url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://archive.lib.msu.edu/crcmath/math/math/s/s129.htm | 1,638,844,809,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363332.1/warc/CC-MAIN-20211207014802-20211207044802-00036.warc.gz | 180,024,360 | 2,887 | ## Second Derivative Test
Suppose is a Function of which is twice Differentiable at a Stationary Point .
1. If , then has a Relative Minimum at .
2. If , then has a Relative Maximum at .
The Extremum Test gives slightly more general conditions under which a function with is a maximum or minimum.
If is a 2-D Function which has a Relative Extremum at a point and has Continuous Partial Derivatives at this point, then and . The second Partial Derivatives test classifies the point as a Maximum or Minimum. Define the Discriminant as
1. If , and , the point is a Relative Minimum.
2. If , , and , the point is a Relative Maximum.
3. If , the point is a Saddle Point.
4. If , higher order tests must be used.
See also Discriminant (Second Derivative Test), Extremum, Extremum Test, First Derivative Test, Global Maximum, Global Minimum, Hessian Determinant, Maximum, Minimum, Relative Maximum, Relative Minimum, Saddle Point (Function)
References
Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 14, 1972. | 276 | 1,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-49 | latest | en | 0.841775 |
https://math.answers.com/Q/How_many_months_are_in_43_years | 1,716,815,793,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00139.warc.gz | 337,657,645 | 47,785 | 0
# How many months are in 43 years?
Updated: 9/16/2023
Wiki User
13y ago
There are exactly three years and seven months.
Wiki User
6y ago
Wiki User
13y ago
There are 12 months in one year. Therefore, 43 years is equal to 43 x 12 = 516 months. | 80 | 253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-22 | latest | en | 0.940845 |
https://kr.mathworks.com/matlabcentral/answers/1756700-reversing-the-digits-into-a-string | 1,679,509,054,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.74/warc/CC-MAIN-20230322180852-20230322210852-00004.warc.gz | 397,751,346 | 27,830 | # reversing the digits into a string
조회 수: 3(최근 30일)
cgo 2022년 7월 10일
댓글: John D'Errico 2022년 7월 10일
I have a list of numbers.
l1 = randi([0,9],[1,10]);
I want to make this into a string without the spaces and then reverse the digits.
댓글을 달려면 로그인하십시오.
### 답변(1개)
Kshittiz Bhardwaj 2022년 7월 10일
Hello cgo,
I understand that you want to convert your list into a string and then reverse it.
S = l1(:)';
This will convert your 2*2 array into a 1*4 array
str = char( S ) converts the input list to a char array
rev = reverse( str ) reverses the order of the characters in str .
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
John D'Errico 2022년 7월 10일
Sorry, but this part is wrong:
S = l1(:)';
since that makes it into a coumn vector. And there is no reason to do so. l1 is a ROW VECTOR ALREADY. It is not a 2x2 array of numbers. You seem to have misunderstood what randi does.
l1 = randi([0,9],[1,10])
l1 = 1×10
5 6 0 6 3 3 9 7 9 1
Now, how do you convert those digits into characters? The use of char is also wrong, but subtly so. You need to convert them into their ascii representation as digits first, if you would use char. Adding '0' does that.
S = char(l1 + '0')
S = '5606339791'
Finally, you can use either flip or reverse to do the order swapping operation.
reverse(S)
ans = '1979336065'
flip(S)
ans = '1979336065'
댓글을 달려면 로그인하십시오.
### 범주
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Translated by | 505 | 1,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-14 | longest | en | 0.756902 |
https://mindcoding.ro/pb/triples | 1,708,873,852,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474617.27/warc/CC-MAIN-20240225135334-20240225165334-00884.warc.gz | 394,191,209 | 4,207 | Triples
Let m and n be two positive integers, 5 ≤ m ≤ 100, 2 ≤ n ≤ 100. Consider the following sets of triples:
Tm,j = {(x, y, z) ∈ ℕ3 | x ≤ y ≤ z ≤ m and xj + yj = zj}, j = 2…n
The problem asks you to compute the sum:
Sm,n = ∑ card(Tm,j), 2 ≤ j ≤ n.
Notes:
• ℕ is the set of non-negative integers (ℕ = {0, 1, 2, …}).
• card(Tm,j) is the number of elements of the set Tm,j.
Input
The input contains several test cases. Each test case occupies exactly two lines of input: the first line contains integer m, and the second line contains integer n. Tests should be read until the end of file is reached.
Output
For each test case, the result will be written to standard output, on a line by itself.
InputOutput
85
95
8128
Questions? | 226 | 740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-10 | latest | en | 0.871266 |
https://www.wiziq.com/tutorials/set-theory | 1,638,463,243,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362230.18/warc/CC-MAIN-20211202145130-20211202175130-00157.warc.gz | 1,127,531,792 | 22,491 | Free Online Set Theory Tutorials
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PROBLEMS in 2008-09 | 289 | 1,096 | {"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-2021-49 | latest | en | 0.803227 |
https://discuss.leetcode.com/topic/104436/time-complexity | 1,513,123,408,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948520042.35/warc/CC-MAIN-20171212231544-20171213011544-00474.warc.gz | 550,246,183 | 8,108 | # Time Complexity!
• Firstly the recursion tree:
n options on 1st layer, n - 1 options on 2nd layer ...
So, the recursion tree has (n!) nodes.
Then, in every recursion node:
we did a traverse on original array to check whether we can take the element or not.
This takes O(n)
And, every time we check an element:
We uses List.contains(), this method cost O(n) too.
So, totally the time complexity should be:
O(n * n * n!) = O(n^3)
If we use a boolean array `boolean[] used`, we can avoid using List.contains() and make this step to O(1), so the total time will be:
O(n * n!) = O(n^2)
Are these correct??? | 164 | 609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-51 | latest | en | 0.766211 |
http://mathhelpforum.com/algebra/206379-simplifying-fractions-exponents.html | 1,519,057,309,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812756.57/warc/CC-MAIN-20180219151705-20180219171705-00089.warc.gz | 232,821,707 | 10,060 | # Thread: Simplifying Fractions with Exponents
1. ## Simplifying Fractions with Exponents
I understand the process of simplifying denominators but im getting stuck with questions like this. I know im trying to get to 4(x)^5/2 but how do i get there. Appreciate the help.
2. ## Re: Simplifying Fractions with Exponents
$\frac{-1}{4x^{\frac{3}{2}}}+\frac{6}{x^{\frac{5}{2}}}$
You have correctly identified the common denominator we want, although this is called the lowest common denominator:
$4x^{\frac{5}{2}}$
Now, we want to multiply both fractions by an expression equal to 1 which makes both terms have this same denominator. If you are unsure what these expressions are, take your lowest common denominator, and divide it by the existing denominators, and the result over itself is the expression we need.
For the first term, this is $\frac{4x^{\frac{5}{2}}}{4x^{\frac{3}{2}}}=x$, so we multiply the first term by $\frac{x}{x}$.
For the second term, this is $\frac{4x^{\frac{5}{2}}}{x^{\frac{5}{2}}}=4$, so we multiply the second term by $\frac{4}{4}$.
Thus, we have:
$\frac{-1}{4x^{\frac{3}{2}}}\cdot\frac{x}{x}+\frac{6}{x^{ \frac{5}{2}}}\cdot\frac{4}{4}=\frac{-x}{4x^{\frac{5}{2}}}+\frac{24}{4x^{\frac{5}{2}}}$
Now, you may combine the terms. | 388 | 1,260 | {"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": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2018-09 | latest | en | 0.839595 |
https://www.sitepoint.com/community/t/mathematical-string-parsing/55271 | 1,723,055,081,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640707024.37/warc/CC-MAIN-20240807174317-20240807204317-00551.warc.gz | 751,433,725 | 6,866 | # Mathematical String Parsing!
Hi all.
I’ve written an application which creates graphs where the values of Y and X are functions of a third variable T. For example, if X = t^2 + 2t + 4, and Y = log2t, a graph like the following is produced:
I have gotten that far.
However, to create such a graph, the following code is used:
``````<?php
include 'Graph.php';
\$Graph = new Graph('t^2 + 2t + 4 against log(2t)', 250, 500);
\$Graph->Axes->X = new Axis("t^2 + 2t + 4", -6, 5);
\$Graph->Axes->Y = new Axis("log2t", -6, 50);
\$Graph->LineThickness = 3;
\$Graph->Inputs = range(-100, 10, 0.01);
\$Graph->MajorGrid = new Grid( new Box(6, 6), Color::Black() );
\$Graph->MinorGrid = new Grid( new Box(0.5, 0.5), Color::Grey() );
\$Graph->addFormula( new Formula( 'x', 'y', Color::Red() ) );
function x(\$t){
return pow(\$t, 2) + 2 * \$t + 4;
}
function y(\$t){
return log(2 * \$t);
}
imagejpeg(\$Graph->Run()->Resource, null, 100);
``````
What I want to do is make this graph application accessible to a few teachers, who would want to enter the X and Y formulae via a form, as a mathematical input.
``````
x = pow(\$t, 2) + 2 * \$t + 4
y = log(2 * t)
``````
They’d want to enter:
``````
x = t^2 + 2t + 4
y = log2t
``````
So I’m interested in looking at the various ways that I can parse the mathematical input into commands that PHP can understand. I am going to be trying a few ideas in the mean time, however I thought I’d ask here to see what you guys could come up with! The theory behind it is what I’m interested in, so code itself isn’t required - I’m just after ideas.
Example Ins/Outs:
``````t^2 + log(t) => pow(\$t, 2) + log(\$t)
(t + sint)^3 => pow( (\$t + sin(\$t) ), 3)
2^t + tan^2(t) => pow( 2, \$t ) + pow( tan( \$t ), 2 )
``````
Take a look at the eval() page…
Eval parses PHP code. I’m looking for a way to parse mathematical language into instructions.
http://www.phpclasses.org/browse/file/11680.html | 623 | 1,929 | {"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-2024-33 | latest | en | 0.830446 |
http://en.wikipedia.org/wiki/Bellard's_formula | 1,432,477,791,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928019.31/warc/CC-MAIN-20150521113208-00071-ip-10-180-206-219.ec2.internal.warc.gz | 80,805,202 | 8,450 | # Bellard's formula
\begin{align} \pi = \frac1{2^6} \sum_{n=0}^\infty \frac{(-1)^n}{2^{10n}} \, \left(-\frac{2^5}{4n+1} \right. & {} - \frac1{4n+3} + \frac{2^8}{10n+1} - \frac{2^6}{10n+3} \left. {} - \frac{2^2}{10n+5} - \frac{2^2}{10n+7} + \frac1{10n+9} \right) \end{align} | 152 | 274 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2015-22 | latest | en | 0.365899 |
https://www.geeksforgeeks.org/count-quadruplets-with-sum-k-from-given-array/?ref=lbp | 1,611,369,309,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703531702.36/warc/CC-MAIN-20210123001629-20210123031629-00359.warc.gz | 797,997,610 | 33,558 | Related Articles
Count quadruplets with sum K from given array
• Last Updated : 04 Dec, 2020
Given an array arr[] of size N and an integer S, the task is to find the count of quadruplets present in the given array having sum S
Examples:
Input: arr[] = {1, 5, 3, 1, 2, 10}, S = 20
Output: 1
Explanation: Only quadruplet satisfying the conditions is arr[1] + arr[2] + arr[4] + arr[5] = 5 + 3 + 2 + 10 = 20.
Input: N = 6, S = 13, arr[] = {4, 5, 3, 1, 2, 4}
Output: 3
Explanation: Three quadruplets with sum 13 are:
1. arr[0] + arr[2] + arr[4] + arr[5] = 4 + 3 + 2 + 4 = 13
2. arr[0] + arr[1] + arr[2] + arr[3] = 4 + 5 + 3 + 1 = 13
3. arr[1] + arr[2] + arr[3] + arr[5] = 5 + 3 + 1 + 4 = 13
Naive Approach: The idea is to generate all possible combinations of length 4 from the given array. For each quadruplet, if the sum equals S, then increment the counter by 1. After checking all the quadruplets, print the counter as the total number of quadruplets having sum S.
Below is the implementation of the above approach:
## C++
`// C++ program for the above approach` `#include ` `using` `namespace` `std;` `// Function to return the number of` `// quadruplets with the given sum` `int` `countSum(``int` `a[], ``int` `n, ``int` `sum)` `{` ` ``// Initialize variables` ` ``int` `i, j, k, l;` ` ``// Initialize answer` ` ``int` `count = 0;` ` ``// All possible first elements` ` ``for` `(i = 0; i < n - 3; i++) {` ` ``// All possible second elements` ` ``for` `(j = i + 1; j < n - 2; j++) {` ` ``// All possible third elements` ` ``for` `(k = j + 1; k < n - 1; k++) {` ` ``// All possible fourth elements` ` ``for` `(l = k + 1; l < n; l++) {` ` ``// Increment counter by 1` ` ``// if quadruplet sum is S` ` ``if` `(a[i] + a[j]` ` ``+ a[k] + a[l]` ` ``== sum)` ` ``count++;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// Return the final count` ` ``return` `count;` `}` `// Driver Code` `int` `main()` `{` ` ``// Given array arr[]` ` ``int` `arr[] = { 4, 5, 3, 1, 2, 4 };` ` ``// Given sum S` ` ``int` `S = 13;` ` ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``// Function Call` ` ``cout << countSum(arr, N, S);` ` ``return` `0;` `}`
## Java
`// Java program for the above approach` `import` `java.util.*;` `class` `GFG{` ` ` `// Function to return the number of` `// quadruplets with the given sum` `static` `int` `countSum(``int` `a[], ``int` `n, ``int` `sum)` `{` ` ` ` ``// Initialize variables` ` ``int` `i, j, k, l;` ` ``// Initialize answer` ` ``int` `count = ``0``;` ` ``// All possible first elements` ` ``for``(i = ``0``; i < n - ``3``; i++) ` ` ``{` ` ` ` ``// All possible second elements` ` ``for``(j = i + ``1``; j < n - ``2``; j++) ` ` ``{` ` ` ` ``// All possible third elements` ` ``for``(k = j + ``1``; k < n - ``1``; k++)` ` ``{` ` ` ` ``// All possible fourth elements` ` ``for``(l = k + ``1``; l < n; l++) ` ` ``{` ` ` ` ``// Increment counter by 1` ` ``// if quadruplet sum is S` ` ``if` `(a[i] + a[j] + ` ` ``a[k] + a[l] == sum)` ` ``count++;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// Return the final count` ` ``return` `count;` `}` `// Driver Code` `public` `static` `void` `main(String args[])` `{` ` ` ` ``// Given array arr[]` ` ``int` `arr[] = { ``4``, ``5``, ``3``, ``1``, ``2``, ``4` `};` ` ``// Given sum S` ` ``int` `S = ``13``;` ` ``int` `N = arr.length;` ` ``// Function Call` ` ``System.out.print(countSum(arr, N, S));` `}` `}` `// This code is contributed by bgangwar59`
## Python3
`# Python3 program for the above approach` `# Function to return the number of` `# quadruplets with the given sum` `def` `countSum(a, n, ``sum``):` ` ` ` ``# Initialize variables` ` ``# i, j, k, l` ` ``# Initialize answer` ` ``count ``=` `0` ` ``# All possible first elements` ` ``for` `i ``in` `range``(n ``-` `3``):` ` ` ` ``# All possible second elements` ` ``for` `j ``in` `range``(i ``+` `1``, n ``-` `2``):` ` ` ` ``# All possible third elements` ` ``for` `k ``in` `range``(j ``+` `1``, n ``-` `1``):` ` ` ` ``# All possible fourth elements` ` ``for` `l ``in` `range``(k ``+` `1``, n):` ` ` ` ``# Increment counter by 1` ` ``# if quadruplet sum is S` ` ``if` `(a[i] ``+` `a[j] ``+` `a[k] ``+` `a[l]``=``=` `sum``):` ` ``count ``+``=` `1` ` ` ` ``# Return the final count` ` ``return` `count` `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` ` ` ` ``# Given array arr[]` ` ``arr ``=` `[ ``4``, ``5``, ``3``, ``1``, ``2``, ``4` `]` ` ``# Given sum S` ` ``S ``=` `13` ` ``N ``=` `len``(arr)` ` ` ` ``# Function Call` ` ``print``(countSum(arr, N, S))` ` ` `# This code is contributed by mohit kumar 29`
## C#
`// C# program for the above approach` `class` `GFG{` ` ` `// Function to return the number of` `// quadruplets with the given sum` `static` `int` `countSum(``int` `[]a, ``int` `n, ``int` `sum)` `{` ` ` ` ``// Initialize variables` ` ``int` `i, j, k, l;` ` ``// Initialize answer` ` ``int` `count = 0;` ` ``// All possible first elements` ` ``for``(i = 0; i < n - 3; i++) ` ` ``{` ` ` ` ``// All possible second elements` ` ``for``(j = i + 1; j < n - 2; j++) ` ` ``{` ` ` ` ``// All possible third elements` ` ``for``(k = j + 1; k < n - 1; k++)` ` ``{` ` ` ` ``// All possible fourth elements` ` ``for``(l = k + 1; l < n; l++) ` ` ``{` ` ` ` ``// Increment counter by 1` ` ``// if quadruplet sum is S` ` ``if` `(a[i] + a[j] + ` ` ``a[k] + a[l] == sum)` ` ``count++;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// Return the final count` ` ``return` `count;` `}` `// Driver Code` `public` `static` `void` `Main()` `{` ` ` ` ``// Given array arr[]` ` ``int` `[]arr = { 4, 5, 3, 1, 2, 4 };` ` ``// Given sum S` ` ``int` `S = 13;` ` ``int` `N = arr.Length;` ` ``// Function Call` ` ``System.Console.Write(countSum(arr, N, S));` `}` `}` `// This code is contributed by SURENDRA_GANGWAR`
Output:
`3`
Time Complexity: O(N4)
Auxiliary Space: O(N)
Better Approach: To optimize the above approach, the idea is to use a Map data structure. Follow the steps below to solve the problem:
• Initialize the counter count with 0 to store the number of quadruplets.
• Traverse the given array over the range [0, N – 3)using the variable i. For each element arr[i], traverse the array again over the range [i + 1, N – 2) using the variable j and do the following:
• Find the value of the required sum(say req) as (S – arr[i] – arr[j]).
• Initialize the count_twice with 0 that will store the count of ordered pairs in the above subarray with sum (S – arr[i] – arr[j]).
• After finding the count_twice, update the count by count_twice / 2.
• After the above steps, print the value of count as the result.
Below is the implementation of the above idea:
## C++
`// C++ program for the above approach` `#include ` `#include ` `using` `namespace` `std;` `// Function to return the number of` `// quadruplets having given sum` `int` `countSum(``int` `a[], ``int` `n, ``int` `sum)` `{` ` ``// Initialize variables` ` ``int` `i, j, k, l;` ` ``// Initialize answer` ` ``int` `count = 0;` ` ``// All possible first elements` ` ``for` `(i = 0; i < n - 3; i++) {` ` ``// All possible second element` ` ``for` `(j = i + 1; j < n - 2; j++) {` ` ``int` `req = sum - a[i] - a[j];` ` ``// Use map to find the` ` ``// fourth element` ` ``unordered_map<``int``, ``int``> m;` ` ``// All possible third elements` ` ``for` `(k = j + 1; k < n; k++)` ` ``m[a[k]]++;` ` ``int` `twice_count = 0;` ` ``// Calculate number of valid` ` ``// 4th elements` ` ``for` `(k = j + 1; k < n; k++) {` ` ``// Update the twice_count` ` ``twice_count += m[req - a[k]];` ` ``if` `(req - a[k] == a[k])` ` ``twice_count--;` ` ``}` ` ``// Unordered pairs` ` ``count += twice_count / 2;` ` ``}` ` ``}` ` ``// Return answer` ` ``return` `count;` `}` `// Driver Code` `int` `main()` `{` ` ``// Given array arr[]` ` ``int` `arr[] = { 4, 5, 3, 1, 2, 4 };` ` ``// Given sum S` ` ``int` `S = 13;` ` ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``// Function Call` ` ``cout << countSum(arr, N, S);` ` ``return` `0;` `}`
## Java
`// Java program for the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to return the number of` `// quadruplets having given sum` `static` `int` `countSum(``int` `a[], ``int` `n, ``int` `sum)` `{` ` ` ` ``// Initialize variables` ` ``int` `i, j, k, l;` ` ``// Initialize answer` ` ``int` `count = ``0``;` ` ``// All possible first elements` ` ``for``(i = ``0``; i < n - ``3``; i++)` ` ``{` ` ` ` ``// All possible second element` ` ``for``(j = i + ``1``; j < n - ``2``; j++) ` ` ``{` ` ``int` `req = sum - a[i] - a[j];` ` ``// Use map to find the` ` ``// fourth element` ` ``HashMap m = ``new` `HashMap<>();` ` ``// All possible third elements` ` ``for``(k = j + ``1``; k < n; k++)` ` ``if` `(m.containsKey(a[k]))` ` ``{` ` ``m.put(a[k], m.get(a[k]) + ``1``);` ` ``}` ` ``else` ` ``{` ` ``m.put(a[k], ``1``);` ` ``}` ` ` ` ``int` `twice_count = ``0``;` ` ``// Calculate number of valid` ` ``// 4th elements` ` ``for``(k = j + ``1``; k < n; k++) ` ` ``{` ` ` ` ``// Update the twice_count` ` ``if` `(m.containsKey(req - a[k]))` ` ``twice_count += m.get(req - a[k]);` ` ``if` `(req - a[k] == a[k])` ` ``twice_count--;` ` ``}` ` ``// Unordered pairs` ` ``count += twice_count / ``2``;` ` ``}` ` ``}` ` ``// Return answer` ` ``return` `count;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` ` ``// Given array arr[]` ` ``int` `arr[] = { ``4``, ``5``, ``3``, ``1``, ``2``, ``4` `};` ` ``// Given sum S` ` ``int` `S = ``13``;` ` ``int` `N = arr.length;` ` ``// Function Call` ` ``System.out.print(countSum(arr, N, S));` `}` `}` `// This code is contributed by Princi Singh`
## Python3
`# Python3 program for the above approach` `# Function to return the number of` `# quadruplets having given sum` `def` `countSum(a, n, ``sum``):` ` ` ` ``# Initialize variables` ` ``# Initialize answer` ` ``count ``=` `0` ` ``# All possible first elements` ` ``for` `i ``in` `range``(n ``-` `3``):` ` ` ` ``# All possible second element` ` ``for` `j ``in` `range``(i ``+` `1``, n ``-` `2``, ``1``):` ` ``req ``=` `sum` `-` `a[i] ``-` `a[j]` ` ``# Use map to find the` ` ``# fourth element` ` ``m ``=` `{}` ` ``# All possible third elements` ` ``for` `k ``in` `range``(j ``+` `1``, n, ``1``):` ` ``m[a[k]] ``=` `m.get(a[k], ``0``) ``+` `1` ` ``twice_count ``=` `0` ` ``# Calculate number of valid` ` ``# 4th elements` ` ``for` `k ``in` `range``(j ``+` `1``, n, ``1``):` ` ` ` ``# Update the twice_count` ` ``twice_count ``+``=` `m.get(req ``-` `a[k], ``0``)` ` ``if` `(req ``-` `a[k] ``=``=` `a[k]):` ` ``twice_count ``-``=` `1` ` ``# Unordered pairs` ` ``count ``+``=` `twice_count ``/``/` `2` ` ``# Return answer` ` ``return` `count` `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` ` ` ` ``# Given array arr[]` ` ``arr ``=` `[ ``4``, ``5``, ``3``, ``1``, ``2``, ``4` `]` ` ``# Given sum S` ` ``S ``=` `13` ` ``N ``=` `len``(arr)` ` ``# Function Call` ` ``print``(countSum(arr, N, S))` `# This code is contributed by ipg2016107`
## C#
`// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{` `// Function to return the number of` `// quadruplets having given sum` `static` `int` `countSum(``int` `[]a, ``int` `n, ` ` ``int` `sum)` `{` ` ` ` ``// Initialize variables` ` ``int` `i, j, k; ` ` ``//int l;` ` ` ` ``// Initialize answer` ` ``int` `count = 0;` ` ``// All possible first elements` ` ``for``(i = 0; i < n - 3; i++)` ` ``{` ` ` ` ``// All possible second element` ` ``for``(j = i + 1; j < n - 2; j++) ` ` ``{` ` ``int` `req = sum - a[i] - a[j];` ` ``// Use map to find the` ` ``// fourth element` ` ``Dictionary<``int``,` ` ``int``> m = ``new` `Dictionary<``int``,` ` ``int``>();` ` ``// All possible third elements` ` ``for``(k = j + 1; k < n; k++)` ` ``if` `(m.ContainsKey(a[k]))` ` ``{` ` ``m[a[k]]++;` ` ``}` ` ``else` ` ``{` ` ``m.Add(a[k], 1);` ` ``}` ` ` ` ``int` `twice_count = 0;` ` ``// Calculate number of valid` ` ``// 4th elements` ` ``for``(k = j + 1; k < n; k++) ` ` ``{` ` ` ` ``// Update the twice_count` ` ``if` `(m.ContainsKey(req - a[k]))` ` ``twice_count += m[req - a[k]];` ` ``if` `(req - a[k] == a[k])` ` ``twice_count--;` ` ``}` ` ``// Unordered pairs` ` ``count += twice_count / 2;` ` ``}` ` ``}` ` ``// Return answer` ` ``return` `count;` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` ` ``// Given array []arr` ` ``int` `[]arr = { 4, 5, 3, 1, 2, 4 };` ` ``// Given sum S` ` ``int` `S = 13;` ` ``int` `N = arr.Length;` ` ``// Function Call` ` ``Console.Write(countSum(arr, N, S));` `}` `}` `// This code is contributed by Princi Singh`
Output:
`3`
Time complexity: O(N3) where N is the size of the given array,
Auxiliary Space: O(N)
Efficient Approach: The idea is similar to the above approach using a map. In this approach, fix the 3rd element, then find and store the frequency of sums of all possible first two elements of any quadruplet of the given array. Follow the below steps to solve the problem:
1. Initialize the counter count to store the total quadruplets with the given sum S and a map to store all possible sums for the first two elements of each possible quadruplet.
2. Traverse the given array over the range [0, N – 1] using the variable i where arr[i] is the fixed 3rd element.
3. Then for each element arr[i] in the above step, traverse the given array over the range [i + 1, N – 1] using the variable j and increment the counter count by map[arr[i] + arr[j]].
4. After traversing from the above loop, for each element arr[i], traverse the array arr[] from j = 0 to i – 1 and increment the frequency of any sum arr[i] + arr[j] of the first two elements of any possible quadruplet by 1 i.e., increment map[arr[i] + arr[j]] by 1.
5. Repeat the above steps for each element arr[i] and then print the counter count as the total number of quadruplets with the given sum S.
Below is the implementation of the above idea:
## C++
`// C++ program for the above approach` `#include ` `#include ` `using` `namespace` `std;` `// Function to return the number of` `// quadruplets having the given sum` `int` `countSum(``int` `a[], ``int` `n, ``int` `sum)` `{` ` ``// Initialize variables` ` ``int` `i, j, k;` ` ``// Initialize answer` ` ``int` `count = 0;` ` ``// Store the frequency of sum` ` ``// of first two elements` ` ``unordered_map<``int``, ``int``> m;` ` ``// Traverse from 0 to N-1, where` ` ``// arr[i] is the 3rd element` ` ``for` `(i = 0; i < n - 1; i++) {` ` ``// All possible 4th elements` ` ``for` `(j = i + 1; j < n; j++) {` ` ``// Sum of last two element` ` ``int` `temp = a[i] + a[j];` ` ``// Frequency of sum of first` ` ``// two elements` ` ``if` `(temp < sum)` ` ``count += m[sum - temp];` ` ``}` ` ``for` `(j = 0; j < i; j++) {` ` ``// Store frequency of all possible` ` ``// sums of first two elements` ` ``int` `temp = a[i] + a[j];` ` ``if` `(temp < sum)` ` ``m[temp]++;` ` ``}` ` ``}` ` ``// Return the answer` ` ``return` `count;` `}` `// Driver Code` `int` `main()` `{` ` ``// Given array arr[]` ` ``int` `arr[] = { 4, 5, 3, 1, 2, 4 };` ` ``// Given sum S` ` ``int` `S = 13;` ` ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``// Function Call` ` ``cout << countSum(arr, N, S);` ` ``return` `0;` `}`
## Java
`// Java program for the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to return the number of` `// quadruplets having the given sum` `static` `int` `countSum(``int` `a[], ``int` `n, ``int` `sum)` `{` ` ` ` ``// Initialize variables` ` ``int` `i, j, k;` ` ``// Initialize answer` ` ``int` `count = ``0``;` ` ``// Store the frequency of sum` ` ``// of first two elements` ` ``HashMap m = ``new` `HashMap<>();` ` ``// Traverse from 0 to N-1, where` ` ``// arr[i] is the 3rd element` ` ``for``(i = ``0``; i < n - ``1``; i++)` ` ``{` ` ` ` ``// All possible 4th elements` ` ``for``(j = i + ``1``; j < n; j++) ` ` ``{` ` ` ` ``// Sum of last two element` ` ``int` `temp = a[i] + a[j];` ` ``// Frequency of sum of first` ` ``// two elements` ` ``if` `(temp < sum && m.containsKey(sum - temp))` ` ``count += m.get(sum - temp);` ` ``}` ` ``for``(j = ``0``; j < i; j++) ` ` ``{` ` ` ` ``// Store frequency of all possible` ` ``// sums of first two elements` ` ``int` `temp = a[i] + a[j];` ` ``if` `(temp < sum)` ` ``if` `(m.containsKey(temp))` ` ``m.put(temp, m.get(temp) + ``1``);` ` ``else` ` ``m.put(temp, ``1``);` ` ``}` ` ``}` ` ` ` ``// Return the answer` ` ``return` `count;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` ` ``// Given array arr[]` ` ``int` `arr[] = { ``4``, ``5``, ``3``, ``1``, ``2``, ``4` `};` ` ``// Given sum S` ` ``int` `S = ``13``;` ` ``int` `N = arr.length;` ` ``// Function Call` ` ``System.out.print(countSum(arr, N, S));` `}` `}` `// This code is contributed by Princi Singh`
## Python3
`# Python3 program for the above approach` `from` `collections ``import` `defaultdict` `# Function to return the number of` `# quadruplets having the given sum` `def` `countSum(a, n, ``sum``):` ` ` ` ``# Initialize answer` ` ``count ``=` `0` ` ``# Store the frequency of sum` ` ``# of first two elements` ` ``m ``=` `defaultdict(``int``)` ` ``# Traverse from 0 to N-1, where` ` ``# arr[i] is the 3rd element` ` ``for` `i ``in` `range``(n ``-` `1``):` ` ``# All possible 4th elements` ` ``for` `j ``in` `range``(i ``+` `1``, n):` ` ``# Sum of last two element` ` ``temp ``=` `a[i] ``+` `a[j]` ` ``# Frequency of sum of first` ` ``# two elements` ` ``if` `(temp < ``sum``):` ` ``count ``+``=` `m[``sum` `-` `temp]` ` ``for` `j ``in` `range``(i):` ` ``# Store frequency of all possible` ` ``# sums of first two elements` ` ``temp ``=` `a[i] ``+` `a[j]` ` ``if` `(temp < ``sum``):` ` ``m[temp] ``+``=` `1` ` ``# Return the answer` ` ``return` `count` `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` ` ``# Given array arr[]` ` ``arr ``=` `[ ``4``, ``5``, ``3``, ``1``, ``2``, ``4` `]` ` ``# Given sum S` ` ``S ``=` `13` ` ``N ``=` `len``(arr)` ` ``# Function Call` ` ``print``(countSum(arr, N, S))` `# This code is contributed by chitranayal`
## C#
`// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{` `// Function to return the number of` `// quadruplets having the given sum` `static` `int` `countSum(``int` `[]a, ``int` `n, ``int` `sum)` `{` ` ` ` ``// Initialize variables` ` ``int` `i, j;` ` ``// Initialize answer` ` ``int` `count = 0;` ` ``// Store the frequency of sum` ` ``// of first two elements` ` ``Dictionary<``int``, ` ` ``int``> m = ``new` `Dictionary<``int``,` ` ``int``>();` ` ``// Traverse from 0 to N-1, where` ` ``// arr[i] is the 3rd element` ` ``for``(i = 0; i < n - 1; i++)` ` ``{` ` ` ` ``// All possible 4th elements` ` ``for``(j = i + 1; j < n; j++) ` ` ``{` ` ` ` ``// Sum of last two element` ` ``int` `temp = a[i] + a[j];` ` ` ` ``// Frequency of sum of first` ` ``// two elements` ` ``if` `(temp < sum && m.ContainsKey(sum - temp))` ` ``count += m[sum - temp];` ` ``}` ` ` ` ``for``(j = 0; j < i; j++) ` ` ``{` ` ` ` ``// Store frequency of all possible` ` ``// sums of first two elements` ` ``int` `temp = a[i] + a[j];` ` ``if` `(temp < sum)` ` ``if` `(m.ContainsKey(temp))` ` ``m[temp]++;` ` ``else` ` ``m.Add(temp, 1);` ` ``}` ` ``}` ` ` ` ``// Return the answer` ` ``return` `count;` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` ` ``// Given array []arr` ` ``int` `[]arr = { 4, 5, 3, 1, 2, 4 };` ` ``// Given sum S` ` ``int` `S = 13;` ` ``int` `N = arr.Length;` ` ` ` ``// Function Call` ` ``Console.Write(countSum(arr, N, S));` `}` `}` `// This code is contributed by shikhasingrajput`
Output:
`3`
Time Complexity: O(N2)
Auxiliary Space: O(N)
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Save
In matters of business and finance, interest rates affect a wide range of issues. One important aspect of interest rates is the impact they have on budgeting and short-term financial stability. A business's gap ratio is a representation of the effects that interest rates have on its short-term finances.
## Aspects
Two variables determine a business's gap ratio. The first is the sum of all assets that are interest-sensitive. Such assets may be debts that other parties owe the business that are affected by interest rate fluctuations. The second is the sum of all liabilities that are interest-sensitive. Such liabilities may be variable-interest loans on which the business must make payments.
## Calculation
To calculate its gap ratio, a business must divide the total value of its interest-sensitive assets by the total value of its interest-sensitive liabilities. Once it has this quotient, the business may represent it as a decimal or as a percentage.
## Application
The purpose of calculating a gap ratio is to gauge how well a business can withstand sudden fluctuations in interest rates. A high number shows financial stability in light of possible interest rate fluctuations, while a low number shows financial instability. For instance, a lender with \$3 million worth of interest-sensitive liabilities and \$5 million worth of interest-sensitive assets is relatively stable because its gap ratio is approximately 1.67. However, if those numbers were reversed, its gap ratio would be 0.6, showing financial instability.
## Limitations
While the gap ratio can be useful as a sign of financial stability, it is not the only aspect of financial stability. For instance, if only a small portion of a company's total assets are heavily interest-sensitive, even if fluctuations occur, this may not have a heavy effect on financial stability or business solvency.
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# What is the difference between the required rate of return and the expected rate of return?...
What is the difference between the required rate of return and the expected rate of return?
According to the scenario using the analysis of the current growth model for the required rate of return and the excepted rate of return we were asked if we could give the investors a 15% return CAPM We used beta as an estimate number which gave us a benchmark TF wanted to know the value of their stock and keep in mind admin changes We used the constant growth model to determine the share price We were given present value of all future ash flows the variables given were: Dividends flat rate \$10 Growth Rate (g) 10% Required Rate on Return 15% Using the Constant Growth Model: (10*(1+10)/ (.15-.10) = \$220 per share of stock compare to the days current trading \$220.65. To determine if stock value is undervalued or overvalued we use CAPM= Rf+B (Rm-Rf) = 3% + .8 (15%-3%) = 12.6 Making it UNDERVALUED
Required rate of return is the return an investor requires for investing in an asset. It is based on inherent risk of the investment. It also depends on the risk aversion of the investor.
Expected rate of return is the expected return an investment would provide. It is calculated on the basis of CAPM. It is the return the market would require on an investment on the basis of the risk of the investment.
As per the calculation price of stock would be 10*(1+0.10)/(0.15-0.10)=\$220
Since the current price of stock is \$220.65 it is slightly OVERVALUED.
If CAPM return is 12.6% and the required return is 15%, then the stock shall be again OVERVALUED because it provides lesser return than required return. | 406 | 1,720 | {"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} | 3.421875 | 3 | CC-MAIN-2024-10 | latest | en | 0.922032 |
http://www.acmerblog.com/hdu-2124-repair-the-wall-3271.html | 1,503,509,750,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886123312.44/warc/CC-MAIN-20170823171414-20170823191414-00395.warc.gz | 478,199,226 | 13,126 | 2013
12-29
# Repair the Wall
Long time ago , Kitty lived in a small village. The air was fresh and the scenery was very beautiful. The only thing that troubled her is the typhoon.
When the typhoon came, everything is terrible. It kept blowing and raining for a long time. And what made the situation worse was that all of Kitty’s walls were made of wood.
One day, Kitty found that there was a crack in the wall. The shape of the crack is
a rectangle with the size of 1×L (in inch). Luckly Kitty got N blocks and a saw(锯子) from her neighbors.
The shape of the blocks were rectangle too, and the width of all blocks were 1 inch. So, with the help of saw, Kitty could cut down some of the blocks(of course she could use it directly without cutting) and put them in the crack, and the wall may be repaired perfectly, without any gap.
Now, Kitty knew the size of each blocks, and wanted to use as fewer as possible of the blocks to repair the wall, could you help her ?
The problem contains many test cases, please process to the end of file( EOF ).
Each test case contains two lines.
In the first line, there are two integers L(0<L<1000000000) and N(0<=N<600) which
mentioned above.
In the second line, there are N positive integers. The ith integer Ai(0<Ai<1000000000 ) means that the ith block has the size of 1×Ai (in inch).
The problem contains many test cases, please process to the end of file( EOF ).
Each test case contains two lines.
In the first line, there are two integers L(0<L<1000000000) and N(0<=N<600) which
mentioned above.
In the second line, there are N positive integers. The ith integer Ai(0<Ai<1000000000 ) means that the ith block has the size of 1×Ai (in inch).
5 3
3 2 1
5 2
2 1
2
impossible
2011-12-16 06:40:40
mark：简单题，贪心。排序即可。
wa了几百次！！！搞了40分钟！！！尼玛没考虑到impossible的时候数组越界啊我日！！！太2了。条件i < n+1 没写。
# include <stdio.h>
# include <stdlib.h>
int a[610] ;
int cmp(const void *a, const void *b)
{
return *(int*)b - *(int*)a ;
}
int main ()
{
int i, L, n ;
while (~scanf ("%d%d", &L, &n))
{
for (i = 1 ; i <= n ; i++)
scanf ("%d", a+i) ;
qsort(a+1, n, 4, cmp) ;
for (i = 1 ; i <= n ;i++)
{
a[i] += a[i-1] ;
if (a[i] >= L)
break ;
}
if (i < n+1 && a[i] >=L) printf ("%d\n", i) ;
else puts ("impossible") ;
}
}
1. 去搜地球史上最震撼的访谈，大卫艾克采访的非洲萨满科瑞多.姆特瓦。他在10几年前就开始揭露蜥蜴人和光明会的阴谋了。 难道你没听说过他？
2. 博主您好，这是一个内容十分优秀的博客，而且界面也非常漂亮。但是为什么博客的响应速度这么慢，虽然博客的主机在国外，但是我开启VPN还是经常响应很久，再者打开某些页面经常会出现数据库连接出错的提示
3. 第23行：
hash = -1是否应该改成hash[s ] = -1
因为是要把从字符串s的start位到当前位在hash中重置
修改提交后能accept，但是不修改居然也能accept | 884 | 2,508 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-34 | longest | en | 0.933477 |
http://docs.mosek.com/8.1/dotnetapi/sensitivity-shared.html | 1,511,272,345,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806388.64/warc/CC-MAIN-20171121132158-20171121152158-00706.warc.gz | 81,976,825 | 10,440 | # 15.3 Sensitivity Analysis¶
Given an optimization problem it is often useful to obtain information about how the optimal objective value changes when the problem parameters are perturbed. E.g, assume that a bound represents the capacity of a machine. Now, it may be possible to expand the capacity for a certain cost and hence it is worthwhile knowing what the value of additional capacity is. This is precisely the type of questions the sensitivity analysis deals with.
Analyzing how the optimal objective value changes when the problem data is changed is called sensitivity analysis.
References
The book [Chv83] discusses the classical sensitivity analysis in Chapter 10 whereas the book [RTV97] presents a modern introduction to sensitivity analysis. Finally, it is recommended to read the short paper [Wal00] to avoid some of the pitfalls associated with sensitivity analysis.
Warning
Currently, sensitivity analysis is only available for continuous linear optimization problems. Moreover, MOSEK can only deal with perturbations of bounds and objective function coefficients.
## 15.3.1 Sensitivity Analysis for Linear Problems¶
### 15.3.1.1 The Optimal Objective Value Function¶
Assume that we are given the problem
(1)$\begin{split}\begin{array}{rclccccl} z(l^c,u^c,l^x,u^x,c) & = & \mbox{minimize} & & & c^Tx & & \\ & & \mbox{subject to} & l^c & \leq & Ax & \leq & u^c, \\ & & & l^x & \leq & x & \leq & u^x, \end{array}\end{split}$
and we want to know how the optimal objective value changes as $$l^c_i$$ is perturbed. To answer this question we define the perturbed problem for $$l^c_i$$ as follows
$\begin{split}\begin{array}{rclcccl} f_{l^c_i}(\beta ) & = & \mbox{minimize} & & & c^Tx & \\ & & \mbox{subject to} & l^c + \beta e_i & \leq & Ax & \leq u^c, \\ & & & l^x & \leq & x \leq & u^x, \end{array}\end{split}$
where $$e_i$$ is the $$i$$-th column of the identity matrix. The function
(2)$f_{l^c_i}(\beta)$
shows the optimal objective value as a function of $$\beta$$. Please note that a change in $$\beta$$ corresponds to a perturbation in $$l_i^c$$ and hence (2) shows the optimal objective value as a function of varying $$l^c_i$$ with the other bounds fixed.
It is possible to prove that the function (2) is a piecewise linear and convex function, i.e. its graph may look like in Fig. 3 and Fig. 4.
Fig. 3 $$\beta =0$$ is in the interior of linearity interval.
Fig. 4 $$\beta =0$$ is a breakpoint.
Clearly, if the function $$f_{l^c_i}(\beta )$$ does not change much when $$\beta$$ is changed, then we can conclude that the optimal objective value is insensitive to changes in $$l_i^c$$. Therefore, we are interested in the rate of change in $$f_{l^c_i}(\beta)$$ for small changes in $$\beta$$ — specifically the gradient
$f^\prime_{l^c_i}(0),$
which is called the shadow price related to $$l^c_i$$. The shadow price specifies how the objective value changes for small changes of $$\beta$$ around zero. Moreover, we are interested in the linearity interval
$\beta \in [\beta_1,\beta_2]$
for which
$f^\prime_{l^c_i}(\beta ) = f^\prime_{l^c_i}(0).$
Since $$f_{l^c_i}$$ is not a smooth function $$f^\prime_{l^c_i}$$ may not be defined at $$0$$, as illustrated in Fig. 4. In this case we can define a left and a right shadow price and a left and a right linearity interval.
The function $$f_{l^c_i}$$ considered only changes in $$l^c_i$$. We can define similar functions for the remaining parameters of the $$z$$ defined in (1) as well:
$\begin{split}\begin{array}{rcll} f_{l^c_i}(\beta ) & = & z(l^c+\beta e_i,u^c,l^x,u^x,c), & i=1,\ldots,m, \\ f_{u^c_i}(\beta ) & = & z(l^c,u^c+\beta e_i,l^x,u^x,c), & i=1,\ldots,m, \\ f_{l^x_{j}}(\beta ) & = & z(l^c,u^c,l^x+\beta e_{j},u^x,c), & j=1,\ldots,n, \\ f_{u^x_{j}}(\beta ) & = & z(l^c,u^c,l^x,u^x+\beta e_{j},c), & j=1,\ldots,n, \\ f_{c_{j}}(\beta ) & = & z(l^c,u^c,l^x,u^x,c+\beta e_{j}), & j=1,\ldots,n. \end{array}\end{split}$
Given these definitions it should be clear how linearity intervals and shadow prices are defined for the parameters $$u^c_i$$ etc.
#### 15.3.1.1.1 Equality Constraints¶
In MOSEK a constraint can be specified as either an equality constraint or a ranged constraint. If some constraint $$e_i^c$$ is an equality constraint, we define the optimal value function for this constraint as
$f_{e_i^c}(\beta ) = z(l^c+\beta e_i,u^c+\beta e_i,l^x,u^x,c)$
Thus for an equality constraint the upper and the lower bounds (which are equal) are perturbed simultaneously. Therefore, MOSEK will handle sensitivity analysis differently for a ranged constraint with $$l^c_i = u^c_i$$ and for an equality constraint.
### 15.3.1.2 The Basis Type Sensitivity Analysis¶
The classical sensitivity analysis discussed in most textbooks about linear optimization, e.g. [Chv83], is based on an optimal basic solution or, equivalently, on an optimal basis. This method may produce misleading results [RTV97] but is computationally cheap. Therefore, and for historical reasons, this method is available in MOSEK.
We will now briefly discuss the basis type sensitivity analysis. Given an optimal basic solution which provides a partition of variables into basic and non-basic variables, the basis type sensitivity analysis computes the linearity interval $$[\beta_1,\beta_2]$$ so that the basis remains optimal for the perturbed problem. A shadow price associated with the linearity interval is also computed. However, it is well-known that an optimal basic solution may not be unique and therefore the result depends on the optimal basic solution employed in the sensitivity analysis. This implies that the computed interval is only a subset of the largest interval for which the shadow price is constant. Furthermore, the optimal objective value function might have a breakpoint for $$\beta = 0$$. In this case the basis type sensitivity method will only provide a subset of either the left or the right linearity interval.
In summary, the basis type sensitivity analysis is computationally cheap but does not provide complete information. Hence, the results of the basis type sensitivity analysis should be used with care.
### 15.3.1.3 The Optimal Partition Type Sensitivity Analysis¶
Another method for computing the complete linearity interval is called the optimal partition type sensitivity analysis. The main drawback of the optimal partition type sensitivity analysis is that it is computationally expensive compared to the basis type analysis. This type of sensitivity analysis is currently provided as an experimental feature in MOSEK.
Given the optimal primal and dual solutions to (1), i.e. $$x^*$$ and $$((s_l^c)^*,(s_u^c)^*,(s_l^x)^*,(s_u^x)^*)$$ the optimal objective value is given by
$z^* := c^T x^*.$
The left and right shadow prices $$\sigma_1$$ and $$\sigma_2$$ for $$l_i^c$$ are given by this pair of optimization problems:
$\begin{split}\begin{array} {rclccl} \sigma_1 & = & \mbox{minimize} & e_i^T s_l^c & & \\ & & \mbox{subject to} & A^T (s_l^c-s_u^c) + s_l^x - s_u^x & = & c, \\ & & & (l^c)^T (s_l^c) - (u^c)^T (s_u^c) + (l^x)^T (s_l^x) - (u^x)^T (s_u^x) & = & z^*, \\ & & & s_l^c,s_u^c,s_l^c,s_u^x \geq 0 & & \end{array}\end{split}$
and
$\begin{split}\begin{array} {rclccl} \sigma_2 & = & \mbox{maximize} & e_i^T s_l^c & & \\ & & \mbox{subject to} & A^T (s_l^c-s_u^c) + s_l^x - s_u^x & = & c, \\ & & & (l^c)^T (s_l^c) - (u^c)^T (s_u^c) + (l^x)^T (s_l^x) - (u^x)^T (s_u^x) & = & z^*, \\ & & & s_l^c,s_u^c,s_l^c,s_u^x \geq 0. & & \end{array}\end{split}$
These two optimization problems make it easy to interpret the shadow price. Indeed, if $$((s_l^c)^*,(s_u^c)^*,(s_l^x)^*,(s_u^x)^*)$$ is an arbitrary optimal solution then
$(s_l^c)_i^* \in [\sigma_1,\sigma_2].$
Next, the linearity interval $$[\beta_1,\beta_2]$$ for $$l_i^c$$ is computed by solving the two optimization problems
$\begin{split}\begin{array} {lclccccl} \beta_1 & = & \mbox{minimize} & & & \beta & & \\ & & \mbox{subject to} & l^c + \beta e_i & \leq & Ax & \leq & u^c, \\ & & & & & c^T x - \sigma_1 \beta & = & z^*, \\ & & & l^x & \leq & x & \leq & u^x, \end{array}\end{split}$
and
$\begin{split}\begin{array} {lclccccl} \beta_2 & = & \mbox{maximize} & & & \beta & & \\ & & \mbox{subject to} & l^c + \beta e_i & \leq & Ax & \leq & u^c,\\ & & & & & c^T x - \sigma_2 \beta & = & z^*, \\ & & & l^x & \leq & x & \leq & u^x. \end{array}\end{split}$
The linearity intervals and shadow prices for $$u_i^c,$$ $$l_j^x,$$ and $$u_j^x$$ are computed similarly to $$l_i^c$$.
The left and right shadow prices for $$c_j$$ denoted $$\sigma_1$$ and $$\sigma_2$$ respectively are computed as follows:
$\begin{split}\begin{array} {lclccccl} \sigma_1 & = & \mbox{minimize} & & & e_j^T x & & \\ & & \mbox{subject to} & l^c + \beta e_i & \leq & Ax & \leq & u^c,\\ & & & & & c^T x & = & z^*,\\ & & & l^x & \leq & x & \leq & u^x, \end{array}\end{split}$
and
$\begin{split}\begin{array} {lclccccl} \sigma_2 & = & \mbox{maximize} & & & e_j^T x & & \\ & & \mbox{subject to} & l^c + \beta e_i & \leq & Ax & \leq & u^c,\\ & & & & & c^T x & = & z^*,\\ & & & l^x & \leq & x & \leq & u^x. \end{array}\end{split}$
Once again the above two optimization problems make it easy to interpret the shadow prices. Indeed, if $$x^*$$ is an arbitrary primal optimal solution, then
$x_j^* \in [\sigma_1,\sigma_2].$
The linearity interval $$[\beta_1,\beta_2]$$ for a $$c_j$$ is computed as follows:
$\begin{split}\begin{array} {rclccl} \beta_1 & = & \mbox{minimize} & \beta & & \\ & & \mbox{subject to} & A^T (s_l^c-s_u^c) + s_l^x - s_u^x & = & c + \beta e_j, \\ & & & (l^c)^T (s_l^c) - (u^c)^T (s_u^c) + (l^x)^T (s_l^x) - (u^x)^T (s_u^x) - \sigma_1 \beta & \leq & z^*, \\ & & & s_l^c,s_u^c,s_l^c,s_u^x \geq 0 & & \end{array}\end{split}$
and
$\begin{split}\begin{array} {rclccl} \beta_2 & = & \mbox{maximize} & \beta & & \\ & & \mbox{subject to} & A^T (s_l^c-s_u^c) + s_l^x - s_u^x & = & c + \beta e_j, \\ & & & (l^c)^T (s_l^c) - (u^c)^T (s_u^c) + (l^x)^T (s_l^x) - (u^x)^T (s_u^x) - \sigma_2 \beta & \leq & z^*, \\ & & & s_l^c,s_u^c,s_l^c,s_u^x \geq 0. & & \end{array}\end{split}$
### 15.3.1.4 Example: Sensitivity Analysis¶
As an example we will use the following transportation problem. Consider the problem of minimizing the transportation cost between a number of production plants and stores. Each plant supplies a number of goods and each store has a given demand that must be met. Supply, demand and cost of transportation per unit are shown in Fig. 5.
Fig. 5 Supply, demand and cost of transportation.
If we denote the number of transported goods from location $$i$$ to location $$j$$ by $$x_{ij}$$, problem can be formulated as the linear optimization problem of minimizing
$\begin{array}{ccccccccccccc} 1x_{11} & + & 2x_{12} & + & 5x_{23} & + & 2x_{24} & + & 1x_{31} & + & 2x_{33} & + & 1 x_{34} \end{array}$
subject to
(3)$\begin{split}\begin{array}{ccccccccccccccl} x_{11} & + & x_{12} & & & & & & & & & & & \leq & 400, \\ & & & & x_{23} & + & x_{24} & & & & & & & \leq & 1200, \\ & & & & & & & & x_{31} & + & x_{33} & + & x_{34} & \leq & 1000, \\ x_{11} & & & & & & & + & x_{31} & & & & & = & 800, \\ & & x_{12} & & & & & & & & & & & = & 100, \\ & & & & x_{23} & + & & & & & x_{33} & & & = & 500, \\ & & & & & & x_{24} & + & & & & & x_{34} & = & 500, \\ x_{11}, & & x_{12}, & & x_{23}, & & x_{24}, & & x_{31}, & & x_{33}, & & x_{34} & \geq & 0. \end{array}\end{split}$
The sensitivity parameters are shown in Table 18 and Table 19 for the basis type analysis and in Table 20 and Table 21 for the optimal partition type analysis.
Table 18 Ranges and shadow prices related to bounds on constraints and variables: results for the basis type sensitivity analysis.
Con. $$\beta_1$$ $$\beta_2$$ $$\sigma_1$$ $$\sigma_2$$
$$1$$ $$-300.00$$ $$0.00$$ $$3.00$$ $$3.00$$
$$2$$ $$-700.00$$ $$+\infty$$ $$0.00$$ $$0.00$$
$$3$$ $$-500.00$$ $$0.00$$ $$3.00$$ $$3.00$$
$$4$$ $$-0.00$$ $$500.00$$ $$4.00$$ $$4.00$$
$$5$$ $$-0.00$$ $$300.00$$ $$5.00$$ $$5.00$$
$$6$$ $$-0.00$$ $$700.00$$ $$5.00$$ $$5.00$$
$$7$$ $$-500.00$$ $$700.00$$ $$2.00$$ $$2.00$$
Var. $$\beta_1$$ $$\beta_2$$ $$\sigma_1$$ $$\sigma_2$$
$$x_{11}$$ $$-\infty$$ $$300.00$$ $$0.00$$ $$0.00$$
$$x_{12}$$ $$-\infty$$ $$100.00$$ $$0.00$$ $$0.00$$
$$x_{23}$$ $$-\infty$$ $$0.00$$ $$0.00$$ $$0.00$$
$$x_{24}$$ $$-\infty$$ $$500.00$$ $$0.00$$ $$0.00$$
$$x_{31}$$ $$-\infty$$ $$500.00$$ $$0.00$$ $$0.00$$
$$x_{33}$$ $$-\infty$$ $$500.00$$ $$0.00$$ $$0.00$$
$$x_{34}$$ $$-0.000000$$ $$500.00$$ $$2.00$$ $$2.00$$
Table 19 Ranges and shadow prices related to bounds on constraints and variables: results for the optimal partition type sensitivity analysis.
Con. $$\beta_1$$ $$\beta_2$$ $$\sigma_1$$ $$\sigma_2$$
$$1$$ $$-300.00$$ $$500.00$$ $$3.00$$ $$1.00$$
$$2$$ $$-700.00$$ $$+\infty$$ $$-0.00$$ $$-0.00$$
$$3$$ $$-500.00$$ $$500.00$$ $$3.00$$ $$1.00$$
$$4$$ $$-500.00$$ $$500.00$$ $$2.00$$ $$4.00$$
$$5$$ $$-100.00$$ $$300.00$$ $$3.00$$ $$5.00$$
$$6$$ $$-500.00$$ $$700.00$$ $$3.00$$ $$5.00$$
$$7$$ $$-500.00$$ $$700.00$$ $$2.00$$ $$2.00$$
Var. $$\beta_1$$ $$\beta_2$$ $$\sigma_1$$ $$\sigma_2$$
$$x_{11}$$ $$-\infty$$ $$300.00$$ $$0.00$$ $$0.00$$
$$x_{12}$$ $$-\infty$$ $$100.00$$ $$0.00$$ $$0.00$$
$$x_{23}$$ $$-\infty$$ $$500.00$$ $$0.00$$ $$2.00$$
$$x_{24}$$ $$-\infty$$ $$500.00$$ $$0.00$$ $$0.00$$
$$x_{31}$$ $$-\infty$$ $$500.00$$ $$0.00$$ $$0.00$$
$$x_{33}$$ $$-\infty$$ $$500.00$$ $$0.00$$ $$0.00$$
$$x_{34}$$ $$-\infty$$ $$500.00$$ $$0.00$$ $$2.00$$
Table 20 Ranges and shadow prices related to the objective coefficients: results for the basis type sensitivity analysis.
Var. $$\beta_1$$ $$\beta_2$$ $$\sigma_1$$ $$\sigma_2$$
$$c_1$$ $$-\infty$$ $$3.00$$ $$300.00$$ $$300.00$$
$$c_2$$ $$-\infty$$ $$\infty$$ $$100.00$$ $$100.00$$
$$c_3$$ $$-2.00$$ $$\infty$$ $$0.00$$ $$0.00$$
$$c_4$$ $$-\infty$$ $$2.00$$ $$500.00$$ $$500.00$$
$$c_5$$ $$-3.00$$ $$\infty$$ $$500.00$$ $$500.00$$
$$c_6$$ $$-\infty$$ $$2.00$$ $$500.00$$ $$500.00$$
$$c_7$$ $$-2.00$$ $$\infty$$ $$0.00$$ $$0.00$$
Table 21 Ranges and shadow prices related to the objective coefficients: results for the optimal partition type sensitivity analysis.
Var. $$\beta_1$$ $$\beta_2$$ $$\sigma_1$$ $$\sigma_2$$
$$c_1$$ $$-\infty$$ $$3.00$$ $$300.00$$ $$300.00$$
$$c_2$$ $$-\infty$$ $$\infty$$ $$100.00$$ $$100.00$$
$$c_3$$ $$-2.00$$ $$\infty$$ $$0.00$$ $$0.00$$
$$c_4$$ $$-\infty$$ $$2.00$$ $$500.00$$ $$500.00$$
$$c_5$$ $$-3.00$$ $$\infty$$ $$500.00$$ $$500.00$$
$$c_6$$ $$-\infty$$ $$2.00$$ $$500.00$$ $$500.00$$
$$c_7$$ $$-2.00$$ $$\infty$$ $$0.00$$ $$0.00$$
Examining the results from the optimal partition type sensitivity analysis we see that for constraint number $$1$$ we have $$\sigma_1 = 3,\ \sigma_2=1$$ and $$\beta_1 = -300,\ \beta_2=500$$. Therefore, we have a left linearity interval of $$[-300,0]$$ and a right interval of $$[0,500]$$. The corresponding left and right shadow prices are $$3$$ and $$1$$ respectively. This implies that if the upper bound on constraint $$1$$ increases by
$\beta \in [0,\beta_1] = [0,500]$
then the optimal objective value will decrease by the value
$\sigma_2 \beta = 1 \beta .$
Correspondingly, if the upper bound on constraint $$1$$ is decreased by
$\beta \in [0,300]$
then the optimal objective value will increase by the value
$\sigma_1 \beta = 3 \beta .$
## 15.3.2 Sensitivity Analysis with MOSEK¶
MOSEK provides the functions Task.primalsensitivity and Task.dualsensitivity for performing sensitivity analysis. The code in Listing 34 gives an example of its use.
Listing 34 Example of sensitivity analysis with the MOSEK Optimizer API for .NET. Click here to download.
using System;
namespace mosek.example
{
class msgclass : mosek.Stream
{
string prefix;
public msgclass (string prfx)
{
prefix = prfx;
}
public override void streamCB (string msg)
{
Console.Write ("{0}{1}", prefix, msg);
}
}
public class sensitivity
{
public static void Main ()
{
const double
infinity = 0;
mosek.boundkey[] bkc = new mosek.boundkey[] {
mosek.boundkey.up, mosek.boundkey.up,
mosek.boundkey.up, mosek.boundkey.fx,
mosek.boundkey.fx, mosek.boundkey.fx,
mosek.boundkey.fx
};
mosek.boundkey[] bkx = new mosek.boundkey[] {
mosek.boundkey.lo, mosek.boundkey.lo,
mosek.boundkey.lo, mosek.boundkey.lo,
mosek.boundkey.lo, mosek.boundkey.lo,
mosek.boundkey.lo
};
int[] ptrb = new int[] {0, 2, 4, 6, 8, 10, 12};
int[] ptre = new int[] {2, 4, 6, 8, 10, 12, 14};
int[] sub = new int[] {0, 3, 0, 4, 1, 5, 1, 6, 2, 3, 2, 5, 2, 6};
double[] blc = new double[] {
-infinity, -infinity,
-infinity, 800, 100, 500, 500
};
double[] buc = new double[] {400, 1200, 1000, 800, 100, 500, 500};
double[] c = new double[] {1.0, 2.0, 5.0, 2.0, 1.0, 2.0, 1.0};
double[] blx = new double[] {0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0};
double[] bux = new double[] {infinity,
infinity,
infinity,
infinity,
infinity,
infinity,
infinity
};
double[] val = new double[] {1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0,
1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0
};
int numcon = 7; /* Number of constraints. */
int numvar = 7; /* Number of variables. */
try
{
using (mosek.Env env = new mosek.Env())
{
using (mosek.Task task = new mosek.Task(env))
{
// Directs the log task stream to the user specified
// method task_msg_obj.streamCB
task.set_Stream(mosek.streamtype.log, new msgclass ("[task]"));
task.inputdata(numcon, numvar,
c,
0.0,
ptrb,
ptre,
sub,
val,
bkc,
blc,
buc,
bkx,
blx,
bux);
/* A maximization problem */
task.putobjsense(mosek.objsense.minimize);
try
{
task.optimize();
}
catch (mosek.Warning w)
{
Console.WriteLine("Mosek warning:");
Console.WriteLine (w.Code);
Console.WriteLine (w);
}
/* Analyze upper bound on c1 and the equality constraint on c4 */
int[] subi = new int [] {0, 3};
mosek.mark[] marki = new mosek.mark[] {mosek.mark.up,
mosek.mark.up
};
/* Analyze lower bound on the variables x12 and x31 */
int[] subj = new int [] {1, 4};
mosek.mark[] markj = new mosek.mark[] {mosek.mark.lo,
mosek.mark.lo
};
double[] leftpricei = new double[2];
double[] rightpricei = new double[2];
double[] leftrangei = new double[2];
double[] rightrangei = new double[2];
double[] leftpricej = new double[2];
double[] rightpricej = new double[2];
double[] leftrangej = new double[2];
double[] rightrangej = new double[2];
task.primalsensitivity( subi,
marki,
subj,
markj,
leftpricei,
rightpricei,
leftrangei,
rightrangei,
leftpricej,
rightpricej,
leftrangej,
rightrangej);
Console.Write("Results from sensitivity analysis on bounds:\n");
Console.Write("For constraints:\n");
for (int i = 0; i < 2; ++i)
Console.Write(
"leftprice = {0}, rightprice = {1},leftrange = {2}, rightrange ={3}\n",
leftpricei[i], rightpricei[i], leftrangei[i], rightrangei[i]);
Console.Write("For variables:\n");
for (int i = 0; i < 2; ++i)
Console.Write(
"leftprice = {0}, rightprice = {1},leftrange = {2}, rightrange ={3}\n",
leftpricej[i], rightpricej[i], leftrangej[i], rightrangej[i]);
double[] leftprice = new double[2];
double[] rightprice = new double[2];
double[] leftrange = new double[2];
double[] rightrange = new double[2];
int[] subc = new int[] {2, 5};
task.dualsensitivity( subc,
leftprice,
rightprice,
leftrange,
rightrange
);
Console.Write("Results from sensitivity analysis on objective coefficients:");
for (int i = 0; i < 2; ++i)
Console.Write(
"leftprice = {0}, rightprice = {1},leftrange = {2}, rightrange = {3}\n",
leftprice[i], rightprice[i], leftrange[i], rightrange[i]);
}
}
}
catch (mosek.Exception e)
{
Console.WriteLine (e.Code);
Console.WriteLine (e);
throw;
}
}
}
} | 6,937 | 19,436 | {"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} | 3.703125 | 4 | CC-MAIN-2017-47 | latest | en | 0.777791 |
https://www.tes.com/news/keep-posted | 1,529,394,448,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861980.33/warc/CC-MAIN-20180619060647-20180619080647-00428.warc.gz | 932,174,579 | 15,757 | # Keep posted
Rachel Cookson adds up the benefits of using colourful counting aids in Reception and key stage 1.
Numbers, Calculations, Measures, Shape, Space and Handling Data 1. Key stage 1. Various prices.
Reception Interactive Pictures. Reception. pound;69.95 + VAT.
Both from Cambridge Mathematics Direct, The Pitt Building, Trumpington Street, Cambridge. Tel: 01223 325588. Email: uksales@cambridge.org
Cambridge Mathematics Direct's interactive picture packs contain 16 write-on, wipe-off, A1 posters and are intended to support whole-class teaching. They are also created to promote discussion and the development of mathematical vocabulary as well as aid the teaching of the main areas of the numeracy strategy - for example, numbers, calculations (including money), shape, space, measures (including time and days of the week) and handling data.
The Reception and key stage 1 packs are both bright and colourful, with simple, clear pictures, and they are large enough for a whole class to use simultaneously. Teachers familiar with Cambridge Mathematics Resources will recognise the range of characters which appear. The pictures feature nursery rhymes and traditional story characters alongside cute animals and cartoon children to stimulate pupils' interest. The posters, which provide a scene for pupils to discuss - for example, "Playground measures" in the Reception pack and "In the classroom" in the key stage 1 pack - try to illustrate real life situations. Many of the pictures could also link to topics such as shopping and literacy themes like the Gingerbread Man, teddy bears and Jack and the Beanstalk.
We used the Reception pack with two classes. The children really enjoyed looking at and discussing the posters, which were excellent for promoting children's knowledge about numbers, position and counting.
One of our teachers and her class of budding mathematicians looked at "The zoo" poster, linking nicely to our topic on animals. The aim of the poster was to stimulate discussion about position, direction and movement, but the class was so engrossed that pupils also counted all the different animals, discovered lots of shapes and sizes and discussed environmental issues. All this from one poster.
Another class loved counting the animals "In the autumn forest" and finding patterns with the Pattern Family in their garden. The "How many" poster was put on a table with a set of dry-wipe pens as an independent activity and the children used it happily all day to count dragons and astronauts and practise writing numbers.
These posters are an excellent resource. They can be used in a variety of ways and genuinely stimulate and support children's interest in and knowledge of mathematics. They offer differentiation, as teachers are able to assess what children know and can do, and the children remain interested and learn from others as they delve deeper into the pictures. Finding different ways to count with younger children can be difficult but these interactive pictures offer a child-friendly option.
Rachel Cookson is assistant head and Reception teacher at Oliver Goldsmith Primary School, Camberwell, London
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18 June 2018 | 762 | 3,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-26 | latest | en | 0.929475 |
https://www.hackmath.net/en/math-problem/81136 | 1,716,646,757,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058829.24/warc/CC-MAIN-20240525131056-20240525161056-00612.warc.gz | 677,842,848 | 8,702 | # Frequencies 81136
In grouped data classes such as 10-15, 16-20, 21-25, 26-30 with the respective frequencies of each class as 3, 5, 4, 3, then the range (range of variation) is:
a. 15
b. 6
c. 20
d. 5
r = 20
## Step-by-step explanation:
Did you find an error or inaccuracy? Feel free to write us. Thank you!
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#### Grade of the word problem:
We encourage you to watch this tutorial video on this math problem: | 146 | 490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-22 | latest | en | 0.781606 |
http://guides.wkbw.com/Approach_to_Solving_Math_Problems_Bayside_NY-r992520-Bayside_NY.html | 1,506,005,929,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687820.59/warc/CC-MAIN-20170921134614-20170921154614-00531.warc.gz | 138,636,404 | 6,588 | # Approach to Solving Math Problems Bayside NY
When learning math, getting stumped without knowing what next is very common. However, before you let go of that pen and paper and scream I hate math to the top of your lungs, take time to read this article and get more information about learning math in Bayside.
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When learning math, getting stumped without knowing what next is very common. However, before you let go of that pen and paper and scream I hate math to the top of your lungs, take time to read this article.
If you have read other articles or books on how to solve math problems, this article will not really provide anything new. However, this provides you with a simple step by step approach to help you through each math problem you encounter.
How do you go about solving a difficult math problem?
Before discussing the steps, you must realize how keeping an open mind about math will improve your performance greatly. You may dislike math, but reminding yourself over and over will make you worse off. So, try assuming even just for a few minutes, that math is manageable. As it actually is.
So, here it goes.
In every math problem, you have to ask yourself 4 very important questions, which are:
1. What information am I given?In this part, you can do any one or more of the following:
• List down the information and facts given in the word problem, along with units and other details
• Point out key words and underline or circle them. There are hundreds of key words possible such as area, speed, factor, per, for every, among many others
• Watch out for units that are mixed. Units are important information, so never neglect them.
• Express the information in the form of math symbols, for example speed is distance over time.
2. What information is being asked?On this step, make sure you do the following:
• Know what the problem expects you to find. Check if it has to be expressed in certain units.
• Express your goal in terms of math symbols, if possible. If you are being asked for the area of a circle, express it as A = pi*r*r
3. What can I do with the given information?
• Play around with the facts you are given. Check what information you can get with the given and see whether it brings you closer to the information you have to solve for.
• Don t hesitate to manipulate the information one step at a time. Simplify the problem if it is relatively complicated.
• Make a picture or diagram, or even a table or chart. if the situation calls for it.
• Look for certain distinct pattern if applicable.
• If the problem seems difficult to understand, find a way to restate the problem which might be easier to grasp.
• Keep manipulating until you arrive at an answer.
4. Does my answer make sense?
• Once you have successfully found a way to solve the problem, check if it makes sense.
• Read the problem again and understand it once more.
• Check if you have the right units, as asked.
• Think of a way to confirm if your answer is right, if there is at all.
Solving math problems should not be tough if you keep your mind and heart open. Start with these 4 questions and see how you ll fare.
John has a site called http://Mathtrench.com , that offers thousands of solved math problems
Provided by ZingArticles.com
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- Reading Enhances Knowledge Bayside NY | 1,120 | 4,647 | {"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} | 3.15625 | 3 | CC-MAIN-2017-39 | longest | en | 0.895465 |
https://www.hackmath.net/en/example/6780?tag_id=74,100 | 1,539,933,843,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512332.36/warc/CC-MAIN-20181019062113-20181019083613-00548.warc.gz | 974,839,810 | 6,372 | # Peaches
There are 20 peaches in the pocket. 3 peaches are rotten. What is the probability that one of the randomly picked two peaches will be just one rotten?
Result
p = 0.067
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
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Same type of medicament produces a number of manufacturers in a variety of packages with different content of active substance. Pack 1: includes 60 pills of 600 mg of active substance per pack cost 9 Eur. Pack 2: includes 150 pills of 500 mg of active sub | 718 | 2,877 | {"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-2018-43 | longest | en | 0.957148 |
https://math.stackexchange.com/questions/255145/is-the-set-of-real-numbers-the-largest-possible-totally-ordered-set | 1,653,308,400,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558015.52/warc/CC-MAIN-20220523101705-20220523131705-00439.warc.gz | 448,315,623 | 72,560 | # Is the set of real numbers the largest possible totally ordered set?
Because I find any totally ordered set can be "lined up" in a straight line, I'm guessing that the set of all the real numbers is the biggest totally ordered set possible. In the sense that any other totally ordered set is isomorphic (order-preserving) to a subset of real numbers. Is that right?
• No. There are arbitrarily large totally ordered sets. See en.wikipedia.org/wiki/Hartogs_number . Dec 10, 2012 at 4:16
• Perhaps it was because OP's name is Voldemort, but I originally read Qiaochu's link as the "Hogwarts number." Dec 10, 2012 at 5:12
• @JavaMan: Yours should be the comment of the day. I wish I could give you +50 for it. Dec 10, 2012 at 5:32
• Related question: Any example of how to order a set of aleph-2 cardinality (for example the set of sets of reals?) May 17, 2015 at 10:25
• The set of real numbers is the largest totally ordered group only in the sense that having no nontrivial convex subgroups. This means any totally ordered group having convex subgroups only zero subgroup is isomorphic to subgroup of R as a ordered group.
– CAA
Jul 1, 2015 at 17:35
Elementary Construction
Without going into too much set theory, it is possible to produce an explicit example of a totally ordered set that cannot be embedded into $(\mathbb{R},\leq_{\mathbb{R}})$ in an order-preserving manner.
Define a total ordering $\preceq$ on $\mathbb{R}^{2}$ as follows: $$\forall a,b,c,d \in \mathbb{R}: \quad (a,b) \prec (c,d) ~ \stackrel{\text{def}}{\iff} ~ (a <_{\mathbb{R}} c) ~~ \text{or} ~~ (a = c ~~ \text{and} ~~ b <_{\mathbb{R}} d).$$ Observe that $\{ \{ r \} \times \mathbb{R} ~|~ r \in \mathbb{R} \}$ is an uncountable collection of pairwise-disjoint non-degenerate $\preceq$-intervals of $\mathbb{R}^{2}$. Hence, $(\mathbb{R}^{2},\preceq)$ cannot be embedded into $(\mathbb{R},\leq_{\mathbb{R}})$ in an order-preserving manner; if this were not the case, then $\mathbb{R}$ would contain uncountably many pairwise-disjoint non-degenerate intervals, so by picking a rational number from each of these intervals, we would end up with uncountably many rational numbers ⎯ contradiction.
Model-Theoretic Construction
There is a neat way to prove the existence of totally ordered sets with arbitrarily large cardinality. It relies on the Compactness Theorem in logic and model theory. Let $\kappa$ be a cardinal and $\mathcal{L}$ the first-order language consisting of
• a set $\mathcal{C}$ of $\kappa$-many constant symbols and
• a single binary relation symbol $\leq$.
Next, define $T$ to be the first-order $\mathcal{L}$-theory consisting of:
1. For distinct $c,d \in \mathcal{C}$, the sentence $\neg(c = d)$.
2. The Axiom of Reflexivity: $(\forall x)(x \leq x)$.
3. The Axiom of Antisymmetry: $(\forall x)(\forall y)(((x \leq y) \land (y \leq x)) \Longrightarrow (x = y))$.
4. The Axiom of Transitivity: $(\forall x)(\forall y)(\forall z)(((x \leq y) \land (y \leq z)) \Longrightarrow (x \leq z))$.
5. The Total-Ordering Axiom: $(\forall x)(\forall y)((x \leq y) \lor (y \leq x))$.
Although $T$ has infinitely many sentences (by (1)), it is easy to see that $(\mathbb{N},\leq_{\mathbb{N}})$ is a model of any finite number of these sentences. Therefore, by the Compactness Theorem, $T$ has a model $(\mathcal{M},\leq^{\mathcal{M}})$. Clearly, if we choose $\kappa > {\frak{c}}$, then we cannot embed $(\mathcal{M},\leq^{\mathcal{M}})$ into $(\mathbb{R},\leq_{\mathbb{R}})$ in an order-preserving manner.
Set-Theoretic Construction
The model-theoretic approach above suffers from a mild drawback, in the sense that the Compactness Theorem is equivalent to the Boolean Prime Ideal Theorem, which is a weak form of the Axiom of Choice. If one wishes to work strictly within the framework of the Zermelo-Fraenkel axioms, then one can use Hartogs’ result that for any set $X$, it is possible to find an ordinal $\alpha$ that cannot be mapped injectively into $X$. In particular, there exists an ordinal $\alpha$ that cannot be mapped injectively into $\mathbb{R}$, much less embedded into $\mathbb{R}$ in an order-preserving manner.
• But $\mathbb{R} \times \mathbb{R}$ is the same size as $\mathbb{R}$. I think the poster is looking for a set strictly larger than $\mathbb{R}$ (For instance if there existed an ordering on $\mathcal{P}(\mathbb{R})$, that would be good.) Dec 10, 2012 at 10:07
• An ordering on $\mathcal{P}(\mathbb{R})$ requires some form of the Axiom of Choice (AC). I wanted to avoid using AC when providing an answer. However, as Qiaochu has mentioned, Hartog's Theorem gives us a way of producing an ordinal that does not inject into a given set. For example, by Hartog's Theorem, there exists an ordinal $\alpha$ that does not inject into $\mathbb{R}$, much less inject into $\mathbb{R}$ in an order-preserving manner. Dec 10, 2012 at 17:30
• The great thing about Hartog's Theorem is that it is a theorem of ZF. Naturally, as Hartog's Theorem is about well orderings, it is a harder result. My example only deals with total orderings, so it is simple to construct. It is clear that my example is not a well-ordered set. Dec 10, 2012 at 17:34
• The point of my construction is that one does not have to resort to larger cardinalities in order to exhibit a counterexample. Dec 10, 2012 at 17:57
• Let me just add that there are certain models of set theory without choice where the reals are maximal in a sense: Define $\mathbb R/E_0$ as the set of equivalence classes of the relation that identifies two reals iff their difference is in $\mathbb Q$. In natural models of the axiom of determinacy, the size of $\mathbb R/E_0$ is a successor of the size of $\mathbb R$: It is strictly larger, and there are no intermediate sizes. This set is not linearly orderable and, if a set $X$ is not linearly orderable, then $\mathbb R/E_0$ can be injected into $X$. Feb 10, 2013 at 3:48
No, absolutely not. Assuming the axiom of choice, let $\kappa$ be any cardinal greater than $2^\omega=|\Bbb R|$; then $\kappa$ in its natural order is a linear order of greater cardinality than $\Bbb R$. If the axiom of choice fails in such a way that $\Bbb R$ cannot be well-ordered, there is still a well-ordered set that cannot be embedded in $\Bbb R$, the Hartogs ordinal of $\Bbb R$.
• Actually, no uncountable ordinal can be embedded in $\mathbb{R}$. I give the argument in my answer. Dec 10, 2012 at 7:20
• @Michael: I know; after all, $\Bbb R$ is ccc, hereditarily separable, etc. I was deliberately concentrating on the cardinality aspect. Dec 10, 2012 at 7:22
The real line is the largest order separable linearly ordered set. That is, if $$(L,\preceq)$$ is a linearly ordered set such that a countable set $$C\subseteq L$$ exists satisfying that for all $$x,y\in L$$ with $$x\prec y$$ there is $$c\in C$$ with $$x\preceq c\preceq y$$, then there exists a function $$f:L\to\mathbb{R}$$ such that $$f(x) iff $$x for all $$x$$ and $$y$$ in $$L$$. The existence of such a set $$C$$ is also necessary. A proof of these facts is given here.
Here is yet another counterexample without order separability: Let $$(W,\preceq)$$ be an uncountable well-ordered set. Let $$W'$$ be $$W$$ without the maximum of $$W$$ if such a maximum exists. For each $$w\in W'$$, let $$S(w)$$ be its immediate sucessor given by $$S(w)=\min\{w'\in W:w'\succ w\}$$. It is easily seen that $$x\prec y$$ implies $$x\prec S(x)\preceq y\prec S(y)$$. So if a strictly increasing function $$f:W\to\mathbb{R}$$ would exist, the uncountable family of intervals $$\{(f(w),f(S(w))):w\in W'\}$$ would be disjoint, which cannot be for the same reason pointed out by Haskel Curry in his answer.
• I think you mean: the uncountable family of intervals $\{(f(w), f(S(x))) : w\in W'\}$. May 26, 2019 at 19:55
• @Anguepa That family of intervals is clearly exactly disjoint when the given family of order intervals is disjoint. May 26, 2019 at 19:59
• I must insist, I think you mean $\{(f(w),f(S(w))) : w\in W'\}$ is place of $\{(w,S(w)): w\in W'\}$. The intervals $(w, S(w))$ are empty. May 26, 2019 at 20:15
• @Anguepa You are right. I'll edit it. May 26, 2019 at 20:19
EDIT: this may be order isomorphic to the reals, but not as a field, as it is not Archimedean.
As an elementary example, the rational functions $\mathbb R(x)$ are an ordered field. The quotient $p(x) / q(x),$ with leading terms $p_m x^m$ and $q_n x^n,$ is called positive if $p_m / q_n$ is positive. Comparison of two rational functions is given by taking the difference and checking the sign as above. So, for example, $1/x$ is positive but is smaller than any positive real number, as with positive $A$ we find $$A - \frac{1}{x} = \frac{Ax-1}{x}.$$
• Regarding your EDIT: Your construction is not order-isomorphic to the reals because $\{(p-1/x,p+1/x):p\in\Bbb R\}$ is an uncountable set of pairwise-disjoint nontrivial intervals, so as in Haskell Curry's answer you get an uncountable set of rational numbers in these sets. Jul 9, 2014 at 16:22
Some linearly ordered sets are not isomorphic to any subset of the reals even though there are not more of them than there are reals. The set of all countable ordinals is an example.
No, in fact it's a famous result that ANY set can be totally ordered--this is known as the Well-Ordering Theorem and is equivalent to the Axiom of Choice. | 2,743 | 9,328 | {"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": 24, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2022-21 | latest | en | 0.8649 |
http://www.sagemath.org/doc/constructions/_sources/groups.txt | 1,429,418,138,000,000,000 | text/plain | crawl-data/CC-MAIN-2015-18/segments/1429246637445.19/warc/CC-MAIN-20150417045717-00037-ip-10-235-10-82.ec2.internal.warc.gz | 764,003,537 | 6,927 | .. _chapter-groups: ****** Groups ****** .. index:: pair: group; permutation .. _section-permutation: Permutation groups ================== A permutation group is a subgroup of some symmetric group :math:S_n. Sage has a Python class PermutationGroup, so you can work with such groups directly:: sage: G = PermutationGroup(['(1,2,3)(4,5)']) sage: G Permutation Group with generators [(1,2,3)(4,5)] sage: g = G.gens()[0]; g (1,2,3)(4,5) sage: g*g (1,3,2) sage: G = PermutationGroup(['(1,2,3)']) sage: g = G.gens()[0]; g (1,2,3) sage: g.order() 3 For the example of the Rubik's cube group (a permutation subgroup of :math:S_{48}, where the non-center facets of the Rubik's cube are labeled :math:1,2,...,48 in some fixed way), you can use the GAP-Sage interface as follows. .. index:: pair: group; Rubik's cube .. skip :: sage: cube = "cubegp := Group( ( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19), ( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35), (17,19,24,22)(18,21,23,20)( 6,25,43,16)( 7,28,42,13)( 8,30,41,11), (25,27,32,30)(26,29,31,28)( 3,38,43,19)( 5,36,45,21)( 8,33,48,24), (33,35,40,38)(34,37,39,36)( 3, 9,46,32)( 2,12,47,29)( 1,14,48,27), (41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39)(16,24,32,40) )" sage: gap(cube) 'permutation group with 6 generators' sage: gap("Size(cubegp)") 43252003274489856000' Another way you can choose to do this: - Create a file cubegroup.py containing the lines:: cube = "cubegp := Group( ( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19), ( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35), (17,19,24,22)(18,21,23,20)( 6,25,43,16)( 7,28,42,13)( 8,30,41,11), (25,27,32,30)(26,29,31,28)( 3,38,43,19)( 5,36,45,21)( 8,33,48,24), (33,35,40,38)(34,37,39,36)( 3, 9,46,32)( 2,12,47,29)( 1,14,48,27), (41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39)(16,24,32,40) )" Then place the file in the subdirectory \$SAGE_ROOT/local/lib/python2.4/site-packages/sage of your Sage home directory. Last, read (i.e., import) it into Sage: .. skip :: sage: import sage.cubegroup sage: sage.cubegroup.cube 'cubegp := Group(( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18) (11,35,27,19),( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37) ( 6,22,46,35),(17,19,24,22)(18,21,23,20)( 6,25,43,16)( 7,28,42,13) ( 8,30,41,11),(25,27,32,30)(26,29,31,28)( 3,38,43,19)( 5,36,45,21) ( 8,33,48,24),(33,35,40,38)(34,37,39,36)( 3, 9,46,32)( 2,12,47,29) ( 1,14,48,27),(41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39) (16,24,32,40) )' sage: gap(sage.cubegroup.cube) 'permutation group with 6 generators' sage: gap("Size(cubegp)") '43252003274489856000' (You will have line wrap instead of the above carriage returns in your Sage output.) - Use the CubeGroup class:: sage: rubik = CubeGroup() sage: rubik The Rubik's cube group with generators R,L,F,B,U,D in SymmetricGroup(48). sage: rubik.order() 43252003274489856000 (1) has implemented classical groups (such as :math:GU(3,\GF{5})) and matrix groups over a finite field with user-defined generators. (2) also has implemented finite and infinite (but finitely generated) abelian groups. .. index:: pair: group; conjugacy classes .. _section-conjugacy: Conjugacy classes ================= You can compute conjugacy classes of a finite group using "natively":: sage: G = PermutationGroup(['(1,2,3)', '(1,2)(3,4)', '(1,7)']) sage: CG = G.conjugacy_classes_representatives() sage: gamma = CG[2] sage: CG; gamma [(), (4,7), (3,4,7), (2,3)(4,7), (2,3,4,7), (1,2)(3,4,7), (1,2,3,4,7)] (3,4,7) You can use the Sage-GAP interface:: sage: gap.eval("G := Group((1,2)(3,4),(1,2,3))") 'Group([ (1,2)(3,4), (1,2,3) ])' sage: gap.eval("CG := ConjugacyClasses(G)") '[ ()^G, (2,3,4)^G, (2,4,3)^G, (1,2)(3,4)^G ]' sage: gap.eval("gamma := CG[3]") '(2,4,3)^G' sage: gap.eval("g := Representative(gamma)") '(2,4,3)' Or, here's another (more "pythonic") way to do this type of computation:: sage: G = gap.Group('[(1,2,3), (1,2)(3,4), (1,7)]') sage: CG = G.ConjugacyClasses() sage: gamma = CG[2] sage: g = gamma.Representative() sage: CG; gamma; g [ ConjugacyClass( SymmetricGroup( [ 1, 2, 3, 4, 7 ] ), () ), ConjugacyClass( SymmetricGroup( [ 1, 2, 3, 4, 7 ] ), (4,7) ), ConjugacyClass( SymmetricGroup( [ 1, 2, 3, 4, 7 ] ), (3,4,7) ), ConjugacyClass( SymmetricGroup( [ 1, 2, 3, 4, 7 ] ), (2,3)(4,7) ), ConjugacyClass( SymmetricGroup( [ 1, 2, 3, 4, 7 ] ), (2,3,4,7) ), ConjugacyClass( SymmetricGroup( [ 1, 2, 3, 4, 7 ] ), (1,2)(3,4,7) ), ConjugacyClass( SymmetricGroup( [ 1, 2, 3, 4, 7 ] ), (1,2,3,4,7) ) ] ConjugacyClass( SymmetricGroup( [ 1, 2, 3, 4, 7 ] ), (4,7) ) (4,7) .. index:: pair: group; normal subgroups .. _section-normal: Normal subgroups ================ If you want to find all the normal subgroups of a permutation group :math:G (up to conjugacy), you can use Sage's interface to GAP:: sage: G = AlternatingGroup( 5 ) sage: gap(G).NormalSubgroups() [ Group( () ), AlternatingGroup( [ 1 .. 5 ] ) ] or :: sage: G = gap("AlternatingGroup( 5 )") sage: G.NormalSubgroups() [ Group( () ), AlternatingGroup( [ 1 .. 5 ] ) ] Here's another way, working more directly with GAP:: sage: print gap.eval("G := AlternatingGroup( 5 )") Alt( [ 1 .. 5 ] ) sage: print gap.eval("normal := NormalSubgroups( G )") [ Group(()), Alt( [ 1 .. 5 ] ) ] sage: G = gap.new("DihedralGroup( 10 )") sage: G.NormalSubgroups() [ Group( of ... ), Group( [ f2 ] ), Group( [ f1, f2 ] ) ] sage: print gap.eval("G := SymmetricGroup( 4 )") Sym( [ 1 .. 4 ] ) sage: print gap.eval("normal := NormalSubgroups( G );") [ Group(()), Group([ (1,4)(2,3), (1,3)(2,4) ]), Group([ (2,4,3), (1,4) (2,3), (1,3)(2,4) ]), Sym( [ 1 .. 4 ] ) ] .. index:: pair: groups; center .. _section-center: Centers ======= How do you compute the center of a group in Sage? Although Sage calls GAP to do the computation of the group center, center is "wrapped" (i.e., Sage has a class PermutationGroup with associated class method "center"), so the user does not need to use the gap command. Here's an example:: sage: G = PermutationGroup(['(1,2,3)(4,5)', '(3,4)']) sage: G.center() Subgroup of (Permutation Group with generators [(3,4), (1,2,3)(4,5)]) generated by [()] A similar syntax for matrix groups also works:: sage: G = SL(2, GF(5) ) sage: G.center() Matrix group over Finite Field of size 5 with 1 generators ( [4 0] [0 4] ) sage: G = PSL(2, 5 ) sage: G.center() Subgroup of (The projective special linear group of degree 2 over Finite Field of size 5) generated by [()] .. NOTE:: center can be spelled either way in GAP, not so in Sage. The group id database ===================== The function group_id requires that the Small Groups Library of E. A. O'Brien, B. Eick, and H. U. Besche be installed (you can do this by typing ./sage -i database_gap-4.4.9 in the Sage home directory). :: sage: G = PermutationGroup(['(1,2,3)(4,5)', '(3,4)']) sage: G.order() 120 sage: G.group_id() # optional - database_gap [120, 34] Another example of using the small groups database: group_id .. skip :: sage: gap_console() GAP4, Version: 4.4.6 of 02-Sep-2005, x86_64-unknown-linux-gnu-gcc gap> G:=Group((4,6,5)(7,8,9),(1,7,2,4,6,9,5,3)); Group([ (4,6,5)(7,8,9), (1,7,2,4,6,9,5,3) ]) gap> StructureDescription(G); "(((C3 x C3) : Q8) : C3) : C2" Construction instructions for every group of order less than 32 =============================================================== AUTHORS: * Davis Shurbert Every group of order less than 32 is implemented in Sage as a permutation group. They can all be created easily. We will first show how to build direct products and semidirect products, then give the commands necessary to build all of these small groups. Let G1, G2, ..., Gn be permutation groups already initialized in Sage. The following command can be used to take their direct product (where, of course, the ellipses are simply being used here as a notation, and you actually must enter every factor in your desired product explicitly). .. skip :: sage: G = direct_product_permgroups([G1, G2, ..., Gn]) The semidirect product operation can be thought of as a generalization of the direct product operation. Given two groups, H and K, their semidirect product, H \ltimes_{\phi} K, (where \phi : H \rightarrow Aut(K) is a homomorphism) is a group whose underlying set is the cartersian product of H and K, but with the operation: .. MATH:: (h_1, k_1) (h_2, k_2) = (h_1 h_2, k_1^{\phi(h_2)} k_2). The output is not the group explicity described in the definition of the operation, but rather an isomorphic group of permutations. In the routine below, assume H and K already have been defined and initialized in Sage. Also, phi is a list containing two sublists that define the underlying homomorphism by giving the images of a set of generators of H. For each semidirect product in the table below we will show you how to build phi, then assume you have read this passage and understand how to go from there. .. skip :: sage: G = H.semidirect_product(K, phi) To avoid unnecessary repitition, we will now give commands that one can use to create the cyclic group of order n, C_n, and the dihedral group on n letters, D_n. We will present one more example of each to ensure that the reader understands the command, then it will be withheld. .. skip :: sage: G = CyclicPermutationGroup(n) sage: G = DihedralGroup(n) Note that exponential notation will be used for the direct product operation. For example, {C_2}^2 = C_2 \times C_2. This table was crafted with the help of *Group Tables*, by AD Thomas and GV Wood (1980, Shiva Publishing). ===== =============================================== =============================================================================================== =========================== Order Group Description Command(s) GAP ID ===== =============================================== =============================================================================================== =========================== 1 The Trivial Group :: [1,1] sage: G = SymmetricGroup(1) 2 C_2 :: [2,1] sage: G = SymmetricGroup(2) 3 C_3 :: [3,1] sage: G = CyclicPermutationGroup(3) 4 C_4 [4,1] 4 C_2 \times C_2 :: [4,2] sage: G = KleinFourGroup() 5 C_5 [5,1] 6 C_6 [6,2] 6 S_3 (Symmetric Group on 3 letters) :: [6,1] sage: G = SymmetricGroup(3) 7 C_7 [7,1] 8 C_8 [8,1] 8 C_4 \times C_2 [8,2] 8 C_2\times C_2\times C_2 [8,5] 8 D_4 :: [8,3] sage: G = DihedralGroup(4) 8 The Quaternion Group (Q) :: [8,4] sage: G = QuaternionGroup() 9 C_9 [9,1] 9 C_3 \times C_3 [9,2] 10 C_{10} [10,2] 10 D_5 [10,1] 11 C_{11} [11,1] 12 C_{12} [12,2] 12 C_6 \times C_2 [12,5] 12 D_6 [12,4] 12 A_4 (Alternating Group on 4 letters) :: [12,3] sage: G = AlternatingGroup(4) 12 Q_6 (DiCyclic group of order 12) :: [12,1] sage: G = DiCyclicGroup(3) 13 C_{13} [13,1] 14 C_{14} [14,2] 14 D_{7} [14,1] 15 C_{15} [15,1] 16 C_{16} [16,1] 16 C_8 \times C_2 [16,5] 16 C_4 \times C_4 [16,2] 16 C_4\times C_2\times C_2 [16,10] 16 {C_2}^4 [16,14] 16 D_4 \times C_2 [16,11] 16 Q \times C_2 [16,12] 16 D_8 [16,7] 16 Q_{8} (Dicyclic group of order 16) :: [16,9] sage: G = DiCyclicGroup(4) 16 Semidihedral Group of order 2^4 :: [16,8] sage: G = SemidihedralGroup(4) 16 Split Metacyclic Group of order 2^4 :: [16,6] sage: G = SplitMetacyclicGroup(2,4) 16 (C_4 \times C_2) \rtimes_{\phi} C_2 :: [16,13] sage: C2 = SymmetricGroup(2); C4 = CyclicPermutationGroup(4) sage: A = direct_product_permgroups([C2,C4]) sage: alpha = PermutationGroupMorphism(A,A,[A.gens()[0],A.gens()[0]^2*A.gens()[1]]) sage: phi = [[(1,2)],[alpha]] 16 (C_4 \times C_2) \rtimes_{\phi} C_2 :: [16,3] sage: C2 = SymmetricGroup(2); C4 = CyclicPermutationGroup(4) sage: A = direct_product_permgroups([C2,C4]) sage: alpha = PermutationGroupMorphism(A,A,[A.gens()[0]^3*A.gens()[1],A.gens()[1]]) sage: phi = [[(1,2)],[alpha]] 16 C_4 \rtimes_{\phi} C_4 :: [16,4] sage: C4 = CyclicPermutationGroup(4) sage: alpha = PermutationGroupMorphism(C4,C4,[C4.gen().inverse()]) sage: phi = [[(1,2,3,4)],[alpha]] 17 C_{17} [17,1] 18 C_{18} [18,2] 18 C_6 \times C_3 [18,5] 18 D_9 [18,1] 18 S_3 \times C_3 [18,3] 18 Dih(C_3 \times C_3) :: [18,4] sage: G = GeneralDihedralGroup([3,3]) 19 C_{19} [19,1] 20 C_{20} [20,2] 20 C_{10} \times C_2 [20,5] 20 D_{10} [20,4] 20 Q_{10} (Dicyclic Group of order 20) [20,1] 20 Hol(C_5) :: [20,3] sage: C5 = CyclicPermutationGroup(5) sage: G = C5.holomorph() 21 C_{21} [21,2] 21 C_7 \rtimes_{\phi} C_3 :: [21,1] sage: C7 = CyclicPermutationGroup(7) sage: alpha = PermutationGroupMorphism(C7,C7,[C7.gen()**4]) sage: phi = [[(1,2,3)],[alpha]] 22 C_{22} [22,2] 22 D_{11} [22,1] 23 C_{23} [23,1] 24 C_{24} [24,2] 24 D_{12} [24,6] 24 Q_{12} (DiCyclic Group of order 24) [24,4] 24 C_{12} \times C_2 [24,9] 24 C_6 \times C_2 \times C_2 [24,15] 24 S_4 (Symmetric Group on 4 letters) :: [24,12] sage: G = SymmetricGroup(4) 24 S_3 \times C_4 [24,5] 24 S_3 \times C_2 \times C_2 [24,14] 24 D_4 \times C_3 [24,10] 24 Q \times C_3 [24,11] 24 A_4 \times C_2 [24,13] 24 Q_6 \times C_2 [24,7] 24 Q \rtimes_{\phi} C_3 :: [24,3] sage: Q = QuaternionGroup() sage: alpha = PermutationGroupMorphism(Q,Q,[Q.gens()[0]*Q.gens()[1],Q.gens()[0].inverse()]) sage: phi = [[(1,2,3)],[alpha]] 24 C_3 \rtimes_{\phi} C_8 :: [24,1] sage: C3 = CyclicPermutationGroup(3) sage: alpha = PermutationGroupMorphism(C3,C3,[C3.gen().inverse()]) sage: phi = [[(1,2,3,4,5,6,7,8)],[alpha]] 24 C_3 \rtimes_{\phi} D_4 :: [24,8] sage: C3 = CyclicPermutationGroup(3) sage: alpha1 = PermutationGroupMorphism(C3,C3,[C3.gen().inverse()]) sage: alpha2 = PermutationGroupMorphism(C3,C3,[C3.gen()]) sage: phi = [[(1,2,3,4),(1,3)],[alpha1,alpha2]] 25 C_{25} [25,1] 25 C_5 \times C_5 [25,2] 26 C_{26} [26,2] 26 D_{13} [26,1] 27 C_{27} [27,1] 27 C_9 \times C_3 [27,2] 27 C_3 \times C_3 \times C_3 [27,5] 27 Split Metacyclic Group of order 3^3 :: [27,4] sage: G = SplitMetacyclicGroup(3,3) 27 (C_3 \times C_3) \rtimes_{\phi} C_3 :: [27,3] sage: C3 = CyclicPermutationGroup(3) sage: A = direct_product_permgroups([C3,C3]) sage: alpha = PermutationGroupMorphism(A,A,[A.gens()[0]*A.gens()[1].inverse(),A.gens()[1]]) sage: phi = [[(1,2,3)],[alpha]] 28 C_{28} [28,2] 28 C_{14} \times C_2 [28,4] 28 D_{14} [28,3] 28 Q_{14} (DiCyclic Group of order 28) [28,1] 29 C_{29} [29,1] 30 C_{30} [30,4] 30 D_{15} [30,3] 30 D_5 \times C_3 [30,2] 30 D_3 \times C_5 [30,1] 31 C_{31} [31,1] ===== =============================================== =============================================================================================== =========================== Table By Kevin Halasz Construction instructions for every finitely presented group of order 15 or less ================================================================================ Sage has the capability to easily construct every group of order 15 or less as a finitely presented group. We will begin with some discussion on creating finitely generated abelian groups, as well as direct and semidirect products of finitely presented groups. All finitely generated abelian groups can be created using the groups.presentation.FGAbelian(ls) command, where ls is a list of non-negative integers which gets reduced to invariants defining the group to be returned. For example, to construct C_4 \times C_2 \times C_2 \times C_2 we can simply use:: sage: A = groups.presentation.FGAbelian([4,2,2,2]) The output for a given group is the same regardless of the input list of integers. The following example yeilds identical presentations for the cyclic group of order 30. :: sage: A = groups.presentation.FGAbelian([2,3,5]) sage: B = groups.presentation.FGAbelian([30]) If G and H are finitely presented groups, we can use the following code to create the direct product of G and H, G \times H. .. skip :: sage: D = G.direct_product(H) Suppose there exists a homomorphism \phi from a group G to the automorphism group of a group H. Define the semidirect product of G with H via \phi, as the Cartesian product of G and H, with the operation (g_1, h_1)(g_2, h_2) = (g_1 g_2, \phi_{h_1}(g_2) h_2) where \phi_h = \phi(h). To construct this product in Sage for two finitely presented groups, we must define \phi manually using a pair of lists. The first list consists of generators of the group G, while the second list consists of images of the corresponding generators in the first list. These automorphisms are similarly defined as a pair of lists, generators in one and images in the other. As an example, we construct the dihedral group of order 16 as a semidirect product of cyclic groups. :: sage: C2 = groups.presentation.Cyclic(2) sage: C8 = groups.presentation.Cyclic(8) sage: hom = (C2.gens(), [ ([C8([1])], [C8([-1])]) ]) sage: D = C2.semidirect_product(C8, hom) The following table shows the groups of order 15 or less, and how to construct them in Sage. Repeated commands have been omitted but instead are described by the following exmples. The cyclic group of order n can be crated with a single command: .. skip :: sage: C = groups.presentation.Cyclic(n) Similarly for the dihedral group of order 2n: .. skip :: sage: D = groups.presentation.Dihedral(n) This table was modeled after the preceding table created by Kevin Halasz. ===== =============================================== =============================================================================================== =========================== Order Group Description Command(s) GAP ID ===== =============================================== =============================================================================================== =========================== 1 The Trivial Group :: [1,1] sage: G = groups.presentation.Symmetric(1) 2 C_2 :: [2,1] sage: G = groups.presentation.Symmetric(2) 3 C_3 :: [3,1] sage: G = groups.presentation.Cyclic(3) 4 C_4 [4,1] 4 C_2 \times C_2 :: [4,2] sage: G = groups.presentation.Klein() 5 C_5 [5,1] 6 C_6 [6,2] 6 S_3 (Symmetric Group on 3 letters) :: [6,1] sage: G = groups.presentation.Symmetric(3) 7 C_7 [7,1] 8 C_8 [8,1] 8 C_4 \times C_2 :: [8,2] sage: G = groups.presentation.FGAbelian([4,2]) 8 C_2\times C_2\times C_2 :: [8,5] sage: G = groups.presentation.FGAbelian([2,2,2]) 8 D_4 :: [8,3] sage: G = groups.presentation.Dihedral(4) 8 The Quaternion Group (Q) :: [8,4] sage: G = groups.presentation.Quaternion() 9 C_9 [9,1] 9 C_3 \times C_3 [9,2] 10 C_{10} [10,2] 10 D_5 [10,1] 11 C_{11} [11,1] 12 C_{12} [12,2] 12 C_6 \times C_2 [12,5] 12 D_6 [12,4] 12 A_4 (Alternating Group on 4 letters) :: [12,3] sage: G = groups.presentation.Alternating(4) 12 Q_6 (DiCyclic group of order 12) :: [12,1] sage: G = groups.presentation.DiCyclic(3) 13 C_{13} [13,1] 14 C_{14} [14,2] 14 D_{7} [14,1] 15 C_{15} [15,1] ===== =============================================== =============================================================================================== =========================== | 6,719 | 18,685 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2015-18 | latest | en | 0.576819 |
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Author Topic: Using analogRead() on floating pin for random number generator seeding/generator (Read 861 times) 0 Members and 1 Guest are viewing this topic.
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Nokia 6100 LCD Library
« on: April 26, 2012, 01:58:29 pm » Bigger Smaller Reset
As a followup of the discussion in this thread, I decided to look at the frequency distribution of values you get by using analogRead() on a floating (unconnected) analog pin.
Background
A "good" random number generator should have a "flat" frequency distribution (equal probability of getting any number in range) - which of course is unlikely in this approach. The better approach is to "seed" the default pseudorandom generator in Arduino library - which does approximate to flat distribution - with some kind of random value - either a timer counter or as discussed in the post, analogRead() on a floating pin. The seed should be a fairly random, wide ranging number.
Test
I generated 1 million values from analogRead() on a floating pin, sent each value to my computer using serial & built a frequency distribution in excel.
Results
99.5% of the 1 million values were distributed between just 38 values out of 1024 possible values. Hmmm.... does not seem to be a wide "seed" range by any means
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Ardgrafix6100 - A fast, full-featured Arduino graphics driver for Nokia 6100 LCDs http://code.google.com/p/ardgrafix6100/
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I don't think you connected the grounds, Dave.
« Reply #1 on: April 26, 2012, 02:03:59 pm » Bigger Smaller Reset
But what if you build your seed from the LSBs of 32 consecutive reads?
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Pete, it's a fool looks for logic in the chambers of the human heart.
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« Reply #2 on: April 26, 2012, 02:55:50 pm » Bigger Smaller Reset
Quote
I decided to look at the frequency distribution of values you get by using analogRead() on a floating (unconnected) analog pin.
With an Uno? Did you also test the Mega? What about the bigger Mega? The Teensys? What about when the Teensy is plugged into a breadboard? What if a neighbor pin is generating a PWM signal? What if a neighbor pin is HIGH? Or LOW? What if the board is powered by USB? What if the board is powered from VIN? What happens when the board is placed close to a desk lamp? Did you try the board near an OttLite? What happens when your hand is close to the pin? What about your mobile phone? Did you test after running the air conditioner a bit so the air is drier?
The point is, an unconnected pin is a terrible way to generate a random number. Even if you get good results some of the time there is no way to guarantee good results all of the time. AWOL's suggestion helps in the case that the voltage is changing but it obviously won't help in the copious cases that the voltage is not changing or is heavily influenced by something near.
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« Reply #3 on: April 26, 2012, 03:04:58 pm » Bigger Smaller Reset
But what if you build your seed from the LSBs of 32 consecutive reads?
Nice & Noisy across the spectrum! :-) i truncated the 32 bit numbers to 16 bits to generate freq dist attached.
@Coding Badly - just an evaluation of an easy alternative that's not PseudoRandom - ie not replicable if u kinda know the seed range.
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« Reply #4 on: April 26, 2012, 03:09:34 pm » Bigger Smaller Reset
Feed the data through this...
http://www.fourmilab.ch/random/
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« Reply #5 on: April 26, 2012, 03:21:03 pm » Bigger Smaller Reset
Feed the data through this...
http://www.fourmilab.ch/random/
interesting.... will try over the weekend. Thanks !
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« Reply #6 on: November 22, 2012, 09:06:50 am » Bigger Smaller Reset
You are entirely correct. I did extensive research on this and essentially debunked this claim made by Arduino. I did not find a stable and viable way to generate randomness on-board the Arduino.
See the full paper here: http://benedikt.sudo.is/ardrand.pdf
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« Reply #7 on: November 22, 2012, 09:11:10 am » Bigger Smaller Reset
Results
99.5% of the 1 million values were distributed between just 38 values out of 1024 possible values. Hmmm.... does not seem to be a wide "seed" range by any means
That's what cryptographers use hash functions for - take a sample of the bottom bits of the values and put them through a hash function (CRC32, MD5...whatever size you need).
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« Reply #8 on: November 22, 2012, 10:50:59 am » Bigger Smaller Reset
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I did extensive research on this and essentially debunked this claim made by Arduino.
Take a look at this thread:
http://arduino.cc/forum/index.php/topic,133261.0.html
The stock analogRead() is essentially incorrectly coded.
I did a Chi-squared test on about 7k data points and it passed with flying color using the revised myanalogRead() provided in that thread. The distribution is also quite uniform.
It also talked about other ways to potentially produce truly random numbers from your arduino. The rc oscillator approach for example can be done with just a resistor + cap + onboard comparator.
The ideal is solid, except it is incorrectly implemented.
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« Reply #9 on: November 22, 2012, 02:24:01 pm » Bigger Smaller Reset
I did a Chi-squared test on about 7k data points and it passed with flying color using the revised myanalogRead() provided in that thread. The distribution is also quite uniform.
So did I. The result was a spectacular failure.
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In theory there is no difference between theory and practice, however in practice there are many...
« Reply #10 on: November 22, 2012, 02:42:13 pm » Bigger Smaller Reset
The floating analog pin is an antenna receiving signals. The ADC converts this to 10 bits. Depending on what the analog pin receives the result will be more or less random.
What does the antenna receive? signals from all electronics in the neighbourhood and beyond ...
True randomness is very difficult to reach imho (although the frequency of this topic popping up in this forum is quite random
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« Reply #11 on: November 22, 2012, 03:00:13 pm » Bigger Smaller Reset
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Depending on what the analog pin receives the result will be more or less random.
That is true but it ignores the issue that the distribution of that randomness is not desirable - which is the point raised by the paper cited in this thread, putting aside 1) the loading by the adc module; and 2) the inherent issue with the analogRead() call, well documented in a separate thread.
Thus, the lsb approach. It actually doesn't rely on the randomness of a signal on the adc pin, but the randomness of the last digit returned by the adc process. You can tie the pin to a known voltage (Vcc or ground), yet the adc's lsb is unstable.
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« Reply #12 on: November 22, 2012, 03:04:13 pm » Bigger Smaller Reset
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« Reply #13 on: November 22, 2012, 03:20:49 pm » Bigger Smaller Reset
Thus, the lsb approach. It actually doesn't rely on the randomness of a signal on the adc pin, but the randomness of the last digit returned by the adc process. You can tie the pin to a known voltage (Vcc or ground), yet the adc's lsb is unstable.
Nope. The technique failed again. Analog input 1 ... failed. Analog input 2 ... failed. Speed up the ADC clock ... failed. Speed it up even more ... failed. Set it to the value used in the core ... failed.
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It actually doesn't rely on the randomness of a signal on the adc pin...
It may not but it most certainly does not work the way you described.
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« Reply #14 on: November 22, 2012, 05:59:48 pm » Bigger Smaller Reset
the inherent issue with the analogRead() call, well documented in a separate thread.
Are you referring to http://arduino.cc/forum/index.php/topic,133261.0.html ?
I might pick this up again..
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Pages: [1] 2 Go Up | 2,358 | 8,918 | {"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-2013-20 | latest | en | 0.869706 |
https://www.physicsforums.com/threads/distance-between-3-islands.182475/ | 1,477,166,538,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719041.14/warc/CC-MAIN-20161020183839-00022-ip-10-171-6-4.ec2.internal.warc.gz | 972,021,073 | 16,354 | # Distance between 3 islands
1. Aug 31, 2007
### kbump
1. The problem statement, all variables and given/known data
A ferryboat transports tourists among three islands. It sails from the first island to the second island, 4.02 km away, in a direction 37.0° north of east. It then sails from the second island to the third island in a direction 69.0° west of north. Finally, it returns to the first island, sailing in a direction 28.0° east of south. (a) Calculate the distance between the second and third islands. (b) Calculate the distance between the first and third islands.
2. Relevant equations
3. The attempt at a solution
Sadly, I don't know where to start. I know I have to find the x and y components, but don't know how since only one side/distance is given. Any help is appreciated..
2. Aug 31, 2007
### rock.freak667
Do you the answers to the question? i got for part a)5.153km if that is correct then i shall help you get out part b)
3. Aug 31, 2007
### bob1182006
I tried but got 5.79 km for part a.
The way I did it:
draw a diagram.
start with the first island, approximate 37 degrees N of E, and do the rest to all the other islands.
Now extend the N/S/E/W lines @ each of the islands. You should get triangles starting to appear. you know 2 of the angles. from the starting island you have 37 degrees, and the right angle of the triangle. So you can find the other angle and from there you can find all other angles.
You can then find the 3 angles within the "triangle" that the ship takes along the islands. You know 1 side 1->2nd island. Use the law of sines A/sin(a) = B/sin(b) = C/sin(c) and you can find all of the distances.
4. Aug 31, 2007
### rock.freak667
Well that is the same way i got my answer, although i most likely made some error in my diagram
5. Aug 31, 2007
### kbump
I don't get the solution. It's actually online, so I type my answer and it lets you know if it was correct. I actually tried the same method and got 24, 107, and 49 for my angles, but didn't know where to go from there. I tried typing in 5.79 for part a and it was incorrect. I only have one more shot at it, so I want to make sure it's right and I know what I'm doing before I submit again.
6. Aug 31, 2007
### P.O.L.A.R
Ok I think I have the trick is to draw it out an use the law of sine a/sin(A)=b/sin(B)=c/sin(C) for part a i got 6.83km much diff than the rest. You have to place the angle across from the side like the picture.
Last edited: Aug 31, 2007
7. Aug 31, 2007
### Yapper
My diagram( very rough) is attached, my angles are 41, 58, 81. And I got 6.05km for my answer,
File size:
5.5 KB
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196 | 746 | 2,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2016-44 | longest | en | 0.943103 |
https://math.stackexchange.com/questions/3002058/is-banach-fixed-point-theorem-a-necessary-and-sufficient-condition-for-the-exist | 1,560,963,140,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999003.64/warc/CC-MAIN-20190619163847-20190619185847-00324.warc.gz | 510,560,985 | 34,501 | # Is Banach fixed point theorem a necessary and sufficient condition for the existence of a fixed point
Banach fixed point theorem requires a contraction mapping from a metric space into itself, but when I was learning some machine learning algorithms, some questions rise above: k-means is an algorithm for clustering, it can be proved that this method will converge, but the proof of convergence of the algorithm doesn't involve fixed point theorem. I feel the iteration of the centering point for each cluster in this method is very similar the iteration of the fixed point iteration steps so I tried to prove this convergence using Banach fixed point theorem. However I couldn't construct a contraction mapping in this problem. So I guess if the iteration steps in k-means is not a contraction at all. In order to test this assumption, I generate some random number on my computer and use the k-means steps to cluster and calculate the norm distance between each iteration point.To my surprise, it is NOT a contraction!but it does converge in finite steps. I think the convergence shows the existence of the fixed point under this iteration. So with these facts, can I say the Bananch Fixed point theorem gives a sufficient condition on the existence of a fixed point, but it might not be necessary?
• If you're asking if contraction in each step is necessary for convergence, then yes you're right - it is not necessary – OnceUponACrinoid Nov 17 '18 at 7:24
• $\mathbb R$ is a Banach space and $f(x)=2x$ is map with a unique fixed point. It is not a contraction. – Kavi Rama Murthy Nov 17 '18 at 12:09 | 348 | 1,608 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-26 | latest | en | 0.904172 |
https://somme2016.org/blog/what-is-the-profitability-index-formula/ | 1,670,504,859,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711336.41/warc/CC-MAIN-20221208114402-20221208144402-00449.warc.gz | 527,023,592 | 9,259 | # What is the profitability index formula?
## What is the profitability index formula?
The profitability index is calculated by dividing the present value of future cash flows that will be generated by the project by the initial cost of the project. A profitability index of 1 indicates that the project will break even. For example, if a project costs \$1,000 and will return \$1,200, it’s a “go.”
## How do we calculate payback period?
To calculate the payback period you can use the mathematical formula: Payback Period = Initial investment / Cash flow per year For example, you have invested Rs 1,00,000 with an annual payback of Rs 20,000. Payback Period = 1,00,000/20,000 = 5 years.
How do you calculate modified IRR?
Take the present value (PV) of the project cash flows from the recovery phase (note not the NPV), divide by the outlay and take the ‘ n th’ root of the result. Multiply the result by one plus the cost of capital (1.1 in this case), deduct one and you have the answer.
How do you calculate net present value?
What is the formula for net present value?
1. NPV = Cash flow / (1 + i)t – initial investment.
2. NPV = Today’s value of the expected cash flows − Today’s value of invested cash.
3. ROI = (Total benefits – total costs) / total costs.
### What is profitability index in capital budgeting?
The profitability index (PI) is a measure of a project’s or investment’s attractiveness. The PI is calculated by dividing the present value of future expected cash flows by the initial investment amount in the project.
### How do I calculate payback period in Excel?
Payback period = Initial Investment or Original Cost of the Asset / Cash Inflows.
1. Payback period = Initial Investment or Original Cost of the Asset / Cash Inflows.
2. Payback Period = 1 million /2.5 lakh.
3. Payback Period = 4 years.
What is MIRR formula?
To calculate the MIRR for each project Helen uses the formula: MIRR = (Future value of positive cash flows / present value of negative cash flows) (1/n) – 1. | 474 | 2,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-49 | latest | en | 0.882415 |
https://roughlydaily.com/tag/games/ | 1,696,355,387,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511170.92/warc/CC-MAIN-20231003160453-20231003190453-00264.warc.gz | 522,113,688 | 25,951 | # (Roughly) Daily
## “Games are a compromise between intimacy and keeping intimacy away”*…
… Maybe, as Greg Costikyan explains, none more so than Rochambeau (or “Rock-Paper-Scissors” as it’s also known)…
Unless you have lived in a Skinner box from an early age, you know that the outcome of tic-tac-toe is utterly certain. At first glance, rock-paper-scissors appears almost as bad. A four-year-old might think there’s some strategy to it, but isn’t it basically random?
Indeed, people often turn to rock-paper-scissors as a way of making random, arbitrary decisions — choosing who’ll buy the first round of drinks, say. Yet there is no quantum-uncertainty collapse, no tumble of a die, no random number generator here; both players make a choice. Surely this is wholly nonrandom?
All right, nonrandom it is, but perhaps it’s arbitrary? There’s no predictable or even statistically calculable way of figuring out what an opponent will do next, so that one choice is as good as another, and outcomes will be distributed randomly over time — one-third in victory for one player, one-third to the opponent, one-third in a tie. Yes? Players quickly learn that this is a guessing game and that your goal is to build a mental model of your opponent, to try to predict his actions. Yet a naïve player, once having realized this, will often conclude that the game is still arbitrary; you get into a sort of infinite loop. If he thinks such-and-so, then I should do this-and-that; but, on the other hand, if he can predict that I will reason thusly, he will instead do the-other-thing, so my response should be something else; but if we go for a third loop — assuming he can reason through the two loops I just did — then . . . and so on, ad infinitum. So it is back to being a purely arbitrary game. No?
No…
Read on for an explanation in this excerpt from veteran game designer Greg Costikyan’s book Uncertainty in Games: “The Psychological Depths of Rock-Paper-Scissors,” from @mitpress.
* Eric Berne
###
As we play, we might send carefully-plotted birthday greetings to Vilfredo Pareto; he was born on this date in 1848. An engineer, mathematician, sociologist, economist, political scientist, and philosopher, he made significant contributions to math and sociology. But he is best remembered for his work in economics and socioeconomics– particularly in the study of income distribution, in the analysis of individuals’ choices, and in his studies of societies, in which he popularized the use of the term “elite” in social analysis.
He introduced the concept of Pareto efficiency (zero-sum situations in which no action or allocation is available that makes one individual better off without making another worse off) and helped develop the field of microeconomics. He was also the first to discover that income follows a Pareto distribution, which is a power law probability distribution. The Pareto principle ( the “80-20 rule”) was built on his observations that 80% of the wealth in Italy belonged to about 20% of the population.
source
Written by (Roughly) Daily
July 15, 2023 at 1:00 am
Posted in Uncategorized
Pradeep Mutalik unpacks the magic and math of how to win games when your opponent goes first…
Most games that pit two players or teams against each other require one of them to make the first play. This results in a built-in asymmetry, and the question arises: Should you go first or second?
Most people instinctively want to go first, and this intuition is usually borne out. In common two-player games, such as chess or tennis, it is a real, if modest, advantage to “win the toss” and go first. But sometimes it’s to your advantage to let your opponent make the first play.
In our February Insights puzzle, we presented four disparate situations in which, counterintuitively, the obligation to move is a serious and often decisive disadvantage. In chess, this is known as zugzwang — a German word meaning “move compulsion.”…
Four fascinating examples: “The Secrets of Zugzwang in Chess, Math and Pizzas,” from @PradeepMutalik.
* Fyodor Dostoyevsky, Notes from Underground
###
As we play to win, we might recall that it was on this date in 2011 that scientists involved in the OPERA experiment (a collaboration between CERN and the Laboratori Nazionali del Gran Sasso) mistakenly observed neutrinos appearing to travel faster than light. OPERA scientists announced the results with the stated intent of promoting further inquiry and debate. Later the team reported two flaws in their equipment set-up that had caused errors far outside their original confidence interval: a fiber optic cable attached improperly, which caused the apparently faster-than-light measurements, and a clock oscillator ticking too fast; accounting for these two sources of error eliminated the faster-than-light results. But even before the sources of the error were discovered, the result was considered anomalous because speeds higher than that of light in a vacuum are generally thought to violate special relativity, a cornerstone of the modern understanding of physics for over a century.
source
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September 22, 2022 at 1:00 am
Posted in Uncategorized
## “Then we got into a labyrinth, and when we thought we were at the end, came out again at the beginning, having still to seek as much as ever.”*…
On the heels of Wordle‘s extraordinary success, there have been a rash of variations: e.g., Crosswordle, Absurdle, Quordle, even the NSFW Lewdle.
Now for the National Gallery of Art, another nifty puzzle: Artle.
Enjoy!
* Plato, Euthydemus
###
As we play, we might recall that it was on this date in 1934 that Mandrake the Magician first appeared in newspapers. A comic strip, it was created by Lee Falk (before he created The Phantom)… and thus its crime-fighting, puzzle-solving hero is regarded by most historians of the form to have been America’s first comic superhero.
source
Written by (Roughly) Daily
June 11, 2022 at 1:00 am
Posted in Uncategorized
## “May all beings have happy minds”*…
But then it’s important to be careful as to how we look for that happiness…
– Games where players either remove pieces from a pile or add pieces to it, with the loser being the one who causes the heap to shake (similar to the modern game pick-up sticks)
– Games of throwing dice
– Ball games
– Guessing a friend’s thoughts
Just a few of the entries in “List of games that Buddha would not play,” from the T. W. Rhys Davids‘ translation of the Brahmajāla Sutta (though the list is duplicated in a number of other early Buddhist texts, including the Vinaya Pitaka).
(TotH to Scott Alexander; image above: source)
* the Buddha
###
As we endeavor for enlightenment, we might recall that it was on this date in 2001 that Wikipedia was born. A free online encyclopedia that is collaboratively edited by volunteers, it has grown to be the world’s largest reference website, attracting 1.7 billion unique-device visitors monthly as of November 2021. As of January 9, 2022, it has more than fifty-eight million articles in more than 300 languages, including 6,436,030 articles in English (serving 42,848,899 active users of English Wikipedia), with 118,074 active contributors in the past month.
source
Written by (Roughly) Daily
January 15, 2022 at 1:00 am
Posted in Uncategorized
## “He’s a pinball wizard”*…
Mark Frauenfelder talks with Tanner Petch, the creator of play-able pieces of art: an arcade full of handmade pinball machines…
Sinkhole is a backwards game that borrows from the aesthetic of early pinball, particularly “wood rail” games from pre-1960s. The fact that it tilts away from you changes your experience a lot more than you’d expect and came from trying to question what were some of the very core aspects of pinball that could be tinkered with. In addition to the wooden components, the art style, playfield design, and overall theme were inspired by the esoteric nature of early games (at least compared to what we expect today)…
Prometheus was the first game I made and is based on the part of the myth where an eagle eats Prometheus’ liver every day after it regenerates. In the game, the player is the eagle, and the only objective is to hit four drop targets which represent four bites of the liver. You do this as many times as you want to, or until you lose. Rather than an individual score, the display shows the cumulative number of livers eaten as long as the machine has existed…
* The Who, “Pinball Wizard,” Tommy
###
As we finesse the flippers, we might recall that it was on this date in 1972 that Hewlett-Packard introduced the first handheld scientific calculator, the HP-35, a calculator with trigonometric and exponential functions. The model name was a reflection of the fact that the unit had 35 keys.
It became known as “the electronic slide rule”– a device that it (and its successors, from both HP and TI) effectively replaced.
source
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January 4, 2022 at 7:23 am
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https://www.mrexcel.com/board/threads/40-iif-statements-efficiently.77504/ | 1,642,690,389,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301863.7/warc/CC-MAIN-20220120130236-20220120160236-00375.warc.gz | 952,433,547 | 20,439 | 40 IIf statements efficiently??
abrown
Board Regular
I have a table with item # and retail price. The retail prices range from \$0.00 to \$400.00
I would like to create a third column called Retail range. Retail range is in increments of \$5. For example:
0-4.99
5-9.99
10-14.99
15-19.99
etc all the way to 400
Can someone suggest the best way to accomplish this task??? IIf's, queries, combo boxes???
Excel Facts
Which lookup functions find a value equal or greater than the lookup value?
MATCH uses -1 to find larger value (lookup table must be sorted ZA). XLOOKUP uses 1 to find values greater and does not need to be sorted.
ChrisUK
Well-known Member
Try a select case statement if this is indeed what you realy want to do.
It is inefficient to store data more than one in a DB . If you have the cost you don't need to store a range separately in a column of the same table all you ned to do is use a range in the query which you could construct using the select case statement
HTH
SydneyGeek
MrExcel MVP
Which version of Access do you use? In XP and higher (not sure about 2000), you can use a Pivot Table view to group the items by increments of (or whatever interval you like).
SydneyGeek
MrExcel MVP
You could enter this expression in a query:
Price Band: (IIf(Round([Retail Price]/5,1)-Int([Retail Price]/5)<0.5,Round([Retail Price]/5,0)+1,Round([Retail Price]/5,0)))*5
This will give the upper limit of each price band. To group on some other interval, change the 5's to your desired interval.
Denis
ChrisUK
Well-known Member
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Thats a really nice feature, i didn't realise you could do it. Thanks
abrown
Board Regular
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Thanks for all of the advice. We tried the IIF in the query that gives you an upper limit and it worked great. BUT the Price column contains some text values so there are #Error messages in the column.
Also what we are really looking for the formula to give us a range and not just the upper limit. So if the price is 14.00 it would give us "10-15" in the price range column.
Is there any way to do this in one formula?
PaulF
New Member
Did you check out the partition function? Use the link above. It is easy and will, I believe, do exactly as you want.
If this gives you a problem post back.
PaulF
New Member
OK: In a new field- Retail_Range:Partition([Retail Price],0,400,5).
This should work.
abrown
Board Regular
The partition formula worked great. Thanks.
The other challenge is that when the price is 5.99 the formula returns the range 6:10. I want 5.99 to fall into 5:9.99 instead.
How can I make the range recognize 2 decimal points? I would like the ranges to layout as follows:
0-4.99
5-9.99
10-14.99
15-19.99
etc all the way to 400
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## Problems with GPS and position and direction distributions?
Keywords: GPS, position and direction distributions, biasing, usr-defined histograms
Forum: Particles
Date: 17 Oct, 2007
From: Shahrokh <Shahrokh>
```Hi, I have some questions about to define position and angular distributions by using GPS, as follows: When I execute my application, the particles traverse a plane. For each particle traversing this plane the position (X and Y coordinates), direction ( X, Y and Z components of momentum direction ) and energy are stored in six 1-D histograms by using AIDA that these horizontal axes are the position coordinates, the components of momentum direction and energy, respectively in which the particle incident upon the plane. Afterwards, I want to define these distributions (positions, directions and energy) by using GPS commands in user-defined histograms or biasing and save these commands in a macro file and then I execute again from this planar source by having these distributions in this macro file. In the following, I divide my works in three steps, the definition of energy, position and angular distributions from these six 1-D histograms by using GPS commands: --------------------------- 1- Energy distribution: I can initialize this energy distribution (1-D histogram) by using the commands following: /gps/ene/type User /gps/hist/type energy /gps/hist/point ... ... /gps/hist/point is followed by its two arguments - the upper boundary of the bin and weight of the bin. I think I do it correctly. In your opinion, is it correct? --------------------------- 2- Direction distribution: For the initialization of the direction distribution by using GPS, firstly I must get angular distribution (Phi and Theta) from momentum direction distribution. For this purpose, after getting X, Y and Z components of momentum direction, I can calculate Phi and Theta angles with the following formulas: Phi = Arctan(Py/Px) Theta = Pi - Arccos(Pz) After getting two 1-D histograms from Phi and Theta distributions, I can use the following commands: For Theta: /gps/ang/type user /gps/hist/type theta /gps/hist/point ... ... ... For Phi: /gps/ang/type user /gps/hist/type phi /gps/hist/point ... ... ... I think I do it correctly. In your opinion, is it correct? --------------------------- 3- Position distribution: The position distribution in this plane is that many particles emanate from a region in the rectangular shape and a few particles emanate from the out of this region. At now, I want to define a planar source by using this position distribution, but I do not know how to do it. I thought that I could do it by using the commands as follows: /gps/pos/type user /gps/hist/type biasx /gps/hist/point ... ... ... Unfortunately, I did not see the command as similar to /gps/pos/type user. Although I do not exactly understand the concept of biasing. How can I do it? Please tell me the role of biasing. In the other hands, is it correct that I bias distributions by using two 1-D histograms of X and Y components, synchronously? Does it superimpose into the rectangular region? Please guide me, Shahrokh. ```
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1 Re: Problems with GPS and position and direction distributions? (Shahrokh - 20 Oct, 2007)
3 Re: Problems with GPS and position and direction distributions? (Fan Lei - 02 Nov, 2007)
to: "Problems with GPS and position and direction distributions?"
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Showing results 1 to 25 of 1000 Search took 0.71 seconds. Search: Posts Made By: gd_barnes
Forum: Riesel Prime Search 2020-10-22, 23:26 Replies: 61 Views: 9,302 Posted By gd_barnes Double checking for k<20 to n=1M is complete. ... Double checking for k<20 to n=1M is complete. No problems found. The links in post #57 are updated. Continuing. Future statuses will be reported with the completion of each 20k range.
Forum: Aliquot Sequences 2020-10-22, 22:47 Replies: 3,343 Views: 261,080 Posted By gd_barnes update 2704542 2704604 2727570 2746900 2748546... update 2704542 2704604 2727570 2746900 2748546 2806860 2835336 update 2854464 2857074 2864896 2871232 2901432 2923764 2941392 2983110
Forum: Conjectures 'R Us 2020-10-21, 19:23 Replies: 388 Views: 41,416 Posted By gd_barnes Will you be testing n>1M? If not I will release... Will you be testing n>1M? If not I will release the base.
Forum: Conjectures 'R Us 2020-10-21, 19:20 Replies: 1,049 Views: 68,597 Posted By gd_barnes Testing the remaining k to n=100K for R3 will be... Testing the remaining k to n=100K for R3 will be one of the goals for 2021. R3 is more interesting than other small-conjectured bases because it is so prime and it is the second smallest base.
Forum: Conjectures 'R Us 2020-10-20, 02:53 Replies: 10 Views: 1,491 Posted By gd_barnes Info for CRUS folks: Sierp base 2 k=90646 and... Info for CRUS folks: Sierp base 2 k=90646 and 101746 have been completed to n=10M and are being released. Stream will send me the residues in a few days. When other k's in our Riesel/Sierp base...
Forum: Conjectures 'R Us 2020-10-19, 08:31 Replies: 2,180 Views: 147,811 Posted By gd_barnes S376 is complete to n=300K; no primes were found... S376 is complete to n=300K; no primes were found for n=250K-300K; 2 k's still remain; base released.
Forum: Conjectures 'R Us 2020-10-18, 02:22 Replies: 161 Views: 35,188 Posted By gd_barnes Pepi, I see that most of the files for bases... Pepi, I see that most of the files for bases in the 1st post have been sent to Yoyo for sieving. Can you go ahead and also send sieve files for R887, S368, and S994 to Yoyo? The target is to...
Forum: Conjectures 'R Us 2020-10-17, 23:25 Replies: 3,740 Views: 231,273 Posted By gd_barnes I did not get results for this one. I did not get results for this one.
Forum: Conjectures 'R Us 2020-10-15, 08:24 Replies: 2,180 Views: 147,811 Posted By gd_barnes I hope you're planning on adding a whole lot of... I hope you're planning on adding a whole lot of resources for n=2500-10K. :-) n=2500-10K is 3 times the n-range of n=1-2500. A test at half the n-range for n=2500-10K (n=6250) would take 25...
Forum: Conjectures 'R Us 2020-10-14, 05:43 Replies: 2,180 Views: 147,811 Posted By gd_barnes S354 is complete to n=300K; no primes were found... S354 is complete to n=300K; no primes were found for n=200K-300K; 3 k's still remain; base released.
Forum: Riesel Prime Search 2020-10-10, 23:40 Replies: 61 Views: 9,302 Posted By gd_barnes Good to know. Thanks MooMoo! :smile: Good to know. Thanks MooMoo! :smile:
Forum: Riesel Prime Search 2020-10-10, 05:54 Replies: 61 Views: 9,302 Posted By gd_barnes I have completed double checking k<10 (not k=1)... I have completed double checking k<10 (not k=1) to n=1M. No problems were found. Below are links to the primes and results. I will continue to add files to the links as I progress upwards. ...
Forum: Conjectures 'R Us 2020-10-10, 04:48 Replies: 2,180 Views: 147,811 Posted By gd_barnes Several CPU years worth of work. I'm sure... Several CPU years worth of work. I'm sure you're aware to remove k's that are squares. After removing k's with trivial factors there are 953 squared k's that need to be removed. There is a...
Forum: Conjectures 'R Us 2020-10-08, 07:13 Replies: 2,180 Views: 147,811 Posted By gd_barnes S394 is complete to n=300K; no primes were found... S394 is complete to n=300K; no primes were found for n=200K-300K; 2 k's still remain; base released.
Forum: Conjectures 'R Us 2020-10-01, 01:35 Replies: 2,180 Views: 147,811 Posted By gd_barnes Reserving S376, S444 and S447 to n=300K. ... Reserving S376, S444 and S447 to n=300K. A few small-range ones...n=250K-300K.
Forum: Conjectures 'R Us 2020-10-01, 01:19 Replies: 2,180 Views: 147,811 Posted By gd_barnes Reserving S354 to n=300K. Reserving S354 to n=300K.
Forum: Conjectures 'R Us 2020-10-01, 01:13 Replies: 3,740 Views: 231,273 Posted By gd_barnes R926 is complete to n=200K; 1 prime was found for... R926 is complete to n=200K; 1 prime was found for n=100K-200K shown below; 6 k's remain; base released. Prime: 70*926^131099-1
Forum: Conjectures 'R Us 2020-09-30, 23:05 Replies: 2,180 Views: 147,811 Posted By gd_barnes A big congrats to BOINC on the proof! :smile: A big congrats to BOINC on the proof! :smile:
Forum: Riesel Prime Search 2020-09-27, 23:53 Replies: 61 Views: 9,302 Posted By gd_barnes Reviving a very old thread... I am starting... Reviving a very old thread... I am starting a double check effort for all k<300 for n<=1M using the well sieved files provided here. I know that most k were at least double checked to n=260K...
Forum: Conjectures 'R Us 2020-09-23, 08:59 Replies: 4 Sticky: 2020 project goals Views: 3,789 Posted By gd_barnes Goal #4 has been completed! :smile: Goal #4 has been completed! :smile:
Forum: Conjectures 'R Us 2020-09-23, 08:57 Replies: 3,740 Views: 231,273 Posted By gd_barnes S612 is complete to n=10K; 1709 primes were found... S612 is complete to n=10K; 1709 primes were found for n=2.5K-10K; 2152 k's remain; base released. All bases with < 4000 k's remaining at n=2500 have been completed to n=10K! :-)
Forum: Conjectures 'R Us 2020-09-21, 20:07 Replies: 2,180 Views: 147,811 Posted By gd_barnes Reserving S394 to n=300K. Reserving S394 to n=300K.
Forum: Conjectures 'R Us 2020-09-21, 08:52 Replies: 868 Sticky: Report top-5000 primes here Views: 69,776 Posted By gd_barnes 80*394^298731-1 is prime!! 775358 digits ... 80*394^298731-1 is prime!! 775358 digits 2 top-5000 primes for R394 in the n=200K-300K range! Base R394 is proven!! My first proof in years!
Forum: Conjectures 'R Us 2020-09-20, 08:55 Replies: 1,414 Views: 107,819 Posted By gd_barnes There are many bases with 2 k's remaining at... There are many bases with 2 k's remaining at n=300K or 400K. Why do you choose to take two k's only out of a base with so many k's remaining? You say you will search them until you find a prime. ...
Forum: Conjectures 'R Us 2020-09-16, 05:55 Replies: 868 Sticky: Report top-5000 primes here Views: 69,776 Posted By gd_barnes 86*394^220461-1 is prime! R394 is now a... 86*394^220461-1 is prime! R394 is now a 1ker. :smile:
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## Thursday, August 16, 2007
### Promiscuity 101
A recent article in the New York Times has brought to the fore a question I'm sure has been bugging many readers of this blog: if men are supposed to be promiscuous and women are supposed to be shy, who, exactly, are the men being promiscuous with?
First, the facts, such as they appear to be: the results of survey after survey are remarkably consistent in indicating that men have more sexual partners, on average, than women. According to a recent study conducted in Britain, the figures for the median are 12.7 and 6.5 respectively (interestingly, the corresponding numbers in the United States are 7 and 4, what's up with that?).
But, argues math professor at the University of California at Berkeley David Gale, this cannot logically be. He even provided the Times with a mathematical proof that should settle the matter definitively :
“By way of dramatization, we change the context slightly and will prove what will be called the High School Prom Theorem. We suppose that on the day after the prom, each girl is asked to give the number of boys she danced with. These numbers are then added up giving a number G. The same information is then obtained from the boys, giving a number B.
Theorem: G=B
Proof: Both G and B are equal to C, the number of couples who danced together at the prom. Q.E.D.”
Well, not so fast. As a smart (female) friend of mine quickly pointed out, there is a huge hole in this proof: it assumes that people “dance” only with one person at a time, and that dancing is balanced throughout the evening, so that if a girl doesn't dance there will be a corresponding boy who doesn't either.
But one wonders in what kind of world professor Gale lives. Biologists have produced plenty of evidence that many animal species, including mammals and particularly primates, easily violate both assumptions. I remember an old Italian joke appropriate to the situation in question: “Every man has, on average, seven women in his life. If I catch that son of a bitch who's got fourteen...”
Of course, Gale has a legitimate point when he cautions social scientists against accepting self-reported figures at face value: critics of self-report surveys point out that the surveys themselves may reinforce the stereotype of promiscuous men and chaste women, which then influences the way respondents answer the question, which reinforces the stereotype, and so on in an increasingly vicious circle. Although difficult, however, it is possible to come up with independent means of checking the reliability of self-reports. One would have to conduct DNA tests in random samples of the population using hospital birth records, comparing genomic markers of newborns with those of the (alleged) fathers. The relatively hard science of molecular biology could then come to the aid of the relatively soft one of sociology. We might finally be able to see whether Woody Allen was right when he said “Sex without love is a meaningless experience, but as far as meaningless experiences go its pretty damn good.”
1. I recall a report of an unpublished study that showed, just on A-B-O blood types, that a startlingly high percentage of children could not have been the offspring of the reported father.
To put a rather cynical viewpoint on it, women could (and sometimes do) marry for resources, and mate for good genes.
2. "One would have to conduct DNA tests in random samples of the population using hospital birth records, comparing genomic markers of newborns with those of the (alleged) fathers."
In line with John's comment, the unspoken premise of the above extract is that all promiscuity results in live births, which is obviously not the case.
3. T "In line with John's comment, the unspoken premise of the above extract is that all promiscuity results in live births, which is obviously not the case."
But what does science actually say? I think some men claim to care what science says, but they don't really care at all.
More and more we are going to realize that the "real world" is not so much what happens between people. The real world(in terms of our continued existence) is what happens between viruses and or bacteria. When people choose to be promiscuous, live births could be the very least of the worries, all the while it is viruses that are exchanging the truly critical(to the rest of your life kinda) information.
A friend of mine worked as a supervisor in the dept at UNM where the vaccine for the HPV (and HIV) was developed. But according to my obgyn that particular vaccine for HPV is not to be thought of as a "cure" as it is being purported to be. It actually only addresses about four strains of HPV, and there are a lot more "out there" than four. As a matter of fact, he also said that over 60% of men and women are already carrying some form of HPV. I think that medical science is really beginning to understand that it is this type of virus which often leaves the door wide open for cancers in the reproductive organs.
Is that what we all wanted and asked for?
Well, yeah, sort of.
Prophylactic and birth control promotions are at least partially to blame for the rise in these types of deadly diseases in women, and even to a certain extent in men. Women should NOT be at all fooled by the propaganda on this vaccine. Babies most definitely are not the only price that women will pay when not opting to protect themselves.
ever the realist..
cal
4. T "In line with John's comment, the unspoken premise of the above extract is that all promiscuity results in live births, which is obviously not the case."
But what does science actually say? I think some men claim to care what science says, but they don't really care at all.
More and more we are going to realize that the "real world" is not so much what happens between people. The real world(in terms of our continued existence) is what happens between viruses and or bacteria. When people choose to be promiscuous, live births could be the very least of the worries, all the while it is viruses that are exchanging the truly critical(to the rest of your life kinda) information.
A friend of mine worked as a supervisor in the dept at UNM where the vaccine for the HPV (and HIV) was developed. But according to my obgyn that particular vaccine for HPV is not to be thought of as a "cure" as it is being purported to be. It actually only addresses about four strains of HPV, and there are a lot more "out there" than four. As a matter of fact, he also said that over 60% of men and women are already carrying some form of HPV. I think that medical science is really beginning to understand that it is this type of virus which often leaves the door wide open for cancers in the reproductive organs.
Is that what we all wanted and asked for?
Well, yeah, sort of.
Prophylactic and birth control promotions are at least partially to blame for the rise in these types of deadly diseases in women, and even to a certain extent in men. Women should NOT be at all fooled by the propaganda on this vaccine. Babies most definitely are not the only price that women will pay when not opting to protect themselves.
ever the realist..
cal
5. Hmmm... I don't think I get the "impossibility" in those numbers at all, so either I'm drooling here (likely) or Dr. Gale is kinda crazy. Well, given he's professor emeritus of math and I am a biologist who can barely count...
But here goes my "reasoning" anyway for you guys to criticize -- maybe it's the same thing Massimo's friend said, just in my verborrhagic style. What is the median? It is the value which is exactly in the middle of all other, i.e. 50% of women have 4 or fewer partners and 50% of women have 4 or more. The median is calculated putting all the number in order and then picking the one in the middle. So, consider the sequence:
1 3 4 5 7 23 98
The median here is 5, because that's the value in the middle (the mean, or average, would be the sum of all those, divided by seven, a.k.a. 20.14)
Where am I going with all this is: the median is as informative as the distribution is relatively uniform, without crazy outliers. So let's (hypothetically) say that 30% of women have 30 partners over their lifetime, while the remaining 70% have the proverbial 4. The median will still be 4, of course. Even if nobody is "dancing" with more than one person at a time. :-)
Where did I go wrong? Did I?
I'd rather be improving my statistics for when those damn survey people show up, actually... :O)
6. Second post (and not a repeat!): the infidelity data John and Thumpalumpacus hinted at. Yes, that sure would be a minimal estimate (given the obvious fact that not all infidelity results in a pregnancy).
An article discussing why women cheat, but the following passage is of special interest:
The first reliable estimate concerning infidelity was made in 1953 by renowned sex researcher Alfred Kinsey, who in his landmark study found that 50 percent of husbands and 26 percent of wives surveyed had cheated by age 40. However, in recent years, women have been catching up to men. A 1997 Ball State University study suggests that young women, those under age 40, are just as likely to commit adultery as men their age. Among older couples, the stereotype of men being more likely to stray holds true.
So, either times'a'changing or women are speaking more the truth or, probably, both.
Another interesting read is this first chapter of a book called "Female Infidelity and Paternal Uncertainty", subtitled "Evolutionary Perspectives on Male Anti-Cuckoldry Tactics"
This is a more standard scholarly thing, with references and all that. They cite the American Association of Blood Banks data that analyzed 280,000 paternity tests (in 1999) and came up with 30% of children fathered by what they call "extra-pair copulations". Gotta love that technical jargon. A poor chap in Texas for example went from having 4 kids to having just one, in the drop of a hat. I mean, test tube.
In spite of this big, impressive 30% number (which could be an underestimate, if you consider also that there are cases where the blood group of the cheated "father" and the biological one will be the same, or not mutually exclusive), they point out that there are estimates of female infidelity ranging from 1% to 30%, with the best estimate at 10%.
Have fun with those texts and numbers, y'all. Just don't get too paranoid, OK? :-)
7. Median is not Mean.
8. "critics of self-report surveys point out that the surveys themselves may reinforce the stereotype of promiscuous men and chaste women, which then influences the way respondents answer the question, which reinforces the stereotype,.."
This I think is an important point. There are cultural pressures for men to exagerate the number of their partners, and for women to minimize the number. And even if the respondent knows that the survey is confidential or anonymous, these cultural pressures may end up influencing what the respondents actually believe about themselves.
Speaking of which. What was the name of that pro basketball player who claimed that he had relations with some ridiculous number of women? Doing the math it worked out to him having to have sex with 3-4 different women per day for 30 years or something like that.
So I rest my case! Men with just a fraction of the bravado and imagination of that basketball player would consistently skew the mean!
And Cal, how exactly is your comment really related to what we are discussing here? Only tangentially is it relevant. As usual you attempt to derail the discussion.
9. Simple explanation: men exaggerate, women understate.
10. @The Ridger, fdc:
Yeah, she is. She won't share.
---
Cal, you can feel guilty about recreational sex if you like, but leave the rest of us alone.
11. Following up with my last comment.
It was Wilt Chamberlain who claimed to have had 20,000 different sex partners throughout his life. (now thats a sports stat!).
http://en.wikipedia.org/wiki/Wilt_Chamberlain
He lived to the age of 63. Assuming that he began his sexual adventures at the age of 15, and continued until he neared death, he had 48 years of sexual activity, requiring him to have relations with 416 women per year, or on average 1.18 women per day. Busy man!
http://espn.go.com/nba/news/1999/1012/110836.html
Apparently Wilt had actually done the math and wrote about it in his auto-biography! And it would seem he actually believed it to be true as suggested by the espn article!
12. Oops correction. Wilt Chamberlain made the claim in 1991, which would give him 40 years of sexual activity, 500 women per year, and 1.42 women per day. I am sure you all are as facinated about this as I am! LOL
13. I think that women and men are definitely socialized in different ways. As the only female in a brood of 5 children, I was definitely constrained sexually--not even allowed to date until just before graduating from highschool. My brothers were freer in their social relations.
Standards of socialization for men and woman, however, are changing. In the past, women were socialized to 'act' shy, especially in response to sexual overtures from men. Sexual aggressiveness in females was considered a faux pas, unfeminine, even b--chy. Its all rather medieval, these social contracts and assumptions.
Now, more women project the same aggressiveness that men have laid claim to for years. I don't have any great edicts to levy against promiscuity. Yet, as a result of the tearing down of constraints, the leveling of high standards (sometimes impossible to maintain), consequences will arise--pregnancies, dieseases, single-parent households, and so forth.
This blog gives a biological edge to the topic, but maybe the social construction of modern society is at the crux of the issue. One can analyze the 'promiscuity' issue with respect to social constraints, various prejudices, stereotypes, religion, and more. A promiscuity explosion may be the result of the medieval code of threading one's animal desires through too rigid an iron grid of moral behavior? Sounds like a set up for a severe neurosis; and, possibly, more promiscuous behavior in the future.
14. sheldon, kimpat,
The topic is promiscuity. (& the science of) And just so you know, it always hurts women the most. so if you think it an effort to derail, that is your problem. In essence, you just don't care about what science really says unless it supports your interests
Not particularly respectable in my estimation.
Of course...
It is great to be loved. And it is especially good to be loved unconditionally even if a person makes an occasional errors along the way. BUT IT IS NOT good to be (so-called) loved by a partner who lets one get away with murder. The HIV rate in Africa proves that this is a lesson that certainly can be learned too late. Love should exist for helping people to have long and healthy lives, not sending them to an early grave.
And do i feel guilt about suggesting any of that? NOT ON YOUR LIFE
cal
15. Cal,
The topic may be "promiscuity", but not neccessarily the moral and health consequences of "promiscuity".
The topic is more specifically the alleged discrepancy between men and women's reported number of sex partners over the course of a lifetime.
I don't think anybody here would disagree that there are negative consequences to irresponsible sexual behaviour. But that is not the topic.
Many would however disagree with your idea of what contsitutes irresponsible sexual behaviour. Anything outside of one partner and marriage, I presume?
You want to derail the discussion by interjecting your religious right wing-nut opinions about HPV virus/vaccine, and birth control methods. This is off topic.
"And just so you know, it always hurts women the most. so if you think it an effort to derail, that is your problem. In essence, you just don't care about what science really says unless it supports your interests. Not particularly respectable in my estimation."
What are your talking about? You also seem to want to suggest and portray women as only victims of any sexual encounter outside of one partner and marriage. That is not the case.
Just so you know, since you are implying something, I have no "interests" in the matter. I have had zero promiscuity since meeting my spouse 8 years ago. Before that, I was serially monogamous, with only a few casual encounters in between. None of those women were passive victims of my sexual aggressions, but willing participants and beneficients.
16. Cal, Having read numerous articles about the HPV vaccine, I have not seen one that said it was a "cure", but rather a preventative measure. And isn't efficacy against four strains better than none, especially if the individual is going to be sexually active regardless of the existence of a vaccine or not.
And how can prophalaxis and promotions birth control, be at blame for the spead of disease among women (or men for that matter). Sounds like the Gospel according to George W. Bush. And look where that dogma as gotten us in other areas that he has spewed on about for the past seven years.
17. Oops correction. Wilt Chamberlain made the claim in 1991, which would give him 40 years of sexual activity, 500 women per year, and 1.42 women per day. I am sure you all are as facinated about this as I am! LOL
Did not Brigitte Bardot claim about a partner a day? I find that more fascinating than Wilt the Stilt. Heh heh.
Then there's David Bowie's wife, Angie (of the Rolling Stones song) who more realistically claimed 90-odd lovers.
18. Dennis,
It doesn't matter what you have read. You see, Cal has a friend who worked in a biology lab, so she clearly knows what she is talking about. Besides, if you give those girls that vaccine, then they will figure that with the reduced chances of cervical cancer they can equalize their number of lifetime sex partners to that of men. That would just be awful! You know, women enjoying a varied sex life free from the worries of that cancer! Civilization would collapse!
19. Very interesting post!! Here's something I put together about ten years ago, which seems germane to this topic:
Consider a village of 9 absolutely faithful couples. They've each had exactly 1.0 partners. Now, another couple moves in. He looks like Tom Cruise and seduces the other 9 guys' wives. (hey - it's just a thought experiment!) Now, considering all 10 guys, their total # of partners is 9x1 + 1x10, or 19, for an average of 1.9 partners each. Every woman has now had 2 partners except Mr. Cool's poor wife, who's had only one. This comes out to 9x2 + 1x1, or 19, giving a women's average of 1.9 as well - still the same as the men. Thus it seems that in any CLOSED group, following ANY kind of behavior always gives EXACTLY the same average number of partners for men as for women.
Mathematically, this is because, while the average for, say, men, is defined as the total # of their partners divided by the # of men (M), it is also equal to the # of unique pairings Np (e.g. Bob+Alice) that has ever happened, divided by the # of men. Assuming that a unique pairing involves one man and one woman, this means the "Np" used to calculate the men's average is the SAME "Np" used to calculate the women's average, so if there are the same number of men as women in the group, then their average numbers of partners MUST be identical. (Note that a threesome & up can be treated as two or more serial 'unique pairings', leading to the same result.) Further, if there are, say, more women than men in the group, then the ratio of the two gender averages must be the same as the ratio of the populations. If there are more women, then the men will average more partners, thusly:
Women's average X = Np / W
Men's average Y = Np / M
Ratio of men's average to women's average = Y/X or (Np/M) / (Np/W) or just W/M.
Therefore, surveys indicating that men have nearly twice as many partners as women do are mathematically proven to be impossible in a closed population.
Now of course no sizable real population is sexually closed, so how does this change things? First, realize there is indeed ONE very large closed population, namely planet Earth, and the above equation MUST hold for that. Within that, it can be proven with a little more math that although there can exist a non-closed subgroup where men do indeed have twice as many partners as women, this forces the existence of another non-closed subgroup wherein women have twice as many as men, to keep the global average at W/M as above. (obviously, this reasoning holds only for heterosexual activities)
-KWFco
20. I wonder how many female prostitutes figure in the questionnaires used in these surveys? One such prostitute could equal the lifetime sexual history of several men in a single evening.
21. The inscriptons on the adjacent tombstones of Alaska frontier hero Frank Reese and a lady of the night known only as Annie K.
"FRANK REESE
HE GAVE HIS LIFE FOR THE HONOR OF THIS TOWN.
ANNIE K.
SHE GAVE HER HONOR FOR THE LIFE OF THIS TOWN."
22. "None of those women were passive victims of my sexual aggressions, but willing participants and beneficients. "
Or, in my case, donors.
23. Massimo, you got the Woody Allen quote slightly wrong.
The actual quote is "Sex without love is an empty experience, but as empty experiences go, it's one of the best."
24. Hi Massimo,
Couple of quick notes -- Gale claims that he was / is well aware that medians are not means, but that since the survey data was given with percentages who answered that they had a # of partners within a particular range, you can generate a high and low bound mean from the data -- so "j" is right of course, but there was more data available than just the median. And given those survey results, the mean number of sexual partners had to be different.
Now of course, the proof sucks -- but as "KWFco" notes, it *would* be correct if:
1) it was specified that the total number of boys and the total number of girls at the dance are equal
and
2) B and G are not the total number of dances, but are generated by dividing the total number of dances by the number of boys and girls.
So it doesn't matter whether 1 boy dances with all 6 girls and the other 5 boys just sit there, or vice-versa. In either case, the average (mean) number of dance partners remains the same for both sexes. Nor does it matter if they dance in groups, etc -- again, the average will stay the same.
So for mean number of sexual partners, the same should be true -- as it must be in all organisms with roughly equal sex ratios (in organisms with unequal sex ratios, the average number of sexual partners in the sexes must vary by the skewing of the ratio). It doesn't matter if one bull elephant seal has sex with dozens of females and other males have sex with almost none, whereas the females are all equally choosy and all have sex with the same smallish number of males -- the averages must be the same, if the number of males and females is the same.
Re: the survey data. Gale recognized as well that what the surveys might, in principle, be missing is the rare very promiscuous woman (prostitutes are the usual suspects here, as Suffenus suggests). But Gale suggests that this is unlikely for a number of reasons (mainly that it is hard to get the numbers to work out in a way that makes sense of the survey data).
So the most likely explanation remains that men "round up" and women "round down" in the surveys (possibly not even deliberately), or count differently -- perhaps, pace Clinton, women have a higher standard for what it takes for someone to count as a sexual partner ;)
What other species show is not that the average number of sexual partners can be different for males and females, but rather that you can't move so quickly from the fact of equal average number of partners to views about the *preferred* number of partners. Male elephant seals ;) no doubt prefer to have lots and lots of mates, it's just that most of them are disappointed in that preference. Female elephant seals may prefer rather fewer mates, and manage to work with that preference. The males and females end up with the same *average* number of mates, but do so in different ways and through the expression of very different preferences.
And again, this points out that paternity doesn't matter to this argument, either. It wouldn't matter if one man was the father of every child born in a particular year -- the average of number of matings resulting in a conception would still be the same for both men and women, if the number of men and women were equal!
best,
jk
25. One more note, re: KWFco note re: closed populations. If you *really* want to defend the different numbers, I think you might be able to by appealing to the (in my view pretty sloppy, but whatever) research suggesting that older men have more sex with younger women than older women do with younger men. The basic idea would be that the population isn't closed, as women are removed from the partnering population sooner than are men, and the population size keeps expanding so ever more young women are added continuously. So the idea here would be to imagine that each woman has sex with e.g., people in her own age group at a 1-1 ratio, but with older men at say a 1:2 ratio, because there are so many more young women than old men, because the population is expanding (that is, if say there are 2 young women for every old man, and every old man has sex with one young woman, only 1/2 of the young women will have had sex with an old man, etc.).
This, I think, probably doesn't really account for the survey data, but if you like that kind of thing, could be made to work...
26. Jonathan,
good points, and thanks for the post. However: the mean and the median are the same only if the distribution is normal (well, and for other unlikely distributions, like a symmetrical bimodal). It would be interesting to see whether that is indeed the case.
Also, I wonder if the relevant statistic here isn't the mode (the most frequent value), rather than the simple arithmetic mean (the average) or the median (the middle of the distribution).
As the joke goes, if I got two chicken and you none, on average we both had a chicken each... :)
27. Means and medians, again.
Massimo, you note that the mean and the median are only the same if the distribution is normal, and that it probably isn't for # of sexual partners. That's right, and Gale (claims that he) knows that.
What Gale got from the original survey data were the percentages of respondents claiming to have had a number of partners within a particular range, e.g., % claiming between 0-1 partners, % claiming between 2-4 partners, between 5-8, etc. So if you know that 20% of the male population (say) claims to have between 9-12 partners, you can estimate the range of possible means first by assuming 9, then by assuming 12, and similarly for all the ranges. That is what Gale did. Of course, you don't get *the same* difference between sexes as you do for the median -- rather, you get a *range* of possible differences, and the minimum difference between the means is about 40% (again, according to Gale).
If what you are interested in is, for example, parenthood over human evolution, than the historical modes would be more useful, yes. If what you are interested in are tendencies or preferences, than none of these is particularly useful, and what you need are data on preferences. If you are interested in how reliable survey data is, means would be best here ;)
jk
28. Problem with your Prom analysis. You're assuming the sample size = the population size.
Fact is, there are a few women out there who sleep with a LOT of guys. If these women are missed in the study, it will skew the average of all women's sexual partners down. And honestly, these are not the type of women who participate in those types of studies.
This is why you have differing numbers.
29. No, I don't think it's that simple. The authors of these kinds of surveys are aware of that problem, but it would take an unlikely high number of "outliers" to significantly skew the results. | 6,403 | 28,814 | {"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} | 2.96875 | 3 | CC-MAIN-2018-05 | longest | en | 0.950941 |
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# Sampling Error Research
## Contents
Types of Simple Random SampleWith replacementWithout replacement 22. With replacementThe unit once selected has the chancefor again selectionWithout replacementThe unit once selected can not beselected again 23. Rosling H. Stratified Random Sampling 29. SELECTION ERROR—This occurs when respondents self select their participation in the study – only those that are interested respond. weblink
My specialties are statistics and operations research. Burgess T. In 1936, many Americans did not own cars or telephones and those who did were largely Republicans. Fitch M.
## How To Reduce Sampling Error
Steps in Sampling ProcessDefine the populationIdentify the sampling frameSelect a sampling design orprocedureDetermine the sample sizeDraw the sample 15. An estimate of a quantity of interest, such as an average or percentage, will generally be subject to sample-to-sample variation.[1] These variations in the possible sample values of a statistic can Harraway J.
These are often expressed in terms of its standard error. Burns, N & Grove, S.K. (2009). Non-sampling error is a catch-all term for the deviations from the true value that are not a function of the sample chosen, including various systematic errors and any random errors that Random Sampling Error Definition See next slide for an example) 66.
Non Probability Sampling Involves non random methods in selection ofsampleAll have not equal chance of being selectedSelection depend upon situationConsiderably less expensiveConvenientSample chosen in many ways 45. Non Sampling Error If you continue browsing the site, you agree to the use of cookies on this website. As a method for gathering data within the field of statistics, random sampling is recognized as clearly distinct from the causal process that one is trying to measure. PopulationThe entire group of people of interest from whom theresearcher needs to obtain informationElement (sampling unit)One unit from a populationSamplingThe selection of a subset of the population through varioussampling techniquesSampling FrameListing
Over the next few articles, we will discuss the several different forms of bias and how to avoid them in your surveys. Sources Of Sampling Error By using this site, you agree to the Terms of Use and Privacy Policy. Sytsma S. Gilmore L.
## Non Sampling Error
Others have moved or are away from home for the period of the survey. https://en.wikipedia.org/wiki/Sampling_error Not-at-home respondents are typically younger with no small children, and have a much higher proportion of working wives than households with someone at home. How To Reduce Sampling Error Arnold Dr. Sampling Error Formula Gibbs G.
Reply RickPenwarden says: April 2, 2014 at 4:44 pm Hey meme! http://ldkoffice.com/sampling-error/sampling-error-marketing-research.html Comments View the discussion thread. . Instead, results are skewed by customers who bought items online. Share Email Sampling Errors byNeeraj Kumar 1458views Errors in research byAbinesh Raja M 15708views Type i and type ii errors byp24ssp 7645views RESEARCH METHOD - SAMPLING byHafizah Hajimia 161803views Sampling Error And Nonsampling Error
This solution is to eliminate the concept of sample, and to test the entire population.In most cases this is not possible; consequently, what a researcher must to do is to minimize Sampling error From Wikipedia, the free encyclopedia Jump to: navigation, search In statistics, sampling error is incurred when the statistical characteristics of a population are estimated from a subset, or sample, Select another clipboard × Looks like you’ve clipped this slide to already. http://ldkoffice.com/sampling-error/sampling-error-in-research-methods.html It follows logic that if the sample is not representative of the entire population, the results from it will most likely differ from the results taken from the entire population. .
Mackrory R. Sampling Error Ppt Advantage Moderate cost; moderate usage External validity high; internal validityhigh; statistical estimation of error Simple to draw sample; easy to verify 39. Start clipping No thanks.
## adult population to gauge their entertainment preferences.
Since sampling is typically done to determine the characteristics of a whole population, the difference between the sample and population values is considered a sampling error.[1] Exact measurement of sampling error Back to Blog Subscribe for more of the greatest insights that matter most to you. Defining Error and Bias In survey research, error can be defined as any difference between the average values that were obtained through a study and the true average values of the this content According to a differing view, a potential example of a sampling error in evolution is genetic drift; a change is a population’s allele frequencies due to chance.
Added: 24/10/12) Search resources Advanced searchSimple search NZC Level 3 4 5 6 7 8 Achievement Standard 1.10 1.11 1.12 1.13 2.8 2.9 2.10 2.11 2.12 2.13 Random sampling (and sampling error) can only be used to gather information about a single defined point in time. Rather, you are trying to get information to project onto a larger population. Download Explorable Now!
Exploring the Importance of Event Feedback 3 Ways to Yield a Higher Survey Response Rate Closing the loop: It’s more than just a survey response Blog Home About Inquisium Categories Contributors Privacy policy About Wikipedia Disclaimers Contact Wikipedia Developers Cookie statement Mobile view Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. For this reason, it is important to understand common sampling errors so you can avoid them. Even if a sampling process has no non-sampling errors then estimates from different random samples (of the same size) will vary from sample to sample, and each estimate is likely to
References Sarndal, Swenson, and Wretman (1992), Model Assisted Survey Sampling, Springer-Verlag, ISBN 0-387-40620-4 Fritz Scheuren (2005). "What is a Margin of Error?", Chapter 10, in "What is a Survey?", American Statistical Measurement Error The question is unclear, ambiguous or difficult toanswer The list of possible answers suggested in the recordinginstrument is incomplete Requested information assumes a frameworkunfamiliar to the respondent The definitions Keep in mind that as the sample size increases, the standard deviation decreases.Imagine having only 10 subjects, with this very little sample size, the tendency of their results is to vary Binnie N.
Sampling process error occurs because researchers draw different subjects from the same population but still, the subjects have individual differences. Notify me of new posts via email. Woodward Jared Hockly Jeanette Saunders Cognition Education Ltd Johnnie Journal of Statistics Education K. Innovation Norway The Research Council of Norway Subscribe / Share Subscribe to our RSS Feed Like us on Facebook Follow us on Twitter Founder: Oskar Blakstad Blog Oskar Blakstad on Twitter
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rmsgrey
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Duck on a lake with unseen predator « on: May 6th, 2023, 3:01am » Quote Modify
In the classic duck on a lake problem, a duck is near the middle of a circular lake, with a fox prowling the shore. The fox can run four times as fast as the duck can swim, but if the duck can make it to shore, it can then take off and escape (note: this is not ornithologically accurate - ducks can generally take off from open water).
With both duck and fox able to see each other at all times, the duck can always escape before the fox can intercept it (while the fox can cover more than half the circumference of the lake in the time it takes the duck to swim the radius, so long as the duck is within a quarter radius of the center, it can manage a higher angular velocity, so can stay opposite the fox until it gets that far out, where it takes less time to swim 3/4 of the radius than the fox does to run half the circumference)
What happens when the duck cannot see the fox, but the fox can see the duck? Does either have a strategy that's guaranteed to always win? (the duck escaping, or the fox preventing the escape) If not, what's the best each can manage?
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rmsgrey
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Re: Duck on a lake with unseen predator « Reply #1 on: May 15th, 2023, 3:10am » Quote Modify
Here's what I've come up with:
The fox cannot guarantee to capture the duck. If the duck swims in circles around the center of the lake just inside the 1/4 radius line, then, no matter what the fox does, the duck will have a greater angular velocity, and, over a long enough period, will sweep through every possible angle from the fox. If the duck circles for long enough, then swims radially to shore at a random angle, there's a chance that the fox is more than 3 radians from the duck and so unable to beat it to its landing point.
Against this strategy, the fox can attempt to minimise its time in the dead zone, by running with the duck whenever it's in range, and against it whenever it's out of range.
The size of the dead zone depends linearly on how close to the r/4 line the duck is circling (zero at 1 - [pi]/4), while the time the fox can stay out of the dead zone is inversely proportional to how close to the r/4 line the duck circles, so the optimum for the duck under this strategy would be to circle at (5-[pi])r/8.
On the other hand, if the fox uses the strategy of heading toward the duck's projected landing point whenever it can beat the duck there, and moving to exit the dead zone as soon as possible on the assumption the duck follows its most obvious projected motion (by whatever method you use to project it), the duck can counter that extremely effectively by hanging out somewhere off center long enough for the fox to take up position directly outside the duck, giving the fox a known starting position that the duck can then escape from provided the fox follows the obvious strategy.
To counter that sort of bait-and-escape strategy, the fox can ignore the duck whenever it's inside the r/4 line, picking a position around the circumference randomly and waiting for the duck to emerge and then pursuing it.
If that is the optimal strategy, then the fox has a flat 3/[pi] chance of catching the duck.
Of course, the duck also has the "flip the table" option of staying in the pond forever.
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Forum software copyright © 2000-2004 Yet another Bulletin Board | 1,062 | 4,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-10 | latest | en | 0.874225 |
https://www.physicsforums.com/threads/particle-acceleration.189296/ | 1,508,747,601,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825812.89/warc/CC-MAIN-20171023073607-20171023093607-00571.warc.gz | 974,162,159 | 14,371 | # Particle Acceleration
1. Oct 5, 2007
### bdb1324
A particle's acceleration is described by the function a_x =(10 -t) m/s^2, where t is in s. Its initial conditions are x_0 =0 m and v_0x =0 m/s at t =0 s.
At what time is the velocity again zero?
I am having a hard time setting this problem up with where to begin if someone could help me with a formula that would be much appreciated.
2. Oct 5, 2007
### bdb1324
Ax = (10-t)m/s^2
Xi = 0 m
Vi = 0 m/s
t = 0 s
since acceleration is A = (Vf-Vi)/t
Last edited: Oct 5, 2007
3. Oct 5, 2007
### hotvette
Hint. You are given a = f(t). How can you get v from this? Think calculus.
4. Sep 20, 2010
### ghostanime2001
integration right? | 234 | 691 | {"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} | 3.21875 | 3 | CC-MAIN-2017-43 | latest | en | 0.902737 |
http://mathcenter.oxford.emory.edu/math117/excelAndPoissonAndHypergeometricDistributions/ | 1,506,255,493,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690016.68/warc/CC-MAIN-20170924115333-20170924135333-00710.warc.gz | 215,411,142 | 1,264 | # Excel Functions Related to Poisson & Hypergeometric Distributions
POISSON.DIST($x,\mu,FALSE$)
The Poisson probability of seeing $x$ occurrences in some well-defined interval when $\mu$ occurrences are expected (i.e., the mean number of occurrences is $\mu$). The last argument indicates that the probability returned should not be cumulative.
POISSON.DIST($x,\mu,TRUE$)
The Poisson probability of seeing $x$ or fewer occurrences in some well-defined interval when $\mu$ occurrences are expected (i.e., the mean number of occurrences is $\mu$). The last argument indicates that the probability is cumulative. That is to say, it gives the sum $P(0) + P(1) + \cdots + P(x)$.
HYPGEOM.DIST($x,k,m,N,FALSE$)
The Hypergeometric probability of seeing exactly $x$ white balls when drawing $k$ balls from an urn containing $N$ balls, $m$ of which are white. The last argument indicates that the probability returned should not be cumulative.
HYPGEOM.DIST($x,k,m,N,TRUE$)
The Hypergeometric probability of seeing exactly $x$ or fewer white balls when drawing $k$ balls from an urn containing $N$ balls, $m$ of which are white. The last argument indicates that the probability is cumulative. That is to say, it gives the sum $P(0) + P(1) + \cdots + P(x)$. | 321 | 1,253 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.125 | 3 | CC-MAIN-2017-39 | latest | en | 0.852926 |
https://physics.stackexchange.com/questions/263921/how-to-calculate-orbit-of-a-sphere-around-a-bigger-one-in-microgravity | 1,718,298,723,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861480.87/warc/CC-MAIN-20240613154645-20240613184645-00153.warc.gz | 408,664,509 | 39,668 | # How to calculate orbit of a sphere around a bigger one in microgravity?
I don't have any knowledge in physics, but in a world building related question, I need to know how can I calculate the orbit or the distance in each point of the orbit around a spherical object in space.
This would be the values of the orbiter 'planet' (A):
Radius: 1645.8294922995256
Diameter: 3291.658984599051
Circumference: 10341.051684139216
Circle area: 8509803.921575
Sphere surface area: 34039215.6863
Sphere volume: 18674248357.085728
And this, the values of the mass that the other would orbit around (B):
Radius: 179665.31384583088
Diameter: 359330.62769166176
Circumference: 1128870.4601659337
Circle area: 101409432758.5
Spere surface area: 405637731034
Sphere volume: 24293010084644784
So my question is, if I want to calculate the longest distance from A to B in the orbit, or the shortest, or basically any point on it, is there an easy formula to do so?
I've researched a little bit and learned about why planets orbit in microgravity. I sourced from a Quora answer:
The earth is constantly trying to fall into the sun, but it keeps missing. That is essentially what an orbit is. The sun exerts an attractive force on the earth, accelerating the earth directly towards the sun. This acceleration is constantly taking place. However, the earth also has some sideways momentum (perpendicular to the direction towards the sun). So as it falls towards the sun, it also moves to the side. As long as that sideways motion is enough to "side-step" the sun, the earth will orbit instead of crashing. You can see this more clearly in a more elliptical orbit:
Is it right to say so?
Also, if I had different values for A and B, how can I calculate the orbit and distances on them? Is there a specific formula?
Edit: Specifying how elliptical would the orbit be and the average distance: As said before it is more on World Building, but it's supposed to simulate Earth's condition but in a smaller area. Using a rule of three I've calculated the 'Sun' size and the orbiter planet area comes from a pixel measure standard (in a map). It should be somehow like Earth, but smaller, what I don't know is if it would be the same orbit or same speed since they are smaller. Also, I'm interested in knowing the factors and formula so I can calculate it myself. The average distance would be an Astronomical Unit reduced to the current scale, which I do not surely know how can I calculate it. The scale is relative to the planet data.
• given two planets there are an infinite number of possible orbits, you need to be more specific, such as the average distance between planets and the ellipticity of the orbit
– user65081
Commented Jun 21, 2016 at 19:20 | 682 | 2,844 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-26 | latest | en | 0.916227 |
http://abroadexport.com/walmart/882026786717aa43824a19-convert-roman-numeral-to-decimal-number-in-java | 1,669,927,292,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710869.86/warc/CC-MAIN-20221201185801-20221201215801-00691.warc.gz | 819,834 | 9,240 | Hence we have tried to do the same through Java code:-. each symbol belongs to a group of unit or magnitude (ones 10^1, hundreds 10^2, thousands 10^3) wich require to be in a specific sequence in order to form a valid roman numeral. Recently I decided to provide all the copyright dates on my sites in Roman numerals rather than decimal, just for something different. 1. The same logic is applied fur values between 10 and 100. The symbols used are I, V, X, C, D and M. Browse other questions tagged java beginner roman-numerals or ask your own question. From the 14th century on, Roman numerals began to be replaced in most contexts by the more convenient Hindu-Arabic numerals; however, this process was gradual, and the use of Roman numerals persists in some minor applications to this day.
100. Theory The numbers between 1 and 10 are composed by a combination between I, V and X. Java 2022-05-14 00:30:17 group all keys with same values in a hashmap java Java 2022-05-14 00:22:08 download csv file spring boot Java 2022-05-14 00:05:59 implementing euclid's extended algorithm Example : XCIX = 99 but my code prints 109. Take a roman number and save it in the array roman_Number. Write a program that converts a number entered in Roman numerals to decimal. 2. An object of type Roman should do the following: a. java converting roman-numerals. Create an instance of the StringBuilder Class. Things to note about roman numerals: When a symbol appears after a larger (or equal) symbol it is added. How to convert a roman number in decimal in Java. else subtract this value by adding . 1000>36 = yes, check with next roman numeral. Example: VI = V + I = 5 + 1 = 6. else subtract this value by adding the value of next symbol to the running total. This could be replaced with parallel arrays. Share. /* Java program for Conversion from Decimal to roman number */ public class Perform { // Display roman value of n public static void result (int n) { switch (n) { // Test Cases case 1: System.out.print ("I"); break; case 4: System.out.print ("IV"); break; case 5: System .
Search: C Program Char To Binary. I would appreciate critical points on everything, . The vocabulary of roman symbols ( I, V) can be translated to numbers (1, 5). But it doesn't work for some exceptional cases. Note that number 4 and 9 are obtained by using the subtractive notation. How do you convert a number into Roman numerals? Similarly, if larger letter appears first it gets added to the smaller number like CL=100+50=150. We can the separate symbol now convert each symbol of Roman Numerals into the value it represents.
Search: C Program Char To Binary. The Overflow Blog WSO2 joins Collectives on Stack Overflow. C#: IEnumerable, yield return, and . 1,000. If current value of symbol is greater than or equal to the value of next symbol, then add this value to the running total. Example: LXX = L + X + X = 50 + 10 + 10 = 70. I have managed to get my code to convert most Roman numerals to its appropriate decimal value. Java answers related to "convert roman numerals to decimal java" Java int to roman; roman numbers equivalent java; Java program to print hexadecimal to decimal number conversion The use of Roman numerals continued long after the decline of the Roman Empire. else subtract this value by adding the value of next symbol to the running total.
Example 1: Input:s = VI Output: 6 Example 2: Input: XOutput: 10XL is a Roman symbol which represents 40 Solution Approach. Convert each symbol of Roman Numerals into the value it represents. Using the while statement, iterate through each digit of the input number.
Here is my code. If possible to convert then: We will append that roman numeral to the new initialized string We will subtract the highest number from the given number We'll follow this checking approach till the smallest possible number which can be converted into Roman numerals i.e 1 = I Throw and catch exceptions. The algorithm for converting from Roman to Decimal is adapted from this post. But it doesn't work for some exceptional cases. Algorithm to convert Roman Numerals to Integer Number: Split the Roman Numeral string into Roman Symbols (character). The StringBuilder will be the corresponding roman numeral. Java string split with "." (dot) You need to escape the dot if you want to split on a literal dot: String extensionRemoved = filename.split ("\\.") [0]; Otherwise you are splitting on the regex ., which means "any character". Now compare the integer with roman numerals and do the following: Your program should consist of a class, say, Roman.
First, split the Roman numeral string into roman symbols. If the smaller letter is written before the larger letter its get subtracted like IV = 5-1 =4. Convert each symbol of Roman Numerals into the value it represents. The write function is invoked to display the character on the screen Create a program that would convert a decimal number to binary, octal or hexadecimal counterpart Normally, when we work with Numbers, we use primitive data types such as int, short, long, float and double, etc All Unicode characters can be represented soly by UTF-8 encoded ones and zeros (binary numbers) You can also . My program makes a conversion between Roman and decimal numbers. 900>36 = yes, check with next roman numeral. For Example: Roman Number III has Integral value is 3 adding value of three I's together (1+1+1=3). Take symbol one by one from starting from index 0: If current value of symbol is greater than or equal to the value of next symbol, then add this value to the running total. Basic rules for writing roman numbers are: 1. else subtract this value by adding the value of next symbol to the running total. Example: IV = V I = 5 1 = 4. Take symbol one by one from starting from index 0: If current value of symbol is greater than or equal to the value of next symbol, then add this value to the running total. Compare the integer with roman values as follows. User interface The decimal (Arabic) number 2063439 converted to a Roman number is (M)(M)(L)(X)MMMCDXXXIX. if it has a larger roman numeral // equivalent than number, then the two letters are counted together as // a roman numeral with value (nextnumber - number). Example Suppose, we have to convert 36 into roman numerals. Follow edited Sep 20, 2015 at . There are 2 more cases to consider: X can be placed before L (50 . Find the highest decimal value v from the table below that is less than or equal to the decimal number x, as well as its corresponding roman numeral n: Write the roman numeral you found and subtract its value v from x; repeat stages 1 and 2 until you get . The grammar consists of rules for adding ( III = 1+1+1) or subtracting ( IV = -1+5) values. We have assigned the first letter as the sum for easy comparison. int nextnumber = lettertonumber (roman.charat (i)); if (nextnumber > number) { // combine the two letters to get one value, and move on to next position in string. 9079009 in Roman numeral and images. 500. Here more solutions. function convertToRoman(num) { var numeral = ""; var arr = [ {number:1, roman: "I"}, {number:4, roman: "IV"}, {number:5, roman: "V"}, {number:9, roman: "IX"}, {number . Note the double backslash needed to create a single backslash in the regex. Roman and decimal number conversions. Java answers related to "convert roman numerals to decimal java" Java int to roman; roman numbers equivalent java; Java program to print hexadecimal to decimal number conversion Usually, roman numbers are written from left to right, which makes it easier for us to scan each roman character and add the corresponding value for each character to get the result.
Convert each symbol of Roman Numerals into the value it represents. Although I have tried to do this but converting a char to int is producing errors. However, the Roman Number for 4 and 9 are not written as IIII and VIIII, respectively. */ import java.util.Map; import java.util.Map.Entry; /** * A class for converting to and from roman numerals and arabic integers. Converting decimal numbers to Roman numerals in JavaScript Back 30 March 2010. Java answers related to "convert decimal to roman in java" java integer to binary string with leading zeros; int to double java; convert int to double with 2 decimal places java To .
About Roman Numerals. Therefore, let's write the code for this function Example But I now want the program to do the opposite and take a char input representing a roman value and convert to decimal.
But if the symbol appears before a larger symbol it is subtracted. C program to convert roman numbers to decimal numbers C Server Side Programming Programming Given below is an algorithm to convert roman numbers to decimal numbers in C language Algorithm Step 1 Start Step 2 Read the roman numeral at runtime Step 3 length: = strlen (roman) Step 4 for i = 0 to length-1 do Step 4.1 switch (roman [i]). Similarly, if larger letter appears first it gets added to the smaller number like CL=100+50=150. First, split the Roman numeral string into roman symbols.
This article will present how to convert a Roman number into a decimal using Java. initialize a string builder, now start checking if input number is >= highest roman numeral then add it to the string builder and reduce its corresponding value from the input number and if input number is < highest roman numeral then check with next highest roman numeral and repeat the process above till input number becomes 0. string builder
Can you give me some advice on how to make it look more readable or to improve in general? We can the separate symbol now convert each symbol of Roman Numerals into the value it represents. Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M. Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000 These symbols can be combined like this to create numbers: Algorithm. I wrote a class that can convert to and fro from Roman Numerals and Decimals. I have managed to get my code to convert most Roman numerals to its appropriate decimal value. Here more solutions. Hence we have tried to do the same through Java code:- 1. eg) input - "Enter Roman value: " X , output - Decimal equivalent - 10 - This case is for letter I. /* Java program for Conversion from Decimal to roman number */ public class Perform { // Display roman value of n public static void result (int n) { switch (n) { // Test Cases case 1: System.out.print ("I"); break; case 4: System.out.print ("IV"); break; case 5: System . From the 14th century on, Roman numerals began to be replaced in most contexts by the more convenient Hindu-Arabic numerals; however, this process was gradual, and the use of Roman numerals persists in some minor applications to this day. JavaScript exercises, practice and solution: Write a JavaScript function that Convert Roman Numeral to Integer. Here is my code. First, create two arrays one for storing the values of roman numerals and second for storing the corresponding letters. An approach to convert integer to roman numeral is simple. A static inner value class is used in the logic class. Separation of the user interface from the logic / model. 2. Step 1: First divide the list of elements in half It is a convention in C to identify the end of a character string with a null (zero) character In this C programming example, you will learn to convert binary numbers to octal and vice versa manually by creating a user-defined function How to convert letters (ASCII characters) to binary and vice versa This is . A simple GUI. Java program for Decimal to roman numeral converter. Weight conversion 2063439 kilograms (kg) = 4549057.6 pounds (lbs) 2063439 pounds (lbs) = 935969.8 kilograms (kg) Length conversion 2063439 kilometers (km) equals to 1282162 miles (mi). The use of Roman numerals continued long after the decline of the Roman Empire. Take symbol one by one from starting from index 0: If current value of symbol is greater than or equal to the value of next symbol, then add this value to the running total. The number 124 is the Roman numeral CXXIV. Within the switch statement of the function digit (), define the value of each digit of the roman number and return the same. It's not a major change I know, but it was something I wanted to do. Let us learn how to convert hexadecimal to decimal number in C programming language World's simplest browser-based hexadecimal to UTF8 converter Assume a virtual MIPS machine (no branch delay slots) 62890625 (d) 0 com ) From the table above, you can see that once you grab each digits value (0-9), just add 48 to it to convert it into the ascii . Convert each symbol of Roman Numerals into the value it represents. First, determine whether the current roman digit's value is less than zero. Algorithm. Similarly for Roman Number: DCLXVI , Its integer value is : 500 + 100 + 50 + 10 +5 + 1= 666. arabic += Java program for Decimal to roman numeral converter. L = 50 C = 100 D = 500 M = 1000. The meaning of the number 9079009: How is 9,079,009 written in letters, facts, mathematics, computer science, numerology, codes. Improve this question. .
If the smaller letter is written before the larger letter its get subtracted like IV = 5-1 =4 2. We can use the following algorithm to convert from Arabic to Roman numerals (iterating through symbols in reverse order from M to I ): LET number be an integer between 1 and 4000 LET symbol be RomanNumeral.values () [0] LET result be an empty String WHILE number > 0: IF symbol's value <= number: append the result with the symbol's name subtract . They are written as IV which is (5-1 = 4), and for 9 it is IX (10 -1=9). Example 1: Input:s = VI Output: 6 Example 2: Input: XOutput: 10XL is a Roman symbol which represents 40 Solution Approach. Example : XCIX = 99 but my code prints 109. */ public class RomanNumeralsConverter {private final Map< String, Integer > numberByNumeral; /** * Constructs an instance capable of converting from a string of roman * numerals into an arabic integer, and vice versa. In this case the logic is a static method that converts from an internal integer to a roman numeral string. convert-to-arabic.js This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. Let's understand the above steps through an example. JavaScript algorithm for converting Roman numbers to decimal numbers Javascript Web Development Object Oriented Programming We are required to write a function that takes in a string of roman number and returns its decimal (base 10) equivalent. Just Paste Binary Value and Press button, get results This is the simple program of java In practice, an 8th bit is added and used as a parity bit to detect transmission errors How to Convert ASCII Text to Binary GitHub Gist: instantly share code, notes, and snippets GitHub Gist: instantly share code, notes, and snippets. | 3,400 | 14,896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-49 | latest | en | 0.821307 |
https://codehs.com/info/standards/explore/WI_6-8/10009 | 1,713,564,219,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817455.17/warc/CC-MAIN-20240419203449-20240419233449-00005.warc.gz | 160,373,089 | 64,968 | # WI 6-8: AP3.b.6.m
## CodeHS Lessons
Compare different algorithms that may be used to solve the same problem in terms of their speed, clarity, and size (e.g., different algorithms solve the same problem, but one might be faster than the other). [Clarification Students are not expected to quantify these differences].
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http://octave.1599824.n4.nabble.com/mesh-plot-td4652572.html | 1,558,574,275,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256997.79/warc/CC-MAIN-20190523003453-20190523025453-00033.warc.gz | 150,072,631 | 14,582 | # mesh plot
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## mesh plot
I am trying to get a mesh plot in a program where I think the N and iter values are very high. So it is taking too long. Is there some way I can change the following code in order to just get a sample of 1 in 5 or 1 in 10 points. That might reduce the density and may speed the whole thing. Please advise. Thanks Asha G colormap(gray)view(30,45)[xx,zz]= meshgrid(x,z);mesh(xx,zz,y)qm = [xx(:) zz(:) y(:)];save -mat-binary cabunbact1doctN41mesh xx zz ysave -ascii cabunbact1doctN41meshascii qm _______________________________________________ Help-octave mailing list [hidden email] https://mailman.cae.wisc.edu/listinfo/help-octave
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## Re: mesh plot
On Fri, 2013-05-03 at 12:46 +0800, asha g wrote: > I am trying to get a mesh plot in a program where I think the N and > iter values are very high. Your script doesn't refer to either 'N' or 'iter', so the following is a *guess* that 'N' is numel(x) and 'iter' is numel(z). Furthermore I *assume* that the quantities 'x', 'y', and 'z' are computed by some external means, have conformable dimensions and Just Exist[tm] in your environment. To pick up every five elements of 'y' in each coordinate direction, you could do something like this: i = 1 : 5 : numel(x); j = 1 : 5 : numel(z); mesh(z(j), x(i), y(i,j)) Adjust the stride (in this case, '5') appropriately if you want more or fewer points in your mesh plot. Note: mesh(z(j), x(i), y(i,j)) is not a misprint. If you use vectors as coordinates, then number of rows of the third parameter must match the number of elements of the *second* parameter and the number of columns of the third parameter must match the number of elements of the first parameter. Sincerely, -- Bård Skaflestad <[hidden email]> _______________________________________________ Help-octave mailing list [hidden email] https://mailman.cae.wisc.edu/listinfo/help-octave
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## Re: mesh plot
Your script doesn't refer to either 'N' or 'iter', so the following is a*guess* that 'N' is numel(x) and 'iter' is numel(z). Furthermore I*assume* that the quantities 'x', 'y', and 'z' are computed by someexternal means, have conformable dimensions and Just Exist[tm] in yourenvironment.To pick up every five elements of 'y' in each coordinate direction, youcould do something like this: i = 1 : 5 : numel(x); j = 1 : 5 : numel(z); mesh(z(j), x(i), y(i,j))Adjust the stride (in this case, '5') appropriately if you want more orfewer points in your mesh plot.Note: mesh(z(j), x(i), y(i,j)) is not a misprint. If you use vectors ascoordinates, then number of rows of the third parameter must match thenumber of elements of the *second* parameter and the number of columnsof the third parameter must match the number of elements of the firstparameter.Am running it right now. Will keep you posted on whether I get the required result.Asha _______________________________________________ Help-octave mailing list [hidden email] https://mailman.cae.wisc.edu/listinfo/help-octave
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## Re: mesh plot
In reply to this post by Bård Skaflestad i = 1 : 5 : numel(x); j = 1 : 5 : numel(z); mesh(z(j), x(i), y(i,j))Adjust the stride (in this case, '5') appropriately if you want more orfewer points in your mesh plot.Note: mesh(z(j), x(i), y(i,j)) is not a misprint. If you use vectors ascoordinates, then number of rows of the third parameter must match thenumber of elements of the *second* parameter and the number of columnsof the third parameter must match the number of elements of the firstparameter.N = 41, iter = 111000;vvvv(iter,:) = V;tmax = 55; x= linspace(0,tmax,niter);z = linspace(0,l,N);y = vvvv';On using the following i = 1 : 5 : numel(x); j = 1 : 5 : numel(z); mesh(z(j), x(i), y(i,j))I get an error message at line mesh(z(j),x(i),y(i,j))error: A(I): Index exceeds matrix dimension.Maybe it needs to be rewritten Asha G _______________________________________________ Help-octave mailing list [hidden email] https://mailman.cae.wisc.edu/listinfo/help-octave
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## Re: mesh plot
On Fri, 2013-05-03 at 17:46 +0800, asha g wrote: > N = 41, iter = 111000; > vvvv(iter,:) = V; Okay. This means that 'vvvv' has (at least) 'iter' (=111,000) rows. You haven't said how many columns it has though. I'm forced to assume that it has N columns and that 'V' (whatever that is) is a 1-by-N (i.e., a 1-by-41) row vector. > > tmax = 55; > x= linspace(0,tmax,niter); I'm confused. What is 'niter'? Is it the maximum value of 'iter' or something else entirely? Whatever it is (presumably a positive integer), at least we know that all(size(x) == [ 1, niter ]) > z = linspace(0,l,N); all(size(z) == [ 1, N ]) > y = vvvv'; Size unknown. In the following, I'm going to assume that all(size(y) == [ N, niter ]) because that seems the most likely given your setup. If that is not the case, then you *really* need to tell us the size of 'y'. > On using the following > i = 1 : 5 : numel(x); > j = 1 : 5 : numel(z); > mesh(z(j), x(i), y(i,j)) > I get an error message at line mesh(z(j),x(i),y(i,j)) > error: A(I): Index exceeds matrix dimension. That's seems consistent with the fact that numel(x) (i.e., 'niter') is (much) larger than numel(z) (i.e., 'N'). > Maybe it needs to be rewritten By 'it' I assume you mean the 'mesh' call, rather than your own calculations. I already said in a previous e-mail that in an invocation like mesh(U, V, W) in which 'U' and 'V' are vectors, then the following conditions must hold rows(W) == numel(V) columns(W) == numel(U) In other words, your 'mesh' invocation must then (assuming, as stated above, that all(size(y) == [ N, niter ])) be written as mesh(z(j), x(i), y(j,i)) Sincerely, -- Bård Skaflestad <[hidden email]> _______________________________________________ Help-octave mailing list [hidden email] https://mailman.cae.wisc.edu/listinfo/help-octave
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## Re: mesh plot
On Fri, 2013-05-03 at 13:39 +0200, Bård Skaflestad wrote: > mesh(z(j), x(i), y(j,i)) Sorry. That should be mesh(x(i), z(j), y(j,i)) Sincerely, -- Bård Skaflestad <[hidden email]> _______________________________________________ Help-octave mailing list [hidden email] https://mailman.cae.wisc.edu/listinfo/help-octave
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## Re: mesh plot
In reply to this post by asha g On 05/03/2013 10:46 AM, asha g wrote: > > Maybe it needs to be rewritten > > Asha G Off-topic: could you please refrain from sending HTML emails? -- Zé _______________________________________________ Help-octave mailing list [hidden email] https://mailman.cae.wisc.edu/listinfo/help-octave | 2,000 | 7,035 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-22 | latest | en | 0.798634 |
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◆ cgees()
subroutine cgees ( character JOBVS, character SORT, external SELECT, integer N, complex, dimension( lda, * ) A, integer LDA, integer SDIM, complex, dimension( * ) W, complex, dimension( ldvs, * ) VS, integer LDVS, complex, dimension( * ) WORK, integer LWORK, real, dimension( * ) RWORK, logical, dimension( * ) BWORK, integer INFO )
CGEES computes the eigenvalues, the Schur form, and, optionally, the matrix of Schur vectors for GE matrices
Download CGEES + dependencies [TGZ] [ZIP] [TXT]
Purpose:
``` CGEES computes for an N-by-N complex nonsymmetric matrix A, the
eigenvalues, the Schur form T, and, optionally, the matrix of Schur
vectors Z. This gives the Schur factorization A = Z*T*(Z**H).
Optionally, it also orders the eigenvalues on the diagonal of the
Schur form so that selected eigenvalues are at the top left.
The leading columns of Z then form an orthonormal basis for the
invariant subspace corresponding to the selected eigenvalues.
A complex matrix is in Schur form if it is upper triangular.```
Parameters
[in] JOBVS ``` JOBVS is CHARACTER*1 = 'N': Schur vectors are not computed; = 'V': Schur vectors are computed.``` [in] SORT ``` SORT is CHARACTER*1 Specifies whether or not to order the eigenvalues on the diagonal of the Schur form. = 'N': Eigenvalues are not ordered: = 'S': Eigenvalues are ordered (see SELECT).``` [in] SELECT ``` SELECT is a LOGICAL FUNCTION of one COMPLEX argument SELECT must be declared EXTERNAL in the calling subroutine. If SORT = 'S', SELECT is used to select eigenvalues to order to the top left of the Schur form. IF SORT = 'N', SELECT is not referenced. The eigenvalue W(j) is selected if SELECT(W(j)) is true.``` [in] N ``` N is INTEGER The order of the matrix A. N >= 0.``` [in,out] A ``` A is COMPLEX array, dimension (LDA,N) On entry, the N-by-N matrix A. On exit, A has been overwritten by its Schur form T.``` [in] LDA ``` LDA is INTEGER The leading dimension of the array A. LDA >= max(1,N).``` [out] SDIM ``` SDIM is INTEGER If SORT = 'N', SDIM = 0. If SORT = 'S', SDIM = number of eigenvalues for which SELECT is true.``` [out] W ``` W is COMPLEX array, dimension (N) W contains the computed eigenvalues, in the same order that they appear on the diagonal of the output Schur form T.``` [out] VS ``` VS is COMPLEX array, dimension (LDVS,N) If JOBVS = 'V', VS contains the unitary matrix Z of Schur vectors. If JOBVS = 'N', VS is not referenced.``` [in] LDVS ``` LDVS is INTEGER The leading dimension of the array VS. LDVS >= 1; if JOBVS = 'V', LDVS >= N.``` [out] WORK ``` WORK is COMPLEX array, dimension (MAX(1,LWORK)) On exit, if INFO = 0, WORK(1) returns the optimal LWORK.``` [in] LWORK ``` LWORK is INTEGER The dimension of the array WORK. LWORK >= max(1,2*N). For good performance, LWORK must generally be larger. If LWORK = -1, then a workspace query is assumed; the routine only calculates the optimal size of the WORK array, returns this value as the first entry of the WORK array, and no error message related to LWORK is issued by XERBLA.``` [out] RWORK ` RWORK is REAL array, dimension (N)` [out] BWORK ``` BWORK is LOGICAL array, dimension (N) Not referenced if SORT = 'N'.``` [out] INFO ``` INFO is INTEGER = 0: successful exit < 0: if INFO = -i, the i-th argument had an illegal value. > 0: if INFO = i, and i is <= N: the QR algorithm failed to compute all the eigenvalues; elements 1:ILO-1 and i+1:N of W contain those eigenvalues which have converged; if JOBVS = 'V', VS contains the matrix which reduces A to its partially converged Schur form. = N+1: the eigenvalues could not be reordered because some eigenvalues were too close to separate (the problem is very ill-conditioned); = N+2: after reordering, roundoff changed values of some complex eigenvalues so that leading eigenvalues in the Schur form no longer satisfy SELECT = .TRUE.. This could also be caused by underflow due to scaling.```
Definition at line 195 of file cgees.f.
197*
198* -- LAPACK driver routine --
199* -- LAPACK is a software package provided by Univ. of Tennessee, --
200* -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
201*
202* .. Scalar Arguments ..
203 CHARACTER JOBVS, SORT
204 INTEGER INFO, LDA, LDVS, LWORK, N, SDIM
205* ..
206* .. Array Arguments ..
207 LOGICAL BWORK( * )
208 REAL RWORK( * )
209 COMPLEX A( LDA, * ), VS( LDVS, * ), W( * ), WORK( * )
210* ..
211* .. Function Arguments ..
212 LOGICAL SELECT
213 EXTERNAL SELECT
214* ..
215*
216* =====================================================================
217*
218* .. Parameters ..
219 REAL ZERO, ONE
220 parameter( zero = 0.0e0, one = 1.0e0 )
221* ..
222* .. Local Scalars ..
223 LOGICAL LQUERY, SCALEA, WANTST, WANTVS
224 INTEGER HSWORK, I, IBAL, ICOND, IERR, IEVAL, IHI, ILO,
225 \$ ITAU, IWRK, MAXWRK, MINWRK
226 REAL ANRM, BIGNUM, CSCALE, EPS, S, SEP, SMLNUM
227* ..
228* .. Local Arrays ..
229 REAL DUM( 1 )
230* ..
231* .. External Subroutines ..
232 EXTERNAL ccopy, cgebak, cgebal, cgehrd, chseqr, clacpy,
234* ..
235* .. External Functions ..
236 LOGICAL LSAME
237 INTEGER ILAENV
238 REAL CLANGE, SLAMCH
239 EXTERNAL lsame, ilaenv, clange, slamch
240* ..
241* .. Intrinsic Functions ..
242 INTRINSIC max, sqrt
243* ..
244* .. Executable Statements ..
245*
246* Test the input arguments
247*
248 info = 0
249 lquery = ( lwork.EQ.-1 )
250 wantvs = lsame( jobvs, 'V' )
251 wantst = lsame( sort, 'S' )
252 IF( ( .NOT.wantvs ) .AND. ( .NOT.lsame( jobvs, 'N' ) ) ) THEN
253 info = -1
254 ELSE IF( ( .NOT.wantst ) .AND. ( .NOT.lsame( sort, 'N' ) ) ) THEN
255 info = -2
256 ELSE IF( n.LT.0 ) THEN
257 info = -4
258 ELSE IF( lda.LT.max( 1, n ) ) THEN
259 info = -6
260 ELSE IF( ldvs.LT.1 .OR. ( wantvs .AND. ldvs.LT.n ) ) THEN
261 info = -10
262 END IF
263*
264* Compute workspace
265* (Note: Comments in the code beginning "Workspace:" describe the
266* minimal amount of workspace needed at that point in the code,
267* as well as the preferred amount for good performance.
268* CWorkspace refers to complex workspace, and RWorkspace to real
269* workspace. NB refers to the optimal block size for the
270* immediately following subroutine, as returned by ILAENV.
271* HSWORK refers to the workspace preferred by CHSEQR, as
272* calculated below. HSWORK is computed assuming ILO=1 and IHI=N,
273* the worst case.)
274*
275 IF( info.EQ.0 ) THEN
276 IF( n.EQ.0 ) THEN
277 minwrk = 1
278 maxwrk = 1
279 ELSE
280 maxwrk = n + n*ilaenv( 1, 'CGEHRD', ' ', n, 1, n, 0 )
281 minwrk = 2*n
282*
283 CALL chseqr( 'S', jobvs, n, 1, n, a, lda, w, vs, ldvs,
284 \$ work, -1, ieval )
285 hswork = int( work( 1 ) )
286*
287 IF( .NOT.wantvs ) THEN
288 maxwrk = max( maxwrk, hswork )
289 ELSE
290 maxwrk = max( maxwrk, n + ( n - 1 )*ilaenv( 1, 'CUNGHR',
291 \$ ' ', n, 1, n, -1 ) )
292 maxwrk = max( maxwrk, hswork )
293 END IF
294 END IF
295 work( 1 ) = maxwrk
296*
297 IF( lwork.LT.minwrk .AND. .NOT.lquery ) THEN
298 info = -12
299 END IF
300 END IF
301*
302 IF( info.NE.0 ) THEN
303 CALL xerbla( 'CGEES ', -info )
304 RETURN
305 ELSE IF( lquery ) THEN
306 RETURN
307 END IF
308*
309* Quick return if possible
310*
311 IF( n.EQ.0 ) THEN
312 sdim = 0
313 RETURN
314 END IF
315*
316* Get machine constants
317*
318 eps = slamch( 'P' )
319 smlnum = slamch( 'S' )
320 bignum = one / smlnum
321 CALL slabad( smlnum, bignum )
322 smlnum = sqrt( smlnum ) / eps
323 bignum = one / smlnum
324*
325* Scale A if max element outside range [SMLNUM,BIGNUM]
326*
327 anrm = clange( 'M', n, n, a, lda, dum )
328 scalea = .false.
329 IF( anrm.GT.zero .AND. anrm.LT.smlnum ) THEN
330 scalea = .true.
331 cscale = smlnum
332 ELSE IF( anrm.GT.bignum ) THEN
333 scalea = .true.
334 cscale = bignum
335 END IF
336 IF( scalea )
337 \$ CALL clascl( 'G', 0, 0, anrm, cscale, n, n, a, lda, ierr )
338*
339* Permute the matrix to make it more nearly triangular
340* (CWorkspace: none)
341* (RWorkspace: need N)
342*
343 ibal = 1
344 CALL cgebal( 'P', n, a, lda, ilo, ihi, rwork( ibal ), ierr )
345*
346* Reduce to upper Hessenberg form
347* (CWorkspace: need 2*N, prefer N+N*NB)
348* (RWorkspace: none)
349*
350 itau = 1
351 iwrk = n + itau
352 CALL cgehrd( n, ilo, ihi, a, lda, work( itau ), work( iwrk ),
353 \$ lwork-iwrk+1, ierr )
354*
355 IF( wantvs ) THEN
356*
357* Copy Householder vectors to VS
358*
359 CALL clacpy( 'L', n, n, a, lda, vs, ldvs )
360*
361* Generate unitary matrix in VS
362* (CWorkspace: need 2*N-1, prefer N+(N-1)*NB)
363* (RWorkspace: none)
364*
365 CALL cunghr( n, ilo, ihi, vs, ldvs, work( itau ), work( iwrk ),
366 \$ lwork-iwrk+1, ierr )
367 END IF
368*
369 sdim = 0
370*
371* Perform QR iteration, accumulating Schur vectors in VS if desired
372* (CWorkspace: need 1, prefer HSWORK (see comments) )
373* (RWorkspace: none)
374*
375 iwrk = itau
376 CALL chseqr( 'S', jobvs, n, ilo, ihi, a, lda, w, vs, ldvs,
377 \$ work( iwrk ), lwork-iwrk+1, ieval )
378 IF( ieval.GT.0 )
379 \$ info = ieval
380*
381* Sort eigenvalues if desired
382*
383 IF( wantst .AND. info.EQ.0 ) THEN
384 IF( scalea )
385 \$ CALL clascl( 'G', 0, 0, cscale, anrm, n, 1, w, n, ierr )
386 DO 10 i = 1, n
387 bwork( i ) = SELECT( w( i ) )
388 10 CONTINUE
389*
390* Reorder eigenvalues and transform Schur vectors
391* (CWorkspace: none)
392* (RWorkspace: none)
393*
394 CALL ctrsen( 'N', jobvs, bwork, n, a, lda, vs, ldvs, w, sdim,
395 \$ s, sep, work( iwrk ), lwork-iwrk+1, icond )
396 END IF
397*
398 IF( wantvs ) THEN
399*
400* Undo balancing
401* (CWorkspace: none)
402* (RWorkspace: need N)
403*
404 CALL cgebak( 'P', 'R', n, ilo, ihi, rwork( ibal ), n, vs, ldvs,
405 \$ ierr )
406 END IF
407*
408 IF( scalea ) THEN
409*
410* Undo scaling for the Schur form of A
411*
412 CALL clascl( 'U', 0, 0, cscale, anrm, n, n, a, lda, ierr )
413 CALL ccopy( n, a, lda+1, w, 1 )
414 END IF
415*
416 work( 1 ) = maxwrk
417 RETURN
418*
419* End of CGEES
420*
subroutine slabad(SMALL, LARGE)
SLABAD
Definition: slabad.f:74
integer function ilaenv(ISPEC, NAME, OPTS, N1, N2, N3, N4)
ILAENV
Definition: ilaenv.f:162
subroutine xerbla(SRNAME, INFO)
XERBLA
Definition: xerbla.f:60
logical function lsame(CA, CB)
LSAME
Definition: lsame.f:53
subroutine ccopy(N, CX, INCX, CY, INCY)
CCOPY
Definition: ccopy.f:81
real function clange(NORM, M, N, A, LDA, WORK)
CLANGE returns the value of the 1-norm, Frobenius norm, infinity-norm, or the largest absolute value ...
Definition: clange.f:115
subroutine cgehrd(N, ILO, IHI, A, LDA, TAU, WORK, LWORK, INFO)
CGEHRD
Definition: cgehrd.f:167
subroutine cgebal(JOB, N, A, LDA, ILO, IHI, SCALE, INFO)
CGEBAL
Definition: cgebal.f:161
subroutine cgebak(JOB, SIDE, N, ILO, IHI, SCALE, M, V, LDV, INFO)
CGEBAK
Definition: cgebak.f:131
subroutine clascl(TYPE, KL, KU, CFROM, CTO, M, N, A, LDA, INFO)
CLASCL multiplies a general rectangular matrix by a real scalar defined as cto/cfrom.
Definition: clascl.f:143
subroutine clacpy(UPLO, M, N, A, LDA, B, LDB)
CLACPY copies all or part of one two-dimensional array to another.
Definition: clacpy.f:103
subroutine ctrsen(JOB, COMPQ, SELECT, N, T, LDT, Q, LDQ, W, M, S, SEP, WORK, LWORK, INFO)
CTRSEN
Definition: ctrsen.f:264
subroutine cunghr(N, ILO, IHI, A, LDA, TAU, WORK, LWORK, INFO)
CUNGHR
Definition: cunghr.f:126
subroutine chseqr(JOB, COMPZ, N, ILO, IHI, H, LDH, W, Z, LDZ, WORK, LWORK, INFO)
CHSEQR
Definition: chseqr.f:299
real function slamch(CMACH)
SLAMCH
Definition: slamch.f:68
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https://www.jiskha.com/questions/568802/A-solid-sphere-of-uniform-density-starts-from-rest-and-rolls-without-slipping-a-distance | 1,571,205,115,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986664662.15/warc/CC-MAIN-20191016041344-20191016064844-00127.warc.gz | 930,976,291 | 4,365 | # Physics
A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 4.4 m down a q = 20° incline. The sphere has a mass M = 4.6 kg and a radius R = 0.28 m. Of the total kinetic energy of the sphere, what fraction is translational?
KEtran/KEtotal = ?
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Security vulnerability when facilitating order
I’m writing a script which facilitates orders. I’m quite new to ethereum, I’m trying to figure out if there are any security vulnerability when facilitating orders.
Here is my script:
contract FacilitateOrder {
return $this->hasMany(MessageParticipant::class); } The Me… In Bundle product order "item status" doesn’t show / Magento 2 Anybody know, why in Bundle Product order, simple products doesn’t has “item status” information/status? It’s quite essential to know if this product is ordered, or backordered. Here is screenshot: https://www.dropbox.com/s/… Order and principal part of$f(x)=\, e^{x^2} \int_{x}^{+\infty}e^{-t^2}\,dt$, infinitesimal as$x\to+\infty$Determine if$f(x)$is an infinite or an infinitesimal as$x\to+\infty$, its order and the principal part $$f(x)=\, e^{x^2} \int_{x}^{+\infty}e^{-t^2}\,dt$$ I can write, and then, using L’Hôpital’s rule:$$\lim_{x \to +\inf… How do I programmatically create a payment for an order? I know how to programmatically update a payment for an order.$payments = \Drupal::entityTypeManager()->getStorage(‘commerce_payment’)
->loadByProperties([ ‘order_id’ => [$order_id] ]); foreach ($payments as \$paym… | 570 | 2,353 | {"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} | 2.8125 | 3 | CC-MAIN-2018-34 | latest | en | 0.846013 |
https://ianmcgregorphotography.com/helping-93 | 1,675,139,087,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499842.81/warc/CC-MAIN-20230131023947-20230131053947-00145.warc.gz | 334,574,507 | 3,605 | # General solution solver
One instrument that can be used is General solution solver. We will also look at some example problems and how to approach them.
## The Best General solution solver
This General solution solver provides step-by-step instructions for solving all math problems. The personal income tax calculator can be viewed through Baidu app. Users can download Baidu app, search for the personal income tax calculator, find the personal income tax wizard calculator applet, and enter information such as pre tax income to calculate after tax income. The steps are as follows: 1 find out the important function keys. Scientific calculators have several function keys, which are very important for learning algebra, trigonometry, geometry, calculus, etc.
Using the annual precipitation data of 2000 to calculate the Rainfall Erosivity Factor: spatial analysis tool → map algebra tool → grid Calculator: input the formula (as shown in the figure) 1 find out the important function keys. Scientific calculators have several function keys, which are very important for learning algebra, trigonometry, geometry, calculus, etc. Find the following function keys on your calculator: Preliminary analytic geometry is to rely on the equations of straight lines and the standard equations of circles, so that students can grasp the basic steps of solving geometric problems with algebraic methods, initially form the ability to solve geometric problems with algebraic methods, and help students understand the basic ideas of analytic geometry. Mathematics is mainly composed of parts that can't use calculators and parts that can use calculators. It mainly examines problem solving and data analysis, core algebra, and basic knowledge of advanced mathematics.
This is mainly because the signal and noise are assumed to be uncorrelated in napc transformation, and the noise is also spatially uncorrelated, which is generally not met in the noise estimation process. For example, when median filtering is used, the resulting noise image usually contains high-frequency components of the original image and contains certain contour information, which has strong spatial correlation. Abstract: the contents of the three mean value theorems in differential calculus are characterized by high degree of abstraction and strong theoretical nature. It is very difficult for students to learn this knowledge, especially the problem of proving the mean value theorem, which makes them at a loss.
The mathematics part is divided into two parts: usable calculators and unusable calculators. It mainly examines algebra, problem solving and data analysis, introduction to higher mathematics, geometry, trigonometry and other related contents. Wherein: IB math exam has strict requirements on the setting of calculators. If the invigilator clears the setting of calculators during the exam, you need to be able to quickly change it to the original mode. Therefore, it is also necessary to spend some time doing calculator exercises. | 565 | 3,022 | {"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} | 3.921875 | 4 | CC-MAIN-2023-06 | latest | en | 0.936628 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=26&t=30518&p=105088 | 1,596,511,219,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735851.15/warc/CC-MAIN-20200804014340-20200804044340-00457.warc.gz | 359,703,917 | 15,976 | ## Accuracy vs Precision
Caroline Crotty 1D
Posts: 32
Joined: Fri Apr 06, 2018 11:02 am
### Accuracy vs Precision
I remember the example we used in class, but I am unsure how to tell which is which. One is when all measurements are close together, but not the desired amount? The other is when all the measurements are on target?
Chem_Mod
Posts: 18400
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times
### Re: Accuracy vs Precision
Accuracy is on target. Precision is how closely you can repeatedly measure a thing.
Allen Chen 1J
Posts: 31
Joined: Wed Nov 15, 2017 3:04 am
### Re: Accuracy vs Precision
Chem_Mod wrote:Accuracy is on target. Precision is how closely you can repeatedly measure a thing.
What is the chemistry context in which these differences will be most important?
Chem_Mod
Posts: 18400
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times
### Re: Accuracy vs Precision
It will be important when you do the experiment. Suppose you have a balance which you use to measure the mass of a chemical. The balance can have poor precision, meaning the numbers fluctuate all the time and you are not very certain. Or it can have good precision, where the numbers are always close if you measure the same thing over and over. I hope this example helps.
Bryan Jiang 1F
Posts: 37
Joined: Fri Apr 06, 2018 11:03 am
Contact:
### Re: Accuracy vs Precision
Caroline Crotty 1D wrote:I remember the example we used in class, but I am unsure how to tell which is which. One is when all measurements are close together, but not the desired amount? The other is when all the measurements are on target?
Accuracy means how close your measurements are to the true/correct value. Precision means how close your measurements are to each other. They are not mututally exclusive, so it is possible, as you said, that all measurements are close together but not the desired amount (very precise, but not very accurate). However, it is also possible that all measurements are close together and are the desired amount (both very precise and very accurate).
Last edited by Bryan Jiang 1F on Sun Apr 15, 2018 6:16 pm, edited 1 time in total.
Posts: 31
Joined: Fri Apr 06, 2018 11:03 am
### Re: Accuracy vs Precision
Precision is how close to each other they are, and accuracy is how close to the true value they are.
Natalie_Martinez_1I
Posts: 30
Joined: Mon Feb 26, 2018 3:00 am
### Re: Accuracy vs Precision
Accuracy tells you how close you are to the the true value, whereas precision indicates how close measurements are to one another.
Haison Nguyen 1I
Posts: 30
Joined: Fri Apr 06, 2018 11:04 am
### Re: Accuracy vs Precision
Accuracy is how close your results are to the true value. Precision is how close your results are to each other.
Paywand Baghal
Posts: 32
Joined: Fri Apr 06, 2018 11:01 am
### Re: Accuracy vs Precision
Allen Chen 1J wrote:
Chem_Mod wrote:Accuracy is on target. Precision is how closely you can repeatedly measure a thing.
What is the chemistry context in which these differences will be most important?
In context of chem, accuracy would be consistency getting the same, or around the same, answer when doing an experiment. Precision would be getting an answer which is correct.
KristinaNguyen_1A
Posts: 30
Joined: Fri Apr 06, 2018 11:04 am
### Re: Accuracy vs Precision
I went to a UA office hour and how she described it was that:
Precision= how CLOSE measurements are to ONE ANOTHER
Accuracy= closeness of measurements to its TRUE VALUES
Maria Zamarripa 1L
Posts: 30
Joined: Fri Apr 06, 2018 11:05 am
Been upvoted: 1 time
### Re: Accuracy vs Precision
Precision is how close the measurements are to one another and accuracy is how close the measurements are to the true value.
breannasung_1K
Posts: 30
Joined: Fri Apr 06, 2018 11:05 am
### Re: Accuracy vs Precision
Precision is how close the measurements/ data are to one another. Accuracy is the closeness of the measurements/ data to their correct values.
When you have high precision and low accuracy, you can multiply all your data by some value to get higher accuracy.
004985802
Posts: 27
Joined: Fri Feb 02, 2018 3:00 am
### Re: Accuracy vs Precision
accuracy measures how close your results are to the true value, and precision measures how exact and consistent your results are
Jessica Lancisi - 1I
Posts: 29
Joined: Fri Apr 06, 2018 11:03 am
### Re: Accuracy vs Precision
@Paywand Baghal, I think that it would be the other way around- In a chemical context, accuracy would be getting the correct value, and precision would be getting the same or close to the same result each time. Precision without accuracy shows that there may be a mistake in the experiment that keeps getting repeated.
Rummel Requerme 1E
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Joined: Fri Sep 25, 2015 3:00 am
Been upvoted: 1 time
### Re: Accuracy vs Precision
Essentially accuracy is how close you are to the actual result, as precision is how close your results are in respect to one another.
Hope_Pham_1G
Posts: 29
Joined: Fri Apr 06, 2018 11:04 am
### Re: Accuracy vs Precision
When visualizing accuracy, think of a bulls-eye target with many shots near the center, but scattered within the radius of the bulls-eye in any direction. The results appear very disbursed. When visualizing precision, think of a bulls-eye target with many shots clumped together, but not necessarily within close range of the bulls eye. The important aspect of precision is that the same value is achieved consistently even if the value is not equal to the true value.
Taizha 1C
Posts: 36
Joined: Fri Apr 06, 2018 11:01 am
### Re: Accuracy vs Precision
According to lecture precision indicates how close measurements are to one another while accuracy is the accuracy to the true value.
Eugene Chung 3F
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Joined: Wed Nov 15, 2017 3:03 am
### Re: Accuracy vs Precision
ran2000
Posts: 66
Joined: Fri Sep 28, 2018 12:15 am
### Re: Accuracy vs Precision
From an experimental chemistry standpoint, an example of accuracy vs. precision would be:
If we conducted an experiment to test the volumes of carbon dioxide released when different concentrations of acid were added to metal carbonates, the study would have-
1. Low accuracy and Low precision if the the volumes of CO2 were found to be very variable (1.3l, 1.7l, 1.8l, 1.0l, 1.6l) and the method used to measure it was wrong (such as the calculating the volume of a balloon after it filled up with CO2, but without maintaining a constant temperature)
2. Low accuracy but high precision if the volumes of CO2 were found to be very close to one another (1.0l, 1.1l, 1,0l, 1.1l, 1.0l) but the method used had a systematic error such as the method of measurement of volume would not provide the actual volume measurement.
3. High accuracy but low precision if the volumes of CO2 were found to be variable (1.7l, 1.9l, 1.1l, 1.4l, 2.0l) but the method of volume measurement did not have any systematic errors, rather the variability in the volume was a problem of random errors.
4. High accuracy and high precision if the volumes of CO2 were found to be close to each other (1.0l, 1.1l, 1.0l, 1.0l, 1.1l) and the method of volume measurement did not have any systematic errors (the method used to measure volume would provide the true values)
ariellasarkissian_3H
Posts: 46
Joined: Fri Sep 28, 2018 12:25 am
### Re: Accuracy vs Precision
Caroline Crotty 1D wrote:I remember the example we used in class, but I am unsure how to tell which is which. One is when all measurements are close together, but not the desired amount? The other is when all the measurements are on target?
Accuracy is how close your value is to the true/exact value. Precision is producing values that are close in proximity. In some cases, if you receive the same value multiple times you are precise, but if it is not close to the exact value, you are not accurate.
Posts: 62
Joined: Fri Sep 28, 2018 12:23 am
### Re: Accuracy vs Precision
So there are four scenarios
1. Low accuracy and low precision
2. Low accuracy and high precision
3. High accuracy and low precision
4. High accuracy and high precision
Obviously we are shooting to get high accuracy and high precision; although, in the event that your data isn't, would it be worse to have high accuracy and low precision or low accuracy and high precision?
Posts: 60
Joined: Fri Sep 28, 2018 12:27 am
### Re: Accuracy vs Precision
Another interesting definition of accuracy - Accuracy can be defined as when the average experimental value is close to the true value. For example, if the three following measurements were taken of a 30 mL sample of water: 20 mL, 35 mL, and 35 mL, the average value of this measurement is 30 mL. By this definition, the sample is considered accurate but not precise.
Rebecca Altshuler 1D
Posts: 61
Joined: Fri Sep 28, 2018 12:15 am
### Re: Accuracy vs Precision
How can values be accurate but not precise? If all the values are accurate (close to the true value) how can they still be not precise in relation to each other?
Elena Maneffa 1E
Posts: 33
Joined: Fri Sep 28, 2018 12:27 am
Been upvoted: 1 time
### Re: Accuracy vs Precision
Rebecca Altshuler 1E wrote:How can values be accurate but not precise? If all the values are accurate (close to the true value) how can they still be not precise in relation to each other
Values can be accurate but not precise because each measurement may be close to the true value (relatively) however not relatively close enough to each other to be considered precise.
### Who is online
Users browsing this forum: No registered users and 1 guest | 2,463 | 9,597 | {"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-2020-34 | latest | en | 0.941237 |
https://assettrading.co.uk/risk-management/ | 1,653,429,583,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662577259.70/warc/CC-MAIN-20220524203438-20220524233438-00114.warc.gz | 154,840,712 | 19,390 | Definition: Identifying potential risks before they happen, analyse them and take action to reduce the risk.
In the financial sector, pretty much everything has an element of risk. Your own bank account, despite you thinking it is the safest place for your money, there are risks holding it there. What happens if the bank collapses and cannot pay out your money?
1. Lehman Brothers famously was the fourth largest investment bank in the United States and collapsed during the financial crisis, leaving its clients scrambling for their money.
2. The Greek banks during their crisis were limiting how much clients could withdraw from their own account.
There are always risks associated to investing your money and understanding those risk is imperative when it comes to trading assets.
## Capital at Risk
Every time you take a trade you are risking your capital and that is the biggest risk of trading any assets. As a trader you therefore have a responsibility to manage your risk, maintaining that any losses are controlled and managed.
Essentially the most important thing is to keep your losses consistent. If you can choose the amount you are going to loser before you lose it, then you should stick to that much every time.
## Risk Reward Ratio
Before we look at risk management strategies, we must first understand what a risk reward ratio (RRR) is. This is the ratio that measures how much you could potentially profit for every dollar you have at risk.
If you were risking £100 to make £100, your RRR would be 1:1. If you were risking £100 to make £300, your RRR would be 1:3.
It is normally advisable to never go below a RRR of 1:1 however this depends on your win rate. If you are hitting 50% win rate, then 1:1 isn’t enough to make profit. However, if your RRR was 1:2 and your win rate was 50%, then you would be making profit. Let’s do the math’s quickly.
Quick Look at the Math's
If you’re risking £100 in order to make £200 and you take the exact same trade 100 times.
50 wins & 50 Losses
(50 x £-100) = £-5,000
+
(50 x £200) = £10,000
= £5,000 Profit
This one example shows how important it is to stick to your strategy and making sure your psychology is correct to make sure you are not cutting your profits short or playing with your stop loss.
If you can get a RRR that works with your strategy and you stick to it then you can see the profits to be made.
In an ideal world then every time you took a trade you would simply take the trade and leave it. Let it either hit the target or stop loss. This is where psychology comes in, people get greedy and people get scared. This is why we would urge new traders to follow a strategy on demo before going live. If you can see it working on demo then it makes it psychologically easier to do when you’re on a live account.
## Risk Management Strategies
Stop Loss & Profit Orders – These are automatic orders that near enough every broker now offers. They are orders that traders can put into the market that once the price gets to that level they will automatically buy or sell the market at that price. If you use these orders every time, you can manage exactly how much you are going to lose every time.
Stop Loss and Take Profit Orders
While that sentence does not sound nice, it is very important that you get it into your head that you will lose and actually managing your losses is the foundation of making consistent profits.
Position Sizing – This is how much you put on each position. Depending on your strategy this may or may not be the same for each trade. Normally this will change for every trade you take, the reason being is because your stop loss will not always be the same distance from your entry.
Before you take a trade you should know exactly:
• How many pips you are will to risk
• How much money you are willing to lose (this should be consistent for every trade)
Once you know these two, only then can you calculate your position size.
Calculate Your Position
Trade Size = £ Risk / Pip Risk
Or
£ Risk = Trade Size x Pip Risk
Trade Size (% of Account at Risk) – The amount you are willing to risk is up to the trader in question. However, you should think about it as a percentage and not as a numerical value.
It is widely recommended in the industry that you should not risk more than 1% of your account on any one trade. If you have an account of £10,000, then no trade should ever lose more than £100. This way it will take 100 bad trades in a row for you to blow your account. The chances of that happening are very slim.
No More Than 1% Risk On Every Trade
There will always be more risk adverse traders out there but the higher your risk goes the less trades it will take to blow your account. Using the example above, if you’re willing to risk £1,000 per trade then it will only take 10 bad trades for you to blow your account.
Loss of Capital % Required to get back to breakeven 10% 11% 20% 25% 30% 43% 40% 67% 50% 100% 60% 150% 70% 233% 80% 400% 90% 900%
Trading is about the long game and not the short term wins. A more impressive portfolio is one that climbs steadily rather than erratically.
If you were a fund manager in charge of millions of other peoples money, who would you rather trust with that money? Someone who has a steady growth with small losses or someone who has huge swings, sometimes finding their balance negative?
Total Risk – Whilst it is said you shouldn’t over 1% on one trade, it is also advised that you shouldn’t risk more than 2% of your account at any one time. This way, should everything go wrong, you will only lose 2% at one time.
If you were to have 10 trades open at 1% risk each, you would have a total of 10% of your account at risk at once, and should the worst happen, you have done your months risk in a day. Absolutely far from ideal!
This does not mean that you can only have two trades on at once, it just means that you need to manage your risk to not go over 2%. You can have 4 trades on at 0.5% or 8 at 0.25%, it is up to you, but ensuring you don’t lose too much in one day is also important.
As we highlighted earlier, trading is a long game and while you should consider each trade, you should also consider your losses in terms of days, weeks and months. Typically the below percentages are the most you should lose in that time period;
Max Losses Per Timeframe
Day – 2%
Week – 6%
Month – 8%
If you start losing more than these amounts then your account is in real danger and realistically you should be going back to the drawing board and assessing what is going wrong because profitable traders do not lose more than that.
A common mistake by new traders is that they are so desperate to make back what they have lost that they end up increasing their risk to try and ‘win back their losses quicker’. Famous last words. If you find yourself saying or doing that, then take a step back and reassess. That is called chasing your losses and more often than not, the trader loses.
Diversification – This is an investment technique which limits a trader or investors portfolio exposure to one asset or asset class.
An example for an FX trader may be that they have 4 trades open and all four have GBP in the currency pair. This means that the trader is exposed to any GBP news or fluctuation and in the likelihood of an extreme move, all trades could come in at a loss at the same time.
Asset managers also adhere to diversification in the asset classes they invest in. They will not like to be overly exposed to any one asset and is why you will often see a percentage split of an investors portfolio across various asset classes.
## Other Risks
Slippage – This is when you enter a market at a price above or below the expected entry. An example may be when you want to buy GBP/USD at 1.3010, but when you execute the trade you in fact have an entry at 1.3030. In this example the trader has been slipped 20 pips.
Example of Slippage (not a real trade)
Slippage can happen in any market, it is perhaps less common in the FX market because of the liquidity however it is still fairly common when you get an extreme move that a trader will be filled at an unexpected price.
Slippage tends to occur during highly volatile moments, in the FX market you would expect to see this during interest decisions or other major announcements.
Gaps – These are when the market simply jumps and does not trade at one price. It is a common occurrence in the stock market because the market closes everyday and will often open at a different price to the close price and hence ‘gapped’ over some prices.
One of the largest risks of gaps pose is that should they not trade at a certain price, then if a trader has a stop loss and the price gaps over it then all of a sudden you have more than you originally planned to risk on a that trade in question.
How to manage gaps in the market:
• Close trades leading up to potential highly volatile events where a gap could occur
• Use a broker who can guarantee your stop loss despite any gaps
• Should your stop loss be ‘gapped’, CLOSE your trade immediately. Suddenly you are running into more risk than you expected and you should cut your losses. | 2,053 | 9,189 | {"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-2022-21 | latest | en | 0.977247 |
https://docs.scipy.org/doc/scipy-1.3.3/reference/generated/scipy.signal.lsim2.html | 1,601,085,032,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400232211.54/warc/CC-MAIN-20200926004805-20200926034805-00376.warc.gz | 343,808,720 | 3,391 | # scipy.signal.lsim2¶
scipy.signal.lsim2(system, U=None, T=None, X0=None, **kwargs)[source]
Simulate output of a continuous-time linear system, by using the ODE solver scipy.integrate.odeint.
Parameters
systeman instance of the lti class or a tuple describing the system.
The following gives the number of elements in the tuple and the interpretation:
• 1: (instance of lti)
• 2: (num, den)
• 3: (zeros, poles, gain)
• 4: (A, B, C, D)
Uarray_like (1D or 2D), optional
An input array describing the input at each time T. Linear interpolation is used between given times. If there are multiple inputs, then each column of the rank-2 array represents an input. If U is not given, the input is assumed to be zero.
Tarray_like (1D or 2D), optional
The time steps at which the input is defined and at which the output is desired. The default is 101 evenly spaced points on the interval [0,10.0].
X0array_like (1D), optional
The initial condition of the state vector. If X0 is not given, the initial conditions are assumed to be 0.
kwargsdict
Additional keyword arguments are passed on to the function odeint. See the notes below for more details.
Returns
T1D ndarray
The time values for the output.
youtndarray
The response of the system.
xoutndarray
The time-evolution of the state-vector.
Notes
This function uses scipy.integrate.odeint to solve the system’s differential equations. Additional keyword arguments given to lsim2 are passed on to odeint. See the documentation for scipy.integrate.odeint for the full list of arguments.
If (num, den) is passed in for system, coefficients for both the numerator and denominator should be specified in descending exponent order (e.g. s^2 + 3s + 5 would be represented as [1, 3, 5]).
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scipy.signal.lsim
#### Next topic
scipy.signal.impulse | 463 | 1,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-40 | latest | en | 0.670719 |
https://www.thenational.academy/pupils/lessons/exploring-missing-lines-of-symmetry-following-a-reflection-70vk6e/video | 1,716,038,002,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057422.43/warc/CC-MAIN-20240518121005-20240518151005-00313.warc.gz | 891,431,467 | 27,840 | # Lesson video
In progress...
Hi there, my name is Miss Darwish, and for today's maths lesson, we are going to be exploring missing lines of symmetry.
But before we get started, if I could just ask you for a favour, if you just want to take yourself to a nice quiet place, ready to start the lesson.
So the agenda for today's lesson is the following.
First of all, we're just going to recap on reflections, and then we're going to be looking at finding the mirror line between a few options, and then we're going to have a look at missing lines of reflection and how to find them.
And then at the end, of course, as always, there will be a quiz for you to do on today's learning.
So, before we start the lesson, you will need a sheet of paper or a notebook, pencil and a ruler.
So if you want to go and get those things, we can start the lesson.
Okay, let's start.
So reflections, in front of us, we have got a grid with some coordinates, and we've got two shapes, and one of the shape has been reflected onto the other.
Can you see that? Now, the only thing that's missing is what? We know these shapes are reflections of each other because we've been told that.
But, what's missing is a mirror line, or a line of reflection.
So we can call it a line of reflection or we could also call it a mirror line, and we can call it a mirror line because if I was to grab a mirror and place it on that line, it would show us what the shape would look like if it's reflected.
Okay, so, is this the correct mirror line, now that I've just added a mirror line? Is it correct? Yes or no, what do you think? Why is it right? Why is it wrong? What do you think? Okay.
Now, is this correct? This one or this one's the correct mirror line, or neither, and why? What about this one? Yeah, that one looks about right.
Why does that mirror line look right? If we look at the first mirror line, it's not really halfway between both shapes, is it? And if we choose one of the points on that shape, and we were to reflect it, it wouldn't reflect the exact same place.
And we wouldn't get that here either.
But if we have a look at this, if you choose one of the points, we can see that there's four squares away from the mirror line, or the line of reflection, and also where it's been reflected is also four squares away.
Can you see that? Okay, so this is the correct mirror line that we're missing.
Okay, below, two of the shapes are reflected.
Okay, again, what's missing? All, so the mirror line's missing, or we can also say the line of reflection.
We can call it the line of reflection or the mirror line.
So where do you think it goes? Have a think.
Maybe you want to grab your ruler and place it on the screen, as to where you think it might go, the line of reflection or the mirror line.
Okay.
Should we have a look together? I've given you three options now.
I'm not going to tell you the answer straight away.
So there's option A, option B and option C.
I wonder, was your ruler a placed on either A, B or C? Okay.
Let's go through each of the options.
So, let's have a look at A first of all.
The triangle has three vertices, correct? I've chosen just one of these, and I've marked it with a black X.
Can you see that? Okay.
So, this point is three squares away from mirror line A.
Can you count to three squares? Okay, good.
Now, it's reflected point, I've also marked with an X.
Is that three squares away from the mirror line? It's not three squares away from the mirror line.
So is A the mirror line? No, we don't think it's A.
So let's check the next one.
Let's check B.
Okay.
We check B.
Again, the first thing I did was point out we know a triangle has three vertices.
I'm just going to pick or choose one of those three.
So I have picked the same one as I did when I was checking A.
How many squares away is it from the mirror line? It's five squares away from the mirror line.
Well done.
And can you see where it's same reflective point is? I want you to put your finger there.
Have you found it? Okay.
Let's have a look.
So this point is five squares away from the mirror line, mirror line B.
It's reflective point is also five squares away from mirror line B.
What does that tell us? Could the line of reflection B, or mirror line B, be the correct one? Okay.
Let's check the other vertices just to be sure.
So we're going to pick another one, so that's another point.
This is also how many squares away from the mirror line? Five squares away from the mirror line.
Well done.
And can you find the same point on the other side? It's reflected point is also five squares away from the mirror line.
Well done.
Okay.
Let's have a look at the last one.
We're checking the last vertices.
How many squares away is it from the mirror line?? Can you count them for me? Three squares away.
Well done.
And it's reflective point is also three squares away from the mirror line.
So have we found our mirror line? We have on hopefully by line B is where you had your ruler.
If you didn't, then don't worry.
Now we know how to check now.
Okay.
Let's look at another example.
So, it says, where is the mirror line? I've given you two shapes.
The ones is a reflection of the other.
Where do you think the mirror line might go? I'm going to give you some thinking time.
Okay.
Should we go through it together now? Giving you some thinking time, maybe you got an idea as to where you might think the mirror line might go.
Let's have a look together.
So, step one, this is my method or my recipe of doing this.
So first of all, step one is to find one of the vertices.
Anyone.
It really doesn't matter which one you pick.
This is the one that I've chosen.
You might choose a different one.
So I've picked it.
Now, you've also got to find where would that point be when it's reflected on the shape? So we look at the reflected shape, and I want you to find the exact same point.
So I've marked both of them.
Okay? Now, step two.
Once I found both of my points, remember it has to be, they're not both random.
I pick one random point, any of the vertices, and then that exact same point on the shape.
I have to find where it is on the second shape.
Okay, on the reflected shape.
So once I've done that, step two is to count how many squares apart are they? Can you count them for me? Okay.
Hopefully you counted eight.
They are eight squares away from each other.
Can you see that? Can you guess what we might do next? Okay, so, we find the halfway point, and half of eight is equal to four, or we can say eight divided by two is equal to four.
And when I found half of it, I've marked it with an X.
Can you see that X? 'Cause I'm finding half way between both of the points and then what have I replaced it with? My line of reflection, or what else do we call it a line of reflection? A mirror line.
Well done.
So I've replaced it with my mirror line or my line of reflection.
Now let's just check.
So I'm going to look for that first point again, and I'm going to count one, two, three.
It is four squares away from the mirror line or the line of reflection.
Okay, so, now I'm going to count exactly the same.
How many squares away from the line of reflection to the reflective point? Four squares away as well.
Good.
Okay.
Let's have another look.
Different example now.
Where is the mirror line? I'm going to give you some thinking time.
Tell me where do you think it is? Okay.
What do you think? So, what was your step number one? What did you do first? Find a random vertice, one of the vertices of many.
Choose any one and then find its reflective point and try and find halfway, after we've counted how many squares.
Okay.
So, have I found the correct mirror line? Was that what you had as well? Okay.
All right, so we can say that this is halfway between both shapes, and both of the shapes, one is a reflection of the other shape.
Well done.
Okay.
Now things get a little bit trickier now.
Where is the mirror line? What do you notice? We've got a shape in the first quadrant, and a shape in the third quadrant.
Where do you think the mirror line might go, again? Just grab your ruler and place it somewhere.
Don't forget to count the squares.
Okay, let's have a look together.
So, what was asked at number one? Tell me.
Tell me what's step number one? Pick any of the vertices.
We've got three of them.
Pick any of them.
And then what do we do? On the reflected shape, I want you to find me right now, I want you to take your finger and point to where that reflective point is on the reflected shape.
Go.
Okay.
Let's see if you're right.
Were you right? Hopefully it's where your finger is.
Okay.
And then what's the next step that we do? What do we count? It's going to be a diagonal count.
Let's have a look.
How many did you count? Do you want to count them now? Let's count together, so one, two, three, four, five, six, seven.
Seven squares between both points.
So, what's the next step I do? I've counted between my original point and its reflected point and there are seven squares, and now I have to divide it by two or find half way.
So what's half of seven? Three and a half or 3.
5.
So that is where my mirror line or my line of reflection would go.
And then again, I'm going to pick the same point and count from that point to the line of reflection.
Should be three and a half squares, but let's count.
One, two, three, and a half, half, one, two, three.
Okay.
Hopefully my nice, easy step method is clear 'cause I like using it when I look at lines of reflection.
Okay, brilliant work you've done today.
Now I would like you to pause the video, and you have got a task to complete.
If you want to go and complete that, do the best that you can.
Stop the video and then come back, and we will go through the answers and check together.
Good luck.
Okay.
Welcome back.
Hopefully you found that okay.
Not teacher key.
Let's have a look at the answers together then.
Okay, before we go for the answers, let's just read through the question together, and then you can check your answers and mark them.
So, shape A has been reflected to shape B.
Draw the missing mirror line.
And then the second part of the question says that shape A has also been reflected to shape C.
Draw the missing mirror line.
So we've got reflection from shape A to shape B, and then shape A to shape C.
So we should have two mirror lines, using a ruler, obviously.
So, let's have a look.
Okay.
So, here is a mirror line, and which mirror lines this whole translating from shape A to shape C.
And again, if you choose one of the points on shape A, hopefully that you did this, or if not, then have a go at doing it now.
Choose any of the vertices and then find the equivalent one on the reflected shape and then count.
It should be the same amount of squares between your original shape and one of the vertices that you've chosen to the mirror line or the line of reflection.
And the same from the corresponding or the same point on the other reflected shape.
So if you want to count the squares and make sure, that's okay.
Okay, and then the second one we're reflecting from shape A to shape B is a vertical line that looks like this.
And again, exactly the same thing.
If you want to choose one or two of the vertices from shape A, find the matching or the corresponding ones onto shape B and, again, count how many squares away it is from the mirror line or the line of reflection.
And it should be exactly the same number of squares as in from the line of reflection to the point that matches the original point.
So make sure you check those, please.
Once you've checked them, make you give yourself a nice big tick.
Okay, so these are the two lines, the lines of reflection or the mirror lines that you should have on your answer sheet.
Well done, if you got those right.
Okay.
Share your work with Oak National, if you would like to, of course. | 2,822 | 12,003 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-22 | latest | en | 0.976208 |
https://projects.ng/project/analysis-of-unbalanced-fixed-effect-non-interactive-model/ | 1,618,792,907,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038862159.64/warc/CC-MAIN-20210418224306-20210419014306-00277.warc.gz | 567,050,626 | 25,147 | #### Project File Details
3,000.00
The Project File Details
• Name: ANALYSIS OF UNBALANCED FIXED-EFFECT NON- INTERACTIVE MODEL
• Type: PDF and MS Word (DOC)
• Size: [276 KB]
• Length: [45] Pages
## ABSTRACT
This study examines the analysis of fixed-effect non-interactive unbalanced data by a method called Intra-Factor Design. And to derive this design for analysis, mathematically, the matrix version of the fixed
7
effect model, Yijk = µ + τi + βj + εijk, was used. This resulted to the definition and formation of many matrices such as the Information Matrix, L; Replication Vector, r; Incidence Matrix, N; Vector of adjusted factor A totals, q; Variance–Covariance Matrix, Q, which is the generalized inverse of the Information Matrix; and other Matrices. The Least Squares Method which gave birth to several normal equations was used to estimate for the parameters, τ and β mathematically. Then, an illustrative example was given to ascertain the workability of this Intra-Factor procedure in testing for the main effects under some stated hypothesis for significance. But, before testing for the variance component of the main effects on the illustrative data, it was necessary to first ascertain that the data is fixedeffect and that interaction is either absent or non-significant since our interest is on “Fixed-Effect Non-interactive Models”. Thereafter, the Analysis of Variance Components for Adjusted Factor A with Unadjusted Factor B effects was carried out. This gave the result that Adjusted Factor A effect, τ, was not significant; whereas, the Unadjusted Factor B effect, β, was significant. Also, the Analysis of Variance Components was performed for Unadjusted Factor A effect with Adjusted Factor B effect and it yielded similar result as that of Adjusted Factor A with Unadjusted Factor B effects. We therefore concluded that for a Fixed Effect, Noninteractive Unbalanced Data Analysis, the Method of Intra-Factor Design can be successfully employed.
Title page…………………………….…….…………………….…i
Certification…………………………………………………….……ii
Dedication………………..………………………………………..iii
Acknowledgement………….……………………………………..iv
List of tables………….…………….…………………………….vii
Abstract……………………..……………………………………viii
CHAPTER ONE
INTRODUCTION
1.0 Introduction 1
1.1 Background of the study 1
1.2 Problem involved in random models 2
1.3 Problem of mixed effect models 3
1.4 Aim and objective of the study 4
1.5 Significance of the study 4
5
1.6 Scope/ Limitations of the study 6
1.7 Organization of the study 6
CHAPTER TWO
LITERATURE REVIEW
2.0 Introduction 7
2.1 The test for interaction effect 7
2.2 Elimination of the interaction effect 11
CHAPTER THREE
METHODOLOGY 3.0 Introduction 13
3.1 Method of data analysis 13
3.2 The Intra-factor analysis of unbalanced data 13
CHAPTER FOUR
PRESENTATION AND ANALYSIS OF DATA
4.0 Introduction 18
4.1 The least squares method of analyzing Intra-factor design 18
4.2 Test of hypothesis 22
4.3 Illustrative example on the application of the Intra-factor
design 24
4.3.1 The hypothesis 25
4.3.2 Test for fixed-effectness of the data 25
4.3.3 Test for absence of interaction 27
4.3.4 Analysis of variance for the main effects 27
CHAPTER FIVE
SUMMARY, CONCLUSION AND RECOMMENDATIONS
5.0 Introduction 30
5.1 Discussion of the main findings 30
5.2 Conclusion 31
5.3 Recommendations 32
## CHAPTER ONE
INTRODUCTION
1.0 INTRODUCTION
Estimating variance component from unbalanced data is not as
straightforward as from balanced data. This is so for two reasons. Firstly,
several methods of estimation are available (most of which reduce to the
analysis of variance method for balanced data), but no one of them has yet
been clearly established as superior to others. Secondly, all the methods
involve relatively cumbersome algebra; discussion of unbalanced data can
therefore easily deteriorate into a welter of symbols, a situation we do our
8
best to minimize here. However, we shall review some works on
unbalanced data.
1.1 GENERAL OUTLAY OF UNBALANCED DATA
Balanced data are those in which every one of the subclasses of the
model has the same number of observations, that is, equal numbers of
observations in all the subclasses. In contrast, unbalanced data are those
data wherein the numbers of observations in the subclasses of the model
are not all the same, that is, unequal number of observations in the
subclasses, including cases where there are no observations in some
classes. Thus unbalanced data refers not only to situations where all
subclasses have some data, namely filled subclasses, but also to cases
where some subclasses are empty, with no data in them. The estimation of
variance components from unbalanced data is more complicated than from
balanced data.
In many areas of research such as this, it is necessary to analyze the
variance of data, which are classified into two ways with unequal numbers
of observations falling into each cell of the classification. For data of this
kind, special methods of analysis are required because of the inequality of
the cell numbers. This we shall attempt to solve in this research work.
1.2 PROBLEM INVOLVED IN RANDOM MODELS
The problem associated with the random effect models has been the
determination of approximate F-test in testing for the main effects say, A
and B using F-ratio. In this case there would be no obvious denominator for testing the hypothesis Ho: σ² = 0; for a level of Factor A crossed with
the level of Factor B in the model such as Xijk = μ + αi +βj + λij + εijk
9
where, Xijk is the kth observation (for k =1,2,…,nij )in the ith level of Factor A and jth level of Factor B; i= 1,2,…,p; j= 1,2,…,q;
µ is the general mean; αi is the effect due to the ith level of Factor A; βj is the effect due to the jth level of Factor B; λij is the interaction between the ith level of Factor A and jth level of Factor
B; εijk is the observation error associated with Xijk.
1.3 PROBLEM OF MIXED EFFECT MODELS
According to Henderson (1953), in the customary unbalanced random model from a crossed classification, the nature of the expected value of the sum of squares is such that: E (SSA) = n.. – Σni.2 σ²A + ∑ Σnij – Σn.j2 σ²B + ∑ Σnij2 – Σ Σnij2 σ2AB + n.. ni. n.. ni. n..
α -1 σ2e ………………………………………………………………..…(1.1)
This means that with unbalanced data from crossed classification in
random model, the expected value of every sum of squares contains every
variance component of the model. But in an unbalanced mixed, with α’s being fixed and not random effects, the expectation would then be:
E (SSA) = ∑ni.αi2 – Σni.αi /n.. + ∑ Σnij2 – Σn.j2 σ2β + ∑ Σnij2 – Σ Σnij2 σ2βα + n.. n.. ni. n..
α -1 σ2e …………………………………………………….…………….(1.2)
j i
b c
j
c
i i
a
j
j
b b j j
b
b
i j j a
a
i a
i
a a
i j i a b b b
10
It is observed that the first term of (1.2) which is a function of the fixed
effect is different from that in (1.1); and this occurs in expected values of
all sum of squares terms (except for SSE). And, more importantly, the
function of the fixed effects is not the same from one expected sum of
squares term to the next. For example, with the α’s fixed, E (SSE) contains the term,
∑ (Σnijαi)2 / n.j – (Σni.αi)2 /n.. which differs from the first term in E (SSA) of (1.2). Thus
E [SSA – SSB] does not get rid of the fixed effects even though it does
eliminate terms in µ. This is true generally in mixed models, expected
values of the sum of squares contain functions of the fixed effects that
cannot be eliminated by considering linear combinations of the sum of squares. This means that the equations E (SS) = Pσ2 + σ2ef of the random model takes the term E (SS) = Pσ2 + σ2ef + q in the mixed model, where q
is a vector of the quadratic functions of the fixed-effects in the model. Hence, σ² cannot be estimated and the analysis of variance method applied
to unbalanced data cannot be used for mixed as well as for fixed-effect
models. It yields biased estimators. Simply put, with unbalanced data, the
analysis of variance method for mixed and fixed effect models lead to
biased estimators of variance components.
Mixed models involve dual estimation problems – estimating both
fixed effects and variance components.
1.4 AIM / OBJECTIVE OF THE STUDY
The aim of this work is to analyze unbalanced fixed effect non
interactive model using the Intra-Factor Design.
1.5 SIGNIFICANCE OF THE STUDY
b
i
a
j
a
i
11
As a result of dual estimation problems of the mixed model with
unbalanced data which accounted for biased estimators of variance
components, Henderson (1953), designed a method to correct this
deficiency. This he does by his method 2 which uses the data first to
estimate fixed effects of the model and then using these estimators to
by the analysis of variance method. This whole procedure was designed so
that the resulting variance estimators were not biased by the presence of
the fixed effects in the model as they were with the analysis of variance
estimators derived from the basic data. So far as the criterion of
unbiasedness was concerned, this was certainly achieved by this method.
But the general method of analyzing data adjusted according to
some estimator of the fixed effects is open to criticism on other grounds
such as: it cannot be uniquely defined, and a simplified form of it, of
which Henderson’s Method 2 is a special case, cannot be used whenever
the model includes interactions between the fixed effects and the random
effects. As such, the need for the birth of this research “Analysis of
Unbalanced Fixed-Effect Non-interactive Model”.
1.6 SCOPE / LIMITATION OF THE STUDY
This study is basically limited to fixed effect non-interactive
models.
1.7 ORGANIZATION OF THE STUDY
The work was organized in five systematic chapters. Chapter one
is made up of the introduction to the study. Other topics considered in this
chapter include: the problem involved in random models, problem of
mixed effect models, the aim and objective of the study, significance of
12
the study, the scope/limitations of the study and the organization of the
study.
In chapter two, the related literatures were reviewed as to see what
consideration. In this chapter, the following subtopics were considered:
the test for interaction effect and elimination of the interaction effect.
In chapter three, the methodology used in the study was
systematically described as to enhance the understanding of the study. The
method of data analysis was also carefully derived mathematically.
The chapter four of this study shows the presentation and analysis
of data. In doing this, the following subtopics were considered: the least
square method of analyzing the unbalanced factor design, test of
hypothesis and illustrative example on the application of the unbalanced
factor design. An example was given to illustrate the analysis of
unbalanced data using the Intra-Factor Design.
Finally, chapter five which is the concluding chapter contains the
discussion of the main findings, conclusion and recommendations.
GET THE FULL WORK
DISCLAIMER: | 2,698 | 10,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-17 | longest | en | 0.916465 |
https://www.electricaltechnology.org/2020/10/inductor-inductance-formula-equations.html | 1,718,485,390,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00840.warc.gz | 666,721,965 | 68,025 | # Formula and Equations For Inductor and Inductance
## Inductor and Inductance Formulas and Equations
The following formulas and equations can be used to calculate the inductance and related quantities of different shapes of inductors as follow.
### Inductance of Inductor:
The inductance of the inductor from the basic formula of inductor:
#### Current of the Inductor:
Where
• V is the voltage across inductor
• L is the inductance of the inductor in Henry
• Di/dt is the instantaneous rate of current change through the inductor.
• ito = current at time t = 0.
#### Reactance of the Inductor:
Inductive reactance is the opposition of inductor to alternating current AC, which depends on its frequency f and is measured in Ohm just like resistance. Inductive reactance is calculated using:
X= ωL = 2πfL
Where
• Xis the Inductive reactance
• f is the applied frequency
• L is the Inductance in Henry
#### Quality Factor of Inductor:
The efficiency of the inductor is known as quality factor & its measured by:
QF = XL/ESR
Where
• Xis the Inductive reactance
• ESR is the equivalent series resistance of the circuit.
#### Dissipation Factor of Inductor:
It is the inverse of the quality factor and it shows the power dissipation inside the inductor & its given by:
DF = tan δ = ESR/XL
Where
• DF is the dissipation factor
• δ is the angle between capacitive reactance victor & negative axis.
• Xis the capacitive reactance
• ESR is the equivalent series resistance of the circuit.
### Energy Stored in an Inductor:
The energy E stored in inductor is given by:
E = ½ Li2
Where
• E is the energy in joules
• L is the inductance in Henry
• i is the current in Amps
Related Posts:
#### Average Power of Inductor
The average power for the inductor is given by:
Pav = Li2 / 2t
Where
• = is the time in seconds.
### Inductor Current During Charge / Discharge:
Just like capacitor, the inductor takes up to 5 time constant to fully charge or discharge, during this time the current can be calculated by:
#### During Charging:
Instantaneous current of the inductor during charging is given by:
#### During Discharging:
The current during the discharging at any time t is given by:
Where
• IC is the current of the inductor
• I0 is the current at time t=0
• t is the time passed after supplying current.
• τ = L/R is the time constant of the RL circuit
Related Posts:
### Inductance Formulas
#### Helical Coil Inductance “Wheeler’s Formula”
Where:
• L is the inductance
• n is the number of turns
• h is the height
#### Spiral Coil Inductance Formula
Where:
• OR is the outermost radius in inches
• IR is the innermost radius in inches
#### Conical Coil Inductance Formula
Where:
• θ is the angle outside the cone, (assume θ ≈ 15°)
Related Formulas and Equations Posts: | 715 | 2,813 | {"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-2024-26 | longest | en | 0.903212 |
https://www.aussiestockforums.com/threads/worked-margin-loan-example-tax-implications.28585/ | 1,579,628,969,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250604849.31/warc/CC-MAIN-20200121162615-20200121191615-00458.warc.gz | 774,236,106 | 16,630 | • Australian (ASX) Stock Market Forum
# Worked Margin Loan Example + Tax Implications
Discussion in 'Stock Market Nuts and Bolts' started by Fluffmeister21, Jun 15, 2014.
1. ### Fluffmeister21
Posts:
6
0
Joined:
Jun 15, 2014
Hi All
I'm just in the process of setting up a Margin Loan and just wanted to clarify the maths on it a bit, especially in regards to the tax conductibility of the interest on the loan. I've based the below on the 37c for each \$1 tax bracket.....Mostly looking at the ability to service a loan with the dividends of a High Yield ETF. Not factoring in any capital gains.. (for those curious I'm using VHY for the maths)
https://static.vgcontent.info/crp/intl/auw/docs/etfs/profiles/VHY_profile.pdf?20140610|110100
They quote a 'grossed up' yield of 7.74% (from what i interpret of the clause it means that it extrapolates out the various franking levels of the companies it reflects to give a pre tax value of the income???)
50/50 Split with the Margin Loan.
Initial Equity Investment 10,000.00
Amount Borrowed 10,000.00
Initial Stock Price 66.80
Shares Purchased 299
Ending Stock Price 66.80
Cash Dividends 1 year \$5.04 per share.
Total Dividends through year \$1505
Margin Loan rate 8%
Interest on Margin Loan \$800
So Net Income =1505-800=705
My question surrounds the tax deductions available on the interest of the loan which in effect from what I understand would reduce my tax bill .37*800...
Net income becomes (.67*705)+(37*800)=768.35
Have I messed up my maths here???
2. ### Hodgie
Posts:
155
3
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Jun 13, 2013
You would also get the franking credits back as a tax off set. The dividends plus franking credits (grossed up dividends) are added on to increase you're taxable income while the interest charges are taken off you're taxable income as deductions. Then after tax has been calculated you add the franking credits back in to off set any tax payable.
3. ### Fluffmeister21
Posts:
6
0
Joined:
Jun 15, 2014
Yea I was working with a grosssed up figure to account for the franking credits... I believe...to treat it as Pre Tax Money for ease of calcuations.
I'm a bit confused by that in fact... they quote a yield at 5.58 but a grossed up yield of 7.54 which I assume is because not everything is fully franked else it would be 7.9 ish.
4. ### Fluffmeister21
Posts:
6
0
Joined:
Jun 15, 2014
I think i was asleep when i first posted this...as i deducted the interest twice...
50/50 Split with the Margin Loan.
Initial Equity Investment 10,000.00
Amount Borrowed 10,000.00
Initial Stock Price 66.80
Shares Purchased 299
Ending Stock Price 66.80
Cash Dividends 1 year \$5.04 per share.
Total Dividends through year \$1505
Margin Loan rate 8%
Interest on Margin Loan \$800
So Net Income =1505-800=705
Net income becomes (.67*705)=514
5. ### VSntchr
Posts:
1,729
46
Joined:
Dec 27, 2010
Income = \$1,500
Deductions = \$800
Taxable Income = \$1,500 - \$800
= \$700
Tax Payable = \$700 x 37%
= \$259
Net Income = \$1,500 - \$259
= \$1,241
I think there is a question relating to whether all of the interest expense can be claimed as a deduction; i.e. have the loan proceeds been used to create enough taxable income to completely offset the interest expense. Perhaps someone can clarify this..
6. ### craft
Posts:
2,000
304
Joined:
May 3, 2008
The 5.58% is a forecast yield and the 7.54% is the grossed up figure based on average franking levels over the previous 12 months.
[table="width: 500, class: grid"]
Equity
\$10,000
\$10,000
Debt
\$10,000
Capital
\$20,000
\$10,000
Net Dividend @ 5.58%
\$1,116
\$558
Imputation @ 7.54%
\$392
\$196
Interest
-\$800
Taxable Income
\$708
\$754
Tax on Taxable Income @39%
-\$276
-\$294
Imputation Credit
\$392
\$196
Net Tax
\$116
-\$98
Net Cash
\$432
\$460
[/table]
It will cost you \$28 in net cash to hold the extra \$10,000 of exposure through the margin loan after taking into account the tax effect. Make a capital gain, you win; Capital loss you loose, the margin loan whilst basically break even on cash flow, ups your exposure to capital growth/loss.
7. ### Fluffmeister21
Posts:
6
0
Joined:
Jun 15, 2014
Just to clarify, one doesn't actually calculate the taxable income based on the net dividend + imputation, this is purely to calculate the net after tax impact as opposed to what one would declare to the ATO?
8. ### Fluffmeister21
Posts:
6
0
Joined:
Jun 15, 2014
Just to clarify as far as the ATO is concerned though when you declare income they want the 558 number, the maths above is just to reflect the full after tax implication?
Posts:
6 | 1,294 | 4,578 | {"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-2020-05 | longest | en | 0.925632 |
https://goprep.co/ex-7.3-q1-abc-and-dbc-are-two-isosceles-triangles-on-the-i-1njg77 | 1,627,857,645,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154277.15/warc/CC-MAIN-20210801221329-20210802011329-00308.warc.gz | 299,514,074 | 30,400 | Δ ABC and &
It is given in the question that:
are two isosceles triangles
(i) In ,
AB = AC (Triangle ABC is isosceles)
BD = CD (Triangle DBC is isosceles)
Three sides equal (SSS)Definition: Triangles are congruent if all three sides in one triangle are congruent to the corresponding sides in the other.
Therefore,
By SSS axiom,
(ii) In
AP = AP (Common)
PAB = PAC (By c.p.c.t)
AB = AC (Triangle ABC is isosceles)
Side-Angle-Side (SASRule. Side-Angle-Side is a rule used to prove whether a given set of triangles are congruent. If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent
Therefore,
By SAS axiom,
(iii) PAB = PAC (By c.p.c.t)
AP bisects A (i)
Also,
In
PD = PD (Common)
BD = CD (Triangle DBC is isosceles)
BP = CP ( so by c.p.c.t)
Side-Side-Side (SSSRule. Side-Side-Side is a rule used to prove whether a given set of triangles are congruent. The SSS rule states that: If three sides of one triangle are equal to three sides of another triangle, then the triangles are congruent.
Therefore,
By SSS axiom,
BDP = CDP (By c.p.c.t) (ii)
By (i) and (ii), we can say that AP bisects A as well as D
(iv) BPD = CPD (By c.p.c.t)
And,
BP = CP (i)
Also,
BPD + CPD = 180o (BC is a straight line)
2BPD = 180o
BPD = 90o (ii)
From (i) and (ii), we get
AP is the perpendicular bisector of BC
Hence, Proved.
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# Problem 5. Triangle Numbers
Solution 143108
Submitted on 27 Sep 2012 by Damien
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
%% n = 1; t = 1; assert(isequal(triangle(n),t))
2 Pass
%% n = 3; t = 6; assert(isequal(triangle(n),t))
3 Pass
%% n = 5; t = 15; assert(isequal(triangle(n),t))
4 Pass
%% n = 30; t = 465; assert(isequal(triangle(n),t)) | 162 | 482 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-51 | latest | en | 0.610828 |
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carresmagiques.blogspot.com | 1,722,972,680,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640508059.30/warc/CC-MAIN-20240806192936-20240806222936-00862.warc.gz | 116,615,524 | 43,227 | ## Saturday 11 May 2024
### 137 Plus-or-Minus Groups of Magic Tori of Order 4, Linked by Plus-and-Minus Operations
Pandiagonal and semi-pandiagonal magic tori of order 4 connected by plus-and-minus links
### Previous Studies
General information about the 255 magic tori of order 4, and about the 880 magic squares that those tori display, can be found in a previous article, published on the 15th June 2023, and entitled "440 Torus-Opposite Pairs of the 880 Frénicle Magic Squares of Order-4".
The 137 complete-torus, same-integer, plus-or-minus groups of magic tori of order 4 have been presented in a previous article, published on the 24th April 2024, and entitled "Plus or Minus Groups of Magic Tori of Order 4".
Further information about the magic tori can be found in the lists of references that are appended to the papers in each of the above-mentioned articles.
### Searching for 2-or-more-integer plus-and-minus links between the 137 ± Groups of Magic Tori
As we are looking for connections between complete-torus, same-integer, plus-or-minus groups of magic tori, it seems logical to search for links with 2-or-more-integer plus-and-minus operations. The closest links between the tori will be those where only 4 out of the 16 numbers are changed, and these will be for example 3/4 & 1/4: (±0, ±1), or 3/4 & 1/4: (±0, ±2) etc., depending on the differences between the cell entries.
A variant of these links will be those where 12 out of the 16 numbers are changed, and these can be written, for example, as 3/4 & 1/4: (±1, ±0), or as 3/4 & 1/4: (±2, ±0) etc., depending on the differences between the cell entries. However, a change of only 1/4 of the cell entries seems less disruptive than a change of 3/4 of these, and we can therefore suppose that the previously-mentioned 3/4 & 1/4: (±0, ±1), or 3/4 & 1/4: (±0, ±2) are closer links.
Close links also include those where only 8 out of the 16 numbers are changed, and these will be, for example, 1/2 & 1/2: (±0, ±1), or 1/2 & 1/2: (±0, ±2) etc., depending on the differences between the torus cell entries.
In some cases 2-integer plus-and-minus operations cannot be found, but more-complex 3-integer links may exist, such as 1/2, 3/8 & 1/8: (±2, ±0, ±1) or 1/2, 1/4 & 1/4: (±4, ±0, ±8). 3-integer links like these are frequent between the paired complementary and self-complementary torus groups.
Sometimes even 3-integer plus-and-minus operations are unavailable, but there exist 4-integer links, such as 1/4, 1/4, 1/4, & 1/4: (±0, ±2, ±4, ±6), or 3/8, 1/4, 1/4 & 1/8: (±3, ±0, ±6, ±9), which, if not particularly close, show interesting patterns. Links like these are common within the paired complementary torus groups that are to be found in the pages that follow.
### Some Precisions Concerning the Enclosed Presentation
Because of its A3 format, this enclosed PDF file can at first seem unwieldy, especially when it is displayed on small-screen mobile phones. However the large display size allows same-page illustrations of complete sets of ± groups, which can be comprised of up to 24 examples. Please bear in mind that, in order to reduce space requirements and facilitate reading, most of the 137 ± groups are represented by single magic square viewpoints of only one of their magic tori. But occasionally, a same group is represented more than once, so as to facilitate the comprehension of multiple links. To find our which other magic tori are within a ± group, please refer to the details given in the paper "Plus or Minus Groups of Magic Tori of Order 4", already mentioned above.
In order to simplify the annotation of the pages of the paper enclosed below, the links between the ± groups are written as ± operations. This is slightly ambiguous, as there must always be a same number of plusses as minuses of any particular integer. Here therefore, the ± sign in front of an integer will mean an equal number of plusses and minuses of that integer. This will become evident as soon as we compare any pair of magic square viewpoints which are connected by the links in question. For full details, please refer to the paper below.
## Wednesday 24 April 2024
### Plus or Minus Groups of Magic Tori of Order 4
As the positive integer entries of normal magic tori, and of their displayed magic squares, are always arranged in same-sum orthogonal arrays, it would seem logical to compensate any additions or subtractions to those entries, by making corresponding subtractions or additions. How can we develop this simple observation?
### Preliminary Definitions
A normal magic square is an N x N array of same-sum rows, columns and main diagonals. If the magic diagonal condition is not satisfied, the square is deemed to be semi-magic. The torus, upon which a magic or semi-magic square is displayed, can be visualised by joining the opposite edges of the magic or semi-magic square in question.
There are N x N square viewpoints on the surface of a torus of order N. A normal magic torus has N x N arrays of same-sum latitudes and longitudes, and at least one magic intersection of magic diagonals on its surface. In his paper "Conformal Tiling on a Torus" published by Bridges in 2011, John M. Sullivan shows that, on a square torus T1, 0, the diagonal grid lines “form (1, ±1) diagonals on the torus, each of which is a round (Villarceau) circle in space.”
Semi-magic tori can also have magic diagonals, but the latter can never produce the magic intersections required for magic squares, either because the magic diagonals are single or parallel, or because they do not intersect correctly. Additional information about magic and semi-magic tori, with further explanations of their diagonals, can be found in the references at the end of the enclosed paper.
In order 4, starting with any magic torus, and examining cases where half of the torus numbers are subjected to equal additions, and the other half are subjected to same-integer subtractions, we notice the following: From its initial state (which we can call 0), each magic square can be transformed by four plus or minus operations that will either produce alternative magic or semi-magic square viewpoints of the same torus, or square viewpoints of other essentially different magic or semi-magic tori. Here's a simple example of two transformations:
3 Magic Tori of Order 4 linked by by Complete-Torus, Same-Integer, Plus or Minus Operations
### Complete-Torus, Same-Integer, Plus or Minus Groups in Order 4
Please note that, in order 4, the orthogonal totals of magic and semi-magic tori are always 34. Therefore, only the diagonal totals, which can vary, are announced here. These totals are interesting because of their symmetries, and in the example illustrated above, there are pairs of diagonals that always sum to 68 (twice the magic sum). Each of the 255 magic tori of order 4 is similarly examined, and the results are given in tables and observations at the end of the following paper: "Groups of Magic Tori of Order 4 Assembled by Complete-Torus, Same-Integer, Plus or Minus Operations"
### Complete-Torus, Same-Integer, Plus or Minus Groups in Orders N > 4
The same method cannot always be applied to higher even orders. A quick look at some examples from even orders N = 6, N = 8, and N = 10 shows that certain magic tori have either no solutions whatsoever, or only one that can be used in a complete-torus same-integer plus or minus matrix to produce another orthogonally magic torus. In singly-even orders, even divisors with odd quotients have to be ruled out. Each case will need to be tested, and systematic computer checks will be necessary for these higher even orders.
An adaptation is of course required for odd orders, as their odd square totals do not have even integer divisors. But partial-torus same-integer divisions produce plus or minus solutions, and there are a variety of approaches.
The following paper, entitled "Examples of Partial Groups of Magic Tori of Orders N > 4 Assembled by Complete-Torus (or Near-Complete in Odd Orders) Same-Integer, Plus or Minus Operations" shows how the method used for order 4 gives some good results in higher orders:
## Thursday 15 June 2023
### Finding the torus-opposite pairs of magic squares in even-orders:
The following diagram, which illustrates the array of essentially-different square viewpoints of a basic magic torus of order-n, starting here with the 4x4 magic square that has the Frénicle index n° 1, shows how a torus-opposite magic square of order-4 can be identified:
Array of 16 essentially-different square viewpoints of a basic magic torus of order-4
This basic magic torus, now designated as type n° T4.05.1.01, can be seen to display a torus-opposite pair of Frénicle-indexed magic squares n° 1-458 (with Dudeney pattern VI), as well as 7 torus-opposite pairs of semi-magic squares that the discerning reader will easily be able to spot.
The relative positions of the numbers of torus-opposite squares can be expressed by a simple plus or minus vector. There are always two equal shortest paths towards the far side of the torus: These are, in toroidal directions, east or west along the latitudes, and in poloidal directions, north or south along the longitudes of the doubly-curved 2D surface. So, if the even-order is n, then:
v = ( ± n/2, ± n/2 )
"Tesseract Torus" by Tilman Piesk CC-BY-4.0 https://commons.wikimedia.org/w/index.php?curid=101975795
Torus-opposite squares are not just limited to basic magic tori, as they also exist on pandiagonal, semi-pandiagonal, partially pandiagonal, and even semi-magic tori of even-orders. Therefore, for order-4, we can either say that there are 880 magic squares, or we can announce that there are 440 torus-opposite pairs of magic squares. Similarly, the 67,808 semi-magic squares of order-4 can be expressed as 33,904 torus-opposite pairs of semi-magic squares, etc.
### Why torus-opposite pairs of magic squares cannot exist in odd-orders
A torus-opposite magic square always exists in even-orders because an even-order magic square has magic diagonals that produce a first magic intersection at a centre between numbers, and another magic intersection at a second centre between numbers on the far side of the torus (at the meeting point of the four corners of the first magic square viewpoint). However, in odd-orders, where a magic square has magic diagonals that produce a magic intersection over a number, then a sterile non-magic intersection always occurs between numbers on the far side of the torus (at the meeting point of the four corners of the initial magic square). This is why torus-opposite pairs of magic squares cannot exist in the odd-orders.
### 255 magic tori of order-4, listed by type, with details of the 440 torus-opposite pairs of the 880 Frénicle magic squares of order-4
The enumeration of the 255 magic tori of order-4 was first published in French on the 28th October 2011, before being translated into English on the 9th January 2012: "255 Fourth-Order Magic Tori, and 1 Third-Order Magic Torus." In this previous article, the corresponding 880 fourth-order magic squares were listed by their Frénicle index numbers, but not always illustrated. The intention of the enclosed paper is therefore to facilitate the understanding of the magic tori of order-4, by portraying each case.
Here, Frénicle index numbers continue to be used, as they have the double advantage of being both well known and commonly accepted for cross-reference purposes. But please also note, that now, in order to simplify the visualisation of the magic tori, their displayed magic squares are not systematically presented in Frénicle standard form.
In the illustrations of the 255 magic tori of order-4, listed by type with the presentation of their magic squares, the latter are labelled from left to right with their Frénicle index number, followed by, in brackets, the Frénicle index number of their torus-opposite magic square, and finally by the Roman numeral of their Dudeney complementary number pattern. Therefore, for the magic square of order-4 with Frénicle index n° 1 (that forms a torus-opposite pair of magic squares with Frénicle index n° 458; both squares having the same Dudeney pattern VI), its label is 1 (458) VI.
To find the "255 magic tori of order-4, listed by type, with details of the 440 torus-opposite pairs of the 880 Frénicle magic squares of order-4" please consult the following PDF:
## Tuesday 7 June 2022
### Polyomino Area Magic Tori
A magic torus can be found with any magic square, as has already been demonstrated in the article "From the Magic Square to the Magic Torus". In fact, there are n^2 essentially different semi-magic or magic squares, displayed by every magic torus of order-n. Particularly interesting to observe with pandiagonal (or panmagic) examples, a magic torus can easily be represented by repeating the number cells of one of its magic square viewpoints outside its limits. However, once we begin to look at area magic squares, it becomes much less evident to visualise and construct the corresponding area magic tori using repeatable area cells, especially when the latter have to be irregular quadrilaterals... The following illustration shows a sketch of an area magic torus of order-3 that I created back in January 2017. I call it a sketch because it may be necessary to use consecutive areas starting from 2 or from 3, should the construction of an area magic torus of order-3, using consecutive areas from 1 to 9, prove to be impossible. And while it can be seen that such a torus is theoretically constructible, many calculations will be necessary to ensure that the areas are accurate, and that the irregular quadrilateral cells can be assembled with precision:
At the time discouraged by the complications of such geometries, I decided to suspend the research of area magic tori. But since the invention of area magic squares, other authors have introduced some very interesting polyomino versions that open new perspectives:
On the 20th May 2021, Morita Mizusumashi (盛田みずすまし @nosiika) tweeted a nice polyomino area magic square of order-3 constructed with 9 assemblies of 5 to 13 monominoes. On the 21st May 2021, Yoshiaki Araki (面積魔方陣がテセレーションみたいな件 @alytile) tweeted several order-4 and order-3 solutions in a polyomino area magic square thread. These included (amongst others) an order-4 example constructed with 16 assemblies of 5 to 20 dominoes. On the 24th May 2021, Yoshiaki Araki then tweeted an order-3 polyomino area magic square constructed using 9 assemblies of 1 to 9 same-shaped pentominoes! Edo Timmermans, the author of this beautiful square, had apparently been inspired by Yoshiaki Araki's previous posts! Since the 22nd June 2021, Inder Taneja has also published a paper entitled "Creative Magic Squares: Area Representations" in which he studies polyomino area magic using perfect square magic sums.
### Intention and Definition
The intention of the present article is to explore the use of polyominoes for area magic torus construction, with the objective of facilitating the calculation and verification of the cell areas, while avoiding the geometric constraints of irregular quadrilateral assemblies. Here, it is useful to give a definition of a polyomino area magic torus:
1/ In the diagram of the torus, the entries of the cells of each column, row, and of at least two intersecting diagonals, will add up to the same magic sum. The intersecting magic diagonals can be offset or broken, as the area magic torus has a limitless surface, and can therefore display semi-magic square viewpoints.
2/ Each cell will have an area in proportion to its number. The different areas will be represented by tiling with same-shaped holeless polyominoes.
3/ The cells can be of any regular or irregular rectangular shape that results from their holeless tiling.
4/ Depending on the order-n of the area magic torus, each cell will have continuous edge connections with contiguous cells (and these connections can be wrap-around, because the torus diagram represents a limitless curved surface).
5/ The vertex meeting points of four cells can only take place at four convex (i.e. 270° exterior angled) vertices of each of the cells.
### Polyomino Area Magic Tori (PAMT) of Order-3
Magic Torus index n° T3, of order-3. Magic sums = 15.Please note that this is not a Polyomino Area Magic Torus,but it is the Agrippa "Saturn" magic square, after a rotation of +90°, in Frénicle standard form.
Polyomino Area Magic Torus (PAMT) of order-3. Magic sums = 15. Tetrominoes.Consecutively numbered areas 1 to 9, in an irregular rectangular shape of 180 units.
Polyomino Area Magic Torus (PAMT) of Order-3. Magic sums = 18. Trominoes.Consecutively numbered areas 2 to 10, in an irregular rectangular shape of 162 units.
Polyomino Area Magic Torus (PAMT) of order-3. Magic sums = 24. Trominoes.Consecutively numbered areas 4 to 12, in an oblong 12 ⋅ 18 = 216 units.
Polyomino Area Magic Torus (PAMT) of Order-3. Magic sums = 36. Trominoes Version 1.Square 18 ⋅ 18 = 324 units.
Polyomino Area Magic Torus (PAMT) of Order-3. Magic sums = 36. Trominoes Version 2.Square 18 ⋅ 18 = 324 units.
Polyomino Area Magic Torus (PAMT) of Order-3. Magic sums = 60. Pentominoes Version 1.Square 30 ⋅ 30 = 900 units.
Polyomino Area Magic Torus (PAMT) of order-3. Magic sums = 24. Monominoes.Consecutively numbered areas 4 to 12, in an irregular rectangular shape of 72 units.
Polyomino Area Magic Torus (PAMT) of order-3. Magic sums = 27. Monominoes Version 1.Consecutively numbered areas 5 to 13, in a square 9 ⋅ 9 = 81 units.
Polyomino Area Magic Torus (PAMT) of order-3. Magic sums = 27. Monominoes Version 2.Consecutively numbered areas 5 to 13, in a square 9 ⋅ 9 = 81 units.
### Polyomino Area Magic Tori (PAMT) of Order-4
Magic Torus index n° T4.198, of order-4. Magic sums = 34.Please note that this is not a Polyomino Area Magic Torus,but it is a pandiagonal torus represented by a pandiagonal square that has Frénicle index n° 107.
The pandiagonal torus above displays 16 Frénicle indexed magic squares n° 107, 109, 171, 204, 292, 294, 355, 396, 469, 532, 560, 621, 691, 744, 788, and 839. It is entirely covered by 16 sub-magic 2x2 squares. The torus is self-complementary and has the magic torus complementary number pattern I. The even-odd number pattern is P4.1. This torus is extra-magic with 16 extra-magic nodal intersections of 4 magic lines. It displays pandiagonal Dudeney I Nasik magic squares. It is classified with a Magic Torus index n° T4.198, and is of Magic Torus type n° T4.01.2. Also, when compared with its two pandiagonal torus cousins of order-4, the unique Magic Torus T4.198 of the Multiplicative Magic Torus MMT4.01.1 is distinguished by its total self-complementarity.
Pandiagonal Polyomino Area Magic Torus (PAMT) of order-4. Magic sums = 34. Pentominoes.Index PAMT4.198, Version 1, Viewpoint 1/16, displaying Frénicle magic square index n° 107. Consecutively numbered areas 1 to 16, in an oblong 34 ⋅ 20 = 680 units.
Pandiagonal Polyomino Area Magic Torus (PAMT) of order-4. Magic sums = 50. Dominoes.Version 1, Viewpoint 1/16.Consecutively numbered areas 5 to 20, in a square 20 ⋅ 20 = 400 units.
Pandiagonal Polyomino Area Magic Torus (PAMT) of Order-4. Sums = 50. Dominoes.Version 1, Viewpoint 16/16.Consecutively numbered areas 5 to 20, in an irregular rectangular shape of 400 units.
### Polyomino Area Magic Tori (PAMT) of Order-5
Pandiagonal Torus type n° T5.01.00X of order-5. Magic sums = 65.Please note that this is not a Polyomino Area Magic Torus.
This pandiagonal torus of order-5 displays 25 pandiagonal magic squares. It is a direct descendant of the T3 magic torus of order-3, as demonstrated in page 49 of "Magic Torus Coordinate and Vector Symmetries" (MTCVS). In "Extra-Magic Tori and Knight Move Magic Diagonals" it is shown to be an Extra-Magic Pandiagonal Torus Type T5.01 with 6 Knight Move Magic Diagonals. Note that when centred on the number 13, the magic square viewpoint becomes associative. The torus is classed under type n° T5.01.00X (provisional number), and is one of 144 pandiagonal or panmagic tori type 1 of order-5 that display 3,600 pandiagonal or panmagic squares. On page 72 of "Multiplicative Magic Tori" it is present within the type MMT5.01.00x.
Pandiagonal Polyomino Area Magic Torus (PAMT) of order-5. Magic sums = 65. Hexominoes.Index PAMT5.01.00X, Version 1, Viewpoint 1/25. Consecutively numbered areas 1 to 25, in an oblong 65 ⋅ 30 = 1950 units.
Pandiagonal Polyomino Area Magic Torus (PAMT) of Order-5. Magic sums = 125. Monominoes.Version 1, Viewpoint 1/25.Consecutively numbered areas 13 to 37, in a square 25 ⋅ 25 = 625 units.
Pandiagonal Polyomino Area Magic Torus (PAMT) of Order-5. Magic sums = 125. Monominoes.Version 2, Viewpoint 1/25.Consecutively numbered areas 13 to 37, in a square 25 ⋅ 25 = 625 units.
### Polyomino Area Magic Tori (PAMT) of Order-6
Partially Pandiagonal Torus type n° T6 of order-6. Magic sums = 111.Please note that this is not a Polyomino Area Magic Torus.
Harry White has kindly authorised me to use this order-6 magic square viewpoint. With a supplementary broken magic diagonal (24, 19, 31, 3, 5, 29), this partially pandiagonal torus displays 4 partially pandiagonal squares and 32 semi-magic squares. In "Extra-Magic Tori and Knight Move Magic Diagonals" it is shown to be an Extra-Magic Partially Pandiagonal Torus of Order-6 with 6 Knight Move Magic Diagonals. This is one of 2627518340149999905600 magic and semi-magic tori of order-6 (total deduced from findings by Artem Ripatti - see OEIS A271104 "Number of magic and semi-magic tori of order n composed of the numbers from 1 to n^2").
Partially Pandiagonal Polyomino Area Magic Torus (PAMT) of order-6. Magic sums = 111.Heptominoes, Version 1, Viewpoint 1/36. Consecutively numbered areas 1 to 36, in an oblong 111 ⋅ 42 = 4662 units.
Partially Pandiagonal Polyomino Area Magic Torus (PAMT) of order-6. Sums = 147.Dominoes, Version 1, Viewpoint 1/36.Consecutively numbered areas 7 to 42, in a square 42 ⋅ 42 = 1764 units.
### Observations
As they are the first of their kind, these Polyomino Area Magic Tori (PAMT) can most likely be improved: The examples illustrated above are all constructed with their cells aligned horizontally or vertically; and though it is convenient to do so, because it allows their representation as oblongs or squares, this method of constructing PAMT is not obligatory. Representations of PAMT that have irregular rectangular contours may well give better results, with less-elongated cells and simpler cell connections.
While the use of polyominoes has the immense advantage of allowing the construction of area magic tori with easily quantifiable units, it also introduces the constraint of the tiling of the cells. It has been seen in the examples above that the PAMT can be represented as oblongs or as squares, while other irregular rectangular solutions also exist. A normal magic square of order-3 displays the numbers 1 to 9 and has a total of 45, which is not a perfect square. As the smallest addition to each of the nine numbers 1 to 9, in order to reach a perfect square total is four (45 + 9 ⋅ 4 = 81), this implies that when searching for a square PAMT with consecutive areas of 1 to 9, in theory the smallest polyominoes for this purpose will be pentominoes.
But to date, in the various shaped examples of PAMT shown above, the smallest cell area used to represent the area 1 is a tetromino, as this gives sufficient flexibility for the connections of a nine-cell PAMT of order-3 with consecutive areas of 1 to 9. Edo Timmermans has already constructed a Polyomino Area Magic Square of order-3 using pentominoes for the consecutive areas of 1 to 9, but it seems that such polyominoes cannot be used for the construction of a same-sized and shaped PAMT of order-3. Straight polyominoes are always used in the examples given above, as these facilitate long connections, but other polyomino shapes will in some cases be possible.
We should keep in mind that the PAMT are theoretical, in that, per se, they cannot tile a torus: As a consequence of Carl Friedrich Gauss's "Theorema Egregium", and because the Gaussian curvature of the torus is not always zero, there is no local isometry between the torus and a flat surface: We can't flatten a torus without distortion, which therefore makes a perfect map of that torus impossible. Although we can create conformal maps that preserve angles, these do not necessarily preserve lengths, and are not ideal for our purpose. And while two topological spheres are conformally equivalent, different topologies of tori can make these conformally distinct and lead to further mapping complications. For those wishing to know more, the paper by Professor John M. Sullivan, entitled "Conformal Tiling on a Torus", makes excellent reading.
Notwithstanding their theoreticality, the PAMT nevertheless offer an interesting field of research that transcends the complications of tiling doubly-curved torus surfaces, while suggesting interesting patterns for planar tiling: For those who are not convinced by 9-colour tiling, 2-colour pandiagonal tiling can also be a good choice for geeky living spaces:
Tiling with irregular rectangular shaped PAMT of order-3. Monominoes. S=24.
Tiling with irregular rectangular shaped PAMT of order-3. Tetrominoes. S=15.
Tiling with irregular rectangular shaped PAMT of order-3. Trominoes. S=18.
Tiling with oblong PAMT of order-3. Trominoes. S=24.
Tiling with oblong pandiagonal PAMT of order-4. Pentominoes. S=34.
Tiling with oblong pandiagonal PAMT of order-5. Hexominoes. S=65.
Tiling with square pandiagonal PAMT of order-5. Monominoes. S=125.
Tiling with oblong partially pandiagonal PAMT of order-6. Heptominoes. S=111.
Tiling with square partially pandiagonal PAMT of order-6. Dominoes. S=147.
There are still plenty of other interesting PAMT that remain to be found, and I hope you will authorise me to publish or relay your future discoveries and suggestions!
## Monday 20 September 2021
### How Dürer's "Melencolia I" is a painful but liberating metamorphosis!
The title of this post may at first seem rather strange, especially when we know that the main subjects of these pages are "Magic Squares, Spheres and Tori." However, the famous "Melencolia I" engraving by Albrecht Dürer does depict, amongst other symbols, a magic square of order-4 (already examined in "A Magic Square Tribute to Dürer, 500+ Years After Melencolia I," and "Pan-Zigzag Magic Tori Magnify the "Dürer" Magic Square").
In the section reserved for correspondence at the end of the post "A Magic Square Tribute to Dürer, 500+ Years After Melencolia I," I have recently received some interesting comments from Rob Sellars. Rob looks at Dürer's engraving from a Judaic point of view and describes the bat-like animal (at the top left) as a flying chimera which has a combination of the Tinshemet features of the "flying waterfowl and the earth mole." Rob's description has made me look harder at this beast, and in doing so, I have noticed some aspects that explain the very essence of "Melencolia I."
## The Historical Context
The year 1514 CE came during a turbulent historical period, just three years before the Protestant Reformation, and the seemingly endless wars of religion which would follow. When Albrecht Dürer created "Melencolia I," he was expressing the philosophical, scientific, and humanist ideas of fifteenth-century Italy, and thus contributing to the beginning of a new phase of the Renaissance. Dürer was one of the first artists of Northern Europe to understand the importance of the Greek classics, and particularly the ideas of Plato and Socrates. The Renaissance idea was revolutionary, as it suggested that everyone was created "in imago Dei," in the image of God, and was capable of developing himself, or herself, to participate in the creation of the universe. This idea was now being gradually transmitted to all classes of society, thanks to the invention of the printing press; but it required a metaphorical language which could be deciphered by all, especially in a largely illiterate population. Although most of Dürer’s prints were intended for this wide public, his three master engravings (“Meisterstiche”), which include "Melencolia I," were aimed instead at a more discerning circle of fellow humanists and artists. The messages were more intellectual, using subtle symbols that would not be evident for common men, but could be decrypted by those initiated in the art.
## The Metamorphosis of the Flying Creature
Detail of the bat-like beast at the top left-hand side of "Melencolia I" which was engraved by Albrecht Dürer in 1514
At first sight, the cartouche at the top left-hand side of Dürer's "Melencolia I" seems to be a flying bat, bearing the title of the engraving on its open wings. The length and thickness of the tail both look oversized, but we can suppose that Dürer was using his artistic licence to amplify the visual impact of the swooping beast. Nearly all species of bats have tails, even if most (if not all) of these, are shorter and thinner than the one that Dürer has depicted.
But looking again with more attention, we can see that, quite weirdly, the body of the animal is placed above its wings, which is impossible unless the bat is flying upside-down! Closer examination suggests that this is not the case, as the mouth and eyes of the beast are clearly those of an animal with an upright head. All the same, we might well ask where the hind feet are, and how the creature can possibly make a safe landing without these!
Looking once again more closely, we can see another, even more troubling detail, in that the “wings,” which carry the title of the engraving, are in fact two large strips of ragged skin, ripped outwards from the belly, as if the animal has disembowelled itself!
Judging from the thickness of its tail and the form of its head, the airborne creature was initially a rat before it began its painful metamorphosis. It has since carried out an auto-mutilation, and is now showing its inner melancholy to the outer world, but at the same time flying free with its hard-earned wings!
Symbolically, the cartouche is telling us that ""Melencolia I" is a painful metamorphosis which precedes a liberating "Renaissance!""
## How Melancholy leads to Renaissance
During 1514 CE the artist's mother, Barbara Dürer (née Holper), passed away, or “died hard” as he described it, and we can therefore suppose that Dürer’s grief would have been a strong catalyst of the very melancholic atmosphere depicted in his “Melencolia I.” The melancholy, referred to in the title of the engraving, is illustrated by an extraordinary collection of symbols that fill the scene. Some of these are tools associated with craft and carpentry. Others are objects and instruments that refer to alchemy, geometry or mathematics. In addition to the bat-like beast, the sky also contains what might be a moonbow and a comet. Further symbols include a putto seated on a millstone, and a robust winged person, also seated, which could well be an allegorical self-portrait. These, and many other symbols, are the object of multiple interpretations by various authors. Some scholars consider the engraving to be an allegory, which can be interpreted through the correct comprehension of the symbols, while others think that the ambiguity is intentional, and designed to resist complete interpretation. I tend to agree with the latter point of view, and think that the confusion symbolises the unfinished studies and works of the main melancholic figure; an apprentice angel, who believes that despite his worldly efforts, he lacks inspiration, and is not making sufficient progress.
Notwithstanding the melancholy that reigns, there is still hope: The 4 x 4 magic square, for example, has the same dimension as Agrippa's Jupiter square, a talisman that supposedly counters melancholy. The intent expression of the main winged person suggests a determination to overcome his doubts, and transcend the obstacles that continue to block his progression. Positive symbols of a resurrection or "Renaissance" are also plainly visible, not only in the hard-earned wings of the flying creature, but also in the growing wings that Dürer gives himself in his portrayal as the apprentice angel.
"Melencolia I" engraved by Albrecht Dürer in 1514
On page 171 of his book entitled "The Life and Art of Albrecht Dürer," Erwin Panofsky considers that "Melencolia I" is the spiritual self-portrait of the artist. There is indeed much resemblance between the features of the apprentice angel, and those of the engraver in previous self-portraits.
Dürer had already adopted a striking religious pose in his last declared self-portrait of 1500 CE, giving himself a strong resemblance to Christ by respecting the iconic pictorial conventions of the time. In other presumed self-portraits, (but not declared as such), Dürer had also presented himself in a Christic manner; in his c.1493 "Christ as a Man of Sorrows;" and in his 1503 "Head of the Dead Christ." What is more, Dürer inserted his self-portraits in altarpieces; in 1506 for the San Bartolomeo church in Venice ("Feast of Rose Garlands"); in 1509 for the Dominican Church in Frankfurt ("Heller Altarpiece"); and in 1511 for a Chapel in Nuremberg's "House of Twelve Brothers" ("Landauer Altarpiece" or "The Adoration of the Trinity"). Thus Dürer was already a master of religious self-portraiture when he engraved "Melencolia I" in 1514, and he might well have continued in the same manner. But this time, probably because the theological, philosophical and humanistic ideas of the Renaissance were not only spiritually, but also intellectually inspiring, he went even further, and gave himself wings!
## Acknowledgement
Passages of "The Historical Context" are inspired by the writings of Bonnie James, in her excellent article "Albrecht Dürer: The Search for the Beautiful In a Time of Trials" (Fidelio Volume 14, Number 3, Fall 2005), a publication of the Schiller Institute.
## Latest Development
After reading this article, Miguel Angel Amela (who like me, is not only interested in magic squares, but also in "Melencolia I") sent me his thanks by email, and enclosed "a paper of 2020 about a painful love triangle..." His paper is entitled "A Hidden Love Story" and interprets the "portrait of a young woman with her hair done up," which was first painted by Albrecht Dürer in 1497, and then reproduced in an engraving by Wenceslaus Hollar, almost 150 years later in 1646. Miguel's story is captivating, and I wish to thank him for kindly authorising me to publish it here. | 8,709 | 34,838 | {"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} | 3.0625 | 3 | CC-MAIN-2024-33 | latest | en | 0.906788 |
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## Errata
• Posts: 14
### shaungandrews created the topic: Errata
Hi Guys
Question on errata published on the website..
"19.08.2019
Page 130. Answer to Question No 4. The GS out was used instead of the GS home. The last line of the calculation should have read "178 ÷ 130 = 82 minutes". The corrrect answer is (b)"
The question asked for the distance and time taken from ETP to Bravo, so we should be using GS out @ 190 and not GS home @ 130. Should the original answer (d - 56min) in the textbook not stand?
• Posts: 17
### SJM replied the topic: Errata
Hi Shaun,
I believe you're correct...
TAS = 160
A->B = 300nm
A->B = 30kt TW
GSout = 160+30 = 190
GShome = 160-30=130
ETP (CP) = Distance x GShome / 2TAS
= 300 x 130/(2(160))
= 122nm
ETP to Bravo distance = 300 - 122 = 178nm
ETP to Bravo time = 178nm @ 190kts = 56 min | 475 | 1,746 | {"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} | 3.15625 | 3 | CC-MAIN-2020-24 | latest | en | 0.923819 |
http://www.downloadmela.com/infosys-nbspplacement-papernbspnbspnbspaptitude-puzzlenbspnbspnbsp--60%3Cfd%3EWw0s%2BNfihpjqSQmkMNdF%2BA%3D%3D | 1,553,325,712,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202728.21/warc/CC-MAIN-20190323060839-20190323082839-00006.warc.gz | 278,833,241 | 7,316 | Infosys Placement Paper Aptitude - Puzzle -
#### Infosys Placement Paper Aptitude - Puzzle -
• Posted by FreshersWorld
7 Jan, 2012
1. There are 4 married couples out ofwhich 3 a group
isneeded . Butther
should not be his of her spouse .How nmany groups are possible ?
Ans 32
2.In the 4 digits 1,2,3,4 how many 4 digited numbers are
possible
which are divisable by 4? Repetetions are
allowed
Ans 64
3. Twow men are goingalong a trackf rail in the opposite
direction.
One goods train crossed the first person in 20 sec.
After 10 min
the train crossed the other person who is commingin
opposite direction
in 18 sec .Afterthe train
haspassed, when thetwo persons will meet?
Approx 72min check it once.
no .of boys .
The no.of boys > no. of girls .The no.of girls >
no.of
familyi conditions
1.No family is without a child
2 Every girl has atleast one brotherand sister .
Ans c>a>b>g>f; 9 6 5 4 3 .
6.There are4 boys Anand ,Anandya ,Madan and Murali with
nicmnames perich ,zomie ,drummy and madeena not in thesame
order
Some com=nditons
Ans Anand : Perich
Anandya: drummy
7. Thereare2diomans ,1 spadeand1 club and 1ace and also 1king ,1
jack
and 1 aceare arranged in a straight line
1.The king is at third place
2.Theleft of jack is a heart and itsright is king
3. No two red colours arein consequtive.
4.The queensareseperated by two cards. Write the orderor which suits
(hearts ,clubs )and names(jacks
queensetc.)
8. Writeeach statementis true or false 8M
1.The sum of the1st three statements and the2nd false statement
givesthe true statement.
2.The no.oftrue statements falsestatement
3. The sum of2nd true statement and 1st
falsestatement gives the first
true statement.
4. Thereareatmost 3 falsestatements
5.There is no two
consequtive true statements
6.If this containsonly 1-5 statements ,theanswer of this is
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9.Question on Venn diagram.
All handsome are also fair skinned
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Some musculars are also handsom All lean are also muscular
Some lean are also fair skinned.
All rich man inot fair skinned but all rich manare handsome
Some questions follows.
10
There are 3 pileseach containe 10 15 20 stones. There are
A,B,C,D,F,G
and h persons .One man can catch
upto four stones from any
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The last manwho takeswill win. If first A starts next B. and
so on
who will
win? Ans May be F | 713 | 2,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2019-13 | latest | en | 0.80464 |
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Abstract
Share and Cite:
Han, C. and Kinnas, S. (2013) Study on the Wake Shape behind a Wing in Ground Effect Using an Unsteady Discrete Vortex Panel Method. Open Journal of Fluid Dynamics, 3, 261-265. doi: 10.4236/ojfd.2013.34032.
1. Introduction
When a wing is flying near the ground, the aerodynamic characteristics of the wing is changed due to the interaction of the wing and the ground. This phenomenon is called as wing-in-ground (WIG) effect. New conceptual vehicles utilizing the WIG effect has been suggested focusing on fuel efficiency and stability [1].
Panel methods have been frequently applied to the conceptual design of wing-in-ground effect vehicles [2]. Many free-wake analysis techniques have been developed in order to represent the complicated wake vortex behavior behind a wing (Hoeijmakers and Vaatstra [3], Ribeiro and Kroo [4], Zhu and Takami [5], for example). Lamarre and Paraschivoiu [6] showed that the large-scale behavior of the roll-up of the aircraft trailing vortices can be simulated using the shape of 2-dimensional vortex sheet at any pseudo time because it resembles the shape of the cross section of the 3-dimensional sheet at a distance behind the wing (see [7] for a comprehensive review). Pullin [8], Krasny [9,10] investigated the roll-up of an initially plane semi-infinite vortex sheet. Morky [11] developed a 2-dimensional algorithm (based on the constant vortex sheet method) for modeling the evolution and propagation of aircraft trailing vortices in a sheared environment near the ground. Han and Cho [12] investigated the unsteady evolution of trailing vortex sheets behind a wing in ground effect using a discrete vortex method.
Han and Cho [12] assumed that the wings in close formation flight near the ground have the elliptic load distributions. Windall and Barrow [13] showed, based on a linear approach, that semi-elliptic wing planform with parabolic wing loading is optimal for a wing in ground effect for all aspect ratios. When a large aspect ratio wing is flying very close to the ground (extreme ground effect: EGE), based on a nonlinear inviscid theory, the optimal spanwise distribution of loading is parabolic rather than elliptic [2].
The objective of the present work is to investigate the effect of wing load distributions on the unsteady wake vortex evolution behind a wing in ground effect (IGE) and extreme ground effect (EGE). A new load distribution is suggested in order to match the change of the load distributions due to the change of the ground height. A lifting line with an initial load distribution is discretized with discrete vortex elements. The trailing wake vortices from each vortex element are represented by free vortices that deform freely by the assumption of a force-free position during the simulation.
2. Numerical Method
For a lifting line solution of a symmetrical thin rectangular wing in ground effect, Tan and Plotkin [14] confirmed that the lifting line solution agrees well with vortex-lattice numerical method. Formal mathematical development of Lifting Line Theory is well described in many text books on aerodynamics [15,16]. Following the notations in [17], solving an integro-differential equation for spanwise circulation distribution in terms of geometrical characteristics of the wing with boundary conditions at both wingtips provides the aerodynamic characteristics of the wing.
(1)
where is the local lift curve slope, is the local wing chord, and and are respectively geometric and zero-lift angles of attack at the spanwise location. It can be easily verified that the solution of Equation (1) for an elliptic planform, untwisted wing out of ground effect (OGE) has an elliptic spanwise distribution of bound circulation and the elliptic wing loading is the optimal value for minimum induced drag within the limitation of flat rigid wake assumption. When a wing is flying in close proximity to the ground, the optimal spanwise distribution of bound circulation is parabolic [13]. Equation (2) shows the equations for bound circulation distributions (for a wing in and out of ground effect) along the spanwise direction, where bound circulations are divided by their maximum values for each case. Thus, it can be said that the nondimensional bound circulation distribution of a wing in ground effect is confined in the envelope with its upper one as an elliptic distribution Equation (2a), and its lower one as a parabolic distribution Equation (2c) (See Figure 1).
(2a)
In Ground Effect:
(2b)
Figure 1. Comparison of spanwise circulation distributions.
(2c)
When applying Equation 2 to the calculation of bound circulation distributions, it is also required to understand the relationship between bound circulation and wing height.
Shigemi et al. [18] performed a wind tunnel test in order to investigate the ground effect on an 8.9% model of HOPE ALFLEX (Automatic Landing Flight Experiment) vehicle. They found an equation for the change of lift coefficient due to the change of the wing height as follows. After modifying the equation in order to match with Equation (2),
(3)
where is the lift coefficient of a wing in free flight (OGE), is the lift coefficient of a wing in extreme ground effect (EGE) and C is the coefficient that makes Equation (3) matching with the experimental data and should have a positive value. h represents the distance between the mid chord to the ground.
As shown in Equation (3), the lift coefficient is a nonlinear function of wing height, which also results in the change of circulation distribution. Thus, in the present paper, it is assumed that p in Equation (2b) changes as follows.
(4)
where is set to 1/2. p has the values of 0 and 1 respectively for h = 1.0 and 0.1.
2.2. Unsteady Discrete Vortex Panel Method
Present method is well described and validated in Ref. where the effect of smoothing schemes or vortex models on the numerical accuracy is shown by comparing computed results with the published data of Krasny [10]. In this approach, a lifting line with an initial load distribution is discretized with discrete vortex panel elements (See Figure 2). Each panel segment is assumed to have a constant-strength vortex. This constant-strength vortex panel element is replaced by a single point vortex with
Figure 2. Representation of a lifting line with elliptic loading with discrete vortex panel elements.
the same vorticity. On each line segment, the point vortex is located at the 3/4 point. The trailing wake vortices from each vortex panel element are represented by free point vortices that can deform freely with an assumption of a force-free position.
The distributions of the dimensionless tangential velocities induced by a point vortex are represented using the Lamb (Oseen) model [19]. The Lamb model has a Gaussian distribution of vorticity. Ling et al. [20] proposed a model that is based on the solution for the actual velocity in a viscous fluid. The core radius, rc, is approximately equal to the radial distance of the point where the maximum velocity is induced. This is expressed
(5)
The second term on the right-hand side of Equation (5) provides an estimate for the growth in the core radius of the vortex. The vortex Reynolds number is related to the time scale over which the laminar diffusion processes in the vortex core region become turbulent [21]. In the present work, an infinite Reynolds number is assumed. The tangential velocity induced by a point vortex with its circulation, Γj, is as follows.
(6)
Several smoothing schemes are used in order to obtain accurate solutions by circumventing the singularity behavior when the distance is very small. Krasny [10] avoids the singularity in the induced velocity of a point vortex through a smoothing factor, δ. The induced velocities because of other vortices can then be written as follows
(7)
where is represented as follows depending on the wing distance to the ground.
(8a)
(8b)
(8c)
In order to enforce the no penetration condition at the ground, an image method is used. Since the wake is force-free, the evolution of each vortex is investigated by moving the positions of point vortices using an Euler convection scheme.
(9)
3. Results and Discussion
Figure 3 shows the validation of the present method by comparing the wake shapes behind a lifting line having an elliptic loading with the published result of Krasny [10]. It is assumed that the wing is placed at h = 0 and is out of ground effect. In both calculations, the in-flow velocity and the maximum circulation at the midspan are assumed to be unity. The smoothing factor for the present calculation is set to 0.05 [the same as Krasny’s]. As shown in Figure 4, an unrealistic vortex sheet crossing is not observed in the present results. A time step size of 0.04 sec is used in the present calculation whereas Krasny [10] used a time step size of 0.01 sec. Thus, it can be said that the present results are in agreement with Krasny’s.
Figure 4 shows the comparison of wake shapes behind wings in and out of ground effect. Both wings are placed at the same ground height of h = 1.0. As shown in the figure, wake vortices behind a wing in ground effect cannot fully develop due to the insufficient distance between the wing and the ground. The positions of wingtip vortices of a wing in ground effect are moved laterally
Figure 3. Comparison of wake shapes between present method and Krasny’s.
Figure 4. Wake shapes behind a wing in and out of ground effect.
more outward and vertically more upward than those of a wing out of ground effect. When the wing is in ground effect, the desingularized point vortices become confined to a small region and coalesce. Present method does not include the viscous diffusion model of interacting point vortices, but it can be said that the interaction of point vortices while merging will decrease further the viscous diffusion. The total circulation of wingtip vortices behind a wing in ground effect will be decreased.
Figure 5 shows the wake shapes behind a wing in close proximity to the ground. When the wing is placed at h = 0.2, from Equation (4),. As shown in Equation (2), p = 0.5 means the elliptic circulation distribution whereas p = 1.0 represents the parabolic circulation distribution. When h = 0.2, the value of exponent p approaches to p = 1.0. Figure 5(a) shows the wake shapes behind a wing at h = 0.2 with three cases of circulation distributions. Figure 5(b) shows the wake shapes behind a wing at h = 0.1 for elliptic and parabolic circulation distributions. As shown in both Figures 5(a) and (b), overall wingtip vortex shapes are similar to each other and does not dependent on the circulation distributions, however, the position of the wingtip vortices for elliptic loading is moved laterally more outward than the other cases. Thus, it can be said that, for a single wing in ground effect, the assumption of elliptic load distributions does not affect much the wake shapes but negligibly changes the position of wingtip vortices.
Figure 6 shows the changes of wake shapes behind a wing in ground effect. As the wing is placed close to the ground, the wingtip vortices departed from the wing is moving vertically downward and laterally outward. As shown in Figure 6(b), when the wing is in close proximity to the ground (h = 0.1), the area of confined vorticity is decreased compared to the case of the wing at h = 0.5.
4. Conclusions
In the present paper, a new formula that matches elliptic
(a)(b)
Figure 5. Wake shapes behind a wing in close proximity to the ground with various wing loadings.
(a)(b)
Figure 6. Wake shapes for different wing heights. (a) Wake shape in the Trefftz plane, T = 4.0; (b) Three dimensional view of unsteady wake evolutions.
From the computed results on the unsteady wake evolution using the suggested model for wing loading, it was found that the overall wingtip vortex shapes were not affected by the wing load distributions. But, as the wing approaches very close to the ground, the parabolic wing load distribution calculation produced more laterally inward wingtip positions than the elliptic load distribution case.
In future, the present model will be applied to the unsteady wake evolution behind wings in formation where the significant interaction between wingtip vortices is expected.
5. Acknowledgements
The research was supported by a grant from the 2011 program for visiting professors overseas in Korea National University of Transportation.
NOTES
Conflicts of Interest
The authors declare no conflicts of interest.
[1] K. V. Rozhdestvensky, “Wing-in-Ground Effect Vehicles,” Progress in Aerospace Sciences, Vol. 42, No. 3, 2006, pp. 183-211. http://dx.doi.org/10.1016/j.paerosci.2006.10.001 [2] K. V. Rozhdestvensky, “Aerodynamics of a Lifting System in Extreme Ground Effect,” Springer, Heidelberg, 2000. http://dx.doi.org/10.1007/978-3-662-04240-3 [3] H. W. M. Hoeijmakers and W. Vaatstra, “A Higher Order Panel Method Applied to Vortex Sheet Roll-up,” AIAA Journal, Vol. 21, No. 4, 1983, pp. 516-523. http://dx.doi.org/10.2514/3.8108 [4] R. S. Ribeiro and I. Kroo, “Vortex-in-Cell Analysis of Wing Wake Roll-Up,” Journal of Aircraft, Vol. 32, No. 5, 1995, pp. 962-969. http://dx.doi.org/10.2514/3.46824 [5] K. Zhu and H. Takami, “Effect of Ground on Wake Roll-Up behind a Lifting Surface,” Proceedings of the 37th Japan National Congress for Applied Mechanics, Tokyo, 1987, pp. 115-123. [6] F. Lamarre and I. Paraschivoiu, “Efficient Panel Method for Vortex Sheet Roll-Up,” Journal of Aircraft, Vol. 29, No. 1, 1992, pp. 28-33. http://dx.doi.org/10.2514/3.46121 [7] T. Sarpkaya, “Computational Methods with Vortices— The 1988 Freeman Scholar Lecture,” Journal of Fluids Engineering, Vol. 111, No. 1, 1989, pp. 5-52. http://dx.doi.org/10.1115/1.3243601 [8] D. I. Pullin, “The Large-scale Structure of Unsteady Self-Similar Rolled-Up Vortex Sheets,” Journal of Fluid Mechanics, Vol. 88, No. 3, 1978, pp. 401-430. http://dx.doi.org/10.1017/S0022112078002189 [9] R. Krasny, “A Study of Singularity Formation in a Vortex Sheet by the Point-Vortex Approximation,” Journal of Fluid Mechanics, Vol. 167, 1986, pp. 65-93. http://dx.doi.org/10.1017/S0022112086002732 [10] R. Krasny, “Computation of Vortex Sheet Roll-Up in the Trefftz Plane,” Journal of Fluid Mechanics, Vol. 184, 1987, pp. 123-155. http://dx.doi.org/10.1017/S0022112087002830 [11] M. Morky, “Numerical Simulation of Aircraft Trailing Vortices Interacting with Ambient Shear or Ground,” Journal of Aircraft, Vol. 38, No. 4, 2001, pp. 636-643. http://dx.doi.org/10.2514/2.2840 [12] C. Han and J. Cho, “Unsteady Trailing Vortex Evolution Behind a Wing in Ground Effect,” Journal of Aircraft, Vol. 42, No. 2, 2005, pp. 429-434. http://dx.doi.org/10.2514/1.6477 [13] S. E. Windall, and T. M. Barrows, “An Analytic Solution for Two-and Three-Dimensional Wings in Ground Effect,” Journal of Fluid Mechanics, Vol. 41, No. 4, 1970, pp. 769-792. http://dx.doi.org/10.1017/S0022112070000915 [14] A. Plotkin and C. H. Tan, “Lifting-Line Solution for a Symmetrical Thin Wing in Ground Effect,” AIAA Journal, Vol. 24, No. 7, 1986, pp. 1193-1194. http://dx.doi.org/10.2514/3.9413 [15] J. Anderson, “Fundamentals of Aerodynamics,” 5th Edition, McGraw Hill, New York, 2010. [16] J. Katz and A. Plotkin, “Low Speed Aerodynamics,” 2nd Edition, Cambridge University Press, Cambridge, 2001. http://dx.doi.org/10.1017/CBO9780511810329 [17] I. G. Sheldon, “Wing Tip Vortices,” In: I. G. Sheldon, Ed., Fluid Vortices, Kluwer Academic Publishers, Berlin, 1995, Chapter X. [18] M. Shigemi, T. Fujita, A. Iwasaki, T. Ohnuki, K. Rinoie, H. Nakayasu and M. Sagisaka, “Experimental Investigation of Static and Dynamic Ground Effect on HOPE ALFLEX Vehicle,” Technical Report of National Aerospace Laboratory TR-1236, National Aerospace Laboratory, Tokyo, 1994. [19] A. Ogawa, “Vortex Flow-CRC Series on Fine Particle Science and Technology,” CRC Press, Boca Raton, 1940. [20] G. C. Ling, P. W. Bearman and J. M. R. Graham, “A Further Simulation of Starting Flow around a Flat Plate by a Discrete Vortex Model,” Internal Seminar on Engineering Applications of the Surface and Cloud Vorticity Methods, Vol. 51, No. 14, 1986, pp. 118-138. [21] J. G. Leishman, “Principles of Helicopter Aerodynamics,” 2nd Edition, Cambridge University Press, Cambridge, 2000. | 4,064 | 16,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-38 | latest | en | 0.894728 |
http://www.physicsforums.com/showthread.php?p=3957602 | 1,369,025,531,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698289937/warc/CC-MAIN-20130516095809-00080-ip-10-60-113-184.ec2.internal.warc.gz | 632,367,821 | 10,478 | ## Why can we see light when looking down a metallic tube? TEM modes can't exist.
Hey everyone, I'm going over my course content for an exam this Saturday and I'm a bit confused by a realisation I've just had.
In a metallic waveguide we know there are no TEM modes. But wouldn't this mean that we couldn't see through metallic waveguides as we see electromagnetic radiation? Obviously this only applies when the waveguides are perfect conductors, but I don't think this is the reason either because we can assume it's at least a 50% perfect conductor and thus should lock 50% of the light.
Any insight?
PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
Recognitions: Homework Help You mean like these things? "a radio wave in a hollow metal waveguide must have zero tangential electric field amplitude at the walls of the waveguide, so the transverse pattern of the electric field of waves is restricted to those that fit between the walls. For this reason, the modes supported by a waveguide are quantized. The allowed modes can be found by solving Maxwell's equations for the boundary conditions of a given waveguide." ... you can think of it as like how standing waves on a string fixed at each end must have zero tengential amplitude at the end - the electric field in a metal tube must show an antinode at the walls. That's all the "no TEM mode" bit means iirc.
I don't think I communicated clearly. Look on that same wikipedia page under Analysis, it says -
"TEM modes (Transverse ElectroMagnetic) have no electric nor magnetic field in the direction of propagation."
This is due to -
If we sub in Ez and Hz = 0 (being transverse Electromagnetic Waves) then Ex = Hx = Ey = Hy = 0. And so there cannot exist a transverse electromagnetic wave in a perfectly conducting waveguide.
The only reason I can think of as to why we can see light going down metallic tubes is that it comes in at a slight angle, but I'm not by any means confident with this explanation.
Recognitions:
Homework Help
## Why can we see light when looking down a metallic tube? TEM modes can't exist.
If we sub in Ez and Hz = 0 (being transverse Electromagnetic Waves) then Ex = Hx = Ey = Hy = 0. And so there cannot exist a transverse electromagnetic wave in a perfectly conducting waveguide.
And yet - clearly, there can - since EM waves go through waveguides all the time. Further, you know that this result is completely compatible with maxwells equations. It follows that you are misunderstanding them, or you are misunderstanding what you are being told when it is asserted that TEM modes are zero.
The requirement that the transverse amplitude is zero at the surface of the guide means that some wavelengths won't make it out the other end of the tube. Somewhere in your notes there should be a derivation of the EM field in a waveguide with conducting walls.
So what are you saying is wrong? Those formulas clearly show that EM waves propagating in z cannot exist - assuming perfectly conducting, propagation solely in z etc. etc.
Recognitions: Science Advisor It's correct that there are no TEM modes in waveguides with simply connected cross sections, but how do you come to the conclusion that no waves can go through it? There are many propagating TE and TM modes!
Recognitions:
Homework Help
Those formulas clearly show that EM waves propagating in z cannot exist
This is what I am saying is wrong - that is not what the formulas are saying. You are making a mistake. I'm trying to get you to elaborate on your reasoning so the nature of the misunderstanding will be exposed.
Of course you could just solve maxwels equations for the EM wave in a metal guide and see that you do, indeed, get waves out the end. But I suspect that won't clear up your confusion.
aside: Thanks vanhees71 :)
Yeah I know that Transverse Electric (TE) and TM modes can exist in these type of waveguides but I was under the impression light we see has to have an Electric and Magnetic component. Does that mean that we can see EM waves with no M or no E field? ---- Edit: I should have been more clear, I meant "Those formulas clearly show that TEM waves..."
Quote by Kyle91 In a metallic waveguide we know there are no TEM modes.
You cannot extrapolate it to arbitrarily high frequency, relative to size of waveguide, and expect the "no TEM mode" rule to make sense.
For example, consider a waveguide the size of a trashcan and a collimated laser beam shining right down the middle. There is no interaction between waveguide walls and light waves.
On the other hand the core of a fiber optic cable propagates infared light with restricted propagation modes in the same way as large metallic waveguides do for microwaves.
The important distinction is whether or not the electromagnetic wave in question is being *guided* by the waveguide or not.
Recognitions:
That reveals your misunderstanding! TE or TM just means that the component of the electric or magnetic field, respectively, in direction of the waveguide's axis (usually chosen as the $z$ axis of the coordinate system) is vanishing. Electromagnetic waves always have non-vanishing electric and magnetic fields. That's already clear from a very qualitative perspective on Maxwell's equations: If there is a time-dependent electric field there must be also a magnetic field and vice versa. Only static fields can be solely electric or magnetic. | 1,183 | 5,505 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2013-20 | latest | en | 0.943747 |
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Quote:
>Hi,
> I am building a report using Crystal Reports 4.5 and I have a
>problem. I hope one of you can solve it. Here it goes:
>I am accessing an access database in which there is a table named RESULT,
>there is a field in that table named resDepth.
>I would like to divide RESULT.resDepth into 6 ranges:
> 1) 0
> 2) 0 to 20
> 3) 20 to 40
> 4) 40 to 60
> 5) 60 to 80
> 6) 80 to 100
>I would then like to plot a pie chart that would contain those 6 ranges and
>the slice would represent the amount of records in each range.
>Is there any way to do this or something close to it?
>Thanks a lot,
>Patrice LaRoche
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> I am building a report using Crystal Reports 4.5 and I have a
> problem. I hope one of you can solve it. Here it goes:
> I am accessing an access database in which there is a table named RESULT,
> there is a field in that table named resDepth.
> I would like to divide RESULT.resDepth into 6 ranges:
> 1) 0
> 2) 0 to 20
> 3) 20 to 40
> 4) 40 to 60
> 5) 60 to 80
> 6) 80 to 100
> I would then like to plot a pie chart that would contain those 6 ranges and
> the slice would represent the amount of records in each range.
> Is there any way to do this or something close to it?
> Thanks a lot,
> Patrice LaRoche
Make 6 formula fields from the Insert menu like
range20to40: IF {Result.resDepth} > 20 AND {Result.resDepth} <= 40 THEN 1 ELSE 0
range40to60: IF {Result.resDepth} > 40 AND {Result.resDepth} <= 60 THEN 1 ELSE 0
etc, etc
Put these fields in the DETAILS section and add them up...
This works quite well and You do not have to rely on the division above...
If You at a later point decide to change Your ranges then You only have to change to IF
statement...
Thu, 05 Nov 1998 03:00:00 GMT
Crystal Reports
Patrice,
The two gentlemen who have answered this previously are right on the
money. Regarding graphing, beware of one thing: Crystal will not graph
on multiple fields at the same time, only summary sections of the same
field (column of numbers). If you want graphs that will do this, I'm
afraid you'll either have to do a lot of data manipulation to get all
your data into the same column, or use another graphing utility
besides Crystal's. I did tech support for Crystal for a long time. If
you were to come up to me on the street and ask me what Crystal's
weakest feature is, I'd have to tell you graphing for the above
reason. Good Luck.
Eric {*filter*}
Atlanta, GA
Quote:
>Hi,
> I am building a report using Crystal Reports 4.5 and I have a
>problem. I hope one of you can solve it. Here it goes:
>I am accessing an access database in which there is a table named RESULT,
>there is a field in that table named resDepth.
>I would like to divide RESULT.resDepth into 6 ranges:
> 1) 0
> 2) 0 to 20
> 3) 20 to 40
> 4) 40 to 60
> 5) 60 to 80
> 6) 80 to 100
>I would then like to plot a pie chart that would contain those 6 ranges and
>the slice would represent the amount of records in each range.
>Is there any way to do this or something close to it?
>Thanks a lot,
>Patrice LaRoche
Thu, 12 Nov 1998 03:00:00 GMT
Page 1 of 1 [ 4 post ]
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# 基于单周控制的新型有源电力 滤波器 - PowerPoint PPT Presentation
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### 基于单周控制的新型有源电力滤波器
2003.11
1.引言
• 电网谐波污染日益受到关注.有源滤波器(APF)与功率因素校正(PFC)已成为电力电子学研究热点之一.PFC需要对产生谐波的电力电子装置本身进行改造,调节其功率因数;APF可以直接与设备相连接,通过谐波抽取与注入来抑制谐波.经典的APF需要较复杂的控制,电路成本较高.
2.UCI-APF工作原理及直流分量问题
• 设有源滤波器开关频率为fs=1/T，全桥部分工作在两个状态：当0<t<DT时，M1、M4导通，当DT<t<T时，M2、M3导通。考虑到开关周期T远小于工频周期,一个开关周期内电感电流基本不变,根据电感伏秒平衡原理,可得到:
How to solve it
?
3.UCI-APF的直流分量抑制策略
• 针对上面所述的直流分量的产生机理，如果将电流直流分量负反馈，对开关占空比进行补偿，稳态情况下可以将直流分量调至零。负反馈可由电流积分得到。
4.实验结果及结论
• 根据上图原理得到实验结果。
(2)
(1)
END
• 谢谢！ | 597 | 1,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-43 | latest | en | 0.420663 |
http://www.traditionaloven.com/building/beach-sand/convert-wt-oz-beach-sand-to-ml-beach-sand.html | 1,521,706,644,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647782.95/warc/CC-MAIN-20180322073140-20180322093140-00719.warc.gz | 476,239,917 | 12,035 | Beach Sand 1 ounce weight to milliliters converter
# beach sand conversion
## Amount: 1 ounce (oz) of weight Equals: 18.54 milliliters (ml) in volume
Converting ounce to milliliters value in the beach sand units scale.
TOGGLE : from milliliters into ounces in the other way around.
## beach sand from ounce to milliliter Conversion Results:
### Enter a New ounce Amount of beach sand to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other beach sand measuring units - complete list.
Conversion calculator for webmasters.
## Beach sand weight vs. volume units
Beach sand has quite high density, it's heavy and it easily leaks into even tiny gaps or other opened spaces. No wonder it absorbs and conducts heat energy from the sun so well. However, this sand does not have the heat conductivity as high as glass does, or fireclay and firebricks, or dense concrete. A fine beach sand in dry form was used for taking these measurements.
Convert beach sand measuring units between ounce (oz) and milliliters (ml) but in the other reverse direction from milliliters into ounces.
conversion result for beach sand: From Symbol Equals Result To Symbol 1 ounce oz = 18.54 milliliters ml
# Converter type: beach sand measurements
This online beach sand from oz into ml converter is a handy tool not just for certified or experienced professionals.
First unit: ounce (oz) is used for measuring weight.
Second: milliliter (ml) is unit of volume.
## beach sand per 18.54 ml is equivalent to 1 what?
The milliliters amount 18.54 ml converts into 1 oz, one ounce. It is the EQUAL beach sand weight value of 1 ounce but in the milliliters volume unit alternative.
How to convert 2 ounces (oz) of beach sand into milliliters (ml)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
18.5387935685 * 2 (or divide it by / 0.5)
QUESTION:
1 oz of beach sand = ? ml
1 oz = 18.54 ml of beach sand
## Other applications for beach sand units calculator ...
With the above mentioned two-units calculating service it provides, this beach sand converter proved to be useful also as an online tool for:
1. practicing ounces and milliliters of beach sand ( oz vs. ml ) measuring values exchange.
2. beach sand amounts conversion factors - between numerous unit pairs variations.
3. working with mass density - how heavy is a volume of beach sand - values and properties.
International unit symbols for these two beach sand measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for ounce is:
oz
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for milliliter is:
ml
### One ounce of beach sand converted to milliliter equals to 18.54 ml
How many milliliters of beach sand are in 1 ounce? The answer is: The change of 1 oz ( ounce ) weight unit of beach sand measure equals = to volume 18.54 ml ( milliliter ) as the equivalent measure within the same beach sand substance type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in oz - ounces for beach sand amount, the rule is that the ounce number gets converted into ml - milliliters or any other beach sand unit absolutely exactly.
Conversion for how many milliliters ( ml ) of beach sand are contained in a ounce ( 1 oz ). Or, how much in milliliters of beach sand is in 1 ounce? To link to this beach sand ounce to milliliters online converter simply cut and paste the following.
The link to this tool will appear as: beach sand from ounce (oz) to milliliters (ml) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 948 | 4,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-13 | latest | en | 0.845508 |
https://socratic.org/questions/what-s-the-difference-between-a-binomial-hypergeometric-and-poisson-probability-#519412 | 1,716,525,124,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00186.warc.gz | 443,807,269 | 6,118 | # What's the difference between a binomial, hypergeometric, and Poisson probability distribution?
Binomial - Random variable $X$ is the number of successes in n independent and identical trials, where each trial has fixed probability of success.
Hypergeometric - Random variable $X$ is the number of objects that are special, among randomly selected $n$ objects from a bag that contains a total of $N$ out of which $K$ are special. If $n$ is much smaller than $N$ then this can be approximated by binomial.
Poisson - Random variable $X$ counts the number of occurrences on an event in a given period, where we know that the concurrences has an average of $\setminus \lambda$ for any period of that length, independent of any other disjoint period. | 169 | 748 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "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} | 3.109375 | 3 | CC-MAIN-2024-22 | latest | en | 0.930528 |
https://brilliant.org/problems/dont-forget-that-energy/?group=w2mVOzgecPha | 1,621,372,202,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991514.63/warc/CC-MAIN-20210518191530-20210518221530-00514.warc.gz | 152,566,163 | 8,692 | # Don't Forget that Energy
A disk of mass $m$ and radius $r$ is pushed up on an inclined plane of angle $\theta$ with initial speed $v$. If $g=10[\frac{m}{s}]$ and the coefficient of friction is such that the disk can roll without slipping, the maximum height reached by the disk can be expressed as $\frac{a}{b}v^2$. Find $a+b$.
× | 95 | 333 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-21 | latest | en | 0.854461 |
kotakenterprise.com | 1,638,375,160,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360803.6/warc/CC-MAIN-20211201143545-20211201173545-00529.warc.gz | 430,568,509 | 16,161 | ## SR Flip Flop Design with NOR Gate and NAND Gate
### Introduction to SR Flip Flop
The SR flip – flop is one of the fundamental parts of the sequential circuit logic. SR flip – flop is a memory device and a binary data of 1 – bit can be stored in it. SR flip – flop has two stable states in which it can store data in the form of either binary zero or binary one. Like all flip – flops, an SR flip – flop is also an edge sensitive device.
SR flip – flop is one of the most vital components in digital logic and it is also the most basic sequential circuit that is possible. The S and R in SR flip – flop means ‘SET’ and ‘RESET’ respectively. Hence it is also called Set – Reset flip – flop. The symbolic representation of the SR Flip Flop is shown below.
#### Working
SR flip – flop works during the transition of clock pulse either from low – to – high or from high – to – low (depending on the design) i.e. it can be either positive edge triggered or negative edge triggered.
For a positive edge triggered SR flip – flop, suppose, if S input is at high level (logic 1) and R input is at low level (logic 0) during a low – to – high transition on clock pulse, then the SR flip – flop is said to be in SET state and the output of the SR flip – flop is SET to
1. For the same clock situation, if the R input is at high level (logic 1) and S input is at low level (logic 0), then the SR flip – flop is said to be in RESET state and the output of the SR flip – flop is RESET to 0.
The SR flip – flops can be designed by using logic gates like NOR gates and NAND gates. Unclocked or simple SR flip – flops are same as SR Latches. The two types of unclocked SR flip – flops are discussed below
### Unclocked S-R Flip-Flop Using NAND Gate
SR flip flop can be designed by cross coupling of two NAND gates. It is an active low input SR flip – flop. The circuit of SR flip – flop using NAND gates is shown in below figure
#### Working
`Case 1:`
When both the SET and RESET inputs are high, then the output remains in previous state i.e. it holds the previous data.
`Case 2:`
When SET input is HIGH and RESET input is LOW, then the flip flop will be in RESET state. Because the low input of NAND gate with R input drives the other NAND gate with 1, as its output is 1. So both the inputs of the NAND gate with S input are 1. This will cause the output of the flip – flop to settle in RESET state.
`Case 3:`
When SET input is LOW and RESET input is HIGH, then the flip flop will be in SET state. Because the low input of NAND gate with S input drives the other NAND gate with 1, as its output is 1. So both the inputs of the NAND gate with R input are 1. This will cause the output of the flip – flop to settle in SET state.
`Case 4:`
When both the SET and RESET inputs are low, then the flip flop will be in undefined state. Because the low inputs of S and R, violates the rule of flip – flop that the outputs should compliment to each other. So the flip flop is in undefined state (or forbidden state).
The table below summarizes above explained working of SR Flip Flop designed with the help of a NAND gates
(or forbidden state).
### Unclocked S R Flip-Flop Using NOR Gate
SR flip flop can also be designed by cross coupling of two NOR gates. It is an active high input SR flip – flop. The circuit of SR flip – flop using NOR gates is shown in below figure.
The operation is same as that of NOR SR Latch.
#### Working
Case 1:
When both the SET and RESET inputs are low, then the output remains in previous state i.e. it holds the previous data
`Case 2:`
When SET input is low and RESET input is high, then the flip flop will be in RESET state. Because the high input of NOR gate with R input drives the other NOR gate with 0, as its output is 0. So both the inputs of the NOR gate with S input are 0. This will cause the output of the flip – flop to settle in RESET state.
`Case 3:`
When SET input is high and RESET input is low, then the flip flop will be in SET state. Because the low input of NOR gate with S input drives the other NOR gate with 1, as its output is 1. So both the inputs of the NOR gate with R input are 1. This will cause the output of the flip flop to settle in SET state.
`Case 4:`
When both the SET and RESET inputs are high, then the flip flop will be undefined state. Because the high inputs of S and R, violates the rule of flip flop that the outputs should complement to each other. So the flip flop is in undefined state (or forbidden state).
The table below summarizes above explained working of SR Flip Flop designed with the help of a NOR gate.
Even though simple SR flip – flops and simple SR latches are same, both the terms are used in their respective contexts.
The problem with simple SR flip – flops is that they are level sensitive to the control signal (although not shown in figure) which makes them a transparent device. In order to avoid this, Gated or Clocked SR flip – flops are introduced (whenever the term SR flip – flop is used, it usually refers to clocked SR flip – flop). Clock signal makes the device edge sensitive (and hence no transparency).
### Clocked SR Flip – Flops
Two types of clocked SR flip – flops are possible: based on NAND and based on NOR. The circuit of clocked SR flip – flop using NAND gates is shown below
This circuit is formed by adding two NAND gates to NAND based SR flip – flop. The inputs are active high as the extra NAND gate inverts the inputs. A clock pulse is given as input to both the extra NAND gates.
Hence the transition of the clock pulse is a key factor in functioning if this device. Assuming it is a positive edge triggered device, the truth table for this flip – flop is shown below.
The same can be achieved by using NOR gates. The circuit of clocked SR flip – flop using NOR gates is shown below.
The figure suggests a structure of RS flip – flop (as R is associated to the output Q), the functionality of SET and RESET remain the same i.e. when S is high, Q is set to 1 and when R is high, Q is reset to 0.
### Applications
SR flip – flops are very simple but are not widely used in practical circuits because of their illegal state where both S and R are high (S = R = 1). But they are used in switching circuits as they provide simple switching function (between Set and Reset). One such application is a Switch de – bounce circuit. The SR flip – flops are used to eliminate mechanical bounce of switches in digital circuits.
#### Mechanical Bounce
Mechanical switches, when pressed or released, often take some time and vibrate several times before settling down. This non – ideal behavior of the switch is called switch bounce or mechanical bounce. This mechanical bounce will tend to fluctuate between low and high voltages which can be interpreted by digital circuit. This can result in variation of pulse signals and these series of unwanted pulses will result in the digital system to work incorrectly.
For example, in this bouncing period of the signal, the fluctuations in the output voltage are very high and therefore the register counts several inputs instead of single input. To eliminate this kind of behavior of digital circuits, we use SR flip – flops.
#### How Does S R Flip-Flop Eliminates the Mechanical Bounce
Based on the present state output, if the set or reset buttons are depressed then the output will change in a manner that it counts more than one signal input i.e. the circuit may receive some unwanted pulse signals and thus because of the mechanical bouncing action of machines, there is no change in outputs at Q.
When the button is pressed, the contact will affect the flip flop input and there will be change in the present state and no further affects on the circuit/machine for any other mechanical switch bounces. If there is any additional input from the switch, there will be no change and SR flip – flop will reset after some small period of time.
So the same switch will come to use only after an SR flip – flop executes a state change i.e. only after receiving the single clock pulse signal.
The circuit of a switch de – bouncing circuit is shown below.
The input to the switch is connected to ground (logic 0). There are two pull up resistors connected to each of the input. They ensure that flip – flop inputs S and R are always 1 when the switch is between contacts.
Another circuit can be constructed with NOR SR flip – flop.
The input to the switch is connected to logic 1. There are two pull down resistors connected to each of the input. They ensure that flip – flop inputs S and R are always 0 when the switch is between the contacts a and b.
Commonly used ICs for eliminating the mechanical switch bounce are MAX6816 – single input, MAX6817 – dual input, MAX6818 – octal input switch de-bouncer ICs. These ICs contain the necessary configuration with the SR flip – flops. | 2,017 | 8,866 | {"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} | 2.796875 | 3 | CC-MAIN-2021-49 | latest | en | 0.936485 |
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# Shivam Singh
### MathWorks
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Hello Anand, “quiver (X, Y, U, V)” plots arrows with directional components U and V at the Cartesian coordinates specified by X...
9 meses ago | 1
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Hello Harin, It is my understanding that you want to know how to add more shapes to the same plot. You can use hold function fo...
10 meses ago | 0 | 1,479 | 6,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-33 | latest | en | 0.891356 |
https://www.jobilize.com/online/course/appendix-a-to-applied-probability-directory-of-m-functions-and-m?qcr=www.quizover.com&page=8 | 1,581,986,585,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143455.25/warc/CC-MAIN-20200217235417-20200218025417-00387.warc.gz | 796,826,412 | 21,585 | # Appendix a to applied probability: directory of m-functions and m (Page 9/24)
Page 9 / 24
trialmatch.m Estimates the probability of matches in n independent trials from identical distributions. The sample size and number of trials mustbe kept relateively small to avoid exceeding available memory.
% TRIALMATCH file trialmatch.m Estimates probability of matches % in n independent trials from identical distributions% Version of 8/20/97 % Estimates the probability of one or more matches% in a random selection from n identical distributions % with a small number of possible values% Produces a supersample of size N = n*ns, where % ns is the number of samples. Samples are separated.% Each sample is sorted, and then tested for differences % between adjacent elements. Matches are indicated by% zero differences between adjacent elements in sorted sample. X = input('Enter the VALUES in the distribution ');PX = input('Enter the PROBABILITIES '); c = length(X);n = input('Enter the SAMPLE SIZE n '); ns = input('Enter the number ns of sample runs ');N = n*ns; % Length of supersample U = rand(1,N); % Vector of N random numbersT = dquant(X,PX,U); % Supersample obtained with quantile function; % the function dquant determines quantile% function values for random number sequence U ex = sum(T)/N; % Sample averageEX = dot(X,PX); % Population mean vx = sum(T.^2)/N - ex^2; % Sample varianceVX = dot(X.^2,PX) - EX^2; % Population variance A = reshape(T,n,ns); % Chops supersample into ns samples of size nDS = diff(sort(A)); % Sorts each sample m = sum(DS==0)>0; % Differences between elements in each sample % -- Zero difference iff there is a matchpm = sum(m)/ns; % Fraction of samples with one or more matches d = arrep(c,n);p = PX(d); p = reshape(p,size(d)); % This step not needed in version 5.1ds = diff(sort(d))==0; mm = sum(ds)>0; m0 = find(1-mm);pm0 = p(:,m0); % Probabilities for arrangements with no matches P0 = sum(prod(pm0));disp('The sample is in column vector T') % Displays of results disp(['Sample average ex = ', num2str(ex),]) disp(['Population mean E(X) = ',num2str(EX),]) disp(['Sample variance vx = ',num2str(vx),]) disp(['Population variance V(X) = ',num2str(VX),]) disp(['Fraction of samples with one or more matches pm = ', num2str(pm),]) disp(['Probability of one or more matches in a sample Pm = ', num2str(1-P0),])
## Distributions
comb.m function y = comb(n,k) Calculates binomial coefficients. k may be a matrix of integers between 0 and n . The result y is a matrix of the same dimensions.
function y = comb(n,k) % COMB y = comb(n,k) Binomial coefficients% Version of 12/10/92 % Computes binomial coefficients C(n,k)% k may be a matrix of integers between 0 and n % result y is a matrix of the same dimensionsy = round(gamma(n+1)./(gamma(k + 1).*gamma(n + 1 - k)));
ibinom.m Binomial distribution — individual terms. We have two m-functions ibinom and cbinom for calculating individual and cumulative terms, $P\left({S}_{n}=k\right)$ and $P\left({S}_{n}\ge k\right)$ , respectively.
$P\left({S}_{n}=k\right)=C\left(n,\phantom{\rule{0.166667em}{0ex}}k\right){p}^{k}{\left(1-p\right)}^{n-k}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left({S}_{n}\ge k\right)=\sum _{r=k}^{n}P\left({S}_{n}=r\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0\le k\le n$
For these m-functions, we use a modification of a computation strategy employed by S. Weintraub: Tables of the Cumulative Binomial Probability Distribution for Small Values of p , 1963. The book contains a particularly helpful error analysis, written by Leo J. Cohen. Experimentation with sums and expectations indicates aprecision for ibinom and cbinom calculations that is better than ${10}^{-10}$ for $n=1000$ and p from 0.01 to 0.99. A similar precision holds for values of n up to 5000, provided $np$ or $nq$ are limited to approximately 500. Above this value for $np$ or $nq$ , the computations break down. For individual terms, function y = ibinom(n,p,k) calculates the probabilities for n a positive integer, k a matrix of integers between 0 and n . The output is a matrix of the corresponding binomial probabilities.
function y = ibinom(n,p,k) % IBINOM y = ibinom(n,p,k) Individual binomial probabilities% Version of 10/5/93 % n is a positive integer; p is a probability% k a matrix of integers between 0 and n % y = P(X>=k) (a matrix of probabilities) if p>0.5 a = [1 ((1-p)/p)*ones(1,n)]; b = [1 n:-1:1]; c = [1 1:n]; br = (p^n)*cumprod(a.*b./c);bi = fliplr(br); elsea = [1 (p/(1-p))*ones(1,n)];b = [1 n:-1:1];c = [1 1:n];bi = ((1-p)^n)*cumprod(a.*b./c); endy = bi(k+1);
#### Questions & Answers
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
how did you get the value of 2000N.What calculations are needed to arrive at it
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A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive | 2,262 | 8,385 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-10 | longest | en | 0.771972 |
http://ebs.cu.edu.tr/?upage=fak&page=drs&f=3&b=140&ch=1&dpage=all&yil=2017&lang=en&dpage=tnm&InKod=6290&dpage=all | 1,607,105,757,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141740670.93/warc/CC-MAIN-20201204162500-20201204192500-00155.warc.gz | 30,174,260 | 18,629 | •
• Information on Degree Programmes
COURSE INFORMATON
Course Title Code Semester L+P Hour Credits ECTS
Statistical Thinking * ISB 152 2 2 2 4
Prerequisites and co-requisites Recommended Optional Programme Components None
Language of Instruction Turkish
Course Level First Cycle Programmes (Bachelor's Degree)
Course Type
Course Coordinator Prof. Dr. Sadullah SAKALLIOĞLU
Instructors
Prof. Dr. SADULLAH SAKALLIOĞLU 1. Öğretim Grup:A Prof. Dr. SADULLAH SAKALLIOĞLU 2. Öğretim Grup:A
Assistants
Goals
The purpose of this course is to provide students an introduction to major concepts and tools of collecting, analyzing and interpreting data.
Content
Testing theories, Types of errors, Decision rule, Producing data, Random Sampling, Types of variables, Displaying distributions, Pie Charts, Bar Graphs, Frequency Plots, Stem and Leaf Plots, Mean, Median Mode, Displaying relationships between variables
Learning Outcomes
-
Course's Contribution To Program
NoProgram Learning OutcomesContribution
12345
1
Explain the essence fundamentals and concepts in the field of Probability, Statistics and Mathematics
X
2
Emphasize the importance of Statistics in life
X
3
Define basic principles and concepts in the field of Law and Economics
4
Produce numeric and statistical solutions in order to overcome the problems
X
5
Use proper methods and techniques to gather and/or to arrange the data
X
6
Utilize computer systems and softwares
7
Construct the model, solve and interpret the results by using mathematical and statistical tehniques for the problems that include random events
X
8
Apply the statistical analyze methods
X
9
Make statistical inference(estimation, hypothesis tests etc.)
X
10
Generate solutions for the problems in other disciplines by using statistical techniques
X
11
Discover the visual, database and web programming techniques and posses the ability of writing programme
12
Construct a model and analyze it by using statistical packages
13
Distinguish the difference between the statistical methods
X
14
Be aware of the interaction between the disciplines related to statistics
X
15
Make oral and visual presentation for the results of statistical methods
X
16
Have capability on effective and productive work in a group and individually
X
17
Professional development in accordance with their interests and abilities, as well as the scientific, cultural, artistic and social fields, constantly improve themselves by identifying training needs
X
18
Develop scientific and ethical values in the fields of statistics-and scientific data collection
X
Course Content
WeekTopicsStudy Materials _ocw_rs_drs_yontem
1 How to make a decision with statistics Source reading
2 Explanation of null hypothesis, alternative hypothesis, decision rule, rejected and accepted regions, types of errors Source reading
3 Explanation of the concetps unit, variable, sample, population, parameter and statistics. Source reading
4 The language of sampling, Sampling methods Source reading
5 Random numbers, Simple Random Sampling, Stratified Random Sampling Source reading
6 Systematic Sampling, Cluster Sampling, Multistage sampling Source reading
7 Introduction to observational studies and experiments. Understanding observational studies Source reading
8 Mid-Term Exam Rewview the topics discussed in the lecture notes and sources
9 Response variable, Explanatory variable, Source reading
10 Principles of planning an experiment Source reading
11 Summarizing data graphically (Types of variables, Distribution of a variable, Pie Charts, Bar Graphs) Source reading
12 Summarizing data graphically (Displaying relationships between two qualitative variables, Frequency Plots, Histogram, Stem-and-leaf plots) Source reading
13 Summarizing data numerically (Mean, Median, Mode) Source reading
14 Summarizing data numerically (Range, Quartiles, Interquartile range, Standart deviation) Source reading
15 Solving problem Rewview the topics discussed in the lecture notes and sources
16-17 Term Exams Written exam | 855 | 4,022 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-50 | latest | en | 0.767251 |
https://math.stackexchange.com/questions/371059/how-does-one-find-a-basis-for-the-orthogonal-complement-of-w-given-w | 1,702,291,372,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103810.88/warc/CC-MAIN-20231211080606-20231211110606-00898.warc.gz | 430,855,923 | 37,492 | How Does One Find A Basis For The Orthogonal Complement of W given W?
I've been doing some work in Linear Algebra for my course at school. I just want to be clear about how to find the orthogonal complement of a subspace. The basis for the subspace, W, is shown below, composed of 3 vectors:$$W = \begin{Bmatrix}\begin{bmatrix}1\\2\\3\\4 \end{bmatrix} \begin{bmatrix}-3\\4\\2\\6\end{bmatrix} \begin{bmatrix}2\\-2\\3\\5\end{bmatrix}\end{Bmatrix}$$
I would like to know if one simply sets $W*W^T$$=0 and takes the columns of the resulting matrix as the basis of the orthogonal complement of W, provided that row reduction has been performed to make sure the remaining columns are linearly independent. Please note that I have created an arbitrary set of vectors above that are not orthogonal, and so if they need to be for W*W^T$$=0$ please point that out. Thank you.
I'd rather put the matrix in this way, here I just give a very simple example, you can solve your matrix in the same way: $$W = \begin{bmatrix} \begin{matrix}1\\0\\0 \end{matrix} \begin{matrix}0\\1\\0\end{matrix} \begin{matrix}0\\0\\1\end{matrix} \begin{matrix}1\\1\\1\end{matrix}\end{bmatrix}$$
Then find the null space of W by solving $$Wx = 0$$
You will get the basis for the nullspace: $$v = \begin{bmatrix} \begin{matrix}-1\\-1\\-1\\1 \end{matrix} \end{bmatrix}$$
the null space is spared by v, $$null(W) = span(v)$$ you can easily find subspace spanned by W is orthogonal to subspace spanned by v, because every basis (each row of W) is orthogonal to v.
Think of a system of linear equations you want to solve that will say your vector $\mathbf x$ is orthogonal to $W$.
• I think I now better understand. x being orthogonal to W, as I just reviewed, means the inner product of x with all vectors in a set spanning W must be zero. So doing a little rearranging, I found that the transpose of W times x must also equal zero. So is it incorrect for me to say W*W^T = 0 when Row W = perp(Nul W^T)? Apr 24, 2013 at 6:00
• EDIT: In above comment, I meant to ask "Therefore is it incorrect for one to say W*W^T = 0 when Col W = perp(Nul W^T)?" (NOT Row W) Apr 24, 2013 at 6:13
• Yes, $\text{Col}(A)^\perp = N(A^T)$. Your $W$ is a subspace, not a matrix, so what you wrote doesn't quite make sense. Apr 24, 2013 at 11:13
• No, you forget that a matrix can be comprised of vectors, and the subspace above is a basis of vectors, which can, in turn, be considered a matrix. I'm afraid you haven't really answered my question, which is specifically leading to "how can one find a vector perpendicular to a set of vectors?" And no, I am not looking for a simple answer such as "come up with a vector with the same number of elements as the vectors in the set whose inner product with each of the set's vectors is zero". I'm afraid it has to be more elegant than that. Apr 25, 2013 at 4:31 | 834 | 2,855 | {"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} | 4.40625 | 4 | CC-MAIN-2023-50 | latest | en | 0.895277 |
saludmed.com | 1,685,903,635,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650201.19/warc/CC-MAIN-20230604161111-20230604191111-00593.warc.gz | 48,720,741 | 2,271 | LABORATORY EXERCISE 2
RELATIONSHIP OF DYNAMIC STRENGTH TO SELECTED GIRTH MEASURES
Objective
To determine if the strength of certain muscle groups may be predicted from girth measures.
Equipment
Gulick tape
Dumbbell with weights
Universal or comparable leg press machine
Procedures
1. The class will divide into groups of four:
Subject
Recorder
Tester (2)
2. Measure the girth of the biceps by use of the Gulick tape. The biceps will be measured with the arm hanging at the side of the body. The measurement will be taken relaxed at the point of greatest circumference between the elbow and the acromion process (Figure 1-2-1).
The quadriceps circumference will be measured with the subject standing (non-flexed). This measure will be taken between the knee and acetabulum at the point of greatest circumference (Figure 1-2-2).
Figure 1-2-1
Method of measuring girth of the biceps.
Figure 1-2-2
Method of measuring girth of the quadriceps.
3. The measurement of dynamic strength will be done as follows:
a. The biceps strength measurement will be taken using a dumbbell with the weights at 2.5 lb. increments. The subject should hold the weights in a supinated position and have the elbow resting on the padded board. Following a warm-up of 5-10 curls at 5.0 lbs., women will begin contractions through a full range of motion until the maximum strength is reached. This will be accomplished using 1 R for each amount of weight added until the maximum is reached. After 5-10 contractions at 20-30 lbs. men will do the same until the maximum strength is reached. At least 1 minute of rest should be allowed between contractions.
b. Quadriceps strength will be measured using the leg press on the Universal Gym. Both male and female subjects should warm up before attemting maximum contractions. Maximum strength measurement will be accomplished by having the subject complete 1 R at increasing weights until the maximum is reached. At least 1 minute of rest should be allowed between contractions.
Results
1. Calculate the mean strength for men and women for both muscle groups.
2. Using the male data only, correlate the strength for the quadriceps with the quadriceps girth.
Conclusions
1. What is the relationship beteen girth measures and strength for the two muscle groups? Can you explain this?
2. What are the mean differences between males and females in strength measures? Is this expected and why?
3. What are the problems associated with using girth measures to predict muscular strength.
Reproduced from:
Vaughn, C., & Johnson, R. (1984). Laboratory experiments in exercise physiology: Measurement / evaluation / application (pp. 14-17). Dubuque, IA: Eddie Bowers Publishing Company. Copyright Ⓒ 1984, by eddie bowers publishing company | 641 | 2,788 | {"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-2023-23 | latest | en | 0.860057 |
http://www.santhoshreddymandadi.com/java/bubble-sort-algorithm-flow-chart.html | 1,550,655,361,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247494694.1/warc/CC-MAIN-20190220085318-20190220111318-00335.warc.gz | 425,480,059 | 6,342 | ## Bubble sort
Bubble sort is a simple and common sorting algorithm. It sorts by iterating through the list to be sorted, comparing each pair of adjacent items and swapping them if they are in the wrong order. This process will be continued until all the elements are being sorted i.e.; no swapping is required in the list.
Bubble sort name came for this algorithm due to - like a bubble comes to the top of the water, each iteration will push one smaller element to the top of the list (if the algorithm is for ascending order).
## Performance
Bubble sort is not the efficient algorithm in terms of the performance because its worst-case and average complexity both ?(n2), where n is the number of items being sorted.
## Algorithm
`START DECLARE n, list ACCEPT n ACCEPT n values into an array list DO swapped := false; FOR EACH i IN 0 to n-1 LOOP IF list[i] > list[i+1] THEN temp := list[i] list[i]:=list[i+1] list[i+1]:=temp END IF END LOOP WHILE swapped = trueEND`
## Java program
`import java.io.*;/** * This class demonstrates the bubble sort. * @author SANTHOSH REDDY MANDADI * @version 1.0 * @since 04th September 2009 */public class BubbleSort{ public static void main(String[] args) throws Exception { int values[]=new int[5]; BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); System.out.println("Enter "+values.length+" values to sort using bubble sort"); for(int i=0;i<values.length;i++) { try{ values[i]=Integer.parseInt(br.readLine()); }catch(NumberFormatException e) { e.printStackTrace(); values[i]=0; } } bubbleSort(values); for(int i=0;i<values.length;i++) { System.out.print(values[i]+" "); } } /** * This method will sort the passed array * @author SANTHOSH REDDY MANDADI * @since 4-Sep-2009 * @param an array of the list of values to be sorted */ public static void bubbleSort(int[] values) { boolean swapped=false; do { //When initializing the each loop assing false to swapped flag swapped = false; //Looping through the list for(int i=0; i<values.length-1; i++) { //Comparing the adjecent elements if(values[i]>values[i+1]) { //Swapping int temp=values[i]; values[i]=values[i+1]; values[i+1]=temp; //Swapped is true swapped=true; } } }while (swapped); }}`
## Explanation:
Let us consider we've executed the above program by inputting the values (43, 25, 21, 56, 4). Here is how the list will be modified in each iteration.
swapped value = false
### First iteration:
43 25 21 56 4 » 25 43 21 56 4 - swapped since 43 > 25
25 43 21 56 4 » 25 21 43 56 4 - swapped since 43 > 21
25 21 43 56 4 » 25 21 43 56 4 - not swapped since 43 < 56
25 21 43 56 4 » 25 21 43 4 56 - swapped since 54 > 4
### Second iteration:
25 21 43 4 56 » 21 25 43 4 56
21 25 43 4 56 » 21 25 43 4 56
21 25 43 4 56 » 21 25 4 43 56
21 25 4 43 56 » 21 25 4 43 56
### Third iteration:
21 25 4 43 56 » 21 25 4 43 56
21 25 4 43 56 » 21 4 25 43 56
21 4 25 43 56 » 21 4 25 43 56
21 4 25 43 56 » 21 4 25 43 56
### Fourth iteration:
21 4 25 43 56 » 4 21 25 43 56
4 21 25 43 56 » 4 21 25 43 56
4 21 25 43 56 » 4 21 25 43 56
4 21 25 43 56 » 4 21 25 43 56
### Fifth iteration:
4 21 25 43 56 » 4 21 25 43 56
4 21 25 43 56 » 4 21 25 43 56
4 21 25 43 56 » 4 21 25 43 56
4 21 25 43 56 » 4 21 25 43 56
Since there're no swaps happened in the fifth operation, the outer loop will be stopped since the condition swapped = true false.
Above program has been tested in JDK 1.4, 1.5, and 1.6. | 1,171 | 3,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-09 | longest | en | 0.574621 |
https://wiki.lazarus.freepascal.org/Solution_3/zh_CN | 1,701,188,809,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099892.46/warc/CC-MAIN-20231128151412-20231128181412-00711.warc.gz | 687,201,484 | 7,459 | # Basic Pascal Tutorial/Chapter 3/Solution/zh CN
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3Ea - 参考答案 (原作者: Tao Yue, 状态: 未更改)
```(* Author: Tao Yue
Date: 19 July 1997
Description:
Find the first 10 Fibonacci numbers
Version:
1.0 - original version
*)
program Fibonacci;
var
Fibonacci1, Fibonacci2 : integer;
temp : integer;
count : integer;
begin (* Main *)
writeln ('First ten Fibonacci numbers are:');
count := 0;
Fibonacci1 := 0;
Fibonacci2 := 1;
repeat
write (Fibonacci2:7);
temp := Fibonacci2;
Fibonacci2 := Fibonacci1 + Fibonacci2;
Fibonacci1 := Temp;
count := count + 1
until count = 10;
writeln;
(* Of course, you could use a FOR loop or a WHILE loop
to solve this problem. *)
end. (* Main *)
```
```(* Author: Tao Yue
Date: 13 July 2000
Description:
Display all powers of two up to 20000, five per line
Version:
1.0 - original version
*)
program PowersofTwo;
const
numperline = 5;
maxnum = 20000;
base = 2;
var
number : longint;
linecount : integer;
begin (* Main *)
writeln ('Powers of ', base, ', 1 <= x <= ', maxnum, ':');
(* Set up for loop *)
number := 1;
linecount := 0;
(* Loop *)
while number <= maxnum do
begin
linecount := linecount + 1;
(* Print a comma and space unless this is the first
number on the line *)
if linecount > 1 then
write (', ');
(* Display the number *)
write (number);
(* Print a comma and go to the next line if this is
the last number on the line UNLESS it is the
last number of the series *)
if (linecount = numperline) and not (number * 2 > maxnum) then
begin
writeln (',');
linecount := 0
end;
(* Increment number *)
number := number * base;
end; (* while *)
writeln;
(* This program can also be written using a
REPEAT..UNTIL loop. *)
end. (* Main *)
```
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The probability that a 1-h rainfall of magnitude 10 cm or more will occur in each oftwo successive years is 0.0004
Return period of rainfall, T 50 years
Therefore Probability of occurrence once in 50 years,
Probability of occurrence in each of 2 successive years
= 0.0004
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## définition - Noncentral F-distribution
voir la définition de Wikipedia
Wikipedia
# Noncentral F-distribution
In probability theory and statistics, the noncentral F-distribution is a continuous probability distribution that is a generalization of the (ordinary) F-distribution. It describes the distribution of the quotient (X/n1)/(Y/n2), where the numerator X has a noncentral chi-squared distribution with n1 degrees of freedom and the denominator Y has a central chi-squared distribution n2 degrees of freedom. It is also required that X and Y are statistically independent of each other.
It is the distribution of the test statistic in analysis of variance problems when the null hypothesis is false. The noncentral F-distribution is used to find the power function of such a test.
## Occurrence and specification
If $X$ is a noncentral chi-squared random variable with noncentrality parameter $\lambda$ and $\nu_1$ degrees of freedom, and $Y$ is a chi-squared random variable with $\nu_2$ degrees of freedom that is statistically independent of $X$, then
$F=\frac{X/\nu_1}{Y/\nu_2}$
is a noncentral F-distributed random variable. The probability density function for the noncentral F-distribution is [1]
$p(f) =\sum\limits_{k=0}^\infty\frac{e^{-\lambda/2}(\lambda/2)^k}{ B\left(\frac{\nu_2}{2},\frac{\nu_1}{2}+k\right) k!} \left(\frac{\nu_1}{\nu_2}\right)^{\frac{\nu_1}{2}+k} \left(\frac{\nu_2}{\nu_2+\nu_1f}\right)^{\frac{\nu_1+\nu_2}{2}+k}f^{\nu_1/2-1+k}$
when $f\ge0$ and zero otherwise. The degrees of freedom $\nu_1$ and $\nu_2$ are positive. The noncentrailty parameter $\lambda$ is nonnegative. The term $B(x,y)$ is the beta function, where
$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.$
The cumulative distribution function for the noncentral F-distribution is
$F(x|d_1,d_2,\lambda)=\sum\limits_{j=0}^\infty\left(\frac{\left(\frac{1}{2}\lambda\right)^j}{j!}e^{-\frac{\lambda}{2}}\right)I\left(\frac{d_1F}{d_2 + d_1F}\bigg|\frac{d_1}{2}+j,\frac{d_2}{2}\right)$
where $I$ is the regularized incomplete beta function.
The mean and variance of the noncentral F-distribution are
$\mbox{E}\left[F\right]= \begin{cases} \frac{\nu_2(\nu_1+\lambda)}{\nu_1(\nu_2-2)} &\nu_2>2\\ \mbox{Does not exist} &\nu_2\le2\\ \end{cases}$
and
$\mbox{Var}\left[F\right]= \begin{cases} 2\frac{(\nu_1+\lambda)^2+(\nu_1+2\lambda)(\nu_2-2)}{(\nu_2-2)^2(\nu_2-4)}\left(\frac{\nu_2}{\nu_1}\right)^2 &\nu_2>4\\ \mbox{Does not exist} &\nu_2\le4.\\ \end{cases}$
## Special cases
When λ = 0, the noncentral F-distribution becomes the F-distribution.
## Related distributions
$Z=\lim_{\nu_2\to\infty}\nu_1 F$
where F has a noncentral F-distribution.
## Implementations
The noncentral F-distribution is implemented in the R language (e.g., pf function), in MATLAB (ncfcdf, ncfinv, ncfpdf, ncfrnd and ncfstat functions in the statistics toolbox) in Mathematica (NoncentralFRatioDistribution function), in NumPy (random.noncentral_f), and in Boost C++ Libraries.[2]
A collaborative wiki page implements an interactive online calculator, programmed in R language, for noncentral t, chisquare, and F, at the Institute of Statistics and Econometrics, School of Business and Economics, Humboldt-Universität zu Berlin.[3]
## Notes
1. ^ S. Kay, Fundamentals of Statistical Signal Processing: Detection Theory, (New Jersey: Prentice Hall, 1998), p. 29.
2. ^ John Maddock, Paul A. Bristow, Hubert Holin, Xiaogang Zhang, Bruno Lalande, Johan Råde. "Noncentral F Distribution: Boost 1.39.0". Boost.org. Retrieved 20 August 2011.
3. ^ Sigbert Klinke (10 December 2008). "Comparison of noncentral and central distributions". Humboldt-Universität zu Berlin.
## References
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un contenu abusif (raciste, pornographique, diffamatoire) | 1,804 | 6,177 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 19, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2020-50 | latest | en | 0.593481 |
http://www.qacollections.com/Did-James-watt-build-a-vacuum-cleaner | 1,524,494,057,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946011.29/warc/CC-MAIN-20180423125457-20180423145457-00054.warc.gz | 489,033,107 | 5,643 | # Did James watt build a vacuum cleaner?
If it has Mayo in it, it must be kept cold!
Top Q&A For: Did James watt build a vacuum cleaner
## Would a 480 watt inverter run a 12 amp vacuum cleaner?
P = V/I or I = P/V or I = 480/120 = 4 amps 12 Amps at 120 volts is 12 * 120 = 1440 W So I'd say no.
## How many amps is your 1000 watt vacuum cleaner?
2-3 days only because of the sour cream and cream cheese.
Related Questions | 127 | 427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-17 | latest | en | 0.925108 |
https://zbook.org/tag/types-of-energy/4 | 1,712,992,081,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816586.79/warc/CC-MAIN-20240413051941-20240413081941-00248.warc.gz | 1,034,575,656 | 6,856 | # Types Of Energy-Page 4
Net-Zero Energy Project Guide 2 A net-zero energy (NZE) building is an extremely energy efficient building that is designed and operated to produce as much energy as it consumes over the course of the year. An ultra-low energy building exemplifies many of the same characteristics of NZE buildings, but may not have renewable energy systems.
conservation of energy tells us that we can never create or destroy energy, but we can change its form. In this lab, you will analyze energy transfer between gravitational potential energy, kinetic energy, and energy lost due to collisions or friction (thermal energy) as a skate boarder rides along a track.
conservation of energy tells us that we can never create or destroy energy, but we can change its form. In this lab, you will analyze energy transfer between gravitational potential energy, kinetic energy, and energy lost due to collisions or friction (thermal energy) as a skate boarder rides along a track.
Energy is a scalar (no direction, not a vector — easy maths!) S.I. unit of energy is the joule, J Examples of energy? Energy of motion - "kinetic energy" Stored energy - "potential energy": gravitational, elastic, chemical Energy in hot o
Conservation of Energy Energy can never be created nor destroyed. The amount of energy in a system will always be the same. Once a coaster starts, the system cannot gain any more energy. However, energy can be transformed from one form to another. Energy is transformed from potential energy to kinetic energy and back again and from kinetic
Chapter 8: Potential Energy and Conservation of Energy Work and kinetic energy are energies of motion. We need to introduce an energy that depends on location or position.This energy is called potential energy.
spiritual. But all of them are rooted in energy. The Magic of Energy Medicine Yoga Energy follows certain rules: 1 Energy wants to move. 2 Energy needs space to move. 3 Energy follows particular patterns in the body. 4 The two most important energy patterns are moving for
Kinetic vs. Potential Energy Potential Energy is the energy an object possesses by virtue of its position or composition. Kinetic Energy is the energy of motion K.E. ½mv 2 where m mass and v velocity Notes 6.1 2c Initial vs. Final Position In the initial position, ball A has a higher potential energy than ball B.
Energy Energy is the ability to do work. It is a scalar quantity. Measured in Joules (J). Energy can exist in: o Sound Energy o Heat Energy o Kinetic Energy o Gravitational Energy Kinetic Energy It is the
heat energy light energy sound energy electrical energy kinetic (m ovement) energy. Stored energy. Some energy has to be stored so that it is ready for use when we need it. Chemical energy is stored in food, fuels and cells. Gravi
Section 16.2 Thermal Energy . Three States of Matter Low Energy High School Dance Slow Song . Three States of Matter Medium Energy . Energy . Thermal Energy The total potential and kinetic energy of all the particles in an object Three things that thermal energy depends on: 1
This principle is called the principle of conservation of energy and is expressed as T 1 V 1 T 2 V 2 Constant T 1 stands for the kinetic energy at state 1 and V 1 is the potential energy function for state 1. T 2 and V 2 represent these energy states at state 2. Recall, the kinetic energy is defined as T ½ mv2. CONSERVATION OF ENERGY . | 740 | 3,427 | {"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-2024-18 | latest | en | 0.938918 |
https://www.physicsforums.com/threads/elastic-collisions-and-drag.70227/ | 1,638,284,101,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359037.96/warc/CC-MAIN-20211130141247-20211130171247-00113.warc.gz | 1,023,206,604 | 19,176 | Elastic Collisions and Drag
Some people argue that a non-viscous gas could not result in a drag for moving objects because the kinetic energy could not be dissipated to the gas. However, this view neglects the fact that even if the gas molecules do not interact with each other (i.e. if the gas is inviscid), they still collide with the object which results in a change of momentum and energy.
Consider a plate moving frontally through a gas with velocity +v. A molecule moving vertically towards the plate with velocity -u in the lab frame has a velocity -u-v in the reference frame of the plate and hence bounces back from the front side with the velocity u+v (as the plate is much heavier than the molecule), i.e. in the lab frame the velocity of the molecule is now u+2v. Correspondingly, at the backside of the plate, a molecule with velocity u in the lab frame has a velocity u-v in the reference frame of the plate and thus after reflection a velocity -u+v, i.e. in the lab frame the velocity is now -u+2v.
This means in the lab frame the kinetic energy of the molecules being reflected from the front has increased from m/2*u² to m/2*(u+2v)² =m/2*u² +2*m*u*v + 2*m*v², i.e. an increase of +2*m*u*v + 2*m*v².
Molecules reflected from the back side on the other hand have changed their energy from m/2*u² to m/2*(-u+2v)² = m/2*u² -2*m*u*v + 2*m*v² i.e. a change of -2*m*u*v + 2*m*v².
Adding and averaging the two contributions one obtains therefore the average increase of the kinetic energy of the molecules with mass m hitting the plate with velocity v as ΔK = 2*m*v². (Note: this value will actually be smaller by about factor 1/2 as the molecules won't be reflected straight back but will be scattered into the whole half-space due to the roughness of the plate's surface).
(One can derive this result also from the energy and momentum conservation equations applied an elastic collision, with the same result if one assumes the plate mass as large compared to the molecular mass).
Let's apply this result to air (although the latter is not strictly inviscid in the above defined sense):
if the plate has a cross section of 1m² and a velocity of 10 m/sec it will collide with 3*10^26 molecules per second, which, assuming a molecule mass of 4.5*10^-26 kg, amounts to a total mass of M=13 kg per second. Replacing m with M in the above result for the energy gain of a molecule, one sees that the kinetic energy gained by the air molecules per second corresponds to the kinetic energy of a plate of about 26 kg (taking the above mentioned correction into account), i.e. a plate of with a mass of 26 kg, a cross section of 1m² and a velocity of 10m/sec would lose its energy within about one second. This result looks actually quite realistic and shows that the assumption of a non-viscous gas can account for observed drag in air, and this obviously suggests also that the same may hold for the aerodynamic lift.
russ_watters
Mentor
You're neglecting the concept of pressure again, Thomas2. If there was no interaction at all between air molecules (that's not what "inviscid" means, btw), then as they hit an object they'd bounce away only on the leading surfaces and there would be a corresponding change in energy, just as you say. But pressure intervenes to make these particles bounce back onto the trailing surfaces of the object, giving back whatever momentum they gained.
edit: once again, there is real data to back this up. No, you can't ever have zero viscocity, but you can have an extremely low viscocity relative to the other components of the Reynolds number. Simply put, you can measure that as viscocity goes down, drag goes down. Once again, you are arguing against reality.
Last edited:
arildno
Homework Helper
Gold Member
Dearly Missed
russ:
One of the major troubles in Thomas2's conception lies here I think:
However, this view neglects the fact that even if the gas molecules do not interact with each other (i.e. if the gas is inviscid)...
That is, he essentially regards an inviscid fluid as a sort of vacuum.
I don't know how to respond to this..
Last edited:
russ_watters
Mentor
arildno said:
russ:
The trouble in Thomas2's conception lies here I think:
However, this view neglects the fact that even if the gas molecules do not interact with each other (i.e. if the gas is inviscid)...
That is, he essentially regards an inviscid fluid as a sort of vacuum.
I don't know how to respond to this..
I mentioned that. It seems his misunderstands always come back to dealing with pressure. The only way to deal with that in this case is to explain to him that an inviscid fluid still has pressure, which I did. Beyond that.... [shrug]
arildno
Homework Helper
Gold Member
Dearly Missed
Oops, I see you did mention it after all, right at the start.Sorry about that..
Last edited:
russ_watters said:
You're neglecting the concept of pressure again, Thomas2. If there was no interaction at all between air molecules (that's not what "inviscid" means, btw), then as they hit an object they'd bounce away only on the leading surfaces and there would be a corresponding change in energy, just as you say. But pressure intervenes to make these particles bounce back onto the trailing surfaces of the object, giving back whatever momentum they gained.
Even if you take the molecular interaction into account, the molecules will lose only some of the velocity 2v that they gained on collision with the plate: if you consider that half of the molecules were actually initially not flying towards the plate and average the velocity gained with these, you still have a velocity gain of +v, i.e. the initially (on average) resting molecules will be imparted a velocity such that they co-move with the plate. Interestingly this is just what is observed in form of the 'stagnation' at the rear and front.
But this was not actually my main point. I merely wanted to show that there is drag in a gas even if the molecules don't interact with each other (whether you call this 'inviscid' or whatever).
russ_watters
Mentor
Thomas2 said:
Even if you take the molecular interaction into account...
Please start using the word "pressure". When you say "molecular interaction" it makes it sound like you are purposefully avoiding taking pressure into account.
...the molecules will lose only some of the velocity 2v that they gained on collision with the plate...
And you base that on what, exactly? Experimental evidence? Mathematical proof? Just saying it doesn't make it true (and the fact is, it isn't true, experimentally or mathematically).
But this was not actually my main point. I merely wanted to show that there is drag in a gas even if the molecules don't interact with each other (whether you call this 'inviscid' or whatever).
Well, if you take away all "molecular interactions", (pressure and viscocity) you are no longer talking about a "fluid" and fluid dynamics/aerodynamics no longer applies. What you are saying may be true of a spacecraft encountering a single hydrogen atom every few kilometers, but it has nothing to do with drag, fluid dynamics, or aerodynamics.
I'm a little incredulous here, Thomas2 - you claim a lot for someone who doesn't even understand what a "gas" is. I am again having doubts about whether you're even serious about this.
russ_watters said:
Please start using the word "pressure". When you say "molecular interaction" it makes it sound like you are purposefully avoiding taking pressure into account. And you base that on what, exactly? Experimental evidence? Mathematical proof? Just saying it doesn't make it true (and the fact is, it isn't true, experimentally or mathematically). Well, if you take away all "molecular interactions", (pressure and viscocity) you are no longer talking about a "fluid" and fluid dynamics/aerodynamics no longer applies. What you are saying may be true of a spacecraft encountering a single hydrogen atom every few kilometers, but it has nothing to do with drag, fluid dynamics, or aerodynamics.
The 'pressure' that the molecules exert on each other is not what causes the drag on an object. It is the one that the molecules exert on the object. If molecules would be point masses, the former would actually be zero as molecules could not collide with each other, but the pressure on macroscopic objects would still be the same.
The point here is that 'pressure' is actually something different for gases on the on hand and fluids or solids on the other. For gases much of a volume is actually empty space: the volume of a molecule is about 10^-24 cm^3 and with 3*10^19 molecules/cm^3 at atmospheric pressure, this means that only a fraction of 3*10^-5 of a given air volume consists of matter, the rest of empty space. In fluids and solids on the other hand, the molecules or atoms are however tightly packed without any empty space between them. The pressure here is is not associated with the kinetic energy of the molecules but with the static potential energy between the atoms (analogously to compressing a spring for instance).
In view of this, it is actually the fluid dynamics approach to gases that needs justification and not the 'particle kinematics' approach. Fact is that within a certain distance from an object in a gas there are no collisions between molecules and hence the latter hit the object in a free flight manner as described in my opening post above.
arildno
Homework Helper
Gold Member
Dearly Missed
Fact is that within a certain distance from an object in a gas there are no collisions between molecules and hence the latter hit the object in a free flight manner as described in my opening post above
No. This is simply false, and you know it.
Stop posting nonsense.
It seems you are making this up as you go along.
Where are you? At an asylum?
Last edited:
enigma
Staff Emeritus
Gold Member
Thomas2 said:
The 'pressure' that the molecules exert on each other is not what causes the drag on an object.
Actually, pressure produces the vast majority (an order of magnitude greater than viscous drag at least) of most ordinary (read: aircraft) flows.
If you pick up an introductory aerodynamics textbook such as:
"Fundamentals of Aerodynamics" by John D. Anderson
You will find that out in chapters 1 and 2 where the basic forces and interactions are derived and explained.
The situation you are describing is called a "Newtonian Flow" and is only applicable for spacecraft and other applications which are in hard vacuum where the molecular mean free path is of the order of magnitude of the cross section of the craft. | 2,389 | 10,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-49 | longest | en | 0.909163 |
https://mathematics-monster.com/lessons/how_to_find_factors.html | 1,627,675,224,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153980.55/warc/CC-MAIN-20210730185206-20210730215206-00003.warc.gz | 395,331,692 | 4,621 | # Finding Factors(KS2, Year 4)
## Finding Factors
The factors of a number can be found.
## A Simple Example of How to Find Factors
Finding the factors of a number is easy. Find pairs of numbers that multiply to make that number.
## Question
What are the factors of the number 18?
# 1
1 and the number itself (18) are factors of 18.
1 × 18 = 18
Write 1 and 18 on the outside of a grid.
# 2
Divide 18 by 2. Does it divide exactly?
18 ÷ 2 = 9
Yes. 2 and the answer of the division (9) are factors of 18.
2 × 9 = 18
Write 2 and 9 in the grid, just inside the outside entries.
# 3
Divide 18 by 3. Does it divide exactly?
18 ÷ 3 = 6
Yes. 3 and the answer of the division (6) are factors of 18.
3 × 6 = 18
Write 3 and 6 in the grid, just inside the previous entries.
# 4
Divide 18 by 4. Does it divide exactly?
18 ÷ 4 = 4.5
No. The answer is not a whole number. 4 is not a factor.
# 5
Divide 18 by 5. Does it divide exactly?
18 ÷ 5 = 3.6
No. The answer is not a whole number. 5 is not a factor.
# 6
Divide 18 by 6. We already know that 6 is a factor of 18, it is already in the grid. We have found all the factors.
## Answer:
The factors of 18 are 1, 2, 3, 6, 9 and 18.
## Lesson Slides
The slider below shows another real example of finding the factors of a number. Open the slider in a new tab
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## See Also
What is a factor? | 469 | 1,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2021-31 | longest | en | 0.922191 |
http://www.chegg.com/homework-help/statistics-for-business-and-economics-11th-edition-solutions-9780321640116 | 1,475,024,006,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661289.41/warc/CC-MAIN-20160924173741-00290-ip-10-143-35-109.ec2.internal.warc.gz | 391,887,854 | 16,222 | # Statistics for Business and Economics (11th Edition) View more editions
• 1570 step-by-step solutions
• Solved by publishers, professors & experts
• iOS, Android, & web
Over 90% of students who use Chegg Study report better grades.
May 2015 Survey of Chegg Study Users
Chapter: Problem:
89% (9 ratings)
What is statistics?
SAMPLE SOLUTION
Chapter: Problem:
89% (9 ratings)
• Step 1 of 1
Statistics is the science of data; it involves collection, evaluation, and interpretation. The science of statistics has infinitely wide applications in almost every sphere of human activity, such as business, government, and physical and social sciences, is not without limitations.
Statistics can be divided into two broad areas:
Descriptive statistics
Inferential statistics.
Descriptive statistics involves classifying, summarizing, and organizing data to determine patterns and to provide meaningful graphical or numerical depictions of information.
Inferential statistics uses sample data to draw conclusions about a larger set of data.
Corresponding Textbook
Statistics for Business and Economics | 11th Edition
9780321640116ISBN-13: 032164011XISBN: Authors:
Alternate ISBN: 9780321641755, 9780321641809, 9780321644213, 9780321644220, 9780321644237, 9780321644695, 9780321734549, 9780321735010, 9780321954121 | 316 | 1,315 | {"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-2016-40 | latest | en | 0.828595 |
https://www.physicsforums.com/threads/rotational-kinetic-energy-question.391623/ | 1,503,344,920,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109525.95/warc/CC-MAIN-20170821191703-20170821211703-00632.warc.gz | 957,692,675 | 14,221 | # Rotational Kinetic Energy Question
1. Apr 1, 2010
### fleabass123
1. The problem statement, all variables and given/known data
A 31.0 kg child runs with a speed of 2.80 m/s tangential to the rim of a stationary merry-go-round . The merry-go-round has a moment of inertia of 520 kg\cdot m^2 and a radius of 2.51 m. When the child jumps onto the merry-go-round, the entire system begins to rotate.
A) Calculate the initial kinetic energy of the system.
B) Calculate the final kinetic energy of the system.
2. Relevant equations
E=1/2*m*v^2+1/2*I*w^2
3. The attempt at a solution
I got the answer to part A by simply doing KE=1/2*m*v^2. The answer was 122 J.
I'm not sure how to approach part B, however. I thought that because of conservation of mechanical energy that the initial energy would equal the final and I could use my answer from part A to solve part B. But this isn't working. Any help would be appreciated. :D
2. Apr 1, 2010
### ehild
The KE energy is not the same as it was initially. There must be some interaction between the child and merry-go-round to set it into rotation and this consumes some energy. You can use conservation of angular momentum to calculate angular velocity, and calculate the rotational energy from that.
ehild | 323 | 1,263 | {"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} | 3.234375 | 3 | CC-MAIN-2017-34 | latest | en | 0.927584 |
https://www.cnblogs.com/bestFy/p/9491887.html | 1,620,953,358,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989616.38/warc/CC-MAIN-20210513234920-20210514024920-00399.warc.gz | 759,332,554 | 7,561 | # AGC009D Uninity
### 题意：
$n\leq 10^5.$
### code:
1 #include<bits/stdc++.h>
2 #define rep(i,x,y) for (int i=(x);i<=(y);i++)
3 #define per(i,x,y) for (int i=(x);i>=(y);i--)
4 #define ll long long
5 #define inf 1000000001
6 #define y1 y1___
7 using namespace std;
8 char gc(){
9 static char buf[100000],*p1=buf,*p2=buf;
11 }
12 #define gc getchar
14 char ch=gc();ll x=0;int op=1;
15 for (;!isdigit(ch);ch=gc()) if (ch=='-') op=-1;
16 for (;isdigit(ch);ch=gc()) x=(x<<1)+(x<<3)+ch-'0';
17 return x*op;
18 }
19 #define N 100005
21 struct edge{int to,nxt;}e[N<<1];
23 void dfs(int u,int pr){
24 for (int i=head[u];i;i=e[i].nxt) if (e[i].to!=pr){
25 int v=e[i].to;
26 dfs(v,u);
27 rep (j,0,20) bit[u][j]+=bit[v][j];
28 }
29 int Min=0;
30 per (i,20,0) if (bit[u][i]>=2){Min=i+1;break;}
31 while (bit[u][Min]) Min++;
32 ans=max(ans,Min);
33 bit[u][Min]=1;
34 rep (i,0,Min-1) bit[u][i]=0;
35 }
36 int main(){
45 } | 396 | 985 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-21 | latest | en | 0.18794 |
chronicengineer.blogspot.com | 1,516,193,934,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886939.10/warc/CC-MAIN-20180117122304-20180117142304-00466.warc.gz | 73,115,983 | 18,508 | ## Saturday, November 19, 2011
### My experiments with toothpaste - II
In part one, we looked at the ubiquitous tube and arrived at an approximate formula for its volume. We had neglected the effect of the increasing major axis on the volume and had taken it as a constant. Here, we shall try to be more accurate by including this tapering as well in our analysis.
For the elliptical cross section of the tube, let the major axis taper linearly from 2R to 3.14R as you move from the bottom of the tube to the top as shown above. The expression for the semi major axis 'a' at any vertical distance 'x' from the top is given by:
The expression for the semi minor axis 'b' at any vertical distance 'x' from the top is given by:
The volume of the tube can be obtained by the following integration.
Substituting the expressions for 'a' and 'b' and working out the integral we get
which is greater than the result obtained earlier.
Applying the result:
For the toothpaste tube under consideration in part one (R=1.6cm, h=13.8cm), the new formula gives a volume of 66.6 ml. Taking a correction factor of 1.5 to account for the bulging profile as opposed to a triangular one, we get a volume of about 100 ml. The choice of an appropriate correction factor is a matter of debate though. For a net weight of 150 gm, the density of the toothpaste would be 1.5 gm/ml as opposed to the earlier result of 2.5 gm/ml. This new result seems more in agreement with literature which says that typically it's in the range of 1.2-1.6 gm/ml.
Thanks to Raghunandan for pointing out an alternative approach to calculating the density of toothpaste coming from the chemical composition data.
Caveat:
Strictly speaking it is not correct to say that both major axis as well as minor axis taper linearly. Whatever be the values of 'a' and 'b', they have to satisfy the constraint that the perimeter of the ellipse has to equal 2 x pi x R (Since our tube was actually formed by pinching one end of a cylindrical tube). Hence, the moment you say that either 'a' or 'b' tapers linearly, the other has to taper non-linearly. But so far as an approximate calculation is concerned we shall not worry about this constraint.
1. Awesome post Lalit.
2. Suggest verifying actual weight of tube/contents
with maufacturer claim
3. @ op741: Thanks :)
@ Anonymous: Nice suggestion. Now that I've used up quite a bit of that tube, I'll try it out when I buy a new tube next time. :)
4. Class! both part-1 and part-2..Very good analysis and wonderfully presented :)
5. @ vinay: Thanks for taking the time to read it. I know how busy you've been...
6. Interesting. Neat work! | 638 | 2,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2018-05 | longest | en | 0.95538 |
http://www.sixsigmacertificationcourse.com/tag/six-sigma-gb-power/ | 1,537,763,830,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160142.86/warc/CC-MAIN-20180924031344-20180924051744-00482.warc.gz | 410,827,917 | 10,454 | ## Six sigma GB Power
Here are my observations based on not just training a few hundred Six Sigma Green Belts across different demographics of geography, industry and age group, but also having mentored them for few months after the training:
• Retention of a subject like Six Sigma is much less than soft-skills or technical training like software languages.
• Ability of Green Belts to relate the application of six sigma concepts to their line of business is quite lesser than soft-skills like team work, time management, negotiation skills, etc.
Here are the 10 things according to me that Six Sigma Green Belts forget within just 10 -days of attending a training program, implying nearly zero retention:
1. Project Charter: How to a write a business case that convinces the management? And also differentiate it from Problem Statement.
2. Fish-Bone: What to do after completing a Fish-Bone diagram? Of course, collect data but on what factors?
3. Gage R&R: Conduct an live Gage R&R study (at least Discrete data).
4. Sampling: Choosing a sampling scheme and deciding the sample size for data collection.
5. Descriptive Statistics: Meaningful and practical interpretation of ‘Standard Deviation’. If there is a process with Standard Deviation of 5 minutes, what does this number mean in real sense?
6. P-value: What does it mean to the business to accept or reject a hypothesis based on P value?
7. Variation: Identify the major sources of variation that impact a project metric?
8. Regression: How to use the regression equation to operate the business efficiently.
9. Control Chart: Explain to a layman (probably a Manager) what an out-of-control data point means in practical sense.
10. Sustenance: How to make sure the project exists after a year. I don’t mean the project deck! Many times, it is considered that sustenance is not in our hands, but actually it is a skill that can be acquired and needs to be taught to all Green Belts.
One of the primary reasons, why Green Belts can’t retain these 10 things, is because most of all this is taught in less than a week with little time for things to settle in, and for the participants to relate.
One of the solutions, that works well for me as a coach, is to limit the class size to 2 or 3 and spread the program over 60 days. I actually found best results during my one-on-one training sessions!
Return-on-Investment of such executive development programs are at most important and so the business & career benefits recoups these investments.
My note will be incomplete if I don’t mention about few things that retain very well after the training, even for several years:
· How to do Fish-Bone Diagram?
· How to play the Gage R&R game?
· Remember and recite the phrase “P is low, Null must go”
· How to map the process? | 609 | 2,784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-39 | longest | en | 0.932127 |
https://www.proprofs.com/quiz-school/story.php?title=kinematics-quiz | 1,553,576,783,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204790.78/warc/CC-MAIN-20190326034712-20190326060712-00036.warc.gz | 858,646,034 | 16,939 | # Introduction To Kinematics (2)
10 Questions
Settings
Another 10 questions over Kinematics 1 (K1).
• 1.
Which of the following options contain an error?
• A.
V_i = 3.2 j (m/s)
• B.
A = -2.5 j (m/s)
• C.
∆x = 500 (m)
• D.
X_f = -2.5 i (m)
• 2.
A car accelerates from 5.0 m/s eastward to 25.0 m/s eastward in a time of 4.0 seconds. What is the acceleration of the car? State your answer to the tenths place.[State your answer in the format: "60.0 i (m/s/s)" or "-15.0 j (m/s/s)", without the quotation marks.]
• 3.
A car accelerates from 5.0 m/s eastward to 25.0 m/s eastward in a time of 4.0 seconds. What is the displacement of the car? State you answer to the tenths place.[State your answer in the format: "10.0 i (m)" or "-45.0 j (m)", without the quotation marks.]
• 4.
A bus begins at xi = -5.0 j (m) and travels for 3.0 seconds at vavg = -10.0 j (m/s).
• A. ∆x
• A.
• B. x_f
• B.
• 5.
A car is driving 30.0 m/s eastward when the driver slams on the brakes. If the rate of acceleration is 5.0 m/s/s westward, then how much space does the car need to come to a complete stop? State your answer to the tenths place; do NOT include direction because I'm asking for distance, not displacement.[State your answer in the format: "20.0 m" or "80.0 m", without the quotation marks.]
• 6.
If a car has a negative acceleration, it might be speeding up.
• A.
True
• B.
False
• 7.
A bicycle is moving at 1.40 m/s westward and begins to slow at the rate of 0.60 m/s/s. What is the velocity of the bicycle after 1.20 seconds? State your answer to the hundredths place.[State your answer in the format: "-6.55 i (m/s)" or "0.82 j (m/s)", without the quotation marks.]
• 8.
A car is driving northward at 18.0 m/s. The driver then begins to accelerate at a constant 2.0 m/s/s northward. Three seconds later, the car has moved 63.0 meters. If we wait an additional 3.0 seconds, will it have moved an additional 63.0 meters (exactly)? That is, will the car move twice the distance in twice the time?
• A.
Yes
• B.
No
• 9.
A car travels 80.0 m southward during a 3.6 s period of time, during which it maintains a constant acceleration of 1.5 m/s/s southward. What was the car's initial velocity? State your answer to the tenths place.[State your answer in the format: "26.0 i (m/s)" or "-8.6 j (m/s)", without the quotation marks.]
• 10.
A car can be at rest but still have an acceleration.
• A.
True
• B.
False | 763 | 2,420 | {"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-2019-13 | latest | en | 0.903899 |
https://gmatclub.com/forum/calling-all-hec-paris-mba-applicants-sept2017-intake-class-of-dec2018-224514.html | 1,726,120,482,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651422.16/warc/CC-MAIN-20240912043139-20240912073139-00590.warc.gz | 255,214,580 | 128,416 | Last visit was: 11 Sep 2024, 22:54 It is currently 11 Sep 2024, 22:54
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# Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018
Intern
Joined: 21 Jun 2015
Posts: 28
Own Kudos [?]: 1 [0]
Given Kudos: 1
GMAT 1: 720 Q50 V38
Intern
Joined: 21 Jun 2015
Posts: 28
Own Kudos [?]: 1 [0]
Given Kudos: 1
GMAT 1: 720 Q50 V38
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Given Kudos: 7
MBA Section Director
Joined: 22 Feb 2012
Affiliations: GMAT Club
Posts: 8792
Own Kudos [?]: 10328 [0]
Given Kudos: 4563
Test: Test
Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
I request all applicants to add HEC MBA program to your MBA timeline so that the applicant stats displayed in the first post reflect true picture of admissions. While adding the program, MAKE SURE you are setting it for graduation year i.e. 'Class of 2019'. (We know the class starting in September 2017 will graduate in December 2018, however since our Bschool system counts the program duration in years, not months, the system classifies them as Class of 2019 students.)
Attachment:
HEC_Paris.png [ 106.64 KiB | Viewed 13812 times ]
MBA Section Director
Joined: 22 Feb 2012
Affiliations: GMAT Club
Posts: 8792
Own Kudos [?]: 10328 [0]
Given Kudos: 4563
Test: Test
Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
paisaj87
Hello Everyone,
I will be submitting my application for the September round. Still waiting for the recommendation letters to finally arrive.
Anyone else in for the sept-17 intake?
Thanks for joining the discussion. As per our forum stats, 19 applicants applying for Sept 2017 intake. Hope others too will join this discussion soon.
Good luck!
MBA Section Director
Joined: 22 Feb 2012
Affiliations: GMAT Club
Posts: 8792
Own Kudos [?]: 10328 [0]
Given Kudos: 4563
Test: Test
Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
Hello Everyone - Please join me in welcoming Ameerkhatri6167 as a Thread Master for HEC Paris (Sept 2017 Intake) MBA Thread. We are thrilled to have him on the site and joining the Thread Master ranks. He joined GMAT Club community in 2015 and has since then demonstrated leadership and helpfulness in forum discussions.
Ameer has been selected into the September 2016 intake for the HEC Paris and is willing to help prospective applicants. If anyone has any questions about HEC application, deadlines, school stats, feel free to PM him. For now, welcome him and wish him best of luck as the ThreadMaster.
Best Regards,
Narenn
Intern
Joined: 19 Nov 2015
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Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
Hello everyone! I got admission offer for Jan 2017 intake but i got it deferred to Sept 2017 intake. So technically, i am the first person to be admitted in the Sept 2017 intake batch.
Let me know if u have any questions
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Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
Narenn
Hello Everyone - Please join me in welcoming Ameerkhatri6167 as a Thread Master for HEC Paris (Sept 2017 Intake) MBA Thread. We are thrilled to have him on the site and joining the Thread Master ranks. He joined GMAT Club community in 2015 and has since then demonstrated leadership and helpfulness in forum discussions.
Ameer has been selected into the September 2016 intake for the HEC Paris and is willing to help prospective applicants. If anyone has any questions about HEC application, deadlines, school stats, feel free to PM him. For now, welcome him and wish him best of luck as the ThreadMaster.
Best Regards,
Narenn
Thank you Narenn.
I will try my best to give back to the community which has served me so well and was incremental in getting me into HEC Paris.
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Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
paisaj87
Hello Everyone,
I will be submitting my application for the September round. Still waiting for the recommendation letters to finally arrive.
Anyone else in for the sept-17 intake?
Best of luck for your Application. If you need any help before that with regards to application or Mock interviews feel free to reach out via PM.
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Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
thescorpianvj
Hello everyone! I got admission offer for Jan 2017 intake but i got it deferred to Sept 2017 intake. So technically, i am the first person to be admitted in the Sept 2017 intake batch.
Let me know if u have any questions
Congratzz on your Admit. You still have more than a year to join in. I suggest you enjoy that to the maximum.
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Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
1
Bookmarks
Hi Everyone,
I am Ameer Khatri. I am your thread master and I will keep you updated on the latest happening in the HEC Campus.
I am thrilled start my MBA at HEC Paris in sept 2016.
I wish everyone the best of luck for their applications. If anyone needs any help with regards to profile evaluations, Application essays or mock interviews for practice please feel free to reach out.
You can PM me here or you can drop me an E-mail at ameer.khatri@hec.edu.
I will try to host webinars and chats with the student council presidents and Adcoms if possible. Keep checking the thread for those updates.
Intern
Joined: 19 Nov 2015
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Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
Ameerkhatri6167
thescorpianvj
Hello everyone! I got admission offer for Jan 2017 intake but i got it deferred to Sept 2017 intake. So technically, i am the first person to be admitted in the Sept 2017 intake batch.
Let me know if u have any questions
Congratzz on your Admit. You still have more than a year to join in. I suggest you enjoy that to the maximum.
Thanks Ameer. I will be in touch with you as probably we will meet in HEC Paris in Sept 2017. I will utilise and enjoy my time to the fullest. :D
Posted from my mobile device
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Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
Hi Ameer,
Welcome to the forum and congratulation on making it to HEC. I am applying for the Masters in International Finance at HEC. Had some queries regarding the program. Wanted to understand if there is any separate group for that discussion, or I can post my queries here?
Thanks!
Intern
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Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
jainshubhams92@gmail.com
Hi Ameer,
Welcome to the forum and congratulation on making it to HEC. I am applying for the Masters in International Finance at HEC. Had some queries regarding the program. Wanted to understand if there is any separate group for that discussion, or I can post my queries here?
Thanks!
Thank you, Shubham
I am not really sure if there is a separate group. Gmatclub and this thread is more of an MBA oriented thread. You can post your queires here but i do't think i will be of much help in regards to MIF. I do help out with application essay consulting and Mock interviews so if you need help with that you can PM me or drop me an E-mail at ameerkhatri6167@gmail.com
Manager
Joined: 27 May 2016
Posts: 65
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Location: India
Concentration: Finance, Strategy
GMAT 1: 690 Q49 V34
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Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
Ameerkhatri6167
jainshubhams92@gmail.com
Hi Ameer,
Welcome to the forum and congratulation on making it to HEC. I am applying for the Masters in International Finance at HEC. Had some queries regarding the program. Wanted to understand if there is any separate group for that discussion, or I can post my queries here?
Thanks!
Thank you, Shubham
I am not really sure if there is a separate group. Gmatclub and this thread is more of an MBA oriented thread. You can post your queires here but i do't think i will be of much help in regards to MIF. I do help out with application essay consulting and Mock interviews so if you need help with that you can PM me or drop me an E-mail at ameerkhatri6167@gmail.com
Thanks Ameer!
Will write out a detailed post here, after my GMAT next week.
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Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
Hi guys, I have sent my CV trough candidate profile assesment and I think the give me an automated reply.. am I right?
they ask me to do another GMAT (mine is 630) so I think I will apply after I retake the test.
Cheers for the admitted and prospective
Intern
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Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
kimtaeyeon89
Hi guys, I have sent my CV trough candidate profile assesment and I think the give me an automated reply.. am I right?
they ask me to do another GMAT (mine is 630) so I think I will apply after I retake the test.
Cheers for the admitted and prospective
The profile assesment is not a automated reply but it is a standard formatted reply where only particular inputs are changed from candidate to candidate.
And yes with 630, i too would suggest you to retake and apply. Best of luck for your second attempt.
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Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
Hello Guys,
Does anyone has already seen the questions for the recomendation letters? I just want to have a general sense of the questions.
Thank you,
Manager
Joined: 12 Aug 2015
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Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
Hi Ameer,
I mailed previously for CV review but got no reply. Can you advise, whom should i mail for CV review.
Regards,
Ameerkhatri6167
Hi Everyone,
I am Ameer Khatri. I am your thread master and I will keep you updated on the latest happening in the HEC Campus.
I am thrilled start my MBA at HEC Paris in sept 2016.
I wish everyone the best of luck for their applications. If anyone needs any help with regards to profile evaluations, Application essays or mock interviews for practice please feel free to reach out.
You can PM me here or you can drop me an E-mail at ameer.khatri@hec.edu.
I will try to host webinars and chats with the student council presidents and Adcoms if possible. Keep checking the thread for those updates.
Manager
Joined: 07 May 2015
Posts: 159
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Given Kudos: 21
Location: India
GMAT 1: 660 Q48 V31
GPA: 3
Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
Hi All,
Thanks for starting this thread. I am applying for Sept'17 intake too. I did submit my profile for quick evaluation and since my score is 630 i was recommended to apply with better score. Their current average is 685.
I am rewriting my test on 4th Oct and there will submit application by Nov deadline. Hopefully I am not too late. Just wanted to check if my assumption that application from all rounds are treated equally. Please let me know
Re: Calling all HEC Paris MBA applicants:(Sept2017 intake)Class of Dec2018 [#permalink]
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234 posts | 3,571 | 13,045 | {"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} | 2.859375 | 3 | CC-MAIN-2024-38 | latest | en | 0.826532 |
https://www.comsol.co.in/blogs/using-radial-basis-functions-for-surface-interpolation/ | 1,590,997,272,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347415315.43/warc/CC-MAIN-20200601071242-20200601101242-00143.warc.gz | 654,034,762 | 19,733 | # Using Radial Basis Functions for Surface Interpolation
March 8, 2016
Have you ever had a set of nonuniformly distributed points in a Cartesian plane that sample a surface height, such as points on the contours of a map or data points representing some material property data? If so, you probably also wanted to reconstruct, or interpolate, a continuous and smooth surface between these points. You can construct such a surface using the core capabilities of COMSOL Multiphysics by using Radial Basis Functions. Let’s find out how…
### An Introduction to Radial Basis Functions
A single Radial Basis Function (RBF) is any function defined in terms of distance (radius) from a point:
(1)
z(x,y)=w \phi \left(\sqrt{(x-x_c)^2+(y-y_c)^2} \right) = w \phi ( || \mathbf {x- c}||) = w \phi(r)
where w is the weight of this RBF; \mathbf{c}=(x_c,y_c) are the coordinates of the point, or center; and r is the distance from any other point in the xy-plane to this center.
The RBF itself can be one of many different types of functions. The family of polyharmonic splines is often used for interpolation, particularly the thin-plate spline function. The thin-plate spline basis function is:
(2)
\phi(r)=r^2 \log (r)
Now, just a single one of these RBFs is not all that interesting, but we can take a sum over a finite number of different centers with different weights and optionally add a linear polynomial term with weights a_0, a_1, a_2, giving us the function:
(3)
z(x,y)= \sum_{i=1}^N w_i \phi(r_i) + a_0 + a_1x + a_2y
If there are enough centers, then this sum of a set of RBFs can be used to represent very complicated single-valued functions. When using the thin-plate spline basis, there is the added advantage that this function is also smooth everywhere and infinitely differentiable.
Let’s now take a look at how to interpolate a smooth surface with these RBFs. If we are given a finite set of center point locations, \mathbf{c}_1, \mathbf{c}_2, … \mathbf{c}_N, and their corresponding known heights, z_1, z_2, … z_N, then we can write a system of linear equations:
(4)
\left[\begin{array}{cccccc}\phi_{1,1} & \dots & \phi_{1,N} & 1 & x_{c,1} & y_{c,1}\\\vdots & \ddots & \vdots & & \vdots & \\\phi_{N,1} & \dots & \phi_{N,N} & 1 & x_{c,N} & y_{c,N}\\1 & & 1 & 0 & 0 & 0 \\x_{c,1} & \dots & x_{c,N} & 0 & 0 & 0 \\y_{c,1} & & y_{c,N} & 0 & 0 & 0\end{array}\right]\left\{\begin{array}{c}w_1\\\vdots\\w_N\\a_0\\a_1\\a_2\end{array}\right\}= \left\{\begin{array}{c}z_1\\\vdots\\z_N\\\\end{array}\right\}
where the terms of the system matrix, \phi_{i,j} = \phi ( || \mathbf {c}_i- \mathbf{c}_j||) , are the Radial Basis Functions evaluated between the centers.
Almost all of the off-diagonal terms will be nonzero when using the thin-plate spline basis, hence this system matrix is quite dense. The linear system can be solved for all of the weights and we can then evaluate the sum of our weighted RBFs at any other point in the xy-plane, giving us a smooth interpolation function. Let’s now look at how to compute these weights and visualize the interpolation function using the core capabilities of COMSOL Multiphysics.
### Surface Interpolation with Radial Basis Functions in COMSOL Multiphysics
We start with a model containing a 3D component with a dimensionless units system. The units system is selected in the settings for Component 1. A dimensionless units system is simpler to use if our data represents material properties rather than a geometry.
The geometry in the model consists of two features. First, a Point feature is used to define the set of points. (The list of coordinates can be copied from a text file.) The Cumulative Selection is used to define a named selection of all of these points, as shown in the screenshot below. There is additionally a Block feature, which has dimensions that are slightly larger than the range of data points and is positioned to enclose all data points.
A screenshot showing the definition of the data points to interpolate and the cumulative selection definition.
The data points and the bounding block.
Once the geometry is defined, we define an Integration Component Coupling operator over the points that we just created. Since the integration is done over a set of points, this operator is equivalent to taking a sum of an expression evaluated over the set of points. Next, we define three variables, as shown in the screenshot below.
First, the variable r = eps+sqrt((dest(x)-x)^2+(dest(y)-y)^2) will be used to compute the distances between all of the centers. Note the usage of the dest() operator, which forces the expression within the operator to be evaluated on the destination points instead of the source points. A very small nonzero term ( eps is machine epsilon) is added so that this expression is never precisely zero.
Next, the variable phi = r^2*log(r) is Equation (2), the thin-plate spline basis function. Note that this function converges to zero for a radius of zero, but we did need to make the radius very slightly nonzero because of the log function so that the basis function can be evaluated at zero. It is also worth remarking that this function could be changed to any other desired basis function.
Lastly, the definition RBF = intop1(w*phi)+a0+a1*x+a2*y is Equation (3), the interpolated surface itself, with weights that are not yet computed. Keep in mind that the integration component coupling operator takes a sum of the expression within the operator over those points.
A screenshot showing the definitions of the variables.
Now that the geometry is set up and all variables are defined, we are ready to solve for the weights for the RBF and the polynomial terms. This is done with a Point ODEs and DAEs interface, defined over the points that we want to interpolate, as shown in the screenshot below. We can set all of the units to dimensionless, since the point locations are also dimensionless. These settings define a set of unknowns, w, which will have a different value for each point.
Settings for the Point ODEs and DAEs interface.
Within this physics interface, only two features need to be modified. Firstly, the settings for the Distributed ODE need to be adjusted as shown below. The Source Term is defined as z-RBF. Since all other terms in the equation are zero when a Stationary study is solved, this term means that RBF=z at all of the selected points. With this one feature, we define rows 1 through N of Equation (4).
The settings for the Source Term at each point.
Secondly, we need to define the last three rows of Equation (4). This is done with a Global Equations feature, as shown in the next screenshot. These three equations solve for the weights a0, a1, and a2. Again, the integration coupling operator takes a sum of the expression over all selected points. With these two features, the problem is completely defined and almost ready to solve.
A screenshot showing the Global Equations for the polynomial weights.
Solving this model requires that we have a mesh on all points, so we apply a free tetrahedral mesh to the bounding box and then solve with a Stationary solver. Once the problem is solved, we can plot our interpolation function, the variable RBF, as shown below.
The smooth and differentiable interpolation surface passes through all of the data points.
### Packaging the Functionality into a Simulation App
If you would like to use this functionality without setting up all of these features in your own models, you are also welcome to download our demonstration app from our Application Gallery, which takes in the xyz data points from a comma-delimited file and computes the interpolation surface. Up to 5000 data points can be interpolated with this demo app.
In addition to computing this surface, the app can write out the complete analytic function describing the surface, and also write out a COMSOL-format CAD file of the surface itself, all within the core capabilities of COMSOL Multiphysics. The CAD data is a NURBS surface and thus only approximately represents the function, but to a very high accuracy for reasonably smooth surfaces. A screenshot of the app’s user interface is shown below.
Screenshot of an app that computes an interpolation function and writes out the function and CAD surface.
### Further Resources
If you’re interested in finding out more about the Application Builder and COMSOL Server™, which can be used to build and run this app, check out the resources below.
What would you like to do with COMSOL Multiphysics? Do you have further questions about the capabilities of Radial Basis Functions? Contact us for help.
#### Categories
##### Dennis Nagy
March 17, 2016
Wow! What goes around comes around. Bill Knudson and I developed software and published a paper on this approach in 1974! I’m sure the above approach must be better after 42 years.
##### Walter Frei
April 1, 2016
Hi Dennis, Nice of you to post this interesting article. I think the advantage here is in the flexibility of using the output in other COMSOL models, either as a geometric surface, or as a function.
##### Peter Oving
December 6, 2018
Hi Walter,
Thanks for your very interesting article!
I had liked to see the results of the application… | 2,197 | 9,234 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2020-24 | latest | en | 0.893522 |
https://studylib.net/doc/17685530/set-up-the-integrals-needed-to-solve-each-of-the...-inclu... | 1,695,672,205,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510085.26/warc/CC-MAIN-20230925183615-20230925213615-00647.warc.gz | 601,837,913 | 10,932 | # SET UP THE INTEGRALS NEEDED TO SOLVE EACH OF THE... INCLUDE AN ILLUSTRATION OF THE VARIABLE.
```SET UP THE INTEGRALS NEEDED TO SOLVE EACH OF THE PROBLEMS BELOW.
INCLUDE AN ILLUSTRATION OF THE VARIABLE.
1. A tank contains 288 ft3 of water. If the density of water is 62.4 lbs/ft3, how much work is needed to
pump all of the water out of the tank? How would your integral change if the tank is on one of its other
sides? Which position would produce the greatest amount of work?
8 ft
4 ft
12 ft
2. A tank is full of water. How much work is needed to pump all of the water out of the tank to a point
3 feet above the tank? How would your integral change if the tank is on its side?
12 ft
5 ft
3. A conical tank filled with kerosene is buried 4 feet underground. The density of kerosene is 51.2
lbs/ft3. The kerosene is pumped out until the level drops 5 feet. How much work is needed to pump the
kerosene to the surface if the variable is given as
A. The distance between the vertex of the cone and the “slice”?
B. The distance between the top of the cone and the “slice”?
C. The distance between the surface and the “slice”?
D. The distance between the final level of the kerosene and the “slice”?
surface
4 ft
6 ft
8 ft
4. A chain is L feet long and weighs W pounds. How much work is needed to pull the chain to the top
of a bridge that is L 5 feet tall?
5. A block of ice weighing 500 pounds will be lifted to the top of a 200 foot building. In the 20 minutes
it will take to do this, the block will lose 12 pounds. How much work is needed to lift the block of ice
to the top of the building?
6. A banner in the shape of an isosceles triangle is hung over the side of a building. The banner has a
base of 25 feet (at the roof line), a height of 20 feet, and weighs 40 pounds. How much work is needed
to lift the banner onto the roof of the building?
``` | 512 | 1,858 | {"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-2023-40 | latest | en | 0.946633 |
https://depedtambayan.net/mathematics-10-quarter-1-module-10-polynomial-equation/ | 1,701,391,302,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100258.29/warc/CC-MAIN-20231130225634-20231201015634-00120.warc.gz | 245,543,890 | 28,785 | # Mathematics 10 Quarter 1 – Module 10: Polynomial Equation
This module was designed and written with you in mind. It is here to help you define and identify a polynomial equation, classify a polynomial equation according to its degree, define root (solution) of a polynomial equation, prove rational root theorem, find the roots of any polynomial equation using the rational root theorem, and solve problems involving polynomial equation. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. The lessons are arranged to follow the standard sequence of the course but the order in which you read and answer this module is dependent on your ability.
This module contains Lesson 1: Illustrating a Polynomial Equation and Lesson 2: Finding the Roots of Polynomial Equation.
After going through this module, you are expected to:
• define a polynomial equation
• identify a polynomial equation
• classify a polynomial equation according to its degree.
• define root (solution) of a polynomial equation,
• prove rational root theorem,
• find the roots of any polynomial equation using the rational root theorem, and
• solve problems involving polynomial equation.
math10_q1_mod10-Polynomial-Equation
## Can't Find What You'RE Looking For?
We are here to help - please use the search box below. | 272 | 1,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-50 | latest | en | 0.905688 |
https://www.business-analysis-made-easy.com/learning-curve-theory1.html | 1,716,301,106,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058484.54/warc/CC-MAIN-20240521122022-20240521152022-00109.warc.gz | 593,131,910 | 8,377 | Learning Curve
by Mahalakshmi
What is the equation for the learning curve prediction calculation?
Dec 16, 2009 Learning Curve Equations by: Dee Reavis Learning Curvey=kx-ay=the time per cyclek=the time for the first cyclex=the number of cyclesa=a constant representing the learning curve percentageLet?s take an example to see how this works. A certain task takes 50 minutes the first time it is done. The second time it takes 40 minutes. The fourth time it takes 32 minutes. Each time the number of cycles is doubled, the time to complete the task is .8 of the time. This would then be called an 80% learning curve.Using the above equation we can solve for the variable a for the 80% learning curve:a=(log k - log y)/log x Using a simple case of second cycle and y=10 this simplifies to:a=(log 10 - log 8)/log 2=.322using the first equation we can verify our time for the 4th cycle:y=50 x 4-.322=32So there you have all the math required to use learning curves.
Combining Two Learning Curves
by Jim Stana
(Orlando FL USA)
If I have a total product made up of several items and subassemblies that each follow a different learning curve but roughly the same quantity, is there a way to derive mathematically the overall learning curve for the sum product of those individual learning curves? I am trying to predict overall product cost (time to manufacture x \$/hr) of a product using what is known about the subunits and convert that to a top level learning curve model.
There probably is a way to derive a combination of 2 learning curves. However, if you don't have a Phd in math, the quick and easy way is to use Monte Carlo Simulation. Since you know the parameters of the individual learning curves, you can just plug them into the simulation and let them run. After a sufficient number of interations, you will have an excellent idea of combined learning curve looks like.
Learning Curve Theory
by Anjala Fernando
(SL)
What is the Important of Learning Curve Theory? | 460 | 1,983 | {"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} | 3.953125 | 4 | CC-MAIN-2024-22 | latest | en | 0.929038 |
http://www.webqc.org/symmetry.php | 1,438,297,080,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042987775.70/warc/CC-MAIN-20150728002307-00213-ip-10-236-191-2.ec2.internal.warc.gz | 822,812,481 | 4,860 | #### Point Group Symmetry Character Tables
Hint: click on a point group symmetry symbol to get irreducible representations, character and product tables
Nonaxial
C1
Cs
Ci
Cn
C2
C3
C4
C5
C6
Cnv
C2v
C3v
C4v
C5v
C6v
Cnh
C2h
C3h
C4h
C5h
C6h
Dn
D2
D3
D4
D5
D6
Dnh
D2h
D3h
D4h
D5h
D6h
Dnd
D2d
D3d
D4d
D5d
D6d
Sn
S4
S6
S8
S10
Higher
Td
Oh
Ih
Linear
C∞v
D∞h
## Point Group Symmetry
Point group symmetry is an important property of molecules widely used in some branches of chemistry: spectroscopy, quantum chemistry and crystallography.
An individual point group is represented by a set of symmetry operations:
• E - the identity operation
• Cn - rotation by 2π/n angle *
• Sn - improper rotation (rotation by 2π/n angle and reflection in the plane perpendicular to the axis)
• σh - horizontal reflection plane (perpendicular to the principal axis) **
• σv - vertical reflection plane (contains the principal axis)
• σd - diagonal reflection plane (contains the principal axis and bisect the angle between two C2 axes perpendicular to the principal axis)
* - n is an integer
** - principal axis is a Cn axis with the biggest n.
Molecule belongs to a symmetry point group if it is unchanged under all the symmetry operations of this group.
Certain properties of the molecule (vibrational, electronic and vibronic states, normal vibrational modes, orbitals) may behave the same way or differently under the symmetry operations of the molecule point group. This behavior is described by the irreducible representation(irrep, character). All irreducible representations of the symmetry point group may be found in the corresponding character table. Molecular property belongs to the certain irreducible representation if it changes undersymmetry operations exactly as it is specified for this irreducible representation in the character table.
If some molecular property A is a product of other properties B and C, the character of A is a product of B and C characters and may be determined from the character product table.
In general assignment of a character (irriducible representation) to a given molecular property depends on the molecular orientation. To make this assignment unambiguous Mulliken developed conventions for symmetry notations which became widely accepted.
By using this website, you signify your acceptance of Terms and Conditions and Privacy Policy. | 592 | 2,371 | {"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-2015-32 | longest | en | 0.827431 |
http://www.panamath.org/wiki/index.php?title=What_is_a_Weber_Fraction%3F&diff=282&oldid=67 | 1,544,628,805,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823895.25/warc/CC-MAIN-20181212134123-20181212155623-00518.warc.gz | 449,077,433 | 32,445 | # What is a Weber Fraction?
(Difference between revisions)
Revision as of 21:06, 28 March 2011 (view source)Panamath (Talk | contribs)← Older edit Latest revision as of 17:35, 6 October 2011 (view source)Sharon (Talk | contribs) (110 intermediate revisions not shown) Line 1: Line 1: + In mathematics and physics, Weber's law states that, as the ratio between the magnitudes of two stimuli increases, the more easily the difference between the two stimuli will be perceived. + + Because the probability of distinguishing whether there are more blue or yellow dots increases as the disparity between the actual number of blue versus yellow dots increases, Weber's law also applies to the Panamath task. Models of the ANS utilize Gaussian curves that represent different numerosities, or mental representations of quantity. The spread of these curves is determined by the Weber fraction, hence allowing analysis and comparison of subject performance. + + As the site is currently under development, Panamath is still collecting all of the necessary citations for this wiki page. No references will be provided until this process is complete. == Model Representations of the ANS == == Model Representations of the ANS == - [[Image:mentalnumberline.jpg|thumb|left|alt=Figures 1a - 1b: Numerosity activations on the mental number line, and the Weber ratio.| Figures 1a - 1b: Numerosity activations on the mental number line, and the Weber ratio.]] + [[Image:mentalnumberline.jpg|thumb|left|250 px|alt=Figures 1a - 1b: Numerosity activations on the mental number line, and the Weber ratio.| Figures 1a - 1b: Numerosity activations on the mental number line, and the Weber ratio.]] - In modeling performance on tasks that engage the ANS, it is necessary first to specify a model for the underlying approximate number representations. It is generally agreed that each numerosity is mentally represented by a distribution of activation on an internal “number line.” These distributions are inherently “noisy” and do not represent number exactly or discretely {{Cite book | last1 = Dechaene | first1 = Stanislas | title = The Number Sense : How the Mind Creates Mathematics | url = http://www.oup.com/us/catalog/general/subject/Psychology/CognitivePsychology/?view=usa&ci=9780195132403 | publisher = Oxford University Press | isbn = 978-0-19-513240-3}}{{Cite journal | author = Gallistel, C., & Gelman, R. | year = 2000 | title = Non-verbal numerical + In modeling performance on tasks that engage the ANS, it is necessary first to specify a model for the underlying approximate number representations. It is generally agreed that each numerosity is mentally represented by a distribution of activation on an internal “number line.” These distributions are inherently “noisy” and do not represent number exactly or discretely. This means that there is some error each time they represent a number; and this error can be thought of as a spread of activation around the number being represented. - cognition: from reals to integers | url = http://eebweb.arizona.edu/faculty/dornhaus/courses/materials/papers/Gallistel%20Gelman%20numbers%20counting%20cognition.pdf | journal = Trends in Neurosciences | volume = 4 | issue = 2 | pages = 59-65}} . This means that there is some error each time they represent a number; and this error can be thought of as a spread of activation around the number being represented. + == The Mental Number Line == == The Mental Number Line == - The mental number line is often modeled as having linearly increasing means and linearly increasing standard deviation . In such a format, the representation for e.g., cardinality seven is a probability density function that has its mean at 7 on the mental number line and a smooth degradation to either side of 7 such that 6 and 8 on the mental number line are also highly activated by instances of seven in the world. In Figure 1a I have drawn idealized curves which represent the ANS representations for numerosities 4-10 for an individual with Weber fraction = .125. You can think of these curves as representing the amount of activity generated in the mind by a particular array of items in the world with a different bump for each numerosity you might experience (e.g., 4 balls, 5 houses, 6 blue dots, etc). Rather than activating a single discrete value (e.g., 6) the curves are meant to indicate that a range of activity is present each time an array of (e.g., 6) items is presented {{Cite journal | doi = 10.1146/annurev.neuro.051508.135550 | author = Nieder, A., & Dehaene, S. | year = 2009 | title = Representation of number in the brain | url = http://www.annualreviews.org/doi/abs/10.1146/annurev.neuro.051508.135550 | journal = Annual Review of Neuroscience | volume = 32 | issue = 1 | pages = 185-208}} . That is, an array of e.g., ''six'' items will greatly activate the ANS numerosity representation of 6, but because these representations are noisy this array will also activate representations of 5 and 7 etc with the amount of activation centered on 6 and gradually decreasing to either side of 6. + The mental number line is often modeled as having linearly increasing means and linearly increasing standard deviation. In such a format, the representation for e.g., cardinality seven is a probability density function that has its mean at 7 on the mental number line and a smooth degradation to either side of 7 such that 6 and 8 on the mental number line are also highly activated by instances of seven in the world. In Figure 1a I have drawn idealized curves which represent the ANS representations for numerosities 4-10 for an individual with Weber fraction = .125. You can think of these curves as representing the amount of activity generated in the mind by a particular array of items in the world with a different bump for each numerosity you might experience (e.g., 4 balls, 5 houses, 6 blue dots, etc). Rather than activating a single discrete value (e.g., 6) the curves are meant to indicate that a range of activity is present each time an array of (e.g., 6) items is presented. That is, an array of e.g., ''six'' items will greatly activate the ANS numerosity representation of 6, but because these representations are noisy this array will also activate representations of 5 and 7 etc with the amount of activation centered on 6 and gradually decreasing to either side of 6. == Neuronal Associations of the Mental Number Line == == Neuronal Associations of the Mental Number Line == Line 18: Line 22: The bell-shaped representations of number depicted in Figure 1a are more than just a theoretical construct; “bumps” like these have been observed in neuronal recordings of the cortex of awake behaving monkeys as they engage in numerical discrimination tasks (e.g., shown an array of six dots, neurons that are preferentially tuned to representing 6 are most highly activated, while neurons tuned to 5 and 7 are also fairly active, and those tuned to 4 and 8 are active above their resting state but less active than those for 5, 6, and 7. These neurons are found in the monkey brain in the same region of cortex that has been found to support approximate number representations in human subjects. This type of spreading, “noisy”, activation is common throughout the cortex and is not specific to representing approximate number. Rather, approximate number representations obey principles that operate quite generally throughout the mind/brain. The bell-shaped representations of number depicted in Figure 1a are more than just a theoretical construct; “bumps” like these have been observed in neuronal recordings of the cortex of awake behaving monkeys as they engage in numerical discrimination tasks (e.g., shown an array of six dots, neurons that are preferentially tuned to representing 6 are most highly activated, while neurons tuned to 5 and 7 are also fairly active, and those tuned to 4 and 8 are active above their resting state but less active than those for 5, 6, and 7. These neurons are found in the monkey brain in the same region of cortex that has been found to support approximate number representations in human subjects. This type of spreading, “noisy”, activation is common throughout the cortex and is not specific to representing approximate number. Rather, approximate number representations obey principles that operate quite generally throughout the mind/brain. - When trying to discriminate one numerosity from another using the Gaussian representations in Figure 1a, the more overlap there is between the two Gaussians being compared the less accurately they can be discriminated. Ratios that are closer to 1 (Ratio = bigger# / smaller#), where the two numbers being compared are close (e.g., 9 versus 10), give rise to Gaussians with greater overlap resulting in poorer discrimination (i.e., “ratio-dependent performance”). Visually, the curve for 5 in Figure 1a looks clearly different in shape from the curve for 4 (e.g., curve 4 is higher and skinnier than curve 5); and discriminating 4 from 5 is fairly easy. As you increase in number (i.e., move to the right in Figure 1a), the curves become more and more similar looking (e.g., is curve 9 higher and skinnier than curve 10?); and discrimination becomes harder. + When trying to discriminate one numerosity from another using the Gaussian representations in Figure 1a, the more overlap there is between the two Gaussians being compared the less accurately they can be discriminated. - But, it is not simply that larger numbers are harder to discriminate across the board. For example, performance at discriminating 16 from 20 (not shown) will be identical to performance discriminating 4 from 5 as these pairs differ by the same ratio (i.e., 5/4 = 1.125 = 20/16); and the curves representing these numbers overlap in the ANS such that the representation of 4 and 5 overlap in area to the same extent that 16 overlaps with 20 (i.e., although 16 and 20 each activate very wide curves with large standard deviations, these curves are far enough apart on the mental number line that their overlap is the same amount of area as the overlap between 5 and 4, i.e., they have the same discriminability). This is ratio-dependent performance. + Ratios that are closer to 1, where the two numbers being compared are close (e.g., 9 versus 10), give rise to Gaussians with greater overlap resulting in poorer discrimination (i.e., “ratio-dependent performance”). The ratio is calculated with this formula, where $N_{smaller} \,$ is the smaller number and $N_{larger} \,$ is the larger number, respectively: + + +
$Ratio = \frac{{N_{larger}}}{N_{smaller}} \,$
+ + + Visually, the curve for 5 in Figure 1a looks clearly different in shape from the curve for 4 (e.g., curve 4 is higher and skinnier than curve 5); and discriminating 4 from 5 is fairly easy. As you increase in number (i.e., move to the right in Figure 1a), the curves become more and more similar looking (e.g., is curve 9 higher and skinnier than curve 10?); and discrimination becomes harder. + + But, it is not simply that larger numbers are harder to discriminate across the board. For example, performance at discriminating 16 from 20 (not shown) will be identical to performance discriminating 4 from 5 as these pairs differ by the same ratio (i.e., $5/4 = 1.125 = 20/16$); and the curves representing these numbers overlap in the ANS such that the representation of 4 and 5 overlap in area to the same extent that 16 overlaps with 20 (i.e., although 16 and 20 each activate very wide curves with large standard deviations, these curves are far enough apart on the mental number line that their overlap is the same amount of area as the overlap between 5 and 4, i.e., they have the same discriminability). This is ratio-dependent performance. == Tasks That Give Rise to the Gaussian Curves == == Tasks That Give Rise to the Gaussian Curves == Line 30: Line 42: When attempting to visualize what the noisy representations of the ANS are like, one can think of the Gaussian activations depicted in Figure 1a, and these representations affect performance in a variety of tasks including discriminating one number for another (e.g., 5 versus 4) and generating number-relevant behaviors (e.g., tapping 9 times). When attempting to visualize what the noisy representations of the ANS are like, one can think of the Gaussian activations depicted in Figure 1a, and these representations affect performance in a variety of tasks including discriminating one number for another (e.g., 5 versus 4) and generating number-relevant behaviors (e.g., tapping 9 times). - == References == + - + == Numerical Discrimination in the ANS == + + [[Image:figure2.JPG|thumb|right|250 px|alt=Figures 2a - 2c: The Panamath task, and Gaussian models of the ANS.| Figures 2a - 2c: The Panamath task, and Gaussian models of the ANS.]] + + To understand how numerical discrimination is possible in the ANS, consider the task of briefly presenting a subject with two arrays, e.g., 5 yellow dots and 6 blue dots, and asking the subject to determine which array is greater in number (Figure 2a). The 5 yellow dots will activate the ANS curve representation of 5 and the 6 blue dots will activate the ANS curve representation of 6 (assume that the subject uses attention to select which dots to send to the ANS for enumerating and then stores and compares those numerosity representations bound to their respective colors) (Figure 2a-b). + + == ANS Modeling and Subtraction == + + An intuitive way to think about ordinal comparison within the ANS is to liken it to a subtraction; this will be mathematically equivalent to other ways of making an ordinal judgment within the ANS and my use of subtraction here should be thought of as one illustration among several mathematically equivalent illustrations. + + Imagine that an operation within the ANS subtracts the smaller (i.e., five-yellow) representation from the larger (i.e., six-blue) representation (Figure 2b). Because the 5 and 6 representations are Gaussian curves, this subtraction results in a new Gaussian representation of the difference which is a Gaussian curve on the mental number line that has a mean of 1 ($6 - 5 = 1) and a standard deviation of: + + + [itex]\sqrt{{{\sigma}_5^2 + {\sigma}_6^2}} \,$
+ + + + In Figure 2c (i.e., when subtracting one Gaussian random variable from another (i.e., $X_6 - X_5 \,$), the result is a new Gaussian random variable with the mean at the difference ($6 - 5 = 1$) and a variance that adds the variances of the original variables: ${\sigma}_5^2 + {\sigma}_6^2 \,$. This results in a Gaussian curve that is centered on 1, but that extends to both the left and right of 0 (Figure 2c). + + One can think of 0 as the demarcation line separating evidence “for” and “against” in that the area under the curve to the right of 0 is the portion of the resulting representation that correctly indicates that ''six'' is greater than ''five'' while the area under the curve to the left of 0 is the portion of the resulting representation that incorrectly indicates that ''five'' is greater than ''six''. This area to the left of 0 results from the overlap between the original Gaussian representations, ''five'' and ''six'', that were being discriminated in which some of the area of ''five-yellow'' is to the right (i.e., greater than) some of the area of six-blue (Figure 2b). + + == Interpreting Gaussian Overlap: Weber's Law == + + One can think of 0 as the demarcation line separating evidence “for” and “against” in that the area under the curve to the right of 0 is the portion of the resulting representation that correctly indicates that six is greater than five while the area under the curve to the left of 0 is the portion of the resulting representation that incorrectly indicates that five is greater than six. This area to the left of 0 results from the overlap between the original Gaussian representations, five and six, that were being discriminated in which some of the area of five-yellow is to the right (i.e., greater than) some of the area of six-blue (Figure 2b). + + Another method would rely on assessing the total evidence for blue and the total evidence for yellow. Either of these ways of making a decision will have the result that, on a particular trial, the probability of the subject getting the trial correct will depend on the relative area under the curve to the left and right of 0 which is itself determined by the amount of overlap between the original Gaussian representations for the numerosities being compared (i.e., ''five'' and ''six''). + + The more overlap there is between the two Gaussian representations being compared, the less accurately they can be discriminated. Consider comparing a subject’s performance on a 5 dots versus 6 dots trial to a trial involving 9 versus 10 dots. Using the curves in Figure 1a as a guide, we see that the overlapping area for the curves representing 5 and 6 is less than the overlapping area for the curves representing 9 and 10, because the curves flatten and spread as numerosity increases. This means that it will be easier for the subject to tell the difference between 5 and 6 than between 9 and 10, i.e., the resulting Gaussian for the subtraction will have more area to the right of 0 for the subtraction of 5 from 6 than for the subtraction of 9 from 10. + + Across multiple trials the subject would give more correct responses on the 5 versus 6 trials than the 9 versus 10 trials owing to the greater evidence for 6 being larger than 5 (i.e., more area to the right of 0 in the subtraction, Figure 2c). The linear increase in the standard deviation of the curves representing the numerosites along the mental number line results in ratio-dependent performance whereby the discriminability of two numerosities increases as the ratio between them (e.g., $bigger number / smaller number$) increases. This is Weber’s law. + + How rapidly performance rises from chance (50%) to near-asymptotic performance (100%) in this kind of dot numerosity discrimination task is controlled by the subject’s Weber fraction (''w''). + + == The Weber Fraction: Overview == + + The Weber fraction indexes the amount of spread in the subject’s ANS number representations and therefore the overlap between any two numbers as a function of ratio (described in a succeeding section). The precision of the ANS varies across individuals with some people having a smaller Weber fraction (i.e., better performance and sharper Gaussian curves) and others having a larger Weber fraction (i.e., poorer performance owing to wider noisier Gaussian curves). + + Numerical discrimination (e.g., determining which color, blue or yellow, has more dots in an array of dots flashed to quickly for explicit counting) is possible in the ANS through a process that attempts to determine which of the two resulting curves (e.g., five-yellow or six-blue) is further to the right on the mental number line. The fullness of these noisy curves is used to make this decision (and not just the mode, or mean or some other metric) and successful discrimination thereby depends on the amount of overlap between the two activated curves (i.e., ratio-dependent performance). + + The amount of overlap is indexed by a subject’s Weber fraction (''w'') with a larger Weber fraction indicating more noise, more overlap, and thereby worse discrimination performance. This model has been found to provide an accurate fit to data from rats, pigeons, and humans of all ages. + + == Relationship to the ANS == + + What does a Weber fraction (''w'') tell us about a subject’s Approximate Number System (ANS) representations? A relationship exists between the Weber fraction (''w''), the standard deviations of the underlying Gaussian numerosity representations, and the discriminability of any two numbers; and a Weber fraction can be thought of in at least three ways, which sometimes leads to confusions. + + These are: 1) the fraction by which a stimulus with numerosity n would need to be increased in order for a subject to detect this change and its direction on 50% of the trials (aka the difference threshold, aka the Just Noticeable Difference, J.N.D.), 2) the midpoint between subjective equality of two numbers and asymptotic performance in discrimination, and 3) the constant that describes the standard deviation of all of the numerosity representations on the mental number line. Option 3 is perhaps the least discussed in the psychophysics literature - it is rarely taught in psychophysics courses - but I will suggest that option 3 is the most productive way to understand the Weber fraction. + + I first describe what a Weber fraction is and how it relates to the underlying number representations and then provide some suggestions for what I take to be a productive way of understanding the Weber fraction. + + == How to Think of the Weber Fraction == + + First, consider Figure 1a to represent the ANS number representations for a particular individual who has a Weber fraction = .125. If one presents the hypothetical subject in Figure 1a with the task of determining which of two colors has the greater number of dots on a trial where there are 16 blue dots and some number of yellow dots, this subject would require an increase of 2 dots from this standard ($16 * .125 = 2$, $N_2 = 16 + 2 = 18$) in order to respond that yellow (18) is more numerous than blue (16) on 75% of the trials that present these two numerosities. That is, a subject’s Weber fraction can be used to determine the amount by which one would need to change a particular number in order for that subject to correctly determine which number was larger on 75% of the trials (chance = 50%). + + Conceived in this way, the Weber fraction describes a relationship between any numerosity and the numerosity that will be consistently discriminated from this standard. This gives one way of understanding why one might choose 75% correct performance as an interesting indicator of performance. + + In order to specify what “consistently discriminated from” might mean one might also choose some other standard (e.g., 66.7% correct, or any other % above chance). From this point of view, often the view taught in courses on Psychophysics, the point is to estimate the steepness of the linear portion of the psychometric function (depicted in Figure 1b) and 66.7% would work for such purposes just as well as 75% or 80%; the choice of 75% correct is presented as more or less an arbitrary one. + + However, as we will see below, 75% correct ''is'' special and the seemingly arbitrary reasons for choosing 75% correct as an index of performance find their non-arbitrariness in a mathematical relationship between correct discrimination, the Weber fraction, and the standard deviations of the underlying Gaussian representations. + + Some readers, more familiar with research on the acuity of the ANS in infants 6-9 months of age and less familiar with the literature on adult psychophysics, may have come to believe that the Weber fraction describes the ratio below which a subject will fail to discriminate two numerosities (e.g., 6-month-olds succeed with a 1:2 ratio and fail with a 2:3 ratio). This suggests a categorical interpretation of the Weber fraction (e.g., a threshold where you will succeed if a numerical difference is “above threshold” and fail if it is “below threshold”). That is, some may have come to believe that performance should be near perfect with ratios easier than a subject’s Weber fraction and at chance for ratios harder than a subject’s Weber fraction. This is not what is seen in typical performance where a large number of trials test a subject’s discrimination abilities across a wide variety of ratios. + + == Modeling Trends == + + [[Image:figure3a.jpg|thumb|left|225 px|alt=Figure 3a: Percent correct as a function of the Weber ratio. This model correlated to over ten thousand subjects.| Figure 3a: Percent correct as a function of the Weber ratio. This model correlated to over ten thousand subjects.]] + + Consider again the simple task of being briefly shown a display that includes some blue and yellow dots and being asked to determine on each flash if there had been more blue or more yellow dots. Percent correct on this numerical discrimination task is not a step function with poor performance “below threshold” and good performance “above threshold”, but rather is a smoothly increasing function from near chance performance to success. + + This performance and the range of individual differences, gathered from over 10,000 subjects between the ages of 8 and 85 years of age participating in this type of blue-yellow dots task, can be seen in Figure 3a-b. Every one of the more than 10,000 participants obeyed this kind of gradual increase in percent correct from a ratio of 1 (where the number of blue and yellow dots are equal) to easier ratios like 2 (where there might be 10 blue dots and only 5 yellow dots; $10 / 5 = 2$). What changes from participant to participant is how steep the left side of the curve is, and these individual differences are shown in the figure by indicating the performance of the 10th and 90th percentile ranking of the more than 10,000 participants. + + + [[Image:figure3b.jpg|thumb|right|225 px|alt=Figure 3b: A graph modeling how the average Weber fraction improves over development.| Figure 3b: A graph modeling how the average Weber fraction improves over development.]] + + Figure 3b shows how the average Weber fraction improves over development. A steeper, quicker rise in the psychometric function (Figure 3a) indicates better sensitivity, better discrimination abilities and is indexed by the subject having a smaller Weber fraction (Figure 3b) (i.e., a ''smaller'' Weber fraction indicates less noise in the underlying ANS representations). + + If one wished to localize the Weber fraction at some point along the smoothly increasing curve in Figure 3a it would be at the midpoint between subjective equality of the two numerosities being compared (typically occurring at a ratio = 1, where $N_1 = N_2$) and asymptotic performance (typically 100% correct, though asymptotic performance may be lower in unskilled subjects resulting in the midpoint falling at a percent correct lower than 75%). + + == The "Difference Threshold" == + + When subjects behave optimally, the Weber fraction is related to the ratio that results in 75% correct performance and not to the first ratio that results in chance, or above chance, performance. Indeed, the actual behavioral data from subjects seen in Figure 3a, and the modeled ideal behavior seen in Figure 1b, suggest that the subjects will ''always'' be above chance no matter how small the difference between $N_1$ and $N_2$. What changes is not whether a participant will succeed or fail to make a discrimination but rather the number of trials an experimenter will have to run in order to find statistically significant performance. + + However, within the practical limits of testing real subjects, the infant literature’s method of looking for a change from above chance performance to below chance performance is a reasonable approach for attempting to roughly locate the Weber fraction of subjects, like infants, who cannot participate in the large number of trials it takes to achieve the smooth data seen in Figure 3a. + + The 75% correct point has been used as an indicator of the “difference threshold” because this point can be translated into a whole-number ratio through a mathematical relationship that holds between the Weber fraction and the ratio at which performance attains 75% correct. + + Researchers occasionally suggest that “human adults can discriminate numerosities that differ by a ratio of $7:8$,” where $8 / 7 = 1.14$ would be the whole number ratio nearest the Weber Ratio that results in 75% correct discrimination. Even this understanding of the Weber fraction as the midpoint between subjective equality and asymptotic performance, while it goes some way towards avoiding the mistaken belief that a difference threshold is a step function, misses the deeper continuous nature of discrimination within the ANS. + + == Understanding Gaussian Spreads == + + Let us consider a third way of understanding the Weber fraction (''w''): as a scaling factor that indexes the amount of noise for every numerical representation of the ANS. Understood in this way, described below, the Weber fraction is not specific to the task of numerical discrimination; indeed, it is wholly independent and prior to discrimination. + + An animal that (bizarrely) could only represent a single numerical value in their ANS (e.g., could only represent approximately-twelve and no other numbers), and who could therefore never discriminate 12 from any other number (i.e., could not even perform a numerical discrimination task), would nonetheless have a Weber fraction and we could measure it. + + Consider the Gaussian curves depicted in Figure 1a. The spread of each successive numerosity from 4 to 10 is steadily wider than the numerosity before it. This means that the discriminability of any two numerosities is a smoothly varying function, dependent on the ratio between the two numerosities to be discriminated. + + In theory, such discrimination is never perfect because any two numerosities no matter how distant from one another will always share some overlap. At the same time, discrimination will never be entirely impossible, so long as the two numerosities are not identical, because any two numerosities, no matter how close (e.g., 67 and 68) will always have some non-overlapping area where the larger numerosity is detectably larger. Correct discrimination may occur on only a small percentage of trials if the two sets are very close in number, but it will never be impossible. This again motivates the intuition that percent correct in a discrimination task should be a smoothly increasing function from the point of subjective equality to asymptotic performance. + + In Figure 1b I have drawn the expected percent correct for the ideal subject in Figure 1a whose w = .125 as derived by a model from classical psychophysics. Those who wish to translate this subject’s Weber fraction into a whole number ratio can determine this ratio from the function in Figure 1b as the nearest whole number ratio to the Weber Ratio that falls at the midpoint between subjective equality (i.e., Weber Ratio = 1, Percent Correct = 50%) and asymptotic performance (i.e., Percent Correct = 100%) and would equal 8:9 (i.e., WR = 1.125) for the subject in Figure 1b. + + Notice, the Weber fraction is equal to this Weber Ratio minus 1 (i.e., $1.125 - 1 = .125$). It is the smoothly increasing spread in the underlying Gaussian representations depicted in Figure 1a that is the source of the smoothly increasing Percent Correct ideal performance in Figure 1b. + + Noting the smoothly increasing spread of the Gaussian representations in Figure 1a motivates one to ask what is the parameter that determines the amount of spread in any each Gaussian representation on the mental number line. + + == Calculating Standard Deviation == + + In fact, it is the Weber fraction that determines the spread of every single representation on the mental number line by the following formula: + + +
$SD = n * w \,$
+ + + The standard deviation (SD) of the Gaussian representing any numerosity is that numerosity multiplied by the Weber fraction. Why is this the case? Intuitively, it is the standard deviations of the underlying Gaussian representations that determines the amount of overlap between the curves that represent any two numerosities, and it is the amount of overlap that determines how well any two numerosities can be discriminated. + + The categorical views of the Weber fraction as a kind of threshold between successful discrimination and failure, or as the midpoint between subjective equality and asymptotic performance, choose to focus on only one particular point of what is actually a continuous and smooth function of increasing success at discrimination, and this entire function is determined by the Weber fraction as this fraction describes the standard deviation of any numerosity representation on the mental number line and thereby the degree of overlap between any two numerosities on the mental number line. + + Viewed in this light, the Weber fraction describes all of the numerosity representations in the ANS. It is not specific to the task of discriminating two numerosities, and not specific to numerosity comparisons near “threshold”. In fact, though this may never be the case in practice, given this understanding of the Weber fraction it would be possible to assess the Weber fraction for an animal who could only represent a single number in the ANS (e.g., can only represent ''approximately-12''). It would be the standard deviation of the Gaussian that represents ''approximately-12'' divided by the mean of this Gaussian. + + == The Weber Fraction as a Scaling Ratio == + + [[Image:figure4.jpg|thumb|left|250 px|alt=Figure 4a-c: Idealized representations of the activation of the mental line for subjects with a Weber fraction of .125 (a), .20 (b), and their performance (c).| Figure 4a-c: Idealized representations of the activation of the mental line for subjects with a Weber fraction of .125 (a), .20 (b), and their performance (c).]] + + The Weber fraction (''w'') is the constant that describes the amount of variability in the underlying ANS number representations. It is a scaling factor by which one could take any one of the curves in Figure 1a and turn it into any of the other curves in Figure 1a. In the linear model in Figure 1a, the analog representation for any numerosity (e.g., n = 7) is a Gaussian random variable with a mean at n (e.g., n = 7) and a standard deviation of (n * w). This means that for a subject who has a Weber fraction of .125, the ANS representation for n = 7 will be a normal curve with a mean of 7 on the mental number line and a standard deviation of $.125 * 7 = .875$. + + By substituting any number you like for n you can easily determine the shape of the underlying ANS representation, without ever having the participant engage in a numerical discrimination task that compares two numbers. This illustrates the power of understanding the Weber fraction as an index of internal noise: rather than simply telling us something about how well a subject will do at discriminating two numbers near their threshold, the Weber fraction (''w'') tells us the shape and overlap of every single number representation inside a person’s Approximate Number System (ANS). The Weber fraction is about all of the representations, not just the ones “near threshold.” + + The inductive power of understanding the Weber fraction (''w'') to be an internal scaling factor is also seen when we compare the Weber fractions of different individuals. Individuals differ in the precision of their ANS representations. Some people have more noise and some people have less. + + In Figure 4a I illustrate some idealized curves which represent the underlying ANS representations for a subject whose w = .125 and in Figure 4b for a subject whose w = .20. Crucially, one can see that the subject in Figure 4b has a greater degree of overlap between the Gaussian curves that represent numerosity than the subject in Figure 4a (recall, a bigger Weber fraction means more noise and fatter curves). It is this overlap that leads to difficulty in discriminating two stimuli that are close in numerosity (i.e., as one nears Weber Ratio = 1). The hypothetical subject in Figure 4b would have poorer discrimination in a numerical discrimination task (e.g., blue & yellow dots task) than the subject in Figure 4a. + + In Figure 4c I have drawn the ideal performance for these two subjects in a numerical discrimination task. The midpoint between subjective equality (Percent Correct = 50%) and asymptotic performance (Percent Correct = 100%) in Figure 4c will fall at WR = 1.125 for the subject in Figure 4a and at WR = 1.20 for the subject in Figure 4b because these are equal to the Weber fraction for each subject plus one. An inspection of Figure 4c reveals that this is the case. This is not an accident; it derives from the relationship described above between the Weber fraction (''w''), the standard deviations of the underlying ANS number representations, and the overlap between any two ANS number representations. + + The values for the Weber fractions in Figure 4 have been chosen so as to illustrate another value of understanding the Weber fraction to be an internal scaling factor; it empowers comparisons across individuals and formal models of individual differences. Converting the Weber fraction for each of these subjects into the nearest whole number fraction reveals that the Weber fraction for the subject in Figure 4a is 8:9 and for the subject in Figure 4b is 5:6 (i.e., $9/8 = 1.125$; $6/5 = 1.20$). Investigating the Gaussian curves in Figure 4a and 4b, which were created by treating the Weber fraction as an internal scaling factor to determine the standard deviations of the ANS Gaussian curves, reveals that the Gaussians for the numerosities 8 and 9 for the subject in Figure 4a are identical to the Gaussians for the numerosities 5 and 6 for the subject in Figure 4b. This is no accident. The only parameter that has been changed in constructing Figures 4a and 4b is the Weber fraction for the subject and this single parameter determines the spread in the Gaussians that represent every possible numerosity in the ANS of each subject. + + In this way, understanding the Weber fraction to be an internal scaling factor that determines the spread of every ANS number representation not only empowers us to compare one number representation to another within a particular subject, it also empowers us to compare across individuals and to create mathematically tractable predictions about how the ANS representations of one person (e.g., the subject in Figure 4a) relate to the ANS representations of another (e.g., the subject in Figure 4b). + + == Deviations From the Panamath Model == + + It is not uncommon for performance in a discrimination task to deviate from the ideal performance depicted in e.g., Figure 1b and seen in actual data in Figure 2a. Two basic deviations capture almost all cases: 1) a “give-up” function wherein subjects give-up on their numerical discrimination abilities on hard ratios and guess at chance, and 2) a “lapse parameter” which tracks participants’ tendency to have a lapse in attention such that they do not attend the e.g., blue and yellow dots on a particular trial. These two forms of deviation from the model presented in this chapter are tractable in that additional parameters can be included in the model to capture these behaviors. + + The model I have described does not take account of participants’ confidence in their answers, nor does it adjust its predictions based on sub-optimal engagement of the Approximate Number System (ANS). But, in practice, sub-optimal performance can occur, especially on difficult trials. + + For example, consider the task of being shown a brief display of blue and yellow dots and being asked to guess which color had more dots. One a fairly easy trial (e.g., 27 yellow versus 7 blue), you would most likely get the correct answer and be highly confident in your answer. This would be true even if the display time was incredibly brief (e.g., 50 msec). + + == Effect of Difficulty on Trials == + + [[Image:figure5.jpg|thumb|right|300 px|alt=Figure 5: Performance of children and adults in a difficiult numerical discrimination task.| Figure 5: Performance of children and adults in a difficiult numerical discrimination task.]] + + But, what will happen on a difficult trial of e.g., 27 yellow and 28 blue dots? On such a trial, a subject can tell fairly rapidly that the trial will be incredibly difficult and they will have very low confidence that their guess will be correct. On such trials, it is not unreasonable that a subject will simply give up on trying to determine whether there are more blue or more yellow dots and guess randomly (e.g., flip a coin). This kind of behavior has been observed in numerical discrimination tasks that present trials that are very difficult for that subject (e.g., presenting 3-year-olds with a brief display of 9 versus 10 objects) and it has also been observed in the discrimination performance of many animal models making a variety of ratio-dependent discriminations (but has yet to be discussed in detail in any literature). When this happens, performance will deviate from the model shown in Figure 1b in that percent correct will drop to chance before the Weber Ratio = 1. Participants will “give-up” on difficult trials and, rather than relying on their ANS representations, will guess randomly. This kind of performance can be seen in the left side of the curves diagramming the performance of children and adults in a difficult numerical discrimination task (Figure 5). + + It is important to note that this kind of drop to chance performance is not predicted by any existing psychophysical model and it appears to be a deviation from the model caused by the subject giving up on the task; it does not appear to be the case, as of yet, that this type of deviation indicates that the underlying model of the ANS itself is flawed. We are currently attempting to specify the exact relationship between difficulty, Weber fraction, and the probability of giving up on a trial. + + == Effect of Attention == + + Another deviation can occur on the right side of the psychometric curve and can also be seen in Figure 5. As has been observed in previous numerical discrimination tasks, performance in unskilled subjects can fail to reach 100% correct even on easy trials, perhaps because of a tendency to suffer a lapse of attention on a subset of trials. As can be seen in the example data in Figure 5, this was the case for the 3-, 4- and 5-year-olds. + + When this type of lapse occurs, it should not be dependent on the ratio that was presented on that trial. Rather, it is an event that occurs because of e.g., fatigue, irrespective of the ratio being presented on that particular trial. As such, this deviation is typically detectable as a flat cut in percent correct across the entire range of ratios presented in an experiment. It is because percent correct is highest on the easiest ratios that the deleterious affects of this lapse in attention are most visible on easier trials. The simplest way to account for this tendency in the model is to include a lapse parameter that is a constant probability of guessing randomly on any particular trial. + + This parameter primarily lowers the model’s asymptotic performance while retaining an accurate estimate of the Weber fraction (''w''). When required, I included this parameter in the model as follows, where $p_{guess} \,$ is the probability of guessing randomly on any particular trial irrespective of ratio, $p_{error} \,$ is the probability of being incorrect given the model above and the chance level is .5 multiplied by 100 to return a percentage: + +
$Percent Correct = [(1 - p_{guess} )(1 - p_{error} ) + p_{guess} * .5 ] * 100 \,$
+ + + This type of guessing has also been observed in children (Figure 5) and in animal and other adult human participants for both numerical discrimination and other ratio-dependent dimensions. + + == What Do These Deviations Mean? == + + Each of these deviations from the model presented in this chapter can be captured by including an additional parameter in the model (i.e., a threshold for “giving up” as a function of ratio, and a lapse parameter; the probability of guessing randomly on any particular trial irrespective of ratio). + + At the moment, the existing datasets do not suggest that deviation from the model presented here is so drastic as to recommend the abandoning of the model. Rather, as with all complex cognition, behaviors that engage the Approximate Number System (ANS) are occurring in a rich context that drives many other considerations that can affect performance. Giving up on a hard trial and being bored such that you fail to attend on a trial are two very basic considerations that can affect performance. + + == Conclusions and Summary == + + If one does some reading in hopes of understanding what a Weber fraction (''w'') is, the most common descriptions turn on the idea of a difference threshold (aka Just Noticeable Difference, J.N.D.). For example, “the fraction by which a stimulus with numerosity n would need to be increased in order for a subject to detect this change and its direction on 50% of the trials” is a common phrase; or, “the midpoint that indicates roughly successful discrimination performance.” + + Thinking about a Weber fraction in these ways promotes confusions and limits our inferences (e.g., the confusion that performance should change from chance performance at difficult ratios to above chance performance at easy ratios; as we’ve seen in Figure 3, the actual performance of subjects does not look this way but instead is a smooth function). It also does not promote the kind of understanding of the Approximate Number System (ANS) that highlights the systematic nature of the noise across various ANS number representations within an individual (e.g., understanding that the noise for any one ANS representation can be easily translated into an understanding of the noise for every single ANS number representation), nor the systematic relationships that exist across individuals (e.g., the comparison of the two subjects in Figure 4). + + In this chapter I have tried to promote a third way of understanding the Weber fraction: it is a scaling factor that enables any ANS number representation to be turned into any other; or, equivalently, it is an index of the amount of noise in every one of the Approximate Number System (ANS) representations. Understood in this way, a Weber fraction does not require the commonsense notion of a “threshold” (i.e., a change from failure to success) and does not generate the same kinds of confusions that this commonsense notion gives rise to. + + The commonsense notion of a “threshold,” as a change from poor to good performance, is a productive one for first coming to understand how discrimination behavior works across many dimensions (e.g., approximate number, loudness, brightness etc). And so, it should not be abandoned. But it does not foster our intuitions about the underlying representations that give rise to this discrimination behavior (i.e., it doesn’t give rise to an accurate picture of the systematic relationships between noise, numerosity, and discrimination performance). An understanding of a Weber fraction (''w'') as an index of the internal noise in the underlying ANS representations promotes productive intuitions and better theorizing about the Approximate Number System (ANS) and other ratio-dependent cognitive dimensions (e.g., line length, weight, brightness, etc). + + A Weber fraction (''w'') indexes the amount of noise in the underlying approximate number representations, it specifies the standard deviations of the underlying Gaussian number curves for every numerosity in a subject’s Approximate Numer System (ANS) and thereby the degree of overlap between any of these representations. For discrimination, the Weber fraction describes a subject’s performance at discriminating any possible combination of numerosities, not just discriminability near their “discrimination threshold”. And, the Weber fraction (''w''), understood as an internal scaling factor, describes all of the representations in the Approximate Number System (ANS) without any reference to a numerical discrimination task. + + == See also == + + *[[Tutorial 1: How To Test A Participant]] + *[[Tutorial 2: How to Determine w]] + *[[Resources]] + *[[Weber Fraction (Beginners)]] + *[[Weber Fraction (Experienced Users)]] + *[[What is a Weber Fraction?]] + *[[Panamath Software Manual]]
## Latest revision as of 17:35, 6 October 2011
In mathematics and physics, Weber's law states that, as the ratio between the magnitudes of two stimuli increases, the more easily the difference between the two stimuli will be perceived.
Because the probability of distinguishing whether there are more blue or yellow dots increases as the disparity between the actual number of blue versus yellow dots increases, Weber's law also applies to the Panamath task. Models of the ANS utilize Gaussian curves that represent different numerosities, or mental representations of quantity. The spread of these curves is determined by the Weber fraction, hence allowing analysis and comparison of subject performance.
As the site is currently under development, Panamath is still collecting all of the necessary citations for this wiki page. No references will be provided until this process is complete.
## Model Representations of the ANS
In modeling performance on tasks that engage the ANS, it is necessary first to specify a model for the underlying approximate number representations. It is generally agreed that each numerosity is mentally represented by a distribution of activation on an internal “number line.” These distributions are inherently “noisy” and do not represent number exactly or discretely. This means that there is some error each time they represent a number; and this error can be thought of as a spread of activation around the number being represented.
## The Mental Number Line
The mental number line is often modeled as having linearly increasing means and linearly increasing standard deviation. In such a format, the representation for e.g., cardinality seven is a probability density function that has its mean at 7 on the mental number line and a smooth degradation to either side of 7 such that 6 and 8 on the mental number line are also highly activated by instances of seven in the world. In Figure 1a I have drawn idealized curves which represent the ANS representations for numerosities 4-10 for an individual with Weber fraction = .125. You can think of these curves as representing the amount of activity generated in the mind by a particular array of items in the world with a different bump for each numerosity you might experience (e.g., 4 balls, 5 houses, 6 blue dots, etc). Rather than activating a single discrete value (e.g., 6) the curves are meant to indicate that a range of activity is present each time an array of (e.g., 6) items is presented. That is, an array of e.g., six items will greatly activate the ANS numerosity representation of 6, but because these representations are noisy this array will also activate representations of 5 and 7 etc with the amount of activation centered on 6 and gradually decreasing to either side of 6.
## Neuronal Associations of the Mental Number Line
The bell-shaped representations of number depicted in Figure 1a are more than just a theoretical construct; “bumps” like these have been observed in neuronal recordings of the cortex of awake behaving monkeys as they engage in numerical discrimination tasks (e.g., shown an array of six dots, neurons that are preferentially tuned to representing 6 are most highly activated, while neurons tuned to 5 and 7 are also fairly active, and those tuned to 4 and 8 are active above their resting state but less active than those for 5, 6, and 7. These neurons are found in the monkey brain in the same region of cortex that has been found to support approximate number representations in human subjects. This type of spreading, “noisy” activation is common throughout the cortex and is not specific to representing approximate number. Rather, approximate number representations obey principles that operate quite generally throughout the mind/brain.
## Interpreting the Gaussian Curves
The bell-shaped representations of number depicted in Figure 1a are more than just a theoretical construct; “bumps” like these have been observed in neuronal recordings of the cortex of awake behaving monkeys as they engage in numerical discrimination tasks (e.g., shown an array of six dots, neurons that are preferentially tuned to representing 6 are most highly activated, while neurons tuned to 5 and 7 are also fairly active, and those tuned to 4 and 8 are active above their resting state but less active than those for 5, 6, and 7. These neurons are found in the monkey brain in the same region of cortex that has been found to support approximate number representations in human subjects. This type of spreading, “noisy”, activation is common throughout the cortex and is not specific to representing approximate number. Rather, approximate number representations obey principles that operate quite generally throughout the mind/brain.
When trying to discriminate one numerosity from another using the Gaussian representations in Figure 1a, the more overlap there is between the two Gaussians being compared the less accurately they can be discriminated.
Ratios that are closer to 1, where the two numbers being compared are close (e.g., 9 versus 10), give rise to Gaussians with greater overlap resulting in poorer discrimination (i.e., “ratio-dependent performance”). The ratio is calculated with this formula, where $N_{smaller} \,$ is the smaller number and $N_{larger} \,$ is the larger number, respectively:
$Ratio = \frac{{N_{larger}}}{N_{smaller}} \,$
Visually, the curve for 5 in Figure 1a looks clearly different in shape from the curve for 4 (e.g., curve 4 is higher and skinnier than curve 5); and discriminating 4 from 5 is fairly easy. As you increase in number (i.e., move to the right in Figure 1a), the curves become more and more similar looking (e.g., is curve 9 higher and skinnier than curve 10?); and discrimination becomes harder.
But, it is not simply that larger numbers are harder to discriminate across the board. For example, performance at discriminating 16 from 20 (not shown) will be identical to performance discriminating 4 from 5 as these pairs differ by the same ratio (i.e., 5 / 4 = 1.125 = 20 / 16); and the curves representing these numbers overlap in the ANS such that the representation of 4 and 5 overlap in area to the same extent that 16 overlaps with 20 (i.e., although 16 and 20 each activate very wide curves with large standard deviations, these curves are far enough apart on the mental number line that their overlap is the same amount of area as the overlap between 5 and 4, i.e., they have the same discriminability). This is ratio-dependent performance.
## Tasks That Give Rise to the Gaussian Curves
The Gaussian curves in Figure 1a are depictions of the mental representations of 4-10 in the ANS. Similar looking curves can be generated by asking subjects to make rapid responses that engage the ANS, such as asking subjects to press a button 9 times as quickly as possible while saying the word “the” repeatedly to disrupt explicit counting. In such tasks, the resulting curves, generated over many trials, represent the number of times the subject pressed the button when asked to press it e.g., 9 times. Because the subject can’t count verbally and exactly while saying “the”, they tend to rely on their ANS to tell them when they have reached the requested number of taps.
When this is the case, the variance in the number of taps across trials is the result of the noisiness of the underlying ANS representations and so can be thought of as another method for determining what the underlying Gaussian representations are. That is, if starting to tap and ending tapping etc did not contribute additional noise to the number of taps (i.e., if the ANS sense of how many taps had been made were the only source of over and under tapping) then the standard deviation of the number of taps for e.g., 9 across trials would be identical to the standard deviation of the underlying ANS number representation of e.g., 9.
When attempting to visualize what the noisy representations of the ANS are like, one can think of the Gaussian activations depicted in Figure 1a, and these representations affect performance in a variety of tasks including discriminating one number for another (e.g., 5 versus 4) and generating number-relevant behaviors (e.g., tapping 9 times).
## Numerical Discrimination in the ANS
To understand how numerical discrimination is possible in the ANS, consider the task of briefly presenting a subject with two arrays, e.g., 5 yellow dots and 6 blue dots, and asking the subject to determine which array is greater in number (Figure 2a). The 5 yellow dots will activate the ANS curve representation of 5 and the 6 blue dots will activate the ANS curve representation of 6 (assume that the subject uses attention to select which dots to send to the ANS for enumerating and then stores and compares those numerosity representations bound to their respective colors) (Figure 2a-b).
## ANS Modeling and Subtraction
An intuitive way to think about ordinal comparison within the ANS is to liken it to a subtraction; this will be mathematically equivalent to other ways of making an ordinal judgment within the ANS and my use of subtraction here should be thought of as one illustration among several mathematically equivalent illustrations.
Imagine that an operation within the ANS subtracts the smaller (i.e., five-yellow) representation from the larger (i.e., six-blue) representation (Figure 2b). Because the 5 and 6 representations are Gaussian curves, this subtraction results in a new Gaussian representation of the difference which is a Gaussian curve on the mental number line that has a mean of 1 (6 − 5 = 1) and a standard deviation of:
$\sqrt{{{\sigma}_5^2 + {\sigma}_6^2}} \,$
In Figure 2c (i.e., when subtracting one Gaussian random variable from another (i.e., $X_6 - X_5 \,$), the result is a new Gaussian random variable with the mean at the difference (6 − 5 = 1) and a variance that adds the variances of the original variables: ${\sigma}_5^2 + {\sigma}_6^2 \,$. This results in a Gaussian curve that is centered on 1, but that extends to both the left and right of 0 (Figure 2c).
One can think of 0 as the demarcation line separating evidence “for” and “against” in that the area under the curve to the right of 0 is the portion of the resulting representation that correctly indicates that six is greater than five while the area under the curve to the left of 0 is the portion of the resulting representation that incorrectly indicates that five is greater than six. This area to the left of 0 results from the overlap between the original Gaussian representations, five and six, that were being discriminated in which some of the area of five-yellow is to the right (i.e., greater than) some of the area of six-blue (Figure 2b).
## Interpreting Gaussian Overlap: Weber's Law
One can think of 0 as the demarcation line separating evidence “for” and “against” in that the area under the curve to the right of 0 is the portion of the resulting representation that correctly indicates that six is greater than five while the area under the curve to the left of 0 is the portion of the resulting representation that incorrectly indicates that five is greater than six. This area to the left of 0 results from the overlap between the original Gaussian representations, five and six, that were being discriminated in which some of the area of five-yellow is to the right (i.e., greater than) some of the area of six-blue (Figure 2b).
Another method would rely on assessing the total evidence for blue and the total evidence for yellow. Either of these ways of making a decision will have the result that, on a particular trial, the probability of the subject getting the trial correct will depend on the relative area under the curve to the left and right of 0 which is itself determined by the amount of overlap between the original Gaussian representations for the numerosities being compared (i.e., five and six).
The more overlap there is between the two Gaussian representations being compared, the less accurately they can be discriminated. Consider comparing a subject’s performance on a 5 dots versus 6 dots trial to a trial involving 9 versus 10 dots. Using the curves in Figure 1a as a guide, we see that the overlapping area for the curves representing 5 and 6 is less than the overlapping area for the curves representing 9 and 10, because the curves flatten and spread as numerosity increases. This means that it will be easier for the subject to tell the difference between 5 and 6 than between 9 and 10, i.e., the resulting Gaussian for the subtraction will have more area to the right of 0 for the subtraction of 5 from 6 than for the subtraction of 9 from 10.
Across multiple trials the subject would give more correct responses on the 5 versus 6 trials than the 9 versus 10 trials owing to the greater evidence for 6 being larger than 5 (i.e., more area to the right of 0 in the subtraction, Figure 2c). The linear increase in the standard deviation of the curves representing the numerosites along the mental number line results in ratio-dependent performance whereby the discriminability of two numerosities increases as the ratio between them (e.g., biggernumber / smallernumber) increases. This is Weber’s law.
How rapidly performance rises from chance (50%) to near-asymptotic performance (100%) in this kind of dot numerosity discrimination task is controlled by the subject’s Weber fraction (w).
## The Weber Fraction: Overview
The Weber fraction indexes the amount of spread in the subject’s ANS number representations and therefore the overlap between any two numbers as a function of ratio (described in a succeeding section). The precision of the ANS varies across individuals with some people having a smaller Weber fraction (i.e., better performance and sharper Gaussian curves) and others having a larger Weber fraction (i.e., poorer performance owing to wider noisier Gaussian curves).
Numerical discrimination (e.g., determining which color, blue or yellow, has more dots in an array of dots flashed to quickly for explicit counting) is possible in the ANS through a process that attempts to determine which of the two resulting curves (e.g., five-yellow or six-blue) is further to the right on the mental number line. The fullness of these noisy curves is used to make this decision (and not just the mode, or mean or some other metric) and successful discrimination thereby depends on the amount of overlap between the two activated curves (i.e., ratio-dependent performance).
The amount of overlap is indexed by a subject’s Weber fraction (w) with a larger Weber fraction indicating more noise, more overlap, and thereby worse discrimination performance. This model has been found to provide an accurate fit to data from rats, pigeons, and humans of all ages.
## Relationship to the ANS
What does a Weber fraction (w) tell us about a subject’s Approximate Number System (ANS) representations? A relationship exists between the Weber fraction (w), the standard deviations of the underlying Gaussian numerosity representations, and the discriminability of any two numbers; and a Weber fraction can be thought of in at least three ways, which sometimes leads to confusions.
These are: 1) the fraction by which a stimulus with numerosity n would need to be increased in order for a subject to detect this change and its direction on 50% of the trials (aka the difference threshold, aka the Just Noticeable Difference, J.N.D.), 2) the midpoint between subjective equality of two numbers and asymptotic performance in discrimination, and 3) the constant that describes the standard deviation of all of the numerosity representations on the mental number line. Option 3 is perhaps the least discussed in the psychophysics literature - it is rarely taught in psychophysics courses - but I will suggest that option 3 is the most productive way to understand the Weber fraction.
I first describe what a Weber fraction is and how it relates to the underlying number representations and then provide some suggestions for what I take to be a productive way of understanding the Weber fraction.
## How to Think of the Weber Fraction
First, consider Figure 1a to represent the ANS number representations for a particular individual who has a Weber fraction = .125. If one presents the hypothetical subject in Figure 1a with the task of determining which of two colors has the greater number of dots on a trial where there are 16 blue dots and some number of yellow dots, this subject would require an increase of 2 dots from this standard (16 * .125 = 2, N2 = 16 + 2 = 18) in order to respond that yellow (18) is more numerous than blue (16) on 75% of the trials that present these two numerosities. That is, a subject’s Weber fraction can be used to determine the amount by which one would need to change a particular number in order for that subject to correctly determine which number was larger on 75% of the trials (chance = 50%).
Conceived in this way, the Weber fraction describes a relationship between any numerosity and the numerosity that will be consistently discriminated from this standard. This gives one way of understanding why one might choose 75% correct performance as an interesting indicator of performance.
In order to specify what “consistently discriminated from” might mean one might also choose some other standard (e.g., 66.7% correct, or any other % above chance). From this point of view, often the view taught in courses on Psychophysics, the point is to estimate the steepness of the linear portion of the psychometric function (depicted in Figure 1b) and 66.7% would work for such purposes just as well as 75% or 80%; the choice of 75% correct is presented as more or less an arbitrary one.
However, as we will see below, 75% correct is special and the seemingly arbitrary reasons for choosing 75% correct as an index of performance find their non-arbitrariness in a mathematical relationship between correct discrimination, the Weber fraction, and the standard deviations of the underlying Gaussian representations.
Some readers, more familiar with research on the acuity of the ANS in infants 6-9 months of age and less familiar with the literature on adult psychophysics, may have come to believe that the Weber fraction describes the ratio below which a subject will fail to discriminate two numerosities (e.g., 6-month-olds succeed with a 1:2 ratio and fail with a 2:3 ratio). This suggests a categorical interpretation of the Weber fraction (e.g., a threshold where you will succeed if a numerical difference is “above threshold” and fail if it is “below threshold”). That is, some may have come to believe that performance should be near perfect with ratios easier than a subject’s Weber fraction and at chance for ratios harder than a subject’s Weber fraction. This is not what is seen in typical performance where a large number of trials test a subject’s discrimination abilities across a wide variety of ratios.
## Modeling Trends
Consider again the simple task of being briefly shown a display that includes some blue and yellow dots and being asked to determine on each flash if there had been more blue or more yellow dots. Percent correct on this numerical discrimination task is not a step function with poor performance “below threshold” and good performance “above threshold”, but rather is a smoothly increasing function from near chance performance to success.
This performance and the range of individual differences, gathered from over 10,000 subjects between the ages of 8 and 85 years of age participating in this type of blue-yellow dots task, can be seen in Figure 3a-b. Every one of the more than 10,000 participants obeyed this kind of gradual increase in percent correct from a ratio of 1 (where the number of blue and yellow dots are equal) to easier ratios like 2 (where there might be 10 blue dots and only 5 yellow dots; 10 / 5 = 2). What changes from participant to participant is how steep the left side of the curve is, and these individual differences are shown in the figure by indicating the performance of the 10th and 90th percentile ranking of the more than 10,000 participants.
Figure 3b shows how the average Weber fraction improves over development. A steeper, quicker rise in the psychometric function (Figure 3a) indicates better sensitivity, better discrimination abilities and is indexed by the subject having a smaller Weber fraction (Figure 3b) (i.e., a smaller Weber fraction indicates less noise in the underlying ANS representations).
If one wished to localize the Weber fraction at some point along the smoothly increasing curve in Figure 3a it would be at the midpoint between subjective equality of the two numerosities being compared (typically occurring at a ratio = 1, where N1 = N2) and asymptotic performance (typically 100% correct, though asymptotic performance may be lower in unskilled subjects resulting in the midpoint falling at a percent correct lower than 75%).
## The "Difference Threshold"
When subjects behave optimally, the Weber fraction is related to the ratio that results in 75% correct performance and not to the first ratio that results in chance, or above chance, performance. Indeed, the actual behavioral data from subjects seen in Figure 3a, and the modeled ideal behavior seen in Figure 1b, suggest that the subjects will always be above chance no matter how small the difference between N1 and N2. What changes is not whether a participant will succeed or fail to make a discrimination but rather the number of trials an experimenter will have to run in order to find statistically significant performance.
However, within the practical limits of testing real subjects, the infant literature’s method of looking for a change from above chance performance to below chance performance is a reasonable approach for attempting to roughly locate the Weber fraction of subjects, like infants, who cannot participate in the large number of trials it takes to achieve the smooth data seen in Figure 3a.
The 75% correct point has been used as an indicator of the “difference threshold” because this point can be translated into a whole-number ratio through a mathematical relationship that holds between the Weber fraction and the ratio at which performance attains 75% correct.
Researchers occasionally suggest that “human adults can discriminate numerosities that differ by a ratio of 7:8,” where 8 / 7 = 1.14 would be the whole number ratio nearest the Weber Ratio that results in 75% correct discrimination. Even this understanding of the Weber fraction as the midpoint between subjective equality and asymptotic performance, while it goes some way towards avoiding the mistaken belief that a difference threshold is a step function, misses the deeper continuous nature of discrimination within the ANS.
Let us consider a third way of understanding the Weber fraction (w): as a scaling factor that indexes the amount of noise for every numerical representation of the ANS. Understood in this way, described below, the Weber fraction is not specific to the task of numerical discrimination; indeed, it is wholly independent and prior to discrimination.
An animal that (bizarrely) could only represent a single numerical value in their ANS (e.g., could only represent approximately-twelve and no other numbers), and who could therefore never discriminate 12 from any other number (i.e., could not even perform a numerical discrimination task), would nonetheless have a Weber fraction and we could measure it.
Consider the Gaussian curves depicted in Figure 1a. The spread of each successive numerosity from 4 to 10 is steadily wider than the numerosity before it. This means that the discriminability of any two numerosities is a smoothly varying function, dependent on the ratio between the two numerosities to be discriminated.
In theory, such discrimination is never perfect because any two numerosities no matter how distant from one another will always share some overlap. At the same time, discrimination will never be entirely impossible, so long as the two numerosities are not identical, because any two numerosities, no matter how close (e.g., 67 and 68) will always have some non-overlapping area where the larger numerosity is detectably larger. Correct discrimination may occur on only a small percentage of trials if the two sets are very close in number, but it will never be impossible. This again motivates the intuition that percent correct in a discrimination task should be a smoothly increasing function from the point of subjective equality to asymptotic performance.
In Figure 1b I have drawn the expected percent correct for the ideal subject in Figure 1a whose w = .125 as derived by a model from classical psychophysics. Those who wish to translate this subject’s Weber fraction into a whole number ratio can determine this ratio from the function in Figure 1b as the nearest whole number ratio to the Weber Ratio that falls at the midpoint between subjective equality (i.e., Weber Ratio = 1, Percent Correct = 50%) and asymptotic performance (i.e., Percent Correct = 100%) and would equal 8:9 (i.e., WR = 1.125) for the subject in Figure 1b.
Notice, the Weber fraction is equal to this Weber Ratio minus 1 (i.e., 1.125 − 1 = .125). It is the smoothly increasing spread in the underlying Gaussian representations depicted in Figure 1a that is the source of the smoothly increasing Percent Correct ideal performance in Figure 1b.
Noting the smoothly increasing spread of the Gaussian representations in Figure 1a motivates one to ask what is the parameter that determines the amount of spread in any each Gaussian representation on the mental number line.
## Calculating Standard Deviation
In fact, it is the Weber fraction that determines the spread of every single representation on the mental number line by the following formula:
$SD = n * w \,$
The standard deviation (SD) of the Gaussian representing any numerosity is that numerosity multiplied by the Weber fraction. Why is this the case? Intuitively, it is the standard deviations of the underlying Gaussian representations that determines the amount of overlap between the curves that represent any two numerosities, and it is the amount of overlap that determines how well any two numerosities can be discriminated.
The categorical views of the Weber fraction as a kind of threshold between successful discrimination and failure, or as the midpoint between subjective equality and asymptotic performance, choose to focus on only one particular point of what is actually a continuous and smooth function of increasing success at discrimination, and this entire function is determined by the Weber fraction as this fraction describes the standard deviation of any numerosity representation on the mental number line and thereby the degree of overlap between any two numerosities on the mental number line.
Viewed in this light, the Weber fraction describes all of the numerosity representations in the ANS. It is not specific to the task of discriminating two numerosities, and not specific to numerosity comparisons near “threshold”. In fact, though this may never be the case in practice, given this understanding of the Weber fraction it would be possible to assess the Weber fraction for an animal who could only represent a single number in the ANS (e.g., can only represent approximately-12). It would be the standard deviation of the Gaussian that represents approximately-12 divided by the mean of this Gaussian.
## The Weber Fraction as a Scaling Ratio
The Weber fraction (w) is the constant that describes the amount of variability in the underlying ANS number representations. It is a scaling factor by which one could take any one of the curves in Figure 1a and turn it into any of the other curves in Figure 1a. In the linear model in Figure 1a, the analog representation for any numerosity (e.g., n = 7) is a Gaussian random variable with a mean at n (e.g., n = 7) and a standard deviation of (n * w). This means that for a subject who has a Weber fraction of .125, the ANS representation for n = 7 will be a normal curve with a mean of 7 on the mental number line and a standard deviation of .125 * 7 = .875.
By substituting any number you like for n you can easily determine the shape of the underlying ANS representation, without ever having the participant engage in a numerical discrimination task that compares two numbers. This illustrates the power of understanding the Weber fraction as an index of internal noise: rather than simply telling us something about how well a subject will do at discriminating two numbers near their threshold, the Weber fraction (w) tells us the shape and overlap of every single number representation inside a person’s Approximate Number System (ANS). The Weber fraction is about all of the representations, not just the ones “near threshold.”
The inductive power of understanding the Weber fraction (w) to be an internal scaling factor is also seen when we compare the Weber fractions of different individuals. Individuals differ in the precision of their ANS representations. Some people have more noise and some people have less.
In Figure 4a I illustrate some idealized curves which represent the underlying ANS representations for a subject whose w = .125 and in Figure 4b for a subject whose w = .20. Crucially, one can see that the subject in Figure 4b has a greater degree of overlap between the Gaussian curves that represent numerosity than the subject in Figure 4a (recall, a bigger Weber fraction means more noise and fatter curves). It is this overlap that leads to difficulty in discriminating two stimuli that are close in numerosity (i.e., as one nears Weber Ratio = 1). The hypothetical subject in Figure 4b would have poorer discrimination in a numerical discrimination task (e.g., blue & yellow dots task) than the subject in Figure 4a.
In Figure 4c I have drawn the ideal performance for these two subjects in a numerical discrimination task. The midpoint between subjective equality (Percent Correct = 50%) and asymptotic performance (Percent Correct = 100%) in Figure 4c will fall at WR = 1.125 for the subject in Figure 4a and at WR = 1.20 for the subject in Figure 4b because these are equal to the Weber fraction for each subject plus one. An inspection of Figure 4c reveals that this is the case. This is not an accident; it derives from the relationship described above between the Weber fraction (w), the standard deviations of the underlying ANS number representations, and the overlap between any two ANS number representations.
The values for the Weber fractions in Figure 4 have been chosen so as to illustrate another value of understanding the Weber fraction to be an internal scaling factor; it empowers comparisons across individuals and formal models of individual differences. Converting the Weber fraction for each of these subjects into the nearest whole number fraction reveals that the Weber fraction for the subject in Figure 4a is 8:9 and for the subject in Figure 4b is 5:6 (i.e., 9 / 8 = 1.125; 6 / 5 = 1.20). Investigating the Gaussian curves in Figure 4a and 4b, which were created by treating the Weber fraction as an internal scaling factor to determine the standard deviations of the ANS Gaussian curves, reveals that the Gaussians for the numerosities 8 and 9 for the subject in Figure 4a are identical to the Gaussians for the numerosities 5 and 6 for the subject in Figure 4b. This is no accident. The only parameter that has been changed in constructing Figures 4a and 4b is the Weber fraction for the subject and this single parameter determines the spread in the Gaussians that represent every possible numerosity in the ANS of each subject.
In this way, understanding the Weber fraction to be an internal scaling factor that determines the spread of every ANS number representation not only empowers us to compare one number representation to another within a particular subject, it also empowers us to compare across individuals and to create mathematically tractable predictions about how the ANS representations of one person (e.g., the subject in Figure 4a) relate to the ANS representations of another (e.g., the subject in Figure 4b).
## Deviations From the Panamath Model
It is not uncommon for performance in a discrimination task to deviate from the ideal performance depicted in e.g., Figure 1b and seen in actual data in Figure 2a. Two basic deviations capture almost all cases: 1) a “give-up” function wherein subjects give-up on their numerical discrimination abilities on hard ratios and guess at chance, and 2) a “lapse parameter” which tracks participants’ tendency to have a lapse in attention such that they do not attend the e.g., blue and yellow dots on a particular trial. These two forms of deviation from the model presented in this chapter are tractable in that additional parameters can be included in the model to capture these behaviors.
The model I have described does not take account of participants’ confidence in their answers, nor does it adjust its predictions based on sub-optimal engagement of the Approximate Number System (ANS). But, in practice, sub-optimal performance can occur, especially on difficult trials.
For example, consider the task of being shown a brief display of blue and yellow dots and being asked to guess which color had more dots. One a fairly easy trial (e.g., 27 yellow versus 7 blue), you would most likely get the correct answer and be highly confident in your answer. This would be true even if the display time was incredibly brief (e.g., 50 msec).
## Effect of Difficulty on Trials
But, what will happen on a difficult trial of e.g., 27 yellow and 28 blue dots? On such a trial, a subject can tell fairly rapidly that the trial will be incredibly difficult and they will have very low confidence that their guess will be correct. On such trials, it is not unreasonable that a subject will simply give up on trying to determine whether there are more blue or more yellow dots and guess randomly (e.g., flip a coin). This kind of behavior has been observed in numerical discrimination tasks that present trials that are very difficult for that subject (e.g., presenting 3-year-olds with a brief display of 9 versus 10 objects) and it has also been observed in the discrimination performance of many animal models making a variety of ratio-dependent discriminations (but has yet to be discussed in detail in any literature). When this happens, performance will deviate from the model shown in Figure 1b in that percent correct will drop to chance before the Weber Ratio = 1. Participants will “give-up” on difficult trials and, rather than relying on their ANS representations, will guess randomly. This kind of performance can be seen in the left side of the curves diagramming the performance of children and adults in a difficult numerical discrimination task (Figure 5).
It is important to note that this kind of drop to chance performance is not predicted by any existing psychophysical model and it appears to be a deviation from the model caused by the subject giving up on the task; it does not appear to be the case, as of yet, that this type of deviation indicates that the underlying model of the ANS itself is flawed. We are currently attempting to specify the exact relationship between difficulty, Weber fraction, and the probability of giving up on a trial.
## Effect of Attention
Another deviation can occur on the right side of the psychometric curve and can also be seen in Figure 5. As has been observed in previous numerical discrimination tasks, performance in unskilled subjects can fail to reach 100% correct even on easy trials, perhaps because of a tendency to suffer a lapse of attention on a subset of trials. As can be seen in the example data in Figure 5, this was the case for the 3-, 4- and 5-year-olds.
When this type of lapse occurs, it should not be dependent on the ratio that was presented on that trial. Rather, it is an event that occurs because of e.g., fatigue, irrespective of the ratio being presented on that particular trial. As such, this deviation is typically detectable as a flat cut in percent correct across the entire range of ratios presented in an experiment. It is because percent correct is highest on the easiest ratios that the deleterious affects of this lapse in attention are most visible on easier trials. The simplest way to account for this tendency in the model is to include a lapse parameter that is a constant probability of guessing randomly on any particular trial.
This parameter primarily lowers the model’s asymptotic performance while retaining an accurate estimate of the Weber fraction (w). When required, I included this parameter in the model as follows, where $p_{guess} \,$ is the probability of guessing randomly on any particular trial irrespective of ratio, $p_{error} \,$ is the probability of being incorrect given the model above and the chance level is .5 multiplied by 100 to return a percentage:
$Percent Correct = [(1 - p_{guess} )(1 - p_{error} ) + p_{guess} * .5 ] * 100 \,$
This type of guessing has also been observed in children (Figure 5) and in animal and other adult human participants for both numerical discrimination and other ratio-dependent dimensions.
## What Do These Deviations Mean?
Each of these deviations from the model presented in this chapter can be captured by including an additional parameter in the model (i.e., a threshold for “giving up” as a function of ratio, and a lapse parameter; the probability of guessing randomly on any particular trial irrespective of ratio).
At the moment, the existing datasets do not suggest that deviation from the model presented here is so drastic as to recommend the abandoning of the model. Rather, as with all complex cognition, behaviors that engage the Approximate Number System (ANS) are occurring in a rich context that drives many other considerations that can affect performance. Giving up on a hard trial and being bored such that you fail to attend on a trial are two very basic considerations that can affect performance.
## Conclusions and Summary
If one does some reading in hopes of understanding what a Weber fraction (w) is, the most common descriptions turn on the idea of a difference threshold (aka Just Noticeable Difference, J.N.D.). For example, “the fraction by which a stimulus with numerosity n would need to be increased in order for a subject to detect this change and its direction on 50% of the trials” is a common phrase; or, “the midpoint that indicates roughly successful discrimination performance.”
Thinking about a Weber fraction in these ways promotes confusions and limits our inferences (e.g., the confusion that performance should change from chance performance at difficult ratios to above chance performance at easy ratios; as we’ve seen in Figure 3, the actual performance of subjects does not look this way but instead is a smooth function). It also does not promote the kind of understanding of the Approximate Number System (ANS) that highlights the systematic nature of the noise across various ANS number representations within an individual (e.g., understanding that the noise for any one ANS representation can be easily translated into an understanding of the noise for every single ANS number representation), nor the systematic relationships that exist across individuals (e.g., the comparison of the two subjects in Figure 4).
In this chapter I have tried to promote a third way of understanding the Weber fraction: it is a scaling factor that enables any ANS number representation to be turned into any other; or, equivalently, it is an index of the amount of noise in every one of the Approximate Number System (ANS) representations. Understood in this way, a Weber fraction does not require the commonsense notion of a “threshold” (i.e., a change from failure to success) and does not generate the same kinds of confusions that this commonsense notion gives rise to.
The commonsense notion of a “threshold,” as a change from poor to good performance, is a productive one for first coming to understand how discrimination behavior works across many dimensions (e.g., approximate number, loudness, brightness etc). And so, it should not be abandoned. But it does not foster our intuitions about the underlying representations that give rise to this discrimination behavior (i.e., it doesn’t give rise to an accurate picture of the systematic relationships between noise, numerosity, and discrimination performance). An understanding of a Weber fraction (w) as an index of the internal noise in the underlying ANS representations promotes productive intuitions and better theorizing about the Approximate Number System (ANS) and other ratio-dependent cognitive dimensions (e.g., line length, weight, brightness, etc).
A Weber fraction (w) indexes the amount of noise in the underlying approximate number representations, it specifies the standard deviations of the underlying Gaussian number curves for every numerosity in a subject’s Approximate Numer System (ANS) and thereby the degree of overlap between any of these representations. For discrimination, the Weber fraction describes a subject’s performance at discriminating any possible combination of numerosities, not just discriminability near their “discrimination threshold”. And, the Weber fraction (w), understood as an internal scaling factor, describes all of the representations in the Approximate Number System (ANS) without any reference to a numerical discrimination task. | 19,132 | 89,985 | {"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": 10, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.901414 |
https://www.programtopia.net/c-programming/examples/c-program-find-factorial-number | 1,669,902,456,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710813.48/warc/CC-MAIN-20221201121601-20221201151601-00060.warc.gz | 1,012,402,601 | 11,109 | # C Program to Find Factorial of a Number
A factorial of a number can be defined as the multiplication of the number itself and its descending natural numbers. Factorial is denoted by ‘!’ symbol. e.g. the factorial of 5 is
`5! = 5 x 4 x 3 x 2 x 1 = 120`
The factorial of 1 is
1! = 1
C program to find the factorial of a number is shown below.
### Program
```#include<stdio.h>
int main()
{
int i,n,fact=1;
printf("Enter a number n");
scanf("%d",&n);
for (i=1;i<=n;i++)
{
fact=fact*i;
}
printf ("The factorial of %d is %d",n,fact);
return 0;
}```
Here, the number entered by the user is stored in variable n. The loop goes on from 1 to the number itself and inside the for loop, the working of the expression can be understood from the following steps.
Let us suppose, the user has entered 4
In first loop,
```i=1 so
fact = 1 * 1 = 1```
In second loop,
```i=2 so
fact = 1 *2 =2```
In third loop,
```i=3 so
fact = 2 * 3 = 6```
In fourth loop,
```i =4 so
fact = 6 * 4 = 24```
which is the final result as 4! = 24.
### Output:
```Enter a number
6
The factorial of 6 is 720
``` | 348 | 1,092 | {"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} | 3.078125 | 3 | CC-MAIN-2022-49 | longest | en | 0.887462 |
https://www.jiskha.com/display.cgi?id=1170943115 | 1,516,124,362,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886476.31/warc/CC-MAIN-20180116164812-20180116184812-00269.warc.gz | 935,213,753 | 4,263 | posted by .
1 four buttons have diameters of 5/8 , 3/4, 9/16, and 7/8 which buttons is smallest?
2. I spent 1/3 of my time doing math homework and 2/7 preparing for a history test . what fraction of study time remain?
3. least common multiple of 3, 6, 9
4. Tom practiced 1 1/2 hrs monday , 3 3/4 hrs tuesday and 2 1/4 hrs wednesday what is the total number of hours practiced?
5. perform the indicated operations
6 2/9 + 1 3/4 - 3 5/6
6. divide
19.41 by 6
7. round to the indicated decimal plavce 26.1978 three decimal places
8. A triangle has a base 13 ft and altitude of 6 ft Find its area.
I was not sure on this i went with the A+L X W
9. Television cost \$815.56 with interest. I paid \$250.00 down and agreeded to pay the remainder over next 18 months what amount will i pay per month?
10. Write a proportion that is equivalent to the statement If 3 gallons of gasoline cost \$5.37 then 8 gallons will cost \$14.32.
3 gal/\$5.37 = 8 gal?/14.32
11. Approximately 5 out of every student at Kenda college lives on campus the college has 5,000 students How many of them live on campus?
12. Find the best buy on cashews (1 lb. is 16 oz.)
A. 1 lb for \$6.09
B. 10 oz for \$4.59
C. 1 lb 4 oz for \$7.09
D. 12 oz for \$5.09
please let me know if i missed any these so i can go back over to see where i messed up thanks
## Similar Questions
1. ### math check
1 four buttons have diameters of 5/8 , 3/4, 9/16, and 7/8 which buttons is smallest?
2. ### would someone please check this math
1 four buttons have diameters of 5/8 , 3/4, 9/16, and 7/8 which buttons is smallest?
3. ### math
Four buttons have diameters of 5/8 3/4, 9/16, and 7/8 inches. Which button is smallest?
4. ### math
just want to chack my answers: 1. Find the least common multiple of 3, 6, and 9. A-36?
5. ### math
peter has 3 different kinds of buttons in one box. there are 124 more ivory buttons than plastic buttons and 1/3 as many wooden buttons as ivory buttons. there are 72 wooden buttons. how many plastic buttons are there?
6. ### math
justine has 162 red buttons,98 blue buttons,and 284 green buttons. she says she knows she has more than 500 buttons without adding.do you agree? | 675 | 2,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-05 | latest | en | 0.917756 |
https://nigerianscholars.com/tutorials/introducing-graphs/identifying-points-on-a-graph/ | 1,653,086,474,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534693.28/warc/CC-MAIN-20220520223029-20220521013029-00453.warc.gz | 486,726,181 | 18,837 | Mathematics » Introducing Graphs » Use the Rectangular Coordinate System
# Identifying Points on a Graph
## Identifying Points on a Graph
In algebra, being able to identify the coordinates of a point shown on a graph is just as important as being able to plot points. To identify the x-coordinate of a point on a graph, read the number on the x-axis directly above or below the point. To identify the y-coordinate of a point, read the number on the y-axis directly to the left or right of the point. Remember, to write the ordered pair using the correct order $$\left(x,y\right).$$
## Example
Name the ordered pair of each point shown:
### Solution
Point A is above $$-3$$ on the $$x\text{-axis},$$ so the $$x\text{-coordinate}$$ of the point is $$-3.$$ The point is to the left of $$3$$ on the $$y\text{-axis},$$ so the $$y\text{-coordinate}$$ of the point is $$3.$$ The coordinates of the point are $$\left(-3,3\right).$$
Point B is below $$-1$$ on the $$x\text{-axis},$$ so the $$x\text{-coordinate}$$ of the point is $$-1.$$ The point is to the left of $$-3$$ on the $$y\text{-axis},$$ so the $$y\text{-coordinate}$$ of the point is $$-3.$$ The coordinates of the point are $$\left(-1,-3\right).$$
Point C is above $$2$$ on the $$x\text{-axis},$$ so the $$x\text{-coordinate}$$ of the point is $$2.$$ The point is to the right of $$4$$ on the $$y\text{-axis},$$ so the $$y\text{-coordinate}$$ of the point is $$4.$$ The coordinates of the point are $$\left(2,4\right).$$
Point D is below $$4$$ on the $$x-\text{axis},$$ so the $$x\text{-coordinate}$$ of the point is $$4.$$ The point is to the right of $$-4$$ on the $$y\text{-axis},$$ so the $$y\text{-coordinate}$$ of the point is $$-4.$$ The coordinates of the point are $$\left(4,-4\right).$$
## Example
Name the ordered pair of each point shown:
### Solution
Point A is on the x-axis at $$x=-4$$. The coordinates of point A are $$\left(-4,0\right)$$. Point B is on the y-axis at $$y=-2$$ The coordinates of point B are $$\left(0,-2\right)$$. Point C is on the x-axis at $$x=3$$. The coordinates of point C are $$\left(3,0\right)$$. Point D is on the y-axis at $$y=1$$. The coordinates of point D are $$\left(0,1\right)$$.
[Attributions and Licenses]
This is a lesson from the tutorial, Introducing Graphs and you are encouraged to log in or register, so that you can track your progress. | 687 | 2,363 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2022-21 | latest | en | 0.721123 |
https://astronomy.stackexchange.com/questions/31756/what-would-happen-if-asteroid-99942-apophis-collides-with-the-earth | 1,716,903,348,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059139.85/warc/CC-MAIN-20240528123152-20240528153152-00188.warc.gz | 85,399,778 | 36,439 | # What would happen if asteroid 99942 Apophis collides with the earth?
In the news recently it's been said that in 2029 the asteroid called Apophis, aka 'The God of Chaos', will fly by very close to earth. To quote Geek.com
In 2029, a massive asteroid called 99942 Apophis will fly past our planet at an approximate distance of 19,000 miles within the distance of some orbiting spacecraft.
It's said the chances of a collision are about 1 in 100,000, which is small but not not completely infeasible either.
The asteroid has a diameter of around 350 meters with a mass of 4×10^10 kg and an average orbital speed of around 30km/s.
What kind of damage could we expect on earth if it were to collide with us?
• I guess the chances given for a collision are based on a uncertainty of the computations 10 years into the future. The closer we get the more certain the projections get, as orbital mechanics is a completely deterministic system. May 3, 2019 at 22:11
• There are various online calculators that can do impact analysis, eg impact.ese.ic.ac.uk/ImpactEarth/index.html May 3, 2019 at 22:41
• Wikipedia's entry for 99942 Apophis has a summary of the rough impact effects it would have. But there are a huge number of variables (e.g. hits the ocean, continental coast, the Antartic, the angle of impact). For this reason I'd call this too broad of a question. Broadly speaking bit crater, possibly lots of dead people (millions) and human life will continue away from the blast. Huge economic impact if it hit some places (e.g. China's silicon valley or a main oil producing area). None if it missed them, Too broad a querstion. May 4, 2019 at 0:51
• You have mis-read part of the article. Apophis will not have a 1/100000 chance of hitting Earth in 2029. The chance it will hit Earth is 0, or at least as close to zero as we can determine. We know where it will be in 2029, and it won't be hitting Earth. It has a very small chance (1 in 100000) of hitting Earth at some point after 2060. May 4, 2019 at 21:29
• It has a very small chance (1 in 100000) of hitting Earth at some point after 2060... I'll be dead or senile by then so I'm not concerned. Apr 28, 2020 at 3:08 | 572 | 2,180 | {"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-2024-22 | latest | en | 0.947748 |
http://bridgewinners.com/article/view/play-problem-from-session-3-of-spingold-r64-2-669z9it3d2/?cj=832963 | 1,585,429,274,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493120.15/warc/CC-MAIN-20200328194743-20200328224743-00126.warc.gz | 30,068,585 | 23,085 | Play problem from session 3 of Spingold R64
(Page of 3)
This was board 15 in the Wilson-Tulin match.
West
North
J76543
KJ105
6
J4
East
South
AQ9863
KJ8
A1083
W
N
E
S
1
X
4
4
5
P
P
P
D
5 South
NS: 0 EW: 0
K
4
2
A
3
1
0
1
What's your plan after (presumably) winning the ace at trick 1? Note that East thought for a bit before passing 5.
You have six hearts, two clubs, certainly a couple ruffs in dummy, and need an eleventh trick. One possibility is to hope that West has a singleton heart. In that case, you can concede a diamond now, win a trump return, eventually ruff two diamonds and a club in dummy with East not begin able to get in.
West
North
J76543
KJ105
6
J4
East
South
AQ9863
KJ8
A1083
W
N
E
S
1
X
4
4
5
P
P
P
D
5 South
NS: 0 EW: 0
K
4
2
A
3
1
0
J
5
6
Q
2
1
1
2
3
7
10
1
2
1
J
7
3
Q
0
2
2
4
You can also play a club to dummy. In that case, if West plays a club you pitch a diamond. Even if that is ruffed, you can still win a trump return and ruff all your diamonds in dummy. However, West can play a trump instead. Now when you play a diamond off of dummy, if you guess wrong and West doubled with 3-2 in the majors you're a trick short (unless a the club 9 is dropping, in which case it's your lucky day). This is better than playing a diamond from hand at trick 2, since you can still play a diamond after West's initial trump return and have a chance to guess right. You also have the potential line of winning West's trump return in hand and trying to pitch your diamond, but that will fail when West is 3-1-4-5 and East ruffs your club winner and returns a spade.
So, what do you do?
There is a much better line available. Considering the bidding, spades are almost certainly 3-4. Unless trumps are 0-3, you can force West to do something for you.
West
North
J76543
KJ105
6
J4
East
South
AQ9863
KJ8
A1083
W
N
E
S
1
X
4
4
5
P
P
P
D
5 South
NS: 0 EW: 0
K
4
2
A
3
1
0
3
7
10
2
1
2
0
3
8
8
2
3
3
0
6
5
K
4
1
4
0
4
10
9
9
3
5
0
3
6
If west wins this trick, a club or diamond return yields an extra trick in that suit. A spade return provides an extra entry to dummy to ruff out its spades. Now you can ruff a spade, ruff a club, ruff another spade, cash your T and dummy is high. If instead West lets the jack of clubs win, that becomes your extra entry to dummy and you ruff the spades good yourself.
The whole deal:
West
Q92
7
A975
KQ965
North
J76543
KJ105
6
J4
East
AK108
42
Q10432
72
South
AQ9863
KJ8
A1083
W
N
E
S
1
X
4
4
5
P
P
P
D
5 South
NS: 0 EW: 0
K
4
2
A
3
1
0
3
7
10
2
1
2
0
3
8
8
2
3
3
0
6
5
K
4
1
4
0
4
10
9
9
3
5
0
3
6 | 990 | 2,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-16 | latest | en | 0.933589 |
https://es.mathworks.com/matlabcentral/answers/470838-comparing-an-array-of-cells-in-a-matrix | 1,726,649,733,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00198.warc.gz | 216,028,774 | 26,495 | # Comparing an array of cells in a matrix
2 visualizaciones (últimos 30 días)
Lev Mihailov el 9 de Jul. de 2019
Comentada: Rik el 9 de Jul. de 2019
Hello! I need to compare values with each other in cell arrays
for redit = 1:length(xintegra)-1
if EE1{redit}-100>= EE1{redit+1};
E1{redit}=0;
else EE1{redit}+100 <= EE1{redit+1};
E1{redit}=EE1{redit};
end
end
You give an error "Index exceeds matrix dimensions."
Help me fix
##### 2 comentariosMostrar NingunoOcultar Ninguno
Rik el 9 de Jul. de 2019
Comment posted as answer by Praveen:
What is the input Xintegra?
Rik el 9 de Jul. de 2019
Reply by Lev Mihailov:
the minimum value of my matrix = 11003, EE1 this is an indicator of where it is located.
Iniciar sesión para comentar.
### Respuestas (1)
Roy Kadesh el 9 de Jul. de 2019
If you want loop through all values of EE1, then you can use the code below.
for redit = 1:numel(EE1)-1
if EE1{redit}-100>= EE1{redit+1}
E1{redit}=0;
elseif EE1{redit}+100 <= EE1{redit+1} %did you mean elseif?
E1{redit}=EE1{redit};
end
end
If you want to catch the case where redit is larger than the numer of elements in EE1, you can fill in the code below.
for redit = 1:numel(xintegra)-1
if (redit+1)>numel(EE1)
%do something else
else
if EE1{redit}-100>= EE1{redit+1}
E1{redit}=0;
elseif EE1{redit}+100 <= EE1{redit+1} %did you mean elseif
E1{redit}=EE1{redit};
end
end
end
##### 1 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
Rik el 9 de Jul. de 2019
If there are indeed only numeric values inside that cell array, it is probably much faster to convert it to a double and use logical indexing to create the output.
Iniciar sesión para comentar.
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Translated by | 600 | 1,903 | {"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} | 3.453125 | 3 | CC-MAIN-2024-38 | latest | en | 0.263657 |
https://www.quesba.com/questions/hurricanes-see-exercise-32-2-average-category-4-on-stafford-simpson-scale-1816908 | 1,725,897,169,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651103.13/warc/CC-MAIN-20240909134831-20240909164831-00867.warc.gz | 913,014,875 | 29,371 | # Hurricanes. (See Exercise 32.2.) On the average, a category 4 (on the Stafford /Simpson scale) or...
Hurricanes. (See Exercise 32.2.) On the average, a category 4 (on the Stafford /Simpson scale) or stronger hurricane strikes the United States once every 6 years. A hurricane of this strength has winds of at least 131 miles per hour and can cause extreme damage. (http://www.aoml.noaa.gov/ hrd/Landsea/deadly/index.html) An insurance agency is considering whether they might want to stop insuring oceanfront homes and wants to assess the risk involved. The president of the company wants to know how long (in years) before the next 3 hurricanes that are category 4 or stronger. a. Why is this a Gamma scenario? b. What does X represent in this scenario? c. What are the parameters?
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Answer- a)As company wants to know the interarrival time between the category 4 or stronger hurricanes and interarrival times are exponentially distributed. b)X represtents the interarriavl time between the hurricanes. c)exponential distribution are defined using parameter called mean interarrival time. mean interaarival time=1/\lambda where \lambda is rate parameter i.e. number of events per unit time d)\lambda=1/6 ,as there is one hurricane in 6 years. mean or expected length=1/(1/6)=6 years e)variance=1/\lambda ^2=6^2=36 f)pdf f(x)=(1/6)e-x/6 g)cdf: P(X<=x)=1-e-x/6 h) P(X>=3)=e-3/6=e-0.5=0.6065 i)P(5<=X<=10)=P(X<=10)-P(X<=5)=e-5/6-e-10/6 =e-0.833-e-1.667=0.4346-0.1889=0.2457 j)As,exponential distribution is memoryless,so,We have to find probability that there are no hurricanes in next 7 years. P(X>7)=e-7/6=e-1.167=0.3114 k) P(X<=x)=0.75 1-e-x/6=0.75 e-x/6=1-0.75=0.25 x/6=-ln(0.25)=1.3863 x=8.32 years...
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# What is a tessellation vertex?
Updated: 11/8/2022
Wiki User
13y ago
I don't know what a tessellation vertex is but I will try to Answer it I think it means the endpoint of a vertex which is also called vertices,which is the pointy ends of the vertex.
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Q: What is a tessellation vertex?
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Related questions
### What is the sum of the angles at any vertex in a tessellation?
It is 360 degrees.
### What is the definition of tessellation vertex mean?
Gee, Hard problem
Hi im bob
3
3
### How do you name a tessellation?
Tessellations are named based on the number of polygons located at a vertex. For example: A regular tessellation, made from only triangles is named 3.3.3
### Does a semi-regular tessellation have a vertex of 360 degress?
Yes. Regular or irregular, the angles at vertices must sum to 360 deg otherwise you will have gaps in the tessellation.
### Can a 12-gon be used to make a tessellation of the plane?
No. You would need 2.4 such shapes to meet at each vertex and since 0.4 of a 12-gon is impossible, so is the tessellation.
### Why doesn't a semi regular tessellation with vertex configuration 346 work?
The question cannot be answered because it is based on the incorrect assertion that a semi-regular tessellation does not work. Sorry, but it does work! | 339 | 1,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-18 | latest | en | 0.950843 |
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# Maths Science Corner
Math and Science for all competitive exams
# Work Book (Swadhyayan Pothi) Solution Video
On maths science corner you can now download new NCERT 2020 Gujarati Medium Textbook Standard 8 Maths Chapter 09 Baisic Padavalio Ane Nityasam (Algebraic Expressions and Identities) in Video format for your easy reference.
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Highlight of the chapter
9.1 What are expressions?
9.1.1 Number line and an expression
9.2 Terms, factors and coefficients
9.3 Types of Expressions : Monomials, binomials, trinomials and Polynomials
9.4 Like and Unlike terms
9.5 Addition and subtraction of algebraic expressions
9.6 Multiplication of algebraic expressions : An Introduction
9.7 Multiplying a monomial by a monomial
9.7.1 Multiplying two monomials
9.7.2 Multiplying three or more monomials
9.8 Multiplying a monomial by a polynomial
9.8.1 Multiplying a monomial by a binomial
9.8.2 Multiplying a monomial by a Trinomial
9.9 Multiplying a Polynomial by a Polynomial
9.9.1 Multiplying a binomial by a binomial
9.9.2 Multiplying a binomial by a trinomial
9.10 What is an Identity?
9.11 Standard Identities
9.12 Applying Identities
9.13 Summary
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Mathematics Standard 8 Chapter 9 Baisic Padavalio Ane Nityasam (Algebraic Expressions and Identities) Work Book (Swadhyayan Pothi) Solution Video
Part 1
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In this Chapter you will be able to learn various types of expressions, term of expression, factor of a term, coefficient of term, addition and subtraction of various expressions, types of expressions such as monomial, binomial, trinomial and polynomial, multiplication of polynomial with polynomial etc..
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