very close match to your population. This means you need a way to access every member of your population to measure stuff about them (variables). Sure, this is not the reality of what happens in the real world in terms of sampling, but remember what I said earlier about what was important 🙂
3. You need to select from your sampling frame using a random sampling method to obtain the members of your sample. It is sufficient (and recommended) to stick to simple random sampling. In some cases, you may be able to make an assumption that what you have can be considered a random sample, but I’d prefer to avoid these kinds of situations where possible at high school level.
4. You need to define and measure variables from each member of the sample/population. We want students working with multivariate data sets, with several options possible for numerical and categorical data (but don’t forget there is the option to create new variables from what was measured).
I’ll try to refer back to these four considerations/requirements when I discuss examples in the posts that will follow.
Just one very relevant NZ NCEA assessment-specific comment before we talk data. For AS91035 and AS91582, the standards state that students are to be provided with the sample multivariate data for the task – so all of (1) (2) (3) and (4) is done by the teacher. Similarly with AS91264, the requirement for the standard is that students select a random sample (3) from a provided population dataset – so (1) (2) and (4) are done by the teacher. This does not mean the students can’t do more in terms of the sampling/collecting processes, just that these are not requirements for the standards and asking students to do more should not limit their ability to complete the task. I’ll try to give some ideas for how to manage any related issues in the examples.
Just one more point. I haven’t made this (5) in the previous section, but something to watch out for is the nature of your “cases”. Tables of data (which we refer to as datasets) that play nicely with statistical software like iNZight are ones where the data is organised so that each row is a case and each column is a variable. Typically at high school level, the datasets we use are ones where each case (e.g. each individual in the defined population) is measured directly to obtain different variables. Things can get a little tricky conceptually when some of the variables for a case are actually measured by grouping/aggregating related but different cases.
For example, if I take five movies from the internet movie database that have “dog” in the title (imdb.com) and another five with “cat” in the title, I could construct a mini dataset like the one below using information from the website:
For this dataset, each row is a different movie, so the cases are the movies. Each column provides a different variable for each movie. The variables Movie title, Year released, Movie length mins, Average rating, Number of ratings, Number photos and Genre were taken straight from the webpage for each movie. I created the variables Number words title, Number letters title, Average letters per word, Animal in title, Years since release and Millennium. [Something I won’t tackle in this post is what to do about the Genre variable to make this usable for analysis.]
The Average rating variable looks like a nice numerical variable to use, for example, to compare ratings of these movies with “dog” in the title and those with “cat”. The thing is, this particular variable has been measured by aggregating individual’s ratings of the movie using a mean (the related but different cases here are the individuals who rated the movies). You can see why this may be an issue when you look at the variable Number of ratings, which again is an aggregate measure (a count) – some of these movies have received less than 200 ratings while others are in the hundreds of thousands. We also can’t see what the distribution of these individual ratings for each movie looks like to decide whether the mean is telling us something useful about the ratings. [For some more really interesting discussion of using online movie ratings, check out this fivethirtyeight article.]
The variable Average letters per word has been measured directly from each case, using the characteristics of the movie title. There are still some potential issues with using the variable Average letters per word as a measure of, let’s say, complexity of words used in the movie title, since the mean is being used, but at least in this case students can see the movie title.
Another example of case awareness can be seen in the mini dataset below, using data on PhD candidates from the University of Auckland online directory:
For this dataset, each row is a different department, so the cases are the departments. Each column provides a different variable for each department. Gender was estimated based on the information provided in the directory and the data may be inaccurate for this reason. The % of PhD candidates that are female looks like a nice numerical variable to use, for example, to compare gender rates between these departments from the Arts and Science faculties. Generally with numerical variables we would use the mean or median as a measure of central tendency. But this variable was measured by aggregating information about each PhD candidate in that department and presenting this measure as a percentage (the related but different cases here are the PhD candidates). Just think about it, does it really make sense to make a statement like: The mean % of PhD candidates that are female for these departments of the Arts faculty is 73% whereas the mean % of PhD candidates that are female for these departments of the Science faculty is 44%, especially when the numbers of PhD candidates varies so much between departments?
Looking at the individual percentages is interesting to see how they vary across departments, but combining them to get an overall measure for each faculty should involve calculating another percentage using the original counts for PhD candidates for each department (e.g. group by faculty). If I want to compare gender rates between the Arts and Science faculties for PhD candidates, I would calculate the proportion of all PhD candidates across these department that are female for each faculty e.g. 58% of the PhD candidates from these departments of the Arts faculty are female, 53% of the PhD candidates from these departments of the Science faculty are female.
[If you’d like to read more about structuring data in the context of creating a dataset, then check out this excellent post by Rob Gould.]
Where to next?
This post was not supposed to deter you from finding and creating your own real datasets! But we do need to think carefully about the data that we provide to students, especially our high school students. Not all datasets are the same and while I’ve seen some really cool and interesting ideas out there for finding/collecting data for investigations, some of these ideas unintentionally produce data that makes it very difficult for students to engage with the core question: What can and can’t I say about the population(s) based on the random sample data?
In the next post, I’ll discuss some examples of finding real data online. Until I find time to write this next post, check out these existing data finding posts:
Using awesome real data
Cat and whisker plots: sampling from the Quick, Draw! dataset
The power of pixels: Modelling with images
## Cat and whisker plots – sampling from the Quick, Draw! dataset
Last night, I saw a tweet announcing that Google had made data available on over 50 million drawings from the game Quick, Draw! I had never played the game before, but it is pretty cool. The idea behind the game is whether a neural network can learn to recognize doodling – watch the video below for more about this (with an example about cats of course!)
For each game, you are challenged to draw a certain object within 20 secs, and you get to see if the algorithm can classify your drawing correctly or not. See my attempt below to draw a trumpet, and the neural network correctly identifying that I was drawing a trumpet.
Since I am clearly obsessed with cats at the moment, I went straight to the drawings of cats. You can see ALL the drawings made for cats (hundred of thousands) and can see variation in particular features of these drawings. I thought it would be cool to be able to take a random sample from all the drawings for a particular category, so after some coding I set up this page: learning.statistics-is-awesome.org/draw/. I’ve included below each drawing the other data provided in the following order:
• the word the user was told to draw
• the two letter country code
• the timestamp
• whether the drawing was correctly classified
• number of individual strokes made for the drawing
[Update: There are now more variables available – see this post for more details]
So, on average, how many whiskers do Quick, Draw! players draw on their cats?
I took a random sample of 50 drawings from those under the cat category using the sampling tool on learning.statistics-is-awesome.org/draw/. Below are the drawings selected 🙂
Counting how many individual whiskers were drawn was not super easy but, according to my interpretation of the drawings, here is my sample data (csv file). Using the awesome iNZight VIT bootstrapping module (and the handy option to add the csv file directly to the URL e.g. https://www.stat.auckland.ac.nz/~wild/VITonline/bootstrap/bootstrap.html?file=http://learning.statistics-is-awesome.org/draw/cat-and-whisker-plots.csv), I constructed a bootstrap confidence interval for the mean number of whiskers on cat drawings made by Quick, Draw! players.
So, turns out it’s a fairly safe bet that the mean number of whiskers per cat drawing made by Quick, Draw! players is somewhere between 2.2 and 3.5 whiskers. Of course, these are the drawings that have been moderated (I’m assuming for appropriateness/decency). When you look at the drawings, with that 20 second limit on drawing time, you can see that many players went for other features of cats like their ears, possibly running out of time to draw the whiskers. In that respect, it would be interesting to see if there is something going on with whether the drawing was correctly classified as being a cat or not – are whiskers a defining feature of cat drawings?
I reckon there are a tonne of cool things to explore with this dataset, and with the ability to randomly sample from the hundreds and hundreds of thousands of drawings available under each category, a good reason to use statistical inference 🙂 I like that students can develop their own measures based on features of the drawings, based on what they are interested in exploring.
After I published this post, I took a look at the drawings for octopus and then for octagon, a fascinating comparison.
I wonder if players of Quick, Draw! are more likely to draw eight sides for an octagon or eight legs for an octopus? I wonder if the mean number of sides drawn for an octagon is higher than the mean number of legs draw for an octopus?
## It’s raining cats and dogs (hopefully)
In April 2017, I presented an ASA K-12 statistics education webinar: Statistical reasoning with data cards (webinar). Towards the end of the webinar, I encouraged teachers to get students to make their own data cards about their cats. A few days later, I then thought that this could be something to get NZ teachers and students involved with. Imagine a huge collection of real data cards about dogs and cats? Real data that comes from NZ teachers and students? Like Census At School but for pets 🙂 I persuaded a few of my teacher friends to create data cards for their pets (dogs or cats) and to get their students involved, to see whether this project could work. Below is a small selection of the data cards that were initially created (beware of potential cuteness overload!)
The project then expanded to include more teachers and students across NZ, and even the US, and I’ve now decided to keep the data card generator (and collection) page open so that the set of data cards can grow over time. Please use the steps below to get students creating and sharing data cards about their pets.
Creating and sharing data cards about dogs and cats
Inevitably, there will be submissions made that are “fake”, silly or offensive (see below).
Data cards submitted to the project won’t automatically be added to any public sets of data cards, and will be checked first. Just like with any surveying process that is based on self-selection, is internet based and relies on humans to give honest and accurate answers, there is the potential for non-sampling errors. To help reduce the quantify of “fake” data cards, if you are keen to have your students involved with this project it would be great if you could do the following:
1. Talk to your students about the project and explain that the data cards will be shared with other students. They will be sharing information about their pet and need to be OK with this (and don’t have to!). The data will be displayed with a picture of their pet, so participation is not strictly anonymous. All of this is important to discuss with students as we need to educate students about data privacy 🙂
2. When students submit their data, they are given the finished data card which they can save. Set up a system where students need to share the data card they have created with you e.g. by saving into a shared Google drive or Dropbox, or by emailing the data card to you. The advantage for you of setting up this system is that you get your class/school set of data cards to use however you want. The advantage for me is that this level of “watching” might discourage silly data cards being created.
Pet data cards
The data collection period for this set of data cards was 1 May 17 to 19 May 17.
The diagram below shows the data included on each data card:
Additional data that could be used from each data card includes:
• Whether the pet photo was taken inside or outside
• Whether the pet photo is rotated (and the angle of rotation)
• The number of letters in the pet name
• The number of syllables in the pet name
## Statistics flowers (data cards)
Inspired by Fisher’s Iris data, this sample of flowers was created through simulation from a carefully designed model. From a student’s perspective, these flowers represent a random sample of flowers from a much bigger population of statistics flowers. The idea is that students get all of the 300 cards and need to measure different features of the flowers and determine other variables to create their sample data.
Designed variables are: type of statistics flower (tictastics, stistactis, or castistist), petal colour (red, orange, blue, green), number of petals, petal length, petal width and stigma diameter. The diagram below shows how the measurements should be taken by students:
I have made the sample size 300 to allow for categorical and distributional exploration e.g. What proportion of all statistics flowers have a black stigma? Does stigma colour appear to be linked to petal colour for statistics flowers? How could the number of petals for statistics flowers be distributed? But I appreciate that it would take a long time for students to measure 300 different flowers and record necessary data! Perhaps students could look at the flowers visually first, sort them by type of flower and see if they can detect any features that appear to differ (e.g. colour, petal length, etc.). Students could then measure some of the flowers and chuck this data into a graph for an initial view before being given access to the digital sample to do some more exploring. Remember these data cards represent a sample and the true population parameters, for example the mean petal length of all statistics flowers, are unknown to you and the students. It is not intended that these cards are used for “population bags”.
Here is the sample data set as a CSV file: flower_power
Here are the data cards as a PDF: flower_power
You will need to print these one to a page if you want the measurements in the CSV file match!
## Auckland Marathon 2015 runners (population data)
The data for each runner entered in the Auckland Marathon 2015 was obtained from https://www.aucklandmarathon.co.nz/. This data is owned by the organisers of the Auckland Marathon and can not be used for commercial purposes unless by prior written permission from the organisers.
For each runner, the following was recorded:
• bib number
• name
• time in hours (this is blank if the runner did not compete in the race)
• place (this is blank if the runner did not compete in the race)
• gender
• division
• age division
• distance in km (this is blank if the runner did not compete in the race)
• mean pace km per hr (this is blank if the runner did not compete in the race)
NB: This data set contains information about the five different races which are part of the Auckland Marathon 2015. It may be necessary to focus on just one of these races for a meaningful investigation, for example if comparing running times for male and female runners (whether as part of a sample-to-population inference or as part of exploring the population data).
Here is the population data set as a CSV file: all_races_auckland_marathon_2015_final
## Rugby World Cup 2015 players (population data)
The data for each player in the Rugby World Cup 2015 was obtained from http://www.rugbyworldcup.com/. This data is owned by the Rugby World Cup Ltd (RWC) and can not be used for commercial purposes unless by prior written permission from the RWC.
Thanks to @cushlat for the idea 🙂
For each player, the following was recorded:
• team played for (team)
• name (name)
• number of international matches played (caps)
• position (position)
• number of years since debuted (years_since_debut)
• date of debut (debut)
• age at Rugby World Cup 2015 (age)
• age minus years_since_debut (approx_age_debuted)
• height in cm (height_cm)
• weight in kg (weight_kg)
NB: This data set should be used with care for sample-to-population inference involving comparison, as both categorical variables (team and position) involve a large number of outcomes (16 teams and 11 positions). This means it is not likely that a random sample of 80 players from the population of Rugby World Cup 2015 players, for example, will contain sufficient numbers of players in any two groups for comparison e.g. England vs New Zealand OR forwards vs backs. If you use all the data for NZ and all the data for England to compare the age of players, for example, you will have used all of the data for this population and so there is no need to “make a call” about what is going on “back in the population” 🙂
My advice would be to use this data set for either single variable sampling investigations OR exploratory data analysis for the entire population. There is also something interesting in using the time variable (debut) to explore other variables 🙂
Here is the population data set as a CSV file: rubgy_world_cup_2015 | 6,927 | 32,744 | {"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-2020-10 | latest | en | 0.954863 |
https://math.stackexchange.com/questions/2626889/finding-the-eigenvalues-between-smallest-and-largest-eigenvalues-of-a-sparse-mat | 1,656,730,717,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103983398.56/warc/CC-MAIN-20220702010252-20220702040252-00108.warc.gz | 422,387,495 | 65,092 | # Finding the eigenvalues between smallest and largest eigenvalues of a sparse matrix without using Shifted Inverse Power Iteration Method
I have a symmetric positive-definite sparse matrix of size 150k X 150k.
I use the Power Iteration Method to find the largest eigenvalue.
I learnt from this post in math.stackexchange.com that if the matrix is positive definite, then by using the shift $B = A - \lambda_{max} I$ and then applying Power Iteration Method to B, I can find the smallest eigenvalue.
EDIT:: It is not numerically stable
However, my question is, if I can apply the same method to calculate eigenvalues between the smallest and largest eigenvalues, placed at regular intervals, by using the same method.
(I need to calculate eigenvalues closest to 300 regularly selected values between the smallest and largest eigenvalue).
If the interval is $s = (\lambda_{max} - \lambda_{min})/301$, then will applying Power Iteration Method on $C = A - sI$ give me the eigenvalue closest to $(\lambda_{max} - s)$?
( 301, since $\lambda_{min} + 301*s = \lambda_{max}$, so I get 300 eigenvalues between $\lambda_{min}$ and $\lambda_{max}$)
I don't want to use Inverse Power Iteration Method because the matrix is too large and the inverse of such a large sparse matrix may or may not be sparse and would be computationally inefficient too.
Shifted power iteration — while theoretically possible — is not very useful since it converges to the eigenvalue farthest away from s.
But, by using the Inverse or Shifted Inverse power iteration algorithm will yield the eigenvalue closest to s. This means by picking appropriate shifts µ, any one eigenvalue of A can be found.
Initialize v(0) with an arbitrary vector such that $||v^{(0)}||_{2} = 1$
for k = 1, 2, . . .
Solve $Aw = v^{(k−1)}$ for $w$
$v(k) = w/||w||_{2}$
$λ(k) = [v^{(k)}]^{T}Av^{(k)}$
end
Inverse of A need not be calculated. Instead for each iteration the eigenvector can be calculated by solving a system of linear equations, which is $Aw = v^{(k-1)}$
Same applies for Shifted Inverse Power Iteration.
So, Inverse Power Iteration doesn't require to calculate inverse of the matrix after all. | 545 | 2,171 | {"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} | 3.890625 | 4 | CC-MAIN-2022-27 | longest | en | 0.817208 |
http://tutor4physics.com/motion2d3d.htm | 1,386,362,019,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163052469/warc/CC-MAIN-20131204131732-00024-ip-10-33-133-15.ec2.internal.warc.gz | 192,406,355 | 4,524 | Scalar or Vector?
To explain the difference we use two words: 'magnitude' and 'direction'. By magnitude we mean how much of the quantity is there. By direction we mean is this quantity having a direction which defines it. Physical quantities which are completely specified by just giving out there magnitude are known as scalars. Examples of scalar quantities are distance, mass, speed, volume, density, temperature etc. Other physical quantities cannot be defined by just their magnitude. To define them completely we must also specify their direction. Examples of these are velocity, displacement, acceleration, force, torque, momentum etc.
If we were to represent two vectors magnitude and direction by two adjacent sides of a parallelogram. The resultant can then be represented in magnitude and direction by the diagonal. This diagonal is the one which passes through the point of intersection of these two sides.
Resolution of a Vector
It is often necessary to split a vector into its components. Splitting of a vector into its components is called resolution of the vector. The original vector is the resultant of these components. When the components of a vector are at right angle to each other they are called the rectangular components of a vector.
In the figure above the green vector has been resolved into two vectors: blue and red. These vectors are at right angles to each other. The are the rectangular components of the green vector.
Rectangular Components of a Vector
As the rectangular components of a vector are perpendicular to each other, we can do mathematics on them. This allows us to solve many real life problems. After all the best thing about physics is that it can be used to solve real world problems.
Note: As it is difficult to use vector notations on the computer word processors we will coin our own notation. We will show all vector quantities in bold. For example 'A' will be scalar quantity and 'A' will be a vector quantity.
Let Ax and Ay be the rectangular components of a vector A
then
A = Ax + Ay this means that vector A is the resultant of vectors Ax and Ay
A is the magnitude of vector A and similarly Ax and Ay are the magnitudes of vectors Ax and Ay
As we are dealing with rectangular components which are at right angles to each other. We can say that:
A = (Ax + Ay)1/2
Similarly the angle Q which the vector A makes with the horizontal direction will be
Q = tan-1 (Ax / Ay) | 500 | 2,440 | {"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} | 4.6875 | 5 | CC-MAIN-2013-48 | longest | en | 0.947367 |
http://lists.contesting.com/_amps/2005-12/msg00198.html | 1,611,470,854,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703547333.68/warc/CC-MAIN-20210124044618-20210124074618-00609.warc.gz | 67,634,085 | 3,525 | Amps
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## Re: [Amps] *** SPAM *** Re: Antenna traps
To: amps@contesting.com Re: [Amps] *** SPAM *** Re: Antenna traps Ian White GM3SEK Ian White GM3SEK Fri, 16 Dec 2005 09:31:29 +0000
```Peter Chadwick wrote: >Rich said: >>** An L-network always has the lowest Q possible. Is mo' Q mo' betta?< >Rubbish. Do the math. The Q is the sq.rt of the impedance ratio -1. >So the bigger the impedance transformation, the higher the working Q. >In any two variable system, Qw is an uncontrolled variable. A 3 >variable network allows Q to be defined. That's the advantage of the T >or pi (which is the dual of the T). I think Rich is more right than Peter on this one. We all agree that the network should have the lowest possible working Q, in order to achieve the lowest possible internal losses and the highest possible power handling capability. When transforming between resistive impedances, the lowest possible working Q is: > the sq.rt of the impedance ratio -1 Any lower value of Q will not achieve a match. Any network with three variables (eg a T) can also achieve a match, but always with higher than the minimum possible values of working Q and internal losses. An L-network only has two variables, so the working Q is automatically determined for you. With a resistive load, this is automatically the lowest that can be achieved - so that is an advantage. The only case where an L-network won't give the lowest possible Q is with certain highly reactive loads... in which case, you need to switch to a totally different configuration of L-network. This last point is the major DISadvantage of L-networks: lack of flexibility. There are a total of 8 different configurations, and they all have a limited matching range. Between them, they can match any impedance (except a 1:1 match requires theoretically zero or infinite component values) but it's a matter of finding which one out of the 8, and most practical L-network tuners can switch between a maximum of 2 configurations. However, I'd conjecture that for any pair of impedances, there will always exist at least one L-network configuration that can match them at the lowest possible working Q. -- 73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek _______________________________________________ Amps mailing list Amps@contesting.com http://lists.contesting.com/mailman/listinfo/amps ```
Current Thread Re: [Amps] *** SPAM *** Re: Antenna traps, Peter Chadwick Re: [Amps] *** SPAM *** Re: Antenna traps, R . Measures Re: [Amps] *** SPAM *** Re: Antenna traps, Steve Thompson Re: [Amps] *** SPAM *** Re: Antenna traps, Ian White GM3SEK <= Re: [Amps] *** SPAM *** Re: Antenna traps, Steve Thompson Re: [Amps] *** SPAM *** Re: Antenna traps, Ian White GM3SEK Re: [Amps] *** SPAM *** Re: Antenna traps, Steve Thompson Re: [Amps] *** SPAM *** Re: Antenna traps, Ian White GM3SEK | 732 | 2,896 | {"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} | 2.703125 | 3 | CC-MAIN-2021-04 | latest | en | 0.913802 |
http://wiki.stat.ucla.edu/socr/index.php?title=AP_Statistics_Curriculum_2007_Gamma&diff=11656&oldid=11154 | 1,531,743,925,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589270.3/warc/CC-MAIN-20180716115452-20180716135452-00420.warc.gz | 396,737,979 | 10,227 | # AP Statistics Curriculum 2007 Gamma
(Difference between revisions)
Revision as of 20:51, 11 July 2011 (view source)TracyTam (Talk | contribs) (→Example)← Older edit Current revision as of 17:34, 23 June 2012 (view source)IvoDinov (Talk | contribs) m (→Normal Approximation to Gamma distribution) (20 intermediate revisions not shown) Line 1: Line 1: + ==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - Gamma Distribution== + ===Gamma Distribution=== ===Gamma Distribution=== '''Definition''': Gamma distribution is a distribution that arises naturally in processes for which the waiting times between events are relevant. It can be thought of as a waiting time between Poisson distributed events. '''Definition''': Gamma distribution is a distribution that arises naturally in processes for which the waiting times between events are relevant. It can be thought of as a waiting time between Poisson distributed events. Line 7: Line 9: - For $X\sim Gamma(k,\theta)\!$, where $k=h$ and $\theta=1/\lambda$, the gamma probability density function is given by + For $X\sim \operatorname{Gamma}(k,\theta)\!$, where $k=h$ and $\theta=1/\lambda$, the gamma probability density function is given by :$\frac{x^{k-1}e^{-x/\theta}}{\Gamma(k)\theta^k}$ :$\frac{x^{k-1}e^{-x/\theta}}{\Gamma(k)\theta^k}$ Line 47: Line 49: The gamma distribution is also used to model errors in a multi-level Poisson regression model because the combination of a Poisson distribution and a gamma distribution is a negative binomial distribution. The gamma distribution is also used to model errors in a multi-level Poisson regression model because the combination of a Poisson distribution and a gamma distribution is a negative binomial distribution. - ===Example=== ===Example=== Line 58: Line 59: The figure below shows this result using [http://socr.ucla.edu/htmls/dist/Gamma_Distribution.html SOCR distributions] The figure below shows this result using [http://socr.ucla.edu/htmls/dist/Gamma_Distribution.html SOCR distributions]
[[Image:Gamma.jpg|600px]]
[[Image:Gamma.jpg|600px]]
+ + + ===Normal Approximation to Gamma distribution=== + + Note that if $$\{X_1,X_2,X_3,\cdots \}$$ is a sequence of independent [[AP_Statistics_Curriculum_2007_Exponential|Exponential(b) random variables]] then $$Y_k = \sum_{i=1}^k{X_i}$$ is a [http://www.math.uah.edu/stat/special/Gamma.html random variable with gamma distribution] with the following shape parameter, '''k''' (positive integer indicating the number of exponential variable in the sum) and scale parameter '''b''' (which is the exponential parameter). By the [[AP_Statistics_Curriculum_2007_Limits_CLT|central limit theorem]], if k is large, then gamma distribution can be approximated by the normal distribution with mean $$\mu=kb$$ and variance $$\sigma^2 =kb^2$$. That is, the distribution of the variable $$Z_k={{Y_k-kb}\over{\sqrt{k}b}}$$ tends to the standard normal distribution as $k\longrightarrow \infty$. + + For the example above, $$\Gamma(k=4, \theta=2)$$, the [http://socr.ucla.edu/htmls/dist/Gamma_Distribution.html SOCR Normal Distribution Calculator] can be used to obtain an estimate of the area of interest as shown on the image below. + +
[[Image:EBook_Gamma_Fig2.png|500px]]
+ + The probabilities of the [http://socr.ucla.edu/htmls/dist/Gamma_Distribution.html real Gamma] and [http://socr.ucla.edu/htmls/dist/Normal_Distribution.html approximate Normal] distributions (on the range [2:4]) are not identical but are sufficiently close. + +
+ {| class="wikitable" style="text-align:center; width:75%" border="1" + |- + ! Summary|| [http://socr.ucla.edu/htmls/dist/Gamma_Distribution.html $$\Gamma(k=4, \theta=2)$$ ] || [http://socr.ucla.edu/htmls/dist/Normal_Distribution.html $$Normal(\mu=8, \sigma^2=4)$$ ] + |- + | Mean||8.000000||8.0 + |- + | Median||7.32||8.0 + |- + | Variance||16.0||16.0 + |- + | Standard Deviation||4.0||4.0 + |- + | Max Density|| 0.112021||0.099736 + |- + ! colspan=3|Probability Areas + |- + | <2|| 0.018988|| 0.066807 + |- + | [2:4]|| 0.123888||0.091848 + |- + | >4|| 0.857123||0.841345 + |} +
+ +
## General Advance-Placement (AP) Statistics Curriculum - Gamma Distribution
### Gamma Distribution
Definition: Gamma distribution is a distribution that arises naturally in processes for which the waiting times between events are relevant. It can be thought of as a waiting time between Poisson distributed events.
Probability density function: The waiting time until the hth Poisson event with a rate of change λ is
$P(x)=\frac{\lambda(\lambda x)^{h-1}}{(h-1)!}{e^{-\lambda x}}$
For $X\sim \operatorname{Gamma}(k,\theta)\!$, where k = h and θ = 1 / λ, the gamma probability density function is given by
$\frac{x^{k-1}e^{-x/\theta}}{\Gamma(k)\theta^k}$
where
• e is the natural number (e = 2.71828…)
• k is the number of occurrences of an event
• if k is a positive integer, then Γ(k) = (k − 1)! is the gamma function
• θ = 1 / λ is the mean number of events per time unit, where λ is the mean time between events. For example, if the mean time between phone calls is 2 hours, then you would use a gamma distribution with θ=1/2=0.5. If we want to find the mean number of calls in 5 hours, it would be 5 $\times$ 1/2=2.5.
• x is a random variable
Cumulative density function: The gamma cumulative distribution function is given by
$\frac{\gamma(k,x/\theta)}{\Gamma(k)}$
where
• if k is a positive integer, then Γ(k) = (k − 1)! is the gamma function
• $\textstyle\gamma(k,x/\theta)=\int_0^{x/\theta}t^{k-1}e^{-t}dt$
Moment generating function: The gamma moment-generating function is
$M(t)=(1-\theta t)^{-k}\!$
Expectation: The expected value of a gamma distributed random variable x is
$E(X)=k\theta\!$
Variance: The gamma variance is
$Var(X)=k\theta^2\!$
### Applications
The gamma distribution can be used a range of disciplines including queuing models, climatology, and financial services. Examples of events that may be modeled by gamma distribution include:
• The amount of rainfall accumulated in a reservoir
• The size of loan defaults or aggregate insurance claims
• The flow of items through manufacturing and distribution processes
• The load on web servers
• The many and varied forms of telecom exchange
The gamma distribution is also used to model errors in a multi-level Poisson regression model because the combination of a Poisson distribution and a gamma distribution is a negative binomial distribution.
### Example
Suppose you are fishing and you expect to get a fish once every 1/2 hour. Compute the probability that you will have to wait between 2 to 4 hours before you catch 4 fish.
One fish every 1/2 hour means we would expect to get θ = 1 / 0.5 = 2 fish every hour on average. Using θ = 2 and k = 4, we can compute this as follows:
$P(2\le X\le 4)=\sum_{x=2}^4\frac{x^{4-1}e^{-x/2}}{\Gamma(4)2^4}=0.12388$
The figure below shows this result using SOCR distributions
### Normal Approximation to Gamma distribution
Note that if $$\{X_1,X_2,X_3,\cdots \}$$ is a sequence of independent Exponential(b) random variables then $$Y_k = \sum_{i=1}^k{X_i}$$ is a random variable with gamma distribution with the following shape parameter, k (positive integer indicating the number of exponential variable in the sum) and scale parameter b (which is the exponential parameter). By the central limit theorem, if k is large, then gamma distribution can be approximated by the normal distribution with mean $$\mu=kb$$ and variance $$\sigma^2 =kb^2$$. That is, the distribution of the variable $$Z_k={{Y_k-kb}\over{\sqrt{k}b}}$$ tends to the standard normal distribution as $k\longrightarrow \infty$.
For the example above, $$\Gamma(k=4, \theta=2)$$, the SOCR Normal Distribution Calculator can be used to obtain an estimate of the area of interest as shown on the image below.
The probabilities of the real Gamma and approximate Normal distributions (on the range [2:4]) are not identical but are sufficiently close.
Summary $$\Gamma(k=4, \theta=2)$$ $$Normal(\mu=8, \sigma^2=4)$$
Mean8.0000008.0
Median7.328.0
Variance16.016.0
Standard Deviation4.04.0
Max Density 0.1120210.099736
Probability Areas
<2 0.018988 0.066807
[2:4] 0.1238880.091848
>4 0.8571230.841345 | 2,246 | 8,226 | {"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": 11, "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} | 4.34375 | 4 | CC-MAIN-2018-30 | latest | en | 0.856615 |
https://stackoverflow.com/questions/34490247/overload-resolution-and-user-defined-conversion | 1,723,542,043,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641075627.80/warc/CC-MAIN-20240813075138-20240813105138-00492.warc.gz | 416,544,063 | 43,674 | # Overload resolution and user defined conversion
Consider the simple code :
``````struct A;
struct B {
B(){}
B(A const&){ }
};
struct A {
operator int() const {return 0;};
};
void func(B){}
void func(char){}
int main()
{
func(A()); //ambiguous call oO
}
``````
First of all I'm not sure if I understand everything correctly, so correct me anytime you find me wrong please.
My understanding was that that `void func(B)` should have been chosen, since argument to `func` is `A` which is class type, hence type of conversion required is "User defined conversion sequence"
Now from IBM C++ ref :
A user-defined conversion sequence consists of the following:
• A standard conversion sequence
• A user-defined conversion
• A second standard conversion sequence
Now there are two user defined conversion present `B::B(const A&)` and `A::operator int (const A&);`
so the sequence are
-> `A()` -> `B::B(const A&)` -> `Standard conversion (identity conversion)`
-> `A()` -> `A::operator int (const A&)` -> `Standard conversion (integral conversion)`
since integral conversion is worse than identity conversion I thought `void func(B)` would called but still the call is ambiguous .
So please help me at which point am I wrong and why the call is ambiguous. Thanks a lot :)
• I think feeding `B` to `func(B)` is identity, `A`->`B::B`->`func(B)` is perhaps not; should check the standard though. Commented Dec 28, 2015 at 8:47
The two conversion sequences here, `A -> B` and `A -> int` are both user-defined because they operate via functions which you defined.
The rule for ranking user-defined conversion sequences is found in 13.3.3.2 (N3797):
User-defined conversion sequence `U1` is a better conversion sequence than another user-defined conversion sequence `U2` if they contain the same user-defined conversion function or constructor or they initialize the same class in an aggregate initialization and in either case the second standard conversion sequence of `U1` is better than the second standard conversion sequence of `U2`
These two conversion sequences don't contain the same user-defined conversion function, and they don't initialize the same class in aggregate initialization (since one initializes `int`).
So it is not true that one sequence ranks above the other, therefore this code is ambiguous.
• So overload resolute is confused just because user defined conversion functions are not same? Commented Dec 28, 2015 at 9:07
• @AngelusMortis yes basically
– M.M
Commented Dec 28, 2015 at 9:10
• You still've a conversion function to `B` in `A`, just removing the constructor in `B` wouldn't do. Commented Dec 28, 2015 at 9:52
• @legends2k gcc and clangs rejects the code but MS C++ accepts the code and function selected was `void func(B)`as expected by me :) Commented Dec 28, 2015 at 9:57
• @M.M Sir doesn't the statement "if they contain the same user-defined conversion function or constructor" limits us in a way that -> object-of-some-class-type as a argument to some overloaded function must contain conversion functions of same type (i.e either constructor one or conversion function), since they have to be same w.r.t overload resolution. Thanks Commented Dec 28, 2015 at 10:12
so the sequence are -> A() -> B::B(const A&) -> Standard conversion (identity conversion)
No! Excerpt from the standard (draft) [over.best.ics] (emphasis mine):
1. If no conversions are required to match an argument to a parameter type, the implicit conversion sequence is the standard conversion sequence consisting of the identity conversion (13.3.3.1.1).
`func(A())` is not identity, it's user-defined. Again from the standard, [[conv]]:
For class types, user-defined conversions are considered as well; see 12.3. In general, an implicit conversion sequence (13.3.3.1) consists of a standard conversion sequence followed by a user-defined conversion followed by another standard conversion sequence.
I think you have a misunderstanding about Standard conversions. They have nothing to do with user-defined types/classes. Standard conversions are only for built-in types: lvalue-to-rvalue conversion, array-to-pointer conversion, function-to-pointer conversion, integral promotions, floating point promotion, integral conversions, floating point conversions, floating-integral conversions, pointer conversions, pointer to member conversions, boolean conversions and qualification conversions. `A` -> `int` is not any of these but a user-defined conversion. The standard on user-defined conversions, [[class.conv]] i.e. 12.3:
Type conversions of class objects can be specified by constructors and by conversion functions. These conversions are called user-defined conversions and are used for implicit type conversions (Clause 4), for initialization (8.5), and for explicit type conversions (5.4, 5.2.9).
You have two user-defined conversion sequences of the same rank (see M.M's answer to know why), so the compiler wants you to disambiguate.
• "You have two user-defined conversion sequences, and thus are of the same rank" -> I still don't understand how did you rank them same. I'd really appreciate if you explain a bit more in simple terms :) Commented Dec 28, 2015 at 9:12
• Conversions can be categorized by ranks: Exact match, Promotion and Conversions. Both your calls fall under the third. In that, both fall under the user-defined conversion category. When two user-defined conversion sequences compete, which one to choose is defined in the standard. This part of the standard is quoted in M.M's answer. Commented Dec 28, 2015 at 9:20 | 1,275 | 5,574 | {"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-2024-33 | latest | en | 0.87672 |
http://topcoder.bgcoder.com/print.php?id=2122 | 1,726,088,185,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651400.96/warc/CC-MAIN-20240911183926-20240911213926-00420.warc.gz | 31,897,231 | 1,955 | ### Problem Statement
NOTE: This problem statement contains superscripts that may not display properly if viewed outside of the applet.
You are given ints n and k. Return the value of the sum 1k + 2k + 3k + ... + nk modulo 1000000007.
### Definition
Class: SumOfPowers Method: value Parameters: int, int Returns: int Method signature: int value(int n, int k) (be sure your method is public)
### Constraints
-n will be between 1 and 109, inclusive.
-k will be between 1 and 50, inclusive.
### Examples
0)
`5` `1`
`Returns: 15`
Here, we have arithmethic progression: 1 + 2 + 3 + 4 + 5 = 15.
1)
`4` `2`
`Returns: 30`
Just a little bit more complicated example here: 12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30.
2)
`13` `5`
`Returns: 1002001`
This one would be harder to check by hand.
3)
`123456789` `1`
`Returns: 383478132`
#### Problem url:
http://www.topcoder.com/stat?c=problem_statement&pm=8725
#### Problem stats url:
http://www.topcoder.com/tc?module=ProblemDetail&rd=12169&pm=8725
mateuszek
#### Testers:
PabloGilberto , Olexiy , ivan_metelsky , ged | 343 | 1,077 | {"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-2024-38 | latest | en | 0.532647 |
https://community.wolfram.com/groups/-/m/t/291196 | 1,726,856,930,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00269.warc.gz | 159,307,496 | 20,809 | 1
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# Using the output on ParametricNDSolve in a Piecewise function
Posted 11 years ago
Hi folks - this is my first post here so I hope I'm putting question in the right place! I'm using ParametricNDSolve to evaluate a function (Eb1r,t]) with four free parameters and two dependent variables between two boundaries rn and ro. However, the function also needs to be defined between 0 and rn - in this case it's just a constant that takes on the values at the boundary rn, given by Eb1[rn,t]. I've outlined the problem on [Stack Exchange here , and in essence I'm unable to use the output of ParametricNDSolve in the same way I'd use output of NDsolve; for example, I try to feed it into piecewise like this; (*Parametric solution for unknowns kme, kmn, j and eo, for functions Ef1 and Eb1*) x = ParametricNDSolve[{eqnBo == 0, Eb1[r, 0] == 0, eqnDe == 0, Ef1[r, 0] == 0, Derivative[1, 0][Ef1][rn, t] == 0, Ef1[ro, t] == eo}, Eb1, {r, rn, ro}, {t, 0, 14400}, {kme, kmn, j, eo}]; which works fine, but my attempts to manipulate output fail miserablely - for example, when I try to define a piecewise function at t = 14400 - Ebound[r_] = Piecewise[{{Eb1[rn, 14400] /. x, r < rn}, {Eb1[r, 14400] /. x, r >= rn}}]; I keep getting the error ParametricNDSolve::fpct: "Too many parameters in {kme,kmn,j,eo} to be filled from {r,14400}." I can't work out if this is just some syntax error, or whether there is a deeper reason why I cannot manipulate the parametric equation. Does anyone have any ideas whether this is possible, or what I'm doing wrong? Thanks! A MWE of the problem is appended to this post for anyone who might be interested. Attachments:
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Posted 11 years ago
ParametricNDSolve is basically the setup part of NDSolve, for use when you want to numerically solve similar differential equations multiple times and they only differ in terms of certain parameters. To get a solution from it, you have to specify the values of the parameters, as NDSolve is a numerical algorithm. It's not like DSolve, which can keep parameters as symbols. You're not providing the values of the parameters.
Posted 11 years ago
Thanks - that's a pain! I was hoping to use nonlinearmodelfit for the parameters {kme,kmn,j,eo} between known limits. The fact that the function is different before and after the boundary rn is complicating my attempts to do so, which is why I was trying to use Piecewise. Is there a clever way around this issue? | 695 | 2,517 | {"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} | 3.03125 | 3 | CC-MAIN-2024-38 | latest | en | 0.865842 |
http://www.maa.org/publications/maa-reviews/measure-for-measure-the-story-of-imperial-metric-and-other-units?device=mobile | 1,438,564,853,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042989331.34/warc/CC-MAIN-20150728002309-00246-ip-10-236-191-2.ec2.internal.warc.gz | 567,966,808 | 8,740 | # Measure for Measure: The Story of Imperial, Metric, and Other Units
Publisher:
Johns Hopkins University Press
Number of Pages:
215
Price:
24.95
ISBN:
0-8018-7072-0
When Bill Berlinghoff and I were working on Math through the Ages, we felt it was important to include a chapter on the history of units, and especially of metric units. After all, this topic is part of the school mathematics curriculum. But we found it difficult to find good references. The best we could find was John Roche's The Mathematics of Measurement: A Critical History, which is useful but too heavy, technical, and expensive for our intended audience. Alex Hebra's Measure for Measure: The Story of Imperial, Metric, and Other Units is an attempt to fill that gap.
In short chapters, Hebra takes us through a plethora of units for all sorts of quantities: length, time, angles, mass and force, temperature, luminosity, etc. He tells us about attempts to create units based on natural phenomena and about ways of formulating natural laws in terms of dimensionless quantities. One of the last chapters takes the reader through an interesting exercise: suppose an alien race, communicating over intergalactic distances, asked us how to build a dam; could we give them directions despite the fact that we have no idea what sort of units of measurement they use?
Despite the subtitle, there is very little history here. Hebra does include lots of stories, but he doesn't really attempt to trace in detail the evolution of the various systems and units of measurement. The organization by kind of unit (a chapter for length, a chapter for time,...) makes it hard to get a sense of historical flow, and the transition from isolated units to systems of units does not get the emphasis it deserves. There are occasional lapses; for example, in the chapter on measuring angles, the radian is defined but no explanation is given as to why one would want to use a unit of measurement that resulted in such strange numbers. (There is also a problem with a graphic in that section: π has come out as p.) Finally, both the discussion of whether the United States should "go metric" and the jokes based on unit conversions (e.g.,"28 to 29 grams of prevention are worth 0.454 kilograms of cure") get old fairly quickly.
Nevertheless, this is a useful book for anyone wanting to know more about units of measurement and their role in science (especially physics). Many of the examples would make for excellent assignments for students, and the many illustrations are very helpful. So: not exactly the book I wanted it to be, but I'll take it.
Fernando Q. Gouvêa is the co-author, with William P. Berlinghoff, of Math through the Ages: A Gentle History for Teachers and Others.
Date Received:
Sunday, May 1, 2005
Reviewable:
Yes
Include In BLL Rating:
No
Alex Hebra
Publication Date:
2003
Format:
Hardcover
Audience:
Category:
General
Fernando Q. Gouvêa
06/1/2004
Publish Book:
Modify Date:
Wednesday, October 19, 2005 | 667 | 2,984 | {"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-2015-32 | latest | en | 0.954702 |
http://s3-us-west-1.amazonaws.com/yummydl/mathematical-miracles-of-quran-19.html | 1,576,120,314,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540536855.78/warc/CC-MAIN-20191212023648-20191212051648-00164.warc.gz | 131,033,310 | 8,619 | Mathematical miracles of the Holy Quran MATHEMATICAL DESIGN OF THE BISMILLAH The four words and the 19 letters of the Bismalah are put together according. Mathematical miracles of the Holy Quran MATHEMATICAL DESIGN OF THE BISMILLAH The four words and the 19 letters of the Bismalah are put together according to a mathematical system which is humanly impossible to compose. Using the data in Tables 5 and 6, we get the following 19-based mathematical facts: FACT 1.
XP Tutorial 4New Perspectives on Creating Web Pages with HTML, XHTML, and XML 1 Designing a Web Page with Tables Tutorial 4 Creating a News Page.
Integers Decimals Exponents Square roots Scientific notation Fractions Percentages Averages (means, medians, and modes) An understanding of these 8. ARITHMETIC SEQUENCES These are sequences where the difference between successive terms of a sequence is always the same number.
Ch 6 Sec 4: Slide #1 Columbus State Community College Chapter 6 Section 4 An Introduction to Applications of Linear Equations. Topics: Put-Call Parity American and European Options Properties Basic Binomial Tree Models. THE ARITHMETIC MEAN The arithmetic mean is the statisticians term for what the layman knows as the average. When the appearance of the letters D-N-A (Dal-Nun-Alif in Arabic) side by side in places in the Qur’an is examined, they appear most frequently in verse 65 of Surat al-Kahf. The letters D-N-A appear side by side three times in this verse, in a most incomparable manner. The number of this exceptional verse in which the term DNA appears so strikingly is 18:65. In Surat al-Kahf, which refers to DNA and the year 1865 when the science of genetics began, DNA is repeated 7 times, as is RNA (the Arabic letters Ra-Nun-Alif). Like DNA, the RNA molecule is a molecule giving rise to genetic structure.
The entire universe, the Earth we live on and all living and inanimate entities are made up of various combinations of elements. The sequence of the Arabic letters in Surat al-Hadid are such that they represent expressions containing our Lord's commands and advice on the one hand and indications concerning atoms on the other. Oxygen is one of the most common elements on Earth, and was discovered in the 1770 by two scientists separate from each other: the Swiss Carl Scheele and Joseph Priestley from Britain.
The French scientist Paul Lecoq de Boisbaudran discovered the element Samarium (Sm) in 1879. The British chemist Sir Humphrey Davy first obtained the element Potassium (K) using electrolysis in 1807.
The element Sulfur (S) is found in high levels in the bodies of living things and the soil.
The element Erbium (Er) was discovered by the Swedish chemist Carl Gustaf Mosander in 1843. The element Vanadium (V) was discovered in 1801 by the Mexican scientist Andres Manuel del Rio and by the Swedish chemist Nils Gabriel Sefstrom in 1830. The French chemist Paul Emile Lecoq de Boisbaudran discovered the element Gallium (Ga) in 1875 using spectral bands.
Nitrogen (N), discovered by the Scottish scientist Daniel Rutherford in 1772, makes up 78% of the atmosphere. The Dutch physicist Dirk Coster and the Hungarian chemist Georg de Hevesy discovered the element Hafnium (Hf) in 1923.
The element Indium (In) was first discovered in 1863 by the scientists Hieronymus Theodor Richter and Ferdinand Reich. Non-radioactive Strontium was first discovered in Scotland in 1790 by Adair Crawford and William Cruikshank.
Discovered in 1899 by the French scientist Andre Debierne, Actinium (Ac) is a rare, radioactive element. Thorium (Th) is a radioactive element discovered in 1828 by the Swedish chemist Jons Jacob Berzelius. The element pure nickel (Ni) was first isolated in 1751 by the Swedish scientist Axel Fredrik Cronstedt.
There are 12 words between the places in Surat al-Hadid where the letter Mg appear together.
Dysprosium (Dy) was discovered in 1886 by the French scientist Paul Emile Lecoq de Boisbaudran.
There are 65 words between the places in Surat al-Hadid where the letters Zn appear together.
The Swedish scientist Carl Wilhelm Scheele first obtained the element Chlorine (Cl) in 1774, centuries after the revelation of the Qur'an. Carbon (C) occupies an extremely important place in the structure of living things and exists in very different substances.
Cerium (Ce) was discovered in 1803 by the geologist Wilhelm von Hisinger and chemist Jons Berzelius, both from Sweden, and, independently of them, by the German chemist Martin Klaproth. This remarkable system is based on the number and the gematrical values of the letters that constitute the four words of the Bismalah.
Insert the sequence number of each letter in the word before its gematrical value in Fact 4. Publishing as Pearson Addison-Wesley Numbers, Variables, and Expressions Natural Numbers and Whole Numbers Prime. Mean Square Estimation Given some information that is related to an unknown quantity of interest, the problem is to obtain a good estimate for the. Publishing as Pearson Addison-Wesley Rules for Exponents Review of Bases and Exponents Zero Exponents The Product.
The beginning of the science of genetics dates back to genetic laws drawn up by the scientist Mendel in 1865. These numbers are an expression of the date when the science of genetics began. This cannot be regarded as a coincidence. For that reason, the appearance of DNA and RNA an equal number of times in this Surat is further proof that these molecules were referred to in the Qur’an hundreds of years ago.
Artificial elements have also been obtained by various experiments in recent times, but these are seldom found on Earth. Reference to this element, used in the cores of nuclear reactors, hundreds of years beforehand in the Qur'an, is a great miracle. Potassium is a requirement for plants and animals and also the human body, and is plentiful in nature.
There are a total of 24 words between the first and second appearances, consecutively, of the letter C and R in Surat al-Hadid.
As with other elements, its atomic number appearing in the Surah is one of the miracles of the Quran. Discovered hundreds of years after the revelation of the Qur'an, Indium is found in very small quantities in nature. There are 71 words from the beginning of Surat al-Hadid to the letters L and U that represent the element. This element, which is rare in nature, is used in nuclear devices, rockets and space vehicles. Sodium atoms are found in a great many objects, from the salt we eat to the stars in space. Magnesium, discovered centuries after the revelation of the Qur'an, is one of the most common elements in the Earth's crust. For example, it performs vitally important functions by serving in enzymes and blood cells in the human body. Carbon atoms are present in very different compounds, from the carbon dioxide released during respiration to diamond.
Let us first summarize the information we need to know about the Bismalah in Table 6 before we review this incredible mathematical system.
The date, a turning point in the history of science, is referred to in verse 65 of Surat al-Kahf, or verse 18:65.
To date, the existence of 110 elements has been verified by the International Union of Pure and Applied Chemistry (IUPAC).
Nitrogen occupies an important place in the structures of living things and is found in such vital components as protein and nucleic acid.
There are 85 letters between the second and third appearances side by side of the letters A and T, which represent Astatine. Lithium is required by industry and used in a wide range of fields, such as manufacturing batteries, freezers and artificial rubber and also in medicine. The way that this element, discovered hundreds of years after the revelation of the Qur'an, was foretold to people so long ago is a sign of the Omniscience of our Lord.
Chlorine is used in a wide range of fields, from drug manufacture to paints, and from petrol products to cleaning products. The sequence number of each word in the Basmalah followed by the number of letters in it forms an 8-digit number which is a multiple of 19: 1 3 2 4 3 6 4 6 = 19 x 19 x 36686 FACT 3. When we insert the sequence numbers of the letters, we get 1 2 2 60 3 40, where the sequence numbers are in italics, the gematrical values are in bold. The presence of a certain amount of sodium in the human body is essential for a normal flow of water between cells and bodily fluids, tissue formation and muscle contraction. The four words of the Basmalah, the English translation, the numberof Arabic letters in each word, and their gematrical values.
Replace the number of letters in each word in Fact 2 by the total geomatrical value of that word.
There are twelve months in a year.Allah in Holy Quran had mentioned word month or Shahar twelve times. This element, discovered centuries after the revelation of the Qur'an, is one of the rarest in nature.
Thus, the sequence number of each word is followed by its total geomatrical value, to form a 15-digit number which is a multiple of 19: 1 102 2 66 3 329 4 289 = 19 x 5801401752331 FACT 4. Replace the total gematrical value of each word in Fact 3 by the gematrical value of every letter in that word.
Similarly, the total gematrical value of the second word, 66, is replaced by 1 30 30 5, and so on. Replace the total gematrical value of each word in Fact 3 by the sum of the gematrical values of the first and the last letter in that word.
The number 42 is the sum of 2 and 40, which are the gematrical values of the first and the last letter in the first word. Similarly, the total gematrical value of the second word, 66, is replaced by 6, the sum of 1 and 5. Repeating this process for the four words of the Basmalah,we get an 11-digit number which is a multiple of 19: 1 42 2 6 3 51 4 41 = 19 x 748755339 (2+40) (1+5) (1+50) (1+40) FACT 7.
In Fact 2, the sequence number of each word is followed by the number of letters (3, 4, 6, and 6) in the word. In Fact 3, we replace the number letters by the gematrical values of the words (102, 66, 329, and 289).
Now, for this case, the sequence number of each word will be followed by the sum of the number of letters and the gematrical value of the word.
It will be70 (4+66) for the second word, 335 (6+329) for the third word, and 295 (6+289) for the fourth word.
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Author: dinamshica | 2,582 | 10,687 | {"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-2019-51 | latest | en | 0.86938 |
https://wiki-helper.com/question/i-will-ask-10-questions-q-2-a-garment-factory-produced-216315-shirts-182736-trousers-and-5-40596350-52/ | 1,638,780,066,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363290.59/warc/CC-MAIN-20211206072825-20211206102825-00391.warc.gz | 670,161,505 | 13,153 | ## I will ask 10 questionsQ.2 A garment factory produced 216315 shirts 182736 trousers and 58704 jackets in a year what i
Question
Q.2 A garment factory produced 216315 shirts 182736 trousers and 58704 jackets in a year what is the total production of all three items in that year
give correct answer to get correct points
in progress 0
5 months 2021-06-23T04:37:04+00:00 1 Answers 0 views 0
Number of shirts products in a year =216315
Number of trouser products in a year =182736
Number of jacket products in a year =58704
Total production in a year =216315+182736+58704=457755 | 169 | 588 | {"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-2021-49 | latest | en | 0.806502 |
https://community.fabric.microsoft.com/t5/DAX-Commands-and-Tips/How-to-average-counts-and-how-to-weight-the-count-based-on/m-p/3724781 | 1,716,629,863,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058789.0/warc/CC-MAIN-20240525065824-20240525095824-00167.warc.gz | 146,030,479 | 117,734 | cancel
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Helper I
## How to average counts, and how to weight the count based on multiple reasons.
Hello,
This is a two part question.
The first part is related to calculating an average count of canceled orders. For this it could be average per month, average per week, average per quarter, average per year. I don't need the calculation hardcoded for a specific period, but an example of how to do so would be helpful in case I decide to put it on a card. My thoughts are to set a time frame and bring in a date field with either the year, year/quarter, year/month, year, year/week, and the calculation would perform correctly based on the level of aggregation.
Thoughts?
The second part of my question is I want to apply a weight to the reasons why someone may have canceled an order. For instance, say I have 5 canceled ordered for the month. The reason column is a concatenated field which I parsed out into multiple rows. So in this example, a canceled order may consist of one row or multiple rows depending on the number of reasons.
My data is currently set up as fact table and a support table.
Here is an example or my original data:
Date Canceled Order Reason(s) 12/3/23 ORD9846 Ordered by accident 12/7/23 ORD9915 Ordered wrong color; Ordered wrong size 12/15/23 ORD9934 Expected arrival too late 12/20/23 ORD9948 Unhappy with shipping costs; Heard bad reviews about product; Product looks cheaply made 12/26/23 ORD9979 Customer changed mind; Order no longer needed
This is after I parse the data into rows:
Canceled Order Reason(s) ORD9846 Ordered by accident ORD9915 Ordered wrong color ORD9915 Ordered wrong size ORD9934 Expected arrival too late ORD9948 Unhappy with shipping costs ORD9948 Heard bad reviews about product ORD9948 Product looks cheaply made ORD9979 Customer changed mind ORD9979 Order no longer needed
This is my model, however I could merge the data back to make one table with multiple rows for each reason:
Basically I want the calculation to return a count of 5 orders, where if I filter the data by a reason, it would calculate 0.5 for each row on orders ORD9915 and ORD9979 and calculate 0.33 for each row on order ORD9948. Currently my calculation shows a total of 9 orders.
Thanks.
4 REPLIES 4
Community Support
According to your description, here are my steps you can follow as a solution.
(1) My test data is the same as yours.
(2) Click "transform data" to go to the power query, delete the date column, and then split the Reason(s).
(3) We can create a date table.
``Date = CALENDAR(MIN('Canceled Orders'[Date]),MAX('Canceled Orders'[Date])) ``
(4) We can create measures.
``````Measure =
var _a=COUNTROWS(ALL('Reasons Canceled'))
var _b=COUNTROWS(FILTER(ALL('Reasons Canceled'),'Reasons Canceled'[Canceled Order]=MAX('Canceled Orders'[Canceled Order])))
RETURN DIVIDE(_b,_a)``````
``````Measure 2 =
var _a= COUNTROWS(ALL('Reasons Canceled'))
var _b= COUNTROWS(ALLSELECTED('Canceled Orders'))
RETURN DIVIDE(_b,_a,0)``````
(5) Then the result is as follows.
If the above one can't help you get the desired result, please provide some sample data in your tables (exclude sensitive data) with Text format and your expected result with backend logic and special examples. It is better if you can share a simplified pbix file. Thank you.
Best Regards,
Neeko Tang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Helper I
Thank you for the attempt, but the two measures do not capture what I am looking for. Unfortunately I currently cannot upload pbix files, but I will provide you with some more information.
I decided to merge the two tables, however I might prefer not to, but I know that would complicate the measure.
I expanded the reasons.
I created three measures:
Here is my canvas with an explaination:
As you can see in the table on the top, the weighted count shows correctly, but when the context of the table changes, as seen in the bottom table, the calculation doesn't work. Thanks.
Community Support
We can create a measure.
``````Measure 3 =
VAR _a =
CALCULATE (
COUNT ( 'Merge Canceled Orders'[Canceled Order] ),
FILTER (
ALL ( 'Merge Canceled Orders' ),
[Canceled Order] = MAX ( 'Merge Canceled Orders'[Canceled Order] )
)
)
VAR _b =
CALCULATE (
DISTINCTCOUNT ( 'Merge Canceled Orders'[Canceled Order] ),
FILTER (
ALL ( 'Merge Canceled Orders' ),
[Canceled Order] = MAX ( 'Merge Canceled Orders'[Canceled Order] )
)
)
RETURN
DIVIDE ( _b, _a)
``````
Best Regards,
Neeko Tang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Helper I
@v-tangjie-msft Aplogize for my delayed response, but I tried to implement this, but the the count doesn't match up.
Measure 3 should match the distinct count of 5 in the example.
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Top Kudoed Authors | 1,305 | 5,381 | {"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-2024-22 | latest | en | 0.916403 |
https://online.stat.psu.edu/stat509/book/export/html/778 | 1,685,271,854,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643663.27/warc/CC-MAIN-20230528083025-20230528113025-00402.warc.gz | 501,579,522 | 4,423 | # 17.3 - Estimating the Probability of Disease
17.3 - Estimating the Probability of Disease
Sensitivity and specificity describe the accuracy of a test. In a clinical setting, we do not know who has the disease and who does not - that is why diagnostic tests are used. We would like to be able to estimate the probability of disease based on the outcome of one or more diagnostic tests. The following measures address this idea.
Prevalence is the probability of having the disease, also called the prior probability of having the disease. It is estimated from the sample as $$\dfrac{\left(a+c\right)}{\left(a+b+c+d\right)}$$.
Positive Predictive Value (PV+) is the probability of disease in an individual with a positive test result. It is estimated as $$\dfrac{a}{\left(a+b\right)}$$.
Negative Predictive Value (PV - ) is the probability of not having the disease when the test result is negative. It is estimated as as $$\dfrac{d}{\left(c+d\right)}$$.
In the FNA study of 114 women with nonpalpable masses and abnormal mammograms,
$$prevalence = \dfrac{15}{114} = 0.13$$
$$PV+ = \dfrac{14}{\left(14+8\right)} = 0.64$$
$$PV - = \dfrac{91}{\left(1+91\right)} = 0.99$$
Thus, a woman's prior probability of having the disease is 0.13 and is modified to 0.64 if she has a positive test result. A women's prior probability of not having the disease is 0.87 and is modified to 0.99 if she has a negative test result.
If the disease under study is rare, the investigator may decide to invoke a case-control design for evaluating the diagnostic test, e.g., recruit 50 patients with the disease and 50 controls. Obviously, prevalence cannot be estimated from a case-control study because it does not represent a random sample from the general population.
Predictive values allow us to determine the usefulness of a test and they vary with the sensitivity and specificity of a test. If all other characteristics held constant, then:
1. as sensitivity of a test increases, PV - increases and
2. as specificity of a test increases, PV+ increases.
Predictive values vary with the prevalence of the disease in the population being tested or the pre-test probability of disease in a given individual.
Sensitivity, specificity, and prevalence can be used in a clinical setting to estimate post-test probabilities (predictive values), even though physicians work with one patient at a time, not entire populations of patients. Three pieces of information are necessary prior to performing the test, namely, (1) either the prevalence of the disease or the prior probability of disease, (2) sensitivity, and (3) specificity.
Then, formulae for PV+ and PV- are:
$$PV+ = \dfrac{\text{Prevalence}\times\text{Sensitivity}}{(\text{Prevalence}\times\text{Sensitivity})+\left\{(1-\text{Prevalence})\times (1-\text{Specificity}) \right\}}$$
$$PV- = \dfrac{(1-\text{Prevalence})\times\text{Specificity}}{\left\{(1-\text{Prevalence})\times\text{Specificity})\right\}+\left\{\text{Prevalence}\times (1-\text{Sensitivity}) \right\}}$$
Although PV+ = 14/(14+8) = 0.64 and PV - = 91/(1+91) = 0.99 can be calculated directly from the 2 × 2 data table because the women constituted a random sample, the above formulae yield the same results:
$$PV+ = \dfrac{(0.13)(0.93)}{{(0.13)(0.93) + (0.87)(0.08)}} = 0.64$$
$$PV- = \dfrac{(0.87)(0.92)}{{(0.87)(0.92) + (0.13)(0.07)}} = 0.99$$
The following example is taken from Sackett et al (1985, Clinical Epidemiology ). Suppose a patient with the following characteristics visits a physician:
• 45-year-old man
• ambulatory with episodic chest pain
• no coronary risk factors except smoking one pack of cigarettes per day
• 3-week history of substernal and precordial pain - stabbing and fleeting
• physical exam shows a single costochondral junction that is slightly tender, but does not reproduce the patient's pain
From this information, the physician estimates an intermediate pre-test (prior) probability of 60% that this patient has significant coronary artery narrowing.
The physician is not sure whether the patient should undergo an exercise electrocardiogram (ECG). How useful would this test be for this patient?
Suppose it is known from the literature that the sensitivity and specificity of the exercise ECG in coronary artery stenosis (as compared to the gold standard of coronary arteriography) are 60% and 91%, respectively.
Then:
$$PV+ = \dfrac{(0.6)(0.6)}{{(0.6)(0.6) + (0.4)(0.09)}} = 0.91$$
$$PV - = \dfrac{(0.4)(0.91)}{{(0.4)(0.91) + (0.6)(0.4)}} = 0.60$$
An additional test characteristic reported in the medical literature is the likelihood ratio, which is the probability of a particular test result (+ or - ) in patients with the disease divided by the probability of the result in patients without the disease. There exists one likelihood ratio for a positive test (LR+) and one for a negative test (LR - ). Likelihood ratios express how many times more (or less) likely the test result is found in diseased versus non-diseased individuals:
$$LR+ = \dfrac{\text{Sensitivity}}{\left(1 - Specificity\right)}$$
$$LR - = \dfrac{\left(1 - \text{Sensitivity}\right)}{\text{Specificity}}$$
From the FNA study in 114 women with nonpalpable masses and abnormal mammograms, LR+ = 0.933/0.081 = 11.52 and LR - = 0.067/0.919 = 0.07. Thus, positive FNA results are 11.52 times more likely in women with cancer as compared to those without, and negative FNA results are .07 times as likely in women with cancer as compared to those without.
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility | 1,465 | 5,548 | {"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} | 4.09375 | 4 | CC-MAIN-2023-23 | latest | en | 0.92893 |
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View Full Version : Course 2 Question-How 2 Pass
Browser
08-20-2002, 12:52 PM
Given that (S 2n) - Accumulated Due value with 2n payments = 300 and (An) = 10 (Simple PV formula) find i.
The hint is to change the accumulated value to a present vlaue with 2n payments. I seem to be getting stuck on good ole Algebra. any help would be great
MathGuy
08-20-2002, 01:19 PM
Clarification Request:
You state that the first annuity is an annuity due, accumulated 2n periods, so the accumulated value is:
S<sub>2n</sub> = (1 + v + ... + v<sup>2n-1</sup>) * (1 + i)<sup>2n</sup> = 300
(This is correct right?) Anyway, the clarification is whether the second annuity you mention (A<sub>n</sub> = 10) is due or immediate. Thanks.
Browser
08-20-2002, 01:26 PM
The first is due, the second is immediate.
Bama Gambler
08-20-2002, 01:31 PM
Mathguy,
How did you use subscripts in your post?
Thanks,
Bama Gambler
MathGuy
08-20-2002, 01:38 PM
Assuming it's what I think it is:
A<sub>n</sub> = 1 + v + ... + v<sup>n-1</sup> = 10
vA<sub>n</sub> = v + v<sup>2</sup> + ... + v<sup>n</sup> = 10v
A<sub>n</sub> - vA<sub>n</sub> = 1 - v<sup>n</sup> = 10 - 10v
1 - v<sup>n</sup> = 10(1 - v) [Formula I]
We'll save this for later.
Next,
S<sub>2n</sub> = (1 + v + ... + v<sup>2n-1</sup>) * (1 + i)<sup>2n</sup> = 300
If we "de-accumulate", which means we just discount back 2n periods using the discount factor v, we get:
1 + v + ... + v<sup>2n-1</sup> = 300v<sup>2n</sup>
1 + v + ... + v<sup>n-1</sup> + v<sup>n</sup> + ... + v<sup>2n-1</sup> = 300v<sup>2n</sup>
but, the bolded terms are the same as A<sub>n</sub> = 10 from above:
10 + v<sup>n</sup> + ... + v<sup>2n-1</sup> = 300v<sup>2n</sup>
10 + v<sup>n</sup>(1 + v + ... + v<sup>n-1</sup>) = 300v<sup>2n</sup>
And there he is again!
10 + v<sup>n</sup>(10) = 300v<sup>2n</sup>
0 = 300v<sup>2n</sup> - 10v<sup>n</sup> - 10
0 = 300(v<sup>n</sup>)<sup>2</sup> - 10v<sup>n</sup> - 10
And use the handy dandy quadratic formula, with v<sup>n</sup> = x:
v<sup>n</sup> = (10 + sqrt(100 - 4(300)(-10)))/(2(300)) = (10 + 110)/600 = 120/600 = .20.
Now, we take Formula I from way back at the beginnning and plug in v<sup>n</sup> = .20:
1 - v<sup>n</sup> = 10(1 - v)
1 - .20 = 10(1 - v)
.8 = 10(1 - v)
.08 = 1 - v
v = .92
(1 + i)<sup>-1</sup> = .92
(1 + i) = .92<sup>-1</sup>
1 + i = 1.087
i = .087
MathGuy
08-20-2002, 01:40 PM
Subscripts: use the html tags <sub> and </sub> around what ever you'd like to have as a subscript. (Same for sup).
If the second is immediate I'll have to think a little more.
Browser
08-20-2002, 02:00 PM
With my inproved notational abilites the problem reads:
..
S<sub>2n</sub> = 300 (Due)
And
a<sub>n</sub>=10 (Immediate)
find i
MathGuy
08-20-2002, 02:10 PM
This solution will be similar to my previous one:
a<sub>n</sub> = v + ... + v<sup>n</sup> = 10
va<sub>n</sub> = v<sup>2</sup> + ... + v<sup>n+1</sup> = 10v
v - v<sup>n+1</sup> = 10 - 10v FORMULA I
The "de-accumulation" of S<sub>2n</sub> works the same as above, and we get to:
1 + (v + v<sup>2</sup> + ... + v<sup>n</sup>) + v<sup>n+1</sup> + ... + v<sup>2n-1</sup> = 300v<sup>2n</sup>
1 + (10) + v<sup>n+1</sup> + ... + v<sup>2n-1</sup> = 300v<sup>2n</sup>
11 +v<sup>n+1</sup> + ... + v<sup>2n-1</sup> = 300v<sup>2n</sup>
11 + v<sup>n</sup>(v + ... + v<sup>n-1</sup>) = 300v<sup>2n</sup>
11 + v<sup>n</sup>(10 - v<sup>n</sup>) = 300v<sup>2n</sup>
11 - 10v<sup>n</sup> - 2v<sup>2n</sup> = 300v<sup>2n</sup>
0 = 302v<sup>2n</sup> - 10v<sup>n</sup> - 11
Again, using the quadratic formula, we get v<sup>n</sup> = .2081
And, using Formula I from above:
v - v<sup>n+1</sup> = 10 - 10v
v(1 - v<sup>n</sup>) = 10 - 10v
v(1 - .2081) = 10 - 10v
.7919v = 10 - 10v
10.7919v = 10
v = 10/10.7919
1/(1+i) = .9266
1 + i = 1.079
i = .079
Browser
08-20-2002, 02:18 PM
c3 taker
08-20-2002, 03:34 PM
And, using Formula I from above:
v - v<sup>n+1</sup> = 10 - 10v
v(1 - v<sup>n</sup>) = 10 - 10v
v(1 - .2081) = 10 - 10v
.7919v = 10 - 10v
10.7919v = 10
v = 10/10.7919
1/(1+i) = .9266
1 + i = 1.079
i = .079
Rather than use formula 1 you could also just say that since a<sub>n</sub> = 10 = (1 - v<sup>n</sup>) / i
then i = (1 - .2081) / 10 = .079
same premise, I find it's easier to remember this way (if you know the formula for an annuity-immediate, which I assume you won't do very well on interest theory without) :D
RiSK kid
08-20-2002, 04:45 PM
Subscripts: use the html tags <sub> and </sub> around what ever you'd like to have as a subscript. (Same for sup).
Amazing, I totally forgot that we could use HTML. This is gonna make the questions a lot more readable.
Thanks
MathGuy
08-21-2002, 08:30 AM
C2 taker: Formulas are for babies. I prefer the all-out guerilla-math style: Everything on first principles!!! For Interest Theory, this is particularly useful because the entire topic is light and fluffy, but "the man" weighs it down with all sorts of formulas and stuff. If you can write down 1 + v + ... + v<sup>n</sup>, and you truly understand what v means, you can answer any question about annuities that comes up. Sure it makes for a crazy four hours come exam time, but it means that there is no studiying required.
c3 taker
08-21-2002, 09:17 AM
C2 taker: Formulas are for babies. I prefer the all-out guerilla-math style: Everything on first principles!!! For Interest Theory, this is particularly useful because the entire topic is light and fluffy, but "the man" weighs it down with all sorts of formulas and stuff. If you can write down 1 + v + ... + v<sup>n</sup>, and you truly understand what v means, you can answer any question about annuities that comes up. Sure it makes for a crazy four hours come exam time, but it means that there is no studiying required.
I understand the concept of 1 + v + ... + v<sup>n</sup>, I just think it's easier to remember a few formulas, but maybe I'm just frightened by your "all-out guerilla-math style" :D | 2,281 | 5,917 | {"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} | 4.0625 | 4 | CC-MAIN-2013-20 | latest | en | 0.691433 |
https://rdrr.io/github/jogrue/jogRu/man/get_CI.html | 1,558,365,203,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256040.41/warc/CC-MAIN-20190520142005-20190520164005-00379.warc.gz | 616,154,670 | 13,534 | # get_CI: Compute confidence interval In jogrue/jogRu: Johann Gründl's collection of R functions
## Description
Computes the confidence interval based on Student's t-distribution or the standard normal distribution. Also reports descriptive statistics (such as the mean). Missings are ignored.
## Usage
```1 2 3``` ```get_CI(x, conf.level = 0.95, dist = "t", incl_n = TRUE, incl_mean = TRUE, incl_SD = TRUE, incl_SE = TRUE, incl_error = TRUE) ```
## Arguments
`x` A numeric (or logical) vector (such as a variable) `conf.level` Confidence level of the interval. Defaults to 0.95. `dist` Distribution the confidence interval should be based on. "t" or "student" (default) for Student's t-distribution, "z" or "normal" for standard normal distribution. `incl_n` A locigal indicating whether the number of observations should be included in the returned results. Defaults to TRUE. `incl_mean` A locigal indicating whether the mean should be included in the returned results. Defaults to TRUE. `incl_SD` A locigal indicating whether the standard deviation should be included in the returned results. Defaults to TRUE. `incl_SE` A locigal indicating whether the standard error should be included in the returned results. Defaults to TRUE. `incl_error` A locigal indicating whether the error based on the confidence level (i.e., the number that is substracted or added to the mean lower and upper bounds of the confidence interval) should be included in the returned results. Defaults to TRUE.
## Value
A numeric vector with at least the lower and upper bounds of the confidence interval.
## Examples
```1 2 3 4 5 6 7``` ```## The height of ten people in cm heights <- c(160, 163, 168, 169, 174, 176, 177, 178, 182, 190) get_CI(heights) # Defaults get_CI(heights, dist = "normal") # Using standard normal distribution get_CI(heights, conf.level = 0.9) # 90%-CI get_CI(heights, incl_SD = FALSE) # Do not include standard deviation ```
jogrue/jogRu documentation built on Sept. 26, 2018, 6:58 a.m. | 510 | 2,003 | {"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-2019-22 | latest | en | 0.801329 |
https://justaaa.com/chemistry/150095-the-rate-constant-for-the-reaction-below-was | 1,723,111,362,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640726723.42/warc/CC-MAIN-20240808093647-20240808123647-00856.warc.gz | 258,495,609 | 10,258 | Question
# The rate constant for the reaction below was determined to be 3.241×10-5 s–1 at 800 K....
The rate constant for the reaction below was determined to be 3.241×10-5 s–1 at 800 K. The activation energy of the reaction is 215 kJ/mol. What would be the value of the rate constant at 9.10×102 K? N2O(g) --> N2(g) + O2(g)
I'm having trouble calculating the rate constant with the arrhenius equation that deals with two temps, could you show me the step by step how to do this?
Ans :- 1.6 x 10-3 s-1
Explanation :-
Given,
rate constant at 800 K = k1 = 3.241 x 10-5 s-1
Temperature = T1 = 800 K
rate constant at 9.10 x 102 K = k2 =?
Temperature = T2 = 9.10 x 102 K
Activation energy = Ea = 215 KJ = 215000 J/mol
From the Arrhenius equation
ln (k2/k1) = (Ea/R) . (1/T1 - 1/T2)
ln (k2/k1) = 215000 J/mol / 8.314 J K-1 mol-1 .(9.10 x 102 - 800 K / (800 K)(9.10 x 102)
ln (k2/k1) = 3.9074
k2 = k1 x exp ( 3.9075)
k2 = 3.241 x 10-5 x 49.7694
k2 = 1.61 x 10-3 s-1 | 377 | 977 | {"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-2024-33 | latest | en | 0.837286 |
https://ubraintv-jp.com/how-much-does-1-l-of-water-weigh/ | 1,638,033,189,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00181.warc.gz | 655,143,891 | 4,790 | ## How plenty of pounds go 1l that water weigh?
2.21 poundsOne liter that water weighs about 2.21 pounds.
You are watching: How much does 1 l of water weigh
## What walk 1l that milk weigh?
The thickness of milk is about 1.03 kilograms per litre so a litre that milk weighs an extremely close to 1 kilogram.
## How to calculation the load of 1L of water?
Example: because that 1L of water at room temperature. 1 transform your 1L number to mL Volume = 1 × 1000 = 1000 mL. 2 weight in grams = volume × density 1000 × 0.99802 = 998.02g. 3 transform grams to pounds.
## Which is much more 1 kg or 1 liter of water?
1 liter of water is not always 1 kg. The liter is one unit the volume and also as temperature fluctuates, 1 liter that water have the right to contain an ext or much less water molecule which have actually a massive of more or less than 1 kg. Let’s occupational backwards native the massive of 1 kg.
See more: How Do You Gift A Song On Itunes, How To Send Itunes Songs And Albums As A Gift
## How lot does one gram of water weigh?
1 gram is same to 0.035274 ounces for this reason to gain a result in ounces just multiply the grams by 0.035274. Friend can likewise use our weight conversion calculators to transform from grams and kilograms come pounds and also ounces. Weight of Water for various Volumes
## How much does one teaspoon of water weigh?
Weight the water by unit the measure. Volume. Weight (oz) load (lb) load (g) weight (kg) 1 teaspoon. 0.1739 oz. 0.0109 lb.
## How much does 1 litre of water sweet in kilograms?
The density of water is 1 kilogram per liter (kg/L) in ~ 39.2°. This way that 1 liter (L) the water weighs 1 kilogram (kg) and 1 milliliter (mL) the water weighs 1 gram (g). In usual US measures, one gallon of water weighs 8.345 pounds.
## What is the molarity that 1L that water?
Molarity is no the moles every litre.1g/ml density means that 1L (=1000 ml) weighs 1000g. So one litre of water has 55.55 mol water, and so molarity if water is 55.55 molar.
## How much weight does 1 gallon the distilled water weigh?
A gallon the distilled/fresh water weighs 8.35 lb. A gallon of sea/ ocean water weighs 8.556 lb. A gallon that saturated sugar water weighs around 8.42 lb. A gallon of ice or frozen water weighs about 7.71 lb. In bespeak to transform Gallon into liters, the complying with are the formulae for both US and Imperial gallon.
## What is the fixed of 1L of water?
One litre the water has a mass of virtually exactly one kilogram as soon as measured at its maximal density, which occurs at about 4 °C. Similarly: 1 millilitre the waterhas a mass of around 1 g; 1,000litres the water has a massive of around 1,000 kg (1 tonne).The weight of Water. | 739 | 2,718 | {"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} | 3.890625 | 4 | CC-MAIN-2021-49 | latest | en | 0.91521 |
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Minimal weight in union-closed families
School of Mathematical Sciences Queen Mary University of London, London E1 4NS, UK.
2011 (English)In: The Electronic Journal of Combinatorics, ISSN 1097-1440, E-ISSN 1077-8926, Vol. 18, no 1, p. P95-Article in journal (Refereed) Published
##### Abstract [en]
Let Omega be a finite set and let S subset of P(Omega) be a set system on Omega. For x is an element of Omega, we denote by d(S)(x) the number of members of S containing x.Along-standing conjecture of Frankl states that if S is union-closed then there is some x is an element of Omega with d(S)(x)>= 1/2|S|. We consider a related question. Define the weight of a family S to be w(S) := A.S|A|.SupposeSisunion-closed. How small can w(S) be? Reimer showed w(S) >= 1/2|S|log(2)|S|, and that this inequality is tight. In this paper we show how Reimer's bound may be improved if we have some additional information about the domain Omega of S: if S separates the points of its domain, then w(S) >= ((vertical bar Omega vertical bar)(2)). This is stronger than Reimer's Theorem when |Omega| > root|S|log(2)|S|. In addition we constructa family of examples showing the combined bound on w(S)istightexcept in the region |Omega| = Theta(root|S|log(2)|S|), where it may be off by a multiplicative factor of 2. Our proof also gives a lower bound on the average degree: if S is a point-separating union-closed family on Omega, then 1/ |Omega|Sigma(x is an element of Omega)d(S)(x)>= 1/2 root|S|log(2)|S| broken vertical bar O(1), and this is best possible except for a multiplicative factor of 2.
##### Place, publisher, year, edition, pages
Electronic Journal of Combinatorics , 2011. Vol. 18, no 1, p. P95-
##### National Category
Discrete Mathematics
##### Identifiers
OAI: oai:DiVA.org:umu-80520DiVA, id: diva2:649983
Available from: 2013-09-19 Created: 2013-09-19 Last updated: 2018-06-08Bibliographically approved
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Falgas-Ravry, Victor
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• rtf | 862 | 2,778 | {"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-2020-16 | latest | en | 0.741152 |
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HP Forum Archive 18
Re: hyperbolic curiosityMessage #2 Posted by Kiyoshi Akima on 14 Oct 2008, 12:44 p.m.,in response to message #1 by Bill Triplett Try this: ```.243 53.5 D->R R->C SINH ``` On my 50g I get (.146, .828) just as in the article. The problem is sinh(u+jt) with u=.243 and t=53.5degrees ~ .934 rads so we want to compute sinh(.243, .934) . kiyoshi
Re: hyperbolic curiosityMessage #3 Posted by Bill Triplett on 14 Oct 2008, 1:26 p.m.,in response to message #2 by Kiyoshi Akima Thanks! That works. I had forgotten that only the second number should be converted from degrees into radians. Now I am curious to see which other calculators can perform the same function. The TI-89 seems like a possible "no" but I am not certain yet. Imagine needing to spend half of an hour, and perform hand calculations using pencil and paper (after consulting tables) in order to do this for a single case.
Re: hyperbolic curiosityMessage #4 Posted by Egan Ford on 14 Oct 2008, 1:49 p.m.,in response to message #3 by Bill Triplett Quote: Now I am curious to see which other calculators can perform the same function. 15C (circa 1985): ```SF 8 ; Set Complex Mode .243 ENTER 53.5 f I ; Enter Complex Number f HYP SIN ; sinh, (use Re<>Im to see both parts) ``` returns: -0.2443, -0.0955 ```SF 8 ; Set Complex Mode .243 ENTER 53.5 ->RAD ; Enter Complex Parts, Convert Im to RAD f I ; Create Complex Number f HYP SIN ; sinh, (use Re<>Im to see both parts) ``` returns: 0.1460, 0.8277 Get a 15C. :-)
Re: hyperbolic curiosityMessage #5 Posted by Bill Triplett on 14 Oct 2008, 2:04 p.m.,in response to message #4 by Egan Ford A beautiful lady surprised me by giving me a 15C recently after she had seen some of the other calculator's in my collection. I will learn how to use it a little more each day. Now I know how to set the flag to enable complex mode. The little 15C could qualify as being the smallest physical machine that can perform the function, even after all these years. The Pickett N4P-ES slide rule either comes close, or it could be the winner in terms of mass. The N4P-ES is a shirt pocket sized hyperbolic slide rule that is similar to the N600-ES that was carried on board five Apollo missions. The N600-ES had slightly less mass, but it did not include hyperbolic scales. On Apollo 13, they just needed to do circular trig functions to navigate, so it worked well enough when they were in a bizarre situation.
Re: hyperbolic curiosityMessage #6 Posted by Bill Triplett on 15 Oct 2008, 12:04 a.m.,in response to message #4 by Egan Ford The test problem is sinh(0.243 + i*0.93375). The answer should be (0.14597 + i*0.82771). I tried reading my HP-41c manual. There does not seem to be any way to compute the sinh(x+iy) function. Anyone more familiar with the machine, please correct me if this is not true. In my collection, so far I have used the following machines to work the example successfully: HP-15c HP-39gs HP-40gs (same machine with CAS) HP-48gx HP-49g+ (good version with excellent keyboard) My TI-89 and TI Voyage 200 can do this, but only in radians mode. Both my TI-nspire machines can do this, with or without CAS. My TI-84+SE refused to accept anything except real numbers as inputs for the sinh function. I was able to track it down, and retrieve it from where it had crawled away to hide under a book, possibly to escape from this problem. My TI-59 battery pack has finally gone flat line, so I have no way to test that until after a bit of surgery. Note. Never turn on a TI-59 while it is connected a power supply with no working battery pack installed to smooth out the rectified AC inside the calculator. Always leave the calculator power switch off (with tape) while charging, and then remove the power supply before turning the power on. This will prevent a boat load of problems from ever developing with the card reader's rate calibration. I have a cheap little Casio FX-115ES that can multiply and divide complex numbers, but it does not know how to use complex numbers as inputs for trig functions, either circular or hyperbolic. It isn't half bad for fifteen bucks, but it would have been nicer if the designers had fully supported complex number operations. So far, it looks as if the least expensive machine that can fully support complex numbers might be the HP-39gs, unless there is some other Casio or similar machine out there that I do not know about.
Re: hyperbolic curiosityMessage #7 Posted by Walter B on 15 Oct 2008, 2:12 a.m.,in response to message #6 by Bill Triplett FWIW, I brought your problem to a Casio fx-991ES, and it caused a Math Error there. Looks like this machine is little better than your "cheap little Casio FX-115ES". Ceterum censeo: HP, launch a 43s. Walter
Cheap little Casio FX-115ES Message #8 Posted by Karl Schneider on 15 Oct 2008, 3:02 a.m.,in response to message #6 by Bill Triplett Quote: I have a cheap little Casio FX-115ES that can multiply and divide complex numbers, but it does not know how to use complex numbers as inputs for trig functions, either circular or hyperbolic. It isn't half bad for fifteen bucks, but it would have been nicer if the designers had fully supported complex number operations. The Casio FX-115ES is a slightly-improved successor to the Casio FX-115MS, which I also bought a few years ago for \$15: -- KS
Re: hyperbolic curiosityMessage #9 Posted by Jeff O. on 14 Oct 2008, 12:56 p.m.,in response to message #1 by Bill Triplett My 42S tells me that ASINH(0.14597 + i 0.82771) = 0.243 + i 0.9338. I noted 0.9338 radians equals 53.5 degrees. The real part matches the real part of the original argument and converted to degrees, it would be 13.923. So is it possible that the author started out with (13.923 + i 53.5) degrees, converted that to (0.243 + i 0.9388) radians, found the result on his slide rule, but then mis-wrote the original argument, mixing the real part in radians with the imaginary part in degrees? Or did the slide rule "expect" the arguments to be input that way? (I'm just speculating, I did not consult the article.)
Re: hyperbolic curiosityMessage #10 Posted by Bill Triplett on 14 Oct 2008, 1:39 p.m.,in response to message #9 by Jeff O. Good point. It looks as if the convention back then was to use polar coordinates. I do not have a hyperbolic slide rule to check it out with yet, but I found this amazing "calculator" emulator: http://www.antiquark.com/sliderule/sim/n4es/virtual-n4es.html When you go to this URL, you can click and drag on the objects to make the simulation work. It looks as if the convention was to provide scales for SINH() from about 0.1 degree to 3.0 degrees. If you needed to perform SINH() for smaller angles, then the angle in degrees would be approximately equal to the SINH value. If you needed to compute SINH() for larger angles, the answer would be approximately equal to e to the x quantity divided by two, so you would just use the logarithm scales for larger angles.
Re: hyperbolic curiosityMessage #11 Posted by Nigel J Dowrick on 14 Oct 2008, 2:27 p.m.,in response to message #1 by Bill Triplett I suspect you are maligning the TI-89! I don't have mine here, but my TI-Nspire (stop laughing!) is quite prepared to find the hyperbolic sine of a complex argument. (It gives the same result as the HP 49/50.) With the TI-89, did you (a) have it in a complex mode, and (b) have it in radians? Sometimes this matters when working with complex numbers on the TI. Whatever you think about TI calculators, the TI-85, 86 and 89 are very capable machines. I would be astonished if the TI-89 couldn't cope with this!
Re: hyperbolic curiosityMessage #12 Posted by Bill Triplett on 14 Oct 2008, 3:37 p.m.,in response to message #11 by Nigel J Dowrick I have an nspire without CAS (blue case) around here somewhere. Perhaps it is hiding from this problem. Is your nspire the blue or the gray CAS machine? I also have an 84+SE, 85, 86 (same machine, more memory) 89, 92+, and Voyage 200 (same machine with more memory and cramped keyboard). I suspect that my venerable old TI-59 might be able to support this function with programs that are included in the standard library module that shipped with every machine. I do not know about the HP-41C standard library (yet), but I would imagine that this complex hyperbolic functionality would have been included in the thousands of programs that have been available for that machine from many sources. I wonder how well the 41C compares to the little 15C in this respect. So far, I have only been able to get my Voyage 200 to give me a domain error, and of course, the same thing happens on the 89. I don't have any of the newer Casio super machines around here, but I wonder whether any of those calculators can do this. Perhaps a Classpad, or an Algebra FX would work. It still blows my mind that the little 15C could do this so long ago, and this was a built in function, not an add-on program.
Re: hyperbolic curiosityMessage #13 Posted by Walter B on 14 Oct 2008, 3:52 p.m.,in response to message #12 by Bill Triplett Quote: It still blows my mind that the little 15C could do this so long ago Those were the days, my friend, where calculators had to do almost everything, since PCs were not capable yet doing reasonable stuff though very expensive, and you wouldn't like to run to a mainframe terminal for every small calculation. Ceterum censeo: HP, launch a 43s. Walter
Those were the days ... 15C/35sMessage #14 Posted by Karl Schneider on 15 Oct 2008, 2:05 a.m.,in response to message #13 by Walter B Hi again, Walter -- Quote: It still blows my mind that the little 15C could do this so long ago Those were the days, my friend, (...) Alas, they did. See Egan's post below. -- KS Edited: 15 Oct 2008, 3:06 a.m.
Re: hyperbolic curiosityMessage #15 Posted by Egan Ford on 14 Oct 2008, 4:24 p.m.,in response to message #12 by Bill Triplett Quote: It still blows my mind that the little 15C could do this so long ago, and this was a built in function, not an add-on program. What blows my mind is that I cannot do this on the state-of-the-art circa 2007 35s. Correct me if I'm wrong people, but I have 0.23400i0.93375 on the stack and HYP SIN returns INVALID DATA.
HP-35s hyperbolic with complexMessage #16 Posted by Karl Schneider on 15 Oct 2008, 2:19 a.m.,in response to message #15 by Egan Ford Hi, Egan -- Quote: What blows my mind is that I cannot do this on the state-of-the-art circa 2007 35s. Correct me if I'm wrong people, but I have 0.23400i0.93375 on the stack and HYP SIN returns INVALID DATA. You're not wrong, and I've mentioned a few times in the past that complex-number functionality on the HP-35s is mathematically incomplete: Hyperbolics, x2 and sqrt, inverse trigonometrics, and other functions are not supported. This shortcoming can be traced historically as follows: HP-41 Math Pac (1980) --> HP-32S (1988) --> HP-32SII (1991) --> HP-33s (2004) --> HP-35s (2007) -- KS Edited: 15 Oct 2008, 3:06 a.m.
Re: HP-35s hyperbolic with complexMessage #17 Posted by Egan Ford on 15 Oct 2008, 10:31 a.m.,in response to message #16 by Karl Schneider Quote: HP-41 Math Pac (1980) --> HP-32S (1988) --> HP-32SII (1991) --> HP-33s (2004) --> HP-35s (2007) I'd assumed (sans Math Pac) that was the case or perhaps 11C -> 32S. Sadly, from my point of view: ```41CX ---. .-----> 28 -----> 48 (RPL Fork) >------> 42S ------< 15C ---' '-----> NONEXISTENT (RPN Fork) ```
Re: HP-35s hyperbolic with complexMessage #18 Posted by Karl Schneider on 16 Oct 2008, 12:02 a.m.,in response to message #17 by Egan Ford Egan -- I failed to clearly state that I was referring only to complex-number functionality, not the developmental lineage of the respective models. I believe that the complex-number functionality of the HP-41 Math Pac was the template for what was implemented in the HP-32S. The similarities -- what's present and what isn't -- are too strong to ignore. It's a usable capability that is not complete like that of the HP-42S. Had to preserve that product differentiation for the considerable difference in price! A small quibble with your tree: The HP-28C and HP-28S preceded the HP-42S by a year or two, so the developmental process overlapped. The HP-42S had the capability of displaying and assembling complex numbers in polar form, while the HP-28 supported only rectangular form, despite its dot-matrix display. The subsequent HP-48 (1990) brought the RPL branch up to par in that respect.
How to do complex hyperbolic functions on calculators that have limited support for complex numbersMessage #19 Posted by Bill Triplett on 16 Oct 2008, 9:43 p.m.,in response to message #15 by Egan Ford Karl covered some of this in an earlier post. I don't have an HP-35s to try this with, but it should cover all of the bases for that and some of the Casio machines: sinh(x + i y) = sinh(x) cos(y) + i cosh(x) sin(y) cosh(x + i y) = cosh(x) cos(y) + i sinh(x) sin(y) tanh(x + i y) = (tanh(x) + i tan(y))/(1 + i tanh(x) tan(y)) arcsinh(x + i y) = ln((x + i y) + sqrt((x + i y) (x + i y) + 1) arccosh(x + i y) = ln((x + i y) + sqrt((x + i y) (x + i y) - 1) arctanh(x + i y) = (1/2)*ln((1 + (x + i y))/(1 - (x + i y))) ln(x + i y) = ln(sqrt(p*p+q*q)) + i ((pi*q)/(2*abs(q)) - arctan(p/q)) I hope the format is readable.
Re: How to do complex hyperbolic functions on calculators that have limited support for complex numbersMessage #20 Posted by Bill Triplett on 16 Oct 2008, 9:49 p.m.,in response to message #19 by Bill Triplett I used incorrect variables for the last entry for complex logarithms. The correct form should be as follows: ln(x + i y) = ln(sqrt(x*x+y*y)) + i ((pi*x)/(2*abs(y)) - arctan(x/y)) No jokes about minding my p's and q's.
Re: How to do complex hyperbolic functions on calculators that have limited support for complex numbersMessage #21 Posted by Bill Triplett on 16 Oct 2008, 9:54 p.m.,in response to message #20 by Bill Triplett Good grief, this is just one more step of correction: ln(x + i y) = ln(sqrt(x*x+y*y)) + i ((pi*y)/(2*abs(y)) - arctan(x/y)) This one seems correct, but I would appreciate it if anyone would try it independently, and check the result, especially using an HP-35s against something like a 49g+, or 50g. Grrr. Now, we can see why it is nice to have this functionality already built into a machine.
LN of complex number, if not built-inMessage #22 Posted by Karl Schneider on 18 Oct 2008, 12:01 a.m.,in response to message #21 by Bill Triplett Bill -- Quote: Good grief, this is just one more step of correction: ln(x + i y) = ln(sqrt(x*x+y*y)) + i ((pi*y)/(2*abs(y)) - arctan(x/y)) This one seems correct, but I would appreciate it if anyone would try it independently, (...) A clearer rearrangement is: ln(x + i y) = ln(sqrt(x*x+y*y)) + i ((pi/2)*(y/abs(y)) - arctan(x/y)) A further simplification is use of the signum function, but it is unavailable on any H-P model that lacks built-in complex-number support, so that's a moot issue. ln(x + i y) = ln(sqrt(x*x+y*y)) + i ((pi/2)*(sign(y)) - arctan (x/y)) Both of these expressions would cause a divide-by-zero error for y = 0. However, the expression is unnecessary for that case, which indicates a real-valued input. Those who are familiar with Fortran and C/C++ language might construct it using the atan2 function as: ln(x + i y) = ln(hypot(x,y)) + i (atan2(y,x)) "hypot" and "atan2" are the functions of rectangular-to-polar conversions. Here, "angle" is returned in radians, and a "ln" is also incorporated. It's not surprising that calculators would use the same code for both. Have a look at this archived post of mine: However, it's not simply conversion per se; the form of the output is still rectangular. So, on the HP-11C, for example, ln(x + i y) would be: ```(y) ENTER (x) RAD ->P LN ``` In summary, on a calculator without complex-number support, rectangular->polar conversions, or signum, I'd use equation 1, with an exception for y = 0 to return ln(x) in place of x. What do you think might be the best way to calculate the polar-format natural logarithm of a polar-format complex number? Don't forget to put the input number into proper form, don't neglect input and output angular mode, and don't do any unnecessary calculations! Quote: Now, we can see why it is nice to have this functionality already built into a machine. Yes, indeed... ;-) -- KS Edited: 20 Oct 2008, 1:28 a.m. after one or more responses were posted
Re: LN of complex number, if not built-inMessage #24 Posted by Pal G. on 19 Oct 2008, 12:46 a.m.,in response to message #23 by Bill Triplett Quote: I am wondering how I would ask my HP-49g+ CAS system to symbolically display the expression See the TRIG command. Hope this helps, PG
Re: LN of complex number, if not built-inMessage #25 Posted by Bill Triplett on 19 Oct 2008, 1:02 a.m.,in response to message #24 by Pal G. Under the TRIG menu, I tried TEXPAND(LN(X+iY)) with no luck. It just responded with LN(X+iY).
Re: Kinda OT: InnovationMessage #26 Posted by Pal G. on 19 Oct 2008, 11:18 a.m.,in response to message #25 by Bill Triplett Quote: Under the TRIG menu, I tried TEXPAND(LN(X+iY)) with no luck. It just responded with LN(X+iY). Try the TRIG *command* which can be found in the Catalog (or under the "Trig" menu). Best, PG Edited: 19 Oct 2008, 12:35 p.m.
Re: LN of complex number, if not built-inMessage #27 Posted by Bill Triplett on 19 Oct 2008, 1:23 a.m.,in response to message #24 by Pal G. I ran this past my Casio Classpad. Like the HP-49g+, it would only respond to ln(x+iy) with ln(x+iy) until I found the classpad's cExpand function. Using cExpand(ln(x+iy)) produces an expression for complex logarithm that also uses arctan(x/y) and a signum function to get the angle into the correct quadrant. It looks sightly different in format, but it is essentially the same expression that is output by both of the TI CAS machines. There probably is something similar for the HP-49g+, HP-50g, or other HP CAS machines. I am just not finding it right now.
Re: LN of complex number, if not built-inMessage #28 Posted by Karl Schneider on 19 Oct 2008, 4:56 p.m.,in response to message #23 by Bill Triplett Bill -- Quote: In the expression for natural logarithm, the angle for the imaginary part would be simply the arctan of y over x, but that would not work in every quadrant. To fix that, and make it work for other quadrants, the expression uses arctan of x over y intentionally. Mea culpa about the input argument to arctan. I have corrected my earlier post. Quote: In the quadrants of interest, (90) degrees minus the arctan of x over y is the same thing as the simpler arctan of y over x, except that when combined with the term for the signum of y the larger expression can work for all four quadrants. Except when y = 0 (for which the expression is unneeded), as my earlier response illustrated. I'd stick with rectangular->polar conversions with internal "atan2" to implement ln(x+ iy) in a program without complex-number support available. About the only H-P scientifics without rectangular<->polar conversions for real-valued inputs -- ironically enough -- are the old HP-35 (which is nonprogrammable), and the new HP-35s (which can compute natural logarithm for complex-valued input). -- KS
Re: LN of complex number, if not built-inMessage #29 Posted by Bill Triplett on 19 Oct 2008, 1:03 a.m.,in response to message #22 by Karl Schneider I should have said 90 degrees. Now I can go back to sleep.
Re: hyperbolic curiosity - TIMessage #30 Posted by Marcus von Cube, Germany on 14 Oct 2008, 4:39 p.m.,in response to message #12 by Bill Triplett My Voyage 200 does it. I had to set it to complex (rectangular) mode and could type in: sinh(.234+i*53.5°) and got the correct result. Marcus
Re: hyperbolic curiosity - TIMessage #31 Posted by Bill Triplett on 14 Oct 2008, 6:29 p.m.,in response to message #30 by Marcus von Cube, Germany I tried that, and it did not work. Then, I also changed from degrees mode to radians mode. That fixed it. The Voyage 200 has a big screen, so there should be plenty of room for an error message that says "I cannot do this unless you put me into radians mode." Instead, the error message says "Domain error." Thanks for helping me to keep trying. I am quite fond of my Voyage 200. Yes, it has many odd quirks, but in general I find it very easy to use, self explanatory, especially in writing quick programs to analyze data without ever needing to worry about whether the program will do what is expected on the first run. The TI-84 does not seem capable of doing this function. I managed to coax my blue nspire out from behind a desk where it had been hiding, and it was able to handle complex numbers in the sinh function. The gray nspire should be equally capable. No word is available yet on whether a Casio can do this. I wonder, what would be the least expensive machine that can do this function, of any brand?
Re: hyperbolic curiosity - TIMessage #32 Posted by Marcus von Cube, Germany on 15 Oct 2008, 3:26 a.m.,in response to message #31 by Bill Triplett I tried it on my classpad 300 plus and it worked. Settings are RAD and COMPLEX. I had to use the approx icon in the upper left corner of the screen to see a numerical result. What bothers me is the reasoning behind a complex number with an angle in degrees as the imaginary part. I can't see any use for such a number.
HP-41 hyperbolics and complex numbersMessage #33 Posted by Karl Schneider on 15 Oct 2008, 2:28 a.m.,in response to message #12 by Bill Triplett Hi again, Bill -- Quote: I do not know about the HP-41C standard library (yet), but I would imagine that this complex hyperbolic functionality would have been included in the thousands of programs that have been available for that machine from many sources. I wonder how well the 41C compares to the little 15C in this respect. The HP-41 Math Pac ROM of 1980 provided rudimentary hyperbolics and complex-number functionality for the HP-41, implemented as RPN keystroke programs. The Math Pac is also available on the combined-module HP-41 Math/Stat Pac. Almost all of the math Pac complex-number functionality was ported to the HP-41 Advantage Pac. Here are several posts of mine from a few years ago: -- KS Edited: 15 Oct 2008, 2:57 a.m.
Re: hyperbolic curiosityMessage #34 Posted by Egan Ford on 15 Oct 2008, 10:45 a.m.,in response to message #12 by Bill Triplett Quote: I wonder how well the 41C compares to the little 15C in this respect. IMHO, the 41C family when initially released was more computer than calculator. And like a computer you need to tailor it for your purpose with the addition of software. That said, take a look at 41Z (2005). This modern module for the 41C family adds 15C/42S-like complex number support. E.g.: ``` 15C 41C/41Z --------- ----------------- OFF SF 8 Insert 41Z Module ON .243 .243 ENTER ENTER 53.5 53.5 ->RAD D-R f I X<>Y f HYP SIN ZSINH ``` 41Z similar to the 42S will display a nicely formatted: ```Z:0.1460+0.8277J ```
HP-41ZMessage #35 Posted by Karl Schneider on 16 Oct 2008, 12:15 a.m.,in response to message #34 by Egan Ford Egan -- Is the 41Z module Angel Martin's "Sandbox ROM" project? If you follow the "Hyperbolics with Advantage" thread linked in my post directly above yours, you'll see where he was working on it in 2004. I'm not sure how feasible it would have been, but maybe an HP-15C style "parallel imaginary stack" solution would have worked for the HP-41. This would have provided a full 4-level stack with complex-valued arguments. Quote: 41Z similar to the 42S will display a nicely formatted: ```Z:0.1460+0.8277J ``` I assume that the displayed value is constructed in the ALPHA register, and that the two components are in the x- and y-stack registers. The "J" ought to precede the imaginary-part component (per electrical engineering and HP-42S convention), lest it be overlooked. "J" stands out better than "I", particularly without good lowercase letters. -- KS
Re: HP-41ZMessage #36 Posted by Egan Ford on 16 Oct 2008, 12:56 a.m.,in response to message #35 by Karl Schneider Quote: Is the 41Z module Angel Martin's "Sandbox ROM" project? No, its a standalone module. Quote: I'm not sure how feasible it would have been, but maybe an HP-15C style "parallel imaginary stack" solution would have worked for the HP-41. This would have provided a full 4-level stack with complex-valued arguments. Angel mentions in the docs that could have been done, but was not. Quote: I assume that the displayed value is constructed in the ALPHA register, and that the two components are in the x- and y-stack registers. Correct. Z functions end with a call to ZVIEW. Read the 41Z PDF from TOS, its quite impressive.
Re: HP-41ZMessage #37 Posted by Walter B on 16 Oct 2008, 1:07 a.m.,in response to message #35 by Karl Schneider Karl, Quote: "J" stands out better than "I", particularly without good lowercase letters. Don't cure the symptoms, fight the root cause. Create good lowercase letters or other suitable layouts to let the "i" show up clearly. Ceterum censeo: HP, launch a 43s. Walter
Re: hyperbolic curiosityMessage #38 Posted by Nigel J Dowrick on 15 Oct 2008, 11:59 a.m.,in response to message #12 by Bill Triplett My Inspire is also the non-CAS variety (blue case). My Casio fx-9860G slim won't do this problem; arithmetic, logarithms, and exponentials are the only functions that accept complex arguments. (Which is rather silly: if you are implementing complex exponentials and logarithms, the extension for trig/hyperbolic functions is simple.) So the TI's do the problem (when set to radians mode); Casio calculators don't seem to (I don't know about the Classpad); some HPs do and some don't. I wonder why TI doesn't make more of this in its advertising? People who buy scientific calculators have probably heard of complex numbers; the ability to handle complex arguments correctly would surely be a selling point, whether or not the ability was actually needed!
Re: hyperbolic curiosityMessage #39 Posted by Bill Triplett on 14 Oct 2008, 8:21 p.m.,in response to message #11 by Nigel J Dowrick I found that the following formula works for any scientific calculator that has hyperbolic functions: Sinh(x+iy) = Sinh(x)Cos(y)+(i)Cosh(x)Sin(y) It requires a few more steps, but it can get the job done. If you find other machines, TI, HP, Casio, or others that do this directly as a built in function, please let me know.
Re: hyperbolic curiosityMessage #40 Posted by Palmer O. Hanson, Jr. on 15 Oct 2008, 2:59 a.m.,in response to message #39 by Bill Triplett Quote: If you find other machines, TI, HP, Casio, or others that do this directly as a built in function, please let me know. I did the specific problem in this thread successfully with the TI-95 Mathematics module. Hyperbolics with complex numbers are not available with the TI-59 Master Library. This subject was addressed earlier this year in a thread "Complex number capability in non-HP machines" dated 5 July 2008. You can access it at http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/archv018.cgi?read=138627#138627
Re: hyperbolic curiosityMessage #41 Posted by Karl Schneider on 15 Oct 2008, 1:44 a.m.,in response to message #1 by Bill Triplett Bill -- Thanks for the interesting link. There are so many comments and observations in this thread about the very topics that I've posted about in the recent and more-distant past, I hardly know where to start. I'll need to make multiple replies... Many people noticed that the H-P models give the same answer in degree or radians mode. Your last post in this thread indicates why: Hyperbolic functions take inputs that are not angles and must be dimensionless. Thus, radians mode is assumed. The basic formulae for computing sinh or cosh of a complex-valued input are quite similar to the familiar "sum of angles" formulae for sine and cosine. In these formulae, hyperbolics are computed for each part of a complex-valued input, but the following identities are substituted to simplify a complex-valued calculation to elementary functions of reals: cosh(ix) = cos(x) sinh(ix) = i*sin(x) I find it curious to enter a complex-valued argument in rectangular coordinates, but with the imaginary part as an angle. However, that's the input form the author clearly indicates. Maybe that's how these slide rules were designed. I'm surprised that the TI-89 wouldn't accept a complex-valued input to sinh or cosh. I'll have to check with my TI-using younger colleagues... Is there a "complex" mode? -- KS Edited: 15 Oct 2008, 2:33 a.m.
Re: hyperbolic curiosityMessage #42 Posted by Marcus von Cube, Germany on 15 Oct 2008, 3:30 a.m.,in response to message #41 by Karl Schneider Karl, you can set the TI machines to either REAL, RECTANGUALAR or POLAR. In REAL mode you get the domain error. RADians mode seems to be the other prerequisite.
Re: hyperbolic curiosityMessage #43 Posted by Bill Triplett on 15 Oct 2008, 5:12 p.m.,in response to message #41 by Karl Schneider References to previous discussion are greatly appreciated, and have been interesting reading. Thanks. I noticed that my old TI-85 can handle this type of problem, and it does not care whether it is set in radians mode or degrees mode. I am looking at the slide rules, and trying to figure out why the imaginary part was an angle. William Robinson's article talks about how the slide rules were made for simplifying solutions to particular types of important problems such as when modeling the stability of a vibrating aircraft, perhaps. His article might contain clues as to why the imaginary part was an angle. I could break down and ask the author about the slide rules. He might be more likely to know why than any other living person. I guess the reason why this topic has me perplexed is that there are many modern calculators that do not support this function, and yet in 1946 people were doing this with slide rules to such an extent that multiple scales on the slide rules were carefully arranged so that a user could do this job with as few physical movements as possible. We tend to forget too much of what had been mastered long ago.
Re: hyperbolic curiosityMessage #44 Posted by Rodger Rosenbaum on 16 Oct 2008, 12:44 a.m.,in response to message #43 by Bill Triplett Quote: His article might contain clues as to why the imaginary part was an angle. I could break down and ask the author about the slide rules. He might be more likely to know why than any other living person. Back in the early part of the 20th century, solution of networks didn't use the matrix methods we use today. The more modern methods were initially due to Feldtkeller in about 1930. Methods developed by the Bell Labs researchers were used, emphasizing the transmission properties of networks. Kennelly (of Kennelly-Heaviside layer fame) published many papers on the use of hyperbolic functions, and he even wrote a book, "Hyperbolic Functions Applied to Electrical Engineering", circa 1912. He showed that the solutions to the voltages and currents along a transmission line (power or telephone) were given by hyperbolic functions. The hyperbolic functions are considered to be analogous to the circular trig functions, sin, cos, tan, etc.; they are just a different type of trig function. The hyperbolics are combinations of exponentials, and as we know, EXP(x + iY) = EXP(x)* (COS(y) + i SIN(y)); that is, the imaginary part of the argument (x + iy) is the argument to sin and cos functions, normally considered to be angles. The real part is just an attenuation constant. And, of course, in engineering work, it's common to measure angles in degrees. Kennelly even referred to the length of a transmission line as its hyperbolic angle. Edited: 17 Oct 2008, 10:30 p.m. after one or more responses were posted
Re: hyperbolic curiosityMessage #45 Posted by Bill Triplett on 16 Oct 2008, 1:00 p.m.,in response to message #44 by Rodger Rosenbaum Thanks for explaining why the functions need angles for the imaginary component. The author of the slide rule article probably still has a copy of Kennelly's book from 1912 safely tucked away, and I imagine it probably has well worn pages.
Re: hyperbolic curiosityMessage #46 Posted by mjcohen on 17 Oct 2008, 3:43 p.m.,in response to message #45 by Bill Triplett Easycalc, on any palm pda, gives 0.1459688 + i*0.927706645. This is my favorite calculator on the pda. It has an amazing amount of power - reals, complex, vectors, matrices, functions up the wazoo (special functions listed below), programmability (I implemented the Lanczos log of gamma function), and it is free! I know it it neither HP nor rpn, but I find it incredibly useful. Features and Specifications: * IEEE-754 double precision math routines, courtesy of MathLib. * Support for color, OS 5.x and DIA (Dynamic Input Area). * Algebraic (not RPN) entry. * Range of values: +/- 1E-308 to +/-1E308 (IEEE 754 double precision); up to 15-digit (user-selectable) decimal precision. * 150 built-in functions including: trigonometry, complex numbers, exponential, probability, statistics, numerical analysis of functions (roots, derivatives, integrals, intersection), discrete math, digital signal processing, list, matrix, and special. * Unlimited user-defined variables and functions. * Multipe levels of nested parentheses limited only by available stack memory. * Multiple screens easily categorize functions. * "ANS" button (last result). * Scrolling history. * Scrolling menu access to user- and built-in functions and variables. * Angular Units: Radian, Degree and Grad. * Integer Base Operations: Decimal, Binary, Hexadecimal, and Octal. * Display Formats: Normal, Scientific and Engineering display modes. * Automatic parentheses matching. * Full PalmOS clipboard support. * Complex number support in all functions. * Graphics: o Normal (Cartesian), polar, and parametric graphs. o Graph up to six functions simultaneously with trace and numerical analysis. o Graphs have user definable colors, axes, axes lables, range, grid, log and log-log. o "Table Mode" of graph function values. o Zoom in/out. o Live screen scrolling. o Adjustable graph resolution. Special functions: Function Name Name in EasyCalc Euler Gamma gamma(z) Beta beta(z:w) Incomplete Gamma igamma(a:x) Error erf(x) Complementary Error erfc(x) Incomplete Beta ibeta(a:b:x) Bessel 1st Kind besselj(n:x) Bessel 2nd Kind bessely(n:x) Mod. Bessel 1st Kind besseli(n:x) Mod. Bessel 2nd Kind besselk(n:x) Incomplete Elliptic Integral 1st kind elli1(m:phi) Incomplete Elliptic Integral 2nd kind elli2(m:phi) Complete Elliptic Integral 1st kind ellc1(m) Complete Elliptic Integral 2nd kind ellc2(m) Jacobi sn sn(m:u) Jacobi cn cn(m:u) Jacobi dn dn(m:u)
Re: hyperbolic curiosityMessage #47 Posted by Rodger Rosenbaum on 17 Oct 2008, 10:38 p.m.,in response to message #46 by mjcohen Quote: Easycalc, on any palm pda, gives 0.1459688 + i*0.927706645. There's a typo here isn't there? Shouldn't the result be 0.1459688 + i*0.827706645? Also, it looks like you're only getting about 5 or 6 correct digits. Can you do the complex SINH calculation in double precision? What does Easycalc get for TAN(1.57079), in radians mode? In both single and double precision, if available.
Re: hyperbolic curiosityMessage #48 Posted by Pal G. on 19 Oct 2008, 9:13 p.m.,in response to message #47 by Rodger Rosenbaum Quote: There's a typo here isn't there? Shouldn't the result be 0.1459688 + i*0.827706645? Yes, there is a typo. EasyCalc reports: 0.145968879271932+0.827706644863301*i Quote: Also, it looks like you're only getting about 5 or 6 correct digits. Can you do the complex SINH calculation in double precision? I think EasyCalc only supports double-precision calculations. It won't run without Mathlib, the Palm shared IEEE-754 double-precision library. Quote: What does Easycalc get for TAN(1.57079), in radians mode? In both single and double precision, if available. EasyCalc: 158,057.913410701003158 Hope this helps, PG
Re: hyperbolic curiosityMessage #49 Posted by Rodger Rosenbaum on 20 Oct 2008, 2:28 a.m.,in response to message #48 by Pal G. Ok, I see the reason for the apparent error in the result for the SINH calculation. You used .243 + i .93375 for the input; the imaginary part was 5 digits. And it looks like Easycalc reduces radian trig arguments with fewer digits of Pi than the Saturn based HP calcs.
Go back to the main exhibit hall | 9,499 | 36,412 | {"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} | 2.875 | 3 | CC-MAIN-2019-43 | longest | en | 0.869444 |
https://jingleyanqiu.wordpress.com/tag/constant/ | 1,618,062,954,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038057142.4/warc/CC-MAIN-20210410134715-20210410164715-00413.warc.gz | 448,647,570 | 21,407 | ## C Is 4 Constant
C Is For C-O-N-S-T-A-N-T, CONSTANT, which can be applied as an adjective or a noun.
Examples of applications:
#1: Billy is a constant believer of capitalism,
#2: Trisha is a constant supporter of all blogging friends.
#3: Poets in Thursday Poets Rally have made constant growth of creativity and productivity.
#4: Given Y=2X+5, we have X and Y as variables, and 5 as a constant.
If Given Y=0X+5=5, then X is a trivial variable, and Y is a constant with its value equal to 5.
Please let me know if you have trouble with the math example. Many Thanks!
*****
This is an entry for
ABC WednesdayC is 4 Constant! | 170 | 636 | {"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-2021-17 | latest | en | 0.931808 |
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Perimeter & Graphing Activity
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Students can struggle with the idea that the perimeter of a shape is the sum of its sides – get your students understanding and thinking of perimeter and units with this fun measuring and graphing worksheet suitable for high school students!
Open the preview to check out the product!
Contents
This double-sided worksheet contains four activities that require students to measure an object with unusual units (thumbs, hand spans, shoes), collect data from other students and then graph the results.
Instructions for use
This is a good lesson to deliver before teaching perimeter to get students thinking about the concept in a non-threatening and fun way. Simply print the worksheet double-sided and distribute to students. You can talk students through the first activity if necessary, but in my year 9 class, my students quickly understood the idea and ran with it, eager to collate data.
Suggestions for use in the classroom
You could print single-sided and create task cards with these activities, or stagger the activities throughout the lesson(s). Students could work in pairs or small groups to complete the tasks. At the end of the lesson, you could collate class data for all tasks and talk about why there was such a range of data (book sizes, shoe sizes etc) and then actually measure the objects with a ruler to see whose estimate was the closest.
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Home / Discussions / Algorithms
# Algorithms
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HOW TO ANSWER A QUESTION Chris Maunder28-Jul-09 2:32 Chris Maunder 28-Jul-09 2:32
How to get an answer to your question Chris Maunder16-Mar-09 10:13 Chris Maunder 16-Mar-09 10:13
Where To Begin - Excel Manipulation (Mike) ipscone 19-May-22 12:14 (Mike) ipscone 19-May-22 12:14
Re: Where To Begin - Excel Manipulation Richard MacCutchan19-May-22 20:53 Richard MacCutchan 19-May-22 20:53
Re: Where To Begin - Excel Manipulation (Mike) ipscone 20-May-22 5:22 (Mike) ipscone 20-May-22 5:22
Re: Where To Begin - Excel Manipulation Richard MacCutchan20-May-22 5:35 Richard MacCutchan 20-May-22 5:35
Re: Where To Begin - Excel Manipulation (Mike) ipscone 20-May-22 10:38 (Mike) ipscone 20-May-22 10:38
Re: Where To Begin - Excel Manipulation Gerry Schmitz20-May-22 5:18 Gerry Schmitz 20-May-22 5:18
Re: Where To Begin - Excel Manipulation (Mike) ipscone 20-May-22 5:39 (Mike) ipscone 20-May-22 5:39
Re: Where To Begin - Excel Manipulation Richard MacCutchan20-May-22 6:09 Richard MacCutchan 20-May-22 6:09
Re: Where To Begin - Excel Manipulation (Mike) ipscone 20-May-22 6:52 (Mike) ipscone 20-May-22 6:52
Re: Where To Begin - Excel Manipulation Gerry Schmitz21-May-22 5:30 Gerry Schmitz 21-May-22 5:30
Creating paths in a 2D array Member 1561542626-Apr-22 1:18 Member 15615426 26-Apr-22 1:18
Re: Creating paths in a 2D array `Randor` 26-Apr-22 17:48 `Randor` 26-Apr-22 17:48
Virtual sports algorithms. How it works? Cryptocurrency3-Apr-22 19:45 Cryptocurrency 3-Apr-22 19:45
Re: Virtual sports algorithms. How it works? Gerry Schmitz26-Apr-22 6:00 Gerry Schmitz 26-Apr-22 6:00
Methods for Solving different forms of Recurrence Relations Dawood Ahmad 20213-Apr-22 10:46 Dawood Ahmad 2021 3-Apr-22 10:46
Re: Methods for Solving different forms of Recurrence Relations `Randor` 26-Apr-22 18:06 `Randor` 26-Apr-22 18:06
Help with allocation algorithm Cynthia Moore30-Mar-22 23:53 Cynthia Moore 30-Mar-22 23:53
Re: Help with allocation algorithm Member 146609423-May-22 4:01 Member 14660942 3-May-22 4:01
1 - I would make people prioritize the records they want - only 1 number, let's say, from 1 to 100. 2 - They would then choose the records they want, in the rank they want. Not allowed to repeat numbers. With that information, I would assign a priority value for each ranking, the higher the priority (1 is highest), the higher this number is. The balancing of these values is going to be a big part on your "justice". I would do a cumulative priority number, and upon collision, the person that asked first for the record would get it. Let's review this, in a slower way: Let's say you use 1 => 400, 2 => 200, 3 => 100, 4 => 80, 5 => 70, 6 => 60 Now we have the first choice round. Everyone gets 400 priority points. Let's say 5 people. But there is a collision. 2 people chose MC Hammer record. The one that asked for it before is going to get it. The other one... well, tough luck. Now, we have the second choice round. Everyone gets 200 priority points. But, the "un-hamnmered" guy now has 600 priority points. Whatever he chose for choice 2, if available, is going to get picked. After that, everyone else can get their choice of number 2, if available. Just repeat until no more records. This should be a fairly fair system - everyone is probably going to be equally pissed... lol... You are seriously underestimating human beings' capacity for being petty - I hope all goes well, but I really think you're going to have some headache.
Very smart pointers Andy Oct202121-Mar-22 9:24 Andy Oct2021 21-Mar-22 9:24
Re: Very smart pointers Greg Utas26-Mar-22 10:14 Greg Utas 26-Mar-22 10:14
Re: Very smart pointers Member 146609423-May-22 4:13 Member 14660942 3-May-22 4:13
Re: Very smart pointers Andy Oct20213-May-22 22:21 Andy Oct2021 3-May-22 22:21
Maximum Sum of Value of Positions Along a Path Jj2093884218-Mar-22 13:30 Jj20938842 18-Mar-22 13:30
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Use Ctrl+Left/Right to switch messages, Ctrl+Up/Down to switch threads, Ctrl+Shift+Left/Right to switch pages. | 1,331 | 4,253 | {"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} | 2.75 | 3 | CC-MAIN-2022-27 | longest | en | 0.570519 |
https://math.stackexchange.com/questions/174945/why-are-sin-and-cos-and-perhaps-tan-more-important-than-their-recip/174957 | 1,718,728,529,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861762.73/warc/CC-MAIN-20240618140737-20240618170737-00264.warc.gz | 348,797,435 | 39,124 | # Why are $\sin$ and $\cos$ (and perhaps $\tan$) "more important" than their reciprocals?
(My personal "feel" is that $\sin$ and $\cos$ are first-class citizens, $\tan$ is "1.5th-class," and the rest are second-class; I'm sure there are others who feel the same.)
Main question(s): From a purely high-school-geometric/"ninth-century-geometer's" standpoint, is there any reason why this should be so? Given the usual elementary knowledge of triangles when one is first introduced to these functions, I think it appears pretty arbitrary. How should I convince a high school student that $\sin$, $\cos$, and $\tan$, instead of their reciprocals, should be our main objects of study? How did history decide on their superiority?
Of course, with real analysis goggles, things look quite a bit different: $\sin$ and $\cos$ are the only ones that are continuous everywhere; $f'' = -f$ characterizes all their linear combinations; they have much nicer series representations; etc. But I suspect this is all hindsight.
(I don't pretend to know enough about complex analysis, but I suspect even more nice things happen there with $\sin$ and $\cos$, and even more ugly things happen with the other four. In any case, I doubt history chose $\sin$ and $\cos$ to be first-class citizens because of their complex properties.)
Secondary questions: Is there any reason why $\sin$ is the "main" function and $\cos$ is "only" its complement, or is this arbitrary as well? Is there any reason why $\tan$ is preferred to $\cot$?
• Circles. ${}{}$
– anon
Commented Jul 25, 2012 at 6:34
• $\tan \theta$ is the slope of a line at angle $\theta$ from horizontal; we prefer $\tan$ to $\cot$ because we prefer $\dfrac{\text{rise}}{\text{run}}$ to $\dfrac{\text{run}}{\text{rise}}$, perhaps. Commented Jul 25, 2012 at 6:41
• I don't think there is any real reason that $\sin$ is the "primary" and $\cos$ is the "co-" function. Indeed, there are ways that $\cos$ often feels "more" primary to me. Commented Jul 25, 2012 at 6:48
• (I've always hated that the inversion of $\cos$ is $\sec$ and the inversion of $\sin$ is the "cosec". Bugs me that there is this inversion of which is primary and which is the "co-". It feels arbitrary.) Commented Jul 25, 2012 at 6:51
• @ThomasAndrews Well, one nice thing about how the co-functions are defined is when it comes to derivatives. The derivatives of the co-functions all sport a minus sign.
– Mike
Commented Jul 25, 2012 at 7:01
Historically, trigonometric functions were originally computed in terms of "chords" (the angles subtended by a circular arc), and the sine was computed from a bisected chord (so a "half-chord"). In fact, the word "sine" originates from a mis-translation (from Arabic) of the Sanskrit word "jyā" which literally means "bow-string".
(The word "jyā" became "jība" in Arabic, and then subsequently "jaib". A cognate of "jaib" in Arabic has the meaning of "bosom", and jaib was mistakenly rendered into Latin as "sinus". So yes, the word "sine" was originally not safe for work).
The importance of the cosine seems also to have been first recognized by Indian mathematicans, who called it "koṭi-jyā" or "kojyā" meaning (roughly) "sine of the extreme angle" ("koṭi" means "the extreme end of a bow" or "extremity" in general).
So the sine as the "main" function, and cosine as the "adjunct" function goes back at least 16 centuries (in the Surya Siddhanta, written some time in the 4th century).
• My goodness, Mr Wheeler, how do you know this stuff? Very impressed!
– user22805
Commented Jul 25, 2012 at 8:31
• Fun fact: In some languages, in particular Spanish, sine still has the same not safe for work meaning. This has lead to uncountable repressed giggles in high school math class. Commented Jul 25, 2012 at 14:48
To elaborate on anon's comment, the classic trigonometric diagram has a right triangle drawn inside a unit circle, with one point at the origin, one point on the circle, and the third point the projection of the second onto the $x$ axis. Sine and cosine then give the lengths of the legs. I suppose it's more "natural" to draw the hypotenuse as having fixed unit length and studying how the legs change as you rotate the point around a circle, than to fix one of the legs.
Of course, which functions are "first-class citizens" really depends on the application; in my area (computer graphics, and in particular physical simulation) the first-class function is $2\tan \frac{\theta}{2}$, since it is easily computed from dot and cross products, bijective between $(-\pi,\pi)$ and the reals, and is the identity to first order near $\theta=0$.
• Even better, $2\tan\frac{\theta}{2}$ is the identity to second order near $\theta=0$ (like $\sin$ and $\tan$). Commented Jul 25, 2012 at 7:11
• Funny, I can't recall encountering $2\tan\frac\theta2$ very much at all in my graphics/simulation code. Can you name some examples?
– user856
Commented Jul 25, 2012 at 8:07
• Here are a few: bending energy of rods and plates, the constraint function for enforcing that two angles are equal, mean value coordinates, some formulations of mean curvature, modeling circular arcs with NURBS (via Weierstrass substitution). On reflection $\cot$ might give $\tan \frac{\theta}{2}$ some competition: the ubiquitous "cotan Laplacian," Wachspress coordinates, etc. Commented Jul 25, 2012 at 23:46 | 1,404 | 5,346 | {"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} | 3.765625 | 4 | CC-MAIN-2024-26 | latest | en | 0.958258 |
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,066 | 3,057 | {"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} | 3.046875 | 3 | CC-MAIN-2018-22 | latest | en | 0.835082 |
https://www.tutorialspoint.com/articles/category/electric-drives | 1,685,386,580,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644907.31/warc/CC-MAIN-20230529173312-20230529203312-00128.warc.gz | 1,159,844,294 | 13,054 | Difference Between Dielectric Constant and Permittivity
Updated on 18-Apr-2023 16:33:16
The study of dielectrics and their behavior in electric fields continue to fascinate physicists and electrical engineers alike. Despite the fact that dielectrics are poor conductors of electricity, they play a fundamental role in electronic circuits, which need a dielectric medium to build the circuit. A basic understanding of dielectrics and their properties is thus required. A dielectric material is nothing but an insulator with a poor conductor of electricity meaning they do not allow current to flow. They are the exact opposite of conductors. Like any other material, a dielectric is an assembly of ions with positive and negative ... Read More
Mechanics of Train Movement and Derivation for Tractive Effort
Updated on 03-May-2022 08:27:28
Mechanism of Train MovementThe essential mechanism of an electric locomotive is shown in the figure below.Here, the armature of the driving motor has a pinion of the diameter d1 attached to it. The tractive effort at the edge of the pinion is transferred to the driving wheel by means of a gearwheel.Let, T = Torque exterted by the motor in NmF1 = Tractive effort at the pinionFt = Tractive effort at the wheel$\gamma$ = Gear ratiod1 = Diameter of piniond2 = Diameter of gear wheelD = Diameter of driving wheel$\eta$ = Efficiency of power transmission from motor to ... Read More
Systems of Track Electrification: AC Electrification System
Updated on 03-May-2022 08:24:50
There are two types of AC system of track electrification −Single-Phase Low-Frequency AC System (15 kV to 25 kV at 16$\frac{2}{3}$ Hz, 25 Hz and 50 Hz)Three-Phase AC System (3.3 kV to 3.6 kV at 16$\frac{2}{3}$ Hz)Single-Phase Low-Frequency AC SystemIn a single-phase low-frequency system of track electrification, an AC voltage of 15 kV at a frequency of 16$\frac{2}{3}$ to 25 Hz is used.The use of high voltage AC in the system reduces the current flowing through the overhead transmission line resulting in the reduced voltage drops and it helps in installing the substations at larger distances, about 50 to 80 ... Read More
Electric Traction: Trapezoidal Speed Time Curve
Updated on 03-May-2022 08:19:33
The speed-time curve of a main line service is best and most easily replaced by trapezoid. The figure shows the simplified trapezoidal speed-time curve for the main line service.Let, $\mathit{V_{m}}$ = Crest speed in kmph$\alpha$ = Acceleration in kmphps$\beta$ = Retardation in kmphpsT = Total time of run in seconds$\mathit{t_{\mathrm{1}}}$ = Acceleration time in seconds$\mathit{t_{\mathrm{2}}}$ = Free running time in seconds$\mathit{t_{\mathrm{3}}}$ = Retardation time in secondsD = Total distance of run in kmTherefore, the time of acceleration in seconds is given by, $$\mathrm{\mathit{t_{\mathrm{1}}}\:=\:\frac{\mathit{V_{m}}}{\alpha }}$$And the time for retardation in seconds is, $$\mathrm{\mathit{t_{\mathrm{3}}}\:=\:\frac{\mathit{V_{m}}}{\beta }}$$Hence, the time for free running in seconds is, ... Read More
Electric Traction: Quadrilateral Speed Time Curve
Updated on 03-May-2022 08:16:12
General Features of Traction Motors: Mechanical Features & Electrical Features
Updated on 28-Apr-2022 08:43:11
An electric motor which is used to provide the primary rotational torque to a machine such as electric locomotives, elevators, conveyors, trolley buses and other electric vehicles etc. is described as a traction motor.General Features of Traction MotorsThe general features of the electric motors used for traction purpose are classified into two classes as −Mechanical FeaturesElectrical FeaturesMechanical Features of Traction MotorsThe mechanical features of electric motors used for traction applications are listed below −Robustness - A traction motor must be mechanically strong and robust so that it can be capable of withstanding severe mechanical vibrations.Small Size - The overall dimensions ... Read More
Updated on 28-Apr-2022 08:41:08
An electric traction system is the type of traction system that uses electrical energy for the movement of vehicle at any stage.The electric traction system is classified into two groups as −Self-contained vehicles or locomotives (Ex. – battery-electric drive and diesel-electric drive)Vehicles or locomotives which receive electric power from a distribution network or suitably placed substations (Ex. - railway electric locomotives, tramways and trolleybuses, etc.)Here, we described different types of electric traction systems along with their advantages and disadvantages −Straight Electric Traction SystemIn the straight electric traction system, either DC series motor or single phase AC series motor or three-phase ... Read More
Systems of Track Electrification: Composite System of Track Electrification
Updated on 28-Apr-2022 08:37:31
There is no system of track electrification which is good in all aspects, i.e. it suffers from some disadvantage.The composite system (as the name implies) combines together any two of the systems of track electrification so as to incorporate good points of both the systems. The composite systems presently being used are as −Single-Phase AC to Three-Phase AC SystemSingle-Phase AC to DC SystemSingle-Phase AC to Three-Phase AC SystemThe single-phase to three-phase system of track electrification is also known as Kando System. As the three phase AC traction system has some disadvantages such as tapping of power supply through two overhead conductors ... Read More | 1,297 | 5,509 | {"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} | 3.171875 | 3 | CC-MAIN-2023-23 | latest | en | 0.901 |
https://edurev.in/chapter/55831_Mathematical-Operations--%e0%a4%97%e0%a4%a3%e0%a4%bf%e0%a4%a4%e0%a5%80%e0%a4%af-%e0%a4%b8%e0%a4%82%e0%a4%95%e0%a5%8d%e0%a4%b0%e0%a4%bf%e0%a4%af%e0%a4%be%e0%a4%8f%e0%a4%81--General-Intelligence-Reasoning--Hindi- | 1,701,718,059,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100534.18/warc/CC-MAIN-20231204182901-20231204212901-00425.warc.gz | 261,855,897 | 63,693 | Mathematical Operations (गणितीय संक्रियाएँ)
# Mathematical Operations (गणितीय संक्रियाएँ) - General Intelligence & Reasoning (Hindi) - Banking Exams - Notes, Videos & Tests
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Mathematical Operations (गणितीय संक्रियाएँ) | General Intelligence & Reasoning (Hindi) | 1,534 | 6,747 | {"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} | 2.625 | 3 | CC-MAIN-2023-50 | latest | en | 0.689069 |
http://blogs.discovermagazine.com/seriouslyscience/2013/06/27/increasing-the-portion-size-of-a-sandwich-increases-energy-intake/ | 1,526,921,744,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864461.53/warc/CC-MAIN-20180521161639-20180521181639-00523.warc.gz | 41,329,834 | 18,174 | # Why foot-long sandwiches are a bad idea.
By Seriously Science | June 27, 2013 1:00 pm
Photo: flickr/_BuBBy_
Many of us are trying to not eat ourselves to death, and we’ve been told that using smaller plates can help us eat less. Other research has proven that we drink more alcohol when drinking giant cocktails. But what about the size of individual snacks– does the size of the snack unit change how much we eat? Could a larger sandwich trick you into eating more than a smaller sandwich? In this study, they had the participants eat a lunch comprised of unlimited numbers of sandwiches of a certain size and measured how many calories they ate. Turns out that yes, the participants ate more in total when eating giant sandwiches. Mmm… giant sandwiches!
Increasing the portion size of a sandwich increases energy intake.
“OBJECTIVE: This study investigated the effect on energy intake of increasing the portion size of a food served as a discrete unit.
DESIGN: A within-subject design with repeated measures was used.
SUBJECTS/SETTING: The sample comprised 75 young adults (37 females and 38 males) from a university community.
INTERVENTION: Individuals ate lunch in the lab once a week for 4 weeks. Each week, they were served one of four sizes of a deli-style sandwich (6, 8, 10, or 12 inches), of which they could eat as much as they wanted.
MAIN OUTCOME MEASURES: Energy intakes were determined for each meal, as were ratings of hunger and satiety before and after each meal.
STATISTICAL ANALYSES PERFORMED: A linear mixed model with repeated measures was used. The influence of subject characteristics was examined using analysis of covariance.
RESULTS: The portion size of the sandwich significantly influenced lunch intake for both males and females
APPLICATIONS/CONCLUSIONS:These results suggest that increasing the portion size of a food served as a discrete unit leads to increased energy intake at a single meal without differentially influencing ratings of hunger and satiety. Dietitians should educate their clients about strategies to moderate the effect on intake of increased portions of high-calorie foods.”
CATEGORIZED UNDER: duh, eat me
• http://www.ghookermls.info/ Geoffrey Hooker
What you do is get a foot-long, cut it in half, eat 6″ today and put the other 6″ in the fridge for tomorrow.
NEW ON DISCOVER
OPEN
CITIZEN SCIENCE
### Seriously, Science?
Seriously, Science?, formerly known as NCBI ROFL, is the brainchild of two prone-to-distraction biologists. We highlight the funniest, oddest, and just plain craziest research from the PubMed research database and beyond. Because nobody said serious science couldn't be silly! | 587 | 2,663 | {"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} | 2.65625 | 3 | CC-MAIN-2018-22 | latest | en | 0.971777 |
https://superbprofessors.com/solving-the-derivative-of-cosx/ | 1,726,332,542,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.73/warc/CC-MAIN-20240914161327-20240914191327-00260.warc.gz | 526,503,932 | 11,326 | # Solving the Derivative of cos(x)
In this lesson, we will see how to find the derivative of cos(x). We will also see the derivatives of all the basic trigonometric functions and discuss why it is useful to be familiar with these derivatives.
Steps to Solve
The derivative of cos(x) is often one that people know by memory, but they don’t know how to show it. Let’s take a look at how we can find the derivative of cos(x) if we can’t remember it off the top of our heads. To do this, we’re going to make use of the fact that cos(x) = 1 / sec(x). We’re going to find the derivative of 1 / sec(x), and in doing so, we will also find the derivative of cos(x), since they’re equal.
cosderv1
Notice that 1 / sec(x) is a quotient. Therefore, we are going to use the quotient rule for derivatives to find the derivative. The quotient rule for derivatives is as follows.
cosderv2
The other facts that we will need to know to find this derivative are as follows:
The derivative of sec(x) is sec(x)tan(x)
The derivative of a constant is 0
tan(x) = sin(x) / cos(x)
(a/b) / (c/d) = (a/b) * (d/c) = (ad / bc)
Okay, now that we have a game plan and all the facts that we’re going to need, let’s dive in! The first thing we want to do is use the quotient rule on 1 / sec(x). In 1 / sec(x), the function in the numerator is f(x) = 1, and the function in the denominator is g(x) = sec(x). We plug these into the quotient rule, which you can see play out here:
cosderv3
Now we’re going to simplify. This is where our facts come into play. During the first part of simplification, we use facts 1 and 2, which you can see here:
cosderv5
And now, we can and will simplify further using facts 3 and 4, like this right here:
cosderv6
We can put all this work in a nice organized and compact form, as follows:
cosderv7
So, with all said and done, we see that the derivative of cos(x) is -sin(x).
If you want to make sure that you get all of the different steps that we’re showing, feel free to pause and take note of them. If this is just a handy review for you, let’s move on.
Trigonometric Function Derivatives
As was said, the derivative of cos(x) is one that is well known. Though it’s always useful to know how to find a derivative, knowing some key derivatives by memory is quite useful. Suppose you were solving a problem and along the way, you had to find the derivative of cos(x). If you had it memorized that the derivative of cos(x) is -sin(x), you wouldn’t have to go through all the steps we just went through. We see that knowing certain derivatives can save us a lot of time and work.
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The post Solving the Derivative of cos(x) appeared first on Superb Professors. | 710 | 2,751 | {"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} | 4.78125 | 5 | CC-MAIN-2024-38 | latest | en | 0.945392 |
https://math.stackexchange.com/questions/1047985/krein-milman-theorem-why-the-closed-convex-hull-is-needed | 1,719,310,381,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865694.2/warc/CC-MAIN-20240625072502-20240625102502-00316.warc.gz | 337,794,210 | 36,776 | # Krein Milman Theorem, why the closed convex hull is needed?
I have a question about the Krein Milman that I'm having difficulty answering. Here is the statement of the theorem.
If ${K}$is a non-empty compact convex subset of a locally convex space ${X}$, then ${\text{ext}\;K\neq\emptyset}$ and ${K=\overline{\text{co}}(\text{ext}\;K)}.$
My question is, why must it be the closed convex hull? Surely when working in a locally convex space with a compact space the convex hull is already closed. I have tried to think of an example in which ${K={\text{co}}(\text{ext}\;K)}$ is not true but haven't been able to think of anything.
Does this problem only occur when we are dealing with infinite dimensional spaces?
I do know that the set of extreme points of a finite dimensional set need not be closed if this is of any help.
Surely when working in a locally convex space with a compact space the convex hull is already closed.
No, that need not be. Consider $\ell^2(\mathbb{N})$, let $(e_n)_{n\in\mathbb{N}}$ be the "standard basis". Define $v_n = \tfrac{1}{n+1}\cdot e_n$ and
$$M := \{ 0\} \cup \left\{v_n : n\in\mathbb{N}\right\}.$$
$M$ is compact, and the convex hull of $M$ is not closed, since every element of the convex hull has only finitely many nonzero coordinates, but
$$x = \sum_{n=0}^\infty 2^{-(n+1)}\cdot v_n \in \overline{\operatorname{co}}(M)$$
has infinitely many nonzero coordinates. Let $K = \overline{\operatorname{co}}(M)$. Then $K$ is a compact convex set, which we can describe as
$$K = \left\{x\in \ell^2(\mathbb{N}) : (\forall k)(x_k \geqslant 0) \land \sum_{n=0}^\infty (n+1)\cdot x_n \leqslant 1\right\}.\tag{1}$$
It is easy to see that the right hand side of $(1)$, let's call that $L$, is a compact convex set, and it is clear that it contains $M$, hence it contains $\overline{\operatorname{co}}(M)$. For $N\in\mathbb{N}$, let $P_N\colon \ell^2(\mathbb{N}) \to \ell^2(\mathbb{N})$ be the orthogonal projection onto the subspace spanned by the $e_n$ with $n\leqslant N$. Then we have $P_N(x) \to x$ for all $x\in \ell^2(\mathbb{N})$, and $P_N(L) \subset \operatorname{co}(M)$, whence $L \subset \overline{\operatorname{co}}(M)$, so indeed $L = K = \overline{\operatorname{co}}(M)$.
Now let's see that $M = \operatorname{ext} K$. The inclusion $M\subset \operatorname{ext} K$ is easy (though if properly done a little tediously) to verify. If $x\in K\setminus M$, then either $x = c\cdot v_m$ for some $m\in\mathbb{N}$ with $0 < c < 1$, then
$$x = c\cdot v_m + (1-c)\cdot 0$$
shows that $x$ is not an extreme point of $K$, or there are (at least) two indices $m < n$ with $x_m > 0$ and $x_n > 0$. Let $w = v_m - v_n$. Then for small enough $\varepsilon > 0$, both $x+\varepsilon w$ and $x-\varepsilon w$ lie in $K$, and $x$ is the midpoint of the segment between these two, hence not an extreme point of $K$.
So we have an example of a compact convex set $K$ in a locally convex space - even a Hilbert space - with
$$\operatorname{co}(\operatorname{ext} K) \subsetneq K = \overline{\operatorname{co}}(\operatorname{ext} K).$$
Does this problem only occur when we are dealing with infinite dimensional spaces?
Yes. In finite-dimensional spaces, Carathéodory's theorem guarantees that the convex hull of a compact set $M$ is again compact, since it puts an upper bound on the number of points that are required in a convex combination. In infinite-dimensional spaces, the number of points of $M$ required to write an $x\in \operatorname{co} (M)$ as a convex combination of points of $M$ can be arbitrarily large, and that allows the existence of points in the closed convex hull that cannot be written as a convex combination of finitely many points of $M$.
• Thanks so much I understand it much better now, I was unaware of Carathéodory's theorem. However I do have a question in how you define K Firstly, stupidly is the upwards arrow in the centre of the set just separating the conditions? But also haven't we used all of v's to describe a single point x above? where do the xk's come from, although I can clearly see why the sum must be less than 1 if i accept that they exist Commented Dec 3, 2014 at 9:45
• The "upward arrow" is a logical "and", it is also common to separate the conditions simply by a comma. $K$ is the set of all $x\in \ell^2$ such that all components of $x$ - the $x_k$ - are non-negative, and the weighted sum of the components is $\leqslant 1$. We could also describe it as $$K = \left\{x =\sum_{m=0}^\infty c_m\cdot v_m : 0 \leqslant c_m\leqslant 1, \sum_{m=0}^\infty c_m \leqslant 1\right\}.$$ The convex hull of $M$ consists of those points where only finitely many of the coefficients are $> 0$. The connection between the two representations is $x_m = c_m\cdot \frac{1}{m+1}$. Commented Dec 3, 2014 at 12:24 | 1,476 | 4,790 | {"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} | 3.328125 | 3 | CC-MAIN-2024-26 | latest | en | 0.873953 |
https://blogs.harvard.edu/samoh/2018/10/29/quantum-computing-and-cryptography/ | 1,632,281,148,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057303.94/warc/CC-MAIN-20210922011746-20210922041746-00565.warc.gz | 198,141,655 | 7,736 | ## Quantum Computing and Cryptography
If you google “Is Quantum Computing Dangerous”, you will find headline after headline about the imminent dangers of quantum computing. Articles instilling fear in readers through the talk of the ways that Quantum computing is a “tool of destruction”, or that it is the “end of cryptography”. Articles such as this one argue that modern cryptography will be defeated and even ends with the following statement “Should the Russian government break all of our encryption before the US develops countermeasures, stolen elections will seem like small potatoes. Welcome to the cyber-battlefield of the 21st century.” Where are these fears coming from, and are they substantiated? Should we actually fear a complete breakdown of cryptographic methods if quantum computing technology advances? Will it advance to that state?
First, the fears about cryptography. Many cryptographic methods and schemes, such as RSA, are built on top of the difficulty of solving one-way mathematical functions, such as prime factorization. These problems are easy to compute in one direction, but take exponential time in the other direction, so the ability to guess and break a method such as RSA are not feasible with modern computers, which at best solve problems in linear time. However, theoretically, quantum computers should be able to compute problems much faster, even problems that would normally take exponential time to solve. On the face of it, this would mean that all our current systems, i.e. banks, national security, etc. would be compromised if someone had a quantum computer that truly worked as a quantum computer (i.e. each qubit, rather than storing two possible states, 0 and 1, store three states, 0 and 1 and 0 and 1. When you multiply out the additional computing power for many hundreds of thousands of bits, then its easy to see the where the additional computing power would come from). Its easy to see the fear in this -> at face value it boils down to an arms race with a crazy powerful weapon that could break all cyper security and cryptographic schemes by solving exponential time algorithms in quadratic time.
But, in reality this fear seems a bit too strong. What is the current progress of quantum computers? Is it even possible to reach create a fully functional quantum computer with more than 50 qubits (the number needed for quantum supremacy, that is a quantum computer that cannot be simulated on a classical computer)? Currently, there are some serious roadblocks in the practical creation of such a machine, with IBM having the closet possibility of it. But, even if such a machine is to come, “it isn’t obvious how useful even a perfectly functioning quantum computer would be. It doesn’t simply speed up any task you throw at it; in fact, for many calculations, it would actually be slower than classical machines. Only a handful of algorithms have so far been devised where a quantum computer would clearly have an edge. And even for those, that edge might be short-lived. The most famous quantum algorithm, developed by Peter Shor at MIT, is for finding the prime factors of an integer. Many common cryptographic schemes rely on the fact that this is hard for a conventional computer to do. But cryptography could adapt, creating new kinds of codes that don’t rely on factorization.” -> https://www.technologyreview.com/s/610250/serious-quantum-computers-are-finally-here-what-are-we-going-to-do-with-them/
The truth is, there are serious difficulties ahead for the advancements of quantum computers to a useful state, and it doesn’t currently seem like there are that many practically useful efforts for quantum computers. Moreover, if a quantum computer was created, cryptography could adapt quickly to rely on a NP-Hard problem that is not easily solvable by a quantum computer (which would need some sort of quantum algorithm to solve the problem in the first place). All in all, the technology is incredibly interesting, but it does not seem like much of the current fear is merited. At least for the time being.
## 1 Comment »
1. ### Cassper Nyovest
October 29, 2018 @ 5:50 pm
1
Quantum Computing hasn’t really been elaborated clearly in my country, I would really love to get more insights in the topic and also cryptocurrency. | 878 | 4,305 | {"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-2021-39 | latest | en | 0.953165 |
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# Is x = 0 ?
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Is x = 0 ? [#permalink] 27 Jan 2009, 19:24
Is $$x = 0$$ ?
1. $$xy = x$$
2. $$x + y = y$$
* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient
My reasoning is the following but OA is different, what am I doing wrong?
Stat1: xy - x = 0 ===> x = 0, y = 1...sufficient
Stat1: x = 0
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Re: Is x = 0 ? [#permalink] 27 Jan 2009, 19:38
x-ALI-x wrote:
Is $$x = 0$$ ?
1. $$xy = x$$
2. $$x + y = y$$
* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient
My reasoning is the following but OA is different, what am I doing wrong?
Stat1: xy - x = 0 ===> x = 0, y = 1...sufficient
Stat1: x = 0
state1: x(y-1)=0 MEANS either x=0 or y=1
its not necessary that both conditions have to satisified
e.g. x=2 y=1 --> x(y-1) = 0
e.g x=0 y=3 -->x(y-1) = 0
x can be 0 or any other number..
multiple solutions
statement 1 is not suffcieint
stat2: straight forward x=0
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Re: Is x = 0 ? [#permalink] 27 Jan 2009, 19:59
aaaah. okay, makes sense. Thanks!
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Re: Is x = 0 ? [#permalink] 28 Jan 2009, 09:35
remains true even if y=0 for if y=0
x+0=0.
x2suresh wrote:
x-ALI-x wrote:
Is $$x = 0$$ ?
1. $$xy = x$$
2. $$x + y = y$$
* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient
My reasoning is the following but OA is different, what am I doing wrong?
Stat1: xy - x = 0 ===> x = 0, y = 1...sufficient
Stat1: x = 0
state1: x(y-1)=0 MEANS either x=0 or y=1
its not necessary that both conditions have to satisified
e.g. x=2 y=1 --> x(y-1) = 0
e.g x=0 y=3 -->x(y-1) = 0
x can be 0 or any other number..
multiple solutions
statement 1 is not suffcieint
stat2: straight forward x=0
_________________
-----------------------
tusharvk
Re: Is x = 0 ? [#permalink] 28 Jan 2009, 09:35
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Display posts from previous: Sort by | 1,372 | 3,992 | {"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} | 3.578125 | 4 | CC-MAIN-2015-27 | longest | en | 0.857415 |
http://mathhelpforum.com/trigonometry/140115-solve-triangle-given-parts-print.html | 1,527,173,098,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866326.60/warc/CC-MAIN-20180524131721-20180524151721-00422.warc.gz | 188,124,930 | 2,971 | # solve the triangle/with given parts
• Apr 19th 2010, 12:32 PM
colie2020
solve the triangle/with given parts
alpha=42.5degree
beta=35.2degree
a=18.0
• Apr 19th 2010, 02:09 PM
skeeter
Quote:
Originally Posted by colie2020
alpha=42.5degree
beta=35.2degree
a=18.0
$\displaystyle \alpha + \beta + \gamma = 180^\circ$
find $\displaystyle \gamma$ , then use the sine law.
• Apr 19th 2010, 02:13 PM
Quote:
Originally Posted by colie2020
alpha=42.5degree
beta=35.2degree
a=18.0
I'm assuming side A is directly opposite angle $\displaystyle \alpha$, there is a side B opposite angle $\displaystyle \beta$, and there is a hypotenuse opposite angle $\displaystyle \gamma$.
If so...
The sum of the interior angles in a triangle equals 180, so:
$\displaystyle \alpha + \beta + \gamma = 180$
$\displaystyle 42.5 + 35.2 + \gamma = 180$
$\displaystyle \gamma = 102.3$
The Law of Sines:
$\displaystyle \frac{18}{\sin (42.5)} = \frac{B}{\sin (35.2)}$
$\displaystyle B = \frac{18\sin (35.2)}{\sin (42.5)}$
$\displaystyle \frac{18}{\sin (42.5)} = \frac{C}{\sin (102.3)}$
$\displaystyle C = \frac{18\sin (102.3)}{\sin (42.5)}$
Check out the wikipedia entry on the law of sines:
Law of sines - Wikipedia, the free encyclopedia
What this problem is asking is known as triangulation:
Triangulation - Wikipedia, the free encyclopedia | 445 | 1,326 | {"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} | 4.4375 | 4 | CC-MAIN-2018-22 | latest | en | 0.694677 |
https://www.biology-online.org/biology-forum/viewtopic.php?p=119397 | 1,516,111,149,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886436.25/warc/CC-MAIN-20180116125134-20180116145134-00413.warc.gz | 873,037,594 | 16,176 | ## Population Genetics (Hardy Weinberg) problem
Genetics as it applies to evolution, molecular biology, and medical aspects.
Moderator: BioTeam
amdugg
Garter
Posts: 2
Joined: Sat Feb 13, 2010 9:28 pm
### Population Genetics (Hardy Weinberg) problem
Hello, this is my first post.
I am having trouble with a homework problem that should be really easy, and I can't figure out what I'm doing wrong.
Here it is:
In eggplants, fruit color is determined by a gene with two alleles: an allele that codes for purple fruit (Z) and an allele that codes for white fruit (W). Heterozygotes have neither purpler nor white fruit; instead ,their fruit is an intermediate color, light violet. A population of 10,000 eggplants is in Hardy-Weinberg equilibrium. In this population, the frequency of individuals with a genotype homozygous for the white allele is .49.
Here is what I did:
Calculate the Allelic frequencies:
p= the frequency of the allele Z = .3 (1-q)
q= the frequency of the allele W = (square root of .49) =.7
Calculate the genotypic frequencies:
genotype ZZ = p^2=.09
genotype WW = q^2 = .49
genotype ZW = 2pq = .0882
This should all add up to 1, but it doesn't and I can't figure out what I'm doing wrong.
I'm sure I'm missing something so basic that I'm going to be embarrassed for even posting this problem. Regardless, any help is much appreciated.
JackBean
Inland Taipan
Posts: 5694
Joined: Mon Sep 14, 2009 7:12 pm
for me 2*0.3*0.7 = 0.42 not 0.0882
and that makes up 1 exactly
http://www.biolib.cz/en/main/
Cis or trans? That's what matters.
amdugg
Garter
Posts: 2
Joined: Sat Feb 13, 2010 9:28 pm
### Re:
JackBean wrote:for me 2*0.3*0.7 = 0.42 not 0.0882
and that makes up 1 exactly
That would be fine, except that it's not 2*0.3*0.7
Its p^2+q^2+2pq
.03^2+.07^2+2[(.03^2)*(.07^2)]
...which does not equal 1.
Am I making this more complicated than I have to?
JackBean
Inland Taipan
Posts: 5694
Joined: Mon Sep 14, 2009 7:12 pm
How that it's not? It is 2*p*q, not 2*p^2*q^2! And p and q are 0.3 and 0.7 respectively.
http://www.biolib.cz/en/main/
Cis or trans? That's what matters.
Haidarh
Garter
Posts: 2
Joined: Tue Jan 26, 2010 6:08 pm
p = 0.3 q = 0.7 2pq = 2 x 0.3 x 0.7 = 0.42
p^2 + q^2 + 2pq = 1
0.9 + 0.49 + 0.42 = 1
### Who is online
Users browsing this forum: No registered users and 0 guests | 778 | 2,334 | {"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} | 3.359375 | 3 | CC-MAIN-2018-05 | latest | en | 0.919629 |
http://erlang.org/pipermail/erlang-questions/2020-November/100181.html | 1,718,407,064,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861575.66/warc/CC-MAIN-20240614204739-20240614234739-00533.warc.gz | 11,730,345 | 3,001 | # Find nearest geographic coordinates
Frank Muller frank.muller.erl@REDACTED
Thu Nov 26 15:13:09 CET 2020
Thanks Mike. Very informative.
I don’t expect an exact answer (2). Even the Euclidean distance will do in
my case.
/Frank
Wed 26 nov. 2020 - 06:14, FRENCH Mike <mike.french@REDACTED> wrote :
> 1. What is the overall extent of the points?
>
>
>
> 2. Do you need the exact answer?
> e.g. will there be many points close together and you *have* to get
> the correct one;
> can you assume a spherical Earth, not model the exact geoid
> shape (WGS84,etc.)
>
>
>
> 3. Does the region include the areas around the poles or crossing the 180
> longitude ‘dateline’?
>
> If (1) is ‘small’ or ‘a few 100 km’, and (2) and (3) are NO, then there
> are some simple methods …
>
>
>
> If it is a localized region near the equator (e.g. within Singapore) you
> can use the Euclidean formula directly.
>
>
>
> If it is a localized region, you can assume the ‘2D tangent plane
> projection’ Euclidean approximation.
> Take the middle value for the latitude, *avlat*, and use *cos(avlat)* to
> scale the longitudinal differences,
> then use the Euclidean formula.
>
>
>
> Regards,
>
> Mike
>
>
>
>
>
> *From:* erlang-questions [mailto:erlang-questions-bounces@REDACTED] *On
> Behalf Of *Ivan Uemlianin
> *Sent:* Thursday, November 26, 2020 6:14 AM
> *To:* Frank Muller; Erlang-Questions Questions
> *Subject:* Re: Find nearest geographic coordinates
>
>
>
> using it for `distance(A, X) < distance(B, X)` you could probably simplify
> the arithmetic.
>
> Ivan
>
> On 25/11/2020 22:09, Frank Muller wrote:
>
> I’m new to GIS field, but this module states that Euclidian distance isn’t
> accurate compared to “Law of Haversines”:
>
>
>
> https://github.com/armon/teles/blob/master/src/teles_geo_query.erl#L87-L102
>
>
>
>
>
> /Frank
>
>
>
> Wed 25 nov. 2020 à 22:45, Ivan Uemlianin <ivan@REDACTED> wrote:
>
> Hi Frank
>
> Do you want to sort the list by Euclidean distance from X? If so, could
> you use lists:sort/2?
>
> Ivan
>
>
> On 25/11/2020 21:38, Frank Muller wrote:
> > Hi guys,
> >
> > I've a list of geographic coordinates:
> >
> > L = [ {{<<"longitude">>,6.1457}, {<<"latitude">>,46.2022}},
> > {{<<"longitude">>,2.3387}, {<<"latitude">>,48.8582}},
> > ... ]
> >
> > and a specific coordinate X = {{<<"longitude">>,-73.5848},
> > {<<"latitude">>,45.4995}}.
> >
> > Question: how can i find the nearest coordinates to X from L (sorted
> > from the nearest to the farest)?
> >
> > /Frank
>
> --
> ============================================================
> Ivan A. Uemlianin PhD
> Llaisdy
>
> Ymchwil a Datblygu Technoleg Lleferydd
> Speech Technology Research and Development
>
> ivan@REDACTED
> @llaisdy
> llaisdy.wordpress.com
> github.com/llaisdy
>
> festina lente
> ============================================================
>
>
>
> --
>
> ============================================================
>
> Ivan A. Uemlianin PhD
>
> Llaisdy
>
>
>
> Ymchwil a Datblygu Technoleg Lleferydd
>
> Speech Technology Research and Development
>
>
>
> ivan@REDACTED
>
> @llaisdy
>
> llaisdy.wordpress.com
>
> github.com/llaisdy
>
>
>
>
> festina lente
>
> ============================================================
>
>
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# MIT2_094S11_lec12 - 2.094 Finite Element Analysis of Solids...
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2.094 Finite Element Analysis of Solids and Fluids Fall ‘08 Lecture 12 - Total Lagrangian formulation Prof. K.J. Bathe MIT OpenCourseWare We discussed: t t X = x x i j d t x = t X d 0 x , d 0 x = t X 1 d t x (12.1) 0 0 0 0 t C = t X T t X (12.2) 0 0 0 d 0 x = 0 t X d t x where 0 t X = 0 t X 1 = t 0 x i (12.3) ∂ x j The Green-Lagrange strain: 1 1 0 t = 0 t X T 0 t X I = 0 t C I (12.4) 2 2 Polar decomposition: 1 2 0 t X = 0 t R 0 t U 0 t = 2 0 t U I (12.5) We see, physically that: where d t t x and d t x are the same lengths the components of the G-L strain do not change. Note in FEA 0 0 k x i = h k x i k for an element (12.6) t t k u i = h k u i k t 0 t x i = x i + u i for any particle (12.7) Hence for the element t 0 k t k x i = h k x i + h k u i (12.8) k k = h k 0 x i k + t u i k (12.9) k = h k t x k k (12.10) k 49
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MIT 2.094 12. Total Lagrangian formulation E.g., k = 4 2nd Piola-Kirchhoff stress 0 ρ 0 t S = t ρ 0 t X t τ 0 t X T components also
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1. Jan 8, 2005
### bomba923
Brachistochrone derivation
I have calculated the parametric equation for it....but why is this path a cycloid? It does seem fast...but how did they determine that it was a cycloid...not any other function?--Hmm-why exactly does the path of a point of a circle as it rolls down a straight line become the fastest distance between the two points in the brachistrone problem??
Last edited: Jan 8, 2005
2. Jan 8, 2005
### da_willem
Why is it a cycloid? Because it makes 'the best use' of the gravitational force. At first the path is very steep, so the object acquires a large acceleration, and because teher is no friction involved the object keeps this large velocity till the end of the track. Ofcourse there are other paths with the same characteristics, and I think it is a cycloid because this follows from the formulas.
If there is a connection between the shape being the one of a piont on a horizontally moving wheel, and the brachistochrone I would definitely want to hear more...!
3. Jan 8, 2005
### dextercioby
Mathematics and only mathematics.It could have been any other curve similar in shape,but it was the equations that led to the solution.
There's no connection between the cycloid followed from a point on rolling wheel and the brahistochrone.The point would describe the same cycliod even in the absence of gravity.
For more detail,check out this wonderful site:you have a model of brahistochrone right on the first page
site
It's in french...But it's math.It's comprehendable.
Daniel.
4. Jan 8, 2005
### bomba923
That's was my question in the first place--! Anyways, I'll try to figure it out
5. Jan 9, 2005
### dextercioby
Please,do tell us if u find it.It would be a really interesting both mathematically and physically result.
Daniel. | 457 | 1,830 | {"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-2018-51 | latest | en | 0.943978 |
https://electronics.stackexchange.com/questions/17045/high-pcb-copper-thickness-what-are-the-pitfalls | 1,721,605,973,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517796.21/warc/CC-MAIN-20240721213034-20240722003034-00637.warc.gz | 196,446,252 | 49,609 | # High PCB copper thickness: What are the pitfalls?
We need to carry high currents on a PCB (~30Amps sustained), so we are likely to order our PCBs with high copper thickness. So far we've only used 35 microns (1 oz) in our designs, so 'high thickness' for us means, 70 (2 oz) or 105 (3 oz).
We do not know what are the things to watch out for with thickness copper. We'd appreciate any experiences. Since this is a very broad topic, I'll go ahead and ask specific questions:
1. It appears that for a lot of manufacturing houses, 105 microns is as high as its gets. Is that correct or are higher thickness possible?
2. Can the copper in the inner layers be as thick as the copper at the top and bottom of the board?
3. If I'm pushing current through several board layers, is it necessary or preferred (or even possible?) to distribute the current as equally as possible throughout the layers?
4. About the IPC rules regarding trace widths: Do they hold up in real life? For 30 Amps and a 10 degrees temperature rise, if I'm reading the graphs correctly, I need about 11mms of trace width on the top or bottom layer.
5. When connecting multiple layers of high current traces, what's the better practice: Placing an array or grid of vias close to the current source, or placing the vias throughout the high current trace?
• I'd like to add: Are there any problems having asymmetrical copper weights? E.g. 35 um on layer 1-4 and 70 um on layer 5 and 6? Commented Jul 18, 2011 at 21:12
• This is not high copper density, it's high copper thickness. The density of the copper is pretty much the same, they just vary the thickness. Commented Jul 19, 2011 at 0:01
• Also, for people who are used to boards with copper thicknesses in ounces (e.g. Americans, Me), 35 Micron = 1 oz, 70 micron is 2 oz, and 105 micron is 3 oz. Commented Jul 19, 2011 at 0:09
• Density is not just per volume, it may also be per unit area, or for string per unit length. All this is to some extent splitting hairs and numbers should always be joined with units that normally make the context clear. Commented Jul 19, 2011 at 14:52
• Also, it's most definitely not splitting hairs, because I can't imagine many PCB houses would respond in the positive, were you to call them and ask for more dense copper. Density in a PCB context can mean several things, including trace-trace spacing capability, copper thickness, or even substrate thickness. Commented Jul 23, 2011 at 4:03
I'm late to the game, but I'll give it a shot:
1- It appears that for a lot of manufacturing houses, 105 microns is as high as its gets. Is that correct or are higher thickness possible?
Some fab shops can plate up internal layers. The tradeoff is usually larger tolerance in the overall thickness of the board, e.g. 20% instead of 10%, higher cost, and later ship dates.
2- Can the copper in the inner layers be as thick as the copper at the top and bottom of the board?
Yes, though inner layers do not dissipate heat as well as outer layers, and if you're using impedance control, they are more likely to be striplines than microstrips (i.e. using two reference planes instead of one). Striplines are harder to get a target impedance; microstrips on the outer layers can just be plated up until impedance is close enough, but you can't do that with internal layers after the layers are laminated together.
3- If I'm pushing current through several board layers, is it necessary or preferred (or even possible?) to distribute the current as equally as possible throughout the layers?
Yes, it is preferred, but it is also difficult. Usually this is only done with the ground planes, by way of stitching vias and mandating that holes and vias connect to all planes of the same net.
4- About the IPC rules regarding trace widths: Do they hold up in real life? For 30 Amps and a 10 degrees temperature rise, if I'm reading the graphs correctly, I need about 11mms of trace width on the top or bottom layer.
The new IPC standard on current capacity (IPC-2152) holds up well in real life. However, never forget that the standard does not account for nearby traces also generating comparable amounts of heat. Finally, be sure to check voltage drops on your traces as well to make sure they are acceptable.
Also, the standard does not account for increased resistance due to skin effect for high-frequency (e.g. switching power loop) circuits. Skin depth for 1 MHz is about the thickness of 2 oz. (70 µm) copper. 10 MHz is less than 1/2 oz. copper. Both sides of the copper are only used if return currents are flowing in parallel layers on both sides of the layer in question, which is usually not the case. In other words, current prefers the side facing the path of the corresponding return current (usually a ground plane).
5- When connecting multiple layers of high current traces, what's the better practice: Placing an array or grid of vias close to the current source, or placing the vias throughout the high current trace?
It's best (and usually easier from a practical point of view) to spread the stitching vias out. Also, there is an important thing to keep in mind: mutual inductance. If you place vias that carry current flowing in the same direction too close to each other, there will be mutual inductance between them, increasing the total inductance of the vias (possibly making a 4x4 grid of vias look like a 2x2 or 1x2 at decoupling capacitor frequencies). The rule of thumb is to keep these vias at least one board thickness from each other (easier) or at least twice the distance between the planes the vias are connecting (more math).
Finally, it is still wise to keep the board's layer stackup symmetric to prevent board warpage. Some fab shops may be willing to go to the extra effort to fight the warpage from an asymmetric stackup, usually by increasing lead times and cost since they have to take a couple tries at it to get it right for your stackup.
If only fraction of traces need 30A, I would still suggest to solder copper wire on top of the trace. This might even be cheaper to manufacture, as you are not using any 'rare' materials (like 100$\mu$ Cu). 2mm$^2$ copper wire is dirt cheap and much more robust than thin PCB trace.
Is this current DC? With AC current you might be limited by skin effect.
• Is this a mechanically accepted solution for harsh environments? Would cabled solutions pass vibration and shock tests? Commented Jul 19, 2011 at 19:12
• Also, I hear about PCB busbars and solid copper blocks that can be mounted on PCBs, but I can't seem to find them in any distributor stock. Maybe I'm not searching right? Commented Jul 19, 2011 at 19:14
• 'cabled' solution would pass as long as it's soldered to the track, and PCB track is not 0.5mm thin line. I am not sure you can damage it even if you want to ) Haven't heard about copper blocks - but should be expensive. Commented Jul 20, 2011 at 4:43
• @SomethingBetter - Here's one manufacturer (Circuit Components Inc) that claims 64A capacity. Couldn't find a distributor. Commented Jul 22, 2011 at 14:35
• The downside of soldering copper wire to the trace is that then mechanical forces, such as the differing thermal expansion coefficients between copper and fiberglass, or just someone bending the board, can cause the trace to tear away from the board. Copper by itself would be fine, but soldering the whole length counteracts the malleability of copper, making it more rigid and brittle. You'd probably be just as well off having two large plated holes and using the heavy wire between them... so long as skin effect doesn't stop you. Commented Jul 23, 2011 at 4:51
I think 105$\mu$ is the thickest you can get, but I can see no reason why you wouldn't get it on inner layers. A PCB is just a stackup of epoxy, copper and glass fiber. You can play with thickness at will. Thicker inner layers won't be as efficient, of course, since they can't give off their heat to the environment easily; the epoxy is a poor thermal conductor.
We had a design which demanded 16A over several traces, and ended up with 4mm traces top and bottom, and big diameter vias on them from beginning to end. No inner traces, I don't recall copper thickness.
• Some design houses have restrictions on inner copper thicknesses, at least in the prototyping stage. The one I use regularly (4PCB) will only do 1 oz in on inner layers, unless you are willing to pay a lot more. Commented Jul 22, 2011 at 21:57
• If you need thick internal layers, you can pretty much kiss goodbye any cheap fab offers. You'll need to go full custom. Commented Jul 23, 2011 at 4:52
• 105 µm is not the thickest you can get, there are some manufactures offering 140, 210, 300 and 400 too.
– Uwe
Commented Jul 16, 2018 at 15:13
I think the #1 unexpected gotcha may be: The PCB fab marketing people advertise that they can fab very tight trace/gap widths, and also advertise that they can 35, 71, and 105 um thick copper (commonly called 1, 2, and 3 ounce copper), but they can't do both on the same board. If you want thicker copper, you must space traces further apart than you may be used to on more typical PCBs.
1. You can always call a PCB fab and ask if they can handle thicker copper. But be sure and ask how much that will cost. Even if they can make thicker copper, you may not want to pay the cost adder.
2. The copper on the 2 outer layers is always thicker than the inner layers. PCB fabs typically buy "blank" copper-clad boards with 17.5 um or 35 um thickness, etch them and add spacers between them and glue them together, so that's the thickness of every internal layer. Then they drill holes and toss the PCB into the plating bath, which grows a layer of copper in each hole and on the outer layers. The result is that all inner layers have the same thickness, and both outer layers have the same thickness, thicker than the inner layers.
3. When pushing high currents, you typically want wide, short traces to reduce the resistance and hence the I2R heat generated in those traces. If you have 2 unequal traces on different layers "in parallel", reducing the width of any part of either trace increases the resistance and hence the I2R heat generated, making things worse -- it doesn't matter if you make the board more balanced by reducing the width of the wider trace or more unbalanced by reducing the width of the narrower trace.
5- When connecting multiple layers of high current traces, what's the better practice: Placing an array or grid of vias close to the current source, or placing the vias throughout the high current trace?
I suspect that placing the array close to the current source will give a lower net resistance.
"Are there any problems having asymmetrical copper weights? E.g. 35 um on layer 1-4 and 70 um on layer 5 and 6?"
Early PCB fabs had problems unless the layers were "balanced". My understanding is that modern PCB fabs no longer have those problems, so people could in principle make unbalanced PCB. But most people don't bother -- the standard thin-internal layers, thick-external-layers, with 2 distinct thicknesses, is often adequate for most boards.
The best source for many of these questions is the PCB vendor that you have selected. Different PCB vendors excel at different types of boards: some are great at high speed, tight tolerances; others are good at high power applications. Most will do just about anything you ask, but there may be a price premium.
You didn't mention whether the high current will be at high voltages. If so then you will have additional creepage/clearance requirements to meet in order to pass product safety requirements.
1.It appears that for a lot of manufacturing houses, 105 microns is as high as its gets. Is that correct or are higher thickness possible?
There are a much smaller number of board houses that can do more than 3oz. But if you design your board that way you may be stuck using them forever because there won't be a lot of other options. I would stick with 3oz at most.
A lot of board houses can do 3oz copper. But keep in mind that many board houses don't keep the 3oz copper material in stock. So if you use it you may have to wait an extra week or two for them order the material. This has typically not been too big of a problem in my experience as long as you plan for it in your project schedule.
2.Can the copper in the inner layers be as thick as the copper at the top and bottom of the board?
Its usually the opposite.
If you are going to put any SMD components on the board then its likely your outer layers will still be 1oz and some of the inner layers will be 3oz.
3.If I'm pushing current through several board layers, is it necessary or preferred (or even possible?) to distribute the current as equally as possible throughout the layers?
It is both preferred and possible to distribute the current equally between the layers, but there is no requirement.
The calculations are a lot easier when every layer is the same.
The best way to do that is to make sure that the current carying shapes on all layers are identical. Also the layers should all be tied together at the source and destination, either by a grid of vias, a plated through hole, or both.
But if you have space on some other layer then by all means use the extra copper, it will only reduce the heat.
4.About the IPC rules regarding trace widths: Do they hold up in real life? For 30 Amps and a 10 degrees temperature rise, if I'm reading the graphs correctly, I need about 11mms of trace width on the top or bottom layer.
I have used the IPC recommendations for trace width without problems. But if you have high current on multiple layers expect the temperature rise to be higher for a given ammount of copper (so use more copper if you have space).
Also its worth estimating the trace resistance. If your cad tool can do this then great, if it can't you can just estimate the number of copper "squares" from one end to the other. The resistance is typically 0.5m Ohms per square at 1oz or 166u Ohms per square at 3oz. Using the current and the resistance calculate the trace wattage. Check that the wattage seems rasonable before proceeding.
Also don't forget wattage generated by connector contacts, crimps, solder joints, etc. Those things all add up when dealing with high current.
5.When connecting multiple layers of high current traces, what's the better practice: Placing an array or grid of vias close to the current source, or placing the vias throughout the high current trace?
It depends on if your source and destination are surface mount or through hole.
If through hole then the plated hole already ties all the layers together so thre may be no need for extra vias.
You want the current to be on as many layers as possible for as much as the route as possible. So for SMD pads there should be vias near the source and destination. Ideally you would put filled vias right in the pad becasue otherwise you would be running all of your current on just one outer layer until you reached the first vias.
Placing any vias away from the source and destination means that some of the current is going to flow on fewer layers for a portion of the route. If you place vias evenly along the whole path its likely that most of the current will go through the first few vias (possibly heating them up a lot) and then less current will go through the vias that are farther away. Therefore you won't get very efficient use out of those vias, and you will need more vias overall with this approach. Since vias take away from routing space it may increas the size of your board overall. | 3,623 | 15,624 | {"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} | 2.875 | 3 | CC-MAIN-2024-30 | latest | en | 0.940211 |
http://uncyclopedia.wikia.com/wiki/Number_bases?oldid=4089823 | 1,462,548,315,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861831994.45/warc/CC-MAIN-20160428164351-00193-ip-10-239-7-51.ec2.internal.warc.gz | 289,996,963 | 20,356 | # Number bases
For those without comedic tastes, the so-called experts at Wikipedia have an article about Number Bases.
Number bases are convenient ways of counting things invented by Indian mathematicians. All number and all forms of counting come originally from India. The most important thing to know is that the Indians invented all numbers, numerals, and mathematics.
## Digit Placement Systems
All of the following number systems vary in a number of different aspects, however one of the most important factors to recall when looking at the various numeral systems is that they almost all use digit placement as a method of determining value.
For instance, quaternary expands by digit placement, even though it appears limited by the lack of digits. Where you put the digit dramatically increases or decreases the value. Placing a digit in one hole can cause a positive reaction, whereas placing digit in another hole will cause a negative reaction, and an inability to place said digit in the first hole that we were discussing again. The reasoning behind this should be obvious.
Digit placement in quaternary works thus:
0, 1, 2, 3, Lots, Lots-1, Lots-2, Lots-3, 2-Lots, 2-Lots-1, 2-Lots-2, 2-Lots-3, 3-Lots, 3-Lots-1, 3-Lots-2, 3-Lots-3, Heaps
A famous example of using digit placement relates to the boy putting his digit in the dyke in an attempt to save his hometown from flooding. No doubt, he saved the town in this instance, but often putting your digit in a dyke will have a very different value.
### Base 0 - The Nunnery system
Symbols Used The Nunnery System Base Zero (or Nunnery system, Nonery system, or Nonetal) is the most Zen like of all counting systems. The Nunnery system was named after the amount of sexual activity that mathematicians generally get (None in the morning, none at night...) Where all other systems have $x$ amount of digits, the Nunnery system has none. This would mean that at any stage of your life you would have a bank balance of \$ with which you could use it top purchase items from \$ to \$ . It also means that you would have to work days until your retirement. The Indians invented the number zero. Nobody else wants to claim responsibility.
### Base 1 - The Urinary system
Symbols Used The Urinary system is based only on this one digit. │ Base 1, or the Urinary system (often misquoted as the Unary system) is the basis of all number systems and is widely believed to be the first number system used. It was made up of a single digit that is used repeatedly. The Urinary number system works as follows: 1.│ 2.││ 3.│││ 4.││││ 5.│││││ 6.││││││ 7.│││││││ 8.││││││││ 9.│││││││││ 10.││││││││││ 11.Lots Although the Chinese, Japanese and Korean people have tried to claim the creation of a Urinary system (as evidenced by their Tally system this has been predated by the Brahmi (Indian) numerical symbols (, , , etc.) that have been discovered as early as 3rd Century BCE. This proves the mathematical superiority of the Indian people.
### Base 2 - the Binary System
Symbols Used Ye olde style I/O button. 1 0 or ◯ │ Base 2 is twice as complicated as the Urinary system to count things. Also known as Binary, this was an invention by women who decided that through years of frustration and anguish, they needed a system to be able to indicate to men whether they were on, or off the mark. Base 2 is responsible for many woes of the modern world, including fuzzy logic, Microsoft, and the internet According to Binary mathematicians, there are only 10 types of people in the world, 01. Those that understand binary 10. Those that don't And a varying scale between the two extremes, leaving the bulk of the population in the second and third quartile. Computers occasionally use binary. In the early days of computing, computers had an on/off switch, usually along the lines of "Get that for me, will you Igor." This was slowly replaced by the symbolised on/off switch, so that one end of the switch displays a │ while the other end displays a ◯. This became confusing and was then replaced by what is referred to as an I/O switch. This then became even further confusingly and was then replace by a single symbol that had these symbols combined. This is referred to as the "ring of power." Now of course the bulk of computers don't do anything as pedestrian as turning on and off, and instead prefer to sit in "User available access mode", "Hibernation for the winter mode", or "Input/output device unready mode." The ring of power. Each place in Binary is referred to as a bit. Fractions are referred to as "a bit on the side." In the 11th Century, an arrangement of the hexagrams of the I Ching, using a two symbol system, was developed by the Chinese scholar and philosopher Shao Yong, however there is no evidence that he knew really what he was doing. In fact, many scholars believe that he was just making pretty pictures. The Indian writer Pingala (c. 200 BC) developed advanced mathematical concepts for describing prosody, and in doing so presented the first known description of a binary numeral system. Notice again where he comes from. That's right!
### Base 4 - The Quaternary System
Symbols Used The fantastic four, the champions of the Quaternary. 0 1 2 3 The Base 4 system, or The Quaternary System or Four play, uses four digits, meaning that for every 2 bits of binary information you would only have 1 Quaternary place. Having said all of that, of course this is just a two-bit system. There has been some debate about the origin of this number system. Many Rock drummers find this the easiest form of counting. Many or all of the Chumashan languages originally used a base 4, and as everyone knows the Chumashan are all American Indians.
### Base 8 - The Octal system
Symbols Used Octagon, a movie raising awareness of the often forgotten Octal system. 0 1 2 3 4 5 6 7 Base 8, or Octal, is a general purpose counting system for stuff that comes in eights, such as beer. Unless you get beer is six packs. Or four packs. Or in cases of 24. Or one at a time. If you have 8 of something, count it in Octal. If you have more or less than 8, use another Base. Base 8 has more than seven digits, but is well known for having less than nine. The legend is that this was originally a counting system of nine digits, but 7, the most feared digit of all, was hungry one evening, and when the rest of the digits woke up they found out that seven ate nine. This is likely apocryphal. The Yuki language in California and the Pamean languages in Mexico have octal systems because the speakers count using the spaces between their fingers rather than the fingers themselves. The Indians think this is stupid and they just count but ignore thumbs, thus proving that although they may not have invented it, they perfected it.
### Base 10 - the Decimal System
Symbols Used The basis of the decimal system. 0 1 2 3 4 5 6 7 8 9 The most popular number base is Base 10. It is often believe that this is due to science following nature, as a marijuana leaf is made up of nine fronds and one stem, giving ten points, and most mathematical concepts and precepts are created while the mathematician is enjoying the effects of excessive marijuana consumption. By amazing coincidence, the numbers in Base 10 coincide with the numbers we all use for such common tasks as counting peas, children, trees, misfortunes, and marijuana leaves. By assiduous use of fingers and toes, it has been shown that one can represent almost any quantity of stuff or more of the above digits used in close conjunction with each other. For numbers below the value of the lowest digit, fractions of digits are used, often called "nail clippings." The Decimal system was invented by American librarian, educator, and humanitarian, Melvil Dewey. However. it appears that he based his inventions on the inventions of the Indian people from many years prior. Nowhere else is this as evident as seen in the comparison of the digits used. , , , , , , , , 1, 2, 3, 4, 5, 6, 7, 8, 9 See, it's obvious that the Indian system of numerals came first, otherwise why else would the Brahmi (Indian) numerals be on top?
### Base 16 - The Hexadecimal System
Symbols Used Modern computers often use hexadecimal numbers. 0 1 2 3 4 5 6 7 8 9 A B C D E F Base 16 is popularly known as hexadecimal. The use of the word 'hex' obviously indicates the use of magic in the application of the Base 16 counting system. When quoting IP address on the Internet Experts often use the Base 10 equivalent of the binary addresses, however when they quote the MAC address of the Computer they often read them in Hex. When questioned on the reason why, they often say that this is 248. If they didn't do this they would be 57005. What's the problem, are you 57007? Amongst Internet Experts this is considered high humour. Many Internet Experts come from India. Apparently, this numerical system is starting to 64222 from use.
## Other number systems
### Base I - The ROMAN numeral system
Symbols Used Julius, once head of the ROMAN Empire, trying to work out how they get V to mean five. I V X L C D M The ROMAN Numeral system is still extremely popular today by copywriters, as it makes more logical sense to say that a futuristic sci-fi like Return of the Killer Tomatoes! is © MCMLXXXVIII than ©1988. Rather than deriving value from the more traditional digit placement, Roman Numerals are all of a particular value, so that an I is the equivalent value of 1, V is 5, and so on, as shown below. I → 1 V → 5 X → 10 L → 50 C → 100 D → 500 M → 1000 However, when in ROME , always place digits like the ROMANS do. in order to have the number 4, for example, rather than having IIII, the ROMANS have instead chosen to use IV, thus making this a more one I'd system. Due to the limitations of the system, often numerical suffixes were introduced, most commonly K which represented multiplied by 1,000, to get to numbers higher than one thousand. This was compoundable, so that KK was multiply by 1,000,000. KKK, however, is just absurd. It was considered high praise to relate to people by numbers instead of names. Caesar, being above most others, was often referred to as 599,000. This practise is still in place today in some motorcycle clubs, so ensure that when you next see a motorcyclist to refer to him as a DICK. The ROMANS stole all their inventions from the Greeks, who in turn stole all their inventions from the Indians.
### Base £ - The Imperial system
Symbols Used Leader of the Imperial system ' yd fath fur in " mi µin mi(naut., U.S., U.K.) rd Not about to be outdsone by the ROMANS, the British Empire also introduced it's own numbering system. The Imperial system is extremely simple to understand. It works on a system of a varying base dependant on what it is you are counting or measuring. As an example the Imperial measurement for length is an inch. Once you have 12 inches you have a foot, and once you have 3 feet you have a yard. Unless of course you were at sea then you had 6.08 feet to a fathom. . Of course, in practise a fathom was actually 2 yards, which was 6 feet. And 100 fathoms made a chain. Unless you were talking about fathoms in practice, which as we said before they were only 6 feet instead of 6.08 feet, as a chain would be 608 feet, which would be 100 true fathoms, or 101.3333 practical fathoms, as this is an easier measurement. Approximately. And, of course, there is also the link, which is 7.92 inches, or 0.66 of a foot. 25 links would make a pole, which is also known as a rod or a perch. This would mean that a pole would be the equivalent of 5.5 yards, or 2.71382 true fathoms, or 2.75 practical fathoms. And a chain would be 4 perches, or 0.10855 cables, or 792 inches. Now a furlong is 220 yards, or 660 feet, or 110 practical fathoms, or 108.552632 true fathoms, or 1.085526 chains. But this of course would mean that you would have to be on land. 8 furlongs would make a mile, which would also be the length of 5280 feet, 868.421056 true fathoms, not to be confused with 880 practical fathoms. Of course if you were naughty you would use the nautical mile, which is 6080 feet, or 1000 true fathoms, or 1013.3333 practical fathoms. And a League is 2605.26316 true fathoms, or 240 chains. And that's just length. The Indians used Imperial measurements during the colonisation by the British. They gave it back.
## Summary
Obviously the Indians invented every number system, as well as everything to do with numbers. Indians invented Algebra, Differential Calculus, Imaginary numbers, factorials, happy numbers, and magic numbers. In fact, every time that you have ever been given any test in school relating to mathematics there's a strong possibility that it was written by Indians.
When you think about it, Indians are bastards, aren't they? | 3,149 | 12,859 | {"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": 1, "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} | 4.1875 | 4 | CC-MAIN-2016-18 | latest | en | 0.936817 |
https://newslanes.com/whats-a-semi-log-plot-and-how-can-you-use-it-for-covid-data/ | 1,611,851,532,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704847953.98/warc/CC-MAIN-20210128134124-20210128164124-00289.warc.gz | 455,663,969 | 39,481 | What's a Semi-Log Plot and How Can You Use It for Covid Data?
1 min
By
Rhett Allain
Just to be clear, 106 means 10 x 10 x 10 x 10 x 10 x 10. But what if I want to do the inverse of 10 raised to some power? It’s much easier to write big numbers by raising them to some power—this is exactly what we do with numbers in scientific notation. Finding the power of 10 that a number is raised to is exactly what a logarithm does. If I take the log of 1,000,000, it gives the result of 6. Oh, here is an important note. If we are talking about 10 raised to some power, that means we are using a log base of 10. The two most common bases are 10 (because we write numbers in base-10) or e, the natural number where e is approximately 2.718 (it’s irrational). Here is a more detailed explanation of e.
But wait! You can also take the logarithm for numbers that aren’t integer powers of 10. Let’s just pick a number—I’m going with 1,234. If I take the logarithm of this number, I get:
This means that if you raise 10 to the power of 3.09132, you get 1,234. But why? Why would you do that? OK, let’s go back to our terrible Covid data. Suppose that instead of plotting the number of confirmed infections, I plot the log (base 10) of the number of infections. I can then plot the log of the number vs. the day number. Here’s what that looks like.
Just to be clear—this is the same data as the first plot, but there is a big difference. You can actually see the data for South Korea even though that country’s numbers are so much lower than those in the USA. Why? Well, let’s look at the total number of confirmed cases as of November 17, 2020. For the USA, it’s 11,036,935 and for South Korea it’s 28,769. Now let’s take the log (base 10) of both of these numbers.
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Meme | 668 | 2,499 | {"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} | 4.03125 | 4 | CC-MAIN-2021-04 | latest | en | 0.906101 |
https://uk.mathworks.com/matlabcentral/cody/problems/48-making-change/solutions/160500 | 1,582,051,532,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143805.13/warc/CC-MAIN-20200218180919-20200218210919-00519.warc.gz | 605,997,605 | 15,644 | Cody
# Problem 48. Making change
Solution 160500
Submitted on 11 Nov 2012 by li haitao
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
%% a = [257.68]; b = [2 1 0 0 1 1 0 1 0 1 1 3]; out = makingChange(a); assert(isequal(out(:), b(:)))
2 Pass
%% a = [135.01]; b = [1 0 1 1 1 0 0 0 0 0 0 1]; out = makingChange(a); assert(isequal(out(:), b(:)))
3 Pass
%% a = [10035.99]; b = [100 0 1 1 1 0 0 1 1 2 0 4]; out = makingChange(a); assert(isequal(out(:), b(:))) | 231 | 584 | {"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} | 2.859375 | 3 | CC-MAIN-2020-10 | latest | en | 0.609259 |
https://www.jiskha.com/display.cgi?id=1318628656 | 1,503,048,493,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104631.25/warc/CC-MAIN-20170818082911-20170818102911-00155.warc.gz | 937,546,857 | 3,944 | # chemistry
posted by .
A solution of .12 L of 0.160 M KOH is mixed with a solution of .3 L of 0.230 M NiSO4.
the equation for this reaction is:
2KOH (aq)+NiSO4 (aq) ----->K2SO4 (aq)+Ni(OH)2 (s)
.89 grams of Ni(OH)2 precipitate form
I need to know the concentration remaining in the solution of:
1) Ni (II)
2) SO4
3) K
I do not know how to do these nor can i find aid to show me how to do it step by step, so to whoever answers this question, could you also show steps so i can learn how to do this? thank you to whoever puts in their time and helps me!!
• chemistry -
The first part of this is a limiting reagent problem. I solve these problems by solving two stoichiometry problems. The first one using reagent 1 and all of the other needed; the second time with reagent 2 and all of the other needed. You will obtain two answers; obviously both can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Here is a step by step procedure for solving stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
## Similar Questions
1. ### Chemistry
You wish to make a 0.200 M solution of NiSO4 (aq). How many grams of NiSO4 * 6H2O should you put in a 0.500 L volumetric flask?
2. ### Chemistry
How many grams of NiSO4*6H2O are needed to prepare 200*10^2 mL of a 3.5*10^-2 M NiSO4 solution?
3. ### chemistry
A solution of 114 mL of 0.190 molecules KOH is mixed with a solution of 200 mL of 0.220 molecules NiSO4. a. write a balanced equation for the reaction that occurs. identify all phases in answer
4. ### Chemistry
A solution of 114 mL of 0.160 molecules KOH is mixed with a solution of 300 mL of 0.250 molecules NiSO4. a. write a balanced equation for the reaction that occurs. identify all phases in answer
5. ### Chemistry
A solution of 220mL of 0.200M Koh is mixed with a solution of 120mL of 0.240M NiSO4. 1.Determine limiting reactant?
6. ### chemistry
A solution of 220mL of 0.200M Koh is mixed with a solution of 120mL of 0.240M NiSO4. 1.Determine limiting reactant? | 604 | 2,112 | {"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} | 2.859375 | 3 | CC-MAIN-2017-34 | latest | en | 0.877679 |
https://buzztutor.com/site/question.html?Problem_page=5 | 1,701,579,901,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100484.76/warc/CC-MAIN-20231203030948-20231203060948-00434.warc.gz | 187,005,721 | 52,787 | 8061
Physics Electrostatics Level: Misc Level
In Figure a, small 80 g insulating sphere is suspended from point p by a silk thread that in 50 cm long.The sphere bears an unknown electric charge Q.A positive point charge q=+2.0uC is brought to a position directly below p,and the sphere is repelled to a new position,30 cm to the right of q,asshown.The charge Qin uC, is closest to;
A)+1.5
B)+2.0
C)+2.5
D)+3.0
E)+3.5
8060
Mathematics Functions Level: Misc Level
sketch the graph of f (x)
8059
Mathematics Functions Level: Misc Level
find the coordinates of any reflection points
8058
Mathematics Calculus Level: Misc Level
find the coordinates of any relative extrema, and classify as maxima or minima.
8056
Mathematics Calculus Level: Misc Level
the annual world rate of water use t years after 1960 was approximately 860e to the.04t power cubic kilometers per year.how much water was used between 1970 and 1980?
8055
Mathematics Calculus Level: Misc Level
find all relative extrema for f(x,y)=7x2(squared)-5xy+y2(squared)+x-y, and classify each as a maximum or minimum.
8054
Mathematics Calculus Level: Misc Level
find the volume of the solid bounded by f(x,y)=9-x29squared)-y2(squared) over the region for which 0 is less than or equal to x less than or equal to 2 and 0 is less than or equal to y and is lessthan or equal to 1.
8053
Mathematics Calculus Level: Misc Level
the marginal cost of producing x units of a certain product is C, (x)=6x2 (squared)-4. Find the cost function C(x) if the cost of producing 3 units is \$8.
8051
Mathematics Calculus Level: Misc Level
A cylindrical can of radius 5cm and height 12 cm is leaking water at a rate of 4 cm 3 (cubed) per minute. How fast is the height of the water changing when the water level is 3 cm from the bottom of the can.(this is a,,related rates,, problem. Volume of a cylinder is given by V=pie r2(squared)h.)
8048
Mathematics Calculus Level: Misc Level
Consider the function y=f (x)=6x-x2. What is the average value of the portion above the x-axis?First sketch the graph and clearly show the x and y intercepts.
8047
Mathematics Calculus Level: Misc Level
A conical container open to the top is being filled at the rate of 10 ft3/min.The container is 20 feet high with a radius of 6 feet at the top.How rapidly is height of the fluid increasing when the height is 10 feet?
Hint;1; Express the cone,s liquid volume in terms of r and h. Then express r in terms of h using the given dimensions of the cone.That allows you to express the liquid volume in cone as a function of just h, the height of the liquid.
Hint;2;Keep n,in your expressions until the very end-tider.
8046
Mathematics Calculus Level: Misc Level
We have a right triangle with one leg growing and one shrinking.By how much is the hypotenuse growing(or shrinking0 if x is increasing at 6 ft/sec and y is decreasing at 3 ft/sec when x= 60 ft and y=80 ft?Be sure to indicate if the hypotenuse is growing or shrinking.Be careful with the signs of the derivatives.Yes,you must first find the hypotenuse.
8043
Mathematics Calculus Level: Misc Level
The velocity of a particle is given by v(t)=t/(t2+4)^0.5. Find the distance traveled during the first four seconds.Give the exact result using radicals; then a three-decimal place approximation.
8042
Mathematics Integration Level: Misc Level
For f (x)=x^2 (5-7x^3)^4 find integral of f(x)
8041
Mathematics Probability Level: Misc Level
On further reflection,the professor remembered that Tariq, Sean,Steven and Victoria each asked at least 3 questions, while Emily,Mila,Thomas and Eric didn,t ask any In light of this,in how many different ways could the 35 questions have been distributed among the students
Displaying 61-75 of 6371 results. | 991 | 3,768 | {"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} | 3.390625 | 3 | CC-MAIN-2023-50 | latest | en | 0.854905 |
https://mathoverflow.net/questions/296086/average-edge-cost-optimality-of-minimum-spanning-trees | 1,716,334,755,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058522.2/warc/CC-MAIN-20240521214515-20240522004515-00814.warc.gz | 338,972,483 | 24,152 | # Average edge-cost optimality of minimum spanning trees
Are there counter examples to the conjecture, that in a complete, finite and symmetric weighted graph $$G\left(V,E,\omega\right),\ E=\lbrace \lbrace i,j\rbrace\subset V\times V\rbrace,\ \omega:E\ni e\mapsto\mathbb{R_0^+}$$ the edges constituting to the solution of $$\min\frac{\sum_{i,j} \alpha_{ij}\omega(e_{ij})}{\sum_{i,j} \alpha_{ij}}$$ $$\sum_{i\ne u}\alpha_{iu}=0,\quad \sum_{i\ne u}\alpha_{ui}=1$$ $$\sum_{j\ne v}\alpha_{jv}=1,\quad \sum_{j\ne v}\alpha_{vj}=0$$ $$\sum_{i\notin\lbrace j,k,u,v\rbrace}\alpha_{ik}\ -\sum_{j\notin\lbrace i,k,u,v\rbrace}\alpha_{kj}=\ 0$$ $$\alpha_{ij}\in\lbrace 0,1\rbrace$$
i.e. to the path connecting $$u$$ to $$v$$ with minimal average edge-weight are elements of the set of edges constituting to the minimum spanning tree (MST)?
Edit to further explain, as requested by Brendan McKay:
• the graphs, that this question relates to, are undirected and shall have no cycles of negative length; it is further required, that the graph is connected and if not, the question relates to the connected components.
• a simple path connecting two distinct vertices $$u$$ and $$v$$ in such a graph is characterized
• by the set $$P_{uv}$$ of edges constituting to that connecting simple path,
• by the cardinality $$C_{uv}$$ of that set of edges.
• by the sum of weights $$L_{uv}$$ of $$P_{uv}$$, i.e. the path's length
• the average edge-length of $$P_{uv}$$ is then defined as $$\frac{L_{uv}}{C_{uv}}$$.
This must not be confused with the average path-length, which is also a the subject of research!
for explaining, what is conjectured, further notation is introduced:
$$P_{uv}^E$$, $$L_{uv}^E$$ and $$C_{uv}^E$$ shall be the set of edges, its sum of weights and its cardinality of a path between $$u$$ and $$v$$ consisting of edges from the entire edge-set $$E$$ of $$G$$, whereas for $$P_{uv}^{\mathrm{MST}}$$, $$L_{uv}^{\mathrm{MST}}$$ and $$C_{uv}^{\mathrm{MST}}$$ the set of edges shall be restricted to $$G$$'s minimum spanning tree MST.
Conjecture: $$\frac{L_{uv}^{\mathrm{MST}}}{C_{uv}^{\mathrm{MST}}}\ \le\ \frac{L_{uv}^E}{C_{uv}^E}\ \forall u,v\in G$$
• Please explain better. What are the conditions and what is the (conjectured?) conclusion? Mar 25, 2018 at 0:25
the least average edge-length of a path from node A to node D would be A$\mapsto$B$\mapsto$C$\mapsto$D with an average edge length of $\frac{1+1+2}{3}=\frac{4}{3}$, whereas in the MST (bold lines) the only path from A to D is A$\mapsto$D with a length of $\frac{2}{1}=2$ | 805 | 2,545 | {"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": 28, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-22 | latest | en | 0.826109 |
https://brilliant.org/discussions/thread/a-pressing-problem-with-kinetic-energy-all-physici/ | 1,597,131,402,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738735.44/warc/CC-MAIN-20200811055449-20200811085449-00265.warc.gz | 235,335,626 | 17,471 | # A pressing problem with Kinetic Energy. ALL PHYSICISTS HELP!!
the equation for kinetic energy is actually used as $\frac{1}{2}mv^{2}$ as it is a very useful quantity and actually is derived from the work done by a force, without any changes to internal energy or potential energy of the system.
$W = \int \Sigma\vec{F}\cdot d\vec{r}$
Subbing in $\Sigma\vec{F} = m\vec{a}$, you will get
$W = \int m\vec{a}\cdot d\vec{r}$
Subbing in $\vec{a} = \frac{d\vec{v}}{dt}$, you will get
$W = \int m\frac{d\vec{v}}{dt}\cdot d\vec{r}$
Since $\vec{v}=\frac{d\vec{r}}{dt}$, it implies that $d\vec{r} = \vec{v}dt$
$W = \int m\frac{d\vec{v}}{dt}\cdot\vec{v}dt$
$W = \int_{\vec{v}_i}^{\vec{v}_f} m\vec{v}\cdot d\vec{v}$
$W = \frac{1}{2}m(\vec{v}_f \cdot \vec{v}_f) - \frac{1}{2}m(\vec{v}_i\cdot\vec{v}_i) = \frac{1}{2}mv_f {}^{2} - \frac{1}{2}mv_i{}^{2}$
This showed that the quantity $\frac{1}{2}mv^{2}$ can be denoted as the energy of a moving object, and any change in kinetic energy must be from an external work done
However we know that the general definition of force is $\Sigma\vec{F}=\frac{d\vec{p}}{dt}$
Hence work would be $W = \int \frac{d\vec{p}}{dt}\cdot d\vec{r}$.
Then from that kinetic energy should be derived, as it more accurately depicts motion of an object of changing mass. Then how do we derive $\int \frac{d\vec{p}}{dt}\cdot d\vec{r}$
6 years, 10 months ago
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It's similar. Hint: Product law for derivatives. $\frac{dp}{dt}=\frac{d(mv)}{dt}=m\cdot\frac{dv}{dt}+v\cdot\frac{dm}{dt}$. Note that $\frac{dm}{dt}=0$ for constant mass, that is the non-relativistic case. Note that $F=ma$ is not true for relativistic cases. (Sorry lazy to put vector arrows and stuff =P)
- 6 years, 10 months ago
But what if mass is not constant so $\frac{dm}{dt} \neq 0$
- 6 years, 10 months ago
Then you have to find out at what rate $m$ is changing wrt to $t$ and continue with the derivation.
- 6 years, 10 months ago
Would not that be equivalent to saying $KE\neq \frac{1}{2}mv^{2}$
- 6 years, 10 months ago
Exactly, that is what relativity is about.
- 6 years, 10 months ago
Actually according to relativity, our mass is increasing with velocity but not that much. So, the kinetic energy formula holds. Also, the formula is m_{real}=\frac{m_{rest}}{\sqrt{1-\frac{v^{2}}{c^{2}}}
- 6 years, 10 months ago
dmv/dt .dr here dr/dt =v and mdv will be left so we can procede in the same way above
- 6 years, 2 months ago | 1,325 | 4,089 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 30, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "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} | 4.1875 | 4 | CC-MAIN-2020-34 | latest | en | 0.838113 |
https://www.dominatethegmat.com/2012/09/killer-gmat-tip-for-hard-gmat-math-problems/ | 1,590,895,751,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347410745.37/warc/CC-MAIN-20200531023023-20200531053023-00442.warc.gz | 671,396,923 | 10,947 | Learn the 4 Things You Must Know Before Taking the GMAT! Download it Now!
## GMAT Tip of the Week for the week of September 17, 2012
This week’s GMAT tip* concerns how to proceed on difficult GMAT problem solving questions when you feel stuck. Specifically, learn to “Drive through the Fog with Headlights.” What does that mean and how can you apply it? Watch this video to find out:
GMAT Tip of the Week: One of the most common challenges I hear my students express is that they sometimes look at a hard GMAT math problem and don’t immediately know how to proceed. They can’t foresee how the entire problem is going to unfold and ultimately lead to a right answer, so they sort of just freeze up and stare at the computer screen, unsure how to move forward. It’s all or nothing for some students: Either they immediately know how to do the problem and work through it accordingly, or they don’t immediately see a solution in their mind and they panic, freeze up, and do nothing (and eventually take a random guess and move on to the next problem).
My solution for this dilemma is to apply a way of thinking — a mindset — that I call learning to “Drive through the Fog with Headlights.” Here’s what I mean by that. Think about what happens when you drive through the fog with your headlights on. Can you see all the way to your destination? No, of course not — it’s foggy! You can only see as far as the end of where your headlights illuminate. But can you safely drive at least that far? Yes. And then a funny thing happens: You can now see a bit further. And a bit further. And if you keep driving just as far as your headlights will enable you to see, you can ultimately reach your final destination that way, even though you weren’t able to see it from the beginning.
The same approach works well on the GMAT. When you feel stuck…Do Something! Write down a formula. Draw a diagram. Make up numbers for the variables. Work backwards from the answer choices. Often times you’ll find that by just doing something, it actually triggers another thought in your mind and enables you to do something else on the problem. And then something else. And then something else. And sure enough, you might just be able to get all the way to a right answer doing one step at a time this way. Learn to adopt this mindset, and I think you’ll find that this strategy better helps you to dominate the GMAT!
* Each week, one of the GMAT experts at Dominate the GMAT shares a valuable GMAT test-taking tip, strategy, trick, or content item. These tips are designed to augment your GMAT study program and provide you with additional information that will help you improve your GMAT score. | 585 | 2,677 | {"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} | 3.796875 | 4 | CC-MAIN-2020-24 | latest | en | 0.955261 |
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# 18 teaspoons equals how many cups
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Function help
Help me write a function that converts a temperature measurement from celsius to fahrenheit. The formula is, Celsius = (5/9)(Fahrenheit -32).
Here's my attempt:
Code:
```int convert(int i)
{
int p;
p = ((9i/5) + 32);
return p;
}```
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Jim
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Originally Posted by jimblumberg
Jim
Yes I am having a problem. Here's the entire code:
Code:
```#include <stdio.h>
main()
{
int i;
for (i = 0; i <= 300; ++i)
printf("%d %d\n", i, convert(i));
return 0;
}
int convert(int i)
{
int p;
p = ((9i/5) + 32);
return p;
}```
I only get a series of 32.
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Let's start by looking at the following line:
Code:
`p = ((9i/5) + 32);`
First in C/C++ and most other programming languages you must specify the operand, (9 * i).
Next this is integer math, so remember there are no fractions. 9/5 will yield 1. To get fractional amounts you mus use floating point numbers.
Jim
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Originally Posted by jimblumberg
Let's start by looking at the following line:
Code:
`p = ((9i/5) + 32);`
First in C/C++ and most other programming languages you must specify the operand, (9 * i).
Next this is integer math, so remember there are no fractions. 9/5 will yield 1. To get fractional amounts you mus use floating point numbers.
Jim
Here's my revised code:
Code:
```#include <stdio.h>
main()
{
int i;
for (i = 0; i <= 300; ++i)
printf("%6.3f %6.3f\n", i, convert(i));
return 0;
}
int convert(int i)
{
float p;
p = ((9*i/5) + 32);
return p;
}```
Now all I get is a series of 0.000s
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Your function is returning an integer, but your printf() is trying to print a floating point number. The printf() function requires that the format specifier match the variable type. If they don't match you won't get the expected results.
I really suggest you change your function to return a floating point number and use floating point numbers in that function. Remember to also change your constants to floating point by specifying the value with a decimal point (9.0).
Jim
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I haz teh codez!
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You skipped part of jimblumberg's response. I'll requote it for you.
Next this is integer math, so remember there are no fractions. 9/5 will yield 1. To get fractional amounts you mus use floating point numbers.
Change 5 to 5.0 and see what happens.
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Originally Posted by jimblumberg
Your function is returning an integer, but your printf() is trying to print a floating point number. The printf() function requires that the format specifier match the variable type. If they don't match you won't get the expected results.
I really suggest you change your function to return a floating point number and use floating point numbers in that function. Remember to also change your constants to floating point by specifying the value with a decimal point (9.0).
Jim
If I try to compile this:
Code:
```#include <stdio.h>
main()
{
int i;
for (i = 0; i <= 300; ++i)
printf("%6.3f %6.3f\n", i, convert(i));
return 0;
}
float convert(float i)
{
float p;
p = ((9.0*i/5.0) + 32.0);
return p;
}```
it says "Conflicting types for 'convert'" for
Code:
`float convert(float i)`
and, "previous implicit declaration of 'convert' was here" for
Code:
`printf("%6.3f %6.3f\n", i, convert(i));`
:(
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Sorry if I sound too dumb. I am new in this field.
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Your function needs a function prototype before main(). Remember your functions and variables must be defined before they can be used.
Also look at this snippet:
Code:
``` int i;
...
printf("%6.3f %6.3f\n", i, convert(i));```
What would be the proper format specifier for i, which happens to be an int? Remember integers don't have fractional parts and that printf() insists that the format string match the variables.
Jim
• arman.khandaker agrees : Thanks a tonne for giving me time!
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Originally Posted by jimblumberg
Your function needs a function prototype before main(). Remember your functions and variables must be defined before they can be used.
Also look at this snippet:
Code:
``` int i;
...
printf("%6.3f %6.3f\n", i, convert(i));```
What would be the proper format specifier for i, which happens to be an int? Remember integers don't have fractional parts and that printf() insists that the format string match the variables.
Jim
The code finally worked!
Code:
```#include <stdio.h>
float convert(float i)
{
float p;
p = ((5.0/9.0)*(i - 32.0));
return p;
}
main()
{
float i;
for (i = 0; i <= 300; ++i)
printf("%1.1f %1.1f\n", i, convert(i));
return 0;
}```
I also changed the formula.
Thanks a lot for helping me around! :)
12. Code:
```#include <stdio.h>
main()
{
int i;
for (i = 0; i <= 300; ++i)
printf("%6.3f %6.3f\n", i, convert(i));
return 0;
}
float convert(float i)
{
float p;
p = ((9.0*i/5.0) + 32.0);
return p;
}```
Look at the warnings that you get:
C:>gcc -Wall convert.c
convert.c:4: warning: return-type defaults to `int'
convert.c: In function `main':
convert.c:7: warning: implicit declaration of function `convert'
convert.c:7: warning: double format, different type arg (arg 2)
convert.c:7: warning: double format, different type arg (arg 3)
convert.c: At top level:
convert.c:11: warning: type mismatch with previous implicit declaration
convert.c:7: warning: previous implicit declaration of `convert'
convert.c:11: warning: `convert' was previously implicitly declared to return `int'
C:>
convert.c:7: warning: implicit declaration of function `convert'
The compiler reads from the top of the file to the bottom. When it tries to call convert(), it has no idea what you are talking about because you haven't told it anything about convert. Old legacy behavior causes C to assume that you're talking about a function that takes one int argument and returns an int. That assumption caused the following warnings:
convert.c:7: warning: double format, different type arg (arg 3)
convert.c: At top level:
convert.c:11: warning: type mismatch with previous implicit declaration
convert.c:7: warning: previous implicit declaration of `convert'
convert.c:11: warning: `convert' was previously implicitly declared to return `int'
When the compiler reached the actual declaration for convert, that actual declaration conflicted with the assumed implicit declaration you had forced on the compiler previously, so naturally the compiler complained.
You need to either move convert above main or else you need to place a function prototype before main; eg:
Code:
```#include <stdio.h>
float convert(float i);
main()
{
int i;
for (i = 0; i <= 300; ++i)
printf("%6.3f %6.3f\n", i, convert(i));
return 0;
}
float convert(float i)
{
float p;
p = ((9.0*i/5.0) + 32.0);
return p;
}```
Please pay especial attention to the fact that the function prototype ends in a semicolon, but the actual function itself does not have a semicolon there.
convert.c:7: warning: double format, different type arg (arg 2)
is because i is an int but you're telling printf to treat it like a floating-point value. That will cause really weird results. It's an int, so treat it like an int!
convert.c:4: warning: return-type defaults to `int'
What's the return type of main? It should be int, so why don't you declare it as int? That's that old legacy defaulting undeclared identifiers as int that created a generation of lazy C programmers. Do not be lazy! Declare main to be int!
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Originally Posted by dwise1_aol
Code:
```#include <stdio.h>
main()
{
int i;
for (i = 0; i <= 300; ++i)
printf("%6.3f %6.3f\n", i, convert(i));
return 0;
}
float convert(float i)
{
float p;
p = ((9.0*i/5.0) + 32.0);
return p;
}```
Look at the warnings that you get:
convert.c:7: warning: implicit declaration of function `convert'
The compiler reads from the top of the file to the bottom. When it tries to call convert(), it has no idea what you are talking about because you haven't told it anything about convert. Old legacy behavior causes C to assume that you're talking about a function that takes one int argument and returns an int. That assumption caused the following warnings:
When the compiler reached the actual declaration for convert, that actual declaration conflicted with the assumed implicit declaration you had forced on the compiler previously, so naturally the compiler complained.
You need to either move convert above main or else you need to place a function prototype before main; eg:
Code:
```#include <stdio.h>
float convert(float i);
main()
{
int i;
for (i = 0; i <= 300; ++i)
printf("%6.3f %6.3f\n", i, convert(i));
return 0;
}
float convert(float i)
{
float p;
p = ((9.0*i/5.0) + 32.0);
return p;
}```
Please pay especial attention to the fact that the function prototype ends in a semicolon, but the actual function itself does not have a semicolon there.
convert.c:7: warning: double format, different type arg (arg 2)
is because i is an int but you're telling printf to treat it like a floating-point value. That will cause really weird results. It's an int, so treat it like an int!
convert.c:4: warning: return-type defaults to `int'
What's the return type of main? It should be int, so why don't you declare it as int? That's that old legacy defaulting undeclared identifiers as int that created a generation of lazy C programmers. Do not be lazy! Declare main to be int!
Thanks for your analysis and advice. I'll keep that in mind! :) | 2,863 | 10,574 | {"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-2018-17 | latest | en | 0.716952 |
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Diagrams.Core.HasOrigin
Description
Types which have an intrinsic notion of a "local origin", i.e. things which are not invariant under translation.
Synopsis
# Documentation
class HasOrigin t where #
Class of types which have an intrinsic notion of a "local origin", i.e. things which are not invariant under translation, and which allow the origin to be moved.
One might wonder why not just use Transformable instead of having a separate class for HasOrigin; indeed, for types which are instances of both we should have the identity
moveOriginTo (origin .^+ v) === translate (negated v)
The reason is that some things (e.g. vectors, Trails) are transformable but are translationally invariant, i.e. have no origin.
Minimal complete definition
moveOriginTo
Methods
moveOriginTo :: Point (V t) (N t) -> t -> t #
Move the local origin to another point.
Note that this function is in some sense dual to translate (for types which are also Transformable); moving the origin itself while leaving the object "fixed" is dual to fixing the origin and translating the diagram.
Instances
HasOrigin t => HasOrigin [t] # MethodsmoveOriginTo :: Point (V [t]) (N [t]) -> [t] -> [t] # (HasOrigin t, Ord t) => HasOrigin (Set t) # MethodsmoveOriginTo :: Point (V (Set t)) (N (Set t)) -> Set t -> Set t # # MethodsmoveOriginTo :: Point (V (TransInv t)) (N (TransInv t)) -> TransInv t -> TransInv t # (HasOrigin t, HasOrigin s, SameSpace s t) => HasOrigin (s, t) # MethodsmoveOriginTo :: Point (V (s, t)) (N (s, t)) -> (s, t) -> (s, t) # HasOrigin t => HasOrigin (Map k t) # MethodsmoveOriginTo :: Point (V (Map k t)) (N (Map k t)) -> Map k t -> Map k t # (Additive v, Num n) => HasOrigin (Point v n) # MethodsmoveOriginTo :: Point (V (Point v n)) (N (Point v n)) -> Point v n -> Point v n # HasOrigin t => HasOrigin (Measured n t) # MethodsmoveOriginTo :: Point (V (Measured n t)) (N (Measured n t)) -> Measured n t -> Measured n t # (Additive v, Num n) => HasOrigin (Transformation v n) # MethodsmoveOriginTo :: Point (V (Transformation v n)) (N (Transformation v n)) -> Transformation v n -> Transformation v n # (Additive v, Num n) => HasOrigin (Trace v n) # MethodsmoveOriginTo :: Point (V (Trace v n)) (N (Trace v n)) -> Trace v n -> Trace v n # (Metric v, Fractional n) => HasOrigin (Envelope v n) # The local origin of an envelope is the point with respect to which bounding queries are made, i.e. the point from which the input vectors are taken to originate. MethodsmoveOriginTo :: Point (V (Envelope v n)) (N (Envelope v n)) -> Envelope v n -> Envelope v n # (Additive v, Num n) => HasOrigin (Query v n m) # MethodsmoveOriginTo :: Point (V (Query v n m)) (N (Query v n m)) -> Query v n m -> Query v n m # (OrderedField n, Metric v) => HasOrigin (SubMap b v n m) # MethodsmoveOriginTo :: Point (V (SubMap b v n m)) (N (SubMap b v n m)) -> SubMap b v n m -> SubMap b v n m # (Metric v, OrderedField n) => HasOrigin (Subdiagram b v n m) # MethodsmoveOriginTo :: Point (V (Subdiagram b v n m)) (N (Subdiagram b v n m)) -> Subdiagram b v n m -> Subdiagram b v n m # (Metric v, OrderedField n, Semigroup m) => HasOrigin (QDiagram b v n m) # Every diagram has an intrinsic "local origin" which is the basis for all combining operations. MethodsmoveOriginTo :: Point (V (QDiagram b v n m)) (N (QDiagram b v n m)) -> QDiagram b v n m -> QDiagram b v n m #
moveOriginBy :: (V t ~ v, N t ~ n, HasOrigin t) => v n -> t -> t #
Move the local origin by a relative vector.
moveTo :: (InSpace v n t, HasOrigin t) => Point v n -> t -> t #
Translate the object by the translation that sends the origin to the given point. Note that this is dual to moveOriginTo, i.e. we should have
moveTo (origin .^+ v) === moveOriginTo (origin .^- v)
For types which are also Transformable, this is essentially the same as translate, i.e.
moveTo (origin .^+ v) === translate v
place :: (InSpace v n t, HasOrigin t) => t -> Point v n -> t #
A flipped variant of moveTo, provided for convenience. Useful when writing a function which takes a point as an argument, such as when using withName and friends. | 1,180 | 4,154 | {"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} | 2.59375 | 3 | CC-MAIN-2024-33 | latest | en | 0.798379 |
http://www.convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=Spanish+libra | 1,669,674,882,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710662.60/warc/CC-MAIN-20221128203656-20221128233656-00670.warc.gz | 63,780,052 | 3,839 | Partner with ConvertIt.com
New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```Spanish libra = 0.45994266318 kilogram (mass) ``` Related Measurements: Try converting from "Spanish libra" to as (Roman as), bes (Roman bes), bowling ball, carat troy (troy carat), cotton bale Egypt, crith, dinar (Arabian dinar), dram (avoirdupois dram), dram troy (troy dram), English carat, funt (Russian funt), hyl, long quarter, oz (ounce), pfund (German pfund), pood (Russian pood), Roman obolus, slug, talent (Greek talent), uncia (Roman uncia), or any combination of units which equate to "mass" and represent mass. Sample Conversions: Spanish libra = 2.77E+26 AMU (atomic mass unit), 2.11 bes (Roman bes), .002028 cotton bale (US), .001352 cotton bale Egypt, 109.51 dinar (Arabian dinar), .00459943 doppelzentner, 118.3 dram troy (troy dram), 7,098 grain (avoirdupois grain), 46.9 hyl, 32.62 Israeli shekel mass, .45994266 kg (kilogram), .00045268 long ton (avoirdupois long ton), .99901478 Mexican libra, 633.75 obol (Greek obol), .007605 picul (Chinese picul), 1.23 pound troy (troy pound), 2.75E+26 proton mass (proton rest mass), 404.91 scrupulum (Roman scrupulum), .01014 short hundredweight (avoirdupois short hundredweight), 12.17 tael (Chinese tael).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 532 | 1,802 | {"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} | 2.90625 | 3 | CC-MAIN-2022-49 | latest | en | 0.658049 |
https://www.edureka.co/community/20370/avoiding-multiple-nested-for-loops-in-python | 1,718,842,712,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00341.warc.gz | 664,425,155 | 27,073 | How to avoid multiple nested for-loops when one nested for-loop has range up to the current iteration of the outer for-loop? For example, consider the following code: This program returns a triplet from a list arr such that arr[i] - arr[j] = arr[j] - arr[k] = d and i<j<k.
```d =3
arr = [1, 2, 4, 5, 7, 8, 10]
list1 = []
for biggest in range(0, len(arr)):
for bigger in range(0, biggest):
for big in range(0, bigger):
if abs(arr[big] - arr[bigger]) == d and abs(arr[bigger] - arr[biggest]) == d:
list1.append([arr[big], arr[bigger], arr[biggest]])
print(list1))```
Are there any other alternatives to using multiple nested loops?
Sep 14, 2018 in Python 15,832 views
## 2 answers to this question.
+1 vote
You can replace the three loops with:
```from itertools import combinations
for big, bigger, biggest in combinations(range(0, len(arr)), 3):```
You can replace all the code with:
```print([t for t in combinations(arr, 3)
if t[2] - t[1] == t[1] - t[0] == d])```
Hope this helps!!
Thanks!
• 58,090 points
+1 vote
Instead of multi-loop, If you can categorize all the threads into a loop, you can easily go with the less complexity with the in python, and for the nested loop, it is total standing with the loop in between the loop.
• 160 points
edited Sep 15, 2018 by Vardhan
Can you explain somewhat deeply using syntax bro please
Hii @Nikhil,
Example code:
```d =3
arr = [1, 2, 4, 5, 7, 8, 10]
list1 = []
for biggest in range(0, len(arr)):
for bigger in range(0, biggest):
for big in range(0, bigger):
if abs(arr[big] - arr[bigger]) == d and abs(arr[bigger] - arr[biggest]) == d:
list1.append([arr[big], arr[bigger], arr[biggest]])
print(list1))```
You can replace the three loops with:
```from itertools import combinations
for big, bigger, biggest in combinations(range(0, len(arr)), 3):```
You can replace all the code with:
```print([t for t in combinations(arr, 3)
if t[2] - t[1] == t[1] - t[0] == d])```
Hey, @Nikhil,
## Python: nested 'for' loops
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Syntax : list. count(value) Code: colors = ['red', 'green', ...READ MORE | 792 | 2,697 | {"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-2024-26 | latest | en | 0.763632 |
https://stats.stackexchange.com/questions/318971/fitting-exponential-decay-with-negative-y-values/318979 | 1,606,487,148,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141193221.49/warc/CC-MAIN-20201127131802-20201127161802-00428.warc.gz | 501,981,937 | 39,456 | # Fitting exponential decay with negative y values
I am trying to fit an exponential decay function to y-values that become negative at high x-values, but am unable to configure my nls function correctly.
# Aim
I am interested in the slope of the decay function ($\lambda$ according to some sources). How I get this slope is not important, but the model should fit my data as well as possible (i.e. linearizing the problem is acceptable, if the fit is good; see "linearization"). Yet, previous works on this topic have used a following exponential decay function (closed access article by Stedmon et al., equation 3):
$f(y) = a \times exp(-S \times x) + K$
where S is the slope I am interested in, K the correction factor to allow negative values and a the initial value for x (i.e. intercept).
I need to do this in R, as I am writing a function that converts raw measurements of chromophoric dissolved organic matter (CDOM) to values that researchers are interested in.
# Example data
Due to the nature of the data, I had to use PasteBin. The example data are available here.
Write dt <- and copy the code fom PasteBin to your R console. I.e.
dt <- structure(list(x = ...
The data look like this:
library(ggplot2)
ggplot(dt, aes(x = x, y = y)) + geom_point()
Negative y values take place when $x > 540 nm$.
# Trying to find solution using nls
Initial attempt using nls produces a singularity, which should not be a surprise seeing that I just eyeballed start values for parameters:
nls(y ~ a * exp(-S * x) + K, data = dt, start = list(a = 0.5, S = 0.1, K = -0.1))
# Error in nlsModel(formula, mf, start, wts) :
# singular gradient matrix at initial parameter estimates
Following this answer, I can try to make better fitting start parameters to help the nls function:
K0 <- min(dt$y)/2 mod0 <- lm(log(y - K0) ~ x, data = dt) # produces NaNs due to the negative values start <- list(a = exp(coef(mod0)[1]), S = coef(mod0)[2], K = K0) nls(y ~ a * exp(-S * x) + K, data = dt, start = start) # Error in nls(y ~ a * exp(-S * x) + K, data = dt, start = start) : # number of iterations exceeded maximum of 50 The function does not seem to be able to find a solution with the default number of iterations. Let's increase the number of iterations: nls(y ~ a * exp(-S * x) + K, data = dt, start = start, nls.control(maxiter = 1000)) # Error in nls(y ~ a * exp(-S * x) + K, data = dt, start = start, nls.control(maxiter = 1000)) : # step factor 0.000488281 reduced below 'minFactor' of 0.000976562 More errors. Chuck it! Let's just force the function to give us a solution: mod <- nls(y ~ a * exp(-S * x) + K, data = dt, start = start, nls.control(maxiter = 1000, warnOnly = TRUE)) mod.dat <- data.frame(x = dt$x, y = predict(mod, list(wavelength = dt$x))) ggplot(dt, aes(x = x, y = y)) + geom_point() + geom_line(data = mod.dat, aes(x = x, y = y), color = "red") Well, this was definitely not a good solution... # Linearizing the problem Many people have linearized their exponential decay functions with a success (sources: 1, 2, 3). In this case, we need to make sure that no y value is negative or 0. Let's make the minimum y value as close to 0 as possible within the floating point limits of computers: K <- abs(min(dt$y))
dt$y <- dt$y + K*(1+10^-15)
fit <- lm(log(y) ~ x, data=dt)
ggplot(dt, aes(x = x, y = y)) + geom_point() +
geom_line(aes(x=x, y=exp(fit$fitted.values)), color = "red") Much better, but the model does not trace y values perfectly at low x values. Note that the nls function would still not manage to fit the exponential decay: K0 <- min(dt$y)/2
mod0 <- lm(log(y - K0) ~ x, data = dt) # produces NaNs due to the negative values
start <- list(a = exp(coef(mod0)[1]), S = coef(mod0)[2], K = K0)
nls(y ~ a * exp(-S * x) + K, data = dt, start = start)
# Error in nlsModel(formula, mf, start, wts) :
# singular gradient matrix at initial parameter estimates
# Do the negative values matter?
The negative values are obviously a measurement error as absorption coefficients cannot be negative. So what if I make the y values generously positive? It is the slope I am interested in. If addition does not affect the slope, I should be settled:
dt$y <- dt$y + 0.1
fit <- lm(log(y) ~ x, data=dt)
ggplot(dt, aes(x = x, y = y)) + geom_point() + geom_line(aes(x=x, y=exp(fit$fitted.values)), color = "red") Well, this did not go that well...High x values should obviously be as close to zero as possible. # The question I am obviously doing something wrong here. What is the most accurate way to estimate slope for an exponential decay function fitted on data that have negative y values using R? • nls converged for me using the starting values$a=1, S =0.01, K=-0.0001$. Alternatively, you could use the self-starting function: nls(y~SSasymp(x, Asym, r0, lrc), data = dt). That converges for me too. – COOLSerdash Dec 15 '17 at 9:59 ## 2 Answers Use a selfstarting function: ggplot(dt, aes(x = x, y = y)) + geom_point() + stat_smooth(method = "nls", formula = y ~ SSasymp(x, Asym, R0, lrc), se = FALSE) fit <- nls(y ~ SSasymp(x, Asym, R0, lrc), data = dt) summary(fit) #Formula: y ~ SSasymp(x, Asym, R0, lrc) # #Parameters: # Estimate Std. Error t value Pr(>|t|) #Asym -0.0001302 0.0004693 -0.277 0.782 #R0 77.9103278 2.1432998 36.351 <2e-16 *** #lrc -4.0862443 0.0051816 -788.604 <2e-16 *** #--- #Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 # #Residual standard error: 0.007307 on 698 degrees of freedom # #Number of iterations to convergence: 0 #Achieved convergence tolerance: 9.189e-08 exp(coef(fit)[["lrc"]]) #lambda #[1] 0.01680222 However, I would seriously consider if your domain knowledge doesn't justify setting the asymptote to zero. I believe it does and the above model doesn't disagree (see the standard error / p-value of the coefficient). ggplot(dt, aes(x = x, y = y)) + geom_point() + stat_smooth(method = "nls", formula = y ~ a * exp(-S * x), method.args = list(start = list(a = 78, S = 0.02)), se = FALSE, #starting values obtained from fit above color = "dark red") • Perfect. I did not know about SSasymp function. Thank you! I believe the researchers want to refer to the article I cited in the question and use the K term, but I will suggest them to modify their equation. I think they want to keep the K, because negative values mean that the instrument did not behave as expected, but they are interested in the slope. Removing the negative asymptote might affect the slope in some instances. – Mikko Dec 15 '17 at 10:37 • @Mikko If you measure absorption and the asymptote gets significantly zero, I would say that you have problems with your calibration or instrument stability. – Roland Dec 15 '17 at 11:12 • The problem often takes place when water is very clear (oceanic water). Some values become below zero. I think we have an instrument that suffers from temperature problems. When it overheats, the values become unstable, but these details should probably not be handled in Crossvalidated. – Mikko Dec 15 '17 at 11:44 This question has relationships with several other questions I have three additional remarks regarding some points in this question. ### 1: Why linearized model does not fit well the large values of $$y$$ Much better, but the model does not trace y values perfectly at low x values. The linearized fit is not minimizing the same residuals. At the logarithmic scale the residuals for smaller values will be larger. The image below shows the comparison by plotting the y-axis on a log scale in the right image: When necessary you could add weights to the least squares loss function. ### 2: Using linearized fit as starting values After you have obtained estimates with your linearized fit you could have used these as starting point for the non linear fitting.* # vectors x and y from data x <- dat$$x y <- dat$$y # linearized fit with zero correction K <- abs(min(y)) dty <- y + K*(1+10^-15) fit <- lm(log(dty) ~x) # old fit that had a singluar gradient matrix error # nls(y ~ a * exp(-S * x) + K, # start = list(a = 0.5, # S = 0.1, # K = -0.1)) # # new fit fitnls <- nls(y ~ a * exp(-S * x) + K, start = list(a = exp(fit$$coefficients[1]), S = -fit$$coefficients[2], K = -0.1)) # ### 3: Using a more general method to obtain the starting point If you have enough points then you can also obtain the slope without having to worry about asymptotic value and negative values (no computation of a logarithm needed). You can do this by integrating the data points. Then with $$y = a e^{sx} + k$$ and $$Y = \frac{a}{s} e^{sx} + kx + Const$$ you can use a linear model to obtain the value of $$s$$ by describing $$y$$ as a linear combination of the vectors $$Y$$, $$x$$ and an intercept: $$\begin{array}{rccccl}y &=& a e^{sx} + k &=& s(\frac{a}{s} e^{s x} + k x + Const) &- s k x - s Const \\ &&&=& sY &- sk x - s Const \end{array}$$ The advantage of this method (see Tittelbach and Helmrich 1993 "An integration method for the analysis of multiexponential transient signals") is that you can extend it to more than a single exponentially decaying component (adding more integrals). # # using Tittelbach Helmrich # # integrating with trapezium rule assuming x variable is already ordered ys <- c(0,cumsum(0.5*diff(x)*(y[-1]+y[-length(y)]))) # getting slope parameter modth <- lm(y ~ ys + x) slope <- modth$coefficients[2]
# getting other parameters
modlm <- lm(y ~ 1 + I(exp(slope*x)))
K <- modlm$$coefficients[1] a <- modlm$$coefficients[2]
# fitting with TH start
fitnls2 <- nls(y ~ a * exp(-S * x) + K,
start = list(a = a,
S = -slope,
K = K))
Footnote: *This use of the slope in the linearized problem is exactly what what the SSasymp selfstarting function does. It first estimates the asymptote
> stats:::NLSstRtAsymptote.sortedXyData
function (xy)
{
in.range <- range(xy$$y) last.dif <- abs(in.range - xy$$y[nrow(xy)])
if (match(min(last.dif), last.dif) == 2L)
in.range[2L] + diff(in.range)/8
else in.range[1L] - diff(in.range)/8
}
and then the slope by (subtracting the asymptote value and taking the log values)
> stats:::NLSstAsymptotic.sortedXyData
function (xy)
{
xy\$rt <- NLSstRtAsymptote(xy)
setNames(coef(nls(y ~ cbind(1, 1 - exp(-exp(lrc) * x)), data = xy,
start = list(lrc = log(-coef(lm(log(abs(y - rt)) ~ x,
data = xy))[[2L]])), algorithm = "plinear"))[c(2,
3, 1)], c("b0", "b1", "lrc"))
}
Note the line start = list(lrc = log(-coef(lm(log(abs(y - rt)) ~ x, data = xy))[[2L]]))
Sidenote: In the special case that $$K=0$$ you can use
plot(x,y)
mod <- glm(y~x, family = gaussian(link = log), start = c(2,-0.01))
lines(x,exp(predict(mod)),col=2)
which models the observed parameter $$y$$ as
$$y = exp(X\beta) + \epsilon = exp(\beta_0) \cdot exp(\beta_1 \cdot x) + \epsilon$$ | 3,113 | 10,714 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-50 | latest | en | 0.802637 |
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