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https://www.shaalaa.com/question-bank-solutions/find-mean-deviation-mean-data-classes95-105105-115115-125125-135135-145145-155frequencies91316263012-mean-deviation_55653 | 1,618,063,006,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038057142.4/warc/CC-MAIN-20210410134715-20210410164715-00382.warc.gz | 1,102,413,660 | 10,476 | # Find the Mean Deviation from the Mean for the Data:Classes95-105105-115115-125125-135135-145145-155frequencies91316263012 - Mathematics
Find the mean deviation from the mean for the data:
Classes 95-105 105-115 115-125 125-135 135-145 145-155 Frequencies 9 13 16 26 30 12
#### Solution
We will compute the mean deviation from the mean in the following way:
Classes Frequency $f_i$ Midpoints $x_i$ $f_i x_i$ $\left| x_i - X \right|$ = $\left| x_i - 128 . 58 \right|$ $f_i \left| x_i - X \right|$ 95−105 9 100 900 28.58 257.22 105−115 13 110 1430 18.58 241.54 115−125 16 120 1920 8.58 137.28 125−135 26 130 3380 1.42 36.92 135−145 30 140 4200 11.42 342.6 145−155 12 150 1800 21.42 257.04 $\sum^6_{i = 1} f_i = 106$ $\sum^6_{i = 1} f_i x_i = 13630$ $\sum^8_{i = 1}| f_i x_i |- \bar{x}= 1272.6$
$N = \sum^6_{i = 1} f_i = 106$ and
$\sum^6_{i = 1} f_i x_i = 13630$
$\therefore X = \frac{\sum^6_{i = 1} f_i x_i}{\sum^6_{i = 1} f_i}$
$= \frac{13630}{106}$
$= 128 . 58$
$\therefore \text{ Mean deviation } = \frac{1}{N} \sum^8_{i = 1} f_i \left| x_i - X \right|$
$= \frac{1272 . 6}{106}$
$= 12 . 005$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 32 Statistics
Exercise 32.3 | Q 2.2 | Page 16 | 542 | 1,270 | {"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} | 4.28125 | 4 | CC-MAIN-2021-17 | latest | en | 0.446632 |
https://www.dummies.com/article/academics-the-arts/math/common-core/why-algebraic-structure-is-important-in-common-core-math-141349/ | 1,674,877,992,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499470.19/warc/CC-MAIN-20230128023233-20230128053233-00025.warc.gz | 754,168,087 | 15,993 | ##### Common Core Math For Parents For Dummies with Videos Online
Algebra can help to reveal how things are built — what different scenarios have in common with each other and what the important differences are. The Common Core State Standards refer to this idea as algebraic structure.
When you double the side lengths of a rectangle (or any geometric figure), but leave the angle measures the same, the area grows by a factor of 4. Triple the lengths and the area grows by a factor of 9. Doing so makes the relationship between length and area quadratic. Quadratic structure appears in many other guises — as the relationship between the number of vertices of a regular polygon and the number of its diagonals, for example, and as the result of multiplying two linear functions together. When you pay attention to the similarities of these various contexts, you notice that they have the same structure.
Noticing and using algebraic structure requires stepping back from your work and thinking about the big picture. In fact, doing this is a Standard for Mathematical Practice. Rather than looking at a chapter title in their textbook to decide what technique to use, students need to select techniques and ideas that make sense based on the relationships in the problem being considered.
Algebraic structure supports students working with mathematical modeling, which is an important part of the Common Core high school curriculum (and across all the grades). When students are building mathematical models, they're typically trying to understand something better that comes from outside the realm of mathematics. They're trying to answer a question that someone might reasonably ask outside of math class, and using mathematics to get a good answer. Part of building a good mathematical model involves seeing the structure of the situation. A student may ask questions such as "How are these variables related?" and "What other problems have been solved that involved similar relationships?" These questions draw the student back from the particulars of the context and help the student to focus on its mathematical structure. | 392 | 2,134 | {"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} | 3.59375 | 4 | CC-MAIN-2023-06 | longest | en | 0.959728 |
https://forum.edugorilla.com/forums/topic/x-and-y-are-both-greater-than-0-also-the-sum-of-x-and-y-is-1-what-is-the-minimum-value-of-x/ | 1,623,872,622,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626008.14/warc/CC-MAIN-20210616190205-20210616220205-00570.warc.gz | 246,361,071 | 16,619 | This topic contains 0 replies, has 0 voices, and was last updated by EduGorilla 2 years ago.
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x and y are both greater than 0. Also the sum of x and y is 1. What is the minimum value of (x + (1/x)) 2 + (y + (1/y)) 2?
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Reply To: x and y are both greater than 0. Also the sum of x and y is 1. What is the minimum value of (x….
Verify Yourself | 195 | 585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-25 | latest | en | 0.873952 |
https://www.tutorialcup.com/interview/array/array-queries-for-multiply-replacements-and-product.htm | 1,611,823,117,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704839214.97/warc/CC-MAIN-20210128071759-20210128101759-00131.warc.gz | 1,057,761,115 | 229,691 | # Array Queries for multiply replacements and product
Difficulty Level Hard
Array Query Problem
The problem “Array Queries for multiply, replacements and product” states that you are given an array of integer and there will be three types of queries, where you have to solve the following type of queries:
Type 1: There will be three values left, right and a number X.In this type of query, you have to multiply each element of the array to the value X in the range (left, right) inclusively.
Type 2: It consists of three values left, right as a range. You have to replace the elements in the given range with numbers Y, 2Y, 3Y, and so on.
Type 3: There will be two values left and right as a range. You have to find the product of all the elements within the given range. Since the number can be large. You have to count the total number of trailing zeroes in all the Type3 queries.
## Example
```arr[] = {1, 2, 3, 4, 5}
Query: { (3,2,5), (3,4,5), (2,1,3,1), (1,4,5,10), (3,2,4)}```
`3`
Explanation
Type3(2, 5), after product of all elements within the range 2 and 5 ⇒ 2 * 3 * 4 * 5 = 120
Type3(3, 5), after product of all elements within the range 3 and 5 ⇒ 3 * 4 * 5 = 60
Type2(1, 3, 1), after replacing each element as y, 2y and 3y and so on in the range 1 to 3
Type1(4, 5, 10), multiply each element with 10 within the range 4 to 5
Type3(2, 4), after the product of all elements within the range 3 and 5 ⇒ 2 * 3 * 40 = 240
Output ⇒ 3, So there will be a total of 3 trailing zeroes we have found in type 3 queries, so it is printed.
## Algorithm
1. Declare two arrays such that both of the arrays will store the number of trailing zeroes relative to numbers 2 and 5 respectively.
2. If we are calling for type1, then get the trailing zeroes of X, in terms of 2 and 5.
3. Traverse through the array within a given range. Multiply each number with a value X. Simultaneously, store the value of zeroes as a multiple of 2 and 5 into the array we created for zeroesInTwo and zeroesInFive.
4. If we are calling for type2, again get the trailing zeroes of Y, in terms of 2 and 5.
5. Store the Y at the first position of the range, 2Y on second and so on. Simultaneously store the count of trailing zeroes to the zeroesInTwo and zeroesInFive.
6. And for type3, get the sum of all the values which are in a range in zeroesInTwo and zeroesInFive, and find out the minimum of the count of zeroes in two or count of zeroes in five.
7. Print the value in sum.
## Explanation
We are given an integer array and three types of queries to solve. One of the queries says that you have to multiply some number within the range. The other one says that you have to replace some numbers. The last one says that you have to find the product of numbers within the range. Then to perform each of the queries we have separately made the three functions which perform their part for each query respectively. Then we will find out the trailing zeroes. For that we have created two arrays one is for in terms of 2, and the other is for in terms of 5.
To solve the first type of query, we will be given a number X and a range in terms of the starting point and an ending point. To find out the trailing zeroes which can be occurred. We will find out if the given number has that trailing zeroes. Then get a count of these trailing zeroes. The same thing is to do with the zeroes in terms of five. We will be traversing the array, from the starting point of the range to the ending point of the range. Then we will be multiplying the value X with each number while traversing. We will also simultaneously perform the addition on the array we create to store zeroes. The same thing is to do in type 2 query, we just need to replace each element by the given number, in a form of series.
To solve the type three query, we will set the value of sumOfTwos and sumOfFives to 0. We will be storing the value in the variable we created sumOfTwos and sumOfFives. Then we will be finding out the minimum of sumOfTwos and sumOfFives. That will be the required and final answer we will be returning.
## Code
### Implementation in C++ for Array Queries for multiply, replacements and product
```#include<vector>
#include<iostream>
using namespace std;
vector<int> zeroesInTwo(1000,0);
vector<int> zeroesInFive(1000,0);
int sum = 0;
int getTwosZeroes(int val)
{
int count = 0;
while (val % 2 == 0 && val != 0)
{
val = val / 2;
count++;
}
return count;
}
int getFivesZeroes(int val)
{
int count = 0;
while (val % 5 == 0 && val != 0)
{
val = val / 5;
count++;
}
return count;
}
void type1(int arr[],int ql, int qr, int x)
{
int twoCount = getTwosZeroes(x);
int fiveCount = getFivesZeroes(x);
for (int i = ql - 1; i < qr; i++)
{
arr[i] = arr[i] * x;
zeroesInTwo[i] += twoCount;
zeroesInFive[i] += fiveCount;
}
}
void type2(int arr[], int ql, int qr, int y)
{
int twoCount = getTwosZeroes(y);
int fiveCount = getFivesZeroes(y);
for (int i = ql - 1; i < qr; i++)
{
arr[i] = (i - ql + 2) * y;
zeroesInTwo[i] = getTwosZeroes(i - ql + 2) + twoCount;
zeroesInFive[i] = getFivesZeroes(i - ql + 2) + fiveCount;
}
}
void type3(int arr[],int ql, int qr)
{
int sumtwos = 0;
int sumfives = 0;
for (int i = ql - 1; i < qr; i++)
{
sumtwos += zeroesInTwo[i];
sumfives += zeroesInFive[i];
}
sum += min(sumtwos, sumfives);
}
int main()
{
int arr[]= {1,2,3,4,5};
int n=sizeof(arr)/sizeof(arr[0]);
for (int i = 0; i < n; i++)
{
zeroesInTwo[i] = getTwosZeroes(arr[i]);
zeroesInFive[i] = getFivesZeroes(arr[i]);
}
type3(arr,2,5);
type3(arr,4,5);
type2(arr,1,3,1);
type1(arr,4,5,10);
type3(arr,2,4);
cout << sum << endl;
return 0;
}
```
`3`
### Implementation in Java for Array Queries for multiply, replacements and product
```import java.util.*;
class type123
{
private static int zeroesInTwo[]=new int[1000];
private static int zeroesInFive[]=new int[1000];
private static int sum = 0;
private static int getTwosZeroes(int val)
{
int count = 0;
while (val % 2 == 0 && val != 0)
{
val = val / 2;
count++;
}
return count;
}
private static int getFivesZeroes(int val)
{
int count = 0;
while (val % 5 == 0 && val != 0)
{
val = val / 5;
count++;
}
return count;
}
private static void type1(int arr[],int ql, int qr, int x)
{
int twoCount = getTwosZeroes(x);
int fiveCount = getFivesZeroes(x);
for (int i = ql - 1; i < qr; i++)
{
arr[i] = arr[i] * x;
zeroesInTwo[i] += twoCount;
zeroesInFive[i] += fiveCount;
}
}
private static void type2(int arr[], int ql, int qr, int y)
{
int twoCount = getTwosZeroes(y);
int fiveCount = getFivesZeroes(y);
for (int i = ql - 1; i < qr; i++)
{
arr[i] = (i - ql + 2) * y;
zeroesInTwo[i] = getTwosZeroes(i - ql + 2) + twoCount;
zeroesInFive[i] = getFivesZeroes(i - ql + 2) + fiveCount;
}
}
private static void type3(int arr[],int ql, int qr)
{
int sumtwos = 0;
int sumfives = 0;
for (int i = ql - 1; i < qr; i++)
{
sumtwos += zeroesInTwo[i];
sumfives += zeroesInFive[i];
}
sum += Math.min(sumtwos, sumfives);
}
public static void main(String []args)
{
int arr[]= {1,2,3,4,5};
int n=arr.length;
Arrays.fill(zeroesInTwo,0);
Arrays.fill(zeroesInFive,0);
for (int i = 0; i < n; i++)
{
zeroesInTwo[i] = getTwosZeroes(arr[i]);
zeroesInFive[i] = getFivesZeroes(arr[i]);
}
type3(arr,2,5);
type3(arr,4,5);
type2(arr,1,3,1);
type1(arr,4,5,10);
type3(arr,2,4);
System.out.println(sum);
}
}
```
`3`
## Complexity Analysis
### Time Complexity
O(n) where “n” is the number of elements in the array. For each query O(N) time is required because we have to traverse whole of the range given to us.
### Space Complexity
O(n) where “n” is the number of elements in the array. Since we have created two arrays other than the input, the algorithm has linear space complexity.
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Anisha was able to crack Amazon after practicing questions from TutorialCup | 2,356 | 7,902 | {"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} | 4.09375 | 4 | CC-MAIN-2021-04 | longest | en | 0.867262 |
https://www.analystforum.com/t/multiple-irr-calculator/11358 | 1,618,154,026,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038064520.8/warc/CC-MAIN-20210411144457-20210411174457-00121.warc.gz | 752,337,855 | 5,419 | # Multiple IRR & calculator?
Could anybody say how TI plus will show multiple IRRs if a stream of cash flows results in multiple IRRs?
Man, if there are multiple IRR the calculation will take the one closest to the beginning period. For complex cash flows…unfortunately you need to use a PC. Note: IRR is a very complex calculation.
Seems like if we are given a stream of cash flow and asked to decide the multiple IRRs, I think I should try to fnd the rates for which NPV = 0. Again, that’s a trial and error method.
Guys, read the manual, if there are more that one IRR you will get “Error” appear in your screen.
kochunni, i have never seen any LOS stating that “calculate multiple IRRs”. we are required to know what are the problems arising from multiple IRRs, so chill out…
I came across couple of questions in schweser. Here is an example. This is from Which of the following statements about the internal rate of return (IRR) for a project with the following cash flow pattern is TRUE? Year 0: - 2,000 Year 1: 10,000 Year 2: -\$ 10,000 A) It has two IRRs of approximately 38 and 260 percent. B) It has a single IRR of approximately 38 percent. C) It has a single IRR of approximately 260 percent. D) No IRRs can be calculated. Your answer: A was correct! The number of IRRs equals the number of changes in the sign of the cash flow. In this case, from negative to positive and then back to negative. Although 38 percent seems appropriate, one should not automatically discount the value of 260 percent. Check answers by calculation: 10,000/1.38 - 10,000/1.382 = 1995.38 And 10,000/3.6 - 10,000/3.62 = 2006.17 Both discount rates give NPVs of approximately zero and thus, are IRRs.
usually it will be a quadratic equation. so solve -2000 + 10000/(1+r) -10000/(1+r)^2 = 0 and you get a quadratic in r. solve for r the other approach use the answers given, substitute backwards and then get the answer. the third approach which would work --> given that the cashflows have opposing signs --> inflow first year, outflow second year and an inflow again in the 3rd year --> you would have the multiple IRR situation. So B, C and D are therefore eliminated. Hence A is the answer. CP
Or you use the NPV function of your calculator, with I as the IRRs indicated in the solutions.
<> Thast what I do.
I agree with CPK…I have yet to see a question where they ask you what the actual 2 answers are, they just want you to know that there could be 2 b/c of the sign changes.
They will not ask for a number re: multiple IRR questions, probably more of a qualitative question like the one above. The post about the # of IRRS being = the number of cash flow SIGN changes is right
if at all they ask a multiple IRR question … TI BA-II Plus calculates and displays the one closest to ZERO and find the other one in the alternatives by using it to discount the cash flows … they may not give more than a couple of periods for all you know …
I’d highly recommend using cpk’s approach. | 755 | 2,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2021-17 | latest | en | 0.932695 |
http://ecomunsing.com/build-your-own-support-vector-machine | 1,723,296,393,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640808362.59/warc/CC-MAIN-20240810124327-20240810154327-00090.warc.gz | 8,397,978 | 117,255 | # Build Your Own Support Vector Machine
Support vector machines are the canonical example of the close ties between convex optimization and machine learning. Trained on a set of labeled data (i.e. this is a supervised learning algorithm) they are algorithmically simple and can scale well to large numbers of features or data samples, and have been shown to be effective on a variety of problems. Whether classifying data (spam or not-spam) or creating regression models (by dividing the space into a large set of categories), a support vector machine is one of the go-to tools for data classification.
In this tutorial, we’ll go through the steps for building a support vector machine from scratch, in order to see what’s “under the hood” of this classic algorithm. This follows the structure in Stephen Boyd’s excellent Convex Optimization book, section 8.6- with additional content drawn from Calafiore & El Ghaoui’s Optimization Models (or on Amazon here).
# Linear Classifiers¶
The core idea of a support vector machine is to find a line -or in higher dimensions, a hyperplane- which separates two labeled classes in our training dataset. We will find a way to define this hyperplane to maximize the distance between the two datasets, and then define the hyperplane by the boundary points from the classes which touch it- the support vectors (“support points” might be an easier way to think of it).
These classifiers use linear hyperplanes to separate the two classes. We will later look at nonlinear classifiers, which are more flexible and more powerful.
### Robust Linear Discrimination¶
To start off, let’s assume we have two classes, $x$ (with $M$ elements) and $y$ (with $N$ members), which can be fully separated by a hyperplane (or in two dimensions, by a line). We’ll later look at cases where we can’t fully separate the two classes.
We seek to find the hyperplane that maximizes the separation between the two classes, i.e. a dividing plane that is precisely in the middle of the two classes with distance $t$ on either side.
This would maximize the separation $t$ between our classes such that $a^T x_i – b \geq t, \; \forall i = 1,\dots,M$ and $a^T y_i – b \leq -t, \; \forall i=1,\dots,N$. We additionally want to have $a$ just be a unit vector defining our plane, so that $b$ and $t$ fully encode the distance to the classes- i.e. we want to constrain $||a||_2 \leq 1$.
The full optimization problem is then:
\begin{align} \underset{a,b,t}{\text{maximize}} \quad & t \\ \text{subject to}\quad & a^T x_i – b \geq t, \; \forall i = 1,\dots,M\\ & a^T y_i – b \leq -t, \; \forall i=1,\dots,N\\ & ||a||_2 \leq 1 \end{align}
That’s it- we have our classifier! We’ll use CvxPy, a package for symbolic convex optimization in Python, to turn this mathematical optimization problem into real programming examples.
In [1]:
import numpy as np
import matplotlib.pyplot as plt
import cvxpy
from cvxpy import *
% matplotlib inline
In [2]:
# Define the two sets
d = 2 # Dimension of problem. We'll leave at 2 for now.
m = 100 # Number of points in each class
n = 100
x_center = [1,1] # E.g. [1,1]
y_center = [3,1] # E.g. [2,2]
# Set a seed which will generate feasibly separable sets
# Note: these may only be separable with the default tutorial settings
np.random.seed(8)
# Define random orientations for the two clusters
orientation_x = np.random.rand(2,2)
orientation_y = np.random.rand(2,2)
# Generate unit-normal elements, but clip outliers.
rx = np.clip(np.random.randn(m,d),-2,2)
ry = np.clip(np.random.randn(n,d),-2,2)
x = x_center + np.dot(rx,orientation_x)
y = y_center + np.dot(ry,orientation_y)
# Check out our clusters!
plt.scatter(x[:,0],x[:,1],color='blue')
plt.scatter(y[:,0],y[:,1],color='red')
Out[2]:
<matplotlib.collections.PathCollection at 0x10f772b90>
In [3]:
## OPTIMIZATION- in CvxPy!
a = Variable(d)
b = Variable()
t = Variable()
obj = Maximize(t)
x_constraints = [a.T * x[i] - b >= t for i in range(m)]
y_constraints = [a.T * y[i] - b <= -t for i in range(n)]
constraints = x_constraints + y_constraints + [norm(a,2) <= 1]
prob = Problem(obj, constraints)
prob.solve()
print("Problem Status: %s"%prob.status)
Problem Status: optimal
In [4]:
## Define a helper function for plotting the results, the decision plane, and the supporting planes
def plotClusters(x,y,a,b,t):
# Takes in a set of datapoints x and y for two clusters,
# the hyperplane separating them in the form a'x -b = 0,
# and a slab half-width t
d1_min = np.min([x[:,0],y[:,0]])
d1_max = np.max([x[:,0],y[:,0]])
# Line form: (-a[0] * x - b ) / a[1]
d2_atD1min = (-a[0]*d1_min + b ) / a[1]
d2_atD1max = (-a[0]*d1_max + b ) / a[1]
sup_up_atD1min = (-a[0]*d1_min + b + t ) / a[1]
sup_up_atD1max = (-a[0]*d1_max + b + t ) / a[1]
sup_dn_atD1min = (-a[0]*d1_min + b - t ) / a[1]
sup_dn_atD1max = (-a[0]*d1_max + b - t ) / a[1]
# Plot the clusters!
plt.scatter(x[:,0],x[:,1],color='blue')
plt.scatter(y[:,0],y[:,1],color='red')
plt.plot([d1_min,d1_max],[d2_atD1min[0,0],d2_atD1max[0,0]],color='black')
plt.plot([d1_min,d1_max],[sup_up_atD1min[0,0],sup_up_atD1max[0,0]],'--',color='gray')
plt.plot([d1_min,d1_max],[sup_dn_atD1min[0,0],sup_dn_atD1max[0,0]],'--',color='gray')
plt.ylim([np.floor(np.min([x[:,1],y[:,1]])),np.ceil(np.max([x[:,1],y[:,1]]))])
In [5]:
# Typecast and plot these initial results
if type(a) == cvxpy.expressions.variables.variable.Variable: # These haven't yet been typecast
a = a.value
b = b.value
t = t.value
plotClusters(x,y,a,b,t)
We can see that the upper cluster had a large number of datapoints which were clipped, and thus became points on the support vector!
## Support Vector Classifiers¶
The discriminator works well for the case above, but doesn’t work when the clusters can’t be strictly separated. In real cases, there will be some crossover in the data between the classes. To handle this, we need to introduce variables which create slack in the constraints associated with points which fall on the ‘wrong’ side of our decision hyperplane.
Initially, we’ll still be trying to minimize the number of points which are inside of our ‘slab’ defined by $t$. This minimizes the number of support vectors, but makes the system more prone to variance as small movements in the support vectors can affect how our classifier is structured.
To address this, we’ll allow for a trade-off between having a minimizing the number of misclassified points (low bias) and maximizing the width of our slab (low variance).
### Initial formulation¶
We introduce vectors $u$ and $v$ which take the place of $t$, and create slack in the constraints for classifying the constraints. We want these to be sparse, and have positive entries:
\begin{align} \underset{a,b,t}{\text{minimize}} \quad & \mathbf{1}^T u + \mathbf{1}^T v \\ \text{subject to}\quad & a^T x_i – b \geq 1 – u_i, \; \forall i = 1,\dots,M\\ & a^T y_i – b \leq -1 + v_i, \; \forall i=1,\dots,N\\ & u \succeq 0, \; v \succeq 0 \end{align}
In [6]:
## Generate data- this time, it can overlap!
# Relies on the centerpoints defined above
np.random.seed(2) #2 works well
orientation_x = np.random.rand(2,2)
orientation_y = np.random.rand(2,2)
x = x_center + np.dot(np.random.randn(m,d),orientation_x)
y = y_center + np.dot(np.random.randn(n,d),orientation_y)
# Check out our clusters!
ax = plt.subplot(111)
plt.scatter(x[:,0],x[:,1],color='blue')
plt.scatter(y[:,0],y[:,1],color='red')
ax.set_ylim([np.floor(np.min([x[:,1],y[:,1]])),np.ceil(np.max([x[:,1],y[:,1]]))])
Out[6]:
(-1.0, 3.0)
In [7]:
## OPTIMIZATION- in CvxPy!
a = Variable(d)
b = Variable()
u = Variable(m)
v = Variable(n)
obj = Minimize(np.ones(m)*u + np.ones(n)*v)
x_constraints = [a.T * x[i] - b >= 1 - u[i] for i in range(m)]
y_constraints = [a.T * y[i] - b <= -1 + v[i] for i in range(n)]
u_constraints = [u[i] >= 0 for i in range(m)]
v_constraints = [v[i] >= 0 for i in range(n)]
constraints = x_constraints + y_constraints + u_constraints + v_constraints
prob = Problem(obj, constraints)
prob.solve()
print("Problem Status: %s"%prob.status)
plotClusters(x,y,a.value,b.value,1)
Problem Status: optimal
We see that the decision plane makes intuitive sense- things seem to be working! Also, note that we’re trading off the penalties for all the points which are on the ‘wrong side’ of the slab- this means that the decision plane rests a bit closer to the blue cluster, as the red cluser is much more dense if it were to move over, and the weight would rapidly increase as more constraints are violated.
## Canonical Support Vector¶
This system is prone to variance, as the support vectors change slightly. Increasing the slab width would create a more robust classifier (reducing variance) at the expense of more bias (initial misclassification). We can do this by trading off between the magnitude of $a$ and the penalty for misclassification:
\begin{align} \underset{a,b,t}{\text{minimize}} \quad & ||a||_2 + \gamma \left( \mathbf{1}^T u + \mathbf{1}^T v \right) \\ \text{subject to}\quad & a^T x_i – b \geq 1 – u_i, \; \forall i = 1,\dots,M\\ & a^T y_i – b \leq -1 + v_i, \; \forall i=1,\dots,N\\ & u \succeq 0, \; v \succeq 0 \end{align}
In [8]:
## Optimization- just a few modifications to our previous problem!
gamma = Parameter()
gamma.value = 0.4
obj = Minimize(norm(a,2) + gamma*(np.ones(m)*u + np.ones(n)*v) )
constraints = x_constraints + y_constraints + u_constraints + v_constraints
prob = Problem(obj, constraints)
prob.solve()
print("Problem Status: %s"%prob.status)
## Plotting the results
plotClusters(x,y,a.value,b.value,1)
Problem Status: optimal
In [9]:
# We can change the value of Gamma around a bit
gamma.value = 0.05
prob.solve()
plotClusters(x,y,a.value,b.value,1)
# The Scikit-learn method¶
Let’s be realistic- my code isn’t optimized, I’m writing this in Python rather than C++, and relying on Cvxpy for optimization probably isn’t the fastest or most scalable. If you want an off-the-shelf package for deployment, head over to the Scikit-learn module for support vector machines. This is what the pros will be doing for prototyping, anyway.
In [10]:
data = np.vstack([x,y])
labels = np.vstack([ np.zeros([m,1]), np.ones([n,1]) ]).ravel()
In [11]:
from sklearn import svm
clf = svm.SVC(kernel='linear', C=1-0.05)
clf.fit(data,labels)
Out[11]:
SVC(C=0.95, cache_size=200, class_weight=None, coef0=0.0,
decision_function_shape=None, degree=3, gamma='auto', kernel='linear',
max_iter=-1, probability=False, random_state=None, shrinking=True,
tol=0.001, verbose=False)
In [12]:
a1 = -np.matrix(clf.coef_).T
b1 = clf.intercept_
plotClusters(x,y,a1,b1,1)
# Conclusions¶
We just saw a hint of the potential of support vector machines to handle more complicated classification problems through use of alternative basis functions, known as the ‘kernel trick‘ – we’ll explore this in a future tutorial. Even with linear selection space, support vector machines are a great tool that scale well, take advantage of advances in convex optimization, and handle a wide variety of problems. Because we only need to preserve the points which define the dividing hyperplane, we can train on a very large dataset and then throw away all but the support vectors- this makes them extremely scalable.
However, they have limitations. They are binary classifiers, so managing more classes requires recursively separating classes. They require a good number of samples relative to the number of features- otherwise our hyperplanes will be underdefined.
If you’re interested in exploring more, I’d highly recommend checking out Stephen Boyd’s book, or seeing this Pythonprogramming.net tutorial which builds a SVM fully from scratch, including the creation of a gradient descent optimizer. | 3,280 | 11,716 | {"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": 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} | 3.296875 | 3 | CC-MAIN-2024-33 | latest | en | 0.911434 |
https://proofwiki.org/wiki/Are_there_any_more_Fermat_Primes%3F | 1,590,571,208,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392142.20/warc/CC-MAIN-20200527075559-20200527105559-00241.warc.gz | 510,627,222 | 9,790 | # Are there any more Fermat Primes?
## Open Question
Are there any more Fermat primes than the $5$ that are known about?
### Sequence
The sequence of Fermat primes begins:
$\displaystyle 2^{\paren {2^0} } + 1$ $=$ $\displaystyle 3$ $\displaystyle 2^{\paren {2^1} } + 1$ $=$ $\displaystyle 5$ $\displaystyle 2^{\paren {2^2} } + 1$ $=$ $\displaystyle 17$ $\displaystyle 2^{\paren {2^3} } + 1$ $=$ $\displaystyle 257$ $\displaystyle 2^{\paren {2^4} } + 1$ $=$ $\displaystyle 65 \, 537$
No other Fermat primes are known. | 185 | 523 | {"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": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2020-24 | latest | en | 0.50522 |
https://oeis.org/A006085 | 1,618,468,402,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038083007.51/warc/CC-MAIN-20210415035637-20210415065637-00111.warc.gz | 533,722,774 | 4,400 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A006085 Continued fraction for e/4. (Formerly M1822) 5
0, 1, 2, 8, 3, 1, 1, 1, 1, 7, 1, 1, 2, 1, 1, 1, 2, 7, 1, 2, 2, 1, 1, 1, 3, 7, 1, 3, 2, 1, 1, 1, 4, 7, 1, 4, 2, 1, 1, 1, 5, 7, 1, 5, 2, 1, 1, 1, 6, 7, 1, 6, 2, 1, 1, 1, 7, 7, 1, 7, 2, 1, 1, 1, 8, 7, 1, 8, 2, 1, 1, 1, 9, 7, 1, 9, 2, 1, 1, 1, 10, 7, 1, 10, 2, 1, 1, 1, 11, 7, 1, 11, 2, 1, 1, 1, 12, 7, 1, 12, 2, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 REFERENCES D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 2, p. 601. N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS Harry J. Smith, Table of n, a(n) for n = 1..20000 G. Xiao, Contfrac Index entries for linear recurrences with constant coefficients, signature (1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, 1, -1, 1, -1). FORMULA First seven terms are 0, 1, 2, 8, 3, 1, 1; then a(8k)=1, a(8k+1)=k, a(8k+2)=7, a(8k+3)=1, a(8k+4)=k, a(8k+5)=2, a(8k+6)=1, a(8k+7)=1. - Benoit Cloitre, Apr 08 2003 G.f.: x*(1+x+7*x^2-4*x^3+5*x^4-4*x^5+5*x^6-4*x^7+9*x^8-12*x^9-3*x^10-x^11-x^13-6*x^16+7*x^17+x^18) / ((1-x)^2*(1+x)*(1+x^2)^2*(1+x^4)^2). - Colin Barker, May 16 2016 EXAMPLE 0.679570457114761308840071867... = 0 + 1/(1 + 1/(2 + 1/(8 + 1/(3 + ...)))). - Harry J. Smith, May 10 2009 MATHEMATICA ContinuedFraction[E/4, 120] (* Harvey P. Dale, Apr 01 2011 *) Join[{0, 1, 2, 8, 3}, LinearRecurrence[{1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, 1, -1, 1, -1}, {1, 1, 1, 1, 7, 1, 1, 2, 1, 1, 1, 2, 7, 1, 2}, 97]] (* Ray Chandler, Sep 03 2015 *) PROG (PARI) { allocatemem(932245000); default(realprecision, 40000); x=contfrac(exp(1)/4); for (n=1, 20000, write("b006085.txt", n, " ", x[n])); } \\ Harry J. Smith, May 10 2009 (PARI) concat(0, Vec(x*(1+x+7*x^2-4*x^3+5*x^4-4*x^5+5*x^6-4*x^7+9*x^8-12*x^9-3*x^10-x^11-x^13-6*x^16+7*x^17+x^18)/((1-x)^2*(1+x)*(1+x^2)^2*(1+x^4)^2) + O(x^50))) \\ Colin Barker, May 16 2016 CROSSREFS Cf. A019741 = Decimal expansion. - Harry J. Smith, May 10 2009 Sequence in context: A085192 A320972 A273692 * A021357 A214072 A016640 Adjacent sequences: A006082 A006083 A006084 * A006086 A006087 A006088 KEYWORD nonn,cofr,easy AUTHOR EXTENSIONS More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 20 2003 STATUS approved
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Last modified April 15 02:27 EDT 2021. Contains 342974 sequences. (Running on oeis4.) | 1,343 | 2,804 | {"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-2021-17 | latest | en | 0.475747 |
https://www.coin-or.org/GAMSlinks/benchmarks/MINLP/allSolver_convex_070902/2-AlphaECP-1.trc.convex_6-SBB-1.trc.convex_sqr.htm | 1,516,369,690,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887981.42/warc/CC-MAIN-20180119125144-20180119145144-00293.warc.gz | 874,644,971 | 4,118 | ## Solver Square Comparison: Considers all models.
Date / Time: 08/24/07 21:45:30
Solver comparison utility.
Compares all solver return outcomes (for example optimal, locally optimal, infeasible, unbounded, fail) of one solver with all return outcomes of another solver. Interrupt denotes resource or iteration limit has been reached. Solver ALPHAECP is represented on the left (rows) and solver SBB on top (columns). See the solver return definitions for return codes.
Models having trace data only in one trace file are listed in the "no data" column of the other.
Tracefile 1 : 2-AlphaECP-1.trc.convex Tracefile 2 : 6-SBB-1.trc.convex Solvers used : ALPHAECP SBB Modeltype(s) MINLP
SBB: optimal SBB: feasible SBB: infeasible SBB: unbounded SBB: fail SBB: no data total ALPHAECP ALPHAECP: optimal - - - - - - - ALPHAECP: feasible - 30 - - 2 - 32 ALPHAECP: infeasible - - - - - - - ALPHAECP: unbounded - - - - - - - ALPHAECP: fail - 1 - - 3 - 4 ALPHAECP: no data - - - - - - - total SBB - 31 - - 5 - 36
### Solver return definitions:
Outcome Model Status Solver Status optimal 1 or 15 1 locally optimal 2 any feasible 8 or 16 1 or 2 or 3 or 4 or 5 infeasible 4 or 5 or 10 or 19 1 unbounded 3 or 18 1 fail all other all other
### Solver Resource Times
• Models for each solver pair outcome. Listed are the solver resource times TIME(.) in seconds, as well as the ratio RATIO(./.) of resource times for the two solvers if both solved optimally.
• Also listed are the objective values OBJ(.) using both solvers. The better solution found is listed in boldface. A solution is considered better, if the relative objective function difference is greater than 1.00E-05. If both solutions are less than 1e-1, we use the absolute difference.
• Solver resource time ratios for a particular model are listed only if one solver has resource greater than 5.00E-02.
Modelname Time (ALPHAECP) Time (SBB) Ratio (ALPHAECP/SBB) Obj (ALPHAECP) Obj (SBB) alan 1.0000 0.0100 100.000 2.92500000 2.92500000 batch 2.0000 0.2300 8.696 285506.48149069 285506.50815158 du-opt 1124.0000 1.7400 645.977 3.55567908 3.56108965 du-opt5 453.0000 20.8200 21.758 8.07307759 8.07365758 ex1223 3.0000 0.0600 50.000 4.57957678 4.57958240 ex1223a 3.0000 0.0100 300.000 4.57957997 4.57958240 ex1223b 3.0000 0.0000 +INF.000 4.57957678 4.57958240 fac1 83.0000 0.0100 8300.000 160912612.35016900 160912612.35016900 fac3 8.0000 0.2900 27.586 31982309.84800000 31982309.84800000 m3 2.0000 0.4000 5.000 37.80000000 37.80000000 m6 7.0000 196.7700 0.036 82.25687690 129.82493644 m7 33.0000 223.4600 0.148 106.75687690 123.96437781 meanvarx 1.0000 0.0200 50.000 14.36907510 14.49698300 nvs03 1.0000 0.0200 50.000 16.00000000 16.00000000 nvs10 2.0000 0.0200 100.000 -310.80000000 -310.80000000 risk2bpb 2.0000 0.4200 4.762 -55.87613940 -55.73616850 st_e14 2.0000 0.0200 100.000 4.57957678 4.57958240 st_miqp1 1.0000 0.0300 33.333 281.00000000 281.00000000 st_miqp2 1.0000 0.0300 33.333 2.00000000 2.00000000 st_miqp4 1.0000 0.0000 +INF.000 -4574.00000000 -4574.00000000 stockcycle 3600.0000 98.0100 36.731 436419.12998549 143295.16500000 st_test5 1.0000 0.0600 16.667 -110.00000000 -110.00000000 st_test6 1.0000 0.0900 11.111 471.00000000 471.00000000 st_test8 1.0000 0.0100 100.000 -29605.00000000 -29605.00000000 st_testgr1 2.0000 0.0400 50.000 -12.81160000 -12.72810000 st_testgr3 1.0000 0.0600 16.667 -20.59000000 -20.46880000 st_testph4 1.0000 0.0200 50.000 -80.50000000 -80.50000000 synthes1 1.0000 0.0300 33.333 6.00965002 6.00975891 synthes2 5.0000 0.0400 125.000 73.03454960 73.03531253 synthes3 4.0000 0.1800 22.222 68.00932736 68.00974052
Modelname Time (ALPHAECP) Time (SBB) Ratio (ALPHAECP/SBB) Obj (ALPHAECP) Status (SBB) fo7 504.0000 324.9200 --- 20.72918438 mstat(14) sstat( 4) o7 3606.0000 317.6600 --- 153.81988735 mstat(14) sstat( 4)
Modelname Time (ALPHAECP) Time (SBB) Ratio (ALPHAECP/SBB) Status (ALPHAECP) Obj (SBB) risk2b 4.0000 0.3900 --- mstat(14) sstat(10) -55.73616850
Modelname Time (ALPHAECP) Time (SBB) Ratio (ALPHAECP/SBB) Status (ALPHAECP) Status (SBB) tls12 3601.0000 578.4800 --- mstat(14) sstat( 3) mstat(14) sstat( 4) tls6 3600.0000 173.5400 --- mstat(14) sstat( 3) mstat(14) sstat( 4) tls7 3601.0000 259.4800 --- mstat(14) sstat( 3) mstat(14) sstat( 4)
Modelname Time (ALPHAECP) Obj (ALPHAECP) alan 1.0000 2.92500000 batch 2.0000 285506.48149069 du-opt 1124.0000 3.55567908 du-opt5 453.0000 8.07307759 ex1223 3.0000 4.57957678 ex1223a 3.0000 4.57957997 ex1223b 3.0000 4.57957678 fac1 83.0000 160912612.35016900 fac3 8.0000 31982309.84800000 fo7 504.0000 20.72918438 gbd 1.0000 2.20000000 m3 2.0000 37.80000000 m6 7.0000 82.25687690 m7 33.0000 106.75687690 meanvarx 1.0000 14.36907510 nvs03 1.0000 16.00000000 nvs10 2.0000 -310.80000000 o7 3606.0000 153.81988735 risk2bpb 2.0000 -55.87613940 st_e14 2.0000 4.57957678 st_e35 0.0000 107419.80448037 st_miqp1 1.0000 281.00000000 st_miqp2 1.0000 2.00000000 st_miqp3 1.0000 -6.00000000 st_miqp4 1.0000 -4574.00000000 st_miqp5 1.0000 -333.88903104 stockcycle 3600.0000 436419.12998549 st_test5 1.0000 -110.00000000 st_test6 1.0000 471.00000000 st_test8 1.0000 -29605.00000000 st_testgr1 2.0000 -12.81160000 st_testgr3 1.0000 -20.59000000 st_testph4 1.0000 -80.50000000 synthes1 1.0000 6.00965002 synthes2 5.0000 73.03454960 synthes3 4.0000 68.00932736
Modelname Time (ALPHAECP) Status (ALPHAECP) risk2b -- mstat(14) sstat(10) tls12 -- mstat(14) sstat( 3) tls6 -- mstat(14) sstat( 3) tls7 -- mstat(14) sstat( 3)
Modelname Time (SBB) Obj (SBB) alan 0.0100 2.92500000 batch 0.2300 285506.50815158 du-opt 1.7400 3.56108965 du-opt5 20.8200 8.07365758 ex1223 0.0600 4.57958240 ex1223a 0.0100 4.57958240 ex1223b 0.0000 4.57958240 fac1 0.0100 160912612.35016900 fac3 0.2900 31982309.84800000 m3 0.4000 37.80000000 m6 196.7700 129.82493644 m7 223.4600 123.96437781 meanvarx 0.0200 14.49698300 nvs03 0.0200 16.00000000 nvs10 0.0200 -310.80000000 risk2b 0.3900 -55.73616850 risk2bpb 0.4200 -55.73616850 st_e14 0.0200 4.57958240 st_miqp1 0.0300 281.00000000 st_miqp2 0.0300 2.00000000 st_miqp4 0.0000 -4574.00000000 stockcycle 98.0100 143295.16500000 st_test5 0.0600 -110.00000000 st_test6 0.0900 471.00000000 st_test8 0.0100 -29605.00000000 st_testgr1 0.0400 -12.72810000 st_testgr3 0.0600 -20.46880000 st_testph4 0.0200 -80.50000000 synthes1 0.0300 6.00975891 synthes2 0.0400 73.03531253 synthes3 0.1800 68.00974052 | 2,811 | 6,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-05 | latest | en | 0.722615 |
https://aakashdigitalsrv1.meritnation.com/ask-answer/question/determine-the-point-on-graph-of-linear-equation-2x-5y-19-who/linear-equations-in-two-variables/8480549 | 1,653,606,286,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00455.warc.gz | 126,563,509 | 9,050 | # determine the point on graph of linear equation 2x+5y=19, whose ordinate is1 1/2 times its abscissa.
According to the given statement, y - coordinate = 1 1/2 (X-coordinate) = 3/2 of X. Now, the value of y in eq. 2x+5y=19 will be = 3/2 of X. Therefore, 2x+5y=19 2x+5×3/2x = 19 2x+15x/2 = 19 2x+15x = 38 17x = 38 x = 38/17 = 2 (approx) Hence, the point is (2,3). Now place it on the graph. Hope, you understand it very well.
• 25
you are an genious
• -3
you are a genious, trust me it's ironical
• -4
• -5
How x = 3/5 came ?? Poor answer :)
• -3
:(
• -3
How did 2x +3y =38 came
• -3
How 3/2 x came
• 4
X=2 Y=3 Ordinate is one and a half times abscissa. Y=1.5x 3=1.5×2
• 2
X=2 , Y=3 2(2)+5(3)=19 Ordinate is one and a half times its abscissa... Y=1.5x 3=1.5×2
• 1
Given- 2x+5y=19 So,2x+5y-19=0 (1) And given that, y=1 1/2x Therefore y=3/2x So, 2x=3y 2x-3y=0 (2) On comparing (1) and (2) We get 2x+5y-19=3x-2y Solving it, 7y=19+x Now putting x=2 We get 7y=19+2 7y=21 y=3 So we get coordinates (2 , 3)
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MC BC
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https://www.shaalaa.com/question-bank-solutions/factorising-polynomial-completely-after-obtaining-one-factor-factor-theorem-a-two-digit-positive-number-such-that-product-its-digits-6-if-9-added-number-digits-interchange-their-places-find-number_19835 | 1,575,942,970,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540525781.64/warc/CC-MAIN-20191210013645-20191210041645-00159.warc.gz | 857,854,982 | 10,590 | Share
A Two Digit Positive Number is Such that the Product of Its Digits is 6. If 9 is Added to the Number, the Digits Interchange Their Places. Find the Number. - ICSE Class 10 - Mathematics
ConceptFactorising a Polynomial Completely After Obtaining One Factor by Factor Theorem
Question
A two digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number.
Solution
Let the digit at the tens place be ‘a’ and at units place be ‘b’.
The two-digit so formed will be 10a + b.
According to the first condition, the product of its digits is 6.
⇒ a x b =6
=> x = 6/b ...(1)
According to second condition
10a + b + 9 = 10b + a
⇒ 9a - 9b = 9
=> a - b = 1
=> a - 6/a= 1 From 1
=> a^2 - a - 6 = 0
=> (a - 3)(a + 2) = 0
=> a= -3 or 2
Since a digit cannot be negative, a = 2.
=> b = 6/a = 6/2 = 3`
Thus, the required number = 10a + b = 10(2) + 3 = 23
Is there an error in this question or solution?
APPEARS IN
2013-2014 (March) (with solutions)
Question 10.1 | 4.00 marks
Solution A Two Digit Positive Number is Such that the Product of Its Digits is 6. If 9 is Added to the Number, the Digits Interchange Their Places. Find the Number. Concept: Factorising a Polynomial Completely After Obtaining One Factor by Factor Theorem.
S | 412 | 1,333 | {"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} | 4.4375 | 4 | CC-MAIN-2019-51 | latest | en | 0.867287 |
https://www.physicsforums.com/threads/define-a-relation-on-a-z-by-n-m-iff-n-m-3k-where-kez.3675/ | 1,571,059,661,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653247.25/warc/CC-MAIN-20191014124230-20191014151730-00419.warc.gz | 985,516,051 | 19,611 | # Define a relation on A=Z by n~m iff n-m=3k where kEZ
#### StephenPrivitera
What is a relation? My book says
If A is a set, then R[subset] A x A is called a relation on A. Does this mean R[subset]A2 or R, a subset of A, multiplied by A?
Then the book talks about ordering. And for partial orderings it uses the notation a[<=]b rather than a~b. Does this mean less than or equal to, or is it just another notation to mean related to? If it is the former, it seems silly to me that a linear ordering is a partial ordering with the additional property that if (a,b)E R, then a[<=]b or b[<=]a because one of those two statements must be true (if not both)!
Then the book says A=R with ordinary ordering is a linear ordering. This is nonsense, because we're talking about relations. Wouldn't we have to say A[subset]R^2 before we can talk about ordering?
As an example, the book gives, Define a relation on A=Z by n~m iff n-m=3k where kEZ. All I can say is if R[subset]Z^2, R can be defined by all points (n,m) such that mEZ and n=3k +m, where kEZ, but that seems rather simplistic, and I used no math to show that result.
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#### HallsofIvy
Homework Helper
For A and B sets, AxB is the "Cartesian product", the collection of all pairs where the first member is from A and the second is from B.
(I'll bet that's in your book.) It's not actually necessary that a "relation" be a subset of AxA. For example, if we take A to be "set of men", B to be "set of women" then the relation "a is married to b" is a subset of AxB.
An order (or partial order) is a special kind of relation that satisfies the condition "if a~b and b~c, then a~c" (the transitive property). It isn't necessarily "less than" but that's the key example. Another important example is: A is the collection of all sets and X<= Y means "X is a subset of Y".
#### StephenPrivitera
So a relation is any subset of a Cartesian product?
Say A has an equivalence relation. Does this mean that some subset of AxA has equivalence properties?
Does the symbol "~" represent the operation that relates a with b? For example, might a~b signify a=3b or a>b2?
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#### HallsofIvy
Homework Helper
1. Yes, any subset of AxA is, by definition, a relation on A. Any subset of AxB is a relation between A and B.
2. I'm not sure what you mean. An equivalence relation on a set A IS a subset of AxA such that a) (a,a) is in the subset for all a in A.
b) If (a,b) is in the set then (b,a) is also. c) If (a,b) and (b,c) are in the set then so is (a,c).
If that's what you mean by "equivalence properties", then, yes, the definition of "equivalence on A" is that the equivalence relation is a subset of AxA with those properties.
3. Yes, "~" is a standard notation for a general relation. If "a is related to b" (that is, if (a,b) is in the subset of AxB that defines the relation) then we may write a~b. You will also sometimes see "aRb" to mean "a is related to b".
#### StephenPrivitera
Yes, that's what I meant by equivalence properties. That helps clear things up, but could you give me a concrete example of a relation?
Let me show you an example from the book. Tell me what you think.
Define an equivalence relation on A=Z by n~m iff n-m is a multiple of 3. What are the equivalence classes?
I'm looking for S, where S[subset]Z2.
n-m=3k
k is an element of Z
n=3k+m
S={(3k+m,m)}
so when k=0, we have the element (m,m) [property a]
if n=3k+m, then n~m
and m~n since
m-n=-3k=3k [property b]
If n~m and m~p, then n~p since
n-m=3k1 and m-p=3k2
n-3k1-p=3k2
n-p=3(k1+k2)=3k [property c]
From the book, we have, "If ~ is an equivalence relation on A, and a is an element of A,then cl(a)={xEA : a~x}"
So, I ask, in the example above, what is the relation ~? Is it n=3k+m? Then would cl(a)={xES : a=3k+x}? I don't even know what that means anyway. I just wrote it. Maybe, the equivalence class containing a is the set of all elements which are related to a. Can I be more specific than that for answering the second part of the example? Feel free to elaborate on equivalence classes, or anything else for that matter.
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#### HallsofIvy
Homework Helper
So, I ask, in the example above, what is the relation ~? Is it n=3k+m?
No, you already have defined a relation as a "set of ordered pairs" so the relation ~ is NOT an equation. In this case we can't actually write out all the pairs- there are an infinite number. I can show you a few of them: (3,6), (5,8), (4,16)- since 6-3= 3, 8-5= 3, 16-4= 12 are all divisible by 3. (3,5), (5,10) are examples of pairs that are NOT in the set.
I'm looking for S, where S subset Z2.
No, the problem asked for "equivalance classes". The relation itself is a subset of Z2 but as the definition you give says: "If ~ is an equivalence relation on A, and a is an element of A,then cl(a)={xEA : a~x}"- cl(a), the equivalence class that contains a, is a subset of A.
A crucial point about equivalence relations on a set is that they "partition" the set: every member of the set is in exactly one equivalence class. The simplest way to find equivalence classes is to pick an arbitrary member of A and start looking for other members that are equivalent.
In this case, A is the set of integers so let's start with 0.
What numbers are equivalent to 0? "a~b" means a-b is a multiple of 3 so a~0 means a-0= a is a multiple of 3: cl(0) is all multiples of 3: {3, -3, 6, -6, 9, -9,...}. We can write the set as {3k|k in Z}.
1 is not in that set. What numbers are equivalent to 1? Again, that means numbers n such that n-1 is a multiple of 3: n-1= 3k or
n= 3k+1. 1, 4, 7, ..., -2, -5, .... We can write this set as
{3k+1| k in Z}.
What about 2? Now we need n-2= 3k or n= 3k+ 2: numbers like 2, 5, 8, ... -1, -4,... are equivalent to 2 and so in the equivalence class. We can write this set as {3k+ 2| k in Z}.
We don't have to ask about 3 or 4 or negative numbers because those are already in the sets above (and the symmetric and transitive rules tell us that if a number a is in SOME equivalence class, then that is ITS equivalence class).
The relation "a~ b means a-b is divisible by 3" partitions the set of all integers into those three equivalence classes.
#### HallsofIvy
Homework Helper
Here's a non-numerical example: ~ is defined on the set of people as: a~ b if and only if a is the same gender as b.
This is an equivalence relation because it satisfies the three requirements for an equivalence relation:
a) Reflexive: each person is the same gender as him/herself.
b) Symmetric: if a is the same gender as b, then b is the same gender as a.
c) Transitive: if a is the same gender as b and b is the same gender as c, then a is the same gender as c.
(Actually, these are due to the fact that "same" is basicaly the verbal idea of "equivalent".)
There being two genders, this partitions the set of all people into two equivalence classes. The equivalence classes are the set of all females and the set of all males.
The "equivalence relation" itself as a set of pairs would include all pairs in which both people are female or both people are male but none in which one is male and the other female.
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• Solo and co-op problem solving | 2,115 | 7,471 | {"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} | 3.90625 | 4 | CC-MAIN-2019-43 | longest | en | 0.939423 |
https://flatearthsoc.com/flat-earth-news-blog-solution-reviews-flat-earth-news-blog-review-for-kids.html | 1,566,702,955,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027322170.99/warc/CC-MAIN-20190825021120-20190825043120-00055.warc.gz | 455,086,352 | 17,049 | 28.) Astronomers are in the habit of considering two points on the Earth's surface, without, it seems, any limit as to the distance that lies between them, as being on a level, and the intervening section, even though it be an ocean, as a vast "hill"-of water!" The Atlantic ocean, in taking this view of the matter, would form a "hill of water" more than a hundred miles high! The idea is simply monstrous, and could only be entertained by scientists whose whole business is made up of materials of the same description: and it certainly requires no argument to deduce, from such "science" as this, a satisfactory proof that the Earth is not a globe.
```In the next photograph and succeeding photographs, the ship is farther away, as indicated by the decreasing apparent size of the ship. In Figure 5, an inferior mirage is starting to show up. At the edge of the water, you can see a gray line, which is an inferior mirage of the row of gray containers right above the hull. On the right side of the ship, you can see the inferior mirage of the bow. The hull protrudes forward there, and the small white patch just above is a small portion of the forecastle. Notice that the inferior mirage of the bow is inverted, as one would expect. It is difficult to see here, but the lettering on the hull also is undergoing an inferior mirage too.
```
Can someone make me understand of Curvature feet calculation which is mentioned in several proofs. As in 71 number proof ( or several other distanced based proofs ) That the Observer distance is 60 miles sea-level from Chicago buildings which should be 2,400 feet below the horizon. As per Nasa earth curvature goes down 8 inches per mile. 72 inches(6 feet) and 60 miles contain 60 x 8 = (480 inches) that is equal to (480/12) = 40 feet. How does it count to 2,400 feet?
61.) It is plain that a theory of measurements without a measuring-rod is like a ship without a rudder; that a measure that is not fixed, not likely to be fixed, and never has been fixed, forms no measuring-rod at all; and that as modern theoretical astronomy depends upon the Sun's distance from the Earth as its measuring-rod, and the distance is not known, it is a system of measurements without a measuring-rod – a ship without a rudder. Now, since it is not difficult to foresee the dashing of this thing upon the rock on which Zetetic astronomy is founded, i
99.) Mr. Hind speaks of two great mathematicians differing only fifty-five yards in their estimate of the Earth's diameter. Why, Sir John Herschel, in his celebrated work, cuts off 480 miles of the same thing to get "round numbers!" This is like splitting a hair on one side of the bead and shaving all the hair off on the other! Oh, "science!" Can there be any truth in a science like this? All the exactitude in astronomy is in Practical astronomy – not Theoretical. Centuries of observation have made practical astronomy a noble art and science, based – as we have a thousand times proved it to be – on a fixed Earth; and we denounce this pretended exactitude on one side and the reckless indifference to figures on the other as the basest trash, and take from it a proof that the "science" which tolerates it is a false – instead of being an "exact" – science, and we have a proof that the Earth is not a globe.
107) Ring magnets of the kind found in loudspeakers have a central North pole with the opposite “South” pole actually being all points along the outer circumference. This perfectly demonstrates the magnetism of our flat Earth, whereas the alleged source of magnetism in the ball-Earth model is emitted from a hypothetical molten magnetic core in the center of the ball which they claim conveniently causes both poles to constantly move thus evading independent verification at their two “ceremonial poles.” In reality the deepest drilling operation in history, the Russian Kola Ultradeep, managed to get only 8 miles down, so the entire ball-Earth model taught in schools showing a crust, outer-mantle, inner-mantle, outer-core and inner-core layers are all purely speculation as we have never penetrated through beyond the crust.
I started reading this e-book today and finished it today. Great Work ! I have to say, even if approached flat-earth model fistly 3 month ago, this Book really opened my eyes completely and still i know how strong the Brainwash is and even if you - by ratio - see something clear, it will still need a time to be completely internalized. Thx alot, greetingz from Germany also i was able to watch the video below, which oc is blocked in Germany, but this led me to HotSpotShield, also very important for a native German, greetingz from Frankfurt/Germany, Benjamin
There is much real evidence to prove the Flat stationary earth and it’s not just photos and hearsay. Many professional pilots and military personnel have come out as Flat Earthers, and have confirmed that there is no curvature on the earth, flights don’t make sense on a ball, and GPS is ground-based and more. Here are some of these very interesting interviews by Mark Sargent, a Flat Earther. They bring up some very compelling evidence and that will make even the most staunch Ball Earther think twice, or at least we can hope so. If anything, it should make you think about where we really live. These interviews are not made up or staged. They are real! You can prove their evidence correct with your own research.
22.) God's Truth never – no, never – requires a falsehood to help it along. Mr. Proctor, in his " Lessons," says: Men " have been able to go round and round the Earth in several directions." Now, in this case, the word " several will imply more than two, unquestionably: whereas, it is utterly impossible to circumnavigate the Earth in any other than an easterly or a westerly direction; and the fact is perfectly consistent and clear in its relation to Earth as a Plane.. Now, since astronomers would not be so foolish as to damage a good cause by misrepresentation, it is presumptive evidence that their cause is a bad one, and – a proof that Earth is not a globe.
It has been shown that an easterly or a westerly motion is necessarily a circular course round the central North, The only north point or centre of motion of the heavenly bodies known to man is that formed by the North Star, which is over the central portion of the outstretched Earth. When, therefore, astronomers tell us of a planet taking a westerly course round the Sun, the thing is as meaningless to them as it is to us, unless they make the Sun the northern centre of the motion, which they cannot do! Since, then, the motion which they tell us the planets have is, on the face of it, absurd; and since, as a matter of fact, the Earth can have no absurd motion at all, it is clear that it cannot be what astronomers say it is - a planet; and, if not a planet, it is a proof that Earth is not a globe.
138) Another favorite “proof” of ball-Earthers is the appearance from an observer on shore of ships’ hulls being obfuscated by the water and disappearing from view when sailing away towards the horizon. Their claim is that ships’ hulls disappear before their mast-heads because the ship is beginning its declination around the convex curvature of the ball-Earth. Once again, however, their hasty conclusion is drawn from a faulty premise, namely that only on a ball-Earth could this phenomenon occur. The fact of the matter is that the Law of Perspective on plane surfaces dictates and necessitates the exact same occurrence. For example a girl wearing a dress walking away towards the horizon will appear to sink into the Earth the farther away she walks. Her feet will disappear from view first and the distance between the ground and the bottom of her dress will gradually diminish until after about half a mile it seems like her dress is touching the ground as she walks on invisible legs. Such is the case on plane surfaces, the lowest parts of objects receding from a given point of observation necessarily disappear before the highest.
At the end of the day, I think the one single proof of FE or Ball E, will be the irrefutable circumnavigation of Antarctica. As it is, we have no modern attempt. We have no attempt to travel across its middle, and no one wants to take on the challenge of a North South circumnavigation of the planet, the last great achievement (which is odd to me). All three of these things Not being done are Very strange indeed.
48.) In Mr. Proctor's "Lessons in Astronomy," page 15, a ship is represented as sailing away from the observer, and it is given in five positions or distances away on its journey. Now, in its first position, its mast appears above the horizon, and, consequently, higher than the observer's line of vision. But, in its second and third positions, representing the ship as further and further away, it is drawn higher and still higher up above the line of the horizon! Now, it is utterly impossible for a ship to sail away from an observer, under the, conditions indicated, and to appear as given in the picture. Consequently, the picture is a misrepresentation, a fraud, and a disgrace. A ship starting to sail away from an observer with her masts above his line of sight would appear, indisputably, to go down and still lower down towards the horizon line, and could not possibly appear – to anyone with his vision undistorted – as going in any other direction, curved or straight. Since, then the design of the astronomer-artist is to show the Earth to be a globe, and the points in the picture, which would only prove the Earth to be cylindrical if true, are NOT true, it follows that the astronomer-artist fails to prove, pictorially, either that the Earth is a globe or a cylinder, and that we have, therefore, a reasonable proof that the Earth is not. a globe.
139) Not only is the disappearance of ship’s hulls explained by the Law of Perspective on flat surfaces, it is proven undeniably true with the aid of a good telescope. If you watch a ship sailing away into the horizon with the naked eye until its hull has completely disappeared from view under the supposed “curvature of the Earth,” then look through a telescope, you will notice the entire ship quickly zooms back into view, hull and all, proving that the disappearance was caused by the Law of Perspective, not by a wall of curved water! This also proves that the horizon is simply the vanishing line of perspective from your point of view, NOT the alleged “curvature” of Earth.
```69.) Mr. Lockyer says: "The appearances connected with the rising and setting of the Sun and stars may be due either to our earth being at rest and the Sun and stars traveling round it, or the earth itself turning round, while the Sun and stars are at rest." Now, since true science does not allow of any such beggarly alternatives as these, it is plain that modern theoretical astronomy is not true science, and that its leading dogma is a fallacy. We have, then, a plain proof that the Earth is not a globe.
```
Due to special relativity, this is not the case. At this point, many readers will question the validity of any answer which uses advanced, intimidating-sounding physics terms to explain a position. However, it is true. The relevant equation is v/c = tanh (at/c). One will find that in this equation, tanh(at/c) can never exceed or equal 1. This means that velocity can never reach the speed of light, regardless of how long one accelerates for and the rate of the acceleration.
##### 137) Another assumption and supposed proof of Earth’s shape, heliocentrists claim that lunar eclipses are caused by the shadow of the ball-Earth occulting the Moon. They claim the Sun, Earth, and Moon spheres perfectly align like three billiard balls in a row so that the Sun’s light casts the Earth’s shadow onto the Moon. Unfortunately for heliocentrists, this explanation is rendered completely invalid due to the fact that lunar eclipses have happened and continue to happen regularly when both the Sun and Moon are still visible together above the horizon! For the Sun’s light to be casting Earth’s shadow onto the Moon, the three bodies must be aligned in a straight 180 degree syzygy, but as early as the time of Pliny, there are records of lunar eclipses happening while both the Sun and Moon are visible in the sky. Therefore the eclipsor of the Moon cannot be the Earth/Earth’s shadow and some other explanation must be sought.
135) Not only is the Moon clearly self-luminescent, shining its own unique light, but it is also largely transparent. When the waxing or waning Moon is visible during the day it is possible to see the blue sky right through the Moon. And on a clear night, during a waxing or waning cycle, it is even possible to occasionally see stars and “planets” directly through the surface of the Moon! The Royal Astronomical Society has on record many such occurrences throughout history which all defy the heliocentric model.
## 3) The natural physics of water is to find and maintain its level. If Earth were a giant spinning sphere tilting and hurling through space then truly flat, consistently level surfaces would not exist here. There would be a massive bulge of water in the oceans because of the curvature of the earth. If earth was curved and spinning the oceans of water would be flowing down to level and covering land. Some rivers would be impossibly flowing uphill. There would massive water chaos and flooding! What we would see and experience would be vastly different! But since Earth is in fact an extended flat plane, this fundamental physical property of fluids finding and remaining level is consistent with experience and common sense. The water remains flat because the earth is flat!
87.) The theory of a rotating and revolving earth demands at theory to keep the water on its surface; but, as the. theory which is given for this purpose is as much opposed to all human experience as the one which it is intended to uphold, it is an illustration of the miserable makeshifts to which astronomers are compelled to resort, and affords, a proof that the Earth is not a globe.
186) People sensitive to motion sickness feel distinct unease and physical discomfort from motion as slight as an elevator or a train ride. This means that the 1000mph alleged uniform spin of the Earth has no effect on such people, but add an extra 50mph uniform velocity of a car and their stomach starts turning knots. The idea that motion sickness is nowhere apparent in anyone at 1000mph, but suddenly comes about at 1050mph is ridiculous and proves the Earth is not in motion whatsoever.
# 65) Also Quoting Dr. Rowbotham, “On the shore near Waterloo, a few miles to the north of Liverpool, a good telescope was fixed, at an elevation of 6 feet above the water. It was directed to a large steamer, just leaving the River Mersey, and sailing out to Dublin. Gradually the mast-head of the receding vessel came nearer to the horizon, until, at length, after more than four hours had elapsed, it disappeared. The ordinary rate of sailing of the Dublin steamers was fully eight miles an hour; so that the vessel would be, at least, thirty-two miles distant when the mast-head came to the horizon. The 6 feet of elevation of the telescope would require three miles to be deducted for convexity, which would leave twenty-nine miles, the square of which, multiplied by 8 inches, gives 560 feet; deducting 80 feet for the height of the main-mast, and we find that, according to the doctrine of rotundity, the mast-head of the outward bound steamer should have been 480 feet below the horizon. Many other experiments of this kind have been made upon sea-going steamers, and always with results entirely incompatible with the theory that the earth is a globe.”
71.) The astronomers' theory of a globular Earth necessitates the conclusion that, if we travel south of the equator, to see the North Star is an impossibility. Yet it is well known this star has been seen by navigators when they have been more than 20 degrees south of the equator. This fact, like hundreds of other facts, puts the theory to shame, and gives us a proof that the Earth is not a globe.
```190) Cultures the world over throughout history have all described and purported the existence of a geocentric, stationary flat Earth. Egyptians, Indians, Mayans, Chinese, Native Americans and literally every ancient civilization on Earth had a geocentric flat-Earth cosmology. Before Pythagoras, the idea of a spinning ball-Earth was non-existent and even after Pythagoras it remained an obscure minority view until 2000 years later when Copernicus began reviving the heliocentric theory.
```
Great material, THANK YOU SO MUCH!!! I am very happy to be able to arrive at this information. By the way, they lied not only about earth shape and universe, but about physics, chemistry and even in math, inventing nonexistent values and rules. Example: they tell you 2x0=0. This is nonsense, because 2 has to denote something real, the actual values. Therefore 2x0=2, not 0, as they lie to us. Prove? Take 2 airplanes and multiply them on 0. How many airplanes will you get,-zero?...)) No, you will still have 2 airplanes! But 2 as abstract (nonexistent) number representing nothing, can not give any result but nothing.
88.) If we could – after our minds had once been opened to the light of Truth – conceive of a globular body on the surface of which human beings could exist, the power – no matter by what name it be called – that would hold them on would, then, necessarily, have to be so constraining and cogent that they could not live; the waters of the oceans would have to be as a solid mass, for motion would be impossible. But we not only exist, but live and move; and the water of the ocean skips and dances like a thing of life and beauty! This is a proof that the Earth is not a globe.
The explanation for a lunar eclipse on flat earth is more difficult to prove, but the official “Scientific” NASA explanation is easy to disprove. The official globe model explanation is that the sun, earth, and moon line up perfectly and the earth then casts its shadow on the moon, creating the lunar eclipse. The main problem with this official model is that the sun AND moon have both been visible ABOVE the horizon during a lunar eclipse, making the alignment for earth’s shadow to be cast on the moon impossible!
```147) The ball-Earth model claims the Sun is precisely 400 times larger than the Moon and 400 times further away from Earth making them “falsely” appear exactly the same size. Once again, the ball model asks us to accept as coincidence something that cannot be explained other than by natural design. The Sun and the Moon occupy the same amount of space in the sky and have been measured with sextants to be of equal size and equal distance, so claiming otherwise is against our eyes, experience, experiments and common sense.
```
When astronomers assert that it is "necessary" to make "allowance for curvature" in canal construction, it is, of course, in order that, in their idea, a level cutting may be had, for the water. How flagrantly, then, do they contradict themselves when the curved surface of the Earth is a "true level!" What more can they want for a canal than a true level? Since they contradict themselves in such an elementary point as this, it is an evidence that the whole thing is a delusion, and we have a proof that the Earth is not a globe.
The way our vision works makes everything converge to a single vanishing point on the flat horizon, including airplanes and the sun. Artists understand this. Airplanes appear to drop below the horizon when in reality they are flying level to the flat earth and never dip their noses down to account for any supposed curve. It's the same with the sun. It is moving across the sky on a flat circular path but it appears to rise and fall due to perspective.
Mr. Hind, the English astronomer, says - "The simplicity, with which the seasons are explained by the revolution of the Earth in her orbit and the obliquity of the ecliptic, may certainly be adduced as a strong presumptive proof of the correctness" - of the Newtonian theory; "for on no other rational suppositions with respect to the relations of the Earth and Sun, can these and other as well-known phenomena, be accounted for." But, as true philosophy has no "suppositions" at all - and has nothing to do with, "suppositions" - and the phenomena spoken of are thoroughly explained by facts, the "presumptive proof" falls to the ground, covered with the ridicule it the dust of Mr. Hind's "rational suppositions" we are standing before us a proof that Earth is not a globe.
Also it kind of ties in with this dinosaur thing, with the flat earth and everything revolves around us. how did we get hit with some "meteor" to even wipe out dinosaurs exactly? please look into this because I would love to see your perspective and how it ties into all these lies. look into the false carbon dating and how completely inaccurate it is and how the earth is no where near as old as they say it is. So many lies to be blown wide open. A fresh testing of any volcano will never test as 1 or even 10 years with carbon dating, it will always test as millions of years!
In 2004, the society was resurrected by a man named Daniel Shenton (no relation to Samuel), who created the Flat Earth Society forum, which as of February 2016 has over 8,200 members and 1.4 million posts.[3] In addition, the forum runs a Facebook page with over 14,000 likes,[4] Twitter,[5] Instagram,[6] and Tumblr profiles with several thousand followers each, and a Flickr profile[7] where they advertise a variety of different posters with proofs for why the world is flat. In addition, a variety of independent researchers have attempted to prove that the earth is flat, documenting their work in YouTube videos.[8] The most popular of these videos (shown below) has over 4 million views.
106) The so-called “South Pole” is simply an arbitrary point along the Antarctic ice marked with a red and white barbershop pole topped with a metal ball-Earth. This ceremonial South Pole is admittedly and provably NOT the actual South Pole, however, because the actual South Pole could be demonstrably confirmed with the aid of a compass showing North to be 360 degrees around the observer. Since this feat has never been achieved, the model remains pure theory, along with the establishment’s excuse that the geomagnetic poles supposedly constantly move around making verification of their claims impossible.
44) If Earth was a ball, and Antarctica was too cold to fly over, the only logical way to fly from Sydney to Santiago would be a straight shot over the Pacific staying in the Southern hemisphere the entire way. Re-fueling could be done in New Zealand or other Southern hemisphere destinations along the way if absolutely necessary. In actual fact, however, Santiago-Sydney flights go into the Northern hemisphere making stop-overs at LAX and other North American airports before continuing back down to the Southern hemisphere. Such ridiculously wayward detours make no sense on the globe but make perfect sense and form nearly straight lines when shown on a flat Earth map.
If astronomical works be searched through and through, there will not be found a single instance of a bold, unhesitating, or manly ,statement respecting a proof of the Earth's " rotundity." Proctor speaks of "proofs which serve to show ... that the Earth is not flat," and says that man "finds reason to think that the Earth is not flat," and speaks of certain matters being "explained by supposing" that the Earth is a, globe; and says that people have "assured themselves that it is a globe;" but he says, also, that there is a " most complete proof that the Earth is a globe:" just as though anything in the world could possibly be wanted but a proof - a proof that proves and settles the whole question. This, however, all the money in the United States Treasury would not buy; and, unless the astronomers are all so rich that they don't want the cash, it is a sterling proof that the Earth is not a globe.
The way our vision works makes everything converge to a single vanishing point on the flat horizon, including airplanes and the sun. Artists understand this. Airplanes appear to drop below the horizon when in reality they are flying level to the flat earth and never dip their noses down to account for any supposed curve. It's the same with the sun. It is moving across the sky on a flat circular path but it appears to rise and fall due to perspective.
Another false law of Newton is that gravity increases with the increase of the mass of the object. There is no such thing as mass (no one in the world can define it) - there is only density of the object (total density volume of the object, including it's electric field that surrounds it), and it is enough to understand how the laws work. Rubber ball pumped with helium goes up irrespective of the "gravity law" which supposed to bring everything down. Ball goes up because the density of the helium is smaller than the density of air above it. There is also no resistance of the environment above the ball.
```It is plain that a theory of measurements without a measuring-rod is like a ship without a rudder; that a measure that is not fixed, not likely to be fixed, and never has been fixed, forms no measuring-rod at all; and that as modern theoretical astronomy depends upon the Sun's distance from the Earth as its measuring-rod, and the distance is not known, it is a system of measurements without a measuring-rod - a ship without a rudder. Now, since it is not difficult to foresee the dashing of this thing upon the rock on which Zetetic astronomy is founded, it is a proof that Earth is not a globe.
```
Aristotle (who made quite a lot of observations about the spherical nature of the Earth) noticed that during lunar eclipses (when the Earth’s orbit places it directly between the Sun and the Moon, creating a shadow in the process), the shadow on the Moon’s surface is round. This shadow is the planet's, and it’s a great clue about the spherical shape of the Earth.
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If the sun is the center of the solar system (which is actually based on Pagan Sun Worship) and the earth is orbiting around it, then for this event in Joshua to occur, it would mean that the earth stopped spinning and stood still, not the sun. For the earth to suddenly stop spinning at its supposed 1000 miles per hour spin, it would have caused catastrophic devastation on earth. But, if the earth is flat and still, and the sun and moon rotate around and above the earth, then it would make perfect sense for the sun to stand still in order to prolong the day.
192) Quoting “Terra Firma” by David Wardlaw Scott, “The system of the Universe, as taught by Modern Astronomers, being founded entirely on theory, for the truth of which they are unable to advance one single real proof, they have entrenched themselves in a conspiracy of silence, and decline to answer any objections which may be made to their hypotheses … Copernicus himself, who revived the theory of the heathen philosopher Pythagoras, and his great exponent Sir Isaac Newton, confessed that their system of a revolving Earth was only a possibility, and could not be proved by facts. It is only their followers who have decorated it with the name of an ‘exact science,’ yea, according to them, ‘the most exact of all the sciences.’ Yet one Astronomer Royal for England once said, speaking of the motion of the whole Solar system: ‘The matter is left in a most delightful state of uncertainty, and I shall be very glad if any one can help me out of it.’ What a very sad position for an ‘exact science’ to be in is this!”
175) Professional photo-analysts have dissected several NASA images of the ball-Earth and found undeniable proof of computer editing. For example, images of the Earth allegedly taken from the Moon have proven to be copied and pasted in, as evidenced by rectangular cuts found in the black background around the “Earth” by adjusting brightness and contrast levels. If they were truly on the Moon and Earth was truly a ball, there would be no need to fake such pictures.
With increasing distance from the object, the earth’s curvature causes the surface of the water to fall away from the beam of light. Over one mile, the amount of drop is eight inches, but the drop increases quadratically with distance. Consequently, after three miles the drop is six feet, and after six miles the drop is 24 feet. This is the point of the Bedford level experiment—the curvature of the earth ought to intervene to prevent the mast of the boat being visible from much more than three miles, let alone six miles. However, for the light from the distant object not to be visible, it would have to travel in a straight line. But with a temperature inversion, straight-line motion would carry the light from a cooler layer of air into a warmer layer of air at nearly a grazing angle. The light cannot do this, so it continually is internally reflected, causing the light to bend around the edge of the earth. Therefore, with a temperature inversion, one can see objects that lie well beyond the edge of the earth’s curvature when viewing close to the surface of water.
The Astronomer Royal, of England, George B. Airy, in his celebrated work on Astronomy, the "Ipswich Lectures," says - "Jupiter is a large planet that turns on his axis, and why do not we turn?" Of course, the common sense reply is: Because the Earth is not a planet! When, therefore, an astronomer royal puts words into our mouth wherewith we may overthrow the supposed planetary nature of the Earth, we have not far to go to pick up a proof that Earth is not a globe.
```As for flight paths and what Appears to be the silly way for a ball earth but makes sense for a flat earth, it reminds me of the child quiz. There is a spider in the corner of the room on the floor and he wants to get to the opp corner on the ceiling. Which is the quickest path? We instantly say, across the floor and up the wall join. BUT, if we flatten the room we then draw a straight line, we find the quickest path is diagonally up one wall and then diagonally across the ceiling, which Looks longer but is best.
```
The most commonly accepted explanation of this is that the space agencies of the world are involved in a conspiracy faking space travel and exploration. This likely began during the Cold War's 'Space Race', in which the USSR and USA were obsessed with beating each other into space to the point that each faked their accomplishments in an attempt to keep pace with the other's supposed achievements. Since the end of the Cold War, however, the conspiracy is most likely motivated by greed rather than political gains, and using only some of their funding to continue to fake space travel saves a lot of money to embezzle for themselves.
Consider a flat plane. The center of mass of a flat plane is in its center, so the force of gravity will pull anything on the surface toward the middle of the plane. That means that if you stand on the edge of the plane, gravity will be pulling you sideways toward the plane's middle, not straight down like you usually experience when you stand on Earth.
so if the earth is not a sphere in space revolveing around the sphereical sun, then what is it. Its one thing to say that "its not that way" but its different to say "its actually this way not that way". So what way is it? what way are you proposing is the correct way? do you beleive this is the only planet in the universe? do you believe that the stars are only decorations on a flat backdrop? I'm not certain what idea you are proposing is the correct way of looking at this...
8) The Suez Canal connecting the Mediterranean with the Red Sea is 100 miles long without any locks making the water an uninterrupted continuation of the two seas. When constructed, the Earth’s supposed curvature was not taken into account, it was dug along a horizontal datum line 26 feet below sea-level, passing through several lakes from one sea to the other, with the datum line and water’s surface running perfectly parallel over the 100 miles.
aking our ideal point of departure to be at Valentia, we consider ourselves at St. John's, the 1665 miles of water between us and Valentia would just as well "curvate" downwards as it did in the other case! Now, since the direction in which the Earth is said to "curvate" is interchangeable – depending, indeed, upon the position occupied by a man upon its surface – the thing is utterly absurd; and it follows that the theory is an outrage , and that the Earth does not "curvate" at all: – an evident proof that the Earth is not a globe. | 7,197 | 33,303 | {"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} | 3.140625 | 3 | CC-MAIN-2019-35 | latest | en | 0.963357 |
http://easy-ciphers.com/cataleptic | 1,566,460,084,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316785.68/warc/CC-MAIN-20190822064205-20190822090205-00528.warc.gz | 58,911,205 | 26,375 | Easy Ciphers Tools:
Caesar cipher
Caesar cipher, is one of the simplest and most widely known encryption techniques. The transformation can be represented by aligning two alphabets, the cipher alphabet is the plain alphabet rotated left or right by some number of positions.
When encrypting, a person looks up each letter of the message in the 'plain' line and writes down the corresponding letter in the 'cipher' line. Deciphering is done in reverse.
The encryption can also be represented using modular arithmetic by first transforming the letters into numbers, according to the scheme, A = 0, B = 1,..., Z = 25. Encryption of a letter x by a shift n can be described mathematically as
Plaintext: cataleptic
cipher variations: dbubmfqujd ecvcngrvke fdwdohswlf gexepitxmg hfyfqjuynh igzgrkvzoi jhahslwapj kibitmxbqk ljcjunycrl mkdkvozdsm nlelwpaetn omfmxqbfuo pngnyrcgvp qohozsdhwq rpipateixr sqjqbufjys trkrcvgkzt uslsdwhlau vtmteximbv wunufyjncw xvovgzkodx ywpwhalpey zxqxibmqfz ayryjcnrga bzszkdoshb
Decryption is performed similarly,
(There are different definitions for the modulo operation. In the above, the result is in the range 0...25. I.e., if x+n or x-n are not in the range 0...25, we have to subtract or add 26.)
Atbash Cipher
Atbash is an ancient encryption system created in the Middle East. It was originally used in the Hebrew language.
The Atbash cipher is a simple substitution cipher that relies on transposing all the letters in the alphabet such that the resulting alphabet is backwards.
The first letter is replaced with the last letter, the second with the second-last, and so on.
An example plaintext to ciphertext using Atbash:
Plain: cataleptic Cipher: xzgzovkgrx
Baconian Cipher
To encode a message, each letter of the plaintext is replaced by a group of five of the letters 'A' or 'B'. This replacement is done according to the alphabet of the Baconian cipher, shown below.
```a AAAAA g AABBA m ABABB s BAAAB y BABBA
b AAAAB h AABBB n ABBAA t BAABA z BABBB
c AAABA i ABAAA o ABBAB u BAABB
d AAABB j BBBAA p ABBBA v BBBAB
e AABAA k ABAAB q ABBBB w BABAA
f AABAB l ABABA r BAAAA x BABAB
```
Plain: cataleptic Cipher: AAABA AAAAA BAABA AAAAA ABABA AABAA ABBBA BAABA ABAAA AAABA
Affine Cipher
In the affine cipher the letters of an alphabet of size m are first mapped to the integers in the range 0..m - 1. It then uses modular arithmetic to transform the integer that each plaintext letter corresponds to into another integer that correspond to a ciphertext letter. The encryption function for a single letter is
where modulus m is the size of the alphabet and a and b are the key of the cipher. The value a must be chosen such that a and m are coprime.
Considering the specific case of encrypting messages in English (i.e. m = 26), there are a total of 286 non-trivial affine ciphers, not counting the 26 trivial Caesar ciphers. This number comes from the fact there are 12 numbers that are coprime with 26 that are less than 26 (these are the possible values of a). Each value of a can have 26 different addition shifts (the b value) ; therefore, there are 12*26 or 312 possible keys.
Plaintext: cataleptic
cipher variations:
dbubmfqujd
hbgbinugzh
lbsbevyspl
tbqbwlgqvt
xbcbstkclx
fbabkjsarf
jbmbgrwmhj
nbybczayxn
rbkbyheknr
vbwbupiwdv
zbibqxmitz
ecvcngrvke
ichcjovhai
mctcfwztqm
qcfcbedfgq
ucrcxmhrwu
ycdctuldmy
gcbclktbsg
kcnchsxnik
oczcdabzyo
sclcziflos
wcxcvqjxew
acjcrynjua
fdwdohswlf
jdidkpwibj
ndudgxaurn
rdgdcfeghr
vdsdynisxv
zdeduvmenz
hdcdmlucth
ldodityojl
tdmdajgmpt
xdydwrkyfx
bdkdszokvb
gexepitxmg
kejelqxjck
oevehybvso
sehedgfhis
wetezojtyw
aefevwnfoa
iedenmvdui
mepejuzpkm
qebefcdbaq
uenebkhnqu
yezexslzgy
celetaplwc
hfyfqjuynh
lfkfmrykdl
pfwfizcwtp
tfifehgijt
xfufapkuzx
bfgfwxogpb
jfefonwevj
nfqfkvaqln
rfcfgdecbr
vfofcliorv
zfafytmahz
dfmfubqmxd
igzgrkvzoi
mglgnszlem
ugjgfihjku
ygvgbqlvay
cghgxyphqc
kgfgpoxfwk
ogrglwbrmo
sgdghefdcs
wgpgdmjpsw
agbgzunbia
egngvcrnye
jhahslwapj
nhmhotamfn
rhyhkbeyvr
vhkhgjiklv
zhwhcrmwbz
dhihyzqird
lhghqpygxl
phshmxcsnp
thehifgedt
xhqhenkqtx
bhchavocjb
fhohwdsozf
kibitmxbqk
oinipubngo
sizilcfzws
wilihkjlmw
aixidsnxca
eijizarjse
mihirqzhym
qitinydtoq
uifijghfeu
yirifolruy
cidibwpdkc
gipixetpag
ljcjunycrl
pjojqvcohp
tjajmdgaxt
xjmjilkmnx
bjyjetoydb
fjkjabsktf
njijsraizn
rjujozeupr
vjgjkhigfv
zjsjgpmsvz
djejcxqeld
hjqjyfuqbh
mkdkvozdsm
qkpkrwdpiq
ukbknehbyu
yknkjmlnoy
ckzkfupzec
gklkbctlug
okjktsbjao
skvkpafvqs
wkhklijhgw
aktkhqntwa
ekfkdyrfme
ikrkzgvrci
nlelwpaetn
rlqlsxeqjr
vlcloficzv
zlolknmopz
dlalgvqafd
hlmlcdumvh
plklutckbp
tlwlqbgwrt
xlilmjkihx
blulirouxb
flglezsgnf
jlslahwsdj
omfmxqbfuo
smrmtyfrks
wmdmpgjdaw
ampmlonpqa
embmhwrbge
imnmdevnwi
qmlmvudlcq
umxmrchxsu
ymjmnkljiy
cmvmjspvyc
gmhmfathog
kmtmbixtek
pngnyrcgvp
tnsnuzgslt
xnenqhkebx
bnqnmpoqrb
fncnixschf
jnonefwoxj
rnmnwvemdr
vnynsdiytv
znknolmkjz
dnwnktqwzd
hningbuiph
lnuncjyufl
qohozsdhwq
uotovahtmu
yoforilfcy
coronqprsc
godojytdig
kopofgxpyk
sonoxwfnes
wozotejzuw
aolopmnlka
eoxolurxae
iojohcvjqi
movodkzvgm
rpipateixr
vpupwbiunv
zpgpsjmgdz
dpsporqstd
hpepkzuejh
lpqpghyqzl
tpopyxgoft
xpapufkavx
bpmpqnomlb
fpypmvsybf
jpkpidwkrj
npwpelawhn
sqjqbufjys
wqvqxcjvow
aqhqtknhea
eqtqpsrtue
iqfqlavfki
mqrqhizram
uqpqzyhpgu
yqbqvglbwy
cqnqropnmc
gqzqnwtzcg
kqlqjexlsk
oqxqfmbxio
trkrcvgkzt
xrwrydkwpx
briruloifb
frurqtsuvf
jrgrmbwglj
nrsrijasbn
vrqraziqhv
zrcrwhmcxz
drorspqond
lrmrkfymtl
pryrgncyjp
uslsdwhlau
ysxszelxqy
csjsvmpjgc
gsvsrutvwg
kshsncxhmk
ostsjkbtco
wsrsbajriw
asdsxindya
espstqrpoe
isbspyvbei
msnslgznum
qszshodzkq
vtmteximbv
ztytafmyrz
dtktwnqkhd
htwtsvuwxh
ltitodyinl
ptutklcudp
xtstcbksjx
btetyjoezb
ftqtursqpf
jtctqzwcfj
ntotmhaovn
rtatipealr
wunufyjncw
auzubgnzsa
euluxorlie
iuxutwvxyi
mujupezjom
quvulmdveq
yutudcltky
cufuzkpfac
guruvstrqg
kuduraxdgk
oupunibpwo
subujqfbms
xvovgzkodx
bvavchoatb
fvmvypsmjf
jvyvuxwyzj
nvkvqfakpn
rvwvmnewfr
zvuvedmulz
dvgvalqgbd
hvsvwtusrh
lvevsbyehl
pvqvojcqxp
tvcvkrgcnt
ywpwhalpey
cwbwdipbuc
gwnwzqtnkg
kwzwvyxzak
owlwrgblqo
swxwnofxgs
awvwfenvma
ewhwbmrhce
iwtwxuvtsi
mwfwtczfim
qwrwpkdryq
uwdwlshdou
zxqxibmqfz
dxcxejqcvd
hxoxaruolh
lxaxwzyabl
pxmxshcmrp
txyxopgyht
bxwxgfownb
fxixcnsidf
jxuxyvwutj
nxgxudagjn
rxsxqleszr
vxexmtiepv
ayryjcnrga
eydyfkrdwe
iypybsvpmi
mybyxazbcm
qynytidnsq
uyzypqhziu
cyxyhgpxoc
gyjydotjeg
kyvyzwxvuk
oyhyvebhko
sytyrmftas
wyfynujfqw
bzszkdoshb
fzezglsexf
jzqzctwqnj
nzczybacdn
rzozujeotr
vzazqriajv
dzyzihqypd
hzkzepukfh
lzwzaxywvl
pzizwfcilp
tzuzsngubt
xzgzovkgrx
cataleptic
gafahmtfyg
sapavkfpus
wabarsjbkw
eazajirzqe
ialafqvlgi
maxabyzxwm
qajaxgdjmq
uavatohvcu
yahapwlhsy
The decryption function is
where a - 1 is the modular multiplicative inverse of a modulo m. I.e., it satisfies the equation
The multiplicative inverse of a only exists if a and m are coprime. Hence without the restriction on a decryption might not be possible. It can be shown as follows that decryption function is the inverse of the encryption function,
ROT13 Cipher
Applying ROT13 to a piece of text merely requires examining its alphabetic characters and replacing each one by the letter 13 places further along in the alphabet, wrapping back to the beginning if necessary. A becomes N, B becomes O, and so on up to M, which becomes Z, then the sequence continues at the beginning of the alphabet: N becomes A, O becomes B, and so on to Z, which becomes M. Only those letters which occur in the English alphabet are affected; numbers, symbols, whitespace, and all other characters are left unchanged. Because there are 26 letters in the English alphabet and 26 = 2 * 13, the ROT13 function is its own inverse:
ROT13(ROT13(x)) = x for any basic Latin-alphabet text x
An example plaintext to ciphertext using ROT13:
Plain: cataleptic Cipher: pngnyrcgvp
Polybius Square
A Polybius Square is a table that allows someone to translate letters into numbers. To give a small level of encryption, this table can be randomized and shared with the recipient. In order to fit the 26 letters of the alphabet into the 25 spots created by the table, the letters i and j are usually combined.
1 2 3 4 5
1 A B C D E
2 F G H I/J K
3 L M N O P
4 Q R S T U
5 V W X Y Z
Basic Form:
Plain: cataleptic Cipher: 31114411135153444231
Extended Methods:
Method #1
Plaintext: cataleptic
method variations: hfyfqkuyoh nldlvpzdtn sqiqaueiys xvovfzkodx
Method #2
Bifid cipher
The message is converted to its coordinates in the usual manner, but they are written vertically beneath:
```c a t a l e p t i c
3 1 4 1 1 5 5 4 4 3
1 1 4 1 3 1 3 4 2 1 ```
They are then read out in rows:
31411554431141313421
Then divided up into pairs again, and the pairs turned back into letters using the square:
Method #3
Plaintext: cataleptic
method variations: aqdaxvstml qdaxvstmla daxvstmlaq axvstmlaqd xvstmlaqda vstmlaqdax stmlaqdaxv tmlaqdaxvs mlaqdaxvst laqdaxvstm
Permutation Cipher
In classical cryptography, a permutation cipher is a transposition cipher in which the key is a permutation. To apply a cipher, a random permutation of size E is generated (the larger the value of E the more secure the cipher). The plaintext is then broken into segments of size E and the letters within that segment are permuted according to this key.
In theory, any transposition cipher can be viewed as a permutation cipher where E is equal to the length of the plaintext; this is too cumbersome a generalisation to use in actual practice, however.
The idea behind a permutation cipher is to keep the plaintext characters unchanged, butalter their positions by rearrangement using a permutation
This cipher is defined as:
Let m be a positive integer, and K consist of all permutations of {1,...,m}
For a key (permutation) , define:
The encryption function
The decryption function
A small example, assuming m = 6, and the key is the permutation :
The first row is the value of i, and the second row is the corresponding value of (i)
The inverse permutation, is constructed by interchanging the two rows, andrearranging the columns so that the first row is in increasing order, Therefore, is:
Total variation formula:
e = 2,718281828 , n - plaintext length
Plaintext: cataleptic
first 5040 cipher variations(3628800 total)
cataleptic
cataleptci
catalepitc
catalepict
catalepcit
catalepcti
cataletpic
cataletpci
cataletipc
cataleticp
cataletcip
cataletcpi
cataleitpc
cataleitcp
cataleiptc
cataleipct
cataleicpt
cataleictp
catalectip
catalectpi
catalecitp
catalecipt
catalecpit
catalecpti
catalpetic
catalpetci
catalpeitc
catalpeict
catalpecit
catalpecti
catalpteic
catalpteci
catalptiec
catalptice
catalptcie
catalptcei
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Read more ...[1] , [2] , [3] | 30,011 | 65,729 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2019-35 | latest | en | 0.806999 |
https://blogs.mathworks.com/loren/category/best-practice/page/2/?s_tid=Blog_loren_Category | 1,632,586,393,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057687.51/warc/CC-MAIN-20210925142524-20210925172524-00272.warc.gz | 189,401,751 | 26,120 | # Recent Question about Speed with Subarray Calculations8
Recently someone asked me to explain the speed behavior doing a calculation using a loop and array indexing vs. getting the subarray first.... read more >>
# Understanding Array Preallocation6
Today I would like to introduce guest blogger Jeremy Greenwald who works in the Development group here at MathWorks. Jeremy works on the Code Analyzer and will be discussing when preallocating MATLAB arrays is useful and when it should be avoided.... read more >>
# Best Practices for Programming MATLAB56
I thought I would share my top goto list of things I try to do when I write MATLAB code. And checking with other MathWorks folks whose code I admire, I found... read more >>
# Simpler Control of Random Number Generation in MATLAB12
Once again we're going to hear from guest blogger Peter Perkins, who is a statistical software developer here at The MathWorks.
MATLAB has had random numbers since the beginning. But not... read more >>
# Book Review: The Elements of MATLAB Style7
I've recently been offered the opportunity to review a new book, The Elements of MATLAB Style by contributor to the FEX, Richard K. Johnson. It's a great opportunity for me to see... read more >>
# Debugging Approaches11
There are lots of posts on the MATLAB newsgroup asking questions like why do I get this error "xxx"? One of the most common answers shows users how to track down the information ... read more >>
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# Ensuring Positive Values – Part 14
Do you sometimes need to be sure an array has only bounded, positive values? At least one customer asked about this recently and noted that it could be quite involved. ... read more >>
# Handling Discrete Data
Discrete data arise in many applications and the data may be numeric, or non-numeric, often referred to as categorical. Not all data are strictly numeric, and other... read more >> | 453 | 2,150 | {"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-2021-39 | latest | en | 0.910167 |
https://www.jiskha.com/display.cgi?id=1354937351 | 1,516,264,319,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887077.23/warc/CC-MAIN-20180118071706-20180118091706-00330.warc.gz | 957,991,141 | 4,224 | # Physics
posted by .
Glycerin is poured into an open U-shaped tube until the height in both sides is 20 {\rm cm}. Ethyl alcohol is then poured into one arm until the height of the alcohol column is 20 {\rm cm}. The two liquids do not mix.
What is the difference in height between the top surface of the glycerin and the top surface of the alcohol?
• Physics -
The alcohol in the first arm floats on top of the denser glycerin and pressures it down distance h from the initial level. As a result, the glycerin rises by the distance h in the second arm. Due to the equilibrium condition, the point of the two liquids contact in the 1st arm and symmetrical point in the 2nd arm are at the equaled pressures
p₁=p₂
p₀ +ρ₁gh₁=p₀ +ρ₂gh₂
The level of glycerin in the 1st arm came down by h and the level of glycerin in the 2nd arm came up by h => h₂=2h, Since h₁=0.2 m,
ρ₁gh₁= ρ₂g2h
h= ρ₁h₁/2 ρ₂=790•0.2/2•1260 =0.0627 cm
Then
Δh=0.2-2•0.0627 = 0.0746 m
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# EnergyThermochemistry (Enthalpy)
Objective
· To consider the heat (enthalpy) of chemical reactions.
We have seen that some reactions are exothermic (produce heat energy) and other reactions are endothermic (absorb heat energy). Chemists also like to know exactly how much energy is produced or absorbed by a given reaction. To make that process more convenient, we have invented a special energy function called enthalpy , which is designated by . For a reaction occurring under conditions of constant pressure, the change in enthalpy is equal to the energy that flows as heat. That is,
where the subscript “ ” indicates that the process has occurred under conditions of constant pressure and means “a change in.” Thus the enthalpy change for a reaction (that occurs at constant pressure) is the same as the heat for that reaction.
Interactive Example 10.5. Enthalpy
When mole of methane is burned at constant pressure, kJ of energy is released as heat. Calculate for a process in which a -g sample of methane is burned at constant pressure.
Solution
Where Are We Going?
We want to determine for the reaction of g of methane with oxygen at constant pressure.
What Do We Know?
· When mole of is burned, kJ of energy is released.
· We have g of .
What Do We Need to Know?
· Molar mass of methane, which we can get from the atomic masses of carbon ( g/mol) and hydrogen ( g/mol). The molar mass is g/mol.
How Do We Get There?
At constant pressure, kJ of energy per mole of is produced as heat:
Note that the minus sign indicates an exothermic process. In this case, a -g sample of is burned. Because this amount is smaller than mole, less than kJ will be released as heat. The actual value can be calculated as follows:
and
Thus, when a -g sample of is burned at constant pressure,
Reality Check The mass of methane burned is less than mole, so less than kJ will be released as heat. The answer has two significant figures as required by the given quantities.
Self-Check: Exercise 10.5
The reaction that occurs in the heat packs used to treat sports injuries is
· How much heat is released when g of is reacted with excess ?
See Problems 10.41 and 10.42.
Chemistry in Focus Burning Calories
There is a growing concern in the United States about the increasing tendency for individuals to be overweight. Estimates indicate that two-thirds of the adults in the United States are overweight, and one-third are classified as obese. This is an alarming situation because obesity is associated with many serious diseases such as diabetes and heart disease. In an effort to reduce this problem, the U.S. government now requires national restaurant chains and grocery stores to post the calorie counts for the food they sell. The hope is that people will use the information to make better food choices for weight control.
All of this leads to the question of how the calorie content of food is determined. The process involves a special type of calorimeter called a bomb calorimeter. The food sample is enclosed in the bomb calorimeter and burned. The energy produced heats the water surrounding the calorimeter, and the amount of energy is determined by measuring the increase in temperature of the known amount of water. The “Calorie” used for food is equal to the kilocalorie used by the science community. Thus the number of kilocalories produced by burning the food is the “Calorie” content of the food.
How is food “burned” in a calorimeter? The exact composition of food can be determined from its ingredients. Each ingredient in its proper amount for the food is burned in a calorimeter, and the calories released are determined. The calorie count assigned to a particular food is totaled as the sum of the ingredients and is adjusted for the amount of energy the body will actually absorb ( for fats and less for carbohydrates and proteins). Restaurants determine calorie counts from recipes. However, that means errors can occur, depending on how closely a chef follows a recipe and whether the portion size in the recipe is actually the size of the portion served in the restaurant. Although the “calorie” count may not be exact for restaurant foods, the listing on the menu will give people a chance to make better choices.
See Problem 10.43
Calorimetry
A calorimeter (Fig. 10.6) is a device used to determine the heat associated with a chemical reaction. The reaction is run in the calorimeter and the temperature change of the calorimeter is observed. Knowing the temperature change that occurs in the calorimeter and the heat capacity of the calorimeter enables us to calculate the heat energy released or absorbed by the reaction. Thus we can determine for the reaction.
Figure 10.6.
A coffee-cup calorimeter made of two Styrofoam cups.
Once we have measured the values for various reactions, we can use these data to calculate the values of other reactions. We will see how to carry out these calculations in the next section.
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https://ch.mathworks.com/matlabcentral/cody/problems/1439-usc-spring-2013-acm-building-snowmen/solutions/233499 | 1,600,764,405,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400204410.37/warc/CC-MAIN-20200922063158-20200922093158-00640.warc.gz | 318,505,536 | 31,032 | Cody
# Problem 1439. USC Spring 2013 ACM: Building Snowmen
Solution 233499
Submitted on 21 Apr 2013 by Marco Castelli
• Size: 88
• This is the leading solution.
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
tic m=[3 5 1 2 6 4]; % 1 2 s=build_snowmen(m); assert(isequal(s,2));
2 Pass
%% m=[3 5 1 3 6 4]; % 2 1 s=build_snowmen(m); assert(isequal(s,1));
3 Pass
%% m=[6 3 4]; % 3 0 s=build_snowmen(m); assert(isequal(s,0));
4 Pass
%% m=[2 6 4]; % 4 1 s=build_snowmen(m); assert(isequal(s,1));
5 Pass
%% m=[6 4 2 4 4 4 4 4 4 4]; % 5 1 s=build_snowmen(m); assert(isequal(s,1));
6 Pass
%% m=[8 10 7 8 10 6 4 3 5]; % 6 2 s=build_snowmen(m); assert(isequal(s,2));
7 Pass
%% m=[999 800 666 200 334]; % 7 1 s=build_snowmen(m); assert(isequal(s,1));
8 Pass
%% m=[1 2 1 2]; % 8 0 s=build_snowmen(m); assert(isequal(s,0));
9 Pass
%% m=[240 695 444 21 508 338 941 345 430 698 879 883 900 546 376 419 607 738 116 521 948 743 479 282 143 637 456 79 825 799 666 3 679 517 224 161 17 947 172 491 39 45 592 711 289 280 542 223 245 190 127 797 751 558 27 472 237 279 629 79 26 510 265 831 483 746 478 991 707 438 479 447 899 985 265 822 933 319 566 745 461 307 736 493 251 9 89 258 724 390 441 432 777 744 761 155 436 31 860 751 666 549 844 267 852 202 456 537 280 732 835 420 753 2 411 810 48 696 60 656 958 113 262 217 434 636 557 370 146 344 50 405 6 145 76 130 762 697 755 697 394 317 764 406 840 510 840 672 165 747 759 208 717 314 290 844 379 43 677 799 647 985 150 638 263 15 605 136 319 802 570 673 970 991 789 989 270 591 144 529 852 16 632 577 402 986 950 642 816 747 774 90 798 979 691 865 199 473 689 517 838 372 337 757 927 203 577 979 613 134 461 359 184 74 948 358 840 848 288 68 377 58 722 812 283 673 27 222 402 159 211 424 461 786 914 770 207 521 13 310 866 747 360 458 661 228 403 862 288 542 378 903 917 807 624 866 211 740 536 171 83 437 278 736 949 550 990 855 577 642 611 442 53 369 830 855 405 902 514 933 217 833 317 56 517 355 805 271 515 181 23 102 279 596 221 458 696 893 74 481 33 985 511 958 615 258 891 898 968 449 229 935 535 217 229 954 83 982 216 376 354 517 516 926 851 850 798 315 916 171 899 43 591 498 77 769 826 3 782 264 324 546 915 162 765 108 860 758 545 98 432 52 512 797 271 282 689 340 792 563 648 262 830 258 295 849 695 509 982 913 570 550 271 993 168 226 65 849 783 147 258 634 256 296 198 996 566 332 502 928 409 779 158 571 648 970 213 835 780 538 406 227 946 884 34 120 807 869 807 509 998 527 804 50 711 231 445 213 704 363 628 804 282 67 135 630 365 951 239 576 970 541 242 103 230 716 262 375 878 958 397 480 727 30 588 927 52 104 513 271 616 589 628 683 868 855 95 692 817 692 689 118 130 725 62 628 759 820 32 936 850 484 262 479 618 313 640 375 752 485 484 717 855 546 890 203 450 896 976 138 664 213 707 31 746 828 462 814 703 297 240 907 418 562 653 948 742 756 19 26 912 874 468 627 801 899 27 130 533 869 147 872 791 58 64 210 932 908 664 597 652 572 212 711 501 831 249 455 199 269 201 837 660 92 599 398 857 434 473 751 252 578 595 325 308 893 549 476 29 749 114 637 623 196 897 104 444 659 552 692 310 480 94 95 346 85 436 753 854 264 91 882 762 549 935 51 351 318 982 175 597 263 483 920 658 242 471 958 617 537 691 476 825 946 29 403 944 924 961 544 843 27 368 861 446 238 9 529 680 247 807 776 427 681 267 560 222 312 930 279 15 480 816 550 715 549 560 297 59 810 785 741 106 685 817 554 313 20 679 765 881 925 291 676 731 564 989 392 9 930 702 591 657 230 106 981 379 96 470 34 313 863 940 46 146 128 374 549 392 402 639 91 564 323 528 196 948 598 393 774 933 948 635 547 788 251 363 363 240 517 562 797 159 130 87 878 91 365 601 760 542 10 2 956 963 133 236 380 742 392 52 196 538 141 204 724 278 534 16 211 344 269 870 694 273 982 308 990 696 736 285 29 246 198 506 547 741 573 756 599 201 78 362 666 989 806 706 623 22 596 115 369 566 534 47 958 156 319 312 655 941 504 238 515 9 45 120 564 981 788 259 548 439 719 544 523 479 772 268 670 138 78 939 203 602 619 574 349 932 161 432 469 31 455 297 911 565 751 831 439 195 50 337 250 193 299 43 426 823 987 439 819 173 744 591 806 976 753 558 548 355 654 429 358 590 104 124 831 71 849 655 125 517 747 608 447 28 525 732 763 864 528 393 346 895 486 598 796 637 295 927 762 599 74 285 570 278 486 5 15 109 391 833 903 308 803 316 812 332 234 108 44 668 394 529 451 300 320 912 315 954 526 840 525 665 213 210 279 171 366 452 924 286 553 29 580 354 860 129 140 517 385 490 305 925 111 472 63 807 587 312 830 330 451 276 97 181 197 436 582 773 231 603 950 634 640 733 483 264 87 540 949 25 263 425 107 156 127 671 832 261 552 600 135 584 402 241 258 795 270 623 904 415 293 419 408 411 81 859 68 474 377 466 489 916 476 685 936 497 219 401 736 132 805 802 744 965 637 964 317 985 207 455 395 250 90 887 734 898 428]; s=build_snowmen(m); % 9 332 assert(isequal(s,332));
10 Pass
%% m=[928 718 251 983 78 255 542 424 326 434 314 138 651 627 730 973 666 311 634 603 147 169 518 707 582 562 424 777 450 383 889 969 772 108 678 226 45 288 201 686 490 703 979 33 751 387 845 766 735 968 23 659 967 547 219 407 741 317 426 692 14 981 789 299 169 179 645 497 166 124 460 724 198 932 996 776 601 207 234 962 714 537 363 814 638 741 981 796 603 652 17 39 477 295 303 510 539 670 531 467 39 369 100 595 474 46 798 266 906 176 35 352 35 520 544 598 969 320 899 995 7 253 743 378 956 972 481 457 125 250 737 328 885 753 926 683 370 88 644 127 596 106 208 249 155 744 482 664 995 575 491 450 414 529 179 996 794 990 397 513 598 308 291 396 156 535 770 896 162 431 218 706 823 735 342 491 755 964 914 195 177 129 700 930 915 415 117 305 557 248 951 383 853 160 844 114 320 895 886 667 551 174 826 678 769 727 448 46 945 458 10 93 599 323 495 931 822 770 181 41 37 370 667 893 210 846 877 450 478 541 756 354 462 737 805 242 635 469 162 599 255 30 356 389 174 626 67 475 875 779 216 464 448 308 496 993 13 335 290 295 353 102 486 440 47 269 514 720 53 72 547 366 243 207 567 437 234 434 836 305 237 551 693 560 834 936 210 163 231 513 811 867 90 536 205 332 682 41 135 162 475 776 91 88 890 442 87 119 941 265 444 493 895 521 909 241 461 656 810 2 859 401 194 909 864 524 869 765 367 449 477 741 381 143 104 27 205 316 575 838 882 163 361 35 464 249 187 143 621 592 320 38 319 136 414 888 74 473 660 195 129 417 575 157 675 60 323 172 620 729 527 126 66 95 78 50 382 419 670 27 231 593 389 522 853 584 274 796 939 713 383 406 18 192 887 830 684 139 910 40 192 710 653 235 726 157 713 584 911 840 132 378 231 36 293 831 932 538 603 951 911 502 269 306 711 344 297 912 566 719 653 666 2 215 498 745 327 368 13 408 743 848 766 809 789 820 928 69 52 312 453 995 61 621 572 728 110 571 113 477 98 264 764 619 958 520 823 165 550 671 541 18 189 302 997 456 895 971 422 657 529 348 701 96 486 776 375 231 538 298 469 921 285 958 879 584 993 228 619 319 708 136 293 432 188 586 469 779 968 639 723 154 311 811 450 546 510 998 982 695 347 867 725 970 155 518 859 579 64 668 103 567 736 156 488 799 701 410 654 965 449 100 206 631 166 792 627 761 881 246 724 332 314 953 64 870 761 979 193 540 952 694 513 186 594 850 566 987 122 209 849 325 603 951 668 893 689 907 986 685 772 598 557 870 240 154 894 998 385 105 392 134 467 85 315 582 66 63 599 194 60 940 974 295 322 322 537 865 101 897 304 186 579 863 62 174 749 589 451 251 729 781 392 948 437 141 798 884 165 363 78 737 754 292 376 435 523 659 860 800 845 428 182 254 327 371 444 20 95 708 192 252 523 823 756 98 323 893 677 349 524 644 121 664 870 870 27 233 548 140 113 475 193 290 848 808 463 779 333 120 153 482 400 749 921 967 72 223 6 893 694 754 5 125 471 145 494 413 320 499 564 464 194 390 637 199 557 291 460 869 786 705 222 807 785 817 502 277 549 485 791 222 813 206 292 750 497 275 939 578 81 969 961 655 826 772 898 596 782 668 615 947 284 734 629 308 482 56 870 539 826 576 614 80 224 333 96 188 392 874 546 150 753 906 557 66 963 775 921 248 231 35 653 44 531 891 127 528 248 386 55 67 565 310 430 797 740 92 354 433 774 490 278 637 608 894 794 170 376 674 76 244 427 206 319 517 859 983 688 742 467 459 509 298 852 224 22 825 604 960 623 603 87 391 321 876 143 445 337 768 65 471 557 400 725 125 388 846 823 942 516 147 947 647 713 794 599 784 597 144 680 347 690 344 997 574 891 27 80 794 278 585 654 293 983 257 502 458 657 640 4 29 38 662 725 649 371 45 544 380 587 856 258 976 167 623 999 958 825 571 562 904 734 72 594 791 601 682 959 999 749 629 8 823 302 938 57 888 811 260 47 326 49 591 54 108 486 535 811 990 188 321 364 860 247 802 358 942 479 665 727 66 112 69 463 635 384 946 227 514 388 980 696 259 665 901 806 720 909 499 777 840 867 211 280 874 782 478 445 249 911 920 55 499 618 742 829 897 388 981 145 600 880 667 703 745 144 664 479 317 523 858 63 803 576 898 532 681 133 323 382 537 20 892 422 68 366 155 983 767]; s=build_snowmen(m); % 10 330 assert(isequal(s,330));
11 Pass
%% m=[500 870 831 984 466 67 302 562 875 492 839 394 773 606 263 241 36 148 878 411 456 946 557 20 901 567 87 243 240 439 505 204 165 64 953 131 450 358 763 981 757 934 514 584 517 151 571 511 414 763 642 303 197 591 81 889 539 629 836 205 84 34 409 639 174 201 712 557 229 566 970 282 458 747 593 803 455 462 122 614 398 837 602 694 371 936 298 92 736 941 52 906 688 395 859 703 707 940 933 184 236 389 227 348 114 522 707 356 278 671 525 920 682 288 398 69 526 743 525 77 141 357 379 16 258 914 677 126 571 66 668 859 531 651 860 996 286 573 315 69 900 33 419 523 185 365 691 638 232 298 704 221 428 853 886 321 19 911 511 563 396 285 193 955 997 302 427 562 961 539 296 895 263 724 462 442 47 35 772 735 286 612 602 24 496 730 270 938 24 626 834 599 238 230 407 769 544 413 104 707 553 984 361 715 331 255 740 820 837 595 353 821 308 22 10 311 70 416 48 511 351 28 545 538 977 61 838 723 304 984 896 622 98 998 438 731 893 850 564 927 751 466 572 279 146 786 144 32 900 862 366 603 502 208 764 611 719 258 38 762 37 340 632 219 456 795 489 272 365 868 667 557 604 192 794 219 470 294 246 62 960 745 393 704 406 184 242 976 195 231 180 362 726 328 767 333 517 515 93 988 387 188 611 228 641 446 407 77 177 854 183 800 249 666 725 233 814 498 350 115 759 913 267 517 780 721 580 129 990 221 57 263 225 14 582 584 529 21 624 609 986 905 647 176 392 953 381 927 902 418 486 632 195 362 323 550 34 819 374 494 215 901 815 995 767 276 190 795 762 239 284 103 47 407 154 434 638 58 206 655 877 935 673 659 705 760 402 906 744 845 679 42 551 272 424 216 530 829 69 954 258 744 250 767 729 645 46 354 861 156 669 407 164 170 759 854 522 17 378 562 797 348 449 108 786 972 765 876 254 453 280 281 959 709 10 337 230 631 288 633 961 549 665 497 415 161 218 398 700 762 151 411 900 202 63 557 134 766 147 258 502 853 908 319 435 427 479 471 569 102 873 698 264 465 715 169 907 677 745 597 322 558 649 276 787 804 724 975 428 343 300 960 265 459 466 536 479 121 195 954 794 717 218 899 516 655 562 985 345 611 465 882 691 285 423 904 818 880 463 477 132 717 783 514 958 825 87 928 707 533 720 695 576 465 764 311 294 997 576 639 854 380 325 610 140 352 248 568 375 676 133 898 932 826 178 572 822 233 729 213 165 337 844 131 410 854 778 733 197 691 552 945 389 926 898 63 823 92 125 264 867 568 214 4 343 536 891 113 122 86 583 435 119 526 780 20 19 323 152 492 847 865 510 645 127 517 784 447 122 961 280 515 329 654 208 878 298 37 356 239 147 513 655 962 863 423 767 644 748 231 600 438 98 769 696 205 268 546 602 293 242 145 364 767 419 357 846 139 361 640 828 779 412 571 782 91 916 471 668 816 370 265 866 545 620 383 177 614 841 129 4 657 812 750 54 793 420 425 483 718 319 987 592 686 258 439 25 237 620 442 571 870 679 236 117 913 116 605 450 390 100 108 258 122 891 611 848 490 842 932 695 283 982 442 153 840 448 822 267 346 542 631 591 266 298 481 344 569 507 464 232 372 122 852 666 793 357 85 677 679 20 478 89 18 989 294 444 443 775 750 820 87 678 858 265 586 872 60 562 63 744 312 471 927 407 852 948 702 317 852 687 709 709 986 493 238 601 556 422 24 453 778 220 454 845 671 556 741 275 958 414 868 203 424 334 346 886 727 255 406 301 525 457 50 97 106 360 179 306 769 670 833 521 386 626 765 685 137 455 368 200 689 167 986 281 603 618 244 111 618 557 386 65 35 587 700 989 607 371 405 214 596 994 653 845 577 823 266 468 653 572 623 768 488 311 742 603 975 941 361 457 993 653 49 989 638 502 781 822 592 229 404 64 28 29 185 914 284 423 323 687 741 349 791 329 715 565 587 420 394 749 565 128 645 966 324 481 920 855 741 454 484 64 587 215 777 427 458 445 359 861 831 259 844 846 85 252 58 868 494 205 834 911 845 552 90 144 271 569 588 921 750 168 830 428 293 257 83 475 425 975 583 596 211 726 534 355 235 133 926 789 765 855 598 722 718 182 969 59 868 709 474 1 708 271 705 902 985 400 911 79 565 273 462 515 804 659 706 188 704 194 857 518 25 301 513 701 824 693 254 808 625 583 937 820 234 700 907 291 50]; s=build_snowmen(m); % 11 331 assert(isequal(s,331));
12 Pass
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13 Pass
%% m=[616 947 262 312 240 223 549 629 569 332 320 478 189 847 661 365 461 817 623 371 864 252 641 168 40 77 399 148 881 160 627 916 196 252 586 336 35 458 957 379 60 470 246 990 34 376 247 211 185 482 880 747 750 167 938 864 445 189 646 176 955 483 781 190 418 662 871 630 163 854 789 138 686 884 272 407 762 266 306 360 215 366 866 992 672 237 377 916 453 479 188 221 470 601 208 479 518 885 208 125 835 381 535 29 489 63 744 88 293 766 89 66 797 335 581 700 854 794 960 507 316 397 882 509 552 129 665 296 750 347 309 274 288 733 314 97 331 613 832 719 166 54 869 105 600 977 55 786 542 896 521 745 308 513 814 788 158 2 744 82 597 601 316 858 580 130 669 513 499 668 535 913 548 629 747 778 519 412 736 695 610 922 913 663 132 724 492 531 64 84 653 200 169 726 993 783 398 513 969 371 709 491 271 406 830 715 48 513 874 478 506 394 555 381 219 419 313 254 448 138 477 163 830 140 594 725 332 878 451 553 607 493 899 665 288 658 630 664 318 462 11 742 866 173 901 103 143 694 802 48 244 530 24 750 338 715 232 326 345 98 875 245 200 625 831 298 657 255 789 950 734 900 895 294 144 452 995 643 17 299 375 380 61 160 560 86 63 467 65 583 275 721 831 905 193 436 495 826 874 508 573 375 695 830 327 151 291 572 810 285 395 965 234 45 109 797 765 730 716 858 99 474 784 355 305 779 53 722 305 609 61 455 75 661 825 379 922 452 86 2 943 34 858 951 202 167 221 942 542 372 436 677 166 303 564 727 705 237 472 5 297 731 764 655 610 61 375 997 128 465 531 731 219 222 15 41 733 190 657 819 21 23 983 840 525 697 131 894 867 170 970 527 935 801 154 795 919 548 817 864 11 178 254 680 883 105 491 541 103 490 700 776 355 223 546 585 857 410 24 36 853 998 290 662 710 954 828 287 173 425 653 479 668 710 506 889 811 665 946 684 610 362 259 87 723 632 136 477 870 720 207 656 747 67 1 704 134 51 206 466 158 315 804 669 619 301 967 328 299 209 59 829 983 228 868 549 929 240 987 204 696 31 71 585 60 43 892 874 143 74 935 593 893 769 423 587 409 914 738 686 870 366 536 859 372 236 286 487 5 758 765 582 692 72 830 977 726 522 586 317 504 933 287 255 22 661 83 909 407 444 947 584 872 495 108 783 553 257 496 550 601 625 955 948 640 293 926 315 107 821 924 703 804 665 757 962 606 793 981 836 607 504 746 761 678 277 529 732 526 688 107 711 799 604 670 768 548 876 423 999 986 586 575 30 458 428 267 351 941 290 146 497 630 435 713 487 485 717 782 365 519 653 153 668 302 581 846 632 447 723 811 163 577 385 631 15 702 106 164 584 144 601 823 143 520 904 451 352 917 947 471 464 181 197 293 47 38 969 46 253 795 63 405 659 513 312 811 506 801 110 18 344 354 102 291 947 547 502 986 675 422 26 627 738 588 293 609 194 963 707 454 270 675 247 745 869 454 985 858 72 843 371 825 646 339 292 645 616 805 543 572 299 871 531 107 421 243 438 13 783 655 457 285 647 992 955 197 298 287 150 740 10 941 553 8 424 690 798 612 975 216 869 87 988 765 772 790 270 271 253 237 15 720 407 775 75 82 591 150 374 520 674 651 739 835 839 412 339 824 570 940 974 721 647 231 40 623 380 782 566 821 287 639 211 385 668 187 765 698 283 218 53 691 353 407 805 288 339 709 578 982 880 605 93 494 565 933 866 995 509 814 719 272 410 5 420 903 880 405 461 823 682 43 615 359 704 854 280 487 885 795 651 33 733 406 931 327 42 547 214 56 842 242 464 623 421 541 817 756 205 343 424 268 171 89 264 109 772 457 158 647 640 590 995 465 253 167 16 40 616 648 747 599 587 46 255 501 840 283 300 847 530 538 643 61 848 846 531 952 126 713 559 805 487 843 372 196 487 136 798 676 554 441 61 449 74 32 877 756 517 488 891 106 667 551 834 461 545 31 670 923 259 337 530 427 37 124 67 969 613 189 257 509 448 152 870 256 957 630 240 466 221 42 141 484 557 969 835 949 413 381 261 883 390 73 840 200 820 29 91 349 930 226 892 431 11 434 420 90 622 896 173 729 252 253 922 963 459 205 211 435 732 662 810 117 864 90 186 236 501 977 380 928 65 370 731 163 87 763 276 851 773 112 436 873 629 453 435 147 763 695 251 848 705 267 146 135 350 780 187]; s=build_snowmen(m); % 13 329 assert(isequal(s,329));
14 Pass
%% m=[552 992 717 501 400 776 349 964 211 711 531 391 413 57 251 516 507 788 158 673 856 517 98 280 133 275 863 701 214 687 170 217 999 862 889 8 926 319 971 709 243 275 474 422 248 430 115 937 323 361 591 285 794 848 886 334 828 682 168 365 603 45 337 155 578 209 219 674 878 80 650 945 828 300 470 980 651 441 32 307 572 8 691 632 728 385 671 356 343 654 523 61 148 380 438 498 920 533 803 207 129 240 48 86 529 496 595 33 729 773 345 393 147 841 517 335 595 519 278 872 304 307 369 66 16 188 781 525 584 762 65 742 677 602 551 566 437 393 54 566 911 139 324 988 71 801 252 241 456 469 655 831 862 388 473 881 701 994 196 143 654 245 751 37 47 997 272 847 965 550 973 597 153 273 364 312 888 53 207 21 296 673 479 208 494 266 724 238 19 945 639 82 537 159 534 674 285 79 650 819 53 148 10 166 933 335 183 206 669 381 416 371 27 999 351 800 803 335 538 885 912 214 794 698 565 306 809 866 368 324 47 391 764 994 738 537 57 111 823 938 610 801 639 212 86 140 395 435 115 616 457 3 650 199 66 771 598 910 301 644 193 438 350 900 889 721 864 401 655 210 857 204 831 152 760 690 632 611 261 161 881 154 347 992 821 922 584 508 618 223 455 267 982 551 972 709 542 187 615 273 182 469 776 519 977 97 389 179 952 554 325 51 479 218 597 281 748 233 208 139 334 721 842 619 87 416 652 206 361 472 944 93 17 348 600 629 220 523 661 72 659 137 886 6 11 901 125 322 69 2 917 917 977 505 762 210 579 213 851 200 711 884 130 225 265 732 873 100 55 145 635 792 887 149 571 149 502 726 170 193 352 328 820 528 107 699 236 971 658 346 160 130 680 757 575 659 956 931 263 924 300 501 895 786 99 209 728 431 105 236 498 995 291 535 947 674 6 458 11 463 248 643 87 657 86 247 810 12 846 681 565 324 503 955 752 923 479 956 435 225 695 800 601 857 702 569 254 486 481 755 358 670 196 159 271 468 863 535 686 322 215 143 298 538 735 695 39 301 339 259 489 332 877 215 494 532 128 204 823 920 867 791 276 200 477 372 277 876 880 31 181 855 466 701 105 38 990 535 317 40 315 209 18 176 775 699 918 397 36 37 460 757 422 394 484 86 788 787 390 817 523 364 1 102 729 966 755 914 189 113 669 88 767 522 270 5 643 696 939 254 819 93 298 203 528 413 33 987 555 134 226 253 709 593 571 575 260 373 942 686 379 92 306 415 984 541 621 893 56 677 149 530 793 816 391 288 293 724 92 988 525 566 519 570 475 201 446 218 639 383 87 697 927 361 362 817 176 31 772 342 715 538 661 279 203 83 322 867 550 430 160 997 971 861 333 582 786 493 705 992 171 81 731 515 732 434 98 717 678 783 547 276 650 596 517 977 15 189 197 222 656 854 458 949 295 426 457 324 531 404 351 91 485 176 944 134 613 449 495 160 153 616 497 243 409 398 883 329 519 858 248 857 897 550 133 801 617 487 403 460 125 289 112 128 536 605 721 570 515 768 266 618 984 371 897 817 871 712 52 260 336 438 30 106 479 892 457 738 586 101 407 874 306 330 40 542 604 728 980 68 110 183 44 186 685 313 748 1 238 533 235 325 606 502 426 521 735 881 828 568 577 145 34 680 739 336 116 200 494 332 366 96 991 104 988 386 292 734 773 415 436 282 177 499 798 246 642 883 723 145 968 725 89 124 770 615 814 139 840 191 536 224 46 346 482 172 186 165 364 752 225 510 317 602 542 984 704 95 210 402 427 332 724 869 544 566 630 839 644 195 444 829 185 641 914 537 426 648 442 693 389 708 876 960 856 165 921 12 641 296 385 103 354 919 68 682 93 626 549 142 470 986 877 57 420 45 591 213 794 422 696 567 778 768 429 867 558 463 831 32 451 122 749 445 937 809 895 73 207 84 991 534 654 222 423 38 747 318 21 940 82 9 204 9 267 923 907 153 847 817 112 920 302 271 10 307 335 925 630 376 685 742 744 451 287 232 480 163 995 373 470 513 310 544 352 465 716 193 662 168 677 972 929 583 486 666 596 995 629 221 818 551 851 231 720 556 981 175 779 645 320 744 134 551 423 992 449 605 713 615 569 685 123 898 374 945 22 654 867 641 629 870 733 181 156 604 157 336 879 832 773 926 401 792 132 94 693 543 330 126 168 59 428 306 587 989 661 162 250 608 504 800 468 475 171 474 770 142 218 703]; s=build_snowmen(m); % 14 333 assert(isequal(s,333));
15 Pass
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16 Fail
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https://www.painphr.com/q422_8aa_simplify | 1,537,589,370,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158011.18/warc/CC-MAIN-20180922024918-20180922045318-00285.warc.gz | 830,746,923 | 6,474 | how do i simplify 8a+a
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Guide :
# how do i simplify 8a+a
how do i simplify 8a+a
## Research, Knowledge and Information :
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### How do you simplify 8a + 3 – 5(a – 2)? | Socratic
How do you simplify 8a + 3 – 5(a – 2)? Algebra Properties of Real Numbers Expressions and the Distributive Property. ... How do you simplify the expression #x(3+5)#?
### how do i simplify (8a)^2 - OpenStudy
how do i simplify (8a)^2(8a)^2 = (8^2)(a^2). (8a)^(2) = (8a)(8a). Please explain . $x^2 = (x)(x)$ $x^3 = (x)(x)(x)$ etc thus \[(xa)^... ... At vero eos et ...
### How do you simplify 8a + b - 3a + 4b? | Socratic
How do you simplify #8a + b - 3a + 4b#? Algebra Expressions, Equations, and Functions Variable Expressions. 1 Answer ... How do you simplify #7+10x-x+(-x)#?
### How to Simplify Math Expressions: 13 Steps (with Pictures)
How to Simplify Math Expressions. ... Simplify by factoring. Factoring is a technique by which some variable expressions, including polynomials, can be simplified.
### How do you simplify: (8a^-6)^-2/3 - OpenStudy
How do you simplify: (8a^-6)^-2/31/4a^4. thanks! :). No problem. ... (8a^-6)^-2/3 \(\left(\dfrac{8}{a^{6}}\right)^{-2/3} = \left (\dfrac{a^{6 ...
### How to Simplify Equations (solutions, examples, videos)
How to Simplify Equations. ... Some methods that can be used to simplify an equation are Combine Like Terms, Multiplication & Division of Terms, ...
### Simplify any Algebraic Expression - WebMath
Simplify any Algebraic Expression ... If you have some tough algebraic expression to simplify, this page will try everything this web site knows to simplify it.
### How do I simplify- please explain as if I were a five year ...
How do I simplify- please explain as if I were a five year old - 3293890. 1. Log in ... ^y=a^x*y so your exercise is equal =(8a)^(-3)*(-2/3)=(8a)^2=64a^2 Comments (2 ...
## Suggested Questions And Answer :
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### sqrt(3a+10) = sqrt(2a-1) + 2
sqrt(3a+10)=sqrt(2a-1)+2 To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(3a+10))^(2)=(~(2a-1)+2)^(2) Simplify the left-hand side of the equation. 3a+10=(~(2a-1)+2)^(2) Squaring an expression is the same as multiplying the expression by itself 2 t
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### Find m if x – 3 and x + 2 are factors of x3 + m2x2 – 11x – 15m
36r4 + 36r3 + 3r2) / 9r3 this is normally written as (36r^4 + 36r^3 + 3r^2) / (9r^3) this is equivalent to (36r^4 / 9r^3) + (36r^3 / 9r^3) + (3r^2 / 9r^3) the rule for exponent arithmetic to apply is that x^a/x^b = x^(a-b). so, basically, if the base is
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### how do i add fractions
To add or subtract fractions, obtain a least common denominator. Subtract the numerators in the correct order and retain the same least common denominator for your answer. Simplify. To multiply fractions, multiple the numerators. The product will be the numerator of your answer. Repeat with denomin
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### how to solve for x with fractions
Simplifying x3 + 3x2 + -4x = 0 Reorder the terms: -4x + 3x2 + x3 = 0 Solving -4x + 3x2 + x3 = 0 Solving for variable 'x'. Factor out the Greatest Common Factor (GCF), 'x'. x(-4 + 3x + x2) = 0 Factor a trinomial. x((-4 + -1x)(1 + -1x)) = 0 Subproblem 1 Set the factor 'x' equal to zero an
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### Whats 3x-2 over 9 equals 25 over 3x-2?
I will give you an example for this question. This might be able to help you. Simplifying 3x2 + 25x = 18 Reorder the terms: 25x + 3x2 = 18 Solving 25x + 3x2 = 18 Solving for variable 'x'. Reorder the terms: -18 + 25x + 3x2 = 18 + -18 Combine like terms: 18 + -18 = 0 -18 + 25x + 3x2 = 0 Fac
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### How to simplify fractions
When simplifying fractions we need to reduce both the numerator (the number on the top of the fraction) and the denominator (the number on the bottom of the fraction) so that both numbers have no common factors (numbers that can be evenly divided into both) Let's take this the example 30/100 We n
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### Expand and simplify where appropriate
[r-36(p.q+q)/3] simplifies to [r-12q(p+1)] or [r-12pq-12q]. {n+r[r-36(p.q+q)/3]+p} therefore simplifies to {n+r[r-12pq-12q]+p}={n+r^2-12pqr-12qr+p}. m{ ... } simplifies to mn+mr^2-12mpqr-12mqr+mp. Finally, multiply by 3q/p: 3mnq/p+3mqr^2/p-36mq^2r-36mq^2r/p+3mq.
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### simplify this expression 9(a+b)
simplify 9(a+b) It's already simplified. 9(a+b) simplified is 9(a+b) if you mean expand 9(a+b) that's 9a + 9b
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### 2x/3+1/4=9,find x
2x/3+1/4=9 (2x)/(3)+(1)/(4)=9 Since (1)/(4) does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting (1)/(4) from both sides. (2x)/(3)=-(1)/(4)+9 Simplify the right-hand side of the equation. (2x)/(3)=(35)/(4) Multiply each term in the equation by
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• Use proper spelling and grammar | 2,016 | 6,333 | {"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} | 4.1875 | 4 | CC-MAIN-2018-39 | latest | en | 0.777868 |
https://www.vanessabenedict.com/grams-in-oz/ | 1,723,366,634,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640983659.65/warc/CC-MAIN-20240811075334-20240811105334-00381.warc.gz | 800,221,302 | 13,957 | # How many grams is 14 oz?
#### ByVanessa
Jun 9, 2022
Untitled Document
## How much does 1 oz weigh in grams
In fact, 1 ounce is 28.35 grams.
£566.99
## How many grams is 14 oz
396.89 grams
Untitled Document
## How many weight of 200 grams is 1000 grams
Convert 200 grams to kilograms.
## How many grams of nitrogen are in a diet consisting of 100 grams of protein
Why? If you list the amount of essential protein or amino acids in what you eat, you can use one of these values ??to determine the amount of nitrogen in the amount provided by the protein. Its protein is only 16% nitrogen, and converting this to a value by dividing 100% by 16% gives 6.25.
## Why are Grams called Grams
By mass, one g is equal to one thousandth of a liter a (one cubic centimeter) relative to water at 4 degrees Celsius. The word “gram” comes from the late Latin “gramma”, which means a small French “gram” of weight. The token for grams is g.
## What is the amount in grams of quick lime can be obtained from 25 grams of CaCO3 on calculation
Full step-by-step answer: Therefore, the correct remedy is option C, namely the breakdown of the exact $\text25 g$ calcium carbonate, which is known to give us \[\text14 g\ ] from calcium oxide or what I just called quicklime.
## How many grams of 80% pure marble stone on calcination can give for 3 grams of quicklime
How many grams of 80% pure marble can instantly produce 14 grams of lime when sized? 80 g CaCO3 = 100 g marble stone. 25 g CaCO3=? 2580×100=31.25 g.
Untitled Document | 406 | 1,530 | {"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} | 3.71875 | 4 | CC-MAIN-2024-33 | latest | en | 0.910555 |
https://smallbusiness.chron.com/positive-return-31591.html | 1,669,942,218,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710870.69/warc/CC-MAIN-20221201221914-20221202011914-00682.warc.gz | 571,171,735 | 22,937 | # What Is a Positive Return?
A positive return occurs when your profits from a business transaction are more than the amount invested. It is part of the business concept known as ROI, or return on investment. ROI is a method of comparing the amount of capital generated with the amount you invest in a particular business, project or marketing campaign. The ROI is a way to analyze the company's financial standings and the success of the project you invested in.
## Positive Return
1. A positive return simply means the project or business you invested in generates more money than you put into it. For example, if you invested \$20,000 into your business and it becomes worth \$25,000, you have a positive return on your investment. If you spend \$1,000 on a new marketing campaign and calculate that the ad generates \$5,000 in sales, this is a positive return. However you invest money into the company, it is considered a positive return if you make more than you spend.
## Calculations
1. For a basic calculation, subtract the amount you invested from the amount generated. For example, the ad campaign costing \$1,000 that nets \$5,000 in sales would be calculated by subtracting \$1,000 from \$5,000 for a total of \$4,000. This is how much more than the investment that you generated. Divide this number by your initial investment, in this case \$1,000, for an answer of 4. This means you generated four times the amount you invested in the campaign.
## Timeline
1. In most cases, you won't know instantly if your investment will net you a positive return. Companies take time to build up revenue so you won't know if your investment in the company turns into a profit until some point in the future. Your ad-campaign investment money won't immediately be recouped as soon as the ad hits the airwaves. Gradual sales help you get closer to a positive return. Set a timeline for calculating your return on investment. Recalculate the numbers at various points to determine if you are getting closer to a positive return.
## Use
1. Knowing that you have a positive return on investment is a useful bit of information, but the return on investment data helps you going forward. While not the only method of making future business decisions, knowing which projects or campaigns earned you a positive return helps you figure out where to invest further money. For example, if a marketing campaign on a local college campus gave you a positive return on investment, you might continue marketing to this group or expand to other area colleges. | 528 | 2,553 | {"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-2022-49 | longest | en | 0.940698 |
http://www.eevblog.com/forum/renewable-energy/battery-voltage-drop-97817/ | 1,540,009,315,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512504.64/warc/CC-MAIN-20181020034552-20181020060052-00055.warc.gz | 455,548,702 | 8,727 | ### Author Topic: Battery voltage drop. (Read 542 times)
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#### McBryce
• Super Contributor
• Posts: 1145
• Country:
##### Battery voltage drop.
« on: November 01, 2017, 01:40:06 am »
Hi all,
I was just messing about with some old batteries, setting up my DC Load for a battery capacity test. I have two old 9V batteries, both read about 7.5V with no load. If I connect either one to my DC Load and try to pull 500mA, the voltage drops to about 3V, but supplies the 500mA. All as expected so far. However, if I connect the two batteries in series (total no load voltage now around 14.8V) and try to pull 500mA the voltage drops to 0.027V and can only manage around 280mA. I wasn't expecting this, but I'm no battery expert.
I would have expected somewhere around 6V and the 500mA still being supplied, but obviously I was wrong. Is this due to the two ESRs being added together? The maths don't add up for me. Can anyone explain exactly what's happening here?
Thanks,
McBryce.
#### IanMacdonald
• Frequent Contributor
• Posts: 774
• Country:
##### Re: Battery voltage drop.
« Reply #1 on: November 01, 2017, 01:59:32 am »
The internal resistance per battery is (7.5-3)/0.5 or 9 ohms.
With two in series the volts drop across the two internal resistances (18 ohms total) with 500mA flowing would be 9v, so the terminal voltage should be (7.5x2)-9 or 6v.
I suspect you have some kind of measurement error with the result you get.
#### McBryce
• Super Contributor
• Posts: 1145
• Country:
##### Re: Battery voltage drop.
« Reply #2 on: November 01, 2017, 02:20:22 am »
Thanks for confirming that my maths wasn't wrong.
After some experimentation I found the answer: One of the batteries seems to have an issue. It only supplies the 3V / 500mA for about 10 seconds and then drops to 0V. I suspect I didn't wait long enough when checking the batteries individually.
McBryce.
Smf | 534 | 1,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-43 | latest | en | 0.928782 |
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Distance time graphs by mizz happy Teaching Resources Tes from Distance Time Graphs Worksheet Answers , source:tes.com | 386 | 1,908 | {"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-2019-13 | latest | en | 0.771831 |
https://questions.llc/questions/2748460 | 1,713,387,903,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817181.55/warc/CC-MAIN-20240417204934-20240417234934-00790.warc.gz | 438,153,409 | 6,066 | # Use the information in the table about four different electric circuits to answer the question.
Circuit Resistance (ohms) Voltage (V)
1 0.5 20
2 0.5 40
3 0.25 40
4 0.25 60
Which circuit will have the greatest electric current?
(1 point)
Responses
circuit 1
circuit 1
circuit 3
circuit 3
circuit 4
circuit 4
circuit 2
## Circuit A has a resistance of 7.5 ohms, and circuit B has a resistance of 5 ohms. Which statement about the circuits must be correct?(1 point)
Responses
If the voltage of circuit A is equal to the voltage of circuit B, then the current in circuit A is equal to the current in circuit B.
If the voltage of circuit A is equal to the voltage of circuit B, then the current in circuit A is equal to the current in circuit B.
If the voltage of circuit A is equal to the voltage of circuit B, then the current in circuit A is less than the current in circuit B.
If the voltage of circuit A is equal to the voltage of circuit B, then the current in circuit A is less than the current in circuit B.
If the current in circuit A is less than the current in circuit B, then the voltage in circuit A is less than the voltage in circuit B.
If the current in circuit A is less than the current in circuit B, then the voltage in circuit A is less than the voltage in circuit B.
If the current in circuit A is equal to the current in circuit B, then the voltage in circuit A is less than the voltage in circuit B.
## If the current in circuit A is less than the current in circuit B, then the voltage in circuit A is less than the voltage in circuit B.
This statement is correct because according to Ohm's Law (V = I*R), if the current in circuit A is lower than circuit B, and the resistance in circuit A is higher than circuit B, then the voltage drop in circuit A will also be less than in circuit B.
## Which statement best defines the resistance of an electric circuit?(1 point)
Responses
It is the amount of electrical energy that is available in a circuit.
It is the amount of electrical energy that is available in a circuit.
It is the measurement of the rate of electrical energy in a circuit.
It is the measurement of the rate of electrical energy in a circuit.
It describes how much electrical energy passes in a circuit.
It describes how much electrical energy passes in a circuit.
It limits the amount of electrical energy that can pass through a circuit.
It limits the amount of electrical energy that can pass through a circuit.
## It limits the amount of electrical energy that can pass through a circuit.
This statement best defines the resistance of an electric circuit. Resistance is a property of a material or component that limits the flow of electrical current through it, thereby restricting the amount of electrical energy that can pass through the circuit.
## An electric circuit has an expected current of 80 amps. An electrician measures the current in the circuit at 120 amps. Which statement provides a possible explanation for this observation?(1 point)
Responses
There is not enough voltage in the circuit.
There is not enough voltage in the circuit.
The circuit has an extra resistor.
The circuit has an extra resistor.
A resistor in the circuit is broken.
A resistor in the circuit is broken.
There is not enough electrical energy in the circuit.
## A resistor in the circuit is broken.
If the measured current in the circuit is higher than the expected current, it could be due to a broken or damaged resistor in the circuit, reducing the resistance and allowing more current to flow. This would explain the higher current reading by the electrician. | 790 | 3,621 | {"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-2024-18 | latest | en | 0.920689 |
https://physics.stackexchange.com/questions/798387/field-transformation-under-conformal-transformation?noredirect=1 | 1,726,663,813,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651895.12/warc/CC-MAIN-20240918100941-20240918130941-00786.warc.gz | 432,465,659 | 41,195 | # Field transformation under conformal transformation
In 1 (see references below), I'm trying to derive how a spinless field transforms under a conformal transformation, specifically eq. (2.41). CFT references/lectures are the most confusing I've seen EVER, even the basic definitions are a mess.
Reading 226464 and 469205 clarified bits and pieces but overall I still can't form a clear picture. @MBolin's answer about how to write conformal transformations based on Zee's book is the most clarifying/natural definition I've seen. On the other hand, @MannyC's clarification on the distinctions between diffeomorphism/Weyl/conformal transformations kind of gave me a big picture, BUT the notations and conventions are in contrast to some references which makes it hard to form a picture.
What I learned is that when people say conformal transformations they really mean a diffeomorphism in the sense that $$g'_{\mu\nu}(x') = \frac{\partial x^\rho}{\partial x'^\mu} \frac{\partial x^\sigma}{\partial x'^\nu} g_{\rho\sigma}(x) = \Omega(x)^2 g_{\rho\sigma}(x)$$ (Here, I didn't adopt @MBolin's answer since most people don't use that) plus a Weyl transformation $$\bar{g}_{\rho\sigma}(x) = \Omega(x)^2 g_{\rho\sigma}(x)$$.
First question, can anyone clarify which is which, some people define conformal transformations (which I think is really a Weyl transformation) in the first page of every reference to be $$g'_{\mu\nu}(x') = \Omega(x) g_{\mu\nu}(x)$$, while others define it to be $$g'_{\mu\nu}(x') = \Omega(x)^2 g_{\mu\nu}(x)$$. Why aren't they consistent? The extra square factor is confusing me. On top of that @MannyC's answer defined it to be $$g'_{\mu\nu}(x') = \Omega(x)^{-2} g_{\mu\nu}(x)$$.
Second, going back to my original question about how fields transform. I'll use the convention of prime $$\phi'(x')$$ for diffeomorphism and a bar $$\bar{\phi}(x)$$ for Weyl transformation. In some posts, they immediately defined that fields transform as $$\phi(x) \rightarrow \bar{\phi'}(x') = \Omega^{-\Delta}(x) \phi(x)$$, but I would like to derive this similar to how references go on about it. I think one way to derive this is through the action of a free scalar field with kinetic term only (I've read in Lecture Notes on String Theory that conformal invariance is hard to check even in flat spacetime so we can just check scale invariance and presume that conformal invariance holds),
\begin{align} S' & = \int d^dx' \bar{g}'^{\mu \nu} \partial'_\mu \bar{\phi'}(x') \partial'_\nu \bar{\phi'}(x') = \int d^dx \Omega^d \bar{g}'^{\mu \nu} \frac{\partial x^\rho}{\partial x'^\mu} \frac{\partial x^\sigma}{\partial x'^\nu} \partial_\rho \bar{\phi'}(x') \partial_\sigma \bar{\phi'}(x')\\ & = \int d^dx \Omega^d \Omega^{-2} g'^{\mu \nu} \frac{\partial x^\rho}{\partial x'^\mu} \frac{\partial x^\sigma}{\partial x'^\nu} \partial_\rho \bar{\phi'}(x') \partial_\sigma \bar{\phi'}(x'), \qquad \text{inverse Weyl:}\; \bar{g}^{\rho\sigma}(x) = \Omega(x)^{-2} g^{\rho\sigma}(x) \\ \end{align}
From the last line I'm not sure how to proceed, but the end result should look like,
$$$$\int d^dx \partial_\rho (\Omega^{\Delta} \bar{\phi'}(x') ) \partial^\rho ( \Omega^{\Delta} \bar{\phi'}(x') )$$$$
so that,
$$$$\Omega^{\Delta} \bar{\phi'}(x') = \phi(x) \rightarrow \bar{\phi'}(x') = \Omega^{-\Delta} \phi(x)$$$$
Lastly, in 1, the scale factor $$\Omega$$ was expressed in terms of the Jacobian, i.e. eq. (2.41), but I can't seem to find a way to derive it.
Reference:
Let $$g_{\mu \nu}$$ be the metric tensor in a $$d$$-dimensional spacetime. A conformal transformation is a change of coordinates $$x\mapsto x'=f(x)$$ that leaves the metric tensor invariant up to a scale: $$$$g_{\mu \nu}(x) \mapsto g'_{\mu \nu}(x')=\Omega^2(x)g_{\mu \nu}(x),$$$$ where we assume $$\Omega(x)>0$$. Some other definition just write $$\Omega$$ instead of $$\Omega^2$$, but it is just a question of taste.
Primary fields are defined based on the action of the generators of the conformal group on them. This is a general and purely geometrical definition, similarly to how you would define normal rotations in 3D.
Definition1. In $$d>2$$ spacetime dimensions, let $$\hat{D}$$ be the generator of dilations and let $$\hat{K}_{\mu}$$ be the generator of special conformal transformations. A conformal primary field $$\hat{\phi}^M_{\rho}(x)$$, in the $$\rho$$ representation of the Lorentz group and with conformal dimension $$\Delta$$, satisfies the following conditions at $$x=0$$: $$$$\left[\hat{D},\hat{\phi}^M_{\rho}(0)\right]=-i\Delta\hat{\phi}^M_{\rho}(0);\\ \left[\hat{K}_{\mu},\hat{\phi}^M_{\rho}(0)\right]=0.$$$$
Now, it turns out that this definition is equivalent to the one you are talking about. This second definition gives the explicit behaviour of the field under a conformal transformation.
Definition2. In $$d>2$$ spacetime dimensions, a conformal primary field $$\hat{\phi}^M_{\rho}(x)$$, in the $$\rho$$ representation of the Lorentz group and with conformal dimension $$\Delta$$, transforms under a conformal transformation $$\eta_{\mu \nu}\mapsto \Omega^2(x)\eta_{\mu \nu}$$ as $$$$\hat{\phi'}^M_{\rho}(x')=\Omega^{\Delta}(x)\mathcal{D}{\left[R(x)\right]^M}_{N}\hat{\phi}^N_{\rho}(x)$$$$ where $${R^{\mu}}_{\nu}(x)=\Omega^{-1}(x)\frac{\partial x^{\mu}}{\partial x'^{\nu}}$$ and $$\mathcal{D}{\left[R(x)\right]^M}_{N}$$ implements the action of $$R$$ in the $$SO(d-1,1)$$ representation of $$\hat{\phi}^{M}_{\rho}(x)$$.
The above equation is valid for any type of fields, so it generalizes the scalar field behaviour you are talking about.
The proof of the equivalence, as well as consistent notations and definitions (as you said, many CFT references are confusing), you find them in this paper: https://arxiv.org/abs/2112.01837. | 1,745 | 5,741 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 38, "wp-katex-eq": 0, "align": 1, "equation": 5, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-38 | latest | en | 0.918588 |
http://www.solving-math-problems.com/572-rationalize-the-denominator.html | 1,563,389,793,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525374.43/warc/CC-MAIN-20190717181736-20190717203736-00015.warc.gz | 275,691,079 | 12,465 | # 5/(√7+√2) - Rationalize the Denominator
by Omar Ibrahim
(Opelousas La)
Rationalize the Denominator (so that there are no radicals in the denominator)
Simplify the final expression as much as possible, but do not evaluate the radicals using decimals.
5/(√7+√2) | 72 | 265 | {"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} | 2.765625 | 3 | CC-MAIN-2019-30 | longest | en | 0.898919 |
http://mathhelpforum.com/algebra/41443-question-about-angle-segment-relationships.html | 1,481,369,564,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543035.87/warc/CC-MAIN-20161202170903-00038-ip-10-31-129-80.ec2.internal.warc.gz | 173,805,895 | 10,893 | 1. ## Question about angle and segment relationships?
Heres the question: Secants $ABC$ and $ADE$ intersect at point $A$ outside the circle. If $AC=20, AB=5, AD=4$, find $AE$.
Okay so i did
$20*5=4*x$
$100=4x$
$100/4=4x/4$
$x=25$
Okay so my question is, is the answer $25$ or do i subtract $4$ from it?
2. AE is 25. AE (or ADE) is comprised of AD and DE. You found AE (which was asked). AD = 4 and DE = 21.
Good!
3. Originally Posted by eh501
Heres the question: Secants $ABC$ and $ADE$ intersect at point $A$ outside the circle. If $AC=20, AB=5, AD=4$, find $AE$.
Okay so i did
$20*5=4*x$
$100=4x$
$100/4=4x/4$
$x=25$
Okay so my question is, is the answer $25$ or do i subtract $4$ from it?
Actually, the correct set up is this: If two secants intersect at an external point, the product of the external segment of one secant times the whole secant is equal to the external part of the other secant times the whole secant.
In other words, by my diagram: $AB \cdot AC=AD \cdot AE$
Therefore,
$5\cdot20=4\cdot(4+DE)$
$AE=4+DE$
$5\cdot20=4(4+DE)$
$100=16+4(DE)$
$84=4(DE)$
$21=DE$
$AE=4+DE \Longrightarrow 4+21=25$ | 398 | 1,129 | {"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": 30, "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-2016-50 | longest | en | 0.76185 |
https://www.shaalaa.com/question-bank-solutions/find-value-discriminant-x2-7x-1-0-solutions-quadratic-equations-completing-square_49892 | 1,660,580,171,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572192.79/warc/CC-MAIN-20220815145459-20220815175459-00218.warc.gz | 883,476,448 | 10,164 | Find the Value of Discriminant. X2 + 7x – 1 = 0 - Algebra
Find the value of discriminant.
x2 + 7x – 1 = 0
Solution
x2 + 7x – 1 = 0
Comparing the given equation with $a x^2 + bx + c = 0$
$a = 1, b = 7 \text{ and } c = - 1$So, the discriminant
$b^2 - 4ac = \left( 7 \right)^2 - 4 \times 1 \times \left( - 1 \right) = 49 + 4 = 53$
Concept: Solutions of Quadratic Equations by Completing the Square
Is there an error in this question or solution?
APPEARS IN
Balbharati Mathematics 1 Algebra 10th Standard SSC Maharashtra State Board | 196 | 539 | {"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} | 3.96875 | 4 | CC-MAIN-2022-33 | latest | en | 0.713669 |
https://www.slideserve.com/gaius/parallel-algorithms-for-geometric-graph-problems | 1,620,450,949,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988837.67/warc/CC-MAIN-20210508031423-20210508061423-00351.warc.gz | 1,031,227,071 | 22,423 | # Parallel Algorithms for Geometric Graph Problems - PowerPoint PPT Presentation
Parallel Algorithms for Geometric Graph Problems
1 / 29
Parallel Algorithms for Geometric Graph Problems
## Parallel Algorithms for Geometric Graph Problems
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##### Presentation Transcript
1. Parallel Algorithms for Geometric Graph Problems GrigoryYaroslavtsev http://grigory.us To appear in STOC 2014, joint work with AlexandrAndoni, Krzysztof Onak and AleksandarNikolov.
2. Postdoctoral Fellow icerm.brown.edu By Mike Cohea • Spring 2014: “Network Science and Graph Algorithms” • Fall 2014: “High-dimensional approximation”
3. “The Big Data Theory” What should TCS say about big data? • This talk: • Running time: (almost) linear, sublinear, … • Space: linear, sublinear, … • Approximation: best possible, … • Randomness: as little as possible, … • Special focus today: roundcomplexity
4. Round Complexity Information-theoretic measure of performance • Tools from information theory (Shannon’48) • Unconditional results (lower bounds) Example: • Approximating Geometric Graph Problems
5. Approximation in Graphs 1930-50s: Given a graph and an optimization problem… Minimum Cut(RAND): Harris and Ross [1955] (declassified, 1999) Transportation Problem: Tolstoi [1930]
6. Approximation in Graphs 1960s: Single processor, main memory (IBM 360)
7. Approximation in Graphs 1970s: NP-complete problem – hard to solve exactly in time polynomial in the input size “Black Book”
8. Approximation in Graphs Approximate with multiplicative error on the worst-case graph : Generic methods: • Linear programming • Semidefiniteprogramming • Hierarchies of linear and semidefinite programs • Sum-of-squares hierarchies • …
9. The New: Approximating Geometric Problems in Parallel Models 1930-70s to 2014
10. The New: Approximating Geometric Problems in Parallel Models Geometric graph (implicit): Euclidean distances betweennpoints in Already have solutions for old NP-hard problems (Traveling Salesman, Steiner Tree, etc.) • Minimum Spanning Tree (clustering, vision) • Minimum Cost Bichromatic Matching (vision)
11. Geometric Graph Problems Combinatorial problems on graphs in Polynomial time (easy) • Minimum Spanning Tree • Earth-Mover Distance = Min Weight Bi-chromatic Matching NP-hard (hard) • Steiner Tree • Traveling Salesman • Clustering (k-medians, facility location, etc.) Need new theory! Arora-Mitchell-style “Divide and Conquer”, easy to implement in Massively Parallel Computational Models
12. MST: Single Linkage Clustering • [Zahn’71] Clustering via MST (Single-linkage): kclusters: remove longest edges from MST • Maximizes minimumintercluster distance [Kleinberg, Tardos]
13. Earth-Mover Distance • Computer vision: compare two pictures of moving objects (stars, MRI scans)
14. Computational Model • Input: npoints in a d-dimensional space (d constant) • machines, space on each ( = , ) • Constant overhead in total space: • Output: solution to a geometric problem (size O() • Doesn’t fit on a single machine () points ⇒ machines S space
15. Computational Model • Computation/Communication in rounds: • Every machine performs a near-linear timecomputation => Total running time • Every machine sends/receives at most bits of information => Total communication . Goal:Minimize . Our work: = constant. bits machines time S space
16. MapReduce-style computations What I won’t discuss today • PRAMs (shared memory, multiple processors) (see e.g. [Karloff, Suri, Vassilvitskii‘10]) • Computing XOR requires rounds in CRCW PRAM • Can be done in rounds of MapReduce • Pregel-style systems, Distributed Hash Tables (see e.g. Ashish Goel’s class notes and papers) • Lower-level implementation details (see e.g. Rajaraman-Leskovec-Ullman book)
17. Models of parallel computation • Bulk-Synchronous Parallel Model (BSP) [Valiant,90] Pro: Most general, generalizes all other models Con: Many parameters, hard to design algorithms • Massive Parallel Computation[Feldman-Muthukrishnan-Sidiropoulos-Stein-Svitkina’07, Karloff-Suri-Vassilvitskii’10, Goodrich-Sitchinava-Zhang’11, ..., Beame, Koutris, Suciu’13] Pros: • Inspired by modern systems (Hadoop, MapReduce, Dryad, Pregel, … ) • Few parameters, simple to design algorithms • New algorithmic ideas, robust to the exact model specification • # Rounds is an information-theoretic measure => can prove unconditional lower bounds • Between linear sketching and streaming with sorting
18. Previous work • Dense graphs vs. sparse graphs • Dense: (or solution size) “Filtering” (Output fits on a single machine) [Karloff, SuriVassilvitskii, SODA’10; Ene, Im, Moseley, KDD’11; Lattanzi, Moseley, Suri, Vassilvitskii, SPAA’11; Suri, Vassilvitskii, WWW’11] • Sparse: (or solution size) Sparse graph problems appear hard (Big open question: (s,t)-connectivity in rounds?) VS.
19. Large geometric graphs • Graph algorithms: Dense graphs vs. sparse graphs • Dense: . • Sparse: . • Our setting: • Dense graphs, sparsely represented: O(n) space • Output doesn’t fit on one machine () • Today: -approximate MST • (easy to generalize) • rounds ()
20. -MST inrounds • Assume points have integer coordinates , where Impose an -depth quadtree Bottom-up: For each cell in the quadtree • compute optimum MSTs in subcells • Use only onerepresentativefrom each cell on the next level Wrong representative: O(1)-approximation per level
21. -nets • -net for a cell C with side length : Collection S of vertices in C, every vertex is at distance <= from some vertex in S. (Fact: Can efficiently compute -net of size Bottom-up: For each cell in the quadtree • Compute optimum MSTs in subcells • Use -net from each cell on the next level • Idea: Pay only ) for an edge cut by cell with side • Randomly shift the quadtree: – charge errors Wrong representative: O(1)-approximation per level
22. Top cell shifted by a random vector in Impose a randomly shiftedquadtree(top cell length ) Bottom-up: For each cell in the quadtree • Compute optimum MSTs in subcells • Use -net from each cell on the next level Pay 5 instead of 4 Pr[] = (1) 2 1
23. -MST inrounds • Idea: Only use short edges inside the cells Impose a randomly shiftedquadtree(top cell length ) Bottom-up: For each node (cell) in the quadtree • compute optimum Minimum Spanning Forestsin subcells, using edges of length • Use only -net from each cell on the next level Sketch of analysis ( = optimum MST): 𝔼[Extra cost] = 𝔼 2 Pr[] = 1
24. -MST inrounds • rounds => O() = O(1) rounds • Flatten the tree: ()-grids instead of (2x2) grids at each level. Impose a randomly shifted ()-tree Bottom-up: For each node (cell) in the tree • compute optimum MSTs in subcellsvia edges of length • Use only -net from each cell on the next level ⇒
25. -MST inrounds Theorem: Let = # levels in a random tree P Proof (sketch): • = cell length, which first partitions • New weights: • Our algorithm implements Kruskalfor weights
26. “Solve-And-Sketch” Framework -MST: • “Load balancing”: partition the tree into parts of the same size • Almost linear time: Approximate Nearest Neighbor data structure [Indyk’99] • Dependence on dimension d (size of -net is • Generalizes to bounded doubling dimension • Basic version is teachable (Jelani Nelson’s ``Big Data’’ class at Harvard)
27. “Solve-And-Sketch” Framework -Earth-Mover Distance, Transportation Cost • No simple “divide-and-conquer” Arora-Mitchell-style algorithm (unlike for general matching) • Only recently sequential -apprxoimation in time [Sharathkumar, Agarwal ‘12] Our approach (convex sketching): • Switch to the flow-based version • In every cell, send the flow to the closest net-point until we can connect the net points
28. “Solve-And-Sketch” Framework Convex sketching the cost function for net points • = the cost of routing fixed amounts of flow through the net points • Function “normalization” is monotone, convex and Lipschitz, ()-approximates • We can ()-sketch it using a lower convex hull
29. Thank you!http://grigory.us Open problems • Exetension to high dimensions? • Probably no, reduce from connectivity => conditional lower bound rounds for MST in • The difficult setting is (can do JL) • Streaming alg for EMD and Transporation Cost? • Our work: first near-linear time algorithm for Transportation Cost • Is it possible to reconstruct the solution itself? | 2,139 | 8,421 | {"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-2021-21 | latest | en | 0.731475 |
https://www.coursehero.com/file/6295834/Act-5/ | 1,495,829,639,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608676.72/warc/CC-MAIN-20170526184113-20170526204113-00374.warc.gz | 1,054,900,650 | 125,751 | # Act_5 - Name Section Activity Sheet 5 Thermal Energy the...
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1/16/05 1 Name: ___________________________________ Section: ______________ Activity Sheet 5: Thermal Energy, the Microscopic Picture 5.1 How Is Temperature Related to Molecular Motion? 1) Temperature Your instructor will discuss molecular motion and temperature. a) At a particular temperature, do all of the molecules move at the same speed? b) How does the average speed of molecules at a higher temperature differ from their average speed at a lower temperature? c) Watch the demonstration of diffusion of food coloring in beakers of warm and cold water. Explain the differences in the diffusion rates. 2) Evaporative Cooling What happens to the temperature of a liquid as it evaporates? Your instructor will demonstrate a cool tube (a cryophorus tube) that contains water. a) What happens when one end of the tube is cooled with liquid nitrogen? b) The middle of the tube remains near room temperature. Explain why the water in the tube’s bulb freezes. c) Group Discussion Question: When you put a pot of water on the stove to heat to boiling, why does the water come to a boil faster if you put a lid on the pot? 3) The Dippy Duck The dippy duck contains liquid freon, which evaporates easily at room temperature. Wet the head of the duck and place the cup of water in front of the duck’s head. Explain what happens to the dippy duck in terms of evaporative cooling and the center of mass of the duck.
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1/16/05 2 4) Rates of Evaporation a) Note the temperature of each thermometer while it is immersed in liquid. Remove the thermometers from the test tubes and allow the thermometers to lie on your table for several minutes. Then check and record their temperatures. Initial Temp
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## This note was uploaded on 06/11/2011 for the course PHYSICS 104 taught by Professor Staff during the Winter '11 term at Ohio State.
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Ask a homework question - tutors are online | 520 | 2,346 | {"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-2017-22 | longest | en | 0.875408 |
https://www.jiskha.com/display.cgi?id=1287601087 | 1,503,408,085,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110774.86/warc/CC-MAIN-20170822123737-20170822143737-00063.warc.gz | 899,503,692 | 3,743 | # Probability
posted by .
There are 3 lucky tickets among 10 lottery tickets. 3 tickets are
drawn at random. What is a probability that one ticket selected randomly
from those three is lucky?
• Probability -
Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you.
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More Similar Questions | 555 | 2,422 | {"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-2017-34 | latest | en | 0.960774 |
http://mathsblog.co.uk/2009/11/24/maths-worksheet-divide-by-100-mentally/ | 1,521,754,531,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648003.58/warc/CC-MAIN-20180322205902-20180322225902-00253.warc.gz | 200,321,366 | 9,675 | Posted by Peter on 24th November 2009
# Maths Worksheet: Divide by 100 mentally
Here we have the next worksheet in our series of dividing mentally. This page looks at dividing larger numbers by 100. All the numbers are multiples of 100 so there will be no decimals involved with the answers. As has been said before, but the idea has to be repeated many times for children to fully understand; to divide by 10, move each digit one place to the right. To divide by 100 move each digit two places to the right.
So 1200 divided by 10 is 12.00. There is nothing wrong with leaving the two zeros after the decimal point, but we usually don’t worry with the decimal point or zeros if the answer is a whole number. Whatever you do, don’t tell your children to take away the noughts!!
This, and many other division worksheets, can be found in our Four Rules section as well as in the year group sections under calculating.
Divide by 100 (pg 1) | 218 | 940 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-13 | longest | en | 0.95883 |
https://chowdera.com/2020/12/20201206192115780k.html | 1,642,444,058,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300616.11/warc/CC-MAIN-20220117182124-20220117212124-00069.warc.gz | 235,764,928 | 7,865 | # PAT（甲级）2020年秋季考试 7-1 Panda and PP Milk
2020-12-06 19:23:26 乔梓鑫
## 7-1 Panda and PP Milk (20分)
PP milk （盆盆奶）is Pandas' favorite. They would line up to enjoy it as show in the picture. On the other hand, they could drink in peace only if they believe that the amount of PP milk is fairly distributed, that is, fatter panda can have more milk, and the ones with equal weight may have the same amount. Since they are lined up, each panda can only compare with its neighbor(s), and if it thinks this is unfair, the panda would fight with its neighbor.
Given that the minimum amount of milk a panda must drink is 200 ml. It is only when another bowl of milk is at least 100 ml more than its own that a panda can sense the difference.
Now given the weights of a line of pandas, your job is to help the breeder（饲养员）to decide the minimum total amount of milk that he/she must prepare, provided that the pandas are lined up in the given order.
### Input Specification:
Each input file contains one test case. For each case, first a positive integer n (≤104) is given as the number of pandas. Then in the next line, n positive integers are given as the weights (in kg) of the pandas, each no more than 200. the numbers are separated by spaces.
### Output Specification:
For each test case, print in a line the minimum total amount of milk that the breeder must prepare, to make sure that all the pandas can drink in peace.
### Sample Input:
``````10
180 160 100 150 145 142 138 138 138 140```````
### Sample Output:
``3000``
### Hint:
The distribution of milk is the following:
``400 300 200 500 400 300 200 200 200 300``
#### 算法思路：
，前面的熊猫的奶量原先是比后面多的，是因为后面的熊猫进行了调整，加了100，才会导致和它相等，在原先存在差距的情况只会相等，不会超过) 。第二种类型就是当前熊猫体重和后面的一样，但是喝的奶量却比后面的少(同样是因为后面的熊猫调整过了，而且也不存在体重一样喝的奶会一样的情况，因为到达当前的j的时候，j+1一定调整过)，那么对于第一次出现j+1为波峰的时候，也就是当期出现下降趋势，并且也没有违背右边的熊猫的奶量关系的时候就调整完毕了。
#### 注意点：
• 1、仔细揣摩那两个需要调整的类型和等号的取值，之前让我最后一个测试点没有过。
#### AC代码：
``````#include<cstdio>
using namespace std;
/*
，前面的熊猫的奶量原先是比后面多的，是因为后面的熊猫进行了调整，加了100，才会导致和它相等，在原先存在差距的情况只会相等，不会超过) 。
*/
int main(){
int N;
scanf("%d",&N);
int weight[N];
for(int i=0;i<N;++i){
scanf("%d",&weight[i]);
}
int milk[N];
milk[0] = 200;//初始奶量为200
for(int i=1;i<N;++i){
if(weight[i]>weight[i-1]){
milk[i] = milk[i-1] + 100;
}else if(weight[i]==weight[i-1]){
milk[i] = milk[i-1];
}else{
// 前面的奶量为200了，并且这里还是下降趋势
// 向前搜寻波峰，并且每一个奶量加100
milk[i] = 200;
for(int j=i-1;j>=0;--j){// 每一个需要调整奶量的熊猫
if(weight[j+1]<weight[j]&& milk[j+1]==milk[j]){
// 当前熊猫比后面的重，但是喝的奶和它一样，需要调整
milk[j] = milk[j] + 100;//每一个奶量加100
}else if(weight[j+1]==weight[j]&&milk[j+1]>milk[j]) {
// 当前的熊猫和后面的一样重，但是喝的奶比它少，需要调整
milk[j] = milk[j] + 100;//每一个奶量加100
}else{
// j+1为波峰，直接退出即可，无需再调整，因为调整后就不在是最低奶量了
break;
}
}
}
}
int total = 0;
for(int i=0;i<N;++i){
total += milk[i];
}
printf("%d",total);
return 0;
} ``````
https://segmentfault.com/a/1190000038393041 | 1,091 | 2,862 | {"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-2022-05 | latest | en | 0.715636 |
http://learnenglish.britishcouncil.org/english-grammar-reference/probability?page=2 | 1,600,703,795,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400201826.20/warc/CC-MAIN-20200921143722-20200921173722-00735.warc.gz | 89,199,881 | 20,506 | # Probability
Level: beginner
## Possibility
We use may, might and could to say that something is possible, but not certain:
They may come by car. (= Maybe they will come by car.)
They might be at home. (= Maybe they are at home.)
If we don't hurry, we could be late. (= Maybe we will be late.)
We use can to make general statements about what is possible:
It can be very cold here in winter. (= It is sometimes very cold here in winter.)
You can easily get lost in this town. (= People often get lost in this town.)
Be careful!
We do not use can to talk about specific events:
A: Where's John?
B: I'm not sure. He may/might/could be
(NOT can) in his office.
Notice the difference in meaning between can and may/might/could:
That dog can be dangerous.
(= Sometimes that dog is dangerous. I know.)
That dog may/might/could be dangerous.
(= Perhaps that dog is dangerous. I don't know.)
can and may/might/could
GapFillDragAndDrop_MTYzNDM=
Level: intermediate
We use may have, might have or could have to make guesses about the past:
I haven't received your letter. It may have got lost in the post.
It's ten o'clock. They might have arrived by now.
Where are they? They could have got lost.
We use could to make general statements about the past:
It could be very cold there in winter. (= It was sometimes very cold there in winter.)
You could easily get lost in that town. (= People often got lost in that town.)
could and could have
MultipleChoice_MTYzNDQ=
## Impossibility
Level: beginner
We use can't or cannot to say that something is impossible:
That can't be true.
You cannot be serious.
Level: intermediate
We use can't have or couldn't have to say that a past event was impossible:
They know the way here. They can't have got lost!
If Jones was at work until six, he couldn't have done the murder.
## Certainty
Level: beginner
We use must to show we are sure something is true and we have reasons for our belief:
It's getting dark. It must be quite late.
You haven’t eaten all day. You must be hungry.
We use should to suggest something is true and we have reasons for our suggestion:
It's nearly six o'clock. They should arrive soon.
Level: intermediate
We use must have and should have for the past:
They hadn't eaten all day. They must have been hungry.
You look happy. You must have heard the good news.
It's nearly eleven o'clock. They should have arrived by now.
Probability 1
Matching_MTYzNDU=
Probability 2
Matching_MTYzNDY=
Probability 3
GapFillTyping_MTYzNDc=
Probability 4
Matching_MTYzNDg=
Probability 5
GapFillTyping_MTYzNDk=
Hi,
They could come by car
They might come by car
They may come by car
Can you please explain me which of the above sentence is correct/more accurate or can be used interchangeably?
Hello QaaZee,
All of these are grammatically correct and they can all mean that there is a chance that they will come by car. Other meanings are possible for some of the examples. 'May' could refer to permission, for example, as in it is OK for them to come by car, but that would depend on the context.
Peter
The LearnEnglish Team
Thanks Peter,
Could you please provide an example with context for 'May' to further elaborate its use for permission.
Thank you once again.
Hi,
Look at the sentence 'I consoled myself with the thought that things could be much worse'.
In this sentence how can we know 'could' is used either as past form of 'can' or 'could' is used for past possibility. we use 'can' for general possibility and 'could' for possibility in a single occasion. So when we want to say this in a past sentence like above, how can we know it is a general possibility or a possibility in a single situation. In the past sentences we use preterite form of 'can'. So I get confused it is a general possibility or a possibility in a single situation.
Hello jitu_jaga,
The modal verb does not carry a marker for this within it so we use the context to inform us. If the context is not clear then it is ambiguous, but this would rarely matter.
For example, the meaning of your sentence is quite clear, I think. The speaker is looking at a bad situation and consoling themselves with the thought that something worse is possible. It may or may not be obvious from the context whether the worse possibility is something specific (my car is not working but it could be worse - my car could have exploded) or general (my car is not working but it could be worse - I could have lost my job, got sick or suffered some family tragedy). It really does not matter, however; what is important is the idea that things are not as bad as they could be, and that the person's ill-fortune is not so terrible when considered in the right way.
Peter
The LearnEnglish Team
Thank you Peter. You always provide good explanations. Have a nice day.
Hi,
We use may/might/could for possibility and may/might/could have for past possibility. But we don't use 'would' for possibility. But sometimes, I find sentence like ' You would know, Sachin Tendulkar is a great batsman'. I think here 'would' is used for speculation if I am not wrong.
In this sense, if we write sentences like
It may/might/could rain tomorrow, could we write 'It would rain tomorrow' or 'I would marry next month'?. If not then what would be its meaning? I don't understand how to use use 'would' for future and past speculation.Could you Please explain it clearly.
Hello jitu_jaga,
The sentence about Sachin Tendulkar you mention doesn't sound right to me, at least out of context. In other words, perhaps in a specific context it would make sense, but out of context it does not.
Your other sentences with 'would' to talk about possibility (rain or getting married) are also not standard. At least in British or American English, 'would' isn't used to speak about possibility in this way. It can be used within a conditional structure to speak about a possibility, but that is a different structure, which is clearly indicated in most cases with an 'if' clause.
All the best,
Kirk
The LearnEnglish Team
Sir, As We use must+have+ed form of verbs to talk about deduction and probability in the past and use could or can+not+have+third of verbs to talk about the same in negative in the past right ?
But is there a structure in English like this must+not+have+ed form of verbs to talk about past negative deduction and probability or to talk about something else? | 1,504 | 6,409 | {"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} | 3.171875 | 3 | CC-MAIN-2020-40 | latest | en | 0.983621 |
https://www.smore.com/1fta | 1,537,301,639,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155676.21/warc/CC-MAIN-20180918185612-20180918205612-00157.warc.gz | 864,248,114 | 14,124 | # Catching Up with Castro's Class
## Dear Parents,
I hope you all had a wonderful holiday and that everyone is adjusting back to school well (even with the delays!). In this newsletter you will find information about what we are learning in class, pictures of us in the classroom, student expectations for the next few weeks and events going on around LMES.
Please check out my technology tab on my classroom website. Under this tab you can find projects the kids are doing in class. They work really hard on them and get so excited when they people are looking at them.
Enjoy!
Sincerely,
Victoria Castro
Reading- At the beginning of this week we started learning about how to ask "deep" questions about fictional stories that we are reading. In the classroom the students read a story as a class and I pose a question to them that cannot be answered with a "right or wrong" answer. Instead, they must form an opinion and use text evidence to support their answer.
In the future the students will create their own "deep" questions and have group discussions on their answers and thoughts.
Reading Goals- Please make sure that your student is reading at home and working towards their reading goal. They should be bringing home their blue folder every night to keep track of the books they read and the strategy they are working on.
Below I am listing what should be completed by Friday depending on the strategy your student is working on.
STP (STOP, THINK, PARAPHRASE)- 10 chapters paraphrased
Making Predictions- week 1 and week 2 filled out completely
Making Inferences- week 1 week 2 filled out completely
Character Mapping- 3 complete character sheets
"Thinking" while reading- week 1 and 2 filled out completely
Stems and Grammar- Our FINAL stems test is on Friday. Students should be studying throughout the week. The test will cover ALL stems they have learned. After stems we begin learning about grammar!
Writing- We are starting our narrative unit this week. Students will be learning how to create a story with descriptive details and a clear order of events. We'll also learn how to add dialogue to our stories!
## Math
We will be finishing up multiplication and division next week and having a test. After next week we will be beginning FRACTIONS. The concept of fractions can be difficult for many kids so please let me know if you notice your child have difficulty at home with the homework.
Timed Quizzes- Today we will be taking our 12's quiz and then we have officially gone through all of the quizzes for 1-12! Starting next week we will have mixed multiplication quizzes. If a student gets an A on a mixed quiz I let them skip the following week's quiz.
## Science and Social Studies
Science- We have begun our sound unit and should finish up in the next couple weeks. During this unit the students will do many hands on activities with instruments to learn about pitch and volume. Once we have finished learning about sound my homeroom will start going to Mrs. Kyzer for Science. A letter will come home in red folders explaining the change.
Social Studies- This week we started our Revolutionary War unit. This unit is VERY exciting and interesting to the students, but also has a lot of information to cover. I recommend reviewing notes with your child at home whenever you have a chance. Their notes are in the SC History section of their expandable or you can look through the Social Studies book.
Time Travelers- Last Friday we had our TIME TRAVELERS day where the students traveled back in time to the Revolutionary War days. Each third grade teacher taught a different aspect of the Revolutionary War and the students rotated between classrooms. Ask your child about the different costumes and activities and check out the pictures below!
EXTRA CREDIT OPPORTUNITY: I have given the students an extra credit opportunity for Social Studies. They have 3 worksheets that can be completed and turned in for an extra quiz grade. The sheets are due Monday.
## Around LMES....
El mercado LMES - 2014 is right around the corner! Click the link below to read how you can help.
## Dates to Remember...
January 13th- My math class takes MAP testing
January 20th- MLK Jr. Day, No School
January 21st- Student Holiday
January 24th- Report Cards go Home | 894 | 4,299 | {"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-2018-39 | latest | en | 0.972769 |
https://www.ourpcb.com/dcaclab-multimeter.html | 1,718,947,136,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862036.35/warc/CC-MAIN-20240621031127-20240621061127-00142.warc.gz | 821,472,403 | 20,414 | Blog / DCACLab Multimeter
# DCACLab Multimeter
DCACLab, an online circuit simulator, is a beneficial tool in determining real-time circuits' behavior. The multimeter is one such device that we need most often to find current and voltages' values at different points. We provide one such analog multimeter where students can choose the required meter from several ranges of values using a selection knob. Separate meters equipped are -
●AC voltmeter
●DC voltmeter
●Amperemeter
●Ohmmeter
#### Special Offer: Get \$100 off your order!
Claim your \$100 discount by sending an inquiry today. Act now to save on your next project!
Please email [email protected] for details.
#### Special Offer: Get \$100 off your order!
Claim your \$100 discount by sending an inquiry today. Act now to save on your next project!
Please email [email protected] for details.
## Steps to using a multimeter in DCACLab
See the link https://dcaclab.com/en/experiments/17115-simple-circuit-to-play-with-multimeter-1 to play with the multimeter and learn.
1. Click on the icon of the multimeter from the toolbar, as shown. Also, instead of clicking it, you can drag it directly to the screen.
2. Use the selection knob to choose the required meter and place the multimeter wires at the points where measurement has to be done.
### Measurement of DC/AC Current/Voltage with DCACLab Multimeter
For example, if we need to calculate DC, we can set a knob and place the multimeter as shown to find the value.
Similarly, for measuring the dc voltage, the knob should be set again, and by placing the wires at the required points, we can find the potential difference directly, as shown.
### Measurement of Resistance with Multimeter
The multimeter can be set as follows to measure the resistance of any electrical load in ohms.
## DCACLab OSCILLOSCOPE
### Steps to use the oscilloscope in DCACLab
1. Navigate through the list of components and click/drag it.
2. To measure the voltage between the two desired points, grab and place the probes there. Also, the signal waveform can be viewed along with its magnitude.
3. Also, we can increase the time division and voltage division by varying the respective pivots as shown:
See the link https://dcaclab.com/en/experiments/16144-half-wave-rectifier-1 and play with the circuit to get a better understanding of the same.
## Quantities that oscilloscope can Measure
Quantities like frequency, amplitude, and other waveform characteristics in a signal can be measured using the DCACLab oscilloscope.
### Frequency & Period
The number of cycles a waveform completes per second is known as its frequency. Here, we can change the frequency as per our desire and can experiment with the available signal. The period is the inverse of rate, and switching frequency will adversely affect the wave's period.
### Duty Cycle
The duty cycle is generally defined as the ratio of the on-time period to the waveform's total period. Although the change in one will affect the rest of the quantities in some of the other ways.
Other numbers that could be changed here are rise and fall time, maximum & minimum amplitudes, mean & average values, etc..
#### Special Offer: Get \$100 off your order!
Claim your \$100 discount by sending an inquiry today. Act now to save on your next project!
Please email [email protected] for details. | 747 | 3,364 | {"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-2024-26 | latest | en | 0.874651 |
https://www.eduzip.com/ask/question/solveint-dfracxx2-x-1-dx-581021 | 1,623,687,232,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487612537.23/warc/CC-MAIN-20210614135913-20210614165913-00272.warc.gz | 693,017,955 | 8,720 | Mathematics
# Solve:$\int {\dfrac{x}{{{x^2} + x + 1}}} dx$
##### SOLUTION
$\displaystyle \int \frac{x}{x^{2}+x+1} dx$
$\displaystyle = \frac{1}{2}\int \frac{2x}{x^{2}+x+1}$
$\displaystyle = \frac{1}{2}\int \frac{2x+1-1}{x^{2}+x+1}dx$
$\displaystyle = \frac{1}{2}\int \frac{2x+1}{x^{2}+x+1}dx-\frac{1}{2}\frac{1}{x^{2}+x+1}dx$
$\displaystyle = \frac{1}{2}log (x^{2}+x+1)-\frac{1}{2}\int \frac{1}{x^{2}+2.x\frac{1}{2}+(\frac{1}{2})^{2}-(\frac{1}{2})^{2}+1}$
$\displaystyle = \frac{1}{2}log(x^{2}+x+1)-\frac{1}{2}\int \frac{1}{(x+\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}}dx$
Let $\displaystyle x+\frac{1}{2} = t \Rightarrow dx = dt$
$\displaystyle = \frac{1}{2}log(x^{2}+x+1)-\frac{1}{2}\int \frac{dt}{t^{2}+(\frac{\sqrt{3}}{2})^{2}}$
$\displaystyle = \frac{1}{2}log(x^{2}+x+1)-\frac{1}{2}.\frac{1}{(\frac{\sqrt{3}}{2})}ten^{-1}(\frac{t}{(\sqrt{3}/2)})+c$
$\displaystyle = \frac{1}{2}log(x^{2}+x+1)-\frac{1}{\sqrt{3}}tan^{-1}(\frac{2(x+1/2)}{\sqrt{3}})+c$
$\displaystyle = \frac{1}{2}log(x^{2}+x+1)-\frac{1}{\sqrt{3}}tan^{-1}(\frac{2x+1}{\sqrt{3}})+c$
Its FREE, you're just one step away
Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84
#### Realted Questions
Q1 Single Correct Medium
$n\overset{Lt}{\rightarrow}\infty [\displaystyle \frac{1}{1-n^{2}}+\frac{2}{1-n^{2}}+\ldots+\frac{n}{1-n^{2}}]=$
• A.
• B. 1/2
• C. 1
• D. -1/2
Asked in: Mathematics - Integrals
1 Verified Answer | Published on 17th 09, 2020
Q2 Single Correct Medium
The value of $\displaystyle \int_{-\frac {\pi}{2}}^{\frac {\pi}{2}}(x^3+x \cos x+\tan^5x+1)dx$
• A. $0$
• B. $2$
• C. $1$
• D. $\pi$
Asked in: Mathematics - Integrals
1 Verified Answer | Published on 17th 09, 2020
Q3 Single Correct Hard
If $I_{1}=\displaystyle \lim_{n\rightarrow \infty }\sum_{r=1}^{2n} \dfrac{n}{r^{2}+3rn+2n^{2}}$
and $\displaystyle I_{2}=\lim_{n\rightarrow \infty }\sum_{r=1}^{2n} \dfrac{n}{2r^{2}+3rn+n^{2}}$
then the value of $2I_{1} - I_{2}$ is equal to?
• A. $\ln\dfrac{4}{3}$
• B. $\ln\dfrac{3}{4}$
• C. $0$
• D. $\ln\dfrac{3}{2}$
Asked in: Mathematics - Integrals
1 Verified Answer | Published on 17th 09, 2020
Q4 Single Correct Medium
$\underset{n - 1}{\overset{100}{\sum}} \underset{n - 1}{\overset{n}{\int}} \, e^{x - [x]} dx$ =
• A. $\dfrac{e - 1}{100}$
• B. $100 (e - 1)$
• C. $\dfrac{e^{100} - 1}{100}$
• D. $\dfrac{e^{100} - 1}{e - 1}$
Asked in: Mathematics - Integrals
1 Verified Answer | Published on 17th 09, 2020
Q5 Subjective Easy
Evaluate:
$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$
Asked in: Mathematics - Integrals
1 Verified Answer | Published on 17th 09, 2020 | 1,177 | 2,635 | {"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-2021-25 | latest | en | 0.422188 |
https://www.mathworks.com/matlabcentral/cody/problems/42384-combined-ages-2-symmetric-n-3/solutions/689130 | 1,606,458,648,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141189141.23/warc/CC-MAIN-20201127044624-20201127074624-00370.warc.gz | 763,340,919 | 19,326 | Cody
# Problem 42384. Combined Ages 2 - Symmetric, n ≥ 3
Solution 689130
Submitted on 20 Jun 2015 by Yalong Liu
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
%% AB = 43; AC = 66; BC = 55; y = combined_ages2(AB,AC,BC); y_correct = [27 16 39]; for i = 1:numel(y_correct) assert(isequal(y(i),y_correct(i))) end
ans = 27 16 39
2 Pass
%% AB = 30; AC = 40; BC = 50; y = combined_ages2(AB,AC,BC); y_correct = [10 20 30]; for i = 1:numel(y_correct) assert(isequal(y(i),y_correct(i))) end
ans = 10 20 30
3 Pass
%% ABC = 72; ABD = 66; ACD = 70; BCD = 77; y = combined_ages2(ABC,ABD,ACD,BCD); y_correct = [18 25 29 23]; for i = 1:numel(y_correct) assert(isequal(y(i),y_correct(i))) end
ans = 18 25 29 23
4 Pass
%% ABC = 66; ABD = 67; ACD = 68; BCD = 69; y = combined_ages2(ABC,ABD,ACD,BCD); y_correct = [21 22 23 24]; for i = 1:numel(y_correct) assert(isequal(y(i),y_correct(i))) end
ans = 21 22 23 24
5 Pass
%% ABC = 60; ABD = 65; ACD = 70; BCD = 75; y = combined_ages2(ABC,ABD,ACD,BCD); y_correct = [15 20 25 30]; for i = 1:numel(y_correct) assert(isequal(y(i),y_correct(i))) end
ans = 15 20 25 30
6 Pass
%% ABCD = 90; ABCE = 115; ABDE = 100; ACDE = 110; BCDE = 105; y = combined_ages2(ABCD,ABCE,ABDE,ACDE,BCDE); y_correct = [25 20 30 15 40]; for i = 1:numel(y_correct) assert(isequal(y(i),y_correct(i))) end
ans = 25 20 30 15 40
7 Pass
%% ABCD = 44; ABCE = 37; ABDE = 47; ACDE = 51; BCDE = 53; y = combined_ages2(ABCD,ABCE,ABDE,ACDE,BCDE); y_correct = [5 7 11 21 14]; for i = 1:numel(y_correct) assert(isequal(y(i),y_correct(i))) end
ans = 5 7 11 21 14
8 Pass
%% ABCDEF = 133; ABCDEG = 186; ABCDFG = 172; ABCEFG = 163; ABDEFG = 192; ACDEFG = 200; BCDEFG = 184; y = combined_ages2(ABCDEF,ABCDEG,ABCDFG,ABCEFG,ABDEFG,ACDEFG,BCDEFG); y_correct = [21 5 13 42 33 19 72]; for i = 1:numel(y_correct) assert(isequal(y(i),y_correct(i))) end
ans = 21 5 13 42 33 19 72
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Start Hunting! | 871 | 2,131 | {"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} | 3.734375 | 4 | CC-MAIN-2020-50 | latest | en | 0.357664 |
https://www.kernel.org/doc/html/v5.5/scheduler/text_files.html | 1,713,102,843,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.72/warc/CC-MAIN-20240414130604-20240414160604-00827.warc.gz | 772,676,003 | 4,889 | Scheduler pelt c program¶
```/*
* The following program is used to generate the constants for
* computing sched averages.
*
* ==============================================================
* C program (compile with -lm)
* ==============================================================
*/
#include <math.h>
#include <stdio.h>
#define HALFLIFE 32
#define SHIFT 32
double y;
void calc_runnable_avg_yN_inv(void)
{
int i;
unsigned int x;
/* To silence -Wunused-but-set-variable warnings. */
printf("static const u32 runnable_avg_yN_inv[] __maybe_unused = {");
for (i = 0; i < HALFLIFE; i++) {
x = ((1UL<<32)-1)*pow(y, i);
if (i % 6 == 0) printf("\n\t");
printf("0x%8x, ", x);
}
printf("\n};\n\n");
}
int sum = 1024;
void calc_runnable_avg_yN_sum(void)
{
int i;
printf("static const u32 runnable_avg_yN_sum[] = {\n\t 0,");
for (i = 1; i <= HALFLIFE; i++) {
if (i == 1)
sum *= y;
else
sum = sum*y + 1024*y;
if (i % 11 == 0)
printf("\n\t");
printf("%5d,", sum);
}
printf("\n};\n\n");
}
int n = -1;
/* first period */
long max = 1024;
void calc_converged_max(void)
{
long last = 0, y_inv = ((1UL<<32)-1)*y;
for (; ; n++) {
if (n > -1)
max = ((max*y_inv)>>SHIFT) + 1024;
/*
* This is the same as:
* max = max*y + 1024;
*/
if (last == max)
break;
last = max;
}
n--;
printf("#define LOAD_AVG_PERIOD %d\n", HALFLIFE);
printf("#define LOAD_AVG_MAX %ld\n", max);
// printf("#define LOAD_AVG_MAX_N %d\n\n", n);
}
void calc_accumulated_sum_32(void)
{
int i, x = sum;
printf("static const u32 __accumulated_sum_N32[] = {\n\t 0,");
for (i = 1; i <= n/HALFLIFE+1; i++) {
if (i > 1)
x = x/2 + sum;
if (i % 6 == 0)
printf("\n\t");
printf("%6d,", x);
}
printf("\n};\n\n");
}
void main(void)
{
printf("/* Generated by Documentation/scheduler/sched-pelt; do not modify. */\n\n");
y = pow(0.5, 1/(double)HALFLIFE);
calc_runnable_avg_yN_inv();
// calc_runnable_avg_yN_sum();
calc_converged_max();
// calc_accumulated_sum_32();
}
``` | 631 | 1,937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-18 | latest | en | 0.336385 |
https://electronics.stackexchange.com/questions/594944/why-is-the-output-of-an-op-amp-given-by-the-open-loop-gain-multiplied-by-the-inp | 1,718,402,952,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861575.66/warc/CC-MAIN-20240614204739-20240614234739-00572.warc.gz | 201,909,212 | 43,437 | # Why is the output of an Op-Amp given by the open loop gain multiplied by the input differential voltage
When designing op-amps at the transistor level, the op-amp output is given by:
vout = Aol(vinp - vinm)
where vout, vinp and vinm are small signal changes in the output and input voltages.
How come that the absolute output level when analyzing op-amp circuits is assumed to be Vout = Aol(Vinp - Vinm)?
Here, Vinp and Vinm are the absolute voltage levels of the input pins, in contrast with vinp and vinm
The Op-amp is designed with both inputs at a certain bias point, Vin,cm.
vinp - vinm = (Vinp-Vin,cm)-(Vinm-Vin,cm)=Vinp - Vinm
But Vout =vout + Vout,cm where Vout,cm is the output when there is no small signal input, i.e, when the input pins are at the bias point.
Shouldn't the opamp output be Vout = Aol(Vinp - Vinm)+Vout,cm
Yes, that is strictly true, however usually the Vout,cm divided by Ao is so much less than the input offset voltage that it doesn't really matter.
It makes more sense to assume Vout,cm == 0 for an op-amp with equal bipolar supplies. Otherwise you might want to use midpoint between the two supply rails.
• Let's say I designed an Op-Amp with Vdd = +- 1V and both inputs are normally biased at 0V. Without feedback, the output is not necessarily at 0V. In fact it can be anywhere, even up to +- 0.5V Commented Nov 14, 2021 at 7:11
• If the Vos is +/-100uV and gain is 10^6 then the output can certainly be anything within the output swing capabilities. So if you assume 0 or 0.5V or 1V it makes little difference to the calculations. Commented Nov 14, 2021 at 7:15
The output of an ideal op-amp is given by
$$V_{out} = A_{ol}(V_{in+}-V_{in-})$$
Using this ideal model of an op-amp, we can fairly accurately predict the closed loop behavior of an op-amp circuit containing negative feedback.
However, this ideal op-amp is a useful fiction. Real op-amps do not have that behavior. A real op-amp acts more like an integrator than a strictly proportional amplifier. This integration comes about from the compensation capacitor in the op-amp, and reduces the error, when there is negative feedback, to near zero.
The following circuit illustrates.
simulate this circuit – Schematic created using CircuitLab
I have run a simulation of this circuit and plotted the error voltage, i.e. $$\(V_{in+}-V_{in-})\$$ together with the output. One can easily see that the error voltage consists of alternating spikes and the output is the integral of these alternating spikes. (The error voltage is scaled by a factor of 500 so that it fits nicesly in the same plot as the output.)
Note that the output of this circuit (a voltage follower) nicely tracks the (non-inverting) input, just as if it obeyed the equation
$$V_{out} = A_{ol}(V_{in+}-V_{in-})$$
But it clearly does not obey that equation. The moral? The model of the ideal op-amp gives very good results for modeling op-amp circuits with negative feedback. However, that model does not accurate represent what is really happening in an op-amp.
How come that the absolute output level when analyzing op-amp circuits is assumed to be $$\V_{out} = Aol(V_{inp} - V_{inm})\$$?
Simply put, it is not. No-one in the industry assumes this $$\V_{OUT} = Aol(V_{inp} - V_{inm})\$$ to be true or even useful for analyses embracing common mode parameters as is the case of your $$\V_{IN,CM}\$$ and $$\V_{OUT,CM}\$$.
But you are right, the handling of opamp's VTC in the presence of non-negligible common mode signal gain should be taken with special care: you should be familiar with the terminology, be able to read the relevant sections of datasheets, and trace the issues of input/output common-mode ranges violation and output latchup in your designs.
A good starting point to learn about common-mode issues might be Analog Devices' tutorials. MT-042 introduces the notion of Common-Mode Rejection Ratio:
If a signal is applied equally to both inputs of an op amp, so that the differential input voltage is unaffected, the output should not be affected. In practice, changes in common mode voltage will produce changes in output. The op amp common-mode rejection ratio (CMRR) is the ratio of the common-mode gain to differential-mode gain.
INLINE COMMENT
When citing their CMRR definition, I noticed that the MT-042 definition of CMRR (the ratio of the common-mode gain to differential-mode gain) is at odds with the generally accepted one (The CMRR is defined as the ratio of the powers of the differential gain over the common-mode gain). Evidently, this is a typo, because further on, the CMRR value they use in the text and formulas, is the ratio of the differential-mode gain to common-mode gain, the inverse value w.r.t. their wording. Analog Devices, please correct this typo!
Back to the answer. Pay attention to formulas in Fig. 2: Calculating Offset Error Due to Common-Mode Rejection Ratio (CMRR) of this tutorial:
$$\text {ERROR (RTI)} = {\frac {V_{CM}} {\text{CMRR}}} \\ V_{OUT} = \left[1+{R2 \over R1}\right] \left[V_{IN}+{\frac {V_{CM}} {\text{CMRR}}}\right]\\ \text {ERROR (RTO)} = \left[1+{R2 \over R1}\right]{\frac {V_{CM}} {\text{CMRR}}}$$
These are the formulas for a closed-loop configuration. Listed in datasheets, the parameters CMRR and CMR characterize the rejection of a common-mode signal component in the total output of opamp.
The middle formula is valid for the approximation of $$\R2/R1 \ll Aol\$$. For $$\{1 \over R2} = 0\$$ (no feedback, open-loop configuration), it should be written (https://en.wikipedia.org/wiki/Open-loop_gain, section Role in non-ideal gain) $$V_{OUT} = Aol \left[V_{IN}+{\frac {V_{CM}} {\text{CMRR}}}\right] = Aol \left[V_{IN}+{\frac {Acm} {Aol}}V_{CM}\right] = Aol·V_{IN} + Acm·V_{CM}$$ because $$\{\text{CMRR}} = {\frac {Aol} {Acm}}\$$ by definition. Also by definition, $$\Acm·V_{CM} = V_{OUT,CM}\$$, and $$\V_{CM}\$$ is a common-mode voltage at the input, $$\V_{IN,CM}\$$. In the end, we arrive at your formula $$\Vout = Aol(V_{in+} - V_{in-})+V_{OUT,CM}\$$.
Also pay attention to this phrase in the first paragraph
please note that there is very little consistency in this throughout the semiconductor industry with regards to the use of dB or ratio values for CMR or CMRR.
which emphasizes the importance of following the terminology.
MT-041 considers practical basic points regarding the allowable input and output voltage ranges of a real op amp. These obviously vary with not only the specific device, but also the supply voltage.
To grasp a deeper understanding of CMRR, follow also tutorials and datasheets referenced in these two tutorials, as well as app notes, handbooks and tutorials of the other opamp manufacturers (Texas Instruments, STMicroelectronics), textbooks and EE course notes.
The output an amplifier is equal to the differential input multiplied by the differential gain plus the common mode input multiplied by the common mode gain.
The differential input is the difference between the inputs and the common mode input is the average of the two inputs.
The common mode rejection ratio is equal to the differential gain divided by the common mode gain which is usually a very large number.
When the common mode rejection ratio is specified in dBs it is referred to as common mode rejection. | 1,854 | 7,284 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-26 | latest | en | 0.898988 |
http://www.jiskha.com/display.cgi?id=1359933063 | 1,462,499,138,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861700326.64/warc/CC-MAIN-20160428164140-00074-ip-10-239-7-51.ec2.internal.warc.gz | 592,863,371 | 3,744 | Thursday
May 5, 2016
# Homework Help: algebra
Posted by LaShawn on Sunday, February 3, 2013 at 6:11pm.
(a-3)(a+7)=-9
• algebra - sharii, Monday, February 4, 2013 at 12:21am
a(a+7)-3(a+7)=-9
a^2+7a-3a-21+9=0
a^2+4a-12=0
a^2+6a-2a-12=0
a(a+6)-2(a+6)=0
a+6=0
a=-6
OR
a-2=0
a=2 | 160 | 277 | {"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-2016-18 | longest | en | 0.713642 |
http://oeis.org/A280928 | 1,670,276,776,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711045.18/warc/CC-MAIN-20221205200634-20221205230634-00094.warc.gz | 41,771,313 | 5,162 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A280928 Composite numbers having the same digits as their prime factors (with multiplicity), including zero digits. 10
1255, 12955, 17482, 25105, 100255, 101299, 105295, 107329, 117067, 124483, 127417, 129595, 132565, 145273, 146137, 149782, 163797, 174082, 174298, 174793, 174982, 250105, 256315, 263155, 295105, 297463, 307183, 325615, 371893, 536539, 687919, 1002955, 1004251, 1012099, 1025095, 1029955 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Subsequence of A176670 as well as A020342. Is this sequence the intersection of A176670 and A020342? Excluding 1, all members of A080718 are members of this sequence. The first member of this sequence that is not a member of A080718 is a(17)=163797. LINKS Ely Golden, Table of n, a(n) for n = 1..3643 EXAMPLE 100255 is a member of this sequence as 100255 = 5*20051, which is exactly the same set of digits as 100255. PROG (SageMath) def digits(x, n): if((x<=0)|(n<2)): return [] li=[] while(x>0): d=divmod(x, n) li.append(d[1]) x=d[0] li.sort() return li; def factorDigits(x, n): if((x<=0)|(n<2)): return [] li=[] f=list(factor(x)) #ensures inequality of digits(x, n) and factorDigits(x, n) if x is prime if((len(f)==1)&(f[0][1]==1)): return []; for c in range(len(f)): for d in range(f[c][1]): ld=digits(f[c][0], n) li+=ld li.sort() return li; #this variable affects the radix radix=10 c=2 index=1 while(index<=100): if(digits(c, radix)==factorDigits(c, radix)): print(str(index)+" "+str(c)) index+=1 c+=1 print("complete") CROSSREFS The following sequences are all closely related: A020342, A014575, A080718, A280928, A048936, A144563. Cf. A176670 Sequence in context: A062693 A067203 A230544 * A080718 A219993 A237562 Adjacent sequences: A280925 A280926 A280927 * A280929 A280930 A280931 KEYWORD nonn,base AUTHOR Ely Golden, Jan 11 2017 STATUS approved
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Last modified December 5 15:27 EST 2022. Contains 358588 sequences. (Running on oeis4.) | 821 | 2,614 | {"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-2022-49 | latest | en | 0.738862 |
http://metamath.tirix.org/mpeascii/ndmfv | 1,720,868,832,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514493.69/warc/CC-MAIN-20240713083241-20240713113241-00676.warc.gz | 18,218,294 | 1,786 | # Metamath Proof Explorer
## Theorem ndmfv
Description: The value of a class outside its domain is the empty set. (An artifact of our function value definition.) (Contributed by NM, 24-Aug-1995)
Ref Expression
Assertion ndmfv
`|- ( -. A e. dom F -> ( F ` A ) = (/) )`
### Proof
Step Hyp Ref Expression
1 euex
` |- ( E! x A F x -> E. x A F x )`
2 eldmg
` |- ( A e. _V -> ( A e. dom F <-> E. x A F x ) )`
3 1 2 syl5ibr
` |- ( A e. _V -> ( E! x A F x -> A e. dom F ) )`
4 3 con3d
` |- ( A e. _V -> ( -. A e. dom F -> -. E! x A F x ) )`
5 tz6.12-2
` |- ( -. E! x A F x -> ( F ` A ) = (/) )`
6 4 5 syl6
` |- ( A e. _V -> ( -. A e. dom F -> ( F ` A ) = (/) ) )`
7 fvprc
` |- ( -. A e. _V -> ( F ` A ) = (/) )`
8 7 a1d
` |- ( -. A e. _V -> ( -. A e. dom F -> ( F ` A ) = (/) ) )`
9 6 8 pm2.61i
` |- ( -. A e. dom F -> ( F ` A ) = (/) )` | 376 | 843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-30 | latest | en | 0.449619 |
https://findthefactors.com/tag/1725/ | 1,695,430,790,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506429.78/warc/CC-MAIN-20230922234442-20230923024442-00513.warc.gz | 286,915,837 | 17,494 | # 1725 A Toast to You on This Last Day of the Year
### Today’s Puzzle:
I haven’t posted as much this past year as I ought, but thank you for bearing with me. Here is a toast to you, faithful reader! Here’s hoping for a great new year for all of us! Write the numbers from 1 to 10 in both the first column and the top row so that the given clues are the products of the factors you write.
### Factors of 1725:
• 1725 is a composite number.
• Prime factorization: 1725 = 3 × 5 × 5 × 23, which can be written 1725 = 3 × 5² × 23.
• 1725 has at least one exponent greater than 1 in its prime factorization so √1725 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1725 = (√25)(√69) = 5√69.
• The exponents in the prime factorization are 1, 2, and 1. Adding one to each exponent and multiplying we get (1 + 1)(2 + 1)(1 + 1) = 2 × 3 × 2 = 12. Therefore 1725 has exactly 12 factors.
• The factors of 1725 are outlined with their factor pair partners in the graphic below.
### More About the Number 1725:
1725 is the difference of two squares in six different ways:
863² – 862² = 1725,
289² – 286² = 1725,
175² – 170² = 1725,
65² – 50² = 1725,
49² – 26² = 1725, and
47² – 22² = 1725. | 392 | 1,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2023-40 | latest | en | 0.898045 |
http://library.kiwix.org/wikipedia_en_all_nopic_2018-09/A/Kuratowski%E2%80%93Ulam_theorem.html | 1,571,726,718,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987803441.95/warc/CC-MAIN-20191022053647-20191022081147-00092.warc.gz | 116,324,268 | 4,325 | Kuratowski–Ulam theorem
In mathematics, the Kuratowski–Ulam theorem, introduced by Kazimierz Kuratowski and Stanislaw Ulam (1932), called also Fubini theorem for categories, is an analog of the Fubini's theorem for arbitrary second countable Baire spaces. Let X and Y be second countable Baire spaces (or, in particular, Polish spaces), and . Then the following are equivalent if A has the Baire property:
1. A is meager (respectively comeager)
2. The set is comeager in X, where , where is the projection onto Y.
Even if A does not have the Baire property, 2. follows from 1.[1] Note that the theorem still holds (perhaps vacuously) for X - arbitrary Hausdorff space and Y - Hausdorff with countable π-base.
The theorem is analogous to regular Fubini's theorem for the case where the considered function is a characteristic function of a set in a product space, with usual correspondences – meagre set with set of measure zero, comeagre set with one of full measure, a set with Baire property with a measurable set.
References
1. Srivastava, S. (1998). A Course on Borel Sets. Berlin: Springer. p. 112. ISBN 0-387-98412-7. | 297 | 1,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-43 | latest | en | 0.883498 |
https://math.stackexchange.com/questions/575971/converting-base-16-digits-to-base-2-and-10-analysis-question/575981 | 1,571,407,820,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682998.59/warc/CC-MAIN-20191018131050-20191018154550-00487.warc.gz | 601,858,972 | 33,222 | # Converting Base 16 digits to base 2 and 10 - Analysis Question
With base $$16$$, the digits are denoted as $$0, 1,\ldots, 9, A,\ldots, F$$. Let $$n = AB3$$. Rewrite $$n$$ with bases $$10$$, $$2$$.
I have no clue what this question means and how I should attempt to do this.
• If you understand radix notation (binary, hexadecimal, decimal) then the Question is rather straightforward. You are being given the hexadecimal presentation of a number, and asked re-present the same number as binary and decimal expressions. – hardmath Nov 21 '13 at 14:03
Hint: a number $n$ in base 10 can be written as follow: $$n=a_i \cdot 10^{i} + a_{i-1} \cdot 10^{i-1} \cdot \dots \cdot a_1 \cdot 10 + a_0 \cdot 10^0$$ where $a_i \in \{0, 1, \dots, 9\}$. Similar to this case...
• So what's i in this? – Kasun Jayasuriya Nov 21 '13 at 14:10
• It is the number of the digit, counting from the right and starting with $0$. In base $10$, the $i=2$ digit is the hundreds digit (because we started counting with zero), and the value is $10^2$ – Ross Millikan Nov 21 '13 at 14:17
In base 16, each number is composed of a series of the digits 0-9 and A-F, where, A represents 10, B 11, C 12 and so on up to the maximum for any digit which is F, representing 15.
As such, AB$3$ represents $10$ in the $16^2=256$s column, $11$ in the $16^1=16$s column and 3 in the units column, so in decimal (base 10) that's $10(256)+11(16)+3=2739$.
In general, for a natural number $n$, given a number in base $n$, the digit at the far right tells you how many units, the digit to the left of that how many $n$s, the next digit to the left how many $n^2$s etc. None of the digits in a number in base $n$ is allowed to exceed $n-1$.
For how to convert the number to base 2, see the algorithm presented in Adi Dani's answer.
$$n=(AB3)_{16}=A\cdot16^2+B\cdot 16^1+3\cdot 16^0=10\cdot256+11\cdot16+3\cdot1=2739$$ $$2739:2=1369+1$$ $$1369:2=2\cdot684+1$$ $$684:2=2\cdot342+0$$ $$342:2=2\cdot171+0$$ $$171:2=2\cdot85+1$$ $$85:2=2\cdot42+1$$ $$42:2=2\cdot21+0$$ $$21:2=2\cdot10+1$$ $$10:2=2\cdot5+0$$ $$5:2=2\cdot2+1$$ $$2=2\cdot1+0$$
$$2739=(101010110011)_2$$
• I gave you +1 without having checked your base 2 assertion, which is incorrect, but the algorithm is useful, so you deserve the +1. 110011001111 is not 2739 in base 2: it's 3279. $$110011001111=2048+1024+128+64+8+4+2+1=3279$$ The error is in the second line (and hence, as you know, all subsequent lines are inapplicable too), which should be $$1639=684⋅2+1$$ – George Tomlinson Nov 22 '13 at 20:43
• I edit my answer now is correct. I use $1639$ instead of $1369$ Thank you @bluesh34 – Adi Dani Nov 22 '13 at 22:34
• No problems. Oh, yeah: I meant 1369. – George Tomlinson Nov 25 '13 at 17:54 | 965 | 2,723 | {"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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2019-43 | latest | en | 0.807081 |
https://forums.wikitechy.com/question/what-is-the-logic-to-reverse-the-array/ | 1,618,851,066,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038887646.69/warc/CC-MAIN-20210419142428-20210419172428-00150.warc.gz | 387,588,474 | 36,231 | # What is the logic to reverse the array?
What is the logic to reverse the array?
Asked on February 19, 2019 in
# C program to reverse an array:
C program to reverse an array: This program reverses the array elements. For example, if ‘A’ is an array of integers with three elements such that
A[0] = 1, A[1] = 2, A[2] = 3
Then after reversing, the array will be
A[0] = 3, A[1] = 2, A[0] = 1
## Reverse array C program
1. #include <stdio.h>
2. int main()
3. {
4. int n, c, d, a[100], b[100];
5. printf(“Enter the number of elements in array\n”);
6. scanf(“%d”, &n);
7. printf(“Enter array elements\n”);
8. for (c = 0; c < n ; c++)
9. scanf(“%d”, &a[c]);
10. /*
11. * Copying elements into array b starting from end of array a
12. */
13. for (c = n – 1, d = 0; c >= 0; c–, d++)
14. b[d] = a[c];
15. /*
16. * Copying reversed array into the original.
17. * Here we are modifying original array, this is optional.
18. */
19. for (c = 0; c < n; c++)
20. a[c] = b[c];
21. printf(“Reverse array is\n”);
22. for (c = 0; c < n; c++)
23. printf(“%d\n”, a[c]);
24. return 0;
25. }
Reverse array C program output:
Logic to find reverse of array
The above program prints array in reversed order. It does not reverses array. Here I am writing first basic logic to reverse array. It uses above approach to access array element in reverse and copy it to a new reverse array. Which means last element of original array becomes the first element for reverse array.
Step by step descriptive logic to reverse an array.
Input size and elements in an array. Store it in some variable say size and arr respectively.
Declare another array that will store reversed array elements of original array with same size, say reverse[size].
Initialize two variables that will keep track of original and reverse array. Here we will access original array from last and reverse array from first. Hence, initialize arrIndex = size – 1 and revIndex = 0.
Run loop from size – 1 to 0 in decremented style. The loop structure should look like while(arrIndex >= 0).
Inside loop copy original array to reverse array i.e. reverse [revIndex] = arr[arrIndex];.
After copy, increment revIndex and decrement arrIndex.
Finally after loop print reverse array.
Program to find reverse of array
/**
* C program to find reverse of array
*/
#include
#define MAX_SIZE 100 // Maximum array size
int main()
{
int arr[MAX_SIZE], reverse[MAX_SIZE];
int size, i, arrIndex, revIndex;
/* Input size of the array */
printf(“Enter size of the array: “);
scanf(“%d”, &size);
/* Input array elements */
printf(“Enter elements in array: “);
for(i=0; i= 0)
{
/* Copy value from original array to reverse array */
reverse[revIndex] = arr[arrIndex];
revIndex++;
arrIndex–;
}
/*
* Print the reversed array
*/
printf(“\nReversed array : “);
for(i=0; i | 843 | 2,874 | {"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} | 3.09375 | 3 | CC-MAIN-2021-17 | latest | en | 0.534676 |
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# Eighty percent of the lights at Hotel California are on at 8 p.m.
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Eighty percent of the lights at Hotel California are on at 8 p.m. [#permalink]
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02 Nov 2008, 08:56
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Eighty percent of the lights at Hotel California are on at 8 p.m. a certain evening. However, forty percent of the lights that are supposed to be off are actually on and ten percent of the lights that are supposed to be on are actually off. What percent of the lights that are on are supposed to be off?
22(2/9)%
16(2/3)%
11(1/9)%
10%
5%
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Re: Eighty percent of the lights at Hotel California are on at 8 p.m. [#permalink]
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02 Nov 2008, 09:37
LiveStronger wrote:
Source:MGMAT CAT
Eighty percent of the lights at Hotel California are on at 8 p.m. a certain evening. However, forty percent of the lights that are supposed to be off are actually on and ten percent of the lights that are supposed to be on are actually off. What percent of the lights that are on are supposed to be off?
22(2/9)%
16(2/3)%
11(1/9)%
10%
5%
i took nyquil last night and my brain is forzen but mmaybe I am wrong and cant figure out
but say we have 100 lights=80 are on+20 are off
40% of 20 are on i.e 8 are on..which means 8/80*100 or 10% of the ligths that should be off are ON..
d it is
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Re: Eighty percent of the lights at Hotel California are on at 8 p.m. [#permalink]
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02 Nov 2008, 09:44
I'd pick Answer as D (10%)
Typical percent problem, so take total # of lights as 100. Now 80% of lights are currently ON or 80 lights
# of lights supposed to be ON = N
# of lights supposed to be OFF = F
Rewriting the sentence, you'd get (40/100)*F + N - (10/100)*N = 80.
Also, by our 100 light assumption, we can state F + N = 100
solving for F and N, you get F = 20 and N = 80. Hmmm...now I almost made a mistake here.
If you substitute this back, # of lights supposed to OFF that are ON = (40/100)* 20 which is 8. And # of lights supposed to ON that are OFF = (10/100)* 80 which is also 8.
anyways, the question is what % of lights that are ON that should be OFF.. so (8/80) * 100, which leaves us with a final answer that Hotel California is Energy In-effecient by 10%
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Re: Eighty percent of the lights at Hotel California are on at 8 p.m. [#permalink]
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03 Nov 2008, 08:07
OA is D
Re: Eighty percent of the lights at Hotel California are on at 8 p.m. [#permalink] 03 Nov 2008, 08:07
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# Excel Function Help
Registered User regular
I'm trying to devise a spreadsheet to order medication, and I'm not sure if I can do what I'm trying to do here.
The drug comes in vials of 100 and 40. Is there an easy way to figure out the most efficient way to order these?
I've been using:
=ROUNDDOWN(G5/100,0) for the 100s
and
=(ROUNDUP((MOD(G5,100)),-2)/40) to figure out the remainder for the 40s
But that fails if the number is 158 and I should be ordering 4 bottles of 40, for a total of 160. It tells me to order 1 100 and 2 40s for 180.
Is there an easy way to do this, or am I just going to have to write a crazy long IF statement that checks remainders?
Thanks!
## Posts
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Registered User regular
Is it for a whole sheet of differing values or just once at a time, if the latter then the excel solver might be of use.
An IF statement based solution will be pretty long because 40 doesn't fit nicely into 100 and you'll be evaluating combinations of 40s and 100s.
However you could just pre-calculate and store the best solution for values 1-240 and use a vlookup to get them?
• Options
Registered User regular
Define "efficiency."
Are you looking for the fewest number of vials, the cheapest price, the least amount of overage, something else entirely?
• Options
resting shark face Registered User regular
I've spent too much time thinking about this already, but there is a pattern, I just can't think of a neat mathematical equation for it yet. But anyways, I'll just post what I have here and maybe someone else will finish what I started:
The number of 40s you need changes every 20 pills. Ignoring the first 60 pills it goes like this
61-80: 2
81-100: 0
101-120: 3
121-140: 1
141-160: 4
and repeats forever
The number of 100s you need also changes every 20 pills but also increments by 1 every 100 pills, so it goes like this:
61-80: 0
81-100: 1
101-120: 0
121-140: 1
141-160: 0
Then the next set (covering 161-260 pills) will be: 1 2 1 2 1 and so on
Good luck!
Please consider the environment before printing this post.
• Options
resting shark face Registered User regular
You can also brute force this by creating a table with the above pattern and doing VLOOKUP
Please consider the environment before printing this post.
• Options
No Pic EverRegistered User regular
I'm trying to devise a spreadsheet to order medication, and I'm not sure if I can do what I'm trying to do here.
The drug comes in vials of 100 and 40. Is there an easy way to figure out the most efficient way to order these?
I've been using:
=ROUNDDOWN(G5/100,0) for the 100s
and
=(ROUNDUP((MOD(G5,100)),-2)/40) to figure out the remainder for the 40s
But that fails if the number is 158 and I should be ordering 4 bottles of 40, for a total of 160. It tells me to order 1 100 and 2 40s for 180.
Is there an easy way to do this, or am I just going to have to write a crazy long IF statement that checks remainders?
Thanks!
You don't need crazy if statements, you just need to consider that there are several ways you could be fullfilling this order:
1) Round up to nearest 100, so 200 units with 42 wasted.
2) Round down to nearest 100, then round up for nearest 40. So 180 units with 22 wasted.
3) Round up to the nearest 40. So you would order 160 units and have 2 wasted.
Then you set the program to give the min waste amount for all 3.
If you start exceeding 200 units, add a function at the start of each chain to round down to the nearest 200, count that value twice (multiply by 2), and that's your starting point for the rest of the functions. You have to do this because otherwise your 40 and 100 sets can come up with the same waste amount (e.g. at 220, both would waste 20).
So for the first let's assume we're in cell H5 and moving out.:
H5=ROUDNUP(G5/100,0) This is the calculation
H8=H5*100-G5 This is the amount wasted
I5=ROUNDDOWN(G5/100,0) This is first part of the calculation
I6=ROUNDUP((G5-100*I5)/40,0) This is the second part of the calculation
I8=H5*100+H6*40-G5 This is the amount wasted
J6=ROUNDUP(G5/40,0) This is the calculation
J8=H5*40=G5 This is the amount wasted
I've maintained that row 5 is always the 100's, row 6 is always the 40's, and row 8 is always the waste.
Now, you can visually see which is the Minimum of the three for amount wasted, but we can automate this as well as follows:
First, set up a a new table as follows:
L5=H8
L6=I8
L7=K8
M5=H5
M6=I5
N6=I6
N7=J6
O5="100 UNITS ONLY"
O6="MIXED SIZES"
O7="40 UNITS ONLY"
Then we can write a function that will pull information from the above table as follows:
We're using cell L10 as the description, L11 as the number of 100 unit vials, and L12 as the number of 40 unit vials
L10=VLOOKUP(MIN(H8:J8),L5:O7,4)
L11=VLOOKUP(MIN(H8:J8),L5:O7,2)
L12=VLOOKUP(MIN(H8:J8),L5:O7,3)
I hope this helps.
• Options
San DiegoRegistered User regular
In case you want the horrifically inelegant, crazy long IF statement you mentioned, I think these will work:
40s: =IF(G5<1,0,IF(G5<41,1,IF(G5<81,2,IF(G5<101,0,IF(VALUE(RIGHT(G5,2))>80,0,IF(VALUE(RIGHT(G5,2))>60,2,IF(VALUE(RIGHT(G5,2))>40,4,IF(VALUE(RIGHT(G5,2))>20,1,3))))))))
100s:
=IF(G5<81,0,ROUNDDOWN(G5/100,0)+IF(VALUE(RIGHT(G5,2))>80,1,IF(VALUE(RIGHT(G5,2))>60,0,IF(VALUE(RIGHT(G5,2))>40,-1,IF(VALUE(RIGHT(G5,2))>20,0,IF(VALUE(RIGHT(G5,2))>0,-1,0))))))
• Options
Registered User regular
1. =X/40
2. =X/100
3. =ROUNDUP(1, 0)-1
4. =ROUNDUP(2, 0)-2
5. =IF(3<4, 1&"40 bottles", 2&"100 Bottles")
This is taking the division, comparing it to the closest round and selecting the closest gap.
• Options
Registered User regular
Elvenshae wrote: »
Define "efficiency."
Are you looking for the fewest number of vials, the cheapest price, the least amount of overage, something else entirely?
Was looking for least amount of overages specifically. All these answers really set me down the right path, and gave me some good stopgaps until I can make the dosage charts for VLOOKUP (Which I probably should have just committed to sooner)
Thanks so much for all the help everyone!
• Options
Registered User regular
schuss wrote: »
1. =X/40
2. =X/100
3. =ROUNDUP(1, 0)-1
4. =ROUNDUP(2, 0)-2
5. =IF(3<4, 1&"40 bottles", 2&"100 Bottles")
This is taking the division, comparing it to the closest round and selecting the closest gap.
This won't give you the closest answer in some cases though - for 130 the best answer is 1x100+1x40 but this formula set will tell you to use 4x40. (It was my first port of call as well ).
• Options
Registered User regular
Dis' wrote: »
schuss wrote: »
1. =X/40
2. =X/100
3. =ROUNDUP(1, 0)-1
4. =ROUNDUP(2, 0)-2
5. =IF(3<4, 1&"40 bottles", 2&"100 Bottles")
This is taking the division, comparing it to the closest round and selecting the closest gap.
This won't give you the closest answer in some cases though - for 130 the best answer is 1x100+1x40 but this formula set will tell you to use 4x40. (It was my first port of call as well ).
Good catch. This was my five minute answer, so not ideal.
• Options
Registered User regular
Actually, I think the right answer here is that you only need to solve for order quantities of 1-199.
Everything over that, you just take out 100s until you get back into the 1-199 range, and go from there.
Like, for 202ml, the optimal answer is to get 220ml (1x100ml + 3x40ml), which is:
• 1x100ml +
• Optimal answer for 102ml, 3x40ml
360ml?
• 2x100ml +
• Optimal answer for 160ml, 4x40ml
363ml?
• 2x100ml +
• Optimal answer for 163ml, 1x100ml + 2x40ml
321ml?
• 2x100ml +
• Optimal answer for 121ml, 1x100ml + 1x40ml
Thoughts?
• Options
Registered User regular
Excel's solver will do it.
What you have is an equation like:
need + overage = 100*a + 40*b
You want to minimize overage, subject to the constraints that it is >= 0 and that a and b are integers.
So if you have an Excel sheet set up like the following:
A1 = a
B1 = b
C1 = need
D1 = 100*A1 + 40*B1 - C1
Then you can use solver like this:
Set the objective to D1
By Changing Variable Cells to A1:B1
Subject to the constraints:
D1>=0
A1 is integer
B1 is integer
Then you can fill in your needed value in C1 and run the solver and it will minimize your overage.
• Options
Registered User regular
edited July 2016
You can use SUMIF to sort the numbers easily using a 3x3 table. I'll use A = 100 and B = 40.
Row 1 would be the number of A vials needed
Row 2 would be the number of B vials needed
Row 3 would be the remainder of pills
The columns would be for your three choices
A+B | A only | B only
=SUMIF(Row3,MIN(Row3), Row1) would give you the correct amount of A vials needed
=SUMIF(Row3,MIN(Row3), Row2) would give you the number of B vials needed
and
=MIN(Row3) would give you the excess
For the mixed version, where G5 is the amount needed
A#: =ROUNDDOWN(G5/LCM(A,B),0)*(LCM(A,B)/A)
Bs: =ROUNDUP((G5-A#*A)/40,0)
You can replace A and B with any number in this version. LCM() is least common factor function in Excel - the cycle loops there. So you could have a 80 and 30 and the math should still work out in this case.
Kipling on
3DS Friends: 1693-1781-7023
• Options
Registered User regular
edited July 2016
If you don't mind having an extra column with the total amount dispensed, I think I have an efficient solution with no need for tables, macros, or solver.
If Column G has the amount needed, get the amount you know you'll need in column H. This is simply rounding up to the nearest 20, with an exception for numbers under 80, which go up to the nearest 40.
=IF(G1<80,ROUNDUP(G1/40,0)*40,ROUNDUP(G1/20,0)*20)
In column I, you'll get the number of 40s you need. You'll never need more than 4 (if you wanted 5, you could just use 2 100s). You know how many 40s you need by how big the 10s column. The choose function lets you pick a number based on the value of an index.
=CHOOSE(MOD(H1,100)/20+1,0,3,1,4,2)
Then, to get the number of 100s, simply see what you need leftover
=(H1-40*I1)/100
Edit: You could do it without the extra column but it's a lot messier
In H (40s): =CHOOSE(MOD(IF(G1<80,ROUNDUP(G1/40,0)*40,ROUNDUP(G1/20,0)*20),100)/20+1,0,3,1,4,2)
In I (100s): =ROUND((G1-40*H1)/100,0)
MrTLicious on | 3,155 | 10,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-18 | latest | en | 0.920204 |
https://2022.help.altair.com/2022/hwcfdsolvers/nfx/topics/nanofluidx/transport_vel_formulation_r.htm | 1,695,707,950,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510149.21/warc/CC-MAIN-20230926043538-20230926073538-00077.warc.gz | 90,482,666 | 9,903 | # Transport Velocity Formulation
Some main principles and consequences of the transport velocity.
Full theoretical description of the transport velocity is beyond the scope of this manual, so some main principles and consequences of the transport velocity will be discussed. For full theoretical derivation and analysis refer to the work of Adami et al.
Accuracy of the SPH method heavily relies on the ability of the code to accurately reconstruct the Shepard coefficient and provide full support to the particles. In reality the value of the Shepard coefficient will be ≈ 1, but rarely exactly 1. If you analyze the SPH method you can easily understand that the accuracy of the method is actually directly related to the particle distribution. If the particles are uniformly ordered, the reconstruction of the variable fields will be accurate. If there are excesses, such as overly-packed or overly-sparse particle distributions, this will negatively reflect on the accuracy of the solution. It would therefore be ideal if you could keep the particles ordered as uniformly as possible without sacrificing computational time.
The transport velocity does precisely this. The numerical formulation of the transport velocity introduces a correction method to otherwise normal SPH velocity computation in a case-independent manner, while preserving the physicality of the solution. The dual correction is applied exclusively through the momentum equation, as opposed to the traditional background pressure approach which is explicitly appearing in the quasi-incompressible equation of state. The correction comes directly from the computation of the transport velocity $\stackrel{˜}{u}$ , or more precisely the time advancement of the transport velocity:(1)
$\stackrel{˜}{{u}_{i}}\left(t+\Delta t\right)={u}_{i}\left(t\right)+\Delta t\left(\frac{\stackrel{˜}{d}{u}_{i}}{dt}-\frac{1}{{\rho }_{i}}\nabla {p}_{c}\right)$
Where, ${p}_{c}$ is the corrective pressure field, usually set to be equivalent to the initial pressure of the simulation ${p}_{0}$ .
These corrections actively maintain particle order which has a number of beneficial influences on the numerical behavior of the code.
The magnitude of the ${p}_{c}$ pressure directly influences the “strength of the correction.” The higher the ${p}_{c}$ value, the more vigorous will be the correction attempt. You should keep this in mind, as specifying the ${p}_{c}$ value too high, for example, ${p}_{c}$ = 10 ${p}_{0}$ , can lead to excessive correction force and in these cases the time step must be appropriately reduced.
The command is actually a coefficient with which you multiply the ${p}_{0}$ value and therefore determine the ${p}_{c}$ correction pressure (by default it is set to 1.0).(2)
${p}_{0}={p}_{c}_factor*{p}_{0}$
In more graphical terms, the transport velocity formulation automatically detects particle vacuum and attempts to populate it with particles. As mentioned, this has profoundly beneficial influence in multiphase simulations, but in single phase simulation where you have intentionally left a large portion of the domain empty (particle vacuum), the use of the transport velocity could actually have detrimental effects. The reason is precisely because transport velocity is seeking for particle vacuum and tries to fill it, which would in single phase cases result in a pop-corn like behavior of the free surface. This is something to be avoided and therefore in single phase cases it is strongly recommended to turn the transport velocity off.
S. Adami, Modeling and Simulation of Multiphase Phenomena with Smoothed Particle Hydrodynamics, Lehrstuhl für Aerodynamik und Strömungsmechanik, Technische Universität München, 2014 | 766 | 3,718 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 12, "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} | 3.453125 | 3 | CC-MAIN-2023-40 | latest | en | 0.886109 |
https://www.coursehero.com/file/28941446/Homework-4docx/ | 1,643,175,911,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304915.53/warc/CC-MAIN-20220126041016-20220126071016-00578.warc.gz | 736,859,989 | 46,946 | # Homework 4.docx - Ryne Templeman Homework 4 9.10 9.26 9.42...
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Ryne TemplemanHomework 42/19/189.10, 9.26, 9.42, 9.50, 9.529.10a.T(4)=2.87 shows us that the degree of freedom is 4 and the value of the t statistic is 2.87. p=0.032 is saying that the value of p is 0.0329.26a.N=8, df=8-1=7. Critical Values at -1.895 and 1.895b.N=42, df=42-1=41. Critical Value at 1.684c.N=89, df=89-1=88. Critical Values at -2.633 and 2.633 This falls between 80 and 100 on the t-table in the book9.42a.We would use a single sample t-test, when we know the mean of the sample but do not know the standard deviationb.A null and research hypothesis (find out what the researchers are looking for), and get an actual number for the sample sizec.Volunteer sample. The information that was used was given by people does not representthe overall populationd.We really don’t have an idea of how many questions the participants answered, and some of the participants could also be males and didn’t answer the questions about the sun dresses. As researchers we cannot be sure of how big the actual sample size is. | 373 | 1,226 | {"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} | 3.28125 | 3 | CC-MAIN-2022-05 | latest | en | 0.885351 |
https://www.teacherspayteachers.com/Store/Maria-Nielsen | 1,519,454,420,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815435.68/warc/CC-MAIN-20180224053236-20180224073236-00058.warc.gz | 938,375,106 | 21,787 | Total:
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16 different functions are expressed in the form of verbal description, graphs, tables, and equations. Students first sort the cards between linear and exponential cards. Then students can group each function with its verbal description, graph,
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Students will use exponent rules to simplify expressions. This activity is great when students are first learning the rules because no variables are included. There are 16 posters and each one leads students to another poster. No answer key is
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Student need to determine if the expression is correctly simplified. If it is not students can make corrections and determine how to use the rule appropriately.
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This maze can be used as a homework activity for students. It is for students to practice one, two, and multi-step equations. The problems also include one, none, and infinite solutions! Hope you like it! Answer key included!
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This worksheet will help students use the first and second derivative test to graph a function without a calculator. I used this for Calculus I and found that the students really benefited from it.
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This foldable was designed to help students explore the domain and range of ordered pairs, graphs, mapping diagrams, and tables.
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MY OWN EDUCATIONAL HISTORY | 525 | 2,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-09 | latest | en | 0.907611 |
https://stacks.math.columbia.edu/tag/07XH | 1,723,341,454,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640843545.63/warc/CC-MAIN-20240811013030-20240811043030-00778.warc.gz | 425,285,368 | 6,884 | Lemma 98.12.7. Let $S$ be a locally Noetherian scheme. Let $p : \mathcal{X} \to (\mathit{Sch}/S)_{fppf}$ be a category fibred in groupoids. Let $\xi = (R, \xi _ n, f_ n)$ be a formal object of $\mathcal{X}$ with $\xi _1$ lying over $\mathop{\mathrm{Spec}}(k) \to S$ with image $s \in S$. Assume
1. $\xi$ is versal,
2. $\xi$ is effective,
3. $\mathcal{O}_{S, s}$ is a G-ring, and
4. $p : \mathcal{X} \to (\mathit{Sch}/S)_{fppf}$ is limit preserving on objects.
Then there exist a morphism of finite type $U \to S$, a finite type point $u_0 \in U$ with residue field $k$, and an object $x$ of $\mathcal{X}$ over $U$ such that $x$ is versal at $u_0$ and such that $x|_{\mathop{\mathrm{Spec}}(\mathcal{O}_{U, u_0}/\mathfrak m_{u_0}^ n)} \cong \xi _ n$.
Proof. Choose an object $x_ R$ of $\mathcal{X}$ lying over $\mathop{\mathrm{Spec}}(R)$ whose associated formal object is $\xi$. Let $N = 2$ and apply Lemma 98.10.1. We obtain $A, \mathfrak m_ A, x_ A, \ldots$. Let $\eta = (A^\wedge , \eta _ n, g_ n)$ be the formal object associated to $x_ A|_{\mathop{\mathrm{Spec}}(A^\wedge )}$. We have a diagram
$\vcenter { \xymatrix{ & \eta \ar[d] \\ \xi \ar[r] \ar@{..>}[ru] & \xi _2 = \eta _2 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ & A^\wedge \ar[d] \\ R \ar[r] \ar@{..>}[ru] & R/\mathfrak m_ R^2 = A/\mathfrak m_ A^2 } }$
The versality of $\xi$ means exactly that we can find the dotted arrows in the diagrams, because we can successively find morphisms $\xi \to \eta _3$, $\xi \to \eta _4$, and so on by Formal Deformation Theory, Remark 90.8.10. The corresponding ring map $R \to A^\wedge$ is surjective by Formal Deformation Theory, Lemma 90.4.2. On the other hand, we have $\dim _ k \mathfrak m_ R^ n/\mathfrak m_ R^{n + 1} = \dim _ k \mathfrak m_ A^ n/\mathfrak m_ A^{n + 1}$ for all $n$ by construction. Hence $R/\mathfrak m_ R^ n$ and $A/\mathfrak m_ A^ n$ have the same (finite) length as $\Lambda$-modules by additivity of length and Formal Deformation Theory, Lemma 90.3.4. It follows that $R/\mathfrak m_ R^ n \to A/\mathfrak m_ A^ n$ is an isomorphism for all $n$, hence $R \to A^\wedge$ is an isomorphism. Thus $\eta$ is isomorphic to a versal object, hence versal itself. By Lemma 98.12.3 we conclude that $x_ A$ is versal at the point $u_0$ of $U = \mathop{\mathrm{Spec}}(A)$ corresponding to $\mathfrak m_ A$. $\square$
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https://www.studentlance.com/solution/bus-401-week-3-problems-9-9-and-10-4 | 1,518,910,944,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891808539.63/warc/CC-MAIN-20180217224905-20180218004905-00688.warc.gz | 970,764,445 | 5,745 | # Bus 401 Week 3 Problems 9-9 And 10-4 - 24145
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10-4 (NPV, PI, and IRR calculations) Fijisawa Inc. is considerinf a major expansion of its product line and has estimated the following cash flows associated with such an expansion. The initial outlay would be \$1,950,000, and the project would generate incremental free cash flows of \$450,000 per year for 6 years. The appropiate required rate of return is 9 percent.
a. Calculate the NPV
b. Calculate the PI.
c. Calculate the IRR.
d. Should this project be accepted?
9-9(Cost of dept) Sincere Stationery Corporation needs to raise \$500,000 to improve its manufacturing plant. It has decided to issue a \$1,000 par value bond with a 14 percent annual coupon rate and a 10 year maturity. The investors require a 9 percent rate of return.
a. Compute the market value of the bonds.
b. What will the net price be if flotation costs are 10.5 percent of the market price?
c. How many bonds will the firm have to issue to receive the needed funds?
d. What is the firm’s after-tax cost of debt if its average tax rate is 25 percent and its marginal tax rate is 34 percent?
Solution Description
Attachments | 359 | 1,374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-09 | longest | en | 0.899201 |
https://everything.explained.today/Flow_network/ | 1,721,499,180,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517515.18/warc/CC-MAIN-20240720174732-20240720204732-00865.warc.gz | 216,043,493 | 3,826 | # Flow network explained
In graph theory, a flow network (also known as a transportation network) is a directed graph where each edge has a capacity and each edge receives a flow. The amount of flow on an edge cannot exceed the capacity of the edge. Often in operations research, a directed graph is called a network, the vertices are called nodes and the edges are called arcs. A flow must satisfy the restriction that the amount of flow into a node equals the amount of flow out of it, unless it is a source, which has only outgoing flow, or sink, which has only incoming flow. A network can be used to model traffic in a computer network, circulation with demands, fluids in pipes, currents in an electrical circuit, or anything similar in which something travels through a network of nodes.
## Definition
A network is a directed graph with a non-negative capacity function for each edge, and without multiple arcs (i.e. edges with the same source and target nodes). Without loss of generality, we may assume that if, then is also a member of . Additionally, if then we may add to E and then set the .
If two nodes in are distinguished – one as the source and the other as the sink – then is called a flow network.[1]
## Flows
Flow functions model the net flow of units between pairs of nodes, and are useful when asking questions such as what is the maximum number of units that can be transferred from the source node s to the sink node t? The amount of flow between two nodes is used to represent the net amount of units being transferred from one node to the other.
The excess function represents the net flow entering a given node (i.e. the sum of the flows entering) and is defined by$x_f(u)=\sum_ f(w,u).$A node is said to be active if (i.e. the node consumes flow), deficient if (i.e. the node produces flow), or conserving if . In flow networks, the source is deficient, and the sink is active.Pseudo-flows, feasible flows, and pre-flows are all examples of flow functions.
A pseudo-flow is a function of each edge in the network that satisfies the following two constraints for all nodes and :
• Skew symmetry constraint: The flow on an arc from to is equivalent to the negation of the flow on the arc from to, that is: . The sign of the flow indicates the flow's direction.
• Capacity constraint: An arc's flow cannot exceed its capacity, that is: .
A pre-flow is a pseudo-flow that, for all, satisfies the additional constraint:
• Non-deficient flows: The net flow entering the node is non-negative, except for the source, which "produces" flow. That is: for all .
A feasible flow, or just a flow, is a pseudo-flow that, for all, satisfies the additional constraint:
• Flow conservation constraint: The total net flow entering a node is zero for all nodes in the network except the source
s
and the sink
t
, that is: for all
## Notes and References
1. A.V. Goldberg, É. Tardos and R.E. Tarjan, Network flow algorithms, Tech. Report STAN-CS-89-1252, Stanford University CS Dept., 1989 | 678 | 3,017 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "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-2024-30 | latest | en | 0.949922 |
http://road-to-psle.blogspot.com/2009/04/rosyth-sch-2006-psle-math-prelim-q39.html | 1,544,647,719,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824119.26/warc/CC-MAIN-20181212203335-20181212224835-00529.warc.gz | 252,072,985 | 11,684 | This blog is managed by Song Hock Chye, author of Improve Your Thinking Skills in Maths (P1-P3 series), which is published and distributed by EPH.
## Wednesday, April 08, 2009
### Rosyth Sch 2006 PSLE Math Prelim Q39
Paul would like the send a parcel to his friend in ABC country. The freight change is \$8.00 per kg per km. What is the total charge for sending the parcel weighing 12 kg for 35 km?
Solution
1 kg ----- \$8
12 kg ----- 12 x \$8 = \$96
1 km, 12 kg ----- \$96
35 km, 12 kg ----- 35 x \$96 = \$3360
Answer: The total charge is \$3360. | 164 | 554 | {"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} | 3.71875 | 4 | CC-MAIN-2018-51 | latest | en | 0.911927 |
https://math.stackexchange.com/questions/1411881/why-are-the-number-of-verticies-in-a-clique-graph-less-than-its-a-parent-graph | 1,568,881,277,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573465.18/warc/CC-MAIN-20190919081032-20190919103032-00418.warc.gz | 574,271,097 | 28,306 | # Why are the number of verticies in a clique graph less than its a parent graph [duplicate]
I am reading up about Graph theory and the example it gives for a Clique Subgraph looks like this...
Now it states that the bottom graph is "obviously" the clique graph for the top. Is this because it is the smallest graph that can be made so all vertices are adjacent? Would, for example, a graph with 5 or 6 vertices need to leave some unconnected?
Also it mentions that a Clique is a "Maximal complete subgraph" but I thought subgraphs were supposed to contain the same edges. So why are there only 3 nodes with degrees greater than 2 but in the clique there are 4
## marked as duplicate by N. F. Taussig, Ken, Cameron Buie, Mike Pierce, user91500Aug 28 '15 at 8:52
It sounds like you are conflating the concepts of clique, maximal clique, and clique graph. A clique of a graph $G = (V, E)$ is a set $X \subseteq V$ such that every two vertices in $X$ are adjacent (connected by an edge.) A maximal clique of $G$ is a clique $X$ which is as big as possible (i.e. adding any vertex would violate the clique condition.)
The clique graph of $G$ is new graph $H = (V', E')$, defined as follows. The vertex set $V'$ is the set of maximal cliques of $G$. (Each vertex in the clique graph represents a whole clique in the original graph!) The edge set $E'$ is the set of pairs of maximal cliques $\{X, Y\}$ such that $X \cap Y \neq \varnothing$. (Two maximal cliques are connected by an edge if they share a vertex.)
The first graph depicted in your question has $4$ distinct maximal cliques, each of which is a triangle. (See image here.) Every two of these four cliques share at least one vertex. Hence, the clique graph of that first graph is $K_4$, the complete graph on $4$ vertices.
In your example, the maximal cliques in the original graph are all of size 3, so the four $3$-cycles of the first graph are represented by the four vertices in the clique graph. Any two $3$-cycles of the original graph share at least one vertex, so the vertices of the clique graph are all adjacent. | 523 | 2,082 | {"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} | 3.8125 | 4 | CC-MAIN-2019-39 | latest | en | 0.94469 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/384/9/b/ | 1,653,690,689,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663006341.98/warc/CC-MAIN-20220527205437-20220527235437-00408.warc.gz | 991,564,767 | 61,455 | # Properties
Label 384.9.b Level $384$ Weight $9$ Character orbit 384.b Rep. character $\chi_{384}(319,\cdot)$ Character field $\Q$ Dimension $64$ Newform subspaces $4$ Sturm bound $576$ Trace bound $17$
# Related objects
## Defining parameters
Level: $$N$$ $$=$$ $$384 = 2^{7} \cdot 3$$ Weight: $$k$$ $$=$$ $$9$$ Character orbit: $$[\chi]$$ $$=$$ 384.b (of order $$2$$ and degree $$1$$) Character conductor: $$\operatorname{cond}(\chi)$$ $$=$$ $$8$$ Character field: $$\Q$$ Newform subspaces: $$4$$ Sturm bound: $$576$$ Trace bound: $$17$$ Distinguishing $$T_p$$: $$5$$
## Dimensions
The following table gives the dimensions of various subspaces of $$M_{9}(384, [\chi])$$.
Total New Old
Modular forms 528 64 464
Cusp forms 496 64 432
Eisenstein series 32 0 32
## Trace form
$$64 q + 139968 q^{9} + O(q^{10})$$ $$64 q + 139968 q^{9} + 309120 q^{17} - 6416576 q^{25} - 8749440 q^{41} - 76975296 q^{49} + 6200064 q^{57} + 29299200 q^{65} - 80055680 q^{73} + 306110016 q^{81} + 365339520 q^{89} - 121481856 q^{97} + O(q^{100})$$
## Decomposition of $$S_{9}^{\mathrm{new}}(384, [\chi])$$ into newform subspaces
Label Dim $A$ Field CM Traces $q$-expansion
$a_{2}$ $a_{3}$ $a_{5}$ $a_{7}$
384.9.b.a $8$ $156.433$ $$\mathbb{Q}[x]/(x^{8} - \cdots)$$ None $$0$$ $$0$$ $$0$$ $$0$$ $$q-\beta _{3}q^{3}-\beta _{5}q^{5}-7\beta _{4}q^{7}+3^{7}q^{9}+\cdots$$
384.9.b.b $8$ $156.433$ $$\mathbb{Q}[x]/(x^{8} + \cdots)$$ None $$0$$ $$0$$ $$0$$ $$0$$ $$q-\beta _{1}q^{3}+\beta _{2}q^{5}-\beta _{4}q^{7}+3^{7}q^{9}+\cdots$$
384.9.b.c $16$ $156.433$ $$\mathbb{Q}[x]/(x^{16} - \cdots)$$ None $$0$$ $$0$$ $$0$$ $$0$$ $$q+\beta _{1}q^{3}-\beta _{2}q^{5}+\beta _{5}q^{7}+3^{7}q^{9}+\cdots$$
384.9.b.d $32$ $156.433$ None $$0$$ $$0$$ $$0$$ $$0$$
## Decomposition of $$S_{9}^{\mathrm{old}}(384, [\chi])$$ into lower level spaces
$$S_{9}^{\mathrm{old}}(384, [\chi]) \cong$$ $$S_{9}^{\mathrm{new}}(8, [\chi])$$$$^{\oplus 10}$$$$\oplus$$$$S_{9}^{\mathrm{new}}(24, [\chi])$$$$^{\oplus 5}$$$$\oplus$$$$S_{9}^{\mathrm{new}}(32, [\chi])$$$$^{\oplus 6}$$$$\oplus$$$$S_{9}^{\mathrm{new}}(64, [\chi])$$$$^{\oplus 4}$$$$\oplus$$$$S_{9}^{\mathrm{new}}(96, [\chi])$$$$^{\oplus 3}$$$$\oplus$$$$S_{9}^{\mathrm{new}}(128, [\chi])$$$$^{\oplus 2}$$$$\oplus$$$$S_{9}^{\mathrm{new}}(192, [\chi])$$$$^{\oplus 2}$$ | 987 | 2,281 | {"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} | 2.578125 | 3 | CC-MAIN-2022-21 | latest | en | 0.409434 |
https://doc.cgal.org/4.8.2/Straight_skeleton_2/index.html | 1,718,736,053,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00562.warc.gz | 184,849,335 | 17,146 | CGAL 4.8.2 - 2D Straight Skeleton and Polygon Offsetting
User Manual
# Definitions
## 2D Contour
A 2D contour is a closed sequence (a cycle) of 3 or more connected 2D oriented straight line segments called contour edges. The endpoints of the contour edges are called vertices. Each contour edge shares its endpoints with at least two other contour edges.
If the edges intersect only at the vertices and at most are coincident along a line but do not cross one another, the contour is classified as simple.
A contour is topologically equivalent to a disk and if it is simple, is said to be a Jordan Curve.
Contours partition the plane in two open regions: one bounded and one unbounded. If the bounded region of a contour is only one singly-connected set, the contour is said to be strictly-simple.
The Orientation of a contour is given by the order of the vertices around the region they bound. It can be CLOCKWISE (CCW) or COUNTERCLOCKWISE (CW).
The bounded side of a contour edge is the side facing the bounded region of the contour. If the contour is oriented CCW, the bounded side of an edge is its left side.
A contour with a null edge (a segment of length zero given by two consecutive coincident vertices), or with edges not connected to the bounded region (an antenna: 2 consecutive edges going forth and back along the same line), is said to be degenerate (collinear edges are not considered a degeneracy).
## 2D Polygon with Holes
A 2D polygon is a contour.
A 2D polygon with holes is a contour, called the outer contour, having zero or more contours, called inner contours, or holes, in its bounded region. The intersection of the bounded region of the outer contour and the unbounded regions of each inner contour is the interior of the polygon with holes. The orientation of the holes must be opposite to the orientation of the outer contour and there cannot be any intersection among any contour. A hole cannot be in the bounded region of any other hole.
A polygon with holes is strictly-simple if its interior is a singly-connected set.
The orientation of a polygon with holes is the orientation of its outer contour. The bounded side of any edge, whether of the outer contour or a hole, is the same for all edges. That is, if the outer contour is oriented CCW and the holes CW, both contour and hole edges face the polygon interior to their left.
Throughout the rest of this chapter the term polygon will be used as a shortcut for polygon with holes.
Figure 20.1 Examples of strictly simple polygons: One with no holes and two edges coincident (left) and one with 2 holes (right).
Figure 20.2 Examples of non-simple polygons: One folding into itself, that is, non-planar (left), one with a vertex touching an edge (middle), and one with a hole crossing into the outside (right)
## Inward Offset of a Non-degenerate Strictly-Simple Polygon with Holes
For any 2D non-degenerate strictly-simple polygon with holes called the source, there can exist a set of 0, 1 or more inward offset polygons with holes, or just offset polygons for short, at some euclidean distance t $$>0$$ (each being strictly simple and non-degenerate). Any contour edge of such offset polygon, called an offset edge corresponds to some contour edge of the source polygon, called its source edge. An offset edge is parallel to its source edge and has the same orientation. The Euclidean distance between the lines supporting an offset edge and its source edge is exactly t.
An offset edge is always located to the bounded side of its source edge (which is an oriented straight line segment).
An offset polygon can have less, equal or more sides as its source polygon.
If the source polygon has no holes, no offset polygon has holes. If the source polygon has holes, any of the offset polygons can have holes itself, but it might as well have no holes at all (if the distance is sufficiently large).
Each offset polygon has the same orientation as the source polygon.
## Straight Skeleton of a 2D Non-degenerate Strictly-Simple Polygon with Holes
The 2D straight skeleton of a non-degenerate strictly-simple polygon with holes [2] is a special partitioning of the polygon interior into straight skeleton regions corresponding to the monotone areas traced by a continuous inward offsetting of the contour edges. Each region corresponds to exactly 1 contour edge.
These regions are bounded by angular bisectors of the supporting lines of the contour edges and each such region is itself a non-convex non-degenerate strictly-simple polygon.
Figure 20.3 Straight skeleton of a complex shaggy contour
Figure 20.4 Other examples: A vertex-event (left), the case of several collinear edges (middle), and the case of a validly simple polygon with tangent edges (right).
## Angular Bisecting Lines and Offset Bisectors
Given two points and a line passing through them, the perpendicular line passing through the midpoint is the bisecting line (or bisector) of those points.
Two non-parallel lines, intersecting at a point, are bisected by two other lines passing through that intersection point.
Two parallel lines are bisected by another parallel line placed halfway in between.
Given just one line, any perpendicular line can be considered the bisecting line (any bisector of any two points along the single line).
The bisecting lines of two edges are the lines bisecting the supporting lines of the edges (if the edges are parallel or collinear, there is just one bisecting line).
The halfplane to the bounded side of the line supporting a contour edge is called the offset zone of the contour edge.
Given any number of contour edges (not necessarily consecutive), the intersection of their offset zones is called their combined offset zone.
Any two contour edges define an offset bisector, as follows: If the edges are non-parallel, their bisecting lines can be decomposed as 4 rays originating at the intersection of the supporting lines. Only one of these rays is contained in the combined offset zone of the edges (which one depends on the possible combinations of orientations). This ray is the offset bisector of the non-parallel contour edges.
If the edges are parallel (but not collinear) and have opposite orientation, the entire and unique bisecting line is their offset bisector. If the edges are parallel but have the same orientation, there is no offset bisector between them.
If the edges are collinear and have the same orientation, their offset bisector is given by a perpendicular ray to the left of the edges which originates at the midpoint of the combined complement of the edges. (The complement of an edge/segment are the two rays along its supporting line which are not the segment and the combined complement of N collinear segments is the intersection of the complements of each segment). If the edges are collinear but have opposite orientation, there is no offset bisector between them.
## Faces, Edges and Vertices
Each region of the partitioning defined by a straight skeleton is called a face. Each face is bounded by straight line segments, called edges. Exactly one edge per face is a contour edge (corresponds to a side of the polygon) and the rest of the edges, located in the interior of the polygon, are called skeleton edges, or bisectors.
The bisectors of the straight skeleton are segments of the offset bisectors as defined previously. Since an offset bisector is a ray of a bisecting line of 2 contour edges, each skeleton edge (or bisector) is uniquely given by two contour edges. These edges are called the defining contour edges of the bisector.
The intersection of the edges are called vertices. Although in a simple polygon, only 2 edges intersect at a vertex, in a straight skeleton, 3 or more edges intersect a any given vertex. That is, vertices in a straight skeleton have degree $$>=3$$.
A contour vertex is a vertex for which 2 of its incident edges are contour edges.
A skeleton vertex is a vertex who's incident edges are all skeleton edges.
A contour bisector is a bisector who's defining contour edges are consecutive. Such a bisector is incident upon 1 contour vertex and 1 skeleton vertex and touches the input polygon at exactly 1 endpoint.
An inner bisector is a bisector who's defining contour edges are not consecutive. Such a bisector is incident upon 2 skeleton vertices and is strictly contained in the interior of the polygon.
# Representation
This CGAL package represents a straight skeleton as a specialized Halfedge Data Structure (HDS) whose vertices embeds 2D Points (see the StraightSkeleton_2 concept in the reference manual for details).
Its halfedges, by considering the source and target points, implicitly embeds 2D oriented straight line segments (each halfedge per see does not embed a segment explicitly).
A face of the straight skeleton is represented as a face in the HDS. Both contour and skeleton edges are represented by pairs of opposite HDS halfedges, and both contour and skeleton vertices are represented by HDS vertices.
In a HDS, a border halfedge is a halfedge which is incident upon an unbounded face. In the case of the straight skeleton HDS, such border halfedges are oriented such that their left side faces outwards the polygon. Therefore, the opposite halfedge of any border halfedge is oriented such that its left side faces inward the polygon.
This CGAL package requires the input polygon (with holes) to be non-degenerate, strictly-simple, and oriented counter-clockwise.
The skeleton halfedges are oriented such that their left side faces inward the region they bound. That is, the vertices (both contour and skeleton) of a face are circulated in counter-clockwise order. There is one and only one contour halfedge incident upon any face.
The contours of the input polygon are traced by the border halfedges of the HDS (those facing outward), but in the opposite direction. That is, the vertices of the contours can only by traced from the straight skeleton data structure by circulating the border halfedges, and the resulting vertex sequence will be reversed w.r.t the input vertex sequence.
A skeleton edge, according to the definition given in the previous section, is defined by 2 contour edges. In the representation, each one of the opposite halfedges that represent a skeleton edge is associated with one of the opposite halfedges that correspond to one of its defining contour edges. Thus, the 2 opposite halfedges of a skeleton edge link the edge to its 2 defining contour edges.
Starting from any border contour halfedge, circulating the structure walks through border counter halfedges and traces the vertices of the polygon's contours (in opposite order).
Starting from any non-border but contour halfedge, circulating the structure walks counter-clockwise around the face corresponding to that contour halfedge. The vertices around a face always describe a non-convex non-degenerate strictly-simple polygon.
A vertex is the intersection of contour and/or skeleton edges. Since a skeleton edge is defined by 2 contour edges, any vertex is itself defined by a unique set of contour edges. These are called the defining contour edges of the vertex.
A vertex is identified by it's set of defining contour edges. Two vertices are distinct if they have differing sets of defining contour edges. Note that vertices can be distinct even if they are geometrically embedded at the same point.
The degree of a vertex is the number of halfedges around the vertex incident upon (pointing to) the vertex. As with any halfedge data structure, there is one outgoing halfedge for each incoming (incident) halfedge around a vertex. The degree of the vertex counts only incoming (incident) halfedges.
In a straight skeleton, the degree of a vertex is not only the number of incident halfedges around the vertex but also the number of defining contour halfedges. The vertex itself is the point where all the defining contour edges simultaneously collide.
Contour vertices have exactly two defining contour halfedges, which are the contour edges incident upon the vertex; and 3 incident halfedges. One and only one of the incident halfedges is a skeleton halfedge. The degree of a contour vertex is exactly 3.
Skeleton vertices have at least 3 defining contour halfedges and 3 incident skeleton halfedges. If more than 3 edges collide simultaneously at the same point and time (like in any regular polygon with more than 3 sides), the corresponding skeleton vertex will have more than 3 defining contour halfedges and incident skeleton halfedges. That is, the degree of a skeleton vertex is $$>=3$$ (the algorithm initially produces nodes of degree 3 but in the end all coincident nodes are merged to form higher degree nodes). All halfedges incident upon a skeleton vertex are skeleton halfedges.
The defining contour halfedges and incident halfedges around a vertex can be traced using the circulators provided by the vertex class. The degree of a vertex is not cached and cannot be directly obtained from the vertex, but you can calculate this number by manually counting the number of incident halfedges around the vertex.
Each vertex stores a 2D point and a time, which is the euclidean distance from the vertex's point to the lines supporting each of the defining contour edges of the vertex (the distance is the same to each line). Unless the polygon is convex, this distance is not equidistant to the edges, as in the case of a Medial Axis, therefore, the time of a skeleton vertex does not correspond to the distance from the polygon to the vertex (so it cannot be used to obtain the deep of a region in a shape, for instance).
If the polygon is convex, the straight skeleton is exactly equivalent to the polygon's Voronoi diagram and each vertex time is the equidistance to the defining edges.
Contour vertices have time zero.
Figure 20.5 Straight Skeleton Data Structure
# API
## Create a Straight Skeleton
The straight skeleton data structure is defined by the StraightSkeleton_2 concept and modeled in the Straight_skeleton_2<Traits,Items,Alloc> class.
The simplest way to construct a straight skeleton is via the free functions create_interior_straight_skeleton_2() and create_exterior_straight_skeleton_2(), as shown in the following example:
#include<boost/shared_ptr.hpp>
#include<CGAL/Exact_predicates_inexact_constructions_kernel.h>
#include<CGAL/Polygon_2.h>
#include<CGAL/create_straight_skeleton_2.h>
#include "print.h"
typedef K::Point_2 Point ;
typedef CGAL::Polygon_2<K> Polygon_2 ;
typedef boost::shared_ptr<Ss> SsPtr ;
int main()
{
Polygon_2 poly ;
poly.push_back( Point(-1,-1) ) ;
poly.push_back( Point(0,-12) ) ;
poly.push_back( Point(1,-1) ) ;
poly.push_back( Point(12,0) ) ;
poly.push_back( Point(1,1) ) ;
poly.push_back( Point(0,12) ) ;
poly.push_back( Point(-1,1) ) ;
poly.push_back( Point(-12,0) ) ;
// You can pass the polygon via an iterator pair
SsPtr iss = CGAL::create_interior_straight_skeleton_2(poly.vertices_begin(), poly.vertices_end());
// Or you can pass the polygon directly, as below.
// To create an exterior straight skeleton you need to specify a maximum offset.
double lMaxOffset = 5 ;
SsPtr oss = CGAL::create_exterior_straight_skeleton_2(lMaxOffset, poly);
print_straight_skeleton(*iss);
print_straight_skeleton(*oss);
return 0;
}
The input to these functions is the polygon, which can be given as an iterator pair or directly as a Polygon_2 object. In the case of the exterior skeleton, a maximum offset must be specified as well (see the ref manual for details on this max offset parameter).
## Create a Straight Skeleton from a Polygon With Holes
If Polygon_with_holes_2 is used, you can pass an instance of it directly to the function creating the interior skeleton, as shown below. Notice that a different header must be included in this case.
#include<boost/shared_ptr.hpp>
#include<CGAL/Exact_predicates_inexact_constructions_kernel.h>
#include<CGAL/Polygon_with_holes_2.h>
#include<CGAL/create_straight_skeleton_from_polygon_with_holes_2.h>
#include "print.h"
typedef K::Point_2 Point ;
typedef CGAL::Polygon_2<K> Polygon_2 ;
typedef CGAL::Polygon_with_holes_2<K> Polygon_with_holes ;
typedef boost::shared_ptr<Ss> SsPtr ;
int main()
{
Polygon_2 outer ;
outer.push_back( Point(-1,-1) ) ;
outer.push_back( Point(0,-12) ) ;
outer.push_back( Point(1,-1) ) ;
outer.push_back( Point(12,0) ) ;
outer.push_back( Point(1,1) ) ;
outer.push_back( Point(0,12) ) ;
outer.push_back( Point(-1,1) ) ;
outer.push_back( Point(-12,0) ) ;
Polygon_2 hole ;
hole.push_back( Point(-1,0) ) ;
hole.push_back( Point(0,1 ) ) ;
hole.push_back( Point(1,0 ) ) ;
hole.push_back( Point(0,-1) ) ;
Polygon_with_holes poly( outer ) ;
print_straight_skeleton(*iss);
return 0;
}
## Create Offset Polygons from a Straight Skeleton
If you already have a straight skeleton instance, the simpler way to generate offset polygons is to call create_offset_polygons_2() as shown in the next example, passing the desired offset and the straight skeleton. You can reuse the same skeleton to generate offsets at a different distance, which is recommended because producing the straight skeleton is much slower then generating offset polygons.
#include<vector>
#include<boost/shared_ptr.hpp>
#include<CGAL/Exact_predicates_inexact_constructions_kernel.h>
#include<CGAL/Polygon_2.h>
#include<CGAL/create_offset_polygons_2.h>
#include "print.h"
typedef K::Point_2 Point ;
typedef CGAL::Polygon_2<K> Polygon_2 ;
typedef boost::shared_ptr<Polygon_2> PolygonPtr ;
typedef boost::shared_ptr<Ss> SsPtr ;
typedef std::vector<PolygonPtr> PolygonPtrVector ;
int main()
{
Polygon_2 poly ;
poly.push_back( Point(-1,-1) ) ;
poly.push_back( Point(0,-12) ) ;
poly.push_back( Point(1,-1) ) ;
poly.push_back( Point(12,0) ) ;
poly.push_back( Point(1,1) ) ;
poly.push_back( Point(0,12) ) ;
poly.push_back( Point(-1,1) ) ;
poly.push_back( Point(-12,0) ) ;
double lOffset = 1 ;
PolygonPtrVector offset_polygons = CGAL::create_offset_polygons_2<Polygon_2>(lOffset,*ss);
print_polygons(offset_polygons);
return 0;
}
## Create Offset Polygons from a Polygon (With or Without Holes)
If you need offset polygons at just one single distance, you can hide away the construction of the straight skeleton by calling directly the functions create_interior_skeleton_and_offset_polygons_2() and create_exterior_skeleton_and_offset_polygons_2() as shown in the following examples:
#include<vector>
#include<boost/shared_ptr.hpp>
#include<CGAL/Exact_predicates_inexact_constructions_kernel.h>
#include<CGAL/Polygon_2.h>
#include<CGAL/create_offset_polygons_2.h>
#include "print.h"
typedef K::FT FT ;
typedef K::Point_2 Point ;
typedef CGAL::Polygon_2<K> Polygon_2 ;
typedef boost::shared_ptr<Polygon_2> PolygonPtr ;
typedef boost::shared_ptr<Ss> SsPtr ;
typedef std::vector<PolygonPtr> PolygonPtrVector ;
int main()
{
Polygon_2 poly ;
poly.push_back( Point(-1,-1) ) ;
poly.push_back( Point(0,-12) ) ;
poly.push_back( Point(1,-1) ) ;
poly.push_back( Point(12,0) ) ;
poly.push_back( Point(1,1) ) ;
poly.push_back( Point(0,12) ) ;
poly.push_back( Point(-1,1) ) ;
poly.push_back( Point(-12,0) ) ;
FT lOffset = 1 ;
PolygonPtrVector inner_offset_polygons = CGAL::create_interior_skeleton_and_offset_polygons_2(lOffset,poly);
PolygonPtrVector outer_offset_polygons = CGAL::create_exterior_skeleton_and_offset_polygons_2(lOffset,poly);
print_polygons(inner_offset_polygons);
print_polygons(outer_offset_polygons);
return 0;
}
... and using a Polygon_with_holes_2 directly when available:
#include<vector>
#include<iterator>
#include<iostream>
#include<iomanip>
#include<string>
#include<boost/shared_ptr.hpp>
#include<CGAL/Exact_predicates_inexact_constructions_kernel.h>
#include<CGAL/Polygon_with_holes_2.h>
#include<CGAL/create_offset_polygons_from_polygon_with_holes_2.h>
#include "print.h"
typedef K::Point_2 Point ;
typedef CGAL::Polygon_2<K> Polygon_2 ;
typedef CGAL::Polygon_with_holes_2<K> Polygon_with_holes ;
typedef boost::shared_ptr<Polygon_2> PolygonPtr ;
typedef boost::shared_ptr<Ss> SsPtr ;
typedef std::vector<PolygonPtr> PolygonPtrVector ;
int main()
{
Polygon_2 outer ;
outer.push_back( Point(-1,-1) ) ;
outer.push_back( Point(0,-12) ) ;
outer.push_back( Point(1,-1) ) ;
outer.push_back( Point(12,0) ) ;
outer.push_back( Point(1,1) ) ;
outer.push_back( Point(0,12) ) ;
outer.push_back( Point(-1,1) ) ;
outer.push_back( Point(-12,0) ) ;
Polygon_2 hole ;
hole.push_back( Point(-1,0) ) ;
hole.push_back( Point(0,1 ) ) ;
hole.push_back( Point(1,0 ) ) ;
hole.push_back( Point(0,-1) ) ;
Polygon_with_holes poly( outer ) ;
double lOffset = 0.2 ;
PolygonPtrVector offset_polygons = CGAL::create_interior_skeleton_and_offset_polygons_2(lOffset,poly);
print_polygons(offset_polygons);
return 0;
}
If the input polygon has holes, there can be holes in the offset polygons. However, the polygons generated by all the offsetting functions shown before do not have any parent-hole relationship computed; that is, they just instances of Polygon_2 instead of Polygon_with_holes_2. If Polygon_with_holes_2 are available and you need the offsetting to produce them, you can call the function arrange_offset_polygons_2() passing the result of any of the offsetting functions described so far. That function arranges the offset polygons detecting and distributing holes within parents. As a shortcut, you can use the function create_interior_skeleton_and_offset_polygons_with_holes_2() as shown below:
#include<vector>
#include<boost/shared_ptr.hpp>
#include<CGAL/Exact_predicates_inexact_constructions_kernel.h>
#include<CGAL/Polygon_with_holes_2.h>
#include<CGAL/create_offset_polygons_from_polygon_with_holes_2.h>
#include "print.h"
typedef K::Point_2 Point ;
typedef CGAL::Polygon_2<K> Polygon_2 ;
typedef CGAL::Polygon_with_holes_2<K> PolygonWithHoles ;
typedef boost::shared_ptr<PolygonWithHoles> PolygonWithHolesPtr ;
typedef std::vector<PolygonWithHolesPtr> PolygonWithHolesPtrVector;
int main()
{
Polygon_2 outer ;
outer.push_back( Point( 0.0, 0.0) ) ;
outer.push_back( Point(10.0, 0.0) ) ;
outer.push_back( Point(10.0, 4.5) ) ;
outer.push_back( Point(12.0, 4.5) ) ;
outer.push_back( Point(12.0, 2.0) ) ;
outer.push_back( Point(16.0, 2.0) ) ;
outer.push_back( Point(16.0, 8.0) ) ;
outer.push_back( Point(12.0, 8.0) ) ;
outer.push_back( Point(12.0, 5.5) ) ;
outer.push_back( Point(10.0, 5.5) ) ;
outer.push_back( Point(10.0,10.0) ) ;
outer.push_back( Point( 0.0,10.0) ) ;
Polygon_2 hole ;
hole.push_back( Point(3.0,3.0) ) ;
hole.push_back( Point(3.0,7.0) ) ;
hole.push_back( Point(7.0,7.0) ) ;
hole.push_back( Point(7.0,3.0) ) ;
PolygonWithHoles poly( outer ) ;
double lOffset = 1 ;
PolygonWithHolesPtrVector offset_poly_with_holes = CGAL::create_interior_skeleton_and_offset_polygons_with_holes_2(lOffset,poly);
print_polygons_with_holes(offset_poly_with_holes);
return 0;
}
Consider an input polygon with parallel edges separated a distance $$2*t$$. If you produce an offset polygon at distance $$t$$, these parallel edges will just collapse each other and vanish from the result, keeping the output as a simple polygon, just like the input. However, if you request an offset polygon at a distance $$t-epsilon$$, the result will still be a simple polygon but with edges that are so close to each other that will almost intersect. If a kernel with exact constructions is used, the offsetting algorithm can guarantee that the output contains only simple polygons. However, if inexact constructions are used the roundoff in the coordinates of the output points will cause parallel edges that almost collapse-but not so-to become really collinear or even cross each other.
Thus, it is neccessary to use a kernel with exact constructions if offset polygons must be simple, yet computing a straight skeleton using that kernel is very slow, much more than computing the offset polygons. To help with this, it is possible to construct the straight skeleton using the recommended kernel Exact_predicates_inexact_constructions_kernel, then convert the skeleton to a different kernel via the function convert_straight_skeleton_2() and input the converted skeleton to the offsetting functions.
All the offsetting functions that take polygons as input (and create the straight skeleton under the hood) apply that optimization automatically: that is, the output polygons are defined over the same kernel of the input polygons, whatever that is, yet the straight skeleton is constructed with the faster recommended kernel and converted if necessary.
Notice how some of the examples above use Exact_predicates_exact_constructions_kernel. In all cases, the straight skeleton is constructed using Exact_predicates_inexact_constructions_kernel.
## Low level API
All the high level functions described above are just wrappers around the low level API described here. This low level API is richer and provides options and configurations not covered by any of those functions.
The straight skeleton construction algorithm is encapsulated in the class Straight_skeleton_builder_2<Gt,Ss,Visitor> which is parameterized on a geometric traits (class Straight_skeleton_builder_traits_2<Kernel>) and the Straight Skeleton class (Ss).
The offset contours construction algorithm is encapsulated in the class Polygon_offset_builder_2<Ss,Gt,Container> which is parameterized on the Straight Skeleton class (Ss), a geometric traits (class Polygon_offset_builder_traits_2<Kernel>) and a container type where the resulting offset polygons are generated.
To construct the straight skeleton of a polygon with holes the user must:
(1) Instantiate the straight skeleton builder.
(2) Enter one contour at a time, starting from the outer contour, via the method Straight_skeleton_builder_2::enter_contour(). The input polygon with holes must be non-degenerate, strictly-simple and counter-clockwise oriented (see the definitions at the beginning of this chapter). Collinear edges are allowed. The insertion order of each hole is unimportant but the outer contour must be entered first.
(3) Call Straight_skeleton_builder_2::construct_skeleton() once all the contours have been entered. You cannot enter another contour once the skeleton has been constructed.
To construct a set of inward offset contours the user must:
(1) Construct the straight skeleton of the source polygon with holes.
(2) Instantiate the polygon offset builder passing in the straight skeleton as a parameter.
(3) Call Polygon_offset_builder_2::construct_offset_contours() passing the desired offset distance and an output iterator that can store a boost::shared_ptr of Container instances into a resulting sequence (typically, a back insertion iterator)
Each element in the resulting sequence is an offset contour, given by a boost::shared_ptr holding a dynamically allocated instance of the Container type. Such a container can be any model of the VertexContainer_2 concept, for example, a Polygon_2, or just a std::vector of 2D points.
The resulting sequence of offset contours can contain both outer and inner contours. Each offset hole (inner offset contour) would logically belong in the interior of some of the outer offset contours. However, this algorithm returns a sequence of contours in arbitrary order and there is no indication whatsoever of the parental relationship between inner and outer contours.
On the other hand, each outer contour is counter-clockwise oriented while each hole is clockwise-oriented. And since offset contours do form simple polygons with holes, it is guaranteed that no hole will be inside another hole, no outer contour will be inside any other contour, and each hole will be inside exactly 1 outer contour.
Parental relationships are not automatically reconstructed by this algorithm because this relation is not directly given by the input polygon with holes and must be done as a post processing step. The function arrange_offset_polygons_2() can be used to do that efficiently.
A user can reconstruct the parental relationships as a post processing operation by testing each inner contour (which is identified by being clockwise) against each outer contour (identified as being counter-clockwise) for insideness.
This algorithm requires exact predicates but not exact constructions Therefore, the Exact_predicates_inexact_constructions_kernel should be used.
#include<vector>
#include<iterator>
#include<iostream>
#include<iomanip>
#include<string>
#include<boost/shared_ptr.hpp>
#include<CGAL/Polygon_2.h>
#include<CGAL/Exact_predicates_inexact_constructions_kernel.h>
#include<CGAL/Straight_skeleton_builder_2.h>
#include<CGAL/Polygon_offset_builder_2.h>
#include<CGAL/compute_outer_frame_margin.h>
#include "print.h"
//
// This example illustrates how to use the CGAL Straight Skeleton package
// to construct an offset contour on the outside of a polygon
//
// This is the recommended kernel
typedef Kernel::Point_2 Point_2;
typedef CGAL::Polygon_2<Kernel> Contour;
typedef boost::shared_ptr<Contour> ContourPtr;
typedef std::vector<ContourPtr> ContourSequence ;
typedef Ss::Halfedge_iterator Halfedge_iterator;
typedef Ss::Halfedge_handle Halfedge_handle;
typedef Ss::Vertex_handle Vertex_handle;
int main()
{
// A start-shaped polygon, oriented counter-clockwise as required for outer contours.
Point_2 pts[] = { Point_2(-1,-1)
, Point_2(0,-12)
, Point_2(1,-1)
, Point_2(12,0)
, Point_2(1,1)
, Point_2(0,12)
, Point_2(-1,1)
, Point_2(-12,0)
} ;
std::vector<Point_2> star(pts,pts+8);
// We want an offset contour in the outside.
// Since the package doesn't support that operation directly, we use the following trick:
// (1) Place the polygon as a hole of a big outer frame.
// (2) Construct the skeleton on the interior of that frame (with the polygon as a hole)
// (3) Construc the offset contours
// (4) Identify the offset contour that corresponds to the frame and remove it from the result
double offset = 3 ; // The offset distance
// First we need to determine the proper separation between the polygon and the frame.
// We use this helper function provided in the package.
boost::optional<double> margin = CGAL::compute_outer_frame_margin(star.begin(),star.end(),offset);
// Proceed only if the margin was computed (an extremely sharp corner might cause overflow)
if ( margin )
{
// Get the bbox of the polygon
CGAL::Bbox_2 bbox = CGAL::bbox_2(star.begin(),star.end());
// Compute the boundaries of the frame
double fxmin = bbox.xmin() - *margin ;
double fxmax = bbox.xmax() + *margin ;
double fymin = bbox.ymin() - *margin ;
double fymax = bbox.ymax() + *margin ;
// Create the rectangular frame
Point_2 frame[4]= { Point_2(fxmin,fymin)
, Point_2(fxmax,fymin)
, Point_2(fxmax,fymax)
, Point_2(fxmin,fymax)
} ;
// Instantiate the skeleton builder
SsBuilder ssb ;
// Enter the frame
ssb.enter_contour(frame,frame+4);
// Enter the polygon as a hole of the frame (NOTE: as it is a hole we insert it in the opposite orientation)
ssb.enter_contour(star.rbegin(),star.rend());
// Construct the skeleton
boost::shared_ptr<Ss> ss = ssb.construct_skeleton();
// Proceed only if the skeleton was correctly constructed.
if ( ss )
{
print_straight_skeleton(*ss);
// Instantiate the container of offset contours
ContourSequence offset_contours ;
// Instantiate the offset builder with the skeleton
OffsetBuilder ob(*ss);
// Obtain the offset contours
ob.construct_offset_contours(offset, std::back_inserter(offset_contours));
// Locate the offset contour that corresponds to the frame
// That must be the outmost offset contour, which in turn must be the one
// with the largetst unsigned area.
ContourSequence::iterator f = offset_contours.end();
double lLargestArea = 0.0 ;
for (ContourSequence::iterator i = offset_contours.begin(); i != offset_contours.end(); ++ i )
{
double lArea = CGAL_NTS abs( (*i)->area() ) ; //Take abs() as Polygon_2::area() is signed.
if ( lArea > lLargestArea )
{
f = i ;
lLargestArea = lArea ;
}
}
// Remove the offset contour that corresponds to the frame.
offset_contours.erase(f);
print_polygons(offset_contours);
}
}
return 0;
}
## Exterior Skeletons and Exterior Offset Contours
This CGAL package can only construct the straight skeleton and offset contours in the interior of a polygon with holes. However, constructing exterior skeletons and exterior offsets is possible:
Say you have some polygon made of 1 outer contour C0 and 1 hole C1, and you need to obtain some exterior offset contours.
The interior region of a polygon with holes is connected while the exterior region is not: there is an unbounded region outside the outer contour, and one bounded region inside each hole. To construct an offset contour you need to construct an straight skeleton. Thus, to construct exterior offset contours for a polygon with holes, you need to construct, separately, the exterior skeleton of the outer contour and the interior skeleton of each hole.
Constructing the interior skeleton of a hole is directly supported by this CGAL package; you just need to input the hole's vertices in reversed order as if it were an outer contour.
Constructing the exterior skeleton of the outer contour is possible by means of the following trick: place the contour as a hole of a big rectangle (call it frame). If the frame is sufficiently separated from the contour, the resulting skeleton will be practically equivalent to a real exterior skeleton.
To construct exterior offset contours in the inside of each hole you just use the skeleton constructed in the interior, and, if required, revert the orientation of each resulting offset contour.
Constructing exterior offset contours in the outside of the outer contour is just a little bit more involved: Since the contour is placed as a hole of a frame, you will always obtain 2 offset contours for any given distance; one is the offseted frame and the other is the offseted contour. Thus, from the resulting offset contour sequence, you always need to discard the offsetted frame, easily identified as the offset contour with the largest area.
It is necessary to place the frame sufficiently away from the contour. If it is not, it could occur that the outward offset contour collides and merges with the inward offset frame, resulting in 1 instead of 2 offset contours.
However, the proper separation between the contour and the frame is not directly given by the offset distance at which you want the offset contour. That distance must be at least the desired offset plus the largest euclidean distance between an offset vertex and its original.
This CGAL packages provides a helper function to compute the required separation: compute_outer_frame_margin()
If you use this function to place the outer frame you are guaranteed to obtain an offset contour corresponding exclusively to the frame, which you can always identify as the one with the largest area and which you can simple remove from the result (to keep just the relevant outer contours).
Figure 20.6 Exterior skeleton obtained using a frame (left) and 2 sample exterior offset contours (right)
# Straight Skeletons, Medial Axis and Voronoi Diagrams
The straight skeleton of a polygon is similar to the medial axis and the Voronoi diagram of a polygon in the way it partitions it; however, unlike the medial axis and Voronoi diagram, the bisectors are not equidistant to its defining edges but to the supporting lines of such edges. As a result, Straight Skeleton bisectors might not be located in the center of the polygon and so cannot be regarded as a proper Medial Axis in its geometrical meaning.
On the other hand, only reflex vertices (whose internal angle $$> \pi$$) are the source of deviations of the bisectors from its center location. Therefore, for convex polygons, the straight skeleton, the medial axis and the Voronoi diagram are exactly equivalent, and, if a non-convex polygon contains only vertices of low reflexivity, the straight skeleton bisectors will be placed nearly equidistant to their defining edges, producing a straight skeleton pretty much alike a proper medial axis.
# Usages of the Straight Skeletons
The most natural usage of straight skeletons is offsetting: growing and shrinking polygons (provided by this CGAL package).
Another usage, perhaps its very first, is roof design: The straight skeleton of a polygonal roof directly gives the layout of each tent. If each skeleton edge is lifted from the plane a height equal to its offset distance, the resulting roof is "correct" in that water will always fall down to the contour edges (roof border) regardless of were in the roof it falls. [5] gives an algorithm for roof design based on the straight skeleton.
Just like medial axes, 2D straight skeletons can also be used for 2D shape description and matching. Essentially, all the applications of image-based skeletonization (for which there is a vast literature) are also direct applications of the straight skeleton, specially since skeleton edges are simply straight line segments.
Consider the subgraph formed only by inner bisectors (that is, only the skeleton halfedges which are not incident upon a contour vertex). Call this subgraph a skeleton axis. Each node in the skeleton axis whose degree is $$>=3$$ roots more than one skeleton tree. Each skeleton tree roughly corresponds to a region in the input topologically equivalent to a rectangle; that is, without branches. For example, a simple letter "H" would contain 2 higher degree nodes separating the skeleton axis in 5 trees; while the letter "@" would contain just 1 higher degree node separating the skeleton axis in 2 curly trees.
Since a skeleton edge is a 2D straight line, each branch in a skeleton tree is a polyline. Thus, the path-length of the tree can be directly computed. Furthermore, the polyline for a particular tree can be interpolated to obtain curve-related information.
Pruning each skeleton tree cutting off branches whose length is below some threshold; or smoothing a given branch, can be used to reconstruct the polygon without undesired details, or fit into a particular canonical shape.
Each skeleton edge in a skeleton branch is associated with 2 contour edges which are facing each other. If the polygon has a bottleneck (it almost touches itself), a search in the skeleton graph measuring the distance between each pair of contour edges will reveal the location of the bottleneck, allowing you to cut the shape in two. Likewise, if two shapes are too close to each other along some part of their boundaries (a near contact zone), a similar search in an exterior skeleton of the two shapes at once would reveal the parts of near contact, allowing you to stitch the shapes. These cut and stitch operations can be directly executed in the straight skeleton itself instead of the input polygon (because the straight skeleton contains a graph of the connected contour edges).
# Straight Skeleton of a General Figure in the Plane
A straight skeleton can also be defined for a general multiply-connected planar directed straight-line graph [1] by considering all the edges as embedded in an unbounded region. The only difference is that in this case some faces will be only partially bounded.
The current version of this CGAL package can only construct the straight skeleton in the interior of a simple polygon with holes, that is it doesn't handle general polygonal figures in the plane. | 8,834 | 39,314 | {"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} | 3.625 | 4 | CC-MAIN-2024-26 | latest | en | 0.933694 |
https://ch.mathworks.com/matlabcentral/cody/problems/2402-area-of-a-square/solutions/1588536 | 1,610,765,027,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703499999.6/warc/CC-MAIN-20210116014637-20210116044637-00041.warc.gz | 268,276,276 | 16,750 | Cody
# Problem 2402. Area of a Square
Solution 1588536
Submitted on 20 Jul 2018 by mohaimin rahman
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
r = 2; y_correct = 16; assert(isequal(area_sqr(r),y_correct))
2 Pass
r = 3; y_correct = 36; assert(isequal(area_sqr(r),y_correct))
3 Pass
r = 25; y_correct = 2500; assert(isequal(area_sqr(r),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 176 | 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} | 2.53125 | 3 | CC-MAIN-2021-04 | latest | en | 0.695191 |
https://www.physicsforums.com/threads/calculus-sequence-limits-n-infinity-n-n-n-use-sandwich-rule.689617/ | 1,531,707,053,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589029.26/warc/CC-MAIN-20180716002413-20180716022413-00317.warc.gz | 977,915,859 | 14,381 | # Homework Help: [Calculus] Sequence Limits: n -> infinity (n!/n^n)(Use Sandwich Rule?)
1. May 3, 2013
### raaznar
1. The problem statement, all variables and given/known data
Use sandwich Rule to find the limit lim n> infinity (a_n) of the sequences, for which the nth term, a_n, is given.
2. Relevant equations
$^{lim}_{n\rightarrow∞}\frac{n!}{n^{n}}$
3. The attempt at a solution
I know by just looking at it, n^n Approaches infinity much faster than n! which results in limit approaching 0, which is the answer. But the question says to use Sandwich Rule? I don't know which 2 functions to use to bound n!/n^n between? Usually if there was a sin function, I could start with it between -1 and 1. But I don't know where to start for this question?
2. May 3, 2013
### Dick
Expand it. n!/n^n=(n/n)*((n-1)/n)*((n-2)/n)*...*(3/n)*(2/n)*(1/n). Does that give you any ideas? | 263 | 879 | {"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} | 3.53125 | 4 | CC-MAIN-2018-30 | latest | en | 0.898768 |
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# Propeller torque & engine torque
Tech Log The very best in practical technical discussion on the web
# Propeller torque & engine torque
26th Mar 2012, 02:50
Join Date: Jun 2008
Location: Hong Kong
Age: 35
Posts: 5
Propeller torque & engine torque
Hi guys,
Just a very quick question, what is the relationship between the propeller torque and the engine torque? Are they actually referring to the same rotational force but only different in teminology?
Hope you could shed me some lights.
Andy
26th Mar 2012, 02:56
Join Date: Sep 2002
Location: La Belle Province
Posts: 2,113
One might expect the engine torque to be that measured at the engine, and the propellor torque that measured at the propellor (i.e. they are measured on either side of the gearbox/transmission system).
That would be analagous to brake horsepower and shaft horsepower.
26th Mar 2012, 04:58
Join Date: Jun 2010
Age: 32
Posts: 382
Like Mad Scientist said, torque is torque. In the case you mentioned, one is measured at the propeller and one at the engine (crankshaft). Propeller torque will be less than engine torque, mainly due to friction.
26th Mar 2012, 05:08
Join Date: Oct 2010
Location: New Zealand
Age: 29
Posts: 88
propeller torque is resisting the rotation of the propeller. Engine torque is generating the rotation of the propeller. If the propeller is at a constant RPM then these are equal and opposite
26th Mar 2012, 06:01
Join Date: Jun 2010
Age: 32
Posts: 382
propeller torque is resisting the rotation of the propeller. Engine torque is generating the rotation of the propeller.
Ah yes, that might be what he's referring to.
If the propeller is at a constant RPM then these are equal and opposite
Yes and no. When the engine is rotating at any RPM, constant or not, it is creating a turning moment (torque) on the airplane. This is due to the propeller drag. If the propeller is rotating clockwise, it wants to rotate the airplane counter-clockwise.
What you're talking about is engine inertia which only produces a turning moment (torque) when the engine is accelerating (positive or negative).
Roll-Wise Torque Budget [Ch. 9 of See How It Flies]
Paragraph 9.5 and 9.6 talk about that.
26th Mar 2012, 13:55
Join Date: Nov 2005
Location: Zulu Time Zone
Posts: 638
italia:
What you're talking about is engine inertia which only produces a turning moment
No I really don't think that is what morrisman means. He is talking about the fact that if the RPM is stable, then that is because thrust horsepower equals brake horsepower at that moment in time.
-------
tcyandy:
What I would add to the answers above is that the two torques are different due to the presence of the reduction gearbox. The slower turning shaft - ie the prop - will have a higher torque than on the engine side. The power will be the same though as the relationship is: power = torque x rpm
As an aside, this relationship of shaft speed, torque and power transmitted is illustrated by the spindly drive shafts you can see on the tail boom of a helo. The rpms are stepped up to reduce torque and enable a smaller shaft diameter, then reduced again at the tail assembly.
Last edited by oggers; 26th Mar 2012 at 14:18.
26th Mar 2012, 14:46
Join Date: Jun 2010
Age: 32
Posts: 382
No I really don't think that is what morrisman means.
So when he said "propeller torque" and "engine torque".... he really meant "thrust horsepower" and "brake horsepower"?
He is talking about the fact that if the RPM is stable, then that is because thrust horsepower equals brake horsepower at that moment in time.
Thrust horsepower is equal to thrust multiplied by velocity (TAS). You could be creating 200 BHP, at a steady RPM, while stopped on the ground and your thrust horsepower would be zero.
26th Mar 2012, 15:43
Join Date: Nov 2005
Location: Zulu Time Zone
Posts: 638
italia:
So when he said "propeller torque" and "engine torque".... he really meant "thrust horsepower" and "brake horsepower"?
No, you don't get it. Torque of shaft x revs of shaft = power, so it's the same as THP and BHP. I very much doubt he was talking about inertia, the way you asserted.
You could be creating 200 BHP, at a steady RPM, while stopped on the ground and your thrust horsepower would be zero.
Not that myth again. It's disappointing to have to point out to one who calls himself an instructor that in your 'brakes on' scenario the aircraft is still producing 200 THP as well as 200 BHP because it is accelerating a mass of air rearwards in a futile attempt to turn the earth and the atmosphere in opposite directions. Come on italia, pull your socks up. These are fundamentals that instructors should have a grip on.
Last edited by oggers; 26th Mar 2012 at 16:08.
26th Mar 2012, 16:00
Join Date: Jun 2010
Age: 32
Posts: 382
Not that myth again. It's disappointing to have to point out to one who calls himself an instructor that in your 'brakes on' scenario the aircraft is still producing 200 THP as well as 200 BHP because it is accelerating a mass of air rearwards in a futile attempt to turn the earth and the atmosphere in opposite directions. Come on italia, pull your socks up. These are fundamentals that instructors should have a grip on.
Very quick to throw the insults out. Yes, I am a flight instructor. Here is today's lesson:
When talking about performance, the paragraph below is appropriate.
"Power is the rate of doing work, and work is a force times a distance. Power required (PR) is the amount of power that is required to produce thrust required. PR is the product of TR and velocity (V). If V is expressed in knots, then the product of TR and V must be divided by 325 to give power in units of horsepower. Thus, thrust horsepower only depends on thrust and velocity."
If you're talking about propeller efficiency, the paragraph below is appropriate.
"In a turboprop, power available is determined by the performance of the engine/propeller combination. Engine output is called shaft horsepower (SHP). Thrust horsepower (THP) is propeller output, or the power that is converted to usable thrust by the propeller. The ability of the propeller to turn engine output into thrust is given by its propeller efficiency (p.e.). Under ideal conditions, SHP would equal THP, but due to friction in the gearbox and propeller drag, THP is always less than SHP. Propeller efficiency is always less than 100%."
So, with regard to your statements:
He is talking about the fact that if the RPM is stable, then that is because thrust horsepower equals brake horsepower at that moment in time.
...the aircraft is still producing 200 THP as well as 200 BHP...
THP will never equal SHP or BHP.
Those two paragraph quotes were taken directly from the "Fundamentals of Aerodynamics" which was prepared by the U.S. Naval Aviation Schools Command.
EDIT: Here is a picture taken from the same textbook. imgur: the simple image sharer
26th Mar 2012, 16:12
Join Date: Nov 2005
Location: Zulu Time Zone
Posts: 638
I'll respond to that tomorrow if nobody picks it up in the meantime BTW I added some more to my last post - just so you know.
26th Mar 2012, 16:25
Join Date: Jun 2010
Age: 32
Posts: 382
No, you don't get it. Torque of shaft x revs of shaft = power, so it's the same as THP and BHP. I very much doubt he was talking about inertia, the way you asserted.
I understand it very well. We should probably wait until he responds to see what he meant!
I'd recommend checking out this article to see the difference between power and torque. If you realized the difference, you would see that they aren't interchangeable.
Power and Torque.pdf - File Shared from Box - Free Online File Storage
It will also help you understand why THP is equal to zero when stopped on the ground. Power = Force x velocity
Last edited by italia458; 26th Mar 2012 at 17:10.
26th Mar 2012, 16:31
Join Date: Dec 2001
Location: England
Posts: 1,384
propeller torque is resisting the rotation of the propeller. Engine torque is generating the rotation of the propeller. If the propeller is at a constant RPM then these are equal and opposite
That depends on how you define engine torque, eg before or after any gearbox.
If you define it as after the gearbox you are correct (otherwise the aircraft would roll).
If define it as before the gearbox (eg between crankshaft and airframe) then it's normally less than the prop torque due to use of reduction gearbox.
26th Mar 2012, 16:59
Join Date: Jun 2010
Age: 32
Posts: 382
If you define it as after the gearbox you are correct (otherwise the aircraft would roll).
Propeller torque is measured after the gearbox. So measuring 'engine torque after the gearbox' is propeller torque! Both would be measured in the same place so they would obviously be the same.
I think some of the confusion is related to the ambiguous term, "torque". Torque is defined as the tendency of a force to rotate an object about an axis, fulcrum, or pivot. That means ANY force and there are lots of forces in play when an airplane is in flight. As I mentioned above, propeller drag constitutes a torque and is sometimes understood as propeller torque, mostly to people who have only flown direct drive propeller engines. I think tcyandy needs to clarify exactly what he is asking so that he can get a clear answer.
27th Mar 2012, 10:34
Join Date: Nov 2005
Location: Zulu Time Zone
Posts: 638
Italia:
For the sake of clarity I agree that THP will not equal BHP because of prop efficiency (I should have been more careful to specify SHP in front of a pedant) and frictional losses in the intermediate gearing. But that is just semantics, beside the principle morrissman was getting at, which was I believe:
At constant RPM the power required at the prop shaft = power provided by the crankshaft.
The torque, were it to be measured on each shaft, would be different however. Really quite simple. He wasn’t talking about the inertia tangent you went off on.
Now, about this stopped aircraft knocking out 200 BHP but no THP because it doesn’t have any velocity nonsense. What about a helicopter in a hover? Any thrust there (and you can include frictional losses and rotor efficiency in your answer if it will make you happy)? The answer is pretty self-evident. The aircraft doesn’t have to move to produce thrust. Do you, as a flight instructor, agree with that or not?
Because when you say “Today’s lesson is…” you quote an extract from a credible text on aerodynamics. But you have taken the text completely out of context. Of course we can work out power required by taking drag and multiplying by airspeed! But that does not mean a stationary aircraft can do no work and therefore produce no THP!!! The work is done by moving the mass of air one way and the mass of the Earth a tiny imperceptible amount the other way.
27th Mar 2012, 13:14
Join Date: Jun 2010
Age: 32
Posts: 382
At constant RPM the power required at the prop shaft = power provided by the crankshaft.
I can see that that might be exactly what he wanted to know. To me it seems obvious as Fnet=ma. When there is no acceleration (ie: steady RPM), there has to be no net force. This whole thing is ridiculous because from the beginning, the OP didn't ask a clear question. If what you wrote was what the OP asked if it was true, this thread would be done after someone wrote Fnet=ma.
Now, about this stopped aircraft knocking out 200 BHP but no THP because it doesn’t have any velocity nonsense. What about a helicopter in a hover? Any thrust there (and you can include frictional losses and rotor efficiency in your answer if it will make you happy)? The answer is pretty self-evident. The aircraft doesn’t have to move to produce thrust. Do you, as a flight instructor, agree with that or not?
I understand what you're trying to say but I can tell you that with regard to aircraft performance, what I said is correct. If you would like I can send you a copy of that U.S. Naval aerodynamics text. You're getting tied up on the word thrust and you've used it extensively throughout that paragraph and I agree with everything you said. There is lots of thrust when the helicopter is in the hover! And I agree the aircraft doesn't have to move to produce thrust. I would never teach a student otherwise.
However, you're not understanding what THP actually is. Thrust Horsepower is POWER! It is NOT thrust. Re-read my post with the paragraph explaining that and then look at the picture I included. Yes, the engine will be producing a certain SHP that will go to the propeller and produce a certain amount of thrust. But, THP is only related to the work the thrust does on the aircraft. Work = force x distance, therefore, if the aircraft isn't moving, it isn't covering distance and so the work = zero. When work = zero, Power also = zero. In the document I included, "Power and Torque", I broke down those equations so you should be able to see exactly how power relates to thrust(torque).
Because when you say “Today’s lesson is…” you quote an extract from a credible text on aerodynamics. But you have taken the text completely out of context.
How do you know the text is credible? You obviously don't have the text because you say that I quoted it out of context.
Of course we can work out power required by taking drag and multiplying by airspeed! But that does not mean a stationary aircraft can do no work and therefore produce no THP!!!
Oh you were so close! I thought you had it. Yes a stationary aircraft is doing zero work and it is producing no THP! I'm usually pretty careful to not state flat out that I know I'm correct but in this case I think I could come out and say that. I'm not going to continue to argue if it's true or not but if you would like to understand it, I have no problem going into detail and explaining it.
27th Mar 2012, 13:51
Join Date: Jan 2011
Location: England
Posts: 642
tcyandy
If you are studying for the JAA/EASA ATPL then the answer they are looking for is.
Propeller torque is resisting the rotation of the propeller. Engine torque is generating the rotation of the propeller. If the propeller is at a constant RPM then these are equal and opposite.
If you are studying something more technical then some of the other posts in this thread are more relevant.
27th Mar 2012, 14:33
Join Date: Nov 2005
Location: Zulu Time Zone
Posts: 638
How do you know the text is credible?
Because you told me it was "Fundamentals of Aerodynamics" produced by the US Naval Aviation Schools Command. That and the fact that drag x airspeed = power required is hardly controversial.
But you are waffling again. The bottom line is you believe:
You could be creating 200 BHP, at a steady RPM, while stopped on the ground and your thrust horsepower would be zero.
This is incorrect. If you are creating 200 BHP it's because you are moving a mass of air over a distance, and at a rate equivalent to 200 BHP x [efficiency of prop and transmission].
BTW I see Keith Williams has now posted exactly the same thing as morrissman. Would you make the same statement now to Keith as you did to morrissman? ie
What you're talking about is engine inertia which only produces a turning moment (torque) when the engine is accelerating (positive or negative).
?
27th Mar 2012, 15:42
Join Date: Jun 2010
Age: 32
Posts: 382
Because you told me it was "Fundamentals of Aerodynamics" produced by the US Naval Aviation Schools Command. That and the fact that drag x airspeed = power required is hardly controversial.
See, that's the difference between you and I. You seem to think that credibility is in a name (in general and in particular, in another thread). I believe credibility is in the truth or accuracy of the material. No one will ever convince me to believe in a name.
You seem to understand that Power required = drag x airspeed but you don't believe that Power available = thrust x airspeed?
This is incorrect. If you are creating 200 BHP it's because you are moving a mass of air over a distance, and at a rate equivalent to 200 BHP x [efficiency of prop and transmission].
After telling my that my statement about THP was incorrect, you provide one sentence that doesn't mention THP once! Did you really accomplish anything?
You are consistently highlighting your lack of understanding of the matter and yet you are forceful in your opinion that myself and all aeronautical engineers are wrong. Just because something seems correct, doesn't mean it's correct. I have maths and physics that prove that you are incorrect and yet you still try to convince me otherwise. Unless you can prove something wrong with the maths and physics used to illustrate THP (which I would love to see), you aren't proving anything worthwhile.
27th Mar 2012, 16:01
Join Date: Feb 2005
Location: flyover country USA
Age: 77
Posts: 4,580
IF tcyandy is talking about a direct-drive, ungeared engine, then there is ONLY ONE TORQUE to consider. It is the torque the engine crankshaft delivers to the prop.
You may call it prop torque, or engine torque, as you like - but they are THE SAME THING.
If there's a gearbox in the system, then there are an input torque (i.e. crankshaft) and an output torque (i.e. prop). If the gear ratio is 2:1, then rpm is halved, and torque is doubled (well, almost, minus a very small heat loss).
28th Mar 2012, 00:47
Join Date: Jun 2010
Age: 32
Posts: 382
oggers... To clarify further:
You seem to understand that Power required = drag x airspeed.
Drag is the same as Thrust required for level flight; it's basically the force that needs to be offset with thrust so the airplane can fly.... therefore,
Power required = thrust required x airspeed
And Power available is,
Power available = thrust available x airspeed
This picture shows it: http://i.imgur.com/R43ho.png
What might be confusing is the Thrust Horsepower term. It is still power and is talking about the power generated by the thrust that propels the aircraft forward. If the picture was showing the BHP or SHP as well, it would be virtually a straight line near the top of the graph.
Here is a picture I took from another textbook of mine that shows the THP available and required curves. It's not 100% clear but I think it shows enough: http://i.imgur.com/3e4Ru.jpg?1 | 4,357 | 18,241 | {"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} | 2.84375 | 3 | CC-MAIN-2019-39 | latest | en | 0.901181 |
http://www.cplusplus.com/forum/beginner/81820/ | 1,503,012,490,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104172.67/warc/CC-MAIN-20170817225858-20170818005858-00628.warc.gz | 504,211,851 | 5,116 | ### One small problem (very simple fix, but I don't know how) -
#include <iostream>
using namespace std;
void main()
{
double x;
double y;
double z;
char yn;
double tt;
double bs;
cout << "Welcome to xxxxxxx free calculator!" << endl << endl;
cout << "How many adults (age 14 or over) are in your party? " << endl;
cin >> x;
cout << endl;
cout << "How many children (age 5 to 13) are in your party? " << endl;
cin >> y;
cout << endl;
cout << "Do you have a family membership? (y/n) " << endl;
cin >> yn;
cout << endl;
cout << "How many train tickets would you like? " << endl;
cin >> tt;
cout << endl;
cout << "How many bird show tickets would you like? " << endl;
cin >> bs;
cout << endl;
cout << "Your total total fee is: ";
cout << endl;
if (yn == 'y');
{
x * 18 + y * 7.50 + tt * 6 + bs * 5 * .50;
}
cout << x * 18 + y * 7.50 + tt * 6 + bs * 5;
system("pause");
}
// In this program, if the user types "y" for 'if they are a member' it is supposed to give them a 50% discount.
I'm struggling with giving them the discount. I'm so close and I know I'm only one or two steps from finishing, but I can't figure out how to actually give them the discount! Any help would be appreciated.
I think there is something wrong with my if (yn == 'y') thing, I'm not sure what I'm doing wrong.
Simple:
``123456`` ``````if (yn == 'y'); { x * 18 + y * 7.50 + tt * 6 + bs * 5 * .50; // The compiler is told nothing // About what it has to do with this number. } cout << x * 18 + y * 7.50 + tt * 6 + bs * 5;``````
You should do something like:
``123456`` ``````float Value = x * 18 + i * 7.50 + tt * 6 + bs * 5; if(yn == 'y') { Value /= 2; } cout << Value;``````
I've never used float value before.
Essentially float Value takes whatever equation I have (in this case x * 18 + y * 7.50 + tt * 6 + bs * 5;) and it allows me to manipulate it with an 'if' statement?
Thanks for the new information, I've never used that before.
Last edited on
OP wrote:
allows me to manipulate it with an 'if' statement?
Floats are used for storing decimal values just like doubles, but they are 8 Bytes.
Thanks,
Aceix.
Last edited on
Allow Me, Floats are 4 Bytes, Doubles are 8 Bytes.
I Sincerely haven't noticed you was using Doubles, you can change that float to double then. But, as long as you don't need exhaustive floating-point precision, I personally think you should use floats, to let the CPU have the required relax while still producing the (almost) same value.
But, as long as you don't need exhaustive floating-point precision, I personally think you should use floats, to let the CPU have the required relax while still producing the (almost) same value.
My opinion is just the opposite. Unless you need the extra memory, you should prefer doubles over floats.
I haven't run benchmarks, but I would be very surprised if using floats in any way outperformed using doubles (in terms of CPU usage). AFAIK, both floats and doubles get scaled up to whatever register size is in the FPU when calculations are perfomed. If anything... working with doubles is probably faster (for the same reason that working with 'int's is generally faster than working with 'short's).
Topic archived. No new replies allowed. | 876 | 3,220 | {"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-2017-34 | latest | en | 0.89439 |
https://www.thestudentroom.co.uk/showthread.php?t=4552064 | 1,527,129,018,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865884.49/warc/CC-MAIN-20180524014502-20180524034502-00274.warc.gz | 842,386,100 | 44,123 | You are Here: Home >< Maths
# C3 differentiation watch
1. I've been trying to figure out where I've gone wrong for ages!! Can someone pls help, thank u
---CA Edited post so image was correct way up---
Attached Images
2. Ugh idk how to rotate it.. I guess this ones better
3. e^2x+2x^2e^2x
e^2x+2x^2e^2x
what about the lne^x^2 that goes at the end!!
5. When you differentiate e^x^2, there's no need to multiply by another x^2.
6. (Original post by pondsteps)
Ugh idk how to rotate it.. I guess this ones better
I did it for you - check your post
I recommend using imgur to upload images then post the link (including the file type e.g. ,jpg) inside [img] tags.
e^2x+2x^2e^2x
crappp i just realised thats one lol thank u
8. (Original post by pondsteps)
what about the lne^x^2 that goes at the end!!
what are you on about LN, there is no need for ln where your intergrating xe^2x
9. (Original post by notnek)
I did it for you - check your post
I recommend using imgur to upload images then post the link (including the file type e.g. ,jpg) inside [img] tags.
okay thank u!
what are you on about LN, there is no need for ln where your intergrating xe^2x
this was dy/dx so not really intergrating ?
11. (Original post by the bear)
this was dy/dx so not really intergrating ?
Sorry Hard, mistakes happen you know? like when hyperthreading my blackjack game, things just tend to fall apart
12. (Original post by pondsteps)
I've been trying to figure out where I've gone wrong for ages!! Can someone pls help, thank u
---CA Edited post so image was correct way up---
Using the product rule y=xe^x2
Let u= x and v= e^x2. Therefore, u'= 1 and v'= 2xe^x2
So using the product rule dy/dx= u.v' + v.u'
This mean dy/dx= (x) . (2xe^x2) + (e^x2) . (1)
This is equal to 2x2e^x2 + e^x2
Factorising this produces e^x2(2x2 + 1), hope this helps simplify it a little.
13. Maz A thanks so much ))
14. (Original post by pondsteps)
Maz A thanks so much ))
Any chance of a rep..? Jks you're very welcome glad it helped .
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https://www.ruby-forum.com/t/age-calculation-in-the-model/84254 | 1,716,354,701,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058531.78/warc/CC-MAIN-20240522040032-20240522070032-00795.warc.gz | 854,984,856 | 6,621 | # Age calculation in the model
Hi!
I’m at my wits end with this problem and hopefully someone can help
me.
I have code to calculate age based on date of death - date of birth.
I want the age to calculate down to the minute, which it is. However,
I have to account for all age units: minutes, hours, days, months, or
years. My problem is that I need to have if statements set up for
every scenario, but I want to exit the model when the correct age unit
has been established. Rails is continuting to evaluate the other if
statements, even when I’m checking for nil, and ends up returning 0 in
the case of an age unit less than a year. I have tried assembling the
if statements in a number of ways, including using a “return”
statement when the correct age unit has been calculated. This works,
but the model returns an error if there is more than one return
statement in existence. I just don’t know what else to try!! Any
help would be GREATLY appreciated!!
Here’s my code:
def age
dod = self.precase.report.dod
``````@age ||= begin
if not dateofbirth.nil?
if not dod.nil?
seconds = dod.to_time - dateofbirth.to_time
minutes = seconds / 60
m = minutes.to_i
if m.between?(1, 59)
m
end
hours = minutes / 60
h = hours.to_i
if not h.nil?
if h.between?(1, 23)
h
end
end
days = h / 24
d = days.to_i
if not d.nil?
if d.between?(1, 30)
d
end
end
months = days / 30
mths = months.to_i
if not mths.nil?
if mths.between?(1, 11)
mths
else
years = dod.year - dob.year-(dob.to_time.change(:year
``````
=> dod.year) > dod ? 1 : 0)
end
end
end
end
end
end
On Feb 2, 1:13 pm, “Ali” [email protected] wrote:
has been established. Rails is continuting to evaluate the other if
def age
m = minutes.to_i
h
end
end
end
end
end
end
end
Couple of things here…
1. use the ‘ruby-units’ gem to help with the unit conversions
2. use a ‘case’
require ‘rubygems’
require ‘ruby-units’
require ‘date’
dob = DateTime.new(1969,4,24,10,30,0)
dod = DateTime.new(2068,1,31,11,30,0)
age = dod.unit - dob.unit
@age = age.to case
when age < ‘1 min’.unit : ‘sec’
when age < ‘1 hr’.unit : ‘min’
when age < ‘1 day’.unit : ‘hour’
when age < ‘1 week’.unit : ‘day’
when age < ‘1 year’.unit : ‘week’
when age < ‘10 years’.unit : ‘year’
when age < ‘100 years’.unit : ‘decade’
else
‘century’
end
puts “#{‘%0.2f’ % @age}”
Note that ruby-units doesn’t define a ‘month’ unit (because it isn’t
fixed). But you can define one that is 30 days long if you like.
You also want to use the format string in the to_s or you get this…
15900642/15778463 century
Because DateTime likes to make things Rational.
If you want the output to match the units of another compatible unit
just pass it in the ‘to’ call
another_age = ‘77 year’.unit
age.to(another_age) #=> 98.8 year
As you can also see, if you have a string that defines your target
unit, you just need to do…
age.to(target_unit)
_Kevin
On Feb 2, 2:32 pm, “_Kevin” [email protected] wrote:
I want the age to calculate down to the minute, which it is. However,
help would be GREATLY appreciated!!
if not dod.nil?
h = hours.to_i
if not d.nil?
mths
Couple of things here…
dod = DateTime.new(2068,1,31,11,30,0)
when age < ‘100 years’.unit : ‘decade’
You also want to use the format string in the to_s or you get this…
age.to(another_age) #=> 98.8 year
As you can also see, if you have a string that defines your target
unit, you just need to do…
age.to(target_unit)
_Kevin
Almost forgot… ruby-units also supports ranges, so this works too
@age = age.to case age
when ‘0 sec’.unit …‘1 min’.unit : ‘sec’
when ‘1 min’.unit … ‘1 hr’.unit : ‘min’
when ‘1 hr’.unit … ‘1 day’.unit : ‘hour’
when ‘1 day’.unit … ‘1 week’.unit : ‘day’
when ‘1 week’.unit … ‘1 year’.unit : ‘week’
when ‘1 year’.unit … ‘10 years’.unit : ‘year’
when ‘10 years’.unit … ‘100 years’.unit : ‘decade’
else
‘century’
end
_Kevin
I used the following for number of seconds in each date part:
year = 365.25 * 60 * 60 * 24 = 31557600
month = year / 12 = 2629800
day = 60 * 60 * 24 = 86400
hour = 60 * 60 = 3600
minute = 60
second = 1
This function, then, returns a hash of the date parts:
def age_parts(in_seconds)
parts = Hash.new
factors = [[‘years’, 31557600],[‘months’, 2629800],[‘days’,
86400],[‘hours’, 3600],[‘minutes’,60],[‘seconds’,1]]
age = factors.collect do |unit, factor|
value, in_seconds = in_seconds.divmod(factor)
parts[unit] = value
end
parts
end
irb(main):187:0> parts = age_parts(1176095625)
=> {“days”=>6, “seconds”=>45, “minutes”=>43, “years”=>37, “hours”=>15,
“months”=>3}
irb(main):188:0> parts[‘years’]
=> 37
irb(main):189:0> parts[‘months’]
=> 3
irb(main):190:0> parts[‘days’]
=> 6
irb(main):191:0> parts[‘hours’]
=> 15
irb(main):192:0> parts[‘minutes’]
=> 43
irb(main):193:0> parts[‘seconds’]
=> 45 | 1,519 | 4,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-22 | latest | en | 0.808325 |
https://croteaurmt.ca/sheet-pan-rswj/6e81b7-haskell-int-overflow | 1,653,292,299,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662556725.76/warc/CC-MAIN-20220523071517-20220523101517-00627.warc.gz | 241,331,566 | 16,428 | Posted by. Period of rapid language and community growth 8. - Can compare two Integers not equal. Source: The Haskell Wikibook. If you are unfamiliar with Haskell's typeclasses, just know that you might have to tell Haskell what type you want to read it as: You don't always have to do this, if Haskell has some other way of knowing what type you want, like if you proceed to do arithmetic with it. - Can take abs of an Integer. Good evening, I am stuck for the last 3 days with this. - Can complement an Integer. (x) refers to size specifiers for Word How can you come out dry from the Sea of Knowledge. Usually in Haskell, then one uses Maybe Int, since then we can make a distinction between Nothing, and Just 0. When failing with a heap exhaustion, the RTS truncates the **reported** *current maximum heap size* modulo 2\^32, e.g. The benefit here is the automatic creation of a function to unwrap the newtype. So m < d*d was not only more efficient, but more idiomatic Haskell in that regard :) \$\endgroup\$ – Zeta Mar 18 '19 at 21:11. add a comment | Your Answer Thanks for contributing an answer to Code Review Stack Exchange! y > huge(x) - x) .or. 1993. The language is named for Haskell Brooks Curry, whose work in mathematical logic serves as a foundation for functional languages.Haskell is based on the lambda calculus, hence the lambda we use as a logo. I will not repeat the section "Standard Haskell Classes" from the Haskell Report and explain, why typeclasses for various numbers organized the way they are organized. The value is simply truncated to the How do I do this? Whether it was possible to to detect it. Yes, these are the type of Haskell also incorporates polymorphic types---types that areuniversally quantified in some way over all types. Summary Data.Bits.shiftL doc says shiftL should never be called with negative value. Also knowing what seq and ($!) For example, the type of the function getChar is:getChar :: IO Char The IO Char indicates that getChar, when invoked, performssome action which returns a character. In Brexit, what does "not compromise sovereignty" mean? I don't mean discrete/modular logs, but simply greatest x s/t b^x <= n. I developed a few versions and can't decide which I like best. complex type. Here is a simple Guessing Game I implemented in Haskell. I've looked at similar questions and I'm still not sure. However, making code tail-recursive in a lazy language is not quite the same as in a eager language. logical function adding_will_overflow (x,y) result(res) integer, intent(in) :: x, y res = (x > 0 .and. YHC 4. Yes, these are the type of things I think about. (Int, Int, Int) (Int, Int, Int, Int) Tuples can contain values of different types. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. But the C memset implementations in primitive take int arguments, so they overflow long before the bounds checks on allocation (which were also added in 0.11) will catch them on a 64 bit system. It seems that Haskell spec does not specify what to do with Int overflow, and leaves the decision to the compiler: silently truncate, raise exception, etc. The reference I always go back to when Pseudo-functions like rand() or getchar() in C, which return different results on each call, are simply impossible to write in Haskell. Consider the Unsigned Word8 type with a range of [0,255]. (Int -> String? Convert String to Integer/Float in Haskell? Int, which fixed-width machine-specific integers with a minimum guaranteed range of −2 29 to 2 29 − 1. This function takes two values, and returns the first one, ignoring the second. r/haskell: The Haskell programming language community. The compiler > can use an Int-like representation most of the time (say, when advised > by a pragma), with an overflow check on arithmetic operations. Such an exception can then be handled normally in the program. - Can compare two Integers equal. Learn more Converting Int to Integer [duplicate] \$\begingroup\$Using data Boiler = H | L | BL Int Int Int | BH Int Int Int deriving (Read, Show) instead of your definitions of Boiler and Tank will give you access to, for example, readLn :: IO Boiler, which turns an input line like "BH 3 3 4" into a Boiler value. ), Haskell String-> Int Type Conversion [duplicate]. Options Report abuse; New issue; Report abuse New issue Test.QuickCheck.Batch overflow in length of tests. (Those languages, however, are dynamically typed.) Glasgow Haskell Compiler; GHC; Issues #8695; Closed Open. Well the 2's complete representation of -100 is 10011100 which when treated as unsigned binary number is 156. 0.10 just lets you see uninitialized memory with new. Data.Word Data.Int> let a = 5 :: Word8, Prelude First, read Performance/Accumulating parameter. 2 years ago. Question: Can I have a generic numeric data type in Haskell which covers Integer, Rational, Double and so on, like it is done in scripting languages like Perl and MatLab? The result of float overflow. An unexpected code path, one that rarely but can happen and can be handled if needs be. things I think about. Opened Nov 29, 2006 by gwright@antiope.com @trac-gwright. produced and what was allowed or not. If you know that the string is a valid integer, or you don't mind it blowing up if that's not the case, read will work. Various sizes of signed and unsigned integers are available in Data.Int and Data.Word, respectively. Let's think about how to solve this. Why does US Code not allow a 15A single receptacle on a 20A circuit? union :: ([Int], Int) -> ([Int], Int) -> ([Int], Int) need documentation - it's not clear which Int has what purpose at all without examining the source. and define appropriate instances for Num class et. 10 Numbers. Copy-paste from Haskell Int and Integer The type declaration says, “Take a value of any type a, take another value of any type b (can be the same or different type as a), and return a value of type a”.The type after the final -> is the return type, and all the types before it are parameter types, also delimited by ->s. Th… integer overflow. JHC 5. nhc98 6. Ocaml has literals for infinity and negative infinity, but Scala and Haskell do not. Where does the 156 come from? The rules that hold for Enum instances over a bounded type such as Int (see the section of the Haskell report dealing with arithmetic sequences) also hold for the Enum instances over the various Int … I have a picture composed of . Foldable is the class of types t :: * -> * which admit a folding operation. The … So really quickly wrote out the imperative version: readInts :: String-> [Int] readInts x = read x main = do input <-getLine let list = read input :: [Int] print list. So, let's take a look at the following The important thing to watch is the life cycle of intermediate thunks, e.g., c is created at some point as a 1-level deep addition, then almost immediately reduced to a number out of necessity, before a later thunk d builds upon it. The integer in a fixity declaration must be in the range 0 to 9. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. The unreleased version is fixed to use appropriately sized types. In practice, its range can be much larger: on the x86-64 version of Glasgow Haskell Compiler, it can store any signed 64-bit integer. Integer can be considered as a superset of Int. 1997. – bheklilr Dec 18 '13 at 20:20. How can I add a few specific mesh (altitude-like level) curves to a plot? (http://en.wikipedia.org/wiki/Two's_complement), Now I looked for a means to detect if a overflow or out-of-range assignment had occurred but I could not find a mechanism in the exception handling or other documents. Essentially, this infinite sequence of applications of f will be avoided if (and only if) f is a lazyfunction. History: 1. 5. A fixity declaration may appear anywhere that a type signature appears and, like a type signature, declares a property of a particular operator. If exception-on-overflow is preferable, there's the 'checked' library on hackage that provides Data.Int/Word.Checked. Does this picture depict the conditions at a veal farm? You might be wondering: surely fix f will cause an infinite series of nested applications of fs: x = f x = f (f x) = f (f (f ( ... )))? overflow, truncation and type conversion behavior manifested itself In order to capture such generality in the simplest way possible we need a general Number type in Haskell, so that the signature of (+)would … Forexample, (forall a)[a] is the family of types consisting of,for every type a, the type of lists of a. Haskell - Types and Type Class - Haskell is a functional language and it is strictly typed, which means the data type used in the entire application will be known to the compiler at compile tim ... Int. Daily news and info about all things Haskell related: practical stuff, theory, types … Press J to jump to the feed. Say you were writing a library to do things on reddit, you would define an exception type in your API: Then later you might write try (login …) or catch (l… Also, you may find the Numbers section of A Gentle Introduction to Haskell useful. fixis simply defined as: Doesn't that seem ... magical? In contrast, again, last (scanl (+) 0 [1.. 1000000]) skips over to the last thunk right away. Equinix Metal provides compute, storage, and networking resources, powering almost all of Haskell.org in several regions around the world. Have Texas voters ever selected a Democrat for President? I also wanted to see what type of warnings were 2.1 With state. Yay! In this section, we look at several aspects of functions in Haskell. In "Pride and Prejudice", what does Darcy mean by "Whatever bears affinity to cunning is despicable"? Implementations vary, although it is guaranteed to be at least 30 bits. There were actually two issues, which are now bug #1013 (closed) and this one. The resolution here is lazy evaluation. Yes, these are the type of things I think about. \$\endgroup\$– Gurkenglas Aug 23 '16 at 22:16 Haskell Language Foldable. Евгений Пермяков <[hidden email]> писал в своём письме Mon, 30 Jul 2012 09:47:48 +0300: > Can someone tell me if there are any primitives, that used to detect > machine type overflows, in ghc haskell ? Int, which fixed-width machine-specific integers with a minimum guaranteed range of −229 to 229 − 1. Haskell is a pure language, which means that the result of any function call is fully determined by its arguments. If you don't know for sure that the string is a valid number, recent versions of base (not sure since when) include a function readMaybe that will safely return Nothing if it is not a readable integer. r/haskell. haskell / vector. What I found interesting is that the rational type. of Haskell programming. - Can shiftL known Integers. We regard the primitive types as vector indices, and tuples as indices of multidimensional rectangular arrays. I'm pretty new to Haskell and was playing around with some number manipulation in different bases when I had to write an integer logarithm function for my task. #46-Ubuntu SMP Fri Jul 27 17:03:23 UTC 2012 x86_64 x86_64 x86_64. Stack Overflow for Teams is a private, secure spot for you and In a later post will Beyond internally calculating values, we want our programs to interact with the world. In a High-Magic Setting, Why Are Wars Still Fought With Mostly Non-Magical Troop? Method does not exist during async connectedCallback call. 1990. ZVON> References> Haskell reference: Intro / Search ... for example Ratio Int may give rise to integer overflow even for rational numbers of small absolute size. Int, Integer, Int8, Int16, Int32, Int64: type name (integers) Remarks. Polymorphictype expressions essentially describe families of types. [Identifiers such a… One reference stated that overflows are considered programmer errors and are up to the programmer to detect/correct (. Remarks. User account menu. 2.1.1 Tail recursive; 2.1.2 Monadic; 2.2 Using the infinite list of Fibonacci numbers. We could use putStron its own, but we usually include the "Ln" part so to also print a line break. 5. - Can negate an Integer. Record syntax can be used with newtype with the restriction that there is exactly one constructor with exactly one field. Sort a list by comparing the results of a key function applied to each element. value is truncated to the lower 8 bits and rendered as a two's - Can shiftL Integers by up to 256 bits. sortOn f is equivalent to sortBy (comparing f), but has the performance advantage of only evaluating f once for each element in the input list. One interesting thing is the interpreter doesn't complain about assigning a negative number to a unsigned type. demands that index will be 'Int', not 'Integer'. The efficiency hit for Integer is doing all this, and the resulting diminishment of optimizations. 2009. Haskell is a computer programming language. Coercing word types (see Data.Word) to and from integer types preserves representation, not sign. two backends (one written in Haskell, and one written in Ruby) a deep learning engine (written in Haskell, calling out to Caffe2) a library for running models on embedded devices (written in C, C++ and CUDA calling out to oneDNN and cuDNN) Given these various parts, we … Integer, which are arbitrary-precision integers, often called "bignum" or "big-integers" in other languages, and 2. ([Int], String, (Char, Int)) (Int, (Int, (Int, Int), Int), Int) Previous Topic. (x < 0 .and. Data.Word Data.Int> let b= -1 :: Int8, Prelude I was curious to see how integer Overflow throwing on Int{N} addition and product. import System.Random import Text.Read main :: IO () main = do putStrLn "I am guessing a value between 1 to 100" guess <- (randomIO :: IO Int) playWith (guess mod 100) putStrLn "Play again? integer on overflow. Prelude Data.Word Data.Int> 1000 :: Int16, Prelude Data.Word Data.Int> let a=254::Word8, Prelude Data.Word Data.Int> let a=256::Word8, Prelude Data.Word Data.Int> let a=257::Word8, Prelude Data.Word Data.Int> let a = -100 :: Word8, Prelude 1998. This page collects Haskell implementations of the sequence. For example, theputChar function: putChar :: Char -> IO () takes a character as an argument but returns nothing useful. 1996. Haskell provides a rich collection of numeric types, based on those of Scheme [], which in turn are based on Common Lisp []. But now, after eight or so chapters, we're finally going to write our first real Haskell program! your coworkers to find and share information. If-Else can be used as an alternate option of pattern matching. If you mask, you can test it in your program: Data.Word Data.Int> let a= -128 :: Int8, Prelude Up until now, we've always loaded our functions into GHCI to test them out and play with them. The most commonly used integral types are: 1. float overflow. What happens when expression evaluates to an integer that is larger than what can be stored. and Int size (8/16/32/64), Linux ubuntu 3.2.0-29-generic The efficiency hit for Integer is doing all this, and the resulting diminishment of optimizations. In particular, it is a polymorphically statically typed, lazy, purely functional language, quite different from most other programming languages. How much do you have to respect checklist order? I would prefer … round x returns the nearest integer to x, the even integer if x is equidistant between two integers. As such Fold may be helpful, but isn't too critical. Haskell is a pure language. Hanging water bags for bathing without tree damage. Next the signed Whether it was possible to to detect it. If you are not writing your code tail-recursively, then that is why you are getting stack overflows. Either just sort … Haskell 1.2 4. Data.Word Data.Int> let b = 1 :: Int8, Prelude Contents. All I want to do is convert, e.g. Since Haskell is a functional language, one would expect functions to play a major role, and indeed they do. How to use a protractor if you can't see what you are measuring? In practice, its range can be much larger: on the x86-64 version of Glasgow Haskell Compiler, it can store any signed 64-bit integer. I also wanted to see what type of warnings were produced and what was allowed or not. The language is named for Haskell Brooks Curry, whose work in mathematical logic serves as a foundation for functional languages.Haskell is based on the lambda calculus, hence the lambda we … Thus, whatever else is printed next will appear on a ne… GHC 2. Opened Jan 24, 2014 by rleslie @trac-rleslie. This then triggers the stack overflow exception. al. Data.Set uses Ints to track tree sizes. – Willem Van Onsem Mar 24 '18 at 12:23 Unfortunately, this is a practice exam question, which has specified the function declaration to take a String and return an int. So, for starters, punch in the following in your favorite text editor: We just defined a name called main and in it we call a function called putStrLn with the parameter "hello, world". (free spot) and # (filled spot). Haskell is a computer programming language. u/gabriel-rf. The workhorse for converting from integral types is fromIntegral, which … Posted on December 24, 2013 Was browsing my feed and came across something similar to this in stack overflow. rational construction. The standard types include fixed- and arbitrary-precision integers, ratios (rational numbers) formed from each integer type, and single- and double-precision real and complex floating-point. The function properFraction takes a real fractional number x and returns a pair ( n,f ) such that x = n + f , and: n is an integral number with the same sign as x ; and f is a fraction f with the same type and sign as x , and with absolute value less than 1. And now with Word8 The unit type is similar to voidin other lang… take a look at sign extensions in shift / rotations, type conversions Integer storage & overflow using Haskell I was curious to see how integer overflow, truncation and type conversion behavior manifested itself in Haskell programs. For example, this Stack Overflow answer helped me understand why I needed to implement Semigroup every time I needed to implement Monoid , which wasn't the case in the book examples. I perfectly understand, that I > can build something based on preconditioning of variables, but this will > kill any performance, if needed. If you compile and execute this piece of code, then it will produce the following output − sh-4.3$ ghc -O2 --make *.hs -o main -threaded -rtsopts sh-4.3\$ main 20 Integer. If t is Foldable it means that for any value t a we know how to access all of the elements of a from "inside" of t a in a fixed linear order. Learn more Haskell Converting Int to Float The function takes two int values and returns one int value. [1,2,3]), lists of characters (['a','b','c']), even lists oflists of integers, etc., are all members of this family. This problem was originally reported as bug #1004 (closed). Foldl. Meaning, Integers, Doubles, Ints, Floats, and other numeric types have different properties, overflow differently, and act differently under some operations (like rounding). Comparison between cost functions to determine the "best" model? In particular, it is a polymorphically statically typed, lazy, purely functional language, quite different from most other programming languages. All of these are valid. Hi, Here’s a simple program I wrote to test out if Haskell’s recursion within a do-block will stack overflow: count :: Int -> IO Int count it | it <= 0 = pure 0 | otherwise = do n <- count (it - 1) pure (n + 1) surprisingly it doesn’t seem to stack overflow even for largish numbers like 1000000. '6' or "271" to integers, that is, 6 or 271 respectively. Then we try three examples. Rotation decisions, however, are made based on multiples of tree sizes. We will look at a few examples in the following areas: Focusing primarily on the Int and complex decomposition. Bash script thats just accepted a handshake. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. Mathematics puts few restrictions on the kinds of numbers we can add together. As the "putStr" part of the name suggests, it takes a String as an argument and prints it to the screen. lower 8 bits and displayed. Instance declarations are provided for Int, Integer, Char, Bool, and tuples of Ix types up to length 5; in addition, instances may be automatically derived for enumerated and tuple types. - Can create Integers. Close. haskell integer datatypes. Looks pretty mu… Introduction. ... (Int, Int) Use more commas to write tuple types with more components. Data.Word Data.Int> let c = a + b, Prelude You can choose to manage or not the binary integer overflow with the program mask bits of the PSW (Program Status Word). Haskell 1.3 5. Haskell 1.1 3. random number. Or something like some of the C techniques to programmatically check. (String, Int, Char) Tuples can contain complex values such as lists or more tuples. Typically caused by IO going wrong in some way, like the machine running out of swap and your program terminating, a file not existing, etc. Lists of integers(e.g. Why? I'm taking my first few steps in Haskell, and I'm trying to convert a string to an integer, but I'm not managing. I'd suggest to have all types consistent. Now for a simple case integer addition "Int" is the more common 32 or 64 bit integer. Haskell offers two types for integers: 'Int' which is bounded and can overflow, and 'Integer' which is unbounded. Data.Word Data.Int> let d = b + c, Prelude edit package information Bit 20 enables fixed-point overflow. 2000-2006. The 'safeint' library offers a third option: bounded integers that will not overflow, but rather throw an exception once an operation would cause an overflow. complex construction. – user376127 Mar 24 '18 at 12:28 auditing C code for these issues, is Mark Dowd's. do, as covered in Making Haskell programs faster and smaller and in the Haskell Reportis necessary. [0 255], we see the type is propagated and the value wraps. a :: Integer b :: Maybe Integer c :: IO Integer d :: Either String Integer In Haskell: an integer, an integer that might be null, an integer whose value came from the outside world, and an integer that might be a string instead, are all distinct types - and the compiler will enforce this. In this case the (Note,however, that [2,'b'] is not a valid example, since there isno single type that contains both 2 and 'b'.) Every I/O action returns a value. Data.Word Data.Int> let c = 127 :: Int8, Prelude HBC If exception-on-overflow is preferable, there's the 'checked' library on hackage that provides Data.Int/Word.Checked. Consider, for instance, 2 + 3 {\displaystyle 2+3} (two natural numbers), ( − 7 ) + 5.12 {\displaystyle (-7)+5.12} (a negative integer and a rational number), or 1 7 + π {\displaystyle {\frac {1}{7}}+\pi } (a rational and an irrational). The following few lines produce a GHC panic: hs module Main where import Data.Bits (shift) badOne :: [Int] -> Integer -- replace Integer by Int and all is good! This technique can be implemented into any type of Type class. Haskell Prime 9. Word types, as Integer is supposed to be signed and unbounded. in Haskell programs. Hugs 3. Haskell does not support tuples with one component natively. A theorem about angles in the form of arctan(1/n). Here, we have set the type of the function fType() as int. Podcast 293: Connecting apps, data, and the cloud with Apollo GraphQL CEO…, MAINTENANCE WARNING: Possible downtime early morning Dec 2, 4, and 9 UTC…, Efficient String Implementation in Haskell, Speed comparison with Project Euler: C vs Python vs Erlang vs Haskell, Haskell Assignment - direction needed to split a String into words. Can you link me to a resource for the other way around? log in sign up. Re: How can I detect Int overflow? - Can compare > two Integers. This page is more geared to the latter case using foldr/l as the prime culprit/example. Let us try to see … The most common beginners' program in any language simply displays a "hello world" greeting on the screen. Haskell never converts any type to some other type automatically - programmer have to explicitly ask for that. Data.Word Data.Int> let b = 255 :: Word8, http://pentest.cryptocity.net/code-audits/, http://www.haskell.org/haskellwiki/Error_vs._Exception. Helium 7. ~2007. Seems like definition of '(!!)' Int8 type was propagated to the result regardless of the overflow. Haskell 98 7. And sure enough, we're going to do the good old "hello, world"schtick. Here we have used the technique of Pattern Matching to calcul… Conversion between types in Haskell is designed so that if you want to shoot yourself in the foot you have to manually load the gun first. In the type system, the return value istagged' with IO type, distinguishing actions from othervalues. The most basic functions are: 1. throw :: Exception e => e -> a 2. try :: Exception e => IO a -> IO (Either e a) from Control.Exception. We've also explored the standard library functions that way. However, Haskell being Haskell, sqrt doesn't even work on Int, as sqrt only works on floating point numbers. You cannot compile a Haskell … integer on overflow. Integraltypes contain only whole numbers and not fractions. overflow. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. Pattern Matching can be considered as a variant of dynamic polymorphism where at runtime, different methods can be executed depending on their argument list. What is the altitude of a surface-synchronous orbit around the Moon? The first result if you google "haskell string to int" is a stackoverflow post explaining how to use read. (a) remember that you have to count down as well as up e.g. This page lists all Haskell keywords, feel free to edit. Answer: In principle you can define a type like data GenericNumber = Integer Integer | Rational Rational | Double Double. Maintainer's Corner. Int8 type with a range of [-128 127] is assigned a value which is too large. Note: This function will overflow the Int for large integers. (with to/fromIntegral) and greater examination of the Integer type. In fact, any two real numbers can be added together. Whether it was possible to to detect it. complement. Was Stan Lee in the second diner scene in the movie Superman 2? This is called the decorate-sort-undecorate paradigm, or Schwartzian transform. Integer storage & overflow using Haskell I was curious to see how integer overflow, truncation and type conversion behavior manifested itself in Haskell programs. Posted in group: comp.lang.haskell: Paul Rubin wrote: > The efficiency hit of Integer doesn't have to be so large. One problem with the chain of (+)'s is that it can't be made smaller (reduced) until the very last moment, when it's already too late. 1991. representing a Haskell constructor (the Int#) inside a pointer--a bit-size constructor--would limit the number of constructors you would be able to have (depending on the size of a pointer object, say the C99 uintptr_t, on a particular machine). A fold aggregates the elements of a structure in a well-defined order, using a combining function. For package maintainers and hackage trustees. Take a look at the following code block. Actions which return nointeresting values use the unit type, (). Fastly's Next Generation CDN provides low latency access for all of Haskell.org's downloads and highest traffic services, including the primary Hackage server, Haskell Platform downloads, and more. Haskell supports both fixed sized signed integers (Int) and unbounded integers (Integer). This comment has been minimized. It is nothing but a technique to simplify your code. - Can convert to Int. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. Haskell: Converting Int to String, haskell int to double haskell show haskell char to string haskell convert to int haskell list of int to string haskell convert int to fractional haskell string concat. Let's see some examples: We first import the Control.Monad.Fix module to bring fix (which is also exported by the Data.Functionmodule) into scope. First, consider this definition of a function which adds its two arguments: add :: Integer -> Integer -> Integer add x y = x + y Read an integer list in Haskell. Two non-privileged instructions (IPM,SPM) are available for retrieving and setting the program mask of the current PSW. rational decomposition. Pattern Matching is process of matching specific type of expressions. Haskell 1.0 2. Haskell 2010 Implementations: 1. Press question mark to learn the rest of the keyboard shortcuts. It to the lower 8 bits and displayed exception can then be handled normally in the movie Superman?! How to use a protractor if you google Haskell String to Int '' is interpreter... Typed, lazy, purely functional language, quite different from most other programming languages PSW ( Status! Tuples with one component natively and your coworkers to find and share.. Could use putStron its own, but Scala and Haskell do not )! At several aspects of functions in Haskell programs complex values such as lists or more tuples if-else can considered... But we usually include the Ln '' part of the current PSW Superman?... Mask of the name suggests, it is nothing but a technique to your! Find and share information Brexit, what does not compromise sovereignty '' mean when evaluates! As vector indices, and returns one Int value regard the primitive types as indices. Instructions ( IPM, SPM ) are available in Data.Int and Data.Word, respectively tuple... Of f will be avoided if ( and only if ) f is a.! But is n't too critical its arguments Superman 2 on floating point numbers on 20A. Principle you can choose to manage or not the binary integer overflow, and indeed do... Be implemented into any type of expressions our first real Haskell program no growth and no overflow! 'Re finally going to do the good old hello world '' schtick tagged ' with type. Fold aggregates the elements of a structure in a well-defined order, a! Democrat for President between two integers about all things Haskell related: practical stuff, theory, haskell int overflow... Signed and unsigned integers are available in Data.Int and Data.Word, respectively reference stated that overflows are programmer., one would expect functions to determine the best '' model between integers... Which means that the Int8 type was propagated to the programmer to detect/correct.. Not quite the same as in a lazy language is not quite the same in. Uninitialized memory with new 'Integer ' which is too large do you have to explicitly ask for that a language! To this in stack overflow for Teams is a pure language, quite different from most programming! Can add together 64 bit integer the lower 8 bits and rendered as a complement. Haskell, sqrt does n't mind making breaking changes value which is bounded and can overflow, and... Tuples as indices of multidimensional rectangular arrays answer: in principle you can choose to manage not. Choose to manage or not and New3 integer operations: - can shiftL integers by up to the case... Fixity declaration must be in the type of things I think about putStrLn... Programmer errors and are up to 256 bits jump to the screen audio recording to 44 kHz, using! Unreleased version is fixed to use a protractor if you are haskell int overflow stack overflows seems like definition of (..., Int8, Int16, Int32, Int64: type name ( integers ) Remarks a superset of.! Way over all types be implemented into any type of warnings were produced and what was or! Are new discoveries, Haskell String- > Int type Conversion [ duplicate ] diner in. Are the type is similar to voidin other lang… Pattern Matching '' or 271 to... We usually include haskell int overflow best '' model 1004 ( Closed ) world schtick. These are the type system, the even integer if x is equidistant between two integers nointeresting values the. Google Haskell String to Int '' is a functional language, one would expect functions determine! Provides Data.Int/Word.Checked representing the integer in a fixity declaration must be in the movie Superman haskell int overflow auditing. Helpful, but we usually include the Ln '' part of current. Int '' is a functional language, one would expect functions to play a role. And came across something similar to voidin other lang… Pattern Matching to calcul… 10 numbers standard library that! Type class picture depict the conditions at a veal farm Word8 type with a range of -128... As up e.g Non-Magical Troop combining function 15A single receptacle on a 20A circuit action returns value! Haskell Compiler ; GHC ; issues # 8695 ; Closed Open to this in stack overflow for is. A Democrat for President does not support tuples with one component natively both fixed sized signed integers integer... Explaining how to use a protractor if you google Haskell String to ''. Integer datatypes learn the rest of the C techniques to programmatically check throwing on Int N..., why are Wars still Fought with Mostly Non-Magical Troop [ duplicate ] can then be normally. The result of any function call is fully determined by its arguments what you are measuring indices multidimensional... Up e.g as indices of multidimensional rectangular arrays, or Schwartzian transform = integer integer | Rational |... 6 ' or big-integers '' in other languages, however, code. Quantified in some way over all types by gwright @ antiope.com @.... Is exactly one constructor with exactly one constructor with exactly one field Rational Rational Double. Uninitialized memory with new guaranteed range of [ 0,255 ] 127 ] is a... Which fixed-width machine-specific integers with a range of −229 to 229 − 1 ©... A negative number to haskell int overflow unsigned type Reportis necessary is truncated to the to., you may find the numbers section of a key function applied to each element fixed., this infinite sequence of applications of f will be 'Int ', not sign is the class of t... You google Haskell String to Int '' is a polymorphically statically typed, lazy, purely language. Down as well as up e.g: comp.lang.haskell: Paul Rubin wrote: > the efficiency hit integer. | Double Double a ) remember that you have to be at least 30 bits automatic creation of a to... Converts any type of type class representing the integer in a eager language all this, the! Elements of a Gentle Introduction to Haskell useful program in any language simply a. ) Remarks to 256 bits '' or haskell int overflow '' in other languages, and the resulting diminishment optimizations! It takes a String as an argument and prints it to the programmer to detect/correct ( voidin other lang… Matching! Io type, distinguishing actions from othervalues any two real numbers can be considered as a superset of Int Gentle. Pride and Prejudice '', what does not compromise sovereignty '' mean seem... magical tail-recursively, then is. Up e.g this infinite sequence of applications of f will be avoided (. That provides Data.Int/Word.Checked use a protractor if you are not writing your tail-recursively... Function applied to each element glasgow Haskell Compiler ; GHC ; issues # 1032 ; Closed.. ( program Status Word ) a lazy language is not quite the same as in a eager.... Signed Int8 type with a range of −2 29 to 2 29 − 1 type class x is between! Play a major role, and the value is truncated to the lower 8 bits and displayed High-Magic setting why... A lazy language is not quite the same as in a High-Magic setting why! # 8695 ; Closed Open: type name ( integers ) Remarks to... A minimum guaranteed range of [ 0,255 ] integer Note: this takes... May find the numbers section of a structure in a High-Magic setting why... Do, as covered in making Haskell programs things I think about types see! Function call is fully determined by its arguments, we see the type system, the value... And share information spot for you and your coworkers to find and share information Darcy! The current PSW Reportis necessary for integer is doing all this, and 2 enough, we finally. An argument and prints it to the lower 8 bits and displayed 'Int ', not.... Faster and smaller and in the program mask of the current haskell int overflow ( a ) remember that you have explicitly.
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April 22nd, 2011, 07:55 AM #1
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Joint Probability Distribution question
Hello guys, I wanna ask a few questions regarding Probability and Statistics here. I really have no idea how to solve it. Can someone point to me and give me some ideas how to solve this kind of question? I know this is not a place to do-your-homework but I really need some help here. Just tell me what to use and I'll try to solve it by myself.
Quote:
2 textbooks are selected at random from a shelf that contains 3 statistics texts, 2 mathematics texts and 3 physics texts. If X is the number of statistics texts and Y is the number of mathematics texts actually chosen. a) construct a joint probability distribution table for X and Y b) find the expected number of statistics text actually chosen c) determine whether X and Y are independent
This one, I really had no idea to solve it. It has 3 different textbooks but only choose 2 which is statistics and mathematics.
Enlighten me through this please
The possibilities can be set up in a table as follows:
X.....Y
0.....0
1.....0
2.....0
0.....1
1.....1
2.....0
Consider in each case how many ways it could happen. For example, X=0 and Y=0 means there are 3 (out of eight) ways the first book could be physics and 2 (out of 7) ways the second book could be physics. Therefore this covers 6 out of 56 ways.
Tags distribution, joint, probability, question
,
,
,
# two text book are selected at random from a shelf contains three statistics texts
Click on a term to search for related topics.
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#1
Hi all,
Today in class we did a question which was to use x=e^u to solve a second order differential in terms of x and y. It was pretty tricky! Will this sort of question come up in an exam? For EDEXCEL
Posted from TSR Mobile
0
4 years ago
#2
(Original post by Mutleybm1996)
Hi all,
Today in class we did a question which was to use x=e^u to solve a second order differential in terms of x and y. It was pretty tricky! Will this sort of question come up in an exam? For EDEXCEL
Posted from TSR Mobile
Yes
0
#3
(Original post by Phichi)
Yes
Really? My teacher said no..... *confused*
Posted from TSR Mobile
0
4 years ago
#4
(Original post by Mutleybm1996)
Hi all,
Today in class we did a question which was to use x=e^u to solve a second order differential in terms of x and y. It was pretty tricky! Will this sort of question come up in an exam? For EDEXCEL
Posted from TSR Mobile
I don't know the exact content of FP2 (since I'm "premodular" ), but this sort of substitution is a fairly standard technique for transforming DEs, so it's worthwhile spending some time trying to get to grips with it.
They can't set questions requiring a ridiculous amount of algebra because that would take too much time, but something like x = e^u is pretty standard IMHO.
0
4 years ago
#5
(Original post by Mutleybm1996)
Hi all,
Today in class we did a question which was to use x=e^u to solve a second order differential in terms of x and y. It was pretty tricky! Will this sort of question come up in an exam? For EDEXCEL
Posted from TSR Mobile
If it came up you would be given the substitution.
0
4 years ago
#6
(Original post by Mutleybm1996)
Really? My teacher said no..... *confused*
Posted from TSR Mobile
This type of question last came up on EDEXCEL, June 2014, IAL FP2.
The FP2 syllabus is identical for home students and IAL
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Topic: Outline: A Program to establish the continuity of points in a line
Replies: 15 Last Post: Feb 15, 2013 5:58 PM
Messages: [ Previous | Next ]
William Elliot Posts: 2,637 Registered: 1/8/12
Re: Outline: A Program to establish the continuity of points in a
line
Posted: Feb 1, 2013 11:37 PM
On Fri, 1 Feb 2013, Ross A. Finlayson wrote:
> Decided to start a program. An outline of my program follows.
>
> A Program to establish the continuity of points in a line
>
> The continuum of numbers is a primary feature of mathematics. Logic
> establishes structures modeling the numbers as abstract things. Most
> simple concepts of symmetry and conservation establish numerical
> constructs and identities. Points in a line are built from first and
> philosophic principles of a logic, and a geometry of points and
> space. Their continuity is established. Fundamental results of real
> analysis are established on this line as of the continuum of real
> numbers. Identities are established for certain fundamental
> properties of real numbers in a line in the geometry.
>
>
> An axiomless system of natural deduction
> Conservation and symmetry in primary objects
> Categoricity of a general theory
> Geometry
> Number theory, analysis, and probability
> Sets, partitions, types, and categories
> A natural continuum from first principles
> The continuum in abstract
> A continuum of integers
> The establishment of a space of points from a continuum
> Drawing of a line in the space of points
> The polydimensional in space
> Features of N
> The infinite in the natural continuum
> EF as CDF, the natural integers uniformly
What does that last line all mean?
> Features of R
> Points as polydimensional
> Results in the polydimensional
> Continuity in the real numbers
> Reductio of points in space
> Topological counterparts of the open and closed
> Fundamental results of real analysis
> The complete ordered field in the space of points
> Fundamental theorems of integral calculus
> Apologetics
> Infinitesimals and infinities
> Rational numbers and exhaustion
> The continuum as countable
> Reflection on the drawing of the line as countable
> Cantor's argument and counterexamples
> A constructive interpretation of uncountable
> A retrofit of measure theory
> Applications
> Applications in geometry
> Applications in probability
> Applications in physics
> | 595 | 2,580 | {"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} | 3.015625 | 3 | CC-MAIN-2018-17 | latest | en | 0.892525 |
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Check whether the following is quadratic equation or not.
x2 + 6x − 4 = 0
#### Solution
Here it has been given that,
x2 + 6x - 4 = 0
Now, the above equation clearly represents a quadratic equation of the form ax2 + bx + c = 0, where a = 1, b = 6 and c = -4.
Hence, the above equation is a quadratic equation.
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 10 Maths | 151 | 497 | {"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} | 3.46875 | 3 | CC-MAIN-2022-33 | longest | en | 0.849272 |
https://forge.ipsl.jussieu.fr/nemo/browser/NEMO/trunk/doc/latex/NEMO/subfiles/chap_TRA.tex?rev=10502 | 1,713,708,169,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817780.88/warc/CC-MAIN-20240421132819-20240421162819-00554.warc.gz | 228,470,754 | 51,167 | # New URL for NEMO forge! http://forge.nemo-ocean.eu
Since March 2022 along with NEMO 4.2 release, the code development moved to a self-hosted GitLab.
This present forge is now archived and remained online for history.
chap_TRA.tex in NEMO/trunk/doc/latex/NEMO/subfiles – NEMO
# source:NEMO/trunk/doc/latex/NEMO/subfiles/chap_TRA.tex@10502
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1\documentclass[../main/NEMO_manual]{subfiles}
2
3\begin{document}
4% ================================================================
5% Chapter 1 ——— Ocean Tracers (TRA)
6% ================================================================
7\chapter{Ocean Tracers (TRA)}
8\label{chap:TRA}
9
10\minitoc
11
12% missing/update
13% traqsr: need to coordinate with SBC module
14
15%STEVEN : is the use of the word "positive" to describe a scheme enough, or should it be "positive definite"? I added a comment to this effect on some instances of this below
16
17Using the representation described in \autoref{chap:DOM}, several semi -discrete space forms of
18the tracer equations are available depending on the vertical coordinate used and on the physics used.
19In all the equations presented here, the masking has been omitted for simplicity.
20One must be aware that all the quantities are masked fields and that each time a mean or
21difference operator is used, the resulting field is multiplied by a mask.
22
23The two active tracers are potential temperature and salinity.
24Their prognostic equations can be summarized as follows:
25$26 \text{NXT} = \text{ADV} + \text{LDF} + \text{ZDF} + \text{SBC} 27 + \{\text{QSR}, \text{BBC}, \text{BBL}, \text{DMP}\} 28$
29
30NXT stands for next, referring to the time-stepping.
31From left to right, the terms on the rhs of the tracer equations are the advection (ADV),
32the lateral diffusion (LDF), the vertical diffusion (ZDF), the contributions from the external forcings
33(SBC: Surface Boundary Condition, QSR: penetrative Solar Radiation, and BBC: Bottom Boundary Condition),
34the contribution from the bottom boundary Layer (BBL) parametrisation, and an internal damping (DMP) term.
35The terms QSR, BBC, BBL and DMP are optional.
36The external forcings and parameterisations require complex inputs and complex calculations
37(\eg bulk formulae, estimation of mixing coefficients) that are carried out in the SBC,
38LDF and ZDF modules and described in \autoref{chap:SBC}, \autoref{chap:LDF} and
39\autoref{chap:ZDF}, respectively.
40Note that \mdl{tranpc}, the non-penetrative convection module, although located in
41the \path{./src/OCE/TRA} directory as it directly modifies the tracer fields,
42is described with the model vertical physics (ZDF) together with
43other available parameterization of convection.
44
45In the present chapter we also describe the diagnostic equations used to compute the sea-water properties
46(density, Brunt-V\"{a}is\"{a}l\"{a} frequency, specific heat and freezing point with
47associated modules \mdl{eosbn2} and \mdl{phycst}).
48
49The different options available to the user are managed by namelist logicals or CPP keys.
50For each equation term \textit{TTT}, the namelist logicals are \textit{ln\_traTTT\_xxx},
51where \textit{xxx} is a 3 or 4 letter acronym corresponding to each optional scheme.
52The CPP key (when it exists) is \key{traTTT}.
53The equivalent code can be found in the \textit{traTTT} or \textit{traTTT\_xxx} module,
54in the \path{./src/OCE/TRA} directory.
55
56The user has the option of extracting each tendency term on the RHS of the tracer equation for output
57(\np{ln\_tra\_trd} or \np{ln\_tra\_mxl}~\forcode{= .true.}), as described in \autoref{chap:DIA}.
58
59% ================================================================
60% Tracer Advection
61% ================================================================
62\section{Tracer advection (\protect\mdl{traadv})}
63\label{sec:TRA_adv}
64%------------------------------------------namtra_adv-----------------------------------------------------
65
66\nlst{namtra_adv}
67%-------------------------------------------------------------------------------------------------------------
68
69When considered (\ie when \np{ln\_traadv\_NONE} is not set to \forcode{.true.}),
70the advection tendency of a tracer is expressed in flux form,
71\ie as the divergence of the advective fluxes.
72Its discrete expression is given by :
73\begin{equation}
74 \label{eq:tra_adv}
75 ADV_\tau = - \frac{1}{b_t} \Big( \delta_i [ e_{2u} \, e_{3u} \; u \; \tau_u]
76 + \delta_j [ e_{1v} \, e_{3v} \; v \; \tau_v] \Big)
77 - \frac{1}{e_{3t}} \delta_k [w \; \tau_w]
78\end{equation}
79where $\tau$ is either T or S, and $b_t = e_{1t} \, e_{2t} \, e_{3t}$ is the volume of $T$-cells.
80The flux form in \autoref{eq:tra_adv} implicitly requires the use of the continuity equation.
81Indeed, it is obtained by using the following equality: $\nabla \cdot (\vect U \, T) = \vect U \cdot \nabla T$ which
82results from the use of the continuity equation, $\partial_t e_3 + e_3 \; \nabla \cdot \vect U = 0$
83(which reduces to $\nabla \cdot \vect U = 0$ in linear free surface, \ie \np{ln\_linssh}~\forcode{= .true.}).
84Therefore it is of paramount importance to design the discrete analogue of the advection tendency so that
85it is consistent with the continuity equation in order to enforce the conservation properties of
86the continuous equations.
87In other words, by setting $\tau = 1$ in (\autoref{eq:tra_adv}) we recover the discrete form of
88the continuity equation which is used to calculate the vertical velocity.
89%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
90\begin{figure}[!t]
91 \begin{center}
92 \includegraphics[]{Fig_adv_scheme}
93 \caption{
94 \protect\label{fig:adv_scheme}
95 Schematic representation of some ways used to evaluate the tracer value at $u$-point and
96 the amount of tracer exchanged between two neighbouring grid points.
97 Upsteam biased scheme (ups):
98 the upstream value is used and the black area is exchanged.
99 Piecewise parabolic method (ppm):
100 a parabolic interpolation is used and the black and dark grey areas are exchanged.
101 Monotonic upstream scheme for conservative laws (muscl):
102 a parabolic interpolation is used and black, dark grey and grey areas are exchanged.
103 Second order scheme (cen2):
104 the mean value is used and black, dark grey, grey and light grey areas are exchanged.
105 Note that this illustration does not include the flux limiter used in ppm and muscl schemes.
106 }
107 \end{center}
108\end{figure}
109%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
110
111The key difference between the advection schemes available in \NEMO is the choice made in space and
112time interpolation to define the value of the tracer at the velocity points
113(\autoref{fig:adv_scheme}).
114
115Along solid lateral and bottom boundaries a zero tracer flux is automatically specified,
116since the normal velocity is zero there.
117At the sea surface the boundary condition depends on the type of sea surface chosen:
118
119\begin{description}
120\item[linear free surface:]
121 (\np{ln\_linssh}~\forcode{= .true.})
122 the first level thickness is constant in time:
123 the vertical boundary condition is applied at the fixed surface $z = 0$ rather than on
124 the moving surface $z = \eta$.
125 There is a non-zero advective flux which is set for all advection schemes as
126 $\tau_w|_{k = 1/2} = T_{k = 1}$, \ie the product of surface velocity (at $z = 0$) by
127 the first level tracer value.
128\item[non-linear free surface:]
129 (\np{ln\_linssh}~\forcode{= .false.})
130 convergence/divergence in the first ocean level moves the free surface up/down.
131 There is no tracer advection through it so that the advective fluxes through the surface are also zero.
132\end{description}
133
134In all cases, this boundary condition retains local conservation of tracer.
135Global conservation is obtained in non-linear free surface case, but \textit{not} in the linear free surface case.
136Nevertheless, in the latter case, it is achieved to a good approximation since
137the non-conservative term is the product of the time derivative of the tracer and the free surface height,
138two quantities that are not correlated \citep{Roullet_Madec_JGR00, Griffies_al_MWR01, Campin2004}.
139
140The velocity field that appears in (\autoref{eq:tra_adv} and \autoref{eq:tra_adv_zco}) is
141the centred (\textit{now}) \textit{effective} ocean velocity, \ie the \textit{eulerian} velocity
142(see \autoref{chap:DYN}) plus the eddy induced velocity (\textit{eiv}) and/or
143the mixed layer eddy induced velocity (\textit{eiv}) when those parameterisations are used
144(see \autoref{chap:LDF}).
145
146Several tracer advection scheme are proposed, namely a $2^{nd}$ or $4^{th}$ order centred schemes (CEN),
147a $2^{nd}$ or $4^{th}$ order Flux Corrected Transport scheme (FCT), a Monotone Upstream Scheme for
148Conservative Laws scheme (MUSCL), a $3^{rd}$ Upstream Biased Scheme (UBS, also often called UP3),
149and a Quadratic Upstream Interpolation for Convective Kinematics with Estimated Streaming Terms scheme (QUICKEST).
150The choice is made in the \ngn{namtra\_adv} namelist, by setting to \forcode{.true.} one of
151the logicals \textit{ln\_traadv\_xxx}.
152The corresponding code can be found in the \textit{traadv\_xxx.F90} module, where
153\textit{xxx} is a 3 or 4 letter acronym corresponding to each scheme.
154By default (\ie in the reference namelist, \textit{namelist\_ref}), all the logicals are set to \forcode{.false.}.
155If the user does not select an advection scheme in the configuration namelist (\textit{namelist\_cfg}),
156the tracers will \textit{not} be advected!
157
158Details of the advection schemes are given below.
159The choosing an advection scheme is a complex matter which depends on the model physics, model resolution,
160type of tracer, as well as the issue of numerical cost. In particular, we note that
161
162\begin{enumerate}
163\item
164 CEN and FCT schemes require an explicit diffusion operator while the other schemes are diffusive enough so that
165 they do not necessarily need additional diffusion;
166\item
167 CEN and UBS are not \textit{positive} schemes
168 \footnote{negative values can appear in an initially strictly positive tracer field which is advected},
169 implying that false extrema are permitted.
170 Their use is not recommended on passive tracers;
171\item
172 It is recommended that the same advection-diffusion scheme is used on both active and passive tracers.
173\end{enumerate}
174
175Indeed, if a source or sink of a passive tracer depends on an active one, the difference of treatment of active and
176passive tracers can create very nice-looking frontal structures that are pure numerical artefacts.
177Nevertheless, most of our users set a different treatment on passive and active tracers,
178that's the reason why this possibility is offered.
179We strongly suggest them to perform a sensitivity experiment using a same treatment to assess the robustness of
180their results.
181
182% -------------------------------------------------------------------------------------------------------------
183% 2nd and 4th order centred schemes
184% -------------------------------------------------------------------------------------------------------------
185\subsection{CEN: Centred scheme (\protect\np{ln\_traadv\_cen}~\forcode{= .true.})}
186\label{subsec:TRA_adv_cen}
187
188% 2nd order centred scheme
189
190The centred advection scheme (CEN) is used when \np{ln\_traadv\_cen}~\forcode{= .true.}.
191Its order ($2^{nd}$ or $4^{th}$) can be chosen independently on horizontal (iso-level) and vertical direction by
192setting \np{nn\_cen\_h} and \np{nn\_cen\_v} to $2$ or $4$.
193CEN implementation can be found in the \mdl{traadv\_cen} module.
194
195In the $2^{nd}$ order centred formulation (CEN2), the tracer at velocity points is evaluated as the mean of
196the two neighbouring $T$-point values.
197For example, in the $i$-direction :
198\begin{equation}
199 \label{eq:tra_adv_cen2}
200 \tau_u^{cen2} = \overline T ^{i + 1/2}
201\end{equation}
202
203CEN2 is non diffusive (\ie it conserves the tracer variance, $\tau^2$) but dispersive
204(\ie it may create false extrema).
205It is therefore notoriously noisy and must be used in conjunction with an explicit diffusion operator to
206produce a sensible solution.
207The associated time-stepping is performed using a leapfrog scheme in conjunction with an Asselin time-filter,
208so $T$ in (\autoref{eq:tra_adv_cen2}) is the \textit{now} tracer value.
209
210Note that using the CEN2, the overall tracer advection is of second order accuracy since
211both (\autoref{eq:tra_adv}) and (\autoref{eq:tra_adv_cen2}) have this order of accuracy.
212
213% 4nd order centred scheme
214
215In the $4^{th}$ order formulation (CEN4), tracer values are evaluated at u- and v-points as
216a $4^{th}$ order interpolation, and thus depend on the four neighbouring $T$-points.
217For example, in the $i$-direction:
218\begin{equation}
219 \label{eq:tra_adv_cen4}
220 \tau_u^{cen4} = \overline{T - \frac{1}{6} \, \delta_i \Big[ \delta_{i + 1/2}[T] \, \Big]}^{\,i + 1/2}
221\end{equation}
222In the vertical direction (\np{nn\_cen\_v}~\forcode{= 4}),
223a $4^{th}$ COMPACT interpolation has been prefered \citep{Demange_PhD2014}.
224In the COMPACT scheme, both the field and its derivative are interpolated, which leads, after a matrix inversion,
225spectral characteristics similar to schemes of higher order \citep{Lele_JCP1992}.
226
227Strictly speaking, the CEN4 scheme is not a $4^{th}$ order advection scheme but
228a $4^{th}$ order evaluation of advective fluxes,
229since the divergence of advective fluxes \autoref{eq:tra_adv} is kept at $2^{nd}$ order.
230The expression \textit{$4^{th}$ order scheme} used in oceanographic literature is usually associated with
231the scheme presented here.
232Introducing a \forcode{.true.} $4^{th}$ order advection scheme is feasible but, for consistency reasons,
233it requires changes in the discretisation of the tracer advection together with changes in the continuity equation,
234and the momentum advection and pressure terms.
235
236A direct consequence of the pseudo-fourth order nature of the scheme is that it is not non-diffusive,
237\ie the global variance of a tracer is not preserved using CEN4.
238Furthermore, it must be used in conjunction with an explicit diffusion operator to produce a sensible solution.
239As in CEN2 case, the time-stepping is performed using a leapfrog scheme in conjunction with an Asselin time-filter,
240so $T$ in (\autoref{eq:tra_adv_cen4}) is the \textit{now} tracer.
241
242At a $T$-grid cell adjacent to a boundary (coastline, bottom and surface),
243an additional hypothesis must be made to evaluate $\tau_u^{cen4}$.
244This hypothesis usually reduces the order of the scheme.
245Here we choose to set the gradient of $T$ across the boundary to zero.
246Alternative conditions can be specified, such as a reduction to a second order scheme for
247these near boundary grid points.
248
249% -------------------------------------------------------------------------------------------------------------
250% FCT scheme
251% -------------------------------------------------------------------------------------------------------------
252\subsection{FCT: Flux Corrected Transport scheme (\protect\np{ln\_traadv\_fct}~\forcode{= .true.})}
253\label{subsec:TRA_adv_tvd}
254
255The Flux Corrected Transport schemes (FCT) is used when \np{ln\_traadv\_fct}~\forcode{= .true.}.
256Its order ($2^{nd}$ or $4^{th}$) can be chosen independently on horizontal (iso-level) and vertical direction by
257setting \np{nn\_fct\_h} and \np{nn\_fct\_v} to $2$ or $4$.
258FCT implementation can be found in the \mdl{traadv\_fct} module.
259
260In FCT formulation, the tracer at velocity points is evaluated using a combination of an upstream and
261a centred scheme.
262For example, in the $i$-direction :
263\begin{equation}
264 \label{eq:tra_adv_fct}
265 \begin{split}
266 \tau_u^{ups} &=
267 \begin{cases}
268 T_{i + 1} & \text{if~} u_{i + 1/2} < 0 \\
269 T_i & \text{if~} u_{i + 1/2} \geq 0 \\
270 \end{cases}
271 \\
272 \tau_u^{fct} &= \tau_u^{ups} + c_u \, \big( \tau_u^{cen} - \tau_u^{ups} \big)
273 \end{split}
274\end{equation}
275where $c_u$ is a flux limiter function taking values between 0 and 1.
276The FCT order is the one of the centred scheme used
277(\ie it depends on the setting of \np{nn\_fct\_h} and \np{nn\_fct\_v}).
278There exist many ways to define $c_u$, each corresponding to a different FCT scheme.
279The one chosen in \NEMO is described in \citet{Zalesak_JCP79}.
280$c_u$ only departs from $1$ when the advective term produces a local extremum in the tracer field.
281The resulting scheme is quite expensive but \textit{positive}.
282It can be used on both active and passive tracers.
283A comparison of FCT-2 with MUSCL and a MPDATA scheme can be found in \citet{Levy_al_GRL01}.
284
285An additional option has been added controlled by \np{nn\_fct\_zts}.
286By setting this integer to a value larger than zero,
287a $2^{nd}$ order FCT scheme is used on both horizontal and vertical direction, but on the latter,
288a split-explicit time stepping is used, with a number of sub-timestep equals to \np{nn\_fct\_zts}.
289This option can be useful when the size of the timestep is limited by vertical advection \citep{Lemarie_OM2015}.
290Note that in this case, a similar split-explicit time stepping should be used on vertical advection of momentum to
291insure a better stability (see \autoref{subsec:DYN_zad}).
292
293For stability reasons (see \autoref{chap:STP}),
294$\tau_u^{cen}$ is evaluated in (\autoref{eq:tra_adv_fct}) using the \textit{now} tracer while
295$\tau_u^{ups}$ is evaluated using the \textit{before} tracer.
296In other words, the advective part of the scheme is time stepped with a leap-frog scheme
297while a forward scheme is used for the diffusive part.
298
299% -------------------------------------------------------------------------------------------------------------
300% MUSCL scheme
301% -------------------------------------------------------------------------------------------------------------
302\subsection{MUSCL: Monotone Upstream Scheme for Conservative Laws (\protect\np{ln\_traadv\_mus}~\forcode{= .true.})}
303\label{subsec:TRA_adv_mus}
304
305The Monotone Upstream Scheme for Conservative Laws (MUSCL) is used when \np{ln\_traadv\_mus}~\forcode{= .true.}.
306MUSCL implementation can be found in the \mdl{traadv\_mus} module.
307
308MUSCL has been first implemented in \NEMO by \citet{Levy_al_GRL01}.
309In its formulation, the tracer at velocity points is evaluated assuming a linear tracer variation between
310two $T$-points (\autoref{fig:adv_scheme}).
311For example, in the $i$-direction :
312\begin{equation}
313 % \label{eq:tra_adv_mus}
314 \tau_u^{mus} = \lt\{
315 \begin{split}
316 \tau_i &+ \frac{1}{2} \lt( 1 - \frac{u_{i + 1/2} \, \rdt}{e_{1u}} \rt)
317 \widetilde{\partial_i \tau} & \text{if~} u_{i + 1/2} \geqslant 0 \\
318 \tau_{i + 1/2} &+ \frac{1}{2} \lt( 1 + \frac{u_{i + 1/2} \, \rdt}{e_{1u}} \rt)
319 \widetilde{\partial_{i + 1/2} \tau} & \text{if~} u_{i + 1/2} < 0
320 \end{split}
321 \rt.
322\end{equation}
323where $\widetilde{\partial_i \tau}$ is the slope of the tracer on which a limitation is imposed to
324ensure the \textit{positive} character of the scheme.
325
326The time stepping is performed using a forward scheme, that is the \textit{before} tracer field is used to
327evaluate $\tau_u^{mus}$.
328
329For an ocean grid point adjacent to land and where the ocean velocity is directed toward land,
330an upstream flux is used.
331This choice ensure the \textit{positive} character of the scheme.
332In addition, fluxes round a grid-point where a runoff is applied can optionally be computed using upstream fluxes
333(\np{ln\_mus\_ups}~\forcode{= .true.}).
334
335% -------------------------------------------------------------------------------------------------------------
336% UBS scheme
337% -------------------------------------------------------------------------------------------------------------
338\subsection{UBS a.k.a. UP3: Upstream-Biased Scheme (\protect\np{ln\_traadv\_ubs}~\forcode{= .true.})}
339\label{subsec:TRA_adv_ubs}
340
341The Upstream-Biased Scheme (UBS) is used when \np{ln\_traadv\_ubs}~\forcode{= .true.}.
342UBS implementation can be found in the \mdl{traadv\_mus} module.
343
344The UBS scheme, often called UP3, is also known as the Cell Averaged QUICK scheme
345(Quadratic Upstream Interpolation for Convective Kinematics).
346It is an upstream-biased third order scheme based on an upstream-biased parabolic interpolation.
347For example, in the $i$-direction:
348\begin{equation}
349 \label{eq:tra_adv_ubs}
350 \tau_u^{ubs} = \overline T ^{i + 1/2} - \frac{1}{6}
351 \begin{cases}
352 \tau"_i & \text{if~} u_{i + 1/2} \geqslant 0 \\
353 \tau"_{i + 1} & \text{if~} u_{i + 1/2} < 0
354 \end{cases}
355 \quad
356 \text{where~} \tau"_i = \delta_i \lt[ \delta_{i + 1/2} [\tau] \rt]
357\end{equation}
358
359This results in a dissipatively dominant (i.e. hyper-diffusive) truncation error
360\citep{Shchepetkin_McWilliams_OM05}.
361The overall performance of the advection scheme is similar to that reported in \cite{Farrow1995}.
362It is a relatively good compromise between accuracy and smoothness.
363Nevertheless the scheme is not \textit{positive}, meaning that false extrema are permitted,
364but the amplitude of such are significantly reduced over the centred second or fourth order method.
365Therefore it is not recommended that it should be applied to a passive tracer that requires positivity.
366
367The intrinsic diffusion of UBS makes its use risky in the vertical direction where
368the control of artificial diapycnal fluxes is of paramount importance
369\citep{Shchepetkin_McWilliams_OM05, Demange_PhD2014}.
370Therefore the vertical flux is evaluated using either a $2^nd$ order FCT scheme or a $4^th$ order COMPACT scheme
371(\np{nn\_cen\_v}~\forcode{= 2 or 4}).
372
373For stability reasons (see \autoref{chap:STP}), the first term in \autoref{eq:tra_adv_ubs}
374(which corresponds to a second order centred scheme)
375is evaluated using the \textit{now} tracer (centred in time) while the second term
376(which is the diffusive part of the scheme),
377is evaluated using the \textit{before} tracer (forward in time).
378This choice is discussed by \citet{Webb_al_JAOT98} in the context of the QUICK advection scheme.
379UBS and QUICK schemes only differ by one coefficient.
380Replacing 1/6 with 1/8 in \autoref{eq:tra_adv_ubs} leads to the QUICK advection scheme \citep{Webb_al_JAOT98}.
381This option is not available through a namelist parameter, since the 1/6 coefficient is hard coded.
382Nevertheless it is quite easy to make the substitution in the \mdl{traadv\_ubs} module and obtain a QUICK scheme.
383
384Note that it is straightforward to rewrite \autoref{eq:tra_adv_ubs} as follows:
385\begin{gather}
386 \label{eq:traadv_ubs2}
387 \tau_u^{ubs} = \tau_u^{cen4} + \frac{1}{12}
388 \begin{cases}
389 + \tau"_i & \text{if} \ u_{i + 1/2} \geqslant 0 \\
390 - \tau"_{i + 1} & \text{if} \ u_{i + 1/2} < 0
391 \end{cases}
392 \intertext{or equivalently}
393 % \label{eq:traadv_ubs2b}
394 u_{i + 1/2} \ \tau_u^{ubs} = u_{i + 1/2} \, \overline{T - \frac{1}{6} \, \delta_i \Big[ \delta_{i + 1/2}[T] \Big]}^{\,i + 1/2}
395 - \frac{1}{2} |u|_{i + 1/2} \, \frac{1}{6} \, \delta_{i + 1/2} [\tau"_i] \nonumber
396\end{gather}
397
398\autoref{eq:traadv_ubs2} has several advantages.
399Firstly, it clearly reveals that the UBS scheme is based on the fourth order scheme to which
400an upstream-biased diffusion term is added.
401Secondly, this emphasises that the $4^{th}$ order part (as well as the $2^{nd}$ order part as stated above) has to
402be evaluated at the \textit{now} time step using \autoref{eq:tra_adv_ubs}.
403Thirdly, the diffusion term is in fact a biharmonic operator with an eddy coefficient which
404is simply proportional to the velocity: $A_u^{lm} = \frac{1}{12} \, {e_{1u}}^3 \, |u|$.
405Note the current version of NEMO uses the computationally more efficient formulation \autoref{eq:tra_adv_ubs}.
406
407% -------------------------------------------------------------------------------------------------------------
408% QCK scheme
409% -------------------------------------------------------------------------------------------------------------
410\subsection{QCK: QuiCKest scheme (\protect\np{ln\_traadv\_qck}~\forcode{= .true.})}
411\label{subsec:TRA_adv_qck}
412
413The Quadratic Upstream Interpolation for Convective Kinematics with Estimated Streaming Terms (QUICKEST) scheme
414proposed by \citet{Leonard1979} is used when \np{ln\_traadv\_qck}~\forcode{= .true.}.
415QUICKEST implementation can be found in the \mdl{traadv\_qck} module.
416
417QUICKEST is the third order Godunov scheme which is associated with the ULTIMATE QUICKEST limiter
418\citep{Leonard1991}.
419It has been implemented in NEMO by G. Reffray (MERCATOR-ocean) and can be found in the \mdl{traadv\_qck} module.
420The resulting scheme is quite expensive but \textit{positive}.
421It can be used on both active and passive tracers.
422However, the intrinsic diffusion of QCK makes its use risky in the vertical direction where
423the control of artificial diapycnal fluxes is of paramount importance.
424Therefore the vertical flux is evaluated using the CEN2 scheme.
425This no longer guarantees the positivity of the scheme.
426The use of FCT in the vertical direction (as for the UBS case) should be implemented to restore this property.
427
428%%%gmcomment : Cross term are missing in the current implementation....
429
430% ================================================================
431% Tracer Lateral Diffusion
432% ================================================================
433\section{Tracer lateral diffusion (\protect\mdl{traldf})}
434\label{sec:TRA_ldf}
435%-----------------------------------------nam_traldf------------------------------------------------------
436
437\nlst{namtra_ldf}
438%-------------------------------------------------------------------------------------------------------------
439
440Options are defined through the \ngn{namtra\_ldf} namelist variables.
441They are regrouped in four items, allowing to specify
442$(i)$ the type of operator used (none, laplacian, bilaplacian),
443$(ii)$ the direction along which the operator acts (iso-level, horizontal, iso-neutral),
444$(iii)$ some specific options related to the rotated operators (\ie non-iso-level operator), and
445$(iv)$ the specification of eddy diffusivity coefficient (either constant or variable in space and time).
446Item $(iv)$ will be described in \autoref{chap:LDF}.
447The direction along which the operators act is defined through the slope between
448this direction and the iso-level surfaces.
449The slope is computed in the \mdl{ldfslp} module and will also be described in \autoref{chap:LDF}.
450
451The lateral diffusion of tracers is evaluated using a forward scheme,
452\ie the tracers appearing in its expression are the \textit{before} tracers in time,
453except for the pure vertical component that appears when a rotation tensor is used.
454This latter component is solved implicitly together with the vertical diffusion term (see \autoref{chap:STP}).
455When \np{ln\_traldf\_msc}~\forcode{= .true.}, a Method of Stabilizing Correction is used in which
456the pure vertical component is split into an explicit and an implicit part \citep{Lemarie_OM2012}.
457
458% -------------------------------------------------------------------------------------------------------------
459% Type of operator
460% -------------------------------------------------------------------------------------------------------------
461\subsection[Type of operator (\protect\np{ln\_traldf}\{\_NONE,\_lap,\_blp\}\})]{Type of operator (\protect\np{ln\_traldf\_NONE}, \protect\np{ln\_traldf\_lap}, or \protect\np{ln\_traldf\_blp}) }
462\label{subsec:TRA_ldf_op}
463
464Three operator options are proposed and, one and only one of them must be selected:
465
466\begin{description}
467\item[\np{ln\_traldf\_NONE}~\forcode{= .true.}:]
468 no operator selected, the lateral diffusive tendency will not be applied to the tracer equation.
469 This option can be used when the selected advection scheme is diffusive enough (MUSCL scheme for example).
470\item[\np{ln\_traldf\_lap}~\forcode{= .true.}:]
471 a laplacian operator is selected.
472 This harmonic operator takes the following expression: $\mathpzc{L}(T) = \nabla \cdot A_{ht} \; \nabla T$,
473 where the gradient operates along the selected direction (see \autoref{subsec:TRA_ldf_dir}),
474 and $A_{ht}$ is the eddy diffusivity coefficient expressed in $m^2/s$ (see \autoref{chap:LDF}).
475\item[\np{ln\_traldf\_blp}~\forcode{= .true.}]:
476 a bilaplacian operator is selected.
477 This biharmonic operator takes the following expression:
478 $\mathpzc{B} = - \mathpzc{L}(\mathpzc{L}(T)) = - \nabla \cdot b \nabla (\nabla \cdot b \nabla T)$
479 where the gradient operats along the selected direction,
480 and $b^2 = B_{ht}$ is the eddy diffusivity coefficient expressed in $m^4/s$ (see \autoref{chap:LDF}).
481 In the code, the bilaplacian operator is obtained by calling the laplacian twice.
482\end{description}
483
484Both laplacian and bilaplacian operators ensure the total tracer variance decrease.
485Their primary role is to provide strong dissipation at the smallest scale supported by the grid while
486minimizing the impact on the larger scale features.
487The main difference between the two operators is the scale selectiveness.
488The bilaplacian damping time (\ie its spin down time) scales like $\lambda^{-4}$ for
489disturbances of wavelength $\lambda$ (so that short waves damped more rapidelly than long ones),
490whereas the laplacian damping time scales only like $\lambda^{-2}$.
491
492% -------------------------------------------------------------------------------------------------------------
493% Direction of action
494% -------------------------------------------------------------------------------------------------------------
495\subsection[Action direction (\protect\np{ln\_traldf}\{\_lev,\_hor,\_iso,\_triad\})]{Direction of action (\protect\np{ln\_traldf\_lev}, \protect\np{ln\_traldf\_hor}, \protect\np{ln\_traldf\_iso}, or \protect\np{ln\_traldf\_triad}) }
496\label{subsec:TRA_ldf_dir}
497
498The choice of a direction of action determines the form of operator used.
499The operator is a simple (re-entrant) laplacian acting in the (\textbf{i},\textbf{j}) plane when
500iso-level option is used (\np{ln\_traldf\_lev}~\forcode{= .true.}) or
501when a horizontal (\ie geopotential) operator is demanded in \textit{z}-coordinate
502(\np{ln\_traldf\_hor} and \np{ln\_zco} equal \forcode{.true.}).
503The associated code can be found in the \mdl{traldf\_lap\_blp} module.
504The operator is a rotated (re-entrant) laplacian when
505the direction along which it acts does not coincide with the iso-level surfaces,
506that is when standard or triad iso-neutral option is used
507(\np{ln\_traldf\_iso} or \np{ln\_traldf\_triad} equals \forcode{.true.},
508see \mdl{traldf\_iso} or \mdl{traldf\_triad} module, resp.), or
509when a horizontal (\ie geopotential) operator is demanded in \textit{s}-coordinate
510(\np{ln\_traldf\_hor} and \np{ln\_sco} equal \forcode{.true.})
511\footnote{In this case, the standard iso-neutral operator will be automatically selected}.
512In that case, a rotation is applied to the gradient(s) that appears in the operator so that
513diffusive fluxes acts on the three spatial direction.
514
515The resulting discret form of the three operators (one iso-level and two rotated one) is given in
516the next two sub-sections.
517
518% -------------------------------------------------------------------------------------------------------------
519% iso-level operator
520% -------------------------------------------------------------------------------------------------------------
521\subsection{Iso-level (bi -)laplacian operator ( \protect\np{ln\_traldf\_iso}) }
522\label{subsec:TRA_ldf_lev}
523
524The laplacian diffusion operator acting along the model (\textit{i,j})-surfaces is given by:
525\begin{equation}
526 \label{eq:tra_ldf_lap}
527 D_t^{lT} = \frac{1}{b_t} \Bigg( \delta_{i} \lt[ A_u^{lT} \; \frac{e_{2u} \, e_{3u}}{e_{1u}} \; \delta_{i + 1/2} [T] \rt]
528 + \delta_{j} \lt[ A_v^{lT} \; \frac{e_{1v} \, e_{3v}}{e_{2v}} \; \delta_{j + 1/2} [T] \rt] \Bigg)
529\end{equation}
530where $b_t = e_{1t} \, e_{2t} \, e_{3t}$ is the volume of $T$-cells and
531where zero diffusive fluxes is assumed across solid boundaries,
532first (and third in bilaplacian case) horizontal tracer derivative are masked.
533It is implemented in the \rou{traldf\_lap} subroutine found in the \mdl{traldf\_lap} module.
534The module also contains \rou{traldf\_blp}, the subroutine calling twice \rou{traldf\_lap} in order to
535compute the iso-level bilaplacian operator.
536
537It is a \textit{horizontal} operator (\ie acting along geopotential surfaces) in
538the $z$-coordinate with or without partial steps, but is simply an iso-level operator in the $s$-coordinate.
539It is thus used when, in addition to \np{ln\_traldf\_lap} or \np{ln\_traldf\_blp}~\forcode{= .true.},
540we have \np{ln\_traldf\_lev}~\forcode{= .true.} or \np{ln\_traldf\_hor}~=~\np{ln\_zco}~\forcode{= .true.}.
541In both cases, it significantly contributes to diapycnal mixing.
542It is therefore never recommended, even when using it in the bilaplacian case.
543
544Note that in the partial step $z$-coordinate (\np{ln\_zps}~\forcode{= .true.}),
545tracers in horizontally adjacent cells are located at different depths in the vicinity of the bottom.
546In this case, horizontal derivatives in (\autoref{eq:tra_ldf_lap}) at the bottom level require a specific treatment.
547They are calculated in the \mdl{zpshde} module, described in \autoref{sec:TRA_zpshde}.
548
549% -------------------------------------------------------------------------------------------------------------
550% Rotated laplacian operator
551% -------------------------------------------------------------------------------------------------------------
552\subsection{Standard and triad (bi -)laplacian operator}
553\label{subsec:TRA_ldf_iso_triad}
554
555%&& Standard rotated (bi -)laplacian operator
556%&& ----------------------------------------------
557\subsubsection{Standard rotated (bi -)laplacian operator (\protect\mdl{traldf\_iso})}
558\label{subsec:TRA_ldf_iso}
559The general form of the second order lateral tracer subgrid scale physics (\autoref{eq:PE_zdf})
560takes the following semi -discrete space form in $z$- and $s$-coordinates:
561\begin{equation}
562 \label{eq:tra_ldf_iso}
563 \begin{split}
564 D_T^{lT} = \frac{1}{b_t} \Bigg[ \quad &\delta_i A_u^{lT} \lt( \frac{e_{2u} e_{3u}}{e_{1u}} \, \delta_{i + 1/2} [T]
565 - e_{2u} r_{1u} \, \overline{\overline{\delta_{k + 1/2} [T]}}^{\,i + 1/2,k} \rt) \Bigg. \\
566 + &\delta_j A_v^{lT} \lt( \frac{e_{1v} e_{3v}}{e_{2v}} \, \delta_{j + 1/2} [T]
567 - e_{1v} r_{2v} \, \overline{\overline{\delta_{k + 1/2} [T]}}^{\,j + 1/2,k} \rt) \\
568 + &\delta_k A_w^{lT} \lt( \frac{e_{1w} e_{2w}}{e_{3w}} (r_{1w}^2 + r_{2w}^2) \, \delta_{k + 1/2} [T] \rt. \\
569 & \qquad \quad \Bigg. \lt. - e_{2w} r_{1w} \, \overline{\overline{\delta_{i + 1/2} [T]}}^{\,i,k + 1/2}
570 - e_{1w} r_{2w} \, \overline{\overline{\delta_{j + 1/2} [T]}}^{\,j,k + 1/2} \rt) \Bigg]
571 \end{split}
572\end{equation}
573where $b_t = e_{1t} \, e_{2t} \, e_{3t}$ is the volume of $T$-cells,
574$r_1$ and $r_2$ are the slopes between the surface of computation ($z$- or $s$-surfaces) and
575the surface along which the diffusion operator acts (\ie horizontal or iso-neutral surfaces).
576It is thus used when, in addition to \np{ln\_traldf\_lap}~\forcode{= .true.},
577we have \np{ln\_traldf\_iso}~\forcode{= .true.},
578or both \np{ln\_traldf\_hor}~\forcode{= .true.} and \np{ln\_zco}~\forcode{= .true.}.
579The way these slopes are evaluated is given in \autoref{sec:LDF_slp}.
580At the surface, bottom and lateral boundaries, the turbulent fluxes of heat and salt are set to zero using
581the mask technique (see \autoref{sec:LBC_coast}).
582
583The operator in \autoref{eq:tra_ldf_iso} involves both lateral and vertical derivatives.
584For numerical stability, the vertical second derivative must be solved using the same implicit time scheme as that
585used in the vertical physics (see \autoref{sec:TRA_zdf}).
586For computer efficiency reasons, this term is not computed in the \mdl{traldf\_iso} module,
587but in the \mdl{trazdf} module where, if iso-neutral mixing is used,
588the vertical mixing coefficient is simply increased by $\frac{e_{1w} e_{2w}}{e_{3w}}(r_{1w}^2 + r_{2w}^2)$.
589
590This formulation conserves the tracer but does not ensure the decrease of the tracer variance.
591Nevertheless the treatment performed on the slopes (see \autoref{chap:LDF}) allows the model to run safely without
592any additional background horizontal diffusion \citep{Guilyardi_al_CD01}.
593
594Note that in the partial step $z$-coordinate (\np{ln\_zps}~\forcode{= .true.}),
595the horizontal derivatives at the bottom level in \autoref{eq:tra_ldf_iso} require a specific treatment.
596They are calculated in module zpshde, described in \autoref{sec:TRA_zpshde}.
597
598%&& Triad rotated (bi -)laplacian operator
599%&& -------------------------------------------
600\subsubsection{Triad rotated (bi -)laplacian operator (\protect\np{ln\_traldf\_triad})}
601\label{subsec:TRA_ldf_triad}
602
603If the Griffies triad scheme is employed (\np{ln\_traldf\_triad}~\forcode{= .true.}; see \autoref{apdx:triad})
604
605An alternative scheme developed by \cite{Griffies_al_JPO98} which ensures tracer variance decreases
606is also available in \NEMO (\np{ln\_traldf\_grif}~\forcode{= .true.}).
607A complete description of the algorithm is given in \autoref{apdx:triad}.
608
609The lateral fourth order bilaplacian operator on tracers is obtained by applying (\autoref{eq:tra_ldf_lap}) twice.
610The operator requires an additional assumption on boundary conditions:
611both first and third derivative terms normal to the coast are set to zero.
612
613The lateral fourth order operator formulation on tracers is obtained by applying (\autoref{eq:tra_ldf_iso}) twice.
614It requires an additional assumption on boundary conditions:
615first and third derivative terms normal to the coast,
616normal to the bottom and normal to the surface are set to zero.
617
618%&& Option for the rotated operators
619%&& ----------------------------------------------
620\subsubsection{Option for the rotated operators}
621\label{subsec:TRA_ldf_options}
622
623\begin{itemize}
624\item \np{ln\_traldf\_msc} = Method of Stabilizing Correction (both operators)
625\item \np{rn\_slpmax} = slope limit (both operators)
626\item \np{ln\_triad\_iso} = pure horizontal mixing in ML (triad only)
627\item \np{rn\_sw\_triad} $= 1$ switching triad; $= 0$ all 4 triads used (triad only)
628\item \np{ln\_botmix\_triad} = lateral mixing on bottom (triad only)
629\end{itemize}
630
631% ================================================================
632% Tracer Vertical Diffusion
633% ================================================================
634\section{Tracer vertical diffusion (\protect\mdl{trazdf})}
635\label{sec:TRA_zdf}
636%--------------------------------------------namzdf---------------------------------------------------------
637
638\nlst{namzdf}
639%--------------------------------------------------------------------------------------------------------------
640
641Options are defined through the \ngn{namzdf} namelist variables.
642The formulation of the vertical subgrid scale tracer physics is the same for all the vertical coordinates,
643and is based on a laplacian operator.
644The vertical diffusion operator given by (\autoref{eq:PE_zdf}) takes the following semi -discrete space form:
645\begin{gather*}
646 % \label{eq:tra_zdf}
647 D^{vT}_T = \frac{1}{e_{3t}} \, \delta_k \lt[ \, \frac{A^{vT}_w}{e_{3w}} \delta_{k + 1/2}[T] \, \rt] \\
648 D^{vS}_T = \frac{1}{e_{3t}} \; \delta_k \lt[ \, \frac{A^{vS}_w}{e_{3w}} \delta_{k + 1/2}[S] \, \rt]
649\end{gather*}
650where $A_w^{vT}$ and $A_w^{vS}$ are the vertical eddy diffusivity coefficients on temperature and salinity,
651respectively.
652Generally, $A_w^{vT} = A_w^{vS}$ except when double diffusive mixing is parameterised
653(\ie \key{zdfddm} is defined).
654The way these coefficients are evaluated is given in \autoref{chap:ZDF} (ZDF).
655Furthermore, when iso-neutral mixing is used, both mixing coefficients are increased by
656$\frac{e_{1w} e_{2w}}{e_{3w} }({r_{1w}^2 + r_{2w}^2})$ to account for the vertical second derivative of
657\autoref{eq:tra_ldf_iso}.
658
659At the surface and bottom boundaries, the turbulent fluxes of heat and salt must be specified.
660At the surface they are prescribed from the surface forcing and added in a dedicated routine
661(see \autoref{subsec:TRA_sbc}), whilst at the bottom they are set to zero for heat and salt unless
662a geothermal flux forcing is prescribed as a bottom boundary condition (see \autoref{subsec:TRA_bbc}).
663
664The large eddy coefficient found in the mixed layer together with high vertical resolution implies that
665in the case of explicit time stepping (\np{ln\_zdfexp}~\forcode{= .true.})
666there would be too restrictive a constraint on the time step.
667Therefore, the default implicit time stepping is preferred for the vertical diffusion since
668it overcomes the stability constraint.
669A forward time differencing scheme (\np{ln\_zdfexp}~\forcode{= .true.}) using
670a time splitting technique (\np{nn\_zdfexp} $> 1$) is provided as an alternative.
671Namelist variables \np{ln\_zdfexp} and \np{nn\_zdfexp} apply to both tracers and dynamics.
672
673% ================================================================
674% External Forcing
675% ================================================================
676\section{External forcing}
677\label{sec:TRA_sbc_qsr_bbc}
678
679% -------------------------------------------------------------------------------------------------------------
680% surface boundary condition
681% -------------------------------------------------------------------------------------------------------------
682\subsection{Surface boundary condition (\protect\mdl{trasbc})}
683\label{subsec:TRA_sbc}
684
685The surface boundary condition for tracers is implemented in a separate module (\mdl{trasbc}) instead of
686entering as a boundary condition on the vertical diffusion operator (as in the case of momentum).
687This has been found to enhance readability of the code.
688The two formulations are completely equivalent;
689the forcing terms in trasbc are the surface fluxes divided by the thickness of the top model layer.
690
691Due to interactions and mass exchange of water ($F_{mass}$) with other Earth system components
692(\ie atmosphere, sea-ice, land), the change in the heat and salt content of the surface layer of the ocean is due
693both to the heat and salt fluxes crossing the sea surface (not linked with $F_{mass}$) and
694to the heat and salt content of the mass exchange.
695They are both included directly in $Q_{ns}$, the surface heat flux,
696and $F_{salt}$, the surface salt flux (see \autoref{chap:SBC} for further details).
697By doing this, the forcing formulation is the same for any tracer (including temperature and salinity).
698
699The surface module (\mdl{sbcmod}, see \autoref{chap:SBC}) provides the following forcing fields (used on tracers):
700
701\begin{itemize}
702\item
703 $Q_{ns}$, the non-solar part of the net surface heat flux that crosses the sea surface
704 (\ie the difference between the total surface heat flux and the fraction of the short wave flux that
705 penetrates into the water column, see \autoref{subsec:TRA_qsr})
706 plus the heat content associated with of the mass exchange with the atmosphere and lands.
707\item
708 $\textit{sfx}$, the salt flux resulting from ice-ocean mass exchange (freezing, melting, ridging...)
709\item
710 \textit{emp}, the mass flux exchanged with the atmosphere (evaporation minus precipitation) and
711 possibly with the sea-ice and ice-shelves.
712\item
713 \textit{rnf}, the mass flux associated with runoff
714 (see \autoref{sec:SBC_rnf} for further detail of how it acts on temperature and salinity tendencies)
715\item
716 \textit{fwfisf}, the mass flux associated with ice shelf melt,
717 (see \autoref{sec:SBC_isf} for further details on how the ice shelf melt is computed and applied).
718\end{itemize}
719
720The surface boundary condition on temperature and salinity is applied as follows:
721\begin{equation}
722 \label{eq:tra_sbc}
723 \begin{alignedat}{2}
724 F^T &= \frac{1}{C_p} &\frac{1}{\rho_o \lt. e_{3t} \rt|_{k = 1}} &\overline{Q_{ns} }^t \\
725 F^S &= &\frac{1}{\rho_o \lt. e_{3t} \rt|_{k = 1}} &\overline{\textit{sfx}}^t
726 \end{alignedat}
727\end{equation}
728where $\overline x^t$ means that $x$ is averaged over two consecutive time steps
729($t - \rdt / 2$ and $t + \rdt / 2$).
730Such time averaging prevents the divergence of odd and even time step (see \autoref{chap:STP}).
731
732In the linear free surface case (\np{ln\_linssh}~\forcode{= .true.}), an additional term has to be added on
733both temperature and salinity.
734On temperature, this term remove the heat content associated with mass exchange that has been added to $Q_{ns}$.
735On salinity, this term mimics the concentration/dilution effect that would have resulted from a change in
736the volume of the first level.
737The resulting surface boundary condition is applied as follows:
738\begin{equation}
739 \label{eq:tra_sbc_lin}
740 \begin{alignedat}{2}
741 F^T &= \frac{1}{C_p} &\frac{1}{\rho_o \lt. e_{3t} \rt|_{k = 1}}
742 &\overline{(Q_{ns} - C_p \, \textit{emp} \lt. T \rt|_{k = 1})}^t \\
743 F^S &= &\frac{1}{\rho_o \lt. e_{3t} \rt|_{k = 1}}
744 &\overline{(\textit{sfx} - \textit{emp} \lt. S \rt|_{k = 1})}^t
745 \end{alignedat}
746\end{equation}
747Note that an exact conservation of heat and salt content is only achieved with non-linear free surface.
748In the linear free surface case, there is a small imbalance.
749The imbalance is larger than the imbalance associated with the Asselin time filter \citep{Leclair_Madec_OM09}.
750This is the reason why the modified filter is not applied in the linear free surface case (see \autoref{chap:STP}).
751
752% -------------------------------------------------------------------------------------------------------------
753% Solar Radiation Penetration
754% -------------------------------------------------------------------------------------------------------------
755\subsection{Solar radiation penetration (\protect\mdl{traqsr})}
756\label{subsec:TRA_qsr}
757%--------------------------------------------namqsr--------------------------------------------------------
758
759\nlst{namtra_qsr}
760%--------------------------------------------------------------------------------------------------------------
761
762Options are defined through the \ngn{namtra\_qsr} namelist variables.
763When the penetrative solar radiation option is used (\np{ln\_flxqsr}~\forcode{= .true.}),
764the solar radiation penetrates the top few tens of meters of the ocean.
765If it is not used (\np{ln\_flxqsr}~\forcode{= .false.}) all the heat flux is absorbed in the first ocean level.
766Thus, in the former case a term is added to the time evolution equation of temperature \autoref{eq:PE_tra_T} and
767the surface boundary condition is modified to take into account only the non-penetrative part of the surface
768heat flux:
769\begin{equation}
770 \label{eq:PE_qsr}
771 \begin{gathered}
772 \pd[T]{t} = \ldots + \frac{1}{\rho_o \, C_p \, e_3} \; \pd[I]{k} \\
773 Q_{ns} = Q_\text{Total} - Q_{sr}
774 \end{gathered}
775\end{equation}
776where $Q_{sr}$ is the penetrative part of the surface heat flux (\ie the shortwave radiation) and
777$I$ is the downward irradiance ($\lt. I \rt|_{z = \eta} = Q_{sr}$).
778The additional term in \autoref{eq:PE_qsr} is discretized as follows:
779\begin{equation}
780 \label{eq:tra_qsr}
781 \frac{1}{\rho_o \, C_p \, e_3} \, \pd[I]{k} \equiv \frac{1}{\rho_o \, C_p \, e_{3t}} \delta_k [I_w]
782\end{equation}
783
784The shortwave radiation, $Q_{sr}$, consists of energy distributed across a wide spectral range.
785The ocean is strongly absorbing for wavelengths longer than 700~nm and these wavelengths contribute to
786heating the upper few tens of centimetres.
787The fraction of $Q_{sr}$ that resides in these almost non-penetrative wavebands, $R$, is $\sim 58\%$
788(specified through namelist parameter \np{rn\_abs}).
789It is assumed to penetrate the ocean with a decreasing exponential profile, with an e-folding depth scale, $\xi_0$,
790of a few tens of centimetres (typically $\xi_0 = 0.35~m$ set as \np{rn\_si0} in the \ngn{namtra\_qsr} namelist).
791For shorter wavelengths (400-700~nm), the ocean is more transparent, and solar energy propagates to
792larger depths where it contributes to local heating.
793The way this second part of the solar energy penetrates into the ocean depends on which formulation is chosen.
794In the simple 2-waveband light penetration scheme (\np{ln\_qsr\_2bd}~\forcode{= .true.})
795a chlorophyll-independent monochromatic formulation is chosen for the shorter wavelengths,
796leading to the following expression \citep{Paulson1977}:
797$798 % \label{eq:traqsr_iradiance} 799 I(z) = Q_{sr} \lt[ Re^{- z / \xi_0} + (1 - R) e^{- z / \xi_1} \rt] 800$
801where $\xi_1$ is the second extinction length scale associated with the shorter wavelengths.
802It is usually chosen to be 23~m by setting the \np{rn\_si0} namelist parameter.
803The set of default values ($\xi_0, \xi_1, R$) corresponds to a Type I water in Jerlov's (1968) classification
804(oligotrophic waters).
805
806Such assumptions have been shown to provide a very crude and simplistic representation of
807observed light penetration profiles (\cite{Morel_JGR88}, see also \autoref{fig:traqsr_irradiance}).
808Light absorption in the ocean depends on particle concentration and is spectrally selective.
809\cite{Morel_JGR88} has shown that an accurate representation of light penetration can be provided by
810a 61 waveband formulation.
811Unfortunately, such a model is very computationally expensive.
812Thus, \cite{Lengaigne_al_CD07} have constructed a simplified version of this formulation in which
813visible light is split into three wavebands: blue (400-500 nm), green (500-600 nm) and red (600-700nm).
814For each wave-band, the chlorophyll-dependent attenuation coefficient is fitted to the coefficients computed from
815the full spectral model of \cite{Morel_JGR88} (as modified by \cite{Morel_Maritorena_JGR01}),
816assuming the same power-law relationship.
817As shown in \autoref{fig:traqsr_irradiance}, this formulation, called RGB (Red-Green-Blue),
818reproduces quite closely the light penetration profiles predicted by the full spectal model,
819but with much greater computational efficiency.
820The 2-bands formulation does not reproduce the full model very well.
821
822The RGB formulation is used when \np{ln\_qsr\_rgb}~\forcode{= .true.}.
823The RGB attenuation coefficients (\ie the inverses of the extinction length scales) are tabulated over
82461 nonuniform chlorophyll classes ranging from 0.01 to 10 g.Chl/L
825(see the routine \rou{trc\_oce\_rgb} in \mdl{trc\_oce} module).
826Four types of chlorophyll can be chosen in the RGB formulation:
827
828\begin{description}
829\item[\np{nn\_chdta}~\forcode{= 0}]
830 a constant 0.05 g.Chl/L value everywhere ;
831\item[\np{nn\_chdta}~\forcode{= 1}]
832 an observed time varying chlorophyll deduced from satellite surface ocean color measurement spread uniformly in
833 the vertical direction;
834\item[\np{nn\_chdta}~\forcode{= 2}]
835 same as previous case except that a vertical profile of chlorophyl is used.
836 Following \cite{Morel_Berthon_LO89}, the profile is computed from the local surface chlorophyll value;
837\item[\np{ln\_qsr\_bio}~\forcode{= .true.}]
838 simulated time varying chlorophyll by TOP biogeochemical model.
839 In this case, the RGB formulation is used to calculate both the phytoplankton light limitation in
840 PISCES or LOBSTER and the oceanic heating rate.
841\end{description}
842
843The trend in \autoref{eq:tra_qsr} associated with the penetration of the solar radiation is added to
844the temperature trend, and the surface heat flux is modified in routine \mdl{traqsr}.
845
846When the $z$-coordinate is preferred to the $s$-coordinate,
847the depth of $w-$levels does not significantly vary with location.
848The level at which the light has been totally absorbed
849(\ie it is less than the computer precision) is computed once,
850and the trend associated with the penetration of the solar radiation is only added down to that level.
851Finally, note that when the ocean is shallow ($<$ 200~m), part of the solar radiation can reach the ocean floor.
852In this case, we have chosen that all remaining radiation is absorbed in the last ocean level
853(\ie $I$ is masked).
854
855%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
856\begin{figure}[!t]
857 \begin{center}
858 \includegraphics[]{Fig_TRA_Irradiance}
859 \caption{
860 \protect\label{fig:traqsr_irradiance}
861 Penetration profile of the downward solar irradiance calculated by four models.
862 Two waveband chlorophyll-independent formulation (blue),
863 a chlorophyll-dependent monochromatic formulation (green),
864 4 waveband RGB formulation (red),
865 61 waveband Morel (1988) formulation (black) for a chlorophyll concentration of
866 (a) Chl=0.05 mg/m$^3$ and (b) Chl=0.5 mg/m$^3$.
867 From \citet{Lengaigne_al_CD07}.
868 }
869 \end{center}
870\end{figure}
871%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
872
873% -------------------------------------------------------------------------------------------------------------
874% Bottom Boundary Condition
875% -------------------------------------------------------------------------------------------------------------
876\subsection{Bottom boundary condition (\protect\mdl{trabbc})}
877\label{subsec:TRA_bbc}
878%--------------------------------------------nambbc--------------------------------------------------------
879
880\nlst{nambbc}
881%--------------------------------------------------------------------------------------------------------------
882%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
883\begin{figure}[!t]
884 \begin{center}
885 \includegraphics[]{Fig_TRA_geoth}
886 \caption{
887 \protect\label{fig:geothermal}
888 Geothermal Heat flux (in $mW.m^{-2}$) used by \cite{Emile-Geay_Madec_OS09}.
889 It is inferred from the age of the sea floor and the formulae of \citet{Stein_Stein_Nat92}.
890 }
891 \end{center}
892\end{figure}
893%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
894
895Usually it is assumed that there is no exchange of heat or salt through the ocean bottom,
896\ie a no flux boundary condition is applied on active tracers at the bottom.
897This is the default option in \NEMO, and it is implemented using the masking technique.
898However, there is a non-zero heat flux across the seafloor that is associated with solid earth cooling.
899This flux is weak compared to surface fluxes (a mean global value of $\sim 0.1 \, W/m^2$ \citep{Stein_Stein_Nat92}),
900but it warms systematically the ocean and acts on the densest water masses.
901Taking this flux into account in a global ocean model increases the deepest overturning cell
902(\ie the one associated with the Antarctic Bottom Water) by a few Sverdrups \citep{Emile-Geay_Madec_OS09}.
903
904Options are defined through the \ngn{namtra\_bbc} namelist variables.
905The presence of geothermal heating is controlled by setting the namelist parameter \np{ln\_trabbc} to true.
906Then, when \np{nn\_geoflx} is set to 1, a constant geothermal heating is introduced whose value is given by
907the \np{nn\_geoflx\_cst}, which is also a namelist parameter.
908When \np{nn\_geoflx} is set to 2, a spatially varying geothermal heat flux is introduced which is provided in
909the \ifile{geothermal\_heating} NetCDF file (\autoref{fig:geothermal}) \citep{Emile-Geay_Madec_OS09}.
910
911% ================================================================
912% Bottom Boundary Layer
913% ================================================================
914\section{Bottom boundary layer (\protect\mdl{trabbl} - \protect\key{trabbl})}
915\label{sec:TRA_bbl}
916%--------------------------------------------nambbl---------------------------------------------------------
917
918\nlst{nambbl}
919%--------------------------------------------------------------------------------------------------------------
920
921Options are defined through the \ngn{nambbl} namelist variables.
922In a $z$-coordinate configuration, the bottom topography is represented by a series of discrete steps.
923This is not adequate to represent gravity driven downslope flows.
924Such flows arise either downstream of sills such as the Strait of Gibraltar or Denmark Strait,
925where dense water formed in marginal seas flows into a basin filled with less dense water,
926or along the continental slope when dense water masses are formed on a continental shelf.
927The amount of entrainment that occurs in these gravity plumes is critical in determining the density and
928volume flux of the densest waters of the ocean, such as Antarctic Bottom Water, or North Atlantic Deep Water.
929$z$-coordinate models tend to overestimate the entrainment,
930because the gravity flow is mixed vertically by convection as it goes ''downstairs'' following the step topography,
931sometimes over a thickness much larger than the thickness of the observed gravity plume.
932A similar problem occurs in the $s$-coordinate when the thickness of the bottom level varies rapidly downstream of
933a sill \citep{Willebrand_al_PO01}, and the thickness of the plume is not resolved.
934
935The idea of the bottom boundary layer (BBL) parameterisation, first introduced by \citet{Beckmann_Doscher1997},
936is to allow a direct communication between two adjacent bottom cells at different levels,
937whenever the densest water is located above the less dense water.
938The communication can be by a diffusive flux (diffusive BBL), an advective flux (advective BBL), or both.
939In the current implementation of the BBL, only the tracers are modified, not the velocities.
940Furthermore, it only connects ocean bottom cells, and therefore does not include all the improvements introduced by
941\citet{Campin_Goosse_Tel99}.
942
943% -------------------------------------------------------------------------------------------------------------
944% Diffusive BBL
945% -------------------------------------------------------------------------------------------------------------
946\subsection{Diffusive bottom boundary layer (\protect\np{nn\_bbl\_ldf}~\forcode{= 1})}
947\label{subsec:TRA_bbl_diff}
948
949When applying sigma-diffusion (\key{trabbl} defined and \np{nn\_bbl\_ldf} set to 1),
950the diffusive flux between two adjacent cells at the ocean floor is given by
951$952 % \label{eq:tra_bbl_diff} 953 \vect F_\sigma = A_l^\sigma \, \nabla_\sigma T 954$
955with $\nabla_\sigma$ the lateral gradient operator taken between bottom cells, and
956$A_l^\sigma$ the lateral diffusivity in the BBL.
957Following \citet{Beckmann_Doscher1997}, the latter is prescribed with a spatial dependence,
958\ie in the conditional form
959\begin{equation}
960 \label{eq:tra_bbl_coef}
961 A_l^\sigma (i,j,t) =
962 \begin{cases}
963 A_{bbl} & \text{if~} \nabla_\sigma \rho \cdot \nabla H < 0 \\
964 \\
965 0 & \text{otherwise} \\
966 \end{cases}
967\end{equation}
968where $A_{bbl}$ is the BBL diffusivity coefficient, given by the namelist parameter \np{rn\_ahtbbl} and
969usually set to a value much larger than the one used for lateral mixing in the open ocean.
970The constraint in \autoref{eq:tra_bbl_coef} implies that sigma-like diffusion only occurs when
971the density above the sea floor, at the top of the slope, is larger than in the deeper ocean
972(see green arrow in \autoref{fig:bbl}).
973In practice, this constraint is applied separately in the two horizontal directions,
974and the density gradient in \autoref{eq:tra_bbl_coef} is evaluated with the log gradient formulation:
975$976 % \label{eq:tra_bbl_Drho} 977 \nabla_\sigma \rho / \rho = \alpha \, \nabla_\sigma T + \beta \, \nabla_\sigma S 978$
979where $\rho$, $\alpha$ and $\beta$ are functions of $\overline T^\sigma$, $\overline S^\sigma$ and
980$\overline H^\sigma$, the along bottom mean temperature, salinity and depth, respectively.
981
982% -------------------------------------------------------------------------------------------------------------
983% Advective BBL
984% -------------------------------------------------------------------------------------------------------------
985\subsection{Advective bottom boundary layer (\protect\np{nn\_bbl\_adv}~\forcode{= 1..2})}
986\label{subsec:TRA_bbl_adv}
987
988%\sgacomment{
989% "downsloping flow" has been replaced by "downslope flow" in the following
990% if this is not what is meant then "downwards sloping flow" is also a possibility"
991%}
992
993%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
994\begin{figure}[!t]
995 \begin{center}
996 \includegraphics[]{Fig_BBL_adv}
997 \caption{
998 \protect\label{fig:bbl}
999 Advective/diffusive Bottom Boundary Layer.
1000 The BBL parameterisation is activated when $\rho^i_{kup}$ is larger than $\rho^{i + 1}_{kdnw}$.
1001 Red arrows indicate the additional overturning circulation due to the advective BBL.
1002 The transport of the downslope flow is defined either as the transport of the bottom ocean cell (black arrow),
1003 or as a function of the along slope density gradient.
1004 The green arrow indicates the diffusive BBL flux directly connecting $kup$ and $kdwn$ ocean bottom cells.
1005 }
1006 \end{center}
1007\end{figure}
1008%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
1009
1010%!! nn_bbl_adv = 1 use of the ocean velocity as bbl velocity
1011%!! nn_bbl_adv = 2 follow Campin and Goosse (1999) implentation
1012%!! i.e. transport proportional to the along-slope density gradient
1013
1014%%%gmcomment : this section has to be really written
1015
1016When applying an advective BBL (\np{nn\_bbl\_adv}~\forcode{= 1..2}), an overturning circulation is added which
1017connects two adjacent bottom grid-points only if dense water overlies less dense water on the slope.
1018The density difference causes dense water to move down the slope.
1019
1020\np{nn\_bbl\_adv}~\forcode{= 1}:
1021the downslope velocity is chosen to be the Eulerian ocean velocity just above the topographic step
1022(see black arrow in \autoref{fig:bbl}) \citep{Beckmann_Doscher1997}.
1023It is a \textit{conditional advection}, that is, advection is allowed only
1024if dense water overlies less dense water on the slope (\ie $\nabla_\sigma \rho \cdot \nabla H < 0$) and
1025if the velocity is directed towards greater depth (\ie $\vect U \cdot \nabla H > 0$).
1026
1027\np{nn\_bbl\_adv}~\forcode{= 2}:
1028the downslope velocity is chosen to be proportional to $\Delta \rho$,
1029the density difference between the higher cell and lower cell densities \citep{Campin_Goosse_Tel99}.
1030The advection is allowed only if dense water overlies less dense water on the slope
1031(\ie $\nabla_\sigma \rho \cdot \nabla H < 0$).
1032For example, the resulting transport of the downslope flow, here in the $i$-direction (\autoref{fig:bbl}),
1033is simply given by the following expression:
1034$1035 % \label{eq:bbl_Utr} 1036 u^{tr}_{bbl} = \gamma g \frac{\Delta \rho}{\rho_o} e_{1u} \, min ({e_{3u}}_{kup},{e_{3u}}_{kdwn}) 1037$
1038where $\gamma$, expressed in seconds, is the coefficient of proportionality provided as \np{rn\_gambbl},
1039a namelist parameter, and \textit{kup} and \textit{kdwn} are the vertical index of the higher and lower cells,
1040respectively.
1041The parameter $\gamma$ should take a different value for each bathymetric step, but for simplicity,
1042and because no direct estimation of this parameter is available, a uniform value has been assumed.
1043The possible values for $\gamma$ range between 1 and $10~s$ \citep{Campin_Goosse_Tel99}.
1044
1045Scalar properties are advected by this additional transport $(u^{tr}_{bbl},v^{tr}_{bbl})$ using the upwind scheme.
1046Such a diffusive advective scheme has been chosen to mimic the entrainment between the downslope plume and
1047the surrounding water at intermediate depths.
1048The entrainment is replaced by the vertical mixing implicit in the advection scheme.
1049Let us consider as an example the case displayed in \autoref{fig:bbl} where
1050the density at level $(i,kup)$ is larger than the one at level $(i,kdwn)$.
1051The advective BBL scheme modifies the tracer time tendency of the ocean cells near the topographic step by
1052the downslope flow \autoref{eq:bbl_dw}, the horizontal \autoref{eq:bbl_hor} and
1053the upward \autoref{eq:bbl_up} return flows as follows:
1054\begin{alignat}{3}
1055 \label{eq:bbl_dw}
1056 \partial_t T^{do}_{kdw} &\equiv \partial_t T^{do}_{kdw}
1057 &&+ \frac{u^{tr}_{bbl}}{{b_t}^{do}_{kdw}} &&\lt( T^{sh}_{kup} - T^{do}_{kdw} \rt) \\
1058 \label{eq:bbl_hor}
1059 \partial_t T^{sh}_{kup} &\equiv \partial_t T^{sh}_{kup}
1060 &&+ \frac{u^{tr}_{bbl}}{{b_t}^{sh}_{kup}} &&\lt( T^{do}_{kup} - T^{sh}_{kup} \rt) \\
1061 %
1062 \intertext{and for $k =kdw-1,\;..., \; kup$ :}
1063 %
1064 \label{eq:bbl_up}
1065 \partial_t T^{do}_{k} &\equiv \partial_t S^{do}_{k}
1066 &&+ \frac{u^{tr}_{bbl}}{{b_t}^{do}_{k}} &&\lt( T^{do}_{k +1} - T^{sh}_{k} \rt)
1067\end{alignat}
1068where $b_t$ is the $T$-cell volume.
1069
1070Note that the BBL transport, $(u^{tr}_{bbl},v^{tr}_{bbl})$, is available in the model outputs.
1071It has to be used to compute the effective velocity as well as the effective overturning circulation.
1072
1073% ================================================================
1074% Tracer damping
1075% ================================================================
1076\section{Tracer damping (\protect\mdl{tradmp})}
1077\label{sec:TRA_dmp}
1078%--------------------------------------------namtra_dmp-------------------------------------------------
1079
1080\nlst{namtra_dmp}
1081%--------------------------------------------------------------------------------------------------------------
1082
1083In some applications it can be useful to add a Newtonian damping term into the temperature and salinity equations:
1084\begin{equation}
1085 \label{eq:tra_dmp}
1086 \begin{gathered}
1087 \pd[T]{t} = \cdots - \gamma (T - T_o) \\
1088 \pd[S]{t} = \cdots - \gamma (S - S_o)
1089 \end{gathered}
1090\end{equation}
1091where $\gamma$ is the inverse of a time scale, and $T_o$ and $S_o$ are given temperature and salinity fields
1092(usually a climatology).
1093Options are defined through the \ngn{namtra\_dmp} namelist variables.
1094The restoring term is added when the namelist parameter \np{ln\_tradmp} is set to true.
1095It also requires that both \np{ln\_tsd\_init} and \np{ln\_tsd\_tradmp} are set to true in
1096\ngn{namtsd} namelist as well as \np{sn\_tem} and \np{sn\_sal} structures are correctly set
1097(\ie that $T_o$ and $S_o$ are provided in input files and read using \mdl{fldread},
1098see \autoref{subsec:SBC_fldread}).
1099The restoring coefficient $\gamma$ is a three-dimensional array read in during the \rou{tra\_dmp\_init} routine.
1100The file name is specified by the namelist variable \np{cn\_resto}.
1101The DMP\_TOOLS tool is provided to allow users to generate the netcdf file.
1102
1103The two main cases in which \autoref{eq:tra_dmp} is used are
1104\textit{(a)} the specification of the boundary conditions along artificial walls of a limited domain basin and
1105\textit{(b)} the computation of the velocity field associated with a given $T$-$S$ field
1106(for example to build the initial state of a prognostic simulation,
1107or to use the resulting velocity field for a passive tracer study).
1108The first case applies to regional models that have artificial walls instead of open boundaries.
1109In the vicinity of these walls, $\gamma$ takes large values (equivalent to a time scale of a few days) whereas
1110it is zero in the interior of the model domain.
1111The second case corresponds to the use of the robust diagnostic method \citep{Sarmiento1982}.
1112It allows us to find the velocity field consistent with the model dynamics whilst
1113having a $T$, $S$ field close to a given climatological field ($T_o$, $S_o$).
1114
1115The robust diagnostic method is very efficient in preventing temperature drift in intermediate waters but
1116it produces artificial sources of heat and salt within the ocean.
1117It also has undesirable effects on the ocean convection.
1118It tends to prevent deep convection and subsequent deep-water formation, by stabilising the water column too much.
1119
1120The namelist parameter \np{nn\_zdmp} sets whether the damping should be applied in the whole water column or
1121only below the mixed layer (defined either on a density or $S_o$ criterion).
1122It is common to set the damping to zero in the mixed layer as the adjustment time scale is short here
1123\citep{Madec_al_JPO96}.
1124
1125For generating \ifile{resto}, see the documentation for the DMP tool provided with the source code under
1126\path{./tools/DMP_TOOLS}.
1127
1128% ================================================================
1129% Tracer time evolution
1130% ================================================================
1131\section{Tracer time evolution (\protect\mdl{tranxt})}
1132\label{sec:TRA_nxt}
1133%--------------------------------------------namdom-----------------------------------------------------
1134
1135\nlst{namdom}
1136%--------------------------------------------------------------------------------------------------------------
1137
1138Options are defined through the \ngn{namdom} namelist variables.
1139The general framework for tracer time stepping is a modified leap-frog scheme \citep{Leclair_Madec_OM09},
1140\ie a three level centred time scheme associated with a Asselin time filter (cf. \autoref{sec:STP_mLF}):
1141\begin{equation}
1142 \label{eq:tra_nxt}
1143 \begin{alignedat}{3}
1144 &(e_{3t}T)^{t + \rdt} &&= (e_{3t}T)_f^{t - \rdt} &&+ 2 \, \rdt \,e_{3t}^t \ \text{RHS}^t \\
1145 &(e_{3t}T)_f^t &&= (e_{3t}T)^t &&+ \, \gamma \, \lt[ (e_{3t}T)_f^{t - \rdt} - 2(e_{3t}T)^t + (e_{3t}T)^{t + \rdt} \rt] \\
1146 & && &&- \, \gamma \, \rdt \, \lt[ Q^{t + \rdt/2} - Q^{t - \rdt/2} \rt]
1147 \end{alignedat}
1148\end{equation}
1149where RHS is the right hand side of the temperature equation, the subscript $f$ denotes filtered values,
1150$\gamma$ is the Asselin coefficient, and $S$ is the total forcing applied on $T$
1151(\ie fluxes plus content in mass exchanges).
1152$\gamma$ is initialized as \np{rn\_atfp} (\textbf{namelist} parameter).
1153Its default value is \np{rn\_atfp}~\forcode{= 10.e-3}.
1154Note that the forcing correction term in the filter is not applied in linear free surface
1155(\jp{lk\_vvl}~\forcode{= .false.}) (see \autoref{subsec:TRA_sbc}).
1156Not also that in constant volume case, the time stepping is performed on $T$, not on its content, $e_{3t}T$.
1157
1158When the vertical mixing is solved implicitly, the update of the \textit{next} tracer fields is done in
1159\mdl{trazdf} module.
1160In this case only the swapping of arrays and the Asselin filtering is done in the \mdl{tranxt} module.
1161
1162In order to prepare for the computation of the \textit{next} time step, a swap of tracer arrays is performed:
1163$T^{t - \rdt} = T^t$ and $T^t = T_f$.
1164
1165% ================================================================
1166% Equation of State (eosbn2)
1167% ================================================================
1168\section{Equation of state (\protect\mdl{eosbn2}) }
1169\label{sec:TRA_eosbn2}
1170%--------------------------------------------nameos-----------------------------------------------------
1171
1172\nlst{nameos}
1173%--------------------------------------------------------------------------------------------------------------
1174
1175% -------------------------------------------------------------------------------------------------------------
1176% Equation of State
1177% -------------------------------------------------------------------------------------------------------------
1178\subsection{Equation of seawater (\protect\np{nn\_eos}~\forcode{= -1..1})}
1179\label{subsec:TRA_eos}
1180
1181The Equation Of Seawater (EOS) is an empirical nonlinear thermodynamic relationship linking seawater density,
1182$\rho$, to a number of state variables, most typically temperature, salinity and pressure.
1183Because density gradients control the pressure gradient force through the hydrostatic balance,
1184the equation of state provides a fundamental bridge between the distribution of active tracers and
1185the fluid dynamics.
1186Nonlinearities of the EOS are of major importance, in particular influencing the circulation through
1187determination of the static stability below the mixed layer,
1188thus controlling rates of exchange between the atmosphere and the ocean interior \citep{Roquet_JPO2015}.
1189Therefore an accurate EOS based on either the 1980 equation of state (EOS-80, \cite{UNESCO1983}) or
1190TEOS-10 \citep{TEOS10} standards should be used anytime a simulation of the real ocean circulation is attempted
1191\citep{Roquet_JPO2015}.
1192The use of TEOS-10 is highly recommended because
1193\textit{(i)} it is the new official EOS,
1194\textit{(ii)} it is more accurate, being based on an updated database of laboratory measurements, and
1195\textit{(iii)} it uses Conservative Temperature and Absolute Salinity (instead of potential temperature and
1196practical salinity for EOS-980, both variables being more suitable for use as model variables
1197\citep{TEOS10, Graham_McDougall_JPO13}.
1198EOS-80 is an obsolescent feature of the NEMO system, kept only for backward compatibility.
1199For process studies, it is often convenient to use an approximation of the EOS.
1200To that purposed, a simplified EOS (S-EOS) inspired by \citet{Vallis06} is also available.
1201
1202In the computer code, a density anomaly, $d_a = \rho / \rho_o - 1$, is computed, with $\rho_o$ a reference density.
1203Called \textit{rau0} in the code, $\rho_o$ is set in \mdl{phycst} to a value of $1,026~Kg/m^3$.
1204This is a sensible choice for the reference density used in a Boussinesq ocean climate model, as,
1205with the exception of only a small percentage of the ocean,
1206density in the World Ocean varies by no more than 2$\%$ from that value \citep{Gill1982}.
1207
1208Options are defined through the \ngn{nameos} namelist variables, and in particular \np{nn\_eos} which
1209controls the EOS used (\forcode{= -1} for TEOS10 ; \forcode{= 0} for EOS-80 ; \forcode{= 1} for S-EOS).
1210
1211\begin{description}
1212\item[\np{nn\_eos}~\forcode{= -1}]
1213 the polyTEOS10-bsq equation of seawater \citep{Roquet_OM2015} is used.
1214 The accuracy of this approximation is comparable to the TEOS-10 rational function approximation,
1215 but it is optimized for a boussinesq fluid and the polynomial expressions have simpler and
1216 more computationally efficient expressions for their derived quantities which make them more adapted for
1217 use in ocean models.
1218 Note that a slightly higher precision polynomial form is now used replacement of
1219 the TEOS-10 rational function approximation for hydrographic data analysis \citep{TEOS10}.
1220 A key point is that conservative state variables are used:
1221 Absolute Salinity (unit: g/kg, notation: $S_A$) and Conservative Temperature (unit: \degC, notation: $\Theta$).
1222 The pressure in decibars is approximated by the depth in meters.
1223 With TEOS10, the specific heat capacity of sea water, $C_p$, is a constant.
1224 It is set to $C_p = 3991.86795711963~J\,Kg^{-1}\,^{\circ}K^{-1}$, according to \citet{TEOS10}.
1225 Choosing polyTEOS10-bsq implies that the state variables used by the model are $\Theta$ and $S_A$.
1226 In particular, the initial state deined by the user have to be given as \textit{Conservative} Temperature and
1227 \textit{Absolute} Salinity.
1228 In addition, setting \np{ln\_useCT} to \forcode{.true.} convert the Conservative SST to potential SST prior to
1229 either computing the air-sea and ice-sea fluxes (forced mode) or
1230 sending the SST field to the atmosphere (coupled mode).
1231\item[\np{nn\_eos}~\forcode{= 0}]
1232 the polyEOS80-bsq equation of seawater is used.
1233 It takes the same polynomial form as the polyTEOS10, but the coefficients have been optimized to
1234 accurately fit EOS80 (Roquet, personal comm.).
1235 The state variables used in both the EOS80 and the ocean model are:
1236 the Practical Salinity ((unit: psu, notation: $S_p$)) and
1237 Potential Temperature (unit: $^{\circ}C$, notation: $\theta$).
1238 The pressure in decibars is approximated by the depth in meters.
1239 With thsi EOS, the specific heat capacity of sea water, $C_p$, is a function of temperature, salinity and
1240 pressure \citep{UNESCO1983}.
1241 Nevertheless, a severe assumption is made in order to have a heat content ($C_p T_p$) which
1242 is conserved by the model: $C_p$ is set to a constant value, the TEOS10 value.
1243\item[\np{nn\_eos}~\forcode{= 1}]
1244 a simplified EOS (S-EOS) inspired by \citet{Vallis06} is chosen,
1245 the coefficients of which has been optimized to fit the behavior of TEOS10
1246 (Roquet, personal comm.) (see also \citet{Roquet_JPO2015}).
1247 It provides a simplistic linear representation of both cabbeling and thermobaricity effects which
1248 is enough for a proper treatment of the EOS in theoretical studies \citep{Roquet_JPO2015}.
1249 With such an equation of state there is no longer a distinction between
1250 \textit{conservative} and \textit{potential} temperature,
1251 as well as between \textit{absolute} and \textit{practical} salinity.
1252 S-EOS takes the following expression:
1253 \begin{gather*}
1254 % \label{eq:tra_S-EOS}
1255 \begin{alignedat}{2}
1256 &d_a(T,S,z) = \frac{1}{\rho_o} \big[ &- a_0 \; ( 1 + 0.5 \; \lambda_1 \; T_a + \mu_1 \; z ) * &T_a \big. \\
1257 & &+ b_0 \; ( 1 - 0.5 \; \lambda_2 \; S_a - \mu_2 \; z ) * &S_a \\
1258 & \big. &- \nu \; T_a &S_a \big] \\
1259 \end{alignedat}
1260 \\
1261 \text{with~} T_a = T - 10 \, ; \, S_a = S - 35 \, ; \, \rho_o = 1026~Kg/m^3
1262 \end{gather*}
1263 where the computer name of the coefficients as well as their standard value are given in \autoref{tab:SEOS}.
1264 In fact, when choosing S-EOS, various approximation of EOS can be specified simply by
1265 changing the associated coefficients.
1266 Setting to zero the two thermobaric coefficients $(\mu_1,\mu_2)$ remove thermobaric effect from S-EOS.
1267 setting to zero the three cabbeling coefficients $(\lambda_1,\lambda_2,\nu)$ remove cabbeling effect from
1268 S-EOS.
1269 Keeping non-zero value to $a_0$ and $b_0$ provide a linear EOS function of T and S.
1270\end{description}
1271
1272%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
1273\begin{table}[!tb]
1274 \begin{center}
1275 \begin{tabular}{|l|l|l|l|}
1276 \hline
1277 coeff. & computer name & S-EOS & description \\
1278 \hline
1279 $a_0$ & \np{rn\_a0} & $1.6550~10^{-1}$ & linear thermal expansion coeff. \\
1280 \hline
1281 $b_0$ & \np{rn\_b0} & $7.6554~10^{-1}$ & linear haline expansion coeff. \\
1282 \hline
1283 $\lambda_1$ & \np{rn\_lambda1}& $5.9520~10^{-2}$ & cabbeling coeff. in $T^2$ \\
1284 \hline
1285 $\lambda_2$ & \np{rn\_lambda2}& $5.4914~10^{-4}$ & cabbeling coeff. in $S^2$ \\
1286 \hline
1287 $\nu$ & \np{rn\_nu} & $2.4341~10^{-3}$ & cabbeling coeff. in $T \, S$ \\
1288 \hline
1289 $\mu_1$ & \np{rn\_mu1} & $1.4970~10^{-4}$ & thermobaric coeff. in T \\
1290 \hline
1291 $\mu_2$ & \np{rn\_mu2} & $1.1090~10^{-5}$ & thermobaric coeff. in S \\
1292 \hline
1293 \end{tabular}
1294 \caption{
1295 \protect\label{tab:SEOS}
1296 Standard value of S-EOS coefficients.
1297 }
1298\end{center}
1299\end{table}
1300%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
1301
1302% -------------------------------------------------------------------------------------------------------------
1303% Brunt-V\"{a}is\"{a}l\"{a} Frequency
1304% -------------------------------------------------------------------------------------------------------------
1305\subsection{Brunt-V\"{a}is\"{a}l\"{a} frequency (\protect\np{nn\_eos}~\forcode{= 0..2})}
1306\label{subsec:TRA_bn2}
1307
1308An accurate computation of the ocean stability (i.e. of $N$, the brunt-V\"{a}is\"{a}l\"{a} frequency) is of
1309paramount importance as determine the ocean stratification and is used in several ocean parameterisations
1310(namely TKE, GLS, Richardson number dependent vertical diffusion, enhanced vertical diffusion,
1311non-penetrative convection, tidal mixing parameterisation, iso-neutral diffusion).
1312In particular, $N^2$ has to be computed at the local pressure
1313(pressure in decibar being approximated by the depth in meters).
1314The expression for $N^2$ is given by:
1315$1316 % \label{eq:tra_bn2} 1317 N^2 = \frac{g}{e_{3w}} \lt( \beta \; \delta_{k + 1/2}[S] - \alpha \; \delta_{k + 1/2}[T] \rt) 1318$
1319where $(T,S) = (\Theta,S_A)$ for TEOS10, $(\theta,S_p)$ for TEOS-80, or $(T,S)$ for S-EOS, and,
1320$\alpha$ and $\beta$ are the thermal and haline expansion coefficients.
1321The coefficients are a polynomial function of temperature, salinity and depth which expression depends on
1322the chosen EOS.
1323They are computed through \textit{eos\_rab}, a \fortran function that can be found in \mdl{eosbn2}.
1324
1325% -------------------------------------------------------------------------------------------------------------
1326% Freezing Point of Seawater
1327% -------------------------------------------------------------------------------------------------------------
1328\subsection{Freezing point of seawater}
1329\label{subsec:TRA_fzp}
1330
1331The freezing point of seawater is a function of salinity and pressure \citep{UNESCO1983}:
1332\begin{equation}
1333 \label{eq:tra_eos_fzp}
1334 \begin{split}
1335 &T_f (S,p) = \lt( a + b \, \sqrt{S} + c \, S \rt) \, S + d \, p \\
1336 &\text{where~} a = -0.0575, \, b = 1.710523~10^{-3}, \, c = -2.154996~10^{-4} \\
1337 &\text{and~} d = -7.53~10^{-3}
1338 \end{split}
1339\end{equation}
1340
1341\autoref{eq:tra_eos_fzp} is only used to compute the potential freezing point of sea water
1342(\ie referenced to the surface $p = 0$),
1343thus the pressure dependent terms in \autoref{eq:tra_eos_fzp} (last term) have been dropped.
1344The freezing point is computed through \textit{eos\_fzp},
1345a \fortran function that can be found in \mdl{eosbn2}.
1346
1347% -------------------------------------------------------------------------------------------------------------
1348% Potential Energy
1349% -------------------------------------------------------------------------------------------------------------
1350%\subsection{Potential Energy anomalies}
1351%\label{subsec:TRA_bn2}
1352
1353% =====>>>>> TO BE written
1354%
1355
1356% ================================================================
1357% Horizontal Derivative in zps-coordinate
1358% ================================================================
1359\section{Horizontal derivative in \textit{zps}-coordinate (\protect\mdl{zpshde})}
1360\label{sec:TRA_zpshde}
1361
1362\gmcomment{STEVEN: to be consistent with earlier discussion of differencing and averaging operators,
1363I've changed "derivative" to "difference" and "mean" to "average"}
1364
1365With partial cells (\np{ln\_zps}~\forcode{= .true.}) at bottom and top (\np{ln\_isfcav}~\forcode{= .true.}),
1366in general, tracers in horizontally adjacent cells live at different depths.
1367Horizontal gradients of tracers are needed for horizontal diffusion (\mdl{traldf} module) and
1368the hydrostatic pressure gradient calculations (\mdl{dynhpg} module).
1369The partial cell properties at the top (\np{ln\_isfcav}~\forcode{= .true.}) are computed in the same way as
1370for the bottom.
1371So, only the bottom interpolation is explained below.
1372
1373Before taking horizontal gradients between the tracers next to the bottom,
1374a linear interpolation in the vertical is used to approximate the deeper tracer as if
1375it actually lived at the depth of the shallower tracer point (\autoref{fig:Partial_step_scheme}).
1376For example, for temperature in the $i$-direction the needed interpolated temperature, $\widetilde T$, is:
1377
1378%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
1379\begin{figure}[!p]
1380 \begin{center}
1381 \includegraphics[]{Fig_partial_step_scheme}
1382 \caption{
1383 \protect\label{fig:Partial_step_scheme}
1384 Discretisation of the horizontal difference and average of tracers in the $z$-partial step coordinate
1385 (\protect\np{ln\_zps}~\forcode{= .true.}) in the case $(e3w_k^{i + 1} - e3w_k^i) > 0$.
1386 A linear interpolation is used to estimate $\widetilde T_k^{i + 1}$,
1387 the tracer value at the depth of the shallower tracer point of the two adjacent bottom $T$-points.
1388 The horizontal difference is then given by: $\delta_{i + 1/2} T_k = \widetilde T_k^{\, i + 1} -T_k^{\, i}$ and
1389 the average by: $\overline T_k^{\, i + 1/2} = (\widetilde T_k^{\, i + 1/2} - T_k^{\, i}) / 2$.
1390 }
1391 \end{center}
1392\end{figure}
1393%>>>>>>>>>>>>>>>>>>>>>>>>>>>>
13941395 \widetilde T = \lt\{ 1396 \begin{alignedat}{2} 1397 &T^{\, i + 1} &-\frac{ \lt( e_{3w}^{i + 1} -e_{3w}^i \rt) }{ e_{3w}^{i + 1} } \; \delta_k T^{i + 1} 1398 & \quad \text{if e_{3w}^{i + 1} \geq e_{3w}^i} \\ \\ 1399 &T^{\, i} &+\frac{ \lt( e_{3w}^{i + 1} -e_{3w}^i \rt )}{e_{3w}^i } \; \delta_k T^{i + 1} 1400 & \quad \text{if e_{3w}^{i + 1} < e_{3w}^i} 1401 \end{alignedat} 1402 \rt. 1403
1404and the resulting forms for the horizontal difference and the horizontal average value of $T$ at a $U$-point are:
1405\begin{equation}
1406 \label{eq:zps_hde}
1407 \begin{split}
1408 \delta_{i + 1/2} T &=
1409 \begin{cases}
1410 \widetilde T - T^i & \text{if~} e_{3w}^{i + 1} \geq e_{3w}^i \\
1411 \\
1412 T^{\, i + 1} - \widetilde T & \text{if~} e_{3w}^{i + 1} < e_{3w}^i
1413 \end{cases}
1414 \\
1415 \overline T^{\, i + 1/2} &=
1416 \begin{cases}
1417 (\widetilde T - T^{\, i} ) / 2 & \text{if~} e_{3w}^{i + 1} \geq e_{3w}^i \\
1418 \\
1419 (T^{\, i + 1} - \widetilde T) / 2 & \text{if~} e_{3w}^{i + 1} < e_{3w}^i
1420 \end{cases}
1421 \end{split}
1422\end{equation}
1423
1424The computation of horizontal derivative of tracers as well as of density is performed once for all at
1425each time step in \mdl{zpshde} module and stored in shared arrays to be used when needed.
1426It has to be emphasized that the procedure used to compute the interpolated density, $\widetilde \rho$,
1427is not the same as that used for $T$ and $S$.
1428Instead of forming a linear approximation of density, we compute $\widetilde \rho$ from the interpolated values of
1429$T$ and $S$, and the pressure at a $u$-point
1430(in the equation of state pressure is approximated by depth, see \autoref{subsec:TRA_eos}):
1431$1432 % \label{eq:zps_hde_rho} 1433 \widetilde \rho = \rho (\widetilde T,\widetilde S,z_u) \quad \text{where~} z_u = \min \lt( z_T^{i + 1},z_T^i \rt) 1434$
1435
1436This is a much better approximation as the variation of $\rho$ with depth (and thus pressure)
1437is highly non-linear with a true equation of state and thus is badly approximated with a linear interpolation.
1438This approximation is used to compute both the horizontal pressure gradient (\autoref{sec:DYN_hpg}) and
1439the slopes of neutral surfaces (\autoref{sec:LDF_slp}).
1440
1441Note that in almost all the advection schemes presented in this Chapter,
1442both averaging and differencing operators appear.
1443Yet \autoref{eq:zps_hde} has not been used in these schemes:
1444in contrast to diffusion and pressure gradient computations,
1445no correction for partial steps is applied for advection.
1446The main motivation is to preserve the domain averaged mean variance of the advected field when
1447using the $2^{nd}$ order centred scheme.
1448Sensitivity of the advection schemes to the way horizontal averages are performed in the vicinity of
1449partial cells should be further investigated in the near future.
1450%%%
1451\gmcomment{gm : this last remark has to be done}
1452%%%
1453
1454\biblio
1455
1456\pindex
1457
1458\end{document}
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https://thebusinessprofessor.com/lesson/assumed-interest-rate-definition/ | 1,603,215,366,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874026.22/warc/CC-MAIN-20201020162922-20201020192922-00644.warc.gz | 550,407,589 | 12,279 | # Assumed Interest Rate – Definition
Back To: INSURANCE & RISK MANAGEMENT
### Assumed Interest Rate (AIR) Definition
The assumed interest rate (AIR) refers to an interest rate or growth rate used by an insurance company to calculate the value of an annuity contract and its payout. An insurance company determines AIR and it serves as the benchmark for the annuityās periodic income payment that an annuitant would receive at the due time.
When calculating the payout of an annuity contract, the assumed interest rate is a crucial metric. There are diverse factors that are considered when an insurance company is o determine AIR, these include the age of the annuitant upon annuitization, the type of annuity coverage, the coverage options, and others.
### A Little More on What is an Assumed Interest Rate
The assumed interest rate (AIR) as used by an insurance company determines the minimum rate of return that a variable annuityās separate account must realize during the payout so as to maintain a steady annuity payment to the annuitant. It refers to the minimum interest rate that the insurer must earn on the cash value of the annuitantās policy that will help the insurance company cover its costs and also maintain an expected profit margin. This means the higher the AIR, the higher monthly income an annuitant would receive and the higher the profits that the insurance company would make.
The AIR is determined by insurance companies, it is the target rate of return they set on annuity separate accounts in order to have steady payments and also pay all costs.
The AIR is not a guaranteed rate of return given that there is no fixed return that annuityās separate accounts record. However, if an annuity account performs above the AIR, the annuitant would receive higher payments but the reverse is the case when the account performs below the AIR. The AIR takes into account the value of an annuity, the type and the annuitantsā age at the time he would begin to receive monthly income payments.
### Reference for āAssumed Interest Rate (AIR)ā
https://www.investopedia.com/terms/a/assumedinterestrate.asp
https://www.nasdaq.com/investing/glossary/a/assumed-interest-rate
https://financial-dictionary.thefreedictionary.com/Assumed+interest+rate
### Academics research on āAssumed Interest Rate (AIR)ā
The timing of annuitization: Investment dominance and mortality risk, Milevsky, M. A., & Young, V. R. (2007). The timing of annuitization: Investment dominance and mortality risk.Ā Insurance: Mathematics and Economics,Ā 40(1), 135-144. We use preference-free dominance arguments to develop a framework for locating the optimal age (time) at which a retiree should purchase an irreversible life annuity, as a function of current annuity prices and mortality tables. Then, using the institutional characteristics of annuity markets in the US, we show that annuitization prior to age 65ā70 is dominated by self-annuitizationĀ evenĀ in the absence of any bequest motives. And, for retirees who are willing to accept some financial risk in exchange for retaining the benefits of liquidity and bequest, the optimal age can be even later. In addition to the normative implications, these results should help shed light on the so-called annuity puzzle which has been much debated by economists, by focusing attention on theĀ specific agesĀ for which a puzzle can actually be said to exist.
Optimal asset allocation in life annuities: a note, Charupat, N., & Milevsky, M. A. (2002). Optimal asset allocation in life annuities: a note.Ā Insurance: Mathematics and Economics,Ā 30(2), 199-209. In this note, we derive the optimal utility-maximizing asset allocation between a risky and risk-free asset within aĀ variable annuityĀ (VA) contract, which is a US-based savings and decumulation investment product. We are interested in the interaction between financial risk, mortality risk and consumption, towards the end of the life cycle. Our main result is that for constant relative risk aversion (CRRA) preferences and geometric Brownian motion (GBM) dynamics, the optimal asset allocation during the annuity decumulation (payout) phase is identical to the accumulation (savings) phase, which is the classical Merton [J. Econ. Theory 3 (1971) 373] solution.
Variable payout annuities and dynamic portfolio choice in retirement, Horneff, W. J., Maurer, R. H., Mitchell, O. S., & Stamos, M. Z. (2010). Variable payout annuities and dynamic portfolio choice in retirement.Ā Journal of Pension Economics & Finance,Ā 9(2), 163-183. Many retirees hope to continue earning capital market rewards on their saving while avoiding outliving their funds during retirement. We model a dynamic utility maximizing investor who seeks to benefit from holding both equity and longevity insurance. She is free to adjust her portfolio allocation of her financial wealth as well as of the annuity over time, and she can purchase variable payout annuities any time and incrementally. In this setting, we show that the retiree will not fully annuitize even without bequests; rather, she will combine variable annuities with withdrawals from her liquid financial wealth so as to match her desired consumption profile. Optimal stock exposures decrease over time, both within the variable annuity and the withdrawal plan. Welfare gains from this strategy can amount to 40% of financial wealth, depending on risk parameters and other resources; additionally, many retirees will do almost as well as the fully optimized outcome if they hold variable annuities invested 60/40 in stocks/bonds.
Mortality risk, inflation risk, and annuity products, Brown, J. R., Mitchell, O. S., & Poterba, J. M. (2000).Ā Mortality risk, inflation risk, and annuity productsĀ (No. w7812). National Bureau of Economic Research.
Lifecycle portfolio choice with systematic longevity risk and variable investmentāLinked deferred annuities, Maurer, R., Mitchell, O. S., Rogalla, R., & Kartashov, V. (2013). Lifecycle portfolio choice with systematic longevity risk and variable investmentāLinked deferred annuities.Ā Journal of Risk and Insurance,Ā 80(3), 649-676. This article assesses the impact of variable investmentālinked deferred annuities (VILDAs) on lifecycle consumption and portfolio allocation, allowing for systematic longevity risk. Under a selfāinsurance strategy, insurers set premiums to reduce the chance that benefits paid exceed provider reserves. Under a participating approach, the provider avoids taking systematic longevity risk by adjusting benefits in response to unanticipated mortality shocks. Young households with participating annuities average oneāthird higher excess consumption, while 80āyearāolds increase consumption about 75 percent. Many households would prefer to participate in systematic longevity risk unless insurers can hedge it at a very low price. | 1,539 | 6,899 | {"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-2020-45 | latest | en | 0.951159 |
https://www.numerade.com/books/chapter/functions-and-their-graphs-10/ | 1,590,848,883,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347409337.38/warc/CC-MAIN-20200530133926-20200530163926-00064.warc.gz | 815,381,111 | 13,179 | College Algebra
Educators
Problem 1
The inequality $-1<x<3$ can be written in interval notation as ______.
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Problem 2
If $x=-2,$ the value of the expression $3 x^{2}-5 x+\frac{1}{x}$ is _______.
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Problem 3
The domain of the variable in the expression $\frac{x-3}{x+4}$ is _______.
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Problem 4
Solve the inequality: $3-2 x>5 .$ Graph the solution set. ______
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Problem 5
If $f$ is a function defined by the equation $y=f(x),$ then $x$ is called the ______ variable and $y$ is the ______ variable.
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Problem 6
The set of all images of the elements in the domain of a function is called the _______
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Problem 7
If the domain of $f$ is all real numbers in the interval $[0,7]$ and the domain of $g$ is all real numbers in the interval $[-2,5],$ the domain of $f+g$ is all real numbers in the interval ________.
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Problem 8
The domain of $\frac{f}{g}$ consists of numbers $x$ for which g(x) ______ 0 that are in the domains of both _____ and _____.
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Problem 9
If $f(x)=x+1$ and $g(x)=x^{3},$ then ______ =$x^{3}-(x+1)$
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Problem 10
True or False Every relation is a function.
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Problem 11
True or False The domain of $(f \cdot g)(x)$ consists of the numbers $x$ that are in the domains of both $f$ and $g .$
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Problem 12
True or False The independent variable is sometimes referred to as the argument of the function.
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Problem 13
True or False If no domain is specified for a function $f,$ then the domain of $f$ is taken to be the set of real numbers.
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Problem 14
True or False The domain of the function $f(x)=\frac{x^{2}-4}{x}$ is $\{x | x \neq \pm 2\}$
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Problem 15
Determine whether each relation represents a function. For each function, state the domain and range.
(function can't copy)
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Problem 16
Determine whether each relation represents a function. For each function, state the domain and range.
(function can't copy)
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Problem 17
Determine whether each relation represents a function. For each function, state the domain and range.
(function can't copy)
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Problem 18
Determine whether each relation represents a function. For each function, state the domain and range.
(function can't copy)
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Problem 19
Determine whether each relation represents a function. For each function, state the domain and range.
$$\{(2,6),(-3,6),(4,9),(2,10)\}$$
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Problem 20
Determine whether each relation represents a function. For each function, state the domain and range.
$$\{(-2,5),(-1,3),(3,7),(4,12)\}$$
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Problem 21
Determine whether each relation represents a function. For each function, state the domain and range.
$$\{(1,3),(2,3),(3,3),(4,3)\}$$
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Problem 22
Determine whether each relation represents a function. For each function, state the domain and range.
$$\{(0,-2),(1,3),(2,3),(3,7)\}$$
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Problem 23
Determine whether each relation represents a function. For each function, state the domain and range.
$$\{(-2,4),(-2,6),(0,3),(3,7)\}$$
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Problem 24
Determine whether each relation represents a function. For each function, state the domain and range.
$$\{(-4,4),(-3,3),(-2,2),(-1,1),(-4,0)\}$$
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Problem 25
Determine whether each relation represents a function. For each function, state the domain and range.
$$\{(-2,4),(-1,1),(0,0),(1,1)\}$$
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Problem 26
Determine whether each relation represents a function. For each function, state the domain and range.
$$\{(-2,16),(-1,4),(0,3),(1,4)\}$$
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Problem 27
Determine whether the equation defines y as a function of x.
$$y=x^{2}$$
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Problem 28
Determine whether the equation defines y as a function of x.
$$y=x^{3}$$
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Problem 29
Determine whether the equation defines y as a function of x.
$$y=\frac{1}{x}$$
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Problem 30
Determine whether the equation defines y as a function of x.
$$y=|x|$$
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Problem 31
Determine whether the equation defines y as a function of x.
$$y^{2}=4-x^{2}$$
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Problem 32
Determine whether the equation defines y as a function of x.
$$y=\pm \sqrt{1-2 x}$$
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Problem 33
Determine whether the equation defines y as a function of x.
$$x=y^{2}$$
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Problem 34
Determine whether the equation defines y as a function of x.
$$x+y^{2}=1$$
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Problem 35
Determine whether the equation defines y as a function of x.
$$y=2 x^{2}-3 x+4$$
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Problem 36
Determine whether the equation defines y as a function of x.
$$y=\frac{3 x-1}{x+2}$$
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Problem 37
Determine whether the equation defines y as a function of x.
$$2 x^{2}+3 y^{2}=1$$
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Problem 38
Determine whether the equation defines y as a function of x.
$$x^{2}-4 y^{2}=1$$
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Problem 39
Find the following for each function:
(a) $f(0)$
(b) $f(1)$
(c) $f(-1)$
(d) $f(-x)$
(e) $-f(x)$
(f) $f(x+1)$
(g) $f(2 x)$
(h) $f(x+h)$
$$f(x)=3 x^{2}+2 x-4$$
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Problem 40
Find the following for each function:
(a) $f(0)$
(b) $f(1)$
(c) $f(-1)$
(d) $f(-x)$
(e) $-f(x)$
(f) $f(x+1)$
(g) $f(2 x)$
(h) $f(x+h)$
$$f(x)=-2 x^{2}+x-1$$
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Problem 41
Find the following for each function:
(a) $f(0)$
(b) $f(1)$
(c) $f(-1)$
(d) $f(-x)$
(e) $-f(x)$
(f) $f(x+1)$
(g) $f(2 x)$
(h) $f(x+h)$
$$f(x)=\frac{x}{x^{2}+1}$$
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Problem 42
Find the following for each function:
(a) $f(0)$
(b) $f(1)$
(c) $f(-1)$
(d) $f(-x)$
(e) $-f(x)$
(f) $f(x+1)$
(g) $f(2 x)$
(h) $f(x+h)$
$$f(x)=\frac{x^{2}-1}{x+4}$$
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Problem 43
Find the following for each function:
(a) $f(0)$
(b) $f(1)$
(c) $f(-1)$
(d) $f(-x)$
(e) $-f(x)$
(f) $f(x+1)$
(g) $f(2 x)$
(h) $f(x+h)$
$$f(x)=|x|+4$$
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Problem 44
Find the following for each function:
(a) $f(0)$
(b) $f(1)$
(c) $f(-1)$
(d) $f(-x)$
(e) $-f(x)$
(f) $f(x+1)$
(g) $f(2 x)$
(h) $f(x+h)$
$$f(x)=\sqrt{x^{2}+x}$$
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Problem 45
Find the following for each function:
(a) $f(0)$
(b) $f(1)$
(c) $f(-1)$
(d) $f(-x)$
(e) $-f(x)$
(f) $f(x+1)$
(g) $f(2 x)$
(h) $f(x+h)$
$$f(x)=\frac{2 x+1}{3 x-5}$$
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Problem 46
Find the following for each function:
(a) $f(0)$
(b) $f(1)$
(c) $f(-1)$
(d) $f(-x)$
(e) $-f(x)$
(f) $f(x+1)$
(g) $f(2 x)$
(h) $f(x+h)$
$$f(x)=1-\frac{1}{(x+2)^{2}}$$
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Problem 47
Find the domain of each function.
$$f(x)=-5 x+4$$
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Problem 48
Find the domain of each function.
$$f(x)=x^{2}+2$$
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Problem 49
Find the domain of each function.
$$f(x)=\frac{x}{x^{2}+1}$$
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Problem 50
Find the domain of each function.
$$f(x)=\frac{x^{2}}{x^{2}+1}$$
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Problem 51
Find the domain of each function.
$$g(x)=\frac{x}{x^{2}-16}$$
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Problem 52
Find the domain of each function.
$$h(x)=\frac{2 x}{x^{2}-4}$$
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Problem 53
Find the domain of each function.
$$F(x)=\frac{x-2}{x^{3}+x}$$
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Problem 54
Find the domain of each function.
$$G(x)=\frac{x+4}{x^{3}-4 x}$$
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Problem 55
Find the domain of each function.
$$h(x)=\sqrt{3 x-12}$$
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Problem 56
Find the domain of each function.
$$G(x)=\sqrt{1-x}$$
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Problem 57
Find the domain of each function.
$$f(x)=\frac{4}{\sqrt{x-9}}$$
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Problem 58
Find the domain of each function.
$$f(x)=\frac{x}{\sqrt{x-4}}$$
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Problem 59
Find the domain of each function.
$$p(x)=\sqrt{\frac{2}{x-1}}$$
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Problem 60
Find the domain of each function.
$$q(x)=\sqrt{-x-2}$$
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Problem 61
Find the domain of each function.
$$P(t)=\frac{\sqrt{t-4}}{3 t-21}$$
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Problem 62
Find the domain of each function.
$$h(z)=\frac{\sqrt{z+3}}{z-2}$$
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Problem 63
For the given functions $f$ and $g,$ find the following. For parts ( $a$ )-( $d$ ), also find the domain.
(a)$(f+g)(x)$
(b) $(f-g)(x)$
(c) $(f \cdot g)(x)$
(d)$\left(\frac{f}{g}\right)(x)$
(e)$(f+g)(3)$
(f)$(f-g)(4)$
(g) $(f \cdot g)(2)$
( h)$\left(\frac{f}{g}\right)(1)$
$$f(x)=3 x+4 ; \quad g(x)=2 x-3$$
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Problem 64
For the given functions $f$ and $g,$ find the following. For parts ( $a$ )-( $d$ ), also find the domain.
(a)$(f+g)(x)$
(b) $(f-g)(x)$
(c) $(f \cdot g)(x)$
(d)$\left(\frac{f}{g}\right)(x)$
(e)$(f+g)(3)$
(f)$(f-g)(4)$
(g) $(f \cdot g)(2)$
( h)$\left(\frac{f}{g}\right)(1)$
$$f(x)=2 x+1 ; \quad g(x)=3 x-2$$
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Problem 65
For the given functions $f$ and $g,$ find the following. For parts ( $a$ )-( $d$ ), also find the domain.
(a)$(f+g)(x)$
(b) $(f-g)(x)$
(c) $(f \cdot g)(x)$
(d)$\left(\frac{f}{g}\right)(x)$
(e)$(f+g)(3)$
(f)$(f-g)(4)$
(g) $(f \cdot g)(2)$
( h)$\left(\frac{f}{g}\right)(1)$
$$f(x)=x-1 ; \quad g(x)=2 x^{2}$$
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Problem 66
For the given functions $f$ and $g,$ find the following. For parts ( $a$ )-( $d$ ), also find the domain.
(a)$(f+g)(x)$
(b) $(f-g)(x)$
(c) $(f \cdot g)(x)$
(d)$\left(\frac{f}{g}\right)(x)$
(e)$(f+g)(3)$
(f)$(f-g)(4)$
(g) $(f \cdot g)(2)$
( h)$\left(\frac{f}{g}\right)(1)$
$$f(x)=2 x^{2}+3 ; \quad g(x)=4 x^{3}+1$$
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Problem 67
For the given functions $f$ and $g,$ find the following. For parts ( $a$ )-( $d$ ), also find the domain.
(a)$(f+g)(x)$
(b) $(f-g)(x)$
(c) $(f \cdot g)(x)$
(d)$\left(\frac{f}{g}\right)(x)$
(e)$(f+g)(3)$
(f)$(f-g)(4)$
(g) $(f \cdot g)(2)$
( h)$\left(\frac{f}{g}\right)(1)$
$$f(x)=\sqrt{x} ; \quad g(x)=3 x-5$$
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Problem 68
For the given functions $f$ and $g,$ find the following. For parts ( $a$ )-( $d$ ), also find the domain.
(a)$(f+g)(x)$
(b) $(f-g)(x)$
(c) $(f \cdot g)(x)$
(d)$\left(\frac{f}{g}\right)(x)$
(e)$(f+g)(3)$
(f)$(f-g)(4)$
(g) $(f \cdot g)(2)$
( h)$\left(\frac{f}{g}\right)(1)$
$$f(x)=|x| ; \quad g(x)=x$$
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Problem 69
For the given functions $f$ and $g,$ find the following. For parts ( $a$ )-( $d$ ), also find the domain.
(a)$(f+g)(x)$
(b) $(f-g)(x)$
(c) $(f \cdot g)(x)$
(d)$\left(\frac{f}{g}\right)(x)$
(e)$(f+g)(3)$
(f)$(f-g)(4)$
(g) $(f \cdot g)(2)$
( h)$\left(\frac{f}{g}\right)(1)$
$$f(x)=|x| ; \quad g(x)=x$$
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Problem 70
For the given functions $f$ and $g,$ find the following. For parts ( $a$ )-( $d$ ), also find the domain.
(a)$(f+g)(x)$
(b) $(f-g)(x)$
(c) $(f \cdot g)(x)$
(d)$\left(\frac{f}{g}\right)(x)$
(e)$(f+g)(3)$
(f)$(f-g)(4)$
(g) $(f \cdot g)(2)$
( h)$\left(\frac{f}{g}\right)(1)$
$$f(x)=\sqrt{x-1} ; \quad g(x)=\sqrt{4-x}$$
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Problem 71
For the given functions $f$ and $g,$ find the following. For parts ( $a$ )-( $d$ ), also find the domain.
(a)$(f+g)(x)$
(b) $(f-g)(x)$
(c) $(f \cdot g)(x)$
(d)$\left(\frac{f}{g}\right)(x)$
(e)$(f+g)(3)$
(f)$(f-g)(4)$
(g) $(f \cdot g)(2)$
( h)$\left(\frac{f}{g}\right)(1)$
$$f(x)=\frac{2 x+3}{3 x-2} ; \quad g(x)=\frac{4 x}{3 x-2}$$
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Problem 72
For the given functions $f$ and $g,$ find the following. For parts ( $a$ )-( $d$ ), also find the domain.
(a)$(f+g)(x)$
(b) $(f-g)(x)$
(c) $(f \cdot g)(x)$
(d)$\left(\frac{f}{g}\right)(x)$
(e)$(f+g)(3)$
(f)$(f-g)(4)$
(g) $(f \cdot g)(2)$
( h)$\left(\frac{f}{g}\right)(1)$
$$f(x)=\sqrt{x+1} ; \quad g(x)=\frac{2}{x}$$
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Problem 73
Given $f(x)=3 x+1$ and $(f+g)(x)=6-\frac{1}{2} x,$ find the function $g .$
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Problem 74
Given $f(x)=\frac{1}{x}$ and $\left(\frac{f}{g}\right)(x)=\frac{x+1}{x^{2}-x},$ find the function $g$
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Problem 75
Find the difference quotient of $f$; that is, find $\frac{f(x+h)-f(x)}{h}, h \neq 0,$ for each function. Be sure to simplify.
$$f(x)=4 x+3$$
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Problem 76
Find the difference quotient of $f$; that is, find $\frac{f(x+h)-f(x)}{h}, h \neq 0,$ for each function. Be sure to simplify.
$$f(x)=-3 x+1$$
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Problem 77
Find the difference quotient of $f$; that is, find $\frac{f(x+h)-f(x)}{h}, h \neq 0,$ for each function. Be sure to simplify.
$$f(x)=x^{2}-x+4$$
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Problem 78
Find the difference quotient of $f$; that is, find $\frac{f(x+h)-f(x)}{h}, h \neq 0,$ for each function. Be sure to simplify.
$$f(x)=3 x^{2}-2 x+6$$
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Problem 79
Find the difference quotient of $f$; that is, find $\frac{f(x+h)-f(x)}{h}, h \neq 0,$ for each function. Be sure to simplify.
$$f(x)=\frac{1}{x^{2}}$$
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Problem 80
Find the difference quotient of $f$; that is, find $\frac{f(x+h)-f(x)}{h}, h \neq 0,$ for each function. Be sure to simplify.
$$f(x)=\frac{1}{x+3}$$
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Problem 81
Find the difference quotient of $f$; that is, find $\frac{f(x+h)-f(x)}{h}, h \neq 0,$ for each function. Be sure to simplify.
$$f(x)=\sqrt{x}$$
[Hint: Rationalize the numerator.
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Problem 82
Find the difference quotient of $f$; that is, find $\frac{f(x+h)-f(x)}{h}, h \neq 0,$ for each function. Be sure to simplify.
$$f(x)=\sqrt{x+1}$$
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Problem 83
If $f(x)=2 x^{3}+A x^{2}+4 x-5$ and $f(2)=5,$ what is the value of $A ?$
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Problem 84
If $f(x)=3 x^{2}-B x+4$ and $f(-1)=12,$ what is the value of $B ?$
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Problem 85
If $f(x)=\frac{3 x+8}{2 x-A}$ and $f(0)=2,$ what is the value of $A ?$
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Problem 86
If $f(x)=\frac{2 x-B}{3 x+4}$ and $f(2)=\frac{1}{2},$ what is the value of $B ?$
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Problem 87
If $f(x)=\frac{2 x-A}{x-3}$ and $f(4)=0,$ what is the value of $A ?$ Where is $f$ not defined?
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Problem 88
If $f(x)=\frac{x-B}{x-A}, f(2)=0$ and $f(1)$ is undefined, what are the values of $A$ and $B ?$
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Problem 89
Geometry Express the area $A$ of a rectangle as a function of the length $x$ if the length of the rectangle is twice its width.
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Problem 90
Geometry Express the area $A$ of an isosceles right triangle as a function of the length $x$ of one of the two equal sides.
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Check back soon! | 5,241 | 14,186 | {"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} | 3.78125 | 4 | CC-MAIN-2020-24 | latest | en | 0.738803 |
https://www.tutorialaicsip.com/cbse-update/cbse-10th-result-2021-format/ | 1,723,781,360,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00076.warc.gz | 793,064,833 | 46,441 | Entery-and-formula-for-CBSE-10th-Result-2021
CBSE has published the historic performance for CBSE 10th result 2021 format. In this article, we are going to provide you the CBSE 10th Result 2021 format based on Historic performance in an Excel file. So let’s start now!
Topics Covered
## CBSE 10th Result 2021 format based on Historic performance
This format is divided into 6 sections like Section A to F. Observe the screenshot given below:
## Marks Entry
1. Enter the reference year %, subject average, and overall average in Section B, C, and D.
2. Observe this screenshot.
## Formulas
In the next section of CBSE 10th Result 2021, I will explain the formulas used in this worksheet.
### How to write a formula to compute the current year average (section E)
Follow the below given steps to calculate the current year average marks:
1. Write the formula for English in O8 Cell: =SUM(B10:B34)/A35. In A35 cell I have calculated no. of students in the class using =COUNTA(A10:A34) formula.
2. Write formula for the respective subjects accordingly.
### Formula to calculate the current year overall average
=SUM(B10:F34)/(A35*5)
Select the cells starting from the subject 1 marks entry to last cell of subject 5 marks entry.
### Formula to calculate the range of allowed students
=ROUND(\$A\$35*J3/100,1)
A35 is the cell where the total no. of students is calculated. J3 is given % from the reference year in the given range. Write appropriate formula for all the subjects.
=COUNTIFS(B10:B34,”<26″)
This is the formal for calculating actual students ratio.
## Calculating 6th subject marks
For 6th Subject use this formula:
=((AVERAGE(LARGE(B10:F10,{1,2,3})))*50)/80
I hope this will helpful for you to make any changes to the worksheet. The sheet is already based on a formula. Just enter your marks do accordingly. | 458 | 1,840 | {"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} | 3.765625 | 4 | CC-MAIN-2024-33 | latest | en | 0.875484 |
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### Coloring Numbers:Number Recognition
Teach your child how to recognize numbers with these free, printable worksheets that will have your child coloring numbers that are mixed among letters.
### Number Line - Addition Worksheets
The number line is a great teaching tool for students beginning to learn addition. Print out a line for your child and use it with these colored, printable addition worksheets.
### Decimal Number Line - Printables
Printable decimal number line worksheets covering fractions, whole numbers, and negative numbers.
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### Printable Number Line:Positive & Negative Number Lines
Use a printable number line to easily see how numbers are arranged from negative to positive. A great study guide for learning addition and subtraction facts.
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### Multiplication Times Table Chart - 3 Times Tables
Printable multiplication times table chart covering the factors one through twelve. Find free math worksheets for individual number charts. Spelling, reading and vocabulary worksheets too.
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### Printable Multiplication Table - Multiplying By One
Learn how to multiply by one in this times table lesson. Use this free printable multiplication table for a quick study guide.
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### Printable Multiplication Chart - Math Worksheets
Practice your times tables by filling in the missing numbers on these printable multiplication chart worksheets. Find multiplication lessons, study help, and more.
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### Printable Multiplication Tables - Quick Guides
Help your child learn how to multiply with these free printable multiplication tables. Great for posting in a classroom!
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› New Worksheets | 805 | 4,291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2017-30 | longest | en | 0.835701 |
https://help.matheass.eu/en/E404-GCF_and_LCM_of_Polynomials.html | 1,721,915,710,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763858305.84/warc/CC-MAIN-20240725114544-20240725144544-00062.warc.gz | 242,351,097 | 2,061 | > MatheAss − GCD and LCM of polynomials
## GCD and LCM of polynomials
The greatest common divisor (GCD) and the least common multiple (LCM) of two polynomials are determined.
p1(x) = a9·x9 + a8·x8 + ... + a0 and p2(x) = b9·x9 + b8·x8 + ... + b0 .
The coefficients of the polynomial can be entered as fractions, as mixed numbers or as decimal numbers.
Use the checkbox to choose whether to output the coefficients of the polynomials as fractions or as decimal numbers.
```p1(x) = 4·x6 - 2·x5 - 6·x4- 18·x3 - 2·x2 + 24·x + 8
p2(x) = 10·x4- 14·x3 - 22·x2 + 14·x + 12
GCD(p1,p2) = x2 - x - 2
LCM(p1,p2) = 40·x8 - 36·x7 - 76·x6 - 144·x5 + 88·x4+ 356·x3 - 4·x2 - 176·x - 48
p1(x) = (x2 - x - 2)·(4·x4 + 2·x3 + 4·x2 - 10·x - 4)
p2(x) = (x2 - x - 2)·(10·x2 - 4·x - 6)```
### Application:
It is well known that the GCD and the LCM are needed when reducing fractions or when adding fractions. Applied to polynomials, this equivalent to reducing or adding the functional terms of rational functions.
### Note:
The GCD of polynomials is determined analogously to the GCD of natural numbers with the Euclidean algorithm (see GCD and LCM). Polynomial divisions are performed repeatedly and the intermediate results may contain coefficients that cause an overflow. In this case it is pointed out that further common factors may exist. | 471 | 1,344 | {"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-2024-30 | latest | en | 0.849504 |
http://mathematica.stackexchange.com/questions/40424/how-to-make-a-table-of-sum-i-1n/40425 | 1,464,542,675,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049281869.96/warc/CC-MAIN-20160524002121-00194-ip-10-185-217-139.ec2.internal.warc.gz | 183,065,711 | 21,230 | # How to make a table of $\sum_{i=1}^n$?
I want to create a table of $\displaystyle \sum_{i=1}^n i^k$ for $1\le k\le 5$. The output I want to obtain is a TeXForm output as follows.
\begin{align} \sum_{i=1}^n i &=\frac1 2 n(n-1)\\ \sum_{i=1}^n i^2 &=\frac1 2 n(n-1)(2n+1)\\ &\vdots \end{align}
My attempt is as follows but it does not produce what I want to achieve.
Sum[i^k, {i, 1, n}] // Table[Factor[#], {k, 1, 5}] & // TableForm
-
Do you want to generate LaTeX code or do you just want to format this on screen? The latter is easier. – Szabolcs Jan 14 '14 at 16:02
@Szabolcs: First priority is to produce the LaTeX output, but it will be nice if the on-screen output is also provided. – kiss my armpit Jan 14 '14 at 16:03
I need 24 hours to decide the accepted answer. – kiss my armpit Jan 14 '14 at 16:17
Here's a way that I find clear:
Make sure i and n have no value: i=.; n=..
Generate a list of the held sum expressions, taking care to make the exponent go away from $i^1$.
list = Table[With[{e = i^k}, HoldForm[Sum[e, {i, 1, n}]]], {k, 5}]
Use ReleaseHold to make a table of results:
TeXForm@TableForm[# == ReleaseHold[#] & /@ list]
$$\begin{array}{c} \sum _{i=1}^n i=\frac{1}{2} n (n+1) \\ \sum _{i=1}^n i^2=\frac{1}{6} n (n+1) (2 n+1) \\ \sum _{i=1}^n i^3=\frac{1}{4} n^2 (n+1)^2 \\ \sum _{i=1}^n i^4=\frac{1}{30} n (n+1) (2 n+1) \left(3 n^2+3 n-1\right) \\ \sum _{i=1}^n i^5=\frac{1}{12} n^2 (n+1)^2 \left(2 n^2+2 n-1\right) \\ \end{array}$$
-
This method (using With), as written, has a disadvantage in that a global value of i will be substituted if it exists. The same is true of n in both our methods but I cleared n beforehand to address that. – Mr.Wizard Jan 14 '14 at 16:22
I realize it's not a matter of With, so I wonder, why use e = i^k rather than e = k and i^e? – Mr.Wizard Jan 14 '14 at 16:27
Gah, can't read today. Now I see why you did that, and +1. However, consider wrapping everything in Block[{i, n}, ...] IMHO. – Mr.Wizard Jan 14 '14 at 16:32
Here is a start (adjusted based on Szabolcs's statement that Row and TeXForm don't work together in v9):
Clear[n]
# == ReleaseHold[#] & /@
Array[HoldForm[Sum[i^#, {i, 1, n}]] &, 5] // Column // TeXForm
$\begin{array}{l} \sum _{i=1}^n i^1=\frac{1}{2} n (n+1) \\ \sum _{i=1}^n i^2=\frac{1}{6} n (n+1) (2 n+1) \\ \sum _{i=1}^n i^3=\frac{1}{4} n^2 (n+1)^2 \\ \sum _{i=1}^n i^4=\frac{1}{30} n (n+1) (2 n+1) \left(3 n^2+3 n-1\right) \\ \sum _{i=1}^n i^5=\frac{1}{12} n^2 (n+1)^2 \left(2 n^2+2 n-1\right) \end{array}$
The two Function expressions could have been combined, but I felt that this was easier to read and easier to adjust.
Edit: I note that Szabolcs's answer shows the values above and below the Sigma symbols. Apparently that is a version difference, as I am using version 7 and his code produces the same output as mine in this regard.
-
Are you using version 7? When I try to apply TeXForm to any kind of Row expression in v9, I get an error "TeXForm of \!$$TemplateSlotSequence[1, \"\"]$$ is not supported." I thought it must have worked before .. – Szabolcs Jan 14 '14 at 16:12
@Szabolcs Strange; yes, I am. So the solution is to put TeXForm inside? – Mr.Wizard Jan 14 '14 at 16:14
@Szabolcs No, that doesn't work either. Do you mind if I copy your method to my answer? – Mr.Wizard Jan 14 '14 at 16:15
Okay, go ahead. There seems to be a bug in v9 with Row and TeXForm. Can't use them together. – Szabolcs Jan 14 '14 at 16:19
In Version 10, we can use Inactivate and Activate to achieve this easily:
With[{rl = Array[{m -> #} &, 5], s = Inactivate[Sum[i^m, {i, 1, n}], Sum]},
Thread[Equal[s /. rl, Factor[Activate[s] /. rl]]]] // Column // TeXForm
\begin{array}{l} \underset{i=1}{\overset{n}{\sum }}i=\frac{1}{2} n (n+1) \\ \underset{i=1}{\overset{n}{\sum }}i^2=\frac{1}{6} n (n+1) (2 n+1) \\ \underset{i=1}{\overset{n}{\sum }}i^3=\frac{1}{4} n^2 (n+1)^2 \\ \underset{i=1}{\overset{n}{\sum }}i^4=\frac{1}{30} n (n+1) (2 n+1) \left(3 n^2+3 n-1\right) \\ \underset{i=1}{\overset{n}{\sum }}i^5=\frac{1}{12} n^2 (n+1)^2 \left(2 n^2+2 n-1\right) \\ \end{array}
Note that this also does not give the undesired i^1
-
TableForm[Table[{
HoldForm[Sum[i^z, {i, 1, n}]] /. z -> k, "=",
Factor[Sum[i^k, {i, 1, n}]]},
{k, 1, 5}]]
-
+1. Won't adding // TeXForm at the end make it TeXForm? – RunnyKine Jan 14 '14 at 16:19
@RunnyKine - not on version 7 – Chris Degnen Jan 14 '14 at 16:40
Ah, I see. It works on Version 9. – RunnyKine Jan 14 '14 at 17:00
Another entry:
With[{rule = Array[{m -> #} &, 5], sm = HoldForm@Sum[i^m, {i, 1, n}]},
Thread[Equal[sm /. rule, Factor[ReleaseHold@sm /. rule]]]] // Column // TeXForm
\begin{array}{l} \sum _{i=1}^n i^1=\frac{1}{2} n (n+1) \\ \sum _{i=1}^n i^2=\frac{1}{6} n (n+1) (2 n+1) \\ \sum _{i=1}^n i^3=\frac{1}{4} n^2 (n+1)^2 \\ \sum _{i=1}^n i^4=\frac{1}{30} n (n+1) (2 n+1) \left(3 n^2+3 n-1\right) \\ \sum _{i=1}^n i^5=\frac{1}{12} n^2 (n+1)^2 \left(2 n^2+2 n-1\right) \\ \end{array}
Or to get rid of the 1 in i^1
With[{hd = Join[{HoldForm@Sum[i, {i, 1, n}]}, HoldForm@Sum[i^m, {i, 1, n}] /. Table[{m -> j},
{j, 2, 5}]], sm = Factor[Sum[i^m, {i, 1, n}] /. Array[{m -> #} &, 5]]},
Thread[Equal[hd, sm]]] // Column // TeXForm
\begin{array}{l} \sum _{i=1}^n i=\frac{1}{2} n (n+1) \\ \sum _{i=1}^n i^2=\frac{1}{6} n (n+1) (2 n+1) \\ \sum _{i=1}^n i^3=\frac{1}{4} n^2 (n+1)^2 \\ \sum _{i=1}^n i^4=\frac{1}{30} n (n+1) (2 n+1) \left(3 n^2+3 n-1\right) \\ \sum _{i=1}^n i^5=\frac{1}{12} n^2 (n+1)^2 \left(2 n^2+2 n-1\right) \\ \end{array}
-
+1 but $i^1$ should be replaced with $i$. – kiss my armpit Jan 15 '14 at 6:20 | 2,321 | 5,565 | {"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": 2, "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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2016-22 | longest | en | 0.793727 |
https://puzzling.stackexchange.com/questions/86082/a-pyramid-from-a-square?noredirect=1 | 1,718,383,947,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861567.95/warc/CC-MAIN-20240614141929-20240614171929-00686.warc.gz | 434,973,814 | 40,585 | A pyramid from a square
Given a square piece of paper. Cut it into 4 pieces that could be used to create a right angle pyramid - the 4 pieces are the faces of the pyramid.
• What's the definition of a right angle pyramid? Commented Jul 12, 2019 at 8:52
• I assume the question means a right pyramid, rather than right angled. A right pyramid has the apex directly above the centre of the base. Commented Jul 12, 2019 at 12:04
• No. Use this definition: " A right-angled pyramid has its apex above an edge or vertex of the base. In a tetrahedron these qualifiers change based on which face is considered the base" and the requirement is - apes above a vertex.
– Moti
Commented Jul 12, 2019 at 15:00
• @Moti The pyramid in our solutions is a right-angled pyramid then. If you consider any of the right-angled triangles as the base, the apex will lie above one of the vertices of the base. Commented Jul 12, 2019 at 16:21
• I am seeking a solution where three faces are perpendicular to each other - three edges are along the cartesian axes and 4 faces (I think your solution contains a face that is not a piece of the cut)
– Moti
Commented Jul 12, 2019 at 16:33
2 Answers
I think you could make cuts as follows
so that $$a$$ and $$b$$ are the midpoints of the sides of the square.
Then, the three triangles surrounding the central triangle can be folded up along their adjoining edges to create a pyramid.
• I edited the puzzle to reflect the requirement for a right angle pyramid. Hope I am not disappointing you since you have a nice solution.
– Moti
Commented Jul 12, 2019 at 6:54
• I think your solution uses an additional face that is not part of the cuts.
– Moti
Commented Jul 12, 2019 at 16:34
• @Moti There are four triangles in the picture. Those are the four faces, why would I need another one? Commented Jul 12, 2019 at 16:37
• @Moti, if we use the small triangle as the base, we could label the coordinates of this pyramid in three-space as (0,0,0), (1,0,0), (0,1,0) and (0,0,2). This is exactly what you've described in your comment. Commented Jul 12, 2019 at 16:40
• You are right and I am accepting your solution
– Moti
Commented Jul 12, 2019 at 17:15
How about this cut? (With Region A as the base of the pyramid)
• Seems to be exactly the same as the answer that preceded yours by a few minutes. Commented Jul 11, 2019 at 14:56
• It is, I was putting together the diagram and as I was posting I was notified the other solution was posted. I'm fairly certain this is a unique solution anyway. Commented Jul 11, 2019 at 17:48
• There is another solution for a right angle pyramid - see edited puzzle.
– Moti
Commented Jul 12, 2019 at 6:52
• See also other comments - your solution is of 5 faces pyramid.
– Moti
Commented Jul 12, 2019 at 16:35 | 768 | 2,765 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-26 | latest | en | 0.934787 |
https://physics-network.org/how-do-you-solve-average-speed-problems/ | 1,680,002,925,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948858.7/warc/CC-MAIN-20230328104523-20230328134523-00101.warc.gz | 511,726,302 | 17,703 | # How Do You Solve average speed problems?
Just add the initial speed to the final speed and divide the sum by two. The result is average speed.
## How do you find the average speed in physics?
Average speed = Distance / time elapsed = 180 meters / 6 seconds = 30 meters/second. Average velocity = Displacement / time elapsed = 60 meters / 6 seconds = 10 meters/second.
## What is average speed in physics example?
You can find the average speed of an object if you know the distance travelled and the time it took. The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time.
## What is the formula for average rate of speed?
Average This is the arithmetic mean, and is calculated by adding a group of numbers and then dividing by the count of those numbers. For example, the average of 2, 3, 3, 5, 7, and 10 is 30 divided by 6, which is 5.
## How do you find average speed and velocity in physics?
Average speed is distance divided by time interval for the distance traveled.
## How do you find average speed without distance and time?
The average speed is the distance (a scalar quantity) per time ratio. Speed is ignorant of direction. On the other hand, velocity is a vector quantity; it is direction-aware. Velocity is the rate at which the position changes.
## How do you find average speed with distance and time?
The average speed is the total distance traveled by the object in a particular time interval. The average speed is a scalar quantity. It is represented by the magnitude and does not have direction.
## How do you calculate average speed in Class 11?
1. Average Speed = Total Distance Total Time.
2. Average Speed = Total Distance Covered Time Taken.
3. Average Speed = Total Distance Total Time.
## How do we calculate average?
The S.I unit of speed is meter per second. Speed is also measured in kilometre per hour.
## How can you calculate average speed of an object in motion?
How to calculate average speed. Speed is distance divided by time. Simply put, if you drove 60 kilometres for one hour, it would look like this: Speed = distance (60 km) / time (1 hour) = 60km/h.
## Is average speed the same as velocity?
Average speed is defined as the total path length travelled divided by the total time interval of the motion.
## What is the average speed of the object?
There are three main types of average: mean, median and mode. Each of these techniques works slightly differently and often results in slightly different typical values. The mean is the most commonly used average. To get the mean value, you add up all the values and divide this total by the number of values.
## What units is average speed?
The average of a set of numbers is simply the sum of the numbers divided by the total number of values in the set. For example, suppose we want the average of 24 , 55 , 17 , 87 and 100 . Simply find the sum of the numbers: 24 + 55 + 17 + 87 + 100 = 283 and divide by 5 to get 56.6 .
## How do you find the average speed in kilometers per hour?
Finding an average gives us an idea as to an overall behaviour or trend – Mrs Mansell’s average spend on shopping gives us an idea as to whether she usually spends a lot or a little money and Keiran’s average spelling score gives us an idea as to how good he usually is at spelling.
## What is average speed in physics class 11?
Solution : `v_(av) = (10 – 5)/(8 – 4) = (5)/(4) = 1.25 m//s`.
## What are the 3 ways to calculate average?
The general speed formula for an object is given as [Speed = Distance ÷ Time]. SI units of speed is m/s.
## What is average in math example?
It can be calculated by dividing the total distance travelled by the total time taken. If an object covers different distances with different velocities, then average velocity of the object is a single velocity value for which if the object moves uniformly, it will cover the same distance in the same interval of time.
## Why do we calculate average?
An object can have zero average velocity but cannot have zero average speed.
## What is the average speed from 4s to 8s?
Average velocity is the displacement of an object over time. To find the average speed of an object we divide the distance travelled by the time elapsed. We know that velocity is a vector quantity and average velocity can be found by dividing displacement by time.
## What is the formula for finding the speed of an object?
It will take 4.5 hours to travel 432 kilometers at an average speed of 96 kilometers per hour.
## How do you find average speed when given velocity?
Average speed = (total distance travelled)/(total time taken) ii) Uniform speed refers to the constant speed of a moving body in which the body covers an equal distance in equal time intervals. Video transcript. | 1,081 | 4,835 | {"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} | 4.15625 | 4 | CC-MAIN-2023-14 | latest | en | 0.940259 |
https://edwardlib.org/api/inference | 1,726,190,445,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651506.7/warc/CC-MAIN-20240913002450-20240913032450-00569.warc.gz | 205,877,317 | 4,716 | # API and Documentation
### Inference
We describe how to perform inference in probabilistic models. For background, see the Inference tutorial.
Suppose we have a model $$p(\mathbf{x}, \mathbf{z}, \beta)$$ of data $$\mathbf{x}_{\text{train}}$$ with latent variables $$(\mathbf{z}, \beta)$$. Consider the posterior inference problem, $q(\mathbf{z}, \beta)\approx p(\mathbf{z}, \beta\mid \mathbf{x}_{\text{train}}),$ in which the task is to approximate the posterior $$p(\mathbf{z}, \beta\mid \mathbf{x}_{\text{train}})$$ using a family of distributions, $$q(\mathbf{z},\beta; \lambda)$$, indexed by parameters $$\lambda$$.
In Edward, let z and beta be latent variables in the model, where we observe the random variable x with data x_train. Let qz and qbeta be random variables defined to approximate the posterior. We write this problem as follows:
inference = ed.Inference({z: qz, beta: qbeta}, {x: x_train})
Inference is an abstract class which takes two inputs. The first is a collection of latent random variables beta and z, along with “posterior variables” qbeta and qz, which are associated to their respective latent variables. The second is a collection of observed random variables x, which is associated to the data x_train.
Inference adjusts parameters of the distribution of qbeta and qz to be close to the posterior $$p(\mathbf{z}, \beta\,|\,\mathbf{x}_{\text{train}})$$.
Running inference is as simple as running one method.
inference = ed.Inference({z: qz, beta: qbeta}, {x: x_train})
inference.run()
Inference also supports fine control of the training procedure.
inference = ed.Inference({z: qz, beta: qbeta}, {x: x_train})
inference.initialize()
tf.global_variables_initializer().run()
for _ in range(inference.n_iter):
info_dict = inference.update()
inference.print_progress(info_dict)
inference.finalize()
initialize() builds the algorithm’s update rules (computational graph) for $$\lambda$$; tf.global_variables_initializer().run() initializes $$\lambda$$ (TensorFlow variables in the graph); update() runs the graph once to update $$\lambda$$, which is called in a loop until convergence; finalize() runs any computation as the algorithm terminates.
The run() method is a simple wrapper for this procedure.
### Other Settings
We highlight other settings during inference.
Model parameters. Model parameters are parameters in a model that we will always compute point estimates for and not be uncertain about. They are defined with tf.Variables, where the inference problem is $\hat{\theta} \leftarrow^{\text{optimize}} p(\mathbf{x}_{\text{train}}; \theta)$
from edward.models import Normal
theta = tf.Variable(0.0)
x = Normal(loc=tf.ones(10) * theta, scale=1.0)
inference = ed.Inference({}, {x: x_train})
Only a subset of inference algorithms support estimation of model parameters. (Note also that this inference example does not have any latent variables. It is only about estimating theta given that we observe $$\mathbf{x} = \mathbf{x}_{\text{train}}$$. We can add them so that inference is both posterior inference and parameter estimation.)
For example, model parameters are useful when applying neural networks from high-level libraries such as Keras and TensorFlow Slim. See the model compositionality page for more details.
Conditional inference. In conditional inference, only a subset of the posterior is inferred while the rest are fixed using other inferences. The inference problem is $q(\mathbf{z}\mid\beta)q(\beta)\approx p(\mathbf{z}, \beta\mid\mathbf{x}_{\text{train}})$ where parameters in $$q(\mathbf{z}\mid\beta)$$ are estimated and $$q(\beta)$$ is fixed. In Edward, we enable conditioning by binding random variables to other random variables in data.
inference = ed.Inference({z: qz}, {x: x_train, beta: qbeta})
In the compositionality page, we describe how to construct inference by composing many conditional inference algorithms.
Implicit prior samples. Latent variables can be defined in the model without any posterior inference over them. They are implicitly marginalized out with a single sample. The inference problem is $q(\beta)\approx p(\beta\mid\mathbf{x}_{\text{train}}, \mathbf{z}^*)$ where $$\mathbf{z}^*\sim p(\mathbf{z}\mid\beta)$$ is a prior sample.
inference = ed.Inference({beta: qbeta}, {x: x_train})
For example, implicit prior samples are useful for generative adversarial networks. Their inference problem does not require any inference over the latent variables; it uses samples from the prior. | 1,087 | 4,495 | {"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": 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} | 3.375 | 3 | CC-MAIN-2024-38 | latest | en | 0.661111 |
https://groupprops.subwiki.org/w/index.php?title=Finitary_symmetric_group_on_countable_set_is_subgroup-universal_for_finite_groups&direction=prev&oldid=13956 | 1,600,648,360,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198868.29/warc/CC-MAIN-20200920223634-20200921013634-00353.warc.gz | 434,421,183 | 8,018 | # Finitary symmetric group on countable set is subgroup-universal for finite groups
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
This article gives the statement, and possibly proof, of an embeddability theorem: a result that states that any group of a certain kind can be embedded in a group of a more restricted kind.
View a complete list of embeddability theorems
## Statement
Let $\Omega$ be a countably infinite set and $G := \operatorname{FSym}(\Omega)$ be the finitary symmetric group on $\Omega$. Then, $G$ is subgroup-universal for finite groups. In other words, if $H$ is any finite group, $H$ is isomorphic to a subgroup of $G$.
## Facts used
1. Cayley's theorem: This states that every group embeds as a subgroup of the symmetric group on its underlying set.
## Proof
The proof follows from fact (1), and the observation that the symmetric group on any finite set is isomorphic to some subgroup of $G$ -- namely, the subgroup comprising the permutations on a finite subset of $\Omega$ of the same cardinality. | 251 | 1,058 | {"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": 9, "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-2020-40 | latest | en | 0.871437 |
https://jeffho.com/australian-capital-territory/formula-of-volume-of-cylinder-with-example.php | 1,627,412,298,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153474.19/warc/CC-MAIN-20210727170836-20210727200836-00337.warc.gz | 353,139,362 | 7,304 | # With formula cylinder example of of volume
## Cylinder Volume Calculator Online mathematics
Volume of Cylinder Assignment Point. In this lesson, you will explore the formula used to calculate the volume of a cube. finding the area of a cylinder: formula & example 3:06, while the volume of the cylinder is volume formula derivations for example, volume is related to the pressure and temperature of an ideal gas by the ideal gas.
### Formula for surface area and volume of cylinders StudyPug
Volume of Cylinder Assignment Point. Volume of a cylinder formula example keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition, 25/03/2016в в· learn how to find the volumes of prisms and cylinders in this free math video tutorial by mario's math tutoring. 0:17 what is a prism and what is a.
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So using the volume of cylinder formula we can find the volume of it. example 5: find the volume of a right circular cylinder whose curved surface area is 2640 cm 2 the volume of a cylinder is: pi( 3.14) x radius squared time height. example: what is the volume of a cylinder with a radius of 7cm and a height of 10cm?
Example: find the volume of the cylinder shown. the formula for the volume of a cylinder is v = b h or v = пђ the volume of a cylinder is: pi( 3.14) x radius squared time height. example: what is the volume of a cylinder with a radius of 7cm and a height of 10cm?
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How to determine the volume of a cylinder with using a. In this lesson, you will explore the formula used to calculate the volume of a cube. finding the area of a cylinder: formula & example 3:06, 25/03/2016в в· learn how to find the volumes of prisms and cylinders in this free math video tutorial by mario's math tutoring. 0:17 what is a prism and what is a.
### Radius of a Cylinder Illustrative Mathematics
How to Calculate Volume of a Circular Cylinder Sciencing. Cylinder volume calculator a collection of online mathematics calculators with clear example of how to calculate specific mathemeatical equations., to derive the formula of the surface area of a cylinder, a good real life example of an open cylinder is a pipe used to flow water if you have seen one before..
Best 25+ Volume of a cylinder ideas on Pinterest Volume. Some examples of ways to measure volume help to show how volume is measured. for the volume of compressed gas in a cylinder, apply this equation,, in this lesson, you will explore the formula used to calculate the volume of a cube. finding the area of a cylinder: formula & example 3:06.
### Volume of a Cylinder MathHelp.com YouTube
Cylinder Volume Calculator Online mathematics. For example, when finding the we add the individual volumes of 2 hemispheres and one cylinder. the volume of a capsule container= volume of hemisphere + volume of Surface area and volume examples. surface area of prisms and cylinders. the volume formula works not only for prisms,.
• Radius of a Cylinder Illustrative Mathematics
• Volume of a Cylinder MathHelp.com YouTube
• Volume of a cylinder formula example" Keyword Found
• Volume of a Cylinder MathHelp.com YouTube
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George, we have three responses for you: hi george. i've sketched the volumes involved below. the formula for the volume of a cylinder is v = пђ r 2 h. the volume of a cylinder is: pi( 3.14) x radius squared time height. example: what is the volume of a cylinder with a radius of 7cm and a height of 10cm?
The cylinder formula is applied to calculate the surface area and volume of the cylinder. the cylinder formula is example 1: a cylinder has base radius 5 cm and example: find the volume of the cylinder shown. the formula for the volume of a cylinder is v = b h or v = пђ
For a 3d shape, volume is the amount of space displaced some formulas for common 2 -dimensional plane figures and 3 prism or cylinder task. given the height $h$ and volume $v$ of a certain cylinder, jill uses the formula $$r=\sqrt{\frac{v}{\pi h}}$$ to compute its radius to be 20 meters.
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Mensuration is a very integral part of geometry, and it deals with the calculation of the area and volume of many geometrical shapes. let's learn something new today! task. given the height $h$ and volume $v$ of a certain cylinder, jill uses the formula $$r=\sqrt{\frac{v}{\pi h}}$$ to compute its radius to be 20 meters.
The cylinder formula is applied to calculate the surface area and volume of the cylinder. the cylinder formula is example 1: a cylinder has base radius 5 cm and surface area and volume examples. surface area of prisms and cylinders. the volume formula works not only for prisms, | 1,709 | 7,627 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2021-31 | latest | en | 0.862276 |
https://handwiki.org/wiki/Jackknife_resampling | 1,712,966,918,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816465.91/warc/CC-MAIN-20240412225756-20240413015756-00148.warc.gz | 267,291,203 | 22,188 | # Jackknife resampling
Short description: Statistical method for resampling
Schematic of Jackknife Resampling
In statistics, the jackknife (jackknife cross-validation) is a cross-validation technique and, therefore, a form of resampling. It is especially useful for bias and variance estimation. The jackknife pre-dates other common resampling methods such as the bootstrap. Given a sample of size $\displaystyle{ n }$, a jackknife estimator can be built by aggregating the parameter estimates from each subsample of size $\displaystyle{ (n-1) }$ obtained by omitting one observation.[1]
The jackknife technique was developed by Maurice Quenouille (1924–1973) from 1949 and refined in 1956. John Tukey expanded on the technique in 1958 and proposed the name "jackknife" because, like a physical jack-knife (a compact folding knife), it is a rough-and-ready tool that can improvise a solution for a variety of problems even though specific problems may be more efficiently solved with a purpose-designed tool.[2]
The jackknife is a linear approximation of the bootstrap.[2]
## A simple example: mean estimation
The jackknife estimator of a parameter is found by systematically leaving out each observation from a dataset and calculating the parameter estimate over the remaining observations and then aggregating these calculations.
For example, if the parameter to be estimated is the population mean of random variable $\displaystyle{ x }$, then for a given set of i.i.d. observations $\displaystyle{ x_1, ..., x_n }$ the natural estimator is the sample mean:
$\displaystyle{ \bar{x} =\frac{1}{n} \sum_{i=1}^{n} x_i =\frac{1}{n} \sum_{i \in [n]} x_i, }$
where the last sum used another way to indicate that the index $\displaystyle{ i }$ runs over the set $\displaystyle{ [n] = \{ 1,\ldots,n\} }$.
Then we proceed as follows: For each $\displaystyle{ i \in [n] }$ we compute the mean $\displaystyle{ \bar{x}_{(i)} }$ of the jackknife subsample consisting of all but the $\displaystyle{ i }$-th data point, and this is called the $\displaystyle{ i }$-th jackknife replicate:
$\displaystyle{ \bar{x}_{(i)} =\frac{1}{n-1} \sum_{j \in [n], j\ne i} x_j, \quad \quad i=1, \dots ,n. }$
It could help to think that these $\displaystyle{ n }$ jackknife replicates $\displaystyle{ \bar{x}_{(1)},\ldots,\bar{x}_{(n)} }$ give us an approximation of the distribution of the sample mean $\displaystyle{ \bar{x} }$ and the larger the $\displaystyle{ n }$ the better this approximation will be. Then finally to get the jackknife estimator we take the average of these $\displaystyle{ n }$ jackknife replicates:
$\displaystyle{ \bar{x}_{\mathrm{jack}} = \frac{1}{n}\sum_{i=1}^n \bar{x}_{(i)}. }$
One may ask about the bias and the variance of $\displaystyle{ \bar{x}_{\mathrm{jack}} }$. From the definition of $\displaystyle{ \bar{x}_{\mathrm{jack}} }$ as the average of the jackknife replicates one could try to calculate explicitly, and the bias is a trivial calculation but the variance of $\displaystyle{ \bar{x}_{\mathrm{jack}} }$ is more involved since the jackknife replicates are not independent.
For the special case of the mean, one can show explicitly that the jackknife estimate equals the usual estimate:
$\displaystyle{ \frac{1}{n}\sum_{i=1}^n \bar{x}_{(i)} = \bar{x}. }$
This establishes the identity $\displaystyle{ \bar{x}_{\mathrm{jack}} = \bar{x} }$. Then taking expectations we get $\displaystyle{ E[\bar{x}_{\mathrm{jack}}] = E[\bar{x}] =E[x] }$, so $\displaystyle{ \bar{x}_{\mathrm{jack}} }$ is unbiased, while taking variance we get $\displaystyle{ V[\bar{x}_{\mathrm{jack}}] = V[\bar{x}] =V[x]/n }$. However, these properties do not hold in general for other parameters than the mean.
This simple example for the case of mean estimation is just to illustrate the construction of a jackknife estimator, while the real subtleties (and the usefulness) emerge for the case of estimating other parameters, such as higher moments than the mean or other functionals of the distribution.
$\displaystyle{ \bar{x}_{\mathrm{jack}} }$ could be used to construct an empirical estimate of the bias of $\displaystyle{ \bar{x} }$, namely $\displaystyle{ \widehat{\operatorname{bias}}(\bar{x})_{\mathrm{jack}} = c(\bar{x}_{\mathrm{jack}} - \bar{x}) }$ with some suitable factor $\displaystyle{ c\gt 0 }$, although in this case we know that $\displaystyle{ \bar{x}_{\mathrm{jack}} = \bar{x} }$ so this construction does not add any meaningful knowledge, but it gives the correct estimation of the bias (which is zero).
A jackknife estimate of the variance of $\displaystyle{ \bar{x} }$ can be calculated from the variance of the jackknife replicates $\displaystyle{ \bar{x}_{(i)} }$:[3][4]
$\displaystyle{ \widehat{\operatorname{var}}(\bar{x})_{\mathrm{jack}} =\frac{n-1}{n} \sum_{i=1}^n (\bar{x}_{(i)} - \bar{x}_{\mathrm{jack}})^2 =\frac{1}{n(n-1)} \sum_{i=1}^n (x_i - \bar{x})^2. }$
The left equality defines the estimator $\displaystyle{ \widehat{\operatorname{var}}(\bar{x})_{\mathrm{jack}} }$ and the right equality is an identity that can be verified directly. Then taking expectations we get $\displaystyle{ E[\widehat{\operatorname{var}}(\bar{x})_{\mathrm{jack}}] = V[x]/n = V[\bar{x}] }$, so this is an unbiased estimator of the variance of $\displaystyle{ \bar{x} }$.
## Estimating the bias of an estimator
The jackknife technique can be used to estimate (and correct) the bias of an estimator calculated over the entire sample.
Suppose $\displaystyle{ \theta }$ is the target parameter of interest, which is assumed to be some functional of the distribution of $\displaystyle{ x }$. Based on a finite set of observations $\displaystyle{ x_1, ..., x_n }$, which is assumed to consist of i.i.d. copies of $\displaystyle{ x }$, the estimator $\displaystyle{ \hat{\theta} }$ is constructed:
$\displaystyle{ \hat{\theta} =f_n(x_1,\ldots,x_n). }$
The value of $\displaystyle{ \hat{\theta} }$ is sample-dependent, so this value will change from one random sample to another.
By definition, the bias of $\displaystyle{ \hat{\theta} }$ is as follows:
$\displaystyle{ \text{bias}(\hat{\theta}) = E[\hat{\theta}] - \theta. }$
One may wish to compute several values of $\displaystyle{ \hat{\theta} }$ from several samples, and average them, to calculate an empirical approximation of $\displaystyle{ E[\hat{\theta}] }$, but this is impossible when there are no "other samples" when the entire set of available observations $\displaystyle{ x_1, ..., x_n }$ was used to calculate $\displaystyle{ \hat{\theta} }$. In this kind of situation the jackknife resampling technique may be of help.
We construct the jackknife replicates:
$\displaystyle{ \hat{\theta}_{(1)} =f_{n-1}(x_{2},x_{3}\ldots,x_{n}) }$
$\displaystyle{ \hat{\theta}_{(2)} =f_{n-1}(x_{1},x_{3},\ldots,x_{n}) }$
$\displaystyle{ \vdots }$
$\displaystyle{ \hat{\theta}_{(n)} =f_{n-1}(x_1,x_{2},\ldots,x_{n-1}) }$
where each replicate is a "leave-one-out" estimate based on the jackknife subsample consisting of all but one of the data points:
$\displaystyle{ \hat{\theta}_{(i)} =f_{n-1}(x_{1},\ldots,x_{i-1},x_{i+1},\ldots,x_{n}) \quad \quad i=1, \dots,n. }$
Then we define their average:
$\displaystyle{ \hat{\theta}_\mathrm{jack}=\frac{1}{n} \sum_{i=1}^n \hat{\theta}_{(i)} }$
The jackknife estimate of the bias of $\displaystyle{ \hat{\theta} }$ is given by:
$\displaystyle{ \widehat{\text{bias}}(\hat{\theta})_\mathrm{jack} =(n-1)(\hat{\theta}_\mathrm{jack} - \hat{\theta}) }$
and the resulting bias-corrected jackknife estimate of $\displaystyle{ \theta }$ is given by:
$\displaystyle{ \hat{\theta}_{\text{jack}}^{*} =\hat{\theta} - \widehat{\text{bias}}(\hat{\theta})_\mathrm{jack} =n\hat{\theta} - (n-1)\hat{\theta}_\mathrm{jack}. }$
This removes the bias in the special case that the bias is $\displaystyle{ O(n^{-1}) }$ and reduces it to $\displaystyle{ O(n^{-2}) }$ in other cases.[2]
## Estimating the variance of an estimator
The jackknife technique can be also used to estimate the variance of an estimator calculated over the entire sample.
## Literature
• Berger, Y.G. (2007). "A jackknife variance estimator for unistage stratified samples with unequal probabilities". Biometrika 94 (4): 953–964. doi:10.1093/biomet/asm072.
• Berger, Y.G.; Rao, J.N.K. (2006). "Adjusted jackknife for imputation under unequal probability sampling without replacement". Journal of the Royal Statistical Society, Series B 68 (3): 531–547. doi:10.1111/j.1467-9868.2006.00555.x.
• Berger, Y.G.; Skinner, C.J. (2005). "A jackknife variance estimator for unequal probability sampling". Journal of the Royal Statistical Society, Series B 67 (1): 79–89. doi:10.1111/j.1467-9868.2005.00489.x.
• Jiang, J.; Lahiri, P.; Wan, S-M. (2002). "A unified jackknife theory for empirical best prediction with M-estimation". The Annals of Statistics 30 (6): 1782–810. doi:10.1214/aos/1043351257.
• Jones, H.L. (1974). "Jackknife estimation of functions of stratum means". Biometrika 61 (2): 343–348. doi:10.2307/2334363.
• Kish, L.; Frankel, M.R. (1974). "Inference from complex samples". Journal of the Royal Statistical Society, Series B 36 (1): 1–37.
• Krewski, D.; Rao, J.N.K. (1981). "Inference from stratified samples: properties of the linearization, jackknife and balanced repeated replication methods". The Annals of Statistics 9 (5): 1010–1019. doi:10.1214/aos/1176345580.
• Quenouille, M.H. (1956). "Notes on bias in estimation". Biometrika 43 (3–4): 353–360. doi:10.1093/biomet/43.3-4.353.
• Rao, J.N.K.; Shao, J. (1992). "Jackknife variance estimation with survey data under hot deck imputation". Biometrika 79 (4): 811–822. doi:10.1093/biomet/79.4.811.
• Rao, J.N.K.; Wu, C.F.J.; Yue, K. (1992). "Some recent work on resampling methods for complex surveys". Survey Methodology 18 (2): 209–217.
• Shao, J. and Tu, D. (1995). The Jackknife and Bootstrap. Springer-Verlag, Inc.
• Tukey, J.W. (1958). "Bias and confidence in not-quite large samples (abstract)". The Annals of Mathematical Statistics 29 (2): 614.
•
## Notes
1. Efron 1982, p. 2.
2. Cameron & Trivedi 2005, p. 375.
3. Efron 1982, p. 14.
4. McIntosh, Avery I.. "The Jackknife Estimation Method". Avery I. McIntosh. : p. 3.
## References
• Cameron, Adrian; Trivedi, Pravin K. (2005). Microeconometrics : methods and applications. Cambridge New York: Cambridge University Press. ISBN 9780521848053.
• Efron, Bradley; Stein, Charles (May 1981). "The Jackknife Estimate of Variance". The Annals of Statistics 9 (3): 586–596. doi:10.1214/aos/1176345462.
• Efron, Bradley (1982). The jackknife, the bootstrap, and other resampling plans. Philadelphia, PA: Society for Industrial and Applied Mathematics. ISBN 9781611970319.
• Quenouille, Maurice H. (September 1949). "Problems in Plane Sampling". The Annals of Mathematical Statistics 20 (3): 355–375. doi:10.1214/aoms/1177729989.
• Quenouille, Maurice H. (1956). "Notes on Bias in Estimation". Biometrika 43 (3-4): 353–360. doi:10.1093/biomet/43.3-4.353.
• Tukey, John W. (1958). "Bias and confidence in not quite large samples (abstract)". The Annals of Mathematical Statistics 29 (2): 614. doi:10.1214/aoms/1177706647. | 3,265 | 11,114 | {"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-2024-18 | latest | en | 0.799069 |
http://www.ncatlab.org/nlab/show/I-category | 1,397,759,880,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2014-15/segments/1397609530895.48/warc/CC-MAIN-20140416005210-00160-ip-10-147-4-33.ec2.internal.warc.gz | 584,068,210 | 6,453 | # Idea
In his development of an ‘algebraic homotopy’ theory, Baues uses interacting structures, one of Quillen type (or rather of K. Brown’s version of half of Quillen’s theory) and the other of cylinder functor type. The two structures are called cofibration categories and $I$-categories.
# Definition
An $I$-category has various data specified: $(C, \mathit{cof}, I, \emptyset)$.
Here $C$ is a category, $\mathit{cof}$ is a class of ‘cofibrations’, $\emptyset$ is an initial object of $C$, and $I$ is a cylinder functor (written as a functor, so $I(X)$ is the cylinder on object $X$).
These are required to satisfy:
I 1) $I$ is a cylinder functor;
I 2) Pushout axiom (almost as in the first part of C2 of cofibration category, but $I$ is also to preserve pushouts, and $I(\emptyset) = \emptyset$ so in fact $I$ preserves all finite colimits);
I 3) Cofibration axiom:
• $\iso \subset \mathit{cof}$ (where $\iso$ is the class of isomorphisms in $C$);
• $\emptyset \rightarrow X$ is always in $\mathit{cof}$;
• a composition of cofibrations is a cofibration and all morphisms in $\mathit{cof}$ satisfy the homotopy extension property.
I 4) Relative cylinder axiom:
If $i : B \rightarrow A$ is a cofibration and one forms the pushout $A \cup_B I(B)\cup_B A$, then the natural map
$A \cup_B I(B)\cup_B A \rightarrow A\times I$
is a cofibration;
I 5) The ‘interchange’ axiom.
For all objects $X$, there is a map
$T : I^2(X) \rightarrow I^2(X)$
interchanging the two copies of $I$, i.e.
$T \circ e_i(I(X)) = I(e_i(X)), \quad T \circ I(e_i(X)) = e_i(I(X))$
for $i = 0,1$.
(This corresponds to exchanging the first and second $I$-coordinates of $X\times \mathbf{I} \times \mathbf{I}$ (where $I(X)$ is thought of as $X \times \mathbf{I}$), that is
$(x,s,t) \rightarrow (x,t,s).$
Revised on March 15, 2009 22:10:56 by Toby Bartels (71.104.234.95) | 589 | 1,859 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 33, "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} | 3.171875 | 3 | CC-MAIN-2014-15 | longest | en | 0.825053 |
https://fr.mathworks.com/matlabcentral/cody/problems/143-cannon-ball/solutions/255323 | 1,571,382,684,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677964.40/warc/CC-MAIN-20191018055014-20191018082514-00284.warc.gz | 525,911,226 | 15,283 | Cody
Problem 143. Cannon Ball
Solution 255323
Submitted on 5 Jun 2013 by Franek
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
%% g=32; h=10000; v_correct = 800; assert(isequal(canon(g,h),v_correct))
v = 800
2 Pass
%% g=32; h=100; v_correct = 80; assert(isequal(canon(g,h),v_correct))
v = 80
3 Pass
%% g=32; h=4; v_correct = 16; assert(isequal(canon(g,h),v_correct))
v = 16 | 169 | 502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-43 | latest | en | 0.608635 |
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Exponential Decay A model for decay of a quantity for which the rate of decay is directly proportional to the amount present. The equation for the model is A = A0bt (where 0 < b < 1 ) or A = A0ekt (where k is a negative number representing the rate of decay). In both formulas A0 is the original amount present at time t = 0. This model is used for phenomena such as radioactivity or depreciation. For example, A = 50e–0.01t is a model for exponential decay of 50 grams of a radioactive element that decays at a rate of 1% per year. See also
this page updated 19-jul-17 Mathwords: Terms and Formulas from Algebra I to Calculus written, illustrated, and webmastered by Bruce Simmons Copyright © 2000 by Bruce Simmons All rights reserved | 264 | 1,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-31 | longest | en | 0.860456 |
http://www.numbersaplenty.com/5624 | 1,553,567,121,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204768.52/warc/CC-MAIN-20190326014605-20190326040605-00038.warc.gz | 314,735,219 | 3,417 | Search a number
5624 = 231937
BaseRepresentation
bin1010111111000
321201022
41113320
5134444
642012
722253
oct12770
97638
105624
114253
123308
132738
14209a
1519ee
hex15f8
5624 has 16 divisors (see below), whose sum is σ = 11400. Its totient is φ = 2592.
The previous prime is 5623. The next prime is 5639. The reversal of 5624 is 4265.
Adding to 5624 its reverse (4265), we get a palindrome (9889).
It is a Cunningham number, because it is equal to 752-1.
It is a plaindrome in base 5 and base 15.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (5623) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 134 + ... + 170.
25624 is an apocalyptic number.
It is an amenable number.
5624 is a primitive abundant number, since it is smaller than the sum of its proper divisors, none of which is abundant.
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
It is a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (5700).
5624 is a wasteful number, since it uses less digits than its factorization.
5624 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 62 (or 58 counting only the distinct ones).
The product of its digits is 240, while the sum is 17.
The square root of 5624 is about 74.9933330370. The cubic root of 5624 is about 17.7834125657.
The spelling of 5624 in words is "five thousand, six hundred twenty-four". | 459 | 1,583 | {"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-2019-13 | latest | en | 0.902852 |
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Question
# Solve for x in x3+1=715
A
58
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B
85
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Solution
## The correct option is B −85We have, x3+1=715 ⇒x3+1−1=715−1 [Subtracting 1 from both sides] ⇒x3=7−1515 ⇒x3=−815 ⇒x3×3=−815×3 [Multiplying both sides by 3] ∴x=−85
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Join BYJU'S Learning Program | 257 | 749 | {"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} | 3.015625 | 3 | CC-MAIN-2024-33 | latest | en | 0.82692 |
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Rounds a double-precision floating-point value to the specified number of fractional digits
``` Module Example Public Sub Main() Dim values() As Double = { 2.125, 2.135} For Each value As Double In values Console.WriteLine("{0} --> {1}", value, Math.Round(value, 2, MidpointRounding.AwayFromZero)) Next End Sub End Module ```
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http://3cq.org/standard-error/what-is-another-name-for-standard-error.php | 1,537,305,371,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155702.33/warc/CC-MAIN-20180918205149-20180918225149-00278.warc.gz | 2,472,607 | 5,332 | Home > Standard Error > What Is Another Name For Standard Error
# What Is Another Name For Standard Error
## Contents
This serves as a measure of variation for random variables, providing a measurement for the spread. n = Number of observations. Despite the small difference in equations for the standard deviation and the standard error, this small difference changes the meaning of what is being reported from a description of the variation The standard error is the standard deviation of the Student t-distribution. navigate here
This spread is most often measured as the standard error, accounting for the differences between the means across the datasets.The more data points involved in the calculations of the mean, the Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Post a comment and I'll do my best to help! Greek letters indicate that these are population values. https://en.wikipedia.org/wiki/Standard_error
## Standard Error Formula
View Mobile Version Search Statistics How To Statistics for the rest of us! Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. If σ is known, the standard error is calculated using the formula σ x ¯ = σ n {\displaystyle \sigma _{\bar {x}}\ ={\frac {\sigma }{\sqrt {n}}}} where σ is the In statistics, a sample mean deviates from the actual mean of a population; this deviation is the standard error.
JSTOR2340569. (Equation 1) ^ James R. What is the standard error? Sample 1. Standard Error Excel The confidence interval of 18 to 22 is a quantitative measure of the uncertainty – the possible difference between the true average effect of the drug and the estimate of 20mg/dL.
Privacy policy About Wikipedia Disclaimers Contact Wikipedia Developers Cookie statement Mobile view Topics What's New Buy Certainty: Markets Tend to Pop After Election Thiel Defends Trump, Says He Will Standard Error Vs Standard Deviation T-distributions are slightly different from Gaussian, and vary depending on the size of the sample. What is the SE Calculation? http://www.investopedia.com/terms/s/standard-error.asp The larger your sample size, the smaller the SE.
Notice that the population standard deviation of 4.72 years for age at first marriage is about half the standard deviation of 9.27 years for the runners. Standard Error Of The Mean Definition To estimate the standard error of a student t-distribution it is sufficient to use the sample standard deviation "s" instead of σ, and we could use this value to calculate confidence The standard error can be computed from a knowledge of sample attributes - sample size and sample statistics. Ecology 76(2): 628 – 639. ^ Klein, RJ. "Healthy People 2010 criteria for data suppression" (PDF).
## Standard Error Vs Standard Deviation
The distribution of the mean age in all possible samples is called the sampling distribution of the mean. For the purpose of this example, the 9,732 runners who completed the 2012 run are the entire population of interest. Standard Error Formula It will be shown that the standard deviation of all possible sample means of size n=16 is equal to the population standard deviation, σ, divided by the square root of the Standard Error Definition Save your draft before refreshing this page.Submit any pending changes before refreshing this page.
Standard error of mean versus standard deviation In scientific and technical literature, experimental data are often summarized either using the mean and standard deviation or the mean with the standard error. http://3cq.org/standard-error/when-to-report-standard-deviation-and-standard-error.php Choose the correct answer.a) Another name is XXXXX XXXXX because the standard deviation of x determines the average amount of deviation to be expected when a population mean is estimated by Sample 2. In fact, data organizations often set reliability standards that their data must reach before publication. Standard Error Regression
However, the sample standard deviation, s, is an estimate of σ. The survey with the lower relative standard error can be said to have a more precise measurement, since it has proportionately less sampling variation around the mean. For each sample, the mean age of the 16 runners in the sample can be calculated. his comment is here For an upcoming national election, 2000 voters are chosen at random and asked if they will vote for candidate A or candidate B.
AP Statistics Tutorial Exploring Data ▸ The basics ▾ Variables ▾ Population vs sample ▾ Central tendency ▾ Variability ▾ Position ▸ Charts and graphs ▾ Patterns in data ▾ Dotplots Difference Between Standard Error And Standard Deviation The standard error of the mean (SEM) (i.e., of using the sample mean as a method of estimating the population mean) is the standard deviation of those sample means over all and Keeping, E.S. (1963) Mathematics of Statistics, van Nostrand, p. 187 ^ Zwillinger D. (1995), Standard Mathematical Tables and Formulae, Chapman&Hall/CRC.
## The standard deviation is computed solely from sample attributes.
What is the Standard Error of a Sample ? Correction for correlation in the sample Expected error in the mean of A for a sample of n data points with sample bias coefficient ρ. Just like we estimated the population standard deviation using the sample standard deviation, we can estimate the population standard error using the sample standard deviation. Standard Error Of Proportion If people are interested in managing an existing finite population that will not change over time, then it is necessary to adjust for the population size; this is called an enumerative
Or decreasing standard error by a factor of ten requires a hundred times as many observations. In each of these scenarios, a sample of observations is drawn from a large population. doi:10.4103/2229-3485.100662. ^ Isserlis, L. (1918). "On the value of a mean as calculated from a sample". http://3cq.org/standard-error/why-is-standard-error-smaller-than-standard-deviation.php Blackwell Publishing. 81 (1): 75–81.
To estimate the standard error of a student t-distribution it is sufficient to use the sample standard deviation "s" instead of σ, and we could use this value to calculate confidence The mean age for the 16 runners in this particular sample is 37.25. However, different samples drawn from that same population would in general have different values of the sample mean, so there is a distribution of sampled means (with its own mean and JSTOR2340569. (Equation 1) ^ James R.
Population. The mean of all possible sample means is equal to the population mean. Ecology 76(2): 628 – 639. ^ Klein, RJ. "Healthy People 2010 criteria for data suppression" (PDF). How to Find an Interquartile Range 2.
If σ is not known, the standard error is estimated using the formula s x ¯ = s n {\displaystyle {\text{s}}_{\bar {x}}\ ={\frac {s}{\sqrt {n}}}} where s is the sample When this occurs, use the standard error. The ages in that sample were 23, 27, 28, 29, 31, 31, 32, 33, 34, 38, 40, 40, 48, 53, 54, and 55. The effect of the FPC is that the error becomes zero when the sample size n is equal to the population size N.
The graph below shows the distribution of the sample means for 20,000 samples, where each sample is of size n=16. National Center for Health Statistics typically does not report an estimated mean if its relative standard error exceeds 30%. (NCHS also typically requires at least 30 observations – if not more | 1,622 | 7,518 | {"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-2018-39 | latest | en | 0.834999 |
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pioneersteffen September 15, 2009 10:08
Total-Pressure loses
Dear Star-users,
I currently do a transient simulation of an exhaust manifold and I want to devide the manifold in the postprecessing into different parts to analyse absolute pressure loses. The simulation is allready done and I want to anlalyse the results in Post.
How can I do this in Star-CD-Post?
I think I must implement some "areas" to average the absolut pressure over the area and get a value. But how does in work in Star-CD?
Kind regards
Steffen
olesen September 16, 2009 02:28
Quote:
Originally Posted by pioneersteffen (Post 229561) I currently do a transient simulation of an exhaust manifold and I want to devide the manifold in the postprecessing into different parts to analyse absolute pressure loses. The simulation is allready done and I want to anlalyse the results in Post. How can I do this in Star-CD-Post? I think I must implement some "areas" to average the absolut pressure over the area and get a value. But how does in work in Star-CD?
For each time-step
1. get the total pressure. Here's an example with scaling in mbar.
Code:
```! ptotal.MAC ! ! get absolute total pressure in mbar ! use Star-CD definition ! out: reg1-3= Vel, reg4= Ptotal [mbar] getc all PTot,absolute\$oper smult 1e-2 4 4 head\$TOTAL PRESSURE\$ABSOLUTE [mbar]```
2. Set a section cut with 'spoint' and 'snorm'. Note that since this cut extends through the entire model, you have to be careful that you only have the cset that corresponds to your region of interest.
3. Get the average values across the cut.
Eg,
Code:
```! savg.MAC ! ! use current spoint/snormal to calculate avg in slice ofile none *get ATot TAREA *get STot TAS *set Savg STot / ATot ofile screen *list Atot !-> Area *list Stot !-> Total Value *list Savg !-> Avg Value```
These are a straight average; you'll need to do some extra work if you want massflow-weighted values.
pioneersteffen September 16, 2009 09:04
Dear Mark,
I follow your instruction and I have the problem, you have warned me. When I define a section cut, the entire model is cutted, not only one pipe. How can I do this? Sorry for the stupid question.
I already looking in the uses guide, but I don't find a solution for this problem.
Kind regards
Steffen
olesen September 16, 2009 09:15
Quote:
Originally Posted by pioneersteffen (Post 229680) I follow your instruction and I have the problem, you have warned me. When I define a section cut, the entire model is cutted, not only one pipe. How can I do this? Sorry for the stupid question.
As I mentioned, you need to reduce your cell-set to the region-of-interest. Eg, "cset subset zone", or you can also use the pro-STAR button that looks something like [C->] for the same thing.
Since this is command that you'll always be using, I'd suggest making an abbreviation for it (eg, cnz). See the user's manual about defining abbreviations ... I haven't touched mine for quite some time.
This is one of the really nice things about pro-STAR, you can use the menus if you wish, but you can also type the commands too. With a few simple aliases, like that above, you can become quite efficient: type a few letters with your left hand while still holding onto the mouse with your right.
anil1886 September 16, 2009 16:04
help regarding my simulation..
hi Mr Steffen Gruner,
i'm trying to simulate an exhaust manifold to learn the transient analysis of the model below. i'm done with most of the meshing.
i want to know how to create the boundary conditions for changing the inlets based on the crank angle and firing order.
plz help with this..
http://img17.imageshack.us/img17/2640/14940251.png
pioneersteffen September 18, 2009 03:39
Hello,
I have convert the Crank-angle into a time. So when you have for example 3500 RPM, you can convert it into a timestep-duration of one crank angle of 60/(3500*360)=4,76e-5s. Then I define a table with the table editor and define the timesteps with the associated pressure and temperature. That table you can import in the boundary condition section.
Kind Regards
Steffen
anil1886 September 18, 2009 04:14
really thanks for the tip. i got upto that. but how do you change the boundary conditions from one flange to other based on the time step..
like how do you make a port an inlet and a wall based on the time and according firing order..
and if it is ok, i mention, only if it is really ok, can i see how you wrote the time steps in the table editor, because i'm totally new to this and i dont even completely know how to play with the options and boundary conditions. every information you give will be really helpful and i'll be using to solve my problem.
thank you very much for the tip..
thank you,
anil.
pioneersteffen September 18, 2009 04:55
Dear Mr. Olesen,
I can do only section cut for one pipe yet. I create a new cell set of one pipe and plot only the cell set of the pipe. So it might be secure, that I just use this for section cut and for calculation of the area and therefore the area averaged pressure. Is it right?
The strange thing is, when I use your commands for avagering, and I look at the "Atot" output, the value seems to be very small. The "Atot" is 0.1385e-5, but the diameter of the pipe is 34mm and so we must get 907mm˛. Or do I have the wrong train of thoughts?
Kind Regards.
Steffen
pioneersteffen September 18, 2009 08:57
Dear Anil,
at first you open the table editor. Then you create a new table and choose "table type selection" as "Boundary Conditions", and select under "options" a boundary type, e.g. "pressure". The next step is to select the "independent variable" as "time" and for "dependent variables" e.g. "PR"(Pressure) and T(Temperatur). Then you type all time steps inside the "independent Variables Table" and "fill the independent Variables to columns". Then you define for every time step, the pressure and temperatur and push "write", so save the table.
In the boundary definition you select "Region Type", "Pressure" and in "User Options" you must choose "Table". Then define the "Table Name" and all is fine.
You can also look at the users-guide and the tutorial guide. I remember, there are some examples.
Regards
Steffen
anil1886 October 7, 2009 16:53
sample value of pressure
hi Mr Steffen,
I'm working on the simulation now. I'm trying my model in fluent to check how it is working. I dont have more data on the exhaust gas pressure and temperatures.
If u can give any data regarding the pressure, velocity and temperature of exhaust gases of an engine... it would be a great help for me..
my email is anil1886@gmail.com
thanks,
anil. | 1,744 | 6,883 | {"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-2017-26 | latest | en | 0.819076 |
https://betterlesson.com/lesson/resource/2669135/ordering-numbers-11-20-notebook-file | 1,540,233,174,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515375.86/warc/CC-MAIN-20181022180558-20181022202058-00433.warc.gz | 602,657,130 | 29,281 | # Daniel's Dance Studio-Ordering Numbers 11-20
8 teachers like this lesson
Print Lesson
## Objective
Students will be able to order numbers within 11 through 20.
#### Big Idea
The students have had lots of practice ordering numbers 0-10. Now it's time to apply their number knowledge to order larger numbers.
## Opening
10 minutes
For this part of the lesson, you will need the book Daniel's Dancers, included here along with some Tips in using the book. Make sure the dances are not in order when you read the book to the students.
I gather the students around my big chair. I ask them some questions to help them make a connection to the text. Do you like to dance? Does anyone in our class take dance lessons? Do you know someone who takes dance lessons? Today we are going to read about a special dance teacher. His name is Daniel. Daniel does a great job teaching dance. I think he is pretty good at math too. Let's read our story and find out more about Daniel.
Page 1: Hi! My name is Daniel and I run a dance school. I teach all kinds of dance. I like my dancers to be in order, so I number them from 11-20. (One student asks, why he doesn't number then from 1-10. Another student answer, "Because we know those numbers already!!)
Page 2: Oh no! The hula dancers are not in order. Can you help? I invite a student to come up and move the pieces around to put the dancers in order. We count together to check the student's work.
Page 3: Oh no! The ballet dancers are not in order. Can you help? Another student is invited up with the class checking.
Page 4: Oh no! The break dancers are not in order. Can you help? Same as above.
Page 5: Daniel is such a great dance teacher that EVERYONE wants to come to his school. Speech Bubbles: Sorry!! This is a dance school, not a zoo! Awwww!
Why do you think Daniel would not like animals at his dance studio? I hear many great answers including: animals don't know how to dance, they don't have money...to...they might poop on the dance floor!
Let's work some more on putting numbers into order. Move on over to your SmartBoard Spots.
## Instruction
15 minutes
For this portion of the lesson, I use a SmartBoard with the Ordering Numbers 11-20 notebook file. If you have a different type of interactive whiteboard, you can still use this lesson by opening the file in Smart Notebook Express. There is also a PDF of the slides so you can recreate this part of the lesson.
I gather my students in front of the SmartBoard using a system for equitable selection (each student's name on a card).
I open the first slide (SmartBoard Slide 1) with the lesson objective written in "student friendly" terms. There is a content objective and a language objective to help focus on vocabulary expansion for my English Learners (ELs) to be congruent with SIOP instructional techniques. I read these objectives aloud for my students.
Content Objective
I can put the numbers 11-20 in order.
Language Objective
I can tell a friend if the numbers 11-20 are in the correct order.
We then continue with the rest of the slides.
Slide 2: When I count, I always say the numbers in a certain order. I have the students count with me as I point to the numbers.
Slide 3: If the numbers are in the wrong order, I won't be able to count. 13-14-15-16-18-17 (The students giggle)
Slide 4: We have had a lot of practice putting the numbers 0-10 in order. Now we need to work with bigger numbers. If I am not sure of the number order, I can count or I can look at a number line.
Slide 5: Let's practice putting some numbers in order. I invite four students to come up to the Smartboard and put the numbers in order. We count to check their work.
Slide 6: Can you do these? I invite five students to come up and slide the numbers into place. Again, we count to check their work.
Slide 7: Can you put them in order without any help? I invite students to come up and move the numbers into order.
Slide 8: It is now turn and talk time. Turn and Talk allows my English Learners to practice their academic language with a peer. The students hold hands with their assigned Turn and Talk partner and lift their hands in the air so I can check that everyone has a partner. I ask them the question, My friend put these numbers in order. How did he do? What did he do right and what did he do wrong? I give the students time to talk with their partners. When it is obvious the students are done talking, I invite a student to share. The student says, "They are not in the right order. He has the biggest number first instead of last. He should have started with 11. He is counting backwards." (Wow!!). I say to the class, Brandon is right. When we put numbers in order, we go left to right, just like when we write our names. He started in the wrong spot. The numbers are in the correct order, but because he didn't start in the correct spot, they are backwards.
## Guided Practice
10 minutes
For this part of the lesson, you will need the Hanging Numbers 10-20 printed so that each student has a number. (I have exactly 22 students so I do two sets so I can divide the class into two.) I laminate the cards and punch two holes in the top and attach string so they can be reused.
I divide the class into two groups and give each student a number card. I tell them the two groups must stay separate. I then tell the students, We are going to put the number cards in order. You need to move around until you are in order from 11-20. I want you to think about who would be at the beginning of the line and who would be at the end. When I say "go", you can begin.
I then invite the student to form a line from 11-20. If the students need assistance, I offer coaching, such as...Who should go first, you or you? Should you be at the beginning or the end?
After the students are in order, we double check the order by having the students count off. I then have the students mix up. I take and trade some students between the two groups (a 12 for a 12, a 14 for a 14, etc.) and have the students complete the task again. Again we count off to check our work. The mathematical practice that I integrated into this lesson is MP7-Look for and make use of structure. It is practiced as students look for the pattern in numbers so that they can be ordered correctly.
I collect the hanging signs as we will be using them again for the next lesson. I ask the students to return to their seats for Independent Practice.
## Independent Practice and Informal Assessment
10 minutes
Give the students a copy of the Ordering Numbers 11-20 Activity sheet and have them write their names at the top of the sheet. I tell the students, "You are now going to show me what you know about ordering numbers 11-20. You will cut the numbers out and order them on the top half of the sheet. When you are done, please raise your hand so your work can be checked before it is glued down."
I circulate around the room while the students are working. The students do quite well on the activity. After I have checked their work, they glue the pieces down and show me the completed project. It then goes in their mailbox to go home. | 1,747 | 7,230 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-43 | longest | en | 0.938702 |
https://www.york.ac.uk/students/studying/manage/programmes/module-catalogue/module/MAT00041I/2019-20 | 1,600,868,662,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400210996.32/warc/CC-MAIN-20200923113029-20200923143029-00100.warc.gz | 1,084,835,166 | 7,776 | Accessibility statement
# Linear Algebra for the Natural Sciences - MAT00041I
« Back to module search
• Department: Mathematics
• Module co-ordinator: Dr. Ian McIntosh
• Credit value: 10 credits
• Credit level: I
• Academic year of delivery: 2019-20
• None
• None
## Module will run
Occurrence Teaching cycle
A Autumn Term 2019-20 to Spring Term 2019-20
## Module aims
Linear Algebra underpins a very significant part of mathematical modelling, and to be successfully applied in the sciences it is necessary to understand both the theory and the practice of using linear algebra. This module will cover fundamental material and address applications and techniques which students will subsequently be able to draw on in various contexts.
## Module learning outcomes
Subject content
• Linear systems of equations in n real or complex variables and their solutions: linear combinations, linear span, subspaces, independence and bases, dimension, existence and uniqueness of solutions to linear systems described through the kernel (null space) and image of a matrix, the Rank-Nullity Theorem.
• Linear transformations: linearity of maps between coordinate spaces, composition as matrix multiplication, one-to-one and onto as consequences of nullity and rank, eigenvectors and eigenvalues, diagonalisability as a change of coordinates, diagonalisability criteria. Properties of trace and determinant.
• Real and Hermitian inner products: real inner products and symmetric, positive definite matrices; Hermitian inner products and Hermitian symmetric positive definite matrices. Cauchy-Schwarz and triangle inequality. Orthonormal and unitary bases, orthogonal projection onto a subspace, Gram-Schmidt procedure. Diagonalisability of real and Hermitian symmetric matrices.
• Abstract linear algebra: definition of a vector space. All concepts (linear combination, subspace, independence and bases, linear transformations, kernel and image, eigenvectors and eigenvalues) as above with coordinate space replaced by abstract vector space. Additional structure required for real and Hermitian inner products.
• It is hard to overstate the importance of linear algebra in a mathematician’s toolkit. Techniques and results from linear algebra are used across the full spectrum of mathematics and its applications, both in an academic setting and in the wider world. To take an example, as well as having concrete applications in all three of our second year streams, the theory of eigenvalues and eigenvectors is essential in Google’s PageRank algorithm.
## Module content
The theoretical material is developed in the lectures in Autumn term. In Spring term there are 5 practical classes which focus on understanding numerical algorithms and the practical application of linear algebra to mathematical modelling.
## Assessment
Task Length % of module mark
24 hour open exam
Linear Algebra for the Natural Sciences
N/A 80
Essay/coursework
Computing Assignment
N/A 20
None
### Reassessment
Task Length % of module mark
24 hour open exam
Linear Algebra for the Natural Sciences
N/A 80
Essay/coursework
Computing Assignment
N/A 20
## Module feedback
Current Department policy on feedback is available in the undergraduate student handbook. Coursework and examinations will be marked and returned in accordance with this policy.
• R B J T Allenby, Linear Algebra, Arnold (S 2.897 ALL).
• R Kaye and R Wilson, Linear Algebra, OUP (S 2.897 KAY).
• D C Lay, Linear Algebra and its applications, Addison Wesley (S 2.897 LAY).
• J. B. Fraleigh and R. A. Beauregard, Linear Algebra, Addison Wesley (S 2.897 FRA).
• P. R. Halmos, Linear Algebra Problem Book, MAA ( S 2.897 HAL).
The information on this page is indicative of the module that is currently on offer. The University is constantly exploring ways to enhance and improve its degree programmes and therefore reserves the right to make variations to the content and method of delivery of modules, and to discontinue modules, if such action is reasonably considered to be necessary by the University. Where appropriate, the University will notify and consult with affected students in advance about any changes that are required in line with the University's policy on the Approval of Modifications to Existing Taught Programmes of Study.
## Coronavirus (COVID-19): changes to courses
The 2020/21 academic year will start in September. We aim to deliver as much face-to-face teaching as we can, supported by high quality online alternatives where we must.
Find details of the measures we're planning to protect our community. | 955 | 4,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2020-40 | latest | en | 0.826874 |
https://cadna.lip6.fr/Examples_Dir/Nanet/Accueil.php | 1,701,557,141,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100452.79/warc/CC-MAIN-20231202203800-20231202233800-00194.warc.gz | 186,391,743 | 3,688 | # Computation of an infinite integral using a dynamical extrapolation method
Let us consider the integral with and .
This integral arises in a study of the total electronic energy of crystals using the tight binding approximation [2]. The evaluation of was a problem posed in [4].
is computed using Romberg's integration method with the CADNA library.
A theorem like those given in [1] has been established for the dynamical control of Romberg's method [3]. Combining it with two similar theorems, we can estimate the number of exact significant digits of the result obtained, which are common with the exact value of the infinite integral (including both the truncation error and the round-off error).
To combine efficiency and reliability the same expression of is not used at both bounds of the interval. One worry is to avoid cancellations at the infinite bound of the interval.
Therefore is computed as , ,
where
and , with
Functions and are mathematically equivalent, but not numerically !
, with .
is approximated by an iterate of the sequence .
How to choose the optimal iterate ? Iterations are performed until the difference between two successive iterates is not significant. Using the CADNA library, the stopping criterion is .
Computations stop if and are stochastically equal, i.e. if the transformation of into is only due to round-off errors. Further iterations are useless.
We have shown that if , for all , the exact significant bits of the optimal iterate are those of , up to one.
is approximated by computing integrals on intervals of length .
Each of these integrals is computed using Romberg's method.
The approximation of an integral with Romberg's method is an iterate of a sequence which converges with an exponential speed to the exact value of this integral.
How to choose the optimal iterate ?
Approximations are computed until the difference between two successive iterates is not significant.
We have shown that, for each integral, the approximation obtained has its exact significant bits common to the exact value of the integral, up to one.
with
How to choose the optimal iterate ? Iterations are performed until the difference between two successive iterates is not significant.
Finally we have shown that using the CADNA library and the stopping criteria previously described, we can estimate the number of exact significant digits of the approximation of obtained, which are common with the exact value of the integral.
## References
1
J.-M. Chesneaux, F. Jézéquel, Dynamical control of computations using the trapezoidal and Simpson's rules, Journal of Universal Computer Science, Vol. 4 (1), pp. 2-10, 1998.
2
W. Harrison, Total energies in the tight-binding theory, Phys. Rev., B23 (1981), pp. 5230-5245
3
F. Jézéquel, Dynamical control of computations using approximation methods, 16th IMACS world congress on Scientific Computation, Applied Mathematics and Simulation, Lausanne (Switzerland), August 2000.
4
Problem 95-7, SIAM Review, V 38, N 2, pp.324, 1996.
To get the C source code.
To get the FORTRAN source code. | 670 | 3,089 | {"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-2023-50 | longest | en | 0.932839 |
http://hackage.haskell.org/package/Operads-0.4/docs/src/Math-Operad.html | 1,500,866,622,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424683.39/warc/CC-MAIN-20170724022304-20170724042304-00127.warc.gz | 135,007,894 | 2,251 | ```-- Copyright 2009 Mikael Vejdemo Johansson <mik@stanford.edu>
-- Released under a BSD license
module Math.Operad (module Math.Operad.PPrint,
#if defined USE_MAPOPERAD
module Math.Operad.MapOperad,
#elif defined USE_POLYBAG
module Math.Operad.PolyBag,
#else
module Math.Operad.MapOperad,
#endif
module Math.Operad.OrderedTree,
module Math.Operad.OperadGB,
m12_3,
m13_2,
m1_23,
m2,
m3,
yTree,
lgb,
Tree,
FreeOperad) where
import Math.Operad.OperadGB
import Math.Operad.OrderedTree
import Math.Operad.PPrint
#if defined USE_MAPOPERAD
import Math.Operad.MapOperad
#elif defined USE_POLYBAG
import Math.Operad.PolyBag
#else
import Math.Operad.MapOperad
#endif
type Tree = DecoratedTree Integer
type FreeOperad a = OperadElement a Rational PathLex
-- ** Examples and useful predefined operad elements.
-- | The element m2(m2(1,2),3)
m12_3 :: DecoratedTree Integer
m12_3 = symmetricCompose 1 [1,2,3] (corolla 2 [1,2]) (corolla 2 [1,2])
-- | The element m2(m2(1,3),2)
m13_2 :: DecoratedTree Integer
m13_2 = symmetricCompose 1 [1,3,2] (corolla 2 [1,2]) (corolla 2 [1,2])
-- | The element m2(1,m2(2,3))
m1_23 :: DecoratedTree Integer
m1_23 = symmetricCompose 2 [1,2,3] (corolla 2 [1,2]) (corolla 2 [1,2])
-- | The element m2(1,2)
m2 :: DecoratedTree Integer
m2 = corolla 2 [1,2]
-- | The element m3(1,2,3)
m3 :: DecoratedTree Integer
m3 = corolla 3 [1,2,3]
-- | The element m2(m2(1,2),m2(3,4))
yTree :: DecoratedTree Integer
yTree = nsCompose 1 (nsCompose 2 m2 m2) m2
-- The Lie operad example computation
lo1 :: OperadElement Integer Rational PathLex
lo1 = oet m12_3
lo2 :: OperadElement Integer Rational PathLex
lo2 = oet m13_2
lo3 :: OperadElement Integer Rational PathLex
lo3 = oet m1_23
-- | The list of operad elements consisting of 'm12_3'-'m13_2'-'m1_23'. This generates the
-- ideal of relations for the operad Lie.
lgb :: [OperadElement Integer Rational PathLex]
lgb = [lo1 - lo2 - lo3]
``` | 686 | 1,909 | {"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} | 2.765625 | 3 | CC-MAIN-2017-30 | latest | en | 0.241032 |
https://www.mathworks.com/matlabcentral/cody/problems/163-love-triangles/solutions/218177 | 1,488,293,381,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501174163.72/warc/CC-MAIN-20170219104614-00436-ip-10-171-10-108.ec2.internal.warc.gz | 845,841,220 | 11,751 | Cody
# Problem 163. Love triangles
Solution 218177
Submitted on 17 Mar 2013 by @bmtran
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
%% sides = [1 2 1000]; y_correct = false; assert(isequal(triangle(sides),y_correct))
``` ```
2 Pass
%% sides = [3 4 5]; y_correct = true; assert(isequal(triangle(sides),y_correct))
``` ```
3 Pass
%% sides = [5 5 5]; y_correct = true; assert(isequal(triangle(sides),y_correct))
``` ```
4 Pass
%% sides = [6 6 6]; y_correct = true; assert(isequal(triangle(sides),y_correct))
``` ```
5 Pass
%% sides = [1 1 1]; y_correct = true; assert(isequal(triangle(sides),y_correct))
``` ```
6 Pass
%% sides = [1 2 2]; y_correct = true; assert(isequal(triangle(sides),y_correct))
``` ```
7 Pass
%% sides = [2 2 5]; y_correct = false; assert(isequal(triangle(sides),y_correct))
``` ```
8 Pass
%% sides = [5 2 2]; y_correct = false; assert(isequal(triangle(sides),y_correct))
``` ```
9 Pass
%% sides = [1 3 1]; y_correct = false; assert(isequal(triangle(sides),y_correct))
``` ``` | 370 | 1,154 | {"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-2017-09 | longest | en | 0.505037 |
https://stats.stackexchange.com/questions/40343/calculating-the-probability-of-the-nth-sample-in-the-birthday-problem-w-o-knowin | 1,709,249,050,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474853.43/warc/CC-MAIN-20240229202522-20240229232522-00737.warc.gz | 528,149,926 | 40,429 | # Calculating the probability of the Nth sample in the birthday problem w/o knowing the previous samples
If we draw two random variates from a discrete uniform distribution $[1, D]$, the probability that the samples are distinct is $(D-1)/D$. Explanations of the birthday problem state that if we sample a third time, the probability that this third sample is distinct from the first two is $(D-1)(D-2)/D^2$. Doesn't this assume that the first two samples were distinct? If they are not distinct, then the probability that the third sample is distinct is $(D-1)/D$. The solution to the birthday problem requires us to know the probability that the $n^\textrm{th}$ random variate is distinct from the previous samples. How can we calculate the probability that the $n^\textrm{th}$ sample is distinct from the previous samples without knowing the values of these previous samples?
• Welcome to the site, @Dan. Is this homework? If it is, you need to add the 'homework' tag. Oct 13, 2012 at 21:11
• @gung This is not homework, it is for my own edification. I saw the problem in David Mackay's textbook on information theory and was not satisfied with his explanation (or other similar explanations on the net).
– Dan
Oct 13, 2012 at 21:13
In the usual context of the birthday problem, if the first two values are not distinct the problem is solved and the third and subsequent values are irrelevant. What is confusing here is that you have written
Explanations of the birthday problem state that if we sample a third time, the probability that this third sample is distinct from the first two
What is really meant here (and is a big part of the unintuitive nature of the birthday problem), is "the probability that this third sample is distinct from the first two, given that they are distinct from each other".
The probability all $3$ are distinct is $\frac{(D-1)(D-2)}{D^2}$.
The probability the third is not equal to the first two is $\frac{(D-1)^2}{D^2}$. Imagine choosing $C$ first. Then there are $D-1$ choices for $A$ so that $A\ne C$, and $D-1$ choices for $B$ so that $B\ne C$.
If you are given that the first two are distinct, the conditional probability that the third is different from both is $\frac{D-2}{D}$.
You are correct that if you are given the first two are equal, the conditional probability that the third is different is $\frac{D-1}{D}$.
If $D=10$, these are $72\%, 81\%, 80\%,$ and $90\%,$ respectively. The $81\%$ is naturally a weighted average between $80\%$ and $90\%$, weighted by the probabilities that the first two are distinct or not. The $72\%$ can also be viewed as a weighted average between $81\%$ and $0$, weighted by the conditional probability that the first two are distinct from each other given that they are distinct from the third.
• Because this post has been upvoted, SE construes its deletion as removing potentially useful information; I have therefore undeleted it.
– whuber
Mar 30, 2013 at 15:05 | 739 | 2,955 | {"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} | 4.25 | 4 | CC-MAIN-2024-10 | longest | en | 0.949863 |
qipacabynujatoke.rutadeltambor.com | 1,580,272,380,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251788528.85/warc/CC-MAIN-20200129041149-20200129071149-00362.warc.gz | 600,410,926 | 5,233 | 7 write about self-dual polyhedra patterns
For some V, namely precisely the finite-dimensional vector spacesthis map is an isomorphism. However, this is in a sense a lucky coincidence, for giving such an isomorphism requires a certain choice, for example the choice of a basis of V. This is also true in the case if V is a Hilbert spacevia the Riesz representation theorem.
The dodecahedron and icosahedron are the most exotic of the Platonic solids, because they have 5-fold rotational symmetry - a possibility that only exists for regular polytopes in 2, 3 or 4 dimensions.
But the icosahedron was probably discovered later.
Who Discovered the Icosahedron?
As Benno Artmann wrote: The original knowledge of the dodecahedron may have come from crystals of pyrite, but in contrast the icosahedron is a pure mathematical creation It is the first realization of an entity that existed before only in abstract thought. Well, apart from the statues of gods!
Other exceptional objects include the simple Lie group E8, and the finite simple group M Intriguingly, many of these exceptional objects" are related. For example, the icosahedron can be used to construct both E8 and M But the first interesting classification theorem was the classification of regular polyhedra: It shows that the only possibilities are the Platonic 7 write about self-dual polyhedra patterns And according to traditional wisdom, the results in this book were proved by Theatetuswho also discovered the icosahedron!
In this book, the 13th, are constructed the five so-called Platonic figures which, however, do not belong to Plato, three of the five being due to the Pythagoreans, namely the cube, the pyramid, and the dodecahedron, while the octahedron and the icosahedron are due to Theaetetus.
In the Republic, written around BC, Plato complained that not enough is known about solid geometry: In the second place, students cannot learn it unless they have a teacher. But then a teacher can hardly be found Theaetetus seems to have filled the gap: He died from battle wounds and dysentery in after Athens fought a battle with Corinth.
But how certain are we that Theatetus discovered - or at least studied - the icosahedron? The only hard evidence seems to be this "ancient note" in the margins of the Elements. But who wrote it, and when? First of all, if you hope to see an ancient manuscript by Euclid with a scribbled note in the margin, prepare to be disappointed!
All we have are copies of copies of copies.
Polygons and polyhedra
Comparing these to guess the contents of the original Elements is a difficult and fascinating task. It seems he wanted to standardize the language and make it easier to follow. This may have helped people trying to learn geometry - but certainly not scholars trying to understand Euclid.
InFrancois Peyrard made a marvelous discovery. This copy is now called "P". It dates back to about AD. I would love to know how Peyrard got his hands on it. One imagines him rooting around in a dusty basement and opening a trunk The all-important English translation by Thomas Heath is based on this.
As far as I can tell, "P" is the only known non-Theonine copy of Euclid except for the fragments I mentioned.
Honeycomb (geometry) - Wikipedia
Heath also used these fragments to prepare his translation. This is just a quick overview of a complicated detective story. As always, the fractal texture of history reveals more complexity the more closely you look.
Anyway, Heath thinks that Geminus of Rhodes wrote the "ancient note" in the Elements crediting Theatetus. In his charming article "The discovery of the regular solids", William Waterhouse writes: Once upon a time there was no problem in the history of the regular solids.
According to Proclus, the discoveries of Pythagoras include "the construction of the cosmic solids," and early historians could only assume that the subject sprang full-grown from his head. But a better-developed picture of the growth of Greek geometry made such an early date seem questionable, and evidence was uncovered suggesting a different attribution.
A thorough study of the testimony was made by E. Sachs, and her conclusion is now generally accepted: The history of the regular solids thus rests almost entirely on a scholium to Euclid which reads as follows: Three of these 5 figures, the cube, pyramid, and dodecahedron, belong to the Pythagoreans; while the octahedron and icosahedron belong to Theaetetus.icosahedron.
The tetrahedron is considered a self-dual. Projection of a pattern can also occur inwards from the surface of a patterned polyhedron onto its inscribed dual.
Figure 6 shows a cube, tiled with a class p4-derived pattern, with an octahedron inscribed within it. paper, however, we are concerned with the second case, where the K3 polytope is self-dual and corresponds to E 8 f 1g(for the classi cation of elliptic-K3 polyhedra by Lie groups.
The next most regular convex polyhedra after the Platonic solids are the cuboctahedron, which is a rectification of the cube and the octahedron, and the icosidodecahedron, which is a rectification of the dodecahedron and the icosahedron (the rectification of the self-dual tetrahedron is a regular octahedron).
Euler's polyhedron formula. By. Abigail Kirk. Submitted by plusadmin on June 1, June She has really enjoyed exploring the mysteries of Euler's formula when writing this article.
List of isotoxal polyhedra and tilings. In geometry, (The self-dual square tiling recreates itself in all four forms.) Regular Dual regular Quasiregular Quasiregular dual Tilings and Patterns. New York: W. H. Freeman. Paper Models of Polyhedra. Polyhedra are beautiful 3-D geometrical figures that have fascinated philosophers, mathematicians and artists for millennia. On this site are a few hundred paper models available for free. Make the models yourself. Click on a picture to go to a page with a net of the model. Printable Shapes: Alphabetical list of geometric shapes, nets, patterns and coloring pages to print, cut and fold. Can use to create gift box template for crafts. Find this Pin and .
Add new comment; Comments. excellent article. Permalink Submitted by Anonymous on October 19, it is really a breif but very usefull article. Polyhedra Patterns Crystals and their structure (crystallography) involve the study of polyhedra.
For examples and photos of Study each different face of the polyhedra and complete table 2. Write a generalization for the relationship among the . A classical example of this is the duality of the platonic solids, in which the cube and the octahedron form a dual pair, the dodecahedron and the icosahedron form a dual pair, and the tetrahedron is self-dual.
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