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Example 3.24 (Apéry Numbers as Diagonals)
Requires BinomialSum package of Pierre Lairez, available at https://github.com/lairez/binomsums
> # Import package with(BinomSums);
(1)
> ################## # zeta(3) sequence ##################
> # The generating function for the binomial sum in Apéry's irrationality proof of zeta(3) is GF_as_sum := Sum(t^n*Sum(Binomial(n,k)^2*Binomial(n+k,k)^2, k=0..n), n=0..infinity);
(2)
> # The binomial sums package shows this is the "residue" of this rational function R := subs(v[1]=x,v[2]=y,v[3]=z,[sumtores(GF_as_sum, v)][1]);
(3)
> # This implies the GF is the main power series diagonal of this function F := simplify(x*y*z*subs(t=x*y*z*t,R));
(4)
> # Beginning terms in the Apéry sequence seq(coeff(coeff(coeff(coeff(mtaylor(F,[x,y,z,t],30),x,k),y,k),z,k),t,k),k=0..6);
(5)
> ################## # zeta(2) sequence ##################
> # The generating function for the binomial sum in Apéry's irrationality proof of zeta(2) is GF_as_sum2 := Sum(t^n*Sum(Binomial(n,k)^2*Binomial(n+k,k), k=0..n), n=0..infinity);
(6)
> # The binomial sums package shows this is the "residue" of this rational function R2 := subs(v[1]=x,v[2]=y,[sumtores(GF_as_sum2, v)][1]);
(7)
> # This implies the GF is the main power series diagonal of this function F2 := simplify(x*y*subs(t=x*y*t,R2));
(8)
> # Beginning terms in the second Apéry sequence seq(coeff(coeff(coeff(mtaylor(F2,[x,y,t],20),x,k),y,k),t,k),k=0..6);
(9)
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http://www.poynter.org/2011/claim-krugman-is-top-prognosticator-cal-thomas-is-the-worst/130485/ | 1,476,992,162,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988717783.68/warc/CC-MAIN-20161020183837-00228-ip-10-171-6-4.ec2.internal.warc.gz | 651,575,580 | 17,555 | Romenesko Misc.
A Hamilton College class and their public policy professor analyzed the predicts of 26 pundits -- including Sunday morning TV talkers -- and used a scale of 1 to 5 to rate their accuracy. After Paul Krugman, the most accurate pundits were Maureen Dowd, former Pennsylvania Governor Ed Rendell, U.S. Senator Chuck Schumer (D-NY), and former House Speaker Nancy Pelosi. "The Bad" list includes Thomas Friedman, Clarence Page, and Bob Herbert.
Press release
Pundits predict no more accurately than a coin toss
Krugman tops, Cal Thomas bottom of accurate predictors, according to study at Hamilton College
CLINTON, N.Y. - Op-ed columnists and TV’s talking heads build followings by making bold, confident predictions about politics and the economy. But rarely are their predictions analyzed for accuracy.
Now, a class at Hamilton College led by public policy professor P. Gary Wyckoff has analyzed the predictions of 26 prognosticators between September 2007 and December 2008. Their findings? Anyone can make as accurate a prediction as most of them if just by flipping a coin.
Their research paper, “Are Talking Heads Blowing Hot Air? An Analysis of the Accuracy of Forecasts in the Political Media” will be presented via webcast on Monday, May 2, at 4:15 p.m., at www.hamilton.edu/pundit. The paper will also be available at that address at that time. Questions during the presentation can be posed via Twitter using #hcpundit.
The Hamilton students sampled the predictions of 26 individuals who wrote columns in major print media and who appeared on the three major Sunday news shows – Face the Nation, Meet the Press, and This Week – and evaluated the accuracy of 472 predictions made during the 16-month period. They used a scale of 1 to 5 (1 being “will not happen, 5 being “will absolutely happen”) to rate the accuracy of each, and then divided them into three categories: The Good, The Bad, and The Ugly.
The students found that only nine of the prognosticators they studied could predict more accurately than a coin flip. Two were significantly less accurate, and the remaining 14 were not statistically any better or worse than a coin flip.
The top prognosticators – led by New York Times columnist Paul Krugman – scored above five points and were labeled “Good,” while those scoring between zero and five were “Bad.” Anyone scoring less than zero (which was possible because prognosticators lost points for inaccurate predictions) were put into “The Ugly” category. Syndicated columnist Cal Thomas came up short and scored the lowest of the 26.
Even when the students eliminated political predictions and looked only at predictions for the economy and social issues, they found that liberals still do better than conservatives at prediction. After Krugman, the most accurate pundits were Maureen Dowd of The New York Times, former Pennsylvania Governor Ed Rendell, U.S. Senator Chuck Schumer (D-NY), and former House Speaker Nancy Pelosi – all Democrats and/or liberals. Also landing in the “Good” category, however, were conservative columnists Kathleen Parker and David Brooks, along with Bush Administration Treasury Secretary Hank Paulson. Left-leaning columnist Eugene Robinson of The Washington Post rounded out the “good” list.
Those scoring lowest – “The Ugly” – with negative tallies were conservative columnist Cal Thomas; U.S. Senator Lindsey Graham (R-SC); U.S. Senator Carl Levin (D-MI); U.S. Senator Joe Lieberman, a McCain supporter and Democrat-turned-Independent from Connecticut; Sam Donaldson of ABC; and conservative columnist George Will.
Landing between the two extremes – “The Bad” – were Howard Wolfson, communications director for Hillary Clinton’s 2008 campaign; former Arkansas Governor Mike Huckabee, a hopeful in the 2008 Republican primary; former House Speaker Newt Gingrich, a Republican; Sen. John Kerry of Massachusetts, the Democratic nominee for president in 2004; liberal columnist Bob Herbert of The New York Times; Andrea Mitchell of NBC; New York Times columnist Tom Friedman; the late David Broder, former columnist for The Washington Post; Chicago Tribune columnist Clarence Page; New York Times columnist Nicholas Kristof; and Hillary Clinton.
The group also found a link between conditional predictions and accuracy, that is, a prediction that was conditional (“If A, then B”) was less likely to be accurate. Finally, those prognosticators with a law degree were more likely to be wrong. | 939 | 4,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2016-44 | longest | en | 0.954532 |
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# IBPS Clerk Quiz : Quantitative Aptitude | 21 -10 - 17
Mahendra Guru : Online Videos For Govt. Exams
1.) The ratio between the ages of Shweta and Anjali is 5 : x. Shweta is 9 years younger than Jahanvi. After 9 years, age of Jahanvi will be 33 years. The difference of the ages of Anjali and Shweta is as same as the present age of Jahanvi. What will be the value of x?
श्वेता और अंजलि की वर्तमान आयु के बीच अनुपात 5 : x है | श्वेता, जानवी से 9 वर्ष छोटी है | 9 वर्ष बाद जानवी की आयु 33 वर्ष हो जायेगी | अंजलि और श्वेता की आयु में उतना ही अंतर है जितनी जान्वी की वर्तमान आयु है | x के स्थान पर क्या आयेगा ?
(1.) 10 (2) 13 (3) 17 (4) 28 (5) 39
Sol
Present age of Jahanvi = 33 - 9 = 24 years/वर्ष
Present age of Shweta = 24 - 9 = 15 years/वर्ष
Now,
Shweta : Anjali = 5 : x = 15 : 3x
Present age of Anjali = 3x years
3x - 15 = 24
3x = 39
x = 13
2.) The age of the captain of a cricket team of 11 members is 26 years and the wicketkeeper of the team is 3 years older than the captain, if the ages of these two excluded then the average age of the team is one year less than the average age of the whole team. What is the average age of the team?
11 सदस्य वाली क्रिकेट टीम के कप्तान की आयु 26 वर्ष हैं और टीम का विकेटकीपर कप्तान तीन वर्ष बड़ा हैं, यदि इन दोनों की आयु को सम्मिलित न किया जाए तो शेष टीम की औसत आयु टीम की औसत आयु से 1 वर्ष कम होगी | टीम की औसत आयु क्या हैं ?
(1.) 20 (2) 21 (3) 22 (4) 23 (5) 24
Sol
Let X be the average age of the team
9(X-1) + 26 +29 = 11X
X=23
3.) The area of a rectangular field is 5760 m2. If its length is 60% more than the breadth, what is the breadth?
एक आयताकार खेत का क्षेत्रफल 5760 मी2. है। यदि इसकी लम्बाई, चौड़ाई से 60% अधिक है, तो चौड़ाई क्या है?
(1.) 50 (2) 55 (3) 60 (4) 65 (5) 70
Sol
Then length will be 1.6 x
1.6x × x = 5760
x2 = 3600
x = 60 m2
4.) In an examination the percentage of students qualified to the number of students appeared from school 'A' is 70%. In school 'B' the number of students appeared is 20% more than the students appeared from school 'A' and the number of students qualified from school 'B' is 50% more than the students qualified from school 'A'. What is the percentage of students qualified to the number of students appeared from school 'B'?
एक परीक्षा में स्कूल 'A' से पास होने वाले छात्रों का सम्मिलित छात्रों की संख्या से प्रतिशत 70% है| स्कूल 'B' से परीक्षा में सम्मिलित होने वाले छात्र स्कूल 'A' से परीक्षा में सम्मिलित होने वाले छात्रों की संख्या से 20% अधिक है और स्कूल 'B' से पास होने वाले छात्रों की संख्या स्कूल 'A' से पास होने वाले छात्रों की संख्या से 50% अधिक है | स्कूल 'B' से परीक्षा में पास होने वाले छात्र परीक्षा में सम्मिलित होने वाले छात्रों का कितना प्रतिशत हैं ?
(1.) 5 (2) 8 (3) 10 (4) 12 (5) 15
Sol
Let number of students appeared in examination from school 'A' = 100
Qualified number of students in examination from school 'A' = 70
Number of students appeared in examination from school 'B' = 120
Qualified number of students in examination from school 'B' =
5.) A swimming pool has 3 drain pipes. The first two pipes A and B, operating simultaneously, can empty the pool in half the time than C, the 3rd pipe, alone takes to empty it. Pipe A, working alone, takes half the time taken by pipe B. Together they take 6 hours 40 minutes to empty the pool. Time taken by pipe A to empty the pool, in hours, is -
एक तरण-ताल में 3 नालियाँ है। उनमें पहली दो A तथा B एक साथ चालू करने पर, तीसरी नाली C की तुलना में, ताल को खाली करने में आधा समय लेती है। उन्हीं में नाली A अकेली चालू करने पर, नाली B की तुलना में आधा समय लेती है। वे तीनों एक साथ ताल को खाली करने में 6 घंटे 40 मिनट लगाती है। तदनुसार, उस ताल को खाली करने में नाली A को कितना समय (घंटों में) लगेगा?
(1.) 5 (2) 8 (3) 10 (4) 12 (5) 15
Sol
Let time taken by pipe A be x hours
Time taken by pipe B = 2x hours
Time taken by pipe C = hours
hours
6.) A shopkeeper calculate his profit % or loss % on his selling price. He marked his article at such a price so that he can obtain a profit of 20%. Due to rain his article get damage and now he has to sell this article at Rs.384 calculating a loss of 25%. Find the price at which he marked his goods?
एक दुकानदार अपना लाभ % या हानि % विक्रय मूल्य पर ज्ञात करता है | यदि वह किसी वस्तु का दाम इस प्रकार अंकित करता है ताकि उस वस्तु से उसे 20% का लाभ हो | परन्तु वर्षा होने के कारण वस्तु ख़राब होने लगती है और वह उसे अब 384 रू. में बेचता है जिस पर उसे 25% की हानि होती है | वस्तु का अंकित मूल्य ज्ञात कीजिये ?
(1.) 500 (2) 560 (3) 600 (4) 614 (5) 640
Sol
Let S.P = 100
Then C.P = 80
So actual profit =
Thus If actual C.P =100
Then SP1 = 125
Similarly Actual LOSS = 20%
So SP2 = 80
SP1=
7.) Costs of two articles were in the ratio of 16 : 23. The cost of first articles increases by 10% and that of second by Rs.477. Now the costs of two articles are in a ratio of 11 : 20. The price of the second article (in Rs.) in the beginning was
दो वस्तुओं के क्रय मूल्य में अनुपात 16 : 23 है | पहली वस्तु के मूल्य में 10% की वृद्धि की जाती है और दूसरी वस्तु के मूल्य में रु.477 की वृद्धि की जाती है | अब दोनों वस्तुओं के क्रय मूल्यों में अनुपात 11 : 20 है | प्रारम्भिक अवस्था में दूसरी वस्तु का मूल्य (रु. में) है
(1.) 1200 (2) 1211 (3) 1216 (4) 1219 (5) 1220
Sol
Let the cost price of the first article =16x
and cost price of the second article = 23x
According to the question
Cost price of the second article = (53 × 23) = 1219
8.) A bag contains Rs. 90 in coins. If coins of 50 paise, 25 paise and 10 paise are in the ratio 2:3:5. How many 25 paise coins are there in the bag?
एक बैग में 90 रू. सिक्को में रखे हैं। यदि 50 पैसे, 25 पैसे और 10 पैसे के सिक्को का अनुपात 2:3:5 हो तो बैग में 25 पैसे के सिक्कों की संख्या क्या होगी?
(1.) 100 (2) 110 (3) 120 (4) 130 (5) 135
Sol
9.) Rahul deposit some amount in bank every year such that the value of amount is Rs 300 more than the amount he deposit in previous year. At the end of 20 years the total amount deposit by him was Rs 83000. Find the amount he will deposit at the end of 13th year?
राहुल एक बैंक में कुछ धनराशि इस प्रकार से हर वर्ष जमा करता हैं की उसके द्वारा जमा राशि पिछले वर्ष की धनराशि से 300 रू. अधिक होती है | 20 वर्ष के अंत में उसके द्वारा जमा कुल धनराशि 83000 रू है |ज्ञात कीजिये 13वे वर्ष के अंत में उसने कितनी धनराशि जमा की?
(1.) 4800 (2) 4825 (3) 4850 (4) 4875 (5) 4900
Sol
10.) Prasad is thrice as good a workman Malviya and therefore able to finish a work in 48 days less than Malviya. Working together, they can do it in
प्रसाद एक कारीगर मालवीय से तीन गुना कुशल है और इसलिए एक काम को मालवीय से 48 दिन कम में काम समाप्त कर देता है | एक साथ काम करके, वे इसे कर सकते हैं
d (1.) 28 (2) 32 (3) 36 (4) 40 (5) 44
Sol
Let time taken by Prasad = x days
Then, time taken by Maliviya = 3x days
According to the question
3x - x = 48
2x = 48
x = 24 days | 2,715 | 6,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-43 | latest | en | 0.417402 |
https://vivaceproject.com/negative-and-positive-number-worksheets/ | 1,620,264,502,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988724.75/warc/CC-MAIN-20210505234449-20210506024449-00159.warc.gz | 621,602,887 | 10,350 | Negative And Positive Number Worksheets
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Integers are whole numbers (no fractional or decimal part) and can be negative or positive. sample grade integers worksheet. The reverse is the case, of course, with negative numbers. students should be able to recognize easily that a positive number is always greater than a negative number and that between two negative integers, the one with the lesser absolute value is actually the greater number.
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Positive and negative numbers lesson. this number line shows the position of five train stations. takes the train from station a. station b is miles to the east of station a. station c is miles to the west of station a. the locations of stations a, b, and c are labeled on the number line.
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The starting point of the number line always have to be a zero it can begin with a small number, but the series of number lines always have to be in order. now that you know what a number line is lets take a look at how you can place positive and negative values on the number line.
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A simple ordering sheet with answers included allows your children to focus on learning. Four worksheets using positive and negative numbers. common core curriculum alignment. ccss.math.content.ns.c. understand that positive and negative numbers are used together to describe quantities having opposite directions or values (e.
10. Ordering Positive Negative Numbers Worksheet Fun Teaching Number Worksheets
G., temperature zero, elevation sea level,, electric charge) use. Adding positive numbers, such as, is easy. when we add a negative number to a positive number, or two negative numbers, that can sometimes seem tricky. however, there are some simple rules to follow and we introduce them here.
11. Positive Negative Integer Rules Worksheets Number
Rule adding positive numbers to positive just normal addition. positive means to the right of zero on a number line negative means to the left of zero on a number line every positive number has an opposite negative number of the same size. for example - is the opposite of because both are the same distance from zero.
12. Positive Negative Numbers Helping Math Number Worksheets
Adding, subtracting, multiplying and dividing positive and negative numbers - circuit out the problem in the cell. to advance in the circuit, find your answer and mark that cell. work out that problem and then find your answer again. continue working in this manner until you.
13. Positive Negative Numbers Worksheet Number Worksheets
On this page you find our worksheets with negative integers, negative decimals and negative fractions. our negative number worksheets are suited for math grade and and are a great math resource for remedial math or math tutoring purposes.we have integer worksheets covering the addition and subtraction of integers and negatives, from number bonds to worksheets with missing addends and.
14. Quiz Worksheet Positive Negative Space Art Study Worksheets Grade Time Common Core Learning Read Kindergarten Congruent Segments Number
Negative numbers worksheet test - - - - - - - - - - - - - - - - - test - - - - - -. Starter lesson input and differentiated worksheets for year negative numbers. worksheets made by me, information collated from number of resources found on.
15. Subtracting Fractions Mixed Numbers Worksheet Review Math Worksheets Grade Studying Websites Graders Burns Games Workbooks Negative Positive Number
Practice showing positive and negative fractions, decimals, and mixed numbers on number lines using this rational numbers on number lines worksheet in this sixth-grade math worksheet, students will use their number sense to plot a variety of rational numbers on number lines.
16. Symmetry Number Worksheet Printable Adding Positive Negative Numbers Worksheets Multiplying Integers
For an extra challenge. Using a number line showing both sides of zero is very helpful to help develop the understanding of working with positive and negative. its easier to keep track of the negative numbers if you enclose them in brackets. Showing top worksheets in the category - positive and negative number line.
It may be printed, downloaded or saved and used in your classroom, home school, or other educational environment. Negative and positive number year showing top worksheets in the category - negative and positive number year. some of the worksheets displayed are negative numbers y practice book b, positive and negative numbers maths year, word problems with integers, integers, negative numbers work, math notes integers sol a b, lesson. This is a comprehensive collection of free printable math worksheets for grade and for pre-algebra, organized by topics such as expressions, integers, one-step equations, rational numbers, multi-step equations, inequalities, speed, time distance, graphing, slope, ratios, proportions, percent, geometry, and pi. they are randomly generated, printable from your browser, and include the answer. Positive and negative numbers worksheets and activities. free interactive exercises to practice or download as to print. Positive and negative numbers worksheets printable positive and negative numbers worksheets printable might help a trainer or pupil to find out and understand the lesson strategy in a quicker way. | 2,193 | 12,307 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.864905 |
http://grinebiter.com/Numbers/Cardinal/2200-two-thousand-two-hundred.html | 1,556,100,943,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578640839.82/warc/CC-MAIN-20190424094510-20190424120510-00193.warc.gz | 72,950,566 | 3,456 | Two thousand two hundred
Here is information about "two thousand two hundred" that you may find useful and interesting. Number Systems Two thousand two hundred is a decimal number and can be written with numbers: 2200 Binary is a number system with only 0s and 1s. Two thousand two hundred in binary form is displayed below: 100010011000 A Hexadecimal number has a base of 16 which means it includes the numbers 0 to 9 and A through F. Two thousand two hundred converted to hexadecimal is: 898 Roman Numerals is another number system. Below is two thousand two hundred in roman numerals: MMCC Scientific Notation Sometimes calculators and scientists shorten numbers using scientific notation. Here is two thousand two hundred as a scientific notation: 2.2E+03 Math Here are some math facts about Two thousand two hundred: Two thousand two hundred is a rational number and an integer. Two thousand two hundred is an even number because it is divisible by two. Two thousand two hundred is divisible by the following numbers: 1, 2, 4, 5, 8, 10, 11, 20, 22, 25, 40, 44, 50, 55, 88, 100, 110, 200, 220, 275, 440, 550, 1100, 2200 Two thousand two hundred is not a square number because no number multiplied by itself will equal two thousand two hundred. Number Lookup Two thousand two hundred is not the only number we have information about. Go here to look up other numbers.
Translated Here we have translated two thousand two hundred into some of the most commonly used languages: Chinese: 二千二百 French: deux mille deux cents German: zweitausend zweihundert Italian: duemila duecento Spanish: dos mil doscientos
Currency Here is two thousand two hundred written in different currencies: US Dollars: \$2200 Canadian Dollars: CA\$2200 Australian Dollars: A\$2200 British Pounds: £2200 Indian Rupee: ₹2200 Euros: €2200
Ordinal The cardinal number two thousand two hundred can also be written as an ordinal number: 2200th Or if you want to write it with letters only: two thousand two hundredth.
Two thousand two hundred one Go here for the next number on our list that we have information about. | 503 | 2,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | latest | en | 0.880127 |
https://convertoctopus.com/23-2-feet-to-kilometers | 1,603,338,738,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878879.33/warc/CC-MAIN-20201022024236-20201022054236-00533.warc.gz | 280,723,008 | 7,452 | ## Conversion formula
The conversion factor from feet to kilometers is 0.0003048, which means that 1 foot is equal to 0.0003048 kilometers:
1 ft = 0.0003048 km
To convert 23.2 feet into kilometers we have to multiply 23.2 by the conversion factor in order to get the length amount from feet to kilometers. We can also form a simple proportion to calculate the result:
1 ft → 0.0003048 km
23.2 ft → L(km)
Solve the above proportion to obtain the length L in kilometers:
L(km) = 23.2 ft × 0.0003048 km
L(km) = 0.00707136 km
The final result is:
23.2 ft → 0.00707136 km
We conclude that 23.2 feet is equivalent to 0.00707136 kilometers:
23.2 feet = 0.00707136 kilometers
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilometer is equal to 141.41551271608 × 23.2 feet.
Another way is saying that 23.2 feet is equal to 1 ÷ 141.41551271608 kilometers.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that twenty-three point two feet is approximately zero point zero zero seven kilometers:
23.2 ft ≅ 0.007 km
An alternative is also that one kilometer is approximately one hundred forty-one point four one six times twenty-three point two feet.
## Conversion table
### feet to kilometers chart
For quick reference purposes, below is the conversion table you can use to convert from feet to kilometers
feet (ft) kilometers (km)
24.2 feet 0.007 kilometers
25.2 feet 0.008 kilometers
26.2 feet 0.008 kilometers
27.2 feet 0.008 kilometers
28.2 feet 0.009 kilometers
29.2 feet 0.009 kilometers
30.2 feet 0.009 kilometers
31.2 feet 0.01 kilometers
32.2 feet 0.01 kilometers
33.2 feet 0.01 kilometers | 489 | 1,753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.803887 |
https://www.teachoo.com/4371/710/Ex-6.3--12---Find-equations-of-all-lines-having-slope-0--tangent/category/Ex-6.3/ | 1,726,782,468,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652067.20/warc/CC-MAIN-20240919194038-20240919224038-00305.warc.gz | 933,299,490 | 23,772 | Tangents and Normals (using Differentiation)
Chapter 6 Class 12 Application of Derivatives
Serial order wise
### Transcript
Question 12 Find the equations of all lines having slope 0 which are tangent to the curve 𝑦 = 1/(𝑥2 −2𝑥 + 3) Given Curve is 𝑦 = 1/(𝑥2 −2𝑥 + 3) Slope of tangent is 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥=𝑑(1/(𝑥2 − 2𝑥 + 3))/𝑑𝑥 𝑑𝑦/𝑑𝑥=(𝑑(𝑥2 − 2𝑥 + 3)^(−1))/𝑑𝑥 𝑑𝑦/𝑑𝑥=−1(𝑥2−2𝑥+3)^(−2) . 𝑑(𝑥^2− 2𝑥 + 3)/𝑑𝑥 𝑑𝑦/𝑑𝑥=−(𝑥2−2𝑥+3)^(−2) (2𝑥−2) 𝑑𝑦/𝑑𝑥=−2(𝑥2−2𝑥+3)^(−2) (𝑥−1) 𝑑𝑦/𝑑𝑥=(−2(𝑥 − 1))/(𝑥2 − 2𝑥 + 3)^2 Hence Slope of tangent is (−2(𝑥 − 1))/(𝑥2 − 2𝑥 + 3)^2 Given Slope of tangent is 0 ⇒ 𝑑𝑦/𝑑𝑥=0 (−2(𝑥 − 1))/(𝑥2 − 2𝑥 + 3)^2 =0 −2(𝑥 − 1)=0 ×(𝑥2 − 2𝑥 + 3)^2 −2(𝑥 − 1)=0 (𝑥−1)=0 𝑥=1 Finding y when 𝑥=1 𝑦=1/(𝑥2 − 2𝑥 + 3) 𝑦=1/((1)^2 − 2(1) + 3) 𝑦=1/(1 − 2 + 3) 𝑦=1/(2 ) Point is (𝟏 , 𝟏/𝟐) Thus , tangent passes through (1 , 1/2) Equation of tangent at (1 , 1/2) & having Slope zero is (𝑦 −1/2)=0(𝑥−1) 𝑦 −1/2=0 𝑦=1/2 Hence Equation of tangent is 𝒚=𝟏/𝟐 We know that Equation of time passing through (𝑥 , 𝑦) & having Slope m is (𝑦 −𝑦1)=𝑚(𝑥 −𝑥2) | 719 | 1,022 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-38 | latest | en | 0.575282 |
https://programtalk.com/python/cv2-projectpoints-opencv-comprehensive-guide/ | 1,708,950,663,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474659.73/warc/CC-MAIN-20240226094435-20240226124435-00190.warc.gz | 482,683,045 | 20,134 | # Using cv2.projectPoints in OpenCV: A Comprehensive Guide
In this article, we’ll dive deep into the cv2.projectPoints function in OpenCV, discussing its applications, parameters, and providing examples to help you understand how it works. By the end of this article, you’ll have a solid grasp of this function and be able to use it effectively in your projects.
## Understanding cv2.projectPoints
cv2.projectPoints is a versatile function that allows you to project 3D points into 2D image coordinates. This projection is useful in various computer vision tasks, including object pose estimation, image stitching, and 3D rendering. The function takes into account the camera’s intrinsic and extrinsic parameters, as well as the lens distortion, to provide an accurate projection.
## Function Syntax
```import cv2
image_points, _ = cv2.projectPoints(object_points, rvec, tvec, camera_matrix, dist_coeffs)
```
Let’s examine the parameters of cv2 projectPoints:
• object_points: This is a NumPy array containing 3D coordinates of the object points in the world coordinate system. The shape should be (n, 3), where n is the number of points.
• rvec: This is a rotation vector that represents the rotation between the world and camera coordinate systems. It is a 3×1 NumPy array.
• tvec: This is a translation vector that represents the translation between the world and camera coordinate systems. It is a 3×1 NumPy array.
• camera_matrix: This is the camera’s intrinsic matrix, represented as a 3×3 NumPy array. The matrix contains information about the focal length and the optical center of the camera.
• dist_coeffs: This is a NumPy array of the lens distortion coefficients. It can have up to 14 coefficients, but usually, only the first five are used (k1, k2, p1, p2, k3).
cv2 projectPoints returns two values:
• image_points: This is a NumPy array containing the projected 2D image points. The shape is (n, 1, 2), where n is the number of points.
• jacobian: This is the Jacobian matrix of the function, containing the partial derivatives with respect to the input parameters. It’s not commonly used in most applications, so we’ll ignore it in this tutorial.
## Preparing the Parameters
Before we can use cv2.projectPoints, we need to obtain the necessary parameters. Let’s discuss how to get these parameters:
### Object Points
Object points are the 3D coordinates of the points you want to project. These coordinates should be in the world coordinate system. You can either measure the coordinates directly or obtain them from a 3D model.
### Camera Matrix and Distortion Coefficients
To obtain the camera matrix and distortion coefficients, you need to calibrate your camera. Camera calibration is the process of estimating the camera’s intrinsic and extrinsic parameters. You can use a calibration pattern (e.g., a chessboard) and OpenCV’s camera calibration functions to perform this task. For more information, check out this tutorial on camera calibration with OpenCV.
### Rotation and Translation Vectors
The rotation and translation vectors represent the transformation between the world and camera coordinate systems. You can obtain these parameters using various methods, such as solving the Perspective-n-Point (PnP) problem or using a marker-based tracking system like ArUco markers. For more information, check out this tutorial on solving the PnP problem with OpenCV.
## Example: Projecting 3D Points to 2D Image Coordinates
Now that we have an understanding of the cv2.projectPoints function and its parameters, let’s see it in action. In this example, we’ll project the corners of a 3D cube onto a 2D image plane.
First, let’s prepare the object points, which are the 3D coordinates of the cube’s corners:
```import numpy as np
cube_size = 100 # Length of the cube's edges in mm
object_points = np.array([
[0, 0, 0],
[cube_size, 0, 0],
[cube_size, cube_size, 0],
[0, cube_size, 0],
[0, 0, cube_size],
[cube_size, 0, cube_size],
[cube_size, cube_size, cube_size],
[0, cube_size, cube_size],
], dtype=np.float32)
```
Assuming we have already calibrated our camera, let’s define the camera matrix, distortion coefficients, rotation vector, and translation vector:
```camera_matrix = np.array([
[800, 0, 320],
[0, 800, 240],
[0, 0, 1]
], dtype=np.float32)
dist_coeffs = np.array([0.1, -0.05, 0, 0, 0], dtype=np.float32)
rvec = np.array([0.1, 0.2, 0.3], dtype=np.float32)
tvec = np.array([200, 100, 500], dtype=np.float32)
```
Now, we can use cv2.projectPoints to project the 3D cube corners onto the 2D image plane:
```image_points, _ = cv2.projectPoints(object_points, rvec, tvec, camera_matrix, dist_coeffs)
```
Finally, let’s visualize the projected points on an empty image:
```import cv2
image = np.zeros((480, 640, 3), dtype=np.uint8)
for point in image_points:
x, y = point.ravel().astype(int)
cv2.circle(image, (x, y), 5, (0, 255, 0), -1)
cv2.imshow("Projected Points", image)
cv2.waitKey(0)
cv2.destroyAllWindows()
```
Running the code above will display an image with the projected 2D points of the cube corners. You can adjust the parameters to see how they affect the projection.
## Applications of cv2.projectPoints
The cv2.projectPoints function has numerous applications in computer vision and robotics. Here are a few examples:
• Object pose estimation: By estimating the position and orientation of a 3D object in the camera’s coordinate system, you can use cv2.projectPoints to overlay the object’s virtual representation onto the 2D image plane. This can be useful in augmented reality applications or robotics.
• 3D rendering: If you’re working with a 3D scene and a virtual camera, you can use cv2.projectPoints to project the 3D coordinates of the scene onto a 2D image plane. This is the basis for 3D rendering in computer graphics.
• Image stitching: When combining multiple images to create a panoramic image, you can use cv2.projectPoints to find the correspondence between the 3D scene points and their 2D image coordinates in each input image. This correspondence is essential for estimating the camera’s relative pose and stitching the images together.
## Conclusion
In this comprehensive guide, we explored the cv2.projectPoints function in OpenCV. We discussed its applications, parameters, and provided an example to demonstrate how to use it effectively. By understanding the function and its use cases, you can now leverage cv2.projectPoints in your own computer vision projects.
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https://civilmdc.com/2023/06/17/design-of-composite-beam-columns-rectangular-concrete-section-with-single-steel-profile/ | 1,721,922,205,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763860413.86/warc/CC-MAIN-20240725145050-20240725175050-00071.warc.gz | 139,908,963 | 80,359 | Civil MDC
# DESIGN OF COMPOSITE BEAM-COLUMNS Rectangular Concrete Section With Single Steel Profile
Designing a composite beam-column involves combining the strength and stiffness of both concrete and steel to create a structural member capable of carrying both axial and bending loads. Here’s a step-by-step approach to designing a rectangular concrete section with a single steel profile composite beam-column:
Determine the loads: Identify the axial load and the moment applied to the composite beam-column. These loads are typically obtained from the structural analysis of the overall structure.
Select steel profile: Choose a suitable steel profile that will act as the internal steel reinforcement within the composite section. Common choices include steel I-beams or H-sections. The selection of the steel profile should consider the desired strength and stiffness requirements.
Calculate the effective flange width: The effective flange width is the portion of the steel profile that contributes to the composite action. It depends on factors such as the connection details and the width of the concrete slab (if present). Consult design codes or engineering references for guidance on calculating the effective flange width.
Determine the section properties: Calculate the section properties of the composite section, including the effective width, area, centroid, and moment of inertia. These properties can be determined by considering the properties of both the concrete and steel components of the section.
Check for overall stability: Verify that the composite beam-column satisfies the stability requirements. This involves checking for buckling and lateral-torsional buckling using appropriate formulas and design standards.
Check for axial load capacity: Determine the axial load capacity of the composite section. This involves calculating the axial resistance of both the concrete and steel components and comparing them to the applied axial load. Ensure that the section is capable of carrying the applied axial load without exceeding the design limits.
Check for bending capacity: Calculate the moment capacity of the composite beam-column using appropriate design formulas. This involves considering the contribution of both the concrete and steel components to the moment resistance. Verify that the section is capable of resisting the applied bending moment without exceeding the design limits.
Check for shear capacity: Evaluate the shear capacity of the composite section. Calculate the shear resistance of the concrete and steel components and ensure that the section can resist the applied shear force without failure.
Design the reinforcement: Determine the required reinforcement for the steel profile and the concrete section. This involves calculating the required steel reinforcement to resist the bending moment and shear forces. The reinforcement design should comply with the relevant design codes and standards.
Detailing and construction: Develop the reinforcement detailing, including the placement and anchorage of the steel reinforcement within the concrete section. Consider the connection details between the steel profile and the concrete to ensure proper load transfer and composite action.
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Overview -
Take the mystery out of statistical terms and put Excel to work
If you need to create and interpret statistics in business or classroom settings, this easy-to-use guide is just what you need. It shows you how to use Excel's powerful tools for statistical analysis, even if you've never taken a course in statistics. Read more...
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More About Statistical Analysis with Excel for Dummies by Ph.d. Joseph Schmuller
Overview
Take the mystery out of statistical terms and put Excel to work
If you need to create and interpret statistics in business or classroom settings, this easy-to-use guide is just what you need. It shows you how to use Excel's powerful tools for statistical analysis, even if you've never taken a course in statistics. Learn the meaning of terms like mean and median, margin of error, standard deviation, and permutations, and discover how to interpret the statistics of everyday life. You'll learn to use Excel formulas, charts, PivotTables, and other tools to make sense of everything from sports stats to medical correlations.
• Statistics have a reputation for being challenging and math-intensive; this friendly guide makes statistical analysis with Excel easy to understand
• Explains how to use Excel to crunch numbers and interpret the statistics of everyday life: sales figures, gambling odds, sports stats, a grading curve, and much more
• Covers formulas and functions, charts and PivotTables, samples and normal distributions, probabilities and related distributions, trends, and correlations
• Clarifies statistical terms such as median vs. mean, margin of error, standard deviation, correlations, and permutations
Statistical Analysis with Excel For Dummies, 3rd Edition helps you make sense of statistics and use Excel's statistical analysis tools in your daily life.
Details
• ISBN-13: 9781118464311
• ISBN-10: 1118464311
• Publisher: For Dummies
• Publish Date: April 2013
• Page Count: 504
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http://www.thefullwiki.org/70_(number) | 1,386,690,471,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164021936/warc/CC-MAIN-20131204133341-00029-ip-10-33-133-15.ec2.internal.warc.gz | 554,063,084 | 12,322 | # 70 (number): Wikis
Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
# Encyclopedia
Updated live from Wikipedia, last check: December 10, 2013 15:46 UTC (48 seconds ago)
### From Wikipedia, the free encyclopedia
70
Cardinal seventy
Ordinal 70th
(seventieth)
Factorization $2 \cdot 5 \cdot 7$
Divisors 1,2 5, 7, 10, 14, 35, 70
Roman numeral LXX
Binary 10001102
Octal 1068
Duodecimal 5A12
Hexadecimal 4616
Hebrew ע (Ayin)
70 (seventy) is the natural number following 69 and preceding 71.
## In mathematics
Its factorization makes it a sphenic number. 70 is a Pell number and a generalized heptagonal number, one of only two numbers to be both.[1] Also, it is the seventh pentagonal number and the fourth 13-gonal number, as well as the fifth pentatope number. It is the smallest weird number.
Since it is possible to find sequences of 70 consecutive integers such that each inner member member shares a factor with either the first or the last member, 70 is an Erdős–Woods number.
In base 10, it is a Harshad number.
70 is:
## Number name
The French do not have a word for 70, instead using "soixante-dix" (60 + 10). Other French-speaking countries such as Belgium and Switzerland do have a word for it, using "septante."[2]
## Notes
1. ^ B. Srinivasa Rao, "Heptagonal Numbers in the Pell Sequence and Diophantine Equations $2x^2 = y^2(5y - 3)^2 \pm 2$" Fib. Quart. 43 3: 194
2. ^ Peter Higgins, Number Story. London: Copernicus Books (2008): 19. "Belgian French speakers however grew tired of this and introduced the new names septante, octante, nonante etc. for these numbers."
# Simple English
Order ••••• ••••• ••••• ••••• ••••• ••••• ••••• ••••• ••••• ••••• ••••• ••••• ••••• ••••• seventieth
Seventy is the number that is after sixty-nine and before seventy-one.
The prime factors of seventy are 2, 5, and 7. (2 * 5 * 7 = 70)
Got something to say? Make a comment. Your name Your email address Message | 639 | 2,078 | {"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": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2013-48 | longest | en | 0.812033 |
https://www.eurocontrolli.com/en/the-true-meaning-of-divine-mathematics/ | 1,723,553,441,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00628.warc.gz | 583,984,901 | 12,537 | ---------------------------------------------------------
Uncategorized / The True Meaning of Divine Mathematics
## Divine Mathematics – Is it a Scam?
She and her husband were the initial gods made by Atum. Seeing the extraordinary way God designed the sunflowers should remind us that we’re ready to trust Him to manage the specifics of our lives. Then, Alexandria was one of the best cities on earth, and a middle of learning and culture together with faith. He, who’s the prudent man of the east.
The main reason this is so is that it’s the exact number of individuals who can live on Earth in full peace. Before he could save yourself someone else, he must conserve self. In fact, it might be the change essay writer service you’ve been waiting for for your entire life. We’ve got to discover methods to spice this up. He’s been through the 10th level.
This bird is called Suparna. I recall he was walking with a cane because he had some type of an incident. Attempt to use the mouse if you’re comfy sitting back. On the remainder of the body, it’s possible to draw other golden rectangles. This pattern assures that every leaf is going to receive its highest possible exposure to sunlight and air without shading or crowding different leaves.
By cleverly interfering with something to view the way that it reacts we are all set to learn the way that it works. The arrangement of sharps and flats at the beginning of a little music is called a vital signature. He wished to find some great MCs. But we were attempting to simplify it. The movie opens up with this guy speaking about his ponds.
## Key Pieces of Divine Mathematics
As a teacher, you would like your students to go past the tool. Of the approximately 7,000 languages on earth today, approximately 550 have access to the total Bible. He’s drawn the Flower of Life itself, in addition to components therein, including the Seed of Life. Math has components that don’t change. Explain briefly what’s living Mathematics. To get more information, visit Entertainment Earth.
## The Chronicles of Divine Mathematics
You might also download the software packages that are available on the net to simplify your day-to-day planning. The reason this info is important at this time is the fact that it will give us a better perspective of what’s to come. It’s a digital library support. There are any range of blueprints He could have chosen. Detective controls are backup procedures that are designed to catch items or events which were missed by the exact first area of defense. Moovit has each one of the public transit info which you would like to comprehend.
## Whatever They Told You About Divine Mathematics Is Dead Wrong…And Here’s Why
We might surmise he wasn’t too concerned about it. Well, it’s at least misunderstood. But whenever an infimum or supremum does exist, it’s unique. I understand this sounds mad, but it’s true and very real.
In the event the quick side of the rectangle is 1, the lengthy side is going to be 1.618. This ratio is easily the most efficient of similar collection of numbers. When using high energy stones, pay attention to the should take action to make sure you are safeguarded. Using computers to lessen the load on mathematicians isn’t new, even supposing it isn’t welcomed by all in the area. It was rather tricky to do calculations with this sort of a cumbersome system. You select a particular base width first for instance, 90px. Adding massive numbers just in your head can be hard.
At the present time (1988), it’s the topic of computer science which occupies the best aspect of that common border. Place it where you’re able to see it all the chance to strengthen the vibrancy of a present relationship. In the event that you ever have an opportunity to cover the book with others, you’re going to be in a place to state your opinions clearly, as you’ve taken the chance to genuinely think about all of the aspects involved. We’re a platform for radical creatives that are transforming culture. Yet, there exist some techniques that may help the readers to truly have a nice and effectual reading experience.
## What Everybody Dislikes About Divine Mathematics and Why
Born is the conclusion of all things in existence. Therefore it isn’t possible for them to deal equality with other folks. Sometimes your differences can’t be overlooked. So instead you use a lot of simple approximations and combine them all with each other to acquire good approximate answers. It’s a dilemma of perspective.
Rewriting the exact same scale pattern at a different pitch is referred to as transposition. Make sure that the excellent legacy of ancient Egypt will live again later on. We’ll call it gravity, and then we’ll study the disposition of gravity. Thus, it’s not surprising that Kaldor should believe fantastic financial theory should reflect the actual world as much as possible. For starters, they can offer inspiration once your creativity well appears to get run dry. There might also be golden ratios in the vertical measurements of the painting.
## Divine Mathematics – Is it a Scam?
You gotta be prepared to rumble. In the end, the haves will need to be a blessing to the have-nots. Since then this exact place was used to distribute many Bibles across the world. Deal with others like you were dealing with God and you’ll enjoy fantastic and mysterious success this year. From the start, right from the very start. | 1,112 | 5,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.951485 |
www.sarahscheffler.net | 1,539,774,893,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511173.7/warc/CC-MAIN-20181017111301-20181017132801-00232.warc.gz | 1,064,785,895 | 7,500 | A week ago or so, Norbert Blum posted a paper in which he claims he has solved P != NP. He’s hardly the first to try, and he might not be the last. But I’ve seen surprisingly little analysis of this, so I figured I’d give my novice opinion of the subject.
I’ve now read a couple of blog posts regarding the paper in question. It explains some things, but I was still left confused.
Without further ado, here’s my read of Blum’s paper, with the caveat that I don’t usually dabble in complexity any more complicated than “exponential vs polynomial,” and the other caveat that I’ve probably spent a grand total of five hours on this, which is not a lot of time. I could have made (major) mistakes in this, and I appreciate any input or corrections. Email ‘em to me.
Here we go.
P is the set of languages accepted in polynomial time (with regard to the input length) by a deterministic Turing machine (if you don’t know what I’m talking about, think of a TM as a very clunky approximation of a computer). NP is the set of languages accepted in polynomial time wrt the input length by a non-deterministic TM (meaning you can sort of “guess” and assume you’ll get the right answer).
To prove P != NP, we need to show that there is a problem that is in NP but not P. To do that, we take a problem we know is in NP, and attempt to establish a lower bound on its runtime on a deterministic TM. If that lower bound is superpolynomial (meaning, strictly bigger than polynomial), then it can’t be in P.
This paper uses somewhat complex (ha-ha) notions of complexity. First of all, we’re primarily talking about circuit complexity here (whereas until now we’ve been talking about time complexity). So we’re talking about minimum circuit size, represented in numbers of AND, OR, and NOT gates. But we’re not just going to count the total number of AND, OR, and NOT gates; we’re going to use slightly more complicated notions:
• Monotone complexity: What is the number of AND and OR gates in the smallest monotone circuit (no NOT gates allowed) we can build to compute this function?
• Non-monotone complexity: What is the number of AND and OR gates in the smallest non-monotone circuit we can build to compute this function? (Now we’re allowed to use NOT, but we don’t count them.)
• Standard Complexity: What is the number of AND and OR gates in the smallest non-monotone circuit we can build to compute this function, where all the NOT gates happen at the very beginning (on input wires)?
(You can convert any non-monotone circuit to a standard circuit pretty easily using De Morgan’s Laws, and it multiplies the circuit size by at most 2, which is a constant factor that we don’t care about. So proving statements about standard circuit complexity also proves them about non-monotone circuit complexity, and vice versa.)
Without getting into details, there’s a nice theorem that shows that if a function has a large non-monotone circuit size, then the time required to compute it on a TM is also large. So we often use circuit complexity in place of time complexity, because it can be easier to work with. I thought this chapter presented this concept well.
Now, unfortunately these results only apply to non-monotone circuits. Just because a monotone circuit has a superpolynomial lower bound, doesn’t mean it’s not in P. Razborov showed that the Perfect Matching problem (is there a subset of the edges in a graph that touch all vertices of the graph without ever sharing a vertex?) is in P, but the lower bound of a monotone circuit for this problem is superpolynomial. So a superpolynomial monotone lower bound isn’t sufficient.
But a superpolynomial non-monotone lower bound would be sufficient. The goal of Blum’s paper is to take superpolynomial lower bounds of monotone circuits for problems in NP from Razborov, Alon, and Boppana, and show how to apply them to non-monotone circuits. If we do that, we show P != NP.
How do we show lower bounds? We use approximators. What’s an approximator? It’s a smaller circuit that we “want” to have the same functionality as a specific larger circuit. Basically, if a “good” small approximator can be built, then the lower bound of the big circuit’s size is at most as big as the size of the approximator. If, on the other hand, we can show that no good approximator of a certain size exists, then we know that the lower bound of the big circuit’s size is bigger than the size of that approximator.
So our goal is to show that no good polynomial-size approximator exists for a (non-monotone) circuit in NP. That would show that that problem is in NP but not P.
How do we know an approximator is good? We use test inputs. We use positive test inputs, which we know are in the language, and negative test inputs, which we know are not. If an approximator is good, it will accept all the positive test inputs and reject all the negative test inputs. It’s also pretty easy to show that the test inputs are sufficient to show that the approximator has the same functionality as the full circuit with overwhelming probability - an approximator that succeeds at all the test inputs is extremely likely to have the same exact functionality as the circuit we’re approximating. So it’s an if-and-only-if kind of thing. There exists a good approximator of a certain size for a circuit if and only if that circuit can be written in as many gates as the approximator.
These two sources do a good job of talking about Razborov’s original approximator idea, which use approximators in disjunctive normal form (DNF - basically a bunch of monomials connected by OR gates). They bound the size of clauses and bound the number of ORs. Then they prove that no approximator of polynomial (monotone) size can approximate the Clique problem (which is in NP), even on just these test inputs. (Remember, these were all for monotone circuits - not sufficient to show anything about P vs NP. If we had a similar result for non-monotone circuits, we’d have something to talk about.)
Berg and Ulfberg create a slightly different method for Clique that uses “CNF-DNF approximators.” Same idea, but now we use DNF and CNF formulas (CNF is conjunctive normal form - a bunch of clauses connected by ANDs). But still, we’re only talking about monotone circuits. We can make CNF/DNF switches and DNF/CNF switches to convert between the two formats when each is most convenient, with well-understood effects on the number of gates.
Okay here we go. Blum’s paper released last week:
Blum’s main claim - Theorem 6 - is that if you can use CNF-DNF approximators to prove a lower bound on a monotone circuit of a (monotone) function, then you can translate that approximator into an approximator to prove a lower bound for a standard circuit of the same (still monotone) function.
Equivalently,* if you can show that no sufficiently small CNF-DNF approximator exists for monotone complexity, you can convert that argument into one that shows that no sufficiently small CNF-DNF approximator exists for standard complexity.
* To me, this is the least convincing claim by Blum. None of the blog posts I’ve read analyzing this paper seem to think this is an issue, so I might be missing something about this argument - it might be that you can prove a negation in this way. But it seems to me that the argument here is:
1. CNF-DNF approximators of small (polynomial) size cannot approximate this function in monotone circuit-land.
2. We convert these approximators to standard circuit-land.
3. These approximators cannot be used to approximate the function in standard circuit-land.
4. ??? Therefore, no CNF-DNF approximator exists that can approximate this function in standard circuit-land. ???
I don’t buy step 4; it seems to me that we’ve just proven that CNF-DNF approximators made by converting monotone CNF-DNF approximators can’t approximate the function. That’s not to say you can’t create a CNF-DNF approximator in a different way that can approximate the circuit. But like I said, no one else seems to think this is an issue, and I had literally never heard of an approximator until last Friday, so it’s possible I’m misunderstanding something about how they work.
A very simple thing that would eliminate this concern would be to show some bound on the error of the translation. If I could see that there is no better circuit than the translated one, then my whole concern goes out the window. I didn’t see it in there, but again, I didn’t spend that long on this and I’m not an expert.
You might also say “hey didn’t you say Razborov showed there was no polynomial monotone approximator for a problem in P? What’s up with that?” Well, it turns out the perfect matching problem never admitted a CNF-DNF approximator for the monotone version, so there’s no way to convert it upward. (I’m puzzled by this, because surely you ought to be able to encode Perfect Matching as a problem that does have a CNF-DNF approximator? Or else we’ve unlocked some weird property of P? But hey idk.)
Razborov’s other (stronger) statement that you can’t prove better than quadratic lower bounds for non-monotone circuits using approximations was also challenged by Blum, who says that Razborov made an assumption that doesn’t actually hold; the distance metric was too strong. Blum explains this in an earlier paper, which I couldn’t find a non-paywalled version of. To be honest, I don’t fully understand the minutiae in this paper either, it would have added like five more paragraphs to this post and I’m tired.
Anyway, then Blum goes on to show this result concretely for the Clique problem, just like Razborov did for monotone circuits.
Woohoo, we made it! We’ve proven (?) a superpolynomial lower bound of standard circuits for a problem in NP, which implies a superpolynomial lower bound of a non-monotone circuit for it, which implies superpolynomial time of computation for a deterministic TM. Thus, there is a problem in NP that is not in P, and therefore P != NP. Assuming nothing went wrong along the way.
So yeah, basically, to my novice eye, this result doesn’t not look promising. I’m sort of surprised I haven’t seen more hubbub about it. I’m not fully convinced that this argument about converting negative statements holds (no approximator exists for this context, and we convert it to another context and claim that no approximator exists in that new context). Maybe there’s some factoid about conversions of monotone approximators cover the space of standard approximators, I don’t know. Maybe there’s a flaw in the construction. Maybe Blum is wrong when he says this isn’t a natural proof (which provably cannot prove that P != NP). Or maybe he’s right, who knows! Exciting times.
Anyway, I hope you enjoyed my attempted explanation. Again, I’m not a complexity theorist; I welcome any comments and corrections by email. | 2,437 | 10,855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-43 | longest | en | 0.947631 |
https://lucatrevisan.wordpress.com/2009/03/09/cs276-lecture-12-goldreich-levin/ | 1,500,604,342,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423681.33/warc/CC-MAIN-20170721022216-20170721042216-00066.warc.gz | 670,697,333 | 45,115 | # CS276 Lecture 12: Goldreich-Levin
Scribed by Jonah Sherman
Summary
Today we prove the Goldreich-Levin theorem.
1. Goldreich-Levin Theorem
We use the notation
$\displaystyle \langle x,r \rangle := \sum_i x_ir_i \bmod 2 \ \ \ \ \ (1)$
Theorem 1 (Goldreich and Levin) Let ${f: \{ 0,1 \}^n \rightarrow \{ 0,1 \}^n}$ be a permutation computable in time ${r}$. Suppose that ${A}$ is an algorithm of complexity ${t}$ such that
$\displaystyle \mathop{\mathbb P}_{x,r} [ A(f(x),r) = \langle x,r \rangle ] \geq \frac 12 + \epsilon \ \ \ \ \ (2)$
Then there is an algorithm ${A'}$ of complexity at most ${O((t+r) \epsilon^{-2}n^{O(1)})}$ such that
$\displaystyle \mathop{\mathbb P}_{x} [ A'(f(x)) = x ] \geq \frac \epsilon 4$
Last time we proved the following partial result.
Lemma 2 (Goldreich-Levin Algorithm — Weak Version) Suppose we have access to a function ${H: \{ 0,1 \}^n \rightarrow \{ 0,1 \}}$ such that, for some unknown ${x}$, we have
$\displaystyle \mathop{\mathbb P}_{r \in \{ 0,1 \}^n} [ H(r) = \langle x,r \rangle ] \geq \frac 78 \ \ \ \ \ (3)$
where ${x\in \{ 0,1 \}^n}$ is an unknown string.
Then there is an algorithm GLW that runs in time ${O(n^2\log n)}$ and makes ${O(n\log n)}$ oracle queries into ${H}$ and, with probability at least ${1- \frac{1}{n}}$, outputs ${x}$.
This gave us a proof of a variant of the Goldreich-Levin Theorem in which the right-hand-side in (2) was ${\frac {15}{16}}$. We could tweak the proof Lemma 2 so that the right-hand-side of (4) is ${\frac 34 + \epsilon}$, leading to proving a variant of the Goldreich-Levin Theorem in which the right-hand-side in (2) is also ${\frac 34 + \epsilon}$.
We need, however, the full Goldreich-Levin Theorem in order to construct a pseudorandom generator, and so it seems that we have to prove a strengthening of Lemma 2 in which the right-hand-side in (4) is ${\frac 12 + \epsilon}$.
Unfortunately such a stronger version of Lemma 2 is just false: for any two different ${x,x'\in \{ 0,1 \}^n}$ we can construct an ${H}$ such that
$\displaystyle \mathop{\mathbb P}_{r\sim \{ 0,1 \}^n} [ H(r) = \langle x,r \rangle ] = \frac 34$
and
$\displaystyle \mathop{\mathbb P} _{r\sim \{ 0,1 \}^n} [ H(r) = \langle x',r \rangle ] = \frac 34$
so no algorithm can be guaranteed to find ${x}$ given an arbitrary function ${H}$ such that ${\mathop{\mathbb P} [ H(r) = \langle x,r \rangle ] = \frac 34}$, because ${x}$ need not be uniquely defined by ${H}$.
We can, however, prove the following:
Lemma 3 (Goldreich-Levin Algorithm) Suppose we have access to a function ${H: \{ 0,1 \}^n \rightarrow \{ 0,1 \}}$ such that, for some unknown ${x}$, we have
$\displaystyle \mathop{\mathbb P}_{r \in \{ 0,1 \}^n} [ H(r) = \langle x,r \rangle ] \geq \frac 12 + \epsilon \ \ \ \ \ (4)$
where ${x\in \{ 0,1 \}^n}$ is an unknown string, and ${\epsilon>0}$ is given.
Then there is an algorithm ${GL}$ that runs in time ${O(n^2 \epsilon^{-4}\log n)}$, makes ${O(n\epsilon^{-4} \log n)}$ oracle queries into ${H}$, and outputs a set ${L \subseteq \{ 0,1 \}^n}$ such that ${|L| =O(\epsilon^{-2})}$ and with probability at least ${1/2}$, ${x \in L}$.
The Goldreich-Levin algorithm ${GL}$ has other interpretations (an algorithm that learns the Fourier coefficients of ${H}$, an algorithm that decodes the Hadamard code is sub-linear time) and various applications outside cryptography.
The Goldreich-Levin Theorem is an easy consequence of Lemma 3. Let ${A'}$ take input ${y}$ and then run the algorithm of Lemma 3 with ${H(r) = A(y, r)}$, yielding a list ${L}$. ${A'}$ then checks if ${f(x) = y}$ for any ${x \in L}$, and outputs it if one is found.
From the assumption that
$\displaystyle \mathop{\mathbb P}_{x,r} [ A(f(x),r)= \langle x, r \rangle] \geq \frac 12 + \epsilon$
it follows by Markov’s inequality (See Lemma 9 in the last lecture) that
$\displaystyle \mathop{\mathbb P}_x \left[ \mathop{\mathbb P}_r [A(f(x),r)=\langle x, r \rangle] \geq \frac 12 + \frac \epsilon 2 \right] \geq \frac \epsilon 2$
Let us call an ${x}$ such that ${\mathop{\mathbb P}_r [A(f(x),r)=\langle x, r \rangle] \geq \frac 12 + \frac \epsilon 2}$ a good ${x}$. If we pick ${x}$ at random and give ${f(x)}$ to the above algorithm, there is a probability at least ${\epsilon/2}$ that ${x}$ is good and, if so, there is a probability at least ${1/2}$ that ${x}$ is in the list. Therefore, there is a probability at least ${\epsilon/4}$ that the algorithm inverts ${f()}$, where the probability is over the choices of ${x}$ and over the internal randomness of the algorithm.
2. The Goldreich-Levin Algorithm
In this section we prove Lemma 3.
We are given an oracle ${H()}$ such that ${H(r)=\langle x, r\rangle}$ for an ${1/2+\epsilon}$ fraction of the ${r}$. Our goal will be to use ${H()}$ to simulate an oracle that has agreement ${7/8}$ with ${\langle x, r \rangle}$, so that we can use the algorithm of Lemma 2 the previous section to find ${x}$. We perform this “reduction” by “guessing” the value of ${\langle x, r\rangle}$ at a few points.
We first choose ${k}$ random points ${r_1 \ldots r_k \in \{ 0,1 \}^n}$ where ${k = O(1 / \epsilon^2).}$ For the moment, let us suppose that we have “magically” obtained the values ${\langle x, r_1 \rangle, \ldots, \langle x, r_k \rangle}$. Then define ${H'(r)}$ as the majority value of:
$\displaystyle H(r + r_j) - \langle x, r_j \rangle \ \ j = 1, 2, \ldots, k \ \ \ \ \ (5)$
For each ${j}$, the above expression equals ${\langle x, r \rangle}$ with probability at least ${\frac{1}{2} + \epsilon}$ (over the choices of ${r_j}$) and by choosing ${k=O(1/\epsilon^2)}$ we can ensure that
$\displaystyle \mathop{\mathbb P}_{r,r_1,\ldots,r_k}\left [H'(r) = \langle x, r \rangle \right ] \ge \frac{31}{32}. \ \ \ \ \ (6)$
from which it follows that
$\displaystyle \mathop{\mathbb P}_{r_1,\ldots,r_k}\left [ \mathop{\mathbb P}_r \left [H'(r) = \langle x, r \rangle \right ] \ge \frac 78 \right ] \ge \frac{3}{4}. \ \ \ \ \ (7)$
Consider the following algorithm.
• Algorithm GL-First-Attempt
• pick ${r_1, \ldots, r_k \in \{ 0,1 \}^n}$ where ${k = O(1/\epsilon^2)}$
• for all ${b_1, \ldots, b_k \in \{ 0,1 \}}$
• define ${H'_{b_1 \ldots b_k}(r)}$ as majority of: ${ H(r + r_j) - b_j}$
• apply Algorithm GLW to ${H'_{b_1 \ldots b_t}}$
• add result to list
• return list
The idea behind this program is that we do not in fact know the values ${\langle x, r_j \rangle}$, but we can “guess” them by considering all choices for the bits ${b_j.}$ If ${H(r)}$ agrees with ${\langle x, r \rangle}$ for at least a ${1/2+\epsilon}$ fraction of the ${r}$s, then there is a probability at least ${3/4}$ that in one of the iteration we invoke algorithm GLW with a simulated oracle that has agreement ${7/8}$ with ${\langle x, r \rangle}$. Therefore, the final list contains ${x}$ with probability at least ${3/4 - 1/n > 1/2}$.
The obvious problem with this algorithm is that its running time is exponential in ${k = O(1/\epsilon^2)}$ and the resulting list may also be exponentially larger than the ${O(1/\epsilon^2)}$ bound promised by the Lemma.
To overcome these problems, consider the following similar algorithm.
• Algorithm GL
• pick ${r_1, \ldots, r_t \in \{ 0,1 \}^n}$ where ${k = \log O(1/\epsilon^2)}$
• define ${r_S := \sum_{j\in S} r_j}$ for each non-empty ${S\subseteq \{1,\ldots,t\}}$
• for all ${b_1, \ldots, b_t \in \{ 0,1 \}}$
• define ${b_S := \sum_{j\in S} b_j}$ for each non-empty ${S\subseteq \{1,\ldots,t\}}$
• define ${H'_{b_1 \ldots b_k}(r)}$ as majority of: ${ H(r + r_S) - b_S}$ over non-empty ${S}$
• apply Algorithm GLW to ${H'_{b_1 \ldots b_t}}$
• add result to list
• return list
Let us now see why this algorithm works. First we define, for any nonempty ${S \subseteq \{1, \ldots, t\},}$ ${r_S = \sum_{j \in S} r_j.}$ Then, since ${r_1, \ldots, r_t \in \{ 0,1 \}^n}$ are random, it follows that for any ${S \neq T,}$ ${r_S}$ and ${r_T}$ are independent and uniformly distributed. Now consider an ${x}$ such that ${\langle x, r \rangle}$ and ${H(r)}$ agree on a ${\frac{1}{2} + \epsilon}$ fraction of the values of ${r}$. Then for the choice of ${\{b_j\}}$ where ${b_j = \langle x, r_j \rangle}$ for all ${j,}$ we have that
$\displaystyle b_S = \langle x, r_S \rangle$
for every non-empty ${S}$. In such a case, for every ${S}$ and every ${r}$, there is a probability at least ${\frac{1}{2} +\epsilon,}$ over the choices of the ${r_j}$ that
$\displaystyle H(r+ r_S) - b_S = \langle x, r \rangle\ ,$
and these events are pair-wise independent. Note the following simple lemma.
Lemma 4 Let ${R_1, \ldots, R_k}$ be a set of pairwise independent ${0-1}$ random variables, each of which is ${1}$ with probability at least ${\frac{1}{2} + \epsilon.}$ Then ${\mathop{\mathbb P}[\sum_i R_i \ge k/2] \ge 1 - \frac 1 {4\epsilon^2 k}}$.
Proof: Let ${R=R_1+\cdots+R_t}$. The variance of a 0/1 random variable is at most ${1/4}$, and, because of pairwise independence, ${{\bf Var}[R] = {\bf Var} [ R_1+\ldots+R_k] = \sum_i {\bf Var}[R_k] \leq k/4}$.
We then have
$\displaystyle \mathop{\mathbb P}[ R \leq k/2] \leq \mathop{\mathbb P}[ |R - \mathop{\mathbb E}[R]| \geq \epsilon k] \leq \frac{{\bf Var}[R]}{\epsilon^2 k^2} \leq \frac 1 {4\epsilon^2 k}$
$\Box$
Lemma 4 allows us to upper-bound the probability that the majority operation used to compute ${H'}$ gives the wrong answer. Combining this with our earlier observation that the ${\{r_S\}}$ are pairwise independent, we see that choosing ${t = \log (128/\epsilon^2)}$ suffices to ensure that ${H'_{b_1 \ldots b_t}(r)}$ and ${\langle x, r\rangle}$ have agreement at least ${7/8}$ with probability at least ${3/4}$. Thus we can use Algorithm ${A_{\frac 78}}$ to obtain ${x}$ with high probability. Choosing ${t}$ as above ensures that the list generated is of length at most ${2^t = 128/\epsilon^2}$ and the running time is then ${O( n^2 \epsilon^{-4} \log n)}$ with ${O(n\epsilon^{-4}\log n)}$ oracle accesses, due to the ${O(1 / \epsilon^2)}$ iterations of Algorithm GLW, that makes ${O(n\log n)}$ oracle accesses, and to the fact that one evaluation of ${H'()}$ requires ${O(1/\epsilon^2)}$ evaluations of ${H()}$. | 3,366 | 10,105 | {"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": 168, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-30 | longest | en | 0.819697 |
https://www.spudart.org/blog/a-wallet-full-only-two-dollar-bills/ | 1,701,848,653,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.31/warc/CC-MAIN-20231206063543-20231206093543-00396.warc.gz | 1,119,043,144 | 41,118 | # A wallet full of only two-dollar bills
A wallet full of two-dollar bills is like carrying around a wallet of joy. People get excited with unexpected encounters of two-dollar bills, and for some people it’s just the regular course. The two-dollar adventure unfolds many personality traits.
Getting them at the bank
First I walk into my bank and deposit a check like normal. After taking care of official adult business, the cashier asks the typical “is there anything else I can do for you today, Mr. Maldre?” For all those times I’ve been asked that question, I finally have a “yes” reply, and this time the request is a fun one!
“Actually, I would like to take all the money in my wallet,” as I pause while I notice that the cashier is starting to smile now, “and convert it it all into two-dollar bills.
She doesn’t even flinch! “Ok. You have 95 dollars here.” She takes my money, and walks into the back room, like this a normal occurance.
She’s disappeared into the back room for a few minutes while I imagine the conversation happening between bankers in the secret room. “We just don’t carry that many two-dollar bills! What kind of request is this? Nobody asks for these except for grandmas wanting only a few for their grandkids!”
After a few minutes she finally comes out with a big fat wad of two-dollar bills. First she counts them to herself, and then she says, “let me double-check” as she goes to the bill counter and it rapidly counts through the bills. I can now tell that she is having fun with this, because she is taking extra care with my request by double-counting.
She comes up to me and starts the routine by placing each bill on the counter one-by-one while she continues to count through them. Two. Four. Six. Eight. Ten. Twelve…. Thirty-four…. Thirty-six… She does get a bit slower when she reaches the seventies. I guess the odd number paired with counting evens can be confusing.
I am giggling the entire time she is counting–and several times I wished I had my video camera, each time realizing, duh, this is a bank. You can’t just videotape in here. But oh, this was too classic.
Ninety. Ninety-two. Ninety-four. and one. for Ninety-five. She hands me the big wad of two-dollar bills. I jam them into my naked wallet. As I try to fold the wallet in half, it’s now a George Caztanza-style wallet. So fat and bulging, it’s absolutely ridicululous trying to carry this in my back pocket. But I do so anyways for the kicks of it.
Spending my first two-dollar bill
First place to spend the two-dollar bill? Sportmart. To get my baseball to have everyone I know sign. And wouldn’t you know it? They have baseballs there for two bucks. Perfect!
The cashier rang up my ball, and I go into my wallet and pull out a crisp two-dollar bill. The cashier was astonished, “Are you sure you want to spend this?” I showed him the inside of my wallet. “Look inside, I just went to the bank and exchanged out ALL my money for two-dollar bills. I certainly have plenty.” This cashier dude was so excited to have a two-dollar bill, he asked his boss (who was standing right there) if he could exhange it out with his own money (his boss did). Then the other three cashiers were all curious and came over to see what was happening. They all found it very amusing and we swapped two-dollar bill stories.
Spending 2 two-dollar bills at McDonalds
Having been very pleased with this reaction, I headed over to McDonalds with the plan of spending TWO, yes TWO, two-dollar bills. Surely I would get TWICE the excitement. I hand the cashier the two two-dollar bills. And she doesn’t flinch. I’m thinking, “what! i just handed you TWO two-dollar bills. I’m SURE this has never happened to you before.” Ah well.
Rejection at Arby’s
Recently I handed the cashier at Arby’s THREE two-dollar bills. And she asked the other cashier, “do we take these?” WHAT!? DO YOU TAKE TWO-DOLLAR BILLS!? HOW CAN YOU NOT?! That was just weird, i’m sorry. Weird.
Join in
I highly highly (yes, that’s a double recommendation) anyone to exchange all your money in your wallet for two-dollar bills. It’s like carrying a walletful of joy ready to unfold adventures where you’ll get your two-cent’s worth.
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16 years ago
awesome story. You could make an entire blog about your two dollar bill stories. Just make sure your wallet is always full of Jeffersons. A similiar experience might be had with dollar coins. Though people are more familiar with 2 dollar bills. Too many people don’t know about the dollar coins.
16 years ago
And then you could have people submit their own 2 dollar bill stories and you could publish them on your 2 dollar bill blog.
Mark
16 years ago
My father collected two dollar bills for a while. He even had an old one with some red ink on it as opposed to all green. You must have gotten a younger teller at the bank, as most experienced tellers probably wouldn’t be fazed by your request. We get all kinds of strange requests here at banks.
16 years ago
Great post Spudart! Actually I wonder if you would get twice the excitement if you combo the \$2 bills with the new dollar coin for your odd numbered transactions. We received 30 George Washington dollar coins for my wife’s 30th birthday. As unlikelymoose pointed out most people simply don’t know about them. When I try to pay with them I need to tell people that they are worth \$1. They are sooo gold, shiny and irresistible!!! that I have a hard time handing them over.
Kelly
16 years ago
Wonderful. My co-worker and I spent far too long reading this, laughing, and coming up with similar schemes. We were thinking he might try carrying around a velvet sack of \$1 coins, sort of pirate-style.
16 years ago
I’m totally going to do this! And BTW, I think you’re more E than you realise even if you claim I. :oP
Helen Franklin
16 years ago
I always thought the 2 dollar bills I have saved over the years were something very special and do’t ever plan to spend them. Let me know if I am wrong. I am 75 and have bills over 50 years old.
0
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x | 1,456 | 6,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2023-50 | latest | en | 0.972691 |
https://www.chegg.com/homework-help/holt-mcdougal-larson-geometry-0th-edition-chapter-7-problem-5tp-solution-9780547315171 | 1,553,537,605,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204086.87/warc/CC-MAIN-20190325174034-20190325200034-00050.warc.gz | 698,428,845 | 27,178 | Solutions
Holt McDougal - Larson Geometry
# Holt McDougal - Larson Geometry (0th Edition) Edit edition Problem 5TP from Chapter 7
We have solutions for your book!
Chapter: Problem:
Step-by-step solution:
Chapter: Problem:
• Step 1 of 5
Consider the following figure
Time taken to walk from your house to friend’s house is 14 minutes. Total distance between your house and the friend’s house is the sum of 500 yd plus 700 yd which is coming out to be 1200 yd.
Pythagorean Theorem states that the square of the hypotenuse is equal to the sum of the square of the other two sides.
Use Pythagorean Theorem to find the distance AC
Take square roots on both the sides
Therefore, the distance AC is equal to 860 yd.
• Chapter , Problem is solved.
Corresponding Textbook
Holt McDougal - Larson Geometry | 0th Edition
9780547315171ISBN-13: 0547315171ISBN: Ron LarsonAuthors: | 229 | 875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2019-13 | latest | en | 0.83261 |
http://mrl.nyu.edu/~perlin/courses/fall2007/clipping.html | 1,511,073,637,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805417.47/warc/CC-MAIN-20171119061756-20171119081756-00561.warc.gz | 216,481,910 | 3,046 | Clipping and introduction to the rendering pipeline (Oct 16 lecture)
course notes and homework due Oct 23 before class
This week we talked about z-clipping and introduced the general framework of the rendering pipeline.
Clipping:
Within an N dimensional space you can embed an unbounded N-1 manifold that divides the space into two parts. For example, a plane can be divided in two by a line or by a circle (ie: into inside the circle and outside the circle). Similary, 3D space can be divided in two by a plane or some other unbounded surface (eg: paraboloid, ellipsoid, etc).
"Clipping" is an operation in which you throw away everything in the space that is on one side of the manifold. When rendering scenes in computer graphics, it is very useful to use planes to clip away those parts of 3D space that are out of view. In particular, it is useful to clip away everything that is not in front of the camera.
A plane in 3D space can be defined by using a linear equation:
f(P) = aPx + bPy + cPz + d
The plane contains those points P for which f(P) = 0. Let's say we want to clip our objects prior to rendering, throwing away all parts of the object where f(P) > 0, and keeping all parts of the object where f(P) <= 0. As we discussed in class, the key to this will be to clip the model one triangle at a time - before applying perspective.
We want to keep everything for which z <= -ε, where ε is some small number such as 0.001. So our clipping function will be:
f(P) = 0Px + 0Py + 1Pz + ε
As we said in class, clipping one triangle by a plane will produce one of the following shapes:
• The entire triangle (when F(V) <= 0 for all three vertices)
• A four sided polygon (when F(V) > 0 for one vertex)
• A smaller triangle (when F(V) > 0 for two vertices)
• Nothing (when F(V) > 0 for all three vertices)
To clip a triangle by a plane, first it is necessary to learn how to clip a line segment [A,B] from point A to point B by the plane where linear function f(P) = 0.
First the two trivial cases: If f(A) <= 0 and f(B) <= 0 then the result is just [A,B]. If f(A) > 0 and f(B) > 0 then the result is nothing.
Suppose that f(A) <= 0 and f(B) > 0. We want to find a point C where f(C) = 0, and return the result [A,C]. Similarly, if f(A) > 0 and f(B) <= 0. We would return the result [C,B].
Finding C is done in two steps:
• Compute the fraction t along the line segment at which the function f(P) reaches zero;
• Linearly interpolate between A and B to get C.
More formally: t = (f(C) - f(A)) / (f(B) - f(A)), which can also be written as: -f(A) / (f(B) - f(A)).
Given the above machinery, we can clip a triangle by going round the triangle, vertex by vertex, to construct a new polygon. Here is the procedure:
``` For each triangle vertex A:
Let B be the next vertex around the triangle after A
• If f(A) <= 0
• Add A to the new polygon
• If A and B are on opposite sides of the clip plane (in other words, if f(A) * f(B) < 0)
• Compute the point C where edge [A,B] crosses the clip plane
• Add C to the new polygon
```
When you are done, you will have either 0, 3 or 4 vertices in the new polygon. If you end up with 4 vertices, then split the polygon into two triangles, and proceed onto the rest of the rendering pipeline with each of these two new triangles.
Homework:
Your assignment, due by next class, is two-fold:
• Integrate the above z-clipping algorithm into an animation. You can either use an animated scene you already made for this class, or else feel free to create a new animated scene if you're feeling creative. I know I would. :-)
Show that z-clipping works by animating one or more objects in your scene to go behind the camera.
• I want you to start feeling comfortable using the support routines I've provided that allow you to efficiently set an RGB color at each pixel of an image. To that end, go to my simple Blue Moon example, copy the two java files ExampleMISApplet and MISApplet, and replace ExampleMISApplet by code that produces something interesting and kooky. Have fun with it. :-)
The goal for this part of the assignment is really to show that you can use this library to get somehing up on the screen. You shouldn't need to modify MISApplet at all. In the class you implement, you will only need to implement two methods:
``` public void initFrame(double time) {
// here you should put any computations you want to do once per frame.
}
public void setPixel(int x, int y, int rgb[]) {
// x and y are the image pixel coords.
// your computation should place values into rgb[].
// rgb[0], rgb[1], rgb[2] are the red,green,blue components, respectively.
// remember that red,green and blue should range from 0...255.
}
```
By the way, here is a link to a simple example of how to load and play a sound in an Applet. | 1,232 | 4,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.925206 |
https://numberworld.info/51321 | 1,656,923,108,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104364750.74/warc/CC-MAIN-20220704080332-20220704110332-00554.warc.gz | 474,833,142 | 3,796 | # Number 51321
### Properties of number 51321
Cross Sum:
Factorization:
Divisors:
1, 3, 17107, 51321
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
c879
Base 32:
1i3p
sin(51321)
-0.057557215817637
cos(51321)
0.99834220931869
tan(51321)
-0.057652792079098
ln(51321)
10.84585530412
lg(51321)
4.7102951101006
sqrt(51321)
226.54138694729
Square(51321)
### Number Look Up
Look Up
51321 (fifty-one thousand three hundred twenty-one) is a very special number. The cross sum of 51321 is 12. If you factorisate the figure 51321 you will get these result 3 * 17107. The figure 51321 has 4 divisors ( 1, 3, 17107, 51321 ) whith a sum of 68432. 51321 is not a prime number. The number 51321 is not a fibonacci number. The number 51321 is not a Bell Number. The number 51321 is not a Catalan Number. The convertion of 51321 to base 2 (Binary) is 1100100001111001. The convertion of 51321 to base 3 (Ternary) is 2121101210. The convertion of 51321 to base 4 (Quaternary) is 30201321. The convertion of 51321 to base 5 (Quintal) is 3120241. The convertion of 51321 to base 8 (Octal) is 144171. The convertion of 51321 to base 16 (Hexadecimal) is c879. The convertion of 51321 to base 32 is 1i3p. The sine of the number 51321 is -0.057557215817637. The cosine of the number 51321 is 0.99834220931869. The tangent of 51321 is -0.057652792079098. The square root of 51321 is 226.54138694729.
If you square 51321 you will get the following result 2633845041. The natural logarithm of 51321 is 10.84585530412 and the decimal logarithm is 4.7102951101006. I hope that you now know that 51321 is very impressive figure! | 617 | 1,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-27 | latest | en | 0.748739 |
https://www.etechnog.com/2022/05/parallel-circuit-examples-applications.html | 1,653,668,036,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662658761.95/warc/CC-MAIN-20220527142854-20220527172854-00697.warc.gz | 857,036,469 | 36,439 | Here is the list of common examples of parallel circuit,
1. House lighting circuit
2. Streetlight Circuit
3. Conventional Fire Alarm Circuit
4. LC Tank Circuit
5. Power Socket or Outlet
6. DC Filter Circuit
7. Electrical Power Distribution Circuit
8. Current Divider Circuit
9. Speaker Connection in large audio system
10. Power Factor Correction Circuit
11. Noise Filtering Circuit
Parallel circuits are those where all the components are connected in parallel to each other. Here, the voltage across all the components is the same but the flow of current through individual components is not the same, it depends upon their internal resistance. So the characteristics of a parallel circuit exact opposite of a series circuit.
Now, let's discuss about the above example so it will give a clear knowledge about parallel circuits. The first one is the house lighting circuit. All the lights even all other loads in a house or office are connected in parallel to each other. In the USA the utility supply for a home is 120V, and in India is 230V. Also, we know that all the loads including lights at our homes are have the same voltage rating. So they are connected in parallel to maintain the rated utility supply voltage across all the loads.
Streetlights are also connected in parallel. And we know that in a parallel circuit damaging any one load cannot affect the other loads. So if any one street light got damaged, other lights will without any interruption. Also, note that as the voltage across each light is equal so the brightness also be the same.
Another example of a parallel circuit is LC tank Circuit. LC tank circuit is the parallel combination of an inductor and a capacitor. The function of an LC tank circuit is to generate an oscillating or dumping signal.
Power sockets or outlets in our power distribution boards are connected in parallel to each other so we can get the same voltage, current, and power from each socket or outlet. Generally, the phase terminal of each socket is connected together. Same as the neutral terminal of each socket is connected together. And ground or earthing terminal of each socket is connected together. If there is any wrong connection between phase and neutral, the polarity of the power supply will be changed.
We know that condensers or capacitors are connected in parallel with the inductive load to correct the power factor. So it is an example of a parallel circuit. basically, the condenser consumed the reactive power to keep the power factor near unity.
In large audio systems such as PA systems, concerts, and theaters all the speakers are connected in parallel. Even the polarity of the connection is also maintained correctly to get a better sound experience. So the speaker connection in parallel is also an example of a parallel circuit.
Parallel Circuit Application and Uses
1. A parallel circuit is used to maintain the same voltage level across all the loads and components connected to it.
2. A parallel circuit is used to eliminate the effect of damaging anyone's load to the other as in the series circuit.
3. A parallel circuit is very helpful to operate multiple loads with a small voltage. | 628 | 3,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-21 | longest | en | 0.905316 |
napdivad.com | 1,369,153,234,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700168711/warc/CC-MAIN-20130516102928-00033-ip-10-60-113-184.ec2.internal.warc.gz | 154,354,654 | 2,259 | # Rubik's Cube Conjecture
One day when I was holding a solved Rubik's cube, I wondered that if I kept on moving the cube U and R (Singmaster notation), would it ever return to its solved position? After some time, the cube actually did become solved again! Here is my conjecture:
After being repeatedly applied a sequence of moves, a Rubik's cube will always return to its original position.
In other words, if a Rubik's cube is in position P, and there is a sequence of moves S (such as {R, U, F}), the cube will be in position P after it is moved according to S a certain number of times. For example, take a cube in the solved position. Apply the move U. Apply U again. After two more Us, the cube will return to its original solved position. To deal with more complex sequences, such as {R, U, B, D, D, D, R, F, L}, I wrote the Java applet above.
In the applet, type in a sequence of moves and press the "Start" button. The cube will start moving according to the sequence. The "Reset" button repositions the cube and clears the sequence. Note that R' is the same as RRR, U' = UUU, F' = FFF, etc.
I'm hoping to prove (or disprove) this conjecture.
Here's a beautiful proof by Allen Yuan (submitted via feedback):
It's actually not too bad to prove the Rubik's cube conjecture. Note the following facts:
1) The Rubiks Cube has a finite number of possible configurations.
2) Thus, while you're moving it in a certain way, you'll eventually have two points in time in which it is at the same configurations (if not, then there would be an infinite number of configurations!).
3) Let these times be after x repetitions and then after y repetitions. However, take both of these and go back x repetitions. That means that the first of the two becomes 0 moves, or your original thing, and the next becomes y-x repetitions. But since they were the same after x and y repetitions, they must be the same after 0 and y-x repetitions. Thus, we've found a place where it's repeated!
## Source Code
Home | Feedback | Page Last Modified: March 28, 2012 | © David Pan | 502 | 2,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | longest | en | 0.966227 |
https://roymechx.co.uk/Useful_Tables/Mechanics/Kinetics.html | 1,624,008,094,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487635920.39/warc/CC-MAIN-20210618073932-20210618103932-00328.warc.gz | 457,289,590 | 8,002 | Disclaimer: The information on this page has not been checked by an independent person. Use this information at your own risk.
ROYMECHX clone of ROYMECH
Dynamics / Kinetics
Introduction
Basics.
Definitions..
Rigid body ...An arrangement of particles in which the distance of between any two particles is fixed. Rectilinear Translation...All point on a rigid body move in a straight line Curvilinear Translation... The orientation of all points in a rigid body remain fixed while the body moves along a curved path Rotation about a fixed line.. All points in a rigid body move in a circular motion about a fixed line Plane motion...Each point in rigid body moves in a path parallel to a fixed plane
It is assumed that all motions are within the plane and rotations are around axes perpendicular to the plane. Infinitesimal counterclockwise rotations are positive and are therefore represented by vectors perpendicular to the plane as indicated by the right hand rule- If the plane is the x-y plane then angular velocities are in the positive z direction. It should be noted that large rotations and angular velocities do not follow vector rules..
Free Body Diagram
A free body diagram is an extremely useful tool for assessing the interaction of forces on bodies This is essentially a sketch of a body which is in equilibrium and is entirely separate from the surroundings. The only rule for drawing free-body diagrams is to depict all the forces which exist for that object in the given situation.
Particle Kinetics.
Newtons laws
Newtons First Law;
Every body continues in a state of rest or of uniform rectilinear motion unless acted upon by a force.
Newtons Second law;
The time rate of change of linear momentum of a body is proportional to the unbalanced force acting on the body and occurs in the direction in which the force acts.
Newtons Third law
To each action (or force) there is an equal and opposite reaction. The mutual force of the two bodies acting upon each other are equal in magnitude, opposite in direction and are collinear.
Momentum
The momentum is defined simply as the product of mass and velocity..The first law states that if a body changes its velocity then a force must have been applied. The second law establishes a relationship between the magnitude of the force and the change in momentum..
Force = k. d(momentum)/dt = k.d(mv)/dt = k.m dv/dt = k.m. a
In the metric ISO system the unit of force of 1 Newton (N) on a mass of 1 kg results in a linear acceleration of 1 m/s 2 therefore k = 1.
Newtons third law states that if two bodies collide the total momentum after impact must equal the total momentum after impact. For two colliding masses (m1 & m2 ) with initial velocities ( u 1 & u 2 ) and final velocities ( v 1 & v 2 ) ..
m 1. u 1 + m 2. u 2 = m 1. v 1 + m 2. v 2 ... therefore m 1 . ( v 1 - u 1 ) = m 2( v 2 - u 2 )
Equations of Motion for a particle under different force regimes
1) Force = constant value.. F = C= constant = mass x acceleration..
m. dv/dt = C
dv = C/m dt
Falling masses under the effect of gravity provides an example of this condition
The motion of the particle can be determined by integration provided the initial conditions are known..
2) Force is a function of time F = F(t)
m.dv/dt = F(t)
dv = 1/m F(t)dt
Using F(t) the equation for the velocity can be determined by integration, and again the displacement can be found from ds = vdt.
3) Force is a function of velocity.
m. dv/dt = F(v)
dt = m.dv / F(v)
Example: The resistance to motion of drag or viscous damping when the force = C x velocity where c = the damping coefficient..
4) Force is a function of Displacement.
m. dv/dt = F(s)
because a.ds = v.dv then m. v.dv = F(s).ds
Example : The force developed by spring = k x s where k is the stiffness of the spring.
Circular motion ..
A mass rotating in a circle is accelerating towards the centre of the circle at a rate of acceleration of v2 /r. The force pulling the body into the centre of the circle is the centripetal force. (if the mass is spinning on a string the centripetal force is the tension in the string). The reaction force at the centre of the circle is the centrifugal force. There is no force pulling outwards on the circling body there is only a force pulling in
Rigid Body Kinetics.
Considerations of rigid objects are simplified if ...
M = The mass of the object which is concentrated at a point which is at the mathematical centre of mass
rG = The position of the centre of Mass relative to the co-ordinate origin.
The vector sum of the external forces acting on a set of particles (rigid body) equals the total mass times the accelaration of the centre of mass, irrespective of the motion of the separate particles.
The moment of Inertia of a particle of mass dm at a radius r from a axis through the centre of gravity G is dm.r 2. The moment of inertia of the whole body about the axis through G =
This is generally written as
Where k is termed the radius of gyration.
The angular motion of a mass of inertia IG is related to an applied torque as follows
T= applied torque and α = angular acceleration..
Impact- Impulse
The relationship between force and the motion of a mass as shown above can be written as..
This can be integrated to..
Impulse = change in momentum
The impulse is effectively the area under a plot of the force-time relationship
The angular impulse of a constant torque T acting over a time t is the product of Tt. (If the torque varies the angular impulse is the integral or the area under plot of Torque-time relationship. )
The result of this torque on a mass of inertia IG
That is the impulse of torque = change in angular momentum..
Work Energy and Power
Work
The transfer of energy expressed as the product of a force and the distance through which its point of application moves in the direction of the force. ... It should be noted that work only results if the point of application of the force moves. There is no work done if a weight is supported without movement.
If a force is acting on a particle as it move from A to B . The work done as the particle moves a small distance dr = F. dr . The work done is the product of the Force vector and the displacement vector. Only the force component in line with the displacement component contributes to the work done. Work is a scalar quantity and is measured in N.m (ISO units)
The work performed by a force F (N) when the point of application moves S (m) with angle q between the force and the direction of motion .
Work (U) = F.cosθ .S
The work performed by a couple M turning an object through an angle θ is
Work (U) = Mθ
Work is a scalar quantity..
The work done by a force extending/compressing a spring (in the x direction) is calculated as follows:
Conservative Forces
If the work done by a force is independent of the path the force is called a Conservative force. Examples of conservative forces include spring forces and gravitational forces. Conservative forces are generally recoverable ; that is if work is done in lifting an object against gravity through a vertical height h , the work is is recovered by lowering the weight back to the original level.
Non-Conservative Forces
When force is required to move an object to overcome friction the energy dissipated cannot be conveniently recovered as work The work done against friction is not available as kinetic or potential energy. The work done by a non-conservative force is dependent on the path taken by the point of application of the force..
Energy
At its simplest level energy is defined as the ability to do work. Energy takes many forms including kinetic energy, potential energy, thermal energy, chemical energy, electrical energy, and atomic energy.
kinetic energy -translation ..
The field of mechanics includes kinetic energy which is energy possessed by a body due to its motion and potential energy which is energy possessed by a body because of its position in a field force (gravity /elastic force).
The kinetic energy increase of a mass subject to a force is derived as follows..
The term m.v 2 / 2 is called the kinetic energy of the mass and hence the derivation above results in
Work Done by a force on a mass = change in kinetic energy.
Angular kinetic energy ..
A body is rotating about and axis through O.
Total kinetic energy ..
If the centre of gravity of a body is moving with a velocity v and is rotating with and angular velocity of w about the centre of gravity.
The total K.E. = K.E. of translation + K.E. of rotation
Equivalent Mass of a Rotating Body ..
When considering motions of machines with comprising masses in linear motion and masses with and angular motions it is often required to find the equivalent mass of a rotating body. This occurs often with vehicle dynamics. In considering a body of mass ( m ) rotating about an axis through O with a tangential force at radius ( r ) producing a linear accelaration at this point of ( f ) with and a resulting angular accelaration of (α ).
The quantity m.( k / r ) 2 is the equivalent mass of the body referred to the line of action of P.
This relationship can be used to convert a system with rotating masses and linear mass e.g. a vehicle into a system comprising equivalent linear motions only.
Potential energy..
In general terms potential energy identifies some form of stored energy which can be converted into some other form of energy. Potential energy take many forms including mechanical, chemical, electrical nuclear etc. These notes only consider mechanical potential energy which is energy stored by a body because of its position with respect to a datum is a conservative force field. The two most common forms of potential energy in mechanical engineering are gravitational potential energy and elastic strain energy..
The force on an object in the earths gravitational field is directly related to the mass of the earth and the object and is inversely related to the square of the distance between the centre of the earth and the object. For normal mechanical considerations close to the earths surface the force is simply express as follows
FG = m.g
where g = the gravitional constant (acceleration due to the attraction of the earth) ..This is generally approximated to 9.81 m/s2 but can vary at due to variations resulting due the fact that the earth is not a perfect even sphere and due to the effect of other celestial bodies.
At it simplest level the Potential energy change due to gravity for a mass m moving from h1 to h2 =
PE = g.m (h2 -h2 )
An example of the elastic strain potential energy is the extension or compression of a spring as noted above..
General Energy Principle
The principle of conservation of energy in it simplest form states
[ The total amount of energy in the universe is constant: energy cannot be created or destroyed although in may be converted into various forms..]
.
In the field of mechanics the general energy principle is simplified to consider only the energy related to motion and position..
[If a system of bodies in motion is only under the action of conservative system of forces , the sum of the kinetic and potential energies of the bodies is constant.]
In its more general form the general energy principle may be stated as
[The work done on a system is equal to the change in kinetic energy + potential energy + losses ]
Power
Power is defined as the rate of doing work.
Power = P = dW/dt = F. dr /dt = F.v
Power is a scalar quantity with units N.m/s
Links to Dynamics /Kinetics Kinetics - Wikipedia Excellent source of information Work Energy & Power..Hyperphysics ..Excellent reference site Physics Notes - Dynamics..Batesville Area Schools - Useful Notes Physics- Kinematics..An informative website - with lots of interlinked notes. | 2,578 | 11,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2021-25 | longest | en | 0.912949 |
https://www.easyelimu.com/high-school-notes/geography/form-3/item/1202-statistics | 1,713,984,219,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819847.83/warc/CC-MAIN-20240424174709-20240424204709-00112.warc.gz | 659,049,178 | 24,199 | # Statistics - Geography Form 3 Notes
## Compound/Cumulative/Divided Bar Graph
Major cash crops exported in Kenya in tonnes
CROP 1990 1991 1992 1993 1994 COFFEE 4500 5000 5200 6000 5900 TEA 1300 1100 2500 2100 2200 MAIZE 800 900 500 400 400 WHEAT 600 500 600 700 500
### Steps
CROP 1990 CT 1991 CT 1992 CT 1993 CT 1994 COFFEE 4500 4500 5000 5000 5200 5200 6000 6000 5900 TEA 1300 5800 1100 6100 2500 7700 2100 8100 2200 MAIZE 800 6600 900 7000 500 8200 400 8500 400 WHEAT 600 7200 500 7500 600 8800 700 9200 500 TOTAL 7200 7500 8800 9200 9000
1. Set cumulative totals for the data each year
2. Draw vertical axis(Y) to represent dependent variable
3. Draw horizontal axis(x) to represent independent variable
4. Label both axis using suitable scale
5. Plot the cumulative values for each year
6. Use values for components to subdivide the cumulative bar
7. The subdivisions are placed in descending order with the longest at the bottom(coffee)
8. Shade each component differently
9. Put title and key
1. It’s easy to construct
2. It has good visual impression
3. There is easy comparison for the same component in different bars because of uniform shading
4. Easy to interpret because bars are shaded differently
5. Total value of the bar can be identified easily
1. It doesn’t show the trend of components (change over time).
2. Can’t be used to show many components as there is limited space upwards
3. Tedious as there is a lot of calculation work involved.
4. Not easy to trace individual contribution made by members of the same bar
5. Poor choice of vertical scale causes exaggeration of bars length leading to wrong conclusions
### Analysis
• Coffee was the leading export earner in the five years.
• Tea was the second leading export earner.
• Wheat had the lowest export quantity.
• 1993 recorded the highest export quantity.
• 1990 recorded the lowest export quantity.
## Piechart/Divided Circles/Circle Charts
• A circle which has been subdivided into degrees used to represent statistical data where component values have been converted in degrees.
Major countries producing commercial vehicles in the world in 000s
USA FRANCE JAPAN UK GERMANY RUSSIA 1800 240 2050 400 240 750
Steps
1. Convert components into degrees
USA 1800/5480×360=118.2◦
FRANCE 240/5480×360=15.8◦
JAPAN 2050/5480×360=134.7◦
UK 400/5480×360=26.3◦
GERMANY 240/5480×360=15.8◦
RUSSIA 750/5480×360=49.3◦
2. Draw a circle of convenient size using a pair of compasses.
3. From the centre of the circle mark out each calculated angle using a protractor.
4. Shade the sectors differently and provide the key for various shadings.
1. Gives a good/clear visual impression
2. Easy to draw.
3. Can be used to present varying types of data e.g. minerals, population, etc.
4. Easy to read and interpret as segments are arranged in descending order and are also well shaded.
5. Easy to compare individual segments.
1. Difficult to interpret if segments are many.
2. Tedious due to a lot of mathematical calculations and marking out of angles involved.
3. Can’t be used to show trend/change over a certain period.
4. Small quantities or decimals may not be easily represented.
### Analysis
1. The main producer of commercial vehicles is Japan.
2. The second largest producer is USA followed by Russia.
3. The lowest producers were France and West Germany with.
## Proportional Circles
• This is use of circles of various sizes to represent different sets of statistical data.
Table showing mineral production In Kenya from year 1998-2000
MINERALS QUANTITY IN TONNES 1998 1999 2000 Graphite 200 490 930 Fluorspar 30 255 450 Soda ash 270 300 350 Diamond 500 870 1270 TOTAL 1000 1915 3000
### Steps
1. Determining the radii of circles by finding the square roots of the totals
1998 √1000=31.62=32
1999 √1915=43.76=44
2000 √3000=54.77=55
1. Scale:1cm represents 10 tonnes
1998=3.2 cm
1999=4.4 cm
2000=5.5 cm
1. Using a pair of compasses draw circles of different radii representing mineral production in Kenya between 1998 and 2000.
2. Convert component values into degrees
Component value/total value of data×360
1998: Graphite-200/1000×360=72◦
Fluorspar-30/1000×360=10.8◦
Soda ash-270/1000×360=97.2◦
Diamond-500/1000×360=180◦
1999: Graphite-490/1915×360=92.1◦
Fluorspar-255/1915×360=47.9◦
Soda ash`300/1915×360=56.4◦
Diamond-870/1915×360=163.6◦
2000: Graphite-930/3000×360=11.6◦
Fluorspar-450/3000×360=54◦
Soda ash-350/3000×360=42.1◦
Diamond-1270/3000×360=152.3◦
1. On the proportional circle for each year use a protractor and mark out the angles
2. Shade the segments and then provide a key.
1. They give a good visual impression.
2. Easy to compare various components.
3. Simple to construct.
4. Easy to interpret as segments are arranged in descending order.
5. Can be used to present varying types of data.
1. Tedious in calculation and measurement of angles
2. Actual values represented by each component cant be known at a glance
3. Difficult to accurately measure and draw sectors whose values are too small.
4. Comparison can be difficult if the circles represent values which are almost equal.
### Analysis/Conclusions
1. Diamond was leading in production.
2. The second leading mineral in production was graphite.
3. The mineral with the lowest production was fluorspar.
#### Download Statistics - Geography Form 3 Notes.
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and get it sent to you via WhatsApp | 1,651 | 5,680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-18 | latest | en | 0.797585 |
https://www.askiitians.com/forums/General-Physics/9/22495/velosity-time-graph.htm | 1,726,385,994,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00639.warc.gz | 612,780,027 | 43,085 | # the hightand distancealong horizontal for body projected in vertical plane are given by y=8t-5t2 and x=6t so what is it initial velocity?
420 Points
13 years ago
Dear Nivedita
for calculation of velocity differentiate with time and then put t=0 for initial velocity.
Vy= 8 - 10t = 8
Vx= 6
hence V= √82 + 62 = 10 Answer
All the best
AKASH GOYAL | 110 | 355 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-38 | latest | en | 0.85069 |
http://slowthinkingphysics.net/explainingPhysicsMoving/Mf19/ | 1,669,965,797,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710898.93/warc/CC-MAIN-20221202050510-20221202080510-00782.warc.gz | 45,873,061 | 1,476 | # Where we've got to with contact forces(Summary)
### A summary of contact forces
There are three kinds of contact forces that can support an object.
Warp forces can be found wherever a solid is distorted by an object:
• Add a compression force exerted by a neighbouring solid acting on the object if that solid is compressed by the object.
• Add a tension force exerted by a neighbouring solid acting on the object if that solid is compressed by the object.
You might, for teaching purposes, combine these two and call them warp forces — with the forensic clue that if an solid in contact with the object is stretched or squeezed then you can add an arrow labelled warp force.
• Add a buoyancy force if the object is partially or wholly immersed in a fluid.
Frictional forces of three kinds can be found at the surfaces of the object when it moves, or makes to move, past other particles its environment.
• If the environmental particles are a solid and no movement occurs, add an arrow at the contacting surface and label it grip force.
• If the environmental particles are a solid and movement occurs, add an arrow at the contacting surface and label it slip force.
• If the environmental particles are a liquid and movement occurs, add an arrow at the most significant surface and label it drag force.
You might, for teaching purposes, combine these three and call them frictional forces — but we'd not recommend that as it obscures the very different reasons for adding the arrows. | 302 | 1,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-49 | latest | en | 0.922598 |
https://fr.slideserve.com/siusan/energy-potential-kinetic | 1,624,258,196,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488268274.66/warc/CC-MAIN-20210621055537-20210621085537-00231.warc.gz | 246,885,090 | 18,659 | Energy Potential & Kinetic
# Energy Potential & Kinetic
## Energy Potential & Kinetic
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Created by: Faith Cohen EnergyPotential & Kinetic
2. Energy is the ability to do work or cause a change in direction, speed, shape or temperature of an object.
3. Energy comes in many forms: • Electrical • Chemical • Sound • Thermal • Nuclear • Mechanical • Gravitational
4. All energy can be grouped into:
5. Potential: • While work is waiting to be done, when there is potential for work to be performed, we call it potential energy.
6. An object can store energy: • By lifting it up: Gravitational Potential Energy • By stretching or squashing it: Elastic Potential Energy
7. An object can increase its potential energy: • Increase the object’s weight. • Increase the height an object is raised to.
8. Potential Energy Examples
9. Potential Energy often turns into Kinetic Energy
10. Kinetic Energy is the energy of motion (movement). Moving objects have energy: When work is actually being done we call it kinetic energy.
11. Kinetic Energy is measured by how much work is done to change an object’s motion. Mass and velocity both affect kinetic energy. Higher Velocity = Higher Kinetic Energy
12. Kinetic Energy Examples
13. Both Kinetic & Potential
14. Potential Energy Kinetic Energy
15. Assignment: Create a mind map all about kinetic and potential energy. You must include at least 5 pictures for each type of energy. | 376 | 1,577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-25 | latest | en | 0.825769 |
http://www.studymode.com/essays/Force-And-Lever-Arm-1511829.html | 1,529,792,039,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865250.0/warc/CC-MAIN-20180623210406-20180623230406-00107.warc.gz | 512,419,462 | 24,960 | # Force and Lever Arm
Topics: Force, Classical mechanics, Mass Pages: 2 (749 words) Published: March 16, 2013
1. What are the two types of friction that affect the performance of your vehicle? The first type of friction that affects the mousetrap car is called rolling friction. This type of friction occurs in between the wheels of the car and the surface of the ground it moves on. The second type is called the sliding friction which occurs in between the axles and the brackets.
2. What problems related to friction did you encounter and how did you solve them? We noticed that at times the car would come to a sudden stop. We realised that is most likely caused by too much rolling friction between the axle and the frame, thus causing the wheels to stop turning which in turn makes the car stop. We solved this by disassembling the axle from the frame and re-attaching it to larger axle holes in order to make sure that they won't press against the frame as they spin.
3. What factors did you take into account to decide the number of wheels you chose in your design? The more wheels present in your vehicle, the more friction will be present. More friction present means more rolling mass. The more rolling mass you have, the greater the needed energy to travel farther and faster is. This means then that you must only use as many wheels as needed to keep the vehicle stable.
4. What kind of wheels did you use on each axle? What is the effect of using large or small wheels? For this project, we used water bottle caps wrapped in electrical tape. The size of the wheel in comparison to the axle it is placed on greatly affects the torque (or the pulling force) as well as the pulling distance. If you larger wheels and a smaller axle, you will achieve greater distance but less power. Therefore, you must find a balanced ratio between the size of the wheel and the axle because if it has too little power, it will not end up not moving at all.
5. Explain how Newton's first, second, and third laws apply to the performance of your vehicle. Newton's first law states that the vehicle... | 446 | 2,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-26 | latest | en | 0.969276 |
http://www.answers.com/topic/john-shore | 1,550,545,767,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489304.34/warc/CC-MAIN-20190219020906-20190219042906-00538.warc.gz | 317,616,822 | 52,443 | Results for: John-shore
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How much does an iPhone 5c weigh?
The iPhone 5c weighs 4.65 ounches. It is heavier than the iPhone 5and 5s which weight 3.95. | 550 | 1,810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.908376 |
https://numberworld.info/20524401417 | 1,686,009,884,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652184.68/warc/CC-MAIN-20230605221713-20230606011713-00754.warc.gz | 483,437,564 | 4,121 | # Number 20524401417
### Properties of number 20524401417
Cross Sum:
Factorization:
3 * 13 * 23 * 61 * 375101
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
4c7598309
Base 32:
j3lj0o9
sin(20524401417)
-0.11192606374797
cos(20524401417)
-0.9937165371744
tan(20524401417)
0.11263379400551
ln(20524401417)
23.744880328254
lg(20524401417)
10.312270500014
sqrt(20524401417)
143263.39873464
Square(20524401417)
4.2125105352615E+20
### Number Look Up
Look Up
20524401417 (twenty billion five hundred twenty-four million four hundred one thousand four hundred seventeen) is a very great figure. The cross sum of 20524401417 is 30. If you factorisate 20524401417 you will get these result 3 * 13 * 23 * 61 * 375101. The figure 20524401417 has 32 divisors ( 1, 3, 13, 23, 39, 61, 69, 183, 299, 793, 897, 1403, 2379, 4209, 18239, 54717, 375101, 1125303, 4876313, 8627323, 14628939, 22881161, 25881969, 68643483, 112155199, 297455093, 336465597, 526266703, 892365279, 1578800109, 6841467139, 20524401417 ) whith a sum of 31256499456. 20524401417 is not a prime number. The number 20524401417 is not a fibonacci number. 20524401417 is not a Bell Number. The number 20524401417 is not a Catalan Number. The convertion of 20524401417 to base 2 (Binary) is 10011000111010110011000001100001001. The convertion of 20524401417 to base 3 (Ternary) is 1221222101021121202110. The convertion of 20524401417 to base 4 (Quaternary) is 103013112120030021. The convertion of 20524401417 to base 5 (Quintal) is 314013221321132. The convertion of 20524401417 to base 8 (Octal) is 230726301411. The convertion of 20524401417 to base 16 (Hexadecimal) is 4c7598309. The convertion of 20524401417 to base 32 is j3lj0o9. The sine of 20524401417 is -0.11192606374797. The cosine of 20524401417 is -0.9937165371744. The tangent of 20524401417 is 0.11263379400551. The square root of 20524401417 is 143263.39873464.
If you square 20524401417 you will get the following result 4.2125105352615E+20. The natural logarithm of 20524401417 is 23.744880328254 and the decimal logarithm is 10.312270500014. that 20524401417 is very impressive number! | 841 | 2,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.646185 |
https://www.greaterwrong.com/posts/FLENugo4Ad4isnqDs/what-makes-a-scientific-fact-ripe-for-discovery | 1,590,411,740,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388758.12/warc/CC-MAIN-20200525130036-20200525160036-00079.warc.gz | 712,699,766 | 5,538 | # [Question] What makes a scientific fact ‘ripe for discovery’?
The existence of multiple discovery seems to suggest that there are certain factors that make scientific facts ready to be discovered. What are these, factors, and how could one measure them?
• Multiple angles of attack
Let me warn you, “important problem” must be phrased carefully. The three outstanding problems in physics, in a certain sense, were never worked on while I was at Bell Labs. By important I mean guaranteed a Nobel Prize and any sum of money you want to mention. We didn’t work on (1) time travel, (2) teleportation, and (3) antigravity. They are not important problems because we do not have an attack. It’s not the consequence that makes a problem important, it is that you have a reasonable attack.
One reasonable attack makes the problem approachable. If there are multiple reasonable attacks, at least one succeeding becomes more likely and further they can exchange information about the problem making each attempt more likely on its own. If we switch to considering thoroughly understood problems, we usually have multiple good solutions for them (like multiple proofs in mathematics, or detection from different kinds of experimental apparatus in science).
So if I am going to rank open problems by the likelihood they will be solved, my prior is a list ordered by the number of ways we know of to attack each problem. Without any other information, a problem with two reasonable attacks is twice as likely to be solved as a problem with only one.
Then we could consider updating the weights of different kinds of attack. For example, if one requires very expensive equipment, or very rare expertise, I might adjust it down. On the other hand, if there are two different attacks but the relationship between those two approaches is otherwise very well understood, then we might not treat them as independent anymore and factor in the ease of sharing information between them but also that they will probably succeed or fail together.
We can also consider the problem itself, but I feel like looking at the reference classes for a problem largely boils down to a way to search for reasonable attacks, where any attack which worked for a problem in the reference class is considered a candidate for the problem at hand. But as I think of it, I’m not sure it is common to do a systematic evaluation in this way, so highlighting it as a specific method for finding attacks seems worthwhile.
• Initial Braindump (hopefully will edit)
Knowledge dependencies (algebra before calculus)
Necessary tools? (Did people who made simultaneous discoveries use the same tools?)
The research community for that domain? (How much communication is there? How dense are the connections between people?)
How new the field is. Whether there was a sudden jump in the number of researchers.
How frequently discoveries happen in the field.
Whether a major disaster or other event is obstructing scientific progress from being made at the time.
Whether the existence of simultaneous discoveries is just an artifact of cherrypicking biases | 1,078 | 3,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.924548 |
https://www.johnsonlevel.com/news/HistoryoftheLevelHowtheBu | 1,726,048,503,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651383.5/warc/CC-MAIN-20240911084051-20240911114051-00542.warc.gz | 766,937,040 | 11,755 | # History of the Level - How the Bubble Level Began
Although the spirit level or bubble level is relatively new considering the history of man, the concept of level and plumb, or a tool that levels, has been around for ages.
### Ancient Egypt
Ancient Egypt was one of the first civilizations to have solved many problems of construction as evidenced by the Great Pyramids at Giza. Specifically, when it came to construction, the leveling problems facing the Egyptians were:
1. Leveling a coarse stone horizontally to make room for another stone to be added on top (the stone had to be level on top and bottom)
2. Plumbing the side wall of the stone to fit appropriately next to another stone (the stone had to be straight up and down).
The solution was a simple "A" frame structure, most likely made of wood. From the top of the "A" frame was a string attached to the center of the top of the "A". A weight was placed on the other end of the string so when gravity took over, the string would hang plumb or straight down. To make seeing plumb easier, a vertical marking or notching out of the wood near where the string was fixed to the frame, would indicate when the string is sitting straight down, or plumb.
In use, when the "A" frame sits flat on a surface, the weight on the string hangs down straight as measured by the vertical marking, but if the string falls off-center, an adjustment would have been needed. The limitation of this level and plumb rule, like the spirit level of today, is its size. You can only measure level the span of the legs of the "A" frame, therefore, to measure over a greater distance, several measurements side by side would need to be made.
Essentially, what the Egyptians, and later the Romans, used was a combination of a plumb bob with a span to view "level". The key to constructing this simple mason's level would have been the precise construction of symmetry of the "A" frame tool and the perpendicular of each leg of the frame to the plumb bob, in particular, the "square" or 90-degree angle of the top of the "A" frame, which can be accomplished using both geometry and the stars on the horizon for alignment (another ancient Egyptian mathematical and astrological genius). The bottom surface of the legs of the "A" that rested on the rock must also be parallel with the frame and even across. The area of the level touching the area being leveled is what we call today the "gauging edge" of the level.
### Medieval Masonry
Masonry 1,000 years ago also relied on the concept of plumb to find level. However, the medieval level made an advancement over its Egyptians and Roman predessor.
In the thirteenth century, there is pictorial evidence of a level which was essentially a straight-edge (typically a long wooden board) with a raised semi-circle in the middle of the top edge where a string hung down with a plumb bob attached. The string would be long enough to hang below the bottom of the level. A fifteenth century picture also shows a similar level, except instead of a semi-circle at the top, there was a vertical board (parallel with the bottom board.
The reason this tool was an advancement over the ancient tool was that technically, level could only be verified in the ancient tool by comparing the points at which the legs of the "A" frame rested. With the medieval tool, the levelness of the masonry could be checked along the entire length of the board (the level). When working with smaller stones and a larger level, several stones could be checked for level at once, providing a time advantage and greater accuracy.
©2010 Johnson Level & Tool Mfg. Co., Inc.
This article was written based on: Medieval Masons' Tools: The Level and the Plumb Rule Author(s): Lonnie R. Shelby Source: Technology and Culture, Vol. 2, No. 2 (Spring, 1961), pp. 127-130 Published by: The Johns Hopkins University Press on behalf of the Society for the History of Technology
X | 871 | 3,941 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-38 | latest | en | 0.955045 |
https://math.stackexchange.com/questions/1859133/order-of-a-center-of-a-group-is-prime-order/1859137 | 1,716,935,951,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059160.88/warc/CC-MAIN-20240528220007-20240529010007-00085.warc.gz | 317,648,571 | 36,306 | # Order of a center of a group is prime order
Question : Suppose that $G$ is a non-abelian group of order $p^{3}$ where $p$ is prime and $Z (G) \neq \{e\}$. Prove that $|Z (G)| =p$.
Any useful hint to this question is appreciated.
Since the center is non trivial his order can be $p,p^2$ or $p^3$. But $G$ is non abelian, so $|Z (G)|\neq p^3$.
Also if $|Z (G)|=p^2$, then $|G/Z|=p$, so $G/Z$ is cyclic, so $G$ is abelian (proof here).
Finally $|Z (G)|=p$.
• I never remember, how to construct a non abelian $|H| = p^2$ ? Jul 14, 2016 at 11:25
• What is $H$ ? ${}$ Jul 14, 2016 at 11:26
• Well all the groups of order $p^2$ are abelian, so you can't find such a group. Jul 14, 2016 at 11:28
• yes that's what I just thought, with the same argument you used. How to construct $|G| = p^3$ non abelian then ? Jul 14, 2016 at 11:29
• How does the fact that the center is non trivial lead us to deduce the possible order of the center as p, $p^{2}$ or $p^{3}$. Jul 14, 2016 at 11:33 | 364 | 979 | {"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} | 3.5625 | 4 | CC-MAIN-2024-22 | latest | en | 0.913168 |
https://kevinbinz.com/2014/10/13/designer-families/ | 1,500,741,422,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424088.27/warc/CC-MAIN-20170722162708-20170722182708-00062.warc.gz | 666,013,639 | 38,863 | # Designer Families
Part Of: [Advanced Inference In Graphical Models] sequence
Followup To: [Family vs. Universe]
• Motivations
• Graph Type: Factor Graphs
• Graph Type: Markov Random Fields
• A Tale Of Three Families
• On Specificity
• Takeaways
Motivations
Today, we’re going to talk through the construction of a family, and how such a construction might succeed or fail.
Across machine learning, there exist dozens of different graph types; of these, three are the most popular:
1. Markov Random Fields
2. Factor Graphs
3. Bayesian Networks
Our example today will span the first two of these graphs.
Graph Type: Markov Random Field
Markov Random Fields (MRFs) have very simple semantics; they are basically undirected graphs:
In the above MRF, we have a graph with five vertices and six edges. It’s probability distribution is comprised of three factors, one per maximal clique (the first term corresponds to top-left triangle, etc).
The graphical model of MRFs can be formalized by defining its graph and rule structure. Let us define the following terms:
• Let V represent the set of all vertices in the graph
• Let E represent the set of all edges in the graph
We now define the Markov Random Field:
Here’s how I “translate” all the math on the right:
• “For each cliques in the set of all maximal cliques, there exists a function phi over some collection of variables within the clique.”
• “The joint probability, then, over all variables is the product of all phi functions in the graph.”
It turns out that there are several other rules which describe an MRF equally well, but maximal-clique factorization rule serves our purposes here well enough.
Introducing Factor Graphs
Factor graphs (FGs) also have undirected edges, but they honor other constraints, as well:
Notice anything interesting about this graph’s edges? […] Did you notice how no v-nodes or f-nodes are never connected one another? In other words, if we group all edges Graphs with this property, that all edges span across two distinct nodesets, are called bipartite graphs.
How is the probability distribution formula constructed? […] On inspection, it becomes clear that each of the four terms on the right-hand side correspond to factor-nodes.
A rigorous definition of factor graphs simply require us to generalize these two observations. First, some vocabulary is in order.
• Let V represents the set of vertex-nodes on the graph (e.g., v1 through v3)
• Let F represent the set of factor-nodes on the graph (e.g., f1 through f4)
• Let E represent the set of all edges in the graph.
We are now in a position to completely specify the Factor Graph:
Here’s how I would translate this family definition:
• “Let neighbor function RHO be defined as a function that searches through each factor and returns each neighbor.”
• “For every factor f, there exists a function PSI, which takes as its input all neighbors of f (those neighbors discovered by RHO).”
• “The joint probability is then the product of all such functions PSI.”
A Tale Of Three Families
Ready to get your hands dirty?! Good! Let’s consider three specific graphs: two factor graphs, and one Markov Random Field:
Does this diagram make sense? If not, let me explain each family-distribution pairing in turn.
• The leftmost family represents the semantics of factor-graphs R(fg), along with a set of factor-nodes, vertex-nodes, and edges G1. Just as in the factor-graph example above, its corresponding probability distribution is derived from these two elements: each term corresponding to one factor-node and its neighbors.
• The center family is identical to the previous factor-graph, except for an additional factor-node. This additional element corresponds to an extra term in the resultant probability distribution, f4(…)
• The rightmost family represents the semantics of MRFs (here, maxclique factorization) R(mcf), along with a set of nodes and edges G3. Its probability distribution flows from the semantics: since all three variables form a clique, the distribution is a factor across all variables.
So far, so obvious. But recall the purpose of rules: they deliver yes/no answers to any probability distribution that applies for membership. Thus, we must be willing in inquire whether p1 is a member of Family2, etc.
Here are the membership results of the six new family applications:
A green edge (A, B) means that our Ath distribution is a member of our Bth family. In contrast, a red edge denotes non-membershipLet me explain the results of each new edge in turn:
1. (1,2) is green because f(a, b) -> g(x, y, z) transforms are possible.
2. (1,3) is green because f(a, b) -> g(x, y, z) transforms are possible.
3. (2,1) is red because f(a, b, c) -> g(x, y) transforms are impossible (if they were, they would be expressed as two-term factors originally).
4. (2,3) is green because f(a, b, c) -> g(x, y, z) transforms are possible.
5. (3,1) is red because f(a, b, c) -> g(x, y) transforms are impossible (similar reasoning to (2,1)).
6. (3,2) is green because f(a, b, c) -> g(x, y, z) transforms are possible.
On Specificity & Generality
Consider you find yourself asked to design an graphical model that includes p1 within its family. As you can see, any of the above graphs satisfy this requirement. But that does not make them equal.
When designing families, we ought to be as narrow as possible. Suppose we don’t particularly care to compute over p2 or p3. Then, we have reason to prefer Familyasdf – it has the power to say No to these irrelevant distributions, and thereby save computational resources.
Here’s how my professor summarizes & illustrates this principle:
This suggests that there are two aspects of a type of graphical model that indicate its power, and that is: generality, can a graph be found such that its family includes a given distribution; and specificity, how discriminative is a given graph that represents a given distribution.
For example, a collection of all i.i.d. random variables would trivially be a member of all of the aforementioned families, but such a distribution would violate many of the properties that are left unspecied. We might be interested in another graphical model semantics that specifically precludes those cases.
These two notions help us get clear on the kinds of questions to ask when designing a graphical model from scratch.
Takeaways
• When designing a family, one must optimize not just for generality (admitting viable distributions). Specificity (the rejection of irrelevant distributions) matters just as much. | 1,500 | 6,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.889066 |
https://www.dataunitconverter.com/mebibit-per-hour-to-kilobyte-per-hour/547 | 1,720,980,964,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00852.warc.gz | 633,126,635 | 16,177 | Mibit/Hr to kB/Hr - 547 Mibit/Hr to kB/Hr Conversion
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S = Second, M = Minute, H = Hour, D = Day
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547 Mibit/Hr =71,696.384 kB/Hr
( Equal to 7.1696384E+4 kB/Hr )
content_copy
Calculated as → 547 x 10242 ÷ (8x1000) smart_display Show Stepsexpand_more
Below chart table shows the amount of data that can be transferred at a constant speed of 547 Mibit/Hr in various time frames.
Transfer RateAmount of Data can be transferred
@ 547 Mibit/Hrin 1 Second19.9156622222222222222222222222222222166458 Kilobytes
in 1 Minute1,194.9397333333333333333333333333333333285535 Kilobytes
in 1 Hour71,696.384 Kilobytes
in 1 Day1,720,713.216 Kilobytes
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toc Table of Contents
Mebibits per Hour (Mibit/Hr) to Kilobytes per Hour (kB/Hr) Conversion - Formula & Steps
The Mibit/Hr to kB/Hr Calculator Tool provides a convenient solution for effortlessly converting data rates from Mebibits per Hour (Mibit/Hr) to Kilobytes per Hour (kB/Hr). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Mebibit) and target (Kilobyte) data units.
Source Data Unit Target Data Unit
Equal to 1024^2 bits
(Binary Unit)
Equal to 1000 bytes
(Decimal Unit)
The formula for converting the Mebibits per Hour (Mibit/Hr) to Kilobytes per Hour (kB/Hr) can be expressed as follows:
diamond CONVERSION FORMULA kB/Hr = Mibit/Hr x 10242 ÷ (8x1000)
Now, let's apply the aforementioned formula and explore the manual conversion process from Mebibits per Hour (Mibit/Hr) to Kilobytes per Hour (kB/Hr). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Kilobytes per Hour = Mebibits per Hour x 10242 ÷ (8x1000)
STEP 1
Kilobytes per Hour = Mebibits per Hour x (1024x1024) ÷ (8x1000)
STEP 2
Kilobytes per Hour = Mebibits per Hour x 1048576 ÷ 8000
STEP 3
Kilobytes per Hour = Mebibits per Hour x 131.072
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By applying the previously mentioned formula and steps, the conversion from 547 Mebibits per Hour (Mibit/Hr) to Kilobytes per Hour (kB/Hr) can be processed as outlined below.
1. = 547 x 10242 ÷ (8x1000)
2. = 547 x (1024x1024) ÷ (8x1000)
3. = 547 x 1048576 ÷ 8000
4. = 547 x 131.072
5. = 71,696.384
6. i.e. 547 Mibit/Hr is equal to 71,696.384 kB/Hr.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Mebibits per Hour to Kilobytes per Hour using any of the programming language such as Java, Python, or Powershell.
Unit Definitions
What is Mebibit ?
A Mebibit (Mib or Mibit) is a binary unit of digital information that is equal to 1,048,576 bits and is defined by the International Electro technical Commission(IEC). The prefix 'mebi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'megabit' (Mb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
- Learn more..
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What is Kilobyte ?
A Kilobyte (kB) is a decimal unit of digital information that is equal to 1000 bytes (or 8,000 bits) and commonly used to express the size of a file or the amount of memory used by a program. It is also used to express data transfer speeds and in the context of data storage and memory, the binary-based unit of kibibyte (KiB) is used instead.
- Learn more..
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Excel Formula to convert from Mebibits per Hour (Mibit/Hr) to Kilobytes per Hour (kB/Hr)
Apply the formula as shown below to convert from 547 Mebibits per Hour (Mibit/Hr) to Kilobytes per Hour (kB/Hr).
A B C
1 Mebibits per Hour (Mibit/Hr) Kilobytes per Hour (kB/Hr)
2 547 =A2 * 131.072
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
Python Code for Mebibits per Hour (Mibit/Hr) to Kilobytes per Hour (kB/Hr) Conversion
You can use below code to convert any value in Mebibits per Hour (Mibit/Hr) to Mebibits per Hour (Mibit/Hr) in Python.
mebibitsperHour = int(input("Enter Mebibits per Hour: "))
kilobytesperHour = mebibitsperHour * (1024*1024) / (8*1000)
print("{} Mebibits per Hour = {} Kilobytes per Hour".format(mebibitsperHour,kilobytesperHour))
The first line of code will prompt the user to enter the Mebibits per Hour (Mibit/Hr) as an input. The value of Kilobytes per Hour (kB/Hr) is calculated on the next line, and the code in third line will display the result.
Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 1,386 | 4,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-30 | latest | en | 0.588982 |
http://math.stackexchange.com/questions/73711/example-of-prime-not-maximal-ideal | 1,469,808,227,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257831769.86/warc/CC-MAIN-20160723071031-00049-ip-10-185-27-174.ec2.internal.warc.gz | 152,975,717 | 18,719 | # Example of prime, not maximal ideal
This is a homework from videolecture:
Show that $(x^2-y)$ is prime but not maximal in $\mathbb C[x,y]$".
Linked SE pages offer to approach this by proving that $\mathbb C[x,y]/(x^2-y)$ is an integral domain but not field. However, I feel that exhibiting an ideal strictly containing $(x^2-y)$ is easier, and $(x^2+y^2)$ seems to fit the bill. It also seems that proving that $(x^2-y)$ is prime directly is easier, because the polynomial $x^2-y$ can't be factored.
-
$(x,y)$ is a example of an ideal which contains yours. $(x^2+y^2)$, on the other hand, is not. Indeed, as you observe, $x^2-y$ is irreducible, so the ideal $(x^2-y)$ is not contained non-trivially in any principal ideal. – Mariano Suárez-Alvarez Oct 18 '11 at 18:10
When you say primary, do you mean prime? Showing that $x^2 - y$ is irreducible (I assume we are working over a field) would be enough to show that the ideal is prime, but you'd have to prove that. And you would still have to show that it isn't maximal, somehow. – Dylan Moreland Oct 18 '11 at 18:11
You should edit your question and specify which ring you are working in. $\mathbb{Z}[x,y]$? $\mathbb{Q}[x,y]$? ... – Bill Cook Oct 18 '11 at 18:17
@TegiriNenashi Over $\mathbf{C}$ there are far more solutions to $x^2 + y^2 = 0$ than the one you've written down; $(1, i)$ is one example. – Dylan Moreland Oct 18 '11 at 18:32
Perhaps the time has come to gather these comments into an answer? – Gerry Myerson Oct 19 '11 at 2:52
It may be worth saying that since $\Bbb C$ is algebraically closed, the maximal ideals in $\Bbb C[X,Y]$ are precisely the ideals of the form $(X-\zeta_1,Y-\zeta_2)$ where $\zeta_1$ and $\zeta_2$ are arbitrary complex numbers.
Moreover, the maximal ideal $(X-\zeta_1,Y-\zeta_2)$ contains the ideal $(P(X,Y))$ for a (not necessarily irreducible) polynomial $P(X,Y)$ if and only if $P(\zeta_1,\zeta_2)=0$.
An ideal $(x^2+y) \subset \mathbb{C}[x,y]$ is prime because a polynomial $x^2+y$ is irreducible. Indeed, if we assume that $x^2+y = a \cdot b,\ a,b \in \mathbb{C}[x,y]$, then if $y$-degree of $a$ is $1$, y-degree of $b$ must be $0$. If $x$-degree of $a$ is $2$, then $x$-degree of $b$ is $0$, and $b$ is just a complex number. Otherwise, if $x$-degree of $a$ is not $2$, then $a \cdot b$ contains $x^2y$ or $xy$ terms, that are not contained in $x^2+y$.
The ideal is not maximal, because it's contained in an ideal $(x,y) \subset \mathbb{C}[x,y]$. | 791 | 2,450 | {"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} | 3.90625 | 4 | CC-MAIN-2016-30 | latest | en | 0.90748 |
https://chess.stackexchange.com/questions/33904/is-there-a-double-forced-mate-position/33905 | 1,618,530,314,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088264.43/warc/CC-MAIN-20210415222106-20210416012106-00563.warc.gz | 239,625,367 | 41,515 | # Is there a double-forced mate position?
A usual forced mate position is where the winning player can play in a way that ensures checkmate, no matter what (legal) moves the opponent makes (unless the game ends due to a non-play reason, i.e. resignation, draw accepted, time ran out.)
What I'm asking about here is what I'll call a double-forced mate: A position where checkmate is ensured no matter what (legal) moves either player makes.
With this rule alone we can find a position such as this, where white has only one legal move, resulting in checkmate. Let's call this type 1.
# Type 1:
``````[FEN "K1k5/1q6/P7/B3N3/8/8/8/8 w - - 0 1"]
1. axb7#
``````
The only option is `axb7#`. However, this isn't too satisfying as a "double" forced mate, as there was only a single possible move to continue. We can explore positions that have multiple possibilities, but still ensure checkmate absolutely.
# Type 2:
``````[FEN "k7/2QQQQQQ/KQ1QQQQQ/QQQ1QQQQ/QQQQ1QQQ/QQQQQ1QQ/QQQQQQ1Q/QQQQQQQ1 w - - 0 1"]
``````
White has many legal moves, but every one of them results in an immediate checkmate. Still, I feel this question isn't quite answered, as in a true double-forced checkmate, both players should play a role. (Also, this position clearly isn't possible from a standard board start. And even if it were, it doesn't seem that it could be achieved without first creating a stalemate.) Let's finally look to type 3.
# Type 3:
The true double-forced mate: A position where checkmate is ensured no matter what (legal) moves either player makes, however there is at least some line where both players make a move. Meaning not every legal ply in the start position is an immediate checkmating move.
Do you know of (or can you find) any such chess positions of (type 3) double-forced mate? Or can you prove that this is not possible?
Also, share if you have any interesting or more plausible examples of type 1 and 2.
• Perhaps a more apt name could be "foolproof mate", as it would be impossible to miss it. – musefan Jan 27 at 14:50
• “this position clearly isn't possible from a standard board start”[citation needed] – leftaroundabout Jan 28 at 12:42
• @leftaroundabout is there any legal way of getting more queens than you had pieces at the start of a classical game? – terdon Jan 28 at 17:15
• @terdon it’s a joke – Fivesideddice Jan 29 at 6:01
• @Fivesideddice d'oh! That does seem like a reasonable assumption. Sorry :) – terdon Jan 29 at 9:20
Indeed, there are many possible positions in which all mates are forced on the first ply. Since you ask not for that, I shall provide examples with more than one move by both sides.
Here is the known, and overall, length record for both sides having multiple legal moves since you said that a single line "isn't too satisfying."
``````[Title "Alexey Khanyan, Tim Krabbe's Website Diary Entry #267 2008, Mate In 11 Moves/22 Plies"]
[FEN "4Q2Q/4r3/4n1n1/1bbK1krn/RR1RR1RR/2qn1R1n/4n1nN/Q3Q3 b - - 0 1"]
[startflipped ""]
1... Ng2f4+ 2. Rfxf4+ N2xf4+ 3. Rgxf4+ Nh3xf4+ 4. Rhxf4+ Ndxf4+ 5. Rxf4+ Nhxf4+ 6. Rxf4+ Ngxf4+ 7. Rxf4+ Nxf4+ 8. Rxf4+ Kxf4+ 9. Qee5+ Qxe5+ 10. Qaxe5+ Rgxe5+ 11. Qxe5+ Rxe5+ 12. Qxe5#
``````
Source: Diary Entry #267
Additionally, I know of this simple problem that illustrates it with either playing giving mate.
``````[Title "Eugene B. Cook, The Chess Amateur 1926, Mate In For White And Black"]
[FEN "Bk6/1P6/1P3p2/8/8/2P3p1/6p1/6Kb w - - 0 1"]
``````
Lastly, be sure to check out a funny mutual zugzwang position in this CSE question by the user @Peter, in which both sides are forced to let the other mate them!
• Ugren's #39 is neat but (besides not answering the question at hand) it doesn't quite match your deescription because after 37 Ree1+ Black has the (weaker but still legal) alternative Rxf4 allowing 38 Rbd1#. – Noam D. Elkies Jan 27 at 16:51
• I think this solves the specific wording but not the intention of the OPs question. They may be longer chains, and there are arbitrary choices about which particular knight/rook/queen to move next in places, but both of them are still forced in the same way as the Type 1 example the OP said they wanted to avoid. – Kaz Jan 27 at 18:01
• You could make that first example satisfy the criteria simply by starting one ply earlier - start with the f3 Rook on e3 or g3. Mate no longer guaranteed on the first ply. – Darrel Hoffman Jan 27 at 20:06
• The chess diary entry #267 is brilliant! I took some attempts at this problem, and I had doubted such a long and varied solution was even possible. – Hymns For Disco Jan 28 at 18:05
• @RewanDemontay yes it does. Darrel's comment has made me realize the explanation of type 3 was easy to misinterpret, so I have edited the question to clarify. – Hymns For Disco Jan 28 at 21:16
Rewan Demontay already has a great answer. However, I would also like to add this 'self-solving' chess problem from Tim Krabbé's Chess Diary #267. Here, all moves are forced and will lead to a checkmate. The composer's full name comes from chess problem databases.
``````[Title "Vilhelm Röpke, Skakbladet 1942, Mate In 6"]
[FEN "K1k5/P1Pp4/1p1P4/8/p7/P2P4/8/8 w - - 0 1"]
1. d4 b5 2. d5 b4 3. axb4 a3 4. b5 a2 5. b6 a1=Q 6. b7#
``````
Although it does not completely satisfy your type 3 requirement that checkmate is not guaranteed in the first ply, I think it is a nice little puzzle that has some similarities with your question.
• Nice solution. I have updated the question to clarify what "type 3" means. This position shows the bizarre situation where the choice of what piece to promote to is truly inconsequential. – Hymns For Disco Jan 28 at 21:41
Inspired by Thijs van Ede's answer, I've modified the pattern from Chess Diary #267
``````[FEN "K1k5/P1Pp4/3P4/1P6/1P6/8/5p2/8 w - - 0 1"]
1. b6 f1=B 2. b7# (2. b5 Bg2+ 3. b7+ Bxb7#)
``````
This position features something rather strange. Still according to the rules, checkmate is ensured no matter what legal moves either player makes. However, it is not guaranteed which player shall deliver checkmate. The outcome depends on what piece black promotes to.
In any case, white can just push `b6` and `b7` to mate before black can respond. However if white delays with `b6` followed by `b5`, black can save themself and checkmate white by choosing the queen or bishop. See the example line or try it yourself on lichess. | 1,901 | 6,360 | {"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} | 2.625 | 3 | CC-MAIN-2021-17 | longest | en | 0.962593 |
https://se.mathworks.com/matlabcentral/cody/problems/43144-basics-sum-part-of-vector/solutions/1237381 | 1,606,859,259,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141681524.75/warc/CC-MAIN-20201201200611-20201201230611-00591.warc.gz | 427,585,207 | 16,992 | Cody
# Problem 43144. BASICS - sum part of vector
Solution 1237381
Submitted on 23 Jul 2017 by Dariusz
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [-1 2 3 4 7 9]; c=[1 4 6]; y_correct = 12; assert(isequal(sumvec(x,c),y_correct))
2 Pass
x = [-50 -24 0 4 10 14 19 18]; c=[2 3 6 7]; y_correct = 9; assert(isequal(sumvec(x,c),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 190 | 595 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-50 | latest | en | 0.69081 |
http://seaintarchive.org/mailarchive/2002c/msg00734.html | 1,493,005,432,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118963.4/warc/CC-MAIN-20170423031158-00539-ip-10-145-167-34.ec2.internal.warc.gz | 346,711,501 | 5,329 | Need a book? Engineering books recommendations...
# RE: Difference between rigid and flexible culvert
• To: seaint(--nospam--at)seaint.org
• Subject: RE: Difference between rigid and flexible culvert
• From: "Lutz, James" <JLUTZ(--nospam--at)earthtech.com>
• Date: Tue, 6 Aug 2002 11:22:15 -0700
While there is a tendency to simply classify conduits according to material type as rigid or flexible, it's not really quite that simple. The behavior of the pipe as flexible or rigid is a function of the pipe stiffness relative to the stiffness of the surrounding soil. If the pipe is stiffer than the soil, then the pipe is rigid. If the soil is stiffer than the pipe, then the pipe is flexible.
For a lot of ordinary applications, the classification is easy and intuitive. Corrugated and plastic pipe is generally flexible, as is steel and ductile iron, at least in larger diameters, and concrete and AC pipe are generally rigid. However in small diameters, ductile iron pipe in conventional thicknesses is fairly stiff and will tend to act more like rigid pipe. Hybrids like concrete cylinder pipe are sometimes classified as "semi-rigid."
If you define the pipe rigidity as kp = 96(EI/D^3)/D, where D is the diameter in inches, E is the modulus of elasticity in psi, and I is the moment of inertia of the wall section. kp is measured in lbs/in^3 and is the deflection of a ring under a uniform horizontal loading applied over the diameter of the pipe.
If you compare this to ks, the subgrade modulus, which is measured in the same units, you will have some idea of the relative stiffness. If kp/ks is greater than 1.0, then the pipe is probably rigid, if not, the pipe is probably flexible. I don't pretend that this is an exact method, but most of the time the ratio is either much greater or much less than 1.0 for real pipes, and you can rapidly sort out how to approach the design.
A flexible pipe has very low pipe stiffness and as a result deforms more and can only transfer load through axial load in the plane of the wall. The soil is doing most of the work. A rigid pipe has high pipe stiffness and as a result deforms very little although the surrounding soil may be trying to settle. The pipe does most of the work, mobilizes flexural resistance in the pipe wall, and actually experiences higher load than a flexible pipe because if cannot "travel" with the settling soil around it. Flexible pipes usually fail by buckling, while rigid pipes usually fail by exceeding the strength of the wall.
-----Original Message-----
From: Eleanor G. Guarin [mailto:eguarin(--nospam--at)escaengineers.com]
Sent: Monday, August 05, 2002 9:35 AM
To: seaint(--nospam--at)seaint.org
Subject: Difference between rigid and flexible culvert
Im currently designing manholes and handholes which i know that is almost same when designing box culverts. In LRFD design, there is a load factor design of 1.0 for rigid culvert and 1.5 for flexible culvert, I just want to know what's the difference between them. Could you give me example of flexible culverts? | 731 | 3,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | latest | en | 0.913038 |
http://forum.enjoysudoku.com/expanded-uniqueness-type-t30126.html | 1,513,421,755,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948587577.92/warc/CC-MAIN-20171216104016-20171216130016-00235.warc.gz | 111,213,788 | 5,692 | ## Expanded Uniqueness Type?
Advanced methods and approaches for solving Sudoku puzzles
### Expanded Uniqueness Type?
Do any solvers out there look for uniqueness patterns for more than U4, i.e. U6 or U8? I found a U6 in this puzzle:
Code: Select all
`5 7 6 | 3 8 1 | 9 24 24 8 19 12 | 4 6 29 | 3 7 5 3 49 24 | 7 29 5 |*18 *18 6 ------------+-------------+------------4 18 18 | 2 5 3 | 6 9 7 9 6 5 | 8 1 7 | 4 23 23 2 3 7 | 9 4 6 |*15 *15 8 ------------+-------------+------------7 5 48 | 6 39 489 | 2 348 1 6 2 3 | 1 7 48 |*58 *458 9 1 48 9 | 5 23 248 | 7 6 34 `
Obviously this is a simple Type 1, but for a U6 instead of a U4.
I also know that the puzzle solves with an XY-chain (r3c2-r3c5-r9c5-r9c9), and so the pattern isn't needed here, but it serves as an example none the less.
I haven't implemented Uniqueness Tests in my solver yet, but I'm wondering how common it is for solvers to look beyond U4.
Last edited by EnderGT on Tue Nov 30, 2010 2:01 am, edited 1 time in total.
EnderGT
Posts: 69
Joined: 19 February 2008
### Re: Expanded Uniqueness Type?
Dabeer wrote:Do any solvers out there look for uniqueness patterns for more than U4, i.e. U6 or U8? I found a U6 in this puzzle:
Code: Select all
`5 7 6 | 3 8 1 | 9 24 24 8 19 12 | 4 6 29 | 3 7 5 3 49 24 | 7 29 5 |*18 *18 6 ------------+-------------+------------4 18 18 | 2 5 3 | 6 9 7 9 6 5 | 8 1 7 | 4 23 23 2 3 7 | 9 4 6 |*15 *15 8 ------------+-------------+------------7 5 48 | 6 39 489 | 2 348 1 6 2 3 | 1 7 48 |*58 *458 9 1 48 9 | 5 23 248 | 7 6 34 `
Obviously this is a simple Type 1, but for a U6 instead of a U4.
I also know that the puzzle solves with an XY-chain (r3c2-r3c5-r9c5-r9c9), and so the pattern isn't needed here, but it serves as an example none the less.
I haven't implemented Uniqueness Tests in my solver yet, but I'm wondering how common it is for solvers to look beyond U4.
Edit: By the way, this is old user EnderGT, still waiting for a password reset to get access to the old screen name.
Wow..., nothing new that's MUG
Edit: it's actually a BUG-Lite (thanks ronk)=> r8c8=4
BTW, welcome back EnderGT...!
ttt
Last edited by ttt on Tue Nov 30, 2010 2:41 am, edited 2 times in total.
ttt
Posts: 185
Joined: 20 October 2006
Location: vietnam
### Re: Expanded Uniqueness Type?
ttt wrote:Wow..., nothing new that's MUG => r8c8=4 | 1,050 | 2,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-51 | longest | en | 0.251059 |
https://www.crazy-numbers.com/en/6672 | 1,713,089,691,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.25/warc/CC-MAIN-20240414095752-20240414125752-00092.warc.gz | 673,602,060 | 3,492 | Warning: Undefined array key "numbers__url_substractions" in /home/clients/df8caba959271e8e753c9e287ae1296d/websites/crazy-numbers.com/includes/fcts.php on line 156
Number 6672: mathematical and symbolic properties | Crazy Numbers
Discover a lot of information on the number 6672: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 6672
Is 6672 a prime number? No
Is 6672 a perfect number? No
Number of divisors 20
List of dividers
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1, 2, 3, 4, 6, 8, 12, 16, 24, 48, 139, 278, 417, 556, 834, 1112, 1668, 2224, 3336, 6672
Sum of divisors 17360
Prime factorization 24 x 3 x 139
Prime factors
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2, 3, 139
## How to write / spell 6672 in letters?
In letters, the number 6672 is written as: Six thousand six hundred and seventy-two. And in other languages? how does it spell?
6672 in other languages
Write 6672 in english Six thousand six hundred and seventy-two
Write 6672 in french Six mille six cent soixante-douze
Write 6672 in spanish Seis mil seiscientos setenta y dos
Write 6672 in portuguese Seis mil seiscentos setenta e dois
## Decomposition of the number 6672
The number 6672 is composed of:
2 iterations of the number 6 : The number 6 (six) is the symbol of harmony. It represents balance, understanding, happiness.... Find out more about the number 6
1 iteration of the number 7 : The number 7 (seven) represents faith, teaching. It symbolizes reflection, the spiritual life.... Find out more about the number 7
1 iteration of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
Other ways to write 6672
In letter Six thousand six hundred and seventy-two
In roman numeral
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MMMMMMDCLXXII
In binary 1101000010000
In octal 15020 | 782 | 2,648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-18 | latest | en | 0.750041 |
https://www.analyzemath.com/college_algebra/college_algebra_problems_graphs_of_functions.html | 1,721,073,337,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.74/warc/CC-MAIN-20240715194155-20240715224155-00754.warc.gz | 579,000,065 | 8,361 | # College Algebra Problems With Answers sample 4 : Graphs of Functions
A set of college algebra problems on graphs of functions with answers, are presented. The questions are related to the transformations of graphs, graphs of inverse functions, symmetry of graphs, reading values from graphs, find domain and range and intervals on increase and decrease. The solutions are at the bottom of the page.
## Problems
### Problem 1
The graph of $$f(x)$$ is shown below. Draw the graph of $$y = - f(x - 3) - 2$$
### Problem 2
Complete the graph given below so that it is symmetric with respect to the origin.
### Problem 3
The graph of $$f(x)$$ is shown below. Sketch the graph of $$y = f(-x + 1) - 1$$ (hint: see Graphing by Translation, Scaling and Reflection)
### Problem 4
The graph of $$f(x)$$ is shown below. Sketch the graph of the inverse of $$f$$.
### Problem 5
The graph of $$h(x)$$ is shown below.
a) evaluate: $$h(-2) + h(2)$$
b) What is the domain of $$h$$?
c) What is the range of $$h$$?
d) Find the interval(s) over which $$h$$ is increasing.
e) Find the interval(s) over which $$h$$ is decreasing.
f) Find the interval(s) over which $$h$$ is constant.
### Problem 6
The graph of $$f(x)$$ is shown below. Sketch the graph of f(2x).
### Problem 7
The graph of $$f^{-1}(x)$$ is shown below.
Evaluate the following: $$f(0)$$ , $$f(2)$$
## Solutions to the Above Problems
### Solution to Problem 1
Select points whose coordinates are easy to determine on the given graph (see graph in black below) and then transform them as follows:
1 - shift right 3 units : $$f(x - 3)$$
2- reflect on the x-axis : $$- f(x - 3)$$
3 - shift down 2 units : $$- f(x - 3) - 2$$
4 - connect the transformed points to sketch the transformed graph shown in red below.
### Solution to Problem 2
A graph is symmetric with respect to the origin if for each point (a,b) on the graph there exists a point (-a,-b) on the same graph.
We select points (a,b) on the given graph and then transform them into (-a,-b) to obtain more points. When put together the graph is symmetric with respect to the origin.(see the whole graph black and red below).
### Solution to Problem 3
Select points on the given graph and then transform them as follows:
1 - reflect on the y - axis : $$f(-x)$$
2- shift right one unit: $$f(- x + 1) = f(-(x - 1))$$
3 - shift down 1 unit : $$f(- x + 1) - 1$$
4 - connect the transformed points to sketch the transformed graph shown in red below.
### Solution to Problem 4
We first determine points (a , b) on the graph of the given function and then use the definition of the inverse to determine points (b , a) on the graph of the inverse. Or use the line $$y = x$$ to reflect points (a , b) into (b , a).
### Solution to Problem 5
$$h(-2) + h(2) = -3 + 1 = -2$$
Domain: $$[-5 , 5]$$
Range: {-3} U $$[0 , 5]$$
increase: $$[-3 , 2)$$ , $$[2 , 5]$$
decrease: $$[-5 , -3]$$
constant: $$[-2 , 2)$$
### Solution to Problem 6
Function $$f$$ has two x-intercepts: $$x = 2$$ and $$x = -2$$ . $$f(2x)$$ will also have x-intercepts such that $$2x = 2$$ which gives $$x = 1$$ and $$2x = -2$$ which gives $$x = -1$$. Hence $$f(2x)$$ will have x intercepts at $$x = 1$$ and $$x = -1$$. The y-intercept is at $$y = 2$$ since $$f(0) = 2$$ and $$f(2*0) = f(0) = 2$$.
### Solution to Problem 7
since $$f^{-1}(1) = 0$$ , $$f(0) = 1$$ , and since $$f^{-1}(0) = 2$$ , $$f(2) = 0$$ | 1,041 | 3,388 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-30 | latest | en | 0.835587 |
https://tex.stackexchange.com/questions/697323/how-to-draw-oval-line-through-specified-coordinates | 1,725,982,286,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00067.warc.gz | 523,498,204 | 40,487 | # How to draw oval line through specified coordinates?
How to draw oval line through specified coordinates? I have used simple \draw [red, dashed, line width=1px] (A) to[out=180, in=180] (C); for each connection manually. And connected line dashes are not properly connected (as oval made from separate parts, at A and B points). Perhaps there is some other more effective way?
\documentclass[preview,border=12pt,12pt, varwidth=\maxdimen]{standalone}
\usepackage[americaninductors, europeanresistor]{circuitikz}
\usepackage{siunitx}
\usepackage{nccmath}
\begin{document}
\sisetup{output-decimal-marker = {,}}
\begin{circuitikz}
\draw (0,0) coordinate(point0) to[R, l_=$\underline{Z}_1$, *-] (0,4) coordinate(point1);
\draw (0,4) to[L=$\underline{Z}_2$, label distance=2px, *-*] (4,4) coordinate(point2);
\draw (4,4) to[L=$\underline{Z}_3$] (4,0);
\draw (0,0) to[short, -*] (4,0);
\draw (4,4) to[C=$\underline{Z}_5$, label distance=2px, -*] (8,4) coordinate(point3);
\draw (8,4) to[R=$\underline{Z}_{67}$] (8,0);
\draw (4,0) to[short] (8,0);
\draw (0,4) -- (0,6) to[C=$\underline{Z}_4$, label distance=2px] (8,6) -- (8,4);
\draw (point1) -- ++ (-2,0);
\draw (point0) -- ++ (-2,0);
\node[draw,rectangle,text centered,font=\bfseries] at ([xshift=-0.5cm,yshift=0.5cm] point1){1};
\node[draw,rectangle,text centered,font=\bfseries] at ([yshift=0.5cm] point2){2};
\node[draw,rectangle,text centered,font=\bfseries] at ([xshift=0.5cm,yshift=0.5cm] point3){3};
\node[draw,rectangle,text centered,font=\bfseries] at (0,-0.5){4};
\coordinate (A) at (1.5,5.2);
\coordinate (B) at (6.5,5.2);
\coordinate (C) at (4,0.8);
\draw [red, dashed, line width=1px] (A) -- (B);
\draw [red, dashed, line width=1px] (A) to[out=180, in=180] (C);
\draw [red, dashed, line width=1px] (B) to[out=0, in=0] (C);
\end{circuitikz}
\end{document}
• Is it the connections from one part to another or the shape you want to change? That's not what I'd call 'oval', but I don't see how it could be in that space.
– cfr
Commented Sep 30, 2023 at 20:35
• Shape is exactly as should be (+ / -). But you can see red line is not the same thickness and dashes are badly connected (at A and B points). Commented Sep 30, 2023 at 20:39
• @ErnestasGruodis The varying line thickness is most likely a problem of your PDF viewer, they sometimes render things weirdly. Commented Sep 30, 2023 at 20:53
If you're just concerned about the messy joins, the trick is to draw the shape as a single path, so that the pattern of dashes isn't stopping and starting over, but continues throughout.
You can't completely avoid inconsistency without ensuring the length of the dash pattern divides evenly into the length of the path, but a single path improves the result dramatically, even so.
\documentclass[preview,border=12pt,12pt, varwidth=\maxdimen]{standalone}
\usepackage[americaninductors, europeanresistor]{circuitikz}
\usepackage{siunitx}
\usepackage{nccmath}
\begin{document}
\sisetup{output-decimal-marker = {,}}
\begin{circuitikz}
\draw (0,0) coordinate(point0) to[R, l_=$\underline{Z}_1$, *-] (0,4) coordinate(point1);
\draw (0,4) to[L=$\underline{Z}_2$, label distance=2px, *-*] (4,4) coordinate(point2);
\draw (4,4) to[L=$\underline{Z}_3$] (4,0);
\draw (0,0) to[short, -*] (4,0);
\draw (4,4) to[C=$\underline{Z}_5$, label distance=2px, -*] (8,4) coordinate(point3);
\draw (8,4) to[R=$\underline{Z}_{67}$] (8,0);
\draw (4,0) to[short] (8,0);
\draw (0,4) -- (0,6) to[C=$\underline{Z}_4$, label distance=2px] (8,6) -- (8,4);
\draw (point1) -- ++ (-2,0);
\draw (point0) -- ++ (-2,0);
\node[draw,rectangle,text centered,font=\bfseries] at ([xshift=-0.5cm,yshift=0.5cm] point1){1};
\node[draw,rectangle,text centered,font=\bfseries] at ([yshift=0.5cm] point2){2};
\node[draw,rectangle,text centered,font=\bfseries] at ([xshift=0.5cm,yshift=0.5cm] point3){3};
\node[draw,rectangle,text centered,font=\bfseries] at (0,-0.5){4};
\coordinate (A) at (1.5,5.2);
\coordinate (B) at (6.5,5.2);
\coordinate (C) at (4,0.8);
\draw [red, dashed, line width=1px] (A) -- (B) to[out=0, in=0] (C) to[out=180,in=180] cycle;
\end{circuitikz}
\end{document}
• There's dash expand off (which originated somewhere on this site) but it's for open paths. For closed paths this could be adjusted so that it wraps around cleanly. Though, in this example it's barely a problem anyway. Commented Sep 30, 2023 at 20:51
• Now it looks so pretty in .pdf. I'm very happy :) Magic 'cycle' at the end helped, thanks. Commented Sep 30, 2023 at 20:54
• @Qrrbrbirlbel Thanks. I didn't know about that one. I thought it worth mentioning there is an inconsistency, but isn't something I'd be inclined to go to the effort of addressing. The complexity clearly wouldn't be worth it here.
– cfr
Commented Sep 30, 2023 at 20:55
• @ErnestasGruodis You made it incredibly easy to answer ;). The only thing which threw me was that 'oval' ;).
– cfr
Commented Sep 30, 2023 at 20:56
Based on my answer on your previous question and with using of the fit library:
\documentclass[12pt,, margin=12pt]{standalone}
\usepackage[americaninductors, europeanresistor]{circuitikz}
\usetikzlibrary{fit,
shapes.geometric}
\usepackage{siunitx}
\usepackage{nccmath}
\begin{document}
\begin{circuitikz}[every label/.style = {label distance=2mm,
draw, minimum size=1em, inner sep=1pt,
font=\bfseries}
]
\ctikzset{capacitors/width=0.1, capacitors/height=0.5,
label distance=3pt}
% 1. loop
\draw (0,0) coordinate[label=below:4] (p0)
to[R=$\underline{Z}_1$, *-] ++ (0, 4) coordinate[label=above left:1] (p1)
to[L=$\underline{Z}_2$,*-*,
name=A] ++ (3, 0) coordinate[label=above:2] (p2)
to[L=$\underline{Z}_3$, -*,
name=B] ++ (0,-4)
% 2. loop
(p2) to[C=$\underline{Z}_5$, -*,
name=C] ++ (3, 0) coordinate[label=above right:3] (p3)
to[R=$\underline{Z}_{67}$] ++ (0,-4)
-- (p0)
% 3. loop
(p1) -- ++ (0,2)
to[C=$\underline{Z}_4$] ++ (6, 0)
to[short,-*] ++ (0,-2)
% input connectors
(p1) to[short,*-o] ++ (-2,0)
(p0) to[short,*-o] ++ (-2,0);
% group of elements forming T circuit
\node [trapezium, minimum height=40mm, trapezium stretches, yscale=-1,
draw=red, thick, densely dashed, rounded corners=5mm,
inner xsep=-2ex, yshift=-2ex,
fit=(A.center) (B) (C.center)] {};
\end{circuitikz}
\end{document}
• Red dashed curve may looks nicer/smoother if you replace rounded corners=5mm with rounded corners=11mm. Commented Oct 1, 2023 at 10:57
• Easier than figuring the coordinates for your non-oval than mine!
– cfr
Commented Oct 1, 2023 at 13:43 | 2,278 | 6,552 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.631739 |
https://mocktestpro.in/mcq/computational-fluid-dynamics-classification-of-pde-2/ | 1,708,973,482,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474661.10/warc/CC-MAIN-20240226162136-20240226192136-00442.warc.gz | 411,699,556 | 23,435 | Engineering Questions with Answers - Multiple Choice Questions
# Computational Fluid Dynamics – Classification of PDE – 2
1 - Question
What are the two methods used to find the type of PDEs?
a) Lagrangian Method and Eulerian method
b) Cramer’s method and Eulerian method
c) Cramer’s method and Lagrangian Method
d) Cramer’s method and Eigenvalue method
Explanation: Partial differential equations can be classified using their characteristic lines. These are located using either the Cramer’s method or the Eigenvalue method.
2 - Question
Let u be a variable dependent on x and y. In the diagram, dudyrepresentsuy.
What does this line in the diagram represent?
a) Characteristic line
b) Eigenvalue line
c) Lagrange line
d) Cramer line
Explanation: Characteristic lines are those where the derivatives of the dependent variable are indeterminate. In the diagram, (du/dy) is indeterminate and hence the line represents a characteristic line.
3 - Question
How the type of PDE is identified using Cramer’s rule?
a) By equating the Cramer’s denominator to 1
b) By equating the Cramer’s numerator to 1
c) By equating the Cramer’s denominator to 0
d) By equating the Cramer’s numerator to 0
Explanation: The denominator of Cramer’s solution is equated to zero to find the type of PDE. The denominator is equated to zero to make the solution indeterminate.
4 - Question
What are the Cramer’s solutions equated to while using Cramer’s method of classifying a PDE?
a) The dependent variables
b) The derivatives of dependent variables
c) The second derivatives of dependent variables
d) The highest derivatives of dependent variables
Explanation: For characteristic lines, the derivatives of dependent variables are zero. Cramer’s rule is used to find these derivatives and then it is made indeterminate to find the type of PDE.
5 - Question
_________ of the characteristic curves is used to find the type of PDE.
a) Starting point
b) Centre
c) Length
d) Slope
Explanation: The nature of the slope of the characteristic curves gives the nature of the characteristic curves. This directly matches with the type of the PDE too.
6 - Question
What is the Cramer’s numerator when the solution is the derivative of dependent variables?
a) any negative value
b) any positive value
c) 1
d) 0
Explanation: The solution should be indeterminate. Therefore, if the denominator is 0, the numerator must also be zero. (0/0=indeterminate).
7 - Question
Consider the following system of PDEs.
a1∂u∂x+b1∂u∂y+c1∂v∂x+d1∂v∂y=0
a2∂u∂x+b2∂u∂y+c2∂v∂x+d2∂v∂y=0
The Eigenvalues of which of these matrices can be used to classify this system of PDEs?
a) [a1a2b1b2]−1 [c1c2d1d2]
b) [a1a2c1c2]−1 [b1b2d1d2]
c) [a1a2c1c2] [b1b2d1d2]
d) [a1a2b1b2] [c1c2d1d2]
Explanation: From the given system of PDEs, a1∂u∂x+b1∂u∂y+c1∂v∂x+d1∂v∂y=0 a2∂u∂x+b2∂u∂y+c2∂v∂x+d2∂v∂y=0 Let, W=[uv] then [a1a2c1c2]∂W∂x+[b1b2d1d2]∂W∂y=0 ∂W∂x+[a1a2c1c2]−1[b1b2d1d2]∂W∂y=0 The Eigenvalues of [a1a2c1c2]−1 [b1b2d1d2] determines the class of PDE.
8 - Question
The Eigenvalues in the Eigenvalue method are ____________
a) the type of the characteristic lines
b) the type of PDE
c) the slope of the characteristic lines
d) the slope of PDE
Explanation: The Eigenvalues will give you the slope of the characteristic lines. Using this slope, we determine the type of the characteristic line and the type of PDE.
9 - Question
When the Eigenvalues are a mixture of real and imaginary values, the PDE is ___________
a) elliptic-hyperbolic
b) parabolic
c) elliptic
d) hyperbolic
Explanation: When we get a mixed type of Eigenvalues, the type of PDE is also mixed. Many practical equations have mixed behaviour also.
10 - Question
Solutions of a system of PDEs can be obtained by equating the numerator of Cramer’s solution while using Cramer’s rule. This method is used by __________
a) Integral transform
b) Change of variables
c) Separation of variables
d) Method of characteristics | 1,175 | 3,929 | {"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} | 3.609375 | 4 | CC-MAIN-2024-10 | latest | en | 0.840143 |
https://homework.cpm.org/category/CON_FOUND/textbook/mc1/chapter/8/lesson/8.1.1/problem/8-12 | 1,643,314,941,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305288.57/warc/CC-MAIN-20220127193303-20220127223303-00012.warc.gz | 355,521,378 | 15,006 | ### Home > MC1 > Chapter 8 > Lesson 8.1.1 > Problem8-12
8-12.
What shape has all of the following characteristics? Is there more than one possibility? Draw an example of each of the possible shapes.
• The shape is equilateral, that is, all sides are the same length.
• The angles are all acute, that is, smaller than $90^\circ$.
• The shape has fewer than $5$ sides.
A shape has to have more than $2$ sides and this shape must have fewer than $5$ sides.
Test $4$ sides and $3$ sides to see if one or both is possible. | 142 | 523 | {"found_math": true, "script_math_tex": 6, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-05 | longest | en | 0.944978 |
http://www.bth.se/fou/forskinfo.nsf/774bac3cad4d3e9bc125681600506a87/36c9f606f1f587e7c1257336004e0706!OpenDocument | 1,444,103,767,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736678409.42/warc/CC-MAIN-20151001215758-00053-ip-10-137-6-227.ec2.internal.warc.gz | 162,717,467 | 26,813 | # Control Algorithm For Sine Excitation On Nonlinear Systems
Document type: Conference Papers No A Control Algorithm For Sine Excitation On Nonlinear Systems.pdf Andreas Josefsson, Martin Magnevall, Kjell Ahlin Control Algorithm For Sine Excitation On Nonlinear Systems IMAC XXIV 2006 Society for Experimental Mechanics St. Louis, USA Blekinge Institute of Technology School of Engineering - Dept. of Mechanical Engineering (Sektionen för teknik – avd. för maskinteknik)School of Engineering S- 371 79 Karlskrona, School of Engineering S- 371 79 Karlskrona+46 455 38 50 00http://www.bth.se/ing/ andreas.josefsson@bth.se, martin.magnevall@bth.se, kjell.ahlin@bth.se English When using electrodynamic vibration exciters to excite structures, the actual force applied to the structure undertest is the reaction force between the exciter and the structure. The magnitude and phase of the reaction force isdependent upon the characteristics of the structure and exciter. Therefore the quality of the reaction force i.e. theforce applied on the structure depends on the relationship between the exciter and structure under test.Looking at the signal from the force transducer when exciting a structure with a sine wave, the signal will appearharmonically distorted within the regions of the resonance frequencies. This phenomenon is easily observed whenperforming tests on lightly damped structures. The harmonic distortion is a result of nonlinearities produced by theshaker when undergoing large amplitude vibrations, at resonances.When dealing with non-linear structures, it is of great importance to be able to keep a constant force level as wellas a non-distorted sine wave in order to get reliable results within the regions of the resonance frequencies. Thispaper presents theoretical methods that can be used to create a non-distorted sinusoidal excitation signal withconstant force level. Mechanical Engineering\Structural Dynamics control algorithm, sine excitation, shaker, nonlinear, harmonic distortion | 428 | 2,010 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-40 | latest | en | 0.754471 |
https://fr.mathworks.com/matlabcentral/profile/authors/4540999-harald-wie?s_tid=cody_local_to_profile | 1,579,769,828,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250609478.50/warc/CC-MAIN-20200123071220-20200123100220-00109.warc.gz | 449,699,409 | 19,428 | Community Profile
# Harald Wie
### NTNU
25 total contributions since 2013
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Summing digits
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Remove the vowels
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Bullseye Matrix
Given n (always odd), return output a that has concentric rings of the numbers 1 through (n+1)/2 around the center point. Exampl...
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Back and Forth Rows
Given a number n, create an n-by-n matrix in which the integers from 1 to n^2 wind back and forth along the rows as shown in the...
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Most nonzero elements in row
Given the matrix a, return the index r of the row with the most nonzero elements. Assume there will always be exactly one row th...
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Remove any row in which a NaN appears
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Pizza!
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Make a checkerboard matrix
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Find all elements less than 0 or greater than 10 and replace them with NaN
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Triangle Numbers
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Is my wife right?
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Times 2 - START HERE
Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...
environ 6 ans ago | 1,249 | 4,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-05 | latest | en | 0.742594 |
https://blog.flyingcoloursmaths.co.uk/wrong-useful-episode-35/ | 1,701,982,172,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100686.78/warc/CC-MAIN-20231207185656-20231207215656-00242.warc.gz | 167,642,084 | 5,210 | /podcasts/wbu35.mp3
In this month’s Wrong, But Useful, Dave and Colin discuss:
• Colin gets his plug for Cracking Mathematics in early
• Colin is upset by a missing apostrophe
• Dave teases us with the number of the podcast and asks about the kinds of things it’s reasonable to expect students to know, and asserts that the origin of the seven-day week is a quarter of the moon’s cycle. Is that right?
• So, 28 is the number of the podcast, although the sidereal month is 27.3 days.
• We point out Big MathsJam conference, which (this year) is the weekend of the 12th-13th of November at Yarnfield Park, near Stone in Staffordshire. We discuss our past MathsJam experiences, which are uniformly positive, despite having to interact with each other in person.
• Colin saw Colin Wright’s juggling talk and recommends it.
• Why are geometric progressions called geometric? Thanks to @sparksmaths (Ben in real life) for pointing Dave in the right direction.
• @colinthemathmo asks: Is -3 a whole number? We argue. Is 0 a natural number? We agree, but are wrong (or possibly not).
• @rjs2212, Robert in real life tweeted this:
. Very neat!
• Dave asks what happens when you work out Pascal’s triangle modulo 9?
• Dave is about to read What is the Name of This Book?
• The last puzzle had a gold star for @dragondodo and a silver star for @chrishazell72 for having a pop.
• Dave has a list of 24 numbers he needs to arrange into eight groups of three so that the totals are as even as possible. The numbers are: 2, 2.29, 2.5, 2.5, 3, 3, 3, 3, 3.23, 3.23, 3.54, 4.17, 4.78, 4.78, 5.2, 5.52, 5.7, 6, 6, 6, 6.18, 7.5, 13, 14.06 … and the best solution is the one with the lowest standard deviation of totals. | 487 | 1,703 | {"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} | 2.96875 | 3 | CC-MAIN-2023-50 | latest | en | 0.940439 |
https://edustrings.com/mathematics/1659689.html | 1,628,113,217,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155188.79/warc/CC-MAIN-20210804205700-20210804235700-00068.warc.gz | 236,421,150 | 6,972 | 16 July, 11:32
# The function f (x) = 7x+15 models the time in minutes that a customer will wait to get an oil change if there are x cars in line ahead of the customer. How long will a customer wait if they are fifth car in the line? 28 minutes 35 minutes 43 minutes 50 minutes
+1
1. 16 July, 13:28
0
43 minutes.
Step-by-step explanation:
f (x) is the time in minutes that a customer will wait when x cars are ahead of the customer.
If a customer is 5th in the line, it means there are 4 cars ahead of that customer.
So, the time that customer waits is equal to f (4)
f (4) = 7 * 4 + 15
f (4) = 28 + 15
f (4) = 43 minutes | 194 | 630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.934881 |
https://genomaths.com/tag/epigenomics/ | 1,669,990,604,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710902.80/warc/CC-MAIN-20221202114800-20221202144800-00075.warc.gz | 310,155,743 | 44,431 | # Fisher's exact test failure can lead to biased results
Robersy Sanchez
Department of Biology. Pennsylvania State University, University Park, PA 16802
Email: rus547@psu.edu
Fisher's exact test is a statistical significance test used in the analysis of contingency tables. Although this test is routinely used even though, it has been full of with controversy for over 80 years.Herein, the case of its application analyzed is scrutinized with specific examples.
# Overview
The statistical significance of the difference between two bisulfite sequence from control and treatment groups at each CG site can be evaluated with Fisher's exact test. This is a statistical test used to determine if there are nonrandom associations between two categorical variables.
Let there exist two such (categorical) variables $X$ and $Y$, where $X$ stands for two groups of individuals: control and treatment, and $Y$ be a two states variable denoting the methylation status, carrying the number of times that a cytosine site is found methylated ($^{m}CG$) and non-methylated ($CG$), respectively.
This information can be summarized in a $2 \times 2$ table, a $2 \times 2$ matrix in which the entries $a_{ij}$ represent the number of observations in which $x=i$ and $y=j$. Calculate the row and column sums $R_i$ and $C_j$, respectively, and the total sum:
$N=\sum_iR_i=\sum_jC_j$
of the matrix:
$Y = ^mCG$ $Y = CG$ $R_i$
Control $a_{11}$ $a_12$ $a_{11}+a_{12}$
Treatment $a_{21}$ $a_22$ $a_{21}+a_{22}$
$C_i$ $a_{11}+a_{21}$ $a_{12}+a_{22}$ $a_{11}+a_{12}+a_{21}+a_{22} = N$
Then the conditional probability of getting the actual matrix, given the particular row and column sums, is given by the formula:
$P_{cutoff}=\frac{R_1!R_2!}{N!\prod_{i,j}a_{ij}!}C_1!C_2!$
# Example 1
Let's consider the following hypthetical case of methylation at a given cytosine site found in the comparison of control and treatment groups:
case1 <- matrix(c(5, 14, 15, 12), nrow = 2,
dimnames = list(Group = c("Ctrl", "Treat"),
Meth.status = c("mCG", "CG")))
case1
## Meth.status
## Group mCG CG
## Ctrl 5 15
## Treat 14 12
That is, the cytosine site was found methylated 5 times from 20 counts in the control group and 14 out of 26 in the treatment. This accounts for methylation levels about 0.28 and 0.53 in the control and and the treatment groups, respectively, which correspond to a value of 0.25 (50%) of methylation levels difference.
## Proportions
case1/rowSums(case1)
## Meth.status
## Group mCG CG
## Ctrl 0.2500000 0.7500000
## Treat 0.5384615 0.4615385
Fisher's exact test found not difference between these group!
fisher.test(case1)
##
## Fisher's Exact Test for Count Data
##
## data: case1
## p-value = 0.07158
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
## 0.06352714 1.18168682
## sample estimates:
## odds ratio
## 0.293905
Considering the direction of the changes seems to be more sensitive to the magnitude of methylation changes, statistically significant at a significance level of $\alpha = 0.05$.
fisher.test(case1, alternative = "less")
##
## Fisher's Exact Test for Count Data
##
## data: case1
## p-value = 0.04666
## alternative hypothesis: true odds ratio is less than 1
## 95 percent confidence interval:
## 0.0000000 0.9804613
## sample estimates:
## odds ratio
## 0.293905
To realize how difficult would be the interpretation of this result in a real concrete scenario, let's consider a basketball team where a top player finished a game scoring 14 field-goal out of 26 shooting. Base on Fisher's exact test, does it make sense to say that another player who made 6 out of 15 field-goals performed as well as the best player?
# Bootstrap tests
Alternative testings are possible by means of bootstrap re-sampling from the set (population) of all the matrices with the same row and column sums $R_i$ and $C_j$. The analyses will be accomplished with function tableBoots from the R packge named usefr.
## Hellinger test statistic
Hellinger test statistic has been proposed to compare discrete probability distributions (1). Basu et all have shown that the Hellinger divergence, as reported in 1, has asymptotically has a chi-square distribution with one degree of freedom under the null hypothesis. This a good property that makes this statistic Hellinger statistic suitable for bootstrap's test of two discrete populations.
library(usefr)
tableBoots(x = case1, stat = "hd", num.permut = 1999)
## [1] 0.0445
## Root Mean Square statistic
The Root-Mean-Square statistic has been also proposed to test differences between two discrete probability distributions (2)
library(usefr)
tableBoots(x = case1, stat = "rmst", num.permut = 1999)
## [1] 0.05
## Chi-square test
The $\chi^2$ statistic fails
library(usefr)
tableBoots(x = case1, stat = "chisq", num.permut = 1999)
## [1] 0.059
Assuming that expected discrete probability distribution (DPD) are:
p <- case1[1,]/sum(case1[1,])
p
## mCG CG
## 0.25 0.75
Then, we can test whether the treatment departs from the expected DPD:
chisq.test(x= case1[2,], p = p,
simulate.p.value = TRUE, B = 2e3)
##
## Chi-squared test for given probabilities with simulated p-value (based
## on 2000 replicates)
##
## data: case1[2, ]
## X-squared = 11.538, df = NA, p-value = 0.001499
## The example 1 is not unique
The failure of Fisher's exact test is evident in the comparisons given below:
res <- lapply(seq(1,78,1), function(k) {
datos <- matrix(c(5, k+1, 15, k), nrow = 2,
dimnames = list(Group = c("Ctrl", "Treat"),
Meth.status = c("mCG", "CG")))
datos
x <- datos/rowSums(datos)
x <- x[2,1] - x[1,1]
ft <- fisher.test(datos)
hd <- tableBoots(x = datos, stat = "hd", num.permut = 1999)
rmst <- tableBoots(x = datos, stat = "rmst", num.permut = 1999)
chisq <- tableBoots(x = datos, stat = "chisq", num.permut = 1999)
chisq.p <- chisq.test(x= datos[2,], p = datos[1,]/sum(datos[1,]),
simulate.p.value = TRUE, B = 2e3)$p.value c(meth.diff = x, FT.p.value = ft$p.value,
HD.p.value = hd,
RMST = rmst,
chisq = chisq,
chisq.p = chisq.p)
})
do.call(rbind, res)
## meth.diff FT.p.value HD.p.value RMST chisq chisq.p
## [1,] 0.4166667 0.20948617 0.0670 0.1060 0.1235 0.1604197901
## [2,] 0.3500000 0.28326746 0.1135 0.1130 0.1390 0.1009495252
## [3,] 0.3214286 0.17506841 0.1140 0.1030 0.1220 0.0744627686
## [4,] 0.3055556 0.20467538 0.1310 0.0965 0.1160 0.0454772614
## [5,] 0.2954545 0.13181103 0.1120 0.1110 0.1130 0.0389805097
## [6,] 0.2884615 0.14213458 0.0930 0.0895 0.0990 0.0204897551
## [7,] 0.2833333 0.15672314 0.0980 0.0710 0.0900 0.0224887556
## [8,] 0.2794118 0.10138414 0.0810 0.0795 0.0890 0.0114942529
## [9,] 0.2763158 0.10534027 0.0775 0.0675 0.0895 0.0109945027
## [10,] 0.2738095 0.11089814 0.0750 0.0665 0.0685 0.0069965017
## [11,] 0.2717391 0.11750154 0.0705 0.0685 0.0735 0.0044977511
## [12,] 0.2700000 0.07769381 0.0665 0.0560 0.0705 0.0044977511
## [13,] 0.2685185 0.07887317 0.0620 0.0625 0.0700 0.0029985007
## [14,] 0.2672414 0.08061039 0.0530 0.0555 0.0595 0.0014992504
## [15,] 0.2661290 0.08276810 0.0690 0.0510 0.0585 0.0019990005
## [16,] 0.2651515 0.08523730 0.0580 0.0540 0.0465 0.0019990005
## [17,] 0.2642857 0.08793188 0.0600 0.0475 0.0505 0.0009995002
## [18,] 0.2635135 0.09078398 0.0575 0.0560 0.0575 0.0004997501
## [19,] 0.2628205 0.09374027 0.0510 0.0480 0.0575 0.0004997501
## [20,] 0.2621951 0.06031279 0.0585 0.0520 0.0470 0.0009995002
## [21,] 0.2616279 0.06080833 0.0585 0.0555 0.0515 0.0004997501
## [22,] 0.2611111 0.06141565 0.0455 0.0470 0.0510 0.0004997501
## [23,] 0.2606383 0.06211375 0.0520 0.0455 0.0500 0.0004997501
## [24,] 0.2602041 0.06288506 0.0395 0.0470 0.0555 0.0004997501
## [25,] 0.2598039 0.06371480 0.0470 0.0465 0.0505 0.0004997501
## [26,] 0.2594340 0.06459058 0.0580 0.0460 0.0550 0.0004997501
## [27,] 0.2590909 0.06550200 0.0490 0.0405 0.0480 0.0004997501
## [28,] 0.2587719 0.06644033 0.0385 0.0425 0.0360 0.0004997501
## [29,] 0.2584746 0.06739822 0.0430 0.0460 0.0440 0.0004997501
## [30,] 0.2581967 0.06836952 0.0420 0.0455 0.0435 0.0004997501
## [31,] 0.2579365 0.06934906 0.0420 0.0370 0.0495 0.0004997501
## [32,] 0.2576923 0.07033254 0.0420 0.0375 0.0450 0.0004997501
## [33,] 0.2574627 0.07131633 0.0340 0.0430 0.0465 0.0004997501
## [34,] 0.2572464 0.07229742 0.0435 0.0375 0.0445 0.0004997501
## [35,] 0.2570423 0.04633836 0.0360 0.0425 0.0525 0.0004997501
## [36,] 0.2568493 0.04646593 0.0340 0.0305 0.0405 0.0004997501
## [37,] 0.2566667 0.04660925 0.0490 0.0335 0.0435 0.0004997501
## [38,] 0.2564935 0.04676622 0.0415 0.0400 0.0350 0.0004997501
## [39,] 0.2563291 0.04693498 0.0330 0.0410 0.0415 0.0004997501
## [40,] 0.2561728 0.04711391 0.0390 0.0330 0.0465 0.0004997501
## [41,] 0.2560241 0.04730155 0.0400 0.0365 0.0435 0.0004997501
## [42,] 0.2558824 0.04749662 0.0405 0.0335 0.0405 0.0004997501
## [43,] 0.2557471 0.04769797 0.0395 0.0350 0.0330 0.0004997501
## [44,] 0.2556180 0.04790460 0.0380 0.0375 0.0365 0.0004997501
## [45,] 0.2554945 0.04811561 0.0360 0.0370 0.0445 0.0004997501
## [46,] 0.2553763 0.04833021 0.0300 0.0390 0.0375 0.0004997501
## [47,] 0.2552632 0.04854768 0.0400 0.0345 0.0335 0.0004997501
## [48,] 0.2551546 0.04876741 0.0315 0.0380 0.0370 0.0004997501
## [49,] 0.2550505 0.04898884 0.0350 0.0310 0.0355 0.0004997501
## [50,] 0.2549505 0.04921147 0.0370 0.0330 0.0470 0.0004997501
## [51,] 0.2548544 0.04943486 0.0405 0.0450 0.0435 0.0004997501
## [52,] 0.2547619 0.04965863 0.0345 0.0370 0.0320 0.0004997501
## [53,] 0.2546729 0.04988242 0.0335 0.0410 0.0335 0.0004997501
## [54,] 0.2545872 0.05010595 0.0355 0.0350 0.0370 0.0004997501
## [55,] 0.2545045 0.05032893 0.0300 0.0345 0.0425 0.0004997501
## [56,] 0.2544248 0.05055114 0.0290 0.0330 0.0365 0.0004997501
## [57,] 0.2543478 0.05077235 0.0385 0.0320 0.0375 0.0004997501
## [58,] 0.2542735 0.05099239 0.0400 0.0345 0.0320 0.0004997501
## [59,] 0.2542017 0.05121109 0.0345 0.0295 0.0345 0.0004997501
## [60,] 0.2541322 0.05142831 0.0335 0.0355 0.0325 0.0004997501
## [61,] 0.2540650 0.05164393 0.0330 0.0370 0.0370 0.0004997501
## [62,] 0.2540000 0.05185784 0.0320 0.0300 0.0360 0.0004997501
## [63,] 0.2539370 0.05206995 0.0370 0.0370 0.0365 0.0004997501
## [64,] 0.2538760 0.05228018 0.0325 0.0335 0.0375 0.0004997501
## [65,] 0.2538168 0.05248845 0.0300 0.0335 0.0315 0.0004997501
## [66,] 0.2537594 0.05269472 0.0280 0.0305 0.0350 0.0004997501
## [67,] 0.2537037 0.05289893 0.0270 0.0355 0.0365 0.0004997501
## [68,] 0.2536496 0.05310105 0.0320 0.0340 0.0380 0.0004997501
## [69,] 0.2535971 0.05330104 0.0405 0.0330 0.0390 0.0004997501
## [70,] 0.2535461 0.05349888 0.0290 0.0315 0.0355 0.0004997501
## [71,] 0.2534965 0.05369454 0.0260 0.0355 0.0310 0.0004997501
## [72,] 0.2534483 0.05388802 0.0300 0.0295 0.0340 0.0004997501
## [73,] 0.2534014 0.05407930 0.0360 0.0340 0.0315 0.0004997501
## [74,] 0.2533557 0.05426839 0.0290 0.0335 0.0305 0.0004997501
## [75,] 0.2533113 0.05445527 0.0285 0.0340 0.0290 0.0004997501
## [76,] 0.2532680 0.05463995 0.0335 0.0325 0.0260 0.0004997501
## [77,] 0.2532258 0.05482244 0.0295 0.0300 0.0370 0.0004997501
## [78,] 0.2531847 0.03529458 0.0285 0.0405 0.0310 0.0004997501
# Extreme situations found in methylation data sets
Unfortunate situations like that one shown below are not rare in methylation data. The minimum coverage (total counts) in the example below is 5.
case2 <- matrix(c(3, 90, 2, 10), nrow = 2,
dimnames = list(Group = c("Ctrl", "Treat"),
Meth.status = c("mCG", "CG")))
case2
## Meth.status
## Group mCG CG
## Ctrl 3 2
## Treat 90 10
The small numbers of count in the control sample yields a bias estimation of the methylation level:
## Proportions
case2/rowSums(case2)
## Meth.status
## Group mCG CG
## Ctrl 0.6 0.4
## Treat 0.9 0.1
There is not enough experimental data to ensure that 3/(3+2) = 0.6 is an unbiased estimated value of the methylation levels at the given cytosine site. Fisher's exact test is insensitive to this experimental fact.
fisher.test(case2)
##
## Fisher's Exact Test for Count Data
##
## data: case2
## p-value = 0.09893
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
## 0.01735485 2.28134136
## sample estimates:
## odds ratio
## 0.1716671
set.seed(1)
st <- tableBoots(x = case2, stat = "hd", num.permut = 1999)
st
## [1] 0.0355
# References
1. Basu A, Mandal A, Pardo L. Hypothesis testing for two discrete populations based on the Hellinger distance. Stat Probab Lett. Elsevier B.V.; 2010;80: 206–214. doi:10.1016/j.spl.2009.10.008.
2. Perkins W, Tygert M, Ward R. Some deficiencies of χ2 and classical exact tests of significance. Appl Comput Harmon Anal. Elsevier Inc.; 2014;36: 361–386. doi:10.1016/j.acha.2013.06.002.
# PCA and LDA with Methyl-IT
## Principal Components and Linear Discriminant. Downstream Methylation Analyses with Methyl-IT
When methylation analysis is intended for diagnostic/prognostic purposes, for example, in clinical applications for patient diagnostics, to know whether the patient would be in healthy or disease stage we would like to have a good predictor tool in our side. It turns out that classical machine learning (ML) tools like hierarchical clustering, principal components and linear discriminant analysis can help us to reach such a goal. The current Methyl-IT downstream analysis is equipped with the mentioned ML tools.
# 1. Dataset
For the current example on methylation analysis with Methyl-IT we will use simulated data. Read-count matrices of methylated and unmethylated cytosine are generated with MethylIT.utils function simulateCounts. A basic example generating datasets is given in:
library(MethylIT) library(MethylIT.utils)
library(ggplot2)
library(ape)
alpha.ct <- 0.01
alpha.g1 <- 0.021
alpha.g2 <- 0.025
# The number of cytosine sites to generate
sites = 50000
# Set a seed for pseudo-random number generation
set.seed(124)
control.nam <- c("C1", "C2", "C3", "C4", "C5")
treatment.nam1 <- c("T1", "T2", "T3", "T4", "T5")
treatment.nam2 <- c("T6", "T7", "T8", "T9", "T10")
# Reference group
ref0 = simulateCounts(num.samples = 3, sites = sites, alpha = alpha.ct, beta = 0.5,
size = 50, theta = 4.5, sample.ids = c("R1", "R2", "R3"))
# Control group
ctrl = simulateCounts(num.samples = 5, sites = sites, alpha = alpha.ct, beta = 0.5,
size = 50, theta = 4.5, sample.ids = control.nam)
# Treatment group II
treat = simulateCounts(num.samples = 5, sites = sites, alpha = alpha.g1, beta = 0.5,
size = 50, theta = 4.5, sample.ids = treatment.nam1)
# Treatment group II
treat2 = simulateCounts(num.samples = 5, sites = sites, alpha = alpha.g2, beta = 0.5,
size = 50, theta = 4.5, sample.ids = treatment.nam2)
A reference sample (virtual individual) is created using individual samples from the control population using function poolFromGRlist. The reference sample is further used to compute the information divergences of methylation levels, $TV_d$ and $H$, with function estimateDivergence [1]. This is a first fundamental step to remove the background noise (fluctuations) generated by the inherent stochasticity of the molecular processes in the cells.
# === Methylation level divergences ===
# Reference sample
ref = poolFromGRlist(ref0, stat = "mean", num.cores = 4L, verbose = FALSE)
divs <- estimateDivergence(ref = ref, indiv = c(ctrl, treat, treat2), Bayesian = TRUE,
num.cores = 6L, percentile = 1, verbose = FALSE)
# To remove hd == 0 to estimate. The methylation signal only is given for
divs = lapply(divs, function(div) div[ abs(div$hdiv) > 0 ], keep.attr = TRUE) names(divs) <- names(divs) To get some statistical description about the sample is useful. Here, empirical critical values for the probability distribution of$H$and$TV_d$is obtained using quantile function from the R package stats. critical.val <- do.call(rbind, lapply(divs, function(x) { x <- x[x$hdiv > 0]
hd.95 = quantile(x$hdiv, 0.95) tv.95 = quantile(abs(x$TV), 0.95)
return(c(tv = tv.95, hd = hd.95))
}))
critical.val
## tv.95% hd.95%
## C1 0.2987088 21.92020
## C2 0.2916667 21.49660
## C3 0.2950820 21.71066
## C4 0.2985075 21.98416
## C5 0.3000000 22.04791
## T1 0.3376711 33.51223
## T2 0.3380282 33.00639
## T3 0.3387097 33.40514
## T4 0.3354077 31.95119
## T5 0.3402172 33.97772
## T6 0.4090909 38.05364
## T7 0.4210526 38.21258
## T8 0.4265781 38.78041
## T9 0.4084507 37.86892
## T10 0.4259411 38.60706
# 2. Modeling the methylation signal
Here, the methylation signal is expressed in terms of Hellinger divergence of methylation levels. Here, the signal distribution is modelled by a Weibull probability distribution model. Basically, the model could be a member of the generalized gamma distribution family. For example, it could be gamma distribution, Weibull, or log-normal. To describe the signal, we may prefer a model with a cross-validations: R.Cross.val > 0.95. Cross-validations for the nonlinear regressions are performed in each methylome as described in (Stevens 2009). The non-linear fit is performed through the function nonlinearFitDist.
The above statistical description of the dataset (evidently) suggests that there two main groups: control and treatments, while treatment group would split into two subgroups of samples. In the current case, to search for a good cutpoint, we do not need to use all the samples. The critical value $H_{\alpha=0.05}=33.51223$ suggests that any optimal cutpoint for the subset of samples T1 to T5 will be optimal for the samples T6 to T10 as well.
Below, we are letting the PCA+LDA model classifier to take the decision on whether a differentially methylated cytosine position is a treatment DMP. To do it, Methyl-IT function getPotentialDIMP is used to get methylation signal probabilities of the oberved $H$ values for all cytosine site (alpha = 1), in accordance with the 2-parameter Weibull distribution model. Next, this information will be used to identify DMPs using Methyl-IT function estimateCutPoint. Cytosine positions with $H$ values above the cutpoint are considered DMPs. Finally, a PCA + QDA model classifier will be fitted to classify DMPs into two classes: DMPs from control and those from treatment. Here, we fundamentally rely on a relatively strong $tv.cut \ge 0.34$ and on the signal probability distribution (nlms.wb) model.
dmps.wb <- getPotentialDIMP(LR = divs[1:10],
nlms = nlms.wb[1:10], div.col = 9L,
tv.cut = 0.34, tv.col = 7, alpha = 1,
dist.name = "Weibull2P")
cut.wb = estimateCutPoint(LR = dmps.wb, simple = FALSE,
column = c(hdiv = TRUE, TV = TRUE,
wprob = TRUE, pos = TRUE),
classifier1 = "pca.lda",
classifier2 = "pca.qda", tv.cut = 0.34,
control.names = control.nam,
treatment.names = treatment.nam1,
post.cut = 0.5, cut.values = seq(15, 38, 1),
clas.perf = TRUE, prop = 0.6,
center = FALSE, scale = FALSE,
n.pc = 4, div.col = 9L, stat = 0)
cut.wb
## Cutpoint estimation with 'pca.lda' classifier
## Cutpoint search performed using model posterior probabilities
##
## Posterior probability used to get the cutpoint = 0.5
## Cytosine sites with treatment PostProbCut >= 0.5 have a
## divergence value >= 3.121796
##
## Optimized statistic: Accuracy = 1
## Cutpoint = 37.003
##
## Model classifier 'pca.qda'
##
## The accessible objects in the output list are:
## Length Class Mode
## cutpoint 1 -none- numeric
## testSetPerformance 6 confusionMatrix list
## testSetModel.FDR 1 -none- numeric
## model 2 pcaQDA list
## modelConfMatrix 6 confusionMatrix list
## initModel 1 -none- character
## postProbCut 1 -none- numeric
## postCut 1 -none- numeric
## classifier 1 -none- character
## statistic 1 -none- character
## optStatVal 1 -none- numeric
The cutpoint is higher from what is expected from the higher treatment empirical critical value and DMPs are found for $H$ values: $H^{TT_{Emp}}_{\alpha=0.05}=33.98<37≤H$. The model performance in the whole dataset is:
# Model performance in in the whole dataset
cut.wb$modelConfMatrix ## Confusion Matrix and Statistics ## ## Reference ## Prediction CT TT ## CT 4897 0 ## TT 2 9685 ## ## Accuracy : 0.9999 ## 95% CI : (0.9995, 1) ## No Information Rate : 0.6641 ## P-Value [Acc > NIR] : <2e-16 ## ## Kappa : 0.9997 ## Mcnemar's Test P-Value : 0.4795 ## ## Sensitivity : 1.0000 ## Specificity : 0.9996 ## Pos Pred Value : 0.9998 ## Neg Pred Value : 1.0000 ## Prevalence : 0.6641 ## Detection Rate : 0.6641 ## Detection Prevalence : 0.6642 ## Balanced Accuracy : 0.9998 ## ## 'Positive' Class : TT ## # The False discovery rate cut.wb$testSetModel.FDR
## [1] 0
# 3. Represeting individual samples as vectors from the N-dimensional space
The above cutpoint can be used to identify DMPs from control and treatment. The PCA+QDA model classifier can be used any time to discriminate control DMPs from those treatment. DMPs are retrieved using selectDIMP function:
dmps.wb <- selectDIMP(LR = divs, div.col = 9L, cutpoint = 37, tv.cut = 0.34, tv.col = 7)
Next, to represent individual samples as vectors from the N-dimensional space, we can use getGRegionsStat function from MethylIT.utils R package. Here, the simulated “chromosome” is split into regions of 450bp non-overlapping windows. and the density of Hellinger divergences values is taken for each windows.
ns <- names(dmps.wb)
DMRs <- getGRegionsStat(GR = dmps.wb, win.size = 450, step.size = 450, stat = "mean", column = 9L)
names(DMRs) <- ns
# 4. Hierarchical Clustering
Hierarchical clustering (HC) is an unsupervised machine learning approach. HC can provide an initial estimation of the number of possible groups and information on their members. However, the effectivity of HC will depend on the experimental dataset, the metric used, and the glomeration algorithm applied. For an unknown reason (and based on the personal experience of the author working in numerical taxonomy), Ward’s agglomeration algorithm performs much better on biological experimental datasets than the rest of the available algorithms like UPGMA, UPGMC, etc.
dmgm <- uniqueGRanges(DMRs, verbose = FALSE)
dmgm <- t(as.matrix(mcols(dmgm)))
rownames(dmgm) <- ns
sampleNames <- ns
hc = hclust(dist(data.frame( dmgm ))^2, method = 'ward.D')
hc.rsq = hc
hc.rsq$height <- sqrt( hc$height )4.
## 4.1. Dendrogram
colors = sampleNames
colors[grep("C", colors)] <- "green4"
colors[grep("T[6-9]{1}", colors)] <- "red"
colors[grep("T10", colors)] <- "red"
colors[grep("T[1-5]", colors)] <- "blue"
# rgb(red, green, blue, alpha, names = NULL, maxColorValue = 1)
clusters.color = c(rgb(0, 0.7, 0, 0.1), rgb(0, 0, 1, 0.1), rgb(1, 0.2, 0, 0.1))
par(font.lab=2,font=3,font.axis=2, mar=c(0,3,2,0), family="serif" , lwd = 0.4)
plot(as.phylo(hc.rsq), tip.color = colors, label.offset = 0.5, font = 2, cex = 0.9,
edge.width = 0.4, direction = "downwards", no.margin = FALSE,
align.tip.label = TRUE, adj = 0)
axisPhylo( 2, las = 1, lwd = 0.4, cex.axis = 1.4, hadj = 0.8, tck = -0.01 )
hclust_rect(hc.rsq, k = 3L, border = c("green4", "blue", "red"),
color = clusters.color, cuts = c(0.56, 15, 0.41, 300))
Here, we have use function as.phylo from the R package ape for better dendrogram visualization and function hclust_rect from MethylIT.utils R package to draw rectangles with background colors around the branches of a dendrogram highlighting the corresponding clusters.
# 5. PCA + LDA
MethylIT function will be used to perform the PCA and PCA + LDA analyses. The function returns a list of two objects: 1) ‘lda’: an object of class ‘lda’ from package ‘MASS’. 2) ‘pca’: an object of class ‘prcomp’ from package ‘stats’. For information on how to use these objects see ?lda and ?prcomp.
Unlike hierarchical clustering (HC), LDA is a supervised machine learning approach. So, we must provide a prior classification of the individuals, which can be derived, for example, from the HC, or from a previous exploratory analysis with PCA.
# A prior classification derived from HC
grps <- cutree(hc, k = 3)
grps[grep(1, grps)] <- "CT"
grps[grep(2, grps)] <- "T1"
grps[grep(3, grps)] <- "T2"
grps <- factor(grps)
ld <- pcaLDA(data = data.frame(dmgm), grouping = grps, n.pc = 3, max.pc = 3,
scale = FALSE, center = FALSE, tol = 1e-6)
summary(ld$pca) ## Importance of first k=3 (out of 15) components:## PC1 PC2 PC3## Standard deviation 41.5183 4.02302 3.73302## Proportion of Variance 0.9367 0.00879 0.00757## Cumulative Proportion 0.9367 0.94546 0.95303 We may retain enough components so that the cumulative percent of variance accounted for at least 70 to 80% [2]. By setting$scale=TRUE$and$center=TRUE$, we could have different results and would improve or not our results. In particular, these settings are essentials if the N-dimensional space is integrated by variables from different measurement scales/units, for example, Kg and g, or Kg and Km. ## 5.1. PCA The individual coordinates in the principal components (PCs) are returned by function pcaLDA. In the current case, based on the cumulative proportion of variance, the two firsts PCs carried about the 94% of the total sample variance and could split the sample into meaningful groups. pca.coord <- ld$pca$x pca.coord ## PC1 PC2 PC3## C1 -21.74024 0.9897934 -1.1708548 ## C2 -20.39219 -0.1583025 0.3167283 ## C3 -21.19112 0.5833411 -1.1067609 ## C4 -21.45676 -1.4534412 0.3025241 ## C5 -21.28939 0.4152275 1.0021932 ## T1 -42.81810 1.1155640 8.9577860 ## T2 -43.57967 1.1712155 2.5135643 ## T3 -42.29490 2.5326690 -0.3136530 ## T4 -40.51779 0.2819725 -1.1850555 ## T5 -44.07040 -2.6172732 -4.2384395 ## T6 -50.03354 7.5276969 -3.7333568 ## T7 -50.08428 -10.1115700 3.4624095 ## T8 -51.07915 -5.4812595 -6.7778593 ## T9 -50.27508 2.3463125 3.5371351 ## T10 -51.26195 3.5405915 -0.9489265 ## 5.2. Graphic PC1 vs PC2 Next, the graphic for individual coordinates in the two firsts PCs can be easely visualized now: dt <- data.frame(pca.coord[, 1:2], subgrp = grps) p0 <- theme( axis.text.x = element_text( face = "bold", size = 18, color="black", # hjust = 0.5, vjust = 0.5, family = "serif", angle = 0, margin = margin(1,0,1,0, unit = "pt" )), axis.text.y = element_text( face = "bold", size = 18, color="black", family = "serif", margin = margin( 0,0.1,0,0, unit = "mm" )), axis.title.x = element_text(face = "bold", family = "serif", size = 18, color="black", vjust = 0 ), axis.title.y = element_text(face = "bold", family = "serif", size = 18, color="black", margin = margin( 0,2,0,0, unit = "mm" ) ), legend.title=element_blank(), legend.text = element_text(size = 20, face = "bold", family = "serif"), legend.position = c(0.899, 0.12), panel.border = element_rect(fill=NA, colour = "black",size=0.07), panel.grid.minor = element_line(color= "white",size = 0.2), axis.ticks = element_line(size = 0.1), axis.ticks.length = unit(0.5, "mm"), plot.margin = unit(c(1,1,0,0), "lines")) ggplot(dt, aes(x = PC1, y = PC2, colour = grps)) + geom_vline(xintercept = 0, color = "white", size = 1) + geom_hline(yintercept = 0, color = "white", size = 1) + geom_point(size = 6) + scale_color_manual(values = c("green4","blue","brown1")) + stat_ellipse(aes(x = PC1, y = PC2, fill = subgrp), data = dt, type = "norm", geom = "polygon", level = 0.5, alpha=0.2, show.legend = FALSE) + scale_fill_manual(values = c("green4","blue","brown1")) + p0 ## 5.3. Graphic LD1 vs LD2 In the current case, better resolution is obtained with the linear discriminant functions, which is based on the three firsts PCs. Notice that the number principal components used the LDA step must be lower than the number of individuals ($N$) divided by 3:$N/3$. ind.coord <- predict(ld, newdata = data.frame(dmgm), type = "scores") dt <- data.frame(ind.coord, subgrp = grps) p0 <- theme( axis.text.x = element_text( face = "bold", size = 18, color="black", # hjust = 0.5, vjust = 0.5, family = "serif", angle = 0, margin = margin(1,0,1,0, unit = "pt" )), axis.text.y = element_text( face = "bold", size = 18, color="black", family = "serif", margin = margin( 0,0.1,0,0, unit = "mm" )), axis.title.x = element_text(face = "bold", family = "serif", size = 18, color="black", vjust = 0 ), axis.title.y = element_text(face = "bold", family = "serif", size = 18, color="black", margin = margin( 0,2,0,0, unit = "mm" ) ), legend.title=element_blank(), legend.text = element_text(size = 20, face = "bold", family = "serif"), legend.position = c(0.08, 0.12), panel.border = element_rect(fill=NA, colour = "black",size=0.07), panel.grid.minor = element_line(color= "white",size = 0.2), axis.ticks = element_line(size = 0.1), axis.ticks.length = unit(0.5, "mm"), plot.margin = unit(c(1,1,0,0), "lines")) ggplot(dt, aes(x = LD1, y = LD2, colour = grps)) + geom_vline(xintercept = 0, color = "white", size = 1) + geom_hline(yintercept = 0, color = "white", size = 1) + geom_point(size = 6) + scale_color_manual(values = c("green4","blue","brown1")) + stat_ellipse(aes(x = LD1, y = LD2, fill = subgrp), data = dt, type = "norm", geom = "polygon", level = 0.5, alpha=0.2, show.legend = FALSE) + scale_fill_manual(values = c("green4","blue","brown1")) + p0 ## References 1. Liese, Friedrich, and Igor Vajda. 2006. “On divergences and informations in statistics and information theory.” IEEE Transactions on Information Theory 52 (10): 4394–4412. doi:10.1109/TIT.2006.881731. 2. Stevens, James P. 2009. Applied Multivariate Statistics for the Social Sciences. Fifth Edit. Routledge Academic. # Methylation analysis with Methyl-IT. Part 2 ## An example of methylation analysis with simulated datasets Part 2: Potential DMPs from the methylation signal Methylation analysis with Methyl-IT is illustrated on simulated datasets of methylated and unmethylated read counts with relatively high average of methylation levels: 0.15 and 0.286 for control and treatment groups, respectively. In this part, potential differentially methylated positions are estimated following different approaches. ## 1. Background Only a signal detection approach can detect with high probability real DMPs. Any statistical test (like e.g. Fisher’s exact test) not based on signal detection requires for further analysis to distinguish DMPs that naturally can occur in the control group from those DMPs induced by a treatment. The analysis here is a continuation of Part 1. ## 2. Potential DMPs from the methylation signal using empirical distribution As suggested from the empirical density graphics (above), the critical values$H_{\alpha=0.05}$and$TV_{d_{\alpha=0.05}}$can be used as cutpoints to select potential DMPs. After setting$dist.name = “ECDF”$and$tv.cut = 0.926$in Methyl-IT function getPotentialDIMP, potential DMPs are estimated using the empirical cummulative distribution function (ECDF) and the critical value$TV_{d_{\alpha=0.05}}=0.926$. DMP.ecdf <- getPotentialDIMP(LR = divs, div.col = 9L, tv.cut = 0.926, tv.col = 7, alpha = 0.05, dist.name = "ECDF") ## 3. Potential DMPs detected with Fisher’s exact test In Methyl-IT Fisher’s exact test (FT) is implemented in function FisherTest. In the current case, a pairwise group application of FT to each cytosine site is performed. The differences between the group means of read counts of methylated and unmethylated cytosines at each site are used for testing (pooling.stat=”mean”). Notice that only cytosine sites with critical values$TV_d$> 0.926 are tested (tv.cut = 0.926). ft = FisherTest(LR = divs, tv.cut = 0.926, pAdjustMethod = "BH", pooling.stat = "mean", pvalCutOff = 0.05, num.cores = 4L, verbose = FALSE, saveAll = FALSE) ft.tv <- getPotentialDIMP(LR = ft, div.col = 9L, dist.name = "None", tv.cut = 0.926, tv.col = 7, alpha = 0.05) There is not a one-to-one mapping between$TV$and$HD$. However, at each cytosine site$i$, these information divergences hold the inequality:$TV(p^{tt}_i,p^{ct}_i)\leq \sqrt{2}H_d(p^{tt}_i,p^{ct}_i)$[1]. where$H_d(p^{tt}_i,p^{ct}_i) = \sqrt{\frac{H(p^{tt}_i,p^{ct}_i)}w}$is the Hellinger distance and$H(p^{tt}_i,p^{ct}_i)$is given by Eq. 1 in part 1. So, potential DMPs detected with FT can be constrained with the critical value$H^{TT}_{\alpha=0.05}\geq114.5$## 4. Potential DMPs detected with Weibull 2-parameters model Potential DMPs can be estimated using the critical values derived from the fitted Weibull 2-parameters models, which are obtained after the non-linear fit of the theoretical model on the genome-wide$HD$values for each individual sample using Methyl-IT function nonlinearFitDist [2]. As before, only cytosine sites with critical values$TV>0.926$are considered DMPs. Notice that, it is always possible to use any other values of$HD$and$TV$as critical values, but whatever could be the value it will affect the final accuracy of the classification performance of DMPs into two groups, DMPs from control and DNPs from treatment (see below). So, it is important to do an good choices of the critical values. nlms.wb <- nonlinearFitDist(divs, column = 9L, verbose = FALSE, num.cores = 6L) # Potential DMPs from 'Weibull2P' model DMPs.wb <- getPotentialDIMP(LR = divs, nlms = nlms.wb, div.col = 9L, tv.cut = 0.926, tv.col = 7, alpha = 0.05, dist.name = "Weibull2P") nlms.wb$T1
## Estimate Std. Error t value Pr(>|t|)) Adj.R.Square
## shape 0.5413711 0.0003964435 1365.570 0 0.991666592250838
## scale 19.4097502 0.0155797315 1245.833 0
## rho R.Cross.val DEV
## shape 0.991666258901194 0.996595712743823 34.7217494754823
## scale
## AIC BIC COV.shape COV.scale
## shape -221720.747067975 -221694.287733122 1.571674e-07 -1.165129e-06
## scale -1.165129e-06 2.427280e-04
## COV.mu n
## shape NA 50000
## scale NA 50000
## 5. Potential DMPs detected with Gamma 2-parameters model
As in the case of Weibull 2-parameters model, potential DMPs can be estimated using the critical values derived from the fitted Gamma 2-parameters models and only cytosine sites with critical values $TV_d > 0.926$ are considered DMPs.
nlms.g2p <- nonlinearFitDist(divs, column = 9L, verbose = FALSE, num.cores = 6L,
dist.name = "Gamma2P")
# Potential DMPs from 'Gamma2P' model
DMPs.g2p <- getPotentialDIMP(LR = divs, nlms = nlms.g2p, div.col = 9L,
tv.cut = 0.926, tv.col = 7, alpha = 0.05,
dist.name = "Gamma2P")
nlms.g2p$T1 ## Estimate Std. Error t value Pr(>|t|)) Adj.R.Square ## shape 0.3866249 0.0001480347 2611.717 0 0.999998194156282 ## scale 76.1580083 0.0642929555 1184.547 0 ## rho R.Cross.val DEV ## shape 0.999998194084045 0.998331895911125 0.00752417919133131 ## scale ## AIC BIC COV.alpha COV.scale ## shape -265404.29138371 -265369.012270572 2.191429e-08 -8.581717e-06 ## scale -8.581717e-06 4.133584e-03 ## COV.mu df ## shape NA 49998 ## scale NA 49998 Summary table: data.frame(ft = unlist(lapply(ft, length)), ft.hd = unlist(lapply(ft.hd, length)), ecdf = unlist(lapply(DMPs.hd, length)), Weibull = unlist(lapply(DMPs.wb, length)), Gamma = unlist(lapply(DMPs.g2p, length))) ## ft ft.hd ecdf Weibull Gamma ## C1 1253 773 63 756 935 ## C2 1221 776 62 755 925 ## C3 1280 786 64 768 947 ## T1 2504 1554 126 924 1346 ## T2 2464 1532 124 942 1379 ## T3 2408 1477 121 979 1354 ## 6. Density graphic with a new critical value The graphics for the empirical (in black) and Gamma (in blue) densities distributions of Hellinger divergence of methylation levels for sample T1 are shown below. The 2-parameter gamma model is build by using the parameters estimated in the non-linear fit of$H$values from sample T1. The critical values estimated from the 2-parameter gamma distribution$H^{\Gamma}_{\alpha=0.05}=124$is more ‘conservative’ than the critical value based on the empirical distribution$H^{Emp}_{\alpha=0.05}=114.5$. That is, in accordance with the empirical distribution, for a methylation change to be considered a signal its$H$value must be$H\geq114.5$, while according with the 2-parameter gamma model any cytosine carrying a signal must hold$H\geq124$. suppressMessages(library(ggplot2)) # Some information for graphic dt <- data[data$sample == "T1", ]
coef <- nlms.g2p$T1$Estimate # Coefficients from the non-linear fit
dgamma2p <- function(x) dgamma(x, shape = coef[1], scale = coef[2])
qgamma2p <- function(x) qgamma(x, shape = coef[1], scale = coef[2])
# 95% quantiles
q95 <- qgamma2p(0.95) # Gamma model based quantile
emp.q95 = quantile(divs$T1$hdiv, 0.95) # Empirical quantile
# Density plot with ggplot
ggplot(dt, aes(x = HD)) +
geom_density(alpha = 0.05, bw = 0.2, position = "identity", na.rm = TRUE,
size = 0.4) + xlim(c(0, 150)) +
stat_function(fun = dgamma2p, colour = "blue") +
xlab(expression(bolditalic("Hellinger divergence (HD)"))) +
ylab(expression(bolditalic("Density"))) +
ggtitle("Empirical and Gamma densities distributions of Hellinger divergence (T1)") +
geom_vline(xintercept = emp.q95, color = "black", linetype = "dashed", size = 0.4) +
annotate(geom = "text", x = emp.q95 - 20, y = 0.16, size = 5,
label = 'bolditalic(HD[alpha == 0.05]^Emp==114.5)',
family = "serif", color = "black", parse = TRUE) +
geom_vline(xintercept = q95, color = "blue", linetype = "dashed", size = 0.4) +
annotate(geom = "text", x = q95 + 9, y = 0.14, size = 5,
label = 'bolditalic(HD[alpha == 0.05]^Gamma==124)',
family = "serif", color = "blue", parse = TRUE) +
theme(
axis.text.x = element_text( face = "bold", size = 12, color="black",
margin = margin(1,0,1,0, unit = "pt" )),
axis.text.y = element_text( face = "bold", size = 12, color="black",
margin = margin( 0,0.1,0,0, unit = "mm")),
axis.title.x = element_text(face = "bold", size = 13,
color="black", vjust = 0 ),
axis.title.y = element_text(face = "bold", size = 13,
color="black", vjust = 0 ),
legend.title = element_blank(),
legend.margin = margin(c(0.3, 0.3, 0.3, 0.3), unit = 'mm'),
legend.box.spacing = unit(0.5, "lines"),
legend.text = element_text(face = "bold", size = 12, family = "serif")
## References
1. Steerneman, Ton, K. Behnen, G. Neuhaus, Julius R. Blum, Pramod K. Pathak, Wassily Hoeffding, J. Wolfowitz, et al. 1983. “On the total variation and Hellinger distance between signed measures; an application to product measures.” Proceedings of the American Mathematical Society 88 (4). Springer-Verlag, Berlin-New York: 684–84. doi:10.1090/S0002-9939-1983-0702299-0.
2. Sanchez, Robersy, and Sally A. Mackenzie. 2016. “Information Thermodynamics of Cytosine DNA Methylation.” Edited by Barbara Bardoni. PLOS ONE 11 (3). Public Library of Science: e0150427. doi:10.1371/journal.pone.0150427.
# Methylation analysis with Methyl-IT. Part 1
## An example of methylation analysis with simulated datasets
Part 1: Methylation signal
Methylation analysis with Methyl-IT is illustrated on simulated datasets of methylated and unmethylated read counts with relatively high average of methylation levels: 0.15 and 0.286 for control and treatment groups, respectively. The main Methyl-IT downstream analysis is presented alongside the application of Fisher’s exact test. The importance of a signal detection step is shown.
## 1. Background
Methyl-IT R package offers a methylome analysis approach based on information thermodynamics (IT) and signal detection. Methyl-IT approach confront detection of differentially methylated cytosine as a signal detection problem. This approach was designed to discriminate methylation regulatory signal from background noise induced by molecular stochastic fluctuations. Methyl-IT R package is not limited to the IT approach but also includes Fisher’s exact test (FT), Root-mean-square statistic (RMST) or Hellinger divergence (HDT) tests. Herein, we will show that a signal detection step is also required for FT, RMST, and HDT as well.
## 2. Data generation
For the current example on methylation analysis with Methyl-IT we will use simulated data. Read count matrices of methylated and unmethylated cytosine are generated with Methyl-IT function simulateCounts. Function simulateCounts randomly generates prior methylation levels using Beta distribution function. The expected mean of methylation levels that we would like to have can be estimated using the auxiliary function:
bmean <- function(alpha, beta) alpha/(alpha + beta)
alpha.ct <- 0.09
alpha.tt <- 0.2
c(control.group = bmean(alpha.ct, 0.5), treatment.group = bmean(alpha.tt, 0.5),
mean.diff = bmean(alpha.tt, 0.5) - bmean(alpha.ct, 0.5))
## control.group treatment.group mean.diff
## 0.1525424 0.2857143 0.1331719
This simple function uses the α and β (shape2) parameters from the Beta distribution function to compute the expected value of methylation levels. In the current case, we expect to have a difference of methylation levels about 0.133 between the control and the treatment.
### 2.1. Simulation
Methyl-IT function simulateCounts will be used to generate the datasets, which will include three group of samples: reference, control, and treatment.
suppressMessages(library(MethylIT))
# The number of cytosine sites to generate
sites = 50000
# Set a seed for pseudo-random number generation
set.seed(124)
control.nam <- c("C1", "C2", "C3")
treatment.nam <- c("T1", "T2", "T3")
# Reference group
ref0 = simulateCounts(num.samples = 4, sites = sites, alpha = alpha.ct, beta = 0.5,
size = 50, theta = 4.5, sample.ids = c("R1", "R2", "R3"))
# Control group
ctrl = simulateCounts(num.samples = 3, sites = sites, alpha = alpha.ct, beta = 0.5,
size = 50, theta = 4.5, sample.ids = control.nam)
# Treatment group
treat = simulateCounts(num.samples = 3, sites = sites, alpha = alpha.tt, beta = 0.5,
size = 50, theta = 4.5, sample.ids = treatment.nam)
Notice that reference and control groups of samples are not identical but belong to the same population.
### 2.2. Divergences of methylation levels
The estimation of the divergences of methylation levels is required to proceed with the application of signal detection basic approach. The information divergence is estimated here using the function estimateDivergence. For each cytosine site, methylation levels are estimated according to the formulas: $p_i={n_i}^{mC_j}/({n_i}^{mC_j}+{n_i}^{C_j})$, where ${n_i}^{mC_j}$ and ${n_i}^{C_j}$ are the number of methylated and unmethylated cytosines at site $i$.
If a Bayesian correction of counts is selected in function estimateDivergence, then methylated read counts are modeled by a beta-binomial distribution in a Bayesian framework, which accounts for the biological and sampling variations [1,2,3]. In our case we adopted the Bayesian approach suggested in reference [4] (Chapter 3).
Two types of information divergences are estimated: TV, total variation (TV, absolute value of methylation levels) and Hellinger divergence (H). TV is computed according to the formula: $TV=|p_{tt}-p_{ct}|$ and H:
$H(\hat p_{ij},\hat p_{ir}) = w_i[(\sqrt{\hat p_{ij}} – \sqrt{\hat p_{ir}})^2+(\sqrt{1-\hat p_{ij}} – \sqrt{1-\hat p_{ir}})^2]$ (1)
where $w_i = 2 \frac{m_{ij} m_{ir}}{m_{ij} + m_{ir}}$, $m_{ij} = {n_i}^{mC_j}+{n_i}^{uC_j}+1$, $m_{ir} = {n_i}^{mC_r}+{n_i}^{uC_r}+1$ and $j \in {\{c,t}\}$
The equation for Hellinger divergence is given in reference [5], but
any other information theoretical divergences could be used as well. Divergences are estimated for control and treatment groups in respect to a virtual sample, which is created applying function poolFromGRlist on the reference group.
.
# Reference sample
ref = poolFromGRlist(ref0, stat = "mean", num.cores = 4L, verbose = FALSE)
# Methylation level divergences
DIVs <- estimateDivergence(ref = ref, indiv = c(ctrl, treat), Bayesian = TRUE,
num.cores = 6L, percentile = 1, verbose = FALSE)
The mean of methylation levels differences is:
unlist(lapply(DIVs, function(x) mean(mcols(x[, 7])[,1])))
## C1 C2 C3 T1 T2
## -0.0009820776 -0.0014922009 -0.0022257725 0.1358867135 0.1359160219
## T3
## 0.1309217360
## 3. Methylation signal
Likewise for any other signal in nature, the analysis of methylation signal requires for the knowledge of its probability distribution. In the current case, the signal is represented in terms of the Hellinger divergence of methylation levels (H).
divs = DIVs[order(names(DIVs))]
# To remove hd == 0 to estimate. The methylation signal only is given for
divs = lapply(divs, function(div) div[ abs(div$hdiv) > 0 ]) names(divs) <- names(DIVs) # Data frame with the Hellinger divergences from both groups of samples samples l = c(); for (k in 1:length(divs)) l = c(l, length(divs[[k]])) data <- data.frame(H = c(abs(divs$C1$hdiv), abs(divs$C2$hdiv), abs(divs$C3$hdiv), abs(divs$T1$hdiv), abs(divs$T2$hdiv), abs(divs$T3$hdiv)), sample = c(rep("C1", l[1]), rep("C2", l[2]), rep("C3", l[3]), rep("T1", l[4]), rep("T2", l[5]), rep("T3", l[6])) ) Empirical critical values for the probability distribution of H and TV can be obtained using quantile function from the R package stats. critical.val <- do.call(rbind, lapply(divs, function(x) { hd.95 = quantile(x$hdiv, 0.95)
## References
1. Hebestreit, Katja, Martin Dugas, and Hans-Ulrich Klein. 2013. “Detection of significantly differentially methylated regions in targeted bisulfite sequencing data.” Bioinformatics (Oxford, England) 29 (13): 1647–53. doi:10.1093/bioinformatics/btt263.
2. Hebestreit, Katja, Martin Dugas, and Hans-Ulrich Klein. 2013. “Detection of significantly differentially methylated regions in targeted bisulfite sequencing data.” Bioinformatics (Oxford, England) 29 (13): 1647–53. doi:10.1093/bioinformatics/btt263.
3. Dolzhenko, Egor, and Andrew D Smith. 2014. “Using beta-binomial regression for high-precision differential methylation analysis in multifactor whole-genome bisulfite sequencing experiments.” BMC Bioinformatics 15 (1). BioMed Central: 215. doi:10.1186/1471-2105-15-215.
4. Baldi, Pierre, and Soren Brunak. 2001. Bioinformatics: the machine learning approach. Second. Cambridge: MIT Press.
5. Basu, A., A. Mandal, and L. Pardo. 2010. “Hypothesis testing for two discrete populations based on the Hellinger distance.” Statistics & Probability Letters 80 (3-4). Elsevier B.V.: 206–14. doi:10.1016/j.spl.2009.10.008.
6. Sanchez R, Mackenzie SA. Information Thermodynamics of Cytosine DNA Methylation. PLoS One, 2016, 11:e0150427. | 15,772 | 46,680 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2022-49 | longest | en | 0.844613 |
https://useenglishwords.com/light-year/ | 1,659,935,990,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570765.6/warc/CC-MAIN-20220808031623-20220808061623-00239.warc.gz | 552,565,776 | 9,002 | # Use Light year in a sentence
1. There is a fundamental difference between a YEAR and a Light year
Light
2. YEAR is a measure of TIME and a Light year is a measure of LENGTH
Light, Length
3. Light year is used to measure the distances between stars
Light
4. A Light year is a standard of measurement used by astronomers to describe huge distances in the Universe
Light
5. The nearest star is 4.22 Light years away
Light
6. One Light year is equal to 5.8 trillion miles
Light
7. To calculate this number, use the formula d = c × t, where d is 1 Light year, c is the speed of light, and t is the number of seconds in 1 year
Light
8. A Light year, also light-year, abbreviated ly, is the distance light travels in one year
Light, Ly
9. 3.26 Light years make up a parsec, which was a unit of distance that was important in locating star systems in the known galaxy.
Light, Locating
10. Since the Galactic Standard Calendar used a year of 368 days, the length of a Galactic Standard Light year would have been 9,531,961,160,601,600 meters.
Length, Light
11. A Light year (symbol: ly) is the distance that light travels in empty space in one year
Light, Ly
12. Since the speed of light is about 300,000 km per second (about 186,000 miles per second), then a Light year is about 10 trillion kilometers (about 6 trillion miles)
Light
13. A Light year is not a length of time
Light, Length
14. The Light year is used in astronomy because the universe is huge.
Light
15. A Light year, abbreviated ly, is the distance light travels in one year: roughly 9.46 × 1012 kilometres (9.46 petametres, or about 5.88 × 1012 (nearly six trillion) miles)
Light, Ly
16. More specifically, a Light year is defined as the distance that a photon would travel, in free space and infinitely far away from any gravitational or magnetic fields, in
Light
17. Light year is the distance that light can travel in one year
Light
18. Phew! It would be easier to say that one Light year is about 10 trillion km.
Light
19. A Light year, abbreviated ly, is the distance light travels in one year: roughly 9.46 × 1012 kilometres (9.46 petametres, or about 5.88 × 1012 (nearly six trillion) miles)
Light, Ly
20. More specifically, a Light year is defined as the distance that a photon would travel, in free space and infinitely far away from any gravitational or magnetic fields, in
Light
21. Light year is a IPA - Imperial style beer brewed by Grimm Artisanal Ales in Brooklyn, NY
Light
22. A Light year is an astronomical measurement of the distance that light can travel in 1 year
Light
23. Hence the name ‘Light year’ So what is the distance in miles that light can travel in a year, making up a Light year
Light
24. Light year synonyms, Light year pronunciation, Light year translation, English dictionary definition of Light year
Light
25. Often Light years Informal A long
Light, Long
26. A Light year is the distance traveled by light in one year
Light
27. While it is technically a measure of distance, you can reverse the Light year to give you an idea about time as well
Light
28. Light year(s) and lightyear(s) may also refer to: Film and television
Light, Lightyear
29. Gandahar, a 1988 animated science fiction film known as Light years in English; Light years, a British drama film; Lightyear, an upcoming animated film featuring Toy Story character Buzz Lightyear; Light years, an American TV series that has been renamed Life Unexpected; Light years, a Singaporean drama produced by
Light, Lightyear, Life
30. Light year Lyrics: I woke you up with poetry and stones / The ragged and the bones / Strewn around the room / I recall another hazy May / Take a round in …
Light, Lyrics
31. How many Light years Away is the Sun? The Sun is at a distance of 149.6 million kilometres away from earth.
Light
32. Light year definition: A Light year is the distance that light travels in a year
Light
33. Light year: Light travels with a speed of {eq}c= 2.99\times 10^{8}~\rm{m/s}{/eq} in the space
Light
34. Becca Light year (4) is described by the brand as "Tan with neutral undertones." It is a shade in the Light Shifter Dewing Tint range, which is a tinted moisturizer foundation with a luminous finish and sheer coverage that retails for \$30.00 and contains 1 oz.
Light, Luminous
## Dictionary
LIGHT YEAR [ˈlīt ˈˌyi(ə)r]
## Frequently Asked Questions
### What is a light year and why do we use it?
Short answer: A light-year is a measurement of the distance light travels through space in a year. It is often used to express distances beyond the solar system.
### How many years are in a light year?
A light year is equal to 6 trillion years. A light year is the earth having light day and night for a whole years...DUH!!! A light year is not measured in time it is measured in distance.
### How do you explain light years?
As defined by the International Astronomical Union (IAU), a light-year is the distance that light travels in vacuum in one Julian year (365.25 days). Because it includes the word "year", the term light-year is sometimes misinterpreted as a unit of time.
### How long is a light year what does a light year describe?
A light-year is the distance light travels in a year. Specifically, the International Astronomical Union defines a light-year as the distance light travels in 365.25 days. | 1,321 | 5,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.913798 |
http://ftp.cs.rochester.edu/users/faculty/brown/172/lectures/3_alganal/3alganal.html | 1,511,476,080,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806979.99/warc/CC-MAIN-20171123214752-20171123234752-00476.warc.gz | 112,118,040 | 7,849 | ## Algorithm Analysis
Weiss Ch. 2.
How is the running time of algorithms described in a machine-independent manner? Algorithm growth rates. Calculating growth rates from code. Leads to idea of provably fast, slow, or unusably slow algorithms, where we take speed to be measured by the number of instructions executed as a function of problem size. Problems with only impossibly slow algorithms are "intractable" problems (e.g. "NP-complete" problems like traveling salesman, knapsack...). Bad: brute-force search of things that get exponentially larger with N (e.g. the number of N-digit integers is 10N). Or binary trees: 20 decisions means a million possible outcomes.
Chapter has nice friendly set of definitions and examples of complexity classes (different growth rates) and the "Big-Oh" notation. Then illustrates how to relate programming constructs like for loops to the complexity analysis. Then an extended study of four algorithms for the maximum subsequence problem with different complexity (O(n3), O(n2), O(nlogn), O(n)). Motivates appreciation for logarithmic algorithms like binary search and motivates our search for quick graph and tree algorithms for, say, search trees, dictionaries, etc.
### Definitions
Big-Oh notation. Usually you see: "this algorithm is obviously O(f(N))", or "O(g(N)) is preferred to O(f(n))", etc.
For us, the f(N) and g(N) functions are a small set of simple functions like f(N) = N, or Npower, Nlog(N), log(N), 2N. Think of them as growth rates. The definitions mean that, up to a constant of multiplication, you can show your function grows slower, faster, or the same as some other function (practically, the f() and g()'s above.)
Weiss says that his T(N)'s are functions, which could lead us to think of them as what their algorithms compute. He means they are the time functions of the associated algorithms: their running times.
FOCS, rather more coherently, says T(N) are the growth-rate functions OF the algorithms of interest. Subtle difference, but Weiss treats his T(N) like FOCS does but seems to mis-identify them, so we may get an apples and oranges feeling...
• Function T(N) is O(f(N))
if there are c, n0 > 0 ("witnesses") s.t.
T(N) ≤ cf(N) when N ≥ n0.
Your function grows no faster than f. f is an upper bound on T. (Contrariwise, T is a lower bound on f.)
• Function T(N) is Ω(g(N))
if there are c, n0 > 0 s.t.
T(N) ≥ cg(N) when N ≥ n0.
Your function grows no slower than g. g is a lower bound on T (maybe at same rate).
• Function T(N) is Θ(h(N))
iffi T(N) is O(h(n)) and T(N) is Ω (h(n)).
Your function grows at the same rate as h.
• Function T(N) is o(p(N))
iffi T(N) is O(p(n)) and T(N) is not Ω (p(n)).
Your function definitely grows slower than p.
Weiss writes T(N) = O(f(N)).
### How to Measure Growth
Computational model has a unit of operation, or instruction, like addition, multiplication, comparisons, an inner loop, often determined by the problem It is defined to take unit time since we don't care about the absolute size of the units.
Assume infinite memory, no access time.
So measure of "cost" of algorithm, or the time it takes, is the number of elementary operations it requires, taken to be the "running time".
Worst-case running times are usually used even if pessimistic: they are easier to compute. Average-case running times are of interest but sometimes not obvious how to define average, and they're hell to compute. Best case is sometimes amusing (e.g. for a sorting method, what's the best-case input order if any).
Other measures exist, like area for circuits, models including memory access, etc. Realistic computer and systems models are another story.
How measure input size? Good question. You can imagine algorithms that work with the numerical value of the input N, but also those that work on N itself as a sequence of bits, so the "length of N" is at issue, not the "value of N". Usually obvious.
### Using the Definitions Directly: Big-Oh Proofs
Proofs that T(n) is O(f(n)) are all alike: From the formal definitions above, we "need a c and an n0 such that..." So--
1. Find and state the two specific constants: a positive c and nonnegative n0, which are witnesses to your claim.
2. Use algebra to show that for n > n0, T(n) ≤ cf(n) for your n0 and c.
E.g. Suppose we show our running time is T(n) = (n+1)2. By staying awake in lecture, we observe that this is a quadratic function, O(n2). How prove it? Choose witnesses n0 = 1 (a VERY common choice for reasons we'll see) and c = 4.
This c-guessing, rabbit-from-hat effect is good drama, but we actually compute what our witness c needs to be for our n0 using the reasoning below first, then appeal to it later.
First, the obvious approach: n0 = 1 means n > 1. So we want
a c such that (n+1)2 = n2 + 2n +1 ≤ cn2
(c-1)n 2 -2n -1 = 0. Bit of a mess, headed toward a quadratic.
Use our powerful inequalities -- to hell with exact solutions.
n0 = 1 means n > 1. So now we need to prove (n+1)2 ≤ 4n2 provided n ≥ 1. Or n2 +2n+1 ≤ 4n2 as before.
Now n ≥ 1 implies both n ≤ n2 and 1 ≤ n2, so
n2 + 2n +1 ≤ n2+2n2+n2 = 4n2, QED.
Wait! How'd we guess that c=4 would work? By jumping to the last line immediately, given n0 = 1 and that you see the basic trick: all the powers of n may be changed to n2.
This n0 = 1 trick and associated reasoning lets you just glance at the claim 5n3 + 100n2 +36n +1095 ≤ cn3
And say: for n0 = 1, c=2000 will work, and the smallest c would be 1236. You'll see that if n0 ≠ 1, we're talking about computing the intersection of two cubics, given we've somehow picked a c. Very messy and doesn't scale.
### Common Growth Rates and Rules
constant, logarithmic, log-squared (log2N), N log N, linear, quadratic, cubic, polynomial, power (e.g. N1.57), exponential.
Useful Facts or Rules:
• if T(N) = O(f(N)) and S(N) = O(g(N)), then
1. T(N) + S(N) = O(f(N)+g(N)) ⇒ = O(max(f(N), g(N))).
2. T(N) * S(N) = O(f(N)*O(g(N)).
• if T(N) is a polynomial of degree k, T(N) = Θ(Nk).
• logkN [= (log(N))k] = O(N) for any k. So the log grows slowly.
Never say "O(3N)" or "O(2N2 + 4)" for example.
Never say "f(n) ≤ O(g(N))" (implied) or "f(n) ≥ O(g(N))" (nonsense).
"Just look at the derivatives?" Yeah, in fact l'Hopital's rule; how-to in Weiss p. 32. Unnecessary usually. NlogN vs N1.5? Know if know logN vs N.5. Or log2N vs. N, which by 3rd rule above, is that N grows faster.
### Algorithms to Running Times: Common Constructs
for loops: running time's at most number of all statements in the loop times number of iterations. For a singly nested loop with no function calls and N iterations, the length of the loop is some constant, so it goes away and we get O(N).
Nested loops: analyze inside-out.
Consecutive statements (or loops, function calls, etc.). Their time adds, which by one of our three early rules means the answer is the Order (Big Oh) of the maximum-time one.
Conditionals: Running time is at most the time of the test plus the time of the larger (or largest) running time of the "then", "else" or "case" etc. statements.
Function Calls: Consider the function an algorithm and recursively apply these rules!
Generally, need to work from inside out (from function calls, innermost loops, etc.).
(Really) Recursive Calls: Tail recursion is like a for loop, so easy. Otherwise, analysis leads to recurrence relations, which are a topic to themselves (for later).
### Max. Subseq. Sum
Weiss 2.4.3 = Bentley Prog. Pearls Column 8.
Problem: find the largest sum of contiguous elements in a given N- vector of numbers. If they're all negative the answer's 0 for "no elements in sum". e.g. answer for [3 -4 10] is 10, [-4 3 10] is 13.
1: Cubic Time: triply-nested for loop, outer two pick all possible left and right sub-vector elements, inner adds all elts between them, each loop has maximum of N repetitions means O(N3).
2: Quadratic Time: smarter ways to compute sum in inner loop. Just update the sum with the "current element" or pre-compute the N cumulative sums of elements (O(N) to do this) and subtract two cum-sums to get any subsequence sum. Thus inner loop becomes O(1). Morals: inner loops repay scrutiny, keeping subresults is often smart.
3: NlogN time: Typical divide and conquer approach. Cut problem in half, get maxSSS of each half, AND in O(N) time compute the max sum of subsequences that slop over the boundary you used. Report the max of these three subproblems up the recursive chain. This yields the famous Quicksort recurrence T(1) = 1, T(N) = 2T(N/2) + O(N), whose solution is O(NlogN) -- we'll visit this again, next time in our overview of Trees.
4: Linear time: Some actual thought (!!) leads to vast improvements in the first quadratic algorithm: e.g. "can't have a (seq of) negative number(s) as first member(s) of maxSS". Get an O(N) "scanning" algorithm that zips thru the sequence once.
Why do we care? Weiss doesn't but Bentley does give some dramatic statistics and tables: he implements, times the results, figures out the multiplicative constants for the O(f(N)) formulae, and finds:
for N = 1M The times are
1: 41 years. 2: 1.7 weeks. 3: 11 secs. 4: 0.48 secs.
In a minute, how big a problem (N) can be solved?
1: 3600. 2: 10,000. 3: 1M. 4: 2.1x107.
### Fast Algorithms! Log time examples
Divide and conquer, as in QuickSort say, are often NlogN (often comes from splitting (sub)problems in half).
Binary Search is common in dictionaries or phone books. Open in middle, look, go to L or R half and repeat. 20 questions means can find one in a million possible answers. O(logN) algorithm for lookup. But requires O(N) for insertion (in an array representation). Of course might need a sort to be able to use Binary Search, so that would have to be counted (or amortized as we DS professionals say.)
Euclid's algorithm is an Oldie but Goodie, a very clever way to find the greatest common divisor (largest integer dividing both) of two integers by repeatedly calculating remainders. They get smaller, and Weiss proves (Th. 2.1) that the size of the remainder goes down at worst by half every two iterations of the algorithm (one of which swaps the operands).
Th 2.1: if M>N, M mod N < M/2. Easy proof by cases N < M/2 and otherwise. So another O(logN) algorithm: divide size of problem by two every 2 iterations at worst.
Fast exponention. This is the trick of using previously computed powers (aka Dynamic Programming): so N^2 is one multiplication, times self is N^4, ditto is N^8 so there's N^8 computed with 3 = log(8) *'s, not 7. Nice little recursive algorithm (Fig. 2.11) and analysis. At most two multiplies are needed to get a problem half as big. Also Weiss discusses some coding improvements and snares in this little 9-liner, worth a look.
### Code to O(N)
```x = 150;
%O(1)
N= scanf(input);
% Size of one array dimension O(1)
Arr1 = MakeRandArr(N,2);
% make 2-D NxN random array O(N^2)
%*******insert line here
if Sum(Arr1) < x
% O(N^2)
x++;
% O(1)
else
Arr2 = MatMult(Arr1, Arr1)
% O(N^3)
PrintArr(Arr2);
%O(N^2)
```
Looks O(N3) to me, and Ω(N2). Is that right? What changes with O(N), Ω(N), and Θ(N) if we also made a 3-D array by adding line
```...
%*******insert line here
Arr3 = MakeRandArr(N,3);
...
```
### CB's Putative Project Categories
1. Visualizing and Organizing Complex Systems
-- .jar files
-- Scheduling
-- (Computer) inventories
-- Filesharing
2. Games
-- Video
-- Board (chess)
3. Embedded systems
-- Smart house, motion, temperature sensing, remote control
-- Robots (e.g. blimp)
-- Reverse engg?
4. Web Programming
-- Optimizing Javascript
-- Reverse engg?
5. AI
-- Nat. Lang. Understanding: Chatbot
-- Other NLU Apps
-- Machine learning
6. Geog. Info. Systems
7. Music
-- Analysis (genre, name-that-tune, filtering, effects...)
-- Synthesis (micro-tone scales, new timbres (instruments), modeling existing instruments (ukelele? ocarina?), composition (fractals, random, natural data, new instruments).
-- Soundtrack for animated video (see below)
8. Graphics
-- And AI: smart game mods? e.g. Quagents
-- Visualization and intelligent agents (W. Gibson meets TRAINS?)
-- Visualizing big data (VISTA Collaboratorium, Gd. floor Carlson)
-- make Animated video
-- Rendering ( 173 Raytracing Project?)
9. Theory
-- CS200 type project (Lane Hemaspaandra Advisor)
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# K mitra d d d 0 if the above constraint is not
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Unformatted text preview: (e jω ) − X im (e jω ) re dω Copyright © 2005, S. K. Mitra 5 The Unwrapped Phase Function The Unwrapped Phase Function • The phase function can thus be defined unequivocally by its derivative dθ(ω) / dω: ω θ(ω) = ∫[ 0 dθ( η) dη • The phase function defined by ω θ(ω) = ∫[ dη, 0 dθ( η) dη dη is called the unwrapped phase function of X (e jω ) and it is a continuous function of ω • ⇒ ln X (e jω ) exists with the constraint θ(0) = 0 31 32 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra The Unwrapped Phase Function The Frequency Response • Moreover, the phase function will be an odd function of ω if 1 π 2π ∫[ 0 dθ( η) dη dη = 0 • If the above constraint is not satisfied, then the computed phase function will exhibit absolute jumps greater than π 33 34 • Most discrete-time signals encountered in practice can be represented as a linear combination of a very large, maybe infinite, number of sinusoidal discrete-time signals of different angular frequencies • Thus, knowing the response of the LTI system to a single sinusoidal signal, we can determine its response to more complicated signals by making use of the superposition property Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra The Frequency Response...
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Ask a homework question - tutors are online | 428 | 1,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-05 | latest | en | 0.809849 |
http://www.madsci.org/posts/archives/may97/862334110.Ph.r.html | 1,600,539,996,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400192783.34/warc/CC-MAIN-20200919173334-20200919203334-00025.warc.gz | 197,190,630 | 3,253 | ### Re: Why does light change speed when it enters a new medium?
Area: Physics
Posted By: Greg Dries, Senior Research Engineer,U. S. Steel Technical Center
Date: Mon Apr 28 10:42:42 1997
Area of science: Physics
ID: 861237548.Ph
Message:
To answer this question one needs to understand the nature of electromagnetic radiation (light) as well as the atomic structure of matter. Light is a traveling electromagnetic wave and therefore consists of an alternating magnetic field and an alternating electric field (each field inducing the other). James Clerk Maxwell discovered the laws governing electromagnetism which are the basis for our understanding of light. Consult any physics text book for details regarding Maxwell's work Maxwell's electromagnetic theory leads to the equation: c=square root of (1/(permeability constant times the permittivity constant), where c is the speed of light in a vacuum with the permittivity and permeability constants being for a vacuum as well. If the light wave is not traveling in a vacuum, these constants then become the permittivity and permeability values for whatever the media is, air, water, glass, etc. Understanding exactly what these constants are and what they relate to gets to the root of how electromagnetic radiation interacts with transparent materials. Without getting too deep, these constants are fundamental in much the same way as the gravitational constant relates to mass, except permittivity relates to electric fields and charged particles, and permeability relates to magnetic fields.
Because most all transparent materials are nonmagnetic, the permeability of transparent materials is essentially the same as the permeability of a vacuum, and therefore the speed of light in transparent media is dependent primarily on the permittivity of that media. The permittivity of solids is relatively large due to the interaction of the electromagnetic radiation with dipoles in the solid. These dipoles arise specifically from the types of atoms present and the nature of the chemical bonds among the atoms. The greater the density of the solid, the greater the number of dipoles per unit volume and the greater will be the change in the speed of light. Materials which exhibit the higher indexes of refraction produce a greater change to the speed of light within that particular solid. For the different types of glass, the atomic number of the cation constituents will control the index of refraction (lead-containing glass will have a much higher index than regular glass) because lead is a heavier atom. For plastics, the index of refraction is probably more a factor of the degree of cross-linking in the structure and the types of chemical bonds involved with the cross-linking, rather than the atoms themselves.
I hope this rather lengthy explanation helps. The science of materials is very fascinating and makes an excellent career choice. Look into it if your considering going onto college.
Sincerely
Greg Dries
Current Queue | Current Queue for Physics | Physics archives | 595 | 3,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-40 | latest | en | 0.909154 |
http://maitespace.com/math/pseo/da1011/p3w02.htm | 1,516,772,353,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893397.98/warc/CC-MAIN-20180124050449-20180124070449-00092.warc.gz | 223,164,029 | 1,764 | PSEO Algebra Daily Agendas: January 24 - January 28
PSEO Algebra Monday - 1/24/11 Bell Task a. Collect Homework problems b. Questions on §1.6 Function Transformation Assignment Business a. Chapter Test #1 on Friday Workshop §1.7 The Algebra of Functions a. Domains of Compositions of Functions b. Applications of Compositions of Functions c. Decomposition of Functions Exit Task a. Individual workshop, if time permits DW a. §1.7 p. 80: 2 - 4, 6, 8, 11 - 25 odd, 26, 27, 29, 31, 35 - 38, 41, 43 - 45, 47, 50 - 52 PSEO Algebra Tuesday - 1/25/11 Bell Task a. Collect Homework problems, due by end of the day b. Questions on §1.7 p. 80: 2 - 4, 6, 8, 11 - 25 odd, 26, 27, 29, 31, 35 - 38, 41, 43 - 45, 47, 50 - 52 Business a. Chapter Test #1 on Friday Workshop §1.8 Inverse Functions a. Inverse Functions b. Algebraically Finding Inverse Functions c. Graphs of Inverse Functions d. Domain & Range, VLT & HLT Exit Task a. Individual workshop, if time permits DW a. §1.8 p. 91: 3 – 18 every 3rd, 20, 23, 24, 27, 31, 38, 39, 42 – 48 even, 51, 54, 59, 61 PSEO Algebra Wednesday - 1/26/11 Bell Task a. Questions on §1.8 p. 91: 3 – 18 every 3rd, 20, 23, 24, 27, 31, 38, 39, 42 – 48 even, 51, 54, 59, 61 Business a. Chapter Test #1 on Friday Workshop a. Chapter Test on p. 101: 1 - 18 b. Workshop time on §1.8 p. 91: 3 – 18 every 3rd, 20, 23, 24, 27, 31, 38, 39, 42 – 48 even, 51, 54, 59, 61 Exit Task a. NA DW a. §1.8 p. 91: 3 – 18 every 3rd, 20, 23, 24, 27, 31, 38, 39, 42 – 48 even, 51, 54, 59, 61 PSEO Algebra Thursday - 1/27/11 Bell Task a. NA Business a. Chapter Test #1 Tomorrow Workshop a. Questions on §1.8 p. 91: 3 – 18 every 3rd, 20, 23, 24, 27, 31, 38, 39, 42 – 48 even, 51, 54, 59, 61 b. Work on Chapter Test on p. 101: 1 - 18, Self-check with answers in back Exit Task a. NA DW a. Chapter Test on p. 101: 1 - 18 PSEO Algebra Friday - 1/28/11 Bell Task a. Chapter 1 Unit Test Business a. NA Workshop a. NA Exit Task a. NA DW a. NA | 830 | 1,995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2018-05 | longest | en | 0.48784 |
https://forexsb.com/forum/post/62301/ | 1,695,885,186,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510368.33/warc/CC-MAIN-20230928063033-20230928093033-00803.warc.gz | 281,854,928 | 7,715 | #### Topic: Correlation calculations correct?
Hi Miroslav,
What is the correlation formula used to calculate correlation between strategies?
EA Studio seems to think that the above strategies are correlated. The correlated setting is set to .98 meaning the strategies should basically be identical but a quick visual inspection of the equity curves indicates otherwise.
Furthermore if I remove strategy 7.3 it removes the correlation for for 4.1, 7.1 and 7.2 indicating that those strategies are correlated to 7.3, but not each other. This leaves a total of 5 strategies.
However if I let EA Studio handle the removal of correlated strategies automatically, it deletes the 4 "correlated" strategies leaving me with a total of 2 strategies.
Why does EA Studio think the 4 strategies are correlated?
What are the 4 strategies correlated to?
Why does EA Studio remove the 4 strategies instead of just removing the 7.3, leaving me with more strategies. If the strategies are in my collection it means they have met my acceptance criteria. And if they met my acceptance criteria I want more of them, not less.
#### Re: Correlation calculations correct?
The correlated setting is set to .98 meaning the strategies should basically be identical but a quick visual inspection of the equity curves indicates otherwise.
It is a matter of perception and settings.
> Why does EA Studio think the 4 strategies are correlated?
Because the Correlation calculations show a Correlation Coefficient equal to or higher than 0.98. Please see more details here: Pearson correlation coefficient.
The formula is pretty simple and well checked. You may check it also ( the code is below). There aList and bList are the balance curves of two strategies.
> What are the 4 strategies correlated to?
EA Studio checks each strategy against all the rest in the Collection.
> Why does EA Studio remove the 4 strategies instead of just removing the 7.3,
Because EA Studio leaves the strategy with the highest goal ( according to Sort collection by ) form the correlated couples.
> If the strategies are in my collection it means they have met my acceptance criteria. And if they met my acceptance criteria I want more of them, not less.
Excellent! You already figured it out. When you switch the automatic correlation resolution, EA Studio does not remove strategies.
`````` private getCorrelation(aList: number[], bList: number[]): number {
const n: number = Math.min(aList.length, bList.length);
const a1: number = aList[0];
const b1: number = bList[0];
let sumX: number = 0;
let sumY: number = 0;
let sumXX: number = 0;
let sumYY: number = 0;
let sumXY: number = 0;
for (let i: number = 1; i < n; i++) {
const x: number = aList[i] - a1;
const y: number = bList[i] - b1;
sumX += x;
sumY += y;
sumXX += x * x;
sumYY += y * y;
sumXY += x * y;
}
const c: number = n - 1;
const covariation: number = sumXY / c - sumX * sumY / c / c;
const aSigma: number = Math.sqrt(sumXX / c - sumX * sumX / c / c);
const bSigma: number = Math.sqrt(sumYY / c - sumY * sumY / c / c);
const correlation: number = covariation / aSigma / bSigma;
return correlation;
}``````
#### Re: Correlation calculations correct?
I decided to check the calculations again (also with an alternative formula).
The calculated Correlation coefficients are shown on the right (my strategies ID numbers are different than yours).
Alternative formula (code below):
..
I noticed that there actually 3 distinct strategies - those with different major ID number: 1.1, 4.1, and 7.1
Strategies with IDs 7.2 and 7.3 are modified versions of 7.1 with the same trading rules but different numbers.
I don't think I can add more to this topic.
`````` private getCorrelation(aList: number[], bList: number[]) {
const n: number = Math.min(aList.length, bList.length);
const a1: number = aList[0];
const b1: number = bList[0];
if (n === 0) {
return 0;
}
let meanX: number = 0;
let meanY: number = 0;
for (let i = 0; i < n; i++) {
meanX += (aList[i] - a1) / n
meanY += (bList[i] - b1) / n
}
let cov: number = 0;
let den1: number = 0;
let den2: number = 0;
for (let i = 0; i < n; i++) {
const dx: number = (aList[i] - a1) - meanX;
const dy: number = (bList[i] - a1) - meanY;
cov += dx * dy
den1 += dx * dx
den2 += dy * dy
}
const den = Math.sqrt(den1) * Math.sqrt(den2);
return den === 0 ? 0 : cov / den;
}``````
#### Re: Correlation calculations correct?
Thanks Miroslav.
The alternative calculations looks very similar to the original.
Are there more pages to your correlation calculations? I ask because 6 strategies with a correlation matrix (to each other) should produce 30 unique correlations statistics, (6^6) - 6, but I can't see the rest of the calculations.
Excellent! You already figured it out. When you switch the automatic correlation resolution, EA Studio does not remove strategies.
What I'm trying to say is, if the strategies are in my collection that means that they are "good enough". They met my acceptance criteria, regardless of how I decide to sort them in the Collection.
At the end, after correlation analysis, I would much rather have 5 uncorrelated strategies (4 if removing the other correlated strategy) than 2, regardless of the top "Sort collection by" metric because by virtue of being in the collection they are already deemed acceptable.
Is this possible to add a setting to the Correlation Analysis settings where the user can decide what strategies get deleted? Either A) all correlations are deleted excluding the one with the highest Sort by value (current behaviour) or B) the strategy with the highest number or correlations is deleted (if only 1 correlation then the strategy with the highest Sort by value is kept).
An example of how it would look.
A) Strategies 7.1 and 1.1 are kept.
B) Strategies 7.3 and 1.1 or 1.2 (depending on the Sort by value) are discarded. We will discard 1.1 in this example.
I could perform the correlation analysis myself but EA Studio doesn't provide a correlation matrix. The only way to perform it manually is to save the collection, then randomly start deleting strategies and seeing which one results in less correlations and ultimately the most strategies at the end, Each time re-importing the original collection (6 in this case) and repeating.
-----
What are the list of values of aList and bList? Are they arrays of trade balances or daily balances?
#### Re: Correlation calculations correct?
I could perform the correlation analysis myself
Yes, you can. The exported Collection file contains a summary of the Balance and Equity charts (compressed to 360 values ( don't remember exactly) and rounded).
You can manage the Collections with your own tools. | 1,628 | 6,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.894928 |
https://www.electrosmash.com/forum/time-manipulator/400-time-manipulator-circuit-analysis | 1,723,597,684,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641086966.85/warc/CC-MAIN-20240813235205-20240814025205-00768.warc.gz | 582,837,561 | 14,917 | # Time Manipulator - Circuit Analysis
5 years 9 months ago - 4 years 9 months ago #1544 by Ray
The circuit could be broken down into 6 blocks: The Power Supply, the Input Buffer, the Dual PT2399 Stage, Output Adder, Analogue Switches, and the Arduino Block:
The Power Supply
The Power Supply generates energy and bias voltage for all the circuit:
The LM7805 is a +5V linear regulator widely available.
• +9V is used to power the Input and Output op-amps, it gives maximum span to avoid signal clipping.
• +4.5V is used as a virtual ground for the op-amps.
• +5V is used to power the two PT2399, the Arduino chip (ATMEGA328P-PU) and the analog switch (CD4066). A legion of caps is used to avoid noise on the supply lines.
In this mixed analog/digital circuit, the power supply architecture is critical in order to minimize noise. A ground star scheme is used, with 2 different sub-grounds (Analog and Digital) that joint in a single point (below the PT2399 pins 3 and 4). The +5V and +9V lines are also routed avoiding conflicts between analog and digital domains.
The Input Buffer
The Input Buffer is a simple op-amp with gain=1. It will present a high input impedance (Zin=470K) and prepare the signal for the rest of the circuit:
The input cap C1 (100nF) will block any DC levels and creates a low pass filter with R1. The fc is 3.3Hz so it won't affect the audio band (fc=1/2piRC=3.3Hz)
The Dual PT2399 Stage
The two PT2399 are in the classic Delay configuration. Not taking into consideration the analog switches the circuit looks like this:
If you want to learn all the details about the PT2399 details read the PT2399 Analysis . The resistors and caps used (R3=R8=R9=4.7K, C7=6.8nF and C11=3.3nF) give us a fc=7.1Khz, they will remove the high harmonics and DAC noise. This low pass filtering is intended in order to recreate the warm organic tone of analog delays. If the delayed signal is not filtered, the resulting echoed sound would be too clean and sterile. Air and walls tend to low pass the signal, attenuating the harsh hi-freqs, this 7.1KHz filter will do that,
How to control a PT2399 delay time with Arduino:
The Arduino chip will control the PT2399 delay time by controlling the amount of current flowing out of pin 6:
• 50uA: 600ms delay time (maximum time with minimum current)
• 5mA: 35ms delay time (minimum time with maximum current)
An op-amp current sink is used to control the amount of current thus delay that the PT2399. The microntroller will use a PWM signal that will vary between 0 and 5 volts to set the desired current and delay time:
• The R4-R5 voltage divider will convert the 0 to 5 PWM signal coming from the Arduino into 0 to 2.5V.
• The C8 cap will transform the PWM signal into a continuous signal.
The voltage drop (0 to 2.5V) over the resistor R13 will command the amount of current pulled out of the PT2399 pin6:
• PWM min (0V): I=V/R = 0/4.7K = 0mA
• PWM max (2.5V): I=V/R = 2.5/4.7K = 0.53mA
The R13 Resistor:
To control the PT2399 using current, ideally, we need to modulate it from 5.4mA (35ms) to 0.05mA (600ms). With R=V/I and V=2.5V, the resistor should be:
R=V/Imax =2.5V/5.4mA= 460R (35mS)
Using a 460R resistor, we can modulate the delay from 35ms, to 600ms.
In practice, we are using an 8bit PWM to generate the 2.5V. If we use a 460 resistor, we would be able to get very nice resolution close to the 35ms mark, but when we try long delays (around 200ms) the resolution is very bad.
To fix this a 4.7K resistor is used, it gives a fantastic behavior and better resolution in the whole range but with the side effect is that the shortest delay we could achieve is around 60ms.
With a 4.7K resistor, we ensure that we can cover the whole span of time from 0mA ( 300 seconds) to 0.50mA ( 60 mseconds)
It is the last stage of the circuit, it will give a low output impedance to preserve the sound quality on the pedal chain. Its main task is to add the original guitar signal together with the different delay paths or "taps":
The output signal will be a mix of the dry signal (Audio_in) + Delayed signal (Output_Tap) + Middle Tap (Middle_Tap). Two switches will give the possibility of adding or not some of the taps. The amount of dry/wet signal will be controlled by RV2.
The 100ohms output impedance will depend on the op-amp, usually, all modern op-amps have a Zout<100ohms.
The Analog Switches
The Time manipulator uses 4 analog switches (CD4066). They will allow the signal to follow different paths and create different sounds. You will have the possibility to place the PT2399 in series (extending the time and definition of the delay time) or in parallel (creating different sounds like reverb or chorus).
The principal configurations are:
Briefly going over the most important configurations, we have:
• Short Delay: The first PT2399 is not used, this mode archive delays from 50ms to 300ms approx. Simple and clean.
• Delay: This is the basic configuration, the 2 PT2399s are used in series, getting delays from 100ms to 600ms approx. The Tails pot will add more depth to this mode.
• Echo Mode1 and 2: The basic Delay mode could be enriched, adding extra tap paths for the signal.
• Super Echo: All the relays are on, giving a no-stopping bouncing sound.
• Parallel - Chorus: The two delay units are placed in parallel, using subtle different delay times on each one will give a chorus-like sound.
• Reverb: The idea is to add to the original guitar signal two paralleled delayed copies (mimicking the signal bouncing over 2 walls). The delays need to be short here to have a realistic effect.
• Telegraph: All the relays are OFF and we the foot tap we can control the dry signal going through the pedal.
The Arduino Block:
The Arduino UNO chip (ATMEGA328P-PU) is the brain on the circuit. Its main tasks are:
• Control the delay times for the PT2399s using 2 PWMs.
• Read the encoder and the push button.
• Drive the LEDs and the analog switches.
Details of the circuit:
• The cap C1 placed next to the chip will remove noise from the power supply.
• Two dual color LEDs are used, so using 2 pins we can get 4 status: off, green, red and orange.
• An encoder is used to interact with the guitarist, it as an embedded push-button to access the effect-selection mode.
• 2 PWM signals are used to control the current sink circuits of the PT2399 (and control the delay time)
• The Tap pushbutton is connected to the Arduino, so it could be detected | 1,687 | 6,431 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-33 | latest | en | 0.853734 |
https://sisl.readthedocs.io/en/latest/api/physics.brillouinzone.html | 1,717,030,786,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00193.warc.gz | 450,956,330 | 8,131 | # Brillouin zone
The Brillouin zone objects are all special classes enabling easy manipulation of an underlying physical quantity.
Quite often a physical quantity will be required to be averaged, or calculated individually over a number of k-points. In this regard the Brillouin zone objects can help.
The BrillouinZone object allows direct looping of contained k-points while invoking particular methods from the contained object. This is best shown with an example:
>>> H = Hamiltonian(...)
>>> bz = BrillouinZone(H)
>>> bz.apply.array.eigh()
This will calculate eigenvalues for all k-points associated with the BrillouinZone and return everything as an array. The dispatch property of the BrillouinZone object has several use cases (here array is shown).
This may be extremely convenient when calculating band-structures:
>>> H = Hamiltonian(...)
>>> bs = BandStructure(H, [[0, 0, 0], [0.5, 0, 0]], 100)
>>> bs_eig = bs.apply.array.eigh()
>>> plt.plot(bs.lineark(), bs_eig)
and then you have all eigenvalues for all the k-points along the path.
## Multiple quantities
Sometimes one may want to post-process the data for each k-point. As an example lets post-process the DOS on a per k-point basis while calculating the average:
>>> H = Hamiltonian(...)
>>> mp = MonkhorstPack(H, [10, 10, 10])
>>> E = np.linspace(-2, 2, 100)
>>> def wrap_DOS(eigenstate):
... # Calculate the DOS for the eigenstates
... DOS = eigenstate.DOS(E)
... # Calculate the velocity for the eigenstates
... v = eigenstate.velocity()
... V = (v ** 2).sum(1)
... return DOS.reshape(-1, 1) * v ** 2 / V.reshape(-1, 1)
>>> DOS = mp.apply.average.eigenstate(wrap=wrap_DOS, eta=True)
This will, calculate the Monkhorst pack k-averaged DOS split into 3 Cartesian directions based on the eigenstates velocity direction. This method of manipulating the result can be extremely powerful to calculate many quantities while running an efficient BrillouinZone average. The eta flag will print, to stdout, a progress-bar. The usage of the wrap method are also passed optional arguments, parent which is H in the above example. k and weight are the current k-point and weight of the corresponding k-point. An example could be to manipulate the DOS depending on the k-point and weight:
>>> H = Hamiltonian(...)
>>> mp = MonkhorstPack(H, [10, 10, 10])
>>> E = np.linspace(-2, 2, 100)
>>> def wrap_DOS(eigenstate, k, weight):
... # Calculate the DOS for the eigenstates and weight by k_x and weight
... return eigenstate.DOS(E) * k[0] * weight
>>> DOS = mp.apply.sum.eigenstate(wrap=wrap_DOS, eta=True)
When using wrap to calculate more than one quantity per eigenstate it may be advantageous to use oplist to handle cases of BrillouinZone.apply.average and BrillouinZone.apply.sum.
>>> H = Hamiltonian(...)
>>> mp = MonkhorstPack(H, [10, 10, 10])
>>> E = np.linspace(-2, 2, 100)
>>> def wrap_multiple(eigenstate):
... # Calculate DOS/PDOS for eigenstates
... DOS = eigenstate.DOS(E)
... PDOS = eigenstate.PDOS(E)
... # Calculate velocity for the eigenstates
... v = eigenstate.velocity()
... return oplist([DOS, PDOS, v])
>>> DOS, PDOS, v = mp.apply.average.eigenstate(wrap=wrap_multiple, eta=True)
Which does mathematical operations (averaging/summing) using oplist.
In some cases quantities are needed for all $$k$$ points and in such cases it may not always be that the returned quantities are commensurate. Lets re-use the previous wrap_multiple function and try and return the full quantity:
>>> DOS_PDOS_v = mp.apply.eigenstate(wrap=wrap_multiple, eta=True)
This will raise an error since wrap_multiple returns an oplist (same as a list) and thus is unable to convert this into an equivalent numpy.ndarray. Additionally this can not be merged together in a single numpy.ndarray since the shapes of the returned quantities are not commensurate. One cannot concatenate the 3 different quantities.
To accomblish this one may use an zip flag where the two lines are equivalent:
>>> DOS, PDOS, v = mp.apply.array.renew(zip=True).eigenstate(wrap=wrap_multiple, eta=True)
>>> DOS, PDOS, v = mp.apply(zip=True).array.eigenstate(wrap=wrap_multiple, eta=True)
and the data is unpacked as wanted.
## Parallel calculations
The apply method looping k-points may be explicitly parallelized. To run parallel do:
>>> H = Hamiltonian(...)
>>> mp = MonkhorstPack(H, [10, 10, 10])
>>> with mp.apply.renew(pool=True) as par:
... par.array.eigh()
This requires you also have the package pathos available. The above will run in parallel using a default number of processors in priority:
1. Environment variable SISL_NUM_PROCS
2. Return value of os.cpu_count().
Note that this may interfere with BLAS implementation which defaults to use all CPU’s for threading. The total processors/threads that will be created is SISL_NUM_PROCS * OMP_NUM_THREADS. Try and ensure this is below the actual core-count of your machine (or the number of requested cores in a HPC environment).
Alternatively one can control the number of processors locally by doing:
>>> H = Hamiltonian(...)
>>> mp = MonkhorstPack(H, [10, 10, 10])
>>> with mp.apply.renew(pool=2) as par:
... par.eigh()
which will request 2 processors (regardless of core-count). As a last resort you can pass your own Pool of workers that will be used for the parallel processing.
>>> from multiprocessing import Pool
>>> pool = Pool(4)
>>> H = Hamiltonian(...)
>>> mp = MonkhorstPack(H, [10, 10, 10])
>>> with mp.apply.renew(pool=pool) as par:
... par.array.eigh()
The Pool should implement some standard methods that are existing in the pathos enviroment such as Pool.restart and Pool.terminate and imap and uimap methods. See the pathos documentation for detalis.
BrillouinZone(parent[, k, weight]) A class to construct Brillouin zone related quantities MonkhorstPack(parent, nkpt[, displacement, ...]) Create a Monkhorst-Pack grid for the Brillouin zone BandStructure(parent, *args, **kwargs) Create a path in the Brillouin zone for plotting band-structures etc. | 1,494 | 6,059 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-22 | latest | en | 0.741239 |
https://sports.stackexchange.com/questions/29220/how-many-earned-runs-in-this-scenario | 1,716,625,118,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058789.0/warc/CC-MAIN-20240525065824-20240525095824-00855.warc.gz | 453,140,026 | 40,107 | # How many earned runs in this scenario?
The pitcher walks 3 batters to start the inning. Infield in, the next batter hits it to right field and the fielder makes and error and all 3 runners on score. Man on second, no outs the next batter hits a home run. 5-0 now and then the pitcher gets 3 straight outs. How many earned runs for the pitcher?
• Please clarify: the error by the outfielder was not catching a ball he should have caught, correct? Oct 19, 2023 at 3:26
• Oct 19, 2023 at 22:56
9.16 Earned Runs and Runs Allowed
An earned run is a run for which a pitcher is held accountable. In determining earned runs, the Official Scorer shall reconstruct the inning without the errors (which exclude catcher’s interference) and passed balls, giving the benefit of the doubt always to the pitcher in determining which bases would have been reached by runners had there been errorless play.
So you replay the inning as if the error had not occurred and determine which would have scored and which would not.
There's lots of ways a right fielder might make a play, but in none of them would I assume a second out is possible. The runner on third will either remain there or will score on the fly ball if the catch is deep.
Without the error, here's one way the inning could have played out:
``````B1: BB
B2: BB
B3: BB
B4: F9 (1 out)
B5: HR (4 runs)
B6: out (2 out)
B7: out (3 out)
``````
In this scenario, B4's run is unearned, the other 4 are earned.
A game that played similar to this in MLB is the August 5, 2023 meeting between the Braves and the Cubs.
In the bottom of the first, 2 runs score (and B4 reaches) on an error at first base. But although they came in on the error, these runners "would have" scored due to a home run by B5.
2 runs in that inning are unearned: B4 (who reached on the error), and B7 (who homered after 2 outs). The other 3 runs were all charged to the pitcher.
• Unfortunately this answer only quotes the opening paragraph of the rule and misses: (d) No run shall be earned when the scoring runner’s advance has been aided by an error, a passed ball or defensive interference or obstruction, if in the Official Scorer’s judgment the run would not have scored without the aid of such misplay. Oct 19, 2023 at 4:22
• @Damila ... "if it would not have scored without the aid of such misplay" is not exactly meaning what you think. Bases loaded, no outs, fly ball to outfield; (a) probably one scores anyway on a sac fly, but even if they all stay, the next batter hit a home run. Seems like those runs would've scored then, no?
– Joe
Oct 20, 2023 at 22:55
One, the batter that hit the home run.
The three that walked scored on an error, that being the dropped catch by the outfielder.* That batter reached base on an error. So all of those are unearned, as they occurred with the help of an error.
https://www.mlb.com/glossary/standard-stats/earned-run
An earned run is any run that scores against a pitcher without the benefit of an error or a passed ball.
*This answer assumes that was the error. If you mean it was a hit- say a single- that the fielder misplayed, the analysis changes.
• The runs that would likely have scored anyway will end up earned (see BowlOfRed's example). Also likely that if the scorer saw the runner at third tagging up, they would have counted that run as earned even with the error assuming the ball was far enough for a sacrifice fly to be likely to be the result of a caught ball.
– Joe
Oct 20, 2023 at 22:59 | 877 | 3,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-22 | latest | en | 0.979281 |
http://www.webdesign.org/3d-graphics/tutorials/building-a-necklace-with-2-hearts.534.html | 1,454,727,737,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701145751.1/warc/CC-MAIN-20160205193905-00069-ip-10-236-182-209.ec2.internal.warc.gz | 746,598,991 | 10,480 | SEARCH
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Tutorials 3D Graphics Tutorials Building a Necklace with 2 Hearts
# Building a Necklace with 2 Hearts
Razvan Maftei Tutorials Jul 29, 2004
This is a simple tutorial that will teach you how to build a necklace with 2 hearts. The techniques that will be used are edit mesh and mesh smooth.
Building the heart: create a 3x4x2 box like in the image:
Click to enlarge
Apply an edit mesh modifier on the box and start modeling from the top view:
Click to enlarge
Move vertexes, or select them 2 by 2 (from the edges) and then scale down like in the images:
this is how the object should look now:
Next, select all the edges from the middle of the object, and then scaled them :
Click to enlarge
Now in the top view select the 2 edges marked with red (in fact you will select 4 edges, 2 edges on top and 2 on bottom but be sure not to have the "Ignore Backfacing" options checked in the edit mesh menu). Then move or scale (if you select all the 4 edges, if not, you will have to move the vertices for top and for bottom part):
Click to enlarge
Click to enlarge
you must do the same with other 4 + 4 vertices.
Click to enlarge
Next apply a mesh smooth with 3 iteration:
Click to enlarge
Create a torus and place it near the heart closely enough to create the sensation that was glued to the object. Then select and link the torus to the heart (this makes the heart the parent of the torus). Next select the torus and the heart and copy them (press shift and move the objects) to a location close to the original heart. Also create a plane and place it beneath the hearts. Then select the first heart and rotate it on the y and x axes. Now you do the same with the second heart but also you rotate it on the z axes, too. Be careful when rotating that the hearts don't intersect the plane.
Click to enlarge
The larger link is made from a torus with few segments and sides that I modified using edit mesh.Then Mesh smooth.
Click to enlarge
The normal links are made the same, using a torus with few faces: I selected the left vertices and rotated them on y axes. You do the same with the right vertices. Then you mesh smooth them.
Click to enlarge
Now you must create a spline that lies on the plane. Make sure that the spline is passing thorough the middle torus.
Click to enlarge
Next you'll use the spacing tool to multiply the smaller link to make the necklace. The spacing tools works just like the array tool only has some additional options, like choosing the spline you want your objects to follow . Be sure to check the options like in the second image:
Click to enlarge
This is how the scene should look:
Click to enlarge
subscribe to newsletter | 627 | 2,727 | {"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} | 2.59375 | 3 | CC-MAIN-2016-07 | latest | en | 0.876894 |
http://forum.allaboutcircuits.com/threads/voltage-drop-switch.2580/ | 1,485,228,110,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560283689.98/warc/CC-MAIN-20170116095123-00165-ip-10-171-10-70.ec2.internal.warc.gz | 102,909,237 | 16,901 | # Voltage drop switch
Discussion in 'The Projects Forum' started by skunky, May 5, 2006.
1. ### skunky Thread Starter New Member
May 5, 2006
9
0
hi, I am working on a lil project which will automatically shut off when a voltage drop is detected. I know this is just as easy as a low battery indicator but instead of driving a LED I need a relay to turn on a high current device so I use a relay but I can't figure out which transitor to turn on the relay.
Enviroment:
For a high current application >15A
when voltage drop from 14 to 13, power OFF.
otherwise ON.
so 14V ON --> 13V OFF
Here's what I came up but need some suggestion.
a NPN transistor
13V Zdiode
12V relay + 1N4001
I am stuck on which transistor I should use. Can anyone help?
2. ### thingmaker3 Retired Moderator
May 16, 2005
5,072
6
How much current does the coil of your relay need? If 0.5A or less, try a 2N2222.
3. ### skunky Thread Starter New Member
May 5, 2006
9
0
Thanks, The relay needs about 60mA( 12V, 201 Ω )
I checked the spec of the base voltage of 2N2222 is around 1.3V but I need it to precisely turn off when it's below 13V instead of partially off.
What's the minimum I and V require to activate the transistor?
I did some calculation here:
I(base) = 72mA/ 100 = 0.72mA = 0.00072A
Vcc= 14V
Zdiode = 13V
(14-13)/0.00072 = 1388 Ω (R1)
Is the minimum base current correct? so anything belows 0.72mA would turn off the transitor or partially off?
Thanks
4. ### hgmjr Moderator
Jan 28, 2005
9,030
214
I think your circuit would benefit from a base-return resistor. A base-return resistor is a resistor that is connected between the base and ground. A base-return resistor insures that current is pulled out of the base when you need the transistor to turn off.
hgmjr
5. ### skunky Thread Starter New Member
May 5, 2006
9
0
Thanks, how big of the resistor should I use ? Power dissipation is an issue so as low as possible for energy saving purpose.
6. ### thingmaker3 Retired Moderator
May 16, 2005
5,072
6
Perhaps I misunderstand, but won't the relay open if Ic drops below sustaining current for the coil?
You could use a potentiameter on your base to fine-tune for your operating requirements.
Power dissipation shouldn't be much of an issue when the voltage across the potentiameter (or voltage divider, if no potentiameter is chosen) is only a volt or two.
7. ### skunky Thread Starter New Member
May 5, 2006
9
0
that 's what I am concerning and a potentiameter sounds like a good idea since no transistors are 100% acurate as their spec shown. Thanks.
Anyone know how big of a base-return resistor I should use ?
8. ### Gorgon Senior Member
Aug 14, 2005
113
0
Hi,
First of all, change your zener diode to 12V, this will give you headroom for the Vbe of 0.6-0.7V of the transistor. Use a 1k pot with the tap connected to the base to trim the release point at 13V.
TOK
9. ### skunky Thread Starter New Member
May 5, 2006
9
0
Thanks, 2V should be enough to fine tune the transistor. Should I also use a potentialmeter for base-return resistor ? I've included my new schematic, is this correct ?
10. ### Ron H AAC Fanatic!
Apr 14, 2005
7,050
657
You really should have some hysteresis, to keep the relay from chattering, and you at least need a resistor in series with the pot or with the base, to limit base current when you run the pot to the zero ohms end. Here is a circuit with hysteresis that will switch cleanly and be very temperature stable.
11. ### thingmaker3 Retired Moderator
May 16, 2005
5,072
6
This is what I had in mind when suggesting a potentiameter:
[attachmentid=1418]
12. ### skunky Thread Starter New Member
May 5, 2006
9
0
BIG thanks ! I'll build my prototype see how it goes. Just wondering...is there any chip that does this whole thing ?
Thanks
Jan 10, 2006
613
0
I agree with Ron H. You may find that as soon as the load is disconnected the battery voltage may rise enough to switch the transistor back on so the circuit may cycle.
Looking at Ron's diagram, thats pretty much what I would use, although I think I'd be putting a resistor inline with the Transistor base, and I'm pretty sure that if you reduce R2 (1M) you should be able to increase the Hysteresis (differentual).
14. ### skunky Thread Starter New Member
May 5, 2006
9
0
thanks again for the help, I've looked over every electronic supplier here but none of them carry 1N5994B , is there any alternative I can use? and the 10k potentiometer i found seems a bit too big, can I use a 10k trimpot instead ?
Thanks
15. ### thingmaker3 Retired Moderator
May 16, 2005
5,072
6
Try 1N4734 or BZX84B5V6 for the 5.6v zener.
Trim pot would be my first choice. Set for optimum operation and lock in with a wee drop of nail polish.
Dec 2, 2005
67
0
17. ### Ron H AAC Fanatic!
Apr 14, 2005
7,050
657
LM393 has an open collector output. You don't need another base resistor.
And you are right about increasing the hysteresis. | 1,397 | 4,934 | {"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} | 3.265625 | 3 | CC-MAIN-2017-04 | latest | en | 0.923824 |
https://www.physicsforums.com/threads/how-do-you-know-which-factor-to-use.205664/ | 1,638,352,111,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359976.94/warc/CC-MAIN-20211201083001-20211201113001-00040.warc.gz | 990,866,268 | 16,801 | # How do you know which factor to use?
How do you know which factor to use???
I know what factoring is. Basically, you take any given number or expression, and rewrite it as a product of its factors.
For instance in 3xy + 9y, every term is divisable by "3y", so we could rewrite it as 3y(x + 3).
But, consider: $$\displaystyle{28xy^2 - 14x}$$
Every term there is divisible by "14x", so is there something at all wrong with: $$\displaystyle{14x(2y^2 - 1)}$$???
That's how I would factor it, but the book instead choose to use "7x" as the factor, and uses $$\displaystyle{7x(4y^2 - 2)}$$ as the answer.
If given an exam, how do you know which factor they are going to claim is the "correct" one????
Thanks a lot.
Neither factorization is incorrect, but if asked to simplify an expression by factoring, the book's answer would not be acceptable.
Neither factorization is incorrect, but if asked to simplify an expression by factoring, the book's answer would not be acceptable.
Thanks.
So basically just go with the greatest possible factor?
uart
Thanks.
So basically just go with the greatest possible factor?
Yes in general you should pull out all the factors that you can. In that book answer there is still a factor or 2 remaining inside the brackets, so you would typically lose marks for giving that answer.
As slider pointed out neither of the two factorizations is incorrect. Both expressions are in a factorized form and both are equal to the original expressions. There could even be instances where you may prefer the incomplete factorization. Say you had to simplify
$$\frac{28 x y^2 - 14x}{4y^2 - 2}$$
In this case the incomplete factorization of the numerator might lead to a slightly quicker solution.
Last edited:
arildno
Homework Helper
Gold Member
Dearly Missed
I know what factoring is. Basically, you take any given number or expression, and rewrite it as a product of its factors.
For instance in 3xy + 9y, every term is divisable by "3y", so we could rewrite it as 3y(x + 3).
But, consider: $$\displaystyle{28xy^2 - 14x}$$
Every term there is divisible by "14x", so is there something at all wrong with: $$\displaystyle{14x(2y^2 - 1)}$$???
That's how I would factor it, but the book instead choose to use "7x" as the factor, and uses $$\displaystyle{7x(4y^2 - 2)}$$ as the answer.
If given an exam, how do you know which factor they are going to claim is the "correct" one????
Thanks a lot.
Factorizations are not unique, and what counts as "simplest" may often be a matter of taste.
One rule of thumb, though:
1. In general, an expression is not simplified by introducing explicit fractions, radicals etc.
For example,
the valid factorization $$28xy^{2}-14x=23\sqrt{x}y(\frac{28}{23}\sqrt{x}y-\frac{14\sqrt{x}}{23y})$$ is not generally regarded as a simplification.
mathwonk
Homework Helper
2020 Award
one does not know which factorization to use until one knows what ring the factorization occurs in. in general one wants to factor somehting into irreducible elements and ignore units.
in the ring Z[X,Y], the units are 1 and -1, so one would have four irreducible factors
2(7)X(2Y^2-1), but if the ring is Q[X,Y], then all non zero costants are units, your element will only have 2 irreducible factors, and hence the 2 and the 7 do not matter. so either of your factorizations is entirely acceptable and equivalent.
i.e. two irreducible factors whose quotient is a unit are equivalent, so 2X, 7X, X, and 14X are all equivalent, so in fact X(28Y^2-14) is just as good as the others.
on the other hand, in the ring Q(X,Y), which is a field, there is no need to factor it at all, since the original element is a unit.
but it seems to me the people who wrote your book are not very knowledgeable and are not following these rules for factorizations but are just following some pattern they think is "nice" and simple. so it is difficult to say what they will prefer as an answer.
Last edited:
Ideally you want to reduce it into its prime factors. Can't get anything simpler than that. So yours would be correct.
mathwonk
Homework Helper
2020 Award
as i said, the meaning of the word "prime" or "irreducible", depends on the containing ring, and what are its units, since units are not prime, but define an equivalence relation on primes.
thus you want to factor into "primes" p , but where u.p and p are considered equivalent if u is a "unit", i.e. is invertible.
thus in Z[X,Y], 2 is a prime, but not in Q[X,Y], and in Q(X,Y) not even X is a prime.
Last edited: | 1,177 | 4,517 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2021-49 | latest | en | 0.958228 |
https://math.answers.com/other-math/What_is_ten_percent_of_sixty_five | 1,708,955,957,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474660.32/warc/CC-MAIN-20240226130305-20240226160305-00159.warc.gz | 369,545,652 | 45,565 | 0
What is ten percent of sixty five?
Updated: 4/28/2022
Wiki User
11y ago
10% of 65
= 10% * 65
= 0.1 * 65
= 6.5
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11y ago
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Q: What is ten percent of sixty five?
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How do you express 0.0565 percent in words?
Five hundred sixty-five ten-thousandths percent.
\$26.50
How do you write in word form 65.6100?
Sixty five and sixty one hundredthsor, if you want to retain the precision,Sixty five and sixty thousand one hundred ten thousandths.Sixty five and sixty one hundredthsor, if you want to retain the precision,Sixty five and sixty thousand one hundred ten thousandths.Sixty five and sixty one hundredthsor, if you want to retain the precision,Sixty five and sixty thousand one hundred ten thousandths.Sixty five and sixty one hundredthsor, if you want to retain the precision,Sixty five and sixty thousand one hundred ten thousandths.
If the price is five hundred and sixty dollars and you have ten percent off how much is it?
The sale price is \$504.00
Ten percent of sixty five thousand?
10% of 65,000= 10% * 65000= 0.1 * 65000= 6,500
How do you write sixty three and five ten thousands?
50,063 is fifty-thousand and sixty three; if you mean sixty three and five ten thousandTHS, this is written as 63.0005
What is seventy-five percent of sixty?
seventy-five percent of sixty = 45= 75% of 60= 75% * 60= 0.75 * 60= 45
6 pounds
3000
75%
15.
What is the Scientific notation for sixty seven thousand five hundred and ten?
The scientific notation for sixty seven thousand five hundred and ten is: 6.751 × 104 | 449 | 1,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.859479 |
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rf505_asciicfg.py
Go to the documentation of this file.
1## \file rf505_asciicfg.py
2## \ingroup tutorial_roofit
3## \notebook -nodraw
4## Organization and simultaneous fits: reading and writing ASCII configuration files
5##
6## \macro_code
7## \macro_output
8##
9## \date February 2018
10## \authors Clemens Lange, Wouter Verkerke (C++ version)
11
12from __future__ import print_function
13import ROOT
14
15
16# Create pdf
17# ------------------
18
19# Construct gauss(x,m,s)
20x = ROOT.RooRealVar("x", "x", -10, 10)
21m = ROOT.RooRealVar("m", "m", 0, -10, 10)
22s = ROOT.RooRealVar("s", "s", 1, -10, 10)
23gauss = ROOT.RooGaussian("g", "g", x, m, s)
24
25# Construct poly(x,p0)
26p0 = ROOT.RooRealVar("p0", "p0", 0.01, 0.0, 1.0)
27poly = ROOT.RooPolynomial("p", "p", x, [p0])
28
29# model = f*gauss(x) + (1-f)*poly(x)
30f = ROOT.RooRealVar("f", "f", 0.5, 0.0, 1.0)
31model = ROOT.RooAddPdf("model", "model", [gauss, poly], [f])
32
33# Fit model to toy data
34# -----------------------------------------
35
36d = model.generate({x}, 1000)
37model.fitTo(d, PrintLevel=-1)
38
39# Write parameters to ASCII file
40# -----------------------------------------------------------
41
42# Obtain set of parameters
43params = model.getParameters({x})
44
45# Write parameters to file
46params.writeToFile("rf505_asciicfg_example.txt")
47
48# Read parameters from ASCII file
49# ----------------------------------------------------------------
50
51# Read parameters from file
52params.readFromFile("rf505_asciicfg_example.txt")
53params.Print("v")
54
55configFile = ROOT.gROOT.GetTutorialDir().Data() + "/roofit/rf505_asciicfg.txt"
56
57# Read parameters from section 'Section2' of file
58params.readFromFile(configFile, "", "Section2")
59params.Print("v")
60
61# Read parameters from section 'Section3' of file. Mark all
62# variables that were processed with the "READ" attribute
63params.readFromFile(configFile, "READ", "Section3")
64
65# Print the list of parameters that were not read from Section3
66print("The following parameters of the were _not_ read from Section3: ", params.selectByAttrib("READ", False))
67
68# Read parameters from section 'Section4' of file, contains
69# 'include file' statement of rf505_asciicfg_example.txt
70# so that we effective read the same
71params.readFromFile(configFile, "", "Section4")
72params.Print("v") | 724 | 2,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.41943 |
https://almazrestaurant.com/what-is-input-gst-and-output-gst/ | 1,674,807,561,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00151.warc.gz | 99,505,416 | 11,942 | What Is Input GST And Output GST?
What is input and output tax GST?
Input credit means at the time of paying tax on output, you can reduce the tax you have already paid on inputs. Say, you are a manufacturer – tax payable on output (FINAL PRODUCT) is Rs 450 tax paid on input (PURCHASES) is Rs 300 You can claim INPUT CREDIT of Rs 300 and you only need to deposit Rs 150 in taxes.
What is output in GST?
Output GST is the GST which is applied on sales of goods or services provided. For example if the CGST is 10% and SGCT is 5% than for a sale of Rs. 10,000/-, the total will be as. Price = Rs. 10,000/-
What do you mean by input GST?
“Input Tax” in relation to a taxable person, means the Goods and Services Tax charged on him for any supply of goods and/or services to him or purchases he makes, which are used or are intended to be used, for the furtherance of his business.
Related Question What is input GST and output GST?
How do you calculate GST output?
• Add GST: GST Amount = (Original Cost x GST%)/100. Net Price = Original Cost + GST Amount.
• Remove GST: GST Amount = Original Cost – [Original Cost x 100/(100+GST%)] Net Price = Original Cost – GST Amount.
• What is input tax credit example?
Input Tax Credit refers to the tax already paid by a person at time of purhase of goods ro services and which is available as deduction from tax payable . For eg- A trader purchases good worth rs 100 and pay tax of 10% on it. Now the trader has to pay Rs. 15 to government but he had already paid Rs.
What is an ITC percentage?
If you are a GST registered business, you must tell us what your entitlement to ITC was for your insurance premium, under the Goods and Services Tax Act. The ITC percentage is the GST paid to you by CCI on your premium and for which you may be able to claim from the Australian Taxation Office.
How GST is calculated with example?
GST can be calculated simply by multiplying the Taxable amount by GST rate. If CGST & SGST/UTGST is to be applied then CGST and SGST both amounts are half of the total GST amount. For example, GST including amount is Rs. 525 and GST rate is 5%.
How do I calculate GST input?
• Step 2 – Go to Services.
• Step 3 – Select the Financial Year and the Return Filing Period from the drop-down.
• Step 4 – Click on the 'View' button in the tile GSTR-2A.
• Step 5 – The GSTR2A – auto drafted details is displayed.
• Step 6 – Under Part A, click on B2B Invoices.
• What is the difference between input and output VAT?
Output VAT is VAT which you must calculate and collect when you sell goods and services, provided that you are registered in the VAT Register. Input VAT is VAT which is included in the price when you purchase vatable goods or services for your business.
Is output tax an expense?
Input and output tax is calculated on revenue or expense items (base amount). The tax amounts are posted to separate tax accounts and refunded by the tax office (input tax) or paid to the tax office (output tax). The input tax can be completely or partially non-deductible.
How are input tax credits calculated?
To calculate your ITCs, you add up the GST/HST paid or payable for each purchase and expense of property and services you acquired, imported, or brought into a participating province. You multiply the amount by the ITC eligibility you can claim. You calculate adjustments for change in use, sales or improvements.
How do I check my GST input credit?
• Step 1: Log in to the Portal. The taxpayer has to login to the official GST Portal.
• Step 2: Enter the Details.
• Step 3: Click Electronic Credit Ledger.
• Step 4: Click Provisional Credit Balance.
• Step 5: Click Save.
• How do I find my HSN number?
For example, if you are selling fresh bananas, you can find your HSN code under Section 02: Vegetable Products, Chapter 08: Edible fruits and nuts; peel of citrus fruits or melons, Heading 03: Bananas including Plantains, Subheading 9010: Bananas, fresh. Your HSN code will be 0803.
What is the GST rate for HSN code?
2018 on FORM GST DRC-07 Notification issued to for waiver of penalty payable for non-compliance of the provisions of notification No. 14/2020 – Central Tax till 30.09. 2021 Fourth Amendment (2021) to CGST Rules SOP for revocation of cancellation of registration DG Systems - Bengaluru has issued Advisory No.
Who is eligible for ITC?
INPUT TAX CREDIT 1) Who is eligible to take Input tax credit (ITC) under SGST & CGST Act? A registered taxable person under GST Act who is paying tax due in the course or furtherance of business can claim and avail ITC credited in electronic ledger [Sec. 16(1)].
What is ITC eligibility?
Eligibility for Availing ITC [Section 16(1)]
Registered Person: As per Section 16(1), Input tax credit is available only to a registered person. When a registered person is supplied with goods or services or both, on which tax has been charged, he is allowed to take credit of the input tax paid.
Is GST calculated on MRP?
MRP is inclusive of all taxes including GST. It must be noted that retailers cannot charge GST over and above the MRP. GST is already included in the MRP printed on the product.
What is e Ledger?
e-Ledger is an electronic form of passbook for GST. These e-ledgers are available to all GST registrants on the GST portal. The e-ledgers contains details of the following: Amount of GST deposited in cash to government in electronic cash ledger. Balance of Input Tax Credit available (ITC) in electronic credit ledger.
How do you calculate input VAT?
Take the gross amount of any sum (items you sell or buy) – that is, the total including any VAT – and divide it by 117.5, if the VAT rate is 17.5 per cent. (If the rate is different, add 100 to the VAT percentage rate and divide by that number.) Multiply the result from Step 1 by 100 to get the pre-VAT total.
Is input VAT asset or liability?
Just to be clear, the VAT is either a current asset or current liability depending on its balance, and the balance changes all the time, sometimes it is positive, sometimes negative.
What is cash ledger?
Electronic Cash Ledger is a cash ledger that contains deposits that a taxpayer has made and any GST payments made through cash. The cash ledger segregates the information head wise such as IGST, CGST, SGST/UTGST, and CESS. Each of these major heads (IGST, SGST, etc.)
How do I calculate GST receivables?
Access the https://www.gst.gov.in/ URL. The GST Home page is displayed. Login to the GST Portal with valid credentials. Click the Services > Ledgers > Electronic Credit Ledger option.
Posted in FAQ | 1,559 | 6,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-06 | latest | en | 0.967218 |
https://www.priory-common.milton-keynes.sch.uk/monday-8/ | 1,624,196,193,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487662882.61/warc/CC-MAIN-20210620114611-20210620144611-00420.warc.gz | 877,442,573 | 14,110 | # Monday
RWI
Practice Start by practising single Sounds using the RWI Sound Sheet
New sound Focus sound of the day: oo
For example; todays sound is oo and is said in the same way as in the words – moo and fool.
Say the rhyme ‘Poo at the zoo’
Word writing - Adult to sound out the following words and allow time for child to write them down a sound at a time. As follows – m-oo-n (encourage child to say the sounds as they write them, remember rhymes for each sound, they may help your child. Make sure the double letter sound stays as one sound – do not be tempted to break it down into individual.)
f-oo-d
p-oo-l
b-oo-t
Key words to learn to read and practice writing today are:
get, he, she, not
ENGLISH
Read/listen to/watch the story of the three Billy Goats Gruff. There is a powerpoint of the story attached.
Think about the story and answer some questions. You can do this by just talking abut the story.
• Who are the 4 characters in the story?
• Which character do you the is the scariest?
• Who do you think is the bravest?
• Can you remember what the troll said when he heard the Billy Goats Gruff?
MATHS
This week is going to be all about counting in 2s and finding pairs. A pair is when there are 2 of something.
Start today by counting up to 20, now do it again. It’s important that you know the order the numbers come in so that you can properly work out how to count in 2s.
Count up to 20 again but this time we’re going to do a whisper and a shout… whisper number 1, shout number 2, whisper 3, shout 4 and so on up to 10. Try it again and see if you can go up to 20.
I wonder if you noticed anything about the numbers you shouted? If you only said the shouted numbers you would be counting in 2s.
Do this one more time but this time write down the numbers you shout, in a line so you can read them again after.
You now have all the numbers you need to count in 2s. Keep practising!
Here’s a song to finish todays maths so you can practise some more…
TOPIC
Follow the link below to find out all about goats. When you have worked through the book, you can do this more than once, write down three facts about goats that you didn’t already know. | 547 | 2,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-25 | longest | en | 0.930087 |
https://math-faq.com/modeling-health-insurance-project-under-development/ | 1,716,778,798,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059028.82/warc/CC-MAIN-20240527021852-20240527051852-00128.warc.gz | 335,211,601 | 10,895 | # Modeling Health Insurance – Project Under Development
This past May, I was faced with a problem regarding health insurance. My employer offers two different plans, the Basic Plus Plan and the Premier. In the past, I was insured under the Premier plan with no cost to me. However, this summer I added my wife and child to my health insurance. I was faced with a decision. Should I cover my family under the Premier plan or change to family coverage under the Basic Plus Plan?
The two plans are described in the summary plan (PDF). Examining the two plans you’ll see that there are several differences for family coverage.
• The out of pocket maximum for the Premier Plan is \$6000 per year versus \$12,000 per year for the Basic Plus Plan.
• The annual deductible for the Premier Plan is \$600 per family versus \$1200 per family for the Basic Plus Plan.
• In general, the Premier Plan covers 80% of medical charges and the Basic Plus Plan covers 60% of medical charges.
The plans differ in cost too. Family coverage under the Premier Plan is \$514 per month compared to \$46 per month under the Basic Plus Plan. There are other difference regarding in network and out of network, but these facts are what I used to make my coverage decision.
Under a number of assumptions (which I’ll detail in later posts), I came up with the following graph.
This graph shows the total annual cost for each plan as a function of the medical charges incurred by the insured. As you can see, each plan is described by a piecewise linear function. If my family incurs less than \$26,880 in medical charges during the year, the Basic Plus Plan is cheaper. For more than \$26,880 in medical charges, the Premier Plan is cheaper.
In this project, I want students to create a graph and accompanying mathematics for other insurance plans. Luckily, it is easy to get health insurance quotes online. Humana One will give you an instance quote when you click on the Get Quote link on their webpage. Although they ask for a lot of personal information, only the state, zip code and birthday information about the insured are required. For a family coverage of a 46 year old in Arizona, this PDF demonstrates that a number of plans are available with different levels of coverage, different out of pocket maximums, and different deductibles. Another great source for health insurance quotes from many different companies is eHealthInsurance. I want my students in College Algebra to enter in their pertinent information and compare two plans like I did in the graph above.
Over the next week, I’ll post the assignments I create to help point students in the proper direction as well as the project letter I come up with. | 556 | 2,702 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-22 | latest | en | 0.954468 |
https://stats.stackexchange.com/questions/390602/correlation-and-p-value/390613 | 1,708,887,316,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474641.34/warc/CC-MAIN-20240225171204-20240225201204-00081.warc.gz | 532,561,604 | 39,563 | # Correlation and P Value
I have run correlation on SPSS for the variables of number of employees in a bank and Net profit margin for the data of 10 years. Results show the correlation coefficient '-0.1649' and p-value '0.649'.
I am unable to interpret the result and need help.
• It just indicates that there is not strong evidence that the correlation is different from 0. To determine if there is a slight negative correlation as the estimate suggests you would need a much larger sample size. Feb 3, 2019 at 19:48
It appears you have data for one bank for 10 years, so N = 10. The low, negative correlation only indicates that for the bank you chose, for the 10 years you have, the linear relationship was small and negative.
Unless you are willing to posit that either a) These 10 years are a random sample of all years from your bank or b) These 10 years are a random sample from those 10 years for some larger population of banks then you won't be doing any inference and the p value isn't important. (And positing either of those things seems unjustified, at least to me).
What I would do is plot the ratio of profit to employees over time and see if anything interesting appears.
• Yes, I have taken data for one bank earlier. Later on, I have run correlation for the 10 years data of 6 banks (60 observations) and found the following results. R = 0.231, Sig (2-tailed) = 0.076. I am actually confused whether p-value will matter over here as per which the above results show that the same are not statistically significant because p-value > 0.05 whereas correlation (R) = 0.231 that means slight positive correlation. Feb 4, 2019 at 20:13
• If you try to do this for 6 banks then your data is not independent and correlation really isn't appropriate. You would want a multilevel model. Feb 5, 2019 at 11:09
Your data shows a slight negative correlation between number of employees and net profit. (correlation coefficient of -0.17).
Your data is consistent with a world in which there was no true correlation between number of employees and net profit, in that it is not unlikely that the data shows a slight negative correlation due to random chance (p-value of 0.65).
In short, almost all rational people would not take this data as proof that there is a negative association between number of employees and net profit. | 538 | 2,341 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-10 | latest | en | 0.949263 |
http://altmetrics.blogspot.com/2016/ | 1,529,940,371,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868135.87/warc/CC-MAIN-20180625150537-20180625170537-00247.warc.gz | 18,795,044 | 12,894 | ## Equalised Mean-based Normalised Proportion Cited
The Equalised Mean-based Normalised Proportion Cited (EMNPC) is an indicator to assess the proportion of documents cited (i.e., with a non-zero citation count) to see whether it is above or below the world average or the values for another group. It is designed for web indicators for which a high proportion is uncited so that average citation indicators, such as MNLCS, are not very accurate. It is particularly suitable for metrics designed by Kayvan Kousha, such as Wikipedia citation counts, and syllabus mentions.
EMNPC has a simple formula. For each field f in which group g publishes, calculate the proportion of articles with at least one citation, pgf. Now for the same set of fields, calculate the proportion of articles published by the world with at least one citation, pwf. Now sum the first set and divide by the sum of the second set to get EMNPC for the group.
If this has a value greater than 1 then a higher proportion of articles by group g is cited than average for the world, so g's research has above average impact in terms of the proportion cited. Note that this calculation is unfair if the group publishes small numbers of articles in some fields because all fields have equal weight in the above formula. Small fields should therefore be removed.
This calculation also gives a higher implicit weight to fields with a relatively high proportion cited because these can dominate the numerator and denominator of the formula, but this is necessary to get narrower confidence intervals (in contrast to the similar MNPC indicator).
Here is a worked example in Excel, showing the formulae used (click to expand).
### Confidence intervals
Confidence intervals can be calculated with the formula below.
Here is a worked example in Excel (click to expand).
### Spreadsheet and original publication
Download a spreadsheet containing the EMNPC calculations. This also has MNPC calculations, and a second worksheet repeating the calculations with a continuity correction for greater accuracy.
Additional details can be found in this article.
Thelwall, M. (in press). Three practical field normalised alternative indicator formulae for research evaluation Journal of Informetrics. 10.1016/j.joi.2016.12.002
## Mean Normalised Log-transformed Citation Score (MNLCS): A field normalised average citation impact indicator for sets of articles from multiple fields and years
The Mean Normalised Log-transformed Citation Score (MNLCS) is a variant of the Mean Normalised Citation Score (MNLCS) to assess the average citation impact of a set of articles. This formula compares the average citation impact of articles within a group to the average citation impact of all articles in the fields and year of the group’s articles. A score of above 1 indicates that the group’s articles have a higher average citation impact than normal for the fields and years in which they were published. The MNLCS uses a log transformation to citation counts before processing them because sets of citation counts are typically highly skewed and this transformation prevents individual articles from having too much influence on the results. The MNLCS calculation is as follows.
1. Log transformation: For each article A in the group to be assessed, replace its citation count c by a log-transformed version, ln(1+c).
2. Field normalisation: Divide the log transformed citation count ln(1+c) of each article A in the group by the average (arithmetic mean) log transformed citation count ln(1+x) for all articles x in the same field and year as A.
3. Calculation: MNLCS is the arithmetic mean of all the field-normalised, log-transformed citation counts of articles in the group.
The MNLCS calculations are illustrated below in Excel for a group publishing articles A,C, H and J in a single field containing articles A to K.
And here is an example of MNLCS calculations for a group that is split between two fields, X and Y, with articles A,C, H and J in field X and O, S, U in field Y.
Confidence intervals can be calculated for MNLCS to assess whether two different MNLCS values are statistically significantly different from each other. The formulae use the means and standard errors of the log transformed citation counts (note: not the field normalised versions) for the articles in the group and the world’s articles in the same field and year as the group. The formula is more complicated if the group publishes in multiple fields and years – this post just discusses the simplest case, with formula below.
The confidence interval calculations are illustrated below in Excel. In this case the group MNLCS is 0.885 but its 95% confidence interval is (0.119, 2.081) so the group average is not statistically significantly different from the world MNLCS value, which is always 1.
All of the above calculations (including confidence intervals) are also built into the free software Webometric Analyst for sets of citation counts from Scopus, the Web of Science as well as for web indicators.
More details are given in this paper:
Thelwall, M. (in press). Three practical field normalised alternative indicator formulae for research evaluation Journal of Informetrics. 10.1016/j.joi.2016.12.002
## Friday, 5 August 2016
What do Wikipedia Citation Counts Mean?
A post on this topic can be found on the 3:AM altmetrics workshop blog. | 1,134 | 5,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-26 | latest | en | 0.923587 |
https://app-wiringdiagram.herokuapp.com/post/melting-and-solidification-in-fluent | 1,563,782,313,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527828.69/warc/CC-MAIN-20190722072309-20190722094309-00002.warc.gz | 317,033,366 | 13,293 | 9 out of 10 based on 844 ratings. 3,326 user reviews.
# MELTING AND SOLIDIFICATION IN FLUENT
ANSYS Fluent Tutorial: Analysis of Melting and
Click to view on Bing37:28Mar 03, 2018From this tutorial, the viewer would be able to learn how to model a PCM and analyse its solidification and melting using ANSYS Fluent. Transient modelling along with solidification and meltingAuthor: Ansys-TutorViews: 23K
Fluent Melting and Solidification Essay. Limitations of the As mentioned above, the phase change formulation in FLUENT Phase Change can be used to model the melting freezing of pure materials, as Model well as alloys. The liquid fraction versus temperature relationship used in FLUENT is the lever rule|i.e., a linear relationship Equation 8.2-3 .
ANSYS Fluent - Solidification/Melting Option
May 30, 2019I am trying to run a solidification case by ANSYS fluent. I have installed the ANSYS with academic student lisence and I added research license that my university has (FENSAP-ICE Release 19.2). I don't see the solidification/melting option among the available options.Multiphase modelling (VOF) of solidification and meltingApr 02, 2019UDF macro to get the liquid fraction when using theNov 19, 2018See more results
Fluent Melting and Solidification , Sample of Essays
Limitations of the As mentioned above, the phase change formulation in FLUENT Phase Change can be used to model the melting freezing of pure materials, as Model well as alloys. The liquid fraction versus temperature relationship used in FLUENT is the lever rule|i.e., a linear relationship Equation 8.2-3 .[PDF]
Chapter 21. Modeling Solidi cation and Melting - ENEA
the turbulence equations to account for reduced porosity in the solid regions. FLUENT provides the following capabilities for modeling solidi cation and melting: Calculation of liquid-solid solidi cation/melting in pure metals as well as in binary alloys c Fluent Inc. November 28, 2001 21-1
ANSYS FLUENT 12.0 Theory Guide - 17. Solidification and
17. Solidification and Melting. This chapter describes how you can model solidification and melting in ANSYS FLUENT information about using the model, see this chapter in the separate User's Guide.
solidification and melting with VOF: how to sink solid PCM
Mar 17, 2015During the melting process of the PCM there will be 2 phases present in the enclosure, the liquid and solid state. The solid PCM will drop to the bottom due to the gravitation force. A 2d mesh was created and exported out from Gambit. In Fluent, I used Solidification and Melting function in my simulation with VOF.
How to define interface in melting and solidification of
How to define interface in melting and solidification of PCM in Ansys Fluent? I'm modelling 2D melting and solidification of a phase change material in a container using Ansys fluent with VOF.
Can anybody share a melting or solidification model file
Popular Answers ( 1) The easiest way to take into account the latent heat of fusion is by including it in the heat capacity coefficient which will then become very large in near the melt temperature. For that, it is essential to introduce a transition interval (temperature range in which the phase change
melting and solidification -- CFD Online Discussion Forums
Apr 18, 2017Re: melting and solidification. The mixture model approach that is implicit in this and many other formulation of the phase change problem are ultimately not very satisfying unless you have vanishing Stokes number and hence all phases have the same local velocity. I have subsequently switched to a full multiphase approach working with my colleague.. | 827 | 3,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | longest | en | 0.873439 |
https://wealthofgeeks.com/coast-fire-easiest-path-to-financial-independence/ | 1,696,473,174,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511717.69/warc/CC-MAIN-20231005012006-20231005042006-00873.warc.gz | 654,629,834 | 36,001 | # Coast FIRE: Easiest Path To FIRE (With Calculator)
CoastFIRE is one of several Financial Independence Retire Early (FIRE) types. Coast FIRE is defined as having enough money invested early in life to stop contributions and still cover a traditional retirement. This allows you to work only to cover current expenses.
Let’s explore CoastFIRE as an alternative route to regular financial independence; retire early (FIRE) and see if it’s right for you.
## What is Coast FIRE?
Coast FIRE is when you aim to invest enough money to let it grow without any further contributions and have that money be sufficient to cover a traditional retirement.
This growth would come from capital gains and dividend reinvestments.
Since you don’t need to contribute towards retirement, you could switch to a part-time job to cover your current expenses.
At that point, you would be “coasting” towards financial independence.
This type of FIRE appeals to many people because you don’t have to save as much money.
CoastFIRE allows you to switch to a part-time job much earlier and not wait until you have accumulated 25x your yearly expenses like traditional FIRE.
## How Is Coast FIRE Calculated?
The easiest way to calculate Coast FIRE is to find your FIRE number first.
If you anticipate having yearly expenses of \$40,000/year in retirement, you would have to save \$1 million.
This is based on a traditional route to FIRE.
FIRE Number = 25 x (Yearly Expenses)
However, for CoastFIRE, you don’t have to wait until you reach \$1 million.
You can invest \$90,000 at an early age and let it grow to \$1 million using compound interest.
The money will be there for you once you retire, even without additional contributions.
Here is what it would look like if you had \$90,000 invested by the age of 30 and you wanted to retire at 65 years old: (assuming a 7% return)
Letting \$90,000 compound over 35 years at a 7% average yearly return would give you \$1 million! This is why compound interest is the 8th wonder of the world.
So in this example, you could stop making contributions at 30 and only worry about your current expenses.
This allows you to design a lifestyle where you work less or spend more now.
You could have more money and time for family, friends, and adventures.
That is the appeal of CoastFIRE.
## Coast FIRE Calculator
Your Coast FIRE number or the amount will depend on two factors:
1. Yearly Expenses
2. Time Until Retirement
Knowing how much you spend in a typical year is the first step to calculating your Coast FIRE Number.
I use Personal Capital to track my expenses. (Get a \$20 Amazon Gift Card with this link when you add at least one investment account containing a balance of more than \$1,000 within 30 days)
The next step is figuring out how much time you have until you want to retire.
If you are 30 and want to retire at 65, you would have 35 years until retirement.
Now we can calculate our Coast FIRE Number:
### Coast FIRE Number = (Yearly Expenses x 25) ÷ (X)
X Depends On Your Time Until Retirement
Here Is An Example:
• Yearly Expenses (\$40,000)
• Time Until Retirement (35 Years)
• X = 11.5
CoastFI Number = (\$40,000 x 25) ÷ (11.5)
CoastFI Number = \$87,000
The Coast FIRE Formula gives us our Coast FIRE Number of \$87,000.
Now we use the Coast FIRE Calculator to determine when we will reach this amount.
It will be much faster than regular FIRE.
Calculate Your Coast FI Number With My Free Coast FIRE Calculator
[convertkit form=2812706]
Using my Coast FIRE Calculator helps you determine how long it will take to reach your Coast FIRE number.
The CoastFIRE Calculator accounts for how much you’ve saved and the amount you currently contribute.
It shows your progress year by year, so you can quickly identify when you will arrive at Coast FIRE.
In this example, it would take 16 years to accumulate \$1 million and achieve Traditional FIRE.
However, that same person would reach CoastFIRE in only 2 to 3 years!
That is a full 13 years before Traditional FIRE.
## Benefits of Coast FIRE
Coast FIRE allows you to switch jobs or achieve part-time work much faster than traditional FIRE.
If you are miserable at work, pursuing Coast FIRE can help you live a happier lifestyle sooner.
Easiest Path To Financial Independence
Coast FIRE is also the easiest path to financial independence.
As the example above showed, it took less than 3 years to reach CoastFIRE.
The majority of the \$1 million was compound interest working for you.
The advantage of CoastFIRE is allowing compound interest to do the heavy lifting.
When using this type of FIRE, time works in your favor.
The earlier you start, the easier it is to achieve Coast FIRE!
Keeps You Engaged
Lastly, by continuing to work, this type of financial independence appeals to those not looking for a fully retired lifestyle.
Perhaps, you enjoy work, just not working for 40 hours a week.
It’s also possible you envision getting bored with early retirement.
If switching to part-time work appeals more than completely retiring, CoastFIRE might be the right FIRE path for you.
The good news is that you can always change your mind later and pursue traditional FIRE.
Regardless, it’s essential to consider this early on in your journey and adjust as needed.
## Different Types Of FIRE
The Financial Independence Retire Early (FIRE) movement is growing and has many FIRE types.
Here are the different types of FIRE in order from easiest to hardest to achieve:
• CoastFIRE (Coast FIRE)
• BaristaFIRE (Barista FIRE)
• LeanFIRE (Lean FIRE)
• FatFIRE (Fat FIRE)
Each type of FIRE has pros and cons and comes down to balancing money and work.
• Coast FIRE is accumulating enough money to stop contributing and still reach FIRE in the future.
• Barista FIRE is having enough money to retire early while still working a part-time job for additional income and health insurance.
• Lean FIRE is the minimalist way of reaching FIRE and means retiring with a “lean” budget.
• Traditional FIRE is accumulating 25 times your annual expenses and retiring early using the 4% rule.
• Fat FIRE is early retirement without embracing frugality and choosing to accumulate a bigger nest egg.
These options for FIRE allow us to pursue the type that best matches our lifestyle goals.
## Negatives of CoastFIRE
A clear negative of Coast FIRE is the need to continue working and depending on an employer for income.
CoastFIRE is not for those who want F-You Money or to design a self-reliant lifestyle.
CoastFIRE also relies on sticking to a budget and embracing some form of frugality.
If you prefer to live budget-free or anticipate increased expenses in the future, CoastFIRE might not be the best option.
## Coast FIRE Financial Independence
As you can see, you can reach FIRE in various ways. CoastFIRE is by far the easiest path to financial independence.
This can be used as a motivational milestone to motivate you to continue towards more challenging types of FIRE.
Alternatively, achieving CoastFIRE by switching from full-time to part-time might be the best option if you enjoy your job.
You can still enjoy the same lifestyle and declare you’ve reached CoastFIRE in only a few years. | 1,577 | 7,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-40 | latest | en | 0.959059 |
http://clay6.com/qa/30044/how-many-optical-isomers-are-possible-for-the-compound-abdc-cabd- | 1,524,604,310,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947328.78/warc/CC-MAIN-20180424202213-20180424222213-00017.warc.gz | 62,490,809 | 25,557 | # How many optical isomers are possible for the compound abdc - cabd?
(a) Two (b) Four (c) Three (d) One
The compound contains 2 chiral carbon atoms and it has a plane of symmetry.
Hence the number of optical isomers = $2^{n-1}+2^{n/2-1}$
$\qquad=2^{2-1}+2^{2/2-1}$
No. of . optical isomers =3
Hence c is the correct answer.
answered Mar 3, 2014 by | 124 | 350 | {"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} | 2.609375 | 3 | CC-MAIN-2018-17 | latest | en | 0.862899 |
https://justaaa.com/economics/163924-aexplain-incidence-of-direct-and-indirect | 1,713,515,270,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817382.50/warc/CC-MAIN-20240419074959-20240419104959-00834.warc.gz | 308,742,412 | 10,819 | Question
# (a)Explain incidence of direct and indirect taxation using appropriate diagrams. (b)Suppose that the demand and supply...
(a)Explain incidence of direct and indirect taxation using appropriate diagrams. (b)Suppose that the demand and supply for a product sold in a perfectly competitive market are given by the equations:
Qd= 130 – 2P
Qs= -50 + P
(i) What are the perfectly competitive equilibrium price and quantity for this product?
(ii) Suppose the industry that produces this product causes environmental damage valued by the government at 0.5 per unit of output. Determine the socially optimal price and output level for this product.
(iii) What tax should the government add to the price of the product to obtain the socially optimal output level?
(c)Suppose the demand function facing a company manufacturing a particular product (X) is given by
Qx = 62 – 2Px + 0.2M + 25A, where
Px is the price of the product
M is the consumer income, and
A is the amount of advertising expenditure.
If Px is 4, M is 150 and A is 4,
(i) Calculate the amount of X purchased.
(ii) Calculate the income elasticity of demand. Is X a normal or an inferior good?
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 285 | 1,234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-18 | latest | en | 0.910712 |
https://www.cut-the-knot.org/pythagoras/TrigPythIMO.shtml | 1,709,141,186,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474737.17/warc/CC-MAIN-20240228143955-20240228173955-00730.warc.gz | 697,354,190 | 6,423 | # Another Face and Proof of a Trigonometric Identity
Problem 2 at the 2007 Irish Mathematical Olympiad required to prove that the identity $\mbox{sin}^{2}\alpha + \mbox{sin}^{2}\beta +\mbox{sin}^{2}\gamma =2$, where $\alpha +\beta +\gamma = 180^{\circ}$, holds if and only if one of the angles $\alpha$, $\beta$, or $\gamma$ is right. The equivalent identity $\cos ^{2}\alpha + \cos ^{2}\beta +\cos ^{2}\gamma =1$ then appears as a form of the Pythagorean theorem. It came to my attention that in the following reincarnation
it was suggested by Poland at the 1967 International Mathematical Olympiad [The IMO Compendium, p. 46].
The equivalence of the three formulations is obvious. Thus, the proof below covers all three.
Assume $\alpha +\beta +\gamma = 180^{\circ}$. Then the identity
$\cos ^{2}\alpha + \cos ^{2}\beta +\cos ^{2}\gamma =1$
Holds if and only if one of the angles $\alpha$, $\beta$, or $\gamma$ is right.
### Proof 1
Multiply the identity by 2 and regroup:
$(2\cos ^{2}\alpha -1) + (2\cos ^{2}\beta -1) +2\cos ^{2}\gamma =0$.
By the double argument formulas, $2\cos ^{2}t -1=\cos (2t)$, this is converted to
$\cos 2\alpha + \cos 2\beta +2\cos ^{2}\gamma =0$.
Next apply the edition formulas $\cos (2s)+\cos (2t)=2\cos (s-t)\cos (s+t)$ to obtain
$\cos (\alpha -\beta)\cos (\alpha +\beta) +\cos ^{2}\gamma =0$.
Taking into account that, by the stipulation of the problem, $\cos \gamma = -\cos (\alpha +\beta)$, we factor the latter
$\cos \gamma\space (\cos (\alpha -\beta)-\cos \gamma) =0$.
It follows that either $\cos \gamma =0$ or $\cos (\alpha -\beta) -\cos \gamma =\cos (\alpha -\beta) +\cos (\alpha +\beta) =0$. In the former case, $\gamma$ is right.
In the latter case, $\cos (\alpha -\beta) +\cos (\alpha +\beta)=2\space\cos \alpha\space\cos \beta =0$, implying that either $\alpha$ or $\beta$ is right.
### Proof 2
As was suggested by a comment below the expression $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1\,$ cann be converted into
$(-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)=0\,$
with a required implication. This follows from a known identity
$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=1\,$
and the Law of Cosines:
$\displaystyle \cos\gamma=\frac{a^2+b^2-c^2}{2ab}, \ldots$
### Proof 3
This proof is by mit Itagi.
If the triangle is right, WLOG let $\alpha=90^o$. Thus,
\displaystyle \begin{align} &\frac{\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma}{\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma} \\ &=\frac{1+\sin^2\beta+\sin^2(90^o-\beta)}{\cos^2\beta+\cos^2(90^o-\beta)} \\ &=\frac{1+\sin^2\beta+\cos^2\beta}{\cos^2\beta+\sin^2\beta}=2 \end{align}
For the converse, let $d$ be the diameter of the circumcircle. Using sine rule,
\displaystyle \begin{align} &\frac{\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma}{\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma}=2 \\ \Rightarrow& \frac{(a^2+b^2+c^2)/d^2}{3-(a^2+b^2+c^2)/d^2}=2 \\ \Rightarrow& (a^2+b^2+c^2)=2d^2 \\ \Rightarrow& (a^2+b^2+c^2)=\frac{a^2+b^2+c^2}{(1+\cos A\cos B\cos C)} \\ \Rightarrow& \cos A\cos B\cos C = 0. \end{align}
Thus, one of the cosine terms goes to zero and that angle is a right angle.
### References
1. D. Djukic et al, The IMO Compendium, Springer, 2011 (Second edition)
### Trigonometry
Copyright © 1996-2018 Alexander Bogomolny
71426612 | 1,175 | 3,302 | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-10 | latest | en | 0.75066 |
https://www.answersarena.com/ExpertAnswers/exercise-1-prove-the-following-for-all-positive-integers-n-a-if-n-is-odd-then-8-n-2-1-b-i-pa156 | 1,701,318,702,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100164.87/warc/CC-MAIN-20231130031610-20231130061610-00647.warc.gz | 735,865,004 | 6,232 | # (Solved): Exercise 1. Prove the following for all positive integers n: a) if n is odd then 8 | n 2 1, b) i ...
Exercise 1. Prove the following for all positive integers n: a) if n is odd then 8 | n 2 − 1, b) if 3̸ | n and n is odd then 24 | n 2 − 1. Hint: 24 = 3 · 8 and gcd(3, 8) = 1.
Exercise 2. Find some integer solution (where possible): a) 3x − 5y = 7 b) 21x − 35y = 24 c) 97x + 127y = 1
Exercise 3. Simplify the fraction 260 712 561 752
Exercise 4. A number l is called a common multiple of m and n if both m and n divide l. There are many such l. The smallest positive one is called the least common multiple of m and n and is denoted by lcm(m, n). For example 30 = lcm(10, 6) because 30 = 3 · 10 = 5 · 6 (so it’s a common multiple), and any smaller multiple of 10 is not a multiple of 6.
(a) Find lcm(8, 12), lcm(20, 30), lcm(51, 68), lcm(23, 18).
(b) Compare the value of lcm(m, n) with the values of m, n and gcd(m, n). In what way are they related? No need to prove this. Just describe it.
(c) Compute lcm(301337, 307829) using the formula you found in (b). You probably need a calculator for this.
Exercise 5. What is the last digit of 72023?
Exercise 6. Find all integer solutions (x, y) to 6x − 13y = 5
We have an Answer from Expert | 433 | 1,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2023-50 | latest | en | 0.858137 |
https://unlearningmath.com/8.5-cm-to-mm/ | 1,726,284,077,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651548.18/warc/CC-MAIN-20240914025441-20240914055441-00545.warc.gz | 552,120,523 | 11,041 | # 8.5 cm to mm (Centimeter to Millimeter)
By / Under Centimeter To Millimeter / Published on
Unfolding the mathematical and practical aspects of the conversion from 8.5 cm to mm using relatable contexts and examples.
## Exploring the Mathematical Conversion from Centimeter to Millimeter
8.5 cm is equal to 85 mm. This mathematical conversion underlines fundamental proficiency in measurement units transfer, often crucial in various fields such as science, engineering, and daily life activities.
In the realm of measurements, centimeters and millimeters are commonly used in numerous arenas and for different reasons. Be it in professional sports, school assignments, or architectural blueprints, accurately converting one unit to another is essential for precise and meaningful results.
Amid the plethora of measurement units, centimeters and millimeters are prominently utilized due to their relative ease of understanding and their applicability across diverse fields. For instance, in physics, the distance between two entities might be calculated in centimeters, but when one requires greater precision, converting these centimeters into millimeters can provide a clearer picture.
An interesting statistic to note here is that 1 cm equals 10 mm. Hence, the conversion of 8.5 cm to 85 mm reiterates this fundamental association. To visualize this conversion, imagine a usual 30 cm scale. It noticeably accommodates a crowded 300 mm, accentuating the accuracy offered by millimeters.
Millimeters play a significant role, especially in contexts requiring detailed precision. For example, in the jewelry industry, diamonds are measured to the nearest millimeter, as a slight miscalculation can result in significant value differences. Further, survey reports reveal that approximately 80% of industries in the manufacturing sector frequently use millimeters for precise measurements.
To convert any centimeter measurement into millimeters, the multiplier '10' is your helper. In essence, converting 8.5 cm to mm isn't as complicated as cooking a three-course meal; it's as simple as multiplying your entire ingredients by 10 to serve more guests.
Try it on other values, and you will realize that the conversion from centimeters to millimeters adheres to a constant change-over rule, enhancing its relevance and accessibility.
Centimeters To Millimeters Conversion
### FAQs
Q: How many mm are there in 8.5 cm? A: There are 85 mm in 8.5 cm.
Q: What is the method to convert 8.5 cm to mm? A: The method to convert 8.5 cm to mm is by multiplying 8.5 with 10. The result is 85 mm.
Q: Why do we need to convert cm to mm? A: We convert cm to mm to get a more precise measurement, especially in contexts where minor variations could lead to significant discrepancies.
Q: What industries require more precise measurements like mm? A: Industries such as jewelry design, engineering, and manufacturing require more precise measurements and thus use millimeters as standard.
This article succinctly captures the crux of converting 8.5 cm to mm and draws upon its utility across various domains. Recognizing the novelty of millimeters and their versatility is central to understanding and appreciating the world around us.
Millimeter: 0 | 674 | 3,248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-38 | latest | en | 0.89927 |
https://en.khanacademy.org/math/algebra/x2f8bb11595b61c86:systems-of-equations/x2f8bb11595b61c86:number-of-solutions-to-systems-of-equations/v/consistent-and-inconsistent-systems | 1,709,414,858,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475897.53/warc/CC-MAIN-20240302184020-20240302214020-00615.warc.gz | 224,747,461 | 128,005 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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# Solutions to systems of equations: consistent vs. inconsistent
A consistent system of equations has at least one solution, and an inconsistent system has no solution. Watch an example of analyzing a system to see if it's consistent or inconsistent. Created by Sal Khan and Monterey Institute for Technology and Education.
## Want to join the conversation?
• Does anyone know why they call it "consistent" or "inconsistent", not some other word?
• "Inconsistent" is because it is not possible for both equations to hold simultaneously. They contradict each other in the sense that if one holds, the other must fail. Thus their graphs never intersect and there is no solution to their system.
"Consistent" is then the opposite. There does exist solution(s) to the system.
• when you graph a system of equation can you have 2 solutions?
• Yes, if the system includes other degrees (exponents) of the variables, but if you are talking about a system of linear equations, the lines can either cross, run parallel or coincide because `line`ar equations represent lines.
If you are graphing a system with a quadratic and a linear equation, these will cross at either two points, one point or zero points.
If you have a quadratic like y = x² + 2x -3 and a linear equation like y = -x + 1 , this example intersects at two points, (-4,5) and (1, 0), so this system does have two solutions.
If you have a quadratic like y = x² - 2x +1 and a linear equation like y = 2x - 3, this example intersects at one point, x = 2. y = 1 so the point (2,1) is the only solution to this system of equations.
If you have a quadratic like y = x² - 2x + 1 and a linear equation like y = (1/5)x - 2
these never cross and so there will be no solution for this system of equations.
.
A straight line and a quadratic will not coincide, because a quadratic equation represents a parabola--a very un-linear curve! They cannot match at every point.
• I really don't get how a consistent system can be overlapped
• Hi Isaiah,
When you say "overlapped" do you mean the lines crossing? That's essentially the definition of a consistent system - that there is a solution, which is that point where the lines cross.
Or when you say "overlapped", maybe you mean that the two lines are the same and they overlap each other from start to finish? That is considered a consistent system too. When your two equations graph to the same line, the solution is all the points on the line, a solution set, rather than just one point. The video "Infinite solutions to systems" has an example of that situation.
• Unless it is the same line, couldn't you just look at the slopes?
If the slope is the same, then it is inconsistent. (unless it's the same line)
If the slope is different, then they have to intersect.
• As Sal said in the video, you can do it. He just wanted to make it clearer to view.
• What do I do if there are 3 variables? Is there a way to figure this out by intuition, other than, of course, generating these vectors in my mind!?
• To solve a system with three variable you need three equations.
Combine them in two sets of two to get rid of one variable. Then combine these two equations to get rid of another variable.
• So a consistent line either has one solution or infinite solutions, right?
• It's not right to say "a consistent line." You need more than 1 line to have either a "consistent" or "inconsistent" system. Then you can say that a consistent system (with at least 2 lines) has one solution or infinite solutions.
• x-y=2 is it inconsistent?
• That is not a system of equations, so a single equation cannot be inconsistent, it is just a linear equation. If you had 2x - 2y = 14 as a second equation, then the two would be inconsistent.
• Technically speaking these equations are being mapped on a 2-dimensinal scale right? Is there such a thing as mapping a 3-dimensional equation, or even a 4-dimensional one? Also I would like to know how those would be used and why.
• Yes these are based in planar geometry. So imagine a square room with 4 walls, a floor, and a ceiling. You see all sorts of lines that could be parallel (if on the same plane which could include planes you do not see like the wall and ceiling on one side of the room and the wall and floor on the opposite side of the room). Similarly, there are a lot of intersecting lines at the corner of the room. Any two lines would still be on a single plane.
The difference in 3d would be that each corner is the intersection of 3 lines which are not on the same plane and there are some lines that are skew to each other, not on the same plane, but also not parallel like the wall and ceiling of one wall and an adjacent wall and floor. So if you have two rooms that are aligned, you could end up with the same like with one long wall divided by an internal wall. I do not know about 4d. They are used in architecture, game design, and anything that needs 3 dimensions.
• But what if we have a system of equations of a plane (ie. 3 variables)? Then how do we determine whether the planes are consistent or inconsistent? | 1,224 | 5,300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.930954 |
https://www.reddit.com/r/AskReddit/comments/khhpl/reddit_what_is_your_favorite_riddle/c2kdlr6/ | 1,670,513,967,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711344.13/warc/CC-MAIN-20221208150643-20221208180643-00379.warc.gz | 1,004,355,494 | 18,727 | ×
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[–] 20 points21 points (4 children)
==BASE CASE==
TO PROVE: F(1) = G(1). If exactly one person had blue eyes, he would leave on day one
1 GIVEN at least one person on the island has blue eyes (from the oracle, who speaks only truth)
2 BASE HYPOTHESIS exactly one person has blue eyes
3 BY EXCLUSION AND 2 The person with blue eyes sees no one else with blue eyes
4 BY ELIMINATION AND 3 AND 1 The person with blue eyes concludes that he must have blue eyes
5 QED: If exactly one person had blue eyes, he would leave on day one. F(1) = G(1)
==INDUCTIVE CASE==
TO PROVE: If F(n) = G(n) then F(n+1) = G(n+1). (If n people have blue eyes, they will leave on day n) implies (If n+1 people have blue eyes, they will leave on day n+1)
6 GIVEN No one leaves until he knows he has blue eyes
7 ASSUMING n+1 people have blue eyes
8 INDUCTIVE HYPOTHESIS n people with blue eyes will leave on day n
9 BY EXCLUSION AND 7 Each person with blue eyes will see n people with blue eyes
10 BY 6 AND 7 AND 8 No one leaves on day n
11 BY 8 AND 10 The number of people with blue eyes cannot be n
12 BY ELIMINATION AND 9 AND 11 Each person with blue eyes reasons that he must have blue eyes (to make the total not equal to n)
13 QED: (If n people have blue eyes, they will leave on day n) implies (If n+1 people have blue eyes, they will leave on day n+1). If F(n) = G(n) then F(n+1) = G(n+1)
==INDUCTIVE CONCLUSION==
14 BY INDUCTION AND 5 AND 13 x people with blue eyes will leave on day x | 452 | 1,531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2022-49 | latest | en | 0.922013 |
https://stat.ethz.ch/pipermail/r-help/2016-December/443723.html | 1,582,141,669,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144167.31/warc/CC-MAIN-20200219184416-20200219214416-00044.warc.gz | 580,184,334 | 2,493 | # [R] Organize regression output
Michael Dewey lists at dewey.myzen.co.uk
Sat Dec 10 10:49:29 CET 2016
```Dear Francesca
i usually do this by collecting the models into a list not a vector
model <- list(ra = ra, rb = rb, and so on
and then I use lapply or sapply to process the model
lapply(mode, function(x) coef(x)[1])
or something like that, not tested
On 10/12/2016 07:32, francesca Pancotto wrote:
> Dear Contributors
>
> I would like to ask some help concerning the automatization process of an analysis, that sounds hard to my knowledge.
> I have a list of regression models.
> I call them models=c(ra,rb,rc,rd,re,rf,rg,rh)
>
> I can access the output of each of them using for example, for the first
>
> ra\$coefficients
>
> and i obtain
>
> (Intercept) coeff1 coeff2 age gender
> 0.62003033 0.00350807 -0.03817848 -0.01513533 -0.18668972
> and I know that ra\$coefficients[1] would give me the intercept of this model.
>
> What i need to do is to collect the coefficients of each regression in models, and calculate and place in a table, the following simple summation:
>
>
> ra rb rc ...
>
> intercept intercept intercept
> intercept+coeff1 intercept+coeff1 intercept+coeff1
> intercept+coeff2 intercept+coeff2 intercept+coeff2
> intercept+coeff1+coeff2 intercept+coeff1+coeff2 intercept+coeff1+coeff2
>
>
> The calculations are trivial(I know how to do it in steps) but what is difficult for me is to invent a procedure that organizes the data in an efficient way.
>
> I tried some step , starting with collecting the coefficients but i think I am going the wrong way
>
> calcolati <- list()
> for (i in c(ra,rb,rc,rd,re,rf,rg,rh))
> {
> calcolati[[i]] <- i\$coefficients[1]
> }
>
>
> f.
> ----------------------------------
> Francesca Pancotto
> Energie:
> http://www.energie.unimore.it/ <http://www.energie.unimore.it/>
> ----------------------------------
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help | 605 | 2,163 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.813817 |
http://www.puzzlesite.nl/math/index_us.html | 1,516,405,598,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888302.37/warc/CC-MAIN-20180119224212-20180120004212-00215.warc.gz | 545,076,378 | 12,452 | A short introduction Cookie policy
# Mathematical Problems
The puzzles are marked with stars (★) that show the degree of difficulty of the given puzzle.
Back to the main page
Copyright © 1996-2017. RJE-productions. All rights reserved. No part of this website may be published, in any form or by any means, without the prior permission of the authors.
## Traveling Toes ★
A school bus travels from Veldhoven to Roosendaal. There are four children in the bus. Each child has four backpacks with him. There are four dogs sitting in each backpack. Every dog has four puppies with her. All these dogs have four legs, with four toes at each leg.
The question: What is the total number of toes in the bus?
The answer: Click here!
## Big Numbers ★
Using the digits 1 up to 9, two numbers must be made. The product of these two numbers should be as large as possible. All digits must be used exactly once.
The question: Which are the requested two numbers?
The answer: Click here!
Another question: There are two numbers that, when multiplied, give 10000, but neither of them contains a zero.
Which numbers are these?
Another answer: Click here!
## Boys and Girls ★
Ronald and Michelle have two children. The probability that the first child is a girl is 50%. The probability that the second child is a girl is also 50%. Ronald and Michelle tell you that they have a daughter.
The question: What is the probability that their other child is also a girl?
A hint : Click here!
The answer: Click here!
## Square Sequence ★★hstar("stt","hstt");
The numbers 1 up to and including 16 can be placed in sequence in such a way, that the sum of each two consecutive numbers is a square.
The question: How should this be done?
The answer: Click here!
## Train Trouble ★★hstar("stt","hstt");
Charles walks over a railway-bridge. At the moment that he is just ten meters away from the middle of the bridge, he hears a train coming from behind. At that moment, the train, which travels at a speed of 90 km/h, is exactly as far away from the bridge as the bridge measures in length. Without hesitation, Charles rushes straight towards the train to get off the bridge. In this way, he misses the train by just four meters! If Charles had rushed exactly as fast in the other direction, the train would have hit him eight meters before the end of the bridge.
The question: What is the length of the railway-bridge?
The answer: Click here!
## Charlie's Chickens ★★hstar("stt","hstt");
Farmer Charlie has a chicken farm. On a certain day, Charlie calculates in how many days he will run out of chicken-food. He notices that if he would sell 75 of his chickens, he could feed the remaining chickens twenty days longer with the chicken-food he has, and that if he would buy 100 extra chickens, he would run out of chicken-food fifteen days earlier.
The question: How many chickens does farmer Charlie have?
The answer: Click here!
Another question: One chicken lays two eggs in three days. How many eggs do three chickens lay in nine days?
Another answer: Click here!
## Rowing Across the River ★★hstar("stt","hstt");
Patrick and Eric are on the two opposite banks of a river. They both have a rowing boat.
They both start at the same time towards the opposite bank. They pass each other at 180 meters from the bank where Patrick departed. When reaching the opposite bank, they both take a rest for the same amount of time before they return. On the way back, they pass each other at 100 meters from the bank from where Patrick returned.
Patrick and Eric both row with a constant speed, but Eric rows faster.
The question: How wide is the river?
The answer: Click here!
## Sneaking Spider ★★hstar("stt","hstt");
A rectangular room measures 7.5 meters in length and 3 meters in width. The room has a height of 3 meters. A spider sits 25 centimeters down from the ceiling at the middle of one of the short walls. A sleeping fly sits 25 centimeters up from the floor at the middle of the opposite wall. The spider wants to walk (i.e., move along the walls, floor, and ceiling only) to the fly to catch it.
The question: How can the spider reach the fly, walking just 10 meters?
The answer: Click here!
## Buying Bitterballs ★★hstar("stt","hstt");
On a nice summer day, two tourists visit the Dutch city of Gouda. During their tour through the center, they spot a cozy terrace. They decide to have a drink and, as an appetizer, a portion of hot "bitterballs" (bitterballs are a Dutch delicacy, similar to croquettes). The waiter tells them that the bitterballs can be served in portions of 6, 9, or 20.
The question: What is the largest number of bitterballs that cannot be ordered in these portions?
The answer: Click here!
## Speedy Sums ★★hstar("stt","hstt");
A salesman drives from Amsterdam to The Hague. The first half of the distance of his journey, he drives at a constant speed of 80 km/h. The second half of the distance of his journey, he drives at a constant speed of 120 km/h.
The question: What is the salesman's average speed for the complete journey?
A hint : The solution is not 100 km/h!
The answer: Click here!
Another question: A racecar driver drove, on a 4 km long racecourse, at an average speed of 120 km/h for the first 2 km. How fast does he have to go the second 2 km to average 240 km/h for the entire course?
Another answer: Click here!
Yet another question: Makkum and Stavoren are two villages. Michael and Donald want to go from Makkum to Stavoren. They leave at the same time. Michael goes by bicycle. Donald goes by car, which is six times as fast as Michael is on his bicycle. Unfortunately, Donald has a car breakdown halfway between Makkum and Stavoren. Fortunately, a passing farmer gives him a lift to Stavoren on his tractor. Unfortunately, the farmer drives only half as fast as Michael drives on his bicycle. Who of the two arrives first in Stavoren?
Yet another answer: Click here!
The fourth question: Normally, the train between Utrecht and Amersfoort drives at an average speed of 90 km/h. One day, the train was delayed a little. Because of this, the average speed of the train between Utrecht and Amersfoort was only 70 km/h, and the train arrived four minutes late in Amersfoort. What is the distance between the stations of Utrecht and Amersfoort?
The fourth answer: Click here!
## Traveling Bird ★★hstar("stt","hstt");
Consider a road with two cars, at a distance of 100 kilometers, driving towards each other. The left car drives at a speed of forty kilometers per hour and the right car at a speed of sixty kilometers per hour. A bird starts at the same location as the right car and flies at a speed of 80 kilometers per hour. When it reaches the left car, it turns its direction, and when it reaches the right car, it turns its direction again to the opposite, etcetera.
The question: What is the total distance that the bird has traveled at the moment that the two cars have reached each other?
The answer: Click here!
## Cork in the Canal ★★hstar("stt","hstt");
A swimmer jumps from a bridge over a canal and swims one kilometer stream up. After that first kilometer, he passes a floating cork. He continues swimming for half an hour, then turns around, and swims back to the bridge. The swimmer and the cork arrive at the bridge at the same time. The swimmer has been swimming with constant effort.
The question: How fast does the water in the canal flow?
The answer: Click here!
## Square and Rectangle ★★hstar("stt","hstt");
The area of the square shown below is 8 × 8 = 64. The square is cut in the four parts A, B, C, and D, which are rearranged into the rectangle shown below. This rectangle has an area of 13 × 5 = 65.
The question: How can you explain the difference in area?
The answer: Click here!
## Two Trains ★★
Between the two towns of Nijmegen and Venlo runs one railroad track, on which one train travels back and forth. Annette, who lives near the railroad, leaves home every morning at a random time between nine and half past nine to walk her dog. Every day she walks the same route to the railroad, plays with her dog until the train has passed, and then walks back home. You would expect Annette to spot the train to Nijmegen as often as the train towards Venlo, but that is not the case! Annette sees the train to Nijmegen five times as often as the train to Venlo.
The question: How is that possible?
The answer: Click here!
## Fint the Fault ★★
Below we prove that 2=1:
Suppose that x = y. Then holds:
2x - x = 2y - y
This we can rewrite to:
2x - 2y = x - y
This we can rewrite to:
2 (x - y) = x - y
If we now divide by x - y, we get:
2 = 1
The question: Where is the fault in this proof?
The answer: Click here!
## Three Taps ★★
There is a water-cask with three different water-taps. With the smallest tap, the water-cask can be filled in 20 minutes. With the middle tap, the water-cask can be filled in 12 minutes. With the largest tap, the water-cask can be filled in 5 minutes.
The question: How long does it take to fill the water-cask with the three taps together?
The answer: Click here!
## Notable Number ★★
There is a unique number of ten digits, for which the following holds:
• all digits from 0 up to 9 occur exactly once in the number;
• the first digit is divisible by 1;
• the number formed by the first two digits is divisible by 2;
• the number formed by the first three digits is divisible by 3;
• the number formed by the first four digits is divisible by 4;
• the number formed by the first five digits is divisible by 5;
• the number formed by the first six digits is divisible by 6;
• the number formed by the first seven digits is divisible by 7;
• the number formed by the first eight digits is divisible by 8;
• the number formed by the first nine digits is divisible by 9;
• the number formed by the ten digits is divisible by 10.
The question: Which number is this?
A hint : Click here!
The answer: Click here!
Another question: There is a unique number of which the square and the cube together use all digits from 0 up to 9 exactly once. Which number is this?
Another answer: Click here!
Yet another question: We are looking for a prime number of ten digits, in which all digits from 0 up to 9 occur exactly once. The number may not start with 0. Which number is this?
Yet another answer: Click here!
The fourth question: We are looking for a number of three digits. It is divisible by both 9 and 11, and when the first and the last digit are interchanged, you get a number which is only two-ninth of the number we are looking for. Which number is this?
The fourth answer: Click here!
## Baffling Birthdays ★★
In Mrs. Melanie's class are twenty-six children. None of the children was born on February 29.
The question: What is the probability that at least two children have their birthdays on the same day?
The answer: Click here!
## William's Whereabouts ★★
William lives in a street with house-numbers 8 up to 100. Lisa wants to know at which number William lives.
She asks him: "Is your number larger than 50?"
William answers, but lies.
Upon this, Lisa asks: "Is your number a multiple of 4?"
William answers, but lies again.
Then Lisa asks: "Is your number a square?"
William answers truthfully.
Upon this, Lisa says: "I know your number if you tell me whether the first digit is a 3."
William answers, but now we do not know whether he lies or speaks the truth.
Thereupon, Lisa says at which number she thinks William lives, but (of course) she is wrong.
The question: What is Williams real house-number?
The answer: Click here!
## The Prince and the Pearls ★★
Long ago, a young Chinese prince wanted to marry a Mandarin's daughter. The Mandarin decided to test the prince. He gave the prince two empty, porcelain vases, 100 white pearls, and 100 black pearls. "You must put all the pearls in the vases", he told the prince. "After this, I will call my daughter from the room next door. She will take a random pearl from one of the two vases. If this pearl is a black one, you are allowed to marry my daughter."
The question: What was the best way in which the prince could divide the pearls over the vases?
The answer: Click here!
Another question: You have three vases: one vase containing two white pearls, one vase containing one white and one black pearl, and one vase containing two black pearls. From one of these vases, a pearl is taken. This pearl turns out to be white. What is the probability that the other pearl in the same vase is also white?
Another answer: Click here!
Yet another question: You have ten vases. Five of the vases contain a white pearl and four of the vases contain a black pearl (note that a vase may contain both a white and a black pearl!). You randomly select one of the ten vases. What is the probability that the vase you chose is empty?
Yet another answer: Click here!
## Plus & Minus ★★
Below is an equation that is not correct yet. By adding a number of plus signs and minus signs between the digits on the left side (without changes the order of the digits), the equation can be made correct.
123456789 = 100
The question: How many different ways are there to make the equation correct?
The answer: Click here!
## Missing Pages ★★
From a book, a number of consecutive pages are missing. The sum of the page numbers of these pages is 9808.
The question: Which pages are missing?
The answer: Click here!
## Postman Pat ★★
Postman Pat delivers the mail in the small village Tenhouses. This village, as you already suspected, has only one street with exactly ten houses, numbered from 1 up to and including 10.
In a certain week, Pat did not deliver any mail at two houses in the village; at the other houses, he delivered mail three times each. Each working day he delivered mail at exactly four houses.
The sums of the house numbers where he delivered mail were:
on Monday: 18
on Tuesday: 12
on Wednesday: 23
on Thursday: 19
on Friday: 32
on Saturday: 25
on Sunday: he never works
The question: Which two houses did not get any mail that week?
The answer: Click here!
Copyright © 2002 E.R. van Veldhoven. All rights reserved.
## All Apples ★★
On the market, Mrs. Jones and Mrs. Smith sell apples. Mrs. Jones sells her apples per two for 0.50 euro. The apples of Mrs. Smith are a bit smaller; she sells hers per three for 0.50 euro. At a certain moment, when both women have the same amount of apples left, Mrs. Smith is being called away. She asks her neighbor to take care of her goods. To make everything not too complicated, Mrs. Jones puts all apples to one big pile, and starts selling them for one euro per five apples. When Mrs. Smith returns at the end of the day, all apples have been sold. However, when they start dividing the money, there appears to be a shortage of 3.50 euro.
The question: Supposing they divide the amount of money equally, how much does Mrs. Jones lose with this deal?
The answer: Click here!
## Camel & Bananas ★★
A banana plantation is located next to a desert. The plantation owner has 3000 bananas that he wants to transport to the market by camel, across a 1000-kilometer stretch of desert. The owner has only one camel, which carries a maximum of 1000 bananas at any moment in time, and eats one banana every kilometer it travels.
The question: What is the largest number of bananas that can be delivered at the market?
The answer: Click here!
## Buying Books ★★
Two friends, Alex and Bob, go to a bookshop, together with their sons Peter and Tim. All four of them buy some books; each book costs a whole amount in shillings. When they leave the bookshop, they notice that both fathers have spent 21 shillings more than their respective sons have. Moreover, each of them paid per book the same amount of shillings as books that he bought. The difference between the number of books of Alex and Peter is five.
The question: Who is the father of Tim?
The answer: Click here!
## Four Flies ★★
Consider four (dimensionless) flies, two males and two females. They are situated at the corners of one square meter. Every fly tries to reach the male/female fly in front of her/him. Their initial situation is visualized in the picture below. Since the flies are flying towards another, they will meet each other at a certain time in the center of the square.
The question: What is the length of the path they have traveled at the moment they reach each other?
The answer: Click here!
## Fabulous Fraction ★★
With all the numbers 0 up to 9 (using each number exactly once), you can make two fractions that add up to exactly 1.
The question: How shall this be done?
The answer: Click here!
## Circling Cyclist ★★
A cyclist drove one kilometer, with the wind in his back, in three minutes and drove the same way back, against the wind in four minutes.
The question: If we assume that the cyclist always puts constant force on the pedals, how much time would it take him to drive one kilometer without wind?
The answer: Click here!
## Odd Oranges ★★
Greengrocer C. Carrot wants to expose his oranges neatly for sale. Doing this he discovers that one orange is left over when he places them in groups of three. The same happens if he tries to place them in groups of 5, 7, or 9 oranges. Only when he makes groups of 11 oranges, it fits exactly.
The question: How many oranges does the greengrocer have at least?
The answer: Click here!
## The Cucumber Case ★★
On a sunny morning, a greengrocer places 200 kilograms of cucumbers in cases in front of his shop. At that moment, the cucumbers are 99% water. In the afternoon, it turns out that it is the hottest day of the year, and as a result, the cucumbers dry out a little bit. At the end of the day, the greengrocer has not sold a single cucumber, and the cucumbers are only 98% water.
The question: How many kilograms of cucumbers has the greengrocer left at the end of the day?
The answer: Click here!
## Palindrome Puzzle ★★
A number is called a palindrome when it is equal to the number you get when all its digits are reversed. For example, 2772 is a palindrome.
We discovered a curious thing. We took the number 461, reversed the digits, giving the number 164, and calculated the sum of these two numbers:
``` 461
164 +
-------
625
```
We repeated the process of reversing the digits and calculating the sum two more times:
``` 625
526 +
-------
1151
1511 +
-------
2662
```
To our surprise, the result 2662 was a palindrome. We decided to see if this was a pure coincidence or not. So we took another 3-digit number, reversed it, which gave a larger number, and added the two. The result was not a palindrome.
We repeated the process, which resulted in another 3-digit number that was still not a palindrome. We had to repeat the process twice more to arrive at a 4-digit number, which was a palindrome finally.
The question: What was the 3-digit number we started with the second time?
The answer: Click here!
## Escalator Exercise ★★
You walk upwards on an escalator, with a speed of 1 step per second. After 50 steps, you are at the end. You turn around and run downwards with a speed of 5 steps per second. After 125 steps, you are back at the beginning of the escalator.
The question: How many steps do you need if the escalator stands still?
The answer: Click here!
Copyright © 1996-2017. RJE-productions. All rights reserved. No part of this website may be published, in any form or by any means, without the prior permission of the authors.
This website uses cookies. By further use of this website, or by clicking on 'Continue', you give permission for the use of cookies. If you want more information, look at our cookie policy. | 4,716 | 19,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-05 | longest | en | 0.952029 |
https://www.sarthaks.com/2856812/plot-the-points-a-2-3-b-6-3-c-4-7-on-a-graph-sheet | 1,659,999,191,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.1/warc/CC-MAIN-20220808213349-20220809003349-00351.warc.gz | 867,371,711 | 15,075 | # Plot the points A(2, 3), B(6, 3), C(4, 7) on a graph sheet.
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closed
i) Plot the points A(2, 3), B(6, 3), C(4, 7) on a graph sheet.
ii) Join the points to form triangle ABC. Identify the type of triangle and justify your answer.
iii) Find its area,
+1 vote
by (44.6k points)
selected
i) given points are A(2, 3), B(6, 3), C(4, 7)
ii) ΔABC is formed, which is an Isosceles triangle.
CD = | 7 – 3 |= 4 units
AB = |16 – 2 | =4 units
iii) Area of ΔABC = 1/2 × b × h .
= 1/2 × AB × CD
= 1/2 × 4 × 4
= 2 × 4 = 8 sq units | 225 | 537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.891881 |
https://www.coursehero.com/file/8771466/1-c-Suppose-that-you-want-to-maximize-the-meanvariance-criterion/ | 1,498,673,735,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323721.80/warc/CC-MAIN-20170628171342-20170628191342-00054.warc.gz | 857,241,584 | 22,618 | Problem Set 1
# 1 c suppose that you want to maximize the
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Unformatted text preview: ent? b) If you invest in N of these projects, what is the mean and the variance of the total profit? [Hint: Use the results from Question 1b]. 1 c) Suppose that you want to maximize the mean variance criterion E [ X ] " Var( X ) 2 (in terms of the class slides, we are assuming parameters α=1, β=½). Write down the mean variance criterion as a function of the number of projects N. ! d) How many projects should you invest if you want to maximize the mean va...
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## This note was uploaded on 02/03/2014 for the course INSR 205 taught by Professor Kent/smetters/nini during the Spring '09 term at UPenn.
Ask a homework question - tutors are online | 234 | 900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-26 | longest | en | 0.91811 |
https://www.fxsolver.com/browse/?like=1823&p=115 | 1,685,981,443,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652149.61/warc/CC-MAIN-20230605153700-20230605183700-00486.warc.gz | 855,105,078 | 66,056 | '
# Search results
Found 1158 matches
Electrical Impedances - In Series
Electrical impedance is the measure of the opposition that a circuit presents to a current when a voltage is applied. The term complex impedance may be ... more
Electrical Impedances - In Parallel
Electrical impedance is the measure of the opposition that a circuit presents to a current when a voltage is applied. The term complex impedance may be ... more
Electrical Impedances - Phase in Series
Electrical impedance is the measure of the opposition that a circuit presents to a current when a voltage is applied. The term complex impedance may be ... more
Electrical Impedances - Phase in Parallel
Electrical impedance is the measure of the opposition that a circuit presents to a current when a voltage is applied. The term complex impedance may be ... more
Triclinic crystal system (Unit cell's volume)
In crystallography, the triclinic crystal system is one of the 7 crystal systems. A crystal system is described by three basis vectors. In the triclinic ... more
Central processing unit power consumption
Central processing unit power dissipation or CPU power dissipation is the process in which central processing units (CPUs) ... more
Electrical Impedances - Magnitude
Electrical impedance is the measure of the opposition that a circuit presents to a current when a voltage is applied. The term complex impedance may be ... more
Worksheet 302
In the wheelbarrow of the following figure the load has a perpendicular lever arm of 7.50 cm, while the hands have a perpendicular lever arm of 1.02 m.(a) What upward force must you exert to support the wheelbarrow and its load if their combined mass is 45.0 kg? (b) What force does the wheelbarrow exert on the ground?
(a) In the case of the wheelbarrow, the output force or load is between the pivot and the input force. The pivot is the wheel’s axle. Here, the output force is greater than the input force. Thus, a wheelbarrow enables you to lift much heavier loads than you could with your body alone. (b) In the case of the shovel, the input force is between the pivot and the load, but the input lever arm is shorter than the output lever arm. The pivot is at the handle held by the right hand. Here, the output force (supporting the shovel’s load) is less than the input force (from the hand nearest the load), because the input is exerted closer to the pivot than is the output.
Strategy
Here, we use the concept of mechanical advantage.
Force (Newton's second law)
Mechanical Advantage - Law of Lever
Subtraction
Discussion
An even longer handle would reduce the force needed to lift the load. The MA here is:
Division
Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
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Category | 651 | 2,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | longest | en | 0.890083 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1-common-core-15th-edition/chapter-1-foundations-for-algebra-1-2-order-of-operations-and-evaluating-expressions-practice-and-problem-solving-exercises-page-14/42 | 1,685,336,786,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644683.18/warc/CC-MAIN-20230529042138-20230529072138-00281.warc.gz | 864,036,356 | 14,395 | ## Algebra 1: Common Core (15th Edition)
$\dfrac{70}{9}$
Using order of operations, the given expression, $\dfrac{2[8+(67-2^6)^3]}{9} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{2[8+(67-64)^3]}{9} \\\\= \dfrac{2[8+(3)^3]}{9} \\\\= \dfrac{2[8+27]}{9} \\\\= \dfrac{2[35]}{9} \\\\= \dfrac{70}{9} .\end{array} | 147 | 319 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-23 | latest | en | 0.468467 |
https://discourse.panda3d.org/t/heightfeild-tessalator-terrain-height/2117 | 1,685,728,740,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648850.88/warc/CC-MAIN-20230602172755-20230602202755-00746.warc.gz | 241,986,898 | 4,903 | # Heightfeild Tessalator terrain height?
Ok well my computer club at school is using panda to make some games for a game competition. This is my first time using panda. But i’m not a noob programmer either. I’ve used other 3D librarys like Irrlicht and Ogre. Python is new to me also but I pretty much got it figured out. I’m having trouble getting the heightfeild tessalator to work in a timely fashion so I came to the forums for help.
``````class Terrain:
def __init__ (self):
self.terrainmap = HeightfieldTesselator ("Heightmap")
self.terrainmap.setHeightfield (Filename("Art/Heightmap.PNG"))
self.UpdateTerrain ()
def UpdateTerrain (self):
self.terrainmap.setHorizontalScale( 3 )
self.terrainmap.setVerticalScale( 100)
self.terrainmap.setFocalPoint (cam.x,cam.y)
self.terrainNode = self.terrainmap.generate ()
self.terrainNode.setTexGen( TextureStage.getDefault( ), TexGenAttrib.MWorldPosition )
self.terrainNode.setTexScale( TextureStage.getDefault( ), 0.05, 0.05 )
self.terrainNode.setPos (0,0,0)
def Draw (self):
self.terrainNode.reparentTo (render)
def GetHeightAt (self,x,y):
nx = x/3
ny = y/3
return self.terrainmap.getElevation (nx,ny)``````
Its crappy at the moment but the problem lies in the GetHeighAt function…
First of all where in my terrain in 3D space is 0,0 in terms of pixels getElevation wants a pixel amount to bad for us were in 3D which doesnt have pixels When I move towards the terrain I get a -y value!!!
but the terrainNode has a position of 0,0,0!!!
So if my rendered heightmap is thus…
``````+-----------+
| |
| |
| |
+-----------+``````
Which corner is the Node at and which corner is 0,0 pixel?
Well I seem to have hit the spot after i posted this so the fix looks like this.
``````class Terrain:
def __init__ (self):
self.terrainmap = HeightfieldTesselator ("Heightmap")
self.terrainmap.setHeightfield (Filename("Art/Heightmap2.PNG"))
self.vScale = 100
self.hScale = 3
self.UpdateTerrain ()
def UpdateTerrain (self):
self.terrainmap.setHorizontalScale( self.hScale )
self.terrainmap.setVerticalScale( self.vScale)
self.terrainmap.setFocalPoint (cam.x,cam.y)
self.terrainNode = self.terrainmap.generate ()
self.terrainNode.setTexGen( TextureStage.getDefault( ), TexGenAttrib.MWorldPosition )
self.terrainNode.setTexScale( TextureStage.getDefault( ), 0.05, 0.05 ) | 624 | 2,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.660612 |
http://hidrobiologie.granturi.ubbcluj.ro/uty/glide-reflection-isometry.html | 1,679,900,256,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948609.41/warc/CC-MAIN-20230327060940-20230327090940-00182.warc.gz | 23,915,183 | 11,891 | translation. In a reflection over the line y = x, the x- and y-coordinates simply switch positions. Which way is the most efficient? Reflections are isometries that have infinitely many fixed points. The isometry h 2 is invertible and its inverse is an isometry, by Corollary2.3. Isometry question. In other words, a frieze pat-tern maps onto itself under translations, and possibly under rotations, reflections, and glide reflections. The Glide Reflection is an isometry as a result of it’s outlined because the composition of two isometries: º Ml, the place P and Q are factors on line l or a vector parallel to line l. A difficulty, in fact, is whether or not this composition is equal to some current isometry — a reflection, rotation, or translation. - The composition of three reflections is either a reflection or a glide reflection. Glide Reflections Good day! Example 1 Find the image of ABunder the glide reflection that translates from (0,0) to (3,4) and reflects about the x-axis. then classify the isometry a. opposite orientation reflection b. opposite orientation translation c. same orientation rotation d. same orientation glide reflection I know that: the conjugate of a reflection by a translation (or by any isometry, for that matter) is another reflection, by some explicitly calculation (haven't try). Exploring Two Reflections with Parallel Lines. Glide Reflections - Concept. An isometry may or may not have invariant, or fixed, sets, i.e., sets S that satisfy S = f(S). Depending on context, we may consider a reflection a special case, where the translation vector is the zero vector. 7.5 Glide Reflections and Compositions 431 USING COMPOSITIONS When two or more transformations are combined to produce a single transformation, the result is called a of the transformations. Just like a translation, a glide reflection doesn't So, yung topic natin ngayon ay glide reflection. Thus the effect of a reflection combined with any translation is a glide reflection, with as special case just a reflection. These are the two kinds of indirect isometries in 2D. For example, there is an isometry consisting of the reflection on the x-axis, followed by translation of one unit parallel to it. Isometries: Reflection. It is a kind of cross between a reflection and a translation: Unlike translation, rotation and reflection, we can't distinguish a glide reflection from the others just by considering how many fixed points there are. The third symbol is m if there is a horizontal line of reflection, a if there is a glide reflection but no horizontal reflection, or 1 if there is horizontal or glide reflection. Its features include: - Digitally signed automatic security updates - The community is always in control of any add-ons it produces - Supports a multi-site architecture out of the … The order of the two constituent transforms (translation and reflection) is not important. To review, open the file in an editor that reveals hidden Unicode characters. In a glide reflection, the order in which the transformations are performed does not affect the final image. Glide Reflection of a Triangle. The intermediate step between reflection and translation can look different from the starting configuration, so objects with glide symmetry are in general, not symmetrical under reflection alone. The product, or composition, of two isometries is the result of applying one and then the other in order. florianmanteyw and 4 more users found this answer helpful. This is somewhat apparent from the name. Exploring Reflections with Intersecting Lines. Proof. a glide reflection is a composition of a rotation and reflection Is reflecting a congruence transformation? A glide reflection is an isometry with no fixed points and one invariant line. List of Amc - Free ebook download as Word Doc (.doc / .docx), PDF File (.pdf), Text File (.txt) or read book online for free. The translation is (x, y) -> (x, y + 3) and the line of reflection is y = -1. A reflection in a line is an isometry. Reflection is isometry: a glide reflection preserves distances. (If we were to allow translations of length zero, then a reflection would be a type of glide reflection. •A Glide Reflection is an isometry. Glide Reflections A glide reflection is transformation created by a translation followed by a reflection. Thus ensuring that a reflection is an isometry, as Math Bits Notebook rightly states. It provides a blog engine and a framework for Web application development. In an inner product space, the above definition reduces to , = , for all , which is equivalent to saying that † =. The initial "move" on the left foot is seen on the left, followed by the "step" on the right foot, which is the "result" of the glide reflection on the left. Glide reflections are a combination of two transformations: reflection and translation. When reflecting over (across) the x-axis, we keep x the same, but make y negative. 4. From the 4 sorts of transformations translation, reflection, glide reflection, and … To remind yourself , an isometry is a transformation that preserves distance. A common example of glide reflections is footsteps in the sand. Exploring Two Reflections with Parallel Lines. either m or 1 indicating the presence or absence of a vertical line of reflection. Reflection is isometry: a glide reflection preserves distances. of ABC for a glide reflection where the translation is (x, y) (x, y +2) and the reflection line is x =1. Because a glide reflection is a composition of a translation and a reflection, this theorem implies that glide reflections are isometries. Information and translations of glide reflection in the most comprehensive dictionary definitions resource on the web. So, yung goals natin today ay ma-identify yung translation at yung mirror reflection components ng isang glind reflection transformation, and then we'll demonstrate glide reflections in software tools, and illustrate glide reflection sa coordinate … Tags: Question 14 . Here is a description of each type of isometry. Reflection Across the X … 2.rotation. Theorem 9-1 The composition of two or more isometries is an isometry. Graph the triangle represented by the polygon matrix (0 -3 2 on top 0 3 6 on bottom). The composition of two (or more) isometries is an isometry. Notification that the direction of translation and also the axis the reflection space parallel. Each isometry is a rigid transformation, ... Glide Reflections. Exploring Glide-Reflections. An isometric transformation (or isometry) is a shape-preserving transformation (movement) in the plane or in space.The isometric transformations are reflection, rotation and translation and combinations of them such as the glide, which is the combination of a … So, ito yung pang-apat na isometry na ididiscuss natin. Rotation Reflection Glide Reflection R R R R R R R R R Can you find more than one way? You can map a left paw print onto a right paw print with a glide reflection. (b) Show that an isometry which is a composition of three reflections is a glide reflec- tion. Reflect the smiley face across line segment r. Create a glide-reflection using line r as the taxis of the glide-reflection (i.e., using it to indicate direction and for reflection). A (6, 7), B (–4, 4), and C (2, 2) Tell whether the orientations are the same or opposite. glide reflection The composite of a reflection and a translation parallel to the reflecting line, also known as a walk. One can easily verify that the same result is obtained by first reflecting and then translating the image. Login . Rotate the smiley face through an angle of 45 degrees about K in a counter clockwise direction. Reflection Reflections Reflector Reflex Refract Refraction Refresh Refuge Regal Regatta Regatta w/Alta Regiment Depending on context, we may consider a reflection a special case, where the translation vector is the zero vector. It is fairly easy to see that a glide-reflection is actually a combination of a reflection and a translation. Two geometric figures related by an isometry are said to be congruent (Coxeter and Greitzer 1967, p. 80). If Lis the line of points equidistant from points P and Q, then re ection in Lexchanges P and Q. Theorem 1.14 (Three Re ections Theorem). For example, if a wallpaper pattern has rotations of order 2 but no higher order rotations, then it can have at … This produces a "footprint" pattern. Sometimes glide-reflections are mentioned as a fourth type of isometry. In 2-dimensional geometry, a glide reflection is a symmetry operation that consists of a reflection over a line and then translation along that line, combined into a single operation. 1. Glide Reflection Cats. tell whether their orientations are the same or opposite. In other words, an isometry is a transformation which preserves distance. There are four types: translations, rotations, reflections, and glide reflections (see below under classification of Euclidean plane isometries).. ;P Glide re ection Re ection r M. Translation t (; ) Rotation R ˚;Q Glide Re ection The table is completed as follows. An isometry can be classified as a reflection, translation, rotation or glide reflection. Let's take some time to prove this! This function: . To say h 1(z) = h 2(z) is the same as saying (h 1 2 h 1)(z) = z. Glide reflections are essential to an analysis of symmetries, as soon as you go beyond rosette symmetry to frieze and wallpaper symmetry. Every isometry of the plane, other than the identity, is either a translation, a rotation, a reße ction , or a glide-r eße ction. Glide reflections are improper isometries because the orientation has shifted. Let's take some time to prove this! (a) Show that the isometry that consists of a rotation about a point followed by a reflection about a line is a glide reflection. The isometric transformations are reflection, rotation and translation and combinations of them such as the glide, which is the combination of a translation and a reflection. Because two reflection axes which meet at an angle $\theta$ produce a rotation symmetry whose angle is $2\theta$, the crystallographic restriction also puts a strong restriction on the possible reflection and glide reflection symmetries. Glide Reflection 1. (b) For each matrix below, identify what type of isometry it is (translation, rotation, mirror reflection or glide reflection) and find its fixed points and stable lines. Glide Reflection MBJH Spartans. Assume P ≠ Q. flections, and glide reflections. – The composition of two reflections is either a translation or a rotation. A reflection is specified by a line of reflection, a.k.a a mirror. The Glide Reflection is an isometry because it is defined as the composition of two isometries: º Ml, where P and Q are points on line l or a vector parallel to line l. An issue, of course, is whether this composition is equivalent to some existing isometry -- a reflection, rotation, or translation. I need to show that every isometry in the plane is a composition of at most three reflections. Reversing the order of combining gives the same result. How do you test to see if a transformation is a reflection? Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. In addition to rotation, translation, reflection, glides, and identity maps, there are other types of isometries. Reflection on a Coordinate Plane Reflection Over X Axis. Therefore, the only required information is the translation rule and a line to reflect over. Glide Reflection 2. #2 is truly a fundamental and phenomenal theorem. Reflections, rotations, translations, and glide reflections are all examples of rigid motions. Four kinds of isometries Translation Rotation Reflection Glide Reflection A reflection is called a rigid transformation or isometry because the image is the same size and shape as the pre-image. What Is Isometry In Geometry Terms? In Euclidean geometry, a rotation is an example of an isometry, a transformation that moves points without changing the ... exchanges left- and right-handed ordering, and a glide reflection does both. The following types of transformations are isometries: translation, rotation, reflection, glide reflection. In thinking about translation and rotation, we discovered that a useful question to ask about an isometry is whether or not it has fixed points, i.e. Distance stays preserved however orientation (or order) adjustments in a glide reflection. Linear isometries are distance-preserving maps in the above sense. A glide reflection is an isometry that results from composing a reflection with a non-trivial translation (aka hyperbolic isometry) along the geodesic corresponding to the reflection. Reflection preserves angles. When the "Execute p1" button is clicked the javascript function p1 is executed. answer choices . A glide reflection is the composition of a translation (a glide) and a reflection across a line parallel to the direction of translation. You will learn about glide reflections later in the lesson. It is a … One more isometry will be important, a glide reflection, a special product of a reflection and a translation along the line of reflection. Glide reflection is a composite transformation which is a translation followed by a reflection in line parallel to the direction of translation.
glide reflection isometry | 2,840 | 13,404 | {"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} | 3.703125 | 4 | CC-MAIN-2023-14 | longest | en | 0.888252 |
http://convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=palm&To=fl+head | 1,585,843,334,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506988.10/warc/CC-MAIN-20200402143006-20200402173006-00111.warc.gz | 41,594,709 | 3,842 | Partner with ConvertIt.com
New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```palm = 0.0762 length (length) ``` Related Measurements: Try converting from "palm" to angstrom, archin (Russian archin), arpentlin, astronomical unit, Biblical cubit, cable length, city block (informal), cloth quarter, earth to moon (mean distance earth to moon), en (typography en), finger, Greek cubit, Greek fathom, Greek span, li (Chinese li), light yr (light year), ri (Japanese ri), soccer field, span (cloth span), yard, or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: palm = 9 barleycorn, 300 caliber (gun barrel caliber), 4.11 digitus (Roman digitus), 3.43 finger, .25 foot, .00037879 furlong (surveyors furlong), .16465423 Greek cubit, .04116356 Greek fathom, .32930845 Greek span, .75 hand, 3 inch, .13761468 Israeli cubit, 8.05E-18 light yr (light year), 36 line, .00000181 marathon, 1.33 nail (cloth nail), .01515152 rod (surveyors rod), .00000579 spindle, 2.51 sun (Japanese sun), .2499995 survey foot.
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 446 | 1,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-16 | longest | en | 0.682838 |
https://www.coursesarchive.com/AnsweredQuestions/assignment-3-p-4-4-theraph-below-depicts-the-supply-and-demand-for-gasoline-in-california-with-price-ca723 | 1,701,636,212,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.53/warc/CC-MAIN-20231203193127-20231203223127-00220.warc.gz | 802,619,545 | 6,986 | #### (Solved): .] Assignment 3, P. 4 4 Theraph Below Depicts The Supply And Demand For Gasoline In California, With ...
.] Assignment 3, p. 4 4 Theraph below depicts the supply and demand for gasoline in California, with prices in dollars per gall, and the quantity in millions of pallons per day - Soda! CO CUW \$ 30 15 20 25 30 35 40 45 Suppose scientific studies determine that every gallon of gasoline used for driving causes \$3 in environmental damage. Draw the social cost curve in the diagram b. What are the equilibrium and optimal quantities of gasoline consumed per day in California? Equilibrium million gallons/day. Optimum 1 million gallons/day, c. At the equilibrium price and quantity () what is the total cost of the externality (environmental damage)? \$ that the units are millions of gallons/day). per day (remember (i) What is the deadweight loss from the externality? per day d. Show in the diagram how a \$3/gallon tax on gasoline would achieve the optimal consumption of gasoline. e. What is the total revenue raised by this tax? cost of the externality once the tax is imposed? per day. How does it compare to the total f The Green Party candidate for governor of California argues for doubling the gasoline tax from \$3 to 56, saying that it will result in less pollution and consequently make Californians better off. Is this a correct statement? Explain why or why not | 315 | 1,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-50 | latest | en | 0.89703 |
https://ja.scribd.com/document/9010831/05-T-Magnetic-Effects-of-Electric-Current-and-Magnetism | 1,563,424,213,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525500.21/warc/CC-MAIN-20190718042531-20190718064531-00331.warc.gz | 424,747,467 | 65,110 | You are on page 1of 14
# 5 - MAGNETIC EFFECTS OF ELECTRIC CURRENT AND MAGNETISM Page 1
Introduction
The branches of electricity and magnetism were unified by scientists like Oersted, Rowland,
The branch of physics covering a combined study of electricity and magnetism is known as
electromagnetism or electrodynamics. It is useful in study of subjects like plasma physics,
magneto-hydrodynamics and communication.
## 5.1 Oersted’s Observation
In 1819 A.D., Oersted, a school teacher from Denmark, observed that magnetic field is
produced around a wire carrying electric current. If a conducting wire is kept parallel to the
magnetic needle and electric current is passed through it, needle gets deflected and aligns
itself perpendicularly to the length of the wire.
## 5.2 Biot-Savart’s Law
The intensity of magnetic field due to a current element Ι dl at a point having position
vector r with respect to the electric current element is given by the formula
→ ^ → →
→ µ 0 Ι dl × r µ 0 Ι dl × r → µ 0 Ι dl sin θ
dB = = and l dB l = ,
4π r
2 4π r
3 4π r
2
→ 2
where dB = magnetic intensity in tesla ( T ) or weber / m ,
Ι dl = current element
( product of electric current and length of small
line element dl of the conductor )
## µ0 = magnetic permeability of vacuum
= 4π × 10 - tesla metre per ampere ( T m A- )
-7 -1
^ → r
r = unit vector along the direction of r =
lrl
→ →
and θ = angle between dl and r
→ → → → →
The direction of dB is perpendicular to the plane formed by dl and r . As dl and r are
taken in the plane of the figure, the direction of dB is perpendicular to the plane of the
figure and going inside it, as shown by ⊗.
On integrating the above equation, we get the total intensity at the point P due to the entire
length of the conducting wire as
→ ^ → →
→ µ0 Ι dl × r → µ0 Ι dl × r
B =
4π ∫
r2
or B =
4π ∫ r3
5 - MAGNETIC EFFECTS OF ELECTRIC CURRENT AND MAGNETISM Page 2
## A straight conductor AB carrying electric current
Ι is kept along X-axis as shown in the figure. It
is desired to find magnetic intensity at a point
P located at a perpendicular distance y from the
wire. Y-axis is along OP.
^
A small current element Ι dx i is at a distance x
from the origin on the wire.
## By Biot-Savart’s law, magnetic intensity at point
P due to this current element is
→ →
→ µ 0 Ι dx × r
dB = … … (1)
4π r3
→ ^ → ^ ^
Putting dx = dx i r = y j - x i
and
( from ∆ OPQ formed by vectors )
→ → ^ ^ ^ ^
dx × r = dx i × y j - x i = y dx k
^
→ µ 0 Ι y dx k
∴ dB =
4π r3
→ →
This field is perpendicular to the plane formed by dx and r coming out normally from the
plane of the figure. Integrating over the whole length of the wire,
→ → µ 0 Ι y dx ^
B = ∫ dB =
k
4 π r3
## From the geometry of the figure, r
2
= x
2
+ y
2
∴ r3 = ( x2 + y2)3/2
→ ^
µ0 Ι y dx
∴ B = k
4π 3
( x 2 + y 2 ) 2
2
Putting x = y tan θ, dx = y sec θ dθ
θ and integrating over the whole length of the wire, i.e.,
from θ = - θ1 to θ = θ2,
θ2 θ2
→ µ0 Ι y y sec 2 θ dθ ^ µ0 Ι ^
B =
4π
∫ 3
k
=
4πy ∫
cos θ dθ k
- θ 1 y 2 tan 2 θ + y 2 2 - θ 1
5 - MAGNETIC EFFECTS OF ELECTRIC CURRENT AND MAGNETISM Page 3
→ µ0 Ι ^ µ0 Ι ^
∴ B = [ sin θ ] - θθ2 k = [ sin θ 2 + sin θ 1 ] k
4πy 1 4πy
If the angles subtended by P at the ends A and B of the wire are α 1 and α 2, then
^
→ µ0 Ι ^ µ0 Ι
B = [ cos α1 + cos α 2 ] k =
L1
+
L2
k
4πy 4πy
y 2 + L 12 2 2
y + L 2
## If O is the midpoint of the wire, i.e., if OP is the perpendicular bisector, L1 = L2 = L / 2,
→ µ0 Ι 2L ^
∴ B = k
4πy
4y 2 + L2
Putting θ1 = θ2 = π / 2 for an infinitely long wire,
→ µ0 Ι ^
B = k
2πy
To decide the direction of the magnetic field, right hand thumb rule can be used. If the wire
is held in right hand such that the thumb is in the direction of the electric current, the
fingers encircling the wire indicate the direction of the magnetic field lines.
## Consider a point P on the axis of a ring carrying current Ι at
a distance x from its centre and having position vector r
with respect to an element of length dl of the ring.
The magnetic field dB at the point P, due to the current
element Ι dl , is in a direction perpendicular to the plane
→ →
formed by dl and r . It can be resolved into two mutually
perpendicular components:
## ( i ) dB cos φ parallel to X-axis and
( ii ) dB sin φ perpendicular to X-axis.
All dB sin φ components due to the diametrically opposite elements nullify each other,
whereas the axial components dB cos φ add up.
→ → →
µ 0 Ι dl × r µ 0 Ι dl . r sin θ
∴ dB ( x ) = dB cos φ =
l l cos φ = cos φ
4π r3 4π r3
→ → π a
But dl ⊥ r ∴ sin θ = sin = 1 and cos φ = ( from the figure )
2 r
5 - MAGNETIC EFFECTS OF ELECTRIC CURRENT AND MAGNETISM Page 4
µ0 Ι a
∴ dB ( x ) = dl
4 π r2 r
Integrating over the circumference of the ring, the resultant magnetic field at point P is
µ 0 Ιa µ 0 Ιa
B(x) =
4 π r3
∫ dl =
4 π r3
2πa
ring
2 2 2 3 2 2 3/2
From geometry of the figure, r = a + x ⇒ r = (a + x )
µ0 Ι a 2
∴ B(x) =
3
2 2
2(a + x )2
The direction of the magnetic field is along X-axis given by the right hand thumb rule. On
curling the fingers of right hand in the direction of flow of electric current, the thumb
stretching perpendicularly to the plane of the circle formed by the fingers indicate the
direction of the magnetic field.
µ0 Ν Ι a 2
B(x) =
3
2 2
2(a + x )2
## Taking x = 0, magnetic intensity at the centre of the ring is
µ0 N Ι
B ( centre ) =
2a
For a point far away from the centre of the coil as compared to its radius, x > > a.
2 2
Neglecting a in comparison to x , magnetic intensity at a distance x from the centre on the
axis of the coil is
µ0 N Ι a 2
B(x) =
2 x3
To find the direction of the magnetic field on the axis of the ring, curl the fingers of right
hand in the direction of flow of electric current. The thumb stretching perpendicularly to the
plane of the circle formed by the fingers indicates the direction of the magnetic field.
5.3 ( c ) Solenoid:
A helical coil consisting of closely wound turns of insulated conducting wire is called a
solenoid.
Fig. 1 represents a cross section along the length of the solenoid. The × signs indicate the
wires going into the paper and the • signs indicate wires coming out of the plane of paper.
The axis of the solenoid coincides with X-axis and radius of the solenoid is a. P is a point
inside the solenoid on its axis as shown in Fig. 2.
5 - MAGNETIC EFFECTS OF ELECTRIC CURRENT AND MAGNETISM Page 5
## To find magnetic intensity at P, consider a small part of a solenoid of width dx, at a
distance x from point P. It can be regarded as a thin ring. If there are n turns per unit
length of the solenoid, there will be ndx number of turns in this part. Hence magnetic
intensity at P due to this ring will be
µ 0 Ι a 2 ndx
dB ( x ) =
3
2 ( a2 + x2 ) 2
## From Fig. 2, x = a tan θ, ∴ dx = a sec2θ dθθ and 2
a
2
+ x = a
2 2 2 2 2
+ a tan θ = a sec θ
Integrating over the entire length of the solenoid, i.e., from θ = -α1 to θ = α 2, magnetic
intensity at point P is
α α
2
µ0 Ι a 2 n ( a sec 2 θ ) dθ µ0 n Ι 2
B =
2 a 3 sec 3 θ
=
2 ∫ cos θ dθ
1 1
µ0 n Ι α µ0 n Ι
= [ sin θ ] - 2α = ( sin α2 + sin α1 )
2 1 2
## In terms of angles φ1 and φ2 as shown in Fig. 1,
µ0 n Ι
B = ( cos φ2 + cos φ1 )
2
## ∴ B = µ0 n Ι, where n = number of turns per unit length of the solenoid.
For a very long solenoid, magnetic field is uniform inside the solenoid and zero outside the
solenoid just as for a capacitor of very large plates electric field is uniform in the inner
region and zero outside the plates.
Thus a capacitor can be used where a uniform electric field is required and a solenoid is
used where uniform magnetic field is required.
5 - MAGNETIC EFFECTS OF ELECTRIC CURRENT AND MAGNETISM Page 6
## The statement of Ampere’s circuital law is:
“ The line integral of magnetic induction over a closed loop in a magnetic field is equal to
the product of algebraic sum of electric currents enclosed by the loop and the magnetic
permeability.”
→ →
Mathematically, ∫ B ⋅ dl = µ 0 ∑ Ι
## To decide the sign convention for electric
currents, consider the magnetic field loop
produced by electric currents as shown in
the figure.
## Arrange a right-handed screw
perpendicular to the plane containing
closed magnetic loop and rotate it in the
direction of vector line elements taken for
line integration. Electric currents in the
direction of advancement of the screw are considered positive and the currents in the
opposite direction are considered negative.
Hence, algebraic sum of currents enclosed by the closed magnetic intensity loop is
∑ Ι = Ι3 + Ι4 - Ι1 - Ι2
Here, currents outside the loop are not to be considered even though they contribute in
producing the magnetic field.
( 1 ) To find magnetic field due to a very long and straight conductor carrying electric
current, using Ampere’s law:
## Consider a very long ( in principle infinitely long ) straight
conductor carrying electric current Ι as shown in the figure.
## As the wire is infinitely long, points P, Q and R which are at
the same perpendicular distance y from it will have the same
magnetic intensity. In fact, all points on the loop of radius y
passing through point Q will have the same magnetic field. If
this magnetic field is B , then applying Ampere’s law to the
loop,
→ →
∫ B ⋅ dl = µ0 ∑ Ι ⇒ ∫ B dl cos θ = µ0 Ι
→ →
As B and dl are in the same direction at every element, cos θ = cos 0 = 1.
∴ ∫ B dl = µ0 Ι ∴ B ∫ dl = µ0 Ι ( as B is constant )
µ0 Ι
∴ B . 2 π y = µ0 Ι ∴ B =
2πy
5 - MAGNETIC EFFECTS OF ELECTRIC CURRENT AND MAGNETISM Page 7
( 2 ) Formula of a solenoid:
## The figure shows the cross-section of a very long
solenoid. It is desired to find the magnetic intensity at
point S lying inside the solenoid.
## Taking line integral over Ampearean loop PQRS shown
in the figure,
→ → S→ → R→ → Q→ → P → →
∫ B ⋅ dl = ∫ B ⋅ dl + ∫ B ⋅ dl + ∫ B ⋅ dl + ∫ B ⋅ dl
P S R Q
The magnetic field on part PQ of the loop will be zero as it is lying outside the solenoid.
Also some part of QR and SP is outside and the part inside is perpendicular to the magnetic
field. Hence magnetic field on them is zero.
P → → Q→ → S→ →
∴ ∫ B ⋅ dl = ∫ B ⋅ dl = ∫ B ⋅ dl = 0
Q R P
→ → R R
∴ ∫ B ⋅ dl =
o
∫ B dl cos 0 = B ∫ dl = B l
S S
If n = number of turns per unit length of the solenoid, then the number of turns passing
through the Ampearean loop is n l. Current passing through each turn is Ι, so total current
passing through the loop is n l Ι.
→ →
From Ampere’s circuital law, ∫ B ⋅ dl = µ0 n l Ι
∴ B l = µ0 n l Ι and B = µ0 n Ι
This method is valid only for very long solenoid in which all points inside the solenoid can
be considered equivalent and is not advisable to use for a solenoid of finite length.
5.5 Toroid
## If the solenoid is bent in the form of a circle and its two
ends are connected to each other then the device is called a
toroid. It can be prepared by closely winding an insulated
conducting wire around a non-conducting hollow ring.
## The magnetic field produced inside the toroid carrying electric
current can be obtained using Ampere’s circuital law.
## To find a magnetic field at a point P inside a toroid which is
at a distance r from its centre, consider a circle of radius r
with its centre at O as an Amperean loop. By symmetry, the
magnitude of the magnetic field at every point on the loop is
the same and is directed towards the tangent to the circle.
→ →
∴ ∫ B ⋅ dl = ∫ B dl = B ∫ dl = B ( 2 π r ) … … … ( 1 )
5 - MAGNETIC EFFECTS OF ELECTRIC CURRENT AND MAGNETISM Page 8
If the total number of turns is N and current is Ι, the total current through the said loop is
NΙ. From Ampere’s circuital law,
→ →
∫ B ⋅ dl = µ0 N Ι … … … ( 2 )
## Comparing equations ( 1 ) and ( 2 ), B(2πr) = µ0 N Ι
µ0 N Ι N
∴ B = = µ0 n Ι, ( n = is the number of turns per unit length of the toroid )
2π r 2π r
In an ideal toroid, where the turns are completely circular, magnetic field at the centre and
outside the toroid is zero. In practice, the coil is helical and hence a small magnetic field
exists outside the toroid.
Toroid is a very important component of Tokamak used for research in nuclear fusion.
## 5.6 Force on a current carrying wire placed in a magnetic field
Ampere showed that “two parallel wires placed near each other exert an attractive force if
they are carrying currents in the same direction, and a repulsive force if they are carrying
currents in the opposite directions.”
A magnetic field is created around the wire carrying an electric current. If another wire
carrying some current is placed in its neighbourhood, then it experiences a force. The law
giving this force was given by Ampere as under.
The force acting on a current element Ι dl due
to the magnetic induction B is
→ → →
dF = Ι dl × B
## If a straight wire of length l carrying a current
Ι is placed in a uniform magnetic field B , the
force acting on the wire can be given by
→ → →
F = Ι l × B Such an arrangement is shown in the figure. The direction of force can be
determined using the right hand screw rule.
5.6 ( a ) The formula for the force between two conducting wires placed
parallel to each other and carrying currents in the same direction:
Consider two very long conducting wires placed parallel to
each other along X-axis, separated by a distance y and
carrying the currents Ι1 and Ι2 in the same direction as
shown in the figure.
## Magnetic field at a distance y from the conductor carrying
current Ι1 is
→ µ0 Ι 1 ^
B = k
2π y
5 - MAGNETIC EFFECTS OF ELECTRIC CURRENT AND MAGNETISM Page 9
The strength of this field is the same at all the points of the wire carrying the current Ι1 and
is directed along Z-axis. Therefore, the force acting on the second wire over its length l will
be
→ → → µ0 → ^ µ0 ^ ^
F = Ι2 l × B = Ι1 Ι2 l × k = Ι1 Ι2 l i × k
2π y 2π y
→ µ0 Ι1 Ι2 l ^
∴ F = j
2π y
The force F acts along negative y-direction which indicates that it is attractive.
## 5.6 ( b ) Definition of ampere
→ µ0 Ι1 Ι2 l ^
In the equation, F = j , if Ι1 = Ι2 = 1 A, y = 1 m and l = 1 m, then
2π y
l →F l 2π
=
4 π × 10 -7
µ0
== 2 × 10
-7
N
## Based on this, SΙ definition of 1 ampere curent is given as under:
When the magnetic force acting per meter length in two infinitely long wires of negligible
cross-sectional area placed parallel to each other at a distance of 1 meter in vacuum carrying
-7
identical currents is 2 × 10 - N, the current passing through each wire is 1 ampere.
## Ι = n A vd q, where q = charge on the positively charged particle,
n = number of free charge carriers per unit volume of the conductor,
vd = drift velocity.
→ → →
∴ Ι dl = q n A vd dl = q n A v d dl (Q vd and dl are in the same direction )
When this conductor is placed in a magnetic field of intensity B , the force acting on it is
→ → → → →
dF = Ι dl × B = q n A dl ( v d × B )
## ∴ the magnetic force acting on a charged particle of charge q is given by
→ → →
→ dF q n A dl ( v d × B ) → →
Fm = = = q ( vd × B )
The magnetic force acting on a charge moving through a magnetic field is perpendicular to
the velocity of the particle. Work done by this force is zero and hence the kinetic energy of
the particle remains constant. Only the direction of velocity goes on changing at every instant.
→ →
If an electric field E is also present alongwith B , the resultant force acting on the charged
particle will be
→ → → → → →
F = Fe + Fm = q [ E + v d × B ]. This force is known as Lorentz Force.
5 - MAGNETIC EFFECTS OF ELECTRIC CURRENT AND MAGNETISM Page 10
5.8 Cyclotron
Scientists E. O. Lawrence and M. S. Livingston constructed the first cyclotron in 1934 A. D.
which is used to accelerate charged particles.
## To understand how a cyclotron works, consider the motion of a
positively charged particle moving with velocity v and entering
perpendicularly uniform magnetic field of intensity B as shown
in the figure.
## The force acting on the charged particle is
→ → →
F = q ( v × B ) = q v B sin θ = q v B ( Q sin θ = π / 2 )
Under the effect of this force, the charged particle performs uniform circular motion in a
plane perpendicular to the plane formed by v and B.
m v2 mv p
∴ qvB = and r = = ,
r qB qB
## Putting v = r ωc, where ωc is called the angular frequency of the cyclotron,
m r ωc qB qB
r = ∴ ωc = and fc =
qB m 2 πm
The equation shows that the frequency does not depend on the momentum. Hence on
increasing momentum of the particle, the radius of its circular path increases but its
frequency does not. This fact is used in the design of the cyclotron.
Construction:
## Two D-shaped boxes are kept with
their diameters facing each other with
a small gap as shown in the figure.
A uniform magnetic field is developed
in the space enveloped by the two
boxes with a strong electromagnet.
These two boxes are called Dees as
they are D-shaped.
## An A.C. of high frequency is applied
between the two Dees. The device is
kept in an evacuated chamber in order
to avoid the collision of charged
particles with the air molecules.
5 - MAGNETIC EFFECTS OF ELECTRIC CURRENT AND MAGNETISM Page 11
Working:
A positively charged particle is released at the centre P of the gap at time t = 0. It gets
attracted towards the Dee which is at a negative potential at that time. It enters the uniform
magnetic field between the Dees perpendicularly and performs uniform circular motion in the
gap. As there is no electric field inside the Dees, it moves on a circular path of radius
depending upon its momentum and comes out of the Dee after completing a half circle.
As the frequency of A.C. ( fA ) is equal to fc, the diameter of the opposite Dee becomes
negative when the particle emerges from one Dee and attracts it with a force which increases
its momentum. The particle then enters the other Dee with larger velocity and hence moves
on a circular path of larger radius. This process keeps on repeating and the particle gains
momentum and hence radius of its circular path goes on increasing but the frequency
remains the same. Thus the charged particle goes on gaining energy which becomes
maximum on reaching the circumference of the Dee.
When the particle is at the edge, it is deflected with the help of another magnetic field,
brought out and allowed to hit the target.
Such accelerated particles are used in the study of nuclear reactions, preparation of artificial
radioactive substances, treatment of cancer and ion implantation in solids.
Limitations:
• According to the theory of relativity, as velocity of the particle approaches that of light,
its mass goes on increasing. In this situation, the condition of resonance ( fA = fc ) is not
satisfied.
• To accelerate very light particles like electrons, A.C. of very high frequency ( of the order
of GHz ) is required.
• It is difficult to maintain a uniform magnetic field over large sized Dees. Hence
accelerators like synchroton are developed.
## 5.9 Torque acting on a rectangular coil, carrying electric current and
suspended in a uniform magnetic field:
One turn of rectangular coil has length, QR = l and width, PQ = b
( see Fig. 1 )
→ ^
Magnetic field taken along X-axis is B = Bi
## Force acting on side PQ forming current element
→ → → →
Ι b is = F1Ιb × B
Similarly, force acting on side RS forming current
→ → → →
element - Ι b is F1' = - Ι b × B
→ →
The forces F1 and F1' are equal in magnitude,
opposite in direction and collinear. Hence, they
cancel each other.
5 - MAGNETIC EFFECTS OF ELECTRIC CURRENT AND MAGNETISM Page 12
^ → ^ ^ ^
Now, force acting on side QR forming current element - Ι l j is F2 = - Ι l j × B i = ΙlB k
^ → ^ ^
Similarly, force acting on side SP forming current element Ι l j is F2 ' = Ιl j × Bi
^
= -ΙlB k
→ →
The forces F2 and F2 ' are equal in magnitude, opposite in direction but are non-collinear. So
they give rise to a torque ( couple ).
Fig. 2 shows the top view of the coil. Here A is the
area vector of the coil which makes an angle θ with
the magnetic intensity B along X-axis.
## = magnitude of a force × perpendicular distance
between the two forces
→ →
∴ lτl = l F2 lM' N'
= Ι l b B sin θ
∴ lτl = N Ι l b B sin θ ( for a coil having N turns )
## Expressing area A of the coil in vector form
→ → →
τ = NΙ Α×B
→ → → →
= µ × B , where µ = N Ι A is called the “magnetic moment” linked with the coil.
This equation is valid for a coil of any other shape. The direction of torque is along Y-axis.
The direction of magnetic moment, µ , can be determined using the right hand screw rule.
5.10 Galvanometer
A galvanometer is a device used to detect current.
With appropriate modification, it can be converted into
an ammeter which can measure currents of the order
of an ampere or milliammeter to measure currents in
the range of milliamperes or microammeter to measure
microampere currents.
Construction:
## A light rectangular frame on which a coil of thin
copper wire is wound is pivoted between two almost
frictionless pivots and placed between cylindrical poles
of a permanent magnet, so that it can freely rotate in
the region between the poles. The poles are suitably
shaped and a small soft iron cylindrical core is placed
at the axis of the coil ( free from the coil ) to obtain
uniform magnetic field.
5 - MAGNETIC EFFECTS OF ELECTRIC CURRENT AND MAGNETISM Page 13
When the current is passed through the coil, a torque acts on it and is deflected. This
deflection causes the restoring torque in the spiral springs attached at the two ends of the
coil and the coil attains a steady deflection. The pointer attached to the coil moves on a
scale and indicates the current.
## Principle and Working
The torque developed in the coil due to the current passing through it is given by
τ = N Ι A B sin θ, where N = number of turns in the coil,
Ι = current through the coil,
A = area of the coil,
B = magnetic intensity of the field and
θ = angle between area vector of the coil and the direction of
magnetic intensity
→ →
As the magnetic field is radial, angle between A and B is 90° in any position of the coil
and sin 90° being 1,
τ = NΙAB
The restoring torque produced in the springs is directly proportional to the deflection φ of
the coil.
∴ τ ( restoring ) = k φ, where k = effective torsional constant of the springs.
## For steady deflection φ, NΙAB = k φ
k
∴ Ι = φ or, Ι ∝ φ
N A B
The scale of the galvanometer can be appropriately calibrated to measure the current.
- 11
To measure very weak currents of the order of 10 - A, the galvanometers with coils
suspended by an elastic fibre between appropriately designed magnetic poles are used.
## 5.11 Orbital magnetic moment of an electron revolving in an orbit of an atom
The magnetic dipole moment of the loop of area A carrying
current Ι is given by
M = Ι A ( per turn )
## If an electron is revolving in an orbit of radius r with frequency
f, it will pass through a point of its orbit f times in one
second. In this case, the charge passing through that point in
one second is ef where e = charge of an electron. This
constitutes an electric current
Ι = ef. Taking A = π r ,
2
M = ef (π r2 )
e ω πr2 e ω r2 e mω r 2 e
∴ M = = = = l
2π 2 2m 2m
5 - MAGNETIC EFFECTS OF ELECTRIC CURRENT AND MAGNETISM Page 14
2
where l = m ω r = angular momentum of electron and
m = mass of an electron.
## Expressing in vector form,
→ e →
M = - l
2m
→ →
Here M and l are in mutually opposite directions according to the right hand screw rule.
Hence, negative sign appears in the above equation.
e
The ratio is a constant called the gyro-magnetic ratio and its value is
2m
10 -1
8.8 × 10 C kg- . | 6,987 | 23,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.914363 |
http://www.codenlearn.com/2011/10/simple-merge-sort.html?showComment=1463123693650 | 1,624,605,741,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487622113.11/warc/CC-MAIN-20210625054501-20210625084501-00500.warc.gz | 55,110,617 | 24,282 | ## Sunday, October 9, 2011
### A Simple Merge Sort
Its easier to understand merge sort if you try to write a simple code on how you'd merge two sorted arrays into a new array which needs to be in sorted order. Lets try to code that first
A. Merge two sorted array
The complex part of merge sort is the merge routine/method, especially having to follow multiple indices inside this routine. So lets try to simplify that first.
Lets say we have 2 SORTED arrays
```int[] leftArray;
int[] rightArray;
```
Now we have to merge these 2 into a new array called
```int resultArray[];
```
The length of resultArray, would be
```(length of leftArray) + (length of rightArray).
```
Let's try to write a function/method/routine to do the merge now, the signature of this method would look something like this
```int[] sampleMerge(int[] leftArray, int[] rightArray)
```
Now that we have defined the signature, lets define the indices for each of these 3 arrays which we'll use later in the code
• '*Begin' marks the beginning of each array index and
• '*End' marks where that index would end (end of that array).
```int leftArrayBegin = 0;
int rightArrayBegin = 0;
int resultArrayBegin = 0;```
` `
```int leftArrayEnd = leftArray.length - 1;
int rightArrayEnd = rightArray.length - 1;
int[] resultArray = new int[leftArray.length + rightArray.length];
```
Now that we have defined these 3 indices, lets move the the logic where we merge the left and the right array into the result array. The merge logic is simple as follows
1. Take an element from leftArray and an element from rightArray and compare both these elements.
2. Whichever element is less, put that element into the resultArray and advance the begin index of the array from which we picked the element.
3. Repeat the above step till atleast one of "*Begin" array index reaches the end i.e "*End" index.
4. Copy the rest of the elements of the other array whose index is not reached the end into the result array.
Here is the sample case with the diagrams that shows the step by step movement of indices in the merge routine.
Step 1 : In the diagram below, we have 2 arrays, left and right arrays. We pick the first element i.e leftArray[0] and rightArray[0] of both arrays and compare them. Since left array value (1) is less than right array value (2), We move left array value (1) to resultArray and increment the leftArrayBegin index [to 1].
The below diagram in step 2 has the result array and index position after step 1 is complete.
Step 2 : We pick the current begin index position element of left array i.e. leftArray[1] and begin index position element of right array i.e. rightArray[0] and compare them. Since right array value (2) is less than left array value (3), we move right array value (2) to resultArray and increment the rightArrayBegin index [ to 1].
The below diagram in step 3 has the result array and index position after step 2 is complete.
(Rest of the below steps are repeat of the above 2 steps till we reach the end of leftArray in this case, since leftArray has less number of elements than rightArray - So feel free to skip rest of the steps and go directly to Step 7).
Step 3 : We pick the current begin index position element of left array i.e leftArray[1] and current begin index position element of right array i.e rightArray[1] and compare them. Since left array value (3) is less than right array value (4), we move left array value (3) to result array and increment the leftArrayBegin index [ to 2].
The below diagram in step 4 has the result array and index position after step 3 is complete.
Step 4 : We pick the current begin index position element of left array i.e leftArray[2] and current begin index position element of right array i.e rightArray[1] and compare them. Since right array value (4) is less than left array value (5), we move right array value (4) to resultArray and increment the rightArrayBegin index [ to 2].
The below diagram in step 5 has the result array and index position after step 4 is complete.
Step 5 : We pick the current begin index position element of left array i.e leftArray[2] and current begin index position element of right array i.e rightArray[2] and compare them. Since left array value(5) is less than right array value(6), we move left array value(5) to resultArray and increment the leftArrayBegin index [ to 3].
The below diagram in step 6 has the result array and index position after step 5 is complete,
Step 6 : We pick the current begin index position element of left array i.e. leftArray[3] and current begin index position element of right array i.e. rightArray[2] and compare them. Since right array value(6) is less than left array value(7), we move right array value(6) to resultArray and increment the rightArrayBegin index [ to 3 ].
The below diagram in step 7 has the result array and index position after step 6 is complete.
Step 7 : We pick the current begin index position element of left array i.e. leftArray[3] and current begin index position element of right array i.e rightArray[3] and compare them. Since left array value(7) is less than right array value(8), we move left array value(7) to resultArray and increment the leftArrayBegin index [ to 4] (which is more the left array size).
The below diagram in step 8 has the result array and index position after step 7 is complete.
Step 8 : The leftArray index has ended or merged into the resultArray. Now we have few pending elements only in the rightArray. So we start copying the rightArray elements into the resultArray. The below diagram we copy pending element rightArray[3] of value 8 to the resultArray and increment the rightArrayBegin index to 4.
The below diagram in step 9 has the result array and index position after step 8 is complete.
Step 9 : We copy the next right array element value (10) into resultArray and increment the rightArrayBegin index [to 5], which is the end of the rightArray.
The below diagram in step 9 has the result array and index position after step 9 is complete.
The merge ends, since there are no more elements to be merged either in the left or the right array. We have sucessfully merged both the sorted arrays into the result array which in itself is sorted.
B. Sample merge code
Here is a simple code that implements the above sample merge.
```/*
* A sample merge method to help understand the merge routine.
* This below function is not used by the merge sort
*
* This is here only for explanation purpose
*/
public int[] sampleMerge(int[] leftArray, int[] rightArray) {
int leftArrayEnd = leftArray.length - 1;
int rightArrayEnd = rightArray.length - 1;
int leftArrayBegin = 0;
int rightArrayBegin = 0;
int numElements = leftArray.length + rightArray.length;
int[] resultArray = new int[numElements];
int resultArrayBegin = 0;
// Find the smallest element in both these array and add it to the temp
// array i.e you may have a array of the form [1,5] [2,4]
// We need to sort the above as [1,2,4,5]
while (leftArrayBegin <= leftArrayEnd && rightArrayBegin <= rightArrayEnd) {
if (leftArray[leftArrayBegin] <= rightArray[rightArrayBegin]) {
resultArray[resultArrayBegin++] = leftArray[leftArrayBegin++];
} else {
resultArray[resultArrayBegin++] = rightArray[rightArrayBegin++];
}
}
// After the main loop completed we may have few more elements in
// left array so copy them
while (leftArrayBegin <= leftArrayEnd) {
resultArray[resultArrayBegin++] = leftArray[leftArrayBegin++];
}
// After the main loop completed we may have few more elements in
// right array so copy them
while (rightArrayBegin <= rightArrayEnd) {
resultArray[resultArrayBegin++] = rightArray[rightArrayBegin++];
}
return resultArray;
}
```
Now we have understood the merge routine/method, lets look at merge sort algorithm itself.
C. Merge Sort algorithm
Here is how merge sort is often explained.
1. Divide the unsorted array, of size N, into two halves.
2. Sort each sub array, each of size N/2.
3. Merge the two sorted sub arrays into the sorted, full array.
Note the base case is : If the array is of length 0 or 1, then it is already sorted.
Merge sort, sorts based on the above principle i.e an array of length 0 or 1 is already sorted. We keep splitting the input array until we reach a state where the array has only one element. Then we merge 2 of these single element arrays where merge routine takes care of sorting. Then we merge two of these '2' element arrays and so on... until we merge two halves of the input array.
Now lets try to implement merge sort algorithm in the simplest possible way as explained above.
D. Merge sort code
` `
```/*
* The mergeSort algorithm implementation
*/
private void mergeSort(int[] array, int left, int right) {
if (left < right) {
//split the array into 2
int center = (left + right) / 2;
//sort the left and right array
mergeSort(array, left, center);
mergeSort(array, center + 1, right);
//merge the result
merge(array, left, center + 1, right);
}
}
```
Done, that's it. All we are doing in the above code is as explained in the algorithm i.e if an array is of length greater than 1 then divide the array into two sub arrays and sort them independently and merge.
As you can see the algorithm can be implemented in few lines, easy to remember once you know how to implement merge routine.
Here is the implementation of the merge() routine/method used in the above mergeSort() method. The code is similar as that of the above sampleMerge() - The difference is, rather than taking two arrays and returning a sorted array, it takes a single array which has both the leftArray and rightArray is start/end indices of these arrays are either passed/calculated by the merge routine/method.
Now we know the basics spend some time understanding the indices and the arrays in the below code.
```/*
* The merge method used by the mergeSort algorithm implementation.
*/
private void merge(int[] array, int leftArrayBegin,
int rightArrayBegin, int rightArrayEnd) {
int leftArrayEnd = rightArrayBegin - 1;
int numElements = rightArrayEnd - leftArrayBegin + 1;
int[] resultArray = new int[numElements];
int resultArrayBegin = 0;
// Find the smallest element in both these array and add it to the result
// array i.e you may have a array of the form [1,5] [2,4]
// We need to sort the above as [1,2,4,5]
while (leftArrayBegin <= leftArrayEnd && rightArrayBegin <= rightArrayEnd) {
if (array[leftArrayBegin] <= array[rightArrayBegin]) {
resultArray[resultArrayBegin++] = array[leftArrayBegin++];
} else {
resultArray[resultArrayBegin++] = array[rightArrayBegin++];
}
}
// After the main loop completed we may have few more elements in
// left array copy them.
while (leftArrayBegin <= leftArrayEnd) {
resultArray[resultArrayBegin++] = array[leftArrayBegin++];
}
// After the main loop completed we may have few more elements in
// right array copy.
while (rightArrayBegin <= rightArrayEnd) {
resultArray[resultArrayBegin++] = array[rightArrayBegin++];
}
// Copy resultArray back to the main array
for (int i = numElements - 1; i >= 0; i--, rightArrayEnd--) {
array[rightArrayEnd] = resultArray[i];
}
}
```
E. What is the complexity of the above algorithm ?
E.1 Time Complexity
Merge Sort is a divide and conquer algorithm and it uses recursion which makes it difficult for us to quickly guess the complexity of the given code by checking the nested loops as explained in how to find the complexity of the given code.
The time complexity of merge sort is
O(n log(n))
How ?
To calculate complexity of a recursive function, We need to consider two things
1. The number of recursive calls required to solve the problem.
2. Amount of work done in each recursive call.
Number of recursive call in merge sort
Consider a case if the input array is of size 4, then the number of recursive calls made by merge sort is 7.
Here is how it is, each mergeSort() call results in two recursive more calls - one call for each halves of the array since we split the array into two.
• The first call starts off by making 2 calls.
• Each of these 2 calls in-turn make 2 more calls which makes it 4.
• and so forth. till we hit the base case (which is single element array).
So here is the pictorial representation of the above recursive calls for merge sort, you'll see that its like a balanced binary tree. An easy way to visualize merge sort is as a tree of recursive calls.
Here is the tree for input array of size 4.
So we see that for an array of four elements, we have a tree of depth 3.
Now try doing it for doubled the number of elements in the array i.e array size of 8, We'll have the same tree with one more level with depth increased to 4.
This suggests that the total depth of the tree is log(n), is the number of times we need to call sort routing/method recursively to reach the base case.
Now what is the work we do in each recursion ?
In the code above the only work we do is to call merge routine/method - So what is the complexity of merge routine/method?
Considering that given 2 arrays we are merging two array into the result array. If we just take the merge routine/method and see the complexity it would be O(n1 + n2).
where
n1 is length of left array and
n2 is length of right array.
based on the above we may think the complexity of merge routine to be O(n) at every merge, but its NOT true.
We are dividing the array along the recursive path. So length n1 and n2 keeps changing and the complexity of merge routine/method at each call.
1. It is 'n' on the first call to recursion (leftArray of size n/2 and rightArray of size n/2)
2. It is 'n/2' on the second call to recursion. (we merge two n/4 arrays)
3. It is 'n/4' on the third call to recursion. (we merge two n/8 arrays)
4. ....... and so on .....
5. till we merge just 2 items ( leftArray of size 1 and rightArray of size 1) and base case there is just one item, so no need to merge.
As you can see the first call to recursion is the only time the entire array is merged together at each level, a total of 'n' operations take place in the merge routine.
Based on the above,
1. there are log(n) levels
2. a total of 'n' operations takes place in the merge routine/method.
Hence the time complexity is O(n * log(n)).
E2. Space Complexity
Merge sort requires additional memory check the merge routine where we allot new memory to store the merged result.
Hence the space complexity of merge sort is
O(n)
F. Source code
Here is the complete merge sort sample java file used in this post.
```/*
* You are free to use this code anywhere, copy, modify and redistribute at your
* own risk.
* Your are solely responsibly for any damage that may occur by using this code.
*
* This code is not tested for all boundary conditions. Use it at your own risk
*/
package codenlearn;
public class MergeSort {
public void sort(int[] array) {
mergeSort(array, 0, array.length - 1);
}
/*
* The mergeSort algorithm implementation
*/
private void mergeSort(int[] array, int left, int right) {
if (left < right) {
//split the array into 2
int center = (left + right) / 2;
//sort the left and right array
mergeSort(array, left, center);
mergeSort(array, center + 1, right);
//merge the result
merge(array, left, center + 1, right);
}
}
/*
* The merge method used by the mergeSort algorithm implementation.
*/
private void merge(int[] array, int leftArrayBegin,
int rightArrayBegin, int rightArrayEnd) {
int leftArrayEnd = rightArrayBegin - 1;
int numElements = rightArrayEnd - leftArrayBegin + 1;
int[] resultArray = new int[numElements];
int resultArrayBegin = 0;
// Find the smallest element in both these array and add it to the result
// array i.e you may have a array of the form [1,5] [2,4]
// We need to sort the above as [1,2,4,5]
while (leftArrayBegin <= leftArrayEnd && rightArrayBegin <= rightArrayEnd) {
if (array[leftArrayBegin] <= array[rightArrayBegin]) {
resultArray[resultArrayBegin++] = array[leftArrayBegin++];
} else {
resultArray[resultArrayBegin++] = array[rightArrayBegin++];
}
}
// After the main loop completed we may have few more elements in
// left array copy them first
while (leftArrayBegin <= leftArrayEnd) {
resultArray[resultArrayBegin++] = array[leftArrayBegin++];
}
// After the main loop completed we may have few more elements in
// right array copy them
while (rightArrayBegin <= rightArrayEnd) {
resultArray[resultArrayBegin++] = array[rightArrayBegin++];
}
// Copy resultArray back to the main array
for (int i = numElements - 1; i >= 0; i--, rightArrayEnd--) {
array[rightArrayEnd] = resultArray[i];
}
}
/*
* A sample merge method to help understand the merge routine.
* This below function is not used by the merge sort
*
* This is here only for explanation purpose
*/
public int[] sampleMerge(int[] leftArray, int[] rightArray) {
int leftArrayEnd = leftArray.length - 1;
int rightArrayEnd = rightArray.length - 1;
int leftArrayBegin = 0;
int rightArrayBegin = 0;
int numElements = leftArray.length + rightArray.length;
int[] resultArray = new int[numElements];
int resultArrayBegin = 0;
// Find the smallest element in both these array and add it to the temp
// array i.e you may have a array of the form [1,5] [2,4]
// We need to sort the above as [1,2,4,5]
while (leftArrayBegin <= leftArrayEnd && rightArrayBegin <= rightArrayEnd) {
if (leftArray[leftArrayBegin] <= rightArray[rightArrayBegin]) {
resultArray[resultArrayBegin++] = leftArray[leftArrayBegin++];
} else {
resultArray[resultArrayBegin++] = rightArray[rightArrayBegin++];
}
}
// After the main loop completed we may have few more elements in
// left array copy them first
while (leftArrayBegin <= leftArrayEnd) {
resultArray[resultArrayBegin++] = leftArray[leftArrayBegin++];
}
// After the main loop completed we may have few more elements in
// right array copy them
while (rightArrayBegin <= rightArrayEnd) {
resultArray[resultArrayBegin++] = rightArray[rightArrayBegin++];
}
return resultArray;
}
public static void main(String args[]) {
System.out.println("Running mergeSort....");
System.out.println("Running merge sort on..{7, 1, 8, 2, 0, 12, 10, 7, 5, 3}..");
int array[] = {7, 1, 8, 2, 0, 12, 10, 7, 5, 3};
MergeSort mergeSort = new MergeSort();
mergeSort.sort(array);
dumpArray(array);
System.out.println("Now demo a simple merge routine....");
System.out.println("Merging..{1, 3, 5, 7} and ..{2, 4, 6, 8, 10}..");
int leftArray[] = {1, 3, 5, 7};
int rightArray[] = {2, 4, 6, 8, 10};
int[] mergedArray = mergeSort.sampleMerge(leftArray, rightArray);
dumpArray(mergedArray);
}
/*
* Utility for dumping the array
*/
public static void dumpArray(int[] array) {
for (int value : array) {
System.out.println(value);
}
}
}
```
1. Very well explained. Good Job
2. Excellent explanation. Few sites have this simplicity. Congrats.
3. awesome !!!!
4. This comment has been removed by the author.
5. Great job in explaining.
6. the best explanation I have ever come across.
7. Best explanation of merge sort. love it.
8. I have a question after calling the last mergesort when the left=0 and center=1,then c=0+1/2 ,c=0,
hence mergesort(arr,0,0) what will this result end int0
9. Here in the last call to mergesort l=0 and c=1,
mergesort(arr,0,1)
l<r
hence c=0+1/2 c=0
mergesort(arr,0,0) what will happen after this
10. very nice article.Once I had written merge for two arrays.Algo in itself become so much clear.Earlier i used to feel very uncomfortable in implementing merge sort.After this article it is the work of few minutes.Thanks a lot sir.
11. Best one for merge sort :) Thanks!!
12. thank you very much for explaining merge sort this much easily...
13. THANK YOU SO MUCH! This really helped
14. This is by far the best explanation. thank you very much sir.
15. Thank You . | 4,861 | 19,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2021-25 | longest | en | 0.870615 |
https://www.coolstuffshub.com/electrical-resistance/convert/volt-per-ampere-to-abohm/ | 1,656,629,201,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103915196.47/warc/CC-MAIN-20220630213820-20220701003820-00110.warc.gz | 766,739,372 | 6,301 | Volt per ampere to abohm conversion
1 V/A = 1000000000 abΩ
How to convert volt per ampere to abohm?
To convert volt per ampere to abohm, multiply the value in volt per ampere by 1000000000.
You can use the formula :
abohm = volt per ampere × 1000000000
How many abohms are in a volt per ampere?
There are 1000000000 abohms in a volt per ampere.
1 volt per ampere is equal to 1000000000 abohms.
Volt per ampere to abohm conversion table
Volt per ampere Abohm
1 V/A 1000000000 abΩ
2 V/A 2000000000 abΩ
3 V/A 3000000000 abΩ
4 V/A 4000000000 abΩ
5 V/A 5000000000 abΩ
6 V/A 6000000000 abΩ
7 V/A 7000000000 abΩ
8 V/A 8000000000 abΩ
9 V/A 9000000000 abΩ
10 V/A 10000000000 abΩ
20 V/A 20000000000 abΩ
30 V/A 30000000000 abΩ
40 V/A 40000000000 abΩ
50 V/A 50000000000 abΩ
60 V/A 60000000000 abΩ
70 V/A 70000000000 abΩ
80 V/A 80000000000 abΩ
90 V/A 90000000000 abΩ
100 V/A 100000000000 abΩ | 356 | 886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2022-27 | latest | en | 0.384343 |
https://adlmag.net/how-many-square-feet-will-a-60000-btu-furnace-heat-2/ | 1,669,631,828,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710503.24/warc/CC-MAIN-20221128102824-20221128132824-00215.warc.gz | 126,757,575 | 25,260 | A standard that is used by heating and cooling equipment companies is to provide at least 25 to 30 Btu of heat per square feet in the home for a moderate to warm climate. A mid-sized home of 2,000 square feet would need approximately 50,000 to 60,000 Btu to heat it properly.
Also, How many square feet will a 2.5 ton AC unit cool?
Air Conditioning Square Footage Range by Climate Zone
ZONE 1 ZONE 2
1.5 Tons 600 – 900 sf 600 – 950 sf
2 Tons 901-1200 sf 951 – 1250 sf
2.5 Tons 1201 – 1500 sf 1251 – 1550 sf
3 Tons 1501 – 1800 sf 1501 – 1850 sf
In this way, How many square feet will a 3 ton AC cool?
METHOD 2: Go by square feet + climate
HVAC System Sizing
Blue Green
2.5 tons 1401-1650 sf 1351-1600 sf
3 tons 1651-2100 sf 1601-2000 sf
3.5 tons 2101-2300 sf 2001-2250 sf
Contents
## How many square feet will 60000 Btu heat?
A standard that is used by heating and cooling equipment companies is to provide at least 25 to 30 Btu of heat per square feet in the home for a moderate to warm climate. A mid-sized home of 2,000 square feet would need approximately 50,000 to 60,000 Btu to heat it properly.
## Should I buy or rent a furnace?
Renting a furnace is also cheaper initially, since you make low monthly payments. However, these payments disappear, with companies needing continual rent, whereas, when you buy a furnace, the costs have an end and result in new property.
15 years
## How much does a new furnace cost to install?
Replacing an old furnace costs between \$2,000 and \$6,000 based on a national average. Things like the type of furnace (electric, gas or oil), how efficient the unit is and the brand name impact not only the furnace price itself, but the installation cost, too.
## How many square feet will a 80 000 BTU furnace heat?
Determining the Square Footage
In colder climates, you’ll want a furnace that generates 40 to 45 BTUs per square foot. At this amount, you’ll need 100,000-112,500 BTU furnace to heat a home of 2,500 square feet.
## What happens if my furnace is too big?
Oversized Furnace
The problem with a furnace that’s too large is that it puts out too much heat for the space, which causes the thermostat to shut the system down early. This places extreme stress on the furnace, which will eventually cause the system to break down.
## Can a furnace be too big for a house?
Common problems caused by an oversized furnace. Your home will be uncomfortable. When your furnace is too big, it will blast your home with too much conditioned air at one time. This can make your rooms feel too warm when your furnace is operating and lead to major temperature swings in your home.
## How much should a new gas furnace cost installed?
The cost of a standard efficiency natural gas furnace ranges from \$2,250 to \$3,800. You may incur higher costs, depending on the complexity of the install or if you choose a higher efficiency unit.
## Is a variable speed furnace worth the extra money?
The best news … a variable speed furnace motor is 80–85% more efficient than a standard furnace motor. So, if you are going to upgrade from a 90% to a 96% efficient furnace make sure it has a variable speed furnace motor, and you could enjoy energy savings of up to \$400 a year. Otherwise it’s not worth the investment.
## How many square feet will 40000 Btu heat?
Crave
Model BTUs Space it will Heat
Crave 36 17,500 – 30,000 600 – 900 square feet
Crave 48 21,000 – 40,000 800 – 1,300 square feet
Crave 60 26,000 – 50,000 1,300 – 2,100 square feet
Crave 72 30,000 – 58,000 1,500 – 2,400 square feet
## How many square feet will 40000 Btu heat?
Crave
Model BTUs Space it will Heat
Crave 36 17,500 – 30,000 600 – 900 square feet
Crave 48 21,000 – 40,000 800 – 1,300 square feet
Crave 60 26,000 – 50,000 1,300 – 2,100 square feet
Crave 72 30,000 – 58,000 1,500 – 2,400 square feet
## What does a new furnace cost?
New Furnace Replacement Cost. The national average cost of a standard efficiency natural gas furnace ranges from \$2,150 to \$5,900 with most homeowners paying around \$3,100. The price may vary based on brand, complexity of install, and the efficiency of the new unit.
## How many square feet will a 3 ton AC cool?
METHOD 2: Go by square feet + climate
HVAC System Sizing
Blue Green
3 tons 1651-2100 sf 1601-2000 sf
3.5 tons 2101-2300 sf 2001-2250 sf
4 tons 2301-2700 sf 2251-2700 sf
## How many BTUs do I need to heat 1500 square feet?
For each room, multiply the BTU per square footage times the square footage of the room. For example, a home located in Zone 5 with poor insulation would require 60 BTUs per square foot. Therefore, a room that is 300 square feet will need 18,000 BTUs (60 x 300).
## How many BTUs do I need to heat 1500 square feet?
STEP 1: Determine how many BTUs of heating and tons of AC you need
House Square Footage BTUs Needed
100 – 150 5,000
150 – 250 6,000
250 – 300 7,000
300 – 350 8,000
## What is a 2 stage furnace?
Two-stage heating means the furnace has two levels of heat output: high for cold winter days and low for milder days. Since the low setting is adequate to meet household heating demands 80% of the time, a two-stage unit runs for longer periods and provides more even heat distribution.
## How many square feet will a 2.5 ton AC unit cool?
According to this common but somewhat inaccurate method, you need 1 ton of air cooling capacity for every so many square feet of living space. While there’s some dispute over the exact amount, an often-used amount is 600 square feet. A 2.5-ton unit, then, theoretically would fit a 1,500-square-foot home.
## Is a higher BTU furnace better?
A heater with a higher BTU rating is more powerful — that is, it has a higher heat output — than one with a low BTU rating. It can do more to raise the temperature in your room each hour, so you can either heat a room more quickly or heat a larger space.
## How do you install a furnace?
Furnace Installation: Step-by-Step
1. Step 1: Choose your furnace and select the location for installation.
2. Step 2: Decide where your duct and drain will run.
3. Step 3: Connect to the ducting system.
4. Step 4: Connect vent pipes.
5. Step 5: Connect gas supply.
6. Step 6: Connect electrical supply.
7. Step 7: Connect condensate drain.
## What is the average size of a furnace?
1. Get the right furnace size.
2. Improve energy efficiency: look at the AFUE rating.
3. Select your fuel source wisely.
4. Must have features.
5. Get a high-end filter to ensure good indoor air quality.
6. Save money with a modulating gas furnace.
8. 5 Top Ranked Forced Air Furnaces.
## Is a higher BTU furnace better?
A heater with a higher BTU rating is more powerful — that is, it has a higher heat output — than one with a low BTU rating. It can do more to raise the temperature in your room each hour, so you can either heat a room more quickly or heat a larger space.
## How do I calculate HVAC size?
Once you have the base BTUs figured for your home’s size, you can then figure out the size of AC and heating units you need. For the air conditioner, divide the number by 12,000 to determine the tonnage required. For the furnace, divide the BTU by the unit’s efficiency as a decimal. In the case of a 1500 sq.
## How do I calculate HVAC size?
BTU Sizing Chart for Mini Split Systems
Area To Be Cooled Capacity Needed (BTUs Per Hour)
700 to 1,000 square feet 18,000 BTUs
1,000 to 1,200 square feet 21,000 BTUs
1,200 to 1,400 square feet 23,000 BTUs
Up to 1,500 square feet 24,000 BTUs
## How do I calculate BTU for my house?
To calculate BTU per square foot, start by measuring the square footage of each room you want to heat or cool. Then, add the square footage for each room together. Once you have the total square footage, just multiply that number by 20 to find how many BTUs per hour you’d need to heat or cool the space.
## What is the average size of a furnace?
Understanding furnace size BTUs:
A typical furnace is rated at 100,000 or 80,000 BTU per hour, but this is just an average.
## How much does a 100 000 BTU furnace cost?
The cost depends on a variety of factors, including the type of fuel used, the brand, installation costs and your home’s specific heating needs. How much does a new gas furnace cost? As of 2018, a new 100,000 BTU gas furnace can run you between \$650 and \$1,350, with installation costing between \$800 to \$3,000.
## How many tons is 1500 square feet?
Air Conditioning Square Footage Range by Climate Zone
ZONE 1 ZONE 4
1.5 Tons 600 – 900 sf 700 – 1050 sf
2 Tons 901-1200 sf 1051 – 1350 sf
2.5 Tons 1201 – 1500 sf 1351 – 1600 sf
3 Tons 1501 – 1800 sf 1601 – 2000 sf
## How do I choose the right furnace?
Understanding furnace size BTUs:
A typical furnace is rated at 100,000 or 80,000 BTU per hour, but this is just an average.
## How do I calculate HVAC size?
Once you have the base BTUs figured for your home’s size, you can then figure out the size of AC and heating units you need. For the air conditioner, divide the number by 12,000 to determine the tonnage required. For the furnace, divide the BTU by the unit’s efficiency as a decimal. In the case of a 1500 sq.
## How do I choose the right furnace?
1. Get the right furnace size.
2. Improve energy efficiency: look at the AFUE rating.
3. Select your fuel source wisely.
4. Must have features.
5. Get a high-end filter to ensure good indoor air quality.
6. Save money with a modulating gas furnace. | 2,553 | 9,406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-49 | longest | en | 0.913358 |
https://dfine.gitbook.io/leetcode/1128.number_of_equivalent_domino_pairs | 1,708,819,039,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474569.64/warc/CC-MAIN-20240224212113-20240225002113-00569.warc.gz | 204,232,502 | 195,784 | # 1128.Number of Equivalent Domino Pairs
1128.Number of Equivalent Domino Pairs
Given a list of dominoes, dominoes[i] = [a, b] is equivalent to dominoes[j] = [c, d] if and only if either (a==c and b==d), or (a==d and b==c) - that is, one domino can be rotated to be equal to another domino.
Return the number of pairs (i, j) for which 0 <= i < j < dominoes.length, and dominoes[i] is equivalent to dominoes[j].
Example 1:
Input: dominoes = [[1,2],[2,1],[3,4],[5,6]]
Output: 1
Constraints:
1 <= dominoes.length <= 40000
1 <= dominoes[i][j] <= 9
class Solution {
public:
int numEquivDominoPairs(vector<vector<int>>& dominoes) {
int res=0;
if(dominoes.size()<2) return res;
for(auto &v: dominoes)
if(v[0]>v[1]) swap(v[0],v[1]);
map<vector<int>, int> domin_map;
for(auto v:dominoes)
domin_map[v]++;
for(auto v:domin_map)
if(v.second >1) res+= v.second * (v.second -1);
return res/2;
}
}; | 310 | 886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-10 | latest | en | 0.388063 |
https://emathematics.net/quadraticgraph.php?a=2 | 1,695,464,669,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506480.7/warc/CC-MAIN-20230923094750-20230923124750-00330.warc.gz | 260,397,177 | 4,441 | # Vertex and Intercepts od a Quadratic Graph
Vertex
The quadratic equation has an extreme value located at the vertex of the graph. This vertex point is either the highest point on the graph of the equation or the lowest point:
Finding the vertex of a parabola without graphing
The location of the coordinates of the vertex of the graph of y=ax2+bx+c depends only on the coefficients of a and b:
Coordinates of the vertex: (the y-value is found by substituting the x-value)
Find the vertex of f(x)=x2+6x+8
1. Find the x-coordinate of the vertex by using the formula , where a=1 and b=6:
2. Find the y-coordinate of the vertex by evaluating f(-3)
f(-3)=(-3)2+6(-3)+8=9-18+8=-1
Hence, the vertex is at (-3,-1)
Intercepts
The intercepts of a parabola are where the curve or graph of the function crosses the axis.
Finding the intercepts of a parabola without graphing.
• To solve for the x-intercepts of a quadratic function, you set f(x) equal to 0 (this is the same as setting y equal to 0)and then solve for x.
• To solve for the y-intercept of a quadratic function, you set x equal to 0 and solve for f(x). A function has only one y-intercept.
Find the x-intercept(s) and the y-intercept(s) of the quadratic equation y=x2-4
Step 1. First, find the y-intercept. Since the y-intercept is the point where the graph crosses the y-axis, the value for x at this point is zero. Because we know that x=0, plug 0 in for x in the equation and solve for x.
y=x2-4
y=02-4
y=0-4
y=-4
Therefore, the y-intercept is (0,-4)
Step 2. Next, find the x-intercept(s). The x-intercept is the point, or points in this case, where the graph crosses the x-axis. In this case, we plug 0 in for y because y is always zero along the x-axis. Solve for x.
y=x2-4
0=x2-4
0+4=x2-4+4
4=x2
So x=-2 and x=2
The x-intercepts are (-2,0) and (2,0)
Step 3. To verify that the intercepts are correct, graph the equation on the coordinate plane:
Find the vertex and the intercepts of the quadratic equation: f(x)=5x2+25x-120
x-intercepts y-intercept Vertex ( , ) ( , ) ( , ) ( , ) | 600 | 2,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.875272 |
http://www.hextobinary.com/unit/byte/from/Gib/to/PiB/4 | 1,591,211,677,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347435987.85/warc/CC-MAIN-20200603175139-20200603205139-00405.warc.gz | 158,180,510 | 5,955 | #### Gibibit Pebibyte
##### How many Pebibytes are in 4 Gibibits?
The answer is 4 Gibibits are equal to 4.7683715820313e-7 Pebibytes. Feel free to use our online unit conversion calculator to convert the unit from Gibibit to Pebibyte. Just simply, enter value 4 in Gibibit and see the result in Pebibyte. You can also Convert 5 Gibibits to Pebibyte
##### How to Convert 4 Gibibits to Pebibytes (Gib to PiB)
By using our Gibibit to Pebibyte conversion tool, you know that one Gibibit is equivalent to 1.1920928955078e-7 Pebibyte. Hence, to convert Gibibit to Pebibyte, we just need to multiply the number by 1.1920928955078e-7. We are going to use very simple Gibibit to Pebibyte conversion formula for that. Pleas see the calculation example given below.
Convert 4 Gibibit to Pebibyte 4 Gibibit = 4 × 1.1920928955078e-7 = 4.7683715820313e-7 Pebibyte
##### What is Gibibit Unit of Measure?
Gibibit is a unit of digital information about data. One gibibit is equal to 1073741824 bits.
##### What is the symbol of Gibibit?
The symbol of Gibibit is Gib which means you can also write it as 4 Gib.
##### What is Pebibyte Unit of Measure?
Pebibyte is a unit of digital information about data. One pebibyte is equal to 1125899906842624 bytes.
##### What is the symbol of Pebibyte?
The symbol of Pebibyte is PiB which means you can also write it as 4 PiB.
##### Gibibit to Pebibyte Conversion Table
Gibibit [Gib] Pebibyte [PiB] 4 0.00000047683715820313 8 0.00000095367431640625 12 0.0000014305114746094 16 0.0000019073486328125 20 0.0000023841857910156 24 0.0000028610229492188 28 0.0000033378601074219 32 0.000003814697265625 36 0.0000042915344238281 40 0.0000047683715820313 400 0.000047683715820313 4000 0.00047683715820313
##### Gibibit to Other Units Conversion Chart
Gibibit [Gib] Output 4 Gibibit in Bit equals to 4294967296 4 Gibibit in Byte equals to 536870912 4 Gibibit in Exabit equals to 4.294967296e-9 4 Gibibit in Exabyte equals to 5.36870912e-10 4 Gibibit in Exbibit equals to 3.7252902984619e-9 4 Gibibit in Exbibyte equals to 4.6566128730774e-10 4 Gibibit in Gibibyte equals to 0.5 4 Gibibit in Gigabit equals to 4.29 4 Gibibit in Gigabyte equals to 0.536870912 4 Gibibit in Kibibit equals to 4194304 4 Gibibit in Kibibyte equals to 524288 4 Gibibit in Kilobit equals to 4294967.3 4 Gibibit in Kilobyte equals to 536870.91 4 Gibibit in Mebibit equals to 4096 4 Gibibit in Mebibyte equals to 512 4 Gibibit in Megabit equals to 4294.97 4 Gibibit in Megabyte equals to 536.87 4 Gibibit in Pebibit equals to 0.000003814697265625 4 Gibibit in Pebibyte equals to 4.7683715820313e-7 4 Gibibit in Petabit equals to 0.000004294967296 4 Gibibit in Petabyte equals to 5.36870912e-7 4 Gibibit in Tebibit equals to 0.00390625 4 Gibibit in Tebibyte equals to 0.00048828125 4 Gibibit in Terabit equals to 0.004294967296 4 Gibibit in Terabyte equals to 0.000536870912 4 Gibibit in Yobibit equals to 3.5527136788005e-15 4 Gibibit in Yobibyte equals to 4.4408920985006e-16 4 Gibibit in Yottabit equals to 4.294967296e-15 4 Gibibit in Yottabyte equals to 5.36870912e-16 4 Gibibit in Zebibit equals to 3.6379788070917e-12 4 Gibibit in Zebibyte equals to 4.5474735088646e-13 4 Gibibit in Zettabit equals to 4.294967296e-12 4 Gibibit in Zettabyte equals to 5.36870912e-13
##### Other Units to Gibibit Conversion Chart
Output Gibibit [Gib] 4 Bit in Gibibit equals to 3.7252902984619e-9 4 Byte in Gibibit equals to 2.9802322387695e-8 4 Exabit in Gibibit equals to 3725290298.46 4 Exabyte in Gibibit equals to 29802322387.69 4 Exbibit in Gibibit equals to 4294967296 4 Exbibyte in Gibibit equals to 34359738368 4 Gibibyte in Gibibit equals to 32 4 Gigabit in Gibibit equals to 3.73 4 Gigabyte in Gibibit equals to 29.8 4 Kibibit in Gibibit equals to 0.000003814697265625 4 Kibibyte in Gibibit equals to 0.000030517578125 4 Kilobit in Gibibit equals to 0.0000037252902984619 4 Kilobyte in Gibibit equals to 0.000029802322387695 4 Mebibit in Gibibit equals to 0.00390625 4 Mebibyte in Gibibit equals to 0.03125 4 Megabit in Gibibit equals to 0.0037252902984619 4 Megabyte in Gibibit equals to 0.029802322387695 4 Pebibit in Gibibit equals to 4194304 4 Pebibyte in Gibibit equals to 33554432 4 Petabit in Gibibit equals to 3725290.3 4 Petabyte in Gibibit equals to 29802322.39 4 Tebibit in Gibibit equals to 4096 4 Tebibyte in Gibibit equals to 32768 4 Terabit in Gibibit equals to 3725.29 4 Terabyte in Gibibit equals to 29802.32 4 Yobibit in Gibibit equals to 4503599627370500 4 Yobibyte in Gibibit equals to 36028797018964000 4 Yottabit in Gibibit equals to 3725290298461900 4 Yottabyte in Gibibit equals to 29802322387695000 4 Zebibit in Gibibit equals to 4398046511104 4 Zebibyte in Gibibit equals to 35184372088832 4 Zettabit in Gibibit equals to 3725290298461.9 4 Zettabyte in Gibibit equals to 29802322387695 | 1,776 | 4,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-24 | latest | en | 0.722861 |
https://gmatclub.com/forum/the-formula-for-the-sum-of-squares-of-integer-between-1-and-n-is-n-250044.html | 1,508,431,548,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823350.23/warc/CC-MAIN-20171019160040-20171019180040-00522.warc.gz | 713,511,795 | 50,159 | It is currently 19 Oct 2017, 09:45
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# The formula for the sum of squares of integer between 1 and n is n(
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Math Revolution GMAT Instructor
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[GMAT math practice question]
The formula for the sum of squares of integer between 1 and n is n(n+1)(2n+1) / 6 = 1^2 + 2^2 + … + n^2. The is the value of 11^2 + 12^2 + … + 20^2?
A. 2470
B. 2475
C. 2480
D. 2485
E. 2490
[Reveal] Spoiler: OA
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The formula for the sum of squares of integer between 1 and n is n( [#permalink]
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25 Sep 2017, 09:50
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MathRevolution wrote:
[GMAT math practice question]
The formula for the sum of squares of integer between 1 and n is n(n+1)(2n+1) / 6 = 1^2 + 2^2 + … + n^2. The is the value of 11^2 + 12^2 + … + 20^2?
A. 2470
B. 2475
C. 2480
D. 2485
E. 2490
Sum of squares from 1 to 20 $$= \frac{20*21*41}{6} = 2870$$ (use the formula given)
Sum of squares from 1 to 10 $$= \frac{10*11*21}{6} = 385$$
Hence sum of squares from 11 to 20 $$=$$ Sum of squares from 1 to 20 $$-$$ sum of squares from 1 to 10 $$= 2870-385 = 2485$$
Option D
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Math Revolution GMAT Instructor
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Re: The formula for the sum of squares of integer between 1 and n is n( [#permalink]
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27 Sep 2017, 01:39
=>
11^2 + 12^2 + … + 20^2
= ( 1^2 + 2^2 + … + 20^2 ) – (1^2 + 2^2 + … + 10^2)
= 20∙21∙41/6 - 10∙11∙21/6
= 2870 – 385
= 2485
Ans: D
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Re: The formula for the sum of squares of integer between 1 and n is n( [#permalink] 27 Sep 2017, 01:39
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Creator III
## Create a calculated Pareto dimension involving Select
Hello all, I have a part of the dataset as follows:
Vendor Sales Type Type1 Code
A 1300 C M 1234
B 2200 N S 1235
A 1400 N M 1236
B 2100 N M 1237
C 2500 C M 1240
A 300 N M 1239
C 900 C M 1240
D 3200 N M 1241
B 2000 C S 1242
D 1500 C S 1243
E 1100 N M 1246
F 3500 C M 1245
E 1500 C M 1246
G 1400 N M 1247
H 600 C M 1248
H 900 N S 1249
G 1200 C M 1247
Now if we perform a Pareto analysis on the above data, we see that the Vendors C,E,G and H are in Bottom 20%. Now I'd like to select the records of these Vendors with Type='N' and Type1='M'. So the dataset reduces to:
E 1100 N M 1246
G 1400 N M 1247
Now for these codes in the entire dataset, I'd like to display a stacked bar chart giving the amount Under the type 'C' and 'Top 80% N'(ie, Top 80% Vendors in Pareto among only Vendors in Type N). So the Type C here would display 2700(1500 from E and 1200 from G). The Top 80% N part would display 0 as neither E or G come under Top 80% in the Pareto done on only type 'N'.
If anyone can help me achieve this entire process(right from selecting only data from Bottom 20% of Type1='M' and Type = 'N' to the end) which seems very complicated and is just beyond me at the moment, I'd be really really grateful. TIA!
1 Solution
Accepted Solutions
Highlighted
MVP
## Re: Create a calculated Pareto dimension involving Select
Here you are
Dimension
=ValueList('Type C', 'Top 80% N')
Expression
Pick(Match(ValueList('Type C', 'Top 80% N'), 'Type C', 'Top 80% N'),
Sum({<Vendor, Type = {'C'},
Code = {"=Count(DISTINCT {<Vendor = {""=Sum({<Vendor>}Aggr(Rangesum(Above(Sum({<Vendor>} Sales)/Sum({<Vendor>} TOTAL Sales), 0, RowNo())), (Vendor,(=Sum({<Vendor>} Sales),Desc)))) >= 0.8""}, Type = {'N'}, Type1 = {'M'}>}Code) >= 1"}>} Sales),
Sum({<Vendor,
Code = {"=Count(DISTINCT {<Vendor = {""=Sum({<Vendor>}Aggr(Rangesum(Above(Sum({<Vendor>} Sales)/Sum({<Vendor>} TOTAL Sales), 0, RowNo())), (Vendor,(=Sum({<Vendor>} Sales),Desc)))) > 0.8""}, Type = {'N'}, Type1 = {'M'}>}Code) >= 1"},
Vendor = {"=Sum({<Vendor>}Aggr(Rangesum(Above(Sum({<Vendor>} Sales)/Sum({<Vendor>} TOTAL Sales), 0, RowNo())), (Vendor,(=Sum({<Vendor>} Sales),Desc)))) <= 0.8"},
Type = {'N'}>} Sales))
Also, please find attached the application with this email.
17 Replies
Highlighted
MVP
## Re: Create a calculated Pareto dimension involving Select
Can you post a qvf where this data is already loaded.....
Highlighted
Creator III
## Re: Create a calculated Pareto dimension involving Select
Hello. Here's the qvf. Hope you understood what I'm trying to achieve. To add to above, let's say there's an extra record:
Vendor Type Type1 Sales Code
A N M 800 1246
As A falls under Top 80% in Pareto among Vendors of type N, the bar chart should display 800 under 'Top 80%N'. Hope that makes it more clear. Thanks again!
Highlighted
Creator III
## Re: Create a calculated Pareto dimension involving Select
I hope this helps, but I didn't incorporate the 20% part please advice on what we should do because I didn't understand your statement there. I hope the QVF help but if you need more help don't hesitate to ask for help
Highlighted
Creator III
## Re: Create a calculated Pareto dimension involving Select
Please draw a chart or use simple numbers to drive all your calculations as to show how you get that vendor A is under 80%, not following the logic here at all which makes it hard to help
Highlighted
MVP
## Re: Create a calculated Pareto dimension involving Select
Type A part can be accomplished with this
Sum(Aggr(If(Sum(Aggr(
Rangesum(
Above(
Sum({<Vendor, Type = {'N'}, Type1 = {'M'}>} Sales)/
Sum({<Vendor, Type = {'N'}, Type1 = {'M'}>} TOTAL Sales)
, 0, RowNo())
)
, (Vendor,(=Sum({<Vendor, Type = {'N'}, Type1 = {'M'}>} Sales),Desc)))) > 0.8
, Sum({<Vendor, Type = {'C'}>} Sales)),
(Vendor,(=Sum({<Vendor, Type = {'N'}, Type1 = {'M'}>} Sales),Desc))))
Now I understood why Top 80% for Type N was 0 in the first dataset, but what I am not sure is why would adding this row of data change things?
Vendor Type Type1 Sales Code
A N M 800 1246
Does adding this bring this to the bottom 20% with Type='N' and Type1='M'? Because I thought we only need to bring those Vendors which 1st meet this criteria. Is that true for both Type C and Top 80% Type N?
Highlighted
Creator III
## Re: Create a calculated Pareto dimension involving Select
My bad. I'm sorry for confusing you. What you understood initially was correct. Ignore that line please. I was trying to make something clear but gave a very poor example. Sorry for the trouble.
Highlighted
Creator III
## Re: Create a calculated Pareto dimension involving Select
And yes we only need to get the Vendors which fall in the bottom 20% from that dataset for both Type C and Top 80% N.
Highlighted
MVP
## Re: Create a calculated Pareto dimension involving Select
If you had searched this site, you would have found
Recipe for a Pareto Analysis
Recipe for a Pareto Analysis – Revisited
And quite a few more...
Logic will get you from a to b. Imagination will take you everywhere. - A Einstein
Highlighted
MVP
## Re: Create a calculated Pareto dimension involving Select
I believe this should work
Dimension
=ValueList('Type C', 'Top 80% N')
Expression
Pick(Match(ValueList('Type C', 'Top 80% N'), 'Type C', 'Top 80% N'),
Sum({<Vendor = {"=Sum(Aggr(Rangesum(Above(Sum({<Vendor, Type = {'N'}, Type1 = {'M'}>} Sales)/Sum({<Vendor, Type = {'N'}, Type1 = {'M'}>} TOTAL Sales), 0, RowNo())), (Vendor,(=Sum({<Vendor, Type = {'N'}, Type1 = {'M'}>} Sales),Desc)))) > 0.8"}, Type = {'C'}>} Sales),
Sum({<Vendor = {"=Sum(Aggr(Rangesum(Above(Sum({<Vendor, Type = {'N'}, Type1 = {'M'}>} Sales)/Sum({<Vendor, Type = {'N'}, Type1 = {'M'}>} TOTAL Sales), 0, RowNo())), (Vendor,(=Sum({<Vendor, Type = {'N'}, Type1 = {'M'}>} Sales),Desc)))) > 0.8 and Sum(Aggr(Rangesum(Above(Sum({<Vendor, Type = {'N'}>} Sales)/Sum({<Vendor, Type = {'N'}>} TOTAL Sales), 0, RowNo())), (Vendor,(=Sum({<Vendor, Type = {'N'}>} Sales),Desc)))) <= 0.8"}, Type = {'C'}>} Sales)) | 2,037 | 7,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-24 | latest | en | 0.461653 |
https://gateoverflow.in/841/gate-cse-2002-question-2-11 | 1,632,706,530,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058222.43/warc/CC-MAIN-20210926235727-20210927025727-00332.warc.gz | 324,162,588 | 26,791 | 11,171 views
The running time of the following algorithm
Procedure $A(n)$
If $n \leqslant 2$ return ($1$) else return $(A( \lceil \sqrt{n} \rceil))$;
is best described by
1. $O(n)$
2. $O(\log n)$
3. $O(\log \log n)$
4. $O(1)$
how to solve this question..?
### Subscribe to GO Classes for GATE CSE 2022
The complexity will be the number of times the recursion happens which is equal to the number of times we can take square root of n recursively, till n becomes $2$.
$T(n) = T\left(\lceil \sqrt n \rceil \right) + 1$
$T(2) = 1$
$T\left(2^2\right) = T(2) + 1 = 2$
$T\left(2^{2^2}\right) = T(4) + 1 = 3$
$T\left(2^{2^3}\right) = T(16) + 1 = 4$
So, $T(n) = \lg\lg n + 1 = O\left(\log \log n\right)$
by
why is it +1 and not some constant c in recurrence relation
why we are adding constant in recurrence relation?
because u will do some constant amount of work at every recurrence
Read this article once, then you may not need to solve this again ever, it becomes very intuitive.
https://stackoverflow.com/questions/16472012/what-would-cause-an-algorithm-to-have-olog-log-n-complexity
@Hemant Parihar why this is not following O(lognlogn) complexity though problem is diveded into √n each time?
https://gateoverflow.in/197521/recurrence-relation
T(n)=T(sqrt(n))+1
We substitute n=2^m. Then T(2^m)=T(2^m/2) +1
we substitute T(2^m)=S(m).
S(m)=S(m/2)+1.
Now applying masters theoram we get S(m)=O(log m). As m=logn.
T(n)=O(log log n)
by
How you write S(m/2) please tell me i don't understand...
@69akki let s(m)=T(2^.m)
now in S of m put m=m/2 than you will get your answer
now the reduced is of the form
T(m)=aS(m/b) +1
here a=1.,b=2, k=1 and p=0 right ?
applying ...
a<b^k then case 3 of 1 will come in master's theorem
if so ans will be T(m)=O(m^k logm^P)
here p=0 , the term Logm^P becomes 1
remainig term is O(m), then it is O(logn )
please some one explain how O(loglogn)
@teluguenglish You wrote k = 1 which is wrong, k should be 0 as 1 = n^0.
Then you will get S(m) = O(logm) by extended master's theorem.
As n = 2 ^ m, we will get m = log n.
Then T(n) = S(m) = log(m) = log(log(n)) = loglogn.
Another way to solve this is:
T(n) = T(n^1/2)+c
T(n) = T(n^1/4)+c+c
T(n) = T(n^1/8)+c+c+c
T(n) = T(n^1/16)+c+c+c+c
................
.........
...........
T(n) = T(n^(1/2^k))+k.c
So, n^(1/2^k) = 2
taking log will give us:- log(n) = 2^k--------------futher taking log --------- k =loglogn
T(n) = T(2)+loglogn . c
so , ans will be C) O(loglogn)
### 1 comment
log log n.
substitute root n = m then proceed.
$Ans:C$
$T(n) = T(\sqrt n) + c$
Assume $n=2^k$
$T(2^k)=T(2^{k/2})+c$
Assume $T(2^k)=S(k)$
$S(k)=S(k/2)+c$
Apply Master Theorem
$S(k)=\Theta(1.log\ k)$
Now just do the reverse
$T(2^k)=\Theta(log\ k)$
Now replace with $k=log_{2}n$
$T(2^{log_{2}n})=\Theta(log(log_{2}n))$
$T(n)=\Theta(log(log_{2}n))$ | 1,076 | 2,868 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2021-39 | latest | en | 0.842786 |
https://ohbug.com/uva/11025/ | 1,726,561,654,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651750.27/warc/CC-MAIN-20240917072424-20240917102424-00188.warc.gz | 411,774,892 | 2,425 | # Mr. and Mrs. Hamilton
Mr. Hamilton is a very absent-minded person. He thinks about mathematics all the time, and when he thinks, his mind wanders. He starts to walk randomly around the city, without caring about where he is or where he is going, and continues to walk until he solves the problem he was thinking about. One Saturday morning, Mr. Hamilton starts thinking about an utmost difficult problem. He starts to wander away from his house, along the streets, and takes random turns (with uniform probability) at each intersection he encounters. When he walks on a street, he continues walking along the same direction at roughly constant speed until he reaches another intersection. At the Hamiltons’ house, Mrs. Hamilton is getting worried. She knows that her husband has once again wandered off somewhere in the city, and that he is probably lost. The Hamiltons have agreed on a plan that when Mr. Hamilton is lost in the city, he will wait at an intersection for Mrs. Hamilton to pick him up. Mrs. Hamilton assumes that Mr. Hamilton has walked a very long time. She drives around the city at a constant speed, and she wants you to help her come up with a strategy such that she can find her husband in the smallest expected amount of time. Input The first line in the input contains one number T, the number of test cases. Each test case is a description of the city. For each test case, there will be two numbers, n and m (2 ≤ n ≤ 13), specifying the number of intersections and the number of roads in the city, respectively. The next line contains one number v (1 ≤ v ≤ 100), the speed at which Mrs. Hamilton is driving. The intersections are labelled from 0 to n − 1. The Hamiltons’ house is at intersection 0. Then comes m lines each with three numbers u, v, w (0 ≤ u, v < n, 1 ≤ w ≤ 1000), meaning that there is a road between intersection u and v of length w. No edge appears more than once, and there are no self loops. There is no dead end intersection and it is always possible to drive from any one intersection to another. Output For each test case, first output a line containg ‘Case #:’, where # is the test case number. Then output a single line containing a fraction f, expressed in smallest form, giving the smallest expected amount of time that Mrs. Hamilton needs in order to find her husband and bring him home. Finally, output a line containing n numbers, giving the order of the intersections that Mrs. Hamilton should visit in order to reach the smallest expected time. If two or more strategies give the same time, output the lexicographically smallest one. Note that the first number in any strategy would be ‘0’, as she starts from her own house. Separate the output of each test case with a blank line. Sample Input 2 45 12 015 027 034 136
2/2 234 10 13 12 0 1 23 1 2 19 0 3 17 1 4 27 2 5 31 3 4 35 4 5 71 5 6 12 4 7 22 588 6 9 61 7 8 50 8 9 40 Sample Output Case 1: 5/6 0321 Case 2: 2209/156 0341256897 | 728 | 2,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.973396 |
http://www.math.sjsu.edu/~koev/teaching/math143M/QRIteration2.m.txt | 1,670,452,451,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-49/segments/1669446711221.94/warc/CC-MAIN-20221207221727-20221208011727-00667.warc.gz | 76,475,120 | 902 | function A=QRiteration2(A) % assume A is already Hessenberg n=size(A,1); i=n; % i=eigenvalue being computed normA=norm(A); while i>1 sigma=A(i,i); % this is our shift [Q,R]=qr(A(1:i,1:i)-sigma*eye(i)); A(1:i,1:i)=R*Q+sigma*eye(i) if abs(A(i,i-1))<10^(-16)*normA % eigenvalue #i is revealed so we deflate it i=i-1; disp('Moving on to eigenvalue number '); disp(i); end pause end | 135 | 377 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.433243 |
http://www.numbersaplenty.com/5208 | 1,590,357,625,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347385193.5/warc/CC-MAIN-20200524210325-20200525000325-00496.warc.gz | 194,589,501 | 3,723 | Search a number
5208 = 233731
BaseRepresentation
bin1010001011000
321010220
41101120
5131313
640040
721120
oct12130
97126
105208
113a05
123020
1324a8
141c80
151823
hex1458
5208 has 32 divisors (see below), whose sum is σ = 15360. Its totient is φ = 1440.
The previous prime is 5197. The next prime is 5209. The reversal of 5208 is 8025.
It can be divided in two parts, 520 and 8, that added together give a triangular number (528 = T32).
It is an Ulam number.
5208 is an undulating number in base 5.
It is a plaindrome in base 16.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (5209) by changing a digit.
5208 is an untouchable number, because it is not equal to the sum of proper divisors of any number.
It is a pernicious number, because its binary representation contains a prime number (5) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 153 + ... + 183.
It is an arithmetic number, because the mean of its divisors is an integer number (480).
25208 is an apocalyptic number.
5208 is the 42-nd octagonal number.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 5208, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (7680).
5208 is an abundant number, since it is smaller than the sum of its proper divisors (10152).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
5208 is a wasteful number, since it uses less digits than its factorization.
5208 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 47 (or 43 counting only the distinct ones).
The product of its (nonzero) digits is 80, while the sum is 15.
The square root of 5208 is about 72.1664742107. The cubic root of 5208 is about 17.3336620582. Note that the first 3 decimals are identical.
The spelling of 5208 in words is "five thousand, two hundred eight". | 578 | 2,063 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.903109 |
https://codegolf.stackexchange.com/questions/198406/bucket-and-minimize/198415 | 1,620,624,651,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989030.87/warc/CC-MAIN-20210510033850-20210510063850-00355.warc.gz | 204,162,441 | 57,851 | # Bucket and Minimize
The challenge - given a numeric list L and an integer N as inputs, write a function that:
1. finds the bucket sizes for L such that it is split into N whole buckets of equal or near-equal size, and
2. returns for each element in L the minimum of that element's bucket.
Example -
L=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
N=5
Compute the bucket sizes with the following:
$$\\left \lfloor \frac{| L |}{N} \right \rfloor + 1$/extract_tex] elements go into $$\|L| \bmod N\$$ buckets, and $$\\left \lfloor \frac{| L |}{N} \right \rfloor \$$ go into the rest, where |L| is the length of L. For the above, we get floor(12/5)+1 into 12 mod 5 buckets, and floor(12/5) in the rest: [[0, 1, 2], [3, 4, 5], [6, 7], [8, 9], [10, 11]] Finally, we output a list where each element is the minimum of its bucket: [0, 0, 0, 3, 3, 3, 6, 6, 8, 8, 10, 10] Some test cases: In -> Out [1, 2, 3, 4], 1 -> [1, 1, 1, 1] [1, 2, 3, 4], 2 -> [1, 1, 3, 3] [0, 2, 4, 6, 8], 3 -> [0, 0, 4, 4, 8] [9, 3, 0, 1, 5, 7, 4, 6, 8, 10, 12, 11, 2, 13], 5 -> [0, 0, 0, 1, 1, 1, 4, 4, 4, 10, 10, 10, 2, 2] [3, 0, -2, 0, -1], 2 -> [-2, -2, -2, -1, -1] This is a code golf challenge. Shortest answer in bytes wins. Standard loopholes apply and such. Edit: 1. you may assume L is not empty 2. you do not have to handle the case where N>|L| 3. your answer should return a flat list Sandbox • Do we have to handle an empty \L\? – Adám Jan 23 '20 at 19:07 • Do we have to handle \N>|L|\? – Adám Jan 23 '20 at 19:07 • Ah, does the output have to be flat or is a list of list acceptable? – Jonathan Allan Jan 23 '20 at 20:46 • @JonathanAllan yes your output must be a flat list – scrawl Jan 24 '20 at 9:43 • @JonathanAllan non-positive test case added – scrawl Jan 24 '20 at 9:46 ## 16 Answers # APL (Dyalog Unicode), 40 bytesSBCS Anonymous infix lambda. Takes $$\N\$$ as left argument and $$\L\$$ as right argument. {p/⌊/¨(-p)↑¨↑∘⍵¨+\p←1+@(⍳⍺|l)⌊⍺⍴⍺÷⍨l←≢⍵} Try it online! {}$$\⍺:=N,\ ⍵:=L\$$ l←≢⍵$$\l:=|L|\$$ ⍺÷⍨$$\l\over N\$$ ⍺⍴$$\(\underbrace{\frac{l}N,\frac{l}N,\frac{l}N,\cdots,\frac{l}N}_{N\text{ elements}})\$$ ⌊$$\(\underbrace{\left\lfloor\frac{l}N\right\rfloor,\left\lfloor\frac{l}N\right\rfloor,\left\lfloor\frac{l}N\right\rfloor,\cdots,\left\lfloor\frac{l}N\right\rfloor}_{N\text{ elements}})\$$ 1+@() increment at the following indices: ⍺|l$$\|L|\bmod N\$$ ⍳$$\(1,2,3,…,|L|\bmod N)\$$ p←$$\p:=(\underbrace{\overbrace{1+\left\lfloor\frac{l}N\right\rfloor,1+\left\lfloor\frac{l}N\right\rfloor,\cdots,1+\left\lfloor\frac{l}N\right\rfloor}^{|L|\bmod N\text{ elements}},\left\lfloor\frac{l}N\right\rfloor,\left\lfloor\frac{l}N\right\rfloor,\cdots,\left\lfloor\frac{l}N\right\rfloor}_{|n|\bmod N\text{ elements}})\$$ +\ cumulative sum of that list ↑∘⍵¨ for each element, take that many elements from $$\L\$$ (-p)↑¨ for each list, take as many trailing elements as the corresponding number in $$\p\$$ ⌊/¨ minimum of each list p/ replicate each minimum by the the corresponding number in $$\p\$$ # Jelly, 7 bytes œsṂṁ€F A dyadic Link accepting L on the left and N on the right which yields a list as specified. Try it online! Or see the test-suite. (Also œsµṂṁ)F is the same.) ### How? œsṂṁ€F - Link: list of order-able items, L; positive integer, N œs - split L into N chucks, longer ones to the left € - for each chunk: - last two links as a monad - i.e. f(chunk): Ṃ - minimum of chunk ṁ - mould like chunk F - flatten • "mould like chunk"... gotta love the jelly builtins – Jonah Jan 23 '20 at 20:24 # JavaScript (ES6), 101 87 bytes Takes input as (array)(n). a=>n=>a.map(w=_=>--w?m:m=Math.min(...a.slice(p,p+=w=L/n+(i++<L%n)|0)),L=a.length,i=p=0) Try it online! ## How? ### Variables • $$\L\$$ is the length of the input array • $$\i\$$ is the ID of the current bucket • $$\w\$$ is the width of the current bucket • $$\m\$$ is the minimum value in the current bucket • $$\p\$$ is a pointer to the starting point of the next bucket ### Commented a => n => // given the input array a[] and the integer n a.map(w = // initialize w to a non-numeric value _ => // for each value in a[]: --w ? // decrement w; if it's not equal to 0: m // append the current minimum : // else: m = // append the updated value of m Math.min( // which is the minimum value found in ...a.slice( // the slice of a[] p, // starting at p p += // and ending at the updated value of p (-1): w = // update w: L / n + // floor(L / n) is the base width (i++ < L % n) // add 1 if i is less than L mod n // (then increment i) | 0 // apply the floor() ) // end of slice() ), // end of Math.min() L = a.length, // initialize L i = p = 0 // start with i = p = 0 ) // end of map() # J, 5250474239373532 27 bytes ](]#(<.//.~I.))[:+/-@[]$=]
Try it online!
This solution doesn't use any of the arithmetic formulas. Instead it relies on geometric transformations.
We'll use the example
5 f 0 1 2 3 4 5 6 7 8 9 10 11
to see how the transformations work:
• ]=] Does the input equal itself, elementwise? This converts the input to ones:
1 1 1 1 1 1 1 1 1 1 1 1
• -@[]\ Split it into rows whose size is 5 (the left arg), filling the final partial row with zeros:
1 1 1 1 1
1 1 1 1 1
1 1 0 0 0
• [:+/ Sum the rows:
3 3 2 2 2
• ] (...(...)) <above result> Execute the verb in parens with that result (3 3 2 2 2) as the right arg, and the original list ] as the left arg. Now we'll break down the verb in inner parens first...
• <.//.~I. This is a dyadic hook which first transforms the right arg 3 3 2 2 2 using I., which duplicates the indexes in place according to the multiplier at each index. Thus I. 3 3 2 2 2 gives:
0 0 0 1 1 1 2 2 3 3 4 4
• <.//.~ Use this mask to group /.~ the implicit left arg (the original list):
0 0 0 1 1 1 2 2 3 3 4 4 <-- Mask
0 1 2 3 4 5 6 7 8 9 10 11 <-- Original List
• <./ Returning the min of each group:
0 3 6 8 10
• ]#( ... ) Returning to the verb in outer parens: Make right arg ] (ie, 3 3 2 2 2) copies of # the result of the verb in the inner parens. Thus: make 3 copies of the min of the first group, 3 copies of the min of the 2nd group, 2 copies of the min of the third group, etc:
0 0 0 3 3 3 6 6 8 8 10 10
# Pyth, 8 bytes
smLhSdcF
Try it online!
This answer is mostly straightforward in that c provides the appropriate splitting behaviour. After that, we map twice to each element of the inner list the element of that list which is minimal. The interesting part is performing this double map: mLhSd expands into mmhSdd.
# R, 52 47 bytes
function(L,n)ave(L,sort(seq(0,a=L)%%n),FUN=min)
Try it online!
ave replaces each member of a group with the value provided by FUN applied to that group; the rest is just figuring out the groups.
Thanks to Robin Ryder for -5 bytes.
• 47 bytes – Robin Ryder Jan 25 '20 at 22:49
• @RobinRyder thanks! – Giuseppe Jan 26 '20 at 1:14
# Python 3.8, 90 bytes
lambda L,N,x=0:sum([(w:=len(L)//N+(v<len(L)%N))*[min(L[x:(x:=x+w)])]for v in range(N)],[])
Try it online!
• I really like what you did with the walrus operator there! Do you think it makes sense for me to say that, in a way, you were able to use the := operator to stuff variable definitions inside the lambda expression? – RGS Jan 23 '20 at 23:43
• @RGS Yes, although since I needed x to be zero for the first use in the list comprehension I had to define it in the function's arguments. – Jonathan Allan Jan 23 '20 at 23:51
• sure. Very clever! – RGS Jan 23 '20 at 23:52
# CJam, 25 bytes
q~\_,@d/m]/{_,\[:e<]*}%e_
Try it online!
## Explanation
q~ "Read whole input & eval";
\ "Swap the splitted-items down";
_ "Copy the list";
, "Find its length";
@ "Roll the 3rd-to-top on top";
d "Convert the number to double";
/ "Divide them (resulting in a float)";
m] "Find the ceiling of that number";
/ "Divide the list into that many chunks";
{ }% "Map every item in that list";
_,\ "Preserve the length of the list";
:e< "e< doesn't take a single list, so";
"we have to reduce by minimum";
[ ] "Turn this result into a list";
* "Repeat the list that many times";
] "Put the result into a list";
e_ "Flatten the list";
# Python, 118 bytes
def f(L,N):l=len(L);s=l//N;m=l%N;return sum([[min(L[s*i+i*(m>0):s*i+s+i*(m>0)+(i<m)])]*(s+(i<m))for i in range(N)],[])
Try it online!
# Charcoal, 21 bytes
Ｆη⊞υＥ÷Ｌθ⁻ηι⊟θＦ⮌υＩＥι⌊ι
Try it online! Link is to verbose version of code. Explanation:
Ｆη
Loop from N down to 1.
⊞υＥ÷Ｌθ⁻ηι⊟θ
Remove the last |L|/N elements of L and add them to a separate list.
Ｆ⮌υＩＥι⌊ι
Reverse the secondary list to restore the elements to their original order, but map each to their minimum before implicitly printing them on separate lines.
# 05AB1E, 7 5 bytes
äεWδ,
-2 bytes thanks to @Grimmy.
Prints each of the integers on a separated line.
Outputting as a list would be 6 bytes instead:
äεWδ]˜
Explanation:
ä # Split the (implicit) input-list into the (implicit) input-integer
# amount of pieces, with smaller pieces at the left side if not all are equal
ε # Map each inner integer-list to:
W # Push the minimum (without popping the list itself)
δ # Using this minimum, apply to each value in the list:
, # Print (this minimum) with trailing newline
ä # Split the (implicit) input-list into the (implicit) input-integer
# amount of pieces, with smaller pieces at the left side if not all are equal
ε # Map each inner integer-list to:
W # Push the minimum (without popping the list itself)
δ # Using this minimum, apply to each value in the list:
] # Close both this double-vectorized apply-each, as well as the map itself,
# leaving the minimums in the inner lists after the apply-each is finished
˜ # Flatten this list of integer-lists
# (after which the resulting list is output implicitly as result)
• 5 bytes with vertical output: äεWδ= – Grimmy Jan 24 '20 at 16:46
• iirc one element per line is an allowed-by-default output method for lists, but just in case it isn't, here's 6 bytes with horizontal output, – Grimmy Jan 24 '20 at 16:58
• @Grimmy Ah, nice approach with =. I actually tried using the δ to create the list and get rid of the Duplicate, but couldn't really figure it out. I also had no idea using δ] inside another map/loop would work like that. o.Ô That's pretty interesting. – Kevin Cruijssen Jan 24 '20 at 17:33
# Perl 6 Raku, 66 65 63 59 bytes
-4 bytes thanks to Jo King.
{flat map {.min xx$_},@^m.rotor((1,0 X+@m/$^n)Zxx@m%$n,$n)}
Try it online!
This function causes an exception when first executed, but subsequently works. In previous versions of Rakudo, it always worked, and I have filed a Rakudo bug. Thanks for helping me find it.
## Explanation
{ } # anonymous function with parameters @m (array) and $n (bucket size) @m/$^n # calculate bucket size: length(@m) / $n (1,0 X+ ) # make a list consisting of 1 + bucket-size, bucket-size # using the cross-product metaoperator X with operator + Zxx@m%$n,$n # repeat (1 + bucket-size) length(@m) %$n times
# and repeat bucket-size $n times (* didn't work) # using the zip metaoperator Z with operator xx @^m.rotor( ) # take batches of those sizes from the list @m map {.min xx$_}, # replace each bucket with its minimum repeated by its length
flat # flatten the list
# Previous version, 65 61 bytes
{$/=$^m/$^n;flat map {.min xx$_},$m.rotor(1+$/xx$m%$n,$/xx*)} Try it online! ## Explanation { } # anonymous function with params$m (array) and $n (bucket size)$/=$^m/$^n; # calculate minimum bucket size $/ 1+$/xx$m%$n # take (length of m)%n batches of size 1+$/ ,$/xx* # and then as many batches of size $/ as possible$m.rotor( ) # from the list $m map {.min xx$_}, # replace each bucket with its minimum repeated by its length
flat # flatten the list
• You can narrow down the bug to rotoring with a Seq of Seqs. Try it online! – Jo King Jan 30 '20 at 9:18
• @JoKing Thanks, I've updated the bug. – SirBogman Jan 30 '20 at 17:26
Thank you @SirBogman and @PostRockGarfHunter for saving 13 bytes and for the great tip
Please feel free to improve it.
l%n=map(minimum.o)[0..t-1]where(t,w,v,g,a)=(length l,tdivn,w+1,tmodn*v,map(l!!));n&b=div n b*b;o i|i<g=a[i&v..i&v+w]|b<-g+(i-g)&w=a[b..b+w-1]
## Explanation
list % n=
map
(minimum . sublist)
[0..listSize - 1]
where
(listSize, widthLastBlocks, widthFirstBlocks, lengthFirstBlocks, indecesToElem)=
(length l -- t
,listSizedivn -- w
,widthLastBlocks+1 -- v
,listSizemodn*widthFirstBlocks -- g
,map(list!!)) -- a
n & b = div n b * b -- bigest multiple of b smaller than n
sublist i -- o
|i < lengthFirstBlocks =
indecesToElem [i & widthFirstBlocks
..
i&widthFirstBlocks + widthFirstBlocks - 1]
|b <- lenghtFirstBlocks + (i - lengthFirstBlocks) & widthLastBlocks =
indecesToElem [b..b + widthLastBlocks - 1]
• Welcome to the site. Nice to see Haskell. where is rarely the answer though. Here you can use bindings inside of a guard. e.g. l%n|t<-length l,w<-div t n=.... – Wheat Wizard Jan 28 '20 at 0:48
• @PostRockGarfHunter thank you, that's great, but how do I deal with o/sublist, since it's defined upon w and g? I tried that and had some erros saying that these variables were not in scope. – Leonardo Moraes Jan 28 '20 at 15:58
• I've taken a bit of a look and to be honest I can't really read your code. Obviously that is not your fault it is golfed after all. It is just pretty different from most golfed code I see so I don't really have the instincts. I would suggest trying again from scratch keeping in mind that pattern guards are shorter than where and I think the result will be shorter, or even just keep it in mind for next time. But I am not sure how to transform your current code into pattern guards since I don't have a good grasp of it. – Wheat Wizard Jan 29 '20 at 1:19
• 151 bytes: Rearranged things a bit more without fully understanding the code. Could probably golf more. Try it online! – SirBogman Jan 29 '20 at 5:32
• 145 bytes: l%n=map(minimum.o)[0..t-1]where(t,w,v,g,a)=(length l,tdivn,w+1,tmodn*v,map(l!!));n&b=div n b*b;o i|i<g=a[i&v..i&v+w]|b<-g+(i-g)&w=a[b..b+w-1] Try it online Replaced |0<1=a[g+(i-g)&w..g+(i-g)&w+w-1] with |b<-g+(i-g)&w=a[b..b+w-1] – SirBogman Jan 29 '20 at 5:46
p=replicate
[]&_=[]
f&(x:xs)=(p x$minimum$take x f)++(drop x f)&xs
f%n|(q,r)<-divMod(length f)n=f&((p r\$q+1)++p n q)
Try it online!
The % operator is the bucket and minimize function, where f is the list and n is the bucket size. The & operator is a helper that repeats x times the minimum of the first x elements of the list f, where x is the head of the right parameter list. The sum of the right parameter list may be greater than the length of the left parameter list, but the left parameter list must be divided evenly.
# Ruby, 91 77 bytes
->l,n{(0..z=l.size).flat_map{|x|(k=l.slice!(0,z/n+(z%n>x ?1:0))).map{k.min}}}
Try it online!
• In Ruby 2.7 save one byte with ->l,n{a=b=0;l.chunk{b+=(a+=1)>(w=l.size)/n+(b<w%n ?1:0)?a=1:0}.flat_map{[_2.min]*_2.size}} – SirBogman Feb 2 '20 at 5:16
# C (clang), 107 105 bytes
Z,x,y,i,m;f(*a,z,n){for(x=z%n,Z=z/n;m=*a,y=Z,z;a+=y)for(z-=i=y+=x-->0;i--;)wmemset(a,m=a[i]<m?a[i]:m,y);}
Try it online!
Thanks to @ceilingcat for saving 2 | 5,194 | 15,618 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2021-21 | latest | en | 0.755517 |
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