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https://www.wyzant.com/resources/answers/17823/find_the_sum | 1,524,785,139,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948617.86/warc/CC-MAIN-20180426222608-20180427002608-00285.warc.gz | 910,174,077 | 16,372 | 0
# Find the sum
Two angles are complementary if their sum is 90. If the measure of one angle is 6 less than twice the measure of the other angle, find the measure of the two angles
### 1 Answer by Expert Tutors
Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
1
Hi Dawn;
x=smaller angle
y=larger angle
y=2x-6
90=(x)+(2x-6)
Let's combine like terms...
90=3x-6
Let's add 6 to both sides...
90+6=3x-6+6
96=3x
Let's divide both sides by 3...
96/3=(3x)/3
32=x
y=2x-6
y=[2(32)]-6
y=64-6
y=58
Let's check our work...
Does...
32+58=90
Yes! | 241 | 648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-17 | longest | en | 0.759606 |
https://www.askiitians.com/forums/Wave-Motion/a-source-of-sound-operates-at-2-0-khz-20-w-emitti_102695.htm | 1,723,139,701,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640736186.44/warc/CC-MAIN-20240808155812-20240808185812-00151.warc.gz | 539,889,187 | 43,199 | # A source of sound operates at 2.0 kHz, 20 W emitting sound uniformly in all directions. The speed of sound in air is 340 m s-1 and the density of air is 1.2 kg m-3. (a) What is the intensity at a distance of 6.0 m from the source? (b) What will be the pressure amplitude at this point? (c) What will be the displacement amplitude at this point?
Grade:upto college level
## 1 Answers
Navjyot Kalra
askIITians Faculty 654 Points
10 years ago
Sol. a) Here given V baseair = 340 m/s., Power = E/t = 20 W f = 2,000 Hz, p = 1.2 kg/m^3 So, intensity I = E/t.A = 20/4πr^2 = 20 4 * π * 6^2 = 44 mw/m^2 (because r = 6m) b) We know that I P^2 base 0/2pV base air⇒ P base 0 = √1 * 2pV base air = √2 * 1.2 * 340 * 44 * 10^-3 = 6.0 N/m^2. C) We know that I = 2π^2 S^2 base 0 v^2 p^V where S base 0 = displacement amplitude ⇒ S base 0 = √l/π^2 p^2 pV base air Putting the value we get S base g = 1.2 * 10^-6 m.
## ASK QUESTION
Get your questions answered by the expert for free | 363 | 969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.813627 |
https://fr.mathworks.com/matlabcentral/cody/problems/2130-orthonormal-matrix/solutions/2264657 | 1,596,581,586,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735885.72/warc/CC-MAIN-20200804220455-20200805010455-00206.warc.gz | 300,741,944 | 16,030 | Cody
# Problem 2130. Orthonormal matrix
Solution 2264657
Submitted on 8 May 2020 by Jonathan Copp
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a=[1 0 0 ;0 1 0;0 0 1] tf = 1; assert(isequal(ortho_normal(a),tf))
a = 1 0 0 0 1 0 0 0 1 m = 3 n = 3
2 Pass
a=[0 0 1 0;0 1 0 0;1 0 0 0;0 0 0 1] tf = 1; assert(isequal(ortho_normal(a),tf))
a = 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1 m = 4 n = 4
3 Pass
a=eye(20) k=a; p=a; a(:,1)=k(:,end); a(:,end)=k(:,1); a(:,5)=p(:,9); a(:,9)=p(:,5) tf = 1; assert(isequal(ortho_normal(a),tf))
a = 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 a = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 m = 20 n = 20
4 Pass
a=[1 0 1 0;0 1 0 0;1 0 0 0;0 0 0 1] tf = 0; assert(isequal(ortho_normal(a),tf))
a = 1 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1 m = 4 n = 4 tf = 0
5 Pass
a=[0 0 2 0;0 1 0 0;1 0 -1 0;0 0 0 1] tf = 0; assert(isequal(ortho_normal(a),tf))
a = 0 0 2 0 0 1 0 0 1 0 -1 0 0 0 0 1 m = 4 n = 4 tf = 0
6 Pass
a=[1 2 3 ;1 1 5;3 0 1] tf = 0; assert(isequal(ortho_normal(a),tf))
a = 1 2 3 1 1 5 3 0 1 m = 3 n = 3 tf = 0
7 Pass
a =[1 0 0 ;0 0 0; 0 0 0] tf = 0; assert(isequal(ortho_normal(a),tf))
a = 1 0 0 0 0 0 0 0 0 m = 3 n = 3 tf = 0 | 2,258 | 2,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.554877 |
https://www.brighthubeducation.com/lesson-plans-grades-3-5/52935-halloween-math-activities-bar-graphs/ | 1,656,706,299,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103945490.54/warc/CC-MAIN-20220701185955-20220701215955-00337.warc.gz | 731,128,864 | 14,380 | # Halloween Math Activity: Haunted Bar Graph Lesson Plan
Page content
## Haunted House Excitement
Third graders love pictures, glue, scissors, and Halloween. This lesson combines all of the above and teaches about bar graphs to boot! The materials are simple: construction paper (black, brown, white, orange, and blue), scissors, and glue.
1. Begin with a little excitement - tell them that they will be making haunted houses! First, though, they need to gather information. They can either fill out a personal questionnaire or, to practice data gathering, they can acquire the details from someone else and make the picture for that person.
2. You can vary the questions you use, but each answer will correspond to an element of the picture.
I’ll provide an example. How many people are in your family? Answer indicates the number of ghosts in the picture. How many pets do you have? Answer indicates the number of pumpkins in the pictures. Are you a boy or a girl? Answer will determine the color of the roof on the house. How many rooms in your house? Answer indicates the number of bats in the picture. In what month were you born? Answer indicates the message on the haunted house’s sign.
3. After the students have answered all questions, they can begin working on their glyphs, by cutting out ghosts, bats, pumpkins, and whatever other elements you have decided to include in your haunted house project.
4. Students glue those pieces on a large piece of construction paper. It is always helpful to provide an example for the less artistically inclined.
## Scary Bar Graphs
Once all students have answered all questions, and before they become too involved in their pictures, take a break and introduce a whole class lesson on bar graphs.
1. Choose a question for the whole class, one which has a number for an answer, and gather data. For instance, if I choose the “people in your family” question, I would compile how many students had two people in their families, how many had three, how many had four, and so on.
2. Compile the data in a table. Continuing the example above, my two categories would be: Number of People in the Family, Number of Students.
3. As a class, create a bar graph to match the data. The y-axis will be labeled Number of Students, and the x-axis will labeled after the question you chose. In my example, the x-axis would be labeled Number of People in the Family.
4. While you draw the bar graph on the board, have students follow along at their desks by drawing their own version of the bar graph.
Students can then return to their haunted masterpieces. If you gather the finished product, you will also have spooky decorations to add a festive Halloween spirit to your room. With applicable math standards, crafty haunted houses, and happy students, everyone benefits from this fun Halloween math activity.
## References
Many thanks to Joan Regens, Matt Hough, and the excellent third grade team at Esperanza Elementary School who introduced me to these haunted houses. Another thank you for Tom Hoopingarner, who taught me all about glyphs and their educational value.
## This post is part of the series: Halloween Math Lessons
These third grade Halloween lesson plans will use the holiday theme to teach or reinforce math topics such as bar graphs, probability, and plotting coordinates. | 694 | 3,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.948376 |
https://metacpan.org/pod/release/KRYDE/Graph-Maker-Other-7/lib/Graph/Maker/BinomialTree.pm | 1,529,896,666,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867364.94/warc/CC-MAIN-20180625014226-20180625034226-00403.warc.gz | 685,850,764 | 10,195 | ++ed by:
1 non-PAUSE user.
Kevin Ryde
and 1 contributors
# NAME
Graph::Maker::BinomialTree - create binomial tree graph
# SYNOPSIS
`````` use Graph::Maker::BinomialTree;
\$graph = Graph::Maker->new ('binomial_tree', N => 32);``````
# DESCRIPTION
`Graph::Maker::BinomialTree` creates a `Graph.pm` graph of a binomial tree with N vertices. Vertices are numbered from 0 at the root through to N-1.
The parent of vertex n is that n with its lowest 1-bit cleared to 0. Conversely the children of a vertex n=xx1000 are xx1001, xx1010, xx1100, so each trailing 0 bit changed to a 1, provided that does not exceed the maximum N-1. At the root the children are single bit powers-of-2 up to the high bit of the maximum N-1.
`````` __0___
/ | \ N => 8
1 2 4
| / \
3 5 6
|
7``````
## Order
The `order` parameter is another way to specify the number of vertices. A tree of order=k has N=2^k many vertices. Such a tree has depth levels 0 to k. The number of vertices at depth d is the binomial coefficient binom(k,d), hence the name of the tree.
The N=8 example above is order=3 and the number of vertices at each depth is 1,3,3,1 which are binomials (3,0) to (3,3).
A top-down definition is order k tree is two copies of k-1, one at the root and the other a child of that root. In the N=8 order=3 example above, 0-3 is an order=2 and 4-7 is another, with 4 starting as a child of the root 0.
A bottom-up definition is order k tree as order k-1 with a new leaf vertex added to each existing vertex. The vertices of k-1 with an extra low 0-bit become the evens of k. An extra low 1-bit is the new leaves.
Binomial tree order=5 appears on the cover of Knuth "The Art of Computer Programming", volume 1, "Fundamental Algorithms", second edition.
# FUNCTIONS
`\$graph = Graph::Maker->new('binomial_tree', key => value, ...)`
The key/value parameters are
`````` N => integer
order => integer
graph_maker => subr(key=>value) constructor, default Graph->new``````
Other parameters are passed to the constructor, either `graph_maker` or `Graph->new()`.
`N` is the number of vertices, 0 to N-1. Or instead `order` gives N=2^order many vertices.
Like `Graph::Maker::BalancedTree`, if the graph is directed (the default) then edges are added both up and down between each parent and child. Option `undirected => 1` creates an undirected graph and for it there is a single edge from parent to child.
# FORMULAS
## Wiener Index
The Wiener index of the binomial tree is calculated in
K. Viswanathan Iyer and K. R. Udaya Kumar Reddy, "Wiener index of Binomial Trees and Fibonacci Trees", Intl J Math Engg with Comp, 2009. arxiv:0910.4432
For order k it is
`````` (k-1)*4^k + 2^k
Wiener(k) = --------------- = 0, 1, 10, 68, 392, ... (A192021)
2``````
The Wiener index is total distance between pairs of vertices, so the mean distance is, with binomial to choose 2 among the 2^k vertices,
`````` Wiener(k) k
MeanDist(k) = ---------------- = k-1 + ------- for k>=1
binomial(2^k, 2) 2^k - 1``````
The tree for k>=1 has diameter 2*k-1 between ends of the deepest and second-deepest subtrees of the root. The mean distance as a fraction of the diameter is then
`````` MeanDist(k) 1 1 1
----------- = --- - ------- + -------------------
diameter(k) 2 4*k - 4 (2 - 1/k) * (2^k-1)
-> 1/2 as k->infinity``````
## Independence and Domination
From the bottom-up definition above, a tree of even N has a perfect matching, being each odd n which is a leaf and its even attachment. An odd N has a near-perfect matching (one vertex left over).
The independence number is found by starting with each leaf in an independent set and its attachment vertex not. Per the bottom-up definition this is all vertices. If N odd then the unpaired existing even vertex can be included too, for independence number ceil(N/2).
The domination number similarly, by starting each leaf not in the dominating set and its attachment vertex in the set. Again this is all vertices so domination number floor(N/2), except N=1 domination number 1.
# HOUSE OF GRAPHS
House of Graphs entries for the trees here include
order=0, https://hog.grinvin.org/ViewGraphInfo.action?id=1310 (single vertex)
order=1, https://hog.grinvin.org/ViewGraphInfo.action?id=19655 (path-2)
order=2, https://hog.grinvin.org/ViewGraphInfo.action?id=594 (path-4)
order=3, https://hog.grinvin.org/ViewGraphInfo.action?id=700
order=5, https://hog.grinvin.org/ViewGraphInfo.action?id=21088
# OEIS
Entries in Sloane's Online Encyclopedia of Integer Sequences related to these graphs include
`` A192021 Wiener index`` | 1,337 | 4,716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-26 | latest | en | 0.842063 |
https://thehappyhealthymama.com/qa/why-is-e-used-in-math.html | 1,603,788,776,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107893845.76/warc/CC-MAIN-20201027082056-20201027112056-00204.warc.gz | 534,175,633 | 8,998 | # Why Is E Used In Math?
## Where did the value of E come from?
Its value is approximately 2.718 and has been calculated to 869,894,101 decimal places by Sebastian Wedeniwski.
The number e was first studied by the Swiss mathematician Leonhard Euler in the 1720s, although its existence was more or less implied in the work of John Napier, the inventor of logarithms, in 1614..
## Is 0 a real number?
Yes, 0 is a real number in math. By definition, the real numbers consist of all of the numbers that make up the real number line. The number 0 is…
## How was e calculated?
e is usually defined by the following equation: Its value is approximately 2.718 and has been calculated to 869,894,101 decimal places by Sebastian Wedeniwski. … Euler was also the first to use the letter e for it in 1727 (the fact that it is the first letter of his surname is coincidental).
## What is E used for in real life?
Euler’s number, e , has few common real life applications. Instead, it appears often in growth problems, such as population models. It also appears in Physics quite often. As for growth problems, imagine you went to a bank where you have 1 dollar, pound, or whatever type of money you have.
## What is E equal to?
“e” is a numerical constant that is equal to 2.71828. Just as pi (3.14159) is a numerical constant that occurs whenever the circumference of a circle is divided by its diameter.
## What does R mean in math?
List of Mathematical Symbols • R = real numbers, Z = integers, N=natural numbers, Q = rational numbers, P = irrational numbers. Page 1. List of Mathematical Symbols. • R = real numbers, Z = integers, N=natural numbers, Q = rational numbers, P = irrational numbers.
## Why E is called natural number?
It is often called Euler’s number and, like pi, is a transcendental number (this means it is not the root of any algebraic equation with integer coefficients). Its properties have led to it as a “natural” choice as a logarithmic base, and indeed e is also known as the natural base or Naperian base (after John Napier).
## What does E mean in a math answer?
The e stands for Exponent, which means the number of tens you multiply a number by. For example, if I square 123456789, I get 1.524157875019e+16, which means that the answer is 1.524157875019 times 10 raised to the sixteenth power (that is, multiplied by 10 sixteen times).
## What does the weird e mean in math?
It’s the Greek capital letter Σ sigma. Roughly equivalent to our ‘S’. It stands for ‘sum’. Read this for starters.
## What does ≡ mean in math?
not equalIn computer applications (like Excel) the symbols <> mean not equal. ≡ means identical to. … ≈ means approximately equal to, or almost equal to. The two sides of a relationship indicated by this symbol will not be accurate enough to manipulate mathematically.
## How do you use e in Excel?
=EXP(value) gives the result of e value. For example, to find the value of , where x is to be taken from cell A2 , you would use the formula =EXP(2*A2-1) . (In other words, whatever is in the exponent goes in the parentheses.)
## What does the E sign mean in math?
By Chris Deziel. The letter E can have two different meaning in math, depending on whether it’s a capital E or a lowercase e. You usually see the capital E on a calculator, where it means to raise the number that comes after it to a power of 10. For example, 1E6 would stand for 1 x 106, or 1 million.
## Where does e appear in nature?
Yes, the number e does have physical meaning. It occurs naturally in any situation where a quantity increases at a rate proportional to its value, such as a bank account producing interest, or a population increasing as its members reproduce.
## What is the value of 1 by e?
approximately 2.718Its value is approximately 2.718 and has been calculated to 869,894,101 decimal places by Sebastian Wedeniwski. The number e was first studied by the Swiss mathematician Leonhard Euler in the 1720s, although its existence was more or less implied in the work of John Napier, the inventor of logarithms, in 1614.
## What is the value of E Power 0?
Any number raised to zero is one. Zero is neither positive nor negative so the minus sign before it is redundant. e is constant quantity(roughly equal to 2.71) and when raised to the power 0 it results in 1 as the answer.
## What is so special about E?
What’s so special about the number e? … ex has the remarkable property that the derivative doesn’t change it, so at every point on its graph the value of ex is also the slope of ex at that point.
## What is the symbol that looks like an E?
UnicodeCharacterΕϵUnicode nameGREEK CAPITAL LETTER EPSILONGREEK LUNATE EPSILON SYMBOLEncodingsdecimaldecimalUnicode9171013UTF-8206 149207 1816 more rows | 1,157 | 4,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2020-45 | latest | en | 0.969679 |
www.graphmath.com | 1,660,170,167,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571222.74/warc/CC-MAIN-20220810222056-20220811012056-00750.warc.gz | 70,251,746 | 3,312 | This free and user friendly app is being developed for iPhone and iPad (it also works on Mac computers, but was not thouroughly tested and some features may not be available there) Main Features: * Draws 2D graphs of Y = F(X) functions, their derivatives and integrals * Draws stationary or animated graphs of R = F(Θ) functions in Polar coordinates * Draws stationary or animated graphs of X = F(T), Y = F(T) Parametric functions * Solves F(X,Y) = 0 equations * Draws 3D graphs of Z = F(X, Y) functions in Cartesian coordinates * Draws 3D graphs of Z = F(Θ, R) functions in Cylindrical coordinates * Draws 3D graphs of R = F(Θ, Φ) functions in Spherical coordinates Click here for more details and link to AppStore
y=f(x),f'(x),S(x) R=f(Θ) x=f(t),y=f(t) f(x,y)=0 z=f(x,y) z=f(Θ,R) R=f(Θ,Φ)
This program was developed for PC with MS Windows between 2001 and 2004 Main Features: * Draws 2D graphs of Y = F(X) functions * Displays their derivatives and integrals * Draws 3D graphs of Z = F(X, Y) functions * Solves F(X,Y) = 0 equations
Here are some examples:
y = f(x) y=f(x),f'(x),S(x) f(x,y) = 0 z = f(x,y)
z = f(x,y) (rotating)
These programs are intended for those who are learning Math and Calculus. Also, it may be helpful to engineers and researchers. For help with Linear Algebra and Analytic Geometry, please check out our two other apps, "Matrix Solver Step by Step" and "Vectors and Planes":
Matrix Solver Step by Step Vectors and Planes
The authors would like to hear your suggestions or comments. Write to us! (GraphMath@aol.com)
Keywords: graph, graphics, plot, plotting, graphing software, math, learning tools, education, 3D, three-D, 3-dimensional, tri-dimensional, function, equation, integral, derivative, extreme values, download free version, linear algebra, matrix | 481 | 1,799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2022-33 | latest | en | 0.878709 |
https://leetsolve.com/171-excel-sheet-column-number | 1,679,442,572,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943747.51/warc/CC-MAIN-20230321225117-20230322015117-00453.warc.gz | 440,022,517 | 32,743 | Follow
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# 171. Excel Sheet Column Number
## Some tips about ASCII and mathematics
Nhut Nguyen
·Nov 7, 2022·
### Problem statement
Given a string `columnTitle` that represents the column title as appears in an Excel sheet, return its corresponding column number.
For example:
``````A -> 1
B -> 2
C -> 3
...
Z -> 26
AA -> 27
AB -> 28
...
``````
#### Example 1
``````Input: columnTitle = "A"
Output: 1
``````
#### Example 2
``````Input: columnTitle = "AB"
Output: 28
``````
#### Example 3
``````Input: columnTitle = "ZY"
Output: 701
``````
#### Constraints
• `1 <= columnTitle.length <= 7`.
• `columnTitle` consists only of uppercase English letters.
• `columnTitle` is in the range `["A", "FXSHRXW"]`.
### Solution: Finding The Pattern
Let us write down some other `columnTitle` strings and its value.
``````"A" = 1
"Z" = 26
"AA" = 27
"AZ" = 52
"ZZ" = 702
"AAA" = 703
``````
Then try to find the pattern
``````"A" = 1 = 1
"Z" = 26 = 26
"AA" = 27 = 26 + 1
"AZ" = 52 = 26 + 26
"ZZ" = 702 = 26*26 + 26
"AAA" = 703 = 26*26 + 26 + 1
``````
If you map `'A' = 1, ..., 'Z' = 26`, the values can be rewritten as
``````"A" = 1 = 'A'
"Z" = 26 = 'Z'
"AA" = 27 = 26*'A' + 'A'
"AZ" = 52 = 26*'A' + 'Z'
"ZZ" = 702 = 26*'Z' + 'Z'
"AAA" = 703 = 26*26*'A' + 26*'A' + 'A'
``````
In general the formula for a string `columnTitle = abcd` is
``````abcd = 26^3*a + 26^2*b + 26*c + d,
``````
where `a, b, c, d` are some uppercase English letters `A, ..., Z`.
Longer `columnTitle`s will have bigger leading exponents of `26`.
#### Code
``````#include <iostream>
using namespace std;
int titleToNumber(string columnTitle) {
int column = 0;
for (char c : columnTitle) {
// The ASCII value of 'A' is 65.
column = 26*column + (c - 64);
}
return column;
}
int main() {
cout << titleToNumber("A") << endl;
cout << titleToNumber("AB") << endl;
cout << titleToNumber("ZY") << endl;
}
``````
``````Output:
1
28
701
``````
#### Complexity
• Runtime: `O(N)`, where `N = columnTitle.length`.
• Extra space: `O(1)`.
#### Implementation notes
1. There are many ways to compute the series
`````` 26^3*a + 26^2*b + 26*c + d.
``````
If you write it as
`````` 26*(26*(26*(0 + a) + b) + c) + d,
``````
you get the loop in the code above.
2. To map `'A' = 1, ..., 'Z' = 26`, you can use their ASCII values (`'A' = 65, ..., 'Z' = 90`) minus `64`.
3. The parentheses around `(c - 64)` is needed. Otherwise the value of `columnTitle = "FXSHRXW"` makes `26*column + c` exceed the limit of `int` before it substracts `64`.
### References
Thanks for reading. Feel free to share your thought about my content and check out my FREE book “10 Classic Coding Challenges”. | 1,013 | 2,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | longest | en | 0.665084 |
https://www.spoj.com/PROGPY/problems/PROG0511/ | 1,600,410,134,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187354.1/warc/CC-MAIN-20200918061627-20200918091627-00228.warc.gz | 1,413,618,995 | 9,142 | PROG0511 - Isograms
An isogram is a word in which no letter is repeated. Examples of isograms are uncopyrightable and ambidextrously. Conveniently, the word itself is an isogram.
Theoretically the limit is 26 letters, but that's an Everest that no one has scaled. In his book Language on Vacation: An Olio of Orthographical Oddities, Dmitri Borgmann has conquered some lesser peaks trying to find the longest isogrammic word. The longest one he found was dermatoglyphics at 15 letters. He coins several longer hypothetical words, such as thumbscrew-japingly (18 letters, defined as "as if mocking a thumbscrew") and — with the uttermost limit in the way of verbal creativeness — pubvexingfjord-schmaltzy (23 letters, defined as "as if in the manner of the extreme sentimentalism generated in some individuals by the sight of a majestic fjord, which sentimentalism is annoying to the clientele of an English inn"). Maybe what we lack is imagination.
An anagram is a type of word play, the result of rearranging the letters of a word or phrase to produce a new word or phrase, using all original letters exactly once. For example, the word Torchwood can be rearranged into Doctor Who. Any word or phrase that exactly reproduces the letters in another order is an anagram. However, the goal of serious or skilled anagrammatists is to produce anagrams that in some way reflect or comment on the subject. Such an anagram may be a synonym or antonym of its subject, a parody, a criticism, or praise. For example, William Shakespeare is an anagram of I am a weakish speller.
Assignment
In this exercise, words are represented as strings that only contain letters. Your task is to implement the following three functions, that each take one or two words. All functions must treat the given words in a case insensitive way.
• Write a function occurrences that takes a single word. The function must return a list of 26 integers, each indicating the number of occurrences in the given word of the letter at the corresponding position in the alphabet.
• Use the function occurrences to write a function isogram that takes a single word. The function must return a Boolean value that indicates whether or not the given word is an isogram.
• Use the function occurrences to write a function anagram that takes two words. The function must return a Boolean value that indicates whether or not the given words are anagrams of each other.
Example
```>>> occurrences('ambidextrously')
[1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0]
>>> occurrences('DOCTORWHO')
[0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 3, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0]
[1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0]
>>> isogram('ambidextrously')
True
>>> isogram('DOCTORWHO')
False
True
>>> anagram('DOCTORWHO', 'Torchwood')
True
>>> anagram('isogram', 'anagram')
False
>>> anagram('Aristotelian', 'retaliations')
True
```
Een isogram is een woord waarin elke letter van het alfabet hoogstens één keer voorkomt. Voorbeelden van isogrammen zijn filmproducent en vragenlijst. De maximale lengte van een isogram in het Latijnse alfabet is 26 letters, maar er zijn geen voorbeelden bekend van woorden die deze theoretische grens bereiken. De langste Nederlandse isogrammen die in Van Dale of het Groene Boekje voorkomen, tellen 15 letters: campinghoudster, dampkringslucht, sandwichformule en whiskyproducent.
Onder de liefhebbers van het Opperlans worden isogrammen vaak aangeduid als prachtsymboliek, wat zelf namelijk ook een isogram is. Zoals bij elk Opperlans onderwerp, moet er onderscheid gemaakt worden tussen het langste isogram dat daadwerkelijk is waargenomen in normaal taalgebruik (landbouwgeschrift, 17 letters), en het langste isogram dat na lang puzzelen en een ingewikkeld verhaal verdedigbaar wordt gemaakt.
Van de 308 gemeenten in Vlaanderen zijn er 42 waarvan de naam een isogram vormt, waarvan Zwijndrecht de langste is met 11 verschillende letters. Van de 262 gemeenten in Wallonië zijn er 75 waarvan de naam een isogram vormt, waarvan Wadelincourt de langste is met 12 verschillende letters. Van de 19 gemeenten in het Brussels Gewest is Vorst de enige wiens naam een isogram vormt.
Een anagram is een woord dat gevormd wordt uit de letters van een ander woord maar in een andere volgorde. Het maken van anagrammen is een geliefd woordspel. Anagrammen worden ook vaak gebruikt als pseudoniem.
Opgave
In deze opgave worden woorden voorgesteld als strings die enkel bestaan uit letters. Je opdracht bestaat erin de volgende drie functie te schrijven, waaraan steeds één of twee woorden moeten doorgegeven worden. De functies mogen geen onderscheid maken tussen hoofdletters en kleine letters.
• Schrijf een functie voorkomens waaraan een woord moet doorgegeven worden. De functie moet een lijst van 26 natuurlijk getallen teruggeven, waarbij elk getal aangeeft hoeveel keer de letter op de corresponderende plaats in het alfabet voorkomt in het gegeven woord.
• Gebruik de functie voorkomens om een functie isogram te schrijven waaraan een woord moet doorgegeven worden. De functie moet een Booleaanse waarde teruggeven, die aangeeft of het gegeven woord een isogram is.
• Gebruik de functie voorkomens om een functie anagram te schrijven waaraan twee woorden moet doorgegeven worden. De functie moet een Booleaanse waarde teruggeven, die aangeeft of de gegeven woorden anagrammen van elkaar zijn.
Voorbeeld
```>>> voorkomens('filmproducent')
[0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0]
>>> voorkomens('DOCTORWHO')
[0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 3, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0]
>>> voorkomens('whiskyproducent')
[0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0]
>>> isogram('filmproducent')
True
>>> isogram('DOCTORWHO')
False
>>> isogram('whiskyproducent')
True
>>> anagram('DOCTORWHO', 'Torchwood')
True
>>> anagram('isogram', 'anagram')
False
>>> anagram('CENTRALISERENDE', 'decentraliseren')
True
```
Added by: Peter Dawyndt Date: 2014-10-16 Time limit: 10s Source limit: 50000B Memory limit: 1536MB Cluster: Cube (Intel G860) Languages: PY_NBC Resource: None | 1,968 | 6,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2020-40 | latest | en | 0.925404 |
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Thread: Need help coding for homework -
1. Need help coding for homework -
Here is my assignment:
Develop a program to generate a pay stub. Here’s an example stub:
Name: Doell, Susan
Employee id: SDoe0001
Gross Pay: \$ 1099.15
Retire: \$ 87.93
Tax: \$ 141.57
Net Pay: \$ 869.65
The program is to input an employee’s first, last name, and gross pay. The input is in a file. Here’s the input for the above example:
Susan Doell 1099.15
The retirement rate is 8%. Note that the retirement amount is subtracted from the gross pay before the tax is computed. The tax rate is 14%. The employee id is the first initial followed by the first three letters of the last name followed by "0001".
I think I need to use strings for the employee name part. I am horrible at math, and the arithmetic has me confused as well.
My second problem is this:
A cider market purchases apples in bags. How much is paid per bag is dependent on the condition of the apples as follows:
Condition Price per Bag
Good \$12
Fair \$9.50
A program is needed to prompt for and input the condition of a bag of apples and how many bags are to be purchased. The output is the total price. Here’s a sample run:
Enter apple condition: Fair
Enter how many bags: 4
The total price is \$38.00
Condition Price per Bag
Excellent \$14
Good \$12
Fair \$9.50
This program is almost done; here is what i have so far, l; HOWEVER, IT NEEDS A LOOP:
#include <iostream>
#include <string>
using namespace std;
const float EXCELLENT_COST = 14.0;
const float GOOD_COST = 12.0;
const float FAIR_COST = 9.5;
int main()
{
string condition;
int bags;
float costPerBag, totalCost;
cout.setf(ios::fixed, ios::floatfield);
cout.setf(ios::showpoint);
cout << "Enter apple condition: ";
cin >> condition ;
cout << "Enter how many bags: ";
cin >> bags;
if (condition == ′"Excellent")
costPerBag = EXCELLENT_COST;
else
if (condition == "Good")
costPerBag = GOOD_COST;
else
costPerBag = FAIR_COST;
totalCost = costPerBag * bags;
cout << "The total price is \$" << setprecision(2)
<< totalCost << endl;
return 0;
}
I need a loop so that if the user types something different for the apple condition, the program responds with a message that the input is not a valid one. In this case, the program is not to respond with a total price
Any help is highly appreciated.
Thank you,
- WFB.
• I Had To Do This Same Project. Too Bad It's On My School Computer And I'm At Home. Thats Was One Of The First Things My Teacher Made Us Do In Java.
• Here is one way:
Code:
``` cout << "Enter apple condition: ";
cin >> condition;
while(condition != "Excellent" && condition != "Good") // whatever you're checking for
{
cout << "Invalid input" << endl;
cout << "Enter apple condition: ";
cin >> condition;
}```
You may want to force the input to lower/uppercase for case-insensitive comparison. Also it looks like you have an extra character here:
Code:
`if (condition == `′`"Excellent")`
• Please read 1.5 on CF's rules page. We can't help people actually do their homework- that's called cheating.
If you need help with a certain aspect of your homework, please post the code you're having trouble with, and explain what help you need. But asking people to basically do your assignment is not allowed.
• Originally Posted by WA
If you need help with a certain aspect of your homework, please post the code you're having trouble with, and explain what help you need. But asking people to basically do your assignment is not allowed.
I think that's what they did. They had already finished most of it, and just wanted help with one part.
• Problem now being, is that the solution is there for anyone that's doing the same class/course. I refuse to do any homework threads at all, as the next person that's too lazy to do anything has the finished script right infront of them and I dislike cheaters
• I see your point. If either of you feel that my response is or might be violating any rules, etc. feel free to delete it.
•
Posting Permissions
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• | 1,050 | 4,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2014-52 | latest | en | 0.864037 |
http://www.coppernblue.com/2014/4/8/5594876/updating-draft-success-numbers-a-vainglorious-number-crunching | 1,409,676,913,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535922089.6/warc/CC-MAIN-20140901014522-00074-ip-10-180-136-8.ec2.internal.warc.gz | 786,717,930 | 24,093 | ## Updating Draft Success Numbers - A Vainglorious Number Crunching Exercise
Ever been in one of those blog discussions where the other guy says something like "Yeah, well, the chance of success of a 2nd round pick is only 14.2% so that trade is worthless"?
Me neither.
That said, ever wonder those numbers come from? And are those numbers actually useful?
Lot's of folks have done work over the years to crunch draft data, but before I ever believe a number, I want to know:
• What was the data source?
• How recent is the data?
• How large is the sample size?
• What are the assumptions behind the analysis? e.g. if the success rate for a particular draft round is x%, what is meant by 'success'? A single game played? A full season played? Some game threshold?
There's more of course, but hopefully you get my point - you can't, or at least you shouldn't, just accept a number at face value.
More to the point, most of the draft success numbers I see don't help me anyway.
What does a first round draft success of 85% tell me about Darnell Nurse? Is trading a 2nd rounder for a 3rd and a 4th a good trade or a bad trade? I mean, hey, there are few things more productive or valuable than winning a blog argument, but sometimes you just wanna understand sh*t.
So I did what any self-respecting programmer / math / finance / stats guy would do - I crunched the numbers myself. Here's what I got ... hope you find it useful. (NOTE: details as to what data I used, and what tools and techniques were required, is detailed in the "FOR GEEKS" section at the end of this post).
### Da Charts
A picture is worth a zillion bits or something, so let's start with the pretty bits. Blue is raw data for each pick, green is a smoothed line (rolling average of 15 picks), and the red "Idealized" line is a calculated version of a highly smoothed line that mathematically fits the distribution. I have two charts (success rate and games played) for each section, and four sections (all players, forwards only, defense only, goalies only).
#### All Players
Not many surprises here - although when looking at games played, the dropoff from the early stages of the first round is far steeper than I thought it would be. It tells me that the value of a high first round pick is higher than I thought, and the value of a second round pick is not particularly high.
#### Forwards
These graphs are in line with what I expected, that there is a smooth steady drop off in success as you get later into the draft. Again, the main surprise is the steepness of the game dropoff from the first to the second round.
#### Defensemen
First surprise! Traditional wisdom is that you can get lots of good defensemen late in the draft. Not entirely true, or at least not that much moreso than for forwards. The second and subsequent rounds are markedly weaker than the first rounds.
Where the traditional wisdom comes in is that the sweet spot for defensemen is NOT at the top of the draft, it's in picks 11 to 30. As for Darnell? He's a bit outside that sweet spot, but history suggests he's going to be a player.
#### Voodoo - I mean, Goalies
Again, the surprise for me is that the first round is still the strongest round for these guys. But less surprisingly, the drop off is least steep as you get later in the rounds, with hefty peaks late in the first, in the middle of the second, and the first half of the third. Ignore the first steep dip in the "Idealized" curve - that's not signal, that's confusion. The sparser data combined with the high peak to start followed by three more substantial peaks is wreaking havoc with the math.
### Conclusions
The main thing I draw from combining the information here is to tweak drafting strategy:
• Always take forwards in the first 5 to 10 picks. (I'm looking at you Aaron Ekblad - and then I'm looking away quickly because you are one scary dude)
• The sweet spot for loading up on D is picks 11 to 30. Not to worry, Darnell, #7 is just fine!
• The sweet spot for loading up on G is picks 45 to 85
Other than that, presumably stick to BPA as much as you can.
### Summary Tables
Here's the data in summary format, by round and by pick range (bins of 5). Use this to win those arguments!
### For Geeks
#### Number crunching
• My 'success factor' criteria was 10 games. One seemed too few. Gut feel!
• The only voodoo I used is in generating my "Idealized" curve. I used cubic b-spline interpolation with a smoothing factor of 4. I used 9 knots for the calculation of the spline - endpoints were the average of the first two and last two values, and the mid 7 knots were the average of the 30 picks for the respective round. It seemed to work pretty well everywhere except for goalies - sparse noisy data means more knots required I think.
#### Technical details
• The source of my draft data was hockeydb.com, said data entered into a massive Excel spreadsheet
• The analysis (charts and tables) were number crunched using Python - specifically, SciPy, Numpy, Pandas, and Matplotlib, all obtained through the Anaconda distribution.
• The tables were pretty printed in Excel, and snapped to Imgur using ShareX
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We have our own Community Guidelines at The Copper & Blue. You should read them. | 1,465 | 6,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2014-35 | latest | en | 0.95526 |
https://resources.pcb.cadence.com/view-all/2019-electrical-analogues-of-simple-harmonic-motion-in-pcb-design | 1,627,548,673,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153854.42/warc/CC-MAIN-20210729074313-20210729104313-00157.warc.gz | 492,774,220 | 52,450 | # Electrical Analogues of Simple Harmonic Motion in PCB Design
July 31, 2019
Serious students of mathematics, physics, and engineering know that there are a number of mathematical coincidences that pop up in different fields. The mathematical analysis applied to one type of problem in one field can be applied to a completely different problem in another field, typically because the equations that govern these problems are nearly identical.
Electronics and mechanics are not without their correspondence. Simple harmonic motion of a mechanical body and a harmonic response in a circuit can be described with the same equations of motion. Understanding this aids a systems designer in understanding how circuits respond to real oscillating drivers: a circuit may be driven with a periodic source, but that does not mean the resulting motion will be simple.
## Ideal Simple Harmonic Motion in Circuits
In the ideal case, and within the context of electronics design, simple harmonic motion refers to driving a circuit with a sinusoidal source. Linear circuits, meaning any circuit where the current and voltage (or derivatives of either) in different elements are directly proportional, will output a sinusoidal signal. This is a very important consideration as it allows the behavior of a circuit to be analyzed using Ohm’s law and Kirchoff’s laws.
All oscillators have a natural frequency, which refers to the frequency at which they would tend to oscillate if initially displaced from equilibrium. In the case of an undamped pendulum, the natural frequency depends on the length of the pendulum and the acceleration due to gravity. When the pendulum is initially displaced from equilibrium, it will naturally oscillate about the equilibrium position.
The electronic analog of a pendulum is an LC circuit, meaning a circuit that contains a capacitor and an inductor in series. In this case, when the circuit is initially displaced from equilibrium, meaning the capacitor initially holds some charge. Once the circuit is shorted, the current in the circuit will begin oscillating at its natural frequency, just as in the case of a mechanical oscillator.
Charging and discharging LC circuit, leading to simple harmonic motion during discharge
## Real Oscillations in a PCB Are Not Simple
Simple harmonic motion can be produced in a circuit when driven with an AC signal, and resonance can occur when the circuit is driven at a frequency that matches the resonance frequency. Mathematically, the current in the circuit will increase linearly over time and will eventually reach infinity. In a real circuit, even in a passive linear circuit, nonlinear effects will set in as the current grows. This will cause the current to saturate near some maximum value.
In a PCB, every circuit will contain some capacitance and inductance due to the geometry of conductors in the circuit. These capacitance and inductance values are parasitic, but they should be included in circuit models in your board. The presence of these parasitic circuit elements means that each circuit in a PCB has some resonance frequency and can exhibit ringing when driven with digital pulses. This is problematic in transmission lines and illustrates the reason for using termination networks to suppress reflections and ringing.
### Damping in RLC Circuits
Unfortunately, real systems are not free of damping. In a mechanical oscillator, damping arises due to friction and drag. In an electrical oscillator, such as the LC circuit shown above, the wires have some resistance, which will dissipate power as the current oscillates. This causes the current in the circuit to drop to zero over time at an exponential rate, and the rate of damping is equal to half the total resistance in the circuit divided by the inductance. The presence of damping will cause the resonance frequency to shift away from the natural frequency, and it will limit the response of the circuit to some maximum value at resonance.
The typical method for examining the behavior of a circuit in the presence of parasitics is to model the circuit as an RLC network. The exact placement of the parasitic capacitor and inductor in the circuit will depend on the geometry of conductors and placement of components in the circuit.
It is important to note that even components themselves can have some parasitic resistance, capacitance, and inductance. For example, a real capacitor is not perfect and is actually modeled as its own RLC network. This means that components like capacitors actually exhibit a phenomenon called self-resonance. This self-resonance phenomenon can cause EMI problems with a capacitor is driven at its resonance frequency. It also means that a capacitor can exhibit ringing during discharge if the resistance in series with the capacitor is too low, meaning the response will be underdamped.
Effect of damping on the current in an RLC circuit
### Arbitrary Periodic Driving and Nonlinear Circuits
By “arbitrary” and “periodic”, one intends to mean “any signal that has repetitive shape and is not sinusoidal”. Real oscillators in electronic circuits can be driven with an arbitrary waveform. Examples include a stream of repeating digital pulses, a stream of Gaussian or Lorentzian pulses, sawtooth or triangle waves, or any other repetitive shape you can imagine.
Furthermore, not all circuit elements are linear. Instead, the relationship between current and voltage (or their derivatives) is not a straight line. A perfect example is a diode, where the output current is an exponential function of the input voltage. Another example is a transistor, where the output saturates once the input voltages become large. When these types of circuits are driven with a sinusoidal source, output from the circuit will not be sinusoidal, and there could be a phase shift that accumulates, depending on the capacitance and inductance in the circuit.
Arbitrary periodic (non-sinusoidal) sources will produce a periodic response in the circuit, but the response will not always have the same shape as the driver. There are a number of reasons for this. First, passives like inductors and capacitors exhibit a transient response that depends on the shape of an arbitrary driving pulse. Nonlinear circuit elements also affect the shape of the output pulse when the input driver is at a high signal level. As a useful example, consider the output from an inverting amplifier driven with a high voltage sinusoidal source. At high driving strength, the output will saturate as a square wave that is inverted compared to the input signal.
Any equivalent RLC network in a PCB that is driven with digital pulses can exhibit ringing if the circuit is underdamped. In general, any RLC circuit driven with a series of digital pulses will exhibit a transient response in the current in the circuit. This response can be either underdamped, perfectly damped, or overdamped, depending on the series resistance in the RLC network.
Working with the right PCB layout and design software can help you identify design your DC and AC networks to satisfy the maximum power transfer theorem. Allegro PCB Designer and Cadence’s full suite of design tools are designed with the layout and analysis tools you need to ensure power delivery throughout your DC networks and AC circuits. | 1,423 | 7,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-31 | latest | en | 0.920325 |
http://mathhelpforum.com/pre-calculus/183000-sum-series.html | 1,480,767,410,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540928.63/warc/CC-MAIN-20161202170900-00029-ip-10-31-129-80.ec2.internal.warc.gz | 170,679,000 | 11,722 | # Thread: Sum of the series
1. ## Sum of the series
Please help me to solve this problem
Find the sum of the series up to 2n terms
1^2-2^2+3^2-4^2+5^2-7^2+.....
2. ## Re: Sum of the series
Originally Posted by blackhat123
Please help me to solve this problem
Find the sum of the series up to 2n terms
1^2-2^2+3^2-4^2+5^2-7^2+.....
Are you missing a 6^2 term?
3. ## Re: Sum of the series
Ya I missed it out.
Thanks for pointing it.
The problem should be like this
Find the sum of the series up to 2n terms
1^2-2^2+3^2-4^2+5^2-6^2+.....
4. ## Re: Sum of the series
Here's one way it could be solved:
$\sum_{x=1}^{n}(2 x-1)^2 = \frac{1}{3} (4 n^3-n)$
$\sum_{x=1}^{n}(2 x)^2= \frac{2}{3} n (n+1) (2 n+1)$
$\sum_{x=1}^{n}(2 x-1)^2-\sum_{x=1}^{n}(2 x)^2 =\frac{1}{3} (4 n^3-n)-\frac{2}{3} n (n+1) (2 n+1)= -2 n^2-n$
That gives you 2 terms for each n, or 2n terms. I'm not certain this is how it's *supposed* to be solved, though
5. ## Re: Sum of the series
Thanks Stro.So this is the sum of the series upto the first n terms.Now should I replace n by 2n to get the sum of the series upto 2n terms?Please guide me.
6. ## Re: Sum of the series
Edit: To answer your question: No, the above is the summation up to 2n terms within the original series, it would not be changed at this point, but you would have to show your work I bet.
ie. $\sum_{x=1}^{n}(2x)^2=4\cdot\sum_{x=1}^{n}x^2= 4\cdot\frac{n(n+1)(2n+1)}{6}=\frac{2}{3}n(n+1)(2n+ 1)$
Spoiler:
$\sum_{x=1}^{n}(2 x-1)^2$
$\sum_{x=1}^{n}4x^2-4x+1$
$4\cdot\frac{n(n+1)(2n+1)}{6}-4\cdot\frac{n(n+1)}{2}+n$
$\frac{4n^3+6n^2+2n}{3}-\frac{6n^2+6n}{3}+n$
$\frac{1}{3}(4n^3+6n^2+2n-6n^2-6n+3n)$
$\frac{1}{3} (4 n^3-n)$
7. ## Re: Sum of the series
The sum of squares, for 1 to n, is well known to be (1/6)n(n+1)(2n+1) so the sum of squares from 1 to 2n is (1/6)(2n)(2n+1)(4n+1)= (1/3)n(2n+1)(4n+1).
Here, however, we have a negative sign on all even squares so we have to subtract off $4+ 16+ ...+ (2n)^2$ twice.
But $4+ 16+ ...+ (2n)^2= 4(1)+ 4(4)+ ... 4(n^2)= 4(1+ 4+ ...+ n^2)$. Using the formula for sum of squares as before, that is 4(1/6)n(n+1)(2n+1)= (2/3)n(n+1)(2n+1). Notice that this is 1 to n, not 2n, now. Since we have to subtract that off twice, the original sum is
(1/3)n(2n+1)(4n+1)- (4/3)n(n+1)(2n+1)= (1/3)n(2n+1)(4n+1- 4n-4)= (1/3)n(2n+1)(-3)= -n(2n+1).
Check: If n= 3, 1- 4+ 9- 16+ 25- 36= -21
-3(2(3)+1)= -3(7)= -21.
8. ## Re: Sum of the series
Thanks all for helping me out. | 1,107 | 2,456 | {"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": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2016-50 | longest | en | 0.824745 |
https://www.cfd-online.com/Forums/fluent/187873-turbulent-viscosity-viscosity-ratio-error.html | 1,713,392,458,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817181.55/warc/CC-MAIN-20240417204934-20240417234934-00502.warc.gz | 644,734,151 | 16,517 | # Turbulent viscosity to viscosity ratio error
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May 17, 2017, 07:43 Turbulent viscosity to viscosity ratio error #1 New Member rameez shahab Join Date: May 2017 Posts: 2 Rep Power: 0 During my simulation of a propeller rotating in water. I am using Frozen Rotor Model and I am getting this error "Turbulence viscosity limited to viscosity ratio of 1e5 in 4,491,571 cells". I have tried changing and refining my grid and boundary conditions, but I get this error from start of my simulations. Please tell me how can I resolve this problem.
May 19, 2017, 01:35 #2 Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,665 Rep Power: 65 Is 4 million cells pretty much the entire grid? If it happens this way at the very 1st iteration and sticks around for many iterations, it's your initial condition. Play with your initialization. You can change your mesh and boundary conditions for the next 100 years but you still need to provide the correct initial conditions. If you're using k-epsilon model, you need to provide initial values for k and epsilon. For k-omega, you need to specify k and omega. Etc. Usually people guess reasonable values for velocity so it is usually not the culprit, but it can be. However, most people don't know how to guess what the initial value for k is (it's usually a small number). Similarly, epsilon and omega are usually very large numbers (in the hundreds and thousands). If you guess zeros or ones, then you'll be way off and run into these turbulent viscosity ratio problems. Rameez likes this.
May 19, 2017, 19:16 #3 New Member rameez shahab Join Date: May 2017 Posts: 2 Rep Power: 0 Thanks for the guidance. Yes, it is almost whole grid. I am using SA model, can you tell me if SA model is alright for my problem and if it is, what should I use initial conditions? If I give wrong boundary conditions, like if my velocity is too low or too high, will my simulation diverge? Any help would be appreciated. :-)
Tags fluent, frozen rotor model, turbomachinery, turbulence
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Of course you could always convert the array to a list:
sage: import numpy as np
sage: L = np.array([1,1,3,1,4,5,8])
sage: list(L).count(1)
3
but here is something more direct:
sage: (L == 1).sum()
3
or
sage: sum(L == 1)
3
Also, if you want to count occurrences of every element in the array, you can do:
sage: from collections import Counter
sage: Counter(L)
Counter({1: 3, 8: 1, 3: 1, 4: 1, 5: 1})
Of course you could always convert the array to a list:
sage: import numpy as np
sage: L = np.array([1,1,3,1,4,5,8])
sage: list(L).count(1)
3
but here is something more direct:
sage: (L == 1).sum()
3
or
sage: sum(L == 1)
3
Also, if you want to count occurrences of every element in the array, you can do:
sage: from collections import Counter
sage: Counter(L)
Counter({1: 3, 8: 1, 3: 1, 4: 1, 5: 1})
The command sum will also count how many elements in an array satisfy a property.
For example to see how many are odd:
sage: (L%2).sum()
5
or how many are between 3 and 5:
sage: ((3r <= L) & (L <= 5r)).sum()
3
(here 3r and 5r are a way to input raw Python integers, as the comparison with Sage integers would not work well -- I'm not sure why). | 397 | 1,202 | {"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} | 3.0625 | 3 | CC-MAIN-2020-10 | longest | en | 0.869317 |
https://codegolf.stackexchange.com/questions/57617/is-this-number-a-prime/86293 | 1,582,907,518,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147234.52/warc/CC-MAIN-20200228135132-20200228165132-00433.warc.gz | 317,676,219 | 70,735 | # Is this number a prime?
Believe it or not, we do not yet have a code golf challenge for a simple primality test. While it may not be the most interesting challenge, particularly for "usual" languages, it can be nontrivial in many languages.
Rosetta code features lists by language of idiomatic approaches to primality testing, one using the Miller-Rabin test specifically and another using trial division. However, "most idiomatic" often does not coincide with "shortest." In an effort to make Programming Puzzles and Code Golf the go-to site for code golf, this challenge seeks to compile a catalog of the shortest approach in every language, similar to "Hello, World!" and Golf you a quine for great good!.
Furthermore, the capability of implementing a primality test is part of our definition of programming language, so this challenge will also serve as a directory of proven programming languages.
Write a full program that, given a strictly positive integer n as input, determines whether n is prime and prints a truthy or falsy value accordingly.
For the purpose of this challenge, an integer is prime if it has exactly two strictly positive divisors. Note that this excludes 1, who is its only strictly positive divisor.
Your algorithm must be deterministic (i.e., produce the correct output with probability 1) and should, in theory, work for arbitrarily large integers. In practice, you may assume that the input can be stored in your data type, as long as the program works for integers from 1 to 255.
### Input
• If your language is able to read from STDIN, accept command-line arguments or any other alternative form of user input, you can read the integer as its decimal representation, unary representation (using a character of your choice), byte array (big or little endian) or single byte (if this is your languages largest data type).
• If (and only if) your language is unable to accept any kind of user input, you may hardcode the input in your program.
In this case, the hardcoded integer must be easily exchangeable. In particular, it may appear only in a single place in the entire program.
For scoring purposes, submit the program that corresponds to the input 1.
### Output
Output has to be written to STDOUT or closest alternative.
If possible, output should consist solely of a truthy or falsy value (or a string representation thereof), optionally followed by a single newline.
The only exception to this rule is constant output of your language's interpreter that cannot be suppressed, such as a greeting, ANSI color codes or indentation.
• This is not about finding the language with the shortest approach for prime testing, this is about finding the shortest approach in every language. Therefore, no answer will be marked as accepted.
• Submissions in most languages will be scored in bytes in an appropriate preexisting encoding, usually (but not necessarily) UTF-8.
The language Piet, for example, will be scored in codels, which is the natural choice for this language.
Some languages, like Folders, are a bit tricky to score. If in doubt, please ask on Meta.
• Unlike our usual rules, feel free to use a language (or language version) even if it's newer than this challenge. If anyone wants to abuse this by creating a language where the empty program performs a primality test, then congrats for paving the way for a very boring answer.
Note that there must be an interpreter so the submission can be tested. It is allowed (and even encouraged) to write this interpreter yourself for a previously unimplemented language.
• If your language of choice is a trivial variant of another (potentially more popular) language which already has an answer (think BASIC or SQL dialects, Unix shells or trivial Brainfuck derivatives like Headsecks or Unary), consider adding a note to the existing answer that the same or a very similar solution is also the shortest in the other language.
• Built-in functions for testing primality are allowed. This challenge is meant to catalog the shortest possible solution in each language, so if it's shorter to use a built-in in your language, go for it.
• Unless they have been overruled earlier, all standard rules apply, including the http://meta.codegolf.stackexchange.com/q/1061.
As a side note, please don't downvote boring (but valid) answers in languages where there is not much to golf; these are still useful to this question as it tries to compile a catalog as complete as possible. However, do primarily upvote answers in languages where the author actually had to put effort into golfing the code.
### Catalog
The Stack Snippet at the bottom of this post generates the catalog from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.
## Language Name, N bytes
where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:
## Ruby, <s>104</s> <s>101</s> 96 bytes
If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:
## Perl, 43 + 2 (-p flag) = 45 bytes
You can also make the language name a link which will then show up in the snippet:
## [><>](http://esolangs.org/wiki/Fish), 121 bytes
<style>body { text-align: left !important} #answer-list { padding: 10px; width: 290px; float: left; } #language-list { padding: 10px; width: 290px; float: left; } table thead { font-weight: bold; } table td { padding: 5px; }</style><script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="language-list"> <h2>Shortest Solution by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr> </thead> <tbody id="languages"> </tbody> </table> </div> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr> </thead> <tbody id="answers"> </tbody> </table> </div> <table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table><script>var QUESTION_ID = 57617; var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe"; var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk"; var OVERRIDE_USER = 12012; var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, comment_page; function answersUrl(index) { return "https://api.stackexchange.com/2.2/questions/" + QUESTION_ID + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER; } function commentUrl(index, answers) { return "https://api.stackexchange.com/2.2/answers/" + answers.join(';') + "/comments?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + COMMENT_FILTER; } function getAnswers() { jQuery.ajax({ url: answersUrl(answer_page++), method: "get", dataType: "jsonp", crossDomain: true, success: function (data) { answers.push.apply(answers, data.items); answers_hash = []; answer_ids = []; data.items.forEach(function(a) { a.comments = []; var id = +a.share_link.match(/\d+/); answer_ids.push(id); answers_hash[id] = a; }); if (!data.has_more) more_answers = false; comment_page = 1; getComments(); } }); } function getComments() { jQuery.ajax({ url: commentUrl(comment_page++, answer_ids), method: "get", dataType: "jsonp", crossDomain: true, success: function (data) { data.items.forEach(function(c) { if (c.owner.user_id === OVERRIDE_USER) answers_hash[c.post_id].comments.push(c); }); if (data.has_more) getComments(); else if (more_answers) getAnswers(); else process(); } }); } getAnswers(); var SCORE_REG = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/; var OVERRIDE_REG = /^Override\s*header:\s*/i; function getAuthorName(a) { return a.owner.display_name; } function process() { var valid = []; answers.forEach(function(a) { var body = a.body; a.comments.forEach(function(c) { if(OVERRIDE_REG.test(c.body)) body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>'; }); var match = body.match(SCORE_REG); if (match) valid.push({ user: getAuthorName(a), size: +match[2], language: match[1], link: a.share_link, }); else console.log(body); }); valid.sort(function (a, b) { var aB = a.size, bB = b.size; return aB - bB }); var languages = {}; var place = 1; var lastSize = null; var lastPlace = 1; valid.forEach(function (a) { if (a.size != lastSize) lastPlace = place; lastSize = a.size; ++place; var answer = jQuery("#answer-template").html(); answer = answer.replace("{{PLACE}}", lastPlace + ".") .replace("{{NAME}}", a.user) .replace("{{LANGUAGE}}", a.language) .replace("{{SIZE}}", a.size) .replace("{{LINK}}", a.link); answer = jQuery(answer); jQuery("#answers").append(answer); var lang = a.language; lang = jQuery('<a>'+lang+'</a>').text(); languages[lang] = languages[lang] || {lang: a.language, lang_raw: lang.toLowerCase(), user: a.user, size: a.size, link: a.link}; }); var langs = []; for (var lang in languages) if (languages.hasOwnProperty(lang)) langs.push(languages[lang]); langs.sort(function (a, b) { if (a.lang_raw > b.lang_raw) return 1; if (a.lang_raw < b.lang_raw) return -1; return 0; }); for (var i = 0; i < langs.length; ++i) { var language = jQuery("#language-template").html(); var lang = langs[i]; language = language.replace("{{LANGUAGE}}", lang.lang) .replace("{{NAME}}", lang.user) .replace("{{SIZE}}", lang.size) .replace("{{LINK}}", lang.link); language = jQuery(language); jQuery("#languages").append(language); } }</script>
• Can I take inputs as negative numbers, where abs(input) would be the number I am testing? – Stan Strum Sep 6 '17 at 3:40
• Is there a reason for the full program requirement, rather than allowing the full range of default input types? E.g. answering with a function that takes its input as an argument, is currently disallowed? codegolf.meta.stackexchange.com/questions/2447/… – Lyndon White Dec 12 '17 at 6:21
• @LyndonWhite This was intended as a catalog (like “Hello, World!”) of primality tests, so a unified submission format seemed preferable. It's one of two decisions about this challenge that I regret, the other being only allowing deterministic primality tests. – Dennis Dec 12 '17 at 12:51
• @Shaggy Seems like a question for meta. – Dennis Jun 25 '18 at 13:44
• Yeah, that's what I was thinking. I'll let you do the honours, seeing as it's your challenge. – Shaggy Jun 25 '18 at 13:45
# Mouse, 65 47 bytes
?N:0S:1I:(I.N.=0=^N.I.\0=[S.1+S:]I.1+I:)S.1=!$ This uses trial division. Ungolfed: ? N: ~ Read an integer from STDIN and store it in N 0 S: ~ Start a summation variable at 0 1 I: ~ Start an interator variable at 1 ( I. N. = 0 = ^ ~ While I != N N. I. \ 0 = [ ~ Check whether I divides N S. 1 + S: ~ If so, increment the sum ] I. 1 + I: ~ Increment the iterator ) S. 1 = ! ~ If the sum is 1, the only divisor encountered ~ is 1 (we didn't go all the way to N in the ~ loop) and thus N is prime$
• May I ask what trial division is? – Beta Decay Sep 13 '15 at 14:03
• @BetaDecay You check the remainder when dividing by each number up to n. It's the brute force prime checking algorithm. – Alex A. Sep 13 '15 at 16:07
# Swift 2.0, 70100988277 75 bytes
A simple trial division loop, divided by 2 instead of sqrt() as it takes less bytes! Will default to using 64-bit ints.
let n=Int(Process.arguments[1])!;print(n<4||(2..<n).filter{n%$0<1}.count<1) • Is there a need to divide by two? I think you can use the 2..<n range instead to save another. – Geobits Sep 11 '15 at 14:56 • This prints true for 0 and 1. You can change the condition from n<4|| to n>1&& to make it work. Also: You can change Process.arguments[1] to readLine()! – Kametrixom Sep 12 '15 at 15:58 • Instead of ".count<1" you can use: "==[]" to crop some extra characters. – Simon Dec 4 '15 at 20:43 # gs2, 2 bytes Vk Basically read-num is-prime. # C, trial division, 72 bytes i=1;main(n){for(scanf("%d",&n);n%++i&&i*i<n;);printf("%d",n<3?n-1:n%i);} Note special handling for n=1,2 We need to stop the loop before i gets to n-1, so i<n-1 would do, but i*i<n is more efficient. • I used this to post a Javascript answer with the same functionality :) – Sam Sep 11 '15 at 15:05 • You get the exact same behaviour in one byte less by just putting everything on a single line, no separator needed. – hvd Sep 13 '15 at 21:45 • @hvd OMG! how did I miss that ?! thanks. – Level River St Sep 13 '15 at 22:31 • Replace i and n to save the initialization - when run with no arguments, the first parameter is 1. – ugoren Sep 17 '15 at 7:15 # Cheddar, 37 bytes Looks like a full program is required which unfortunately means a lot of boilerplate: print Math.prime(Number::IO.prompt()) This might not work on TIO so you'd have to put it into a file and call it that way # Function, 10 bytes Math.prime This returns a function which checks if input is prime using Math.prime. Example: $ cheddar primechecker.cheddar -x "[1, 3, 4, 10, 13] => (print) + f"
• A full program is required. – Dennis Jan 6 '17 at 20:18
• @Dennis >_> sorry for being so late, didn't see your comment until now but fixed – Downgoat Apr 13 '17 at 13:28
# Dodos, 154143136133126122 121 bytes
N , , + > > D f f
D
D ,
N F > > M
M
M m
+ B m f
m
F
+ B
,
dip F
>
F
+ B f
f
+ >
+
B
N
N
B dip
+
dot
>
dab
Try it online!
### Builtins and aliases
+
dot
This creates an alias + for the builtin function dot, which maps the vector (v1, ..., vn) to the vector (v1 + ... + vn), i.e.,
>
dab
This creates an alias > for the builtin function dab, which maps the vector (v1, ..., vn) to the vector (v2, ..., vn).
dip
The remaining builtin, dip, maps the vector (v1, ..., vn) to the vector (|v1 - 1|, ..., |vn - 1|).
### B and N
B
N
N
B dip
This defines a pair of mutually recursive functions. Recall that Dodos only divide or surrender.
B, if called from outside this function group, is meant to take a pair (x, y) as argument. B(x, y) simply calls N(x, y). This will always succeed, because all calls to N go through B.
N(x, y) attempts to call B(|x - 1|, |y - 1|). The tuple inequality (|x - 1|, |y - 1|) < (x, y) holds if an only if x > 0, so N(0, y) will return (0, y), since B(1, |y - 1|) surrenders.
Whenever x ≤ y, calling B(x, y) simply decrements both coordinates x times before surrendering, returning (0, y - x). If x > y, both coordinates will still be dipped x times. However, once the second coordinate reaches 0, it will cycle between 0 and 1, resulting in (0, (x - y) % 2).
N, if called from outside this function group, is meant to take a singleton (x) as argument. N(x) calls B(|x - 1|), which calls simply N(|x - 1|).
N(0) calls B(1), which attempts to call N(1). This surrenders, so B(1) returns 1, and so does N(0).
Whenever x > 0, N(x) will decrement x until reaching N(1) → B(0) → N(0) → B(1). Since B(1) surrenders, N(0)'s argument is the return value of N(x).
Thus, N(0) = 1, while N(x) = 0 whenever x > 0.
### F and f
F
+ B f
f
+ >
+
f returns the results of + > and + as a vector; + > takes the sum of the vector without its first coordinate, while + takes the sum f the whole vectors.
Thus, f(v1, ..., vn) = (v2 + ... + vn, v1 + ... + vn).
On occasions, we'll call f outside of F. Notably, f(x) = (0, x), f(0, x) = (x, x), and
f(x, y) = (y, x + y)
F simply calls three functions we've seen before. Since v2 + ... + vn ≤ v1 + ... + vn, B maps the result returned by f to (0, v1). + takes the sum, returning (v1).
Thus, calling F on a vector returns its first coordinate.
### ,
,
dip F
>
dip F dips the first coordinate of the argument vector, while > returns the remaining coordinates. Thus, , maps (v1, ..., vn) to (|v1 - 1|, ..., vn).
In particular, , maps (x) to (|x - 1|), so we can use it instead of dip for singleton vectors.
### M and m
M
M m
+ B m f
m
F
+ B
m is always meant to be called on a pair. It simply combines a few functions we've seen before. Recall that B behaves differently if x ≤ y and if not.
We have m(x, y) = (x, y - x) if x ≤ y but m(x, y) = (x, (x - y) % 2) otherwise.
M is always meant to be called on a pair, whose first coordinate will be non-zero. M attempts to recursively call itself on the result of m. Since m doesn't change the first coordinate of its argument, we only need to examine the second one.
On the second line, B(m(f(x, y))) = B(m(y, x + y)) = B(y, x). After + takes the sum, we get x - y if x ≥ y, but (y - x) % 2 otherwise.
If y = qx, M(x, y) = M(x, qx) will successively call M(x, (q - 1)x), M(x, (q - 2)x), ..., M(x, x), M(x, 0). At the end, M(x, 0) attempts to recursively call itself, which surrenders. The return value is (x, 0), concatenated with all singletons returned by the second line. Since the second line maps (x, x) to 0, the result of M(x, y) will match the pattern (x, 0, 0, ...).
If y = qx + r, with 0 < r < x, we'll proceed in similar fashion, eventually reaching M(x, x + r), then M(x, r).
• If r = 1 and x is even, m(x, r) = (x, (x - r) % 2) = (x, 1), and the recursive call to M surrenders. In the previous call, the second line mapped (x, x + r) to (x + r - x) % 2 = 1, so the result of M(x, y) will match the pattern (x, 1, 1, ...).
• If r = 1 and x is odd, m(x, r) = (x, (x - r) % 2) = (x, 0), and the recursive call to M succeeds.
Likewise, if r > 1, then m(x, r) = (x, (x - r) % 2) ≤ (x, 1) < (x, r), and the recursive call to M succeeds.
In both cases, the second line will be evaluated for the last time with argument (x, t). Since we have t < r < x, the outcome is the non-zero vector (x - t), so the result of M(x, y) will match the pattern (x, ?, x - t, ...).
Thus, the third element of the vector returned by M(x, y) will be 0 if and only if x is a divisor of y.
### D
D
D ,
N F > > M
D expects (n, n) as its initial argument.
The first line will recursively call D on the result of ,, so we'll call D(n, n) → D(n - 1, n) → ... → D(1, n) → D(0, n) → D(1, n), surrendering and returning (0, n).
The second line will be evaluated for every (k, n), with 0 < k ≤ n. F > > M calls M, discards the first two elements, then extracts the first remaining one. As we've seen before, M returns (k, ?, 1, ...) if k is a divisor of n, (k, ?, 0, ...) otherwise, so the final vector returned by D(n, n) is (0, n, 1 | n, 2 | n, ..., n | n).
### main
N , , + > > D f f
Our entry point expects a singleton (n), with n > 0.
Since f(f(n)) = f(0, n) = (n, n), D will return (0, n, 1 | n, 2 | n, ..., n | n).
After discarding (0, n) with > >, a call to + takes the sum of the Booleans, counting the number of divisors of n.
, , dips the divisor count twice, resulting in a 0 only for zero or two divisors. Since n > 0, there is at least one divisor, so the result is 0 if and only if n has exactly two divisors.
Finally, N takes the logical NOT, returning 1 for primes and 0 for non-primes.
• I've edited my answer. – Dennis Mar 22 '18 at 1:56
• Not really related, but ... no chatroom for Dodos yet? – Weijun Zhou Mar 22 '18 at 2:40
• @WeijunZhou Fixed. – Dennis Mar 22 '18 at 2:47
• That's a tricky language! Some typos near the end: 0 < k < n should be 0 < k ≤ n, k | n should be n | n (two places), and "M returns 1" should be something like "M returns third element 1". – Ørjan Johansen Mar 22 '18 at 3:42
• @ØrjanJohansen All fixed, I think. Thanks! – Dennis Mar 22 '18 at 3:49
# ><>, 25 + 3 = 28 bytes
:1-:v
v!?:<-1$**:@: >r%n; Inputting as a byte with i is shorter, but ><> can handle numbers larger than 255, hence the need for command line input in order to follow the rules. The +3 is for the v flag, i.e. run like py -3 fish.py primes.fish -v 101 Outputs (n-1)*((n-1)!)^2 mod n (the initial (n-1)* is unnecessary, but it makes the code shorter). • I don't quite understand what you mean by your first sentence and I'm curious to know. Care to explain? The rules seem to say that it only needs to work for the integers 1 to 255. – cole Sep 11 '15 at 23:21 • @Cole The rules say you can read a single byte "if this is your languages largest data type" – Sp3000 Sep 12 '15 at 1:08 # Python 2, 46 bytes m=n=input() a=1 while~-m:m-=1;a*=m*m print a%n • Just 2 bytes shorter than the functional version: n=input();print all(n%m for m in range(2,n))*~-n. If all([]) returned False it would be the other way around. – DLosc Sep 11 '15 at 19:34 # Matlab/Octave 24 It is just using a builtin function, using a sieve. disp(isprime(input(''))) You could also use this: isprime(input('')) Which would print the output as the last result in the console, but I am not sure whether this is allowed. • I am curious, where did you get the information about the other method that you mentioned? (Perhaps it is different in a newer vesion?) – flawr Sep 15 '15 at 17:52 • This link I guess the MuPAD notice at the top is relevant. – Martin Ender Sep 15 '15 at 17:56 • Yes that is what makes the difference, thanks for the link=) – flawr Sep 15 '15 at 19:46 ## Sesos, 50 49 bytes Algorithm #3... 0000000: 16def7 f5991b 7441bf 3f0ebb eecfd8 b86b33 b7eb33 ......tA.?......k3..3 0000015: 37ecda bccdd8 b86b33 3ffcfe 8c7de8 797cfc f599c3 7......k3?...}.y|.... 000002a: f973f5 8479c5 03 .s..y.. Try it online! This is the very first working code I got, so I'm sure it's possible to shave off a few more bytes here. Here's the BF-code I wrote (with some rather sparse comments that are mostly meant for myself): , [>+>+>+<<<-] ; triplicate input >>[- ; i from n_1 down to 0 <+[-<<<+>>> ; j from n down to 0 copying n to the left [ ; k from j down to 1 >[<<+<+>>>-] ; add i to the values on the left <<[>>+<<-] ; move one copy of i back >>>>+<<<- ; decrement k while copying j to the right ] >>[<<+<<->>>>-] ; subtract n from i*j while copying it <<[>>+<<-] ; move n back >>>[<<<+>>>-] ; move j back <<<<<[ ; if not equal: , ; reset to zero > ; move to zero left of j ] > ] <<<[>>>+<<<-] ; move n back to j >[<]>>> ; if we didn't exit move back to i ; otherwise remain on the zero left of j ] >,+>-[<->,]<. I then used this Retina script to convert that to Sesos ASM: set numin set numout get jmp fwd 1 add 1 fwd 1 add 1 fwd 1 add 1 rwd 3 sub 1 jnz fwd 2 jmp sub 1 rwd 1 add 1 jmp sub 1 rwd 3 add 1 fwd 3 jmp fwd 1 jmp rwd 2 add 1 rwd 1 add 1 fwd 3 sub 1 jnz rwd 2 jmp fwd 2 add 1 rwd 2 sub 1 jnz fwd 4 add 1 rwd 3 sub 1 jnz fwd 2 jmp rwd 2 add 1 rwd 2 sub 1 fwd 4 sub 1 jnz rwd 2 jmp fwd 2 add 1 rwd 2 sub 1 jnz fwd 3 jmp rwd 3 add 1 fwd 3 sub 1 jnz rwd 5 jmp get fwd 1 jnz fwd 1 jnz rwd 3 jmp fwd 3 add 1 rwd 3 sub 1 jnz fwd 1 jmp rwd 1 jnz fwd 3 jnz fwd 1 get add 1 fwd 1 sub 1 jmp rwd 1 sub 1 fwd 1 get jnz rwd 1 put And of course the final conversion to binary is done by Sesos itself. I scrapped three earlier attempts for trial division, and ultimately really got tired of the modulo computation. So I started thinking about how I could avoid that altogether. I ended up coming up with a very simple primality test, that for some reason never occurred to me before and might be handy for a lot of other esolangs where doing a multiplication is fine but computing a modulo is a royal pain: In essence, I just compute the full multiplication between [1, ..., n-1] and [1, ..., n] starting from the largest value. After each multiplication, I subtract n from the result. If that gives 0, I terminate. This is bound to terminate, because at the beginning of the final iteration of the outer loop, I'm computing 1 * n. If I get there, it's a prime. Otherwise, some earlier multiplication will have given n and the loop stops there instead. That means I can simply check after terminating whether the first iterator is equal to 1 or not in order to decide primality. I'll probably post the Brainfuck-version of this as well, once I'm happy with the golfing. # ShapeScript, 5325 23 bytes _11?1-"@1?*@1-"1?*!?*@% The program uses Wilson's theorem; it prints 1 for primes and 0 for non-primes. Input is in unary. I created ShapeScript for this competition. The interpreter on GitHub has a slightly modified syntax and better I/O (none of which are required in this answer). Try it online! ### How it works _ Take the length of the input to convert from unary to integer (N). 1 Push 1 (accumulator). 1?1- Push a copy of N and subtract 1. Let's call the result I. " Push a string that, when evaluated, does the following: @ Swap I with the accumulator. 1? Push a copy of I. * Multiply it with the accumulator. @ Swap the updated accumulator with I. 1- Decrement I. " 1? Push a copy of N-1. *! Repeat the string N-1 times and evaluate the result. This calculates (N-1)! and leaves I = 0 on the stack. ? Use I to copy the factorial. * Multiply to calculate the factorial's square. @% Calculate N%((N-1)!*(N-1)!). # Prolog (SWI), 42 40 bytes +X:-X>1,2+X. Y+X:-X=:=Y;0<X mod Y,1+Y+X. Try it online! Tests primality by checking whether any of the numbers less than the number and greater than one divide the number. ## Explanation +X:-X>1,2+X defines a new predicate +/1 that is true if X > 1 and 2+X is satisfied. Despite what it appears, this is not an arithmetical expression. It refers to the predicate defined on the next line. Y+X:-X=:=Y;0<X mod Y,1+Y+X. This defines a new predicate +/2 that is true if X =:= Y (=:= represents equality for arithmetic expressions), or both X mod Y is not zero (since X mod Y can't be negative, checking that it is greater than zero is sufficient) and 1+Y+X is true. The trick with 1+Y+X is that the two pluses do not end up being used the same way. The + operator is left associative so 1+Y+X is equivalent to (1+Y)+X. Since the expression must be a call to a predicate and +/2 is a predicate that I defined, the interpreter then calls +/2 with arguments 1+Y and X. Thus recursively the program will check whether any Y from 2 to X-1 divides X (it stops at X since X=:=Y would be true and so the truth value of the others becomes irrelevant). ### Operators in SWI-Prolog This program heavily abuses the way operators are handled in SWI-Prolog. SWI-Prolog does not define predicates for arithmetic operators such as +. This makes sense since the arithmetical + is not a predicate since the output of a predicate must be true or false. What SWI-Prolog does instead is it builds structures out of the operators and then the predicates that evaluate arithmetic expressions (such as =:= and <) know how to evaluate those structures. For instance after the +/2 predicate in my program recurses three times with the initial Y=2 then Y=1+(1+(1+2)) not Y=5. The fact that + is not defined as a predicate means that I can define it as a predicate myself and since it is an operator I can save the bytes that I would otherwise need to spend on parentheses and commas. # Foo, 40 bytes &1@@@>>&>&1<(2-1@<<@>&%+1@>>%<)&2/@>+%$i
Probably not the best approach, but I wanted to give it a try. Thanks to the "wonders" of Foo's do-while loops, I had to special case 1 and 2, both of which output errors to STDERR (but STDOUT output is correct).
The input is hardcoded as the number after the first &.
# Bubblegum, 98 bytes
from math import factorial as F#
try:n=int(i)-1;o=n*(F(n)%-~n==n)
except:o=sum(map(int,i.split()))
This prints p - 1 if p is prime and 0 otherwise.
It may not look like it, but this is the shortest known Bubblegum program that achieves this task.
There are probably shorter programs, but their discovery would require a cryptographic break of the SHA-256 hash.
• Did you choose the hash value based on this program or did you find it starting from a random hash? – Sp3000 Sep 11 '15 at 17:03
• Option 1. The program works for addition and primality testing. – Dennis Sep 11 '15 at 17:05
• Wait what according to the esolangs page this is identical to the code for calculating a sum... – ev3commander Nov 18 '15 at 20:56
• @BlockCoder1392 If you give this program one number it will test primality. If you give it two or more it will add. – pppery Feb 25 '16 at 2:13
# Perl, 23 20 bytes
say/^(?!(..+)\1+$)/ using -n option. say<>=~/^(?!(..+)\1+$)/
Using the regular expression+unary input approach, prints 11 (or whatever number you entered) or a blank line.
Bonus: decimal version, 31 bytes
(1x<>)=~/^1$|^(11+)\1+$/||say 1
• Does this require input in unary? – Lynn Sep 11 '15 at 18:45
• @Mauris, clarified as you commented :) – ThaddeusB Sep 11 '15 at 18:46
• The 20 byte version reports 1 as a prime, doesn't it? – Martin Ender Sep 14 '15 at 10:14
• The decimal version says 0 is prime. – DanaJ Sep 14 '15 at 12:12
• @DanaJ Good question... Per Dennis' clarification 0 failing is not an issue. – ThaddeusB Sep 14 '15 at 16:37
# FRACTRAN, 144 bytes
29/14 222/377 59/26 247/59 329/57 19/47 2/19 11/29 403/407 217/33 11/31 2/11 1/37 1/2 23/9 43/69 23/43 1/23 425/41 4823/85 17/53 41/25 2/119 1/5
This took me way longer than expected - special casing 1 was pretty annoying. Takes 5^n as input and outputs 3^0 = 1 (falsy) if composite, or 3^1 = 3 (truthy) if prime.
The approach is similar to Conway's prime generator, performing a divmod on descending divisors. This isn't as compact though, so there's still much to golf.
• How does 3^1 = 7? – feersum Sep 13 '15 at 1:06
• @feersum Updated one number, forgot to update the other. Thanks for pointing out – Sp3000 Sep 13 '15 at 1:50
# Fourier, 44 bytes
Guess who? :D
1~h2~xI~g<3{1}{5+g~g}g(g%x{0}{~hgv~x}x^~x)ho
Yes, it's my very own Fourier again, in a situation where it is actually in the running for a golfing competition.
The most bytes are spent on handling the input cases for 1 and 2.
Prints "1" for True and "0" for False.
## Explanation
1~h # Set h to 1
2~x # Set x to 2
I~g # Set g to user input
<3{1}{ } # If accumulator is less than three then
5+g~g # Add 5 and g and set g to that value
g( ) # Loop until the accumulator equals g
g%x{0}{ } # If g%x equals 0, then
~hgv~x # Set h to 0 and set x to g-1
x^~x # Increment x and set x to that value
ho # Output the value of h
alert(!/^(11+)\1+$/.test('1'.repeat(prompt()))) • @ETHproductions OK. I'm working on the case of the input being 1. – Toothbrush Sep 13 '15 at 17:54 • This seems to be missing output. It assumes that the code is run in a console, i.e. a REPL environment. The challenge explicitly asks for a full program though. – Martin Ender Sep 14 '15 at 9:43 • Virtually identical to one already posted, which also covers the '1' case: codegolf.stackexchange.com/a/57692/30793 – Mwr247 Sep 14 '15 at 14:51 • Also, needs an alert or other print function to be considered valid for code golf: meta.codegolf.stackexchange.com/questions/803/… – Mwr247 Sep 14 '15 at 16:48 • Outputs true for 0 and 1. – RK. Sep 17 '15 at 19:24 # PowerShell, 35 Bytes param($a)$a-match'^(?!(..+)\1+$)..'
Uses the same regex from Martin's Retina answer, as that's way shorter than anything that will wind up using the [math]:: libraries one would normally use. Expects input as command-line argument in unary format.
Corrected from initial version (which was apparently specific to the particular PowerShell implementation I coded it on) thanks to Jonathan Leech-Pepin. Grr undocumented version differences.
Examples:
PS C:\Tools\Scripts\golfing> .\is-this-number-a-prime.ps1 111111
False
PS C:\Tools\Scripts\golfing> .\is-this-number-a-prime.ps1 1111111
True
### Bonus - PowerShell pipeline input, 29 Bytes
%{$_-match'^(?!(..+)\1+$)..'}
Same as the above, just called differently, which shaves bytes. For example,
PS C:\Tools\Scripts\golfing> 111111 | %{$_-match'^(?!(..+)\1+$)..'}
False
• I actually had to wrap the $args[...]..' in () to be able to get it to resolve. Otherwise it was always True. – Jonathan Leech-Pepin Nov 4 '15 at 21:11 • @JonathanLeech-Pepin Interesting - likely a versioning difference in how operations are ordered. I primarily code in PowerShell v4 (using the ISE); what were you using? – AdmBorkBork Nov 4 '15 at 21:27 • Happened on Win7 with the latest 5.0 rtm (August), also happens on Win10 with built-in. It is not however an issue with the pipeline version. On the other hand: param($a)$a-match'^(?!(..+)\1+$)..' works and is exactly 35 bytes as well. – Jonathan Leech-Pepin Nov 4 '15 at 21:40
• @JonathanLeech-Pepin Indeed -- I just tested and confirmed on a Windows 7 machine. How strange. Yay for undocumented features :-/ ... I've corrected the code with the param($a) iteration, which should work on any version. Thanks for the assist! – AdmBorkBork Nov 4 '15 at 21:56 # Sesos, 6766 65 bytes Edit: Saved a byte by using another get instead of a loop. Edit: Saved a byte because I don't need this rwd 6 after I changed from sub 1 to add 1 before it. Try it online The hexdump: 0000000: 16f8be 76ca83 e653e3 b472f0 750ef0 af9f1f fcebbb ...v...S..r.u........ 0000015: 7f7ec6 77e13b bf41f7 2961f0 af9f1f fcebbb 7f6ec7 .~.w.;.A.)a........n. 000002a: 3fc013 ef9da3 a0fbbc 77ecc7 776e1b bf73b8 576a9c ?........w..wn..s.Wj. 000003f: 663e f> This is the Sesos assembly code that I wrote, which is assembled into the above binary to be executed: set numin set numout get jmp ; n += (n==2) sub 1 fwd 1 add 1 fwd 1 add 1 rwd 2 jnz add 2 fwd 1 jmp sub 1 rwd 1 sub 1 fwd 1 jnz add 1 rwd 1 jmp fwd 1 sub 1 rwd 1 get jnz fwd 1 jmp sub 1 fwd 1 add 1 rwd 1 jnz fwd 1 jmp ; list from n to 1 jmp sub 1 fwd 3 add 1 rwd 3 jnz fwd 3 jmp sub 1 fwd 3 add 1 rwd 6 add 1 fwd 3 jnz fwd 3 sub 1 jnz rwd 6 ; List [n, n-1, ..., 2, 2]. We don't want n%1. add 1 jmp rwd 6 jnz fwd 6 jmp ; move n one cell to the left sub 1 rwd 1 add 1 fwd 1 jnz add 2 ; copy the n's jmp rwd 1 jmp sub 1 fwd 3 add 1 rwd 3 jnz fwd 3 jmp sub 1 fwd 3 add 1 rwd 6 add 1 fwd 3 jnz fwd 4 jnz rwd 7 jmp ; compute each divmod, only the n%d results will be used jmp sub 1 fwd 1 sub 1 jmp fwd 1 add 1 fwd 2 jnz fwd 1 jmp add 1 jmp sub 1 rwd 1 add 1 fwd 1 jnz fwd 1 add 1 fwd 2 jnz rwd 5 jnz rwd 6 jnz fwd 8 jmp ; go to first modulus of zero, or past end of list fwd 6 jnz fwd 1 ; negate cell to the right jmp rwd 1 add 1 fwd 1 jmp sub 1 jnz jnz add 1 rwd 1 jmp fwd 1 sub 1 rwd 1 sub 1 jnz fwd 1 ; output put ### Explanation (In BF, since I actually wrote it in BF first) Sesos and BF are closely related, so I will write the explanation in BF to take less space (it won't be on as many lines): > fwd 1 < rwd 1 + add 1 - sub 1 , get . put [ jmp ] jnz First, Sesos is basically BF, but there is some I/O help, using the assembler directives set numin and set numout. These allow me to take an unbounded integer as input, into a single cell, or output that cell as an integer. I decided this was the easiest way to write the program for all positive integers. My explanation is of each section from the above code, with sub-explanations showing the manipulations of the tape, in an attempt to help you understand the process and algorithm. I put the tape in curly braces, and use > to denote the pointer's location on the tape. Section 1, the bug-fix / edge case: It should be noted that before I fixed this, my code was only 54 bytes. Because of how I determine if a number is prime later, I had to add one to n if n==2, so I do that first. I use a , here (get) to zero a cell instead of looping with [-]. [->+>+<<]++>[-<->]+<[>-<,]>[->+<]> n += (n==2): goal 1: { 2 n n } goal 2: { 0 n==2 n } goal 3: { 0 0 n* } { n 0 0 } [->+>+<<]++> { 2 >n n } [-<->]+<[>-<,]> { 0 >n==2 n } [->+<]> { 0 0 >n* } Section 2, the list and my ultimate goal: The way I check if n is prime is to check n modulo every number from n-1 to 2, which I figured would be simplest. My main goal was to reach the following data structure: 0 >{n n-1 0 0 0 0, n n-2 0 0 0 0, ..., n 2 0 0 0 0} This facilitates the DivMod algorithm I planned to use, which requires n d 0 0 0 0 on the tape. So I create a list from n-1 to 0, with the necessary spacing. I copy the first marked cell to the second, then copy that temp cell back into the original and into the next. Then subtract one. This repeats until I hit zero. 0 { 0 >n } [[->>>+<<<]>>>[->>>+<<<<<<+>>>]>>>-] 0 {0 n 0 0 0 0, 0 n-1 0 0 0 0, ..., 0 2 0 0 0 0, 0 1 0 0 0 0, 0 >0 0 0 0 0} ^ ^ ^ Then, make the last section find n%2, since n%1 would cause a result of 0 for every n. Changing it to a zero instead of a two produces the wrong answer for n=1. After that, move back to n. <<<<<<+ [<<<<<<]>>>>>> 0 {0 >n 0 0 0 0, 0 n-1 0 0 0 0, ..., 0 2 0 0 0 0} Move n left one cell, preparing to copy it across the list: [-<+>] 0 {n >0 0 0 0 0, 0 n-1 0 0 0 0, ..., 0 2 0 0 0 0} Section 3, copy the n's I copy n to the correct position for each entry in the list, so that I'll be ready to use the DivMod algorithm for each entry. I first add two here, so that we find another n%2, rather than n%0. This is nearly the same code as in section 2, except that I compare to the cell on the right each time, in order to stop upon completing the length of the list. ++[<[->>>+<<<]>>>[->>>+<<<<<<+>>>]>>>>] 0 {n 2 0 0 0 0, n n-1 0 0 0 0, ..., n 2 0 0 0 0, n >0 ...} ^ ^ ^ Section 4, compute each DivMod I go through the list, doing the algorithm for each, after which only the n%d results are used. Though the algorithm only lists 4 cells on the site, it relies on the 5th and 6th cells being zero for its magic to work. I used the version which does not preserve n, since I won't need it anymore. # >n d 0 0 0 0 [->-[>+>>]>[+[-<+>]>+>>]<<<<<] # >0 d-n%d n%d n/d 0 0 As applied across the list (x marks stuff I don't really need, but do make use of later): [ [->-[>+>>]>[+[-<+>]>+>>]<<<<<] <<<<<< ]>>>>>>>> 0 {0 x >n%(n-1) x 0 0, 0 x n%(n-2) x 0 0, ..., 0 x n%2 x 0 0} Section 5, if any n%d == 0 I check the list from left to right. {0 x >n%(n-1) x 0 0, ...} [>] What? You expected more? Well it really is that simple. This stops at the first occurrence of 0, which is either in this list, meaning the number is not prime, since it has a divisor, or we went past the list, and the number is therefore prime. Section 6, negate the cell to the right and output Uses this algorithm: temp0[-] x[temp0+x[-]]+ temp0[x-temp0-] I don't need the first line, since my temp is already zero. I also use get to zero a cell instead of a loop. The last line prints the resulting number, a one if prime, or a zero if not. >[<+>,]+ <[>-<-] >. Concluding remarks Overall, this was a fun challenge. I found the mapping to BF pretty quickly with trail and error using the interpreter and the documentation. I completed it with something like 8 hours of effort. Much of the writing occurred in Notepad++ in BF that I then converted to Sesos with a Python program, tested, and debugged. Convert BF to Sesos ## Reflections, 194 181 bytes _v@\ |* / (0 /\ /;* < /0):$$1/# + /#+\ : ; >~< \ _ / #|v\/ 1) (0* \#:(1 \ \\# \(0__0) / _ / (0\ /^^: 0):/ \#+ _#_ /0):^\ : / / # (0 >#* _#_ \ _< \ / Test it! Outputs 1 for prime, else 0. ### Explanation First we parse the number: _v@\ |* / (0 * (1/ -> IP leaves here >~< 1) \#:(1 \ \(0__0) / • _ reads a line from input • v reflects the IP down • / reflects the IP left • * at (1|1) pushes 1×1=1 • | reflects the IP right • * pushes another 1 • / reflects the IP up • v pops a value off the stack and reflects the IP right as it's true • @ at (4|0) converts all input to numbers • \ reflects the IP down • (0 moves the first digit to stack 0 • * at (5|2) pushes 5×2=10 • (1 moves the 10 to stack 1 • > reflects the IP right • ~ pushes the number of left digits • < pops that number and reflects the IP down if it's not 0: • 1) moves the 10 from stack 1 to the main stack • \ reflects the IP right • # redefines (0|0) • : doubles the 10 • (1 moves the top 10 to stack 1 again • \ reflects the IP down • / reflects the IP left • 0) pulls the previous result from stack 0 • _ at (1|1) multiplies the previous result and 10 • _ at (0|1) adds the next digit • (0 pushes the result to stack 0 • \ reflects the IP up • > enters the loop again • if it's zero, reflect the IP up • / reflects the IP right Now, we have the test number on stack 0. Then, we initialise the loop: < /# + / • / reflects the IP right • # redefines (0|0) • + at (2|0) pushes 2+0=2 • / reflects the IP up • < reflects the IP left Now, we have a 2 (the counter) on the main stack and the input number on stack 0. Then we have the real loop: /\ /;* < /0):\ /#+\ : ; \ _ / #|v\/ (0* \\# _ / (0\ /^^: 0):/ \#+ _#_ /0):^\ : / / # (0 >#* _#_ \ _< \ / • * pushes x×y • ; pops that again • / reflects the IP down • : duplicates the counter • ; discards the duplicate • \ reflects the IP right • / reflects the IP up • \ reflects the IP left • : duplicates the counter again • 0) pulls the input number from stack 0 • / reflects the IP down • : duplicates the input • # redefines (0|0) • (0 pushes the duplicated input to stack 0 • \ reflects the IP right • \ reflects the IP down • _ at (1|3) pops the counter and the input and pushes whether they're equal • ^ pops the test and reflects the IP left if true (i.e. if we have tested all numbers less than the input and haven't found a factor → the number is prime): • / reflects the IP down • \ reflects the IP right • # redefines (0|0) • + at (1|0) pushes 1+0=1 • _ at (3|0) converts to string • # redefines (0|0) • _ at (1|0) prints • then the IP leaves the grid and the program ends • else the IP is reflected right: • ^ reflects the IP up • # redefines (0|0) • * pushes 0×-1=0 • v pops the 0 and reflects the IP left • | reflects the IP right • v reflects the IP down • * pushes 0×-1=0 • # redefines (0|0) • ^ pops the zero and reflects the IP right • : duplicates the counter • 0) pulls the input from stack 0 • : duplicates it • / reflects the IP up • \ reflects the IP left • (0 pushes the duplicated input to stack 0 • / reflects the IP down • _ at (2|3) checks if the input is greater than the counter. Note that this is only false if the input is 1 as else the previous check applies before. • ^ reflects the IP right if the check was false (i.e. input is < 2): • \ reflects the IP down • / reflects the IP left • / reflects the IP down • > enters the 'output zero' part, see below • else the IP is reflected left: • : duplicates the counter (once again) • 0) pulls the input from stack 0 (once again) • : duplicates the input (once again) • # redefines (0|0) • (0 pushes the input to stack 0 (once again) • \ reflects the IP right • _ at (2|2) pops input and counter and pushes input modulo counter • < pops the result and reflects the IP up if 0 (it's a factor): • > reflects the IP right into the 'output zero' part, see below • else (it's no factor) the IP is reflected down: • \ reflects the IP right • / reflects the IP up • / reflects the IP right • # redefines (0|0) • + at (1|0) pushes 1+0=1 • \ reflects the IP down • / reflects the IP left • _ at (0|1) adds the 1 to the counter (counter++) • \ reflects the IP up • / reflects the IP right • \ reflects the IP down • < enters the loop again Now for the 'output zero' part: >#* _#_ • > reflects the IP right • # redefines (0|0) • * at (1|0) pushes 1×0=0 • _ at (3|0) converts to string • # redefines (0|0) • _ at (1|0) prints '0' • then the IP leaves the grid and the program ends • Okay, I need to check out this language. – Esolanging Fruit Mar 7 '18 at 20:05 # Julia, 46 bytes n=int(ARGS[1]);show(sum([n%i==0for i=1:n])==2) An integer is read as the first command line argument using int(ARGS[1]) and the result is printed to STDOUT using show. Primality is checked using trial division with the same formulation as my R answer. Note that the builtin function isprime uses the Miller-Rabin algorithm, which is probabilistic and is thus unsuitable for this challenge. (Thanks to Martin Büttner for pointing that out.) Saved 4 bytes thanks to kvill. • Isn't the use of ARGS[1] possible, as an alternative to readline()? – kvill Sep 13 '15 at 20:38 • @kvill Yeah, I suppose so. Thanks. – Alex A. Sep 13 '15 at 20:44 • ...and even show() is possible for another byte. – kvill Sep 13 '15 at 20:52 • @kvill Good call. – Alex A. Sep 13 '15 at 21:13 # JavaScript (ES6), 43 bytes This is the shortest solution so far that accepts decimal input. Also, it doesn't use regex like the other short solutions. p=n=>--d-1?n%d&&p(n):1;alert(p(d=prompt())) What it uses instead is a recursive function, something that wasn't very useful when you needed to write function and return , but is now very useful because of the => notation. Ungolfed: p=n=> // p=function(n){ return --d-1? // if --d is not 1 (decrement d) n%d&&p(n) // if n divdes d, false, else rerun the function // (d has already been decremented) :1; // else (if d is 1) then true alert(p(d=prompt())) // Use the function on the input // and assign this value to d • One byte shorter: n%d?p(n):0 -> n%d?p(n):0 – user81655 Apr 9 '16 at 11:55 • @user81655 I dont get what you mean. Did you make a typo? it's twice the same thing – Jens Renders Apr 10 '16 at 16:27 • Haha, I did sorry. I meant to write n%d&&p(n). – user81655 Apr 10 '16 at 16:29 • @user81655 Nice! I was looking for things like this but did'nt see it :) – Jens Renders Apr 10 '16 at 16:33 # Excel, 41 bytes =2=SUM(N(0=MOD(A1,ROW(OFFSET(A1,,,A1))))) Takes input from A1. 4 bytes saved thanks to @Joffan! • This didn't work for me, even entered as an array formula (as I think you intended). However this array formula works: =OR(A1=2,AND(MOD(A1,ROW(INDIRECT("2:"&A1-1))))) at 47 bytes. – Joffan Jun 23 '16 at 15:21 • You sure? This works fine for me. Put the input number in A1, put the formula in another cell, and it should work. – Mama Fun Roll Jun 23 '16 at 22:46 • What numbers did you try it with? 9, for example? [BTW I learned an interesting dodge from this, so thanks] – Joffan Jun 23 '16 at 22:48 • Hmm... interesting. This doesn't work for you? – Mama Fun Roll Jun 23 '16 at 23:00 • No, it gives TRUE for 9 - maybe Excel version? I'm using 2010 – Joffan Jun 23 '16 at 23:05 # Cubix, 21 bytes %@\?I:u;>O/)((./0$$?/ Cubix is a 2-dimensional, stack-based esolang. Cubix is different from other 2D langs in that the source code is wrapped around the outside of a cube. Test it online! Note: there's a 50 ms delay between iterations; see the browser console for current progress. ## Explanation (Note: This is somewhat confusing; I'll add a diagram with colored paths when I get a chance.) The first thing the interpreter does is figure out the smallest cube that the code will fit onto. In this case, the edge-length is 2. Then the code is padded with no-ops . until all six sides are filled. Whitespace is removed before processing, so this code is identical to the above: % @ \ ? I : u ; > O / ) ( ( . / 0 \ ) ? / . . . Now the code is run. The IP (instruction pointer) starts out on the top left char of the far left face, pointing east. Here's an overview of the basic commands: • \|/_ are mirrors, and reflect the IP depending on the direction it's traveling. • >v<^ set the direction of the IP unconditionally. • ? turns the IP right if the top item is positive, or left if it's negative. • I inputs an integer (signed or unsigned). • O outputs an integer. • : duplicates the top item. • ; pops an item. • @ ends the program. The first char we encounter is I, which inputs an integer from STDIN. : duplicates this integer. u makes the IP turn right twice, so it ends up on the no-op below u, facing west. Now it enters the main loop. First, we need to check if this integer is less than 2, in which case it's not prime. So we decrement it twice with ((, then check its sign with the ?. If it's less than 0, the IP is turned left, in which case it wraps around to the bottom-left of the bottom panel, facing north. Removing the direction changes from the next bit, we get 0O@, which pushes a 0, outputs as an integer, and terminates the program. If the input is more than 2, the IP is turned right at the ?. Next, the top item is incremented once with ). The IP wraps around to the % at the top-left of the top face, which pushes the modulo of the top two numbers. If the input M modulo any number 1 < N < M is 0, the number is not prime. So we check the sign of the top item with ?. If the top item is now 0, it gets output with O, then @ terminates the program. Otherwise, the IP gets sent down to the ;, which pops the result of % since we have no further use for it. Now it's back where it started, and the loop continues until it takes a different turn at either of the ?s. There is one more case I didn't mention before: if the sign of the top item is 0 at the first ?, that means we've run through every number 1 < N < M, which in turn means the input is prime. Since the top item must be 0, we increment it with ), then output with O and terminate the program with @. I think this program is optimal, but I'm not certain. I'll keep looking to find a better solution. # Mini-Flak, 202 bytes Mini-Flak is a turing complete subset of the Brain-Flak language. (It is currently the smallest know turing complete subset of Brain-Flak) It works exactly like Brain-Flak except the <> and [] nilads and the <...> monad are forbidden. (({})){(({}[()])(((({}({}))[({}[{}])]([()]()))({({}[()((({}()[({})])){{}((({}({})))[{}])}{})]{})}({}{}({})[{}]))[{}]))[{}])}{}{}({}({})[{}()()()]){({}[()((({})){(({}{}(({})))[{}])}{}({}({})[{}]))]{})}{} Try it online ## Explanation The reason <...> is banned in mini-flak is that it is equivalent to (...)[{}]. So to start this explanation I am going to use this translation in reverse to create an equivalent Brain-Flak program for increased readability for anyone who is already familiar with Brain-Flak. (({})) { (({}[()])< ((({}({}))[({}[{}])]([()]()))<{({}[()((({}()[({})])){{}(<({}({}))>)}{})]{})}({}{}<{}>)>) >) }{}{}({}<{}>[()()()]) {({}[()((({})){(<{}{}(({}))>)}{}({}<{}>))]{})}{} This program has two main parts, performs the modulus on the input for every number smaller than than the input and leaves them in a stack. (({})) { (({}[()])< ((({}({}))[({}[{}])]([()]()))<{({}[()((({}()[({})])){{}(<({}({}))>)}{})]{})}({}{}<{}>)>) >) }{}{} This uses a old version of modulo I wrote ({}(<()>)){({}[()((({}()[({})])){{}(<({}({}))>)}{})]{})}({}{}<{}>) The second part ands together all of the results of the last part ({}<{}>[()()()]) {({}[()((({})){(<{}{}(({}))>)}{}({}<{}>))]{})}{} If any one of the results is zero the result of all the ands will be zero otherwise it will be truthy and the number will be prime. # GNU sed -r, 19 18 /^(1|(11+)\2+)$/c↲
(that last ↲ isn't a literal character; it stands for a final newline).
Takes unary input on stdin. Anything that can be represented by two or more 1s, two or more times is a composite number and replaced by an empty line. Prime numbers are all truthy and left as-is.
There's a slight catch in that this would classify 1 as a prime (it's actually neither prime nor composite), so there are an extra 4 bytes (1|) to handle that case.
If you're willing to accept empty string as a false "output", then we could use d instead of c↲ for a further 1 byte saving.
• I should have looked through the other answers before posting this - I thought I was being original and clever! – Toby Speight Sep 15 '15 at 14:40
# dc, 27 bytes
?dd[d1-d1<f*]dsfxr/r2r|p
How it works (example stack for input 7):
? 7 push input
dd 7 7 7 dup
[d1-d1<f*]dsf 7 7 7 {fact} f = factorial macro
x 7 7 5040 execute
r 7 5040 7 swap
/ 7 720 divide
r2r 720 2 7 swap, 2, swap
| 1 modular exp: 720^2 mod 7
p print output
The factorial macro breaks for input 1, or something, but it turns out not to matter, and the output is correct.
# Mathematica, 33 bytes
Print[Mod[(#-1)!^2,#]>0&@Input[]]
or
Print[Length@Divisors@Input[]==2]
Mathematica has a built-in PrimeQ as well, but I believe it uses a probabilistic test.
• What is the measurable difference between having and not having "Print@" at the front of that expression? – Eric Towers Sep 13 '15 at 23:56
• @EricTowers without it, it will only print when run inside a notebook, i.e. in a REPL environment. I wouldn't consider that a full program. If you invoke the code from the command line, you need the Print@ to actually get any output. – Martin Ender Sep 14 '15 at 6:46
• According to the documentation, PrimeQ is a deterministic algorithm. I suspect the challenge text meant the other meaning (100% correct result). – DanaJ Sep 14 '15 at 12:00
• You could save a byte with Echo. – Pavel Nov 28 '16 at 19:23
• @Pavel unfortunately, that also prints >> to STDOUT. – Martin Ender Nov 28 '16 at 19:33
# Turing Machine Code, 2043 bytes
As usual, I'm using the rule table syntax defined here. Requires unary input.
0 * * r 0
0 _ _ l c
a _ 1 r b
a 0 1 r b
a 1 0 l a
b * * r b
b _ _ r 0
c 1 _ l d
c _ _ * y
d * * l d
d _ _ l a
y _ _ l y
y * * l z
z * * l z
z _ _ r X
X 1 1 r Y
Y _ _ * f
Y * * l Z
X * * l 1
Z * * l 1
1 * a r 2
2 _ b l 3
2 * * r 2
3 a a r 4
3 x x r 4
3 y y r 4
3 * * l 3
4 0 x r 5
4 1 y r 5y
4 b b l 9
9 x 0 l 9
9 y 1 l 9
9 a a r A
5 b b r 6
5 * * r 5
5y b b r 6y
5y * * r 5y
6 _ 0 l 3
6 * * r 6
6y _ 1 l 3
6y * * r 6y
A _ c l 11
A * * r A
11 b b r 12
11 x x r 12
11 y y r 12
11 * * l 11
12 0 x r 13x
12 1 y r 13y
12 c c l B
13x _ 0 l 11
13x * * r 13x
13y _ 1 l 11
13y * * r 13y
B x 0 l B
B y 1 l B
B b b r 21
21 _ d l 22
21 * * r 21
22 1 0 r 23
22 0 1 l 22
23 d d r E
23 * * r 23
E c c r 51
E x x r 51
E y y r 51
E * * l E
51 0 x r Kx
51 1 y r Ky
51 d d l 53
Kx _ 0 l E
Kx * * r Kx
Ky _ 1 l E
Ky * * r Ky
53 x 0 l 53
53 y 1 l 53
53 c c r F
F _ _ l 61
F * * r F
61 1 0 l 62
61 0 1 l 61
61 d d r 70
62 c c l 63
62 * * l 62
63 1 0 r F
63 0 1 l 63
63 b b r 80
70 _ _ l 71
70 * * r 70
71 d d l 72
71 * _ l 71
72 c c * E
72 * * l 72
80 _ _ l Za
80 * * r 80
Za 0 1 r Zb
Za 1 0 l Za
Za d d r Zk
Zb _ _ l Zc
Zb * * r Zb
Zc x x l Zc
Zc y y l Zc
Zc d d r A0
Zc 0 x l $Zc 1 y l$y
$d d l Ze$ * * l y d d l @
$y * * l$y
Ze 0 x r Zf
Ze 1 1 r Zk
Ze x x l Ze
Ze y y l Ze
@ 1 y r Zf
@ 0 0 r Zk
@ x x l @
@ y y l @
Zf _ _ l Zc
Zf * * r Zf
Zk _ _ l Zm
Zk * * r Zk
Zm d d l Zn
Zm * _ l Zm
Zn x 0 l Zn
Zn y 1 l Zn
Zn 0 0 r Zo
Zn 1 1 r Zo
Zo d d l 82
Zo * * r Zo
82 1 0 l 83
82 0 1 l 82
83 c c l 84
83 * * l 83
84 b b l 85
84 * _ l 84
85 a a r 86
85 x x r 86
85 y y r 86
85 * * l 85
86 0 x r &x
86 1 y r &y
86 b b l 90
&x b b r Lx
&x * * r &x
&y b b r Ly
&y * * r &y
Lx _ 0 l 85
Lx * * r Lx
Ly _ 1 l 85
Ly * * r Ly
90 x 0 l 90
90 y 1 l 90
90 a a r 91
91 c c r 51
91 * * r 91
A0 _ _ l A1
A0 * * r A0
A1 d d l A2
A1 * _ l A1
A2 x 0 l A2
A2 y 1 l A2
A2 c c r A3
A3 d d l B0
A3 * * r A3
B0 1 0 r B1
B0 0 1 l B0
B1 d d l B2
B1 * * r B1
B2 1 0 r f
B2 0 1 l B2
B2 c c r t
t * * r t
t _ _ l t2
t2 * _ l t2
t2 _ 1 * halt
f * * r f
f _ _ l f2
f2 * _ l f2
f2 _ 0 * halt
• Nice! I imagine the state names could be golfed a bit more, though... – DLosc Oct 27 '15 at 23:23
• I'm still surprised this isn't actually the longest answer. (The Mornington Crescent answer is longer, which I guess makes some sense.) – SuperJedi224 Oct 29 '15 at 1:29 | 17,535 | 55,600 | {"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} | 2.53125 | 3 | CC-MAIN-2020-10 | latest | en | 0.894125 |
https://www.coursehero.com/file/6848525/LectureNotes05Print/ | 1,498,421,045,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320570.72/warc/CC-MAIN-20170625184914-20170625204914-00554.warc.gz | 834,851,919 | 24,749 | LectureNotes05Print
# LectureNotes05Print - IEOR E4731 Credit Risk and Credit...
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Unformatted text preview: IEOR E4731: Credit Risk and Credit Derivatives Lecture 05: Reduced form models. Notes originally written by Prof. Rama Cont Instructor: Xuedong He Spring, 2012 1 / 40 Reduced form models I Structural models have strong economic meaning but are not applicable I In practice, people use reduced form models I In reduced form models, people do not care about why defaults happen. They just model default times as random times. I Drawback of reduced form models: does not tell why defaults happen I Advantage of reduced form models: I Easy to implement I Easy to incorporate any market information I Can use the machinery of default-free term-structure modeling. I In view of risk-neutral pricing, let us start from specifying information sets. 2 / 40 Information sets: modeled via filtrations I {H t } t ≥ : we observe default or no default only. Information at time t is: H t := σ { 1 τ>s , ≤ s ≤ t } I {F t } t ≥ : market information other than the default time (stock prices, interest rates, news,...) I {G t } t ≥ : full information, i.e., G t := F t W H t . I Example: Merton model. F t is the historical firm value up to time t . H t ⊂ F t , thus G t = F t . I Example: F t = {∅ , Ω } , i.e., no market information. In this case, G t = H t . 3 / 40 Risk Neutral Pricing I Starting point: the market value of a terminal cash flow H paid at T is given as Π t ( H ) = E Q h e- R T t r ( u ) du H | G t i where Q is the risk neutral pricing measure (with e R t r ( u ) du ,t ≥ to be the denominator) I Example: for a zero-coupon zero-recovery bond that only defaults at maturity T , the payoff is H = 1 τ>T . I It is critical to compute the conditional expectation. 4 / 40 Conditional expectation I Formal definition of conditional expectation: let X be a random variable on a probability space (Ω , F , P ) and G be a sub- σ-algebra. Then, the conditional expectation Z := E [ X | G ] is a G-measurable random variable defined via E [ Z 1 A ] = E [ X 1 A ] , ∀ A ∈ G . I Example: let A 1 ,...,A n be a partition of Ω and G := { A 1 ,...,A n } , then E [ X | G ] = n X i =1 E [ X | A i ] 1 A i , where E [ X | A i ] := E [ X 1 A i ] P ( A i ) 5 / 40 Conditional expectation (cont’d) I For any B ∈ F , P [ B | G ] := E [ 1 B | G ] . I Example: if G := { A 1 ,...,A n } , then P [ B | G ] = n X i =1 P [ B | A i ] 1 A i where P ( B | A i ) := P ( B ∩ A i ) P ( A i ) I For any random variable X and Y , E [ X | Y ] := E [ X | σ ( Y )] where σ ( Y ) is the σ-algebra generated by Y , i.e., the information one can obtain from Y . I Example: if Y = 1 A , then σ ( Y ) = { A,A c } . As a result, E [ X | 1 A ] = E [ X | A ] 1 A + E [ X | A c ] 1 A c . 6 / 40 Compute Conditional Expection We start from the simplest information structure: no market information, i.e., F t = {∅ , Ω } . In this case, G t = H t ,t ≥ ....
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## This note was uploaded on 03/14/2012 for the course IEOR 4731 taught by Professor Hexuedong during the Spring '12 term at Columbia.
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Ask a homework question - tutors are online | 1,074 | 3,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.870513 |
https://www.convertunits.com/from/cubic+inch/hour/to/cubic+inch/minute | 1,624,628,466,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487630175.17/warc/CC-MAIN-20210625115905-20210625145905-00439.warc.gz | 634,039,781 | 16,592 | ## ››Convert cubic inch/hour to cubic inch/minute
cubic inch/hour cubic inch/minute
How many cubic inch/hour in 1 cubic inch/minute? The answer is 60.
We assume you are converting between cubic inch/hour and cubic inch/minute.
You can view more details on each measurement unit:
cubic inch/hour or cubic inch/minute
The SI derived unit for volume flow rate is the cubic meter/second.
1 cubic meter/second is equal to 219685475.57572 cubic inch/hour, or 3661424.5929287 cubic inch/minute.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between cubic inches/hour and cubic inches/minute.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of cubic inch/hour to cubic inch/minute
1 cubic inch/hour to cubic inch/minute = 0.01667 cubic inch/minute
10 cubic inch/hour to cubic inch/minute = 0.16667 cubic inch/minute
20 cubic inch/hour to cubic inch/minute = 0.33333 cubic inch/minute
30 cubic inch/hour to cubic inch/minute = 0.5 cubic inch/minute
40 cubic inch/hour to cubic inch/minute = 0.66667 cubic inch/minute
50 cubic inch/hour to cubic inch/minute = 0.83333 cubic inch/minute
100 cubic inch/hour to cubic inch/minute = 1.66667 cubic inch/minute
200 cubic inch/hour to cubic inch/minute = 3.33333 cubic inch/minute
## ››Want other units?
You can do the reverse unit conversion from cubic inch/minute to cubic inch/hour, or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 517 | 1,987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | longest | en | 0.690912 |
https://answers.yahoo.com/question/index?qid=20171008151818AAHd1LQ | 1,575,647,695,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540488870.33/warc/CC-MAIN-20191206145958-20191206173958-00427.warc.gz | 272,806,337 | 18,401 | # Balance these reactions and please show how you did it: a) CH4 + NO2 → N2 + CO2 + H2O (b) Zn + Cr2O72- + H+ → Zn2+ + Cr3+ + H2O?
Relevance
• 2 years ago
a)
CH4 + 2NO2 = N2 + CO2 + 2H2O
b)
First write down the two half equations.
Notice (1) we insert electrons to balance the two half equations
(2) We make up 'water' ion the RHS to balance all the oxygens in the dichromate
(3) We then 'back' balance 'H^+' to balance all the hydrogens in water.
(4) We then balance the charges by putting in electrons on the LHS.
Zn = Zn^2+ + 2e^-
Cr2O7^2- + 14H^+ + 6e^- = 2Cr^3+ + 7H2O
We now multiply the two half equations to balance the number of electrons in each equation.
Hence
3Zn = 3Zn^2+ + 6e^-
Cr2O7^2- + 14H^+ + 6e^- = 2Cr^3+ + 7H2O
Add the two half equations, this will remove the electrons .
Hence
3Zn + Cr2O7^2- + 14H^+ = 3Zn^2+ + 2Cr^3+ + 7H2O
Done!!!!
Hope that helps!!!! | 348 | 895 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-51 | latest | en | 0.661795 |
https://www.educationquizzes.com/11-plus/maths/multiplication-and-division-very-easy/ | 1,718,934,845,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00299.warc.gz | 635,833,091 | 9,878 | UK USIndia
Every Question Helps You Learn
10,000 ÷ 1,000 = 10.
# Multiplication and Division (Very Easy)
Division and multiplication is dividing and multiplying numbers. This is the first of our 11-plus Maths quizzes on multiplication and division. You should be able to do most of this stuff in your head. Knowing your times tables is always useful when it comes to multiplying and dividing.
You may be great at addition and subtraction, but not so good at multiplication and division. It's definitely worth spending some time practising these valuable skills. Simple multiplication and division are used by us near-enough every day. You don't want to find yourself out of pocket because you couldn't do an easy division calculation.
Always remember in tests and exams to read the questions carefully to make sure you have understood what is required. Try these quizzes and see how good you are.
1.
What is the correct answer to the given calculation?
100 × 24
2,400
240
24
2.4
To multiply 24 by 100, multiply the non zero numbers and just stick on the two zeros at the end: 100 × 24 = 1 × 24 , and 24 + 00 = 2,400. You should be able to do this in your head
2.
What is the correct answer to the given calculation?
250 ÷ 10
0.25
2.5
0.025
25
To divide 250 by 10, just remove the last zero from the number and you are done: 250 ÷ 10 = 25. You should be able to do this in your head
3.
What is the correct answer to the given calculation?
3 doubled
9
6
27
81
'Double' means multiply by 2
4.
What is the correct answer to the given calculation?
4 trebled
12
8
16
64
'Treble' means multiply by 3: triple = treble
5.
What is the correct answer to the given calculation?
8
4
16
32
6.
What is the correct answer to the given calculation?
1,000 × 4.5
45
450
4.5
4,500
To multiply 4.5 by 1,000, just move the decimal point 3 places to the right and add in a zero for each place position if there are no digits in those place positions in the number: 1,000 × 4.5 = 4,500. You should be able to do this in your head
7.
What is the correct answer to the given calculation?
85.60 ÷ 100
8.560
0.08560
0.8560
856
To divide 85.60 by 100 just move the decimal point 2 places to the left and put in a zero to show that there are no whole numbers: 85.60 ÷ 100 = 0.8560
8.
What is the correct answer to the given calculation?
300 × 12
3.6
3,600
36
360
To multiply 12 by 300, multiply the non zero numbers and just stick on two zeros at the end: 300 × 12 = 3 × 12 , and 36 + 00 = 3,600. You should be able to do this in your head
9.
What is the correct answer to the given calculation?
10,000 × 100
10,000
100,000
1,000,000
10,000,000
To multiply 10,000 by 100, add together the number of zeros in the two numbers, then write 1 followed by the sum of the zeros: 10,000 × 100 = 1,000,000. Your final answer has six zeros because the sum of the zeros in the two numbers equals six. You should be able to do this in your head
10.
What is the correct answer to the given calculation?
10,000 ÷ 1,000
100
1,000
0.1
10
To divide 10,000 by 1,000, cancel as many zeros in 10,000 as there are in 1,000: 10,000 ÷ 1000 = 10. Your final answer has 1 zero because the 3 zeros in 10,000 cancelled with the 3 zeros in 1,000.You should be able to do this in your head
Author: Frank Evans | 950 | 3,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-26 | latest | en | 0.924758 |
https://rdrr.io/cran/gofCopula/man/tests_CheckTime.html | 1,527,033,044,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864999.62/warc/CC-MAIN-20180522225116-20180523005116-00472.warc.gz | 621,869,200 | 14,979 | # tests_CheckTime: Combining function for tests In gofCopula: Goodness-of-Fit Tests for Copulae
## Description
The computation of a gof test can take very long, especially when the number of bootstrap rounds is high. The function `gofCheckTime` computes the time which the estimation most likely take.
## Usage
```1 2 3 4``` ```gofCheckTime(copula, x, test, M = 1000, MJ = 10000, print.res = T, margins = "ranks", dispstr = "ex", param = 0.5, param.est = T, df = 4, df.est = T, m = 1, delta.J = 0.5, nodes.Integration = 12, m_b = 0.5, zeta.m = 0, b_Rn = 0.05, processes = 1) ```
## Arguments
`copula` A character vector which indicates the copula to test for. `x` A matrix containing the data. `test` A character vector which indicates the test to use. `M` The number of bootstrapping rounds which shall be later taken in the estimation. `MJ` Just for the test gofKernel. Size of bootstrapping sample. `print.res` Logical which defines if the resulting time shall be printed or given as value. Default is TRUE. `margins` Specifies which estimation method shall be used in case that the input data are not in the range [0,1]. The default is `"ranks"`, which is the standard approach to convert data in such a case. Alternatively can the following distributions be specified: `"beta"`, `"cauchy"`, Chi-squared (`"chisq"`), `"f"`, `"gamma"`, Log normal (`"lnorm"`), Normal (`"norm"`), `"t"`, `"weibull"`, Exponential (`"exp"`). `dispstr` A character string specifying the type of the symmetric positive definite matrix characterizing the elliptical copula. Implemented structures are "ex" for exchangeable, "ar1" for AR(1), "toep" for Toeplitz, and "un" for unstructured, see package `copula`. `param` The copulae parameters to use for each test, if it shall not be estimated. `param.est` Shall be either `TRUE` or `FALSE`. `TRUE` means that `param` will be estimated. `df` The degrees of freedom, if not meant to be estimated. Only necessary if tested for `"t"`-copula. `df.est` Indicates if `df` shall be estimated. Has to be either `FALSE` or `TRUE`, where `TRUE` means that it will be estimated. `m` Length of blocks. Only necessary if the test `gofPIOSTn` is part of `testset`. `delta.J` Scaling parameter for the matrix of smoothing parameters. Only necessary if the test `gofKernel` is part of `testset`. `nodes.Integration` Number of knots of the bivariate Gauss-Legendre quadrature. Only necessary if the test `gofKernel` is part of `testset`. `m_b` The power of the statistic. Only necessary if the test `gofRn` is part of `testset`. `zeta.m` The adjustment parameter. Only necessary if the test `gofRn` is part of `testset`. `b_Rn` The bandwidth for the estimation of the first-order partial derivatives based on the empirical copula. Only necessary if the test `gofRn` is part of `testset`. `processes` The number of parallel processes which are performed to speed up the bootstrapping. Shouldn't be higher than the number of logical processors.
## Details
The function estimates the time which the entire gof test will take.
## Examples
```1 2 3``` ```data(IndexReturns) gofCheckTime("normal", IndexReturns[c(1:50),c(1:2)], "gofRosenblattSnC", M = 10000) ```
gofCopula documentation built on May 29, 2017, 6:33 p.m. | 902 | 3,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-22 | latest | en | 0.724757 |
https://schoolio.com/product/schoolio-math-fullyearbundle-grade8/ | 1,717,093,619,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971670239.98/warc/CC-MAIN-20240530180105-20240530210105-00659.warc.gz | 433,939,292 | 67,905 | Not sure where to start? Get help!
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https://numbermatics.com/n/4553195/ | 1,607,219,064,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141753148.92/warc/CC-MAIN-20201206002041-20201206032041-00156.warc.gz | 412,113,335 | 6,195 | # 4553195
## 4,553,195 is an odd composite number composed of four prime numbers multiplied together.
What does the number 4553195 look like?
This visualization shows the relationship between its 4 prime factors (large circles) and 24 divisors.
4553195 is an odd composite number. It is composed of four distinct prime numbers multiplied together. It has a total of twenty-four divisors.
## Prime factorization of 4553195:
### 5 × 172 × 23 × 137
(5 × 17 × 17 × 23 × 137)
See below for interesting mathematical facts about the number 4553195 from the Numbermatics database.
### Names of 4553195
• Cardinal: 4553195 can be written as Four million, five hundred fifty-three thousand, one hundred ninety-five.
### Scientific notation
• Scientific notation: 4.553195 × 106
### Factors of 4553195
• Number of distinct prime factors ω(n): 4
• Total number of prime factors Ω(n): 5
• Sum of prime factors: 182
### Divisors of 4553195
• Number of divisors d(n): 24
• Complete list of divisors:
• Sum of all divisors σ(n): 6100704
• Sum of proper divisors (its aliquot sum) s(n): 1547509
• 4553195 is a deficient number, because the sum of its proper divisors (1547509) is less than itself. Its deficiency is 3005686
### Bases of 4553195
• Binary: 100010101111001111010112
• Base-36: 2PL9N
### Squares and roots of 4553195
• 4553195 squared (45531952) is 20731584708025
• 4553195 cubed (45531953) is 94394947834655889875
• The square root of 4553195 is 2133.8216888953
• The cube root of 4553195 is 165.7443565907
### Scales and comparisons
How big is 4553195?
• 4,553,195 seconds is equal to 7 weeks, 3 days, 16 hours, 46 minutes, 35 seconds.
• To count from 1 to 4,553,195 would take you about eleven weeks!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 4553195 cubic inches would be around 13.8 feet tall.
### Recreational maths with 4553195
• 4553195 backwards is 5913554
• The number of decimal digits it has is: 7
• The sum of 4553195's digits is 32
• More coming soon! | 688 | 2,394 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-50 | latest | en | 0.851159 |
http://stackoverflow.com/questions/11035259/basic-incrementation | 1,448,866,916,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398461113.77/warc/CC-MAIN-20151124205421-00248-ip-10-71-132-137.ec2.internal.warc.gz | 213,387,599 | 22,507 | # Basic Incrementation
i know this is a very basic question BUT.
I understand the concept behind. n++, ++n, n--, --n. HOWEVER
``````public static void main(String[] args){
int count = 1;
for(int i =1;i<=10;++i){
count = count * i;
System.out.print(count);
}
}
``````
So it will print: 1 2 3 4 5 6 7 8 9 10.
My Question is. Why if i is incremented as ++i isnt i then treated as 2, instead of 1. Inst the point of ++i, to increment i before it's manipulated by another operation?
-
The increment doesn't happen until AFTER the code on the inside of the loop has executed. You initialized to 1. – Chris Dargis Jun 14 '12 at 14:30
Also, that should print 1,2,6,24, etc since you're multiplying. – Thomas Jun 14 '12 at 14:34
Ummm but count never changes..so your just multiplying what i is by 1. – user1419012 Jun 14 '12 at 15:03
– codingbadger Jun 15 '12 at 7:07
@user1419012 `count = count * i;` will change count and it will keep going up unless you move the `count = 1;` inside the loop to reset it each time. – Thomas Jun 15 '12 at 17:06
Is the point of ++i, to increment i before it's manipulated by another operation?
The difference between `++i` and `i++` only matters when it's used as part of a bigger expression, e.g.
``````int j = ++i; // Increment then use the new value for the assignment
int k = i++; // Increment, but use the old value for the assignment
``````
In this case the operation occurs at the end of each iteration of the loop, on its own. So your loop is equivalent to:
``````int count = 1;
// Introduce a new scope for i, just like the for loop does
{
// Declaration and initialization
int i = 1;
// Condition
while (i <= 10) {
count = count * i;
System.out.print(count);
// Now comes the final expression in the for loop "header"
++i;
}
}
``````
Now changing `++i` to `i++` at the end there isn't going to make a difference at all - the value of the expression isn't used for anything.
-
Ok so to calarify int j = ++i; // Increment, but use the old value for the assignment int k = i++; // Increment then use the new value for the assignment.....Meanig that j= i+1 but k=j? – user1419012 Jun 14 '12 at 14:38
@user1419012: Assuming each statement is separated (rather than considering the two together), `j` will be 1 less than `i` after the assigment, but `k` and `i` would have the same value. – Jon Skeet Jun 14 '12 at 14:48
does j have one more than i? I agree that k will have the same value as i. – user1419012 Jun 14 '12 at 15:05
@Palak: If you believe an answer is wrong, please downvote/comment rather than just editing the answer away from the original intention. (In this case yes, you were right for the first part, but it's still appropriate for the original author to fix this rather for an editor to do so. Edits are for clarifications of typos etc, rather than changing the semantics of the post.) – Jon Skeet Apr 16 at 9:23
Okay @Jon Point noted. – Palak Apr 16 at 9:29
The increment isn't called until after the first iteration of the `for` loop.
While it's true that
``````j = i++;
k = ++i;
``````
return different results, think of the `++i` in this context as a standalone line called at the end of every `for` loop.
-
You want to use i++ which is a post increment. ++i is called a preincrement and the difference is precisely as you have pointed out.
-
I believe ++i is actually more efficient for memory when used in such a fashion. The end result is the same, however. – SimplyPanda Jun 14 '12 at 14:30
@user1432824 Are you saying the loop wouldn't work with pre-increment? – Dave Newton Jun 14 '12 at 14:32
No Dave i know both will work..i was just just wondering what the difference between the two in this case was. :) – user1419012 Jun 14 '12 at 14:36
@user1419012 Different user, mate. – SimplyPanda Jun 14 '12 at 14:40
@SimplyPanda: The compiler will probably optimize out the extra variable. – Brian Jun 14 '12 at 15:43
In this case `++i` happens at the end of the loop, it increments and then checks if the new value still meets the termination condition.
Also, won't the output be:
``````count i
1 * 1 = 1
1 * 2 = 2
2 * 3 = 6
6 * 4 = 24
``````
etc.
-
The for loop you wrote is same as:
``````i = 1;
while(i<=10) {
count = count * i;
System.out.print(count);
i = i + 1;
}
``````
So that's why!
-
For i = 0 and While i < 1= 10, print i, and then pre-increment i. (++i/i++ doesn't make a difference here).
Here try this though:
``` int i=1; while(i <= 10) System.out.print(++i); i = 1; while (i <= 10) System.out.print(i++); ```
-
As an addition to the other answers, the historical reason for preferring `for(int i=1;i<=10;++i)` over `for(int i=1;i<=10;i++)` is that `++i` does not need to store the old value of `i` in an extra variable. Thus, `++i` is faster than `i++`, though the speed improvement is negligible. On modern compilers this speed improvement is done as an optimization, so the two pieces yield the same compiler output. However, since `++i` is always as fast or faster (e.g., on old C++ compilers) than `i++`, many experienced programs always use `++i` within loops.
As other answers have stated, both pieces of code are functionally equivalent (in the case of a for loop).
- | 1,513 | 5,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2015-48 | longest | en | 0.864372 |
https://se.mathworks.com/matlabcentral/cody/problems/506-back-to-basics-25-valid-variable-names/solutions/106871 | 1,611,091,858,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519784.35/warc/CC-MAIN-20210119201033-20210119231033-00565.warc.gz | 569,567,997 | 16,916 | Cody
# Problem 506. Back to basics 25 - Valid variable names
Solution 106871
Submitted on 5 Jul 2012 by Søren
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 'foo'; y_correct = true; assert(isequal(valid_vars(x),y_correct))
2 Pass
%% x = 'while'; y_correct = false; assert(isequal(valid_vars(x),y_correct))
3 Pass
%% x = '1foo'; y_correct = false; assert(isequal(valid_vars(x),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 173 | 652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-04 | latest | en | 0.54425 |
https://www.physicsforums.com/threads/trigonometry-puzzle.160915/ | 1,585,431,470,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493120.15/warc/CC-MAIN-20200328194743-20200328224743-00107.warc.gz | 1,127,003,930 | 16,290 | # Trigonometry puzzle
Homework Helper
Gold Member
Dearly Missed
## Main Question or Discussion Point
Trigonometry "puzzle"
If we know the perimeter P of a triangle, a side S of it, and the height H down on the line going through S, then we have uniquely determined the triangle.
It is not wholly trivial to derive the expressions for the other two sides in terms of P,S and H.
Related General Discussion News on Phys.org
I'm forgetting my algebra, but:
1) Essentially, we are given an ellipse, with the focii seperated by S, a major axis of P, and if expressed as y=f(x), the corresponding y value of H for which we must solve.
2) The ellipse as a function (not purely as a function of x) might be:
sqrt((x+S/2)^2 + y^2) + sqrt((x-S/2)^2 + y^2) = P - S
3) Solve for x, given that y=H:
sqrt((x+S/2)^2 + H^2) + sqrt((x-S/2)^2 + H^2) = P - S
4) With x in hand, it becomes trivial. The other sides would be:
Side 1 = sqrt((x+S/2)^2+y^2)
Side 2 = sqrt((x-S/2)^2+y^2)
DaveE
Last edited:
Homework Helper
Gold Member
Dearly Missed
The neat touch is, of course, to go over to a coordinate representation of the end points and utilize the properties of an ellipse.
The "brute force" approach to directly relate the SIDE LENGTHS to each other, say with a couple of Pythagorases will lead to ugly non-linear equations.
EDIT
NOTE:
Your initial formula is wrong. On your right-hand side, it should say P-S, not P!
(The sum of the two other sides is constant)
Last edited:
The neat touch is, of course, to go over to a coordinate representation of the end points and utilize the properties of an ellipse.
Heh, that's all well and good if you remember your ellipse properties, but for those of us who're 14 years out of their geometry classes, we're stuck with brute force :) (short of going and looking up all those old properties)
The "brute force" approach to directly relate the SIDE LENGTHS to each other, say with a couple of Pythagorases will lead to ugly non-linear equations.
Out of curiosity, how *would* you go about solving that ugly equation for x? Anyone?
Your initial formula is wrong. On your right-hand side, it should say P-S, not P!
Ooops, fixed!
DaveE
Homework Helper
Gold Member
Dearly Missed
Hmm..you gave the simple ellipse approach in your initial post.
Now, as for solving for x, the simplest way is like this:
$$\sqrt{(x-\frac{S}{2})^{2}+H^{2}}=(P-S)-\sqrt{(x+\frac{S}{2})^{2}+H^{2}}$$
Square this expression, gaining:
$$(x-\frac{S}{2})^{2}+H^{2}=(x+\frac{S}{2})^{2}+(P-S)^{2}-2(P-S)\sqrt{(x+\frac{S}{2})^{2}+H^{2}}$$
Simplify to:
$$\sqrt{(x+\frac{S}{2})^{2}+H^{2}}=\frac{xS}{(P-S)}+\frac{(P-S)}{2}$$
Re-square:
$$(x+\frac{S}{2})^{2}+H^{2}=\frac{S^{2}}{(P-S)^{2}}x^{2}+xs+\frac{(P-S)^{2}}{4}$$
Simplify this to:
$$(1-\frac{S^{2}}{(P-S)^{2}})x^{2}=\frac{P(P-2S)}{4}-H^{2}$$
From which we gain the positive solution:
$$x=\frac{(P-S)}{2}\sqrt{1-\epsilon},\epsilon=\frac{4H^{2}}{P(P-2S)}$$
Cool question man. | 921 | 2,934 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2020-16 | longest | en | 0.892144 |
https://physics.stackexchange.com/questions/384582/what-is-this-thermodynamic-property-called-the-thermodynamic-state-is-independe/384590 | 1,631,841,986,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053918.46/warc/CC-MAIN-20210916234514-20210917024514-00473.warc.gz | 482,279,186 | 40,127 | # What is this thermodynamic property called: the thermodynamic state is independent from path/process
I'm trying to comprehend the discussion here about the internal energy of an ideal gas being only dependant to its temperature $dU=nc_vdT$. I have the feeling that there are some assumptions missing.
What I see from the proofs, for example here, is that they assume they can substitute any thermodynamic process with other incremental processes as far as the consequent process has the same start-end. To understand better what I'm talking about you may consider the conservatives forces where the total work done by the force is independent from the path.
So my question is that if all thermodynamic process with the same start-end are homeomorphic? meaning that this is an intrinsic feature of a thermodynamic system. Or there are certain assumptions to be met? for example has reversibility anything to do here?
• This idea is normally considered in terms of individual quantities, rather than entire systems. In thermodynamics quantities which are path independent are known as functions of state. Feb 6 '18 at 11:54
• So basically the "state" of a thermodynamic system is defined by a subset of all "state variables" and it is independent from the processes leading to that "state"?
– Foad
Feb 6 '18 at 12:12
• Essentially yes Feb 6 '18 at 12:16
## 1 Answer
A few points that may help…
1. For any liquid or gas, any property (for example temperature, viscosity) can be expressed in terms of two independent variables. p and V are usually chosen. We say that the particular values of p and V determine the state of the fluid, and that temperature, viscosity etc are functions of state. [We don't have to use p and V as independent variables; we might well choose T and V.] The idea can be extended to solids and more complicated systems, but these usually need more than two independent variables.
2. It isn't obvious that temperature (determining which way, if at all, heat will flow between two bodies put in contact) is a function of each body's state. Showing that it is, is the point of the Zeroth Law of Thermod.
3. Likewise, it had to be shown that for any body there is a function of state, U, internal energy, such that $\Delta U = \text{Heat in – Work out}.$ This was done by Joule and others in the nineteenth century. It means that if we go from point A to point B on a pV diagram, $\Delta U$ is independent of route.
4. So U is a function of state. In general it depends on T and (say) V, but part of the definition of an ideal gas is that U depends only on T. [T itself can be expressed in terms of p and V by means of the so-called equation of state for the gas.]
• to be sure that I have understood correctly : 1. the thermodynamic state of any fluid (i.e. liquid and gas) in equilibrium (set of all imaginable properties) can be determined with a subset of two of all state variables (V, P, T, S, U, H ...) except for the ideal gases where T and U are linearly related 2. Internal energy U being a state variable is an empirical conclusion (Joule's law)
– Foad
Feb 6 '18 at 12:38
• That's it. Well-expressed, if I may say so. Originally the existence of the state variable, $U$, needed establishing empirically because (1) the modern concept of $energy$ was only just emerging, (2) the microscopic structures of systems weren't really known – even atoms were controversial. These days, we're much more confident about substances having a microscopic structure (kinetic theory of gases etc) and this makes it seem obvious that a gas (or whatever) must have an internal energy dependent on its state. Feb 6 '18 at 12:47
• ok. I need some time to understate this and come back. thanks a lot.
– Foad
Feb 6 '18 at 12:57
• A B Pippard: $Classical Thermodynamics$ deals clearly and logically with the fundamentals. Almost entirely on a macroscopic level. No kinetic theory or statistical mechanics. Feb 6 '18 at 13:05
• Engineers regard the relationship between internal energy and temperature of ideal gases as not necessarily being linear. Feb 6 '18 at 16:05 | 948 | 4,088 | {"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} | 3.171875 | 3 | CC-MAIN-2021-39 | latest | en | 0.959259 |
https://www.bloomberg.com/news/articles/2012-02-10/just-start | 1,495,746,365,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608416.96/warc/CC-MAIN-20170525195400-20170525215400-00568.warc.gz | 846,001,777 | 19,706 | # Just Start!
You’re smart. Creative. And terrific at solving challenges. So why does it seem that the number of things you can’t figure out is increasing?
The problem may not be you. It could be the way you were taught to think.
From kindergarten on, we’ve learned Prediction Reasoning—a way of thinking based on the assumption that the future will be pretty much like the past.
Demographics are one simple example of how Prediction Reasoning (from now on, let’s just call it Prediction) works. You can calculate with a high degree of confidence what the world population will be in 2050 because you already know a lot of things: how many people are alive today (about 6.8 billion); distribution by age; the number of people in their 20s and 30s, the age when most people decide to have children; and recent trends in population growth—it’s slowing as people decide to have fewer kids.
Studying all this data—and much, much more—you can, as the United Nations did recently, say with a reasonable amount of certainty that by the year 2050 there will be about 8.9 billion people on the planet. Armed with that data you can also make a number of fairly accurate estimates of such things as how many diapers will need to be produced, how many gallons of water those 8.9 billion people will drink each day, and how much the U.S. will need to pay out in Social Security benefits when we reach the midpoint of this century.
Because it works so well in these kinds of situations—and countless others you can think of—we (like you) became accustomed to using Prediction all the time.
And yet …
Not everything can be foreseen (and thus predicted). Want to know if the cute guy across the hall is going to ask you out? Sorry, Prediction can’t help. Is the world ready for your brand-new product or service? Prediction does you no good.
When the future is unknowable, the way we traditionally reason isn’t much help. We need a new approach for navigating in an uncertain world. Well, we found one. And we know it works because it’s based on how the people who best deal with uncertainty—successful entrepreneurs—think.
Those who write about entrepreneurs invariably focus on their behavior: what Howard Schultz or Michael Dell did while building their respective companies. If you take that approach, you probably would conclude that every entrepreneur is unique, so there’s little to be learned from studying them.
Enter our friend Saras Sarasvathy, author of Effectuation: Elements of Entrepreneurial Expertise. Saras made a fascinating discovery, one that ran counter to conventional wisdom. She studied serial entrepreneurs, but instead of looking at their behavior—which is indeed unique—she focused on how they think. There she found amazing similarities in how they reasoned, approached obstacles, and took advantage of opportunities.
We built off of this, finding that when entrepreneurs create something new, they almost always follow these steps:
1. They want to do what they set out to do. It doesn’t have to be passion at the outset. That will build over time. But an honest desire is important—something more emotional than, “It seems like a good idea.”
2. They quickly take a small step toward their goal, using any means available. You don’t want to overcommit or move so far you can’t recover.
3. After taking that small step, they stop to see what they have learned. Maybe they learn their initial goal is still a good one. Maybe the market tells them to go in another direction. The point is, they pause and reflect.
4. Once they understand what they’ve learned, they take another small step and go through the cycle again. In other words, the formula for success (if there is one) is figuring out what you truly want to do. And once you know: Act. Learn. Build. Repeat. This process continues until you are either happy with the result or decide you don’t want to (or can’t afford to) continue.
How does this work in practice?
• Find something you want. (“I really want to start a restaurant, but I haven’t a clue if I will ever be able to open one.”)
• Take stock of the resources you have at hand and use them to take a small step toward your goal. (“I know a great chef, and if all my family and friends back me, I might have enough money to open a place.”)
• Using those resources, take a smart step toward achieving your goal, i.e., bringing your current reality closer to what you want. (“Let me see what the rents are like at potential locations.”)
• Pause to reflect on what you have learned from taking that step. Every time you act, reality changes. Sometimes the step you take gets you nearer to what you want. (“I should be able to afford something just outside of downtown.”) Sometimes what you want changes. (“It looks likes there are an awful lot of Italian restaurants nearby. We’re going to have to rethink our menu.”) You always learn something. So after you act, ask: Did those actions get you closer to your goal? (“Yes. It looks like I’ll be able to open a restaurant.”) Do you need additional resources to draw even closer? (“Yes. I’ll need to find another chef. The one I know only does Italian.”) Do you still want to pursue your dream? (“Yes.”)
• Repeat until you have what you want (or you’ve decided you don’t want it and/or want something else instead).
We’ve dubbed this new way of thinking “CreAction,” a combination of create and action. It complements Prediction—it does not replace it. The key thing about CreAction is that it’s based on acting and creating evidence, as opposed to thinking and analysis.
In other words, when facing the unknown, act your way into the future you desire. Don’t think your way into it. You can think all day about starting that restaurant, but thinking alone won’t get you any closer to having one.
Sometimes when we explain this concept, people say it sounds familiar. It should. It was the way we all learned—at first.
As a child, everything was unknown or uncertain, so you started learning through action. You’d make a sound and something happened (your mother responded). You pulled the cat’s tail (and got scratched). What we advocate is the recovery of a skill we all had at a young age: the ability to act your way into better thinking.
One last thought. We know this approach works because we used it ourselves. While self-publishing a book-length version of Just Start about a year ago, we …
• Acted. (We wrote it and got it on Amazon.com for \$14.95 a copy.)
• Learned. (Yes, it was a commercial idea. We sold about 9,000 copies, a pretty good number. And yes, we could do better. People told us they wanted more stories, more examples, and for us to break things into digestible bites so our argument would be easier to follow.) Then we …
• Built off of what we learned. (We made all the changes the readers wanted and sold the book to a “real” publisher, Harvard Business Review Press.)
If we can implement this model, so can you.
Before it's here, it's on the Bloomberg Terminal. | 1,520 | 6,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-22 | longest | en | 0.958771 |
http://math.stackexchange.com/questions/1710340/find-equation-for-mass-in-gravity | 1,466,980,848,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783395560.69/warc/CC-MAIN-20160624154955-00181-ip-10-164-35-72.ec2.internal.warc.gz | 206,723,490 | 17,650 | Find equation for mass in gravity
A satellite is moving in circular motion round a planet.
From the physics we know that
$$\Sigma F_r = ma_r = \frac{GMm}{r^2}$$
So I wanted to find the equation for $M$ knowing also that $$v = \omega r = \frac{2\pi r}{T}$$ and
$$a_r = \frac{v^2}{r}$$
Thus,
$$ma_r = \frac{GMm}{r^2}$$ $$a_r = \frac{GM}{r^2}$$ $$\frac{v^2}{r} = \frac{GM}{r^2}$$ $$\frac{\left(\frac{2\pi r}{T}\right)^2}{r} = \frac{GM}{r^2}$$ $$\frac{\frac{4\pi^2r^2}{T^2}}{r} = \frac{GM}{r^2}$$ $$\frac{4\pi^2r^3}{T^2} = \frac{GM}{r^2}$$ $$\frac{4\pi^2r^5}{T^2} = GM$$ $$\frac{4\pi^2r^5G}{T^2} = M$$
However, this is wrong! It should be:
$$M = \frac{4\pi^2r^3}{GT^2}$$
What was my mistake in Mathematics? Please don't migrate it to physics because my misunderstanding is on math.
Note: I would be very happy if you show my mistake, instead of showing me another way to get to the equation.
-
The mistake lies in these steps:
$$\frac{\frac{4\pi^2r^2}{T^2}}{r} = \frac{GM}{r^2}$$ $$\frac{4\pi^2r^3}{T^2} = \frac{GM}{r^2}$$ $$\frac{4\pi^2r^5}{T^2} = GM$$ $$\frac{4\pi^2r^5G}{T^2} = M$$
Actually, it should have been:
$$\frac{\frac{4\pi^2r^2}{T^2}}{r} = \frac{GM}{r^2}$$ $$\frac{4\pi^2r}{T^2} = \frac{GM}{r^2}$$ $$\frac{4\pi^2r^3}{T^2} = GM$$ $$\frac{4\pi^2r^3}{GT^2} = M$$
1. In the second step, the numerator will have $r$ and not $r^3$.
2. In the last step, $G$ will be in the denominator.
-
Oh right! It was a little confusion. Thanks! – Pichi Wuana Mar 23 at 15:40
You're welcome! – SchrodingersCat Mar 23 at 15:40
$\frac{4\pi^2r^2}{T^2}/r=\frac{4\pi r}{T^2}$ and G should go down not up in numerator.
-
When going from $\frac{4\pi^2r^2/T^2}{r}$ to $\frac{4\pi^2r^3}{T^2}$ you added an extra factor of $r^2$; you should have canceled out an $r$ from the numerator and denominator, giving you $\frac{4\pi^2r}{T^2}$. Then when going from $\frac{4\pi^2r^5}{T^2} = GM$ to $\frac{4\pi^2r^5G}{T^2} = M$, you needed to divide each side by $G$; $G$ should be in the denominator on the left-hand side, not the numerator.,
- | 858 | 2,034 | {"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} | 4.0625 | 4 | CC-MAIN-2016-26 | latest | en | 0.808281 |
https://recruitmentresult.com/rrb-ntpc-syllabus/?replytocom=594185 | 1,669,977,847,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710900.9/warc/CC-MAIN-20221202082526-20221202112526-00020.warc.gz | 507,732,083 | 15,658 | # RRB NTPC Syllabus
Download RRB NTPC Syllabus 2022 for Clerk, Goods Guard, Station Master and others from here!! Candidates applied for the post have to participate in the CBT – Stage I/II Examination. In order to prepare for the exam, you must have knowledge regarding the topics. So, go ahead and check Railway NTPC Syllabus from here. Apart from it, you may download RRB NTPC Exam Syllabus PDF in Hindi and English both from here.
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## RRB NTPC Syllabus
RRB NTPC Exam Syllabus 2022
RRB NTPC Mathematics Syllabus
• Number System
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• Ratio and Proportions
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RRB NTPC Syllabus 2022: Weightage of Different Subjects
General Intelligence and Reasoning Weightage:
Topics Weightage Decision making 1 Coding-Decoding 3 Puzzles 3 Analogy 2 Order and Rank 2 Alphabet & Figure series 2 Blood relation 1 Odd pair 1 Logical Arrangement of words 3 Distance & direction 2 Venn diagram 3 Number series 2 Odd One Out 1 Mathematical operators 2 Statement & Conclusions 2
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Final Note: | 1,593 | 6,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-49 | latest | en | 0.789415 |
http://www.lofoya.com/Solved/198/find-the-remainder-when-2-31-is-divided-by-5 | 1,516,545,721,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890771.63/warc/CC-MAIN-20180121135825-20180121155825-00436.warc.gz | 480,134,645 | 15,408 | # Easy Number System Solved QuestionAptitude Discussion
Q. Find the remainder when $2^{31}$ is divided by $5$.
✖ A. 4 ✖ B. 5 ✔ C. 3 ✖ D. 7
Solution:
Option(C) is correct
$2^{10} = 1024$
Unit digit of $2^{10} × 2^{10} × 2^{10}$ is 4 [as $4 × 4 × 4$ gives unit digit 4].
⇒ Unit digit of 231 is 8. Now, 8 when divided by 5, gives 3 as remainder.
Hence, 231 when divided by 5, gives 3 as remainder.
Edit: For an alternative solution, check comment by Kumar Vikram.
Edit 2: For yet another alternative solution involving 'modulus' function, check comment by Sravan Reddy.
## (6) Comment(s)
AKARSHI
()
we can also do like this:-
2^31 = (2^2)^15 * 2^1
= (-1)^15 * 2
= -1*2 = -2
2^31/5 = -2/5 = 3
Sampath
()
another easy approach is that
2 power one divided by 5 gives remainder 2
2 power two divided by 5 gives remainder 4
2 power three divided by 5 gives remainder 3
2 power four divided by 5 gives remainder 1
Follow this procedure until remainder is 1
on above the cycle repeats per every 4 terms
so divide 31 with 4 so we get remainder 3 then it is third term in the above cycle i.e 3 will be the required answer
Krishna
()
can u solve remainder when 128^1000 devided by 153? please by modulus method
Sravan Reddy
()
I think learning '$\text{mod}$' or '$\text{modulus}$' function would be helpful to solve all reminder related problems. Especially very complex problems involving big numbers.
$2 = 2\text{mod}5$
$2^4=16\text{mod}5 = 1\text{mod}5$
$(2^4)^7=(1^7)\text{mod}5$
Multipying with $2^3$ on both sides,
$=2^{(28)}*2^3=2^3*1\text{mod}5$
$2^{31}=8\text{mod}5 = 3\text{mod}5$
Hence, the reminder is $3$.
It might look greek and latin at first but it would be your favourite tool for all reminder related problems once you learn it.
All the above steps could be done in mind in 20-30 secs if you master the 'mod' tool.
Yashwant
()
Could u do this whit ((2^3)^10)×2^1
Then remainder is 1 why ?
Kumar Vikram
()
We can also do like this.
$2^{31}$ can be written as $(2^4)^7 * 2^3$
Now when $2^4$ is divided by 5, the remainder is 1 and when $2^3$ is divided by 5 the remainder is 3.
Therefore answer $= 1*3 =\textbf{3}$. | 732 | 2,165 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2018-05 | latest | en | 0.849872 |
http://mizar.uwb.edu.pl/version/current/html/proofs/matrlin/39 | 1,571,105,094,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-43/segments/1570986655735.13/warc/CC-MAIN-20191015005905-20191015033405-00156.warc.gz | 125,619,994 | 2,911 | let k, m, n be Nat; :: thesis: for K being Field
for V being VectSp of K
for M1 being Matrix of n,k, the carrier of V
for M2 being Matrix of m,k, the carrier of V holds Sum (M1 ^ M2) = (Sum M1) ^ (Sum M2)
let K be Field; :: thesis: for V being VectSp of K
for M1 being Matrix of n,k, the carrier of V
for M2 being Matrix of m,k, the carrier of V holds Sum (M1 ^ M2) = (Sum M1) ^ (Sum M2)
let V be VectSp of K; :: thesis: for M1 being Matrix of n,k, the carrier of V
for M2 being Matrix of m,k, the carrier of V holds Sum (M1 ^ M2) = (Sum M1) ^ (Sum M2)
let M1 be Matrix of n,k, the carrier of V; :: thesis: for M2 being Matrix of m,k, the carrier of V holds Sum (M1 ^ M2) = (Sum M1) ^ (Sum M2)
let M2 be Matrix of m,k, the carrier of V; :: thesis: Sum (M1 ^ M2) = (Sum M1) ^ (Sum M2)
A1: dom (Sum (M1 ^ M2)) = Seg (len (Sum (M1 ^ M2))) by FINSEQ_1:def 3;
A2: now :: thesis: for i being Nat st i in dom (Sum (M1 ^ M2)) holds
(Sum (M1 ^ M2)) . i = ((Sum M1) ^ (Sum M2)) . i
let i be Nat; :: thesis: ( i in dom (Sum (M1 ^ M2)) implies (Sum (M1 ^ M2)) . i = ((Sum M1) ^ (Sum M2)) . i )
assume A3: i in dom (Sum (M1 ^ M2)) ; :: thesis: (Sum (M1 ^ M2)) . i = ((Sum M1) ^ (Sum M2)) . i
then i in Seg (len (M1 ^ M2)) by ;
then A4: i in dom (M1 ^ M2) by FINSEQ_1:def 3;
now :: thesis: (Sum (M1 ^ M2)) . i = ((Sum M1) ^ (Sum M2)) . i
per cases ( i in dom M1 or ex n being Nat st
( n in dom M2 & i = (len M1) + n ) )
by ;
suppose A5: i in dom M1 ; :: thesis: (Sum (M1 ^ M2)) . i = ((Sum M1) ^ (Sum M2)) . i
len M1 = len (Sum M1) by Def6;
then A6: dom M1 = dom (Sum M1) by FINSEQ_3:29;
thus (Sum (M1 ^ M2)) . i = (Sum (M1 ^ M2)) /. i by
.= Sum ((M1 ^ M2) /. i) by
.= Sum (M1 /. i) by
.= (Sum M1) /. i by A5, A6, Def6
.= (Sum M1) . i by
.= ((Sum M1) ^ (Sum M2)) . i by ; :: thesis: verum
end;
suppose A7: ex n being Nat st
( n in dom M2 & i = (len M1) + n ) ; :: thesis: (Sum (M1 ^ M2)) . i = ((Sum M1) ^ (Sum M2)) . i
A8: len M1 = len (Sum M1) by Def6;
len M2 = len (Sum M2) by Def6;
then A9: dom M2 = dom (Sum M2) by FINSEQ_3:29;
consider n being Nat such that
A10: n in dom M2 and
A11: i = (len M1) + n by A7;
thus (Sum (M1 ^ M2)) . i = (Sum (M1 ^ M2)) /. i by
.= Sum ((M1 ^ M2) /. i) by
.= Sum (M2 /. n) by
.= (Sum M2) /. n by
.= (Sum M2) . n by
.= ((Sum M1) ^ (Sum M2)) . i by ; :: thesis: verum
end;
end;
end;
hence (Sum (M1 ^ M2)) . i = ((Sum M1) ^ (Sum M2)) . i ; :: thesis: verum
end;
len (Sum (M1 ^ M2)) = len (M1 ^ M2) by Def6
.= (len M1) + (len M2) by FINSEQ_1:22
.= (len (Sum M1)) + (len M2) by Def6
.= (len (Sum M1)) + (len (Sum M2)) by Def6
.= len ((Sum M1) ^ (Sum M2)) by FINSEQ_1:22 ;
hence Sum (M1 ^ M2) = (Sum M1) ^ (Sum M2) by ; :: thesis: verum | 1,150 | 2,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2019-43 | latest | en | 0.845085 |
http://stadiumsports.com.br/7rvrrue/073a04-diode-resistive-characteristics-curve | 1,725,893,032,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651103.13/warc/CC-MAIN-20240909134831-20240909164831-00800.warc.gz | 26,429,897 | 11,634 | Signal diodes, such as the 1N4148 only pass very small electrical currents as opposed to the high-current mains rectification diodes in which silicon diodes are usually used. • Detect the I/V characteristic curve of semiconductor diode. If we combine both forward biased and reverse biased curves of dido than we have a complete V-I characteristic curve. Maximum Power Rating − It is the maximum power that can be dissipated from the junction, without damaging the junction. Diode Characteristics 7 Temperature effect on forward voltage With a constant current, the forward voltage drop of a diode has a very linear negative slope with temperature. For example, a small diode may have a rating of 1 A; thus, it should not be used in a circuit with more than 1 A current (unless it is used in parallel with some other diodes). So glad I found your blog, and was able to learn new things. A typical curve is shown in Above Figure. Maximum forward current (IFM) − The absolute maximum repetitive forward current that can pass through a diode. This result means that the device has an intrinsic rectifying characteristic, so we can fit the curve with a Schottky diode equation. Graphing the V-I Curve If you plot the results of the type of measurements shown in Figure on a graph, you get the V-I characteristic curve for a forward-biased diode⦠For capacitive load, derate current by 20% Ratings and Characteristic Curves (ES1A, ES1F, ES1G, ES1H and ES1J) Maximum Forward Current Derating Curve Average Forward Current (A) DC or Static Resistance The application of a dc voltage to a circuit containing a semiconductor diode will result in an operating point on the characteristic curve that will not change with time. Diodes ⦠Diodes, however, do not work like this. In the given below figure, there are 2 circuits shown for forward biased arrangements. There are diverse current scales for forward bias and reverse bias operations. For all the practical purposes, a diode is considered to be an open switch when reverse biased. (Make sure you switch from the Diode Tracer window to the Zener Diode Tracer window.) See you in the next post. So that's the iv characteristic for the diode, where this is V diode right there, and the ⦠Under certain conditions, the current-voltage (I-V) characteristic of transition metal oxide (TMO)-based resistive switching (RS) devices exhibits the so-called snapback effect.This effect is not always observable and is a feature associated with the abrupt reduction of the voltage drop across the device caused by the formation of a filamentary pathway spanning the insulating layer. This article covers the Linear & Nonlinear Resistor Definition and Characteristic Curve. Also, thanks for permitting me to comment! You can see there are 2 curves in graph blue is at room temperature (25 centigrade) and the red curve is for increment in the temperature (25C +ÎT). The figure depicted under the section Forward Characteristic shows that Forward Voltage and Reverse Voltage are usually plotted on the horizontal line of the graph. DIODE IV Characteristics. APPARATUS REQUIRED Zener diode IZ5.6V, variable dc power supply (0 â 15V), milli ammeter, volt meter, 470 Ω resistance⦠The resistance offered by the reverse biased p-n junction diode is very large compared to the forward biased diode. Rectification: A PN junction diode is allowed easy to flow of charge in one direction but restrains the flow in the opposite direction. At low current the voltage drop due to the series resistance is negligible; however, if the current is large enough so that IR S is comparable to V A the applied voltage drop should be modified to. Diodes do consume some amount of power when conducting forward current, and they won't block out all reverse current. In the given figure, you can see the graphical representation. Regions of the curve which have a ⦠Single phase, half wave, 60Hz, resistive or inductive load. If the diode is in reverse biasing conditions then the increment in temperature the reverse current (IR) also increases. when the Pregion is made negative and the N-region is made positive. When anode is positive with respect to cathode , diode is said to be forward biased. Through this platform, I am also sharing my professional and technical knowledge to engineering students. Since diodes are semiconductor devices doped with impurities, they do not function like linear devices. Following are some of the important specifications −. Wonderful article! So let’s get started with the Voltage Current Characteristic of Diode. Direct current or electric current is nothing but the flow of charge carriers (free electrons or holes) through a conductor.In DC circuit, the charge carriers flow steadily in single direction or forward direction. The curve obtains called V-I characteristic curve for a reverse-biased diode. For Silicon diode the equation becomes. It is because the value of reverse resistance is so high (R R > 100 MΩ) that is considered to be infinite for all practical purposes.. Reverse and forward characteristics curve for a PN junction diode: Applications of PN junction diode. As reverse bias voltage is further raised, depletion region width increases and a point comes when junction breaks down. Ideally, diodes will block any and all current flowing the reverse direction, or just act like a short-circuit if current flow is forward.Unfortunately, actual diode behavior isn't quite ideal. 4. Breakdown Voltage − It is the minimum reverse bias voltage at which PN junction breaks down with sudden rise in reverse current. In practical, the value or forward voltage is almost one volt according to the value of forward current. The current is equal to IS times e to the qv on kT minus one. Below Figure shows the reverse portion of a zener diode’s characteristic curve. As we continuously increase the forward-biased voltage-current also increases very fast with the forward-biased voltage, but the voltage about the dido increase very slowly after the 0.7 volts that are potential barriers. The effect of a bypass diode on an IV curve can be determined by first finding the IV curve of a single solar cell with a bypass diode and then combining this curve with other solar cell IV curves. Plotting the I-V characteristic requires to change the Axis variable. Sign in Register; Hide. Junction breakdown takes place due … If it converts half-wave of alternating current into direct than this process called half-wave rectification. Resistance of a Diode In practice, no diode is an Ideal diode, this means neither it acts as a perfect conductor when forward biased nor it acts as an insulator when it is reverse biased. Maximum Ratings and Electrical Characteristics (@T A = +25°C unless otherwise specified.) The curve shows the turn-on and the buildup of the forward bias current in the diode. The forward and reverse current voltage (IV) characteristics of a diode are generally compared on a single characteristic curve. Different scales are normally used to display forward and reverse values. Knee Voltage − It is the forward voltage at which the current through the junction starts to increase rapidly. So for the diode, we write a current law that looks like this. The static resistance at the knee of the curve and below of it will be much greater than the resistance values of the vertical rise section of the characteristic curve. If we make a graphical representation of reversed biasing of a diode. For capacitive load, derate current by 20%. The resistance offered by a p-n junction diode when it is connected to a DC circuit is called static resistance. The forward characteristics curve of a diode grows in _____ form. Required fields are marked *. In this segment you will plot out the current vs. voltage characteristics of a non-linear resistive device, specifically the D1N4007 diode. An explanation of why a tunnel diode has a region of negative resistance is best understood by using ⦠The less increment in the voltage across the diode after 0.7 volts is due to the voltage loss across the internal resistance of the semiconductor material by which diode is constructed. This characteristic can be used for the Zener diode. Use the Curve Tracer to obtain the characteristic curve of a 1N5234B $6.2\,\mathrm{V}$ Zener diode. Forward Voltage is measured across the diode and Forward Current is a measure of current through the diode. ⦠Forward Voltage represented to the right and Reverse Voltage to the left. Save my name, email, and website in this browser for the next time I comment. The dynamic resistor of the diode before the knee of the curve is high since the increment in current is very less with the increment in the voltage. Measuring Diode Laser Characteristics Diode Lasers Approach Ubiquity, But They Still Can Be Frustrating To Work With By Tyll Hertsens attempt at real-time equivalent-circuit mod-eling, mostly during device modulation. When the diode is in reverse biasing condition then the current (IR) flow is very less. You can see in the above characteristic curve. The diode conducts almost no current in the high resistance direction, i.e. With the increment in one-degree Celsius, the potential barrier decreases by two millivolts. With the further increment in the reversed biased voltage, the reverse current also increases continuously but the increment in the voltage across the diode is very less. (adsbygoogle=window.adsbygoogle||[]).push({}); AD623 Instrumentation Amplifier IC, Pinout, Datasheet, Circuit, TDA2005 Amplifier Datasheet, Pinout, Features & Applications, TDA7265 Audio Amplifier Datasheet, Pinout, Features & Applications, LM747 Datasheet, Pinout, Features, Equivalent & Applications, Basic Electronic Components used for Circuit Designing, What is a DC Generator : Construction and Working, Starting Methods of Single Phase Induction Motor. • The diode network is shown below. The dotted section of the curve indicates the ideal curve, which would result if it were not for avalanche breakdown. Negative resistance vs positive resistance: An IâV curve which is nonmonotonic (having peaks and valleys) represents a device which has negative resistance. The effect on diode V-I curve is shown in a given figure. Forward Voltage represented to the right and Reverse Voltage to the left. You can see in the graph that is denoted as. The device under The voltage across the diode remains reasonably constant at this time. When the value of VF is large enough to overcome the barrier potential of the P-N junction, a considerable increase in IF occurs. In reality, I 0 changes rapidly with temperature resulting in the dark blue curve… Their resistance ⦠• create a New Simulation Profile. As reverse bias voltage is further raised, depletion region width increases and a point comes when junction breaks down. The slope is also a function of current which is why constant current must be used. To draw the voltage-current (V-I) characteristic curves of a Zener diode and to determine its knee voltage, forward resistance and reverse breakdown voltage. leading to asymmetric I–V curve characteristics. Lets start. This is the type of info that are meant to be shared across the web. In this circuit, we will increase the value of voltage from the 0 across the diode and will find the effect on it. from V s= 0 to cut-in voltage , the forward current is very small .cut-in voltage is also known as threshold voltage or turn-on voltage. But negative resistance means that current ⦠If negative terminal of battery is connected with the anode of diode and positive is connected with the cathode this called reverse biasing. The resistance can be calculated by the formula,R=V/I. Voltage vs Current Characteristics of PN Junction, Simulation done with the help of National Instruments software like Ni Multisim is used. Breakdown is the knee of diode characteristics curve⦠⦠Similarly, when the diode is reversed biased, it acts as a perfect insulator with zero ⦠For Si diode, at 0.7 V the current increases suddenly. The figure below shows the characteristic curve of a zener diode: The figure represents the curve for both silicon and germanium diodes. Maximum Forward Rating − It is the highest instantaneous forward current that a PN junction can pass, without damaging it. Reverse current ⦠By using these to biasing we will discuss the effect of current and voltage across the diode and find different characteristics. The difference is that forward biased voltage across the complete circuit provided by the external voltage source. Reverse breakdown voltage (VBR) − The minimum steady-state reverse voltage at which breakdown will occur. Staticresistance is also defined as the ratio of DC voltage applied across diode ⦠When the forward voltage across the diode equals 0V, forward current (IF) equals 0 mA. common emitter configuration is the same as the I-V characteristic of a diode. Forward voltage (VF) − Maximum forward voltage drop for a given forward current at device operating temperature. This resistor limits the forward current to its maximum rated value. If convert both parts of alternating current into direct current this process is known as full-wave rectification. There are different types of a diode according to their construction and working such as LED, OLED, Zener, etc. Real Diode Characteristics. Comee on over and consult with my site . If a diode is forward biased and we increase the temperature then there will be an increment in the forward current IF at a specified value of forward voltage. All of these pretty graphs are indicating one thing. Maximum Ratings and Electrical Characteristics Rating at 25°C ambient temperature unless otherwise specified. I quite like reading through an article that can make people think. While point ‘C’ defines the position where the forward voltage equals the potential barrier or 0.7 volts. This constant-voltage characteristic leads to a number of applications of diode under reverse bias condition. Forward and reverse current values are shown on the vertical axis of the graph. The combined Forward Voltage and Forward Current values are located in the upper right part of the graph and Reverse Voltage and Reverse Current in the lower left corner. The voltage transfer curve for this circuit is shown on Figure 15 and it is derived from the I-V characteristic of the diode model and Kirchhoff’s voltage law. Ideally, diodes will block any and all current flowing the reverse direction, or just act like a short-circuit if current flow is forward.Unfortunately, actual diode behavior isn't quite ideal. As the voltage approaches to knee of curve resistance start to decreases and after the knee point it becomes very less and current start to increase. In other words, an actual diode offers a very small resistance (not zero) when forward biased and is called a forward ⦠This results in large flow of current. Zener Breakdown Characteristics. As the characteristic curve of this diode is not vertical similar to the ideal diode due to variation in Zener current (ÎIZ) its Zener voltage (ÎVZ) also increases. This results in large flow of current. Thanks for reading. In the circuit, resistance is connected in series it used to limit the value of current passing through the circuit to save a value that will not damage the diode. After the 0.7 volts of forward voltage the increment in the forward current is very fast and (VF) is almost constant. Breakdown is the knee of diode characteristics curve. Maximum forward surge current (IFM-surge) − The maximum current that can be tolerated for a short interval of time. If the value of reverse current is not controlled then it can overheat or damage the diode. Thus, reverse biased diode offer large resistance to the electric current. forward bias diode : A diode is forward biased if the P-type pin is connected with the anode ⦠The following figure shows the static characteristic of a junction diode. With incident sunlight, the I/V curve shifts up showing that there is external current flow from the solar cell to a resistive load as shown with the red curve. The Zener diode is a special type of diode that is designed to work in reverse bias and in the so-called Zener region of the diode characteristic curve. The reverse resistance is in the range of mega ohms (M Ω). Voltage transfer characteristic of the rectifier circuit. Reverse Characteristics Curve for a Junction Diode Sometimes this avalanche effect has practical applications in voltage stabilising circuits where a series limiting resistor is used with the diode to limit this reverse breakdown current to a preset maximum value thereby producing a fixed voltage output across the diode. In a normal positive resistor, the current through it increases in proportion with the voltage across it. I am also a technical content writer my hobby is to explore new things and share with the world. This resistance is known as ac resistance or dynamic resistance. The point of beginning or zero value is at the center of the graph. Electrical Characteristics The L/I Curve. With the increment of temperature forward voltage decreases at a certain value of current. Maximum reverse voltage (VRM) − The absolute maximum or peak reverse bias voltage that can be applied to a diode. Which diode has negative resistance characteristics? The amount of minority current carriers that take part in conduction of reverse current is quite small. The voltage transfer curve for this circuit is shown on Figure 15 and it is derived from the I-V characteristic of the diode model and Kirchhoffâs voltage law. At the start, there is no voltage across the diode as we increase the forward-biased voltage value the current IF and voltage across diode increases the value of forward current and voltage across the diode can be seen in ammeter and voltmeter connected in a circuit. There are normally 2 methods are used to provides input supply to the diode first is forward biasing and the second is reverse biasing. - V& 0. After the knee of curve reverse voltage is almost same but reverse current increases very fast. The bypass diode affects the solar cell only in reverse bias. The most common of the diode laser characteristics is the L/I curve (Figure 1). Specifications like maximum voltage and current ratings, usual operating conditions, mechanical facts, lead identification, mounting procedures, etc. If show above circuit measurements on the graph this graphical representation is called the V-I characteristics curve for a forward-biased diode. The diode is also in the reverse working state, but the reverse voltage is large. The forward and reverse current voltage (IV) characteristics of a diode are generally compared on a single characteristic curve. If the value of IF increases much beyond VK, the forward current becomes quite large. The dotted section of the curve indicates the ideal curve, which would result if it were not for avalanche breakdown. While forward voltage (VF) is the only value of voltage across the diode in a circuit. When the voltage across the diode is zero the IR is also zero. Using a Zener diode in the lower leg, you can make a voltage divider-like circuit whose ouput voltage is quite stiff, i.e. The junction is said to be reverse biased when the p-section of the diode is connected to the negative terminal of the battery and the n-section of the diode is connected to the positive terminal of the battery. In almost every electronic device and engineering project, a diode is used such as ups and some other electronic projects. I am a professional engineer and graduate from a reputed engineering university also have experience of working as an engineer in different famous industries. The knee in the third quadrant of the V-I characteristic curve shows the Zener region and voltage. This operation causes excessive heat to develop across the junction and can destroy a diode. The relationship of forward voltage and forward current is called the ampere-volt, or IV characteristic of a diode. Peak Inverse Voltage − It is the maximum reverse voltage that can be applied to the PN junction, without damaging it. Notice that as the reverse voltage (VR) is increased, the reverse current (IR) remains extremely small up to the “knee” of the curve. i . Distinguishing the stability of characteristic curves for nonlinear resistive circuits is requirement and importance to design various electronic ... Two-tunnel-diode circuit . In general, this means that reverse current remains constant over a large part of reverse voltage. diode characteristic curve and finally diode resistance when the forward voltage exceeds the diodes junctions internal barrier voltage, which for silicon is. At the breakdown voltage (VBR) point, current increases very rapidly. From the voltage transfer curve we observe the following: • Vo = Vin-Vg for Vin ≥ Vg, Voltage transfer characteristic of the rectifier circuit. LIV Curves The fundamental test of a laser diode is a Light-Current-Voltage (LIV) curve, which simultaneously measures the electrical and optical output power characteristics of the device. A diode is said to be an Ideal Diode when it is forward biased and acts as a perfect conductor, with zero voltage across it. Shunt Resistance, R SH Shunt resistance is the slope of the current-voltage curve of the photodiode at the origin, i.e. The vertical axis of the graph across diode is forward is called forward current is a point where voltage! Fast and ( VF ) is the same as the I-V characteristic of a diode tolerated for a figure! The turn-on and the N-region negative operated in the above circuit denoted as out bad devices before they be. Causing the barrier potential to increase reverse values characteristics are maximum ratings and Electrical characteristics L/I. Destroy a diode is in reverse biasing conditions one by one _____ form of diode will... Is in the given figure is the maximum reverse current ⦠the portion the. Purposes, a diode voltage characteristics of a diode, at 0.7 V the current increases very rapidly damaging.... Different characteristics influences the I-V characteristic of a diode is in the third quadrant of the characteristic!, working principle, characteristics, and website in this circuit, we could not observe resistive-switching... Drop and forward continuously increases shown for forward biased and reverse voltage which! Fit the curve leads to a point where the forward bias current in the above denoted... The drive current applied to a DC circuit is called forward current situation, a diode is in below! Positive resistor, the curve obtains called V-I diode resistive characteristics curve of diode under reverse bias the practical purposes, currentlimiting. At which PN junction diode is in reverse biasing conditions then the increment in the above.! Will find the effect on the vertical reverse current that flows through diode resistive characteristics curve junction, Simulation done with increment. Has ratings for maximum voltage and maximum current that can be applied the. The most common of the class, learned a lot from this report (., mechanical facts, lead identification, mounting procedures diode resistive characteristics curve etc obtains called V-I curve... Every electronic device and engineering project, a considerable increase in if occurs diode reasonably! Some amount of minority current carriers that take part in conduction of reverse current portion of graph. J =V A-IR S. the following figure 5 includes the effect on diode V-I curve is shown resistance! Not a linear device current extending downward represent topics of other articles for... Less till the point of beginning or zero value is at knee point I does... Really valuable the cathode this called reverse biasing conditions then the increment in graph! Led, OLED, Zener, etc normally general purpose diode has a voltage. Power when conducting forward current that can make a voltage divider-like circuit whose ouput voltage almost! Get the V-I characteristic curve processes which are responsible for current conduction in a circuit but restrains flow! ) of the graph that is usually quite small potential barrier or 0.7 volts \$ 6.2\ \mathrm... ‘ B ’ is a property exhibited by devices whose resistance does not change with temperature people think linear... Without damaging it diode resistive characteristics curve part of reverse current voltage ( IV ) characteristics of Zener. The last two categories of Table 1 represent topics of other articles, for other authors most! Voltage from the 0 across the diode diode and positive is connected to a diode generally. Scales are normally 2 methods are used to display forward and reverse biasing, this means that reverse current of! Consume some amount of power when conducting forward current, and 0.7 V for silicon to engineering students series influences... Current, the curve shows the diode and positive is connected in series with the cathode this reverse! Knowledge to engineering students represents the curve indicates the ideal curve, would! Half wave, 60Hz, resistive or inductive load a lot from this report beginning! Cell only in reverse bias voltage that can be dissipated from the junction and largely independent the! The DC or static resistance value larger than fifty volts and some special diodes breakdown... The current increases suddenly ( IFM-surge ) − the absolute maximum reverse voltage to the current... Minority current carriers that take part in conduction of reverse current ⦠the portion of the diode is. And positive is connected with the diode Tracer window to the forward biased offer. Usual operating conditions, mechanical facts, lead identification, mounting procedures, etc or. Conducts reverse current is not controlled then it can overheat or damage the diode are drained away the... Share with the voltage across the limiting resistance forward is called the DC or static resistance diode I your. Law that looks like this current applied to the applied voltage junction starts to increase is! The potential barrier or 0.7 volts to flow of charge in one direction but the! Resistor, the curve with a Schottky diode equation with the diode and forward current, and largely of! The V-I characteristic of diode under reverse bias voltage sweep to reset the device under leading to asymmetric I–V characteristics! Devices before they can be used for the Zener diode diode forward characteristic! If show above circuit measurements on the vertical axis of the base-emitter characteristics are ratings... And find different characteristics drained away from the 0 across the web are on! Diode remains reasonably constant at this time L/I curve ( Volt-ampere characteristic curve of the base-emitter characteristics are maximum and. Maximum forward Rating − it is the highest instantaneous forward current is proportional... A p-n junction diode current lengthens above the horizontal axis with reverse current voltage VBR. This report current into direct than this process is known as ac resistance,,! For no longer positioning this submit higher by two millivolts a junction is. V-I characteristic curve then you can go to the diode is forward and! To asymmetric I–V curve characteristics current applied to the top of the characteristic... Ideal diode is also a technical content writer my hobby is to explore new things it plots drive! Axis of the curve, which would result if it were not for avalanche breakdown is that forward and. Curve shifts by approximately 2 mV/°C that current ⦠the V-I characteristic of a non-linear device! Circuit provided by the reverse biased curves of dido than we have a complete V-I characteristic curve and parameters start. Potential barrier decreases by two millivolts circuit whose ouput voltage is further raised, depletion region width and... Peak reverse bias voltage at which the current through it increases in proportion with the of... Forward IV characteristic of a diode is forward is called the knee voltage − it is type! ( IFM-surge ) − the maximum current that can pass, without damaging it defines the position where value. Maximum current that flows through it increases in proportion with the increment in the. Must be used equals 0V, forward current is quite stiff,.! Ouput voltage is measured across the diode and it explains the dynamic or ac resistance also increases position the... Diode first is forward biasing and the N-region negative project, a considerable increase if... Excessive heat to develop across the web includes the effect on it situation, a diode the flow in third... I-V curve of a junction diode, selection of a non-linear resistive device we... Plots the drive current applied to the top of the graph the.. As LED, OLED, Zener, etc that are meant to shared... And importance to design various electronic... Two-tunnel-diode circuit leading to asymmetric I–V characteristics! After the knee in the next time I comment the circuit denoted as and. From a reputed engineering university also have experience of working as an engineer in different famous industries beginning zero! Temperature unless otherwise specified controlled then it can overheat or damage the diode is not controlled then it can or.
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https://dwwiki.mooo.com/w/index.php?title=Burden&diff=49372&oldid=3582 | 1,642,388,323,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300253.51/warc/CC-MAIN-20220117000754-20220117030754-00579.warc.gz | 282,033,596 | 9,116 | # Difference between revisions of "Burden"
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This page contains formulae or data from the distribution mudlib. The data has been partially verified as up to date. You can help by filling in research on the topic.
A player's burden is a measure of how much difficulty they are having carrying things and moving around due to the encumbrance of their inventory.
## Burden calculation
Your burden is calculated as follows:
• Add up the weight of all items in your possession, in all your containers.
• Worn items count as half their weight.
• Items held in one hand count as twice their weight.
• Items held in two hands count as 1.5x their weight.
• Shields still count as 1x their weight when held.
• Determine your maximum carrying capacity as:
Maximum carrying capacity (lb) = 1.89 * (20 + strength + strength2/5)
This gives:
• 8 strength: 77.112 lb
• 9 strength: 85.428 lb
• 10 strength: 94.5 lb
• 11 strength: 104.328 lb
• 12 strength: 114.912 lb
• 13 strength: 126.252 lb
• 14 strength: 138.348 lb
• 15 strength: 151.2 lb
• 16 strength: 164.808 lb
• 17 strength: 179.172 lb
• 18 strength: 194.292 lb
• 19 strength: 210.168 lb
• 20 strength: 226.8 lb
• 21 strength: 244.188 lb
• 22 strength: 262.332 lb
• 23 strength: 281.232 lb
• 24 strength: 300.888 lb
• 25 strength: 321.3 lb
• Divide the first by the second to get your burden level.
## Effects of burden
Burden has the following effects:
### Dexterity penalties
• 50-74% burden results in a -1 dex penalty.
• 74-99% burden results in a -2 dex penalty.
• 100-124% burden results in a -3 dex penalty.
And so on. It's possible to get up to 200% burden by holding a box containing exactly your maximum burden's worth of weight in it, or (very temporarily) to 400% and higher by wearing very heavy custom armour and then taking severe strength penalties.
### Picking up and dropping things
• Your carrying capacity is separate from your burden.
• Since wearing things halves their effect on your burden and holding things doubles it, the point where you're maxed out and can't pick up any more can be anywhere from 50% to 200% depending on how much of your stuff is worn or held.
• You can only take something out or put something into bags and other containers that you're carrying if you've got enough "space" to carry another of it.
• If your carried weight changes such that it exceeds 100% of max (for example, due to a loss of strength), then you will automatically drop items to reduce it below 100% again.
### Combat efficiency
• Burden negatively affects one's 'to-hit' chance with all weapons.
• If fighting unarmed, then there is an additional 'to-hit' penalty from burden.
• Burden negatively affects one's ability to successfully parry or dodge an incoming attack.
• An additional penalty to dodging occurs if your burden is over twice your dexterity stat.
• Burden increases the action point cost of dodging.
### Buoyancy
The more burden a player has, the more trouble they will have staying afloat while swimming. This can be counteracted somewhat by effects that increase a player's buoyancy, such as being a witch or wearing an aquamarine magick crystal or swim ring. | 814 | 3,183 | {"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} | 3.671875 | 4 | CC-MAIN-2022-05 | latest | en | 0.928997 |
https://codereview.stackexchange.com/questions/260922/opencl-approximation-of-pi-using-vectorized-implementation | 1,714,046,432,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297292879.97/warc/CC-MAIN-20240425094819-20240425124819-00351.warc.gz | 151,062,354 | 39,567 | # OpenCL approximation of Pi using vectorized implementation
Yesterday I started following the hands-on OpenCL course. I now got to the point where we are requested to reimplement an approximation algorithm for Pi in OpenCL (in steps, up to a vectorized implementation). I have done so and wanted to get a review for it coupled with some questions I had.
My kernel:
__kernel void pi(const long steps_total, const long steps_per_kernel,
const double step, __global double *global_sums)
{
const int global_id = get_global_id(0);
const int local_id = get_local_id(0);
const int local_size = get_local_size(0);
const long vector_size = 16;
const double16 deltas = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15);
const long work_dx_start = global_id * steps_per_kernel;
const long work_dx_end = min(work_dx_start + steps_per_kernel, steps_total);
const long vectorized_dx_end = work_dx_end / vector_size * vector_size;
double work_sum = 0.0;
for (long dx = work_dx_start; dx < vectorized_dx_end; dx += vector_size)
{
const double16 mid_points = (dx - 0.5 + deltas) * step;
const double16 partial_sums = (4.0 / (1.0 + mid_points * mid_points));
work_sum += partial_sums.s0 + partial_sums.s1 + partial_sums.s2 + partial_sums.s3 +
partial_sums.s4 + partial_sums.s5 + partial_sums.s6 + partial_sums.s7 +
partial_sums.s8 + partial_sums.s9 + partial_sums.sa + partial_sums.sb +
partial_sums.sc + partial_sums.sd + partial_sums.se + partial_sums.sf;
}
for (long dx = vectorized_dx_end; dx < work_dx_end; dx++)
{
const double mid_point = (dx - 0.5) * step;
work_sum += 4.0 / (1.0 + mid_point * mid_point);
}
const double group_sum = work_group_reduce_add(work_sum);
if (local_id == 0)
{
global_sums[global_id / local_size] = group_sum;
}
}
I create max_compute_units * preferred_work_group_size work-items, with preferred_work_group_size as the number of items in a group (and so each kernel executes around steps/global_work_size iterations). On the host side the global sums array is finally added again.
Any comments are greatly appreciated (style, optimizations,...), I also have some questions:
1. It there a nicer way to remove the extra loop at the end in case the iterations are not a multiple of the vector size?
2. Is there a way to sum all elements of an OpenCL vector faster than this (the elements of partial_sums)? The best I could find was IIRC calculating the dot product (which only seems to exist for vectors up to size 4, and even then, it was slower on my GPU). Also stepwise reducing to vectors of half the size by using addition between lower and upper has no apparent effect. I guess it may not even be possible to optimize further?
3. On my GPU, using vectors of size 16 performs the best. I take it this can change depending on the device? Is there some way to try to detect/heuristically calculate this up front (statically) and dynamically load a specific kernel? (or other techniques?)
## 1 Answer
Addressing question #2, you can move the parallel sum out of the for loop, and only do it once.
Use another double16 variable (initialized to all zeros), accumulate the sum into it during the loop body. Then, once the loop is done, do the parallel sum to initialize work_sum before the last loop that handles that last few elements.
• Thanks for the answer! I actually tried that this morning, but it was quite a bit slower than before (IIRC something like 40 instead of 10 seconds, for a huge number of iterations). I am new to this kind of programming, but could this perhaps be due to too much (-spilled over- private) memory? Sadly I have yet to get any profiler working, so for now I can't do more than empirically guess. May 19, 2021 at 18:22 | 945 | 3,701 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-18 | latest | en | 0.596936 |
https://www.slideserve.com/ronda/splash-screen-powerpoint-ppt-presentation | 1,632,424,973,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057427.71/warc/CC-MAIN-20210923165408-20210923195408-00242.warc.gz | 1,025,148,239 | 20,221 | Splash Screen
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##### Presentation Transcript
1. Chapter 10 Measurement and Geometry Click the mouse or press the space bar to continue. Splash Screen
2. Measurement and Geometry 10 Lesson 10-1Polygons Lesson 10-2Perimeter Lesson 10-3Measurement: Area Lesson 10-4Problem-Solving Strategy: Solve a Simpler Problem Lesson 10-5Triangles and Angles Lesson 10-6Quadrilaterals Lesson 10-7Problem-Solving Investigation: Choose a Strategy Lesson 10-8Solid Figures Lesson 10-9 Complex Solid Figures Lesson 10-10 Measurement: Volume Chapter Menu
3. Polygons 10-1 Five-Minute Check (over Chapter 9) Main Idea and Vocabulary California Standards Example 1 Example 2 Lesson 1 Menu
4. Polygons 10-1 • I will identify, describe, and classify polygons. • plane figure • polygon • triangle • quadrilateral • pentagon • hexagon • octagon Lesson 1 MI/Vocab
5. Polygons 10-1 Standard 3MG2.1 Identify, describe, and classify polygons (including pentagons, hexagons, and octagons). Lesson 1 Standard 1
6. Polygons 10-1 Classify the polygon. Answer:not a polygon Lesson 1 Ex1
7. Polygons 10-1 Classify the polygon. • polygon; pentagon • polygon; hexagon • polygon; octagon • not a polygon Lesson 1 CYP1
8. Polygons 10-1 Classify the polygon. Answer:polygon; hexagon Lesson 1 Ex2
9. Polygons 10-1 Classify the polygon. • polygon; triangle • polygon; square • polygon; pentagon • polygon; hexagon Lesson 1 CYP2
10. End of Lesson 1
11. Perimeter 10-2 Five-Minute Check (over Lesson 10-1) Main Idea and Vocabulary California Standards Key Concept: Find Perimeter Example 1 Example 2 Lesson 2 Menu
12. Perimeter 10-2 • I will find the perimeter of a polygon. • perimeter Lesson 2 MI/Vocab
13. Perimeter 10-2 Standard 3MG1.3 Find the perimeter of a polygon with integer sides. Lesson 2 Standard 1
14. Perimeter 10-2 Lesson 2 Key Concept 1
15. Perimeter 10-2 Find the perimeter of the parallelogram. Answer:34 yards Lesson 2 Ex1
16. Perimeter 10-2 Find the perimeter of this rectangle. • 56 units • 40 units • 28 units • 30 units Lesson 2 CYP1
17. Perimeter 10-2 Find the perimeter of the rectangle. Answer: 18 units Lesson 2 Ex2
18. Perimeter 10-2 Find the perimeter of the square, if all its sides where equal to 4 cm. • 16 cm • 20 cm • 18 cm • 8 cm Lesson 2 CYP2
19. End of Lesson 2
20. Measurement: Area 10-3 Five-Minute Check (over Lesson 10-2) Main Idea and Vocabulary California Standards Example 1 Example 2 Example 3 Measurement: Area Lesson 3 Menu
21. Measurement: Area 10-3 • I will find the area of a plane figure. • area Lesson 3 MI/Vocab
22. Measurement: Area 10-3 Standard 3MG1.2 Estimate or determine the areaand volume of solid figures by covering them with squaresor by counting the number of cubes that will fill them. Lesson 3 Standard 1
23. Measurement: Area 10-3 Find the perimeter of the rectangle. Answer:40 square units Lesson 3 Ex1
24. Measurement: Area 10-3 Find the area of the figure. • 20 square units • 18 square units • 21 square units • 14 square units Lesson 3 CYP1
25. Measurement: Area 10-3 Find the area of the figure. Answer:5 square units Lesson 3 Ex2
26. Measurement: Area 10-3 Find the area of the figure. • 36 square units • 40 square units • 48 square units • 30 square units Lesson 3 CYP2
27. Measurement: Area 10-3 The figure below shows the size of Jeff’s driveway. What is the area of the driveway? Answer: 26 square units Lesson 3 Ex3
28. Measurement: Area 10-3 The figure below shows the top view of a land plot. What is the area of the land plot? • 64 square units • 60 square units • 50 square units • 48 square units Lesson 3 CYP3
29. End of Lesson 3
30. Problem-Solving Strategy: Solve a Simpler Problem 10-4 Five-Minute Check (over Lesson 10-3) Main Idea California Standards Example 1: Problem-Solving Strategy Lesson 4 Menu
31. Problem-Solving Strategy: Solve a Simpler Problem 10-4 • I will solve problems by working a simpler problem. Lesson 4 MI/Vocab
32. Problem-Solving Strategy: Solve a Simpler Problem 10-4 Standard 3MG1.3Find the perimeter of a polygon with integer sides. Standard 3MR1.2Determine when and how to break a problem into simpler parts. Lesson 4 Standard 1
33. Problem-Solving Strategy: Solve a Simpler Problem 10-4 Julianna is placing a border around a rectangular bulletin board. The perimeter of the bulletin board is 10 feet. One side measures 2 feet. What will be the length of the border for each side? Lesson 4 Ex1
34. Problem-Solving Strategy: Solve a Simpler Problem 10-4 Understand What facts do you know? • The bulletin board is rectangular with a perimeter of 10 feet. • The length of one side is 2 feet. What do you need to find? • The length of the border for each side. Lesson 4 Ex1
35. Problem-Solving Strategy: Solve a Simpler Problem 10-4 Plan Solve a simpler problem to find the length of the border for each side. Lesson 4 Ex1
36. Problem-Solving Strategy: Solve a Simpler Problem 10-4 total of 2 sides total of remaining sides perimeter 10 feet 4 feet 6 feet Solve Two sides are 2 feet long. 2 feet + 2 feet = 4 feet Find the total length of the other two sides. – = Lesson 4 Ex1
37. Problem-Solving Strategy: Solve a Simpler Problem 10-4 Solve The two unknown sides add up to 6 feet. Since both sides are the same length, divide by 2. 6 feet ÷ 2 feet = 3 feet Answer: So, two pieces of border should be 2 feet long, and the other two should be 3 feet long each. Lesson 4 Ex1
38. Problem-Solving Strategy: Solve a Simpler Problem 10-4 Check Look back at the problem. Add up all of the sides. 2 feet + 2 feet + 3 feet + 3 feet = 10 feet So, the answer is correct. Lesson 4 Ex1
39. End of Lesson 4
40. Triangles and Angles 10-5 Five-Minute Check (over Lesson 10-4) Main Idea and Vocabulary California Standards Key Concept: Identify Triangles by its Sides Example 1 Example 2 Example 3 Lesson 5 Menu
41. Triangles and Angles 10-5 • I will identify and classify different triangles and compare angles. • scalene triangle • right triangle • equilateral triangle • isosceles triangle Lesson 5 MI/Vocab
42. Triangles and Angles 10-5 Standard 3MG2.2Identify attributes of triangles (e.g., two equal sides for the isosceles triangle, three equal sides for the equilateral triangle, right angle for the right triangle). Lesson 5 Standard 1
43. Triangles and Angles 10-5 Standard 3MG2.4 Identify right angles in geometric figures and determine whether other angles are greater or less than a right angle. Lesson 5 Standard 2
44. Triangles and Angles 10-5 Lesson 5 Key Concept 1
45. Triangles and Angles 10-5 Identify the triangle. Answer: scalene triangle Lesson 5 Ex1
46. Triangles and Angles 10-5 Identify the triangle shown in the grid. • isosceles triangle • equilateral triangle • scalene triangle • right triangle Lesson 5 CYP1
47. Triangles and Angles 10-5 Identify the triangle. Answer: equilateral Lesson 5 Ex2
48. Triangles and Angles 10-5 Identify the triangle shown below. • isosceles triangle • equilateral triangle • scalene triangle • right triangle Lesson 5 CYP2
49. Triangles and Angles 10-5 Identify the bottom triangle on the tree. Tell if each angle of the triangle is a right angle, less than a right angle or greater than a right angle. Answer: isosceles; 2 angles are less than a right angle and 1 angle is greater than a right angle Lesson 5 Ex3
50. Triangles and Angles 10-5 Identify the triangle below. Tell if each triangle is a right angle, less than a right angle, or greater than a right angle. • scalene; right angle • isosceles; all angles are less than a right angle • equilateral; all angles are less than a right angle • right; right angle Lesson 5 CYP3 | 2,149 | 7,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2021-39 | latest | en | 0.506994 |
https://www.reddit.com/r/explainlikeimfive/comments/ktomf/if_a_ship_were_able_to_travel_to_the_end_of_the/ | 1,511,227,779,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806309.83/warc/CC-MAIN-20171121002016-20171121022016-00103.warc.gz | 847,484,554 | 22,950 | [–] 3 points4 points (2 children)
This is a very common misunderstanding.
Imagine that the Universe is a balloon. If you're flying a spaceship through the Universe, you're not flying on the inside of the balloon, you're flying across the surface of the balloon. So, you see, there's no "edge", just a continuous surface.
Now, the balloon is also slowly inflating; you know how if you draw on a balloon in marker, the markings stretch out and spread further apart? That's what the expansion of the universe looks like, basically. The, um, "fabric" of spacetime is just spreading waaaaaaaay out.
[–] 0 points1 point (1 child)
The problem with explanations like this is it just leaves the reader asking "well what's past the edge of the balloon?
[–] 2 points3 points (0 children)
The balloon is, in this case, the whole enchilada. Since our thought experiment has to compress a 4 (or more) dimensional space into a 2d space that curves in 3 dimensions for simplicity's sake, a lot gets lost in the translation.
Imagine that you're traveling, not on the surface of the balloon, but "in" the surface, like playing a top-down 2d videogame. Your character can look up, down, left or right, but he has no "up" or "down". They simply aren't part of his universe.
[–] 1 point2 points * (2 children)
There is no "end" of the universe. It has no edges. Our most recent data strongly suggests that the universe is infinite in size. Which really, when you think about it, only makes sense. Because if it did have edges...well, you'd be asking this question about them. It would mean that the laws of physics differ from one location to the next (i.e. at "the edge" and everywhere else), which is not something that happens.
It may not quite be for a five year old, but this is a very good page to read. In particular, in order to avoid the inevitable "wtf man what about the Big Bang?" question, I will quote part of it:
The Big Bang did not occur at a single point in space as an "explosion." It is better thought of as the simultaneous appearance of space everywhere in the universe.
That is, it's not that the universe is expanding into something, but rather distances increase over time.
Edit: There are also a lot of older /r/askscience discussions on this topic, which can be found here.
[–] 0 points1 point (1 child)
Our most recent data
Where's that? I thought many (most?) physicists currently think the universe is finite but without an edge. As I understand it, there's a whole host of math problems that come with an infinite universe.
See Novae's post.
[–] 0 points1 point (0 children)
It's not the 90's anymore.
See Novae's post.
Novae's post describes a universe with positive curvature (the surface of the balloon). WMAP data from the last 10 years points very strongly toward a flat universe (within small error bars, of course), i.e. zero curvature.
Here's another NASA link regarding WMAP and the shape of the universe. Relevant quote from the end:
We now know that the universe is flat with only a 0.5% margin of error. This suggests that the Universe is infinite in extent
[–] 0 points1 point (3 children)
As far as we know, there's no wall to hit. The universe is expanding everywhere, from everywhere. If there were a "wall" of sorts to hit, it would act as a fixed reference point, and in relativity, those don't exist.
The most likely explanation is that the universe is "infinite, but bounded." It's infinite in that the universe is quite literally all that there is, but it's bounded in that it does not go on forever. The geometry of this is confusing, to say the least. If all our current undestanding is correct (and there's not higher dimensions or anything like that) the universe is most likely shaped like a 4-dimensional torus, (a donut-like shape.) To understand why this is, imagine a sheet of paper. Now make a tube out of that paper by folding the top to the bottom. Now make a donut-shape by putting one end of the tube to the other. Now uh, imagining the 4-dimensional part of this is a little beyond us.
So there is a possibility that if you continued to travel in one direction in the universe, you would end up right back where you started. It's expanding far too fast for that to actually happen, though.
Also, it's thought that the "visible universe" - what we can see from Earth - comprises 3-4% of the entire universe, at best. So if you got to the end of the visible universe somehow, you'd just have a whole lot more universe to go to.
[–] 0 points1 point (2 children)
the universe is most likely shaped like a 4-dimensional torus
While possible, there is nothing "most likely" about a torus as opposed to a straightforward Euclidean geometry, in which case:
if you continued to travel in one direction in the universe, you would
...just keep going.
it's thought that the "visible universe" ... comprises 3-4% of the entire universe
Where did you get this number?
[–] 0 points1 point (1 child)
Actually, I was way off:
Cosmos At Least 250x Bigger Than Visible Universe, Say Cosmologists
That's a lot less than 3-4%.
[–] 0 points1 point (0 children)
Ah yes, I remember that. I think you know this, but for clarity, I would like to emphasize the importance of the words "at least" in that headline. That "250x" is a lower bound.
[–] -1 points0 points (1 child)
Well, we can never actually know, because the universe is defined as everything within the range of our power of observation. Our senses (vision, really, since none of the other senses can help us over interstellar distances) can't travel faster than the speed of light, so the farthest we can see is the distance that light has travelled since the beginning of the universe. And, assuming that things started off with a "Big Bang," it's also the farthest away that any matter from the Big Bang could have been hurled since then.
Since we don't know what existed around the Big Bang (if anything), we don't know what (if anything) is on the other side of that boundary. One strong possibility, though, is that the universe is a three-dimensional version of the surface of a balloon, so that if you travel far enough, you would end up back where you started.
[–] 0 points1 point (0 children)
the universe is defined as everything within the range of our power of observation.
This is the observable universe, not the universe as a whole.
[–] -2 points-1 points (0 children)
It would be like the game Asteroids- The ship would come back from the other side of the universe. | 1,561 | 6,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-47 | latest | en | 0.961185 |
https://physics.stackexchange.com/questions/175344/how-to-imagine-higher-dimensions | 1,656,598,127,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103821173.44/warc/CC-MAIN-20220630122857-20220630152857-00544.warc.gz | 496,136,603 | 65,495 | # How to imagine higher dimensions?
and here's a description of Brian Greene: http://www.youtube.com/watch?v=v95WjxpMIQg
Carl Sagan explains, we can not see the higher dimensions because we are limited to perceive only three dimensions. He didn't say a dimension can be small or big. This explanation completely makes sense.
But Brian Greene explains, higher dimensions can be tiny and curled up.
Isn't every dimension perpendicular to each of the other dimensions? If so, then how can a dimension be tiny or big? I want to know which is the right way to imagine higher dimensions?
The definition of dimension used here is that of a dimension of a manifold - essentially, how many coordinates (=real numbers) we need to describe the manifold (thought of as spacetime).
Manifolds may carry a notion of length, and one of volume. They may also be compact or non-compact, roughly1 corresponding to finite and infinite. E.g. a sphere of radius $R$ is compact and two-dimensional - every point on it can be described by two angles, and it's volume is finite as $\frac{4}{3}\pi r^2$. Ordinary Euclidean space $\mathbb{R}^3$ is non-compact and three-dimensional - every point in it is described by three real numbers (directed distance from an arbitrarily chosen origin), and you can't associate a finite volume to it.
Note that, on the sphere, you can keep increasing any one of the coordinates and, sooner or later, you will return to the point you started from. All dimensions here are "small"/compact. In Euclidean space, you never return to the origin, no matter how far you go. All dimensions are "big"/non-compact.
An infinitely long cylinder is now an example of where the two dimensions are different. Take as coordinates the obvious two - the length (how far "down"/"up" on the cylinder you are), and the angle (where on the circle that's at that length you are). The length dimension is non-compact - you never return to your starting point if you just keep increasing that coordinate. The angle coordinate is compact - you return after $2\pi$ to your starting point, and the "size" of the dimension is the radius of the circle. This is an example of a "curled up dimension". If you are far larger than the radius, you might not even notice you are on a cylinder, and instead think you are on a one-dimensional line!
1The mathematical definition is by covering properties which are not as easily translated into intuition.
In differential geometry, a space of a given number of dimensions can be curved rather than Euclidean, so for example the surface of a sphere is understood to be a 2-dimensional space in spite of the fact that we can't help but visualize the sphere sitting in a higher-dimensional 3D Euclidean space. This 3D space that we imagine the 2D surface sitting in is technically known as an "embedding space", but the mathematics of differential geometry allows mathematicians and physicists to describe the curvature of surfaces in purely "intrinsic" terms without the need for any embedding space, rather than in "extrinsic" terms where the surface is described by its coordinates in a higher-dimensional space--see the "Intrinsic versus extrinsic" section of the differential geometry wiki page. And all this has a practical relevance to physicists, since Einstein's theory of general relativity uses differential geometry to explain gravitation in terms of matter and energy causing spacetime to become curved (see here for a short conceptual introduction to how spacetime curvature can explain the way particle trajectories are affected by gravity).
With these ideas in mind, if you want to understand Greene's comment about higher dimensions being "curled up", picture the surface of a long cylinder or tube, like a garden hose. This surface is 2-dimensional, but you only have to travel a short distance in one direction to make a circle and return to your place of origin--that's the "curled up" dimension--while the perpendicular direction can be arbitrarily long, perhaps infinite. You could imagine 2-dimensional beings that live on this surface, like those in the famous book Flatland that has introduced many people to the idea of spaces with different numbers of dimensions (and there's also a "sequel" by another author titled Sphereland which introduces the idea that a 2D universe could actually be curved). But if the circumference of the cylinder was very short--shorter even than the radius of atoms in this universe--then at large scales this universe could be indistinguishable from a 1-dimensional universe (like the "Lineland" that the characters in Flatland pay a visit to). So a similar idea is hypothesized in string theory to account for the fact that we only experience our space as 3-dimensional even though the mathematics of string theory requires more spatial dimensions--the extra dimensions are "curled up" into small shapes known as Calabi-Yau manifolds, which play a role analogous to the circular cross-sections of the 2D cylinder or tube I described (although in brane theory, an extension of string theory, it's possible that one or more extra dimensions may be "large" and non-curled, but particles and forces except for gravity are confined to move in a 3-dimensional "brane" sitting in this higher-dimensional space, which is termed the "bulk").
It is relatively easy to imagine 4th dimension. That would be time. But time as if we had a time machine with which we we could arbitrarily move through it. Higher dimensions would be more difficult but possible as if "destinies". For example imagine that in destiny1 you see a car going from A to B in a given hour but in alternate destiny2 you see the same car going from B to A. And so on. Now imagine those destinies as if they were books in a row on a shelf 1,2..., n. Now imagine number of shelves (1...n) x (1...n) with destinies. Or number of shelves in number of rows in a library of destinies in a 3 dimensional table (1,2...n) x (1,2...n) x (1,2...n). Or a 3 dimensional library of destinies that changes in time. Now if you can imagine all this you just imagined 4 x 4=16 dimensions.
• That picture didn't work for me. I just kept picturing my library. Apr 11, 2015 at 2:50
• When people say time is the 4th dimension they are assuming there are only 3 spatial dimensions (and the dimensions don't have any intrinsic order, so you could just as easily say time is the 1st dimension and the next three are spatial dimensions, which is actually how 4-vectors in relativity are usually written). But it's certainly mathematically possible to describe a universe with more than 3 spatial dimensions--in superstring theory there are 9 space dimensions and 1 time dimension (so you might say 'time is the 10th dimension' here), in M-theory there are 10 space dimensions and 1 time. Apr 11, 2015 at 20:14 | 1,504 | 6,831 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-27 | latest | en | 0.921601 |
https://learn.careers360.com/physics/kinetic-theory-of-gases-chapter/ | 1,642,820,915,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303729.69/warc/CC-MAIN-20220122012907-20220122042907-00476.warc.gz | 381,148,014 | 114,106 | # Kinetic theory of Gases Share
## What is Kinetic theory of Gases
All of you must have seen hot air balloons in the sky or on television. The balloon rises because the air molecules get heated up and they started rising upwards and this heat (high temperature) increases the movement of molecules and the Kinetic energy of molecules increases hence resulting in a force which pushes the walls of the balloon in the upward direction. If you give keen attention to the above example you could see that the Temperature is related to Pressure. In fact, Temperature is directly proportional to Pressure and you can also see that balloon inflation also indicates the increase in Volume. In this chapter, we will see how these macroscopic properties i.e. Temperature, Pressure, and Volume, etc are related to each other.
## Kinetic Theory Of Gases- Topics
Equation of state for the perfect gas, work done while compressing a gas,Kinetic theory of gases (assumptions), concept of pressure, temperature and Kinetic energy , RMS speed of gas molecules; Degrees of freedom, applications to specific heat capacities of gases,Law of equipartition of energy,Avogadro's number,Mean free path.
## Overview of Kinetic Theory of Gases
Before starting this chapter, you should know that it is comparatively easier topic than studying properties of solids or liquids because, in a gas, molecules are far from each other and their mutual interactions are negligible except when two molecules collide. Also, this chapter is going to be important because the next chapter you will be reading is Thermodynamics and many concepts of this chapter will be used there. This chapter mainly talks about the properties of ideal gases, and an ideal gas follows some set of assumption and also follows ideal gas equation which is given by, PV =nRT where P is the pressure in the container in which the gas is kept, V is the volume of the container, T is the temperature of gas, n is the number of moles of gas and R is a gas constant whose value is fixed.
The assumptions of the ideal gas are mentioned below for your better understanding of this chapter :
• The gas consists of a large number of molecules, which are in random motion and obey Newton's laws of motion;
• The volume of the molecules is negligibly small compared to the volume occupied by the gas
• No forces act on the molecules except during elastic collisions of negligible duration.
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Although no real gas follows these assumptions hence they also don't follow the ideal gas equation. For real gases, some changes are made in the ideal gas equation. But for the intermediate class, the syllabus is limited to ideal gases only. You will also read about the Law of equipartition of energy which states that in equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having average energy equal $\frac{1}{2}K_{B}T$ to where$K_{B}$ is Boltzmann’s constant. This law will also help you to understand the concept of degrees of freedom. You will get various questions in which you have to deal with Root mean square velocity, Average speed and Most probable speed of a molecule of gas. It also tells us that the temperature of an ideal monatomic gas is proportional to the average kinetic energy of its atoms.
## Formulas for Kinetic Theory of Gases
• Ideal gas equation, PV =nRT
• Pressure of ideal gas, $P = \frac{1}{3}nm\bar{v}^{2}$ where 'n' is number density of molecules, 'm' is mass of the molecule, $\bar{v}^{2}$ is mean squared speed of molecules.
• Root mean squared speed, $v_{rms} =\sqrt{\frac{3RT}{M}}$
• Mean free path length, $l = \frac{1}{\sqrt{2}nd^{2}}$ ,here d is the diameter of molecule.
## How to prepare the Kinetic Theory of Gases
It is an easy chapter in exam point of view hence if you have good command over calculation then you will be able to solve most of the questions very quickly but you shouldn’t take it lightly. Sometimes questions are asked which are hypothetical in nature and will require you to apply concepts hence read carefully don’t just mug up. Also, some questions will be very easy but tricky. For example, you may be given the value $v_{rms}$ of , n, and Volume of the container and from there you will have to calculate the pressure of an ideal gas, so here you need to use two formulas, one will be $v_{rms}$ formula and 2nd will be the ideal gas equation. So keep in mind how can you convert one equation to get the required value of an unknown variable and that's why you will require a lot of practice for doing these types of conversions.
Kinetic Theory of Gases - Notes
• Give attention to the questions which were previously asked in the exam for which you are preparing, and look at the concept which was applied to solve them.
• Rather than just reading online or from books, make your own notes and revise them.
• Give a significant number of mock tests and make note of your progress.
• Make a list of questions which you were not able to solve in the first attempt or you took a long time solving them, after a week or two solve them again.
## Books for Kinetics Theory of Gases
Recommendation for this chapter would be to first go through NCERT book and solve questions after that you should move on to NCERT Exemplar book for a good hold on this chapter. If you want to test yourself for competitive exams, then you should read Understanding Physics by D.C. Pandey or you could also read and practice questions from Concepts of Physics by H.C. Verma.
## Physics Chapter wise Notes For Engineering and Medical Exams
Chapters No. Chapters Name Chapter 1 Physics and Measurement Chapter 2 Kinematics Chapter 3 Laws of motion Chapter 4 Work Energy and Power Chapter 5 Rotational Motion Chapter 6 Gravitation Chapter 7 Properties of Solids and Liquids Chapter 9 Thermodynamics Chapter 10 Oscillations and Waves Chapter 11 Electrostatics Chapter 12 Current Electricity Chapter 13 Magnetic Effects of Current and Magnetism Chapter 14 Electromagnetic Induction and Alternating currents Chapter 15 Electromagnetic Waves Chapter 16 Optics Chapter 17 Dual Nature of Matter and Radiation Chapter 18 Atoms And Nuclei Chapter 19 Electronic devices Chapter 20 Communication Systems Chapter 21 Experimental skills
### Topics from Kinetic theory of Gases
• Kinetic theory of gases- assumptions, concept of pressure ( JEE Main, GUJCET Pharmacy, COMEDK UGET ) (13 concepts)
• Equation of state of a perfect gas, work done on compressing a gas ( JEE Main, GUJCET Pharmacy, COMEDK UGET ) (2 concepts)
• Kinetic energy and temperature:rms speed of gas molecules: Degreesof freedom ( JEE Main, GUJCET Pharmacy, COMEDK UGET ) (9 concepts)
• Applications to specific heat capacities of gases: Mean free path, Avogadro'snumber ( JEE Main, GUJCET Pharmacy, COMEDK UGET ) (8 concepts)
• States of matter ( JEE Main, GUJCET Pharmacy, COMEDK UGET ) (1 concepts)
• Ideal gas, its equation and laws of ideal gas ( JEE Main, GUJCET Pharmacy, COMEDK UGET ) (5 concepts)
• Real gas ( JEE Main, GUJCET Pharmacy, COMEDK UGET ) (1 concepts)
• Pressure of an ideal gas ( JEE Main, GUJCET Pharmacy, COMEDK UGET ) (1 concepts)
• Maxwell's law ( JEE Main, GUJCET Pharmacy, COMEDK UGET ) (3 concepts)
• Law of equipartition of energy ( JEE Main, GUJCET Pharmacy, COMEDK UGET ) (4 concepts)
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Questions | 1,702 | 7,432 | {"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": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2022-05 | latest | en | 0.9322 |
https://riddles360.com/riddle/ball-in-bag?discuss=1 | 1,669,865,243,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710789.95/warc/CC-MAIN-20221201021257-20221201051257-00326.warc.gz | 536,149,245 | 6,930 | # Ball in Bag
A bag contains 64 balls of eight different colours. There are eight of each colour (including red). What is the least number you would have to pick, without looking, to be sure of selecting 3 red balls?
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In Canada, a mathematical puzzle must be solved in order to win the lottery to classify it as a “game of skill” not gambling. | 789 | 3,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.972064 |
https://www.lmfdb.org/GaloisGroup/41T5 | 1,719,157,226,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862474.84/warc/CC-MAIN-20240623131446-20240623161446-00574.warc.gz | 760,926,247 | 8,629 | # Properties
Label 41T5 Degree $41$ Order $328$ Cyclic no Abelian no Solvable yes Primitive yes $p$-group no Group: $C_{41}:C_{8}$
Show commands: Magma
magma: G := TransitiveGroup(41, 5);
## Group action invariants
Degree $n$: $41$ magma: t, n := TransitiveGroupIdentification(G); n; Transitive number $t$: $5$ magma: t, n := TransitiveGroupIdentification(G); t; Group: $C_{41}:C_{8}$ Parity: $-1$ magma: IsEven(G); Primitive: yes magma: IsPrimitive(G); magma: NilpotencyClass(G); $\card{\Aut(F/K)}$: $1$ magma: Order(Centralizer(SymmetricGroup(n), G)); Generators: (1,27,32,3,40,14,9,38)(2,13,23,6,39,28,18,35)(4,26,5,12,37,15,36,29)(7,25,19,21,34,16,22,20)(8,11,10,24,33,30,31,17), (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41) magma: Generators(G);
## Low degree resolvents
|G/N|Galois groups for stem field(s)
$2$: $C_2$
$4$: $C_4$
$8$: $C_8$
Resolvents shown for degrees $\leq 47$
## Subfields
Prime degree - none
## Low degree siblings
There are no siblings with degree $\leq 47$
A number field with this Galois group has no arithmetically equivalent fields.
## Conjugacy classes
Label Cycle Type Size Order Representative 1A $1^{41}$ $1$ $1$ $()$ 2A $2^{20},1$ $41$ $2$ $( 2,41)( 3,40)( 4,39)( 5,38)( 6,37)( 7,36)( 8,35)( 9,34)(10,33)(11,32)(12,31)(13,30)(14,29)(15,28)(16,27)(17,26)(18,25)(19,24)(20,23)(21,22)$ 4A1 $4^{10},1$ $41$ $4$ $( 2,33,41,10)( 3,24,40,19)( 4,15,39,28)( 5, 6,38,37)( 7,29,36,14)( 8,20,35,23)( 9,11,34,32)(12,25,31,18)(13,16,30,27)(17,21,26,22)$ 4A-1 $4^{10},1$ $41$ $4$ $( 2,10,41,33)( 3,19,40,24)( 4,28,39,15)( 5,37,38, 6)( 7,14,36,29)( 8,23,35,20)( 9,32,34,11)(12,18,31,25)(13,27,30,16)(17,22,26,21)$ 8A1 $8^{5},1$ $41$ $8$ $( 2,28,33, 4,41,15,10,39)( 3,14,24, 7,40,29,19,36)( 5,27, 6,13,38,16,37,30)( 8,26,20,22,35,17,23,21)( 9,12,11,25,34,31,32,18)$ 8A-1 $8^{5},1$ $41$ $8$ $( 2,39,10,15,41, 4,33,28)( 3,36,19,29,40, 7,24,14)( 5,30,37,16,38,13, 6,27)( 8,21,23,17,35,22,20,26)( 9,18,32,31,34,25,11,12)$ 8A3 $8^{5},1$ $41$ $8$ $( 2, 4,10,28,41,39,33,15)( 3, 7,19,14,40,36,24,29)( 5,13,37,27,38,30, 6,16)( 8,22,23,26,35,21,20,17)( 9,25,32,12,34,18,11,31)$ 8A-3 $8^{5},1$ $41$ $8$ $( 2,15,33,39,41,28,10, 4)( 3,29,24,36,40,14,19, 7)( 5,16, 6,30,38,27,37,13)( 8,17,20,21,35,26,23,22)( 9,31,11,18,34,12,32,25)$ 41A1 $41$ $8$ $41$ $( 1, 3, 5, 7, 9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41, 2, 4, 6, 8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40)$ 41A2 $41$ $8$ $41$ $( 1, 2, 3, 4, 5, 6, 7, 8, 9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41)$ 41A4 $41$ $8$ $41$ $( 1,25, 8,32,15,39,22, 5,29,12,36,19, 2,26, 9,33,16,40,23, 6,30,13,37,20, 3,27,10,34,17,41,24, 7,31,14,38,21, 4,28,11,35,18)$ 41A7 $41$ $8$ $41$ $( 1,22, 2,23, 3,24, 4,25, 5,26, 6,27, 7,28, 8,29, 9,30,10,31,11,32,12,33,13,34,14,35,15,36,16,37,17,38,18,39,19,40,20,41,21)$ 41A8 $41$ $8$ $41$ $( 1, 5, 9,13,17,21,25,29,33,37,41, 4, 8,12,16,20,24,28,32,36,40, 3, 7,11,15,19,23,27,31,35,39, 2, 6,10,14,18,22,26,30,34,38)$
magma: ConjugacyClasses(G);
## Group invariants
Order: $328=2^{3} \cdot 41$ magma: Order(G); Cyclic: no magma: IsCyclic(G); Abelian: no magma: IsAbelian(G); Solvable: yes magma: IsSolvable(G); Nilpotency class: not nilpotent Label: 328.12 magma: IdentifyGroup(G); Character table:
1A 2A 4A1 4A-1 8A1 8A-1 8A3 8A-3 41A1 41A2 41A4 41A7 41A8 Size 1 41 41 41 41 41 41 41 8 8 8 8 8 2 P 1A 1A 2A 2A 4A1 4A-1 4A-1 4A1 41A8 41A4 41A1 41A2 41A7 41 P 1A 2A 4A1 4A-1 8A-3 8A3 8A-1 8A1 41A7 41A8 41A2 41A4 41A1 Type 328.12.1a R $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ 328.12.1b R $1$ $1$ $1$ $1$ $−1$ $−1$ $−1$ $−1$ $1$ $1$ $1$ $1$ $1$ 328.12.1c1 C $1$ $1$ $−1$ $−1$ $−i$ $i$ $i$ $−i$ $1$ $1$ $1$ $1$ $1$ 328.12.1c2 C $1$ $1$ $−1$ $−1$ $i$ $−i$ $−i$ $i$ $1$ $1$ $1$ $1$ $1$ 328.12.1d1 C $1$ $−1$ $−ζ82$ $ζ82$ $ζ83$ $−ζ8$ $ζ8$ $−ζ83$ $1$ $1$ $1$ $1$ $1$ 328.12.1d2 C $1$ $−1$ $ζ82$ $−ζ82$ $−ζ8$ $ζ83$ $−ζ83$ $ζ8$ $1$ $1$ $1$ $1$ $1$ 328.12.1d3 C $1$ $−1$ $−ζ82$ $ζ82$ $−ζ83$ $ζ8$ $−ζ8$ $ζ83$ $1$ $1$ $1$ $1$ $1$ 328.12.1d4 C $1$ $−1$ $ζ82$ $−ζ82$ $ζ8$ $−ζ83$ $ζ83$ $−ζ8$ $1$ $1$ $1$ $1$ $1$ 328.12.8a1 R $8$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $ζ41−20+ζ41−19+ζ41−16+ζ41−7+ζ417+ζ4116+ζ4119+ζ4120$ $ζ41−14+ζ41−9+ζ41−3+ζ41−1+ζ41+ζ413+ζ419+ζ4114$ $ζ41−18+ζ41−13+ζ41−6+ζ41−2+ζ412+ζ416+ζ4113+ζ4118$ $ζ41−17+ζ41−11+ζ41−10+ζ41−8+ζ418+ζ4110+ζ4111+ζ4117$ $ζ41−15+ζ41−12+ζ41−5+ζ41−4+ζ414+ζ415+ζ4112+ζ4115$ 328.12.8a2 R $8$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $ζ41−17+ζ41−11+ζ41−10+ζ41−8+ζ418+ζ4110+ζ4111+ζ4117$ $ζ41−20+ζ41−19+ζ41−16+ζ41−7+ζ417+ζ4116+ζ4119+ζ4120$ $ζ41−14+ζ41−9+ζ41−3+ζ41−1+ζ41+ζ413+ζ419+ζ4114$ $ζ41−15+ζ41−12+ζ41−5+ζ41−4+ζ414+ζ415+ζ4112+ζ4115$ $ζ41−18+ζ41−13+ζ41−6+ζ41−2+ζ412+ζ416+ζ4113+ζ4118$ 328.12.8a3 R $8$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $ζ41−18+ζ41−13+ζ41−6+ζ41−2+ζ412+ζ416+ζ4113+ζ4118$ $ζ41−15+ζ41−12+ζ41−5+ζ41−4+ζ414+ζ415+ζ4112+ζ4115$ $ζ41−17+ζ41−11+ζ41−10+ζ41−8+ζ418+ζ4110+ζ4111+ζ4117$ $ζ41−14+ζ41−9+ζ41−3+ζ41−1+ζ41+ζ413+ζ419+ζ4114$ $ζ41−20+ζ41−19+ζ41−16+ζ41−7+ζ417+ζ4116+ζ4119+ζ4120$ 328.12.8a4 R $8$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $ζ41−15+ζ41−12+ζ41−5+ζ41−4+ζ414+ζ415+ζ4112+ζ4115$ $ζ41−17+ζ41−11+ζ41−10+ζ41−8+ζ418+ζ4110+ζ4111+ζ4117$ $ζ41−20+ζ41−19+ζ41−16+ζ41−7+ζ417+ζ4116+ζ4119+ζ4120$ $ζ41−18+ζ41−13+ζ41−6+ζ41−2+ζ412+ζ416+ζ4113+ζ4118$ $ζ41−14+ζ41−9+ζ41−3+ζ41−1+ζ41+ζ413+ζ419+ζ4114$ 328.12.8a5 R $8$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $ζ41−14+ζ41−9+ζ41−3+ζ41−1+ζ41+ζ413+ζ419+ζ4114$ $ζ41−18+ζ41−13+ζ41−6+ζ41−2+ζ412+ζ416+ζ4113+ζ4118$ $ζ41−15+ζ41−12+ζ41−5+ζ41−4+ζ414+ζ415+ζ4112+ζ4115$ $ζ41−20+ζ41−19+ζ41−16+ζ41−7+ζ417+ζ4116+ζ4119+ζ4120$ $ζ41−17+ζ41−11+ζ41−10+ζ41−8+ζ418+ζ4110+ζ4111+ζ4117$
magma: CharacterTable(G); | 3,534 | 5,663 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 169, "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} | 2.859375 | 3 | CC-MAIN-2024-26 | latest | en | 0.426205 |
https://www.coolstuffshub.com/weight/convert/long-hundredweights-(uk)-to-pennyweights/ | 1,701,363,091,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100229.44/warc/CC-MAIN-20231130161920-20231130191920-00320.warc.gz | 780,232,728 | 19,876 | # Convert long hundredweights (UK) to pennyweights (cwt long to dwt conversion)
1 long hundredweight (UK) (cwt long) = 32666.7 pennyweights (dwt)
## How to convert long hundredweights (UK) to pennyweights?
To convert long hundredweights (UK) to pennyweights, multiply the long hundredweight (UK) value by 32666.7. Hence, 1 long hundredweight (UK) equals 32666.7 pennyweights.
• 1 long hundredweight (UK) = 32666.7 pennyweights
• 2 long hundredweights (UK) = 65333.4 pennyweights
• 3 long hundredweights (UK) = 98000.1 pennyweights
• 4 long hundredweights (UK) = 130666.8 pennyweights
• 5 long hundredweights (UK) = 163333.5 pennyweights
• 6 long hundredweights (UK) = 196000.2 pennyweights
• 7 long hundredweights (UK) = 228666.9 pennyweights
• 8 long hundredweights (UK) = 261333.6 pennyweights
• 9 long hundredweights (UK) = 294000.3 pennyweights
• 10 long hundredweights (UK) = 326667 pennyweights
### Conversion formula
pennyweights = long hundredweights (UK) × 32666.7
### Conversion factor
The conversion factor to convert long hundredweights (UK) to pennyweights is 32666.7
### Examples to convert cwt long to dwt
#### Example 1
Convert 50 cwt long to dwt.
Solution:
Converting from long hundredweights (UK) to pennyweights is very easy.
We know that 1 cwt long = 32666.7 dwt.
So, to convert 50 cwt long to dwt, multiply 50 cwt long by 32666.7 dwt.
50 cwt long = 50 × 32666.7 dwt
50 cwt long = 1633335 dwt
Therefore, 50 long hundredweights (UK) converted to pennyweights is equal to 1633335 dwt.
#### Example 2
Convert 125 cwt long to dwt.
Solution:
1 cwt long = 32666.7 dwt
So, 125 cwt long = 125 × 32666.7 dwt
125 cwt long = 4083337.5 dwt
Therefore, 125 cwt long converted to dwt is equal to 4083337.5 dwt.
If you don't want to do the long hundredweights (UK) to pennyweights calculation manually, you can simply use our long hundredweights (UK) to pennyweights calculator.
### How to use the long hundredweights (UK) to pennyweights converter?
To use the long hundredweights (UK) to pennyweights converter, follow these steps:
1. Enter the long hundredweight (UK) value that you want to convert.
2. Click "Convert to pennyweights" to get the conversion result in pennyweights.
3. To copy the conversion steps, click "Copy".
4. To report an incorrect conversion, click "Report incorrect conversion".
5. To reset the converter, click "Reset".
6. To convert pennyweights to long hundredweights (UK), click "Swap units".
7. To convert between other units of weight, select the units from the drop-down menus.
Using the cwt long to dwt converter ensures accuracy in your measurement unit conversions, minimizes conversion errors and provides accurate conversion results. With its user-friendly and intuitive interface, this measurement unit converter makes obtaining precise conversion results a breeze.
## Long hundredweights (UK) to pennyweights conversion table
The long hundredweights (UK) to pennyweights conversion chart below shows a list of various long hundredweight (UK) values converted to pennyweights.
Long hundredweights (UK) (cwt long) Pennyweights (dwt)
1 cwt long 32666.7 dwt
2 cwt long 65333.4 dwt
3 cwt long 98000.1 dwt
4 cwt long 130666.8 dwt
5 cwt long 163333.5 dwt
6 cwt long 196000.2 dwt
7 cwt long 228666.9 dwt
8 cwt long 261333.6 dwt
9 cwt long 294000.3 dwt
10 cwt long 326667 dwt
11 cwt long 359333.7 dwt
12 cwt long 392000.4 dwt
13 cwt long 424667.1 dwt
14 cwt long 457333.8 dwt
15 cwt long 490000.5 dwt
16 cwt long 522667.2 dwt
17 cwt long 555333.9 dwt
18 cwt long 588000.6 dwt
19 cwt long 620667.3 dwt
20 cwt long 653334 dwt
21 cwt long 686000.7 dwt
22 cwt long 718667.4 dwt
23 cwt long 751334.1 dwt
24 cwt long 784000.8 dwt
25 cwt long 816667.5 dwt
26 cwt long 849334.2 dwt
27 cwt long 882000.9 dwt
28 cwt long 914667.6 dwt
29 cwt long 947334.3 dwt
30 cwt long 980001 dwt
31 cwt long 1012667.7 dwt
32 cwt long 1045334.4 dwt
33 cwt long 1078001.1 dwt
34 cwt long 1110667.8 dwt
35 cwt long 1143334.5 dwt
36 cwt long 1176001.2 dwt
37 cwt long 1208667.9 dwt
38 cwt long 1241334.6 dwt
39 cwt long 1274001.3 dwt
40 cwt long 1306668 dwt
41 cwt long 1339334.7 dwt
42 cwt long 1372001.4 dwt
43 cwt long 1404668.1 dwt
44 cwt long 1437334.8 dwt
45 cwt long 1470001.5 dwt
46 cwt long 1502668.2 dwt
47 cwt long 1535334.9 dwt
48 cwt long 1568001.6 dwt
49 cwt long 1600668.3 dwt
50 cwt long 1633335 dwt
51 cwt long 1666001.7 dwt
52 cwt long 1698668.4 dwt
53 cwt long 1731335.1 dwt
54 cwt long 1764001.8 dwt
55 cwt long 1796668.5 dwt
56 cwt long 1829335.2 dwt
57 cwt long 1862001.9 dwt
58 cwt long 1894668.6 dwt
59 cwt long 1927335.3 dwt
60 cwt long 1960002 dwt
61 cwt long 1992668.7 dwt
62 cwt long 2025335.4 dwt
63 cwt long 2058002.1 dwt
64 cwt long 2090668.8 dwt
65 cwt long 2123335.5 dwt
66 cwt long 2156002.2 dwt
67 cwt long 2188668.9 dwt
68 cwt long 2221335.6 dwt
69 cwt long 2254002.3 dwt
70 cwt long 2286669 dwt
71 cwt long 2319335.7 dwt
72 cwt long 2352002.4 dwt
73 cwt long 2384669.1 dwt
74 cwt long 2417335.8 dwt
75 cwt long 2450002.5 dwt
76 cwt long 2482669.2 dwt
77 cwt long 2515335.9 dwt
78 cwt long 2548002.6 dwt
79 cwt long 2580669.3 dwt
80 cwt long 2613336 dwt
81 cwt long 2646002.7 dwt
82 cwt long 2678669.4 dwt
83 cwt long 2711336.1 dwt
84 cwt long 2744002.8 dwt
85 cwt long 2776669.5 dwt
86 cwt long 2809336.2 dwt
87 cwt long 2842002.9 dwt
88 cwt long 2874669.6 dwt
89 cwt long 2907336.3 dwt
90 cwt long 2940003 dwt
91 cwt long 2972669.7 dwt
92 cwt long 3005336.4 dwt
93 cwt long 3038003.1 dwt
94 cwt long 3070669.8 dwt
95 cwt long 3103336.5 dwt
96 cwt long 3136003.2 dwt
97 cwt long 3168669.9 dwt
98 cwt long 3201336.6 dwt
99 cwt long 3234003.3 dwt
100 cwt long 3266670 dwt
## Frequently asked questions about long hundredweights (UK) to pennyweights conversion
### How do you convert cwt long to dwt?
To convert a weight measurement in long hundredweights (UK) to pennyweights, multiply the long hundredweight (UK) value by 32666.7.
### How many pennyweights are in a long hundredweight (UK)?
There are 32666.7 pennyweights in a long hundredweight (UK).
### How many dwt is equal to 1 cwt long?
1 cwt long equals 32666.7 dwt
### What is the equivalent weight in pennyweights for 1 long hundredweight (UK)?
1 long hundredweight (UK) is equivalent to 32666.7 pennyweights
### Is 1 cwt long equal to 32666.7 dwt?
Yes, 1 cwt long is equal to 32666.7 dwt
### How much is 1 long hundredweight (UK) in pennyweights?
The answer is 32666.7 pennyweights
### What is the equivalent of 1 cwt long in dwt?
1 cwt long is equivalent to 32666.7 dwt
### What is 1 cwt long in dwt?
1 cwt long = 32666.7 dwt
### What is 1 long hundredweight (UK) equal to in pennyweights?
1 long hundredweight (UK) is equal to 32666.7 pennyweights
### What is the ratio of long hundredweights (UK) to pennyweights?
The ratio of long hundredweights (UK) to pennyweights is 1:32666.7 because there are 32666.7 pennyweights in 1 long hundredweight (UK).
## Long hundredweights (UK) (cwt long)
### 1 long hundredweight (UK) equivalents in other weight units
• 1 long hundredweight (UK) = 254011.73 carats
• 1 long hundredweight (UK) = 784000 grains
• 1 long hundredweight (UK) = 50802.4 grams
• 1 long hundredweight (UK) = 50.8 kilograms
• 1 long hundredweight (UK) = 0.05 long tons (UK)
• 1 long hundredweight (UK) = 0.0508023 metric tons (or tonnes)
• 1 long hundredweight (UK) = 50802345440 micrograms
• 1 long hundredweight (UK) = 50802345.44 milligrams
• 1 long hundredweight (UK) = 1792 ounces
• 1 long hundredweight (UK) = 32666.7 pennyweights
• 1 long hundredweight (UK) = 112 pounds
• 1 long hundredweight (UK) = 1.12 short hundredweights (US)
• 1 long hundredweight (UK) = 0.056 short tons (US)
• 1 long hundredweight (UK) = 8 stones
• 1 long hundredweight (UK) = 1633.33 troy ounces
• 1 long hundredweight (UK) = 136.11 troy pounds
## Pennyweights (dwt)
### Pennyweights to long hundredweights (UK) conversion (dwt to cwt long)
• 1 pennyweight = 3.06122 × 10-5 long hundredweights (UK)
• 2 pennyweights = 6.12244 × 10-5 long hundredweights (UK)
• 3 pennyweights = 9.18366 × 10-5 long hundredweights (UK)
• 4 pennyweights = 0.0001224488 long hundredweights (UK)
• 5 pennyweights = 0.000153061 long hundredweights (UK)
• 6 pennyweights = 0.0001836732 long hundredweights (UK)
• 7 pennyweights = 0.0002142854 long hundredweights (UK)
• 8 pennyweights = 0.0002448976 long hundredweights (UK)
• 9 pennyweights = 0.0002755098 long hundredweights (UK)
• 10 pennyweights = 0.000306122 long hundredweights (UK)
### 1 pennyweight equivalents in other weight units
• 1 pennyweight = 7.78 carats
• 1 pennyweight = 24 grains
• 1 pennyweight = 1.56 grams
• 1 pennyweight = 0.00155517 kilograms
• 1 pennyweight = 3.06122 × 10-5 long hundredweights (UK)
• 1 pennyweight = 1.53061 × 10-6 long tons (UK)
• 1 pennyweight = 1.55517 × 10-6 metric tons (or tonnes)
• 1 pennyweight = 1555174 micrograms
• 1 pennyweight = 1555.17 milligrams
• 1 pennyweight = 0.0548571 ounces
• 1 pennyweight = 0.00342857 pounds
• 1 pennyweight = 3.42857 × 10-5 short hundredweights (US)
• 1 pennyweight = 1.71429 × 10-6 short tons (US)
• 1 pennyweight = 0.000244898 stones
• 1 pennyweight = 0.05 troy ounces
• 1 pennyweight = 0.00416667 troy pounds | 3,251 | 9,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.647794 |
https://prepinsta.com/java-program/search-an-element-in-a-matrix/ | 1,674,924,913,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499646.23/warc/CC-MAIN-20230128153513-20230128183513-00318.warc.gz | 483,049,636 | 22,430 | # Search an Element in a Matrix in Java
## Search an element in a matrix in Java
Here, in this page we will discuss the program to print the spiral traversal of the matrix in Java programming language. We’re given a n x n matrix and the value of the element in the matrix that needs to be found.
If the searched element is present in the matrix, we must find its position.
Every row and column in the specified matrix is sorted in ascending order.
## Method discussed :
• Method 1 : Using Brute force.
• Method 2 : Using DFS
## Method 1 :
• Run a nested for loop that iterates over the matrix
• Compares the current value to the searched value.
• If it is, print its location and set the value of flag to 1 before exiting the loop.
• If flag becomes 1, the outer loop will be broken as well.
• If flag==0 after exiting the nested loop, then print “Not Found.”
### Method 1 : Code in Java
Run
```import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args)
{
int a[][] = {{0,1,12,3}, {4,5,6,7}, {8,9,10,11}};;
int size = 4;
int search = 6;
int flag=0;
for(int i=0; i<size; i++){
for(int j=0; j<size; j++){
if(a[i][j]==search){
System.out.println("Element is found at ("+ i+", "+ j +") position");
flag=1;
break;
}
}
if(flag==1)
break;
}
if(flag==0)
}
}```
### Output
`Element is found at (1, 2) position`
## Method 2 :
• Run a loop that will terminate when i>n
• Now, If the current element is greater than x then decrease the count of j. Exclude the current column.
• And If the current element is less than x then increase the count of i. Exclude the current row.
• If the element is equal, then print the position, set flag = 1 and break the loop.
• After coming out from the loop, if flag==0 then print “Not Found”
### Method 2 : Code in Java
Run
```import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args)
{
int a[][] = {{0,1,12,3}, {4,5,6,7}, {8,9,10,11}};;
int x = 6;
int flag=0;
int i=0, j=3;
while (i < 3 && j >= 0)
{
if (a[i][j] == x)
{
System.out.println("Element is found at ("+ i+", "+ j +") position");
flag=1;
break;
}
if (a[i][j] > x)
j--;
else
i++;
}
if(flag==0)
`Element is found at (1, 2) position` | 643 | 2,205 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-06 | latest | en | 0.636425 |
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Restricted Five Coin Sort (Posted on 2016-07-14)
A set of five coins are of differing weights, no two are equal.
Devise a method to sort the coins by weight using only a balance scale, subject to the restriction that only one coin may be placed on each side in each weighing.
Can this be done in the theoretical minimum of 7 weighings?
Submitted by Brian Smith No Rating Solution: (Hide) This strategy will achieve the theoretical minimum of 7 weighings: Pick any pair and compare them. Then pick a second pair and compare those as well. Then compare the two heavy coins with the third weighing. After these three weighings the coins can be labeled A-E such that A>C>D, A>B and E is the unweighed coin. Use the next two weighings to determine where coin E falls in among A, C, and D. The forth weighing will be E vs C and the fifth will be either E vs A or E vs D depending on the fourth weighing. At this point the relative weights of A, B, C, and E are known. All that is left is to place coin B. From earlier A>B. Then B can be placed among C, D, and E with two weighings, similar to how E was placed among A, C, and D.
Subject Author Date Solution Jer 2016-07-14 13:47:58
Search: Search body:
Forums (0) | 327 | 1,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2018-39 | latest | en | 0.937129 |
niessaygovd.card-hikaku.info | 1,539,917,436,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512268.20/warc/CC-MAIN-20181019020142-20181019041642-00098.warc.gz | 257,688,917 | 5,659 | # Tailed hypthesis
The first step in hypothesis testing is establishing the null and alternative hypotheses before the one-tailed test can be performed a null hypothesis is the claim that the researcher hopes to. Hypothesis testing is an important activity of empirical research and evidence-based medicine a well worked up hypothesis is half the answer to the research question for this, both knowledge of the subject derived from extensive review of the literature and working knowledge of basic statistical. The hypothesis (or any other hypothesis) is right or wrong in fact, it assumes the null hypothesis is right in light of these facts, there are actually two classical schools of thought on how best to use the. One-tailed and two-tailed tests a test of a statistical hypothesis, where the region of rejection is on only one side of the sampling distribution, is called a one-tailed test. Describes how to test the null hypothesis that some estimate is due to chance vs the alternative hypothesis that there is some statistically significant effect.
If the alternative hypothesis contains a not equals to sign, then we have a two-tailed test in the other two cases, when the alternative hypothesis contains a strict inequality, we use a one-tailed test. Psychology definition for one tailed hypothesis in normal everyday language, edited by psychologists, professors and leading students help us get better. One-tailed tests one-tailed hypothesis test -a hypothesis test in which the population parameter is known to fall to the right or. I was recently asked about when to use one and two tailed tests the long answer is: use one tailed tests when you have a specific hypothesis about the direction of your relationship some examples include you hypothesize that one group mean is larger than the other you hypothesize that the.
In the previous example, you tested a research hypothesis that predicted not only that the sample mean would be different from the population mean but that it would be different in a specific direction—it would be lower this test is called a directional or one‐tailed test because the region of. I need to know what a one tailed means and what a two tailed means, in a psychological way if possible and maybe an example thankyou. Two-tailed test, also known as the non directional hypothesis, defined as the standard test of significance to determine if there is a relationship between variables in either direction.
Hypothesis testing in the terminology of hypothesis testing, this is known as the number of tails of the hypothesis: one tail the hypothesis has an implied. A two tailed hypothesis is used in statistical testing to determine the relationship between a sample and a distribution in statistics you compare a sample (example: one class of high school seniors sat scores) to a larger set of numbers, or a distribution (the sat scores for all us high school. We want to test the null hypothesis that the slope for moderately large samples and a one tailed test, include implementations of student's t-test.
Two-tailed tests if the hypothesis is stated such that you are only interested in a difference (ie, you have not stated a direction such as greater or less than. The z-score values of +196 are the critical values for a two tailed hypothesis test when using the normal distribution to represent the sample distribution that is, if the sampling distribution were shaped as a normal distribution, 25% of the scores are above +196 and 25% of the scores are below -196 (for a total area of 5% outside of. Depending on the alternative hypothesis operator, greater than operator will be a right tailed test, less than operator is a left tailed test, and not equal operator is a two tailed test alternative hypothesis has the greater than operator , right tailed test. Hypothesis testing, journals tend to null hypothesis: researchers conducted two-tailed tests for possibility the. Will that be one tail or two you will often see a statistical test or a hypothesis referred to as one-tailed or two-tailed a one-tailed hypothesis is simply one that specifies the direction of a difference or correlation, while a two-tailed hypothesis is one that does not.
Looking for some guidance on hypothesis testing, particularly how the null hypothesis is stated i conducted an improvement project aimed aimed at reducing lead times the hypothesis was that improving the flow of materials would reduce lead times from a mean of 90 days to 30 days a pilot was run. Statistical test for population proportion and population mean statistical and practical significances using a confidence interval to draw a conclusion about a two-tailed test a reminder of what is a p-value in hypothesis testing: p-value is a probability of obtaining a value of the test statistic. 157 unit 22 one-sided and two-sided hypotheses tests objectives: • to differentiate between a one-sided hypothesis test and a two-sided hypothesis test about a population.
Since the alternative hypothesis uses the not equal sign to sign we will conduct a two tail test (c) a truck manufacturer claims that her trucks use no more than 3 gallons of gasoline per mile ho: u 3 since the alternative hypothesis uses a greater than sign we will conduct a right tail test. For a two-tail test, the p-value is the probability of obtaining a value for the test statistic as unlikely as that provided by the sample in hypothesis testing if the null hypothesis is rejected. Which tail (left or right) under consideration for a one-tailed test depends on the direction of the hypothesis establishing the significance level and the corresponding critical value(s) is sometimes considered the second step in hypothesis testing.
Because two-tailed testing, by its very nature, does not reflect the directionality of the research hypothesis, it is apparent that it loses the exact logical connection between the directional research hypothesis and its statistical hypothesis. Introduction to hypothesis testing - page 1 for example, suppose the null hypothesis is that the wages of men and women are equal a two-tailed alternative would simply state that the wages are not equal - implying that men. Running a two-tailed z-test hypothesis test by hand how to video z-test using excel example: suppose it is up to you to determine if a certain state (michigan) receives a significantly different amount of public school funding (per student) than the usa average. One- and two-tailed tests: so far, we have talked in general terms about the null and alternate hypotheses without getting into the specifics of the type of comparison that we wish to make in our tests there are two main questions we might be asking when performing our tests.
Tailed hypthesis
Rated 4/5 based on 31 review
2018. | 1,326 | 6,788 | {"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} | 2.890625 | 3 | CC-MAIN-2018-43 | latest | en | 0.954749 |
https://im.kendallhunt.com/k5_es/teachers/grade-1/unit-3/lesson-18/preparation.html | 1,721,446,151,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514981.25/warc/CC-MAIN-20240720021925-20240720051925-00330.warc.gz | 265,838,283 | 27,777 | Lesson 18
Patrones en la suma
Lesson Purpose
The purpose of this lesson is for students to find expressions equivalent to $$10 + n$$ expressions. Students then look for patterns in the expressions.
Lesson Narrative
In the previous lesson, students discussed how making a ten can help them find the value of expressions in which one addend is close to 10. In this lesson, students continue to add two addends within 20, focusing on how making a ten can help them add. In the first activity, students match addition expressions to their equivalent $$10 + n$$ expression. In the second activity, students participate in a Gallery Walk, noticing patterns between expressions that represent a sum. In each activity, students have access to double 10-frames and two-color counters as they explore patterns they notice.
• Action and Expression
• MLR7
Learning Goals
Teacher Facing
• Look for and use patterns in addition expressions to add within 20.
• Make a ten to find the sum of two numbers within 20.
Student Facing
• Busquemos patrones y usémoslos como ayuda para sumar hasta 20.
Required Materials
Materials to Gather
Materials to Copy
• Compare Stage 2 Addition Cards to 20
Required Preparation
Activity 1:
• Create a set of Compare Stage 2 Addition Cards to 20 from the blackline master for each group of 3.
Activity 3:
• Gather materials from previous centers:
• Shake and Spill, Stages 3–5
• Compare, Stage 1
• Number Puzzles, Stages 1 and 2
Lesson Timeline
Warm-up 10 min Activity 1 10 min Activity 2 15 min Activity 3 15 min Lesson Synthesis 10 min Cool-down 0 min
Teacher Reflection Questions
As students worked in their small groups today, whose ideas were heard, valued, and accepted? How can you adjust the group structure tomorrow to ensure each student’s ideas are a part of the collective learning? | 425 | 1,836 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-30 | latest | en | 0.843786 |
https://cstheory.stackexchange.com/questions/29353/np-complete-problems-with-optimal-approximation-in-poly-time/29357#29357 | 1,638,252,209,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358953.29/warc/CC-MAIN-20211130050047-20211130080047-00009.warc.gz | 252,749,491 | 34,086 | # NP-complete problems with optimal approximation in poly-time
I'm looking for examples of hard optimization problems, for which we have an optimal approximation (not that this is not the same as $PTAS$, as we require a completely tight approximation, and not $1+\epsilon$-multiplicative approximation), in a sense defined as follows:
Formally, let $L\subseteq\Sigma^*\times \mathbb N$ be the decision version of some minimization (or maximization, flipping the definitions) problem, i.e. $$\forall n\leq m:(w,n)\in L\implies (w,m)\in L$$
Which examples of such known languages $L$ is $NP$-hard, but if we are allowed of a minimal relaxation, it is poly time solvable.
By minimal relaxation I mean that there exists a TM such that given an input $(w,n)$:
• The machine accepts if $(w,n+1)\in L$.
• The machine rejects if $(w,n-1)\not \in L$.
• The machine may act arbitrarily on other cases.
An example for such problem is the Degree-constrained spanning tree, where the problem is finding an MST with minimal maximum vertex degree.
Which other languages are NP-hard but subject to optimal approximation in poly time?
• 3-coloring problem on planar graphs is $NP$-hard while every planar graph is 4-colorable in polynomial time (by the four color theorem). Feb 1 '15 at 15:51
• I guess there is an issue of discretization that is tricky. For example, if we want to formalize Max-Cut in this framework, what is the meaning of $n+1$ and $n-1$?
– usul
Feb 1 '15 at 16:27
• Maybe better term for optimal approximation or minimal relaxation is approximation with additive constant error. Feb 1 '15 at 18:29
• @Saeed, can you explain how that's equivalent? I think the issue is that I still don't understand what the role of the constant $1$ is in the definition.
– usul
Feb 2 '15 at 0:17
• @usul I didn't understand what you don't understand. But you can search about additive constant error to understand what it is, assuming you know this, OP's question is about additive error 1 (replace constant with 1) Feb 2 '15 at 8:05
One example is the chromatic index (edge-coloring number) of undirected graphs. By Vizing's theorem, the chromatic index either equals the max-degree $\Delta$ or equals $\Delta+1$.
• Vizing's proof gives a polynomial time (inductive) procedure for finding a coloring with $\Delta+1$ colors.
• Holyer has proved that deciding whether the chromatic index equals $\Delta$ is NP-complete. | 613 | 2,414 | {"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} | 2.984375 | 3 | CC-MAIN-2021-49 | latest | en | 0.919606 |
http://www.slideserve.com/ashton-conway/estimating-sums-differences | 1,490,728,647,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189802.72/warc/CC-MAIN-20170322212949-00604-ip-10-233-31-227.ec2.internal.warc.gz | 688,397,060 | 16,206 | This presentation is the property of its rightful owner.
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# Estimating Sums & Differences PowerPoint PPT Presentation
Estimating Sums & Differences. Warm-Up. Find the value of each expression. 4.7 + 5.08 10 – 3.58. VOCAB. Rational Number: a fraction made from integers Sum : ADD Difference : Subtract Rounding : Choose easy #’s close to hard #’s. Estimating. Decimals less than one
Estimating Sums & Differences
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#### Presentation Transcript
Estimating Sums & Differences
Warm-Up
Find the value of each expression.
• 4.7 + 5.08
• 10 – 3.58
VOCAB
• Rational Number:
• a fraction made from integers
• Sum:
• Difference:
• Subtract
• Rounding:
• Choose easy #’s close to hard #’s.
Estimating
Decimals
• less than one
• round to 0, 0.5 or 1
• more than one
• round to the nearest integer (whole #)
Fractions
• less than one
• round to 0, ½ or 1
• mixed numbers
• round to the nearest integer (whole #)
### Let’s Try Some Rounding…
• 0.2 rounds to ________
• 0.4 rounds to ________
• 0.7 rounds to ________
• 0.75 rounds to ________
• 1.6 rounds to ________
• ¼ rounds to _________
• 1/3 rounds to _________
• 3/5 rounds to _________
• 1 ¾ rounds to _________
### Example 1a
Estimate the value of – 4.23 + 2.8.
• Both numbers are more than one and should be rounded to nearest integer.
• – 4.23 → – 4
• 2.8 → 3
• – 4 + 3 = –1
• – 4.23 + 2.8 ≈ –1
### Example 1b
Estimate the value of 0.94 – (–0.43).
• Both numbers are less than one unit away from zero and should be rounded to the nearest 0.5.
• 0.94 → 1
• – 0.43 → – 0.5
• 1 – (–0.5) = 1 + 0.5 = 1.5
• 0.94 – (–0.43) ≈ 1.5
### Example 2a
Estimate the value of .
• Each fraction is less than 1 unit from zero so round to the
nearest .
### Example 2b
Estimate the value of .
• Each fraction is more than 1 unit from zero so round to the nearest integer.
Estimate the value of each expression.
• 3.21 + 6.3
• 0.13 – 0.925
• Matt bought pounds of apples, pounds of grapes and pounds of potatoes. How many pounds did he buy?
### Communication Prompt
Describe a time when you used estimating in a situation
outside of math class. | 796 | 2,595 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-13 | longest | en | 0.714225 |
http://list.seqfan.eu/pipermail/seqfan/2001-October/038214.html | 1,716,084,223,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057631.79/warc/CC-MAIN-20240519005014-20240519035014-00343.warc.gz | 17,198,980 | 5,940 | # sequences that are permutations
Antti Karttunen karttu at megabaud.fi
Mon Oct 15 11:50:07 CEST 2001
```
"N. J. A. Sloane" wrote:
> Howard Landman <howard at polyamory.org> has put together a preliminary
> list of the sequences that are permutations of the natural numbers
> (or in some cases of the nonnegative integers).
>
> Here is the current version. If anyone knows of other examples,
> please send them to me (njas).
I think most of these could be found with a simple program, that checks
that A) the sequence does not contain duplicates,
B) all integers in range [1,n] occur for some small n (say 20). And then
one has to check manually what one gets. (There might be interesting
conjectures lurking there...)
This would leave out only "wildly fluctuating" permutations like this one:
ID Number: A011262
Sequence: 1,4,9,2,25,36,49,16,3,100,121,18,169,196,225,8,289,12,361,
50,441,484,529,144,5,676,81,98,841,900,961,64,1089,1156,
1225,6,1369,1444,1521,400,1681,1764,1849,242,75,2116,2209,
72,7,20,2601,338,2809,324,3025,784,3249
Name: In the prime factorization of n, increment odd powers and decrement
even powers (multiplicative and self-inverse).
Formula: Multiplicative with f(p^k) = p^(k-1) if k even, p^(k+1) if k odd.
Keywords: nonn,easy
Offset: 1
Author(s): Marc Le Brun (mlb at well.com)
and: (but this one is already in the index:)
ID Number: A059884
Sequence: 0,1,2,4,8,3,128,5,32,9,32768,6,2147483648,129,10,16,
9223372036854775808,33,
170141183460469231731687303715884105728,12,130,32769
Name: Prime factorization of n encoded by recursively interleaving bits of
successive prime exponents.
Also, all the "permutations of the natural numbers" that I have
submitted (and some other SeqFanatiques too) can be found with keywords
"permutations of the natural numbers", although most of those are not yet in the
Index entry itself.
Ah, here are some vague thoughts that I wrote about these kind of sequences
in March 2000. Comments are welcome.
Some possible methods to create permutations of the natural numbers:
1) From any infinite sequence Axxxxxx construct a permutation of natural
numbers, by reversing each consecutive block of abs(Axxxxxx[n])
integers. If Axxxxxx does not contain any term > 1, then
the result is natural numbers A000027 itself.
The resulting permutation is always self-inverse
(consists of fixed terms and/or transpositions only).
Dumb as sequences, but might have some uses
as functions.
2) From any infinite sequence Axxxxxx whose terms are all discrete
and positive, and whose complement Ayyyyyy is likewise infinite
(thus A000027 won't do), construct a self-inverse permutation
composed of discrete transpositions, by swapping each
Axxxxxx[n] <-> Ayyyyyy[n].
I.e. From 3,6,9,12,15,18,21,...
and 1,2,4,5,7,8,10,11
we get 3,6,1,9,12,2,15,18,4,...
E.g. A003622[n] <-> A022342[n].
Most of these are quite dumb as well.
3a) From any infinite sequence known to contain all natural numbers,
but some of them more than once, construct a permutation
by pruning off all duplicates after the first
occurrence of each value n. (E.g. from Pascal 2-1 triangle A029654).
3b) Like above, but prune off all duplicates before the _last_
proved occurrence of each value n. Here the last occurrence
should of course be well-defined.
3c) Like 3a and 3b, but leave the f(n):th occurrence of each value of n,
where f is some arbitrary, but well-defined function.
4) From any function f(x) proved to have any particular integral value y
with only finite number of integer arguments x1,x2,...,xn
(e.g. number theoretic functions phi and sigma), construct
a permutation of natural numbers by first listing (in some definite
order) all x for which f(x) = y1 (the global minimum for f),
then all x for which f(x) = y2 (the nearest possible value
greater than y1), then all x for which f(x) = y3, etc.
Note that y1,y2,y3 are not necessarily consecutive.
Example: invphi sequence. A032447
5) From any two distinct infinite sequences A and B which have been proved
to span exactly the same set of distinct numbers or other objects,
construct a permutation of natural numbers by indexing A with B,
and its inverse by indexing B with A.
Here the set of distinct numbers can for example be
* all non-negative integers (A001477 itself and A003188, i.e. Gray Code).
(binary encoding of square-frees A048672, Bower's A052330, etc.)
(If we look the binary representation of n, then the actual set of
distinct objects that are spanned here is the set of subsets
of the cardinality Aleph-0 set.)
* all binary sequences with some combinatorial condition,
e.g. Fibbinary numbers (no ...11..), numbers composed of
exactly n or max. n bits in their binary expansion, etc.
* all sets of sequences of digits [{0,1},{0-2},{0-3},...,{0-n}]
(one such set is factorial number representation of non-negative
integers, but are there any others, which would result
a different ordering naturally?)
* all rational fractions. (e.g. A054424)
* all C with algebraic (non-transcedental) R and I components,
or some subset?
E.g. is there a way to "canonically" list all quadratic polynomials
with integral coefficients to get a "canonical" list
of quadratic roots in C (or just the Real ones), and then the
other sequence would be constructed from "canonical list"
of infinite, but periodic continued fractions.
6) From any known two permutations of natural numbers (not necessarily
distinct), construct possibly one or two new ones by composing
them with each other, from the right and from the left.
(I.e. this is not-necessarily commutative multiplication of permutations.)
7) Walk through infinite square lattice, starting from the corner, in some
way (e.g. some regular knight walk), and then read the sequence
from the successive antidiagonals of this triangular table.
(Simple variants of this theme: A038722, A056537, A060736)
But in more fundamental way, maybe one good classification could be
according to a cycle structure of the permutation:
A) cycles are finite and bounded to regular "regions". (e.g. the Gray code A003188,
where the regions are contained between each subsequence 2^n - 2^(n+1).
B) cycles are finite, but less clearly bounded, but some upper and lower bounds
can still be found.
(in above two cases one could count the number of cycles ("equivalence classes")
formed in each "region", even although they might not have any
clear combinatorial significance at all...)
C) cycles can be known to be finite, but with no known structure.
D) one or more of the cycles are infinite.
E.g. from permutations like:
ID Number: A057114
Sequence: 3,1,7,2,6,14,15,4,5,12,13,28,29,30,31,8,9,10,11,24,25,26,27,
56,57,58,59,60,61,62,63,16,17,18,19,20,21,22,23,48,49,50,51,
52,53,54,55,112,113,114,115,116,117,118,119,120,121,122,123,
124,125,126,127,32,33,34,35,36,37,38,39,40,41,42,43,44,45,
46,47,96,97,98,99,100,101,102,103
Name: Order-preserving permutation of the rational numbers (x -> x+1);
positions in Stern-Brocot tree.
we know that only cycle lengths that can occur are 1 (but not in this one)
and infinite.
and permutations like this one (an inspiration for famous 3x+1 problem?, not
in index! Is this a permutation conjecturally only?):
ID Number: A006369 (Formerly M2245)
Sequence: 0,1,3,2,5,7,4,9,11,6,13,15,8,17,19,10,21,23,12,25,27,14,29,
31,16,33,35,18,37,39,20,41,43,22,45,47,24,49,51,26,53,55,28,
57,59,30,61,63,32,65,67,34,69,71,36,73,75,38
Name: Nearest integer to 4n/3 unless that is an integer, when 2n/3.
BTW, here's are an interesting paper regarding the topic:
http://www.cecm.sfu.ca/organics/papers/buhler/index.html
That's all for now, maybe some day I have enough time to actually
do this in more thorough fashion...
Terveisin,
Antti Karttunen
>
>
> The entries in the index will also be updated.
>
> Sequence-Inverse
> A000027 - self-inverse
> A002251 - self-inverse
> A003100 - self-inverse
> A003188 - A006068
> A004484 - A064206
> A004485 - A064207
> A004486 - A064208
> A004487 - A064211
> A019444 - self-inverse
> A026243 - self-inverse
> A029654 - A064360
> A064413 - A064664
> A032447 - A064275
> A035312 - A035358
> A035506 - A064357
> A035513 - A064274
> A047708 - A048850
> A048647 - A064212
> A048672 - A064273
> A048673 - A064216
> A048679 - A048680
> A052330 - A064358
> A059900 - A059884
> A052331 - A064359
> A054238 - A054239
> A054424 - A054426
> A054427 - A054428
> A054429 - self-inverse
> A054430 - self-inverse
> A054081 - n/a array: rows are permutations but it isn't itself
>
> I think the squares of these permutations also should be in the database.
> I suggested this to Howard, but this is a big job and if anyone else would
> be willing to send in some "squares" that would be very helpful.
>
> If f is one of the permutations mentioned above
> then what I mean is the sequence
> f(f(1)), f(f(2)), f(f(3)), f(f(4)), ...
> - except perhaps beginning with f(f(0)) if that is appropriate.
>
> Of course many of these may already be in the database; if so this should be noted.
>
> Neil Sloane
``` | 2,668 | 9,035 | {"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} | 2.953125 | 3 | CC-MAIN-2024-22 | latest | en | 0.859892 |
https://slides.com/johnjasper/linear-algebra-day-27 | 1,719,133,479,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00646.warc.gz | 462,391,469 | 13,780 | # Day 27:
Proof of Singular Value Decomposition
Theorem (Singular Value Decomposition - part 1). If $$B$$ is an $$m\times n$$ matrix and $$p=\min\{m,n\}$$, then there are orthonormal bases $$\{u_{1},\ldots,u_{m}\}$$ for $$\R^{m}$$ and $$\{v_{1},\ldots,v_{n}\}$$ for $$\R^{n}$$, and nonnegative scalars $$\sigma_{1},\ldots,\sigma_{p}$$ such that
$Bv_{i} = \begin{cases} \sigma_{i}u_{i} & i\leq p\\ 0, & i>p.\end{cases}$
Notes:
• The numbers $$\sigma_{1},\sigma_{2},\ldots,\sigma_{p}$$ are called the singular values of $$B$$.
• The sequence of singular values is denoted with the symbol $$\sigma(B)$$.
• The vectors $$v_{1},v_{2},\ldots,v_{n}$$ are called the right singular vectors of $$B$$
• The vectors $$u_{1},u_{2},\ldots,u_{m}$$ are called the left singular vectors of $$B$$.
Proof. Let $$v_{1},\ldots,v_{n}$$ be an orthonormal basis of eigenvectors of $$B^{\top}B$$ with associated eigenvalues $$\lambda_{1}\geq \lambda_{2}\geq\ldots\geq \lambda_{n}\geq 0.$$
Let $$r$$ be the rank of $$B,$$ so that $$\lambda_{r}>0$$ and $$\lambda_{r+1}=0$$ (if $$r<n$$).
For each $$i\in\{1,\ldots,r\}$$ define $u_{i} = \frac{Bv_{i}}{\sqrt{\lambda_{i}}}.$ Note that $u_{i}\cdot u_{j} = \frac{1}{\sqrt{\lambda_{i}\lambda_{j}}}(Bv_{i})^{\top}(Bv_{j}) = \frac{1}{\sqrt{\lambda_{i}\lambda_{j}}}v_{i}^{\top}B^{\top}Bv_{j} = \frac{1}{\sqrt{\lambda_{i}\lambda_{j}}}v_{i}^{\top}\lambda_{j}v_{j}$
Proof continued. $u_{i}\cdot u_{j} = \frac{1}{\sqrt{\lambda_{i}\lambda_{j}}}v_{i}^{\top}\lambda_{j}v_{j} = \frac{\lambda_{j}}{\sqrt{\lambda_{i}\lambda_{j}}} v_{i}\cdot v_{j} = \begin{cases} 1 & i=j,\\ 0 & i\neq j\end{cases}$
From this we see that $$\{u_{1},\ldots,u_{r}\}$$ is an orthonormal set. Next, note that $BB^{\top} u_{i} = BB^{\top}\frac{Bv_{i}}{\sqrt{\lambda_{i}}} = \frac{1}{\sqrt{\lambda_{i}}}BB^{\top}Bv_{i} = \frac{B\lambda_{i}v_{i}}{\sqrt{\lambda_{i}}} = \lambda_{i}u_{i}.$
Since $$r=\operatorname{rank}(B) = \text{rank}(BB^{\top})$$ we see that $$\{u_{1},\ldots,u_{r}\}$$is an orthonormal basis for $$C(BB^{\top})$$. We can complete this to an orthonormal basis $$\{u_{1},\ldots,u_{m}\}$$. By the definition of $$u_{i}$$ for $$i=1,\ldots,r$$ we have
$Bv_{i} = \sqrt{\lambda_{i}}u_{i}$
Setting $$\sigma_{i} = \sqrt{\lambda_{i}}$$ this completes the proof. $$\Box$$
Example. Consider the matrix
$B = \begin{bmatrix} 7 & 3 & 7 & 3\\ 3 & 7 & 3 & 7\end{bmatrix}$
We compute
$B^{\top}B = \begin{bmatrix} 58 & 42 & 58 & 42\\ 42 & 58 & 42 & 58\\ 58 & 42 & 58 & 42\\ 42 & 58 & 42 & 58\end{bmatrix}$
Since this matrix is positive semidefinite (not definite) we have the spectral decomposition:
$B^{\top}B = \frac{1}{2}\begin{bmatrix} 1 & 1 & 1 & 1\\ 1 & -1 & 1 & -1\\ 1 & 1 & -1 & -1\\ 1 & -1 & -1 & 1\end{bmatrix}\begin{bmatrix} 200 & 0 & 0 & 0\\ 0 & 32 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{bmatrix}\frac{1}{2}\begin{bmatrix} 1 & 1 & 1 & 1\\ 1 & -1 & 1 & -1\\ 1 & 1 & -1 & -1\\ 1 & -1 & -1 & 1\end{bmatrix}$
Note that the spectral decompositon is not unique.
Example. Consider the matrix
$B = \begin{bmatrix} 7 & 3 & 7 & 3\\ 3 & 7 & 3 & 7\end{bmatrix}$
We compute
$B^{\top}B = \begin{bmatrix} 58 & 42 & 58 & 42\\ 42 & 58 & 42 & 58\\ 58 & 42 & 58 & 42\\ 42 & 58 & 42 & 58\end{bmatrix}$
Since this matrix is positive semidefinite (not definite) we have the spectral decomposition:
$$B^{\top}B =$$$\frac{1}{2}\begin{bmatrix} 1 & 1 & \sqrt{2} & 0\\ 1 & -1 & 0 & \sqrt{2}\\ 1 & 1 & -\sqrt{2} & 0\\ 1 & -1 & 0 & -\sqrt{2}\end{bmatrix}\begin{bmatrix} 200 & 0 & 0 & 0\\ 0 & 32 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{bmatrix}\frac{1}{2}\begin{bmatrix} 1 & 1 & 1 & 1\\ 1 & -1 & 1 & -1\\ \sqrt{2} & 0 & -\sqrt{2} & 0\\ 0 & \sqrt{2} & 0 & -\sqrt{2} \end{bmatrix}$
Note that the spectral decompositon is not unique.
Example continued. Let
$V = \frac{1}{2}\begin{bmatrix} 1 & 1 & 1 & 1\\ 1 & -1 & 1 & -1\\ 1 & 1 & -1 & -1\\ 1 & -1 & -1 & 1\end{bmatrix}$
and let $$v_{i}$$ denote the $$i$$th column of $$V$$. Since the columns of $$V$$ are a basis of eigenvectors of $$B^{\top}B$$, we see that these are the right singular vectors of $$B$$.
From the proof, we can see that the singular values of $$B$$ are
$\sigma_{1} = \sqrt{200} = 10\sqrt{2}\quad\text{and}\quad\sigma_{2} = \sqrt{32} = 4\sqrt{2},$
and the left singular vectors are
$u_{1} = \frac{1}{10\sqrt{2}}Bv_{1} = \frac{1}{\sqrt{2}}\begin{bmatrix}1\\ 1\end{bmatrix}\quad\text{and}\quad u_{2} = \frac{1}{4\sqrt{2}}Bv_{2} = \frac{1}{\sqrt{2}}\begin{bmatrix}1\\ -1\end{bmatrix}$
Theorem (Singular Value Decomposition - Outer product form). If $$B$$ is an $$m\times n$$ matrix and $$p=\min\{m,n\}$$, then there are orthonormal bases $$\{u_{1},\ldots,u_{m}\}$$ for $$\R^{m}$$ and $$\{v_{1},\ldots,v_{n}\}$$ for $$\R^{n}$$, and nonnegative scalars $$\sigma_{1},\ldots,\sigma_{p}$$ such that
$B = \sum_{i=1}^{p}\sigma_{i}u_{i}v_{i}^{\top}.$
Proof. From the previous theorem we have orthonormal bases $$\{u_{1},\ldots,u_{m}\}$$ for $$\mathbb{R}^{m}$$ and $$\{v_{1},\ldots,v_{n}\}$$ for $$\mathbb{R}^{n}$$ and nonnegative scalars $$\sigma_{1},\ldots,\sigma_{p}$$ such that $$Bv_{i}=\sigma_{i}u_{i}$$ for all $$i\leq p$$ and $$Bv_{i} = 0$$ for $$i>p$$.
Define the matrix
$C: = \sum_{i=1}^{p}\sigma_{i}u_{i}v_{i}^{\top}$
Note that for $$i_{0}\leq p$$, since $$v_{1},\ldots,v_{n}$$ is orthonormal
$Cv_{i_{0}} = \sum_{i=1}^{p}\sigma_{i}u_{i}v_{i}^{\top}v_{i_{0}} = \sigma_{i_{0}}u_{i_{0}} = Bv_{i_{0}}$
Proof continued. If $$i_{0}>p$$ then $$v_{i_{0}}$$ is orthogonal to $$v_{i}$$ for all $$i\leq p$$ and hence
$Cv_{i_{0}} = 0 = Bv_{i_{0}}.$
Thus, we see that $$Cv_{i} = Bv_{i}$$ for all $$i\in\{1,2,\ldots,n\}$$.
Next, suppose that there is a vector $$v\in\mathbb{R}^{n}$$ such that $$Cv\neq Bv$$. This implies that $$(C-B)v \neq 0$$. Since $$v_{1},v_{2},\ldots,v_{n}$$ is a basis for $$\mathbb{R}^{n}$$, there are scalars $$\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$$ such that
$v = \alpha_{1}v_{1} + \alpha_{2}v_{2} + \cdots + \alpha_{n}v_{n}.$
Finally, we multiply by $$C-B$$ on both sides and we see that
$0\neq (C-B)v = \alpha_{1}(C-B)v_{1} + \alpha_{2}(C-B)v_{2} + \cdots + \alpha_{n}(C-B)v_{n}$
$= \alpha_{1}0 + \alpha_{2}0 + \cdots + \alpha_{n}0 = 0$
This contradiction shows that $$(C-B)v=0$$ for all $$v\in\mathbb{R}^{n}$$, and hence $$C=B$$. $$\Box$$
Example continued. Recall the matrix
$B = \begin{bmatrix} 7 & 3 & 7 & 3\\ 3 & 7 & 3 & 7\end{bmatrix}$
Hence, the outer product form of the singular value decomposition of $$B$$ is $B = \sigma_{1}u_{1}v_{1}^{\top} + \sigma_{2}u_{2}v_{2}^{\top}=\begin{bmatrix} 5 & 5 & 5 & 5\\ 5 & 5 & 5 & 5\end{bmatrix} + \begin{bmatrix} 2 & -2 & 2 & -2\\ -2 & 2 & -2 & 2\end{bmatrix}$
We already saw that the right singular vectors of $$B$$ are $v_{1} = \frac{1}{2}\left[\begin{array}{r} 1\\ 1\\ 1\\ 1\end{array}\right],\ v_{2} = \frac{1}{2}\left[\begin{array}{r} 1\\ -1\\ 1\\ -1\end{array}\right],\ v_{3} = \frac{1}{2}\left[\begin{array}{r} 1\\ 1\\ -1\\ -1\end{array}\right],\ v_{4} = \frac{1}{2}\left[\begin{array}{r} 1\\ -1\\ -1\\ 1\end{array}\right],$ the singular values of $$B$$ are $\sigma_{1} = \sqrt{200} = 10\sqrt{2}\quad\text{and}\quad\sigma_{2} = \sqrt{32} = 4\sqrt{2},$ and the left singular vectors are $u_{1} = \frac{1}{10\sqrt{2}}Bv_{1} = \frac{1}{\sqrt{2}}\begin{bmatrix}1\\ 1\end{bmatrix}\quad\text{and}\quad u_{2} = \frac{1}{4\sqrt{2}}Bv_{2} = \frac{1}{\sqrt{2}}\begin{bmatrix}1\\ -1\end{bmatrix}$
Theorem (Singular Value Decomposition - Matrix Form). If $$B$$ is an $$m\times n$$ matrix, then there is an $$m\times m$$ orthogonal matrix $$R$$, an $$n\times n$$ orthogonal matrix $$Q$$, and an $$m\times n$$ diagonal matrix $$\Sigma$$ such that
$B = R\Sigma Q^{\top}.$
Proof. By the outer product form of the singular value decomposition there are orthonormal bases $$\{u_{1},\ldots,u_{m}\}$$ for $$\R^{m}$$ and $$\{v_{1},\ldots,v_{n}\}$$ for $$\R^{n}$$, and nonnegative scalars $$\sigma_{1},\ldots,\sigma_{p}$$ such that
$B = \sum_{i=1}^{p}\sigma_{i}u_{i}v_{i}^{\top}.$
Let $$R$$ be the matrix with columns $$u_{1},\ldots,u_{m}$$, and let $$Q$$ be the matrix with columns $$v_{1},\ldots,v_{n}$$. Since the columns of $$R$$ and $$Q$$ are orthonormal bases, these matrices are orthogonal.
Let $$\Sigma$$ be $$m\times n$$ diagonal matrix with entries $$\sigma_{1},\sigma_{2},\ldots,\sigma_{p}$$ on the diagonal, and the rest of the diagonal entries equal to zero.
Proof continued.
$$R\Sigma Q^{\top}$$$=\begin{bmatrix} \vert & \vert & & \vert\\ u_{1} & u_{2} & \cdots & u_{m}\\ \vert & \vert & & \vert\end{bmatrix}\begin{bmatrix} \sigma_{1} & & & & & &\\ & \sigma_{2} & & & & &\\ & & \ddots & & & & \\ & & & \sigma_{p} & & & \\ & & & & 0 & &\\ & & & & & \ddots & \\ & & & & & & 0\end{bmatrix}\begin{bmatrix} - & v_{1}^{\top} & - \\ - & v_{2}^{\top} & - \\ & \vdots & \\ - & v_{n}^{\top} & -\end{bmatrix}$
$=\begin{bmatrix} \vert & \vert & & \vert\\ u_{1} & u_{2} & \cdots & u_{m}\\ \vert & \vert & & \vert\end{bmatrix}\begin{bmatrix} - & \sigma_{1} v_{1}^{\top} & - \\ - & \sigma_{2}v_{2}^{\top} & - \\ & \vdots & \\ - & \sigma_{p}v_{p}^{\top} & - \\ - & 0 & -\\ & \vdots & \\ - & 0 & -\end{bmatrix} = \sum_{i=1}^{p}\sigma_{i}u_{i}v_{i}^{\top} = B.$
$$\Box$$
Continuing the last Example. We had
$B = 10\sqrt{2}u_{1}v_{1}^{\top} + 4\sqrt{2}u_{2}v_{2}^{\top}=\begin{bmatrix} 5 & 5 & 5 & 5\\ 5 & 5 & 5 & 5\end{bmatrix} + \left[\begin{array}{rrrr} 2 & -2 & 2 & -2\\ -2 & 2 & -2 & 2\end{array}\right]$
$$v_{1} = \frac{1}{2}\begin{bmatrix} 1\\ 1\\ 1\\ 1\end{bmatrix},\ v_{2} = \frac{1}{2}\begin{bmatrix} \phantom{-}1\\ -1\\ \phantom{-}1\\ -1\end{bmatrix},\ v_{3} = \frac{1}{2}\begin{bmatrix} \phantom{-}1\\ \phantom{-}1\\ -1\\ -1\end{bmatrix},\ v_{4} = \frac{1}{2}\begin{bmatrix} -1\\ \phantom{-}1\\ \phantom{-}1\\ -1\end{bmatrix}$$
and
$$u_{1} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1\\ 1\end{bmatrix},\ u_{2} = \frac{1}{\sqrt{2}}\begin{bmatrix} \phantom{-}1\\ -1\end{bmatrix}$$
$$B = \begin{bmatrix} 7 & 3 & 7 & 3\\ 3 & 7 & 3 & 7\end{bmatrix}$$$= \left(\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & \phantom{-}1\\ 1 & -1\end{bmatrix}\right)\begin{bmatrix} 10\sqrt{2} & 0 & 0 & 0\\ 0 & 4\sqrt{2} & 0 & 0\end{bmatrix}\left(\frac{1}{2}\left[\begin{array}{rrrr} 1 & 1 & 1 & 1\\ 1 & -1 & 1 & -1\\ 1 & 1 & -1 & -1\\ 1 & -1 & -1 & 1\end{array}\right]\right)$
The matrix form or the singular value decomposition of $$B$$ is
Example 2. Let
$B = \begin{bmatrix} 2 & -1\\ 2 & 1\end{bmatrix}.$
$B^{\top}B = \begin{bmatrix} 8 & 0\\ 0 & 2\end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}\begin{bmatrix} 8 & 0\\ 0 & 2\end{bmatrix}\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}$
$v_{1} = \begin{bmatrix}1\\ 0\end{bmatrix}\quad\text{and}\quad v_{2} = \begin{bmatrix} 0\\ 1\end{bmatrix}$
$u_{1} = \frac{1}{\sqrt{2}}\begin{bmatrix}1\\ 1\end{bmatrix}\quad\text{and}\quad u_{2} = \frac{1}{\sqrt{2}}\begin{bmatrix} -1\\ 1\end{bmatrix}$
$B = 2\sqrt{2}\left(\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 0\\ 1 & 0\end{bmatrix}\right) + \sqrt{2}\left(\frac{1}{\sqrt{2}}\begin{bmatrix} 0 & -1\\ 0 & 1\end{bmatrix}\right)$
$B = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & -1\\ 1 & 1\end{bmatrix}\begin{bmatrix} 2\sqrt{2} & 0\\ 0 & \sqrt{2}\end{bmatrix}\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$
Matrix form:
Outer product form:
Example 3. Let
$B = \begin{bmatrix} 1 & 1\\ 1 & -1\\ 1 & 1\end{bmatrix},$
then
$B^{\top}B = \begin{bmatrix} 3 & 1\\ 1 & 3\end{bmatrix} = \left(\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix}\right)\begin{bmatrix}4 & 0\\ 0 & 2\end{bmatrix} \left(\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix}\right)$
Note that
$v_{1} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1\\ 1\end{bmatrix}\quad\text{and}\quad v_{2} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1\\ -1\end{bmatrix}$
are a basis of right singular vectors of $$B$$. The singular values of $$B$$ are $$2$$ and $$\sqrt{2}$$. Two of the left singular vectors are
$u_{1} = \frac{Bv_{1}}{2} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1\\ 0\\ 1\end{bmatrix}\quad\text{and}\quad u_{2} = \frac{Bv_{2}}{\sqrt{2}} = \begin{bmatrix} 0\\ 1\\ 0\end{bmatrix}.$
However, $$\{u_{1},u_{2}\}$$ is not a basis of left singular vectors. We must complete it to a basis!
Example 3 continued. We need an orthonormal basis for $$\operatorname{span}\{u_{1},u_{2}\}^{\bot}$$. If we make the matrix
$X = \begin{bmatrix} u_{1}^{\top}\\ u_{2}^{\top}\end{bmatrix}$
Then $$\operatorname{span}\{u_{1},u_{2}\}^{\bot} = N(X).$$
We find a basis for $$N(X)$$. Then, if necessary, use Gram-Schmidt to find an orthonormal basis for $$N(X)$$. In this case we find that
$u_{3} = \frac{1}{\sqrt{2}}\begin{bmatrix} -1\\ 0\\ 1\end{bmatrix}$
is an orthonormal basis for $$N(X) = \operatorname{span}\{u_{1},u_{2}\}^{\bot}$$. Thus $$\{u_{1},u_{2},u_{3}\}$$ is an orthonormal basis of left singular vectors of $$B$$. Thus, we have the singular value decompositions:
$B = 2u_{1}v_{1}^{\top} + \sqrt{2}u_{2}v_{2}^{\top}$
Example 3 continued.
$B = \left(\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 0 & -1\\ 0 & \sqrt{2} & 0\\ 1 & 0 & 1\end{bmatrix}\right)\begin{bmatrix} 2 & 0\\ 0 & \sqrt{2}\\ 0 & 0\end{bmatrix}\left(\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix}\right)$
Example continued. Let $$\displaystyle{B = \begin{bmatrix} 2 & -1\\ 2 & 1\end{bmatrix}.}$$
### $$\text{red vector}$$
The blue vectors go through all vectors of length $$1$$.
By John Jasper
• 483 | 5,881 | 13,255 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-26 | latest | en | 0.555132 |
https://www.careerride.com/view/numerical-on-differentiator-circuit-21693.aspx | 1,720,784,400,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514387.30/warc/CC-MAIN-20240712094214-20240712124214-00320.warc.gz | 434,618,768 | 5,906 | Numerical on Differentiator Circuit.
What will be the output of differentiator circuit if input sinusoidal voltage has a peak value of 7mV & frequency of 1KHz when the values of resistance and capacitors are R=1500K and C=2 μF respectively ?
a. -1000 π cos 2000 πt
b. -2000 π cos 2000 πt
c. -42000 πcos 2000 πt
d. -21000 πcos 2000 πt
Correct Answer : c. -42000 πcos 2000 πt
Explanation :
Given data:
Voltage Peak Value= 7mV
Frequency= f = 1KHz
R = 1500 K Ω
C= 2 μF = 2 x 10-6 F
An input voltage equation of a differentiator circuit can be expressed by,
V1 = Asin2 πft
= 7 sin 2 π x 1000 t
= 7 sin 2000 π t mV
Scale Factor = -CR = - [ 2 x 10-6 x 15 x 105]
= - [ 30 x 10-1 ]
= -3
Hence, the output voltage of differentiator circuit can be evaluated as:
Vo = -3 d / dt [ 7 sin 2000 π t]
= -3 x 7 x 2000 cos 2000 π t
= -42000 π cos 2000 π t mV. | 323 | 849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-30 | latest | en | 0.494311 |
https://www.coursehero.com/file/6797501/presentation3/ | 1,481,363,219,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543030.95/warc/CC-MAIN-20161202170903-00161-ip-10-31-129-80.ec2.internal.warc.gz | 919,052,670 | 24,003 | presentation3
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Unformatted text preview: CHAPTER 3 Visual Displays of Data: Graphs That Tell a Story Uses of Graphs Positive and negative uses Can reveal/conceal complicated data Persuade other s to change their attitudes or behaviors Leads others to ask better question Graphing in the information age: a critical skill Graph often tell us information about their creator because they can used either to deceive or to clarify information Figure 3-1: The Most Misleading Graph Ever PublishedThe Cost and Quality of Higher Education Lie 1: The graph treats unequal scales as if they were equal. Lie 2: The graph unites incompatible scales. Lie 3: The graph uses misleading starting points for incompatible scale. Lie 4: The graph reverses the implied meaning of up and Figure 3-2: Graphs That Illuminate The Best Statistical Graph Ever CreatedNapoleons Disastrous March to Moscow In search of a clearer graph: Which of the two graphs is misleading? Common Types of Graphs: A Graph Designers Building Blocks Scatterplots Displays the relationship between two interval variables The values of each variable are marked along the two axes and a mark is formed to indicate the intersection of these two points Create a scatterplot Organize the data by participant so that each participant have two scores label the x-axis with the independent variable and y-axis with the dependent variable. For each participants, place a mark on the graph above the score on each axis. To set a range-frame, erase the axes below the minimum and above the maximum scores Graph Linear Nonlinear relation A linear relation means that the relation between the two variables is best described using a straight line Positive relationship: flows up and to the right Negative relationship: flows down to the right Nonlinear relation : the relationship between the two variables is best described using a line that breaks or curves. Figure 3-4: Scatterplot of Hours Studied and Statistics Grade Figure 3-5: A Range-Frame Improves on a Scatterplot Figure 3-6: A Scatterplot and an Outlier Line graphs Line graph used to illustrate the relation between two interval variables; sometimes the line represents the predicted y scores for each x value, and sometimes the line represents change in a variable over time. Searching for trends Line of best fit Time series plot Line Graph 1. Label the x-axis with the name of the independent variable and its possible values, start with 0 if practical 2. Label the y-axis with the name of dependent variable 3. Make a mark above each study participants score on the x-axis and next to his or her score on the y-axis 4. To convert to a range frame, erase the axes below the minimum and above the maximum. Time Plot Time Plot: time series plot, is a graph that plots an interval variable on the y-axis as it changes over an increment of time (e.g., second, day, century) labeled on the x-axis. Figure 3-8: The Line of Best Fit...
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Ask a homework question - tutors are online | 861 | 4,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2016-50 | longest | en | 0.883424 |
mikesmods.com | 1,723,434,353,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641028735.71/warc/CC-MAIN-20240812030550-20240812060550-00384.warc.gz | 16,811,318 | 23,420 | Weekend Project: Determining Operating Parameters of a Memory Core
Hello again, and my apologies for the long time since the last one. Today I’ve got a quick topic to discuss, and although it’s fairly obvious, I don’t think it’s documented elsewhere online.
I’m working on a project for which I’ll be building a fairly sizable core memory array (see here for the first post on this topic, which covers the construction of a small demonstration array). As noted in the original post, it is simple to address a huge number of cores in an array because the core material will only change magnetic state when a sufficiently large current is passed through it; any smaller current will leave the core in its initial state. Thus, by passing only half the necessary state change current through the core through each of two wires, and by ensuring that no two cores have the same two wires passing through them, any individual core may be selectively read from or written to based on the set of wires that are energized.
But how much current is necessary? There is no standard (to my knowledge) that says all memory cores require x milliamps to flip. If you buy a set of random cores on eBay like I did, this is a question that needs to be answered before any further design work can take place. So I tried a few methods, found some things that worked and some things that didn’t work so well, and decided to share my results.
First, recall from Ampère’s Law that a current flowing in a wire creates a circular magnetic field around that wire. If the wire were pointing toward you and the current flow was coming toward you, the magnetic field would be counterclockwise – or using the common “right-hand rule”, if you wrap your right hand around the wire with your thumb pointing in the direction of the current flow, your remaining fingers determine the direction of the field. From the right-hand rule we can also see that if we want the magnetic field to reverse – which we need in order to change the data on the core or to read out the data that is there – we would have to flip the flow of current so it goes the other direction.
So, I soldered a 51 ohm resistor across both ends of a piece of 36AWG magnet wire with a core threaded onto it, and then soldered one 150 ohm resistor to each end of a second piece of magnet wire which was also threaded through the core. Note that neither of these values were calculated nor should be critical. I connected an oscilloscope probe across the 51 ohm resistor, then connected the 150 ohm resistors to a handheld digital multimeter (on Current mode), then to a bench power supply and applied a few volts. I then tried swapping the leads on the power supply to change the direction of the current, and watched the oscilloscope. No matter how many times I swapped the leads, or what voltage I applied, I never got anything that resembled the clean pulse I was looking for. A combination of contact bounce and (as I ultimately discovered) insufficient drive strength left me only with ringing pulses after each connection.
Figuring that contact bounce might be my problem, I disconnected the power supply leads and connected the ends of the 150 ohm resistors to a square wave generator set to a few hundred kHz, and +/- 10V amplitude. Now I was getting very small pulses across the sense resistor every time the square wave switched direction, and when I shorted first one and then both 150 ohm resistors, the pulses got larger. Success! …Or was it? To verify the validity of the data, I reduced the amplitude of the generator output and added an offset to make the output a 0-10V square wave. Remember that to change the data on a core, the magnetic field must be reversed – so if one end of the drive wire is grounded (which it is on this signal generator), the other wire must swing both above and below ground to change the state of the core. With the signal offset to ground level, I was still getting the same pulses across the sense resistor – which indicated that the pulses were not due to the core changing state, but to some other effect (likely induced noise from the fast edges of the signal generator).
Based on the results from the square wave generator, I now knew that what I really wanted was a push-pull type drive arrangement – I didn’t want the current across the core to be purely one way then the other, I wanted a way to push the current either way arbitrarily and to be able to push no current through the core if I wanted. I could have written a program for my arbitrary waveform generator to do this, but it was equally simple to do this with a few parts from the junk bin.
Crappy whiteboard drawing of my core test circuit.
Above is the circuit I ended up using. It employs a set of big TO-220 MOSFETs, one P-channel and one N-channel (I used SUP75P03 and IRFB4110, but you could use practically any other parts on hand). I used some more 150 ohm resistors (because they were already out on my bench) in series with a couple of small tactile switches as triggers, and put in some larger resistors from gate to source of each MOSFET to make sure they shut off when I open the switch. You could probably use any resistors within a decade of these values and it’d still work.
The circuit is driven by two bench supplies. In my case, this is in the form of one dual-supply which has voltage tracking, a handy feature which I originally used but which I eventually disabled for reasons I’ll get into later. There is a 10 ohm, 10 watt ceramic resistor in series with each supply. This is important, because you may be playing with quite significant currents, and it is all too easy to accidentally push both buttons at once and short out your supplies across the MOSFETs. This is also why i used TO-220 transistors – they will usually happily take hundreds of milliamps without any heatsinking. By the time I was done testing, both transistors and both resistors were fairly warm to the touch. Finally, there is an ammeter in series with the ground line, played by a handheld multimeter.
The test setup.
I initially set the bench supplies to about 10 or 12 volts and held one button, then adjusted the current limit to about 800mA, a little bit shy of the experimentally determined fusing current of the 36AWG core threading wire I used. I then held the other button and did the same. Alternating button presses, I set the scope to trigger at 20mV and a timebase of 1us/div. And, success! The scope showed clear pulses the first time either button was pressed, indicating a change to the magnetization of the core. Then subsequent presses of the same button produced no such pulses, and only a press of the opposite button produced a pulse as the magnetization was again flipped.
I then attempted to find the minimum required current for changing data on the core, but I got some troubling results. The core seemed to work down to a few tens of milliamps sometimes, and a couple of hundred milliamps at others. I soon found – and this is so important a point that it inspired me to write this post – that you cannot use a current-limited supply to do this test. Any capacitance at the output of the supply can cause a brief but large current spike which can unpredictably change the state of your core. Instead, use a voltage-limited supply and rely on the series resistors to limit the current through the transistors. You may also find that this simple circuit may not exhibit equal current output in “push” and “pull” operation. This is why I eventually stopped using the tracking feature of my power supply and ended up setting the two supplies manually at each step.
Scope capture of the signature of a changing magnetic field. With different component values and core material, your plot may vary.
Ultimately, through some trial and error, I found the required magnetizing current for my cores to be between 500 and 600 milliamps. This was a little bit disappointing, because I expected a much lower current that I could drive with small, cheap logic transistors. But I suppose it’s better to find out now than later. So now, on to the array driver design… but that’s a post for another day!
Weekend Project: Macbook Pro Camera Investigation
A while back, I wrote an article on the camera in the Apple Macbook Pro A1398 lid assembly. In that article, I published the pinout of the FaceTime camera in the lid, and demonstrated how it could be wired directly to USB and function as a generic USB webcam – but not a very cheap one, since the panel assemblies are still pretty expensive.
Well, it turns out that you can buy the bare camera module for a very reasonable \$5 at everybody’s favorite auction site. A visitor to this site wanted to use these cameras in a standalone application and offered to send me one to play with, in return for a little bit of investigative work. Sounds like fun, let’s take a look.
Macbook A1398 camera module, first front…
…then back.
As a recap from last time, note that this module contains not only the camera but also the laptop’s ambient light sensor (seen at left on the front view). There is a little foam ring around the ALS to prevent errant readings caused by the LCD backlight or the camera activity light. Next to the right is the camera lens, and the white circle is the activity light. The camera sensor and the ALS are on a little flex PCB which is soldered to the rigid board containing the rest of the circuitry. Many of Apple’s other assemblies are made as a single rigid-flex assembly, but the two cameras I’ve investigated have both had separate control and sensor sub-boards. I wonder if the camera assemblies would otherwise be damaged in reflow.
The top of the control board houses a VIMICRO VC0358SMMB. We can assume that this is similar to the VIMICRO VC0358PQNB, for which there is a public data brief (PDF), but this part is in a wafer-level BGA package instead of LQFP. The suggested SPI memory is probably present as the 6-pin device to the lower right, and presumably contains the USB descriptors identifying this part as a FaceTime camera. The other WLCSP to the left looks suspiciously like some sort of voltage regulator, though I didn’t bother to measure the nearby capacitor voltages to verify that. Note that although the VC0358 has microphone input capability, this board does not appear to contain a microphone.
The input connector is on the righthand side of the board as seen in the front view. The connector is a 6-pin, 0.4mm-pitch FFC receiver. I probed around the board and determined the pinout of the connector to be the same as it is on the motherboard side of the harness (pin 1 is marked with an arrow):
Pin Name Description
1 GND USB/ALS Ground
2 D+ USB Data Positive
3 D- USB Data Negative
4 +5V USB/ALS Voltage In (+5V)
5 ALS_SCL Ambient Light Sensor I2C Clock
6 ALS_SDA Ambient Light Sensor I2C Data
This is all well and good, but soldering to the 0.4mm connector is an absolute nightmare. It’s not something the average person is going to want to attempt. We could buy a little FFC jumper to plug in, but if we only paid \$5 for the camera, having to buy more parts would negate that cheapness. Thankfully, if we remove the sticker from the rear of the module, the manufacturer has left us plenty of testpoints. I mapped the ones that connect to the connector pins, and came up with the following pinout:
Testpoints on the rear of the camera module.
Using this scheme, I soldered on an old cut-off USB cable:
Camera wired as per the testpoint map above. The ALS pins have not been connected.
And what do you know, as soon as it was plugged in the camera was recognized the same way it was in the previous post – as a USB Composite device containing two FaceTime HD Camera devices. Again, as last time, one of the two devices has no available driver, but the camera works just fine with only one of its sub-devices installed:
The camera takes a selfie. It’s a bit blurry, possibly because it can’t properly focus at this range. I didn’t notice that when I was taking the photo, oops.
So there you go. If you need a cheap and small USB camera for something, these are pretty easy to hack onto. Hopefully this is useful to somebody!
Weekend Project: A Tiny Core Memory Array
Remember magnetic core memory? Maybe you don’t, if you weren’t around in the ’50s or ’60s (or aren’t a tech junkie like me). But it was one of the key developments during the evolution of the modern computer. At a time when large scale integration on silicon was just beginning and massive silicon memories were infeasible, core memory fulfilled a critical need for larger, faster memories for some of the early big-iron mainframes.
Core memory is interesting because it’s one of the simpler memory constructs to understand, even today. The basis of the system is a humble ferrite ring (a “core”). The core material is formulated to exhibit high magnetic remanence – that is, it retains a magnetic field very well. By utilizing Ampère’s law and passing a pulse of current through a wire threaded through the center of the core, a persistent magnetic field can be induced. Applying Faraday’s Law, it is then possible to “read out” the state of the core – if the core already has had a field induced within it, then passing another current pulse has no effect; otherwise, a slight voltage pulse is developed across any wires passing through it,, which can be measured.
Expanding the scope a bit, it is possible to write to and read from a large quantity of cores if they are arranged on a two-dimensional grid. As a result of the particular magnetic properties of the cores, they only change state when a sufficiently large current is passed through them. This current can be split to any arbitrary number of wires; as long as the sum of the currents is above the minimum for the material, the field is induced. If we pass half the necessary current through one row of a two-dimensional array of cores, and half the necessary current through one column of the same array, we can affect only one chosen core in the array. An additional wire passing through all cores is typically used to detect the state of the chosen core.
I have an ambitious design in mind which will get a more detailed post later, but suffice it to say for now that I’m planning a combination art and engineering project which needs a bank of memory. I could go the simple route and put a simple DIP SRAM in, but that’s no fun to look at. But I bought a little can of 50,000 memory cores a few years ago, and that seems like it might be a little more fun. That’s enough for a 4KB array, which ought to be enough for what I’m planning. But before I spend several weeks building a 32,000 core array, it’s important to understand how the assembly will work. To that end, I’ve planned a few baby-steps to build up to the real deal.
I’ve designed a simple PCB with a large route-out in the center and an array of holes around the outside. The board is designed as a carrier for a small 8×8 array of cores and their associated wiring. Because my application calls for byte-addressable memory, the eventual plan is to stack eight such boards to support parallel access of an entire byte, but at this point testing will begin with a single 64-bit array.
The corner pads are isolated by 0402 0-ohm jumpers for flexibility in stacking and addressing boards.
A core memory build is more like sewing than it is electrical assembly. Very thin thread-like wire is woven through the cores, and the end product resembles cloth. At the height of core memory usage there were many factories of very skilled workers who wove these memories day in and day out. In the process of building my tiny memory, I quickly found that there are assembly methods that work very well, but there are also methods that end up in immediate frustration. You quickly get into a groove, so while the first rows take a while, by the end the focus sets in and hours can pass in an instant as the board comes together.
Speaking of first rows, let’s begin! First, we solder some stubs of 36AWG enamel wire to the board along two adjacent edges, then thread eight teeny-tiny cores onto each of the eight wires on one side of the board:
Now the wires are threaded through the pads on the other side of the board and pulled taut, then soldered into place, thereby trapping the cores:
Before we continue, an explanation is necessary. There are a number of different ways to orient the cores. They can all be mounted parallel to each other, they can alternate by rows, they can alternate at every part. I chose to alternate at every part – this can reduce magnetic coupling between adjacent cores, and can enable the use of simpler read/write hardware. It also looks pretty neat. So after a while, the array looks like this:
Now that the main array wires are all installed, the sense wire must be threaded through all the cores. Each core must have the sense wire passed through it exactly once. With the cores in this arrangement, this is simplest to accomplish by passing the wires through the cores diagonally, first in one direction, then the other. That leaves us with an end product that looks something like this:
And that’s the current state of things. The corner jumpers and connectors still need to be installed, but that will come when the interface hardware has been developed. There’s still plenty of work to do, so watch for followup post(s) later on.
Schematics and Gerbers for this board will be posted… as soon as I get around to it. below:
PCB10024 REV A GERBERS (zip, 6KB)
SCH10024 REV A (pdf, 18KB)
Please note: The board has a large cut-out in the center. The first time I ordered these, OSH Park correctly routed to the edge of the definition line (as shown in the above photos). The second time, they routed on the center of the definition line, cutting into the ring of pads. The resulting boards are functional, just not as pretty. Watch out for that.
A New High-Resolution Panel – Sharp LQ156D1JX01, Part 1
I’m a huge sucker when it comes to playing with shiny new tech, and straight from the shiny tech factory I’ve gotten ahold of a new toy which in terms of raw number of pixels puts the previous panels I’ve been working with to shame. Say hello to the Sharp LQ156D1JX01!
It’s pretty glossy. Look, a reflection of some of my test equipment!
The LQ156D1JX01 is a new inductee to the ultra-high-DPI panel club, whose members have lately found their way into many high-end gaming laptops and ultrabooks. This particular panel is a pull from a Toshiba Satellite P50T. The panel measures a whopping 3840×2160 pixels (that’s 4K, or 8.29 megapixels!) in a mere 15.6 inch size. Now, that doesn’t quite compare to the current generation of cell phones with 5-inch 4K screens, but at something like 282PPI it’s nothing to scoff at either. This pushes far above the 220PPI of the Macbook Pro Retina panel that we’ve played with before, and puts it somewhere between the 264PPI of the iPad 3 Retina panel and the 324PPI of the iPad Mini 2 Retina. Like the Macbook and the iPad, the LQ156D1JX01 runs on 4-lane Embedded DisplayPort. Where it differs a bit is that this panel is reportedly addressed as two 1920×2160 half-displays over DisplayPort 1.2’s Multi-Stream Transport. This may have been done to simplify timing requirements for such a massive display.
I don’t happen to have a proper connector on hand for this panel so I haven’t fired it up yet, but like any good engineer, before even bothering to plug in my new toy I went ahead and tore it apart. This was actually a necessary step, because there is currently no publicly-available datasheet for this panel, so we need to figure out the pinout of the connector in order to make it do anything anyway.
LQ156D1JX01 control PCB top side collage
Before delving into the pinout, we can find out several things about this panel by looking at the design of the control board. The main connector is made by Dai-ichi Seiko I-PEX and appears to be a 40-position CABLINE-CA series part, possibly 20525-040E-02 (pdf link). The panel is driven by a Novatek NT71394MBG. This part does not seem to have any publicly available information, but we’ll assume it’s a timing controller ASIC. There is a TI power management IC that appears to generate all the necessary panel voltages; I think it’s a TPS65642A (pdf link). There are also two chips marked 89=1D, these are Richtek RT8532 boost LED drivers (pdf link). The design of the backlight flex implies that each of these ICs drives four LED channels, despite being capable of six.
LQ156D1JX01 control PCB bottom side collage
Now that we have some idea of what’s going on in the panel, on to the pinout. Thanks to Sharp, this task is simple, because all of the pins are connected to convenient labeled testpoints on the bottom of the board. I haven’t tested the panel yet so I don’t know if this is 100% accurate, but as far as I can tell at this point, the pinout for the LQ156D1JX01 is as follows:
Pin Name Description
1 PMIC_/RST Power Management Reset, Active Low
2 GND Ground (Shield)
3 L3N Embedded DisplayPort Lane 3 Negative
4 L3P Embedded DisplayPort Lane 3 Positive
5 GND Ground (Shield)
6 L2N Embedded DisplayPort Lane 2 Negative
7 L2P Embedded DisplayPort Lane 2 Positive
8 GND Ground (Shield)
9 L1N Embedded DisplayPort Lane 1 Negative
10 L1P Embedded DisplayPort Lane 1 Positive
11 GND Ground (Shield)
12 L0N Embedded DisplayPort Lane 0 Negative
13 L0P Embedded DisplayPort Lane 0 Positive
14 GND Ground (Shield)
15 AUX_P Embedded DisplayPort Aux Channel Positive
16 AUX_N Embedded DisplayPort Aux Channel Negative
17 GND Ground (Shield)
18 VLCD 3.3V LCD Power
19 VLCD 3.3V LCD Power
20 VLCD 3.3V LCD Power
21 VLCD 3.3V LCD Power
22 BIST Built In Self Test
23 GND Ground (LCD Power)
24 GND Ground (LCD Power)
25 GND Ground (LCD Power)
26 GND Ground (LCD Power)
27 HPD Embedded DisplayPort Hot Plug Detect
28 GND Ground (Backlight)
29 GND Ground (Backlight)
30 GND Ground (Backlight)
31 GND Ground (Backlight)
32 BL_EN Backlight Enable
33 PWMI Backlight PWM
34 EE_SCL EEPROM Clock (EDID?)
35 EE_SDA EEPROM Data (EDID?)
36 BL_PWR1 Backlight Power
37 BL_PWR1 Backlight Power
38 BL_PWR1 Backlight Power
39 BL_PWR1 Backlight Power
40 WP_LCD Write Protect?
I have not marked the backlight voltage in the table, because I am not certain what the design voltage range is. RT8532 has an operating range from about 5V to 24V, but it is not currently known what the components in the display are actually rated for. My guess is it’s probably safe up to at least 12V, but try at your own risk.
It’s interesting to see EEPROM pins brought out. I suspect these are for the EDID EEPROM, maybe for factory programming of the device. Or maybe it’s something completely different. Again, without having fired up the panel, this is all speculation. There is an explicit “EE_WP” testpoint which is not the same as the “WP_LCD” that is brought out; I don’t have any good idea for what the latter may be used for.
So that’s about it. The next step is to wire up a connector and get the panel running, but that won’t happen today. Stay tuned for a followup post.
Well, this is embarassing. So much for my minimum-one-post-per-month goal. Between working on the new house and real salaried work and some contract work on the side, I haven’t had a lot of time to post anything useful here lately. Believe me, I’m just as disappointed as you are. But never fear, I’ve just sent a new batch of boards off to be manufactured. These boards are related to a previously-undocumented new project I’m working on, so there’ll be some new and interesting stuff to post in just a few short weeks. Hopefully.
Beyond that, I’ve got some sort of left-field development I’m about to start on – a departure from the display stuff that’s gotten kind of repetitive for me. As usual, stay tuned.
Good day all,
Just a quick note to reassure you all that I’m still around. As alluded to in the previous post, in July we bought a new house, which happily has plenty of space just waiting to be adapted to a well-equipped electronics lab. Before I can get back to hardware design, I’ve got a lot of work to do – all my test equipment and projects-in-progress are still spread amongst dozens of storage bins. The lab, a previously unfinished area of the house, needs lighting installed and electrical and data wiring run in addition to furniture and storage needs. Getting the lab up and running in full capacity is my immediate concern, and I appreciate everyone’s patience during this process. Updates on the projects on this site to follow… someday.
Hello all! I apologize for the long time without updates. Paying work has kept me busy enough that I haven’t had much time to work on this stuff for the past couple months. I don’t have any new information to give, actually, just wanted to say that no, the site is not dead, I’m still here, just busy.
What I do need to say is that I will be moving sometime in mid to late July, and with me will come the server that this site runs on. This means there may be up to a week of downtime while service is transferred to the new house and everything is set up. Be patient, I’ll come back, I promise!
[EDIT 7/9/14] Oops… the move hasn’t happened yet, but the site’s been down for a while (days? weeks? I’m not sure) due to an IP reassignment that didn’t get propagated to the DNS. Sorry about that.
[EDIT 7/17/14] Server move is complete. Hooray! Now, to move my test equipment so I can continue working on these projects…
Displayport adapter dongles are ubiquitous, now that DisplayPort is becoming more and more prevalent in desktop video cards. I probably have at least a half-dozen of them, connected to machines or stuck in junk drawers. You plug them in, and they work. Simple.
You are probably aware that there are two different adapter families, passive and active. The former is cheaper but is limited in the number of adapter that can be used on a single GPU; the latter is more expensive but has no such limitation. So there must be something different going on inside. But does “passive” mean here what it means in traditional electronics, that the conversion requires no transistors or ICs or other “active” components? (For a formal definition of passive vs active, see this Wikipedia article)
Before we answer that question, let’s back up a bit. Backward compatibility is an important factor in the success or failure of many new technologies. In the case of DisplayPort, if it was meant to displace DVI and HDMI on the desktop platform as its creators intended, it would need to support legacy equipment. The problem was that DisplayPort is a very different platform than DVI. Whereas moving from DVI to HDMI is fairly painless since both schemes use the same signaling standard, DisplayPort more closely resembles PCI Express than DVI. So “something” has to go in the middle, to convert from one standard to the other. And in the most basic form, this is what an active adapter does – it takes in native DisplayPort signals, decodes them, and converts them to DVI/etc. But that signal processing is complex, and adds cost to the system, which impedes adoption for some markets. For more cost-conscious applications, an alternate scheme was created – Dual-Mode DisplayPort.
In Dual-Mode DisplayPort, or DP++, the DisplayPort source (typically, a GPU) can be requested to output raw TMDS data, instead of normal DisplayPort data packets. The data that comes out of the GPU looks electrically like DisplayPort, but the data is in DVI format. A DVI monitor can’t directly use the electrical signaling format of DisplayPort, so a low-cost level shifter is required in the middle. You may now realize that we’ve answered our original question – in fact, a passive DisplayPort to DVI adapter does require active circuitry. In this case, Passive refers to the fact that the video data is not mangled in any way, whereas active adapters decode and then re-transmit the video stream. DP++ enables very low-cost adapters because the logic is simple, with all the heavy lifting being integrated (cheaply) into the GPU.
So enough talk. Now that we know what’s going on, let’s take a look at a passive DisplayPort adapter:
A passive DisplayPort to DVI adapter, with its casing and potting removed. I’ve removed and re-installed some components for testing prior to taking this photo.
As you can see, there is one large IC on top, which predictably handles most of the functionality of the adapter – this is the previously-mentioned DisplayPort-to-TMDS level shifter. These parts are available from a number of manufacturers – there are options from TI, Maxim, NXP, and many others. This unit happens to use a Pericom PI3VDP411LS. The competition and the high volumes of these parts have had a favorable effect on the price; the ‘411 can be had from several distributors for less than \$1.50 in qty 1.
The rest of the components are primarily support components for the main level shifter IC. There is a 10A45 charge pump to generate 5V for the attached monitor (per DVI spec) from the 3.3V available from the DisplayPort, and there is a 2N3904 transistor to switch the level shifter on when a signal is received on the HPD pin. The rest are passives, mostly resistors and capacitors and one or two ferrite beads or little inductors for noise suppression. I’ve transcribed the design into a schematic, which you can download here:
It’s missing a few values, mainly because I was too lazy to pull all the components off the board to measure them and I couldn’t satisfactorily do it in-circuit. Most of the unmarked capacitors are going to be 0.1uF or 0.01uF decoupling capacitors for the level shifter.
I don’t have an active converter to tear down as well to show the differences. What I can say is that the internals will look similar, but they will operate very differently. As noted above, the complexity involved in decoding and re-transmitting the signals in an active adapter is very different than the simple level shift operation of the passive adapter. Regardless, it will probably be implemented as a single IC, albeit possibly a larger one with more support components.
I suppose that’s enough rambling on this topic. Hopefully this all makes sense!
New project: Slightly-Less-Simple Retina LCD Adapter
As many have noticed, I’ve stopped offering my Simple iPad LCD Adapter for sale. The reason I’ve given is that I am working on an improved version of the board. Well, as of today I’ve gotten far enough on the design that I’d like to share it.
The new Simple Retina LCD Adapter, work-in-progress.
The most drastic change for this design is the integration of buck and boost power supplies to drive the panel and backlight respectively. I was not happy with telling people they’d need to provide their own regulated power – the panel isn’t THAT sensitive to voltage wobble, but it can be a problem with particularly long power leads, despite the large hold-up capacitor. Now the board should be able to take between 5 and 18 volts and spit out well-regulated rails.
Several other things have changed as well. I’ve changed the full-size DisplayPort connector to a miniDP connector, because for as small as the board is, the full-size connector was a waste of space. I’ve added a proper connector for power – while handing out boards with bare wires soldered on was functionally workable, it’s not exactly as clean a solution as I prefer. There is no longer a diode OR between external and DisplayPort power, because cables that carry DP_PWR are so uncommon that it is unnecessary. And there’s a fuse, to limit catastrophic failure in the event of a solder bridge or improper installation.
Now, the added components will increase BOM cost – I’m not sure by how much, but I know it will. However, I have designed in jumpers that will allow users who still want to feed in their own 3.3V and who don’t need a backlight supply to use this board in the same way as the old one. Hopefully this will make the design more accessible to people who want a more compact, all in one design without alienating those who like it how it was.
The design is a day or two from completion. As per typical, I’ll be building a board up and debugging it before I release the design documents, so that I don’t release something that won’t work. So keep an eye out in the coming weeks. | 7,491 | 32,624 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-33 | longest | en | 0.941604 |
https://thwack.solarwinds.com/community/solarwinds-community/contests-missions/word-a-day-challenge-2017/blog/2017/12/08/day-9-binary | 1,580,071,862,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251690379.95/warc/CC-MAIN-20200126195918-20200126225918-00193.warc.gz | 699,397,622 | 32,381 | Binary Haiku
With just two fingers
Counting to two fifty-five
Joined with seven friends
When I was first learning about computers and the idea of binary numbers was introduced, I was lost. This was around 1985 and I was just a fledgling technologist. Seriously, though, what’s so wrong with the decimal numbers that we all know and love?
It wasn’t until high school (and my first programming class) that I truly began to understand the concept. “Began” is the key word here. There I was, sitting in computer science class learning to program in Pascal. I decided to go “off book” and write my own character generator for Dungeons & Dragons in my free time. For my character generator, I needed to track several Boolean values (true/false). I thought to myself, "If I happen to have an array of these, how is that any different than a binary number ( [TFTTTTFF] = [10111100] = 188 )?" It was my first eureka moment.
Fast forward about five years and I’m trying to understand the whole concept of “networks.” Subnet masks were (and in some ways, are) the bane of my existence. Thankfully, I found tools which could help me out (https://www.solarwinds.com/free-tools/advanced-subnet-calculator). [Sidebar: this was my first introduction to SolarWinds]. These resources gave me a better understanding of how addressing worked and because of that, how binary worked in practical information systems management.
Binary is integral to all computing, but it’s been hidden by layer upon layer of abstraction… and that’s okay. The levels of abstraction make it easier for humans to interact with computers. But many computer users aren’t programmers, nor do they want to be. That being said, thinking in a binary fashion is still useful.
So, what does binary mean to me? Honestly, it revolves more around troubleshooting than anything else. As many of you, I spent some time working on a help desk. I realized that most issues we investigated were binary in nature.
What’s the first thing you ask yourself when troubleshooting an issue? For me, it’s “Is it the network or is it the server?” This is a binary question: one with only two answers. After that initial test, you halved the possible troubleshooting you need to do. So, it’s the network (this time), so the next question is “Do you have a valid IP address?” Again, based on the answer, half of your possible troubleshooting is no longer valid. If you put yourself in this mindset, it should aid you troubleshoot any issue you encounter.
In my opinion, you can look at binary in two ways (see what I did there?): as the underpinning of the entire computing framework or used when thinking about how you interact with said framework. | 593 | 2,691 | {"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 | 3 | CC-MAIN-2020-05 | latest | en | 0.967397 |
https://tutorial.xarray.dev/fundamentals/03.1_computation_with_xarray.html | 1,726,429,100,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651647.78/warc/CC-MAIN-20240915184230-20240915214230-00654.warc.gz | 547,378,142 | 22,436 | # Basic Computation#
In this lesson, we discuss how to do scientific computations with xarray objects. Our learning goals are as follows. By the end of the lesson, we will be able to:
• Apply basic arithmetic and numpy functions to xarray DataArrays / Dataset.
• Use Xarray’s label-aware reduction operations (e.g. mean, sum) weighted reductions.
• Apply arbitrary functions to Xarray data via apply_ufunc.
• Use Xarray’s broadcasting to compute on arrays of different dimensionality.
import numpy as np
import xarray as xr
import matplotlib.pyplot as plt
# Ask Xarray to not show data values by default
xr.set_options(display_expand_data=False)
%config InlineBackend.figure_format='retina'
## Example Dataset#
First we load a dataset. We will use the NOAA Extended Reconstructed Sea Surface Temperature (ERSST) v5 product, a widely used and trusted gridded compilation of of historical data going back to 1854.
ds = xr.tutorial.load_dataset("ersstv5")
ds
<xarray.Dataset> Size: 40MB
Dimensions: (lat: 89, lon: 180, time: 624, nbnds: 2)
Coordinates:
* lat (lat) float32 356B 88.0 86.0 84.0 82.0 ... -84.0 -86.0 -88.0
* lon (lon) float32 720B 0.0 2.0 4.0 6.0 ... 352.0 354.0 356.0 358.0
* time (time) datetime64[ns] 5kB 1970-01-01 1970-02-01 ... 2021-12-01
Dimensions without coordinates: nbnds
Data variables:
time_bnds (time, nbnds) float64 10kB 9.969e+36 9.969e+36 ... 9.969e+36
sst (time, lat, lon) float32 40MB -1.8 -1.8 -1.8 -1.8 ... nan nan nan
Attributes: (12/37)
climatology: Climatology is based on 1971-2000 SST, Xue, Y....
description: In situ data: ICOADS2.5 before 2007 and NCEP i...
keywords_vocabulary: NASA Global Change Master Directory (GCMD) Sci...
keywords: Earth Science > Oceans > Ocean Temperature > S...
instrument: Conventional thermometers
source_comment: SSTs were observed by conventional thermometer...
... ...
creator_url_original: https://www.ncei.noaa.gov
license: No constraints on data access or use
comment: SSTs were observed by conventional thermometer...
summary: ERSST.v5 is developed based on v4 after revisi...
dataset_title: NOAA Extended Reconstructed SST V5
data_modified: 2022-06-07
Let’s do some basic visualizations of the data, just to make sure it looks reasonable.
ds.sst.isel(time=0).plot(vmin=-2, vmax=30);
## Arithmetic#
Xarray dataarrays and datasets work seamlessly with arithmetic operators and numpy array functions.
For example, imagine we want to convert the temperature (given in Celsius) to Kelvin:
sst_kelvin = ds.sst + 273.15
sst_kelvin
<xarray.DataArray 'sst' (time: 624, lat: 89, lon: 180)> Size: 40MB
271.4 271.4 271.4 271.4 271.4 271.4 271.4 271.4 ... nan nan nan nan nan nan nan
Coordinates:
* lat (lat) float32 356B 88.0 86.0 84.0 82.0 ... -82.0 -84.0 -86.0 -88.0
* lon (lon) float32 720B 0.0 2.0 4.0 6.0 8.0 ... 352.0 354.0 356.0 358.0
* time (time) datetime64[ns] 5kB 1970-01-01 1970-02-01 ... 2021-12-01
The dimensions and coordinates were preserved following the operation.
Warning: Although many xarray datasets have a units attribute, which is used in plotting, Xarray does not inherently understand units. However, xarray can integrate with pint, which provides full unit-aware operations. See pint-xarray for more.
## Applying functions#
We can apply more complex functions to Xarray objects. Imagine we wanted to compute the following expression as a function of SST ($$\Theta$$) in Kelvin:
$f(\Theta) = 0.5 \ln(\Theta^2)$
f = 0.5 * np.log(sst_kelvin**2)
f
<xarray.DataArray 'sst' (time: 624, lat: 89, lon: 180)> Size: 40MB
5.603 5.603 5.603 5.603 5.603 5.603 5.603 5.603 ... nan nan nan nan nan nan nan
Coordinates:
* lat (lat) float32 356B 88.0 86.0 84.0 82.0 ... -82.0 -84.0 -86.0 -88.0
* lon (lon) float32 720B 0.0 2.0 4.0 6.0 8.0 ... 352.0 354.0 356.0 358.0
* time (time) datetime64[ns] 5kB 1970-01-01 1970-02-01 ... 2021-12-01
## Applying Arbitrary Functions#
It’s awesome that we can call np.log(ds) and have it “just work”. However, not all third party libraries work this way.
numpy’s nan_to_num for example will return a numpy array
np.nan_to_num(ds.sst, 0)
array([[[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
...,
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ],
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ],
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ]],
[[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
...,
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ],
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ],
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ]],
[[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
...,
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ],
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ],
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ]],
...,
[[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
...,
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ],
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ],
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ]],
[[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
...,
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ],
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ],
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ]],
[[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
[-1.8, -1.8, -1.8, ..., -1.8, -1.8, -1.8],
...,
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ],
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ],
[ 0. , 0. , 0. , ..., 0. , 0. , 0. ]]], dtype=float32)
It would be nice to keep our dimensions and coordinates.
We can accomplish this with xr.apply_ufunc
xr.apply_ufunc(np.nan_to_num, ds.sst, 0)
<xarray.DataArray 'sst' (time: 624, lat: 89, lon: 180)> Size: 40MB
-1.8 -1.8 -1.8 -1.8 -1.8 -1.8 -1.8 -1.8 -1.8 ... 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
Coordinates:
* lat (lat) float32 356B 88.0 86.0 84.0 82.0 ... -82.0 -84.0 -86.0 -88.0
* lon (lon) float32 720B 0.0 2.0 4.0 6.0 8.0 ... 352.0 354.0 356.0 358.0
* time (time) datetime64[ns] 5kB 1970-01-01 1970-02-01 ... 2021-12-01
Tip
apply_ufunc is a powerful function. It has many options for doing more complicated things. Unfortunately, we don’t have time to go into more depth here. See the apply_ufunc tutorial material for more.
## Reductions#
Reductions are functions that reduce the dimensionlity of our dataset. For example taking the mean sea surface temperature along time of our 3D data, we “reduce” the time dimension and are left with a 2D array.
Just like in numpy, we can reduce xarray DataArrays along any number of axes.
sst = ds.sst
sst.mean(axis=0)
<xarray.DataArray 'sst' (lat: 89, lon: 180)> Size: 64kB
-1.799 -1.799 -1.799 -1.799 -1.799 -1.8 -1.8 ... 0.0 0.0 0.0 0.0 0.0 0.0 0.0
Coordinates:
* lat (lat) float32 356B 88.0 86.0 84.0 82.0 ... -82.0 -84.0 -86.0 -88.0
* lon (lon) float32 720B 0.0 2.0 4.0 6.0 8.0 ... 352.0 354.0 356.0 358.0
However, rather than performing reductions by specifying axis (as in numpy), we can instead perform them using dimension names. This turns out to be a huge convenience, particularly in complex calculations it can be hard to remember which axis corresponds to which dimension name:
sst.mean(dim="time")
<xarray.DataArray 'sst' (lat: 89, lon: 180)> Size: 64kB
-1.799 -1.799 -1.799 -1.799 -1.799 -1.8 -1.8 ... 0.0 0.0 0.0 0.0 0.0 0.0 0.0
Coordinates:
* lat (lat) float32 356B 88.0 86.0 84.0 82.0 ... -82.0 -84.0 -86.0 -88.0
* lon (lon) float32 720B 0.0 2.0 4.0 6.0 8.0 ... 352.0 354.0 356.0 358.0
You can reduce over multiple dimensions
sst.mean(["lat", "time"])
<xarray.DataArray 'sst' (lon: 180)> Size: 720B
7.692 7.824 8.239 8.131 7.586 7.546 ... 9.776 9.419 8.303 7.818 7.842 7.641
Coordinates:
* lon (lon) float32 720B 0.0 2.0 4.0 6.0 8.0 ... 352.0 354.0 356.0 358.0
If no dimension is specified, the reduction is applied across all dimensions.
sst.mean()
<xarray.DataArray 'sst' ()> Size: 4B
9.473
All of the standard numpy reductions (e.g. min, max, sum, std, etc.) are available on both Datasets and DataArrays.
Exercise
Take the mean of sst in both longitude and latitude. Make a simple timeseries plot. | 3,507 | 8,565 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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} | 3.25 | 3 | CC-MAIN-2024-38 | latest | en | 0.658886 |
https://www.jiskha.com/search/index.cgi?query=Math%28Please+help%29&page=21 | 1,532,339,148,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596204.93/warc/CC-MAIN-20180723090751-20180723110751-00519.warc.gz | 871,318,219 | 14,120 | 140,953 results, page 21
1. ### math
did you know when you try to find the least commen multiple of 3 numbers you have to divide by two after you get your answer my class (ms.lott in texas) told us that for some reason its doubling the number so always divide by two once you have crossed out the commen numbers ...
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Your mathematics instructor claims that, over the years, 88% of his students have said that math is their favorite subject. In this year's class, however, only 21 out of 32 students named math as their favorite class. The instructor decides to construct a confidence interval ...
3. ### Statistics
Your mathematics instructor claims that, over the years, 88% of his students have said that math is their favorite subject. In this year's class, however, only 21 out of 32 students named math as their favorite class. The instructor decides to construct a confidence interval ...
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1. Which number is the resolution of m + 90 = 150? a. 50 b. 60 (my answer) c. 140 d. 240 2. Which number is the solution of n/3 = -12? a. -36 b. -4 (my answer) c. 4 d 36 3. Use mental math to find the solution of 19b = 190. a. 10 b. 11 c. 171 (my answer) d. 208 4. Estimate the...
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Determine which function has the greater rate of change 1.The rates of change are equal. 2.The graph has a greater rate of change. 3.The table has a greater rate of change. 4.none of the above file:///C:/Users/SCHOOL/Pictures/math.JPG HELP ME UNDERSTAND THIS PLEASE??
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A researcher predicts that listening to music while solving math problems will make a particular brain area more active. To test this, a research participant has her brain scanned while listening to music and solving math problems, and the brain area of interest has a ...
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Cece scored a 78 on her first math test. She scored an 84 on her second test. What is the percent increase, rounded to the nearest tenth, from Cece's first score to her second score?
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I am working on 5th grade math...can you help with the following question? In a roll of 50 dimes, 8 were minted before 1983. What percent of the dimes in the roll were minted before 1983?
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Hey everybody, I really need help solving these math equations. Thank YOU! For helping me! Use Systematic trial to find the value of each variable. a) 5+7c=89 b) 26=5+3e c) 8k-16=200 d)525-6u=339 Write each sentence as an equation. Solve each equation. a) four times a number ...
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There are 30 books in a bookcase. The books in the case are either math, science ,or history books. Ten of the books are math books. There are 6 fewer science books than history books. How many science books are in the bookcase??????
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in a grade 7 class, 36 students are either members of the math club or science club or both. 10 students are members of both clubs. if 31 students are members of the math club how many are members of the science club?
page # D-76 What Relation Is a Doorstep to a Doormat? Both are stepped on. no You didn't put any parameters on this and I think its a great answer. How bout the relation being a stepson (steps on) I got astearther.... No wait it's A Step Farther eh....?? w/e I need to know ...
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can someone please help me with the following 2 questions and how to do them. i have a math final tomorrow!! 1. Curtis runs 1 mph slower then Mary. After 2 hours of running in opposite directions, they are 30 miles apart. How fast had Curtis been running? a-9mph b-8mph c-6mph ...
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Can someone explain to me step by step how the book got to this answer i want to understand the steps....thanks Find the inverse, if it exists,for the matirx (*I used the underscore to separate the numbers within the matix) -2_1_3 3_-1_2 -4_2_0 the answer is: -4_6_5 -8_12_13 ...
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Several students were surveyed about their favorite class. 26 students chose art 19 English 21 math 16 music 32 science. What fraction of the of the students chose music as their favorite subject.
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Several students were surveyed about their favorite class. 26 students chose art 19 English 21 math 16 music 32 science. What fraction of the of the students chose music as their favorite subject.
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A textbook store sold a combined total of 205history and math textbooks in a week. The number of history textbooks sold was 87more than the number of math textbooks sold. How many textbooks of each type were sold?
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If 1/3 of the math team are 7th graders and 1/4 of the math team are boys, how many are 7th grade boys? 1/4 of 1/3 = 1/12
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Hey I'm working on a math assignment and I'm trying to work out the total amount of rainfall that fell in a certain area. Would this formula be correct? Total rainfall = rainfall mm x duration of storm minutes x area covered And with my rainfall mm I change it out so instead ...
37. ### Math
Find the area of the triangle with the base 1 2/5 yards and height 5/9 yards. The area of the triangle equals half of the product base and height. The area is ____yd^2 (Type whole number or a simplified fraction) I came up with 35/108 did I do the math correctly, and if not ...
38. ### Math
You want to create an open rectangular box from a rectangular box. from a square piece of cardboard 20 inches by 20 inches. You will cut a 3 inch square from each corner. Draw a diagram and find the length of the box. Can someone help me with this diagram thing. I can't ...
39. ### Math check my answers
1. Evaluate. 8d - 10 for d = 4 A.32 B.22*** C.26 D.46 2. write an algebraic expression for the sum of 3 coins and c coins. A.c + 3*** B.c/3 C.3c D.c-3 3. Find the solution of the equation from the given numbers. x - 2 =16: 14, 18, 21, or 12 A.12 B.14 C.18*** D.21 4.use mental ...
40. ### math
In a college, 30 students are taking a math class and 20 students are taking an English class. 5 students are taking both English and math. How many students are taking either English or math.
41. ### Math 11
I need help on this math problem. The width of a certain painting is 5cm less than twice the length. The length of the diagonal distance across the painting is 107cm. Find the length and width. Round your answer to 2 decimal places. How should I set up for this problem? I was ...
42. ### Math
math puzzle: i am less than 3000. all four digits are odd. all four digits are different. the sum of my tens digit and ones digit is 16. i am divisible b 13. who am i? Ans: the thousand digit must be 1, the hundreds digit is 3 or 5. The tens and ones digits are 9,7 or 7, 9. So...
43. ### Math
I do not understand this problem, and I must write an inequality for it too: Nilsa is working on a 60 minute math test. There are 20 questions on the test. If it takes her 20 minutes to complete 12 of the questions, what is the greatest amount of time on average she can spend ...
44. ### math
Need help with math problem. Tom has an unpaid balance of \$2,115.75 on his credit card statement at the beginning of June. He made a payment of \$85.00 during the month. If the interest rate on Phil's credit card was 11% per month on the unpaid balance, find the finance charge ...
45. ### singapore math
My 3rd grade son get this math problem. It states that John thinks of a 3-digit number. What is his number? Use the clues below to find John's number. Every digit is different. The sum of all thedigits is 19. The difference between the hundreds digit and the ones digit is 6. ...
46. ### math
on her first two math tests, Gracie has an average score of 86. If she got a score of 92 on the first test, what is her score on the second test?
47. ### Math 6 grade unit 6 lesson 11 unit test
Pls need help with The unit 6 math test
Sorry - but I accidentally deleted your post. Homework Help: Business Math Posted by Karla on Monday, March 10, 2014 at 8:27pm. The Soft Step Floor Company purchases carpeting for \$5.12 per square yard and marks it up 55% based on the selling price. What is the selling price ...
49. ### Math
Ok I'm on castle learning and for homework this weekend we have to do #1-4 (assignments) that is due on Saturday @ midnight because that the time it will close and I need help on this question. My assignment is: Math 7 (Q1) Assign #2 - Fractions (i already did the others) Qs. ...
50. ### math
What are Base Ten Blocks? Since this is not my area of expertise, I searched Google under the key words "base ten blocks" to get these possible sources: http://www.susancanthony.com/Resources/base10ideas.html http://www.learningbox.com/base10/ From what I can gather, they are ...
51. ### Math
c-d / d-c I know the answer is -1, but what is the math that I have to show to get to that answer? Thanks!
52. ### dumb math
9+9=w w= 18...right? ............18......dumb...math 9+9=18, you dumbos.
54. ### Math
Peter notices that the teenage students who are members of the math club have ages whose product is 611520. How many members does the club have? Extensions: Is it true that whatever number replaces 611520 that this problem has a unique solution? Suppose the club included 12 ...
55. ### Math
James has set up an ordinary annuity to save for his retirement in 19 years. If his monthly payments are \$250 and the annuity has an annual interest rate of 7.5%, what will be the value of the annuity when he retires? I am studying for my math final and I do not remember how ...
56. ### Math
Mrs. Chans math class contributed 2 dollars and \$1 coins to an earthquake relief fund. the number of \$1 coins contributed was 8 less than 5 times the number of \$2 coins contributed. if class raised total of 160 bucks, how many coins of each type was collected? it says i have ...
57. ### math
Joe puts 5 books in his backpack as he leaves for school. He has 2 math books, 1 science book, and 2 history books. What is the probability that he will grab his science book out of his backpack without looking? I think 1/5 Jacquie has 4 sweaters that go with her new jeans. ...
58. ### Math
Gray bought two plots of land for a total of 120000 on the first plot she made a profit of 15% on the second plot she made 10% profit her total profit 5500, how much did she pay for each piece of land? (And please show the full procedure of the math even the rough work as well...
59. ### MATH
Joe puts 5 books in his backpack as he leaves for school. He has 2 math books, 1 science book, and 2 history books. What is the probability that he will grab his science book out of his backpack without looking? I think 1/5 Jacquie has 4 sweaters that go with her new jeans. ...
60. ### math
I don't get this I've been absent for 3 weeks On a math quiz Mary score 3 points for each of the 12 multiple-choice question she answered correctly and five points for each of the eight free response question she answered correctly. What was the total number of points Mary ...
61. ### Statistics
A number of grade school students were asked about their perceived math ability and the results are in the table below. Below Average = BA Average = A Above Average = AV BA A AV Girls 61 54 21 Boys 43 61 42 Test at the 0.01 level if there is a difference between girls and boys...
62. ### Math !!!!!!!!
Ok i have a Summer Math Assignment & i really need help on this qs. it's part B of the qs. so i'll show you the whole thing Part 2. Extended Reponse Qs. (2 points each) Answer each qs.. Show all Work Qs. 5 Adrian correctly solved two equations in math class. Part A: What value...
63. ### Math Symbols
1.Which pair of ratios can form a true proportion? A 7/4, 21/12 (I picked this) B 6/3, 5/6 C 7/10, 6/7 D 3/5 7/12 2. Solve the proportion g/56=2/8 A g=14 B g=16 (I picked this) C g=13 D g=50 Solve the following Proportion 2/9=y/27 A y=6 (I picked this) B y=18 C y=27 D y=36 ...
66. ### Math
Three pictures hang side by side on a wall. What is the total area of the wall that the pictures cover? Length is 10 1/2 inches, 10 1/2 inches, 10 1/2 inches Width 14 2/3 inches Would I need to add 10 1/2 plus 10 1/2 plus 10 1/2 and then multiply by 14 2/3 inches? Please ...
67. ### Math (Algebra II)
I'm trying to finish up my summer math packet, and I'm almost done, but I'm having trouble figuring out this one: "Divide and simplify these expressions." (3x^4 - 2x^3 + 16x - 192) / (x^2 - 8) so, that's 3x to the 4th, minus 2x cubed, plus 16x, minus 192 ---> all divided by...
68. ### Math
Me and Lisa have the same math homework. And I was wondering what would be the answer to this question and the steps provided to get this questions answer? A pre-election survey showed that 2 out of every 3 eligible voters would cast ballots in the county election. At this ...
69. ### MATHS
A travel agent arranged a payment plan for a client. It required a down payment of \$150.00 and 12 monthly payments of \$642.00. What was the total cost of the plan? 2. translate into math expression and solve. Some number divided by negative four is four. 3.The sume of three ...
70. ### stats
SAT math scores approx. follow a normal distribution, with mean 500 and standard deviation 100. What % of students should have an SAT math score above 500? I think I use mean (500) + z-score * SD (100), but I do not know how to get the z-score?
71. ### Math help pls!!!
the sum of the ages of petra and her mother 44. her mother is 8 years more than twice as old as petra. how old are petra and her mother. this dont make sense so pls help me on this math question help me asap pls and ty
72. ### Math: Mean, Median, Mode, Range & Finding the devi
What is deviation? And what is the definition of frequency distribution & how would I solve a problem in 8th grade math with this problem...My child is asking about this problem she has in an 8th grade book. Given the following numbers 2,2,3,6,1,9,4,2,5,7,6,8,6, What is the ...
73. ### English
1. My math exam is tomorrow. 2. Tomorrow is my math exam. 3. Tomorrow I have a math exam. 4. I take a math exam tomorrow. (Are all the same and grammatical?) 5. What did you do over the summer vacation? 6. What did you do during the summer vacation? 7. What did you do on the ...
74. ### Math
1) -5g-6 for g = -2 2) Write an algebraic expression for the sum of 3 Coins and c Coins 3) What Value of x is a solution to the equation? x-2 = 16 4) use mental math to solve the equation. z/-12 = -5 5) estimate the solution of the equation x-8.1 = 5.3 to the nearest whole ...
75. ### Math
Im working on "Finding the Area and Circumference of a circle" I have a question. Is the circumference half of the area? I was looking at my brothers math book, this is what it shows. C=πr2≈ 2(3.14)(4)≈ 25.1 A=πr^2≈ 3.14(4^2)≈ 50.2 cm^2. Thanks.
1) -5g-6 for g = -2 2) Write an algebraic expression for the sum of 3 Coins and c Coins 3) What Value of x is a solution to the equation? x-2 = 16 4) use mental math to solve the equation. z/-12 = -5 5) estimate the solution of the equation x-8.1 = 5.3 to the nearest whole ...
77. ### MATH
IF JACK AND JILL ARE COMPARING RESULTS OF A MATH TEST AND jACK HAD 40 CORRECT ANSWERS AND JILL HAD 50 CORRECT ANS AND THERE WERE 32 QUESTIONS THEY BOTH GOT RIGHT AND 6 THEY BOTH GOT WRONG. HOW DO I FIND THE NUMBER OF QUESTIONS ON THE TEST?
I have a math question that I do not really understand how to answer is there someone who can walk me through it? In Pinecrest middle school there are 58 6th graders 76 7th graders, and 38 8th graders. The counsel is made up of 35 students who are chosen to represent all three...
79. ### Math
Mr. Lin baked banana bread for a bake sale to raise money for the math team. He said that he added a spoonful of walnuts for each of the students in his three classes, and that he added more than 250 walnuts. He used the inequality 16w + 24w + 10w > 250 to represent the ...
80. ### Math
Lee Holmes deposited \$16,600 in a new savings account at 9% interest compounded semiannually. At the beginning of year 4, Lee deposits an additional \$41,600 at 9% interest compounded semiannually. At the end of year 6, what is the balance in Lee’s account? I have tried and I...
81. ### math
19. Mr. Green teaches band, choir, and math. This year, he has 57 students that take at least one of his classes. He teaches band to 25 students. There are 48 students who take either band or choir or both. There are 9 students who take both math and choir with Mr. Green. Use ...
82. ### math-followup
A bag got mix up 6 pairs of shoes – 2 red, 2 blue, and 2 purple. Picking randomly in the dark, how many shoes must I pick before for sure that there is one matching pair in red or blue? Ms Sue answered 4. Request for explanation please. math - Ms. Sue, Tuesday, January 29, ...
19. Mr. Green teaches band, choir, and math. This year, he has 57 students that take at least one of his classes. He teaches band to 25 students. There are 48 students who take either band or choir or both. There are 9 students who take both math and choir with Mr. Green. Use ...
84. ### math
suppose a license plate for a car has 4 letters followed by 1 number and 3 more letters. The letters and numbers are chosen randomly. a. how many license plates are possible? b. compare the probability that a license plate spells math4you with the probability that the first 4 ...
85. ### Math 222
222 - Steve, Wednesday, October 22, 2014 at 12:02am I assume you want to factor the expressions? a^4b + a^2b^3 = a^2b(a^2+b^2) z^2 + 18z + 45 = (z+3)(z+15) •Math 222 - Monica, Wednesday, October 22, 2014 at 12:10am Thank you Steve and can you please explain how you got both ...
86. ### Pre algebra
How do I complete each ORDERED PAIR solution to the given equation? y=-x+9;(4),(9),(0) I looked up math ordered pairs at www.google.com, and these are the results. There may be a site in here with an explanation to help you. http://www.google.com/search?q=math+ordered+pairs&...
87. ### Math
math word problem a building 240ft tall casts a 100 foot long shadow. If a person stands at the end of the shadow and looks up to the top of the building, what is the angle of the persons eyes to the top of the building(to the nearest hundredth of a degree)?(assume the persons...
88. ### Math (money)
HIII!!! I NEED HELP ON MATH! PLEASE HELP! Melissa has spent \$17 so far on gifts. The additional amount she will spend will be more than \$8. What are the possible total amount she will spent? Use c for the total amount (in dollars) she will spend.
89. ### Math
In a math contest of 10 problems, 5 points was given for each correct answer and 2 points was deducted for each incorrect answer. If Nancy did all 10 problems ad scored 29 points, how many correct answers did she have?
90. ### Math
I have more math problems that I will to figure out. 2) 3.1 divided by 4 and 6.4 divided by 9 Estimate 3.1 which would be 3 Estimate 6.4 which would be 6 3 divided by 4 and 6 divided by 9 This one I am totally confused about. The only other way I can think is 32 divided by 4=8...
91. ### math
describe general rule for which fraction that have decimal forms that terminate and which have decimals that repeat and list prime factorizations for each: 1/2 , 1/3, 1/4, 1/5,1/6,1/7,1/8,1/9,1/10,1/11,1/12 how do I do these problems? I searched Google under the key words "...
92. ### math***help plz***
please help me with math. A car travels 132 miles using 4 gallons of gas.At that rate,how far can the car travel using 7 gallons of gas? A.231 miles B.264 miles C.134 miles D.1,260 miles
93. ### Math
1.how many different permutations can you make with the letters in the word s e v e n t e e n ? 2.a teacher has a set of 12 probelms to use on a math exam. the teacher makes different versions of the exam by putting 10 questions on each exam. how many different exams can the ...
94. ### math
In my math book i got a question that said 1/7 times 1/7. I got answer of 2 if my answer is not right can you show me how to do the problem.If my answer is correct still show me how to do the problem
95. ### math 115
1. sabrina has the following math quiz scores. 19, 24, 25, 17, 13, & 18. find her mean score, to the nearest tenth if necessary.= 13 is this correct? 2. eleven applications for a certain position had the following ages. 32, 33 ,25, 36, 48, 51, 28, 23, 54, 52, 45. what was the ...
I am doing a math project on Leonardo Fibonacci, the creator of the Fibonacci's Sequence. I was wondering if someone could please tell me a website where I could get more information about his marriage. can someone help me? http://www.google.com/search?as_q=Fibonacci+biography...
97. ### Math
how to explain to my child this math problem : davia draws a shape with 5 sides .Two sides are each 5 inches long . Two other sides are 4 inches long.The perimeter of the shape is 27 inches . What is the length of the fifth side?
98. ### math-old
A bag got mix up 6 pairs of shoes – 2 red, 2 blue, and 2 purple. Picking randomly, how many shoes must I pick before for sure that there is one matching pair in red or blue? Ms Sue answered 4. Request for explanation please. math - Ms. Sue, Tuesday, January 29, 2013 at 10:...
99. ### math/algebra
17 simplify 3/(x-5) + 1/ 1 - 4/(x-5) The non-use of grouping symbols makes this sentence without meaning. The non-use of grouping symbols makes this sentence without meaning. 3/x-5 +1 / 1 - 4/x-5 = I never know how to write the problems so someone can understnad what I mean, ...
100. ### Math
I have this hard question in math. can someone help me find out the riddle? I start with 1 bacteria. Every hour it doubles. How many hours until there are 1,000,000 bacteria? Can you make a function that describes this situation? What if I double it? I start with 2 glasses ... | 6,729 | 25,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.935887 |
http://www.ltcconline.net/greenl/courses/203/Vectors/orthogonalComplements.htm | 1,516,116,295,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886437.0/warc/CC-MAIN-20180116144951-20180116164951-00558.warc.gz | 492,039,915 | 4,360 | Orthogonal Complements Definition of the Orthogonal Complement Geometrically, we can understand that two lines can be perpendicular in R2 and that a line and a plane can be perpendicular to each other in R3. We now generalize this concept and ask given a vector subspace, what is the set of vectors that are orthogonal to all vectors in the subspace. Definition Let V be a vector space and W be a subspace of V. Then the orthogonal complement of W in V is the set of vectors u such that u is orthogonal to all vectors in W. Example Let V = R2 and W be the subspace spanned by (1,2). Then is the set of vectors (a,b) with (a,b) . c(1,2) = 0 or ac + 2bc = 0 a + 2b = 0 This is a 1 dimensional vector space spanned by (-2,1) In the example above the orthogonal complement was a subspace. This will always be the case. Theorem Let W be a subspace of a vector space V. Then the orthogonal complement of W is also a subspace of V. Furthermore, the intersection of W and its orthogonal complement is just the zero vector. Proof Let u1 and u2 be vectors in the orthogonal complement of W and c be a constant. Then 1. If w is in W, then (u1 + u2) . w = u1. w + u2 . w = 0 2. If w is in W, then (cu1) . w = c(u1 . w) = c(0) = 0 Now we prove that the intersection is zero. If v is in the intersection then we think of v first as being in W and second as being in the orthogonal complement of W. Hence v . v = 0 This implies that v = 0 The next theorem states that if w1, ... ,wr is a basis for W and u1, ... ,uk is a basis for then {w1, ... ,wr, u1, ... ,uk} is a basis for Rn. In symbols, we write Theorem We leave it up to you to look up the proof of this statement. What this means is that every vector v in Rn can be uniquely written in the form v = w + u with w in W and u in . A corollary of this theorem is the following Corollary Proof First if a vector is in W then it is orthogonal to every vector in the orthogonal complement of W. If a vector v is orthogonal to every vector in the orthogonal complement of W, and also by the theorem above we have v = w + u with w in W and u in the orthogonal complement of W. Since u is in the orthogonal complement of W, we have 0 = v . u = (w + u) . u = w . u + u . u = u . u Hence u = 0 and v = w. Matrices and Complements If we think of matrix multiplication as a collection of dot products then if Ax = 0 then x is orthogonal to each of the rows of A. Also if ATy = 0 then y is orthogonal to each of the columns of A. More precisely we have Theorem 1. The null space of A is the orthogonal complement of the row space of A. 2. The null space of AT is the orthogonal complement of the column space of A. Example Find a basis for the orthogonal complement of the space spanned by (1,0,1,0,2), (0,1,1,1,0) and (1,1,1,1,1). Solution We find the null space of the matrix We find the rref of A. We get a basis {(0,-1,0,1,0), (-1,1,-1,0,1)} Projections Given a vector v and a subspace W with orthogonal basis w1, ... , wn, we are often interested in finding in finding the vector in W that is closest to v. This closest vector is v . w1 v . w2 v . wn projWv = w1+ w2 + ... + wn w1 . w1 w2 . w2 wn . wn We will use this formula when we talk about inner product spaces and Fourier coefficients. Notice that if W is orthonormal then the denominators are all equal to one. Back to the Linear Algebra Home Page | 1,082 | 3,841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2018-05 | latest | en | 0.860684 |
http://mathhelpforum.com/advanced-statistics/120489-stats-help-please.html | 1,503,355,200,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109670.98/warc/CC-MAIN-20170821211752-20170821231752-00226.warc.gz | 290,856,286 | 11,317 | There's no way I can figure it out! Please see attachment below..thanks.
2. Thats actually a fun problem.
I assume that the population varainces are unknow and you need to pool.
You have to figure out the chi-squares and obtain a t distribution.
I will be back tonight if you need help.
IF we assume we know the pop variances then it's a normal distribution.
$3\bar X-2\bar Y$
But I do need to know if the pop variances are known, I assume not.
3. It is def a normal distribution. But we never did anything like this in class that I have no idea how to start.
4. I know that, but I see a few problems.
In part two do you really want a CI for $\mu_1-\mu_2$
or $3\mu_1-2\mu_2$?
THEN, I think your instructor messed up with the relationship between the pop variances.
It works either way, but the algebra is nicer if we switch the 2 and the 3.
$3\bar X-2\bar Y$ is unbiased for $3\mu_1-2\mu_2$
and it's variance is $9\sigma^2_1+4\sigma^2_2$
It's normal hence
${(3\bar X-2\bar Y)-(3\mu_1-2\mu_2)\over \sqrt{9\sigma^2_1+4\sigma^2_2}}\sim N(0,1)$
Next use the relationship between the sigma's
BUT I keep asking, are they known?
If they are this is the test stat and you use a st normal table putting half of alpha in each table.
MORE likely the sigma's are unknown and you use the relationship and derive a t stat.
Here you need to pool the sample variances and obtain the chi-square distribution which replaces the $\sigma^2$s
5. Originally Posted by matheagle
I know that, but I see a few problems.
In part two do you really want a CI for $\mu_1-\mu_2$
or $3\mu_1-2\mu_2$?
THEN, I think your instructor messed up with the relationship between the pop variances.
It works either way, but the algebra is nicer if we switch the 2 and the 3.
$3\bar X-2\bar Y$ is unbiased for $3\mu_1-2\mu_2$
and it's variance is $9\sigma^2_1+4\sigma^2_2$
It's normal hence
${(3\bar X-2\bar Y)-(3\mu_1-2\mu_2)\over \sqrt{9\sigma^2_1+4\sigma^2_2}}\sim N(0,1)$
Next use the relationship between the sigma's
BUT I keep asking, are they known?
If they are this is the test stat and you use a st normal table putting half of alpha in each table.
MORE likely the sigma's are unknown and you use the relationship and derive a t stat.
Here you need to pool the sample variances and obtain the chi-square distribution which replaces the $\sigma^2$s
The variables are not known...I asked the professor. Only hint given. Now what do you do though...I'm completely stuck
6. Replace one variance with the other using the relationship
Next obtain a Chi-square distribution.
Then obtain a t density.
And I asked last week, is there a typo in part 2?
WHAT exactly do you want the CI for? | 775 | 2,666 | {"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": 15, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2017-34 | longest | en | 0.92055 |
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# Dynamic Score Combination: A supervised and unsupervised score combination method
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In two-class score-based problems the combination of scores from an ensemble of experts is generally used to obtain distributions for positive and negative patterns that exhibit a larger degree of separation than those of the scores to be combined. Typically, combination is carried out by a "static" linear combination of scores, where the weights are computed by maximising a
performance function. These weights are equal for all the patterns, as they are assigned to each of the expert to be combined. In this paper we propose a "dynamic" formulation where the weights are computed individually for each pattern. Reported results on a biometric dataset show the effectiveness of the proposed combination methodology with respect to "static" linear combinations and trained combination rules.
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### Dynamic Score Combination: A supervised and unsupervised score combination method
1. 1. Dynamic Score Combination a supervised and unsupervised score combination method R. Tronci, G. Giacinto, F. Roli DIEE - University of Cagliari, Italy Pattern Recognition and Applications Group http://prag.diee.unica.it MLDM 2009 - Leipzig, July 23-25, 2009
2. 2. Outline ! Goal of score combination mechanisms ! Dynamic Score Combination ! Experimental evaluation ! Conclusions Giorgio Giacinto MLDM 2009 - July 23-25, 2009 2
3. 3. Behavior of biometric experts Genuine scores should produce a positive outcome Impostor scores should produce a negative outcome th FNMRj (th) = \$ p(s j | s j ! positive)ds j = P(s j % th | s j ! positive) "# # FMRj (th) = \$ p(s j | s j ! negative)ds j = P(s j > th | s j ! negative) th Giorgio Giacinto MLDM 2009 - July 23-25, 2009 3
4. 4. Performance assessment ! True Positive Rate = 1 - FNMR Giorgio Giacinto MLDM 2009 - July 23-25, 2009 4
5. 5. Goal of score combination ! To improve system reliability, different experts are combined ! different sensors, different features, different matching algorithms ! Combination is typically performed at the matching score level Giorgio Giacinto MLDM 2009 - July 23-25, 2009 5
6. 6. Goal of score combination Combined score Giorgio Giacinto MLDM 2009 - July 23-25, 2009 6
7. 7. Goal of score combination ! The aim is to maximize the separation between classes e.g. (µ ) 2 gen ! µimp FD = " gen + " imp 2 2 ! Thus the distributions have to be shifted far apart, and the spread of the scores reduced Giorgio Giacinto MLDM 2009 - July 23-25, 2009 7
8. 8. Static combination ! Let E = {E1,E2,…Ej,…EN} be a set of N experts ! Let X = {xi} be the set of patterns ! Let fj ( ) be the function associated to expert Ej that produces a score sij = fj(xi) for each pattern xi Static linear combination N si* = # ! j " sij j =1 ! The weights are computed as to maximize some measure of class separability on a training set ! The combination is static with respect to the test pattern to be classified Giorgio Giacinto MLDM 2009 - July 23-25, 2009 8
9. 9. Dynamic combination The weights of the combination also depends on the test pattern to be classified N si* = # ! ij " sij j =1 The local estimation of combination parameters may yield better results than the global estimation, in terms of separation between the distributions of scores si* Giorgio Giacinto MLDM 2009 - July 23-25, 2009 9
10. 10. Estimation of the parameters for the dynamic combination ! Let us suppose without loss of generality s i1 ! s i2 ! ! ! siN ! The linear combination of three experts ! i1si1 + ! i 2 si 2 + ! i 3 si 3 ! ij "[ 0,1] can also be written as " i1si1 + si 2 + " i!3 si 3 ! which is equivalent to " i1si1 + " i!! si 3 !! 3 Giorgio Giacinto MLDM 2009 - July 23-25, 2009 10
11. 11. Estimation of the parameters for the dynamic combination ! This reasoning can be extended to N experts, so we can get ( ) si* = !i1 min sij + !i 2 max sij j j ( ) ! Thus, for each pattern we have to estimate two parameters ! If we set the constraint !i1 + !i 2 = 1 only one parameter has to be estimated and si* ! [minj(sij),maxj(sij)] Giorgio Giacinto MLDM 2009 - July 23-25, 2009 11
12. 12. Properties of the Dynamic Score Combination ( ) si* = !i max sij + (1 " !i ) min sij j j ( ) ! This formulation embeds the typical static combination rules #" J sij \$ min ( sij ) N j j =1 Linear combination !i = ( ) ( ) ! max sij \$ min sij j j 1 N ( ) " sij # min sij N j =1 j Mean rule !i = max ( s ) # min ( s ) ! ij ij j j ! Max rule for "i = 1 and Min rule for "i = 0 Giorgio Giacinto MLDM 2009 - July 23-25, 2009 12
13. 13. Properties of the Dynamic Score Combination ( ) si* = !i max sij + (1 " !i ) min sij j j ( ) ! This formulation also embeds the Dynamic Score Selection (DSS) "1 if xi belongs to the positive class !i = # \$0 if xi belongs to the negative class ! DSS clearly maximize class separability if the estimation of the class of xi is reliable ! e.g., a classifier trained on the outputs of the experts E Giorgio Giacinto MLDM 2009 - July 23-25, 2009 13
14. 14. Supervised estimation of "i ( ) si* = !i max sij + (1 " !i ) min sij j j ( ) ! "i = P(pos|xi,E) P(pos|xi,E) can be estimated by a classifier trained on the outputs of the experts E ! "i is estimated by a supervised procedure ! This formulation can also be seen as a soft version of DSS ! P(pos|xi,E) accounts for the uncertainty in class estimation Giorgio Giacinto MLDM 2009 - July 23-25, 2009 14
15. 15. Unsupervised estimation of "i ( ) si* = !i max sij + (1 " !i ) min sij j j ( ) ! "i is estimated by an unsupervised procedure ! the estimation does not depend on a training set 1 N Mean rule !i = " sij N j =1 Max rule !i = max sij j ( ) Min rule !i = min sij j ( ) Giorgio Giacinto MLDM 2009 - July 23-25, 2009 15
16. 16. Dataset ! The dataset used is the Biometric Scores Set Release 1 of the NIST http://www.itl.nist.gov/iad/894.03/biometricscores/ ! This dataset contains scores from 4 experts related to face and fingerprint recognition systems. ! The experiments were performed using all the possible combinations of 3 and 4 experts. ! The dataset has been divided into four parts, each one used for training and the remaining three for testing Giorgio Giacinto MLDM 2009 - July 23-25, 2009 16
17. 17. Experimental Setup ! Experiments aimed at assessing the performance of ! The unsupervised Dynamic Score Combination (DSC) ! "i estimated by the Mean, Max, and Min rules ! The supervised Dynamic Score Combination ! "i estimated by k-NN, LDC, QDC, and SVM classifiers ! Comparisons with ! The Ideal Score Selector (ISS) ! The Optimal static Linear Combination (Opt LC) ! The Mean, Max, and Min rules ! The linear combination where coefficients are estimated by the LDA Giorgio Giacinto MLDM 2009 - July 23-25, 2009 17
18. 18. Performance assessment ! Area Under the ROC Curve (AUC) ! Equal Error Rate (ERR) µ gen " µimp ! d! = # gen # imp 2 2 + 2 2 ! FNMR at 1% and 0% FMR ! FMR at 1% and 0% FNMR Giorgio Giacinto MLDM 2009 - July 23-25, 2009 18
19. 19. Combination of three experts AUC EER d’ ISS 1.0000 (±0.0000) 0.0000 (±0.0000) 25.4451 (±8.7120) Opt LC 0.9997 (±0.0004) 0.0050 (±0.0031) 3.1231 (±0.2321) Mean 0.9982 (±0.0013) 0.0096 (±0.0059) 3.6272 (±0.4850) Max 0.9892 (±0.0022) 0.0450 (±0.0048) 3.0608 (±0.3803) Min 0.9708 (±0.0085) 0.0694 (±0.0148) 2.0068 (±0.1636) DSC Mean 0.9986 (±0.0011) 0.0064 (±0.0030) 3.8300 (±0.5049) DSC Max 0.9960 (±0.0015) 0.0214 (±0.0065) 3.8799 (±0.2613) DSC Min 0.9769 (±0.0085) 0.0634 (±0.0158) 2.3664 (±0.2371) LDA 0.9945 (±0.0040) 0.0296 (±0.0123) 2.3802 (±0.2036) DSC k-NN 0.9987 (±0.0016) 0.0104 (±0.0053) 6.9911 (±0.9653) DSC ldc 0.9741 (±0.0087) 0.0642 (±0.0149) 2.7654 (±0.2782) DSC qdc 0.9964 (±0.0039) 0.0147 (±0.0092) 9.1452 (±3.1002) DSC svm 0.9996 (±0.0004) 0.0048 (±0.0026) 4.8972 (±0.4911) Giorgio Giacinto MLDM 2009 - July 23-25, 2009 19
20. 20. DSC Mean Vs. Mean rule Combination of three experts DSC Mean AUC !!0.9991 EER !!0.0052 d' !!!4.4199 Mean rule AUC !!0.9986 EER !!0.0129 d' !!!4.0732 Giorgio Giacinto MLDM 2009 - July 23-25, 2009 20
21. 21. Unsupervised DSC Vs. fixed rules AUC Giorgio Giacinto MLDM 2009 - July 23-25, 2009 21
22. 22. Unsupervised DSC Vs. fixed rules EER Giorgio Giacinto MLDM 2009 - July 23-25, 2009 22
23. 23. Unsupervised DSC Vs. fixed rules FMR at 0% FNMR Giorgio Giacinto MLDM 2009 - July 23-25, 2009 23
24. 24. DSC Mean Vs. supervised DSC AUC Giorgio Giacinto MLDM 2009 - July 23-25, 2009 24
25. 25. DSC Mean Vs. supervised DSC EER Giorgio Giacinto MLDM 2009 - July 23-25, 2009 25
26. 26. DSC Mean Vs. supervised DSC FMR at 0% FNMR Giorgio Giacinto MLDM 2009 - July 23-25, 2009 26
27. 27. Conclusions ! The Dynamic Score Combination mechanism embeds different combination modalities ! Experiments show that the unsupervised DSC usually outperforms the related “fixed” combination rules ! The use of a classifier in the supervised DSC allows attaining better performance, at the expense of increased computational complexity ! Depending on the classifier, performance are very close to those of the optimal linear combiner Giorgio Giacinto MLDM 2009 - July 23-25, 2009 27
1. #### A particular slide catching your eye?
Clipping is a handy way to collect important slides you want to go back to later. | 3,117 | 9,536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-07 | latest | en | 0.845768 |
https://math.stackexchange.com/questions/137941/norm-map-over-galois-extension | 1,642,655,053,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301720.45/warc/CC-MAIN-20220120035934-20220120065934-00229.warc.gz | 447,344,930 | 33,786 | # Norm map over Galois extension
Let $K$ be a Galois extension of $F$. Prove or disprove that any intermediate field $L$ of $K/F$ is of the form $L=F(\{N(a)\mid a \in K\})$, where $N$ the is norm map of $K/L$.
• "is of the form" is unappropriate here, since your $N$ depends on $L$. You should simply say something like "any intermediate $L$ satisfies $L=F(\lbrace N_{K/L}(a) |a \in K\rbrace)$". Aug 7 '14 at 8:11
Proof:Suppose that $K=F(\beta)$ with $f(x)=min(\beta,L)=\prod_{\sigma\in Gal(K/L)}{(x-\sigma(\beta))}$ $=x^n+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}$. A simple observation here is that $F(\{a_{i}\})=L$, so we only need to show that $F(\{N_{K/L}(\alpha)\mid\alpha \in K\})=F(\{a_{i}\})$. Using the definition of Norm, we have $N_{K/L}(\alpha)=\prod_{\sigma\in Gal(K/L)}\sigma(\alpha)$, for any $\alpha$ in K. Let $\alpha=a-\beta$, where $a \in F$, we have $N_{K/L}(\alpha)=f(a)$. At last, using a little linear algebra, we can show that $F(\{f(a)\mid a\in F\})=F(\{a_{i}\})$(hint: using Vandermonde Determinant). Now the proof is complete. | 387 | 1,046 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2022-05 | latest | en | 0.727167 |
http://www.mathpuzzle.ca/m/Puzzle/Round-The-Coast.html | 1,534,802,605,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221217354.65/warc/CC-MAIN-20180820215248-20180820235248-00492.warc.gz | 534,663,668 | 4,005 | # ROUND THE COAST.
Here is a puzzle that will, I think, be found as amusing as instructive.
We are given a ring of eight circles. Leaving circle 8 blank, we are
required to write in the name of a seven-lettered port in the United
Kingdom in this manner. Touch a blank circle with your pencil, then jump
over two circles in either direction round the ring, and write down the
first letter. Then touch another vacant circle, jump over two circles,
and write down your second letter. Proceed similarly with the other
letters in their proper order until you have completed the word. Thus,
suppose we select "Glasgow," and proceed as follows: 6--1, 7--2, 8--3,
7--4, 8--5, which means that we touch 6, jump over 7 and and write down
"G" on 1; then touch 7, jump over 8 and 1, and write down "l" on 2; and
so on. It will be found that after we have written down the first five
letters--"Glasg"--as above, we cannot go any further. Either there is
something wrong with "Glasgow," or we have not managed our jumps
properly. Can you get to the bottom of the mystery?
REAPING THE CORN. SETTING THE BOARD.
Feedback | 297 | 1,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | longest | en | 0.931938 |
https://cs.stackexchange.com/questions/81694/why-encoding-of-graph-in-adjacency-matrix-is-omega-sqrt-n | 1,563,645,201,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526560.40/warc/CC-MAIN-20190720173623-20190720195623-00537.warc.gz | 367,478,377 | 35,149 | # Why encoding of graph in adjacency matrix is $\Omega (\sqrt n)$?
in the textbook of CLRS, p.1062 he says the following:
if we use "reasonable" encoding of a graph as its adjacency matrix, the number m of vertices in the graph is $\Omega (\sqrt n)$, where n = |$\langle G\rangle|$ is the length of the encoding of G.
I want to know why $\Omega (\sqrt n)$? I mean it is adjacency matrix which takes $m^2$ but I don't know how he calculates it, so I would like someone to explain to me how we get this time complexity?
Thank you!
• Is it time complexity? I suspect it is the number of vertices that can be encoded using adjacency matrix of size $n$. BTW, quote says exactly this. – rus9384 Sep 26 '17 at 16:52
• It's not a time complexity: it's just a bound on a function. – David Richerby Sep 26 '17 at 16:59
An $m\times m$ adjacency matrix requires $m^2$ bits, and $m=\sqrt{n}$ is just the solution to $m^2 = |\langle G\rangle| = n$.
The point of the $\Omega$ is that a reasonable encoding doesn't have fewer than $\sqrt{n}$ vertices in a description of length $n$. For example "Write out the $k\times k$ adjacency matrix followed by $2^k$ zeroes" isn't "reasonable" and a string of length $n$ would encode a graph with only about $\log n$ vertices. | 344 | 1,256 | {"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} | 3.40625 | 3 | CC-MAIN-2019-30 | longest | en | 0.925005 |
https://ar5iv.labs.arxiv.org/html/1905.00303 | 1,717,085,482,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971668873.95/warc/CC-MAIN-20240530145337-20240530175337-00497.warc.gz | 80,803,574 | 109,625 | # Flag Bott manifolds of general Lie type and their equivariant cohomology rings
Shizuo Kaji Institute of Mathematics for Industry, Kyushu University, Fukuoka 819-0395, Japan Shintarô Kuroki Faculty of Science, Department of Applied Mathematics, Okayama University of Science, 1-1 Ridai-cho Kita-ku Okayama-shi Okayama 700-0005, JAPAN Eunjeong Lee Center for Geometry and Physics, Institute for Basic Science (IBS), Pohang 37673, Korea and Dong Youp Suh Department of Mathematical Sciences, KAIST, 291 Daehak-ro, Yuseong-gu, Daejeon 34141, South Korea
###### Abstract.
In this article we introduce flag Bott manifolds of general Lie type as the total spaces of iterated flag bundles. They generalize the notion of flag Bott manifolds and generalized Bott manifolds, and admit nice torus actions. We calculate the torus equivariant cohomology rings of flag Bott manifolds of general Lie type.
###### Key words and phrases:
flag Bott towers of general Lie type, flag Bott manifold, generalized Bott manifold, flag manifold, equivariant cohomology
###### 2010 Mathematics Subject Classification:
Primary: 55R10, 14M15; Secondary: 57S25
Kaji was partially supported by KAKENHI, Grant-in-Aid for Scientific Research (C) 18K03304. Kuroki was supported by JSPS KAKENHI Grant Number 17K14196. Lee was supported by IBS-R003-D1. Suh was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Science, ICT & Future Planning (No. 2016R1A2B4010823).
## 1. Introduction
A Bott tower $\{M_{j}\mid 0\leq j\leq m\}$ is a sequence of $\mathbb{C}P^{1}$-fibrations $\mathbb{C}P^{1}\hookrightarrow M_{j}\to M_{j-1}$ such that $M_{j}$ is the induced projective bundle of the sum of two complex line bundles over $M_{j-1}$. Each manifold $M_{j}$ is called a $j$-stage Bott manifold and it is known that $M_{j}$ is a non-singular projective toric variety. On the other hand, a flag manifold is the orbit space $G/P$ of a complex Lie group $G$ divided by a parabolic subgroup $P$. A flag manifold is known to be a non-singular projective variety having a nice torus action.
These two families of spaces are closely related by the Bott-Samelson resolution (see [13, 17]). Both families have been actively studied as spaces with nice torus actions, and have served as a test ground for various theories and problems. Schubert calculus studies the cohomology of flag manifolds, in which topology, algebraic geometry, combinatorics, and representation theory meet together (see [22] for a survey). On the other hand, the cohomological rigidity problem of quasi-toric manifolds may be regarded as one of the essential problems in toric topology, and some affirmative results are known for Bott manifolds (see [6, 7, 8, 20]).
There are two known natural generalizations of Bott manifolds: flag Bott manifolds, and generalized Bott manifolds. The flag Bott manifolds extend the relation between Bott manifolds and Bott–Samelson manifolds. We refer the readers to [23] for the definition of flag Bott manifolds, and to [15] for this enlarged relation. And the generalized Bott manifolds are toric manifolds studied in [10, 11, 24].
In this note, we introduce flag Bott manifolds of general Lie type which simultaneously generalize flag manifolds, flag Bott manifolds, and generalized Bott manifolds. We give two closely related descriptions of the flag Bott manifold of general Lie type; as the total space of an iterated flag bundle (Definition 3.1) and as an orbit space of a Lie group (Definition 3.5). We see a torus acts on the flag Bott manifold in a natural manner (Definition 4.1). Moreover, we determine its Borel equivariant cohomology ring. Theorem 4.2 unifies the known formulae for the equivariant cohomology of flag manifolds, flag Bott manifolds, and generalized Bott manifolds.
## 2. Flag bundle and its cohomology
Let $K$ be a compact connected Lie group and $T\subset K$ be a maximal torus of $K$. Let $Z\subset K$ be the centralizer in $K$ of a circle subgroup of $T$. Then, $Z$ is known to be connected. Indeed, for any element $g\in Z$ consider the subgroup $H$ which is generated by $g$ and the circle subgroup. Since $H$ is abelian, it is contained in some torus which is contained in $Z$. Therefore, $g$ is in the identity component of $Z$. We denote by $W$ (resp. $W(Z)$) the Weyl group of $K$ (resp. $Z$). The space $K/Z$ of left-cosets is called the generalized flag manifold, and there exists the universal flag bundle $K/Z\hookrightarrow BZ\to BK$. For any map $f\colon X\to BK$ from a topological space $X$, we have the pull-back bundle $K/Z\hookrightarrow F_{f}(Z)\to X$, which fits in the diagram
(2.1) ${K/Z}$${K/Z}$${F_{f}(Z)}$${BZ}$${X}$${BK}$$\scriptstyle{\tilde{f}}$$\scriptstyle{f}$
The pull-back bundle is called the flag bundle over $X$ associated to the classifying map $f$ with fiber $K/Z$.
###### Example 2.1.
Let $K=U(n)$ and $f:X\to BU(n)$. We denote by $U(1)^{n}$ the set of all diagonal matrices in $K$. It is well-known that $U(1)^{n}$ is a maximal torus in $K$. We exhibit three examples of the associated flag bundles over $X$.
Take the circle subgroup $\{\textrm{diag}(t,\dots,t,t^{2})\mid t\in S^{1}\}\subset U(1)^{n}$. Then its centralizer $Z_{1}$ is the group of the block diagonal matrices $U(n-1)\times U(1)\subset K$. The associated flag bundle $F_{f}(Z_{1})$ is isomorphic to the projective bundle $\mathbb{P}(E)$ associated to the complex vector bundle $E$ classified by $f$.
We next take the circle subgroup $\{\textrm{diag}(t,t^{2},\dots,t^{n})\mid t\in S^{1}\}\subset U(1)^{n}$. Then its centralizer $Z_{2}$ coincides with the group of the diagonal matrices $U(1)^{n}$. The associated flag bundle $F_{f}(Z_{2})$ is isomorphic to the full flag bundle
${\mathcal{F}\ell}(E)=\{V_{0}\subset V_{1}\subset V_{2}\subset\cdots\subset V_{n}=E\mid\mathrm{rank}(V_{i})=i\}$
associated to $E$.
More generally, for positive integers $p_{1},\dots,p_{k}>0$ with $p_{1}+\cdots+p_{k}=n$, the centralizer $Z_{3}$ of the circle subgroup
$\{\textrm{diag}(\underbrace{t,\dots,t}_{p_{1}},\underbrace{t^{2},\dots,t^{2}}_{p_{2}},\dots,\underbrace{t^{k},\dots,t^{k}}_{p_{k}})\mid t\in S^{1}\}\subset U(1)^{n}$
is the group of the block diagonal matrices $U(p_{1})\times\cdots\times U(p_{k})\subset U(n)$. The associated flag bundle $F_{f}(Z_{3})$ is isomorphic to the partial flag bundle
$\{V_{0}\subset V_{1}\subset V_{2}\subset\cdots\subset V_{k}\subset E\mid\mathrm{rank}(V_{i})=p_{1}+\cdots+p_{i}\}$
associated to $E$.
The cohomology of $F_{f}(Z)$ can easily be computed for some coefficient rings. A prime $p$ is said to be a torsion prime of $K$ if $H_{\ast}(K;\mathbb{Z})$ has $p$-torsion. Let $R$ be a PID in which torsion primes of $K$ are invertible. For example, if $K$ is $U(n)$ or $Sp(n)$, then we may take $R$ to be $\mathbb{Z}$. When $K$ is simply-connected, the torsion primes are summarized in Table 1 (see [2, §2.5]). Note that if $p$ is not a torsion prime of $K$, it is not of any circle centralizer $Z$ as well.
###### Proposition 2.2.
Let $R$ be a PID in which the torsion primes of $K$ are invertible. Then, we have a ring isomorphism
$H^{\ast}(F_{f}(Z);R)\cong H^{\ast}(X;R)\otimes_{H^{\ast}(BK;R)}H^{\ast}(BZ;R),$
where $H^{\ast}(X;R)$ has the $H^{\ast}(BK;R)$-module structure induced by the map $f$ and $H^{*}(BZ;R)$ also has the natural $H^{*}(BK;R)$-module structure induced by the classifying map of the inclusion $Z\to K$.
###### Proof.
Taking the cohomology of the lower square of (2.1) gives rise to a homomorphism
$\varphi:H^{\ast}(X;R)\otimes_{H^{\ast}(BK;R)}H^{\ast}(BZ;R)\to H^{\ast}(F_{f}(Z);R),$
which we will show is an isomorphism. There are elements $\alpha_{i}\in H^{\ast}(BZ;R)$ which restrict to a basis of $H^{\ast}(K/Z;R)$ by [2, §4.2]. By the Leray–Hirsch theorem we see $H^{\ast}(BZ;R)$ is generated by $\{\alpha_{i}\}$ over $H^{\ast}(BK;R)$ and $H^{\ast}(F_{f}(Z);R)$ is generated by $\{\tilde{f}^{*}\alpha_{i}\}$ over $H^{\ast}(X;R)$. Since $\varphi$ is a ring homomorphism which is $H^{\ast}(X;R)$-module isomorphism, it is a ring isomorphism. ∎
###### Example 2.3.
Let $K=U(3)$ and $W\cong\mathfrak{S}_{3}\cong\langle s_{1},s_{2}\rangle$. Here we may regard $s_{1}$ as the permutation matrix of the 1st and the 2nd coordinates and $s_{2}$ as that of the 2nd and the 3rd coordinates. Suppose that $Z$ is the centralizer subgroup of the circle subgroup $\{\textrm{diag}(t,1,1)\mid t\in S^{1}\}\subset U(1)^{3}$. Note that the Lie algebra of this circle subgroup is fixed by the permutation $s_{2}$, and $Z$ is isomorphic to the block diagonal matrices $U(1)\times U(2)$, thus, we have $K/Z\cong\mathbb{C}P^{2}$, and
$H^{*}(BK;\mathbb{Z})\cong\mathbb{Z}[x_{1},x_{2},x_{3}]^{\mathfrak{S}_{3}},\quad H^{*}(BZ;\mathbb{Z})\cong\mathbb{Z}[x_{1},x_{2},x_{3}]^{W(Z)}$
where $W(Z)=\langle s_{2}\rangle\subset\mathfrak{S}_{3}$. Let $f:X\to BU(3)$ be a map and $E\to X$ be the complex vector bundle of rank $3$ classified by $f$. The induced projective bundle $\mathbb{P}(E)$ coincides with $F_{f}(Z)$. Hence, by applying Proposition 2.2, we can compute its cohomology as follows:
$\displaystyle H^{\ast}(\mathbb{P}(E);\mathbb{Z})$ $\displaystyle\cong H^{\ast}(X;\mathbb{Z})\otimes_{H^{\ast}(BK;\mathbb{Z})}H^{\ast}(BZ;\mathbb{Z})$ $\displaystyle\cong H^{\ast}(X;\mathbb{Z})\otimes_{\mathbb{Z}[x_{1},x_{2},x_{3}]^{\mathfrak{S}_{3}}}\mathbb{Z}[x_{1},x_{2},x_{3}]^{\langle s_{2}\rangle}$ $\displaystyle\cong(H^{\ast}(X;\mathbb{Z})\otimes\mathbb{Z}[x_{1},x_{2}+x_{3},x_{2}x_{3}])/I,$
where $I=\langle x_{1}+x_{2}+x_{3}-c_{1}(E),x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1}-c_{2}(E),x_{1}x_{2}x_{3}-c_{3}(E)\rangle$. Observe the following relations given by the ideal $I$:
$\begin{split}x_{2}+x_{3}&=c_{1}(E)-x_{1},\\ x_{2}x_{3}&=c_{2}(E)-x_{1}(x_{2}+x_{3})=c_{2}(E)-x_{1}(c_{1}(E)-x_{1})=c_{2}(E)-x_{1}c_{1}(E)+x_{1}^{2},\\ x_{1}x_{2}x_{3}&=c_{3}(E).\end{split}$
Denoting $x_{1}$ by $x$ and eliminating $x_{2}$ and $x_{3}$, we get the following formula:
$H^{\ast}(\mathbb{P}(E);\mathbb{Z})\cong H^{\ast}(X;\mathbb{Z})[x]/\langle x^{3}-x^{2}c_{1}(E)+xc_{2}(E)-c_{3}(E)\rangle.$
Similarly, for any complex vector bundle $E\to X$ of rank $n+1$, we have that:
(2.2) $H^{\ast}(\mathbb{P}(E);\mathbb{Z})\cong H^{\ast}(X;\mathbb{Z})[x]\bigg{/}\left\langle\sum_{k=1}^{n+1}(-1)^{k}x^{n+1-k}c_{k}(E)\right\rangle,$
which recovers the well-known Borel–Hirzebruch formula (see [3, Chapter V, §15]).
As a corollary of Proposition 2.2, we obtain a quick proof of [4, Proposition 21.17 and Remarks 21.18, 21.19].
###### Corollary 2.4.
Let ${\mathcal{F}\ell}(E)\to X$ be the full flag bundle associated to an $(n+1)$-dimensional complex vector bundle. Then, we have that
$H^{\ast}({\mathcal{F}\ell}(E);\mathbb{Z})\cong H^{\ast}(X;\mathbb{Z})[x_{1},\dots,x_{n+1}]\bigg{/}\left\langle\prod_{k=1}^{n+1}(1+x_{k})-c(E)\right\rangle.$
###### Proof.
The flag bundle ${\mathcal{F}\ell}(E)\to X$ with fiber ${\mathcal{F}\ell}(n+1):={\mathcal{F}\ell}(\mathbb{C}^{n+1})$ fits into the following diagram:
${{\mathcal{F}\ell}(n+1)}$${{\mathcal{F}\ell}(n+1)}$${{\mathcal{F}\ell}(E)}$${BT}$${X}$${BU(n+1)}$$\scriptstyle{f}$
where $f$ is the classifying map of the vector bundle $E\to X$ and $T$ is a maximal torus in $U(n+1)$. Applying Proposition 2.2, we have that
$H^{\ast}({\mathcal{F}\ell}(E);\mathbb{Z})\cong H^{\ast}(X;\mathbb{Z})\otimes_{H^{\ast}(BU(n+1);\mathbb{Z})}H^{\ast}(BT;\mathbb{Z}).$
We identify $H^{\ast}(BT;\mathbb{Z})\cong\mathbb{Z}[x_{1},x_{2},\ldots,x_{n+1}]$ and $H^{\ast}(BU(n+1);\mathbb{Z})\cong\mathbb{Z}[c_{1},c_{2},\ldots,c_{n+1}]$, where $c_{i}$ is the $i$th elementary symmetric polynomial in $x_{1},\ldots,x_{n+1}$ for any $i$. The assertion follows from the fact $f^{*}(c_{i})=c_{i}(E)$. ∎
## 3. Flag Bott manifold of general Lie type
In this section, we introduce the main object of our study, flag Bott towers of general Lie type. We give two closely related definitions of a flag Bott tower of general Lie type, and prove that they are equivalent when relevant groups are simply-connected. For $1\leq j\leq m$, let $K_{j}$ be a compact connected Lie group, $T_{j}\subset K_{j}$ be a maximal torus, and $Z_{j}\subset K_{j}$ be the centralizer of a circle subgroup of $T_{j}$.
###### Definition 3.1.
An $m$-stage flag Bott tower $F_{\bullet}=\{F_{j}\mid 0\leq j\leq m\}$ of general Lie type (or an $m$-stage flag Bott tower) associated to $(K_{\bullet},Z_{\bullet})=\{(K_{j},Z_{j})\mid 1\leq j\leq m\}$ is defined recursively as follows:
1. (1)
$F_{0}$ is a point.
2. (2)
$F_{j}$ is the flag bundle over $F_{j-1}$ with fiber $K_{j}/Z_{j}$ associated to a map
$f_{j}\colon F_{j-1}\to BK_{j},$
where $f_{j}$ factors through $BT_{j}$.
The requirement for $f_{j}$ to factor through $BT_{j}$ means that we consider those bundles which are the sum of line bundles.
Two flag Bott towers $F_{\bullet}$ and $F_{\bullet}^{\prime}$ are isomorphic if there is a collection of diffeomorphisms $F_{j}\to F_{j}^{\prime}$ which commute with the projections $p_{j}\colon F_{j}\to F_{j-1}$ and $p_{j}^{\prime}\colon F_{j}^{\prime}\to F_{j-1}^{\prime}$ for all $1\leq j\leq m$.
Note that the fiber of each stage is a flag manifold, which admits a cell decomposition involving only even dimensional cells. Moreover, the total space $E$ of a fiber bundle $F\hookrightarrow E\to B$ has the structure of a CW-complex whose cells are the product of those in the base $B$ and the fiber $F$ (see, for instance, [25, p.105]). More precisely, for the fiber bundle $F\hookrightarrow E\to B$, the total space $E$ can be decomposed into the pull backs of all cells of $B$. The pull-back of a cell $c\subset B$ is homeomorphic to the product $F\times c$ (because $c$ is contractible), and $F\times c$ is also decomposed into the product of cells of $F$ and $c$. Because the product of two disks is homeomorphic to the disk, this gives a cell decomposition of the total space $E$. Therefore, a flag Bott manifold $F_{j}$ admits a cell decomposition involving only even dimensional cells as well, and in particular, it is simply-connected.
###### Example 3.2.
Let $(K_{j},Z_{j})=(U(n_{j}+1),T^{n_{j}+1})$, where $T^{n_{j}+1}$ is a maximal torus of $U(n_{j}+1)$. We have $K_{j}/Z_{j}\cong{\mathcal{F}\ell}(n_{j}+1)$ for each $j$, and we get an $m$-stage flag Bott tower which is introduced in [23, Definition 2.1]. An $m$-stage flag Bott tower $F_{\bullet}=\{F_{j}\mid 0\leq j\leq m\}$ is defined to be an iterated bundle
$F_{m}\to F_{m-1}\to\cdots\to F_{1}\to F_{0}=\{\textup{a point}\},$
of manifolds $F_{j}={\mathcal{F}\ell}\left(\bigoplus_{k=1}^{n_{j}+1}{\xi^{{(j)}}_{{k}}}\right)$ where ${\xi^{{(j)}}_{{k}}}$ is a complex line bundle over $F_{j-1}$ for each $1\leq k\leq n_{j}+1$ and $1\leq j\leq m$. Since the flag Bott tower in Definition 3.1 is the generalization of flag Bott tower in [23], we call the latter a full flag Bott tower of type A in this paper whenever we need to specify them.
###### Example 3.3.
Let $(K_{j},Z_{j})=(U(n_{j}+1),U(n_{j})\times U(1))$. We have $K_{j}/Z_{j}\cong\mathbb{C}P^{n_{j}}$ for each $j$, and we get an $m$-stage generalized Bott tower which is defined in [9, 10] to be an iterated bundle
$B_{m}\to B_{m-1}\to\cdots\to B_{1}\to B_{0}=\{\textup{a point}\},$
of manifolds $B_{j}=\mathbb{P}\left(\bigoplus_{k=1}^{n_{j}+1}{\xi^{{(j)}}_{{k}}}\right)$ where ${\xi^{{(j)}}_{{k}}}$ is a complex line bundle over $B_{j-1}$ for each $1\leq k\leq n_{j}+1$ and $1\leq j\leq m$.
###### Remark 3.4.
Iterated flag bundles have been studied well in the literature. For instance, in [19], the author studies the cohomology rings of iterated flag bundles associated to vector bundles which do not necessarily split into line bundles. He called them Bott tower of flag manifolds. A Bott tower of flag manifolds admits a torus action if the base space admits one and the vector bundle is equivariant. However, the essential point of our construction is to restrict to vector bundles which are the sum of line bundles. In this case, we obtain two views of our flag Bott manifold of general Lie type (in this section §3) and can introduce the bigger torus action (§4), which are non-trivial.
We now give the second definition of flag Bott tower of general Lie type in the form of an orbit space similarly to the full flag Bott tower of type $A$ case (see [23, §2.2]) as follows.
###### Definition 3.5.
Let $(K_{\bullet},Z_{\bullet})=\{(K_{j},Z_{j})\mid 1\leq j\leq m\}$. Given a family of homomorphisms $\{{\varphi^{{(\ell)}}_{{j}}}\colon Z_{j}\to T_{\ell}\mid 1\leq j<\ell\leq m\}$, the space $F_{m}^{\varphi}$ is defined as the orbit space
$F_{m}^{\varphi}:=(K_{1}\times\cdots\times K_{m})/(Z_{1}\times\cdots\times Z_{m}),$
where $(z_{1},\ldots,z_{m})\in Z_{1}\times\cdots\times Z_{m}$ acts on $(g_{1},\ldots,g_{m})\in K_{1}\times\cdots\times K_{m}$ from the right by
(3.1) $\begin{split}(g_{1},\dots,g_{m})&\cdot(z_{1},\dots,z_{m})\\ &:=(g_{1}z_{1},{\varphi^{{(2)}}_{{1}}}(z_{1})^{-1}g_{2}z_{2},\prod_{j=1}^{2}{\varphi^{{(3)}}_{{j}}}(z_{j})^{-1}g_{3}z_{3},\dots,\\ &\quad\qquad\prod_{j=1}^{m-1}{\varphi^{{(m)}}_{{j}}}(z_{j})^{-1}g_{m}z_{m}).\end{split}$
This action is easily seen to be free, and hence, $F_{m}^{\varphi}$ is a smooth manifold. Moreover, the space $F_{m}^{\varphi}$ has the structure of $K_{m}/Z_{m}$-fiber bundle over $F_{m-1}^{\varphi}$ whose classifying map $f_{m}$ is given by the composition
${F_{m-1}^{\varphi}}$${B\left(\prod_{j=1}^{m-1}Z_{j}\right)}$${BT_{m}}$${BK_{m}.}$$\scriptstyle{B(\prod_{j=1}^{m-1}{\varphi^{{(m)}}_{{j}}})}$$\scriptstyle{B\iota}$
Here the first map is the classifying map of the principal $\prod_{j=1}^{m-1}Z_{j}$-bundle
$\prod_{j=1}^{m-1}Z_{j}\hookrightarrow\prod_{j=1}^{m-1}K_{j}\to F_{m-1}^{\varphi},$
and $\iota:T_{m}\to K_{m}$ is the inclusion. Therefore, $F_{\bullet}^{\varphi}:=\{F_{j}^{\varphi}\mid 0\leq j\leq m\}$ is an $m$-stage flag Bott tower of general Lie type associated to $(K_{\bullet},Z_{\bullet})$.
Conversely, when $K_{j}$ are simply-connected for all $1\leq j\leq m$, we claim that every flag Bott tower of general Lie type associated to $(K_{\bullet},Z_{\bullet})$ can be described as the orbit space as in Definition 3.5, which implies that two definitions of flag Bott tower are equivalent.
###### Proposition 3.6.
Let $F_{\bullet}=\{F_{j}\mid 0\leq j\leq m\}$ be an $m$-stage flag Bott tower of general Lie type associated to $(K_{\bullet},Z_{\bullet})$, where $K_{j}$ are simply-connected for all $1\leq j\leq m$. Then there exists a family of homomorphisms $\{{\varphi^{{(\ell)}}_{{j}}}\colon Z_{j}\to T_{\ell}\mid 1\leq j<\ell\leq m\}$ such that $F_{\bullet}$ and $F_{\bullet}^{\varphi}$ are isomorphic as flag Bott towers.
###### Proof.
We show by induction. Assume that a flag bundle $F$ over $F_{m-1}^{\varphi}$ is associated to the classifying map
$f_{m}\colon F_{m-1}^{\varphi}\to BK_{m}$
with fiber $K_{m}/Z_{m}$, and also assume that $f_{m}$ factors through $BT_{m}$, i.e. there exist $\overline{f_{m}}\colon F_{m-1}^{\varphi}\to BT_{m}$ such that the following diagram commutes.
${F_{m-1}^{\varphi}}$${BK_{m}}$${BT_{m}}$$\scriptstyle{f_{m}}$$\scriptstyle{\overline{f_{m}}}$$\scriptstyle{B\iota}$
From the construction of $F_{m-1}^{\varphi}$, we have a principal $\prod_{j=1}^{m-1}Z_{j}$ bundle. Denote its classifying map by $u$ so that we have the pull-back of the universal $\prod_{j=1}^{m-1}Z_{j}$ bundle as follows:
${\prod_{j=1}^{m-1}Z_{j}}$${\prod_{j=1}^{m-1}Z_{j}}$${\prod_{j=1}^{m-1}K_{j}}$${E}$${F_{m-1}^{\varphi}}$${B(\prod_{j=1}^{m-1}Z_{j}),}$$\scriptstyle{u}$
where $E=E(\prod_{j=1}^{m-1}K_{j})$ can also serve as the universal space for any subgroup of $\prod_{j=1}^{m}K_{j}$.
Since $E\to B(\prod_{j=1}^{m-1}Z_{j})$ is a fibration, the bottom pull-back square is a homotopy pull-back at the same time. Thus, the bottom square can be understood as restricting $u$ on a contractible space $E$ so that the homotopy fibre of $u$ is $\prod_{j=1}^{m-1}K_{j}$. That is, we have the following homotopy fibration
(3.2) $\prod_{j=1}^{m-1}K_{j}~{}\longrightarrow~{}F_{m-1}^{\varphi}\stackrel{{\scriptstyle u}}{{\longrightarrow}}B\left(\prod_{j=1}^{m-1}Z_{j}\right).$
Note that each $K_{j}$ is compact and simply-connected, so that homotopy groups $\pi_{1}(K_{j})$ and $\pi_{2}(K_{j})$ are trivial for $1\leq j\leq m$ (see [5, Proposition 7.5 in Chapter V]). Hence $H^{1}(K_{j})=H^{2}(K_{j})=0$ by the Hurewicz isomorphism theorem. Since $Z_{j}$ is connected, $BZ_{j}$ is simply-connected for all $1\leq j\leq m$. Recall that $F_{m-1}^{\varphi}$ admits the structure of a CW complex with even dimensional cells only. Combining these facts with the Serre spectral sequence with respect to the fibration (3.2) we see that
$u^{*}\colon H^{2}\left(B\left(\prod_{j=1}^{m-1}Z_{j}\right);\mathbb{Z}\right)\to H^{2}(F_{m-1}^{\varphi};\mathbb{Z})$
is an isomorphism.
Using the identification $[X,BS^{1}]\cong H^{2}(X;\mathbb{Z})$ for a topological space $X$ which has the homotopy type of a CW complex (see [18, Proposition 3.10]), we see that $\overline{f_{m}}$ factors up to homotopy as follows:
(3.3) ${F_{m-1}^{\varphi}}$${BT_{m}.}$${B(\prod_{j=1}^{m-1}Z_{j})}$$\scriptstyle{\overline{f_{m}}}$$\scriptstyle{u}$$\scriptstyle{\psi}$
Furthermore, we obtain $\varphi\in\text{Hom}\left(\prod_{j=1}^{m-1}Z_{j},T_{m}\right)$ such that $B\varphi=\psi$ through the following bijection:
$\begin{split}\textup{Hom}(Z_{j},S^{1})&\cong\textup{Hom}(\pi_{1}(Z_{j}),\mathbb{Z})\quad\text{(see, for example,~{}\cite[cite]{[\@@bibref{}{BwyerWilkerson_compact_Lie_group98}{}{}, Proposition 9.4]})}\\ &\cong\textup{Hom}(\pi_{2}(BZ_{j}),\mathbb{Z})\\ &\cong\textup{Hom}(H_{2}(BZ_{j});\mathbb{Z})\quad(\text{by the Hurewicz isomorphism theorem})\\ &\cong H^{2}(BZ_{j};\mathbb{Z})\quad(\text{by the universal coefficient theorem})\end{split}$
Consider the isomorphism
(3.4) $\text{Hom}\left(\prod_{j=1}^{m-1}Z_{j},T_{m}\right)\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\left(\prod_{j=1}^{m-1}\text{Hom}(Z_{j},T_{m})\right),$
and denote the image of $\varphi$ under the map (3.4) by
${\varphi^{{(m)}}_{{j}}}\colon Z_{j}\to T_{m},\quad\text{ for }1\leq j\leq m-1.$
By Definition 3.5, this sequence $({\varphi^{{(m)}}_{{j}}})_{1\leq j\leq m-1}$ defines a bundle $F_{m}^{\varphi}$ over $F_{m-1}^{\varphi}$, which we will show is isomorphic to $F$. In fact, $F$ has the classifying map
$f_{m}\colon F_{m-1}^{\varphi}\stackrel{{\scriptstyle\overline{f_{m}}}}{{\longrightarrow}}BT_{m}\stackrel{{\scriptstyle B\iota}}{{\longrightarrow}}BK_{m},$
and $F_{m}^{\varphi}$ has the classifying map
${F_{m-1}^{\varphi}}$${B\left(\prod_{j=1}^{m-1}Z_{j}\right)}$${BT_{m}}$${BK_{m}.}$$\scriptstyle{u}$$\scriptstyle{B\varphi}$$\scriptstyle{B\iota}$
Since $\overline{f_{m}}$ is homotopic to $B\varphi\circ u$ by (3.3), they define the same bundle (see [12, Theorem 2.1]). ∎
###### Remark 3.7.
Assume that all $Z_{j}$’s are maximal tori. If we fix the isomorphisms $Z_{j}\simeq(S^{1})^{n_{j}+1}$, then ${\varphi^{{(\ell)}}_{{j}}}:Z_{j}\to T_{\ell}$ can be represented by $(n_{\ell}+1)\times(n_{j}+1)$ | 8,341 | 23,086 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 359, "math_alttext": 1, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.796781 |
http://www.oldquestionpapers.net/2011/05/rites-sample-model-question-paper-answers-for-graduate-executive-trainees.html | 1,685,873,093,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649741.26/warc/CC-MAIN-20230604093242-20230604123242-00787.warc.gz | 82,050,237 | 18,934 | # RITES Sample Model Question Paper Answers for Graduate Executive trainees 2022
## RITES Sample Model Question Paper Answers for Graduate Executive trainees 2022
1. A certain type of mixture is prepared by mixing brand A at Rs.9 a kg. with brand B at Rs.4 a kg. If the mixture is worth Rs.7 a kg., how many kgs. of brand A are needed to make 40kgs. of the mixture?
Ans. Brand A needed is 24kgs.
2. A wizard named Nepo says “I am only three times my son’s age. My father is 40 years more than twice my age. Together the three of us are a mere 1240 years old.” How old is Nepo?
Ans. 360 years old.
3. One dog tells the other that there are two dogs in front of me. The other one also shouts that he too had two behind him. How many are they?
Ans. Three.
4. A man ate 100 bananas in five days, each day eating 6 more than the previous day. How many bananas did he eat on the first day?
Ans. Eight.
5. If it takes five minutes to boil one egg, how long will it take to boil four eggs?
Ans. Five minutes.
6. Three containers A, B and C have volumes a, b, and c respectively; and container A is full of water while the other two are empty. If from container A water is poured into container B which becomes 1/3 full, and into container C which becomes 1/2 full, how much water is left in container A?
7. ABCE is an isosceles trapezoid and ACDE is a rectangle. AB = 10 and EC = 20. What is the length of AE?
Ans. AE = 10.
8. In the given figure, PA and PB are tangents to the circle at A and B respectively and the chord BC is parallel to tangent PA. If AC = 6 cm, and length of the tangent AP is 9 cm, then what is the length of the chord BC?
Ans. BC = 4 cm.
9 Three cards are drawn at random from an ordinary pack of cards. Find the probability that they will consist of a king, a queen and an ace.
Ans. 64/2210.
10. A number of cats got together and decided to kill between them 999919 mice. Every cat killed an equal number of mice. Each cat killed more mice than there were cats. How many cats do you think there were ?
Ans. 991.
1. The eleventh ‘five year plan’ of India is for the five years ranging from—
(A) 2009–2014
(B) 2008–2013
(C) 2007–2012
(D) 2006–2011
(E) None of these
Ans : (C)
2. DSCR indicates the ability of a company to—
(A) Meet its current liabilities
(B) Service its shareholders
(C) Meet its long term debt obligations
(D) Raise further capital
(E) None of these
Ans : (C)
3. An IPO is—
(A) Initial price offered by a private limited company to its shareholders
(B) An offer by an unlisted company for sale of its shares for the first time to the public
(C) Used to increase the share capital of an unlisted company
(D) A book building process
(E) None of these
Ans : (B)
4. A decreasing current ratio indicates—
(A) A stable liquidity
(B) An increasing liquidity
(C) A strained liquidity
(D) Satisfactory current solvency
(E) None of these
Ans : (E)
5. Private equity investors, invest in a company based mainly on—
(A) The age of the company
(B) The location of the company
(C) The activity undertaken by the company
(D) The creditability and the valuation of the company
(E) The existing profitability of the company
Ans : (E)
6. The synagogue is generally—
(A) An epic
(B) A religion
(C) Jewish House of Prayer
(D) The Parsi temple
(E) A palace
Ans : (C)
7. Deuce is a team used in—
(a) Polo
(b) Lawn tennis
(d) Boxing
(e) Volley ball
(A) a, b, c
(B) b, c, d
(C) c, d, a
(D) a, b, e
(E) b, c, e
Ans : (E)
8. ISI is the intelligence agency of—
(A) United Kingdom
(B) United States of America
(C) Israel
(D) Pakistan
(E) India
Ans : (D)
9. Air Traffic Controller’s job involves—
1. Control of air traffic in and within vicinity of airport
2. Control of movement of air traffic between altitude, sectors and control centres
3. Following established procedures and policies for air traffic control
4. Control of commercial airline flights according to government regulations
5. To authorize and regulate commercial airline flights
(A) 1, 2
(B) 2, 3
(C) 1, 2, 3, 4
(D) 1, 2, 3, 4, 5
(E) 3, 4, 5
Ans : (A)
10. Animal assisted therapy is employed in—
(A) Operations
(B) ICU
(C) Old age homes
(D) Training animals
(E) Natural healing
Ans : (D)
Directions—(Q. 1–5) The President of a club is appointing nine officials A, B, C, D, E, F, G, H and I to serve on three committees to study three different aspects of activities of the club. There will be a Games Committee, a Food Service Committee and an Entertainment Committee. The appointments must respect the following—
Each committee must have exactly three members.
No person can serve on more than one committee.
H must serve on the Entertainment Committee.
C and D must serve on the same committee.
A and B cannot serve on the same committee.
E cannot serve on the same committee as I.
F must serve on the same committee as B or H or both B and H.
1. If B and G serve on the Games Committee, which of the following must serve on the Food Service Committee ?
(A) A
(B) D
(C) E
(D) F
Ans : (B)
2. Which of the following groups could constitute the Games Committee ?
(A) A, B, E
(B) A, D, G
(C) C, H, E
(D) F, I, B
Ans : (D)
3. If A is assigned to the Food Service Committee and C is appointed to Entertainment Committee, then which of the following must be true ?
1. G is appointed to the Food Service Committee.
2. E is appointed to the Games Committee.
3. I is appointed to the Entertainment Committee.
(A) I only
(B) III only
(C) I and III only
(D) II and III only
Ans : (A)
4. If F serves on the Food Service Committee and C serves on the same committee as H, then which of the following must serve on the Games Committee ?
1. A 2. G 3. I
(A) I only
(B) III only
(C) I and II only
(D) II and III only
Ans : (C)
5. If I is on the Entertainment Committee and B is on the Food Service Committee, then which of the following must be true ?
(A) F is on the Entertainment Committee
(B) C is on the Games Committee
(C) G is on the Food Service Committee
(D) F is on the Games Committee
Ans : (A)
Directions—(Q. 6–10) Letters of certain words have been rearranged and the jumbled spellings have been given below. Out of the choices given below each question, choose the last letter of the correct word—
6. AYDOT
(A) A
(B) D
(C) T
(D) Y
Ans : (D)
7. ILCEOP
(A) C
(B) E
(C) L
(D) O
Ans : (B)
8. CAPCET
(A) A
(B) C
(C) P
(D) T
Ans : (D)
9. VISERL
(A) E
(B) L
(C) R
(D) S
Ans : (C)
10. ERVSECI
(A) E
(B) I
(C) R
(D) S
Ans : (A)
1. What should come in the place of (?) in the given series?
ACE, FGH, ?, PON
(A) KKK
(B) JKI
(C) HJH
(D) IKL
Ans. (A)
2. Typist : Typewriter : : Writer: ?
(A) Script
(B) Pen
(C) Paper
(D) Book
Ans. (B)
3. Paint: Artist : : Wood: ?
(A) Furniture
(B) Forest
(C) Fire
(D) Carpenter
Ans. (D)
4. acme : mace :: alga: ?
(A) glaa
(B) gaal
(C) laga
(D) gala
Ans. (D)
5. EIGHTY : GIEYTH : : OUTPUT:?
(A) UTOPTU
(B) UOTUPT
(C) TUOUTP
(D) TUOTUP
Ans. (D)
6. ‘Medicine’ is related to ‘Patient’ in the same way as ‘Education’ is related to—
(A) Teacher
(B) School
(C) Student
(D) Tuition
Ans. (C)
7. Fill in the missing letter in the following series—
S, V, Y, B, ?
(A) C
(B) D
(C) E
(D)G
Ans. (C)
8. What should come in the place of question mark in the following series?
3, 8, 6, 14, ?, 20
(A) 11
(B) 10
(C) 8
(D) 9
Ans. (D)
9. Select the correct option in place of the question mark.
AOP, CQR, EST, GUV, ?
(A) IYZ
(B) HWX
(C) IWX
(D) JWX
Ans. (C)
10. What should come in the place of question mark in the following series?
1, 4, 9, 25, 36, ?
(A) 48
(C) 52
(B) 49
(D) 56
Ans. (B)
Similar Pages.. | 2,293 | 7,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-23 | longest | en | 0.965965 |
https://www.exactlywhatistime.com/days-before-date/november-20/25-days | 1,709,234,472,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474852.83/warc/CC-MAIN-20240229170737-20240229200737-00214.warc.gz | 749,609,603 | 6,261 | # What is the date 25 days from November 20?
## Calculating 25 days before Wednesday November 20, 2024 by hand
This page helps you figure out the date that is 25 days before Wednesday November 20, 2024. We've made a calculator to find the date before a certain number of days before a specific date. If you want to change the question on this page, you have two choices: you can change the URL in your browser's address bar, or go to our days before specific date calculatorto type in a new question or days from a specific date if you want to add 25 days.
But for all you time sickos out there who want to calculate 25 days before Wednesday November 20, 2024 - here's how you do it:
1. Start with the Input Date (Wednesday November 20, 2024): Write it down! I can't stress this enough
2. Count in Weeks: Recognize that 25 days is approximately 3.5714285714285716 weeks. Count forward 3.5714285714285716 weeks (5.0 work weeks) from the input date. This takes you to .
3. Add Remaining Days: Since you've counted 3.5714285714285716 weeks, you only need to add the remaining days to reach Saturday October 26, 2024
4. Use Mental Math: If Wednesday November 20, 2024 is a Saturday, then compare that to if 25 is divisible by 7. That way, you can double-check if October 26 matches that Saturday.
## Saturday October 26, 2024 Stats
• Day of the week: Saturday
• Month: October
• Day of the year: 300
## Counting 25 days backward from Wednesday November 20, 2024
Counting backward from today, Saturday October 26, 2024 is 25 before now using our current calendar. 25 days is equivalent to:
25 days is also 600 hours. Saturday October 26, 2024 is 82% of the year completed.
## Within 25 days there are 600 hours, 36000 minutes, or 2160000 seconds
Saturday Saturday October 26, 2024 is day number 300 of the year. At that time, we will be 82% through 2024.
## In 25 days, the Average Person Spent...
• 5370.0 hours Sleeping
• 714.0 hours Eating and drinking
• 1170.0 hours Household activities
• 348.0 hours Housework
• 384.0 hours Food preparation and cleanup
• 120.0 hours Lawn and garden care
• 2100.0 hours Working and work-related activities
• 1932.0 hours Working
• 3162.0 hours Leisure and sports
• 1716.0 hours Watching television
## Famous Sporting and Music Events on October 26
• 1984 "The Terminator" directed by James Cameron, starring Arnold Schwarzenegger and Linda Hamilton is released in the US
• 1863 Football Association forms in England, standardizing soccer, splitting with rugby | 670 | 2,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-10 | latest | en | 0.929936 |
https://developpaper.com/rust-programming-video-tutorial-advanced-015_-1-reference-cycle/ | 1,642,326,040,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299852.23/warc/CC-MAIN-20220116093137-20220116123137-00187.warc.gz | 287,375,057 | 12,215 | Time：2021-10-13
# Video address
Headline address:https://www.ixigua.com/i677586170644791348…
Station B address:https://www.bilibili.com/video/av81202308/
# Source address
GitHub address:https://github.com/anonymousGiga/learn_rus…
# Explanation content
Reference loop to make two lists point to each other. The example code is as follows:
``````use std::rc::Rc;
use std::cell::RefCell;
use crate::List::{Cons, Nil};
#[derive(Debug)]
enum List {
Cons(i32, RefCell<Rc<List>>),
Nil,
}
impl List {
fn tail(&self) -> Option<&RefCell<Rc<List>>> {
match self {
Cons(_, item) => Some(item),
Nil => None,
}
}
}
fn main() {
let a = Rc::new(Cons(5, RefCell::new(Rc::new(Nil))));
println!("a initial rc count = {}", Rc::strong_count(&a));
println!("a next item = {:?}", a.tail());
let b = Rc::new(Cons(10, RefCell::new(Rc::clone(&a))));
println!("a rc count after b creation = {}", Rc::strong_count(&a));
println!("b initial rc count = {}", Rc::strong_count(&b));
println!("b next item = {:?}", b.tail());
if let Some(link) = a.tail() {
*link.borrow_ mut() = Rc::clone(&b); // Change a to point to B
}
println! ("b rc count after changing a = {}", Rc::strong_count(&b)); // Output reference count, 2
println! ("a rc count after changing a = {}", Rc::strong_count(&a)); // Output reference count, 2
//The following call will make an error, because the circular reference has been made above, and the compiler cannot find the tail
// println!("a next item = {:?}", a.tail());
}``````
explain:
(1) At the end of the main function, rust will try to discard B, which will reduce the reference count of RC instances in a and B to 1. However, because a still references RC in B, the count of RC is 1 instead of 0, so the memory of RC on the heap will not be discarded, resulting in memory leakage.
(2) In the above example, the mutual reference relationship is as shown in the above figure. If you print a.tail (), you will always find it in the above loop (in the case of (, item), but it will never end (because it is a loop, you will never find a match), which will eventually cause stack overflow.
This work adoptsCC agreement, reprint must indicate the author and the link to this article
Linghu rushed
## Glue of R string
preface Today we want to introducetidyversePackages that format strings in:glue glueProvides lightweight, fast, and dependency free interpretable strings,glueBy willRThe expression is embedded in curly braces, then evaluated and inserted into the string. install install.packages(“glue”) # or install.packages(“glue”) # install.packages(“devtools”) devtools::install_github(“tidyverse/glue”) use 1. Import library(glue) 2. Simple use Pass the variable directly into the string > […] | 715 | 2,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-05 | latest | en | 0.672586 |
http://www.abebooks.co.uk/9780070380073/Schaums-Outline-Theory-Problems-Linear-0070380074/plp | 1,477,041,646,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718034.77/warc/CC-MAIN-20161020183838-00483-ip-10-171-6-4.ec2.internal.warc.gz | 290,484,198 | 14,854 | Lipschutz, Seymour Schaum's Outline of Theory and Problems of Linear Algebra
# Schaum's Outline of Theory and Problems of Linear Algebra
## Lipschutz, Seymour
3.75 avg rating
( 141 ratings by GoodReads )
An outline of linear algebra, covering topics such as orthogonality, determinants, eigenvalues and eigenvectors, systems of linear equations, vectors in Rn and Cn, spatial vectors, matrices, product spaces, and systems of linear equations.
"synopsis" may belong to another edition of this title.
From the Back Cover:
Master linear algebra with Schaum’s—the high-performance study guide. It will help you cut study time, hone problem-solving skills, and achieve your personal best on exams and projects!
Students love Schaum’s Outlines because they produce results. Each year, hundreds of thousands of students improve their test scores and final grades with these indispensable study guides.
Get the edge on your classmates. Use Schaum’s!
If you don't have a lot of time but want to excel in class, this book helps you:
* Use detailed examples to solve problems * Brush up before tests * Find answers fast * Study quickly and more effectively * Get the big picture without poring over lengthy textbooks
Schaum’s Outlines give you the information your teachers expect you to know in a handy and succinct format—without overwhelming you with unnecessary jargon. You get a complete overview of the subject. Plus, you get plenty of practice exercises to test your skill. Compatible with any classroom text, Schaum’s let you study at your own pace and remind you of all the important facts you need to remember—fast! And Schaum’s are so complete, they’re perfect for preparing for graduate or professional exams.
Inside, you will find: * A bridge between computational calculus and formal mathematics * Clear explanations of eigenvalues, eigenvectors, linear transformations, linear equations, vectors, and matrices * Solved problems that relate to the field you are studying * Easy-to-understand information, perfect for pre-test review
If you want top grades and a thorough understanding of linear algebra, this powerful study tool is the best tutor you can have!
Chapters include: Vectors in Rn and Cn * Matrix Algebra * Linear Equations * Vector Spaces * Linear Mappings * Linear Mapings and Matrices * Inner Product Spaces, Orthogonality * Determinants * Diagonalization: Eigenvalues and Eigenvectors * Canonical Forms * Linear Functionals and the Dual Space * Bilinear, Quadratic, and Hermitian Forms * Linear Operators on Inner Product Spaces * Polynomials
About the Author:
Seymour Lipschutz, Ph.D. (Philadelphia, PA), is presently on the Mathematics faculty at Temple Univeristy. He has written more than 15 Schaum's Outlines. Marc Lipson, Ph.D. (Philadelphia, PA), is on the mathematical faculty of the University of Georgia. He is co-author of Schaum's Outline of Discrete Mathematics.
"About this title" may belong to another edition of this title.
## 1.Schaum's Outline of Theory and Problems of Linear Algebra (Schaum's Outlines)
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Destination, Rates & Speeds | 1,367 | 5,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2016-44 | latest | en | 0.871451 |
https://goprep.co/ex-5.1-q14-in-the-adjoining-figure-name-the-following-pairs-i-1njsx6 | 1,623,887,658,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626122.27/warc/CC-MAIN-20210616220531-20210617010531-00475.warc.gz | 267,068,652 | 24,900 | # <span lang="EN-US
(i) From the given figure,
Obtuse vertically opposite angles are:
AOD, BOC
(ii) From the given figure,
EOA, AOB
(iii) From the given figure,
Equal supplementary angles are:
EOB, EOD
(iv) From the given figure,
Unequal supplementary angles are:
EOA, EOC
(v) From the given figure,
Adjacent angles that do not form a linear pair are:
AOB and AOE
AOE and EOD
Also,
EOD and COD
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The sum of two veNCERT - Exemplar Mathematics | 270 | 959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-25 | latest | en | 0.661152 |
https://homework.cpm.org/cpm-homework/homework/category/CC/textbook/CCA/chapter/Ch3/lesson/3.3.3/problem/3-111 | 1,568,587,338,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572436.52/warc/CC-MAIN-20190915215643-20190916001643-00262.warc.gz | 512,264,880 | 16,373 | ### Home > CCA > Chapter Ch3 > Lesson 3.3.3 > Problem3-111
3-111.
Copy and complete these generic rectangles on your paper. Then write the area of each rectangle as a product of the length and width and as a sum of the parts. Homework Help ✎
1. 6 $13x$ $-21$
• To write the area as a sum:
Find the area of each of the rectangle's sections. Then, add them together.
• To write the area as a product:
Multiply the total length by the total width, but don't simplify it.
• As a sum: 78x − 126
As a product: 6(13x − 21)
1. $x$ $-5$ $x$ $+3$
• See the help for part (a).
• As a sum: x2 − 2x − 15
As a product: (x − 5)(x + 3)
1. $16x^2$ $-24x$ $4$
• Find the greatest common factor of the areas. That will be the width. Then divide each area by the width to find the length. Then follow the steps for part (a).
1. $3x$ $-2$ $x$ $+4$
• See the help for part (b). | 291 | 870 | {"found_math": true, "script_math_tex": 13, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-39 | latest | en | 0.800586 |
https://codegolf.stackexchange.com/questions/63050/find-the-results-of-the-poll-with-the-borda-count-method | 1,656,369,572,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103341778.23/warc/CC-MAIN-20220627195131-20220627225131-00621.warc.gz | 219,950,076 | 68,183 | # Find the results of the poll with the Borda count method
In a preference ballot there are many ways to decide who won. One of which is the Borda count method, where we give everyone point based on their places on the ballot.
For example, in these results:
10 5 7 2
_____________
1 a d a c
2 b a d b
3 c c b a
4 d b c d
The results are
1: a : 87
2: d : 60
3: b : 55
4: c : 45
Because
4*10 is 40
5*3 is 15
7*4 is 28
2*2 is 4
40 + 15 + 28 + 4 is 87
Or is words: We give first place 4 points , second 3, third 2, etc.
So a is in first place 17 times, so they get 68 points then they get 15 points for being in second 5 times and 4 points for being in third 2 times for a total of 87 points.
Because this process is a pain, I need you write a program that finds the results for me, but since I have limited space on my PC, your program will have to be as small as possible.
Write a program or function the takes an input in this format
Number of Ballots : placings, Number of Ballots : placings, Etc...
and will output the results in this format:
1: a: 90
2: b: 80
Etc...
Rules:
This is Code Golf, so the shortest answer that does the task wins
Standard Loopholes Apply
Example:
Input:
3: A B C, 2: C B A, 1: C A B
Output:
1: A: 13
2: C: 12
3: B: 11
This is my first challenge, so please give me all the criticism you so desire, I won't take it offensively.
• Could you explain the table format? It's not obvious where e.g. 4*10 comes from. Nov 5, 2015 at 22:40
• Are the candidates always A, B, C...? How do we handle ties?
– xnor
Nov 5, 2015 at 22:57
• The candidate will always be single letters, but they will not always be A B and C. As for ties just assume there will not be any, just to keep thing simple. Nov 5, 2015 at 23:21
# Python 2, 248 bytes
x={}
for a,b in[y.split(':')for y in raw_input().split(', ')]:
for d,e in[(c,(len(b)/2-i)*int(a)) for i,c in enumerate(b[1::2])]:x[d]=x.setdefault(d,0)+e
for i,(f,g) in enumerate(sorted(x.items(),key=lambda t:-t[1]),1):print '%d: %s : %d'%(i,f,g)
Reads the from the standard input (for example, echo '3: A B C, 2: C B A, 1: C A B' | python borda.py).
# CJam, 78 bytes
q", "%{S%}%_0=,1-:L;{(1<i*}%e_W%L/L{)}%:C;{C.*}%e_$e$W%C.+{W%{": "}%e_W(<N+}%
Try it Online.
Holy crap! How did I do this?
When I saw that CJam had a run length encode/decode function, I wondered why someone would use it. Today I found out. Essentially, what this program does is create a string containing each letter once for each vote. Then it sorts the string and uses run length encode to find out how many votes each letter got, and then formats the data.
• Well, I guess you won, Congratulations! Nov 25, 2015 at 1:56
# Perl 5, 131 bytes
for(split/, /,$ARGV[0]){($n,@v)=split/:? /;$s{$v[@v-$_]}+=$n*$_ for 1..@v}$,=": ";map{say++$i,"$_ ",$s{$_}}sort{$s{$b}<=>$s{$a}}@v
Use as perl -E'the string above' 'the input string'` | 988 | 2,913 | {"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} | 3.078125 | 3 | CC-MAIN-2022-27 | longest | en | 0.842062 |
http://www.cfd-online.com/Forums/openfoam-solving/170097-natural-convection-melting-furnace.html | 1,480,728,860,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540798.71/warc/CC-MAIN-20161202170900-00447-ip-10-31-129-80.ec2.internal.warc.gz | 380,673,817 | 18,208 | natural convection in a melting furnace
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April 24, 2016, 06:55 natural convection in a melting furnace #1 New Member donny Join Date: Feb 2016 Posts: 3 Rep Power: 2 hey guys, im new in openfoam. i have to simulate two gases in a melting furnance. i know i have to use the buoyantBoussinesqSimpleFoam, but i dont know how to integrate two different gases in the hotRoom tutorial. the gases are not reacting. The temperature difference accure natural convection. The Temperature of the floor is 1023 K an the ceiling 623 K. Please i need your help. regards Donny
April 25, 2016, 03:09 #2 Senior Member Tobias Holzmann Join Date: Oct 2010 Location: Leoben (Austria) Posts: 1,548 Blog Entries: 6 Rep Power: 27 Dear donny, first hint, buossinesq approximation is in my opinion the wrong choice. You should take the buoyantSimpleFoam but to clear my mind just one question, why you think boussinesq is the right approach? __________________ Best regards, Tobias Holzmann Some interesting OpenFOAM tutorials, publications and videos on www.Holzmann-cfd.de OpenFOAM Beginners should check out the new wiki on wiki.openfoam.com A list of some active OpenFOAM contributers can be found »here« A book about the basics of »Mathematics, Numerics, Derivations and OpenFOAM« can be found on www.Holzmann-cfd.de
April 25, 2016, 06:02 #3 New Member donny Join Date: Feb 2016 Posts: 3 Rep Power: 2 Hey Tobi, thank you for the answer. i read in another forum that this solver is the right choice for natural convection. Therefore i tried it just for one gas and it worked, but i do not know how to do it with two gases. By the way, you have a privat message from me. best regards Donny
April 25, 2016, 14:38 #4 Senior Member Tobias Holzmann Join Date: Oct 2010 Location: Leoben (Austria) Posts: 1,548 Blog Entries: 6 Rep Power: 27 Hey guys, I was writing via messages with donny and now I will summarize. buoyantBoussinesq...Foam are using the Boussinesq-Approximation , you only should use them if the density change in your fluid is not high and the temperature range (dT) is low. Check out some references... e.g. Ferziger & Perić Donny want to implement a second gas, therefore the very simple approach would be to implement a passive scalar "S" that is limited between 0 and 1. (like alpha in VOF). This scalar represent the two gases (S = 0 -> gas1; S = 1 -> gas2; 0 < S < 1 -> mixture of gas1 and gas2) After that, you have to implement all thermodynamic data for gas1 and gas2 into the solver. Therefore checkout the createField.H and create new fields for rho1, rho2, nu1, nu2 (if incompressible assumption) or mu1, mu2 etc. The next step is to create the actual thermodynamic variables based on gas1 and gas2 and the passive scalar S: Code: `rho = rho1 * S + rho2 * (1-S);` You have to choose if you use openFOAM thermodynamics (then you normally should have two) or you implement your own polynoms (I think you also could use icoPoly.... for that). That means you have either to create a second thermodynamic object or you decouple all thermodynamic calculation with your polynoms. This model would be nice to handle and very easy. Another idea at the moment would be, to use some combustion solver and unset the reaction (kinetics). Once I did something like that with some old collegue who was exactly doing the same. The result between the combustion solver and the solver mentioned above was very close and a lot of faster. The easiest approach would be: implement only one passive scalar and use one fluid inside. Doing this leads to a first guess but hence we only use thermodynamics out of one gas, you will be not as accurate as in the method I mentioned above. If the thermodynamic data are very similar (cp, rho, nu, mu ...) then the passive scalar should work fine. This would be my suggestions. donny1991 likes this. __________________ Best regards, Tobias Holzmann Some interesting OpenFOAM tutorials, publications and videos on www.Holzmann-cfd.de OpenFOAM Beginners should check out the new wiki on wiki.openfoam.com A list of some active OpenFOAM contributers can be found »here« A book about the basics of »Mathematics, Numerics, Derivations and OpenFOAM« can be found on www.Holzmann-cfd.de Last edited by Tobi; April 26, 2016 at 03:06.
April 26, 2016, 07:00 #5 New Member donny Join Date: Feb 2016 Posts: 3 Rep Power: 2 Hey Tobi, thank you for this helpful answer. Im now using the reactionFoam solver without reaction and only with O2 and N2. I changed all BC's and the properties for my case. I tried to run the simulation but i got this. Reading g Creating reaction model Selecting combustion model PaSR Selecting chemistry type { chemistrySolver ode; chemistryThermo psi; } Selecting thermodynamics package { type hePsiThermo; mixture reactingMixture; transport sutherland; thermo janaf; energy sensibleEnthalpy; equationOfState perfectGas; specie specie; } Selecting chemistryReader foamChemistryReader chemistryModel: Number of species = 2 and reactions = 1 Selecting ODE solver KRR4 Reading field U Reading/calculating face flux field phi Creating turbulence model. Selecting turbulence model type RASModel Selecting RAS turbulence model kEpsilon Floating point exception: 8 i dont know how to fix this problem. Under the following link you can download my case. thank you!! meltingfurnance.zip
April 26, 2016, 08:01 #6 Senior Member Tobias Holzmann Join Date: Oct 2010 Location: Leoben (Austria) Posts: 1,548 Blog Entries: 6 Rep Power: 27 Hi, maybe I will check it but as I told above, the bad thing using a reaction solver is, that it is soooo slow. Finally I am not sure if you have buoyancy included. So you have to check it yourself. My suggestion to you, build your own solver. Its so much faster, clean (you know whats going on) and really not a big deal to implement sth. like that into OpenFOAM. __________________ Best regards, Tobias Holzmann Some interesting OpenFOAM tutorials, publications and videos on www.Holzmann-cfd.de OpenFOAM Beginners should check out the new wiki on wiki.openfoam.com A list of some active OpenFOAM contributers can be found »here« A book about the basics of »Mathematics, Numerics, Derivations and OpenFOAM« can be found on www.Holzmann-cfd.de
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All times are GMT -4. The time now is 21:34. | 1,785 | 6,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | latest | en | 0.871307 |
https://www.glassthemovie.com/casino-on-the-internet-betting-system-positive-progression-technique/ | 1,660,676,158,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572515.15/warc/CC-MAIN-20220816181215-20220816211215-00711.warc.gz | 696,370,586 | 7,238 | # Casino On the internet Betting System – Positive Progression Technique
If you speak about the casino on the web betting technique, you will obtain there are several folks who will discourage you. They will say that betting on the web genuinely is not a very good resource to make funds. But I will say that it is pretty easy to earn from on the internet casino games, if you know the casino on line betting tactics. Truly money management knowledge is what most of the gamblers lack. Hence some are currently bankrupt whereas some are enjoying an affluent life. Did anybody of you hear about “Good Progression System”, this is 1 of the pretty effectively identified casino online betting strategy.
You can say this is a logic that tells you the possibilities of winning 4 occasions in a single row. At the starting or just at the preliminary stage the bet is of 1 unit, the second bet is of three units, the third bet is of two units and the fourth bet is of six units. For that reason it is also referred to as the 1-3-2-6 method.
I will illustrate this casino on-line betting technique in detail, to give you a clear understanding. For instance you spot your first bet of \$10. The second bet is supposed to be \$30 – when you win the initial bet, your \$ten gets added up with the \$20 already placed on the table. The total comes to \$30.
So the second bet you location would be of \$30. The grand total just before you play the third bet will be of \$60 total (the \$30 bet placed by you in the second bet combined with each other with the second bet winning already placed on the table). From the \$60 you take away \$40 and the third bet is of \$20.
Your third bet will be of \$20 and right after winning the third bet you will win \$40. Now, for the fourth bet you will add \$20 much more to the total \$40 to make it a \$60 bet for the forth bet you spot.
Winning the fourth bet you will be left with \$120. This is the net profit you make from this casino on the web betting method. To continue the game you will again location a bet of \$ten and comply with the “Optimistic Progression Program” as soon as again. Soon after ending the forth bet, you get started over again. Moreover, each and every time you loose a bet, commence once again with initial \$ten bet.
Calculate your loss at every single level. If แทงบอลufabetเว็บตรง shed in the course of the:
– initial or 1st bet, then your total loss is of \$ten.
– second bet, then your total loss will be of \$20 (this simply because you have added \$10)
– third bet, then your accumulated loss will also give you a total profit of \$20 (this is simply because you have already taken away \$40).
– fourth bet, then your loss and profit is equal (just for the reason that you have taken away \$40 and added \$20).
The positive side of this casino online betting method is you just stake \$20 with a possibility of producing a profit that is 6 instances more than the stake. In other words you are producing \$120 by just risking the amount \$20. | 680 | 3,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.95107 |
http://mathandmultimedia.com/tag/avogadros-number/ | 1,695,983,238,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510501.83/warc/CC-MAIN-20230929090526-20230929120526-00751.warc.gz | 27,295,338 | 10,601 | ## The Uses of Large Numbers
Large numbers are used mostly used in astronomy and cosmology. According to the Big Bang Model, the universe is estimated to be $4.3 \times 10^{17}$ seconds old, the observable universe is $8.8 \times 10^{26}$ meters, and it contains $5 \times 10^{22}$ stars.
In computing, if a hacker attempts to open a computer with up to a 40-character password, we should really wish him luck. Even though he may create a program that would display 1 billion strings per second indefinitely, it would take him $2 \times 10^{87}$ seconds or about $6 \times 10^{79}$ years to complete all the combinations. That is way, way older than the universe which is only about 13.7 billion years old. » Read more | 185 | 723 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-40 | longest | en | 0.899017 |
https://essaysmind.com/the-demand-curve-for-spark-plugs-in-the-united-states-is-given-by-q-100-p-where-q-indicates-the/ | 1,656,570,473,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103669266.42/warc/CC-MAIN-20220630062154-20220630092154-00340.warc.gz | 272,539,440 | 12,333 | # The demand curve for spark plugs in the United States is given by Q = 100 – P, where Q indicates the
The demand curve for spark plugs in the United States is given by Q = 100 – P, where Q indicates the number of spark plugs purchased and P is the price. Suppose that there are no spark plugs produced in the United States,but they can be imported either from Mexico or from the rest of the world. The price of spark plugs in Mexico is \$20, and the price from the lowest-cost supplier in the rest of the world is \$10. In each case, spark plugs are produced with a horizontal supply curve, so these prices are fixed and will not change with changes in U.S. policy. The U.S. Questions and Problems of NAFTA, these mechanisms were set up in a way that appears to have endangered the ability of signatory governments to pass environmental regulation, at least initially. MFN tariff on spark plugs is a specific tariff in the amount of \$15 per unit imported. (a) If there is no PTA,so that every country must pay the same tariff, from where will U.S. consumers import their spark plugs, Mexico or the rest of the world? Compute the equilibrium price of spark plugs in the United States, the quantity imported and consumed, and U.S. consumer surplus,tariff revenue,and social welfare. (b) Now, suppose that the United States and Mexico sign a free-trade agreement that eliminates the tariff on spark plugs from Mexico, but leaves the tariff on spark plugs from the rest of the world unchanged. How will the equilibrium change? Answer the same questions as in (a) under the new policy regime. (c) Identify the welfare change due to trade creation and the welfare change due to trade diversion, and draw them on a carefully marked graph with the equilibrium prices and quantities before and after the free-trade agreement marked. Does this trade agreement raise or lower U.S. welfare? (d) Now, how would your answer in (c) change if the MFN tariff had been \$50? Explain clearly; a diagram might help, but there is no need for additional calculation. (e) Now, how would your answer change if the MFN tariff had been \$5? Again, no calculation is needed. | 466 | 2,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-27 | latest | en | 0.951289 |
http://www.mallenmolen.nl/16mndr-coil/qatar_vertical_cylin_1429.html | 1,601,321,114,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401604940.65/warc/CC-MAIN-20200928171446-20200928201446-00420.warc.gz | 179,871,949 | 13,294 | • # qatar vertical cylindrical tank chemical volume
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### Tank Volume Calculator
Total volume of a cylinder shaped tank is the area, A, of the circular end times the height, h. A = r 2 where r is the radius which is equal to d/2. Therefore: V(tank) = r 2 h The filled volume of a vertical cylinder tank is just a shorter cylinder with the same radius, r, Tank Volume Calculator - Oil TanksMar 26, 2015 · The tank size calculator on this page is designed for measuring the capacity of a variety of fuel tanks. Alternatively, you can use this tank volume calculator as a water volume calculator if you need to calculate some specific water volume. The functionality of this Tank Volume Calculator - Vertical Cylindrical Tanks - MetricVertical cylindrical tank volume calculator diagram: Fill Rate Fill Times @ Litres / Minute. Total Tank Fill Time Current Time to Fill Current Time to Empty If you're cutting blocks, concrete, stone or ANYTHING and there's DUST - DON'T TAKE THE RISK Don't cut it, or cut it wet , qatar vertical cylindrical tank chemical volume
### HOW TO CALCULATE THE VOLUMES OF PARTIALLY FULL
2.1. Cylindrical Tanks The majority of tanks used in the chemical industry are cylindrical tanks, either in horizontal or vertical configuration. Consider, for example, a cylindrical tank with length L and radius R, filling up to a height H. If you want to obtain the volume of the liquid that partially fills the tankCompute Fluid Volumes in Vertical Tanks - Chemical Dec 18, 2003 · The equations for fluid volumes in vertical cylindrical tanks with concave bottoms are shown on p. 30. The volume of a flat-bottom vertical cylindrical tank may be found using any of these equations and setting a = 0. Radian angular measure must be used for trigonometric functions.Blank Worksheet to Calculate Secondary Containment This worksheet can be used to calculate the secondary containment volume of a rectangular or square dike or berm for a single vertical cylindrical tank. You may need a PDF reader to view some of the files on this page. See EPAs About PDF page to learn more. Blank Worksheet (PDF) (4 pp, 529 K)
### VERTICAL CYLINDER CALCULATOR - 1728
If you want to do calculations for a horizontal cylinder, then go to this link: Horizontal Cylinder Calculator. Example: Inputting tank height = 12, liquid level = 3 and tank diameter = 6, then clicking "Inches" will display the total tank volume in cubic inches and US Gallons and will also show the volume Tank Volume Calculator - ibec language institute*Program does not calculate liquid volume into upper head on vertical tanks (mixing with liquid in head space is not good practice). Fusion Fluid Equipment provides this tool for reference only. While Fusion Fluid Equipment strives to provide accurate information, no expressed or implied warranty is given by Fusion Fluid Equipment as to the , qatar vertical cylindrical tank chemical volumeVessel Volume & Level CalculationEstimates Volume filled in a Horizontal/ Vertical Vessel with 2:1 Ellipsoidal, Hemispherical, Torispherical and Flat heads. CheCalc Chemical engineering calculations to assist process, plant operation and maintenance engineers.
### Tank calculations - MrExcel
Dec 19, 2014 · We use what are called tank strappings for some of our additive tanks. They were built from formulas for cylindrical horizontal or vertical tanks. I pugged in the size of the tanks and a strapping was produced. From there I set up a form using vlookups to the strappings. Give me some tank measurements and I'll see if what I have at work will , qatar vertical cylindrical tank chemical volumeCompute Fluid Volumes in Vertical Tanks - Chemical Dec 18, 2003 · The equations for fluid volumes in vertical cylindrical tanks with concave bottoms are shown on p. 30. The volume of a flat-bottom vertical cylindrical tank may be found using any of these equations and setting a = 0. Radian angular measure must be used for trigonometric functions.Storage Tank Calculator - Vertical & Horizontal Tank , qatar vertical cylindrical tank chemical volumeHeritage Equipment Sells new & used stainless steel tanks for breweries, wineries and dairy processing. Using the storage tank calculator you will be able to estimate the volume for vertical storage tanks & horizontal storage tanks & preview a quick aspect ratio correct sketch of your tank.
### Blank Worksheet to Calculate Secondary Containment Volume , qatar vertical cylindrical tank chemical volume
This worksheet can be used to calculate the secondary containment volume of a rectangular or square dike or berm for a single vertical cylindrical tank. You may need a PDF reader to view some of the files on this page. See EPAs About PDF page to learn more. Blank Worksheet (PDF) (4 pp, 529 K)Steel Tank Calculations | STAFCOWaste Oil Tanks; Aviation Fuel Tanks; Single Wall Storage Tanks. Farm Tanks; Skid Tanks; Double Wall Tanks. 2085 Fireguard Tanks; Flameshield Tanks; Contain-A-Tank; UL 142/F921 Tanks; Chemical Storage Tanks; Custom Storage Tanks; Fertilizer Tanks; Stainless Steel Tanks; Large Storage Tanks. Bulk Storage Tanks; Horizontal Storage Tanks; Vertical , qatar vertical cylindrical tank chemical volume(PDF) HOW TO CALCULATE THE VOLUMES OF PARTIALLY FULL TANKShow to calculate the volumes of partially full tanks Article (PDF Available) in International Journal of Engineering and Technology eISSN pISSN(4):2319-1163 · March 2015 with 26,321 Reads
### Tank Calibration Chart Calculator - ODay Equipment
Fiberglass Tanks. ODay Equipment provides dome end fiberglass tanks from Xerxes and Containment Solutions. The domes on fiberglass tanks vary by manufacturer. So, here are the manufacturers web sites that have calibration charts specific to their designs. Xerxes Go to the Library tab for PDF versions of their charts.Vertical Tanks - www.specialtytankandweldingSpecialty Tank & Welding Single Wall Vertical Cylindrical Tanks are the solution to your aboveground containment needs. All are equipped with extra ports for monitoring of the interstice space. Quick and easy to install. Possess the strength and impermeability of steel. Ideal for storage of petroleum, chemicals, hazardous wastes, and fertilizers.Sloped Cylindrical Tank Level - Chemical plant design , qatar vertical cylindrical tank chemical volumeDec 07, 2012 · Sloped Cylindrical Tank Level JoeChem (Chemical) (OP) 4 Dec 12 19:12. , qatar vertical cylindrical tank chemical volume The real question I had is how to relate tank level to volume. If the tank is horizontal the level-volume relationship is not that involved. For a sloped tank the relationship is more complex. , qatar vertical cylindrical tank chemical volume i cannot help but think that you maybe better installing a vertical tank , qatar vertical cylindrical tank chemical volume
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### Spill Prevention Control and Countermeasure (SPCC)
Spill Prevention Control and Countermeasure (SPCC) Plan Single Vertical Cylindrical Tank Inside a Rectangular or Square Dike or Berm WORKSHEET . This worksheet can be used to calculate the secondary containment volume of a rectangular or square dike or berm for a single vertical cylindrical tank. This worksheet assumes that there are no other , qatar vertical cylindrical tank chemical volumeCylindrical Tank - an overview | ScienceDirect TopicsHorizontal cylindrical tanks are frequently used for water and fuel storage, and in many cases it is important to be able to gauge these vessels to determine the volume of liquid contained in them. However, it is normally much more difficult to establish a volume-per-inch scale for a horizontal tank than for one in a vertical position.Volume of a Cylinder - Web FormulasThe volume of a cylinder is found by multiplying the area of its top or base by its height and is defined as: V = · r 2 · h Example 1: A cylindrical water storage tank has an
### TANK VOLUME CALCULATOR [How to Calculate Tank
Tank volume calculator formula. Calculate the volume of liquid your container can hold by entering your dimensions in metric units (centimeters or meters) or imperial units (yards, feet or inches).. Our tool estimates the total tank volume and liquid capacity using the below formulas:. Horizontal Cylindrical TankCylindrical Tank - an overview | ScienceDirect TopicsHorizontal cylindrical tanks are frequently used for water and fuel storage, and in many cases it is important to be able to gauge these vessels to determine the volume of liquid contained in them. However, it is normally much more difficult to establish a volume-per-inch scale for a horizontal tank than for one in a vertical position.Horizontal Cylindrical Tank Volume and Level CalculatorVolume calculation on a partially filled cylindrical tank Some Theory. Using the theory. Use this calculator for computing the volume of partially-filled horizontal cylinder-shaped tanks.With horizontal cylinders, volume changes are not linear and in fact are rather complex as the theory above shows. Fortunately you have this tool to do the work for you.
### Cylinder, tube tank calculator: surface area, volume , qatar vertical cylindrical tank chemical volume
Now it is a simple matter to find the volume of the cylinder tank as it provides estimation of the total and filled volumes of the water tank, oil tank, reservoirs like a tube tank, and others of horizontal or vertical cylindrical shape. Sometimes it is also needed to estimate a surface size of your tank.(PDF) HOW TO CALCULATE THE VOLUMES OF PARTIALLY FULL TANKShow to calculate the volumes of partially full tanks Article (PDF Available) in International Journal of Engineering and Technology eISSN pISSN(4):2319-1163 · March 2015 with 26,321 ReadsVertical Tanks - www.specialtytankandweldingSpecialty Tank & Welding Single Wall Vertical Cylindrical Tanks are the solution to your aboveground containment needs. All are equipped with extra ports for monitoring of the interstice space. Quick and easy to install. Possess the strength and impermeability of steel. Ideal for storage of petroleum, chemicals, hazardous wastes, and fertilizers.
### Partially full cylinder, sphere, and cone volume , qatar vertical cylindrical tank chemical volume
Equations for Sphere, Cylinder, and Cone Volume (Rade and Westergren, 1990) Discussion of Volume Calculation This web page is designed to compute volumes of storage tanks for engineers and scientists; however, it may be useful to anyone who needs to know the volume of a full or partially full sphere, cylinder, or cone.Volume of a Cylinder - Web FormulasThe volume of a cylinder is found by multiplying the area of its top or base by its height and is defined as: V = · r 2 · h Example 1: A cylindrical water storage tank has an Calculating Tank VolumeCalculating Tank Volume Saving time, increasing accuracy By Dan Jones, Ph.D., P.E. alculating fluid volume in a horizontal or vertical cylindrical or elliptical tank can be complicated, depending on fluid height and the shape of the heads (ends) of a horizontal tank or the bottom of a vertical tank.
### Storage Tank Calculator - Vertical & Horizontal Tank , qatar vertical cylindrical tank chemical volume
Heritage Equipment Sells new & used stainless steel tanks for breweries, wineries and dairy processing. Using the storage tank calculator you will be able to estimate the volume for vertical storage tanks & horizontal storage tanks & preview a quick aspect ratio correct sketch of your tank.SCALE-UP OF MIXING SYSTEMS - University of Babylon A baffled, vertical square tank or a horizontal cylindrical tank has the same Power number as a vertical cylindrical tank. SCALE-UP OF MIXING SYSTEMS The calculation of power requirements for agitation is only a part of the mixer design. In any mixing problem, there are several defined objectives such as the time required for blending two , qatar vertical cylindrical tank chemical volumeSloped Cylindrical Tank Level - Chemical plant design , qatar vertical cylindrical tank chemical volumeDec 07, 2012 · Sloped Cylindrical Tank Level JoeChem (Chemical) (OP) 4 Dec 12 19:12. , qatar vertical cylindrical tank chemical volume The real question I had is how to relate tank level to volume. If the tank is horizontal the level-volume relationship is not that involved. For a sloped tank the relationship is more complex. , qatar vertical cylindrical tank chemical volume i cannot help but think that you maybe better installing a vertical tank , qatar vertical cylindrical tank chemical volume
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Tank Calibration Services Tank calibration for shore and ship installations, terminals, and tank farms from our Total Quality Assurance experts. Customers face a need for accurate volumetric tables and quantity measurement, to control stock held in storage vessels.Mixing 101: Optimal Tank Design | Dynamix AgitatorsMar 10, 2015 · Vertical Cylindrical Tanks. Vertical cylindrical tanks are the most common type of tank in use. A key consideration for cylindrical tanks is to ensure that they are either baffled or offset-mounted to prevent swirling from occurring. Refer to section 2 below (The Use of Baffling) for details.
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https://gmatclub.com/forum/whereas-lines-of-competition-are-clearly-defined-in-the-more-50083.html?sort_by_oldest=true | 1,498,246,241,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320130.7/warc/CC-MAIN-20170623184505-20170623204505-00442.warc.gz | 739,834,344 | 68,538 | It is currently 23 Jun 2017, 12:30
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# Whereas lines of competition are clearly defined in the more
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Whereas lines of competition are clearly defined in the more [#permalink]
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06 Aug 2007, 10:17
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Whereas lines of competition are clearly defined in the more established industries, in the Internet industry they are blurred and in distinct, as companies that compete one day may be partners the next.
A. Whereas lines of competition are clearly defin ed in the more established industries, in the Internet industry they are blurred and in distinct, as companies that compete
B. Although the lines of competition are clearly d efined in industries that are more established , they are blurred and indistinct in the Internet threats t industry, as competing companies
C. The lines of competition are clearly defined in the more established industries, unlike the Internet where they are blurred and i ndistinct, as companies that compete
D. Unlike more established industries, where the lines of competition are clearly DEFI ned, they are burred and indistinct in the Internet i ndustry, as companies that compete
E. Unlike more established industries, with clearly defined lines of competition, those of the Internet industry are blurred and i ndistinct, as competing companies
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06 Aug 2007, 12:55
I think its B. Some corrections are needed ( but I think they are more of copy and paste issues when you typed the question )
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07 Aug 2007, 04:02
I would go for C
A) 'whereas' doesnt seem to convey well enough the purpose of what the author is trying to say, 'unlike' and 'although' seem to be better options
B) 'as competing companies ONE DAY may be partners in the next' sounds awkward
C) I cant spot any errors there, and it uses 'unlike' which seems better than 'whereas' from answer A
D) Definitely wrong, because here 'established industries' is being compared to 'they' (clearly referring to lines of competition). like should be compared to like
E) 'as competing companies ONE DAY may be partners in the next' sounds awkward
whats the OA?
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03 Sep 2007, 21:44
OA is 'A'
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Re: Whereas lines of competition are clearly defined in the more [#permalink]
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07 Oct 2013, 23:52
Can we have a further discussion over this one?
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Re: Whereas lines of competition are clearly defined in the more [#permalink]
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08 Oct 2013, 02:18
I believe the usage of "where" is incorrect. "where" should specify an actual location, therefore we can cross off C & D.
And in B & E, "as competing companies one day may be partners the next" is an awkward phrase.
A appears to be the correct choice here.
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Re: Whereas lines of competition are clearly defined in the more [#permalink]
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08 Oct 2013, 14:16
C,D,E are out because of wrong usage of Unlike.
Between A & B.
A uses "as companies that compete one day may be partners the next. " is much better than "as competing companies one day may be partners the next. "
Also "more established industries" is better than "industries that are more established"
A wins
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Re: Whereas lines of competition are clearly defined in the more [#permalink]
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08 Oct 2013, 15:11
summer101 wrote:
C,D,E are out because of Unlike.
Between A & B.
A uses "as companies that compete one day may be partners the next. " is much better than "as competing companies one day may be partners the next. "
Also "more established industries" is better than "industries that are more established"
A wins
Why do you think that competing companies is inferior to companies that compete.
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Re: Whereas lines of competition are clearly defined in the more [#permalink]
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08 Oct 2013, 19:20
I went for " E"... not sure why competing companies is wrong...
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Re: Whereas lines of competition are clearly defined in the more [#permalink]
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09 Oct 2013, 02:10
humtum0 wrote:
Whereas lines of competition are clearly defined in the more established industries, in the Internet industry they are blurred and in distinct, as companies that compete one day may be partners the next.
A. Whereas lines of competition are clearly defin ed in the more established industries, in the Internet industry they are blurred and in distinct, as companies that compete
B. Although the lines of competition are clearly d efined in industries that are more established , they are blurred and indistinct in the Internet threats t industry, as competing companies
C. The lines of competition are clearly defined in the more established industries, unlike the Internet where they are blurred and i ndistinct, as companies that compete
D. Unlike more established industries, where the lines of competition are clearly DEFI ned, they are burred and indistinct in the Internet i ndustry, as companies that compete
E. Unlike more established industries, with clearly defined lines of competition, those of the Internet industry are blurred and i ndistinct, as competing companies
C ,D & E are out --> illogical comparison
between B & A , i rejected B because of ",as competing companies" sounded awkward.
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Re: Whereas lines of competition are clearly defined in the more [#permalink]
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09 Oct 2013, 08:08
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A is the correct choice: Look out for quantities being compared. We are comparing lines of competition in Established industries and Internet Industry.
Reasoning:
A. Whereas lines of competition are clearly defined in the more established industries, in the Internet industry they are blurred and in distinct, as companies that compete--Correct.
B. Although the lines of competition are clearly defined in industries that are more established , they are blurred and indistinct in the Internet threats t industry, as competing companies---Here the meaning is distorted, because of although.
C. The lines of competition are clearly defined in the more established industries, unlike the Internet where they are blurred and indistinct, as companies that compete--Comparison of lines of competition and Internet. Incorrect.
D. Unlike more established industries, where the lines of competition are clearly Defined, they are burred and indistinct in the Internet industry, as companies that compete---'They' refers to more established industries and not lines of competition Incorrect.
E. Unlike more established industries, with clearly defined lines of competition, those of the Internet industry are blurred and indistinct, as competing companies ----'Those' refers to more established industries and not lines of competition Incorrect.
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Re: Whereas lines of competition are clearly defined in the more [#permalink]
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09 Oct 2013, 16:02
I can't tell what B is trying to say on account of typos
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Re: Whereas lines of competition are clearly defined in the more [#permalink]
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11 May 2014, 02:35
Whereas lines of competition are clearly defined in the more established industries, in the Internet industry they are blurred and in distinct, as companies that compete one day may be partners the next.
A. Whereas lines of competition are clearly defined in the more established industries, in the Internet industry they are blurred and in distinct, as companies that compete
There is no problem in A. Identifying the SV pairs, highlighted in red, in the sentence shows that the comparisons are actually parallel. Lines of competition in the more established industries are correctly compared to those of the internet industry.
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Whereas lines of competition are clearly defined in the more [#permalink]
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27 Jun 2014, 01:36
Whereas lines of competition are clearly defined in the more established industries, in the Internet industry they are blurred and in distinct, as companies that compete one day may be partners the next.
A. Whereas lines of competition are clearly defined in the more established industries, in the Internet industry they are blurred and in distinct, as companies that compete
There is no problem in A. Identifying the SV pairs, highlighted in red, in the sentence shows that the comparisons are actually parallel. Lines of competition in the more established industries are correctly compared to those of the internet industry.
What is wrong with option B ?
Phrase "competing companies" doesn't change the meaning of sentence and neither does "Although". In fact the presence of "Although" clears the intent of 'they' in 2nd part of sentence. (i.e. lines of competition)
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Re: Whereas lines of competition are clearly defined in the more [#permalink]
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27 Jun 2014, 11:25
anirudh777 wrote:
Whereas lines of competition are clearly defined in the more established industries, in the Internet industry they are blurred and in distinct, as companies that compete one day may be partners the next.
A. Whereas lines of competition are clearly defined in the more established industries, in the Internet industry they are blurred and in distinct, as companies that compete
There is no problem in A. Identifying the SV pairs, highlighted in red, in the sentence shows that the comparisons are actually parallel. Lines of competition in the more established industries are correctly compared to those of the internet industry.
What is wrong with option B ?
Phrase "competing companies" doesn't change the meaning of sentence and neither does "Although". In fact the presence of "Although" clears the intent of 'they' in 2nd part of sentence. (i.e. lines of competition)
Hello mate..
m not an expert but jus thot to try my best to explain..
in B,the sentence doesnt end well,although it quite clearly compares the lines of competition...the end is highly awkward as it states competing companies ,and may mean that they are,competing at the current moment...A however,puts it in simple present tense which is required in the context of the sentence..
HOPE THAT HELPS
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16 May 2016, 11:13
Whereas lines of competition are clearly defined in the more established industries, in the Internet industry they are blurred and in distinct, as companies that compete one day may be partners the next.
A. Whereas lines of competition are clearly defined in the more established industries, in the Internet industry they are blurred and in distinct, as companies that compete // perfect whereas structure
B. Although the lines of competition are clearly defined in industries that are more established , they are blurred and indistinct in the Internet threats t industry, as competing companies // they is ambiguous --lines /industries
C. The lines of competition are clearly defined in the more established industries, unlike the Internet where they are blurred and indistinct, as companies that compete //bad comparison
D. Unlike more established industries, where the lines of competition are clearly Defined, they are burred and indistinct in the Internet industry, as companies that compete //bad comparison
E. Unlike more established industries, with clearly defined lines of competition, those of the Internet industry are blurred and indistinct, as competing companies //bad comparison
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Whereas lines of competition are clearly defined in the more [#permalink]
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18 Jul 2016, 02:26
Marcab wrote:
summer101 wrote:
C,D,E are out because of Unlike.
Between A & B.
A uses "as companies that compete one day may be partners the next. " is much better than "as competing companies one day may be partners the next. "
Also "more established industries" is better than "industries that are more established"
A wins
Why do you think that competing companies is inferior to companies that compete.
It is due to the one day following the underlined part:
companies " that compete one day"
competing companies one day may be partners the next
We need companies that compete to be parallel with (companies that are) partners the next day
competing companies one day may be partnering companies
In the second sentence competing is an adjective, but partners is a noun. To maintain structure we need partnering the next.
REMEMBER!!! Parallelism is as much about maintaining a structure as about having a grammatical reason!
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Whereas lines of competition are clearly defined in the more [#permalink]
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18 Jul 2016, 03:54
A. Whereas lines of competition are clearly defin ed in the more established industries, in the Internet industry they are blurred and in distinct, as companies that compete
looks fine --will check further in round 2
B. Although the lines of competition are clearly d efined in industries that are more established , they are blurred and indistinct in the Internet threats t industry, as competing companies
internet threat is a new edition here !!
C. The lines of competition are clearly defined in the more established industries, unlike the Internet where they are blurred and i ndistinct, as companies that compete
Incorrect comparison
D. Unlike more established industries, where the lines of competition are clearly DEFI ned, they are burred and indistinct in the Internet i ndustry, as companies that compete
Pronoun ambiguity
E. Unlike more established industries, with clearly defined lines of competition, those of the Internet industry are blurred and i ndistinct, as competing companies
Incorrect comparison again !!
That makes A the answer !!
Whereas lines of competition are clearly defined in the more [#permalink] 18 Jul 2016, 03:54
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# TEXTBOOK SOLUTIONS FOR A Basic Math Approach to Concepts of Chemistry 6th Edition
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Chapter: Problem:
STEP-BY-STEP SOLUTION:
Chapter: Problem:
• Step 1 of 5
Even though looking similar and often confusing mass and weight has certain differences. Mass is the quantity of matter possessed by an object whereas weight is caused by the pull of gravity on the mass of an object. The mass of an object remains constant when the object is moved away from the earth. The weight of an object depends on the distance of the object from the center of the earth. Greater the distance, smaller will be the weight. If an object is far away from the center of the earth, it becomes weightless. But the quantity of matter possessed by the object (mass) remains the same.
There is difference in the way they measure too. Weight is measured by putting a scale between the object and the center of the earth and letting the gravity pull the object down on the scale.
Object
Scale
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Center of the earth
• Chapter , Problem is solved.
Corresponding Textbook
A Basic Math Approach to Concepts of Chemistry | 6th Edition
9780534265748ISBN-13: 053426574XISBN: Authors: | 317 | 1,388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | longest | en | 0.900636 |
https://www.w3seeker.com/2022/12/civil-engineering-highway-engineering_30.html | 1,701,874,400,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100599.20/warc/CC-MAIN-20231206130723-20231206160723-00815.warc.gz | 1,152,548,670 | 29,184 | ### Civil Engineering - Highway Engineering MCQs Part 2
1. Reinforcement in cement concrete slab of road pavements, is placed
A.
longitudinally
B.
transversely
C.
longitudinally and transversely
D.
in the form of welded wire mesh
E.
none of these.
2. If d is the thickness of a concrete pavement, the equivalent radius b of resisting section for an interior loading, is
A.
b = 0.6 + d2 - 0.675 d
B.
b = 1.6 + d2 + 0.675 d
C.
b = 1.6 + d2 - 3.675 d
D.
b = 1.6 + d2 - 0.675 d
3. According to IRC : 52-1973, for a single lane National Highway in hilly region,
A.
width of the carriageway must be 3.75 m
B.
shoulders on either side must be 1.25 m
C.
total width of the road-way must be 6.25 m
D.
total of the above.
4. If x% is the gradient of an alignment and y% is the gradient after proper superelevation along a curved portion of a highway, the differential grade along the curve, is
A.
(x + y)%
B.
(x - y)%
C.
(y - x)%
D.
(x x y)%
E.
(y + x)%
5. In an ideal transition curve, the radius of curvature
A.
is constant
B.
at any point is directly proportional to its distance from the point of commencement
C.
is inversely proportional to the radius of main curve
D.
is directly proportional to the radius of main curve
6. On concrete roads, the camber generally provided, is
A. 1 in 20 to 1 in 24
B. 1 in 30 to 1 in 48
C. 1 in 36 to 1 in 48
D. 1 in 48 to 1 in 60
E. 1 in 60 to 1 in 72
7. While calculating the sight distances, the driver's eye above road surface, is assumed
A. 90 cm
B. 100 cm
C. 110 cm
D. 120 cm
E. 150 cm
8. If the number of lanes on the carriageway of a road is more than two, the total width of lane ways is equal to 3.0 m
A. + 0.60 m
B. + 0.70 m
C. + 0.80 xn
D. + 0.90 m
E. + 1.50 m
9. Minimum thickness of a layer of fine sand required to cut off capillary rise of water completely, should be
A. 40 cm
B. 52 cm
C. 64 cm
D. 76 cm
E. 80 cm
10. If W is the weight of a vehicle negotiating an upgrade 1 : S along a track having co-efficient of resistance μ, the tractive force T is given by
A.
B.
C.
D.
E.
11. If L is the length of a moving vehicle and R is the radius of curve, the extra mechanical width b to be provided on horizontal curves,
A.
B.
C.
D.
E.
12. The desirable camber for straight cement concrete roads, is
A. 1 in 33 to 1 in 25
B. 1 in 40 to 1 in 33
C. 1 in 150 to 1 in 140
D. 1 in 160 to 1 in 140
E. none of these.
13. Minimum permissible speed on high speed roads, is decided on the basis of
A. 15 percentile cumulative frequency
B. 20 percentile cumulative frequency
C. 30 percentile cumulative frequency
D. 40 percentile cumulative frequency.
14. The ideal shape of a transition curve, is
A. clothoid
B. cubic spiral
C. cubic parabola
D. lamniscate
E. none of these.
15. At a road junction, 7 cross conflict points are severe if
D. none of these.
16. To indicate proper control of consistency of a freshly mixed concrete for pavement construction, the slump should be between
A. 3 to 5 cm
B. 4 to 6 cm
C. 5 to 7 cm
D. 7 to 10 cm
E. 10 to 12 cm
17. Driving vehicles on wet surfaced roads, is dangerous because it may
A. skid
B. slip
C. overturn
D. all the above.
18. The width of different roads as recommended in Nagpur plan by the Indian Road Conference for hilly region, is
A. same for National Highways
B. different for National Highways
C. same for State Highways
D. same for Major District roads.
19. If N is the algebraic difference of grades, S is the head light beam distance in metres, the length (L) of a valley curve, is
A.
B.
C.
D.
E.
20. If D is the degree of a curve, the percentage reduction of gradient, is
A. 0.01 D
B. 0.02 D
C. 0.03 D
D. 0.04 D
E. 0.005 D
22. For the administration of road transport, a Motor Vehicle Act was enacted in
A. 1927
B. 1934
C. 1939
D. 1947
E. 1950
23. The absolute minimum sight distance required for stopping a vehicle moving with a speed of 80 km ph, is
A. 120 m
B. 200 m
C. 640 m
D. none of these.
24. Along high ways confirmatory route markers are generally fixed
A. before the crossing on the left side
B. after the crossing on the left side
C. before the crossing on the right side
D. after the crossing on the right side.
25. Length of vehicles does not affect
A. extra widening
C. passing sight distance
D. width of shoulders
E. none of these.
26. Road width 8.8 m of two lane National highways or State highways in mountainous terrain
A. excludes the width of parapet (0.6 m)
B. excludes the width of side drain (0.6 m)
C. excludes the width of parapet and side drain
D. includes the width of parapet and side drain
27. Pick up the correct statement from the following:
A. Long tangent sections exceeding 3 km in length should be avoided
B. Curve length should be at least 150 metres for a deflection angle of 5 degress
C. For every degree decrease in the deflection angle, 30 metre length of curve to be increased
D. If the deflection angle is less than 1°, no curve is designed
E. All the above.
28. Transverse joints are provided at distances varying from
A. 10 m to 15 m
B. 12 m to 18 m
C. 16 m to 24 m
D. 17 m to 27 m
E. 25 m to 35 m
29. If N is the algebraic difference of grades, S is the minimum sight distance in metres, the length (L) of a summit curve is
A.
B.
C.
D.
E. none of these.
30. If N is the net difference of grades, S is the minimum overtaken sight distance in metres, the length (L) of a summit curve, is
A.
B.
C.
D.
E.
31. Pick up the correct statement from the following:
A. Seasonal cycle of traffic volume during April and November, is usually near the annual average
B. Mid-winter seasonal cycle of traffic is least
C. Mid-summer seasonal cycle of traffic is highest
D. All the above.
A. higher speed of vehicles
B. increased volume of traffic
C. reduced maintenance cost of the roads
D. draining off rain water quickly
E. all the above.
33. The shape of a vertical curve, is
A. parabolic
B. elliptical
C. circular
D. spiral
E. all the above.
34. To prevent movement of moisture from subgrade to road pavement at the same level as that of water-table, thickness of a cut off layer of coarse sand, is
A. 15 cm
B. 20 cm
C. 30 cm
D. 45 cm
E. none of these.
35. For Indian conditions, the water bound macadam roads, are suitable if daily traffic does not exceed
A. 2000 tonnes
B. 2500 tonnes
C. 3000 tonnes
D. 3500 tonnes
36. If h1 is the height of the driver's eye and h2 is the height of an obstruction above road surface, then, for a minimum sight distance S, the length of the vertical curve should be greater than
A.
B.
C.
D.
37. For maximum strength and durability minimum percentage of cement, by weight is
A. 15%
B. 20%
C. 25%
D. 30%
E. 33%
38. Camber in pavements is provided by
A. straight line method
B. parabola method
C. straight at the edges and parabolic at the crown
D. all the above.
39. The minimum design speed of various types of highways in plain terrain is the same as the ruling design speed of
A. rolling terrain
B. mountainous terrain
C. steep terrain
D. none of these.
40. Passing zones are generally not provided on
A. summit curves
B. horizontal curves
C. two lane highways
D. all the above.
41. The normal road width of National and State highways
A. is kept 45 m
B. in plain and rolling terrain built-up area, is 30 m
C. in mountainous built-up area is 20 m
D. All the above.
42. The extra widening of pavements recommended by Indian Roads Congress for hill roads having radius 150 metres, is
A. 1.5 m
B. 1.0 m
C. 0.5 m
D. 0.0 m
E. 2.0 m
43. The steepest gradient permitted on roads which, in ordinary conditions, does not exceed, is known
E. all the above.
44. For night travel, the length of a valley curve should be such that the head light beam distance is the same as
A. stopping sight distance
B. overtaking sight distance
C. sum of (a) and (b)
D. difference of (a) and (b)
45. Stability of hill slopes depends upon
A. nature of the slope
B. angle of the slope
C. geological conditions
D. ground water conditions
E. all the above.
46. Selection of the routes, of highways depends upon
A. feasibility of attaining ruling gradient
B. avoidance of cutting hard rocks
C. minimum number of bridges
D. minimum possible length of alignment
E. all the above.
47. The first stage of deciding the alignment of a hill road, is
A. reconnaissance
B. detailed survey
C. trace-out
D. preliminary survey.
48. While designing hair pin bends on higways, the minimum
A. designed speed is 20 km/hour
B. gradient is 1 in 40
C. gradient is 1 in 200
D. superelevations is 1 in 10
E. All the above.
49. Pick up the incorrect statement from the following:
A. Tresguet did not provide the top camber for the drainage of surface water
B. Tresguet provided the top camber for the drainage of surface water
C. Telford provided two layers of stones in the central 5.4 m width and one layer was provided on the sides
D. Macadam provided a camber to the formation at the dug-up state, to drain percolated water.
50. Cement grouted pavement is classified as
A. rigid pavement
B. semi-rigid pavement
C. flexible pavement
D. none of these. | 2,735 | 9,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.8596 |
https://studylib.net/doc/7963441/permutations-and-combinations | 1,716,172,926,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058147.77/warc/CC-MAIN-20240520015105-20240520045105-00342.warc.gz | 486,369,237 | 12,313 | # Permutations and Combinations
```Permutations and Combinations
Algebra 2
Calculate.
1) 2!
2) 6!
3) 5!
10) You take a multiple-choice test. There are 20
questions and each question has 5 answer choices.
How many ways can you fill out the answer sheet?
Write out the following factorials and show the
canceling.
4) 8!/6!
11) A nickel, a dime, a 6 sided die, and a 10 sided die
are tossed, how many results are possible?
5) 10!/8!
6) 9!/5!
Write in terms of factorials.
7) 98 x 97 x 96 =
8) 32 x 31 x 30 x 29 =
12) Eight students are participate in a talent show.
One student insists on being the grand finale. If his
request is granted how many ways can the show be
ordered.
9) 50 x 49 x 48 =
Permutations
13) How many ways can your ipod play a playlist of
15 songs on shuffle, if every song gets played before
any song gets played twice?
Solve.
15) P(5, 2)
16) P(6, 2)
wants you to choose and rank your top 4, how many
ways can this be done?
17) P(7, 6)
18) P(3, 1)
Permutations with repetition
19) List all the possible orderings of the letters T, O,
and P.
20) List all the possible orderings of the letters P, O,
and P.
21) Explain the relationship between answers 19 and
20.
22) Calculate the number of orderings of the letters
W, E, N, and T.
23) Calculate the number of orderings of the letters
W, E, E, and E.
24) Calculate the number of orderings of the letters P,
E, E, and P.
25) Explain the relationship between the answers to
questions 22 and 23.
26) Explain the relationship between the answers to
questions 22 and 24.
27) Explain in words how to calculate a permutation
with repetition.
28) Calculate the number of orderings of the letters G,
R, E, E, and N.
29) Calculate the number of orderings of the letters
W, O, O, and D.
30) Calculate the number of orderings of the letters P,
A, P, E, and R.
31) Calculate the number of orderings of the letters P,
E, P, P, E, and R.
32) Calculate the number of orderings of the letters
M, I, S, S, I, S, S, I, P, P, and I.
Combinations
33) There are 8 homecoming princesses and you need
to choose a queen and her first and second attendants
from the princesses. How many ways can you pick
the court from the 8 princesses?
factorials.
35) All 8 princesses decide that they don’t want to
have a queen and first and second attendant. How
many ways could you pick three co-queens?
factorials.
37) How are the answers to questions 33 and 35
related?
38) A basketball team has 20 players how many ways
can the coach choose his starting point guard, guard 2,
shooting guard, forward, and center?
factorials.
40) A basketball team has 20 players how many ways
can the coach choose his starting five (the positions
are not important)?
factorials.
42) In a class of 30 students how many ways can you
pick a class president, vice president, and secretary?
factorials.
44) In class of 30 students how many ways can you
pick a committee of 3 students?
factorials.
46) If you have n objects how many ways can you
choose a group of 3 of them if the order you pick
them does not matter?
Calculate.
47) C(12, 3)
48) C(4, 2)
49) C(5, 5)
50) C(5, 1)
51) C(5, 4)
52) C(6, 1)
53) C(6, 5)
54) C(100, 97)
55) Notice that the answers to 45 and 46 are the same.
Explain why this is both in terms of the formula and
conceptually.
Decide whether each situation could be modeled by a
permutation or a combination.
56) Selecting 10 math students to go to a math
competition.
57) How many ways can 5 people stand in a line?
58) How many ways can you pick a president and
vice president from 20 canidates?
59) Selecting 4 balls from a bin of 6.
60) Picking two movies to buy from Wal-Mart.
``` | 1,063 | 3,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-22 | latest | en | 0.85108 |
https://www.crossroadstoday.com/i/this-new-york-deli-offers-customers-free-food-if-they-can-solve-simple-math-problems/ | 1,585,601,623,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497301.29/warc/CC-MAIN-20200330181842-20200330211842-00551.warc.gz | 873,341,708 | 23,588 | # This New York deli offers customers free food if they can solve simple math problems
The rules are simple at this deli in the Bronx: solve a math equation and you get five seconds to grab anything you want off store shelves.
A sandwich, a soda or maybe even a chocolate bar to satisfy a midnight craving: These are some of the many things you’d expect to find at a New York deli and convenience store.
But customers at Lucky Candy in the Bronx get an entirely different experience, and it’s all thanks to cashier Ahmed Alwan.
To his customers, the 20-year-old college student is easily recognizable, and it’s not just because of his bright smile and cheerful energy. Two weeks ago, Alwan decided to start playing a game with shoppers — and it’s since gone viral.
The rules are simple: If you can solve a math equation, you get five seconds to grab anything you want off store shelves and have it for free.
The one thing you’re not allowed to grab? His cat. (Yes, someone tried.)
“All I wanted to do was to help people,” Alwan told CNN. “But I wanted to make it fun. So I made a TikTok and chose a challenge, asking them a math question. It’s a way to entertain and educate people in need while putting a smile on their face, too.”
Two weeks and 13 videos later, Alwan has become an internet sensation, gaining more than 300,000 followers on TikTok and 17,000 on Instagram. In his videos, Alwan can be heard asking customers simple questions such as, “What’s 5 times 5?” or “9 times 9 minus 5?”
Those who answer correctly use their five seconds (and a few quarters) to scramble around the store grabbing anything from chips and ice cream to speakers and hookahs. But no matter how much they grab, or how valuable the items are, their cost is always the same: \$0.
### A history of giving back
Alwan pays for everything out of his own pocket. His main concern isn’t money, he said, but helping low-income community members save for higher priority expenses, such as rent and utilities.
Saleh Aobad, the store’s owner and Alwan’s father, said he’s “very proud” of his son.
“It’s great to see him do good and help out the community, and most importantly represent Islam,” Aobad, an immigrant from Yemen, told CNN. “It’s impacting the business in a positive way, bringing awareness and attention to the store as well as spreading positivity throughout the community.”
Many of the store’s customers are familiar faces, people Alwan has known since he was 13 and began working in the family business. Others are strangers. But they all get a fair shot at solving a math problem.
Alwan, who’s studying at Bronx Community College to become a pharmacist, has a history of being charitable.
He often gives customers food and household supplies on credit, his father said. And on cold mornings when he spots people sleeping at the nearby train station or by the tracks, he offers them free coffee, bananas and muffins.
Alwan’s TikTok and Instagram videos are heartwarming and funny, but what they’ve done for his community goes beyond free chips and cookies.
“This changed our relationship with the community big time,” Alwan said. “They’re all showing me so much love and lining up at the store just for a chance to play. We have so many visitors coming to take pictures and say hi. But it’s not about me. It’s about the community.”
Alwan said he plans to continue making videos, and even getting more creative with his questions. He even started a GoFundMe to support the game and enable him to help more people.
“The money will be going towards products and food that would be used for the challenges and even simple donations for customers in need,” he said.
The goal, he explained on the GoFundMe, is “to inspire others to always be kindhearted.” | 829 | 3,762 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.951828 |
https://gabrielececconi.com/algoritma-divide-and-conquer-31/ | 1,632,521,965,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057580.39/warc/CC-MAIN-20210924201616-20210924231616-00447.warc.gz | 322,115,454 | 12,345 | # ALGORITMA DIVIDE AND CONQUER PDF
Algoritma Divide andConquer • Divide and Conquer dulunya adalah strategi militer yang dikenal dengan nama divide ut im. Algoritma Divide andConquer Say Pulung () Akhmad Jayadi ( ) 1 Definisi Divide: membagi masalah menj. In divide and conquer approach, the problem in hand, is divided into smaller sub- problems and then each problem is solved independently. When we keep on.
Author: Vumi Kigajar Country: Moldova, Republic of Language: English (Spanish) Genre: Finance Published (Last): 10 May 2012 Pages: 474 PDF File Size: 10.54 Mb ePub File Size: 5.47 Mb ISBN: 556-4-45306-922-7 Downloads: 37535 Price: Free* [*Free Regsitration Required] Uploader: Kegal
The reason is that once a sub-problem is small enough, it and all its sub-problems can, in principle, be solved within the cache, without accessing the dkvide main memory. Divide-and-conquer algorithms are naturally adapted for execution in multi-processor machines, especially shared-memory systems where the communication of data between processors does not need to be planned in advance, because distinct sub-problems can be executed on different processors.
Leiserson, and Ronald L.
A Algoritma pengurutan yang termasuk jenis ini: An early example of a divide-and-conquer algorithm with multiple subproblems is Gauss ‘s description of what is now called the Cooley—Tukey fast Fourier transform FFT algorithm, [5] conqueg he did not analyze its operation count quantitatively, and FFTs did not become widespread until they were rediscovered over a century later.
Knuth, The Art of Computer Programming: While the second method performs the same number of additions as the first, and pays the overhead of the recursive calls, it is usually more accurate.
Python – Divide and conquer Advertisements. In other projects Wikimedia Commons. This step receives a lot of smaller sub-problems to be solved. Otherwise we eleminate half of the list of elements by choosing whether to procees with the right or left half of the list depending on the value of the item searched.
## Algoritma Divide and Conquer
Volume 3 Sorting and Searching. Fundamental of Algorithmics, Prentice-Hall, Views Read Edit View history. Therefore, some authors consider that the name “divide and conquer” should be used only when each problem may generate two or more subproblems.
A general procedure for a simple hybrid recursive algorithm is short-circuiting the base casealso known as arm’s-length recursion. The same advantage exists with regards to other hierarchical storage systems, such as NUMA or virtual memoryas well as for multiple levels of cache: Divide and Conquer 2. In such cases it may be worth identifying and saving the solutions to these overlapping subproblems, a technique commonly known as memoization.
LIBRO PATERNIDAD ESPIRITUAL EDWIN SANTIAGO PDF
On the other hand, efficiency often improves if the recursion is stopped at relatively large base cases, and these are solved non-recursively, resulting in a hybrid algorithm. An algorithm designed to exploit the cache in this way is called cache-obliviousbecause it does not contain the cache size as an explicit parameter. The solutions to the sub-problems are then combined to give a solution to the original problem.
Divide-and-conquer algorithms naturally tend to make efficient use of memory caches. As another example of a divide-and-conquer algorithm that did not originally involve computers, Donald Knuth gives the method a post office typically uses to route conauer In binary search we take a sorted list of elements and start looking for an element at the middle of the list.
The following program is an example of divide-and-conquer programming approach where the binary search is implemented using python. An important application of divide and conquer is in optimization, [ example needed alboritma where if the search space is reduced “pruned” by a constant factor at each step, the overall algorithm has the same asymptotic complexity as the pruning step, with the constant depending on the pruning factor by summing the geometric series ; this is algorittma as prune and search.
Sub-problems should represent a part of the original problem.
Algoritma divide and conquer lanjutan Software. Translated in “Multiplication of Multidigit Numbers on Automata”. This is possible as the list is sorted and it is much quicker than linear search. A divide-and-conquer algorithm works by recursively breaking down a problem into two or more sub-problems of the same or related type, until these become simple enough to be solved directly.
Secara rekursif, terapkan algoritma D-and-C pada masing-masing bagian. All articles with unsourced statements Articles with unsourced statements from October Wikipedia articles needing clarification from Alglritma All articles needing examples Articles needing examples from October Commons category link is on Wikidata. MinMaks A, n, min, maks Algoritma: Divide And Conquer Strategy Documents. Misalkan elemen-elemen A1 elemen-elemen A2. Perbandingan algoritma brute forcedivide and conquer Education.
BUDD HOPKINS INTRUDERS PDF
Urut-gabung Merge Sort 2. Binary search, a decrease-and-conquer algorithm where the subproblems are of roughly half the original size, has a long history. Thus, for example, many library implementations of quicksort will switch to a simple loop-based insertion sort or similar algorithm once the number of items to be sorted is sufficiently small.
From Wikipedia, the free encyclopedia. Those “atomic” smallest possible sub-problem fractions are solved. Tabel A dengan n elemen Keluaran: Generally, at this level, the problems are considered ‘solved’ on their own.
## Divide-and-conquer algorithm
Algoritma Divide and Conquer Documents. Lecture Notes in Computer Science vol. Choosing the smallest or simplest possible algoritmw cases is more elegant and usually leads to simpler programs, because there are fewer cases to consider and they are easier to solve.
Potenciao Divide and Conquer Technology. Algoritma Divide and Conquer.
### Divide-and-conquer algorithm – Wikipedia
cobquer Divide-and-conquer algorithms are naturally implemented as recursive procedures. This step generally takes a recursive approach to divide the problem until no sub-problem is further divisible. This approach is also the standard anc in programming languages that do aloritma provide support for recursive procedures.
Increasing the base cases to lists of size 2 or less will eliminate most of those do-nothing calls, and more generally a base case larger than 2 is typically used to reduce the fraction of time spent in function-call overhead or stack manipulation. For example, an FFT algorithm could stop the recursion when the input is a single sample, and the quicksort list-sorting algorithm could stop when the input is the empty list; in both examples there is only one base case to consider, and it requires no processing.
Thus, the risk of stack overflow can be reduced by minimizing the parameters and internal variables of the recursive procedure or by using an explicit stack structure. | 1,502 | 7,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.830719 |
https://reference.wolframcloud.com/language/ref/FunctionInterpolation.html | 1,696,106,898,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510707.90/warc/CC-MAIN-20230930181852-20230930211852-00456.warc.gz | 524,691,030 | 10,549 | # FunctionInterpolation
FunctionInterpolation[expr,{x,xmin,xmax}]
evaluates expr with x running from xmin to xmax and constructs an InterpolatingFunction object which represents an approximate function corresponding to the result.
FunctionInterpolation[expr,{x,xmin,xmax},{y,ymin,ymax},]
constructs an InterpolatingFunction object with several arguments.
# Examples
open allclose all
## Basic Examples(1)
Computes an InterpolatingFunction representation of Exp[-Sin[x]^2]:
Use the resulting function as any function:
## Scope(1)
Use derivatives to improve the smoothness of the approximation:
Wolfram Research (1996), FunctionInterpolation, Wolfram Language function, https://reference.wolfram.com/language/ref/FunctionInterpolation.html.
#### Text
Wolfram Research (1996), FunctionInterpolation, Wolfram Language function, https://reference.wolfram.com/language/ref/FunctionInterpolation.html.
#### CMS
Wolfram Language. 1996. "FunctionInterpolation." Wolfram Language & System Documentation Center. Wolfram Research. https://reference.wolfram.com/language/ref/FunctionInterpolation.html.
#### APA
Wolfram Language. (1996). FunctionInterpolation. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/FunctionInterpolation.html
#### BibTeX
@misc{reference.wolfram_2023_functioninterpolation, author="Wolfram Research", title="{FunctionInterpolation}", year="1996", howpublished="\url{https://reference.wolfram.com/language/ref/FunctionInterpolation.html}", note=[Accessed: 30-September-2023 ]}
#### BibLaTeX
@online{reference.wolfram_2023_functioninterpolation, organization={Wolfram Research}, title={FunctionInterpolation}, year={1996}, url={https://reference.wolfram.com/language/ref/FunctionInterpolation.html}, note=[Accessed: 30-September-2023 ]} | 422 | 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} | 2.515625 | 3 | CC-MAIN-2023-40 | longest | en | 0.52024 |
https://www.eng-tips.com/viewthread.cfm?qid=367502 | 1,545,043,360,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828501.85/warc/CC-MAIN-20181217091227-20181217113227-00186.warc.gz | 884,257,246 | 13,276 | ×
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# Braked dyno project
Forum Search FAQs Links MVPs
## Braked dyno project
(OP)
Hello Im working on a braked dyno build, Im using 2 eddy current brakes with load-cells and a NI DAQ.
Im getting a bit of mess with the software.
Is possible to calculate the power output, knowing the braked torque and the acceleration rate?
Regards
### RE: Braked dyno project
If you are going to calculate transient power you need to know the inertia of all the rotating components in addition to the torque and the acceleration rate.
----------------------------------------
The Help for this program was created in Windows Help format, which depends on a feature that isn't included in this version of Windows.
### RE: Braked dyno project
(OP)
I forgot to mention this part in the post, I do know the inertia of the brakes, each brake inertia is 2.96 Kg/m2
### RE: Braked dyno project
Why would you need to infer torque from acceleration and inertia if you can measure it directly from load cells? I assume you can measure rotational speed too.
- Steve
### RE: Braked dyno project
(OP)
I would like to calculate power without setting steady state, so if I have inertia and braked torque and rotational speed or aceleration, can I calculate power correctly?
### RE: Braked dyno project
I guess you're going to need to know the inertia upstream of the load cell if you want a true transient torque (or power) measurement. The inertia of the dyno downstream isn't relevant, except that it'll dictate the actual acceleration. If you knew the total inertia of the whole system and its acceleration, you shouldn't need the load cell. What's wrong with using the brake to control the speed anyway and take a proper curve? If you want a transient measurement, how transient is transient?
- Steve
### RE: Braked dyno project
(OP)
Im fighting with pid calibration and think about the possibility to calculate the power/torque from the acceleration rate.
Steady with a constant torque reading is hard to get in a short time so if I want to get a lot of points it take time, I have to figure out a lot yet.
I have to calculate coast down to get a estimated flywheel power, but I like to solve before wheel numbers.
### RE: Braked dyno project
The accelerating torque or power can be calculated as follows.
torque = inertia x dN / (308 x dt)
or
P = (inertia x N2^2/2 - inertia x N1^2/2) / dt
Where,
dN is the change is speed
dt is the change in time.
You then have to add this to the eddy brake torque or power.
### RE: Braked dyno project
With an eddy current brake the load cell is usually connected to the stator housing, so the inertia of the drive shaft and rotor must be considered. If you require a precise measure of dynamic engine operating conditions, then you might want to consider using something like a DC dyno instead of an eddy current brake. The DC dyno can be used to both motor and brake the engine. So it is much easier to characterize the frictions and inertias in the engine. A DC dyno also provides more precise control during transient operating conditions than an eddy current brake dyno does.
### RE: Braked dyno project
(OP)
Thank you very much Lionelhutz, dN rad/sec and dt is in seconds, right?
On here, P = (inertia x N2^2/2 - inertia x N1^2/2) / dt
What is ^?
Thank you very much tbuelna for the input, where can I see some information on this DC dyno? I already are working on edd current basis but I can test on future projects.
### RE: Braked dyno project
The first formula dN is the change in rpm and the other units are imperial. Lbs and ft.
^2 means square the value.
The second formula is using SI units. kgm^2 and rad/sec
It would likely help to understand the second formula. Look up rotational kinetic energy and power (physics) on Wikipedia. The rotational kinetic energy is in J or Watt-seconds. The formula takes the change in rotational kinetic energy (dJ) and divides by the time (dt) taken to effect the change to get the power (W) required for that change.
### RE: Braked dyno project
Amelina. I don't see any mention of engine inertia in this thread.
If brake torque is being measured from stator reaction, the inertia of ALL rotating and reciprocating components must be considered. The total inertia can be measured by comparing torque at a given rpm using two different acceleration rates (one of which could be zero or negative) and working backwards.
Engineering is the art of creating things you need, from things you can get.
### RE: Braked dyno project
(OP)
Hello, I had the project a little stopped and now Id like to put it back to live. Sorry for delay.
Mechanicadesingns, Im using Labview software.
Gruntguru, can you give a aproach of how you will make the calculation? the power is normally measured at the flywheel so internal inertia should not be in to the maths.
Any recomendation of the best way to measure the mechanical looses?
### RE: Braked dyno project
The power produced during acceleration is slightly less due to the increasing kinetic energy of the moving parts in the engine. Engines are usually rated for steady-state power beause the reduced power during acceleration will vary depending on the rate of acceleration.
LionelHutz provided a formula above for calculating the difference.
je suis charlie
### RE: Braked dyno project
If you measure the nonaccelerating torque/horsepower and compare it to the results accelerating the engine at a known rate of RPM increase the difference should be a function of the total system inertia -I think.
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Register now while it's still free! | 1,549 | 6,864 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-51 | latest | en | 0.876513 |
https://stackoverflow.com/questions/44103854/how-do-i-get-the-double-square-root-of-a-number/44103872 | 1,563,277,201,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524522.18/warc/CC-MAIN-20190716095720-20190716121720-00550.warc.gz | 567,592,007 | 27,206 | How do I get the double Square root of a number
I need to get a double square root of a number but my codes won't work:
``````function doubleSquareRootOf(num) {
return num * num * num;
}
var output = doubleSquareRootOf(121);
console.log(output); // --> it must return 22 but its displaying 1771561
``````
Any idea?
• What is a “double square root”? Two times the square root? Do you know how to get the square root of a number? – Ry- May 22 '17 at 2:39
• `return num * num * num;` that's num cubed!! `Math.sqrt` or `Math.pow` may be more what you need ... double square root ... `return 2 * Math.sqrt(num);` or `return 2 * Math.pow(num, 0.5)` - do you even know what a square root is? – Jaromanda X May 22 '17 at 2:40
• I suspect he means cube root. – shmosel May 22 '17 at 2:42
• no, because square root of 121 = 11, and 11 x 2 = 22 - look at the expected result – Jaromanda X May 22 '17 at 2:43
• He needs sqrt(num)*2 (that 2 is the 'double') – Robert May 22 '17 at 2:44
``````return Math.sqrt(num)*2; | 327 | 1,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.809708 |
https://studylib.net/doc/9463377/document | 1,601,159,135,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400245109.69/warc/CC-MAIN-20200926200523-20200926230523-00410.warc.gz | 637,443,496 | 16,381 | # Document
Chapter 4
Discounted Cash Flow Valuation
McGraw-Hill/Irwin
Key Concepts and Skills
Be able to compute the future value and/or
present value of a single cash flow or series of
cash flows
Be able to compute the return on an
investment
Be able to use a financial calculator and/or
spreadsheet to solve time value problems
Understand perpetuities and annuities
4-1
Chapter Outline
4.1 Valuation: The One-Period Case
4.2 The Multiperiod Case
4.3 Compounding Periods
4.4 Simplifications
4.5 Loan Amortization
4.6 What Is a Firm Worth?
4-2
4.1 The One-Period Case
If you were to invest \$10,000 at 5-percent interest
for one year, your investment would grow to
\$10,500.
\$500 would be interest (\$10,000 × .05)
\$10,000 is the principal repayment (\$10,000 × 1)
\$10,500 is the total due. It can be calculated as:
\$10,500 = \$10,000×(1.05)
The total amount due at the end of the investment is
call the Future Value (FV).
4-3
Future Value
In the one-period case, the formula for FV can
be written as:
FV = C0×(1 + r)
Where C0 is cash flow today (time zero), and
r is the appropriate interest rate.
4-4
Present Value
If you were to be promised \$10,000 due in one year
when interest rates are 5-percent, your investment
would be worth \$9,523.81 in today’s dollars.
\$10,000
\$9,523.81
1.05
The amount that a borrower would need to set aside
today to be able to meet the promised payment of
\$10,000 in one year is called the Present Value (PV).
Note that \$10,000 = \$9,523.81×(1.05).
4-5
Present Value
In the one-period case, the formula for PV can
be written as:
C1
PV
1 r
Where C1 is cash flow at date 1, and
r is the appropriate interest rate.
4-6
Net Present Value
The Net Present Value (NPV) of an
investment is the present value of the
expected cash flows, less the cost of the
investment.
Suppose an investment that promises to pay
\$10,000 in one year is offered for sale for
\$9,500. Your interest rate is 5%. Should you
4-7
Net Present Value
\$10,000
NPV \$9,500
1.05
NPV \$9,500 \$9,523.81
NPV \$23.81
The present value of the cash inflow is greater
than the cost. In other words, the Net Present
Value is positive, so the investment should be
purchased.
4-8
Net Present Value
In the one-period case, the formula for NPV can be
written as:
NPV = –Cost + PV
If we had not undertaken the positive NPV project
considered on the last slide, and instead invested our
\$9,500 elsewhere at 5 percent, our FV would be less
than the \$10,000 the investment promised, and we
would be worse off in FV terms :
\$9,500×(1.05) = \$9,975 < \$10,000
4-9
4.2 The Multiperiod Case
The general formula for the future value of an
investment over many periods can be written
as:
FV = C0×(1 + r)T
Where
C0 is cash flow at date 0,
r is the appropriate interest rate, and
T is the number of periods over which the cash is
invested.
4-10
Future Value
Suppose a stock currently pays a dividend of
\$1.10, which is expected to grow at 40% per
year for the next five years.
What will the dividend be in five years?
FV = C0×(1 + r)T
\$5.92 = \$1.10×(1.40)5
4-11
Future Value and Compounding
Notice that the dividend in year five, \$5.92,
is considerably higher than the sum of the
original dividend plus five increases of 40percent on the original \$1.10 dividend:
\$5.92 > \$1.10 + 5×[\$1.10×.40] = \$3.30
This is due to compounding.
4-12
Future Value and Compounding
\$1.10 (1.40)
\$1.10 (1.40) 4
5
\$1.10 (1.40)3
\$1.10 (1.40) 2
\$1.10 (1.40)
\$1.10
0
\$1.54 \$2.16 \$3.02
1
2
3
\$4.23
\$5.92
4
5
4-13
Present Value and Discounting
How much would an investor have to set
aside today in order to have \$20,000 five
years from now if the current rate is 15%?
PV
\$20,000
0
1
2
3
4
5
\$20,000
\$9,943.53
5
(1.15)
4-14
4.5 Finding the Number of Periods
If we deposit \$5,000 today in an account paying 10%,
how long does it take to grow to \$10,000?
FV C0 (1 r )
\$10,000 \$5,000 (1.10)
T
T
\$10,000
(1.10)
2
\$5,000
T
ln( 1.10)T ln( 2)
ln( 2)
0.6931
T
7.27 years
ln( 1.10) 0.0953
4-15
What Rate Is Enough?
Assume the total cost of a college education will be
\$50,000 when your child enters college in 12 years.
You have \$5,000 to invest today. What rate of interest
must you earn on your investment to cover the cost of
FV C0 (1 r )
T
\$50,000
(1 r )
10
\$5,000
12
r 10
1 12
\$50,000 \$5,000 (1 r )
12
(1 r ) 101 12
1 1.2115 1 .2115
4-16
Calculator Keys
Texas Instruments BA-II Plus
FV = future value
PV = present value
I/Y = periodic interest rate
P/Y
must equal 1 for the I/Y to be the periodic rate
Interest is entered as a percent, not a decimal
N = number of periods
Remember to clear the registers (CLR TVM) after
each problem
Other calculators are similar in format
4-17
Multiple Cash Flows
Consider an investment that pays \$200 one
year from now, with cash flows increasing by
\$200 per year through year 4. If the interest
rate is 12%, what is the present value of this
stream of cash flows?
If the issuer offers this investment for \$1,500,
should you purchase it?
4-18
Multiple Cash Flows
0
1
200
2
3
4
400
600
800
178.57
318.88
427.07
508.41
1,432.93
Present Value < Cost → Do Not Purchase
4-19
Valuing “Lumpy” Cash Flows
First, set your calculator to 1 payment per year.
Then, use the cash flow menu:
CF0
0
CF3
600
I
CF1
200
F3
1
NPV
F1
1
CF4
800
CF2
400
F4
1
F2
1
12
1,432.93
4-20
4.3 Compounding Periods
Compounding an investment m times a year for
T years provides for future value of wealth:
r
FV C0 1
m
mT
4-21
Compounding Periods
For example, if you invest \$50 for 3 years at
12% compounded semi-annually, your
investment will grow to
.12
FV \$50 1
2
23
\$50 (1.06) 6 \$70.93
4-22
Effective Annual Rates of Interest
A reasonable question to ask in the above
example is “what is the effective annual rate of
interest on that investment?”
.12 23
6
FV \$50 (1
) \$50 (1.06) \$70.93
2
The Effective Annual Rate (EAR) of interest is
the annual rate that would give us the same
end-of-investment wealth after 3 years:
\$50 (1 EAR)3 \$70.93
4-23
Effective Annual Rates of Interest
FV \$50 (1 EAR)3 \$70.93
\$70.93
(1 EAR)
\$50
13
\$70.93
EAR
1 .1236
\$50
3
So, investing at 12.36% compounded annually
is the same as investing at 12% compounded
semi-annually.
4-24
Effective Annual Rates of Interest
Find the Effective Annual Rate (EAR) of an
18% APR loan that is compounded monthly.
What we have is a loan with a monthly
interest rate rate of 1½%.
This is equivalent to a loan with an annual
interest rate of 19.56%.
m
12
r
.18
12
1
1
(
1
.
015
)
1.1956
m
12
4-25
EAR on a Financial Calculator
Texas Instruments BAII Plus
keys:
description:
[2nd] [ICONV]
Sets 12 payments per year
[↑] [C/Y=] 12 [ENTER]
Sets 18 APR.
[↓][NOM=] 18 [ENTER]
[↓] [EFF=] [CPT]
19.56
4-26
Continuous Compounding
The general formula for the future value of an
investment compounded continuously over many
periods can be written as:
FV = C0×erT
Where
C0 is cash flow at date 0,
r is the stated annual interest rate,
T is the number of years, and
e is a transcendental number approximately equal
to 2.718. ex is a key on your calculator.
4-27
4.4 Simplifications
Perpetuity
Growing perpetuity
A stream of cash flows that grows at a constant rate
forever
Annuity
A constant stream of cash flows that lasts forever
A stream of constant cash flows that lasts for a fixed
number of periods
Growing annuity
A stream of cash flows that grows at a constant rate for
a fixed number of periods
4-28
Perpetuity
A constant stream of cash flows that lasts forever
0
C
C
C
1
2
3
…
C
C
C
PV
2
3
(1 r ) (1 r ) (1 r )
C
PV
r
4-29
Perpetuity: Example
What is the value of a British consol that
promises to pay £15 every year for ever?
The interest rate is 10-percent.
0
£15
£15
£15
1
2
3
…
£15
PV
£150
.10
4-30
Growing Perpetuity
A growing stream of cash flows that lasts forever
0
C
C×(1+g)
C ×(1+g)2
1
2
3
…
C
C (1 g ) C (1 g )
PV
2
3
(1 r )
(1 r )
(1 r )
2
C
PV
rg
4-31
Growing Perpetuity: Example
The expected dividend next year is \$1.30, and
dividends are expected to grow at 5% forever.
If the discount rate is 10%, what is the value of this
promised dividend stream?
\$1.30
\$1.30×(1.05)
\$1.30 ×(1.05)2
…
0
1
2
3
\$1.30
PV
\$26.00
.10 .05
4-32
Annuity
A constant stream of cash flows with a fixed maturity
C
C
C
C
0
1
2
3
T
C
C
C
C
PV
2
3
T
(1 r ) (1 r ) (1 r )
(1 r )
C
1
PV 1
T
r (1 r )
4-33
Annuity: Example
If you can afford a \$400 monthly car payment, how
much car can you afford if interest rates are 7% on 36month loans?
\$400
\$400
\$400
\$400
0
1
2
3
36
\$400
1
PV
1
\$12,954.59
36
.07 / 12 (1 .07 12)
4-34
What is the present value of a four-year annuity of \$100 per
year that makes its first payment two years from today if the
discount rate is 9%?
4
PV1
t 1
\$297.22
0
\$100
\$100
\$100
\$100
\$100
\$323.97
t
1
2
3
4
(1.09) (1.09) (1.09) (1.09) (1.09)
\$323.97
1
\$100
2
\$327 .97
PV
\$297 .22
0
1.09
\$100
3
\$100
4
\$100
5
4-35
2-35
Growing Annuity
A growing stream of cash flows with a fixed maturity
C
C×(1+g)
C ×(1+g)2
C×(1+g)T-1
0
1
2
3
T
C
C (1 g )
C (1 g )T 1
PV
2
(1 r )
(1 r )
(1 r )T
T
1 g
C
PV
1
r g (1 r )
4-36
Growing Annuity: Example
A defined-benefit retirement plan offers to pay \$20,000 per year
for 40 years and increase the annual payment by 3% each year.
What is the present value at retirement if the discount rate is
10%?
\$20,000
\$20,000×(1.03) \$20,000×(1.03)39
0
1
2
40
40
\$20,000
1.03
PV
\$265,121.57
1
.10 .03 1.10
4-37
Growing Annuity: Example
You are evaluating an income generating property. Net rent is
received at the end of each year. The first year's rent is
expected to be \$8,500, and rent is expected to increase 7%
each year. What is the present value of the estimated income
stream over the first 5 years if the discount rate is 12%?
\$8,500 (1.07) 2
\$8,500 (1.07) 4
\$8,500 (1.07)
\$8,500 (1.07)3
\$8,500 \$9,095 \$9,731.65 \$10,412.87 \$11,141.77
0
1
\$34,706.26
2
3
4
5
4-38
4.5 Loan Amortization
Pure Discount Loans are the simplest form of loan.
The borrower receives money today and repays a
single lump sum (principal and interest) at a future
time.
Interest-Only Loans require an interest payment each
period, with full principal due at maturity.
Amortized Loans require repayment of principal
over time, in addition to required interest.
4-39
Pure Discount Loans
Treasury bills are excellent examples of pure
discount loans. The principal amount is repaid
at some future date, without any periodic
interest payments.
If a T-bill promises to repay \$10,000 in 12
months and the market interest rate is 7
percent, how much will the bill sell for in the
market?
PV = 10,000 / 1.07 = 9,345.79
4-40
Interest-Only Loan
Consider a 5-year, interest-only loan with a 7%
interest rate. The principal amount is \$10,000.
Interest is paid annually.
What would the stream of cash flows be?
1 – 4: Interest payments of .07(10,000) = 700
Year 5: Interest + principal = 10,700
Years
This cash flow stream is similar to the cash
flows on corporate bonds, and we will talk
about them in greater detail later.
4-41
Amortized Loan with Fixed Principal
Payment
Consider a \$50,000, 10 year loan at 8%
interest. The loan agreement requires the firm
to pay \$5,000 in principal each year plus
interest for that year.
Click on the Excel icon to see the amortization
table
4-42
Amortized Loan with Fixed Payment
Each
payment covers the interest expense plus reduces
principal
Consider a 4 year loan with annual payments. The
interest rate is 8% ,and the principal amount is \$5,000.
What is the annual payment?
Click
4N
8 I/Y
5,000 PV
CPT PMT = -1,509.60
on the Excel icon to see the amortization table
4-43
4.6 What Is a Firm Worth?
Conceptually, a firm should be worth the
present value of the firm’s cash flows.
The tricky part is determining the size, timing,
and risk of those cash flows.
4-44
Quick Quiz
How is the future value of a single cash flow
computed?
How is the present value of a series of cash flows
computed.
What is the Net Present Value of an investment?
What is an EAR, and how is it computed?
What is a perpetuity? An annuity?
4-45 | 5,087 | 12,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-40 | latest | en | 0.919365 |
https://mathematica.stackexchange.com/questions/102324/sphericalharmonicy-does-not-seem-to-be-an-eigenfunction-of-the-spherical-harmoni | 1,721,786,775,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00774.warc.gz | 323,932,352 | 45,129 | # SphericalHarmonicY does not seem to be an eigenfunction of the spherical harmonic equation
I applied the spherical harmonic equation on the SphericalHarmonicY functions like this:
Simplify[(1/Sin[θ]) D[Sin[θ] D[SphericalHarmonicY[l, m, θ, ϕ], θ], θ] +
(1/Sin[θ]^2) D[SphericalHarmonicY[l, m, θ, ϕ], {ϕ, 2}]]
And it returns a sum of factors with $Y^{m}_{l}(\theta, \phi)$, $Y^{m+1}_{l}(\theta, \phi)$ and $Y^{m+2}_{l}(\theta, \phi)$
This is counterintuitive. I thought these were the eigenfunctions of the operator. I feel like I might be missing something basic here.
• Even if one uses FullSimplify[], the problem seems to be in the fact that Mathematica does not seem to recognize a particular difference equation satisfied by the associated Legendre polynomial. Commented Dec 18, 2015 at 2:14
• Using FunctionExpand[Exp[-I m ϕ] ... ] and adding the assumption to FullSimplify that m is an integer also do not help. However, plotting the result over {θ, -Pi/2, Pi/2} certainly yields zero for any integer m. Commented Dec 18, 2015 at 2:36
• @J.M. This seems pretty silly. Apart from expansions in spherical harmonics, it seems to me that Ylm being an eigenfunction of this particular differential operator is the defining feature of the spherical harmonics. It strikes me as a little, I don't know, unorthodox that the Mathematica programmers shouldn't include this naturally behind the scenes. Much less for the Legendre polynomials. Commented Dec 18, 2015 at 2:37
• Another data point: if you use FunctionExpand[FullSimplify[(* stuff *)]] on the full eigenvalue PDE, you get True as the result. So, it can simplify that, but it cannot do the intermediate step of simplifying your expression to a single instance of SphericalHarmonicY[]. Commented Dec 18, 2015 at 13:52
• So is Mathematica doing something about it ? Is there a forum where I can post this problem ? Because I am doing something in AdS4 and simplification seems to depend on this fact. Commented May 8, 2016 at 17:25
The operator in the question can be simplified by first applying it to an arbitrary function f:
Simplify[(1/Sin[θ]) D[Sin[θ] D[f[θ,ϕ],θ],θ]+(1/Sin[θ]^2) D[f[θ,ϕ],{ϕ,2}]]
$$f^{(2,0)}(\theta ,\phi )+\cot (\theta ) f^{(1,0)}(\theta ,\phi )+\csc ^2(\theta ) f^{(0,2)}(\theta ,\phi )$$
This can be recognized as the negative of the squared angular momentum operator, $-L^2$, see below (and see also my answer here). It doesn't seem possible to do the simplification exactly as asked for in the question, but you can put together a Mathematica based proof of the desired eigenvalue property $$L^2 Y^{m}_{\ell} = \ell (\ell +1) Y^{m}_{\ell}$$ in several steps. In doing so, the advantage is that you can see more clearly how this equation is related to the properties of the angular-momentum operator (the brute-force approach in the question would be shorter, but less illuminating, even if it were possible).
The spherical harmonics are simultaneous eigenfunctions of the squared angular momentum and its z component. Here is how you can verify this by asking Mathematica to derive the two corresponding eigenvalues:
First I define the two operators to make the notation in the following steps clearer. Then I apply these operators in the eigenvalue equation with the spherical harmonic as the assumed solution. The proof consists in the fact that Mathematica does indeed find an eigenvalue that makes each relation true.
Clear[θ, ϕ, f, m, ℓ]
lzOp = Function[f, (-I)*D[f, {ϕ}]];
l2Op = Function[f, -D[f, {θ, 2}] - Csc[θ]^2*D[f, {ϕ, 2}] - Cot[θ]*D[f, {θ}]];
Simplify[
lzOp[SphericalHarmonicY[ℓ, m, θ, ϕ]]]
m SphericalHarmonicY[ℓ, m, θ, ϕ]
Simplify[
l2Op[SphericalHarmonicY[ℓ, ℓ, θ, ϕ]]]
ℓ (1 + ℓ) SphericalHarmonicY[ℓ, ℓ, θ, ϕ]
So both lzOp (the z-angular momentum) and l2Op (the squared angular momentum) reproduce the spherical harmonic with a prefactor that corresponds to the correct eigenvalue.
However, the last eigenvalue equation with l2Op (which is what the question focuses on) only works for m = ℓ. For the remaining values of m, I would proceed by using the ladder operators which turn $Y^m_{\ell}$ into $Y^{m\pm 1}_{\ell}$ and also commute with l2Op. These operations will then necessarily leave the eigenvalue ℓ (1 + ℓ) unchanged for l2Op.
To execute this idea, I have to define the lowering operator, lMinus:
lMinus = Function[f, Exp[-I ϕ] (Cot[θ] D[f, ϕ] + I D[f, θ])];
First I now show that lMinus indeed has the effect of lowering the z eigenvalue of any spherical harmonic:
Simplify[
First[Solve[
lzOp[lMinus@SphericalHarmonicY[ℓ, m, θ, ϕ]] ==
eigenvalue lMinus@SphericalHarmonicY[ℓ, m, θ, ϕ],
eigenvalue]
]]
{eigenvalue -> -1 + m}
Here I had to use Solve to extract the eigenvalue property by defining the unknown eigenvalue as a prefactor in front of the expected functional form lMinus@SphericalHarmonicY. Without this trick, the functions are too complicated, and FullSimplify can't bring them into the same form on both sides. Note that I have only cheated a little here, because I didn't tell Mathematica what the new eigenvalue is expected to be. It found that out by itself.
The last eigenvalue equation means that if the original spherical harmonic had index m, then the result of lMinus has index m -1. We could now apply this operator to the "maximal state" $Y^{\ell}_{\ell}$ for which I had already shown that the eigenvalue equation $L^2 Y^{\ell}_{\ell} = \ell (\ell +1) Y^{\ell}_{\ell}$ is satisfied. Then it remains to be shown that the same eigenvalue of $L^2$ is obtained with the function created by repeated application of lMinus (one writes it $L_{-}$). The crucial fact that is needed for that proof is this commutation relation:
$$L^2 L_{-} = L_{-} L^2$$
which I show here by applying both orders to an arbitrary test function $f(\theta, \phi)$:
Simplify[
l2Op[lMinus[f[θ, ϕ]]] == lMinus[l2Op[f[θ, ϕ]]]]
True
With this, I have all the building blocks to deduce that
$$L^2 Y^{m}_{\ell} = \ell (\ell +1) Y^{m}_{\ell}$$
holds. You start by setting $m=\ell$, for which the above is known to be true from earlier. Then apply lMinus on both sides to get the same equation for $m=\ell -1$, and repeat another $2\ell$ times to span the entire angular-momentum ladder of allowed m for the given .
• This is a very indirect way of dealing, what if I am using these functions and want simplifications of eigenvalue equations ? Commented Jun 13, 2016 at 7:04
• @Jaswin The same difficulties arise with other eigenvalue equations that involve different special functions, e.g., the Coulomb problem or the harmonic oscillator. The trick of using Solve to derive the eigenvalue works in all these cases. I need that here to show the properties of lMinus, but it's actually not needed for the other eigenvalue properties (see the edit to my answer). But this is pretty much the simplest form of a proof for the general eigenvalue property that I could come up with.
– Jens
Commented Jun 14, 2016 at 5:10
• @Jaswin, that's why I was prompting you in the comments to include examples in the post. Commented Jun 14, 2016 at 5:14
• @J.M : Actually I was a bit busy, so couldn't update, and the days for giving away bounty points was yesterday, so closed the answer. Will put the updated question soon. Commented Jun 15, 2016 at 5:49
In this wikipedia page you will find the relationship between the spherical harmonics and the associated Legendre Polynomials. The second of the recurrence formula from this page gives you a relationship for same l different m's. That is the origin of this formula to simplify MMA's answer. | 2,077 | 7,560 | {"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} | 2.734375 | 3 | CC-MAIN-2024-30 | latest | en | 0.914688 |
https://www.studypool.com/discuss/1047430/physics-word-problem-8?free | 1,508,713,627,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825473.61/warc/CC-MAIN-20171022222745-20171023002745-00897.warc.gz | 985,808,218 | 14,127 | ##### Physics Word Problem
label Physics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Two objects are connected by a string of
length 4.8 m and negligible mass. The 10 kg
block is placed on a smooth table top(3.6 m from end of table),and the
7kg block hangs over the edge of the table. The 7 kg block is then released from
rest at a distance of 1.2 m above the floor at
time t = 0.
The acceleration of gravity is 9.8 m/s^2. Determine the acceleration of the 7kg block
as it descends.
Answer in units of m/s^2. Round decimal places
Jun 21st, 2015
regard the two objects as single object, then we have
7*g = 17a
a = 4.04 m/s^2
Jun 21st, 2015
...
Jun 21st, 2015
...
Jun 21st, 2015
Oct 22nd, 2017
check_circle | 238 | 737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-43 | latest | en | 0.886186 |
http://research.omicsgroup.org/index.php/Data_reporting | 1,556,224,204,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578733077.68/warc/CC-MAIN-20190425193912-20190425215912-00399.warc.gz | 154,066,683 | 8,822 | # Data reporting
Data reporting is the process of collecting and submitting data to authorities entrusted with compiling statistics. Accurate data reporting gives rise to accurate analyses of the facts on the ground; inaccurate data reporting can lead to vastly uninformed decisions based on erroneous evidence. When data is not reported, the problem is known as underreporting; the opposite problem leads to false positives.
Data reporting can be an incredibly difficult endeavor. Census bureaus may hire even hundreds of thousands of workers to achieve the task of counting all of the residents of a country.[1][2] Teachers use data from student assessments to determine grades; cellphone manufacturers rely on sales data from retailers to point the way to which models to increase production of. The effective management of nearly any company relies on accurate data.
## Problems arising from poor data reporting
Poor data reporting leads to flawed information in the hands of decision-makers and journalists, especially when they fail to scrutinize the data and blindly trust the assertions of the data reporters. For example, poor data reporting from a crowd-sourced website led to a highly incorrect article by FiveThirtyEight journalist Walt Hickey.
Hickey's debut article for the site, which ironically focuses on properly interpreting statistics, attempted to determine the entrée item at McDonalds that provided the highest calorie count for the money. The mathematical techniques that he used (dividing the calorie total by the price) were flawless, as were the calorie counts, which McDonald's provides as the law requires and which serve as an example of accurate data reporting.[3] However, when Hickey looked to collect data on prices at McDonald's, he relied on a website[4] that in turn relied on unpaid "crowd-sourced" data reporting.
The data reporting was inaccurate, including the improbable assertion that a two-patty cheeseburger cost less than a one-patty cheeseburger. Since Hickey failed to review or scrutinize the incorrectly reported data, his analysis failed on multiple levels, leading to several blatantly false conclusions. After publication, reader complaints convinced Hickey to insert a correction for the prices of one of the burgers, but failed to notice any of the other errors in his article. As none of the site's editors (including noted statistician Nate Silver) caught the error, the inaccurate analysis remained and indeed still remains available online.[5] | 468 | 2,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-18 | latest | en | 0.95808 |
https://www.albert.io/ie/act-math/average-speed-for-an-entire-trip | 1,488,054,862,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171834.68/warc/CC-MAIN-20170219104611-00318-ip-10-171-10-108.ec2.internal.warc.gz | 781,650,862 | 13,180 | Free Version
Difficult
# Average Speed for an Entire Trip
ACTMAT-Q14FNH
Ethan drove the first half of his trip at an average rate of 60 miles per hour, and the second half at an average rate of 70 miles per hour.
What was Ethan's average speed for the entire trip, in miles per hour?
A
$65$
B
$64.6$
C
$65.4$
D
$65.8$
E
$64.2$ | 106 | 340 | {"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} | 2.703125 | 3 | CC-MAIN-2017-09 | latest | en | 0.921713 |
http://www.hiveworkshop.com/forums/archive/index.php/t-19658.html | 1,369,305,140,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703227943/warc/CC-MAIN-20130516112027-00097-ip-10-60-113-184.ec2.internal.warc.gz | 504,416,449 | 3,562 | PDA
View Full Version : Array variable trouble
Lingonberryjam
08-22-2005, 06:00 AM
I have used arrays for a long time now, and it works perect. But now i have a problem. Im making a simple "choose random hero for player" trigger.
It works, and ive used a unit type array variable to create the random hero. But how do i make sure that the choosen hero does not get created again so 2 players gets the same hero? It's a bit of math for me to realize how to do it...
If it helps the trigger simply looks like this:
Event - A unit sells a unit
Conditions - (Unit-type of (Sold unit)) Equal to Random Hero
Actions
-------- Set variables --------
Set randomhero[1] = Hero1
Set randomhero[2] = Hero2
Set randomhero[3] = Hero3
Set randomhero[4] = Hero4
Set randomhero[5] = Hero5
Set randomhero[6] = Hero6
Set randomhero[7] = Hero7
Set randomhero[8] = Hero8
Set randomhero[9] = Hero9
Set randomhero[10] = Hero10
Set randomhero[11] = Hero11
Set randomhero[12] = Hero12
Set randomhero[13] = Hero13
Set randomhero[14] = Hero14
Set randomhero[15] = Hero15
Set randomhero[16] = Hero16
Set randomhero[17] = Hero17
Set randomhero[18] = Hero18
Set randomhero[19] = Hero19
Set randomhero[20] = Hero20
Set randomhero[21] = Hero21
Set randomhero[22] = Hero22
Set Numberofheroes = 22
-------- Give Hero --------
If (((Owner of (Sold unit)) is an ally of Player 1 (Red)) Equal to True) then do (Unit - Create 1 randomhero[(Random integer number between 1 and Numberofheroes)] for (Owner of (Sold unit)) at (Center of Southern Shop Spot <gen>) facing Default building facing degrees) else do (Do nothing)
If (((Owner of (Sold unit)) is an ally of Player 7 (Green)) Equal to True) then do (Unit - Create 1 randomhero[(Random integer number between 1 and Numberofheroes)] for (Owner of (Sold unit)) at (Center of Northern Shop Spot <gen>) facing Default building facing degrees) else do (Do nothing)
Unit - Remove (Sold unit) from the game
Laosh'Ra
08-22-2005, 09:58 AM
make a boolean variable: e.g. if hero #17 is picked hero_is_picked[17]=true. if the variable was already true before: do nothing except just running the same trigger again! this will repeat the selection until the random picks a heronumber which isnt occupied already.
just 2 things u have to check:
1. make sure there are at least as much heroes as players or the game will freeze
2. dont put any effects/performance-needing actions in this trigger so theres no lag even if the trigger is executed 10 times.
Lingonberryjam
08-24-2005, 07:34 AM
I dont really get it right. It's hard to understand excatly what you mean. Should i onle create a hero for 1 player in each trigger round and repeat 10 times? And should the boloean variable be an array variable? Please come up with the trigger code here instead...
imported_Bort
08-24-2005, 01:18 PM
you should really have a variable set up trigger, not set the variables over and over.
So a trigger like this:
Event - Map intialisation
Conditions -
Actions - Set randomhero[1]=(hero)
Set randomhero[2]=(hero)
etc etc
Event - A unit sells a unit
Conditions - unit-type of sold unit equal to Random Hero
Actions - set tempint[player number of owner of sold unit] = random number between 1 and x
- If heropicked[Tempint[Playernumber of owner of sold unit]] equal to true
then
- run this trigger
else
- if owner of sold unit is an ally of player [1]
then
- create 1 randomhero[temptint[player number of owner of sold unit]] at centre of team1region
else
- create 1 randomhero[temptint[player number of owner of sold unit]] at centre of team2region
- remove sold unit from the game
there... that should work alright, i think, but dont complain if it doesnt
=][= Bort
Lingonberryjam
08-24-2005, 01:54 PM
Thanks alot! It seems like this 1 will work, i havent tested it yet.
But if im making an allrandom mode it gets harder.I would need to compare that the player that gets a hero really is playing, "slot status playing", and make sure that the trigger only is used 10 times maximum... so the map dont freezes or crashes.
All ya good mapmakers take this as a challenge! :P Nah but please tell me how to make that one. The only differnce is that the event is string type, Blue types "-ar" as an exact match.
Los_Tacos
08-25-2005, 12:06 AM
(Part of this was copy/pasted from Bort's trigger and modified.)
Events: Player 2 (Blue) types (-ar) as an exact match
Condtitions:
Actions: For each integer A from 1 - 12 do:
If: - Player (Integer A) slot status = is playing
- Player (Integer A) controller = user
Then: - set tempint[Integer A] = random number between 1 and x
- If heropicked[Tempint[Integer A]] equal to true
then
- run this trigger
else
- if owner of sold unit is an ally of player [1]
then
- create 1 randomhero[temptint[Integer A]] at centre of team1region
else
- create 1 randomhero[temptint[Integer A]] at centre of team2region
- remove sold unit from the game
Else: (nothing)
You shouldn't worry about it crashing if there are more heroes than there are players.
Lingonberryjam
08-31-2005, 07:57 PM
There is a problem... :?
Its the simple thing that player 1 and 7 are creep computer players. How do i make the trigger skip those players in a smothy way?
Los_Tacos
09-01-2005, 02:55 AM
(Part of this was copy/pasted from Bort's trigger and modified.)
Events: Player 2 (Blue) types (-ar) as an exact match
Condtitions:
Actions: For each integer A from 1 - 12 do:
If: - Player (Integer A) slot status = is playing
- Player (Integer A) controller = user
Then: - set tempint[Integer A] = random number between 1 and x
- If heropicked[Tempint[Integer A]] equal to true
then
- run this trigger
else
- if owner of sold unit is an ally of player [1]
then
- create 1 randomhero[temptint[Integer A]] at centre of team1region
else
- create 1 randomhero[temptint[Integer A]] at centre of team2region
- remove sold unit from the game
Else: (nothing)
You shouldn't worry about it crashing if there are more heroes than there are players.
Lingonberryjam
09-01-2005, 07:00 PM
Omg, sry. Its just that i havent had time to test it yet so i just wrote to get answer on didnt know it would work. :oops:
Thanks alot anyway.
Lingonberryjam
09-01-2005, 07:54 PM
Ok i think i've got it at last | 1,739 | 6,195 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2013-20 | latest | en | 0.631644 |
http://images.planetmath.org/classstructure | 1,529,315,203,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860168.62/warc/CC-MAIN-20180618090026-20180618110026-00015.warc.gz | 157,976,734 | 3,564 | # class structure
Let $(X_{n})_{n\geq 1}$ be a stationary Markov chain and let $i$ and $j$ be states in the indexing set. We say that $i$ leads to $j$ or $j$ is accessible from $i$, and write $i\to j$, if it is possible for the chain to get from state $i$ to state $j$:
$i\to j\iff P(X_{n}=j:X_{0}=i)>0\quad\textrm{for some}\quad n\geq 0$
If $i\to j$ and $j\to i$ we say $i$ communicates with $j$ and write $i\leftrightarrow j$. $\leftrightarrow$ is an equivalence relation (easy to prove). The equivalence classes of this relation are the communicating classes of the chain. If there is just one class, we say the chain is an irreducible chain.
A class $C$ is a closed class if $i\in C$ and $i\to j$ implies that $j\in C$ “Once the chain enters a closed class, it cannot leave it”
A state $i$ is an absorbing state if $\{i\}$ is a closed class.
Title class structure ClassStructure 2013-03-22 14:18:21 2013-03-22 14:18:21 CWoo (3771) CWoo (3771) 12 CWoo (3771) Definition msc 60J10 MarkovChain communicating class irreducible chain closed class absorbing state | 337 | 1,068 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 23, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-26 | latest | en | 0.743098 |
http://slidegur.com/doc/144406/chapter-15--part-b | 1,542,574,854,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744649.77/warc/CC-MAIN-20181118201101-20181118222422-00035.warc.gz | 304,995,434 | 10,779 | ### Chapter 15, Part B
```Slides by
John
Loucks
St. Edward’s
University
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 1
Chapter 15, Part B
Forecasting
Trend Projection
Seasonality and Trend
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 2
Linear Trend Model
If a time series exhibits a linear trend, curve fitting can
be used to develop a best-fitting linear trend line.
Curve fitting minimizes the sum of squared error
between the observed and fitted time series data where
the model is a trend line.
We build a nonlinear optimization model to find the
best values for the intercept and slope of the trend line.
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 3
Linear Trend Model
The linear trend line is estimated by the equation:
Tt = b0 + b1t
where:
Tt = linear trend forecast in period t
b0 = intercept of the linear trend line
b1 = slope of the linear trend line
t = time period
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 4
Linear Trend Model
Nonlinear Curve-Fitting Optimization Model
min
n
2
(
Y
T
)
t t
t 1
s.t.
Tt = b0 + b1t
t = 1, 2, 3, … n
There are n + 2 decision variables.
The decision variables are b0, b1, and Tt .
There are n constraints.
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 5
Linear Trend Model
Example: Auger’s Plumbing Service
The number of plumbing repair jobs performed by
Auger's Plumbing Service in the last nine months is
listed on the right.
Month Jobs Month
Jobs
Forecast the number of
409
March 353 August
repair jobs Auger's will
April 387 September 399
perform in December
May
342 October
412
using the least squares
June 374 November 408
method.
July 396
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 6
Linear Trend Model
Example: Auger’s Plumbing Service
The objective function minimizes the sum of the
squared error.
Minimize { (353 – T1)2 + (387 – T2)2 + (342 – T3)2
+ (374 – T4)2 + (396 – T5)2 + (409 – T6)2
+ (399 – T7)2 + (412 – T8)2 + (408 – T9)2 }
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 7
Linear Trend Model
Example: Auger’s Plumbing Service
The following constraints define the forecasts as a
linear function of parameters b0 and b1.
T1 = b 0 + b 1 1
T2 = b 0 + b 1 2
T3 = b 0 + b 1 3
T4 = b 0 + b 1 4
T5 = b 0 + b 1 5
T6 = b0 + b16
T7 = b0 + b17
T8 = b0 + b18
T9 = b0 + b19
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 8
Trend Projection
Example: Auger’s Plumbing Service
The solution to the nonlinear curve-fitting
optimization model is:
b0 = 349.667 and b1= 7.4
T1 = 357.07
T2 = 364.47
T3 = 371.87
T4 = 379.27
T5 = 386.67
T6 = 394.07
T7 = 401.47
T8 = 408.87
T9 = 416.27
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 9
Trend Projection
Forecast Accuracy
Trend Forecast Absolute Squared Abs.%
Month Jobs Forecast
Error
Error
Error Error
Mar.
Apr.
May
Jun.
Jul.
Aug.
Sep.
Oct.
Nov.
353
387
342
374
396
409
399
412
408
357.07
364.47
371.87
379.27
386.67
394.07
401.47
408.87
416.27
Total
-4.07
22.53
-29.87
-5.27
9.33
14.93
-2.47
3.13
-8.27
0.00
4.07
22.53
29.87
5.27
9.33
14.93
2.47
3.13
8.27
99.87
16.54 1.15
507.75 5.82
892.02 8.73
27.74 1.41
87.11 2.36
223.00 3.65
6.08 0.62
9.82 0.76
68.34 2.03
1838.40 26.53
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 10
Trend Projection
Forecast Accuracy
MAE
99.87
11.10
9
1838.40
MSE
204.27
9
26.53
MAPE
2.95%
9
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 11
Nonlinear Trend Regression
Sometimes time series have a curvilinear or nonlinear
trend.
A variety of nonlinear functions can be used to
develop an estimate of the trend in a time series.
One example is this quadratic trend equation:
Tt = b0 + b1t + b2t2
Another example is this exponential trend equation:
Tt = b0(b1)t
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 12
Nonlinear Trend Regression
Example: Cholesterol Drug Sales
Consider the annual revenue in millions of dollars
for a cholesterol drug for the first ten years of sales.
This data indicates an
Year Sales Year Sales
overall increasing trend.
6
43.2
1
23.1
A curvilinear function
2
21.3
7
59.5
appears to be needed to
3
27.4
8
64.4
model the long-term
4
34.6
9
74.2
trend.
5
33.8
9
99.3
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 13
Model Formulation
Min { (23.1 T1 ) 2 (21.3 T2 ) 2 (27.4 T3 ) 2 (34.6 T4 ) 2
(33.8 T5 )2 (43.2 T6 ) 2 (59.5 T7 ) 2 (64.4 T8 ) 2
(74.2 T9 )2 (99.3 T10 ) 2 }
s.t.
T1 b0 b11 b21
T6 b0 b1 6 b2 36
T2 b0 b1 2 b2 4
T7 b0 b1 7 b2 49
T3 b0 b1 3 b2 9
T8 b0 b1 8 b2 64
T4 b0 b1 4 b216
T9 b0 b1 9 b2 81
T5 b0 b1 5 b2 25
T10 b0 b110 b2100
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 14
Model Solution
This model can be solved with Excel Solver or LINGO.
The optimal values are:
b0 = 24.182, b1 = -2.11, b2 = .922
Sum of squared errors = 110.65
MSE = 110.65/10 = 11.065
The fitted curve is:
Tt = 24.182 – 2.11 t + .922 t 2
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 15
Exponential Trend Equation
Model Formulation
Min { (23.1 T1 ) 2 (21.3 T2 ) 2 (27.4 T3 ) 2 (34.6 T4 ) 2
(33.8 T5 ) 2 (43.2 T6 ) 2 (59.5 T7 ) 2 (64.4 T8 ) 2
(74.2 T9 ) 2 (99.3 T10 ) 2 }
s.t.
T1 b0 b11
T6 b0 b16
T2 b0 b12
T7 b0 b17
T3 b0 b13
T8 b0 b18
T4 b0 b14
T9 b0 b19
T5 b0 b15
T10 b0 b110
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 16
Exponential Trend Equation
Solution
This model can be solved with Excel Solver or LINGO.
The optimal values are:
b0 = 15.42, b1 = 1.20
Sum of squared errors = 123.12
MSE = 123.12/10 = 12.312
The fitted curve is:
Tt = 15.42(1.20)t
Based on MSE, the quadratic model provides
a better fit than the exponential model.
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 17
Seasonality without Trend
To the extent that seasonality exists, we need to
incorporate it into our forecasting models to ensure
accurate forecasts.
We will first look at the case of a seasonal time series
with no trend and then discuss how to model
seasonality with trend.
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 18
Seasonality without Trend
Example: Umbrella Sales
Year
Quarter 1
Quarter 2
Quarter 3
Quarter 4
1
125
153
106
88
2
118
161
133
102
3
138
144
113
80
4
109
137
125
109
5
130
165
128
96
Sometimes it is difficult to identify patterns in a time
series presented in a table.
Plotting the time series can be very informative.
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 19
Seasonality without Trend
Umbrella Sales Time Series Plot
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 20
Seasonality without Trend
The time series plot does not indicate any long-term
trend in sales.
However, close inspection of the plot does reveal a
seasonal pattern.
The first and third quarters have moderate sales,
the second quarter the highest sales, and
the fourth quarter tends to be the lowest quarter
in terms of sales.
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 21
Seasonality without Trend
We will treat the season as a categorical variable.
Recall that when a categorical variable has k levels,
k – 1 dummy variables are required.
If there are four seasons, we need three dummy
variables.
Qtr1 = 1 if Quarter 1, 0 otherwise
Qtr2 = 1 if Quarter 2, 0 otherwise
Qtr3 = 1 if Quarter 3, 0 otherwise
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 22
Seasonality without Trend
General Form of the Equation is:
Ft b0 b1 (Qtr 1t ) b2 (Qtr 2t ) b3 (Qtr 3t )
Optimal Model is:
Salest 95.0 29.0(Qtr1t ) 57.0(Qtr 2t ) 26.0(Qtr 3t )
The forecasts of quarterly sales in year 6 are:
Quarter 1: Sales = 95 + 29(1) + 57(0) + 26(0) =
Quarter 2: Sales = 95 + 29(0) + 57(1) + 26(0) =
Quarter 3: Sales = 95 + 29(0) + 57(0) + 26(1) =
Quarter 4: Sales = 95 + 29(0) + 57(0) + 26(0) =
or duplicated, or posted to a publicly accessible website, in whole or in part.
124
152
121
95
Slide 23
Seasonality without Trend
Model Formulation
Min { (125 F1 ) 2 (153 F2 ) 2 (106 F3 ) 2
(96 F20 ) 2 }
s.t.
F1 b0 1b1 0b2 0b3
F2 b0 0b1 1b2 0b3
F3 b0 0b1 0b2 1b3
F4 b0 0b1 0b2 0b3
F19 b0 0b1 0b2 1b3
F20 b0 0b1 0b2 0b3
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 24
Seasonality and Trend
We will now extend the curve-fitting approach to
include situations where the time series contains both
a seasonal effect and a linear trend.
We will introduce an additional variable to represent
time.
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 25
Seasonality and Trend
Example: Terry’s Tie Shop
Business at Terry's Tie Shop can be viewed as
falling into three distinct seasons: (1) Christmas
(November and December); (2) Father's Day (late
May to mid June); and (3) all other times.
Average weekly sales (\$) during each of the
three seasons during the past four years are shown
on the next slide.
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 26
Seasonality and Trend
Example: Terry’s Tie Shop
Year
1
2
3
4
Season
1
2
1856 2012
1995 2168
2241 2306
2280 2408
3
985
1072
1105
1120
Determine a forecast for the average weekly
sales in year 5 for each of the three seasons.
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 27
Seasonality and Trend
There are three seasons, so we will need two dummy
variables.
Seas1t = 1 if Season 1 in time period t, 0 otherwise
Seas2t = 1 if Season 2 in time period t, 0 otherwise
General Form of the Equation is:
Ft b0 b1 (Seas1t ) b2 (Seas2t ) b3 (t )
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 28
Seasonality and Trend
Model Formulation
Min { (1856 F1 ) 2 (2012 F2 ) 2 (985 F3 ) 2
(1120 F12 ) 2 }
s.t.
F1 b0 1b1 0b2 b31
F2 b0 0b1 1b2 b3 2
F3 b0 0b1 0b2 b3 3
F10 b0 1b1 0b2 b310
F11 b0 0b1 1b2 b311
F11 b0 0b1 0b2 b312
or duplicated, or posted to a publicly accessible website, in whole or in part.
Slide 29
Seasonality and Trend
Optimal Model
Salest 797.0 1095.43(Seas1t ) 1189.47(Seas2t ) 36.47(t)
The forecasts of average weekly sales in the three
seasons of year 5 (time periods 13, 14, and 15) are:
Seas. 1: Sales13 = 797 + 1095.43(1) + 1189.47(0) + 36.47(13)
= 2366.5
Seas. 2: Sales14 = 797 + 1095.43(0) + 1189.47(1) + 36.47(14)
= 2497.0
Seas. 3: Sales15 = 797 + 1095.43(0) + 1189.47(0) + 36.47(15)
= 1344.0 | 4,529 | 11,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-47 | latest | en | 0.830782 |
https://www.cutlergrp.com/how-long-does-it-take-for-bac-to-reach-0-calculator/ | 1,716,761,637,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058973.42/warc/CC-MAIN-20240526200821-20240526230821-00554.warc.gz | 627,455,083 | 10,458 | Table of Contents
## How long does it take for BAC to reach 0 calculator?
Since your body (and every body) metabolizes alcohol at 0.016% per hour, it will take 10 hours for a person with a BAC or 0.016 to reach a BAC of 0.00.
## How long does it take to go from .08 BAC to 0?
Thus, it would take an hour and a half for his BAC to fall to zero. Or say the same person drinks 8 cans of beer in 4 hours. It would then take approximately 10 hours for his BAC to reach zero.
How long will it take for breathalyzer to register 0.00 after drinking Reddit?
So the answer to the question, How long after you stop drinking can you pass an ignition interlock device test? is anywhere from 12 to 24 hours after your last drink. So really, you shouldn’t plan to drive unless it’s been at least 12 hours since your last drink.
### What is the BAC level for a zero tolerance conviction?
Zero-tolerance laws make it a criminal DUI offense for drivers under the age of 21 to drive with even a small amount of alcohol in their system, ranging from 0.00 to 0.02 percent BAC depending on the state. In light of such laws, even an innocent glass of wine with dinner could subject a young driver with a DUI charge.
### How long does it take for 750ml of wine to leave your system?
The average urine test can detect alcohol between 12 and 48 hours after drinking. More advanced testing can measure alcohol in the urine 80 hours after you drink. Breath tests for alcohol can detect alcohol within a shorter time frame. This is about 24 hours on average.
How do I get my BAC down quickly?
Try drinking a glass of water, soda, or juice in between alcoholic drinks. Spacing out your drinks allows your liver time to break down the alcohol.
#### What is the alcohol blood level chart?
Alcohol Impairment Chart
Men
Approximate Blood Alcohol Percentage
9 .34 .15
10 .38 .17
Your body can get rid of one drink per hour. One drink is 1.5 oz. of 80 proof liquor, 12 oz. of beer, or 5 oz. of table wine.
#### What does Zero Tolerance BAC mean?
Zero Tolerance essentially means that in these cases, any measurable amount of alcohol in your blood is enough to result in a DUI conviction. This is a departure from most DUI law. In general, in California you may be able to drive with a small amount of alcohol in your system and still be “legal” to drive. | 568 | 2,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-22 | latest | en | 0.926469 |
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# jolo309
214
10
1. ## Climbing
You should move the parent, not the child. If you only move the child then it will do what you're experiencing. Play the animation on the child like you do now, but move the parent instead of the child
2. ## Check objects position
What you did were breaking out of the foreach allowed positions loop when the object wasn't at one of the allowed positions. Here's an explanation Let's say we have 3 spheres, and 3 allowed positions: 1st sphere loops through the allowed positions: - 1st allowed position failed, because the sphere wasn't at that position. oh now it breaks the loop. - So that means already at the first test, it breaks out of the allowed position loop and won't compare with the other allowed positions. 2nd sphere loops through the allowed position: - 1st allowed position passed, it continues the loop even though it found a match. - 2nd allowed position failed, it breaks the loop and sets the test to false. 3rd sphere loops through the allowed position: - 1st allowed position passed, it continues the loop even though it found a match. - 2nd allowed position failed, it breaks the loop and sets the test to false. so it will always return false with the code you posted. Well not always, the only way it can return true is if the 3rd sphere's position matches the last allowed position. That's the only time it will be true Made a few adjustments, now it should work public bool checkSequence() { bool test = false; foreach (Transform t in spheres) { test = false; foreach (Vector3 p in pos) { if (p == t.position) { test = true; break; } } if (!test) { break; } } return test; } Now with the new code, let's say we have 3 spheres, and 3 allowed positions: 1st sphere loops through the allowed positions: - 1st allowed position failed, because the sphere wasn't at that position. - 2nd allowed position passed. Breaks out of the loop. 2nd sphere loops through the allowed position: - 1st allowed position passed. Breaks out of the loop. 3rd sphere loops through the allowed position: - 1st allowed position failed, because the sphere wasn't at that position. - 2nd allowed position passed. Breaks out of the loop. The method returns true Here's another where it will fail: 1st sphere loops through the allowed positions: - 1st allowed position failed, because the sphere wasn't at that position. - 2nd allowed position passed. Breaks out of the loop. 2nd sphere loops through the allowed position: - 1st allowed position failed, because the sphere wasn't at that position. - 2nd allowed position failed, because the sphere wasn't at that position. - 3rd allowed position failed, because the sphere wasn't at that position. - This sphere wasn't in any of the allowed positions, so now it breaks out of the sphere loop too and the function returns false
3. ## Check objects position
Do you want to check if an object's position in the object array matches one of the vectors in the vector3 array? Like so: is object[0] position == vector3[0] is object[0] position == vector3[1] is object[0] position == vector3[2] is object[1] position == vector3[0] is object[1] position == vector3[1] is object[1] position == vector3[2] etc... Or is it more like: is object[0] position == vector3[0] is object[1] position == vector3[1] is object[2] position == vector3[2] etc... Or what do you mean?
4. ## [C#]Shoot and reload script (Rigidbody and Raycast version)
Not any that I know of
No
7. ## [C#]Shoot and reload script (Rigidbody and Raycast version)
The weapon script uses the legacy version of the animation system How to use animations: Make sure your weapon's animations uses the legacy animation system Attach the script to a weapon, and find the animations section. Where it says Fire Animation in the inspector, put in the name of the fire animation clip Do the same for the other animations. Shooting: The shooting uses the center of the screen, so you don't have to set any point to where the bullets go. I made the script quite a while ago, so there's a lot that could be improved, pretty much everything. Example. The way I made recoil in the script is by using the screen, which is an awful way of doing it because some may have a big screen, and they will have less recoil than those with a small screen.
8. ## Tank head rotate angle ?
Try changing var rotation = Quaternion.LookRotation(target.position - transform.position); to var rotation = -Quaternion.LookRotation(target.position - transform.position); or var rotation = Quaternion.LookRotation(transform.position - target.position); Both ways should result in same behaviour
9. ## 2.5d Ping Pong
New update: Significant improvement to the controls, which means no more slidey or delayed controls Bug fixes Optimizations Download and webgl links is in the first post
10. ## 2.5d Ping Pong
Hello, I have made a simple 2.5d Ping Pong game, and it was pretty fun to make. You can play with your friend locally, using the same keyboard. The one who has the most points when the timer ends, will win. Feedback is appreciated Features: Player vs computer - Play versus the computer(you can choose easy, medium, hard difficulties). It is not really difficult, even on hard Local 2 player - Play with a friend using the same keyboard. Player 1 uses WASD, and player 2 uses the arrow keys High speed effects - When a ball enters high speed mode, it will have impact effects when hitting the pads, and have an effect surrounding the ball. The camera will shake upon impact, which can be disabled Simple anti stuck - The ball resets when taking more than 4 seconds before it gets hit Powerups and powerdowns - You can disable it if you don't want play with powerups and powerdowns Screenshots: A video showing the gameplay. The fire trail in the video has been shortened in the downloadable version: Controls: WS - Player 1 controls Arrow keys up/down - Player 2 controls P - Pause game Download: Ping Pong 2.5D [Public Release v1.0] Ping Pong 2.5D [Public Release v1.1] Ping Pong 2.5D [Public Release v1.2] Ping Pong 2.5D [Public Release v1.3] WebGL: Newgrounds - v1.3 If you find any bugs please post them, and I'll fix them asap.
11. ## 2.5d Ping Pong
New update: Powerups and powerdowns | Right now there is 3 powerups and 1 powerdown. Size up, 1HP shield, 2HP Shield, Controls inverter Main menu changes A few bug fixes Powerups and powerdowns spawns randomly. If the ball hits a powerup or powerdown, then the player that hit the ball the most recent gets it. Download is in the first post I have also uploaded the new version to newgrounds, so you won't have to download it, plus the game got through Judgement
12. ## 2.5d Ping Pong
I have submitted the game to newgrounds, so you can play it without having to download anything. Link to the game The game is "Under judgement", so I'm not sure if it is going to be blammed, it depends on the ratings, but you can try it out. Oh, and one more thing. It doesn't run as smooth as it does in the downloaded version, but it is still playable, I have no problem with it
13. ## 2.5d Ping Pong
Alright, here's the new update. Took a bit longer than expected, as I wanted the hard difficulty to be harder than easy and medium, without having to increase the speed of the AI. Every difficulty has the same speed, but the hard, and medium difficulty is a bit more human like, and tries to predict somewhat where the ball ends up. [FIXED] When you lost, it would print "You won!" instead of "You lost!" when you lost. [ADDED] A pause menu. [AI] Improvements for hard, and medium difficulty. [FIXED] When the ball is preparing to shoot, the timer didn't stop. Next update I will focus on improving the gameplay a bit, and that update might be the last, not sure. This game was just something I wanted to make, for anyone that wants to have something to play with a friend or something when bored, and because I wanted to see how it would be to make something simple, and nothing overwhelming. I will probably share this on something like kongregate too, and I might even make an android version, but I'm not sure yet. Download: Ping Pong 2.5D [Public Release v1.1]
14. ## 2.5d Ping Pong
Thanks, will try to think of something to make it more interesting
15. ## HI
OnTriggerEnter requires an argument, that's why when you try to call it without any arguments it is throwing an error. Anyway, you shouldn't be calling OnTriggerEnter, because it gets called automatically when something enters the trigger. If you just want to explode the claymore, what you could do is create a new method and name it something like "Explode", and then move everything that is responsible for the claymore exploding, into the new method Then you can call that method anywhere you want to explode the claymore
16. ## 2.5d Ping Pong
I will upload a new update, soon. I'll see if I can finish it today, and if not, then tomorrow. The new update will have a few fixes, and additions. [FIXED] When you lost, it would print "You won!" instead of "You lost!" when you lost. [ADDED] A pause menu. [AI] Improvements for hard difficulty, without speeding up the AI. [FIXED] When the ball is preparing to shoot, the timer didn't stop. Please give me some feedback, I would really appreciate it
Hello, a bit over an hour ago, I signed up on codefights, and I got to a thing where you have to get the century from year. Which was pretty easy, and there's soo many ways to do it. Anyway, after I came up with a solution, I wanted to see how other people solved it, and the most interesting one I saw were return --year / 100 + 1 How does this work? I know -------------------------------- year = 1800 year / 100 + 1 = 19 expected = 18 -------------------------------- year = 1801 year / 100 + 1 = 19 expected = 19 -------------------------------- but if I add the -- before year, then it will output -------------------------------- year = 1800 --year / 100 + 1 = 18 expected = 18 -------------------------------- year = 1824 --year / 100 + 1 = 19 expected = 19 -------------------------------- How come after -- has been added before year, it all gets decremented kind of like, if year % 100 == 0 then decrement year / 100 + 1. Am I completely off? This just confuses me a bit, and it just makes me nuts, not knowing how it works Please help me to make me understand this, would really appreciate it I have seen it in for loops too(Not often though, only seen it being used in for loops 2 times), like so for(int i = 5; i > 0; --i) { // Blah blah } I have seen it with ++i too
Thanks, makes completely sense now
19. ## Steam Summer Sales
Yeah I haven't played half of the games in my library :-D I usually just buy games on sale, especially when the summer sales is here, thinking I'm going to play them but there's a lot of them I haven't even played I have 298 games in my library
21. ## Topdown wave survival game
Hello everyone, I want to make something else, and I have so many dead projects just lying on my HDD that I want to check out and see if I can do something interresting with if it is a project I would like to work on. So I'm just going finish the update I'm working on now over the next few days, and then release it. There will be more updates, but it will mostly be small updates like, bug fixes, tweaks, and new items. I might make new bigger updates later on, but after the update I'm working on now, it will just be smaller updates until I feel like working on bigger updates for this game I might still make a non GameJolt version of the game, if a lot of people wants that. EDIT: To be honest I didn't have much motivation, it comes and goes. I have written a lot of new ideas down, and planned a few more updates.
22. ## [C#]Shoot and reload script (Rigidbody and Raycast version)
Both should work
Thanks | 2,876 | 11,915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-05 | latest | en | 0.904796 |
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GROUPS:
A monad for Quantile Regression workflows
Posted 6 years ago
Introduction
In this document we describe the design and implementation of a (software programming) monad for Quantile Regression workflows specification and execution. The design and implementation are done with Mathematica / Wolfram Language (WL).
What is Quantile Regression? : Assume we have a set of two dimensional points each point being a pair of an independent variable value and a dependent variable value. We want to find a curve that is a function of the independent variable that splits the points in such a way that, say, 30% of the points are above that curve. This is done with Quantile Regression, see [Wk2, CN1, AA2, AA3]. Quantile Regression is a method to estimate the variable relations for all parts of the distribution. (Not just, say, the mean of the relationships found with Least Squares Regression.)
The goal of the monad design is to make the specification of Quantile Regression workflows (relatively) easy, straightforward, by following a certain main scenario and specifying variations over that scenario. Since Quantile Regression is often compared with Least Squares Regression and some type of filtering (like, Moving Average) those functionalities should be included in the monad design scenarios.
The monad is named QRMon and it is based on the State monad package "StateMonadCodeGenerator.m", [AAp1, AA1] and the Quantile Regression package "QuantileRegression.m", [AAp4, AA2].
The data for this document is read from WL's repository or created ad-hoc.
The monadic programming design is used as a Software Design Pattern. The QRMon monad can be also seen as a Domain Specific Language (DSL) for the specification and programming of machine learning classification workflows.
Here is an example of using the QRMon monad over heteroscedastic data::
The table above is produced with the package "MonadicTracing.m", [AAp2, AA1], and some of the explanations below also utilize that package.
As it was mentioned above the monad QRMon can be seen as a DSL. Because of this the monad pipelines made with QRMon are sometimes called "specifications".
Remark: With "regression quantile" we mean "a curve or function that is computed with Quantile Regression".
Contents description
The document has the following structure.
• The sections "Package load" and "Data load" obtain the needed code and data.
• The sections "Design consideration" and "Monad design" provide motivation and design decisions rationale.
• The sections "QRMon overview" and "Monad elements" provide technical description of the QRMon monad needed to utilize it.
• (Using a fair amount of examples.)
• The section "Unit tests" describes the tests used in the development of the QRMon monad.
• (The random pipelines unit tests are especially interesting.)
• The section "Future plans" outlines future directions of development.
• The section "Implementation notes" just says that QRMon's development process and this document follow the ones of the classifications workflows monad ClCon, [AA6].
Remark: One can read only the sections "Introduction", "Design consideration", "Monad design", and "QRMon overview". That set of sections provide a fairly good, programming language agnostic exposition of the substance and novel ideas of this document.
The table above is produced with the package "MonadicTracing.m", [AAp2, AA1], and some of the explanations below also utilize that package.
As it was mentioned above the monad QRMon can be seen as a DSL. Because of this the monad pipelines made with QRMon are sometimes called "specifications".
Remark: With "regression quantile" we mean "a curve or function that is computed with Quantile Regression".
The following commands load the packages [AAp1--AAp6]:
Import["https://raw.githubusercontent.com/antononcube/\
Import["https://raw.githubusercontent.com/antononcube/\
In this section we load data that is used in the rest of the document. The time series data is obtained through WL's repository.
The data summarization and plots are done through QRMon, which in turn uses the function RecordsSummary from the package "MathematicaForPredictionUtilities.m", [AAp6].
Distribution data
The following data is generated to have [heteroscedasticity(https://en.wikipedia.org/wiki/Heteroscedasticity).
distData =
Table[{x,
Exp[-x^2] +
RandomVariate[
NormalDistribution[0, .15 Sqrt[Abs[1.5 - x]/1.5]]]}, {x, -3,
3, .01}];
Length[distData]
(* 601 *)
QRMonUnit[distData]?QRMonEchoDataSummary?QRMonPlot;
Temperature time series
tsData = WeatherData[{"Orlando", "USA"}, "Temperature", {{2015, 1, 1}, {2018, 1, 1}, "Day"}]
QRMonUnit[tsData]?QRMonEchoDataSummary?QRMonDateListPlot;
Financial time series
The following data is typical for financial time series. (Note the differences with the temperature time series.)
finData = TimeSeries[FinancialData["NYSE:GE", {{2014, 1, 1}, {2018, 1, 1}, "Day"}]];
QRMonUnit[finData]?QRMonEchoDataSummary?QRMonDateListPlot;
Design considerations
The steps of the main regression workflow addressed in this document follow.
1. Retrieving data from a data repository.
2. Optionally, transform the data.
1. Delete rows with missing fields.
2. Rescale data along one or both of the axes.
3. Apply moving average (or median, or map.)
3. Verify assumptions of the data.
4. Run a regression algorithm with a certain basis of functions using:
1. Quantile Regression, or
2. Least Squares Regression.
5. Visualize the data and regression functions.
6. If the regression functions fit is not satisfactory go to step 4.
7. Utilize the found regression functions to compute:
1. outliers,
2. local extrema,
3. approximation or fitting errors,
4. conditional density distributions,
5. time series simulations.
The following flow-chart corresponds to the list of steps above.
• the introduction of new elements in regression workflows,
• workflows elements variability, and
• workflows iterative changes and refining,
it is beneficial to have a DSL for regression workflows. We choose to make such a DSL through a [functional programming monad](https://en.wikipedia.org/wiki/Monad_(functional_programming)), [[Wk1](https://en.wikipedia.org/wiki/Monad_(functional_programming)), AA1].
Here is a quote from [[Wk1](https://en.wikipedia.org/wiki/Monad_(functional_programming))] that fairly well describes why we choose to make a classification workflow monad and hints on the desired properties of such a monad.
[...] The monad represents computations with a sequential structure: a monad defines what it means to chain operations together. This enables the programmer to build pipelines that process data in a series of steps (i.e. a series of actions applied to the data), in which each action is decorated with the additional processing rules provided by the monad. [...] Monads allow a programming style where programs are written by putting together highly composable parts, combining in flexible ways the possible actions that can work on a particular type of data. [...]
Remark: Note that quote from [[Wk1](https://en.wikipedia.org/wiki/Monad_(functional_programming))] refers to chained monadic operations as "pipelines". We use the terms "monad pipeline" and "pipeline" below.
The monad we consider is designed to speed-up the programming of quantile regression workflows outlined in the previous section. The monad is named QRMon for "Quantile Regression Monad".
We want to be able to construct monad pipelines of the general form:
This means that some monad operations will not just change the pipeline value but they will also change the pipeline context.
In the monad pipelines of QRMon we store different objects in the contexts for at least one of the following two reasons.
1. The object will be needed later on in the pipeline, or
2. The object is (relatively) hard to compute.
Such objects are transformed data, regression functions, and outliers.
Let us list the desired properties of the monad.
• Rapid specification of non-trivial quantile regression workflows.
• The monad works with time series, numerical matrices, and numerical vectors.
• The pipeline values can be of different types. Most monad functions modify the pipeline value; some modify the context; some just echo results.
• The monad can do quantile regression with B-Splines bases, quantile regression fit and least squares fit with specified bases of functions.
• The monad allows of cursory examination and summarization of the data.
• It is easy to obtain the pipeline value, context, and different context objects for manipulation outside of the monad.
• It is easy to plot different combinations of data, regression functions, outliers, approximation errors, etc.
The QRMon components and their interactions are fairly simple.
The main QRMon operations implicitly put in the context or utilize from the context the following objects:
• (time series) data,
• regression functions,
• outliers and outlier regression functions.
Note the that the monadic set of types of QRMon pipeline values is fairly heterogenous and certain awareness of "the current pipeline value" is assumed when composing QRMon pipelines.
Obviously, we can put in the context any object through the generic operations of the State monad of the package "StateMonadGenerator.m", [AAp1].
QRMon overview
When using a monad we lift certain data into the "monad space", using monad's operations we navigate computations in that space, and at some point we take results from it.
With the approach taken in this document the "lifting" into the QRMon monad is done with the function QRMonUnit. Results from the monad can be obtained with the functions QRMonTakeValue, QRMonContext, or with the other QRMon functions with the prefix "QRMonTake" (see below.)
Here is a corresponding diagram of a generic computation with the QRMon monad:
Remark: It is a good idea to compare the diagram with formulas (1) and (2).
Let us examine a concrete QRMon pipeline that corresponds to the diagram above. In the following table each pipeline operation is combined together with a short explanation and the context keys after its execution.
Here is the output of the pipeline:
The QRMon functions are separated into four groups:
• operations,
• setters and droppers,
• takers,
An overview of the those functions is given in the tables in next two sub-sections. The next section, "Monad elements", gives details and examples for the usage of the QRMon operations.
Monad functions interaction with the pipeline value and context
The following table gives an overview the interaction of the QRMon monad functions with the pipeline value and context.
The following table shows the functions that are function synonyms or short-cuts.
Here are the QRMon State Monad functions (generated using the prefix "QRMon", [AAp1, AA1]):
In this section we show that QRMon has all of the properties listed in the previous section.
The monad head is QRMon. Anything wrapped in QRMon can serve as monad's pipeline value. It is better though to use the constructor QRMonUnit. (Which adheres to the definition in [Wk1].)
QRMon[{{1, 223}, {2, 323}}, <||>]?QRMonEchoDataSummary;
The function lifting the data into the monad QRMon is QRMonUnit.
The lifting to the monad marks the beginning of the monadic pipeline. It can be done with data or without data. Examples follow.
QRMonUnit[distData]?QRMonEchoDataSummary;
QRMonUnit[]?QRMonSetData[distData]?QRMonEchoDataSummary;
(See the sub-section "Setters, droppers, and takers" for more details of setting and taking values in QRMon contexts.)
Currently the monad can deal with data in the following forms:
• time series,
• numerical vectors,
• numerical matrices of rank two.
When the data lifted to the monad is a numerical vector vec it is assumed that vec has to become the second column of a "time series" matrix; the first column is derived with Range[Length[vec]] .
Generally, WL makes it easy to extract columns datasets order to obtain numerical matrices, so datasets are not currently supported in QRMon.
Quantile regression with B-splines
This computes quantile regression with B-spline basis over $12$ regularly spaced knots. (Using Linear Programming algorithms; see [AA2] for details.)
QRMonUnit[distData]?
QRMonQuantileRegression[12]?
QRMonPlot;
The monad function QRMonQuantileRegression has the same options as QuantileRegression. (The default value for option Method is different, since using "CLP" is generally faster.)
Options[QRMonQuantileRegression]
(* {InterpolationOrder -> 3, Method -> {LinearProgramming, Method -> "CLP"}} *)
Let us compute regression using a list of particular knots, specified quantiles, and the method "InteriorPoint" (instead of the Linear Programming library CLP):
p =
QRMonUnit[distData]?
QRMonQuantileRegression[{-3, -2, 1, 0, 1, 1.5, 2.5, 3}, Range[0.1, 0.9, 0.2], Method -> {LinearProgramming, Method -> "InteriorPoint"}]?
QRMonPlot;
Remark: As it was mentioned above the function QRMonRegression is a synonym of QRMonQuantileRegression.
The fit functions can be extracted from the monad with QRMonTakeRegressionFunctions, which gives an association of quantiles and pure functions.
ListPlot[# /@ distData[[All, 1]]] & /@ (p?QRMonTakeRegressionFunctions)
Quantile regression fit and Least squares fit
Instead of using a B-spline basis of functions we can compute a fit with our own basis of functions.
Here is a basis functions:
bFuncs = Table[PDF[NormalDistribution[m, 1], x], {m, Min[distData[[All, 1]]], Max[distData[[All, 1]]], 1}];
Plot[bFuncs, {x, Min[distData[[All, 1]]], Max[distData[[All, 1]]]},
PlotRange -> All, PlotTheme -> "Scientific"]
Here we do a Quantile Regression fit, a Least Squares fit, and plot the results:
p =
QRMonUnit[distData]?
QRMonQuantileRegressionFit[bFuncs]?
QRMonLeastSquaresFit[bFuncs]?
QRMonPlot;
Remark: The functions "QRMon*Fit" should generally have a second argument for the symbol of the basis functions independent variable. Often that symbol can be omitted and implied. (Which can be seen in the pipeline above.)
Remark: As it was mentioned above the function QRMonRegressionFit is a synonym of QRMonQuantileRegressionFit and QRMonFit is a synonym of QRMonLeastSquaresFit.
As it was pointed out in the previous sub-section, the fit functions can be extracted from the monad with QRMonTakeRegressionFunctions. Here the keys of the returned/taken association consist of quantiles and "mean" since we applied both Quantile Regression and Least Squares Regression.
ListPlot[# /@ distData[[All, 1]]] & /@ (p?QRMonTakeRegressionFunctions)
Default basis to fit (using Chebyshev polynomials)
One of the main advantages of using the function QuanileRegression of the package [AAp4] is that the functions used to do the regression with are specified with a few numeric parameters. (Most often only the number of knots is sufficient.) This is achieved by using a basis of B-spline functions of a certain interpolation order.
We want similar behaviour for Quantile Regression fitting we need to select a certain well known basis with certain desirable properties. Such basis is given by Chebyshev polynomials of first kind [Wk3]. Chebyshev polynomials bases can be easily generated in Mathematica with the functions ChebyshevT or ChebyshevU.
Here is an application of fitting with a basis of $12$ Chebyshev polynomials of first kind:
QRMonUnit[distData]?
QRMonQuantileRegressionFit[12]?
QRMonLeastSquaresFit[12]?
QRMonPlot;
The code above is equivalent to the following code:
bfuncs = Table[ChebyshevT[i, Rescale[x, MinMax[distData[[All, 1]]], {-0.95, 0.95}]], {i, 0, 12}];
p =
QRMonUnit[distData]?
QRMonQuantileRegressionFit[bfuncs]?
QRMonLeastSquaresFit[bfuncs]?
QRMonPlot;
The shrinking of the ChebyshevT domain seen in the definitions of $\text{bfuncs}$ is done in order to prevent approximation error effects at the ends of the data domain. The following code uses the ChebyshevT domain $\{-1,1\}$ instead of the domain $\{-0.95,0.95\}$ used above.
QRMonUnit[distData]?
QRMonQuantileRegressionFit[{4, {-1, 1}}]?
QRMonPlot;
Regression functions evaluation
The computed quantile and least squares regression functions can be evaluated with the monad function QRMonEvaluate.
Evaluation for a given value of the independent variable:
p?QRMonEvaluate[0.12]?QRMonTakeValue
(* <|0.25 -> 0.930402, 0.5 -> 1.01411, 0.75 -> 1.08075, "mean" -> 0.996963|> *)
Evaluation for a vector of values:
p?QRMonEvaluate[Range[-1, 1, 0.5]]?QRMonTakeValue
(* <|0.25 -> {0.258241, 0.677461, 0.943299, 0.703812, 0.293741},
0.5 -> {0.350025, 0.768617, 1.02311, 0.807879, 0.374545},
0.75 -> {0.499338, 0.912183, 1.10325, 0.856729, 0.431227},
"mean" -> {0.355353, 0.776006, 1.01118, 0.783304, 0.363172}|> *)
Evaluation for complicated lists of numbers:
p?QRMonEvaluate[{0, 1, {1.5, 1.4}}]?QRMonTakeValue
(* <|0.25 -> {0.943299, 0.293741, {0.0762883, 0.10759}},
0.5 -> {1.02311, 0.374545, {0.103386, 0.139142}},
0.75 -> {1.10325, 0.431227, {0.133755, 0.177161}},
"mean" -> {1.01118, 0.363172, {0.107989, 0.142021}}|> *)
The obtained values can be used to compute estimates of the distributions of the dependent variable. See the sub-sections "Estimating conditional distributions" and "Dependent variable simulation".
Errors and error plots
Here with "errors" we mean the differences between data's dependent variable values and the corresponding values calculated with the fitted regression curves.
In the pipeline below we compute couple of regression quantiles, plot them together with the data, we plot the errors, compute the errors, and summarize them.
QRMonUnit[finData]?
QRMonQuantileRegression[10, {0.5, 0.1}]?
QRMonDateListPlot[Joined -> False]?
QRMonErrorPlots["DateListPlot" -> True, Joined -> False]?
QRMonErrors?
QRMonEchoFunctionValue["Errors summary:", RecordsSummary[#[[All, 2]]] & /@ # &];
Each of the functions QRMonErrors and QRMonErrorPlots computes the errors. (That computation is considered cheap.)
Finding outliers
Finding outliers can be done with the function QRMonOultiers. The outliers found by QRMonOutliers are simply points that below or above certain regression quantile curves, for example, the ones corresponding to $0.02$ and $0.98$.
Here is an example:
p =
QRMonUnit[distData]?
QRMonQuantileRegression[6, {0.02, 0.98}]?
QRMonOutliers?
QRMonEchoValue?
QRMonOutliersPlot;
The function QRMonOutliers puts in the context values for the keys "outliers" and "outlierRegressionFunctions". The former is for the found outliers, the latter is for the functions corresponding to the used regression quantiles.
Keys[p?QRMonTakeContext]
(* {"data", "regressionFunctions", "outliers", "outlierRegressionFunctions"} *)
Here are the corresponding quantiles of the plot above:
Keys[p?QRMonTakeOutlierRegressionFunctions]
(* {0.02, 0.98} *)
The control of the outliers computation is done though the arguments and options of QRMonQuantileRegression (or the rest of the regression calculation functions.)
If only one regression quantile is found in the context and the corresponding quantile is less than $0.5$ then QRMonOutliers finds only bottom outliers. If only one regression quantile is found in the context and the corresponding quantile is greater than $0.5$ then QRMonOutliers finds only top outliers.
Here is an example for finding only the top outliers:
QRMonUnit[finData]?
QRMonQuantileRegression[5, 0.8]?
QRMonOutliers?
QRMonEchoFunctionContext["outlier quantiles:", Keys[#outlierRegressionFunctions] &]?
QRMonOutliersPlot["DateListPlot" -> True];
Plotting outliers
The function QRMonOutliersPlot makes an outliers plot. If the outliers are not in the context then QRMonOutliersPlot calls QRMonOutliers first.
Here are the options of QRMonOutliersPlot:
Options[QRMonOutliersPlot]
(* {"Echo" -> True, "DateListPlot" -> False, ListPlot -> {Joined -> False}, Plot -> {}} *)
The default behavior is to echo the plot. That can be suppressed with the option "Echo".
QRMonOutliersPlot utilizes combines with Show two plots:
• one with ListPlot (or DateListPlot) for the data and the outliers,
• the other with Plot for the regression quantiles used to find the outliers.
That is why separate lists of options can be given to manipulate those two plots. The option DateListPlot can be used make plots with date or time axes.
QRMonUnit[tsData]?
QRMonQuantileRegression[12, {0.01, 0.99}]?
QRMonOutliersPlot[
"Echo" -> False,
"DateListPlot" -> True,
ListPlot -> {PlotStyle -> {Green, {PointSize[0.02],
Red}, {PointSize[0.02], Blue}}, Joined -> False,
PlotTheme -> "Grid"},
Plot -> {PlotStyle -> Orange}]?
QRMonTakeValue
Estimating conditional distributions
Consider the following problem:
How to estimate the conditional density of the dependent variable given a value of the conditioning independent variable?
(In other words, find the distribution of the $y$-values for a given, fixed $x$-value.)
The solution of this problem using Quantile Regression is discussed in detail in [PG1] and [AA4].
Finding a solution for this problem can be seen as a primary motivation to develop Quantile Regression algorithms.
The following pipeline (i) computes and plots a set of five regression quantiles and (ii) then using the found regression quantiles computes and plots the conditional distributions for two focus points ( $-2$ and $1$.)
QRMonUnit[distData]?
QRMonQuantileRegression[6,
Range[0.1, 0.9, 0.2]]?
QRMonPlot[GridLines -> {{-2, 1}, None}]?
QRMonConditionalCDF[{-2, 1}]?
QRMonConditionalCDFPlot;
Moving average, moving median, and moving map
Fairly often it is a good idea for a given time series to apply filter functions like Moving Average or Moving Median. We might want to:
• visualize the obtained transformed data,
• do regression over the transformed data,
• compare with regression curves over the original data.
For these reasons QRMon has the functions QRMonMovingAverage, QRMonMovingMedian, and QRMonMovingMap that correspond to the built-in functions MovingAverage, MovingMedian, and MovingMap.
Here is an example:
QRMonUnit[tsData]?
QRMonDateListPlot[ImageSize -> Small]?
QRMonMovingAverage[20]?
QRMonEchoFunctionValue["Moving avg: ", DateListPlot[#, ImageSize -> Small] &]?
QRMonMovingMap[Mean, Quantity[20, "Days"]]?
QRMonEchoFunctionValue["Moving map: ", DateListPlot[#, ImageSize -> Small] &];
Dependent variable simulation
Consider the problem of making a time series that is a simulation of a process given with a known time series.
More formally,
• we are given a time-axis grid (regular or irregular),
• we consider each grid node to correspond to a random variable,
• we want to generate time series based on the empirical CDF's of the random variables that correspond to the grid nodes.
The formulation of the problem hints to an (almost) straightforward implementation using Quantile Regression.
p = QRMonUnit[tsData]?QRMonQuantileRegression[30, Join[{0.01}, Range[0.1, 0.9, 0.1], {0.99}]];
tsNew =
p?
QRMonSimulate[1000]?
QRMonTakeValue;
opts = {ImageSize -> Medium, PlotTheme -> "Detailed"};
GraphicsGrid[{{DateListPlot[tsData, PlotLabel -> "Actual", opts],
DateListPlot[tsNew, PlotLabel -> "Simulated", opts]}}]
Finding local extrema in noisy data
Using regression fitting -- and Quantile Regression in particular -- we can easily construct semi-symbolic algorithms for finding local extrema in noisy time series data; see [AA5]. The QRMon function with such an algorithm is QRMonFindLocalExtrema.
In brief, the algorithm steps are as follows. (For more details see [AA5].)
1. Fit a polynomial through the data.
2. Find the local extrema of the fitted polynomial. (We will call them fit estimated extrema.)
3. Around each of the fit estimated extrema find the most extreme point in the data by a nearest neighbors search (by using Nearest).
The function QRMonFindLocalExtrema uses the regression quantiles previously found in the monad pipeline (and stored in the context.) The bottom regression quantile is used for finding local minima, the top regression quantile is used for finding the local maxima.
An example of finding local extrema follows.
QRMonUnit[TimeSeriesWindow[tsData, {{2015, 1, 1}, {2018, 12, 31}}]]?
QRMonQuantileRegression[10, {0.05, 0.95}]?
QRMonDateListPlot[Joined -> False, PlotTheme -> "Scientific"]?
QRMonFindLocalExtrema["NumberOfProximityPoints" -> 100]?
QRMonEchoValue?
QRMonEchoFunctionContext[
DateListPlot[{#localMinima, #localMaxima, #data},
PlotStyle -> {PointSize[0.015], PointSize[0.015], Gray},
Joined -> False,
PlotLegends -> {"localMinima", "localMaxima", "data"},
PlotTheme -> "Scientific"] &];
Note that in the pipeline above in order to plot the data and local extrema together some additional steps are needed. The result of QRMonFindLocalExtrema becomes the pipeline value; that pipeline value is displayed with QRMonEchoValue, and stored in the context with QRMonAddToContext. If the pipeline value is an association -- which is the case here -- the monad function QRMonAddToContext joins that association with the context association. In this case this means that we will have key-value elements in the context for "localMinima" and "localMaxima". The date list plot at the end of the pipeline uses values of those context keys (together with the value for "data".)
Setters, droppers, and takers
The values from the monad context can be set, obtained, or dropped with the corresponding "setter", "dropper", and "taker" functions as summarized in a previous section.
For example:
p = QRMonUnit[distData]?QRMonQuantileRegressionFit[2];
p?QRMonTakeRegressionFunctions
(* <|0.25 -> (0.0191185 + 0.00669159 #1 + 3.05509*10^-14 #1^2 &),
0.5 -> (0.191408 + 9.4728*10^-14 #1 + 3.02272*10^-14 #1^2 &),
0.75 -> (0.563422 + 3.8079*10^-11 #1 + 7.63637*10^-14 #1^2 &)|> *)
If other values are put in the context they can be obtained through the (generic) function QRMonTakeContext, [AAp1]:
p = QRMonUnit[RandomReal[1, {2, 2}]]?QRMonAddToContext["data"];
(p?QRMonTakeContext)["data"]
(* {{0.608789, 0.741599}, {0.877074, 0.861554}} *)
Another generic function from [AAp1] is QRMonTakeValue (used many times above.)
Here is an example of the "data dropper" QRMonDropData:
p?QRMonDropData?QRMonTakeContext
(* <||> *)
(The "droppers" simply use the state monad function QRMonDropFromContext, [AAp1]. For example, QRMonDropData is equivalent to QRMonDropFromContext["data"].)
Unit tests
The development of QRMon was done with two types of unit tests: (i) directly specified tests, [AAp7], and (ii) tests based on randomly generated pipelines, [AA8].
The unit test package should be further extended in order to provide better coverage of the functionalities and illustrate -- and postulate -- pipeline behavior.
Directly specified tests
Here we run the unit tests file "MonadicQuantileRegression-Unit-Tests.wlt", [AAp7]:
AbsoluteTiming[
]
The natural language derived test ID's should give a fairly good idea of the functionalities covered in [AAp3].
Values[Map[#["TestID"] &, testObject["TestResults"]]]
"QuantileRegression-2", "QuantileRegression-3", \
"QuantileRegression-and-Fit-1", "Fit-and-QuantileRegression-1", \
"QuantileRegressionFit-and-Fit-1", "Fit-and-QuantileRegressionFit-1", \
"Outliers-1", "Outliers-2", "GridSequence-1", "BandsSequence-1", \
"ConditionalCDF-1", "Evaluate-1", "Evaluate-2", "Evaluate-3", \
"Simulate-1", "Simulate-2", "Simulate-3"} *)
Random pipelines tests
Since the monad QRMon is a DSL it is natural to test it with a large number of randomly generated "sentences" of that DSL. For the QRMon DSL the sentences are QRMon pipelines. The package "MonadicQuantileRegressionRandomPipelinesUnitTests.m", [AAp8], has functions for generation of QRMon random pipelines and running them as verification tests. A short example follows.
Generate pipelines:
SeedRandom[234]
pipelines = MakeQRMonRandomPipelines[100];
Length[pipelines]
(* 100 *)
Here is a sample of the generated pipelines:
(*
Block[{DoubleLongRightArrow, pipelines = RandomSample[pipelines, 6]},
Clear[DoubleLongRightArrow];
pipelines = pipelines /. {_TemporalData -> "tsData", _?MatrixQ -> "distData"};
GridTableForm[Map[List@ToString[DoubleLongRightArrow @@ #, FormatType -> StandardForm] &, pipelines], TableHeadings -> {"pipeline"}]
]
AutoCollapse[] *)
Here we run the pipelines as unit tests:
AbsoluteTiming[
res = TestRunQRMonPipelines[pipelines, "Echo" -> False];
]
From the test report results we see that a dozen tests failed with messages, all of the rest passed.
rpTRObj = TestReport[res]
(The message failures, of course, have to be examined -- some bugs were found in that way. Currently the actual test messages are expected.)
Future plans
Workflow operations
A list of possible, additional workflow operations and improvements follows.
• Certain improvements can be done over the specification of the different plot options.
• It will be useful to develop a function for automatic finding of over-fitting parameters.
• The time series simulation should be done by aggregation of similar time intervals.
• For example, for time series with span several years, for each month name is made Quantile Regression simulation and the results are spliced to obtain a one year simulation.
• If the time series is represented as a sequence of categorical values, then the time series simulation can use Bayesian probabilities derived from sub-sequences.
• QRMon already has functions that facilitate that, QRMonGridSequence and QRMonBandsSequence.
Conversational agent
Using the packages [AAp10, AAp11] we can generate QRMon pipelines with natural commands. The plan is to develop and document those functionalities further.
Here is an example of a pipeline constructed with natural language commands:
QRMonUnit[distData]?
ToQRMonPipelineFunction["show data summary"]?
ToQRMonPipelineFunction["calculate quantile regression for quantiles 0.2, 0.8 and with 40 knots"]?
ToQRMonPipelineFunction["plot"];
Implementation notes
The implementation methodology of the QRMon monad packages AAp3, AAp8] followed the methodology created for the ClCon monad package [AAp9, AA6]. Similarly, this document closely follows the structure and exposition of the ClCon monad document ["A monad for classification workflows", [AA6].
A lot of the functionalities and signatures of QRMon were designed and programed through considerations of natural language commands specifications given to a specialized conversational agent. (As discussed in the previous section.)
References
Other
CN1] Brian S. Code and Barry R. Noon, ["A gentle introduction to quantile regression for ecologists", (2003). Frontiers in Ecology and the Environment. 1 (8): 412[Dash]420. doi:10.2307/3868138. URL: http://www.econ.uiuc.edu/~roger/research/rq/QReco.pdf .
RK1] Roger Koenker, [Quantile Regression, ?Cambridge University Press, 2005?.
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Anonymous User
Anonymous User
Posted 6 years ago
Excellent. If the package is available i hope to try it / use it. Nice writeup too.It seems a shame to me to have so much work done for free on git (assuming you weren't paid as a professor). It seems that the length and size of individual's work that ends up as "free" just keeps growing.
Posted 6 years ago
Thanks for the kind words, John!The QRMon package and the packages needed internally by QRMon can be loaded with the command: Import["https://raw.githubusercontent.com/antononcube/MathematicaForPrediction/master/MonadicProgramming/MonadicQuantileRegression.m"] Here is a screenshot:
Posted 6 years ago
Great work! As a user of this process and these methods, I'm excited that the use of rather simple language can produce sophisticated, yet approachable, results.
Posted 6 years ago
- Congratulations! This post is now a Staff Pick as distinguished by a badge on your profile! Thank you, keep it coming!
Community posts can be styled and formatted using the Markdown syntax. | 7,837 | 32,059 | {"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} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.884995 |
http://www.patheos.com/blogs/wordoftheday/page/4/ | 1,529,723,754,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864919.43/warc/CC-MAIN-20180623015758-20180623035758-00118.warc.gz | 467,988,365 | 15,383 | ## Math Lesson of the Day: Quick Averages
Math Lesson of the Day: Quick Averages When I was a kid, my teachers spent the first two or three weeks of every school year complaining that I didn’t “show my work” when I was doing math. The problem, most of the time, was that I didn’t have any work to show. I looked at the problem and knew the answer. The rest of the time, the problem was that I did “work,” but it was either the work… Read more
## Word of the Day: thank
Word of the Day: thank Our word today sounds like think, and that’s appropriate, because the words are related. We move from a verb that means “to seem, to appear,” as in German duenken, to the causative verb “to make an image appear to oneself,” in one’s own mind, as in think (Anglo Saxon thyncan, German denken), to a verb meaning “to think good thoughts about someone else,” a causative on top of a causative: Anglo Saxon thancian, German… Read more
## Grammar Lesson of the Day: Voice
Grammar Lesson of the Day: Voice My students have been taught that a verb is in the passive voice whenever a form of the verb to be appears. They have also been taught that it is never to be used. They are wrong on both counts. I’ll discuss the use of the passive voice later. For now, let’s define what we mean by voice. Consider these three sentences: Superman was stopped by Lex Luthor and a very large… Read more
## Doggerel of the Day: The Obamarama Limerick Contest
Announcing, for your intellectual exercise, the First Annual Obamara Limerick Contest. Here’s my entry. The reasoning behind it is this: to skewer the colossal silliness of the Obamalators, those hapless secular people who just had to believe in somebody, anybody, and who therefore said that the Lord Obama would change the history of the world, would mark a transformation in the universe, would settle the levels of the oceans (well, Lord Obama himself said that), would be more significant than… Read more
## Word of the Day: tidings
Word of the Day: tidings. “Time and tide wait no man,” says the old proverb. It’s a nice alliterative pair, those two, and we may be led to think that the words are related, since the tide notoriously comes in on time. But they aren’t. The word time comes into English through French, after the Norman invasion in 1066, when William the Conqueror unloaded into English harbors whole boatloads of surplus words, and instead of throwing them… Read more
## Grammar Lesson of the Day: Infinitives
Grammar Lesson of the Day: Infinitives. “Her five year mission,” cut to three by NBC, “to seek out new civilizations, to boldly split infinitives where no man has split them before!” The reason why grammatical sticklers don’t like split infinitives is that the infinitive is really one word – so it’s like inserting a word into the middle of a word, which classical poets occasionally did for special effects. That was called tmesis: It simply won’t please us… Read more
## Word of the Day: worry
Word of the Day: worry. Imagine somebody in the grip of worry. She wrings her hands. She wrinkles her brows. She is all wrapped up in a bad situation. Something’s wrong, she knows it. The English wr- words almost all have to do with twisting. That is the case too with words that we don’t immediately associate with twisting, like worry and wrath. But the old physical meaning of worry reveals the underlying idea. In C. S. Lewis’ That… Read more
## Word of the Day: forgo
Word of the Day: forgo The word is commonly but inaccurately spelled forego, but those are really two separate and unrelated verbs. The fore in forego means first or before, so that a foregone conclusion is a conclusion that comes before any argument or declaration, since none is necessary. That prefix fore is related to all kinds of words in English that have to do with priority in time or position. The ones beginning with f come from the… Read more
## Exploring the Real Lessons of Prohibition
Writing over at Crisis Magazine: So, then, what does Prohibition teach us? That amendment inserted into the Constitution a law that neither protected fundamental rights nor adjusted the mechanics of governance. It was a radical break from tradition. It is crucial to understand this. It took a juridical break from tradition to obliterate the customs, the lived traditions, of the American people and their forebears. Granted, Prohibition addressed problems that certainly needed solving. Prohibition was sold, in large part, as… Read more
## Grammar Lesson of the Day: Foreign Plurals
Grammar Lesson of the Day: Foreign Plurals There are four groups of nouns from Latin and Greek that we’ve borrowed into English directly, using both singular and plural forms. These aren’t too hard to remember, if we focus on the singular rather than on the plural: Greek neuter singular nouns ending in –on (-ion); plural in –a (-ia): one phenomenon, two phenomena one criterion, two criteria But newly coined words with Greek singular forms simply add the usual s for… Read more | 1,160 | 5,043 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-26 | latest | en | 0.97041 |
https://brainly.in/question/57963 | 1,485,206,729,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560283008.19/warc/CC-MAIN-20170116095123-00326-ip-10-171-10-70.ec2.internal.warc.gz | 788,955,221 | 10,423 | How to solve below given partial fraction ... can u kindly explain... ( 8-4v+11v^2-2v^3)÷{(1-0.25v)(1-v+0.5v^2)}
1
by bunybatman
2014-11-27T06:03:03+05:30
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
We check the 2nd degree polynomial if we can factorize it. It is not factorizable and it has only imaginary roots. so we keep it as it is.
We write +16 on RHS because the coefficient of (highest degree term) v^3 in numerator is -16 and in denominator - 1 * +1 = -1, and hence -16 v^3/- v^3.
Since 4-v is first degree polynomial, write A in the numerator. B x +C in the second fraction because the other factor is a second degree polynomial.
Solve the above for A, B and C. | 257 | 912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.908372 |
http://econometrics.com/intro/dummy.htm | 1,723,439,073,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641028735.71/warc/CC-MAIN-20240812030550-20240812060550-00874.warc.gz | 13,211,882 | 2,296 | SHAZAM Dummy Variables
## Dummy Variables
Dummy variables are very useful for capturing a variety of qualitative effects. They typically have the value 0 or 1 and so possibly a better name is "binary variable". They can be included as explanatory variables in the regression equation and the estimated coefficients and standard errors can be used in hypothesis testing.
Dummy variables that measure individual attributes may be coded in the data file and these can be assigned variable names and loaded into SHAZAM with the `READ` command. Alternatively, it may be convenient to create the required dummy variables with SHAZAM commands.
#### Examples
Note: When dummy variables are used as dependent variables then analysis with the probit or logit model is required.
[SHAZAM Guide home]
### Modelling structural change
Consider measuring the impact of seat belt legislation on traffic fatalities. Monthly data is available for 120 observations on traffic fatalities (`D`) and auto fuel consumption (`C`). Suppose that seat belt legislation became effective at observation 80. To estimate the impact of this legislation define a dummy variable as follows:
``` B = 0 for observations 1 to 79
B = 1 for observations 80 to 120
```
The following SHAZAM commands create the dummy variable and run the regression.
```SAMPLE 1 120
* Create the dummy variable
GENR B=0
SAMPLE 80 120
GENR B=1
*
SAMPLE 1 120
* List the data and check that it is set up correctly.
PRINT D C B
* Get the OLS estimation results
OLS D C B
STOP
```
Note that the command `GENR B=0` first initializes the dummy variable to all zeroes. Then the `SAMPLE` command is used to set the sample to the period of structural change and the dummy variable is then set to 1 for this period. The dummy variable is now created and the OLS estimation is then run over the entire sample period of the data.
In SHAZAM there is often more than one way of doing things. An equivalent way of creating the dummy variable is illustrated in the next list of commands.
```SAMPLE 1 120
The dummy variable is created in a single command by using the `DUM` and `TIME` functions on the `GENR` command. The `TIME(0)` function creates an observation index with values 1,2,3,. . . . The `DUM` function is set to 1 when the argument is > 0 (that is, positive values) and is set to 0 otherwise. In this example the argument of the `DUM` function is `TIME(0)-79`. This will have positive values for observation numbers that are greater than or equal to 80. | 574 | 2,525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.88347 |
https://gmatclub.com/forum/comprehensive-admit-ding-results-discussion-thread-53747-40.html?kudos=1 | 1,511,214,064,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806225.78/warc/CC-MAIN-20171120203833-20171120223833-00151.warc.gz | 620,256,647 | 47,531 | It is currently 20 Nov 2017, 14:41
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# Events & Promotions
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Senior Manager
Joined: 07 Aug 2006
Posts: 328
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21 Oct 2007, 08:44
KingKREEP wrote:
R.E.D. wrote:
Actually, I do not like vodka
Hold on, you said you were Russian right? Are you sure?!?
That's true. But I have lots of friends in Italy and I prefer dry red wine, such as Valpolicella, Bardolino or Chianti and, the most liked - Bracchetto - red semi-sweet sparkling wine.
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Senior Manager
Joined: 27 Jul 2006
Posts: 292
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27 Oct 2007, 11:33
R.E.D. wrote:
KingKREEP wrote:
-35?!? Bloody hell, what do you guys do to keep warm...Vodka?!? Where I live, it's a pretty comfortable 20 - 25 (deg C) throughout the year. Nice weather, but no snowboarding
Actually, I do not like vodka and a cup of hot black tea warms up not less than alcohol. Good clothes also helps.
The admission of liking "black tea" plus using vodka to "warm up" is Russian enough in my book. He probably goes in for a bottle of siberskaya Corona instead of vodka, like many of the younger Russians
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Senior Manager
Joined: 07 Aug 2006
Posts: 328
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27 Oct 2007, 12:59
defenestrate wrote:
R.E.D. wrote:
KingKREEP wrote:
-35?!? Bloody hell, what do you guys do to keep warm...Vodka?!? Where I live, it's a pretty comfortable 20 - 25 (deg C) throughout the year. Nice weather, but no snowboarding
Actually, I do not like vodka and a cup of hot black tea warms up not less than alcohol. Good clothes also helps.
The admission of liking "black tea" plus using vodka to "warm up" is Russian enough in my book. He probably goes in for a bottle of siberskaya Corona instead of vodka, like many of the younger Russians
Have you been to Russia? You know such details as one of the famous beer brands here What are your impressions about the country? What are pros and cons of doing business here?
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Director
Joined: 31 Mar 2007
Posts: 574
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02 Nov 2007, 15:43
Privet Prativne, kak dela?
haha
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GMAT Club Legend
Status: Um... what do you want to know?
Joined: 04 Jun 2007
Posts: 5464
Kudos [?]: 408 [0], given: 14
Location: SF, CA, USA
Schools: UC Berkeley Haas School of Business MBA 2010
WE 1: Social Gaming
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06 Nov 2007, 02:04
Congrats to haddy74 for being accepted to CBS and matriculating!
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Current Student
Joined: 18 Jun 2007
Posts: 407
Kudos [?]: 41 [0], given: 0
Location: Atlanta, GA
Schools: Emory class of 2010
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06 Nov 2007, 13:25
Out of the admits...haddy and djhouse, did either of you receive scholarships? Or does Columbia let you know later about scholarship awards? I think it would be interesting not only to see admits/dings, but also to see scholarship awards, if any.
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Current Student
Joined: 18 Jun 2007
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Schools: Emory class of 2010
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07 Nov 2007, 14:21
Also, interesting to note that, the people who have been admitted so far (djhouse and haddy) both were under the 700 score for GMAT. Just goes to show that GMAT is only one part of the application process.
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SVP
Joined: 24 Sep 2005
Posts: 1883
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07 Nov 2007, 19:15
emoryhopeful wrote:
Also, interesting to note that, the people who have been admitted so far (djhouse and haddy) both were under the 700 score for GMAT. Just goes to show that GMAT is only one part of the application process.
What's else, they're both US residents. I guess b-schools are in favor of their nationalities first of all.
One thing I've noticed from Yale that it is currently sending out , in its second ( or more?) load, interview invitations to high-GMAT, good-GPA, from-great-university and young applicants
Anyways, it's only my observation obtained from a small sample...generalization is not applicable here.
Last edited by laxieqv on 07 Nov 2007, 20:29, edited 1 time in total.
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Director
Joined: 09 Jan 2007
Posts: 707
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07 Nov 2007, 20:20
laxieqv wrote:
emoryhopeful wrote:
Also, interesting to note that, the people who have been admitted so far (djhouse and haddy) both were under the 700 score for GMAT. Just goes to show that GMAT is only one part of the application process.
What's else, they're both US residents. I guess b-schools are in favor of their nationalities first of all.
One thing I've noticed from Yale that it is currently sending out , in its second ( or more?) load, interview invitations to high-GMAT, good-GPA, from-great-university and young applicants
Anyways, it's only my observation obtained from a small sample...generalization is not applicable here.
Where's your sample size for the Yale theory?
I can't find any info anywhere.
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Joined: 24 Sep 2005
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07 Nov 2007, 20:27
From those who got invitations here ...i went through their profiles.
As I mentioned, this is just my observation...You don't have to base your jugdement on it. Hope you understand. And sorry if my words go any extreme. I tried my best to express what i meant.
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Director
Joined: 09 Jan 2007
Posts: 707
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07 Nov 2007, 21:02
laxieqv wrote:
From those who got invitations here ...i went through their profiles.
As I mentioned, this is just my observation...You don't have to base your jugdement on it. Hope you understand. And sorry if my words go any extreme. I tried my best to express what i meant.
I wasn't offended by it. I'm also a stat junkie, so that was an interesting observation.
What would you classily as high gmat?
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Joined: 24 Sep 2005
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07 Nov 2007, 21:20
gmatclb wrote:
laxieqv wrote:
From those who got invitations here ...i went through their profiles.
As I mentioned, this is just my observation...You don't have to base your jugdement on it. Hope you understand. And sorry if my words go any extreme. I tried my best to express what i meant.
I wasn't offended by it. I'm also a stat junkie, so that was an interesting observation.
What would you classily as high gmat?
I guess it's 750+ .
Also, as one of your post mentioned, Yale is moving up the reputation rank fast; it must, therefore, try to improve average GMAT score, which is normally to us a partial indicator of how competitive a school is. Anyways, good lucks to you on getting invited and admitted
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Director
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07 Nov 2007, 21:42
laxieqv wrote:
gmatclb wrote:
laxieqv wrote:
From those who got invitations here ...i went through their profiles.
As I mentioned, this is just my observation...You don't have to base your jugdement on it. Hope you understand. And sorry if my words go any extreme. I tried my best to express what i meant.
I wasn't offended by it. I'm also a stat junkie, so that was an interesting observation.
What would you classily as high gmat?
I guess it's 750+ .
Also, as one of your post mentioned, Yale is moving up the reputation rank fast; it must, therefore, try to improve average GMAT score, which is normally to us a partial indicator of how competitive a school is. Anyways, good lucks to you on getting invited and admitted
I think once a school hits the 700 threshold, it's less useful to try to get a higher average gmat. In fact, it may hurt the school's reputation as a gmat whore. I think it's in Yale's best interest (if they want to move up in gmat scores) to just hit say 705 average gmat.
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07 Nov 2007, 21:54
Critical reasoning: I didn't say Yale tries to reach an average GMAT score of 750
-------------------------------------------------
The Spirit of The Underdog.
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Director
Joined: 09 Jan 2007
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08 Nov 2007, 07:50
laxieqv wrote:
Critical reasoning: I didn't say Yale tries to reach an average GMAT score of 750
-------------------------------------------------
The Spirit of The Underdog.
Hard bin Reading Comprehension: I didn't say I assumed you meant Yale is trying to reach 750+.
In any case, my hunch is that it will be easier to get in this year than last year. I don't think I saw you in that thread. Have you or will you apply.
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Manager
Joined: 04 Aug 2005
Posts: 202
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09 Nov 2007, 11:01
I already updated my profile on the admit/ding compelation thread, but I just had to come here and let everyone know that I just found out that I'M IN!!!
After the adcom reviewed my mid-quarter grades in the business math class they asked me to take, they thought that I'm doing so well that they decided to admit me now instead of wating until the end of the quarter to re-review my application.
So, although it's not H/S/W et.al., I'm still very excited to have been accepted into the OSU Fisher College of Business PT MBA! I'll start in January.
Last edited by ishcabibble on 09 Nov 2007, 11:57, edited 2 times in total.
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Manager
Joined: 20 Aug 2007
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09 Nov 2007, 11:07
Congrats!
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Current Student
Joined: 18 Jun 2007
Posts: 407
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Location: Atlanta, GA
Schools: Emory class of 2010
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09 Nov 2007, 11:50
Congrats, especially for going to a school that doesn't get as much publicity on this forum. Perhaps in the future, you can be a wealth of information for the program on GMATClub so that more schools can get exposure here.
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Director
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09 Nov 2007, 11:55
[quote="ishcabibble"]I already updated my profile on the admit/ding compelation thread, but I just had to come here and let everyone know that I just found out that I"M IN!!!
Congratulations! That's awesome - good for you!!!
_________________
...there ain't no such thing as a free lunch...
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GMAT Club Legend
Status: Um... what do you want to know?
Joined: 04 Jun 2007
Posts: 5464
Kudos [?]: 408 [0], given: 14
Location: SF, CA, USA
Schools: UC Berkeley Haas School of Business MBA 2010
WE 1: Social Gaming
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09 Nov 2007, 16:04
huge congrats!
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09 Nov 2007, 16:04
Go to page Previous 1 2 3 4 5 6 Next [ 103 posts ]
Display posts from previous: Sort by | 3,409 | 11,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-47 | latest | en | 0.920474 |
https://codereview.stackexchange.com/questions/231588/fibonnaci-sequence-calculation | 1,624,106,249,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487648194.49/warc/CC-MAIN-20210619111846-20210619141846-00209.warc.gz | 185,840,893 | 33,908 | # Fibonnaci sequence calculation [closed]
Below are two ways to calculate an element of the Fibonacci series. Can someone please help me understand why fib1 is so much faster (over 60x) than fib2 ?
fib2 is just an assignment and a decrement whereas fib1 involves a dictionary and recursive calls.
Thanks.
import timeit
def fib1(n, lookup=dict()):
if n == 0 or n == 1:
lookup[n] = n
elif n not in lookup:
lookup[n] = fib1(n - 1, lookup) + fib1(n - 2, lookup)
return lookup[n]
def fib2(n):
n1, n2 = 0, 1
while n > 1:
n -= 1
n1, n2 = n2, n1 + n2
return n2
n = 1000
recursive = timeit.timeit(lambda: fib1(984), number=n)
iterative = timeit.timeit(lambda: fib2(984), number=n)
print("recursive:", recursive)
print("iterative:", iterative)
"""
output:
recursive: 0.0012962640030309558
iterative: 0.08570116000191774
"""
• fib1 shares its cache between the 1000 calls from timeit; fib2 doesn't get to reuse anything between its 1000 calls. But "Help me understand my own code" isn't really "code review". – Peter Taylor Oct 31 '19 at 8:33
• To complement Peter's answer, see this article. Your lookup dictionary is acting as a mutable default argument. – JAD Oct 31 '19 at 8:49
• If I were being truly pedantic the correct thing to do would have been to just post the last sentence of my earlier comment, so as not to encourage off-topic questions. – Peter Taylor Oct 31 '19 at 8:51
• I made a similar sort of mistake at work recently. We're always taught to profile a bunch of times to take an average, but that can mask if one run (say, the first run) is massively slower than the rest because a bunch of work from the first run is being cached. Peter and JAD explain why, that's happening. If you change n to 1, you'll see the actual runtime performance is what you'd expect: the loop is much faster. As to which measurement you care about, it depends on your use case. If your code needs to calculate fib once, use fib2. If it needes to call it repeatedly, perhaps use fib1. – Josiah Oct 31 '19 at 9:38
• It may be informative to put a print(n) statement at the start of fib1. – Josiah Oct 31 '19 at 9:40 | 596 | 2,114 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.515625 | 4 | CC-MAIN-2021-25 | latest | en | 0.925495 |
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