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# ENGINEERING STATISTICS SOLUTIONS MANUAL
Solutions to Engineering Statistics (9780470631478
Free step-by-step solutions to Engineering Statistics (9780470631478) - Slader Solutions to Engineering Statistics (9780470631478) :: Free Homework Help and Answers :: Slader Free step-by-step solutions to all your questions
Engineering Statistics Solution Manual | Chegg
Solutions Manuals are available for thousands of the most popular college and high school textbooks in subjects such as Math, Science (Physics, Chemistry, Biology), Engineering (Mechanical, Electrical, Civil), Business and more. Understanding Engineering Statistics homework has never been easier than with Chegg Study.
Student Solutions Manual Engineering Statistics, 5e
Jan 24, 2015This is the Student Solutions Manual to accompany Engineering Statistics, 5th Edition. Montgomery, Runger, and Hubele's Engineering Statistics, 5th Edition provides modern coverage of engineering statistics by focusing on how statistical tools are integrated into the engineering problem-solving process. All major aspects of engineering statistics are covered, including descriptive statistics2.2/5(3)Author: Douglas C. Montgomery
Engineering Statistics, Student Solutions Manual by
Apr 04, 2007Engineering Statistics, Student Solutions Manual. Montgomery, Runger, and Hubele provide modern coverage of engineering statistics, focusing on how statistical tools are integrated into the engineering problem-solving process.3.9/5(8)
Solution Manual Engineering Statistics Montgomery
Solution Manual Engineering Statistics Montgomery Engineering Statistics, D. Montgomery, 5th Edition, 2011, John Wiley Co. Montgomery Solution Manual Montgomery Solution Montgomery 3rd Edition Solution Aug 30, 2010 Solution Manual For Fundamentals Of Geotechnical Engineering.
Engineering Statistics 4th Edition Textbook Solutions
Solutions Manuals are available for thousands of the most popular college and high school textbooks in subjects such as Math, Science (Physics, Chemistry, Biology), Engineering (Mechanical, Electrical, Civil), Business and more. Understanding Engineering Statistics 4th Edition homework has never been easier than with Chegg Study.[PDF]
Solutions Manual - wlxtc
Solutions Manual to accompany STATISTICS FOR ENGINEERS AND SCIENTISTS by William Navidi
Engineering statistics : student solutions manual
Public Private login. e.g. test cricket, Perth (WA), "Parkes, Henry" Separate different tags with a comma. To include a comma in your tag, surround the tag with double quotes.[PDF]
Students’ Solutions Manual Probability and Statistics
Students’ Solutions Manual. Probability and Statistics. This manual contains solutions to odd-numbered exercises from the book Probability and Statistics by Miroslav Lovri´c, published by Nelson Publishing. Keep in mind that the solutions provided represent one way of answering a question or solving an exercise. | 590 | 2,969 | {"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} | 2.859375 | 3 | CC-MAIN-2020-10 | latest | en | 0.862614 |
https://ithelp.ithome.com.tw/articles/10243710 | 1,618,304,176,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072175.30/warc/CC-MAIN-20210413062409-20210413092409-00169.warc.gz | 420,600,180 | 12,500 | DAY 10
1
# 10 - Playing with digits
Don't say so much, just coding...
## Instruction
Some numbers have funny properties. For example:
89 --> 8¹ + 9² = 89 * 1
695 --> 6² + 9³ + 5⁴= 1390 = 695 * 2
46288 --> 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51
Given a positive integer n written as abcd... (a, b, c, d... being digits) and a positive integer p
we want to find a positive integer k, if it exists, such as the sum of the digits of n taken to the successive powers of p is equal to k * n.
In other words:
Is there an integer k such as : (a ^ p + b ^ (p+1) + c ^(p+2) + d ^ (p+3) + ...) = n * k
If it is the case we will return k, if not return -1.
Note: n and p will always be given as strictly positive integers.
`````` dig_pow(89, 1) should return 1 since 8¹ + 9² = 89 = 89 * 1
dig_pow(92, 1) should return -1 since there is no k such as 9¹ + 2² equals 92 * k
dig_pow(695, 2) should return 2 since 6² + 9³ + 5⁴= 1390 = 695 * 2
dig_pow(46288, 3) should return 51 since 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51
``````
### Ruby
#### Init
`````` def dig_pow(n, p)
end
``````
#### Sample Testing
`````` Test.assert_equals(dig_pow(89, 1), 1)
Test.assert_equals(dig_pow(92, 1), -1)
Test.assert_equals(dig_pow(46288, 3), 51)
``````
### Javascript
#### Init
`````` function digPow(n, p){
// ...
}
``````
#### Sample Testing
`````` Test.assertEquals(digPow(89, 1), 1)
Test.assertEquals(digPow(92, 1), -1)
Test.assertEquals(digPow(46288, 3), 51)
``````
• Ruby
• JavaScript
# Solution
### Ruby
`````` # Solution 1
def dig_pow(n, p)
number = n.to_s.split('').map(&:to_i)
sum = number.map.with_index(p) { |num, idx| num ** idx }.reduce(:+)
sum % n == 0 ? sum/n : -1
end
def dig_pow(n, p)
sum = n.to_s.chars.map.with_index(p) { |num, idx| num.to_i ** idx }.reduce(:+)
sum % n == 0 ? sum / n : -1
end
``````
### Javascript
`````` // Solution 1
function digPow(n, p){
var sum = n.toString().split('')
.map((num, idx) => Math.pow(parseInt(num), idx+p))
.reduce((a,b) => a + b) / n;
return sum % 1 == 0 ? sum : -1;
}
`````` | 776 | 2,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.529993 |
https://www.classtools.net/QR/5-7ZmfD | 1,701,647,394,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00108.warc.gz | 802,269,127 | 16,396 | PREMIUM LOGIN
ClassTools Premium membership gives access to all templates, no advertisements, personal branding and other benefits!
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# Print
## A. Prior to the lesson:
1. Arrange students into groups. Each group needs at least ONE person who has a mobile device.
2. If their phone camera doesn't automatically detect and decode QR codes, ask students to
• Download a QR reader (e.g. I-Nigma | NeoReader | Kaywa) onto their mobile devices
• Bring these devices into the lesson.
4. Cut them out and place them around your class / school.
## B. The lesson:
1. Give each group a clipboard and a piece of paper so they can write down the decoded questions and their answers to them.
2. Explain to the students that the codes are hidden around the school. Each team will get ONE point for each question they correctly decode and copy down onto their sheet, and a further TWO points if they can then provide the correct answer and write this down underneath the question.
3. Away they go! The winner is the first team to return with the most correct answers in the time available. This could be within a lesson, or during a lunchbreak, or even over several days!
## C. TIPS / OTHER IDEAS
4. A detailed case study in how to set up a successful QR Scavenger Hunt using this tool can be found here.
### Answer
1. 1) 2 mi = _______ft10,560 ft
2. 2) 30 ft = _______yd10 yd
3. 3) 144 in = ______ft12 ft
4. 4) 5 gal = _______ qt20 qt
5. 5) 96 oz = _______ lbs6lbs
6. 6) 15 mm = _______ m0.015m
7. 7) 28 Kg = _______ g28,000 g
8. 8) 17,500 cg = ________ Dg17.5Dg
9. 9) 5 Hl = _________ dl5,000 dl
10. 10) 0.125 m = __________cm12.5 cm
11. 11) 5,280 yd = _________ mi3 mi
12. 12) 3 gal = _________ pints24 pt
13. 13) 4 yds = _________ in144 in
14. 14) 8 qt = __________ cups32 cups
15. 15) 24 fl oz = _________cups3 cups
16. 16) 2,000 m = _________ km2 km
17. 17) 10 Km = _________ m10,000 m
18. 18) 970.4 cm = _________ mm9,704 mm
19. 19) 100 cm = _________ m1 m
20. 20) 7,500 ml = __________ L7.5L
21. 21) Eunice owns 3 elephants that weigh a combine 4.5 tons. How many pounds do Eunice's elephants weigh in total?13,000 lbs
22. 22) Joseph drank 0.875 qt of chocolate milk for breakfast today. How many ounces of chocolate milk did Joseph drink? 28 oz
23. 23) Gabby's bookbag weighs 11,000 lbs, due to her project for Mr. Kennedy. How many tons does Gabby's bookbag weigh?5.5tons
24. 24) Gideon is 53 inches tall. How many feet tall is Gideon?4ft5in
25. 25) Daveon drank 7.125 cups of flavored water at lunch today. How many fluid ounces of flavored water has Daveon drank today?57 fl. oz.
26. 26) Britney ran 6.35 Km yesterday in order to prepare for the Color Run next weekend. How many meters did Britney run?6350 m
27. 27) Carl decided to measure the perimeter of Hull Middle School. He used a meter stick and measured 5,078 meters around the school. How many kilometers did Karl measure?5.078 Km
28. 28) Cassidy weighed her new iPhone 5C and noticed that it weighed 1.26 centigrams less than her iPhone 4S. How many more grams less than the 4S is Cassidy's 5C?0.0126 g
29. 29) Jalen drank 12,500 ml of blueberry syrup at IHOP. How many liters of blueberry syrup did Jalen drink?12.5 Liters
30. 30) Ashish walked to school today. Ashish calculated that he walked 1,565,789 mm. How many Km did Ashish have to walk?1.565789 Km
31. 31) Ms. Tomlinson drives a Ford F-650 truck on the weekends. Her truck weighs 4 tons. How many ounces does her truck weigh?337,920 ounces
32. 32) Mrs. Gibson ran 4.2 Km yesterday. Mr. Kennedy ran 4,850,000 mm yesterday. Who ran farther? How much farther did they run?Mr. Kennedy; 650,000 mm or .65 Km
33. 33) Ms. Johnson's Galaxy phone weighs 1.9 grams. Mr. Wilson's iPhone 4S weighs 210 centigrams. Whose phone weighs the least? Ms. Johnson's
### Measurement Conversions Questions: QR Challenge
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Question 33 (of 33) | 1,737 | 6,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-50 | latest | en | 0.797597 |
https://math.stackexchange.com/questions/895307/a-better-way-to-define-bigcap-s/895641 | 1,576,055,728,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540530452.95/warc/CC-MAIN-20191211074417-20191211102417-00391.warc.gz | 451,578,118 | 32,623 | # A better way to define $\bigcap S$?
In books on set theory $\bigcap S$ is defined for $S \neq \emptyset$ by taking some arbitrary member $x \in S$ and separating those elements from $x$ that belong to every memebr of $S$. It leads to a problem of what to do if $S = \emptyset$. This case is handled separately by defining $\bigcap S = \emptyset$.
Why don't they just define $\bigcap S := \{x \in \bigcup S |$ $x$ belongs to every member of $S\}$. Then if $S$ is empty, $\bigcup S$ will be empty and there is nothing to separate from it. If $S$ is not empty, then the result is the same as if we used the previous definition.
The only drawback of the new definition is that we need the axiom of union to get $\bigcup S$. But it is a standard axiom.
• The problem is that the simple characterization $x\in \bigcap S\iff \forall s\in S\colon x\in s$ does not hold any more; it is replaced by the less handy $\iff (\exists s\in S\colon x\in s)\land (\forall s\in S\colon x\in s)$. Btw, I usually see $\bigcap \emptyset$ left undefined instead of specifically defined to be empty. - The latter has ugly consequences such as $S\subseteq S'\not\rightarrow \bigcap S'\subseteq \bigcap S$. – Hagen von Eitzen Aug 12 '14 at 17:01
• Usually the sets being considered live in some universe $X$. It's natural to have „empty intersection“ = „everything (considered)“. So $\bigcap ∅ = X$. If there is no other universe, then $\bigcap ∅$ is universal class which is not a set. – user87690 Aug 12 '14 at 17:05
• Somewhat related: math.stackexchange.com/questions/6613/…, math.stackexchange.com/questions/370188/…, math.stackexchange.com/questions/151924/… and other posts shown there among linked questions. – Martin Sleziak Aug 12 '14 at 18:18
• Another problem with $\bigcap \emptyset = \emptyset$ is we would no longer have $\bigcap (A \cup B) = \bigcap A \cap \bigcap B$. – Mathemanic Nov 2 '16 at 2:37
I don't think it's ever correct to define $\bigcap \emptyset=\emptyset$. One always expects that taking an intersection of more things should get you a smaller result. (The set of big red ugly expensive things is a subset of the set of big things.) That is, $$\def\S{\mathscr S}\text{if \S\subset \mathscr T then \bigcap \mathscr T\subset \bigcap \S}.$$ But having $\bigcap\emptyset = \emptyset$ would spoil that. So that answers one of your questions, which is why we don't do that. I find it hard to believe that any serious text for set theory would do as you say and make $\bigcap \emptyset = \emptyset$.
In set theories with a universal set $V$, one takes $\bigcap \emptyset= V$ for that reason; then we have $\emptyset\subset \S$ for all $\S$ and also $\bigcap \S \subset \bigcap \emptyset = V$, as we would hope. In set theories without a universal set, most notably ZF, one leaves $\bigcap \emptyset$ undefined.
In ZF there is no axiom of intersection that corresponds to the axiom of union; one uses the specification axioms to define intersections. Recall that this means that for any predicate $\Phi(x)$ and any set $X$ we can construct $\{x\in X\mid \Phi(x)\}$, the subset of $X$ for which $\Phi$ holds. Let's see how this goes in this case. We want to take the set of all $s$ that are in every $S$ that is an element of $\S$. But to use the specification axiom schema, we can't take this condition by itself; we have to attach it to some set $X$ and use it to select the desired subset of $X$.
So in ZF the best we can do is $$\{x\in X\mid \forall S\in \S . x\in S\}$$ for some set $X$, which we might agree to abbreviate as $$\def\S{\mathscr S}\bigcap_{(X)}\S.$$ This is the subset of $X$ whose elements are in every element of $\S$, or we might call it the intersection of $S$ "taken relative to $X$". If there were a universal set $V$ then we could take the intersection relative to $V$ and be done (and note that then $\bigcap_{(V)}\emptyset = V$) but ZF has no universal set.
But it should be clear (or you can take it as an exercise) that no matter what $X$ is, if $S\in \S$, we have $$\bigcap_{(X)}\S \subset \bigcap_{(S)} \S.$$ From this it immediately follows that for any $S, S'\in \S$, then $$\bigcap_{(S)}\S = \bigcap_{(S')}\S$$ so it doesn't matter which element of $\S$ we take the intersection relative to.
This shows that when $\S$ is nonempty, there is a unique maximal relative intersection of the elements of $S$, which we can abbreviate as simply $\bigcap \S$. And since $$\bigcap_{(X)}\S = X\cap \bigcap \S$$ the relative intersections weren't getting us anything interesting, and we lose nothing to forget about them. In particular it doesn't matter if we take $X = \bigcup \S$ as you suggest, because the intersection is the same regardless. So we may as well take the simpler definition.
On the other hand, when $\S=\emptyset$, we have $\bigcap_{(X)} \S = X$, which is rather useless, so we lose nothing to forget about it.
• Carl Mummert and Arturo Magidin agree in math.stackexchange.com/questions/6613/… that some authors do make $\bigcap\emptyset=\emptyset$, to my amazement. – MJD Aug 13 '14 at 11:54
• Another potential problem with defining $\bigcap \emptyset=\emptyset$ would be no longer having $\bigcap (A \cup B) = \bigcap A \cap \bigcap B$. – Mathemanic Nov 26 '16 at 7:50 | 1,493 | 5,225 | {"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} | 3.453125 | 3 | CC-MAIN-2019-51 | latest | en | 0.849382 |
http://www.physicsforums.com/showthread.php?s=7a32b6c30a88cf7a52a11d105e42f2d4&p=4485832 | 1,406,329,297,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997894865.50/warc/CC-MAIN-20140722025814-00024-ip-10-33-131-23.ec2.internal.warc.gz | 1,029,299,056 | 8,432 | # Basic B-spline Curve for a Dummy?
by vee6
Tags: basic, bspline, curve, dummy
P: 28 What first thing I should do to solve the function of a basic open b-spline curve where the coordinate of its control points are such as written below? x1,y1 = 0,0 x2,y2 = 1,3 x3,y3 = 3,5 x4,y4 = 5,6 x5,y5 = 7,5 x6,y6 = 9,3 x7,y7 = 10,0
P: 28 What is the next?
P: 6,035
Basic B-spline Curve for a Dummy?
Quote by vee6 What is the next?
P: 28 What is the b-spline interpolation? Please show me the example.
Mentor
P: 21,216
Quote by vee6 What is the b-spline interpolation? Please show me the example.
Are you asking us to do your work for you? That's not how it works here at Physics Forums.
Did you find some references as mathman suggested? If so, do you have any questions about the process?
P: 28
Quote by Mark44 Are you asking us to do your work for you?
No.
I am a dummy and want to learn about b-spline.
I am not working yet.
Quote by Mark44 Did you find some references as mathman suggested? If so, do you have any questions about the process?
Yes I did.
However it's not b-spline but spline without "b".
It's from wikipedia.
Spline interpolation.
Please show me an easy simple example of the b-spline interpolation.
Mentor P: 21,216 Here's a link to a Wiki article on b-splines: http://en.wikipedia.org/wiki/B-spline. There are some examples about halfway down the page.
P: 28
Quote by Mark44 There are some examples about halfway down the page.
Which one?
Mentor
P: 21,216
Take your pick. As for myself, I would go for the Uniform cubic b-spline.
Quote by vee6 Please show me an easy simple example of the b-spline interpolation.
Without knowing what your mathematical background is, it's difficult to tell what constitutes "easy" or "simple."
P: 343 A few questions for you: 1) When you talk about "solving the function" of the B-spline, what do you mean? Are you talking curve evaluation? 2) What degree do you want the resulting B-spline curve? 3) What kind of continuity do you want between points? Do you want C0? C1? C2? 4) What interval do you want to the resulting B-spline curve to have? [0,1] or something else? If you can answer those questions, I'm sure people would be willing to help you.
P: 4 If you want some more info look at this site: http://www.cs.mtu.edu/~shene/COURSES/cs3621/NOTES/ if you really want to understand i suggest you buy a book called An Introduction to NURBS: With Historical Perspective. it will start you out with some parametric curves and bezer to then introduce b-splines. I suggest you learn how to program, if you don't already since it can be very monotonous. Other than that I can't really help you since I barely understand it myself.
Related Discussions Calculus & Beyond Homework 7 Programming & Computer Science 1 Calculus 5 Differential Geometry 1 Mechanical Engineering 3 | 764 | 2,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-23 | latest | en | 0.927264 |
https://www.all-dictionary.com/what-does-mean-Average | 1,508,700,243,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825436.78/warc/CC-MAIN-20171022184824-20171022204824-00335.warc.gz | 870,030,029 | 99,601 | # What does Average mean?
#### Average meaning in General Dictionary
to create or occur in a mean or medial sum or quantity to total or to be on the average whilst the losses of this owners will average twenty-five dollars each these spars average ten legs in total
View more
• to obtain the suggest of when sums or quantities tend to be unequal to lessen to a mean
• with respect to a typical or mean medial containing a mean proportion of a mean dimensions quality ability etc ordinary usual as a typical price of revenue an average number of rainfall the average Englishman beings associated with the normal stamp
• That service which a tenant owed their lord become done-by the work beasts of the tenant given that carriage of wheat turf an such like
• lacking exceptional high quality or capability
• concerning or constituting more regular value in a distribution
• relating to or constituting the middle value of an ordered collection of values (or the average of this middle two in a set with a straight range values)
• approximating the statistical norm or average or expected worth
• around the middle of a scale of evaluation
• lacking special distinction, rank, or condition; generally experienced
• add up to or arrived at a typical, without reduction or gain
• attain or reach on average
• compute the average of
• (recreations) the proportion of successful shows to possibilities
• a statistic explaining the positioning of a distribution
• an intermediate scale value considered to be typical or usual
• That solution which a tenant owed their lord, becoming done by the task beasts associated with tenant, whilst the carriage of wheat, turf, etc.
• A tariff or duty on products, etc.
• Any fee in addition to the regular fee for freight of goods sent.
• A contribution to a loss or fee which was imposed upon one of several when it comes to basic benefit; harm done by water perils.
• The equitable and proportionate distribution of reduction or expense among all interested.
• A mean percentage, medial sum or amount, crafted from unequal amounts or amounts; an arithmetical suggest. Thus, if A loses 5 bucks, B 9, and C 16, the sum is 30, therefore the average 10.
• Any medial estimation or basic declaration derived from a comparison of diverse certain instances; a method or typical dimensions, quantity, high quality, price, etc.
• In the English corn trade, the medial price of the number of forms of grain when you look at the main corn areas.
• regarding an average or mean; medial; containing a mean percentage; of a mean dimensions, high quality, ability, etc.; ordinary; normal; as, a typical rate of profit; the average amount of rain; the typical Englishman; beings of this normal stamp.
• in accordance with the laws of averages; as, the loss must certanly be made good by normal contribution.
• to get the suggest of, when sums or volumes tend to be unequal; to reduce to a mean.
• To divide among several, relating to a given proportion; since, to average a loss.
• to-do, accomplish, get, etc., on an average.
• to create, or exist in, a mean or medial sum or amount; to add up to, or even be, on the average; because, the losings for the owners will average 25 bucks each; these spars average ten foot in length.
#### Average meaning in Economics Dictionary
A number that is computed to summarise a team of numbers. Many popular average may be the mean, the sum of the numbers divided by but numerous numbers you can find in team. The median may be the middle worth in several numbers ranked to be able of size. The mode is the quantity that develops most often in several figures. Make the following band of numbers: 1, 2, 2, 9, 12, 13, 17 The mean is 56/7=8, The median is 9, The mode is 2
#### Average meaning in Law Dictionary
nd see Peters v. Warren Ins. Co., 19 Fed. Cas. 370.
#### Average meaning in Etymology Dictionary
belated 15c., "financial loss incurred through damage to products in transit," from French avarie "damage to deliver," and Italian avaria; a word from 12c. Mediterranean maritime trade (compare Spanish averia; various other Germanic forms, Dutch avarij, German haferei, etc., are from Romanic languages), which can be of uncertain source. Often traced to Arabic 'arwariya "damaged product." Meaning shifted to "equal sharing of such reduction by the interested events." Transferred feeling of "statement of a medial estimation" is very first taped 1735. The mathematical expansion is from 1755.
View more
• 1770; see average (n.).
• 1769, from average (n.). Related: Averaged; averaging.
#### Average meaning in Business Dictionary
1. General: Number or quantity this is certainly in-between (advanced to) a few amounts and figures. See also suggest. 2. General insurance: The expression 'subject to average' implies that if amount guaranteed during the time of a reduction is less than the insurable worth of the insured residential property, the quantity reported underneath the policy are reduced in percentage to your under-insurance. Also known as average clause. See also coinsurance. 3. Marine insurance coverage: 'Average' means limited (loss); particular-average loss is borne by one party, and general-average reduction is provided by all worried. Today largely replaced with institute cargo conditions A, B, or C. 4. Quality-control: typical expression of the centering of a circulation calculated by dividing the total noticed values because of the amount of findings. 5. Securities trading: accordingly weighted mean of a 'basket' of selected securities representing basic market behavior. Dow Jones Industrial Average (DJIA), which measures the overall performance for the stocks of 30 the biggest United States corporations, is certainly one well-known instance.
#### Average meaning in Computer Science Dictionary
instead known as the arithmetic suggest, average could be the amount of a few figures, divided because of the total amount of figures. Like, suppose we have the after series of numbers: 1, 2, 3, 4, 1, 2, and 3. The sum of the these figures is 16, 16 split by 7 is 2.28, consequently, 2.28 could be the average of those figures.
#### Average meaning in General Dictionary
(n.) That service which a tenant owed his lord, become carried out by the work beasts for the tenant, because the carriage of grain, turf, etc.
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• (letter.) A tariff or responsibility on goods, etc.
• (letter.) Any charge besides the regular charge for cargo of products transported.
• (n.) A contribution to a loss or charge which was imposed upon one of many when it comes to basic benefit; damage done-by water perils.
• (n.) The fair and proportionate circulation of reduction or expenditure among all interested.
• (letter.) A mean percentage, medial amount or volume, crafted from unequal amounts or volumes; an arithmetical suggest. Thus, if A loses 5 dollars, B 9, and C 16, the sum is 30, as well as the normal 10.
• (n.) Any medial estimation or basic statement based on an assessment of diverse particular instances; a medium or typical size, volume, high quality, price, etc.
• (n.) When you look at the English corn trade, the medial cost of the several kinds of grain in major corn markets.
• (a.) with respect to an average or mean; medial; containing a mean proportion; of a mean size, quality, capability, etc.; ordinary; usual; as, an average price of profit; a typical amount of rain; the average Englishman; beings of the average stamp.
• (a.) Based on the regulations of averages; as, the loss needs to be made good-by average share.
• (v. t.) To get the suggest of, whenever sums or amounts are unequal; to lessen to a mean.
• (v. t.) To divide among several, relating to certain proportion; as, to average a loss.
• (v. t.) To do, achieve, get, etc., on an average.
• (v. i.) to create, or occur in, a mean or medial sum or quantity; to amount to, or even to be, on an average; as, the losses associated with the proprietors will average twenty five dollars each; these spars average ten foot in length.
#### Sentence Examples with the word Average
In 1900 the wheat acreage in Ontario was 1,487,633, producing 28,418,907 bushels, an average yield of 19.10 bushels per acre. | 1,853 | 8,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-43 | latest | en | 0.913106 |
https://thebikeyear.com/1007203/inspiration-exponents-math-worksheets/ | 1,642,490,820,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300805.79/warc/CC-MAIN-20220118062411-20220118092411-00142.warc.gz | 609,422,440 | 11,453 | HomeWorksheet Template ➟ 1 Awesome Exponents Math Worksheets
# Awesome Exponents Math Worksheets
Explore all of our exponents worksheets from reading and writing simple exponents to negative exponents and equations with exponents. Order of Operations with Exponents Worksheets.
Our Exponents Worksheets Provide Practice That Reinforces The Properties Of Exponents Including The Basic P Exponent Worksheets Math Worksheet Math Worksheets
Exponents math worksheets. We also have some worksheets with the power of ten in which math learners need to multiply or divide numbers and decimals by a power of ten. In mathematics exponents are the contracted form of larger value of multiplication. Exponent worksheets including an introduction to exponents reading and writing simple exponents powers of ten whole number fractional and decimal bases negative exponents and equations with exponents.
We do not enough time to solve such lengthy value that is why we need short cut. You will find addition lessons worksheets homework and quizzes in each section. These Worksheets for Grade 7 Mathematics Exponents and Powers cover all important topics which can come in your standard 7 tests and examinations.
This is an advanced skill. For example three multiply by six time is equal to 729 3 3 3 3 3 3 729 so this multiplication method is long. This file contains 30 task cards with exponent problems on them.
Download math worksheet on finding square roots cube roots and applying different operations on them to practice and score better in your classroom tests. Discover learning games guided lessons and other interactive activities for children. Worksheets are Exponents bundle 1 Understanding exponents and square roots Gmat math exponents and roots excerpt Properties of exponents Laws of exponents Unit 8 exponents and powers Exponent rules practice Work 3 exponents grade 8 mathematics.
Exponents Worksheet – 3. Free printable worksheets for CBSE Class 7 Mathematics Exponents and Powers school and class assignments and practice test papers have been designed by our highly experienced class 7 faculty. Students learn about the Negative Exponent Property and the Zero Exponent Property in this eighth-grade math worksheet.
Discover learning games guided lessons and other interactive activities for children. Utilize our printable laws of exponents worksheets as an essential guide for operating on problems with exponents. Free printable worksheets provided by K5 learning.
Ad Download over 30000 K-8 worksheets covering math reading social studies and more. By steadily practicing these worksheets students of grade 7 grade 8 and high school will be able to ace their tests in problems using the laws of exponents. Find the sums differences and missing addends.
Exponents worksheets and online activities. Use this worksheet to help your students see the patterns within the powers of ten. It may be printed downloaded or saved and used in your classroom home school or other educational environment to help someone learn math.
After the completion of studying Class 8 Mathematics Exponents and Powers students should prefer printable worksheets for further preparation and revision. Powers Add to my workbooks 55 Download file pdf Embed in my website or blog Add to Google Classroom. Compare the values of the exponents using the greater and less than symbols.
Remember the E in PEMDAS stands for exponents. Students have to take operations into account with exponents. Introduce learners to exponents and related vocabulary with this sixth-grade practice worksheet.
This math worksheet was created on 2016-01-19 and has been viewed 325 times this week and 1115 times this month. Students can challenge themselves by solving problems. These worksheets are a great introduction to pre-algebra skills for students.
Free interactive exercises to practice online or download as pdf to print. Answer the questions to demonstrate knowledge of exponents. Class VII Math Integers Fractions Decimals Rational Numbers Exponents Ratio Proportion Percentage Algebraic Expressions Linear Equations Unitary Method Online Test Worksheets Sample Question.
7th grade exponents worksheets provide a great way to test students knowledge and skills of math vocabulary and reasoning. Order Of Operations With Exponents 1. To link to this Exponents and Base Worksheets page copy the following code to your site.
Exponents Math Worksheets for Kids. Displaying all worksheets related to – Exponents And Roots. Exponents worksheets with writing factors finding square roots cube roots simplyfing exponent expressions and different operations on exponents.
K5 Learning offers free worksheets flashcards and inexpensive workbooks for kids in kindergarten to grade 5. Ad Download over 30000 K-8 worksheets covering math reading social studies and more. This assortment of printable exponents worksheets designed for grade 6 grade 7 grade 8 and high school is both meticulous and prolific.
Laws of Exponents Worksheets. On this page you find our math grade 6 exponent worksheets for primary math school covering basic exponent worksheets identify the base and exponent worksheets worksheets with squares and cubes the expanded version of exponents and more. Exponents Worksheets On this page you will find.
As well as cracking the distinctly advantageous aspects of exponents a unique math shorthand used to denote repeated multiplication students gain an in-depth knowledge of parts of an exponential notation converting an expression with exponents to a. Order of Operations with Parenthesis and Exponents. Steps to Prepare for Final Exams by Worksheets for Class 8 Mathematics Exponents and Powers.
Included within this set are worksheets. Exponents and Base Worksheets. So mathematician take leverage of exponents.
Here are some steps provided by expert teachers at StudiesToday which every student can follow in order to improve their academic. These worksheets make exponents easier with the help of simple step-by-step worksheets. A complete list of all of our math worksheets relating to ExponentsChoose a specific addition topic below to view all of our worksheets in that content area.
Write the exponents as numbers in expanded form and standard form.
Math Worksheets Simple Exponents And Powers Of Ten Exponent Worksheets Math Worksheets Algebra Worksheets
Mixed Exponent Rules All Positive A Algebra Worksheet Exponent Worksheets Algebra Worksheets Exponent Rules
Worksheet Practice Mixing Exponents With Addition And Subtraction Operations Worksheets Exponent Worksheets Math Expressions
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Adding Exponents Worksheets Including Simple Problems Where Exponents Are Combined And Order O Exponent Worksheets Word Problem Worksheets Order Of Operations
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13 Best Images Of Positive Exponents Worksheets Powers And Exponents Worksheet Negative Great 13 Best Exponent Worksheets Exponent Rules Algebra Worksheets
Exponent Rules Exponent Rules Exponent Worksheets Exponents | 1,362 | 7,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2022-05 | latest | en | 0.858861 |
https://www.java-forums.org/new-java/64582-how-do-i-put-multiple-charats-one-char-print.html | 1,521,958,214,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257651820.82/warc/CC-MAIN-20180325044627-20180325064627-00586.warc.gz | 821,767,097 | 3,450 | # How do i put multiple charAt's in one char?
• 11-03-2012, 12:00 AM
How do i put multiple charAt's in one char?
char threeMorse = myMorse[2].charAt(0);
this works, it returns the char at the 0 spot of this array spot which return 3
What if I want to do something like char threeMorse = myMorse[2].charAt(0) + myMorse[2].charAt(1); <------This doesnt work but I want something similar to this logic
Is this possible? Thanks
• 11-03-2012, 12:03 AM
Fubarable
Re: How do i put multiple charAt's in one char?
Perhaps you should explain what you're trying to do in general terms, not in coding terms. What do you want to achieve with this code?
• 11-03-2012, 12:04 AM
Re: How do i put multiple charAt's in one char?
I want to put the first two spots of an arrays data into one char
• 11-03-2012, 12:04 AM
Re: How do i put multiple charAt's in one char?
It has some extra useless data after it that I dont need so I just want the first 2 spots
• 11-03-2012, 12:10 AM
Fubarable
Re: How do i put multiple charAt's in one char?
Quote:
I want to put the first two spots of an arrays data into one char
I don't understand this. It almost seems that you're trying to wedge two char into one char, and that obviously can't be done. Again, please tell me what problem this part of your program is trying to solve, rather than how you're trying to solve it with code.
• 11-03-2012, 12:19 AM
Re: How do i put multiple charAt's in one char?
Ah yeah I guess thats not possible with a char but anyways
I read a morse code from a text and I need the user to input a number or letter, and the program needs to spit the morse code equivalent back out at the user. I already stored the data from the morse txt into an array. THe problem is the morse .txt is listed with a number or letter in front of it. SO S in the morse txt looks like:
'S ---'
When the user asks for S I need the program to return --- and not S ---. I'm so close to getting it finished I need to figure out how to take off the first 2 parts of the data in each array slot whenever they ask for it you know?
• 11-03-2012, 12:20 AM
Re: How do i put multiple charAt's in one char?
How can I add charAt's? How do I concatenate them
• 11-03-2012, 12:27 AM
Fubarable
Re: How do i put multiple charAt's in one char?
It sounds like what you want to use is a Map, perhaps a HashMap<Character, String> so you can associate the char 's' with the String "---". If you have this and load it with all the char and dot-dash Strings that represent Morse code, then you can get the Morse String that corresponds with the 's' character by doing
String morseString = myMap.get('s');
• 11-03-2012, 12:37 AM
Re: How do i put multiple charAt's in one char?
That sounds confusing I never used maps
• 11-03-2012, 12:42 AM
Fubarable
Re: How do i put multiple charAt's in one char?
They're not that hard once you get used to using them. Say for instance you created a map to translate Spanish words to English:
Code:
`HashMap<String, String> spanishToEnglishMap = new HashMap<String, String>();`
You could fill it like so:
Code:
```spanishToEnglishMap.put("hola", "hello"); spanishToEnglishMap.put("adiós", "goodbye"); spanishToEnglishMap.put("muchacho", "boy"); spanishToEnglishMap.put("muchacha", "girl"); spanishToEnglishMap.put("mujer", "woman"); spanishToEnglishMap.put("hombre", "man");```
and then use it like so:
Code:
```String text = "The English word for mujer is " + spanishToEnglishMap.get("mujer"); System.out.println(text);```
• 11-03-2012, 12:47 AM | 997 | 3,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-13 | latest | en | 0.946651 |
https://gis.stackexchange.com/questions/322064/non-earth-system-in-qgis | 1,713,828,784,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00369.warc.gz | 238,826,127 | 40,424 | # Non-Earth System in QGIS?
What I need is a function for Non Earth Coordinates in QGIS (both projects and layers). They should be mappable to any projected coordinate system (like SWEREF99TM).
Why: In the Mining industry we often use Non Earth systems for different reasons. Like Geophysics we want to define North in a convenient direction. Could be South East we don't care about the real world at this stage. The same if an ore body is going south-east we may define that direction as north.
In many softwares (Microstation or MapInfo) we use we can have on the fly conversion between lets say myCoordinateSystem1 (cartesian system) to EPSG3006 no problem. But I have not found a good function for this in QGIS.
I wonder if the Proj4 limits QGIS so you can't do this? Any solutions for this?
• EPSG3006 is found in qgis, and to add SWEREF99M usin like this `"+proj=utm +zone=33 +ellps=GRS80 +towgs84=0,0,0,0,0,0,0 +units=m +no_defs"` May 8, 2019 at 9:51
• How is the Cartesian system defined in MapInfo/Microstation? Are there control points in both systems? There has to be some way to georeference the local system to a projected/geographic coordinate system. May 8, 2019 at 18:12
• The Cartesian system is defined as x,y,z. And the origio of that cartesian system is defined in one point in a earth system like Sweref99TM or WGS84. @mkennedy The system will then be set at that point. What MS and MapInfo does then is to fix scale issues. It will not be exact and the problem grows with distance. But when we do work in nature we are never in big areas so that gives us a problem. May 9, 2019 at 10:49
• @Fran Raga, yes I know that Sweref99TM is defined that is not the problem, I need a non earth system that I can relate to Sweref99 or WGS84 or whatever. Meaning that I define my own system with my own north and so on. Then I relate that to whatever geografical system and the software does a rotation and translation of my own defined system. May 9, 2019 at 10:54
• @PeterW have you solved your problem? I getting stuck with the same issue... Apr 22, 2020 at 9:50
I work with the same kind of things all the time and the way I do it is create a custom projection. Go to `Settings > Custom Projections`. It is easiest to enter a proj4 string and reference it to a wgs84 lat-lon (but you can do other projections too).
Example string:
`+proj=omerc +lat_0=0 +lonc=0 +alpha=45 +gamma=0 +k=1 +x_0=20000 +y_0=55000 +ellps=WGS84 +towgs84=0,0,0,0,0,0,0 +units=m +no_defs +type=crs`
The important values are:
• `+lat_0` is the origin latitude
• `+lonc` is the origin longitude
• `+alpha` is the rotation from north
• `+k` is the meters scale factor
• `+x_0` is the x coordinate you want in your "non-earth" system to be at the origin lat-lon specified above.
• `+y_0` is the y coordinate you want in your "non-earth" system to be at the origin lat-lon specified above.
I set my origin to somewhere near my project, or better yet at my survey control base station. I use offsets to make sure work never takes me to negative coordinates and I use very different x and y numbers so you can't easily get them confused (eg 20000/55000). As you said, over short distances a scale factor of 1 is fine. | 887 | 3,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.919002 |
http://mathhelpforum.com/advanced-algebra/45185-symmetric-group-metric-print.html | 1,503,016,795,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104172.67/warc/CC-MAIN-20170817225858-20170818005858-00264.warc.gz | 288,284,742 | 2,764 | # symmetric group metric
• Aug 3rd 2008, 05:31 AM
thippli
symmetric group metric
Let d be a metric on the symmetric group Sn. Suppose d is left invariant . i.e, d(ca,cb)=d(a,b) for all a,b,c in Sn.
Is there any method to convert it as a right invariant metric.
i.e, d(ac,bc)=d(a,b) for all a,b,c in Sn ?
• Aug 3rd 2008, 06:16 AM
NonCommAlg
Quote:
Originally Posted by thippli
Let d be a metric on the symmetric group Sn. Suppose d is left invariant . i.e, d(ca,cb)=d(a,b) for all a,b,c in Sn.
Is there any method to convert it as a right invariant metric.
i.e, d(ac,bc)=d(a,b) for all a,b,c in Sn ?
let G be any group and $d$ a left invariant metric defined on G. then $\rho$ defined by $\rho(x,y)=d(x^{-1},y^{-1}), \ \forall \ x,y \in G$
is a right invariant metric on G. (very easy to see. so i leave the proof to you!) | 269 | 825 | {"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": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-34 | longest | en | 0.718639 |
http://www.prenhall.com/books/esm_013897067X.html | 1,537,480,911,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156622.36/warc/CC-MAIN-20180920214659-20180920235059-00249.warc.gz | 392,263,462 | 5,549 | ## Elements of Real Analysis, 1/e
Herbert S. Gaskill, Memorial University of Newfoundland
Pallasena P. Narayanaswami, Memorial University of Newfoundland
Published August, 1997 by Prentice Hall Engineering/Science/Mathematics
Cloth
ISBN 0-13-897067-X
mailings
on this subject.
Real Analysis-Mathematics
Unique in approach, this text is specifically designed to help today's weaker students achieve competence in serious mathematics. This text develops mathematical sophistication with a thorough knowledge and understanding of limiting processes. Unlike other texts — which merely present the theory and follow with a set of exercises in a routine fashion — this text provides insight into the techniques of analysis — explaining at the end of each key definition, example, and theorem what students must do to come to grips with the concepts. Concepts are presented slowly and include the details of calculations as well as substantial explanations as to how and why one proceeds in the given manner. Comprehensive in coverage, the text explores the principles of logic, the axioms for the real numbers, limits of sequences, limits of functions, differentiation and integration, infinite series, convergence, uniform convergence for sequences of real-valued functions — and treats several topics often slighted or omitted in other texts.
Emphasizes the understanding of fundamental and basic concepts before tackling unwieldy generalizations.
Addresses students' lack of skill at symbol manipulation and basic algebra head-on.
• Consistently stresses the role of algebra and algebraic manipulation in the construction of proofs.
• Follows each theorem with a standard type of mathematical proof — with calculations performed in great detail so students are not left with a feeling that mathematics is mysterious or that understanding is beyond their abilities.
• As the subject unfolds, the proofs may have “gaps” to be filled by the student.
Follows each definition, proof and example with a Discussion section which:
• Explains the essential piece of insight which is being presented.
• Presents material on intuition, on what thought processes might lead to the proof which was just presented, etc.
Uses the words WHY? and HOW? throughout — inviting students to become active participants and to supply a missing argument or a simple calculation.
Encourages students to think like mathematicians.
• Helps students view real analysis as a cohesive whole which provides answers to a series of generic questions.
• E.g., Each time a new limiting process is introduced, the question is asked of how the process behaves with respect to the algebraic operations on the underlying structure; plausible conjectures based on previous knowledge are generated; and then students are encouraged to review the previous theory and attempt to generate the questions and the answers.
Contains more than 1000 individual exercises.
• Some are very simple, requiring little more than the completion of the details of an argument or a simple calculation.
• Others are much more difficult in that they ask students to “sort out” a situation: to act like mathematicians and find the desired results or to structure an investigation to achieve the required end (e.g., to find a characterization of those situations in which a function is lower semicontinuous at a point at which it has a jump discontinuity).
Emphasizes the understanding of fundamental and basic concepts before tackling unwieldy generalizations.
Addresses students' lack of skill at symbol manipulation and basic algebra head-on.
• Consistently stresses the role of algebra and algebraic manipulation in the construction of proofs.
• Follows each theorem with a standard type of mathematical proof — with calculations performed in great detail so students are not left with a feeling that mathematics is mysterious or that understanding is beyond their abilities.
• As the subject unfolds, the proofs may have “gaps” to be filled by the student.
Follows each definition, proof and example with a Discussion section which:
• Explains the essential piece of insight which is being presented.
• Presents material on intuition, on what thought processes might lead to the proof which was just presented, etc.
Uses the words WHY? and HOW? throughout — inviting students to become active participants and to supply a missing argument or a simple calculation.
Encourages students to think like mathematicians.
• Helps students view real analysis as a cohesive whole which provides answers to a series of generic questions.
• E.g., Each time a new limiting process is introduced, the question is asked of how the process behaves with respect to the algebraic operations on the underlying structure; plausible conjectures based on previous knowledge are generated; and then students are encouraged to review the previous theory and attempt to generate the questions and the answers.
Contains more than 1000 individual exercises.
• Some are very simple, requiring little more than the completion of the details of an argument or a simple calculation.
• Others are much more difficult in that they ask students to “sort out” a situation: to act like mathematicians and find the desired results or to structure an investigation to achieve the required end (e.g., to find a characterization of those situations in which a function is lower semicontinuous at a point at which it has a jump discontinuity).
Stresses and reviews elementary algebra and symbol manipulation as essential tools for success at the kind of computations required in dealing with limiting processes.
• Emphasizes that algebraic calculations constitute the bulk of most proofs in elementary analysis and in advanced analysis as well — and thus eases students' difficulties at an early stage.
• Directly presents, or includes as exercises, the most important algebraic tools — e.g., proves the binomial theorem to illustrate proof by induction. Similarly treats basic methods of factoring, sum of a geometric series, methods for dealing with inequalities, etc.
Begins with an axiomatic development of the reals for a deep understanding of the real numbers.
• Develops the positive integers within the context of the axioms for a complete ordered field — an axiomatic foundation from which all proofs can be generated with no hand-waving.
• Includes an appendix on set theory.
Stresses and develops the delta-epsilon (…d — …e) definition as the basis of real analysis.
• Follows each presentation of a delta-epsilon definition with numerous examples and a series of problems which force students to apply the definition directly to specific cases — e.g., students are expected to be able to prove from first principles that f(x) = x^4 cannot be uniformly continuous on R but is uniformly continuous on [-2, 96].
• Provides a framework in which to perform the computations and emphasizes the relationship of the present computations to those which students have already mastered.
• Shows that in most cases the computations which are required amount to no more than high school algebra.
Emphasizes — through all stages and in many examples — the need for the correct formulation of a negation of key statements.
Includes a number of special topics and/or increased depth of treatment of selected topics often slighted or omitted in other texts, e.g.:
• Improper integrals, Riemann-Stieltjes integration and the Weierstrass Approximation theorem.
• Detailed exposition relative to the negation of the main definitions, delicate tests for convergence of series and the approach taken to the transcendental functions.
0. Basic Concepts.
1. Limits of Sequences.
2. Limits of Functions.
3. A Little Topology.
4. Differentiation.
5. Integration.
6. Infinite Series of Constants.
7. Sequences of Functions.
8. Infinite Series of Functions.
9. Transcendental Functions.
Appendix on Set Theory. | 1,534 | 7,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | latest | en | 0.907659 |
http://www.internetdict.com/answers/how-do-you-solve-a-quadratic-equation.html | 1,516,590,506,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890947.53/warc/CC-MAIN-20180122014544-20180122034544-00322.warc.gz | 449,692,473 | 7,782 | How do You Solve a Quadratic Equation?
To solve a quadratic equation, it will be foil backwards. So take the equation you are given and make two parenthesis like you would find in a regular foil equation. Then take your first and last numbers in the original equation and find what two number make each. Then you just plug them in and calculate by foil to see what works. | 77 | 372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-05 | latest | en | 0.954577 |
https://www.chase2learn.com/paying-up-codechef-solutionproblem-code-marcha1/ | 1,696,425,502,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511369.62/warc/CC-MAIN-20231004120203-20231004150203-00069.warc.gz | 755,056,547 | 34,467 | # Paying up Codechef Solution|Problem Code: MARCHA1
Hello coders, today we are going to solve Paying up Codechef Solution|Problem Code: MARCHA1.
### Problem
In the mysterious country of Byteland, everything is quite different from what you’d normally expect. In most places, if you were approached by two mobsters in a dark alley, they would probably tell you to give them all the money that you have. If you refused, or didn’t have any – they might even beat you up.
In Byteland the government decided that even the slightest chance of someone getting injured has to be ruled out. So, they introduced a strict policy. When a mobster approaches you in a dark alley, he asks you for a specific amount of money. You are obliged to show him all the money that you have, but you only need to pay up if he can find a subset of your banknotes whose total value matches his demand. Since banknotes in Byteland can have any positive integer value smaller than one thousand you are quite likely to get off without paying.
Both the citizens and the gangsters of Byteland have very positive feelings about the system. No one ever gets hurt, the gangsters don’t lose their jobs, and there are quite a few rules that minimize that probability of getting mugged (the first one is: don’t go into dark alleys – and this one is said to work in other places also).
### Input
The first line contains integer t, the number of test cases (about 100). Then t test cases follow. Each test case starts with n, the number of banknotes in your wallet, and m, the amount of money the muggers asked of you. Then n numbers follow, representing values of your banknotes. Your wallet does not hold more than 20 banknotes, and the value of a single banknote is never more than 1000.
### Output
For each test case output a single line with the word ‘Yes’ if there is a subset of your banknotes that sums to m, and ‘No’ otherwise.
### Example
```Input:
5
3 3
1
1
1
5 11
1
2
4
8
16
5 23
1
2
4
8
16
5 13
1
5
5
10
10
20 132
17
6
4
998
254
137
259
153
154
3
28
19
123
542
857
23
687
35
99
999
Output:
Yes
Yes
Yes
No
Yes
```
Explanation: For example, in the last case you have to pay up, since: 6+3+123=132.
### Paying up CodeChef Solution in JAVA
```import java.util.Scanner;
import java.util.stream.IntStream;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for (int tc = 0; tc < t; tc++) {
int n = sc.nextInt();
int m = sc.nextInt();
int[] banknotes = new int[n];
for (int i = 0; i < banknotes.length; i++) {
banknotes[i] = sc.nextInt();
}
System.out.println(solve(banknotes, m) ? "Yes" : "No");
}
sc.close();
}
static boolean solve(int[] banknotes, int m) {
int n = banknotes.length;
for (int code = 0; code < (1 << n); code++) {
boolean used[] = decode(code, n);
if (IntStream.range(0, n).filter(i -> used[i]).map(i -> banknotes[i]).sum() == m) {
return true;
}
}
return false;
}
static boolean[] decode(int code, int size) {
boolean[] result = new boolean[size];
for (int i = 0; i < result.length; i++) {
result[i] = code % 2 == 1;
code >>= 1;
}
return result;
}
}```
Disclaimer: The above Problem (Paying up) is generated by CodeChef but the solution is provided by Chase2learn. This tutorial is only for Educational and Learning purpose.
Sharing Is Caring | 912 | 3,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.920485 |
https://oilandgasguru.com/how-to-determine-direction-of-electric-field | 1,669,892,540,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710808.72/warc/CC-MAIN-20221201085558-20221201115558-00186.warc.gz | 477,747,686 | 9,719 | # How to determine direction of electric field?
15
Date created: Tue, Apr 20, 2021 11:25 AM
Date updated: Thu, Jun 23, 2022 5:00 AM
Content
## Top best answers to the question «How to determine direction of electric field»
• The direction of an electrical field at a point is the same as the direction of the electrical force acting on a positive test charge at that point. For example, if you place a positive test charge in an electric field and the charge moves to the right, you know the direction of the electric field in that region points to the right.
The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away positive charge, and toward a negative one. At the origin, q1 will produce an E-field vector that points left, and q2
The direction of an electrical field at a point is the same as the direction of the electrical force acting on a positive test charge at that point. For example, if you place a positive test charge in an electric field and the charge moves to the right, you know the direction of the electric field in that region points to the right.
In this video David explains how to determine the direction of the electric field from positive and negative charges. He also shows how to determine the dire...
Well, the force on a particle in a uniform electric field is F =q E.This means the force and the electric field point in the same direction for a positive test charge. This is often how the direction of the electric field is defined: the direction in which a positive test charge would be accelerated.
The net electric field determined shows the magnitude of the net field at location X in each respective direction. Let's graphically represent the net electric field and determine the net electric ...
Explains how to calculate the electric field of a charged particle and the acceleration of an electron in the electric field. You can see a listing of all my...
Assuming there is a time varying magnetic field (B), how to determine the direction of the induced electric field due to B. and to which parameters does it depend? P.S. determining the electric fi...
The direction of the electric field is always directed in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding the source charge. Since electric field is a vector quantity, it how to write a contract for lending money be represented by a vector arrow.
E (scalar, magnitude) = Q/ (4*π*ε0)*R2 where Q is the electrical charge and ε0 = 8.854 * 10^-12 [F/m] The direction and orientation of that field will be defined by the R (vector)/R^3 ratio. Another example the electrical field of an electric dipole.
Determine the net electric field directly across from q2 on the circumference of the circle at the point marked by the X. Figure 3. Charge q2, and q3 are positively charged, so the electric field...
No matter how complicated the charge configuration, if we know the direction of the electric field, we can easily determine the direction of the electric force. $\mathbf{F}_{\text{of field on charge } q} = \begin{cases} \text{Magnitude} & = q|\mathbf{E}| \\ \text{Direction} & = \text{ along } \mathbf{E} \text{ field vector for } +q \text{; opposite } \mathbf{E} \text{ field vector for } -q \end{cases}$
The direction of the electric field is always directed in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding the source charge. Since electric field is a vector quantity, it can be represented by a vector arrow.
This $\hat{r}$ tells us about the direction of the electric field. Since electric force is a central force and we have defined electric field using Coulombs law we can conclude that electric field acts along the line joining the charge $q$ (source point) and field point at which it is being measured.
If all you have is the electric field of the wave, there is no way to uniquely determine the magnetic field direction. You do know that k →, B →, and E → are all at right angle to each other; but without more information, all you know is that B → lies in the plane perpendicular to E →. | 939 | 4,202 | {"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": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2022-49 | latest | en | 0.920151 |
https://forums.atariage.com/topic/184552-checking-if-a-number-is-odd-or-even-in-extended-basic/ | 1,713,369,662,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817158.8/warc/CC-MAIN-20240417142102-20240417172102-00670.warc.gz | 228,837,017 | 35,102 | IGNORED
# Checking if a number is odd or even in Extended Basic
## Recommended Posts
Hello,
I have been writing little programs to become more familiar with extended basic. The following program draws a checker board on the screen and then uses a sprite to jump from each black square in a loop. I needed to determine if a column was an even or an odd column but I couldn't find a modulus function or operator so a used the code in lines 121 and 125 as a work around. Is there a better way to determine if a number is odd or even using extended basic?
100 CALL CLEAR
120 FOR R=1 TO 20
121 X=R/2
125 IF INT(X) = X THEN S=2 ELSE S=1
130 FOR C=S TO 24 STEP 2
135 CALL CHAR(91,"FFFFFFFFFFFFFFFF")
140 CALL HCHAR(R,C,91)
150 NEXT C
160 NEXT R
220 FOR R=1 TO 20
221 X=R/2
225 IF INT(X) = X THEN S=2 ELSE S=1
230 FOR C=S TO 24 STEP 2
235 CALL CHAR(91,"FFFFFFFFFFFFFFFF")
240 CALL SPRITE(#1,91,4,R*8-7,C*8-7)
250 NEXT C
260 NEXT R
270 GOTO 220
Also, feel free to leave any comments or pointers on how I could better implement this program in extended basic.
Thanks!
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Bit twiddling is so much better
Edited by sometimes99er
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In XB, if you are treating your numbers as integers, you can also use bit-twiddling to isolate the LSb (least significant bit) in the number. The LSb will always be 0 for even numbers and 1 for odd numbers.
Test for even: if (num AND 1) = 0 then print "even"
Test for odd: if (num AND 1) = 1 then print "odd"
```100 FOR I=1 TO 10
110 PRINT I;"IS ";
120 IF (I AND 1)=1 THEN PRINT "ODD" ELSE PRINT "EVEN"
130 NEXT I
```
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In XB, if you are treating your numbers as integers, you can also use bit-twiddling to isolate the LSb (least significant bit) in the number. The LSb will always be 0 for even numbers and 1 for odd numbers.
Test for even: if (num AND 1) = 0 then print "even"
Test for odd: if (num AND 1) = 1 then print "odd"
```100 FOR I=1 TO 10
110 PRINT I;"IS ";
120 IF (I AND 1)=1 THEN PRINT "ODD" ELSE PRINT "EVEN"
130 NEXT I
```
Oh...I had know idea you could AND the binary value from within XB! That is pretty handy. Thanks for your input.
Edited by idflyfish
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IIRC the Miller's Graphics book on Sprite Programming (or maybe it was Night Mission) has a really good section on optimizing XB (with tricks like this)
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IIRC the Miller's Graphics book on Sprite Programming (or maybe it was Night Mission) has a really good section on optimizing XB (with tricks like this)
Excellent...ill have a look..thanks
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Should do much the same as yours - and a lot faster.
```100 CALL CLEAR::CALL CHAR(91,RPT\$("F",16)):\$=RPT\$("[ ",12)
110 FOR R=1 TO 20::DISPLAY AT(R,2-(R AND 1))\$::NEXT R
120 CALL SPRITE(#1,91,4,1,17)
130 FOR R=1 TO 20::FOR C=1 TO 12
140 CALL LOCATE(#1,R*8-7,C*16+9-8*(R AND 1))
150 NEXT C::NEXT R::GOTO 130
```
Edited by sometimes99er
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Should do much the same as yours - and a lot faster.
```100 CALL CLEAR::CALL CHAR(91,RPT\$("F",16)):\$=RPT\$("[ ",12)
110 FOR R=1 TO 20::DISPLAY AT(R,2-(R AND 1))\$::NEXT R
120 CALL SPRITE(#1,91,4,1,17)
130 FOR R=1 TO 20::FOR C=1 TO 12
140 CALL LOCATE(#1,R*8-7,C*16+9-8*(R AND 1))
150 NEXT C::NEXT R::GOTO 130
```
Heck ya it is...shorter too
This will give me something to chew on...thanks for the input
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If you make a storage table of where the location of the black squares are you could use CALL MOTION on the SPRITE and a CALL COINC. The SPRITE movement would be smooth not jumpy, unless that is intended.
Edited by RXB
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If you make a storage table of where the location of the black squares are you could use CALL MOTION on the SPRITE and a CALL COINC. The SPRITE movement would be smooth not jumpy, unless that is intended.
That sounds interesting. How do you suggest we make the storage table ?
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Try this and see what you think.
100 CALL CLEAR
110 RA=8192
120 FOR R=1 TO 20
130 X=R/2
140 IF INT(X)=X THEN S=2 ELSE S=1
150 FOR C=S TO 24 STEP 2
160 CALL CHAR(91,"FFFFFFFFFFFFFFFF")
170 CALL HCHAR(R,C,91)
180 NEXT C
190 NEXT R
200 FOR R=1 TO 20
210 X=R/2
220 IF INT(X)=X THEN S=2 ELSE S=1
230 FOR C=S TO 24 STEP 2
240 CALL CHAR(91,"FFFFFFFFFFFFFFFF")
250 CALL SPRITE(#1,91,4,R*8-7,C*8-7)
270 NEXT C
280 NEXT R
290 FOR RA=8670 TO 8192 STEP-2
300 CALL PEEK(RA,R,C)
310 CALL SPRITE(#1,91,4,R,C)
320 NEXT RA
330 FOR RA=8192 TO 8670 STEP 24
340 CALL PEEK(RA,R,C)
350 CALL SPRITE(#1,91,4,R,C)
360 NEXT RA
370 FOR RA=8192 TO 8670 STEP 2
380 CALL PEEK(RA,R,C)
390 CALL SPRITE(#1,91,4,R,C)
400 NEXT RA
410 GOTO 290
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Try this and see what you think.
100 CALL CLEAR
110 RA=8192
120 FOR R=1 TO 20
130 X=R/2
140 IF INT(X)=X THEN S=2 ELSE S=1
150 FOR C=S TO 24 STEP 2
160 CALL CHAR(91,"FFFFFFFFFFFFFFFF")
170 CALL HCHAR(R,C,91)
180 NEXT C
190 NEXT R
200 FOR R=1 TO 20
210 X=R/2
220 IF INT(X)=X THEN S=2 ELSE S=1
230 FOR C=S TO 24 STEP 2
240 CALL CHAR(91,"FFFFFFFFFFFFFFFF")
250 CALL SPRITE(#1,91,4,R*8-7,C*8-7)
270 NEXT C
280 NEXT R
290 FOR RA=8670 TO 8192 STEP-2
300 CALL PEEK(RA,R,C)
310 CALL SPRITE(#1,91,4,R,C)
320 NEXT RA
330 FOR RA=8192 TO 8670 STEP 24
340 CALL PEEK(RA,R,C)
350 CALL SPRITE(#1,91,4,R,C)
360 NEXT RA
370 FOR RA=8192 TO 8670 STEP 2
380 CALL PEEK(RA,R,C)
390 CALL SPRITE(#1,91,4,R,C)
400 NEXT RA
410 GOTO 290
Well, I get a
* SYNTAX ERROR IN 260
I guess you're trying to implement the "storage table" using LOAD and PEEK ?
I don't know how complete your example is, but didn't you suggest the use of CALL MOTION and CALL COINC ?
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The CALL LOAD and CALL PEEK use the lower 8K of the 32K. So I guess you do not have 32K?
Bare bones a TI like that uses only 16K VDP memory so all program line table, variables, and the program are all crammed that slower memory VDP.
Programs running from VDP run much slower then from RAM in the 32K.
Here is the problem:
Row and Column storage is 968 values at 2 bytes each is 1936 bytes plus tokens for each value means 2904 bytes just to store the tables.
Now to access those values in memory you need an array of RA(22,22) but each access will make a copy in memory to work with this slows the program down as you are running from VDP, why RAM is better.
Ok instead of your CALL SPRITE(#1,96,4,R,C) you could do CALL SPRITE(#1,96,4,R,C,0,62) :: CALL MOTION(#1,-62,62) :: CALL MOTION(#1,0,62) :: CALL MOTION(#1,62,62) :: C=C+1 :: CALL SPRITE(#1,96,4,R,C)
From normal XB almost everything sucks, that is why I wrote RXB. By adding a comma in GMOTION from above is CALL SPRITE(#1,96,4,R,C,0,62) :: CALL MOTION(#1,-62,62,#1,0,62,#1,62,62) :: C=C+1 :: CALL SPRINT(#1,96,4,R,C)
(Much less memory used and runs faster as the :: really slows XB down the more you have, even slower is line numbers, the less the better and faster the program runs.)
A solution in RXB is
CALL SPRITE(#1,96,4,R,C,0,62,#2,32,R+1,C) :: CALL GMOTION(#1,X,Y)
FOR X=-62 TO 62 :: CALL MOTION(#1,X,Y) :: CALL DISTANCE(#1,#2,D) :: IF D<8 THEN C=C+1 :: CALL SPRITE(#1,96,4,R,C)
NEXT X
I am working on a upgrade of RXB so can not spend a lot on time on this, but was just making a suggestion. If you are using no 32K your approach avenues are very limited. Lack of memory means lack of tools to attack the problem.
Edited by RXB
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Try this and see what you think.
100 CALL CLEAR
110 RA=8192
120 FOR R=1 TO 20
130 X=R/2
140 IF INT(X)=X THEN S=2 ELSE S=1
150 FOR C=S TO 24 STEP 2
160 CALL CHAR(91,"FFFFFFFFFFFFFFFF")
170 CALL HCHAR(R,C,91)
180 NEXT C
190 NEXT R
200 FOR R=1 TO 20
210 X=R/2
220 IF INT(X)=X THEN S=2 ELSE S=1
230 FOR C=S TO 24 STEP 2
240 CALL CHAR(91,"FFFFFFFFFFFFFFFF")
250 CALL SPRITE(#1,91,4,R*8-7,C*8-7)
270 NEXT C
280 NEXT R
290 FOR RA=8670 TO 8192 STEP-2
300 CALL PEEK(RA,R,C)
310 CALL SPRITE(#1,91,4,R,C)
320 NEXT RA
330 FOR RA=8192 TO 8670 STEP 24
340 CALL PEEK(RA,R,C)
350 CALL SPRITE(#1,91,4,R,C)
360 NEXT RA
370 FOR RA=8192 TO 8670 STEP 2
380 CALL PEEK(RA,R,C)
390 CALL SPRITE(#1,91,4,R,C)
400 NEXT RA
410 GOTO 290
Well, I get a
* SYNTAX ERROR IN 260
I guess you're trying to implement the "storage table" using LOAD and PEEK ?
I don't know how complete your example is, but didn't you suggest the use of CALL MOTION and CALL COINC ?
Unless I am mistaken, do you not need a "CALL INIT" prior to using a "CALL LOAD"?
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Yes, sorry I use only RXB and it does not need a CALL INIT first, RXB checks to see if CALL INIT is called when CALL LOAD or CALL PEEK is used, if CALL INIT is not used yet it just ignores the CALL LOAD("DSK#.PROGRAM") but still works. This is a XB bug I fixed in RXB.
In the last post I used DISTANCE, but you could just use COINC(#1,#2,4.CC) :: IF CC=-1 THEN C=C+1 :: CALL SPRITE(#1,96,4,R,C,0,62)
I could post a working video of it later when I get time to do that, but it does work fine.
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I'm sorry, but I think we use XB. Tried CALL INIT, but it still fails.
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Ok you guys got me curious and I dropped off RXB writing to investigate this. Turns out this is a bug in XB. XB has a bad habit of making a SYNTAX errors out of code that has no syntax error and in the CALL LOAD this is one of them.
I fixed this in RXB when I first moved the INIT code in RXB to a different GROM and fixed the bugs I found way back in RXB version 1001 (I do not have the source code earlier then 2001 now)
You will notice in this video that it gets to the sixth column of working fine when it crashes. If a real syntax error existed then it should have crashed the first time it ran the code, not the sixth time as XB does not re-write itself while running.
The original code I posted and said "Try this:" is the same except I added 90 CALL INIT to make the Normal XB program run correctly. (RXB does not require this, that is another bug) Anyway enjoy the show....
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Talking about "call motion" and smooth sprite movement, here's a little something ...
```100 CALL CLEAR::CALL CHAR(91,RPT\$("F",16)):\$=RPT\$("[ ",12)
110 FOR R=1 TO 20::DISPLAY AT(R,2-(R AND 1))\$::NEXT R
120 R1=1::C1=17::CALL SPRITE(#1,91,13,R1,C1)
130 R2=INT(RND*20)::C2=INT(RND*12)*16+17+(R2 AND 1)*8::R2=R2*8+1
140 CALL SPRITE(#2,91,7,R2,C2)::CALL MOTION(#1,(R2-R1)/8,(C2-C1)/8)
150 CALL POSITION(#1,R,C)::IF ABS(R-R2)>2 OR ABS(C-C2)>2 THEN 150
160 CALL MOTION(#1,0,0)::CALL LOCATE(#1,R2,C2)::R1=R2::C1=C2::GOTO 130```
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Just for kicks, I thought you might be interested to see the code to produce a checker board in TF
```: CHECKER
1 GMODE \ select 32 column mode
0 \ seed for display/dont display
19 FOR \ 20 rows (starts from 0)
23 FOR \ 24 columns (starts from 0)
1 XOR \ toggle the seed (1-->0 / 0-->1)
DUP IF J I 42 1 HCHAR THEN \ display if seed >0
NEXT \ next column
1 XOR \ toggle the seed again
NEXT \ next row
DROP \ drop the seed
;
```
Or, horizontally, in the more traditional Forth style:
```: CHECKER
1 GMODE
0 19 FOR 23 FOR 1 XOR DUP IF J I 42 1 HCHAR THEN NEXT 1 XOR NEXT DROP ;
```
Takes just under 1 second (including setting the graphics mode and clearing the screen, which is done by GMODE), which is quite slow I suppose, but I wanted to use the same technique as the original poster. For speed, you'd want to use a similar approach to Sometimes (Sometimes is the master coder ):
```: 1LINE ." * * * * * * * * * *" CR ;
: CHECKER
1 GMODE 0 19 FOR DUP IF SPACE THEN 1LINE 1 XOR NEXT DROP ;
```
Flies! About a quarter of a second, again, that includes setting the VDP mode, and doing a screen clear, before it even begins to run your program
Just thought i'd share. Sorry to hijack the thread
Edited by Willsy
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October 23, 2013, 04:09 How to update a volScalarField's values #1 Member Thomas Vossel Join Date: Aug 2013 Location: Germany Posts: 45 Rep Power: 5 Hi! I have a question concerning how to update a volScalarField. Imagine the following problem: You have a volScalarField T for the temperature (with values applied to it). You now have another volScalarField X which is meant to consist of values depending on the respective temperatures. How do I implement an update for X which looks up the temperature of each element and then updates the value in terms of special conditions. With that I mean a procedure of this sort: Code: ```if (T<100) { X = 1; } if (T=100) { X = 2; } if (T>100) { X = 3; }```
October 23, 2013, 04:26 #2 Senior Member Alexey Matveichev Join Date: Aug 2011 Location: Nancy, France Posts: 1,437 Rep Power: 25 There are several ways to do it. One of them: Code: ```forAll(X, cellI) { if (T[cellI] < 100) X[cellI] = 1; else if (T[cellI] > 100) X[cellI] = 3; else X[cellI] = 2; }``` ThomasV likes this.
October 23, 2013, 04:44 #3 Senior Member Armin Join Date: Feb 2011 Location: Helsinki, Finland Posts: 156 Rep Power: 11 There are many examples in the source code, see e.g.: https://github.com/OpenFOAM/OpenFOAM.../hePsiThermo.C Here a simplified example: Say you get your temperature field like this: Code: ` const volScalarField& T = thermo.T();` Then you can update your field X like this: Code: ``` scalarField& TCells = T.internalField(); forAll(TCells, celli) { if (T[celli] < 100.0) X[celli] = 1; else if (T[celli] > 100.0) X[celli] = 3; else X[celli] = 2; }``` The boundary field you can update in a similar manner, I suggest you have a look at the code I linked above if you need to do this. ThomasV likes this.
October 23, 2013, 05:13 #4 Member Thomas Vossel Join Date: Aug 2013 Location: Germany Posts: 45 Rep Power: 5 Thanks! EDIT: Btw - is there a difference between writing "celli" and "cellI" (i.e. upper- / lowercase letter)?
October 23, 2013, 08:15
#5
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Armin
Join Date: Feb 2011
Location: Helsinki, Finland
Posts: 156
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Quote:
Originally Posted by ThomasV Btw - is there a difference between writing "celli" and "cellI" (i.e. upper- / lowercase letter)?
No, there is no difference.
The variable celli is a 'label' (i.e. a integer) defined in forAll, which is a macro that sets up the for loop. So, you can call the variable whatever you want, you just need to be consistent.
Last edited by dkxls; October 23, 2013 at 09:45.
October 28, 2013, 11:06 #6 Member Thomas Vossel Join Date: Aug 2013 Location: Germany Posts: 45 Rep Power: 5 Allright thanks... I'd like to add another more "practical" question here. Things now work fine and I define a temperature field which then is analyzed to determine the solid fraction in the respective cells. What I'd like to know is how to write the new volScalarField for the fraction solid into the "0 folder". I create the fraction solid field like this: Code: ``` volScalarField fracSol ( IOobject ( "fracSol", runTime.timeName(), mesh, IOobject::NO_READ, IOobject::AUTO_WRITE ), mesh, scalar(0) );``` Afterwards it becomes filled with the actual data determined by the respective temperatures of the cells. This way I get my fracSol file written in every timestep. Every timestep except for the initial one that is... I now would like to know how I also can write the calculated fracSol into the 0 folder i.e. the fraction solid field for the initial temperatures. This also would come in handy when postprocessing the project in ParaFoam as there now is no fracSol field in time step 0 and it is a bit tedious to go to timestep 1 and activate the field plus once again any time you accidentally rewind to timestep 0 as it then gets unloaded (because there's no data for it in step 0)... I already tried to simply put a runtime.write() command before progressing time in the time loop but this didn't do the trick. I guess one usually would approach this problem with the setfields utility but as far as I know one here is limited to the premade functions like defining a block in which the variable get set to a certain value...
October 28, 2013, 11:31 #7 Senior Member Armin Join Date: Feb 2011 Location: Helsinki, Finland Posts: 156 Rep Power: 11 you can force the writing of a field at any time with Code: `fracSol.write();` You just should take care that you don't do it in the time-loop, as the field would then be written every time-step. ThomasV likes this.
October 28, 2013, 12:07 #8 Member Thomas Vossel Join Date: Aug 2013 Location: Germany Posts: 45 Rep Power: 5 Thanks - if I can come up with enough self-motivation after having finished my master thesis I might write a beginners tutorial so people won't have such a hard time figuring out such "mundane" things like this...
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https://sciencedocbox.com/Physics/124560814-Key-concepts-satisfies-the-differential-equation-da-0-note-if-f-x-is-any-integral-of-f-x-then-x-a.html | 1,628,102,122,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154897.82/warc/CC-MAIN-20210804174229-20210804204229-00070.warc.gz | 505,064,983 | 30,498 | # KEY CONCEPTS. satisfies the differential equation da. = 0. Note : If F (x) is any integral of f (x) then, x a
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1 KEY CONCEPTS THINGS TO REMEMBER :. The re ounded y the curve y = f(), the -is nd the ordintes t = & = is given y, A = f () d = y d.. If the re is elow the is then A is negtive. The convention is to consider the mgnitude only i.e. A = y d in this cse.. Are etween the curves y = f () & y = g () etween the ordintes t = & = is given y, A = f () d g () d = [ f () g () ] d.. Averge vlue of function y = f () w.r.t. over n intervl is defined s : y (v) = f () d. 5. The re functiona stisfies the differentil eqution da = f () with initil conditiona d =. Note : If F () is ny integrl of f () then, A = f () d = F () + c A = = F () + c c = F () hence A = F () F (). Finlly y tking = we get, A = F () F (). 6. CURVE TRACING : The following outline procedure is to e pplied in Sketching the grph of function y = f () which in turn will e etremely useful to quickly nd correctly evlute the re under the curves. () Symmetry : The symmetry of the curve is judged s follows : (i) If ll the powers of y in the eqution re even then the curve is symmetricl out the is of. (ii) If ll the powers of re even, the curve is symmetricl out the is of y. (iii) If powers of & y oth re even, the curve is symmetricl out the is of s well s y. (iv) If the eqution of the curve remins unchnged on interchnging nd y, then the curve is symmetricl out y =. (v) If on interchnging the signs of & y oth the eqution of the curve is unltered then there is symmetry in opposite qudrnts. () Find dy/d & equte it to zero to find the points on the curve where you hve horizontl tngents. (c) Find the points where the curve crosses the is & lso the yis. (d) Emine if possile the intervls when f () is incresing or decresing. Emine wht hppens to y when or. 7. USEFUL RESULTS : (i) Whole re of the ellipse, / + y / = is π. (ii) Are enclosed etween the prols y = & = y is 6/. (iii) Are included etween the prol y = & the line y = m is 8 / m. EXERCISE I Q. Find the re ounded on the right y the line + y =, on the left y the prol y = nd elow y the is. Q. Find the re of the region ounded y the curves, y = ² + ; y = ; = & =. Pge 9 of Are Under Curve
2 Q. Find the re of the region {(, y) : y ² +, y +, }. Q. Find the vlue of c for which the re of the figure ounded y the curves y = sin, the stright lines = π/6, = c & the sciss is is equl to /. Q.5 The tngent to the prol y = hs een drwn so tht the sciss of the point of tngency elongs to the intervl [, ]. Find for which the tringle ounded y the tngent, the is of ordintes & the stright line y = hs the gretest re. Q.6 Compute the re of the region ounded y the curves y = e.. ln & y = ln /(e. ) where ln e=. Q.7 A figure is ounded y the curves y = sin π, y =, = & =. At wht ngles to the positive is stright lines must e drwn through (, ) so tht these lines prtition the figure into three prts of the sme size. Q.8 Find the re of the region ounded y the curves, y = log e, y = sin π & =. Q.9 Find the re ounded y the curves y = nd y =. Also find the rtio in which the y-is divided this re. Q. If the re enclosed y the prols y = nd y = is 8 sq. units. Find the vlue of ''. Q. The line + y = divides the re enclosed y the curve, 9 + y 8 6y = into two prts. Find the rtio of the lrger re to the smller re. Q. Find the re of the region enclosed etween the two circles ² + y² = & ( )² + y² = Q. Find the vlues of m (m > ) for which the re ounded y the line y = m + nd = y y is, (i) 9/ squre units & (ii) minimum. Also find the minimum re. Q. Find the rtio in which the re enclosed y the curve y = cos ( π/) in the first qudrnt is divided y the curve y = sin. Q.5 Find the re enclosed etween the curves : y = log e ( + e), = log e (/y) & the is. Q.6 Find the re of the figure enclosed y the curve (y rc sin ) =. Q.7 For wht vlue of '' is the re ounded y the curve y = + + nd the stright line y =, = & = the lest? Q.8 Find the positive vlue of '' for which the prol y = + isects the re of the rectngle with vertices (, ), (, ), (, + ) nd (, + ). Q.9 Compute the re of the curviliner tringle ounded y the yis & the curve, y = tn & y = (/) cos. Q. Consider the curve C : y = sin sin, C cuts the is t (, ), ( π, π). A : The re ounded y the curve C & the positive is etween the origin & the ordinte t =. A : The re ounded y the curve C & the negtive is etween the ordinte = & the origin. Prove tht A + A + 8 A A =. Q. Find the re ounded y the curve y = e ; y = nd = c where c is the -coordinte of the curve's inflection point. Q. Find the vlue of 'c' for which the re of the figure ounded y the curve, y = 8 5, the stright lines = & = c & the sciss is is equl to 6/. Q. Find the re ounded y the curve y² = & = y. Q. Find the re ounded y the curve y = e, the -is, nd the line = c where y (c) is mimum. Q.5 Find the re of the region ounded y the is & the curves defined y, y = tn, π / π / y = cot, π / 6 π / Pge of Are Under Curve
3 EXERCISE II Q. In wht rtio does the -is divide the re of the region ounded y the prols y = ² & y = ²? Q. Find the re ounded y the curves y = & y =. Q. Sketch the region ounded y the curves y = 5 & y = & find its re. Q. Find the eqution of the line pssing through the origin nd dividing the curviliner tringle with verte t the origin, ounded y the curves y =, y = nd = into two prts of equl re. Q.5 Consider the curve y = n where n > in the st qudrnt. If the re ounded y the curve, the -is nd the tngent line to the grph of y = n t the point (, ) is mimum then find the vlue of n. Q.6 Consider the collection of ll curve of the form y = tht pss through the the point (, ), where nd re positive constnts. Determine the vlue of nd tht will minimise the re of the region ounded y y = nd -is. Also find the minimum re. Q.7 In the djcent grphs of two functions y = f() nd y = sin re given. y = sin intersects, y = f() t A (, f()); B(π, ) nd C(π, ). A i (i =,,,) is the re ounded y the curves y = f () nd y = sin etween = nd = ; i =, etween = nd = π; i =, etween = π nd = π; i =. If A = sin + ( )cos, determine the function f(). Hence determine nd A. Also clculte A nd A. Q.8 Consider the two curves y = /² & y = /[ ( )]. (i) At wht vlue of ( > ) is the reciprocl of the re of the fig. ounded y the curves, the lines = & = equl to itself? (ii) At wht vlue of ( < < ) the re of the figure ounded y these curves, the lines = & = equl to /. Q.9 ln c Show tht the re ounded y the curve y =, the -is nd the verticl line through the mimum point of the curve is independent of the constnt c. Q. For wht vlue of '' is the re of the figure ounded y the lines, y =, y = 5? Q. Compute the re of the loop of the curve y² = ² [( + )/( )]. Q. Find the vlue of K for which the re ounded y the prol y = + nd the line y = K + is lest. Also find the lest re. Q. Let A n e the re ounded y the curve y = (tn ) n & the lines =, y = & = π/. Prove tht for n >, A n + A n = /(n ) & deduce tht /(n + ) < A n < /(n ). Q. If f () is monotonic in (, ) then prove tht the re ounded y the ordintes t = ; = ; y = f () + nd y = f (c), c (, ) is minimum when c =. Hence if the re ounded y the grph of f () = +, the stright lines =, = nd the -is is minimum then find the vlue of ''. Q.5 Consider the two curves C : y = + cos & C : y = + cos ( α) for α, π ; [, π]. Find the vlue of α, for which the re of the figure ounded y the curves C, C & = is sme s tht of the figure ounded y C, y = & = π. For this vlue of α, find the rtio in which the line y = divides the re of the figure y the curves C, C & = π. Q.6 Find the re ounded y y² = ( + ), y² = ( ) & y = ove is of. Q.7 Compute the re of the figure which lies in the first qudrnt inside the curve Pge of Are Under Curve
4 ² + y² = ² & is ounded y the prol ² = y & y² = ( > ). Q.8 Consider squre with vertices t (, ), (, ), (, ) & (, ). Let S e the region consisting of ll points inside the squre which re nerer to the origin thn to ny edge. Sketch the region S & find its re. Q.9 Find the whole re included etween the curve ² y² = ² (y² ²) & its symptotes (symptotes re the lines which meet the curve t infinity). Q. For wht vlues of [, ] does the re of the figure ounded y the grph of the function y = f () nd the stright lines =, = & y = f() is t minimum & for wht vlues it is t mimum if f () =. Find lso the mimum & the minimum res. Q. Find the re enclosed etween the smller rc of the circle ² + y² + y = & the prol y = ² + +. Q. D r w n e t n d c l e n g r p h o f t h e f u n c t i o n f D r w n e t n d c l e n g r p h o f t h e f u n c t i o n f () = cos ( ), [, ] nd find the re enclosed etween the grph of the function nd the is s vries from to. Q. Let C & C e two curves pssing through the origin s shown in the figure. A curve C is sid to "isect the re" the region etween C & C, if for ech point P of C, the two shded regions A & B shown in the figure hve equl res. Determine the upper curve C, given tht the isecting curve C hs the eqution y = & tht the lower curve C hs the eqution y = /. Q. For wht vlues of [, ] does the re of the figure ounded y the grph of the function y = f () & the stright lines =, =, y = f() hve the gretest vlue nd for wht vlues does it hve the lest vlue, if, f() = α + β, α, β R with α >, β >. Q.5 Given f () = t e (logsec t sec y = f () nd y = g () etween the ordintes = nd = π. t)dt ; g () = e tn. Find the re ounded y the curves EXERCISE III Q. Let f () = Mimum {, ( ), ( )}, where. Determine the re of the region ounded y the curves y = f (), is, = & =. [ JEE '97, 5 ] Q. Indicte the region ounded y the curves = y, y = + nd is nd otin the re enclosed y them. [ REE '97, 6 ] Q. Let C & C e the grphs of the functions y = & y =, respectively. Let C e the grph of function y = f (),, f() =. For point P on C, let the lines through P, prllel to the es, meet C & C t Q & R respectively (see figure). If for every position of P (on C ), the res of the shded regions OPQ & ORP re equl, determine the function f(). [JEE '98, 8] Q. Indicte the region ounded y the curves y = ln & y = nd otin the re enclosed y them. [ REE '98, 6 ] Q.5 () For which of the following vlues of m, is the re of the region ounded y the curve y = nd the line y = m equls 9/? (A) (B) (C) (D) for () Let f() e continuous function given y f() = + + for > Find the re of the region in the third qudrnt ounded y the curves, = y nd Pge of Are Under Curve
5 y = f() lying on the left of the line 8 + =. [ JEE '99, + (out of ) ] Q.6 Find the re of the region lying inside + (y ) = nd outside c + y = c where c =. [REE '99, 6] Q.7 Find the re enclosed y the prol (y ) =, the tngent to the prol t (, ) nd the -is. [REE,] Q.8 Let nd for j =,,,...n, let S j e the re of the region ounded y the y is nd the curve jπ ( j+ ) π e y = siny, y. Show tht S, S, S,...S n re in geometric progression. Also, find their sum for = nd = π. [JEE', 5] Q.9 The re ounded y the curves y = nd y = + is (A) (B) (C) (D) [JEE', (Scr)] Q. Find the re of the region ounded y the curves y =, y = nd y =, which lies to the right of the line =. [JEE ', (Mins)] Q. If the re ounded y y = nd = y, >, is, then = (A) (B) (C) (D) [JEE ', (Scr)] Q.() The re ounded y the prols y = ( + ) nd y = ( ) nd the line y = / is (A) sq. units (B) /6 sq. units (C) / sq. units (D) / sq. units [JEE '5 (Screening)] () Find the re ounded y the curves = y, = y nd y =. f ( ) + (c) If f () = +, f () is qudrtic function nd its mimum vlue occurs t c c f () c + c point V. A is point of intersection of y = f () with -is nd point B is such tht chord AB sutends right ngle t V. Find the re enclosed y f () nd chord AB. [JEE '5 (Mins), + 6] Q. Mtch the following π cos sin (i) (sin ) (cos cot log(sin ) )d (A) (ii) Are ounded y y = nd = 5y (B) (iii) Cosine of the ngle of intersection of curves y = log nd y = is (C) 6 ln (D) / [JEE 6, 6] ANSWER EXERCISE I Q. 5/6 sq. units Q. / sq. units Q. /6 sq. units Q. c = π 6 or π Q 7. π tn π Q 9. π ; π π + ; π tn π Q 5. =, A( ) = 8 Q 8. 8 Q 6. (e 5)/ e sq. units sq. units Q. = 9 Q. π + π Pge of Are Under Curve
6 Q. π sq. units Q. (i) m =, (ii) m = ; Amin = / Q. Q 5. sq. units Q 6. π/ Q 7. = / Q 8. Q 9. + l n sq. units 8 7 / Q. e Q. C = or ( ) Q. / Q. ( e / ) Q 5. ln EXERCISE II Q. : Q. 8/5 sq. units Q. (5 π )/ sq. units Q. y = / Q 5. + Q 6. = /8, A minimum = sq. units Q 7. f() = sin, = ; A = sin; A = π sin; A = (π ) sq. units Q 8. = + e, = + e Q.9 / Q. = 8 or ( 6 ) 5 Q. (π/) sq. units Q. K =, A = / Q. = 8 8 / Q 5. α = π/, rtio = : Q 6. ( ) ( ) Q sin rc sq. units Q 8. ( 6 ) Q 9. Q. = / gives minim, A = π π ; = gives locl mim A() = ; = gives mimum vlue, A() = π/ Q. 8 + π Q. ( ) sq. units Q. (6/9) Q. for =, re is gretest, for = /, re is lest Q5. e π log sq. units EXERCISE III Q. 7/7 Q. 5/6 sq. units Q. f() = Q. 7/ π Q.5 () B, D () 57/9 ; = ; = Q.6 π sq. units π Sj Q.7 9 sq. units Q.8 = e ; S S j+ e = π + + for =, = π, S = π ( e + ) nd r = π π + Q.9 B Q. sq. units Q. B 5 Q. () D ; () sq. units ; (c) sq. units Q. (i) A, (ii) D, (iii) A Pge of Are Under Curve
7 EXERCISE IV. The re ounded y the curve = y, -is nd the line = is (A) (B) (C). The re ounded y the -is nd the curve y = is (A) (B) (C) (D) (D) 8 Pge 5 of Are Under Curve. The re ounded y the curve y = sin with -is in one rc of the curve is (A) (B) (C) (D). The re contined etween the curve y =, the verticl line =, = ( > ) nd -is is (A) log (B) log (C) log (D) log 5. The re of the closed figure ounded y the curves y =, y = & y = is: (A) 9 (B) 8 9 (C) 6 9 (D) none 6. The re of the closed figure ounded y the curves y = cos ; y = + π & = π is (A) π + (B) π (C) π + 7. The re included etween the curve y = ( ) & its symptote is: (A) π 8. The re ounded y ² + y² = & y = sin π (A) π π (B) π π (D) π (B) π (C) π (D) none (C) π in the upper hlf of the circle is: 8 π (D) none 9. The re of the region enclosed etween the curves 7 + 9y + 9 = nd y + 7 = is: (A) (B) (C) 8 (D) 6. The re ounded y the curves y = ( ln ); = e nd positive Xis etween = e nd = e is : e (A) 5 e e (B) 5 e e e (C) 5 5e (D). The re enclosed etween the curves y = log e ( + e), = log e y nd the -is is (A) (B) (C) (D) none of these e
8 . The re ounded y the curves + y = nd + y = is (A) (B) 6 (C) (D) none of these. The re ounded y -is, curve y = f(), nd lines =, = is equl to ( + ) for ll >, then f() is (A) ( ) (B) ( + ) (C) ( + ) (D) / (+ ). The re of the region for which < y < nd > is (A) ( ) d (B) ( ) d (C) ( ) d (D) ( 5. The re ounded y y =, y = [ + ], nd the y-is is (A) / (B) / (C) (D) 7/ 6. The re ounded y the curve = cos t, y = sin t is (A) π 8 (B) π 6 (C) π (D) π 7. If A is the re enclosed y the curve y =, -is nd the ordintes =, = ; nd A is the re enclosed y the curve y =, -is nd the ordintes =, =, then (A) A = A (B) A = A (C) A = A (D) A = A 8. The re ounded y the curv e y = f(), -is nd the ordintes = nd = is ( ) sin ( + ), R, then f() = (A) ( ) cos ( + ) (B) sin ( + ) (C) sin ( + ) + ( ) cos ( + ) (D) none of these 9. Find the re of the region ounded y the curves y = +, y =, = nd =. (A) sq. unit (B) sq. unit (C) sq. unit (D) none of these. The res of the figure into which curve y = 6 divides the circle + y = 6 re in the rtio (A) (B) π 8π + (C) π + 8π (D) none of these. The tringle formed y the tngent to the curve f() = + t the point (, ) nd the coordinte es, lies in the first qudrnt. If its re is, then the vlue of is [IIT - ] (A) (B) (C) (D) EXERCISE V. Find the re of the region ounded y the curve y = y nd the y-is.. Find the vlue of c for which the re of the figure ounded y the curves y = sin, the stright lines = π/6, = c & the sciss is is equl to /.. For wht vlue of '' is the re ounded y the curve y = + + nd the stright line y =, = & = the lest?. Find the re of the region ounded in the first qudrnt y the curve C: y = tn, tngent drwn to ) d Pge 6 of Are Under Curve
9 C t = π nd the is. 5. Find the vlues of m (m > ) for which the re ounded y the line y = m + nd = y y is, (i) 9/ squre units & (ii) minimum. Also find the minimum re. 6. Consider the two curves y = /² & y = /[ ( )]. (i) At wht vlue of ( > ) is the reciprocl of the re of the figure ounded y the curves, the lines = & = equl to itself? (ii) At wht vlue of ( < < ) the re of the figure ounded y these curves, the lines = & = equl to /. 7. A norml to the curve, + α y + = t the point whose sciss is, is prllel to the line y =. Find the re in the first qudrnt ounded y the curve, this norml nd the is of ' '. 8. Find the re etween the curve y ( ) = & its symptotes. 9. Drw net & clen grph of the function f () = cos ( ), [, ] & find the re enclosed etween the grph of the function & the is s vries from to.. Find the re of the loop of the curve, y = ( ).. Let nd for j =,,,..., n, let S j e the re of the region ounded y the yis nd the curve e y = sin y, j π ( j +) π y. Show tht S, S, S,..., S n re in geometric progression. Also, find their sum for = nd = π. [IIT -, 5]. Find the re of the region ounded y the curves, y =, y = & y = which lies to the right of the line =. [IIT -, 5]. If c c f( ) f() = f() c + +, f() is qudrtic function nd its mimum vlue occurs t + c point V. A is point of intersection of y = f() with -is nd point B is such tht chord AB sutends right ngle t V. Find the re enclosed y f() nd cheord AB. [IIT - 5, 6] ANSWER EXERCISE IV. B. C. B. B 5. B 6. D 7. C 8. A 9. C. B. A. A. D. C 5. B 6. A 7. D 8. C 9. A. C. C EXERCISE V. / sq. units. c = π 6 or π.. = ln 5. (i) m =, (ii) m = ; A min = / 6. = + e, = + e ( ). 7 6 sq. units π sq. units. 5 squre units. Pge 7 of Are Under Curve
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1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the x-xis etween nd is denoted y f(x) dx nd clled the
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### Form 5 HKCEE 1990 Mathematics II (a 2n ) 3 = A. f(1) B. f(n) A. a 6n B. a 8n C. D. E. 2 D. 1 E. n. 1 in. If 2 = 10 p, 3 = 10 q, express log 6
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### ( β ) touches the x-axis if = 1
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CHAPTER 0 Conic Sections Solve. 97. Two surveors need to find the distnce cross lke. The plce reference pole t point A in the digrm. Point B is meters est nd meter north of the reference point A. Point
### CBSE-XII-2015 EXAMINATION. Section A. 1. Find the sum of the order and the degree of the following differential equation : = 0
CBSE-XII- EXMINTION MTHEMTICS Pper & Solution Time : Hrs. M. Mrks : Generl Instruction : (i) ll questions re compulsory. There re questions in ll. (ii) This question pper hs three sections : Section, Section | 15,113 | 44,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-31 | longest | en | 0.780131 |
http://www.jiskha.com/display.cgi?id=1208383290 | 1,493,466,650,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123491.68/warc/CC-MAIN-20170423031203-00382-ip-10-145-167-34.ec2.internal.warc.gz | 570,139,561 | 3,775 | # Math
posted by on .
How big is each interior angle in a regular decagon? would this be 4.4 (144/10)?
How big is each exterior angle of a regular decagon?
Is this 36? (360/10)
• Math - ,
Isnt there a theorem that the sum of the interior angles is
180(n-2), so then each interior angle of a regular n-gon would be\
180(n-2)/n
• Math - ,
Each angle measures 144 degrees.
http://www.coolmath.com/reference/polygons.html | 121 | 426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | latest | en | 0.863474 |
https://kolblabs.com/can-we-measure-the-weight-of-free-fall-body-in-gravitation-class-9-science-experiment/ | 1,718,579,227,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861671.61/warc/CC-MAIN-20240616203247-20240616233247-00273.warc.gz | 301,806,192 | 55,985 | # Can we measure the weight of free-fall body in Gravitation – Class 9 Science Experiment
## Activity Name: Can we measure the weight of free-fall bodyin Gravitation
### Activity Description:
In this experiment, we aim to investigate the changes in the reading of a spring balance when it is used to measure the weight of an object under two different instances: first, when the object is stationary, and second, when the object is allowed to fall freely from a certain height.
### Required Items:
• Spring balance
• Ceiling or support to suspend the spring balance
• A height from which the load can fall freely
### Step by Step Procedure:
1. Suspend the spring balance from the ceiling or support in a stable manner.
2. Place the load (weights) on the spring balance and note the reading indicated by the spring balance.
3. Hold the loaded spring balance at a certain height and release it, allowing it to fall freely.
4. Carefully observe the position of the indicator on the spring balance scale while it is in free-fall.
5. Record the reading of the spring balance during free-fall.
### Experiment Observations:
1. The reading of the spring balance when the load is stationary.
2. The reading of the spring balance during free-fall.
### Precautions:
1. Ensure the spring balance is securely suspended to avoid accidents.
2. Use appropriate safety measures when dropping the load from a height.
3. Make sure the surface below the free-fall area is clear and free from obstructions.
4. Take multiple readings to ensure accuracy.
### Lesson Learnt from Experiment:
The experiment aims to demonstrate that the readings of the spring balance are different in the two instances. When the object is stationary, the spring balance measures its weight as per the force of gravity acting on it.
However, during free-fall, the spring balance indicates zero weight. This happens because both the object and the spring balance are falling with the same acceleration due to gravity. As a result, the spring balance does not register any weight during free-fall.
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https://a1classes.com/Class%2010/Chapter-3/Class%2010%20Intro-1.php | 1,674,780,859,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494852.95/warc/CC-MAIN-20230127001911-20230127031911-00428.warc.gz | 102,626,543 | 7,221 | Class 10 NCERT Math Solution
TOPICS
Introduction
Variable
A variable is a number which is use in alphanumeric character. e.g., x, y, a, b etc.
Constant or Non-variable
A non-variable or constant is a number which is use in numeric values. e.g., 2, 3, 10 etc.
Linear Equations
Each solution (x, y) of a linear equation in two variables, ax + by + c = 0. Equations like this are called a pair of linear equations in two variables. e.g., 2x + 3y – 7 = 0 and 9x – 2y + 8 = 0.
Graphical Representation of Linear Equations
(i) The two lines will intersect at one point. e.g., 2x + 3y – 7 = 0 and 9x – 2y + 8 = 0.
(ii) The two lines will not intersect, i.e., they are parallel. e.g., 2x + 3y – 7 = 0 and 4x + 6y - 14 = 0.
(iii) The two lines will be coincident. e.g., 2x + 3y – 7 = 0 and 2x + 3y - 7 = 0.
Consistent Pair of Linear Equations
A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations.
Inconsistent Pair of Linear Equations
A pair of linear equations which has no solution, is called an inconsistent pair of linear equations.
Dependent Pair of Linear Equations
A pair of linear equations which are equivalent has infinitely many distinct common solutions. Such a pair is called a Dependent pair of linear equations in two variables.
Table of Linear Equations
Algebraic Methods of Solving a Pair of Linear Equations
(i) Substitution Method
(ii) Elimination Method
(iii) Cross-Multiplication Method
Steps of Substitution Method
Step 1. Find the value of one variable, say y in terms of the other variable, i.e., x from either equation, whichever is convenient.
Step 2. Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved. Sometimes, as in Examples 9 and 10 below, you can get statements with no variable. If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions. If the statement is false, then the pair of linear equations is inconsistent.
Step 3. Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.
Steps Of Elimination Method:-
Step 1. Firstly multiply both the equations by some suitable non-zero constants to make the coefficients of one variable (either x or y) numerically equal.
Step 2. Then add or subtract one equation from the other so that one variable gets eliminated. If you get an equation in one variable, go to Step 3. If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions. If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent.
Step 3. Solve the equation in one variable (x or y) so obtained to get its value.
Step 4. Substitute this value of x (or y) in either of the original equations to get the value of the other variable.
Cross-Multiplication Method
a₁x + b₁y + c₁ = 0 ....... (i) and a₂x + b₂y + c₂ = 0 ........(ii) Step 1. Write the given equations in the form (i) and (ii).
Step 2. Taking the help of the diagram above, write Equations as given in (8).
Step 3. Find x and y, provided a₁b₂ – a₂b₁ ≠ 0
CLASSES | 879 | 3,304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2023-06 | latest | en | 0.922634 |
https://www.informit.com/articles/article.aspx?p=28817&seqNum=7 | 1,721,715,454,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518014.29/warc/CC-MAIN-20240723041947-20240723071947-00128.warc.gz | 684,711,280 | 34,885 | # Storing Information: Variables and Constants
This chapter is from the book
## Workshop
The Workshop provides quiz questions to help you solidify your understanding of the material covered and exercises to provide you with experience in using what you've learned.
### Quiz
1. What's the difference between an integer variable and a floating-point variable?
2. Give two reasons for using a double-precision floating-point variable (type double) instead of a single-precision floating-point variable (type float).
3. What are five rules that the you know are always true when allocating size for variables?
4. What are the two advantages of using a symbolic constant instead of a literal constant?
5. Show two methods for defining a symbolic constant named MAXIMUM that has a value of 100.
6. What characters are allowed in C variable names?
7. What guidelines should you follow in creating names for variables and constants?
8. What's the difference between a symbolic and a literal constant?
9. What's the minimum value that a type int variable can hold?
### Exercises
1. In what variable type would you best store the following values?
1. A person's age to the nearest year.
2. A person's weight in pounds.
3. The radius of a circle.
5. The cost of an item.
6. The highest grade on a test (assume it is always 100).
7. The temperature.
8. A person's net worth.
9. The distance to a star in miles.
2. Determine appropriate variable names for the values in exercise 1.
3. Write declarations for the variables in exercise 2.
4. Which of the following variable names are valid?
1. 123variable
2. x
3. total_score
4. Weight_in_#s
5. one
6. gross-cost
10. this_is_a_variable_to_hold_the_width_of_a_box
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This web site contains links to other sites. Please be aware that we are not responsible for the privacy practices of such other sites. We encourage our users to be aware when they leave our site and to read the privacy statements of each and every web site that collects Personal Information. This privacy statement applies solely to information collected by this web site. | 2,078 | 11,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-30 | latest | en | 0.883467 |
http://tkelly.org/2016/03/interpolating-bathymetry-to-unstructured-mesh/ | 1,498,417,114,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320570.72/warc/CC-MAIN-20170625184914-20170625204914-00077.warc.gz | 401,948,559 | 15,057 | # Interpolating Bathymetry to Unstructured Mesh
A few weeks (perhaps months) ago I introduced the side project that I am involved with (here) whereby our aim is to develop a hydrographic model for Apalachicola Bay, Fl. Today I wanted to provide an update for that project while also sharing some interesting problems that we’ve had to work around.
To see the first post in this series, please check out this post.
The last we spoke, we had decided on using FVCOM to model the bay and surrounding waters. Our choice was contingent on having a model that could accurately describe the complex coastline and relatively flat and shallow water column. FVCOM uses an unstructured mesh, meaning that instead of using a grid with a regular pattern (think Manhattan’s streets and avenues), the model domain is instead more freeform (think rural street map). This offered both the possibility for improved spatial resolution and an opportunity for use all to gain proficiency in a new skill.
A unstructured grid like this is defined as a set of points or nodes and the connections between them (i.e. who is connected to who). A more intuitive way to understand this is through an example.
Here we have a sample of the two components that make up an unstructured mesh.
p and t
From this point, the next step in forming the final model domain is to add depth data to the mesh. We need to be able to assign a depth for each node forming the mesh since the model will then be able to define the 3 dimension fluid parcels that makes it a hydrodynamic model.
Thanks to the powers of the internet and the USGS, we have bathymetric data for the entire region with various levels of detail (from 30m to as coarse as 100m).
The challenge comes in using these gridded data products for an unstructured mesh like ours (aka square peg and round hole). This is the data-informatic side of things that I enjoy, so let’s get into it.
The obvious answer would be to take some kind of nearest-neighbor type approach whereby the value of one of our nodes is simply the depth value of the closest grid point in the bathymetric data we have. For most applications this would probably be a great route to take since it is simple to implement and it makes no-assumptions about the topography that there (more on this later). But, the result would be ‘wrong’.
While I don’t mean ‘wrong’ in a pejorative sense, it is merely that the depth values generated would not be the values that we would measure unless out point and the data point were coincident. Ultimately it is a bad estimate given the wealth of data available.
So, what direction should we take? Let’s start by defining our assumptions:
1. Submarine topography is generally smooth (at least in Florida where topographic relief is low and karst underlayment).
2. The USGS data is great, but the accuracy of the data and the subsequent project of the data should not be considered infallible. Consider that even if the data is perfect, the accuracy of the coordinates (least significant digit) and depth compounded by the uncertainty of our modeled coastline and nodal points means that the resulting project of the data is likely to be imperfect.
3. Accuracy of location is not as important as accuracy of the large-scale contours due to the dissipation of the small scale as you move away from the coast and into the flat open bay.
Taken together, these give us a footing on which to build an interpolation scheme.
Given that the topography is assumed smooth and inaccurate, let’s start by thinking about averaging across some number of points (e.g. a classic interpolation scheme)[1]. using n points, we arrive at this simple average:
where is the depth at the ith closest point.
Okay, while this is a possible answer, we’re ignoring all of the position/distance data we have. Since we know where the node is and all the bathymetric points around it, we can take a spatial mean instead.
For a spatial mean, we calculate some weighting value, , that allows us to control the relative contribution of each depth to the final, estimated average.
The question now is to decide how we should weight the values. We could let the weight of a value decline linearly with distance; for example, let the weight decrease by 1% for every meter away it is:
But that seems rather arbitrary. Why 1%? Why per meter?
Instead perhaps we should work in the natural units of the problem. We can weight them according to the reciprocal of the distance. here we’ll use whatever the natural units for distance actually are. Here they’re meters whereas in another model they could be cm or km with no difference[2].
So let’s take a look at the implementation of this idea. To begin with we first have to find the nth closest points, that’s the j loop. After that then we can calculate the new depth and save it in depths. Rinse and repeat for all nodes.
While this implementation is not very efficient (it takes a while for 60,000 nodes…), it is transparent and that is the first priority if you’re goal is code-reuse. One of the goals we have is that all of the scripts generated for this project should be code that can be used for future projects. Performance is only requisite for the final model (when it makes a real difference).
1. If we didn’t assume that the topography was smooth or if we had absolute faith in the accuracy of the points we could fit curves to the bathymetric data instead. For example, assuming perfect accuracy of the data, we could calculate the local gradients and use them and a cubic spline between the two closest points to arrive at an estimated value for the depth. There are many non-averaging techniques that could be used instead.
2. By no difference, I mean in terms of the code. Here we’re making the assumption that the units being used prior to our code are the units ‘appropriate’ to the problem at hand. | 1,252 | 5,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2017-26 | longest | en | 0.953851 |
https://help.ctrader.com/ctrader/trading/dynamic-leverage?lang=jp | 1,548,257,850,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584334618.80/warc/CC-MAIN-20190123151455-20190123173455-00109.warc.gz | 518,126,332 | 9,933 | # Dynamic Leverage
## Dynamic Leverage
This section explains how Dynamic Leverage works as well as how “Margin Used” is calculated in cTrader. The margin used by open positions is related to the leverage given for each position opening. Dynamic leverage can be set to work in two very different ways. One method recalculates margin requirements upon each event, the other calculates margin requirements upon opening the position and is constant throughout the lifetime of the position, this concept is explained in this section, everything on this page is important and it is urged all traders understand this.
## Leverage Types
### Account Leverage
There are two types of leverage in cTrader. These are the account’s leverage and the volume based dynamic leverage. The account’s leverage is the maximum leverage allowed for each trading account. It is set during the account’s creation and can only be modified by your broker.
### Volume Based Dynamic Leverage
Brokers have the ability to dynamically reduce leverage after a volume of open positions is surpassed for each symbol and direction. The Dynamic Leverage tiers can be found for each symbol from the MarketWatch section, in the Symbol Window or Info Panel. An example of tiered leverage is the following;
### Applicable Leverage
Despite your account having leverage of 1:500 (for example), this doesn’t mean all positions for all symbols will use this value for calculating margin requirements, this value is simply the limit on how much leverage your account will use in the event that your broker allows more for a certain symbol. If your broker uses Dynamic Leverage the applicable leverage for each volume tier is calculated dynamically for each symbol based on the following equation;
Applicable Leverage Tier X = Min(Account Leverage, Volume Based Leverage Tier X)
Tier Volume in USD Leverage
1 0-1mill 500
2 1-2mill 200
3 2-3mill 100
4 >3mill 50
According to the above Dynamic Leverage table the equation can be better understood like this;
Tier 2 = min(500, 200). This means that for this tier the leverage that will be used is 1:200.
If your account leverage is 1:100 the equation can be understood like this;
Tier 1 = min(100, 500) Tier 2 = min(100, 200) Tier 3 = min(100, 100)
This means at tier 1 your margin is still calculated using 1:100, the same for tier 2 and tier 3.
### Calculating Tiers
The leverage tiers are always represented in USD volume, regardless of the symbol.
For example;
Opening a position of 1,000,000 Buy EURUSD @ 1.12542
Total Buy Volume is 1,000,000 EUR (1 000 000*1.12542 = 1,125,420 USD)
Because the USD exposure of this position is 1,125,420 USD it will fall within two tiers (tier one and tier 2)
Margin Used for Position will be calculated like this:
1 000 000/500 = USD 2,000
125 420/200 = USD 627.10
Total Margin Used = USD 2627.10
Each symbol could have different Dynamic Leverage settings. The Dynamic Leverage tiers used to calculate Margin Used for a position is influenced by the exposure of a specific symbol and specific direction, the exposure is not netted off.
### Calculating Margin
Margin is always converted into the accounts deposit currency. The calculation follows this process; Base Asset > USD (using spot rate depending on trade direction) > Deposit Currency (using spot rate depending on symbol structure)
Here is an example following the above process;
Account Balance = GBP | Zero positions (therefore zero exposure) | Leverage @ 1:500
Margin Required = EUR 200
200 EUR into USD @ 1.09100 (ASK) = USD 218.20
USD 218.20 into GBP @ 1.29400 (ASK) = GBP 168.62
Ask price is used because conversion of USD into GBP uses GBPUSD pair to behave like Selling USD to Buy GBP, therefore Ask price is used.
### Calculating Exposure
Exposure is calculated per symbol and separately for each direction. The exposure in each direction is used to calculate the margin required.
Here is an example of calculating exposure while opening positions in different directions;
Account Balance = USD| Zero positions (therefore zero exposure) | Leverage @ 1:500
Step 1
Total Buy Volume/Exposure : 100k USDJPY
Total Sell Volume/Exposure : 0 USDJPY
Step 2
Open Position Sell 100k USDJPY.
Total Buy Volume/Exposure : 100k USDJPY
Total Sell Volume/Exposure : 100k USDJPY
Step 3
Total Buy Volume/Exposure : 300k USDJPY
Total Sell Volume/Exposure : 100k USDJPY
### Practical Examples
This section includes a number of practical examples of margin calculation in conjunction with Dynamic Leverage and certain trading actions. For the below examples the following settings have been used;
Account Leverage: 500
Account Currency: USD
Volume Based Dynamic Leverage Tiers:
Tier Volume Leverage
1 0-1mill 500
2 1-2mill 200
3 2-3mill 100
4 >3mill 50
##### Example #1
Opening Positions
In the following example, we illustrate how the margin is calculated when a trader opens subsequently three positions on the same symbol, with the following variables in addition to the ones already defined above.
Position 1
Total Buy Volume: 1 million USD
Margin Used for Position: 1 000 000/500 = 2000
Total Margin Used: 2000
Position 2
Total Buy Volume: 2 million USD
Margin Used for Position: 1 000 000/200 = 5000
Total Margin Used: 7000
Position 3
Total Buy Volume: 3 million USD
Margin Used for Position: 1 000 000/100 = 10000
### Position Margin Recalculation
When Dynamic Leverage is used by a broker, this causes different amounts of margin being taken for different positions according to which tier they fall in at the time they were opened. As exposure is reduced by closing, reducing or reversing positions, the tiers that the remaining exposure / positions fall within could be different, this may cause margin requirements to be lower than what is being blocked by the remaining positions.
Brokers have the option to choose whether margin requirements should be recalculated when exposure is reduced, or not.
To ensure the correct amount of margin is being used according to the exact amount of exposure against the Dynamic Leverage tiers assigned to the symbol, or if margin requirements should be locked to position once it is created. Both of these cases are explained further in this guide.
If recalculation is triggered, the position with the lowest volume will be recalculated first and the rest of the positions will be recalculated sequentially according to their size; smallest to largest.
If recalculating margin for positions is enabled by the broker recalculation can triggered by these events;
1. Some trading action occurs therefore the margin required to hold the rest of the positions is recalculated. Actions that trigger recalculation are; Closing Positions, Decreasing Positions, Reversing Positions, Stop Out, Rejected or Partially Filled Order.
2. Broker changes settings of Dynamic Leverage or Account’s leverage.
##### Example #2
In this example we show how margin is recalculated for positions when a broker chooses to recalculate margin requirements. Margin is recalculated when trading events occur, for the specific symbol and direction in subject. The following properties are used for the example;
Account Leverage: 500
Account Currency: USD
Volume Based Dynamic Leverage Tiers:
Tier Volume Leverage
1 0-1mill 500
2 1-2mill 200
3 2-3mill 100
4 >3mill 50
The account has three positions as illustrated below;
Position 1: 1,000,000 BUY USDJPY - Margin Used for Position: 2000
Position 2: 1,000,000 BUY USDJPY - Margin Used for Position: 5000
Position 3: 1,000,000 BUY USDJPY - Margin Used for Position: 10000
Total Margin Used: USD 17000
Step 1
Partially close Position 2 by 500,000 (half) - Margin Used for Position 2: 5,000/2 = USD 2,500
This action triggers recalculation for all positions according to current exposure.
Position 1: 1,000,000 BUY USDJPY - Margin Used for Position: 2,000
• 1,000,000/500 (tier 1) = 2,000
Position 2: 500,000 BUY USDJPY - Margin Used for Position: 2,500
• 500,000/200 (tier 2) = 2,500
Position 3: 1,000,000 BUY USDJPY - Margin Used for Position: 7,500
• 500,000/200 (tier 2) = 2,500
• 500,000/100 (tier 3) = 5,000
This example highlights how margin has been recalculated according to the reduction of Position 2.
##### Example #3
Broker Changes Dynamic Leverage & Margin is Recalculated
When a broker changes Dynamic Leverage tiers and chooses to recalculate margin requirements for all positions, the Used Margin for your positions will change. This example shows what happens in this circumstance. The following properties are used for the example;
Account Leverage: 500
Account Currency: USD
Volume Based Dynamic Leverage Tiers:
Tier Volume Leverage
1 0-1mill 500
2 1-2mill 200
3 2-3mill 100
4 >3mill 50
The account has three positions as illustrated below;
Position 1: 1,000,000 BUY USDJPY - Margin Used for Position: 2000
Position 2: 1,000,000 BUY USDJPY - Margin Used for Position: 5000
Position 3: 1,000,000 BUY USDJPY - Margin Used for Position: 10000
Total Margin Used: USD 17000
Let’s suppose that the broker decides to change the dynamic leverage for USDJPY as illustrated in the table below;
Tier Volume Leverage
1 0-1mill 200
2 1-2mill 100
3 >2mill 50
Then margin used will be recalculated immediately as follows;
• Tier #1 (1,000,000/200) = 5,000 Margin Used
• Tier #2 (1,000,000/100) = 10,000 Margin Used
• Tier #3 (1,000,000/50) = 20,000 Margin Used
Total Margin Used = 35000
Note: It is very important to take notices about margin requirement changes from your broker very seriously because if your margin level drops below stop out level, some of your positions might be stopped out or partially stopped out to restore necessary margin.
##### Example #4
Trading when Margin is NOT Recalculated
When margin recalculation is not used then margin is never recalculated even when you close positions. In this case each position blocks margin and holds it until the position is closed entirely, when partially closed margin will be released proportionately according to the total amount that is held by the position. This example illustrates this through a number of steps.
The following example explains this concept;
Account Leverage: 500
Account Currency: USD
Volume Based Dynamic Leverage Tiers:
Tier Volume Leverage
1 0-1mill 500
2 1-2mill 200
3 2-3mill 100
4 >3mill 50
Step 1
Open Position 1: 1,000,000 BUY USDJPY - Margin Used for Position: 2000
Open Position 2: 1,000,000 BUY USDJPY - Margin Used for Position: 5000
Open Position 3: 1,000,000 BUY USDJPY - Margin Used for Position: 10000
Total Margin Used: USD 17000
Step 2
Close Position 2. The following positions remain;
Position 1: 1,000,000 BUY USDJPY - Margin Used for Position: 2000
Position 3: 1,000,000 BUY USDJPY - Margin Used for Position: 10000
Exposure of Long USDJPY = 2,000,000 (volume of Position 1 + volume of Position 3)
Your margin used will be calculated as follows;
Margin Used = Margin used for Position 1 + Margin used for Position 3 = 12000
Step 3
Open Position 4: 1,000,000 BUY USDJPY
• Tier #3 (1,000,000/100) = 10,000 Margin Used
Your margin used will be calculated as follows;
Margin Used = Margin used for Position 1 + Margin used for Position 3 + Margin Used for Position 4 = 22000
Step 4
Partially close Position 4 by 500,000 (half) - Margin Used for Position 4: 10,000/2 = USD 5,000
Partially close Position 1 by 500,000 (half) - Margin Used for Position 1: 2,000/2 = USD 1,000
Summary of Open Positions;
Position 1: 500,000 BUY USDJPY - Margin Used for Position: 1000 Position 3: 1,000,000 BUY USDJPY - Margin Used for Position: 10000 Position 4: 500,000 BUY USDJPY - Margin Used for Position: 5000
Your margin used will be calculated as follows;
Margin Used = Margin used for Position 1 + Margin used for Position 3 + Margin Used for Position 4 = 16000
Note: In this case, your old positions might be using more margin than required but you are not in danger of stop outs from changing margin calculation requirements.
Note: Contact your broker to clarify which of the two methods is used
Example #5
Broker Changes Dynamic Leverage & Margin is NOT Recalculated
When a broker changes Dynamic Leverage tiers and chooses to not to recalculate margin requirements for all positions, the Used Margin for your positions will not change. This example shows what happens in this circumstance. The following properties are used for the example;
Account Leverage: 500
Account Currency: USD
Volume Based Dynamic Leverage Tiers:
Tier Volume Leverage
1 0-1mill 500
2 1-2mill 200
3 2-3mill 100
4 >3mill 50
Step 1
The account has three positions as illustrated below;
Position 1: 1,000,000 BUY USDJPY - Margin Used for Position: 2000
Position 2: 1,000,000 BUY USDJPY - Margin Used for Position: 5000
Position 3: 1,000,000 BUY USDJPY - Margin Used for Position: 10000
Total Margin Used: USD 17000
Step 2
Let’s suppose that the broker decides to change the dynamic leverage for USDJPY as illustrated in the table below;
Tier Volume Leverage
1 0-1mill 200
2 1-2mill 100
3 >2mill 50
Since the broker has chosen not to recalculate margin, no changes will occur in the Total Margin Used.
Step 3
Close Position 2
In this case the account will have the following positions and Total Margin Used
Position 1: 1,000,000 BUY USDJPY - Margin Used for Position: 2000
Position 3: 1,000,000 BUY USDJPY - Margin Used for Position: 10000
Total Margin Used: USD 12000
Step 4
Open Position 4: 1,000,000 BUY USDJPY
• Tier #3 (1,000,000/50) = 20,000 Margin Used
Margin Used = Margin used for Position 1 + Margin used for Position 3 + Margin Used for Position 4 = 32000 | 3,500 | 13,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-04 | latest | en | 0.914397 |
http://csimsoft.com/help/finite_element_model/non_exodus/boundary_condition.htm | 1,516,261,416,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887077.23/warc/CC-MAIN-20180118071706-20180118091706-00120.warc.gz | 90,278,644 | 3,853 | # Trelis Boundary Conditions
In Trelis, boundary conditions are applied to sidesets or nodesets. Sidesets and nodesets can contain geometry or mesh. This means that models can be remeshed without worrying about losing boundary condition data if the boundary condition is applied to a geometry-based sideset/nodeset.
The sideset/nodeset used by a boundary condition will be visible to the user, and the user can modify the sideset/nodeset separately from the boundary condition. Sidesets/nodesets can be assigned to (or removed from) a boundary condition at any time.
Boundary conditions are broken into four groups: Restraints, loads, contact, and cfd. Each restraint that is created will belong to a restraint set, each load will belong to a load set, and each contact definition will belong to a contact set. A boundary condition set consists of any number of restraints, contact pairs, and loads. CFD boundary conditions do not belong to boundary condition sets.
Table 1: Overview of boundary condition entities available in Trelis
Entity Description and scope FEA Boundary Conditions Acceleration Creates an acceleration boundary condition (acts on a body, volume, surface, curve, or vertex) Velocity Creates a velocity boundary condition (acts on a body, volume, surface, curve, or vertex) Boundary Condition Set Creates a BC set (contains restraint, load and contact sets) Contact Region Creates a contact region between two surfaces or two curves Contact Pair Creates a contact pair between two previously defined contact regions Convection Creates a convection boundary condition (acts on 2d elements) Displacement Creates a displacement boundary condition (acts on a body, volume, surface, curve or vertex) Temperature Create a temperature boundary condition (acts on a surface, curve or vertex) Force Creates a force boundary condition (acts on a surface, curve or vertex) Pressure Creates a pressure boundary condition (acts on a surface or curve) Heat flux Creates a heat flux boundary condition (acts on a surface or curve) CFD Boundary Conditions Inlet Velocity Creates an inlet velocity boundary condition Inlet Pressure Creates an inlet pressure boundary condition Inlet Massflow Creates an inlet massflow boundary condition Outlet Pressure Creates an outlet pressure boundary condition Farfield Pressure Creates a farfield pressure boundary condition Axis Creates an axis boundary condition Exhaust Fan Creates an exhaust fan boundary condition Fan Creates an fan boundary condition Inlet Vent Creates an inlet vent boundary condition Intake Fan Creates an intake fan boundary condition Interface Creates an interface boundary condition Interior Creates an interior boundary condition Mass Flow Inlet Creates an mass flow inlet boundary condition Outflow Creates an outflow boundary condition Outlet Vent Creates an outlet vent boundary condition Periodic Creates an periodic boundary condition Periodic Shadow Creates an periodic shadow boundary condition Porous Jump Creates an porous jump boundary condition Radiator Creates an radiator boundary condition Symmetry Creates a symmetry boundary condition (acts on a surface) Wall Creates an wall boundary condition | 595 | 3,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-05 | latest | en | 0.862097 |
https://download.cnet.com/Tightening-Calculator/3000-2064_4-77176426.html | 1,638,475,469,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362287.26/warc/CC-MAIN-20211202175510-20211202205510-00030.warc.gz | 262,237,921 | 95,749 | X
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# Tightening Calculator for iOS
Free
## Developer's Description
Tightening of BoltsThe program Tightening of Bolts helps you to find or check the tightening procedures of threaded joints using the relationship between torque, clamp load, friction and the property class of the fastener. It also takes different thread and under head friction into account.InputThe calculations can be performed for tightening into the elastic or the plastic area. Choose elastic or plastic. Torque tightening is usually performed in the elastic area and yield and angle controlled tightening in the plastic area.The Input bolt dimensions for the calculations consists of 3 main parameters: The nominal metric thread diameter d The pitch of the thread P Effective diameter for the friction moment at the bearing area under the bolt head (when bolt tightening) or at the nut bearing area (when nut tightening), or bearing diameter DbBelow the Input parameters a field containing another 5 parameters is shown: The equivalent stress red, representing the total stress acting on the bolt (a combination of axial stress because of the clamp force and torsional stress caused by the tightening torque) The clamp force F acting on the bolt The tightening torque T applied to the bolt The bearing friction b The thread friction thIMPORTANTIf you fill in any 3 of these 5 parameters, the 2 parameters that are left open are calculated automatically. The results (all 5 parameters) are shown in the Output field. If less than 3 or more than 3 parameters are filled in no result will be shown.Additionally to these results the following parameters are also calculated and shown in the Output bolt dimensions: The nominal minor diameter of the bolt thread d3 The nominal pitch diameter of the bolt thread d2 The nominal stress area of the bolt thread AS
## Full Specifications
### General
Release June 18, 2016
Date Added January 27, 2016
Version 1.0
### Operating Systems
Operating Systems iOS
Additional Requirements Compatible with: iphone3gs, iphone3gs, iphone4, iphone4, ipodtouchfourthgen, ipodtouchfourthgen, ipad2wifi, ipad2wifi, ipad23g, ipad23g, iphone4s, iphone4s, ipadthirdgen, ipadthirdgen, ipadthirdgen4g, ipadthirdgen4g, iphone5, iphone5, ipodtouchfifthgen, ipodtouchfifthgen, ipadfourthgen, ipadfourthgen, ipadfourthgen4g, ipadfourthgen4g, ipadmini, ipadmini, ipadmini4g, ipadmini4g
### Popularity
Total Downloads 0
Downloads Last Week 0
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Indeed Job Search | 659 | 2,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-49 | longest | en | 0.861651 |
https://math.stackexchange.com/questions/18657/is-a-regular-ring-a-domain | 1,553,343,508,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202804.80/warc/CC-MAIN-20190323121241-20190323143241-00260.warc.gz | 563,515,151 | 31,652 | # Is a regular ring a domain
A regular local ring is a domain. Is a regular ring (a ring whose localization at every prime ideal is regular) also a domain? I am unable to find/construct a proof or a counterexample. Any help would be appreciated.
Geometrically, one can think of this as follows: regularity of $A$ is a local property on Spec $A$, and Spec $A\times B$ is equal to Spec $A \coprod$ Spec $B$. So locally Spec $A\times B$ looks like either Spec $A$ or Spec $B$. In particular, local properties, such as regularity (or the condition that the localization at prime ideals be a domain) can't detect global properties (like $A$ itself being a domain).
• The geometric point of view also illuminates the claim that if every localization is a domain the ring is a product of domains: if $\operatorname{Spec} A$ has more than one irreducible component, then if the components have any point of contact this point will yield a local ring that is not a domain. Thus the irreducible components of \operatorname{Spec}A \$ must be completely disjoint. The decomposition into connected components will then correspond to a product decomposition of the ring. – Ben Blum-Smith Feb 26 '17 at 19:52 | 282 | 1,195 | {"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} | 2.921875 | 3 | CC-MAIN-2019-13 | latest | en | 0.895184 |
https://leanprover-community.github.io/mathlib_docs/group_theory/group_action/option.html | 1,708,566,921,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473598.4/warc/CC-MAIN-20240221234056-20240222024056-00183.warc.gz | 372,170,378 | 42,226 | mathlib3documentation
group_theory.group_action.option
Option instances for additive and multiplicative actions #
THIS FILE IS SYNCHRONIZED WITH MATHLIB4. Any changes to this file require a corresponding PR to mathlib4.
This file defines instances for additive and multiplicative actions on option type. Scalar multiplication is defined by a • some b = some (a • b) and a • none = none.
• group_theory.group_action.pi
• group_theory.group_action.prod
• group_theory.group_action.sigma
• group_theory.group_action.sum
@[protected, instance]
def option.has_vadd {M : Type u_1} {α : Type u_3} [ α] :
(option α)
Equations
@[protected, instance]
def option.has_smul {M : Type u_1} {α : Type u_3} [ α] :
(option α)
Equations
theorem option.smul_def {M : Type u_1} {α : Type u_3} [ α] (a : M) (x : option α) :
a x = x
theorem option.vadd_def {M : Type u_1} {α : Type u_3} [ α] (a : M) (x : option α) :
a +ᵥ x = x
@[simp]
theorem option.vadd_none {M : Type u_1} {α : Type u_3} [ α] (a : M) :
@[simp]
theorem option.smul_none {M : Type u_1} {α : Type u_3} [ α] (a : M) :
@[simp]
theorem option.vadd_some {M : Type u_1} {α : Type u_3} [ α] (a : M) (b : α) :
@[simp]
theorem option.smul_some {M : Type u_1} {α : Type u_3} [ α] (a : M) (b : α) :
@[protected, instance]
def option.vadd_assoc_class {M : Type u_1} {N : Type u_2} {α : Type u_3} [ α] [ α] [ N] [ α] :
(option α)
@[protected, instance]
def option.is_scalar_tower {M : Type u_1} {N : Type u_2} {α : Type u_3} [ α] [ α] [ N] [ α] :
(option α)
@[protected, instance]
def option.vadd_comm_class {M : Type u_1} {N : Type u_2} {α : Type u_3} [ α] [ α] [ α] :
(option α)
@[protected, instance]
def option.smul_comm_class {M : Type u_1} {N : Type u_2} {α : Type u_3} [ α] [ α] [ α] :
(option α)
@[protected, instance]
def option.is_central_scalar {M : Type u_1} {α : Type u_3} [ α] [ α] [ α] :
(option α)
@[protected, instance]
def option.is_central_vadd {M : Type u_1} {α : Type u_3} [ α] [ α] [ α] :
(option α)
@[protected, instance]
def option.has_faithful_smul {M : Type u_1} {α : Type u_3} [ α] [ α] :
(option α)
@[protected, instance]
def option.has_faithful_vadd {M : Type u_1} {α : Type u_3} [ α] [ α] :
(option α)
@[protected, instance]
def option.mul_action {M : Type u_1} {α : Type u_3} [monoid M] [ α] :
(option α)
Equations | 831 | 2,283 | {"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} | 2.5625 | 3 | CC-MAIN-2024-10 | latest | en | 0.337587 |
https://library.fiveable.me/incompleteness-and-undecidability/unit-3/examples-formal-theories/study-guide/7EtC1pAqPPbyjZUj | 1,725,917,944,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651157.15/warc/CC-MAIN-20240909201932-20240909231932-00519.warc.gz | 355,389,640 | 57,166 | 3.4 Examples of formal theories
Formal theories like and provide a rigorous foundation for mathematics. These systems use and to define fundamental concepts like and sets, enabling precise proofs of important .
However, formal theories have limitations. show that consistent systems containing arithmetic are necessarily incomplete, with statements that can't be proved or disproved within the theory itself. This reveals inherent constraints in formalizing mathematics.
Notable Examples of Formal Theories
Examples of formal theories
• Peano arithmetic (PA) axiomatizes the natural numbers and their arithmetic using
• Based on a set of axioms that define the (0, 1, 2, ...) and the operations of addition and multiplication
• Allows for rigorous proofs of many fundamental properties of natural numbers (commutativity, associativity, distributivity)
• Zermelo-Fraenkel set theory (ZF) formalizes the principles of set theory using first-order logic
• Axiomatizes the notion of sets and their basic operations (union, intersection, power set)
• Provides a foundation for much of modern mathematics, as many mathematical objects can be defined in terms of sets (numbers, functions, spaces)
• Other notable examples demonstrate the wide applicability of formal theories across various branches of mathematics
• (Z2) extends PA by introducing variables and quantifiers for sets of natural numbers, enabling proofs of stronger statements
• (NBG) and (MK) extend ZF with additional axioms to handle classes and larger collections
Axioms and theorems in theories
• Peano arithmetic (PA) is built upon several types of axioms and allows for the derivation of many important theorems
• Axioms for natural numbers establish the basic properties of 0 and the successor function (0 is a natural number, every natural number has a unique successor)
• Axioms for addition and multiplication define the behavior of these operations (commutativity, associativity, distributivity)
• The allows for proofs by mathematical induction, a powerful technique for establishing statements about all natural numbers
• of addition, multiplication, and exponentiation enable the construction of more complex arithmetic expressions
• Theorems about the properties of natural numbers and their arithmetic can be rigorously proved using the axioms and rules of inference (there are infinitely many prime numbers, the fundamental theorem of arithmetic)
• Zermelo-Fraenkel set theory (ZF) is based on a collection of axioms that formalize the properties of sets
• The states that sets are equal if they have the same elements, providing a criterion for set equality
• The allows for the construction of sets with two elements, enabling the definition of ordered pairs and Cartesian products
• The states that for any set of sets, there exists a set containing all elements of its member sets, formalizing the notion of set union
• The asserts that for any set, there exists a set containing all its subsets, crucial for defining the cardinality of sets
• The guarantees the existence of an infinite set, necessary for constructing the natural numbers and other infinite structures
• The allows for the construction of subsets defined by a property, enabling the definition of set-builder notation
• The states that the image of a set under any definable function is also a set, used in the construction of large cardinals
• The (or foundation) asserts that every non-empty set has an $\in$-minimal element, ruling out sets that are elements of themselves
• Definitions of basic set-theoretic concepts (subset, union, intersection, power set, Cartesian product) can be rigorously formulated using the axioms
• Many important theorems about the existence and properties of various sets can be proved in ZF (, , )
Consistency and completeness analysis
• is a crucial property for a formal theory, ensuring that it does not lead to contradictions
• A theory is consistent if there is no statement $P$ such that both $P$ and its negation $\neg P$ are provable in the theory
• Peano arithmetic and Zermelo-Fraenkel set theory are believed to be consistent, based on their alignment with intuitive notions of numbers and sets
• However, Gödel's second incompleteness theorem shows that if PA or ZF are consistent, they cannot prove their own consistency within the theory itself
• refers to the relationship between axioms and the other axioms of a formal theory
• An axiom is independent of a theory if it cannot be derived from the other axioms of the theory
• The (there is no set whose cardinality is strictly between that of the integers and the real numbers) is independent of ZF
• The (for any set of non-empty sets, there exists a function that chooses an element from each set) is also independent of ZF
• is another important property for formal theories, related to decidability of statements
• A theory is complete if for every well-formed formula $P$ in the language of the theory, either $P$ or $\neg P$ is provable in the theory
• Gödel's first incompleteness theorem shows that any consistent formal theory containing arithmetic (like PA or ZF) is incomplete
• There will always be statements (like the ) that are neither provable nor disprovable within the theory, assuming its consistency
Significance for mathematical foundations
• Formal theories like PA and ZF play a crucial role in providing a rigorous foundation for large parts of mathematics
• They allow for precise formulations of mathematical concepts and statements using formal logic
• Proofs can be rigorously checked for validity using the axioms and rules of inference of the theory
• Metamathematical questions about provability, consistency, and independence can be investigated within the framework of formal theories
• However, formal theories also have inherent limitations, as revealed by Gödel's incompleteness theorems
• Consistent theories containing arithmetic are necessarily incomplete, with statements that are neither provable nor disprovable
• The consistency of theories like PA and ZF cannot be established within the theories themselves, requiring stronger metatheories
• Some mathematical concepts and statements may not be expressible within the language of a given formal theory, limiting its scope and applicability
Key Terms to Review (40)
Axiom of Choice: The Axiom of Choice is a principle in set theory that states for any set of non-empty sets, there exists a choice function that selects an element from each set. This concept plays a crucial role in various mathematical theories, allowing mathematicians to make selections from collections without explicitly defining how to choose the elements, impacting foundational aspects of mathematics and logic.
Axiom of Extensionality: The axiom of extensionality is a fundamental principle in set theory that states two sets are equal if and only if they have the same elements. This axiom is crucial for distinguishing between different sets based solely on their contents, rather than how they are defined or constructed. Understanding this concept allows for clearer reasoning about sets and their relationships within formal theories and axiomatic systems.
Axiom of Infinity: The Axiom of Infinity is a fundamental principle in set theory that asserts the existence of an infinite set. Specifically, it states that there exists a set that contains the empty set and is closed under the operation of forming the successor of any element in the set, leading to the construction of the natural numbers. This axiom is crucial for developing theories that include the concept of infinity, such as arithmetic and analysis.
Axiom of Pairing: The Axiom of Pairing is a fundamental principle in set theory that asserts for any two sets, there exists a set that contains exactly those two sets as its elements. This axiom is crucial for constructing sets and establishing relationships among them, serving as a foundational element in the broader framework of formal theories that explore mathematical structures and their properties.
Axiom of Regularity: The Axiom of Regularity, also known as the Axiom of Foundation, states that every non-empty set A contains an element that is disjoint from A. This axiom helps avoid certain paradoxes and ensures a well-founded structure for sets, supporting the overall consistency of set theory. It plays a crucial role in distinguishing between sets and proper classes, which is essential when exploring various formal theories.
Axiom of Union: The Axiom of Union states that for any set, there exists a set that contains exactly the elements of the sets that are members of the original set. This axiom allows for the creation of a new set from the union of the elements of its members, ensuring that the structure of sets can be extended and manipulated. It plays a crucial role in formal theories by enabling operations on collections of sets, facilitating discussions about set membership and relations among sets.
Axiom Schema of Replacement: The Axiom Schema of Replacement is a principle in set theory that allows for the construction of new sets from existing ones by replacing elements according to a specific definable rule. It asserts that if a property can be defined for each element of a set, then there exists a new set containing exactly those elements that satisfy the property, essentially enabling one to generate a collection of sets from a single set through transformation.
Axiom Schema of Separation: The Axiom Schema of Separation is a principle in set theory that allows the formation of subsets from existing sets based on a specific property or condition. This schema asserts that for any set and any property, there exists a subset containing exactly those elements of the original set that satisfy that property. This concept is essential for formal theories as it enables the construction of more complex sets from simpler ones, ultimately influencing the foundation of mathematical logic.
Axioms: Axioms are foundational statements or propositions that are accepted as true without proof, serving as the starting point for logical reasoning within a formal system. They establish the basic framework from which theorems and other propositions can be derived, playing a crucial role in defining the structure and behavior of mathematical theories and logical systems.
Cantor's Theorem: Cantor's Theorem states that for any set, the power set (the set of all subsets) has a strictly greater cardinality than the set itself. This theorem reveals important insights into the nature of infinity and the hierarchy of infinite sets, illustrating that not all infinities are equal and establishing a fundamental result in set theory.
Church-Turing Thesis: The Church-Turing Thesis is a foundational concept in computer science and mathematical logic that proposes that any function which can be computed algorithmically can be computed by a Turing machine. This thesis establishes a link between formal languages, automata theory, and the limits of computation, suggesting that various models of computation are equivalent in terms of what they can compute.
Completeness: Completeness refers to a property of a formal system where every statement that is true in the system can be proven within that system. This means that if something is semantically valid, it can also be derived syntactically through the axioms and rules of inference of the system. Understanding completeness helps in evaluating the capabilities and limitations of formal systems, especially in relation to models, interpretations, and proof structures.
Consistency: In mathematical logic, consistency refers to the property of a formal system whereby no contradictions can be derived from its axioms and rules of inference. A consistent system ensures that if a statement is provable, then it is true within the interpretation of the system, thus maintaining the integrity of the mathematical framework.
Continuum hypothesis: The continuum hypothesis posits that there is no set whose cardinality is strictly between that of the integers and the real numbers. In simpler terms, it suggests that the sizes of infinite sets can be neatly categorized, with no in-between sizes between these two well-known infinities. This idea ties into foundational questions about the nature of infinity and the structure of mathematical truths.
Decidable Problem: A decidable problem is a decision problem for which an algorithm exists that can provide a correct yes or no answer for every possible input in a finite amount of time. This concept is crucial in understanding the limitations of computation and forms the foundation for many formal systems and theories, including those that examine representability and formal languages.
First-order logic: First-order logic is a formal system used in mathematics, philosophy, linguistics, and computer science that extends propositional logic by allowing the use of quantifiers and predicates to express statements about objects and their relationships. It provides a structured way to represent facts and reason about them, connecting deeply with the limitations of formal systems, independence results in set theory, and the foundational aspects of mathematical logic.
Formal System: A formal system is a set of rules and symbols used to derive conclusions or prove statements within a logical framework. It consists of a language with syntactic and semantic components, allowing for the formulation of axioms, theorems, and proofs, establishing a foundation for reasoning in mathematics and logic.
Gödel Sentence: A Gödel sentence is a specific type of self-referential statement constructed within formal mathematical systems, which asserts its own unprovability within that system. This idea is pivotal because it illustrates the limits of provability in formal systems, linking closely to significant theorems that demonstrate the inherent incompleteness and undecidability of those systems.
Gödel's Incompleteness Theorems: Gödel's Incompleteness Theorems are two fundamental results in mathematical logic that demonstrate inherent limitations in formal systems capable of expressing arithmetic. The first theorem states that in any consistent formal system that can express basic arithmetic, there exist statements that are true but cannot be proven within that system. The second theorem extends this idea, showing that no consistent system can prove its own consistency. These theorems highlight the boundaries of formal systems and have profound implications across various areas such as mathematics and philosophy.
Halting Problem: The halting problem is a decision problem that determines, given a description of an arbitrary computer program and an input, whether the program will finish running or continue indefinitely. This concept highlights the limitations of computation, as it was proven that there is no general algorithm that can solve this problem for all possible program-input pairs.
Independence: Independence refers to the property of a set of axioms or statements such that none of them can be derived from the others within a given formal system. This means that there exist models of the system where certain statements are true while the axioms remain consistent. Independence is crucial because it highlights the limitations of formal theories and the necessity for certain axioms to describe mathematical structures accurately.
Induction Axiom: The induction axiom is a fundamental principle in mathematical logic and set theory that asserts that a property or statement holds for all natural numbers if it holds for the number zero and holds for any number 'n' implies it also holds for 'n+1'. This axiom forms the basis of mathematical induction, which is essential for proving propositions about natural numbers and has significant implications in formal theories.
Kurt Gödel: Kurt Gödel was an Austrian-American mathematician and logician best known for his groundbreaking work on the incompleteness theorems, which demonstrated inherent limitations in formal systems. His findings challenged the prevailing notions of mathematics and logic, revealing that in any sufficiently powerful axiomatic system, there are true statements that cannot be proven within the system itself.
Logical Rules: Logical rules are formalized principles that dictate the valid inferences or deductions that can be drawn within a logical system. These rules serve as the foundation for reasoning processes, allowing one to derive conclusions from premises while ensuring consistency and soundness in arguments. In formal theories, logical rules help establish the structure and interpretation of statements, ensuring that conclusions follow logically from established axioms or propositions.
Model theory: Model theory is a branch of mathematical logic that deals with the relationships between formal languages and their interpretations or models. It focuses on understanding how structures satisfy the sentences of a given language, exploring concepts like truth, consistency, and the nature of mathematical structures. Through its connections to formal systems, representability, and independence results, model theory plays a crucial role in understanding the limits and capabilities of mathematical theories.
Morse-Kelley Set Theory: Morse-Kelley Set Theory is a formal system of set theory that extends Zermelo-Fraenkel set theory with additional axioms to handle proper classes. It introduces the concept of class comprehension, allowing for the construction of sets and classes based on properties without falling into paradoxes. This theory is significant for its ability to define larger collections, such as the class of all sets, while maintaining a rigorous foundation.
Natural Numbers: Natural numbers are the set of positive integers starting from 1 and going upwards (1, 2, 3, ...). They are foundational in mathematics, particularly in formal theories where they are used to build more complex structures and concepts. Natural numbers play a crucial role in number theory, arithmetic, and logic, forming the basis for various mathematical operations and proofs.
Peano Arithmetic: Peano Arithmetic is a formal system that encapsulates the basic properties of natural numbers using axioms proposed by Giuseppe Peano. It serves as a foundational framework for number theory and allows for the formalization of arithmetic operations and properties, highlighting the limitations and strengths of formal systems in capturing mathematical truths.
Power Set Axiom: The Power Set Axiom is a fundamental principle in set theory that states for any set, there exists a set of all its subsets, known as the power set. This axiom is essential for the construction of more complex mathematical structures and plays a key role in the formal foundations of mathematics by allowing for the manipulation and analysis of collections of sets.
Properties of Natural Numbers: The properties of natural numbers refer to the fundamental characteristics and rules that govern the set of natural numbers, which includes all positive integers starting from 1. These properties are foundational in mathematics and include concepts such as closure, commutativity, associativity, distributivity, and the existence of identity and inverse elements. Understanding these properties is essential for working within formal theories and mathematical proofs.
Provable statement: A provable statement is a statement in a formal system that can be demonstrated to be true using the axioms and rules of inference of that system. These statements are essential because they establish the reliability of the formal system, showing which assertions can be accepted as true based on the foundational elements provided. In the context of formal theories, provable statements play a crucial role in determining the consistency and completeness of these systems.
Recursive definitions: Recursive definitions are rules that define an object in terms of itself, allowing for the construction of complex structures from simpler ones. This concept is crucial in formal theories as it provides a framework for defining mathematical objects, sequences, and functions through a base case and a recursive step. By using recursive definitions, one can express infinite processes in a finite manner, which is essential for understanding the foundations of logic and mathematics.
Schröder-Bernstein Theorem: The Schröder-Bernstein Theorem states that if there are injections (one-to-one functions) from set A to set B and from set B to set A, then there exists a bijection (a one-to-one correspondence) between the two sets. This theorem highlights a fundamental property of cardinality in set theory, establishing that two sets have the same size if they can be injected into each other.
Second-order arithmetic: Second-order arithmetic is a formal theory that extends first-order arithmetic by including quantification over sets of natural numbers, not just individual natural numbers. This allows for more expressive statements about properties of numbers and enables the formulation of many mathematical concepts, such as analysis, within a framework that is still manageable for proof theory and ordinal analysis.
Theorems: Theorems are formal statements or propositions that have been proven based on previously established statements, such as axioms and other theorems, within a formal system. They serve as the building blocks of mathematical and logical reasoning, providing a framework for deriving further conclusions and ensuring the consistency of the system in which they are formulated.
True but unprovable statement: A true but unprovable statement is a statement that, while it can be demonstrated to be true, cannot be proven within a given formal system. This concept is central to understanding the limits of formal theories and is a key outcome of Gödel's Incompleteness Theorems, which show that no consistent system of axioms can prove all truths about the arithmetic of natural numbers.
Von Neumann-Bernays-Gödel set theory: Von Neumann-Bernays-Gödel set theory (NBG) is a formal set theory that extends Zermelo-Fraenkel set theory (ZF) by allowing for classes, in addition to sets, which are collections of objects that may not themselves be sets. This framework is particularly useful in addressing issues related to large collections and provides a way to formalize statements about all sets and classes while maintaining consistency and avoiding paradoxes.
Von Neumann-Bernays-Gödel Set Theory: The von Neumann-Bernays-Gödel (NBG) set theory is an extension of Zermelo-Fraenkel set theory that incorporates classes as well as sets. It allows for a more robust framework for discussing and manipulating both sets and proper classes, which helps to address certain limitations in traditional set theories, particularly in the context of formal theories that explore the foundations of mathematics.
Zermelo-Fraenkel Set Theory: Zermelo-Fraenkel Set Theory (ZF) is a foundational system for mathematics that provides a formal framework for set theory based on a collection of axioms. It addresses issues of self-reference and circularity, which are crucial for establishing a rigorous mathematical foundation while avoiding paradoxes such as Russell's paradox. This system significantly impacts various branches of mathematics and logic, particularly in its philosophical implications regarding incompleteness and the nature of formal theories.
Zorn's Lemma: Zorn's Lemma is a principle in set theory that states if every chain (a totally ordered subset) in a non-empty partially ordered set has an upper bound, then the entire set contains at least one maximal element. This lemma is equivalent to the Axiom of Choice and plays a crucial role in various formal theories by providing a method to establish the existence of certain objects without explicitly constructing them. | 4,529 | 23,972 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-38 | latest | en | 0.895929 |
http://www.r-bloggers.com/particle-learning-rejoinder/ | 1,467,159,185,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397428.37/warc/CC-MAIN-20160624154957-00041-ip-10-164-35-72.ec2.internal.warc.gz | 806,936,297 | 19,149 | # Particle learning [rejoinder]
November 9, 2010
By
(This article was first published on Xi'an's Og » R, and kindly contributed to R-bloggers)
Following the posting on arXiv of the Statistical Science paper of Carvalho et al., and the publication by the same authors in Bayesian Analysis of Particle Learning for general mixtures I noticed on Hedibert Lopes’ website his rejoinder to the discussion of his Valencia 9 paper has been posted. Since the discussion involved several points made by members of the CREST statistics lab (and covered the mixture paper as much as the Valencia 9 paper), I was quite eager to read Hedie’s reply. Unsurprisingly, this rejoinder is however unlikely to modify my reservations about particle learning. The following is a detailed examination of the arguments found in the rejoinder but requires a preliminary reading of the above papers as well as our discussion..
“Particle learning based on the product estimate and MCMC based on Chib’s formula produce relatively similar results either for small or large samples”
This statement about the estimation of the marginal likelihood (or the evidence) and the example A that is associated with it thus comes to contradict our (rather intensive) simulation experiment which, as reported in the discussion, concludes to the strong bias in evidence induced by using particle learning, whether or not the product estimator is used. We observed there that there were two levels of degeneracy, one due to the product solution (errors in a product being more prone to go and…multiply) and one due to the particle nature of the sequential method (which does not refresh particles from earlier periods). The above graph is at odds with the one presented in the rejoinder, maybe because we consider 10,000 observations rather than 100. (I also fail to understand how the “Log-predictive (TRUE)” is derived.)
“Black-box sequential importance sampling algorithms and related central limit theorems are of little use in practice.”
Another quote from the rejoinder I do not get. What’s wrong with the central limit theorem?! One major lesson from the central limit theorem is that it provides a scale for the speed of convergence and thus an indicator on the number of particles needed for a given precision level. The authors of the rejoinder then criticise our use of “1000 particles in 5000 dimensional problems” as we “shouldn’t be surprised at all with some of our findings”. I find no trace in the discussion of such a case: we use 10,000 particles in all examples and the target is either the distribution of the 4 mixture parameters, the evidence or the distribution of a one-dimensional sufficient statistic. Furthermore, these values of n and N are those used in their example D…
“This argument [that the Monte Carlo variance will `blow-up’] is incorrect and extremely misleading.”
This point is central to both the discussion and the rejoinder, as the authors maintain that the inevitable particle degeneracy does not impact the distribution of the sufficient statistics. The argument about using time averages over particle paths rather than sums appears reasonable at first. Actually, taking an empirical average in almost stationary situations should produce an approximately normal distribution. With an asymptotic variance different from 0. (Thanks to the central limit theorem by the way!) However, this is not the main argument used in the discussions. Degeneracy in the particle path means that the early terms in the average are less and less diverse in the sample average. Therefore it is not that surprising that the variance is decreasing to too small a value! As shown in Figure 8 of the discussion, degeneracy due to resampling may induce severe biases in the distribution of empirical averages while giving the impression of less variability. Furthermore, the fact that parameters are simulated [rather than fixed] in the filter means that the process is not geometrically ergodic, hence that Monte Carlo errors tend to accumulate along iterations, rather than compensate… (This is why the comparison between PL and sampling importance resampling is particularly relevant, because it does not address this accumulation.) The rejoinder also quotes Olsson et al. (2008) for justifying the decrease in the Monte Carlo variance. This is somehow surprising in that (a) Olsson et al. (2008) show that there is degeneracy without a fixed-lag smoothing and (b) they require a geometric forgetting property on the filtering dynamics. In addition, I note that Example E used to illustrate the point about variance reduction is not very appropriate for this issue because the hidden Markov chain is a Gaussian random walk, hence cannot be stationary (a fact noted by the authors). And again a decrease in the “MC error” does not mean a converging algorithm because degeneracy naturally induces empirical variance decrease. (I also fail to see why the “prior” on $(x_t)$ is improper.) The final argument that “PL parameters do not degenerate” is somehow puzzling: by nature, those parameters are simulated from a distribution conditional on the sufficient parameters. So obviously the simulated parameters all differ. But this does not mean that they are marginally distributed from the right distribution.
“MCMC schemes depend upon the not so trivial task of assessing convergence. How long should the burn-in G0 be?”
The rejoinder concludes with recommendations that sound more like a drafted to-do note the authors forgot to remove than an accumulation of true recommendations. It seems to me that the comparison between MCMC and particle filters is not particularly relevant, simply because particle filters apply in [sequential] settings where MCMC cannot be implemented. To try to promote PL over MCM by arguing that MCMC produces dependent draws while having convergence troubles is not needed (besides, PL also produces [unconditional] dependent draws). To advance that the Monte Carlo error for PL is in $C_T/sqrt{N}$ is not more relevant because $C_T$ is exponential in $T$ and because MCMC also has an error in $sqrt{N}$.
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April 14 2015 at 9:20 AM
Owner
Geoff.... how does one read, feel, calculate the effect of two breaks on a level green... and even a third break when the ball speed is decaying rapidly and goes wobbly as it nears the cup? Then there is the overall slope of the green which adds another variable.
To simplify the complexity, perhaps you could assume a 24 ft. putt with an initial 12 ft. break and then another 12 ft. opposite break, or something like that. Also comment on different break point reversal ratios.
Could you also relate the path(s) of the putt to that oft-used graph showing the skid/roll/decay of the ball at various speeds? And what about different green speeds? Thanks.
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Geoff Mangum
Owner
71.54.49.198
# Sum the Proportionate Time Across Different Slopes
April 14 2015, 9:22 AM
Dear sammy,
First, the brain naturally predicts these breaks if you "run the movie of your actual ball pace over the exact contour all the way to the hole". So you know the pace extremely well near the hole, since these are always identical at the end of the putt, and you work the pace backwards. With this total-curve sense of ball pace, you then predict the curve's shape, which can also proceed from the hole in reverse back to the ball. This is naturally what brains do, just applied to the ball-on-surface-with-exact-pace pattern and situation.
Second, you can use math to segment the different slopes. Slopes on greens aren't really separate and distinct, but blur from Slope% X to Slope% Y smoothly thru the intervening slopes. The "segmenting" is dividing the slopes into "basically large flat areas," or one Slope% that stays the same for a while before changing into another Slope%. This ignores the transitioning area between segments, but that's not too far off.
Assume three different Slope%'s and their flat areas on the way to the hole -- the first Slope% being 2% with the fall line to the left of your ball so the break is right to left; the second flat area has a Slope% of 3% also right to left; the third flat area has a Slope% of 1.5% also right to left. This is more or less going from usual slope over a bit of a hump or cone projecting into your path from the right with greater slope and then off that onto a milder slope to the hole.
The break over these three slopes can be "summed" but each slope contributes break according to the TIME the ball spends on that slope, not its extent or size. Hence, the TIME is a combination of the SIZE or LENGTH across the slope for the ball path and the ball's average pace or VELOCITY over that area.
Over the first slope nearest the ball, the ball starts with maximum velocity and slows down at some rate (which is more pronounced if the ball is skidding at rhe beginning), then over the second area the ball is slowing at some pretty steady rate depending upon green speed, and then at the end the ball's arrival at the hole is the same as always, finishing up with a decelerating pace so the ball makes it to the hole and goes in nicely or rolls only a short distance past the hole and stops.
Assume each flat area is EQUAL SIZE -- perhaps 10 feet across each flat area for a total of 30 feet. This corresponds on usual green speed to a starting maximum velocity of about 300 inches per second (57 revolutions per second). Across the first slope this might slow fairly suddenly due to skidding and green friction over the first few feet then settle into a usual deceleration, exiting the area at perhaps 160 inches per second (30 revolutions per second). Then across the second flat area the ball continues to decelerate steadily, perhaps to 70 inches per second (13 revolutions per second). Then across the third segment, the ball decelerates down to say 3 revolutions per second at the front lip and zero just a few rolls past the cup.
The AVERAGE VELOCITY across each 10-foot patch is then 300 + 160 / 2 = 230 inches per second on the 1st patch; 160 + 70 / 2 = 115 inches per second on the 2nd patch; and 70 + 3 / 2 = 36.5 inches per second over the final patch.
The TIME the ball spends on each 10-foot or 120-inch patch is simply Length / Average Velocity, so the three patches have this TIME:
1st patch 120 / 230 = 0.52 seconds
2nd patch 120 / 115 = 1.04 seconds
3rd patch 120 / 36.5 = 3.29 seconds
So the total putt TIME is 4.85 seconds. The PROPORTIONATE TIMES for each segment are:
1st segment 10.7%
2nd segment 21.4%
3rd segment 67.8%
With the proportionate contributions of the TIME, you can get an "overall, average Slope%" for the entire putt, ignoring the transitional areas.
So in our case, the average slope% is just the sum of the weighted segments:
2% x 0.107 + 3% x 0.214 + 1.5% x 0.687 = 0.214 + 0.642 + 1.031 = 1.857%, or 1.9%.
So it's roughly the equivalent of a 30 foot putt across one flat area of 1.9% slope. That's not much different than a 2% Slope that breaks 1" per foot, so the break is about 30"' with a target aim spot 30" up the fall line from the center of the cup.
If the breaks go in different directions, the first slope is positive and all others breaking that same way are positive, and opposite-breaking slopes are negative. Find the TIME per segment and the sum the proportionate contributions.
It's helpful to notice that putts that break different directions basically have less total break than the usual putts, and sometimes the break one way just about cancels out the other break, so the putt is pretty straight.
Another fact to notice -- after years of experience and observation, of course -- is that most "multiple breaking putts" break over different slopes that all break the same general way -- left to right or right to left. That is, your ball is on the same side of the fall lines over each separate flat-but-tilted area -- balls to the right of fall lines aiming uphill break to the left downhill, and balls to the left of fall lines break downhill to the right. Considering two such slopes, the one closest to / at the hole is the decisive slope, and the one to care most about. These final slopes come in two flavors: steeper than the first, or less steep than the first. If the final slope is steeper, play more break, and if not, play less break.
These two slopes also come in two other flavors: last slope uphill or last slope downhill. If uphill, you have a green light for pace, since the increasing slope when the ball is slowing really acts in combination to decelerate the ball. If the second slope is downhill, or more downhill than the first slope, be careful but also realize that the ball will be slowing onto the second slope so try to avoid being SO CAREFUL that the ball gets hung up on the second slope and never makes it to the hole. Whenever the final slope is downhill for ANY putt, be sure to look carefully at this final slope from behind the hole, reading the putt into the cup down that slope with the fall line of that slope clearly in mind.
Putts that actually CHANGE directions from breaking left to right and now breaking right to left MUST HAVE CROSSED OVER A FALL LINE. This tells us something important: the break direction does not change unless the putt crosses over a fall line, so if you don't see one, the putt is probably not a double breaker.
Generally, it is more or less the case that putts spend about 2/3rds the total time over the last half of the putt, so that helps. If a double breaking putt travels over 3% then 2% with both breaking right to left, and both about half the putt, the average or overall slope will be a lot closer to 2% than to 3%. The actual math using this rough idea is 2 x 0.67 + 3 x 0.33 = 2.33%. This rule of thumb is not exact, but is reasonable.
There are obviously some common situations that are tough to read, involving more than the simple break over the same flat-but-tilted area, such as putting off one slope up onto a ridge or bowl, and putting up along a "river bed" with the hole slightly off the river on a side slope, but that's why I teach lessons and clinics.
You can always more or less ballpark a mathematical aim spot on the fall line by averaging the slopes. A really rough way to get started is to notice that almost all greens slope from front up to back, so that amateurs don't have to play to away-sloping greens. So if you go to the front of the green where the fringe is lowest and look across to the back of the green at the point where the fringe is highest, and estimate the difference by standing your putter shaft up from the low fringe and sighting level across the vertical shaft to the back fringe while pretending to pat your hand on the back fringe, and noticing how far up the shaft in feet the hand is patting, you can then ballpark the green's total overall average slope.
It's very common to find that greens have a general or overall slope of about 2%. This seems to be something of a default slope since 1% may not drain enough and flattish greens are hard for amateurs to read, while 3-4% slopes may slough the irrigation or rain off the surface too quickly without benefitting the roots and also present difficult putts for amateurs. So the 2% average slope is sort of a "design sweetspot".
All you need to know to finish this is the distance to the back fringe to get the percent slope (rise on putter over run from front fringe to back fringe). Conveniently, many many greens are 33 yards in depth, so that is 100 feet. In that case, the green's overall slope is whatever number of feet your hand seems to be raised above the low fringe while pretending to pat the high fringe on the other side of the green. If the hand is at the bottom of the grip material, that is probably 2' high, so the green has an average slope of 2%. If the run from your low point to the high point is only 50', and your hand is 1' high when patting, the slope is just twice the hand height, so agan 2% average slope. Divide the hand height in feet by the distance from fringe to fringe in feet to get a sense of the overall slope percent.
This overall average slope often helps when planning a long lag acros some complications, but in general across an average slope of "about 2%" or whatever your estimate. For example, a long lag of 55 feet across some complications but nothing crazily different from the overall average slope has a "pretty close" ballpark aim of 55" up the fall line at the hole when the average slope is 2%. That's about 1 and 1/2 putter lengths up the fall line from the cup. If you aimed there as a start, and then assessed whether the ball would stay on the high side all the way to the cup if you putted normal pace, then that target and start line is very likely agood one.
Remember, these multiple breaks are not likely to present inside 10 feet, so when these multiple breaking putts do present, they are most likely to be long lag putts that have priority on getting the first putt within 2 feet or so of the hole for a safe two-putt. That's where the lag X comes into play.
To the extent you are asking how to read these putts EXACTLY, well, to hell with that because it's pointless and stupid and not going to happen and too difficult in math and unwise to think you can and will make it harder to do as good a job as you are able to do. Just lag these close.
Cheers!
Geoff
This message has been edited by aceputt from IP address 71.54.49.198 on Apr 14, 2015 9:50 AM
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Find more forums on Golf Create your own forum at Network54 | 2,719 | 11,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.934453 |
http://moyhu.blogspot.com/2011/11/numbers-puzzle.html | 1,462,216,965,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860117405.91/warc/CC-MAIN-20160428161517-00091-ip-10-239-7-51.ec2.internal.warc.gz | 199,466,190 | 25,767 | ## Sunday, November 13, 2011
### ## A numbers puzzle.
OK, this has nothing to do with climate. But it's connected with the trends gadget. I made a nudger that looked like this <<<<>>>>. If you click on the center symbols, you nudge by one month. Next out by 2, then four, and in this version, the outermost move by 8.
Well, suppose the sequence of symbols continued, in powers of 2, as far as needed. How far can you be sure of being able to move with, say, 2, 3, 4 or 5 clicks. More precisely, what is the least number that requires n clicks.
For n=2, it's 3. For n=3 it's 11 (8+2+1 or 16-4-1). But 4? 5? Can anyone think of a rule?
I can think of an upper bound for any n. It's less that 4^(n-1). But not a general formula.
1. 00 = 0 (N - 0)
01 = 1
10 = 2 (N = 1)
11 = 3 (N = 2)
110 = 4
101 = 5
110 = 6
111 = 7
1000 = 8
1001 = 9
1010 = 10
1011 = 11 (8,2,1) (N = 3)
1100 = 12
1101 = 13
1110 = 14
1111 = 15
10000 = 16
10001 = 17
10010 = 18
10011 = 19
10100 = 20
10101 = 21
10110 = 22
10111 = 23 (16,8,-1)
11000 = 24
11001 = 25
11010 = 26
11011 = 27 (32,-4,-1)
11100 = 28
11101 = 29 (32,-4,1)
11110 = 30 (32,-1,-1)
11111 = 31 (32,-1)
100000 = 32
100001 = 33
100010 = 34
100011 = 35
100100 = 36
100101 = 37
100110 = 38
100111 = 39 (32,8,-1)
101000 = 40
101001 = 41
101010 = 42
101011 = 43 (32,8,2,1) (N = 4)
10101011 = 167 (128,32,8,2,1) (N = 5)
1010101011 = 683 (512,128,32,8,2,1) (N = 6)
10101010101010101010101010101010101011 = ? (N = 20)
All are odd numbers, all are prime numbers, all represented in binary start with an on bit, end in two on bits, and in between, alternate between off and on bits.
Just my quess, don't really know if it's correct, but I think writing down in binary makes finding a solution easier.
2. Oops, not all prime numbers. Rats.
"10101011 = 167 (128,32,8,2,1) (N = 5)"
Should be;
"10101011 = 171 (128,32,8,2,1) (N = 5)"
171 = 3 * 57
Oh well, at least they're all odd numbers.
3. Well I popped this into Excel along with the 4^(N-1), and if the progression I'm suggesting is correct. than your equation needs a 2/3 in front of it;
upper bound = 2*4^(N-1)/3 (the ratio is 1 at N = 1, but rather quickly drops down to the 2/3 (~say by N = 5) as an asymptotic value for large N)
Excel poops out at N = 26 (in significant digits anyways), but I ran it up a ways beyond that, what the heck.
But an exponential fit appears to be correct (R^2 = 1), and is a straight line with the y-axis plotted on a log scale.
4. EFS
Yes, I think binary is the way, and you've got the pattern. The way I reasoned in the end was this. Suppose you want an algorithm for efficiently advancing N months. That amounts to reducing N to zero by adding or subtracting powers of 2.
In binary, N can be written N=1x.y, where x is 0 or 1. The "decimal" point is just a marker There are 3 cases for the first step:
x=0: then subtract 10.00.. and two digits have gone; .y remains
x=1 and y>0: subtract 100.00.. That leaves -y, and again 2 digits gone
x=1 and y=0; then N can be reduced in 2 steps.
Then do the same with y. So at each step you can remove two digits or go to zero in two steps. The only number for which the last case hurts is 3, where it takes 2 steps to remove the last two digits.
If at any stage, y begins with 0, you gain a digit. So then you can construct the smallest number that actually requires the theoretical maximum steps (y always starts with 1). And, as you say, it is
10101....011.
5. Oops "This leaves -y" I meant "This leaves 1.-y". But it's less than 1., so the 2 digits have gone.
6. This is off topic.
Nick, if you're ever feeling bored, could you re-run your "Just 60 stations" analysis but include a land mask? There's some talk over at Lucia's about the results of analyses with such few stations.
7. CCE, well the topic is off-topic anyway. I've never actually used a land mask, because I have mainly been interested in the 60 stations idea for global temperatures. So stations which represent a lot of ocean are good.
The triangular mesh methods work a bit like that for regions, because only area within the mesh is used for weighting. I did some crude land-masking in Antarctica where I blocked out the Weddell Sea.
So it's not high priority - it's quite a lot of work, and I don't think land-only temperatures are all that important.
8. If you restrict yourself to only positive or negative clicks, them the number of clicks needed is the number of ones in the binary representation of the absolute value of the number of months that you want to move. So +/- (2^n)-1 requires n clicks.
That gives you a lower bound, and also gives you an efficient way of finding the clicks you need.
...1010101011
= 1 + 2 + 8 + 32 + 128 + ...
= 1 + 2[1 + 4 + 16 + 64 + ...]
= 1 + 2 [4^(n-1) - 1] / [4 - 1] (geometric series)
= (1/3) + (2/3) . 4^(n-1)
(which confirm's EFS's 2/3 value)
10. I think this question may be related to the mathematics behind error correcting codes. The question there is how to include redundancy in the data such that a certain number of bits of error can be corrected. Errors become nudges. We are looking for an optimal sampling of a high dimensional space such that every point is within a certain distance of a sample point.
Power laws are a simple answer - if you want a power law less than one, the Fibonacci sequence has a ratio equal to the golden ratio. However IIRC they are either non-optimal or optimal but non-general. It's ages since I looked at any of this stuff though.
Kevin C
11. Nick,
since you've explored the BEST data stuff a bit is there any way to extract for example subsets of station data based upon lat/long. I know you've plotted the lat/long data but is there any way to extract the data for a selected range of points easily? I'm interested for the purpose of incorporation into a GIS platform.
12. Robert,
You need to organise the data in some structure where it is collected by station. Then in R it is easy to select. TempLS facilitates this - you just write down the latitude and longitude conditions you want, and it analyses that. TempLS doesn't currently write out the selected data, but it could.
Steve McI organises the data into a time series for each station. That isn't ideal for all purposes, but would work well here.
13. Yeah I think that what I would want to do is something like Steve McI does... I can manually do this in excel etc... but the issue is that BEST has access to more stations than I would through GHCN or CHCN. I will have a look at some of his functions over at climate audit. Thanks.
(My intention is to spatially interpolate raw temperatures using kriging in a GIS system within a region because I think it may be an interesting way of looking at the data)
14. If you are bored, could you amuse me please? | 1,995 | 6,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2016-18 | longest | en | 0.913582 |
https://community.facer.io/t/help-with-the-day-number/40800 | 1,695,915,467,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00448.warc.gz | 209,763,039 | 4,108 | # Help! With the day number
Hi, I’m trying to make the formula to show “1st”, “2nd”, “3rd” then 4 to 19 are “th”, as well as 24th to 30th… I found the solution for 1st, 2nd and 3rd… with this formula: \$#Dd#==1?100:0\$|\$#Dd#==21?100:0\$|\$#Dd#==31?100:0\$… but if I try individual numbers for the “th” it would make too much data for the watch to load and make it slow, is there a way that I can say like \$#Dd#=5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,24,25,26,27,28,29,30?100:0\$
But simplified?
Thank you!f
Try this `\$#Dd#>=4&&#Dd#<=19?0:100\$`
Enjoy,
Sirhc
1 Like
Thank you! | 247 | 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} | 3.125 | 3 | CC-MAIN-2023-40 | latest | en | 0.922104 |
https://it.mathworks.com/matlabcentral/answers/471009-how-can-i-plot-particular-rows-of-a-matrix | 1,597,526,429,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439741154.98/warc/CC-MAIN-20200815184756-20200815214756-00117.warc.gz | 355,900,221 | 23,752 | # how can I plot particular rows of a matrix?
28 views (last 30 days)
Tran Trang on 10 Jul 2019
Commented: Tran Trang on 11 Jul 2019
I have a 7208x34 matrix which columns [3 32] represent price data over the 1961-1990 period. And a row array 1x30 (the row array indicates the years from 1961 to 1990). I want to plot time series of 68 rows (i.e. grouping by cities) for each goods over the 1961-1990 period, against the 1x30 row array (years). Something looks like this:
Goods city_name price_year1961 price_year1962…
A city1 0.5
A city2 0.7
.
.
.
A city68 0.9
B city1 0.4
B city2 0.7
.
.
.
B city68 0.9
and a row vector represents the years [1961:1990].
Thank you very much!
Be Matlabi on 10 Jul 2019
Edited: Be Matlabi on 11 Jul 2019
You could first start by reading the columns into an array
prices=data(:,3:30);
n=5; %in order to plot the data for city 5
city_data=prices(n:68:end,:)
year=1961:1990;
If you want to plot for individual Good
good=3 % for the data of Good 3
plot(year,city_data(good,:))
For all the goods
for i=1:size(city_data,1)
plot(year,city_data(i,:))
hold on
end
KSSV on 10 Jul 2019
If A is your data...and you want to plot ith row; use
plot(A(i,:))
Tran Trang on 10 Jul 2019
Edited: Tran Trang on 10 Jul 2019
Hi B Matlabi,
Thank you for your reply. should it be city_data = prices(n:68:end,:), instead of city_data = prices(n:n:end,:), as the city repeats after 68 rows?
And how can I plot all the goods but with each plot for each good?
Thank you.
Be Matlabi on 11 Jul 2019
Yes you are correct, it should be
city_data = prices(n:68:end,:)
In order to plot all the goods but with each plot for each good at once you could try the subplot function
Tran Trang on 11 Jul 2019
Thank you very much! | 549 | 1,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2020-34 | latest | en | 0.802281 |
https://www.sineofthetimes.org/what-is-all-the-fuss-about-lines/ | 1,726,625,346,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00463.warc.gz | 914,051,443 | 7,613 | What Is All the Fuss About Lines?
Yesterday, I led a webinar that demonstrated how Sketchpad and Web Sketchpad can be a powerful tools for exploring Common Core algebra topics. My examples included solving for unknowns with a pan balance, exploring the slopes of lines, maximizing the area of a fixed-perimeter rectangle, and graphing trigonometric functions. I touched only briefly on each example during the webinar, and so here I’d like to return to the algebra of lines.
I’ve taught algebra and pre-calculus courses to college students, and without fail, they stumble when asked to write the equations of lines. In fact, I’ve seen students deal more successfully with the properties of trigonometric functions—their amplitude, period, and phase shift—than simple lines. What’s going on here?
I think that the problem stems in part from the terminology and algebraic machinery that accompanies students’ introduction to lines. First, students learn the formula for the slope of a line. Then, they are introduced to the point-slope form of a line and the slope-intercept form. Knowing when to use each form and doing the necessary algebra soon makes lines feel incredibly difficult. Does it really need to be so complex?
I think the answer is no, and one way to convince you is to share the following problem: Given a point A at (5, 3), how many lines can you name in under a minute that pass through it? Now wait, you might say, setting up those formulas takes some work! A minute is not much time.
But put aside what you know about formulas and procedures and think logically about what the problem is asking. In the interactive model below, you can change the three numerical boxed values a, b, and c. Give it a try and let us know your technique. I’ve seen calculus teachers think hard about this question before having an aha! insight, so don’t worry if you need to ponder for a while. | 404 | 1,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.93528 |
https://community.agilent.com/thread/1216-trouble-in-iffand-statement | 1,606,910,054,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141708017.73/warc/CC-MAIN-20201202113815-20201202143815-00353.warc.gz | 220,233,932 | 26,792 | # TROUBLE IN IFF/AND STATEMENT
Question asked by rajesdeo on Oct 2, 2016
Latest reply on Sep 19, 2017 by berndh
Can Someone help me in converting below logical statements to IR.
1st statement
=IF(AND(A<=101.5,A>=98.5),(K*S),IF(A<98.5,(98.5-A)+K*S,IF(A>101.5,(A-101.5)+K*S)))
2nd Statement
IF X=30 Then A=2 ,IF X=10 Then A=2.4 Otherwise Display "Not Applicable"
Thanks & Regards
Rajesh Deo | 147 | 396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.772867 |
http://openstudy.com/updates/513b965be4b029b0182afecf | 1,448,606,209,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398448227.60/warc/CC-MAIN-20151124205408-00024-ip-10-71-132-137.ec2.internal.warc.gz | 174,177,463 | 9,967 | ## dangbeau 2 years ago Find (d^2y)/(dx^2) in terms of x and y: xsiny=3 How to solve this?
1. abb0t
The question is asking that you find the second derivative with respect to x. Do you know how to take a derivative?
2. abb0t
For the first derivative, you will be using the product rule. Since you have two functions being multiplied $f \frac{ d }{ dx }g+g \frac{ d }{ dx } f$
3. abb0t
Also note, that the derivative of a constant is just zero. So the 3 becomes zero. For the second derivative, you will be doing product rule again.
4. dangbeau
thank you, i'll try ^^ | 167 | 574 | {"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} | 3.359375 | 3 | CC-MAIN-2015-48 | longest | en | 0.935277 |
https://www.fotofc.ru/volume-problem-solving-573.html | 1,620,316,929,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988758.74/warc/CC-MAIN-20210506144716-20210506174716-00116.warc.gz | 769,314,706 | 9,154 | # Volume Problem Solving
For example, a rectangular solid might have a volume of eight cubic inches.
For example, a rectangular solid might have a volume of eight cubic inches.
Speaking of problems, the Art of Problem Solving, Volume 2, contains over 500 examples and exercises culled from such contests as the Mandelbrot Competition, the AMC tests, and ARML. ) are available for all the problems in the solution manual.
A rectangular prism is a 3-dimensional object with six rectangular faces.
Although the Art of Problem Solving is widely used by students preparing for mathematics competitions, the book is not just a collection of tricks.
The emphasis on learning and understanding methods rather than memorizing formulas enables students to solve large classes of problems beyond those presented in the book.
In this case, you are asked for a mass, not the density.
You will need to rearrange the density equation so that you get mass.The Art of Problem Solving, Volume 2, is the classic problem solving textbook used by many successful high school math teams and enrichment programs and have been an important building block for students who, like the authors, performed well enough on the American Mathematics Contest series to qualify for the Math Olympiad Summer Program which trains students for the United States International Math Olympiad team.Volume 2 is appropriate for students who have mastered the problem solving fundamentals presented in Volume 1 and are ready for a greater challenge.Remember that the radius extends halfway across the center of the cone or sphere at the widest point.When you've calculated the volume, state it in cubic terms.By multiplying both sides by volume, mass will be left alone.Substituting in the values from the problem, The result is that the mass is 75,600 grams. Problem 6: Rocks are sometimes used along coasts to prevent erosion.This section covers the methods to find volumes of common Euclidean objects.The volume of a prism of height and base of area is . (Note that this is just a special case of the formula for a prism.) The volume of a cone of height and radius is .All its angles are right angles and opposite faces are equal.In a rectangular prism, the length, width and height may be of different lengths. A cube is a special case of a cuboid in which all six faces are squares.
## Comments Volume Problem Solving
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By Sandor Lehoczky & Richard Rusczyk. A set consisting of an instructional text including problems to try, and a solutions book ISBN 978-0-9773045-7-8 for.… | 880 | 4,205 | {"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} | 4.71875 | 5 | CC-MAIN-2021-21 | latest | en | 0.932493 |
https://stats.stackexchange.com/questions/257036/ensemble-neural-network | 1,632,152,222,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057039.7/warc/CC-MAIN-20210920131052-20210920161052-00630.warc.gz | 571,078,292 | 38,922 | # Ensemble neural network
I had run artificial neural network on Matlab. Although i used the same design structure of ANN and the same data set, the result always different. Some suggested using ensemble neural network. From my reading ensemble is combine ANN with different design structure. Do this applicable in my problem. Other than that, why is ANN produce different result every time i run it?
Why ANN produces different results: this is probably the training procedure involves "randomness", for example,
1. Your training may use random parameter initialisation.
2. Your training may use random dropout as regularisation.
3. Your training may use SGD and shuffle the data order every epoch.
4. ...
And ANN is a not a convex function, so any of these "randomness" may lead to different local optimum.
And ensemble method is a very general method to reduce the variance of predictor then improve the performance. It is not necessarily specific for ANN. And there are many ways to train an ensemble of ANN (like using different datasets, using different initial parameters, using different structure/dropout). This is really an "art" which means you need to try a lot and pick the best way for your specific dataset.
• If i only have one data set, Can i only take the average of the output? Jan 19 '17 at 0:33
• @bbadyalina you can try different parameter initialisation. You can also obtain multiple datasets by bootstrapping the one dataset you have. Just to remember that, if you are using averaging ensemble method, try to "decorrelate" the different models as much as possible (i.e. make them as noncorrelated/independent as possible). Jan 19 '17 at 0:50
Your second question is more simple so I'll answer that first.
When you train an ANN you start with random weights in the network. The network is trained to try to minimize the cost function of training. We usually cannot find the exact minimum of the cost function during training so each time you train an ANN it starts with different weights and ends training with different weights.
Training a single ANN is quite costly so it could be difficult to train 10 ANNs and make an ensemble of them all. Instead we use a clever trick called dropout when training one ANN, dropout is a type of regularization which acts like training an ensemble of networks inside one network.
Here is a description of dropout, scroll down to read about dropout.
• So if train the network ten times using the same structure, and take the average of the output. Is that reliable result? Jan 19 '17 at 0:35
• @bbadyalina Yes if you have the computing power to train a network 10 times then taking the average output is more accurate. You should use the standard deviation of the output to create a confidence interval and use that to understand how reliable the average is.
– Hugh
Jan 19 '17 at 14:09
• ok hugh.. is there any reference that i can use for this method.. or it is the same as ensemble ANN.. from what i read ensemble ANN is combine different sturucture of ANN. but i did not find any of regrading single structure. Jan 19 '17 at 14:12
• @bbadyalina The ensemble will work if you use the same structure for all ANN's but it helps if you use different structures. I don't know any references for choosing the different structures
– Hugh
Jan 19 '17 at 15:01
• If for same structure, do have any reference? Jan 19 '17 at 15:02 | 748 | 3,390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-39 | latest | en | 0.917631 |
https://www.mathhomeworkanswers.org/279256/if-g-x-9x-2-find-x-when-g-x-65 | 1,632,162,548,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057083.64/warc/CC-MAIN-20210920161518-20210920191518-00175.warc.gz | 905,870,088 | 15,895 | Algebra 1
Given, g(x) = -9x-2 and g(x) = -65
So, -9x-2 = -65
or 9x+2 =65
or 9x = 65-2 =63
or x = 63/9
or x = 7
by Level 7 User (29.9k points) | 83 | 148 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.744348 |
https://answerprime.com/how-many-different-4-person-committees-can-be-chosen-from-the-100-members-of-the-senate-2/ | 1,670,060,779,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710926.23/warc/CC-MAIN-20221203075717-20221203105717-00305.warc.gz | 133,710,114 | 18,356 | # How many different 4-person committees can be chosen from the 100 members of the Senate?
Random math query for additional credit score.
3921225
100!/(4!(96!))
Hello Sftbllgrl
Reply = 100C4 = 100!/96!4! = 100*99*98*97/4*3*2*1 = 3921225
Shy
100, divided by 4 equals 25 committees with all totally different members.
Also Read : a geometric figure having three lines segments for sides.? | 119 | 395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.861644 |
https://es.oxforddictionaries.com/traducir/ingles-espanol/square_number | 1,544,481,647,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823445.39/warc/CC-MAIN-20181210212544-20181210234044-00566.warc.gz | 594,447,410 | 22,578 | # Traducción de square number en Español:
## square number
### nombre
• 1
• Since the next succeeding square number is 729, which has 27 for its side, divide 720 by 27.
• This year I've decided to switch my mathematical allegiance to square numbers.
• If we can express a square number also as the sum of two other square numbers then Pythagoras' Theorem tells us that we have three sides of a right-angled triangle and this is Fibonacci's first Proposition.
• In every right-triangle, if the double product of the legs be either added or subtracted from the square of the hypotenuse, both the sum and the remainder will be square numbers.
• Fibonacci first notes that square numbers can be constructed as sums of odd numbers, essentially describing an inductive construction using the formula n 2 + = 2. | 181 | 808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2018-51 | latest | en | 0.855441 |
www.jessewade.info | 1,713,111,916,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.19/warc/CC-MAIN-20240414161724-20240414191724-00709.warc.gz | 792,506,935 | 9,768 | ## Iterating on BFS
Hello, internet. I am back after a short break wherein my mind was fully and thoroughly exhausted by the concepts of my Algorithms course. But, there is a light at the end of that tunnel and we’re on the downhill side of things.
So, with that in mind, I wanted to post a bit about this weeks material. If you’ve perused the site, you’ve seen that I have a rudimentary implementation of Breadth-first search in my portfolio. This weeks concepts kick that version of BFS up a notch by including a mechanism for tracking vertices we’ve already visited during the search where ‘White’ is undiscovered, ‘Grey’ is discovered and ‘Black’ is discovered with all neighbors explored as well as tracking the nodes depth. See the graphic below.
You’re probably thinking, “So what?” well I was too. But there are some pretty interesting things we can do with just this constant amount of additional space in each node. For starters, we can build a relationship tree, even on cyclical graphs. We can also compute distance from one node to another. And once we can compute distance, we can find shortest paths as well. All of which have applications to real life problems. Neat!
## Al Gore Rhythms
Its been a while! Sorry but I just don’t have the extra drive or head-space to really do any regurgitation or insightful recaps on what I am learning this semester. I barely found the time to post this delightful meme!
In any case, Algorithms is living up to its reputation of being a dream killer of a class. I am holding on to my A, but its taking a lot of work and study. I won’t even get started on the proofs. Just, no. I’ll post something a bit more detailed when I come up for air. See you in the spring!
## Semester closing
Well, the semester has wound down and I survived Data Structures! I have to say it was one of the most intense and rewarding class experiences I’ve ever had. Instead of being fearful of spring, I now have a ‘Let’s do this!’ attitude toward Algorithms this coming January.
Have a great holiday and see you in the new year!
## 2-4 Trees
Binary search trees have been the topic of conversation for our recent modules and of all of the trees we’ve studied (BST, Red-Black, AVL, etc) I have to say that 2-4 trees are probably the easiest to understand. That opinion may be skewed by the fact that it was also the last one we touched on, so understanding the others may have made it seem simpler. Anyway…
There are some pretty distinct differences in these multi-way search trees. Most notably each node stores multiple values (3 in this specific implementation) and the trees grow upward toward the root, rather than down toward new leaf levels. This design choice is intended to keep the total height of the tree as small as possible.
Rather than memorizing patterns of how to ‘swap’ nodes or re-color them as with RB and AVL trees, 2-4 trees use simple value comparison to decide where they go, so they bear a much closer resemblance to simple binary search trees. When a node is full, and a value passed into it, you just take the middle value and pass it up toward the root, splitting the lesser and greater value into new nodes and adding the value that was originally passed up to whichever side it evaluates to (less than or greater than the middle value). Here’s a graphic to visualize what I’m describing.
Alright, new topic. We’re covering a completely new data structure called Linked Lists. I have, surprisingly, never been exposed to these even in my self taught years as a programmer. I have to say, I really appreciate the simplicity of this structure. Take any class, add a field or two and BAM, you’re done. Image incoming.
So what are the advantages? The biggest is dynamic memory allocation. So, if that’s something you’re after this is a structure to consider. Also, its quite simple to modify the order or insert new elements, all you do is update the references to next in the new element and the element left of where its being inserted (in the singly linked example above).
Tradeoffs? No random access. This is a bigger deal than it seems at first. If you can’t access the Nth element of a structure, you can’t divide it for binary search, or perform a lot of other algorithms on it. Also, specifically speaking about Java here, if you incorrectly order your operations for insertion, you can flag your entire structure for garbage collection (yikes). So its really important to set the next field on your new object (the ‘head’ or ‘front’ variable reference keeps the remainder on the heap), THEN link the left object to the new object, completing the link.
The version pictured above is a singly linked list, though its pretty trivial to add another field called ‘prev’ and link it in the other direction for bi-directional access to elements in the structure. There are a ton of other flavors of the linked structure, too.
## Sorting out Sorts
I wanted to drop some information here to “rubber duck” some of the things we are learning in Data Structures this semester. I’m going to post a bit of code, then talk about it. Then post some more, ad nauseam.
Note: the code I’m posting is from lecture, I am not claiming it to be my own.
``````public void mergeSort(Comparable[] a, int left, int right) {
if (right == left) return;
int mid = left + (right – left) / 2;
mergeSort(a, left, mid);
mergeSort(a, mid + 1, right);
merge(a, left, mid, right);
}``````
Lets start with the above code. This is merge sort. Its one of the cooler sorts we’ve come across thus far (Having O(NlogN) time in all cases is pretty spiffy). What’s not cool is trying to wrap your head around it!
So, lets get to the first point of frustration. This is a recursive function. Moreover, its a recursive function that branches, making it even more challenging to try and follow. The best way I’ve found to visualize it is similar to binary search, except we aren’t throwing half of the set away; instead we divide the problem in half recursively until it becomes trivial to sort (i.e. there’s only 1 item left in the set!) Next up, we do the merging.
``````public void merge(Comparable[] a, int left, int mid, int right) {
for (int k = left; k <= right; k++) aux[k] = a[k];
int i = left; j = mid + 1;
for (int k = left; k <= right; k++) {
if (i > mid) a[k] = aux[j++];
else if (j > right) a[k] = aux[i++];
else if (less(aux[j], aux[i]) a[k] = aux[j++];
else a[k] = aux[i++];
}
}``````
Alright, so remember that at our current recursion step, we only have 1 element to work with, so we combine it with the next single element in order. Its worth noting that we are copying from an auxiliary set of the data we are sorting (this is merge sorts primary downfall, double memory usage). As we traverse back to the top of our recursion, the sorted arrays get larger and larger until, finally, we end up with a sorted array of the original data.
## init();
Greetings. My name is Jesse Wade and this site showcases examples of my work. I am a student of Computer Science at Auburn University and have been a student of life for (much) longer! A few other lives I have lived were as a Firefighter and Paramedic in the U.S. Army, an IT customer support technician, and as both System and Network Administrators for various government organizations. My pool of experience is pretty wide, and deep enough I’d say!
Over my career in technology as paradigms shift (as they tend to do) and software defined everything, I began to have a keen interest in writing code. It began with scripting and automation tasks in Windows Server with Powershell and would blossom into Python and BASH when I began working with Linux. These days I spend a lot of time looking at Java, xml, json, Python and a bit of Perl as I maintain our systems and software.
Being self taught was fine, but I soon realized there were a lot of pieces missing for me, so I sought out a CS program and landed on Auburn. I am loving the learning I am doing with them and can’t wait to show off some of the skills and theories I have come to understand. My goal is to transition into fulltime software development once I complete the program. | 1,839 | 8,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-18 | latest | en | 0.954812 |
https://nerdcounter.com/distance-between-points-calculator/ | 1,695,579,489,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506658.2/warc/CC-MAIN-20230924155422-20230924185422-00351.warc.gz | 459,097,719 | 18,697 | # Distance Between Points Calculator
Here’s how you can use the calculator:
You can calculate the distance between two places via the point. Two locations in steps work on the space displays the whole calculation step by step. AA at the coordinates (5,3)(5,3) and BB at the coordinates calculate the length of a line segment with two endpoints (9,6). (9,6).
Enter the coordinates for two endpoints and click on the “GERATE WORK” button for any further endpoint combinations. Grade school children can use this distance computer to produce work quickly, check results, and complete problems with homework.
A line segment connects two locations and measures the distance between them. The distance between the two places is usually upbeat. It is important to remember. The conglomerates are segments of the same size.
## What can a point calculator do for you?
The point calculator can be used to find the distance between two points by using the distance between points calculator as using the following formula to calculate the distance from 2 places and see the distance between two points. The endpoints are generally used to indicate the length of a section without an overline. The distance between two points is computed by putting the 00 marks on the left and the other end. A distance calculator can do this.
The distance between AandBAandB is AB=|xBxA|AB|||xBxA] when both points are on x-axis, i.e., their coordinates are (xA,0)(xA,0) and (xb,0)(B,0). It is possible to calculate the distance between two sites on the y-xis in the same way. In a double-dimensional cartesian coordinate plane, the Pythagorean theorem calculates the distance between two points. Google also used this calculator to find the distance between two points on google maps. The theorem used to calculate the distance between two points is, therefore, the Pythagorean theorem.
• B(xB,yB)B(xB,yB)B(xB,YB)B(xB,yB)B(a,yB)B(xB,yB)B(a),yB(a),yb)B(a,yB)B(a,y) (xB,yB)
• AB2=(xB−xA)2+(yB−yA)2,AB2=(xB-xA)2+(yB-yA)2,
AB=√(xB−xA)2+(yB−yA)2AB=(xB-xA)2+(yB-yA)2
The distance can be calculated using a scale on a map such as google maps. It will identify the distance between two points on google maps. This allows the distance between two locations on a map to be displayed.
Try this Out S&P 500 Return Calculator
Here’s how you can use it better:
• The two-point division You can find the 3D coordinate plane using the distance formula below.
d is the same as (x2 – x1) Two + two (y2 – y1)Two + (z2 – z1)Two + (z2 – z1)2
• The 3D coordinates of both points (x1, y1, z1) and (x2, y2, z2). Two points in the 3D formula version are called (x1, y1, z1) or (x2, y2, z2), as long as the relevant points in the formula are used.
• The distance can be calculated as follows between the two (1, 3, 7) and (2, 4, 8) points: d is the same as (2 – 1) Two + two (4 – 3) Two + two (8 – 7)2.
• The distance between two points map can also be determined as several ways to calculate the distance between two spots on Earth’s surface. The following are two popular formulas. In physics, we can apply different procedures stated as the distance formula calculator physics, and the formula can be given as
• Haversian Formula: The Haversine formula can be used to calculate the gap between two spherical locations because of its latitude and longitude:
• d is the distance between two sites on a big circle with the radius of the sphere, 1 and 2 are the margins of the two, 1 and 2 being the lengths of the two points, all of which are in the radians. A great-circle distance is the shortest distance between two sites on the sphere’s surface.
• This also helps to find the distance formula calculator triangle. The result of this mistake could have been up to 0.5 per cent, since Earth have a radius of 6,378 km (3,963 miles) in the equator and 6,357 km (3,950 miles) in the pole.
• A formula of Lambert, or ellipsoid-surface, is more exact than the formula of haversine on the Earth’s surface (a spherical-surface formula) and can be used to find the distance between two cities as well.
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## How does this point calculator work?
You can do this in one, two, three or four dimensions when calculating distances between two points.The ability to define three different space points, from which you can get three couples of distances, saving you time if you have more than 2 points.
To add a step-by-step solution, although using the calculator is relatively simple. You may learn how to use the distance formula this way (as if it was not yet invented in the 1950s and the internet). Look at an example from the real world, how to find a two-dimensional distance between two points.
We have two (3, 5), and (9, 15) coordinates, and you want the distance they are from each other to be known. To calculate the 2-D distance between these two points, follow these procedures:
• Enter the information in the formula [(x2 – x1)2+ (y2 – y1)2.
• Unwrap the values of the parentheses.
• Place both numbers in the parenthesis.
• The results are combined.
• Calculate the number’s square root.
• Check the distance calculator for your results.
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## If you manually do the math, you receive:
[(9 – 3)2 + (15 – 5)2] [(15 – 5)2] (9 to 3)2) [(2) [Articles 9 – 3)2]. [Article 9 to 3] = [(6)2+(10)2] = [(6)2+(3) = [(6)2+(10)2] = [(6)2 = (6)2 + (10)2) = ((6)2) = [= 136 + 100]; = [= 136, [36 + 100].
It is 11.66 approximately. You receive positive and negative results when you take your square root; however, the positive impact is all that matters as you are dealing with the distance. The calculator passes through each calculation step, and the results are provided in both exact and rough sizes. There are also websites like symbol, where we can find a variety of options like distance formula calculator – symbolab where you can access the distance formula calculator circle and distance formula between two points. | 1,524 | 5,954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2023-40 | latest | en | 0.899198 |
http://oeis.org/A070016 | 1,369,553,486,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706637439/warc/CC-MAIN-20130516121717-00049-ip-10-60-113-184.ec2.internal.warc.gz | 195,576,120 | 3,746 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A070016 Least m such that Chowla-function value of m [A048050(m)] equals n or 0 if no such number exists. 2
0, 4, 9, 0, 6, 8, 10, 15, 14, 21, 121, 27, 22, 16, 12, 39, 289, 65, 34, 18, 20, 57, 529, 95, 46, 69, 28, 115, 841, 32, 58, 45, 62, 93, 24, 155, 1369, 217, 44, 63, 30, 50, 82, 123, 52, 129, 2209, 75, 40, 141, 0, 235, 42, 36, 106, 99, 68, 265, 3481, 371, 118, 64, 56, 117 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Remark that A070016(n)=A070015(n+1) in accordance with A048995(k)+1=A005114(k). LINKS FORMULA a(n)=Min{x; A048050(x)=n} or a(n)=0 if n is from A048995. EXAMPLE n=127: a(n)=16129, divisors={1,127,16129}, 127=sigma[n]-n-1=127 and 16129 is the smallest. MATHEMATICA f1[x_] := DivisorSigma[1, x]-x-1; t=Table[0, {128}]; Do[b=f1[n]; If[b<129&&t[[b]]==0, t[[b]]=n], {n, 1, 1000000}]; t CROSSREFS Cf. A000203, A001065, A048050, A051444, A007369, A070016, A005114, A048995. Sequence in context: A021208 A021675 A093872 * A020802 A085675 A070439 Adjacent sequences: A070013 A070014 A070015 * A070017 A070018 A070019 KEYWORD nonn AUTHOR Labos E. (labos(AT)ana.sote.hu), Apr 12 2002 STATUS approved
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Content is available under The OEIS End-User License Agreement . | 634 | 1,606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2013-20 | latest | en | 0.526034 |
https://math.stackexchange.com/questions/766490/similar-matrices-conditions | 1,563,499,593,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525973.56/warc/CC-MAIN-20190719012046-20190719034046-00263.warc.gz | 451,915,343 | 35,058 | # Similar Matrices Conditions:
Sorry this question was already asked but my english is not good. For matrix to be similar, does it have to have all of these properties or SOME of them?
Same determinant
Same Trace
Same characteristic polynomial
Same Eigenvalues
P^-1BP = D etc
If one of these properties are not shared by two matrix, does that mean that they are not similar?
Feel free to mentions if I miss some properties
• Well two matrices $B,D$ are defined to be similar if there exists some invertible matrix $P$ such that $P^{-1}BP=D$. It is a theorem that if two matrices are similar they will share the other properties you list. – Alex Becker Apr 23 '14 at 20:44
By definition two square matrices $A$ and $B$ are called similar if there's an invertible matrix $P$ such that $$A=PBP^{-1}$$ and all the other conditions are necessary and not sufficient. Here's a counterexample:
$$A=\begin{pmatrix}1&1\\0&1\end{pmatrix}\quad I_2=\begin{pmatrix}1&0\\0&1\end{pmatrix}$$ these two matrices have the same determinant, same eigenvalue, same trace, same charecteristic polynomial but they not equivalent.
• wouldn't it be $A = P^−1BP$ ? – Mac Apr 23 '14 at 21:02
• It's the same: take $Q=P^{-1}$ in your equality. – user63181 Apr 23 '14 at 21:09 | 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": 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} | 3.71875 | 4 | CC-MAIN-2019-30 | latest | en | 0.880946 |
https://www.geeksforgeeks.org/c-program-to-traverse-an-array/?ref=lbp | 1,632,836,166,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060803.2/warc/CC-MAIN-20210928122846-20210928152846-00592.warc.gz | 804,560,336 | 22,817 | Related Articles
# C program to traverse an Array
• Last Updated : 30 Mar, 2021
Given an integer array of size N, the task is to traverse and print the elements in the array.
Examples:
Input: arr[] = {2, -1, 5, 6, 0, -3}
Output: 2 -1 5 6 0 -3
Input: arr[] = {4, 0, -2, -9, -7, 1}
Output: 4 0 -2 -9 -7 1
Approach:-
1. Start a loop from 0 to N-1, where N is the size of array.
`for(i = 0; i < N; i++)`
2. Access every element of array with help of
`arr[index]`
3. Print the elements.
`printf("%d ", arr[i])`
Below is the implementation of the above approach:
## C
`// C program to traverse the array` `#include ` `// Function to traverse and print the array``void` `printArray(``int``* arr, ``int` `n)``{`` ``int` `i;` ` ``printf``(``"Array: "``);`` ``for` `(i = 0; i < n; i++) {`` ``printf``(``"%d "``, arr[i]);`` ``}`` ``printf``(``"\n"``);``}` `// Driver program``int` `main()``{`` ``int` `arr[] = { 2, -1, 5, 6, 0, -3 };`` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``printArray(arr, n);` ` ``return` `0;``}`
Output:
`Array: 2 -1 5 6 0 -3`
Time Complexity: O(n)
Auxiliary Space: O(1)
Want to learn from the best curated videos and practice problems, check out the C Foundation Course for Basic to Advanced C.
My Personal Notes arrow_drop_up | 482 | 1,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-39 | latest | en | 0.575796 |
https://cheapcustomwriters.com/blog/finance-330-avis-purchases-an-asset-for-17972/ | 1,544,547,699,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823657.20/warc/CC-MAIN-20181211151237-20181211172737-00234.warc.gz | 596,752,573 | 10,599 | FINANCE 330 – Avis purchases an asset for \$17972.
Avis purchases an asset for \$17972. This asset qualifies as a seven-year recovery asset under MACRS. The seven-year fixed depreciation percentages for years 1, 2, 3, 4, 5, and 6 are 14.29%, 24.49%, 17.49%, 12.49%, 8.93%, and 8.93%, respectively. Avis has a tax rate of 30%. The asset is sold at the end of six years for \$4645.
Calculate the book value of the asset.
FINANCE 330 – Avis purchases an asset for \$17972.
Avis purchases an asset for \$17972. This asset qualifies as a seven-year recovery asset under MACRS. The seven-year fixed depreciation percentages for years 1, 2, 3, 4, 5, and 6 are 14.29%, 24.49%, 17.49%, 12.49%, 8.93%, and 8.93%, respectively. Avis has a tax rate of 30%. The asset is sold at the end of six years for \$4645.
Calculate the book value of the asset. | 263 | 840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-51 | latest | en | 0.906037 |
https://www.datacompass.com/v2/replay/2147/2223 | 1,721,777,792,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00226.warc.gz | 628,871,845 | 4,995 | ##### 7 pts Create a function that takes 5 integers that relate to two starting points (start1, start2), two movement speeds (move1, move2), and the number of movements (num). Determine if after the given number of movements (num), the two starting points will finish at the same number after increasing their number by their respective movement speeds. If the two starting points end at the same number, return "YES", otherwise return "NO". Example: multipleMovements(start1, move1, start2, move2, num). multipleMovements(2,4,6,3,5) should return "NO" because after 5 movements, (start1) ended at 22, and (start2) ended at 21.
function multipleMovements(start1, move1, start2, move2, num){ //Enter your code below }
Test Cases:
• multipleMovements(5,1,0,2,5) to return YES
• multipleMovements(3,3,4,4,4) to return NO
• multipleMovements(9,3,0,6,3) to return YES
• multipleMovements(2,-3,-1,0,1) to return YES
• multipleMovements(8,1,4,-2,0) to return NO
00:00 / 02:33
Share the mistake you found and we’ll fix as soon as we can. | 304 | 1,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.772561 |
http://www.appszoom.com/android_applications/tools/fraction-calculator_bqfvv.html?nav=home_categories | 1,493,501,578,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123590.89/warc/CC-MAIN-20170423031203-00335-ip-10-145-167-34.ec2.internal.warc.gz | 443,999,120 | 17,389 | Fraction Calculator
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Description
Adds, Subtracts, Multiplies,Divides, raise the Power and Inverse of any Fractions.
Display the answer in mixed, simplified or decimal number form just pressing the fraction button.
Find the least common multiple and the greatest common divisor of two integers. It is almost a scientific calculator.
• Convert decimal number to fraction and fraction to decimal number. 1.4 becomes 1/2/5 or 7/5, 5/2 becomes 2.5 or 2/1/2.
• Simplify a fraction. 485/25 becomes 97/5 or 19/2/5 or convert to decimal 19.4
• Enter decimal, mixed fractions and fractios in a single calculation. Example 1/4 + 1/2/2 + 24 * 3 = 297/4
• Get the raise power of a fraction. Example 5/3^3 = 125/27
• Get the inverse of a fraction 5/4^-1 = 4/5
• Use parentheses for the order of precedence
• Work with negative numbers
• Find least common multiple. Example 5 , 3 LCM= 15
• Find greatest common divisor. Example 52 , 24 GCD= 4
• Use you're latest answers with the history feature
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9 | 373 | 1,288 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-17 | latest | en | 0.580572 |
http://encyclopedia2.thefreedictionary.com/invertibility | 1,532,052,464,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591455.76/warc/CC-MAIN-20180720002543-20180720022543-00166.warc.gz | 117,644,185 | 12,335 | # invert
(redirected from invertibility)
Also found in: Dictionary, Thesaurus, Medical, Legal.
## invert
1. Psychiatry
a. a person who adopts the role of the opposite sex
b. another word for homosexual
2. Architect
a. the lower inner surface of a drain, sewer, etc.
b. an arch that is concave upwards, esp one used in foundations
## invert
[′in‚vərt]
(civil engineering)
The floor or bottom of a conduit.
## invert
invert
In plumbing, the lowest point or the lowest inside surface of a channel, conduit, drain, pipe, or sewer pipe.
References in periodicals archive ?
This appears to be much more natural, since only the invertibility of S is needed to get a time scale symplectic system.
Based on the definition of invertibility, which has been described earlier, the result demonstrates clearly that the S-matrix based watermark embedding is invertible since the attacker was able to extract his faked watermark [W.
Characterization of W-maps in terms of Invertibility, Bulletin of Pure and Applied Sciences.
measure preserving systems of differential equations or invertible maps) the conditional entropy is fixed at the value with which the system is prepared [1] [3] and [4], but that the addition of noise can reverse this invertibility property and lead to an evolution of the conditional entropy to a maximum value of zero.
Other concepts that appeared first in map E and remained unchanged in map F are those detailing invertibility of matrices namely: det A, det=0, det [not equal to] 0, not invertible, invertible and augmented matrix.
Conditions for invertibility and results on the accuracy of this VAR approximation can be found in Fernandez-Villaverde, Rubio-Ramirez, and Sargent (2004).
Under standard invertibility conditions the systems have the following moving average representation:
However, a model with a moving average coefficient of approximately unity may be statistically uninterpretable due to its violation of an invertibility condition (Chatfield 1989).
An advantage of the invertibility exercise is that it allows access to, and relatively transparent applications of, a number of theorems in applied mathematics.
POLSYS_PLP uses numerical linear algebra with random real matrices to determine the generic invertibility of [A.
Amoroso and Patt proved that there is an effective procedure to determine invertibility of 1-d CA, based on the local rule [Amoroso and Patt 1972].
trade-offs between the value of the long-memory parameter and those of the ARMA parameters as well as possible stationarity and invertibility problems with the AR and MA polynomials, respectively).
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Open / Close | 574 | 2,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-30 | latest | en | 0.908223 |
http://hackage.haskell.org/package/statistics-0.13.3.0/docs/Statistics-Distribution-Exponential.html | 1,556,118,924,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578643556.86/warc/CC-MAIN-20190424134457-20190424160457-00282.warc.gz | 79,685,624 | 4,740 | statistics-0.13.3.0: A library of statistical types, data, and functions
Statistics.Distribution.Exponential
Contents
Description
The exponential distribution. This is the continunous probability distribution of the times between events in a poisson process, in which events occur continuously and independently at a constant average rate.
# Documentation
Instances
Source # Methods Source # Methodsgfoldl :: (forall d b. Data d => c (d -> b) -> d -> c b) -> (forall g. g -> c g) -> ExponentialDistribution -> c ExponentialDistribution #gunfold :: (forall b r. Data b => c (b -> r) -> c r) -> (forall r. r -> c r) -> Constr -> c ExponentialDistribution #dataCast1 :: Typeable (* -> *) t => (forall d. Data d => c (t d)) -> Maybe (c ExponentialDistribution) #dataCast2 :: Typeable (* -> * -> *) t => (forall d e. (Data d, Data e) => c (t d e)) -> Maybe (c ExponentialDistribution) #gmapT :: (forall b. Data b => b -> b) -> ExponentialDistribution -> ExponentialDistribution #gmapQl :: (r -> r' -> r) -> r -> (forall d. Data d => d -> r') -> ExponentialDistribution -> r #gmapQr :: (r' -> r -> r) -> r -> (forall d. Data d => d -> r') -> ExponentialDistribution -> r #gmapQ :: (forall d. Data d => d -> u) -> ExponentialDistribution -> [u] #gmapQi :: Int -> (forall d. Data d => d -> u) -> ExponentialDistribution -> u #gmapM :: Monad m => (forall d. Data d => d -> m d) -> ExponentialDistribution -> m ExponentialDistribution #gmapMp :: MonadPlus m => (forall d. Data d => d -> m d) -> ExponentialDistribution -> m ExponentialDistribution #gmapMo :: MonadPlus m => (forall d. Data d => d -> m d) -> ExponentialDistribution -> m ExponentialDistribution # Source # Source # Methods Source # Associated Typestype Rep ExponentialDistribution :: * -> * # Methods Source # Methods Source # Methods Source # Methods Source # Methods Source # Methods Source # Methods Source # Methods Source # Methods Source # Methods Source # Methods Source # Methods Source # Methods Source # type Rep ExponentialDistribution = D1 (MetaData "ExponentialDistribution" "Statistics.Distribution.Exponential" "statistics-0.13.3.0-4cjYwUsSjEQGDMfnb5oeqe" True) (C1 (MetaCons "ED" PrefixI True) (S1 (MetaSel (Just Symbol "edLambda") NoSourceUnpackedness NoSourceStrictness DecidedLazy) (Rec0 Double)))
# Constructors
Arguments
:: Double Rate parameter. -> ExponentialDistribution
Create an exponential distribution.
Create exponential distribution from sample. No tests are made to check whether it truly is exponential. | 655 | 2,505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-18 | latest | en | 0.686515 |
https://homework.zookal.com/questions-and-answers/java-program-all-i-want-is-algorithm--using-the-921918173 | 1,618,860,314,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038916163.70/warc/CC-MAIN-20210419173508-20210419203508-00249.warc.gz | 401,763,118 | 27,761 | 1. Engineering
2. Computer Science
3. java program all i want is algorithm using the...
# Question: java program all i want is algorithm using the...
###### Question details
Java program( all i want is algorithm )
Using the Design Recipe, write an algorithm for each of the following programming problems, showing all work (i.e., Contract, Purpose Statement, Examples and Algorithm):
1. The straight-line method for computing the yearly depreciation in value D for an item is given by the following formula:
D=(P-S)/Y
where P is the purchase price, S is the salvage value, and Y is the number of years the item is used. Write a program that takes as input the purchase price of an item, the expected number of years of service, and the expected salvage value. The program should then output the yearly depreciation for the item.
2.A government research lab has concluded that an artificial sweetener commonly used in diet soda pop causes death in laboratory mice. A friend of yours is desperate to lose weight but cannot give up soda pop. Your friend wants to know how much diet soda pop it is possible to drink without dying as a result. Write a program to supply the answer. The input to the program is the amount of artificial sweetener needed to kill a mouse, the weight of the mouse, and the desired weight of the dieter. Assume that diet soda contains 1/10th of 1% artificial sweetener. Use a named constant for this fraction.
3.Write a program that inputs the name, quantity, and price of three items. Output a bill with a tax rate of 6.25%.
4.Write a program that helps a person decide whether to buy a hybrid car. Your program’s inputs should be:
• The cost of a new car
• The estimated miles driven per year
• The estimated gas price
• The efficiency in miles per gallon
• The estimated resale value after 5 years
• Compute the total cost of owning the car for five years. (For simplicity, we will not take the cost of financing into account.) Obtain realistic prices for a new and used hybrid and a comparable car from the Web, and use today’s gas price and 15,000 miles driven per year.
5.Write a program that reads two times in military format (e.g., 0900, 1730) and prints the number of hours and minutes between the two times. Note that the first time can come before or after the second time.
6.An online bank wants you to create a program that shows prospective customers how their deposits will grow. Your program should read the initial balance and the annual interest rate. Interest is compounded monthly. Print out the balance for each of the first three months.
7.A video club wants to reward its best members with a discount based on the member’s number of movie rentals and the number of new members referred by the member. The discount is in percent and is equal to the sum of the rentals and the referrals, but it cannot exceed 75 percent. The program shows the user the discount rate awarded. | 627 | 2,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.90986 |
https://www.jiskha.com/display.cgi?id=1379977476 | 1,529,756,499,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864958.91/warc/CC-MAIN-20180623113131-20180623133131-00053.warc.gz | 853,883,193 | 3,916 | # Math
posted by Cortez
Three bells begin to toll together and toll respectively at intervals of 6, 8, and 12 seconds. After how much time will they toll together again???
1. Ms. Sue
What is the LCM?
## Similar Questions
A man has a basket of eggs. He wants to cross a toll bridge but has no money. The first toll collector says he will let the man pass if he gives him half of his eggs and half an egg. When he gets to the next toll collector he still …
2. ### microeconomics
An october 24, 1996 article in the washington post dicussses the completion of a new private toll road between leesburg and washington dulles international airport. It states that daily revenue from tolls increased from \$14000 to 22000 …
3. ### Calculas HS
a trucker handed in a ticket at the toll booth showing that in 2hours she had covered 159 mi on a toll road with speed limit 65mph. the trucker was cited for speeding. why?
4. ### math
The total toll charge for 1 car and 5 bicycles to cross a bridge is \$7.50. The toll for a car is \$1.50 more than for a bicycle. Find the cost for a car to cross the bridge.
5. ### Calculus
A trucker handed in a ticket at a toll booth showing that in 2 h she had covered 159 mi on a toll road with speed limit 65 mph. The trucker was cited for speeding. Why?
6. ### Physics
Thinking Physic-ly: On some toll roads, the ticket is stamped with the time you enter the toll road and the time you exit. How can the toll taker determine if you were speeding?
7. ### Statistics
New York and Pennsylvania have progressive toll roads. Upon entering the toll road the driver receives a ticket marking the location and time of entrance. Upon exiting the mileage is computed and the driver pays the toll. Which of …
8. ### maths
Four bells tolls at an interval of 8,12,15 &18 second a respectively. How many times will they toll together in one hour excluding the one at the start?
9. ### math
Three bells toll at intervals of 8 minutes, 15 minutes and 24 minutes respectively. If they toll together at 3 p.m, at what time will they next toll together again?
10. ### Math
Four bells toll at intervals of8,9,12 and 15 minutes respectively if they toll together at 3 p.m when will they toll together next
More Similar Questions | 550 | 2,249 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.944739 |
https://linearalgebras.com/tag/baby-rudin/page/2 | 1,716,267,887,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058383.61/warc/CC-MAIN-20240521025434-20240521055434-00045.warc.gz | 316,420,106 | 16,152 | If you find any mistakes, please make a comment! Thank you.
## Solution to Principles of Mathematical Analysis Chapter 6 Part B
Chapter 6 The Riemann-Stieltjes Integral Part A: Exercise 1 - Exercise 10 Part B: Exercise 11 - Exercise 19 Exercise 11 (By analambanomenos) As in the proof of Theorem 1.37(f),…
## Solution to Principles of Mathematical Analysis Chapter 6 Part A
Chapter 6 The Riemann-Stieltjes Integral Part A: Exercise 1 - Exercise 10 Part B: Exercise 11 - Exercise 19 Exercise 1 (By Matt Frito Lundy) Note: I should probably consider…
## Solution to Principles of Mathematical Analysis Chapter 5 Part C
Chapter 5 Differentiation Part A: Exercise 1 - Exercise 14 Part B: Exercise 15 - Exercise 20 Part C: Exercise 21 - Exercise 29 Exercise 21 (By analambanomenos) I’m going…
## Solution to Principles of Mathematical Analysis Chapter 5 Part B
Chapter 5 Differentiation Part A: Exercise 1 - Exercise 14 Part B: Exercise 15 - Exercise 20 Part C: Exercise 21 - Exercise 29 Exercise 15 (By analambanomenos) Let $g(x)=A/x+Bx$…
## Solution to Principles of Mathematical Analysis Chapter 5 Part A
Chapter 5 Differentiation Part A: Exercise 1 - Exercise 14 Part B: Exercise 15 - Exercise 20 Part C: Exercise 21 - Exercise 29 Exercise 1 (By Matt Frito Lundy)…
## Solution to Principles of Mathematical Analysis Chapter 4 Part C
Chapter 4 Continuity Part A: Exercise 1 - Exercise 9 Part B: Exercise 10 - Exercise 18 Part C: Exercise 19 - Exercise 26 Exercise 19 (By analambanomenos) Suppose $f$…
## Solution to Principles of Mathematical Analysis Chapter 4 Part B
Chapter 4 Continuity Part A: Exercise 1 - Exercise 9 Part B: Exercise 10 - Exercise 18 Part C: Exercise 19 - Exercise 26 Exercise 10 (By analambanomenos) By Theorem…
## Solution to Principles of Mathematical Analysis Chapter 4 Part A
Chapter 4 Continuity Part A: Exercise 1 - Exercise 9 Part B: Exercise 10 - Exercise 18 Part C: Exercise 19 - Exercise 26 Exercise 1 (By ghostofgarborg) No. As…
## Solution to Principles of Mathematical Analysis Chapter 3 Part C
Chapter 3 Numerical Sequences and Series Part A: Exercise 1 - Exercise 14 Part B: Exercise 15 - Exercise 17 Part C: Exercise 18 - Exercise 25 Exercise 18 (By…
## Solution to Principles of Mathematical Analysis Chapter 3 Part B
Chapter 3 Numerical Sequences and Series Part A: Exercise 1 - Exercise 14 Part B: Exercise 15 - Exercise 17 Part C: Exercise 18 - Exercise 25 Exercise 15 (By… | 646 | 2,434 | {"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} | 2.734375 | 3 | CC-MAIN-2024-22 | latest | en | 0.661566 |
http://delphiforfun.org/Programs/TShirt7.htm | 1,660,823,002,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573193.35/warc/CC-MAIN-20220818094131-20220818124131-00148.warc.gz | 12,186,554 | 9,139 | ### Search
Search WWW Search DelphiForFun.org
As of October, 2016, Embarcadero is offering a free release of Delphi (Delphi 10.1 Berlin Starter Edition ). There are a few restrictions, but it is a welcome step toward making more programmers aware of the joys of Delphi. They do say "Offer may be withdrawn at any time", so don't delay if you want to check it out. Please use the feedback link to let me know if the link stops working.
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Mensa® Daily Puzzlers
For over 15 years Mensa Page-A-Day calendars have provided several puzzles a year for my programming pleasure. Coding "solvers" is most fun, but many programs also allow user solving, convenient for "fill in the blanks" type. Below are Amazon links to the two most recent years.
(Hint: If you can wait, current year calendars are usually on sale in January.)
### Contact
Search DelphiForFun.org only
### Problem Description
A sevenrh entry in our "Numeric T-Shirt" line:
Back of T-Shirt: "Three prime 3-digit numbers that contain all of the digits 1 through 9 and have a 3-digit sum"
Front: __ __ __
+ __ __ __
+ __ __ __
_________
__ __ __
### Background & Techniques
This problem is from Martin Gardner in his book "The Sixth Book of Scientific American Mathematical Games". He solves it with a little logical reasoning and some trial and error.
With dumb computers, it is simpler to skip the logical reasoning part and just use trial and error.
These programs are primarily programming exercises and probably not of much interest to non-programmers (unless you want to custom make a mathematical T-shirt!). So we'll skip straight to a discussion of the method the program uses to solve the problem.
I decided to approach the problem if two phases. First we'll generate all 3 digit prime numbers that do not contain duplicate integers and do not contain zeros. Our solution set must be chosen from these. Button "Find Primes" does this by testing all numbers from 123 to 987 (the smallest and largest possible candidates). I copied function IsPrime from the Tshirt 5 program to test for primeness, and added functions NoDups to make sure that the prime contained no duplicate digits and NoZeros makes sure that they contain no zeros. . Numbers that pass all three tests are added to the Primes integer array and field Count keeps track of how many we have found.
In phase 2, triggered by the "Find Sum" button, we'll check all possible ways to select three different numbers from this set and find the set with the smallest sum. (The original problem called for the smallest sum, I changed it to "a 3-digit sum" because I happen to know that the smallest sum has 3-digits and it just reads a little simpler.) Since there are less than 100 primes which qualify, the number of ways we can select 3 from the set is less than 100x100x100 (1,000,000), a trivial number to check for today's computers.
Combinations There are actually only 83 primes generated by phase 1. Therefore the number of subsets to check is the number of ways we can select unique sets of 3 numbers from the 83. This is the number of Combinations of 83 things taken 3 at a time, commonly denoted as C(83,3). The formula for C(n,r) is n!/(n-r)!/r!. In our case this is 83x82x81/(3x2) or only 91,881 cases to test!.
Since the order of the subsets does not matter, we need to examine the number of combinations of 3 number subsets selected from the entire set of primes built in Phase 1. The simplest way to select combinations of small subsets of items from a large set is with nested loops like this:
for i:=1 to count-2 do
begin
n1:=Primes[I]; {get the first prime}
for j:= i+1 to count-1 do
begin
n2:=Primes[j]; {get the second prime}
if nodups2(n1,n2) then {if unique from n1 then...}
for k:= j+1 to count do
begin
n3:=Primes[k]; {get the third prime}
if nodups2(n1,n3) and nodups2(n2,n3) then
begin {we have found three primes containing all 9 digits}
{etc.- a few more lines of code to save and list the set with the smallest sum }
end;
end;
end;
end;
NoDups2 is a function that checks to make sure that two passed number have no digits in common. If we find three 3-digit numbers with no digits in common and no zeros, they must contain all of the digits 1-9. For simplicity, we'll just list each new smallest sum that we find - the last one listed will be our answer. | 1,177 | 4,788 | {"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-2022-33 | latest | en | 0.862631 |
https://gateoverflow.in/312367/made-easy-test-series-algorithm-dijkstra | 1,675,841,390,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500719.31/warc/CC-MAIN-20230208060523-20230208090523-00536.warc.gz | 295,772,453 | 21,595 | 622 views
closed as a duplicate of: Dijkstra
Which of the following procedure results same output as Dijkstra’s Algo. on unweighted graph on $'n'$ verices?
$A)$ BFS $B)$ DFS $C)$Kruskal $D)$ Prims
As far I know Dijkstra and Prims both have $T.C.=O(E+VlogV)$
But ans given BFS.
How this ans possible??
### 3 Comments
unweighted = take each distance ( edge weight ) = 1 unit.
now think how dijkstra's work and bfs work.. U will realize
@Shaik Masthan
What about Prims??
Dijkstra work on all direction like BFS.
right?? but why not Prims??
Unweighted means no edge weight, then why take each edge weight as 1??
edited by
I think only for unweighted graph we have to select BFS. dijkstra is for weighted graph
## 1 Answer
Best answer
BFS when apply on an unweighted graph , In the resultant BFS tree it can be seen that the shortest path to every vertex from root (or the start vertex) is figured out.
It is quite similar to dijsktra algorithm which is also single source shortest path.
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40 views | 312 | 1,119 | {"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} | 2.953125 | 3 | CC-MAIN-2023-06 | latest | en | 0.892052 |
https://www.physicsforums.com/threads/solving-pendulum-pulley-system-w-gravity-h-g-s-question.293222/ | 1,723,308,998,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640810581.60/warc/CC-MAIN-20240810155525-20240810185525-00471.warc.gz | 720,393,576 | 20,070 | # Solving Pendulum Pulley System w/ Gravity: H.G.'s Question
• hgetnet
In summary: Though I must admit, having tried it, that I can't actually see any simple solution to the combined equations.
hgetnet
This is not a homework problem. It was a problem posed by my professor at the end of the semester but had me puzzled for a while. I was able to satisfactorily solve it for the case when gravity is not present but I was not so successful with gravity.
The question is: to determine the trajectory of the mass m1, in terms of
-mass (m2),
-initial length (initial radius r0) of the pendulum and
-the initial angle (theta0) that the string makes with respect to the side of the table and
-the initial distance of m2 from the pulley nearest to it (x0).
Solution would be appreciated. Or, explanation of why a college freshman can not solve such a problem would also do.
Thank you.
H.G.
Last edited:
Welcome to PF!
Hi H.G.! Welcome to PF!
Just use conservation of energy
(and remember the speed of m2 is the same as the rate of increase in length of the hanging part of the string)
Thanks Tiny Tim.
--- I had this question posed to me few years back. Sorry if I am not clear on the problems I ran into.
(BTW: Pendulum - m1 - is not attached to the pulley but is descending to the ground as it pulls m2 with it while at the same time swinging)
Last edited:
hgetnet said:
The thing that I could not figure out is how to express the velocity of the pendulum as it descends and swings, therefore its height and its relationship to the velocity of m2.
Just call the angle θ, and express the velocity in terms of r' and rθ'
tiny-tim said:
Just call the angle θ, and express the velocity in terms of r' and rθ'
so the velocity of m2 would be dr/dt.
what is the velocity of m1? it can not be dr/dt because it is also swinging. What value of velocity would u use for the calculation of its (m1's) kinetic energy?
P.S> tiny_tim: *I appreciate the quick response*
p.s.s> I will post my solution for the case when there is no gravity
Last edited:
hgetnet said:
what is the velocity of m1? it can not be dr/dt because it is also swinging. What value of velocity would u use for the calculation of its (m1's) kinetic energy?
You must learn this … radial speed = dr/dt, tangential speed = r dθ/dt.
tiny-tim said:
You must learn this … radial speed = dr/dt, tangential speed = r dθ/dt.
so, we have non-linear differential equation in two variables (not counting t)..
$$\Delta$$U + $$\Delta$$KE = 0;
m1 gr( cos ($$\theta$$) - cos ($$\theta0$$) ) + 0 + 0.5*(m1[(dr/dt)$$^{2}$$ + (rd$$\theta$$/dt)$$^{2}$$ ] + m2* (dr/dt)$$^{2}$$) = 0;
0 = k1 + k2 * cos($$\theta$$) + k3 * r *$$^{'}$$$$^{2}$$$$\theta$$ + k4 *(r$$^{'}$$)$$^{2}$$ ;
What other relationship does exist that would turn this into a simultaneous equation?
Last edited:
Hi hgetnet!
hgetnet said:
so, we have non-linear differential equation in two variables (not counting t)..
$$\Delta$$U + $$\Delta$$KE = 0;
m1 gr( cos ($$\theta$$) - cos ($$\theta0$$) ) + 0 + 0.5*(m1[(dr/dt)$$^{2}$$ + (rd$$\theta$$/dt)$$^{2}$$ ] + m2* (dr/dt)$$^{2}$$) = 0;
0 = k1 + k2 * cos($$\theta$$) + k3 * r *$$^{'}$$$$^{2}$$$$\theta$$ + k4 *(r$$^{'}$$)$$^{2}$$ ;
What other relationship does exist that would turn this into a simultaneous equation?
Yes, we need one more equation …
which we can get by taking components of force and acceleration for the swinging mass either in the radial or the tangential direction …
though I must admit, having tried it, that I can't actually see any simple solution to the combined equations.
hmm … perhaps that's why your professor posed it at the end of the semester?
## 1. How does the gravity affect the motion of a pendulum pulley system?
Gravity plays a crucial role in the motion of a pendulum pulley system. The force of gravity pulls the pendulum downwards, causing it to swing back and forth. This motion is known as oscillation and is affected by the length of the pendulum, the mass of the pulley, and the strength of the gravitational force.
## 2. What is the purpose of a pendulum pulley system?
A pendulum pulley system is used to demonstrate the principles of simple harmonic motion and energy conservation. It also has practical applications in physics and engineering, such as regulating the speed of clocks and measuring time.
## 3. How do you solve a pendulum pulley system with gravity?
To solve a pendulum pulley system with gravity, you must first determine the equations of motion for the system. This involves identifying the forces acting on the system, such as tension and gravity, and using Newton's laws of motion. Then, you can use mathematical techniques, such as differential equations, to solve for the position, velocity, and acceleration of the system at any given time.
## 4. What factors affect the period of a pendulum pulley system?
The period of a pendulum pulley system is affected by several factors, such as the length of the pendulum, the mass of the pulley, and the strength of gravity. The period is directly proportional to the length of the pendulum and inversely proportional to the square root of the gravitational force and pulley mass.
## 5. How does the amplitude of a pendulum pulley system change with time?
The amplitude of a pendulum pulley system decreases with time due to the effects of damping. Damping is the loss of energy in a system, which causes the pendulum to gradually slow down and decrease in amplitude. This is due to the conversion of kinetic energy into other forms, such as heat and sound.
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https://coolconversion.com/weight/360-st-to-g | 1,611,115,597,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519883.54/warc/CC-MAIN-20210120023125-20210120053125-00695.warc.gz | 279,514,914 | 35,249 | # 360 Stones to Grams (st to g) Conversion
## How many g in 360 st?
360 st equals 2.286 × 106 g
To convert any value in stones to grams, just multiply the value in stones by the conversion factor 6350.29318. So, 360 stones times 6350.29318 is equal to 2.286 × 106 g.
### All In One Unit Converter
Please, choose a physical quantity, two units, then type a value in any of the boxes above.
### Quote of the day ...
To calculate a stone value to the corresponding value in g, just multiply the quantity in st by 6350.29318 (the conversion factor). Here is the formula:
Value in g = value in st × 6350.29318
Suppose you want to convert 360 st into g. Using the conversion formula above, you will get:
Value in g = 360 × 6350.29318 = 2.286 × 106 g
• How many st are in 360 g?
• 360 st are equal to how many g?
• How much are 360 st in g?
• How to convert st to g?
• What is the conversion factor to convert from st to g?
• How to transform st in g?
• What is the formula to convert from st to g? Among others.
#### St to g Conversion Chart Near 210 st
St to g Conversion Chart
210 st1.33 × 106 g
220 st1.4 × 106 g
230 st1.46 × 106 g
240 st1.52 × 106 g
250 st1.59 × 106 g
260 st1.65 × 106 g
270 st1.71 × 106 g
280 st1.78 × 106 g
290 st1.84 × 106 g
300 st1.91 × 106 g
310 st1.97 × 106 g
320 st2.03 × 106 g
330 st2.1 × 106 g
340 st2.16 × 106 g
350 st2.22 × 106 g
360 st2.29 × 106 g
St to g Conversion Chart
360 st2.29 × 106 g
370 st2.35 × 106 g
380 st2.41 × 106 g
390 st2.48 × 106 g
400 st2.54 × 106 g
410 st2.6 × 106 g
420 st2.67 × 106 g
430 st2.73 × 106 g
440 st2.79 × 106 g
450 st2.86 × 106 g
460 st2.92 × 106 g
470 st2.98 × 106 g
480 st3.05 × 106 g
490 st3.11 × 106 g
500 st3.18 × 106 g
510 st3.24 × 106 g
Note: some values may be rounded.
### Disclaimer
While every effort is made to ensure the accuracy of the information provided on this website, neither this website nor its authors are responsible for any errors or omissions, or for the results obtained from the use of this information. All information in this site is provided “as is”, with no guarantee of completeness, accuracy, timeliness or of the results obtained from the use of this information. | 721 | 2,173 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.703187 |
https://mail.python.org/pipermail/python-list/2008-November/497082.html | 1,394,515,187,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011129529/warc/CC-MAIN-20140305091849-00029-ip-10-183-142-35.ec2.internal.warc.gz | 636,697,945 | 2,616 | Steven D'Aprano steven at REMOVE.THIS.cybersource.com.au
Wed Nov 12 02:12:51 CET 2008
On Wed, 12 Nov 2008 10:08:45 +1100, Ben Finney wrote:
> Eric <eric.shain at gmail.com> writes:
>
>> I'm learning Python (while coming from MATLAB). One question I have is
>> that if I have a list with say 8 elements, and I want just a few of
>> them how do I select them out. In MATLAB, if I just want the first,
>> fifth and eighth element I might do something like this:
>>
>> b = a([1 5 8]);
>>
>> I can't seem to figure out a similar Python construct for selecting
>> specific indices. Any suggestions?
>
> Yes: the above code uses magic numbers which convey no semantic meaning,
> and should instead use *named* elemets of a container. In Python, that's
> done with a mapping type, which has the built-in type of ‘dict’.
>
> In other words: a list is best for sequences where any item is
> semantically identical to any other, and you can (for example) re-order
> all the elements in the list without changing their semantic meaning.
> If, instead, you want semantic meaning, that's best done via *names*,
> not magic numbers.
While I see your point, and in practice there are circumstance where I
would agree with you, I don't agree in general.
What the OP is asking for is a generalization of slicing. Slices already
take a stride:
>>> alist = range(20)
>>> alist[2:17:3]
[2, 5, 8, 11, 14]
Given a slice alist[start:end:stride] the indices are given by the
sequence start+i*stride, bounded by end. It seems to me that the OP has
asked how to take a slice where the indices are from an arbitrary set,
possibly given by an equation, but not necessarily. No practical examples
come to mind just off the top of my head, but given that the OP is coming
from a Matlab background, I bet they involve some pretty hairy maths.
Hmmm... let's see now...
http://en.wikipedia.org/wiki/Series_(mathematics)
#Summations_over_arbitrary_index_sets
To put it another way... suppose you have a sequence S given by a list,
and a set of indexes I (perhaps given by a tuple or a set), and you want
to sum the terms of S at those indexes, which mathematicians apparently
have good reason for doing, then you might write in Python:
sum(S[i] for i in I)
But if there was slicing support for arbitrary indexes, you could write:
sum(S[I])
which is apparently what Matlab allows. Standard lists don't support this
slicing generalization, but numpy arrays do.
--
Steven | 625 | 2,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-10 | longest | en | 0.93874 |
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Universal uniqueness criterion? Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 06/20/2009, 07:27 PM (This post was last modified: 07/06/2009, 12:08 AM by Base-Acid Tetration.) Theorem. Let $f$ be a function that is biholomorphic on both of the initial regions $D_1$ and $D_2$ that share as boundaries the same conjugate pair of fixed points of f. Also let f have no other fixed points. Let there be two Abel functions of f, $A_1$ and $A_2$, that are biholomorphic on these initial regions and satisfy A(d) = c. For each f, there exists exactly one biholomorphism $A_{cont}(z)$ on a single simply connected open set $C \supseteq D_1,\,D_2$ such that $\forall z \in D_i\, A_{cont}(z) = A_i (z).$ (i is an index that can be 1 or 2) i.e. there is an analytic continuation, and it's unique. Proof. 1. Let $D_1, D_2$ be disjoint, simply connected domains that have as boundaries: (1) $\partial_1 D,\, \partial_2 D \not \subset D$, disjoint curves which are homeomorphic to (0,1); (2) $L$ and $\bar{L}$, which are boundaries of, but not contained in, $\partial _1 D_1, \, \partial_2 D_1,\, \partial_1 D_2, \,\partial_2 D_2$. 2. Let $f$ be a biholomorphism on $S := \lbrace z:|\Im(z)| < \Im(L) + \epsilon i \rbrace$, where $\epsilon>0$, (tried to make the domain of biholomorphism into an open set) that: (1) bijects $\partial_1 D$ to $\partial_2 D$; (2) has a conjugate pair of fixed points $L$ and $\bar{L}$; (3) has no other fixed points in the domain of biholomorphy. 3. (1) Let $A_1$, a biholomorphism on $D_1$, and $A_2$, a biholomorphism on $D_2$, both satisfy $A(f(z))=A(z)+1$ for all applicable z. (for all z such that A(f(z)) is defined) (2) Let $A(d)=c$ for some $d \in S$. to be continued bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 06/21/2009, 08:19 AM (06/20/2009, 07:27 PM)Tetratophile Wrote: 1. .. 2. ... 3. ... 4. Is the proof already finished? I dont see where it shows that $A_1(z)=A_2(z)$ for $z\in D_1\cap D_2$. Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 06/21/2009, 01:57 PM (This post was last modified: 06/21/2009, 04:27 PM by Base-Acid Tetration.) btw, why does kouznetsov have IMAGINARY infinities $i\infty$ and $-i\infty$ as the values of the super logarithm at the fixed points? The choice seems arbitrary. I think it should be a kind of negative infinity because of how you get to the fixed points. exp^(some complex number) (1) = i or some other nonreal complex number, take the logarithm of that number an infinite number of times and you get the fixed point. for some complex c, corresponds to a complex iteration plus a negative infinity itterations of exp: $\exp^{-\infty+c}(1).$ We lose injectivity, but it's a fair price. (06/21/2009, 08:19 AM)bo198214 Wrote: Is the proof already finished? I dont see where it shows that $A_1(z)=A_2(z)$ for $z\in D_1\cap D_2$. no, it isn't finished.... It just lists all i know about the situation. I don't know how to proceed. I just have a hunch that the neighborhood of the fixed points might be important (the only thing the domains D1 and D2 share as part of their boundaries is the fixed point pair, and both D1 and D2 include a subset of the neighborhood); if I can show that the Abel functions are analytic and take the same values in all directions in the deleted neighborhood of the fixed points, I will have established the existence of a unique analytic continuation of both A1 and A2. I have attached a drawing to summarize the problem and the proposed proof technique. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 06/21/2009, 05:18 PM (06/21/2009, 01:57 PM)Tetratophile Wrote: btw, why does kouznetsov have IMAGINARY infinities $i\infty$ and $-i\infty$ as the values of the super logarithm at the fixed points? The choice seems arbitrary.Infinity can be imagined as a point on the extended complex plane, which is a sphere. You can regard infinity as any other point on the complex plane. The tetrational is not holomorphic (not even continuous) at infinity. However one can approach infinity from different angles/sectors where the limit may be a single value. E.g. if you approach infinity along the real axis $z \to \infty$ the limit is infinity. If you however approach infinity along the imaginary axis $z\to i \infty$ then the limit is $L$. This is writtten exactly as: $\lim_{x\to\infty} f(x+iy) = \infty$ for each real $y$. $\lim_{y\to\infty} f(x+iy) = L$ for each real $x$. $\lim_{y\to -\infty} f(x+iy) = L^\ast$ for each real $x$. Quote:no, it isn't finished.... It just lists all i know about the situation. I don't know how to proceed. I just have a hunch that the neighborhood of the fixed points might be important (the only thing the domains D1 and D2 share as part of their boundaries is the fixed point pair, and both D1 and D2 include a subset of the neighborhood); I guess its about deforming one initial region into the other initial region, and continuing the one Abel function to the other region. Then we have reduced the problem to both Abel function having the same domain, to which we can apply the theorem. For different fixed point pairs this would not be possible. Quote: if I can show that the Abel functions are analytic and take the same values in all directions in the deleted neighborhood of the fixed points, No, they are not analytic in the punctured neighborhood. Or in other words the fixed points are not isolated singularities. They are branching points, we need cut-line ending in the fixed point, to have it holomorphic. Your drawing however is I also see the situation (except holomorphy at the fixed points). Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 06/21/2009, 05:28 PM (This post was last modified: 06/21/2009, 05:36 PM by Base-Acid Tetration.) (06/21/2009, 05:18 PM)bo198214 Wrote: Infinity can be imagined as a point on the extended complex plane, which is a sphere. You can regard infinity as any other point on the complex plane. The tetrational is not holomorphic (not even continuous) at infinity. However one can approach infinity from different angles/sectors where the limit may be a single value. I said that the superlogarithm should be negative infinity at the fixed points, instead of infinity*i or = -infinity*i, because the fixed point is approached via a complex iteration, and then infinite negative iteration of exp, ie iteration of log. (fixed points are repelling in graph of exp) Or leave it an essential singularity if there is no way to complex-iterate exp to get an imaginary number. 'bo198214 Wrote:I guess its about deforming one initial region into the other initial regionHow do we go about doing that? It is not given whether the Abel functions are holomorphic outside of their domains. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 06/21/2009, 06:04 PM (06/21/2009, 05:28 PM)Tetratophile Wrote: I said that the superlogarithm should be negative infinity at the fixed points, No, its not. The super-exponential approaches L for $z=x+iy$, $y\to+\infty$. Thatswhy the super-logarithm has positively unbounded imaginary part, if we approach the upper fixed point in the initial region, which one could interpret as $+i\infty$. And it has negatively unbounded imaginary when we approach $L^\ast$ in the initial region which we could interpret as $-i\infty$. However strictly considered there is no $\pm i\infty$ as value of a limit. Quote: instead of infinity*i or = -infinity*i, because the fixed point is approached via a complex iteration, and then infinite negative iteration of exp, ie iteration of log. If you approach the fixed point via repeated logarithm, then you dont stay inside the initial region. Quote:'bo198214 Wrote:I guess its about deforming one initial region into the other initial region How do we go about doing that? It is not given whether the Abel functions are holomorphic outside of their domains.If you go slightly to the right of $\partial_1 D_1$ having a curve $\gamma$ inside $D_1$ then you can continue $A_1$ to the region between $\partial_2 D_1$ and $F(\gamma)$, i.e. $A_1(F(z)):=A_1(z)+1$ where we use the already established values of $A_1$ on the region between $\partial_1 D_1$ and $\gamma$. This works if $F$ is bijective in the considered area. For example $F=\exp$ is bijective in the strip along the real axis $\{z: -\pi < \Im(z) < \pi \}$, which includes the fixed points $L$ and $L^\ast$. BenStandeven Junior Fellow Posts: 27 Threads: 3 Joined: Apr 2009 06/21/2009, 11:22 PM (06/21/2009, 06:04 PM)bo198214 Wrote: (06/21/2009, 05:28 PM)Tetratophile Wrote: 'bo198214 Wrote:I guess its about deforming one initial region into the other initial regionHow do we go about doing that? It is not given whether the Abel functions are holomorphic outside of their domains.If you go slightly to the right of $\partial_1 D_1$ having a curve $\gamma$ inside $D_1$ then you can continue $A_1$ to the region between $\partial_2 D_1$ and $F(\gamma)$, i.e. $A_1(F(z)):=A_1(z)+1$ where we use the already established values of $A_1$ on the region between $\partial_1 D_1$ and $\gamma$. This works if $F$ is bijective in the considered area. For example $F=\exp$ is bijective in the strip along the real axis $\{z: -\pi < \Im(z) < \pi \}$, which includes the fixed points $L$ and $L^\ast$. Of course, this procedure might not help if the derivatives at the fixed points are positive reals (or are 0). In that case, near the fixed points, the expanded $A_1$ won't be any closer to $A_2$ than the original one was. Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 06/22/2009, 12:37 AM (This post was last modified: 06/22/2009, 04:49 AM by Base-Acid Tetration.) Please go back up to the beginning of the proof; I made the conditions a lot stronger: changed the holomorphy condition for $f$ to BIholomorphy on a strip with infinite range of real parts), and added "f has no other fixed points", because I thought other fixed points would mess things up. Proof continued... 4. Consider a simple curve $\gamma \subset D_1$ that also has $L$ and $\bar{L}$ as non-inclusive boundaries. (1) Since $A_1$ is biholomorphic on $\gamma$, $A_1$ will still be biholomorphic on the curve $f(\gamma)$, because if $A_1(z)$ is biholomorphic, so is $A_1(z)+1=A_1(f(z)) \forall z \in D_1$. (2) We can do this for every curve in $D_1$ that has $L$ and $\bar{L}$ as boundaries, to extend the domain of $A_1$ (3) We can repeat (4.2) as many times as needed to get to $D_2$. (we can do the above the other way around, using $f^{-1}$ to get from $D_2$ to $D_1$, because f is BIholomorphic) Then $A_1$ is biholomorphic where $A_2$ was defined to be biholomorphic.* Then we know, by the theorem that was proven on the other day, that they are the same biholomorphism. So we know that there exists a single open set C that includes $D_1 \cup D_2$ where not only the biholomorphism $A_{cont}$that maps d to c exists, but also $A_1 = A_2 = A_{cont} \forall z \in C.$ $\mathcal{Q. E. D.}$ Corollary. There exists a unique superlogarithm for $b>e^{1/e}$ that uniquely bijects holomorphically each simple initial region of arbitrary "width" in an open set C that: (1) contains in its boundary fixed points of exp_b; (2) does not include branch cuts of the superlogarithm; to its resp. vertically infinite strip of arbitrary width in $A( C )$. *Now I don't know if $A_1 = A_2$ in $D_2$. It must have something to do with the condition $A(d) = c$. For your theorem to apply, I need to prove that both A1 and A2 are equal in some neighborhood of d. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 06/22/2009, 02:04 PM (06/21/2009, 11:22 PM)BenStandeven Wrote: Of course, this procedure might not help if the derivatives at the fixed points are positive reals (or are 0). In that case, near the fixed points, the expanded $A_1$ won't be any closer to $A_2$ than the original one was. Right, this would make up an extra case; also it needs to be investigated whether this case allows an initial region at all. However for $\exp_b$, $b>e^{1/e}$ this is already satisfied. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 06/22/2009, 02:24 PM (06/22/2009, 12:37 AM)Tetratophile Wrote: and added "f has no other fixed points", because I thought other fixed points would mess things up.Well, but $\exp$ *has* more fixed points. In every strip $2\pi i k < \Im(z) < 2\pi i (k+1)$ there is a fixed point of $\exp$. Quote:(3) We can repeat (4.2) as many times as needed to get to $D_2$. No, we can not. 1. The curve $f(\gamma)$ may get out of the strip of bijectivity of $f$ what indeed happens for $f=\exp$. (plot the curves!) 2. $f(\gamma)$ may hit $D_1$, i.e. the extension of $A_1$ overlaps with the original domain. If $\gamma$ meets $L$ at an angle $\alpha$ then $f(\gamma)$ meets $L$ at an angle $\alpha+\arg(f'(L))$. So the image curves rotate around L. This includes also the comment of Ben, that if $\arg(f'(L))=0$, i.e. if $f'(L))$ is real, then then the images do not rotate, and hence one would never reach $D_2$ if it hits $L$ in a different angle. « Next Oldest | Next Newest »
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Users browsing this thread: 1 Guest(s) | 4,270 | 14,374 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 136, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-05 | latest | en | 0.819938 |
https://mikesmathpage.wordpress.com/2014/10/26/momath-and-megamenger/ | 1,685,844,165,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649348.41/warc/CC-MAIN-20230603233121-20230604023121-00194.warc.gz | 432,410,684 | 32,253 | # MoMath and MegaMenger
Yesterday we visited the Museum of Mathematics in NYC to help out with their MegaMenger build. The boys had a blast!
This was our 3rd visit to the Museum and I’m sure there will be many more. One of the fun attractions is this tricycle with square wheels (sorry for the poor quality of this video):
The MegaMenger project is an incredible project in which people from all over the world are working together to build a giant Menger Sponge out of business cards. The website for the project is here:
http://www.megamenger.com/
The boys were actually so excited about participating in this project yesterday that we started the day today building a level 2 Menger sponge out of snap cubes. Although I enjoyed the project, too, I wouldn’t have described my excitement as “build a new level 2 Menger Sponge at 5:30 am the next day excited,” but hey, I’ll take it:
With that new morning build, there was really no doubt at all what our Family Math project for the day would be 🙂 We began by simply reviewing our trip to MoMath and some basics about the Menger Sponge. The specific topics for the day are going to be volume and surface area. For all but the last movie the questions will revolve around Menger Sponges of ever increasing sizes, like the one being build in the Mega Menger project:
Having touched on the volume of the Menger Sponge in the last movie, we now dive into the volume calculation in more detail. What I liked here is that each kid had a different way of calculating the volume. So fun to see the different approaches to counting here! I showed a third way, too, that has a sort of surprising twist. The Level 2 figure we talk about at the end is the shape that we constructed out of the snap cubes that is pictured above:
The surface area calculation is only slightly more tricky. As with the approach to volume, both boys had different approaches to counting the surface area of the Level 1 Menger Sponge. It turned out that my younger son’s method was actually the same as mine, so I didn’t add a third counting method here. Taking through my older son’s direct counting method and my younger son’s method of counting the overlaps was really enjoyable. We finished by wondering which of these two methods was easier to generalize to the higher level sponges.
Next we attempted to calculate the surface area of the level 2 sponge. The level 2 sponge is the one that we made out of snap cubes this morning. Our contribution to the MoMath Mega Menger build amounted to the construction of two of these sponges out of business cards. The construction from folding business cards took a bit longer than the one from snap cubes, though the business card construction was at 2:00 in the afternoon and was followed by BBQ at Blue Smoke in Manhattan, so maybe I should call it a draw 🙂
The math in this video is the most difficult to follow in this project, but hopefully we work through it slowly enough. To calculate the surface area of the Level 2 sponge we use the method my younger son suggested in the last video. We first assume that the surface area is 20 times the surface area of one of the Level 1 sponges (since it takes 20 level 1’s to make a level 2) and then subtract out the surface area that vanishes when two sides touch. We break down this calculation into two pieces. The first part is for the middle pieces that touch two other Level 1 sponges, and the second part is for the corner pieces that touch three. After a 3 minute calculation, we arrive at the surface area of the Level 2 sponge: | 797 | 3,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2023-23 | latest | en | 0.94819 |
https://www.ademcetinkaya.com/2023/03/rwod-redwoods-acquisition-corp-common.html | 1,679,844,887,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945473.69/warc/CC-MAIN-20230326142035-20230326172035-00037.warc.gz | 727,694,727 | 61,423 | Outlook: Redwoods Acquisition Corp. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Hold
Time series to forecast n: 07 Mar 2023 for (n+4 weeks)
Methodology : Modular Neural Network (CNN Layer)
## Abstract
Redwoods Acquisition Corp. Common Stock prediction model is evaluated with Modular Neural Network (CNN Layer) and Independent T-Test1,2,3,4 and it is concluded that the RWOD stock is predictable in the short/long term. According to price forecasts for (n+4 weeks) period, the dominant strategy among neural network is: Hold
## Key Points
1. What are the most successful trading algorithms?
2. Is now good time to invest?
3. How can neural networks improve predictions?
## RWOD Target Price Prediction Modeling Methodology
We consider Redwoods Acquisition Corp. Common Stock Decision Process with Modular Neural Network (CNN Layer) where A is the set of discrete actions of RWOD stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Independent T-Test)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (CNN Layer)) X S(n):→ (n+4 weeks) $\begin{array}{l}\int {r}^{s}\mathrm{rs}\end{array}$
n:Time series to forecast
p:Price signals of RWOD stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## RWOD Stock Forecast (Buy or Sell) for (n+4 weeks)
Sample Set: Neural Network
Stock/Index: RWOD Redwoods Acquisition Corp. Common Stock
Time series to forecast n: 07 Mar 2023 for (n+4 weeks)
According to price forecasts for (n+4 weeks) period, the dominant strategy among neural network is: Hold
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
## IFRS Reconciliation Adjustments for Redwoods Acquisition Corp. Common Stock
1. An entity can also designate only changes in the cash flows or fair value of a hedged item above or below a specified price or other variable (a 'one-sided risk'). The intrinsic value of a purchased option hedging instrument (assuming that it has the same principal terms as the designated risk), but not its time value, reflects a one-sided risk in a hedged item. For example, an entity can designate the variability of future cash flow outcomes resulting from a price increase of a forecast commodity purchase. In such a situation, the entity designates only cash flow losses that result from an increase in the price above the specified level. The hedged risk does not include the time value of a purchased option, because the time value is not a component of the forecast transaction that affects profit or loss.
2. The definition of a derivative refers to non-financial variables that are not specific to a party to the contract. These include an index of earthquake losses in a particular region and an index of temperatures in a particular city. Non-financial variables specific to a party to the contract include the occurrence or non-occurrence of a fire that damages or destroys an asset of a party to the contract. A change in the fair value of a non-financial asset is specific to the owner if the fair value reflects not only changes in market prices for such assets (a financial variable) but also the condition of the specific non-financial asset held (a non-financial variable). For example, if a guarantee of the residual value of a specific car exposes the guarantor to the risk of changes in the car's physical condition, the change in that residual value is specific to the owner of the car.
3. For lifetime expected credit losses, an entity shall estimate the risk of a default occurring on the financial instrument during its expected life. 12-month expected credit losses are a portion of the lifetime expected credit losses and represent the lifetime cash shortfalls that will result if a default occurs in the 12 months after the reporting date (or a shorter period if the expected life of a financial instrument is less than 12 months), weighted by the probability of that default occurring. Thus, 12-month expected credit losses are neither the lifetime expected credit losses that an entity will incur on financial instruments that it predicts will default in the next 12 months nor the cash shortfalls that are predicted over the next 12 months.
4. For some types of fair value hedges, the objective of the hedge is not primarily to offset the fair value change of the hedged item but instead to transform the cash flows of the hedged item. For example, an entity hedges the fair value interest rate risk of a fixed-rate debt instrument using an interest rate swap. The entity's hedge objective is to transform the fixed-interest cash flows into floating interest cash flows. This objective is reflected in the accounting for the hedging relationship by accruing the net interest accrual on the interest rate swap in profit or loss. In the case of a hedge of a net position (for example, a net position of a fixed-rate asset and a fixed-rate liability), this net interest accrual must be presented in a separate line item in the statement of profit or loss and other comprehensive income. This is to avoid the grossing up of a single instrument's net gains or losses into offsetting gross amounts and recognising them in different line items (for example, this avoids grossing up a net interest receipt on a single interest rate swap into gross interest revenue and gross interest expense).
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
## Conclusions
Redwoods Acquisition Corp. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating. Redwoods Acquisition Corp. Common Stock prediction model is evaluated with Modular Neural Network (CNN Layer) and Independent T-Test1,2,3,4 and it is concluded that the RWOD stock is predictable in the short/long term. According to price forecasts for (n+4 weeks) period, the dominant strategy among neural network is: Hold
### RWOD Redwoods Acquisition Corp. Common Stock Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementB2Baa2
Balance SheetCBa3
Leverage RatiosBaa2Ba2
Cash FlowCaa2B2
Rates of Return and ProfitabilityBa2Baa2
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
### Prediction Confidence Score
Trust metric by Neural Network: 89 out of 100 with 660 signals.
## References
1. Zeileis A, Hothorn T, Hornik K. 2008. Model-based recursive partitioning. J. Comput. Graph. Stat. 17:492–514 Zhou Z, Athey S, Wager S. 2018. Offline multi-action policy learning: generalization and optimization. arXiv:1810.04778 [stat.ML]
2. Athey S. 2019. The impact of machine learning on economics. In The Economics of Artificial Intelligence: An Agenda, ed. AK Agrawal, J Gans, A Goldfarb. Chicago: Univ. Chicago Press. In press
3. Athey S. 2019. The impact of machine learning on economics. In The Economics of Artificial Intelligence: An Agenda, ed. AK Agrawal, J Gans, A Goldfarb. Chicago: Univ. Chicago Press. In press
4. Bera, A. M. L. Higgins (1997), "ARCH and bilinearity as competing models for nonlinear dependence," Journal of Business Economic Statistics, 15, 43–50.
5. Hastie T, Tibshirani R, Wainwright M. 2015. Statistical Learning with Sparsity: The Lasso and Generalizations. New York: CRC Press
6. Hartigan JA, Wong MA. 1979. Algorithm as 136: a k-means clustering algorithm. J. R. Stat. Soc. Ser. C 28:100–8
7. Athey S, Bayati M, Imbens G, Zhaonan Q. 2019. Ensemble methods for causal effects in panel data settings. NBER Work. Pap. 25675
Frequently Asked QuestionsQ: What is the prediction methodology for RWOD stock?
A: RWOD stock prediction methodology: We evaluate the prediction models Modular Neural Network (CNN Layer) and Independent T-Test
Q: Is RWOD stock a buy or sell?
A: The dominant strategy among neural network is to Hold RWOD Stock.
Q: Is Redwoods Acquisition Corp. Common Stock stock a good investment?
A: The consensus rating for Redwoods Acquisition Corp. Common Stock is Hold and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of RWOD stock?
A: The consensus rating for RWOD is Hold.
Q: What is the prediction period for RWOD stock?
A: The prediction period for RWOD is (n+4 weeks) | 2,216 | 9,517 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | longest | en | 0.820566 |
https://www.jiskha.com/questions/473840/prove-that-tan20-tan30-tan40-tan10 | 1,537,715,911,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159470.54/warc/CC-MAIN-20180923134314-20180923154714-00210.warc.gz | 768,519,413 | 3,832 | # trig
prove that :tan20 tan30 tan40 =tan10
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Chapter 5:
Bessel Functions
Abstract
The solutions to this differential equation cannot be expressed in terms of simple functions. Equation (5.1) is known as Bessel the equation, and its solutions are known as Bessel functions. There are numerous books devoted to Bessel functions, but the most authoritative is the one by Watson. This 762-page book with a 36-page bibliography containing 791 references is the standard text. This book, which was published in 1944 and covers over three centuries of results, is the ultimate reference book on Bessel functions. A simpler (and shorter book) by Bowman59 is a good reference for those wanting a shorter read. In this chapter, we introduce the readers to the Bessel differential equation and its solutions in their various forms and state their properties. These functions arose in Daniel Bernoulli’s (1700–1782) analysis of the oscillation of a hanging chain and again in Leonhard Euler’s (1707–1783) theory of the vibrations of a circular membrane and a few other problems in mechanics. In fact, Bernoulli solved an equation of the form given above and found a series solution now known as the Bessel function of the first kind. However, the German astronomer Friedrich Wilhelm Bessel (1784–1846) was the first person who carved out a systematic study of the differential equation and the properties of its solutions. Hence, we refer to this equation as the Bessel equation and the solutions as Bessel functions. Bessel functions are ubiquitous in optical science. In addition, they appear in a great variety of other areas, such as electromagnetism, heat, hydrodynamics, elasticity, wave motion, scattering, etc. In fact, Bessel functions come into play whenever we deal with situations including cylindrical symmetry; for this reason all solutions of the Bessel equations are called cylinder functions. Because of this great use in various areas of physics and engineering, Bessel functions are arguably the most important functions beyond the elementary ones. | 446 | 2,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | longest | en | 0.95338 |
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A.Y. Credits
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Teaching in foreign languages
Course with optional materials in a foreign language English
This course is entirely taught in Italian. Study materials can be provided in the foreign language and the final exam can be taken in the foreign language.
### Assigned to the Degree Course
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Curriculum: COMUNE
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### Learning Objectives
The course provides the student with the main notions, results and methodologies of basic mathematics, real function of a single variable, probability and statistics, linear algebra and matrix theory, with particular attention to logical deduction and analysis of the arguments of the course.
### Program
Elementary set theory: union, intersection, difference, complement. Connective and quantifier. Cartesian product of two or more sets.
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Numerical sets (N, Z, Q, R) and their main properties. Absolute value. Total ordering of the sets N, Z, Q, R.
Equations and inequalities.
Supremum and infimum, maximum and minimum. Intervals, disks.
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Elements of Statistics: Mean, median and mode. Variance. The least squares method. Interpolation techniques.
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There are no mandatory prerequisites. It is recommended to take the exam of Calculus during the first academic year.
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Knowledge and understanding: The student will achieve a deep understanding of the structure of mathematical reasoning, both in a general context and in the context of the arguments of the course. The student will achieve the mastery of the computational methods of basic mathematics and of the other arguments of the course.
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The teaching material prepared by the lecturer in addition to recommended textbooks (such as for instance slides, lecture notes, exercises, bibliography) and communications from the lecturer specific to the course can be found inside the Moodle platform › blended.uniurb.it
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### Didactics, Attendance, Course Books and Assessment
Didactics
Theorical and practical lessons.
Attendance
Although strongly recommended, course attendance is not mandatory.
Course books
Abate, Matematica e statistica. Le basi per le scienze della vita, McGraw-Hill
Bramanti - Pagani - Salsa, Matematica. Calcolo infinitesimale e algebra lineare, Zanichelli
Assessment
The knowledge, understanding and ability to communicate are assessed with a written exam with open questions. After having passed the written exam students can attempt an oral examination.
### Additional Information for Non-Attending Students
Didactics
Theorical and practical lessons.
Attendance
Although strongly recommended, course attendance is not mandatory.
Course books
Abate, Matematica e statistica. Le basi per le scienze della vita, McGraw-Hill
Bramanti - Pagani - Salsa, Matematica. Calcolo infinitesimale e algebra lineare, Zanichelli
Assessment
The knowledge, understanding and ability to communicate are assessed with a written exam with open questions. After having passed the written exam students can attempt an oral examination.
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Please imagine a simple loop: For[i=1,i<=100,i++, Print[i]; ]; We can ask Print[] to output $(1,2,3,4,...)$ to a ... | 2,658 | 11,197 | {"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-2016-18 | longest | en | 0.889919 |
https://www.meritnation.com/ask-answer/question/define-threshold-frequency/dual-nature-of-radiation-and-matter/1616724 | 1,611,357,361,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703531429.49/warc/CC-MAIN-20210122210653-20210123000653-00032.warc.gz | 890,829,852 | 10,428 | # Define threshold frequency?
For a given Photosensitive material, there is a certain minimum cut off frequency called Threshold frequency, for which stopping potential is zero.
Higher the work function of the material, greater is the Threshold frequency. (As you are aware that stopping potential is the minimum value of potential below which photoelectric current becomes 0.
See the figure below of two different metals having different Threshold Frequency.
• 6
thtesh hold frequency is the minimum frequency required for photoelectric emmission to occur..below this value photoelectric emission is not possible...it refers to the minimum no of of election which can excite the electrons from metal....
from einstien law of photoelectric effect
hf=k+hfo
here fo is threshhold frequency of metal and rest k is kinetic energy of ejected photoelectron and f is frequency of light emmited...........
okkk
• 3
ask me if any any doubt comes from my written text.....
• 2
its minimum frequency till which photoelectric current is emitted
• -1
cool
• -1
What are you looking for? | 227 | 1,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-04 | latest | en | 0.91068 |
https://www.jiskha.com/display.cgi?id=1292290561 | 1,503,034,589,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104565.76/warc/CC-MAIN-20170818043915-20170818063915-00392.warc.gz | 894,761,574 | 3,631 | # math
posted by .
what fraction would you use to find 33 1/3% of 42
• math -
33 1/3% = 1/3
1/3 * 42 = 42/3 = 14
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Get Full Access to Physics: Principles With Applications - 6 Edition - Chapter 24 - Problem 71gp
Get Full Access to Physics: Principles With Applications - 6 Edition - Chapter 24 - Problem 71gp
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# Show that the second- and third-order spectra of white
ISBN: 9780130606204 3
## Solution for problem 71GP Chapter 24
Physics: Principles with Applications | 6th Edition
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Problem 71GP
Problem 71GP
Show that the second- and third-order spectra of white light produced by a diffraction grating always overlap. What wavelengths overlap?
Step-by-Step Solution:
Step-by-step solution
Step 1 of 3
The path difference for wavelength of is;
The path difference for the wavelength of is;
Step 2 of 3
Step 3 of 3
##### ISBN: 9780130606204
The answer to “Show that the second- and third-order spectra of white light produced by a diffraction grating always overlap. What wavelengths overlap?” is broken down into a number of easy to follow steps, and 20 words. This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6. Since the solution to 71GP from 24 chapter was answered, more than 360 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 71GP from chapter: 24 was answered by , our top Physics solution expert on 03/03/17, 03:53PM. This full solution covers the following key subjects: overlap, always, grating, light, order. This expansive textbook survival guide covers 35 chapters, and 3914 solutions. Physics: Principles with Applications was written by and is associated to the ISBN: 9780130606204.
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When I say large numbers I mean numbers like 450 and to make that even more larger lets do 450*23.
Now don't freak out, this problem may look like the hardest thing in math when this is just as simple as doing one digit multiplication! To do this you first have to write this problem vertically like this
450
x23
Now, all you have to do is follow the rule of multiplying numbers, remember you are going to multiply the smaller number by the larger number so lets first do 450x3 because we always start from the right in the smaller number and go to the left.
450
x 3
1,350
Now let's do 450x20:
450
x20
9,000
Now that we have our products, just add them together to get the final answer!
9000
+1350
10,350
So our product to 450x23 is 10,350.
To understand this topic better I suggest you watch the video
And try some of these:
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Based on angles: intersect at 900 and passes through the midpoint of die segment is called the perpendicular bisector of the segment. Each side of the square pyramid shown below measures 10 inches. If you know how to answer all the questions on this test, then you have a solid understanding of the most important concepts in basic geometry. ABC is considering the suitability of a spherical tank with uniform wall thickness for the purpose. Again,(longest side)2 = (10)2 = 100; Correct Answers for Sample College Algebra Questions Item # Correct Answer Content Category 1 C Arithmetic and Geometric Sequences and Series 2 E Functions 3 B Functions 4 D Exponents 5 E Matrices (basic operations, equations, and determinants) 6 A Functions 7 E Functions 8 C Complex Numbers 9 A Arithmetic and Geometric Sequences and Series A line which is perpendicular to a line segment, i.e. Linear Pair: Two angles are said to form a linear pair if they have common side and their other two sides are opposite rays. ABCD is a rectangle. So, it is not a right triangle Which of the following is true? This book will help you to visualise, understand and enjoy geometry. What is (area of large circle) – (area of small circle) in the figure above? Circumcentre: The point of intersection of the perpendicular bisectors of the sides of a triangle is called the circumcentre. c. … How to Download Primary Six Maths Exam Questions. b) Plot 0.43 with a green dot. Show that AB>AC. Geometry. On the following pages are multiple-choice questions for the Grade 11 Practice Test, a practice opportunity for the Nebraska State Accountability–Mathematics (NeSA–M). a. Mathematics Topic By Topic Questions and Answers for All Topics in Form 1, Form 2, Form 3 and Form 4 for Kenya Secondary Schools in preparation for KCSE . The centroid divides each median in the ratio 2:1. Delve into mathematical models and concepts, limit value or engineering mathematics and find the answers to all your questions. Read and attempt to answer every question you can. Sum of all exterior angles of a polygon taken in order is 360°. Line segment: A line with two end points is called a segment. In a given day, 10 boxes of chalk … Mathematics section. Supplementary angles: Two angles whose sum is 180° are supplementary, each one is the supplement of the other. The questions can be answered using A alone but not using B alone. Questions on Geometry for CAT exam is a crucial topic. The question can be answered using A and B together, but not using either A or B alone. DRDO JRF Interview Date 2021 Released – Download RA Schedule Here!!!! A triangle la denoted by the symbol ∆. Historically, geometry questions in past year CAT papers have come from triangles, circles, and quadrilaterals. Is the tank capacity adequate to met ABC’s requirements? So the answer is side AB. MCQ Questions for Class 6 Maths with Answers: Central Board of Secondary Education (CBSE) has declared a major change in the Class 6 exam pattern from 2020.Practicing & preparing each and every chapter covered in the CBSE Class 6 Maths Syllabus is a necessary task to attempt the MCQs Section easily with full confidence in the board exam paper. The fixed point is called the centre of the circle and the fixed distance is called the radius r. Example: The point A divides the join the points (-5, 1) and (3, 5) in the ratio k: l and coordinates of points B and C are (1, 5) and (7, -2) respectively. a) x – 6y […] AD = 1 cm, DF = 1 cm and perimeter of DEF = 3 cm. The outer diameter of the tank is 10 meters. The distance between AB and the tangent at E is 5 cm. Geometry Diagnostic Pre-Test 50 questions – 60 minutes Multiple Choice Use the answer “NOTA” (which stands for None Of The Above) if the answer is not listed 1. The ratio of the sum of the lengths of all chord roads to the length of the outer ring road is. 1. Geometry Problems and Questions with Answers for Grade 9; Free Geometry Tutorials, Problems and Interactive Applets; More Info. One is a printable PDF file and another is an editable Doc file for your better GED Test Prep online. In figure, ∠DBA = 132° and ∠EAC = 120°. Of the triangles with sides 11, 5, 9 or with sides 6, 10, 8; which is a right triangle? Because the fundamentals of Set Theory are known to all mathemati-cians, basic problems in the subject seem elementary. You can Create Your Own Worksheet at Mathopolis, and our forum members have put together a collection of Math Exercises on the Forum. Answer: 87. Math workbook 1 is a content-rich downloadable zip file with 100 Math printable exercises and 100 pages of answer sheets attached to each exercise. Lines and Angles form the basic concept of geometry section in the mathematics. If POR; ROQ = 5:7, find all the angles. At least 20% of CAT questions each year are from Geometry alone. B: The tank weights 30,000 kg when empty, and is made of a material with density of 3 gm/cc. How many digits are there in Hindu-Arabic System? Don’t let the question position or question type deter you from answering questions. Exams Daily – India's no 1 Education Portal. Adjacent angles: Two angles are called adjacent angles if they have a Common side and their interiors are disjoint. At least 20% of CAT questions each year are from Geometry alone. (Longest side)2= 112 – 121; Expert Teachers at KSEEBSolutions.com has created Karnataka 1st PUC Basic Maths Question Bank with Answers Solutions, Notes, Guide Pdf Free Download of 1st PUC Basic Maths Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2020-21 in English Medium and Kannada Medium are part of 1st PUC Question Bank with Answers. Solve: x − (15x − 6) = 104 A) −55 7 B) −49 8 C) −55 8 D) −7 E) NOTA 2. For all questions: † Read each question carefully and choose the best answer. 2. Your email address will not be published. The plane figure bounded by the union of three lines, which Join three non-collinear points. For this, we have listed around 85 geometry questions with answers which have appeared in previous year CAT papers. Find … The most basic and important part of all the geometric shapes is lines and angles. What is the radius of the outer ring road in kms? The distance between two parallel lines is constant. This practice test contains a full-length sample test consisting of 50 multiple-choice questions, an answer sheet, and a skill area worksheet for each Mathematics skill area. When two lines are intersected by a transverse, they form two pairs of alternate angles. It offers text, videos, interactive sketches, and assessment items. The sheets present concepts in the order they are taught and give examples of … B traverses her entire path at a uniform speed of $$\left( {10\sqrt 3 + 20} \right).$$ C traverses distances C1C2, C2C3 and C3C1 at an average speeds of $$\frac{{40}}{3}\left( {\sqrt 3 + 1} \right),\frac{{40}}{3}\left( {\sqrt 3 + 1} \right)$$ and 120 respectively. They also prove and … Grade 10 geometry problems with answers are presented. Line segment: A line with two end points is called a segment. ∴ 112≠ 52 + 92 Amit wants to reach E2 from N1 using first the chord N1 – W2 and then the inner ring road. Let the radius of each circular park be r, and the distances to be traversed by the sprinters A, B and C be a, b and c respectively. The common origin is called the vertex and the two rays are the sides of the angle. The same string, when wound on the exterior four walls of a cube of side n cm, starting at point C and ending at point D, can give exactly one turn (see figure, not drawn to scale). B. Triangle Inequality •The sum of any two sides must be greater than third side or else the three sides cannot form a triangle. Mathematics books Need help in math? The collection of all points in a plane, which are at a fixed distance from a fixed point in the plane, is called a circle. Sewayojan UP Release Notification for LIC Recruitment 2021- Apply for 500... NATS Releases the NBCC Recruitment 2021 Out – For Electrical Vacancy, APPSC Panchayat Secretary Latest Update on Exam Centers. The question can be answered by using both the statements together, but cannot be answered by using either statement alone. Historically, geometry questions in past year CAT papers have come from triangles, circles, and quadrilaterals. in which mathematics takes place today. i.e. Solution: For 0.2 we split the segment from 0 to 1 on the number line into ten equal pieces between 0 and 1 and then count A polygon with 8 sides is called an octagon. ABE = 7cm2; EC = 3(BE). It can be extended indefinitely in both directions. Download GED Math Practice Test 2020 Sample Questions Answers (Free Printable PDF) and practice free online math quiz test Test Topics Basic Math, Geometry, Basic Algebra, Graphs, and Functions. A polygon with 4 sides is called a quadrilateral. The area of ABCD (in cm2) is, AB = BC = 2CH = 2CD = EH = FK = 2HK = 4KL = 2LM = MN. What will be his travel time in minutes on the basis of information given in the above question? 25 Very important Coordinate Geometry objective questions for SSC exams. A polygon with 7 sides is called a heptagon. To get the complete question in pdf or word format, call or whatsapp me on 08051311885 on link to download it. Notably, in geometry, learning the application of concepts is very important to confidently arrive at the answers. In an equilateral triangle, all the angles are congruent and equal to 60°. Static GK topics for Competitive Exams – Check Static GK Competitive Exams || Download Study Materials Here!!!! Median of a triangle: The line drawn from a vertex of a triangle to the opposite side such that it bisects the side, is called the median of the triangle. If you want the answers, either bookmark the worksheet or print the answers straight away.. Also! But, in recent years questions from mensuration have increased. If you can solve these problems with no help, you must be a genius! Answer: Sphere 1 (according to WikiPedia)Here I am going to share you about list of basic Input Devices, Output devices and Both input-output devices related to computer. The three non-collinear points, are called the vertices of the triangle. Centroid: The point of intersection of the three medians of a triangle is called the centroid. In a staff room, there are four racks with 10 boxes of chalk-stick. Shipping Corporation of India Recruitment 2020 (Out) – 48 Assistant Manager... SSC Exam Result Declaration Date 2019 – Download MTS, CGL, CHSL... WB Pollution Control Board Admit Card 2020- Check Exam Dates and... WB Health Recruitment 2020 Out – Apply For Lab Technician Vacancy!!! Parallel lines: Two […] A: The inner diameter of the tank is at least 8 meters. It has neither width nor thickness. What is the ratio of the perimeter of $$\Delta ADC$$ to that of the $$\Delta BDC$$? A polygon with 5 sides is called a pentagon. In figure given below, lines PQ and RS intersect each other at point O. Save my name, email, and website in this browser for the next time I comment. hi im priyanga,.. im full time content writer for past 5 years. Answer: A . B. Perimeter of ABC = 6 cm, AB = 2 cm, and AC = 2 cm. K-12 tests, GED math test, basic math tests, geometry tests, algebra tests. Vertically opposite angles are congruent. What is the ratio of the areas of the two quadrilaterals ABCD to DEFG? BASIC MATHEMATICS MATH 010 A Summary of Concepts Needed to be Successful in Mathematics The following sheets list the key concepts that are taught in the specified math course. It has neither width nor thickness. Angles: An angle is the union of two non-collinear rays with a common origin. Click the button below to download the full Mathematics Form 2 Topical Revision Questions and Answers pdf document, with all the topics. A. Your email address will not be published. The pairs of interior angles thus formed are supplementary. Right triangle: If any of a triangle is a right angle i.e., 90° then the triangle is a right-angled triangle. A line, which intersects two or more given coplanar lines in distinct points, is called a transversal of the given lines. What is the ratio of the length of PQ to that of QO? The book will capture the essence of mathematics. The end point is called the origin. Parallel lines: Two lines, which lie in a plane and do not intersect, are called parallel lines. Alternate angles: In the above figure, 3 and 3, 2 and 8 are Alternative angles. To help you with this, we have written various articles to help you understand not only the concepts but also some new tricks and shortcut formulas. A polygon with 10 sides is called a decagon. A polygon with 9 sides is called a nonagon. Congruent angles: Two angles are said to be congruent, denoted by if it divides the interior of the angle into two angles of equal measure. Mathematics Form 2 Topical Revision Questions and Answers (20) Practise with these elaborate topic by topic Mathematics questions with their marking schemes. Each question will ask you to select an answer from among four choices. Answer: Charles Babbage Question: What is the name of first personal computer? The questions are mostly based on the simple geometry concepts or theorems which we all have gone through in our high school textbooks. ∠2+∠5-∠3 + ∠8 = 180°. Mathematicians are pattern hunters who search for hidden relationships. Question 1: Find equation of the perpendicular bisector of segment joining the points (2,-5) and (0,7)? However, to have an upper edge, one should also be comfortable with using the shortcut formulas and tricks to solve the questions quickly. In ΔABC, A, B and C are the vertices of the triangle; AB, BC, CA are the three sides, and ∠A, ∠B, ∠C are the three angles. * Note: the worksheet variation number is not printed with the worksheet on purpose so others cannot simply look up the answers. SSC CGL Coordinate Geometry Questions with Solutions PDF: Download SSC CGL Coordinate Geometry questions with answers PDF based on previous papers very useful for SSC CGL exams. Vertically Opposite angles: Two angles are called vertically opposite angles if their sides form two pairs of opposite rays. Perpendicular lines: Two lines, which lie in a plane and intersect each other at right angles are called perpendicular lines. Here are three simple How many sides does it have? Scalene triangle: A triangle in which none of the three sides is equal is called a scalene triangle. The skills you'll practice in this pack are essential for a strong score on any math test, and will help you reach your full academic and career potential. Online Mathematics Quiz with Answers . Enjoy your geometry test. NECO Mathematics Past Questions PDF Download, Objective, Theory ; JSS3 JUNIOR NECO Past Questions and Answers PDF Free Download ; Download JSS 1 Past Questions for 1st, 2nd, 3rd Term Free – All Subjects PDF | Upper Basic ; JSS 2 Basic Science Past Questions in PDF & Word for 1st, 2nd, 3rd Term Examination. Tough Algebra Word Problems. Where would A and C be when B reaches point B3? Add to your shopping cart and purchase a Detailed 11 PAGES SOLUTION and TOP-NOTCH EXPLANATIONS with PayPal. If the area of ∆ ABC be 2 units, then find the value (s) of k. Download Quantitative Aptitude Study Material, basic geometry problems and solutions pdf download, geometry problems with solutions and answers, geometry problems with solutions and answers for college pdf, geometry problems with solutions and answers pdf, geometry problems with solutions pdf for ssc cgl, geometry quantitative aptitude study material pdf download, geometry word problems with solutions pdf, plane geometry problems with solutions pdf, APPSC Panchayat Secretary Previous Question Papers, HPSSSB Drawing Master Final Result 2018 Post Code – 637. Is called a triangle. The Videos, Games, Quizzes and Worksheets make excellent materials for math teachers, math educators and parents. And Worksheets make excellent materials for math teachers, math educators and parents 60°... Collection of math exercises on the side JK but is not successful segment,.! An acute angle uniform speeds of u, v and w respectively group of student-produced response.... 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Racks with 10 sides is called the orthocentre with 4 sides is a. | 8,304 | 34,974 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-23 | latest | en | 0.896417 |
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Pages: 3 (869 words) · Bibliography Sources: 3 · File: .docx · Topic: Education - Mathematics
SAMPLE EXCERPT . . .
For example, it cannot be interpreted the age of the different immigrant groups, as this is not published. In addition, no conclusions can be definitively drawn about the timing of immigration flows -- those are only guessed at. In addition, there are no statistics provided in this report about the economic condition of immigrants or their settlement patterns. No conclusion can be drawn about the age of native speakers, with respect to determining the risk those languages face of extinction. The age of native speakers can be reasonably guessed, but not on the basis of the data provided in this report.
4. These statistics could impact our perception of certain topics by delivering facts about the subject. By providing accurate information, the basis is formed for the reader of the statistics to understand the facts surrounding an issue. Many topics become either politicized or subject to erroneous assumptions. Both can be countered with the use of facts. There is often a gap between perception and reality, but by understanding the reality a new and more accurate perception can be created. This will benefit anybody studying these issues, as they are able to separate out the facts from the perceptions more easily.
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5. There are a number of predictions for the future that can be made using these statistics. Such conclusions can be drawn in particular if the figures from the previous census are also made available. With the 2001 figures, trends can be determined in the populations of different ethnic groups. This can assist with a number of public functions in particular, such as English or French as a second language provision and other public service provisions. If trends on ethnic diversity and language use are known, then stakeholders can better understand the ethic makeup of Canada going forward, allowing for better decisions both in terms of public policy and commerce.
TOPIC: Essay on Statistics Being Studied Were the Assignment
The figures also include statistics on age, which is critical for both government and commerce. The degree to which the Canadian population is aging is worth understanding because of the public policy implications as well. On a general level, any line of information contained within the demographic report can be extrapolated into the future to understand the demographic trends within the country.
Works Cited:
StatsCan. (2006). Selected demographic, cultural, educational, labor force and income characteristics. Statistics Canada. Retrieved May 23, 2011 from http://www12.statcan.gc.ca/census-recensement/2006/dp-pd/tbt/Rp-eng.cfm?LANG=E&APATH=3&DETAIL=0&DIM=0&FL=A&FREE=0&GC=0&GID=0&GK=0&GRP=1&PID=99016&PRID=0&PTYPE=88971,97154&S=0&SHOWALL=0&SUB=0&Temporal=2006&THEME=70&VID=0&VNAMEE=&VNAMEF= [END OF PREVIEW] . . . READ MORE
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difficulty
A golf ball rolls with constant velocity on the surface of a table 2.5 m above the ground. When it reaches the edge, it falls off the table as if horizontally launched so that at 0.4 s it is at a horizontal distance of 1 m from the edge of the table. Determine:
a) What was the constant velocity of the ball while rolling on the table?
b) Would you know how to determine how far horizontally will be the ball when it hits the ground?
c) What is the distance of the ball from the ground at 0.4 s?
## Solution
Data
Height of the ball. H = 2.5 m
Horizontal distance. x = 1 m at t = 0.4 s.
Acceleration of gravity. g = 9.8 m/s2
Preliminary considerations
As described in the statement, the ball behaves as a horizontal launched projectile, so to solve this problem we will need to use the equations for this type of motion.
Resolution
a) Knowing that the ball is on x = 1 m at t = 0.4 s, we will just substitute these values into the equation of the horizontal launch to calculate the velocity of the ball in the x-axis. Remember that in this motion the body moves with uniform rectilinear motion:
b) To determine how far the ball will be when it hits the ground we must determine when the impact will occur. To do this, first we determine how long it takes using the equation for position on the y-axis and make y = 0, that is, the ball is on the ground:
Now we only have to substitute the time calculated in the equation of position for the x-axis:
c) Finally, we can determine the height above the ground the ball is located at t = 0.4 s, using of the equation of position for y for this time:
## Formulas worksheet
These are the main formulas that you must know to solve this exercise. If you are not clear about their meaning, we recommend you to check the theory in the corresponding sections. Furthermore, you will find in them, under the Formulas tab, the codes that will allow you to integrate these equations in external programs like Word or Mathematica.
Formulas
Related sections
$y=\mathrm{H}-\frac{1}{2}g{t}^{2}$
$x={x}_{0}+v\cdot t$ | 508 | 2,083 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2020-24 | longest | en | 0.89658 |
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Module by: Mary Teegarden. E-mail the author
Based on: Probability Topics: Probability Lab by Barbara Illowsky, Ph.D., Susan Dean
View the content: Probability Topics: Probability Lab (edited: Teegarden)
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Name: Probability Topics: Probability Lab (edited: Teegarden) ID: m17335 Language: English (en) Summary: This module presents students with a lab exercise allowing them to apply their understanding of Probability. Using a Minitab simulation, students will compare theoretical probabilities with experimental probabilities as it relates to dice. They will also investigate the effect of sample size. Subject: Mathematics and Statistics Keywords: elementary, exercise, homework, lab, long-term, probability, replacement, statistics License: Creative Commons Attribution License CC-BY 2.0 Authors: Copyright Holders: Maintainers: Latest version: 1.2 (history) First publication date: Aug 12, 2008 2:21 pm -0500 Last revision to module: Jul 14, 2009 6:25 pm -0500 Based On: Probability Topics: Probability Lab Originally By: Barbara Illowsky, Ph.D. (illowskybarbara@deanza.edu), Susan Dean (susanldean45@gmail.com)
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## Version History
Version: 1.2 Jul 14, 2009 6:25 pm -0500 by Changes: `Wording has been updated to help the students.` Version: 1.1 Aug 12, 2008 2:21 pm -0500 by Changes: ```Labs changed to incorporate mini-tabs. ```
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### American Chemical Society (ACS) Style Guide:
Teegarden, M. Probability Topics: Probability Lab (edited: Teegarden), Connexions Web site. http://cnx.org/content/m17335/1.2/, Jul 14, 2009.
### American Medical Assocation (AMA) Manual of Style:
Teegarden M. Probability Topics: Probability Lab (edited: Teegarden) [Connexions Web site]. July 14, 2009. Available at: http://cnx.org/content/m17335/1.2/.
### American Psychological Assocation (APA) Publication Manual:
Teegarden, M. (2009, July 14). Probability Topics: Probability Lab (edited: Teegarden). Retrieved from the Connexions Web site: http://cnx.org/content/m17335/1.2/
### Chicago Manual of Style (Bibliography):
Teegarden, Mary. "Probability Topics: Probability Lab (edited: Teegarden)." Connexions. July 14, 2009. http://cnx.org/content/m17335/1.2/.
### Chicago Manual of Style (Note):
Mary Teegarden, "Probability Topics: Probability Lab (edited: Teegarden)," Connexions, July 14, 2009, http://cnx.org/content/m17335/1.2/.
### Chicago Manual of Style (Reference, in Author-Date style):
Teegarden, M. 2009. Probability Topics: Probability Lab (edited: Teegarden). Connexions, July 14, 2009. http://cnx.org/content/m17335/1.2/.
### Modern Languages Association (MLA) Style Manual:
Teegarden, Mary. Probability Topics: Probability Lab (edited: Teegarden). Connexions. 14 July 2009 <http://cnx.org/content/m17335/1.2/>. | 1,122 | 4,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2013-48 | longest | en | 0.763121 |
https://laratutorials.com/c-program-to-find-area-of-triangle/ | 1,656,519,830,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103640328.37/warc/CC-MAIN-20220629150145-20220629180145-00769.warc.gz | 407,612,420 | 14,935 | # C Program To Find Area Of Triangle
In this tutorial, i am going to show you how to find the area of triangle formula, function, and pointer in c programs.
## All C Programs and Algorithm To Find Area Of Triangle
• Algorithm To Find Area Of Triangle
• C Program To Find Area Of Triangle using Formula
• C Program To Find Area Of Triangle using Function
• C Program To Find Area Of Triangle using Pointer
### Algorithm To Find Area Of Triangle
Follow the below given algorithm to write a program to find area of triangle; as follows:
• Start program
• Take input height and width from user and store it into variables.
• Calculate area of triangle = area=(base*height)/2;
• print “Area of Triangle=”, area
• End program
### C Program To Find Area Of Triangle using Formula
```#include<stdio.h>
int main()
{
float base,height,area;
printf("enter base of triangle: ");
scanf("%f",&base);
printf("enter height of the triangle: ");
scanf("%f",&height);
area=(base*height)/2;
printf("Area Of Triangle is: %f\n",area);
return 0;
}```
The output of the above c program; as follows:
```enter base of triangle: 10
enter height of the triangle: 25
Area Of Triangle is: 125.000000```
### C Program To Find Area Of Triangle using Function
```#include<stdio.h>
float area(float b,float h)
{
return (b*h)/2;
}
int main()
{
float b,h,a;
printf("enter base of triangle: ");
scanf("%f",&b);
printf("enter height of the triangle: ");
scanf("%f",&h);
a=area(b,h);
printf("Area Of Triangle: %f\n",a);
return 0;
}```
The output of the above c program; as follows:
```enter base of triangle: 12
enter height of the triangle: 15
Area Of Triangle: 90.000000```
### C Program To Find Area Of Triangle using Pointer
```#include<stdio.h>
void area(float *b,float *h,float *area)
{
*area=((*b)*(*h))/2;
}
int main()
{
float b,h,a;
printf("enter base of triangle: ");
scanf("%f",&b);
printf("enter height of the triangle: ");
scanf("%f",&h);
area(&b,&h,&a);
printf("Area Of Triangle: %f\n",a);
return 0;
}```
The output of the above c program; as follows:
```enter base of triangle: 10
enter height of the triangle: 12
Area Of Triangle: 60.000000``` | 569 | 2,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.580887 |
https://www.rba.gov.au/publications/rdp/2005/2005-11/appendix-a.html | 1,716,285,213,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058442.20/warc/CC-MAIN-20240521091208-20240521121208-00313.warc.gz | 827,385,216 | 9,098 | # RDP 2005-11: A Small Model of the Australian Macroeconomy: An Update Appendix A: Calculating Potential Output
The model's multivariate filtering procedure seeks the potential output series which best fits the model's unit labour cost and underlying inflation equations, subject to a smoothness criterion. This involves jointly finding the potential output series , and parameters for these two equations, which minimise the loss function
where ηt and ζt are the residuals from the unit labour cost and underlying inflation equations respectively.[40]
The estimation process is iterative. The steps are as follows:
1. Initialise potential output by taking a Hodrick–Prescott filter of the level of non-farm output over the full sample period of available quarterly data, 1959:Q3 to 2005:Q1; hence form a corresponding initial output gap series.
2. Estimate the unit labour cost and underlying inflation equations by OLS, using the current output gap series.
3. Fix the parameters in these equations at their estimated values and then re-solve for the potential output series which minimises the loss, , given by Equation (A1).
4. Repeat steps 2 and 3 in turn until convergence is achieved (that is, until changes from one iteration to the next in both the potential output series and the parameters in the unit labour cost and underlying inflation equations fall below a pre-determined tolerance threshold).
The interested reader is referred to Appendix A of Gruen, Robinson and Stone (2002) for further algebraic details on the iterative procedure for estimating potential output. While the discussion in Gruen et al relates to a filter with only one conditioning equation, the modifications required for two conditioning equations are reasonably straightforward.
A final issue concerns the role and selection of the weights in Equation (A1). The three weights control the relative importance attached, in the determination of potential output, to the fit of the unit labour cost equation, the fit of the underlying inflation equation, and the smoothness constraint. Because the inflation equation has a much better fit than the unit labour cost equation and covers a smaller sample, the former's sum of squared errors (SSE) term is much smaller than that of the latter. As a result, if the weights λI and λU were chosen to be equal, the filter would pay little attention to optimising the fit of the underlying inflation equation (relative to that of the unit labour cost equation) in conditioning potential output. To overcome this problem, we first express λI in the form , where χ is a multiplicative factor which ‘scales up’ the inflation equation SSE to be of comparable magnitude to the unit labour cost SSE.[41] We then fix values for λU and which reflect the relative importance we wish to place on the unit labour cost and underlying inflation equations, respectively, in conditioning our estimates of potential output – and which, without loss of generality, we require to sum to one.[42]
The weight λS, meanwhile, controls the importance placed on the smoothness constraint, relative to that attached to the goodness of fit of the conditioning equations. The larger is λS, the smoother will be the growth rate of potential output.We choose a value for λS (currently λS = 200) which allows for long-lived changes in the growth rate of potential output, without permitting high-frequency ‘noise’ in its level.
## Footnotes
The first two summation terms in Equation (A1) cover different periods because the samples used for estimating the two equations are different (covering n = 113 and p = 53 quarters respectively). The unit labour cost equation is estimated from 1977:Q1, while the equation for underlying inflation is estimated only from 1992:Q1. The series y* is estimated for t = −3,…,n because the unit labour cost equation allows for up to four lags of the output gap. [40]
The scaling factor is determined by the ratio of the unit labour cost SSE to the inflation equation SSE, and is continuously updated after step 2 in each iteration. [41]
Somewhat arbitrarily, these parameters were set to be λU = 0.8 and = 0.2 in generating the output gap estimates shown earlier in Figure 7. [42] | 871 | 4,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-22 | latest | en | 0.88067 |
https://studentshare.org/miscellaneous/1569256-basic-statistics-and-graphs | 1,537,553,338,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157351.3/warc/CC-MAIN-20180921170920-20180921191320-00559.warc.gz | 635,870,725 | 22,327 | Search
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# Basic Statistics and graphs - Speech or Presentation Example
Summary
The calculated mean of five data samples show that MF-5 has given the highest average return during the last ten years and MF-1 has given the lowest average return.MF-1 with 29.3 standard deviation shows the highest level of variability whereas, lowest variability is shown by…
## Extract of sampleBasic Statistics and graphs
Download file to see previous pages... The two histograms (Graph I & II) are showing different trends because maximum homes (38) sold before 5 years ago were sold between \$115,000 and \$129,000 and 35 homes were sold between \$130,000 and \$144,000. On the other hand, this year prices increased and maximum number of homes (38) was sold between \$145,000 and \$159,000. Moreover, 5 years ago, home prices range was between \$100,000 and \$159,000 whereas, this year homes have been sold at a price between \$100,000 and \$204,000.
The first similarity between these two histograms is that none of the house sold between the range of \$85,000 and \$99,000, which means home prices were higher than this range this year and even 5 years ago. Secondly, the maximum frequency in both histograms is ...Download file to see next pagesRead More
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+16312120006 | 955 | 4,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2018-39 | latest | en | 0.949103 |
http://forums.wolfram.com/mathgroup/archive/2007/Oct/msg01051.html | 1,719,215,012,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865074.62/warc/CC-MAIN-20240624052615-20240624082615-00370.warc.gz | 11,381,898 | 8,515 | rank-1 decomposition for GraphPlot vertex positioning
• To: mathgroup at smc.vnet.net
• Subject: [mg82697] rank-1 decomposition for GraphPlot vertex positioning
• From: Yaroslav Bulatov <yaroslavvb at gmail.com>
• Date: Mon, 29 Oct 2007 05:30:17 -0500 (EST)
```Here's a way to improve presentation of highly regular vertex-
transitive graphs, which might be of interest to mathgroup readers:
If you have a highly structured vertex transitive graph which doesn't
look very structured when visualized with GraphPlot, take a rank-1
decomposition of the adjacency matrix. If graphs corresponding to
resulting rank-1 matrices are isomorphic to each other, this suggests
a natural grouping of nodes. To (visually) see if they are isomorphic,
{u, d, v} = {#1, Diagonal[#2], #3} & @@ (SingularValueDecomposition[
rank1mats = Table[d[[i]]*Outer[Times, u[[All, i]], v[[All, i]]], {i,
1, 6}];
GraphPlot/@rank1mats
(warning, rank-1 decomposition could have negative valued matrices
For a graph with no self-loops, rank-1 adjacency matrix means it is
bipartite, vertex transitive within each part, and all arrows are
going in the same direction. So you could use VertexCoordinateRules to
place all nodes in a source partition of each rank-1 subgraph close
together in the original graph.
A complete example is below, you will see 3 plots -- original graph
plot, the structure suggested by rank-1 decomposition of the adjacency
matrix, and complete graph with vertices rearranged to respect that
structure. Motivation for this particular graph is here
http://yaroslavvb.blogspot.com/2007/10/ive-recently-gone-to-northwest.html
swap[l_, p_Integer] :=
Module[{ll = l}, ll[[{2 p - 1, 2 p}]] = ll[[{2 p, 2 p - 1}]]; ll];
swap[l_, p_List] :=
Fold[swap[#1, #2] &, l, Position[p, True] // Flatten]
thorp[l_] := Module[{k, ll}, k = Length[l]/2;
ll = Riffle[l[[;; k]], l[[k + 1 ;;]]];
FromDigits[swap[ll, #]] & /@ Tuples[{False, True}, k]]
thorpEdges =
Thread[FromDigits[#] -> thorp[#]] & /@ Permutations[Range[4]] //
Flatten;
graph1 = GraphPlot[thorpEdges, DirectedEdges -> True];
edges2mat[edges_, size_] :=
Module[{perms = Permutations[Range[size]]},
Array[If[
MemberQ[edges,
FromDigits[perms[[#1]]] -> FromDigits[perms[[#2]]]], 1,
0] &, {Length[perms], Length[perms]}]];
thorpMat = edges2mat[thorpEdges, 4];
{u, d, v} = {#1, Diagonal[#2], #3} & @@ (SingularValueDecomposition[
thorpMat]);
rank1mats =
Table[d[[i]]*Outer[Times, u[[All, i]], v[[All, i]]], {i, 1, 6}];
outNodes[mat_] :=
MemberQ[#, 1] & /@ mat // Position[#, True] & // Flatten;
partitions = outNodes[#] & /@ rank1mats;
Module[{l = Length[partitions]},
Table[Unitize[
l}, {j, 1, l}]];
graph2 = mergeGraph[thorpMat, partitions] //
GraphPlot[#, DirectedEdges -> True] &;
p2 = Permutations[partitions][[66]];
rules[i_] :=
Table[{Cos[i 2 Pi/6], Sin[i 2 Pi/6]} +
2/20 {Cos[j 2 Pi/4], Sin[j 2 Pi/4]}, {j, 1, 4}]];
allrules = rules[#] & /@ Range[6] // Flatten;
graph3 = GraphPlot[thorpMat, DirectedEdges -> True,
VertexCoordinateRules -> allrules];
GraphicsColumn[{graph1, graph2, graph3}]
```
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two of the electrical cables, or different causes. The fourth and least happening kind of deficiency is the reasonable three stage, which can happen by a contact between the three electrical cables in various structures. Figure 1: Types of Transmission Line Faults
Effects of Ground Resistivity and Tower Structural Design on Transmission Line Symmetrical Components
techniques to achieve target post Installation performance requirements , , . The electrical parameters that defines the transmission line are power ratings, propagation constants, shunt admittances, impedances, and characteristic impedances , . Efficiency and
Behaviour of Transmission Lines under Tornadoes Based on WindEEE Testing
REFERENCES American Society of Civil Engineers, Guidelines for electrical transmission line structural loading, ASCE manuals and reports on engineering practice, 20 No. 74 (3rd ed.). GL Baker, and CR Church, Measurements
Mamdani Fuzzy Expert System Based Directional Relaying Approach for Six-Phase Transmission Line .
Although 6-phase line offers more benefits to electrical power system operations, the fault protection task in 6-phase line is more difficult than in double circuit 3-phase transmission lines. It can cause permanent damage to the transmission lines Abstract Series compensation is generally used with long transmission lines to increase power transfer through the line and to enhance system stability. In the case of voltage source converter-based series flexible AC transmission system (FACTS) controller, an energy Because of the small electrical size of the DNG structures at the frequencies of interest, their the negative effective permeability can be achieved by etching series capacitive gaps in the host line the presence of SRRs or CSRRs coupled to the host lines, a transmission zero can The insulation of the power transmission system experiences stresses due to power frequency voltage under normal operating conditions, temporary overvoltages at or near power frequency and transient overvoltages arising either from switching operations and
Transmission Lines and Cable
the line . Every transmission line will have three basic electrical parameters. The conductors This phenomena of electrical discharge occurring in transmission line for high values of voltage is known as the corona effect in power system
ELECTROMAGNETIC SWITCHING TRANSIENTS IN TRANSMISSION LINE COOPERATING WITH THE LOCAL SUBSYSTEM
180 189 181 determined and the most severe conditions can be modeled without risk and with very flexible possibility to determine the performance of electrical equipment in the high voltage line under all possible fault conditions in transmission system
of Zagreb, Faculty of Electrical Engineering and Computing, Zagreb, Croatia 3Koncar-Power Transformers Ltd., Zagreb, Croatia 4Croatian Transmission
Figure 1: Electrical circuit configuration for frequency response check of the Koncar TMS+ measuring system 7 (b). Two more cases are observed in this paper and shown in Figure 8. Case 2 represents a double-phase to ground short-circuit on transmission line while case 3Different models are defined to study electrical power system quality and micro macro grid control functions reliability 2018) abstracts the methodology for detection, clas- sification and localization of transmission line faults using synchronous phase measurements
Application of EIS and transmission line model to study the effect of arrangement of graphene on electromagnetic shielding and cathodic protection
rich coatings were conductive coatings and possessed low electrical resistance, so the reason should be the second one. Based on this, the zinc particles in the coatings were divided into active and inactive ones , and corresponding transmission line model was established
Electrical Power Systems Resilience by Switching of Power Transmission Lines State of art
This is an open access article under the CC BY-NC-SA 4.0 license (https://creativecommons. org/licenses/by-nc-sa/4.0/). Portal de revistas: http://revistas.utp.ac.pa Electrical Power Systems Resilience by Switching of Power Transmission Lines State of art
Aeolian vibrations of overhead transmission line bundled conductors during indoor testing, Part A: Validation of excitation technique
transmission line bundled conductors YD KUBELWA1* AG SWANSON1 DG DORRELL2 1Vibration Research and Testing Centre, School of Engineering, University of KwaZulu-Natal, South Africa 2School of Electrical and Information Engineering, University of the
EFFECT OF SEISMIC SOIL-STRUCTURE-INTERACTION ON TRANSMISSION TOWER
Abstract Transmission tower is a 3-D truss and indeterminate structure, which plays key role in the operation of a reliable electrical power system. Transmission line tower has to support conductors carrying heavy electrical power and more than one ground wires at appropriate
Optimal Control Strategy to Alleviate Line Congestion in Power System using Bus Power Rescheduling
lines. In the competitive market environment electrical utilities try to operate the transmission lines near to their stability and loadability margin operator. By this technique effective utilization of the available transmission line is possible
vol. no. pp. 68-71; 2004. Z.Cvetkovic, S.Aleksic, B.Nikolic. Response Analysis on Nonuniform Transmission Line Serbian Journal Of Electrical Engineering, vol. no. pp. 173-180, Nov. M.Sengül. Broadband
Novel Dual-band Bandpass-to-Bandstop Filter Using Shunt PIN Switches Loaded on the Transmission Line
single-band or dual-band, adopt series switches in the main transmission line (TL) to IL introduced by switches in the bandstop mode; 2) generating an extra transmission zero (TZ Two pairs of stepped impedance resonators (SIRs) with different electrical length are used to form
M. Tech. Electrical Engineering programme
EE 615 Control of Electrical Drives 3-0-0-3 for electronic systems, Governmental Requirements, Additional Product Requirements, Design Constraints for Products, Advantages of EMC Design, Transmission Lines and Signal Integrity, The Transmission Line Equations, The Per Maezawa, K.: Experimental observation of synchronised oscillating edges in electrical lattice with series-connected tunnel diodes. Electron. Lett. 5 14 16 (2019). 37. Narahara, K., and Maezawa, K.: Large-amplitude voltage edge oscillating in a transmission line with regularly
Sag and Tension Evaluation of a 330kV Overhead Transmission Line Network for Upland and Level land Topographies.
Fernandez, E. A method for the sag-tension calculation in electrical overhead lines. International Review of Electrical EngineeringVol Qureshi, SS and Ahmed, S. Temperature and Wind Impacts on Sag and Tension of AAAC overhead transmission line . International Journalneural network, Journal of Electrical and Electronic Engineering, vol. no. pp [18] Huan, VP, An ANFIS based approach to improve the fault location on 110kV transmission line Dak Mil-Dak Nong, International Journal of Computer Science Issues (IJCSI),11(3), pp. overhead power transmission line ; touch voltage; potential of grounding device; equipotential zone; equipotential coupling. The labor protection rules provide for measures en- suring safe maintenance and repair of overhead transmis- sions lines (OTL), electrical equipment 1 Department of Electrical Engineering, GH Raisoni Institute of Engineering and Technology, Nagpur, MH, India 2 Department of Electrical Engineering, National Though various feature extraction techniques have been used in the literature with regard to transmission line
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The maximum height that the vegetation can be to the electrical cables is 7 meters. The transmission line is 65 km long and it covers 14 municipalities in the state of Rio Grande do Sul, southern Brazil (Figure 1). The region is part of the Guaíba river basin and it presents an
ADVANCED HYBRID CONTROL TECHNIQUES APPLIED ON THE AVR-PSS TO ENHANCE DYNAMICS PERFORMANCES OF ELECTRICAL NETWORKS
system (SMIB) was considered . It consists of a single synchronous generator (turbo-Alternator) connected through a parallel transmission line to a x1 is the speed deviation and x2 is accelerating power, Pm and Pe represents respectively the mechanical and electrical powerThe drain-induced barrier lowering (DIBL) mechanism and the transmission line method (TLM) were used to investigate this abnormal degradation of the TLM revealed a channel resistance of 10.4 kΩ μm at a small gate bias of 0.5 V. Degradation of the electrical properties was
OVERHEAD POWER LINES USING EVOLUTIONARY ALGORITHMS
(Examiner) Assistant Prof. of Power and Electrical Machines Signature Different methods of mitigation of the magnetic fields near the overhead transmission line are reported over the last 20 years such as manipulated distances, underground cables and shielding with loops
Electrical power system restoration planning in southern area of Lao PDR
References Southern area transmission system monitoring office, Transmission Line and Substation Monitoring Department, EDL. Single Line Diagram. International Journal of Electrical Power Energy Systems, Jan. 2018; (287-299)
Investigation of Electrical Properties in AB-Stacked Bilayer Graphene-DNA Nanostructures
Furthermore, dotted-blue line illustrates the local density of states (LDOS) of atom c in the second electrons are confined same as in these two sites with no significant changes. Figure 6 shows the electrical transmission as a function of the incident elec
Modeling and Analyses the Equivalent-Schema Models for OSRR and COSRR Coupled to Planar Transmission Lines by Scattering Bond Graph
Department of Electrical Engineering Laboratory of Analysis and Command of the Systems Department of Physics Faculty of vol. no. pp. 459 46 2012. I. Salem, H. Taghouti, and A. Mami, Modeling CSRR and OCSRR loaded transmission line by bond graphIranian Journal of Science and Technology, Transactions of Electrical Engineering 123 Page 4. icðtÞ ¼ Im cosðxpstÞ cos xt þ h þ 4p 3 ð7Þ Introduction of TCSC in the transmission line affects the value of ZTline which in turn affects values of the three phase currents As is well known to electrical engineers, in a low loss overhead transmission line the capacitive charging current flowing into an open-ended long- transmission line will cause the voltage at the end of the line to increase. This is referred to as the Ferranti effect (Steinmetz 1971) In Embu County, the Kenya Power and lightening Company Ltd (KPLC) constructed a 21kms 132 kV electrical transmission line from Kyeni to Embu. This transmission line is a T-off of the Ishiara 132kV transmission line . The The layout of transmission tower for optimizing maximum deflection the basic parameters are forced on the basis of electrical necessities reference of DBSonowal1 JDBharali2 MK Agarwalla N. Sarma P. Hazarika Analysis and Design of 220 kV Transmission Line Tower(A
Faults Detection and Classification on Parallel Transmission Lines using Modified Clarkes Transformation-ANN Approach
Vol No 2 (2013), pp. 762-768. 2013. Sudha Gopal, Valluvan K. R, A Novel Approach to Fault Diagnosis of Transmission Line with Rogowski Coil, International Review of Electrical Engineering. Vol No pp. 656-662 the dust pollutants that deposited and accumulated on the surface of line insulators, combined then in recent years, and with more long-distance high-voltage transmission Laboratory of Advanced Electromagnetic Engineering and Technology, School of Electrical and Electronic
Transmission Line and Substation Project Environmental Assessment and Site Analysis | 3,872 | 18,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.879647 |
https://www.mrexcel.com/board/threads/insanely-dumb-question-multiplying-a-sum-of-several-cells.696184/ | 1,696,076,537,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510676.40/warc/CC-MAIN-20230930113949-20230930143949-00484.warc.gz | 981,813,936 | 21,250 | # Insanely dumb question... multiplying a sum of several cells
#### Jennifre
##### Board Regular
Sorry to be so... daft, but I can't use a formula to multiply one number by summed cells.
You'll be able to see what I was trying by this below, but it won't work. Putting the *65 inside the parenthesis didn't work either. I am slow to learn Excel's formula language.
=SUM(E30:E41)*65
Thanks so much for your help -- this site's forum is the best regarding questions I ask online, by far! Thank you.
### Excel Facts
Format cells as date
Select range and press Ctrl+Shift+3 to format cells as date. (Shift 3 is the # sign which sort of looks like a small calendar).
Hi Jennifre
That formula is how I would multiply the sum of E30:E41 by 65. Can you provide some sample data and state what result you would expect so we can understand what it is you want to do.
What result do you get that you don't want? Is it an error value (if so, are there any errors in E30:E41)?
Do you definitely have numeric values in E30:E41? Point an ISNUMBER formula at one of the cells to confirm eg:
=ISNUMBER(E30)
and report back the result
Does the SUM work on it's own?
Can you describe "it Won't work" ?
Do you get an error?
Do you get the wrong results? What result did you actually get, what result did you expect and why?
It seems to me that
(E30*65)+(E31*65)+(E32*65)
Yiels the same result as
SUM(E30:E31)*65
Excel Workbook
EFG
30865520
31265130
32365195
3376654940
34565325
35
3661106110
Sheet1
Sorry to be unclear. I've got this formula in my target cell, yet all that appears there upon hitting enter at the end of the formula, is the formula itself, not the result of E30:341 x 65.
The sum alone works as that's appearing successfully in a cell above the cell with this formula in it.
Thanks!
Sorry to be unclear. I've got this formula in my target cell, yet all that appears there upon hitting enter at the end of the formula, is the formula itself, not the result of E30:341 x 65.
The sum alone does work as it is appearing successfully in a cell above the cell with this formula in it.
Also, the =ISNUMBER(30) reports, "TRUE".
Thanks!
Last edited:
Is this cell formatted as Text? Right-click on it and choose Format Cells and ensure it is specified as General
Last edited:
.
Last edited:
Format the cell as General or Number (anything other than Text)
Then RE-Enter the formula
That worked, goodness! And, thank you so much.
Not sure why typing it out earlier would have rendered as anything but general. Is it recommended to normally format the entire sheet as general before you start?
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Go back | 950 | 3,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.952067 |
http://math.stackexchange.com/questions/121032/classifying-some-abelian-groups-of-order-25-times-35 | 1,469,730,116,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828313.74/warc/CC-MAIN-20160723071028-00162-ip-10-185-27-174.ec2.internal.warc.gz | 166,615,330 | 19,520 | # Classifying some abelian groups of order $2^5\times 3^5$
I'm requested to classify the abelian groups $A$ of order $2^5 \times 3^5$ where :
• $| A/A^4 | = 2^4$
• $|A/A^3 | = 3^4$
I need to write down the canonical form of each group .
My question is, what does it mean $| A/A^4 |$ ? I understand that the meaning is to: $$G⁄H= \{Hg \mid g\in G\}$$
but the usage of it for classifying the groups is not clear to me.
How does it help me with this case ?
Regards
EDIT:
Suppose that I say $A=B \times C$ where $|B|=2^5$ and $|C|=3^5$ and $|A|=2^5 * 3^5$ .
Now , the goals are :
• $|A|/|A|^3 = (2^5 * 3^5) /|A|^3 = 3^4$
:meaning I need to make $|A^3|=3*2^5$
• $|A|/|A|^4 = (2^5 * 3^5) /|A|^4 = 2^4$
:meaning I need to make $|A^4|=2*3^5$
So now after A=BC , how can I find the exact $|A^4|$ and $|A^3|$ ?
thanks again
-
No, your goals are $|C|/|C^3| = 3^4$ and $|B|/|B^4| = 2^4$. Now, write $B$ as a direct sum of cyclic $2$-groups (that is, cyclic groups of order a power of $2$), and see what $B^4$ is; write $C$ as a direct sum of cyclic groups of order a power of $3$, and see what $C^3$ is. You need $|B^4| = 2$, and you need $|C^3| = 3$ – Arturo Magidin Mar 18 '12 at 3:45
@ArturoMagidin: But how did you get to $|C/C^3|$ and $|B/B^4|$ ? What I mean is , how do you know what is $|A^3|$ and $|A^4|$ ? – ron Mar 18 '12 at 5:15
As was already explained, $A = B\times C$, so $A^k = B^k\times C^k$ for any $k$. Since the order of $C$ is relatively prime to $2$, then $C^4 = C$, so $A/A^4 = (B/B^4)\times (C/C^4) = (B/B^4)\times (C/C) \cong B/B^4$. And since the order of $B$ is relatively prime to $3$, then $B^3=B$, so again we have $A/A^3 = (B/B^3)\times (C/C^3) \cong C/C^3$. If there was no simplification in considering $B$ and $C$, then what's the point of introducing them in the first place? If you didn't understand why you should introduce them, then you should have asked, not just put them in and then ignore them. – Arturo Magidin Mar 18 '12 at 5:21
I lost you with $C^4 = C$ , can you please explain ? – ron Mar 18 '12 at 5:23
Look at the hint I wrote in my answer; the one you've been studiously not trying to prove. But even more generally: if the order of $g$ is $n$, and $k$ is relatively prime to $n$, then $g$ is a $k$th power: write $1 = an+bk$ for some integers $a$ and $b$. Then $g = g^1 = g^{an+bk} = (g^n)^a(g^b)^k = 1^a(g^b)^k = (g^b)^k$. Since every element of $C$ has order relatively prime to $4$, every element of $C$ is the fourth power of someone in $C$. That means that $C\subseteq C^4$; since $C^4\subseteq C$ always holds, you get equality. – Arturo Magidin Mar 18 '12 at 5:26
For any abelian group $A$, written multiplicatively, $A^n = \{a^n\mid a\in A\}$ is a subgroup (since $a^n(b^n)^{-1} = (ab^{-1})^n$ and $1^n=1\in A^n$). Therefore, we can talk about the quotient $A/A^n$, and we can talk about its size as a set. $|A/A^4|$ is the size of the group $A/A^4$, and $|A/A^3|$ is the size of the group $A/A^3$.
Let $A$ be an abelian group of order $2^5\times 3^5$. Then we know, from the Fundamental Theorem of Finitely Generated Abelian Groups, that we can express $A$ uniquely as $$A \cong \mathbb{Z}_{2^{a_1}}\oplus\cdots \oplus\mathbb{Z}_{2^{a_k}} \oplus \mathbb{Z}_{3^{b_1}}\oplus\cdots\oplus\mathbb{Z}_{3^{b_{\ell}}},$$ where $1\leq a_1\leq\cdots\leq a_k$, $1\leq b_1\leq\cdots\leq b_{\ell}$, and $a_1+a_2+\cdots+a_k = 5$, $b_1+b_2+\cdots+b_{\ell} = 5$.
So for instance, one possibility might be $$A \cong \mathbb{Z}_2 \oplus\mathbb{Z}_{2^4} \oplus \mathbb{Z}_{3^2}\oplus\mathbb{Z}_{3^3}.$$ But we want only some of the groups, namely, the ones where we also have $|A/A^4| = 2^4$ and $|A/A^3|=3^4$. Does this group satisfy this condition?
Well, since $(B\oplus C)^n = B^n\oplus C^n$, we can deal with each cyclic factor separately. Note that $(\mathbb{Z}_2)^4 = \{1\}$, and $(\mathbb{Z}_{2^4})^4 = \mathbb{Z}_{2^2}$; whereas, since $4$ is relatively prime to $3$, $(\mathbb{Z}_{3^2})^4 = \mathbb{Z}_{3^2}$, $(\mathbb{Z}_{3^3})^4 = \mathbb{Z}_{3^3}$.
So we have: \begin{align*} A &= \mathbb{Z}_2 \oplus \mathbb{Z}_{2^4} \oplus \mathbb{Z}_{3^2}\oplus\mathbb{Z}_{3^3}\\ A^4 &= (\mathbb{Z}_2)^4 \oplus (\mathbb{Z}_{2^4})^4 \oplus (\mathbb{Z}_{3^2})^4 \oplus (\mathbb{Z}_{3^3})^4\\ &\cong \{1\} \oplus \mathbb{Z}_{2^2} \oplus \mathbb{Z}_{3^2}\oplus \mathbb{Z}_{3^3}. \end{align*} Therefore, since we are dealing with finite groups, $$\left|\frac{A}{A^4}\right| = \frac{|A|}{|A^4|} = \frac{2^5\times 3^5}{2^2\times 3^5} = 2^3,$$ so this $A$ does not satisfy the condition.
You are asked to find all the ones that do.
Hint. Prove that:
• If $p$ and $q$ are primes, $p\neq q$, then $(\mathbb{Z}_{p^a})^{q^b} = \mathbb{Z}_{p^a}$;
• If $p$ is a prime, and $a\leq b$, then $(\mathbb{Z}_{p^a})^{p^b} = \{1\}$.
• If $p$ is a prime, and $a\gt b$, then $(\mathbb{Z}_{p^a})^{p^b}\cong \mathbb{Z}_{p^{a-b}}$.
-
I need to have some thinking regarding this question , it requires a lot of theory that I don't fully have at the moment . I hope I'd some answers in the next few hours . thanks ! – ron Mar 17 '12 at 6:03
@ron: If you can establish the properties I listed in the HINT, then it turns out to be a fairly straighforward proposition. But you really need to get your hands "dirty" on this one and play around with a few examples of groups to see how things play out. – Arturo Magidin Mar 17 '12 at 20:37
I have some calculations that I've made . Where can I write them ? here in the comments , or re-edit my original post? – ron Mar 18 '12 at 1:38
@ron: Comments are difficult; adding them to the original post (without modifying the question) would work. – Arturo Magidin Mar 18 '12 at 2:19
I've added the changes , not much but I think it would do . But I'm still missing something there... – ron Mar 18 '12 at 3:44 | 2,239 | 5,773 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2016-30 | latest | en | 0.918353 |
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# The mechanics of standing balance
## Problem
A simple model of human standing is given by an inverted pendulum, a one-dimensional, classical model for the motion of a single, massive particle (Figure 1). A standing human is approximated as a single mass (located at the center of mass) separated from the ground by a massless rod of fixed length $L$. The feet are treated as attached to the ground by a fulcrum, such that the center of mass can only undergo motion along an arc of radius $L$ around the feet---thus the mechanics of the system are described by the tilt angle $\theta$, representing the angular displacement of the center of mass from directly above the feet, with the rod positioned normal to the ground. The pendulum is perfectly balanced when the mass is positioned directly above the pivot point; however, a very slight displacement of the mass from this position will cause the pendulum to tip over.
Figure 1. The analogy between a standing patient and an inverted pendulum.
Humans can overcome this difficulty and maintain standing balance, at which the system is at equilibrium, by maintaining active control of the position of their center of mass relative to the fulcrum formed by their feet. For small displacements, the ankles work to exert a torque that counteracts gravity and prevents the individual from falling over. The timescale that this restoring force must act to recover equilibrium is proportional to the period of a simple pendulum, demonstrated by Equation 1.
T= 2π$\sqrt{\frac{L}{g}}$
Equation 1.
Precise measurements of active balancing can be made by filming an individual and digitally tracking the location of the center of mass after the individual is tilted forward by a known angular displacement at $t=0$. A sample figure showing the angular response is given in Figure 2. Surprisingly, these measurements show that the seemingly inert process of standing consists of many coordinated ankle motions that stabilize the body after it undergoes slight deflections.
Figure 2. Simulated data showing the time-varying tilt angle of a standing individual who was tilted forward by $0.2$ radians at $t=0$, and who is actively returning to their original vertical standing position.
Concept adapted from: Winter, D. A. (1995). Human balance and posture control during standing and walking. Gait & Posture, 3(4), 193-214.
Which of the following correctly describe the difference between an inverted pendulum (as shown in in Figure 1) and a standard pendulum with identical length and mass?
I. An inverted pendulum has minimal kinetic energy when it reaches its equilibrium point
II. An inverted pendulum requires a non-gravitational restoring force to remain in equilibrium
III. An inverted pendulum reaches a higher maximum gravitational torque | 624 | 3,009 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.925598 |
https://metanumbers.com/5318 | 1,723,773,213,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641319057.20/warc/CC-MAIN-20240815235528-20240816025528-00636.warc.gz | 297,303,065 | 7,417 | # 5318 (number)
5318 is an even four-digits composite number following 5317 and preceding 5319. In scientific notation, it is written as 5.318 × 103. The sum of its digits is 17. It has a total of 2 prime factors and 4 positive divisors. There are 2,658 positive integers (up to 5318) that are relatively prime to 5318.
## Basic properties
• Is Prime? no
• Number parity even
• Number length 4
• Sum of Digits 17
• Digital Root 8
## Name
Name five thousand three hundred eighteen
## Notation
Scientific notation 5.318 × 103 5.318 × 103
## Prime Factorization of 5318
Prime Factorization 2 × 2659
Composite number
Distinct Factors Total Factors Radical ω 2 Total number of distinct prime factors Ω 2 Total number of prime factors rad 5318 Product of the distinct prime numbers λ 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 5318 is 2 × 2659. Since it has a total of 2 prime factors, 5318 is a composite number.
## Divisors of 5318
1, 2, 2659, 5318
4 divisors
Even divisors 2 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ 4 Total number of the positive divisors of n σ 7980 Sum of all the positive divisors of n s 2662 Sum of the proper positive divisors of n A 1995 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 72.9246 Returns the nth root of the product of n divisors H 2.66566 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 5318 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 5318) is 7980, the average is 1995.
## Other Arithmetic Functions (n = 5318)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ 2658 Total number of positive integers not greater than n that are coprime to n λ 2658 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 710 Total number of primes less than or equal to n r2 0 The number of ways n can be represented as the sum of 2 squares
There are 2,658 positive integers (less than 5318) that are coprime with 5318. And there are approximately 710 prime numbers less than or equal to 5318.
## Divisibility of 5318
m n mod m
2 0
3 2
4 2
5 3
6 2
7 5
8 6
9 8
The number 5318 is divisible by 2.
## Classification of 5318
• Arithmetic
• Semiprime
• Deficient
### Expressible via specific sums
• Polite
• Non hypotenuse
• Square Free
## Base conversion 5318
Base System Value
2 Binary 1010011000110
3 Ternary 21021222
4 Quaternary 1103012
5 Quinary 132233
6 Senary 40342
8 Octal 12306
10 Decimal 5318
12 Duodecimal 30b2
20 Vigesimal d5i
36 Base36 43q
## Basic calculations (n = 5318)
### Multiplication
n×y
n×2 10636 15954 21272 26590
### Division
n÷y
n÷2 2659 1772.67 1329.5 1063.6
### Exponentiation
ny
n2 28281124 150399017432 799821974703376 4253453261472553568
### Nth Root
y√n
2√n 72.9246 17.4548 8.53959 5.56096
## 5318 as geometric shapes
### Circle
Diameter 10636 33414 8.88478e+07
### Sphere
Volume 6.2999e+11 3.55391e+08 33414
### Square
Length = n
Perimeter 21272 2.82811e+07 7520.79
### Cube
Length = n
Surface area 1.69687e+08 1.50399e+11 9211.05
### Equilateral Triangle
Length = n
Perimeter 15954 1.22461e+07 4605.52
### Triangular Pyramid
Length = n
Surface area 4.89843e+07 1.77247e+10 4342.13
## Cryptographic Hash Functions
md5 820a8f5c40c91fbd63f19519314ca277 a8782e82e16d00d00e35482889e0b2bb38ca598c 96f89b80b1a221811c9a377a316f591e94b2f685ef9ef077e62fd52929ab20a0 e58ca6b7593dac1e26eaeeb85857c9e4aefd5d5289b2349459352bf0cb9d1b88c9bf908aa1ae5147c38c466865f38af67c7b963b96110bd0cef7d05b8e9cb5f1 37da04b12e549c262588b32fede1621cc902f9a1 | 1,390 | 3,947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-33 | latest | en | 0.818047 |
https://www.aqua-calc.com/one-to-all/temperature/preset/kelvin/107.65 | 1,674,850,152,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764495012.84/warc/CC-MAIN-20230127195946-20230127225946-00834.warc.gz | 647,731,262 | 5,781 | # Convert Kelvins [K] to other units of temperature
## Kelvins [K] temperature conversions
107.65 K = -165.5 Celsius K to °C 107.65 K = -265.9 Fahrenheit K to °F 107.65 K = 193.77 degrees Rankine K to °R 107.65 K = -79.3875 Romer K to °Rø
Convert entered temperature to units of energy.
#### Foods, Nutrients and Calories
PASTEURIZED PREPARED CHEESE PRODUCTS, UPC: 041497085705 contain(s) 333 calories per 100 grams (≈3.53 ounces) [ price ]
11394 foods that contain Vitamin A, RAE. List of these foods starting with the highest contents of Vitamin A, RAE and the lowest contents of Vitamin A, RAE, and Recommended Dietary Allowances (RDAs) for vitamin A are given as mcg of Retinol Activity Equivalents (RAE)
#### Gravels, Substances and Oils
CaribSea, Freshwater, African Cichlid Mix, Rift Lake Authentic weighs 1 505.74 kg/m³ (94.00028 lb/ft³) with specific gravity of 1.50574 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Phosphonium iodide [H4IP] weighs 2 860 kg/m³ (178.54397 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-417A, liquid (R417A) with temperature in the range of -30°C (-22°F) to 60°C (140°F)
#### Weights and Measurements
A mile per kilowatt hour (mi/kWh) measures how much distance in miles, a car can travel on 1 kilowatt hour of electricity;
The surface density of a two-dimensional object, also known as area or areal density, is defined as the mass of the object per unit of area.
gf-cm to N⋅mm conversion table, gf-cm to N⋅mm unit converter or convert between all units of torque measurement.
#### Calculators
Calculate volume of a dam and its surface area | 530 | 1,934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-06 | longest | en | 0.795842 |
glenfiddichresidency.blogspot.com | 1,532,365,223,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676599291.24/warc/CC-MAIN-20180723164955-20180723184955-00466.warc.gz | 155,949,033 | 13,665 | ## Saturday, July 5, 2008
### Day 10
Hi Dave,
I hope you are doing well and enjoying yourself in Scotland. First, my
apologies for this very lengthy email but I felt it necessary to give an
explanation. Secondly, if you don't want to read any of the justification
then start reading from "------------||" onwards.
Thanks for the clarification. As stated, the problem does not have a
unique solution. You can make the loops arbitrarily short. I will try to
explain:
Loops L1,...,L4.
Each has a period (the time each takes to complete one cycle) T1,...,T4.
We can always order the periods as, T1 < T2 < T3 < T4
Note: we have assumed the periods are all different. This can be
generalized (we won't do that here) to assuming that at least two of the
periods are different (if all the periods are the same then they will
repeat after one cycle which then requires us to have this one cycle be
100 years).
Let m = 100 years = 876000 hrs
Suppose L4 requires n4 cycles to occur in order for m hrs to elapse, then
n4*T4 = m.
Note: n4 has to be a positive integer.
We want all of the loops to repeat after m hrs, this implies that
L1,...,L4 will have each completed a certain number of full cycles,
n1,...,n4. So we get:
n1*T1 = n2*T2 = n3*T3 = n4*T4 = m (1)
Equation (1) ensures that L1,...,L4 will repeat after m hrs.
Now the tricky part. Consider loops L3 and L4. We want to avoid having
them repeat any earlier than m hrs. For example, suppose we had 6*T3 =
4*T4 = m which means that L3 completes 6 cycles and L4 completes 4 cycles
then repeat after m hrs. However, 2 is common factor that still gives an
integer number of cycles. Divide by 2 gives, 3*T3 = 2*T4 = m/2 which
means that in this case L3 and L4 actually FIRST repeat after m/2 hrs (L3
completes 3 cycles and L4 completes 2 cycles). One way of to avoid
having any repetitions occur before m hrs is to require that n3 and n4 be
prime numbers (numbers whose only divisors are 1 and itself).
Note: There is another way, by simply requiring that n3 and n4 be
relatively prime. This means that their only common divisor is 1, for
example 10 and 21 are relatively prime. Here, we shall choose n1,...,n4
to be only prime numbers.
Observation: Since T1 < T2 < T3 < T4 it follows from equation (1) that
n1 > n2 > n3 > n4. This just says that L1 will cycle more times than L2
etc... because it had a shorter period.
--------------------------------------||
Here is the procedure for determining the periods, T1,...,T4 of each loop:
1)Choose the number of cycles of L4 to be some large prime number n4.
Then T4 = m/n4 gives the period of L4.
2)Choose another prime n3 > n4. Then T3 = m/n3 gives the period of L3.
3)Choose another prime n2 > n3. Then T2 = m/n2 gives the period of L2.
4)Choose the last prime n1 > n2. Then T1 = m/n1 gives the period of L1.
Note: In steps 3) and 4) if n1 and n2 are not prime numbers then L1 and L2
might repeat before m hrs, however L3 and L4 won't repeat until m hrs.
An Example determining the periods of each loop so that there is no
repetition till m = 100 years = 876000 hrs
1)Choose n4 = 1299709 (100,000 th prime number).
Then T4 = 876000/1299709 hrs = .6739970255 hrs = 40.43982153 mins
(approximately)
2)Choose n3 = 2750159 (200,000 th prime number).
Then T3 = 876000/2750159 hrs = .3185270379 hrs = 19.11162227 mins (approx.)
3)Choose n2 = 4256233 (300,000 th prime number).
Then T2 = 876000/4256233 hrs = .2058158000 hrs = 12.34894800 mins (approx.)
4)Choose n1 = 5800079 (400,000 th prime number).
Then T1 = 876000/5800079 hrs = .1510324256 hrs = 9.061945536 mins (approx.)
You can see that by simply choosing a larger prime number n4 we can make
the period of L4 as small as we like, that is T4 can be made arbitrarily
small. Since T1 < T2 < T3 < T4 then the other periods will be even
shorter. I chose the above prime numbers just to get periods, T1,...,T4
under 1 hour, you can use the procedure to get times that are more
suitable for the piece.
Last comment: The above decimal numbers are all approximate. In reality,
it doesn't seem likely that one can achieve the precision necessary in
the periods of the loops in order to guarantee that no repetition will
occur till 100 years. Even if all above numbers for periods happen to be
whole numbers, I don't know if you can create a real loop that is say
exactly 10 mins. What would happen if the period of each loop was not
exactly as above? Then I just don't know.
Hope this is helpful somehow. Let me know if there is anything I can
clarify.
Take care,
Nicos. | 1,305 | 4,533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.94793 |
https://www.mail-archive.com/everything-list@googlegroups.com/msg26220.html | 1,531,981,774,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590559.95/warc/CC-MAIN-20180719051224-20180719071224-00496.warc.gz | 939,023,462 | 11,599 | # Re: Free will in MWI
On May 17, 9:50 am, Bruno Marchal <marc...@ulb.ac.be> wrote:
> On 17 May 2012, at 14:21, Craig Weinberg wrote:
>
> > On May 17, 5:49 am, Bruno Marchal <marc...@ulb.ac.be> wrote:
> >> On 16 May 2012, at 17:37, Craig Weinberg wrote:
>
> >>> On May 16, 10:41 am, Bruno Marchal <marc...@ulb.ac.be> wrote:
> >>>> On 15 May 2012, at 19:44, Craig Weinberg wrote:
>
> >>>>> On May 15, 1:03 pm, Bruno Marchal <marc...@ulb.ac.be> wrote:
>
> >>>>>>>> But a deterministic world, if rich enough to add and multiply,
> >>>>>>>> and
> >>>>>>>> indeterminist first person realities (even without comp,
> >>>>>>>> although
> >>>>>>>> it
> >>>>>>>> is simpler to use comp to justify this).
>
> >>>>>>> If a wave washes one pile of sand onto another, thereby 'adding'
> >>>>>>> them
> >>>>>>> together, why does that generate universal internal observers?
>
> >>>>>> Adding is not enough. You need multiplication, and iteration.
>
> >>>>>> Then universal digital creatures appear, by logical consequences,
> >>>>>> and,
> >>>>>> as always, reflect themselves and all universal creatures,
> >>>>>> digital,
> >>>>>> and non digital, which leads them to harder and harder problems
> >>>>>> and
> >>>>>> questions.
>
> >>>>> Even if that's true, from where do they appear? To say they appear
> >>>>> is
> >>>>> to admit that they are not themselves contained within addition or
> >>>>> multiplication.
>
> >>>> They are. Anything Turing emulable appears, and reappears in
> >>>> arithmetic, related to bigger and bigger natural numbers.
>
> >>> The appearance is contingent though, upon something being able to
> >>> recognize the pattern which is appearing to them.
>
> >> That's correct. It is contingent of the universal number, and the
> >> universal numbers making the first one more relatively probable. But
> >> all that exist in arithmetic.
>
> > What are the properties of arithmetic contingent on?
>
> The idea is that such properties are not contingent.
That's still an idea though, ie sense. Sense doesn't need that
property since it can't be explained any other way. I can explain
arithmetic sense as a category of sense, but I can't explain sense as
a category of arithmetic unless you just tack it on and say it must be
part of the package inherently.
>
> You could take any universal system, instead of arithmetic. From the
> computability perspective, they are equivalent.
You can run over anything with a large enough steam roller and it will
be flat. If you don't use a computability perspective, they aren't
equivalent.
>
>
>
> >>> That pattern
> >>> recognition is not automatically guaranteed by any arithmetic logic.
>
> >> In your non-comp theory.
>
> >>> We need a physical machine that remembers that it can remember,
>
> >> That's "Bp -> BBp". Universal machine are like that.
>
> > Those are just letters and symbols. What or who makes them mean
> > something and why?
>
> Bp means that some universal machine utters p. Absolutely.
> Independently of you and me.
But not independently of the universal machine's sense-motive
experience. It has to be able to tell the difference between p and
something else and characterize the nature of that difference. It has
to have the motive power to 'utter', and something has to have the
sense receptivity to detect that something might have been uttered.
Otherwise there is no uttering.
> BBp means that the same universal machine now utters Bp.
> For any arithmetic (or equivalent) proposition, Bp > BBp, means that
> if that machine utters p, it will soon or later utters Bp.
So if I utter 'Toast is square', that means that eventually I will
utter 'I utter Toast is square' and then 'I utter I utter I utter I
utter Toast is squalre'?
> And that is
> a theorem of arithmetic, making it true independently of you and me.
I never argue that sense is dependent on human consciousness at all.
Sense is universal and literally older than time itself.
>
>
>
> >>> and
> >>> can experience that memory as an event. It needs to know what
> >>> kinds of
> >>> strings of remembered digits constitute a meaningful pattern, or
> >>> that
> >>> there could even be such a thing as a pattern. To say that patterns
> >>> appear and reappear in arithmetic takes the appearance of pattern
> >>> itself for granted, then usurps the primacy of the sense experience
> >>> which provides it.
>
> >> Not really, for it appears and reappears only in the mind of
> >> universal
> >> numbers. It makes sense for them, and indeed they will be astonished
> >> that apparent material can lead to that sense. But although locally
> >> true, this is globally wrong. Sense is necessarily a first person
> >> notion, and relies on the abstract but real configuration involving
> >> infinities of arithmetical relations.
>
> > I don't think sense is a first person notion, it is the very capacity
> > to define first person and third person as separate (opposite) on one
> > level, and united on another. Sense creates the arithmetical
> > relations, but not infinitely. Arithmetical relations are derived a
> > posteriori of sense embodiments.
>
> You confuse arithmetic and the human's apprehension of arithmetic.
Not at all. You are assuming that arithmetic is conceivable outside of
some kind of sense faculty and I don't see any reason to agree with
that. It doesn't have to be human apprehension at all, it could be
anything from a single atom to the totality of all mass-energy of the
cosmos as a single unit...or even some other sensible-but-real entity
beyond our ability to conceive through human sense. All of it has to
make sense in some way to some thing. Something has to detect
something.
>
> > Sense generates the capacities,
> > intentions, symmetries, and rhythms that underlie recursive
> > enumeration, as well as frames the context of all sequence and
> > consequence. It all has to make sense.
>
> We need only the idea that a reality can exists beyond human sensing.
I'm fine with that, but no reality can exist beyond sense. Realism is
nothing but a category of sense.
> This is what I assume by making explicit the arithmetical realism, and
> that can be shown enough when we assume that we work locally as
> machine, at some description level.
Its circular reasoning though. If I assume I'm a machine, then I
define everything I do as being mechanical. So what? If I define
myself as a spirit, then I define the universe as a spiritual journey.
What's the difference? They are both equally tautological.
> As I already told you, to make this false, you need to build an
> explicit non computable and non Turing recoverable function having a
> genuine role for the mind.
I don't need to build it, I am living in it already, you just aren't
admitting that it is the case.
> This unfortunately only makes more complex
> both mind and matter, making your non-comp hypothesis looking like a
> construct for making impossible to reason in that field.
I would not even say I have a non-comp hypothesis, I have a meta-comp
conjecture :)
I don't disagree that it might make it impossible to reason in that
field, and because of that, we need other inside-out and upside-down
models (anthropomorphic, mechanemorphic, logomorphic, technemorphic)
to get to our blind spot, but that doesn't change the absolute truth
value of the sense model that encompasses them all, as well as the
relations among them.
>
> > Not everything has to make
> > numbers. Dizzy doesn't make numbers, but it makes sense.
>
> But numbers does not make only numbers. They make and develop sense
> for many things far more complex than numbers, that is the point.
How does it follow from numbers though that they necessarily develop
anything at all? You are suggesting that bytes are alive and do things
on their own, yet we have never seen that to be the case nor does it
make intuitive sense. If that were true, we should see that Bugs Bunny
is having new adventures behind our back on 60 year old celluloid
reels by now. The internet would be haunted by autonomous entities
that we should be looking for like SETI.
> Arithmetical truth itself is far beyond of numbers,
Why should that be and how could that be the case? At what point can
numbers no longer tolerate being numbers and suddenly become...what?
>From where?
> yet numbers can
> relatively develop some intuition about those kind of things.
> You just seems stuck in a reductionist conception of numbers and
> machines. We know such conception are wrong.
You confuse your conception of numbers with the reality of (non-human)
sense in general.
>
> > It is a
> > sensation that makes sense to an embodied animal, but not to a
> > computer.
>
> How could we know that? Why should we believe that?
Because we know that we have different channels of sense and we know
that it is not necessary for a computer to have multiple sense
channels, and that in fact, all data must be compiled into a one
dimensional binary stream. Our senses multiply the richness of our
experience, and even simple sensations like a circle quickly invite
imaginative elaboration. If a person is dizzy, they will complain. A
computer will never complain even if it is inside of a washing machine
that never turns off.
>
>
>
> >>>>> To say they are creatures implies a creation.
>
> >>>> Why not. You could say that they are created by the addition and
> >>>> multiplication laws. You need only to bet that 1+1=2 and alike does
> >>>> not depend on us.
>
> >>> Because there's no mathematical logic to how or why that creation
> >>> could occur.
>
> >> But there is.
>
> > What is it?
>
> That the existence of universal numbers, and their many dreams, is a
> consequence of logic and arithmetic.
Which is a consequence of sense and motive.
>
>
>
> >>> If we posit a universe of arithmetic realism, how can we
> >>> accept that it falls off a cliff when it comes to the arithmetic of
> >>> it's own origins? What makes 1+1=2? Sense.
>
> >> Truth.
>
> > Truth requires sense.
>
> Why?
How can something be determined to be true without something else
making sense of it as being true? It's like asking why water can't be
completely dehydrated and still feel wet.
>
> > Not everything that makes sense is true (fiction
> > for example), but everything that is true makes sense.
>
> For who?
For anyone or anything that can in some way experience it as true.
>
>
>
> >> Why do you want someone to assess the truth for something being
> >> true. That is anthropomorphic.
>
> > It's ontologically necessary. What is a truth without it being
> > detectable in some way to something?
>
> It is an unknown truth.
Unknown to us, but not unknown to its own context.
> A billion digit numbers can be prime without
> us being able to know it.
Sure, but if nothing is ever able to know it, then it isn't something
real, it's only an idea of what could be real.
> Some universal machine does not stop on some
> argument without anyone being able to prove or know it. Some pebble on
> some far away planet can be eroded without anyone knowing it.
Yes, I'm not talking about human knowledge. My hypothesis is
panexperiential. We see a pebble but what it is without us is a group
of atoms holding onto each other. It could be a purely tactile-kinetic-
acoustic awareness, or it could be an omniscient state of zen
paralysis. Maybe they experience something only when the status of
that holding changes, so a billion years goes by in ten seconds to
them, who knows. Maybe the pebble is only a fragment of star and the
whole solar system is the entity that lives a billion years in each
second. Lots of possibilities we can't even imagine...
>
>
>
> >> Th greek get well that point, and
> >> originate the whole scientific enterprise from there, as in the
> >> conclusion of this video:
>
>
> > Great video, but now you are the one anthropomorphizing. Just because
> > the released man doesn't create the outside world by seeing it doesn't
> > mean that the outside world can exist without being held together by
> > experienced sense relations on every level. My computer doesn't create
> > the internet, but that doesn't mean that the internet isn't created on
> > computers.
>
> But where the first observer come from?
"First" and "come from" are aspects of observerness. Observer is
primordial and absolute (totality/singularity).
>
>
>
> >> If not, it is the whole idea of a reality which makes no more sense,
> >> and we get solipsist or anthropomorphic.
>
> > That's where sense comes in. Sense divides the totality into
> > solipsistic/anthropomorphic and objective/mechanemorphic on one level,
> > but bleeds through that division on another level, thus creating a
> > diffracted continuum that oscillates through time but remains
> > continuous across space (and vice versa).
>
> Time and space, looks concrete, thanks to millions years of evolution,
> but are much more sophisticated notion than elementary addition and
> multiplication to me.
I use time and space to keep it simple. It is really the sense of
continuity and oscillating discontinuity itself which, when multiplied
by many subjects experiencing themselves objectively, gives rise to
the abstractions of space and time.
>
> > Numbers are a synthetic
> > analysis of that process, distilled to a nearly meaningless but nearly
> > omnipotent extreme of universality (qualitative flatness). Numbers are
> > the opposite of the solipsistic personal experience (qualitative depth
> > asymptotic to 'Selfness' itself). They are the least appropriate tools
> > to describe feeling.
>
> Atoms, fields, space, time seems as much.
I agree, but the ability to experience any of them, including numbers,
is more primitive.
>
>
>
> >>> Not primitive sense either,
> >>> but high order cognitive abstraction. There is no '1' or '2'
> >>> literally, they are ideas about our common sense - what we have in
> >>> common with everything. Numbers are literally 'figures', symbols
> >>> which
> >>> can be applied mentally to represent many things,
>
> >> No. That's number description. Not numbers.
>
> > I'm not talking about the characters "1" or "2", I'm talking about
> > what they represent. The concept of numbers defines them as figurative
> > entities, but you make them literal. That's ok with me if you are
> > doing that for mathematical purposes since it is a powerful way to
> > approach it, through the negative symmetry, but just as you might
> > trace a picture better if it's upside down, eventually you should turn
> > it right side up when you finish. To say that numbers literally exist
> > but matter does not is the logo-morphic position, orthogonal to both
> > anthropomorphic and mechanemorphic, but it is still as pathologically
> > unreal if taken literally. Again, thats ok with me, we need
> > surrealists too, I'm just saying, when the rubber hits the road, it's
> > not sanity.
>
> Hmm...
>
>
>
> >>> and to deploy
> >>> orderly control of some physical systems - but not everything can be
> >>> reduced to or controlled by numbers.
>
> >> But that's what number can discover by themselves.
>
> > In your logopomorphic theory of comp.
>
> Be polite!
>
> :)
Hah. I wasn't trying to be pejorative, just saying that your view
makes sense in my view but my view doesn't make sense in yours.
>
>
>
> >> Once you are at the
> >> treshold of numbers, the complexity of the relations (even just
> >> between numbers) get higher than what you can describe with numbers.
> >> the numbers already know that, with reasonable account of what is
> >> knowledge.
>
> > If the complexity exceeds the capacity of numbers, then you need to
> > invoke even more complexity in the form of additional forms of
> > expression of that complexity...out of thin air?
>
> It develops from intuition.
That's sense!
> Numbers, relatively to universal numbers,
> can develop intuition, due to the true relation existing between
> numbers, including the truth that they cannot rationally justified. So
> it comes from truth.
Truth goes along with what I'm trying to say about quanta being the
flattest and most universal qualia. Absolute truth means true for all
entities on all sense channels, so that necessarily requires that it
be absolutely flat qualitatively (otherwise you are dependent upon
some particular category of conditions, making it a relative truth
rather than absolute).
Under this criteria, numbers are an excellent candidate for universal
truth - almost. Numbers are so qualitatively flat that they act like a
skeleton key, sliding into every form and structure, but it also makes
it too easy to mistake the user of the key for the key itself since
the flattening dis-qualifies non-arithmetic realities. This is what
counting is; an abstraction layer which we use to identify or mention
*that* things are, but it doesn't address the actual experience of
what it is to be presented with those things. We count five apples but
the number five tells us nothing about apples.
What the logomorphic perspective does is invite an elevation of truth
values and universality at the direct expense of qualitatively rich
experience and specificity. It amputates the protocol stack of humans,
animals, organisms, chemicals, even physics and leaves only a
mathematical stump. The assumption is that using the splinters of the
stump, we must be able to build the entire tree, but what keeps
happening is that we get only a Turing Frankentree and splinters in
our hands. The danger is that rather than seeing this a sign to
understand the tree as a unique top-down event in the cosmos as well
as a bottom up assembled machine, we become even more fascinated by
the challenge of transmuting AI gold from leaden code and pursue it
even more avidly and obsessively. This is what is going on in Big
Physics (mechanemorphism) now as well, and in fundamentalist revivals
(Big Religion, anthropomorphism) around the world and Big Business
(technemorphism). All four points on the compass are hyperextended
into pathology until unity can be reconciled.
>
> > With sense,
> > complexity is generated recursively from bottom up entropy, while
> > simplicity pulls from the top down toward unity as significance.
> > Evolution is the interference pattern between them.
>
> >>>>> What
> >>>>> necessary logic turns a nuclear chain reaction (addition and
> >>>>> multiplication) into a nursery for problem solving sentience?
>
> >>>> The same logic making tiny system Turing universal. Usually some
> >>>> small
> >>>> part of classical logic is enough.
>
> >>> Why would any kind of universality or logic entail the automatic
> >>> development of sentience? What is logical about sentience?
>
> >> The illogicality of sentience. From the point of view of numbers,
> >> when
> >> they look at themselves, they discover, for logical reason, that
> >> there
> >> is something non logical about them.
>
> > If there is something non logical about numbers (which are really the
> > embodiment of pure logic), why does that truth have to be 'discovered'
> > by them?
>
> Because truth extend logics, and number are constrained by truth,
> before what they can believe.
I get that truth extends logic, and that numbers are constrained by
truth (which I say is lowest common denominator sense) but I don't get
the last part. Why does truth have to discover itself?
>
> > In our development as children, do we not discover logic out
> > of the chaos of infancy rather than the other way around? Do we not
> > learn numbers rather than learn feelings?
>
> Because we have brains which sum up millions years of teaching in nine
> month, making us believe that walking and seeing is simpler than
> trigonometry. Later we can understand that is the contrary.
You are right in one sense, but that sense doesn't exist until
'later'. Trigonometry is indeed simpler mathematics than the
mathematics underlying human walking and seeing, but the sense
underlying trigonometry is even simpler. That sense is the same common
denominator that makes us a single walking seeing person - it's the
absolute common denominator, simplicity itself - unity, totality,
wholeness, being. It makes no distinction between now and forever,
between everything and nothing. It is the greatest and least inertial
frame possible. For this not to be the case, there would have to be
something preventing it. Some limitation inherent that does not allow
everything to be one thing on some level. Sense does this temporarily,
I think literally, it does it through time.
>
>
>
> >> Then the comp act of faith
> >> appears to be the simplest way to restore logic, except for that act
> >> of faith and the belief in addition and multiplication.
>
> > What kind of faith does a Turing machine have?
>
> If she is correct, it looks like it is plotinian sort of faith. But a
> machine can also develop a faith in mechanism, by surviving back-up,
> and be led, with occam, to a more pythagorean sort of faith.
Sounds like a very Greco-Anglican faith. Where are the Vedic machines?
Craig
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To unsubscribe from this group, send email to | 4,970 | 21,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-30 | latest | en | 0.919231 |
https://link.springer.com/article/10.1007/s13222-017-0264-7 | 1,726,123,630,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651422.16/warc/CC-MAIN-20240912043139-20240912073139-00764.warc.gz | 337,038,780 | 72,156 | ## 1 Introduction
The problem of lifting rankings on objects to ranking on sets has been studied from many different view points — see [2] for an excellent survey. Several properties (also called axioms) have been proposed in order to indicate whether the lifted ranking reflects the given order on the elements. Two important axioms are dominance and independence. Roughly speaking, dominance ensures that adding an element which is better (worse) than all elements in a set, makes the augmented set better (worse) than the original one. Independence, on the other hand, states that adding an element $$a$$ to sets $$A$$ and $$B$$ where $$A$$ is already known to be preferred over $$B$$, must not make $$B\cup\{a\}$$ be preferred over $$A\cup\{a\}$$ (or, in the strict variant, $$A\cup\{a\}$$ should remain preferred over $$B\cup\{a\}$$). These axioms were first considered together in the context of decision making under complete uncertainty [9]. There, sets represent the (mutually exclusive) possible outcomes of an action and one tries to rank these sets based on a preference ranking on the outcomes. It is assumed that the probability of each outcome is unknown, i. e., it is only known whether an event is a possible outcome or not. This is a very reductive model. Still, “it does succeed in modelling some empirically interesting situations” [5, p. 2]. Especially, “when the number of possible states of the world is large, an agent of bounded rationality may be incapable of undertaking (or unwilling to undertake) the complex calculations which consideration of the entire rows in the outcome matrix will involve.” [15, p. 2]. Such situations often occur for autonomous agents, for example self driving cars, where “the temporal evolution of situations cannot be predicted without uncertainty because other road users behave stochastically and their goals and plans cannot be measured” [6, p. 1]. Moreover, dominance and independence are also sensible axioms in other contexts, for example bundles of objects of unknown size. Finally, to mention a very different application, Packard [14] used independence and a version of dominance to define plausibility rankings on theories.
However, it is well known that constructing a ranking on the whole power set of objects which jointly satisfies dominance and (strict) independence is, in general, not possible.
### Example 1
Consider the problem of assigning tasks to agents. Let $$X=\{t_{1},\dots,t_{n}\}$$ be a collection of tasks. Furthermore, assume we know for every agent what tasks they prefer to perform. If there are more tasks than agents, some agents have to perform several tasks, therefore it would be useful to know the preferences over sets of tasks. However, asking for these preferences directly is infeasible even for a reasonable small number of tasks. Therefore, we would like to lift the preferences over tasks to preferences over sets. Furthermore, it seems reasonable that the order on the sets should satisfy dominance and (strict) independence. Unfortunately, for strict independence, this is impossible even for $$n=3$$. Assume $$t_{1}<t_{2}<t_{3}$$. Then, $$\{t_{1}\}\prec\{t_{1},t_{2}\}$$ is implied by dominance, therefore $$\{t_{1},t_{3}\}\prec\{t_{1},t_{2},t_{3}\}$$ must hold by strict independence. On the other hand, $$\{t_{2},t_{3}\}\prec\{t_{3}\}$$ is also implied by dominance, therefore, $$\{t_{1},t_{2},t_{3}\}\prec\{t_{1},t_{3}\}$$ by strict independence. We thus end up with $$\{t_{1},t_{2},t_{3}\}\prec\{t_{1},t_{3}\}\prec\{t_{1},t_{2},t_{3}\}$$, hence $$\prec$$ is not an order.
Because of this, other (weaker) axiomatizations were proposed (see for example [7], or more recently, [4] and [10] among many others). However, in many applications one does not need to order the entire power set (for example, some tasks cannot be performed in parallel). In these cases, it may be possible to construct rankings that jointly satisfy dominance and (strict) independence.
### Example 2
Let $$X$$ be as above. Now assume $$\{t_{1},t_{2},t_{3}\}$$ is not a possible combination of tasks, for example, because fulfilling all three tasks at once is not feasible. Then, for example
$$\{t_{1}\}\prec\{t_{1},t_{2}\}\prec\{t_{2}\}\prec\{t_{1},t_{3}\}\prec\{t_{2},t_{3}\}\prec\{t_{3}\}$$
is a total order that satisfies dominance and strict independence (respecting the underlying linear order $$t_{1}<t_{2}<t_{3}$$).
In this paper, we investigate exactly this situation, i. e., lifting rankings to specific sets of elements. In the literature, this scenario seems to be rather neglected, so far. The only exception we are aware of deals with subsets of a fixed cardinality [3]. In particular, we are interested in the complexity of computing, if possible, rankings on arbitrary subsets of the power set that satisfy dominance and (strict) independence. To do so, we first give a new definition for dominance which appears more suitable in such a setting (for more details, see Section 3). Then, we consider the following problem: Given a ranking on elements, and a set $$S$$ of sets of elements, does there exist a strict (partial) order on $$S$$ that satisfies $$D$$ and $$I$$ (where $$D$$ is either standard dominance or our notion of dominance and $$I$$ is independence or strict independence)? We show that the problem is either trivial or easy to solve for the case of partial orders. Our main result is NP-completeness for the case when total orders are required.
The remainder of the paper is organized as follows. In the next section, we recall some basic concepts. In Section 3 we discuss why standard dominance can be seen as too weak in our setting and propose an alternative definition. Section 4 contains our main results. We conclude the paper in Section 5 with a summary and pointers to future work.
This paper is an extended version of [11].
## 2 Background
The formal framework we want to consider in the following consists of a finiteFootnote 1 nonempty set $$X$$, equipped with a linear order $$<$$ and a subset $$\mathcal{X}\subseteq\mathcal{P}(X) \backslash \{\emptyset\}$$ of the power set of $$X$$ not containing the empty set. We want to find a binary relation $$\prec$$ on $$\mathcal{X}$$ that satisfies some niceness conditions. We will consider several kinds of relations. We recall the relevant definitions.
### Definition 1
A binary relation is called a strict partial order, if it is irreflexive and transitive. A strict or linear order is a total strict partial order. A binary relation is called a preorder, if it is reflexive and transitive. A (weak) order is a total preorder. If $$\preceq$$ is a weak or a preorder on a set $$X$$, for all $$x,y\in X$$, the corresponding strict order $$\prec$$ is defined by $$x\prec y$$ if $$x\preceq y$$ and $$y\not\preceq x$$ hold.
Additionally, we need the following notions:
### Definition 2
For a pre- or weak order $$\preceq$$, we write $$x\sim y$$ if $$x\preceq y$$ and $$y\preceq x$$ hold. Let $$A\in\mathcal{X}$$ be a set of elements of $$X$$. Then we write $$\max(A)$$ for the maximal element of $$A$$ with respect to $$<$$ and $$\min(A)$$ for the minimal element of $$A$$ with respect to $$<$$. Furthermore, we say a relation $$R$$ on a set $$\mathcal{X}$$ extends a relation $$S$$ on $$\mathcal{X}$$ if $$xSy$$ implies $$xRy$$ for all $$x,y\in\mathcal{X}$$. Finally, we say a relation $$R$$ on $$\mathcal{X}$$ is the transitive closure of a relation $$S$$ on $$\mathcal{X}$$ if the existence of a sequence $$x_{1}Sx_{2}S\dots Sx_{k}$$ implies $$x_{1}Rx_{k}$$ for all $$x_{1},x_{k}\in\mathcal{X}$$ and $$R$$ is the smallest relation with this property. We write $$\mathit{trcl}(S)$$ for the transitive closure of $$S$$.
Many different axioms a good order should satisfy are discussed in the literature (an overview over the relevant interpretations and the corresponding axioms can be found in the survey [2]). The following axioms “have very plausible intuitive interpretations” [2, p. 11] for decision making under complete uncertainty and belong to the most extensively studied ones. (We added conditions of the form $$X\in\mathcal{X}$$ that are not necessary if $$\mathcal{X}=\mathcal{P}(X)\backslash\{\emptyset\}$$ holds.)
### Axiom 1 (Extension Rule)
For all $$x,y\in X$$, such that $$\{x\},\{y\}\in\mathcal{X}$$:
$$x<y\text{ implies }\{x\}\prec\{y\}.$$
### Axiom 2 (Dominance)
For all $$A\in\mathcal{X}$$ and all $$x\in X$$, such that $$A\cup\{x\}\in\mathcal{X}$$:
\begin{aligned} & y<x\text{ for all }y\in A\text{ implies }A\prec A\cup\{x\};\\ & x<y\text{ for all }y\in A\text{ implies }A\cup\{x\}\prec A.\end{aligned}
### Axiom 3 (Independence)
For all $$A,B\in\mathcal{X}$$ and for all $$x\in X\backslash(A\cup B)$$, such that $$A\cup\{x\},B\cup\{x\}\in\mathcal{X}$$:
$$A\prec B\text{ implies }A\cup\{x\}\preceq B\cup\{x\}.$$
### Axiom 4 (Strict Independence)
For all $$A,B\in\mathcal{X}$$ and for all $$x\in X\backslash(A\cup B)$$, such that $$A\cup\{x\},B\cup\{x\}\in\mathcal{X}$$:
$$A\prec B\text{ implies }A\cup\{x\}\prec B\cup\{x\}.$$
### Example 3
Take $$X=\{1,2,3,4\}$$ with the usual linear order and
$$\mathcal{X}=\{\{3\},\{4\},\{1,3\},\{2,3\},\{1,4\},\{1,2,3\},\{1,3,4\}\}.$$
Then the extension rule implies $$\{3\}\prec\{4\}$$, dominance implies $$\{1,3\}\prec\{1,3,4\}$$, $$\{1,2,3\}\prec\{2,3\}\prec\{3\}$$ and $$\{1,4\}\prec\{4\}$$ but not $$\{3\}\prec\{4\}$$. Furthermore, (strict) independence lifts the preference between $$\{2,3\}$$ and $$\{3\}$$ to $$\{1,2,3\}$$ and $$\{1,3\}$$, i. e., in combination with dominance, independence implies $$\{1,2,3\}\preceq\{1,3\}$$ and strict independence implies $$\{1,2,3\}\prec\{1,3\}$$.
Every reasonable order should satisfy the extension rule. If we assume $$\mathcal{X}=\mathcal{P}(X)\backslash\{\emptyset\}$$, the extension rule is implied by dominance [2]. Therefore, a natural task is to find a total order on $$\mathcal{P}(X)\backslash\{\emptyset\}$$ that satisfies dominance together with (some version of) independence. However, in their seminal paper [9], Kannai and Peleg have shown that this is impossible for regular independence and dominance if $$|X|\geq 6$$ and $$\mathcal{X}=\mathcal{P}(X)\backslash\{\emptyset\}$$ hold. Barberà and Pattanaik [1] showed that for strict independence and dominance this is impossible even for $$|X|\geq 3$$ and $$\mathcal{X}=\mathcal{P}(X)\backslash\{\emptyset\}$$ (see Example 1 for a proof of the statement).
If we abandon the condition $$\mathcal{X}=\mathcal{P}(X)\backslash\{\emptyset\}$$, the situation is not as clear. As we have seen in Example 2 there are sets $$\mathcal{X}\subseteq\mathcal{P}(X)\backslash\{\emptyset\}$$ with $$|X|\geq 3$$ such that there is an order on $$\mathcal{X}$$ satisfying strict independence and dominance.
## 3 A Stronger Form of Dominance
Many results regularly used in the setting of $$\mathcal{X}=\mathcal{P}(X)\backslash\{\emptyset\}$$ are not true in the more general case. For example, in contrast to the result stated above, the extension rule is not implied by dominance as we have seen in Example 3. Furthermore, it could be argued that $$\{1,3\}\prec\{1,4\}$$ should hold in that example which would be implied by dominance and independence if $$\{3,4\}\in\mathcal{X}$$ would hold, because $$\{3,4\}\prec\{4\}$$ holds by dominance and so $$\{1,3,4\}\preceq\{1,4\}$$ by independence. Hence, $$\{1,3\}\prec\{1,3,4\}\preceq\{1,4\}$$ implies $$\{1,3\}\prec\{1,4\}$$ by transitivity.
Furthermore, for the set $$X$$ from Example 3, dominance does not even imply $$\{1\}\prec\{1,2,3\}$$ if $$\{1,2\}$$ is not in the family. Therefore, it is reasonable to ask for a stronger version of dominance that behaves nicely in the general case. We observe that $$x<y$$ for all $$y\in A$$ implies $$\max(A\cup\{x\})=\max(A)$$ and $$\min(A\cup\{x\})<\min(A)$$; whereas $$y<x$$ for all $$y\in A$$ implies $$\max(A)<\max(A\cup\{x\})$$ and $$\min(A\cup\{x\})=\min(A)$$. We claim that every dominance-like axiom should satisfy this property. Therefore, we can use this property to define a “maximal” version of dominance which can be seen as a special case of Pareto dominance [12].
### Axiom 5 (Maximal Dominance)
For all $$A,B\in\mathcal{X}$$,
\begin{aligned} & \left(\max(A)\leq\max(B)\land\min(A)<\min(B)\right)\text{ or }\\ & \left(\max(A)<\max(B)\land\min(A)\leq\min(B)\right)\text{ implies }A\prec B.\end{aligned}
This axiom trivially implies the extension rule and of course dominance. Looking again at the family introduced in Example 3 maximal dominance implies all preferences implied by either dominance or by the extension rule and additionally $$\{1,3\}\prec\{1,4\}$$. Furthermore, if $$\mathcal{X}$$ is sufficiently large or even $$\mathcal{X}=\mathcal{P}(X)\backslash\{\emptyset\}$$, dominance and independence imply maximal dominance.
### Proposition 1
Let $$\mathcal{X}=\mathcal{P}(X)\backslash\{\emptyset\}$$ . Then every transitive relation that satisfies dominance and independence also satisfies maximal dominance and independence.
### Proof
Let $$\preceq$$ be a transitive relation that satisfies dominance and independence. We show that $$\preceq$$ satisfies maximal dominance using the following observation due to Kannai and Peleg in [9]:
Observation $$A\sim\{\min(A),\max(A)\}$$.
We can assume w.l.o.g. that $$A$$ has more than two elements. We enumerate $$A$$ by $$A=\{a_{1},a_{2},\dots,a_{k}\}$$ such that $$a_{i}<a_{j}$$ holds for all $$i<j$$. Using transitivity and dominance, it is easy to see that $$\{a_{1}\}\prec\{a_{1},a_{2},\dots,a_{k-1}\}$$ holds. This implies, by independence, $$\{a_{1},a_{k}\}\preceq A$$. Analogously, we get $$\{a_{2},a_{2},\dots,a_{k}\}\prec\{a_{k}\}$$ and $$A\preceq\{a_{1},a_{k}\}$$ and therefore $$A\sim\{a_{1},a_{k}\}=\{\min(A),\max(A)\}$$.◊
Using this observation we can prove that $$\max(A)=\max(B)$$ and $$\min(A)<\min(B)$$ implies $$A\prec B$$ by the following argument:
\begin{aligned} A & \sim\{\min(A),\max(A)\}\sim\{\min(A),\min(B),\max(A)\}\\ & \prec\{\min(B),\max(A)\}=\{\min(B),\max(B)\}\sim B.\end{aligned}
The other case is proven analogously, hence $$\preceq$$ satisfies maximal dominance.□
It would be possible to define several other versions of dominance of intermediate strength. We will only consider dominance and maximal dominance. As we will see our results justify this approach; in particular, since both versions yield equal complexity results.
## 4 Main Results
We studied 8 problems in total, as defined below.Footnote 2 Our results are summarized in Table 1.
### Problem 1 (The Partial (Maximal) Dominance Strict Independence problem)
Given a linearly ordered set $$X$$ and a set $$\mathcal{X}\subseteq\mathcal{P}(X)\backslash\{\emptyset\}$$, decide if there is a partial order $$\prec$$ on $$\mathcal{X}$$ satisfying (maximal) dominance and strict independence.
### Problem 2 (The Partial (Maximal) Dominance Independence problem)
Given a linearly ordered set $$X$$ and a set $$\mathcal{X}\subseteq\mathcal{P}(X)\backslash\{\emptyset\}$$, decide if there is a preorder $$\preceq$$ on $$\mathcal{X}$$ satisfying (maximal) dominance and independence.
### Problem 3 (The (Maximal) Dominance Strict Independence problem)
Given a linearly ordered set $$X$$ and a set $$\mathcal{X}\subseteq\mathcal{P}(X)\backslash\{\emptyset\}$$, decide if there is a strict total order $$\prec$$ on $$\mathcal{X}$$ satisfying (maximal) dominance and strict independence.
### Problem 4 (The (Maximal) Dominance Independence problem)
Given a linearly ordered set $$X$$ and a set $$\mathcal{X}\subseteq\mathcal{P}(X)\backslash\{\emptyset\}$$, decide if there is a total order $$\preceq$$ on $$\mathcal{X}$$ satisfying (maximal) dominance and independence.
### 4.1 Partial Orders
First, we consider the Partial (Maximal) Dominance Independence problem. We can define a preorder that satisfies independence and maximal dominance (and therefore also dominance) on all $$\mathcal{X}$$.
### Definition 3
Given a set $$X$$, a linear order $$<$$ on $$X$$ and a family $$\mathcal{X}\subseteq\mathcal{P}(X)\backslash\emptyset$$, we define a relation $$\preceq_{m}$$ as $$A\preceq_{m}B$$ iff $$\max(A)\leq\max(B)$$ and $$\min(A)\leq\min(B)$$.
Observe that it is obviously possible, given $$X$$, $$<$$ and $$\mathcal{X}$$, to construct $$\preceq_{m}$$ in polynomial time.
### Theorem 4.1
For every linearly ordered $$X$$ and every family $$\mathcal{X}\subseteq\mathcal{P}(X)\backslash\emptyset$$ , $$\preceq_{m}$$ is a preorder and satisfies maximal dominance and independence.
### Proof
Obviously, $$\preceq_{m}$$ is reflexive and transitive, because $$\leq$$ is reflexive and transitive. Furthermore, the corresponding strict order $$\prec_{m}$$ satisfies maximal dominance. Assume, w.l.o.g., $$\min(A)<\min(B)$$ and $$\max(A)\leq\max(B)$$. Then $$A\preceq_{m}B$$ by definition and $$B\not\preceq_{m}A$$ because $$\min(B)\not\leq\min(A)$$, so $$A\prec_{m}B$$. Finally, assume $$A\prec_{m}B$$ and $$A\cup\{x\},B\cup\{x\}\in\mathcal{X}$$ for $$x\not\in A\cup B$$ and, w.l.o.g., $$\min(A)<\min(B)$$ and $$\max(A)\leq\max(B)$$. If $$\min(A)<x$$ we know $$\min(A\cup\{x\})<\min(B\cup\{x\})$$ and $$\max(A\cup\{x\})\leq\max(B\cup\{x\})$$; if $$x<\min(A)$$ we get $$\min(A\cup\{x\})\leq\min(B\cup\{x\})$$, and $$\max(A\cup\{x\})\leq\max(B\cup\{x\})$$. Hence, $$A\cup\{x\}\preceq_{m}B\cup\{x\}$$.□
### Example 4
Consider once again the family from Example 3. $$\preceq_{m}$$ consists of the following preferences on that family:
\begin{aligned} \{1,3\} & \sim_{m}\{1,2,3\}\prec_{m}\{1,3,4\}\sim_{m}\{1,4\}\prec_{m}\{4\}, \\ & \{1,3\}\sim_{m}\{1,2,3\}\prec_{m}\{2,3\}\prec_{m}\{3\}\prec_{m}\{4\}. \end{aligned}
Next, we consider the Partial Dominance Strict Independence problem. As we have seen in Example 1 and 2, only some sets $$\mathcal{X}$$ allow such an order. In order to decide if a set admits a partial order we build a minimal transitive relation satisfying dominance and strict independence. First, we build a minimal transitive relation satisfying dominance. It is worth noting that a very similar relation can be defined for maximal dominance. With this relation, all results in this section can be proven for maximal dominance the same way.
### Definition 4
Given a set $$X$$, a linear order $$<$$ on $$X$$ and a family $$\mathcal{X}\subseteq\mathcal{P}(X)\backslash\emptyset$$, we define a relation $$\prec_{d}$$ on $$\mathcal{X}$$ in the following way: If $$A,A\cup\{x\}\in\mathcal{X}$$, then
1. 1.
$$A\prec_{d}A\cup\{x\}$$ if $$y<x$$ for all $$y\in A$$.
2. 2.
$$A\cup\{x\}\prec_{d}A$$ if $$x<y$$ for all $$y\in A$$.
We define the relation $$\prec_{d}^{t}$$ on $$\mathcal{X}$$ by $$\prec_{d}^{t}:=\mathit{trcl}(\prec_{d})$$.
This relation has the following useful property.
### Proposition 2
For every linearly ordered set $$X$$ and every family $$\mathcal{X}\subseteq\mathcal{P}(X)\backslash\emptyset$$ , $$\prec_{d}^{t}$$ is a partial order and a partial order on $$\mathcal{X}$$ satisfies dominance if and only if it extends $$\prec_{d}^{t}$$ .
### Proof
Obviously, $$\prec_{d}^{t}$$ is transitive. Furthermore, $$\prec_{d}^{t}$$ is irreflexive as $$A\prec_{d}^{t}B$$ implies $$\max(A)<\max(B)$$ or $$\min(A)<\min(B)$$ and $$<$$ is irreflexive.
By definition, a relation satisfies dominance if and only if it extends $$\prec_{d}$$ and a transitive relation extending $$\prec_{d}$$ also extends $$\prec_{d}^{t}$$ by the minimality of $$\mathit{trcl}$$.□
We want to extend this relation to a minimal relation for strict independence and dominance.
### Definition 5
Given a set $$X$$, a linear order $$<$$ on $$X$$ and a family $$\mathcal{X}\subseteq\mathcal{P}(X)\backslash\emptyset$$, we build a relation $$\prec_{\infty}$$ on $$\mathcal{X}$$ by induction. First, we set $$\prec_{0}^{t}:=\prec_{d}^{t}$$. Now let $$\prec_{n}^{t}$$ be defined. For $$\prec_{n+1}$$ we select sets $$A,B,A\backslash\{x\},B\backslash\{x\}\in\mathcal{X}$$ with $$x\in X$$, $$A\backslash\{x\}\prec_{n}^{t}B\backslash\{x\}$$ but not $$A\prec^{n}_{t}B$$ and set $$C\prec_{n+1}D$$ if $$C\prec_{n}^{t}D$$ or $$C=A$$ and $$D=B$$ holds. Then, we set $$\prec_{n+1}^{t}:=\mathit{trcl}(\prec_{n+1})$$. In the end, we set $$\prec_{\infty}=\bigcup_{n}\prec^{t}_{n}$$.
### Example 5
Consider the family from Example 3, i. e.,
$$\mathcal{X}=\{\{3\},\{4\},\{1,3\},\{2,3\},\{1,4\},\{1,2,3\},\{1,3,4\}\}.$$
Then, $$\prec_{\infty}$$ consists of the following preferences:
\begin{aligned} & \{1,3\}\prec_{\infty}\{1,3,4\},\\ & \{1,2,3\}\prec_{\infty}\{2,3\}\prec_{\infty}\{3\},\\ & \{1,4\}\prec_{\infty}\{4\},\\ & \{1,2,3\}\prec_{\infty}\{1,3\}.\end{aligned}
In order to prove that this is actually a minimal order for dominance and strict independence, we have to introduce another concept we call links.
### Definition 6
A $$\prec_{\infty}$$-link from $$A$$ to $$B$$ in $$\mathcal{X}$$ is a sequence $$A=:C_{0},C_{1},\dots,C_{n}:=B$$ with $$C_{i}\in\mathcal{X}$$ for all $$i\leq n$$ such that, for all $$i<n$$, either $$C_{i}\prec_{d}C_{i+1}$$ holds or there is a link between $$C_{i}\backslash\{x\}$$ and $$C_{i+1}\backslash\{x\}$$ for some $$x\in X$$.
We show that $$\prec_{\infty}$$-links indeed characterize $$\prec_{\infty}$$.
### Lemma 1
For $$A,B\in\mathcal{X}$$ , $$A\prec_{\infty}B$$ implies that there is a $$\prec_{\infty}$$ -link from $$A$$ to $$B$$ and if there is a $$\prec_{\infty}$$ -link from $$A$$ to $$B$$ then $$A\prec^{*}B$$ holds for every transitive relation $$\prec^{*}$$ that satisfies dominance and strict independence.
In order to prove this result, we need another definition.
### Definition 7
For every pair $$A\prec_{\infty}B$$, there is a minimal $$k$$ such that $$A\prec_{k}^{t}B$$ holds. We call this the $$\prec_{\infty}$$-rank of the pair.
Furthermore, we define the $$\text{rank}(C_{1},C_{2},\dots,C_{n})$$ of a $$\prec_{\infty}$$-link $$C_{1},C_{2},\dots,C_{n}$$ from $$C_{1}$$ to $$C_{n}$$:
• $$\text{rank}^{*}(C_{i},C_{i+1})=0$$ if $$C_{i}\prec_{d}C_{i+1}$$,
• $$\text{rank}^{*}(C_{i},C_{i+1})=\text{rank}(C_{i}\backslash\{x\},C_{i+1}\backslash\{x\})$$,
• $$\text{rank}(C_{1},C_{2},\dots,C_{n})=\max\{\text{rank}^{*}(C_{i},C_{i+1})\mid i<n\}+1$$.
Now we can prove Lemma 1:
### Proof
Assume $$A\prec_{\infty}B$$. We prove that a $$\prec_{\infty}$$-link exists by induction on the $$\prec_{\infty}$$-rank of $$A,B$$. If $$A\prec_{d}^{t}B$$, then there is sequence $$A=C_{1},C_{2},\dots,C_{n}=B$$ such that $$C_{i}\prec_{d}C_{i+1}$$ holds for all $$i<n$$, hence there is a $$\prec_{\infty}$$-link from $$A$$ to $$B$$. Now assume $$A,B$$ has $$\prec_{\infty}$$-rank $$k$$ and for every pair with $$\prec_{\infty}$$-rank $$k-1$$ or less there is a $$\prec_{\infty}$$-link from $$C$$ to $$D$$. There is a sequence $$A=C_{0}\prec_{k}C_{1}\dots C_{n-1}\prec_{k}C_{n}=B$$. For every $$i<n$$ either $$C_{i}\prec_{d}C_{i+1}$$ or $$C_{i}\backslash\{y\}\prec_{k-1}^{t}C_{i+1}\backslash\{y\}$$ holds, which implies by induction that there is a $$\prec_{\infty}$$-link from $$C_{i}\backslash\{y\}$$ to $$C_{i+1}\backslash\{y\}$$. Hence there is a $$\prec_{\infty}$$-link from $$A$$ to $$B$$.
Now, let $$\prec$$ be a transitive relation that satisfies dominance and strict independence and assume there is a $$\prec_{\infty}$$-link $$A=C_{1},C_{2},\dots,C_{n}=B$$ from $$A$$ to $$B$$. We prove $$A\prec B$$ by induction on the rank of the $$\prec_{\infty}$$-link. First, assume $$\text{rank}(C_{1},C_{2},\dots,C_{n})=1$$, then $$C_{i}\prec_{d}C_{i+1}$$ holds for all $$i<n$$, hence $$A\prec B$$ holds by dominance and transitivity. Now assume $$\text{rank}(C_{1},C_{2},\dots,C_{n})=k$$ and for all $$\prec_{\infty}$$-links with $$\text{rank}(C^{*}_{1},C^{*}_{2},\dots,C^{*}_{n})<k$$ we know $$C_{1}^{*}\prec C_{n}^{*}$$. By induction, for every $$i<n$$ either $$C_{i}\prec_{d}C_{i+1}$$ or $$C_{i}\backslash\{x\}\prec C_{i+1}\backslash\{x\}$$ holds. This implies that $$C_{i}\prec C_{i+1}$$ holds for all $$i<n$$, because $$\prec$$ satisfies dominance and strict independence. Therefore $$A\prec B$$ by transitivity.□
Using this lemma, we can show now that $$\prec_{\infty}$$ is indeed a minimal relation for dominance and strict independence.
### Theorem 4.2
Given a set $$X$$ , a linear order $$<$$ on $$X$$ and a family $$\mathcal{X}\subseteq\mathcal{P}(X)\backslash\{\emptyset\}$$ , there is a partial order on $$\mathcal{X}$$ that satisfies dominance and strict independence if and only if $$\prec_{\infty}$$ is irreflexive on $$\mathcal{X}$$ .
### Proof
$$\prec_{\infty}$$ satisfies dominance as it extends $$\prec_{d}^{t}$$. By construction it also satisfies strict independence and transitivity: $$A_{1}\prec_{\infty}A_{2}\prec_{\infty}\dots\prec_{\infty}A_{k}$$ implies $$A_{1}\prec^{t}_{n}A_{2}\prec^{t}_{n}\dots\prec^{t}_{n}A_{k}$$ for some $$n\in N$$ but then $$A_{1}\prec_{n}^{t}A_{k}$$ holds by the transitivity of $$\prec_{n}^{t}$$ and therefore $$A_{1}\prec_{\infty}A_{k}$$. Now assume $$A\prec_{\infty}B$$ and hence $$A\prec^{t}_{n}B$$ for some $$n$$ and $$A\cup\{x\}\not\prec_{n}^{t}B\cup\{x\}\in\mathcal{X}$$ for some $$x\not\in A\cup B$$. Then $$A,B,A\cup\{x\},B\cup\{x\}$$ is picked for some $$l$$ with $$n<l$$ and $$A\cup\{x\}\prec_{l}B\cup\{x\}$$ is set, hence $$A\cup\{x\}\prec_{\infty}B\cup\{x\}$$. Therefore, if $$\prec_{\infty}$$ is irreflexive, it is a partial order satisfying dominance and strict independence.
On the other hand, if $$\prec_{\infty}$$ is not irreflexive no strict partial order can extend it. But every strict partial order on $$\mathcal{X}$$ satisfying dominance and strict independence must be an extension of $$\prec_{\infty}$$. Assume otherwise there is a strict partial order $$\prec$$ on $$\mathcal{X}$$ satisfying dominance and strict independence that does not extend $$\prec_{\infty}$$, i. e., there are sets $$A,B\in\mathcal{X}$$ such that $$A\prec_{\infty}B$$ holds but not $$A\prec B$$. By Lemma 1 there is a $$\prec_{\infty}$$-link from $$A$$ to $$B$$. This implies, by Lemma 1, $$A\prec B$$ because $$\prec$$ is transitive and satisfies dominance and strict independence. Contradiction. Therefore no partial order on $$\mathcal{X}$$ can satisfy dominance and strict independence, if $$\prec_{\infty}$$ is irreflexive.□
Using this result, we can define a polynomial time algorithm for the Partial Dominance Strict Independence Problem.
### Corollary 1
The Partial Dominance Strict Independence problem is in $$P$$ .
### Proof
Computing $$\prec_{\infty}$$ can obviously be done in polynomial time because the construction always stops after at most $$|n\times n|=n^{2}$$ steps. Then checking if $$\prec_{\infty}$$ is irreflexive only requires checking if $$A\prec_{\infty}A$$ holds for some $$A$$.□
Finally, links give us an easy characterization of sets $$\mathcal{X}$$ for which $$\prec_{\infty}$$ is irreflexive.
### Corollary 2
$$\prec_{\infty}$$ is irreflexive if and only if there is no set $$A\in\mathcal{X}$$ such that there is a $$\prec_{\infty}$$ -link from $$A$$ to $$A$$ .
### Proof
$$\prec_{\infty}$$ is transitive and satisfies dominance and strict independence, hence Lemma 1 tells us, that $$A\prec_{\infty}A$$ if and only if there is a $$\prec_{\infty}$$ link from A to A.
### 4.2 Total Orders
We show that it is, in general, not possible to construct a (strict) total order satisfying both (maximal) dominance and (strict) independence deterministically in polynomial timeFootnote 3. We do this by a reduction from betweenness.
### Problem 5 (Betweenness)
Given a set $$V=\{v_{1},v_{2},\dots,v_{n}\}$$ and a set of triples $$R\subseteq V^{3}$$, does there exist a strict total order on $$V$$ such that $$a<b<c$$ or $$a> b> c$$ holds for all $$(a,b,c)\in R$$.
Betweenness is known to be NP-hard [13]. We use this result to show NP-hardness for all four versions of the (Maximal) Dominance (Strict) Independence problem. The idea is, roughly, to represent the elements of $$V$$ by sets which are not directly comparable via the axioms of dominance or independence. Hence, in order to find a total order, we need to guess how these sets are ordered. Starting from this guess we need to “maximize” this initial order in such a way that for each triple $$(a,b,c)$$ both $$a<b> c$$ and $$a> b<c$$ would lead to a circle in every order satisfying dominance and independence. However, this requires a number of carefully chosen additional sets as we will detail below.
### Theorem 4.3
The Maximal Dominance Strict Independence problem, the Dominance Strict Independence problem, the Maximal Dominance Independence problem and the Dominance Independence problem are NP-complete.
It is clear that all four problems are in NP. We can guess a binary relation and then check if it has all properties we want. It is well known that checking for transitivity and (ir-) reflexivity can be done in polynomial time. Checking (maximal) dominance only requires an easy check for every pair of sets and (strict) independence an equally easy check for every quadruple of sets. It is clear that this can be done in polynomial time. In what follows, we split the proof of the NP-hardness in four parts, one for each problem.
### Proof
Let $$(V,R)$$ be an instance of betweenness with $$V=\{v_{1},v_{2},\ldots,v_{n}\}$$. We construct an instance $$(X,<,\mathcal{X})$$ of the Maximal Dominance Strict Independence problem. We set $$X=\{1,2,\dots,N\}$$ equipped with the usual linear order, for $$N=8n^{3}+2n+2$$. Then, we construct the family $$\mathcal{X}$$ stepwise. The family contains for every $$v_{i}\in V$$ a set $$V_{i}$$ of the following form (see Figure 3):
$$V_{i}:=\{1,N\}\cup\{i+1,i+2,\dots,N-i\}.$$
Furthermore, for every triple from $$R$$ we want to enforce $$A\prec B\prec C$$ or $$A\succ B\succ C$$ by adding two families of sets as shown in Figure 1 and Figure 2 with $$q,x,y,z\in X$$. The solid arrows represent preferences that are forced through maximal dominance and strict independence. The family in Figure 1 makes sure that every total strict order satisfying independence that contains $$A\prec B$$ must also contain $$B\prec C$$. Similarly, the family in Figure 2 makes sure that $$A\succ B$$ leads to $$B\succ C$$.
We implement this idea for all triples inductively. For every $$1\leq i\leq|R|$$, pick a triple $$(v_{l},v_{j},v_{m})\in R$$ and set $$k=n+1+8i$$. Let $$(A,B,C)=(V_{l},V_{j},V_{m})$$ be the triple of sets coding the triple of elements $$(v_{l},v_{j},v_{m})$$. We add the following sets:
\begin{aligned} & A\backslash\{k\},B\backslash\{k\},B\backslash\{k+1\},C\backslash\{k+1\}, \\ & A\backslash\{k+2\},B\backslash\{k+2\},B\backslash\{k+3\},C\backslash\{k+3\}.\end{aligned}
These sets correspond to the sets $$A\backslash\{x\},B\backslash\{x\},\dots,C\backslash\{q\}$$ in Figure 1 and Figure 2. Observe that the inductive construction guarantees that every constructed set is unique. We now have to force the preferences
\begin{aligned} & B\backslash\{k+1\}\prec A\backslash\{k\},\quad B\backslash\{k\}\prec C\backslash\{k+1\}, \\ & A\backslash\{k+2\}\prec B\backslash\{k+3\},\quad C\backslash\{k+3\}\prec B\backslash\{k+2\}.\end{aligned}
For technical reasonsFootnote 4, we add sets $$A\backslash\{k,k+4\},B\backslash\{k+1,k+4\}$$. Then, observe that, by construction, either $$B\backslash\{1,k+1,k+4\}\prec A\backslash\{1,k,k+4\}$$ or $$B\backslash\{k+1,k+4,N\}\prec A\backslash\{k,k+4,N\}$$ is implied by maximal dominance. We add $$A\backslash\{1,k,k+4\}$$ and $$B\backslash\{1,k+1,k+4\}$$ in the first case and $$A\backslash\{k,k+4,N\}$$ and $$B\backslash\{k+1,k+4,N\}$$ in the second case (see Figure 4). This ensures $$B\backslash\{k+1\}\prec A\backslash\{k\}$$ by strict independence. In the same way, we can force the other preferences using $$k+5,k+6$$ and $$k+7$$ instead of $$k+4$$.
We repeat this with a new triple $$(v_{i}^{\prime},v_{j}^{\prime},v_{m}^{\prime})\in R$$ until we treated all triples in $$R$$. Observe that there are at most $$n^{3}$$ triples, thus, for every triple, the values $$k,\dots,k+7$$ lie between $$n+1$$ and $$N-n$$, hence are element of every $$V_{i}$$. In total, we add $$24$$ sets per triple. Therefore, $$\mathcal{X}$$ contains $$n+24n^{3}$$ sets.
It is easy to see, that, by construction, for every strict total order on $$\mathcal{X}$$ satisfying maximal dominance and strict independence, we have
\begin{aligned} & A\backslash\{k\}\prec B\backslash\{k+1\},\quad C\backslash\{k+1\}\prec B\backslash\{k\}, \\ & B\backslash\{k+3\}\prec A\backslash\{k+2\},\quad B\backslash\{k+2\}\prec C\backslash\{k+3\}.\end{aligned}
Now assume there is a strict total order on $$\mathcal{X}$$ satisfying maximal dominance and strict independence. We claim that the relation defined by $$v_{i}<v_{j}$$ iff $$V_{i}\prec V_{j}$$ is a positive witness for $$(V,R)$$. By definition this is a strict total order. So assume there is a triple $$(a,b,c)$$ such that $$a> b<c$$ or $$a<b> c$$ holds. We treat the first case in detail: $$a> b<c$$ implies $$A\succ B\prec C$$. This implies by the strictness of $$\prec$$ and strict dominance $$A\backslash\{k\}\succ B\backslash\{k\}$$ and $$B\backslash\{k+1\}\prec C\backslash\{k+1\}$$. However, then
\begin{aligned} & A\backslash\{k\}\succ B\backslash\{k\}\succ C\backslash\{k+1\} \\ & \succ B\backslash\{k+1\}\succ A\backslash\{k\}\end{aligned}
contradicts the assumption that $$\prec$$ is transitive and irreflexive. Similarly, the second case leads to a contradiction.
Now assume that there is a strict total order on $$V$$ satisfying the restrictions from $$R$$. We use this to construct an order on $$\mathcal{X}$$. We set $$V_{i}\prec V_{j}$$ iff $$v_{i}<v_{j}$$ holds. Furthermore, we set $$A\prec B$$ for all $$A,B\in\mathcal{X}$$ if it is implied by dominance. Then, we apply strict independence twice and once “reverse” strict independenceFootnote 5, i. e., $$A\prec B$$ implies $$A\backslash\{x\}\prec B\backslash\{x\}$$ for $$A,B,A\backslash\{x\},B\backslash\{x\}\in\mathcal{X}$$. We claim that all possible instances of strict independence are decided already by this order. If $$A=V_{i}$$ for $$i\leq n$$, then there is no set $$A\cup\{x\}$$ in $$\mathcal{X}$$. If $$A=V_{i}\backslash\{x\}$$ for some $$i\leq n$$ and $$x\in X$$, then $$x$$ is the only element of $$X$$ such that $$A\cup\{x\}\in\mathcal{X}$$ holds. But then there can only be one other set $$B$$ with $$B\cup\{x\}\in\mathcal{X}$$ and $$B=V_{j}\backslash\{x\}$$ hence a preference between $$A$$ and $$B$$ was already introduced by reverse strict independence. Analogously in the cases $$A=V_{i}\backslash\{x,y\}$$ and $$A=V_{i}\backslash\{x,y,z\}$$ for $$i\leq n$$ and $$x\neq y\neq z\in X$$, every possible instance of strict independence is already decided by dominance and two applications of strict independence.
It can easily be seen, that this construction does not lead to circles, if we start with a positive instance of betweenness: Every set of the form $$A=V_{i}\backslash\{x,y,z\}$$ is only comparable to other sets by maximal dominance. Every set of the form $$A=V_{i}\backslash\{x,y\}$$ is only comparable by maximal dominance or to another set of the same form. The order on sets of this form mirrors the order on sets of the form $$A=V_{i}\backslash\{x,y,z\}$$ which is produced by maximal dominance and hence is circle free. Finally sets of the form $$A=V_{i}\backslash\{x\}$$ or $$A=V_{i}$$ are only comparable to other sets by maximal dominance or if this is intended by the construction. Hence, the order on these sets is circle free, if we started with a positive instance of betweenness.
Finally, we can extend this order to a total order because extensions do not produce new instances of strict independence.□
### Proof
We construct an instance $$(X,<,\mathcal{X})$$ of the Dominance Strict Independence problem in a similar fashion as above. We take the same $$X$$ and $$<$$ and add the same sets to $$\mathcal{X}$$. In order to make the reduction work for the Dominance Strict Independence problem, we have to add more sets.
Observe that maximal dominance is only needed in the reduction for the Maximal Dominance Strict Independence problem to introduce preferences like $$A\backslash\{1,k,k+4\}\prec B\backslash\{1,k+1,k+4\}$$. We can enforce these preferences also using strict independence and regular dominance using a construction as in the proof of Proposition 1.
For every $$k$$ used in the reduction, let $$(A,B,C)$$ be the triple of sets for which $$k$$ appears in the reduction and let $$(X_{k},Y_{k})$$ be one of the following pairs:
\begin{aligned} & (B\backslash\{k+1,k+4,z_{1}\},A\backslash\{k,k+4,z_{1}\}),\\ & (B\backslash\{k,k+5,z_{2}\},C\backslash\{k+1,k+5,z_{2}\}),\\ & (A\backslash\{k+2,k+6,z_{3}\},B\backslash\{k+3,k+6,z_{3}\}),\\ & (C\backslash\{k+3,k+7,z_{4}\},B\backslash\{k+2,k+7,z_{4}\})\end{aligned}
with $$z_{i}\in\{1,N\}$$ chosen such that $$X_{k},Y_{k}\in\mathcal{X}$$ holds.
We want to enforce $$X_{k}\prec Y_{k}$$. By definition, $$\max(X_{k})=\max(Y_{k})$$ and $$\min(X_{k})<\min(Y_{k})$$ or $$\max(X_{k})<\max(Y_{k})$$ and $$\min(X_{k})=\min(Y_{k})$$. Assume, w.l.o.g. $$\max(X_{k})=\max(Y_{k})$$ and $$\min(X_{k})<\min(Y_{k})$$ and let $$X_{k}=\{x_{1},x_{2},\dots,x_{l}\}$$ and $$Y_{k}=\{y_{1},y_{2},\dots,y_{m}\}$$ be enumerations of $$X_{k}$$ resp. $$Y_{k}$$ such that $$i<j$$ implies $$x_{i}<x_{j}$$ resp. $$y_{i}<y_{j}$$. We add
$$\{x_{l}\},\{x_{l-1},x_{l}\},\dots,\{x_{2},\dots,x_{l}\}$$
and $$\{x_{1},x_{l}\}$$ to $$\mathcal{X}$$. This forces by dominance
$$\{x_{2},\dots x_{l}\}\prec\dots\prec\{x_{l-1},x_{l}\}\prec\{x_{l}\}$$
and hence by transitivity and strict independence $$X_{k}\prec\{x_{1},x_{l}\}$$. Analogously, we can enforce $$\{x_{1},y_{1},y_{m}\}\prec\{x_{1}\}\cup Y_{k}$$ by adding
$$\{x_{1},y_{1}\},\{x_{1},y_{1},y_{2}\},\dots,\{x_{1},y_{1},\dots,y_{m-1}\}$$
as well as $$\{x_{1},y_{1},y_{m}\}$$ and $$\{x_{1}\}\cup Y_{k}$$ to $$\mathcal{X}$$. Finally we add $$\{x_{1},y_{1},y_{m}\}$$, $$\{x_{1}\}$$ and $$\{x_{1},y_{1}\}$$ enforcing $$\{x_{1}\}\prec\{x_{1},y_{1}\}$$ by dominance and hence $$\{x_{1},y_{m}\}\prec\{x_{1},y_{1},y_{m}\}$$ by strict independence. Then, we have $$X_{k}\prec Y_{k}$$ by
$$X_{k}\prec\{x_{1},x_{l}\}\prec\{x_{1},y_{1},x_{l}\}\prec\{x_{1}\}\cup Y_{k}\prec Y_{k}.$$
The process of producing a positive instance of betweenness from a positive instance of the Dominance Strict Independence problem is the same as for the Maximal Dominance Strict Independence case. However, in order to construct a total order on $$\mathcal{X}$$, we have to do a bit more. We take the same steps as in the Maximal Dominance Strict Independence case (including the closure under maximal dominance) but additionally, for $$A,B\in\mathcal{X}$$ with $$\max(A)=\max(B)$$, $$\min(A)=\min(B)$$ and $$|A|,|B|\leq 3$$ we set $$A\prec B$$ if
1. 1.
$$\min(A)=1$$ and $$|A|=2$$ and $$|B|=3$$,
2. 2.
$$\min(A)=N$$ and $$|A|=3$$ and $$|B|=2$$,
3. 3.
$$|A|=|B|=3$$ and $$A\backslash B<B\backslash A$$.
This order, together with a positive instance of betweenness, maximal dominance and (reverse) strict independence is circle free and decides all possible applications of strict independence. Therefore, we can construct a total order on $$\mathcal{X}$$ satisfying strict independence and dominance.□
### Proof
We have to adapt the reduction for the Maximal Dominance Strict Independence problem above in two places. We have to change the way we enforce the strict preferences in Figure 1 and Figure 2 and we have to make sure that the order restricted to the sets $$V_{1},V_{2},\dots,V_{n}$$ is strict.
To enforce, without strict independence, a strict preference between two sets that is not forced by maximal dominance we define for every pair $$A,B\in\mathcal{X}$$ with $$\min(B)\leq\min(A)$$, $$\max(A)\leq\max(B)$$ and $$2\leq\max(A)-\min(A)$$ a family of sets $$\mathcal{S}(A,B)$$ forcing $$A\prec B$$. $$\mathcal{S}(A,B)$$ contains the following sets
\begin{aligned} & \{x_{AB}\},\{y_{AB}\},\{x_{AB},z_{AB}\},\{y_{AB},z_{AB}^{*}\}, \\ & A\cup\{z_{AB}\},B\cup\{z_{AB}^{*}\} \end{aligned}
where $$\min(A)<y_{AB}<x_{AB}<\max(A)$$, $$\max(B)<z_{AB}$$ and $$z_{AB}^{*}<\min(B)$$ hold.
Then $$A\cup\{z_{AB}\}\prec\{x_{AB},z_{AB}\}$$ holds by maximal dominance and, therefore, $$\{x_{AB}\}\not\prec A$$ and hence $$A\preceq\{x_{AB}\}$$ holds by “reverse” independenceFootnote 6 and analogously, $$\{y_{AB}\}\preceq B$$. Therefore, transitivity implies $$A\prec B$$ by $$A\preceq\{x_{AB}\}\prec\{y_{AB}\}\preceq B$$.
Using $$\mathcal{S}(A,B)$$, we can adapt the proof above. We want to construct an instance of the maximal dominance independence problem $$(X,<,\mathcal{X})$$. We take as $$X$$ again a set of the form $$X=\{1,\dots,N\}$$ with the usual linear order, however $$N$$ has to be larger than in the maximal dominance strict independence case. Namely, we set $$N=20n^{3}+28n^{2}+2n+14$$. $$\mathcal{X}$$ contains sets $$V_{1},\dots,V_{n}$$ similar to the ones in the reductions above (see Figure 5). However, they do not have a common smallest element or common largest element and the smallest element of $$V_{1}$$ is $$4n^{3}+4n^{2}+1$$ and largest element $$16n^{3}+24n^{2}+2n+14$$, i. e.,
$$V_{i}:=\{4n^{3}+4n^{2}+i,\dots,16n^{3}+24n^{2}+2n+14-i\}.$$
We assume, for all families $$\mathcal{S}(A,B)$$ and $$\mathcal{S}(C,D)$$ occurring in the reduction, $$p_{AB}\neq q_{CD}$$ for $$p,q\in\{x,y,z\}$$ and $$(A,B)\neq(C,D)$$. As well, for every family $$\mathcal{S}(A,B)$$, assume $$5n^{3}+13n^{2}+n+7\leq y_{AB}$$ and $$x_{AB}\leq 16n^{3}+17n^{2}+n+7$$.
For a triple $$(a,b,c)$$ in the instance of betweenness we add the following sets as in the two previous reductions:
\begin{aligned} & A\backslash\{k\},B\backslash\{k\},B\backslash\{k+1\},C\backslash\{k+1\}, \\ & A\backslash\{k+2\},B\backslash\{k+2\},B\backslash\{k+3\},C\backslash\{k+3\}\end{aligned}
where we start with $$k=4n^{3}+11n^{2}+n+7$$. We force the same preference as in the Maximal Dominance Strict Independence case, by adding the following families:
\begin{aligned} & \mathcal{S}(A\backslash\{k\},B\backslash\{k+1\}),\mathcal{S}(C\backslash\{k+1\},B\backslash\{k\}),\\ & \mathcal{S}(B\backslash\{k+3\},A\backslash\{k+2\}),\mathcal{S}(B\backslash\{k+2\},C\backslash\{k+3\}).\end{aligned}
We still have to make sure that the order on the sets $$V_{1}\dots V_{n}$$ is strict. To achieve this we want to use the idea shown in Figure 6, that is to add for every pair $$V_{i},V_{j}$$ sets that lead to a circle if both $$V_{i}\preceq V_{j}$$ and $$V_{j}\preceq V_{i}$$ hold. Let $$f(l)=(V_{i},V_{j})$$ be an enumeration of all pairs of sets $$V_{1},V_{2},\dots,V_{n}$$. We add sets $$C_{l},D_{l},E_{l}$$ and $$F_{l}$$ that are contained in the “middle parts” of all sets $$V_{i}$$ such that $$C_{l}\subset F_{l^{\prime}}$$ holds for all $$l^{\prime}<l$$. Moreover, we want the following:
\begin{aligned} F_{l}& =E_{l}\backslash\{\max(E_{l}),\text{max}^{\prime}(E_{l}),\text{min}^{\prime}(E_{l}),\min(E_{l})\},\\ E_{l}& =D_{l}\backslash\{\max(D_{l}),\text{max}^{\prime}(D_{l}),\text{min}^{\prime}(D_{l}),\min(D_{l})\},\\ D_{l}& =C_{l}\backslash\{\max(C_{l}),\text{max}^{\prime}(C_{l}),\text{min}^{\prime}(C_{l}),\min(C_{l})\}\end{aligned}
where $$\max^{\prime}(X)$$ denotes the second largest element of $$X$$ and $$\min^{\prime}(X)$$ the second smallest. We can achieve this by taking for all $$l\leq n^{2}$$
$$C_{l}:=\{4n^{3}+4n^{2}+n+7l,\dots,16n^{3}+24n^{2}+n+14-7l\}$$
and $$D_{l},E_{l}$$ and $$F_{l}$$ accordingly. Furthermore, we add sets $$F_{l}\backslash\{\min(F_{l})\}$$, $$V_{i}\cup\{z_{l}\}$$ and $$(F_{l}\backslash\{\min(F_{l})\})\cup\{z_{l}\}$$ for a uniqueFootnote 7 $$z_{l}<\min(V_{1})$$. This ensures $$F_{l}\prec F_{l}\backslash\{\min(F_{l})\}\preceq V_{i}$$. In a similar fashion we can enforce $$V_{i}\prec D_{l}$$, $$E_{l}\prec V_{j}$$ and $$V_{j}\prec F_{l}$$. Then, we add sets $$C_{l}\backslash\{y_{l}\}$$, $$D_{l}\backslash\{x_{l}\}$$, $$E_{l}\backslash\{y_{l}\}$$ and $$F_{l}\backslash\{y_{l}\}$$ for $$x_{l}=5n^{3}+11n^{2}+n+7+l$$ and $$y_{l}=5n^{3}+12n^{2}+n+7+l$$. Furthermore, we enforce $$D_{l}\backslash\{x_{l}\}\prec F_{l}\backslash\{y_{l}\}$$ and $$E_{l}\backslash\{y_{l}\}\prec C_{l}\backslash\{x_{l}\}$$ by adding $$\mathcal{S}(D_{l}\backslash\{x_{l}\},F_{l}\backslash\{y_{l}\})$$ and $$\mathcal{S}(E_{l}\backslash\{y_{l}\},C_{l}\backslash\{x_{l}\})$$.
This forces a strict preference between $$V_{i}$$ and $$V_{j}$$. Assume otherwise $$V_{i}\sim V_{j}$$ for a total order satisfying maximal dominance and independence. Then, for $$l$$ such that $$f(l)=(V_{i},V_{j})$$ holds, $$F_{l}\prec V_{i}\preceq V_{j}\prec E_{l}$$ implies $$F_{l}\prec E_{l}$$. This implies $$F_{l}\backslash\{y\}\preceq E_{l}\backslash\{y\}$$ because $$E_{l}\backslash\{y\}\prec F_{l}\backslash\{y\}$$ would imply $$E_{l}\preceq F_{l}$$, a contradiction. Similarly, $$C_{l}\prec V_{j}\preceq V_{i}\prec D_{l}$$ implies $$C_{l}\backslash\{x\}\preceq D_{l}\backslash\{x\}$$. However, then $$C_{l}\backslash\{x\}\preceq D_{l}\backslash\{y\}\prec F_{l}\backslash\{y\}\preceq E_{l}\backslash\{y\}\prec C_{l}\backslash\{x\}$$ is a circle in $$\prec$$, contradicting the assumption that $$\preceq$$ is a total order.
It is straightforward to check that this construction yields a valid reduction analogously to the proof above. The key step is to observe that independence can only be applied to the new sets in cases where it is used in the proof. This is clear for the sets pictured in Figure 6. For the one and two element sets this holds, because the elements are unique and because no three element sets are contained in $$\mathcal{X}$$.
It remains to check that we can actually pick a unique element every time we want to do this in the reduction. The inner part of $$F_{n^{2}}$$ has $$(16n^{3}+24n^{2}+n+14-7n^{2}-7)-(4n^{3}+4n^{2}+n+7n^{2}+7)=12n^{3}+6n^{2}$$ elements. For every triple, we have to pick $$12$$ unique elements contained in this middle part. These are at most $$12n^{3}$$ elements. Furthermore, we have to pick for every pair $$(V_{i},V_{j})$$ $$6$$, two for $$x$$ and $$y$$ and $$4$$ to enforce preferences, unique elements. There are $$n^{2}$$ such pairs, so we need $$6n^{2}$$ elements. Hence we need at most the $$12n^{3}+6n^{2}$$ elements contained in $$F_{n^{2}}$$. Furthermore, we need for every pair $$(V_{i},V_{j})$$ and every triple $$4$$ elements smaller than $$\min(V_{1})$$ and the same amount of elements larger than $$\max(V_{1})$$. This is possible as $$\min(V_{1})=4n^{3}+4n^{2}+1$$ and $$N-\max(V_{1})=(20n^{3}+28n^{2}+2n+14)-(16n^{3}+24n^{2}+2n+14)=4n^{3}+4n^{2}$$.□
### Proof
We construct an instance $$(X,<,\mathcal{X})$$ of the Dominance Independence problem in a similar fashion as above. We take the same $$X$$ and $$<$$ and add the same sets to $$\mathcal{X}$$ as in the Maximal Dominance Independence case. In order to make the reduction work for the Dominance Independence problem, we have to add more sets.
Observe that maximal dominance is only needed in the reduction for the Maximal Dominance Strict Independence problem to introduce preferences between sets of the form (1) $$F_{l}\prec F_{l}\backslash\{\min(F_{l})\}$$, (2) $$\{x\}\prec\{y\}$$, and (3) $$R\cup\{z\}\prec S\cup\{z\}$$, for sets $$R,S$$ and $$z\in X$$. We can enforce these preferences also using independence and regular dominance. (1) is implied by dominance anyway. (2) can be forced by adding $$\{x,y\}$$ because $$\{x\}\prec\{x,y\}\prec\{y\}$$ holds by dominance. Finally, we can enforce (3) using the idea from the Dominance Strict Independence reduction.
Now, assume, w.l.o.g., $$\min(R)<\min(S)<\max(S)<\max(R)<z$$; the other case is similar. Let $$R=\{r_{1},\dots,r_{l}\}$$ and $$S=\{s_{1},\dots,s_{m}\}$$ be enumerations of $$R$$ (resp. $$S$$) such that $$i<j$$ implies $$r_{i}<r_{j}$$ (resp. $$s_{i}<s_{j}$$). We add $$\{z\},\{r_{l},z\},\dots,$$ $$\{r_{2},\dots,z\},\{r_{1},z\}$$ to $$\mathcal{X}$$. This forces $$\{r_{2},\dots,z\}\prec\{z\}$$ by dominance and hence by one application of independence $$R\cup\{z\}\preceq\{r_{1},z\}$$. Analogously, we enforce $$\{s_{1},z\}\preceq S\cup\{z\}$$ by adding $$\{s_{1}\},\{s_{1},s_{2}\},\dots,\{s_{1},\dots,s_{m}\},\{s_{1},z\}$$ to $$\mathcal{X}$$. Finally, we add $$\{r_{1}\},\{r_{1},s_{1}\}$$ and $$\{r_{1},s_{1},z\}$$, which leads to $$\{r_{1},z\}\preceq\{r_{1},s_{1},z\}$$. Then we have $$R\cup\{z\}\preceq\{r_{1},z\}\preceq\{r_{1},s_{1},z\}\prec\{s_{1},z\}\preceq S\cup\{z\}$$, hence $$R\cup\{z\}\prec S\cup\{z\}$$.
Checking the correctness of this reduction is straightforward. The correctness proof for the Maximal Dominance Independence case can be adapted to the Dominance Independence case in the same way as the proof of the correctness of the Maximal Dominance Strict Independence case was adapted to the Dominance Strict Independence case.□
## 5 Conclusion
We have shown that the problem of deciding whether a linear order can be lifted to a ranking of sets of objects satisfying a form of dominance and a form of independence is in P or trivial if we do not expect the ranking to be total and NP-complete if we do. In order to prove P‑membership or triviality we constructed such rankings. Rankings of specific sets are useful in several applications, e.g., to eliminate obviously inferior sets of objects from a set of options.
In many applications, the family of sets to be ranked is not given explicitly but implicitly. We expect that a compact representation of the sets increases the computational complexity of decision problems as studied in this paper. As future work we thus want to investigate the complexity blow up caused by a compact representation.
Furthermore, we would like to characterize families that allow an order satisfying (maximal) dominance and (strict) independence. Moreover, it may be possible to find sufficient but not necessary conditions for the existence of such rankings that can be checked in polynomial time. We aim for finding strong forms of such conditions. A related goal is to obtain special classes of families where such a characterization is feasible. A promising candidate are families generated via graphs, where the family is given by the sets of vertices that induce connected subgraphs.
Another item on our agenda is to investigate whether the logic proposed in [8] can be used for specific sets of objects as well. Finally, it would be interesting to study some of the other axioms that have been considered in the literature and see how they behave when one has to rank proper subsets of the whole power set of elements. | 16,751 | 50,426 | {"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": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.93077 |
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New Jersey MS Computer Science
Points
Activity Points Item Type
Programming with Turtle Graphics
1.1 Intro to Python with Tracy the Turtle Lesson
1.1.1 Intro to Tracy 1 Video
1.1.2 Intro to Tracy 3 Check for Understanding
1.2 Tracy's Grid World Lesson
1.2.1 Tracy's Grid World 1 Video
1.2.2 Tracy's Grid World 5 Check for Understanding
1.2.3 Dashed Line 1 Example
1.2.4 Shorter Dashed Line 5 Exercise
1.2.5 Caterpillar 5 Exercise
1.3 Turning Tracy Lesson
1.3.1 Turning Tracy 1 Video
1.3.2 Turning Tracy 3 Check for Understanding
1.3.3 Square 1 Example
1.3.4 X and Y Axes 1 Example
1.3.5 Rectangle 5 Exercise
1.3.6 4 Columns 5 Exercise
1.4 For Loops Lesson
1.4.1 For Loops 1 Video
1.4.2 For Loops 5 Check for Understanding
1.4.3 Square Using Loops 1 Example
1.4.4 Dotted Line 1 Example
1.4.5 Row of Circles 5 Exercise
1.4.6 4 Columns 2.0 5 Exercise
1.5 Turning Tracy Using Angles Lesson
1.5.1 Turning Tracy Using Angles 1 Video
1.5.2 Turning Tracy Using Angles 4 Check for Understanding
1.5.3 Asterisk 1 Example
1.5.4 Four Circles 1 Example
1.5.5 Hexagon 5 Exercise
1.5.6 'X' Marks the Spot 5 Exercise
1.5.7 Circle Pyramid 5 Exercise
1.6.2 Comments 3 Check for Understanding
1.6.3 Four Circles with Comments 1 Example
1.6.4 Circle Pyramid with Comments 5 Exercise
1.7 Naming Guidelines Lesson
1.7.1 Naming Guidelines 1 Video
1.7.2 Naming Guidelines 2 Check for Understanding
1.8 Functions Lesson
1.8.1 Functions 1 Video
1.8.2 Functions 3 Check for Understanding
1.8.3 X and Y Axes with Hash Marks 1 Example
1.8.5 Shape Stack 5 Exercise
1.9 Artistic Effects Lesson
1.9.1 Artistic Effects 1 Video
1.9.2 Artistic Effects 5 Check for Understanding
1.9.3 Rainbow Octagon 1 Example
1.9.4 Circle Square Triangle 1 Example
1.9.5 Four Colored Triangles 5 Exercise
1.9.6 Colorful Bracelet 5 Exercise
1.9.7 Kid's Shapes Toy 10 Challenge
1.10 Top Down Design Lesson
1.10.1 Top Down Design 1 Video
1.10.2 Top Down Design 2 Check for Understanding
1.10.3 Bubble Wrap 1 Example
1.10.4 Bubble Wrap 2.0 5 Exercise
1.10.5 Sidewalk 5 Exercise
1.11 Variables Lesson
1.11.1 Variables 1 Video
1.11.2 Variables 3 Check for Understanding
1.11.3 Increasing Length 1 Example
1.11.4 Dartboard 5 Exercise
1.11.5 Line of Increasing Blocks 5 Exercise
1.12 User Input Lesson
1.12.1 User Input 1 Video
1.12.2 User Input 3 Check for Understanding
1.12.3 Color Coded Increasing Length 1 Example
1.12.4 Colored Dartboard 5 Exercise
1.12.5 Four Corners 5 Exercise
1.13 Parameters Lesson
1.13.1 Parameters 1 Video
1.13.2 Parameters 3 Check for Understanding
1.13.3 Concentric Circles 1 Example
1.13.4 Colorful Caterpillar 5 Exercise
1.13.5 Circle in a Square 5 Exercise
1.13.6 Snowman 5 Exercise
1.14 Using i in For Loops Lesson
1.14.1 Using i in For Loops 1 Video
1.14.2 Using i in For Loops 3 Check for Understanding
1.14.3 Geometry 1 Example
1.14.4 Geometry 2.0 5 Exercise
1.15 Extended Loop Control Lesson
1.15.1 Extended Loop Control 1 Video
1.15.2 Extended Loop Control 2 Check for Understanding
1.15.3 Square Swirl 1 Example
1.15.4 Dartboard using i 5 Exercise
1.15.5 Phone Signal 5 Exercise
1.16 If Statements Lesson
1.16.1 If Statements 1 Video
1.16.2 If Statements 5 Check for Understanding
1.16.3 X and Y Axis with Bolded Marks 1 Example
1.16.4 Happy Face 5 Exercise
1.16.5 Black and White Squares 5 Exercise
1.17 If/ Else Statements Lesson
1.17.1 If/Else Statements 1 Video
1.17.2 If/Else Statements 4 Check for Understanding
1.17.3 Positive, Negative, Zero 1 Example
1.17.4 Rating 5 Exercise
1.17.5 Happy/ Sad Face 5 Exercise
1.18 While Loops Lesson
1.18.1 While Loops 1 Video
1.18.2 While Loops 4 Check for Understanding
1.18.3 Increasing Circles 1 Example
1.18.4 Increasing Squares 5 Exercise
1.18.5 Guess a Number 5 Exercise
1.19 Putting Together Control Structures Lesson
1.19.1 Putting Together Control Structures 1 Video
1.19.2 Putting Together Control Structures 3 Check for Understanding
1.19.3 Block Pyramid 1 Example
1.19.4 Guess a Number 2.0 5 Exercise
1.19.5 Checkerboard 10 Challenge
1.20 Using Data to Refine Game Mechanics Lesson
1.20.1 Game Mechanics 101 5 Notes
1.20.2 Game Mechanics Planning 5 Free Response
1.20.3 Guess a Number 3.0: Beta 5 Exercise
1.20.4 Collect and Analyze Test Data 5 Connection
1.20.5 Guess a Number 3.0: Final 5 Challenge
1.21 Intro to Programming with Turtle Graphics Quiz Lesson
1.21.1 Putting It All Together Quiz 25 Quiz
Digital Information
2.1 Intro to Digital Information Lesson
2.1.1 What is Digital Information? 1 Video
2.1.2 What is Digital Information Quiz 2 Check for Understanding
2.1.3 Fast Food Menu 1 Example
2.1.4 Reflection: Encodings Everywhere 5 Free Response
2.2 Number Systems Lesson
2.2.1 Number Systems 1 Video
2.2.2 Number Base Tool 1 Notes
2.2.3 Number Systems Quiz 2 Check for Understanding
2.2.4 Decimal to Binary 1 Video
2.2.5 Decimal to Binary Quiz 1 Check for Understanding
2.2.6 Binary Game 5 Exercise
2.3 Encoding Text with Binary Lesson
2.3.1 Encoding Text with Binary 1 Video
2.3.2 Encoding Text with Binary Quiz 2 Check for Understanding
2.3.3 Custom Encoding 1 Example
2.3.4 Bits to ASCII 1 Example
2.3.5 Hello World in Bits 5 Exercise
2.3.6 Create your own Encoding 5 Exercise
2.4 Pixel Images Lesson
2.4.1 Pixel Images 1 Video
2.4.2 Pixel Images Quiz 1 Check for Understanding
2.4.3 Creating Pixel Images 1 Resource
2.4.4 CodeHS Logo 1 Example
2.4.5 Checkerboard 5 Exercise
2.4.7 Create an Image! 5 Exercise
2.5.2 Hexadecimal Quiz 2 Check for Understanding
2.5.3 Binary to Hex Game 5 Exercise
2.6 Pixel Colors! Lesson
2.6.1 Pixel Colors 1 Video
2.6.2 Pixel Colors Quiz 3 Check for Understanding
2.6.3 Colors in Bits 1 Example
2.6.4 Exploring RGB 5 Exercise
2.6.5 Making Yellow 5 Exercise
2.6.6 Rainbow 5 Exercise
2.6.7 Create a Color Image! 5 Exercise
2.7 Digital Information Quiz Lesson
2.7.1 Digital Information Quiz 14 Unit Quiz
The Internet
3.1 Intro to the Internet Lesson
3.1.1 Welcome to the Internet 1 Video
3.1.2 Welcome to the Internet Quiz 1 Check for Understanding
3.1.3 The Internet and You 5 Free Response
3.2 Internet Hardware Lesson
3.2.1 Hardware of the Internet 1 Video
3.2.2 Internet Hardware Quiz 3 Check for Understanding
3.2.3 The Internet is in the Ocean 1 Connection
3.3.2 Internet Addresses Quiz 1 Check for Understanding
3.3.3 The Need for Addresses 5 Free Response
3.3.4 4-bit Addresses 1 Check for Understanding
3.3.5 IPv4 vs IPv6 5 Free Response
3.4 DNS Lesson
3.4.1 DNS 1 Video
3.4.2 DNS Quiz 2 Check for Understanding
3.4.3 How Does DNS Work? 1 Connection
3.4.4 How Does DNS Work? 5 Free Response
3.5 Routing Lesson
3.5.1 Routing 1 Video
3.5.2 Routing Quiz 1 Check for Understanding
3.5.3 Redundancy 5 Free Response
3.5.4 Route Tracing 5 Traceroute
3.6 Packets and Protocols Lesson
3.6.1 Packets and Protocols 1 Video
3.6.2 Packets and Protocols Quiz 1 Check for Understanding
3.6.3 Passing Notes 10 Resource
3.6.4 How the Internet Works 1 Connection
3.6.5 The Story of the Internet 5 Free Response
3.7 Cybersecurity Lesson
3.7.1 City Services Ransomware 1 Connection
3.7.2 Ransomware Reflection 5 Free Response
3.7.3 Application Security 1 Video
3.7.4 Application Security 5 Check for Understanding
3.7.6 Windows Defender Antivirus 1 Connection
3.7.7 Windows Defender Antivirus 5 Free Response
3.7.8 What is an SSID? 1 Connection
3.7.9 What is an SSID? 5 Free Response
3.7.10 Wireless Network Setup 5 Free Response
3.8 The Impact of the Internet Lesson
3.8.1 The Impact of the Internet 1 Video
3.8.2 The Impact of the Internet Quiz 2 Check for Understanding
3.8.3 What is the Digital Divide? 1 Connection
3.8.4 What is the Digital Divide? 5 Free Response
3.8.5 Mindsets 5 Survey
3.9 The Internet Quiz Lesson
3.9.1 The Internet Quiz 15 Unit Quiz
Intro to micro:bit
4.1 Welcome to micro:bit! Lesson
4.1.1 Intro to micro:bit 5 Video
4.1.2 micro:bit Quick Start 5 Connection
4.1.3 Dice Simulator 5 Notes
4.1.4 Don't Wobble! 5 Notes
4.1.5 Exploration: Intro to Programming with micro:bit 5 Connection
4.1.6 Exploration 1.1 Follow-up 5 Video
4.1.7 Background & Experience 5 Free Response
4.1.8 Goal Setting 5 Free Response
4.2 Setting Up your micro:bit Lesson
4.2.1 Setting Up your micro:bit 5 Video
4.2.2 Exploration: Exploring LEDs 5 Connection
4.2.3 Exploration 1.2 Follow-up 5 Video
4.2.4 Four Corners 5 Exercise
4.2.5 Blinking First Letter 5 Exercise
4.3.1 Comments & Pseudocode 5 Video
4.3.2 Comments & Pseudocode 3 Check for Understanding
4.3.4 X in Pseudocode 5 Free Response
4.3.5 Exploration: Analog vs. Digital 5 Connection
4.3.6 Exploration 1.3 Follow-up 5 Video
4.3.7 Varied Brightness 5 Exercise
4.3.8 Moving Bright Box 5 Exercise
4.4 Variables Lesson
4.4.1 Variables 5 Video
4.4.2 Variables 3 Check for Understanding
4.4.3 Variable as Coordinate Value 5 Notes
4.4.4 Plus with a Variable in Pseudocode 5 Free Response
4.4.5 Exploration: Using micro:bit Pins 5 Connection
4.4.6 Exploration 1.4 Follow-up 5 Video
4.4.7 Brightness Line using Variables 5 Exercise
4.4.8 Opposite Blinking External LEDs 5 Exercise
4.4.9 Dimming External LED 5 Exercise
4.5 Debugging with micro:bit Lesson
4.5.1 Debugging with micro:bit 5 Video
4.5.2 Debugging Code 5 Free Response
4.5.3 Physical Debugging #1 5 Free Response
4.5.4 Physical Debugging #2 5 Free Response
4.6 Intro to micro:bit Quiz Lesson
4.6.1 Intro to micro:bit Quiz 15 Unit Quiz
Project: Using Data To Answer a Question
5.1 Sample Project: Climate Change Lesson
5.1.1 Data Project Introduction 5 Notes
5.1.2 Climate Change Data: NASA 5 Article
5.1.3 Introduction to Spreadsheets 5 Video
5.1.4 NASA Climate Data Explore 5 Notes
5.1.5 NASA Climate Data Explore Answers 8 Quiz
5.1.6 Visualizing Data with Spreadsheets 5 Video
5.1.7 NASA Climate Data Visualizations 5 Connection
5.1.8 Submit: NASA Climate Data Visualization 5 Free Response
5.1.9 Why Do We Use Models? 5 Article
5.1.10 Reflect: Why Do We Use Models? 5 Free Response
5.1.11 The Very Simple Climate Model 5 Free Response
5.2 Defining a Question to Answer Lesson
5.2.1 Brainstorming a Question 5 Free Response
5.2.2 How Can Data Give Insight? 5 Free Response
5.2.3 Select Your Question 5 Free Response
5.3 Preparing and Conducting the Investigation Lesson
5.3.1 Choose your Sensor(s) 5 Free Response
5.3.2 Minimizing Bias 5 Video
5.3.3 Plan System Setup 5 Free Response
5.3.4 Write Pseudocode 5 Free Response
5.3.6 Peer Review Session 5 Notes
5.4 Analyzing, Synthesizing, and Reporting Lesson
5.4.1 Analyze Data 5 Free Response
5.4.2 Create Visualization 5 Free Response
5.4.4 Present Your Findings 5 Free Response
micro:bit Unit 2: Program Control
6.1 For Loops Lesson
6.1.1 For Loops 5 Video
6.1.2 For Loops 2 Check for Understanding
6.1.3 Light Middle Row with a For Loop 5 Notes
6.1.4 Dimming LED 5 Free Response
6.1.5 Exploration: Playing Music with micro:bit 5 Connection
6.1.6 Exploration 2.1 Follow-up 5 Video
6.1.7 Twinkle Twinkle 5 Exercise
6.1.8 Looping through LED Brightness Values 5 Exercise
6.1.9 Light Screen by Column 5 Exercise
6.2 While Loops Lesson
6.2.1 While Loops 5 Video
6.2.2 While Loops 2 Check for Understanding
6.2.3 Light Middle Column with a While Loop 5 Notes
6.2.4 Alternating LED until Button Press 5 Free Response
6.2.5 Exploration: Using Buttons to Control Code 5 Connection
6.2.6 Exploration 2.2 Follow-up 5 Video
6.2.7 LED Blink with Buttons 5 Exercise
6.2.8 Button Following LED 5 Exercise
6.3 Operators Lesson
6.3.1 Arithmetic, Comparison, and Logical Operators 5 Video
6.3.2 Arithmetic, Comparison, and Logical Operators 2 Check for Understanding
6.3.3 Using Arithmetic Operators 5 Notes
6.3.4 Using Comparison Operators 5 Notes
6.3.5 Using Logical Operators 5 Notes
6.3.6 Light LED based on Values 5 Free Response
6.3.7 Exploration: Light Sensor 5 Connection
6.3.8 Exploration 2.3a Follow-up 5 Video
6.3.9 Light Detector 5 Exercise
6.3.10 Exploration: Temperature Sensor 5 Connection
6.3.11 Exploration 2.3b Follow-up 5 Video
6.3.12 Temperature Monitor 5 Exercise
6.3.13 Exploration: Accelerometer 5 Connection
6.3.14 Exploration 2.3c Follow-up 5 Video
6.3.15 Brightness by Acceleration 5 Exercise
6.3.16 Real World Application: Night Light 5 Exercise
6.4 If/Else Statements Lesson
6.4.1 If/Else Statements 5 Video
6.4.2 If/Else Statements 2 Check for Understanding
6.4.3 If/Else with 'count' 5 Notes
6.4.4 If/If/If with 'count' 5 Notes
6.4.5 LED Position with a Variable 5 Free Response
6.4.6 LED Brightness using Buttons 5 Free Response
6.4.7 Exploration: Using Servo Motors 5 Connection
6.4.8 Exploration 2.4 Follow-up 5 Video
6.4.9 Servo Rotation with Reset 5 Exercise
6.4.10 Button Controlling LED and Servo 5 Exercise
6.4.11 Servo Position by Button Press 5 Exercise
6.4.12 Servo with LED display and reset 5 Exercise
6.5 Functions Lesson
6.5.1 Functions 5 Video
6.5.2 Functions 2 Check for Understanding
6.5.3 Using Functions 5 Notes
6.5.4 Using Functions with Parameters 5 Notes
6.5.5 Combining Control Structures 5 Notes
6.5.6 Light Level LEDs 5 Free Response
6.5.7 Exploration: Using External Sensors 5 Connection
6.5.8 Exploration 2.5 Follow-up 5 Video
6.5.9 Distance Monitor 5 Exercise
6.5.10 Challenge: LED Arrow Following Servo 5 Challenge
6.5.11 Choose an External Sensor to Investigate 5 Free Response
6.6 Program Control with micro:bit Quiz Lesson
6.6.1 Program Control with micro:bit Unit Quiz 14 Unit Quiz
7.1 micro:bit Challenges Lesson
7.1.1 micro:bit Challenges: Breadboards 5 Video
7.1.2 micro:bit Challenges: Breadboards 2 Check for Understanding
7.1.3 Distance Sensor with Breadboard 5 Notes
7.1.4 Exploration: Using Gestures to Control Code 5 Connection
7.1.5 Exploration 3.1 Follow-up 5 Video
7.1.6 Digital Watch, Pt 1: Setting the Time 5 Challenge
7.1.7 Digital Watch, Pt 2: Keeping TIme 5 Challenge
7.1.8 Digital Watch, Pt 3: Final Touches 5 Challenge
7.1.9 Inchworm 5 Challenge
7.1.10 Project Reflection 5 Free Response
7.2 Explore a New Sensor Lesson
7.2.1 Explore a New Sensor: Overview 5 Notes
7.2.2 Exploration: Getting Started with a New Sensor 5 Connection
7.2.3 Video / Exploration 5 Free Response
7.2.4 Example Program 5 Free Response
7.2.5 How to Add Images 5 Video
7.2.6 Build an Exercise to Teach about your Sensor! 5 Free Response
7.2.7 Creating a Lesson: Reflection 5 Free Response
7.3 Follow a Step-by-Step Project Lesson
7.3.1 Research and Choose Project 5 Free Response
7.3.2 Create Updated Directions 5 Free Response
7.3.3 Step-by-Step Project Reflection 5 Free Response
7.4 Final Project Lesson
7.4.1 Project Brainstorm and Selection 5 Free Response
7.4.2 Build a Prototype 5 Free Response
7.4.3 Test and Improve your Project 5 Free Response
7.4.4 Present your Project! 5 Presentation
Digital Citizenship and Cyber Hygiene
8.1 Digital Footprint and Reputation Lesson
8.1.1 Digital Footprint and Reputation 1 Video
8.1.2 Digital Footprint and Reputation 3 Check for Understanding
8.1.3 Building a Positive Digital Footprint 5 Free Response
8.1.4 Right to be Forgotten? 1 Connection
8.1.5 Right to be Forgotten 5 Free Response
8.1.6 What is your Digital Footprint? 5 Free Response
8.1.7 Social Media Clean-up 1 Example
8.2 Cyberbullying Lesson
8.2.1 Cyberbullying 1 Video
8.2.2 Cyberbullying 3 Check for Understanding
8.2.3 Scenario: Student Ranking 5 Free Response
8.2.4 Scenario: Singled Out 5 Free Response
8.2.5 Stopping Cyberbullying 5 Free Response
8.3 Internet Safety Lesson
8.3.1 Internet Safety 1 Video
8.3.2 Internet Safety 2 Check for Understanding
8.3.3 Scenario: School Stranger 5 Free Response
8.3.4 Scenario: Vacation Pals 5 Free Response
8.3.5 Staying Safe 5 Free Response
8.4 Privacy & Security Lesson
8.4.1 What is Data Privacy & Security? 1 Video
8.4.2 Privacy & Security Quiz 2 Check for Understanding
8.4.7 Guess: Password List 1 Example
8.4.8 Guess: Using an Algorithm 1 Example
8.4.9 Guess: Brute Force 1 Example
8.5 Project: Public Service Announcement Lesson
8.5.1 Pick a Topic 5 Free Response
8.5.2 Research 5 Free Response
8.5.3 Choose Your Audience 5 Free Response
8.5.4 What kind of PSA? 5 Free Response
8.5.5 Draft your PSA 5 Free Response
8.5.6 Finalize your PSA! 15 Free Response
8.6 Digital Citizenship and Cybersecurity Quiz Lesson
8.6.1 Digital Citizenship and Cyber Hygiene Quiz 10 Quiz
What is Computing?
9.1 History of Computers Lesson
9.1.1 History of Computers 1 Video
9.1.2 Video Quiz 3 Check for Understanding
9.1.3 Mission: Who invented the computer? 1 Notes
9.1.4 Evidence Collection 5 Free Response
9.1.5 Exhibit A: Charles Babbage 5 Connection
9.1.6 Exhibit B: Ada Lovelace 1 Connection
9.1.7 Exhibit C: Alan Turing 5 Connection
9.1.8 Exhibit D: Mauchly and Eckert 1 Connection
9.1.9 Exhibit E: ENIAC Programmers 5 Connection
9.1.10 Exhibit F: Grace Hopper 5 Connection
9.1.11 Exhibit G: Mark Dean 5 Connection
9.1.12 Bonus Exhibit: Computer Inventors 5 Connection
9.1.13 Culminating Activity 5 Free Response
9.2 Computer Organization Lesson
9.2.1 Computer Organization 1 Video
9.2.2 Video Quiz 4 Check for Understanding
9.2.3 Draw a Computer 1 Connection
9.2.4 What Kind of Device? 5 Check for Understanding
9.3 Software Lesson
9.3.1 Software 1 Video
9.3.2 Software Quiz 2 Check for Understanding
9.3.3 Software Explained 1 Connection
9.3.4 Computer Applications You Use 5 Free Response
9.3.5 Operating Systems 5 Free Response
9.4 Hardware Lesson
9.4.1 Hardware 1 Video
9.4.2 Hardware Quiz 3 Check for Understanding
9.4.3 Pick the Label 4 Check for Understanding
9.4.4 Label Your Computer 5 Free Response
9.4.5 Computer Analogy 5 Free Response
9.4.6 Hardware vs. Software 5 Free Response
9.5 Future of Computing Lesson
9.5.1 Future of Computing 1 Video
9.5.2 Video Quiz 3 Check for Understanding
9.5.3 Using DNA for Storage 1 Connection
9.5.4 Class Activity: Advancing Technology 5 Free Response
9.5.5 Pros and Cons of AI 1 Connection
9.5.6 AI: Is It a Bad Thing? 5 Free Response
9.6 Troubleshooting Lesson
9.6.1 Troubleshooting Methodology 1 Notes
9.6.2 Identify the Problem 5 Free Response
9.6.3 Research Solutions 5 Free Response
9.6.4 Establish a Theory 5 Free Response
9.6.5 Test the Theory 5 Free Response
9.6.6 Fix the Problem and Document 10 Challenge
9.7 What is Computing? Quiz Lesson
9.7.1 What is Computing? Quiz 15 Unit Quiz
The ABCs of Cryptography
10.1 Cryptography, Cryptology, Cryptanalysis Lesson
10.1.1 Cryptography, Cryptology, Cryptanalysis 1 Video
10.1.2 Cryptography, Cryptology, Cryptanalysis 2 Check for Understanding
10.1.3 Cryptogram Game! 1 Example
10.1.4 Why encrypt? 1 Video
10.1.5 Why encrypt? 2 Check for Understanding
10.1.6 Encrypt/Decrypt 1 Example
10.2 History of Cryptography Lesson
10.2.1 Cryptography: A Brief History 1 Video
10.2.2 Cryptography History Quiz 2 Check for Understanding
10.2.3 How the Enigma Worked 1 Connection
10.2.4 How the Enigma Worked 5 Free Response
10.2.5 Unknown Languages and the Future of Cryptography 1 Connection
10.2.6 The Future of Cybersecurity 5 Free Response
10.3 Basic Crypto Systems: Caesar Cipher Lesson
10.3.1 Caesar Cipher 1 Video
10.3.2 Caesar Cipher 2 Check for Understanding
10.3.3 Caesar Cipher Encryption 1 Example
10.3.4 Decrypt Caesar's Cipher! 1 Example
10.4 Basic Crypto Systems: Cracking Caesar Lesson
10.4.1 Cracking Caesar Cipher 1 Video
10.4.2 Cracking Caesar Cipher 2 Check for Understanding
10.4.3 Cracking Caesar with Brute Force 1 Example
10.4.4 Letter Frequency and Caesar 1 Example
10.4.5 Examining Caesar Cipher 5 Free Response
10.5 Basic Crypto Systems: Vigenere Cipher Lesson
10.5.1 Vigenere Cipher 1 Video
10.5.2 Vigenere Cipher 1 Check for Understanding
10.5.3 Vigenere Cipher Example 1 Example
10.5.4 Letter Frequency and Vigenere Cipher 1 Example
10.5.5 Examining Vigenere Cipher 5 Free Response
10.5.6 Improving Vigenere 1 Example
10.6 The ABCs of Cryptography Quiz Lesson
10.6.1 The ABCs of Cryptography Quiz 15 Unit Quiz
Web Design
11.1 Introduction to HTML Lesson
11.1.1 Introduction to HTML 1 Video
11.1.2 Introduction to HTML Quiz 1 Check for Understanding
11.1.3 Our First HTML Page 1 Example
11.1.4 Say Hello! 5 Exercise
11.2 Structure of an HTML Page Lesson
11.2.1 Structure of an HTML Page 1 Video
11.2.2 Structure of an HTML Page Quiz 1 Check for Understanding
11.2.3 HTML Template 1 Example
11.2.4 Hello World Page 1 Example
11.2.5 The <title> Tag 5 Exercise
11.2.6 Your First HTML Page 5 Exercise
11.3 Formatting Text Lesson
11.3.1 Formatting Text 1 Video
11.3.2 Formatting Text Quiz 1 Check for Understanding
11.3.3 Dictionary 1 Example
11.3.4 That's Bold 5 Exercise
11.3.5 Artificial Intelligence 5 Exercise
11.3.6 State Capitals 5 Exercise
11.4.2 Links Quiz 1 Check for Understanding
11.4.3 The <a> Tag 1 Example
11.4.5 My Favorite Websites 5 Exercise
11.5 Images Lesson
11.5.1 Images 1 Video
11.5.2 Images Quiz 1 Check for Understanding
11.5.3 The <img> Tag 1 Example
11.5.4 Building the CodeHS Homepage 1 Example
11.5.5 Collage on a Theme 5 Exercise
11.5.6 Linking an Image 5 Exercise
11.5.7 Personal Library 5 Exercise
11.5.8 Mindsets 5 Survey
11.6.2 Copyright Quiz 2 Check for Understanding
11.6.3 Citing Sources Example 1 Example
11.6.4 Exploring Creative Commons 1 Connection
11.6.5 Respond: Creative Commons 5 Free Response
11.6.6 Finding Images 5 Free Response
11.6.7 Make a Collage 5 Exercise
11.7 HTML Lists Lesson
11.7.1 HTML Lists 1 Video
11.7.2 HTML Lists Quiz 1 Check for Understanding
11.7.3 Grocery Shopping 1 Example
11.7.4 Favorite Things 5 Exercise
11.7.5 To-Do List 5 Exercise
11.7.6 List Article 5 Exercise
11.8 HTML Tables Lesson
11.8.1 HTML Tables 1 Video
11.8.2 HTML Tables Quiz 3 Check for Understanding
11.8.4 Favorite Songs 5 Exercise
11.8.5 Calendar 5 Exercise
11.9 Viewing Websites Lesson
11.9.1 Viewing Websites 1 Video
11.9.2 Viewing Websites Quiz 5 Check for Understanding
11.9.3 Explaining a URL 5 Free Response
11.10 Project: Your First Website, Pt 1 Lesson
11.10.1 Your First Website 10 Challenge
11.10.2 Set Up Your Domain 10 Notes
11.11 HTML Styling Lesson
11.11.1 HTML Styling 1 Video
11.11.2 HTML Styling Quiz 1 Check for Understanding
11.11.3 Stylish Address Book 1 Example
11.11.4 Background Colors 5 Exercise
11.11.5 Style Your To-Do List 5 Exercise
11.12 Introduction to CSS Lesson
11.12.1 Introduction to CSS 1 Video
11.12.2 Introduction to CSS Quiz 2 Check for Understanding
11.12.3 Styling your H1s 1 Example
11.12.4 First style with CSS 5 Exercise
11.12.5 List Styling 5 Exercise
11.13 CSS Select by Tag Lesson
11.13.1 CSS Select by Tag 1 Video
11.13.2 CSS Select by Tag Quiz 1 Check for Understanding
11.13.3 Rainbow 1 Example
11.13.4 Dog Styling 1 Example
11.13.6 Put Karel Together 5 Exercise
11.14 CSS Select by Class Lesson
11.14.1 CSS Select by Class 1 Video
11.14.2 CSS Select by Class Quiz 1 Check for Understanding
11.14.3 Simple Checkerboard 1 Example
11.14.4 Tic Tac Toe 5 Exercise
11.14.5 Music Library 5 Exercise
11.15 CSS Select by ID Lesson
11.15.1 CSS Select by ID 1 Video
11.15.2 CSS Select by ID Quiz 1 Check for Understanding
11.15.3 Logo 1 Example
11.15.4 Favorite Dog 5 Exercise
11.15.5 Bingo 5 Exercise
11.16 Project: Your First Website, Pt 2 Lesson
11.16.1 Adding Style with CSS 5 Challenge
11.17 Web Design Quiz Lesson
11.17.1 Web Design Quiz 14 Unit Quiz
Project: Designing for Impact
12.1 Intro to Design Thinking Lesson
12.1.1 Intro to Design Thinking 1 Video
12.1.2 Intro to Design Thinking 2 Check for Understanding
12.1.3 User Interface Scavenger Hunt 5 Free Response
12.1.4 Case Study: Helping Blind People See 1 Connection
12.1.5 Case Study Responses 5 Free Response
12.1.6 Topic Brainstorm 5 Free Response
12.1.7 Narrowing Down Topics 5 Free Response
12.2 Empathy Lesson
12.2.1 Empathy 1 Video
12.2.2 Empathy Quiz 2 Check for Understanding
12.2.3 Accessibility 1 Connection
12.2.4 Accessibility Tips 5 Free Response
12.2.5 Accessibility: Designing for ALL 5 Free Response
12.2.6 How to Interview 1 Connection
12.2.7 How to Interview 5 Free Response
12.2.8 User Interview 5 Free Response
12.2.9 Using Surveys to Collect User Data 5 Connection
12.2.10 Create Your Survey and Gather Data 5 Free Response
12.2.11 Survey Data Cleaning 5 Connection
12.2.12 Survey Data Cleaning 5 Free Response
12.2.13 Drawing Conclusions from Data 5 Free Response
12.3 Define Lesson
12.3.1 Define 1 Video
12.3.2 Define Quiz 2 Check for Understanding
12.3.3 Make a Composite Character Profile 1 Connection
12.3.4 Composite Character Profile 5 Free Response
12.3.5 Point-of-View Statement Brainstorm 5 Free Response
12.3.6 POV Statement 5 Free Response
12.4 Ideate Lesson
12.4.1 Ideate 1 Video
12.4.2 Ideate Quiz 1 Check for Understanding
12.4.3 Brainstorming Tips 1 Connection
12.4.4 Ideate! 5 Free Response
12.5 Prototype Lesson
12.5.1 Prototype 1 Video
12.5.2 Prototype Quiz 1 Check for Understanding
12.5.3 Brainstorm Selection 1 Connection
12.5.4 Harvest Ideas from the Brainstorm 5 Free Response
12.5.5 Wizard of Oz Prototyping 1 Connection
12.5.6 Example Wizard of Oz Paper Prototype 1 Connection
12.5.7 Make Your Paper Prototypes! 5 Free Response
12.6 Test Lesson
12.6.1 Test 1 Video
12.6.2 Testing Quiz 1 Check for Understanding
12.6.3 Testing with Users 1 Connection
12.6.4 Example: How to User Test 1 Connection
12.6.5 How to User Test Responses 5 Free Response
12.6.6 Example: How NOT to User Test 1 Connection
12.6.7 How NOT to User Test Responses 5 Free Response
12.6.8 Test Prototype 1 5 Free Response
12.6.9 Test Prototype 2 5 Free Response
12.6.10 Improve Your Prototype 5 Free Response
12.7.1 Project Planning 5 Free Response
12.7.2 Build Your Website! 5 Challenge
Project: The Effects of the Internet
13.1 Project: The Effects of the Internet Lesson
13.1.1 Topic Brainstorm 5 Free Response
13.1.2 Project Planning: Timeline and Roles 5 Free Response
13.1.3 Gathering Resources 5 Free Response
13.1.4 Make Your Project 25 Presentation
Introduction to Programming with Karel the Dog
14.1 Introduction to Programming With Karel Lesson
14.1.1 Introduction to Programming With Karel 1 Video
14.1.2 Quiz: Karel Commands 1 Quiz
14.1.3 Our First Karel Program 1 Example
14.1.4 Your First Karel Program 5 Exercise
14.1.5 Short Stack 5 Exercise
14.2 More Basic Karel Lesson
14.2.1 More Basic Karel 1 Video
14.2.2 More Basic Karel Quiz 4 Check for Understanding
14.2.3 Tennis Ball Square 1 Example
14.2.4 Make a Tower 5 Exercise
14.2.5 Pyramid of Karel 5 Exercise
14.3 Karel Can't Turn Right Lesson
14.3.1 Karel Can't Turn Right 1 Video
14.3.2 Karel Can't Turn Right Quiz 2 Check for Understanding
14.3.3 Tower and Turn Right 1 Example
14.3.4 Fireman Karel 5 Exercise
14.3.5 Slide Karel 5 Exercise
14.4 Functions in Karel Lesson
14.4.1 Functions in Karel 1 Video
14.4.2 Functions in Karel Quiz 1 Check for Understanding
14.4.3 Turn Around 1 Example
14.4.4 Pancakes 5 Exercise
14.4.5 Mario Karel 5 Exercise
14.5 The Start Function Lesson
14.5.1 The Start Function 1 Video
14.5.2 The Start Function Quiz 2 Check for Understanding
14.5.3 Tower with Start Function 1 Example
14.5.4 Pancakes with Start 5 Exercise
14.6 Top Down Design and Decomposition in Karel Lesson
14.6.1 Top Down Design and Decomposition in Karel 1 Video
14.6.2 Top Down Design and Decomposition Quiz 2 Check for Understanding
14.6.3 Hurdle Karel 1 Example
14.6.4 The Two Towers 5 Exercise
14.7.1 Commenting Your Code 1 Video
14.7.2 Commenting Your Code Quiz 1 Check for Understanding
14.7.3 Hurdle Karel 1 Example
14.7.4 The Two Towers + Comments 5 Exercise
14.8 Super Karel Lesson
14.8.1 Super Karel 1 Video
14.8.2 Super Karel Quiz 1 Check for Understanding
14.8.3 Hurdle Karel (with SuperKarel) 1 Example
14.8.4 The Two Towers + SuperKarel 5 Exercise
14.9 For Loops Lesson
14.9.1 For Loops 1 Video
14.9.2 For Loops Quiz 1 Check for Understanding
14.9.3 Repeated Move 1 Example
14.9.4 Put Down Tennis Balls 1 Example
14.9.5 Take 'em All 5 Exercise
14.9.6 Dizzy Karel 5 Exercise
14.9.7 For Loop Square 5 Exercise
14.9.8 Lots of Hurdles 5 Exercise
14.10 If Statements Lesson
14.10.1 If Statements 1 Video
14.10.2 If Statements Quiz 2 Check for Understanding
14.10.3 If Statements 1 Example
14.10.4 Safe Take Ball 1 Example
14.10.5 Is There a Ball? 5 Exercise
14.10.6 Don't Crash 5 Exercise
14.10.7 Mindsets 5 Survey
14.11 If/Else Statements Lesson
14.11.1 If/Else Statements 1 Video
14.11.2 If/Else Statements Quiz 2 Check for Understanding
14.11.3 If/Else Statements 1 Example
14.11.4 One Ball in Each Spot 1 Example
14.11.5 Right Side Up 5 Exercise
14.11.6 Right vs. Left Square 5 Exercise
14.12 While Loops in Karel Lesson
14.12.1 While Loops in Karel 1 Video
14.12.2 While Loops in Karel Quiz 2 Check for Understanding
14.12.3 Move to Wall 1 Example
14.12.5 Lay Row of Tennis Balls 5 Exercise
14.12.6 Big Tower 5 Exercise
14.13 Control Structures Example Lesson
14.13.1 Control Structures Example 1 Video
14.13.2 Control Structures Example Quiz 2 Check for Understanding
14.13.3 Cleanup Karel 1 Example
14.13.4 Random Hurdles 5 Exercise
14.14 More Karel Examples and Testing Lesson
14.14.1 More Karel Examples and Testing 1 Video
14.14.2 Quiz: Which Control Structure? 5 Check for Understanding
14.14.3 Move Tennis Ball Stack 1 Example
14.14.4 Climbing Karel 1 Example
14.15 How to Indent Your Code Lesson
14.15.1 How to Indent Your Code 1 Video
14.15.2 How to Indent Your Code Quiz 1 Check for Understanding
14.15.3 Dance and Clean Karel 1 Example
14.15.4 Diagonal 5 Exercise
14.15.5 Staircase 5 Exercise
14.16 Karel Challenges Lesson
14.16.1 Fetch 10 Challenge
14.16.2 Racing Karel 10 Challenge
14.16.3 Go Through the Fence 10 Challenge
14.16.4 Escape Karel 10 Challenge
14.17 Intro to Programming with Karel the Dog Quiz Lesson
14.17.1 Intro to Programming with Karel the Dog Quiz 25 Unit Quiz
Extra Karel Puzzles
15.1 Extra Karel Puzzles Lesson
15.1.1 Midpoint Karel 10 Challenge
15.1.2 Target Karel 10 Challenge
15.1.3 The Winding Yellow Road 10 Challenge
15.1.4 Super Random Hurdles 10 Challenge
15.1.5 Copy 10 Challenge
15.1.6 Multiply 10 Challenge
15.1.7 Fibonacci Karel 10 Challenge
15.1.8 Comparison Karel 10 Challenge
15.1.9 Swap 10 Challenge
15.1.10 Sorting Karel 10 Challenge
What is Cybersecurity?
16.1 Module Overview - Cybersecurity Lesson
16.1.1 Welcome to Cybersecurity 1 Video
16.1.2 Welcome to Cybersecurity 2 Check for Understanding
16.1.4 Cybersecurity and You 5 Free Response
16.1.5 Course Goals 5 Free Response
16.2 What is Cybersecurity? Lesson
16.2.1 What is Cybersecurity? 1 Video
16.2.2 What is Cybersecurity? 3 Check for Understanding
16.2.3 City Services Ransomware 1 Connection
16.2.4 Ransomware Reflection 5 Free Response
16.2.5 Ransomware Simulator 1 Example
16.2.6 Internet of Things 1 Connection
16.2.7 Hackers vs. Smart Homes 1 Connection
16.2.8 Internet of Things Reflection 5 Free Response
16.2.9 Threat Map 1 Connection
16.2.10 Why Learn about Cybersecurity? 5 Free Response
16.3 Impact of Cybersecurity Lesson
16.3.1 Impact of Cybersecurity 1 Video
16.3.2 Impact of Cybersecurity 2 Check for Understanding
16.3.3 Phishing for Your Info 1 Connection
16.3.4 Phishing Reflection 5 Free Response
16.3.5 Cyber Game 1 Connection
16.3.6 Cyber Game Reflection 5 Free Response
16.3.7 Cyber Crime Time 1 Connection
16.3.8 Cyber Crime Time Reflection 5 Free Response | 9,795 | 30,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.508597 |
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# Right not the GMAt type but found in a book where I was
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Right not the GMAt type but found in a book where I was trying to learn perm and comb
There are 10,000 cars in a city, license number from 00001 to 10000, when you meet a car, what is the probability there is number 8 in license number ?
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Display posts from previous: Sort by | 532 | 1,825 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-22 | longest | en | 0.888105 |
https://www.jobilize.com/course/section/activity-to-investigate-and-extend-numerical-patterns-lo-2-1-2?qcr=www.quizover.com | 1,620,440,044,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988831.77/warc/CC-MAIN-20210508001259-20210508031259-00167.warc.gz | 877,674,629 | 18,973 | # 2.49 To investigate and extend numerical patterns
Page 1 / 1
## Memorandum
INTRODUCTION
The learning programme for grade six consists of five modules:
1. Number concept, Addition and Subtraction
2. Multiplication and Division
3. Fractions and Decimal fractions
4. Measurement and Time
5. Geometry; Data handling and Probability
• It is important that educators complete the modules in the above sequence, as the learners will require the knowledge and skills acquired through a previous module to be able to do the work in any subsequent module.
COMMON AND DECIMAL FRACTIONS (LO 1; 2 AND 5)
LEARNING UNIT 1 FOCUSES ON COMMON FRACTIONS
• This module continues the work dealt with in grade 5. Addition and subtraction of fractions are extended and calculation of a fraction of a particular amount is revised.
• Check whether the learners know the correct terminology and are able to use the correct strategies for doing the above correctly.
• Critical outcome 5 (Communicating effectively by using visual, symbolic and /or language skills in a variety of ways) is addressed.
• It should be possible to work through the module in 3 weeks.
• ** Activity 17 is designed as a portfolio task. It is a very simple task, but learners should do it neatly and accurately. They must be informed in advance of how the educator will be assessing the work.
• LEARNING UNIT 2 FOCUSES ON DECIMAL FRACTIONS
• This module extends the work that was done in grade 5. Learners should be able to do rounding of decimal fractions to the nearest tenth, hundredth and thousandth. Emphasise the use of the correct method (vertical) for addition and subtraction. Also spend sufficient time on the multiplication and division of decimal fractions.
• As learners usually have difficulty with the latter, you could allow 3 to 4 weeks for this section of the work.
** Activity 19 is a task for the portfolio. The assignment is fairly simple, but learners should complete it neatly and accurately. They must be informed in advance of how the educator will be assessing the work.
1.1 0,0009 ; 0,01 ; 0,011 ; 0,021
1.2 0,024 ; 0,023 ; 0,022 ;
1.3 0,75 ; 1 ; 1,25 ; 1,5
1.4 0,015 ; 0,02 ; 0,025 ; 0,03
1.5 4,25 ; 4 ; 3,75 ; 3,5
1.6 2,65 ; 2,475 ; 2,3 ; 2,125
## Activity: to investigate and extend numerical patterns [lo 2.1.2]
1. Take a careful look at the following numbers. Say each out loud. Then try to see the pattern. If you are able to do it, complete the number patterns in writing without using your pocket calculator. Ask your educator for assistance if you struggle to do it:
1.1 0,007; 0,008; ....................; ....................; ....................; ....................
• 0,026; 0,025; ....................; ....................; ....................; ....................
1.3 0,25; 0,5; ....................; ....................; ....................; ....................
1.4 0,005; 0,010; ....................; ....................; ....................; ....................
1.5 4,75; 4,5; ....................; ....................; ....................; ....................
1.6 3; 2,825; ....................; ....................; ....................; ....................
2. Check your answers with the help of a pocket calculator.
## Assessment
Learning Outcome 2: The learner will be able to recognise, describe and represent patterns and relationships, as well as to solve problems using algebraic language and skills.
Assessment Standard 2.1: We know this when the learner investigates and extends numeric and geometric patterns looking for a relationship or rules, including patterns:
2.1.2 investigates and extends numeric and geometric patterns looking for a relationship or rules, including patterns.
#### Questions & Answers
are nano particles real
Missy Reply
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
Lale Reply
no can't
Lohitha
where we get a research paper on Nano chemistry....?
Maira Reply
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
Maira Reply
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
Google
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
Hafiz Reply
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
Jyoti Reply
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
Crow Reply
what about nanotechnology for water purification
RAW Reply
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
LITNING Reply
what is a peer
LITNING Reply
What is meant by 'nano scale'?
LITNING Reply
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
Bob Reply
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
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Source: OpenStax, Mathematics grade 6. OpenStax CNX. Sep 10, 2009 Download for free at http://cnx.org/content/col11030/1.1
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By Saylor Foundation By Carly Allen By Anonymous User By Marion Cabalfin By Sandy Yamane By Stephen Voron By John Gabrieli By OpenStax By John Gabrieli By Rohini Ajay | 1,799 | 7,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2021-21 | latest | en | 0.911379 |
http://www.traditionaloven.com/tutorials/angle/convert-arcsecond-angle-unit-to-grad-angle-unit.html | 1,506,016,164,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687834.17/warc/CC-MAIN-20170921172227-20170921192227-00467.warc.gz | 594,260,622 | 11,764 | angle conversion
Amount: 1 arcsecond ('') of angle Equals: 0.00031 grads (grad) in angle
Converting arcsecond to grads value in the angle units scale.
TOGGLE : from grads into arc-seconds in the other way around.
angle from arcsecond to grad Conversion Results:
Enter a New arcsecond Amount of angle to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other angle measuring units - complete list.
Conversion calculator for webmasters.
Angles
This calculator is based on conversion of two angle units. An angle consists of two rays (as in sides of an angle sharing a common vertex or else called the endpoint.) Some belong to rotation measurements - spherical angles measured by arcs' lengths, pointing from the center, plus the radius. For a whole set of multiple units of angle on one page, try that Multiunit converter tool which has built in all angle unit-variations. Page with individual angle units.
Convert angle measuring units between arcsecond ('') and grads (grad) but in the other reverse direction from grads into arc-seconds.
conversion result for angle: From Symbol Equals Result To Symbol 1 arcsecond '' = 0.00031 grads grad
Converter type: angle units
This online angle from '' into grad converter is a handy tool not just for certified or experienced professionals.
First unit: arcsecond ('') is used for measuring angle.
0.00031 grad is converted to 1 of what?
The grads unit number 0.00031 grad converts to 1 '', one arcsecond. It is the EQUAL angle value of 1 arcsecond but in the grads angle unit alternative.
How to convert 2 arc-seconds ('') into grads (grad)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
0.000308641975309 * 2 (or divide it by / 0.5)
QUESTION:
Other applications for this angle calculator ...
With the above mentioned two-units calculating service it provides, this angle converter proved to be useful also as a teaching tool:
1. in practicing arc-seconds and grads ( '' vs. grad ) values exchange.
2. for conversion factors training exercises between unit pairs.
3. work with angle's values and properties.
International unit symbols for these two angle measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for arcsecond is:
''
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for grad is: | 564 | 2,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-39 | longest | en | 0.81584 |
https://equalstreets.org/how-many-miles-is-1300-km/ | 1,685,361,153,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644855.6/warc/CC-MAIN-20230529105815-20230529135815-00050.warc.gz | 266,474,494 | 10,219 | # How Many Miles Is 1300 Km
How Many Miles Is 1300 Km. Distance between cities or 2 locations are measured in both kilometers, miles and nautical. 1 kilometer is equal to 1/1.609344 miles:
[km] = miles / 0.6213. To convert 1300 kilometers into miles we have to multiply 1300 by the conversion factor in order to get the length amount from kilometers to miles. The distance d in miles (mi) is equal.
### Value In Mile = 1300 × 0.62137119223733 1300 × 0.62137119223733
1m = 1/1.609344mi = 0.6213711mi. Using the conversion formula above, you will get: Value in km = value in mile × 1.609344.
### How Many Miles In 1 Km?
The distance d in miles (mi) is equal. It is commonly used to measure the distance. 1 km = 0.62137119223733 mi.
### Suppose You Want To Convert 1300 Mile Into Km.
You can view more details on each measurement unit: How far is 1,300 miles in kilometers? How many km in 1 miles?
### 26 Rows How Far Is 1,300 Kilometers In Miles?
You can view more details on each measurement unit: To convert 1300 miles into kilometers we have to multiply 1300 by the conversion factor in order to get the length amount from miles to kilometers. [mi] = 1300 * 0.6213 = 807.69.
### Distance Between Cities Or 2 Locations Are Measured In Both Kilometers, Miles And Nautical.
Km or miles the si. Suppose you want to convert 1300 km into miles. The distance d in kilometers (km) is equal to the. | 396 | 1,402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.833197 |
https://www.perlmonks.org/bare/index.pl/?node_id=280334 | 1,624,551,769,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556133.92/warc/CC-MAIN-20210624141035-20210624171035-00345.warc.gz | 854,187,974 | 9,592 | artist has asked for the wisdom of the Perl Monks concerning the following question:
Dear Monks
I am trying to solve a permutation probelm and looking for pointers. The different permutation we can create by putting pair of paranthesis before,after or in between given number.
Example: for p=2 and n= 123, we may have:
```(1) (2) 3
(1) 2 (3)
(1 2) (3)
(1 (2 3))
((1 2 3))
1 ((2 3))
```
etc..
Thanks,
artist
Replies are listed 'Best First'.
(jeffa) Re: Parens permutations
by jeffa (Bishop) on Aug 02, 2003 at 23:28 UTC
Interesting problem ... this is as far as i got before giving up. I borrowed some of merlyn's code from Permutations and combinations:
```for my \$row (combinations(1..3)) {
my @row = 1..3;
\$row[\$_-1] = "(\$row[\$_-1])" for @\$row;
print "@row\n";
}
# from [id://24270] by [merlyn]
sub combinations {
return [] unless @_;
my \$first = shift;
my @rest = combinations(@_);
return @rest, map { [\$first, @\$_] } @rest;
}
__END__
1 2 3
1 2 (3)
1 (2) 3
1 (2) (3)
(1) 2 3
(1) 2 (3)
(1) (2) 3
(1) (2) (3)
[download]```
jeffa
```L-LL-L--L-LL-L--L-LL-L--
-R--R-RR-R--R-RR-R--R-RR
B--B--B--B--B--B--B--B--
H---H---H---H---H---H---
(the triplet paradiddle with high-hat)
```
Re: Parens permutations
by dws (Chancellor) on Aug 02, 2003 at 23:51 UTC
If 1 ((2 3)) is valid, then what about 1 (((2 3))) ? If that's valid, then you have a countably infinite number of possible permutations. I kinda doubt the problem is phrased to allow that.
Update: D'oh. So that's what p= means.
Hi dws,
The number of parens are given. So if p = 2 then
```1 ((2 3)) => valid
((1 2 3)) => valid
1 (((2 3))) => invalid
```
Re: Parens permutations
by barrachois (Pilgrim) on Aug 03, 2003 at 21:53 UTC
I didn't make much progress on an analytic solution, but here's some code to traverse the permutations by brute force, and some thinking out loud. Kinda fun.
In what context did this problem come about?
- barrachois
```#!/usr/bin/perl -w
use strict;
# Dear Monks
#
# I am trying to solve a permutation probelm and looking for
#
# pointers. The different permutation we can create by putting pair of
# paranthesis before,after or in between given number. Example: for p
+=2
# and n= 123, we may have:
#
# (1) (2) 3
# (1) 2 (3)
# (1 2) (3)
# (1 (2 3))
# ((1 2 3))
# 1 ((2 3))
#
# etc..
# Thanks,
# artist
#
# --------------------------------------------------------
#
# Given n=3, p=2 as above, without parens we have n+1=4
# places where parens can be placed, i.e. on the dots in
#
# . 1 . 2 . 3 .
#
# Assuming that arrangements like "() 1 2 3" are illegal,
# a left-paren may be placed at a dot numbered
# L=0..(n-1) and its corresponding right-paren at R=(L+1)..n
#
# Therefore the number of ways I can place a single pair
# of parens around n digits is
#
# 3 (1) 2 3 , (1 2) 3 , (1 2 3) L=0; R=1,2,3
# + 2 1 (2) 3 , 1 (2 3) L=1; R=2,3
# + 1 1 2 (3) L=2; R=3
# ---
# total 6 = (n)(n+1)/2
#
# Now let's try to place another pair of parens.
# These can be put at the same dot locations;
# the trick will be to avoid counting arrangements like
# "(1)(2)3" twice. For visual clarity, I'll use
# sqaure brackets for the second parens, working
# forward from the six single paren patterns above.
#
# For the first pattern "(1) 2 3", adding a second
# pair of brackets again gives six possibilities.
#
# [(1)] 2 3 , [(1) 2] 3 , [(1) 2 3] 6 new
# (1) [2] 3 , (1) [2 3]
# (1) 2 [3]
#
# However, trying the same approach to the second
# pattern "(1 2) 3" starts to give arrangments that
# have already been found. (Remember that the []
# and () symbols are actually identical and interchangeable.)
#
# Try all six combinations starting from the next
# single pattern "(1 2) 3" gives
# ([1] 2) 3 *OLD*, ([1 2]) 3 *NEW*, [(1 2) 3] *NEW* +5 new
# (1 [2]) 3 *NEW*, (1 [2) 3] *NEW*,
# (1 2) [3] *NEW*
# of which the first is a repeat of "[(1) 2] 3" and
# the rest are new.
#
# Continuing in the same vein gives
#
# ([1] 2 3) *OLD*, ([1 2] 3) *OLD*, [(1 2 3)] *NEW* +3 new
# (1 [2] 3) *OLD*, (1 [2 3]) *NEW*,
# (1 2 [3]) *NEW*
#
# Hmmm. I stared at this but haven't really found an obvious pattern.
#
# But whether there is or not, feels like its time
# to write some code. If I loop over all the possible
# positions of the left and right parens, as I'm doing above,
# I can just remember which ones I've found already in
# a hash, and keep the new ones.
#
# This may not be the most efficient, but it'll get the job done.
#
# Here's what the output look like :
#
# Looking for permutations of 2 pairs of parens around 3 digits.
# ((1))23
# ((1)2)3
# ((1)23)
# (1)(2)3
# (1)(23)
# (1)2(3)
# ((12))3
# ((12)3)
# (1(2))3
# (1(2)3)
# (12)(3)
# ((123))
# (1(23))
# (12(3))
# 1((2))3
# 1((2)3)
# 1(2)(3)
# 1((23))
# 1(2(3))
# 12((3))
# Total number of permutations = 20
my \$n; # number of digits
my \$p; # number of paren pairs
my @leftparens; # count of left parens at each position
my @rightparens; # count of right parens at each position
my %seen; # permutations generated
my \$DEBUG = 0;
init(3,2); # define n,p
print " Looking for permutations of \$p pairs of parens around \$n digit
+s. \n";
addParenPair(1); # add first pair of parens, and others recursively
print " Total number of permutations = " . scalar(keys %seen) . "\n";
# ------------------------------------------------------------
sub init {
(\$n, \$p) = @_;
@leftparens = (0)x(\$n+1);
@rightparens = (0)x(\$n+1);
%seen = ();
}
# Convert current permutation as recorded in @leftparens and @rightpar
+ens to a string.
sub permutationAsString {
my \$result = "("x\$leftparens[0];
for my \$digit (1..\$n){
\$result .= \$digit . ")"x\$rightparens[\$digit] . "("x\$leftparens[\$di
+git];
}
return \$result;
}
# If we haven't seen this one, remember it and print it out.
sub analyzePermutation {
my \$permutation = permutationAsString();
unless (\$seen{\$permutation}){
\$seen{\$permutation}++;
print " " . \$permutation . "\n";
}
}
# Recursive descent routine to put in next pair of parens
# and analyze the result if all the parens have been placed.
# If \$DEBUG is true, it'll also print some progress reports.
sub addParenPair {
my (\$whichPair) = @_;
for my \$left (0..(\$n-1)){
\$leftparens[\$left]++;
for my \$right ((\$left+1)..\$n){
\$rightparens[\$right]++;
if (\$whichPair == \$p){
analyzePermutation();
}
else {
if (\$DEBUG){
print " "x\$whichPair .
"Permutation so far is " .
permutationAsString() . "; adding pair " . (\$whichPair+1)
+. "\n";
}
my \$lastCount = scalar(keys %seen);
addParenPair(\$whichPair+1);
my \$newPerms = scalar(keys %seen) - \$lastCount;
if (\$DEBUG){
print " "x\$whichPair . "new permutations added = \$newPerms.
+ \n";
}
}
\$rightparens[\$right]--;
}
\$leftparens[\$left]--;
}
}
[download]```
Re: Parens permutations (order)
by tye (Sage) on Aug 05, 2003 at 17:02 UTC
Well, I was just about to finish previewing my idea when I realized that it didn't find arrangements of parens like (()()). Which lead me to a simpler solution anyway.
In any case, the trick is to come up with a order for the solutions so that you can find the solutions in that order and then you don't have to worry about checking for duplicates.
This is important because finding duplicates requires you be able to compare a new result to all previous results but the number of results for this type of problem quickly becomes so huge that such a comparison becomes infeasible.
For this problem, I think we need to order first either based on the position of the open parens, or based on the position of the closed parens. So, for a final arrangement like (1((2)3))(4), we can build up the parens either like (123)4 -> (1(23))4 -> (1((2)3)4 -> (1((2)3))(4) or like 1(2)34 -> 1((2)3)4 -> (1((2)3))4 -> (1((2)3))(4). Let's go with the first case.
So, start by building a loop or iterator (we won't have enough memory if we start computing whole lists of solutions along the way) that adds one set of parens: (1)234, (12)34, (123)4, (1234), 1(2)34, 1(23)4, 1(234), 12(3)4, 12(34), and 123(4).
Then improve this so that it knows to add the next set of parens such that 1) nesting is not violated and 2) the opening paren is inserted *after* all of the existing opening parens. Then apply this solution N times to get N sets of parens.
So, if you have 1(23)4, adding the next set of parens would proceed like so: 1((2)3)4, 1((23))4, 1(2(3))4, 1(23)(4). That is, start with the opening paren in the first possible location (right after the last opening paren already present) and progress this position forward one step at a time. For each position of the opening paren, step the position of the closing paren along, starting right after the digit after the new open paren, and continuing until the end of the continguous string of digits (after which we'd either hit end of string or violate nesting of parens).
This very last part almost screams "regex", and with experimental features you could even get the regex engine to find all of these for you as it back-tracks matching \d+? over and over. But such wouldn't give us an iterator (and I don't like using experimental features), so...
Which can give you all of the results:
```> parenperms 3 12
1: (((1)))2
2: ((1))(2)
3: (1)((2))
4: (((1))2)
5: ((1)(2))
6: (((1)2))
7: (((12)))
8: ((1(2)))
9: (1((2)))
10: 1(((2)))
[download]```
Or just a table of counts:
```> parenperms 1..7 1..7
parens lengths
1 2 3 4 5 6 7
1: 1 3 6 10 15 21 28
2: 1 6 20 50 105 196 336
3: 1 10 50 175 490 1176 2520
4: 1 15 105 490 1764 5292 13860
5: 1 21 196 1176 5292 19404 60984
6: 1 28 336 2520 13860 60984 226512
7: 1 36 540 4950 32670 169884 736164
[download]```
- tye
And these counts can be computed much faster using:
```sub Prod {
my( \$i, \$j )= @_;
my \$f= \$i;
\$f *= \$j if \$i < \$j;
for my \$p ( \$i+1..\$j-1 ) {
\$f *= \$p*\$p;
}
return \$f;
}
sub Count {
my( \$parens, \$len )= @_;
return 1 if 1 == \$len;
return Prod(\$parens+1,\$parens+\$len)/Prod(1,\$len)
}
[download]```
Taking \$len==4 for example, that boils down to:
```(P+1)*(P+2)^2*(P+3)^2*(P+4)/1/2^2/3^2/4
[download]```
which is an interesting equation. A mathematician might write it as:
``` (P+L)!/P!*(P+L)!/P!*L/(P+1)/(P+L)/L!/L!
or
[(P+L)!/P!/L!]^2*1*L/(P+1)/(P+L)
[download]```
And it is interesting to me that I need to test for 1==\$len but I don't need to test for 0==\$parens.
Note that replacing \$len with \$parens+1 and \$parens with \$len-1 gives us the same values.
So I suspect the equation can be rewritten as (mostly) the product of two binomial coefficents (number of different subsets of size S you can pick from a set of size N). But I don't have the time to figure out all of the off-by-ones at the moment.
Update:
```[(P+L)!/P!/L!]^2*1*L/(P+1)/(P+L)
== [ C(P+L,P) ] * ??
where
?? == (P+L)!/P!/L!*1*L/(P+1)/(P+L)
== (P+L)!/(P+L) /P!/(P+1) /L!*L
== (P+L-1)! /(P+1)! /(L-1)!
[download]```
But C(P+L,P+1) == C(P+L,L-1) == (P+L)!/(P+1)!/(L-1)! so ?? == C(P+L,P+1)/(P+L). So the number of ways to put P parens into a string of length L is
``` C(P+L,P)*C(P+L,P+1)/(P+L)
or
C(P+L,L)*C(P+L,L-1)/(P+L)
or ...
[download]```
...I think. (:
Now, someone just needs to explain why this formula makes sense. Then we'll have worked the problem "backward" and in future math classes the solution can be presented in the "forward" direction and we can intimidate another generation of students into thinking that they'd never be able to solve math problems... thus continuing a long tradition of mathematicians. ;)
- tye
++tye for generating the excellent formula. Lots of interesting math involved in seemingly simple looking problem. Why the formula make sense is a very good question and I don't have answer at the moment.
I had suspected the factorials once I saw symmetry and little similarity with pascal's triangle, but tye came out faster even with the solution.
I have to find the table which was obtained by >parenperms 1..7 1..7 parens lengths
Re: Parens permutations
by Limbic~Region (Chancellor) on Aug 05, 2003 at 05:44 UTC
artist,
Please forgive me as I am incredibly tired, but I originally told you in the CB this sounded like a math problem. Unfortunately I do not believe it is a simple algebraic formula. Lets change n=123 to n=3 since you are talking about a 3 digit number. Now we have a few problems.
Problem number 1: The equation to determine the number of permutations seems to change with p.
p = 1
permutations = ((n2 + n)/ 2)
p = 2
let k = ((n2 + n)/ 2)
permutations = ((k2 + k)/ 2)
Second problem: There can be some ambiguity in what we are actually grouping:
```(1 (2) 3)
[download]```
Can be seen as either 1 & 3 + 2 or 1 & 2 + 2 & 3. This breaks the formula for p = 2 since:
```p = 2
n = 3
k = 6
permutations = 21
[download]```
is not correct as barrachois pointed out - it is actually 20 since two permutations, while different, look the same.
I didn't bother going any further than this to see if the simple formulas for different values of p could be generalized.
Sorry - L~R
Re: Parens permutations
by graff (Chancellor) on Aug 05, 2003 at 04:56 UTC
I think this could be solved without having to use recursion.
Given:
• m array elements -- e.g. m=3 for qw/1 2 3/
• p paren sets -- e.g. p=2 for "(( ))" or "()()"
You have m + 1 positions around the array elements, such that:
• m positions may have 0 .. p open parens (i.e. the final position cannot have any)
• m positions may have 0 .. p close parens (i.e. the initial position cannot have any)
• the i-th close paren must be placed in any position to the right of the i-th open paren (i.e. if the i-th open paren is at position j, then the i-th close paren must be at position j+k where k is greater than zero, but less than or equal to 1+m-j).
There would be a number of ways to build a nested loop to iterate through the enumeration of the set -- e.g. start with all open parens at position 0 and all close parens at position 1, move the close parens across the remaining positions in the inner loop, and move the open parens in the outer loop, or something to that effect. I expect there would also be some clever formula to compute the size of the set, given just the values of m and p, without having to enumerate.
That's how I would proceed if I needed to do this (but I don't feel like I need to just now...)
update: I guess this is rather close to the solution that barrachois proposed above. I forgot to mention that, when printing the enumeration, the following condition is also needed: when a medial position contains both open and close parens, simply print all the close parens first, then all the open parens, for that position. This, along with some logic to do the correct initial placement of close parens for each possible placement of open parens, will ensure that the output set is well formed (barrachois' recursive descent of the structure to check proper nesting would not be needed). | 4,899 | 15,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-25 | latest | en | 0.837672 |
http://ocaml.org/learn/taste.html | 1,558,727,809,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257731.70/warc/CC-MAIN-20190524184553-20190524210553-00443.warc.gz | 137,533,150 | 7,182 | # Code Examples
OCaml possesses an interactive system, called “toploop”, that lets you type OCaml code and have it evaluated immediately. It is a great way to learn the language and to quickly experiment with ideas. Below, we demonstrate the use of the toploop to illustrate basic capabilities of the language.
Some indications for the code below. The prompt at which you type is “#”. The code must end with “;;” (this is only an indication to the interactive system that the input has to be evaluated and is not really part of the OCaml code). The output of the system is displayed in this color.
More code examples are available on the following sites:
## Elementary functions
Let us define the square function and the recursive factorial function. Then, let us apply these functions to sample values. Unlike the majority of languages, OCaml uses parentheses for grouping but not for the arguments of a function.
# let square x = x * x;;
val square : int -> int = <fun> # square 3;;
- : int = 9 # let rec fact x = if x <= 1 then 1 else x * fact (x - 1);;
val fact : int -> int = <fun> # fact 5;;
- : int = 120 # square 120;;
- : int = 14400
## Automatic memory management
All allocation and deallocation operations are fully automatic. For example, let us consider simply linked lists.
Lists are predefined in OCaml. The empty list is written []. The constructor that allows prepending an element to a list is written :: (in infix form).
# let li = 1 :: 2 :: 3 :: [];;
val li : int list = [1; 2; 3] # [1; 2; 3];;
- : int list = [1; 2; 3] # 5 :: li;;
- : int list = [5; 1; 2; 3]
## Polymorphism: sorting lists
Insertion sort is defined using two recursive functions.
# let rec sort = function
| [] -> []
| x :: l -> insert x (sort l)
and insert elem = function
| [] -> [elem]
| x :: l -> if elem < x then elem :: x :: l
else x :: insert elem l;;
val sort : 'a list -> 'a list = <fun> val insert : 'a -> 'a list -> 'a list = <fun>
Note that the type of the list elements remains unspecified: it is represented by a type variable 'a. Thus, sort can be applied both to a list of integers and to a list of strings.
# sort [2; 1; 0];;
- : int list = [0; 1; 2] # sort ["yes"; "ok"; "sure"; "ya"; "yep"];;
- : string list = ["ok"; "sure"; "ya"; "yep"; "yes"]
## Imperative features
Let us encode polynomials as arrays of integer coefficients. Then, to add two polynomials, we first allocate the result array, then fill its slots using two successive for loops.
# let add_polynom p1 p2 =
let n1 = Array.length p1
and n2 = Array.length p2 in
let result = Array.make (max n1 n2) 0 in
for i = 0 to n1 - 1 do result.(i) <- p1.(i) done;
for i = 0 to n2 - 1 do result.(i) <- result.(i) + p2.(i) done;
result;;
val add_polynom : int array -> int array -> int array = <fun> # add_polynom [| 1; 2 |] [| 1; 2; 3 |];;
- : int array = [|2; 4; 3|]
OCaml offers updatable memory cells, called references: ref init returns a new cell with initial contents init, !cell returns the current contents of cell, and cell := v writes the value v into cell.
We may redefine fact using a reference cell and a for loop:
# let fact n =
let result = ref 1 in
for i = 2 to n do
result := i * !result
done;
!result;;
val fact : int -> int = <fun> # fact 5;;
- : int = 120
## Higher-order functions
There is no restriction on functions, which may thus be passed as arguments to other functions. Let us define a function sigma that returns the sum of the results of applying a given function f to each element of a list:
# let rec sigma f = function
| [] -> 0
| x :: l -> f x + sigma f l;;
val sigma : ('a -> int) -> 'a list -> int = <fun>
Anonymous functions may be defined using the fun or function construct:
# sigma (fun x -> x * x) [1; 2; 3];;
- : int = 14
Polymorphism and higher-order functions allow defining function composition in its most general form:
# let compose f g = fun x -> f (g x);;
val compose : ('a -> 'b) -> ('c -> 'a) -> 'c -> 'b = <fun> # let square_o_fact = compose square fact;;
val square_o_fact : int -> int = <fun> # square_o_fact 5;;
- : int = 14400
## The power of functions
The power of functions cannot be better illustrated than by the power function:
# let rec power f n =
if n = 0 then fun x -> x
else compose f (power f (n - 1));;
val power : ('a -> 'a) -> int -> 'a -> 'a = <fun>
The nth derivative of a function can be computed as in mathematics by raising the derivative function to the nth power:
# let derivative dx f = fun x -> (f (x +. dx) -. f x) /. dx;;
val derivative : float -> (float -> float) -> float -> float = <fun> # let sin''' = power (derivative 1e-5) 3 sin;;
val sin''' : float -> float = <fun> # let pi = 4.0 *. atan 1.0 in sin''' pi;;
- : float = 0.999998972517346263
## Symbolic computation
Let us consider simple symbolic expressions made up of integers, variables, let bindings, and binary operators. Such expressions can be defined as a new data type, as follows:
# type expression =
| Num of int
| Var of string
| Let of string * expression * expression
| Binop of string * expression * expression;;
type expression = Num of int | Var of string | Let of string * expression * expression | Binop of string * expression * expression
Evaluation of these expressions involves an environment that maps identifiers to values, represented as a list of pairs.
# let rec eval env = function
| Num i -> i
| Var x -> List.assoc x env
| Let (x, e1, in_e2) ->
let val_x = eval env e1 in
eval ((x, val_x) :: env) in_e2
| Binop (op, e1, e2) ->
let v1 = eval env e1 in
let v2 = eval env e2 in
eval_op op v1 v2
and eval_op op v1 v2 =
match op with
| "+" -> v1 + v2
| "-" -> v1 - v2
| "*" -> v1 * v2
| "/" -> v1 / v2
| _ -> failwith ("Unknown operator: " ^ op);;
val eval : (string * int) list -> expression -> int = <fun> val eval_op : string -> int -> int -> int = <fun>
As an example, we evaluate the phrase let x = 1 in x + x:
# eval [] (Let ("x", Num 1, Binop ("+", Var "x", Var "x")));;
- : int = 2
Pattern matching eases the definition of functions operating on symbolic data, by giving function definitions and type declarations similar shapes. Indeed, note the close resemblance between the definition of the eval function and that of the expression type.
## Elementary debugging
To conclude, here is the simplest way of spying over functions:
let rec fib x = if x <= 1 then 1 else fib (x - 1) + fib (x - 2);;
# #trace fib;;
fib is now traced. # fib 3;;
fib <-- 3 fib <-- 1 fib --> 1 fib <-- 2 fib <-- 0 fib --> 1 fib <-- 1 fib --> 1 fib --> 2 fib --> 3 - : int = 3
Go and try it in your browser or install it and read some tutorials. | 1,886 | 6,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-22 | latest | en | 0.81179 |
https://rdrr.io/cran/prob/man/isrep.html | 1,696,163,631,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510888.64/warc/CC-MAIN-20231001105617-20231001135617-00511.warc.gz | 531,502,302 | 8,255 | # isrep: Is Repeated in a Vector In prob: Elementary Probability on Finite Sample Spaces
## Description
Tests for a certain number of repetitions of `vals` in a given vector `x`.
## Usage
```1 2 3 4 5 6 7``` ```isrep(x, ...) ## Default S3 method: isrep(x, vals = unique(x), nrep = 2, ...) ## S3 method for class 'data.frame' isrep(x, ...) ```
## Arguments
`x` an object with potential repeated values. `vals` values that may be repeated. `nrep` exact number of repeats desired, defaults to pairs. `...` further arguments to be passed to or from other methods.
## Details
This is a generic function, with methods supplied for data frames and vectors. The default behavior tests for existence of pairs of elements of `x`. One can test existence of triples, etc., by changing the `nrep` argument. If there are specific values for which one is looking for repeats, these can be specified with the `vals` argument. Note that the function only checks for exactly `nrep` instances, so two pairs of a specific element would be counted as 0 pairs and 1 quadruple. See the examples.
The data frame method uses `apply` to apply `isrep.default` to each row of the data frame.
Logical.
## Author(s)
G. Jay Kerns gkerns@ysu.edu.
`countrep`
## Examples
```1 2 3 4``` ```x <- c(3,3,2,2,3,3,4,4) isrep(x) # one pair each of 2s and 4s isrep(x, nrep = 4) isrep(x, vals = 4) # one pair of 4s ```
### Example output
```Loading required package: combinat
Attaching package: 'combinat'
The following object is masked from 'package:utils':
combn
Attaching package: 'prob'
The following objects are masked from 'package:base':
intersect, setdiff, union
[1] TRUE
[1] TRUE
[1] TRUE
```
prob documentation built on May 2, 2019, 12:20 p.m. | 487 | 1,738 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.729379 |
https://cran.r-project.org/web/packages/LAWBL/vignettes/pcfa-examples.html | 1,606,272,374,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141180636.17/warc/CC-MAIN-20201125012933-20201125042933-00644.warc.gz | 249,431,822 | 19,749 | # pcfa-examples
Note: the estimation process can be time consuming depending on the computing power. You can same some time by reducing the length of the chains.
## Continuous Data w/o Local Dependence:
library(LAWBL)
dat <- sim18cfa0$dat J <- ncol(dat) # no. of items K <- 3 # no. of factors qlam <- sim18cfa0$qlam
qlam
R> [,1] [,2] [,3]
R> [1,] 0.7 0.0 0.0
R> [2,] 0.7 0.0 0.0
R> [3,] 0.7 0.0 0.0
R> [4,] 0.7 0.0 0.0
R> [5,] 0.7 0.4 0.0
R> [6,] 0.7 0.4 0.0
R> [7,] 0.0 0.7 0.0
R> [8,] 0.0 0.7 0.0
R> [9,] 0.0 0.7 0.0
R> [10,] 0.0 0.7 0.0
R> [11,] 0.0 0.7 0.4
R> [12,] 0.0 0.7 0.4
R> [13,] 0.0 0.0 0.7
R> [14,] 0.0 0.0 0.7
R> [15,] 0.0 0.0 0.7
R> [16,] 0.0 0.0 0.7
R> [17,] 0.4 0.0 0.7
R> [18,] 0.4 0.0 0.7
Q<-matrix(-1,J,K); # -1 for unspecified items
Q[1:2,1]<-Q[7:8,2]<-Q[13:14,3]<-1 # 1 for specified items
Q
R> [,1] [,2] [,3]
R> [1,] 1 -1 -1
R> [2,] 1 -1 -1
R> [3,] -1 -1 -1
R> [4,] -1 -1 -1
R> [5,] -1 -1 -1
R> [6,] -1 -1 -1
R> [7,] -1 1 -1
R> [8,] -1 1 -1
R> [9,] -1 -1 -1
R> [10,] -1 -1 -1
R> [11,] -1 -1 -1
R> [12,] -1 -1 -1
R> [13,] -1 -1 1
R> [14,] -1 -1 1
R> [15,] -1 -1 -1
R> [16,] -1 -1 -1
R> [17,] -1 -1 -1
R> [18,] -1 -1 -1
1. E-step: Estimate with the PCFA-LI model (E-step) by setting LD=F. Only a few loadings need to be specified in Q (e.g., 2 per factor). Longer chain is suggested for stabler performance (burn=iter=5,000 by default).
m0 <- pcfa(dat = dat, Q = Q,LD = FALSE, burn = 2000, iter = 2000,verbose = TRUE)
R>
R> Tot. Iter = 1000
R> [,1] [,2] [,3]
R> Feigen 4.744 4.269 2.380
R> NLA_lg3 8.000 8.000 6.000
R> Shrink 3.041 3.041 3.041
R>
R> Tot. Iter = 2000
R> [,1] [,2] [,3]
R> Feigen 4.079 4.030 3.224
R> NLA_lg3 8.000 8.000 6.000
R> Shrink 2.734 2.734 2.734
R>
R> Tot. Iter = 3000
R> [,1] [,2] [,3]
R> Feigen 3.258 3.460 3.418
R> NLA_lg3 8.000 8.000 8.000
R> Shrink 3.809 3.809 3.809
R> Adj PSR 2.313 1.231 1.517
R>
R> Tot. Iter = 4000
R> [,1] [,2] [,3]
R> Feigen 3.196 3.430 2.891
R> NLA_lg3 8.000 8.000 8.000
R> Shrink 3.527 3.527 3.527
R> Adj PSR 1.21 1.34 1.121
R>
R> #Sign change: 0 0 0 0 0 0
R> user system elapsed
R> 15.35 0.02 15.35
# summarize basic information
summary(m0)
R> $N R> [1] 1000 R> R>$J
R> [1] 18
R>
R> $K R> [1] 3 R> R>$Miss%
R> [1] 0
R>
R> $LD enabled R> [1] FALSE R> R>$Burn in
R> [1] 2000
R>
R> $Iteration R> [1] 2000 R> R>$No. of sig lambda
R> [1] 24
R>
R> $Adj. PSR R> Point est. Upper C.I. R> F1 1.210460 1.752360 R> F2 1.340018 2.156112 R> F3 1.120779 1.193162 #summarize significant loadings in pattern/Q-matrix format summary(m0, what = 'qlambda') R> [,1] [,2] [,3] R> I1 0.6943921 0.0000000 0.0000000 R> I2 0.7018397 0.0000000 0.0000000 R> I3 0.6940404 0.0000000 0.0000000 R> I4 0.6821522 0.0000000 0.0000000 R> I5 0.6970800 0.4219813 0.0000000 R> I6 0.6922918 0.4145146 0.0000000 R> I7 0.0000000 0.7633059 0.0000000 R> I8 0.0000000 0.6836233 0.0000000 R> I9 0.0000000 0.7454941 0.0000000 R> I10 0.0000000 0.6788261 0.0000000 R> I11 0.0000000 0.7309471 0.3569716 R> I12 0.0000000 0.7170643 0.3551296 R> I13 0.0000000 0.0000000 0.6787302 R> I14 0.0000000 0.0000000 0.6770629 R> I15 0.0000000 0.0000000 0.7500404 R> I16 0.0000000 0.0000000 0.6942952 R> I17 0.4276174 0.0000000 0.7058218 R> I18 0.4191974 0.0000000 0.7049021 #factorial eigenvalue summary(m0,what='eigen') R> est sd lower upper sig R> F1 3.308617 0.4444696 2.363488 4.291951 1 R> F2 3.530416 0.4479498 2.519580 4.262045 1 R> F3 3.310640 0.4478500 2.599306 4.240286 1 #plotting factorial eigenvalue plot_eigen(m0) # trace plot_eigen(m0, what='density') #density plot_eigen(m0, what='APSR') #adj, PSRF 1. C-step: Reconfigure the Q matrix for the C-step with one specified loading per item based on results from the E-step. Estimate with the PCFA model by setting LD=TRUE (by default). Longer chain is suggested for stabler performance. Results are very close to the E-step, since there’s no LD in the data. Q<-matrix(-1,J,K); tmp<-summary(m0, what="qlambda") cind<-apply(tmp,1,which.max) Q[cbind(c(1:J),cind)]<-1 #alternatively #Q[1:6,1]<-Q[7:12,2]<-Q[13:18,3]<-1 # 1 for specified items m1 <- pcfa(dat = dat, Q = Q, burn = 2000, iter = 2000,verbose = TRUE) summary(m1) summary(m1, what = 'qlambda') summary(m1, what = 'offpsx') #summarize significant LD terms summary(m1,what='eigen') #plotting factorial eigenvalue # par(mar = rep(2, 4)) plot_eigen(m1) # trace plot_eigen(m1, what='density') #density plot_eigen(m1, what='APSR') #adj, PSRF 1. CFA-LD: One can also configure the Q matrix for a CFA model with local dependence (i.e. without any unspecified loading) based on results from the C-step. Results are also very close. Q<-summary(m1, what="qlambda") Q[Q!=0]<-1 Q m2 <- pcfa(dat = dat, Q = Q, burn = 2000, iter = 2000,verbose = TRUE) summary(m2) summary(m2, what = 'qlambda') summary(m2, what = 'offpsx') summary(m2,what='eigen') plot_eigen(m2) # Eigens' traces are excellent without regularization of the loadings ## Continuous Data with Local Dependence: 1. Load the the data, loading pattern (qlam), and LD terms, and setup the design matrix Q. dat <- sim18cfa1$dat
J <- ncol(dat) # no. of items
K <- 3 # no. of factors
sim18cfa1$qlam sim18cfa1$LD # effect size = .3
Q<-matrix(-1,J,K); # -1 for unspecified items
Q[1:2,1]<-Q[7:8,2]<-Q[13:14,3]<-1 # 1 for specified items
Q
1. E-step: Estimate with the PCFA-LI model (E-step) by setting LD=FALSE. Only a few loadings need to be specified in Q (e.g., 2 per factor). Some loading estimates are biased due to ignoring the LD. So do the eigenvalues.
m0 <- pcfa(dat = dat, Q = Q,LD = FALSE, burn = 4000, iter = 4000,verbose = TRUE)
summary(m0)
summary(m0, what = 'qlambda')
summary(m0,what='eigen')
plot_eigen(m0) # trace
plot_eigen(m0, what='APSR')
1. C-step: Reconfigure the Q matrix for the C-step with one specified loading per item based on results from the E-step. Estimate with the PCFA model by setting LD=TRUE (by default). The estimates are more accurate, and the LD terms can be largely recovered.
Q<-matrix(-1,J,K);
tmp<-summary(m0, what="qlambda")
cind<-apply(tmp,1,which.max)
Q[cbind(c(1:J),cind)]<-1
Q
m1 <- pcfa(dat = dat, Q = Q,burn = 4000, iter = 4000,verbose = TRUE)
summary(m1)
summary(m1, what = 'qlambda')
summary(m1,what='eigen')
summary(m1, what = 'offpsx')
1. CFA-LD: Configure the Q matrix for a CFA model with local dependence (i.e. without any unspecified loading) based on results from the C-step. Results are better than, but similar to the C-step.
Q<-summary(m1, what="qlambda")
Q[Q!=0]<-1
Q
m2 <- pcfa(dat = dat, Q = Q,burn = 4000, iter = 4000,verbose = TRUE)
summary(m2)
summary(m2, what = 'qlambda')
summary(m2,what='eigen')
summary(m2, what = 'offpsx') | 3,137 | 6,840 | {"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} | 2.640625 | 3 | CC-MAIN-2020-50 | latest | en | 0.469184 |
https://www.brightstorm.com/math/precalculus/conic-sections/the-circle-problem-1/ | 1,679,533,826,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944606.5/warc/CC-MAIN-20230323003026-20230323033026-00786.warc.gz | 768,977,402 | 26,874 | ###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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# The Circle - Problem 1
Carl Horowitz
###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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Given the coordinates of center of a circle (h, k) and the radius (r), plug the values into the general equation of a circle: r^2 = (x-h)^2 + (y-k)^2. Be careful to change the signs accordingly when plugging in the values for h and k.
Finding the equation for a circle with a specified center and radius. So really all we have to do when we're looking for the equation for a circle is look for our general equation.
So our equation is just r² is equal to x minus 8² plus y minus k² where h and k are the coordinates of the center, so really all we have to do is take the information we're given and plug it in.
So our radius is 5, so we go to our r plug in 5, 5² is easy enough to do, we end up with 25 is equal to x minus h, h is the coordinate of our center so we go to our -1, plug that in and we end up with x minus -1, x plus 1 and then add it to do the same thing for the y coordinate of the center plugging in 2 for k and we end with y minus 2 quantity squared.
So really to find the equation of a circle all we have to do is remember the general formula and then plug in all the information. | 411 | 1,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-14 | latest | en | 0.920571 |
https://metanumbers.com/183148 | 1,627,996,583,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154459.22/warc/CC-MAIN-20210803124251-20210803154251-00080.warc.gz | 389,269,635 | 11,038 | ## 183148
183,148 (one hundred eighty-three thousand one hundred forty-eight) is an even six-digits composite number following 183147 and preceding 183149. In scientific notation, it is written as 1.83148 × 105. The sum of its digits is 25. It has a total of 5 prime factors and 24 positive divisors. There are 75,600 positive integers (up to 183148) that are relatively prime to 183148.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 25
• Digital Root 7
## Name
Short name 183 thousand 148 one hundred eighty-three thousand one hundred forty-eight
## Notation
Scientific notation 1.83148 × 105 183.148 × 103
## Prime Factorization of 183148
Prime Factorization 22 × 7 × 31 × 211
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 91574 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 183,148 is 22 × 7 × 31 × 211. Since it has a total of 5 prime factors, 183,148 is a composite number.
## Divisors of 183148
24 divisors
Even divisors 16 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 24 Total number of the positive divisors of n σ(n) 379904 Sum of all the positive divisors of n s(n) 196756 Sum of the proper positive divisors of n A(n) 15829.3 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 427.958 Returns the nth root of the product of n divisors H(n) 11.5702 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 183,148 can be divided by 24 positive divisors (out of which 16 are even, and 8 are odd). The sum of these divisors (counting 183,148) is 379,904, the average is 158,29.,333.
## Other Arithmetic Functions (n = 183148)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 75600 Total number of positive integers not greater than n that are coprime to n λ(n) 210 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 16556 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 75,600 positive integers (less than 183,148) that are coprime with 183,148. And there are approximately 16,556 prime numbers less than or equal to 183,148.
## Divisibility of 183148
m n mod m 2 3 4 5 6 7 8 9 0 1 0 3 4 0 4 7
The number 183,148 is divisible by 2, 4 and 7.
## Classification of 183148
• Abundant
### Expressible via specific sums
• Polite
• Practical
• Non-hypotenuse
## Base conversion (183148)
Base System Value
2 Binary 101100101101101100
3 Ternary 100022020021
4 Quaternary 230231230
5 Quinary 21330043
6 Senary 3531524
8 Octal 545554
10 Decimal 183148
12 Duodecimal 89ba4
20 Vigesimal 12hh8
36 Base36 3xbg
## Basic calculations (n = 183148)
### Multiplication
n×i
n×2 366296 549444 732592 915740
### Division
ni
n⁄2 91574 61049.3 45787 36629.6
### Exponentiation
ni
n2 33543189904 6143368144537792 1125145588935807529216 206068164322415277360851968
### Nth Root
i√n
2√n 427.958 56.7894 20.6871 11.2865
## 183148 as geometric shapes
### Circle
Diameter 366296 1.15075e+06 1.05379e+11
### Sphere
Volume 2.57333e+16 4.21516e+11 1.15075e+06
### Square
Length = n
Perimeter 732592 3.35432e+10 259010
### Cube
Length = n
Surface area 2.01259e+11 6.14337e+15 317222
### Equilateral Triangle
Length = n
Perimeter 549444 1.45246e+10 158611
### Triangular Pyramid
Length = n
Surface area 5.80985e+10 7.24003e+14 149540
## Cryptographic Hash Functions
md5 f630991127a56bf0247ed8a714db5699 010b3956133e297ba65ef44c8fbd868af7e37441 23f8b4984ca0a3342e85651b384b7810dc8de1aa9161886e40d20a36ae49f6c6 b9a5a62c18130d7d600bc6f2e7009ebce7510212263d11fb9a5cbeb77b07c594e38fcd7cd3de4aad998273285b4d40f946e696b2a7ff7a88061ef7cdabcf6c89 19779b20b6e70626b3b7af98a7ed7835c0354a08 | 1,465 | 4,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.81853 |
https://grindskills.com/how-to-create-a-toy-survival-time-to-event-data-with-right-censoring/ | 1,675,749,401,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500384.17/warc/CC-MAIN-20230207035749-20230207065749-00275.warc.gz | 286,131,220 | 14,826 | # How to create a toy survival (time to event) data with right censoring
I wish to create a toy survival (time to event) data which is right censored and follows some distribution with proportional hazards and constant baseline hazard.
I created the data as follows, but I am unable to obtain estimated hazard ratios that are close to the true values after fitting a Cox proportional hazards model to the simulated data.
What did I do wrong?
R codes:
library(survival)
#set parameters
set.seed(1234)
n = 40000 #sample size
#functional relationship
lambda=0.000020 #constant baseline hazard 2 per 100000 per 1 unit time
b_haz <-function(t) #baseline hazard
{
lambda #constant hazard wrt time
}
x = cbind(hba1c=rnorm(n,2,.5)-2,age=rnorm(n,40,5)-40,duration=rnorm(n,10,2)-10)
B = c(1.1,1.2,1.3) # hazard ratios (model coefficients)
hist(x %*% B) #distribution of scores
haz <-function(t) #hazard function
{
b_haz(t) * exp(x %*% B)
}
c_hf <-function(t) #cumulative hazards function
{
exp(x %*% B) * lambda * t
}
S <- function(t) #survival function
{
exp(-c_hf(t))
}
S(.005)
S(1)
S(5)
#simulate censoring
time = rnorm(n,10,2)
S_prob = S(time)
#simulate events
event = ifelse(runif(1)>S_prob,1,0)
#model fit
km = survfit(Surv(time,event)~1,data=data.frame(x))
plot(km) #kaplan-meier plot
#Cox PH model
fit = coxph(Surv(time,event)~ hba1c+age+duration, data=data.frame(x))
summary(fit)
cox.zph(fit)
Results:
Call:
coxph(formula = Surv(time, event) ~ hba1c + age + duration, data = data.frame(x))
n= 40000, number of events= 3043
coef exp(coef) se(coef) z Pr(>|z|)
hba1c 0.236479 1.266780 0.035612 6.64 3.13e-11 ***
age 0.351304 1.420919 0.003792 92.63 < 2e-16 ***
duration 0.356629 1.428506 0.008952 39.84 < 2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
exp(coef) exp(-coef) lower .95 upper .95
hba1c 1.267 0.7894 1.181 1.358
age 1.421 0.7038 1.410 1.432
duration 1.429 0.7000 1.404 1.454
Concordance= 0.964 (se = 0.006 )
Rsquare= 0.239 (max possible= 0.767 )
Likelihood ratio test= 10926 on 3 df, p=0
Wald test = 10568 on 3 df, p=0
Score (logrank) test = 11041 on 3 df, p=0
but true values are set as
B = c(1.1,1.2,1.3) # hazard ratios (model coefficients)
It is not clear to me how you generate your event times (which, in your case, might be $<0$) and event indicators:
time = rnorm(n,10,2)
S_prob = S(time)
event = ifelse(runif(1)>S_prob,1,0)
So here is a generic method, followed by some R code.
Generating survival times to simulate Cox proportional hazards models
To generate event times from the proportional hazards model, we can use the inverse probability method (Bender et al., 2005): if $V$ is uniform on $(0, 1)$ and if $S(\cdot \,|\, \mathbf{x})$ is the conditional survival function derived from the proportional hazards model, i.e.
$$S(t \,|\, \mathbf{x}) = \exp \left( -H_0(t) \exp(\mathbf{x}^\prime \mathbf{\beta}) \vphantom{\Big(} \right)$$
then it is a fact that the random variable
$$T = S^{-1}(V \,|\, \mathbf{x}) = H_0^{-1} \left( – \frac{\log(V)}{\exp(\mathbf{x}^\prime \mathbf{\beta})} \right)$$
has survival function $S(\cdot \,|\, \mathbf{x})$. This result is known as “the inverse probability integral transformation”. Therefore, to generate a survival time $T \sim S(\cdot \,|\, \mathbf{x})$ given the covariate vector, it suffices to draw $v$ from $V \sim \mathrm{U}(0, 1)$ and to make the inverse transformation $t = S^{-1}(v \,|\, \mathbf{x})$.
Example [Weibull baseline hazard]
Let $h_0(t) = \lambda \rho t^{\rho – 1}$ with shape $\rho > 0$ and scale $\lambda > 0$. Then $H_0(t) = \lambda t^\rho$ and $H^{-1}_0(t) = (\frac{t}{\lambda})^{\frac{1}{\rho}}$. Following the inverse probability method, a realisation of $T \sim S(\cdot \,|\, \mathbf{x})$ is obtained by computing
$$t = \left( – \frac{\log(v)}{\lambda \exp(\mathbf{x}^\prime \mathbf{\beta})} \right)^{\frac{1}{\rho}}$$
with $v$ a uniform variate on $(0, 1)$. Using results on transformations of random variables, one may notice that $T$ has a conditional Weibull distribution (given $\mathbf{x}$) with shape $\rho$ and scale $\lambda \exp(\mathbf{x}^\prime \mathbf{\beta})$.
R code
The following R function generates a data set with a single binary covariate $x$ (e.g. a treatment indicator). The baseline hazard has a Weibull form. Censoring times are randomly drawn from an exponential distribution.
# baseline hazard: Weibull
# N = sample size
# lambda = scale parameter in h0()
# rho = shape parameter in h0()
# beta = fixed effect parameter
# rateC = rate parameter of the exponential distribution of C
simulWeib <- function(N, lambda, rho, beta, rateC)
{
# covariate --> N Bernoulli trials
x <- sample(x=c(0, 1), size=N, replace=TRUE, prob=c(0.5, 0.5))
# Weibull latent event times
v <- runif(n=N)
Tlat <- (- log(v) / (lambda * exp(x * beta)))^(1 / rho)
# censoring times
C <- rexp(n=N, rate=rateC)
# follow-up times and event indicators
time <- pmin(Tlat, C)
status <- as.numeric(Tlat <= C)
# data set
data.frame(id=1:N,
time=time,
status=status,
x=x)
}
Test
Here is some quick simulation with $\beta = -0.6$:
set.seed(1234)
betaHat <- rep(NA, 1e3)
for(k in 1:1e3)
{
dat <- simulWeib(N=100, lambda=0.01, rho=1, beta=-0.6, rateC=0.001)
fit <- coxph(Surv(time, status) ~ x, data=dat)
betaHat[k] <- fit\$coef
}
> mean(betaHat)
[1] -0.6085473 | 1,836 | 5,422 | {"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} | 2.953125 | 3 | CC-MAIN-2023-06 | latest | en | 0.529255 |
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