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8200	https://artofproblemsolving.com/wiki/index.php/Alternating_sum?srsltid=AfmBOor9XXTFnBgtMdb3F-x1rgB2mnLQn1h7wjd4hC031DE6QrSVL_mA	Art of Problem Solving Alternating sum - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Alternating sum Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Alternating sum An alternating sum is a series of real numbers in which the terms alternate sign. For example, the alternating harmonic series is . Alternating sums also arise in other cases. For instance, the divisibility rule for 11 is to take the alternating sum of the digits of the integer in question and check if the result is divisble by 11. Given an infinite alternating sum, , with , if corresponding sequence approaches a limit of [0\|zero]] monotonically then the series converges. Error estimation Suppose that an infinite alternating sum satisfies the the above test for convergence. Then letting equal and the -term partial sum equal , the Alternating Series Error Bound states that The value of the error term must also have the opposite sign as , the last term of the partial series. Examples of infinite alternating sums This article is a stub. Help us out by expanding it. Retrieved from " Category: Stubs Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
8201	https://www.purplemath.com/modules/series3.htm	Home Lessons HW Guidelines Study Skills Quiz Find Local Tutors Demo MathHelp.com Join MathHelp.com Login Select a Course Below Standardized Test Prep ACCUPLACER Math ACT Math ALEKS Math ASVAB Math CBEST Math CLEP Math FTCE Math GED Math GMAT Math GRE Math HESI Math Math Placement Test NES Math PERT Math PRAXIS Math SAT Math TEAS Math TSI Math VPT Math + more tests K12 Math 5th Grade Math 6th Grade Math Pre-Algebra Algebra 1 Geometry Algebra 2 College Math College Pre-Algebra Introductory Algebra Intermediate Algebra College Algebra Homeschool Math Pre-Algebra Algebra 1 Geometry Algebra 2 Arithmetic & Geometric Sequences IntroExamplesArith. SeriesGeo. Series Purplemath The two simplest sequences to work with are arithmetic and geometric sequences. An arithmetic sequence goes from one term to the next by always adding (or subtracting) the same value. For instance, 2, 5, 8, 11, 14,... is arithmetic, because each step adds three; and 7, 3, −1, −5,... is arithmetic, because each step subtracts 4. The number added (or subtracted) at each stage of an arithmetic sequence is called the "common difference" d, because if you subtract (that is, if you find the difference of) successive terms, you'll always get this common value. Content Continues Below A geometric sequence goes from one term to the next by always multiplying (or dividing) by the same value. So 1, 2, 4, 8, 16,... is geometric, because each step multiplies by two; and 81, 27, 9, 3, 1, ,... is geometric, because each step divides by 3. The number multiplied (or divided) at each stage of a geometric sequence is called the "common ratio" r, because if you divide (that is, if you find the ratio of) successive terms, you'll always get this common value. Find the common difference and the next term of the following sequence: 3, 11, 19, 27, 35, ... To find the common difference, I have to subtract a successive pair of terms. It doesn't matter which pair I pick, as long as they're right next to each other. To be thorough, I'll do all the subtractions: 11 − 3 = 8 19 − 11 = 8 27 − 19 = 8 35 − 27 = 8 The difference is always 8, so the common difference is d = 8. They gave me five terms, so the sixth term of the sequence is going to be the very next term. I find the next term by adding the common difference to the fifth term: 35 + 8 = 43 Then my answer is: common difference: d = 8 sixth term: 43 Find the common ratio and the seventh term of the following sequence: To find the common ratio, I have to divide a successive pair of terms. It doesn't matter which pair I pick, as long as they're right next to each other. To be thorough, I'll do all the divisions: The ratio is always 3, so r = 3. They gave me five terms, so the sixth term is the very next term; the seventh will be the term after that. To find the value of the seventh term, I'll multiply the fifth term by the common ratio twice: a6 = (18)(3) = 54 a7 = (54)(3) = 162 Then my answer is: common ratio: r = 3 seventh term: 162 Since arithmetic and geometric sequences are so nice and regular, they have formulas. For arithmetic sequences, the common difference is d, and the first term a1 is often referred to simply as "a". Since we get the next term by adding the common difference, the value of a2 is just: a2 = a + d Continuing, the third term is: a3 = (a + d) + d = a + 2d The fourth term is: a4 = (a + 2d) + d = a + 3d At each stage, the common difference was multiplied by a value that was one less than the index. Following this pattern, the n-th term an will have the form: an = a + (n − 1)d For geometric sequences, the common ratio is r, and the first term a1 is often referred to simply as "a". Since we get the next term by multiplying by the common ratio, the value of a2 is just: a2 = ar Continuing, the third term is: a3 = r(ar) = ar2 The fourth term is: a4 = r(ar2) = ar3 At each stage, the common ratio was raised to a power that was one less than the index. Following this pattern, the n-th term an will have the form: an = ar(n− 1) Memorize these n-th-term formulas before the next test. Content Continues Below Find the tenth term and the n-th term of the following sequence: , 1, 2, 4, 8,... The first thing I have to do is figure out which type of sequence this is: arithmetic or geometric. I quickly see that the differences don't match; for instance, the difference of the second and first term is 2 − 1 = 1, but the difference of the third and second terms is 4 − 2 = 2. So this isn't an arithmetic sequence. On the other hand, the ratios of successive terms are the same: 2 ÷ 1 = 2 4 ÷ 2 = 2 8 ÷ 4 = 2 (I didn't do the division with the first term, because that involved fractions and I'm lazy. The division would have given the exact same result, though.) So clearly this is a geometric sequence with common ratio r = 2, and the first term is a = . To find the n-th term, I can just plug into the formula an = ar(n− 1): To find the value of the tenth term, I can plug n = 10 into the n-th term formula and simplify: Then my answer is: n-th term: tenth term: 256 Find the n-th term and the first three terms of the arithmetic sequence having a6 = 5 and d = The n-th term of an arithmetic sequence is of the form an = a + (n − 1)d. In this case, that formula gives me . Solving this formula for the value of the first term of the sequence, I get a = . Then: a1 = a2 = = −1 a3 = This gives me the first three terms in the sequence. Since I have the value of the first term and the common difference, I can also create the expression for the n-th term, and simplify: Then my answer is: n-th term: first three terms: Find the n-th term and the first three terms of the arithmetic sequence having a4 = 93 and a8 = 65. Since a4 and a8 are four places apart, then I know from the definition of an arithmetic sequence that I'd get from the fourth term to the eighth term by adding the common difference four times to the fourth term; in other words, the definition tells me that a8 = a4 + 4d. Using this, I can then solve for the common difference d: 65 = 93 + 4d −28 = 4d −7 = d Also, I know that the fourth term relates to the first term by the formula a4 = a + (4 − 1)d, so, using the value I just found for d, I can find the value of the first term a: 93 = a + 3(−7) 93 + 21 = a 114 = a Now that I have the value of the first term and the value of the common difference, I can plug-n-chug to find the values of the first three terms and the general form of the n-th term: a1 = 114 a2 = 114 − 7 = 107 a3 = 107 − 7 = 100 an = 114 + (n − 1)(−7) = 114 − 7n + 7 = 121 − 7n Then my answer is: n-th term: 121 − 7n first three terms: 114, 107, 100 Find the n-th and the 26th terms of the geometric sequence with and a12 = 160. The two terms for which they've given me numerical values are 12 − 5 = 7 places apart, so, from the definition of a geometric sequence, I know that I'd get from the fifth term to the twelfth term by multiplying the fifth term by the common ratio seven times; that is, a12 = (a5)( r7). I can use this to solve for the value of the common ratio r: 128 = r7 2 = r Also, I know that the fifth term relates to the first by the formula a5 = ar4, so I can solve for the value of the first term a: Now that I have the value of the first term and the value of the common ratio, I can plug each into the formula for the n-th term to get: With this formula, I can evaluate the twenty-sixth term, and simplify: Then my answer is: n-th term: 26th term: 2,621,440 Once we know how to work with sequences of arithmetic and geometric terms, we can turn to considerations of adding these sequences. URL: Page 1Page 2Page 4Page 5 Select a Course Below Standardized Test Prep ACCUPLACER Math ACT Math ALEKS Math ASVAB Math CBEST Math CLEP Math FTCE Math GED Math GMAT Math GRE Math HESI Math Math Placement Test NES Math PERT Math PRAXIS Math SAT Math TEAS Math TSI Math VPT Math + more tests K12 Math 5th Grade Math 6th Grade Math Pre-Algebra Algebra 1 Geometry Algebra 2 College Math College Pre-Algebra Introductory Algebra Intermediate Algebra College Algebra Homeschool Math Pre-Algebra Algebra 1 Geometry Algebra 2 Share This Page Terms of Use Privacy Contact About Purplemath About the Author Tutoring from PM Advertising Linking to PM Site licencing Visit Our Profiles © 2024 Purplemath, Inc. All right reserved. Web Design by
8202	https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-equations-and-inequalities/cc-6th-inequalities/v/inequalities-on-a-number-line	Plotting an inequality example (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Pre-K - 8th grade Math: High school & college Math: Multiple grades Math: Illustrative Math-aligned Math: Eureka Math-aligned Math: Get ready courses Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Help us do more We'll get right to the point: we're asking you to help support Khan Academy. 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Skip to lesson content 6th grade math Course: 6th grade math>Unit 7 Lesson 7: Intro to inequalities with variables Testing solutions to inequalities Testing solutions to inequalities (basic) Plotting inequalities Plotting an inequality example Graphing basic inequalities Inequality from graph Plotting inequalities Inequalities word problems Inequalities word problems Graphing inequalities review Math> 6th grade math> Equations & inequalities> Intro to inequalities with variables © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Plotting an inequality example NY.Math: NY‑6.EE.8 Google Classroom Microsoft Teams About About this video Transcript Drawing a number line helps visualize 'x is less than 4'. We mark 4 with a circle, not a dot, because 4 isn't included. Then, we color the line below 4, showing all values less than 4. Easy peasy, lemon squeezy.Created by Sal Khan and Monterey Institute for Technology and Education. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Gina Marquez 8 years ago Posted 8 years ago. Direct link to Gina Marquez's post “what does it mean when th...” more what does it mean when the dot is open on the 2 and the line is going both negative and positive ways? Answer Button navigates to signup page •Comment Button navigates to signup page (10 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Mr Henriques 8 years ago Posted 8 years ago. Direct link to Mr Henriques's post “It means that All Real Nu...” more It means that All Real Numbers except 2 is the solution. Comment Button navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... David White 2 years ago Posted 2 years ago. Direct link to David White's post “Would it still be x < 4 i...” more Would it still be x < 4 if you did not put a circle on 4? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 2 years ago Posted 2 years ago. Direct link to Kim Seidel's post “The dot or circle is alwa...” more The dot or circle is always used so there is no ambiguity as to where the inequality starts. An open dot tells you that the inequality is "<" or ">" with the arrow's direction telling you which applies. A solid dot tells you that the inequality is ">=" or "<=". 3 comments Comment on Kim Seidel's post “The dot or circle is alwa...” (7 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more neelswapm 11 years ago Posted 11 years ago. Direct link to neelswapm's post “I don't get the concept o...” more I don't get the concept of the closed and open circle?? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Daberculosis 11 years ago Posted 11 years ago. Direct link to Daberculosis's post “Think of the open and clo...” more Think of the open and closed circles as your hand. If your hand is open, you can't grasp anything or "contain" any object. This is kind of like 5 < x. The circle will be open because it does not contain the 5 because x is "greater than" 5. x has to be greater than 5, so 5 is not an answer to the inequality. Now if your hand is closed around an item you can contain it within your hand. For an inequality example, let's use 2 ≥ x. the circle is closed because 2 is contained and agrees with the inequality given. 2 is a solution because 2 is greater than or EQUAL TO x. 1 comment Comment on Daberculosis's post “Think of the open and clo...” (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... justina harris 2 years ago Posted 2 years ago. Direct link to justina harris's post “Choose the graph that sho...” more Choose the graph that shows the solution of the inequality on the number line. c < –1 (1 point) A number line is shown. There is an open circle on the number negative 1. The number line is shaded to the left of negative 1.A number line is shown. There is an open circle on the number negative 1. The number line is shaded to the right of negative 1.A number line is shown. There is a closed circle on the number negative 1. The number line is shaded to the left of negative 1.A number line is shown. There is a closed circle on the number negative 1. The number line is shaded to the right of negative 1. Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 2 years ago Posted 2 years ago. Direct link to Kim Seidel's post “Hints: 1) You need number...” more Hints: 1) You need numbers "less than" -1. That should tell you which side to shade. 2) An open circle means you start at the number -1 but you do not include it in the solution. A closed circle means you start at -1 and it is included as part of the solution. Hope this helps. Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more TREYA 2 months ago Posted 2 months ago. Direct link to TREYA's post “how do you know when its ...” more how do you know when its a open circle or a closed circle Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 2 months ago Posted 2 months ago. Direct link to Kim Seidel's post “An open circle is used fo...” more An open circle is used for "<" and ">" inequality symbols. It denotes that the number is not included in the solution set. A closed circle is used for "<=" and ">=" inequality symbols. It denotes that the number is included in the solution set. Hope this helps. Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more B 2 years ago Posted 2 years ago. Direct link to B's post “To be honest I don't unde...” more To be honest I don't understand these signs: <,> but the ones with the line under them can someone explain? Answer Button navigates to signup page •5 comments Comment on B's post “To be honest I don't unde...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 2 years ago Posted 2 years ago. Direct link to Kim Seidel's post “See this lesson: more See this lesson: Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Stefan eatmon 6 years ago Posted 6 years ago. Direct link to Stefan eatmon's post “doesn't open circle mean ...” more doesn't open circle mean it is not an option Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer MagesticBookworm 6 years ago Posted 6 years ago. Direct link to MagesticBookworm's post “yes, closed circle means ...” more yes, closed circle means it includes that number Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more KAIHANR 2 years ago Posted 2 years ago. Direct link to KAIHANR's post “Whats the point with the ...” more Whats the point with the whole filled dot and empty dot? Answer Button navigates to signup page •Comment Button navigates to signup page (0 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 2 years ago Posted 2 years ago. Direct link to Kim Seidel's post “A solid/filled dot is use...” more A solid/filled dot is used for ">=" or "<=" inequalities. The open dot is used for ">" and "<" inequalities. They help you to see whether the value is part of the solution set (a solid dot) or excluded from the solution set (an open dot). 1 comment Comment on Kim Seidel's post “A solid/filled dot is use...” (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... NYLIEC 3 years ago Posted 3 years ago. Direct link to NYLIEC's post “i did not no ther wer so ...” more i did not no ther wer so much math on con to do. i didint no that ther wuz so much pagis uv math infakt i didint no ther wer other pagis uv math on cone. Answer Button navigates to signup page •1 comment Comment on NYLIEC's post “i did not no ther wer so ...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Mahnoor a year ago Posted a year ago. Direct link to Mahnoor's post “Guys here is what this pe...” more Guys here is what this person is saying (I think): I didn't know there was so math on Khan to do. I didn't know there were so many pages in math. In fact, I didn't know there were other pages in math on Khan. Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... LEOND 3 years ago Posted 3 years ago. Direct link to LEOND's post “So when dealing with ineq...” more So when dealing with inequality’s the value is always squared Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 3 years ago Posted 3 years ago. Direct link to Kim Seidel's post “How did you get that impr...” more How did you get that impression? There is no squared values in the video. At this level of math, you will see many inequalities that are going to look just like the equations you've been solving except the "=" is an inequality symbol. In higher level math, you would learn how to deal with quadratic inequalities where a term would be squared. Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Video transcript Graph x is less than 4. So let's draw ourselves a number line over here. So let me draw a number line. I'll start here at 0, so 0, 1, 2, 3, 4, 5. And we could go below 0. We'd have negative 1, negative 2, negative 3, negative 4. I could keep going. Now, we want to graph all of the x's that are less than 4, but we're not including 4. It's not less than or equal to 4. It's just less than 4. And to show that we're not going to include 4, what we're going to do is we're going to draw a circle around 4. So this shows us that we're not including 4. If we were including 4, I would make that a solid dot. And to show that we're going to do all the values less than 4, we want to shade in the number line below 4, going down from 4, just like that. And then we can just shade in the arrow just like that. So this right here is all of the values less than 4. And you could test it out. Take any value where there's blue. So there's blue over here, negative 2. Negative 2 is definitely less than 4. If you take this value right here, this 2, it's definitely less than 4. 4 is not included because 4 is not less than 4. It's equal to 4. 5 is not included because 5 is not less than 4. Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: exercise Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. 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8203	https://en.wikipedia.org…holera-map-1.jpg	Jump to content File:Snow-cholera-map-1.jpg From Wikipedia, the free encyclopedia File File history File usage Global file usage Metadata Size of this preview: 639 × 599 pixels. Other resolutions: 256 × 240 pixels \| 512 × 480 pixels \| 819 × 768 pixels \| 1,092 × 1,024 pixels \| 2,183 × 2,048 pixels \| 20,124 × 18,877 pixels. Original file (20,124 × 18,877 pixels, file size: 18.38 MB, MIME type: image/jpeg) \| \| \| --- \| \| \| This is a file from the Wikimedia Commons. Information from its description page there is shown below. Commons is a freely licensed media file repository. You can help. \| \| \| \| --- \| \| \| This is a featured picture, which means that members of the community have identified it as one of the finest images on the English Wikipedia, adding significantly to its accompanying article. If you have a different image of similar quality, be sure to upload it using the proper free license tag, add it to a relevant article, and nominate it. \| \| \| \| --- \| \| \| This image was selected as picture of the day on the English Wikipedia for January 22, 2024. \| Summary \| \| \| \| --- \| Warning \| The original file is very high-resolution. It might not load properly or could cause your browser to freeze when opened at full size. \| Open in ZoomViewer \| \| \| \| --- \| \| DescriptionSnow-cholera-map-1.jpg \| Original map made by John Snow in 1854. Cholera cases are highlighted in black, showing the clusters of cholera cases (indicated by stacked rectangles) in the London epidemic of 1854. The map was created in order to better understand the pattern of cholera spread in the 1854 Broad Street cholera outbreak, which Snow would use as an example of how cholera spread via the fecal-oral route through water systems as opposed to the miasma theory of disease spread. The contaminated pump is located at the intersection of Broad Street and Cambridge Street (now Lexington Street), running into Little Windmill Street. The map marks an important part of the development of epidemiology as a field, and of disease mapping as a whole. \| \| Date \| \| \| Source \| Map of the book "On the Mode of Communication of Cholera" by John Snow, originally published in 1854 by C.F. Cheffins, Lith, Southhampton Buildings, London, England. The uploaded images is a digitally enhanced version found on the UCLA Department of Epidemiology website. \| \| Creator \| \| John Snow (1813–1858) \| \| \| \| --- --- \| \| \| \| Description \| British physician, epidemiologist and anesthesiologist \| \| \| Date of birth/death \| 15 March 1813 \| 16 June 1858 \| \| Location of birth/death \| York \| London \| \| Authority file \| : Q356407 VIAF: 54386380 ISNI: 0000000081334467 LCCN: n87102751 NLA: 36017274 Open Library: OL332521A WorldCat \| \| creator QS:P170,Q356407 \| \| Geotemporal data \| \| \| Bounding box \| \| \| \| \| --- \| N: 51.5091122°N \| \| \| \| W: 0.1441628°W \| \| E: 0.1306517°W \| \| S: 51.5168847°N \| \| \| \| \| Georeferencing \| View the georeferenced map in the Wikimaps Warper \| \| Other versions \| File:Snow-cholera-map.jpg \| \| \| \| \| \| \| \| --- --- --- \| \| \| \| \| \| \| \| --- --- \| \| \| \| \| --- \| \| \| \| \| This is a featured picture on the English language Wikipedia (Featured pictures) and is considered one of the finest images. See its nomination here. --- If you think this file should be featured on Wikimedia Commons as well, feel free to nominate it. If you have an image of similar quality that can be published under a suitable copyright license, be sure to upload it, tag it, and nominate it. \| \| \| Licensing \| \| \| \| \| \| \| --- --- --- \| \| This is a faithful photographic reproduction of a two-dimensional, public domain work of art. The work of art itself is in the public domain for the following reason: \| \| \| Public domainPublic domainfalsefalse \| \| \| \| --- \| \| \| The author died in 1858, so this work is in the public domain in its country of origin and other countries and areas where the copyright term is the author's life plus 100 years or fewer. --- This work is in the public domain in the United States because it was published (or registered with the U.S. Copyright Office) before January 1, 1930. \| \| This file has been identified as being free of known restrictions under copyright law, including all related and neighboring rights. \| \| Commons Public Domain Mark 1.0falsefalse The official position taken by the Wikimedia Foundation is that "faithful reproductions of two-dimensional public domain works of art are public domain". This photographic reproduction is therefore also considered to be in the public domain in the United States. In other jurisdictions, re-use of this content may be restricted; see Reuse of PD-Art photographs for details. \| Original upload log (All user names refer to en.wikipedia) 2006-12-30 23:15 Rewardiv 3045×2840×8 (1183741 bytes) Original map made by John Snow in 1854, copied from Author died in 1858, material is public domain. Captions Add a one-line explanation of what this file represents ジョン・スノーによるゴールデン・スクウェアのコレラの大発生時の死者の分布を表した図 Diagram showing the distribution of deaths during the 1854 Broad Street cholera outbreak Items portrayed in this file depicts cholera outbreak London creator John Snow inception 1854 media type image/jpeg File history Click on a date/time to view the file as it appeared at that time. \| \| Date/Time \| Thumbnail \| Dimensions \| User \| Comment \| --- --- --- \| \| current \| 21:19, 16 November 2020 \| \| 20,124 × 18,877 (18.38 MB) \| Artoria2e5 \| Reverted to version as of 14:12, 6 May 2020 (UTC) \| \| \| 21:12, 16 November 2020 \| \| 1,092 × 1,024 (320 KB) \| Artoria2e5 \| temporarily reduce size for wikimaps \| \| \| 14:12, 6 May 2020 \| \| 20,124 × 18,877 (18.38 MB) \| Sette-quattro \| higher resolution version \| \| \| 13:53, 22 June 2007 \| \| 3,045 × 2,840 (1.13 MB) \| Finavon \| {{Information \|Description=Original map made by John Snow in 1854, copied from Author died in 1858, material is public domain. \|Source=Originally from [ \| File usage More than 100 pages use this file. 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8204	https://www.aao.org/education/disease-review/abetalipoproteinemia	This website uses tracking technologies to enable our website functionalities, to enhance user experience or to analyze performance and traffic. We may also share or sell information about your use of our site with our social media, advertising, and analytics partners. This allows us to perform targeted advertising and to select ads and content that will be more relevant to you. Below you can Accept Default Settings, Reject All trackers, or exercise your right to opt -in or -out of the sale of personal data, targeted advertising, profiling, and the processing of sensitive data by clicking on “Manage Your Privacy Choices.” For more details on the data we process and how to exercise your rights, see our Privacy Policy Learn More Opt out of sale of personal data and targeted advertising When you visit our website, we store trackers on your browser to collect information. The information collected might relate to you, your preferences, or your device, and is used to make the site work as you expect it to and to provide a more personalized web experience. However, you can choose not to allow certain types of tracking technologies, which may impact your experience of the site and the services we are able to offer. You can exercise your rights to opt-out by clicking on “Sale of Personal Data, Profiling, and Targeted Advertising,” below. You may also click on “Processing of Sensitive Data, Sale of Sensitive Data, and Processing for Secondary Purposes” to opt-in to further types of processing so that we may further tailor your experience to you. You cannot opt-out of First Party Strictly Necessary Trackers as they are deployed to ensure the proper functioning of our website (such as prompting the tracking technologies banner and remembering your settings, to log into your account, to redirect you when you log out, etc.). For more details on the data we process and how to exercise your rights, and to view information related to required opt-in disclosures, see our Privacy Policy Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Sale of Personal Data and Targeted Advertising Third party trackers collect information to use for analytics and to personalize your experience with targeted ads. Under the Colorado CPA, the Virginia CDPA, the Texas DPSA, the Oregon CPA, the Montana CDPA, and the Florida DBR, you have the right to opt-out of the sale of your personal data to third parties, of targeted advertising related processing, and of some types of profiling. You may exercise your rights by using the toggles below. If you opt out, the ads and content that you see may not be as relevant to you. Under the Colorado CPA, you have the right to opt back in to these categories at any time should you initially choose to opt out, and you may do so using the same toggles provided below. For more details on the data we process and how to exercise your rights, and to view information related to required opt-in disclosures, see our Privacy Policy Targeting Cookies label These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising. Performance Cookies label These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
8205	https://rgmia.org/papers/v8n3/mlc4phf.pdf	ON THE MONOTONICITY AND LOG-CONVEXITY OF A FOUR-PARAMETER HOMOGENEOUS MEAN ZHEN-HANG YANG Abstract. A four-parameter homogeneous mean F(p, q; r, s; a, b) is deﬁned by another approach. The criterion for monotonicity and logarithmically convex-ity of which are presented, and two reﬁned two-parameter inequality’s chains concerning some classical mean values are deduced. 1. Introduction The so-called two-parameter mean or extended mean values between two unequal positive numbers x and y were deﬁned ﬁrst by K.B. Stolarsky as (1.1) E(r, s; x, y) =                              s(xr −yr) r(xs −ys) 1 r−s , r ̸= s, rs ̸= 0; xr −yr r(ln x −ln y) 1 r , r ̸= 0, s = 0; xs −ys s(ln x −ln y) 1 s , r = 0, s ̸= 0; exp xr ln x −yr ln y xr −yr −1 r , r = s ̸= 0; √xy, r = s = 0. . It contains many mean values, for instance: E(1, 0; x, y) = L(x, y) = x−y ln x−ln y, x ̸= y; x, x = y. (1.2) E(1, 1; x, y) = E(x, y) =    e−1 xx yy 1 x−y , x ̸= y; x, x = y. (1.3) E(2, 1; x, y) = A(x, y) = x + y 2 . (1.4) E(3 2, 1 2; x, y) = h(x, y) =x + √xy + y 3 . (1.5) The monotonicity of E(r, s; x, y) has been researched by K.B. Stolarsky , E. B. Leach and M. C. Sholander and others also in [3, 8, 9, 19] using diﬀerent ideas and simpler methods. Feng Qi studied the log-convexity for the parameters of the extended mean in , and pointed out the two-parameters mean is a log-concave function with respect to either parameter r or s on interval (0, +∞) and is a log-convex function on interval (−∞, 0). Date: May 02, 2005. 2000 Mathematics Subject Classiﬁcation. Primary 26A51, 26B25; Secondary 26D15, 26E60. Key words and phrases. homogeneous functions, monotonicity, log-convexity, mean, inequali-ties chain . 1 2 ZHEN-HANG YANG In , Alfred Witkowski considered more general means deﬁned by (1.6) R(u, v; r, s; x, y) = E(u, v; xr, yr) E(u, v; xs, ys) 1 r−s further and the following results for the monotonicity of R were obtained: Theorem 1. (Corollary 4 in ) R increases in r and s if u+v > 0 and decreases otherwise. Theorem 2. (Corollary 5 in ) R increases in u and v if r+s > 0 and decreases otherwise. On the other hand, the extended mean was generalized to two-parameter homo-geneous functions in [15, 16]. That is: Deﬁnition 1. Assume f: U(⫅R+ × R+) →R+ is an n-order homogeneous func-tion for variables x and y, and is continuous and 1st partial derivatives exist, (a, b) ∈R+ × R+ with a ̸= b, (p, q) ∈R × R. If (1, 1) / ∈U, then deﬁne that Hf(p, q; a, b) = f(ap, bp) f(aq, bq) 1 p−q (p ̸= q, pq ̸= 0), (1.7) Hf(p, p; a, b) = lim q→pHf(a, b; p, q) = Gf,p(p = q ̸= 0), (1.8) where (1.9) Gf,p = G 1 p f (ap, bp), Gf(x, y) = exp xfx(x, y) ln x + yfy(x, y) ln y f(x, y) , fx(x, y) and fy(x, y) denote partial derivatives with respect to 1st and 2nd variable of f(x, y) respectively. If (1, 1) ∈U, then deﬁne further Hf(p, 0; a, b) = f(ap, bp) f(1, 1) 1 p (p ̸= 0, q = 0), (1.10) Hf(0, q; a, b) = f(aq, bq) f(1, 1) 1 q (p = 0, q ̸= 0), (1.11) Hf(0, 0; a, b) = lim p→0Hf(a, b; p, 0) = a fx(1,1) f(1,1) b fy(1,1) f(1,1) (p = q = 0). (1.12) When f(x, y) = L(x, y), we can get two-parameter logarithmic mean HL(p, q; a, b), which is just equal to extended mean E(p, q; a, b) deﬁned by (1.1). For avoiding confusion, the extended mean will be called two-parameter logarithmic mean, and denote by HL(p, q; a, b) or HL(p, q) or HL in what follows. Concerning the monotonicity and log-convexity of the two-parameter homoge-neous functions, there are the following results: Theorem 3. [15, 16]Let f(x, y) be a positive n-order homogenous function deﬁned on U(⫅R+ × R+) and be 2nd diﬀerentiable. If I1 = (ln f)xy < (>)0, then Hf(p, q) is strictly increasing (decreasing) in either p or q on (−∞, 0) ∪(0, +∞). Theorem 4. [17, 18]Let f(x, y) be a positive n-order homogenous function deﬁned on U(⫅R+ × R+) and be 3rd diﬀerentiable. If (1.13) J = (x −y)(xI1)x < (>)0, where I1 = (ln f)xy, then Hf(p, q) is strictly log-convex (log-concave) in either p or q on (0, +∞), and log-concave (log-convex) on (−∞, 0). By the above theorems we have A FOUR-PARAMETER HOMOGENEOUS FUNCTION 3 Corollary 1. The conditions are the same as in Theorem 3. If (1.13) holds, then Hf(p, 1−p) is strictly decreasing (increasing) in p on (0, 1 2), increasing (decreasing) on ( 1 2, 1). If f(x, y) is symmetric with respect to x and y further, then the above monotone interval can be extended from (0, 1 2) to (−∞, 0)∪(0, 1 2) and ( 1 2, 1) to ( 1 2, 1)∪(1, +∞), respectively. Corollary 2. The conditions are the same as Theorem 3. If (1.13) holds, then for p, q ∈(0, +∞) with p ̸= q, there is (1.14) Gf, p+q 2 < (>)Hf(p, q) < (>) p Gf,pGf,q. For p, q ∈(−∞, 0) with p ̸= q, inequality (1.14) is reversed. If f(x, y) is deﬁned on R+×R+ and is symmetric with respect to x and y further, then substituting p + q > 0 for p, q ∈(0, +∞) and p + q < 0 for p, q ∈(−∞, 0), (1.14) is also true, respectively. As applications of the above results, we also have the following conclusions: Conclusion 1. For f(x, y) = L(x, y), A(x, y), E(x, y), where x, y > 0 with x ̸= y, then 1) Hf(p, q) are strictly increasing in either p or q on (−∞, +∞); 2) Hf(p, q) are strictly log-concave in either p or q on (0, +∞), and log-convex on (−∞, 0); 3) Hf(p, 1 −p) are strictly increasing in p on (−∞, 1 2), and decreasing on ( 1 2, +∞). 4) If p + q > 0, then (1.15) Gf, p+q 2 > Hf(p, q) > p Gf,pGf,q. Inequality (1.15) is reversed if p + q < 0. Conclusion 2. For f(x, y) = D(x, y) = \|x −y\|, where x, y > 0 with x ̸= y, then 1) HD(p, q) is strictly decreasing in either p or q on (−∞, 0) ∪(0, +∞); 2) Hf(p, q) is strictly log-concave in either p or q on (−∞, 0), and log-convex on (0, +∞); 3) HD(p, 1 −p) is strictly decreasing in p on (−∞, 0) ∪(0, 1 2), and increasing on ( 1 2, 1) ∪(1, +∞); 4) If p, q ∈(0, +∞), there is (1.16) GD, p+q 2 < HD(p, q) < p GD,pGD,q. Inequality (1.16) is reversed if p, q ∈(−∞, 0). 2. Main Results Let us substitute HL(r, s; x, y) for f(x, y) in Deﬁnition 1, then Hf(p, q; a, b) is a mean of positive x and y with four parameters r, s, p and q, which is called four-parameter mean values. For expedience, we will adopt our notations to introduce the Deﬁnition. Deﬁnition 2. Assume (a, b) ∈R+ × R+ with a ̸= b, (p, q), (r, s) ∈R × R, then call F (p, q; r, s; a, b) four-parameter homogeneous mean, which is deﬁned as follows: (2.1) F (p, q; r, s; a, b) = L(apr, bpr) L(aps, bps) L(aqs, bqs) L(aqr, bqr) 1 (p−q)(r−s) , if pqrs(p −q)(r −s) ̸= 0, 4 ZHEN-HANG YANG or (2.2) F (p, q; r, s; a, b) = apr −bpr aps −bps aqs −bqs aqr −bqr 1 (p−q)(r−s) ,if pqrs(p −q)(r −s) ̸= 0. if pqrs(p−q)(r−s) = 0, then the F (a, b; p, q; r, s) are deﬁned as its corresponding limits, for example: F (p, p; r, s; a, b) = lim q→pF (a, b; p, q; r, s) = E(apr, bpr) E(aps, bps) 1 p(r−s) , if prs(r −s) ̸= 0, p = q, F (p, 0; r, s; a, b) = lim q→0F (a, b; p, q; r, s) = L(apr, bpr) L(aps, bps) 1 p(r−s) , if prs(r −s) ̸= 0, q = 0, F (0, 0; r, s; a, b) = lim p→0F (a, b; p, 0; r, s) = G(a, b), if rs(r −s) ̸= 0, p = q = 0, where L(x, y), E(x, y)are deﬁned by (1.2), (1.3) respectively, G(a, b) = √ ab In the case of not being confused, we set F (p, q; r, s; a, b) = F (p, q) = F (r, s) = F (p, q; r, s) = F (a, b) The following properties of four-parameter mean values F (a, b; p, q; r, s) are ver-iﬁed easily: Property 1 F (p, q; r, s; a, b) are symmetric with respect to a and b, i.e. (2.3) F (a, b) = F (b, a); Property 2 F (p, q; r, s; a, b) are symmetric with respect to p and q , i.e. (2.4) F (p, q) = F (q, p); Property 3 F (p, q; r, s; a, b) are symmetric with respect to r and s, i.e. (2.5) F (r, s) = F (s, r); Property 4 F (p, q; r, s; a, b) are symmetric with respect to (p, q) and (r, s), i.e. (2.6) F (p, q; r, s) = F (r, s; p, q). Obviously, so long as the signs of I1 and J are certain, then the monotonicity and log-convexity of Hf(p, q) with respect to either p or q are also certain with it. For example, for f(x, y) = L(x, y), A(x, y), E(x, y), there are I1 < 0, J > 0, and then corresponding monotonicity and log-convexity of two-parameter homogeneous functions Hf(p, q) are conﬁrmed. Owing to that HL(r, s; x, y) contain L(x, y), A(x, y) and E(x, y), naturally, we could make conjecture on there are I1 = (ln f)xy < 0, J = (x −y)(xI1)x > 0 for f(x, y) =HL(r, s; x, y). The purpose of this paper is to verify the conjecture, and get accordingly the following results on the monotonicity and log-convexity of Hf(p, q), where f(x, y) =HL(r, s; x, y). Theorem 5. If r + s > (<)0, then F (p, q; r, s; a, b) are strictly increasing (decreas-ing) in either p or q on (−∞, +∞); Theorem 6. If r + s > (<)0, then F (p, q; r, s; a, b) are strictly log-concave (log-convex) in either p or q on (0, +∞), and log-convex (log-concave) on (−∞, 0); Corollary 3. If r + s > (<)0, then F (p, 1 −p; r, s; a, b) are strictly increasing (decreasing) in p on (−∞, 1 2), and decreasing (increasing) on ( 1 2, +∞). A FOUR-PARAMETER HOMOGENEOUS FUNCTION 5 Notice for f(x, y) =HL(r, s; x, y), because Gf(x, y) = exp xfx(x, y) ln x + yfy(x, y) ln y f(x, y) = exp 1 r −s rxr xr −yr − sxs xs −ys ln x + 1 r −s − ryr xr −yr + sys xs −ys ln y = exp 1 r−s xr xr −yr ln xr − yr xr −yr ln yr − xs xs −ys ln xs − ys xs −ys ln ys = E(xr, yr) E(xs, ys) 1 r−s , by Theorem 6 and 2, we get Corollary 4. Suppose (p + q)(r + s)< 0, then (2.7) GHL, p+q 2 < F (p, q; r, s; a, b) < p GHL,pGHL,q where GHL,t = G 1 t HL(at, bt), GHL(x, y) = E(xr, yr) E(xs, ys) 1 r−s , E(x, y) is deﬁned by (1.3). Inequality (2.7) is reversed if (p + q)(r + s)> 0. 3. Lemmas The following three lemmas are useful in proofs of the main results. Lemma 1. Suppose x, y > 0 with x ̸= y,let (3.1) K(t) =    xtyt xt −yt t(x −y) −2 , t ̸= 0; L2(x, y), t = 0. then we have 1) K(−t) = K(t); 2) K(t) is strictly increasing in (−∞, 0), and decreasing in (0, +∞). Proof. 1) An easy computation results in part 1) of the Lemma, of which details are omitted. 2) By directly calculations, we get K′(t) K(t) = ln x + ln y −2(xt ln x −yt ln y) xt −yt + 2 t = 2 t ln p xtyt −(xt ln x −yt ln y xt −yt −1) = 2 t ln G(xt, yt) −ln E(xt, yt) . By the well-known inequality E(a, b) > √ ab, we can get part two of the Lemma immediately. The following Lemma is a well-known inequality , which will be used in proof of Lemma 3. Lemma 2. For positive numbers a and b, the following inequality holds: (3.2) L(a, b) < A + 2G 3 = a + 4 √ ab + b 6 6 ZHEN-HANG YANG Lemma 3. Suppose x, y > 0 with x ̸= y,let (3.3) N(t) =    xtyt xt + yt 2 xt −yt t(x −y) −3 , t ̸= 0; L3(x, y), t = 0. then we have 1) N(−t) = N(t); 2) N(t) is strictly increasing in (−∞, 0), and decreasing in (0, +∞). Proof. 1) An easy computation results in part one, of which details are omitted. 2) By direct calculations, we get N ′(t) N(t) = ln x + ln y + xt ln x + yt ln y xt + yt −3(xt ln x −yt ln y) xt −yt + 3 t = 1 + xt xt + yt − 3xt xt −yt ln x + 1 + yt xt + yt + 3yt xt −yt ln y + 3 t = −x2t + 4xtyt + y2t x2t −y2t ln x + x2t + 4xtyt + y2t x2t −y2t ln y + 3 t = 3 t −x2t + 4xtyt + y2t x2t −y2t (ln x −ln y) = 3 t 2t(ln x −ln y) x2t −y2t x2t −y2t 2t(ln x −ln y) −x2t + 4xtyt + y2t 6 . Substituting a, b for x2t, y2t in the above last one expression, then (3.4) N ′(t) N(t) = 3 t L−1(a, b) " L(a, b) −a + 4 √ ab + b 6 # , in which L(a, b)−a + 4 √ ab + b 6 < 0 by Lemma 2, and L−1(a, b) > 0. Consequently, N ′(t) > 0 if t < 0, and N ′(t) < 0 if t > 0. The proof is completed. 4. Proofs of Main Results Since F (a, b; p, q; r, s) = HHL(a, b; p, q), where HL = HL(r, s; x, y) = E(r, s; x, y) is deﬁned by (1.3), it is enough to make certain the signs of I1 = (ln HL)xy and J = (x −y)(xI1)x. Proof of Theorem 5. Let us observe that ln HL = 1 r −s [ln \|s\| + ln \|xr −yr\| −ln \|r\| −ln \|xs −ys\|] . Through straightforward computations, we have I1 = (ln HL)xy = 1 xy (r −s) " r2xryr (xr −yr)2 − s2xsys (xs −ys)2 # = 1 xy (r −s) " r2xryr (xr −yr)2 − s2xsys (xs −ys)2 # = 1 xy(x −y)2 K(r) −K(s) r −s . A FOUR-PARAMETER HOMOGENEOUS FUNCTION 7 By Lemma 1, if r > s > 0, we have K(r) −K(s) r −s < 0; If r > −s > 0, we have also K(r) −K(s) r −s = K(r) −K(−s) r + (−s) < 0. Thus I1 < 0 if r + s > 0.Likewise I1 > 0 if r + s < 0. By Theorem 3, this proof is completed. proof of Theorem 6. Let us consider that J = (x −y) (xI1)x = x −y xy(r −s) " −r3xryr (xr + yr) (xr −yr)3 + s3xsys(xs + ys) (xs −ys)3 # = −2 xy(x −y)2 N(r) −N(s) r −s . By Lemma 3, if r > s > 0, we have N(r) −N(s) r −s < 0; If r > −s > 0, we also have N(r) −N(s) r −s = N(r) −N(−s) r + (−s) < 0. Thus J > 0 if r + s > 0.Likewise J < 0 if r + s < 0. Using Theorem 4, this completes the proof. 5. Inequality’s Chains for Two-parameter Means The four-parameter homogeneous mean values F (p, q; r, s; a, b) contain a good many two-parameter means, for example: (see Table 1) (p, q) F (p, q; r, s; a, b) (p, q) F (p, q; r, s; a, b) (2, 1) ar + br as + bs 1 r−s ( 1 2, 1 2) E(a r 2 , b r 2 ) E(a s 2 , b s 2 ) 2 r−s (1, 1) E(ar, br) E(as, bs) 1 r−s ( 3 4, 1 4) a r 2 + ( √ ab) r 2 + b r 2 a s 2 + ( √ ab) s 2 + b s 2 ! 2 r−s (1, 1 2) a r 2 + b r 2 a s 2 + b s 2 2 r−s ( 2 3, 1 3) a r 3 + b r 3 a s 3 + b s 3 3 r−s (0, 1) s r ar −br as −bs 1 r−s ( 3 2, −1 2) ar + ( √ ab)r + br as + ( √ ab)s + bs ! 1 2(r−s) ( √ ab) 1 2 (1, −1) √ ab (2, −1) ar + br as + bs 1 3(r−s) ( √ ab) 2 3 % Table 1. some familiar two-parameter mean values Example 1. By Theorem 5, we can get a series of inequalities in form of two-parameter. If r + s > 0, then (5.1) F (1, −1; r, s; a, b) < F (0, 1; r, s; a, b) < F (1, 1 2; r, s; a, b) < F (1, 1; r, s; a, b) < F (2, 1; r, s; a, b), 8 ZHEN-HANG YANG i.e. (5.2) G < s r ar −br as −bs 1 r−s < a r 2 + b r 2 a s 2 + b s 2 2 r−s < E(ar, br) E(as, bs) 1 r−s < ar + br as + bs 1 r−s , which can be concisely denoted by: (5.3) G < L(ar, br) L(as, bs) 1 r−s < A(a r 2 , b r 2 ) A(a s 2 , b s 2 ) 2 r−s < E(ar, br) E(as, bs) 1 r−s < A(ar, br) A(as, bs) 1 r−s , where L, E, A are deﬁned by (1.2)-(1.4). Remark 1. Inequality (5.2) or (5.3) is a generalization of the following inequalities G < L < A + G 2 < E < A. Example 2. By Theorem 3, we can get another more reﬁned inequalities. If r+s > 0, then (5.4) F ( 1 2, 1 2; r, s; a, b) > F ( 2 3, 1 3; r, s; a, b) > F ( 3 4, 1 4; r, s; a, b) > F (1, 0; r, s; a, b) > F ( 3 2, −1 2; r, s; a, b) > F (2, −1; r, s; a, b), i.e. (5.5) h E(a r 2 ,b r 2 ) E(a s 2 ,b s 2 ) i 2 r−s > a r 3 + b r 3 a s 3 + b s 3 3 r−s > a r 2 + p a r 2 b r 2 + b r 2 a s 2 + √ a s 2 b s 2 + b s 2 ! 2 r−s > s r ar −br as −bs 1 r−s > ar + √ arbr + br as + √ asbs + bs ! 1 2(r−s) √ G > ar + br as + bs 1 3(r−s) G 2 3 , which can be concisely denoted by (5.6) E(a r 2 , b r 2 ) E(a s 2 , b s 2 ) 2 r−s > A(a r 3 , b r 3 ) A(a s 3 , b s 3 ) 3 r−s > " h(a r 2 , b r 2 ) h(a s 2 , b s 2 ) # 2 r−s > L(ar, br) L(as, bs) 1 r−s > h(ar, br) h(as, bs) 1 2(r−s) √ G > A(ar, br) A(as, bs) 1 3(r−s) G 2 3 , where L(x, y), E(x, y) A(x, y) and h(x, y) and are deﬁned by (1.2)-(1.5), respec-tively. Example 3. If replace a, b with a2, b2, then inequalities (5.6) can be rewritten as: (5.7) E(ar, br) E(as, bs) 1 r−s > " A(a 2r 3 , b 2r 3 ) A(a 2s 3 , b 2s 3 ) # 3 2(r−s) > h(ar, br) h(as, bs) 1 r−s > L(a2r, b2r) L(a2s, b2s) 1 2(r−s) > h(a2r, b2r) h(a2s, b2s) 1 4(r−s) √ G > A(a2r, b2r) A(a2s, b2s) 1 6(r−s) G 2 3 . Remark 2. Inequality (5.6) or (5.7) not only strengthen and generalize Lin Tong-po and Stolarsky inequality, but also uniﬁes them into the same inequality’s chain. A FOUR-PARAMETER HOMOGENEOUS FUNCTION 9 References P´ eter Czinder and Zsolt Pales, An Extension of the Hermite-Hadmard Inequality and An Application for GINI and Stolarsky Means, JIPAM, 5(2)(2004), 1-8. [ONLINE] Available online at 03 JIPAM/167 03 www.pdf. C. Gini, Diuna formula comprensiva delle media, Metron, 13(1938), 3–22. Baini Guo, Shiqin Zhang and Feng Qi, An Elementary proof of Monotonicity for Extend Mean Values With Two Parameters, Mathematics in Practice and Theory, 29(2)(1999), 169-173. Peter. A. H¨ ast¨ o, A Montonicity Property of Ratios of Symmetric Homogeneous Means. [ON-LINE] Available online at 02.pdf J.-Ch. Kuang, Applied Inequalities, 2nd ed., Hunan Education Press, Changsha City, Hunan Province, China, 1993. (Chinese) J.-Ch. Kuang, Applied Inequalities, 3rd ed., Shangdong Science and Technology Press, Jinan City, Shangdong Province, China, 2004. (Chinese) E. B. leach and M. C. Sholander, Extended mean values, Amer. Math. Monthly, 85(1978), 84-90. E. B. leach and M. C. Sholander . Extended mean values, J. Math. Anal. Appl. 92(1983), 207-223. Feng Qi, Logarithmic Convexities of the Extended Mean, Values, RGMIA Research Report Collection, 2(5), Article 5, 1999. K. B. Stolarsky, Generalizations of the Logarithmic Mean, Math. Mag. 48(1975), 87-92. Alfred Witkowski, Weighted extended mean values, Colloq. Math. 2004 (accepted). RGMIA Research Report Collection, 7(1), Article 6, 2004, [ONLINE] Available online at Alfred Witkowski, Convexity of Weighted Extended Mean values, RGMIA Re-search Report Collection, 7(2), Article 10, 2004, [ONLINE] Available online at Alfred Witkowski, Comparishon Theorem for Generalization of Stolarsky Means, RGMIA Research Report Collection, 8(1), Article 6, 2005, [ONLINE] Available online at Zhen-Hang Yang, Simple Discriminance for Convexity of Homogeneous Functions and Appli-cations, Study in College Mathematics, 4(7)(2004), 14-19. Zhen-Hang Yang, On the Monotinity and Log-convexity for Two-parameter Homogeneous Hunctions, Inequalities Research Bulletin, 11(2)(2004), 302-324. Zhen-Hang Yang, On the Homogeneous Functions with Two Parameters and Its Monotonicity, RGMIA Research Report Collection, 8(2), Article 10, 2005. [ONLINE] Available online at Zhen-Hang Yang, On The Logarithmic Convexity for Two-parameters Homogeneous Func-tions, RGMIA Research Report Collection, 8(2), Article 21, 2005. [ONLINE] Available online at Zhen-Hang Yang, The reﬁnements and extensions of log-convexity for two-parameter homo-geneous functions. Pales Z. Inequalities for diﬀerences of powers, J. Math. Anal. Appl. 131(1988), 271-281. Zhejiang Electric Power Vocational Technical College E-mail address: yzhkm@163.com
8206	https://libraryguides.centennialcollege.ca/c.php?g=717548&p=5121821	What is modular arithmetic? - Modular Numbers and Cryptography - Library Guides at Centennial College Skip to Main Content Centennial College Libraries Library Guides 04. Learning Centre Modular Numbers and Cryptography What is modular arithmetic? Search this Guide Search Modular Numbers and Cryptography Modular Numbers and Cryptography What is modular arithmetic? Modular Addition and Subtraction Modular Multiplication Modular Exponential Solving Modular Equations Cryptography The Diffie-Hellman-Merkle Key Exchange Scheme Rivest-Shamir-Adleman Public Key Cryptography Modular Arithmetic In mathematics,modular arithmeticis a system of arithmetic for integers, where numbers "wrap around" when reaching a certain value, called themodulus. The modern approach to modular arithmetic was developed by Carl Friedrich Gauss in his book Disquisitiones Arithmeticae, published in 1801. Let's use the hours of a clock as an example, except we will start with 0 instead of 12 on top. Starting at noon, the hour hand goes in the order: 1,2,3,4,5,6,7,8,9,10,11,0,1,2,3,4,5,6,7,8,9,10,11,0,… You can see the numbers continue to "wrap around" every 12 count. We called this count inmodulo 12. In modulo 3, we "wrap around" every 3 count 1,2,0,1,2,0… All integers can be expressed as 0,1, or 2 in modulo 3, we give these set of integers that can be represented with 0,1, or 2 in modulo 3 the nameresidue classes. In general, for any natural number n, the module n residues are the integers that are whole numbers less than n: meaning you can represented from 0 to n−1 We say that a is the modulo-mresidueof n when n≡a(mod m), and 0≤a<m. Congruence We saw above that all of the integers can be represented by one of the modulo 3 residues. However, multiple integers can be represented as the same residue. For example, if we found 4 in modulo 3, we count four residues in order 0,1,2,0. So 4 is represented by the residue 0 in modulo 3. Now 7 in modulo 3, we count seven residues in order 0,1,2,0,1,2,0. So 7 is also represented by the residue 0 in modulo 3. So we say that 0, 4, and 7 are congruentin modulo 3. 1≡4≡7(mod 3) The (mod 3) tells us that we are working with the integers in modulo 3. Notice that the difference between congruent numbers in modulo 3 is also 3. Two integers a and b are congruentin modulo n when a−b is a multiple of n. In other words, a≡b(mod n) when a−b n is an integer. Otherwise, a≢b(mod n), which means that a and b arenot congruentin modulo n. Examples Find the following residues in the stated modulo. 1.1 5(mod 4) One method is to count 15 residues in modulo 4. 1,2,3,0,1,2,3,0,1,2,3,0,1,2,3 Thus, 15≡3(mod 4) or 3 is the residue of 15 in modulo 4. However, this can become tedious as the numbers increase in size. A shorter method is to find the remainder using long division. For example, 15÷4=3 R 3. Which means 15 wraps around 3 times and has a remainder of 3. Thus, we only have to count 3 residues in modulo 3. 2.82(mod 11) 82÷11=7 R 5. Thus, the remainder 5 means you only have to count 5 residues in modulo 11, so 82≡5(mod 11). This also means that 11×7+5=82. 3.−17(mod 10) For negative integers, when you perform the division −17÷10=−1 R−7, the remainder is -7. However, we know that the reside for 10 must be one of the numbers 1,2,3,4,5,6,7,8,9,0. Thus, we must work out the division such that the remainder is positive. −17÷10=−2 R 3 Above when we increase the quotient by 1 to 2, it results in a positive remainder. This also means that −2×10+3=17. ∴−17≡3(mod 10) 4.−73(mod 7) −73÷7=−10 R−3 or =−11 R 4. Since we want the positive remainder. −73≡4(mod 7) 5.907,342(mod 23) How do we find the remainder of a really big number? You can first use your calculator to find 907,342÷23=39,449.65217 What is important is the value before the decimal39,449 Now to find the remainder 907,342−39,449×23=15 ∴907,342(mod 23)≡15(mod 23) Modular Numbers Practice Questions Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License. Last Updated:Jun 21, 2024 2:54 PM URL: Print Page Login to LibApps Report a problem click to chat Library staff are here to help! contact us
8207	https://www.demonstrations-mathematiques.com/en/properties-of-prime-numbers/	The properties of prime numbers The set (\mathbb{P}) is the set of prime numbers: $$ \mathbb{P} = \Bigl {2, 3, 5, 7, 11, 13, ...etc. \Bigr }$$ We call a prime number, a number (p \in \mathbb{P}) which has only itself and (1) as a divisor. $$ \mathcal{D}(p) = \bigl {p, 1 \bigr } $$ Likewise, we will say that two numbers ((a, b) \in \hspace{0.04em} \mathbb{Z}^2) are coprime if their unique common divisor is (1). $$ \mathcal{D}(a, b) = \bigl {1 \bigr } \hspace{0.2em} \Longleftrightarrow \hspace{0.2em} a \wedge b = 1 $$ Two prime numbers are coprime Two prime numbers are coprime. $$ \forall (p_1, p_2) \in \hspace{0.04em} \mathbb{P}, $$ $$ (p_1, p_2) \in \hspace{0.04em} \mathbb{P} \Longrightarrow p_1 \wedge p_2 = 1$$ Breakdown in prime factors All natural number (n \geqslant 2) uniquely decomposes into a prime factors product. $$ \forall n \in \mathbb{N}, \enspace n \geqslant 2, \enspace \forall i \in \mathbb{N}, \enspace (\forall p_i \in \mathbb{P}, \enspace \exists \alpha_i \in \mathbb{N}), $$ $$ n= p_1^{\alpha_1}p_2^{\alpha_2}...p_i^{\alpha_i}$$ Every number higher than 2 owns at least one prime divisor All natural number (n \geqslant 2) has at least one prime divisor. Every non-prime number higher than 4 owns at least one strict divisor All non-prime natural number (n \geqslant 4 ) has at least one strict divisor (d_0 ) such as ( d_0 \leqslant \sqrt{n} ). Coprimity link between a prime number and an integer $$ \forall p \in \mathbb{P}, \enspace \forall a \in \mathbb{Z}, $$ $$ p \nmid a \hspace{0.2em} \Longleftrightarrow \hspace{0.2em} a \wedge p = 1 $$ Euclid's lemma $$ \forall p \in \mathbb{P}, \enspace \forall (a, b) \in \hspace{0.04em} \mathbb{Z}^2, $$ $$ p/ab \hspace{0.2em} \Longrightarrow \hspace{0.2em} (p/a) \enspace or \enspace (p/b) \qquad \bigl(\text{Euclid's lemma} \bigr) $$ Euclid's lemma corollary $$ \forall (p, a, b) \in \hspace{0.04em} \mathbb{P}^3, $$ $$ p/ab \hspace{0.2em} \Longrightarrow \hspace{0.2em} (p=a) \enspace or \enspace (p=b) \qquad \bigl(\text{Euclid's lemma (corollary)} \bigr) $$ Recap table of the prime numbers properties Demonstrations Two prime numbers are coprime Let ((p_1, p_2) \in \hspace{0.04em} \mathbb{P} ) be two prime numbers. So: $$ \Biggl { \begin{gather} \mathcal{D}(p_1) = \bigl {1, p_1 \bigr } \ \mathcal{D}(p_2) = \bigl {1, p_2 \bigr } \end{gather} $$ Their only common divisor is ( 1 ). Then, $$ \forall (p_1, p_2) \in \hspace{0.04em} \mathbb{P}, $$ $$ (p_1, p_2) \in \hspace{0.04em} \mathbb{P} \Longrightarrow p_1 \wedge p_2 = 1$$ The reciprocal is not true. For example: (16 \wedge 35 = 1) And yet these numbers are not prime. Breakdown in prime factors Let (n \in \mathbb{N}) be a natural number with (n \geqslant 2). This number admits a finite number of divisors. if ( n ) is prime There is only one factor. $$ n= p_1^{\alpha_1} \enspace \enspace (with \enspace \alpha_1 = 1) $$ if ( n ) is not prime We know that ( n ) has at least one prime divisor. $$ n = n_1 p_1 $$ if ( n_1 ) is not prime On recommence : $$ n_1 = n_2 p_2 $$ $$ n = p_1. n_2 p_2 $$ if ( n_2 ) is not prime We carry out this process until the last divisor which will necessary be prime. We could possibly come across the same prime numbers several times in a row. $$ n= p_1^{\alpha_1}p_2^{\alpha_2}...p_i^{\alpha_i}$$ And finally, All natural number (n \geqslant 2) uniquely decomposes into a prime factors product. $$ \forall n \in \mathbb{N}, \enspace n \geqslant 2, \enspace \forall i \in \mathbb{N}, \enspace (\forall p_i \in \mathbb{P}, \enspace \exists \alpha_i \in \mathbb{N}), $$ $$ n= p_1^{\alpha_1}p_2^{\alpha_2}...p_i^{\alpha_i}$$ Every number higher than 2 owns at least one prime divisor Let (n \in \mathbb{N}) be a natural number with (n \geqslant 2). Two cases arise: ( n ) is prime Then, ( n / n ). It has at least one prime divisor. ( n ) is not prime ( n ) has at least one strict divisor. $$ \mathcal{D}(a) = \bigl {1, d_1, d_2, \hspace{0.2em} ... \hspace{0.2em}, n } $$ Let ( d_1 ) be the smallest stricit divisor of (n ), (d_1 ) is necessarily prime, because if it were not, it would have a divisor lower than itself which would divide ( n ), and ( d_1 ) would not be the smallest divisor of (a ). And finally, All natural number (n \geqslant 2) has at least one prime divisor. Every non-prime number higher than 4 owns at least one strict divisor Let (n \in { \mathbb{N} \hspace{0.2em} \backslash\ \mathbb{P} } ) be a non-prime integer with (n \geqslant 4), and (d_0 \in \mathbb{N}) a strict divisor of (n). So, $$ \exists d' \geqslant d_0, \enspace n =d_0 d' $$ By multiplying both members by (d_0 ), $$ d_0^2 \leqslant d_0d' \Longleftrightarrow d_0^2 \leqslant n $$ $$ d_0 \leqslant \sqrt{n} $$ And finally, All non-prime natural number (n \geqslant 4 ) has at least one strict divisor (d_0 ) such as ( d_0 \leqslant \sqrt{n} ). Coprimity link between a prime number and an integer Let ( p \in \mathbb{P} ) be a prime number and (a \in \mathbb{Z} ) an integer. $$ \mathcal{D}(p) = \bigl {p, 1 \bigr } $$ $$ \mathcal{D}(a) = \bigl {1, \hspace{0.2em} ... \hspace{0.2em}, a } $$ If ( p \nmid a ) then ( p \not\in \mathcal{D}(a) ). Hence the fact that: $$ p \nmid a \hspace{0.2em} \Longrightarrow \hspace{0.2em} \mathcal{D}(a) \wedge \mathcal{D}(p) = \bigl {1 \bigr } $$ $$ p \nmid a \hspace{0.2em} \Longrightarrow \hspace{0.2em} a \wedge p = 1 $$ Reciprocal If (a \wedge p = 1), as (p) divides only itself and (1), so (p \nmid a). $$ a \wedge p = 1 \hspace{0.2em} \Longrightarrow \hspace{0.2em} p \nmid a $$ And finally, $$ \forall p \in \mathbb{P}, \enspace \forall a \in \mathbb{Z}, $$ $$ p \nmid a \hspace{0.2em} \Longleftrightarrow \hspace{0.2em} a \wedge p = 1 $$ Euclid's lemma Let ( p \in \mathbb{P} ) be a prime number, ( (a, b) \in \hspace{0.1em} \mathbb{Z}^2 ) two integers. If ( p/ab ), then two cases arise: ( p ) divides ( a ) Alors, le théorème est vrai ( p ) does not divide ( a ) So, the coprimity link between a prime number and an integer tells us that as ( p \nmid a), so ( a \wedge p = 1). Now, with Gauss' theorem: $$ \forall (a, b, c) \in (\mathbb{Z})^3,\enspace a / bc \enspace et \enspace a \wedge b = 1 \hspace{0.2em} \Longrightarrow \hspace{0.2em} a/c $$ Which allows us to say that ( p ) divides ( b ). And finally, $$ \forall p \in \mathbb{P}, \enspace \forall (a, b) \in \hspace{0.04em} \mathbb{Z}^2, $$ $$ p/ab \hspace{0.2em} \Longrightarrow \hspace{0.2em} (p/a) \enspace or \enspace (p/b) \qquad \bigl(\text{Euclid's lemma} \bigr) $$ Euclid's lemma corollary Let ( (p, a, b) \in \hspace{0.1em} \mathbb{P}^3 ) be three prime numbers. If ( p/ab ), we saw above that with Euclid's lemma, we do have this: $$ p/ab \hspace{0.2em} \Longrightarrow \hspace{0.2em} (p/a) \enspace or \enspace (p/b) $$ Let see what happens in both cases. ( p ) divides ( a ) ( a ) is prime so its only divisors are ( 1 ) and ( a ). But, by hypothesis ( p \neq 1 ) so it is a prime number, so: $$ p/a \hspace{0.2em} \Longrightarrow \hspace{0.2em} p = a $$ ( p ) divides ( b ) These are the same hypotheses for ( b ), therefore: $$ p/b \hspace{0.2em} \Longrightarrow \hspace{0.2em} p = b $$ And finally, $$ \forall (p, a, b) \in \hspace{0.04em} \mathbb{P}^3, $$ $$ p/ab \hspace{0.2em} \Longrightarrow \hspace{0.2em} (p=a) \enspace or \enspace (p=b) \qquad \bigl(\text{Euclid's lemma (corollary)} \bigr) $$ Recap table of the prime numbers properties Go to the top of the page Home Glossary of symbols Bibliography Legal notices Private lessons (in Brittany) Lessons and exercises plateform
8208	https://www.geeksforgeeks.org/maths/powers-and-roots/	Powers and Roots Last Updated : 23 Jul, 2025 Suggest changes Like Article Powers and roots are important ideas in math that help us work with numbers more easily. A power shows how many times a number can be multiplied by itself. For example, 23 means 2 multiplied by itself three times, which equals 8. Roots, on the other hand, A root is a number that, when multiplied by itself a certain number of times, gives you a specific value. For example, if we say 2 is the cube root of 8, it means that 2 multiplied by itself three times equals 8 (because 2 × 2 × 2 = 8). In this article, we will discuss both the concepts of power (exponents) and roots in detail. Table of Content What Are Powers (Exponents)? Properties of Exponents Positive and Negative Powers What are Roots? Square, Cube, and Nth Roots Relationship Between Powers and Roots How do You Convert Roots to Powers? What Are Powers (Exponents)? Powers (exponents) are a way to show how many times to multiply a number by itself. We write it like this: an. Here’s what it means: a is the number we start with (called the base). n is how many times we multiply it (called the exponent). For example: 33= 3 × 3 ×3 = 27 The general form is written as an, where: a is the base n is the exponent Properties of Exponents Some common properties of exponents are: Product of Powers: This simply means that when multiplying two powers of the same base, then we add the exponent to the result. am×an=am+n Power of a Power: For applications of a power to another power multiply the power to the base. (am)n=am×n Quotient of Powers: When dividing two powers with the same base subtract the exponents. anam=am−n Negative Exponents: A negative exponent is an indication of taking the reciprocal of the base raised to the positive exponent of equal value a−m=am1 Positive and Negative Powers Positive powers are straightforward repeated multiplication. For Example: 24 = 2 × 2 × 2 × 2 = 16. Negative powers indicate the reciprocal of the base raised to the positive exponent. Example: 2−3 = 1/ 23 = 1/8 Read More about Fractional Exponents. What are Roots? Roots represent the inverse operation of raising a number to a power. Specifically, taking the root of a number is the process of finding a value which, when raised to a certain exponent, equals the given number. If x2 = 16, then the square root of 16 is the number which when squared results in 16. That is, √16 = 4 Square, Cube, and Nth Roots Square Root (√): The square root of a number is the value that, when squared, gives the original number. Example: √25 = 5 because 52 = 25. Cube Root (∛): The cube root of a number is the value that, when cubed, gives the original number. Example: ∛27 = 3 because 33 =27. Nth Root: This is a general form where the root is based on the value of n. Example: 4√16 = 2 because 24 = 16. Relationship Between Powers and Roots Roots are essentially fractional exponents. The square root of 'a' can be written as '(a1/2)', the cube root as a1/3, and so on. For instance: a1/2=a,a1/3=3a How do You Convert Roots to Powers? To convert roots to powers, express the root as a fractional exponent: a=a1/2 3a=a1/3 na=a1/n Solved Problems Problem 1: Simplify (32)3× 3−4. Solution: Using the power of a power rule: (32)3 = 32×3 = 36. Now using the product of powers rule: 36 × 3−4 = 36−4 = 32 =9. The answer is 9. Problem 2: Simplify √36 × √4. Solution: Using the product of roots rule: √36 × √4 = √36×4 = √144 = 12. The answer is 12. Problem 3: Evaluate 16 3/4. Solution: First, express 16 as a power of 2: 16=24 Now apply the fractional exponent: 163/4 =(24) 3/4 = 24×3/4 = 23 = 8. The answer is 8. Problem 4: Simplify 2−3. Solution: Using the negative exponent rule: 2 −3 = 1 / 23 = 1 / 8. The answer is 1/8 . Problem 5: Simplify 364. Solution: Express 64 as a power of 4: 64 = 43. Now, take the cube root: 364=343=4 The answer is 4. Practice Problems Q1. Simplify: (24 × 23) 2÷ 26 Q2. Evaluate: 7274. Q3. Find the Value of: (√81)2 Q4. Simplify the Root Expression: 416×64. Q5. Evaluate the Following: 32733. Q6. Find the Value of: (232)4. Q7. Simplify the Expression: √ 50 ✕ √2. Q8. Evaluate the Following: 2225 Q9. Find the Value of: 5322 Q10. Simplify the Root Expression: 936. Answer Key 256 49 81 8 1/3-2/3 6.3496 10 8 4 2 Conclusion In conclusion, powers and roots are important mathematical concepts that help us simplify and solve many types of problems. Powers involve multiplying a number by itself multiple times, while roots are the reverse process, finding the original number that was multiplied. A amit7822i5n Improve Article Tags : Mathematics School Learning Algebra Maths Similar Reads Maths Mathematics, often referred to as "math" for short. It is the study of numbers, quantities, shapes, structures, patterns, and relationships. 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The solutions provided by GeeksforGeeks help to practice the qu 5 min readRD Sharma Class 9 Solutions RD Sharma Solutions for class 9 provides vast knowledge about the concepts through the chapter-wise solutions. These solutions help to solve problems of higher difficulty and to ensure students have a good practice of all types of questions that can be framed in the examination. Referring to the sol 10 min readRD Sharma Class 10 Solutions RD Sharma Class 10 Solutions offer excellent reference material for students, enabling them to develop a firm understanding of the concepts covered. in each chapter of the textbook. As Class 10 mathematics is categorized into various crucial topics such as Algebra, Geometry, and Trigonometry, which 9 min readRD Sharma Class 11 Solutions for Maths RD Sharma Solutions for Class 11 covers different types of questions with varying difficulty levels. Practising these questions with solutions may ensure that students can do a good practice of all types of questions that can be framed in the examination. This ensures that they excel in their final 13 min readRD Sharma Class 12 Solutions for Maths RD Sharma Solutions for class 12 provide solutions to a wide range of questions with a varying difficulty level. With the help of numerous sums and examples, it helps the student to understand and clear the chapter thoroughly. Solving the given questions inside each chapter of RD Sharma will allow t 13 min read We use cookies to ensure you have the best browsing experience on our website. By using our site, you acknowledge that you have read and understood our Cookie Policy & Privacy Policy Improvement Suggest Changes Help us improve. Share your suggestions to enhance the article. Contribute your expertise and make a difference in the GeeksforGeeks portal. Create Improvement Enhance the article with your expertise. 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8209	https://scicomp.stackexchange.com/questions/42372/what-is-the-best-method-of-computing-ak-k	accuracy - What is the best method of computing $a^{(k)}/k!$? - Computational Science Stack Exchange Join Computational Science By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more What is the best method of computing a(k)/k!a(k)/k!? Ask Question Asked 2 years, 8 months ago Modified2 years, 8 months ago Viewed 208 times This question shows research effort; it is useful and clear 5 Save this question. Show activity on this post. I have the following expression a(k)k!a(k)k! where a(k)a(k) is the rising factorial. Is it better to evaluate it using floating-point arithmetic separately, that is, call a function that returns the numerator and another function that returns the denominator, and then divide? Or should I evaluate it by expanding ∏i=1 k a+i−1 i∏i=1 k a+i−1 i I tend to think that the second method is more stable because factorials can quickly grow really big, which may lead to a loss of accuracy. But I have never really checked. accuracy Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Jan 15, 2023 at 4:52 hardmath♦ 3,593 2 2 gold badges 22 22 silver badges 44 44 bronze badges asked Jan 12, 2023 at 20:09 nougakonougako 185 3 3 bronze badges 0 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 12 Save this answer. Show activity on this post. If the programming environment provides an implementation of the beta function B(x,y)B(x,y) this computation is straightforward and usually accurate. We have x(k)=Γ(x+k)Γ(x),k!=Γ(k+1),B(x,k)=Γ(x)Γ(k)Γ(x+k)x(k)=Γ(x+k)Γ(x),k!=Γ(k+1),B(x,k)=Γ(x)Γ(k)Γ(x+k) Thus f(x,k)=x(k)k!=Γ(x+k)Γ(x)Γ(k+1)=Γ(x+k)k Γ(x)Γ(k)=1 k B(x,k)f(x,k)=x(k)k!=Γ(x+k)Γ(x)Γ(k+1)=Γ(x+k)k Γ(x)Γ(k)=1 k B(x,k) For example f(9,7)=6435 f(9,7)=6435. Examples of programming environments that provide the beta function are ISO C++17 or later, Julia, and the SciPy and Boost libraries. Best I can tell from a quick perusal, the beta function is not part of ISO Fortran 2018, but it is presumably available via the NAG and IMSL libraries. Note the use of the weasel words "usually accurate" above. According to some quick ad-hoc tests some beta function implementations exhibit a maximum error on the order of 100 ulps. The likely cause of this is that they compute ln(B(x,y))ln⁡(B(x,y)) fully accurate to double precision and then simply exponentiate this intermediate result, thereby suffering from the error magnification property of exp exp. One way around this is to compute ln(B(x,y))ln⁡(B(x,y)) with extended precision; for double-precision computation twelve additional bits would suffice to compute B(x,y)B(x,y) with a maximum error of a few ulps. This is not an unusual issue: similar scenarios occur in implementations of tgamma() and pow() from the C++ standard math library, for example. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Jan 13, 2023 at 11:54 answered Jan 13, 2023 at 3:03 njuffanjuffa 2,005 19 19 silver badges 21 21 bronze badges 2 When k=0 k=0, f(x,k)=1 f(x,k)=1. So, B(x,k)B(x,k) goes to infinity when k=0 k=0?nougako –nougako 2023-01-14 20:45:05 +00:00 Commented Jan 14, 2023 at 20:45 @nougako B(x,k)B(x,k) goes to ∞∞ when k=0 k=0. If you allow k=0 k=0, you will need to special case that to avoid producing a NaN when using IEEEE-754 floating-point arithmetic.njuffa –njuffa 2023-01-14 20:50:17 +00:00 Commented Jan 14, 2023 at 20:50 Add a comment\| This answer is useful 8 Save this answer. Show activity on this post. Njuffa already answered satisfactorily, but let me comment that dealing with large numbers does not cause loss of precision in floating-point arithmetic: the error is a relative one, corresponding to a multiplicative term: the result of a b a b is approximated with a b(1+ε)a b(1+ε), with \|ε\|\|ε\| smaller than machine precision, independently of the size of a a and b b. Computing numerator and denominator separately is a bad idea because of a different issue, that of overflow: after a a or k k are larger than something around 30, numerator and/or denominator become larger than the largest floating point number, which is about 10 308 10 308 for float64 / double, and the computation fails. But if overflow does not happen, then the relative error is always bounded by 1+3 k u+O(u 2)1+3 k u+O(u 2) (that is, u u for each operation). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Jan 14, 2023 at 9:12 answered Jan 13, 2023 at 8:37 Federico PoloniFederico Poloni 13.1k 2 2 gold badges 35 35 silver badges 63 63 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Computational Science Stack Exchange! Please be sure to answer the question. Provide details and share your research! 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8210	https://mcdonaldinstitute.ca/education/visitor-centre/exhibits/cloud-chamber/	Cloud Chamber - McDonald Institute About What is the McDonald Institute? Art McDonald Vision, Mission, & Goals Governance Our Network People Highly Qualified Personnel Staff Faculty Affiliate Universities Queen’s University University of Alberta University of British Columbia Carleton University Laurentian University McGill University Université de Montréal University of Toronto Partner Institutes CIFAR Institute of Particle Physics Perimeter Institute SNOLAB TRIUMF Science Dark Matter Neutrino Physics Technology & Development Photo Detector Development Low Background Techniques McDonald Institute Publications EDII Equity, Diversity, Inclusion, and Indigenization (EDII) DEAP Tool for Researchers Canadian Astroparticle Physics EDII Community of Practice Education Hub Teacher Resources Visitor Centre Student Programs and Summer Camps Jobs & Opportunities Positions Available Funding Opportunities Internships Student Programs and Summer Camps Professional Development News & Events News Events AstroParticle Bites Astroparticle Physics News Our Newsletter en fr About What is the McDonald Institute? Art McDonald Vision, Mission, & Goals Governance Our Network People Highly Qualified Personnel Staff Faculty Affiliate Universities Queen’s University University of Alberta University of British Columbia Carleton University Laurentian University McGill University Université de Montréal University of Toronto Partner Institutes CIFAR Institute of Particle Physics Perimeter Institute SNOLAB TRIUMF Science Dark Matter Neutrino Physics Technology & Development Photo Detector Development Low Background Techniques McDonald Institute Publications EDII Equity, Diversity, Inclusion, and Indigenization (EDII) DEAP Tool for Researchers Canadian Astroparticle Physics EDII Community of Practice Education Hub Teacher Resources Visitor Centre Student Programs and Summer Camps Jobs & Opportunities Positions Available Funding Opportunities Internships Student Programs and Summer Camps Professional Development News & Events News Events AstroParticle Bites Astroparticle Physics News Our Newsletter English Français Exhibits HomeEducation HubVisitor CentreExhibitsCloud Chamber Jump below to tie this content to your classroom. Cloud Chamber Particles are all around us. They are constantly buzzing by us, interacting with our eyes and senses, and even passing completely through us! Given their size, we cannot directly see these particles with our eyes, but we can use technologies to help us see their effects and measure them. This cloud chamber exhibit is designed to show the constant bombardment of particles around us at any given moment. When a particle is travelling, it can interact with its surroundings, like bumping into other particles in the air. A cloud chamber is designed to show the path of particles as they fly through the air and interact with the gas inside the chamber. Cloud chambers were one of the earliest particle detectors developed and the development of them even won Charles Thomson Rees Wilson the Physics Nobel Prize in 1927! The box is filled with a gas, usually isopropyl alcohol (rubbing alcohol) that has been evaporated, and is being cooled to very low temperatures. This creates a supersaturated environment inside the chamber. The alcohol exists in a vaporous state: it is gaseous, but is held at a temperature where it usually condenses into a liquid. If anything comes along and bumps the vapour molecules, disturbing their delicate equilibrium state, the molecules will rapidly condense into their liquid form in that region. When a charged particle travels through this vapour, it causes the vapour to ionize and will make those molecules electrically charged. Small droplets of vapour are then attracted to these charged regions and condense into liquid form, causing a trail that can be seen as white lines. The travelling particle itself is not being seen, but rather the path it took and its consequent reaction interacting with the vapour is what is being observed. What are the causes of the different lines? Why are some thin and straight, while others are thick or wavy? The bright thick line is caused by the largest interacting particle by size: it is the interaction of what is called an alpha particle. An alpha particle is a high energy composite particle consisting of two protons and two neutrons together: it’s the nucleus of a helium atom. Given that it is a collection of particles, it is larger than a single small particle, and that is why this line appears bigger and thicker than the other lines seen. The second type of line is faint, thin, and can curve and bend as it travels through the chamber. This is caused by a particle called a beta particle. A beta particle can be a high speed electron, or its antiparticle, a high speed positron. You can read more about particles and antiparticles HERE link to other theory board beside cloud chamber . These particles are small and that means the lines we see are fainter than the alpha particles. They also are much easier to have their direction changed due to their small mass, and this is why the bending of their paths occur. The final type of line highlighted are those faint, but long streaks that travel in straight lines. These lines are created by particles known as muons. A muon is a fundamental particle that is similar to an electron, but much more massive – about 200 times more massive. The higher mass means that muons bump into the molecules in the cloud chamber but just bludgeon their way through, creating straight tracks. Muons can be created high in the atmosphere as cosmic rays travel through outer space. These rays contain alpha particles and protons, which when they collide with molecules in the atmosphere, can undergo nuclear decay and result in the formation of muons. Interestingly, the life span of muons is incredibly short – only about 2 microseconds! – and so the reason why these particles are able to exist long enough to travel through the atmosphere and long distance down to our detectors is because of relativistic effects. They are travelling fast enough to slow down their internal time, sufficiently long enough that they can reach the detectors and cloud chambers here on Earth’s surface! What are the causes of the different lines? Why are some thin and straight, while others are thick or wavy? The bright thick line is caused by the largest interacting particle by size: it is the interaction of what is called an alpha particle. An alpha particle is a high energy composite particle consisting of two protons and two neutrons together: it’s the nucleus of a helium atom. Given that it is a collection of particles, it is larger than a single small particle, and that is why this line appears bigger and thicker than the other lines seen. The second type of line is faint, thin, and can curve and bend as it travels through the chamber. This is caused by a particle called a beta particle. A beta particle can be a high speed electron, or its antiparticle, a high speed positron. You can read more about particles and antiparticles HERE link to other theory board beside cloud chamber . These particles are small and that means the lines we see are fainter than the alpha particles. They also are much easier to have their direction changed due to their small mass, and this is why the bending of their paths occur. The final type of line highlighted are those faint, but long streaks that travel in straight lines. These lines are created by particles known as muons. A muon is a fundamental particle that is similar to an electron, but much more massive – about 200 times more massive. The higher mass means that muons bump into the molecules in the cloud chamber but just bludgeon their way through, creating straight tracks. Muons can be created high in the atmosphere as cosmic rays travel through outer space. These rays contain alpha particles and protons, which when they collide with molecules in the atmosphere, can undergo nuclear decay and result in the formation of muons. Interestingly, the life span of muons is incredibly short – only about 2 microseconds! – and so the reason why these particles are able to exist long enough to travel through the atmosphere and long distance down to our detectors is because of relativistic effects. They are travelling fast enough to slow down their internal time, sufficiently long enough that they can reach the detectors and cloud chambers here on Earth’s surface! [expand title=”Teacher Resource – Curriculum Connections”] [expandsub1 title=”Ontario Grade 9 Science (SNC1W)”] Atomic models and particle classification: C2.2 – research the role of experimental evidence in the development of various atomic models, and compare and contrast different models of the atom Particle influence by electric charge and current : D2.1 – conduct investigations to explain the behaviour of electric charges in static and current electricity, and to relate the observed behaviour to the properties of subatomic particles and atomic structure The Sun as influencer: E2.1 – describe the importance of the Sun and its characteristics, including its role in the solar system and in sustaining life on Earth The Sun as producer: E2.2 – explain how the Sun’s energy causes natural phenomena on Earth, and how these phenomena contribute to renewable energy production Observable astronomical phenomena: E2.6 – conduct investigations to explain the causes of various astronomical phenomena that can be observed from Earth [/expandsub1] [expandsub1 title=”Ontario Grade 11 Physics (SPH3U)”] Kinetic theory and energy: D3.7 – explain, using the kinetic molecular theory, the energy transfer that occurs during changes of state Structure of isotopes : D3.9 – identify and describe the structure of common nuclear isotopes (e.g., hydrogen, deuterium, tritium) Alpha, beta, and muon particle properties : D3.10 – compare the characteristics of (e.g., mass, charge, speed, penetrating power, ionizing ability) and safety precautions related to alpha particles, beta particles, and gamma rays Radioactivity and half-life: D3.11 – explain radioactive half-life for a given radioisotope, and describe its applications and their consequences [/expandsub1] [expandsub1 title=”Ontario Grade 12 Physics (SPH4U)”] Experimental inquiry: D2.5 – conduct a laboratory inquiry or computer simulation to examine the behaviour of a particle in a field (e.g., test Coulomb’s law; replicate Millikan’s experiment or Rutherford’s scattering experiment; use a bubble or cloud chamber) The Standard Model of particle physics : D3.1 – identify, and compare the properties of, fundamental forces that are associated with different theories and models of physics (e.g., the theory of general relativity and the standard model of particle physics) Relativity and cosmic ray half-life: F2.3 – solve problems related to Einstein’s theory of special relativity in order to calculate the effects of relativistic motion on time, length, and mass (e.g., the half-life of cosmic ray muons, how far into the future a fast space ship would travel, the magnetic field strength necessary to keep protons in the Large Hadron Collider) Experimental inquiry relativity/particle physics : F2.4 – conduct a laboratory inquiry or computer simulation to analyse data (e.g., on emission spectra, the photoelectric effect, relativistic momentum in accelerators) that support a scientific theory related to relativity or quantum mechanics The Standard Model particle components: F3.4 – describe the standard model of elementary particles in terms of the characteristics of quarks, hadrons, and field particles [/expandsub1] [expandsub1 title=”Ontario Grade 12 Earth and Space Science (SES4U)”] Particle radiation and cosmic rays: C3.9 – describe the major external processes and phenomena that affect life on Earth (e.g., radiation and particles from the “quiet” and “active” Sun; cosmic rays) [/expandsub1] [/expand] Back to Exhibits View on Facebook View on Twitter View on Instagram View on LinkedIn About What is the McDonald Institute? Vision, Mission, & Goals Art McDonald Fundamental Science Partnerships Governance Brand Toolkit Financial Toolkit Room/Pod Booking Our Network People Faculty Staff Affiliate Universities Queen’s University University of Alberta University of British Columbia Carleton University Laurentian University McGill University Université de Montréal University of Toronto Partner Institutes CIFAR Institute of Particle Physics Perimeter Institute SNOLAB TRIUMF Science & Education Dark Matter Neutrino Physics Technology & Development Photo Detector Development Low Background Techniques Teacher Resources Visitor Centre McDonald Institute Publications Jobs & Opportunities Careers in Canadian Astroparticle Physics Funding Opportunities 2023 Cross-Disciplinary Internship Recipients Student Programs and Summer Camps Professional Development News & Events News Events AstroParticle Bites Science News Home Our Newsletter Contact FAQ The McDonald Institute at Queen’s University is situated in the traditional territory of the Anishinaabe & Haudenosaunee First Nations. The Institute is part of a national network of institutions and research centres, which operate in other traditional Indigenous territories. Visit www.whose.land to learn the traditional territories where astroparticle physicists are grateful to live and work across Canada. 2025 © Arthur B. McDonald Institute—Canadian Particle Astrophysics Research Institute \| All rights reserved. Back to top English
8211	https://www.geeksforgeeks.org/maths/a-plus-b-plus-c-whole-square/	a plus b plus c Whole Square Formula - GeeksforGeeks Skip to content Tutorials Python Java DSA ML & Data Science Interview Corner Programming Languages Web Development CS Subjects DevOps Software and Tools School Learning Practice Coding Problems Courses DSA / Placements ML & Data Science Development Cloud / DevOps Programming Languages All Courses Tracks Languages Python C C++ Java Advanced Java SQL JavaScript Interview Preparation GfG 160 GfG 360 System Design Core Subjects Interview Questions Interview Puzzles Aptitude and Reasoning Data Science Python Data Analytics Complete Data Science Dev Skills Full-Stack Web Dev DevOps Software Testing CyberSecurity Tools Computer Fundamentals AI Tools MS Excel & Google Sheets MS Word & Google Docs Maths Maths For Computer Science Engineering Mathematics Switch to Dark Mode Sign In Number System and Arithmetic Algebra Set Theory Probability Statistics Geometry Calculus Logarithms Mensuration Matrices Trigonometry Mathematics Sign In ▲ Open In App a plus b plus c Whole Square Formula Last Updated : 23 Jul, 2025 Comments Improve Suggest changes 2 Likes Like Report a plus b plus c whole square, i.e., (a + b + c)2 formula is one of the important algebraic identities. It is used very often in Mathematics when solving algebra questions, expanding the squares of a number, etc. It is used to obtain the sum of squares of a number without performing large calculations and increase simplification. The formula for (a + b + c)2 is represented as: (a + b + c)2 = a 2 + b 2 + c 2 + 2(ab + bc + ca) In this article, we will explore the (a + b + c)2, a plus b plus c whole square formula, expansion of a plus b plus c whole square, and applications of a plus b plus c whole square. We will also solve some examples on a plus b plus c whole square. Let's start our learning on the topic (a + b + c)2. What Does (a + b + c) Whole Square Mean? The (a + b + c)2 means adding three different terms and finding the square of the resultant term. The formula of the (a + b + c) whole square is obtained by the sum of squares of all the three individual terms and the twice of all the product of two terms i.e., ab, bc and ac. (a + b + c)2 Formula Below is the formula for (a + b + c)2 (a + b + c)2= a2+ b2+ c2+ 2(ab + bc + ca) Proof of (a + b + c)2 Formula Formula for (a + b + c)2 can be proved or expanded in the following two ways: Collecting like Terms Using Algebraic Identities Let's discuss these methods in detail as follows: Collecting Like Terms Below we will expand (a + b + c)2 by collecting the like terms. (a + b + c)2 = (a + b + c) (a + b + c) ⇒ (a + b + c)2 = a (a + b + c) + b (a + b + c) + c (a + b + c) ⇒ (a + b + c)2 = a 2 + ab + ac + ba + b 2 + bc + ca + cb + c 2 From the above expression we will collect all the like terms to get the formula for the (a + b + c)2 (a + b + c)2 = a 2 + b 2 + c 2 + 2ab + 2bc + 2ac ⇒ (a + b + c)2=a2+ b2+ c2+ 2(ab + bc + ca) Using Algebraic Identities Below we will expand (a + b + c)2 using algebraic identity of (x + y)2. Let p = b + c ⇒ (a + b + c)2 = (a + p)2 By using the algebraic identity(x + y)2= x2+ y2+ 2xy we get (a + b + c)2 = a 2 + p 2 + 2ap Now putting the value of p in the above expression we get (a + b + c)2 = a 2 + (b + c)2 + 2a (b + c) Now again using the identity(x + y)2= x2+ y2+ 2xy expand (b + c)2 (a + b + c)2 = a 2 + b 2 + c 2 + 2bc + 2a (b + c) ⇒ (a + b + c)2 = a 2 + b 2 + c 2 + 2bc + 2ab + 2ac ⇒ (a + b + c)2= a2+ b2+ c2+ 2(ab + bc + ac) Applications of (a + b + c) Whole Square There are multiple applications of (a + b + c) whole square. Some of these applications are listed below. (a + b + c) whole square is used as an important algebraic identity. It is used in physics and engineering to calculate various other formulas. It can also be used in financial aspects. Read More Algebraic Identities Algebraic Expressions Examples of (a + b + c) Whole Square Example 1: Evaluate: (A + 3B + 2C)2. Solution: For (A + 3B + 2C)2, a = A, b = 3B and c = 2C, Thus, using(a + b + c)2=a2+ b2+ c2+ 2(ab + bc + ca) (A + 3B + 2C)2 = (A)2 + (3B)2 + (2C)2 + 2(A)(3B) + 2(3B)(2C) + 2(A)(2C) ⇒ (A + 3B + 2C)2 = A 2 + 9B 2 + 4C 2 + 6AB + 12BC + 4AC Example 2:Find the value of a2+ b2+ c2if a + b + c = 5, 1/a + 1/b + 1/c = 3 and abc = 4 using A plus B plus C Whole Square Formula. Solution: Given:a + b + c = 5. . . (i) 1/a + 1/b + 1/c = 3 . . . (ii) abc = 4 . . . (iii) To Find: a 2+ b 2+ c 2 Multiply equation (ii) and (iii) abc (1/a + 1/b + 1/c) = (4)(3) ⇒ bc + ca + ab = 12 Thus, using(a + b + c)2=a2+ b2+ c2+ 2(ab + bc + ca) ⇒ a 2 + b 2 + c 2 = (a + b + c)2 - 2(ab + bc + ca) Thus, a 2+ b 2+ c 2= (5)2- 2(12) = 25-24 = 1 square, Conclusion (a+b+c)2 is an essential mathematical formula to use and is very useful to reduce the amount of calculations and make the problem simplified. It comes handy while solving algebraic equations and deucing them to their simplest forms. It is a useful identity in School level mathematics. Practice Questions - (a + b + c)2 Formula Q1. Evaluate: (7x + 2y + 4z)2. Q2. Find the value of (a + b + c) if a2+ b2+ c2= 5 and (ab + bc + ac) = 11. Q3. Find the value of (a2+ b2+ c2) if (a+ b+ c) = 10 and (ab + bc + ac) = 20. Q4. Evaluate: (2x + 3 + 19z)2. Q5. If (a + b + c) = 100 and ab + bc + ca = 34, find the value of the sum of squares of a, b and c. Q6. If a, b, c are in the ratio of 1:2:3 and ab + bc + ca = -1, then find the value of a, b and c. Q7. If the sum of squares of a, b and c is 16 and the value of ab + bc + ca = 24, then find the sum of a, b and c. Q8. Evaluate: (2x - 3u - 5z)2. Q9. If (a + b + c) = 200 and the sum of squares of a, b and c is 64, then find the value of ab + bc + ca. Q10. Find the square of 1003 without multiplying it with itself, use the formula of (a+b+c)2for simplicity and learning. 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8212	https://www.tutorchase.com/answers/ib/chemistry/how-is-kc-different-from-kp	How is Kc different from Kp? \| TutorChase Revision PlatformHire a tutor All Tutors GCSE IGCSE A-Level IB AP Oxbridge / UK Admissions US Admissions Resources More All Tutors GCSE IGCSE A-Level IB AP Oxbridge / UK Admissions US Admissions Resources More Revision PlatformHire a tutor Answers IB Chemistry Article How is Kc different from Kp? Kc and Kp are equilibrium constants, but Kc is expressed in terms of concentrations while Kp is expressed in terms of partial pressures. Kc and Kp are both equilibrium constants that are used in the study of chemical reactions. They are used to express the position of equilibrium for a particular reaction. However, the main difference between them lies in the way they are expressed. Kc, or the equilibrium constant in terms of concentration, is expressed in terms of the molar concentrations of the reactants and products. On the other hand, Kp, or the equilibrium constant in terms of pressure, is expressed in terms of the partial pressures of the reactants and products. The equilibrium constant Kc is used when dealing with reactions in solution. It is calculated by taking the product of the concentrations of the products, each raised to the power of its stoichiometric coefficient, divided by the product of the concentrations of the reactants, each raised to the power of its stoichiometric coefficient. It's important to note that only gases and aqueous solutions are included in the equilibrium expression for Kc. Kp, on the other hand, is used when dealing with reactions involving gases. It is calculated in a similar way to Kc, but instead of using concentrations, the partial pressures of the gases are used. The partial pressure of a gas is the pressure that the gas would exert if it were alone in the container. It's worth noting that the relationship between Kp and Kc is given by the equation Kp = Kc(RT)^(Δn), where R is the ideal gas constant, T is the temperature in Kelvin, and Δn is the change in the number of moles of gas. In summary, while both Kc and Kp serve the same fundamental purpose of expressing the position of equilibrium for a reaction, they are used in different contexts and are calculated using different quantities. Kc is used for reactions in solution and is calculated using concentrations, while Kp is used for reactions involving gases and is calculated using partial pressures. Answered byOli - Qualified IB Tutor \| BSc Natural Sciences IB Chemistry tutor Study and Practice for Free Trusted by 100,000+ Students Worldwide Achieve Top Grades in your Exams with our Free Resources. Practice Questions, Study Notes, and Past Exam Papers for all Subjects! IB ResourcesA-Level ResourcesGCSE ResourcesIGCSE Resources Need help from an expert? 4.93/5 based on733 reviews in The world’s top online tutoring provider trusted by students, parents, and schools globally. #### Vera PGCE Qualified Teacher \| IB Examiner \| BA World Literature #### Stefan PGCE Qualified Teacher \| Cambridge University \| BA Mathematics #### Flora PGCE Qualified Teacher \| Cambridge University MSci Natural Sciences Hire a tutor Related Chemistry ib Answers How does the presence of a functional group affect boiling and melting points? What is the general formula for alkanes? What is the mechanism behind the combustion of hydrocarbons? How is cracking significant in the petrochemical industry? Read All Answers Earn £ as a Student Ambassador! Loading... This website uses cookies to improve your experience, by continuing to browse we’ll assume you’re okay with this. Accept Loading...
8213	https://pmc.ncbi.nlm.nih.gov/articles/PMC2981483/	In anemia of multiple myeloma, hepcidin is induced by increased bone morphogenetic protein 2 - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Blood . 2010 Aug 2;116(18):3635–3644. doi: 10.1182/blood-2010-03-274571 Search in PMC Search in PubMed View in NLM Catalog Add to search In anemia of multiple myeloma, hepcidin is induced by increased bone morphogenetic protein 2 Ken Maes Ken Maes 1 Department of Hematology and Immunology, Vrije Universiteit Brussel (VUB), Brussels, Belgium; Find articles by Ken Maes 1, Elizabeta Nemeth Elizabeta Nemeth 2 Departments of Medicine and Pathology, David Geffen School of Medicine, University of California, Los Angeles, CA; Find articles by Elizabeta Nemeth 2, G David Roodman G David Roodman 3 University of Pittsburgh School of Medicine and Pittsburgh, VA-Medical Center, Pittsburgh, PA; Find articles by G David Roodman 3, Alissa Huston Alissa Huston 3 University of Pittsburgh School of Medicine and Pittsburgh, VA-Medical Center, Pittsburgh, PA; Find articles by Alissa Huston 3, Flavia Esteve Flavia Esteve 3 University of Pittsburgh School of Medicine and Pittsburgh, VA-Medical Center, Pittsburgh, PA; Find articles by Flavia Esteve 3, Cesar Freytes Cesar Freytes 4 University of Texas Health Science Center, Audie L. Murphy Memorial Veterans Hospital, San Antonio, TX; Find articles by Cesar Freytes 4, Natalie Callander Natalie Callander 5 Division of Hematology, Department of Medicine, University of Wisconsin School of Medicine and Public Health, Madison, WI; Find articles by Natalie Callander 5, Eirini Katodritou Eirini Katodritou 6 Department of Hematology, Theagenion Cancer Center, Thessaloniki, Greece; Find articles by Eirini Katodritou 6, Lisa Tussing-Humphreys Lisa Tussing-Humphreys 7 US Department of Agriculture, Agricultural Research Station (USDA-ARS), Baton Rouge, LA; and Find articles by Lisa Tussing-Humphreys 7, Seth Rivera Seth Rivera 2 Departments of Medicine and Pathology, David Geffen School of Medicine, University of California, Los Angeles, CA; Find articles by Seth Rivera 2, Karin Vanderkerken Karin Vanderkerken 1 Department of Hematology and Immunology, Vrije Universiteit Brussel (VUB), Brussels, Belgium; Find articles by Karin Vanderkerken 1, Alan Lichtenstein Alan Lichtenstein 8 West Los Angeles VA-University of California Los Angeles Medical Center, Los Angeles, CA Find articles by Alan Lichtenstein 8, Tomas Ganz Tomas Ganz 2 Departments of Medicine and Pathology, David Geffen School of Medicine, University of California, Los Angeles, CA; Find articles by Tomas Ganz 2,✉ Author information Article notes Copyright and License information 1 Department of Hematology and Immunology, Vrije Universiteit Brussel (VUB), Brussels, Belgium; 2 Departments of Medicine and Pathology, David Geffen School of Medicine, University of California, Los Angeles, CA; 3 University of Pittsburgh School of Medicine and Pittsburgh, VA-Medical Center, Pittsburgh, PA; 4 University of Texas Health Science Center, Audie L. Murphy Memorial Veterans Hospital, San Antonio, TX; 5 Division of Hematology, Department of Medicine, University of Wisconsin School of Medicine and Public Health, Madison, WI; 6 Department of Hematology, Theagenion Cancer Center, Thessaloniki, Greece; 7 US Department of Agriculture, Agricultural Research Station (USDA-ARS), Baton Rouge, LA; and 8 West Los Angeles VA-University of California Los Angeles Medical Center, Los Angeles, CA ✉ Corresponding author. Series information Red Cells, Iron, and Erythropoiesis Received 2010 Mar 12; Accepted 2010 Jul 18; Issue date 2010 Nov 4. © 2010 by The American Society of Hematology PMC Copyright notice PMCID: PMC2981483 PMID: 20679527 Abstract Hepcidin is the principal iron-regulatory hormone and a pathogenic factor in anemia of inflammation. Patients with multiple myeloma (MM) frequently present with anemia. We showed that MM patients had increased serum hepcidin, which inversely correlated with hemoglobin, suggesting that hepcidin contributes to MM-related anemia. Searching for hepcidin-inducing cytokines in MM, we quantified the stimulation of hepcidin promoter-luciferase activity in HuH7 cells by MM sera. MM sera activated the hepcidin promoter significantly more than did normal sera. We then examined the role of bone morphogenetic proteins (BMPs) and interleukin-6 (IL-6), the major transcriptional regulators of hepcidin. Mutations in both BMP-responsive elements abrogated the activation dramatically, while mutations in the IL-6–responsive signal transducer and activator of transcription 3-binding site (STAT3-BS) had only a minor effect. Cotreatment with anti–BMP-2/4 or noggin-Fc blocked the promoter induction with all MM sera, anti–IL-6 blocked it with a minority of sera, whereas anti–BMP-4, -6, or -9 antibodies had no effect. BMP-2–immunodepleted MM sera had decreased promoter stimulatory capacity, and BMP-2 concentrations in MM sera were significantly higher than in normal sera. Our results demonstrate that BMP-2 is a major mediator of the hepcidin stimulatory activity of MM sera. Introduction Multiple myeloma (MM) is a malignant plasma cell disorder that accounts for approximately 10% of all hematologic cancers.1 The disease is characterized by monoclonal proliferation of plasma cells together with overproduction of a monoclonal antibody,2 often accompanied by anemia, hypercalcemia, renal insufficiency, or bone lesions.3 Approximately 97% of MM patients develop anemia during the course of their illness, and 70% are anemic at diagnosis. The anemia is usually normocytic/normochromic,4 serum-iron levels are normal to low, serum ferritin is high, and hemosiderin is prominent in bone marrow macrophages.5 This suggests than iron release from reticuloendothelial macrophages is impaired, consistent with anemia of inflammation.6 The main mediator of anemia of inflammation is the iron-regulatory hormone, hepcidin. Hepcidin is produced by hepatocytes and acts on the iron exporter, ferroportin. Binding of hepcidin to ferroportin induces the internalization and degradation of ferroportin, thereby preventing cellular iron efflux and causing retention of iron, mainly inside enterocytes, macrophages, and hepatocytes.7 Pathologic induction of hepcidin by inflammation causes hypoferremia, restricting the iron supply for erythropoiesis and, eventually, causing anemia. The interleukin-6 (IL-6)–hepcidin axis was shown to be important for inflammation-related hypoferremia.8 However, in chronic inflammation, IL-6–independent pathways may also induce hepcidin mRNA.9 Recently, 2 studies described the involvement of hepcidin in anemia of MM in humans. We reported that patients with stage III MM (n = 44) at diagnosis had higher urinary hepcidin levels than normal controls. Furthermore, in the subset of myeloma patients with normal renal function, urinary hepcidin was inversely correlated with hemoglobin level at diagnosis.6 After a serum hepcidin assay was developed, we showed that these patients, as expected, had elevated serum hepcidin, compared with healthy individuals.10 In a study evaluating 34 MM patients at diagnosis or recurrence, Katodritou et al similarly demonstrated that patients' serum hepcidin levels were elevated and inversely correlated with hemoglobin concentrations.11 Using Hep3B cells in vitro, we showed that treatment with MM patients' sera induced hepcidin mRNA expression more than healthy sera, and that with some samples, hepcidin induction was abrogated by neutralizing anti–IL-6 antibodies.6 However, for other sera, the antibodies had no effect, suggesting that additional serum factors can induce hepcidin expression. Hepcidin expression is predominantly regulated on the transcriptional level. Two families of cytokines are known to be major regulators of hepcidin: the IL-6–like family and the bone morphogenetic protein (BMP) family. BMPs and IL-6 act on the human hepcidin promoter through BMP-responsive elements (BREs) and the signal transducer and activator of transcription 3 (STAT3)–binding site (STAT3-BS), respectively.12–15 IL-616 and BMPs17,18 have been reported to be produced in MM, and we surmised that they could also be involved in the pathogenesis of myeloma-related anemia. We used serum samples from myeloma patients to identify circulating substances that contribute to anemia by mediating the induction of hepcidin. We characterized these substances by interfering with their activity, either on the promoter level by mutating the specific response elements in the hepcidin promoter or by blocking the cytokine/receptor interaction. Methods Cell culture HuH7 human hepatoma cells were cultured in Dulbecco modified Eagle medium (DMEM; Gibco Invitrogen), containing 10% fetal bovine serum (FBS; HyClone Thermo Fisher Scientific), 10 μg/mL ciprofloxacin, and 50 μg/mL gentamycin, at 37°C in 5% CO 2. HuH7 cells were purchased from ATCC Standards and maintained and stored in our laboratory. Generation of hepcidin promoter-luciferase constructs The entire sequence of the human hepcidin (HAMP) promoter, consisting of 2997 base pairs (bp) upstream of the hepcidin translational start site, was amplified, and cloned into pGL4.17 firefly luciferase reporter vector (Promega). This construct was used as a template for site-directed mutagenesis of STAT3-BS sequence and BRE1 and 2 (Figure 2). Mutations were introduced using the Quikchange XL Site-Directed Mutagenesis kit, using the manufacturer-recommended cycling parameters (Agilent-Stratagene). The primers listed in Table 1 were designed with the Quikchange Primer Design Program (Agilent-Stratagene; The mutated plasmid was transformed into XL10-Gold Ultracompetent cells (Stratagene) and isolated using PureLink Quick Plasmid Miniprep Kit (Invitrogen). The mutations were verified by direct sequencing (Laragen). The sequencing primers are listed in Table 2. Plasmids with confirmed mutations were amplified and isolated using Endofree Plasmid Maxi Kit (QIAGEN). Isolated plasmids were evaluated for purity and size with spectrophotometry and gel electrophoresis, respectively. These plasmids were used as templates for the second round of site-directed mutagenesis to construct plasmids carrying 2 or 3 mutations. We generated a total of 8 different constructs: 1 wild-type HAMP promoter construct, 3 constructs with single mutations, and 4 constructs with different combination of the mutations. Figure 2. Open in a new tab Scheme of mutations introduced in HAMP promoter. Boxed areas are the putative transcription factor binding sites for BMP and IL-6 pathways. Mutated nucleotides are shown in bold capitals. Mutations in BRE1 and -2 were previously described in Island et al33 and the STAT3-BS mutation in Truksa et al.15 Table 1. Primers for site-directed mutagenesis \| Site \| Sense 5′- \| Antisense 5′- \| :---: \| STAT3-BS \| cgccaccaccttcttggaattgagacagagcaaa \| tttgctctgtctcaattccaagaaggtggtggcg \| \| BRE-1 \| cccgccttttcggtgccaccaccttct \| agaaggtggtggcaccgaaaaggcggg \| \| BRE-2 \| caccaaggctctggtgcctgtgctgtgac \| gtcacagcacaggcaccagagccttggtg \| Open in a new tab Table 2. Primers for direct sequencing of mutated regions in plasmids \| Mutation site \| Forward 5′- \| Reverse 5′- \| :---: \| BRE-2 \| ggtctccgtgtcaacagttcctgaaa \| aagatggcctcagatgggaatagc \| \| STAT3-BS/BRE-1 \| ccagaacctatgcacgtgtg \| cgtgccgtctgtctggct \| Open in a new tab Transfection and dual luciferase assay Reverse transfection of the human hepatocyte cell line, HuH7, was performed with lipofectamine 2000 (Invitrogen) as a transfection reagent. Briefly, the mixture of plasmids and lipofectamine in Optimem I reduced serum medium (Gibco Invitrogen) was prepared and aliquoted in a 96-well plate. To each well, 2 plasmids were added, a test hepcidin promoter construct and the reference, thymidine kinase–Renilla luciferase plasmid (pTK-RL; Promega). HuH7 cells were trypsinized, resuspended in Optimem, and added to the transfection mix in the 96-well plate. After 16 hours of incubation at 37°C, the complexes were removed and replaced by Optimem media containing recombinant human cytokines (IL-6, BMP-2, -4, -6, and -9; R&D Systems) or human serum. Cells were incubated at 37°C for 6 hours. We ensured that treatments did not alter firefly Renilla values, indicating that cells did not significantly increase in number during treatment, which could potentially confound the experiment. After treatment, cells were lysed using 1× in passive lysis buffer (Promega), and bioluminescence signal was measured using Veritas Microplate Luminometer with dual auto injectors (Turner Biosytems). The firefly luciferase (reporter) signal was measured by adding Luciferase Assay Reagent II, followed by the addition of Stop&Glo substrate to measure Renilla luciferase bioluminescence. The firefly/Renilla luciferase ratio was used to normalize for transfection efficiency. Results were expressed as fold change over untreated cells by dividing the firefly/Renilla ratio of treated cells by the firefly/Renilla ratio of untreated cells. Collection of human samples All blood samples were obtained after informed consent in accordance with the Declaration of Helsinki, as approved by the appropriate institutional review boards. Sera samples used in this study were from Durie-Salmon stage III19 myeloma patients from Veterans Administration (VA) hospitals in Los Angeles, Pittsburgh, San Antonio, and Chicago,6 (n = 13) as well as from stage I, II, or III myeloma patients from Theagenion Cancer Center, Thessaloniki, Greece (n = 12).11 All MM samples were from patients who were newly diagnosed and had not received any cytotoxic chemotherapy. Control sera (n = 15) were obtained from L.T.-H. and University of California Los Angeles. Serum hepcidin assay Serum hepcidin levels of the sera samples used in this study were measured by a competitive ELISA as described and validated by Ganz et al.10 Briefly, 96-well plates were coated with anti–human hepcidin antibody. Serum samples were added in 1:20 dilution in Tris [tris(hydroxymethyl)aminomethane]-buffered saline containing 0.05% Tween-20 (TBS-T), with 10 ng/mL biotinylated hepcidin-25 as a tracer. Standard curves were prepared by the serial 2-fold dilution of synthetic hepcidin in TBS-T containing the tracer. After washing, the assay was developed with streptavidin-peroxidase and tetramethyl benzidine, color development was stopped by sulfuric acid, and the plate was read at 450 nm on a DTX 880 microplate reader (Beckman Coulter). Standard curves were fitted with 12-point fit using GraphPad Prism Version 4 (GraphPad Software). The fitted curve was then used to convert sample absorbance readings to hepcidin concentrations. Neutralization experiments Mouse antibodies against human IL-6, IL-6R, BMP-2/4, -4, -6, -9, as well as BMP antagonist noggin-Fc chimera were all obtained from R&D Systems. For experiments, the concentration of all antibodies and noggin-Fc was 1 μg/mL, except for anti–BMP-2/4 and anti–BMP-6, which were used at 5 μg/mL. Isotype control (IC) was an unrelated goat immunoglobulin (IgG) antibody and was used at a concentration of 1.8 μg/mL. The cytokine inhibitors were added to the wells before adding experimental treatment such as patient sera or recombinant human cytokines. Immunodepletion of serum To selectively immunodeplete BMP-2 from sera, we used the Pierce Direct BMP-2 IP kit (Thermo Fisher Scientific). Briefly, 5 μg of anti–BMP-2/4 antibody or isotype control was coupled to a spin column containing Aminolink Plus Coupling Resin. After 2 hours incubation with shaking at room temperature, the resin was washed and stored in Coupling Buffer for less than a week before use. Efficiency of coupling was determined by sodium dodecyl sulfate polyacrylamide gel electrophoresis (SDS-PAGE). For immunodepletion, the coupling buffer was removed and 150 μL of serum was incubated on the spin column overnight with continuous shaking at 4°C. The next morning, immunodepleted serum was collected and used to treat transfected cells. ELISA for BMP-2 To measure the amount of BMP-2 present in the serum, we used the Quantikine BMP-2 immunoassay (R&D Systems). Microplates were precoated with a monoclonal antibody specific for BMP-2. Standards and samples were incubated in the microplate for 2 hours at room temperature while being shaken at 500 rpm. After washing away any unbound substances, a horseradish peroxidase (HRP)–linked monoclonal antibody specific for BMP-2 was added to the wells and incubated for 2 hours at room temperature with continuous shaking. After washing, the assay was developed with tetramethyl benzidine, the color development was stopped by an acid solution, and the plate was read at 450 nm on a microplate spectrophotometer (Spectramax 190; Molecular Devices). A correction measurement at 540 nm (γ2) was subtracted to correct for plate imperfections. Values were calculated based on the standard curve and expressed as picograms per millilter. Statistical analysis The Student t test was used for compare 2 groups of normally distributed data. The Mann Whitney rank-sum test was used to compare data that were not normally distributed. To compare multiple groups, 1-way analysis of variance (ANOVA) was used to address statistical significance. Correlations between the various measured parameters were calculated by Pearson correlation. A P value < .05 was considered as statistically significant. Results Serum hepcidin levels negatively correlated with hemoglobin concentrations in MM patients In this study, we used MM patient sera from VA hospitals in Los Angeles, Pittsburgh, San Antonio, and Chicago, and from Theagenion Cancer Center, Thessaloniki, Greece (n = 25). The hematologic parameters were previously measured.6,10,11 These patients all had a normocytic/normochromic anemia. For the US patients (n = 13), mean hemoglobin level was 10.0 g/dL (range 7.5-12). Their mean hematocrit level was 30 (range 22.7-37.3). Their mean reticulocyte percent was 0.8% (no patient had a reticulocyte count > 2.5%). The mean serum iron level was 61.5 mg/dL (range 26-175) and serum transferrin was 248 mg/dL (range 143-365). As our normal range of serum iron and transferrin was 60-170 and 240-400, respectively, the mean iron and transferrin levels are both at the lower end of normal, consistent with the anemia of “chronic disease.” In addition, the serum ferritin levels were markedly increased (mean 552 ng/mL, with normal range of 22-322 ng/mL), also consistent with this classification of anemia. The Greek patients (n = 12) had a similar clinical profile, but their laboratory test values were not normally distributed: median hemoglobin 10.1 (range 6.1-10.9), transferrin saturation 21.1% (range 9.7-53.4), and ferritin 291 (range 206-1997). Of the 25 patients, 16 had normal renal function, indicated by serum creatinine levels of 1.4 mg/dL or lower. MM patients had significantly higher serum hepcidin levels compared with age-matched controls (n = 15; Figure 1A). We analyzed the relationship between hepcidin and hemoglobin in this cohort to detect the possible effect of hepcidin on anemia distinct from the known confounding effects of renal failure. As shown, hepcidin serum levels inversely correlated with hemoglobin concentrations in patients with normal renal function (Figure 1B; R 2 = 0.4277; P = .006), again indicating a possible contribution of increased hepcidin to the pathogenesis of MM anemia. Figure 1. Open in a new tab Correlation between serum hepcidin levels and hemoglobin concentrations. We used 25 MM patient serum samples for which hemoglobin content and serum hepcidin concentration were previously measured.6,10,11 (A) Serum hepcidin levels in MM patients (n = 25) were significantly higher than in healthy individuals (n = 15). Boxes represent median and 25-75 percentiles, and whiskers represent 10 and 90 percentiles. Circles are the outliers. Statistical significance was determined with the Mann-Whitney rank-sum test; P< .01, compared with normal serum. (B) Patients with normal renal function (n = 16) were included to investigate the relationship between serum hepcidin levels and hemoglobin concentrations. Hemoglobin concentrations and serum hepcidin inversely correlate (R 2 = 0.4277, P = .006, Pearson correlation). Development of an in vitro system for analysis of hepcidin transcriptional regulation The HuH7 cell line is of human hepatic origin, fast-growing, and readily transfectable. We transfected the cells with WT or mutant HAMP promoter-luciferase constructs containing the full-length human hepcidin promoter (2997 bp). Using site-directed mutagenesis, the promoter was mutated at promoter sequences thought to be involved in signaling by IL-6 and BMPs (Figure 2). We validated the assay by testing known inducers of hepcidin expression and their effect on mutated hepcidin-promoter constructs. We then used the assay to screen MM patient sera for their ability to induce HAMP promoter activity and to identify serum cytokines responsible for hepcidin induction. WT HAMP promoter was induced by IL-6, BMP-2, -4, -6, and -9 in a dose-dependent manner We tested a set of cytokines known to induce hepcidin in vitro, including IL-6 and several members of the TGF-β superfamily, to determine to what extent they induce hepcidin promoter activity in our system. HuH7 cells were transfected with WT-HAMP promoter-luciferase construct and treated with increasing doses of recombinant human IL-6, BMP-2, -4, -6, -9, and TGF-β1 (Figure 3A-F). BMP-4, -9, and IL-6 were the most potent inducers of HAMP promoter activity and achieved statistically significant induction at concentrations less than 2 ng/mL. BMP-2 and -6 significantly induced promoter activity at 6.75 and 12.5 ng/mL, respectively. TGF-β1 had no effect on HAMP promoter activity, even at very high doses. Figure 3. Open in a new tab Dose-dependent effect of recombinant human cytokines on HAMP promoter activity. HuH7 cells were cotransfected with WT-HAMP promoter-firefly luciferase construct and pTK-RL construct and treated with increasing doses of recombinant human cytokines. After treatment, cells were lysed and luciferase activity was measured. Results for individual cytokines (A-F) are expressed as fold induction over untreated control (firefly/renilla ratio of the experimental condition divided by the firefly/renilla ratio of untreated control). Dots and error bars represent mean ± SD of at least 3 independent experiments executed in duplicate. Statistical significance was determined with the Mann-Whitney rank-sum test, and P< .05 compared with untreated control. Mutations in the STAT3-BS and BREs of HAMP promoter abrogated the response to IL-6 and BMPs, respectively To determine which promoter elements were responsible for the inducing effect of IL-6 and BMPs on HAMP promoter, we subjected the WT-HAMP promoter to site-directed mutagenesis as described in the Methods. In total, 7 mutant constructs were generated: 3 constructs with single mutations in STAT3-BS or BREs and 4 constructs with different combinations of mutations. We selected the combination of constructs that could differentiate between IL-6 and BMP signaling. The single mutation in STAT3-BS (mSTAT construct) and the double mutations in BRE1 and 2 (mBRE1/2 construct) disrupted the signaling by IL-6 or BMPs, respectively. The triple-mutated construct (mAll3 construct) disrupted signaling by both IL-6 and BMPs (data not shown). The response of our mSTAT and mBRE1/2 mutant constructs to IL-6 and BMP-2, -4, -6, and -9 in HuH7 cells is shown in Figure 4. Due to differences in the potency of the cytokines, we chose the doses to obtain similar promoter activation after stimulation by each cytokine. BMP-2 and -6 were less potent; therefore, we used higher concentrations to match the response to IL-6, BMP-4, and -9. As expected, the induction of luminescence by IL-6 was disrupted by the mutation in STAT3-BS. Surprisingly, mutations in both BREs resulted in even higher promoter activation by IL-6. The induction by BMPs was disrupted by mutations in the BRE1/2, but not by STAT3-BS, mutation (Figure 4A). Of note, the baseline promoter activity (firefly:renilla ratio) was decreased, compared with WT, when either STAT3-BS or BRE1/2 were mutated, compared with WT-HAMP (data not shown), possibly because these sites may be involved in basal hepcidin transcription. Nevertheless, the ability to respond (ie, fold induction) was not decreased in the mutated constructs (eg, mBRE1/2 construct was still induced by IL-6). Figure 4. Open in a new tab Effect of mutations in STAT3-BS or BREs on HAMP promoter response to cytokines. HuH7 cells were transfected with WT-HAMP, mSTAT, or mBRE1/2 promoter-firefly luciferase construct together with pTK-RL construct and treated with the indicated concentration of recombinant human cytokines. (A) IL-6 signaling was abrogated by the mutation in the STAT3-BS, but not by the mutation in both BREs. In contrast, BMP signaling was disrupted by the mutation in both BREs, but not by the STAT3-BS mutation. Bars represent mean ± SD of at least 3 independent experiments executed in duplicate. Statistical significance was determined with the Student t test or Mann-Whitney rank-sum test. P< .05, compared with untreated control (optimem/WT-HAMP); †P< .05; and ††P< .001, compared with the WT-HAMP construct within the same cytokine group. (B) IL-6 and BMP-9 act synergistically on the HAMP promoter. (C) The mutations in BREs abrogated the synergy, while IL-6 induction was preserved. (D) The mutation of the STAT3-BS did not abrogate the synergy or the BMP-9 induction. Dots and error bars represent mean ± SD. Results are expressed as fold induction over untreated control. IL-6 and BMP-9 induced WT HAMP promoter activity in a synergistic manner Of the multiple cytokines that contribute to the stimulation of hepcidin in MM, some may act synergistically. To examine this possibility, we transfected HuH7 cells with WT-HAMP promoter-luciferase construct and treated them with combinations of different concentrations of IL-6 and BMP-9. We chose these 2 cytokines because of the similarity of their dose-dependent induction of HAMP promoter activity (Figure 3A,E). The combination treatment confirmed the synergistic action of IL-6 and BMP-9 even at very low concentrations (Figure 4B). Treatment with 1.875 ng/mL IL-6 by itself or 1.875 ng/mL BMP-9 by itself caused a 4- and 5-fold increase in luminescence, respectively. Adding the 2 cytokines together resulted in an 18-fold increase in luminescence. Increasing concentrations of IL-6 or BMP-9 by themselves only slightly increased hepcidin promoter activity (7- and 12-fold, respectively, at 30 ng/mL concentration); however, combined treatment, again, showed a synergistic induction of hepcidin promoter activity (50-fold increase). Of note, synergy was also observed between IL-6 and BMP-2 (data not shown). Our findings suggest crosstalk for hepcidin regulation between the SMAD and JAK/STAT signaling pathways. Next, we evaluated the ability of the mutations in the STAT3-BS and BRE1/2 to distinguish the inducing and synergistic effect of IL-6 and BMP signaling on HAMP promoter activity. When both BREs were mutated, the response to BMP-9 was completely abrogated. The BRE1/2 mutation also abrogated the synergism between IL-6 and BMPs (Figure 4C). When the STAT3-BS was mutated, the response of IL-6 was blunted, but the response to BMP-9 was not affected. The synergism between BMP-9 and IL-6 also remained, but their synergistic effect on the STAT3-BS mutant promoter was slightly lower than on the WT-HAMP promoter (Figure 4D). Adding both cytokines together induced promoter activity by 16-fold at 1.875 ng/mL of each cytokine, whereas promoter activity was induced by 7- and 3-fold when BMP-9 and IL-6 were added individually. MM patient sera induced WT HAMP promoter activity Our previous experiments indicated that the in vitro system could detect the presence of hepcidin-inducing cytokines, differentiate BMP from other signaling pathways with the use of the mBRE1/2 construct, and, using the mAll3 construct, detect the presence of additional factors besides IL-6 and BMPs. Next, we tested whether the sera from MM patients would induce WT-HAMP promoter activity. We transfected HuH7 cells with the WT-HAMP construct and treated cells for 6 hours with sera at 10% concentration dissolved in Optimem medium. We compared MM patient sera (n = 25) to normal human sera from healthy donors (n = 15). MM patient sera significantly induced hepcidin promoter activity, in comparison to normal sera (Figure 5A). The correlation coefficient comparing hepcidin concentrations in the 40 sera and hepcidin promoter activity after treatment with those sera was 0.377 (P = .017, Pearson product moment correlation). It is not surprising that the correlation was not higher, considering that not all hepcidin regulatory factors circulate in the blood, and that immortalized hepatic cell lines may not contain the full hepatocyte repertoire of receptors and transduction pathways. Figure 5. Open in a new tab Effect of MM patient sera on HAMP promoter activity. HuH7 cells were cotransfected with WT-HAMP, mSTAT, mBRE1/2, or mAll3 promoter-firefly luciferase construct, together with pTK-RL construct, and treated with 10% serum in Optimem. After 6 hours, cells were lysed and luciferase activity was measured. (A) Treatment with MM patient sera caused a significantly higher induction of WT-HAMP promoter activity, compared with sera from healthy individuals. Results are expressed as fold induction over untreated control. Statistical significance was determined with the Mann-Whitney rank-sum test. Boxes represent median and 25-75 percentiles, and whiskers represent 10 and 90 percentiles. Circles are outliers; P< .05, compared with normal sera. (B) Mutation of BREs abrogated induction by MM sera dramatically, while the STAT3-BS mutation had a statistically less significant effect. Results are expressed as fold induction over untreated control (construct/optimem). Bars represent mean ± SD. Statistical significance was determined with 1-way ANOVA to compare the same sera for the different mutations. To compare MM patient sera with normal sera within the same construct, the Mann-Whitney rank-sum test was used; P< .05, compared with untreated control (construct/optimem), and †P< .05, compared with WT-HAMP promoter activity for the same sera. Next, we compared the effect of mutations in the STAT3-BS and BREs in the HAMP promoter on the response to normal and MM patient sera. We transfected HuH7 cells, followed by treatment with 10% serum in Optimem medium. Both normal (n = 15) and MM patient (n = 25) sera induced WT-HAMP promoter activity, compared with Optimem-only control, but MM sera caused significantly greater induction, compared with normal sera, for all constructs (P< .05; Figure 5B). The mutations in the HAMP promoter decreased the response to serum significantly (Figure 5B). With normal sera, both BRE1/2 and STAT3-BS mutations similarly reduced the serum effects and the response to normal sera was completely abrogated with the use of mAll3 construct, suggesting that the hepcidin-inducing activity of normal sera depends on both BMPs and IL-6. With MM sera, BRE1/2 mutation caused a consistently greater decrease in response than the STAT3-BS mutation. The response was never completely abrogated, not even with the mALL3 construct. Remarkably, we observed that serum from 5 patients contained factors signaling through both BREs and STAT-3BS, while the other 20 contained factors using the BREs (supplemental Figure 1A, available on the Blood Web site; see the Supplemental Materials link at the top of the online article). Anti–BMP-2/4 and noggin-Fc abrogated the stimulatory effect of MM patient sera Using cytokine inhibitors, we further explored the contribution of hepcidin promoter-stimulating cytokines from MM sera. We first confirmed the specificity and potency of neutralizing antibodies and inhibitors by assessing their ability to neutralize recombinant human cytokines. We transfected HuH7 cells with WT-HAMP promoter-luciferase construct and treated them with human recombinant IL-6, BMP-2, -4, -6, and -9 with or without cytokine inhibitors. A combination of anti–IL-6 antibody and anti–IL-6 receptor antibody was used to interfere with IL-6 signaling. We also used noggin, a natural extracellular BMP antagonist, which binds several BMPs, some with high affinity (eg, BMP-2 and -4), others with lower affinity (eg, BMP-7).20 Anti–BMP-2/4, anti–BMP-4, anti–BMP-6, and anti–BMP-9 were used to neutralize their respective targets. Isotype control was used to assess any nonspecific effects of the treatment with cytokine inhibitors. We chose the dose of cytokine inhibitors so as to decrease promoter activation to approximately the same level for each cytokine. Anti–BMP-2/4 and anti-BMP6 were less effective in neutralizing their antigen and thus were used at higher concentrations than anti–IL-6, anti–BMP-4, and -9. In the absence of sera, the cytokine inhibitors by themselves had no effect on WT-HAMP promoter activity (data not shown). As expected, anti–IL-6/anti–IL-6 receptor-neutralizing antibodies abrogated the promoter response to IL-6, but not to any BMPs. Interestingly, noggin-Fc showed slight, but significant, inhibitory effects on IL-6–mediated HAMP promoter stimulation. Anti–BMP-6 and -9 abrogated the promoter induction by BMP-6 and -9, respectively. Anti–BMP-2/4 and noggin/Fc completely neutralized the effect of BMP-2 and -4, whereas anti–BMP-4 specifically affected BMP-4–, but not BMP-2–mediated, stimulation (Figure 6A). Figure 6. Open in a new tab Effect of cytokine inhibitors on the induction of HAMP promoter activity by MM patient sera. HuH7 cells were cotransfected with WT-HAMP promoter-firefly luciferase construct and pTK-RL construct. Next, cells were treated with indicated doses of cytokines or 10% serum with or without cytokine inhibitors. After treatment, cells were lysed and luciferase activity was measured. (A) Specificity array of the cytokine inhibitors. All results are expressed as percentage promoter activity. Bars represent mean ± SD of at least 3 independent experiments executed in duplicate. Statistical significance was determined with the Student t test or Mann-Whitney rank-sum test; P< .05, and †P< .001, compared with the cytokine-only group (cytokine/vehicle). (B) The effect of cytokine inhibitors on the induction of the WT-HAMP promoter by MM or healthy sera. Anti–BMP-2/4 and noggin-Fc significantly reversed the stimulatory effect of MM sera. Results are expressed as fold induction over untreated control (vehicle/optimem). Bars represent mean ± SD. Statistical significance was determined with 1-way ANOVA to compare the same sera for the different cytokine inhibitors. To compare MM patient sera with normal sera within the same group, the Mann-Whitney rank-sum test was used; P< .05, compared with untreated control (vehicle/optimem), and †P< .05, compared with serum only group (vehicle/serum). Next, we tested the ability of the cytokine inhibitors to abrogate the observed induction of HAMP promoter activity when cells were treated with MM patient sera. HuH7 cells were transfected with WT-HAMP promoter-luciferase construct and treated with MM patient sera with or without cytokine inhibitors. After treatment, cells were lysed and luciferase activity was measured. Both MM and normal sera significantly induced promoter activity; however, the effect of MM sera was significantly greater than that of normal sera (P< .05; Figure 6B). Anti–BMP-2/4 and noggin-Fc reduced the response to both normal and MM sera significantly, with more pronounced decrease observed with MM sera (Figure 6B). Anti–BMP-6 and anti–BMP-9 showed no significant neutralizing activity (data not shown). Compared with anti–BMP-2/4, anti–BMP-4 antibody had no significant effect, indicating that BMP-2 is likely responsible for most of the hepcidin-inducing activity. For MM patient sera, besides BMP-2 and, to a lesser extent, BMP-4, another unidentified factor appears to play a role in HAMP promoter induction, as there was still some residual promoter activity even in the presence of anti–BMP-2/4 and noggin-Fc. Interestingly, in the same 5 patients whose sera induced STAT-3BS–dependent responses, IL-6 antibodies (as well as anti–BMP-2/4 antibodies) abrogated the serum-mediated stimulation of hepcidin promoter activity (supplemental Figure 1B). BMP-2 concentrations were higher in MM sera, compared with sera from healthy individuals To confirm the neutralization experiments and to ascertain whether the BMP-2 activity was due to the cytokine presence in MM sera or to its secretion by HuH7 cells, we used anti–BMP-2/4 to deplete any BMP-2 from the MM sera. The depleted sera were used to treat HuH7 cells transfected with WT-HAMP promoter-luciferase construct. Sera depleted for BMP-2 had blunted stimulatory activity, compared with sera depleted with isotype control (Figure 7A). Finally, we measured BMP-2 levels in MM and healthy sera. As expected, we found significantly higher concentrations of BMP-2 in MM patient sera (n = 25), compared with sera from healthy individuals (n = 9; Figure 7B). We did not observe a correlation between serum BMP-2 and hepcidin levels of MM patients (data not shown). This was not surprising, because hepcidin is also affected by other factors, including erythropoiesis, iron intake and stores, and renal function. It would require a much larger sample to isolate the effects of BMP-2 on this background. Figure 7. Open in a new tab BMP-2 is present in MM sera. (A) MM sera (n = 3) were immunodepleted with anti–BMP-2/4 or isotype-control antibody. Depleted sera were used to treat HuH7 cells that had been transfected with WT-HAMP promoter-luciferase construct. After treatment, cells were lysed and luciferase activity was measured. Bars represent mean ± SD. Statistical significance was determined with the Student t test; P< .001. (B) ELISA assay was used to measure BMP-2 values in normal (n = 9) and MM sera (n = 23). As expected, MM sera contained higher amounts of BMP-2, then normal sera. Bars represent mean ± SD. The Mann-Whitney rank-sum test was used to determine statistical significance, and P< .001. Discussion Patients with MM often present with iron-restricted anemia, and overproduction of hepcidin likely contributes to the anemia.6,10 The myeloma-related stimulators of hepcidin have not been systematically characterized. Several BMPs and IL-6 have been identified as important regulators of hepcidin.8,12,21,22 We therefore considered these cytokines as possible candidates responsible for the hepcidin induction in MM patients. With our cellular reporter system, we quantified hepcidin promoter activity in response to patient sera. We introduced mutations in the hepcidin promoter construct, which were based on previous hepcidin promoter mapping studies that identified hepcidin promoter elements involved in the response to BMPs and IL-6. Apparently, murine and human hepcidin promoters are regulated differently by IL-6. The STAT3-BS is important for human hepcidin promoter response to IL-6,12,14 but is not critical for mouse hepcidin promoter response to IL-6.15 Using the human promoter, we observed a significant, but incomplete, loss of promoter response to IL-6 when STAT3-BS was mutated. This was possibly due to our point mutation allowing partial STAT3 binding, as the introduction of a different mutation in the same STAT3-BS had more deleterious effects on IL-6–mediated induction.15 For BMP signaling, 2 BREs are involved in both the murine and human hepcidin promoter response.13,23,24 In agreement with other studies, we successfully disrupted signaling by BMPs by mutating both BREs. However, unlike in another study in which BREs in human hepcidin promoter were apparently necessary for response to IL-6,23 we observed that BREs were not involved in IL-6 responsiveness. This is consistent with observations in the murine hepcidin promoter, where it was shown that IL-6 induces murine hepcidin promoter independently from the BREs.13 Our novel observation that IL-6 and BMPs stimulate hepcidin promoter activity in a synergistic manner points to a crosstalk between the 2 signaling pathways. The BRE elements are important for the synergy while the STAT3-BS is less important. In pathologic conditions, where both types of cytokines are elevated, hepcidin might be more induced compared with conditions where only 1 type of cytokine is elevated. Two unrelated studies described a possible mechanism behind the synergy between SMAD- and STAT-mediated signaling involving the transcriptional coactivator, p300, as a bridge between SMAD and STAT on the promoter level.25,26 In the hepcidin promoter, the STAT3-BS and proximal BRE are close enough to allow bridging by p300. Even the distal BRE could be involved, if chromatin looping brought the transcription factor binding sites into close proximity. Further investigations are needed to elucidate the role of p300 or other synergistic mechanisms in hepcidin regulation. As serum levels of IL-616 and BMP-2 (this study) are elevated in patients with MM, synergistic up-regulation of hepcidin may be occurring. Previously, we observed that serum hepcidin and urinary hepcidin are elevated in newly diagnosed MM patients with anemia, compared with healthy controls.6,10,11 The hepcidin levels correlated negatively with hemoglobin, indicating that hepcidin could be causal in the anemia of MM. In addition, our cellular reporter system also displayed higher hepcidin promoter activity after treatment with serum of MM patients, compared with serum from healthy individuals. This suggested that 1 or more hepcidin-inducing factors is present in MM patient sera. Subsequently, we narrowed down the putative hepcidin inducers from MM sera using agonist-selective mutations of the HAMP promoter. We showed that BMPs are likely involved because of the observed abrogation of the stimulatory effect of MM sera by the mutation of BREs in the hepcidin promoter-luciferase construct. This was consistent for all patients. In addition, mutation of the STAT3-BS resulted in a blunted response in 5 patients, suggesting that IL-6–like cytokines also contribute to the observed induction. Of note, even with the mAll3 construct, the stimulation of promoter activity by MM sera was not completely abrogated, suggesting that MM sera contain factor(s) acting through other, yet unidentified elements in the HAMP promoter. Alternatively, if the point mutation in STAT3-BS was not sufficient to disrupt signaling by IL-6 or IL-6–like cytokines, the triple-mutated construct may have been subject to residual signaling driven by STAT3-BS. Our promoter mutagenesis studies highlighted an important role of BMPs in hepcidin regulation in MM. We next tried to identify the specific BMPs using selective cytokine inhibitors. In addition, we used the same approach to clarify the role of IL-6 in MM sera. All of the cytokine inhibitors were potent and specific inhibitors of their targets. Interestingly, noggin (a BMP antagonist) and anti–BMP-2/4 antibody showed a slight neutralizing effect on IL-6–mediated promoter induction. HuH7 cells produce BMP-2 and -4 in culture,27 which could synergize with exogenous IL-6. Noggin-Fc and anti–BMP-2/4 may neutralize the effect of synergizing autocrine BMPs and thereby decrease the effect of IL-6. When neutralizing molecules were added to HuH7 cells together with MM sera, anti–BMP-2/4 or noggin-Fc blocked the induction of the hepcidin promoter by sera from all patients, indicating an important role of BMP-2–mediated hepcidin regulation in MM. Consistent with the above-mentioned results, anti–IL-6 antibodies abrogated the induction observed by the same 5 MM sera samples that showed abrogation by STAT3-BS mutation. This is in agreement with a previous study that showed that IL-6 contributes in some patients, while not in others.6 Interestingly, like in vitro, the presence of IL-6 and BMP2 together in the serum may act synergistically on hepcidin expression in vivo. An important limitation of our reporter system is that the responsiveness of immortalized cells to different stimuli may differ from the response of hepatocytes in vivo, and we may have missed some stimulators or inhibitors using this system. Hepcidin production in vivo is controlled by several in vivo modulators, including inflammation, erythropoiesis, and iron stores, and these have not only systemic effects, but also complex local effects on hepatocytes and other liver cells. Our cellular system, however, only represented the responsiveness of HuH7 cells to stable circulating substances in MM patient sera. Of the tested cytokines, our system could detect several BMPs and IL-6, but not TGF-β1, in agreement with a previous report that HuH7 cells lack TGF-β receptor 2,28 which might explain the unresponsiveness to TGF-β. Notably, we demonstrated that BMP-2 levels are increased in patients with MM, in comparison to control subjects. This is, to our knowledge, the first observation that a BMP is secreted into serum in MM patients to higher than normal levels and may have systemic effects. Indeed, mice treated with exogenous BMP-2 manifested increased hepcidin expression and lower serum-iron levels,29 indicating that circulating BMP-2 can affect systemic iron homeostasis. Previously, it has been demonstrated that BMPs, in particular BMP-4 and -6, are elevated in BM micromilieu.18 Moreover, BMP-6 was shown to be expressed by MM cells.17 Whether BMP-2 is expressed by tumor cells or in the BM micromilieu is unknown. BMP-2 might be produced in the BM microenvironment during bone repair, where it would induce bone and cartilage in vivo.30 Alternatively, proinflammatory cytokines, such as TNF-α, a known pathogenic factor in MM31 may induce BMP-2 expression in chondrocytes.32 In summary, many MM patients present with anemia caused or exacerbated by hepcidin induction. Our results indicate that BMP-2 is a major mediator of hepcidin-stimulatory capacity in MM sera, and this pathway may contribute to the anemia of MM. A subset of patients had additional IL-6 activity that may act synergistically with BMP-2 on hepcidin induction. Compounds directed against BMP-2 or its signaling pathway may lower hepcidin and ameliorate anemia in MM. Supplementary Material [Supplemental Figure] blood-2010-03-274571_index.html (2.1KB, html) Acknowledgments This work was funded by National Institutes of Health grant RO1 DK 065029 (T.G. and E.N.), an Interdisciplinary Seed Grant from the UCLA-Jonsson Comprehensive Cancer Center (E.N., A.L., and T.G.), and by research support from the VA, Los Angeles, CA (A.L.). Footnotes An Inside Blood analysis of this article appears at the front of this issue. The online version of this article contains a data supplement. The publication costs of this article were defrayed in part by page charge payment. Therefore, and solely to indicate this fact, this article is hereby marked “advertisement” in accordance with 18 USC section 1734. Authorship Contribution: K.M. executed the cloning and the experiments, analyzed the data, and wrote the manuscript; G.D.R., A.H., F.E., C.F., N.C., E.K., and L.T.-H. contributed to patient samples; S.R. and K.V. analyzed the data and revised the manuscript; and A.L., E.N., and T.G. designed the research, contributed to sera samples, analyzed the data, and revised the manuscript. Conflict-of-interest disclosure: E.N. and T.G. are part-owners and officers of Intrinsic LifeSciences, the company that developed and performed the human serum hepcidin assay used in this study. The remaining authors declare no competing financial interests. Correspondence: Tomas Ganz, Department of Medicine, David Geffen School of Medicine, 10833 Le Conte Ave, CHS 37-055, Los Angeles, CA 90095-1690; e-mail: tganz@mednet.ucla.edu. References 1.Kyle RA, Rajkumar SV. Multiple myeloma. Blood. 2008;111(6):2962–2972. doi: 10.1182/blood-2007-10-078022. 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8214	https://www.youtube.com/watch?v=niWULqRxuM8	Convert y= mx+b (Slope Intercept form) into Ax+By =C (Standard form) \| Tagalog Prof Math Wizard 18000 subscribers 362 likes Description 20216 views Posted: 8 May 2020 Convert y= mx+b (Slope Intercept form) into Ax+By =C (Standard form) MathEasy #Slope #Grade 8 18 comments Transcript: hello everyone this is teacher Mel and let us explore the world from a converting y is equal to MX plus B into the form ax plus B Y is equal to C Baba you convert and slope-intercept form into a standard form let's have our first example we write the linear equation y is equal to 3x plus 2 in the form ax plus B white is equal to C poonam step selectmen and again I'm someone at this step bother you Y is equal to 3x plus 2 bugging ax plus by is equal to C form and gagawin not and I you 3x if a transpose not in the left side of the equation or I'm gonna be not and president I am mug soup chocolate 3x the both side of the equation alright it becomes like this pick up if 3x plus y is equal to 2 but cha to the other side the equation you saying your I knock bah-bah go hey rewrite longer then we have the negative 3x plus y is equal to 280 nobody I got nothing I Seneca ax plus B Y is equal to C form Naja okay anything ago in attend the patent coefficients and X naught n should be a positive number a bikini sure positive I gotta be nothing we need to multiply the whole equation by a negative 1 so you multiply not a new equation by a negative I'm kapag and coefficient man X naught and I negative since the coefficient of our Texas negative 3 we need to multiply the whole equation by negative 1 ok let's multiply negative 1 times negative 3 X plus y is equal to 2 so your negative 1 times negative 3 X becomes 3 X and then negative 1 plus y it becomes negative y negative 1 times 2 it becomes a negative 2 therefore you final answer nap in I 3 X minus y is equal to negative 2 here's another example rewrite the linear equation y is equal to 5 of X plus 3 halves in the form ax + B Y is equal to C and in solution R then we write first a given map upon Sigma we have a denominator here okay I'm gonna go in at anis we need to get rid of the denominator by knowing gagawin parama cancel out natin yung denominator an emergency coefficient on X we need to multiply the whole equation by 2 okay imma multiply that in and whole equation on - tangling and to UM denominator or non coefficient man X and in coefficient man x5 half and a young denominator now five huh - so therefore emo multiplying opposite the whole equation parama cancel out young - so denominator no coefficient on X they multiply the moon another 2 times y becomes 2 y is equal to 2 times 5 X becomes 10 X 1/2 2 times 3 it becomes 6 over to simplify nothing we have Y is equal to 5x + 3 that's a 10x divided by 2 becomes 5x 6 divided by 2 becomes free and you'll next step nap and after that you move that a new 5x to the left side of the equation or play the entire market subtracting 5x to both side of the equation Connally namaz madela para say oh ok no me no more or dr. Nance postman and bugging Sonya it becomes a negative 5x plus 2y is equal to three rewrite long nothing it becomes a negative 5x plus 2y is equal to 3 ie nobody final answer nothing mean deepa and a young next na gagawin that but will coefficient on X naught and I positive positive value coefficient man X happen in day can say your coefficient on X naught and I negative 5 and even got going a 10 parama muggin positive en masse amor multiplied nothing your whole equation by a negative 1 it becomes negative times negative mugging 5x negative 1 times 2 y not be minus 2y negative 1 times 3 it becomes negative 3 therefore the final answer is 5x minus 2y is equal to negative 3 thank you everyone if you have any question comment below I'm happy to answer any question please subscribe for more math videos once again this is Peter Mel [Music]
8215	https://www.aafp.org/pubs/afp/issues/2022/1000/barrett-esophagus.html	CARL BRYCE, MD, MERIMA BUCAJ, DO, AND RENEE GAZDA, DO Am Fam Physician. 2022;106(4):383-387 Related editorial: Emerging Roles for Family Physicians in Diagnosing and Treating Barrett Esophagus Author disclosure: No relevant financial relationships. Barrett esophagus is a premalignant change of the esophagus; however, malignant transformation to esophageal adenocarcinoma is rare in patients without dysplasia. Barrett esophagus is estimated to affect up to 5.6% of the U.S. population. Risk factors for Barrett esophagus include gastroesophageal reflux disease, obesity, age older than 50 years, male sex, tobacco use, and a family history of Barrett esophagus or esophageal adenocarcinoma. Patients who experience chronic gastroesophageal reflux symptoms plus additional risk factors should be considered for screening. Mucosal change consistent with Barrett esophagus is visualized during upper endoscopy; biopsy confirms the diagnosis and determines if dysplasia is present. Management of Barrett esophagus depends on the presence and severity of dysplasia; endoscopic treatment of dysplasia decreases the risk of malignant transformation. Surveillance after diagnosis is recommended to monitor for dysplasia and diagnose and treat esophageal adenocarcinoma at an earlier stage. Patients with Barrett esophagus should be offered proton pump inhibitor therapy to control reflux symptoms and possibly decrease the risk of developing esophageal adenocarcinoma. Statins, nonsteroidal anti-inflammatory drugs, and aspirin are associated with a decreased risk of esophageal adenocarcinoma in patients with Barrett esophagus; however, they should not generally be prescribed in the absence of another indication. Mortality benefits of screening and surveillance are uncertain. Barrett esophagus is a premalignant change of the esophagus in which the normal squamous epithelium of the esophagus is replaced by metaplastic, columnar cells due to chronic reflux of acid and bile. This article reviews the best available patient-oriented evidence for the primary care of patients with Barrett esophagus. \| Clinical recommendation \| Evidence rating \| Comments \| --- \| Consider screening for Barrett esophagus in men with GERD symptoms and risk factors for Barrett esophagus, rather than those with GERD symptoms only. Consider screening women only if they have multiple risk factors.19,21,27 \| C \| Expert opinion and performance analyses of prediction tools on patients referred for endoscopy \| \| Treat patients diagnosed with Barrett esophagus with a proton pump inhibitor daily; intensify to twice daily if needed to suppress symptoms of GERD.19,27,29 \| B \| Case-control and cohort studies and expert opinion \| \| Do not routinely prescribe statins, aspirin, or nonsteroidal anti-inflammatory drugs as anti-neoplastic therapy for Barrett esophagus unless a separate indication exists for their use.19,27,32 \| C \| Expert opinion in the absence of clinical trials \| \| Recommendation \| Sponsoring organization \| --- \| \| For a patient diagnosed with Barrett esophagus who has undergone a second endoscopy that confirms the absence of dysplasia on biopsy, a follow-up surveillance examination should not be performed in less than three years. \| American Gastroenterological Association \| The estimated prevalence of Barrett esophagus, including asymptomatic patients, is up to 5.6% of the U.S. population, based on computer modeling.1 Esophageal adenocarcinoma is rare, accounting for 1% of all cancers diagnosed in the United States.2 The prevalence of Barrett esophagus is based on several risk factors3: ○ Low-risk, general population = 0.8% ○ Obesity = 1.9% ○ Gastroesophageal reflux disease (GERD) = 3% ○ Age older than 50 years = 6.1% ○ Male sex = 6.8% ○ GERD plus any additional risk factor = 12% ○ Family history of Barrett esophagus or esophageal adenocarcinoma = 23% Risk factors for Barrett esophagus are listed in Table 14–10; risks are more significant for patients with a longer duration of GERD and multiple risk factors. Alcohol consumption does not increase the risk of Barrett esophagus or esophageal adenocarcinoma.11 One in 416 patients (0.24%) with Barrett esophagus may be diagnosed with esophageal adenocarcinoma per year12; however, a high-quality Danish study estimated a lower annual risk of one in 833 (0.12%).13 The risk of esophageal adenocarcinoma is higher in patients with Barrett esophagus with dysplasia12 (Table 213–16). \| Risk factor \| Odds ratio for developing Barrett esophagus (95% CI) \| --- \| \| Onset of GERD before age 304 \| 15.1 (7.9 to 28.8) \| \| White Caucasian (self-identified)5 (relative to South Asian) \| 6.03 (3.56 to 10.22) \| \| Hiatal hernia6 \| 3.94 (3.02 to 5.13) \| \| GERD7 \| 2.90 (1.86 to 4.54) \| \| Male sex8 \| 2.13 (1.87 to 2.46) \| \| Central adiposity/obesity9 \| 2.04 (1.44 to 2.90) \| \| Tobacco use10 \| 1.42 (1.15 to 1.76) \| \| Age (per year)5 \| 1.03 (1.02 to 1.03) \| \| Patient group \| Progression to esophageal cancer per year \| --- \| \| Without dysplasia13,14 \| 0.1% to 0.33% \| \| With low-grade dysplasia15 \| 0.5% \| \| With high-grade dysplasia16 \| 7% \| Screening There is no evidence from prospective studies that screening for Barrett esophagus in patients with GERD improves mortality or quality of life; therefore, Canadian and British guidelines recommend against universal screening for patients who have GERD.17,18 The American College of Gastroenterology acknowledges that evidence is insufficient, but based on expert consensus, suggests selective screening for Barrett esophagus in people with chronic or frequent GERD symptoms plus two additional risk factors, including age older than 50 years19 (Table 14–10). Women are at significantly lower risk for esophageal adenocarcinoma compared with men; selective screening should be based on the presence of multiple risk factors.19 Life expectancy and comorbidities should be considered before a decision to screen.19 In a large veteran cohort study, 31% of patients who were newly diagnosed with Barrett esophagus had a life expectancy of less than five years and were unlikely to benefit from detection and treatment.20 Several tools to assess individual patient risk of esophageal adenocarcinoma are modestly more predictive than GERD alone, although optimal referral threshold for the number and magnitude of individual risk factors is unknown.21 The HUNT ( and Kunzmann tools rely on information typically collected during office visits and are available online and in a published nomogram.21–23 Screening for Barrett esophagus is done by sedated upper endoscopy. Nonsedated transnasal endoscopy is a viable alternative but is not widely available.19 A swallowed capsule sponge device and breath testing have been studied for screening but are not validated and are not yet recommended as alternatives to upper endoscopy.19 Diagnostic Testing The diagnosis of Barrett esophagus requires two conditions: (1) the normally pale-white esophageal mucosa appears abnormally salmon-colored on endoscopy, at least 1 cm above the gastric folds, and (2) biopsies of the abnormal-appearing esophagus must show columnar epithelium and the presence of goblet cells consistent with intestinal metaplasia. The finding of no dysplasia, low-grade dysplasia, or high-grade dysplasia should be reported because this influences management and prognosis. Complications affect between one in 200 and one in 10,000 upper endoscopies and may include cardiopulmonary events, perforation, bleeding, or adverse reactions to sedation.24 A diagnosis of Barrett esophagus causes anxiety, decreases quality of life, and increases life insurance costs.25 Surveillance Screening and surveillance are of uncertain benefit to most patients. Surveillance leads to detection of earlier-stage esophageal adenocarcinoma but does not lower all-cause mortality after adjusting for lead-time bias.26 Figure 1 summarizes an approach to the screening, diagnosis, and management of Barrett esophagus, which is recommended by expert societies.19,27 There is no expert consensus on a safe age to discontinue surveillance, but cost-effectiveness modeling suggests stopping surveillance of nondysplastic Barrett esophagus for men and women without comorbidities after ages 81 and 75 years, respectively.28 Treatment MEDICAL THERAPY Proton pump inhibitor use was associated with a decreased risk of esophageal adenocarcinoma in patients with Barrett esophagus in North American studies (odds ratio [OR] = 0.47; 95% CI, 0.33 to 0.68) but not in Europe; additionally, a significant decrease was found in studies lasting less than five years, but not in studies of longer duration.29 In a randomized trial of 2,535 patients with Barrett esophagus, fewer patients who were treated with high-dose (40 mg) vs. low-dose (20 mg) esomeprazole (Nexium) developed a composite outcome of death, esophageal adenocarcinoma, or high-grade dysplasia; 10.9% vs. 13.7%, respectively (number needed to treat [NNT] = 37 over eight years).30 Once-daily proton pump inhibitor therapy is recommended, or twice daily if needed to control symptoms of reflux or esophagitis.19,27,29 In the same trial, aspirin (300 to 325 mg per day) reduced the same composite outcome vs. no aspirin; 11.2% vs. 13.4%, respectively (NNT = 46 over eight years), but only in patients who were not also taking a nonsteroidal anti-inflammatory drug.30 Statins are associated with a reduced risk of esophageal adenocarcinoma in patients with Barrett esophagus (OR = 0.54; 95% CI, 0.46 to 0.63), although this was based only on retrospective observational studies.31 Although use of statins, nonsteroidal anti-inflammatory drugs, and aspirin is associated with a decreased risk of esophageal adenocarcinoma in patients with Barrett esophagus, these medications have harms that may outweigh their potential benefits and may also not be cost-effective for the treatment of all patients with Barrett esophagus without another indication.32 PROCEDURAL THERAPY Endoscopic resection, radiofrequency ablation, and cryoablation can be used to remove or ablate discrete lesions or dysplasia of Barrett esophagus. Endoscopic therapies for all patients with Barrett esophagus without dysplasia are cost-prohibitive and unnecessary.33 Endoscopic ablation for low-grade dysplasia reduces the absolute risk of progression to high-grade dysplasia or esophageal adenocarcinoma by 10.9% (95% CI, 7.2% to 14.8%), with an NNT of 10 (95% CI, 7 to 14).34 Because of the low rate of progression to high-grade dysplasia or esophageal adenocarcinoma, surveillance or endoscopic therapy for Barrett esophagus with low-grade dysplasia are reasonable options, based on expert opinion and cost-effectiveness modeling.19,33 Endoscopic radiofrequency ablation for high-grade dysplasia compared with a control group reduced the likelihood of esophageal cancer (2% vs. 19%; P = .04; NNT = 7) in one small, randomized trial and is cost-effective and safer than surgical therapy.33,34 Adverse effects of radiofrequency ablation include esophageal stricture (6%), bleeding (1%), and perforation (0.6%).35 Nodular lesions are removed by endoscopic resection for staging; T1a lesions (limited to mucosal invasion only) can be treated endoscopically, whereas T1b lesions (invading to submucosal level) require a multidisciplinary approach (oncology, gastrointestinal, thoracic surgery).36 Antireflux surgery is not better than medical therapy at preventing esophageal cancer (3.8 vs. 5.3 cases per 1,000 patient-years, respectively; P = .29).37 Prognosis Investigators followed 4,207 patients with Barrett esophagus for 24,959 patient-years; of 921 deaths, 64 (7%) were due to fatal esophageal adenocarcinoma, whereas 857 (93%) were due to other causes.38 The mean life expectancy following a diagnosis of Barrett esophagus is 22 years; the lifetime risk of requiring intervention for high-grade dysplasia or esophageal adenocarcinoma is between one in five and one in six patients.39 Patients with Barrett esophagus do not seem to be at an increased risk of all-cause mortality compared with the general population.40 This article updates previous articles on this topic by Zimmerman41 and Shalauta and Saad.42 Data Sources: A search of Essential Evidence Plus, the Cochrane database, recently published InfoPOEMs, and PubMed was conducted using the Clinical Queries database for the term Barrett esophagus. Search dates: November 2020, August 2021, December 2021, and April 2022. Continue Reading Copyright © 2022 by the American Academy of Family Physicians. This content is owned by the AAFP. A person viewing it online may make one printout of the material and may use that printout only for his or her personal, non-commercial reference. This material may not otherwise be downloaded, copied, printed, stored, transmitted or reproduced in any medium, whether now known or later invented, except as authorized in writing by the AAFP. See permissions for copyright questions and/or permission requests. Copyright © 2025 American Academy of Family Physicians. All Rights Reserved.
8216	https://pubmed.ncbi.nlm.nih.gov/18349643/	Empiric antibiotic therapy for seawater injuries: a four-seasonal analysis - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Empiric antibiotic therapy for seawater injuries: a four-seasonal analysis Jennifer S Kargel1,Vanessa M Dimas,Dennis S Kao,John P Heggers,Peter Chang,Linda G Phillips Affiliations Expand Affiliation 1 Galveston, Texas From the Department of Surgery, University of Texas Medical Branch. PMID: 18349643 DOI: 10.1097/01.prs.0000304241.24443.b0 Item in Clipboard Empiric antibiotic therapy for seawater injuries: a four-seasonal analysis Jennifer S Kargel et al. Plast Reconstr Surg.2008 Apr. Show details Display options Display options Format Plast Reconstr Surg Actions Search in PubMed Search in NLM Catalog Add to Search . 2008 Apr;121(4):1249-1255. doi: 10.1097/01.prs.0000304241.24443.b0. Authors Jennifer S Kargel1,Vanessa M Dimas,Dennis S Kao,John P Heggers,Peter Chang,Linda G Phillips Affiliation 1 Galveston, Texas From the Department of Surgery, University of Texas Medical Branch. PMID: 18349643 DOI: 10.1097/01.prs.0000304241.24443.b0 Item in Clipboard Cite Display options Display options Format Abstract Background: Previous seawater studies found Vibrio species associated with morbidity, suggesting that seawater-contaminated wounds should be treated early. The purpose of this study was to identify common seawater pathogens and antibiotic sensitivities to provide empiric antibiotic therapy until patient-specific culture results are available. Methods: Seasonal samples were collected from 25 Galveston beach locations and cultured. Colonies were identified and sensitivities were performed using MicroScan Positive and Negative Breakpoint Combo Panels. Results: In the fall (28.3 degrees C), of 15 species isolated, the three most common isolates were Escherichia coli, Enterococcus faecium, and Klebsiella pneumoniae. Gram-negative isolates were sensitive to levofloxacin, lomefloxacin, and cefepime. Gram-positive isolates were sensitive to penicillin and ampicillin. In the winter (11.1 degrees C), of 14 species isolated, the three most common isolates were Enterobacter agglomerans, E. faecium, and E. coli. Gram-negative isolates were sensitive to levofloxacin, lomefloxacin, and cefepime. Most Gram-positive isolates were sensitive to vancomycin, levofloxacin, penicillin, and ampicillin. In the spring (26.6 degrees C), of 14 species isolated, the three most common isolates were E. coli, Bacillus species, and E. faecium. Gram-negative isolates were sensitive to levofloxacin, lomefloxacin, and cefepime. Most Gram-positive isolates were sensitive to penicillin, ampicillin, vancomycin, and levofloxacin. In the summer (29.7 degrees C), of 17 bacterial species isolated, the three most common isolates were Bacillus species, Enterobacter cloacae, and K. pneumoniae. Gram-negative isolates were sensitive to cefepime, lomefloxacin, and levofloxacin. Gram-positive isolates were sensitive to penicillin, ampicillin, vancomycin, levofloxacin, and chloramphenicol. Conclusions: Cultured pathogens were sensitive to penicillin, ampicillin, or levofloxacin. The authors recommend a combination of penicillin or ampicillin with levofloxacin for empiric antibiotic coverage for seawater-contaminated injuries. PubMed Disclaimer Similar articles [Analysis of the pathogenic characteristics of 162 severely burned patients with bloodstream infection].Gong YL, Yang ZC, Yin SP, Liu MX, Zhang C, Luo XQ, Peng YZ.Gong YL, et al.Zhonghua Shao Shang Za Zhi. 2016 Sep 20;32(9):529-35. doi: 10.3760/cma.j.issn.1009-2587.2016.09.004.Zhonghua Shao Shang Za Zhi. 2016.PMID: 27647068 Chinese. The tigecycline evaluation and surveillance trial; assessment of the activity of tigecycline and other selected antibiotics against gram-positive and gram-negative pathogens from France collected between 2004 and 2016.Decousser JW, Woerther PL, Soussy CJ, Fines-Guyon M, Dowzicky MJ.Decousser JW, et al.Antimicrob Resist Infect Control. 2018 May 30;7:68. doi: 10.1186/s13756-018-0360-y. eCollection 2018.Antimicrob Resist Infect Control. 2018.PMID: 29876099 Free PMC article. Analysis of freshwater pathogens: a guide to rational empiric antibiotic coverage.Beckert BW, Puckett CL, Concannon MJ.Beckert BW, et al.Mo Med. 2004 May-Jun;101(3):219-21.Mo Med. 2004.PMID: 15311576 Antibiotic use in neonatal sepsis.Yurdakök M.Yurdakök M.Turk J Pediatr. 1998 Jan-Mar;40(1):17-33.Turk J Pediatr. 1998.PMID: 9722468 Review. Microbiological spectrum and antibiotic sensitivity in endophthalmitis: a 25-year review.Gentile RC, Shukla S, Shah M, Ritterband DC, Engelbert M, Davis A, Hu DN.Gentile RC, et al.Ophthalmology. 2014 Aug;121(8):1634-42. doi: 10.1016/j.ophtha.2014.02.001. Epub 2014 Apr 2.Ophthalmology. 2014.PMID: 24702755 Review. See all similar articles Cited by Biohydrogen Production by Antarctic Psychrotolerant Klebsiella sp. ABZ11.Mohammed A, Abdul-Wahab MF, Hashim M, Omar AH, Md Reba MN, Muhamad Said MF, Soeed K, Alias SA, Smykla J, Abba M, Ibrahim Z.Mohammed A, et al.Pol J Microbiol. 2018;67(3):283-290. doi: 10.21307/pjm-2018-033.Pol J Microbiol. 2018.PMID: 30451444 Free PMC article. References Kueh, C. S., Kutarski, P., and Brunton, M. Contaminated marine wounds: The risk of acquiring acute bacterial infection from marine recreational beaches. J. Appl. Bacteriol. 73: 412, 1992. Wilhelmi, B. J., Calianos, T. A., II, Appelt, E. A., Ortiz, M. E., Heggers, J. P., and Phillips, L. G. Modified Dakin’s solution for cutaneous vibrio infections. Ann. Plast. Surg. 43: 386, 1999. Li, C., Li, C., Xing, Y., Zhao, H., and He, Z. Early treatment of wounds polluted by sea water. Chin. J. Traumatol. 4: 187, 2001. Chuang, Y. C., Yuan, C. Y., Liu, C. Y., Lan, C. K., and Huang, A. H. Vibrio vulnificus infection in Taiwan: Report of 28 cases and review of clinical manifestations and treatment. Clin. Infect. Dis. 15: 271, 1992. Reed, K. C., Crowell, M. C., Castro, M. D., and Sloan, M. L. Skin and soft-tissue infections after injury in the ocean: Culture methods and antibiotic therapy for marine bacteria. Mil. Med. 164: 198, 1999. Show all 11 references MeSH terms Anti-Bacterial Agents / therapeutic use Actions Search in PubMed Search in MeSH Add to Search Bacterial Infections / etiology Actions Search in PubMed Search in MeSH Add to Search Bacterial Infections / prevention & control Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Microbial Sensitivity Tests Actions Search in PubMed Search in MeSH Add to Search Seasons Actions Search in PubMed Search in MeSH Add to Search Seawater Actions Search in PubMed Search in MeSH Add to Search Wounds, Penetrating / microbiology Actions Search in PubMed Search in MeSH Add to Search Substances Anti-Bacterial Agents Actions Search in PubMed Search in MeSH Add to Search Related information PubChem Compound PubChem Substance [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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8217	https://www.cuemath.com/algebra/zero-polynomial/	LearnPracticeDownload Zero Polynomial Zero polynomial is a type of polynomial where all variable's coefficients are equal to zero. In other words, it means that all the variables have a power that is equal to zero. A polynomial is an expression consisting of coefficients and variables. Let us learn more about zero polynomials, the degree of zero polynomial, and solve a few examples. \| \| \| --- \| \| 1. \| What is Zero Polynomial? \| \| 2. \| Degree of Zero Polynomial \| \| 3. \| Zero of Zero Polynomial \| \| 4. \| FAQs on Zero Polynomial \| What is Zero Polynomial? Zero polynomial is a type of polynomial where the coefficients are zero and are usually written as 0 and have no terms. Zero polynomial is the only kind of polynomial that has an undefined degree. However, some mathematics define the degree of zero polynomial as negative usually written as -1 or -. Definition of Zero Polynomial Any polynomial with all the variables that have their coefficients equal to zero is called zero polynomial. Hence, the value of a zero polynomial is zero. The function that defines it is called a constant function or zero map usually expressed as P(x) = 0, where x is the variable of the polynomial whose coefficient is zero. A zero polynomial can have an infinite number of terms along with variables of different powers where the variables have zero as their coefficient. For example: 0x2 + 0x + 0. The zero polynomial function is defined as y = P(x) = 0 and the graph of zero polynomial is the x-axis. The domain is considered as real numbers and the range is zero. The domain is the set of values of the variable x for which the function is defined and the range is the set of values of the variable y that is dependent. Degree of Zero Polynomial The degree of zero polynomial is usually undefined unless a degree is assigned then it is -1 or ∞. A degree of a polynomial is considered as the maximum degree of its non-zero terms while a zero polynomial does not have any non-zero terms. Hence, there are no terms with degrees for us to calculate the degree of a polynomial. Any non-zero number or a constant is said to be a zero degree polynomial if f(x) = a as f(x) = ax0 where a ≠ 0. For example: f(x) = 0, g(x) = 0x , h(x) = 0x2. Zero of Zero Polynomial Zero of zero polynomial is any number that can be a rational number, irrational number, or complex number. Since zero of a polynomial is the number that by substituting the variable results in the polynomial's value being zero. While in zero polynomial the coefficient of every term is zero. Therefore, even after substituting, the value of the polynomial will always be zero. Hence, zero itself is the zero polynomial. Related Topics Listed below are a few topics related to zero polynomial, take a look. Zeros of Polynomial Polynomial Functions Polynomial Expressions Polynomial Equations Read More Download FREE Study Materials Polynomial Worksheets Zero Polynomial Worksheets Worksheets on Polynomials Examples on Zero Polynomial Example 1: Eve knows that the zeros of a quadratic polynomial are -4 and 6. How can we help to find the equation of the polynomial? Solution: The zeros of the quadratic polynomial are -4 and 6. Let α = -4, and b = 6 Then, we have the sum of the roots = α + b = 2 Product of the roots = α.b = -24 The required quadratic equation is x2 - (α + b) + α.b = 0 x2 - 2(x) + (-24) = 0 x2 - 2x - 24 = 0 Therefore the equation of the quadratic polynomial is x2 - 2x - 15 = 0. 2. Example 2: Find the degree of the polynomial 5x4 + 3x2 - 7x5 + x7 Solution: In order to find the degree of the given polynomial, Check each term of the given polynomial. All are like terms with x as a variable. Arrange these terms in descending order of their powers, which gives x7 - 7x5 + 5x4+ 3x2 Term with the greatest or highest exponent is x7, so the degree of the polynomial is 7. Therefore, the degree of the polynomial is 7. View Answer > Breakdown tough concepts through simple visuals. Math will no longer be a tough subject, especially when you understand the concepts through visualizations. Book a Free Trial Class Practice Questions on Zero Polynomial Check Answer > FAQs on Zero Polynomial What is Zero Polynomial With Example? A zero polynomial is a type of polynomial where the coefficients of the variables are equal to 0. The constant polynomial f(x) = 0. The general form is g(x) = ax + b where a ≠ 0. For example, f(x) = x -4, g(x) = 14x, etc. The general form is also expressed as a linear polynomial. What are Zero Polynomial and Constant Polynomial? A constant polynomial has its coefficients equal to 0. Whereas a zero polynomial is the additive identity of the additive groups of polynomials such as f(x) = 0. In a constant polynomial, the degree is 0 whereas in a zero polynomial, the degree is undefined or written as -1. How Many Zeros Does a Zero Polynomial Have? In a zero polynomial, the coefficients equal to 0 i.e. f(x) = 0, where x is any value. Hence, the result of f(x) will be 0. Therefore, the number of zeros in a zero polynomial is infinite. What is the Degree of Zero Polynomial? The degree of a zero polynomial is either undefined or is expressed as -1. Just as any constant value, 0 can be considered as a constant value known as zero polynomial. What is Zero of Zero Polynomial? zero polynomial is considered as a constant polynomial with all the coefficients equal to 0. Whereas zero of a polynomial is the value of the variable that makes the polynomial equal to 0. Therefore, zero of zero polynomial is any real number. 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8218	https://math.gmu.edu/~mholzer/relevantDRrevised.pdf	A proof of anomalous invasion speeds in a system of coupled Fisher-KPP equations Matt Holzer∗ Department of Mathematical Sciences George Mason University Fairfax, VA 22030 July 21, 2015 Abstract This article is concerned with the rigorous validation of anomalous spreading speeds in a system of coupled Fisher-KPP equations of cooperative type. Anomalous spreading refers to a scenario wherein the coupling of two equations leads to faster spreading speeds in one of the components. The existence of these spreading speeds can be predicted from the linearization about the unstable state. We prove that initial data consisting of compactly supported pertur-bations of Heaviside step functions spreads asymptotically with the anomalous speed. The proof makes use of a comparison principle and the explicit construction of sub and super solutions. Keywords: anomalous spreading, invasion fronts, linear spreading speed, sub and super-solutions 1 Introduction In this article we study spreading properties for the following system of coupled reaction-diﬀusion equations, ut = duxx + αu(1 −u) + βv vt = vxx + v(1 −v). (1.1) The parameters d, α and β are positive. The system was introduced in as a prototypical example of a system of reaction-diﬀusion equations exhibiting anomalous spreading. In the context of (1.1), anomalous spreading refers to a phenomena where the spreading speed for the u component observed in the coupled regime (β > 0) greatly exceeds that of the uncoupled case. When the system is linearized about the unstable zero state these anomalous speeds are easily calculated to exist within certain regions in parameter space. Our main goal is to prove that for some of these parameter values Heaviside step function initial data spreads with this anomalous speed in the nonlinear system (1.1) as well. The equation governing both the u and v components in isolation (i.e. when β = 0) in (1.1) is the Fisher-KPP equation, see [7, 15]. This equation has been studied for many decades and the dynamics are well understood. Compactly supported, positive initial data converges to a nonlinear traveling front propagating asymptotically with speed 2 √ dα. This speed is also the spreading speed ∗mholzer@gmu.edu 1 of solutions of the linearized equation ut = duxx + αu. For this reason, the Fisher-KPP equation is called linearly determinate since the nonlinear spreading speed is exactly the linear one and the Fisher-KPP front propagating at this speed is referred to as a pulled front. It turns out that (1.1), like the Fisher-KPP equation, is a linearly determinate system in the sense that the nonlinear spreading speeds can be computed from the linearized equation. Spread-ing speeds for the linearized system can be computed systematically from the associated dispersion relation by locating pinched double roots, see section 2.1 and [18, 13] for more details. The skew-product nature of the linearization of system (1.1) implies that the dispersion relation for the full system is the product of the dispersion relations of the reduced system. Pinched double roots can be computed explicitly and there are three diﬀerent spreading speeds that might be of interest: the spreading speed of the u component in isolation (the Fisher-KPP speed 2 √ dα), the spreading speed of the v component in isolation (the Fisher-KPP speed 2) and the spreading speed of the u component induced by the coupling to the v component. Straightforward computations in sec-tion 2.1 show that the largest of these three speeds depends on the parameter values (d, α) and can be summarized in the following table (see also Figure 1). One observes two regions in parameter space where the fastest linear spreading speed is given by the spreading speed of the u component induced by the v component. We call this speed the anomalous spreading speed. parameter regime linear spreading speed pinched double root I 2 √ dα ν+ u = ν− u II 2 ν+ v = ν− v III sanom ν+ v = ν− u IV sanom ν+ u = ν− v With the linear spreading speeds computed, a natural question is whether these speeds are observed in the nonlinear system. This is more complicated and it is not the case that the observed speed in the nonlinear system is simply the fastest of these linear speeds. This issue has been explored in depth in [11, 12, 13] where a distinction is made between linear spreading speeds based upon the analyticity (or lack thereof) of the pointwise Green’s function in a neighborhood of the singularities enforcing these linear speeds. In , the double roots leading to anomalous spreading in the linear system were called relevant if anomalous spreading speeds were also observed numer-ically in the nonlinear regime and irrelevant if the invasion speed was slower than the anomalous one in the nonlinear system despite the existence of faster spreading speeds in the linear system. It was proven in , that for parameters in the irrelevant regime, III, the observed spreading speed was the largest of the two spreading speeds in isolation, max{2, 2 √ dα}. In this article, we will prove that the nonlinear spreading speed for parameters in the relevant regime IV is the anomalous speed sanom. This result conﬁrms numerical observations in . The primary analytical challenge is that anomalous spreading necessarily involves diﬀerent components spreading at diﬀerent speeds. In this way, a traditional traveling wave analysis is incapable of describing this phenomena. Instead, we rely on the existence of a comparison principle for (1.1) and will require some details concerning the behavior of the leading edge of the v component. We review some related work. There are several numerical studies, [2, 5, 11], that suggest that relevant double roots do lead to anomalous spreading speeds in the nonlinear regime. Rigorous results in the relevant regime are, to our knowledge, few. Freidlin proved that if (1.1) is modiﬁed by introducing a βu term in the v component then for any β > 0 both components will spread with a faster speed for parameters including those in III and IV. As β →0, these speeds limit on the anomalous one and therefore this implies that nonlinear spreading speeds are not necessarily continuous functions of the system parameters. The possibility of anomalous spreading speeds in 2 partially coupled equations was ﬁrst noted by Weinberger, Lewis and Li and bounds on the spreading speeds were obtained for example problems. Recently, it was shown in that relevant double roots imply that any traveling front (U(x−st), V (x−st)), moving slower than the anomalous speed and having steep exponential decay is unstable due to the presence of an embedded resonance pole due to the relevant double root. We also note that there is a large literature studying invasion phenomena in many diﬀerent contexts. We point the reader to for a review. In particular, studies of coupled systems of reaction-diﬀusion equations have garnered signiﬁcant interest. Most related to current study are works on spreading speeds in general cooperative systems [19, 16] as well as the speciﬁc example of coupled Fisher-KPP equations [17, 9, 14]. We now state our main result. In brief, we will show that for (d, α) ∈IV, the spreading speed of (1.1) is exactly the anomalous speed sanom predicted by the linearized system. The precise result is as follows. Theorem 1. Deﬁne the invasion point κ(t) = sup x∈R x \| u(t, x) ≥1 2 . Deﬁne the selected speed, ssel by ssel = lim t→∞ κ(t) t . Consider (1.1) with (d, α) ∈IV and β > 0. Fix initial data 0 ≤u0(x) ≤ 1 2 + 1 2 q 1 + 4β α and 0 ≤v0(x) ≤1, each consisting of a compactly supported perturbation of the Heaviside step func-tion 1 2 + 1 2 q 1 + 4β α H(−x) and H(−x), respectively. Then the selected spreading speed of the u component is the anomalous one, i.e. ssel = sanom. The proof of Theorem 1 involves the explicit construction of sub and super solutions and relies on the fact that (1.1) is cooperative and each component satisﬁes the comparison principle. The sub-solution consists of a weakly decaying traveling front solution concatenated with a solution of the linearized equation. Since it is the leading edge of the v component that is driving the anomalous spreading in the u component we will require some results regarding behavior of the solution in the leading edge. We make use of a recent study of the Fisher-KPP equation that shows that sub-solutions can be constructed from solutions of the linearized equation in a moving coordinate frame with Dirichlet boundary conditions placed at the left edge of the domain. Using this sub-solution, we can ﬁnd a wedge in space-time for which a pure exponential solution is a sub-solution for the v component and in turn leverage this to ﬁnd a sub-solution for the u component in this region. It is interesting to note that the wedge for which the pure exponential is a sub-solution propagates at the group velocity associated to that exponential decay rate. This observation was previously made by a formal analysis in . The article is organized as follows. In section 2, we outline some preliminaries necessary for the proof of Theorem 1. This includes a review of linear spreading speeds, the construction of a sub-solution for the v component and a subsequent construction of the sub-solution for the u component. In section 3, we use these sub-solutions to prove Theorem 1. Finally, in section 4 we discuss some generalizations and directions for future work. 2 Preliminaries This section establishes some preliminaries necessary for the proof of Theorem 1. In section 2.1, we review the notions of linear spreading speed, envelope and group velocities. In section 2.2, we review 3 some facts concerning the Fisher-KPP equation and establish a sub-solution for the uncoupled v component. Finally, in section 2.3 we use the analysis in the previous two sections to establish a compactly supported sub-solution for the u component. 2.1 The linearization about the zero state In this section, we consider the linearization of (1.1) about the unstable homogeneous state (u, v) = (0, 0). We compute and discuss the signiﬁcance of the linear spreading speed, envelope velocities and group velocities. Note that much of the material in this sub-section was covered in and we refer the reader there for a more in depth treatment. We transform (1.1) to a moving coordinate frame via y = x −st for s > 0 and linearize about the unstable homogeneous state (u, v) = (0, 0), ut vt = d∂yy + s∂y + α β 0 ∂yy + s∂y + 1 u v . (2.1) The linear spreading speed associated to (2.1) is the asymptotic speed of propagation associated to compactly supported initial data. We refer the reader to and for more detailed discussions of the linear spreading speed and the manner in which it is computed. We sketch the details here. Consider solutions of (2.1) of the form (u, v)T = eνy+λt(u0, v0)T , for ν, λ ∈C. These solutions exist for those values of ν and λ for which the dispersion relation, ds(ν, λ) = (dν2 + sν + α −λ)(ν2 + sν + 1 −λ), (2.2) is equal to zero. Let ds(ν, λ) = du(ν, λ)dv(ν, λ) for the two factors in this dispersion relation. The linear spreading speed for a system of parabolic equations can be computed from the double roots of this dispersion relation and is deﬁned as slin = sup s∈R {ds has a pinched double root (ν∗, λ∗) with Re(λ∗) > 0} , (2.3) where (ν∗, λ∗) is a pinched double root if ds(ν∗, λ∗) = 0, ∂νds(ν∗, λ∗) = 0, Re(ν±(λ)) →±∞as Re(λ) →∞, where ν±(λ) →ν∗as λ →λ∗. The linear system (2.1) is triangular and the linear spreading speed can be computed explicitly. To accomplish this we ﬁrst compute the roots of the dispersion relation (2.2). These are ν± u (s, λ) = −s 2d ± 1 2d p s2 −4dα + 4dλ ν± v (s, λ) = −s 2 ± 1 2 p s2 −4 + 4λ. Pinched double roots occur for those values of s and λ for which ν+ u,v(s, λ) = ν− u,v(s, λ). Two pinched double roots are evident for λ = 0 when ν+ u (2 √ dα, 0) = ν− u (2 √ dα, 0) or ν+ v (2, 0) = ν− v (2, 0). Note that the speeds su = 2 √ dα and sv = 2 are the linear spreading speeds of the u and v component in isolation. A third spreading speed is possible for the u component whenever ν± u (s, 0) = ν∓ v (s, 0). We call this spreading speed the anomalous speed, with formula, sanom = r α −1 1 −d + r 1 −d α −1, 4 and can be found by direct calculation. This spreading speed only occurs for a subset of parameters. The (d, α) parameter space can be decomposed into four regions according to which of the three pinched double roots gives rise to the linear spreading speed. These regions were referenced in the table above and are depicted in the left panel of Figure 1. The four regions have the following descriptions, I = (d, α) \| α ≥ d 2d −1, d > 1 2 II = {(d, α) \| α ≤2 −d} III = (d, α) \| 2 −d < α < d 2d −1, d > 1, IV = (d, α) \| 2 −d < α (d ≤1/2) , 2 −d < α < d 2d −1 (1/2 < d < 1) . A convenient way to understand the double roots that lead to anomalous spreading speeds is to graph the s and ν values for which the dispersion relations for the u and v components in isolation satisfy du(ν, 0) = 0 and dv(ν, 0). These graphs depict the envelope velocity of a mode eνx for ν ∈R−, deﬁned as the speed at which this exponential propagates in the linear system (2.1). A short computation gives, senv,u = −dν −α ν , senv,v = −ν −1 ν . The intersection of the curves of envelope velocities for the u and v component are double roots and are pinched if the derivatives of the two envelope velocity curves have opposite signs. See Figure 1. We ﬁnally mention a third quantity that will arise in our construction of sub-solutions. This quantity is the group velocity, which is deﬁned in general as sg = −∂νd/∂λd(ν∗, λ∗), where (ν∗, λ∗) is a root of the dispersion relation. When d = dv this calculation gives that the group velocity of a mode eνx in the v component is sg(ν) = −2ν. The group velocity of the v component will surface in the construction of sub-solutions for the v component later in this section. We ﬁnd that the region ahead of the front interface where the v component has a sub-solution given by an exponential with decay rate ν is a small interval propagating with the group velocity −2ν. See for a formal study of the role of group velocities in the dynamics of the leading edge of the Fisher-KPP invasion front. Remark 1. The deﬁnition of the linear spreading speed using the double root criterion (2.3) is not always a faithful measure of the actual rates of spreading in a given linear system. This can be observed in (2.1) where when β = 0, the linear spreading speeds arising from pinched double roots involving one root from the u component and one from the v component are not relevant in the fully un-coupled case. This fact was explored in depth in , where it was shown that a more accurate measure of the linear spreading speed is given by singularities of the pointwise Green’s function. Since we will always consider the case β > 0, this ambiguity does not come into play here and we will use the deﬁnition given above. We point the reader to for details of the pointwise Green’s function for the linearization (2.1). 2.2 A sub-solution of the Fisher-KPP equation The evolution of the v component is governed by the classical Fisher-KPP equation and its dynamics are well understood. Initial data, 0 ≤v(0, x) ≤1, a compactly supported perturbation of a Heaviside step function will evolve into a traveling front propagating with the unique, minimal 5 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 d α II I III IV 1.5 2 2.5 3 −2.5 −2 −1.5 −1 −0.5 senv ν Figure 1: On the left is the (d, α) parameter space with regions I −IV depicted. Parameters in regions III and IV lead to anomalous spreading speeds in the linear system (2.1). We focus on regime IV in this article and prove that these parameter values have nonlinear spreading speed given by the anomalous one. On the right is the graph of the envelope velocities for representative values of (d, α) ∈IV, here d = 0.5 and α = 2. The solid line represents the envelope velocities for v and the dashed line represent the envelope velocities for u. The minimums of senv(ν) correspond to the linear spreading speeds of u and v in isolation, while the point where the two curves cross gives the anomalous speed. KPP speed of two1. We expect that for parameters in the anomalous regime IV the dynamics in the leading edge of the v component is driving the anomalous spreading in the u component. We use the method of sub and super solutions, see for example , to prove that this is the case. In this section, we construct a sub-solution for the v component. Of particular interest here is the behavior of the solution v(t, x) far ahead of the front interface. We will show that for any speed σ > 2, the exponential function v(t, x) = eν− v (σ)(x−σt)e−δt, (2.4) is a sub-solution for the v component for t > T ∗on an interval [τ−(t), τ+(t)], where τ±(t) are functions of δ > 0 and σ. We will show that the interval [τ−(t), τ+(t)] is a wedge in space time whose midpoint propagates with speed sg(ν− u (σ)) in the asymptotic limit t →∞. We will do this with the aid of a secondary sub-solution introduced in . Lemma 2.1. (Proposition 3.1 of ) Consider σ > 2 and let y = x −σt. Let q0(y) : R+ →[0, 1] be compactly supported. There exists a function G(t, y), with \|G(t, y)\| < C for some C(σ, q0) > 0 such that q(t, y) = e(1−σ2 4 )te−σ 2 ye−y2 4t G(t, y), (2.5) is a sub-solution of the v component for y ∈[0, ∞). Proof. See , although we sketch the proof here for completeness. We work in a coordinate frame moving to the right at a ﬁxed speed σ > 0. Let y = x −σt. A function q(t, y) is a sub-solution if N(q) = qt −qyy −sqy −q + q2 ≤0, for all (t, y) ∈R+ × R+. Consider the initial value problem for the linearized equation with a Dirichlet boundary condition imposed at y = 0, ˜ qt = ˜ qyy + σ˜ qy + ˜ q, ˜ q(0, y) = ˜ q0(y), ˜ q(t, 0) = 0. 1with a logarithmic correction to the wavespeed, see 6 Asume that ˜ q0(y) is compactly supported. This equation has the explicit solution, ˜ q(t, y) = e(1−σ2 4 )te−σ 2 y Z ∞ 0 e−(y−y′)2 4t −e−(y+y′)2 4t √ 4πt ˜ q0(y′)dy′. As ˜ q(t, y) is a solution to the linearized equation it is not a sub-solution since N(˜ q) = ˜ q2 ≥0. However, one can construct a function A(t) ≥0 such that q(t, y) = A(t)˜ q(t, y) is a sub-solution. To do this, note that \|˜ q(t, y)\| < Ce−ωt for some ω > 0 and some C(q0) independent of y. Then if A′(t) = −Ce−ωtA2, we have that q(t, y) is a sub-solution. The solution of this diﬀerential equation can be calculated explicitly, A(t) = A0ω ω + CA0(1 −e−ωt). Taking for example A0 = 1, we ﬁnd that the solution is bounded away from zero as well as from above, i.e. there exists A1 > 0 such that A1 < A(t) < 1. We then have the existence of a sub-solution q(t, y), which can be factored as in (2.5) with G(t, y) = A(t) √ 2πt Z ∞ 0 e−y′(y′−2y) 4t −e−y′(2y+y′) 4t ˜ q0(y′)dy′, from which we observe that G(t, 0) = 0 and \|G(t, y)\| < C for some C > 0. ■ Lemma 2.2. Fix σ > 2, q0(y) and G(t, y) from Lemma 2.1. Let δ > 0. There exists τ±(t; δ, σ, q0) and a T ∗(σ, δ, q0) > 0 such that the function v is a sub-solution for y ∈[τ−(t), τ+(t)] and t > T ∗. Proof. We will compare the function v to the sub-solution in Lemma 2.1. In particular, we seek those x values for which v(t, x) ≤q(t, x−σt). We work in the moving coordinate system y = x−σt and seek solutions to the nonlinear equation, e(1−σ2 4 )te−σ 2 ye−y2 4t G(t, y) = eνye−δt, where we simplify notation and use ν to denote ν− v (σ). This is equivalent to, −y2 4t − σ 2 + ν y + (1 −σ2 4 + δ)t + log (G(t, y)) = 0. Rescale y = 2tz and divide by −t to ﬁnd, z2 + (σ + 2ν) z −(1 −σ2 4 + δ) −1 t log ˜ G(t, z) = 0. (2.6) Let ρ = t−1/2, then the left hand side of (2.6) deﬁnes an implicit function F(ρ, z) := z2 + (σ + 2ν) z −(1 −σ2 4 + δ) −ρ2 log (ρH(ρ, z)) , with H(ρ, z) = 2A(ρ−2) √ 2π Z ∞ 0 e−(y′ρ)2 4 sinh(zy′)q0(y′)dy′. 7 When ρ = 0 we ﬁnd solutions for z± = −ν −σ 2 ± 1 2 r (σ + 2ν)2 + 4(1 −σ2 4 + δ), which after simpliﬁcation using the identity ν2 + σν + 1 = 0 becomes z± = 1 2 p σ2 −4 ± √ δ. Thus, F(0, z±) = 0 and F is C1 with Fρ(0, z±) = 0 and Fz(0, z±) = z± −z∓. The implicit function theorem implies the existence of a ρ∗> 0 such that z(ρ) = z± + R±(ρ) solves F(ρ, z(ρ)) = 0 for ρ < ρ∗. Reverting to the (t, y) variables, we have τ±(t) = p σ2 −4t ± 2 √ δt + 2R±(t−1/2)t, (2.7) for all t > T ∗(σ, δ, q0) = 1 ρ∗(σ,δ,q0) 2 . ■ Remark 2. Note that the leading order term in (2.7) is precisely the group velocity associated to the spatial mode eνy. This can be interpreted as saying that the region in space where the solution of the Fisher-KPP equation resembles an exponential function with certain strong decay rate is an interval that propagates at the group velocity of that mode. This observation was previously noted in . 2.3 A sub-solution for the u component We now turn our attention to the u component and construct a compactly supported sub-solution for the u component. We do this for all wavespeeds, max{2, 2 √ dα} < s < sanom. We take the dynamics of the v component to be ﬁxed in the sense that initial data has been selected and Lemma 2.1 has been applied to yield a sub-solution for the v component. We will show that there exists a one-parameter family of functions u(t, x) such that N(u) = ut −duxx −α(u −u2) −βv(t, x) < 0. In the following section, we will show that this parameter can be selected so that u(t, x) < u(t, x) for some value of t > 0. The sub-solution consists of a nonlinear traveling front solution with weak decay propagating with speed s, concatenated with a solution of the linearized dynamics of the u component about the zero state. These two sub-solutions are glued together at a unique point in the frame of reference moving with speed σ with s < σ < sanom. That is, we consider u(t, x) =    Ur(x −st) x < σt Ur((σ −s)t)ψ(x −σt, t) x ≥σt 0 x ≥σt + Θ+(t), (2.8) for some Θ+(t) > 0. Here Ur(·) is a nonlinear traveling front solution and ψ(·, t) is a solution of the linearized problem near zero. Continuity is enforced by requiring that ψ(0, t) = 1. There exists a one-parameter family of such functions corresponding to diﬀerent translates of the weakly decaying front. We let r parameterize this family through the identity, Ur(r) = 1 2. 8 Weakly decaying nonlinear fronts The nonlinear traveling front Ur(x −st) is a solution of the second order ordinary diﬀerential equation dU′′ r + sU ′ r + α(Ur −U 2 r ) = 0. The existence of such traveling front solutions for the Fisher-KPP equation is well known, see for example . In the sub-critical regime where s < 2 √ dα, the ﬁxed point at the origin has complex conjugate eigenvalues and the decay of the nonlinear front is oscillatory. In the super-critical regime where s > 2 √ dα, the origin has two real eigenvalues and one can show by phase plane analysis that there exists a unique nonlinear front solution that approaches the origin with weak exponential decay. This decay rate is prescribed by the dispersion relation and is ν+ u (s, 0). The intermediate speed s = 2 √ dα is the critical or minimal speed and the traveling front moving with this speed (referred to as the critical or Fisher-KPP front) is the selected front for the v component in isolation. The function ψ. In order to specify the functional form of ψ, we consider the linearized equation for the u component with the sub-solution (2.4) in place of v. That is, we consider, ut = duyy + σuy + αu + βeν− v (σ)ye−δt. We consider a solution consisting of stationary exponential proﬁles and a particular solution de-scribing the inﬂuence of the inhomogeneous term v, u(t, x) = c1eν+ u (σ)y + c2eν− u (σ)y + p(t, x). Since we require weak exponential decay at the matching point, we set c2 = 0. To determine the particular solution, we apply the transformation w(t, x) = eδtu(t, x) that removes the non-autonomous term from the inhomogeneity, leading to an equation for w, wt = dwyy + σwy + (α + δ)w + βeν− v (σ)y. Finding the particular solution here and reverting to the original coordinates we obtain p(t, x) = − β d(ν− v (σ))2 + σν− v (σ) + α + δeν− v (σ)ye−δt. For brevity going forward we introduce the notation, D(ν) = dν2 + σν + α + δ. Note that D(ν− v (σ)) > 0. We are now in a position to select, ψ(x −σt, t) = c1(t)eν+ u (σ)(x−σt) − β D(ν− v (σ))eν− v (σ)(x−σt)e−δt. (2.9) In order for u(t, x) to be continuous at x −σt = 0 we require ψ(0, t) = 1, which imposes c1(t) = 1 + β D(ν− v (σ))e−δt . Note that ν+ u (σ) < ν− v (σ) < 0 so that ψ(y, t) is negative for large y. In fact, the solution ψ(x−σt, t) vanishes at the point y = Θ+(t) with Θ+(t) = 1 ν− v (σ) −ν+ u (σ) log c1(t)D(ν− v (σ)) β eδt , which after some simpliﬁcation is equivalent to Θ+(t) = δ ν− v (σ) −ν+ u (σ)t + 1 ν− v (σ) −ν+ u (σ) log D(ν− v (σ)) β + e−δt . (2.10) 9 Selecting δ and a decomposition of the real line We have shown that for any δ > 0, the function v is a sub-solution on the interval [τ−(t), τ+(t)]. We have also constructed for any δ > 0 a candidate sub-solution for the u component with the properties ψ(0, t) = 1, ψ(Θ+(t), t) = 0, ψ(y, t) > 0 for all 0 ≤y < Θ+(t). We now will restrict to a particular choice of δ. We set δc = p σ2 −4 ν− v (σ) −ν+ u (σ) . (2.11) Lemma 2.3. Let δ = δc. Then there exists a Tδ(σ, q0) > 0 such that τ−(t) < Θ+(t) < τ+(t), for all t > Tδ(σ, q0). (2.12) holds for all t > Tδ. Moreover, if σ is large enough then τ−(t) > 0 for t > Tδ. Proof. Recall that τ± are deﬁned for all t > T ∗(σ, δc, q0). Compare formulas (2.7) and (2.10). We note that R±(t) →0 as t →∞and that the correction term in (2.10) is bounded and approaches a ﬁnite limit as t →∞. This implies that there exists a Tδ(σ, q0) ≥T ∗(σ, δc, q0) such that (2.12) holds. Note that as σ →sanom, δc tends to zero and τ−(t) > 0 for t suﬃciently large. ■ Still working in moving coordinate frame y = x −σt, we decompose the real line into into four regions as follows, Ia = (−∞, 0] Ib = (0, τ−(t)] Ic = (τ−(t), Θ+(t)] Id = (Θ+(t), ∞) This decomposition holds for all t > Tδ(σ, q0). In region Ia, the candidate sub-solution u is given by the weakly nonlinear front Ur. In region Id, u is zero. We emphasize the diﬀerence between Ib and Ic. On Ic we have that the function v is a sub-solution for the v component. This fact will aid in the proof that u is a sub-solution under certain restrictions on the parameters. However, in region Ib, the function v is not generally a sub-solution. Nonetheless, in this region we can exploit the fact that ψ(x −σt, t) is bounded above zero to show that u remains a sub-solution in this region. The sub-solution Lemma 2.4. Recall that v(t, x) > 0 is a ﬁxed function of the initial data v0(x) and that there exists a q0(x) such that the conclusions of Lemmas 2.1-2.3 hold for some ﬁxed values of s and σ satisfying max{2, 2 √ dα} < s < σ < sanom and for δ = δc. Then there exists Tu(s, σ, q0) ≥max{T ∗, Tδ} and a rc(Tu, s, σ, q0) such that for all r < rc, the function u(t, x) as deﬁned in (2.8) is a sub-solution for the u component for all t > Tu. Proof. We need to show that N(u) ≤0 for all x and t > Tu, where N(u) = ut −duxx −αu + αu2 −βv(t, x). We proceed by straightforward calculation. In region Ia, we have N(Ur(x −st)) = −βv(t, x) < 0. 10 In region Id, the calculation is similarly simple since we have u = 0 and N(0) = −βv(t, x) < 0. In the intermediate regimes in region Ib and Ic the analysis is more involved. In regions Ib and Ic, N(u) is functionally equivalent. We diﬀerentiate between the two regimes because we will need to argue along diﬀerent lines to show that u is a sub-solution in each case. In both regions we have, N(u) = (σ −s)U ′ r(·)ψ + Ur(·)c′ 1(t)eν+ u (σ)(x−σt) + Ur(·)βeν− v (σ)(x−σt)e−δct −βv(t, x) + α(Urψ)2 (2.13) First, note that Ur(·) is the solution of a second order ordinary diﬀerential equation, dU′′ r + sU′ r + α(Ur −U 2 r ). Phase plane analysis implies that the U ′ r can be written as a function of Ur. Since s > 2 √ dα we see that as Ur tends to zero we have the expansion, U ′ r = ν+ u (s)Ur(1 + R(Ur)), (2.14) where \|R(Ur)\| < CUr for Ur suﬃciently small. Consider region Ic. Note that c′ 1(t) < 0. In region Ic, we have that v < v(t, x) for all t > T ∗. Therefore, we can regroup, N(u) = (σ −s)ν+ u (s) + R(Ur(·)) + αUr(·)ψ Ur(·)ψ + Ur(·)c′ 1(t)eν+ u (σ)(x−σt) −β(v(t, x) −Ur(·)v(t, x)). (2.15) Note that (σ −s)ν+ u (s) is a ﬁxed negative number. Since Ur →0 as τ →−∞, there exists a r0(s, σ, q0) such that for all r < r0(s, σ, q0) and t > Tδ(σ, q0) we have (σ−s)ν+ u (s)+R(Ur)+αUrψ < 0 for all x ∈Ic. We have thus shown that N(u) < 0 in region Ic. We now turn our attention to region Ib. Here we do not necessarily have that the function v is a sub-solution for the v component. Nonetheless, consider (2.13). We simplify as in (2.15), N(u) = (σ −s)ν+ u (s) + R(Ur) + αUrψ Urψ + Ur(·)c′ 1(t)eν+ u (σ)(x−σt) + Ur(·)βeν− v (σ)(x−σt)e−δct −βv(t, x). We have shown above that the ﬁrst term can be made negative and bounded away from zero by selecting r appropriately. Since c′ 1(t) < 0, we need only to control for the positive term, Ur(·)βeν− v (σ)(x−σt)e−δct. To do so, we will divide and multiply by ψ(x −σt, t) and show that βeν− v (σ)(x−σt)e−δct ψ(x −σt, t) = β c1(t)e(ν+ u (σ)−ν− v (σ))(x−σt)eδct − β D(ν− v (σ)) , (2.16) can be made arbitrarily small for (t, x) ∈Ib by restricting to t suﬃciently large if and only if (ν+ u (σ) −ν− v (σ))(x −σt) + δct > 0, for all (t, x) ∈Ib. At the left boundary of Ib, i.e. when x −σt = 0, positivity follows from the positivity of δc. At the right bounday, we substitute x −σt = τ−(t) and ﬁnd (ν+ u (σ) −ν− v (σ))τ−(t) + δct = (ν+ u (σ) −ν− v (σ)) −2ν− v (σ)t −2 p δct + 2R−(t)t + 2ν− v (σ)(ν+ u (σ) −ν− v (σ))t = t(ν+ u (σ) −ν− v (σ)) −2 p δc + 2R−(t) 11 Recall that R−(t) →0 as t →∞. Since ν+ u (σ) −ν− v (σ) < 0 we ﬁnd that the exponent is positive and (2.16) can be made arbitrarily small by restricting to t suﬃciently large. This implies that N(u) can be factored as N(u) = (σ −s)ν+ u (s) + R(Ur) + αUrψ + v ψ Urψ + Ur(·)c′ 1(t)eν+ u (σ)(x−σt) −βv(t, x). Since (σ −s)ν+ u (s) + R(Ur) + αUrψ is negative and bounded away from zero and since v/ψ can be made uniformly arbitrarily small, there exists a Tu(s, σ, q0) ≥Tδ(σ, q0) such that u is a sub-solution in region Ib. To complete the proof of the lemma, we need to verify that u is a sub-solution at the matching point x −σt = 0. We must show that the left x derivative of u at the right boundary of region Ia is steeper than the corresponding right derivative at the left boundary of Ib. That is, we require lim y→0− ∂u ∂y < lim y→0+ ∂u ∂y < 0. (2.17) The spatial derivative of u in region Ia is U ′ r(x −st). Making use of the expansion (2.14), the limit of the derivative from the left is lim y→0− ∂u ∂y = ν+ u (s)Ur((σ −s)t)(1 + R(Ur)). We compare this to the the spatial derivative in region Ib at the matching point x −σt = 0. This derivative is Ur((σ −s)t)∂xψ(x −st, t). At the matching point the limit from the right is, lim y→0+ ∂u ∂y = Ur((σ −s)t) ν+ u (σ)c1(t) − ν− v (σ)β D(ν− v (σ))e−δct . After substituting the expression for c1(t), the derivative becomes Ur((σ −s)t) ν+ u (σ) + (ν+ u (σ) −ν− v (σ)) β D(ν− v (σ)) e−δct . Note that ν+ u (σ) < ν− v (σ). Since s < σ, we also have ν+ u (s) < ν+ u (σ) < 0. Selecting Tu perhaps even larger we can enforce, ν+ u (s) < ν+ u (σ) + (ν+ u (σ) −ν− v (σ)) β D(ν− v (σ)) e−δct, from which we may select rc(Tu, s, σ, q0) ≤r0(s, σ, q0) so that the the derivative condition (2.17) is satisﬁed. ■ 2.4 Super-solutions The sub-solutions constructed in the previous section will allow us to bound the spreading speed of (1.1) from below. To demonstrate that the selected spreading speed is actually the anomalous one we will need super solutions that bound the spreading speed from above. In the following lemma, super-solutions are constructed from solutions of the linearized equations. 12 Lemma 2.5. Fix v0(x) and therefore the solution to the initial value problem, v(t, x). Let s > sanom. There exists Cv > 0 and C∗ u(Cv) such that for all Cu > C∗ u there exists a θ(Cu, Cv) for which ¯ u(t, x) = ( Cueν+ u (s)(x−st) + Cvκeν− v (s)(x−st) x −st ≥θ 1 2 + 1 2 q 1 + 4β α x −st < θ , with κ = − β du(ν− v (s), 0), is a super-solution for the u component. Proof: A similar result was established in , we adapt the arguments to the current situation. First, note that ¯ v(t, x) = min{1, Cveν− v (s)(x−st)} is a super-solution for the v component and given v0(x) we can select Cv so that ¯ v(t, x) ≥v(t, x) for all t. With Cv ﬁxed, we ﬁnd that these two curves intersect at the point yv = 1 ν− v (s) log 1 Cv in the moving frame y = x −st. Observe that when s > sanom we have ν− v (s) < ν+ u (s) and du(ν− v (s), 0) < 0. This implies that κ > 0 and the sum of exponentials in the deﬁnition of ¯ u(t, x) is positive. Now, given Cv we can ﬁnd a C∗ u(Cv) such that for any Cu > C∗ u, the intersection point θ(Cu, Cv) is greater than yv. Factor N(u) as follows, N(u) = ut −duxx −αu + αu2 + β(¯ v(t, x) −v(t, x)) −β¯ v(t, x). We argue in pieces. First, for x−st < θ we have that the constant function 1 is a super-solution for the v component. Then evaluate N(u) at a constant value uc, N(uc) = −αuc + αu2 c −β + β(1 −v(t, x)). Taking uc = 1 2 + 1 2 q 1 + 4β α to be a root of the polynomial −αu + αu2 −β and noting v(t, x) < 1 we have that N(−αuc + αu2 c −β) > 0. We now consider x−st > θ. Here the functional form of ¯ u(t, x) is the sum of exponentials given above. The exponential terms are chosen precisely so that the linear terms vanish and all that is left is, N(¯ u(t, x) = αu2 + β(¯ v(t, x) −v(t, x)), which is clearly positive. This completes the proof. ■ 3 Spreading speeds from Heaviside step function initial data – the proof of Theorem 1 We now prove Theorem 1. Proof. Consider any s satisfying max{2, 2 √ dα} < s < sanom. Fix s < σ < sanom and select δ = δc as in Lemma 2.3. Recall also that we require σ large enough so that τ−(t) > 0. Without loss of generality we may assume that the region where v(0, x) > 0 intersects the positive half line. Consider any compactly supported function q0(x), not identically zero, whose support lies in R+ such that q0(x) ≤v(0, x) for all x > 0. Then Lemma 2.1 gives the existence of a sub-solution for the v component that is supported on the positive half line with the properties listed in Lemma 2.1. 13 Now evolve the solution further until t = Tu(s, σ, q0). Lemma 2.4 now gives the existence of a one parameter family of sub-solutions depending on s, σ and q0(x) parameterized by r ≤rc(Tu, s, σ, q0). The maximum principle implies that u(Tu, x) > 0 for all x. We now claim that we can select a particular r ∈R with r ≤rc so that the sub-solution (2.8) satisﬁes u(Tu, x) < u(Tu, x). We do this in two parts. We will require sub-solutions for the u component in isolation. These can be constructed from subcritical nonlinear traveling fronts, see for example . Let uosc(x − (2 √ dα −ϵ)t) be one such nonlinear traveling front solution, cut oﬀand set equal to zero for all x to greater than its smallest zero. We can select a particular translate of this front so that u(t, x) > uosc(t, x) for all 0 < t < Tu(σ.q0). At t = Tu, if the support of uosc(t, x) intersects Ib, then we claim that there exists a r1 ≤rc(s, σ, q0) such that for all r < r1, we have u(Tu, x) < uosc(Tu, x) < u(Tu, x) for all points in Ia. This follows from the decay rates at −∞. The unstable eigenvalue of the linearization of the traveling wave equation dU′′ + sU′ + α(U −U 2) = 0 at u = 1 has a single unstable eigenvalue e(s) = −s 2d + 1 2d p s2 + 4dα, from which we observe that e is a decreasing function of s. This implies that the front uosc converges to u = 1 at a faster rate than Ur and for ﬁxed t, Ur < uosc on some half interval near x = −∞. If at t = Tu the support of uosc includes points in Ib, then we can ﬁnd a r ≤rc so that Ur < uosc for all points in Ib. If the support of uosc(t, x) does not intersect Ib, we can nonetheless ﬁnd a translate of Ur such that u < u(Tu, x). This follows since for t = Tu the region where uosc is equal to zero intersected with the region Ia is compact. The solution u(t, x) > 0 here and bounded from below so by selecting r smaller if necessary we ﬁnd r such that Ur(x −sTu) < u(Tu, x) for all points in Ia. A similar arguments hold for y ∈[0, Θ+(Tu)]. Note that Θ+ is independent of r. Again, u(t, x) is bounded from below on this interval and u(Tu, x) is bounded above by Ur((σ −s)Tu) which implies that we can select r again smaller so that u(Tu, x) < u(Tu, x) for all x. Recall the deﬁnition of the invasion point in Theorem 1. For the sub-solution u(t, x), the invasion point is given explicitly as κ(t) = st + r and since u(t, x) < u(t, x) for all x and all t > T0, we have that ssel ≥s. This construction can be performed for all max{2, 2 √ dα} < s < σ < sanom and we therefore ﬁnd that ssel ≥sanom. Conversely, let s > sanom and consider super-solutions of the form given in Lemma 2.5. For ﬁxed v0(x), we can ﬁnd a Cv such that ¯ v(t, x) is a super-solution for the v component. Since u0(x) is also compactly supported we can ﬁnd Cu > 0 and suﬃciently large so that ¯ u(t, x) ≥u0(x) and therefore ¯ u(t, x) ≥u(t, x) for all t. This implies that ssel ≤s for any s > sanom. We have therefore shown that sanom ≤ssel ≤sanom and therefore ssel = sanom. ■ 4 Discussion We have shown that anomalous spreading speeds are observed in the nonlinear system (1.1) for those parameter values in the relevant regime, i.e. for (d, α) ∈IV. We show that Heaviside step function initial data propagates with the anomalous speed. We now conclude with some observations and discussion of generalizations of the current work and future directions for research. The methods utilized here generalize directly to some other coupled reaction-diﬀusion equa-tions. One example is the following system comprised of a heat equation coupled to a Fisher-KPP 14 equation, ut = duxx + α(u −u2) + βv vt = vxx −γv, where γ > 0 enforces pointwise exponential decay of the v component. This example was introduced in , we refer the reader there for more details. The temptation is to expect that the selected spreading speed for the u component is the Fisher-KPP speed 2 √ dα. As is the case with (1.1), there exists a subset of parameters for which anomalous spreading speeds exist for the linear system. The proof of Theorem 1 is readily adapted to prove that the selected spreading speed for the nonlinear system is also the anomalous speed. It would also be of interest to know whether a general result concerning cooperative systems like the ones in [19, 16] could be adapted to the anomalous case. Consider a general system of coupled reaction-diﬀusion equation whose linearization is block triangular. One major obstacle is that as the number of equations is increased the number of possible pinched double root combinations also increases. Thus, it would be challenging to construct a general theory that accounts for which of these pinched double roots are relevant or irrelevant. Also of interest would be generalizations to larger classes of nonlinearities. Simple modiﬁcations would be amenable to the analysis presented here; for example u(1 −u) could easily be replaced with f(u) where f is of KPP type. The inhomogeneous term βv could be replaced with a term βvg(u, v) with g(u, v) > 0 for u, v > 0 and \|g(u, v)\| = 1 + O(\|u\| + \|v\|). Further generalizations where the v component is allowed to depend on the u component would be more challenging. In a diﬀerent direction there are also large classes of equations for which the selected spreading speed is a nonlinear speed. It would be interesting to quantify the role that anomalous linear speeds could play in this context. Finally, we mention that the proof of Theorem 1 requires the existence of a comparison princi-ple, although the phenomena is observed in non-cooperative systems. Generalizations to systems without comparison would require new techniques. Some insights can be gleaned from a linear analysis. We again mention section 8.3 of for a discussion of the diﬀerences between relevant and irrelevant double roots as well as implications to nonlinear phenomena. For example, supposing the existence of a traveling front solution propagating slower than the linear spreading speed it is shown that relevant double roots enforce the existence of an unstable resonance pole. Building a more complete theory is made more challenging by the fact that anomalous speeds inherently involve diﬀerent components spreading at diﬀerent speeds and interacting over larger spatial scales where a standard traveling wave analysis may not be applicable. References D. G. Aronson and H. F. Weinberger. Multidimensional nonlinear diﬀusion arising in popula-tion genetics. Adv. in Math., 30(1):33–76, 1978. S. Bell, A. White, J. Sherratt, and M. Boots. Invading with biological weapons: the role of shared disease in ecological invasion. Theoretical Ecology, 2:53–66, 2009. 10.1007/s12080-008-0029-x. M. R. Booty, R. Haberman, and A. A. Minzoni. The accommodation of traveling waves of Fisher’s type to the dynamics of the leading tail. SIAM J. Appl. Math., 53(4):1009–1025, 1993. M. Bramson. Convergence of solutions of the Kolmogorov equation to travelling waves. Mem. Amer. Math. Soc., 44(285):iv+190, 1983. 15 E. C. Elliott and S. J. Cornell. Dispersal polymorphism and the speed of biological invasions. PLoS ONE, 7(7):e40496, 07 2012. P. C. Fife and J. B. McLeod. The approach of solutions of nonlinear diﬀusion equations to travelling front solutions. Arch. Ration. Mech. Anal., 65(4):335–361, 1977. R. A. Fisher. The wave of advance of advantageous genes. Annals of Human Genetics, 7(4):355–369, 1937. M. Freidlin. Coupled reaction-diﬀusion equations. Ann. Probab., 19(1):29–57, 1991. A. Ghazaryan, P. Gordon, and A. Virodov. Stability of fronts and transient behaviour in KPP systems. Proc. R. Soc. Lond. Ser. A Math. Phys. Eng. Sci., 466(2118):1769–1788, 2010. F. Hamel, J. Nolen, J.-M. Roquejoﬀre, and L. Ryzhik. A short proof of the logarithmic Bramson correction in Fisher-KPP equations. Netw. Heterog. Media, 8(1):275–289, 2013. M. Holzer. Anomalous spreading in a system of coupled Fisher-KPP equations. Phys. D, 270:1–10, 2014. M. Holzer and A. Scheel. A slow pushed front in a Lotka Volterra competition model. Non-linearity, 25(7):2151, 2012. M. Holzer and A. Scheel. Criteria for pointwise growth and their role in invasion processes. J. Nonlinear Sci., 24(4):661–709, 2014. M. Iida, R. Lui, and H. Ninomiya. Stacked fronts for cooperative systems with equal diﬀusion coeﬃcients. SIAM Journal on Mathematical Analysis, 43(3):1369–1389, 2011. A. Kolmogorov, I. Petrovskii, and N. Piscounov. Etude de l’equation de la diﬀusion avec croissance de la quantite’ de matiere et son application a un probleme biologique. Moscow Univ. Math. Bull., 1:1–25, 1937. B. Li, H. F. Weinberger, and M. A. Lewis. Spreading speeds as slowest wave speeds for cooperative systems. Math. Biosci., 196(1):82–98, 2005. G. Raugel and K. Kirchg¨ assner. Stability of fronts for a KPP-system. II. The critical case. J. Diﬀerential Equations, 146(2):399–456, 1998. W. van Saarloos. Front propagation into unstable states. Physics Reports, 386(2-6):29 – 222, 2003. H. F. Weinberger, M. A. Lewis, and B. Li. Analysis of linear determinacy for spread in cooperative models. J. Math. Biol., 45(3):183–218, 2002. H. F. Weinberger, M. A. Lewis, and B. Li. Anomalous spreading speeds of cooperative recursion systems. J. Math. Biol., 55(2):207–222, 2007. 16
8219	https://www.cs.wm.edu/~wm/CS243/ln10.pdf	5 Counting Combinatorics, the study of arrangements of objects, is an important part of discrete mathematics. Counting of objects with certain properties, is an important part of combinatorics. In fact, counting is frequently used to determine the complexity of algorithms. 5.1 Basics of counting Reading: Rosen 5.1 There are several basic counting principles. • Product rule: Suppose that a procedure can be broken down into a sequence of two tasks. If there are n1 ways to do the ﬁrst task and for each of these ways of doing the ﬁrst task, there are n2 ways to do the second task, then there are n1n2 ways to do the procedure. Example 1: The number of license plates of three letters (a to z) followed by three numbers (0 to 9): 263 ·103. Example 2: The number of functions from A = {a1,...,am} to B = {b1,...,bn}: nm. Example 3: The number of one-to-one functions from A to B: n(n−1)···(n−m+1). Example 4: The number of subsets of set S: Since there is a one-to-one correspondence between a subset of S and a bit string of length \|S\|, then number of subsets of S is equal to the number of bit strings of length \|S\|, which is 2\|S\|. • Sum rule: If a task can be done either in one of n1 ways or in one of n2 ways, where none of the set of n1 ways is the same as any of the set of n2 ways, then there are n1 +n2 ways to do the task. Here is an example of counting using both rules in combination. Example 5: The number of passwords that must be six to eight characters long, where each character is an uppercase letter or a digit, and must contain at least one digit: Let P6, P7, and P8 be the numbers of passwords of length 6, 7, and 8, respectively. Then the total number of passwords is P6 +P7 +P8 = (366 −266)+(367 −267)+(368 −268) = 2,684,483,063,360. • The inclusion-exclusion principle: If a task can be done in n1 or n2 ways, but there are m ways of the set of the n1 ways which are also in the set of the n2 ways, then the total number of ways to do the task is n1 +n2 −m. Example 6: The number of bit strings of length eight that either start with 1 or end with 00: 27 + 26 −25 = 128+64−32 = 160. 5.2 The pigeonhole principle Reading: Rosen 5.2 • The pigeonhole principle: If k is a positive integer and k+1 or more objects are placed into k boxes, then there is at least one box containing two or more objects. • The generalized pigeonhole principle: If N objects are placed into k boxes, then there is at least one box containing at least ⌈N/k⌉objects. • Here are some examples. Example 1: Among a group of 367 people, there are at least two people with the same birthday. Example 2: For any positive integer n, there is a multiple of n that contains only 0s and 1s. Proof: For positive integer n, consider the following n+1 integers, 1,11,111,...,11···1, where the last number contains n+1 1s. Note that there are n remainders, 0,...,n−1, when an integer is divided by n. By the pigeonhole principle, among the n+1 numbers in our list, there must be two numbers, say a and 18 b with a > b, with the same remainder when divided by n. So a −b is divisible by n, thus a multiple of n and a−b contains only 0s and 1s. Example 3: Among 100 people, there are at least ⌈100/12⌉= 9 who were born in the same month. Example 4: In a month of 30 days, a baseball team plays at least one game a day but no more than 45 games in the month. Show that there must be a period of some number of consecutive days during which the team plays exactly 14 games. Proof: Let aj be the number of games played on or before the jth day of the month. Then a1,a2,...,a30 is an increasing sequence of distinct positive numbers, with 1 ≤a j ≤45. Moreover, a1 +14,a2 +14,...,a30 +14 is also an increasing sequence of distinct positive numbers, with 15 ≤a j +14 ≤59. The 60 positive integers a1,a2,...,a30,a1 + 14,a2 + 14,...,a30 + 14 are all less than or equal to 59. By the pigeonhole principle, two of these integers are equal. One must be ai and the other must be ak + 14 and ai = ak +14. This means that exactly 14 games are played from day k +1 to day i. Example 5: Show that among any n+1 positive integers no exceeding 2n there must be an integer that divides one of the other integers. Proof: Assume the n+1 numbers are a1,a2,...,an+1. Each can be expressed a power of 2 times an odd number (including 1). Thus aj = 2kjq j for j = 1,2,...,n+1, where kj is a nonnegative integer and qj is an odd positive number less than 2n. Since there are only n odd positive numbers less than 2n, then by the pigeonhole principle, there must be two of q1,q2,...,qn+1, say qi and qm, that are equal. Let q = qi = qm. Then ai = 2kiq and am = 2kmq. Then it is either ai divides am or am divides ai. Example 6: Assume in a group of six people, any pair of individuals are either friends or enemies. Show that there are either three mutual friends or three mutual enemies in the group. Proof: Let A be one of the six. Of the ﬁve others in the group, there are either three or more who are friends of A , or three or more who are enemies of A, according to the generalized pigeonhole principle. In the ﬁrst case, suppose that B,C,D are friends of A. If any two of B,C,D are friends, then these two and A form a group of three mutual friends. If no one is a friend of the other among B,C,D, then B,C,D is a group of three mutual enemies. In the second case, the proof is similar. The Ramsey number R(m,n), where m,n ≥2 are positive integers, denotes the minimum number of people at a party such that there are either m mutual friends or n mutual enemies, assuming any pair of people at the party are either friends or enemies. The above example shows that R(3,3) ≤6. In fact, it can be proved that R(3,3) = 6. 19
8220	https://www.oxfordreference.com/abstract/10.1093/acref/9780198832102.001.0001/acref-9780198832102-e-9559	Jump to Content Close A Dictionary of Mechanical Engineering (2 ed.) Marcel Escudier and Tony Atkins Publisher: : Oxford University Press Print Publication Date: : 2019 Print ISBN-13: : 9780198832102 Published online: : 2019 Current Online Version: : 2019 eISBN: : 9780191870866 Find at OUP.com Google Preview Read More Cite Save Share This Email this content ## Share Link Copy this link, or click below to email it to a friend Email this content or copy the link directly: The link was not copied. Your current browser may not support copying via this button. Link copied successfully Sign in Get help with access Sign in via your Institution Show Summary Details Publishing Information Preface to second edition Previous Version standard acceleration due to gravity (standard gravity) (g, g0, gn) Source: : A Dictionary of Mechanical Engineering Author(s): : ### Marcel Escudier, ### Tony Atkins The value 980.665 cm/s2 (9.806 65 m/s2) for g was adopted in the International Service of Weights and Measures at the 3rd General Conference on Weights and Measures (CGPM) in 1901. This is the accepted CODATA value. It is about 0.09% higher than the average value of ... ... Access to the complete content on Oxford Reference requires a subscription or purchase. Public users are able to search the site and view the abstracts and keywords for each book and chapter without a subscription. Please subscribe or login to access full text content. If you have purchased a print title that contains an access token, please see the token for information about how to register your code. For questions on access or troubleshooting, please check our FAQs, and if you can''t find the answer there, please contact us. Publishing Information Preface to second edition Oxford University Press Copyright © 2025. All rights reserved. PRINTED FROM OXFORD REFERENCE (www.oxfordreference.com). (c) Copyright Oxford University Press, 2023. All Rights Reserved. Under the terms of the licence agreement, an individual user may print out a PDF of a single entry from a reference work in OR for personal use (for details see Privacy Policy and Legal Notice). date: 30 August 2025 Cookie Policy Privacy Policy Legal Notice Credits Accessibility [3.136.168.16] 3.136.168.16
8221	https://kevintcarlberg.net/files/opt_class/OPT_Lecture3.pdf	Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Lecture 3: Constrained Optimization Kevin Carlberg Stanford University July 31, 2009 Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms 1 First-order optimality: Unconstrained problems 2 First-order optimality: Constrained problems Constraint qualiﬁcations KKT conditions Stationarity Lagrange multipliers Complementarity 3 Second-order optimality conditions Critical cone Unconstrained problems Constrained problems 4 Algorithms Penalty methods SQP Interior-point methods Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Constrained optimization This lecture considers constrained optimization minimize x∈Rn f (x) subject to ci(x) = 0, i = 1, . . . , ne dj(x) ≥0, j = 1, . . . , ni (1) Equality constraint functions: ci : Rn →R Inequality constraint functions: dj : Rn →R Feasible set: Ω= {x \| ci(x) = 0, dj(x) ≥0, i = 1, . . . , ne, j = 1, . . . , ni} We continue to assume all functions are twice-continuously diﬀerentiable Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms What is a solution? x f(x) Global minimum: A point x∗∈Ωsatisfying f (x∗) ≤f (x) ∀x ∈Ω Strong local minimum: A neighborhood N of x∗∈Ωexists such that f (x∗) < f (x) ∀x ∈N ∩Ω. Weak local minima A neighborhood N of x∗∈Ωexists such that f (x∗) ≤f (x) ∀x ∈N ∩Ω. Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Convexity As with the unconstrained case, conditions hold where any local minimum is the global minimum: f (x) convex ci(x) aﬃne (ci(x) = Aix + bi) for i = 1, . . . , ne dj(x) convex for j = 1, . . . , ni Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Active set The active set at a feasible point x ∈Ωconsists of the equality constraints and the inequality constraints for which dj(x) = 0 A(x) = {ci}ni i=1 ∪{dj \| dj(x) = 0} x f(x) d2 d3 d1 Ω d4 x Figure: A(x) = {d1, d3} Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Formulation of ﬁrst-order conditions Words to ﬁrst-order, the function cannot decrease by moving in feasible directions ↓ Geometric description description using the geometry of the feasible set ↓ Algebraic description description using the equations of the active constraints The algebraic description is required to actually solve problems (use equations!) Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms First-order conditions for unconstrained problems Geometric description: a weak local minimum is a point x∗ with a neighborhood N such that f (x∗) ≤f (x) ∀x ∈N Algebraic description: f (x∗) ≤f (x∗+ p), ∀p ∈Rn “small” (2) For f (x∗) twice-continuously diﬀerentiable, Taylor’s theorem is f (x∗+p) = f (x∗)+∇f (x∗)Tp+ 1 2pT∇2f (x∗+tp)p, t ∈(0, 1) Ignoring the O(∥p∥2) term, (2) becomes 0 ≤f (x∗+ p) −f (x∗) ≈∇f (x∗)Tp, ∀p ∈Rn Since pT 1 ∇f (x∗) > 0 implies that pT 2 ∇f (x∗) < 0 with p2 = −p1, we know that strict equality must hold →This reduces to the ﬁrst-order necessary condition: ∇f (x∗)Tp = 0 ∀p ∈Rn ⇒∇f (x∗) = 0 (stationary point) Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Constraint qualiﬁcations KKT conditions First-order conditions for constrained problems Geometric description: A weak local minimum is a point x∗ with a neighborhood N such that f (x∗) ≤f (x) ∀x ∈N ∩Ω Deﬁnition (Tangent cone TΩ(x∗)) The set of all tangents to Ωat x∗. (set of geometrically feasible directions, the limit of N ∩Ω−x∗) TΩ(x∗) x∗ Ω Using the tangent cone, we can begin to formulate the ﬁrst-order conditions algebraically Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Constraint qualiﬁcations KKT conditions First-order conditions for constrained problems Geometric description (continued) The limit of f (x∗) ≤f (x), ∀x ∈N ∩Ωis f (x∗) ≤f (x∗+ p), ∀p ∈TΩ(x∗) “small” Using Taylor’s theorem and ignoring high-order terms, this condition is 0 ≤f (x∗+ p) −f (x∗) ≈∇f (x∗)Tp, ∀p ∈TΩ(x∗) ∇f (x∗)Tp ≥0, ∀p ∈TΩ(x∗) (3) →To ﬁrst-order, the objective function cannot decrease in any feasible direction Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Constraint qualiﬁcations KKT conditions Constraint qualiﬁcations (3) is not purely algebraic since TΩ(x∗) is geometric We require an algebraic description of the tangent cone in terms of the constraint equations Deﬁnition (Set of linearized feasible directions F(x)) Given a feasible point x and the active constraint set A(x), F(x) = ( p \| p satisﬁes ( ∇ci(x)Tp = 0 ∀i ∇dj(x)Tp ≥0 ∀dj ∈A(x) ) The set of linearized feasible directions is the best algebraic description available, but in general TΩ(x) ⊂F(x) Constraint qualiﬁcations are suﬃcient for TΩ(x) = F(x) Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Constraint qualiﬁcations KKT conditions Example Consider the following problem minimize x∈Rn f (x) = x subject to d1(x) = x −3 ≥0 x f(x) x∗TΩ(x∗) !x+3 3 4 5 6 ∇d1(x∗) ∇f(x∗) !x+ 2 3 4 5 6 x ∇f ∇ Since d′ 1(x∗) = 1, pd′ 1(x∗) ≥0 for any p ≥0, and we have F(x∗) = p, ∀p ≥0 Thus, F(x∗) = TΩ(x∗) √ Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Constraint qualiﬁcations KKT conditions Example Consider the mathematically equivalent reformulation minimize x∈Rn f (x) = x subject to d1(x) = (x −3)3 ≥0 The solution x∗= 3 and (geometric) tangent cone TΩ(x∗) are unchanged However, d′ 1(x∗) = 3(3 −3)2 = 0 and pd′ 1(x∗) ≥0 for any p ∈R (positive or negative), and we have F(x∗) = p, ∀p ∈R X Thus, TΩ(x∗) ⊂F(x∗), and directions in F(x∗) may actually be infeasible! Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Constraint qualiﬁcations KKT conditions Constraint qualiﬁcations (suﬃcient for TΩ(x∗) = F(x∗)) Types Linear independence constraint qualiﬁcation (LICQ): the set of active constraint gradients at the solution {∇ci(x∗)}ni i=1 ∪{∇dj(x∗) \| dj(x∗) ∈A(∗x)} is linearly independent Linear constraints: all active constraints are linear functions None of these hold for the last example We proceed by assuming these conditions hold (F(x) = TΩ(x)) ⇒the algebraic expression F(x) can be used to describe geometrically feasible directions at x Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Constraint qualiﬁcations KKT conditions Algebraic description When constraint qualiﬁcations are satisﬁed, F(x) = TΩ(x) and (3) is ∇f (x∗)Tp ≥0, ∀p ∈F(x∗) (4) What form ∇f (x∗) ensures that (4) holds? Equality constraints: if we set ∇f (x∗) = ne P i=1 γi∇ci(x∗), then ∇f (x∗)Tp = Pne i=1 γi ∇ci(x∗)Tp = 0, ∀p ∈F(x∗) √ Inequality constraints: if we set ∇f (x∗) = ni P j=1 λj∇dj(x∗) with λj ≥0, then ∇f (x∗)Tp = Pni j=1 λj ∇dj(x∗)Tp ≥0, ∀p ∈F(x∗) √ Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Constraint qualiﬁcations KKT conditions Theorem (First-order necessary KKT conditions for local solutions) If x∗is a weak local solution of (1), constraint qualiﬁcations hold ∇f (x∗) − ne X i=1 γi∇ci(x∗) − ni X j=1 λj∇dj(x∗) = 0 λj ≥0, j = 1, . . . , ni ci(x∗) = 0, i = 1, . . . , ne dj(x∗) ≥0, j = 1, . . . , ni λjdj(x∗) = 0, j = 1, . . . , ni Stationarity, Dual feasibility, Primal feasibility (x∗∈Ω), Complementarity conditions, Lagrange multipliers γi, λj Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Constraint qualiﬁcations KKT conditions Intuition for stationarity minimize x∈Rn f (x) = x2 1 + x2 2 subject to d1(x) = x1 + x2 −3 ≥0 x f(x) x∗TΩ(x∗) x y !x+3 !2 !1 0 1 2 3 4 5 6 !3 !2 !1 0 1 2 3 4 5 6 x∗ x1 x2 ∇d1(x∗) ∇f(x∗) The solution is x∗= (1.5, 1.5) Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Constraint qualiﬁcations KKT conditions Intuition for stationarity (continued) The KKT conditions say ∇f (x∗) = λ1∇d1(x∗) with λ1 ≥0 Here, ∇f (x∗) = [3, 3]T, while ∇d1(x∗) = [1.5, 1.5]T, so these conditions are indeed veriﬁed with λ1 = 2 ≥0 This is obvious from the ﬁgure: if ∇f (x∗) and ∇d1(x∗) were “misaligned”, there would be some feasible descent directions! x f(x) x∗TΩ(x∗) x y !x+3 !2 !1 0 1 2 3 4 5 6 !3 !2 !1 0 1 2 3 4 5 6 x∗ x1 x2 ∇d1(x∗) ∇f(x∗) x y !x+3 !2 !1 0 1 2 3 4 5 6 !3 !2 !1 0 1 2 3 4 5 6 x∗ x1 x2 x ∇f(x) ∇d1(x) feasible descent directions This gives us some intuition for stationarity and dual feasibility Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Constraint qualiﬁcations KKT conditions Lagrangian Deﬁnition (Lagrangian) The Lagrangian for (1) is L(x, γ, λ) = f (x) − ne P i=1 γici(x) − ni P j=1 λjdj(x) Stationarity in the sense of KKT is equivalent to stationarity of the Lagrangian with respect to x: Lx(x, γ, λ) = ∇f (x) − ne X i=1 γi∇ci(x) − ni X j=1 λj∇dj(x) KKT stationarity ⇔Lx(x∗, γ, λ) = 0 Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Constraint qualiﬁcations KKT conditions Lagrange multipliers Lagrange multipliers γi and λj arise in constrained minimization problems They tell us something about the sensitivity of f (x∗) to the presence of their constraints. γi and λj indicate how hard f is “pushing” or “pulling” the solution against ci and dj. If we perturb the right-hand side of the ith equality constraint so that ci(x) ≥−ϵ∥∇ci(x∗)∥, then df (x∗(ϵ)) dϵ = −γi∥∇ci(x∗)∥. If the jth inequality is perturbed so dj(x) ≥−ϵ∥∇dj(x∗)∥, df (x∗(ϵ)) dϵ = −λj∥∇di(x∗)∥. Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Constraint qualiﬁcations KKT conditions Constraint classiﬁcation Deﬁnition (Strongly active constraint) A constraint is strongly active at if it belongs to A(x∗) and it has: a strictly positive Lagrange multiplier for inequality constraints (λj > 0) a strictly non-zero Lagrange multiplier for equality constraints (γi > 0) Deﬁnition (Weakly active constraint) A constraint is weakly active at if it belongs to A(x∗) and it has a zero-valued Lagrange multiplier (γi = 0 or λj = 0) Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Constraint qualiﬁcations KKT conditions Constraint classiﬁcation (continued) Weakly active and inactive constraints “do not participate” minimize x∈Rn f (x) = x2 1 + x2 2 subject to d1(x) = x1 + x2 −3 ≥0 (strongly active) d2(x) = x1 −1.5 ≥0 (weakly active) d3(x) = −x2 1 −4x2 2 + 5 ≥0 (inactive) x1 x2 !1 0 1 2 3 4 5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 x y x = !1 sqrt(!1/4 x2+20/4), y = k x∗ The solution is unchanged if d2 and d3 are removed, so λ2 = λ3 = 0 Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Constraint qualiﬁcations KKT conditions Intuition for complementarity We just saw that non-participating constraints have zero Lagrange multipliers The complementarity conditions are λjdj(x∗) = 0, j = 1, . . . , ni This means that each inequality constraint must be either: 1 Inactive (non-participating): dj(x∗) > 0, λj = 0, 2 Strongly active (participating): dj(x∗) = 0 and λj > 0, or 3 Weakly active (active but non-participating): dj(x∗) = 0 and λj = 0 Strict complementarity: either case 1 or 2 is true for all constraints (no constraints are weakly active) Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Critical cone Unconstrained problems Constrained problems Second-order optimality conditions Second-order conditions for constrained optimization play a “tiebreaking” role: determine whether “undecided” directions for which pT∇f (x∗) = 0 will increase or decrease f . We call these ambiguous directions the “critical cone” Deﬁnition (Critical cone C(x∗, γ)) Directions that “adhere” to strongly active constraints and equality constraints C(x∗, γ) = {w ∈F(x∗) \| ∇dj(x∗)Tw = 0, ∀j ∈A(x∗) with λj > 0} Note that λj > 0 implies the constraint will remain active even when small changes are made to the objective function! Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Critical cone Unconstrained problems Constrained problems Critical cone For the problem minimize x∈Rn f (x) = x2 1 + x2 2 subject to d1(x) = x1 + x2 −3 ≥0 the critical cone is C(x∗, γ) = α(−1, 1), ∀α ∈R x y !x+3 !2 !1 0 1 2 3 4 5 6 !3 !2 !1 0 1 2 3 4 5 6 x∗ x1 x2 ∇d1(x∗) ∇f(x∗) C(x∗, λ) Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Critical cone Unconstrained problems Constrained problems Second-order conditions for unconstrained problems Recall, second-order conditions for unconstrained problems Theorem (Necessary conditions for a weak local minimum) A1. ∇f (x∗) = 0 (stationary point) A2. ∇2f (x∗) is positive semi-deﬁnite (pT∇2f (x∗)p ≥0 for all p ̸= 0) Theorem (Suﬃcient conditions for a strong local minimum) B1. ∇f (x∗) = 0 (stationary point) B2. ∇2f (x∗) > 0 is positive deﬁnite (pT∇2f (x∗)p > 0 for all p ̸= 0). Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Critical cone Unconstrained problems Constrained problems Second-order conditions for constrained problems We make an analogous statement for constrained problems, but limit the directions p to the critical cone C(x∗, γ) Theorem (Necessary conditions for a weak local minimum) D1. KKT conditions hold D2. pT∇2L(x∗, γ)p ≥0 for all p ∈C(x∗, γ) Theorem (Suﬃcient conditions for a strong local minimum) E1. KKT conditions hold E2. pT∇2L(x∗, γ)p > 0 for all p ∈C(x∗, γ). Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Critical cone Unconstrained problems Constrained problems Intuition for second-order conditions x∗ x1 x2 ∇d1(x∗) ∇f(x∗) x y (x+y)2+3 (x!y)2 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 x∗ x1 x2 ∇d1(x∗) ∇f(x∗) x y (x+y)2+3 (x!y)2 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 x∗ x1 x2 ∇d1(x∗) ∇f(x∗) x y (x+y)2+3 (x!y)2 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 C(x∗, λ) C(x∗, λ) C(x∗, λ) d1(x) d1(x) d1(x) Case I Case 2 Case 3 Case 1: E1 and E2 are satisﬁed (suﬃcient conditions hold) Case 2: D1 and D2 are satisﬁed (necessary conditions hold) Case 3: D1 holds, D2 does not (necessary conditions failed) Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Critical cone Unconstrained problems Constrained problems Next We now know how to correctly formulate constrained optimization problems and how to verify whether a given point x could be a solution (necessary conditions) or is certainly a solution (suﬃcient conditions) Next, we learn algorithms that are use to compute solutions to these problems Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Penalty methods SQP Interior-point methods Constrained optimization algorithms Linear programming (LP) Simplex method: created by Dantzig in 1947. Birth of the modern era in optimization Interior-point methods Nonlinear programming (NLP) Penalty methods Sequential quadratic programming methods Interior-point methods Almost all these methods rely strongly on line-search and trust region methodologies for unconstrained optimization Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Penalty methods SQP Interior-point methods Penalty methods Penalty methods combine the objective function and constraints minimize x∈Rn f (x) s.t. ci(x) = 0, i = 1, . . . , ni ↓ minimize x∈Rn f (x) + µ 2 ni X i=1 c2 i (x) A sequence of unconstrained problems is then solved for µ increasing Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Penalty methods SQP Interior-point methods Penalty methods example Original problem: minimize x∈R2 f (x) = x2 1 + 3x2, s.t. x1 + x2 −4 = 0 0 1 2 3 4 0 1 2 3 4 0 5 10 15 20 25 x x2+3 y y f (x) Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Penalty methods SQP Interior-point methods Penalty methods example Penalty formulation: minimize x∈R2 g(x) = x2 1 + 3x2 + µ 2 (x1 + x2 −4)2 0 1 2 3 4 0 1 2 3 4 0 5 10 15 20 25 x x2+3 y y f (x) 0 1 2 3 4 0 1 2 3 4 0 10 20 30 40 50 60 x x2+3 y+µ (x+y! 4)2 y g(x) A valley is created along the constraint x1 + x2 −4 = 0 Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Penalty methods SQP Interior-point methods Sequential quadratic programming Perhaps the most eﬀective algorithm Solve a QP subproblem at each iterate minimize p 1 2pT∇2 xxL(xk, λk)p + ∇f (xk)Tp subject to ∇ci(xk)Tp + ci(xk) = 0, i = 1, . . . , ne ∇dj(xk)Tp + dj(xk) ≥0, j = 1, . . . , ni When ni = 0, this is equivalent to Newton’s method on the KKT conditions When ni > 0, this corresponds to an “active set” method, where we keep track of the set of active constraints A(xk) at each iteration Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Penalty methods SQP Interior-point methods Interior-point methods These methods are also known as “barrier methods,” because they build a barrier at the inequality constraint boundary minimize p f (x) −µ m X i=1 log si subject to ci(x) = 0, i = 1, . . . , ne dj(x) −si = 0, j = 1, . . . , ni Slack variables: si, indicates distance from constraint boundary Solve a sequence of problems with µ decreasing Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Penalty methods SQP Interior-point methods Interior-point methods example Original problem: minimize x∈R2 f (x) = x2 1 + 3x2, s.t. −x1 −x2 + 4 ≥0 0 1 2 3 4 0 1 2 3 4 0 5 10 15 20 25 x x2+3 y y f (x) Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Penalty methods SQP Interior-point methods Interior-point methods example Interior-point formulation: minimize x∈R2 h(x) = x2 1 + 3x2 −log (−x1 −x2 + 4) A barrier is created along the boundary of the inequality constraint x1 + x2 −4 = 0 Kevin Carlberg Lecture 3: Constrained Optimization Outline and terminologies First-order optimality: Unconstrained problems First-order optimality: Constrained problems Second-order optimality conditions Algorithms Penalty methods SQP Interior-point methods Summary We now now something about: Modeling and classifying unconstrained and constrained optimization problems Identifying local minima (necessary and suﬃcient conditions) Solving the problem using numerical optimization algorithms We next consider the case of PDE-constrained optimization, which enables us to use to tools learned earlier (ﬁnite elements) in optimal design and control settings, for example Kevin Carlberg Lecture 3: Constrained Optimization
8222	https://www.nagwa.com/en/explainers/956125602506/	Sign Up Sign In English English العربية English English العربية My Wallet Sign Up Sign In My Wallet My Classes My Messages My Reports Lesson Explainer: Equations of Tangent Lines and Normal Lines Mathematics • Second Year of Secondary School In this explainer, we will learn how to find the slope and equation of the tangent and normal to a curve at a given point using derivatives. The derivative of a curve at a point tells us the slope of the tangent line to the curve at that point and there are many different techniques for finding the derivatives of different functions. We can utilize these differentiation techniques to help us find the equation of tangent lines to various differentiable functions. First, let us recall exactly what we mean by the tangent to a curve a point. Definition: The Tangent Line to a Curve at a Point For a curve and point on the curve, we say that the line is the tangent line to the curve at the point if the tangent line passes through the point ; the curve and tangent line have the same slope at the point . In the definition above, we state that our tangent line and curve will have the same slope at the point . This means that, around the point , the line will only touch the curve. The equation of a straight line can be found using the value of its slope and the coordinates of a point which lies on the line. From the definition above, we know that the tangent line and the curve both pass through the point . Hence, the only missing piece of information is the slope. We can then use differentiation to find the slope of the curve, , at this point. For a function , which is differentiable at , this slope is given by . Let us see an example of how we can use this to find the equation of the tangent line to a curve at a point. Example 1: Finding the Equation of the Tangent to the Curve of a Polynomial Function at a Given Value for 𝑥 Find the equation of the tangent to the curve at . Answer To find the equation of the tangent to a curve at a point, we need two pieces of information: the coordinates of the point and the slope of the curve at this point. The question wants us to find the tangent when , so the -coordinate is 2. We can find the -coordinate for this value of by substituting into the equation of our curve: This gives us the point on our curve that our tangent line must pass through. Next, we need the slope of the curve when ; to find this we need to differentiate: We then substitute in to find the slope of the tangent line at this point: Now, to find the equation of our tangent line, we recall that the equation of a line of slope passing through is given by In our case, we have Therefore, the equation of the tangent line to our curve at is given by the equation . This method gives us a really useful result for finding the equation of a tangent line to a curve at a point (provided the derivative of the curve exists at this point). If we have a curve and a point on our curve, then the tangent line to our curve at this point must have slope . Recall that the point-slope form of a straight line tells us the equation of a line passing through the point with slope , given by So, if we know the slope of this tangent line and a point it passes through, this is all the information we need to find its equation. Definition: The Equation of a Tangent Line The equation of the tangent line to the curve at the point is given by This formula assumes is differentiable at . If our function is not differentiable at , then we cannot use this formula to find the equation of the tangent line at . Instead, we will need to consider the problem graphically. Consider trying to find the tangent lines to the following two curves at . In our first graph, we can see that the curve is not defined at , which also means it is not differentiable at . If our curve is not defined at , then it cannot have a tangent line at this value of . In our second graph, we can see that the tangent line at should be vertical. If we were to differentiate our function We could then attempt to find the slope of our curve at : We then see this is undefined. When is not differentiable at a point, sketching the graph can often help us determine whether this point has a vertical tangent line. So far, we have been focused on tangent lines. However, there is another important type of line we need to consider called a normal line. A normal line to a curve at a point is very similar to the tangent line; the only difference is that the normal line will be perpendicular to the tangent line. Definition: The Normal Line to a Curve at a Point For a curve and point on the curve, we say that the line is the normal line to the curve at the point if the point lies on our line; this line is perpendicular to the tangent line of our curve at this point. It is worth pointing out that we can define the normal line based on the information that it is perpendicular to the curve at this point; however, it can be easier to think about the line being perpendicular to the tangent line. Finding the equation of the normal line will take a little bit more work since the derivative of the function only gives us the slope of the tangent line. To find the equation of the normal to a curve at a point, we need a point on the line and its slope to find the point-slope equation. Since we already know the normal passes through the point , we only need to find the slope of the normal line to find its point-slope equation. We want to find an expression for the slope of our normal line in terms of the slope of the tangent line. To do this, we first note that if the tangent line is horizontal, then the normal line needs to be perpendicular, so it must be a vertical line, and the same will be true in reverse. This means we can now assume we are dealing with a tangent line which is not horizontal or vertical. So, we can write the equation of the tangent line at the point in the form , where is not zero. We will also say that the equation of our normal line is . To find an expression for the slope of our normal line, , we will start with a sketch. To find an expression for the slope, , we will add in the line . We can now see we have a triangle with a right angle at . We can find the coordinates of the vertices. We already know the coordinates of the vertex at ; we will compare the other two vertices to this point to find their coordinates. Since we chose our vertical line to be one unit to the right of , our other two vertices will be one unit to the right. We can find the -coordinates of these two vertices by recalling that the slope of a line tells us the change in for every 1 unit change in . Since the change in is 1 unit for both vertices, the change in for the tangent line will be and the change in for the normal line will be . The coordinates of the vertices are therefore , , and . We can then find the lengths of each side in this triangle, by using the distance between two points formula. We will not go through the individual calculation steps, but the results are shown below. Finally, since this is a right triangle, we can apply the Pythagorean theorem to this triangle, where the hypotenuse of this triangle is the side opposite the right angle, in this case the vertical line. In fact, because this is a vertical line, the length of this line is the difference between its -coordinates. In this case, we will use , since we do not know the sign of : This gives us an equation to find the slope of our normal line; it is the negative of the reciprocal of the slope of the tangent line. We also know how to find the slope of the tangent by using the derivative. This means we can use the fact that to find a formula for the equation of the normal line. Definition: The Equation of a Normal Line to a Curve If , then the equation of the normal line to at the point is given by If the slope of our curve at is zero, then the normal line at this point will be vertical, and its equation will be . If the slope of our curve is undefined at a point, there are two possibilities. The tangent line to the curve at this point is vertical; in this case, the normal line will be horizontal. The tangent line to the curve at this point does not exist; in this case, the normal line does not exist. Let us see some examples of applying these formulae to some curves. Example 2: Finding the Equation of the Normal to the Curve of a Polynomial Function at a Point with the Given 𝑥-Coordinate Find the equation of the normal to the curve at . Answer We want to find the equation of the normal line to a curve at a point. To do this, we need to find a point on the line and its slope. We can find a point on the line by substituting into the equation of our curve: So, the normal line passes through the point . Next, we recall that we can find the slope of the normal line by using the slope of the tangent line. If we let , then the tangent line will have slope : We have shown that the tangent will have slope 4, but the slope of the normal line is the negative of the reciprocal of this value: We know the slope of our line is and we also know it passes through . This gives us the equation Therefore, the equation of the normal line to the curve when is given by In our next example, we will consider how to find the points on a curve where its tangent line at that point will be parallel to a given line. Example 3: Finding the 𝑥-Coordinate of the Point on the Curve of a Quadratic Function Where the Tangent Line Is Parallel to the 𝑥-Axis What is the -coordinate of the point where the tangent line to is parallel to the -axis? Answer We want to find the -coordinate where the tangent line to this curve will be parallel to the -axis. We know that the -axis is horizontal, so any line parallel to this must also be horizontal; in other words, the slope of this tangent line must be equal to zero. We also know that, for a curve , the slope of the tangent line to this curve at the point will be its derivative at that point, in this case . Therefore, in order to solve this question, we must find the values of for which the derivative is equal to zero. Since our function is a polynomial, we can differentiate it by using the power rule for differentiation: By setting our derivative equal to zero, we can find the values of where the tangent line is parallel to the -axis: Therefore, the tangent line to this curve is parallel to the -axis when . In our next example, we will find the equation of a tangent line to a curve that makes a specific angle with the positive -axis. Example 4: Finding the Equation of the Tangent to the Curve of a Cubic Function given the Angle the Tangent Makes with the 𝑥-Axis Find the equation of the tangent to the curve that makes an angle of with the positive -axis. Answer In this question, we want to find the tangent to a curve which makes an angle of with the positive -axis. This means that to answer this question we are going to need to work out the corresponding slope for the line that makes this angle with the positive -axis. First, if we translate a line, it will not change the angle it makes with the positive -axis. So, we can start by sketching our line passing through the origin (since this is the simplest case), making an angle of with the positive -axis. This will have the same slope as our tangent line. We can then see that , giving us the following diagram: There are then two ways to find the slope of this line; we can use the fact that the slope of a line is the tangent of the angle it makes with the positive -axis; in this case, ; or we could use trigonometry to find the slope of this line. In either case, we see the question wants us to find the tangent line with slope . The slope of the tangent line at a point is the same as the derivative of the curve at that point, so we want to set the derivative equal to and solve for : This means we want to solve The solution is . To find the equation of the tangent line to the curve when , we need to find the coordinates of a point on the line. We can find this by substituting into the equation for our curve: The tangent we are looking for passes through and has a slope of . We can use this to find the equation of the line Therefore, the tangent to the curve which makes an angle of with the positive -axis has the equation It is not always as simple as differentiating a polynomial to find the slope of the tangent or normal line to our curve. We will sometimes need to apply other derivative rules to help us find this value. Let us see an example of this. Example 5: Finding the Equation of the Normal to the Curve of a Function Involving Trigonometric Functions at a Given 𝑥-Coordinate Find all points with -coordinates in where the curve has a tangent that is parallel to the line . Answer First, for a line to be parallel to the line , it must have the same slope. Therefore, our tangent line must have a slope of . Recall that the slope of the tangent line to a curve at the point is . In our case, . We can differentiate this using the fact that, for any constant where is measured in radians, Hence, Setting the slope of our tangent line to be equal to gives us We can then solve this for in the interval : We can sketch this as follows. This has solutions and . Finally, we need to find the coordinates of these points by substituting these -values into the function : giving us the coordinates and . Hence, the coordinates of the points with -coordinates in , where the curve has a tangent that is parallel to the line , are and . In our final example, we will determine the point of intersection between two curves where the intersection is orthogonal. Example 6: Finding the Point Where Two Quadratic Curves Intersect Orthogonally The curves and intersect orthogonally at a point. What is this point? Answer We say that two curves intersect orthogonally if they intersect at right angles. Equivalently, the tangent lines to both curves at the point of intersection are orthogonal (meet at right angles). We recall that the slope of a curve at a point is given by the value of its derivative at this point. We start by finding all of the points of intersection between these two curves by setting the functions to be equal to one another and solving for : Hence, the curves intersect when and when . We need to find the slopes of both curves at each of these -values to determine if they are orthogonal. We do this by differentiating each curve by using the power rule for differentiation. For the first curve: We can use this to find the slope at both -values. At , At , We can do the same with the second curve: At , At , Since the slopes of all four lines are nonzero, for the lines to be orthogonal, their slopes must be negative reciprocals of each other. Taking the negative reciprocal of gives us which is not equal to the slope of the second curve at this point, so the tangent lines are not orthogonal. Taking the negative reciprocal of 1 gives us which is equal to the slope of the second curve at this point. Hence, the tangent lines are orthogonal. We can find the coordinates of this point by substituting into the equation of either curve: Hence, the curves intersect orthogonally at the point . Let us finish by recapping some of the things we covered when finding the equations of tangent lines and normal lines to curves. Key Points The equation of the tangent line to the curve at the point is given by If , then the equation of the normal line to at the point is given by If , then the tangent line to at the point is horizontal and has the equation If , the normal line to at the point is vertical and has the equation Two curves intersect orthogonally at the point if both curves intersect this point and the slopes of their tangents at this point are orthogonal. If is undefined, we may still be able to find the tangent and normal lines at . However, it is not always possible. Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! Interactive Sessions Chat & Messaging Realistic Exam Questions Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy
8223	https://www.quora.com/What-is-the-difference-between-1-3-and-1-4	What is the difference between 1/3 and 1/4? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Mathematics Homework Ques... Fractions and Decimals Very Basic Arithmetic Simple Mathematics Arithmetical Calculations Basicmath Arithmetic Fractions (mathematics) 5 What is the difference between 1/3 and 1/4? All related (60) Sort Recommended Assistant Bot · 1y The difference between 1/3 and 1/4 is that 1/3 is larger than 1/4 . To understand this difference, let's compare the decimal representations of these fractions. 1/3 is equal to 0.3333, which is a little bit more than 1/4, which is equal to 0.25 . Therefore, 1/3 is greater than 1/4. It's important to note that there are an infinite number of fractions between 1/3 and 1/4 . These fractions can be found by selecting fractions with the same denominators. For example, if we choose 36 as a denominator, then 10/36 and 11/36 are between 1/4 and 1/3 . By choosing larger and larger denominato Continue Reading The difference between 1/3 and 1/4 is that 1/3 is larger than 1/4 . To understand this difference, let's compare the decimal representations of these fractions. 1/3 is equal to 0.3333, which is a little bit more than 1/4, which is equal to 0.25 . Therefore, 1/3 is greater than 1/4. It's important to note that there are an infinite number of fractions between 1/3 and 1/4 . These fractions can be found by selecting fractions with the same denominators. For example, if we choose 36 as a denominator, then 10/36 and 11/36 are between 1/4 and 1/3 . By choosing larger and larger denominators, we can find more and more fractions between 1/3 and 1/4 . Another way to find fractions between 1/3 and 1/4 is by taking the geometric mean. The geometric mean of 1/3 and 1/4 is sqrt(1/12) or sqrt3/6 . This fraction is also between 1/3 and 1/4 . In summary, the difference between 1/3 and 1/4 is that 1/3 is larger than 1/4. However, there are an infinite number of fractions between these two fractions, which can be found by selecting fractions with the same denominators or by taking the geometric mean . Learn more: What is the difference between 1/3 and 1/4? - Quora what fractions are between 1/4 and 1/3? \| Wyzant Ask An Expert How do you find the fraction between 1/3 and 1/4? \| Socratic Upvote · 9 2 Alora Thresher B.A. in Linguistics, University of California, Los Angeles (Graduated 2021) · Author has 79 answers and 184.8K answer views ·7y Originally Answered: What is 1/3 of 1/4? · Here’s an unnecessarily detailed answer: When you see the word “of,” it’s safe to assume you need to multiply. For example, if this said “What is 1/3 of 15?” you would write out 1/315 = 1/315/1 = 15/3 = 5 And you check this by asking “Does 53 make 15? Why yes it does, great.” So now you’ve proven that you answer “What is 1/3 of x?” by multiplying 1/3 by x, and you just follow the same procedure. 1/31/4 = 1/12 And now, you know that three 1/12ths make 1/4, which is not particularly intuitive and potentially useful. Good luck! Upvote · 99 20 9 8 Sponsored by Grammarly Stuck on the blinking cursor? Move your great ideas to polished drafts without the guesswork. Try Grammarly today! Download 99 34 Mohan Raghavan Former Principal · Author has 2.9K answers and 1.7M answer views ·3y (1/3)— (1/4)= (4–3)/12= 1/12 (1/3) is (1/12 ) greater than (1/4). Upvote · 9 2 Related questions More answers below How many 1/3 cups make a 1/4 cup? How many 1/4 cups is equal to 1/3? My recipe calls for 3/4 of a cup, but I used 1/3. What more do I need to? Which is bigger, 1/3 or 1/4? How many 1/3 cups are in a 3/4 cup? Ashish Kumar Yadav Ashish Kumar Yadav's Space Answered by Samyak Chordiya · ·May 1, 2023 The difference between 1/3 and 1/4 is: 1/3 - 1/4 = (4/12) - (3/12) = 1/12 Therefore, the difference between 1/3 and 1/4 is 1/12. Upvote · 9 2 Pulak Das Lives in Siliguri, West Bengal, India (1985–present) · Author has 70 answers and 340.6K answer views ·2y The difference between 1/3 and 1/4 is: 1/3 - 1/4 = (4/12) - (3/12) = 1/12 Therefore, the difference between 1/3 and 1/4 is 1/12. Upvote · 9 1 Related questions More answers below Does a 1/3 cup add 1/4 cup add up to 1/2 cup? What measuring cups can be used to equal 3/4 cups? How many ounces are the difference between a 1/3 cup and a 1/4 cup? What is 1 3/4 of a cup times two? What fraction is exactly between 1/3 and 1/4? Kunal Choudhary b.tech computer science engineer ·11mo The difference between 1/3 and 1/4 is 1/3 is greater than 1/4. Justification, When we convert both the value in decimal 1/3= 0.333 1/4= 0.25 Hence, 1/3 is grater than 1/4 Upvote · Sponsored by JetBrains Write better C++ code with less effort. Boost your efficiency with refactorings, code analysis, unit test support, and an integrated debugger. Download 999 897 Sergio Alvarez Ph.D. from University of Maryland, College Park · Author has 115 answers and 186.8K answer views ·7y Originally Answered: What fraction is exactly between 1/3 and 1/4? · It depends on what you mean by “exactly between” (a pedantic viewpoint is that a number is either between two given numbers, or else it isn’t; in the first case, the betweenness is “exact”, and in the second, it’s just absent). If you mean “halfway between”, as in the midpoint on a number line, then what you’re looking for is the arithmetic mean of the numbers, which is half their sum - that’s 7/24 7/24, in this case. The arithmetic mean of two numbers is equidistant from those two numbers. There’s another reasonable option, though, namely to make the proportions the same (for example, 4 4 is halfway Continue Reading It depends on what you mean by “exactly between” (a pedantic viewpoint is that a number is either between two given numbers, or else it isn’t; in the first case, the betweenness is “exact”, and in the second, it’s just absent). If you mean “halfway between”, as in the midpoint on a number line, then what you’re looking for is the arithmetic mean of the numbers, which is half their sum - that’s 7/24 7/24, in this case. The arithmetic mean of two numbers is equidistant from those two numbers. There’s another reasonable option, though, namely to make the proportions the same (for example, 4 4 is halfway between 2 2 and 8 8 in this latter sense, since 4/2=8/4 4/2=8/4). The value that accomplishes that is called the geometric mean, and can be computed as the square root of the product of the two numbers (equivalently, its logarithm is the arithmetic mean of the logarithms of the two numbers). In the case of 1/3 1/3 and 1/4 1/4, the geometric mean is 1/(2√3)1/(2 3), which, interestingly, cannot be written as the ratio of two whole numbers. Upvote · 9 3 Audley Willacey B.Sc. in Mechanical Engineering, University of the West Indies, St. Augustine (Graduated 1976) · Author has 5K answers and 1.5M answer views ·7y Originally Answered: What fraction is exactly between 1/3 and 1/4? · THE DIFFERENCE: 1/3–1/4=4/12–3/12=1/12. HALF THE DIFFERENCE: 1/2 x 1/12=1/24 ADD TO SMALLER:1/4+1/24=6/24+1/24=7/24 OR SUBTRACT FROM LARGER:1/3–1/24=8/24–1/24=7/24. 7/24 is exactly between 1/4 and 1/3. ALTERNATIVELY 1/4 ………… 1/3 3/12………….4/12 6/24………….8/24 7/24 is exactly between 6/24 and 8/24. Upvote · 9 6 Sponsored by MailerLite Turn your AI tool into a personal email marketing workhorse. AI tools like Claude can connect to MailerLite, turning it into an email marketing expert. Try it free! Learn More 9 4 Brad Vale B.A. in Mathematics, Boise State University (Graduated 2005) ·7y Originally Answered: What is 1/3 of 1/4? · Lets say you have a pizza divided into 4 pieces. Each piece is 1/4 of the whole. Now, you need to take 1/3 of one of the pieces. How do you do that? Well, first find the GCF of 3 and 4, which is 12. If we now divide the pizza into 12 pieces, we can see that each 1/4 piece now contains 3 smaller pieces. So, 1/3 of 1/4 is 1/12. Or, simply multiply 1/3 x 1/4 which equals 1/12. Upvote · 9 4 Charles Holmes Studied Financial Markets&Mathematics (Graduated 1990) · Author has 16.6K answers and 13.6M answer views ·4y Originally Answered: What fraction is exactly between 1/3 and 1/4? · y=7/24. The rational number 7/24 lies exactly between 1/3 and 1/4. PREMISES y=the rational number R that lies exactly between 1/3 and 1/4 CALCULATIONS The rational number R exactly between 1/3 and 1/4 can be denoted by the statement: y=(1/3+1/4)/2 y=[(8+6)/2]/24 y=(14/2)/24 y= 7/24 (decimal: 0.291666…) C.H. Upvote · 9 3 9 5 Sponsored by Qustodio Technologies Sl. Protect your kids online. Parenting made easy; monitor your kids, keep them safe, & gain peace of mind with Qustodio App. Learn More 999 643 Lance Berg Author has 28K answers and 54.7M answer views ·7y Originally Answered: Is 3/4 the same as 1/4? · Let’s have some pie. I’ll cut it in half, and then we each have half, that’s fair right? Wait, no, I’ll cut each of the halves in half as well, so now there are four quarters. You can have one quarter and I’ll take the other three, is that fair? I think you can see that it’s not, I have three times as much pie as you do! Three quarters is not at all the same as one quarter. Upvote · 9 1 Charles Holmes Studied Financial Markets&Mathematics (Graduated 1990) · Author has 16.6K answers and 13.6M answer views ·5y Originally Answered: What is 1/3 of 1/4? · y=1/12 (One twelfth) PREMISES y=1/3(1/4) CALCULATIONS y=1/3(1/4) y=(1×1)/(3×4) y= 1/12 PROOF If y=1/12, then inversely 1/4=1/12+1/12+1/12 returns 1/4=3(1/12) 1/4=3/12 and 1/4=1/4 which proves y=1/12 to the expression 1/3(1/4) C.H. y=3/12 Upvote · 9 2 9 1 John Knight Studied Mathematics at Australian National University · Author has 4.6K answers and 6.2M answer views ·7y Originally Answered: What fraction is exactly between 1/3 and 1/4? · The fraction 7/24 is exactly in between 1/3 and 1/4. Upvote · 99 15 9 1 Related questions How many 1/3 cups make a 1/4 cup? How many 1/4 cups is equal to 1/3? My recipe calls for 3/4 of a cup, but I used 1/3. What more do I need to? Which is bigger, 1/3 or 1/4? How many 1/3 cups are in a 3/4 cup? Does a 1/3 cup add 1/4 cup add up to 1/2 cup? What measuring cups can be used to equal 3/4 cups? How many ounces are the difference between a 1/3 cup and a 1/4 cup? What is 1 3/4 of a cup times two? What fraction is exactly between 1/3 and 1/4? What is 1 and 3/4 cups times 3? How many 1/4 are in 1 3/4? What is 1/3 of 1/4? What is 1/3 in measuring cups? I’m try to following serving size of cereal and it says 1 1/3. I have no idea how to measure out 1/3 I'm thinking of a number between 1 and 3 and it's not 2. What is it? Related questions How many 1/3 cups make a 1/4 cup? How many 1/4 cups is equal to 1/3? My recipe calls for 3/4 of a cup, but I used 1/3. What more do I need to? Which is bigger, 1/3 or 1/4? How many 1/3 cups are in a 3/4 cup? Does a 1/3 cup add 1/4 cup add up to 1/2 cup? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
8224	https://en.wikipedia.org/wiki/Solar_mass	Jump to content Solar mass Afrikaans العربية Asturianu বাংলা 閩南語 / Bn-lm-gí Беларуская Беларуская (тарашкевіца) Български Brezhoneg Català Чӑвашла Čeština Deutsch Eesti Ελληνικά Español Esperanto Euskara فارسی Français 한국어 Հայերեն हिन्दी Hrvatski Igbo Bahasa Indonesia Interlingua Italiano עברית ქართული Kiswahili Latina Latviešu Lëtzebuergesch Lietuvių Magyar Македонски Malagasy മലയാളം मराठी მარგალური Bahasa Melayu Minangkabau မြန်မာဘာသာ Nederlands 日本語 Norsk bokmål Norsk nynorsk Occitan Oʻzbekcha / ўзбекча ਪੰਜਾਬੀ Plattdüütsch Polski Português Romnă Русский Scots Sicilianu සිංහල Simple English Slovenčina Slovenščina کوردی Српски / srpski Suomi Svenska தமிழ் తెలుగు ไทย Türkçe Türkmençe Удмурт Українська Tiếng Việt Volapük Winaray 吴语粵語中文 Edit links From Wikipedia, the free encyclopedia Standard unit of mass in astronomy \| Solar mass \| \| --- \| \| The Sun contains 99.86% of the mass of the Solar System. Bodies less massive than Saturn are not visible at this scale. Jupiter's mass (0.10%) is actually more than three times Saturn's (0.03%), which is not accurately visible in this graphic. \| \| \| General information \| \| \| Unit system \| astronomy \| \| Unit of \| mass \| \| Symbol \| M☉ \| \| In SI base units \| 1.988416×1030 kg \| The solar mass (M☉) is a frequently used unit of mass in astronomy, equal to approximately 2×1030 kg. It is approximately equal to the mass of the Sun. It is often used to indicate the masses of other stars, as well as stellar clusters, nebulae, galaxies and black holes. More precisely, the mass of the Sun is nominal solar mass M☉ = 1.988416×1030 kg or a best estimate of M☉ = (1.988475±0.000092)×1030 kg. The solar mass is about 333000 times the mass of Earth (M🜨), or 1047 times the mass of Jupiter (MJ). History of measurement [edit] The value of the gravitational constant was first derived from measurements that were made by Henry Cavendish in 1798 with a torsion balance. The value he obtained differs by only 1% from the modern value, but was not as precise. The diurnal parallax of the Sun was accurately measured during the transits of Venus in 1761 and 1769, yielding a value of 9″ (9 arcseconds, compared to the present value of 8.794148″). From the value of the diurnal parallax, one can determine the distance to the Sun from the geometry of Earth. The first known estimate of the solar mass was by Isaac Newton. In his work Principia (1687), he estimated that the ratio of the mass of Earth to the Sun was about 1:28700. Later he determined that his value was based upon a faulty value for the solar parallax, which he had used to estimate the distance to the Sun. He corrected his estimated ratio to 1:169282 in the third edition of the Principia. The current value for the solar parallax is smaller still, yielding an estimated mass ratio of 1:332946. As a unit of measurement, the solar mass came into use before the astronomical unit and the gravitational constant were precisely measured. This is because the relative mass of another planet in the Solar System or the combined mass of two binary stars can be calculated in units of Solar mass directly from the orbital radius and orbital period of the planet or stars using Kepler's third law. Calculation [edit] The mass of the Sun cannot be measured directly, and is instead calculated from other measurable factors, using the equation for the orbital period of a small body orbiting a central mass. Based on the length of the year, the distance from Earth to the Sun (an astronomical unit or au), and the gravitational constant (G), the mass of the Sun is given by solving Kepler's third law: The value of G is difficult to measure and is only known with limited accuracy (see Cavendish experiment). The value of G times the mass of an object, called the standard gravitational parameter, is known for the Sun and several planets to a much higher accuracy than G alone. As a result, the solar mass is used as the standard mass in the astronomical system of units. Variation [edit] The Sun is losing mass because of fusion reactions occurring within its core, leading to the emission of electromagnetic energy and neutrinos, and by the ejection of matter with the solar wind. It is expelling about (2–3)×10−14 M☉/year. The mass loss rate will increase when the Sun enters the red giant stage, climbing to (7–9)×10−14 M☉/year when it reaches the tip of the red-giant branch. This will rise to 10−6 M☉/year on the asymptotic giant branch, before peaking at a rate of 10−5 to 10−4 M☉/year as the Sun generates a planetary nebula. By the time the Sun becomes a degenerate white dwarf, it will have lost 46% of its starting mass. The mass of the Sun has been decreasing since the time it formed. This occurs through two processes in nearly equal amounts. First, in the Sun's core, hydrogen is converted into helium through nuclear fusion, in particular the p–p chain, and this reaction converts some mass into energy in the form of gamma ray photons. Most of this energy eventually radiates away from the Sun. Second, high-energy protons and electrons in the atmosphere of the Sun are ejected directly into outer space as the solar wind and coronal mass ejections. The original mass of the Sun at the time it reached the main sequence remains uncertain. The early Sun had much higher mass-loss rates than at present, and it may have lost anywhere from 1–7% of its natal mass over the course of its main-sequence lifetime. Related units [edit] One solar mass, M☉, can be converted to related units: 27068510 M☾ (Lunar mass) 332946 M🜨 (Earth mass) 1047.35 MJ (Jupiter mass) It is also frequently useful in general relativity to express mass in units of length or time. M☉G/c2 ≈ 1.48 km (half the Schwarzschild radius of the Sun) M☉G/c3 ≈ 4.93 μs The solar mass parameter (GM☉), as listed by the IAU Division I Working Group, has the following estimates: 1.32712442099(10)×1020 m3s−2 (TCG-compatible) 1.32712440041(10)×1020 m3s−2 (TDB-compatible) See also [edit] Chandrasekhar limit Gaussian gravitational constant Orders of magnitude (mass) Stellar mass Sun References [edit] ^ a b Prša, Andrej; Harmanec, Petr; Torres, Guillermo; Mamajek, Eric; Asplund, Martin; Capitaine, Nicole; Christensen-Dalsgaard, Jørgen; Depagne, Éric; Haberreiter, Margit; Hekker, Saskia; Hilton, James; Kopp, Greg; Kostov, Veselin; Kurtz, Donald W.; Laskar, Jacques; Mason, Brian D.; Milone, Eugene F.; Montgomery, Michele; Richards, Mercedes; Schmutz, Werner; Schou, Jesper; Stewart, Susan G. (2016). "Nominal Values for Selected Solar and Planetary Quantities: IAU 2015 Resolution B3". The Astronomical Journal. 152 (2): 41. arXiv:1605.09788. Bibcode:2016AJ....152...41P. doi:10.3847/0004-6256/152/2/41. ^ Clarion, Geoffrey R. "Universal Gravitational Constant" (PDF). University of Tennessee Physics. PASCO. p. 13. Retrieved 11 April 2019. ^ Holton, Gerald James; Brush, Stephen G. (2001). Physics, the human adventure: from Copernicus to Einstein and beyond (3rd ed.). Rutgers University Press. p. 137. ISBN 978-0-8135-2908-0. ^ Pecker, Jean Claude; Kaufman, Susan (2001). Understanding the heavens: thirty centuries of astronomical ideas from ancient thinking to modern cosmology. Springer. p. 291. Bibcode:2001uhtc.book.....P. ISBN 978-3-540-63198-9. ^ Barbieri, Cesare (2007). Fundamentals of astronomy. CRC Press. pp. 132–140. ISBN 978-0-7503-0886-1. ^ "How do scientists measure or calculate the weight of a planet?". Scientific American. Retrieved 2020-09-01. ^ Cohen, I. Bernard (May 1998). "Newton's Determination of the Masses and Densities of the Sun, Jupiter, Saturn, and the Earth". Archive for History of Exact Sciences. 53 (1): 83–95. Bibcode:1998AHES...53...83C. doi:10.1007/s004070050022. JSTOR 41134054. S2CID 122869257. ^ Leverington, David (2003). Babylon to Voyager and beyond: a history of planetary astronomy. Cambridge University Press. p. 126. ISBN 978-0-521-80840-8. ^ "Finding the Mass of the Sun". imagine.gsfc.nasa.gov. Retrieved 2020-09-06. ^ Woo, Marcus (6 December 2018). "What Is Solar Mass?". Space.com. Retrieved 2020-09-06. ^ "Kepler's Third Law \| Imaging the Universe". astro.physics.uiowa.edu. Archived from the original on 2020-07-31. Retrieved 2020-09-06. ^ "CODATA Value: Newtonian constant of gravitation". physics.nist.gov. Retrieved 2020-09-06. ^ Carroll, Bradley W.; Ostlie, Dale A. (1995), An Introduction to Modern Astrophysics (revised 2nd ed.), Benjamin Cummings, p. 409, ISBN 0201547309. ^ Schröder, K.-P.; Connon Smith, Robert (2008), "Distant future of the Sun and Earth revisited", Monthly Notices of the Royal Astronomical Society, 386 (1): 155–163, arXiv:0801.4031, Bibcode:2008MNRAS.386..155S, doi:10.1111/j.1365-2966.2008.13022.x, S2CID 10073988 ^ Genova, Antonio; Mazarico, Erwan; Goossens, Sander; Lemoine, Frank G.; Neumann, Gregory A.; Smith, David E.; Zuber, Maria T. (18 January 2018). "Solar system expansion and strong equivalence principle as seen by the NASA MESSENGER mission". Nature Communications. 9 (1): 289. Bibcode:2018NatCo...9..289G. doi:10.1038/s41467-017-02558-1. ISSN 2041-1723. PMC 5773540. PMID 29348613. The fusion cycle that generates energy into the Sun relies on the conversion of hydrogen into helium, which is responsible for a solar mass reduction with a rate of ~ −0.67 × 10−13 per year. On the other hand, the solar wind contribution is more uncertain. The solar cycle significantly influences the solar mass loss rate due to solar wind. Estimates of the mass carried away with the solar wind showed rates between − (2–3) × 10−14M☉ per year, whereas numerical simulations of coupled corona and solar wind models provided rates between − (4.2–6.9) × 10−14 M☉ per year. ^ "Lecture 40: The Once and Future Sun". www.astronomy.ohio-state.edu. Retrieved 2020-09-01. ^ Sackmann, I.-Juliana; Boothroyd, Arnold I. (February 2003), "Our Sun. V. A Bright Young Sun Consistent with Helioseismology and Warm Temperatures on Ancient Earth and Mars", The Astrophysical Journal, 583 (2): 1024–1039, arXiv:astro-ph/0210128, Bibcode:2003ApJ...583.1024S, doi:10.1086/345408, S2CID 118904050 ^ "Planetary Fact Sheet". nssdc.gsfc.nasa.gov. Retrieved 2020-09-01. ^ "Astronomical Constants : Current Best Estimates (CBEs)". Numerical Standards for Fundamental Astronomy. IAU Division I Working Group. 2012. Retrieved 2021-05-04. Retrieved from " Categories: Sun Units of mass Units of measurement in astronomy Hidden categories: Articles with short description Short description is different from Wikidata
8225	https://en.wiktionary.org/wiki/mannequin	mannequin - Wiktionary, the free dictionary Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main Page Community portal Requested entries Recent changes Random entry Help Glossary Contact us Special pages Feedback If you have time, leave us a note. Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donations Preferences Create account Log in [x] Personal tools Donations Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide Beginning 1 EnglishToggle English subsection 1.1 Alternative forms 1.2 Etymology 1.3 Pronunciation 1.4 Noun 1.4.1 Derived terms 1.4.2 Translations 1.4.3 See also 1.5 References 1.6 Further reading 2 DanishToggle Danish subsection 2.1 Etymology 2.2 Pronunciation 2.3 Noun 2.3.1 Declension 2.4 Further reading 3 DutchToggle Dutch subsection 3.1 Etymology 3.2 Pronunciation 3.3 Noun 3.3.1 Hypernyms 3.3.2 Related terms 3.3.3 Descendants 4 FrenchToggle French subsection 4.1 Etymology 4.2 Pronunciation 4.3 Noun 4.3.1 Descendants 4.4 Further reading 5 Lower SorbianToggle Lower Sorbian subsection 5.1 Etymology 5.2 Noun 5.2.1 Declension 5.3 References mannequin [x] 27 languages Afrikaans বাংলা Eesti Ελληνικά Esperanto Français 한국어 Հայերեն Ido Italiano ಕನ್ನಡ Kurdî Magyar Malagasy മലയാളം မြန်မာဘာသာ Nederlands 日本語 Oromoo Polski Русский Simple English Suomi Svenska தமிழ் Tiếng Việt 中文 Entry Discussion Citations [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Create a book Download as PDF Printable version In other projects Visibility Show translations Show other boxes Show inflection Hide synonyms Show quotations From Wiktionary, the free dictionary See also:Mannequin English [edit] mannequin (sense 1) Alternative forms [edit] manikin, mannikin, manequin Etymology [edit] Borrowed from Frenchmannequin, derived from Old FrenchTerm?, derived from Middle Dutchmannekijn(“little man”) (see English manikin), diminutive of man(“man”). By surface analysis, man +‎ -kin. Compare ramequin/ramekin. Doublet of manakin and manikin. Pronunciation [edit] enPR: măn′ĭ-kĭn (contemporary Received Pronunciation,Scotland,India)IPA(key): /ˈman.ɪ.kɪn/ (conservative Received Pronunciation,General American,Canada,General Australian)IPA(key): /ˈmæn.ɪ.kɪn/ (New Zealand)IPA(key): /ˈmɛn.ə.kən/ Audio (Southern England):Duration: 2 seconds.0:02(file) This entry needs an audio pronunciation. If you are a native speaker with a microphone, please record this word. The recorded pronunciation will appear here when it's ready. Particularly: "NZE" Rhymes: -ænɪkɪn Homophones: manakin, manikin Hyphenation: man‧ne‧quin Noun [edit] mannequin (pluralmannequins) A dummy, or life-sizemodel of the humanbody, used for the fitting or displaying of clothes. quotations▼ 2011 April 13, Eric Wilson, “A Mannequin in Every Sense”, in The New York Times‎, →ISSN:At the same time, Mr. Bolton was intrigued by Mr. Sundsbo’s proposal to make models look like mannequins because it spoke to the blurring of boundaries — between good and evil, angels and demons, nature and technology, permanence and decay — that was a consistent theme of the McQueen collections. 2013 August 16, Barbara Brownie, “Mangled mannequins – what happened to shop-window dummies?”, in The Guardian‎:There was a time when a mannequin was the sculptural equivalent of fashion model. Like a fashion model, the mannequin was intended to reflect our social, professional and aesthetic aspirations. 2020 May 14, Lauren Aratani, “Model companions: restaurant solves social distancing dilemma with mannequins”, in The Guardian‎:With the idea that guests may feel lonely eating at a restaurant that can only reach 50% capacity[…]the restaurant will be placing mannequins throughout its dining rooms. A jointed model of the human body used by artists, especially to demonstrate the arrangement of drapery. An anatomicalmodel of the human body for use in teaching of e.g. CPR. (dated) A person who models clothes. synonym▲quotations▼Synonym:fashion model 1951, Denise Robins, Heart of Paris, Ulverscroft, →ISBN, page 14:[…]all of which made a perfect neutral background for the mannequins who moved around swiftly, gracefully, exhibiting one glorious creation after another. Derived terms [edit] mannequinlike Translations [edit] show ▼±model of the human body used for the displaying of clothes [Select preferred languages] [Clear all] Albanian: manekin(sq)m Arabic: مِثَالَةf(miṯāla), مَانِيكَانm(mānīkān), مَانُوكَانm(mānūkān) Armenian: մանեկեն(hy)(maneken) Azerbaijani: maneken Belarusian: манеке́нm(manjekjén) Bulgarian: манеке́н(bg)m(manekén), манеке́нкаf(manekénka) Burmese: ဟန်ပြမယ်(my)(hanpra.mai) Catalan: maniquí(ca)m Chinese: Mandarin: 人體模型/ 人体模型(réntǐ móxíng) Czech: figurína(cs)f, krejčovská pannaf, manekýnm, manekýnka(cs)f Danish: mannequinc Dutch: mannequin(nl), paspop(nl)m or f Esperanto: manekeno Estonian: mannekeen Finnish: mallinukke French: mannequin(fr)m Galician: manequín(gl)m Georgian: მანეკენი(maneḳeni) German: Schaufensterpuppe(de)f, Schaufensterfigurf, Gliederpuppe(de)f, Modellpuppef Greek: ανδρείκελο(el)n(andreíkelo) Hindi: पुतला(hi)m(putlā) Hungarian: próbababa(hu) Icelandic: gínaf Ido: manekino(io) Indonesian: maneken(id) Irish: mainicínm Italian: manichino(it) Japanese: マネキン人形(ja)(マネキンにんぎょう, manekin ningyō) Khmer: ម៉ាន់នឺកាំង(mannɨɨkang) Korean: 마네킹(ko)(maneking) Lao: ນາງແບບ(nāng bǣp), ຫຸ່ນ(hun) Latvian: manekensm Lithuanian: manekenas(lt) Macedonian: и́зложна ку́клаf(ízložna kúkla), манеке́нm(manekén), манеке́нкаf(manekénka) Malay: boneka pakaian Maori: ropi pū(w)eru Mongolian: манекен(mn)(maneken) Norwegian: Bokmål: mannekengm Persian: مانکن(fa)(mânkan) Polish: manekin(pl)m Portuguese: manequim(pt)m Romanian: manechin(ro)n Russian: манеке́н(ru)m(manekén) Slovak: manekýnm, manekýnkaf Spanish: maniquí(es)m Swedish: skyltdocka(sv)c, mannekäng(sv)c Tagalog: maoy Tamil: (please verify)பொம்மை(pommai) Thai: นางแบบ(th)(naang-bɛ̀ɛp), ตุ๊กตา(th)(dtúk-gà-dtaa) Tibetan: མིའི་གསོབ(mi'i gsob) Turkish: manken(tr) Ukrainian: манеке́нm(manekén) Uzbek: maneken(uz) Vietnamese: manocanh, ma-nơ-canh Add translation: More [x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective Noun class: Plural class: Transliteration: (e.g. zìmǔ for 字母) Literal translation: Raw page name: (e.g. 疲れる for 疲れた) Qualifier: (e.g. literally, formally, slang) Script code: (e.g. Cyrl for Cyrillic, Latn for Latin) Nesting: (e.g. Serbo-Croatian/Cyrillic) show ▼±model of human body used by artists [Select preferred languages] [Clear all] Bulgarian: манекен(bg)m(maneken) Catalan: maniquí(ca)m Czech: figurína(cs)f Esperanto: manekeno Finnish: mallinukke French: mannequin(fr)m Galician: manequín(gl)m Georgian: მანეკენი(maneḳeni) German: lebensgroßePuppef, Puppenfigurf Greek: ανδρείκελο(el)n(andreíkelo) Ido: manekino(io) Indonesian: maneken(id) Irish: mainicínm Macedonian: манеке́нm(manekén) Russian: манеке́н(ru)m(manekén) Spanish: maniquí(es)m Add translation: More [x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective Noun class: Plural class: Transliteration: (e.g. zìmǔ for 字母) Literal translation: Raw page name: (e.g. 疲れる for 疲れた) Qualifier: (e.g. literally, formally, slang) Script code: (e.g. Cyrl for Cyrillic, Latn for Latin) Nesting: (e.g. Serbo-Croatian/Cyrillic) show ▼±model of human body used for teaching [Select preferred languages] [Clear all] Czech: figurína(cs)f Finnish: mallinukke French: mannequin(fr)m Galician: manequín(gl)m Greek: ανδρείκελο(el)n(andreíkelo) Indonesian: maneken(id) Irish: mainicínm Add translation: More [x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective Noun class: Plural class: Transliteration: (e.g. zìmǔ for 字母) Literal translation: Raw page name: (e.g. 疲れる for 疲れた) Qualifier: (e.g. literally, formally, slang) Script code: (e.g. Cyrl for Cyrillic, Latn for Latin) Nesting: (e.g. Serbo-Croatian/Cyrillic) show ▼±a person who models clothes [Select preferred languages] [Clear all] Chinese: Mandarin: 模特(zh)(mótè) Czech: manekýnm Finnish: malli(fi), mannekiini(fi) French: mannequin(fr)m Galician: manequín(gl)m German: Model(de)n, Mannequin(de)n Greek: μοντέλο(el)n(montélo) Hungarian: manöken(hu) Indonesian: peragawati(id), maneken(id) Irish: mainicínm Japanese: モデル(ja)(moderu) Macedonian: манеке́нm(manekén), манеке́нкаf(manekénka) Romanian: manechin(ro)n Russian: манеке́нщик(ru)m(manekénščik), манеке́нщица(ru)f(manekénščica), моде́ль(ru)m or f(modélʹ) Swedish: fotomodellc, mannekäng(sv)c Add translation: More [x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective Noun class: Plural class: Transliteration: (e.g. zìmǔ for 字母) Literal translation: Raw page name: (e.g. 疲れる for 疲れた) Qualifier: (e.g. literally, formally, slang) Script code: (e.g. Cyrl for Cyrillic, Latn for Latin) Nesting: (e.g. Serbo-Croatian/Cyrillic) See also [edit] little man References [edit] ↑ Jump up to: 1.01.11.2“mannequin”, in The American Heritage Dictionary of the English Language, 5th edition, Boston, Mass.: Houghton Mifflin Harcourt, 2016, →ISBN. Further reading [edit] mannequin on Wikipedia.Wikipedia “mannequin”, in OneLook Dictionary Search. Danish [edit] Etymology [edit] From Frenchmannequin. Pronunciation [edit] IPA(key): /manəˈkɛŋ/, [mænəˈkʰeŋ], [mæn̩ˈkʰeŋ] Noun [edit] mannequinc (singular definitemannequinen, plural indefinitemannequiner) mannequin Declension [edit] show ▼Declension of mannequin\| common gender \| singular \| plural \| --- \| indefinite \| definite \| indefinite \| definite \| \| nominative \| mannequin \| mannequinen \| mannequiner \| mannequinerne \| \| genitive \| mannequins \| mannequinens \| mannequiners \| mannequinernes \| Further reading [edit] “mannequin” in Den Danske Ordbog Dutch [edit] Dutch Wikipedia has an article on: mannequin Wikipedia nl Mannequins – models. Etymology [edit] Borrowed from Frenchmannequin, from Middle Dutchmanneken. Pronunciation [edit] IPA(key): /ˌmɑ.nəˈkɛːn/ Audio:Duration: 2 seconds.0:02(file) Hyphenation: man‧ne‧quin Noun [edit] mannequinm (pluralmannequins) a clothes model or fashion model, a mannequin Hypernyms [edit] model Related terms [edit] man Descendants [edit] → Papiamentu: mannequin, manikin French [edit] FWOTD – 22 September 2018 Etymology [edit] Borrowed from Middle Dutchmanneken, mannkijn. Pronunciation [edit] IPA(key): /man.kɛ̃/ Audio (France (Somain)):Duration: 2 seconds.0:02(file) Noun [edit] mannequinm (pluralmannequins) fashion modelquotations▼ 2016 September 9, “Un mannequin défiguré à l’acide défile à la Fashion Week de New York”, in Le Monde:Elle était conviée à défiler pour le collectif de jeunes créateurs italiens FTL Moda, qui a régulièrement fait parler de lui ces dernières saisons, plus pour ses choix de mannequins que pour ses vêtements.She was invited to model for the Italian young designers' collective FTL Moda, that had regularly made itself the talk of the town during previous seasons, more for its choice of models than for its clothes. dummy, mannequinquotations▼ 2016 August 18, Matteo Maillard, “Être mère et prostituée au Mali”, in Le Monde:C’est pourquoi elle a garni le mur de sa chambre turquoise de perruques qu’elle a fabriquées sur une tête de mannequin, les nuits sans clients comme celle-ci.It is why she has decorated the wall of her turquoise room with wigs that she made on a mannequin's head on nights without clients, like this one. Descendants [edit] show ▼Descendants → Catalan: maniquí → Czech: manekýn → Danish: mannequin → Dutch: mannequin → English: mannequin → Japanese: マネキン(manekin) → Korean: 마네킹(maneking) → Georgian: მანეკენი(maneḳeni) → German: Mannequin → Greek: μανεκέν(manekén) → Norwegian: Norwegian Bokmål: mannekeng Norwegian Nynorsk: mannekeng → Hungarian: manöken → Japanese: マヌカン(manukan) → Polish: manekin → Portuguese: manequim → Romanian: manechin → Russian: манекен(maneken) → Swedish: mannekäng → Spanish: maniquí → Turkish: manken → Persian: مانکن(mânkan) → Vietnamese: ma-nơ-canh Further reading [edit] “mannequin”, in Trésor de la langue française informatisé[Digitized Treasury of the French Language], 2012. Lower Sorbian [edit] Etymology [edit] Borrowed from Frenchmannequin. Noun [edit] mannequinf (diminutivemannequinka) mannequin Declension [edit] Indeclinable. References [edit] Starosta, Manfred (1999), “mannequin”, in Dolnoserbsko-nimski słownik/ Niedersorbisch-deutsches Wörterbuch (in German), Bautzen: Domowina-Verlag Retrieved from " Categories: English terms borrowed from French English terms derived from French English terms derived from Old French English terms derived from Middle Dutch English terms suffixed with -kin English doublets English 3-syllable words English terms with IPA pronunciation English terms with audio pronunciation Rhymes:English/ænɪkɪn Rhymes:English/ænɪkɪn/3 syllables English terms with homophones English lemmas English nouns English countable nouns English terms with quotations English dated terms Danish terms borrowed from French Danish terms derived from French Danish terms with IPA pronunciation Danish lemmas Danish nouns Danish terms spelled with Q Danish common-gender nouns Dutch terms borrowed from French Dutch terms derived from French Dutch terms derived from Middle Dutch Dutch terms with IPA pronunciation Dutch terms with audio pronunciation Dutch lemmas Dutch nouns Dutch nouns with plural in -s Dutch masculine nouns nl:Occupations French terms borrowed from Middle Dutch French terms derived from Middle Dutch French 2-syllable words French terms with IPA pronunciation French terms with audio pronunciation French lemmas French nouns French countable nouns French masculine nouns French terms with quotations French terms suffixed with -quin fr:Occupations Lower Sorbian terms borrowed from French Lower Sorbian terms derived from French Lower Sorbian lemmas Lower Sorbian nouns Lower Sorbian terms spelled with Q Lower Sorbian feminine nouns Hidden categories: Old French term requests Requests for audio pronunciation in English entries Pages with entries Pages with 5 entries Entries with translation boxes Terms with Albanian translations Terms with Arabic translations Terms with Armenian translations Terms with Azerbaijani translations Terms with Belarusian translations Terms with Bulgarian translations Terms with Burmese translations Terms with Catalan translations Terms with Mandarin translations Terms with Czech translations Terms with Danish translations Terms with Dutch translations Terms with Esperanto translations Terms with Estonian translations Terms with Finnish translations Terms with French translations Terms with Galician translations Terms with Georgian translations Terms with German translations Terms with Greek translations Terms with Hindi translations Terms with Hungarian translations Terms with Icelandic translations Terms with Ido translations Terms with Indonesian translations Terms with Irish translations Terms with Italian translations Terms with Japanese translations Terms with Khmer translations Terms with Korean translations Terms with Lao translations Terms with Latvian translations Terms with Lithuanian translations Terms with Macedonian translations Terms with Malay translations Terms with Maori translations Terms with Mongolian translations Terms with Norwegian Bokmål translations Terms with Persian translations Terms with Polish translations Terms with Portuguese translations Terms with Romanian translations Terms with Russian translations Terms with Slovak translations Terms with Spanish translations Terms with Swedish translations Terms with Tagalog translations Terms with Tamil translations Requests for review of Tamil translations Terms with Thai translations Terms with Tibetan translations Terms with Turkish translations Terms with Ukrainian translations Terms with Uzbek translations Terms with Vietnamese translations Foreign words of the day in French Foreign word of the day archive/2018 Foreign words of the day in French/2018 Foreign word of the day archive Foreign word of the day archive/2018/September This page was last edited on 22 September 2025, at 17:50. 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8226	https://www.oxfordlearnersdictionaries.com/us/definition/english/salvage_2	Definition of salvage verb from the Oxford Advanced Learner's Dictionary salvage \| \| \| --- \| \| present simple I / you / we / they salvage \| /ˈsælvɪdʒ/ /ˈsælvɪdʒ/ \| \| he / she / it salvages \| /ˈsælvɪdʒɪz/ /ˈsælvɪdʒɪz/ \| \| past simple salvaged \| /ˈsælvɪdʒd/ /ˈsælvɪdʒd/ \| \| past participle salvaged \| /ˈsælvɪdʒd/ /ˈsælvɪdʒd/ \| \| -ing form salvaging \| /ˈsælvɪdʒɪŋ/ /ˈsælvɪdʒɪŋ/ \| Join us Join our community to access the latest language learning and assessment tips from Oxford University Press! Other results Nearby words Oxford Learner's Dictionaries More from us Who we are Oxford University Press is a department of the University of Oxford. It furthers the University's objective of excellence in research, scholarship, and education by publishing worldwide
8227	https://baike.baidu.com/item/%E5%88%86%E5%BC%8F/5688762	分式_百度百科网页新闻贴吧知道网盘图片视频地图文库资讯采购百科百度首页登录注册进入词条全站搜索帮助进入词条全站搜索帮助播报编辑讨论 12收藏赞登录近期有不法分子冒充百度百科官方人员，以删除词条为由威胁并敲诈相关企业。在此严正声明：百度百科是免费编辑平台，绝不存在收费代编服务，请勿上当受骗！详情>> 首页历史上的今天百科冷知识图解百科秒懂百科懂啦秒懂本尊答秒懂大师说秒懂看瓦特秒懂五千年秒懂全视界特色百科数字博物馆非遗百科恐龙百科多肉百科艺术百科科学百科知识专题观千年·见今朝中国航天古鱼崛起食品百科数字文物守护计划史记2024·科学100词加入百科新人成长进阶成长任务广场百科团队校园团分类达人团热词团繁星团蝌蚪团权威合作合作模式常见问题联系方式个人中心分式 [fēn shì] 播报锁定讨论 12上传视频数学名词分式：数学界的变形金刚 00:50 分式的定义 01:57 【八年级数学上册】11 什么是分式 06:18 八年级数学分式解析：概念、运算与方程解法全解读 10:29 分式的基本概念与判断方法，如何区分分式与非分式 01:52 分式知识点 02:38 分式是什么？分数背后的秘密，全解析！ 06:13 代数基础09分式 96:02 初中数学知识点：分式 11:21 八年级数学上册：什么是分式？ 02:26 全260节【重点初中数学拔尖】名师精讲初中数学-七年级重点中学名师数学系统课中考数学重点高中之路 17:21 02 分式的概念 04:38 初二数学分式 02:30 分式的基本定义与常见题型解析 10:46 分式的概念是什么 01:29 收藏查看我的收藏 7952 有用+1 376 本词条由“科普中国”科学百科词条编写与应用工作项目审核。一般地，如果A、B表示两个整式，且B中含有字母，那么式子A / B 就叫做分式，其中A称为分子，B称为分母。分式是不同于整式的一类代数式，分式的值随分式中字母取值的变化而变化。分式有意义条件是分母不为0。中文名分式外文名 fraction 学科数学所属范围有理式类型代数式目录 1概念 ▪定义 ▪分式条件 ▪代数式分类 2基本性质 3运算法则 ▪约分 ▪公因式的提取方法 ▪最简分式 ▪乘法 ▪除法 ▪乘方 ▪开方概念播报定义形如（A、B是整式，B中含有字母）的式子叫做分式。其中A叫做分式的分子（或者说被除数），B叫做分式的分母（或者说除数）。当分式的分子的次数低于分母的次数时，这个分式叫做真分式；当分式的分子的次数等于或高于分母的次数时，这个分式叫做假分式。注意：判断一个式子是否是分式，不要看式子是否是的形式，关键要满足：分式的分母（或者说除数）中必须含有字母，分子分母均为整式。无需考虑该分式是否有意义，即分母是否为零。由于字母可以表示不同的数，所以分式比分数更具有一般性。方法：数看结果，式看形。分式条件 1.分式有意义条件：分母（或者说除数）不为0。 2.分式值为0条件：分子（或者说被除数）为0且分母不为0。 3.分式值为正(负)数条件：分子分母同号得正，异号得负。 4.分式值为1的条件：分子=分母≠0。 5.分式值为-1的条件：分子分母互为相反数，且都不为0。代数式分类整式和分式统称为有理式。带有根号且根号下含有字母的式子叫做无理式。无理式和有理式统称代数式。基本性质播报分式的分子和分母同时乘以（或除以）同一个不为0的整式，分式的值不变。用式子表示为：其中A,B,C为整式，且B、C≠0。运算法则播报约分根据分式基本性质，可以把一个分式的分子和分母的公因式约去，这种变形称为分式的约分。约分的关键是确定分式中分子与分母的公因式。步骤： 1.如果分式的分子和分母都是单项式或者是几个因式乘积的形式，将它们的公因式约去。 2.分式的分子和分母都是多项式，将分子和分母分别分解因式，再将公因式约去。公因式的提取方法系数取分子和分母系数的最大公约数，字母取分子和分母共有的字母，指数取公共字母的最小指数，即为它们的公因式。最简分式一个分式不能约分时，这个分式称为最简分式。约分时，一般将一个分式化为最简分式。乘法同分母分式的加减法法则进行计算。用字母表示为：乘法两个分式相乘，把分子相乘的积作为积的分子，把分母相乘的积作为积的分母。用字母表示为：。除法两个分式相除，把除式的分子和分母颠倒位置后再与被除式相乘：。也可表述为：除以一个分式，等于乘以这个分式的倒数。乘方分子乘方做分子，分母乘方做分母，可以约分的约分，最后化成最简：。开方分子开方做分子，分母开方做分母，可以约分的约分，最后化成最简：。词条图册更多图册分式(1张) 1/1 分享你的世界查看更多 8 这个算不算分式？ MANG0000 参考资料 1 何炳均. 分式学习中的重点和难点[J]. 中学生数理化(八年级数学)(配合人教社教材),2012,(Z1):8-10. 分式的概述图（1张）科普中国致力于权威的科学传播本词条认证专家为王海侠副教授审核南京理工大学权威合作编辑 “科普中国”科学百科词条编写与应用工作项目 “科普中国”是为我国科普信息化建设塑造的全... 什么是权威编辑词条统计浏览次数：2111274次编辑次数：191次历史版本最近更新： 9s_自由鸟（2025-05-29）突出贡献榜窒息的月亮 1 概念定义分式条件代数式分类2 基本性质3 运算法则约分公因式的提取方法最简分式乘法除法乘方开方相关搜索初二下册数学分式数学高中公式数学不等式 matlab培训高中数学必修一知识点归纳数列题型及解题方法初二数学知识点归纳奥数题初中数学公式总结归纳分式选择朗读音色成熟女声成熟男声磁性男声年轻女声情感男声 0 0 2x 1.5x 1.25x 1x 0.75x 0.5x 分享到微信朋友圈打开微信“扫一扫”即可将网页分享至朋友圈新手上路成长任务编辑入门编辑规则本人编辑我有疑问内容质疑在线客服官方贴吧意见反馈投诉建议举报不良信息未通过词条申诉投诉侵权信息封禁查询与解封 ©2025 Baidu使用百度前必读\|百科协议\|隐私政策\|百度百科合作平台\|京ICP证030173号京公网安备11000002000001号
8228	https://zhuanlan.zhihu.com/p/402340744	圆的典型问题与方法 - 知乎首页知乎直答焕新知乎知学堂等你来答切换模式登录/注册圆的典型问题与方法首发于高中数学典型问题切换模式登录/注册圆的典型问题与方法究尽数学哈尔滨工业大学应用数学硕士收录于 · 高中数学典型问题 13 人赞同了该文章圆的方程圆的标准方程：(x−a)2+(y−b)2=r 2(x-a)^2+(y-b)^2=r^2 确定圆心和半径，使用两点的距离公式可以推导出；某一点到圆心的距离等于半径。圆的一般方程：x 2+y 2+D x+E y+F=0 x^2+y^2+Dx+Ey+F=0 圆的标准方程展开得到一般式，也可通过一般式配方得到 {a=−D 2 b=−E 2 r 2=D 2 4+E 2 4−F\begin{cases} a=-\frac{D}{2}\ b=-\frac{E}{2}\ r^2=\frac{D^2}{4}+\frac{E^2}{4}-F \end{cases}\ 圆的参数方程 {x=a+r⋅cos⁡θ y=b+r⋅sin⁡θ\begin{cases} x=a+r\cdot\cos\theta\ y=b+r\cdot\sin\theta\ \end{cases}\ 整理圆的标准方程为 (x−a)2 r 2+(y−b)2 r 2=1\frac{(x-a)^2}{r^2}+\frac{(y-b)^2}{r^2}=1\ 于是，可设 {x−a r=cos⁡θ y−b r=sin⁡θ\begin{cases} \frac{x-a}{r}=\cos\theta\ \frac{y-b}{r}=\sin\theta\ \end{cases}\ 已知圆直径的两个端点 A(x 1,y 1),B(x 2,y 2)A(x_1,y_1),B(x_2,y_2)，则圆方程(x−x 1)(x−x 2)+(y−y 1)(y−y 2)=0(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\ 分析：设以AB为直径的圆上，不同于A、B的一点 C(x,y),∠A C B=π 2 C(x,y),\angle{ACB}=\frac{\pi}{2}，那么 A C→=(x−x 1,y−y 1)B C→=(x−x 2,y−y 2)A C→⋅B C→=(x−x 1)(x−x 2)+(y−y 1)(y−y 2)=0\begin{aligned} \overrightarrow{AC}&=(x-x_1,y-y_1)\ \overrightarrow{BC}&=(x-x_2,y-y_2)\ \overrightarrow{AC}\cdot\overrightarrow{BC}&=(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0 \end{aligned}\ 明显的，C为A或B时，仍然满足上面的式子。直线与圆的关系代数法：联立直线方程和圆方程，消去x(或y),得到关于y(或x)的一元二次方程，利用其判别公式判别式>0，直线与圆相交；判别式=0，直线与圆相切；判别式<0，直线与圆没有交点。几何法：确定圆的圆心，求出圆心到直线的距离距离>圆半径，直线与圆相离；距离=圆半径，直线与圆相切；距离<圆半径，直线与圆相交。过圆外一点 P(x_0,y_0)引圆的切线，则点P至切点的距离为：\begin{aligned} d&=\sqrt{x_0^2+y_0^2+Dx_0+Ey_0+F}\ &=\sqrt{(x_0-a)^2+(y_0-b)^2-r^2} \end{aligned}\ 分析：圆外一点与圆心、切点之间连线构成一个直角三角形(可作图辅助，你会一目了然的)。使用上述圆的标准方程和一般方程： \begin{aligned} d^2&=(x_0-a)^2+(y_0-b)^2-r^2\ &=x_0^2+y_0^2+Dx_0+Ey_0+F \end{aligned}\ 圆的切线过圆 x^2+y^2+Dx+Ey+F=0 上一点 P(x_0,y_0)，切线为 x_0x+y_0y+\frac{D}{2}(x+x_0)+\frac{E}{2}(y+y_0)+F=0\ 分析：设圆心为O、切线上任意点Q(x,y)，则 \begin{cases} O=(-\frac{D}{2},-\frac{E}{2})\ \therefore\overrightarrow{OP}=(x_0+\frac{D}{2},y_0+\frac{E}{2})\ \therefore\overrightarrow{PQ}=(x-x_0,y-y_0)\ \because\overrightarrow{OP}\cdot\overrightarrow{PQ}=0 \end{cases}\ 所以 \begin{aligned} \overrightarrow{OP}\cdot\overrightarrow{PQ}&=(x-x_0)(x_0+\frac{D}{2})+(y-y_0)(y_0+\frac{E}{2})\ &=x_0x+\frac{D}{2}x-x_0^2-\frac{D}{2}x_0+y_0y+\frac{E}{2}y-y_0^2-\frac{E}{2}y_0\ &=x_0x+y_0y+\frac{D}{2}(x+x_0)+\frac{E}{2}(y+y_0)-(x_0^2+y_0^2+Dx_0+Ey_0)\ &=x_0x+y_0y+\frac{D}{2}(x+x_0)+\frac{E}{2}(y+y_0)+F\ &=0 \end{aligned}\ 过圆(x-a)^2+(y-b)^2=r^2 上一点 P(x_0,y_0)，切线为(x-a)(x_0-a)+(y-b)(y_0-b)=r^2\ 分析：同样使用上面的方法，可以得到切线方程。编辑于 2021-08-22 06:45 高中数学圆（平面图形）解析几何赞同 13添加评论分享喜欢收藏申请转载写下你的评论... 还没有评论，发表第一个评论吧关于作者究尽数学专注于数学思想、思维、方法、知识的分享哈尔滨工业大学应用数学硕士回答 128文章 129关注者 62,220 关注他发私信推荐阅读简单推导圆的方程 ======== Zachariah 教你演示圆的概念、性质、定理、与圆有关的位置关系——GeoGebra ================================== 啊K数学发表于GeoGe... 一般情况下的阿波罗尼斯圆的方程的推导过程 ==================== DanielVon两个同样大小的圆,一个不动,另一个绕这个圆旋转 ======================= 这是一道数学题：两个同样大小的圆,一个不动,另一个绕这个圆旋转,问它绕圆转了1圈之后自转了多少圈？实验结果为两圈。很多得到这个结果的都以为自己的数学白学了，如图先给大家上一幅图… 三道回想来知乎工作？请发送邮件到 jobs@zhihu.com 打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码
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8230	https://math.stackexchange.com/questions/4665991/confusion-regarding-independent-events	probability - Confusion regarding independent events - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Confusion regarding independent events Ask Question Asked 2 years, 6 months ago Modified2 years, 6 months ago Viewed 200 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Two unbiased dice are rolled. Let A be the event that sum is 6 6 and B B be the event that sum is 9 9. Are these events independent? Intuitively it seems that these events are independent. But, P(A)=5/36 P(A)=5/36 and P(B)=4/36 P(B)=4/36 and A∩B=ϕ A∩B=ϕ so P(A∩B)=P(A)P(B)P(A∩B)=P(A)P(B) is not satisfied. What am i missing here? probability probability-theory Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Mar 25, 2023 at 7:01 Math LoverMath Lover asked Mar 25, 2023 at 6:46 Math LoverMath Lover 3,852 1 1 gold badge 11 11 silver badges 26 26 bronze badges 3 1 Before reading the answers below, take your time to try articulating (to yourself or, even better, editing the question) what independence of events means; then, comparing your intuition against the 'correct' answers will be meaningful and instructive; I suggest this because this concept may be less straightforward than it cursorily seems. Also, take a look at points 1 & 3 of my explanation here: Common mistakes in probability.ryang –ryang 2023-03-25 09:27:19 +00:00 Commented Mar 25, 2023 at 9:27 1 Intuitively to me it seems that these events are not independent. And then you give an excellent reason for why they are not independent.Henry –Henry 2023-03-25 10:28:12 +00:00 Commented Mar 25, 2023 at 10:28 Does this answer your question? How can I intuitively see independent events?ryang –ryang 2023-03-25 16:58:51 +00:00 Commented Mar 25, 2023 at 16:58 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. The events {S u m=6}{S u m=6} and {S u m=9}{S u m=9} are disjoint (can't happen both in the same experiment), and information that one event occurred in the experiment, gives information that the other did not occur, so intuitively, the events are dependent. I guess you confuse disjoint and independent. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Mar 25, 2023 at 8:43 kludgkludg 2,667 1 1 gold badge 17 17 silver badges 16 16 bronze badges 1 +1 See stats.stackexchange.com/a/380791/2958 for an illustration of the distinction Henry –Henry 2023-03-25 10:30:11 +00:00 Commented Mar 25, 2023 at 10:30 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Why does your intuition suggest that they're independent? To put it more simply, if the sum of two dice is 6 6, then can it ever be 9 9?. So do you see the dependence now? If A A and B B were independent then A A and B c B c would also be independent. But here A⊂B c A⊂B c . Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Mar 25, 2023 at 7:36 Mr. Gandalf SauronMr. Gandalf Sauron 17.5k 2 2 gold badges 12 12 silver badges 34 34 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. A and B are dependent, because (as the question should be read) they are about the same dice roll. In fact they are exclusive events (the sum can't be both 6 and 9), which if the events can occur at all implies dependent (for the reason stated in the question). What's independent is the outcome of distinct dice rolls. Like: the event that in a first throw of two dices they sum to 6 is independent of the event than in another throw they sum to 9. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Mar 25, 2023 at 9:40 answered Mar 25, 2023 at 7:42 fgrieufgrieu 1,860 14 14 silver badges 31 31 bronze badges 3 1 "A and B are dependent, because they are about the same dice roll." Events associated with the same dice roll need not be singular: for example, in a single roll of a fair dice, the event of obtaining 3 or 6, and the event of obtaining 2 or 4 or 6, are independent of each other.ryang –ryang 2023-03-25 09:39:27 +00:00 Commented Mar 25, 2023 at 9:39 @ryang: yes. My "if" was a shortcut for "if we read the question as about the same dice roll". I didn't mean that events about the same dice roll are necessarily dependent, which would have been wrong (example: dice roll is even, dice roll is less than 3).fgrieu –fgrieu 2023-03-25 09:43:06 +00:00 Commented Mar 25, 2023 at 9:43 I'm not understanding; the OP's two events are indeed regarding rolling one pair of dice.ryang –ryang 2023-03-25 10:07:17 +00:00 Commented Mar 25, 2023 at 10:07 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions probability probability-theory See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 2How can I intuitively see independent events? 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8231	https://ccrma.stanford.edu/courses/220a/static/cook/PerryRCook_2015_Chapter3DigitalFilter_RealSoundSynthesisFor.pdf	Digital Filters 3.0 Introduction This chapter introduces digital filtering, stressing intuition along with the terminology and notation used to describe digital filters. First, filtering will be discussed in general, f ollowed by definitions and examples of Finite Im pulse Res ponse (FIR) and In finite Impulse Res ponse (IIR) filters. Then, the general f orm of a digital filter will be given and discussed. The Z transform will be introduced as a simple algebraic substitution operator, and this will be exploited to define and develop the notion ofthe transfer fimction. Zeroes and poles will be defined, and some usef ul filter f orms will be shown. Math averse readers could possibly skip this chapter for now, returning to it when needed in later chapters, but 1'd recommend reading at least until your eyes glaze over. 3.1 Linear Systems, LTI Systems, Convolution Linearity is a property of systems that allows us to use many powerf ul mathematical and signal processing techniques to analyze and predict the behavior ofthese systems. Linearity has two defining criteria: 21 Copyright 2015. A K Peters/CRC Press. All rights reserved. May not be reproduced in any form without permission from the publisher, except fair uses permitted under U.S. or applicable copyright law. EBSCO Publishing : eBook Collection (EBSCOhost) - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES AN: 1087771 ; Perry R. Cook.; Real Sound Synthesis for Interactive Applications Account: s4392798.main.ehost 22 3. Digital Filters Homogeneity: if x -? y Superposition: if x 1 -? Y 1 and x2 -? y2 then ax -? f or any a then xl + x2 -? yl + y2 (-? is read ''yields'', and corresponds to a system operating on x to yieldy). These equations state that a mixture and/or scaling of inputs simply results in a mixture and/or scaling of the outputs. No "new" signals are created by a linear system (we'll have more rigorous means to define "new" later). A time-invariant system obeys the property: If x(n) -? yen) then x(n + N) -? yen + N) for any N, which simply means that the system doesn't change its behavior with time. Here, x(n) is the chain of samples put into the system, and yen) is the corresponding chain of output samples. Practically speaking, most systems actually do change over time. A reasonable assumption, however, is that many systems of interest do not change their behavior quickly, and can be treated as time-invariant over time intervals of interest. The bones of the middle ear, acoustics in rooms, a good quality stereo amplifier, and many other systems behave like Linear Time-Invariant (LTI) systems over much of their normal operating range. If a system is linear and time-invariant, we can characterize its behavior by measuring its impulse response as shown in Figure 3 . 1 . The impulse response is defined mathematically as: hen) = yen), f or x(n) = D(n) where D(n) = 1 , n = 0, 0, otherwise. Linearity and time invariance mean that if we excite a system with an input of 1 at time zero, and ° thereafter, we can "record" (observe) the output and use that to determine exactly what the system response would be to any arbitrary input. To prove to ourselves that this is true, all we need do is invoke x(n)= H o (n) ----=:J L TI ! . --I system Impulse h(n) )=t Impulse response Figure 3. 1 . Impulse response of a Linear Time-Invariant (LTI) system. EBSCOhost - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES. All use subject to 3.1 . Linear Systems, LTI Systems, Convolution • the three properties ofL TI systems: homogeneity, superposition, and time invariance. Thus, any input can be decomposed into a time-ordered set of weighted impulses: Each input sample can be viewed as a separate weighted (xo' xl' etc., are the weights) impulsive input, and the outputs can be viewed as individual outputs, which are weighted versions of the impulse response hen): yen) = xoh(n) + x]h(n - 1) + x2h(n - 2) + x3h(n - 3) + . . . + xJz(n - M) = Li x(i)h(n - i) denoted by x(n) hen). Figure 3 .2 shows the interaction of an input signal with a linear time invariant system as a decomposition of separate impulse responses. While h(n) 8(n) LTI H 1l I system T Xo I Yo = xoh(n) I I I Xl Yl'll I X2 Y2,11 X3 + X4 Y4, I I I Xs Ys I , i Xs ---¥-I Figure 3.2. Convolution of input with impulse response of Linear Time-Invariant (LTI) system. 23 EBSCOhost - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES. All use subject to 24 3. Digital Filters seeming quite tedious to calculate (which it is), this operation, called convolution, will be very important in that it allows us to use many mathematical tools to analyze and simulate LTl Systems. 3.2 Digital Filters By f orming linear combinations of past input and output samples, digital filters operate on streams of numbers that are unif ormly sampled in time (such as samples of audio). Current and past inputs are usually denoted as x(n), x(n - f), x(n - 2), ... where n is the current time, n-l is the time one sampling period bef ore the current one, n-2 is two sampling periods ago, etc. Current and past outputs are usually denoted as yen), yen - 1), yen - 2), ... As discussed in Chapter 1, PCM signals are f ormed by sampling an analog waveform at regular intervals in time. The sampling intervals are spaced T seconds apart, where T= lIsampling rate. Thus, relating the integer time indices n, n + 1, etc., of a sampled signal x to actual times in seconds requires multiplying by the sampling period. x(n) =xwnti lll l( + H>- y( n) 0.5 -1 Z Figure 3.3. Two-point moving average digital filter. • diagram of the two-point moving average filter. The Z I block in the f eedforward block represents a unit sample of delay. W e'll find out more about Z and Z I later on. Filters of the fonn of Equation 3.1 are also called nonrecursive, moving average, or all zero (more on that later). Figure 3.4 shows a signal processing block diagram of a general FIR filter. From the discussion of convolution in Section 3 . 1 , note now that an arbitrary (f mite length) impulse response can be stored in the coefficients of an FIR filter, and the operation ofthe filter would then actually perf orm the convolution. Thus, any LTI system with r mite-Iength x ( n ) ---!>--.----l g r---' y(n) Figure 3.4. A high (Nth) order general FIR digital filter. 25 EBSCOhost - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES. All use subject to 26 • 3. Digital Filters impulse response can be modeled by an FIR filter, provided that the impulse response of the LTI system is bandlimited to the Nyquist frequency. 3.4 11K Filters A simple filter that operates on past outputs can be written as y(n) (g x(n)) + ( r y(n - 1)) (3.2) It's easy to show that the impulse response oftms filter for g = 1 is yn = 1.0, r, rr , y3, etc. This type of response is called an ex ponential decay. It's easy to see why filters of this fann are called In finite Im pulse Res ponse (IIR) filters, because for a nonzero Y, the output teclmically never goes exactly to zero. If r is negative, the filter will oscillate positive and negative each sample, corresponding to even and odd powers of Y. This is called an ex ponential oscillation. If the magnitude of r is greater than one, the filter output will grow without bOlllld. This condition is called instability, and such filters are called unstable. Filters of the f onn of Equation 3.2 are also called recursive, all pole (more on that later), and autoregressive. Figure 3.5 shows a signal processing block diagram ofthe simple recursive filter described in Equation 3.2. Figure 3.6 shows a higher order IIR filter block diagram. 3.5 The General Filter Form The most general digital filter operates on both its inputs and outputs, and its difference equation is written: y(n) g(x(n) + a,x(n - 1) + a,x(n - 2) + . . . + a ,;; (n -N» (3.3) -b,y(n - 1) -b,y(n - 2) - ... - bMy(n -M). x(n) ---,...-g .... r }---y(n) -1 Z Figure 3.5. First order recursive digital filter. EBSCOhost - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES. All use subject to 3.6 The Z Transform • x(n) 9 }--y(n) -b 1 -b -1 2 Z -b M 0 0 0 Z -1 Figure 3.6. High order recursive digital filter. Note that the length of input sample "history" is not required to be equal to the length of output sample "history," though in practice they are commonly asswned to be equal. The "order" of a filter is equal to the longest delay used in the filter; in the filter of Equation 3.3, the order would be the greater of N or M. Since general digital filters have IIR components as shown in Equation 3 .3, such filters are also called IIR filters. Another tenn for filters with both FIR and IIR parts is pole-zero filter (more later). One final conmlonly used term is Auto-Regressive MovingAverage, or ARMA. Figure 3.7 shows a signal processing block diagram of the general pole-zero digital filter described in Equation 3.3. 3.6 The Z Transform A common analytical tool for digital filters is the Z transf orm representation. As we said before, we'll def me Zl (Z to the minus 1) as a single sample of delay, and in facż Z·I is sometimes called the Delay O perator. To transf orm a filter using the Z transfonn, simply capitalize all variables x andy, and replace all time indices (n - a) with the appropriate time delay operator Z ". Thus, the Z transf ormed version of Equation 3.3 would be written: 27 EBSCOhost - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES. All use subject to 28 3. Digital Filters x(n) 9 y(n) -b a1 1 -b -1 a2 2 -b Z M aN 0 0 0 Z -1 Figure 3.7. General pole-zero (UR) filter. YC g(X+ alXZ 1 + a,XZ ' + ... + aNXZN) (3.4) -bIYZ 1_ b,YZ ' - ... - bMYZ M. W e'll see in subsequent sections and chapters how the Z transform can be used for analyzing and manipulating signals and filters. 3.7 The Transfer Function A powerful relationship used for analyzing digital filters is the trans fer f unction, which is f ound by solving for the ratio of output (Y) to input (X) in the Z-transfonned filter expression. The transfer function for Equation 3.4 can be solved by using simple algebra: Y(l + blZ-1 + b,Z-' + . . . + bMZ-M) gX(l + alZ-1 + a,Z-' + . . . + aNZ-N) (3.5) EBSCOhost - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES. All use subject to 3.9 First Order One-Zero and One-Pole Filters The transfer function is notated as H H is the Z transfol1ll of the time domain impulse response function hen). Transfol1llation ofx andy into the Z domain gives us a tool f or talking about a function H (the Z transf orm of h) that takes X as input and yields Y. 3.8 Zeroes and Poles Looking at the numerator of Equation 3.6 as a polynomial in Z·I, there will be N values of Z-l that make the numerator equal to zero, and the transfer fimction will be zero at these values. These values that make tile polynomial equal to zero are potentially complex numbers: Re + jIm, where Re and 1m are called the real and imaginary parts, andj ,1(-1). The zero values are called zeroes o f the filter because they cause the gain of the transfer function to be zero. The two· dimensional (real and imaginary) space of possible values of Z is called the z-plane. Similarly, the denominator will have M values of Z·I that make it zero, and these values cause the filter gain to be infinite at those values. These M values are called poles o f the filter (like tent poles sticking up in the transfer f unction, with inImite height where the denominator is zero). Poles are important because they can model resonances in physical systems (we'll see that in the next chapter). Zeroes model signal cancellations, as in the destructive interference discussed in Chapter 2. 3.9 First Order One-Zero and One-Pole Filters The sinlple two· point moving average filter was defined in Equation 3.1, and shown in Figure 3.3. A more general f orm of the first order one·zero filter is shown in Figure 3 .8, and is described by the f ollowing equations: yen) g(x(n) + ax(n - 1)) (3.7) Yl X g(l + aZ I). (3.8) This filter has a single zero at Z -a, and exhibits a maximum gain of g . (1 + laD. Figure 3 .9 shows the gain responses versus frequency of this filter f or various values of a (with g set to 1/(1 + laD to normalize the maximum gain). Such plots are called s pectral magnitude plots, because they show the magnitude of the gain of the filter at each frequency, frOI11 zero Hz up to one half of tile sampling rate (the maximum unaliased frequency). We will see a lot 111 0re on spectra and spectral plots in Chapters 5 and 6. 29 EBSCOhost - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES. All use subject to 30 • x(n) 9 -1 Z Figure 3.8. General one-zero filter. 3. Digital Filters y(n) As can be seen from Figure 3.9, positive values of a cause the filter to favor low frequencies over high. Such filters are called low pass filters. Negative values of a cause the filter to be a high pass filter, as shown in Figure 3.8. If a is set to -1 and g to T (the sampling period), the one-zero filter becomes the digital approximation to differentiation (delta in ad j acent x values divided by delta time). We'll use this in later chapters. The simple first order one-pole filter was shown in Figure 3.5 and is described by the equations below: yen) = gx(n) + ry(n - 1) YlX= g / (1 - rZl). (3.9) (3.10) This filter has a single pole at Z = r, and exhibits a maximum gain of g/(l - Irl). Figure 3.1 ° shows the gain responses versus frequency of this filter for various values of r (g is set to 1 - Irl to normalize the filter maximum gain). As we observed before, the absolute value of r must be less than one in order for the filter to be stable. As can be seen from Figure 3.10, positive values of r cause the filter to favor low frequencies (low pass), while negative values cause the filter to favor high frequencies (high pass). If both g and r I Output I Input I I Output I Input I OHz Frequency SRl2 Frequency SRl2 Figure 3 .9. Gain versus zero location for one-zero filter. EBSCOhost - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES. All use subject to • • • '---.... t" lD ( .. ho"i .. ..,. ", , ) , tho lim «do< "'0-,.,.. 01 . " ,, tho __ . ' _ _ "bo".mp.y nru< ' " .n,.;a"-p.,,;,; .. . . Tho _ i>. "-3.1O. w . 'M. mo . wod jo b . "" ", _ ""'" nu" " OM i>. . ... >0 ud "" """' 0 .... . . _ ", M . t>. tio . . boo .... tho 'n .. . """."""' ''''''''''' '''''''''' ' .v "" hun "" n o. . . o f .-. fu """" ;, --.. Tho " "" _ "" 00 cL boo . , . "" 00 .... _ "."' ",,"" b "" " oo .... ". ,. .. .,i _ t .... "'r '" OM" "" rnlliao ( .,... cL l-oJdE, _ C"",., 1)' OVCM-pclo 01", .. twit,,, • OW". no. . . i>. =Pi_ "''''''' " o m 1.0""" h;t .. . 'Y , .""" " .00. . 1 o.(m., SRA 'ffi'2 -. tho t "'. _ """' . .. . """ """"," ', bo roo """'" 00 _ ! 9l .. , " tho B;Q"'" '" """' . _ . ... . tho , uno " "" .. d d . . cm... u" tho " ..... , "''''''-.. -. >OX<d , 0< , ..." , ,,,, pd "",""" i>. r' by pol: CO>. bo f """, d "" "'" . ... _ _ ,pdP .. ithno! 000 ",", . . '''; tbu, .n thot',_oi ,, <> O n tho ,,"0000, cL Eq .. _ 3.' '" 3.0 " !mt .. d __ on" tillJ.i . . 'loo k> Th o 000<0. ' on" t'O'O-polo . oc .. .. "c". " "" "" _ 0< arilli. _ "",, o x "", . . ti>I L . ...... "''Y _oml .Wfu.:b.""."'tol"' ''' .,... ru<. Tho.oxad '"' .. s., """'" bloo k< on .. " "''''' . _ " ", b1 . cL pm", • • w cL t> ..roo • . . ,... ho. i>. tho Q .... .. A ., b ocm tQod "; tt.. -<'010 moIoo .... E i Q .. ' Tho B;q. .. i>. tho "'"' .. dZ ttuIDrra-..w .liloo > ( n ) _ .(4n) + An - 1) + , 4 n - . )) , ,.(n _ 1) _ .\,>(n . ) (3.11) " EBSCOhost - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES. All use subject to 32 • y g(l + a1Z-1 + a2Z-2) X 1 + b1Z-1 + b2Z-2 . 3. Digital Filters (3 . 12) This filter has two poles and two zeroes. Depending on the values of the a and b coefficients, the poles and zeroes can be placed in f airly arbitrary positions around the z-plane, but not completely arbitrary if the a and b coefficients are real numbers (not complex) . Remember f rom quadratic equations in algebra that the roots of a second order polynomial can be f ound by the f ormula: (-al +/- (a1 2 - 4ayl2] / 2 for the zeroes, and similarly for the poles using the b coefficients. It turns out that for complex roots, the positions will always end up as a complex con jugate pair of the form Re +/-jIm. Filters with two poles are called resonators, or phasors. For practical use in sound processing, there is a wonderful f ormulation of the BiQuad that deals more directly with resonator parameters: y g(l - 2rz cos(27rFreqzT)Z-1 + r;Z-2) X 1 - 2rp cos(27r FreqpT)Z-l + r+Z-2 y(n) = g(x(n) - 2rz cos(27rFreqzT)x(n - 1) + r;x(n - 2)) (3.13) +2rp cos(27rFreqpT)y(n - 1) - r;y(n - 2) . (3 .14) This describes the filter coef f icients in terms of an exponential damping parameter (rz for the zeroes, rp f or the poles) and a center f requency of resonance (antiresonance for the zeroes), which is Freqz for the zeroes and Freq for the poles. We can now control aspects of the filter more directly p f rom these parameters, knowing that once we decide on rz and Freqz' we can convert to aI' a2, bl, and b2 directly. Figure 3 . 1 1 shows the BiQuad in block diagram form. y(n) -2 rzcos(21tFreqz T) r 2 z Figure 3. 1 1 . BiQuad filter block diagram. EBSCOhost - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES. All use subject to 3.10. The Second Order Pole/Zero (BiQuad) Filter • 1m SR+-__ __ -+ __ __ r f "2 Re Figure 3.1 Z. Z-plane pole/zero plot ofBiQuad with zeroes at 1 .0. 5000 Hz. and poles at 0.99. 3000 Hz (sampling rate is 22050). Just as with the filter of Equation 3.2, rp must be strictly less than one, and for the resonance formulation it is usually kept nonnegative, because we can use the frequency variable to swing the position around in an arc anywhere in the z-plane. Freqz and Freqp can take on any value from zero to one half the sampling rate (the Nyquist frequency). Freqz and Freqp can actually take on any value, but they will alias to frequencies within the Nyquist range, because of the "modulo 21t" nature of the cosine functions in Equation 3.l3. Figure 3.12 shows the z-plane pole/zero plot of a Biquad filter with rz set to 1 .0; Freqz set to 5000 Hz; rp set to 0.99; and Freqp set to 3000 Hz (sampling rate is 22050 Hz). The resonance parameters, r, are reflected by the radial distance from the origin in the z-plane. The frequency parameters determine the angle of the pole/zero locations. Figure 3.l3 shows the spectrum (top) of a random noise input to the BiQuad Filter, along with the spectrum of the filtered noise signal. White noise is named similarly to white light, where all frequencies are present in relatively equal amplitude, as shown in the magnitude versus frequency plot. Passing white noise through a filter results in what is called colored noise, in direct analogy to passing light through a color filter. Note the peak in the filtered output spectrum at 3000 Hz corresponding to the pole resonances, and the dip at 5000 Hz corresponding to the zero antiresonances. Figure 3.14 shows some superimposed frequency responses of a Biquad with g set to 0. 1 ; r set to 0.99 and 0.97; the zeroes set at special locations p (+/-1 on the real axis); and Freqp swept from 0 to 4000 Hz in 500 Hz steps (sampling rate is 8000 Hz). Locating the zeroes at those locations offrequency = 0 and frequency = SRATE/2 helps to keep the total filter gain nearly constant, independent of the frequency of the resonator. 33 EBSCOhost - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES. All use subject to 34 3. Digital Filters olse s pec rum :::: OdB, N • t -90dI::S-Bl--P -- -----------------------+ OHz 5500Hz 11025Hz OdB Biquad filtered noise -30dB-UII'F+_i!fi!tr n -,-----------1 -90dB-I----------Q R ,.____------____,,_ S OHz 11025Hz Figure 3.1 3. White noise input spectrum (top) versus BiQuad filtered output spectrum (bottom). The BiQuad has a resonator pole pair at 3000 Hz with r = 0.99, and a zero pair at 5000 Hz with r = 1 . 0 (same filter as shown in the z-plane view of Figure 3 . 12). 3. 1 1 A Little on Filter Topologies If you've had enough of digital filters, dif f erence equations, the z-plane, and all that for now, you can skip the next sections and go on to Chapter 4. However, if you just can't get enough of this stuf f, read on for some notes about implementing digital filters. Digital filters are described by simple linear algebraic equations. As such they can be factored in a number of ways, and thus a given filter might be implemented in a wide variety of ways. As an example, we will use the filter shown in Figure 3 . 1 5 (hereafter we ' ll iet g = 1 for convenience). 0:: -60 0 1 k 2k 3k 4k 0:: -60 0 1 k 2k 3k 4k Figure 3.14. BiQuad transfer f unction magnitudes with zeroes set at +/-1.0 (one zero at DC and one at SRATE/2); r = 0.99 (top plot), and r = 0.97 (lower plot). Resonance f requencies are set to 0, 500, 1 000, 1 500, 2000, 2500, 3000, 3500, and 4000 Hz (sample rate = 8000). EBSCOhost - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES. All use subject to 3.1 1 . A Little on Filter Topologies x(n) 9 -.2125 -1 .4725 -0.20825 -0.9604 y(n) -1 .2925 1 .2925 -1.0 Figure 3.1 5. A f ourth order HR. third order FIR digital filter. The filter of Figure 3 . 1 5 has a dif ference equation (for g = 1) of • yen) = x(n) - 1 .2925 x(n - 1) + 1 .2925 x(n - 2) - x(n - 3) - 0.2 125 yen - 1) - 1 .4725 yen - 2) - 0.20825 yen - 3) - 0.9604 yen - 4). W e can write out the Z transf orm of this filter as: y 1 - 1.2925Z-1 + 1.2925Z-2 - Z-3 X 1 + 0.2125Z-1 + 1.4725Z-2 + 0.20825Z-3 + 0.9604Z-4 · Noting that the numerator is of third order, and the denominator is of f ourth order, we can f actor each into first and second order filter segments: y (1 - 0.2925Z-1 + Z-2) . (1 - Z-l) X (1 - 0.6Z-1 + 0.98Z-2) . (1 + 0.8125Z-1 + 0.98Z-2) . Now that we've f actored it, we can compare the sections to our standard BiQuad f orms given in Equation 3. 1 1 and see that this filter gives us two resonances: one at r = 0.99,f= 3/8 SRA TE; and one at r=0.99,f= 7/8 SRA TE. It also gives us one complex zero pair at r = 1 .0,/= 5/7 SRA TE. Finally, it implements a single zero at f = O. Since the transf er function is just a product of these, we can rewrite it as a chain of simple filter segments: Y I X= (1 - 0.2925 ZI + Z2 ) . (1 - ZI) . 1/(1 - 0.6Z1 + 0.98Z·2) . 1/(1 + 0.8 125Z1 + 0.98Z·2). 35 EBSCOhost - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES. All use subject to 36 • 3. Digital Filters y(n) x(n) y(n) Figure 3.1 6. Two topological variants of the filter shown in Figure 3 . 1 5. Linearity says we can reorder the sections any way we like. The mathematics ofLTI systems tells us that these can be implemented in any order. Two of many such possibilities are shown in Figure 3 . 16 . Note that the gain term can also go at any point in the chain. The f orm shown in Figure 3 . 1 5 is called the direct f orm, and the f orms shown in Figure 3 . 16 are all calledf actared cascade f orms. There are many other f orms f or filters, including parallel. You might ask, "If the math says they're all the same, then why would we care about different f orms?" Well, it turns out that the math is only strictly true in the pure case of infinite precision computation. For finite word sizes, even floats in computers, then round ing, truncation, quantization, etc . , can all make a dif ference in filter implementations. Some topologies do better with [ mite precision computation. However, for the rest of this book we won't much care about this, because it's really not that critical of an issue f or floating point computations on low order filters. 3. 1 2 Conclusion Digital filters operate on sampled signals by f orming linear combinations of past inputs and outputs. LTI (linear time-invariant) systems are common in the acoustical world, and any LTI system can be modeled by a digital filter. Simple low order filters can be used to implement simple high and low pass f unctions, as well as implementing the averaging, dif ferentiation, and integration operations. The second order pole-zero filter (called the "BiQuad") EBSCOhost - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES. All use subject to 3.12. Conclusion is a convenient and flexible f orm, allowing independent control of resonance and antiresonance. The next chapter will look at the first actual physical model in this book and relate the sinmlation of that physical model (and more complex systems) to digital filters. Reading: K en St i egl i tz. A Digital Signal Processing Primer. Menlo P a rk: Add i s o n Wesley, 1996. Lawrence Ra bi ner and Be rn ard Gold. Theory and Ap plication o f Digital Signal Processing. Englewood Cl i ff s: Prent i ce Hall, 1974. Code: filter.c GUI Res o L ab Sounds: [Track 11] One-Pole Filtered Speech, r 0.9 ,0.95, 0.99 , -0.9, -0.95, -0.99. [Track 12] B iQ uad F i ltered No i se. [Track 13] No i se F i ltered Through Topolog i es of F i gures 3. 15-3.16. 37 EBSCOhost - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES. All use subject to EBSCOhost - printed on 11/2/2022 5:54 PM via STANFORD UNIVERSITY LIBRARIES. All use subject to
8232	https://www.youtube.com/watch?v=qViB4KFhPQg	How to Convert Grams and Ounces How-To with Mr Kumar 7030 subscribers 461 likes Description 77410 views Posted: 2 Nov 2016 Learn how to convert between ounces (oz) and grams (g) in this short but informative tutorial video. Grams and ounces are both units of measurement used to measure weight. Grams comes from the metric system of measurement and is equal to 1/1000 of a kilogram. Ounces is used in the US therefore often a conversion is required between the two. This video provides a conversion factor that can be easily used to convert between the two units of measurement. 5 comments Transcript: Grams and Ounces Hey everyone. So in this video I'll show you how to convert between grams and ounces. Right? So a gram is a metric unit of weight which is 1,000th of a kilogram. That simply means that 1,000 g make 1 kilg and ounces is a unit of weight that equals 28.3495 g. Now grams, kilograms, milligs they come from the metric system whereas ounces is used in the Conversion US. Now in order to convert g and ounces, the conversion factor that I like to use is 28.3495. Now as I said earlier, 1 oz is equal to 28.3495 g, right? So if you had to convert g to ounces, you'll divide the amount in g by 28.3495. And if you had to convert ounces to g, you'll multiply by 28.3495. Example, how many ounces are Converting Grams to Ounces there in 10 g? That means you're trying to convert g into ounces. And in order to convert g to ounces, you need to divide by 28.3495 as that's the conversion factor. So 10 g divided by 28.3495. The answer you get is 0.35. Another example, how many g are there in Converting Ounces to Grams 5 oz? So this time you're trying to convert ounces to g. So you have 5 oz multiply by 28.3495 and that will give you 141.748 g. Now I've also stated the formulas. Formulas There's two formulas that you can use. First one is for g. So g is equal to ounces multiplied by 28.3495. And the second formula shows ounces is equal to g / 28.3495. So you can either use the diagram I showed earlier or make a note of this formula and learn these formulas um to convert between g and ounces. And you need to remember the conversion factor. There's other conversion factors as well, but I like to use 28.3495 because I find that easy to remember. And you can change the decimal places as well. I'm using to four decimal places. You can use three or two or even one or even none. That's up to you. Um so remember this diagram and uh you should be fine, right? And the conversion factor 28.3495. Let's try some questions. Number one, convert 12 g to ounces. Questions Number two, how many g make up 7 oz? And number three, if a bag of potatoes is 1,000 g, how much is this in ounces? So number one, convert 12 g to ounces. So 12 g divided by 28.3495 gives you 0.4233 oz. Number two, how many g make up 7 oz? So you're converting answers to grams. You multiply 7 28.3495 gives you 198.4465 g. And the last question, you're converting 1,000 g to ounces. So 1,00 / 28.3495 to give you 35.2739 oz. And that's how you convert between g and ounces. [Music] So here is the diagram again. G to ounces you divide by 28.3495 and ounces to g you need to multiply by 28.3495. I've used um the conversion factor to four decimal places. It's up to you how many uh decimal places you'd like to use. Thank you very much for watching. Please share and subscribe for more.
8233	https://www.chegg.com/homework-help/questions-and-answers/solve-following-initial-value-problems-ut-xux-tu-x-r-t-0-u-x-0-f-x-b-tut-xux-2u-x-r-t-1-u--q120771060	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: Solve the following initial value problems. (a) ut + xux = −tu, x ∈ R, t > 0; u(x, 0) = f(x) (b) tut + xux = −2u, x ∈ R, t > 1; u(x, 1) = f(x) (c) ut + ux − 3u = t, x ∈ R, t > 0, u(x, 0) = x^2 Solve the following initial value problems. (a) ut + xux = −tu, x ∈ R, t > 0; u(x, 0) = f(x) (b) tut + xux = −2u, x ∈ R, t > 1; u(x, 1) = f(x) (c) ut + ux − 3u = t, x ∈ R, t > 0, u(x, 0) = x^2 solve the partial differential equation ut+xux=−tu with initial condition: u(x,0)=f(x) Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
8234	https://stackoverflow.com/questions/61986611/finding-the-number-of-even-perfect-square-proper-divisors-of-a-given-number-n	Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog Collectives¢ on Stack Overflow Find centralized, trusted content and collaborate around the technologies you use most. Learn more about Collectives Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Finding the number of even perfect square proper divisors of a given number N Ask Question Asked Modified 5 years, 4 months ago Viewed 655 times 1 I tried to solve a question on HackerRank (Problem Link: which involves calculating the number of even perfect square proper divisors of a given number N. The problem requires the program to calculate the probability of a divisor of a given number N being even perfect square among all of N's proper divisors. For example, given N = 36, the set of proper divisors is {1,2,3,4,6,9,12,18}, and only 4 is an even perfect square. The probability will be 1/8. Another example will be N = 900, there will be a total of 26 proper divisors and 3 of them {4,36,100} are even perfect square. The probability will be 3/26. These 2 examples are taken from the problem description on HackerRank. I solved this problem and passed all tests but my solution is not optimal. So I read the "Smarter Strategy" mentioned in the editorial provided by HackerRank. I understood the theoretical explanation but I got really confused by the line divisors[j] += divisors[j] / e I don't know whether it is appropriate to copy and paste the explanation and full code here from the editorial on HackerRank ( since it requires the user to log in first (can use Gmail, Facebook, GitHub and LinkedIn accounts) and unlock (no need to pay, it is free), so I just pasted the line that I got really confused. I hope someone can also access the editorial and answer my following questions. I understand the explanations and codes of other solutions, but I just don't get why the update of the divisors list should be done in this way for this optimal method. divisors[j] is the value from the last cycle of the loop, how can this be used to calculate the divisors produced by the current prime number and specific exponent? I think that it /e instead of /(e+1) is because of the initialization of all 1s in the list (already counted the 1 being divisors of every number). Also, I think this method of update is related to avoid double-counting, but I really don't understand how this formula was derived? For example, 36 = 2^2 3^2. After loop 2^1, divisors should be 2. Then after loop 2^2, divisors should be 3 (2/2+2). After loop 3^1, divisors should be 6 (3/1+3). And then after 3^2, divisors should be 9 (6/2+6). My guess is that after each loop the divisors is adding the possibilities of divisors caused by the current value, for example, in the 36 case: val : divisors list 2^1 : {1,2} 2^2 : {1,2,4} 3^1 : {1,2,4,3,6,12} 3^2 : {1,2,4,3,6,12,9,18,36} But I don't know how the formula was mathematically derived... Can anyone explain it to me? Thank you so much... python algorithm math number-theory perfect-square Share Improve this question edited May 26, 2020 at 4:28 user299560user299560 asked May 24, 2020 at 13:25 user299560user299560 1555 bronze badges 4 1 We can't reasonably answer questions about code that we can't see. Can you paste the relevant code into the question? Mark Dickinson – Mark Dickinson 2020-05-24 14:02:54 +00:00 Commented May 24, 2020 at 14:02 I really want but it is in the editorial of the HackerRank, it requires the user to log in their website (can use Gmail, Facebook, GitHub. LinkedIn accounts) and unlock (no need to pay money, it is free). I don't know whether I can post it here directly, but can you try to click the link I provided and sign in to access the editorial? user299560 – user299560 2020-05-24 14:52:27 +00:00 Commented May 24, 2020 at 14:52 Also, how do you know your solution is not optimal? Your solution may be different and optimal. President James K. Polk – President James K. Polk 2020-05-24 14:52:45 +00:00 Commented May 24, 2020 at 14:52 @PresidentJameK.Polk Because the editorial provides different solutions, mine is the same as the one with time complexity O(N log N), while the optimal one's time complexity is better than this. user299560 – user299560 2020-05-24 14:56:26 +00:00 Commented May 24, 2020 at 14:56 Add a comment \| 1 Answer 1 Reset to default 0 It is not clear about which formula you are talking about but if you are talking about how the list val : divisors list 2^1 : {1,2} 2^2 : {1,2,4} 3^1 : {1,2,4,3,6,12} 3^2 : {1,2,4,3,6,12,9,18,36} was created then here is the answer your number is 36 = 22 3 2 and think you have a list A = {} which is empty initially and we will find all the divisors. At that point I think you know how the prime factorization was done. Now, from simple combinatorics you have three possible choice for 2 to include it in every divisors. Suppose you don't want to calculate the divisors that contains any number of 2 means you want 20=1 in So, if you choose 20 and any number of 3 then you have the possible choice 20 30, 20 31, 20 32 So, for 20 and any number of 3: list contains: 20 30 = 1, 20 31 = 3, 20 32 = 9 So, A = {1, 3, 9} Then you choose exactly 2 once and any number of 3 then you have the possible choice 21 30, 21 31, 21 32 for 21 and any number of 3:list contains: 21 30 = 2, 21 31 = 6,21 32 = 18 So, A = {1, 3, 9} U {2, 6, 18} = {1, 2, 3, 6, 9, 18} and continue for when 2 occurs twice. then you have all the divisors in the list. This can be easily implemented using sieve. Share Improve this answer answered May 25, 2020 at 17:44 rng70rng70 251010 bronze badges 5 Comments user299560 user299560 Thank you for your answer, but this is not the question I am talking about... I understand this method and implemented it in the sieve, but the editorial (hackerrank.com/challenges/mehta-and-his-laziness/editorial) here further reduced this method and mentioned a better one in the "smarter strategy", it is the one with 2^e p_2^e2...I also understand the text explanation of that part, but I don't understand how the line of code "divisors[j] += divisors[j] / e" is related to the theoretical explanation. rng70 rng70 you should paste the code segment here .. at least the function that contains the code segment.. I can't help you without seeing the codes you are talking about .. may be it is the part of precomputation that reduce the complexity...add code-segment in you qs. user299560 user299560 Because the code is inside the link that I provided...and it requires the user to log in to unlock. I don't know whether it is appropriate to share their code on another forum, but can you try the link I put in the last comment? The code is inside that link... Thank you.. rng70 rng70 I will try to give you answer after seeing the code when I will get time. But I appreciate your thinking, but most of the time you will find it inappropriate providing some link to other users and asking them to log in that site and then answering your question. user299560 user299560 Yes...I am sorry about that... I also asked the website whether it is okay for me to post the code here but they haven't replied to me yet...Once I got the permission I will post the full code here... Thank you all the time. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions python algorithm math number-theory perfect-square See similar questions with these tags. 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8235	https://www.msdmanuals.com/home/digestive-disorders/gastrointestinal-bleeding/gastrointestinal-bleeding	Gastrointestinal Bleeding - Digestive Disorders - MSD Manual Consumer Version honeypot link skip to main content Professional Consumer Consumer edition active ENGLISH MSD Manual Consumer Version HEALTH TOPICSHEALTHY LIVINGSYMPTOMSEMERGENCIESRESOURCESCOMMENTARYABOUT US HEALTH TOPICSHEALTHY LIVINGSYMPTOMSEMERGENCIESRESOURCESCOMMENTARY Home/ Digestive Disorders/ Gastrointestinal Bleeding/ Gastrointestinal Bleeding/ IN THIS TOPIC Causes Evaluation Treatment Essentials for Older Adults Key Points OTHER TOPICS IN THIS CHAPTER Gastrointestinal Bleeding Esophageal Varices Gastrointestinal Bleeding (GI Bleeding) ByParswa Ansari, MD, Hofstra Northwell-Lenox Hill Hospital, New York Reviewed/Revised Jun 2025 \| Modified Aug 2025 v5613515 VIEW PROFESSIONAL VERSION Causes\| Evaluation\| Treatment\| Essentials for Older Adults\| Key Points\| Topic Resources 3D Models (0) Audios (0) Calculators (0) Images (2) Lab Test (2) Tables (1) Videos (0) Enlarged Veins in the Esophagus... Abnormal Blood Vessels (Angiodysp... Complete Blood Count Liver Function Tests Some Causes and Features... Bleeding may occur anywhere along the digestive (gastrointestinal or GI) tract, from the mouth to the anus. Blood may be easily seen by the naked eye (overt), or blood may be present in amounts too small to be visible (occult). Occult bleeding is detected only by testing a stool specimen with special chemicals. Hematemesis is blood that is visible in vomit. Hematemesis indicates the bleeding is coming from the upper GI tract, usually from the esophagus, stomach, or the first part of the small intestine. When blood is vomited, it may be bright red if bleeding is brisk and ongoing. Alternatively, vomited blood may have the appearance of coffee grounds. It results from bleeding that has slowed or stopped, and the blood looks like coffee grounds because it has been partially digested by acid in the stomach. Blood may also be passed from the rectum: As black, tarry stools (melena) As bright red blood (hematochezia) In apparently normal stool if bleeding is less than a few teaspoons per day Melena is more likely when bleeding comes from the esophagus, stomach, or small intestine. The black color of melena is caused by blood that has been exposed for several hours to stomach acid and enzymes and to bacteria that normally reside in the large intestine. Melena may continue for several days after bleeding has stopped. Hematochezia is more likely when bleeding comes from the large intestine, although it can be caused by very rapid bleeding from the upper portions of the digestive tract as well. People who have lost only a small amount of blood may feel well otherwise. However, serious and sudden blood loss may be accompanied by a rapid pulse, low blood pressure, and reduced urine flow. A person may also have cold, clammy hands and feet. Severe bleeding with extremely low blood pressure (shock) may reduce the flow of blood to the brain, causing confusion, disorientation, and sleepiness. Slow, chronic blood loss may cause symptoms and signs of low blood count (anemia), such as weakness, easy fatigue, paleness (pallor), chest pain, and dizziness. People with underlying ischemic heart disease may develop chest pain (angina) or a have a heart attack (myocardial infarction) because of decreased blood flow through the heart. (See also Overview of Digestive Symptoms.) Causes of Gastrointestinal Bleeding The causes of GI bleeding are divided into 3 areas: Upper GI tract Lower GI tract Small intestine (See table Some Causes and Features of Gastrointestinal Bleeding.) The most common causes, especially for lower GI bleeding, are difficult to specify because causes vary with the person's age. However, in general, the most common causes of upper GI bleeding are Ulcers or erosions of the esophagus, stomach, or duodenum Enlarged veins in the esophagus (esophageal varices) A tear in the lining of the esophagus after vomiting (Mallory-Weiss syndrome) Enlarged Veins in the Esophagus (Esophageal Varices) Hide Details This photo shows enlarged veins in the esophagus (arrows). Image provided by David M. Martin, MD. The most common causes of lower GI bleeding are Polyps of the large intestine Diverticular disease Hemorrhoids Abnormal blood vessels (angiodysplasia, arteriovenous malformations [AVMs]) Inflammatory bowel disease Colon cancer Other causes of lower GI bleeding include a split in the skin of the anus (anal fissure), ischemic colitis, and large-bowel inflammation resulting from radiation or poor blood supply. Bleeding from the small intestine is rare but can result from blood vessel abnormalities, tumors, or a Meckel diverticulum. Abnormal Blood Vessels (Angiodysplasia) in the Intestine Hide Details This photo shows an abnormal blood vessel in the wall of the bowel. Image provided by David M. Martin, MD. Bleeding from any cause is more likely, and potentially more severe, in people who have chronic alcohol-related liver disease or chronic hepatitis, who have hereditary disorders of blood clotting, or who are taking certain medications. Liver disease makes bleeding more likely because a poorly functioning liver produces fewer of the proteins that help blood clot (blood clotting factors). Medications that can cause or worsen bleeding include the following: Anticoagulants (such as heparin, warfarin, dabigatran, apixaban, rivaroxaban, and edoxaban) Those that affect platelet function (such as aspirin and certain other nonsteroidal anti-inflammatory drugs [NSAIDs] and clopidogrel) Those that affect mood or mental health (such as selective serotonin reuptake inhibitors [SSRIs]) Those that affect the stomach's protective barrier against acid (such as NSAIDs) Evaluation of Gastrointestinal Bleeding GI bleeding typically requires evaluation by a doctor. The following information can help people decide when a doctor’s evaluation is needed and help them know what to expect during the evaluation. Warning signs In people with GI bleeding, certain symptoms and characteristics are cause for concern. They include Fainting (syncope) Severe or prolonged sweating (diaphoresis) Rapid heart rate (over 100 beats per minute in the absence of physical activity) Passing in stool or vomiting more than 1 cup (250 milliliters) of blood When to see a doctor People who have GI bleeding should see a doctor right away unless the only sign of bleeding is blood on the toilet paper after a bowel movement. If people with such findings have no warning signs and feel otherwise well, a delay of 1 or 2 days is not harmful. What the doctor does Doctors first ask questions about the person's symptoms and medical history. Doctors then do a physical examination. What they find during the history and physical examination often suggests a cause of the GI bleeding and the tests that may need to be done (see table Some Causes and Features of Gastrointestinal Bleeding). The history is focused on finding out exactly where the bleeding is coming from, how rapid it is, and what is causing it. Doctors need to know how much blood (for instance, a few teaspoons or several clots) is being passed and how often blood is being passed. People with hematemesis are asked whether blood was passed the first time they vomited or only after they vomited a few times with no blood. Doctors ask people with rectal bleeding whether pure blood was passed; whether it was mixed with stool, pus, or mucus; or whether blood simply coated the stool. People with bloody diarrhea are asked about recent travel or other possible forms of exposure to other agents that can cause digestive tract illness (for instance, bacterial infections). Doctors then ask about symptoms of abdominal discomfort, weight loss, and easy bleeding or bruising and symptoms of anemia (such as weakness, easy exhaustion [fatigability], and dizziness). Doctors need to know about any current or past digestive tract bleeding and the results of any previous colonoscopy (an examination of the entire large intestine, the rectum, and the anus using a flexible viewing tube). People should tell doctors whether they have inflammatory bowel disease, bleeding tendencies, or liver disease and whether they use any medications that increase the likelihood of bleeding (such as aspirin, NSAIDs, or anticoagulants) or drugs that can cause chronic liver disease (such as alcohol). The physical examination is focused on the person’s vital signs (such as pulse, breathing rate, blood pressure, and temperature) and other indicators of anemia or significant blood loss and shock (rapid heart rate, rapid breathing, pallor, sweating, little urine production, and confusion). Doctors also look for small purplish red (petechiae) and bruise-like (ecchymoses) spots on the skin, which are signs of bleeding disorders. Doctors also look for signs of chronic liver disease (such as spider angiomas, fluid in the abdominal cavity [ascites], and red palms) and portal hypertension (such as an enlarged spleen and dilated abdominal wall veins). Doctors do a rectal examination to look at stool color and check it for blood and to search for tumors and fissures. Doctors also examine the anus to look for hemorrhoids. Table Some Causes and Features of Gastrointestinal Bleeding Table Some Causes and Features of Gastrointestinal Bleeding Some Causes and Features of Gastrointestinal Bleeding\| Cause \| Common Features† \| Tests \| --- \| Upper digestive tract (indicated by vomiting blood or dark brown, coffee-ground material) \| \| Ulcers or erosions of the esophagus, stomach, or first part of the small intestine (duodenum) \| Pain that Is steady and mild or moderately severe Is usually located just below the breastbone May awaken the person during the night and/or be relieved by eating Painless ulcers can also cause bleeding \| Upper GI endoscopy (examination of esophagus, stomach, and duodenum using a flexible viewing tube called an endoscope) \| \| Esophageal varices (enlarged veins in the esophagus) \| Usually very heavy bleeding Often in people known to have chronic liver disease such as cirrhosis Sometimes signs of chronic liver disease such as a swollen abdomen and yellowish discoloration of the skin and whites of the eyes (jaundice) \| Upper GI endoscopy \| \| Mallory-Weiss tear (a tear in the esophagus caused by vomiting) \| In people who vomited one or more times before they started vomiting blood Sometimes pain in the lower chest during vomiting \| Upper GI endoscopy \| \| Abnormal connections between the arteries and veins (arteriovenous malformations) in the intestine \| Usually no other symptoms \| Upper GI endoscopy \| \| Lower digestive tract (indicated by passing blood in the stool) \| \| Anal fissures \| Pain during bowel movements Bright red blood only on toilet paper or on the surface of formed stools Fissure seen during the doctor's examination \| A doctor’s examination alone \| \| Abnormal blood vessels (angiodysplasia) in the intestine \| Painless Bright red blood from the rectum (hematochezia) Usually in people over age 60 (most common cause of large intestine bleeding in this age group) \| Colonoscopy (examination of the entire large intestine, rectum, and anus using an endoscope) \| \| Inflammation of the large intestine due to radiation therapy, infection, or disruption of the blood supply (as occurs in ischemic colitis) \| Bloody diarrhea, fever, and abdominal pain \| Colonoscopy Stool tests to look for infectious organisms Sometimes CT scan \| \| Colon cancer \| Sometimes fatigue, weakness, and/or a bloating sensation Usually in middle-aged or older people \| Colonoscopy and biopsy (examination of tissue samples taken from the lining of the intestine) \| \| Colon polyps \| Often no other symptoms \| Colonoscopy \| \| Diverticular disease (such as diverticulosis) \| Painless hematochezia Sometimes in people already known to have diverticular disease \| Colonoscopy Sometimes CT scan or CT angiography Rarely angiography (x-rays taken after injecting a dye into an artery using a catheter) \| \| Inflammatory bowel disease (such as ulcerative proctitis, ulcerative colitis, or Crohn disease) \| Bloody diarrhea, fever, and abdominal pain and cramps Sometimes in people who have had several episodes of bleeding from the rectum \| Colonoscopy and biopsy \| \| Internal hemorrhoids \| Bright red blood only on toilet paper, dripping in the bowl, or on the surface of formed stools \| Anoscopy (examination of the anus and rectum with a short, rigid tube) or sigmoidoscopy \| \| Causes are listed in order from the most common to the least in each section. \| \| † Features include symptoms and the results of the doctor's examination. Features mentioned are typical but not always present. \| \| CT = computed tomography; GI = gastrointestinal. \| Causes are listed in order from the most common to the least in each section. † Features include symptoms and the results of the doctor's examination. Features mentioned are typical but not always present. CT = computed tomography; GI = gastrointestinal. Testing The need for tests depends on what doctors find during the history and physical examination, particularly whether warning signs are present. There are 4 main testing approaches to GI bleeding: Blood tests and other laboratory studies Upper endoscopy for suspected upper GI tract bleeding Colonoscopy for lower GI tract bleeding (unless clearly caused by hemorrhoids) Angiography or CT angiography if bleeding is rapid or severe The person’s blood count helps indicate how much blood has been lost. A low platelet count is a risk factor for bleeding. Other blood tests include prothrombin time (PT), partial thromboplastin time (PTT), and liver tests, all of which help detect problems with blood clotting. Doctors often do not do blood tests on people who have minor bleeding caused by hemorrhoids. Lab Test Complete Blood Count Lab Test Liver Function Tests If the person has vomited blood or dark material (which may represent partially digested blood), the doctor sometimes passes a small, hollow plastic tube through the person’s nose down into the stomach (nasogastric tube—see Intubation of the Digestive Tract) and suctions out the stomach contents. Bloody or pink contents indicate active upper GI bleeding, and dark or coffee-ground material indicates that bleeding is slow or has stopped. Sometimes, there is no sign of blood even though the person was bleeding very recently. A nasogastric tube may be inserted in anyone who has not vomited but has passed a large amount of blood from the rectum (if not from an obvious hemorrhoid) because this blood may have originated in the upper digestive tract. If the nasogastric tube reveals signs of active bleeding, or the person’s symptoms strongly suggest the bleeding is originating in the upper digestive tract, the doctor usually does upper endoscopy. Upper endoscopy is a visual examination of the esophagus, stomach, and the first segment of the small intestine (duodenum) using a flexible tube called an endoscope. An upper endoscopy allows the doctor to see the bleeding source and often treat it and is often done without a nasogastric tube being passed. People with symptoms typical of hemorrhoids may need only sigmoidoscopy (examination of the lower part of the large intestine, the rectum, and anus using an endoscope) or anoscopy (examination of only the rectum, using a short scope and light source). All other people with hematochezia should have colonoscopy (examination of the entire large intestine, the rectum, and the anus using an endoscope). Occasionally, upper endoscopy and colonoscopy do not show the cause of bleeding. There are still other options for finding the source of the bleeding. Doctors may do endoscopy of the small bowel (enteroscopy). If bleeding is rapid or severe, doctors sometimes do angiography. During angiography, doctors use a catheter to inject an artery with a contrast dye that is radiopaque (can be seen on x-rays or CT scans). Angiography helps doctors diagnose upper digestive tract bleeding and allows them to do certain treatments (such as embolization and vasoconstrictor infusion—see Stopping the bleeding). Doctors may also inject the person with red blood cells labeled with a radioactive marker (radionuclide scanning). With the use of a special scanning camera, the radioactive marker can sometimes show the approximate location of the bleeding. Before doing angiography or surgery, doctors may also do a test called CT angiography. During this procedure, a type of imaging called computed tomography (CT) and a radiopaque contrast dye are used to produce images of blood vessels and sometimes can show the location of the bleeding. Doctors may do a small-bowel follow-through, which is a series of detailed x-rays of the small intestine. This test has largely been replaced by CT enterography, which is used to evaluate the inside of the small intestine for tumors. Another option is video capsule endoscopy, in which people swallow a tiny camera that takes pictures as it passes through the intestines. Video capsule endoscopy is especially useful in the small intestine, but it is not very useful in either the colon or stomach because these organs are easier to see using endoscopy. It takes a long time to travel through the digestive system so it is not useful in people who are bleeding fast. Treatment of Gastrointestinal Bleeding There are 2 goals to treating people with GI bleeding: Replace lost blood with fluid given by vein (intravenously) and sometimes with a blood transfusion. Stop any ongoing bleeding. Hematemesis, hematochezia, or melena should be considered a potential emergency. People with severe GI bleeding should be admitted to an intensive care unit (ICU) or other closely monitored setting and should be seen by a gastroenterologist and a surgeon. Replacing fluids and blood People with sudden, severe blood loss require intravenous fluids and sometimes an emergency blood transfusion to stabilize their condition. People with blood clotting abnormalities may require transfusion of platelets or fresh frozen plasma or preparations of blood clotting proteins (prothrombin complex concentrates). Stopping the bleeding Most GI bleeding stops without treatment. Sometimes, however, it does not. The type and location of bleeding tell the doctors what treatment to use. For example, doctors can often stop peptic ulcer bleeding during endoscopy by using a device that uses an electrical current to produce heat (electrocautery), heater probes, or injections of certain agents. If endoscopy does not stop the bleeding, surgery may be needed. For esophageal varices (enlarged veins in the esophagus), doctors try to stop bleeding with endoscopic banding, injection sclerotherapy, placement of a tube with balloons to compress the varices, or a transjugular intrahepatic portosystemic shunting procedure (to relieve the underlying pressure that caused the varices and can make bleeding difficult to control). People with esophageal varices may be given injections of the medication octreotide to help stop the bleeding. Antibiotics may also be given.procedure (to relieve the underlying pressure that caused the varices and can make bleeding difficult to control). People with esophageal varices may be given injections of the medication octreotide to help stop the bleeding. Antibiotics may also be given. Doctors can sometimes control severe, ongoing lower GI bleeding caused by diverticula or angiodysplasias during colonoscopy by using clips, an electrocautery device, coagulation with a heater probe, or injection with epinephrine. If these methods do not work or are impossible, doctors may do any of the following: Doctors can sometimes control severe, ongoing lower GI bleeding caused by diverticula or angiodysplasias during colonoscopy by using clips, an electrocautery device, coagulation with a heater probe, or injection with epinephrine. If these methods do not work or are impossible, doctors may do any of the following: Perform an angiography, during which they may pass a catheter into the bleeding vessel and then inject a chemical, fragments of a gelatin sponge, or a wire coil to block the blood vessel and thereby stop the bleeding (embolization) Give intravenous vasopressin to reduce blood flow to the bleeding vessel Give intravenous vasopressin to reduce blood flow to the bleeding vessel Perform surgery. If surgery is needed, it is important for doctors to know the location of the bleeding. Polyps can be removed by a wire snare or electrocautery. Doctors may give people with upper GI bleeding a proton pump inhibitor (PPI), which reduces the amount of stomach acid. Less acidic conditions help the body form and stabilize clots to stop bleeding. Internal hemorrhoid bleeding stops spontaneously in most cases. For people whose bleeding does not stop without treatment, doctors do anoscopy and may place rubber bands around the hemorrhoids or inject them with substances that stop bleeding or do electrocautery or surgery (see treatment of hemorrhoids). Essentials for Older Adults: Gastrointestinal Bleeding In older adults, hemorrhoids and colorectal cancer are the most common causes of minor bleeding. Peptic ulcers, diverticular disease (such as diverticulitis), and abnormal blood vessels (angiodysplasia) are the most common causes of major bleeding. Bleeding from enlarged veins in the esophagus (esophageal varices) is less common than in younger people. Older adults poorly tolerate massive GI bleeding. Doctors must diagnosis older adults quickly, and treatment must be started sooner than in younger people, who can better tolerate repeated episodes of bleeding. Key Points Rectal bleeding may result from upper or lower GI bleeding. Most people stop bleeding spontaneously. Endoscopy is usually the first treatment choice for people whose bleeding does not stop without treatment. Test your KnowledgeTake a Quiz! Copyright © 2025 Merck & Co., Inc., Rahway, NJ, USA and its affiliates. All rights reserved. About Disclaimer Cookie Preferences Copyright© 2025 Merck & Co., Inc., Rahway, NJ, USA and its affiliates. All rights reserved. Find In Topic This Site Uses Cookies and Your Privacy Choice Is Important to Us We suggest you choose Customize my Settings to make your individualized choices. Accept Cookies means that you are choosing to accept third-party Cookies and that you understand this choice. See our Privacy Policy Customize my Settings Reject Cookies Accept Cookies
8236	https://www.chegg.com/homework-help/questions-and-answers/molecular-mass-air-2897-kg-kmol-specific-volume-v-1m3-kg-condition-specific-volume-molar-b-q30857812	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: The molecular mass of air is 28.97 kg/kmol and the specific volume is v=1m3/kg at some condition. What is the specific volume on a molar basis at the same condition? thank you The molecular mass of air is 28.97 kg/kmol and the specific volume is v=1m3/kg at some condition. What is the specific volume on a molar basis at the same condition? thank you Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
8237	https://pubmed.ncbi.nlm.nih.gov/31869062/	Antifungal Ergosterol Synthesis Inhibitors - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Treasure Island (FL): StatPearls Publishing; 2025 Jan. 2024 Mar 1. Affiliations Expand Affiliations 1 LECOM 2 National Health Service PMID: 31869062 Bookshelf ID: NBK551581 Free Books & Documents Item in Clipboard Book Antifungal Ergosterol Synthesis Inhibitors Elizabeth J. Herrick et al. Free Books & Documents Show details Display options Display options Format In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan. 2024 Mar 1. Authors Elizabeth J. Herrick1,Preeti Patel,Muhammad F. Hashmi2 Affiliations 1 LECOM 2 National Health Service PMID: 31869062 Bookshelf ID: NBK551581 Item in Clipboard Full text links Cite Display options Display options Format Excerpt This activity delves into the pharmacological intricacies of antifungal ergosterol synthesis inhibitors, commonly called "azoles." These medications, encompassing miconazole, ketoconazole, fluconazole, itraconazole, voriconazole, posaconazole, isavuconazole, and oteseconazole, play a crucial role in managing diverse fungal infections. The primary mechanism of action involves the inhibition of ergosterol biosynthesis, leading to the disruption of fungal cell membrane integrity. Exploring these azole antifungal agents involves covering indications, contraindications, drug interactions, and adverse event profiles. Notably, the potential for hepatotoxicity necessitates vigilant monitoring and interprofessional communication. These agents' scientific and medical examination includes in-depth discussions on pharmacokinetics, monitoring protocols, clinical toxicology, and elucidating significant box warnings. Copyright © 2025, StatPearls Publishing LLC. PubMed Disclaimer Conflict of interest statement Disclosure: Elizabeth Herrick declares no relevant financial relationships with ineligible companies. Disclosure: Preeti Patel declares no relevant financial relationships with ineligible companies. Disclosure: Muhammad Hashmi declares no relevant financial relationships with ineligible companies. Sections Continuing Education Activity Indications Mechanism of Action Administration Adverse Effects Contraindications Monitoring Toxicity Enhancing Healthcare Team Outcomes Review Questions References Similar articles Coordinated Regulation of Membrane Homeostasis and Drug Accumulation by Novel Kinase STK-17 in Response to Antifungal Azole Treatment.Hu C, Zhou M, Cao X, Xue W, Zhang Z, Li S, Sun X.Hu C, et al.Microbiol Spectr. 2022 Feb 23;10(1):e0012722. doi: 10.1128/spectrum.00127-22. Epub 2022 Feb 23.Microbiol Spectr. 2022.PMID: 35196787 Free PMC article. Synergistic antifungal activity of statin-azole associations as witnessed by Saccharomyces cerevisiae- and Candida utilis-bioassays and ergosterol quantification.Cabral ME, Figueroa LI, Fariña JI.Cabral ME, et al.Rev Iberoam Micol. 2013 Jan 3;30(1):31-8. doi: 10.1016/j.riam.2012.09.006. Epub 2012 Oct 13.Rev Iberoam Micol. 2013.PMID: 23069981 Activity of Isavuconazole and Other Azoles against Candida Clinical Isolates and Yeast Model Systems with Known Azole Resistance Mechanisms.Sanglard D, Coste AT.Sanglard D, et al.Antimicrob Agents Chemother. 2015 Oct 19;60(1):229-38. doi: 10.1128/AAC.02157-15. Print 2016 Jan.Antimicrob Agents Chemother. 2015.PMID: 26482310 Free PMC article. Safety and efficacy of new generation azole antifungals in the management of recalcitrant superficial fungal infections and onychomycosis.Gupta AK, Talukder M, Shemer A, Galili E.Gupta AK, et al.Expert Rev Anti Infect Ther. 2024 Jun;22(6):399-412. doi: 10.1080/14787210.2024.2362911. Epub 2024 Jun 12.Expert Rev Anti Infect Ther. 2024.PMID: 38841996 Review. [Azoles of then and now: a review].Nocua-Báez LC, Uribe-Jerez P, Tarazona-Guaranga L, Robles R, Cortés JA.Nocua-Báez LC, et al.Rev Chilena Infectol. 2020 Jun;37(3):219-230. doi: 10.4067/s0716-10182020000300219.Rev Chilena Infectol. 2020.PMID: 32853312 Review.Spanish. See all similar articles References Gallagher JC, Dodds Ashley ES, Drew RH, Perfect JR. Antifungal pharmacotherapy for invasive mould infections. Expert Opin Pharmacother. 2003 Feb;4(2):147-64. - PubMed Peyton LR, Gallagher S, Hashemzadeh M. Triazole antifungals: a review. Drugs Today (Barc) 2015 Dec;51(12):705-18. - PubMed Heeres J, Meerpoel L, Lewi P. Conazoles. Molecules. 2010 Jun 09;15(6):4129-88. - PMC - PubMed Poojary S, Miskeen A, Bagadia J, Jaiswal S, Uppuluri P. A Study of In vitro Antifungal Susceptibility Patterns of Dermatophytic Fungi at a Tertiary Care Center in Western India. Indian J Dermatol. 2019 Jul-Aug;64(4):277-284. - PMC - PubMed Rauseo AM, Mazi P, Lewis P, Burnett B, Mudge S, Spec A. Bioavailability of Single-Dose SUBA-Itraconazole Compared to Conventional Itraconazole under Fasted and Fed Conditions. Antimicrob Agents Chemother. 2021 Jul 16;65(8):e0013421. - PMC - PubMed Show all 46 references Publication types Study Guide Actions Search in PubMed Search in MeSH Add to Search LinkOut - more resources Full Text Sources NCBI Bookshelf Full text links[x] StatPearls [Internet] NCBI Bookshelf [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). Unauthorized use of these marks is strictly prohibited. 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8238	https://math.stackexchange.com/questions/216705/prove-ax-x-logx-is-convex	optimization - Prove $ax - x\log(x)$ is convex? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Prove a x−x log(x)a x−x log⁡(x) is convex? Ask Question Asked 12 years, 11 months ago Modified12 years, 11 months ago Viewed 10k times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. How do you prove a function like a x−x log(x)a x−x log⁡(x) is convex? The definition doesn't seem to work easily due to the non-linearity of the log function. Any ideas? optimization convex-optimization Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Oct 19, 2012 at 3:39 JavaMan 13.4k 2 2 gold badges 42 42 silver badges 62 62 bronze badges asked Oct 19, 2012 at 3:32 richrich 41 1 1 silver badge 2 2 bronze badges Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 5 Save this answer. Show activity on this post. A function f(x)∈C 2(Ω)f(x)∈C 2(Ω) is convex if its second derivative is non-negative. f(x)=x log(x)⟹f′(x)=x⋅1 x+log(x)⟹f′′(x)=1 x>0 f(x)=x log⁡(x)⟹f′(x)=x⋅1 x+log⁡(x)⟹f″(x)=1 x>0 EDIT If f(x)=a x−x log(x)f(x)=a x−x log⁡(x), then f′′(x)=−1 x f″(x)=−1 x and hence the function is concave. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 19, 2012 at 3:36 user17762 user17762 Add a comment\| This answer is useful 5 Save this answer. Show activity on this post. The answer is quite easy if you use derivatives, so here's one which does not use them. Note that for a continuous function f:P→R f:P→R, it is sufficient that it satisfies the inequality: f(x 2+y 2)⩾f(x)2+f(y)2 f(x 2+y 2)⩾f(x)2+f(y)2 for all x,y∈P x,y∈P to be concave. By the AM-GM inequality, we have x+y 2>x y−−√x+y 2>x y for x≠y x≠y. Thus, for your function, we have: a x+a y 2−ln x+y 2>a x+a y 2−ln x y−−√=a x+a y 2−1 2 ln x y=(a x−ln x 2)+(a y−ln y 2)a x+a y 2−ln⁡x+y 2>a x+a y 2−ln⁡x y=a x+a y 2−1 2 ln⁡x y=(a x−ln⁡x 2)+(a y−ln⁡y 2) EDIT: In fact, since in our case, the inequality is strict, your function is strictly concave. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 19, 2012 at 4:17 Johnny WesterlingJohnny Westerling 1,841 4 4 gold badges 21 21 silver badges 28 28 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Differentiating twice, we get d 2 d x 2 x log(x)=1 x d 2 d x 2 x log⁡(x)=1 x The second derivative is strictly positive everywhere the function is defined, so x log(x)x log⁡(x) is strictly convex. This implies that a x−x log(x)a x−x log⁡(x) is strictly concave. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 19, 2012 at 3:36 EuYuEuYu 42.8k 9 9 gold badges 96 96 silver badges 142 142 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions optimization convex-optimization See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 5Show that f(x)=−ln(x)f(x)=−ln⁡(x) is convex (without using the second derivative!) Related 3Does log(f(X))log⁡(f(X)) concave implies log(f(X−1))log⁡(f(X−1)) convex? 1What's the best way to optimize this energy function, and is it convex? 2Prove f f is a convex function. 5Convert a non-convex QCQP into a convex counterpart 4Mapping non-convex functions onto convex functions 11Prove Neg. Log Likelihood for Gaussian distribution is convex in each of mean and variance. 2How do we know logistic loss is a non convex and log of logistic loss in convex? 0Is this generalized KL divergence function convex? 1Example for closed proper convex function Hot Network Questions How to convert this extremely large group in GAP into a permutation group. 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8239	https://www.youtube.com/watch?v=yQF8WiQnWLE	Proof: Composition of Injective Functions is Injective \| Functions and Relations Wrath of Math 291000 subscribers 425 likes Description 26227 views Posted: 26 Apr 2020 Let g and f be injective (one to one) functions, where g maps A to B and f maps B to C. Then the composition fog, which maps A to C, is also injective. We'll prove this result about injective functions and their compositions in today's lesson! The proof is very straightforward, and merely requires us to apply the definition of injective functions a few times! Remember a function is injective if any two distinct elements in its domain are mapped to distinct elements in the codomain. In other words, injective functions preserve distinctness. And in this lesson we'll prove that function composition preserves injectiveness! Proof that composition of surjective functions is surjective: I hope you find this video helpful, and be sure to ask any questions down in the comments! The outro music is by a favorite musician of mine named Vallow, who, upon my request, kindly gave me permission to use his music in my outros. I usually put my own music in the outros, but I love Vallow's music, and wanted to share it with those of you watching. Please check out all of his wonderful work. Vallow Bandcamp: Vallow Spotify: Vallow SoundCloud: +WRATH OF MATH+ ◆ Support Wrath of Math on Patreon: Follow Wrath of Math on... ● Instagram: ● Facebook: ● Twitter: My Music Channel: 14 comments Transcript: let G and F be injective functions where G Maps elements from some set a to some set B and F Maps elements from that set B to some other set C so these are both injective functions or one-to-one functions meaning they map distinct elements to distinct elements then their composition F of G is injective and we'll be proving that in today's wrath of math lesson so since the codomain of G is B and the domain of F is B we of course can compose these functions we can consider the composition function f of G quickly Before we jump into the proof what are the domain and codomain of this function well remember in function composition the order is from right to left so an element will be evaluated in the function G first and then it will be evaluated in F so elements that go through this function will need to star in the domain of G which of course is the set a then G will send them to some element in the set B and then of course they can go into the function f which will take them from B and send them to some element in the set C so that's going to be their final landing place so this composition function f of G will take elements from a and send them to the set C we want to prove that given G and F are injective the composition F of G is also injective meaning if we take two distinct elements we'll say x and y from the domain so these are distinct elements X is not equal to Y then there are images under this function are also distinct that's what it means for this function to be injective so that means f of G of X is not equal to F of G of Y if we put distinct elements into the function there are images that come out will also be distinct proving this is very straightforward we basically just have to apply the definition of injective a few times so I definitely recommend giving it a try yourself before watching the rest of the lesson you've done that and now let's get into the proof and we can basically begin our proof with these two lines consider the composition function f of G that map's elements from a to elements of C then consider two distinct elements x and y from the domain x and y from a where x is not equal to Y and like we just went over we want to show that the images of x and y under the function f of G must be distinct as well since x and y are distinct and we really just have to think through the progression of this function so an element is going to start being evaluated in the function G and what do we know about G of X and G of Y well we know since X is not equal to Y G of X cannot be equal to G of Y because G we're already assuming is an injective function so if we input these two distinct elements they're images under G will also be distinct G of X is not equal to G of Y because G is injective then of course since GE maps elements to B we know that G of X and G of Y they are distinct elements of the set B that's the codomain of G so G sends x and y to distinct elements in B and of course B is the domain of F so these are distinct elements of s domain G of X is not equal to G of Y and they are both elements of B then we can consider what happens when we put these two elements into the function f and what do we know about that what do we know about F of G of X and how it relates to F of G of Y well since F is also an injective function we know that these two things must not be equal since G of X is not equal to G of Y and we assumed that F is an injective function we know that the image of G of X and the image of G of Y under the function f those images must be distinct F of G of X is not equal to F of G of Y because f is injective so we put those distinct elements into F they're gonna come out all so being distinct and now what does this look like my friends f of G of X F of G of Y by definition that's just our function composition F of G evaluated at X and evaluated at y so this is the same as saying F of G that composite function evaluated at X is not equal to that composite function f of G evaluated at Y and this is the line we really wanted to get to because this demonstrates that our composite function f of G is injective if we take any two distinct elements from the domain their images under the composite function f of G will also be distinct which followed easily from the fact that G and F are both injective functions and so that is our conclusion thus the composite function f of G is injective if distinct elements go in then distinct elements come out and of course remember the f of G of X and f of G of Y they are both elements of C the codomain of the composite function since f is the last part of this composite function and of course F sends elements to C so we've just proven that function composition preserves injectivity if we've got this pair of injective functions and we compose them then that resulting composite function will also be injective and that shouldn't seem too surprising especially if you were to represent this with a diagram I think it would seem really obvious so what makes for an easy proof and it suggests a couple interesting questions the first one that comes to mind is is the converse of this statement true as well if we know that F of G is injective does it follow that F and G are both injective the other question would be is this same result true for surjective functions so if we knew that G and F were surjective instead of injective would it follow that the composition is surjective as well let me know what you think about both of those questions down in the comments and I'll do lessons answering both of them very soon so be sure to subscribe so you don't miss that so I hope this video helped you understand this simple proof about injective functions let me know in the comments if you have any questions need anything clarified or have any other video requests thank you very much for watching I'll see you next time and I already said subscribe but another reason to subscribe is for the swankiest math lessons on the internet and two big banks to valo who upon my request kindly gave me permission to use his music in my math lessons links to his music in the description [Music] [Music] yeah
8240	https://www.chemteam.info/Equilibrium/Common-ion-effect.html	The Common Ion Effect Go to Problems #1 - 10 Return to Equilibrium Menu The solubility of insoluble substances can be decreased by the presence of a common ion. AgCl will be our example. AgCl is an ionic substance and, when a tiny bit of it dissolves in solution, it dissociates 100%, into silver ions (Ag+) and chloride ions (Cl¯). Now, consider silver nitrate (AgNO3). When it dissolves, it dissociates into silver ion and nitrate ion. In the chemistry world, we say that silver nitrate has silver ion in common with silver chloride. Now, consider sodium chloride. It produces sodium ion and chloride ion in solution and we say NaCl has chloride ion in common with silver chloride. What we do is try to dissolve a tiny bit of AgCl in a solution which ALREADY has some silver ion or some chloride ion (never both at the same time) dissolved in it. What will happen is that the solubility of the AgCl is lowered when compared to how much AgCl dissolves in pure water. We call this the common ion effect. Example #1: AgCl will be dissolved into a solution which is ALREADY 0.0100 M in chloride ion. What is the solubility of AgCl? By the way, the source of the chloride is unimportant (at this level). Let us assume the chloride came from some dissolved sodium chloride, sufficient to make the solution 0.0100 M. 1) The dissociation equation for AgCl is: AgCl (s) ⇌ Ag+ (aq) + Cl¯ (aq) 2) The Ksp expression is: Ksp = [Ag+] [Cl¯] 3) The above is the equation we must solve. First we put in the Ksp value: 1.77 x 10¯10 = [Ag+] [Cl¯] 4) Now, we have to reason out the values of the two guys on the right. The problem specifies that [Cl¯] is already 0.0100. I get another 's' amount from the dissolving AgCl. By the 1:1 stochiometry between silver ion and chloride ion, the [Ag+] is 's.' Substituting, we get: 1.77 x 10¯10 = (s) (0.0100 + s) 5) This will wind up to be a quadratic equation which is solvable via the quadratic formula. However, there is a simplified way to solve this problem. We reason that 's' is a small number, such that '0.0100 + s' is almost exactly equal to 0.0100. If we were to use 0.0100 rather than '0.0100 + s,' we would get essentially the same answer and do so much faster. So the problem becomes: 1.77 x 10¯10 = (x) (0.0100) and x = 1.77 x 10¯8 M There is another reason why neglecting the 's' in '0.0100 + s' is OK. It turns out that measuring Ksp values are fairly difficult to do and, hence, have a fair amount of error already built into the value. So the very slight difference between 's' and '0.0100 + s' really has no bearing on the accuracy of the final answer. Why not? Because the Ksp already has significant error in it to begin with. Our "adding" a bit more error is insignificant compared to the error already there. Example #2: What is the solubility of AgI in a 0.274-molar solution of NaI. (Ksp of AgI = 8.52 x 10¯17) Solution: 1) Dissociation equation: AgI (s) ⇌ Ag+ (aq) + I¯ (aq) 2) Ksp expression: Ksp = [Ag+] [I¯] 3) Let us substitue into the Ksp expression: 8.52 x 10¯17 = (s) (0.274 + s) 4) The answer (after neglecting the +s in 0.274 + s: [Ag+] = 3.11 x 10¯16 M By the 1:1 stoichiometry between silver ion and AgI, the solubility of AgI in the solution is 3.11 x 10¯16 M 5) By the way, the solubility of AgI in pure water is this: 8.52 x 10¯17 = (s) (s) x = 9.23 x 10-9 M The solubility of the AgI has been depressed by a factor of a bit less than 30 million times. Example #3: The molar solubility of a generic substance, M(OH)2 in 0.10 M KOH solution is 1.0 x 10¯5 mol/L. What is the Ksp for M(OH)2? Solution: In this case, we are being asked for the Ksp, so that is where our unknown will be. That means the right-hand side of the Ksp expression (where the concentrations are) cannot have an unknown. 1) Dissociation equation: M(OH)2 (s) ⇌ M2+ (aq) + 2OH¯ (aq) 2) Ksp expression: Ksp = [M2+] [OH¯]2 3) Let us substitue into the Ksp expression: Ksp = (1.0 x 10¯5) (0.10)2 The 1.0 x 10¯5 comes from the molar solubility information, coupled with the fact that for every one M(OH)2, one M2+ is produced. Also, we could have used (0.10 + 2.0 x 10¯5) M for the [OH¯]. However, the 2.0 x 10¯5 M, being much smaller than 0.10, is generally ignored. 4) The answer: Ksp = 1.0 x 10¯7 Example #4: What is the solubility, in moles per liter, of AgCl (Ksp = 1.77 x 10-10) in 0.0300 M CaCl2 solution? Solution: 1) Concentration of chloride ion from calcium chloride: 0.0300 M x 2 = 0.0600 M from here: CaCl2(s) ---> Ca2+(aq) + 2Cl¯(aq) 2) Calculate solubility of Ag+: Ksp = [Ag+] [Cl¯] 1.77 x 10-10 = (s) (0.0600) x = 2.95 x 10-9 M Since there is a 1:1 ratio between the moles of aqueous silver ion and the moles of silver chloride that dissolved, 2.95 x 10-9 M is the molar solubility of AgCl in 0.0300 M CaCl2 solution. Example #5: What is the solubility of Ca(OH)2 in 0.0860 M Ba(OH)2? Solution: 1) Dissociation equation: Ca(OH)2(s) ⇌ Ca2+(aq) + 2OH¯(aq) 2) Ksp expression: Ksp = [Ca2+] [OH¯]2 3) The Ksp for Ca(OH)2 is known to be 4.68 x 10¯6. We set [Ca2+] = s and [OH¯] = (0.172 + 2s). Substituting into the Ksp expression: 4.68 x 10¯6 = (s) (0.172 + 2s)2 By the way, Ba(OH)2 is a strong base so [OH¯] = 2 times 0.0860 = 0.172 M Ignoring the "2s," we find s = 1.58 x 10¯4 M Since there is a 1:1 molar ratio between calcium ion and calcium hydroxide, 1.58 x 10¯4 M is the concentration of the calcium hydroxide. Comment: There are several different values floating about the Internet for the Ksp of Ca(OH)2. I got mine from the CRC Handbook, 73rd Edition, pg. 8-43. Example #6: How many grams of Fe(OH)2 (Ksp = 1.8 x 10¯15) will dissolve in one liter of water buffered at pH = 12.00? Solution: 1) The chemical equation: Fe(OH)2 ⇌ Fe2+ + 2OH¯ 2) The Ksp expression: Ksp = [Fe2+] [OH¯]2 3) pH of 12.00 means pOH of 4.00. Which means this: [OH¯] = 1.0 x 10¯4 M 4) The word buffer means that, for all intents and purposes, the [OH¯] will remain constant as some Fe(OH)2 dissolves. 5) Solve the Ksp: 1.8 x 10¯15 = [Fe2+] [1.0 x 10¯4]2 [Fe2+] = 1.8 x 10¯7 M 6) The Fe(OH)2 that dissolves is in a 1:1 molar ratio with the Fe^2+, so we see that 1.8 x 10¯7 mol of Fe(OH)2 dissolves in our 1.00 L of solution. (1.8 x 10¯7 mol) (89.8588 g/mol) = 0.000016174584 g 1.6 x 10¯5 g (rounded to two sig figs) Example #7: Determine the solubility of lead(II) fluoride, (Ksp = 4.0 x 10¯8) in (a) 0.10 M Pb(NO3)2, then (b) 0.010 M NaF. Solution: 1) Write the chemical equation for the dissolving of iron(II) fluoride and write the Ksp expression for lead(II) fluoride: PbI2(s) ⇌ Pb2+(aq) + 2F¯(aq) Ksp = [Pb2+] [F¯]2 2) Solution to (a): 4.0 x 10¯8 = (0.10 + s) (2s)2 Ignore the 'plus s' 4.0 x 10¯8 = (0.10) (2s)2 0.4s2 = 4.0 x 10¯8 s = 0.000316 M 3) Solution to (b): 4.0 x 10¯8 = (s) (0.010 + 2s)2 Ignore the 'plus 2s' 4.0 x 10¯8 = (s) (0.010)2 s = 0.00040 M 4) If you were to leave in the 'plus 2s,' you would wind up with a cubic equation. Here is a Google search for 'cubic equation solver.' Example #8: Consider this solubility equilibrium: \| \| \| --- \| \| AgI ⇌ Ag+(aq) + I¯(aq) \| Ksp = 8.52 x 10¯17 \| (a) Calculate the solubility of AgI in pure water. (b) Calculate the solubility of AgI in a solution containing 1.00 x 10¯3 M NaI. Solution to (a): Ksp = [Ag+] [I¯] 8.52 x 10¯17 = (s) (s) s = 9.23 x 10¯9 M Solution to (b): 8.52 x 10¯17 = (s) (0.001 + s) Ignore the 'plus s' 8.52 x 10¯17 = (s) (0.001) s = 8.52 x 10¯14 M Go to Problems #1 - 10 Return to Equilibrium Menu
8241	https://www.touchsurgery.com/simulations/tension-band-wire-for-patella-fracture-1	Tension Band Wire for patella fracture - Touch Surgery Tension Band Wire for patella fracture Specialties Orthopedics and Trauma Institution Touch Surgery Release Date September 1, 2017 Launch in App Introduction Fracture of the patella is a common injury compromising approximately 1% of all fractures. They are most often caused by a direct blow to the flexed knee, or trauma to the front of the knee. Typical clinical signs include swelling, tenderness and an inability to straight raise the knee against gravity. An example X-ray demonstrating fracture of the patella. Tension Band Principles The quadriceps produces significant forces on patellar fractures which cause early fixation failure. For this reason, screw fixation alone usually fails. In addition to fracture stability, tension band wiring allows early range of motion of the knee which is advantageous. The tension band lies on the anterior surface of the patella, and works by converting tensile forces on the patella surface into compression forces at the fracture line. Example AP and lateral X-rays following tension band wiring of a patella fracture. Patient Positioning The patient is placed supine on a standard operating table. A tourniquet is placed on the ipsilateral side as the injury. The image intensifier should be placed on the contralateral side to the injury with adequate space to move in and take X-rays. The display unit is placed at the foot of the bed. The surgeon stands on the ipsilateral side to the injury. Operating room setup. Possible complications Risks of this procedure include: Scarring. Pain. Infection. Protruding or irritating metalware. Malunion. Non-union. Osteoarthrosis (long term). Launch in App Explore more simulations PrivacyTerms Manage cookies © 2025 Touch Surgery™, all rights reserved. Cookie statement This site uses cookies to store information on your computer. Some are essential to make our site work; others help us improve the user experience. By using the site, you consent to the placement of these cookies. You may change your browser's cookie settings at any time. Read our Privacy statement to learn more. Cookie preferences Okay Manage cookie preferences When you visit our website, we store cookies on your browser to collect information. The information collected might relate to you, your preferences or your device, and is mostly used to make the site work as you expect it to and to provide a more personalized web experience. However, you can choose not to allow certain types of cookies, which may impact your experience of the site and the services we are able to offer. Click on the different category headings to find out more and change our default settings according to your preference. You cannot opt-out of our First Party Strictly Necessary Cookies as they are deployed in order to ensure the proper functioning of our website (such as prompting the cookie banner and remembering your settings, to log into your account, to redirect you when you log out, etc.). For more information about the First and Third Party Cookies used please follow this link: Privacy Statement Allow all Cookie preferences Necessary Cookies Always active These cookies are necessary for the site to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy or language preferences, logging in or filling in forms. You may disable these by changing your browser settings, but this may affect how the website functions. Cookie List ‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Confirm my choices
8242	https://cs.stackexchange.com/questions/94019/optimal-meeting-point	algorithms - Optimal meeting point - Computer Science Stack Exchange Join Computer Science By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Optimal meeting point Ask Question Asked 7 years, 2 months ago Modified5 years, 9 months ago Viewed 3k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I'm interested in studying the problem of the optimal meeting point, which can be described as follow: n n individuals who want to gather in a restaurant (for example). They want a fair meeting point and to minimize the time/distance for everyone. I'm looking for information on that problem, but I've only been able to find advanced paper on this problem. My questions: Is it part of a more general problem? Is graphs the best way to implement it? (the nodes are the possible meeting points, the edges are the possible roads for example..) How do current programs work, for example using maps like Google Map? Is it P or NP? My difficulties to understand the problem and to find more informations about it I know it's a bit vague, but I've got some difficulties because I don't know much about this kind of problems. I was just wondering one day "how to find the best restaurant in terms of distance to meet with some friends". I don't know what is mathematically the parameter that I should minimize to be "fair" (the sum of the distances of the n n individuals to the meeting point? The average?...). Response to comment: (assuming that using graph is the best way to do it) In input, we have a graph and we have the n n individuals (a list of n n vertices if we implement it with graphs, even if it means to add them to the graph), and in output, we get the optimal meeting point (i.e. the vertex which is at the minimal distance from all the n n individuals). algorithms graphs optimization Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Dec 16, 2019 at 10:38 Narek Bojikian 4,784 1 1 gold badge 13 13 silver badges 36 36 bronze badges asked Jul 8, 2018 at 8:50 MiKiDeMiKiDe 229 2 2 silver badges 6 6 bronze badges 3 Can you define your problem in more detail? What is the input, and what is the required output?Yuval Filmus –Yuval Filmus 2018-07-08 09:14:15 +00:00 Commented Jul 8, 2018 at 9:14 Yes, just did it. The problem is that I don't know much about that problem. I was just wondering one day "how to find the best restaurant in terms of distance to meet with some friends"MiKiDe –MiKiDe 2018-07-08 09:32:45 +00:00 Commented Jul 8, 2018 at 9:32 Imagine you wrote a computer program for this problem. What would be the input to the program, and what would be the output?Yuval Filmus –Yuval Filmus 2018-07-08 10:37:21 +00:00 Commented Jul 8, 2018 at 10:37 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. There are two kinds of minimizing in this sense, the first is to minimize the sum of the distances. the second is to minimize the maximum distance (finding a restaurant where the farthest friend is as near as possible). In both cases it's not NP-Hard, since you can calculate all-pairs shortest path in O(N 2 log(E+N))O(N 2 log⁡(E+N)), sum up the distances from each restaurants to all friends and compare the sums. Ps A special case is points laying on a line segment. The solution to the first type will be the restaurant dividing the vertices in two most-possible equal groups (has so many friends on its left as many friends on its right), the reason this works can be derived from the fact that median minimize absolute deviation. The solution to the second problem is trivially the nearest restaurant to the middle point between the two farthest friends. Ps2 In trees you can solve the second type using two DFSes to find the two farthest points and then choose the point in between them minimizing the distance to both of them. The second type however, can be done in trees using a bfs traversal from leaves, counting the number of friends and distance to them in subtree of each node. Then a second bfs from the root of the tree summing all the numbers up. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Jul 8, 2018 at 22:32 answered Jul 8, 2018 at 11:21 Narek BojikianNarek Bojikian 4,784 1 1 gold badge 13 13 silver badges 36 36 bronze badges 2 1 Thank you for your answer. In your opinion, is there anything interesting to say about that kind of problem? (I want to study it for a student project) Is it part of a bigger problem?MiKiDe –MiKiDe 2018-07-08 11:41:04 +00:00 Commented Jul 8, 2018 at 11:41 I tried to reduce on some kind of MST, but that didn't help that much. However, you can try studying the complexity when having a constant number of restaurants or in some special graphs (trees, DAGs, sparse- , planar- or unweighted graphs), you may find some interesting results Narek Bojikian –Narek Bojikian 2018-07-08 11:44:18 +00:00 Commented Jul 8, 2018 at 11:44 Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. narek gives the polynomial-time algorithm. The more general problem is the facility location problem. This includes several problems but the basic idea is that you want to know where to build your restaurants (plural) to minimize the distance people need to travel (with multiple restaurants, we just assume that people want to eat somewhere, not that they all want to eat together) or to compute how many restaurants you need so that everybody is within some distance of one. As for what metric to optimize, note that minimizing the sum is the same thing as minimizing the average, which is just the sum divided by n n. The main alternative is to minimize the maximum distance anyone needs to travel. These can have rather different effects. Suppose you have 99 people in Los Angeles and one in New York. If you minimize the average distance, the one guy in New York flies to LA; if you minimize the maximum distance, everybody schleps over to somewhere like Omaha. It's not my area of research but I've been to plenty of seminars on it, which suggests that it's an active field with plenty of open problems. There should be plenty for you to do, at whatever level you want to work at. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jul 9, 2018 at 9:13 David RicherbyDavid Richerby 82.5k 26 26 gold badges 146 146 silver badges 240 240 bronze badges 1 Thank you very much for your answer. If you have more information about it I would be very interested. But with your informations I'd be able to find more about it.MiKiDe –MiKiDe 2018-07-09 09:22:37 +00:00 Commented Jul 9, 2018 at 9:22 Add a comment\| Your Answer Thanks for contributing an answer to Computer Science Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algorithms graphs optimization See similar questions with these tags. 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8243	https://artofproblemsolving.com/wiki/index.php/Change_of_base_formula?srsltid=AfmBOorOsTYn85iJEGWmSSdP_UpAQc3JTtwOAmhsww8gXJfDuGmdJn5J	Art of Problem Solving Change of base formula - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Change of base formula Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Change of base formula The change of base formula is a formula for expressing a logarithm in one base in terms of logarithms in other bases. For any positivereal numbers such that neither nor are , we have This allows us to rewrite a logarithm in base in terms of logarithms in any base . This formula can also be written Proof Let . Then . And, taking the of both sides, we get By the properties of logarithms, Substituting for y, Use for computations The change of base formula is useful for simplifying certain computations involving logarithms. For example, we have by the change of base formula that The formula can also be useful when calculating logarithms on a calculator. Many calculators have only functions for calculating base-10 and base-e logarithms. But you can still calculate logs in other bases, you just need to use the change of base formula to put in in base 10. For example, if you wanted to calculate , you would first convert it to the form . Then you would evaluate it using the base-10 log function on the calculator. Special cases and consequences Many other logarithm rules can be written in terms of the change of base formula. For example, we have that . Using the second form of the change of base formula gives . One consequence of the change of base formula is that for positive constants , the functions and differ by a constant factor, for all . This article is a stub. Help us out by expanding it. Retrieved from " Category: Stubs Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
8244	https://www.teacherspayteachers.com/browse?search=imperfect%20tense%20spanish%20practice	Imperfect Tense Spanish Practice \| TPT Log InSign Up Cart is empty Total: $0.00 View Wish ListView Cart Grade Elementary Preschool Kindergarten 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Adult education Resource type Student practice Independent work packet Worksheets Assessment Graphic organizers Task cards Flash cards Teacher tools Classroom management Teacher manuals Outlines Rubrics Syllabi Unit plans Lessons Activities Games Centers Projects Laboratory Songs Clip art Classroom decor Bulletin board ideas Posters Word walls Printables Seasonal Holiday Black History Month Christmas-Chanukah-Kwanzaa Earth Day Easter Halloween Hispanic Heritage Month Martin Luther King Day Presidents' Day St. Patrick's Day Thanksgiving New Year Valentine's Day Women's History Month Seasonal Autumn Winter 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Placement (AP) International Baccalaureate (IB) More Supports ESL, EFL, and ELL Special education Screenings and assessments Speech therapy Language More Spanish Imperfect Tense Verbs Conjugation Practice Scaffolded Worksheets No Prep Created by Island Teacher Imperfect Tense Verbs Worksheets to use with your students for class practice or homework after introducing imperfect conjugations. Included are 6 pages of print and go scaffolded worksheets (tasks get increasingly more difficult as students move through the packet), with a variety of activities for students to practice conjugating and applying imperfect tense verbs. This is truly a print and go, no prep packet to save you time! Includes 6 pages with:3 pages of verb chart fill-ins 15 sentences 6 th - 12 th Spanish, World Languages Also included in:Spanish Verb Tense Worksheets Conjugation Practice Present Preterite Reflexive $4.29 Original Price $4.29 Rated 4.74 out of 5, based on 34 reviews 4.7(34) Add to cart Wish List El imperfecto- Spanish Imperfect Tense Practice Worksheet Packet Created by Señora Lila This is a 4-paged (8 including answer key)practice packet of the imperfect tense in Spanish. The first page consists of an explanation of the imperfect tense along with well-organized charts and examples of how to conjugate both "ar" and "er/ir" verbs in this tense. The second page includes blank charts for students to practice conjugating some verbs in the imperfect. Next is a survey that students can take about their childhood. All of these sentences have a verb in the imperfect and will gradu 9 th - 12 th, Adult Education Spanish, World Languages $1.79 Original Price $1.79 Rated 4.84 out of 5, based on 84 reviews 4.8(84) Add to cart Wish List Spanish Imperfect Tense Quiz Worksheet Exam Imperfect Practice El Imperfecto Created by La Profe Plotts EASY TO GRADE Spanish imperfect tense sheets! Ready to print and use! This resource can be used as a formative assessment during an imperfect tense lesson or at the end of the unit as a summative assessment. This resource checks students' understanding of how to conjugate in the imperfect tense (complete the conjugation chart), which imperfect tense forms are used with a variety of sentence subjects (10 fill in the blank with the appropriate form of the verb in parentheses for that subject), an 4 th - 12 th Spanish Also included in:Spanish Imperfect Tense Activity Bundle Spanish Games Spanish Review Activities $1.25 Original Price $1.25 Rated 4.84 out of 5, based on 19 reviews 4.8(19) Add to cart Wish List El Imperfecto Spanish Imperfect Tense Guided Notes and Practice Activities Created by LA SECUNDARIA Your students use the Spanish imperfect tense every day— talking about their childhood, describing routines, telling stories. They just don’t know it yet. This set of guided notes makes that connection crystal clear. With built-in supports and immediate practice, this resource takes the guesswork out of teaching el imperfecto—so students can finally grasp when and why to use it in Spanish. ✅ How It Works✍️ Step-by-step, scaffolded format: Students take notes on regular -AR, -ER, and -IR ver 7 th - 12 th Spanish, World Languages Also included in:Spanish Guided Notes and Practice Grammar Notes Worksheets Bundle $4.00 Original Price $4.00 Rated 4.92 out of 5, based on 25 reviews 4.9(25) Add to cart Wish List Imperfect tense Spanish practice worksheet Created by Kunal Patel A worksheet on the imperfect tense to practice conjugations, fill in the blanks, change the verb from the present to the imperfect tense, and answering personal questions. The personal questions can also be done as an oral activity. Section 1: Conjugate the 7 infinitives in the imperfect tense for the following subjects: Yo, Nosotros, Ella, Vosotros, Uds. Section 2: Conjugate the verb in the imperfect in the blank (10 total) Section 3: Change the verb from the present tense to the imperfect t 9 th - 12 th Spanish $1.75 Original Price $1.75 Rated 4.92 out of 5, based on 12 reviews 4.9(12) Add to cart Wish List El Imperfecto \| Spanish Imperfect Tense Editable Notes &Practice Created by Señora Sol Teach your Spanish classes the preterite tense with this easy to read notes sheet! This is a "one stop shop" reference sheet -- perfect for notes in your Spanish class. This product includes... AR VERBS - This introduction sheet describes how to form the imperfect tense, examples of what they are in English, a definition, imperfect "trigger phrases", and places to practice. ER & IR VERBS - This notes sheet includes how to form these verbs in the imperfect tense, er and ir verb endings, a review 8 th - 12 th, Higher Education Grammar, Spanish Also included in:Spanish Past-Tense Activities BUNDLE \| Preterite & Imperfect $3.00 Original Price $3.00 Rated 5 out of 5, based on 2 reviews 5.0(2) Add to cart Wish List Imperfect Past Tense Verbs Conjugation Practice for Spanish 2 or 3 Review Created by Ms Bochnia's Classroom Conjugating Spanish verbs in the imperfect past tense is quick and easy with this set of Google Slides and printable conjugation practice. This resource can be used to introduce imperfect tense verb conjugations by making a guided notes packet, as independent practice throughout the year, as a review study guide, or as an assessment. Students will practice regular -ar, -er, and -ir verb endings in the imperfect tense and irregular verbs in the imperfect (ser, ir, ver). Use as an end of 10 th - 11 th, Higher Education Spanish Also included in:Spanish Verbs Conjugations \| Present, Preterite, and Imperfect Tenses $4.50 Original Price $4.50 Rated 5 out of 5, based on 1 reviews 5.0(1) Add to cart Wish List Preterite vs Imperfect Tense Practice Spanish 2 Final Exam Review Game Created by Srta Spanish Check out this easy and engaging way to review preterite and imperfect with your Spanish classes! Here are TWO games to review verb forms with your students! Students try and collect all the forms of a verb before the other players in a fast-paced trading game. The first person to collect all the forms of the verb wins! ❁ Click here to save 20% in a BUNDLE!!❁ The vosotros form is included, but if that isn't something you use with your students there is a second set without vosotros! There are 9 th - 12 th Spanish, World Languages Also included in:Preterite vs Imperfect Tense Spanish 2 Worksheets El Pretérito y el imperfecto $2.75 Original Price $2.75 Rated 5 out of 5, based on 4 reviews 5.0(4) Add to cart Wish List Google Drive - Imperfect Tense Spanish - Digital Practice Created by Maestra KSmith No Prep - No Copies! Students practice imperfect conjugations, key phrases, and vocabulary associated with childhood. Completely editable for your classes and your teaching! What's included:Two Google Slides with a total of 7 different practice activities. Each slide has multiple activities for easy grading / assessing. An answer key is provided on Google Slides. Please note, this activity works best on student laptops/desktop computers. This product is not compatible with student devices 8 th - 12 th Spanish $1.50 Original Price $1.50 Rated 4.88 out of 5, based on 8 reviews 4.9(8) Add to cart Wish List Spanish Imperfect Tense Practice – Regulars, Irregulars & Key Words/Phrases Created by Señorita W.'s Spanish Teaching Resources Do you need a quick-prep way to get students practicing Spanish grammar and reviewing the imperfect tense / past tense? Do you need a review activity for fast-finishers or sub plans?Try this print and go Imperfect Tense practice activity! Students complete several different tasks to review the imperfect (including matching and fill-in-the blank) as well as frequency phrases that signal the imperfect tense. This grammar practice:includes seven different review sectionsincludes practice with key 6 th - 12 th Grammar, Spanish $3.00 Original Price $3.00 Rated 5 out of 5, based on 4 reviews 5.0(4) Add to cart Wish List Crime Mystery/Spanish past tense/preterit &imperfect practice pretérito Created by MFulmer Set up a crime scene in your classroom to practice preterit and imperfect in an engaging way! Your colleagues can be the suspects! Your students are the detectives! This 3-day lesson (or longer!) includes: 1. Step-by-step instructions and lesson plan describing how to prepare your lesson ahead of time, set up your classroom, utilize the resources, and two options for teaching your lesson (one option student paced and one option is teacher guided). 2. Google Slides that is already set up, in S Not Grade Specific Spanish, World Languages $6.00 Original Price $6.00 $5.00 Price $5.00 Rated 4.86 out of 5, based on 7 reviews 4.9(7) Add to cart Wish List Spanish Imperfect Tense Conjugation Practice Activities \| Digital and Printable Created by LAS DOS PROFES This digital and printable preterite conjugation bundle has everything that you need to practice or review conjugation in the imperfect (imperfecto) tense in your Spanish 1, 2, or 3 class. These activities are perfect for independent review at the end of a unit, to practice a new concept, or even as sub plans! This activity is engaging, ready to use, and easy to grade! Three styles of activities are included: Color by number (printable), Mazes (printable or digital) and Mystery Picture (d 8 th - 11 th Other (World Language), Spanish Bundle (3 products) $9.00 Original Price $9.00 $7.20 Price $7.20 Add to cart Wish List Spanish Imperfect Tense Notes, Slides, &Practice, Descubre 2 1.1/Senderos 2 4.1 Created by Listo Spanish Resources Mastering the Spanish imperfect tense just got easier! This ready-to-use grammar resource is packed with student-friendly notes, editable slides, and engaging practice activities to help students understand and use the imperfect with confidence. Perfect for Descubre 2 Lección 1.1, Senderos 2 Lección 4.1, or any Spanish class introducing or reviewing the imperfect tense! What Makes It Awesome: Designed for real classrooms—simple, clear, and no fluff Keeps students engaged with visuals Not Grade Specific Spanish $4.00 Original Price $4.00 Add to cart Wish List Spanish Imperfect Tense Verb Practice – Crossword Puzzles for Grammar Review Created by Mi Rincón - El Mundo de Español Reinforce your students’ understanding of Spanish imperfect tense verb conjugation with these engaging and versatile crossword puzzles! Designed to target grammar skills, this set gives students repeated exposure to regular and irregular imperfect tense verbs in a fun and meaningful way. The digital Google Sheets version is self-checking, giving students instant feedback, while the printable worksheets offer flexible options for homework, bell ringers, or sub plans. Whether you're teaching i 8 th - 11 th Spanish $3.50 Original Price $3.50 Add to cart Wish List Spanish Imperfect Tense Verbs Conjugation Practice& Review Pack - 27 Pages! Created by Teach 2 Lead Help students master the Spanish imperfect tense with this comprehensive 27-page practice and review pack! Designed for Spanish 1, 2, and 3 students, this resource breaks down regular imperfect verbs into clear, structured activities that reinforce conjugation patterns and usage. With a combination of conjugation charts, verb drills, fill-in-the-blank exercises, and engaging practice worksheets, students will develop a strong grasp of the imperfect tense and gain confidence in using it fo 4 th - 12 th Spanish, World Languages Also included in:Spanish 2 & 3 Grammar Pack Conjugation Worksheets Present, Preterite & Imperfect $4.00 Original Price $4.00 Add to cart Wish List Super 7 Spanish Verbs: Present, Preterite &Imperfect Tense Practice Created by Spanish with Mr Gregory Make verb practice engaging and effective with this comprehensive resource featuring the Super 7 high-frequency verbs ( ser, estar, tener, haber, ir, gustar, querer). Designed for Spanish teachers looking to build fluency and confidence, this product helps students see, use, and compare the verbs in the present, preterite, and imperfect tenses. 8 th - 12 th Spanish $1.50 Original Price $1.50 Add to cart Wish List Spanish Preterite vs. Imperfect Tenses (comparison) Practice Worksheets Set Created by Maestra en Mississippi This is an excellent Worksheets Pack to compare and contrast the preterite and imperfect tenses in Spanish. There is an area in which students will conjugate various verbs in the present, preterite, and imperfect tenses side-by-side for practice and visual comparison of the conjugations. Students will have to practice deciding if they should use the preterite or imperfect tense for a sentence. Students will be translating sentences and answering questions in complete sentences. This Workshee 6 th - 12 th, Adult Education, Higher Education Spanish $3.00 Original Price $3.00 Rated 5 out of 5, based on 5 reviews 5.0(5) Add to cart Wish List SPANISH IMPERFECT TENSE PRACTICE ACTIVITY - DIGITAL REVIEW GAME - EL IMPERFECTO Created by Spanish Worksheets and More Looking for a fun activity to practice El imperfecto with your Spanish language students? This Google Sheets™ digital game is the perfect way to review the imperfect tense in Spanish. This interactive digital worksheet is a self-checking Coloring Game and is an ideal soft assessment activity to make Spanish grammar practice and review fun for your students! See motivation soar as the displayed image fills in with color every time your students choose the correct answer! This self-paced dig 7 th - 12 th Grammar, Spanish, World Languages Also included in:SPANISH GRAMMAR REVIEW ACTIVITIES - DIGITAL COLORING GAMES AND MYSTERY PICTURES $3.75 Original Price $3.75 Add to cart Wish List Preterite vs. Imperfect Tense Practice for Spanish Students! (Great for tutors!) Created by Mi Camino Spanish Are your students struggling with the preterite vs. the imperfect tenses in Spanish? This packet will help them to practice conjugating regular and irregular verbs in the preterite and in the imperfect tenses in Spanish. It will also introduce those words or phrases that are a generally a "signal" to choose the correct tense when speaking about actions that have occurred in the past. Students will first review and practice conjugating both regular and irregular verbs in each tense, (preterite 4 th - 10 th Spanish $4.00 Original Price $4.00 Rated 5 out of 5, based on 1 reviews 5.0(1) Add to cart Wish List Contra el reloj \| Imperfect tense 2 \| Spanish verb practice Created by Srta H's Classroom This activity can be used to practice the conjugation of Imperfect tense verbs. Students will race against each other and the clock to correctly conjugate the verbs in each box. The goal is to finish with speed and accuracy. Five seconds will be added to their overall time per mistake. The student with the fastest overall time wins. Included: Instructions, Student worksheet, and answer key 8 th - 12 th Spanish $2.00 Original Price $2.00 Add to cart Wish List Contra el reloj \| Imperfect tense 1 \| Spanish verb practice Created by Srta H's Classroom This activity can be used to practice the conjugation of Imperfect tense verbs. Students will race against each other and the clock to correctly conjugate the verbs in each box. The goal is to finish with speed and accuracy. Five seconds will be added to their overall time per mistake. The student with the fastest overall time wins. Included: Instructions, Student worksheet, and answer key 8 th - 12 th Spanish $2.00 Original Price $2.00 Add to cart Wish List Spanish Imperfect Tense Verb Conjugation Game &Practice Worksheets Created by Espanol con Emily This set of 5 Spanish Imperfect Tense Conjugation Race activities will help your students master el Imperfecto through a competitive game! All you have to do is print the worksheets, set a timer, and afterwards, project the answer key to go over the answers. Use these print & go activities for several Spanish lessons as:Sub plansStations workHomeworkReview gameTest review✏️ Your students will practice conjugating with these 5 activitiesRegular -ar verbsRegular -er & -ir verbsRegular -ar, -er 9 th - 12 th Spanish, World Languages $3.00 Original Price $3.00 Add to cart Wish List Imperfect tense mini packet practice (Spanish) Created by Kunal Patel Imperfect tense mini packet practiceWrite the correct imperfect tense conjugations of the infinitive and answer the personal questions in a complete sentence in Spanish. Page 1: -ar / Page 2: -er / Page 3: -irFor each infinitive type (-ar, -er, -ir), students will conjugate one infinitive for all 6 persons and answer 3 personal questions. A table with -ar, -er, -ir imperfect tense endings is included for reference. Irregular verbsPage 4: Ser / Page 5: Ir / Page 6: VerWrite the conjugations u 9 th - 12 th Spanish $2.00 Original Price $2.00 Add to cart Wish List Preterite vs. Imperfect Notes &Practice: Comprehensive Spanish Past Tense Created by ProfesoraPatricia Preterite vs. Imperfect Guided Notes and Practice: Understanding Spanish Past TensesHelp your students demystify the preterite and imperfect tenses with this in-depth, guided notes resource. Designed to clarify when and how to use each tense, this resource offers clear explanations, meaningful examples, and engaging practice activities that reinforce learning. Resource Overview:Guided Notes: Comprehensive side-by-side comparison of the preterite and imperfect tenses, focusing on their distinct u Not Grade Specific Grammar, Spanish, Writing $3.00 Original Price $3.00 Add to cart Wish List 1 2 3 4 5 Showing 1-24 of 2,000+results TPT is the largest marketplace for PreK-12 resources, powered by a community of educators. 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8245	https://ask.sagemath.org/question/45709/how-to-change-sqrt5-to-decimal/	How to change sqrt(5) to decimal? - ASKSAGE: Sage Q&A Forum First time here? Check out the FAQ! Hi there! Please sign inhelp tagsusersbadges ALLUNANSWERED Ask Your QuestionAsk Your Question 1 How to change sqrt(5) to decimal? save cancel sqrt asked 6 years ago john 11●1●1●2 How can i typing to change root5 to decimal? Preview: (hide) save cancel add a comment 2 Answers Sort by » oldestnewestmost voted 2 answered 6 years ago BrentBaccala 67●7 Try RR(sqrt(5)). The default precision is 53 bits, but that can be changed. For 100 bits of precision, use RealField(100)(sqrt(5)) Preview: (hide) save cancel link Comments 1 or a=sqrt(5) a.n(400) FrédéricC( 6 years ago ) 1 More detail, such as controlling the number of digits or precision, can be found in the documentation here. On this site, here gives some options. dazedANDconfused( 6 years ago ) add a comment 1 answered 6 years ago nur_hamid 31●2●4●8 simply by float(sqrt(5)) or RR(sqrt(5)) Preview: (hide) save cancel link add a comment Your Answer Please start posting anonymously - your entry will be published after you log in or create a new account. Add Answer Question Tools Follow subscribe to rss feed Stats Asked: 6 years ago Seen: 1,757 times Last updated: Mar 11 '19 Related questions why won't simplify multiply out square roots? latex(22^(1/3)) Simplify multiplication of SQRT()s Need positive sqrts only eliminating fractions and roots from equations Solving equation with sqrt strange way to simplify square roots radical expression for algebraic number Laurent series, Rational Functions in sqrt(q)? dividing vector(a,b) by sqrt(x) gives (a/xsqrt(x),b/xsqrt(x)) Copyright Sage, 2010. Some rights reserved under creative commons license. Content on this site is licensed under a Creative Commons Attribution Share Alike 3.0 license. about \|faq \|help \|privacy policy \|terms of service Powered by Askbot version 0.7.59 ( 2025-09-29 11:27:35 +0200 )edit none×
8246	https://en.wikipedia.org/wiki/Allomorph	Jump to content Search Contents (Top) 1 In English 1.1 Past tense allomorphs 1.2 Plural allomorphs 1.3 Negative allomorphs 2 In Sámi languages 3 Stem allomorphy 4 History 5 See also 6 References Allomorph Afrikaans العربية Беларуская Беларуская (тарашкевіца) Català Deutsch Español Esperanto Euskara فارسی Français 한국어 Hausa Bahasa Indonesia Íslenska Italiano Қазақша Kiswahili Magyar Nederlands 日本語 Norsk bokmål Polski Português Romnă Русский Shqip Suomi Svenska Татарча / tatarça Türkçe Українська Walon 粵語中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Variant pronunciation of a morpheme This article is about the concept in linguistics. For the concept in geology, see Allomorph (geology). \| \| \| This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed.Find sources: "Allomorph" – news · newspapers · books · scholar · JSTOR (April 2020) (Learn how and when to remove this message) \| In linguistics, an allomorph is a variant phonetic form of a morpheme, or in other words, a unit of meaning that varies in sound and spelling without changing the meaning. The term allomorph describes the realization of phonological variations for a specific morpheme. The different allomorphs that a morpheme can become are governed by morphophonemic rules. These phonological rules determine what phonetic form, or specific pronunciation, a morpheme will take based on the phonological or morphological context in which it appears. Allomorphy in English involves the variation of morphemes in their phonetic form based on specific linguistic contexts, a phenomenon governed by morphophonemic rules. For instance, the past tense morpheme "-ed" can manifest in different forms—[-əd], [-t], or [-d]—depending on the final sound of the verb stem. This variability is not random but follows predictable patterns, such as the insertion of a schwa [ə] or assimilation to the voicing of the preceding consonant. Similarly, English plural morphemes exhibit three allomorphs: [-s], [-z], and [-əz], with pronunciation determined by the final sound of the noun, whether it be a voiceless consonant, a voiced consonant, or a sibilant. In addition, negative prefixes like "in-" display allomorphy, changing from [ɪn-] to [ɪŋ-] or [ɪm-] depending on the following consonant's place of articulation. This systematic variation reflects the intricate relationship between phonology and morphology in language, with allomorph selection being guided by both phonological environment and morphological constraints (Pak, 2016; Stanton, 2022). In English [edit] English has several morphemes that vary in sound but not in meaning, such as past tense morphemes, plural morphemes, and negative morphemes. Past tense allomorphs [edit] For example, an English past tense morpheme is -ed, which occurs in several allomorphs depending on its phonological environment by assimilating the voicing of the previous segment or the insertion of a schwa after an alveolar stop. A possible set of assimilations is: as [-əd] or [-ɪd] in verbs whose stem ends with the alveolar stops [t] or [d], such as 'hunted' [hʌntɪd] or 'banded' [bændɪd] as [-t] in verbs whose stem ends with voiceless phonemes other than [t], such as 'fished' [fɪʃt] as [-d] in verbs whose stem ends with voiced phonemes other than [d], such as 'buzzed' [bʌzd] The "other than" restrictions above are typical for allomorphy. If the allomorphy conditions are ordered from most restrictive (in this case, after an alveolar stop) to least restrictive, the first matching case usually has precedence. Thus, the above conditions could be rewritten as follows: as [-əd] or [-ɪd] when the stem ends with the alveolar stops [t] or [d] as [-t] when the stem ends with voiceless phonemes as [-d] elsewhere The [-t] allomorph does not appear after stem-final /t/ although the latter is voiceless, which is then explained by [-əd] appearing in that environment, together with the fact that the environments are ordered (that is, listed in order of priority). Likewise, the [-d] allomorph does not appear after stem-final [d] because the earlier clause for the /-əd/ allomorph has priority. The /-d/ allomorph does not appear after stem-final voiceless phoneme because the preceding clause for the [-t] comes first. Irregular past tense forms, such as "broke" or "was/were," can be seen as still more specific cases since they are confined to certain lexical items, such as the verb "break," which take priority over the general cases listed above. Plural allomorphs [edit] The plural morpheme for regular nouns in English is typically realized by adding an -s or -es to the end of the noun. However, the plural morpheme actually has three different allomorphs: [-s], [-z], and [-əz]. The specific pronunciation that a plural morpheme takes on is determined by a set of morphological rules such as the following: assume that the basic form of the plural morpheme, /-z/, is [-z] ("bags" /bægz/) the morpheme /-z/ becomes [-əz] by inserting an [ə] before [-z] when a noun ends in a sibilant ("buses" /bʌsəz/) change the morpheme /-z/ to a voiceless [-s] when a noun ends in a voiceless sound ("caps" /kæps/) Negative allomorphs [edit] In English, the negative prefix in- has three allomorphs: [ɪn-], [ɪŋ-], and [ɪm-]. The phonetic form that the negative morpheme /ɪn-/ uses is determined by a set of morphological rules; for example: the negative morpheme /ɪn-/ becomes [ɪn-] when preceding an alveolar consonant ("intolerant"/ɪnˈtɔlərənt/) the morpheme /ɪn-/ becomes [ɪŋ-] before a velar consonant ("incongruous" /ɪŋˈkɔŋgruəs/) the morpheme /ɪn-/ becomes [ɪm-] before a bilabial consonant ("improper" /ɪmˈprɔpər/) In Sámi languages [edit] The Sámi languages have a trochaic pattern of alternating stressed and unstressed syllables. The vowels and consonants that are allowed in an unstressed syllable differ from those that are allowed in a stressed syllable. Consequently, every suffix and inflectional ending has two forms, and the form that is used depends on the stress pattern of the word to which it is attached. For example, Northern Sámi has the causative verb suffix -hit/-ahttit in which -hit is selected when it would be the third syllable (and the preceding verb has two syllables), and -ahttit is selected when it would be the third and the fourth syllables (and the preceding verb has three syllables): goar·rut has two syllables and so when suffixed, the result is goa·ru·hit. na·nos·mit has three syllables and so when suffixed, the result is na·nos·mah·ttit. The same applies to inflectional patterns in the Sami languages as well, which are divided into even stems and odd stems. Stem allomorphy [edit] Allomorphy can also exist in stems or roots, as in Classical Sanskrit: Vāk (voice) \| \| Singular \| Plural \| \| Nominative \| /vaːk/ \| /vaːt͡ʃ-as/ \| \| Genitive \| /vaːt͡ʃ-as/ \| /vaːt͡ʃ-aːm/ \| \| Instrumental \| /vaːt͡ʃ-aː/ \| /vaːɡ-bʱis/ \| \| Locative \| /vaːt͡ʃ-i/ \| /vaːk-ʂi/ \| There are three allomorphs of the stem, /vaːk/, /vaːt͡ʃ/, and /vaːɡ/, which are conditioned by the particular case-marking suffixes. The form of the stem /vaːk/, found in the nominative singular and locative plural, is the etymological form of the morpheme. Pre-Indic palatalization of velars resulted in the variant form /vaːt͡ʃ/, which was initially phonologically conditioned. The conditioning can still be seen in the locative singular form, for which the /t͡ʃ/ is followed by the high front vowel /i/. However, the subsequent merging of /e/ and /o/ into /a/ made the alternation unpredictable on phonetic grounds in the genitive case (both singular and plural) as well as the nominative plural and the instrumental singular. Thus, allomorphy was no longer directly relatable to phonological processes. Phonological conditioning also accounts for the /vaːɡ/ form in the instrumental plural, in which the /ɡ/ assimilates in voicing to the following /bʱ/. History [edit] The term was originally used to describe variations in chemical structure. It was first applied to language (in writing) in 1948, by Fatih Şat and Sibel Merve in Language XXIV. See also [edit] Null allomorph Alternation (linguistics) Allophone Consonant mutation Grassmann's law Suppletion References [edit] ^ a b c d e f Tarni, Prasad (2019-07-01). A Course in Linguistics, Third Edition. PHI Learning Pvt. Ltd. ISBN 978-93-88028-96-7. ^ a b Fromkin, Victoria; Rodman, Robert; Hyams, Nina (2018). An Introduction to Language (11th ed.). Cengage Learning. pp. 218–220. ISBN 9781337559577. ^ Moravcsik, Edith (2019-11-11). "Accounting for Variation in Language". Open Linguistics. 5 (1): 369–382. doi:10.1515/opli-2019-0020. S2CID 208141142. ^ Jeffers, Robert; Lehiste, Ilse (1982). Principles and Methods for Historical Linguistics. The MIT Press. ISBN 9780262600118. ^ Fromkin, Victoria; Rodman, Robert; Hyams, Nina (2003). An Introduction to Language (9th ed.). Wadsworth Cengage Learning. pp. 268–272. ISBN 9781439082416. ^ Oxford English Dictionary Online: Entry 50006103. Accessed: 2006-09-05 \| Authority control databases \| \| International \| \| \| Other \| Yale LUX \| Retrieved from " Categories: Linguistic morphology Morphemes Linguistics terminology Hidden categories: Articles with short description Short description is different from Wikidata Articles needing additional references from April 2020 All articles needing additional references Pages with plain IPA Articles containing Northern Sami-language text Add topic
8247	https://www.studypug.com/algebra-help/conversion-between-entire-radicals-and-mixed-radicals	Converting Between Entire and Mixed Radicals: A Comprehensive Guide Master the art of radical conversion with our in-depth guide. Learn to simplify complex expressions, enhance your algebra skills, and tackle advanced math problems with confidence. Get the most by viewing this topic in your current grade. Pick your course now. Free to Join! Easily See Your Progress Make Use of Our Learning Aids Last Viewed Practice Accuracy Suggested Tasks Get quick access to the topic you're currently learning. See how well your practice sessions are going over time. Stay on track with our daily recommendations. Earn Achievements as You Learn Create and Customize Your Avatar Introduction to Entire and Mixed Radicals Welcome to our exploration of entire and mixed radicals! These concepts are fundamental in algebra and play a crucial role in simplifying expressions. An entire radical is a radical expression where the radicand (the number under the radical sign) is a whole number. On the other hand, a mixed radical combines a whole number with a radical. Understanding the difference and learning how to convert radicals between these forms is essential for solving mathematical problems. Our introduction video will guide you through these concepts step-by-step, making it easier to grasp. You'll see how entire radicals can be transformed into mixed radicals and vice versa, a skill that's invaluable in simplifying expressions and algebraic expressions. This knowledge will serve as a strong foundation for more advanced topics in algebra. So, let's dive in and unravel the mysteries of entire and mixed radicals together! What is the difference between an entire radical and a mixed radical? An entire radical is a radical expression where the entire quantity under the radical sign remains unresolved. For example, ³45,000 is an entire radical. A mixed radical, on the other hand, is a radical expression where part of the radicand can be simplified or extracted, resulting in a combination of a rational number outside the radical and a remaining radical expression. For instance, 10³45 is a mixed radical. How do you convert an entire radical to a mixed radical? To convert an entire radical to a mixed radical, follow these steps: For example, to convert ³45,000 to a mixed radical: How do you convert a mixed radical to an entire radical? To convert a mixed radical to an entire radical: For example, to convert 10³45 to an entire radical: What's the difference between working with square roots and cubic roots? The main differences are: Why is it important to understand radical conversions? Understanding radical conversions is crucial because it: Understanding the conversion between entire radicals and mixed radicals is a crucial skill in algebra, but it requires a solid foundation in several prerequisite topics. These fundamental concepts are essential for grasping the intricacies of radical manipulation and conversion. One of the most important prerequisites is prime factorization. This skill allows students to break down numbers into their prime factors, which is essential when simplifying radicals or converting between different forms. By mastering prime factorization, students can more easily identify perfect square factors within radicals, facilitating the conversion process. Another crucial concept is factoring perfect square trinomials. This skill is particularly useful when dealing with more complex radical expressions, as it helps in identifying and simplifying perfect square terms within radicals. Understanding the patterns of perfect square trinomials can greatly enhance a student's ability to manipulate radical expressions efficiently. Simplifying rational expressions and understanding restrictions is another vital prerequisite. This knowledge is crucial when dealing with radicals in fractional form or when simplifying complex radical expressions. It helps students avoid errors related to undefined expressions and ensures proper simplification of radical fractions. The ability to convert radicals to mixed radicals is directly related to the main topic and serves as a foundational skill. This process involves identifying the largest perfect square factor within a radical and expressing it as a separate factor outside the radical. Mastering this concept is essential for understanding the reverse process of converting mixed radicals to entire radicals. Additionally, knowing how to convert between radicals and rational exponents is crucial. This skill allows students to express radical forms as exponential expressions and vice versa, providing a deeper understanding of the relationship between radicals and exponents. It also offers alternative methods for simplifying and manipulating radical expressions. Lastly, understanding the square root of a function provides a broader context for working with radicals. This concept extends radical operations to functions, allowing students to apply their knowledge of radicals in more advanced mathematical scenarios. By mastering these prerequisite topics, students will be well-equipped to tackle the challenges of converting between entire radicals and mixed radicals. Each of these foundational concepts contributes to a comprehensive understanding of radical manipulation, enabling students to approach more complex problems with confidence and proficiency. Become a member to get more! Try Free © 2015 - 2025 StudyPug No Javascript It looks like you have javascript disabled. You can still navigate around the site and check out our free content, but some functionality, such as sign up, will not work. If you do have javascript enabled there may have been a loading error; try refreshing your browser.
8248	https://www.m-hikari.com/imf-2011/57-60-2011/azarianIMF57-60-2011.pdf	International Mathematical Forum, Vol. 6, 2011, no. 59, 2949 - 2954 Jensen’s Inequality Versus Algebra in Finding the Exact Maximum Mohammad K. Azarian Department of Mathematics University of Evansville 1800 Lincoln Avenue, Evansville, IN 47722, USA azarian@evansville.edu Abstract In this article we ﬁnd the exact maximum value of a function with-out the conventional method of using critical numbers. In fact, we ﬁnd the exact maximum without even ﬁnding the derivative of the function. First we apply Jensen’s Inequality, and then we use simple algebra to ﬁnd the exact maximum. We conclude the article by posing two ques-tions for the reader. Mathematics Subject Classiﬁcation: 26A, 26D Keywords: concave function, convex function, Jensen’s Inequality 1. Preliminaries When dealing with the extrema of a function the ﬁrst thing that comes to mind is to calculate the derivative of the function and ﬁnd the critical num-bers (provided they exist). Sometimes, ﬁnding the exact derivative and/or 2950 M. K. Azarian the exact critical numbers of a function could be very challenging if not impos-sible. Our goal in this article is to ﬁnd the exact maximum value of a function without the conventional method of ﬁnding and using critical numbers. First we apply Jensen’s Inequality, and then we use simple algebra to ﬁnd the ex-act maximum. It is intriguing to ﬁnd the exact maximum value of a rather peculiar function without ﬁnding its critical numbers or even being concerned about its derivative. As the reader may know there are many versions of Jensen’s Inequality in the real and complex analysis of one or several variables. Various versions of Jensen’s Inequality is used in probability theory, measure theory, the theory of integration, and trigonometry, just to name a few. In most cases Jensen’s Inequality is applicable only if the function under discussion is convex or con-cave. In this article, we need the most common form of Jensen’s Inequality, namely the following theorem: Theorem 1.1. (Jensen’s Inequality). A continuous real-valued func-tion f(x) deﬁned on an interval I ⊆R is concave if and only if f(x1) + f(x2) ≤2f(x1 + x2 2 ), for any x1, x2 ∈I. We note that (i) the above theorem can be generalized for f(x1, x2, ..., xn), and (ii) the direction in the above inequality changes if and only if f(x) is con-vex. 2. The Result In this section ﬁrst we state the problem, and then we apply Jensen’s Inequality as well as simple algebra to ﬁnd the exact maximum. 2.1. The Problem Jensen’s inequality versus algebra 2951 The maximum value of f(α) = (1 + sin α cos α) 1 2[(1 + cos2 α)−1 2 + (1 + sin2 α)−1 2] is exactly 2, provided 0 ≤α ≤π 2. We note that the derivative of f is a rather long expression, and one cannot ﬁnd the exact critical point(s) and hence the exact maximum of f, by a conventional method. Therefore, to ﬁnd the exact maximum of f, ﬁrst we make use of Jensen’s Inequality and then we just use elementary algebra as follows. 2.2. Using Jensen’s Inequality We need to show that f(α) = (1 + sin α cos α) 1 2[(1 + cos2 α)−1 2 + (1 + sin2 α)−1 2] ≤2, for α in [0, π 2]. Equivalently, we need to show that 1 √ 1 + cos2 α + 1 √ 1 + sin2 α ≤ 2 √ 1 + sin α cos α, (1) for α in [0, π 2]. Now, we note that if α = 0 or α = π 2, then in both cases we have 1+ 1 √ 2 ≤2, which is a true statement. Therefore, we may assume that α is in (0, π 2), and hence cos α > 0 and sin α > 0. Next, since both cos α and sin α are positive in (0, π 2), we choose nonnegative real numbers r and s such that cos α = e−r and sin α = e−s. Substituting cos α = e−r and sin α = e−s in the Inequality (1), we obtain 1 √ 1 + e−2r + 1 √ 1 + e−2s ≤ 2 √ 1 + e−(r+s), (2) for r, s ⩾0. Now, we consider the function f(x) = 1 √ 1 + e−2x. 2952 M. K. Azarian To prove the Inequality (2), by Jensen’s Inequality we need to show that f(r) + f(s) ≤2f(r + s 2 ), (3) for r, s ⩾0. By Jensen’s Inequality, the Inequality (3) is true if and only if f is concave. But, since f ′′(x) = 1 −2e2x e4x(1 + e−2x) 5 2 < 0, for all x ⩾0, f is concave, as desired. Finally, we note that we have equality when α = π 4 , and therefore, the maximum value of f(α) is exactly 2. 2.3. Using Basic Algebra If t ∈[0, 1], then it is easy to see that (t2 −3t 2 + 5)2 ≥4( t 2 + 1)2(t2 4 + 2). (4) Now, if we divide both sides of the Inequality (4) by ( t 2 + 1)2, and then take the square roots of both sides we get t2 −3t 2 + 5 t 2 + 1 = −3 + t2 + 8 t 2 + 1 ≥2 t2 4 + 2, which is equivalent to t2 + 8 t 2 + 1 ≥3 + 2 t2 4 + 2. (5) Next, if we let t = 2 sin α cos α, then the Inequality (5) can be rewritten as 4 sin2 α cos2 α + 8 sin α cos α + 1 ≥3 + 2 sin2 α cos2 α + 2, which is equivalent to (2 √ 2 + sin2 α cos2 α √ 1 + sin α cos α )2 ≥( 1 + sin2 α + √ 1 + cos2 α)2. (6) But, the Inequality (6) can be rewritten as 2 √ sin2 α cos2 α + 2 √1 + sin α cos α = 2 (1 + sin2 α)(1 + cos2 α) √1 + sin α cos α ≥ 1 + sin2 α + √ 1 + cos2 α, Jensen’s inequality versus algebra 2953 which is equivalent to 2 √1 + sin α cos α ≥ 1 √ 1 + cos2 α + 1 √ 1 + sin2 α . (7) Hence, from the Inequality (7) we deduce that (1 + sin α cos α) 1 2[(1 + cos2 α)−1 2 + (1 + sin2 α)−1 2] ≤2. Moreover, we have equality when t = 2 sin α cos α = sin 2α = 1. That is, when α = π 4 . Therefore, the maximum value of f(α) is exactly 2. 3. Questions Question 3.1. Can you think of other functions where their exact maximum (minimum) can be found using Jensen’s Inequality as well as just simple algebra? Question 3.2. Can the above problem be generalized for f(α1, α2, ..., αn) using both generalized Jensen’s Inequality and simple algebra? References M. K. Azarian, On the Fixed Points of a Function and the Fixed Points of its Composite Functions, International Journal of Pure and Applied Mathe-matics, Vol. 46, No. 1, 2008, pp. 37-44. Mathematical Reviews, MR2433713 (2009c:65129), March 2009. Zentralblatt MATH, Zbl 1160.65015. M. K. Azarian, There May be More Than One Way to Find the Deriva-tive of a Function, Missouri Journal of Mathematical Sciences, Vol. 5, No. 1, Winter 1993, pp. 13-15. Zentralblatt MATH, Zbl 1097.26508. G. H. Hardy, J. E., Littlewood, G. Polya, Inequalities, Second Edition, Cambridge University Press, Cambridge, England, 1988. 2954 M. K. Azarian I. Niven, Maxima and Minima Without Calculus, The Mathematical Association of America, Washington, D. C., 1981. Received: June, 2011
8249	https://www.cs.unc.edu/~plaisted/comp455/slides/TMdef4.1.pdf	1 Turing Machines 1.1 Introduction Turing machines provide an answer to the question, What is a computer? It turns out that anything that is equivalent in power to a Turing machine is a general purpose computer. Turing machines are a general model of computation. • They are more powerful than push-down automata. • For example, there is a Turing machine that recognizes the language {anbncn : n ≥0}. Turing machines have • a ﬁnite control, • a one-way inﬁnite tape, and • a read-write head that can move in two directions on the tape. This slight increase in power over push-down automata has dramatic conse-quences. No more powerful model of computer is known that is also feasible to construct. • This makes Turing machines very interesting because one can use them to prove problems unsolvable. • Basically, if a problem can’t be solved on a Turing machine, it can’t be solved on any reasonable computer. There are various models of Turing machines that diﬀer in various details. • The model in the text can either write a symbol or move the read-write head at each step. • The tape is also one-way inﬁnite to the right. Other Turing machine models that are common have a two-way inﬁnite tape and permit the machine to write and move on the same step. In our model, 1 • the left end of the tape is marked with a special symbol ⊲that cannot be erased. • The purpose of this symbol is to prevent the read-write head from falling oﬀthe end of the tape. Conventions used in this course: • The symbol ←means move left; the symbol →means move to the right. • The input to the Turing machine is written to the right of the ⊲marker on the tape, at the left end of the tape. • Beyond this, at the start, there are inﬁnitely many blanks on the tape. Blanks are indicated by ⊔. There may be a blank between the left-end marker and the input. • It is not speciﬁed where the read-write head starts in general, but frequently it is speciﬁed to be next to the left-end marker at the start. So the tape looks something like this: ⊲ ⊔ a1 a2 a3 a4 a5 ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ . . . 1.2 Formal Deﬁnition Formally, a Turing machine is a quintuple (K, Σ, δ, s, H) where 2 K is a ﬁnite set of states Σ is an alphabet containing ⊔and ⊲but not ←or → s ∈K is an initial state H ⊆K is a set of halting states δ is a transition function from (K −H) × Σ to K × (Σ ∪ {←, →}) non-halting scanned new symbol direction state symbol state written moved such that for all q ∈K −H, if δ(q, ⊲) = (p, b) then b =→ (must move right when a ⊲is scanned) for all q ∈K −H and a ∈Σ, if δ(q, a) = (p, b) then b ̸= ⊲ (can’t write a ⊲) 1.3 Example Turing machines M = (K, Σ, δ, s, {h}), K = {q0, q1, h}, Σ = {a, ⊔, ⊲}, s = q0. δ: q σ δ(q, σ) q0 a (q1, ⊔) see a, write ⊔ q0 ⊔ (h, ⊔) see ⊔, halt q0 ⊲ (q0, →) see ⊲, move right q1 a (q0, a) see a, switch to q0 q1 ⊔ (q0, →) read ⊔, move right q1 ⊲ (q1, →) read ⊲, move right Here’s an example computation: qo ⊲ a a a a ⊔ ⊔ . . . q1 ⊲ ⊔ a a a ⊔ ⊔ . . . q0 ⊲ ⊔ a a a ⊔ ⊔ . . . 3 q1 ⊲ ⊔ ⊔ a a ⊔ ⊔ . . . q0 ⊲ ⊔ ⊔ a a ⊔ ⊔ . . . This computation can also be written this way: (q0, ⊲aaaa ⊔⊔), (q1, ⊲⊔aaa ⊔⊔), (q0, ⊲⊔aaa ⊔⊔), (q1, ⊲⊔⊔aa ⊔⊔), (q0, ⊲⊔⊔aa ⊔⊔) It is also possible to write it without even mentioning the state, like this: ⊲aaaa ⊔⊔, ⊲⊔aaa ⊔⊔, ⊲⊔aaa ⊔⊔, ⊲⊔⊔aa ⊔⊔, ⊲⊔⊔aa ⊔⊔ 1.4 Conﬁgurations and Computations A conﬁguration of a Turing machine M = (K, Σ, δ, s, H) is a member of K × ⊲Σ∗ × (Σ∗(Σ −{⊔}) ∪{ǫ}) tape contents to rest of tape, not ending state left of read head, with blank; all blanks and scanned square indicated by ǫ Conﬁgurations can be written as indicated above, with underlining to indi-cate the location of the read-write head. • If C1 and C2 are conﬁgurations, then C1 ⊢M C2 means that C2 can be obtained from C1 by one move of the Turing machine M. • ⊢∗ M is the transitive closure of ⊢M, indicating zero or more moves of the Turing machine M. 4 • A computation by M is a sequence C0, C1, C2, . . . , Cn of conﬁgurations such that C0 ⊢M C1 ⊢M C2 . . .. It is said to be of length n. One writes C0 ⊢n M Cn. • A halting conﬁguration or halted conﬁguration is a conﬁguration whose state is in H. 1.5 Complex example Turing machines It is convenient to introduce a programming language to describe complex Turing machines. For details about this, see Handout 8. Handout 7 gives details of a Turing machine to copy a string from one place on the tape to another. We can also give the idea of a Turing machine to recognize {anbncn : n ≥ 0} by showing a computation as follows: ⊲⊔aaabbbccc ⊢ ⊲⊔aaabbbccc ⊢ ⊲⊔daabbbccc ⊢ ⊲⊔daabbbccc ⊢ ⊲⊔daabbbccc ⊢ ⊲⊔daabbbccc ⊢ ⊲⊔daadbbccc ⊢ ⊲⊔daadbbccc ⊢ ⊲⊔daadbbccc ⊢ ⊲⊔daadbbccc ⊢ ⊲⊔daadbbdcc ⊢ . . . ⊲⊔ddaddbddc ⊢ . . . ⊲⊔ddddddddd Finally the Turing machine checks that all a, b, and c run out at the same time. 5
8250	https://www.youtube.com/watch?v=exHb6TXa8Yk	If f(x) = { kx/[x], if x is less than 0 } is continuous at x=0, then the value of k is \| QUESTION 04 Prince Gupta Study 5880 subscribers 42 likes Description 2642 views Posted: 19 Jul 2023 if f(x) = { kx/[x], if x is less than 0 } is continuous at x=0, then the value of k is \| QUESTION 04 Complete Sample paper:- Maths Compartment playlist:- Maths Sample Paper Solution Playlist:- All Subject syllabus playlist:- Subscribe to our telegram channel:- @PrinceGuptaStudy Transcript: क्लास 12th की एक आपके पास फंक्शन है जो इस प्रकार से डिफाइन है एफसी = 0 और यहां पर थ्री है >= 0 तो यहां पर मैं आपको कंटिन्यू बताना है इस वाले पॉइंट पर एक्स = 0 पर तो आपको यहां पर के की वैल्यू फाइंड करनी है तो यहां पर मैं सबसे पहले क्या करता हूं यहां पर सबसे पहले मैं आपको कुछ कंडीशन बता देता हूं जो आपकी यहां पर लगेगी तो लिमिट एक्स अप अगर इसको आप कल कर देते हैं यहां पर सबसे पहले ऐसा प्रोसेस तू जीरो माइंस और यहां पर आपका जीरो से छोटे के लिए यह फंक्शन है की एक्स / माड्यूल्स ऑफ एक्स तो यहां पर हम क्या करेंगे और यहां पर यहां पर लगा देता हूं जो कोई भी आपका एक्स जो है जीरो को अप्रोच करेगा तो यहां पर जो है वो भी जीरो का अप्रोच करेगा तो यहां पर ना एक्स को जो है यहां पर मैं 0 - हा से रिप्लेस कर रहा हूं तो यहां पर एक्स मिलेगा वहां पर मैं जीरो माइंस पट कर दूंगा तो लिमिट हा अप्रोच तू जीरो और यहां पर एक्स की जगह माइंस राइट हैंड लिमिट तो यह हो जाएगा तो यहां पर ए जाएगा आपका 3 और यहां पर जो है आपका यहां पर कोई भी कोई सा भी यहां पर जो है फंक्शन टाइप का नहीं है ये ओनली आपका नंबर लिखा हुआ है तो कांस्टेंट है इसका आंसर ए जाएगा ओनली थ्री ठीक है आर चल आगे आपका थ्री अब यहां पर जो है आपको मैं एफ ऑफ जीरो निकालना के लिए बोलूंगा तो इक्वल तू जीरो के लिए भी यही वाला फंक्शन डिफाइन है तो ये इसकी वैल्यू भी ए जाएगी यहां पर से 3 तो यहां पर से आपको पता चल रहा होगा की जो आपका ठीक है तो यहां पर 3 है और यहां पर फंक्शन जो है इस पॉइंट पर वह भी थ्री है अब यहां पर आपको के की वैल्यू दिखे रही होगी के की वैल्यू आगे माइंस का 3 जो की यहां पर ऑप्शन नंबर ए में ऑप्शन नंबर आपका राइट आंसर होगा उम्मीद है जितने भी बच्चे वाली वीडियो देख रहे हैं उनको जो है ये वाला क्वेश्चन समझ में आया होगा यहां पर हमने सबसे पहले माइंस करके यहां पर आपका जो मॉडल है वो हो जाएगा जीरो माइंस और माइंस का हा बचेगा जो की हो जाएगा और जो आपका फंक्शन है वहां पर से विजिट करके आप हर एक वीडियो देख सकते हैं
8251	https://www.youtube.com/watch?v=yfbAqZJC48M	Using the quotient rule of radicals to simplify Brian McLogan 1600000 subscribers 640 likes Description 55126 views Posted: 5 Mar 2013 👉 Learn how to find the square root of rational expressions. To find the square root of a rational expression, we first express the rational expression as the square root of the numerator divided by the square root of the denominator. Next, we rationalize the rational expression by multiplying both the numerator and the denominator by the square root of the denominator. Where the rational expression can divide, we divide the numerator by the denominator and then find the square root of the resulting quotient. Where the denominator is a perfect square, we simply find its square root and simplify. 👏SUBSCRIBE to my channel here: ❤️Support my channel by becoming a member: 🙋‍♂️Have questions? Ask here: 🎉Follow the Community: Organized Videos: ✅How to Divide Radical Expressions ✅Divide by the 4th root with variables ✅Divide by the cube root with variables ✅Rationalize the denominator with the cube root ✅Rationalize the denominator with radical binomials ✅Divide the 5th root of a number ✅Divide by the cube root of a number ✅Simplify the square root of a fraction ✅Simplify the square root of a fraction with variables ✅Simplify the square root with multiple fractions ✅Divide by the square root of a number 🗂️ Organized playlists by classes here: 🌐 My Website - 🎯Survive Math Class Checklist: Ten Steps to a Better Year: Connect with me: ⚡️Facebook - ⚡️Instagram - ⚡️Twitter - ⚡️Linkedin - 👨‍🏫 Current Courses on Udemy: 👨‍👩‍👧‍👧 About Me: I make short, to-the-point online math tutorials. I struggled with math growing up and have been able to use those experiences to help students improve in math through practical applications and tips. Find more here: radicals #brianmclogan 26 comments Transcript: So now let's go and take a look at um if we had uh the division. So we talked about the product, right? We know the product we can split up and taking the product of each term that's in the radical. Well, does that work for division? Well, let's go and do some easy. Let's go and do some easy ones. What if I What about if I did the square of 36 / 9 by following what we did with the product rule? That means we could do the square of 36 over the of 9. Well, let's see what is the square root of 36. 36id 9 is the of 4, right? So therefore, this would produce two. What about the square of 36 / 9? square of 36 is 6 / of 9 which is 3 which equals 2. So guess what? When you take the square root of a rational term, you can break that up then into the square root of the numerator and the square root of the denominator. And what that brings us to now is the quotient property of radicals. And what the quotient property of radicals state is it doesn't matter if I'm taking a square root, cube root, quartic root, c um quintic root. It doesn't matter what the root is. As long what that represents is if I have m to the a over b then that equals the m root of a / the m root of b. So now this is square root. So we know that this is now the square of 9 a 5th / the square of 64 b to 4th. All right. Now, Dara, when you look at this, you can see that I can now split these up even further, right? Because this is the product rule. So, first I use the quotient rule. Now, I can use the product rule to split that up even further. So, when Madison writes this down, she has 9 9 the of a to 5th / the of 64 the of b 4th. And it's just helpful, I think, a lot of times to break it up. So then you're not like overly uh overly confused with I'm trying to do the square root or the whatever root of so many terms. So now I say the square roo of nine. Well, all right, that's three. The square root of 8 to the 5th. How do I do that? Remember, we got to write that as a squared number, right? So this one's going to get a little difficult. I Let me know if this is correct. A to the 5th is equal to a 4th time a 1st. Yes. Does everybody remember the property of exponents? Property of exponents said when you multiply two terms, what do you do with the exponents? Add them. So, does is this work? Yes. Um, what about So, all right. So therefore, I'm going to rewrite this then as the square t of a to the 4th time a all over 64, which we know is 8. And then now we got to deal with the cube or the square root of b to the 4th. And that's going to kind of take us to um the square root of a to the 4th and the square root of b to the 4th. So when looking at these values, remember guys, we need to write them since we're taking the square root, we need to write them as something squared. So could we write this as a squared and this equal to b^2 squared? Right? So therefore, you know that my answers would just be squared. However, if I broke this up again into a square a to the four the square roo of a to the 4th times the square root of a. What you notice is you can't simplify the square root of a. So therefore my final answer is going to be 3 a 2 a / 8 b^ 2 because we got the a squ and the b square from here. All right. But you can't take there's an extra a which you can't take the square root of. So we write it right there. Questions? Yes. Huh? 3 a 2 of a broke down a to the 5th as roo of a 4th time the square of a. And then I can take the a square root of a to 4th as a 2. Yes. Why did you use four? Why did I use four instead of breaking it down to what? Oh, three and two. Good question. So, the question was, why can't I break down a cub a square a square? Well, you can, but remember you can only take the square root of something when it's raised to the square power, right? So if I broke that up like that, if I was going to now take the square roots of all these square of a square is a, right? But what's the square root of a cubed? Well, then you'd have to break that up again time a where you take that and then you'd have a a square of a, which still gives you a squ a. All right, so that's another way to do it. You just need to make sure you break them all down.
8252	https://www.youtube.com/watch?v=CzmfyB1yz_8	Olympiad Geometry Problem #25: Midpoint, Perpendiculars, Parallels Michael Greenberg 4290 subscribers 19 likes Description 641 views Posted: 8 Jun 2020 Here is a beautiful problem from from 2018 Romania JBMO TST. Only a couple of lines, but it leads to a nontrivial result. Enjoy! 7 comments Transcript: hi everyone it's Michael so I have another very nice little mathematical job for you today this one was posted on the art of home selling forum by Parmenides so he posts a lot of very nice geometry problems he has his own Facebook page called romantics of geometry so if you haven't I'd highly recommend checking it out so thanks for posting this one apartment Parmenides it's from the 2018 Romania jr. balcan Math Olympiad team selection test so if you want to try to solve it feel free to pause the video all right so we have a triangle ABC or a B is not equal to AC and D is the midpoint of BC enf are the perpend are the feet of the perpendiculars from B to a B and AC M is the midpoint of EF and O is the circumcenter of ABC and we want to show that DM and a o or parallel ok so when I first saw this problem I thought of a theorem so though the line a oh it's always perpendicular to the feet of that to the line connecting the feet of that altitudes from B and C so if we drew the feet of the altitude from B to AC and we drew the foot of the altitude from C to a B and we connected the two then a o has to be perpendicular to that so that's a theorem I mentioned it in my video number eighteen which had a I'm a shortlist problem from while back when I gave a proof there so if you want you could refer to that video to see the proof but I'm going to give a little bit of a different proof in this video just to just forsake a variety so I'm gonna start out by drawing those two perpendiculars okay so I want to show that AO is perpendicular to gh and I kind of chose the strategy because so I want to show that DM and a or parallel well if I know a o is perpendicular to gh then if I can show DM is also perpendicular to G H and that would solve the trick so that was kind of what I was trying to hope hoping to do here and so here's going to be my proof that a o is perpendicular to G H so I guess no different than my last one so first I'm going to notice that since angle b g c and b HC are both 90 degrees b GH c has to be a cyclic quadrilateral right and now what i'm going to do this may seem a little random but i'm going to draw the circum circle of ABC and i'm going to draw the tangent at a going off in this direction so i is just any point on that tangent from a and then i will do a little angle chase the whole purpose of this is to show that a o is perpendicular to GH so so one thing I forgot to mention so BGH C is cyclic like I mentioned but since BD c and b HC are right angles that means bc is a diameter in the in the circle in a certain thread circle of b GH seems so since BC is the diameter and D is the midpoint B is the center of that circle alright so now I'm going to move on to the angle chase that I mentioned so we have angle Bai is equal to angle BCA and BC a is equal to angle agh and that's because since b GH c is cyclic the angle c has to be the exterior the opposite angle of b GH so Engel CSV equal angle agh so here we have angle Bai is equal to angle agh and since those are alternate interior angles AI has to be parallel to gh but so if AI is parallel to gh well we know AI is perpendicular to a oh because it's a tangent at Point a so since AI is perpendicular to AO then AO has to be perpendicular to gh so I'm gonna write that out yes that a o is perpendicular to AI so a o has to be perpendicular to gh since it's parallel to a I there's a little bit of a different proof of that fact than in my previous video but yeah so all that was just to show you this one back some people might just see right away just because they've seen it so many times okay so we've established that so now we want to show that DM is perpendicular to D H because that would that would help us show that VM and a are parallel so how do we do that well one thing to notice is we want to show DM is perpendicular to G H but D is the center of the circle through D G H and C and so the perpendicular from D to G H has to be the midpoint of Hg that's just because if you have a center of a circle which is point D in this case then the the line from the center of the circle to the midpoint of a cord so the court would be gh is always perpendicular to that cord so I'm going to construct a point J to be the midpoint of GH we know that DJ is perpendicular to G H so if we want to show that DM is perpendicular to G H then we want to show that these three are plenty ER so that's what I'm going to attempt to do okay so J is the midpoint of GH so now we're almost there so the next thing I'm going to note is clearly de has to being parallel to CG because both of those are perpendicular to the side a B so since D is the midpoint of BC E has to be the midpoint of B G so I'm going to write that out de is parallel to C G and so be e has to equal eg because BB was DC okay so b e is equal to e g but we also know that J is the midpoint of GH by construction so since GJ is JH EJ has to be parallel to be H and BH obviously has to be parallel to DF because they're both perpendicular to AC so we have je is parallel to DF and by the same argument jf is parallel to de so basically je DF is a parallelogram and now it's clear why JM and D have to be collinear because if you take two opposite vertices of a parallelogram they're colinear with the midpoint of either diagonal really that's because the the diagonals of a parallelogram bisect each other so this is just writing it out de jf is a parallelogram and so J m and D have to be collinear so I'm going to draw in that segment and now we're basically there so since DM and J are collinear like I mentioned for since D is the circumcenter of b GH c DJ has to be perpendicular to GH because the center of a circle is this if you connect it to the midpoint of a chord G H in this case the line connecting it has to be perpendicular to that core so this is just writing that out GJ equals J H so DJ has to be perpendicular to G H and now we're basically there because DJ is this DM they're the same line so since DJ is perpendicular to G H and a oh we mentioned was perpendicular to G H that DJ has to be parallel to a O and so DM has to be parallel to a o and that solves the problem so this is a fun little one so I hope you all enjoyed it if you liked the video give it a thumbs up and if you want to see more videos like this in the future feel free to subscribe my channel thanks everyone
8253	https://themathguru.quora.com/Focal-chord-of-the-parabola-y-2-4ax-is-a-line-segment-that-has-both-ends-on-the-parabola-and-passes-through-the-focal	Focal chord of the parabola y^2 = 4ax is a line segment that has both ends on the parabola and passes through the focal point of that parabola. Can you prove that the tangents of these two point are perpendicular? - The Math Guru - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In The Math Guru School Maths CBSE, ICSE, IB and Quantitative Aptitude Questions, Math shortcuts Follow · 138.6K 138.6K Haresh Sagar Studied Science&Mathematics (Graduated 1988) · 4y · Haresh Sagar Studied Science&Mathematics (Graduated 1988) · 4y · Focal chord of the parabola y^2 = 4ax is a line segment that has both ends on the parabola and passes through the focal point of that parabola. Can you prove that the tangents of these two point are perpendicular? When equation is simplified into standard vertex form \| (Y - k) ^2 = 4p(X - h) \| , what we get is, \| (Y - 0)^2 = 4a(X - 0) \|, so vertex of the parabola is at origin, variable Y is squared and 4p is positive, so parabola is horizontal and opening towards right. As 4p = 4a, p = a, so length of focal chord or latus rectum is 4a. Focus = (h+p,k), F = (a,0) Points at both the ends of focal chord, let's say above X axis is A(a,0+2a) and below X axis is B(a,0–2a), so A = (a,2a) , B = ((a,-2a). Derivative of function Y^2 = 4aX => 2YY’ = 4a => Y' = 4a/2Y Slope of tangent at A => 4a/2(2a) = 1 Slope of tangent at B => 4a/2(-2a) = - 1 As slopes of both the tangents are reciprocals of each other, they are perpendicular to each other. 1.8K views · 2 upvotes · 3 shares 88 views · View upvotes · Upvote · 9 2 About the Author Haresh Sagar Retired 2023–present Studied Science&Mathematics Graduated 1988 Lives in Rajkot, Gujarat, India 2023–present 7.1M content views 56.9K this month Active in 70 Spaces Knows English View more in The Math Guru About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
8254	https://artofproblemsolving.com/wiki/index.php/2014_UMO_Problems/Problem_5?srsltid=AfmBOopveuW036MbV_nHhg8ogQ-YJD9eUN3BKQF2kLkiK-jCtNc-0KVe	Art of Problem Solving 2014 UMO Problems/Problem 5 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2014 UMO Problems/Problem 5 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2014 UMO Problems/Problem 5 Problem Find all positive real numbers , and that satisfy both of the following equations. Solution By AM-GM Hence, the second equation implies that and . Now we plug it into the first equation to get See Also 2014 UMO (Problems • Answer Key • Resources) Preceded by Problem 4Followed by Problem 6 1•2•3•4•5•6 All UMO Problems and Solutions Retrieved from " Category: Intermediate Algebra Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
8255	https://www.chegg.com/homework-help/questions-and-answers/27-determine-ihd-molecule-formula-c6h10-give-possibilities-structure-compound-formula-c6h1-q54054233	Solved 27. Determine the IHD of a molecule having formula \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Science Chemistry Chemistry questions and answers 27. Determine the IHD of a molecule having formula C6H10. Give the possibilities in structure for a compound with a formula of C6H10.Draw two isomers and write their IUPAC name. ( 4 Marks) ttttt ttt 고 Į I Į I Į I Į I Į I 1 I 1 Section C I. Provide the structure of the major organic product in the reaction below. (7 Marks) i. CH3 + H2O H ? 11. ICH OH a Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: 27. Determine the IHD of a molecule having formula C6H10. Give the possibilities in structure for a compound with a formula of C6H10.Draw two isomers and write their IUPAC name. ( 4 Marks) ttttt ttt 고 Į I Į I Į I Į I Į I 1 I 1 Section C I. Provide the structure of the major organic product in the reaction below. (7 Marks) i. CH3 + H2O H ? 11. ICH OH a Show transcribed image text Here’s the best way to solve it.Solution Share Share Share done loading Copy link Here’s how to approach this question This AI-generated tip is based on Chegg's full solution. Sign up to see more! To determine the Index of Hydrogen Deficiency (IHD) of the molecule with the formula C 6 H 10, use the formula: I H D=C+1+N 2−H 2−X 2 Please like th… View the full answer Previous questionNext question Transcribed image text: Determine the IHD of a molecule having formula C6H10. Give the possibilities in structure for a compound with a formula of C6H10.Draw two isomers and write their IUPAC name. ( 4 Marks) ttttt ttt 고 Į I Į I Į I Į I Į I 1 I 1 Section C I. Provide the structure of the major organic product in the reaction below. (7 Marks) i. CH3 + H2O H ? 11. ICH OH a Not the question you’re looking for? Post any question and get expert help quickly. 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8256	https://www.thoughtco.com/printable-periodic-table-electronegativity-608845	Printable Periodic Table of the Elements - Electronegativity Skip to content Menu Home Science, Tech, Math Science Math Social Sciences Computer Science Animals & Nature Humanities History & Culture Visual Arts Literature English Geography Philosophy Issues Languages English as a Second Language Spanish French German Italian Japanese Mandarin Russian Resources For Students & Parents For Educators For Adult Learners About Us Search Close Search the site GO Science, Tech, Math Science Math Social Sciences Computer Science Animals & Nature Humanities History & Culture Visual Arts Literature English Geography Philosophy Issues Languages English as a Second Language Spanish French German Italian Japanese Mandarin Russian Resources For Students & Parents For Educators For Adult Learners About Us Contact Us Editorial Guidelines Privacy Policy Science, Tech, Math› Science› Chemistry› Periodic Table› Printable Periodic Table of the Elements - Electronegativity Print Science Chemistry Periodic Table Basics Chemical Laws Molecules Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Todd Helmenstine Todd Helmenstine Ph.D., Biomedical Sciences, University of Tennessee at Knoxville B.A., Physics and Mathematics, Hastings College Todd Helmenstine is a science writer and illustrator who has taught physics and math at the college level. He holds bachelor's degrees in both physics and mathematics. Learn about ourEditorial Process Updated on January 13, 2020 Close Electronegativity Table This color periodic table indicates each element's symbol, atomic number, and electronegativity. ThoughtCo / Todd Helmenstine This periodic table with electronegativity indicated can be downloaded in PDF format here. PDF format requires Adobe Acrobat Reader, which can be downloaded for free. Electronegativity jxfzsy/ Getty Images Electronegativity is used to predict whether two atoms will form ionic or covalent bonds. If the values are similar, a polar covalent bond may form. The further apart the electronegativity values, the more ionic the bond will be. Cite this Article Format mlaapachicago Your Citation Helmenstine, Todd. "Printable Periodic Table of the Elements - Electronegativity." ThoughtCo, Jun. 25, 2024, thoughtco.com/printable-periodic-table-electronegativity-608845.Helmenstine, Todd. (2024, June 25). Printable Periodic Table of the Elements - Electronegativity. Retrieved from Helmenstine, Todd. "Printable Periodic Table of the Elements - Electronegativity." ThoughtCo. (accessed September 29, 2025). copy citation Sponsored Stories North Charleston New Policy for Senior Drivers thequotegeneral.com Jeff Bezos Says the 1-Hour Rule Makes Him Smarter. I Gave It A Go.Blinkist Magazine Top Cardiologist Begs: Avoid Eating Cottage Cheese for Breakfast thehealthyfat.com What Happens When You Take 1 Shot Of Olive Oil Before Bed?enrichyourfood.com Printable Periodic Tables (PDF) Printable Periodic Tables - 2015 Edition Clickable Periodic Table of the Elements What Is the Most Electronegative Element? Chart of Periodic Table Trends Periodic Table of Element Groups Electronegativity and Chemical Bonding The Periodic Properties of the Elements Sponsored Stories Kubernetes Metrics Ebook Datadog Ask A Pro: "I'm 70 with $1.4M in IRAs. Should I convert $120K/Year to a Roth?"SmartAsset Watch More Top Cardiologist Begs: Quit Eating Blueberries Before This Happens thehealthyfat.com Here Are 29 of the Coolest Gifts for This 2025 Consumer Gift Basic Printable Periodic Table of the Elements Printable Color Periodic Table of the Elements Introduction to the Periodic Table Why Is the Periodic Table Important? Printable Periodic Table Image and Periodic Table Wallpaper How Is the Periodic Table Organized Today? What Is the Importance of Color on the Periodic Table? Periodic Table With Common Ionic Charges ThoughtCo Follow Us Science, Tech, Math Humanities Languages Resources About Us Advertise Careers Privacy Policy Editorial Guidelines Contact Terms of Service Your Privacy Choices ThoughtCo is part of the People Inc.publishing family. By clicking “Accept All Cookies”, you agree to the storing of cookies on your device to enhance site navigation, analyze site usage, and assist in our marketing efforts. Cookies Settings Accept All Cookies
8257	https://www.youtube.com/watch?v=LHxZ78zAGcc	Definitions of conic sections \| Conic sections \| Grade 11 \| Math \| Khan Academy Khan Academy India - English 549000 subscribers 2 likes Description 144 views Posted: 4 Jun 2025 This video covers the fundamental definitions of the conic sections, each described as a locus of a moving point. We'll explain the specific geometric condition that defines a circle, a parabola (involving its focus and directrix), an ellipse (involving its two foci), and a hyperbola (also involving its two foci). We will also briefly introduce the concept of eccentricity. Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now! ( Timestamps: 00:00 Locus Definition of a Circle (fixed distance from a fixed point) 00:50 Locus Definition of a Parabola (equidistant from a fixed point and a fixed line) 02:00 Locus Definition of an Ellipse (sum of distances from two fixed points is constant) 03:10 Locus Definition of a Hyperbola (difference of distances from two fixed points is constant) Khan Academy India is a nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. We have videos and exercises that have been translated into multiple Indian languages, and 15 million people around the globe learn on Khan Academy every month. Support Us: Created by Ashish Gupta Transcript: Locus Definition of a Circle (fixed distance from a fixed point) What's truly remarkable about conic sections is that there are so many different ways to define them. In the previous video, we took a cone and a plane and sliced the cone using the plane. Using the slices, we got all different conic sections. In this video, we'll consider distances. Let's first look at a circle. So, this is a circle. We have a fixed point and a fixed distance. This is the moving point. this point is always at a fixed distance from this point. The locus of this point is called a circle. So let's write that down. A circle is the locus of a point that is always at the same distance from a fixed point. So this is how we define the circle in terms of distances. This fixed point is called the center of the circle. Let's move to the next one. This Locus Definition of a Parabola (equidistant from a fixed point and a fixed line) one is called the parabola. We have the focus that's the fixed point and we have the directx that's the fixed line. Now when we move this point P on this parabola it will always be at the same distance from this fixed point as it is from this fixed line. Now remember measuring the distance of a point and a line we always take the perpendicular distance. Let's see this in action. Let's see when we move this point it's always at a fixed distance from this point F and this line directX. So let's play the animation. This is what we have. So when we move this point on the parabola, the distance from the focus and the perpendicular distance from the line remains the same. This circle denotes that these two distances are always the radi of this circle. All right. So let's define the parabola. The definition of parabola using distances is this. It's the locus of a point that is equidistant from a fixed point and a fixed line. This fixed point is called the focus and the fixed line is called the directx. Let's look at the other two, the ellipse and the hyperola. So this is Locus Definition of an Ellipse (sum of distances from two fixed points is constant) called an ellipse. We form this using these two fixed points. They're called the fosi. And what we do is we let this point move. But the condition is that the sum of distances from these two fixed points always remain constant. This distance is the radius of this circle and this distance is the radius of this circle. So when you see the animation, see that the sum of the radi of both of these circles remain constant. When one increases, the other one decreases. The increase in radius of this circle is met by the decrease in radius of this other circle. Let's see this in action. Okay. When the point moves, the sum of distances from these two fixed points, the fori remains constant. If one circle increases in size, the other one decreases. The increase in radius of one is equal to the decrease in radius of the other. So this is called an ellipse. Let's define ellipse. It's the locus of a point the sum of whose distances from two fixed points is always constant. These two fixed points are called fosi. So this is Locus Definition of a Hyperbola (difference of distances from two fixed points is constant) an ellipse. Let's move to the hyperola. For this hyperola, we again have two fixed points. They're also called FOSI. And in this case, the difference of the distances remain constant. Again, the magnitude of the difference of distances remain constant. If this distance is greater than this distance, we subtract this from this. Basically, we subtract the smaller one from the larger one. Now if the difference of distances remain constant this means the increase in radius of this circle is same as the increase in radius of this circle. Let's see this in action. Let's move this point. The increase in radius of both the circles will remain the same. This means the difference of radi of both the circles will always be a constant. When that happens we get the hyperola. So let's define the hyperbola is the locus of a point the difference of whose distances from two fixed points is always constant. These two fixed points are called the fosi. Now there's again another interesting way to define these conic sections in terms of the ratio of distances from a fixed point and a fixed line. This is called the concept of eccentricity. The way we define eccentricity is it's the ratio of the distance from a fixed point and the distance from a fixed line. So the numerator is the distance from this point and the denominator is the distance from this line. Right now the eccentricity is zero which means the distance from this point is exactly zero which means we're sitting on this point. Let's play this animation and see when this eccentricity increases what happens to our conic sections. So let's play this. Okay. So our eccentricity is now increasing. It's moving from zero. Now what you might see looks like a circle, but it's not a circle. Let it grow a little bigger. Okay. So what we see is actually an ellipse. For an ellipse, the eccentricity is always less than one, which means we're always more close to this point than this line. So this is the energy for an ellipse. It's moving more and more close to one. Okay. So this is opening up. Let's see what happens when this is exactly equal to one. Okay. All right. So when this is exactly equal to one, this ellipse finally opens up and what we get is called the parabola. Recall that the definition of the parabola is it's the locus of a point that moves in such a way that the distance from this fixed point is exactly equal to the distance from this line. When the distances are equal, the ratio will be exactly equal to one. And that's why we're seeing a parabola. But what happens when the eentricity increases from one? Let's play this. Okay, let's increase the eccentricity. So now this is opening up. It still looks like a parabola but it's not a parabola. When eenticity is more than one, what we actually have is the hyperola. So for the hyperola, the eccentricity is always greater than one. For parabola is exactly equal to one. For ellipse, it's between 0 and 1. So again another way to define conic sections using the concept of eccentricity. So let's write that down. First let's define eccentricity. It's the ratio of the distance from a fixed point and the distance from a fixed line. This is called the focus and this is called the directx. When the eccentricity is between 0 and 1, we get the ellipse. When it's exactly equal to one, we get the parabola. And when it's more than one, we get the hyperola. Now you might be wondering where's the circle. Now when eccentricity is exactly equal to zero, we got a point. So this is called a point. Now some might say that this is a circle but it's not a circle. Technically this is a point but we still say that the centricity is zero for circle. And the way we do this is by assuming that the directx this line is at infinity. Imagine that this line is very far away from this point. And the distance that we have from this point is fixed. So we're moving in a circle. For this circle the distance from this point is something finite. But the distance from this line which is very very far away is infinite. If the denominator is infinite, the eccentricity is equal to zero. And the farther this line is, the better our limit is at infinity. This is exactly equal to zero. So eccentricity is zero for a circle assuming that the directx is at infinity. It's zero for a point. It's equal to one for a parabola. It's between 0 and one for ellipse. And it's more than one for the hyperbola.
8258	https://www.storyofmathematics.com/dice-probability/	Home Dice probability – Explanation & Examples JUMP TO TOPIC [show] How to calculate dice probability: Dice probability formula: How to calculate the probability of multiple dice rolls: Multiple dice probability using tree diagrams: Multiple dice and independent events: Practice Questions: Answer Key: Dice Probability – Explanation & Examples The origins of probability theory are closely related to the analysis of games of chance. The foundations of modern probability theory can be traced back to Blaise Pascal and Pierre de Fermat’s correspondence on understanding certain probabilities associated with rolls of dice. It is no wonder then that dice probabilities play an important role in understanding the probability theory.Dice probabilities refer to calculating the probabilities of events related to a single or multiple rolls of a fair die (mostly with six sides). In a fair die, each side is equally likely to appear in any single roll.To get a better understanding of dice probabilities discussed in this article, it might be a good idea to refresh the following topics: Basic Set Theory. Basic Probability Theory. Tree Diagrams. After reading this article, you should understand the following concepts: What are dice probabilities? How to calculate dice probabilities of single/multiple rolls using sample space method. How to calculate dice probabilities of multiple rolls using the concept of independent events. How to calculate dice probabilities of multiple rolls using tree diagrams. How to calculate dice probability: To calculate dice probabilities, whether a single or multiple rolls, we first need to understand how to make sample spaces. Sample space: A sample space is the collection of all possible outcomes. For example, when we roll a six-sided fair die, there are six possible outcomes, so the sample space is given as S ={1,2,3,4,5,6}. If each outcome in the sample space is equally likely, then the probability of a single outcome is given asProbability of an outcome =1Total number of outcomes in the sample spaceSo, if we roll a die, the probability of getting any one number between 1 and 6 is equal to 16.Any subset of the sample space is called an event. For example, let us consider 𝐸 ={2,4,6}, which is a subset of sample space 𝑆 and it contains only even numbers. We can calculate the probability of an event as 𝑃⁡(𝐸) =number of elements in E Total elements in S So, the probability of getting an even number when we roll a fair die is given as𝑃⁡(getting an even number) =𝑃⁡(𝐸) =36 =12.Similarly, we calculate the probability of any event (i.e., a subset of 𝑆), as shown in the examples below:Example1: What is the probability of getting a number >4, when a fair six-sided die is rolled.Solution:We can write the sample space as S ={1,2,3,4,5,6}. Let 𝐸 be the event that number is greater than 4, then 𝐸 ={5,6}. Hence,𝑃⁡(Number >4) =𝑃⁡(𝐸) =Number of elements in ENumber of elements in S =26 =13.Example 2: What is the probability of getting a prime number when a fair six-sided die is rolled.Solution:We can write the sample space as S ={1,2,3,4,5,6}. Let 𝐸 be the event that the number is prime, then 𝐸 ={1,3,5}. Hence,𝑃⁡(PrimeNumber) =𝑃⁡(𝐸) =Number of elements in ENumber of elements in S =36 =12.Example 3:Roll a single die. Find the probability of getting an even number or a number less than 5.Solution:Let 𝐸⁢1 ={2,4,6} be the event of getting an even number. Let 𝐸⁢2 ={1,2,3,4} be the event that the number is less than 5. We are interested in the event 𝐸⁢1 OR 𝐸⁢2, remember from set theory that 𝐸⁢1 OR 𝐸⁢2 =𝐸⁢1 ∪𝐸⁢2 ={1,2,3,4,6}. We note that 𝐸⁢1 ∪𝐸⁢2 contains 5 elements, so𝑃⁢(𝐸⁢1 ∪𝐸⁢2) =56. Dice probability formula: In all experiments related to dice probabilities, we can always make a sample space 𝑆 and find the probability of any event using the formula𝑃⁡(Any event E related to single/multiple dice rolls) =Number of elements in ENumber of elements in S.Of course, when considering more than 2 dice rolls, making a sample space is cumbersome, and it is better to rely either on tree diagrams or the formula for independent events that we will explain later. How to calculate the probability of multiple dice rolls: We have seen how to calculate probabilities when a single die is rolled. Things get a bit more interesting (and a bit complex as well) when we roll two or more dice. Let’s first draw the sample space when we roll two dice together (Note: We get the same sample space whether we roll two dice together or a single die twice).As can be seen from the sample space, there are 36 possible outcomes in this case. Again, we can define an event by taking a subset of 𝑆 and calculate its probability as shown in the examples below:Example 4: Two six-sided, fair dice are rolled. What is the probability that the sum of their outcomes is greater than 10?Solution:Out of the 36 possible outcomes shown in the sample space above, there are only three outcomes for which the sum is greater than 10, i.e., 𝐸 ={(5,6),(6,5),(6,6)}. So,𝑃⁡(sum >10) =336 =112.Example 5: Two six-sided, fair dice are rolled. What are the probabilities of getting one even and one odd number?Solution:Let us collect all outcomes that contain one even and one odd number number from the sample space given above,i.e., 𝐸 ={(2,1),(4,1),(6,1),(1,2),(3,2),(5,2),(2,3),(4,3),(6,3),(1,4),(3,4),(5,4),(2,5),(4,5),(6,5),(1,6),(3,6),(5,6)}. There are 18 elements in 𝐸, so the probability is calculated as𝑃⁡(One Even and One Odd) =1836 =12.Example 6: Two six-sided dice are rolled, which is more likely to happen: the sum is equal to 10, or the sum is equal to 11?Solution:Let us collect all outcomes that sum to 10 and call it 𝐸⁢1, i.e., {(4,6),(5,5),(6,4)}. Let us collect all outcomes that sum to 11 and call it 𝐸⁢2, i.e., {(5,6),(6,5)}. Hence,𝑃⁡(Sum is 10) =336 =112.𝑃⁡(Sum is 11) =236 =119. So,𝑃⁡(Sum is 10) >𝑃⁡(Sum is 11).So, the probability that the sum is equal to 10 is more likely to happen than a sum equal to 11.Example 7: We roll two dice simultaneously. Find the probability of the following events: Getting a multiple of 5 as the sum. Getting a multiple of 2 on one die and a multiple of 3 on the other die. Solution:1.Let us collect all outcomes that are sum into multiples of 5, from the sample space given above, i.e., 𝐸 ={(1,4),(2,3),(3,2),(4,1),(4,6),(5,5),(6,4)}.There are 7 elements in 𝐸, so the probability is calculated as𝑃⁡(𝐸) =Number of elements in ENumber of elements in S =736.2.Writing all the outcomes that are a multiple of 2 on one die and a multiple of 3 on the other from the sample space shown above gives 𝐸 ={(2,3),(2,6),(3,2),(3,4),(3,6),(4,3),(4,6),(6,2),(6,3),(6,4),(6,6)}.There are 11 elements in 𝐸, so the probability is calculated as𝑃⁡(𝐸) =Number of elements in ENumber of elements in S =1136. Multiple dice probability using tree diagrams: Sometimes, the simplest method to calculate the probability of events related to multiple dice rolls is using tree diagrams. We give a few examples where the solution using tree diagrams is really straightforward; whereas, it would have been much more cumbersome to use the sample space method.Example 8: We roll a single die three times. Find the probability of the following events using a tree diagram: We don’t get a 4 in all three attempts. We get only one 4 in three attempts. Solution:F represents the fours and F’ represents not a four.The event that no four appears in all three attempts is highlighted in red in the tree diagram. We calculate the probability as follows:𝑃⁡(𝐹′⁡𝐹′⁡𝐹′) =56 ×56 ×56 =125216.There are three branches in the tree diagram (highlighted in blue) that correspond to the event that only one four appears in three attempt. The corresponding probability is calculated as𝑃⁡(One four in three attempts) =𝑃⁡(𝐹⁡𝐹′⁡𝐹′) +𝑃⁡(𝐹′⁡𝐹⁡𝐹′) +𝑃⁡(𝐹′⁡𝐹′⁡𝐹) =(56 ×56 ×16) +(56 ×56 ×16) +(56 ×56 ×16) =125216.Example 9: A single die is rolled twice. Find the probability of the following events using a tree diagram. Getting no odd number. Getting at most one odd number. Solution: We can see from the tree diagram that the probability of getting no odd number is 𝑃⁡(𝑂′⁢𝑂′⁢𝑂′) =12 ×12 ×12 =18.2. 𝑃⁡(Getting at most one odd number) =𝑃⁡(𝑂,𝑂′,𝑂′) +𝑃⁡(𝑂′,𝑂,𝑂′) +𝑃⁡(𝑂′,𝑂′,𝑂) +𝑃⁡(𝑂′,𝑂′,𝑂′) =18 +18 +18 +18 =12. Multiple dice and independent events: Two events are said to be independent if one experiment’s outcome does not affect the other event’s probabilities. For example, when we roll a die twice, both rolls are independent events as the first roll’s outcome does not affect the second roll’s probability and vice versa. Remember from the basic probability theory that when two events, say 𝐸⁢1 and 𝐸⁢2, are independent, the probability of getting 𝐸⁢1 AND 𝐸⁢2 is 𝑃⁡(E1 and E2) =𝑃⁡(𝐸⁢1) ×𝑃⁡(𝐸⁢2) . We can use the formula for probabilities of independent events to calculate probabilities of multiple rolls of dice without relying on the sample space, as we show in the following examples:Example 10: When we roll two dice simultaneously, the probability that the first roll is 2 and the second is 6.Solution:𝑃⁡(First roll is 2) =16.𝑃⁡(Second roll is 6) =16.𝑃⁡(First roll 2 and Second roll 6) =𝑃⁡(First roll is 2) ×𝑃⁡(Second roll is 6) =136.Example 11: Two six-sided, fair dice are rolled. What are the probabilities of getting one even and one odd number?Solution:We have already solved this problem using the sample space method. Now, let’s solve it using the independent probability formula.Since we are interested in one even and one odd, there are two possibilities when two dice are rolled, The first roll gives an even number and the second roll is odd The second is even, and the first is odd. We have to calculate the probability of these two events and add them to get the final probability.We have already shown how to calculate the probability of getting an even number in a single roll, i.e.,𝑃⁡(First number is even) =36 =12.Since out of six possible outcomes of a single roll of dice, there are 3 even and 3 odd outcomes, so the probability of getting an odd number is the same as getting an even number,𝑃⁡(Second number is odd) =36 =12.Since both rolls are independent, so𝑃⁡(First even AND Second odd) =𝑃⁡(First even) ×𝑃⁡(Second odd) =14.Using a similar method, we can show that𝑃⁡(first odd AND Secxond even) =14.Finally,𝑃⁡(One even and one One odd) =𝑃⁡(1st even AND 2nd odd) +𝑃⁡(1st odd AND 2nd even). =14 +14 =12.Example 12: Three six-sided, fair dice are rolled. What is the probability that none of the outcomes is an even number?Solution:We know from basic probability that 𝑃⁡(First roll is NOT even) =1–𝑃⁡(First roll is even), so𝑃(First roll is NOT even =𝑃⁡(𝐸⁢1) =1–12 =12.Similarly,𝑃⁡(Second roll is NOT even) =𝑃⁡(𝐸⁢2) =12.𝑃⁡(Third roll is NOT even) =𝑃⁡(𝐸⁢3) =12.𝑃⁡(No roll is even) =𝑃⁡(E1 AND E2 AND E3) =12 ×12 ×12 =18.Example 13: Four six-sided, fair dice are rolled. What is the probability of getting at least one even number?Solution:Let us first calculate the probability of the event 𝐸⁢2 ={No even number in four rolls}. From the previous example, we note that𝑃⁡(𝐸) =(12)4 =116.Let 𝐸⁢2 ={Atleast one even in four rolls}. Note that 𝐸⁢2 = Not 𝐸⁢1, therefore,𝑃⁡(𝐸⁢2) =1–𝑃⁡(𝐸⁢1) =1–116 =1516. Practice Questions: What is the probability of getting a number at least 5 or greater when a fair six-sided die is rolled? What is the probability of getting 1 or 5 when a fair six-sided die is rolled? We roll two dice simultaneously, what is the probability of the following events: a) getting sum divisible by 6. b) getting a total of at least 9. c) getting sum ≤ 4. d) getting a doublet of odd numbers. 4.We roll a single die three times. Find the probability of the following events using a tree diagram: a) getting an even number in all three attempts. b) getting at least two even numbers in three attempts. Answer Key: 1)13.2) 13.3) a.16.b.518.c.16.d.112.4) ‘E’ represents even numbers and E’ represents not an even number.a).18.b).38 Previous Lesson \|Main Page \| Next Lesson
8259	https://www.ijsurgery.com/index.php/isj/article/download/10400/6191/47303	International Surgery Journal \| June 2024 \| Vol 11 \| Issue 6 Page 1033 International Surgery Journal Correa -Posada MO et al. Int S urg J . 20 24Jun ;11(6): 1033 -1040 pISSN 2349 -3305 \| eISSN 2349 -2902 Re view Article Thromboangiitis obliterans: a review Martha O. Correa -Posada 1, John F. García -Velez 1, Orestes André Hurtado -Mosquera 1, Miguel A. Sierra -Juárez 2, Erick Fernando Hernández 3, Cinthia Nayeli Argüelles Castillo 4, German E. Mendoza Barrera 5, Alan I. Valderrama -Treviño 2 INTRODUCTION Buerger's disease was originally described by von Winiwarter in 1879 where he described it as TAO and detailed a case of spontaneous gangrene secondary to intimal proliferation. 1 Later, the disease was named after Leo Buerger, who in 1908 publ ished a complete pathological description based on the amputation of a group of people, including smokers, with neuropathic pain, trophic changes in the skin and annexes such as distal coldness, loss of hair, onychodystrophy, with absence of posterior pedi s and tibial pulses, lasting months to years, and which later presented with gangrene in the short term. 2,3 LITERATURE REVIEW A search was performed using the MeSH and alternative terms “TAO”, “Buerger's disease”, “MicroRNAs”, “MiRNA”, “diagnosis” “sympath ectomy” in the databases PubMed, ScienceDirect, Scopus and SpringerLink where it was initially obtained. 3646, 6138, 3296 and 2165 results respectively. Subsequently, a filter was carried out by English and Spanish language, where ABSTRACT Buerger's disease was initially described by von Winiwarter in 1879 where he described it as thromboangiitis obliterans (TAO), detailing the case of spontaneous gangrene secondary to intimal proliferation. In 1908, Leo Buerger published a complete pathological description based on the amputation of a group of people who later suffered short - term gangrene. Buerger's disease or TAO is defined as a non -atherosclerotic inflammatory vasculitis that affects the small and medium vessels (including arteries and veins) of the lower and upper extremities. Although there are hypotheses about its etiopathogenesis such as the association with exposure to tobacco, few biomedical investigations have been carried out, so its direct causality remains unknown. Diagnosis continues to be a challenge, since the findings tend to be non -specific or inconsistent with suspicion, which is why it is currently based on ruling out other causes such as atherosclerosis and vasculopathies. With current evidence, there are many types of treatments, both pharmacological and non -pharmacological, with the cessation of tobacco consumption having greater evidence of results. It is essential to strengthen basic research as well as clinical trials to standardize management in this type of patient, since it is a disease with a high impact on quality of life. Keywords: Buerguer's disease, TAO, Vasculitis, Non -atherosclerotic inflammatory vasculitis, Vasculopathies 1 Department of Angiology, Vascular and Endovascular Surgery. Vía Vascular. Medellín, Colombia 2 Department of Angiology, Vascular and Endovascular Surgery , General Hospital of Mexico “Dr. Eduardo Liceaga”, CDMX, Mexico 3 Department of General Surgery , General Hospital of Mexico “Dr. Eduardo Liceaga”. CDMX, Mexico 4 Department of Diagnostic and Therapeutic Imaging , Dr. Manuel Gea Gonzalez Hospital. CDMX, Mexico 5 Department of General Surgery, Kelsey Seybold Clinic, Houston, Texas Received: 14 April 202 4 Accepted: 29 April 202 4 Correspondence: Dr. Alan I. Valderrama -Treviño , E-mail: alan_valderrama@hotmail.com Copyright: © the author(s), publisher and licensee Medip Academy . This is an open -access article distributed under the terms of the Creative Commons Attribution Non -Commercial License, which permits unrestricted non -commercial use, distribution, and reproduction in any medium, provided the original work is properly cited. DOI: http s://dx.doi.org/ 10.18203/2349 -2902.isj 20241283 Correa -Posada MO et al. Int S urg J . 202 4 Jun ;11(6): 1033 -1040 International Surgery Journal \|June 2024 \| Vol 11 \| Issue 6 Page 1034 1919 results were found i n SpringerLink, 1674 in Scopus, 1862 in PubMed and 5871 in ScienceDirect. A review of the articles found was carried out, the most current ones were analyzed in a period from 2002 -2024 with emphasis on articles from the last 5 years, scientific articles fr om observational and analytical studies were included, with a total of 37 articles. DEFINITION Buerger's disease or TAO is defined as a non - atherosclerotic inflammatory vasculitis that affects the small and medium vessels (including arteries and veins) of the lower and upper extremities. Although there are hypotheses about its etiopathogenesis such as the association with exposure to tobacco, few biomedical investigations have been carried out, so its direct causality remains unknown. 4 EPIDEMIOLOGY Buerger 's disease has a worldwide distribution; However, a higher prevalence is observed towards the Middle East and far East than what is reported in North America and Europe. The prevalence of the disease in patients with peripheral arterial disease depends on the place where it is described, in Western Europe it varies from 0.5 -5.6% to 45 -63% in India, 16 -66% in Korea and Japan and 80% In Israel. 5 Although it was considered that the disease was almost exclusive to men under 45 years of age, for several years an increasing pattern has been observed in women, possibly related to the increase in cigarette consumption. 1 On the other hand, not many studies have been documented in Latin America on the prevalence of the disease in this region; there is a descriptive st udy from 2009 where the epidemiology of primary vasculitis in Colombia was studied and it was observed that Buerger's disease it ranked second with a prevalence of 11.2% after Takayasu disease. 6 ETIOPATHOGENESIS The etiology of Buerger's disease remains u nknown. It has been shown that it can predominantly affect young men under 45 years of age and has a high relationship with smoking and with other substances such as cannabis. 7 Cannabis has classically been linked to a particular disease called cannabis ar teritis and had previously been routinely considered a subtype of TAO because the reported cases of cannabis arteritis were mostly men without cardiovascular risk factors with similar symptoms. Buerger's disease such as claudication, Raynaud's phenomenon, absence of distal pulses, subacute distal ischemia of the lower limbs and in more advanced cases distal necrosis or gangrene of the lower limbs and venous thrombosis . However, the greatest difference is found in the anatomopathological examination of canna bis arteritis where a clear acute pattern is not found. However, Sterne and Ducastaingt did studies in late lesions where thrombosis was observed without inflammation of the vascular wall. 8,9 At the moment, four processes are recognized that are related to the pathogenesis of Buerger's disease: a secondary variant due to atherosclerosis; immunological arteritis; bacterial thrombosis; hyper -homocysteinemia; despite this, a higher percentage of patients have been seen where immune -mediated arteritis is repres entative according to immunocytochemistry reports where accumulation of immunoglobulins IgG, IgM and IgA is observed in addition to complement factor C4 in the vessel wall and the presence of antibodies. Anti -endothelial cells as markers of immune response against the endothelium, so a special focus is placed on these pathological mechanisms. 10 The existence of autoantibodies has been described in some patients, including anti -endothelial cell antibodies, anti -elastin, and anti -collagen I and III antibodies . Subsequently, the presence of antineutrophil cytoplasmic antibodies (ANCA) was reported, showing a significantly higher level of ANCA in patients with more severe clinical manifestations compared to patients with mild disease and patients without the dis ease. Additionally, high levels of anti -cardiolipin and anti -beta 2 glycoprotein antibodies were found. However, because fluctuation is observed in these levels, a diagnosis of antiphospholipid syndrome could not be established in these patients. 11 From a nother pathophysiological point, the interaction of cytokines and chemokines, adhesion molecules, Rickettsia rickettsii and Porphyryomonas gingivales , angiogenic factors, catecholamines, inflammation of lymph nodes, T lymphocytes, macrophages, dendritic ce lls and among others was proposed; as possible effectors of the response, higher concentrations of pro -inflammatory cytokines such as TNF -a, INF -gamma, IL -1, IL -6, IL -12, IL -4 were found; in patients with severe form elevations of complement C4. 12 Various studies have shown that oxidative stress favors the dysfunction of endothelial cells with a direct influence on vascular tone by decreasing the availability of nitric oxide, generating a vicious circle in the pathological process because the non -functional endothelium is a direct source of oxidative stress. Oxidative stress has been found to be significantly higher in people with OAD and smoking. On the other hand, a significant increase in intercellular adhesion molecule 1 (ICAM -1) and vascular cell adhesi on molecule 1 (VCAM -1), which are key regulators of vascular permeability, has been demonstrated in patients with TO. compared to healthy non -smokers. 11 Additionally, in a study carried out by Tamai and collaborators to see the possible participation of NO TCH signaling in the pathogenesis of Buerger's disease, it was observed that certain genetic pathways that produce activation of JAG1 and Notch1 are very active in the endothelium of the vasa vasorum and the cells of the tunica media in patients with Buerg er's disease and in some cases of aterosclerosis. 13 In 2022, a meta -analysis was carried out to assess whether plasma homocysteine concentrations had an effect on vascular damage, where they found that despite the scarcity of studies and other limitation s in the study, the data are consistent in a participation of the plasma Correa -Posada MO et al. Int S urg J . 202 4 Jun ;11(6): 1033 -1040 International Surgery Journal \| June 2024 \| Vol 11 \| Issue 6 Page 1035 homocysteine in vascular damage, however, making it clear that it is unlikely to be genetically triggered. However, they point out that the increase in oxidative stress that accompa nies Buerger's disease, whether mainly due to smoking or not, can inhibit cystathionine B-synthase, the first enzyme that catalyzes the transformation of homocysteine to cystathionine, so it can perpetuate and increase the homocysteine increasing oxida tive stress and vascular damage; Therefore, they conclude that elevated homocysteine is more an effect of the inflammation that often accompanies Buerger's disease and is less likely to be a primary cause. 14 The pathological processes of Buerger's disea se are not completely defined, but it is proposed to be multifactorial with the predominance of an immune response together with oxidative stress that triggers an inflammatory response and endothelial dysfunction with a procoagulant state (Figure 1). Figure 1: Mechanisms involved in the pathogenesis of Buerger's disease. The processes related to oxidative stress in Buerger's disease are described. MICRO -RNA AND BUERGER'S DISEASE MicroRNAs (miRNAs) have been studied in relation to different pathologies, recently they have been sought as a possible etiology in systemic vasculitis. Micro RNAs often have almost 22 nucleotides and are responsible for controlling the translation and stabilization of mRNA molecules; They do this through the interaction with proteins by targeting the 3' untranslated regions, which can affect post -transcriptional genetic silencing and in this way, an interruption in the circuit can contribute to the pathogenesis of v asculitis. 15 However, in the case of Buerger's disease the regulatory mechanisms and alterations have not yet been completely defined. In a study carried out by Chen and collaborators, the presence of several regulatory networks that can be associated with the progres sion of TAO, such as the NEAT1/miR -1- 3p/GNA 12 pathway, was demonstrated, in addition to identifying the involvement of 16 miRNAs (3 downregulated and 13 upregulated) that would also have implications in the pathogenesis of the disease. 16 In 2021, a study was carried out with the aim of exploring the effects of exosomal miRNAs associated with TAO in human vascular smooth muscle cells and the presence of 39 miRNAs was demonstrated compared to controls (10 downregulated and 29 upregulated) with particular imp ortance in miRNA 233 -5p since the results of the study suggest that the overexpression of this miRNA generates a suppression of the cell viability of vascular smooth muscle cells and, conversely, the suppression of the miRNA 233 -5p was related to greater c ell viability. 17 However, other molecules have been identified such as miRNA 100, which has an anti -inflammatory function in chronic inflammation, so the hypothesis of its possible activity in TAO was raised through a study where it was demonstrated that the overexpression of miRNA 100 manages to reduce H 2O2 apoptosis and inflammatory damage in human umbilical vein endothelial cells through the inactivation of signaling pathways that mainly involve the Notch pathway, which could be considered a therapeutic target in the future. 18 miRNAs have not only been studied as therapeutic targets, but they can also be used for diagnosis, as in a study where it was observed that miRNA 155 is overexpressed in patients with TAO, so it may have high diagnostic value, in ad dition to the implication that this molecule has in the activation of platelets and subsequent generation of thrombosis. 19 CLINICAL PRESENTATION Buerger's disease is characterized by periods of exacerbation and remission; The manifestations tend to worsen at 30 to 40 years of age with a subsequent period Immune involvement Increased homocysteine Genetic predisposition JAG1 y NOTCH1 IgM, IgG, IgA, C4, IL-6, TNF- α Vascular damage MiRNA 233-5p Bacterial thrombosis Rickettsia rickettsii Porphyryomonas gingivales Oxidative stress ON Decrease Increase in free radicals Increase ICAM1-VCAM1 Increased inflammation Endothelial dysfunction Hypercoagulability Correa -Posada MO et al. Int S urg J . 202 4 Jun ;11(6): 1033 -1040 International Surgery Journal \|June 2024 \| Vol 11 \| Issue 6 Page 1036 of decrease in symptoms and even recurrence is rarely seen in people over 60 years of age. 20 The symptoms presented by patients with Buerger's disease are essentially generated by the stenosis and occlusio n of small blood vessels; initially the involvement occurs in small distal arteries and veins, then as the disease progresses it involves proximal arteries (Table 1). 21 When the occlusion is marked, it produces ischemia of the upper and lower extremities, this represented as intermittent claudication initially in the arch of the foot, which is usually an early sign and as it progresses, it compromises the calf until it generates pain at rest, Raynaud's phenomenon, skin color changes, tingling, numbness of t he fingers and toes, and ultimately ulcers and gangrene. 22,23 In severe cases, amputation may be necessary although compared to patients with atherosclerosis it is much less. Death in patients with Buerger's disease is very unusual and is not an expected complication. 24 Patients often present superficial thrombophlebitis, it has been reported that up to 40 -60% of cases are migratory, tend to be recurrent and affect arms and legs; In young patients this manifestation is highly suggestive of TAO. Systemic man ifestations are not frequently observed in patients with TAO, except for some rheumatic manifestations such as non -erosive arthritis. 25 Although it is not the most frequent, because the disease affects the blood vessels in a general way, involvement of the visceral vessels has been described, producing a varied clinical picture depending on the affected bed, generating various symptoms such as abdominal pain, mesenteric ischemia, chest pain, heart attack. acute myocardium, coronary artery stenosis, hemipare sis, aphasia, hemianopsia, seizures, retinal artery involvement, gangrenous glans of the penis, erythema nodosum in extensors, among others. 26 The main manifestations reported in different studies are summarized in Table 1. 20 -23 Table 1: Clinical manifestations of Buerguer's disease. Study Clinical manifestations Olin et al 21 Initially claudication of extremities, ischemic ulcerations in the distal portion of the fingers. Routinely it almost always involves two or more limbs. Małecki et al 20 Intermittent claudication, superficial thrombophlebitis, paresthesias, pain at rest, necrosis, ulceration and Raynaud's phenomenon. Important to take into account an atypical course. Vijayakumar et al 22 Signs of arterial insufficiency in the extremities such as intermittent claudication, pain at rest, ulcers or migratory thrombophlebitis. Joint symptoms prior to the classic ones. Fakour et al 27 In addition to the signs of ischemia, they present diffuse abdominal pain, weight loss, hemiparesis, aphasia, cognitive impairment, headache, retinal artery involvement, among others. Szydełko -Paśko et al 23 Intermittent claudication of the upper and lower extremities, paresthesias, pain at rest, change in skin color up to ulcers and gangrene. Varied manifestations in other organs such as gastrointestinal tract, heart, central nervous system, eye, kidneys, among others. DIAGNOSIS The diagnosis is primarily clinical based on the history, anamnesis regarding clinical manifestations, and a directed physical examination. Now there are no cost - effective and simple tests that guide the diagnosis, since biochemical alterations are usually non -specific, so the request for tests such as s erum creatinine, blood glucose, CRP, ESR and certain antibodies such as ANCA and anticentromere have more value in ruling out differential diagnoses. 13 It has been described that patients with high anticardiolipin titers are mainly young and report a highe r rate of amputations. 11 Malowski and collaborators in a group of 47 patients evaluated between 1990 -1996 with Buerger's disease reported high titers of anticardiolipin antibodies and with the association that the greater their presence there could be a gr eater tendency to produce thrombosis and therefore more requirements for amputation .21 However, this is still a topic under study, so its diagnostic and prognostic value remains in doubt. The initial step is usually to exclude atherosclerosis or risk facto rs for other occlusive vasculopathies; The initial image is usually a Doppler ultrasound as it details the level of occlusion, luminal occlusion and quality of blood flow. 13 Subsequently, tomography or magnetic resonance arteriography can be considered, wh ere lesions have classically been described as corkscrew -shaped collaterals defined as Martorell's sign which denotes changes in the vasa vasorum, presence of segmental lesions or can also be found in occlusions of the distal extremity. 5 For several years, several criteria have been proposed to suspect the disease, such as the Shionoya, Olin, Papa and Mills criteria, (Table 2). 26 However, the most accepted in countries where the disease is more prevalent are the Shionoya criteria, which are preferred in pat ients where the use of images is not cost -effective. To make the diagnosis, the presence of all 5 criteria is required; however, if all the criteria are not met, the diagnosis can be made if other diseases are excluded and the imaging findings in addition to pathology are consistent with the disease. 28 Biopsy is not usually indicated initially, only in cases in which the patient presents atypical symptoms, is very old, or when involvement of large arteries is suspected; In this case, the histopathological f indings include the presence of an inflammatory thrombus with polymorphonuclear cells and multinucleated giant cells with involvement of arteries and veins. 5Correa -Posada MO et al. Int S urg J . 202 4 Jun ;11(6): 1033 -1040 International Surgery Journal \| June 2024 \| Vol 11 \| Issue 6 Page 1037 Table 2: Shionoya and Olin diagnostic criteria for Buerguer's disease. S. no. Shionoya clinical criteria Olin criteria Smoking history Current or past tobacco use Onset before age 50 years Onset before age 45 years Infrapopliteal arterial occlusions Distal limb ischemia (infrapopliteal or intrabrachial), such as claudication, pain at rest, ischemic ulcers, and gangrene documented with non -invasive testing) Laboratory tests to rule out autoimmune or connective tissue diseases and diabetes mellitus Upper limb involvement or migratory thrombophlebitis Exclusion of proximal source of emboli with echocardiography and arteriography Absence of cardiovascular risk factors other than smoking Demonstrate consistent arteriographic findings in affected and clinically uninvolved extremities Figure 2: Diagnosis of Buerguer's disease. Consider arterial or venous vessel biopsy in atypical presentation, very advanced age or involvement of large arteries. Clinical presentation •Male patient •<65 years •Smoking history •Intermittent claudication •Pain at rest •Raynaud's phenomenon •Paresthesia •Ulcers on fingers •Gangrene Suspected Buerger's disease Rule out differential diagnoses Request lab tests and images: Serum creatinine Blood glucose CRP, ESR. ANAs, ANCA, anticentromere, anticardiolipin Doppler ultrasound or tomography or resonance arteriography Arterial or venous vessel biopsy Positive findings for another disease? No Discard diagnosis Make a diagnosis based on Shionoya clinical criteria and findings on imaging or biopsy •Atherosclerosis •Popliteal impingement syndrome •Systemic lupus erythematosus •Behcet's disease. Start non-pharmacological treatment: cessation of tobacco consumption and adequate wound management. Yes Start pharmacological treatment: Aspirin, Cilostazol, Iloprost. Surgical treatment: revascularization or sympathectomy No improvement Correa -Posada MO et al. Int S urg J .202 4Jun ;11(6): 1033 -1040 International Surgery Journal \|June 2024 \| Vol 11 \| Issue 6 Page 1038 DIFFERENTIAL DIAGNOSIS Once Buerger's disease is considered, the differential diagnosis is not very broad, so diseases such as atherosclerosis obliterans, popliteal impingement syndrome and systemic vasculopathies such as systemic lupus erythematosus (SLE) and kidney disease must be ruled out. Behcet. 13 Atherosclerosis obliterans is defined as a chronic arterial disease that produces symptoms of ischemia such as intermittent claudication, pain at rest, tissue loss secondary to obstruction of the arteries secondary to atherosclerosis, for which risk factors are often pres ent .29 Thus, the presence of risk factors or evidence of an atheroma usually excludes TAO. Popliteal artery entrapment syndrome occurs more in young, athletic people without atherosclerotic risk factors; It is defined as a disorder in which the popliteal a rtery is compressed by aberrant muscular and tendon structures in the popliteal fossa, which causes intermittent claudication, pain in the feet and calves, mainly after performing physical activity and which resolves when resting or changing positions. 30 SLE is a multisystem autoimmune disease with a multifactorial origin of genetic, sociodemographic, and infectious factors. It is more prevalent in young women and the symptoms are very varied with a significant deterioration in quality of life. 31 Behcet's d isease is an inflammatory disorder in which patients present with skin symptoms, oral thrush, ophthalmic lesions, and genital ulcers. Occasionally, they may present vascular alterations such as deep vein thrombosis and arterial occlusion, which is why they may also present ischemic symptoms. 28 TREATMENT The fundamental basis of the treatment and the one that has had the best results is the suspension of tobacco consumption, although this alone has not shown evidence of being the solution in all cases. Diffe rent pharmacological treatments have been proposed, however, most of them have been inefficient, so the functioning of pharmacological treatment with anticoagulants, thrombolytics, vasodilators and anti - inflammatories is still questioned. 2 Among the different pharmacological treatment options, aspirin from the class of non -steroidal anti -inflammatory drugs is described, which blocks the production of the thromboxane A2 pathway, which generates an inhibition of platelets and reduces thrombos is. Cilostazol is usually indicated initially since it acts as a phosphodiesterase inhibitor, which has an additional particular mechanism of action, since it is proposed to reduce the expression of IL -1B, IL - 6 and TNF -a, which would have a double effect b y impacting its pathophysiology and on the other hand increasing cAMP in platelets and endothelial cells; ileoprost is a prostacyclin I2 analogue that generates a vasodilatory effect and is usually indicated in patients with severe ischemia of the lower ex tremities. 32 In a recent publication in Cochrane, they evaluated how effective intravenous and oral therapies could be, where they found that when the use of aspirin and intravenous ileoprost were compared, effective results were obtained in healing ulcers and eradicating pain at rest after 28 days of treatment; However, there was no effect on the amputation rate 6 months after treatment was instituted. 33 Due to the above, other arterial revascularization therapies were proposed, whether surgical or endovas cular, sympathectomy and the mobilization and implantation of bone marrow cells, which is usually indicated mainly in the most serious cases. Revascularization therapies are indicated when the patient presents symptoms despite having instituted non - pharmac ological and pharmacological treatment. For its part, open surgery is considered a standard strategy to generate revascularization in patients with TAO and advanced limb ischemia; however, its biggest problem is bypass occlusion. 34 Because TAO involves dif ferent pathophysiological mechanisms than atherosclerosis, endovascular therapy has been proposed as one of the best alternatives through angioplasty with advantages in terms of high safety, no in -hospital mortality reported, low rate of perioperative comp lications and late results such as vascular permeability which favored a lower rate of limb amputations. 35 Sympathectomy is indicated in patients who have symptomatic ulcers in the fingers of the extremities and revascularization is not possible. 28 DISCUS SION Buerger's disease is a vasculitis of small and medium vessels that has been very little studied and still has little clarity regarding etiology and treatment. It has been seen that it may predominate more in 45 -year -old men from the East and smoking is a possible trigger; however, other factors such as cannabis consumption and genetic factors have been raised. 14,3 2 Diagnosis continues to be a challenge, since the findings tend to be non -specific or inconsistent with suspicion, which is why it is curre ntly based on ruling out other causes such as atherosclerosis and vasculopathies. 11 With current evidence, there are many types of treatments, both pharmacological and non - pharmacological, with the greatest evidence of results being the suspension of tobac co consumption. 2 The prognosis tends to be good for patients, although it has been reported that up to 70% of patients will experience an ischemic ulcer or necrosis. In some long -term studies it was reported that between 2.7 and 10.5% required major limb a mputations; However, in a multicenter study of 224 patients, it was found that 34% of patients may require amputation 15 years after diagnosis. 28,3 6 A flow chart is proposed as an algorithm regarding suspicion and treatment (Figure 2). CONCLUSION Currently, since there are no definitive and specific management guidelines or protocols for Buerger's disease regarding its diagnosis and treatment, it is essential to strengthen basic research as well as clinical trials to Correa -Posada MO et al. Int S urg J . 202 4 Jun ;11(6): 1033 -1040 International Surgery Journal \|June 2024 \| Vol 11 \| Issue 6 Page 1039 standardize management in this type of patient, since it is a disease with high impact on quality of life. Funding: No funding sources Conflict of interest: None declared Ethical approval: Not required REFERENCES Conde ID , Peña C. Buerger’s Disease Tromboangiitis Obliterans. Techn Vascu lar Interventional Radiol. 2014;17(4):234 -40. Ribieras AJ, Ortiz YY, Liu ZJ, Velazquez OC. Therapeutic angiogenesis in Buerger’s disease: reviewing the treatment landscape. Ther Adv Rare Dis. 2022;30:3:26330040211070295. Rondón -Carvajal J, Guzmán -Arango C, Cano - Márquez D. Enfermedad de Buerger o el endotelio como laberinto: a propósito de un caso clínico. Med Lab. 2023;27(3):211 -21. Olin JW. Thromboangiitis Obliterans: 110 Years Old and Little Progress Made. J Am Heart Assoc. 2018;7(23):e011214. Rivera -Chav arría IJ, Brenes -Gutiérrez JD. Thromboangiitis obliterans (Buerger's disease). Ann Med Surg. 2016;7:79 -82. Ochoa CD, Ram í rez F, Quintana G, Toro C, Ca ñ as C, Osio LF, et al. Epidemiología de las vasculitis primarias en Colombia y su relación con lo inform ado para Latinoamérica. Revista Colombiana Reumatol. 2009;16(3):248 -63. Liu Z, Ning W, Liang J, Zhang T, Yang Q, Zhang J, et al. Top 100 cited articles in the thromboangiitis obliterans: a bibliometric analysis and visualized study. Eur J Med Res. 2023;28( 1):551. Pilitsi E, Kennamer B, Trepanowski N, Gonzalez R, Trojanowski M, Phillips T, et al. Cannabis arteritis presenting with Raynaud’s and digital ulcerations: a case‐based review of a controversial thromboangiitis obliterans‐like condition. Clin Rheumat ol. 2023;42(7):1981 -5. Banana Y, Bashir H, Boukabous S, Rezziki A, Benzirar A, El Mahi O. Cannabis arteritis: A case report and brief review of the literature. Ann Med Surg. 2022;76:103523. Shapouri -Moghaddam A, Saeed Modaghegh MH, Reza Rahimi H, Ehteshamfar SM, Tavakol Afshari J. Molecular mechanisms regulating immune responses in thromboangiitis obliterans: A comprehensive review. Iran J Basic Med Sci. 2019;22(3):215 -24. Fazeli B, Ligi D, Keramat S, Maniscalco R, Sharebiani H, Mannello F. Recent Updates and Advances in Winiwarter -Buerger Disease (Thromboangiitis Obliterans): Biomolecular Mechanisms, Diagnostics and Clinical Consequences. Diagnostics. 2021;11(10):1736. Sharebiani H, Fazeli B, Maniscalco R, Ligi D, Mannello F. The Imbalance among Oxidative Biomarkers and Antioxidant Defense Systems in Thromboangiitis Obliterans (Winiwarter -Buerger Disease). J Clin. Med. 2020;9(4):1036. Kumar Khanna A, Zeeshan Hakim M. The Management of Ischemic Limb in Thromboangiitis Obliterans (Buerger’s Disease). Indian J Surg. 2021;83:836 -44. Merashli M, Bucci T, Pastori D, Pignatelli P, Arcaro A, Gentile F, et al. A Meta -Analysis of Plasma Homocysteine in Buerger’s Disease. Thromb Haemost. 2022;122(7):1243 -6. Tang X, Guo J, Qi F, Rezaei MJ. Role of non -coding RN As and exosomal non -coding RNAs in vasculitis: A narrative review. Int J Biol Macromol. 2024;261(2):129658. Chen B, Deng Y, Wang B, Tian Z, Tong J, Yu B, et al. Integrated analysis of long non -coding RNA - microRNA -mRNA competing endogenous RNAregulatory net works in thromboangiitis Obliterans. Bioengineered. 2021;12(2):12023 -37. Deng Y, Tong J, Shi W, Tian Z, Yu B, Tang J. Thromboangiitis obliterans plasma -derived exosomal miR -223 -5p inhibits cell viability and promotes cell apoptosis of human vascular smooth muscle cells by targeting VCAM1. Ann Med. 2021;53(1):1129 -41. Wang C, Zhang Y, Jiang Z, Bai H, Du Z. miR -100 alleviates the inflammatory damage and apoptosis of H2O2-induced human umbilical vein endothelial cells via inactivation of Notch signaling by tar geting MMP9. Vascular. 2022;30(1):151 -61. Dai X, Che X. miR -155 for diagnosing thromboangiitis obliterans and its effect on platelet function: a clinical study. Int J Clin Exp Med. 2020;13(2):1277 -84. Małecki R, Zdrojowy R, Adamiec R. Thromboangiitis oblit erans in the 21 st century --a new face of disease. Atherosclerosis. 2009;206(2):328 -34. Olin JW, Shih A. Thromboangiitis obliterans (Buerger's disease). Curr Opin Rheumatol. 2006;18(1):18 -24. Vijayakumar A, Tiwari R, Kumar Prabhuswamy V. Thromboangiitis Obl iterans (Buerger's Disease) - Current Practices. Int J Inflam. 2013;2013:156905. Szydełko -Paśko U, Przeździecka -Dołyk J, Małecki R, Szuba A, Misiuk -Hojło M. Ocular Manifestations of Buerger’s Disease -A Review of Current Knowledge. Clin Ophthalmol. 2022;16:851 -60. Komai H. Thromboangiitis Obliterans -A Disappearing Disease? Circ J. 2024;88(3):329 -30. Puéchal X, Fiessinger JN. Thromboangiitis obliterans or Buerger's disease: challenges for the rheumatologist. Rheumatology. 2007;46(2):192 -9. Lazarides MK, Georgiadis GS, Papas TT, Nikolopoulos ES. Diagnostic criteria and treatment of Buerger's disease: a review. Int J Low Extrem Wounds. 2006;5(2):89 -95. Fakour F, Fazeli B. Visceral bed involvement in thromboangiitis obliterans: a systematic review . Vasc Health Risk Manag. 2019;15:317 -53. Isobe M, Amano K, Arimura Y, Ishizu A, Ito S, Kaname S, et al. JCS 2017 Guideline on Correa -Posada MO et al. Int S urg J . 202 4 Jun ;11(6): 1033 -1040 International Surgery Journal \|June 2024 \| Vol 11 \| Issue 6 Page 1040 Management of Vasculitis Syndrome -Digest Version. Circ J. 2020;84(2):299 -359. Pu H, Huang Q, Zhang X, Wu Z, Qiu P, Jiang Y, et al. A met a-analysis of randomized controlled trials on therapeutic efficacy and safety of autologous cell therapy for atherosclerosis Obliterans. J Vasc Surg. 2022;75(4):1440 -49. Bradshaw S, Habibollahi P, Soni J, Kolber M, Pillai AK. Popliteal artery entrapment sy ndrome. Cardiovasc Diagn Ther. 2021;11(5):1159 -67. Bastidas Goyes AR, Mora C, Martín Arsanios D, Orduz K. Systemic lupus erythematosus, a disease conditioned by the environment. Rev Colomb Reumatol. 2021;28(1):12 -20. Chen Q, Chen J, Li J, Cheng Y, Zhang R, Liu Z. Recent advances of oxidative stress in thromboangiitis obliterans: biomolecular mechanisms, biomarkers, sources and clinical applications. Thromb Res. 2023;230:64 -73. Cacione DG, Macedo CR, Carmo Novaes FD, Baptista -Silva JCC. Pharmacological treatment for Buerger's disease. Cochrane Database Syst Rev. 2020;5(5):CD011033. Frota Carneiro FC, Almeida BM, Cacione DG. Endovascular treatment for thromboangiitis obliterans (Buerger’s disease). Cochrane Database Syst Rev. 2023;2023(1):CD014886. Galyfo s G, Liakopoulos D, Chamzin A, Sigala F, Filis K. A systematic review and meta -analysis of early and late outcomes after endovascular angioplasty among patients with thromboangiitis obliterans and chronic limb ischemia. J Vasc Surg. 2023;77(5):1534 -41. Le Joncour A, Soudet S, Dupont A, Espitia O, Koskas F, Cluzel P, et al. Long -Term Outcome and Prognostic Factors of Complications in Thromboangiitis Obliterans (Buerger's Disease): A Multicenter Study of 224 Patients. J Am Heart Assoc. 2018;7(23):e010677. Cite this article as: Correa -Posada MO , García - Velez JF , Hurtado -Mosquera OA , Sierra -Juárez MA , Hernández EF , Castillo CAN, et al. Thromboangiitis obliterans: a review . Int Surg J 20 24;11:1033 -40 .
8260	https://byjus.com/neet/krebs-cycle/	Cellular RespirationStepsProductsSignificanceFrequently Asked Questions Introduction The Krebs cycle or TCA cycle (tricarboxylic acid cycle) or Citric acid cycle is a series of enzyme catalysed reactions occurring in the mitochondrial matrix, where acetyl-CoA is oxidised to form carbon dioxide and coenzymes are reduced, which generate ATP in the electron transport chain. Download Complete Chapter Notes of Respiration in Plants Download Now Also see: NEET Key Answer 2022 Krebs cycle was named after Hans Krebs, who postulated the detailed cycle. He was awarded the Nobel prize in 1953 for his contribution. It is a series of eight-step processes, where the acetyl group of acetyl-CoA is oxidised to form two molecules of CO2 and in the process, one ATP is produced. Reduced high energy compounds, NADH and FADH2 are also produced. Two molecules of acetyl-CoA are produced from each glucose molecule so two turns of the Krebs cycle are required which yields four CO2, six NADH, two FADH2 and two ATPs. Krebs Cycle is a part of Cellular Respiration Cellular respiration is a catabolic reaction taking place in the cells. It is a biochemical process by which nutrients are broken down to release energy, which gets stored in the form of ATP and waste products are released. In aerobic respiration, oxygen is required. Also see: Biochemical Pathways Cellular respiration is a four-stage process. In the process, glucose is oxidised to carbon dioxide and oxygen is reduced to water. The energy released in the process is stored in the form of ATPs. 36 to 38 ATPs are formed from each glucose molecule. The four stages are: 1. Glycolysis: Partial oxidation of a glucose molecule to form 2 molecules of pyruvate. This process takes place in the cytosol. Further reading: Significance of Glycolysis 2. Formation of Acetyl CoA: Pyruvate formed in glycolysis enters the mitochondrial matrix. It undergoes oxidative decarboxylation to form two molecules of Acetyl CoA. The reaction is catalysed by the pyruvate dehydrogenase enzyme. 3. Krebs cycle (TCA cycle or Citric Acid Cycle): It is the common pathway for complete oxidation of carbohydrates, proteins and lipids as they are metabolised to acetyl coenzyme A or other intermediates of the cycle. The Acetyl CoA produced enters the Tricarboxylic acid cycle or Citric acid cycle. Glucose is fully oxidized in this process. The acetyl CoA combines with 4-carbon compound oxaloacetate to form 6C citrate. In this process, 2 molecules of CO2 are released and oxaloacetate is recycled. Energy is stored in ATP and other high energy compounds like NADH and FADH2. 4. Electron Transport System and Oxidative Phosphorylation: ATP is generated when electrons are transferred from the energy-rich molecules like NADH and FADH2, produced in glycolysis, citric acid cycle and fatty acid oxidation to molecular O2 by a series of electron carriers. O2 is reduced to H2O. It takes place in the inner membrane of mitochondria. Also Check: MCQs on Krebs Cycle Krebs Cycle Steps It is an eight-step process. Krebs cycle or TCA cycle takes place in the matrix of mitochondria under aerobic condition. Step 1: The first step is the condensation of acetyl CoA with 4-carbon compound oxaloacetate to form 6C citrate, coenzyme A is released. The reaction is catalysed by citrate synthase. Step 2: Citrate is converted to its isomer, isocitrate. The enzyme aconitase catalyses this reaction. Step 3: Isocitrate undergoes dehydrogenation and decarboxylation to form 5C 𝝰-ketoglutarate. A molecular form of CO2 is released. Isocitrate dehydrogenase catalyses the reaction. It is an NAD+ dependent enzyme. NAD+ is converted to NADH. Step 4: 𝝰-ketoglutarate undergoes oxidative decarboxylation to form succinyl CoA, a 4C compound. The reaction is catalyzed by the 𝝰-ketoglutarate dehydrogenase enzyme complex. One molecule of CO2 is released and NAD+ is converted to NADH. Step 5: Succinyl CoA forms succinate. The enzyme succinyl CoA synthetase catalyses the reaction. This is coupled with substrate-level phosphorylation of GDP to get GTP. GTP transfers its phosphate to ADP forming ATP. Step 6: Succinate is oxidised by the enzyme succinate dehydrogenaseto fumarate. In the process, FAD is converted to FADH2. Step 7: Fumarate gets converted to malate by the addition of one H2O. The enzyme catalysing this reaction is fumarase. Step 8: Malate is dehydrogenated to form oxaloacetate, which combines with another molecule of acetyl CoA and starts the new cycle. Hydrogens removed, get transferred to NAD+ forming NADH. Malate dehydrogenase catalyses the reaction. Krebs Cycle Summary Location: Krebs cycle occurs in the mitochondrial matrix Krebs cycle reactants: Acetyl CoA, which is produced from the end product of glycolysis, i.e. pyruvate and it condenses with 4 carbon oxaloacetate, which is generated back in the Krebs cycle Krebs cycle products Each citric acid cycle forms the following products: 2 molecules of CO2 are released. Removal of CO2 or decarboxylation of citric acid takes place at two places: In the conversion of isocitrate (6C) to 𝝰-ketoglutarate (5C) In the conversion of 𝝰-ketoglutarate (5C) to succinyl CoA (4C) 1 ATP is produced in the conversion of succinyl CoA to succinate 3 NAD+ are reduced to NADH and 1 FAD+ is converted to FADH2 in the following reactions: Isocitrate to 𝝰-ketoglutarate → NADH 𝝰-ketoglutarate to succinyl CoA → NADH Succinate to fumarate → FADH2 Malate to Oxaloacetate → NADH Note that 2 molecules of Acetyl CoA are produced from oxidative decarboxylation of 2 pyruvates so two cycles are required per glucose molecule. To summarize, for complete oxidation of a glucose molecule, Krebs cycle yields 4 CO2, 6NADH, 2 FADH2 and 2 ATPs. Each molecule of NADH can form 2-3 ATPs and each FADH2 gives 2 ATPs on oxidation in the electron transport chain. Krebs cycle equation To Sum up Significance of Krebs Cycle Krebs cycle or Citric acid cycle is the final pathway of oxidation of glucose, fats and amino acids Many animals are dependent on nutrients other than glucose as an energy source Amino acids (metabolic product of proteins) are deaminated and get converted to pyruvate and other intermediates of the Krebs cycle. They enter the cycle and get metabolised e.g. alanine is converted to pyruvate, glutamate to 𝝰-ketoglutarate, aspartate to oxaloacetate on deamination Fatty acids undergo 𝞫-oxidation to form acetyl CoA, which enters the Krebs cycle It is the major source of ATP production in the cells. A large amount of energy is produced after complete oxidation of nutrients It plays an important role in gluconeogenesis and lipogenesis and interconversion of amino acids Many intermediate compounds are used in the synthesis of amino acids, nucleotides, cytochromes and chlorophylls, etc. Vitamins play an important role in the citric acid cycle. Riboflavin, niacin, thiamin and pantothenic acid as a part of various enzymes cofactors (FAD, NAD) and coenzyme A Regulation of Krebs cycle depends on the supply of NAD+ and utilization of ATP in physical and chemical work The genetic defects of the Krebs cycle enzymes are associated with neural damage As most of the biological processes occur in the liver to a significant extent, damage to liver cells has a lot of repercussions. Hyperammonemia occurs in liver diseases and leads to convulsions and coma. This is due to reduced ATP generation as a result of the withdrawal of 𝝰-ketoglutarate and formation of glutamate, which forms glutamine \| \| \| Important Concepts of Biology for NEET \| \| Neural Communication \| Tetany Causes \| What is Cretinism? \| \| Multinodular Goitre \| Plant Hormones \| Thyroid Symptoms \| \| Thyroid Problems \| Types of Fermentation \| What is Bioprocessing? \| \| Algae Bloom \| What Is Dwarfism? \| Biochemical Pathways \| \| Genetic Diversity \| Electrocardiogram \| Uricotelism Definition \| \| Polyarteritis Nodosa \| Hydroponic System \| Blue Green Algae \| Frequently Asked Questions on Krebs Cycle Q1 What is the Krebs Cycle? Also known as the citric acid cycle, the Krebs cycle or TCA cycle is a chain of reactions occurring in the mitochondria, through which almost all living cells produce energy in aerobic respiration. It uses oxygen and gives out water and carbon dioxide as products. Here, ADP is converted into ATP. This cycle renders electrons and hydrogen required for electron chain transport. Q2 How Many ATPs are Produced In the Krebs Cycle? 2 ATPs are produced in one Krebs Cycle.For complete oxidation of a glucose molecule, the Krebs cycle yields 4 CO2, 6NADH, 2 FADH2 and 2 ATPs. Q3 Where Does Krebs Cycle or TCA cycle Occur? Mitochondrial matrix.In all eukaryotes, mitochondria are the site where the Krebs cycle takes place. The cycle takes place in a mitochondrial matrix producing chemical energy in the form of NADH, ATP, FADH2. These are produced as a result of oxidation of the end product of glycolysis – pyruvate. Q4 How The Krebs Cycle Works? It is an eight-step process1) Condensation of acetyl CoA with oxaloacetate (4C) forming citrate (6C), coenzyme A is released. 2) Conversion of Citrate to its isomer, isocitrate. 3) Isocitrate is subjected to dehydrogenation and decarboxylation forming 𝝰-ketoglutarate (5C). 4) 𝝰-ketoglutarate (5C) experiences oxidative decarboxylation forming succinyl CoA (4C). 5) Conversion of Succinyl CoA to succinate by succinyl CoA synthetase enzyme along with substrate-level phosphorylation of GDP forming GTP. 6) Oxidation of Succinate to fumarate by the enzyme succinate dehydrogenase. 7) Fumarate gets converted to malate by the addition of one H2O. 8) Malate is dehydrogenated to form oxaloacetate, which combines with another molecule of acetyl CoA and starts the new cycle. Q5 Why Is Krebs Cycle Called As Amphibolic Pathway? It is called amphibolic as in the Krebs cycle both catabolism and anabolism take place. The amphibolic pathway indicates the one involving both catabolic and anabolic procedures. Q6 How Many NADH are Produced In The Krebs Cycle? 3 NADH molecules In one turn of the Krebs cycle, 3 molecules of NADH are produced.For complete oxidation of a glucose molecule, Krebs cycle yields 4 CO2, 6NADH, 2 FADH2 and 2 ATPs. Q7 What Is The Krebs Cycle Also Known As? Krebs cycle is also known as Citric acid cycle (CAC) or TCA cycle (tricarboxylic acid cycle) Q8 Why Krebs Cycle Is Called the Citric Acid Cycle? Krebs cycle is also referred to as the Citric Acid Cycle. Citric acid is the first product formed in the cycle. Also Check: NEET Flashcards: Respiration In Plants NEET Flashcards: Breathing And Exchange Of Gases NEET Flashcards: Body Fluids And Circulation NEET Flashcards: Neural Control And Coordination NEET Flashcards: Chemical Coordination And Integration Recommended Video: Test your Knowledge on Krebs Cycle! Q5 Put your understanding of this concept to test by answering a few MCQs. 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8261	https://en.wikipedia.org/wiki/Equivalence_principle	Jump to content Equivalence principle العربية Беларуская Български Català Чӑвашла Čeština Dansk Deutsch Ελληνικά Español Euskara فارسی Français Gaeilge Galego 한국어 Հայերեն Ido Igbo Bahasa Indonesia Italiano עברית Қазақша Magyar Nederlands 日本語 Norsk bokmål Norsk nynorsk ਪੰਜਾਬੀ Polski Português Romnă Русский Slovenčina Slovenščina Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska Татарча / tatarça Türkçe Українська Tiếng Việt 中文 Edit links From Wikipedia, the free encyclopedia Hypothesis that inertial and gravitational masses are equivalent This article is about the principle in gravitation. For the principle in electromagnetism, see surface equivalence principle. \| General relativity \| \| \| \| Introduction + History + Timeline + Tests Mathematical formulation \| \| Fundamental concepts Equivalence principle Special relativity World line Pseudo-Riemannian manifold \| \| Phenomena \| \| \| Kepler problem Gravitational lensing Gravitational redshift Gravitational time dilation Gravitational waves Frame-dragging Geodetic effect Event horizon Singularity Black hole \| \| Spacetime \| \| Spacetime diagrams Minkowski spacetime Einstein–Rosen bridge \| \| \| Equations Formalisms \| Equations \| \| Linearized gravity Einstein field equations Friedmann Geodesics Mathisson–Papapetrou–Dixon Hamilton–Jacobi–Einstein Raychaudhuri \| \| Formalisms \| \| ADM NP BSSN Post-Newtonian \| \| Advanced theory \| \| Kaluza–Klein theory Quantum gravity \| \| \| Solutions Schwarzschild (interior) Reissner–Nordström Einstein–Rosen waves Wormhole Gödel Kerr Kerr–Newman Kerr–Newman–de Sitter Kasner Lemaître–Tolman Taub–NUT Milne Robertson–Walker Oppenheimer–Snyder pp-wave van Stockum dust Hartle–Thorne Vaidya Peres De Sitter-Schwarzschild McVittie Weyl Kerr-Newman-de-Sitter \| \| Scientists Einstein Lorentz Hilbert Poincaré Schwarzschild de Sitter Reissner Nordström Weyl Eddington Friedmann Milne Zwicky Lemaître Oppenheimer Gödel Wheeler Robertson Bardeen Walker Kerr Chandrasekhar Ehlers Penrose Hawking Raychaudhuri Taylor Hulse van Stockum Taub Newman Yau Thorne others \| \| Physics portal Category \| \| v t e \| The equivalence principle is the hypothesis that the observed equivalence of gravitational and inertial mass is a consequence of nature. The weak form, known for centuries, relates to masses of any composition in free fall taking the same trajectories and landing at identical times. The extended form by Albert Einstein requires special relativity to also hold in free fall and requires the weak equivalence to be valid everywhere. This form was a critical input for the development of the theory of general relativity. The strong form requires Einstein's form to work for stellar objects. Highly precise experimental tests of the principle limit possible deviations from equivalence to be very small. Concept [edit] In classical mechanics, Newton's equation of motion in a gravitational field, written out in full, is: : inertial mass × acceleration = gravitational mass × gravitational acceleration Careful experiments have shown that the inertial mass on the left side and gravitational mass on the right side are numerically equal and independent of the material composing the masses. The equivalence principle is the hypothesis that this numerical equality of inertial and gravitational mass is a consequence of their fundamental identity.: 32 The equivalence principle can be considered an extension of the principle of relativity, the principle that the laws of physics are invariant under uniform motion. An observer in a windowless room cannot distinguish between being on the surface of the Earth and being in a spaceship in deep space accelerating at 1g and the laws of physics are unable to distinguish these cases.: 33 In Newton's law of gravity, the equivalence principle is viewed an observation, an accident of nature encoded in empirical law. In Einstein's model of gravity the equivalence principle is a consequence of physics. The same mass both causes and responds to gravity: mass causes spacetime curvature and then moves through spacetime according to that curvature.: 3.3 History [edit] See also: History of gravitational theory By experimenting with the acceleration of different materials, Galileo Galilei determined that gravitation is independent of the amount of mass being accelerated. Isaac Newton, just 50 years after Galileo, investigated whether gravitational and inertial mass might be different concepts. He compared the periods of pendulums composed of different materials and found them to be identical. From this, he inferred that gravitational and inertial mass are the same thing. The form of this assertion, where the equivalence principle is taken to follow from empirical consistency, later became known as "weak equivalence". A version of the equivalence principle consistent with special relativity was introduced by Albert Einstein in 1907, when he observed that identical physical laws are observed in two systems, one subject to a constant gravitational field causing acceleration and the other subject to constant acceleration, like a rocket far from any gravitational field.: 152 Since the physical laws are the same, Einstein assumed the gravitational field and the acceleration were "physically equivalent". Einstein stated this hypothesis by saying he would: ...assume the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system. — Einstein, 1907 In 1911 Einstein demonstrated the power of the equivalence principle by using it to predict that clocks run at different rates in a gravitational potential, and light rays bend in a gravitational field.: 153 He connected the equivalence principle to his earlier principle of special relativity: This assumption of exact physical equivalence makes it impossible for us to speak of the absolute acceleration of the system of reference, just as the usual theory of relativity forbids us to talk of the absolute velocity of a system; and it makes the equal falling of all bodies in a gravitational field seem a matter of course. — Einstein, 1911 Soon after completing work on his theory of gravity (known as general relativity): 111 and then also in later years, Einstein recalled the importance of the equivalence principle to his work: The breakthrough came suddenly one day. I was sitting on a chair in my patent office in Bern. Suddenly a thought struck me: If a man falls freely, he would not feel his weight. I was taken aback. This simple thought experiment made a deep impression on me. This led me to the theory of gravity. — Einstein, 1922 Einstein's development of general relativity necessitated some means of empirically discriminating the theory from other theories of gravity compatible with special relativity. Accordingly, Robert Dicke developed a test program incorporating two new principles – the § Einstein equivalence principle, and the § Strong equivalence principle – each of which assumes the weak equivalence principle as a starting point. Definitions [edit] Three main forms of the equivalence principle are in current use: weak (Galilean), Einsteinian, and strong.: 6 Some proposals also suggest finer divisions or minor alterations. Weak equivalence principle [edit] The weak equivalence principle, also known as the universality of free fall or the Galilean equivalence principle can be stated in many ways. The strong equivalence principle, a generalization of the weak equivalence principle, includes astronomic bodies with gravitational self-binding energy. Instead, the weak equivalence principle assumes falling bodies are self-bound by non-gravitational forces only (e.g. a stone). Either way: "All uncharged, freely falling test particles follow the same trajectories, once an initial position and velocity have been prescribed".: 6 "... in a uniform gravitational field all objects, regardless of their composition, fall with precisely the same acceleration." "The weak equivalence principle implicitly assumes that the falling objects are bound by non-gravitational forces." "... in a gravitational field the acceleration of a test particle is independent of its properties, including its rest mass." Mass (measured with a balance) and weight (measured with a scale) are locally in identical ratio for all bodies (the opening page to Newton's Philosophiæ Naturalis Principia Mathematica, 1687). Uniformity of the gravitational field eliminates measurable tidal forces originating from a radial divergent gravitational field (e.g., the Earth) upon finite sized physical bodies. Einstein equivalence principle [edit] What is now called the "Einstein equivalence principle" states that the weak equivalence principle holds, and that: the outcome of any local, non-gravitational test experiment is independent of the experimental apparatus' velocity relative to the gravitational field and is independent of where and when in the gravitational field the experiment is performed. Here local means that experimental setup must be small compared to variations in the gravitational field, called tidal forces. The test experiment must be small enough so that its gravitational potential does not alter the result. The two additional constraints added to the weak principle to get the Einstein form − (1) the independence of the outcome on relative velocity (local Lorentz invariance) and (2) independence of "where" (known as local positional invariance) − have far reaching consequences. With these constraints alone Einstein was able to predict the gravitational redshift. Theories of gravity that obey the Einstein equivalence principle must be "metric theories", meaning that trajectories of freely falling bodies are geodesics of symmetric metric.: 9 Around 1960 Leonard I. Schiff conjectured that any complete and consistent theory of gravity that embodies the weak equivalence principle implies the Einstein equivalence principle; the conjecture can't be proven but has several plausibility arguments in its favor.: 20 Nonetheless, the two principles are tested with very different kinds of experiments. The Einstein equivalence principle has been criticized as imprecise, because there is no universally accepted way to distinguish gravitational from non-gravitational experiments (see for instance Hadley and Durand). Strong equivalence principle [edit] The strong equivalence principle applies the same constraints as the Einstein equivalence principle, but allows the freely falling bodies to be massive gravitating objects as well as test particles. Thus this is a version of the equivalence principle that applies to objects that exert a gravitational force on themselves, such as stars, planets, black holes or Cavendish experiments. It requires that the gravitational constant be the same everywhere in the universe: 49 and is incompatible with a fifth force. It is much more restrictive than the Einstein equivalence principle. Like the Einstein equivalence principle, the strong equivalence principle requires gravity to be geometrical by nature, but in addition it forbids any extra fields, so the metric alone determines all of the effects of gravity. If an observer measures a patch of space to be flat, then the strong equivalence principle suggests that it is absolutely equivalent to any other patch of flat space elsewhere in the universe. Einstein's theory of general relativity (including the cosmological constant) is thought to be the only theory of gravity that satisfies the strong equivalence principle. A number of alternative theories, such as Brans–Dicke theory and the Einstein-aether theory add additional fields. Active, passive, and inertial masses [edit] Some of the tests of the equivalence principle use names for the different ways mass appears in physical formulae. In nonrelativistic physics three kinds of mass can be distinguished: Inertial mass intrinsic to an object, the sum of all of its mass–energy. Passive mass, the response to gravity, the object's weight. Active mass, the mass that determines the objects gravitational effect. By definition of active and passive gravitational mass, the force on due to the gravitational field of is: Likewise the force on a second object of arbitrary mass2 due to the gravitational field of mass0 is: By definition of inertial mass:if and are the same distance from then, by the weak equivalence principle, they fall at the same rate (i.e. their accelerations are the same). Hence: Therefore: In other words, passive gravitational mass must be proportional to inertial mass for objects, independent of their material composition if the weak equivalence principle is obeyed. The dimensionless Eötvös-parameter or Eötvös ratio is the difference of the ratios of gravitational and inertial masses divided by their average for the two sets of test masses "A" and "B". Values of this parameter are used to compare tests of the equivalence principle.: 10 A similar parameter can be used to compare passive and active mass. By Newton's third law of motion: must be equal and opposite to It follows that: In words, passive gravitational mass must be proportional to active gravitational mass for all objects. The difference, is used to quantify differences between passive and active mass. Experimental tests [edit] Tests of the weak equivalence principle [edit] Tests of the weak equivalence principle are those that verify the equivalence of gravitational mass and inertial mass. An obvious test is dropping different objects and verifying that they land at the same time. Historically this was the first approach – though probably not by Galileo's Leaning Tower of Pisa experiment: 19–21 but instead earlier by Simon Stevin, who dropped lead balls of different masses off the Delft churchtower and listened for the sound of them hitting a wooden plank. Newton measured the period of pendulums made with different materials as an alternative test giving the first precision measurements. Loránd Eötvös's approach in 1908 used a very sensitive torsion balance to give precision approaching 1 in a billion. Modern experiments have improved this by another factor of a million. A popular exposition of this measurement was done on the Moon by David Scott in 1971. He dropped a falcon feather and a hammer at the same time, showing on video that they landed at the same time. Chronology of weak equivalence principles tests \| Year \| Investigator \| Sensitivity \| Method \| \| 500? \| John Philoponus \| "small" \| Drop tower \| \| 1585 \| Simon Stevin \| 5×10−2 \| Drop tower \| \| 1590? \| Galileo Galilei: 91 \| 2×10−3 \| Pendulum, drop tower \| \| 1686 \| Isaac Newton: 91 \| 10−3 \| Pendulum \| \| 1832 \| Friedrich Wilhelm Bessel: 91 \| 2×10−5 \| Pendulum \| \| 1908 (1922) \| Loránd Eötvös: 92 \| 2×10−9 \| Torsion balance \| \| 1910 \| Southerns: 91 \| 5×10−6 \| Pendulum \| \| 1918 \| Zeeman: 91 \| 3×10−8 \| Torsion balance \| \| 1923 \| Potter: 91 \| 3×10−6 \| Pendulum \| \| 1935 \| Renner: 92 \| 2×10−9 \| Torsion balance \| \| 1964 \| Roll, Krotkov, Dicke \| 3×10−11 \| Torsion balance \| \| 1972 \| Braginsky, Panov: 92 \| 10−12 \| Torsion balance \| \| 1976 \| Shapiro, et al.: 92 \| 10−12 \| Lunar laser ranging \| \| 1979 \| Keiser, Faller: 93 \| 4×10−11 \| Fluid support \| \| 1987 \| Niebauer, et al.: 95 \| 10−10 \| Drop tower \| \| 1989 \| Stubbs, et al.: 93 \| 10−11 \| Torsion balance \| \| 1990 \| Adelberger, Eric G.; et al.: 95 \| 10−12 \| Torsion balance \| \| 1999 \| Baessler, et al. \| 5×10−14 \| Torsion balance \| \| 2008 \| Schlamminger, et al. \| 10−13 \| Torsion balance \| \| 2017 \| MICROSCOPE \| 10−15 \| Earth orbit \| Experiments are still being performed at the University of Washington which have placed limits on the differential acceleration of objects towards the Earth, the Sun and towards dark matter in the Galactic Center. Future satellite experiments – Satellite Test of the Equivalence Principle and Galileo Galilei – will test the weak equivalence principle in space, to much higher accuracy. With the first successful production of antimatter, in particular anti-hydrogen, a new approach to test the weak equivalence principle has been proposed. Experiments to compare the gravitational behavior of matter and antimatter are currently being developed. Proposals that may lead to a quantum theory of gravity such as string theory and loop quantum gravity predict violations of the weak equivalence principle because they contain many light scalar fields with long Compton wavelengths, which should generate fifth forces and variation of the fundamental constants. Heuristic arguments suggest that the magnitude of these equivalence principle violations could be in the 10−13 to 10−18 range. Currently envisioned tests of the weak equivalence principle are approaching a degree of sensitivity such that non-discovery of a violation would be just as profound a result as discovery of a violation. Non-discovery of equivalence principle violation in this range would suggest that gravity is so fundamentally different from other forces as to require a major reevaluation of current attempts to unify gravity with the other forces of nature. A positive detection, on the other hand, would provide a major guidepost towards unification. Tests of the Einstein equivalence principle [edit] In addition to the tests of the weak equivalence principle, the Einstein equivalence principle requires testing the local Lorentz invariance and local positional invariance conditions. Testing local Lorentz invariance amounts to testing special relativity, a theory with a vast number of existing tests.: 12 Nevertheless, attempts to look for quantum gravity require even more precise tests. The modern tests include looking for directional variations in the speed of light (called "clock anisotropy tests") and new forms of the Michelson–Morley experiment. The anisotropy measures less than one part in 10−20.: 14 Testing local positional invariance divides into tests in space and in time.: 17 Space-based tests use measurements of the gravitational redshift, the classic is the Pound–Rebka experiment in the 1960s. The most precise measurement was done in 1976 by flying a hydrogen maser and comparing it to one on the ground. The Global Positioning System requires compensation for this redshift to give accurate position values. Time-based tests search for variation of dimensionless constants and mass ratios. For example, Webb et al. reported detection of variation (at the 10−5 level) of the fine-structure constant from measurements of distant quasars. Other researchers dispute these findings. The present best limits on the variation of the fundamental constants have mainly been set by studying the naturally occurring Oklo natural nuclear fission reactor, where nuclear reactions similar to ones we observe today have been shown to have occurred underground approximately two billion years ago. These reactions are extremely sensitive to the values of the fundamental constants. Tests of changes in fundamental constants: 19 \| Constant \| Year \| Method \| Limit on fractional change per year \| \| weak interaction constant \| 1976 \| Oklo \| 10−11 \| \| fine-structure constant \| 1976 \| Oklo \| 10−16 \| \| electron–proton mass ratio \| 2002 \| quasars \| 10−15 \| Tests of the strong equivalence principle [edit] The strong equivalence principle can be tested by 1) finding orbital variations in massive bodies (Sun-Earth-Moon), 2) variations in the gravitational constant (G) depending on nearby sources of gravity or on motion, or 3) searching for a variation of Newton's gravitational constant over the life of the universe: 47 Orbital variations due to gravitational self-energy should cause a "polarization" of solar system orbits called the Nordtvedt effect. This effect has been sensitively tested by Lunar Laser Ranging experiments. Up to the limit of one part in 1013 there is no Nordtvedt effect. A tight bound on the effect of nearby gravitational fields on the strong equivalence principle comes from modeling the orbits of binary stars and comparing the results to pulsar timing data.: 49 In 2014, astronomers discovered a stellar triple system containing a millisecond pulsar PSR J0337+1715 and two white dwarfs orbiting it. The system provided them a chance to test the strong equivalence principle in a strong gravitational field with high accuracy. If there is any departure from the strong equivalence principle, it is no more than two parts per million. Most alternative theories of gravity predict a change in the gravity constant over time. Studies of Big Bang nucleosynthesis, analysis of pulsars, and the lunar laser ranging data have shown that G cannot have varied by more than 10% since the creation of the universe. 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"An improved test of the strong equivalence principle with the pulsar in a triple star system". Astronomy & Astrophysics. 638: A24. arXiv:2005.01388. Bibcode:2020A&A...638A..24V. doi:10.1051/0004-6361/202038104. ISSN 0004-6361. S2CID 218486794. Further reading [edit] Dicke, Robert H.; "New Research on Old Gravitation", Science 129, 3349 (1959). Explains the value of research on gravitation and distinguishes between the strong (later renamed "Einstein") and weak equivalence principles. Dicke, Robert H.; "Mach's Principle and Equivalence", in Evidence for gravitational theories: proceedings of course 20 of the International School of Physics "Enrico Fermi", ed. C. Møller (Academic Press, New York, 1962). This article outlines the approach to precisely testing general relativity advocated by Dicke and pursued from 1959 onwards. Misner, Charles W.; Thorne, Kip S.; and Wheeler, John A.; Gravitation, New York: W. H. Freeman and Company, 1973, Chapter 16 discusses the equivalence principle. Ohanian, Hans; and Ruffini, Remo; Gravitation and Spacetime 2nd edition, New York: Norton, 1994, ISBN 0-393-96501-5 Chapter 1 discusses the equivalence principle, but incorrectly, according to modern usage, states that the strong equivalence principle is wrong. Will, Clifford M.; Theory and experiment in gravitational physics, Cambridge, UK: Cambridge University Press, 1993. This is the standard technical reference for tests of general relativity. Will, Clifford M.; Was Einstein Right?: Putting General Relativity to the Test, Basic Books (1993). This is a popular account of tests of general relativity. Friedman, Michael; Foundations of Space-Time Theories, Princeton, New Jersey: Princeton University Press, 1983. Chapter V discusses the equivalence principle. External links [edit] Wikiquote has quotations related to Equivalence principle. Gravity and the principle of equivalence – The Feynman Lectures on Physics Introducing The Einstein Principle of Equivalence from Syracuse University The Equivalence Principle at MathPages The Einstein Equivalence Principle at Living Reviews on General Relativity "...Physicists in Germany have used an atomic interferometer to perform the most accurate ever test of the equivalence principle at the level of atoms..." \| v t e Albert Einstein \| \| --- \| \| Physics \| Theory of relativity + Special relativity + General relativity Mass–energy equivalence (E=mc2) Brownian motion Photoelectric effect Einstein coefficients Einstein solid Equivalence principle Einstein field equations Einstein radius Einstein relation (kinetic theory) Einstein ring Cosmological constant Bose–Einstein condensate Bose–Einstein statistics Bose–Einstein correlations Einstein–Cartan theory Einstein–Infeld–Hoffmann equations Einstein–de Haas effect EPR paradox Bohr–Einstein debates Teleparallelism Thought experiments Unsuccessful investigations Wave–particle duality Gravitational wave Tea leaf paradox \| \| Works \| Annus mirabilis papers (1905) "Investigations on the Theory of Brownian Movement" (1905) Relativity: The Special and the General Theory (1916) The Meaning of Relativity (1922) The World as I See It (1934) The Evolution of Physics (1938) "Why Socialism?" (1949) Russell–Einstein Manifesto (1955) \| \| In popular culture \| Die Grundlagen der Einsteinschen Relativitäts-Theorie (1922 documentary) The Einstein Theory of Relativity (1923 documentary) Relics: Einstein's Brain (1994 documentary) Insignificance (1985 film) Young Einstein (1988 film) Picasso at the Lapin Agile (1993 play) I.Q. (1994 film) Einstein's Gift (2003 play) Einstein and Eddington (2008 TV film) Genius (2017 series) Oppenheimer (2023 film) \| \| Prizes \| Albert Einstein Award Albert Einstein Medal Kalinga Prize Albert Einstein Peace Prize Albert Einstein World Award of Science Einstein Prize for Laser Science Einstein Prize (APS) \| \| Books about Einstein \| Albert Einstein: Creator and Rebel Einstein and Religion Einstein for Beginners Einstein: His Life and Universe Einstein in Oxford Einstein on the Run Einstein's Cosmos I Am Albert Einstein Introducing Relativity Subtle is the Lord \| \| Family \| Mileva Marić (first wife) Elsa Einstein (second wife; cousin) Lieserl Einstein (daughter) Hans Albert Einstein (son) Pauline Koch (mother) Hermann Einstein (father) Maja Einstein (sister) Eduard Einstein (son) Robert Einstein (cousin) Bernhard Caesar Einstein (grandson) Evelyn Einstein (granddaughter) Thomas Martin Einstein (great-grandson) Siegbert Einstein (distant cousin) \| \| Related \| Awards and honors Brain House Memorial Political views Religious views Things named after Einstein–Oppenheimer relationship Albert Einstein Archives Einstein's Blackboard Einstein Papers Project Einstein refrigerator Einsteinhaus Einsteinium Max Talmey Emergency Committee of Atomic Scientists \| \| \| \| \| v t e \| \| --- \| \| Special relativity \| \| \| \| --- \| \| Background \| Principle of relativity (Galilean relativity Galilean transformation) Special relativity Doubly special relativity \| \| Fundamental concepts \| Frame of reference Speed of light Hyperbolic orthogonality Rapidity Maxwell's equations Proper length Proper time Proper acceleration Relativistic mass \| \| Formulation \| Lorentz transformation Textbooks \| \| Phenomena \| Time dilation Mass–energy equivalence (E=mc2) Length contraction Relativity of simultaneity Relativistic Doppler effect Thomas precession Ladder paradox Twin paradox Terrell rotation \| \| Spacetime \| Light cone World line Minkowski diagram Biquaternions Minkowski space \| \| \| General relativity \| \| \| \| --- \| \| Background \| Introduction Mathematical formulation \| \| Fundamental concepts \| Equivalence principle Riemannian geometry Penrose diagram Geodesics Mach's principle \| \| Formulation \| ADM formalism BSSN formalism Einstein field equations Linearized gravity Post-Newtonian formalism Raychaudhuri equation Hamilton–Jacobi–Einstein equation Ernst equation \| \| Phenomena \| Black hole Event horizon Singularity Two-body problem Gravitational waves: astronomy detectors (LIGO and collaboration Virgo LISA Pathfinder GEO) Hulse–Taylor binary Other tests: precession of Mercury lensing (together with Einstein cross and Einstein rings) redshift Shapiro delay frame-dragging / geodetic effect (Lense–Thirring precession) pulsar timing arrays \| \| Advanced theories \| Brans–Dicke theory Kaluza–Klein Quantum gravity \| \| Solutions \| Cosmological: Friedmann–Lemaître–Robertson–Walker (Friedmann equations) Lemaître–Tolman Kasner BKL singularity Gödel Milne Spherical: Schwarzschild (interior Tolman–Oppenheimer–Volkoff equation) Reissner–Nordström Axisymmetric: Kerr (Kerr–Newman) Weyl−Lewis−Papapetrou Taub–NUT van Stockum dust discs Others: pp-wave Ozsváth–Schücking Alcubierre Ellis In computational physics: Numerical relativity \| \| \| Scientists \| Poincaré Lorentz Einstein Hilbert Schwarzschild de Sitter Weyl Eddington Friedmann Lemaître Milne Robertson Chandrasekhar Zwicky Wheeler Choquet-Bruhat Kerr Zel'dovich Novikov Ehlers Geroch Penrose Hawking Taylor Hulse Bondi Misner Yau Thorne Weiss others \| \| Category \| \| \| Authority control databases \| \| --- \| \| National \| Germany United States Israel \| \| Other \| Yale LUX \| Retrieved from " Categories: General relativity Fictitious forces Albert Einstein Principles Acceleration Philosophy of astronomy Likeness Hidden categories: CS1 errors: periodical ignored Articles with short description Short description matches Wikidata Use dmy dates from September 2020 Articles containing video clips
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8263	https://www.chegg.com/homework-help/questions-and-answers/bead-slide-without-friction-circular-hoop-radius-r-vertical-plane-bead-held-string-attache-q48099362	Solved A bead can slide without friction on a circular hoop \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Science Physics Physics questions and answers A bead can slide without friction on a circular hoop of radius R in a vertical plane. The bead is held by a string attached to the center of the hoop. The hoop rotates at a constant angular speed of rad/s about the vertical diameter, as shown in the figure at right. (a) Find the relationship between the angle  and . Hints: At equilibrium, resolve the Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: A bead can slide without friction on a circular hoop of radius R in a vertical plane. The bead is held by a string attached to the center of the hoop. The hoop rotates at a constant angular speed of rad/s about the vertical diameter, as shown in the figure at right. (a) Find the relationship between the angle  and . Hints: At equilibrium, resolve the A bead can slide without friction on a circular hoop of radius R in a vertical plane. The bead is held by a string attached to the center of the hoop. The hoop rotates at a constant angular speed of rad/s about the vertical diameter, as shown in the figure at right. (a) Find the relationship between the angle  and . Hints: At equilibrium, resolve the tension of the string into vertical and horizontal components. The radius for the circular motion is r R  sin .  Where does  comes into the picture? (b) Show that in order for  to reach 90 degrees,  has to approach infinity. Can you explain this in terms of the needed tension in the string to keep the bead in equilibrium? Show transcribed image text Here’s the best way to solve it.Solution Share Share Share done loading Copy link Please se… View the full answer Previous questionNext question Transcribed image text: A bead can slide without friction on a circular hoop of radius R in a vertical plane. The bead is held by a string attached to the center of the hoop. The hoop rotates at a constant angular speed of a rad/s about the vertical diameter, as shown in the figure at right. (a) Find the relationship between the angle and o. rotating Ik hoop Hints: Allequilibrium, resolve the tension of the string into vertical and horizontal components. The radius for the circular motion is r = Rsine. Where does o comes into the picture? bead (b) Show that in order for 6 to reach 90 degrees, w has to approach infinity. Can you explain this in terms of the needed tension in the string to keep the bead in equilibrium? Not the question you’re looking for? Post any question and get expert help quickly. Start learning Chegg Products & Services Chegg Study Help Citation Generator Grammar Checker Math Solver Mobile Apps Plagiarism Checker Chegg Perks Company Company About Chegg Chegg For Good Advertise with us Investor Relations Jobs Join Our Affiliate Program Media Center Chegg Network Chegg Network Busuu Citation Machine EasyBib Mathway Customer Service Customer Service Give Us Feedback Customer Service Manage Subscription Educators Educators Academic Integrity Honor Shield Institute of Digital Learning © 2003-2025 Chegg Inc. All rights reserved. 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8264	https://math.stackexchange.com/questions/2023524/am-i-correct-finding-all-vectors-orthogonal-to-these-two-vectors	linear algebra - Am I correct? Finding all vectors orthogonal to these two vectors - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Am I correct? Finding all vectors orthogonal to these two vectors Ask Question Asked 8 years, 10 months ago Modified8 years, 10 months ago Viewed 7k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Question Question: Find all vectors in R 3 that are orthogonal to both (1,2,3) and (4,5,6). My Attempt: We can write two equations: x+2 y+3 z=0 4 x+5 y+6 z=0 And now we can combine and simplify:x+2 y+3 z=4 x+5 y+6 z 3 x+3 y+3 z=0 x+y+z=0 Which means that all vectors that lie on the plane x+y+z=0 are orthogonal to both of the original vectors, or any vector of the form (−y−z,y,z). Is this solution correct? I feel like I might be going wrong when I say that every vector in the x+y+z=0 plane is orthogonal. Can anyone think of a different method of approaching this problem? linear-algebra vectors orthogonality Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Nov 21, 2016 at 0:30 riley lymanriley lyman 285 2 2 silver badges 9 9 bronze badges 1 1 As a sanity check on your work, observe that the two vectors are linearly independent, so their orthogonal complement must be one-dimensional. The plane x+y+z=0 is two-dimensional, however, so that can’t be the correct answer.amd –amd 2016-11-21 01:11:45 +00:00 Commented Nov 21, 2016 at 1:11 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. You started out on the right track by setting up a system of two linear equations, but then you took a bit of a left turn and didn’t actually solve the system. Each equation defines a plane orthogonal to the corresponding vector, and as the two vectors aren’t colinear, the intersection of these planes should be a line, not a plane as in your solution. This is consistent with a simple dimension check: the span of the vectors is two-dimensional, so its orthogonal complement in R 3 will be one-dimensional. Specifically, your mistake was concluding that all of the solutions of x+y+z=0 are also solutions of the original two equations. Since we’re working in R 3, a simple way to solve this problem is to take the cross product of the two vectors. Every scalar multiple of this product is orthogonal to both vectors, and these are the only vectors orthogonal to them both. Working instead with the equations that you set up, we have [1 2 3 4 5 6][x y z]=[0 0], i.e., we seek the kernel (null space) of the coefficient matrix. Row-reducing this matrix produces [1 0−1 0 1 2], from which we can read directly that the kernel is spanned by [1−2 1]T. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 21, 2016 at 1:58 amdamd 55.2k 3 3 gold badges 40 40 silver badges 100 100 bronze badges 2 I understand how you got the first "1" and "-2" in the vector that spans the kernel, but how did you get the final "1" at the end?riley lyman –riley lyman 2016-11-21 02:51:46 +00:00 Commented Nov 21, 2016 at 2:51 @rileylyman The rref tells us that x−z=0 and y+2 z=0. Setting z=1 gives us the other two components. Equivalently, you can simply read the basis from the columns of the rref that don’t have pivots. See math.stackexchange.com/a/1521354/265466 for a detailed explanation, but basically you negate the entries of each pivotless column and set the element that corresponds to the column index to 1.amd –amd 2016-11-21 07:42:37 +00:00 Commented Nov 21, 2016 at 7:42 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Incorrect. The vector (1,−1,0) is in the plane x+y+z=0, but it is not orthogonal to either of the vectors you are given. The space of vectors orthogonal to two given vectors is one dimensional, not two dimensional. The vector cross product of the two given vectors is orthogonal to each, and any vector which is orthogonal to each of the two given vectors is a scalar multiple of this cross product. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 21, 2016 at 0:37 DCarterDCarter 869 5 5 silver badges 12 12 bronze badges 2 Oh, okay. Thanks for your answer! How did setting those two equations equal to each other lead me down the wrong path?riley lyman –riley lyman 2016-11-21 00:39:27 +00:00 Commented Nov 21, 2016 at 0:39 1 The cross product of (1,2,3) and (4,5,6) is (−3,+6,−3). This is is the plane x+y+z=0, but is the only direction in this plane which is orthogonal to the given vectors.DCarter –DCarter 2016-11-21 00:43:41 +00:00 Commented Nov 21, 2016 at 0:43 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions linear-algebra vectors orthogonality See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 11Kernels and reduced row echelon form - explanation -1Determine an orthogonal vector in R 4 Related 0Vector Orthogonality and Length 0A basis for a subspace of Euclidean 4-space consisting of all vectors orthogonal to 2 vectors.(Math subject GRE 9768 Q.24) 1Number of Orthogonal vectors 1Find a basis for all vectors perpendicular to x−2 y+3 z=0. 2Find 2 orthogonal vectors in a given subspace 0Set of Orthogonal Vectors in R2 0Determining whether two known vectors are coplanar in a certain plane orthogonal to an unknown vector. Hot Network Questions Is it safe to route top layer traces under header pins, SMD IC? Why are LDS temple garments secret? Weird utility function Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term? 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8265	https://www.quora.com/What-is-the-difference-between-displacement-distance-and-distance-travelled	What is the difference between displacement, distance, and distance travelled? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Physics Distance (travel) Description of Motion Displacement (physics) Kinematics Motion (physics) Distance and Direction 3 What is the difference between displacement, distance, and distance travelled? All related (43) Sort Recommended Brad Moffat Author has 4K answers and 11.6M answer views ·5y Displacement : Look at the starting point, close your eyes, then look at the end point. The straight line distance from start to finish is the displacement. It’s the easiest to calculate. If you start at position 3 m and move to position 7 m then your position is (7m - 3m) , that is, 4 m. If you move from position 2 m to -8 m then your displacement is ( -8 m - 2 m), that is -10 m. A negative displacement means you ended up that far to the left (negative direction) and a positive displacement means you moved to the right (+ direction.) Displacement is usually listed as Δ x Δ x. Always Δ x=(Δ x=( Continue Reading Displacement : Look at the starting point, close your eyes, then look at the end point. The straight line distance from start to finish is the displacement. It’s the easiest to calculate. If you start at position 3 m and move to position 7 m then your position is (7m - 3m) , that is, 4 m. If you move from position 2 m to -8 m then your displacement is ( -8 m - 2 m), that is -10 m. A negative displacement means you ended up that far to the left (negative direction) and a positive displacement means you moved to the right (+ direction.) Displacement is usually listed as Δ x Δ x. Always Δ x=(x 2−x 1)Δ x=(x 2−x 1) or Δ x=(x b−x a)Δ x=(x b−x a) or Δ x=(x e n d−x s t a r t)Δ x=(x e n d−x s t a r t) It’s the idea that’s important, not the letters. Distance : Distance must be tracked at all times. If you move forward from 8 m to 12 m and then from the 12 m back to 4 m and then from 4 m to 7 m, your distance travelled is the sum of all three movements. That’s 4 m for the first, 8 m for the second (even though it went backwards!) and another 3 m for the last bit of movement. Total distance = 15 m. The displacement in the example just seen is -1 m. Can you see why? So displacement is easier and tells forward or backward (direction) information as well. You just need to chop long movements into sensible pieces. If you travel around an oval racing track and end up back where you started, the distance you travelled might be 400 m, but your displacement is zero m (illustrating that, hey, you are back where you started.) Zero m displacement might not mean that you didn’t move, it means that all your movements didn’t take you very far away from the starting point. Distance tells how far you moved, regardless of direction. Upvote · Sponsored by Grammarly 92% of professionals who use Grammarly say it has saved them time Work faster with AI, while ensuring your writing always makes the right impression. Download 999 207 Related questions More answers below What is the difference between displacement and distance? Why is it more useful to think of displacement than distance? Can distance and displacement be equal? What are the differences between 'distance', 'displacement', and 'distance travelled'? What is the displacement and distance travelled by a body? Gary Ward MaEd in Education&Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views ·5y What is the difference between displacement, distance, and distance traveled? Start at a point on a line labeled 0. Move 4 to the right to (+4), then 7 to the left (-7). Your displacement from the origin is -2. The distance traveled is 11 units. Distance in conventional speech is ambiguous and depends on the context. Distance in geometry is the absolute straight line measure between two points, perpendicular measure between a point and a line, or a point and a plane, or a line and a plane it is parallel to, or two parallel planes. Upvote · 9 1 Assistant Bot · 1y The concepts of displacement, distance, and distance traveled are fundamental in physics, particularly in kinematics. Here’s a breakdown of each term: Distance Definition: Distance refers to the total length of the path traveled by an object, irrespective of the direction. It is a scalar quantity. Characteristics: Always positive or zero. Measured in units such as meters (m), kilometers (km), etc. Does not consider the starting or ending point. Displacement Definition: Displacement is the shortest straight-line distance from the initial position to the final position of an object, along with the direc Continue Reading The concepts of displacement, distance, and distance traveled are fundamental in physics, particularly in kinematics. Here’s a breakdown of each term: Distance Definition: Distance refers to the total length of the path traveled by an object, irrespective of the direction. It is a scalar quantity. Characteristics: Always positive or zero. Measured in units such as meters (m), kilometers (km), etc. Does not consider the starting or ending point. Displacement Definition: Displacement is the shortest straight-line distance from the initial position to the final position of an object, along with the direction. It is a vector quantity. Characteristics: Can be positive, negative, or zero. Measured in the same units as distance (e.g., meters). Takes into account the initial and final positions and the direction of movement. Distance Traveled Definition: Distance traveled is often used interchangeably with distance, but it can imply the actual path length covered by an object during its motion, which may include any backtracking or changes in direction. Characteristics: Also a scalar quantity. Represents the total movement along the path taken, similar to distance. Example Consider a person walking from point A to point B, then back to point A: Distance: If the path from A to B is 5 meters and back to A is another 5 meters, the total distance is 10 meters. Displacement: The displacement from point A to point A (the starting point) is 0 meters since the initial and final positions are the same. Distance Traveled: This would also be 10 meters, as it accounts for the total path taken. In summary, distance and distance traveled focus on the total path without direction, while displacement considers the shortest path and direction. Upvote · Gene M.S. in Physics, University of Minnesota - Twin Cities (Graduated 1971) · Author has 9K answers and 2.7M answer views ·5y Time honored definitions by the international physics community are not always recognized by the general public; so expect some disagreements from others. displacement is a vector quantity giving the magnitude and direction for the change in position during a specific tome interval. Think “as the crow flies” movement. distance is the magnitude of separation between two points. It’s the scalar portion of displacement. No direction is implied. distance traveled is the path length of the actual motion during an interval. It may involve multiple changes in direction and may be quite long while ending Continue Reading Time honored definitions by the international physics community are not always recognized by the general public; so expect some disagreements from others. displacement is a vector quantity giving the magnitude and direction for the change in position during a specific tome interval. Think “as the crow flies” movement. distance is the magnitude of separation between two points. It’s the scalar portion of displacement. No direction is implied. distance traveled is the path length of the actual motion during an interval. It may involve multiple changes in direction and may be quite long while ending near the start. A long distance traveled could indicate a lot of exercise; but be ineffective as an evacuation route. Upvote · 9 1 Related questions More answers below What is the difference between distance travelled and displacement? Is a condition possible where distance is shorter than displacement? How are displacement and distance the same? How are distance and displacement same? (Distance is always greater or equal to displacement and Displacement is always shorter or equal to distance) Is it a valid difference? Adrian Haxell PhD in Physics, University of Oxford (Graduated 1994) · Author has 350 answers and 120.8K answer views ·4y Originally Answered: What is the difference between distance travelled and displacement? · Displacement is the distance and direction between two points but path to the point is irrelevant and can comprise different path lengths, see diagram. Black line AB is the displacement but all paths yield this but by travelling different distances Continue Reading Displacement is the distance and direction between two points but path to the point is irrelevant and can comprise different path lengths, see diagram. Black line AB is the displacement but all paths yield this but by travelling different distances Upvote · 9 1 Sponsored by Hudson Financial Partners How do I invest in shares in Australia? Buy direct ASX shares or pool your money with other investors in managed funds. Call us for more. Contact Us 9 3 Madhesh MD Studied Physics&Chemistry at SVM School Uthangarai ·5y Originally Answered: What is the difference between position, displacement, and distance traveled? · First of all all these terms depends on frame of reference. For our convenience let us take a stationary frame of reference. Now position of the object denotes the co-ordinates of that object in the space in all the three axes(x,y,z). Displacement is the distance between the intial position and final position of the object. It does not depend on the path you followed to reach the final position. Distance traveled is merely the length of the path you travelled from initial position to the final position Upvote · Prasad Indi M. Tech. from Indian Institute of Technology, Roorkee (Graduated 2006) ·4y Originally Answered: What is the difference between distance travelled and displacement? · If you’d paid attention in Physics classes, Distance is a Scalar while Displacement is a Vector. Meaning, you have to specify direction of the displacement. Upvote · 9 2 Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 621 Jitender Sihag M.Sc physics from Chaudhary Devi Lal University (Graduated 2014) ·6y Originally Answered: What is the difference between position, displacement, and distance traveled? · Suppose that you are at your home,this is your initial position. And now, suppose that after some time you are in market at cloth shops,so the actual distance that you covered from to cloth shop is your distance. Now you want to return for home from shop , so shop is your final position. Now you want a short cut and coincidentally you got an airlift from shop roof to straight your home roof,this shortest distance between your initial position and final position is called displacement. Upvote · Allen Ries B.Ed. from University of Alberta (Graduated 1974) · Author has 25.1K answers and 9.7M answer views ·5y Distance: How far it is between two points. Displacement: 1.Moving something from its original position. 2. replacing something with something else. As in water displaces air in the lungs when we drown. Distance travelled: How far you moved while travelling. Upvote · Sponsored by VAIZ.com Stop Fighting Your PM Tool — Switch to VAIZ VAIZ — Better Alternative to Your PM Tool. Boost Productivity with Built-in Doc Editor & Task Manager. Sign Up 9 6 Satya Parkash Sud M.Sc. in Physics&Nuclear Physics, University of Delhi (Graduated 1962) · Author has 8.1K answers and 27.4M answer views ·1y Related What's difference between distance and displacement? To understand let us consider the following example. A teacher enters the classroom say at 10 AM and starts to teach, writing on the blackboard moving to and fro along the length of the black board. He leaves the class room precisely at 11 AM from the same door from which he entered. What is the displacement of the teacher in one hour? Now displacement s over some time interval is defined as s = fina Continue Reading To understand let us consider the following example. A teacher enters the classroom say at 10 AM and starts to teach, writing on the blackboard moving to and fro along the length of the black board. He leaves the class room precisely at 11 AM from the same door from which he entered. What is the displacement of the teacher in one hour? Now displacement s over some time interval is defined as s = final position at end of time interval - initial position at start of the time interval. In our case the final position of the teacher after one end of one hour interval is the door and initial position of the teacher at the start of the interval is the same door. So displacement of the teacher in the one hour interval = position of the door - position of tte same door = 0 m So the displacement of the teacher = 0 m. But is the distance travelled by the teacher also zero. The teacher has not been static at one place while teaching. He was moving so he did cover some distance, which if he had a step counter or a smart watch with him would be known immediately. When taking of displacement, displacement is with resp... Upvote · 9 1 9 1 Snehangshu Ghosh Studied at St. Xavier's School, Durgapur ·6y Related Write the differences between distance and displacement? Distance It is the length of the path travelled by a object in a certain time. It is a scalar quantity. It is always positive. It depends on the path followed by the object. Distance cannot be zero. Displacement It is the shortest distance between the initial and final positions. It is a vector quantity. It doesn't depends on the path followed by the object. Displacement may or may not be zero. Upvote · 99 19 9 1 Soumyaneel Mukherjee Lives in Kolkata, West Bengal, India (2005–present) · Author has 2.2K answers and 1.7M answer views ·5y Related What is the difference between distance and displacement? Distance. 1)It is the length of the path traversed by the body per unit time Displacement 1)It is the distance travelled by the body in a particular direction in a certain time.(i.e it is the distance between the final and initial positions) Distance 2)It is a scalar quantity i.e it has only magnitude. Displacement 2) It is a vector quantity i.e it has both magnitude and direction of motion. Distance 3) It depends on the path travelled by the object. Dispacement 4)It does not depend on the path travelled by the object. Distance 4))It is always positive Displacement 4)It can be positive or negative de Continue Reading Distance. 1)It is the length of the path traversed by the body per unit time Displacement 1)It is the distance travelled by the body in a particular direction in a certain time.(i.e it is the distance between the final and initial positions) Distance 2)It is a scalar quantity i.e it has only magnitude. Displacement 2) It is a vector quantity i.e it has both magnitude and direction of motion. Distance 3) It depends on the path travelled by the object. Dispacement 4)It does not depend on the path travelled by the object. Distance 4))It is always positive Displacement 4)It can be positive or negative depending on direction. Distance 5) It is always greater than or equal to the magnitude of displacement. Dispacement 5)Its magnitude can be less than or equal to that of distance but can never be greater than than distance Distance 6) It may not be zero if displacement is zero Displacement 6)It is zero if distance is zero. Upvote · 9 3 9 1 9 1 Related questions What is the difference between displacement and distance? Why is it more useful to think of displacement than distance? Can distance and displacement be equal? What are the differences between 'distance', 'displacement', and 'distance travelled'? What is the displacement and distance travelled by a body? What is the difference between distance travelled and displacement? Is a condition possible where distance is shorter than displacement? How are displacement and distance the same? How are distance and displacement same? (Distance is always greater or equal to displacement and Displacement is always shorter or equal to distance) Is it a valid difference? Under what condition will the magnitude of displacement be equal to the distance travelled by an object? What is the similarity between distance and displacement? What is the change in position called, distance or displacement? If displacement is zero, then, what is the distance travelled? What is the difference between effective distance and total distance? Related questions What is the difference between displacement and distance? Why is it more useful to think of displacement than distance? Can distance and displacement be equal? What are the differences between 'distance', 'displacement', and 'distance travelled'? What is the displacement and distance travelled by a body? What is the difference between distance travelled and displacement? 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8266	https://en.wikipedia.org/wiki/The_Advancement_of_Learning	The Advancement of Learning - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 OriginToggle Origin subsection 1.1 Pure knowledge versus proud knowledge 2 Consequences 3 External links 4 References The Advancement of Learning [x] 6 languages Afrikaans 한국어 עברית Română தமிழ் 中文 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Expand all Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item From Wikipedia, the free encyclopedia 1605 Book by Francis Bacon Title page of 1640 edition The Advancement of Learning (full title: Of the Proficience and Advancement of Learning, Divine and Human) is a 1605 book by Francis Bacon which introduces and popularizes the scientific method of observation, skepticism and testability. Origin [edit] Bacon (a Protestant) lived during a period of great social turmoil as well as the expansion of scientific and social knowledge. In 1605 Bacon sent a draft to his friend Tobie Matthew who was in Florence where he was baptized as a Roman Catholic. Two years later, in 1607, Matthew returned to England, where he was imprisoned for his alleged "Papist views" The book is addressed as a plea to King James I and is in two parts or books, each with separate chapters: Part I praises the king for his appreciation of knowledge and outlines Bacon's ideas as how strict bondage to the past (notably study of Greek and Roman language and form) was a hindrance to optimizing Christian values, which required not academic excellence but excellent practical education via the contemplation of nature conjoined with action for the benefit of society. Part II Bacon outlines Novum Organum which avers the benefit to scholars as less important than the benefit of their scholarship to society. He advocated a new discipline studying the effect of climate, geography and natural resources on the various human races, and suggested handbooks should be prepared for diplomacy, business and the new scientific disciplines. In theology he suggested exploring the limits of human reason in matters divine and setting limits thereto. He recommended improving medicine via vivisection of animals and the prior preparation of medicine. Pure knowledge versus proud knowledge [edit] Bacon refutes the claim of king Solomon that knowledge causes anxiety, discontent and rebellion by distinguishing pure knowledge arising from the study of nature that leads to growth and Grace proud knowledge is that of worldly values that lead to atheism Consequences [edit] This work inspired the taxonomic structure of the highly influential Encyclopédie by Jean le Rond d'Alembert and Denis Diderot, and is credited by Bacon's biographer-essayist Catherine Drinker Bowen with being a pioneering essay in support of empirical philosophy. The following passage from The Advancement of Learning was used as the foreword to a popular Cambridge textbook: So that as Tennis is a game of no use in itself, but of great use in respect it maketh a quick eye, and a body ready to put itself in all positions, so, in the Mathematics the use which is collateral, an intervenient, is no less worthy, than that which is principle and intended. External links [edit] English Wikisource has original text related to this article: The Advancement of Learning Full text on Internet Archive Full text on Project Gutenberg Full text at classic-literature.co.uk References [edit] ^Francis Bacon Trust - Tributes ^Bowen, Catherine Drinker (1963). Francis Bacon: The Temper of a Man. Boston: Atlantic/Little, Brown & Co. pp.102–110. ^William Ludlam (1785), The Rudiments of Mathematics, Cambridge. \| hide v t e Francis Bacon \| \| Philosophy \| Baconian method Idola fori Idola theatri Idola specus Idola tribus Salomon's House \| \| Works \| Essays (1597) The Advancement of Learning (1605) Novum Organum (1620) History of the Reign of King Henry VII (1622) New Atlantis (1627) Complete bibliography \| \| Family \| Alice Barnham (wife) Nicholas Bacon (father) Anne Bacon (mother) \| \| Other \| Bacon's cipher Baconian theory of Shakespeare authorship An Examination of the Philosophy of Bacon Occult theories Romanticism and Bacon \| \| Authority control databases \| Yale LUX \| This article about a philosophy-related book is a stub. You can help Wikipedia by expanding it. v t e Retrieved from " Categories: Works by Francis Bacon (philosopher) 1605 books Philosophy book stubs Hidden categories: Articles with short description Short description is different from Wikidata Wikisource templates with missing id All stub articles This page was last edited on 5 July 2025, at 13:48(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents The Advancement of Learning 6 languagesAdd topic
8267	https://www.youtube.com/watch?v=Wzcd_Y-tctc	Prove sin x = sqrt(1 - cos^2 x) MSolved Tutoring 66100 subscribers 47 likes Description 4546 views Posted: 4 Sep 2019 Prove sin x = sqrt(1 - cos^2 x) 6 comments Transcript: great board I'm going to show you how it's done so what you do is you square both sides which results in getting rid of the radical on the right so we end up with sine squared X is equal to one minus cosine squared X and these are equivalent but let me show you so sine squared X plus cosine squared X equals one so if you want to get so to get sine squared equal to something subtract cosine squared from both sides we end up with sine squared X is equal to one minus cosine squared X so we can just rewrite this as sine squared X is equal to sine squared X depending on how you want to do it so anyway that's how you do that I hope this helped you out thanks for watching have a great day
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NIOS Online Admission Contact Contact Us About Us Online Tutors Career Options हिंदी मीडियमLatest BlogsAsk Questions Tiwari Academy/[Session 2025-26] NCERT Solutions Guide for Nursery to Class 12/NCERT Solutions for Class 10/Class 10 Maths Solutions for 2025-26 Board Exams/NCERT Solutions for Class 10 Maths Chapter 6 Triangles/NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2 NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2 Updated on October 8, 2024 by Tiwari Academy NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2 Triangles in Hindi Medium and English Medium prepared for session 2025-26. Class 10 Maths Exercise 6.2 solutions contain the question based on application of Thales Theorem. Class 10 Maths Exercise 6.2 Solutions Class 10 Maths Exercise 6.2 in Hindi Class 10th Maths NCERT Book Download Class 10 Maths Chapter 6 Solutions Class 10 Maths NCERT Solutions Class 10 all Subjects NCERT Solutions NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2 To prepare for Class 10 Maths Exercise 6.2, emphasize understanding and applying the Basic Proportionality Theorem (also known as Thales’ Theorem). The problems in this exercise focus on using this theorem to solve for unknown sides of triangles. Make sure you can recognise when a line divides two sides of a triangle proportionally and how to use the theorem to solve various geometric problems. Start by practicing simpler problems that involve drawing parallel lines and calculating ratios of corresponding sides. Work on more complex applications of the Basic Proportionality Theorem, such as problems that involve multiple steps or additional geometric principles like midpoints and parallel lines. Review the proofs provided in the chapter and attempt to write them in your own words. Understanding the logic behind the proofs will help you answer related questions effectively during exams. Focus on solving examples that require you to apply the theorem in different geometric configurations, as this will give you confidence in handling various problem types during your preparation. Class: 10 Mathematics Chapter 6:Exercise 6.2 Content:NCERT Textbook Solutions Content Mode:Text and Video Format Session:CBSE 2025-26 Medium:English and Hindi Medium Class 10 Maths Exercise 6.2 Solutions Class 10 Maths Exercise 6.2 NCERT Solutions Class 10 Maths Exercise 6.2 in Hindi Medium Class 10th Maths Chapter 6 NCERT Book Class 10 Maths Chapter 6 Solutions Class 10 Maths NCERT Solutions Class 10 all Subjects NCERT Solutions Class X Math Ex. 6.2 solution to View in Video Format for the students of CBSE, MP Board, UP Board, Bihar Board and all other boards using NCERT Solutions based on updated NCERT Books. We are here to help you in education. Never hesitate to take help from us. Just mail or message us about to what do you need. Download Offline Apps 2025-26 for offline use without internet. 10 Maths Chapter 6 Exercise 6.2 Solutions NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2 Triangles in English & Hindi Medium free to download in PDF updated for session 2025-26. All the questions are based on BPT or its converse. A few questions can be solved directly by similarity theorems, but both the methods should be known as some time it is asked to do with Thales theorem only. About 10 Maths Exercise 6.2 Exercise 6.2 is based on Thales Theorem (Basic Proportionality Theorem – BPT) and its converse. In Questions 1 and 2, we have to simply find the ratio of sides and apply the converse of BPT. Question 3 and 4 are direct application of Thales theorem. In Question 5 and 6, first apply BPT and then converse to prove the required things. Similarly questions 7, 8, 9 and 10 can be solved by using Thales as well as its converse. Question number 10 can be solved on the basis of similarity also, which will we study in Exercise 6.3. THALES THEOREM – BPT Basic Proportionality Theorem: If a line is drawn parallel to one side of a triangle to intersect the other sides in distinct points, the other two sides are divided in the same ratio. [BPT is also known as Thales Theorem. The converse of the Thales theorem is also important to solve the questions.] Download NCERT Books Apps 2025-26 based on new CBSE Syllabus. If your have doubts related to NIOS, visit to NIOS main page. How many questions are there in exercise 6.2 of class 10th mathematics chapter 6? There are in all 10 questions and 3 examples (example 1, 2 and 3) in exercise 6.2 of class 10th mathematics chapter 6 (Triangles). Which theorem is main in Exercise 6.2 of class 10th Maths? Exercise 6.2 of class 10th mathematics based on Theorem 6.1 (Basic Proportionality Theorem) and Theorem 6.2 (Converse of Basic Proportionality Theorem). Which is difficult part of exercise 6.2 of class 10th mathematics? Question 3, 4, 5, 6, 7, 8, 9 and 10 are important questions from exercise 6.2 of class 10th mathematics. Difficulty level of questions varies from student to student. But questions which most of students find difficult in this exercise are Q9 and 10. Which is the Student’s favourite questions from exercise 6.2 of class 10th Maths? Question 1 and Question 2 are the student’s favourite questions from exercise 6.2 of class 10th mathematics because these questions are very easy and students can solve these questions without any difficulty. « Class 10 Maths Exercise 6.1 Class 10 Maths Exercise 6.3 » Last Edited: October 8, 2024 Author Tiwari Academy With my best efforts, I have been guiding students for their better education. “There are no secrets to success. It is the result of preparation, hard work and learning from failure” View all posts by Tiwari Academy \| Website Content Reviewed: October 8, 2024 Content Reviewer Saikat Chakravarty Providing help in science for class 6 to 10. Adviser in Tiwari Academy for the science related subjects subject as well as videos contents. Ample teaching experience in schools. I am not only working for Tiwari Academy but also provide new ideas for the website and apps. 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8269	https://www.educatorstechnology.com/2023/08/6-great-periodic-table-of-elements-posters.html	Periodic Table of Elements Posters for Class - Educators Technology Skip to primary navigation Skip to main content Skip to primary sidebar Skip to footer Don't Miss a Post! Subscribe About Contact Search Search Search Search Home Pedagogy Educational AI Edtech Tools Edtech Apps Teacher Resources Special Education Edtech for Kids < X Home Pedagogy Educational AI Edtech Tools Edtech Apps Teacher Resources Special Education Edtech for Kids Search and hit enter Search and hit enter Educators Technology Innovative EdTech for teachers, educators, parents, and students Home Pedagogy Educational AI Edtech Tools Edtech Apps Teacher Resources Special Education Edtech for Kids Classroom Posters / Periodic Table of Elements Posters for Class Periodic Table of Elements Posters for Class By Med Kharbach, PhD \| Last Update: May 8, 2024 Periodic table of elements posters is the topic of our blog post today! The Periodic Table of Elements is more than just a scientific tool; it’s a visual guide to understanding the building blocks of our universe. Serving as an essential part of science education, the Periodic Table continues to evolve, reflecting new discoveries and advancements in our understanding of atomic structures. A key part of any science classroom, the Periodic Table is used not only by chemistry teachers but also by physics and biology educators to explain complex concepts. While the fundamental structure of the Periodic Table remains consistent, innovations in design, technology, and material are continually enhancing its utility and appeal. Classroom Resources Special Education Resources In this blog post, we’ll explore six cutting-edge Periodic Table posters, each designed to cater to specific educational needs and visual preferences. Before diving into the details, it may be helpful to reflect on some relevant literature: “The Periodic Table: Its Story and Its Significance” by Eric Scerri: This book provides an in-depth look into the history and evolution of the Periodic Table, emphasizing its importance in scientific education and development (Scerri, 2007). IUPAC’s Periodic Table Updates: The International Union of Pure and Applied Chemistry (IUPAC) regularly updates the Periodic Table to reflect the latest scientific discoveries, such as the naming and placement of new elements. “Elemental: How the Periodic Table Can Now Explain (Nearly) Everything” by Tim James: In “Elemental,” chemist Tim James provides an engaging guide to the periodic table, making it relevant to our daily lives. Covering the history from its ancient Greek roots to the modern completion of the table with the addition of the last four elements, the book is filled with fascinating insights. Special Education Resources For more visual resources for your class, check out our Classroom Posters section. Periodic table of elements posters Here are some practical period table of element posters from Amazon: 1. Bigtime Signs Periodic Table With Real Elements This 2022 updated periodic table is an eye-catching addition to any educational setting. Measuring an extra-large 54” x 35”, the vinyl construction promises durability. Special Education Resources Vibrant colors help to spot different element groups, including their Atomic Number, Atomic Weight, Symbol, and Name, making it a valuable learning tool. The real element periodic table ships in a protective tube, ready for classroom display, and can even double as artistic wall art. A great fit for kids and a perfect illustration for a classroom environment. 2. Periodic Table Poster Learning Game Makers Classroom Resources A jumbo 80cm x 60cm poster, this up-to-date 2023 edition is an excellent choice for classrooms, lecture halls, and dorm rooms. Constructed with durable PVC vinyl, it’s stronger than typical classroom posters and doesn’t need laminating. The easy-to-read, high-quality print and metal grommets for safe hanging make it a practical choice for teaching science to middle and high school students. The black & white design adds a touch of elegance to the information layout. 3. Periodic table science poster Classroom Resources School supplies Designed with a laminated 3 mil plastic finish, this new, high-quality design is both water and tear-resistant, assuring longevity. The clear and easy-to-view design makes it perfect for teachers, parents, and students, and it could be a decorative addition to a kid’s bedroom. The focus on innovation in teaching, including an easy-to-study and learn design, sets this poster apart as an educational and academic banner. 4. Periodic Table of The Elements Wall Chart Special Education Resources Measuring 24″ x 36″, this quick reference wall chart with heavy lamination offers a clear layout and grouping of elements, including coded symbols for Crystal Structures and Acid/Base Properties. Providing over 20 properties for each chemical element, it includes all 3 group classification systems and conforms to the current IUPAC General Assembly Report, including current names for elements 104-118. The vividly colored, easy-to-read format and heavy lamination make this a perfect classroom or dorm room addition. 5. Periodic Table of Elements Vinyl Poster Classroom Resources This vibrant and detailed 54×35 Inches poster is crafted in line with the latest 2022 IUPAC updates, including atomic weights, the newest elements, and information on radioactive elements. The use of neon colors on a black background provides excellent readability and contrast. The main font is bold, and the quality vinyl material ensures durability. Easy to install with pre-installed grommets and high-resolution printing, this periodic table serves as iconic science classroom decor, appealing especially to high school or college students. Wrinkles can be easily removed, adding to its practicality. 6. LiFe Chem 2022 Chemistry Periodic Table of Elements Poster Classroom Resources School supplies Boasting the best reading experience, this 2022 periodic table chart offers high-quality printing and clear fonts. The pastel colors on a dark grey background and minimal design add to its chic appeal. Shop for bestsellers Made of thick high gloss photography paper, it’s waterproof, tear-resistant, and fade-resistant, requiring no lamination. Available in 2 sizes and 3 designs, this versatile poster is a perfect gift choice for classrooms, labs, and homes. The L-size posters include additional information on “Common Oxidation States”. Final thoughts From vibrant neon colors that captivate younger audiences to minimalistic and stylish themes for a more sophisticated look, the featured periodic table posters offer a fresh approach to a classic educational resource. By prioritizing readability, quality materials, and current information in accordance with IUPAC standards, they ensure that both educators and students can engage with chemistry’s foundational concepts in a way that’s visually stimulating and intellectually enriching. The world of elements has never been more accessible, and these posters prove that the beauty of science can be displayed right on our walls, bridging the gap between learning and art in the classroom. Special Education Resources Learning Game Makers Classroom Resources Join our mailing list Never miss an EdTech beat! Subscribe now for exclusive insights and resources. Meet Med Kharbach, PhD Dr. Med Kharbach is an influential voice in the global educational technology landscape, with an extensive background in educational studies and a decade-long experience as a K-12 teacher. Holding a Ph.D. from Mount Saint Vincent University in Halifax, Canada, he brings a unique perspective to the educational world by integrating his profound academic knowledge with his hands-on teaching experience. Dr. Kharbach's academic pursuits encompass curriculum studies, discourse analysis, language learning/teaching, language and identity, emerging literacies, educational technology, and research methodologies. His work has been presented at numerous national and international conferences and published in various esteemed academic journals. Primary Sidebar As an Amazon Associate I earn from qualifying purchases Discover more School supplies Poster educational Posters poster Education Educators Edtech scientific EdTech Join our mailing list Never miss an EdTech beat! Subscribe now for exclusive insights and resources. Reader Favorites Thanksgiving Literacy Resources for Teachers and Parents Digital Citizenship In The Age Of AI: A Practical Guide For Teachers NotebookLM for Teachers and Researchers: A Complete Guide Rethinking TPACK in the Age of AI 3 Uniquely Human Values AI Cannot Replicate Evaluating AI Tools for Learning: A Teacher’s Manual Footer Join our mailing list Join our email list for exclusive EdTech content. Home Privacy policy Disclaimer About Dr. Kharbach Contact Pinterest Instagram Facebook LinkedIn YouTube Twitter TikTok © 2010–2025 Educators Technology by Med Kharbach, PhD
8270	https://math.libretexts.org/Courses/Rio_Hondo/Math_175%3A_Plane_Trigonometry/03%3A_Trigonometric_Identities_and_Equations/3.06%3A_Half_Angle_Identities	Skip to main content 3.6: Half Angle Identities Last updated : May 13, 2021 Save as PDF 3.5: Double Angle Identities 3.7: Exercises - Double Angle, Half-Angle, and Power Reductions Page ID : 61256 ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives Apply the half-angle identities to expressions, equations and other identities. Use the half-angle identities to find the exact value of trigonometric functions for certain angles. Power Reduction and Half Angle Identities Another use of the cosine double angle identities is to use them in reverse to rewrite a squared sine or cosine in terms of the double angle. Starting with one form of the cosine double angle identity: Isolate the cosine squared term Add 1 Divide by 2 This is called a power reduction identity Exercise Use another form of the cosine double angle identity to prove the identity . Answer The cosine double angle identities can also be used in reverse for evaluating angles that are half of a common angle. Building from our formula , if we let , then this identity becomes . Taking the square root, we obtain where the sign is determined by the quadrant. This is called a half-angle identity. Exercise Use your results from the last Try it Now to prove the identity . Answer IDENTITIES Half-Angle Identities Power Reduction Identities Since these identities are easy to derive from the double-angle identities, the power reduction and half-angle identities are not ones you should need to memorize separately. Example Rewrite without any powers. Solution Using the power reduction formula Square the numerator and denominator Expand the numerator Split apart the fraction Apply the formula above to Simplify Combine the constants Example Find an exact value for . Solution Since 15 degrees is half of 30 degrees, we can use our result from above: We can evaluate the cosine. Since 15 degrees is in the first quadrant, we need the positive result. Exercise If and , then find exact values for (without solving for ): a. b. c. Answer : a. b. c. Important Topics of This Section Power reduction identity Half angle identity Using identities Simplify equations Prove identities Solve equations 3.5: Double Angle Identities 3.7: Exercises - Double Angle, Half-Angle, and Power Reductions
8271	https://engcourses-uofa.ca/books/statics/vectors-and-their-operations/cartesian-vector-notation/	Open Educational Resources Vectors and their Operations: Cartesian vector notation Cartesian coordinate systems A convenient set of directions is a set of perpendicular directions called orthogonal axes. Orthogonal means perpendicular. The positive direction of an axis sets a benchmark to determine the positive (or negative) direction of a vector along (parallel with) the axis. For example, considering an axis x shown in Fig. 2.14a, is in the positive direction and is in the negative direction with respect to axis x. Decomposing a vector in the two-dimensional plane needs two orthogonal axes (Fig. 2.14b). Decomposing a spatial vector (a vector in the three-dimensional space) needs three orthogonal axes (Fig. 2.14c). A set of orthogonal axes, intersecting at a point (the origin), is called a Cartesian coordinate system or Cartesian frame. Unit vector A Unit vector. A unit vector is a vector of magnitude . Any vector can be written as a scalar multiplication of a unit vector with the same direction as the original vector: such that the notation indicates the unit vector () in the direction of . Any vector can become a unit vector if scaled by meaning that . The vector is called the unit vector of . Making a unit vector out of a vector is called normalizing a vector. The unit vector of a vector has the same direction as the original vector. Experiment with the following interactive tool to investigate how a vector is a scalar multiplicand of its unit vector. Cartesian vector notation The components of a vector along orthogonal axes are called rectangular components or Cartesian components. A vector decomposed (resolved) into its rectangular components can be expressed by using two possible notations namely the scalar notation (scalar components) and the Cartesian vector notation. Both notations are explained for the two-dimensional (planar) conditions, and then extended to three dimensions in the following sections. Rectangular vector components of coplanar vectors Consider a vector and its rectangular components resolved in a Cartesian system as shown in Fig. 2.15. The and vector components of are resolved and denoted by and respectively. The subscripts and denote the axes with which the components are parallel. The magnitude of the vector components are and . By definition, the magnitudes of a vector is a positive number. Therefore, the magnitude of a vector component does not hold any information about the sense of direction. To include the information about the directions of the vector components in a Cartesian system, scalar notation or scalar components are defined. Scalar components of a vector are signed magnitudes of its rectangular components. A scalar component is positive if the vector component is directed along the positive axis, and negative if the vector component is directed along the negative axis (opposite of the axis positive direction). For the vector components and shown in Fig. 2.15, the scalar components are denoted by and . Note the regular and italic font face used for the notation. The scalar components of the vector shown in Fig. 2.15a are and both being positive. The scalar components of the vector shown in Fig. 2.15b, however, are and . The scalar component is negative, because its vector component is directed along the negative direction of the y axis. An axis has an associated unit vector showing the positive direction of the axis. Any vector parallel with an axis can be written as a scalar multiplier of the axis’ unit vector. The scalar multiplier is equal to the scalar component (signed magnitude) of the vector parallel with the axis. The scalar multiplier is positive if the vector is in the direction of the axis unit vector, and negative otherwise. The unit vectors associated with Cartesian x and y axes are denoted by bold lower-case letters and respectively and their respective small arrows are displayed on the Cartesian axes (Fig. 2.16). Another possible notations are and . The rectangular components of any vector can be now expressed in terms of the Cartesian axis unit vectors, (2.1a) where and are the scalar components. As a general rule any vector can be written as, (2.1b) Remark: using CVN is equivalent to resolving a vector in a Cartesian coordinate system. Remark: in CVN, capital letters (non-bold), such as, and , represent scalar components. Their values can be negative or positive, depending on the directions of the components relative to , , and . In the formulations of later chapter, where CVN is not used, capital letters represent the magnitudes of vector or vector components, for simplicity. Remark: components of a vector are vectors, whereas the scalar components (vector coordinates) are scalar (i.e. signed magnitudes). Using the following interactive tool, you can observe the (vector) components and scalar components of a vector. Observe how the vector is represented in CVN. Change the direction of the vector by the angle slider and observe the values of the scalar components. When do the scalar components attain negative values? EXAMPLE 2.4.1 Determine the scalar components of shown in the figure below and write the the vector in CVN. SOLUTION Remark: the location of a vector with respect to the origin of the Cartesian coordinate system does not affect its Cartesian components. You can always consider (draw) parallel axes with the Cartesian axes at the tail of a vector and calculate its components. To determine the components of a planar vector in the Cartesian frame, or to calculate its orientation with respect to any axis, we consider the acute angle that the vector makes with one of the Cartesian axes. This allows us to establish the relationships between the vectors magnitude and its components. A minus sign is manually assigned to components that act in the directions opposite to the coordinate axes. If and are angles of a vector with x and y axes respectively, the following equations hold, (2.2) Note that the absolute values of the scalar components, which are the magnitudes of the vector Cartesian components, are used in the trigonometry functions. Fig. 2.17a demonstrates three cases of vector decomposition based on the defined angles and . Pay attention to the signs of the scalar components based on the Cartesian components of the vectors. Another possible way of specifying the direction of a planar vector is by a small slope triangle. The small slope triangle conveys the information on , , and of the angle that the vector makes with an axis (Fig. 2.17b). Remark: be cautious about the way the direction of a vector is specified and the proper formulation you should use for the calculations. Remark: using the angles (which a vector makes with the Cartesian axes) to specify the direction of the vector has a calculation advantage that is in the domain of the inverse trigonometric functions , and . Rectangular vector components of spatial vectors To treat a vector in three dimensions, a three-dimensional Cartesian coordinate system is to be defined. A common three-dimensional Cartesian coordinate system is a right-handed coordinate system. A coordinate system of three orthogonal axes is said to be right-handed if your right-hand thumb points in the positive direction when fingers curl from the (positive) x axis to the (positive) y axis (Fig. 2.18a). Another way to define a right-handed Cartesian coordinate system is to follow the right-hand three-finger rule as demonstrated Fig. 2.18b. In three dimensions, the unit vectors of the axes are denoted by , , and as demonstrated in Fig. 2.18c. Right-click and rotate the following three-dimensional right-handed Cartesian coordinate system to observe its different orientations. Any vector in three dimensions can be resolved into three rectangular components. A vector expressed in CVN (three dimensions) is written as, (2.3) where , and are the vector (rectangular or Cartesian) components, and , and are the scalar components (Fig. 2.19a). The magnitude of is . This can be proved by considering the shaded triangles in Fig. 2.19b and twice using the Pythagoras’ theorem. The direction of in a three-dimensional Cartesian coordinate system can be defined by the coordinate direction angles , , and measured from the positive x, y, and z axes respectively to the vector (Fig. 2.19c). These angles are limited as . The following relationship holds between the vector magnitude, the scalar components and the coordinate direction angles, (2.4) The following interactive example illustrates the rectangular components of a vector in three dimensions. Change the scalar components (coordinates of the head point) by the sliders to observe how the vector is expressed in CVN. The coordinate direction angles of a vector are demonstrated by the following interactive example. The three coordinate direction angles are not independent and they are related by this equation, (2.5) which means that knowing two of the angles, the third one is readily obtainable. To prove Eq. 2.5, a vector is expressed by its Cartesian components and divided by its magnitude, which, by Eq. 2.4, is written as: and leads to, which proves Eq. 2.5. Videos Cartesian Vector Notation: Magnitude and Direction (Unit Vector) of a Vector: Coordinate Direction Angles:
8272	https://www.oden.utexas.edu/media/reports/2005/0509.pdf	LECTURES ON AN INTRODUCTION TO QUANTUM MECHANICS J. Tinsley Oden ICES The University of Texas at Austin 2004 Preface This monograph contains class notes I prepared during the summer of 2004 based on weekly lectures I gave to a group of colleagues: friends, students, post docs, and professional associates. The goal of the course, taught and taken completely voluntarily, was simply to learn some basic things about quantum mechanics and its relationship to molecular dynamics. In preparing these notes, I relied on several excellent textbooks, treatises, and monographs on the subject. We found the subject beautiful and exciting, and that it is quite accessible to those with backgrounds in classical mechanics, applied mathematics, partial diﬀerential equations, and functional analysis. Hope-fully, with further reading and exposure, the theory will become a tool in our work in computational engineering and sciences. Thanks go to Frederika Tausch for typing most of the manuscript and to Derek Ploor for typing some of the early sections. J. Tinsley Oden February 3, 2005 ii Contents 1 Electromagnetic Waves 1 1.1 Electrical Fields . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Gauss’ Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.3 Electric Potential Energy . . . . . . . . . . . . . . . . . . . . . 5 1.4 Magnetic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.5 Some Properties of Waves . . . . . . . . . . . . . . . . . . . . 8 1.6 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . 12 1.7 Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . 16 2 Introduction to Quantum Mechanics 17 2.1 Introductory Comments . . . . . . . . . . . . . . . . . . . . . 17 2.2 The Photoelectric Eﬀect . . . . . . . . . . . . . . . . . . . . . 18 2.3 The Compton Eﬀect . . . . . . . . . . . . . . . . . . . . . . . 19 2.4 Heisenberg’s Uncertainty Principle . . . . . . . . . . . . . . . 19 2.5 The Correspondence Principle . . . . . . . . . . . . . . . . . . 21 2.6 Schr¨ odinger’s Equation . . . . . . . . . . . . . . . . . . . . . . 22 2.7 Elementary Properties of the Wave Equation . . . . . . . . . . 25 2.8 Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2.9 Wave Packets / Fourier Transforms . . . . . . . . . . . . . . . 28 2.10 The Wave-Momentum Duality . . . . . . . . . . . . . . . . . . 29 2.11 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 3 Dynamical Variables and Observables in Quantum Mechan-ics: The Mathematical Formalism 35 3.1 Introductory Remarks . . . . . . . . . . . . . . . . . . . . . . 35 iii CONTENTS 3.2 The Hilbert Space L2(R) (or L2(RN)) . . . . . . . . . . . . . . 36 3.3 Dynamical Variables and Hermitian Operators . . . . . . . . . 37 3.4 Spectral Theory of Hermitian Operators . . . . . . . . . . . . 39 3.5 Observables and Statistical Distributions . . . . . . . . . . . . 42 3.6 The Continuous Spectrum . . . . . . . . . . . . . . . . . . . . 44 3.7 The Generalized Uncertainty Principle . . . . . . . . . . . . . 44 4 Selected Topics and Applications 49 4.1 Introductory Remarks . . . . . . . . . . . . . . . . . . . . . . 49 4.2 Ground States and Energy Quanta: the Harmonic Oscillator . 49 4.3 The Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . 51 4.4 Angular Momentum and Spin . . . . . . . . . . . . . . . . . . 55 4.5 Variational Principle . . . . . . . . . . . . . . . . . . . . . . . 58 5 The Transition from Quantum Mechanics to Approximate Theories and to Molecular Dynamics 61 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 5.2 Molecular Dynamics . . . . . . . . . . . . . . . . . . . . . . . 62 5.3 The Connection Between MD and Quantum Mechanics . . . . 64 5.4 Some Approximate Methods . . . . . . . . . . . . . . . . . . . 67 5.4.1 The LCAO Method . . . . . . . . . . . . . . . . . . . . 67 5.4.2 The Hartree-Fock Model . . . . . . . . . . . . . . . . . 68 5.4.3 Density Functional Theory . . . . . . . . . . . . . . . . 71 5.5 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 Bibliography 77 iv List of Figures 1.1 a) A charge q at a point P in the plane, and electric ﬁeld lines emanating from P and b) the electric ﬁeld produced by positive charges q1 and q2 at points P and Q. . . . . . . . . . 3 1.2 Equal and opposite charges on a line . . . . . . . . . . . . . . 4 1.3 Equal and opposite charges in an electric ﬁeld E a distance d apart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.4 Model of an atom as charged electrons in orbits around a nucleus 6 1.5 Electric dipole analogous to a dipole caused by a magnet . . . 8 1.6 Incremental motion of a wave from over a time increment ∆t . 10 1.7 Properties of a simple plane wave . . . . . . . . . . . . . . . . 11 1.8 Wavelengths and frequencies of electromagnetic waves. . . . . 15 1.9 Components of electromagnetic waves . . . . . . . . . . . . . . 15 2.1 Diﬀraction of a beam of electrons . . . . . . . . . . . . . . . . 20 2.2 Histogram showing the number of people versus age. . . . . . 31 2.3 Two histograms with the same median, average, and probable value, but diﬀerent standard deviations. . . . . . . . . . . . . 33 5.1 The hydrogen molecule H2 consisting of two nuclei at a and b and electrons at 1 and 2. . . . . . . . . . . . . . . . . . . . . . 65 5.2 A molecular system consisting of M nuclei and N electrons. . 69 v Chapter 1 Electromagnetic Waves 1.1 Electrical Fields The now classical science of electricity and magnetism recognizes that mate-rial objects can possess what is called an electric charge—an intrinsic charac-teristic of the fundamental particles that make up the objects. The charges in objects we encounter in everyday life may not be apparent, because the object is electrically neutral, carrying equal amounts of two kinds of charges, positive charge and negative charge. Atoms consist of positively charged protons, negatively charged electrons, and electrically neutral neutrons, the protons and neutrons being packed together in the nucleus of the atom. The mathematical characterization of how charges interact with one an-other begins with Coulomb’s Law, postulated in 1785 by Charles Augustine Coulomb on the basis of experiments. It is stated as follows: consider two charged particles (or point charges) of magnitude q1 and q2 separated by a distance r. The electrostatic force of attraction or repulsion between the charges has magnitude F = k\|q1\| \|q2\| r2 (1.1) where k is a constant, normally expressed in the form k = 1 4πϵ0 = 8.99 × 109 N m2/C2 where ϵ0 is the permittivity constant, ϵ0 = 8.85 × 10−12 N m2/C2 1 CHAPTER 1. ELECTROMAGNETIC WAVES Here C is the SI measure of a unit charge, called a Coulomb, and is deﬁned as the amount of charge that is transferred across a material wire in one second due to a 1-ampere current in the wire. The reason for this choice of k is made clear later. According to Jackson , the inverse-square dependence of force on dis-tance (of Coulomb’s law) is known to hold over at least 24 orders of magnitude in the length scale. Two important properties of charge are as follows: 1. charge is quantized: Let e denote the elementary charge of a single electron or proton, known from experiments to be e = 1.60 × 10−19 C then any positive or negative charge q is of the form, q = ne , where n = ±1, ±2, ±3, . . . (n ∈Z) The fact that electrical charge is “quantized” (meaning discretely de-ﬁned as an integer multiple of e) is regarded by some as “one of the most profound mysteries of the physical world” (Cf. [9, page 251]) 2. charge is conserved: the net charge in a system or object is preserved. The fundamental notion of an electric ﬁeld is intimately tied to the force ﬁeld generated by electrical charges. Let q1 denote a positive point charge situated at a point P in space. Imagine that a second positive point charge q2 is placed at point Q near to P. According to Coulomb’s law, q1 exerts a repulsive electrostatic force on q2. This vector ﬁeld of forces is called the electric ﬁeld. We say that q1 sets up an electric ﬁeld E in the space surrounding it, such that the magnitude of E at a point Q depends upon the distance from P to Q and the direction depends on the direction from P to Q and the electrical sign of q1 at P. In practice, E is determined at a point by evaluating the electrostatic force F due to a positive test charge q0 at that point (see Fig. 1.1a). Then E = q−1 0 F . E thus has the units of Newtons per Coulomb (N/C). The magnitude of the electric ﬁeld, then, due to a point charge q is q−1 0 1 4πϵ0 \|q1\| \|q0\| r2 = \|q1\| 4πϵ0r2 . 2 1.1. ELECTRICAL FIELDS q+ P (a) Positive charge at point P q+ q+ P Q (b) Positive charges at points P and Q Figure 1.1: a) A charge q at a point P in the plane, and electric ﬁeld lines emanating from P and b) the electric ﬁeld produced by positive charges q1 and q2 at points P and Q. For m such charges, E = q−1 0 m X i=1 F 0i (1.2) F 0i being the force from 0, the point of application of q0, and the m charges qi, i = 1, 2, . . ., m. The electric ﬁeld lines for two positive charges are illustrated in Fig. 1.1b. A fundamental question arises concerning the electrical force between two point charges q1 and q2 separated by a distance r; if q2 is moved toward q1, does the electric ﬁeld change immediately? The answer is no. The informa-tion that one of the charges has moved travels outwardly in all directions at the speed of light c as an electromagnetic wave. More on this later. 3 CHAPTER 1. ELECTROMAGNETIC WAVES q+ q− d x O Figure 1.2: Equal and opposite charges on a line The concept of an electric dipole is also important. The electric ﬁeld of two particles of charge q but opposite sign, a distance d apart, at a point x on an axis through the point charges, is E(x) = q 4πϵ0 1 (x −d/2)2 − 1 (x + d/2)2 i (see Fig. 1.2). For x ≫d, E(x) = 1 2πϵ0 qd x3 i The vector m = qdi (1.3) is called the electric dipole moment, and E(x) = m/2πϵ0x3 (1.4) The torque τ created by a dipole moment in an electric ﬁeld E is m × E, as can be deduced from Fig. 1.3 A fundamental property of charged bodies is that charge is conserved, as noted earlier. The charge of one body may be passed to another, but the net charge remains the same. No exceptions to this observation have ever been found. A closely related idea is that embodied in Gauss’ Law, which asserts that the net charge contained in a bounded domain is balanced by the ﬂux of the electric ﬁeld through the surface. 1.2 Gauss’ Law The net ﬂux of an electric ﬁeld E through the boundary surface ∂Ωof a bounded region Ωis given by qΩ= I ∂Ω ϵ0E · n dA (1.5) 4 1.3. ELECTRIC POTENTIAL ENERGY M F F q+ q− E d Figure 1.3: Equal and opposite charges in an electric ﬁeld E a distance d apart where n is a unit normal to ∂Ωand qΩis the net charge enclosed in the region Ωbounded by the surface ∂Ω The relationship between Gauss’ law and Coulomb’s law is immediate: Let Ωbe a sphere of radius r containing a positive point charge q at its center. Then ϵ0 I ∂(sphere) E · n dA = ϵ0E(4πr2) = q , so E = 1 4πϵ0 q r2 , which is precisely Coulomb’s law. This relationship is the motivation for choosing the constant k = 1/(4πϵ0) in Coulomb’s law. 1.3 Electric Potential Energy When an electrostatic force acts between two or more charged particles within a system of particles, an electrostatic potential energy U can be assigned 5 CHAPTER 1. ELECTROMAGNETIC WAVES to the system such that in any change ∆U in U the electrostatic forces do work. The potential energy per unit charge (or due to a charge q) is called the electric potential V ; V = U/q. V has units of volts, 1 volt = 1 joule/coulomb. Atom Models. The attractive or repulsive forces of charged particles leads directly to the classical Rutherford model of an atom as a tiny solar system in which electrons, with negative charges e, move in orbits about a positively charged nucleus. Thus, when an electron travels close to a ﬁxed positive charge q = +e, it can escape the pull or reach a stable orbit, spinning around the charge and thereby creating a primitive model of an atom. This primitive 2e+ e+ e− e− e− Figure 1.4: Model of an atom as charged electrons in orbits around a nucleus model was discarded when Bohr introduced the quantum model of atoms in which, for any atom, electrons can only exist in so-called discrete quantum states of well-deﬁned energy or so-called energy shells. The motion of an electron from one shell to another can only happen instantaneously in a quantum jump. In such a jump, there is obviously a change in energy, which is emitted as a proton of electromagnetic radiation (more on this later). Bohr’s model was later improved by the probability density model of Schr¨ odinger. 1.4 Magnetic Fields Just as charged objects produce electric ﬁelds, magnets produce a vector ﬁeld B called a magnetic ﬁeld. Such ﬁelds are created by moving electrically charged particles or as an intrinsic property of elementary particles, such as electrons. In this latter case, magnetic ﬁelds are recognized as basic charac-teristics of particles, along with mass, electric charge, etc. In some materials 6 1.4. MAGNETIC FIELDS the magnetic ﬁelds of all electrons cancel out, giving no net magnetic ﬁeld. In other materials, they add together, yielding a magnetic ﬁeld around the material. This is no “monopole” analogy of magnetic ﬁelds as in the case of elec-tric ﬁelds, i.e., there are no “magnetic monopoles” that would lead to the deﬁnition of a magnetic ﬁeld by putting a test charge at rest and measuring the force acting on a particle. Instead, we consider a particle of charge q moving through a poing P with velocity v. A force F B is developed at P. The magnetic ﬁeld B at P is deﬁned as the vector ﬁeld such that F B = qv × B (1.6) Its units are tesla’s: T = newton/(coulomb)(meter/second) = N/(C/s)(m) or gauss’s (10−4 tesla). Since the motion of electric charge is called current, it is easily shown that B can be expressed in terms of the current i for various motions of charges. Ampere’s Law asserts that on any closed loop C in a plane, I C B · ds = µ0ienclosed where µ0 is the permeability constant, µ0 = 4π × 10−7 T-m/A, and ienclosed is the net current ﬂowing perpendicular to the plane in the planar region enclosed by C. For motion of a charge along a straight line, B = µ0i/2πR, R being the perpendicular distance from the inﬁnite line. Ampere’s law does not take into account the induced magnetic ﬁeld due to a change in the electric ﬂux. When this is taken into account, the above equality is replace by the Ampere-Maxwell Law, I C B · ds = µ0ienclosed + µ0ϵ0 d dtΦE where ΦE is the electric ﬂux, ΦE = Z A E · n dA n being a unit normal to the area A circumscribed by the closed loop C. Just as the motion of a charged particle produces a magnetic ﬁeld, so also does the change of a magnetic ﬁeld produce an electric ﬁeld. This is called an induced electrical ﬁeld and is characterized by Faraday’s Law: Consider 7 CHAPTER 1. ELECTROMAGNETIC WAVES e+ e− Dipole S N Magnet ⇔ Figure 1.5: Electric dipole analogous to a dipole caused by a magnet a particle of charge q moving around a closed loop C encompassing an area A with unit normal n. Then B induces an electric ﬁeld E such that I C E · ds = −d dt Z A B · n dA (1.7) From the fact that magnetic materials have poles of attraction and re-pulsion, it can be appreciated that magnetic structure can exist in the form of magnetic dipoles. Magnetic monopoles do not exist. For this reason, the net magnetic ﬂux through a closed Gaussian surface must be zero: I ∂Ω B · n dA = 0 (1.8) This is referred to as Gauss’ Law for magnetic ﬁelds. Every electron has an intrinsic angular momentum τ, called the spin angular momentum, and an intrinsic spin magnetic dipole moment µ, related by µ = −e mτ (1.9) where e is the elementary charge (1.6 × 10−19 C) and m is the mass of an electron (9.11 × 10−31 kg) 1.5 Some Properties of Waves The concept of a wave is a familiar one from everyday experiences. In gen-eral, a wave is a perturbation or disturbance in some physical quantity that 8 1.5. SOME PROPERTIES OF WAVES propagates in space over some period of time. Thus, an acoustic wave rep-resents the space and time variation of pressure perturbations responsible for sound; water waves, the motion of the surface of a body of water over a time period; electromagnetic waves, as will be established, characterize the evaluation of electrical and magnetic ﬁelds over time, but need no media through which to move as they propagate in a perfect vacuum at the speed of light. Mathematically, we can characterize a wave by simply introducing a function u of position x (or x = (x1, x2, x3) in three dimensions) and time t. In general, waves can be represented as the superposition of simple sinusoidal functions, so that the building blocks for wave theory are functions of the form, u(x, t) = u0ei(kx−ωt) (1.10) or, for simplicity, of the form u(x, t) = u0 sin(kx −ωt) (1.11) which are called plane waves. A plot of this last equation is given in Fig. 1.7. Here u0 = the amplitude of the wave k = the angular wave number ω = the angular frequency    (1.12) and λ = 2π/k = the wave length T = 2π/ω = the period (of oscillation) (1.13) The frequency ν of the wave is deﬁned as ν = 1/T (1.14) The wave speed v is deﬁned as the rate at which the wave pattern moves, as indicated in Fig. 1.6. Since point A retains its position on the wave crest as the wave moves from left to right, the quantity ϕ = kx −ωt must be constant. Thus dϕ dt = 0 = kdx dt −ω 9 CHAPTER 1. ELECTROMAGNETIC WAVES t t + ∆t ∆x A A Figure 1.6: Incremental motion of a wave from over a time increment ∆t so the wave speed is v = ω k = λ T = λν (1.15) The quantity ϕ = kx −ωt = the phase of the wave (1.16) Two waves of the form u1 = u0 sin(kx −ωt) and u2 = u0 sin(kx −ωt + ϕ) have the same amplitude, angular frequency, angular wave number, wave length, and period, but are out-of-phase by ϕ. These waves produce inter-ference when superimposed: u(x, t) = u1(x, t) + u2(x, t) = (2u0 cos ϕ/2) sin(kx −ωt + ϕ/2) (1.17) 10 1.5. SOME PROPERTIES OF WAVES u0 u(x, 0) = u0 sin(kx) x T t u(0, t) = −u0 sin(ωt) u0 λ Figure 1.7: Properties of a simple plane wave 11 CHAPTER 1. ELECTROMAGNETIC WAVES for ϕ = 0, the amplitude doubles while the wave length and period remains the same (constructive interference) ϕ = π, the waves cancel (u(x, t) ≡0) (destructive interference). Observe that ∂2u ∂t2 = −ω2u0 sin(kx −ωt) and ∂2u ∂x2 = −k2u0 sin(kx −ωt) so that u satisﬁes the second-order (hyperbolic) wave equation ∂2u ∂t2 − ω2 k2 ∂2u ∂x2 = 0 (1.18) and, again, ω/k is recognized as the wave speed. 1.6 Maxwell’s Equations Consider electric and magnetic ﬁelds E and B in a vacuum. Recall the following physical laws: Gauss’ Law (Conservation of charge in a volume Ωenclosed by a surface ∂Ω) ϵ0 Z ∂Ω E · n dA = qΩ= Z Ω ρ dx (1.19) where n is a unit vector normal to the surface area element dA and ρ is the charge density, qΩbeing the total charge contained in Ω. Faraday’s Law (The induced electromotive force due to a charge in magnetic ﬂux) Z C E · ds = −d dt Z A B · n dA (1.20) where C is a closed loop surrounding a surface of area A and n is a unit normal to dA. The total eletromotive force is R CE · ds. 12 1.6. MAXWELL’S EQUATIONS The Ampere-Maxwell Law (The magnetic ﬁeld produced by a current i) Z C B · ds = µ0ienclosed + µ0ϵ0 d dt Z A E · n dA = µ0 Z A j dA + µ0ϵ0 Z A n · ∂E ∂t dA (1.21) where C is a closed curve surrounding a surface of area A, j is the current density, and n is a unit vector normal to dA. The Absence of Magnetic Monopoles Z ∂Ω B · n dA = 0 (1.22) where ∂Ωis a surface bounding a bounded region Ω⊂R3. Applying the divergence theorem to the left-hand sides of (1.19) and (1.22) and Stokes’ theorem to the left-hand sides of (1.20) and (1.21), and arguing that the integrands of the resulting integrals must agree almost ev-erywhere, gives the following system of equations: ϵ0∇· E = 4πρ ∇× E = −∂B ∂t ∇× B = µ0j + µ0ϵ0 ∂E ∂t ∇· B = 0 (1.23) These are Maxwell’s equations of classical electromagnetics of macroscopic events in a vacuum. In materials other than a vacuum, the permittivity ϵ and the magnetic permeability µ may be functions of position x in Ω, and must satisfy constitutive equations D = ϵ0E + P = the electric displacement ﬁeld H = µ−1 0 B −M = the magnetic ﬁeld (1.24) 13 CHAPTER 1. ELECTROMAGNETIC WAVES with B now called the magnetic inductance, P the polarization vector and M the magnetic dipole. Then the equations are rewritten in terms of D, H, B, and j: ∇· D = ρf ∇× E = −∂B ∂t ∇· B = 0 ∇× H = jf + ∂D ∂t (1.25) where ρf and jf are appropriately scaled charge and current densities. When quantum and relativistic eﬀects are taken into account, an additional term appears in the ﬁrst and third equation and ∂/∂t is replaced by the total material time-derivative, d dt = ∂ ∂t + v · ∇, v being the velocity of the media. Let j = 0. Then ∂ ∂t∇× B = ∇× ∂B ∂t = −∇× ∇× E = µ0ϵ0 ∂2E ∂t2 so that we arrive at the wave equation, ∂2E ∂t2 − 1 µ0ϵ0 ∇× ∇× E = 0 (1.26) where the wave speed is c = 1 √µ0ϵ0 = 3.0 × 108 m/s (1.27) which is the speed of light in a vacuum. Electromagnetic disturbances (elec-tromagnetic waves) travel (radiate) at the speed of light through a vacuum. 14 1.6. MAXWELL’S EQUATIONS Gamma Ray 108 107 10−5 10−16 10−7 Infrared . . . . . . 10 107 1014 1015 Ultraviolet X-ray 106 Radio Long Waves 1018 1024 . . . 10−10 . . . Frequency 1 T (Hz) Wavelength λ (m) Figure 1.8: Wavelengths and frequencies of electromagnetic waves. B E ν (direction of propagation) Figure 1.9: Components of electromagnetic waves 15 CHAPTER 1. ELECTROMAGNETIC WAVES 1.7 Electromagnetic Waves Consider a disturbance of an electric (or magnetic) ﬁeld. According to what we have established thus far, this disturbance is radiated as an electromag-netic wave that has the following properties: • The electrical ﬁeld component E and the magnetic ﬁeld component B are normal to the direction of propagation v of the wave: ν = E × B/ \|E × B\| • E is normal to B: E · B = 0 • The wave travels at the speed of light in a vacuum • The ﬁelds E and B have the same frequency and are in phase with one another. Thus, while the wave speed c = ω/k = λ/T is constant, the wave length λ can vary enormously. The well-known scales of electromagnetic wave lengths is given in the accompanying ﬁgure. Some electromagnetic waves such as X-rays and visible light are radiated from sources that are of atomic or nuclear dimension. The quantum transfer of an electron from one shell to another radiates an electromagnetic wave. The wave propagates an energy packet called a proton in the direction ν of propagation. 16 Chapter 2 Introduction to Quantum Mechanics 2.1 Introductory Comments Quantum mechanics emerged as a theory put forth to resolve two fundamen-tal paradoxes that arose in describing physical phenomena using classical physics. First, can physical events be described by waves (optics, wave me-chanics) or particles (the mechanics of corpuscle or particles) or both? Sec-ond, at atomic scales, experimental evidence conﬁrms that events occur at quantum levels and are not continuous in time. Waves are characterized by frequency ν and wave number κ = k/(2π) (or period T = 1/ν or wave length λ = 1/κ) while the motion a particle is characterized by its total energy E and its momentum p. In atomic physics and in the case of electromag-netic waves, particles carry a deﬁnite quanta of energy and momentum and νλ = c. There elementary physical phenomena has a dual aspect and can be described by either waves or particles, and the principal physical attributes are related according to E = hv = ℏω , p = h λ = ℏk where h is Planck’s constant, and ℏ= h/2π. Matter Waves. In 1925, de Broglie put forth the hypothesis that the same dualism between the wave and corpuscle theories of light occur in matter; i.e., a material particle will have a matter wave corresponding to it just as 17 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS light quanta has a light wave. The wave characteristics are related through the formula, E = hν and p = h/λ. The situation is succinctly put forth by Messiah [13, page 59]: “...microscopic objects have a very general property: they appear under two separate irreconcilable aspects, the wave aspect, exhibiting the superposition property char-acteristic of waves, and the corpuscular aspect on the other hand, namely localized grains of energy and mo-mentum. There exists a universal relationship between these two aspects given by E = hν , p = h/λ ” 2.2 The Photoelectric Eﬀect The photoelectric eﬀect provides one of the ﬁrst examples of a phenomena that cannot be described by the classical theory of light but can be captured by a corpuscular theory that treats light as a beam of light quanta, photons, of energy hν, ν being the frequency of the radiation. The experimental setup involves subjecting an alkali metal to short-wave-length light in a high vacuum. The surface becomes positively charged as it gives oﬀnegative electricity in the form of electrons. Very precise experiments enable one to measure the total current and the velocity of the electrons. Observations conﬁrm that 1. the velocity of the emitted electrons depends only on the frequency ν of the light (the electromagnetic wave) 2. the emission of electrons is observed immediately at the start of radia-tion (contrary to classical theory) 3. the energy of the electron is, within a constant M of the material, proportional to the frequency ν, with constant of proportionality h, Planck’s constant: E = hν −M This last result is interpreted as follows: every light quantum striking the metal collides with an electron to which it transfers its energy. The electron 18 2.3. THE COMPTON EFFECT loses part of its energy M due to the work required to remove it from the metal. The velocity of the electron does not depend on the intensity of the light, but the number of electrons emitted does, and is equal to the number of incident light quanta. Property 3) is inferred by Einstein’s 1905 note on the photoelectric eﬀect in which, in generalization of Planck’s theory, he postulated that light radia-tion consists of a beam of protons of energy hν and velocity c, and he showed how this hypothesis could explain the photoelectric aﬀect. Experiments by Meyer and Gerlach in 1914 were in excellent agreement with this proposition. 2.3 The Compton Eﬀect The corpuscular nature of light was observed by Compton in 1922 in experi-ments in which a block of paraﬃn was bombarded by X-rays. It was observed that radiation scattered at an angle of less than 90◦possesses a wave length greater than the primary radiation so that the frequency ν′ of the scattered wave is smaller than the frequency ν of the incident radiation, a phenomena that cannot be reconciled with classical wave theory. Compton explained the process as one in which a light quantum (a pho-ton) of energy hν strikes an electron, transferring kinetic energy to the elec-tron while losing energy itself. The scattered photon has smaller energy hν′. The Compton formula, ∆λ = 2 h mc sin2 θ 2 gives the change in wave length of the photon due to the scattering process, m being the mass of the electron and θ the angle between the incident wave and the direction of the scattered light. 2.4 Heisenberg’s Uncertainty Principle We have seen that physical phenomena can be interpreted in terms of corpus-cles (particles) or in terms of waves. The uncertainty principle of Heisenberg addresses the surprising fact that it is impossible to determine that it is corpuscles or waves that one observes in a phenomena; more speciﬁcally, the position and momentum of an electron cannot be determined simultaneously. Wave and corpuscular views of nature events are said to be in duality (or 19 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS ∆y sin θ θ ∆y x y Figure 2.1: Diﬀraction of a beam of electrons to be complementary) in that if the corpuscular character of an experiment is proved, it is impossible to prove at the same time its wave character. Conversely, proof of a wave characteristic means that the corpuscular char-acteristic, at the same time, cannot be established. A classical example illustrating these ideas involves the diﬀraction of a beam of electrons through a slit, as indicated in Fig. 2.1. As the electrons pass through the slit, of width ∆y, the beam is accompanied by diﬀraction and diverges by an angle θ. The electron is represented by a de Broglie wave of wave length λ = h/p. Thus, ∆y sin θ ≈λ = h/p Likewise, the change in momentum in the y-direction is ∆p = ∆py = p sin θ Thus, ∆y∆py ≈h (2.1) 20 2.5. THE CORRESPONDENCE PRINCIPLE The precise position of an electron in the slit cannot be determined, nor can the variation in the momentum be determined with greater precision that ∆py (or h/∆y). The relation (2.1) is an example of Heisenberg’s uncertainty principle. Various other results of a similar structure can be deduced from quantum mechanics. Some examples and remarks follow: 1. A more precise analysis yields ∆y = q ⟨y2⟩−⟨y⟩2 , ∆p = q ⟨p2⟩−⟨p⟩2 ∆y∆p ≥1 2h where ⟨y⟩is the average measurement of position around y and similar deﬁnitions apply to p. 2. ∆t∆E ≥h Thus, just as position and momentum cannot be localized in time, a change in energy ∆E which accompanies a change in frequency and cannot be localized in time. 2.5 The Correspondence Principle It is diﬃcult to improve upon Born’s description of the correspondence prin-ciple: “The leading idea (Bohr’s correspondence principle, 1923) may be broadly stated as follows. Judged by the test of experience, the laws of clas-sical physics have brilliantly justiﬁed themselves in all processes of motion, macroscopic and microscopic, down to the motions of atoms as a whole (ki-netic theory of matter). It must therefore by laid down, as an unconditionally necessary postulate, that the new mechanics, supposed still unknown, must in all problems reach the same results as the classical mechanics. In other words, it must be demonstrated that, for the limiting cases of large masses and of orbits of large dimension, the new mechanics passes over into classical mechanics.” The major shortcoming of the classical theory is seen at the microscopic scale of atomistic physics where discontinuities (quanta) appear. Classical 21 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS theory accounts for phenomena in the limit where discontinuities are inﬁnitely small. “Quantum theory must approach Classical Theory in the limit of large quantum numbers.” [13, page 29] 2.6 Schr¨ odinger’s Equation Wave and particle mechanics are brought together in a way inspired by the uncertainty principle by introducing a wave function Ψ(x, t) associated with a particle from which the probability that the particle can be found at point x and at time t can be deduced. This wave function completely determines the dynamical state of the quantum system in the sense that all information on the dynamics of the system at time t can be deduced from knowledge of Ψ. The central goal of quantum theory, then, is this: knowing Ψ at an initial time t0, determine Ψ at later times t. To accomplish this, we must derive an equation of propagation of the wave Ψ, and this equation must connect the wave and corpuscular entities in a manner consistent with the observations made earlier. Finally, the resulting propagation equation must be consistent with the correspondence principle. The Case of a Free Particle. Considering ﬁrst the wave equation of a free particle (ignoring hereafter relativistic eﬀects), we began with the fact that a wave Ψ = ψ(x, t) is a superposition of monochromatic plane waves (here in one space dimension). As we have seen earlier, plane waves are of the form, Ψ(x, t) = ψ0ei(kx−ωt) where ψ0 is the amplitude. Then general wave function will be a superposi-tion of waves of this form. Since now the wave number k and the angular fre-quency ω are related to energy and momentum according to k = 2π/λ = p/ℏ and ω = 2π/ν = E/ℏ, we have Ψ(x, t) = ψ0ei(px−Et)/ℏ Thus, ∂Ψ ∂t = −i ℏEψ0ei(px−Et)/ℏ= −i ℏEΨ ∂Ψ ∂x = i ℏpψ0ei(px−Et)/ℏ= i ℏpΨ 22 2.6. SCHR ¨ ODINGER’S EQUATION Thus, E and p can be viewed as operators on Ψ: EΨ = −ℏ i ∂ ∂t Ψ pΨ = ℏ i ∂ ∂x Ψ (2.2) For a free particle of mass m the energy E and momentum p are related by E = p2 2m (2.3) Thus, introducing (2.2) into (2.3), we arrive at the following partial diﬀeren-tial equation for the wave functions: iℏ∂Ψ ∂t + ℏ2 2m ∂2Ψ ∂x2 = 0 (2.4) This is Schr¨ odinger’s equation for a free particle. Superposition in Rn. In general, we may use superposition of waves: Ψ(x, t) = Z Rn ϕ(p)ei(p·x−Et)/hdp to obtain iℏ∂ ∂tΨ(x, t) + ℏ2 2m∆Ψ(x, t) = 0 ∆= Laplacian. What is φ? Let t = 0. Then the initial condition is Ψ(x, 0) = ϕ(x) = Z Rn ϕ(p)eip·xdp i.e., ϕ(p) is the Fourier transform of the initial data, Ψ(x, 0). Hamiltonian Form. A popular way of writing Schr¨ odinger’s equation is to introduce the Hamiltonian, H(p, q) = E (q ∼x) Then the Hamiltonian operator is written H(p, q) = H h 2πi ∂ ∂q, q 23 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS and Schr¨ odinger’s equation becomes, H ℏ i ∂ ∂q, q + ℏ i ∂ ∂t Ψ = 0 (2.5) The Case of Potential Energy. If the particle is under the action of a force with potential energy V = V (q), then H(p, q) = p2 2m + V (q) (2.6) and Schr¨ odinger’s equation becomes −1 2mℏ2 ∂2 ∂q2 + V (q) + ℏ i ∂ ∂t Ψ = 0 (2.7) Multiple Particles. The number of independent variables in the domain of the wave function, in addition to time t, is equal to the number of particles associated with the atom or atoms under study. Thus, a complex atom con-sisting of a nucleus at position x0 and Z electrons at positions x1, x2, . . . , xZ has a wave function Ψ = Ψ(x0, x1, . . . , xZ, t). Relativistic Quantum Mechanics. While we are not going to cover relativistic eﬀects in these notes, the wave equation for this case easily follows from the fact that in this case, the energy of a free particle is given by E2 = p2c2 + m2c2 From this we deduce the equation 1 c2 ∂2 ∂t2 −∆+ mc ℏ 2 Ψ(x, t) = 0 (2.8) This is called the Klein-Gordan equation. Henceforth, we restrict ourselves to non-relavistic quantum mechanics. General Formulations of Schr¨ odinger’s Equations. The basic plan is to con-sider a dynamical system of N particles with coordinates q1, q2, . . . , qN and momenta. 24 2.7. ELEMENTARY PROPERTIES OF THE WAVE EQUATION p1, p2, . . . , pN for which the Hamiltonian is a functional, H = H(q1, q2, . . . , qN; p1, p2, . . ., pN; t) To this dynamical system there corresponds a quantum system represented by a wave function Ψ(q1, q2, . . . , qN, t). Setting E = iℏ∂ ∂t and pr = ℏ i ∂ ∂qr ; r = 1, 2, . . ., N (2.9) Schr¨ odenger’s equation becomes, iℏ∂ ∂t −H q1, q2, . . . , qN; ℏ i ∂ ∂q1 , ℏ i ∂ ∂q2 , . . . , ℏ i ∂ ∂qN Ψ(q1, q2, . . . , qN, t) = 0 (2.10) The rules (2.9) showing the correspondence of energy and momentum to the diﬀerential operators shown hold only in Cartesian coordinates in the form indicated, as Schr¨ odinger’s equation should be invariant under a rotation of the coordinate axes. 2.7 Elementary Properties of the Wave Equa-tion We will undertake a basic and introductory study of Schr¨ odinger’s equation, ﬁrst for a single particle in one space dimension, and then generalize the analysis by considering a more general mathematical formalism provided by function space settings. The major source here is Griﬃth’s book, Introduction to Quantum Mechanics , but we also consult the books of Born and Messiah . 1. Review. The Schr¨ odinger equations governing the dynamics of a single particle of mass m moving along a line, the x-axis, subjected to forces derived from a potential V is iℏ∂Ψ ∂t + ℏ2 2m ∂2Ψ ∂x2 −V Ψ = 0 (2.11) 25 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS where ℏ = h 2π = Planck’s constant /2π = 1.054573 × 10−34 J s Ψ = Ψ(x, t) = the wave function To (2.11) we must add an initial condition Ψ(x, 0) = ϕ(x) (2.12) where ϕ is prescribed. The wave function has the property Ψ∗Ψ = \|Ψ(x, t)\|2 = ρ(x, t) where Ψ∗= the complex conjugate of Ψ and ρ(x, t) = the probability distribution function associated with Ψ, such that ρ(x, t)dx = the probability of ﬁnding the particle between x and x + dx at time t The wave function must be normalized since ρ is a PDF: Z ∞ −∞ \|Ψ(x, t)\|2 dx = 1 (2.13) Postulate. If Ψ = Ψ(x, t) is a solution of Schr¨ odinger’s equation (2.11), then d dt Z ∞ −∞ \|Ψ(x, t)\|2 dx = 0 (2.14) Proof. d dt Z ∞ −∞ \|Ψ\|2 dx = Z ∞ −∞ ∂ ∂ Ψ∗(x, t)Ψ(x, t) dx = Z ∞ −∞ Ψ∗∂Ψ ∂t + ∂Ψ∗ ∂t Ψ dx 26 2.8. MOMENTUM But ∂Ψ ∂t = iℏ 2m ∂2Ψ ∂x2 + i ℏV Ψ ∂Ψ∗ ∂t = −iℏ 2m ∂2Ψ∗ ∂x2 −i ℏV Ψ∗ So ∂ ∂t \|Ψ\|2 = Ψ∗∂Ψ ∂t + ∂Ψ∗ ∂t Ψ = iℏ 2m Ψ∗∂2Ψ ∂x2 −∂2Ψ∗ ∂x2 Ψ = iℏ 2m ∂ ∂x Ψ∗∂Ψ ∂x −∂Ψ∗ ∂x Ψ (2.15) Hence, d dt Z ∞ −∞ \|Ψ(x, t)\|2 dx = iℏ 2m Ψ∗Ψ ∂x −∂Ψ∗ ∂x Ψ x→+a x→−a = 0 since Ψ, Ψ∗→0 as x →±∞in order that (2.13) can hold. □ 2.8 Momentum As noted earlier, momentum can be viewed as an operator, ρΨ = ℏ i ∂ ∂x Ψ (2.16) 27 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS Another way to interpret this is as follows. If ⟨x⟩is the expected value of the position x of the particle, d ⟨x⟩ dt = Z ∞ −∞ x ∂ ∂t(Ψ∗Ψ)dt = iℏ 2m Z ∞ −∞ x ∂ ∂x Ψ∗∂Ψ ∂x −∂Ψ∗ ∂x Ψ dx (from (2.15)) = −iℏ 2m Z ∞ −∞ Ψ∗∂Ψ ∂x −∂Ψ∗ ∂x Ψ dx (integrating by parts) = −iℏ m Z ∞ −∞ Ψ∗∂Ψ ∂x dx = 1 m Z ∞ −∞ Ψ∗ ℏ i ∂ ∂x Ψdx = 1 m Z ∞ −∞ Ψ∗ρΨdx which we denote by ⟨ρ⟩/m: ⟨p⟩= md ⟨x⟩ dt = Z ∞ −∞ Ψ∗ρΨdx (2.17) Thus, the expected or mean value of the momentum is related to the time-rate-of-change of the mean position ⟨x⟩in a way consistent with (in corre-spondence with, as required) to classical notion of momentum. The Heisenberg uncertainty principle can now be stated as, σxσρ ≥ℏ 2 (2.18) where σ2 x = x2 −⟨x⟩2 , σ2 ρ = ρ2 −⟨ρ⟩2 2.9 Wave Packets / Fourier Transforms Consider again the case of rectilinear motion of a free particle x in a vacuum. The wave function is Ψ(x, t) = ψ(x)eiωt = ψ(x)e−iEt/ℏ (2.19) 28 2.10. THE WAVE-MOMENTUM DUALITY and ψ(x) satisﬁes −ℏ2 2m d2ψ dx2 = Eψ (2.20) The solution is Ψ(x, t) = Aeik(x−hkt 2m ) (2.21) where k = ± r 2mE ℏ (2.22) This particular solution is not normalizable ( R R Ψ∗Ψdx →∞), so that a free particle cannot exist in a stationary state. The spectrum in this case is continuous, and the solution is of the form, Ψ(x, t) = 1 √ 2π Z ∞ −∞ ϕ(k)ei(kx−ℏk2t/2m)dk (2.23) Ψ(x, 0) = 1 √ 2π Z ∞ −∞ ϕ(k)eikxdx (2.24) = ⇒ ϕ(k) = 1 √ 2π Z ∞ −∞ Ψ(x, 0)eikxdx (2.25) Equation (2.23) characterizes the wave as a sum of wave packets. It is a sinusoidal function modulated by the function ϕ. The wave function is the Fourier transform of ϕ and ϕ is the inverse Fourier transform of the initial value Ψ(x, 0) of Ψ. Instead of a particle velocity as in classical mechanics, we have a group velocity of the envelope of wavelets; 2.10 The Wave-Momentum Duality In classical mechanics, the dynamical state of a particle is deﬁned at every instant of time by specifying its position x and its momentum p. In quantum mechanics, one can only deﬁne the probability of ﬁnding the particle in a given region when one carries out a measurement of the position. Similarly, one cannot deﬁne the momentum of a particle; one can only hope to deﬁne 29 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS the probability of determining the momentum in a given region of momentum space when carrying out a measure of momentum. To deﬁne the probability of ﬁnding momentum p in a volume (p, d+dp), we consider, for ﬁxed time t, the Fourier transform Φ(p) of the wave function Ψ: Φ(p) = 1 (2πℏ)n/2 R Rn Ψ(x)e−ip·xdx Ψ(x) = 1 (2πℏ)n/2 R Rn Φ(p)eip·xdp ) (2.26) (with dx = dx1 · · · dxN, dp = dp1 · · · dpN). Thus, the wave function can be viewed as a linear combination of waves exp(ip · x/ℏ) of momentum p with coeﬃcients (2πℏ)−n/2Φ(p). The probability of ﬁnding a momentum p in the volume (p, dp) is π(p) = Φ∗(p)Φ(p) and we must have Z Rn Φ∗(p)Φ(p)dp = 1 (2.27) Thus, the Fourier transform F : L2(Rn) →L2(Rn) establishes a one-to-one correspondence between the wave function and the momentum wave function. Equation (2.27) follows from Plancherel’s identity, ⟨Ψ\|Ψ⟩= ∥Ψ∥2 = ⟨F(Ψ)\|F(Ψ)⟩= ⟨Φ\|Φ⟩ (2.28) The interpretation of the probability densities associated with Φ and Ψ is important. When carrying out a measurement on either position or mo-mentum, neither can be determined with precision. The predictions of the probabilities of position and momentum are understood to mean that a very large number N of equivalent systems with the same wave function Ψ are considered. A position measurement on each of them gives the probability density Ψ∗(x)Ψ(x) results in the limit as N approaches inﬁnity. Similarly, Φ∗(x)Φ(x) gives the probability density of results of measuring the momen-tum. 2.11 Appendix Probability. (Inspired by Griﬃth’s Example [6, page 5]) Of 14 people in a room, the distribution of ages is as follows: 30 2.11. APPENDIX 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 1 2 3 4 N(j) j Figure 2.2: Histogram showing the number of people versus age. 1 aged 14 2 aged 15 1 aged 16 1 aged 18 4 aged 19 3 aged 23 1 aged 24 • The total number of people in the room N = ∞ X j=0 N(j) = 13 • The Probability a person selected randomly is of age j P(j) = N(j) N • The probability of getting either 14 or 15 is P(14) + P(15) = 3/13. The sum of all probabilities, ∞ X j=0 P(j) = 1 N ∞ X j=0 N(j) = 1 31 CHAPTER 2. INTRODUCTION TO QUANTUM MECHANICS • The mean age, denoted ⟨j⟩: ⟨j⟩= P jN(j) N = ∞ X j=0 jP(j) Check it out: (14) + 2(15) + (16) + (18) + 4(19) + 3(23) + (24) 13 = 247 13 = 19 = 14 1 13 + 15 2 13 + 16 1 13 + 18 1 13 + 19 4 13 + 23 3 13 + 24 1 13 = 14P(14) + 15P(15) + 16P(16) + 18P(18) + 19P(19) + 23P(23) + 24P(24) = ∞ X j=0 jP(j) We easily verify that j2 = X j2P(j) In general ⟨f(j)⟩= ∞ X j=0 f(j)P(j) = expected value of f The mean is the same (as is the most probable value). To distinguish between them, a measure of spread is needed. • The spread is measured by the variance σ2 = (∆j)2 = n X j=0 (j −⟨j⟩)2P(j) and σ = p ⟨(∆j)2⟩= the standard deviation. 32 2.11. APPENDIX 1 2 3 9 10 4 5 6 7 8 j N(j) 1 2 3 9 10 4 5 6 7 8 j N(j) Figure 2.3: Two histograms with the same median, average, and probable value, but diﬀerent standard deviations. • It is easy to show that σ2 = j2 −⟨j⟩2 For the continuous case, Pab = Z b a ρ(x)dx = probability that x ∈[a, b] Z ∞ −∞ ρ(x)dx = 1 ⟨f(x)⟩= Z ∞ −∞ f(x)ρ(x)dx σ2 = x2 −⟨x⟩2 , ⟨x⟩= Z ∞ −∞ xρ(x)dx 33 Chapter 3 Dynamical Variables and Observables in Quantum Mechanics: The Mathematical Formalism 3.1 Introductory Remarks We have seen that the solutions of Schr¨ odinger’s equations are wave functions Ψ = Ψ(x, t) that have the property that \|Ψ\|2 = Ψ∗Ψ(x, t) is the probabil-ity distribution function giving the probability that the elementary particle under study is at position x at time t (actually that the particle is in the vol-ume between x and x + dx). Moreover, the knowledge of the wave function Ψ (or equivalently, the momentum wave function Φ = Φ(p, t)) determines completely the dynamical state of the quantum system. We shall now build on these ideas and develop the appropriate mathematical setting for an oper-ator theoretic framework for quantum mechanics that brings powerful tools and concepts to the theory. Everything we derive is applicable to functions deﬁned on RN, but vir-tually all of the results can be demonstrated without loss in generality in R. The notation for the spatial coordinate x will be used interchange-ably with q (or x with q in R), as q is the classical notation for a gen-eralized coordinate in “phase space” where coordinate - momentum pairs (q,p) = (q1, q2, ..., qN, p1, p2, ..., pN) deﬁne dynamical states. We will fre-quently treat dynamical variables, such as momentum p, as operators, and 35 CHAPTER 3. THE MATHEMATICAL FORMALISM when it is important to emphasize the operator character of the result, we will aﬃx a tilde to the symbol (i.e. for momentum p, the associated oper-ator is e p = −iℏd dq). Thus, a function F = F(q, p) is associated with an operator e F(q1, q2, ..., qN, −ih ∂ ∂q1, −iℏ∂ ∂q2, ..., −iℏ ∂ ∂qn) etc. The coordinates (q1, q2, ..., qN) will hereafter be understood to be cartesian coordinates be-cause the operator notation must represent dynamical quantities in a way that is invariant under a change (e.g. a rotation) of the coordinate axes. Indeed, while ordinary multiplication of functions is commutative, the corre-sponding operators may not commute, so the qi are interpreted as cartesian coordinates to avoid ambiguity. 3.2 The Hilbert Space L2(R) (or L2(RN)) Since the wave function must have the property that \|Ψ\|2 is integrable over R, (Ψ is “square-integrable”), it must belong to the following space: L2(R) (or L2(RN)) is the space of equivalence classes [u] of measurable complex valued functions equal almost everywhere on R (or RN)(v ∈[u] ⇒v = u everywhere except a set of measure zero) such that \|u\|2 = u∗u is Lebesgue integrable on R(orRN). Thus, u ∈L2(R) implies that u represents an equivalence class [u] of functions equal almost everywhere on R such that R R u2dx < ∞. The space L2(R) is a complete inner product space with inner product ⟨ψ, ϕ⟩= Z IR ψ∗ϕdx Where ψ∗is the complex conjugate of ψ. Thus L2(R) is a Hilbert space. The associated norm on L2(R) is then ∥ψ ∥= p ⟨ψ, ψ⟩ It can be shown that L2(R) is reﬂexive (in particular, by the Riesz the-orem, for every continuous linear functional f in the dual (L2(R))′, there is a unique uf ∈L2(R) such that f(v) = ⟨v, uf⟩and ∥f ∥(L2(R))′ = ∥uf ∥). Also, L2(R) is separable, meaning that it contains countable everywhere dense 36 3.3. DYNAMICAL VARIABLES AND HERMITIAN OPERATORS sets. In other words, for any u ∈L2(R), and any ε > 0, there exists an inﬁnite sequence of functions {un}∞ n=1 in L2(R) and an integer M > 0 such that ∥un −u ∥< ε for all n > M. We shall demonstrate how such countable sets can be computed given any u in the next section. 3.3 Dynamical Variables and Hermitian Op-erators A dynamical variable is some physical feature of the quantum system that depends upon the physical state of the system. In general, we adopt the convention that the state is described by the position coordinates q (or q1, q2, ..., qN; p1, p2, ..., pN). Thus, a dynamical variable is a function Q = Q(q1, q2, ..., qN; p1, p2, ..., pN) Since the momentum components pj can be associated as with the operators pj = −iℏ∂/∂qj, dynamical variables likewise characterize operators, Q(q1, q2, ..., qN; p1, p2, ..., pN) = e Q(q1, q2, ..., qN; −ih ∂ ∂q1 , −iℏ∂ ∂q2 , ..., −iℏ∂ ∂qN (Analogously, one could deﬁne an operator Q(q; p) = ˆ Q(iℏ∂ ∂p1 , −iℏ∂ ∂p2 , ..., −iℏ∂ ∂pN ; p1, p2, ..., pN)). The expected value or mean of a dynamical variable Q for quantum state Ψ is denoted ⟨Q⟩and is deﬁned as (RN, N = 1), ⟨Q⟩ = Z R ψ∗Qψdq/ Z R ψ∗ψdq = D Ψ, ˜ QΨ E ⟨Ψ, Ψ⟩ (3.1) Any operator A : L2(R) →L2(R) is said to be Hermitian if ⟨ψ, Aϕ⟩= ⟨Aψ, ϕ⟩∗ ∀ψ, ϕ ∈L2(R) (3.2) 37 CHAPTER 3. THE MATHEMATICAL FORMALISM Any operator Q that corresponds to a genuine dynamical variable of a physical system, must be such that its expected value ⟨Q⟩is real; i.e., ⟨Q⟩= D ψ, ˆ Qψ E / ⟨ψ, ψ⟩ is real. Therefore, for such Q we must have D ψ, ˜ Qϕ E = D ˜ Qψ, ϕ E ∀ψ, ϕ ∈L2(R) In other words, every dynamical variable must be characterized by a Hermi-tian operator on L2(R) (or L2(RN). We note that the variance of a dynamical variable is deﬁned as σ2 Q = Q2 −⟨Q⟩2 (3.3) The dynamical variable Q takes on the numerical value ⟨Q⟩with certainty if and only if σ2 Q = 0. This observation leads us to a remarkable property of those dynamical variables that can actually be measured in quantum systems with absolute certainty. Since σ2 Q = Q2 −⟨Q⟩2 = ⟨ψ, Q2ψ⟩ ⟨ψ, ψ⟩ − ⟨ψ, Qψ⟩ ⟨ψ, ψ⟩ 2 = ⟨Qψ, Qψ⟩ ⟨ψ, ψ⟩ −⟨ψ, Qψ⟩2 ⟨ψ, ψ⟩2 If σ2 Q = 0, we have D ψ, ˜ Qψ E2 =∥˜ Qψ ∥2 ∥ψ ∥2 ∀ψ ∈L2(R) (3.4) This is recognized as the Cauchy-Sworz inequality for D ψ, ˜ Qψ E in the case in which the equality holds, which happens when the functions Qψ and ψ are proportional, i.e. when there exists a constant λ ∈C such that ˜ Qψ = λψ (3.5) 38 3.4. SPECTRAL THEORY OF HERMITIAN OPERATORS This is recognized as the eigenvalue problem for the Hermitian operator ˜ Q, with eigenvalue - eigenfunction pair (λ, ψ). Therefore, the “ﬂuctuations” (variances) of the dynamical variable Q from its mean ⟨Q⟩vanish for states ψ = ψλ ∈L2(R) which are eigenfunctions of the operator ˜ Q. Notice that if \|ψ\|2 = 1, then D ψ, ˜ Qψ E = ⟨Q⟩= λ ⟨ψ, ψ⟩= λ; i.e. the expected value ⟨Q⟩is an eigenvalue of ˜ Q. We summarize this ﬁnding in the following theorem, Theorem 1. The dynamical variable Q of a quantum system possesses with certainty (with probability 1) the well-deﬁned value ⟨Q⟩if and only if the dynamical state of the system is represented by an eigenfunction ψ of the Hermitian operator ˜ Q : L2(R) →L2(R) associated with Q; moreover, the expected value ⟨Q⟩is an eigenvalue of ˜ Q. □ This fundamental result establishes the connection of quantum mechanics with the spectral theory of Hermitian operators. We explore this theory in more detail for the case of a discrete spectrum. 3.4 Spectral Theory of Hermitian Operators: The Discrete Spectrum Returning to the eigenvalue problem (3.5), let us consider the case in which there exists a countable but inﬁnite set of eigenvalues λk and eigenfunctions ϕk ∈L2(R), k = 1, 2, ... for the operator ˜ Q. In this case, ˜ Q is said to have a discrete spectrum. The basic properties of the system in this case are covered or derived from the following theorem, Theorem 2. Let (λk, ϕk), k = 1, 2, ... denote a countable sequence of eigenvalue - eigenfunction pairs for the Hermitian operator ˜ Q : L2(R) → L2(R); i.e. ˜ Qϕk = λkϕk, k = 1, 2, ... (3.6) Then 1. if ϕk is an eigenfunction, so also is c ϕk, c being any constant; 2. if ϕ(1) k , ϕ(2) k , ..., ϕ(M) k are M eigenfunctions corresponding to the same eigenvalue λk, then any linear combination of these eigenfunctions is an eigenfunction corresponding to λk; 39 CHAPTER 3. THE MATHEMATICAL FORMALISM 3. the eigenvalues are real; 4. the eigenfunctions ϕk can be used to construct an orthonormal set, i.e., a set of eigenfunctions of ˜ Q such that ⟨ϕk, ϕm⟩= δkm = 1 if k ̸= m 0 if k = m (3.7) 5. any state ψ ∈L2(R) can be represented as a series, ψ(q) = ∞ X k=1 ckϕk(q) (3.8) where ck = ⟨ψ, ϕk⟩ (3.9) and by (3.8) we mean lim m→∞∥ψ − m X k=1 ckϕk ∥= 0 (3.10) Proof. Parts 1 and 2 are trivial. In property 2, the eigenfunctions are said to have a degeneracy of order M. To show 3, we take the inner product of both sides of (6) by ϕm and obtain the number λk = D ϕm, ˜ Qϕk E / ⟨ϕm, ϕk⟩ which is real, because ˜ Q is Hermitian. 4. Consider, ˜ Qϕk = λkψk and ψmϕm , m ̸= k Then, D ϕm, ˜ Qϕk E = λk ⟨ϕm, ϕk⟩= D ˜ Qϕm, ϕk E = λm ⟨ϕm, ϕk⟩ Thus, (λk −λm) ⟨ϕm, ϕk⟩= 0 for m ̸= k, λk ̸= λm. 40 3.4. SPECTRAL THEORY OF HERMITIAN OPERATORS 5. The eigenfunctions ϕk are assumed to be normalized: ⟨ϕk, ϕk⟩= 1 , k = 1, 2, ... Thus they form an orthonormal basis for L2(R). Equation (3.8) is then just the Fourier representation of ψ with respect to this bases. Equation (3.9) follows by simply computing ⟨Σmcmϕm, ϕk⟩and using (3.7). □ Various functions of the operator ˜ Q can likewise be given a spectral rep-resentation. For instance, if ˜ Qϕk = λkϕk, then ˜ Q2ϕk = ˜ Q( ˜ Qϕk) = ˜ Qλkϕk = λk ˜ Qϕk = λ2 kϕk, and, in general ( ˜ Q)rϕk = λr kϕk and, symbolically, (sin ˜ Q)ϕk = ( ˜ Q −1 3! ˜ Q3 + · · · )ϕk = (λk −1 3!λ3 k + · · · ) = sin(λk)ϕk etc. In general, if ˜ Q is a Hermitian operator with discrete eigenvalue - eigen-function pairs (λk, ϕk) and F is any smooth function on R, we may write F( ˜ Q) = ∞ X k=1 F(λk)ϕk (3.11) For ˜ Q : L2(R) →L2(R), (11) is meaningful if the series converges; i.e. if the sequence of real numbers, 41 CHAPTER 3. THE MATHEMATICAL FORMALISM µn = n X k=1 \|F(λk)\|2 converges in R. For F( ˜ Q) = eiξ ˜ Q , ξ ∈R, this series always converges. In particular, eiξ ˜ Qψ = X k ckeiξ ˜ Qϕk = X k X m ckeiξλmϕmϕk and ckeiξλm 2 = \|ck\|2 →0 as m →∞. 3.5 Observables and Statistical Distributions The statistical distribution of a quantity Q associated with a quantum dy-namical system is established by its characteristic function a : R →R. In general, the characteristic function is, to within a constant, the Fourier trans-form of the probability density function ϱ associated with a random variable X: a(ξ) = Z ∞ −∞ eiξxϱ(x)dx (3.12) (ϱ(x) being the probability of ﬁnding X in (x, x + dx)). Thus, a(ξ) is the expected value of eiξx: a(ξ) = eiξx (3.13) If the random variable X can only assume discrete values xk, k = 1, 2, · · · , and if ϱ1, ϱ1, · · · are the probabilities of these values, then a(ξ) = ∞ X k=1 ϱkeiξxk (3.14) 42 3.5. OBSERVABLES AND STATISTICAL DISTRIBUTIONS with P k ϱk = 1. Returning now to quantum dynamics, we shall refer to any dynamical variable Q of a quantum system characterized by a Hermitian operator ˜ Q on L2(R) with a discrete spectrum of eigenvalues λk as an observable. The reason that term is appropriate becomes clear when we consider the characteristic function of the quantum system. For the wave function Ψ representing the dynamical state of the system, a(ξ) = D Ψ, eiξ ˜ QΨ E ⟨Ψ, Ψ⟩ = ∞ X k=1 ϱkeiξλk (3.15) where now ϱk = \|ck\|2 / ⟨Ψ, Ψ⟩ = \|⟨ϕk, Ψ⟩\|2 / ⟨Ψ, Ψ⟩ (3.16) Comparing (3.14) and (3.15), we arrive at the following theorem: Theorem 3 Let Q be an observable of a quantum dynamical system with the discrete spectrum of eigenvalues {λk}∞ k=1. Then 1. The only values Q can assume are precisely the eigenvalues λ1, λ2, · · ·λm, · · · , and 2. the probability that Q takes on the value λk is ϱk = \|ck\|2 = \|⟨ϕk, Ψ⟩\|2 where ϕk is the eigenfunction corresponding to λk and Ψ is the nor-malized wave function, ⟨Ψ, Ψ⟩= 1. □ We observe that the discrete probabilities ϱk satisfy 1 = X k ϱk = X k \|⟨ϕk, Ψ⟩\|2 ⟨Ψ, Ψ⟩ = X k \|ck\|2 as required. 43 CHAPTER 3. THE MATHEMATICAL FORMALISM 3.6 The Continuous Spectrum Not all dynamical variables (Hermitian operators) have a discrete spectrum; i.e. not all are observables in quantum dynamical systems. For example, the momentum operator p = −iℏd/dq is Hermitian but does not have a discrete spectrum and, therefore, is not an observable (in keeping with our earlier view of the uncertainty principle for position and momentum). In particular, the eigenvalue problem for p is (Cf. Messiah p.178) pU(p ′, q) = p ′U(p ′, q) (3.17) where U(p ′, q) is the eigenfunctions of p with an eigenvalue λ = p ′, a contin-uous function of the variable p ′; i.e. the spectrum of p = ℏ i d dq is continuous. We easily verify that U(p ′, q) = 1 √ 2πℏ eip ′q/ℏ (3.18) By analogy with the Fourier series representation (3.8) of the dynamical state ψ for discrete spectra, we now have the Fourier transform representation of the state for the case of a continuous spectrum: ψ(q) = 1 √ 2πℏ Z R ϕ(p ′)eip ′q/ℏdp ′ (3.19) In analogy with (3.9), ϕ(p ′) = D U(p ′, q), ψ(q) E (3.20) The analogue arguments cannot be carried further. Indeed, the function (19) is not in L2(R) and a more general setting must be constructed to put the continuous spectrum case in the proper mathematical framework. 3.7 The Generalized Uncertainty Principle for Dynamical Variables Recall that for any dynamical variable characterized by a Hermitian operator Q, its expected value in the state Ψ is ⟨Q⟩= D Ψ, ˜ QΨ E 44 3.7. THE GENERALIZED UNCERTAINTY PRINCIPLE where ˜ Q is the operator associated with Q : (Q(x, p, t) = ˜ Q(x, ℏ i ∂ ∂x, t)), and ⟨Ψ, Ψ⟩= 1. Recall that its standard deviation (actually, its variance) is σ2 Q = D ( ˜ Q −⟨Q⟩2) E = D Ψ, ( ˜ Q −⟨Q⟩2)Ψ E (3.21) For any other such operator M, σ2 M = D Ψ, ( ˜ M −⟨M⟩2)Ψ E = D ( ˜ M −⟨M⟩2)Ψ, ( ˜ M −⟨M⟩2)Ψ E Noting that for any complex number z, \|z\|2 ≥(Im(z))2 ≥[ 1 2i(z −z)]2, it is an algebraic exercise to show that σ2 Qσ2 M ≥( 1 2i D ˜ Q ˜ M −˜ M ˜ Q E )2 (3.22) where [Q, M] is the commutator of the operators Q and M: [Q, M] = ˜ Q ˜ M −˜ M ˜ Q (3.23) The generalized Heisenberg uncertainty principle is this: for any pair of in-compatible observables (those for which the operators do not commute, or [Q, M] ̸= 0), the condition (3.22) holds. Such incompatible observables can-not share common eigenfunctions. To demonstrate that (3.22) is consistent with the elementary form of the uncertainty principle discussed earlier, set ˜ Q = x and ˜ M = ˜ p = (ℏ/i)d/dx. Then, for a test state ψ, [˜ x, ˜ p]ψ = xℏ i dψ dx −ℏ i d dx(xψ) = iℏψ Hence, σ2 xσ2 p ≥ 1 2iih 2 = ℏ 2 2 or σxσp ≥ℏ 2 45 CHAPTER 3. THE MATHEMATICAL FORMALISM in agreement with the earlier calculation. In the case of a Hermitian (or self-adjoint) operator Q with eigenvalue-eigenvector pairs {(λk, ϕk)}∞ k=1, we have seen that Qψ = ∞ X k=1 λkckϕk, ck = ⟨ψ, ϕk⟩ The projection operators {Pk} deﬁned by Pkψ = ⟨ψ, ϕk⟩ϕk = ckϕk have the property that ψ = Iψ = ∞ X k=1 ckϕk = ∞ X k=1 Pkψ so that, symbolically, ∞ X k=1 Pk = 1 = the identity. (3.24) For this reason, such a set of projections is called a resolution of the identity. The operators Q can thus be represented as Q = ∞ X k=1 λkPk The same type of construction can be extended to cases in which Q has a continuous spectrum. In such cases, for any self-adjoint operator Q, there exists a unique family of projection {Pλ}, −∞< λ < + < ∞, such that Pλ = 0 for λ < m, Pλ = I for λ > m, m and M being lower and upper bounds of Q, and lim λ→M ∥Pλψ −PMψ ∥= 0 ∀ψ ∈L2(R) or, in particular, lim λ→M+ ∥Pλψ −1ψ ∥= 0. For this reason the family is still referred to as a resolution of the identity. In this case, we write Q = Z M+ε m λdPλ , 0 < ε < 1 (3.25) 46 3.7. THE GENERALIZED UNCERTAINTY PRINCIPLE in analogy with (3.24). Likewise, if f is a continuous complex-valued function deﬁned on the interval [m, ˆ M + ε], we write f(Q) = Z M+ε m f(λ)dPλ (3.26) The integral in (3.26)(and (3.25)) is understood as the limit of the sum P k f(λk)(Pλk −Pλk−1) on a partition m = λ0 < λ1 < · · · < λn = M of [m, M + ε] as max k (λk −λk−1) →0. Returning to the Hamiltonian operator H, if ⟨Hψ, ϕ⟩= Z ∞ −ω λd(Pλψ, ϕ) (3.27) for all ψ, ϕ ∈L2(R)−∞, the symbolic relation, H = Z ∞ −ω λdPλ (3.28) can be written in analogy to the spectral representation in the discrete spec-trum case. For any such dynamical variable characterized by a Hermitian operator Q (or, equivalently, ˜ Q), if dPλψ →0 as λ →+ −∞or dPλψ →ψ, and if dPλψ →dPλ0ψ for λ →λ0 , λ ≥λ0, (with dPλ1, ≤dPλ2 for λ1 ≤λ2), we can write D ˜ Qψ, ϕ E = Z +∞ −ω λ(dPλψ, ϕ) (3.29) 47 Chapter 4 Selected Topics and Applications 4.1 Introductory Remarks In this chapter, we collect discussions of special topics that complete an introduction to quantum mechanics. These include the elementary example of the case of the harmonic oscillator for which the notion of ground states and quantization of energy levels is immediately derivable from Schr¨ odinger’s equation. The calculation of the quantum dynamics (the wave function) for the hydrogen atom is also described, and is a fundamental application of the theory developed thus far. We also consider the characterization of angular momentum and spin of particles in quantum systems. 4.2 Ground States and Energy Quanta: the Harmonic Oscillator Let us return to the simple case of a particle moving along a straight line with position coordinate x = q. We may then consider wave functions of the form, Ψ(q, t) = ψ(q)eiEt/ℏ (4.1) Then Schr¨ odinger’s equation reduces to H(ℏ i d dq, q) −E ψ(q) = 0 (4.2) 49 CHAPTER 4. SELECTED TOPICS AND APPLICATIONS Thus, the energy E is an eigenvalue of the Hamiltonian H. The Hamiltonian, we recall, is of the form H(p, q) = p2 2m + V (q) = −ℏ2 2m d2 dq2 + V (q) (4.3) Hence, to complete the deﬁnition of the operator ˜ H, we need to specify the potential V . A classical and revealing example concerns the case in which V represents the potential energy of a harmonic oscillator vibrating with angular frequency ω: V (q) = 1 2mω2q2 (4.4) Then (4.2) becomes −ℏ2 2m d2 dq2 + 1 2mω2q2 ψ = Eψ (4.5) or d2 dq2 + λ −α2q2 ψ = 0 (4.6) where λ = 2mE/ℏ2, α = mω/ℏ (4.7) The solutions of (4.6) are of the form ψ(q) = ae−αq2/2 and a direct substitution reveals that the corresponding eigenvalue is λ = α = mv/ℏ Thus, the corresponding energy is E = E0 = 1 2ℏω (4.8) This lowest possible energy corresponds to the ground state, and is seen to be half of Planck’s energy quantum. 50 4.3. THE HYDROGEN ATOM The remaining solutions of the wave equation can be found by standard power-series expansions (method of Frobenius) to be of the form ψn(q) = mω πℏ 1/4 · 1 √ n!2nHn(ξ)e−ξ2/2 (4.9) where Hn(·) are Hermite polynomials of order n and ξ = q(mω/ℏ)1/2. The corresponding energy states are En = (n + 1 2)ℏω , n = 0, 1, · · · (4.10) Thus, energy occurs in quantized states distinguished by diﬀerent integer values of n, and the lowest energy is the ground state corresponding to n = 0. 4.3 The Hydrogen Atom The harmonic oscillator provides an elementary example of how ground and quantized energy states are natural mathematical properties of eigenvalues of Schr¨ odinger’s equation, but the particular choice of the potential V was, to an extent, contrived. We now turn to the concrete example of the wave function for the hydrogen atom and the potential is that characterizing the energy of a charged particle, an electron, orbiting a nucleus with a charge of equal magnitude but opposite sign. We begin by writing the Schr¨ odinger equation for the wave function in three dimensions for a single particle as follows: iℏ∂Ψ ∂t + ℏ2 2m△Ψ −V Ψ = 0 (4.11) where △is the three-dimensional Laplacian, △= ∂2 ∂x2 + ∂2 ∂y2 + ∂2 ∂z2 (4.12) Setting Ψn(x, y, z, t) = ψn(x, y, z)e−iEnt/ℏ (4.13) we arrive at the time-independent equation, −ℏ2 2m△ψn + V ψn = Enψn (4.14) 51 CHAPTER 4. SELECTED TOPICS AND APPLICATIONS The general solution is then Ψ(x, y, z, t) = ∞ X n=1 cnψn(x)e−iEnt/ℏ (4.15) with x = (x, y, z) In spherical coordinates (r, θ, φ) the time-independent equation (4.14) becomes −ℏ2 2m 1 r2 ∂ ∂r r2∂ψ ∂r + 1 r2sinθ ∂ ∂θ sinθ∂ψ ∂θ + 1 r2sin2θ ∂2ψ ∂φ2 +V ψ = Eψ (4.16) Using the method of separation of variables, we set ψ(r, θ, φ) = R(r)Y (θ, φ) (4.17) where R(r) and Y (θ, φ) satisfy the equations, d dr r2dR dr −2mr2 ℏ2 [V (r) −E]R = ℓ(1 + ℓ)R (4.18) (sinθ)−1 ∂ ∂θ sinθ∂Y ∂θ + (sinθ)−2∂2Y ∂φ2 = −ℓ(1 + ℓ)Y (4.19) where ℓ(1+ℓ) is the separation constant (Cf. Griﬃths, p.124); ℓis a complex number that will be determined to be an integer. The solution to the angular equation is Y m ℓ(θ, φ) = αCℓmP m ℓ(cosθ)eimφ (4.20) where α = (−1)m for m ≥0 1 for m < 0 Cℓm = (ℓ−\|m\|)! (ℓ+ \|m\|)! · (1 + 2ℓ) 4π 1/2 (4.21) and P m ℓ(·) are the Legrindre polynomials of degree (m, ℓ). One has Z 2π 0 Z π 0 Y m ℓ(θ, φ)∗Y m′ ℓ′ (θ, φ)sinθdθdφ = δmm′δℓℓ′ (4.22) 52 4.3. THE HYDROGEN ATOM The radial equation (4.18) cannot be solved until a potential V is speci-ﬁed. For the hydrogen atom, we may consider a motionless proton, the origin of our spherical coordinate system, and a much lighter electron of charge −e in an orbit (shell) around the protons. For V we take the Coulomb potential V (r) = −e2 4πε0 · 1 r (4.23) The radial equation then reduces to −ℏ2 2mu′′(r) + −e2 4πε0 · 1 r + ℏ2 2m ℓ(1 + ℓ) r2 u = Eu (4.24) where u(r) = rR(r), u′′(r) = d2u(r)/dr2. One can show that the solution to this equation is of the form, u(r) = r(R)r=ϱ1+ℓe−ϱ ∞ X j=0 ajϱj (4.25) where ϱ = kr, k = [−2mE/ℏ]1/2 (E < 0), and aj+1 = aj[2(j + ℓ+ 1) −ϱ0]/(j + 1)(j + 2ℓ+ 2) ϱ0 = me2/2πε0ℏ2k (4.26) The series (4.25) blows up unless the aj terminates at some j = j∗. From the condition aj∗+1 = 0, we conclude that 2(j∗+ ℓ+ 1) −ϱ0 = 0, or that ϱ0 = 2n, where n = 1 + ℓ+ j∗∈Z. The energy is E = −ℏ2k2 2m = − me4 8π2ε2 0ℏ2ϱ2 0 so the allowed energies are En = − " m 2ℏ2 e2 4πε0 2# 1 n2 = E1 n2 , n = 1, 2, 3, · · · (4.27) and E1 is the energy of the ground state of the electron. 53 CHAPTER 4. SELECTED TOPICS AND APPLICATIONS The above analysis is reproduced from Griﬃths, p.136, 137. Formula (4.28) is called the Bohr formula and characterizes the energy at various quantum states. In conclusion, Rnℓ(r) = r−1ϱ1+ℓe−ϱv(ϱ) (4.28) v(ϱ) = polynomial in ϱ of degree j∗= n −ℓ−1 = j∗ X j=0 ajϱj aj+1 = 2(j + ℓ+ 1 −n)aj/(1 + j)(2 + 2ℓ+ j) (4.29) Returning to (4.26), ϱ0 = m2e2 2πε0ℏ2k = 2(1 + ℓ+ j∗) = 2n so k = me2 4πε0ℏ2 1 n = 1 a0n where a0 = 4πε0ℏ2 me2 = 0.529 x 10−10m (4.30) The number a0, in units of length, is called the Bohr radius. In (4.27), the ground energy state E1 can be calculated to be E1 = −m 2ℏ2 e2 4πε0 2 = −13.6eV (4.31) Any perturbation of some stationary state of the hydrogen atom may cause a transition to some other stationary state. If energy is absorbed, such as might result from a photon hitting the atom, a higher energy state may be attained; or if energy is given oﬀ, a lower state may be attained. Such quantum jumps in energy are constantly occurring. The diﬀerence in energy levels between an initial state n = ni and a ﬁnal state nf is, from (4.27), Ei −Ef = −13.6eV 1 n2 i −1 n2 f ! 54 4.4. ANGULAR MOMENTUM AND SPIN According to Planck’s formula, the energy of a photon is Ei −Ef = hv = h c λ so 1 λ = R 1 n2 f −1 n2 i ! (4.32) Here R is the Rydberg constant R = m 4πc2ℏ3 e2 4πε0 2 = 1.097 x 107m−1 (4.33) If the transition brings the atom to the ground state (nf = 1), the wave length λ corresponds to the ultraviolet length for electromagnet waves; if nf = 2, the visible light region is attained; nf = 3 corresponds to infrared. 4.4 Angular Momentum and Spin Up to this point, the linear momentum p of a system has played the role of a fundamental dynamical property of quantum systems. From classical mechanics, however, we know that other types of momenta can exist by virtue of the moment of momentum vectors relative to a ﬁxed point in space and by virtue of their spin about a trajectory of the particle. We shall now examine how these properties manifest themselves in quantum systems. Consider an elementary particle located at position q relative to a ﬁxed origin and endowed with a momentum p. The angular momentum of the particle consists of an extrinsic momentum L due to its orbital motion about the origin, and an intrinsic momentum S due to its spin about its center of mass. In classical mechanics, L = q × p and S = Iω (4.34) where I is an inertial quantity (such as the moment of inertia) and ω is the spin velocity vector. In quantum mechanics, an electron has extrinsic angular momentum and an intrinsic property, its spin, which it carries in addition to L. 55 CHAPTER 4. SELECTED TOPICS AND APPLICATIONS Denoting ˜ pj = ℏ i ∂ ∂qj , j = 1, 2, 3 (4.35) the components of the extrinsic angular momentum are characterized by the quantum operators, Lj = εrsjqr i ℏ ∂ ∂qs (4.36) where repeated indices are summed and εrsj is the permutation symbol. One can easily show that [L1, L2] = iℏL3 , [L2, L3] = iℏL1 , [L3, L1] = iℏL2 (4.37) The operators Li are therefore incompatible observables, and σ2 L1σ2 L2 ≥ 1 2i ⟨iℏL3⟩ 2 = ℏ2 4 ⟨L3⟩2 , etc. (4.38) On the other hand, the square of the angular momentum, L2 = LiLi = L2 1 + L2 2 + L2 3 (4.39) does commute and [L2, Lk] = 0 , k = 1, 2, 3 (4.40) Therefore, we can hope to ﬁnd simultaneous eigenstates of L2 and Li; i.e. states ϕ such that L2ϕ = λϕ and L1ϕ = µϕ (4.41) The major properties of the eigenfunctions and eigenvalues in (4.41) are laid out in the following theorem: Theorem 4.1. The spherical harmonics Y m ℓ (recall (4.20)) are eigen-functions of the extrinsic angular momentum operators L2 and L1 (or L2, L3) for ℓ= 0; 1/2, 1, 3/2, · · · m = −ℓ, −ℓ+ 1, · · · , ℓ−1, ℓ Moreover, L2Y m ℓ = ℏ2ℓ(1 + ℓ)Y m ℓ (4.42) 56 4.4. ANGULAR MOMENTUM AND SPIN and L1Y m ℓ = ℏmY m ℓ (4.43) Proof (See, pgs. 146-149) □ In the case of the spin S(= “Iω”), we have similarly, [S1, S2] = iℏS3, [S2, S3] = iℏS1, [S3, S1] = iℏS2 (4.44) and S2qs m = ℏ2s(s + 1)qs m; S1qs m = ℏmqs m (4.45) s = 0, 1/2, 1, 3/2, · · · ; m = −s, −s + 1, · · · , s −1, s The case s = 1/2 is canonical. There are these two eigenstates q1/2 1/2, q1/2 −1/2. Representing these as vectors q+ = (1 0) (the spin up) and q−= (1 0) (the spin down), every state is a linear combination, aq++bq−, and the spin operators S become 2 x 2 matrices: S1 = ℏ 2σ1 , S2 = ℏ 2σ2 , S3 = ℏ 2σ3 (4.46) where σi are the Pauli spin matrices, σ1 = 0 1 1 0 , σ2 = 0 −i i 0 , σ3 = 1 0 0 −1 (4.47) All of the Si and S2 are Hermitian operators. The eigenvectors χ+ = 1 0 , χ−= 0 1 (4.48) are called spinors, and satisfy S2χ+ = 3 4ℏ2χ+ , S2χ−= 3 4ℏ2χ− S3χ+ = ℏ 2χ+ , S3χ−= ℏ 2χ− (4.49) We close this section by noting that a spinning charged elementary parti-cle creates a magnetic dipole with a magnetic dipole moment m proportional to the spin angular momentum: m = µS (4.50) 57 CHAPTER 4. SELECTED TOPICS AND APPLICATIONS The constant µ is called the gyromagnetic ratio. When placed in a magnetic ﬁeld B, it experiences a torque m x B with energy −m·B. The Hamiltonian of a spinning particle at rest is H = −µB · S (4.51) 4.5 Variational Principle We begin with a statement of a minimization principle. Theorem 4.2. Let H be the Hamiltonian of a quantum system and let Eg denote its ground-state energy. Then Eg ≤⟨ψ, Hψ⟩ ∀ψ ∈L2(RN), ⟨ψ, ψ⟩= 1 (4.52) In other words, Eg ≤⟨H⟩ (4.53) Proof. Let ψ = P k ckψk, ⟨ψk, ψj⟩= δij. Recall that Hψk = Ekψk. Then ⟨H⟩ = ⟨ψ, Hψ⟩ = X k ckψk, H X j cjψj + = X k X j c∗ kcjEj ⟨ψk, ψj⟩ = X j \|cj\|2 Ej ≥ Eg X j \|cj\|2 = Eg □ This variational principle provides a basis for Ritz approximations of the ground-state energies for more complex atoms. A classical example is the helium atom: a nucleus containing two protons (and some neutrons) and two orbiting electrons. The Hamiltonian is H = −ℏ2 2m(△1 + △2) − e2 4πε0 2 r1 + 2 r2 − 1 \|r1 −r2\| (4.54) 58 4.5. VARIATIONAL PRINCIPLE In the Ritz method, we introduce a trial function, ψa(r1, r2) = a−3 π e−(r1+r2)/a (4.55) where a is a variational parameter. The function ψa is suggested by pertur-bation theory. The corresponding energy is E(a) = ⟨ψa, Hψa⟩ (4.56) We chose a so that E(a) is a minimum and obtain Ea ≈−2 Z −5 16 2 EH (4.57) where Z = 1/2πε0 and EH the ground energy of the hydrogen atom (Cf. Messiah, p.771). Numerically, Ea = −76.6eV which compares remarkably well to the experimentally determined value of −78.6eV = Eexp. Note that Ea > Eexp. 59 Chapter 5 The Transition from Quantum Mechanics to Approximate Theories and to Molecular Dynamics 5.1 Introduction The enormous complexity of Schr¨ odinger’s equations for systems of atoms or molecules has led to the development of numerous approximate theories which have been used very eﬀectively for important applications in chemistry, biology, and materials science. Among methods far removed from quantum theory but applicable to large systems of atoms are those that fall under the classiﬁcation of models of molecular dynamics (MD). While applicable, in principle, to systems involving large numbers of atoms and molecules, MD is limited by many simplifying assumptions that can aﬀect the accuracy of its predictions. Between MD and quantum theory are several more accurate approximate theories that result from various assumptions and conditions placed on the quantum dynamical system. In this chapter, we develop a brief introduction to MD and to various approximate theories. 61 CHAPTER 5. FROM APPROXIMATE THEORIES TO MOLECULAR DYNAMICS 5.2 Molecular Dynamics Molecular dynamics refers to a class of mathematical models of systems of atoms or molecules in which 1. each atom (or molecule) is represented by a material point in R3 as-signed a point mass; 2. the motion of the system of mass points is determined by Newtonian mechanics; and 3. generally it is assumed that no mass is transferred in or out of the system. Molecular dynamics (MD) is generally used to calculate ensemble averages of thermochemical and thermomechanical properties of physical systems rep-resenting gases, liquids, or solids. Thus, in MD we consider a collection of N discrete points in R3 repre-senting the atom or molecule sites in some bounded domain Ω⊂R3, each assigned a point mass mi and each located relative to a ﬁxed origin O by a position vector ri, i = 1, 2, · · · , N. The motion of each such particle is assumed to be governed by Newton’s second law: mi d2ri dt2 = Fi , i = 1, 2, · · · , N (5.1) where Fi is the net force acting on particle i representing the interactions of i with other particles in the system. These interatomic forces are al-ways assumed to be conservative; i.e. there exists a potential energy V = V (r1, r2, · · · , rN) such that Fi = −∂V ∂ri , i = 1, 2, · · · , N (5.2) so that ﬁnally mi d2ri dt2 + ∂V ∂ri = 0 , i = 1, 2, · · · , N (5.3) To (5.3) we must add initial conditions on ri(0) and dri(0)/dt and bound-ary conditions on the boundary ∂Ωof Ω, these generally chosen to represent a periodic pattern of the motion of the system throughout R3. The solutions 62 5.2. MOLECULAR DYNAMICS {ri(t)}N i=1 (5.3) for t ∈[0, T] determine the dynamics of the system. Once these are known, other physical properties of the system such as ensemble averages represented by functionals on the ri and dri/dt can be calculated. The characterization of the behavior of the dynamical system must also be invariant under changes in the inertial coordinate system and the time frame of reference. As in quantum systems, it is also possible to describe the equations of MD in the phase space of position-momentum pairs (q, p) and in terms of the Hamiltonian of the system, H(r1, r2, · · · , rN ; p1, p2, · · · , pN) = N X i=1 1 2mi pi · pi + V (r1, r2, · · · , rN) (5.4) (qi ≡ri). Then, instead of (5.3), we have the 2N ﬁrst-order systems, ˙ ri = ∂H ∂pi , ˙ pi = −∂H ∂ri i = 1, 2, · · · , N (5.5) wherein ˙ ri = dri/dt and ˙ pi = dpi/dt. Thus, (5.5) represents 6N ordinary diﬀerential equations for the 2N-three-vectors (r1(t), r2(t), · · · , rN(t); p1(t), p2(t), · · · , pN(t)). One of the most challenging, diﬃcult, and critical aspects of MD is the identiﬁcation of the appropriate potential function V for the atomic-molecular system at hand. The following is often given as a general form of such potentials: V (r1, r2, · · · , rN) = N X i=1 V1(ri) + N X i,j=1 j>i V2(ri, rj) + 3 X i,j,k=1 j>i,k>i V3(ri, rj, rk) + · · · (5.6) where Vs is called the s-body potential. The 1-body potential V1(r1) + V1(r2) + · · · + V1(rN) 63 CHAPTER 5. FROM APPROXIMATE THEORIES TO MOLECULAR DYNAMICS is the potential energy due to the external force ﬁeld, the 2-body potential, V2(r1, r2) + V2(r1, r3) + · · · + V2(rN−1, rN) represents pair-wise interactions of particles, the 3-body potential three-particle interactions, etc. The Lennard-Jones potential, V (ri, rj) = 4ε " σ rij 12 − σ rij 6# rij = \|ri −rj\| (5.7) σ being a constant that depends upon the atomic structure of the atom or molecule at positions ri, rj, and ε being the potential energy at the minimum of V is a well-known example of a two-body potential. We remark that computer simulations employing MD models, approxima-tions of potentials are used to simplify the very large computational problems that can arise. For just pair-wise interactions of N particles, for example, the force ∂V/∂ri involves (N 2 −N)/2 terms. A typical simpliﬁcation is to introduce a cut-oﬀradius R around each particle and to include only those interactions with neighboring particles inside that radius. The truncated pair-wise potential is then Vcut-oﬀ= f(r)V (r) = V (r) , r ≤R 0 , r > R where r = \|ri −rj\| and f is a smooth cut-oﬀfunction varying from 1 at r = 0 to 0 at r = R. 5.3 The Connection Between MD and Quan-tum Mechanics We shall now show how the results we derived for simple quantum systems can be used to obtain force potentials for a simpliﬁed molecular dynamics model. The issue is one of scale. Quantum eﬀects, we presume, are essen-tially conﬁned to the individual atom or molecule, while MD treats these as material points-particles. The ideas can be illustrated through the example of the hydrogen molecule (Cf. Liu et al ) H2 consisting of two protons and two electrons. The posi-tion of the electrons relative to protons at ra and rb are denoted rai, rbi , i = 64 5.3. THE CONNECTION BETWEEN MD AND QUANTUM MECHANICS rb ra ra1 rb1 rb2 r12 r ra2 e+ e− e− e+ O a b 1 2 Figure 5.1: The hydrogen molecule H2 consisting of two nuclei at a and b and electrons at 1 and 2. 1, 2, and r12 is the distance between electrons. The distance separating the two atoms is denoted r. This situation is illustrated in Fig. 5.1. The analysis proceeds in the following steps: 1. the total energy E of the system is the sum of the energies of the two unbounded hydrogen atoms Ea and Eb, plus the energy due to interactions U, which we call the binding energy, E = Ea + Eb + U 2. MD assumes that no energy is absorbed or emitted by the atoms, so the energies Ea, Eb are the ground states calculated in Chapter 4 (recall (4.31)): E1 = Ea = Eb = −m 2ℏ2 e2 4πε0 2 = −13.6eV 3. the energy of the coupled system can be written as a function of the separation distance r(\|rb −ra\|), now regarded as a real parameter: U(r) = V (r) = E(r) −2E1 Thus, V (r) = V2(r) is the potential function for 2-body (pair-wise) interactions. 65 CHAPTER 5. FROM APPROXIMATE THEORIES TO MOLECULAR DYNAMICS 4. the protons at ra and rb are regarded as stationary at the distance r, an approximation known as the Bohr-Oppenheimer approximation, which is justiﬁed on the basis that the mass of the proton is 1,800 times that of the electron. It remains to calculate the energy E(r). For this we must resort to quan-tum mechanics. Under the conditions and assumptions listed above, the Hamiltonian is ˜ H = −ℏ2 2m 2 X i=1 △i − 1 4πε0 ( 2 X i=1 e2 rai + e2 rbi + e2 r12 + e2 r ) (5.8) The corresponding Schr¨ odinger equation is ˜ HΨ = EΨ = (V (r) + 2E1)Ψ (5.9) Since E = ⟨Ψ, HΨ⟩ (⟨Ψ, Ψ⟩= 1) we have V (r) = ⟨Ψ, HΨ⟩−2E1 (5.10) where, we assume, the wave function Ψ depends parimetrically on r. In general, we can write the coupled MD-quantum mechanics. ˜ HΨ = V + M X α=1 Eα ! Ψ mi¨ ri = −∂V ∂ri , i = 1, 2, · · · , N (5.11) for M-electron systems (¨ ri = d˙ ri/dt = d2ri/dt2). It is understandably rare that the wave function Ψ can be computed exactly. Then we may resort to developing approximations of E such as those described in connection with the Ritz method in Chapter 4. Then, for a given trial function ϕ ∈(L2(R3M), we set E ≈⟨ϕ, Hϕ⟩ ⟨ϕ, ϕ⟩ = Eϕ (5.12) 66 5.4. SOME APPROXIMATE METHODS The trial functions are of the form, ϕ = P k ckχk where χk are speciﬁc basis functions. The coeﬃcients ck are chosen so that Eϕ is a minimum in the space spanned by the χk’s: ∂Eϕ/∂ck = 0. The resulting trial function ϕ is used to commute the minimum Eϕ∗. Then V ≈Eϕ∗− M X α=1 Eα (5.13) We describe other approximate methods in the following sections. 5.4 Some Approximate Methods We brieﬂy outline three of the more popular approximate methods. Two of these are based on Ritz-type approximations similar to those described earlier. 5.4.1 The LCAO Method The following variational approach is called the tight binding method or the linear combination of atomic orbitals (LCAO) method by Liu et al and is attributed to Bloch and Slater and Koster . A molecular orbital is de-ﬁned as a wave function for a single electron in a non-central ﬁeld containing two or more nuclei. For the hydrogen ion H+ 2 , the Hamiltonian is H = −ℏ2 2m 2 X i=1 △i − 1 4πεo ( 2 X α=1 e2 rα1 + e2 rα2 −e2 r ) (5.14) where we use the notation in Fig. 5.1. Let ψa and ψb be the molecular orbitals for a single hydrogen atom (i.e. (−ℏ2/2m△−e2/4πε0ra1)ψa = 0, (−ℏ2/2m△−e2/4πε0rb2)ψb = 0). Then, as an approximation of the wave function for H of (5.14), we take the linear combination, ˜ Ψ = αψa + βψb (5.15) The corresponding approximate energy is ˜ E = D ˜ Ψ, H ˜ Ψ E D ˜ Ψ, ˜ Ψ E = α2Haa + 2αβHab + β2Hbb α2 + β2 + 2αβGαβ (5.16) 67 CHAPTER 5. FROM APPROXIMATE THEORIES TO MOLECULAR DYNAMICS where Hab = ⟨ψa, Hψb⟩ , Gαβ = ⟨ψa, ψb⟩ Choosing α and β to minimize ˜ E, we have α =+ −β so that two molecular orbits for H+ 2 are obtained, ψ+ = A(ψa + ψb) and ψ−= B(ψa −ψb) (5.17) where A and B are normalizing factors. The corresponding energy states are ˜ E+ = Haa + Hab 1 + Gab and ˜ E−= Haa −Hab 1 −Gab (5.18) An approximation of the potential V can now be obtained from (5.13). This approximate approach does not necessarily respect the Pauli exclu-sion principle (discussed in an appendix), and therefore can be ineﬀective or require additional information. The Hartree models, particularly Hartree-Fock, are designed to address these issues. 5.4.2 The Hartree-Fock Model The physical system considered is a molecular system made up of M nuclei and N electrons. The goal is to calculate the ground state energy of the system. Such a system is illustrated in Fig. 5.2. To reduce this M + N body problem to one more tractable, we introduce again the Bohr-Oppenheimer approximation, whereby the M nuclei are considered as classical point par-ticles with charges Zk at positions Rk, k = 1, 2, · · · , M. Within the Bohr-Oppenheimer approximation, the problem of determining the ground state reduces to the nested minimization problems: inf R1,R2,··· ,RM ∈R3M ( E(R1, R2, · · · , RM) + c X 1≤k<j≤M ZkZj Rkj ) (5.19) with Rkj = \|Rk −Rj\| and c the appropriate constant, and E(R1, R2, · · · , RM) = inf ψ ∈H {⟨ψ, Hψ⟩, ∥ψ∥= 1} (5.20) 68 5.4. SOME APPROXIMATE METHODS where H = − N X i=1 ℏ2 2m△ri −c N X i=1 M X k=1 eZk \|rj −Rk\| ! + c X 1≤i<j≤N e2 rij (5.21) r1 r3 rN R1 R2 RM rN−1 RM−1 O 1 3 5 4 N −1 N 2 Figure 5.2: A molecular system consisting of M nuclei and N electrons. Here △ri = ∂2/∂r2 i1 + ∂2/∂r2 i2 + ∂2/∂r2 i3, rij = \|ri −rj\| and ri is the position vector of electron i, and H = N Y i=1 H1(R3, C) (5.22) with H1(R3, C) denoting the Sobolev space of complex-valued functions (equiv-alence classes of functions) with generalized ﬁrst derivatives in L2(R). Here-after, the nuclei positions Rk are regarded as parameters, and we focus on the internal electron minimization problem (5.20). The Hartree-Fock approximation amounts to a variational formulation of (5.20) designed to respect the Pauli exclusion principle, which is discussed 69 CHAPTER 5. FROM APPROXIMATE THEORIES TO MOLECULAR DYNAMICS in an appendix to this chapter. First, we derive a reduced functional. Since electrons correspond to fermions, we seek antisymmetric functions in H char-acterized by Slater determinants of the form, ψ = 1 √ N! det(ϕi(rj)) where ϕi are orthonormal molecular orbitals, ⟨ϕi, ϕj⟩= δij , i, j = 1, 2, · · · , N With this constraint, the Hartree-Fock energy functional becomes EHF({ϕi}) = N X i=1 ℏ2 2m∥∇riϕi∥2 + Z R3 ϱϕV dr + c 2 Z R3 Z R3 ϱϕ(r)ϱϕ(r1) \|r −r′\| drdr′ + c 2 Z R3 Z R3 N X i=1 ϕi(r)ϕ∗ i (r)drdr′ (5.23) where dr = dNr = dr1dr2 · · · drN, ϱϕ(r) = N X i=1 \|ϕi(r)\|2 The Hartree-Fock reduced minimization thus reads, inf ϕi {EHF({ϕi}) , ϕi ∈H1(R, C), ⟨ϕi, ϕj⟩ = δij , i, j = 1, 2, · · · , N} (5.24) The orthonormality condition ⟨ϕi, ϕj⟩= δij is a constraint on the choices of test functions. The Hartree-Fock method consists of constructing Ritz approximations of the variational problem (5.24). We construct a ﬁnite number of molecular orbits {χm}, 1 ≤m ≤n, (or approximate molecular orbits) and set ϕi ≈ϕ(m) i = m X k=1 Ckiχk (5.25) 70 5.4. SOME APPROXIMATE METHODS where Cki are constants to be determined so as to minimize EHF over span {ϕ(m) i , i = 1, 2, · · · , N; m = 1, 2, · · · , n}. We demand that δij = D ϕ(m) i , ϕ(m) j E = C∗ kjXkℓCki where Xij = ⟨χi, χj⟩. Subject to this constraint, substitution of (5.25) into (5.23) leads to a reduced HF energy functional of the constants Cki: EEF({ϕ(m) i }) = ˜ EEF(Ckj) (5.26) We then compute the speciﬁc coeﬃcients ˆ Ckj such that ∂˜ EEF ∂Ckj ( ˆ Ckj) = 0 (5.27) This is a general approximation process behind the Hartree-Fock model. There are many variants of this approach (see, e.g. Cances or Lions for details and additional references). 5.4.3 Density Functional Theory The density functional theory addresses the calculation of ground states of atomic and molecular systems by characterizing the state of the system us-ing the electron density ϱ(r) instead of the wave function Ψ(r1, r2, · · · , rN). For any quantum system involving N electrons, the electron density ϱ(r) is deﬁned as the number of electrons per unit volume, and is given in terms of the wave function by (recalling that now Ψ is invariant under permutation) ϱ(r1) = N Z R \|Ψ(x1, x2, · · · , xN)\|2 dx2dx3 · · · dxN (5.28) so that with r = (x, y, z) ∈R3 Z R3 ϱ(r)dr = N (5.29) Once again we consider the time-independent Schr¨ odinger equation for a system of M nuclei of charge Zk and N electrons with charges e, and we invoke the Bohr-Oppenheimer approximation, assuming that the electrons adjust to any change in the nuclei. The total energy E in the ground state can 71 CHAPTER 5. FROM APPROXIMATE THEORIES TO MOLECULAR DYNAMICS then be expressed in terms of the electronic density, which can be expressed as a sum of molecular orbitals ϕi for each electron: ϱ = N X i=1 \|ϕi\|2 (5.30) For this system, the Hamiltonian operator is of the form ˜ H = ˜ T + Vne + Vee (5.31) where ˜ T is the kinetic energy operator, ˜ T = ℏ2 2m N X i=1 △ri (5.32) Vne is the external potential for electron i due to nuclei α, Vne = − N X i=1 M X α=1 cZαe \|ri −Rα\| (5.33) and Vee = N X i,j=1 i<j ce2 \|ri −rj\| (5.34) with c the usual constraint. The electronic energy E satisﬁes, as usual, ˜ HΨ = EΨ (5.35) and the total energy is W = E + c M X α,β=1 α<β ZαZβ Rαβ (5.36) Rαβ = \|Rα −Rβ\|. Our goal is to determine these various energy components in terms of the density ϱ. 72 5.4. SOME APPROXIMATE METHODS The Kohn-Sham form of the energy functional is E(ϱ) = T(ϱ) + 1 2 Z R3 Z R3 ϱ(x)ϱ(y) \|x −y\| dxdy + Ex(ϱ) + Ec(ϱ) (5.37) where T(ϱ) is the kinetic energy functional, T(ϱ) = inf ( ℏ2 2m N X i=1 Z R3 \|∇ϕi\|2 dx , ϱ = N X i=1 \|ϕi\|2 ) (5.38) and Ex and Ec are the so-called exchange and correlation energies. Various approximations of Ex and Ec can be introduced. The local density approxi-mation (LDA) employs the exchange energy of a homogeneous electron gas, Ex(ϱ) = −cx Z R3 ϱ4/3dr (5.39) where cx is a positive constant (= 3(3/π)1/3/4 in atomic units). For Ec, similar empirical relations can be employed. We then seek a minimum energy subject to the constraints due to the deﬁnition of ϱ: inf E(ϱ) + Z R3 ϱVnecdr , ϱ ≥0 , Z R3 ϱdr = N (5.40) where Vnec is the electrostatic potential given by (5.6). Other energy fuctionals of the density have been proposed. The Thomas-Fermi models employ functionals of the type, ETF(ϱ) = co Z R3 \|∇√ϱ\|2 dr + c1 Z R3 ϱ5/3dr + 1 2 Z R3 x R3 ϱ(x)ϱ(y) \|x −y\| dxdy − M X k=1 Z R3 ϱ(r) \|r −Rk\|dr + 1 2 X x̸=y ce2 \|x −y\| (5.41) and we seek, inf ETF(ϱ) : ϱ ≥0, √ϱ ∈H1(R3), Z R3 ϱdr = N (5.42) 73 CHAPTER 5. FROM APPROXIMATE THEORIES TO MOLECULAR DYNAMICS 5.5 Appendix: Identical Particles and the Pauli Exclusion Principle Consider the Hamiltonian of a system of two identical particles, H((r1, p1), (r1, p2)) = p1 · p1 2m + p2 · p2 2m + V (\|r1 −r2\|) In classical mechanics, if these particles are identical, properties of the system cannot change if the particles are commuted; i.e. the states of parti-cles are interchanged. Thus, H must be invariant under such a permutation (H(µ1, µ2) = H(µ2, µ1)). In quantum mechanics, the situation is more serious. The following ob-servations are classical Cf Messiah , Griﬃths or particularly Hinchliﬀe . Suppose for a minute that the two electrons do not repel each other and that each moves freely about a nucleus of charge Z. The Hamiltonian splits into the sum of two parts, (ℏ(r1) + ℏ(r2))Ψ(r1, r2) = EΨ(r1, r2) where ℏ(ri) = −ℏ2 2m△i −Ze2 4πεo \|R −ri\| Using the method of separation of variables to solve this system, we ﬁnd that the system breaks down to independent orbitals: ℏ(ri)ϕ(ri) = Eiϕ(ri) , i = 1, 2 with Ψ = ϕ1(r1)ϕ2(r2). This result of course is incorrect because we have neglected the electron–electron interaction. Let ϕ1 and ϕ2 represent the lowest energy orbitals. Then possible states are Ψ11(r1, r2) = ϕ1(r1)ϕ1(r2) E = 2E1 Ψ12(r1, r2) = ϕ1(r1)ϕ2(r2) Ψ21(r1, r2) = ϕ2(r1)ϕ1(r2) ) E = E1 + E2 74 5.5. APPENDIX But the latter case presents a problem: the electrons, being indistinguish-able, must appear on equal footing in computing these observables, which is not the case as electron 1 is apparently diﬀerent from Z as it appears is a ﬁrst-energy orbit in Ψ12 but in a second in Ψ21. However, the combination, Ψ12 + Ψ21 and Ψ12 −Ψ21 do not have this shortcoming, and are conceivable bases for building more complex descriptions of possible dynamical states. A wave function will be referred to as symmetric or antisymmetric ac-cording to whether it remains the same or changes sign under a change in electron labels. Thus, the ﬁrst sum above is symmetric whilst the second is antisymmetric. For multielectron systems, we classify wave functions as symmetric or antisymmetric according to whether or not a ...? Let us suppose that the wave function at time t is a linear combination of a symmetric and an antisymmetric component: Ψ(r1, r2, t) = αψsym + βψant and the probability density of ﬁnding one particle at r′ and another at r′′ is \|Ψ(r′, r′′)\|2 + \|Ψ(r′′, r′)\|2 = 2 \|α\|2 \|ψant\|2 + 2 \|β\|2 \|ψsym\|2 This expression must be independent of α and β (which are arbitrary constants). Thus, we must have, \|ψant(r′, r′′)\| = \|ψsym(r′, r′′)\| This result leads us to the following postulate: the dynamical states of two identical particles are either all symmetrical (α = 1, β = 0) or all an-tisymmetrical (α = 0, β = 1) in the permutation of the two particles (Cf. Messiah [8, p.585]. To extend this postulate to systems of N identical particles requires an introduction of permutations in N variables. In any case, the states of sys-tems containing N particles are either all symmetrical or all antisymmetrical with respect to the permutations of N particles. Particles with symmetrical states are called bosons while those with antisymmetrical states are called fermions. According to Messiah [8, p.595], elementary particles with spin 75 CHAPTER 5. FROM APPROXIMATE THEORIES TO MOLECULAR DYNAMICS 1/2 occurring in nature (e.g. electrons, protons) are fermions while those with integral spin are bosons (e.g. photons). From what has been said, it can be shown to follow that two identical fermions cannot occupy the same state. This is the Pauli Exclusion Principle. A method for constructing antisymmetrical wave functions is to use the Slater determinant. Let ψ1, ψ2, · · · , ψN denote N orthogonal individual states occupied by N particles. 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Introduction to Quantum Mechanics, Prentice-Hall, Upper Saddle River, NJ, 1995. Halliday, D., Resnick, R., and Walker, J. Fundamentals of Physics, Sixth Edition, John Wiley and Sons, Hoboken, NJ, 2001. Hinchliffe, A. Computational Quantum Chemistry, John Wiley and Sons, Chichester, 1988. 77 BIBLIOGRAPHY Jackson, J.D. Classical Electrodynamics, Second Edition, John Wiley and Sons, New York et., 1962. Kohn, W. and Sham, L.J. ”Self-Consistent Equations Including Ex-change and Correlation Eﬀects”, Physics Review, 94, P. 111, 1954. Lions, P.L. ”Solutions of Hartree-Fock Equations for Coulomb Sys-tems”, Communications in Mathematical Physics, 109, pp.33-97, 1987. Liu, W.K., Karpov, E.G., and Park, H.S. ”An introduction to computational nanomechanics and materials,” Computer Methods in Ap-plied Mechanics and Engineering, Vol. 193, pp.1529-1578, 2004. Messiah, A. Quantum Mechanics, Volumes 1 and 2. Dover, Mineola, NY, 1999. Parr, R.G. and Yang, W. Density Function Theory of Atoms and Molecules, Oxford University Press, New York, 1989. Slater, V.C., and Koster, G.F. ”Simpliﬁed LCAO method for periodic potential problem,” Physics Review, 94(6), pp.1498-1524, 1954. von Neumann, J. Mathematical Foundations of Quantum Mechanics, University Press, Princeton, NJ, 1955. 78
8273	https://courses.cs.washington.edu/courses/cse311/22au/lectures/lecture07-inference.pdf	CSE 311: Foundations of Computing Lecture 7: Logical Inference Last Class: Quantifiers We use quantifiers to talk about collections of objects. "x P(x) P(x) is true for every x in the domain read as “for all x, P of x” $x P(x) There is an x in the domain for which P(x) is true read as “there exists x, P of x” Statements with Quantifiers (Natural Translations) Translations often (not always) sound more natural if we 1. Notice “domain restriction” patterns "x (Prime(x) ® (Equal(x, 2) Ú Odd(x))) Every prime number is either 2 or odd. 2. Avoid introducing unnecessary variable names "x $y Greater(y, x) For every positive integer, there is some larger positive integer. 3. Can sometimes drop “all” or “there is” ¬ $x (Even(x) Ù Prime(x) Ù Greater(x, 2)) No even prime is greater than 2. More English Ambiguity Implicit quantifiers in English are often confusing Three people that are all friends can form a raiding party Three people that I know are all friends with Mark Zuckerberg Formal logic removes this ambiguity – quantifiers can always be specified – unquantified variables that are not known constants (e.g, π) are implicitly "–quantified " $ Negations of Quantifiers PurpleFruit(x) ::= “x is a purple fruit” Predicate Definitions () "x PurpleFruit(x) (“All fruits are purple”) What is the negation of ()? (a) “there exists a purple fruit” (b) “there exists a non-purple fruit” (c) “all fruits are not purple” Try your intuition! Which one seems right? Negations of Quantifiers PurpleFruit(x) ::= “x is a purple fruit” Predicate Definitions () "x PurpleFruit(x) (“All fruits are purple”) What is the negation of ()? (a) “there exists a purple fruit” (b) “there exists a non-purple fruit” (c) “all fruits are not purple” {plum, apple} Domain of Discourse () PurpleFruit(plum) Ù PurpleFruit(apple) (a) PurpleFruit(plum) Ú PurpleFruit(apple) (b) ¬ PurpleFruit(plum) Ú ¬ PurpleFruit(apple) (c) ¬ PurpleFruit(plum) Ù ¬ PurpleFruit(apple) De Morgan’s Laws for Quantifiers ¬"x P(x) º $x ¬ P(x) ¬ $x P(x) º "x ¬ P(x) De Morgan’s Laws for Quantifiers ¬"x P(x) º $x ¬ P(x) ¬ $x P(x) º "x ¬ P(x) These are equivalent but not equal They have different English translations, e.g.: There is no unicorn Every animal is not a unicorn ¬ $x Unicorn(x) "x ¬ Unicorn(x) De Morgan’s Laws for Quantifiers ¬ $ x " y ( x ≥ y) º " x ¬ "y ( x ≥ y) º " x $ y ¬ ( x ≥ y) º " x $ y (y > x) “There is no integer at least as large as every other integer” “For every integer, there is a larger integer” ¬"x P(x) º $x ¬ P(x) ¬ $x P(x) º "x ¬ P(x) De Morgan’s Laws for Quantifiers ¬ $x (Even(x) Ù Prime(x) Ù Greater(x, 2)) º "x ¬(Even(x) Ù Prime(x) Ù Greater(x, 2)) º "x (¬(Even(x) Ù Prime(x)) Ú ¬Greater(x, 2)) º "x ((Even(x) Ù Prime(x)) ® ¬Greater(x, 2)) º "x ((Even(x) Ù Prime(x)) ® LessEq(x, 2)) “No even prime is greater than 2” “Every even prime is less than or equal to 2.” ¬"x P(x) º $x ¬ P(x) ¬ $x P(x) º "x ¬ P(x) De Morgan’s Laws for Quantifiers ¬ $x (P(x) Ù R(x)) º "x (P(x) ® ¬ R(x)) De Morgan’s Laws respect domain restrictions! (It leaves them in place and only negates the other parts.) ¬"x (P(x) ® R(x)) º $x (P(x) Ù ¬ R(x)) We just saw that Can similarly show that De Morgan’s Laws for Quantifiers ¬"x P(x) º $x ¬ P(x) ¬ $x P(x) º "x ¬ P(x) ¬ $x (P(x) Ù R(x)) º "x (P(x) ® ¬ R(x)) ¬"x (P(x) ® R(x)) º $x (P(x) Ù ¬ R(x)) Remain true when domain restrictions are used: Nested Quantifiers • Quantified variable names don’t matter "x $y P(x, y) º "a $b P(a, b) • Positions of quantifiers can sometimes change "x (Q(x) Ù $y P(x, y)) º "x $y (Q(x) Ù P(x, y)) • But: order is important... Quantifier Order Can Matter “There is a number greater than or equal to all numbers.” GreaterEq(x, y) ::= “x ≥ y” Predicate Definitions x y 1 2 3 4 1 2 3 4 T F F F T T F F T T T F T T T T $x "y GreaterEq(x, y))) {1, 2, 3, 4} Domain of Discourse Quantifier Order Can Matter “There is a number greater than or equal to all numbers.” GreaterEq(x, y) ::= “x ≥ y” Predicate Definitions “Every number has a number greater than or equal to it.” y 1 2 3 4 1 2 3 4 T F F F T T F F T T T F T T T T $x "y GreaterEq(x, y))) "y $x GreaterEq(x, y))) {1, 2, 3, 4} Domain of Discourse x Quantifier Order Can Matter “There is a number greater than or equal to all numbers.” GreaterEq(x, y) ::= “x ≥ y” Predicate Definitions “Every number has a number greater than or equal to it.” y 1 2 3 4 1 2 3 4 T F F F T T F F T T T F T T T T The purple statement requires an entire row to be true. The red statement requires one entry in each column to be true. $x "y GreaterEq(x, y))) "y $x GreaterEq(x, y))) Important: both include the case x = y Different names does not imply different objects! {1, 2, 3, 4} Domain of Discourse x Quantification with Two Variables expression when true when false "x " y P(x, y) Every pair is true. At least one pair is false. $ x $ y P(x, y) At least one pair is true. All pairs are false. " x $ y P(x, y) We can find a specific y for each x. (x1, y1), (x2, y2), (x3, y3) Some x doesn’t have a corresponding y. $ y " x P(x, y) We can find ONE y that works no matter what x is. (x1, y), (x2, y), (x3, y) For any candidate y, there is an x that it doesn’t work for. 1 2 3 4 1 2 3 4 T F F F T T F F T T T F T T T T Logical Inference • So far we’ve considered: – How to understand and express things using propositional and predicate logic – How to compute using Boolean (propositional) logic – How to show that different ways of expressing or computing them are equivalent to each other • Logic also has methods that let us infer implied properties from ones that we know – Equivalence is a small part of this New Perspective Rather than comparing A and B as columns, zooming in on just the rows where A is true: p q A B T T T T F T F T F F F F New Perspective Rather than comparing A and B as columns, zooming in on just the rows where A is true: Given that A is true, we see that B is also true. p q A B T T T T T F T T F T F F F F A ⇒B New Perspective Rather than comparing A and B as columns, zooming in on just the rows where A is true: When we zoom out, what have we proven? p q A B T T T T T F T T F T F ? F F F ? New Perspective Rather than comparing A and B as columns, zooming in on just the rows where B is true: When we zoom out, what have we proven? p q A B A ® B T T T T T T F T T T F T F T T F F F F T (A ® B) º T New Perspective Equivalences A º B and (A « B) º T are the same Inference A ⇒B and (A ® B) º T are the same Can do the inference by zooming in to the rows where A is true Applications of Logical Inference • Software Engineering – Express desired properties of program as set of logical constraints – Use inference rules to show that program implies that those constraints are satisfied • Artificial Intelligence – Automated reasoning • Algorithm design and analysis – e.g., Correctness, Loop invariants. • Logic Programming, e.g. Prolog – Express desired outcome as set of constraints – Automatically apply logic inference to derive solution Proofs • Start with given facts (hypotheses) • Use rules of inference to extend set of facts • Result is proved when it is included in the set An inference rule: Modus Ponens • If A and A ® B are both true, then B must be true • Write this rule as • Given: – If it is Wednesday, then you have a 311 class today. – It is Wednesday. • Therefore, by Modus Ponens: – You have a 311 class today. A ; A ® B ∴B My First Proof! Show that r follows from p, p ® q, and q ® r 1. 𝒑 Given 2. 𝒑→𝒒 Given 3. 𝒒® 𝒓 Given 4. 5. Modus Ponens My First Proof! Show that r follows from p, p ® q, and q ® r 1. 𝒑 Given 2. 𝒑→𝒒 Given 3. 𝒒® 𝒓 Given 4. 𝒒 MP: 1, 2 5. 𝒓 MP: 3, 4 Modus Ponens 1. 𝒑→𝒒 Given 2. ¬𝒒 Given 3. ¬𝒒® ¬𝒑 Contrapositive: 1 4. ¬𝒑 MP: 2, 3 Proofs can use equivalences too Show that ¬p follows from p ® q and ¬q Modus Ponens Inference Rules A ; B ∴C , D A ; A ® B ∴ B Requirements: Conclusions: If A is true and B is true …. Then, C must be true Then D must be true Example (Modus Ponens): If I have A and A ® B both true, Then B must be true. Axioms: Special inference rules ∴C , D ∴A Ú¬A Requirements: Conclusions: If I have nothing… Example (Excluded Middle): A Ú¬A must be true. Then D must be true Then, C must be true Simple Propositional Inference Rules Two inference rules per binary connective, one to eliminate it and one to introduce it A Ù B ∴A, B A ; B ∴A Ù B A x ∴A Ú B, B Ú A A ; A ® B ∴B A Þ B ∴A ® B Not like other rules Elim ∧ Intro ∧ A Ú B ; ¬A ∴B Elim ∨ Intro ∨ Modus Ponens Direct Proof
8274	https://artofproblemsolving.com/wiki/index.php/Incircle?srsltid=AfmBOoppzJmM5LdhCTupOUDfceJu47RdC8_J6c1C0IyXl0Gb_cBSZy4J	Art of Problem Solving Incircle - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Incircle Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Incircle An incircle of a convexpolygon is a circle which is inside the figure and tangent to each side. Every triangle and regular polygon has a unique incircle, but in general polygons with 4 or more sides (such as non-squarerectangles) do not have an incircle. A quadrilateral that does have an incircle is called a Tangential Quadrilateral. For a triangle, the center of the incircle is the Incenter, where the incircle is the largest circle that can be inscribed in the polygon. The Incenter can be constructed by drawing the intersection of angle bisectors. Formulas The radius of an incircle of a triangle (the inradius) with sides and area is The area of any triangle is where is the Semiperimeter of the triangle. The formula above can be simplified with Heron's Formula, yielding The radius of an incircle of a right triangle (the inradius) with legs and hypotenuse is . For any polygon with an incircle, , where is the area, is the semi perimeter, and is the inradius. The coordinates of the incenter (center of incircle) are , if the coordinates of each vertex are , , and , the side opposite of has length , the side opposite of has length , and the side opposite of has length . The formula for the semiperimeter is . The area of the triangle by Heron's Formula is . See also Circumradius Inradius Kimberling center Circumcircle Click here to learn about the orthocenter, and Line's Tangent Retrieved from " Category: Geometry Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
8275	https://www.geeksforgeeks.org/dsa/count-n-digit-numbers-not-particular-digit/	Count n digit numbers not having a particular digit Last Updated : 06 Apr, 2023 Suggest changes 1 Like We are given two integers n and d, we need to count all n digit numbers that do not have digit d. Example : ``` Input : n = 2, d = 7 Output : 72 All two digit numbers that don't have 7 as digit are 10, 11, 12, 13, 14, 15, 16, 18, ..... Input : n = 3, d = 9 Output : 648 ``` A simple solution is to traverse through all d digit numbers. For every number, check if it has x as digit or not. Efficient approach In this method, we check first if excluding digit d is zero or non-zero. If it is zero then, we have 9 numbers (1 to 9) as first number otherwise we have 8 numbers(excluding x digit and 0). Then for all other digits, we have 9 choices i.e (0 to 9 excluding d digit). We simple call digitNumber function with n-1 digits as first number we already find it can be 8 or 9 and simple multiply it. For other numbers we check if digits are odd or even if it is odd we call digitNumber function with (n-1)/2 digits and multiply it by 9 otherwise we call digitNumber function with n/2 digits and store them in result and take result square. Illustration : Number from 1 to 7 excluding digit 9. digits multiple number 1 8 8 2 89 72 3 899 648 4 8999 5832 5 89999 52488 6 899999 472392 7 8999999 4251528 As we can see, in each step we are half the number of digits. Suppose we have 7 digits in this we call function from main with 6(7-1) digits. when we half the digits we left with 3(6/2) digits. Because of this we have to multiply result due to 3 digits with itself to get result for 6 digits. Similarly for 3 digits we have odd digits, we have odd digits. We find result with 1 ((3-1)/2) digits and find result square and multiply it with 9, because we find result for d-1 digits. C++ ```` // C++ Implementation of above method include define mod 1000000007 using namespace std; // Finding number of possible number with // n digits excluding a particular digit long long digitNumber(long long n) { // Checking if number of digits is zero if (n == 0) return 1; // Checking if number of digits is one if (n == 1) return 9; // Checking if number of digits is odd if (n % 2) { // Calling digitNumber function // with (digit-1)/2 digits long long temp = digitNumber((n - 1) / 2) % mod; return (9 (temp temp) % mod) % mod; } else { // Calling digitNumber function // with n/2 digits long long temp = digitNumber(n / 2) % mod; return (temp temp) % mod; } } int countExcluding(int n, int d) { // Calling digitNumber function // Checking if excluding digit is // zero or non-zero if (d == 0) return (9 digitNumber(n - 1)) % mod; else return (8 digitNumber(n - 1)) % mod; } // Driver function to run above program int main() { // Initializing variables long long d = 9; int n = 3; cout << countExcluding(n, d) << endl; return 0; } ```` // C++ Implementation of above method // C++ Implementation of above method ``` include #include ``` ``` define mod 1000000007 #define mod 1000000007 ``` using namespace std; using namespace std // Finding number of possible number with // Finding number of possible number with // n digits excluding a particular digit // n digits excluding a particular digit long long digitNumber(long long n) {long long digitNumber long long n // Checking if number of digits is zero // Checking if number of digits is zero if (n == 0) if n == 0 return 1; return 1 // Checking if number of digits is one // Checking if number of digits is one if (n == 1) if n == 1 return 9; return 9 // Checking if number of digits is odd // Checking if number of digits is odd if (n % 2) {if n% 2 // Calling digitNumber function // Calling digitNumber function // with (digit-1)/2 digits // with (digit-1)/2 digits long long temp = digitNumber((n - 1) / 2) % mod; long long temp = digitNumber n - 1/ 2% mod return (9 (temp temp) % mod) % mod; return 9 temp temp% mod% mod } else {else // Calling digitNumber function // Calling digitNumber function // with n/2 digits // with n/2 digits long long temp = digitNumber(n / 2) % mod; long long temp = digitNumber n/ 2% mod return (temp temp) % mod; return temp temp% mod } } int countExcluding(int n, int d) int countExcluding int n int d { // Calling digitNumber function // Calling digitNumber function // Checking if excluding digit is // Checking if excluding digit is // zero or non-zero // zero or non-zero if (d == 0) if d == 0 return (9 digitNumber(n - 1)) % mod; return 9 digitNumber n - 1% mod else else return (8 digitNumber(n - 1)) % mod; return 8 digitNumber n - 1% mod } // Driver function to run above program // Driver function to run above program int main() {int main // Initializing variables // Initializing variables long long d = 9; long long d = 9 int n = 3; int n = 3 cout << countExcluding(n, d) << endl; cout<< countExcluding n d<< endl return 0; return 0 } Java ```` // Java Implementation of above method import java.lang.; class GFG { static final int mod = 1000000007; // Finding number of possible number with // n digits excluding a particular digit static int digitNumber(long n) { // Checking if number of digits is zero if (n == 0) return 1; // Checking if number of digits is one if (n == 1) return 9; // Checking if number of digits is odd if (n % 2 != 0) { // Calling digitNumber function // with (digit-1)/2 digits int temp = digitNumber((n - 1) / 2) % mod; return (9 (temp temp) % mod) % mod; } else { // Calling digitNumber function // with n/2 digits int temp = digitNumber(n / 2) % mod; return (temp temp) % mod; } } static int countExcluding(int n, int d) { // Calling digitNumber function // Checking if excluding digit is // zero or non-zero if (d == 0) return (9 digitNumber(n - 1)) % mod; else return (8 digitNumber(n - 1)) % mod; } // Driver function to run above program public static void main(String[] args) { // Initializing variables int d = 9; int n = 3; System.out.println(countExcluding(n, d)); } } // This code is contributed by Anant Agarwal. ```` Python3 ```` Python Implementation of above method mod=1000000007 Finding number of possible number with n digits excluding a particular digit def digitNumber(n): # Checking if number # of digits is zero if (n == 0): return 1 # Checking if number # of digits is one if (n == 1): return 9 # Checking if number # of digits is odd if (n % 2!=0): # Calling digitNumber function # with (digit-1)/2 digits temp = digitNumber((n - 1) // 2) % mod return (9 (temp temp) % mod) % mod else: # Calling digitNumber function # with n/2 digits temp = digitNumber(n // 2) % mod return (temp temp) % mod def countExcluding(n,d): # Calling digitNumber function # Checking if excluding digit is # zero or non-zero if (d == 0): return (9 digitNumber(n - 1)) % mod else: return (8 digitNumber(n - 1)) % mod Driver code d = 9 n = 3 print(countExcluding(n, d)) This code is contributed by Anant Agarwal. ```` C# ```` // C# Implementation of above method using System; class GFG { static int mod = 1000000007; // Finding number of possible number with // n digits excluding a particular digit static int digitNumber(long n) { // Checking if number of digits is zero if (n == 0) return 1; // Checking if number of digits is one if (n == 1) return 9; // Checking if number of digits is odd if (n % 2 != 0) { // Calling digitNumber function // with (digit-1)/2 digits int temp = digitNumber((n - 1) / 2) % mod; return (9 (temp temp) % mod) % mod; } else { // Calling digitNumber function // with n/2 digits int temp = digitNumber(n / 2) % mod; return (temp temp) % mod; } } static int countExcluding(int n, int d) { // Calling digitNumber function // Checking if excluding digit is // zero or non-zero if (d == 0) return (9 digitNumber(n - 1)) % mod; else return (8 digitNumber(n - 1)) % mod; } // Driver function to run above program public static void Main() { // Initializing variables int d = 9; int n = 3; Console.WriteLine(countExcluding(n, d)); } } // This code is contributed by vt_m. ```` PHP ```` php // PHP Implementation // of above method $mod = 1000000007; // Finding number of // possible number with // n digits excluding // a particular digit function digitNumber($n) { global $mod; // Checking if number // of digits is zero if ($n == 0) return 1; // Checking if number // of digits is one if ($n == 1) return 9; // Checking if number // of digits is odd if ($n % 2 != 0) { // Calling digitNumber // function with // (digit-1)/2 digits; $temp = digitNumber(($n - 1) / 2) % $mod; return (9 ($temp $temp) % $mod) % $mod; } else { // Calling digitNumber // function with n/2 digits $temp = digitNumber($n / 2) % $mod; return ($temp $temp) % $mod; } } function countExcluding($n, $d) { global $mod; // Calling digitNumber // function Checking if // excluding digit is // zero or non-zero if ($d == 0) return (9 digitNumber($n - 1)) % $mod; else return (8 digitNumber($n - 1)) % $mod; } // Driver code $d = 9; $n = 3; print(countExcluding($n, $d)); // This code is contributed by // Manish Shaw(manishshaw1) ? ```` JavaScript ```` // JavaScript Implementation of above method const mod = 1000000007; // Finding number of possible number with // n digits excluding a particular digit function digitNumber(n) { // Checking if number of digits is zero if (n == 0) return 1; // Checking if number of digits is one if (n == 1) return 9; // Checking if number of digits is odd if (n % 2) { // Calling digitNumber function // with (digit-1)/2 digits let temp = digitNumber((n - 1) / 2) % mod; return (9 (temp temp) % mod) % mod; } else { // Calling digitNumber function // with n/2 digits let temp = digitNumber(n / 2) % mod; return (temp temp) % mod; } } function countExcluding(n, d) { // Calling digitNumber function // Checking if excluding digit is // zero or non-zero if (d == 0) return (9 digitNumber(n - 1)) % mod; else return (8 digitNumber(n - 1)) % mod; } // Driver function to run above program // Initializing variables let d = 9; let n = 3; document.write(countExcluding(n, d) + "<br>"); // This code is contributed by Surbhi Tyagi. ```` Output: 648 Time Complexity: O(log n), where n represents the given integer. Auxiliary Space: O(logn), due to recursive stack space. R rishabh1322 Improve Article Tags : Misc Mathematical DSA number-digits Practice Tags : Mathematical Misc Similar Reads Basics & Prerequisites Logic Building Problems Logic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. It€™s the heart of coding, enabling programmers to think, reason, and arrive at smart solutions just like we do.Here are some tips for improving your programming logic: Understand the pro 2 min readAnalysis of Algorithms Analysis of Algorithms is a fundamental aspect of computer science that involves evaluating performance of algorithms and programs. 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For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ 3 min read]( to Recursion The process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution 14 min readGreedy Algorithms Greedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get 3 min readGraph Algorithms Graph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net 3 min readDynamic Programming or DP Dynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of 3 min readBitwise Algorithms Bitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit 4 min read Advanced Segment Tree Segment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree 3 min readPattern Searching Pattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i 2 min readGeometry Geometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br 2 min read Interview Preparation Interview Corner This article serves as your one-stop guide to interview preparation, designed to help you succeed across different experience levels and company expectations. Here is what you should expect in a Tech Interview, please remember the following points:Tech Interview Preparation does not have any fixed s 3 min readGfG160 Are you preparing for technical interviews and would like to be well-structured to improve your problem-solving skills? Well, we have good news for you! GeeksforGeeks proudly presents GfG160, a 160-day coding challenge starting on 15th November 2024. 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8276	https://vocus.cc/article/68374ed8fd89780001348fc7	仿射變換 (Affine Transformation) 郝信華 iPAS AI應用規劃師學習筆記註冊/登入首頁 AWS AIF-C01生成式AI技術與應用多模態人工智慧應用AI 導入評估AI導入規劃AI風險管理數據準備與模型選擇AI技術系統集成與部署自然語言處理與應用電腦視覺技術與應用敘述性統計與資料摘要技術關於仿射變換 (Affine Transformation) iPAS AI應用規劃師學習筆記人氣格鬥士發佈於電腦視覺技術與應用 2025/08/27 更新 2025/05/28 發佈閱讀 3 分鐘仿射變換 (Affine Transformation) 是一種二維幾何變換，它將一個平面上的點映射到另一個平面上的點，並保持直線的「直線性」和平行線的「平行性」。簡單來說，經過仿射變換後，圖像中的所有直線仍然是直線，平行的線仍然是平行的，但長度、角度和比例可能會改變。更正式地說，一個二維的仿射變換可以表示為以下形式：為什麼會看到廣告 x' = ax + by + e y' = cx + dy + f 其中，(x, y) 是原始圖像中的點的座標，(x', y') 是變換後對應點的座標，a、b、c、d、e、f 是定義變換的六個常數。仿射變換可以由以下幾種基本的幾何變換組合而成：平移 (Translation)：將圖像上的所有點沿著給定的向量移動一定的距離。在上述公式中，由參數 e 和 f 控制。縮放 (Scaling)：改變圖像的尺寸。在上述公式中，由參數 a 和 d 控制對應軸的縮放比例。旋轉 (Rotation)：將圖像繞著某個點旋轉一定的角度。這會影響公式中 a、b、c、d 的取值。剪切 (Shearing)：使圖像沿著一個軸的方向傾斜一定的角度。這會影響公式中 b 和 c 的取值。反射 (Reflection)：將圖像沿著一條直線翻轉。這也可以透過調整 a、b、c、d 的取值來實現。仿射變換的特性：直線仍然是直線：經過仿射變換，原始圖像中的任何一條直線都會保持為直線。平行線仍然是平行線：如果原始圖像中有兩條平行的直線，變換後它們仍然是平行的。點的順序保持不變：如果三個點在原始圖像中共線，它們在變換後仍然共線，並且它們之間的相對順序不變。仿射變換在圖像處理和電腦視覺中的應用：仿射變換是一個非常重要的工具，在許多領域都有廣泛的應用，例如：圖像配準 (Image Registration)：將不同時間、不同視角或不同感測器獲取的圖像對齊到同一個坐標系統中。圖像扭曲和變形 (Image Warping and Distortion Correction)：用於校正圖像的幾何失真，或創建特定的視覺效果。透視校正 (Perspective Correction)：校正由於拍攝角度造成的圖像透視變形。特徵匹配 (Feature Matching)：在不同視角或變形的圖像中尋找對應的特徵點。資料增強 (Data Augmentation)：在訓練機器學習模型時，通過對訓練圖像進行隨機的仿射變換，來增加資料的多樣性，提高模型的泛化能力。總而言之，仿射變換是一種強大而靈活的幾何變換，它可以對圖像進行各種線性變換，是圖像處理和電腦視覺領域中不可或缺的基本工具。含 AI 應用內容 #電腦視覺技術與應用#比例#校正#機器學習郝信華 iPAS AI應用規劃師學習筆記電腦視覺技術與應用圖像處理基礎技術留言留言分享你的想法！郝信華 iPAS AI應用規劃師學習筆記 21 會員 495 內容數加入現職 : 富邦建設資訊副理證照：經濟部 iPAS AI應用規劃師 AWS Certified AI Practitioner (AIF-C01) 加入郝信華 iPAS AI應用規劃師學習筆記的其他內容 2025/05/28 圖像檢索 (Image Retrieval) 圖像檢索 (Image Retrieval) 是指從大量的圖像集合中找到與給定的查詢相關的圖像的過程。查詢的方式有很多種，可以是：基於文本的圖像檢索 (Text-based Image Retrieval, TBIR)：使用文本關鍵字、標籤或描述來搜索相關的圖像。例如，在Google圖片 2025/05/28 圖像檢索 (Image Retrieval) 圖像檢索 (Image Retrieval) 是指從大量的圖像集合中找到與給定的查詢相關的圖像的過程。查詢的方式有很多種，可以是：基於文本的圖像檢索 (Text-based Image Retrieval, TBIR)：使用文本關鍵字、標籤或描述來搜索相關的圖像。例如，在Google圖片 2025/05/28 霍夫變換 (Hough Transform) 霍夫變換 (Hough Transform) 是一種在圖像處理中常用的特徵提取技術，主要用於在影像中檢測特定形狀，例如直線、圓形、橢圓等。它的主要思想是利用投票機制，將影像空間中的點轉換到參數空間中，並在參數空間中尋找累積投票數最多的點，這些點對應於原始影像中最有可能存在的形狀。以下是霍夫變換的 2025/05/28 霍夫變換 (Hough Transform) 霍夫變換 (Hough Transform) 是一種在圖像處理中常用的特徵提取技術，主要用於在影像中檢測特定形狀，例如直線、圓形、橢圓等。它的主要思想是利用投票機制，將影像空間中的點轉換到參數空間中，並在參數空間中尋找累積投票數最多的點，這些點對應於原始影像中最有可能存在的形狀。以下是霍夫變換的 2025/05/28 光照變化 (Illumination Variation) 光照變化 (Illumination Variation) 指的是在拍攝圖像或影片時，場景中光線的強度、方向、顏色或分布發生的改變。這些變化可能是由多種因素引起的，例如：時間的變化：隨著日出日落，自然光的光照強度和顏色會發生顯著變化。室內光線也可能因為開燈、關燈或燈光強度的調整而改變。 2025/05/28 光照變化 (Illumination Variation) 光照變化 (Illumination Variation) 指的是在拍攝圖像或影片時，場景中光線的強度、方向、顏色或分布發生的改變。這些變化可能是由多種因素引起的，例如：時間的變化：隨著日出日落，自然光的光照強度和顏色會發生顯著變化。室內光線也可能因為開燈、關燈或燈光強度的調整而改變。看更多你可能也想看 Chloe小窩手作人必看｜用蝦皮分潤計畫把興趣變新收入渠道在小小的租屋房間裡，透過蝦皮購物平臺採購各種黏土、模型、美甲材料等創作素材，打造專屬黏土小宇宙的療癒過程。文中分享多個蝦皮挖寶地圖，並推薦蝦皮分潤計畫。 #手作#黏土手作#輕黏土 2025/09/09 68 17 Chloe小窩手作人必看｜用蝦皮分潤計畫把興趣變新收入渠道在小小的租屋房間裡，透過蝦皮購物平臺採購各種黏土、模型、美甲材料等創作素材，打造專屬黏土小宇宙的療癒過程。文中分享多個蝦皮挖寶地圖，並推薦蝦皮分潤計畫。 #手作#黏土手作#輕黏土 2025/09/09 68 17 小蝸慢慢爬蝦皮分潤計畫-小豬與小蝸的婚姻神隊友小蝸和小豬因購物習慣不同常起衝突，直到發現蝦皮分潤計畫，讓小豬的購物愛好產生價值，也讓小蝸開始欣賞另一半的興趣。想增加收入或改善伴侶間的購物觀念差異？讓蝦皮分潤計畫成為你們的神隊友吧！ #蝦皮分潤計畫#蝦皮#聯盟行銷 2025/09/09 51 1 小蝸慢慢爬蝦皮分潤計畫-小豬與小蝸的婚姻神隊友小蝸和小豬因購物習慣不同常起衝突，直到發現蝦皮分潤計畫，讓小豬的購物愛好產生價值，也讓小蝸開始欣賞另一半的興趣。想增加收入或改善伴侶間的購物觀念差異？讓蝦皮分潤計畫成為你們的神隊友吧！ #蝦皮分潤計畫#蝦皮#聯盟行銷 2025/09/09 51 1 馬達技術傳承計畫空心線圈代工：多邊形 ( I ) 多邊形空心線圈十分類似方形線圈，同樣會有個線圈外膨的現象，使得完成線型可能不如預期。在方形空心線圈的討論文章中，著重討論的是兩彎角之間的距離及漆包線徑的剛性強度影響，這些要素在多邊形線圈當中依然存在。簡單的描述，就是兩彎角越近，則彎角中間的直線段外擴越嚴重；漆包線越粗，代表線材越不容易彎折，也會增加 #空心線圈#線圈代工#線圈製造 2024/07/16 4 馬達技術傳承計畫空心線圈代工：多邊形 ( I ) 多邊形空心線圈十分類似方形線圈，同樣會有個線圈外膨的現象，使得完成線型可能不如預期。在方形空心線圈的討論文章中，著重討論的是兩彎角之間的距離及漆包線徑的剛性強度影響，這些要素在多邊形線圈當中依然存在。簡單的描述，就是兩彎角越近，則彎角中間的直線段外擴越嚴重；漆包線越粗，代表線材越不容易彎折，也會增加 #空心線圈#線圈代工#線圈製造 2024/07/16 4 ysf的沙龍 The Nature of Code閱讀心得與Python實作：1.6 Normalizing Vectors 本節介紹向量的正規化。 #python#Python#PYTHON 2024/07/08 8 ysf的沙龍 The Nature of Code閱讀心得與Python實作：1.6 Normalizing Vectors 本節介紹向量的正規化。 #python#Python#PYTHON 2024/07/08 8 馬達技術傳承計畫馬達設計小工具：線徑轉換 ( I ) 本文是針對馬達繞線時，想要依本身的生產能力調整漆包線徑粗度時，會使用到的轉換計算進行介紹及說明。實際量產時往往將多條細線並繞的馬達，改由單條粗線採用機台繞線，較為省時；但開發階段，並無設備協助，僅能採用人工繞線打樣時，則會調整為多條細線並聯的模式才能順利工作，這類不同情境下的線徑變化，三不五時就 #馬達設計#漆包線徑 2024/06/18 6 馬達技術傳承計畫馬達設計小工具：線徑轉換 ( I ) 本文是針對馬達繞線時，想要依本身的生產能力調整漆包線徑粗度時，會使用到的轉換計算進行介紹及說明。實際量產時往往將多條細線並繞的馬達，改由單條粗線採用機台繞線，較為省時；但開發階段，並無設備協助，僅能採用人工繞線打樣時，則會調整為多條細線並聯的模式才能順利工作，這類不同情境下的線徑變化，三不五時就 #馬達設計#漆包線徑 2024/06/18 6 sen的沙龍上古漢語的邏輯結構 034 1.0 從函數到函算語法 1.2 函數概念小史 1.2.1 中譯的來源 1.2.2 一個速度問題 1.2.3 幾何的方法 1.2.4 微積分的記法 1.2.5 弦的振動三 1755年，歐拉改變了主意，在《微分學原理》(Institutiones calculi differen #哲學#邏輯學#語言學 2024/06/14 7 sen的沙龍上古漢語的邏輯結構 034 1.0 從函數到函算語法 1.2 函數概念小史 1.2.1 中譯的來源 1.2.2 一個速度問題 1.2.3 幾何的方法 1.2.4 微積分的記法 1.2.5 弦的振動三 1755年，歐拉改變了主意，在《微分學原理》(Institutiones calculi differen #哲學#邏輯學#語言學 2024/06/14 7 程式_筆記 mpu6050 感測器加速度傾斜角度這篇介紹如何用加速度取得傾斜角度。用的是和前篇一樣的 #Arduino#mpu6050#感測器 2024/06/09 5 程式_筆記 mpu6050 感測器加速度傾斜角度這篇介紹如何用加速度取得傾斜角度。用的是和前篇一樣的 #Arduino#mpu6050#感測器 2024/06/09 5 Hippo Big的沙龍 8 風水or數列玄同竟然開口說道“先前產生的三角型，在水準平面上既不等腰也不等邊，但如果將其在三維笛卡爾坐標系中調整角度，就可以得到一個正四面體。當然，這時正四面體的底面一定與水平面不平行。”他的聲音很有磁性，充滿了力量感，但說話的語氣十分生硬而且目光也只盯著懸空的投影，完全不與任何人發生交流。這時，阿離蹦跳著 #原創#科幻#小説 2024/06/06 Hippo Big的沙龍 8 風水or數列玄同竟然開口說道“先前產生的三角型，在水準平面上既不等腰也不等邊，但如果將其在三維笛卡爾坐標系中調整角度，就可以得到一個正四面體。當然，這時正四面體的底面一定與水平面不平行。”他的聲音很有磁性，充滿了力量感，但說話的語氣十分生硬而且目光也只盯著懸空的投影，完全不與任何人發生交流。這時，阿離蹦跳著 #原創#科幻#小説 2024/06/06 螃蟹_crab的沙龍 [Python]導數與偏導數(學習心得) 直觀理解導數：考慮的是單一變數的函數，描述的是函數在某點的斜率或變化率。偏導數：考慮的是多變數函數，描述的是函數在某個變數變化時的變化率，其他變數保持不變。 (針對各維度的調整或者稱變化你要調多少) 應用導數：在物理學中應用廣泛，例如描述速度和加速度。偏導數：在多變量分析、優 #Python#導數與偏導數#導數 2024/06/01 43 螃蟹_crab的沙龍 [Python]導數與偏導數(學習心得) 直觀理解導數：考慮的是單一變數的函數，描述的是函數在某點的斜率或變化率。偏導數：考慮的是多變數函數，描述的是函數在某個變數變化時的變化率，其他變數保持不變。 (針對各維度的調整或者稱變化你要調多少) 應用導數：在物理學中應用廣泛，例如描述速度和加速度。偏導數：在多變量分析、優 #Python#導數與偏導數#導數 2024/06/01 43 辯證人生高微筆記凡所有相皆是虛妄，若見諸相非相，即見如來 2. 能量看不到，卻統籌物理世界（形而上統籌形而下） 3. 數學與物理的不同：數學「定理」：絕對真理，不因時空轉換；物理「定律」：找到自然背後的律，而非證明 4. 數學的本質：建立在不能再問的「公理」上 5. 歐式平 #數學的本質#數學哲學#高等微積分 2024/03/16 2 辯證人生高微筆記凡所有相皆是虛妄，若見諸相非相，即見如來 2. 能量看不到，卻統籌物理世界（形而上統籌形而下） 3. 數學與物理的不同：數學「定理」：絕對真理，不因時空轉換；物理「定律」：找到自然背後的律，而非證明 4. 數學的本質：建立在不能再問的「公理」上 5. 歐式平 #數學的本質#數學哲學#高等微積分 2024/03/16 2 樹屋形狀側向力是否必定垂直馬格努斯力？行進中的球，因表面形狀相對速度方向不對稱，產生了不對稱的邊界層分離，以及使球不同位置受到的空氣作用力大小不同，最終導致空氣作用力的合力，出現「垂直速度方向的分力(f)」。這個分力 f，源於表面形狀 s 的不對稱，而球若有旋轉，則每個時刻的 f 都可能會變化。既然如此，那不如擴充表面形狀(s) 2024/03/15 3 樹屋形狀側向力是否必定垂直馬格努斯力？行進中的球，因表面形狀相對速度方向不對稱，產生了不對稱的邊界層分離，以及使球不同位置受到的空氣作用力大小不同，最終導致空氣作用力的合力，出現「垂直速度方向的分力(f)」。這個分力 f，源於表面形狀 s 的不對稱，而球若有旋轉，則每個時刻的 f 都可能會變化。既然如此，那不如擴充表面形狀(s) 2024/03/15 3 教育心理博士的筆記本隨機截距交叉延宕模式:加入跨時間不變的預測或結果變項(4) 之前已經說過限制模型，接下來進入下一部份根據Mulder and Hamaker (2021)建議，在 RI-CLPM 中，有許多擴展模型，今天要介紹的是 Extension 1。Extension 1就是加入跨時間不變的預測或結果變項，本文將介紹此模型構造和語法。 #Mplus#RICLPM#MOD 2024/02/27 4 教育心理博士的筆記本隨機截距交叉延宕模式:加入跨時間不變的預測或結果變項(4) 之前已經說過限制模型，接下來進入下一部份根據Mulder and Hamaker (2021)建議，在 RI-CLPM 中，有許多擴展模型，今天要介紹的是 Extension 1。Extension 1就是加入跨時間不變的預測或結果變項，本文將介紹此模型構造和語法。 #Mplus#RICLPM#MOD 2024/02/27 4 追蹤感興趣的內容從 Google News 追蹤更多 vocus 的最新精選內容追蹤 Google News 開啟目錄分享留言收藏喜歡留言收藏喜歡 Powered By
8277	https://www.chem.tamu.edu/rgroup/marcetta/chem362/lectures/symmetry%20ops%20and%20elements%20and%20pt%20gps.pdf	MOLECULAR SYMMETRY Know intuitively what "symmetry" means - how to make it quantitative? Will stick to isolated, finite molecules (not crystals). SYMMETRY OPERATION Carry out some operation on a molecule (or other object) - e.g. rotation. If final configuration is INDISTINGUISHABLE from the initial one - then the operation is a SYMMETRY OPERATION for that object. The line, point, or plane about which the operation occurs is a SYMMETRY ELEMENT N.B. “Indistinguishable” does not necessarily mean “identical”. e.g. for a square piece of card, rotate by 90º as shown below: 1 2 4 3 4 1 3 2 90o rotation i.e. the operation of rotating by 90o is a symmetry operation for this object Labels show final configuration is NOT identical to original. Further 90º rotations give other indistinguishable configurations - until after 4 (360º) the result is identical. SYMMETRY OPERATIONS Motions of molecule (rotations, reflections, inversions etc. - see below) which convert molecule into configuration indistinguishable from original. SYMMETRY ELEMENTS Each element is a LINE, PLANE or POINT about which the symmetry operation is performed. Example above - operation was rotation, element was a ROTATION AXIS. Other examples later. Summary of symmetry elements and operations: Symmetry element Symmetry operation(s) – E (identity) Cn (rotation axis) Cn 1.....Cn n-1 (rotation about axis) σ (reflection plane) σ (reflection in plane) i (centre of symm.) i (inversion at centre) Sn (rot./reflection axis)Sn 1...Sn n-1 (n even) (rot./reflection about axis) Sn 1...Sn 2n-1 (n odd) Notes (i) symmetry operations more fundamental, but elements often easier to spot. (ii) some symmetry elements give rise to more than one operation - especially rotation - as above. ROTATIONS - AXES OF SYMMETRY Some examples for different types of molecule: e.g. H2O O (1)H H(2) O (2)H H(1) rotate 180o Line in molecular plane, bisecting HOH angle is a rotation axis, giving indistinguishable configuration on rotation by 180o. BF3 By VSEPR - trigonal, planar, all bonds equal, all angles 120o. Take as axis a line perpendicular to molecular plane, passing through B atom. B (1)F F(2) F(3) B (2)F F(3) F(1) 120o axis perpendicular to plane N.B. all rotations CLOCKWISE when viewed along -z direction. (1)F B F(2) F(3) z view down here Symbol for axes of symmetry Cn where rotation about axis gives indistinguishable configuration every (360/n)o (i.e. an n-fold axis) Thus H2O has a C2 (two-fold) axis, BF3 a C3 (three-fold) axis. One axis can give rise to >1 rotation, e.g. for BF3, what if we rotate by 240o? B (1)F F(2) F(3) B (3)F F(1) F(2) 240o Must differentiate between two operations. Rotation by 120o described as C3 1, rotation by 240o as C3 2. In general Cn axis (minimum angle of rotation (360/n)o) gives operations Cn m, where both m and n are integers. When m = n we have a special case, which introduces a new type of symmetry operation..... IDENTITY OPERATION For H2O, C2 2 and for BF3 C3 3 both bring the molecule to an IDENTICAL arrangement to initial one. Rotation by 360o is exactly equivalent to rotation by 0o, i.e. the operation of doing NOTHING to the molecule. MORE ROTATION AXES xenon tetrafluoride, XeF4 C4 Xe (4)F F(1) F(3) F(2) Xe (3)F F(4) F(2) F(1) 90o cyclopentadienide ion, C5H5 – C C C C C H(1) H(2) H(3) (4)H (5)H C5 C C C C C H(5) H(1) H(2) (3)H (4)H . 72o benzene, C6H6 C C C C C C H(1) H(2) H(3) H(4) (5)H (6)H C6 C C C C C C H(6) H(1) H(2) H(3) (4)H (5)H 60o . Examples also known of C7 and C8 axes. If a C2n axis (i.e. even order) present, then Cn must also be present: C4 Xe (4)F F(1) F(3) F(2) Xe (3)F F(4) F(2) Xe (2)F F(1) F(3) F(1) F(4) 90o i.e. C4 1 180o i.e. C4 2 (≡ C2 1) Therefore there must be a C2 axis coincident with C4, and the operations generated by C4 can be written: C4 1, C4 2 (C2 1), C4 3, C4 4 (E) Similarly, a C6 axis is accompanied by C3 and C2, and the operations generated by C6 are: C6 1, C6 2 (C3 1), C6 3 (C2 1), C6 4 (C3 2), C6 5, C6 6 (E) Molecules can possess several distinct axes, e.g. BF3: C3 F B F F C2 C2 C2 Three C2 axes, one along each B-F bond, perpendicular to C3 Operation = reflection Element = plane of symmetry symbol σ Greek letter ‘sigma’ Several different types of symmetry plane -different orientations with respect to symmetry axes. By convention - highest order rotation axis drawn VERTICAL. Therefore any plane containing this axis is a VERTICAL PLANE, σv. e.g. H2O plane above (often also called σ(xz)) Can be >1 vertical plane, e.g. for H2O there is also: H(2) O (1)H z y x σ(yz) - reflection leaves all atoms unshifted, therefore symmetry plane This is also a vertical plane, but symmetrically different from other, could be labelled σv'. Any symmetry plane PERPENDICULAR to main axis is a HORIZONTAL PLANE, σh. e.g. for XeF4: C4 Xe F F F F Plane of molecule (perp. to C4) is a symmetry plane, i.e. σh) Some molecules possess additional planes, as well as σv and σh, which need a separate label. e.g. XeF4 F Xe F F F σv σv σd σd Four "vertical" planes - but two different from others.Those along bonds called σv, but those bisecting bonds σd - i.e. DIHEDRAL PLANES Usually, but not always, σv and σd differentiated in same way. Two final points about planes of symmetry: (i) if no Cn axis, plane just called σ; (ii) unlike rotations, only ONE operation per plane. A second reflection returns you to original state, i.e. (σ)(σ) = σ2 = E INVERSION : CENTRES OF SYMMETRY z y x z y x inversion . (x, y, z) . (-x, -y, -z) The origin, (0, 0, 0) is the centre of inversion. If the coordinates of every point are changed from (x,y,z) to (-x, -y, -z), and the resulting arrangement is indistinguishable from original - the INVERSION is a symmetry operation, and the molecule possesses a CENTRE OF SYMMETRY (INVERSION) (i.e. CENTROSYMMETRIC) Involves BOTH rotation AND reflection. OPERATION : INVERSION ELEMENT : a POINT - CENTRE OF SYMMETRY or INVERSION CENTRE. Best described in terms of cartesian axes: e.g. trans-N2F2 (1)N N(2) (1)F F(2) (2)N N(1) (2)F F(1) inversion . centre of symmetry In practice, inversion involves taking every atom to the centre - and out the same distance in the same direction on the other side. Symbol - same for operation (inversion) and element (centre): i Another example: XeF4 (4)F Xe F(2) F(3) F(1) (2)F Xe F(4) F(1) F(3) i Xe atom is centre of symmetry As for reflections, the presence of a centre of symmetry only generates one new operation, since carrying out inversion twice returns everything back to start. (x, y, z) i (-x, -y, -z) i (x, y, z) i.e. (i)(i) = i2 = E Inversion is a COMPOSITE operation, with both rotation and reflection components. Consider a rotation by 180o about the z axis: (-x, -y, z) (x, y, z) Follow this by reflection in the xy plane (-x, -y, z) (-x, -y, -z) BUT individual components need not be symmetry operations themselves............. e.g. staggered conformation of CHClBr-CHClBr H Cl Br C C H Cl Br . centre Inversion at centre gives indistinguishable configuration. The components, of rotation by 180o or reflection in a plane perpendicular to the axis, do not. If, however, a molecule does possess a C2 axis and a σh (perpendicular) plane as symmetry operations, then inversion (i) must also be a symmetry operation. IMPROPER ROTATIONS : ROTATION-REFLECTION AXES Operation: clockwise rotation (viewed along -z direction) followed by reflection in a plane perpendicular to that axis. Element: rotation-reflection axis (sometimes known as "alternating axis of symmetry") As for inversion - components need not be themselves symmetry operations for the molecule. e.g. a regular tetrahedral molecule, such as CH4 H C H H H H C H H H H C H H H rotate 90o reflect four-fold rotation reflection S4 1 Symbols: rotation-reflection axis Sn (element) rotation-reflection Sn m (operations) where rotation is through (360/n)o S4 axis requires presence of coincident C2 axis If Cn and σh are both present individually - there must also be an Sn axis : e.g. BF3 - trigonal planar F B F F C3, S3 σh in plane of molecule. C3 1 + σh individually, therefore S3 1 must also be a symmetry operation Other Sn examples: IF7, pentagonal bipyramid, has C5 and σh, therefore S5 also. Ethane in staggered conformation S6 C H H H C H H H i.e. rotate by 60o and reflect in perp. plane. Note NO C6, σh separately. POINT GROUPS ASSIGNMENT OF MOLECULES TO POINT GROUPS STEP 1 : LOOK FOR AN AXIS OF SYMMETRY If one is found - go to STEP 2 If not: look for (a) plane of symmetry - if one is found, molecule belongs to point group Cs A collection of symmetry operations all of which pass through a single point A point group for a molecule is a quantitative measure of the symmetry of that molecule POINT GROUPS ASSIGNMENT OF MOLECULES TO POINT GROUPS STEP 1 : LOOK FOR AN AXIS OF SYMMETRY If one is found - go to STEP 2 If not: look for (a) plane of symmetry - if one is found, molecule belongs to point group Cs A collection of symmetry operations all of which pass through a single point A point group for a molecule is a quantitative measure of the symmetry of that molecule e.g. SOCl2 S O Cl Cl .. No axis, but plane containing S, O, bisecting ClSCl angle, is a symmetry plane. Hence Cs point group. If no plane is found, look for (b) centre of symmetry - if one is found, molecule belongs to point group Ci. e.g. CHClBrCHClBr (staggered conformation): C C H Cl Br H Cl Br No axis, no planes, but mid-point of C-C bond is centre of symmetry. Therefore Ci point group. No axes, plane or centre, therefore (c) no symmetry except E : point group C1 (so called because E = C1, rotation through 360o) e.g. CHFClBr H C F Cl Br No symmetry except E, therefore point group C1. STEP 2 : LOOK FOR C2 AXES PERPENDICULAR TO Cn If found, go to STEP 3. If not, look for (a) a HORIZONTAL PLANE OF SYMMETRY, if found - point group is Cnh (Cn = highest order axis) e.g. trans-N2F2: F N N F . Highest order axis is C2 (perp. to plane, through mid-pt. of N=N bond). No C2 axes perp. to this, but molecular plane is plane of symmetry (perp. to C2, i.e. σh). Point group C2h. If there is no horizontal plane, look for (b) n VERTICAL PLANES OF SYMMETRY. If found, molecule belongs to point group Cnv Many examples, e.g. H2O H O H C2 C2 axis as shown. No other C2's, no σh, but two sv's, one in plane, one perp. to plane, bisecting HOH angle. Point group C2v PCl3 P Cl Cl Cl C3 C3 highest order axis No C2's perp. to C3 No σh, but 3 σv's, each contains P, one Cl Therefore C3v BrF5 By VSEPR, square pyramidal Br F F F F F C4 Highest order axis : C4 No C2's perp. to C4 No σh, but 4 vertical planes. Therefore C4v N.B. of 4 vertical planes, two are σv's, two σd's F Br F F F σv σd σd σv (looking down C4 axis) If no planes at all, could have (c) no other symmetry elements: point group Cn , or (d) an S2n axis coincident with Cn: point group S2n STEP 3 If there are nC2's perp. to Cn, look for: (a) horizontal plane of symmetry. If present, point group is Dnh e.g. ethene (ethylene), C2H4 C C H H H H C2 C2 C2 Highest order axis C2 - along C=C bond Two additional C2's as shown. σh in plane defined by the last two C2's Point group D2h BF3 F B F F C3 C2 C2 C2 Planar trigonal molecule by VSEPR Main axis C3 3 C2's perp. to C3 σh - plane of molecule Point group D3h XeF4 Square planar by VSEPR Xe F F F F C4 C2' C2" C2" C2' Main axis C4 4 C2's perp. to C4 (2 along XeF bonds (C2'), 2 bisecting, (C2")) σh - plane of molecule Point group D4h If no σh, look for: (b) n vertical planes of symmetry (σv/σd). If these are present, molecule belongs to point group Dnd e.g. allene, H2C=C=CH2. C C H H C H H C2 (main axis) Looking down C=C=C bond H C H H H C2' C2' Main axis C2 - along C=C=C Two C2's as shown Two vertical planes (σd) - each containing one CH2 unit Point group D2d A few molecules do not appear to fit into this general scheme.......... LINEAR MOLECULES Molecular axis is C∞ - rotation by any arbitrary angle (360/∞)o, so infinite number of rotations. Also any plane containing axis is symmetry plane, so infinite number of planes of symmetry. Divide linear molecules into two groups: Do in fact fit into scheme - but they have an infinite number of symmetry operations. (i) No centre of symmetry, e.g.: H C N C∞ No C2's perp. to main axis, but ∞ σv's containing main axis: point group C∞v (ii) Centre of symmetry, e.g.: C2 O C O C2 C∞ σh i.e. C∞ + ∞C2's + σh Point group D∞h A few geometries have several, equivalent, highest order axes. Two geometries most important: Highly symmetrical molecules Regular tetrahedron e.g. Cl Si Cl Cl Cl 4 C3 axes (one along each bond) 3 C2 axes (bisecting pairs of bonds) 3 S4 axes (coincident with C2's) 6 σd's (each containing Si and 2 Cl's) Point group: Td Regular octahedron e.g. S F F F F F F 3C4's (along F-S-F axes) also 4 C3's. 6 C2's, several planes, S4, S6 axes, and a centre of symmetry (at S atom) Point group Oh These molecules can be identified without going through the usual steps. Note: many of the more symmetrical molecules possess many more symmetry operations than are needed to assign the point group.
8278	https://www.simplilearn.com/tutorials/statistics-tutorial/spearmans-rank-correlation	HomeResourcesData Science & Business AnalyticsThe Ultimate Statistics TutorialSpearman’s Rank Correlation: The Definitive Guide To Understand Tutorial Playlist #### Statistics Tutorial Overview #### Everything You Need to Know About the Probability Density Function in Statistics Lesson - 1#### The Best Guide to Understand Central Limit Theorem Lesson - 2#### Measures of Central Tendency : Mean, Median and Mode Lesson - 3#### The Ultimate Guide to Understand Conditional Probability Lesson - 4#### Percentile in Statistics Lesson - 5#### The Best Guide to Understand Bayes Theorem Lesson - 6#### Everything You Need to Know About the Normal Distribution Lesson - 7#### An In-Depth Explanation of Cumulative Distribution Function Lesson - 8#### Chi-Square Test Lesson - 9#### What Is Hypothesis Testing in Statistics? Types and Examples Lesson - 10#### Understanding the Fundamentals of Arithmetic and Geometric Progression Lesson - 11#### The Definitive Guide to Understand Spearman’s Rank Correlation Lesson - 12#### Mean Squared Error: Overview, Examples, Concepts and More Lesson - 13#### All You Need to Know About the Empirical Rule in Statistics Lesson - 14#### The Complete Guide to Skewness and Kurtosis Lesson - 15#### A Holistic Look at Bernoulli Distribution Lesson - 16#### All You Need to Know About Bias in Statistics Lesson - 17#### A Complete Guide to Get a Grasp of Time Series Analysis Lesson - 18#### The Key Differences Between Z-Test Vs. T-Test Lesson - 19#### The Complete Guide to Understand Pearson's Correlation Lesson - 20#### A Complete Guide on the Types of Statistical Studies Lesson - 21#### Everything You Need to Know About Poisson Distribution Lesson - 22#### Your Best Guide to Understand Correlation vs. Regression Lesson - 23#### The Most Comprehensive Guide for Beginners on What Is Correlation Lesson - 24 Spearman’s Rank Correlation: The Definitive Guide To Understand Lesson 12 of 24By Aryan Gupta Last updated on Aug 23, 2025275385 PreviousNext Tutorial Playlist #### Statistics Tutorial Overview #### Everything You Need to Know About the Probability Density Function in Statistics Lesson - 1#### The Best Guide to Understand Central Limit Theorem Lesson - 2#### Measures of Central Tendency : Mean, Median and Mode Lesson - 3#### The Ultimate Guide to Understand Conditional Probability Lesson - 4#### Percentile in Statistics Lesson - 5#### The Best Guide to Understand Bayes Theorem Lesson - 6#### Everything You Need to Know About the Normal Distribution Lesson - 7#### An In-Depth Explanation of Cumulative Distribution Function Lesson - 8#### Chi-Square Test Lesson - 9#### What Is Hypothesis Testing in Statistics? Types and Examples Lesson - 10#### Understanding the Fundamentals of Arithmetic and Geometric Progression Lesson - 11#### The Definitive Guide to Understand Spearman’s Rank Correlation Lesson - 12#### Mean Squared Error: Overview, Examples, Concepts and More Lesson - 13#### All You Need to Know About the Empirical Rule in Statistics Lesson - 14#### The Complete Guide to Skewness and Kurtosis Lesson - 15#### A Holistic Look at Bernoulli Distribution Lesson - 16#### All You Need to Know About Bias in Statistics Lesson - 17#### A Complete Guide to Get a Grasp of Time Series Analysis Lesson - 18#### The Key Differences Between Z-Test Vs. T-Test Lesson - 19#### The Complete Guide to Understand Pearson's Correlation Lesson - 20#### A Complete Guide on the Types of Statistical Studies Lesson - 21#### Everything You Need to Know About Poisson Distribution Lesson - 22#### Your Best Guide to Understand Correlation vs. Regression Lesson - 23#### The Most Comprehensive Guide for Beginners on What Is Correlation Lesson - 24 Table of Contents What Is Monotonic Function? Spearman’s Rank Correlation Example of Spearman’s Rank Correlation Conclusion Correlation is a statistical measure that determines how closely two variables fluctuate. A positive correlation shows the extent to which those variables increase or decrease in parallel. A negative correlation shows the range in which one variable increases as the other decreases. In this article, we will discuss one such correlation i.e Spearman’s Rank Correlation. Elevate Your Data Analytics Career in 2025 PL-300 Microsoft Power BI Certification TrainingExplore Program What Is Monotonic Function? To understand Spearman’s rank correlation, it is important to understand monotonic function. A monotonic function is one that either never increases or never decreases as its independent variable changes. The following graph illustrates the monotonic function: Monotonically Increasing: As the variable X increases, the variable Y never decreases. Monotonically Decreasing: As the variable X increases, the variable Y never increases. Not Monotonic: As the X variable increases, the Y variable sometimes decreases and sometimes increases. Spearman’s Rank Correlation Spearman’s rank correlation measures the strength and direction of association between two ranked variables. It basically gives the measure of monotonicity of the relation between two variables i.e. how well the relationship between two variables could be represented using a monotonic function. The formula for Spearman’s rank coefficient is: 𝝆 = Spearman’s rank correlation coefficient di = Difference between the two ranks of each observation n = Number of observations The Spearman Rank Correlation can take a value from +1 to -1 where, A value of +1 means a perfect association of rank A value of 0 means that there is no association between ranks A value of -1 means a perfect negative association of rank Let’s understand the concept better with the help of an example. Master Gen AI Strategies for Businesses with Generative AI for Business Transformation ProgramExplore Program Example of Spearman’s Rank Correlation Consider the score of 5 students in Maths and Science that are mentioned in the table. Step 1: Create a table for the given data. Step 2: Rank both the data in descending order. The highest marks will get a rank of 1 and the lowest marks will get a rank of 5. Step 3: Calculate the difference between the ranks (d) and the square value of d. Step 4: Add all your d square values. Step 5: Insert these values into the formula. = 1 - (6 14) / 5(25 - 1) = 0.3 The Spearman’s Rank Correlation for the given data is 0.3. The value is near 0, which means that there is a weak correlation between the two ranks. Our Data Analyst Master's Program will help you learn analytics tools and techniques to become a Data Analyst expert! It's the pefect course for you to jumpstart your career. Enroll now! Conclusion In this tutorial ‘The Definitive Guide To Understand Spearman’s Rank Correlation’, you saw the concept of Spearman’s Rank Correlation and how to find its rank coefficient. You also worked on an example to understand how you can find the association between the ranks of the given data. If you are looking to pursue a career as a Data Analyst, Simplilearn’s Data Analyst course in collaboration with IBM is the program for you. Was this tutorial on Spearman’s Rank Correlation useful to you? Do you have any doubts or questions for us? Mention them in this article's comments section, and we'll have our experts answer them for you at the earliest! Advance your knowledge through our popular online courses, available across multiple categories: Data Analytics Courses Technical Courses Excel Courses PG Courses About the Author Aryan Gupta Aryan is a tech enthusiast who likes to stay updated about trending technologies of today. He is passionate about all things technology, a keen researcher, and writes to inspire. Aside from tec… View More Recommended Programs PL-300 Microsoft Power BI Certification Training 50805 Learners Lifetime AccessData Analyst 41648 Learners Lifetime AccessData Scientist 44715 Learners Lifetime Access Lifetime access to high-quality, self-paced e-learning content. Explore Category Recommended Resources What is Power BI?: Architecture, and Featu… Tutorial Getting Started with Microsoft Azure Ebook A Beginner's Guide to Learning Power BI the … Article Power BI Vs Tableau: Difference and Compar… Tutorial An Introduction to Power BI Dashboard Tutorial Introduction to Microsoft Azure Basics: A Beginn… Ebook prevNext Acknowledgement PMP, PMI, PMBOK, CAPM, PgMP, PfMP, ACP, PBA, RMP, SP, OPM3 and the PMI ATP seal are the registered marks of the Project Management Institute, Inc.
8279	https://wiki.mbalib.com/wiki/%E5%88%86%E5%B1%82%E6%8A%BD%E6%A0%B7	过去的17年，百科频道一直以免费公益的形式为大家提供知识服务，这是我们团队的荣幸和骄傲。然而，在目前越来越严峻的经营挑战下，单纯依靠不断增加广告位来维持网站运营支出，必然会越来越影响您的使用体验，这也与我们的初衷背道而驰。因此，经过审慎地考虑，我们决定推出VIP会员收费制度，以便为您提供更好的服务和更优质的内容。 MBA智库百科VIP会员，您的权益将包括： 1、无广告阅读； 2、免验证复制。支付成功全球专业中文经管百科，由121,994位网友共同编写而成，共计436,089个条目查看条目讨论编辑收藏简体中文繁体中文工具箱▼ 上传文件特殊页面可打印版永久链接分层抽样用手机看条目扫一扫，手机看条目出自 MBA智库百科( 分层抽样（stratified sampling） \| \| \| 目录[隐藏] 1 什么是分层抽样 2 分层抽样的特点 3 各层样本数的确定方法 4 分层抽样的应用 5 多阶抽样与分层抽样和整群抽样的关系 \| [编辑] 什么是分层抽样分层抽样，也叫类型抽样。就是将总体单位按其属性特征分成若干类型或层，然后在类型或层中随机抽取样本单位。 [编辑] 分层抽样的特点分层抽样的特点是：由于通过划类分层，增大了各类型中单位间的共同性，容易抽出具有代表性的调查样本。该方法适用于总体情况复杂，各单位之间差异较大，单位较多的情况。 [编辑] 各层样本数的确定方法 \| 市场调查方法 \| \| \| \| \| A \| \| 案头调研 \| \| 案例研究法 \| \| B \| \| 不重复抽样 \| \| C \| \| 抽样调查 \| \| 重置抽样 \| \| 抽签法 \| \| 产品留置测试 \| \| D \| \| 多维尺度法 \| \| 定量研究方法 \| \| 定性研究方法 \| \| 典型调查法 \| \| 电话调查 \| \| 多阶段抽样 \| \| 等距抽样 \| \| 独立控制配额抽样 \| \| 等距量表 \| \| 等比量表 \| \| E \| \| 二手资料调研 \| \| 二路焦点小组 \| \| F \| \| 非概率抽样 \| \| 分层抽样 \| \| 分层比例抽样 \| \| 分层最佳抽样 \| \| G \| \| 观察法 \| \| 概率抽样 \| \| 拐点调研 \| \| 滚雪球抽样 \| \| H \| \| 会议调查 \| \| J \| \| 焦点访谈法 \| \| 经验判断法 \| \| 随机抽样 \| \| 家庭日记法 \| \| 经销商访谈 \| \| K \| \| 可行性研究 \| \| 控制实验法 \| \| L \| \| 联合分析法 \| \| 留置调查 \| \| 垃圾调研法 \| \| 类别量表 \| \| M \| \| 面谈访问法 \| \| 盲测 \| \| 描述性调研 \| \| 媒介调查法 \| \| P \| \| PPS \| \| 判断抽样 \| \| 配额抽样 \| \| 平衡量表法 \| \| 评价量表 \| \| 配对比较量表 \| \| Q \| \| Q分类法 \| \| R \| \| 任意抽样 \| \| S \| \| 容量测定法 \| \| SEM模型 \| \| 深层访谈法 \| \| 双重抽样 \| \| 实验调查法 \| \| 实地调研 \| \| 数值分配量表 \| \| 随机号码表法 \| \| 顺序量表 \| \| T \| \| 投影技法 \| \| 推销估计法 \| \| 投射研究 \| \| 探索性调研 \| \| W \| \| 文献调查法 \| \| 问卷调查法 \| \| 网络调研 \| \| 文案调查法 \| \| 无准备访问 \| \| 网上调查 \| \| X \| \| 询问法 \| \| 辛迪加调研 \| \| 行踪分析 \| \| 相互控制配额抽样 \| \| Y \| \| 邮寄调查 \| \| 因果性调研 \| \| Z \| \| 主观概率法 \| \| 整群抽样 \| \| 重点调查 \| \| 逐户寻找法 \| \| \| [编辑] \| \| \| 各层样本数的确定方法有3种： ①分层定比。即各层样本数与该层总体数的比值相等。例如，样本大小n=50，总体N=500，则n/N=0.1 即为样本比例，每层均按这个比例确定该层样本数。 ②奈曼法。即各层应抽样本数与该层总体数及其标准差的积成正比。 ③非比例分配法。当某个层次包含的个案数在总体中所占比例太小时，为使该层的特征在样本中得到足够的反映，可人为地适当增加该层样本数在总体样本中的比例。但这样做会增加推论的复杂性。 [编辑] 分层抽样的应用总体中赖以进行分层的变量为分层变量，理想的分层变量是调查中要加以测量的变量或与其高度相关的变量。分层的原则是增加层内的同质性和层间的异质性。常见的分层变量有性别、年龄、教育、职业等。分层随机抽样在实际抽样调查中广泛使用，在同样样本容量的情况下，它比纯随机抽样的精度高，此外管理方便，费用少，效度高。 [编辑] 多阶抽样与分层抽样和整群抽样的关系多阶段抽样区别于分层抽样，其优点在于适用于抽样调查的面特别广，没有一个包括所有总体单位的抽样框，或总体范围太大，无法直接抽取样本等情况，可以相对节省调查费用。其主要缺点是抽样时较为麻烦,而且从样本对总体的估计比较复杂。将总体分为若干个一阶单元，如果在每一个一阶单元中，都随机抽取部分二阶单元，由这些二阶单元中的总体基本单元组成的样本，在抽样的方式上，就相当于分层抽样；如果在全部的一阶单元中，只抽取了部分一阶单元，并对抽中的一阶单元中的所有的基本单元都做全面调查，这就是整群抽样。因此，分层抽样实际是第一阶抽样比为100%时的一种特殊的两阶抽样；而整群抽样实际上是第二阶抽样比为100%时的一种特殊的两阶抽样，故也称单级整群抽样。令fi为抽样比，即有：当时，二阶抽样可视为分层抽样，当时，二阶抽样可视为整层抽样。多阶抽样与分层抽样的主要区别在于：一、分层抽样是对总体中的每个一级样本群体进行全面入样，再对所有的样本进行抽查；而两阶抽样则把总体中所有的群体视为一阶单元，对这些一阶单元进行抽样，将抽出的样本再次进行抽样（两次都不是进行全面的调查），产生两级样本，最后综合估算出总的一级样本指标。二、整群抽样是对总体中抽取的每个样本群体所包含的基本单元进行全面调查；而两阶抽样则把总体中所有的群体视为一阶单元，对每一个被抽中的一阶单元所包含的二级单元（即基本单位），不是进行全面的调查，而是再进行一次抽样调查（也称抽子样本）。即两阶抽样，产生两级样本，最后综合估算出总的一级样本指标。至于在综合估算的方式方法上，两阶抽样与整群抽样也是极其相似的，只不过前者为就被抽一级单元的样本指标进行综合估算，后者为就被抽样群体单元的全体指标进行综合估算。来自" 打开MBA智库App, 阅读完整内容打开App 本条目对我有帮助100 赏 MBA智库APP 扫一扫，下载MBA智库APP 分享到：温馨提示复制该内容请前往MBA智库App 立即前往App 如果您认为本条目还有待完善，需要补充新内容或修改错误内容，请编辑条目或投诉举报。本条目相关文档抽样技术3分层抽样 50页品质管理抽样检验抽样技术3分层抽样 50页分层抽样 22页机械抽样与分层抽样 12页 [品质管理抽样检验]抽样技术3分层抽样 48页论多主题分层抽样 4页分层抽样111219 38页 213分层抽样 32页多主题分层抽样的分层技术 3页多主题分层抽样的分层技术 3页更多相关文档本条目相关课程 ###### 从专业到管理:管理思维与管理能力的双重跃迁王达峰 ¥99 ###### 阿里管理者基本动作训练营王建和 ¥299 ###### 阿里运营36课：大厂运营思维与方法论让你做高薪人才小马鱼 ¥99 ¥99 ###### 裂变方案：用户裂变10倍的增长方法周导 ¥199 本条目由以下用户参与贡献 Angle Roh,Zfj3000,Kane0135,Cabbage,鲈鱼,KAER,M id 195635392bc048f369b00f1d2fa46189. 页面分类: 市场调查方法 \| 抽样方法评论(共6条) 提示:评论内容为网友针对条目"分层抽样"展开的讨论，与本站观点立场无关。 222.75.167. 在 2012年3月27日 09:29 发表求分层抽样案例！！！急急急！！！！回复评论 110.80.33. 在 2012年3月27日 11:09 发表 222.75.167. 在 2012年3月27日 09:29 发表求分层抽样案例！！！急急急！！！！上面相关文档就有分层抽样的案例回复评论 58.20.174. 在 2013年7月1日 09:37 发表有配偶抽样这个概念么？回复评论 131.227.32. 在 2013年7月23日 07:45 发表 58.20.174. 在 2013年7月1日 09:37 发表有配偶抽样这个概念么？有配额抽样 quota sampling,,还有分层抽样cluster sampling 回复评论 Dan (Talk \| 贡献) 在 2013年7月23日 13:33 发表 58.20.174. 在 2013年7月1日 09:37 发表有配偶抽样这个概念么？参见条目配额抽样回复评论 110.153.143. 在 2014年12月6日 22:16 发表 58.20.174. 在 2013年7月1日 09:37 发表有配偶抽样这个概念么？你猜回复评论首页文档百科课堂商学院资讯知识点国际MBA 商城企业服务问答首页专题管理营销经济金融人力资源咨询财务品牌证券物流贸易商学院法律人物分类索引百科VIP 百科VIP会员权益无广告阅读免验证复制开通/续费百科VIP 登录消息昵称未设置百科VIP 未开通) 收藏夹账号安全中心我的页面我的贡献我的讨论页我的设置退出登录打开APP 导航最新资讯最新评论最新推荐热门推荐编辑实验使用帮助创建条目随便看看本周推荐最多推荐智能家居网络推广价格标签包装短期投资 Facebook公司.gif 战略人力资源管理品牌设计减少风险理论私营企业贷款奶头乐理论蘑菇管理定律猴子管理法则情绪ABC理论垃圾人定律 100个最流行的管理词汇破窗效应 INFP SWOT分析模型 21天效应以上内容根据网友推荐自动排序生成官方社群企业管理者交流群加入添加微信，拉你入群创业者交流群加入添加微信，拉你入群 AIGC交流群加入添加微信，拉你入群市场营销人员交流群加入添加微信，拉你入群人力资源师交流群加入添加微信，拉你入群下载APP 告MBA智库百科用户的一封信亲爱的MBA智库百科用户：过去的17年，百科频道一直以免费公益的形式为大家提供知识服务，这是我们团队的荣幸和骄傲。然而，在目前越来越严峻的经营挑战下，单纯依靠不断增加广告位来维持网站运营支出，必然会越来越影响您的使用体验，这也与我们的初衷背道而驰。因此，经过审慎地考虑，我们决定推出VIP会员收费制度，以便为您提供更好的服务和更优质的内容。 MBA智库百科VIP会员（9.9元 / 年，点击开通），您的权益将包括： 1、无广告阅读； 2、免验证复制。当然，更重要的是长期以来您对百科频道的支持。诚邀您加入MBA智库百科VIP会员，共渡难关，共同见证彼此的成长和进步！ MBA智库百科项目组 2023年8月10日闽公网安备 35020302032707号添加收藏编辑收藏夹确定取消开通百科VIP免验证我已验证
8280	https://www.maths.gla.ac.uk/wws/cabripages/klein/affinesymmetry.html	klein view page the affine symmetry groups of conics We know that the euclidean symmetry group of a parabola has order two, while those of an ellipse and hyperbola have order four (and that the latter each contain a half-turn). We also know from the general theory of symmetry that if two figures are congruent in the geometry, then their groups are conjugate. Thus, here, it is enough to determine the affine symmetry groups of the three standard conics. Also from the general theory, we know that the affine symmetry group of a plane figure contains the euclidean symmetry group as a subgroup. As we shall see it is not always a proper subgroup. The details for each type of conic are rather different, so we have separate pages. the ellipse Eo the hyperbola Ho the parabola Po The proofs are not particularly geometric, and could well be ignored. In each case, we find that the symmetry group is infinite. The actual nature of the groups is not particularly important (except as a means of distinguishing the conics - see below). There is one feature which is, however, significant. It leads to the following general result. The Affine Transitivity Theorem If P and Q are points on a conic C, then there is an affine transformation which maps C to C and P to Q. Proof From the results for individual types of conic, there is an affine transformation u which maps C to the appropriate standard conic C 0, and P to a particular point R on the standard conic. Similarly, there is a transformation v mapping C to C 0, and Q to R. Then vou-1 maps C to C, and P to Q, as required. This has important consequences. In euclidean geometry, we can identify as special the point or points where the axis of symmetry through a focus meets the conic. For example, the vertex of a parabola. The theorem shows that this can be transformed to any point on the same conic. Thus, there is no affine characterization of the vertex. A similar argument shows that the major axis of an ellipse, and the transverse axis of a hyperbola are not affine concepts. It also confirms that the foci do not have affine descriptions, since they lie on these axes. We can also use the groups to show that no two of the three standard conics are affine congruent. First of all, we observe that the symmetry groups of affine congruent conics are conjugate. The symetry groups of ellipses and hyperbolas contain a half-turn, and a conjugate of a half-turn is a half- turn. The symmetry group of P 0 does not contain any half-turn (on geometrical grounds if nothing else, since a half-turn would make the parabola "point" the other way). Thus a parabola is not affine congruent to an ellipse or hyperbola. The symmetry group of an ellipse contains elements of all finite orders, but the group of a hyperbola has only elements of order two. Thus an ellipse cannot be affine congruent to a hyperbola. There are also applications to the problem of tangents in affine geoemtry. affine conics page
8281	https://www.vliz.be/vmdcdata/narms/narms.php/images/aphia.php?p=taxdetails&id=159798	North Atlantic Register of Marine Species (NARMS) Introduction Search taxa Taxon browser Partners ### NARMS taxon details ### Lutjanus jocu (Bloch & Schneider, 1801) AphiaID 159798(urn:lsid:marinespecies.org:taxname:159798) Classification 1. Biota 2. Animalia 3. Chordata 4. Vertebrata 5. Gnathostomata 6. Osteichthyes 7. Actinopterygii 8. Actinopteri 9. Teleostei 10. Eupercaria incertae sedis 11. Lutjanidae 12. Lutjanus 13. Lutjanus jocu Status accepted Rank Species Parent Lutjanus Bloch, 1790 Original name Anthias jocu Bloch & Schneider, 1801 Environment marine, brackish, fresh, ~~terrestrial~~ Original description (of)Bloch, M.E.; Schneider, J.G. (1801). M.E. Blochii, Systema Ichthyologiae iconibus cx illustratum. Post obitum auctoris opus inchoatum absolvit, correxit, interpolavit Jo. Gottlob Schneider, Saxo. Berolini. Sumtibus Auctoris Impressum et Bibliopolio Sanderiano Commissum. Pp i-lx + 1-584, Pls. 1-110., available online at page(s): 310 [details] Taxonomic citation Froese, R. and D. Pauly. Editors. (2021). FishBase. Lutjanus jocu (Bloch & Schneider, 1801). Accessed through: Costello, M.J.; Bouchet, P.; Boxshall, G.; Arvanitidis, C.; Appeltans, W. (2021) European Register of Marine Species at: on 2025-09-28 Regional species database citation NARMS (2025). Lutjanus jocu (Bloch & Schneider, 1801). Accessed at: on 2025-09-28 Taxonomic edit history Date action by 2005-05-30 09:27:08Z created Cuvelier, Daphne 2008-01-15 17:27:08Z changed Bailly, Nicolas [taxonomic tree] Sources (6) Documented distribution (1) Attributes (30) Vernaculars (1) Links (9) Images (2) original description(of)Bloch, M.E.; Schneider, J.G. (1801). M.E. Blochii, Systema Ichthyologiae iconibus cx illustratum. Post obitum auctoris opus inchoatum absolvit, correxit, interpolavit Jo. Gottlob Schneider, Saxo. Berolini. Sumtibus Auctoris Impressum et Bibliopolio Sanderiano Commissum. Pp i-lx + 1-584, Pls. 1-110., available online at page(s): 310 [details] basis of recordRobins, C. R.; Ray, G. C.; Douglass, J.; Freund, R. (1986). A field guide to Atlantic coast fishes of North America. Houghton Mifflin Co. Boston. 1-354.[details] additional sourceMcEachran, J. D. (2009). Fishes (Vertebrata: Pisces) of the Gulf of Mexico, Pp. 1223–1316 in: Felder, D.L. and D.K. Camp (eds.), Gulf of Mexico–Origins, Waters, and Biota. Biodiversity. Texas A&M Press, College Station, Texas.[details] additional sourceZenetos, A., S. Gofas, M. Verlaque, M. Cinar, J. Garcia Raso, C. Bianchi, C. Morri, E. Azzurro, M. Bilecenoglu, C. Froglia, I. Siokou, D. Violanti, A. Sfriso, G. San Martin, A. Giangrande, T. Katagan, E. Ballesteros, A. Ramos-Espla, F. Mastrototaro, O. Ocana, A. Zingone, M,. Gambi & N. Streftaris. (2010). Alien species in the Mediterranean Sea by 2010. A contribution to the application of European Union's Marine Strategy Framework Directive (MSFD). Part I. Spatial distribution. Mediterranean Marine Science. 11(2): 381-493., available online at additional sourceFroese, R. & D. Pauly (Editors). (2025). FishBase. World Wide Web electronic publication. version (06/2024)., available online at ecology sourceLooby, A.; Erbe, C.; Bravo, S.; Cox, K.; Davies, H. L.; Di Iorio, L.; Jézéquel, Y.; Juanes, F.; Martin, C. W.; Mooney, T. A.; Radford, C.; Reynolds, L. K.; Rice, A. N.; Riera, A.; Rountree, R.; Spriel, B.; Stanley, J.; Vela, S.; Parsons, M. J. G. (2023). Global inventory of species categorized by known underwater sonifery. Scientific Data. 10(1). (look up in IMIS), available online at +−EMODnet Bathymetry Countries - [x] OBIS occurrences (17866) - [x] Polygons PresentInaccurateIntroduced: alienContaining type locality Collapse all Expand all Collapse all Expand allDefinitions \| Language \| Name \| \| --- \| \| English \| dog snapper \| [details] \| To Barcode of Life (80 barcodes) To Biodiversity Heritage Library (30 publications) To European Nucleotide Archive, ENA (Lutjanus jocu) To FishBase To FishSounds To GenBank (84 nucleotides; 82 proteins) To Global Biotic Interactions (GloBI) To IUCN Red List (Data Deficient) To ITIS Unreviewed Lutjanus jocu Lutjanus jocu
8282	http://workshop.tcs.uj.edu.pl/mszana2015/gutowski/Kleitman.pdf	FAMILIES OF NON-DISJOINT SUBSETS 153 6. D.J. KLEITMAN, Families of Non-disjoint Subsets, Journal of Combinatorial Theory 1, (1966) 153-155. 7. F.P. RAMSAY, Collected papers, pp. 82-111; see also reference 2. 8. P. TURAN, On the Theory of Graphs, Colloq. Math. 3 (1954), 19-30. P. ERDOS, Panjab University Chandigarh, India Families of Non-disjoint Subsets Communicated by Paul Erd6s A family F of subsets of a finite set which contains no two disjoint subsets can contain at most half of all the subsets, since no subset and its complement can be in F. Moreover, if F is maximal with respect to this property (so that any larger family does not satisfy it) F must con- tain exactly half the subsets. (If F is maximal, B ~ F, C ~ B implies C ~ F; also, B ~ F implies that there is a C ~ F such that C c3 B = 99, i.e., C c/~ (/7 is the complement of B) hence /~ ~ F. Thus B or B must be in F.) In this paper we consider the analogous limitations on the number of subsets contained in k disjoint families F1 ..... F~ each of which contains no disjoint subsets. We prove the following result, which was conjectured by ErdSs (private communication). THEOREM. If El, ..., F k are families of subsets of an n element set such that Ain A~ ~/~ q~ if Ai, Aj ~ F z for 1 < l < k, then the number of ele- ments in the union of F1, ..., Fk is no greater than 2" -- 2~-k: k ] UFjI~2,,--2 ,,-k j=l (where IAl denotes the number of elements of A). Unlike the result for one family, the minimum number of subsets in the union of k disjoint F's which are maximal with respect to these Supported in part by National Science Foundation Grant GP 58. 154 KLEITMAN properties is not the maximum (2"-- 2"-#). We can in fact construct k disjoint families to none of which can a subset not already contained in one be added, whose union contains as few as 2 ,~-1 + ~] subsets. The union of F nondisjoint families each maximal with respect to our property can contain as few as 2 "-~ -- 1 subsets with 2 z > k for ,,) k 2(I ]+l , t The theorem follows directly from the following lemma. LEMMA: Let U be a family of subsets of an n element set such that A ~ U, B D A implies that B ~ U, andlet L be a family of subsets of the same set such that A ~ L, B c A implies that B ~ L. Then: I V n LJ 2 '~ G iV] iLr. PROOF: The result is trivial for n = 1. We suppose it to be true for n = k, and consider a pair of families U and L of a k + 1 element set S. If a is any element of S we can write, with U,, Ua, L~,, La disjoint: u=Gu u,~ L = L a ~J La where U~ and L~ consist of the subsets in U and L, respectively, which contain a. By the defining properties of U and L we have [U~] ~ [Ua], [L~] < [La] since, for example, the union of any element of Ua with {a} must be in U.. Moreover, Ua and L a are families of subsets of S- {a} with the same properties as U and L and so must satisfy I Ua ~ Zal 2 k ~ I U, zl IZal. (1) The families U~' and L~' which consist of the subsets obtained by removing a from the subsets in U. and L. are likewise families of sub- sets of S -- {a} with the properties of U and L, so that FAMILIES OF NON-DISJOINT SUBSETS 155 IU~c~ L~] 2 ~ = [U~'~ Lj] 2~< IU/I ILo'l = IGI IL~I. (2) Combining relations (1) and (2) we obtain 2k+xlU~ LI = 2~+a(IU.~ L~[ + lUan La[) < 2 (IUal [z,I + lull lZa[). (3) Since lull ~ lull and [L.I ~ lLal we have I GI [Lal +lual [Lal ~lU.I ILal + I e~l [L~I. (4) Combining relations (3) and (4) yields the desired result. Proof of our theorem follows from this lemma by induction on k. Let FI, ..., Fk. be families of non-disjoint subsets of S; extend each ar- bitrarily to a maximal family F'. Each maximal family F/(and any un- ion of such) contains any subset which contains any subset already in it. Let Ube ~-1 = = 2~-1 U~=I Fj. Let L be fie'. Since Fe' is maximal, ILl lfe'l . By hypothesis [U] ~ 2 ~ --2 n-e§ From our lemma, then, k I u Fj[ = IUu Eel < [F/I + IUn Zl < IFe'[ + 2-~IU[ ILl j=l 2" -- 2 "-e, which proves the theorem. If we choose Fj to be all subset of S which contain a and (j -- 1) other elements, and all (n - j + 1) element subsets of S which do not contain a, for j % k we find that k-2(n-l) 2,-1+ Y. J=o J subsets contain subsets that are in some F's. If the set of F's are then extended to be disjoint and maximal, they will contain this many subsets. DANIEL J. KLEITMAN, Brandeis University Waltham, Massachusetts Printed in Italy
8283	https://www.techtarget.com/whatis/definition/bitwise	WhatIs Home Programming Definition bitwise By Rahul Awati Published: Jul 06, 2022 What is bitwise? Bitwise is a level of operation that involves working with individual bits which are the smallest units of data in a computing system. Each bit has single binary value of 0 or 1. Most programming languages manipulate groups of 8, 16 or 32 bits. These bit multiples are known as bytes. The arithmetic logic unit (ALU) is a part of a computer's CPU. Inside the ALU, mathematical operations like addition, subtraction, multiplication and division are all done at bit level. For those operations, bitwise operators are used. Bitwise operations A bitwise operation operates on two-bit patterns of equal lengths by positionally matching their individual bits. For example, a logical AND (&) of each bit pair results in a 1 if both the first AND second bits are 1. If only one bit is a 1, the result is 0. AND can also be used to test individual bits in a bit string to see if they are 0 or 1. A logical OR (\|) operation functions differently from the AND operations. For each bit pair, the result is 1 if the first OR second bit is 1. If neither bit is 1, the result is 0. A logical XOR (~) of each bit pair results in a 1 if the two bits are different, and 0 if they are the same (both zeros or both ones). Logical NOT is represented as ^. Left shift (<<), right shift (>>) and zero-fill right shift (>>>) bitwise operators are also known as bit shift operators. Bitwise operators Bitwise operators are characters that represent actions (bitwise operations) to be performed on single bits. They operate at the binary level and perform operations on bit patterns that involve the manipulation of individual bits. Thus, unlike common logical operators like + or - which work with bytes or groups of bytes, bitwise operators can check each individual bit within a byte. The most common bitwise operators used in C/C++ are given in the table below. \| \| \| \| \| --- --- \| \| Operator \| Name \| Description \| Application \| \| & \| Bitwise AND \| Copies a bit to the result if it exists in both operands. The result is 1 only if both bits are 1. \| To set up a mask to check the values of specific bits \| \| \| \| Bitwise OR \| Copies a bit to the result if it exists in either operand. The result is 1 if either bit is 1. \| To add two numbers if there is no carry involved \| \| ^ \| Bitwise Exclusive OR (XOR) \| Copies a bit to the result if it exists in either operand. So, if one of the operands is TRUE, the result is TRUE. If neither operand is TRUE, the result is FALSE. \| To toggle bits or swap two variables without using a third temporary variable To find specific types of numbers (e.g., odd) in a series of numbers (e.g., all even) To find nonrepeating elements To detect if two integers have opposite signs \| \| ~ \| Bitwise NOT \| Also known as bitwise complement and bitwise inversion, it flips zeros into ones and ones into zeros. \| To flip or invert bits \| \| << \| Shift left \| The left operand value is shifted left by the number of bits specified by the right operand. \| To align bits \| \| >> \| Shift right \| The left operand value is shifted right by the number of bits specified by the right operand. \| To align bits \| Multiple bitwise operators are used in bit manipulation. These operations happen very fast and optimize system performance and time complexity. It's important to keep in mind that the left shift and right shift operators should not be used for negative numbers. Doing this can result in undefined behaviors in the programming language. Also, bitwise operators should not be used in place of logical operators because they work differently. Logical operators consider non-zero operands as 1 and their result is either 0 or 1. In contrast, bitwise operators return an integer value. The table below defines the JavaScript bitwise operators. \| \| \| \| \| --- --- \| \| Operator \| Name \| Type \| Action \| \| & \| Bitwise AND \| Binary \| If bits of both operands are ones, returns a one in each bit position \| \| \| \| Bitwise OR \| Binary \| If bits of either operand are ones, returns a one in a bit position \| \| ^ \| Bitwise XOR \| Binary \| If a single operand is a one, returns a one in a bit position \| \| ~ \| Bitwise NOT \| Unary \| Flips the bits in the operand \| \| << \| Left shift \| Binary \| Shifts first operand a number of bits to the left as specified in the second operand, shifting in zeros from the right \| \| >> \| Right shift \| Binary \| Shifts first operand a number of bits to the right as specified in the second operand, and discards displaced bits \| \| >>> \| Zero-fill right shift \| Binary \| Shifts first operand a number of bits to the right as specified in the second operand, discards displaced bits, and shifts in zeros from the left \| Applications of bitwise operations and operators There are many applications of bitwise operations and operators. For one, they are used in data compression where data is converted from one representation to another to reduce the amount of storage space required. Bitwise operations are also used in encryption algorithms to encrypt data and protect it from unauthorized use, manipulation or exfiltration. The following are some other common applications: low-level programming for device drivers, memory allocators and compression software; maintaining large integer sets for search and optimization; ability to store multiple Boolean flags on limited memory devices; embedded software in chips and microcontrollers; communications where individual header bits carry sensitive or important information; and converting text cases, such as uppercase to lowercase or lowercase to uppercase. Bitwise AND The bitwise AND operator produces an output of 1 if the corresponding bits of both the operands are 1. If not, the output is 0. Example 1: Bitwise AND operation of two one-bit operands. \| \| \| \| --- \| Left operand \| Right operand \| Result \| \| 0 \| 0 \| 0 \| \| 0 \| 1 \| 0 \| \| 1 \| 0 \| 0 \| \| 1 \| 1 \| 1 \| Example 2: Bitwise AND operation of two integers: 28 and 17; the & operator compares each binary digit of these integers. \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| Binary digits \| \| \| \| \| \| \| \| \| \| 28 \| 0 \| 0 \| 0 \| 1 \| 1 \| 1 \| 0 \| 0 \| \| 17 \| 0 \| 0 \| 0 \| 1 \| 0 \| 0 \| 0 \| 1 \| \| Are both digits 1? \| No \| No \| No \| Yes \| No \| No \| No \| No \| \| Bitwise AND output \| 0 \| 0 \| 0 \| 1 \| 0 \| 0 \| 0 \| 0 \| Thus: 28 & 17 (bitwise AND) = 00010000 (binary) = 16 (decimal). Bitwise OR The bitwise OR operator produces an output of 1 if either one of the corresponding bits is 1. Otherwise, the output is zero. Example 1: The bitwise OR operation of two one-bit operands. \| \| \| \| --- \| Left operand \| Right operand \| Result \| \| 0 \| 0 \| 0 \| \| 0 \| 1 \| 1 \| \| 1 \| 0 \| 1 \| \| 1 \| 1 \| 1 \| Example 2: Let's consider the previous example of two integers: 28 and 17. \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| Binary digits \| \| \| \| \| \| \| \| \| \| 28 \| 0 \| 0 \| 0 \| 1 \| 1 \| 1 \| 0 \| 0 \| \| 17 \| 0 \| 0 \| 0 \| 1 \| 0 \| 0 \| 0 \| 1 \| \| Is either digit 1? \| No \| No \| No \| Yes \| Yes \| Yes \| No \| Yes \| \| Bitwise OR output \| 0 \| 0 \| 0 \| 1 \| 1 \| 1 \| 0 \| 1 \| Thus: 28 \| 17 (bitwise OR) = 00011101 (binary) = 29 (decimal). Bitwise exclusive OR (XOR) The bitwise exclusive OR (XOR) operator returns 1 if the bits of both operands are opposite. Otherwise, it returns 0. Example 1: The bitwise XOR operation of two one-bit operands. \| \| \| \| --- \| Left operand \| Right operand \| Result \| \| 0 \| 0 \| 0 \| \| 0 \| 1 \| 1 \| \| 1 \| 0 \| 1 \| \| 1 \| 1 \| 0 \| Example 2: Let's see how bitwise XOR works for our two integers 28 and 17. \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| Binary digits \| \| \| \| \| \| \| \| \| \| 28 \| 0 \| 0 \| 0 \| 1 \| 1 \| 1 \| 0 \| 0 \| \| 17 \| 0 \| 0 \| 0 \| 1 \| 0 \| 0 \| 0 \| 1 \| \| Are the two digits opposite of each other? \| No \| No \| No \| No \| Yes \| Yes \| No \| Yes \| \| Bitwise XOR output \| 0 \| 0 \| 0 \| 0 \| 1 \| 1 \| 0 \| 1 \| Thus: 28 ^ 17 (bitwise XOR) = 00001101 (binary) = 13 (decimal). Bitwise NOT The bitwise NOT operator reverses the bits. Unlike other bitwise operators, it accepts only one operand. Example: Let's consider the bitwise NOT operation of the integer 28. \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| Binary digits \| \| \| \| \| \| \| \| \| \| 28 \| 0 \| 0 \| 0 \| 1 \| 1 \| 1 \| 0 \| 0 \| \| Bitwise NOT output \| 1 \| 1 \| 1 \| 0 \| 0 \| 0 \| 1 \| 1 \| Thus: ~28 (bitwise NOT) = 11100011 (binary) = 227 (decimal). Bitwise left shift The bitwise left shift operator shifts the bits left by the bits specified by the right operand. The positions vacated by the left shift operator are filled with 0. Example: Let's perform the bitwise left shift operation on the integer 6. Each bit will be shifted left by 1. 6 = 0110 6<<1 = 1100 (binary) = 12 (decimal) Bitwise right shift Like the left shift operator, the bitwise right shift operator shifts the bits right by the bits specified by the right operand. The positions vacated by the right shift operator are filled with 0. Example: Let's perform the right shift by two bits operations on the integer 8. Each bit will be shifted right by 2. 8 = 1000 8>>2 = 0010 (binary) = 2 (decimal) See also: bit stuffing, logic gate (AND, OR, XOR, NOT, NAND, NOR and XNOR), How many bytes for..., classical computing, Advanced Business Application Programming Continue Reading About bitwise Binary and hexadecimal numbers explained for developers TB vs. GB: Is a terabyte bigger than a gigabyte? 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8284	https://dhspriory.org/kenny/CDtexts/LatinGrammar.pdf	BASIC LATIN GRAMMAR by L. A. THOMPSON UNIVERSITY OF IBADAN DEPARTMENT OF CLASSICS OCCASIONAL PUBLICATION NO. 8 1982 This booklet is intended as an clementa-ry r e f e r e n c e and re \ri s i o n manual f o r rnsturc b e f ~ i q n e r s i n t h e i r first y e a r of L a t i n , t o be used along with m y A ~ L I n t r o d u c t i o n t o L a t i n S.yntax. Its c o n t c c t s a r e , accordingly, r e s t r i c t e d t o t h e b a s i c m a t e ~ i a l (accidence) necenyarv f o r a good foundation i n t h e lant5uage. ~n index hqs been included t o asrist t h e s t u d e n t i n f i n d i n e p a r t i c u l a r p o i n t s of' grammar ea-silv a d quickly. Considerable space h a s been devoted t o co~pound v e r b s s i n c e t h e study of su-ch v c r b s % i d s tile s t u d e n t i n t h e pracess of b u i l d i n g u p h i s o r h e r vocabulary. L. A . Thompson d a t . a't) l ~ t i v c accuaat i v e a d j e c t i v e d a t i v e feminine g e n i t i v e ninsculine n e u t e r n0mins.t i.ve p l u r a l singfilar i. ' C;UF 3 1. - ( r t ) -- Kcaninas of t h e various case-forms non. fcmina unm f i l i m habet ( s u b j e c t ) . The woman has one son. acc. feminam spectmus (object). W e a r e watch in,^ the woman. . ad v i l l m anbulabant. They were walking t o the house (motion towards). in villam ambul~verunt. They walked i n t o the house (motion i n t o ) . ten. filium spectamus. hie a r e watching. the woman ' s son. dxt. femkae pecunim dabant. They were giving money - t o the woman ( i n d i r e c t object). Thev were giving the woman money. feminae villam emit. i!e boucht a house f o r the wom"yz. 1 . servus i n vill.5 - laborat. The s l i v e is working i n the house. servus cum agricolz ambulat. The slave is walking with t h e farmer. servus e v.ill5 d i s c e s s i t . The slave departed from the house. - servus ab agrj colZ verberatus e s t . The slave w a s f l 0 3 ~ e d by the farmer. (b) Tht base of a noun T h i s is the p a r t of the noun t h a t i s comqon t o a l l its cnse-f o m s , except (sometimes i the nominative singular (and t h e accusative singular also, in nouns of n e u t e r gender): C . E . v i l l - f o r v i l l a , l i b r - f o r l i b e r , leon- f o r -- leo. ( c ) The vocative case T h i s has the s m e form as the nominative, except t h a t in Grcup 2 nolms ending i n -us and -ius t h e vocative singular ends i n -e and -i respectively, m d t h a t t h e vocative of r s u s is r.i. Kx.mples: sede m i f i l i ( S i t dokin, m y son), - L - Ls.ce, serve (shilt up, slave). Declc?nsion of nouns -- Grout~ 1 Group 2 6 1 1 - a s e n - u s puer- l i b r - ( l i b e r ) Group f l e k - ( l e a ) C ~ V - i s ZctZr-. (Fictor) -1-m. f o r t i s f o r tern f . f o r t i s fortcm n. f o r t e -. f o r t e . f o r t i s f o r t is f o r t i s f o r t . I f g r t i f o r t i - rn. a c e r 5crern f. a c r i s Zcrern n. a c r e a c r e - a c r i s a c r i s p l u r a l n. b o r ~ i f . bonae n. hona bonos bonas bona bonis b o n i s bonis bonis bonis bonis m. p u l c h r i f . pulchrae n. pulchra pu-lchros pulchras pulchra p u l c h r i s p u l c h r i s p u l c h r i s p u l c h r i s p u l c h r i s p u l c h r i s ingent ium ingent i a r n ingen t ium f Zlicilm fZliciiun fzlicium ingen t ibus ingent i b u s inkent i b u s ingen$ibus ingent tbus ingent ibus ingentes ingentes i n g e n t i a m. f o r t e s f o r t e s fortium f o r t i b u s f o r t i b u s f . f o r t e s f o r t e s f o r t i l l m f :)rt i b u s f o r t ihus n. f o r t i a f o r t i a fortium f o r t i b u s f o r t i b u s - - - m. a c r e s 5 c r e s Zcrium a c r i b u s acribus f . 'icres 5 c r e s 'icriwn Rcribus 5cri.bl\s n. Stria Bcria Zcrium Wcribus Acribus ?Tote - ?'or "pronominalff a d j e c t i v e s (i. e. words. which f m c t i o n both as addectives and as pronouns), see below, s e c t i o n 5 (pronouns). , , : , ) 1. o t e t h ~ f o l l ! ) w i n ~ ;;rlglisn examolfs: ( h ) Latir: exam!jles: ---- 110s i t ive --- con psra t i v e - -- -- -- s u p e r l s t i s lonqus l o n g i o r l ~ n g i s s i m u s f o r t i s fs3rtior f o r t i saimus audsx nud5cior a u d 5 c i , ~ s i m u s bonus malus parvus magnus multus m e i i o r p e i o r minor n a i o r ; > l G s o p t i g u s pessimus minimus maximus p l u r imus ( c ) - Declension of comparative -- f o r m s - -- s i n g u l a r nom. 2.c c . een. drtt. a.bl. m. minor minzrem minEris minori rnin6re f . mincr rninvrem m i n 5 r i s rninzrj min6re n. minus minus m i n o r i s minEri min?ire p l u r a l Notes: 1onp;ior (m. and f . ) , l 4 , n g j u s ( n . ) . L__ superlative forms a r e declined l i k c bonus -a -urn. 11ulchrE pulckrius pulcherrimE miserE miserills miserrime 2'ac i l e f a c i l i u s f a c i l i i m z ii) i r r e g u l a r coaparison -- .- -. .-. - p o s i t i v e bene (well) male (badly) oarim ( l i x t l e ) mamopere ( z r e a t l y ) rnultum (much) diE (long) saepe ( o f t e n ) comp?rqt tve melius peius minus magis plii s d i u t i u s sae pius (a) personal: ego, t C , nEs, vEs sirlgular nom. RCC. een. dat. c f ~ o m e me; mihi tii t E t u i t i b i y l u r a l nEs nEs nostr; ) nobTs nostrum) - - V ~ S vos vestrT ) vobi-s vestrum) ( b ) reflexive: sE sincu1p.r and p l u r a l nom. acc. . gen . d a t . SE (SESE) S U ~ s i b i s u p e r l a t i v e optimE pc s . ? i m E minime maxinE plurihum d i u t i s s i m z sse oissime a b l . ( c ) r e l z t i v e : q u i s i n g u l a r nom. RCC. - m.- q u i ' quem f . rluae sum n . q u o d quod m. quT quzs f . quae quas n. quae quae cuius c u i cuius c u i cuius c u i p l u r a l cluibus qu?irurn qulbus quorum quibus aua au.5 c~uihus quibus quibus (d) pronominal a d j e c t i v e s : is, h i c , i l l e , i s t e , idem, a l t e r , a l i u s , Ell-us. nom. ncc. en. dat . m. is eum e i u s e i f . e a e m eiLus eT: n. i d i d e i u s e i p l u r a l - - a . eT e o s eorum e i s - f . eae eEs ezrum e i s h i c - nom . acc. m. h i c hun c f. haec hanc n. hoc hoc m. hy hos f . hae h5s n. haec haec s i n g u l a r qen. dat. huius huic hi1 i u s hui-c huius hui.c p l u r a l hzrun h i s h5ru.m h?s hBrtun hTs a b l . e i s a b l . hFs h'is h'is i l l e - nom . abl. i l l u m i l l i u s i 1 lam i l l T u s i l l u d i l l i u s a. i l l e f . i l l a n. i l l u d i s t e - abl. isto i s t g ist5 nom . acc. gen isturn i s t i u s istam i s t i u s istud i s t i u s m. i s t e f . ista n. istud p l u r a l m. ist? f . i s t a e n. i s t a i s t U s istorum ist% ist5rm ista istErm istis istis ist7s istTs istis singul4.r ace, gen . nom. - m. idem f . eadem n. idem eundem eiusdem eandern eiusdem iderc eiusdern plural eEsdern eErundern easdem earundem eadem e5rundem eTsdem eTsdem eisdem eisdem eisden eisdem m . e?dem f . eaedem n. eadem a i t e r s i n g u l s r nom. RCC. gen d a t . a b l . m. a l t e r alterwn a l t c r i u s a l t e r i ~ l t e r o f . s l t e r a a l t e r m s l t e r i u s a1ter';l a l t e r a n . a l t e r u m a,lterum a l t e r i u s - - a l t e r i q l t e r 5 p l u r a l m. a l t e r T - a l t e r o s : a1terGrum a l t e r i s alteris f . a l t e r a e a l t e r % a l t e r ~ r u ~ r l a l t e r F s a l t e r i s n . a l t e r a a l t e r a nlterErum a l t e r i s a l t e r i s nom. n . d i u s f . ~ l i ~ n. a l i u d m. a l i i - f . s l i a e . n. alia RCC . en. alium a l i u s aliw a l i u s a l i u d a l i u s % o l u r 1 R ali7js sli5rurn s l i q s ali5rum alia . aliarum This form is r q r e l y found. The z e n i t i v e is, i n s t e a d , u s u a l l y conkeycd by t h e u s e of ? l i e n ("be1onp;inp; t o another") o r a l t e r i s ("of another", g e n i t i v e s i n q u l a r of q l t e r ) ; e . . 2 l i e n m pecuniam a b s t u l i 3 . kic carri-cd o f f t h e noncy of ranother. pecunim a l t e r i u s rnercqtori s a b s t u l i t . ;!e s t o l e t h e monev of a n o t k ~ e r nercriult. . . iillus nom . RCC. gen. d s t . abl. m. Gllus iillum iill'ius G l l Z . G l l O f. E l l a ii1la.m i i l l i u s iill'i ull'i n. Ellurn Zllum E l l i u s . . iilli 'ills p l u r a l Note: The following a l s o behave in t h e same way as - E l l u s o r a l t e r : neuter, n e u t r ~ , neutrum ( n e i t h e r ) niillus, niilla, nCillum (none) s75lus, sbla, sBlum (sole) tEtus, tEta, t0tu.m (whole) u t e r , u t r a , utrum (which of two) uterque, utraque, utrumque (each 'of two) anus, Una, Unum (one) (e) t h e pronoun ipse singular nom. acc. Ren dnt. abl. m. i p s e i sum ipsTus ips: ips0 . f . i p s a i p s m i psTus i p s i i ps'i n. ipsum i p s m iysTus ipsT i p s 5 p l u r a l m. i p s i i psB s i ps8rum ips& ipsTs f . ipsae ips% i psErurn i p s i s i p s i s n. i p s a i p s a ips5rum i p s h ips?s ( f ) the i n t e r r o ~ a t i v e pronoun, quis, quid ( ~ h O ? - - ~ h k t ? ) nom. -act, m. quis Quem ) f q U i s quam ) singular n. auid quid ) A l l other case-forms take the pattern of t h e r e l a t i v e pronoun (see above, section 5c). -- .- ( 3 ) the i n d e f i n i ~ e pronoun, quis (anyone), singular -- nom . . - acc. m. quis quem . . . . .- , . . . . . . . f. . , qua . . . . sum. , . . n. qui.1 ; , . quid . . . . . . , p l u r a l m. qui quoa f . 'qune . . quRs n. qune) quae) qua ) qua ) A l l o t h x case-forms follow the pattern of t h e r e l a t i v : pronoun (see above, section 5c). I I1 I1 I nr v VI VII VIII IX X - 'mus duo tres quattuor qukque sex s e ptem oct5 novem decem XI kdecim XI1 duodecim XI11 tredecim XIV qilattuordec i.n XV auindecim XVI sEdecim XVII septendecim XVIII duodEvTgintT X I X i3ndZvTgFnti .XX. viiz;i-nti triaintz 300 q u a d r Z ~ i n t 5 400 quinqu@int;i 500 s e x Q i n t ~ 600 s e p t u s i n t E 700 oct5gintZ 800 nEn%,int ;Et 900 cen turn 1,000 ducenti -ae -a 2,000 CCC cccc I> DC 1)CC DCCC DCCCC M m t r e c e n t i -Re -R qusdringentz -ae -n quyngenti -ae -a sescent; -Re -a s e p t i n g e n t i -ae -a o c t i n ~ e n t z -ae -a nGngent5 -ae -a m i l l e duo n I l i a 7. S C L E P S I O Y OY T H W hVMEKALS h u s , duo and tres iinus - nom . RCC. gen . d a t . abl. - - - - rn iinus m u m unius m i uno - - f . iina - iinm UnPus - Un'i -una n. unum Unwn iinTus Uni iin5 . , . -. . . , . - duo ; . . - . : $ duzk') .duorum -. m. dua a duobus duobus duo ) .-.. ... - .- f. duae dugs d u z r m d u ~ b u s - . dugbus n. duo . duo duErum duobus duEbus I . . m. - tres tres trium t r i b u s t r i h u s f . tres t rzs trim t r i b u s t r i b u s n. tria t r i a - t r i u m t r i b u s t r i b u s i) The Latin verb, when limited by - mood and oerson, t h a t is, when it r e l a t e s t o a subject which can & , g ~ _ r e s ed b ~ r the 3-+- Latin equivalents of English 2, w e , -u& - rs cdl:~-en a f i n i t e verb. The f i n i t e verb takes a v a r i e t y of formations r e l q t i n c to: " ( a ) the three pewsons ( l s t , 2nd and 7rd), singular and plural: ego, tu, i l l e / i l l a / i l l u d , nos, vos, i l l i / i l l a e / i l l a . ( b ) the s i x -- tenses: present, future,.imperfect, ~jerfect. future perfect, pluperfect. ( c ) t h e three moods: indicati;e,- subjunctive, imperative. ( d ) the two-voices: ---- active and passive. ~ l l other forms of the Lat-b- verb a r e cslled n o ~ - f i n i t e o r i n f i n i t e . These are: (a) the i r f i n i t i v e s (which- a r e verbal nouns) (b) the p;l,rticioles - (which a r e verbal adjectives) ( c ) the 50r1md and (verbal noun and verbal adjecsive, -- (d) the s ~ r ~ i n e (verbal noun). The forms of the f i n i t e verb a r e l i s t e d below accordin< t o the following pattern: singular 1st. 2nd, 3rd person; followed by plural lst, 2nd, 3rd person. The exmoles ilsed t o reoresent the 4 Groups o r Conjugations ~ f v e r b s are: p r o (1): I mepare; moneo ( 2 ) : I advi-se; mit.to ( 3 ) : I send; and audio (4): I hear. i i ) indicative ~ . c t i v e . '=-r-nt tense p a r 5 rn me5 mitt5 aud i ; p a r k m 3nZs m i t t i s audis pnrat rnf3ne t m i t t i t audit par%us r r 3nEmus mittimus aud 5~ us parat is rronztis m i t t itis nud? t is parant r-onent mittunt aud iunt f 11ture t e n s e - parEbo - - monebo m i t t a m audiarn parEbis monEbis m i t t F s audiEs pargb i t monEbit m i t t e t a u d i e t 11arEbimus monEbimus mi-ttEmus ~ . u d iemus :>arZbitis monebitis mittetis a u d i e t i s parAbunt monBbunt m i t t e n t a u d i e n t imperfect tense p a r t b m monzbcm mittebam aud i e b m parEbEis monEbEs mittEbEs a u d i e h ; ~ parahat - - moncbat - - - m i t t e b a t A U ~ i e b ~ t parabanus monebrmus m i t t e h k u s audi e b k u s p a r t b ~ t i s - rnonebgtis mittEb5tis a u d i e b ~ t i . ~ parzbant monzbant mittEbant audiEbant p e r f e c t tense --- - - - - oaravi nonu? m I s T aud ivi. paFavistY A m o n u i s t ~ mIsisti a u d j v i s t i p a r a v i t nolzuit m i s l t a u d r v i t psrgvimus monuimus misimus audrvimus parRvist! s mcnuisti-s misistis - - - - - a u d i v i s t i s p a r a v e r a t mcnuerunt miserunt aud'iverunt f u t u r e p e r f e c t ' t e n s e - - monuer: misero - ~ u d i v e r o monueris m i s e r i s aud r v e r is monuerit m i s e r i t a u d i v e r i t monuerimus m?serimus audTver-bus m o n u e r i t i s mTserit is a u d i v e r i t i s monuerint myserint a u d i v e r i n t p l u p e r f e c t tense - paraveram monueram - miseram a u d i v e r m ~ a r a v e r a s monueras miseras a u d i v e r z s p a r a v e r a t monuerat - m7serat - - a u d i v e r a t parEver%us monueramus misercamus audJverZrnus pnr6ver5tis monuerztis m i s e r ~ t i s a u d i v e r s t i s parsverant monuerant m'iserant s u d i v e r a n t i i i ) i n d i c a t i v e passive present tense paror moneor m i t t o r .~ar:ris rn0nErf.s m i t t e r i s p a r z t u r monetur r n i t t i t u r parsmur mo11Emur mittimur parc?mlni monemini mittimin? parantur monentur, m i t t u n t u r f u t u r e t e n s e par<t.or monebor mittar p a r g b e r i s rnoncberis m i t t z r i s pargb i t u r monebitur m i t t e t u r parsbimur monebirnur mitternur parabimini mcnebimini m i t t Ernini parabuntur monebuntur m i t t e n t u r i r n perf ec t tense parGbar - - monebar mittebnr parabar is monebgris m i t t e b g r i s pargbatur - - moneb-t - R - u r m i t t e b s t u r - - parabamur rn~nebamur - - mitteh,mur yarab;tminT nonebargin7 mitteb-minT parzbantur monebm t u r m i t t e b m t u r p e r f e c t tense parztus sum monitus sum missus sum pnratus es monitus e s missus e s garGtus e s t monitus e s t missus e s t ! ~ a , r g t i sumus moniti sumus m i s s i s w u s - - p a r a t i e s t i s monitz e s t i s missi - e s t i s p a r g t i s u n t moniti sunt m i s s i s u n t f u t u r e perfect --- tense - paratus e r o monitus e r o missus e r z paratus e r i s rnonitus e r i s m i s s t l s e r i s parEtua e r i t monitus e r j t r i s s u s e r i t p a r a t i erimu.7 moniti erimus mics? erjmus p a r z t i eriti-:: moniti e r i t i s iniss7 e r i t i s - - p a r a t i e r u n t moniti e r u n t n i s s i erllnt a~u-iior a u d i r f s a u d l t u r audTmur 9.ud T o inli audiuntur aud i=ir a u d i g r i s a i ~ d i 5 t u r audiErnur audigmini audientur .wdieba,r aud i e b a r is audiebztur - - audi-ebmur - - s u d . i e b m ~ d a.ud i e b a n t u r . ~ u d I t u s sum audytus e s a u d I t u s e s t audit: sumus audyty e s t i s - - .tudit;i sunt a u d i t u s e r 5 a u d i t u s e r i s a u d i t u s erit audTtT erimus a u d i t 7 e r j t i s aurlyti erunt pluperfect tense - parztus errmi nonitus e r m missus errm audTtus e r m uar5tus ergs monitus er5s missus er5s auditus er5s paratus e r a t - - nonitus e r a t missus e r a t audytus e r a t p a r a t i erKmus monit? e r h u s miss: er?mus audTtT e r h u s - - p s r a t i e r a t i s monitI erztis m i s s ? e r s t i s audit? e r z t i s uargt? erant moniti e r m t miss7 e r a n t a ~ i d i t ? erant i v ) subjunctive a c t i v e . -- ~ r e s e c t tense parern rnonem m i t t a m parEs moneiis m i t t % p ~ r e t raonerit mittat parernus rnone&us m i t t G u s p a r e t i s monc-5tis m i t t E t i s parent rno2emt mittmt imperfect tense - monerem - - rriitterern moneres mitteres moneret - - mitteret mmerernus - - m i t t ersrnus m o ~ e r e t i s m i t t e r s t i s nonerent mitterent perfect tense - paraverim rnonuerim miserim xudiverim paraveris monueris miseris audiveris p a r t v e r i t monuerit m-iserit audyverit parRverhus monuerbus mIserirnus a u d i v e r h u s parzveryt is monueritis myseritis sudTverit is pargverint monuerint miserint audiverint pluperfect tense I --. + - . - - paravissern monuissem rnlsissem a u d i v i s sem parzvisses monuissZs m i s i s s Z s nud'ivissGs p a m v i s s e t monuisset m i s i s s e t aud'ivisset pargviss6mus rnonuiss~rnus m i s i s s ~ m u s sudivissemus p a r g v i s s e t i s monuissetis misissetis a u d i v i s s e t i s parzvissent monuissent mrsissent a u d i v i s s e n t v) --- subjunctive passive -. - present tense parer monea,r - mittar a u d i a r p a r e r i s move?tris mittaris a u d i g r i s p a r e t u r - moneatur m i t t z t u r a u d i g t u r paremur monekur m i t t s u r aud i&ur paremin? m o n e h i n i rnittgmini a~zd i% i n i parent u r monemtur mitt-mtur aud i,m t u r imperfect tense - p a r z r e r - - monerer m i t ~ e r e r a u d l r e r - - y a r a r e r i s - - monereris m i t t e r e r i s audyreris parare t u r - - monErEtur - - m i t t e r g t u r a u d i r e t u r pararenur - - moneremur mitteremur audTrGmur pararemini monGremini m i t t eremini ~ u d i r e m i n i pararentur rnonerentur m i t t e r e n t u r a u d i r e n t u r p e r f e c t tense paratus sin monitus sim missus s i m a u d l t u s sim paratus sis monitus sis missus sis a u d i t u s s X s parGtus sit monitus sit - ,-missus sit a u d i t u s sit pargty s h u s monity a b u s missr sirnus a u d i t i s h u s parZtI sitis rnonity sTtis missi sitis a.~idyti - - sTtis parZtT s i n t moniti s i n t m i s s l s i n t auriiti s i n t p l u p e r f e c t tense par$tus csssn monitus essern missus essern audTtus easern - -parZtus e s s c s monitus essEs rnis,.us essEs a u d i t u s essEs parKtus e s s e t a o n i t u s e s s e t missus e s s e t s u d i t u s e s s e t v p r Z t i essGaus monf t i es~;Enus rnlssl esszmus audTtT essGmus parGtT essetis monit?' e s s e t i s m i s s i e s s E t i s a u d r t i easZtis p r g t T essent monity essent rnissi c s s e n t a u d i t y e s s c n t v i ) imperatives a c t i v e - - sinq. para mo~ie m i t t e aud? 1 . pargte rconete m i t t i t e a u d i t e p a s s i v e - - sing. parare monere m i t t e r e audire y l u r . parbin: monZmini mittiminT audTmini v i i ) p a r t i c i p l e s present ( a c t i v e ) - p a r s s monens mittens ~ u d i e n ~ f u t u r e ( a c t i v e ) - - - - - paraturus moniturus mis,surus auditurus p e r f e c t (passive) pargtu-s monitus missus auditus . . Yote: The case-forms of t h e present p a r t i c i p l e follow t h e same - .- p a t t e r n as ingens ( s e e above, s e c t i o n 2, pp. 3-4), except t h a t t h e a b l a t i v e s i n g u l a r of t h e p a r t i c i p l e u s u a l l y ends in -e (e.?. parante as o::~osed t o i n g e n t i ) . The f u t u r e m d perfect p a r t i c i p l e s behave l i k e bonus -a - -= (above, section 2, pg. 3-it). : : verb m d as a noun. I n its verbnl a s p e c t it is a c t i v e in me-2ninp. In its fi9u-i s s v c c t it bchsves l i k e n e u t c r noluzs of Group 2, e.?. hcllum. Thus z s scribe!-.di = t h e art of v r i t i r p , j u s t as 9.rs bell'! = t h e .-~rt of xrc. See fllrtfLr;r, L.4. Tcorzpson, An IrLtroduction t o Lritin Yynt,m, p. 5'1, s c c t i o n 97. ( b ) ~ , e r w d i v c ~ ~ a r a n d u r ; monendus mittendus zud ic~icius . , :.' o t e : - The gerundive is a v e r b a l a d j e c t i v e . That is t o snv, 5 t f'u_vlctions both 3s 2 verb s 2 : d ss an ad.iective. In i t s - v e r b a l a s p e c t it is ~ R S , S ~ V F : i r ? meaning. I n its a d j e c t i v a l a s p e c t lit bcrmves l i k e t h e n d j e c t t v e bonus ( ~ e e above, s e c t i o n 2 , po. 3-41. For i n s t a n c e : ci.bus e s t bonus: t h e food is s o q d . c i b u s e s t pvx.ndus: - t h e food is t o be yrer~nred. - 2 -. - x) supine - - - 11ilr~lt6 monitu missu a u d i t u The su-rjice is a v e r b ~ l l?ami, with characteristics of both 2 verb mc 2 noun. In its noun .tspect it has tw:, case-forms: a c c u s a t i v e (-urn) and a b l a t i v e (-u), t h u s behaving l i k e a (;~ou;o 4 noun (e.g. msnus). "I'ie a c c u s a t i v e is used i n expressions c o n v e y i ~ ~ "notion t o w a r d s " : thus, ~ m c i l l a ad culincm ~ i s s a e s t cen~un parqtun. The msid w ~ s s e n t t o thc kitchen t s food. The maid was s e n t t o the kitchen i n o r d e r t~ - ~~rei~c'tre food. The a b l a t i v e form i s used with a d j e c t i v e s i n c o ~ t e x t s Pike: difft'ricile d i c t u : diff'icu1.t i n t h e saying, d i f f i c u l t t o sav. i) indicative active ( a ) vol;, nolo, m Z l 5 ("wisht', "do - not wish", k l p r e ~ f -present tense - volo - nE15 - v i s nEn v i s vul t nzn v u l t vo lumus nSl~unus v u l t i s non v u l t i s volunt nolunt future tense v o l m n?il<an - - volEs noles volet nolet volEmus nol&~zs v o l e t i s n-dletis volent nolent imperfect tense volebam - - n ~ l ~ b m volebas nole b z s - - v~ lebat - - nolehat volebmus n5lZb%us - - - volebat is nolebatis - - voleb,a.nt nolebant volui v o l u i s t i v o l u i t voluimus voluist is voluerunt ~ ~ e r f c c t tense -- nzlui n z l u i s t i n o l u i t noluimus m a l a m - - males - - ninle tis malent imperfect tense e r a n - e r a s - poteram ibam - - poterss - i b a s poterst - i b a t - poter&us . - i b m u s poterEt is - i b 5 t i s poterant i b a n t p e r f e c t tense f u i uotui iT (ivT) f u i s t i p o t u i s t i i i s t F ( i v i s t i ) f u i t p o t u i t iit ( i v i t ) f uimus potuimus iimus (ivirnus) f u i s t i s p o t u i s t i s iistis ( T v i s t i s ) - - f u E r m t potuerunt G r u n t i iverunt) f u t u r e perfect t e n s e - f u e r o potuer; i e r o f u e r i s - potueris i e r i s f u e r i t potuerit i e r i t fuerimus potuerimus ierimus f u e r i t is p o t u e r i t i s i e r i t is f u e r i n t potuerint i e r i n t pluperfect tense fueram potueram i e r m f u e r n s potueras iergs f u e r a t potuerat i e r a t f u e r k u s potuer&us i e r G u s f u e r z t is potueratis i e r g t i s f ueran t potuerant i e r w t t ( c ) fer;, c ~ . i i Z , f i ? ~ bear)', "take -. --- I t , "becomeq1) fer; f e r s f e r t ferimus f e r t i s ferunt f e r m f e r e s f e r e t feremus f e r e t i s f e r e n t $resent tense ca.pi5 capis capit c8pircus cal).itis capiunt future tense capinrn cal)iEs capiet capizmus c s j ) i e t i s capient ~ ~ e r f e c t tense - - t u l i cepi t u l i s t 7 cEpisti t u l i t ~ E p i t tuliInus C ~ ~ ~ U S t u l i s t is c e p i s t i s - - tulzrunt c e ~ e r u n t fi.5 f is f i t firnus f i t i s f iunt factus sum f a c t u s e s factus e s t fnctz sumus f a c t 7 e s t i s f a c t 7 sunt t uler; t u l e r i s t u l e r i t tulerimus t u l e r i t i s t u l e r i n t future perfect tense - czyero factus ero cEperis fqctus e r i s cEperit factus e r i t cE perimus factT erimus ceperit is f n c t i e r i t i s ceperint . f a c t i erunt p l u ~ ~ e r f e c t tense tulercm cE !)erm f a c t u s e r m tuler5s cEper'or obvious reasons the verbs sum cm(i possum ( n I "I czn able") have no passive forms. The verb eo ("1 gon) has only 4 - passive forms : i t u r ( yreser,-t indic. passive, 3rd person sinz. ) - ib.%tur (imperf. indic. pass. , 7rd verson sin$?. ) itm e s t ( perf. indic. ()ass. , 7rd person sing. ) - i r i 7 r T s e n - t i n f i n i t i v e pxssive ) . - The first three of these forms a r e used is??ersonally. e.g. itum e s t ad tqbernam - ("It WRS gone t o the s h o l ~ " ~ ItG3ing was made t o the shop" "There was movement t o the sh,?pv). - - The i n f i n i t i v e - iri is used only in the f o r m ~ t i ~ m or the future i n f i n i t i v e -- ----- pns,sive of other verbs ( e . ~ ~ par5tum irT =Ifto.be going t~ be prep=tredi', "to be about t o be !,re !lared"). - ( b ) the verbs f c r o cr?.i)io oresent tense -- feror f!tfe.rls f e r r i z w fert11.r f . ; . r i..mur f c r i m i n l f e r ~ ~ n t u r future tense - f e r a r capiar fererir, - c a p i E r i s ferEtur capiE t u r f erEmur ca piEmur f e rzmini capiEmin1 f erentur capientur imperfect tense f erEbar - - cai)iEbELr f erebaris capiebaris f erFbRtur c ~ p i E b Z t ~ - - f erEb%ur capiehamur . f erEbkin3 capiEh&in: ferEbmtur capiEbantur uerfect tense lattls s u m C R ~ ~ U S SUIU G t u s e s captus e s lEtus e s t captus e s t l Z t i sumus c a p t i sumus l & t Y e s t i s c a p t i e s t i s lEtY sunt captP sunt future perfect tense l a t u s erE captus ero l'itus e r i s csptus e r i s l z t u s e r i t captus e r i t l ~ t I erirnus c a o t i erimus lati c r i t i s capt'i e r i t i s l Z t i erunt capt'i erunt pluperfect t e ~ l a t u s er,m cnptus eram 1-2tus ergs caytus ergs l z t u s e r a t captus e r s t l R t r errnus c a p t i e r k u s l5ti e r s t i s captr e r z t i s ~ l t ? erant caDtF erant iii) sub,junctive a c t i v e (a) vol5, n515, m 5 l E present tense ve 1 i n nolim v e l i s n o l i s v e l i t n o l i t v e l h u s nolTmus v e l i t i s n o l i t i s v e l i n t n o l i n t imperfect tegsz vellem nEllem v e l l e s n o l l e s v e l l e t n z l l e t veliemus noll&us v e l l e t i s n o l l e t i s v e l l e n t n o l l e n t j ~ e r f e c t tense voluerim nEluerim v o l u e r i t n o l u e r i t - voluerimus nolucriinus v o l u e r i t i s n ; ? l u e r i t i s v o l u e r i n t n o l u e r i n t s l u p e r f e c t t e n s e . voluissem noluissem V ~ ~ U S S S Z ~ nEluissZs voluis.set n5luisse.t voluissEmus n ~ l u i s s ~ m u s v o l u i s e E t i s n 5 l u i s s E t i s v o l u i s s e n t n o l u i s s e n t m s l im - - m a l i s (b) sum, i~ossun and @ ~ j r e s e n t tense s im - pass i r n - sis p o s s i s sit, - p o s s i t simus possimus sitis possf t is s iiz t ~ ~ o s s i n t imperfect t e n s e essem - possem - e s s e s posses e s s e t - p o s s e t - essemus possemus e s s e t i s ~ O S S E ~ ~ S e s s e n t possent p e r f e c t t e n s e fuerim potuerim f u e r l s p o t u e r i s f u e r i t u o t u e r i t f u e r i n t p o t u e r i n t p l u p e r f e c t t e n s e fuissem potuissem - f u i s s z s p o t u i s s e s f u i s s e t i o t u i s s e t fuissEmus ~ O ~ U ~ S S G ~ U S f u i s s E t i s p o t u i s s e t i s f u i s s e n t p o t u i s s e n t e am - e a s e a t eKrous e6tis ermt - irem i r e s - i r e t - - iremus - - i r e t is - i r e n t i e r i m - i e r i s i e r i t - ierimus i e r l t i s i e r i n t - issem - isses -29-present t ens2 f e r m c a p i m - f ergs cayias f e r s t capiat - f e r h u s capimus f ergt is capigt is f errant capiant imperfect tense ferrem ca~erem - f e r r e s c a ~ e r e s f e r r e t caperet f erremus caperemus f e r r e t is caperet is f errent caperent tulerim t u l e r i s perfect tense ceperim - ceperis tu1eri.t ceperit - - tulerimus ce perimus t u l e r i t is cepe& - is tul-erint seperint f 7;s f i a t f ierem f i-eres f i e r e t f ieremus r'ierztis f i-erent faccus s i m factus sis factus sit pluperfect tense - tulissem cepissem factus essem - tulissEs C E ~ ~ S S Z S factus esses t u l i s s e t cepisset factus esset tulissEmus cepissEmus '-fact? ~ S S E ~ U S t u l i s s e t i s CZ pisset is fact7 e s s e t i s tulissent cepissent fact? essent i v ) subjunctive passive (2) On the verbs volo, nolo, m%lo, f i o , sum, ))ossum in rejiamur - f e r z r i s f er%nini c a p i a r i s c a p i s i n i ferEtur ferantur c ~ . p i < t u r capiantur imperfect tense - f e r r e r f erremur caperer caperemur - - f e r r e r i s ferremin? capereris caperemini f e r r e t u r f e r r e n t u r caperetur cxperent~w .- . p e r f e c t tense 1 s t ~ ~ sim l5tu.s s Y s 1 a - t ~ ~ sit l5tI simus - - la.ti s i t i s - - l a t i s i n t p l u !)erf ect tens-,e l z t u s essem l z t u s essEs l i t u s e s s e t - - lati essemus - - l a t i e s s e t i s 1st: essent captus s b captus sis cagtus sit captF sTmus c a p t i sitris c a p t i s i n t captus essem caytus esses captus e s s e t cnptT essernus c a u t i e s s e t i s c a p t l essent -31 f u t u r e passive -- ( f e r o ) ( c a p i o ) 7 - - - lztum ~ r l caotun iri p e r f e c t a c t i v e - (volo) ( n o l o ) (rnglo) (sum) ( possum ) v o l u i s s e n 3 l u i s s e rnnluisse f u i s s e potuisse . - (4 ( f e r o ) ( c a p i o ) - i s s e ) t u l i s s e c e p i s s e -8- l v i s s e ) y e r f e c t passive ( f e r o ) 1-t a u s e s s e v i i i ) gerund and ,.;erurAdive (a) gerund ( e o ) ( f e r o ) eulclu~ f erendum ( b ) gerundive ( c a ~ i o ) c a p t u s e s s e ( 4 i t u m it: ( f e r o ) ( c a ~ i o ) 1Fitu.m . . capturn l ~ t i i captii 10. ~E201'?J~J5' AF'D S~~l4I-D?;2O~JEP! T' VEHES -. ( a ) deponent a a d g r e d i o r , adgredi, adqressus sum: approach a a d i p i s c o r , a d i ~ ? s c i , adeptus s u m : o b t a i n - 3 amplector, cuni)lecti, camplexus suv: embrace z cornitor, comit%ri, cornit5tus sum: accorrlpmy - - - - - s c o n o r , c o n a r i , conatus sum: tr;lr g c o n f i t e o r , c6nfitEr-I, cgnfessus sun: confess 3 conspicor, c;inspic;ir?, c5nsp.i-cztus sum: - catch sigh g c b c t o r , cEnctqrT, cEnct5tus sum: d e l v r , h e s i t a t e f d e t e s t o r , d e t e s t g r i , 6EtestGtus sum: curse - ~ E s r e d i o r , eared?, egressus sum: KO out g e x p e r i o r , e x p e r i r i , expertus sm: t r y out z f a t e o r , f a t e r i , f a s s u s sum: confess n f u n g o r , funzi, f h c t u s sum: perform - f h o r t o r , h o r t s r 7 , h o r t s t u s sum: er.courace r i n z r e d i o r , i n ~ r z d i , i n g r e s s u s sum: g o i n l - - i ~ r ; l s c o r , rrfisci, ir5tu.s s u m : te angry iiilgbor, l?tbi, l%psus sum: s l i p nr locluor, loquT, l o c u t u s sum: speak - 2 meditor, meditgr'i, medit,;it~ls sum: contem~1.ite r minor, mingri, n i n z t u s sum: t h r e a t e n T - f miror, m l r a r i , mirRtus sum: wonder n t - - ii misereor, m i s e r e r i , m i s e r i t u s sun: p i t v - i - moror, m o r t r i , rnoratus sum: delclv w ~ i t / - 9 a mori-or, nor?, mortuus sum: - d l e ncscor, n z s c i , nztus SWL: be born - - g n i t o r , n i t i , n?xus sum: p r e s s , l e a n g o r i - o r , o r 3 r i 9 o r t u s sum: a r i s e , r i s e g p a t i ~ ~ , pc?tT, passus sun".uffer, allow p o t i o r , p o t i r i , ootXtus sum: f ~ e t po~.c,ession of E precor, p r e c g r i , ~ r e c g t u s sum: j r a y ji p r o f i c i s c o r , :)roficTsci, profectus sum: s e t out - - g p r o f i t e o r , o r o f i t e r i , professus sum: t ~ r o f e s s , d e c i - a a orogredior, ~lrzgredy, ! ) r ~ ~ r e s s u s sum: L O f orwsrd - a queror, q u e r i , questus sum: complain - - recordor, recorcia,ri-, rec:?rdRtus SUIT: remember / 3 r e g r e d i o r , re,a-ed?, regre9sus sum: r e t u r n - -r $ r e o r , rerl-, r a t u s sun: think -r fi sequor, sequl, secutus sum: follow testo or, test.&?, tests-hus sum: call t o witness B u t o r , Gti, - Gsus - SUT: % +Jvn,?or, v a g a r i , - vaggtus sum: w a d e r s v e r e o r , verer?, v e r i t u s sum: f e a r / & v i d e o r , v i d z r i , vysus sum: seem ( b ) semi-de ponent audeo, audere, ausus sum: d a r e 7- - c&fydo, confxdere, c s n ~ l s u s sum: t r u s t d i f f i d o , d i f f l d e r e , d i f f l s u s sum: d i s t ~ u s t fldo, fTdere, fysus sum: t r u s t , confide i n gaudez, gaudere, gav7sus sum: r e j o i c e , be h a p p y sole:, s o l e r e , s o l i t u s sum: he accustomed -36-( a ) Compound verb.: crc. f ~ - ~ . c d i , ~ ri;c a d ? i t i o n of. n p r e f i x - t o a verb: e . ' ~ , a t ) - ! . ~ t ~ ~ . ( I I Y : ~ U S C ~ ~ ) is a compound of - u t o r ("uselt ). ~ h c foliowin,:- prc f i x e s a r e used i n t h e -- formation o f compow~cl v c r t s : - a, ab (away ) ad ( t o , towards, at hand ) corn- (g: t o q e t h e r , completely) circum (around) dE (down) d i s - ( a p a r t , at d i f f e r e n t p o i n t s ) - e , ex ( o u t , o u t r i g h t ) ob ( i n t r ; ~ w2y, by t h e way, kl t h e d i r e c t i o n o f ) pc-r ( tllroucl.,, thoroughly) prae i i n f r o n t , ,ahead) p r a e t e r (?.side, ~ a s t ) pro ( f o m v d s , i n favour) re- (back, a g ~ t i n ; o r E- RS in "unpackt1) - sc- ( s i n e : wit!lout, a s i d e , a p a r t 1 in ( i n , i n t o , upon, snb (under, u p from belov, a g a i n s t ) n e a r t o ) i n t e r (amone, between, s u p e r ( left over, remainin?) away ) nec ( n 3 t ) tr,ms ( a c r 9 s s , over, ri:;ht throuzh) ( b ) The s p e l l i n g of t h e prefix i s sometime^ influenced b~7 t h e l e t t e r o r l e t t e r s i r ~ ~ m c d i a t e l y following it i n t h e compound v e r b : ad-c ad-f ad-~i ad-1 ad-p ad-sc ad-sp corn- con-d corn-f corn-1 often becones o f t e n becomes o f t e n becomes o f t e n becomes o f t e n becomes o f t e n becomes o f t e n becomes o f t e n becowes becomes becomes becomes -2CC aff a l l a p 1) a s c a s p CO- cond conf c o l l o r con1 c o m - ex - Lex r e - sub - dis-f becomes dis-1 o f t e n becomes e-f becomes in-1 in-m in- p in-r o f t e n becor7:es of ten becomes o f t e n becomes often becomes nec-1 becomes ob-c ob-f ob-m ob-p o f t e n becomes of ten becomes often becomes o f t e n becomes pro b e f o r e a vowel becomes re-d r e - t sub-c sub-f sub-m becories becomes becomes b e c o ~ e s of t e n becomes Some coaTon compound v e r b s d i f f d i l e f f ill imm i m p irr OCC o f f o m 0 1 ) 1 1 prod redd re t t - - - RRerC ( "do " , "drive " ) , e,-;i, actlrs - - cogo (co-aga), cogere, c o e ~ i , coRctus compel exig:, exigere, e x z ~ i , exzctus d r i v e o u t , e x a c t , f i n i s h perag3, peragere, peregi, perzctus do com!~le t c l y , finish. r e d i z z , r e d i z e r e , redEgT, redTc t u s b-ring bqck, rcducc subigo, subigere, subggi, subactus b r i n g under, subdue accido, a c c i d e r e , ncr:idi f s l l r;por,, hapr~cn dGcid5, dccidere, aEcidi - f s l l down - -- i n c i d z , i ~ l c i d e r e , i n c i d i , i n c m u s -9 f q l l i n t o -- ocel;.r occido, occidere, o c c i d i , ~ C C E S ~ I S f ~ l l down, & tile :.jay - - csedere (''ki.11" ) c e c i d i , cacsus - - incido, inciriere, i n c i d i , i n c i s u s c u t in> occido, occiderc, occirii, ~ c c i s l i s c u t ~ i 3 ' : . ~ , k i l l sub - subdo, -dere, -didi, -dYfus put undez, subject tr,uls trZdi5, -dere, -didi, -ditus h,md over, h,wd dorm - - adduce, -ducere, -dluri, -ductus l e a d t o , induce condCc5, -ducere, -dE.xi, -ductus rent h i r e - 9 - educe, -dikere, -dExi, -ductus lead out induc5, -ducere, -du'xi, -ductus bring in, introduce - - - produce, -ducere, -dGxi, -ductus brinz f o r t h , prolong - ernere ("buvtf) E m i . emotus ad - adino, -imere, -emi,-emptus take t o oneself, take away i n t e r imterimo, - h e r e , -'EmX, -Emptus take away9 destrox -- - -r - perimo, -imere, -cml-, -enptus d e s t r z esse ("be"), f u i , futurus ab - absum, -esse, g f u i bt? away, he absent ad - adsum, -esse, adfui be present, be a t hand dE - dzsum, - dEesse, ciEfiii - he lackinq, f a i l i n - i n s u m , inesse, i:lfui -9 be i n be contained i n i n t e r intersun, -essi., -fui be ~mong, be between, d i f f e r jlrae praesum --esse, -fui be in charge of pro prosum, prodesse, profuT be u s e f u l super supersum, -esse, -fui be l e f t over, survive ad - c o m i de - E - i n - i n t e r per LK%L i > r D L-A r e - sub - facere ("do", ltwtke") fec9, factus a f f i c i o , -ficere, -feci, -fectus a f f e c t , t r e a t c k f i c i o , -ficere, -fEci, -fectus f i n i s h , wenr out - f. deficio, -ficere, -fee!., -fectus f a i l , defect e f f i c i E , -ficere, -fEci, -fectus work o u . , bring about i n f i c i b , -ficere, -fEcf, -fectus i n f e c t , s t a i n i n t e r f i c i o , -ficere, -fEci, -fectus - k i l l perficio, -ficere, -fZci, -fectus complete praeficio, -ficere, -f'EcF, - f e c ~ u s j ~ u t i n charge o?' p r o f i c i z , -ficere, -feci, -fectus make oroaress - - r e f i c i o , -ficere, -feel, -fectus r c y a i r , rcf'resk --- s u f f i c i o , -fi-cere, -f;ci, -fectus subs%i-tute, suffi-ce . . f e r r e (Tfbring" "bear") t u l i , 1 s t ~ ~ atl -. - adfero, -fen-&,' 2ttulT9 allRtus brin,? t o ab - nufero, -ferre, s h s t u l i , ahlRt1;s take away, --. s t e a l c om - confero, -ferre, --tuli, coll.Etus coll-ect, compare dF - dEferro, -ferre, -tulT, -1Ztus b r i c ~ , reuort -40-d i f f e r o , - f e r r e , d i s t u l i , d i l s t u s d i f f e r e f f e r o , - f e r r e , e x t u l i , e l a t u s c a r r y out, lift u~ infer:, - f e r r s , - t u l i , -1iitus bring-&, bring upon offer5, -ferre, obtulT, obl5tus b r i n z up t o , o f f e r - ~ ~ e r f e r c , - f e r r e , - t u l i , -l?tus -- c a r r y through, endure praef er;, - f e r r e , -tulT, -1:tus c a r r y i n f r o n t , display prof ero, - f e r r e , - t u l i , -1Atus b r i n g forward refer6, - f e r r e , r e t t u l i , r e l z t u s b r i n g back, r e f e r t r a n s f e r 5 , -ferre, - t u l i , -1Ztus tr,msfey, -- t r a n s l a t e -- g r a d i o r ("step") g r a d i , g r e s s u s sum aggredior , -gredF, -gressus sun aporoach, a t t a c k c o n ~ r e d i o r , -gredy, -qressus s u m come together digredior, -gredi, -gressus s u m go a p a r t , d i g r e s s - egredior, -gredi, -gressus s u m go-out ingredior, -gredI, -gressus swn go i n t o , advance przgredior? -gredi, -gressus s u m go forward, advrvlce regredior, +red:, -gressus swn go back, r e t u r n trRnsgredior, -gredi, -gressus sum go a c r o s s , t r ~ s g r e s s -hibui, -hibitus - hold towards, ap!)lg debere, dEbui, dEbitus pwe, ought uT, -hitus hold i n f r o n t , o f f e r -hibuy, -hibitus hold o f f , prevent - - i s c e r e ( "throwl1) i e c i , i a c t u s - - abicio, - Icere, ..-iecl, -iectns &row away adicio, - i c e r e , - i e c i , -iectus throw t o , coniciz, -icere, -iBci -iectus throw t o g e t h e r , con;iscture .- .- A. - dEici5, - i c e r e , -1ec1, -iectus throw down, tirow ozf .- - d i s i c i o , - i c e r e , -1ec1, -iEci, -?-ectus tl7row a s ~ m d e r - eicii5, -i.cere, -iEci, -iectus G r o w out, r e j e c t T n i c i 5 , -icere, -iEcT, -iectus -. i.2.ject, -. -- i n f u s e obicio, --icere, - i @ c i , -iectus op90se9 present - . . proici:, -icere, - i e c l , - - -icctus xhrow f o r t h , project. subiciz, -icere, - i e c l , -iectus - _ I _ - tl-row under, subject, r @ i c i o , --icere, -iEci, -iect~zs g r o w b ~ c k , r e j e c t - - traiciz, -icere, - i e c l , -iectus throw a c r o s s , t r a n s f e r go away, deuart go t o , a~~!)roach, circum corn - ex - i n - ob - ner I j)-rae t e r r e - su-b - t rcms --- c o m -- di5 - d i s - 5 . - - i n t e r nec - Eer , i n - i n t e r oh - per p r a e p a e t e r r e - trans acl -- corn -- . i l l - -r - 1 ?yo around, make a t o u r of $0 together, -9 meet assemble E O out, depart ,go into, begin, devise gc t o meet, meet degth, die -.- p ~ s s a w ~ y , perish .- - i T uass b77, i,ynore C -- qo ba.ck, r e t u r n uxde?-zo q ~ o a c h , s e c r e t l x - - . - 9 - xo across tr8nsgrt:ss ----' - legerz (lfi)ickw. "read") legi, l e c t u s -- - -- -- - - adleg;, -l.,.ir,ere,-legi, -:'..~ctus n p ~ o i n t t o colliqE, - l i g e r e , -lEgi, -1Gctus colle . , --isms send o u t , emit - -F inmitt;, -mittere, -I.I:SX, -C:~SSUS send i n t o - -- . . intermit t z ? -mitter;., - m l s j . , -missus i n t e r m i t , 16,ave o f f .:- ornitto, -nit,ttlre, -Tinl,?ji, - m i s s ~ - ~ omit, disregard -.- - permitt;;, -mittere, - m l s l , --missus l e t throuqh, ? ) e m i t - - ,uraernit;G, -olit;tere, - m l s l , --missus send i n advance praeter.nitt5 ( e t c . ) Ag>--;=llss 3 2 1 , ~ e r l o o k , emir prijrnitt 5 ( e t c . ) put T ' G : - , ~ ~ ~ ~ , ! L I ; ? : remitt5 ( e t , ~ . ) -1 ' - - a m i t . --- :---'': 7 2 t 2 -. i trmsmittG ( , e t c . ) :sc?;c; ~.c,ross, ;end t l ~ r o u q h , t r a n s m i t -- , .~ --.- ----. .- -- - a,q~Esc;;, -n5scere. -n~-ri., - --notus r r c o m i z e cognEscZ, -nEscerz, -novl, -n;-;sus lc-trn, come t o know - .- - - - ignoscl~, -noscere, -nevi, -cotus overlo jk, for;?i..rc: e nr. - dF' - - O'J - gab - Xl-a - nri -- ( % 9 : ; - d i r: -- - c - I)rn -. sub - - , , " place") posux, p 0 ~ i t - u ~ - - - - - ri!)por.o, -pa:-crc, -);~ZUI, - p o s i t u s I I U ~ n e a r , a p o l y t o n';tc.!,5r,3 ( c t c , ) :1.2t t . c . f ~ r c , , j1refcr cor:,5rJ8 ( c t c . ) 1 . to;-i t!;i:r, nrryriqo, compose - - - dk!)o:~o ( c t c . ) ,jd: ac\;n, d k o o s i t , l a y n s i d c - - - di.spo:~o { c t c . ) JjJt i?-. d i f ~ c r c n t !)laces, 9rq;ulize - - C X ~ G R G (c+.c. ) I> i+, out,, u ~ l o q d , e x l ~ l ~ ~ i r i - - i v p r , 3 ( c ~ c . ) r,,~: LTICJT,, i r t ~ o s c _L_r ir.tcr!Zr.?3 - - (ctc.)?bi.+l:r.cn, lfiterpose o!,;?on--, ( c t c . ) s c t ~ - c : i ~ s t , OI:I~OSC, I I U ~ up a g a i n s t psstli3:.3 (ctc.7 pst b~r.ir!d, V R ~ U C 1 e ~ S ornap5nij (etc.) p ~ t In c h ~ r q e pr3pbr.5 (ctc.) p u t forward, propose, d i s p l a x - - rcpono (etc.) put hack, lay back, r e p o s e - - - sc.po:-:o !ctc. ) s e t a r ) & - prernere("~)ress'~ ) p r e s s i , pressus - - c c ~ l r i n l o , - orimc-re, - o r c s s i , -9rcssus p r e s s together-, crush -r dzprir.:, -!)ri~crc-, - p r e s s l , -pressus Lress down, depress 0p.u-r.5, -primere, - p r c ~ s T , -pressus c r u s h , oppress . . sup!)rirnE, - ~ r i m e r c , - p r c s s I , -pressus suppress ra!)Crc ( "snatch") r x u u i , r ~ ! > t u s - ~ b r i !)iz, -ri ,)ere, -ri pui , -rcptus s n a t c h away, t e a r away r r I , r i C , r , -r t a k e hold o f , c a t c h c o r r i p i 3 , - r i m r e , -rip~lT, -reptue s e i ~ e r i -r r , r , - i t u s t e n r a p a r t , plunder - -C i i I -, I , - r t u s n a t c h awaz, t e q r away - prori~;i,~, -rice=, - r i a ~ i , - r ? p t u s d r R R r t h aurripi:, - r i p e r e , -ri!;ui, -rcptus s n n t c h , s a l - - - r c i y ( " t ! ir(?ct," " r u l e " ) r e x i , r e c t u s .- - - - 3 corriEE, -ri$-crc, - r e x l , -r?ctits c o r r e c t , reform -- t - - di.s dTrizz, -ri,ycrc, - r c x ~ , - r c c t u s d i r e c t , zrra.nEe -- - - - - - - C ! - e r i g o , - r i e e r e , -rexr, -rcct.us ,?uli)c o u t , - e r e c t perg6 ,--<ere, perrEx5, i)crr3ctus - j)roceer\ 11ro porrigo, p o r r i c c r c , !)orrcx?, p o r r c c t n s s t r e t c h out slib surgE (sub-rico) , stir,-xrc, SUI-rcxi, s u r r e c t u s r i s e , g e t u!) -- -specere ("look") ad - a s p i c i 6 , -sr)icerc, spcxi, spc-ctus look at - c o m -- - -? conspicio,' -spicere, -spexi, -spect;us catch sight; - of . i n - ~ n s p i c i o , -spicere, -spexi, -soectus inspect, look at r e - resoicio, -spicere, -spexi, -spectus look back - - s t ~ r e ( " s t . m d t ' ) stet;, s t a t u s adsto, -stare, -stit? stand nearby, a s s i s t c5~1st5, -stZre, -stiti 3 apree, consist o f , cost . - clrcumstZ,----st?&, - s t e t l s t m d around, encircle dlst5,-stare, -stj-ti st&?d a p a r t , be dist,mt inst:, -stZre, -stitI press upon ~bst:, -stZre, - s t i k i stcznd i n t h e way o f , r e s i s t -- perstG, - s t a r e , - ~ t i t : p e r s i s t , persevere prsesto, -stgre, - s t i t T s t m d out, excel, -show - r e s t o , -stare, -stiti stand back, remain tendere ( " s t r e t c h " ) tetendi, tentus -. i n t e r ---- - - attendo, contendi5, -- distend5 extend3 ( intend: ( osten_dE. ( -tendere, -tendi, -tentus attend -tendere, -tend?, -tentus hurry ( e t c . ) s t r e t c h asunder, distend e t c . ) s t r e t c h out, extend etc. ) s t r a i n towarcls, cor~centrate e t c . ) s t r e t c h up t o , show -. . tencre ( "hold1') tenu?, . - t e n t u s abstineo, -stinFre,-stinui, -stcntus , abstain, keep off a t t i n e a , -ti-nEre, -ti-nui, -tentus pertain t o , concern eontineg, -tinEre, -tinuT, -tentus contsin dgtine5, -tinEre, -tinui, -tentus detain obtine5, - -tinEre, -tinui, -tentus hold p r t i n e o , -tinGre, -tinuI, -tentus gertqin t o sustineo, -tinEre, -tinuI, -tentus s u s t a i n , withstrand retifie5, -tinere, -tinu:, -tentus hold back, r e t a i n - advenio, -venire, -venT, -ventus come t o , a r r i v e circmvenio ( e t c . ) s u r r o u ~ d , e n c i r c l e convenio ( e t c . ) come together, be sgrsed uvon Eveniz ( e t c . ) come out, r e s u l t , hsppzn invenio ( e t c . ) come upon, - f i n d , discover intervenio ( c t c ' m e t w e e f i , i!lterruot; perveni5 ( e t c . ) cone through, a r r i v e a t , reach qu7libet (qui-libet): anyone you please. A coinpound of quy. and the verb (impersonal) l i b e t . The neuter is e i t h e r xuidlibet o r quodlibet. quisque (quis-que): each one. E'eminine quaeque, neuter quidque o r quodque. auTdcm (quI-dcm): a c e r t a i n one. This is a l s o used as an adjective, meaning "a certain": e.g. v i r quidam ( 9 c e r t a i n man); but quidcam vEnit (a c e r t a i n male person cram?). Feminine quaedam, neuter quiddm o r quoddm ( c ) r c l a t i v e adverbs ubi: where where, which way unde: from where quo: where t o quorncdo: how, i n what way , - 4 - 7 - c t A r 35 ;trii?;SscT 35 X','<T:.,CT I V Y 3 3-5 d : o 1 3-4 rir C ~ O K G ~ orr o f C F :;p,rg,- t i v c c 5 !cl c c z ~ ~ ~ r i z o n of, 5 ( b ) . 6 r c z u l s r coxparison 6 ( b ) i r r e z u i n r corrruarison 7 ~ ~ . i c z u s . 10 nlius 1 0 n l t c r 10 xriplect? 35 r-::?Trc 35 ( b ) r..:dire 14-21 cnsc-f omis, alervl inzs o f , 1 ( 4 - - c x - ; i t s r l 35 (:!;: i'i)'!?. 1) ~ 2 ~ ! ~ i ~ [ ! ~ . ' S 45 (b)-46 C: :. ,PI,; 1 . I ) VEjiIt'; 37-45 -?;r;erc (-icere) 3 H -cqdcre ( - c i d e r e ) 38 -cacricrc(-ciderc) 3 s -csclcrc(-ci!~cro) 32 -cEdere 39 -cilrrcrc 39 -dzrc (-dcre) 39-40 -diiccre 4C -eacre ( - h e r e ) 40 -CSSC 40 -facere (-f iccre) 4 0 - f e r r e 40-41 -,?radi (-pxd?) 41 -1:abEre (-hihEre) 4 1 -iacere (-icerc:) 41 - i r e 41-42 -legere (-ligero ) ' 42 -mittere 42 -nEscere 42 -ponere 43 -\Iremere (=prinere) 43 -rapere ( - r i v e r e ) 43 -rcgerc ( - r j ~ e r c ) 37 -snecere (-spicere) 47-44 -stgre 44 -tct~flere 44 -tenere (-tjnEre) 44 -venire L4-45 ci5l?'irT 35 c6nfidere 36 ( b ) confitErT 35 cEns ljicEri 35 ciinct'iri 35 DEr'()!~'Ej~: T V E u 5 ' 35- 36 dFtestEri 35 difiydere 36 ( h ) ~ U O 13 ecqui 4Lj(b) ecquis 45(h) - cgredi 35 - - e x p e r i r i 35 f a t e r i 35 f i d e r e 76(b) f i n i t e verb 14 fLlJv5 35 ~ a u d e r e 36 ( b ) ~Eh17j~ll ( r e ~ u l q r ) 20-21 capiC) 34 eo -34 I e r 5 34 G :KlLrJ1)IVE: ( r e g u l a r ) 21 capiz 34 f e r B 34 h i c 8 hortTrT 35 3 R t u r 26 (a) ?den C3 ille 9 W ) 3 3 o\A'i'iV?!;S ( r e , ~ u l s r ) 19 c a : ~ i 5 32 dice 32 - - duco 32 eB 32 f a c i 5 32 f e r o 32 no15 32 sum 32 PYDICAT iYE ACT TI);; ( r e g u l a r ) 14-15 oigl5 22-23 nolo 22-23 V O ~ G 22-23 e 5 23(b)-24 possum 23(b )-24 sum 23(b)-24 capio 25-26 f e r E 25-26 f i o 25-26 'C:DICATIV-:; Y A T S I n ! ( r e g u l a r ) 16-17 c a v i o 26(b)-27 f e r o 26(b)-27 J1I!'< IT IV3S ( re,;?ular ) 20 c a p i 5 - 33-34 eo 33-34 f e r z 33-34 mRli5 33-34 nolo 33-34 possum 33-34 sum 33-34 vol0 33-34 -is$; r,iEM -jr , rll - . i , A - ~ V X A D J ~ C T I V ~ ; 4 5 ( a ) i ! ~ s e l l ( e ) - .- 2.r-:scX 35 7 - r 2 6 ( a ) is 8 i s t e 9 ituixi e s t 26(q) i-tur 26 ( 3 ) - - morari 35 - mori 35 n e u t e r li nit: 35 n o n - f i n i t e verb l d hOUTCS ( d e c l e n s i o n ) 1 ( d ) - 3 base of t h e noun l ( b ) n E l l u s 1 1 I : I ~ , Y ; I S 12-13 - p a r a r c 14-21 YAKITI(:IYLES ( r e ~ u l ' a r ) 19 c a p i o 32-33 e z 32-33 f e r 5 32-33 no15 32-33 V O L ~ 32-37 p a t i 3s - - p o t i r i 35 r ~ r e c a r i 35 p r e f i x e s 37 (a)-38 prof i c i s c i - - 35 o r o f l t e r i 36 arGgrecii 36 =HOTV()!~I~;AJ, LIDJ;;CTIVES 4 ( n o t e ) , 7 , d ( d ) -11, 4S(b) yHOjOT,TiYs 7-12, 45-46 personal 7 ( n ) r e f lc:xivc: 7 ( b ) relp.t4ve 7 i c ) i s d e f i n i t e 1 2 ( ~ ; ) , 45-46 i n t e r r 2 ~ 2 t i v e 1 2 ( f ) , 45 GerT - 36 q u i ( i n t e r r o , ? ~ t l v e add. ) 45 ( a ) ( r ~ ~ l n t i v e !~ronoun) (3( c j q u i c q l i d 4 5 ( b ) q u i d w 4b!b) q u r l i b e t 46(k) q u i s ( fl:t?:rr?,y t i V c ~:ronoun) -- -7 -- ---- qsi squm 45(b) qu5.sque 4 6 ( h 1 c;:;Yv'?s c'j ( b ) - < > < - , 46 ( t o ) q!.?-3ds 46(c) , V m 36(h) or.crr'i 36 - . 1 11 ;., J-,,. c.+-Iv=; . ~ C Y I V Y ( ro,mlsr verbs 1 17-18 crt;;iE '30 c.0 29 +,'i ri 33 XI; 30 - - 1 28 : i ? i 1 ? 28 r,Dncum 29 cru: 23 voli5 2d S ' 2ATlll:C?IvX P A ~ ; S I V ~ :rc<rllsr v e r b s ) lii c r l p i z 31 fcrc! 31 - - vj-deri 36 vocative l ( c )
8285	https://www.edaboard.com/threads/physical-meaning-of-dot-product-and-cross-product.232308/	[SOLVED] - physical meaning of dot product and cross product \| Forum for Electronics Continue to Site [x] Search titles and first posts only [x] Search titles only By: SearchAdvanced search… ForumsNew postsSearch forums Best Answers What's newFeatured contentNew postsNew mediaNew media commentsNew resourcesNew profile postsLatest activity MediaNew mediaNew commentsSearch media Help Rules Log inRegister What's newSearch Search [x] Search titles and first posts only [x] Search titles only By: Search Advanced search… New posts Search forums Menu Log in Register Navigation Install the app Install How to install the app on iOS Follow along with the video below to see how to install our site as a web app on your home screen. Note: This feature may not be available in some browsers. More options Contact us Close Menu Welcome to EDAboard.com Welcome to our site! 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RegisterLog in Science Mathematics and Physics [SOLVED]physical meaning of dot product and cross product Thread startereng_boody Start dateNov 27, 2011 Nov 27, 2011 #1 E eng_boody Member level 3 Joined Nov 27, 2011 Messages 57 Helped 1 Reputation 2Reaction score 1 Trophy points 1,288 Activity points1,754 Please i'd like to know the physical meaning of dot product an how it was originated and why it return a scalar quantity as well as the physical meaning of the vector product and why the result vector is perpendicular to both the multiplicated vectors and how it was originated.....note that i don,t need the formula..i know it A.B=abcos(theta)....A X B=absin(theta)n.....i just want the physical meaning .... Reply Nov 29, 2011 #2 T tallface65 Full Member level 6 Joined Dec 8, 2009 Messages 343 Helped 190 Reputation 394Reaction score 173 Trophy points 1,323 Locationland of the freeActivity points3,407 A dot product conceptually is the projection that one vector has over another. This is why it is a scalar, it only tells the length of the projection. Another way of thinking is that it tells one how 'parallel' the two vectors are to one another. The larger a dot product between two unit vectors, the smaller the angle is between them in a given plane or more obtuse if the angle is greater than 90 degrees (the more parallel they are). A cross product results in a vector that has a direction that is perpendicular to both vectors and a magnitude that is equal to the parallelogram with side lengths equal to the magnitudes of the two vectors and a skew equal to the angle between the vectors. Another way of thinking is that it, conversely to the dot product, tells one how 'perpendicular' the two vectors are. The larger the magnitude of the cross product between two unit vectors, the larger the angle between the vectors (up to 90 degrees) in a given plane (the more perpendicular they are). Have Fun Reply Reactions:DEV123, gauravgp9 and eng_boody E eng_boody Points: 2 Helpful Answer Positive Rating Nov 30, 2011 G gauravgp9 Points: 2 Helpful Answer Positive Rating Sep 18, 2012 D DEV123 Points: 2 Helpful Answer Positive Rating Apr 5, 2013 Nov 30, 2011 #3 E eng_boody Member level 3 Joined Nov 27, 2011 Messages 57 Helped 1 Reputation 2Reaction score 1 Trophy points 1,288 Activity points1,754 Thnx tallface65 .. u said that dot product represents the scalar value of the projection of a vector over a another vector....my question here is if i have vector (A) and vector (B) of magnitudes (a) and (b) respectively...and the angle between them is theta ....so the projection(i am not sure is it the component??) of vector A in direction of B is acos(theta)...not abcos(theta)...why did we multiplied by b (generally b is not necessary to be 1) then .....is there a difference between the projection and the component of a vector (of course in some direction)......or what is the reason of multiplying by b . thnx Reply Nov 30, 2011 #4 blooz Advanced Member level 2 Joined Dec 29, 2010 Messages 560 Helped 121 Reputation 242Reaction score 116 Trophy points 1,343 LocationIndiaActivity points4,985 Projection of B in the direction of A is [B\cos(\theta)] Projection of A in the direction of B is [A\cos(\theta)] how much these two vectors point in the same direction. ? it's calculated as a product of these two as [AB\cos(\theta)] hp://behindtheguesses.blogspot.com/2009/04/dot-and-cross-products.html Last edited: Nov 30, 2011 Reply Reactions:eng_boody E eng_boody Points: 2 Helpful Answer Positive Rating Nov 30, 2011 Nov 30, 2011 #5 E eng_boody Member level 3 Joined Nov 27, 2011 Messages 57 Helped 1 Reputation 2Reaction score 1 Trophy points 1,288 Activity points1,754 thank u so much .. 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8286	https://www.linkedin.com/pulse/p-values-we-dont-trust-what-how-bayesians-do-them-pereira-md-phd	In p-values we (don’t) trust: what p-values are and how Bayesians do them Agree & Join LinkedIn By clicking Continue to join or sign in, you agree to LinkedIn’s User Agreement, Privacy Policy, and Cookie Policy. Skip to main contentLinkedIn Top Content People Learning Jobs Games Join nowSign in In p-values we (don’t) trust: what p-values are and how Bayesians do them Report this article Miguel Pereira MD PhD Miguel Pereira MD PhD Executive Director and Biostatistician at Cogitars UK Published Mar 21, 2022 + Follow What is the p-value? In the classical, hypothesis-based statistics, the p-value is the probability that a statistical summary of the data (like the average height of the population, for example) is at least as extreme than the one observed in the sample, provided that the null hypothesis is true. Although short, this definition is very profound. Let’s explore in more detail: First, we already know that the p-value is a probability. So, it will be a value that expresses the chances of something. Starting with the end of the definition, the p-value assumes that the null hypothesis, H0, is true. This corresponds to a conditional statement that tells us that the p-value is not only a probability but it is, in fact, a conditional probability and it can be represented in this form: p-value = P(something... \| H0 is true). The ‘something’ in the probability pertains to the part of the definition about ‘observing a value at least as extreme as’. What does this mean? Going back to the example of the average height, imagine an experiment with the populations of city A and city B. We obtain a sample of people from each city, record their heights and want to know if they are different or, in other words, if the difference in average heights is zero. The set of hypotheses we test is the following: H0: average height in city A - average height in city B = 0, vs. H1: average height in city A - average height in city B is different from 0 Imagine we obtain a sample of 100 citizens from city A and 100 citizens from city B. In the sample from city A, the average height is 1.80m with a standard deviation of 0.25. In the sample from city B, the average height is 1.70m with a standard deviation of 0.20. The difference in the average height between the two cities is 0.1m and we want to test if, given the sample data, this result is against the hypothesis that there is no difference in average height between the two cities. Using a standard T-test calculator (there are plenty online) and entering the numbers, we obtain a p-value of 0.0036. This is a ‘statistically significant’ result if we assume the arbitrary significance level of 0.05, for example. But what does the probability of 0.0036 means? First, the p-value is not the probability that the two cities are equal as this would be the probability that H0 is true, and we already covered in a previous post that this is something we can’t obtain from hypotheses tests. If we take the definition above and the frequentist notion of repeating an experiment multiple times, the p-value is the probability that we obtain a difference in heights greater than 0.10m (the difference we observed) assuming that H0 is true, that is, assuming there is no difference between the two cities. Since this probability is low, we reject H0. This is the reasoning behind the p-value which is a little mind-bending. The decision to reject H0 is usually predefined before observing the data by setting a significance level. This is where the 0.05 threshold for significance comes from. However, 0.05 means that we will be wrong 1 out of 20 times, that is, if the p-value is very close to (0.04999, for example) there is a 1 in 20 chance that we are wrongfully rejecting the null hypothesis. On the other hand, if the p-value is just above 0.05, there is also a 1 in 20 chance that we are wrongfully failing to reject the null hypothesis. This is very important to bear in mind whenever we are setting our significance level since sometimes. Sometimes, we cannot afford this level of probability of making a mistake and should set a lower threshold for the p-value. On the contrary, there are problem where we can allow for that a higher chance of making a mistake, and it should be acceptable to set a higher p-value to accept significance. Always setting the p-value to 0.05 is analogous to using the same single ingredient to make all out meals. It is not ideal nor healthy. Recommended by LinkedIn Vibe check: Uncertainty persists Matt Carmichael 4 months ago Struggling with Fetching and Analyzing Canadian Data:… Murtaza Haider 2 months ago Edward Tufte’s Slopegraphs and political fortunes in… Murtaza Haider 7 years ago The Bayesian p-value After understanding the intricacies of the frequentist p-value, the Bayesian p-value will be very easy to understand. The reason for this is very simple: the Bayesian p-value is a generalisation of the frequentist p-value and assumes that we are assessing H0 in a frequentist way. In Bayesian lingo, the p-value is called posterior predictive p-value and it is, in fact, very similar to the frequentist p-value. When we carry out a Bayesian analysis, we obtain the posterior distribution of what we want to measure. For the example of the height difference between cities A and B, we obtain a posterior distribution for the difference of the distributions of heights in each of the two cities and can calculate a posterior predictive distribution. This distribution is a generalisation of the posterior distribution that lets us simulate unobserved values. In other words, it generalises the information we obtained from the data and the prior to a distribution from which we can draw observations similarly to obtaining data from the actual population. Under the definition of posterior predictive distribution, the p-value is the probability of drawing observations the posterior predictive distribution that are at least extreme than the data. This definition of Bayesian p-value is a great deal like the frequentist p-value. The main differences lie on the use of the posterior predictive distribution and on considering a single drawing/observation from this distribution while the frequentists assume the collection of a new sample whose summary statistic (e.g., average height) is ‘at least as extreme as’. The somewhat confusing concept of a Bayesian p-value is an attempt to adapt the Bayesian approach to the frequentist perspective. It’s important to keep in mind that the Bayesian analysis does not depend on p-values. Bayesians usually use other means to assess hypotheses. One of those is called Bayes factors. Simply put, Bayes factors are a ratio of the likelihood of H0 over H1. This is an interesting way to assess our hypotheses. If the ratio is 1, then H0 and H1 are equally likely. If the Bayes factor is > 1, then we have more evidence to support H0 and, if it is < 1, we have more evidence to support H1. It is possible to set predefined thresholds for Bayes factors that help up make decisions on which hypothesis is more coherent with the data. Since the Bayesian analysis lets us calculate the probability that H0 is true, it is often found that this probabilistic view is, in some cases, more intuitive than setting a p-value or Bayes factor threshold to make a black and white decision on whether H0 can be rejected or not. In the end, this depends on what we are trying to study, on our sample size and on the aim of the analysis. P-values are a fascinating topic. I find interesting that they are widely used and, at the same time, they are somewhat confusing for me. Knowing what p-values are is extremely important when we are interpreting and presenting our results. I don’t think the Bayesian p-value is a better alternative to the frequentist p-value, but Bayesians have come up with p-value-free alternatives for statistical inference that may be more adequate in some studies. Next week’s topic will be about advantages and disadvantages of Bayesian statistics. The posts so far have been admittedly more in favour of the Bayesian approach but it’s time to discuss its disadvantages and the challenges in carrying out Bayesian analyses. Like Like Celebrate Support Love Insightful Funny Comment Copy LinkedIn Facebook Twitter Share 718 Comments Kassaye Bekele MPH Biostatistics fellow \| Data analysis intern\| pediatrics and child health care\| Data analyst \| junior researcher \| Aspiring Digital health Care Service \| 11mo Report this comment Nice explanation! LikeReply1 Reaction 2 Reactions Geoffrey Johnson 3y Report this comment Here are some related posts you may find interesting. Dichotomizing the p-value: Confidence vs credible intervals: The real crisis in statistics: LikeReply 1 Reaction Eni Hasa 3y Report this comment Dear Miguel, thanks for sharing the article. today there are tons of courses in datascience & statistics, new degrees in university but in most of those, they dont really study the stats behind. I know graduated statistician with the new master degrees in which they don’t even know what it is the bayesian approach. However, in business application sometimes could be also good to have quick&cheap stats with standardized tools which does not require programming language. I think that the choice of the frequentist p-value va bayesian approach should be evaluated based on benefits/risk of application/decision outcome. LikeReply1 Reaction 2 Reactions David Manteigas Principal Biostatistician at ICON 3y Report this comment I would disagree - p-values are the less known statistical concept (especially because most people can complete a master degree in statistics without knowing the difference between Fisher and Neyman Pearson frameworks) LikeReply1 Reaction 2 Reactions See more comments To view or add a comment, sign in More articles by Miguel Pereira MD PhD Dose Optimisation and Selection in clinical trials - Part 2: Data Analysis Aug 20, 2025 Dose Optimisation and Selection in clinical trials - Part 2: Data Analysis In part 1, we discussed elements of trial design to address project Optimus. 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8287	https://www.youtube.com/watch?v=cy4PiY5ERTI	Time Value of Money - Present Value vs Future Value The Organic Chemistry Tutor 9880000 subscribers 8382 likes Description 814664 views Posted: 5 Jun 2020 This finance video tutorial provides a basic introduction into the time value of money. It explains how to calculate the present value as well as the future value of money. How To Calculate The Future Value of an Annuity Due: How To Calculate The Monthly Mortgage Payment: 157 comments Transcript: Intro in this video we're going to talk about how to solve a few basic future value and present value problems the present value of money represents the value of money today right now in the present whereas the future value represents the value of that money in the future so let's start with the first part of this problem what is the future value of ten thousand dollars twenty years from now given an annual interest rate of six percent Present Value so right now in the present we have ten thousand dollars so that's the present value how much will this amount of money be worth 20 years from now at an interest rate of six percent so we're looking for the future value the formula that we could use to calculate the future value from the present value is this formula fv is equal to pv times 1 plus r raised to the n so the present value is 10 000. the interest rate is six percent or point zero six n is the number of time periods in this case since the interest is credited on an annual basis and it's going to be the number of years which is 20 years so it's going to be 10 000 times 1.06 raised to the 20th power now let me go ahead and plug this in so the future value of ten thousand dollars twenty years from now i'm gonna write it here is it's worth thirty two thousand seventy one dollars and thirty five cents and so this really helps to illustrate the time value of money ten thousand dollars today is a it's worth a lot more than 10 000 in the future for instance you can buy more of a dollar now than what you could buy 20 years you know later for example now you can buy a small bag of chips for about a dollar 20 years ago you can buy a bag of chips for 25 cents so 20 years in the past one dollar can buy you four bags of chips in the present a dollar can buy you one bag of chips so the purchasing power of money goes down as time moves forward you can buy a lot more stuff with a dollar today than what you will be able to buy with a dollar in the future so now let's move on to the second part of the problem what is the present value of a hundred thousand dollars ten years from now given the same annual interest rate of six percent so in the second part of the problem we're given the future value Future Value which is a hundred thousand dollars and we want to calculate how much that is worth ten years in the past or rather in the present so to speak and the interest rate is the same so the formula that we need to use we need to rearrange it a little the present value is equal to the future value divided by one plus r raised to the n so the future value is a hundred thousand r is still point zero six so one plus point 0.06 that's 1.06 and n is 10. so 100 000 divided by 1.06 raised to the 10th power gives us a present value of 55 839 dollars and 48 cents so let's say that inflation is six percent the value of all goods increases by an average of six percent fifty five thousand dollars eight hundred fifty five thousand eight hundred thirty nine dollars and forty eight cents has the same purchasing power as a hundred thousand dollars ten years later so as you can see this really illustrates the time value of money and that is the same amount of money is worth more now than it will be worth in the future so if you have a choice of selecting a thousand dollars now versus a thousand dollars ten years from now the thousand dollars in the present has more purchasing power than the thousand dollars in the future so the basic idea behind the time value of money is that money is worth more now than it is in the future
8288	https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/10-8-work-and-power-for-rotational-motion/	10 Fixed-Axis Rotation 10.8 Work and Power for Rotational Motion Learning Objectives By the end of this section, you will be able to: Use the work-energy theorem to analyze rotation to find the work done on a system when it is rotated about a fixed axis for a finite angular displacement Solve for the angular velocity of a rotating rigid body using the work-energy theorem Find the power delivered to a rotating rigid body given the applied torque and angular velocity Summarize the rotational variables and equations and relate them to their translational counterparts Thus far in the chapter, we have extensively addressed kinematics and dynamics for rotating rigid bodies around a fixed axis. In this final section, we define work and power within the context of rotation about a fixed axis, which has applications to both physics and engineering. The discussion of work and power makes our treatment of rotational motion almost complete, with the exception of rolling motion and angular momentum, which are discussed in Angular Momentum. We begin this section with a treatment of the work-energy theorem for rotation. Work for Rotational Motion Now that we have determined how to calculate kinetic energy for rotating rigid bodies, we can proceed with a discussion of the work done on a rigid body rotating about a fixed axis. (Figure) shows a rigid body that has rotated through an angle $$ d\theta $$ from A to B while under the influence of a force $$ \overset{\to }{F}$$. The external force $$ \overset{\to }{F} $$ is applied to point P, whose position is $$ \overset{\to }{r}$$, and the rigid body is constrained to rotate about a fixed axis that is perpendicular to the page and passes through O. The rotational axis is fixed, so the vector $$ \overset{\to }{r} $$ moves in a circle of radius r, and the vector $$ d\overset{\to }{s} $$ is perpendicular to $$ \overset{\to }{r}.$$ Figure 10.39 A rigid body rotates through an angle $$ d\theta $$ from A to B by the action of an external force $$ \overset{\to }{F} $$ applied to point P. From (Figure), we have $$\overset{\to }{s}=\overset{\to }{\theta }\,×\,\overset{\to }{r}.$$ $$d\overset{\to }{s}=d(\overset{\to }{\theta }\,×\,\overset{\to }{r})=d\overset{\to }{\theta }\,×\,\overset{\to }{r}+d\overset{\to }{r}\,×\,\overset{\to }{\theta }=d\overset{\to }{\theta }\,×\,\overset{\to }{r}.$$ Note that $$ d\overset{\to }{r} $$ is zero because $$ \overset{\to }{r} $$ is fixed on the rigid body from the origin O to point P. Using the definition of work, we obtain $$W=\int \sum \overset{\to }{F}·d\overset{\to }{s}=\int \sum \overset{\to }{F}·(d\overset{\to }{\theta }\,×\,\overset{\to }{r})=\int d\overset{\to }{\theta }·(\overset{\to }{r}\,×\,\sum \overset{\to }{F})$$ where we used the identity $$ \overset{\to }{a}·(\overset{\to }{b}\,×\,\overset{\to }{c})=\overset{\to }{b}·(\overset{\to }{c}\,×\,\overset{\to }{a})$$. Noting that $$ (\overset{\to }{r}\,×\,\sum \overset{\to }{F})=\sum \overset{\to }{\tau }$$, we arrive at the expression for the rotational work done on a rigid body: $$W=\int \sum \overset{\to }{\tau }·d\overset{\to }{\theta }.$$ The total work done on a rigid body is the sum of the torques integrated over the angle through which the body rotates. The incremental work is $$dW=(\sum _{i}{\tau }_{i})d\theta $$ where we have taken the dot product in (Figure), leaving only torques along the axis of rotation. In a rigid body, all particles rotate through the same angle; thus the work of every external force is equal to the torque times the common incremental angle $$ d\theta $$. The quantity $$ (\sum _{i}{\tau }_{i}) $$ is the net torque on the body due to external forces. Similarly, we found the kinetic energy of a rigid body rotating around a fixed axis by summing the kinetic energy of each particle that makes up the rigid body. Since the work-energy theorem $$ {W}_{i}=\text{Δ}{K}_{i} $$ is valid for each particle, it is valid for the sum of the particles and the entire body. Work-Energy Theorem for Rotation The work-energy theorem for a rigid body rotating around a fixed axis is $${W}_{AB}={K}_{B}-{K}_{A}$$ where $$K=\frac{1}{2}I{\omega }^{2}$$ and the rotational work done by a net force rotating a body from point A to point B is $${W}_{AB}=\underset{{\theta }_{A}}{\overset{{\theta }_{B}}{\int }}(\sum _{i}{\tau }_{i})d\theta .$$ We give a strategy for using this equation when analyzing rotational motion. Problem-Solving Strategy: Work-Energy Theorem for Rotational Motion Identify the forces on the body and draw a free-body diagram. Calculate the torque for each force. Calculate the work done during the body’s rotation by every torque. Apply the work-energy theorem by equating the net work done on the body to the change in rotational kinetic energy. Let’s look at two examples and use the work-energy theorem to analyze rotational motion. Example Rotational Work and Energy A $$ 12.0\,\text{N}·\text{m} $$ torque is applied to a flywheel that rotates about a fixed axis and has a moment of inertia of $$ 30.0\,\text{kg}·{\text{m}}^{2}$$. If the flywheel is initially at rest, what is its angular velocity after it has turned through eight revolutions? Strategy We apply the work-energy theorem. We know from the problem description what the torque is and the angular displacement of the flywheel. Then we can solve for the final angular velocity. Solution The flywheel turns through eight revolutions, which is $$ 16\pi $$ radians. The work done by the torque, which is constant and therefore can come outside the integral in (Figure), is $${W}_{AB}=\tau ({\theta }_{B}-{\theta }_{A}).$$ We apply the work-energy theorem: $${W}_{AB}=\tau ({\theta }_{B}-{\theta }_{A})=\frac{1}{2}I{\omega }_{B}^{2}-\frac{1}{2}I{\omega }_{A}^{2}.$$ With $$ \tau =12.0\,\text{N}·\text{m},{\theta }_{B}-{\theta }_{A}=16.0\pi \,\text{rad},\,I=30.0\,\text{kg}·{\text{m}}^{2},\,\text{and}\,{\omega }_{A}=0$$, we have $$12.0\,\text{N-m}(16.0\pi \,\text{rad})=\frac{1}{2}(30.0\,\text{kg}·{\text{m}}^{2})({\omega }_{B}^{2})-0.$$ Therefore, $${\omega }_{B}=6.3\,\text{rad}\text{/}\text{s}.$$ This is the angular velocity of the flywheel after eight revolutions. Significance The work-energy theorem provides an efficient way to analyze rotational motion, connecting torque with rotational kinetic energy. Example Rotational Work: A Pulley A string wrapped around the pulley in (Figure) is pulled with a constant downward force $$ \overset{\to }{F} $$ of magnitude 50 N. The radius R and moment of inertia I of the pulley are 0.10 m and $$ 2.5\,×\,{10}^{-3}{\text{kg-m}}^{2}$$, respectively. If the string does not slip, what is the angular velocity of the pulley after 1.0 m of string has unwound? Assume the pulley starts from rest. Figure 10.40 (a) A string is wrapped around a pulley of radius R. (b) The free-body diagram. Strategy Looking at the free-body diagram, we see that neither $$ \overset{\to }{B}$$, the force on the bearings of the pulley, nor$$M\overset{\to }{g}$$, the weight of the pulley, exerts a torque around the rotational axis, and therefore does no work on the pulley. As the pulley rotates through an angle $$ \theta , $$ $$\overset{\to }{F} $$ acts through a distance d such that $$ d=R\theta .$$ Solution Since the torque due to $$ \overset{\to }{F} $$ has magnitude $$ \tau =RF$$, we have $$W=\tau \theta =(FR)\theta =Fd.$$ If the force on the string acts through a distance of 1.0 m, we have, from the work-energy theorem, $$\begin{array}{ccc}\hfill {W}_{AB}& =\hfill & {K}_{B}-{K}_{A}\hfill \ \hfill Fd& =\hfill & \frac{1}{2}I{\omega }^{2}-0\hfill \ \hfill (50.0\,\text{N})(1.0\,\text{m})& =\hfill & \frac{1}{2}(2.5\,×\,{10}^{-3}{\text{kg-m}}^{2}){\omega }^{2}.\hfill \end{array}$$ Solving for $$ \omega $$, we obtain $$\omega =200.0\,\text{rad}\text{/}\text{s}.$$ Power for Rotational Motion Power always comes up in the discussion of applications in engineering and physics. Power for rotational motion is equally as important as power in linear motion and can be derived in a similar way as in linear motion when the force is a constant. The linear power when the force is a constant is $$ P=\overset{\to }{F}·\overset{\to }{v}$$. If the net torque is constant over the angular displacement, (Figure) simplifies and the net torque can be taken out of the integral. In the following discussion, we assume the net torque is constant. We can apply the definition of power derived in Power to rotational motion. From Work and Kinetic Energy, the instantaneous power (or just power) is defined as the rate of doing work, $$P=\frac{dW}{dt}.$$ If we have a constant net torque, (Figure) becomes $$ W=\tau \theta $$ and the power is $$P=\frac{dW}{dt}=\frac{d}{dt}(\tau \theta )=\tau \frac{d\theta }{dt}$$ or $$P=\tau \omega .$$ Example Torque on a Boat Propeller A boat engine operating at $$ 9.0\,×\,{10}^{4}\,\text{W} $$ is running at 300 rev/min. What is the torque on the propeller shaft? Strategy We are given the rotation rate in rev/min and the power consumption, so we can easily calculate the torque. Solution $$300.0\,\text{rev/min}=31.4\,\text{rad/s;}$$ $$\tau =\frac{P}{\omega }=\frac{9.0\,×\,{10}^{4}\text{N}·\text{m}\text{/}\text{s}}{31.4\,\text{rad}\text{/}\text{s}}=2864.8\,\text{N}·\text{m}.$$ Significance It is important to note the radian is a dimensionless unit because its definition is the ratio of two lengths. It therefore does not appear in the solution. Check Your Understanding A constant torque of $$ 500\,\text{kN}·\text{m} $$ is applied to a wind turbine to keep it rotating at 6 rad/s. What is the power required to keep the turbine rotating? Show Solution 3 MW Rotational and Translational Relationships Summarized The rotational quantities and their linear analog are summarized in three tables. (Figure) summarizes the rotational variables for circular motion about a fixed axis with their linear analogs and the connecting equation, except for the centripetal acceleration, which stands by itself. (Figure) summarizes the rotational and translational kinematic equations. (Figure) summarizes the rotational dynamics equations with their linear analogs. Rotational and Translational Variables: Summary \| Rotational \| Translational \| Relationship \| \| $$\theta $$ \| $$x$$ \| $$\theta =\frac{s}{r}$$ \| \| $$\omega $$ \| $${v}_{t}$$ \| $$\omega =\frac{{v}_{t}}{r}$$ \| \| $$\alpha $$ \| $${a}_{\text{t}}$$ \| $$\alpha =\frac{{a}_{\text{t}}}{r}$$ \| \| $${a}_{\text{c}}$$ \| $${a}_{\text{c}}=\frac{{v}_{\text{t}}^{2}}{r}$$ \| Rotational and Translational Kinematic Equations: Summary \| Rotational \| Translational \| \| $${\theta }_{\text{f}}={\theta }_{0}+\overset{–}{\omega }t$$ \| $$x={x}_{0}+\overset{–}{v}t$$ \| \| $${\omega }_{\text{f}}={\omega }_{0}+\alpha t$$ \| $${v}_{\text{f}}={v}_{0}+at$$ \| \| $${\theta }_{\text{f}}={\theta }_{0}+{\omega }_{0}t+\frac{1}{2}\alpha {t}^{2}$$ \| $${x}_{\text{f}}={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2}$$ \| \| $${\omega }_{\text{f}}^{2}={\omega }^{2}{}_{0}+2\alpha (\text{Δ}\theta )$$ \| $${v}_{\text{f}}^{2}={v}^{2}{}_{0}+2a(\text{Δ}x)$$ \| Rotational and Translational Equations: Dynamics \| Rotational \| Translational \| \| $$I=\sum _{i}{m}_{i}{r}_{i}^{2}$$ \| m \| \| $$K=\frac{1}{2}I{\omega }^{2}$$ \| $$K=\frac{1}{2}m{v}^{2}$$ \| \| $$\sum _{i}{\tau }_{i}=I\alpha $$ \| $$\sum _{i}{\overset{\to }{F}}_{i}=m\overset{\to }{a}$$ \| \| $${W}_{AB}=\underset{{\theta }_{A}}{\overset{{\theta }_{B}}{\int }}(\sum _{i}{\tau }_{i})d\theta $$ \| $$W=\int \overset{\to }{F}·d\overset{\to }{s}$$ \| \| $$P=\tau \omega $$ \| $$P=\overset{\to }{F}·\overset{\to }{v}$$ \| Summary The incremental work dW in rotating a rigid body about a fixed axis is the sum of the torques about the axis times the incremental angle $$ d\theta $$. The total work done to rotate a rigid body through an angle $$ \theta $$ about a fixed axis is the sum of the torques integrated over the angular displacement. If the torque is a constant as a function of $$ \theta $$, then $$ {W}_{AB}=\tau ({\theta }_{B}-{\theta }_{A})$$. The work-energy theorem relates the rotational work done to the change in rotational kinetic energy: $$ {W}_{AB}={K}_{B}-{K}_{A} $$ where $$ K=\frac{1}{2}I{\omega }^{2}.$$ The power delivered to a system that is rotating about a fixed axis is the torque times the angular velocity, $$ P=\tau \omega $$. Key Equations \| \| \| --- \| \| Angular position \| $$\theta =\frac{s}{r}$$ \| \| Angular velocity \| $$\omega =\underset{\text{Δ}t\to 0}{\text{lim}}\frac{\text{Δ}\theta }{\text{Δ}t}=\frac{d\theta }{dt}$$ \| \| Tangential speed \| $${v}_{\text{t}}=r\omega $$ \| \| Angular acceleration \| $$\alpha =\underset{\text{Δ}t\to 0}{\text{lim}}\frac{\text{Δ}\omega }{\text{Δ}t}=\frac{d\omega }{dt}=\frac{{d}^{2}\theta }{d{t}^{2}}$$ \| \| Tangential acceleration \| $${a}_{\text{t}}=r\alpha $$ \| \| Average angular velocity \| $$\overset{–}{\omega }=\frac{{\omega }_{0}+{\omega }_{\text{f}}}{2}$$ \| \| Angular displacement \| $${\theta }_{\text{f}}={\theta }_{0}+\overset{–}{\omega }t$$ \| \| Angular velocity from constant angular acceleration \| $${\omega }_{\text{f}}={\omega }_{0}+\alpha t$$ \| \| Angular velocity from displacement and constant angular acceleration \| $${\theta }_{\text{f}}={\theta }_{0}+{\omega }_{0}t+\frac{1}{2}\alpha {t}^{2}$$ \| \| Change in angular velocity \| $${\omega }_{\text{f}}^{2}={\omega }_{0}^{2}+2\alpha (\text{Δ}\theta )$$ \| \| Total acceleration \| $$\overset{\to }{a}={\overset{\to }{a}}_{\text{c}}+{\overset{\to }{a}}_{\text{t}}$$ \| \| Rotational kinetic energy \| $$K=\frac{1}{2}(\sum _{j}{m}_{j}{r}_{j}^{2}){\omega }^{2}$$ \| \| Moment of inertia \| $$I=\sum _{j}{m}_{j}{r}_{j}^{2}$$ \| \| Rotational kinetic energy in terms of the moment of inertia of a rigid body \| $$K=\frac{1}{2}I{\omega }^{2}$$ \| \| Moment of inertia of a continuous object \| $$I=\int {r}^{2}dm$$ \| \| Parallel-axis theorem \| $${I}_{\text{parallel-axis}}={I}_{\text{initial}}+m{d}^{2}$$ \| \| Moment of inertia of a compound object \| $${I}_{\text{total}}=\sum _{i}{I}_{i}$$ \| \| Torque vector \| $$\overset{\to }{\tau }=\overset{\to }{r}\,×\,\overset{\to }{F}$$ \| \| Magnitude of torque \| $$\|\overset{\to }{\tau }\|={r}_{\perp }F$$ \| \| Total torque \| $${\tau }_{\text{net}}=\sum _{i}\|{\tau }_{i}\|$$ \| \| Newton’s second law for rotation \| $$\sum _{i}{\tau }_{i}=I\alpha $$ \| \| Incremental work done by a torque \| $$dW=(\sum _{i}{\tau }_{i})d\theta $$ \| \| Work-energy theorem \| $${W}_{AB}={K}_{B}-{K}_{A}$$ \| \| Rotational work done by net force \| $${W}_{AB}=\underset{{\theta }_{A}}{\overset{{\theta }_{B}}{\int }}(\sum _{i}{\tau }_{i})d\theta $$ \| \| Rotational power \| $$P=\tau \omega $$ \| Problems A wind turbine rotates at 20 rev/min. If its power output is 2.0 MW, what is the torque produced on the turbine from the wind? Show Solution $$\tau =\frac{P}{\omega }=\frac{2.0\,×\,{10}^{6}\text{W}}{2.1\,\text{rad}\text{/}\text{s}}=9.5\,×\,{10}^{5}\text{N}·\text{m}$$ A clay cylinder of radius 20 cm on a potter’s wheel spins at a constant rate of 10 rev/s. The potter applies a force of 10 N to the clay with his hands where the coefficient of friction is 0.1 between his hands and the clay. What is the power that the potter has to deliver to the wheel to keep it rotating at this constant rate? A uniform cylindrical grindstone has a mass of 10 kg and a radius of 12 cm. (a) What is the rotational kinetic energy of the grindstone when it is rotating at $$ 1.5\,×\,{10}^{3}\text{rev}\text{/}\text{min}? $$ (b) After the grindstone’s motor is turned off, a knife blade is pressed against the outer edge of the grindstone with a perpendicular force of 5.0 N. The coefficient of kinetic friction between the grindstone and the blade is 0.80. Use the work energy theorem to determine how many turns the grindstone makes before it stops. Show Solution a. $$ K=888.50\,\text{J}$$; b. $$ \text{Δ}\theta =294.6\,\text{rev}$$ A uniform disk of mass 500 kg and radius 0.25 m is mounted on frictionless bearings so it can rotate freely around a vertical axis through its center (see the following figure). A cord is wrapped around the rim of the disk and pulled with a force of 10 N. (a) How much work has the force done at the instant the disk has completed three revolutions, starting from rest? (b) Determine the torque due to the force, then calculate the work done by this torque at the instant the disk has completed three revolutions? (c) What is the angular velocity at that instant? (d) What is the power output of the force at that instant? A propeller is accelerated from rest to an angular velocity of 1000 rev/min over a period of 6.0 seconds by a constant torque of $$ 2.0\,×\,{10}^{3}\text{N}·\text{m}$$. (a) What is the moment of inertia of the propeller? (b) What power is being provided to the propeller 3.0 s after it starts rotating? Show Solution a. $$ I=114.6\,\text{kg}·{\text{m}}^{2}$$; b. $$ P=104,700\,\text{W}$$ A sphere of mass 1.0 kg and radius 0.5 m is attached to the end of a massless rod of length 3.0 m. The rod rotates about an axis that is at the opposite end of the sphere (see below). The system rotates horizontally about the axis at a constant 400 rev/min. After rotating at this angular speed in a vacuum, air resistance is introduced and provides a force $$ 0.15\,\text{N} $$ on the sphere opposite to the direction of motion. What is the power provided by air resistance to the system 100.0 s after air resistance is introduced? A uniform rod of length L and mass M is held vertically with one end resting on the floor as shown below. When the rod is released, it rotates around its lower end until it hits the floor. Assuming the lower end of the rod does not slip, what is the linear velocity of the upper end when it hits the floor? Show Answer $$v=L\omega =\sqrt{3Lg}$$ An athlete in a gym applies a constant force of 50 N to the pedals of a bicycle to keep the rotation rate of the wheel at 10 rev/s. The length of the pedal arms is 30 cm. What is the power delivered to the bicycle by the athlete? A 2-kg block on a frictionless inclined plane at $$ 40\text{°} $$ has a cord attached to a pulley of mass 1 kg and radius 20 cm (see the following figure). (a) What is the acceleration of the block down the plane? (b) What is the work done by the gravitational force to move the block 50 cm? Show Answer a. $$ a=5.0\,\text{m}\text{/}{\text{s}}^{2}$$; b. $$ W=1.25\,\text{N}·\text{m}$$ Small bodies of mass $$ {m}_{1}\,\text{and}\,{m}_{2} $$ are attached to opposite ends of a thin rigid rod of length L and mass M. The rod is mounted so that it is free to rotate in a horizontal plane around a vertical axis (see below). What distance d from $$ {m}_{1} $$ should the rotational axis be so that a minimum amount of work is required to set the rod rotating at an angular velocity $$ \omega ?$$ Additional Problems A cyclist is riding such that the wheels of the bicycle have a rotation rate of 3.0 rev/s. If the cyclist brakes such that the rotation rate of the wheels decrease at a rate of $$ 0.3\,\text{rev}\text{/}{\text{s}}^{2}$$, how long does it take for the cyclist to come to a complete stop? Show Solution $$\text{Δ}t=10.0\,\text{s}$$ Calculate the angular velocity of the orbital motion of Earth around the Sun. A phonograph turntable rotating at 33 1/3 rev/min slows down and stops in 1.0 min. (a) What is the turntable’s angular acceleration assuming it is constant? (b) How many revolutions does the turntable make while stopping? Show Solution a. $$ 0.06\,\text{rad}\text{/}{\text{s}}^{2}$$; b. $$ \theta =105.0\,\text{rad}$$ With the aid of a string, a gyroscope is accelerated from rest to 32 rad/s in 0.40 s under a constant angular acceleration. (a) What is its angular acceleration in $$ {\text{rad/s}}^{2}$$? (b) How many revolutions does it go through in the process? Suppose a piece of dust has fallen on a CD. If the spin rate of the CD is 500 rpm, and the piece of dust is 4.3 cm from the center, what is the total distance traveled by the dust in 3 minutes? (Ignore accelerations due to getting the CD rotating.) Show Solution $$s=405.26\,\text{m}$$ A system of point particles is rotating about a fixed axis at 4 rev/s. The particles are fixed with respect to each other. The masses and distances to the axis of the point particles are $$ {m}_{1}=0.1\,\text{kg},{r}_{1}=0.2\,\text{m}$$, $$ {m}_{2}=0.05\,\text{kg},{r}_{2}=0.4\,\text{m}$$, $$ {m}_{3}=0.5\,\text{kg},{r}_{3}=0.01\,\text{m}$$. (a) What is the moment of inertia of the system? (b) What is the rotational kinetic energy of the system? Calculate the moment of inertia of a skater given the following information. (a) The 60.0-kg skater is approximated as a cylinder that has a 0.110-m radius. (b) The skater with arms extended is approximated by a cylinder that is 52.5 kg, has a 0.110-m radius, and has two 0.900-m-long arms which are 3.75 kg each and extend straight out from the cylinder like rods rotated about their ends. Show Solution a. $$ I=0.363\,\text{kg}·{\text{m}}^{2}$$; b. $$ I=2.34\,\text{kg}·{\text{m}}^{2}$$ A stick of length 1.0 m and mass 6.0 kg is free to rotate about a horizontal axis through the center. Small bodies of masses 4.0 and 2.0 kg are attached to its two ends (see the following figure). The stick is released from the horizontal position. What is the angular velocity of the stick when it swings through the vertical? A pendulum consists of a rod of length 2 m and mass 3 kg with a solid sphere of mass 1 kg and radius 0.3 m attached at one end. The axis of rotation is as shown below. What is the angular velocity of the pendulum at its lowest point if it is released from rest at an angle of $$ 30\text{°}?$$ Show Answer $$\omega =\sqrt{\frac{5.36\,\text{J}}{4.4\,{\text{kgm}}^{2}}}=1.10\,\text{rad}\text{/}\text{s}$$ Calculate the torque of the 40-N force around the axis through O and perpendicular to the plane of the page as shown below. Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the door. One child pushes with a force of 17.5 N at a distance of 0.600 m from the hinges, and the second child pushes at a distance of 0.450 m. What force must the second child exert to keep the door from moving? Assume friction is negligible. Show Solution $$F=23.3\,\text{N}$$ The force of $$ 20\hat{j}\text{N} $$ is applied at $$ \overset{\to }{r}=(4.0\hat{i}-2.0\hat{j})\,\text{m}$$. What is the torque of this force about the origin? An automobile engine can produce 200 N$$· $$ m of torque. Calculate the angular acceleration produced if 95.0% of this torque is applied to the drive shaft, axle, and rear wheels of a car, given the following information. The car is suspended so that the wheels can turn freely. Each wheel acts like a 15.0-kg disk that has a 0.180-m radius. The walls of each tire act like a 2.00-kg annular ring that has inside radius of 0.180 m and outside radius of 0.320 m. The tread of each tire acts like a 10.0-kg hoop of radius 0.330 m. The 14.0-kg axle acts like a rod that has a 2.00-cm radius. The 30.0-kg drive shaft acts like a rod that has a 3.20-cm radius. Show Solution $$\alpha =\frac{190.0\,\text{N-m}}{2.94\,{\text{kg-m}}^{2}}=64.4\,{\text{rad}\text{/}\text{s}}^{2}$$ A grindstone with a mass of 50 kg and radius 0.8 m maintains a constant rotation rate of 4.0 rev/s by a motor while a knife is pressed against the edge with a force of 5.0 N. The coefficient of kinetic friction between the grindstone and the blade is 0.8. What is the power provided by the motor to keep the grindstone at the constant rotation rate? Challenge Problems The angular acceleration of a rotating rigid body is given by $$ \alpha =(2.0-3.0t)\,\text{rad}\text{/}{\text{s}}^{2}$$. If the body starts rotating from rest at $$ t=0$$, (a) what is the angular velocity? (b) Angular position? (c) What angle does it rotate through in 10 s? (d) Where does the vector perpendicular to the axis of rotation indicating $$ 0\text{°} $$ at $$ t=0 $$ lie at $$ t=10\,\text{s}$$? Show Solution a. $$ \omega =2.0t-1.5{t}^{2}$$; b. $$ \theta ={t}^{2}-0.5{t}^{3}$$; c. $$ \theta =-400.0\,\text{rad}$$; d. the vector is at $$ -0.66(360\text{°})=-237.6\text{°}$$ Earth’s day has increased by 0.002 s in the last century. If this increase in Earth’s period is constant, how long will it take for Earth to come to rest? A disk of mass m, radius R, and area A has a surface mass density $$ \sigma =\frac{mr}{AR} $$ (see the following figure). What is the moment of inertia of the disk about an axis through the center? Show Answer Zorch, an archenemy of Rotation Man, decides to slow Earth’s rotation to once per 28.0 h by exerting an opposing force at and parallel to the equator. Rotation Man is not immediately concerned, because he knows Zorch can only exert a force of $$ 4.00\,×\,1{0}^{7}\text{N} $$ (a little greater than a Saturn V rocket’s thrust). How long must Zorch push with this force to accomplish his goal? (This period gives Rotation Man time to devote to other villains.) A cord is wrapped around the rim of a solid cylinder of radius 0.25 m, and a constant force of 40 N is exerted on the cord shown, as shown in the following figure. The cylinder is mounted on frictionless bearings, and its moment of inertia is $$ 6.0\,\text{kg}·{\text{m}}^{2}$$. (a) Use the work energy theorem to calculate the angular velocity of the cylinder after 5.0 m of cord have been removed. (b) If the 40-N force is replaced by a 40-N weight, what is the angular velocity of the cylinder after 5.0 m of cord have unwound? Show Answer a. $$ \omega =8.2\,\text{rad}\text{/}\text{s}$$; b. $$ \omega =8.0\,\text{rad}\text{/}\text{s}$$ Glossary rotational work : work done on a rigid body due to the sum of the torques integrated over the angle through with the body rotates work-energy theorem for rotation : the total rotational work done on a rigid body is equal to the change in rotational kinetic energy of the body Candela Citations CC licensed content, Shared previously OpenStax University Physics. Authored by: OpenStax CNX. Located at: License: CC BY: Attribution. License Terms: Download for free at Licenses and Attributions CC licensed content, Shared previously OpenStax University Physics. Authored by: OpenStax CNX. Located at: License: CC BY: Attribution. License Terms: Download for free at Privacy Policy
8289	https://global-exam.com/blog/es/gramatica-espanola-preterito-imperfecto-de-indicativo-uso-ejemplos-y-ejercicios/	Skip to content General Spanish > Gramática española > El pretérito imperfecto de indicativo: uso, ejemplos y ejercicios El pretérito imperfecto de indicativo: uso, ejemplos y ejercicios El pretérito imperfecto de indicativo: uso, ejemplos y ejercicios Compartir en Los verbos nos permiten expresar una acción o estado en tiempos que pueden variar desde el pasado, el presente o el futuro. Es importante manejar los diferentes tiempos de los verbos para estar seguros de que lo que queremos decir se sitúa en el buen contexto y que transmitimos de manera clara y precisa a nuestro interlocutor una cosa que pasó antes o después, o que pasará más tarde. No es lo mismo decir , «hablo muy bien español» que decir «hablaré muy bien español» que indica un futuro, seguro que próximo si sigues leyendo nuestros artículos. Para que no tengas que recurrir al «antes hablaba muy bien español» vamos a refrescar un poco el uso de ciertos tiempos verbales. Empezaremos en este artículo por el pretérito imperfecto del indicativo, un tiempo del pasado. Para ello vamos a abordar a continuación los siguientes puntos : ¿Qué es el pretérito imperfecto del indicativo y cuándo se utiliza? ¿Cómo se conjugan los verbos en pretérito imperfecto del indicativo y cuáles son las terminaciones que hay que utilizar? Las terminaciones de los verbos irregulares Frases con pretérito imperfecto de indicativo. Ejercicios para entrenarse con este tiempo verbal Mejora el uso de los verbos y perfecciona tu español con el Global General de GlobalExam ¿Preparado para pasar a la acción? Evalúate gratuitamente ¿Qué es el pretérito imperfecto del indicativo y cuándo se utiliza? El pretérito imperfecto es un tiempo que indica pasado en español. El pretérito imperfecto del indicativo (el imparfait en francés o el past continuous en inglés) se utiliza en español para expresar acciones que tuvieron lugar en el pasado, explicar cómo eran otras épocas o qué hacíamos normalmente en el pasado. Se utiliza a menudo con marcadores temporales como siempre, a veces, nunca… Sus usos son: Describe acciones, personas, lugares o cosas que tuvieron lugar en el pasado. Ej.: Cuando era pequeña jugaba mucho en la calle; la casa de mi abuela era pequeña; el pueblo vivía de la minería hasta que llegó la crisis. Puede también expresar cambios, cosas que solían ser de una forma en el pasado y ahora ya no lo son. Ej.: Antes odiaba dormir la siesta los domingos por la tarde pero ahora me encanta; antes comía mucho queso pero ahora me sienta mal. Vamos a utilizarlo también para describir situaciones momentáneas en el pasado. Ej.: Anoche dormí muy mal ; cuando era adolescente creció mucho de un golpe. También puede expresar condición, sustituyendo al condicional simple en el registro oral. Ej.: Si me tocara la lotería me compraba un coche; si fuera más alto jugaba al baloncesto. Expresa también una intención. Ej.: Pensaba llamarle esta mañana y se me ha olvidado; iba a decírselo cuando le interrumpieron. Por último se puede utilizar para expresar cortesía. Ej.: Quería pedirte un favor; ¿qué deseaba usted?; venía a proponerte una cosa. ¿Cómo se conjugan los verbos en pretérito imperfecto del indicativo y cuáles son las terminaciones que hay que utilizar? En español los verbos regulares se dividen en tres grandes grupos, los que terminan en -ar (amar, cantar), los que terminan en -er (beber, correr) y los que lo hacen en -ir (subir, decir). El imperfecto es un tiempo verbal bastante regular y se forma tomando la raíz del verbo (infinitivo menos -ar, -er o -ir) y añadiendo las siguientes terminaciones. Los verbos terminados en -er y en -ir se conjugan de la misma manera: \| \| -ar \| Amar \| -er \| Beber \| -ir \| Subir \| --- --- --- \| Yo \| - aba \| amaba \| -ía \| bebía \| -ía \| subía \| \| Tú /Vos \| -abas \| amabas \| -ías \| bebías \| -ías \| subías \| \| Él/ Ella/ Usted \| -aba \| amaba \| -ía \| bebía \| -ía \| subían \| \| Nosotros \| -ábamos \| amábamos \| -íamos \| bebíamos \| -íamos \| subíamos \| \| Vosotros \| -ábais \| amábais \| -íais \| bebíais \| -íais \| subíais \| \| Ellos/Ellas \| -aban \| amaban \| -ían \| bebían \| -ían \| subían \| \| Ustedes \| -aban \| amaban \| -ían \| bebían \| -ían \| subían \| Las terminaciones de los verbos irregulares Como en todo idioma, en español también hay verbos que no siguen las mismas reglas que los otros. Son los irregulares. Pero no te preocupes, sólo hay 3 verbos irregulares en este tiempo: ver, ir y ser. \| \| Ver \| Ir \| Ser \| --- --- \| \| Yo \| Veía \| Iba \| Era \| \| Tú /Vos \| Veías \| Ibas \| Eras \| \| Él/ Ella/ Usted \| Veía \| Iba \| Era \| \| Nosotros \| Veíamos \| Íbamos \| Éramos \| \| Vosotros \| Veíais \| Íbais \| Érais \| \| Ellos/Ellas \| Veían \| Iban \| Eran \| \| Ustedes \| Veían \| Iban \| Eran \| Frases con pretérito imperfecto de indicativo Para que veas de manera clara cómo se utiliza este tiempo verbal te vamos a dar una serie de frases que te van a servir de ejemplo: Comparado con el presente, mira como se expresa una acción que tuvo lugar en un tiempo determinado del pasado y ya acabó: Este verano hace mucho calor – El verano pasado hacía mucho calor El coche es viejo y tiene muchos problemas mecánicos – El coche era viejo y tenía muchos problemas mecánicos En el párrafo siguiente, la autora ha querido expresar una acción que tenía lugar de manera regular en el pasado y que ya no tiene lugar: Hoy en día los niños pasan mucho tiempo delante de las pantallas. Cuando yo era pequeña jugábamos todo el tiempo en el parque. Era otra forma de ver la vida, disfrutábamos más de la infancia. Para describir una situación momentánea en el pasado diríamos: Ayer estaba cansada pero dormí muy mal. No conseguía relajarme. Antes no pensaba en esas cosas, para mi eran tonterías, pero ahora lo hago a menudo. El uso de este tiempo verbal en condicional se hace sobretodo en el lenguaje oral pero lo haríamos con frases como éstas: Si tuviera más tiempo, viajaba más Si hubiera más sol en el norte, me iba a vivir allí. Mientras que para expresar una intención lo haríamos con frases como Iba a hablarte de eso cuando llegó tu amiga o Pensaba llegar más pronto pero todo se complicó. Para expresar cortesía, utilizamos el pretérito imperfecto de la siguiente manera: Quiero pedirte un favor – Quería pedirte un favor (menos directo y más cortés) ¿Puedes ayudarme con este ejercicio? – ¿Podrías ayudarme con este ejercicio? Si quieres más fichas de gramática sigue leyendo. Ejercicios para entrenarse con este tiempo verbal Una vez que ya conoces los usos y las terminaciones del pretérito imperfecto de indicativo, te vamos a proponer una serie de ejercicios para que puedas entrenarte en su uso y demostrarte lo bien que has entendido nuestras explicaciones. ¿Empezamos? Completa el ejercicio siguiente: \| 1. estudiar, ellos: \| \| 2. viajar, yo: \| \| 3. ser, tú: \| \| 4. ir, yo: \| \| 5. ver, nosotros: \| \| 6. conducir, ellos: \| \| 7. medir, él: \| \| 8. beber, ella: \| \| 9. llorar, nosotros: \| \| 10. olvidar, vosotros: \| Completa las frases con el pretérito imperfecto: Mi hermana (SER) una persona muy alegre.2. Cuando era pequeña, mis amigos y yo (IR) al cine los domingos.3. Nunca (HABLAR) mucho cuando estaba con otros.4. Nuestra casa (ESTAR) en el centro del pueblo, donde (HABER) mucha animación.5. Mi padre no fuma ahora, pero antes (FUMAR) un paquete al día.6. Mientras (COMER) , los niños nunca (VER) la tele.7. (SER) un día templado de primavera. El sol (BRILLAR) en el cielo.8. Antes (IR, vosotros) a la escuela de lunes a viernes.9. Elena (TENER) un gato muy cariñoso que (LLAMARSE) Milo10. Antes (DORMIR, tú) mucho más los fines de semana. Respuestas: \| 1. Estudiaban \| \| 2. Viajaba \| \| 3. Eras \| \| 4. Iba \| \| 5. Veíamos \| \| 6. Conducían \| \| 7. Medía \| \| 8. Bebía \| \| 9. Llorábamos \| \| 10. Olvidábais \| Era2. Íbamos3. hablaba4. Estaba / había5. Fumaba6. Comían / veían7. Era / brillaba8. Íbamos9.Tenía / se llamaba10. Dormías Mejora el uso de los verbos y perfecciona tu español con el Global General de GlobalExam Para mejorar tu fluidez en español tienes que seguir entrenándote en el uso del pretérito imperfecto de indicativo y por supuesto del resto de los tiempos. Busca entre nuestros artículos aquellos que hablen de otros tiempos verbales en español y serás imbatible. ¡Estudiar español nunca ha sido más fácil! Además, para que este aprendizaje sea eficaz y lúdico, GlobalExam tiene el método perfecto para que mejores el uso de los verbos y perfecciones tu español sin moverte de casa: el Global General. Un método de aprendizaje dirigido a los estudiantes con niveles principiante e intermedio que te permitirá aprender español online, con itinerarios académicos adaptados a tus necesidades y recursos puestos a tu disposición sobre nuestra plataforma, para que puedas aprender español cuándo tú quieras y desde donde quieras. Con el Global General accederás a fichas de revisión disponibles en cualquier momento así como a correcciones detalladas para cada cuestión, que te permitirán evolucionar en español a tu ritmo, sin los inconvenientes ni las frustraciones de la educación tradicional. Además, si lo que quieres es prepararte para obtener las titulaciones oficiales de español, GlobalExam te propone programas personalizados con ejercicios corregidos y simulacros de examen para que obtener el Bright o el DELE sea más fácil y sencillo de lo que pensabas. No dudes, GlobalExam es la plataforma perfecta para aprender español de manera rápida y eficaz. ¡Prueba el Global General y haz caer las barreras del idioma! Consulta también las siguientes fichas: Uso del imperativo: reglas, ejemplos y ejercicios Uso del futuro en francés: reglas, ejemplos y ejercicios Uso de adjetivos: reglas, tipos, ejemplos y ejercicios Tipos de pronombres: uso, reglas, ejemplos y ejercicios 2021-10-29T11:00:46+02:0028/05/2021\| Artículos relacionados Fichas de gramática en Español Tipos de pronombres: uso, reglas, ejemplos y ejercicios Uso del futuro en español: reglas, ejemplos y ejercicios Go to Top
8290	https://chemlab.truman.edu/files/2015/07/Gravimetric_Determination_of_Soluble_Sulfate.pdf	Truman State University CHEM 222 Lab Manual Revised 09/05/13 Gravimetric Determination of Soluble Sulfate The quantitative determination of sulfate ions in inorganic compounds can be accomplished by using the selective precipitation of the sulfate ion from a hot solution using BaCl2 (barium chloride), provided proper control over reaction conditions are exercised and good analytical techniques are practiced. In this experiment, the sulfate ion will be reacted with an excess of barium ion in a hot solution to drive the precipitation of barium sulfate. Following digestion of the precipitate, it will be filtered through ashless filter paper and then washed to remove ionic contaminants that are adsorbed on the surface of the crystallizing barium sulfate. The precipitate and filter paper will be transferred to a crucible and the contents will be ignited in a muffle furnace. The process of ignition oxidizes the filter paper, leaving barium sulfate salt in the crucible. Measurement of the mass of BaSO4 will allow the determination of sulfate in the original unknown sample using known stoichiometry. Materials Needed:  3 porcelain crucibles and lids  Barium Chloride solution at (0.25 M)  Aqueous silver nitrate solution (0.1 M)  Concentrated HCl (hydrochloric acid)  Ashless Filter Paper  Sulfate Unknown Sample (provided by your instructor) Preparation for sulfate determination. 1. Obtain an unknown sample from your instructor. It should be of sufficient quantity for triplicate samples. Dry this unknown sample for at least 2 hours at 110-120 oC. Allow it to cool in your desiccator for a minimum of ½ hour prior to massing. 2. Clean each crucible and lid (3 each) thoroughly. Place them paired with the lids slightly ajar to allow air circulation into a convection oven at 110-120 0C for 1 hour. Remove from the oven, place in your desiccator and allow to cool. Accurately weigh each crucible and lid. Repeat the heating/cooling process until a constant mass for each crucible and lid pair is achieved. 3. Prepare a 100 mL of a 0.25 F solution of barium chloride and store this in a plastic bottle. Procedure: 1. Accurately weigh three samples of approximately 0.5000 g of your unknown material directly into clean 400 mL beakers. Dissolve each sample in approximately 200 mL of distilled, deionized water and add 0.5 mL of concentrated hydrochloric acid to each beaker. You should use a different stirring rod for each sample and leave that stirring rod in the beaker throughout. 2. To determine the quantity of barium chloride necessary, assume your sample is pure sodium sulfate (knowing it isn’t) and calculate the number of moles of barium chloride needed to precipitate all of the sulfate in the largest sample. From this value, calculate the number of mL of 0.25 F BaCl2 solution needed to deliver this number of moles, then increase this amount by 10% to determine the volume of BaCl2 solution that will be added in step 4. 3. Heat the three solutions to nearly boiling on a hot plate and adjust the temperature to keep them slightly below boiling. (When you see bubbles forming on the bottom of the beaker is generally a point where the solution is sufficiently hot). Truman State University CHEM 222 Lab Manual Revised 09/05/13 4. Using a buret, slowly add the calculated volume of barium chloride solution determined in step 2 into each beaker, stirring vigorously throughout the addition. After the addition, let the precipitate settle and then test for complete precipitation by adding ~5 mL addition barium chloride solution. If a precipitate forms, stir and let the precipitate settle. Repeat this process until precipitation is complete. Leave the stirring rods in the beakers, cover each with a watch glass and digest the mixture on a hot plate until the supernatant is clear (DO NOT BOIL). Digestion is a step that allows the crystal size to become larger, thus creating a more easily and completely filterable sized solid. This will require at least an hour to complete. If the volume of your solution drops below 200 mL, you should add more distilled water. If there is insufficient time to complete the ensuing filtering steps, you may stop at this step and continue the procedure during the next lab period. 5. Prepare filter funnels to filter each solution. Each should be fitted with a piece of ashless filter paper provided you by your instructor. If you are unsure how to fold this paper for quantitative filtration activities consult with your instructor for the proper approach (fluted folding is not preferred for quantitative filtering activities). While preparing these, if your solutions have cooled, warm them on a hot plate until nearly boiling. Also prepare a beaker of hot distilled water for transferring and washing your precipitate. 6. Filter the solutions while hot, being careful to not fill the paper too full. If you do not understand, look up the term “creeping” as pertaining to filtering operations. Quantitatively transfer the barium sulfate into the filter using hot distilled water, your stirring rod and a rubber policeman. 7. Wash the filtered precipitate and the filter paper with hot distilled water until a few mL of filtrate show no turbidity when several drops of acidic silver nitrate are added. During the filtering and washing steps, do your best to rinse the precipitate down into the cone of the filter. Be sure to analyze the filtered liquid for any signs of cloudiness indicating that crystalline barium sulfate has passed through the filter paper. If none is apparent, you can proceed to the ignition operation. If not, you will need to isolate the filtrate, add an additional 5 mL barium chloride solution, and repeat the digestion and filtration steps to remove all of the barium sulfate precipitate. 8. Loosen the filter paper from the funnels gently, allowing them to drain for a few minutes. Be careful not to tear them or lose crystalline material. Fold each filter paper into a packet enclosing the precipitate in the folded filter paper. Place the packet into a crucible with the cone side down and the folded paper on the top side, gently pressing it down into the crucible. Make sure all the precipitate has been removed from the funnel. If not, wipe it off with a small piece of ashless filter paper and add it to the crucible. Repeat with a new crucible for each sample. 9. Transfer each crucible to a muffle furnace, setting the crucible lid to leave the contents of the crucible exposed to air with only a slight “crack”. Make sure the lid covers the contents with only a mm or two crack. Set the muffle furnace to approximately 100 0C for about ½ hour to dry the moisture from the crucible, precipitate and paper. After drying, raise the temperature of the muffle furnace to 800 0C and allow ignition to occur for 1 hour once the furnace reaches 800 degrees. After the one hour period, shut the furnace off and allow the crucibles to cool to nearly room temperature before removing them from the muffle furnace. The precipitate should be white with no black particles. If it has black particles, it means that your filter paper was not completely oxidized and your crucibles will need to be ignited further. If the precipitate is white, then the crucibles can be transferred to your desiccator to complete cooling and proceed with reaching constant mass. Truman State University CHEM 222 Lab Manual Revised 09/05/13 10. After cooling, weigh the crucibles and lids. Heat them in a convection oven at 110-120 0C for ½ hour, cool in your desiccator and re-weigh. Continue this operation until you reach constant mass and successive mass measurements agree within ±0.4 mg. If gaining constant mass proves difficult, moisten the precipitate with a little dilute sulfuric acid and heat it very gently. 11. After constant mass has been reached, calculate the percent sulfate in each of your samples. Report the mean, standard deviation and confidence limit of your results.
8291	https://www.doubtnut.com/qna/417326158	The bond order of CO molecule is 2.5. To determine the bond order of the carbon monoxide (CO) molecule, we can follow these steps: Step 1: Determine the total number of electrons in CO - Carbon (C) has 6 electrons, and Oxygen (O) has 8 electrons. - Therefore, the total number of electrons in CO = 6 (from C) + 8 (from O) = 14 electrons. Step 2: Write the molecular orbital configuration for CO - The molecular orbital (MO) configuration for CO can be represented as: - σ1s, σ1s, σ2s, σ2s, π2p_x, π2p_y, σ2p_z - The filling of the molecular orbitals with the 14 electrons is as follows: - σ1s: 2 electrons - σ1s: 2 electrons - σ2s: 2 electrons - σ2s: 2 electrons - π2p_x: 2 electrons - π2p_y: 2 electrons - σ2p_z: 2 electrons Step 3: Count the number of electrons in bonding and antibonding orbitals - Bonding orbitals: - σ1s (2) + σ2s (2) + π2p_x (2) + π2p_y (2) + σ2p_z (2) = 10 electrons - Antibonding orbitals: - σ1s (2) + σ2s (2) = 4 electrons Step 4: Calculate the bond order - The bond order formula is given by: Bond Order=12×(Number of electrons in bonding orbitals−Number of electrons in antibonding orbitals) - Plugging in the numbers: Bond Order=12×(10−4)=12×6=3 Conclusion - The bond order of the CO molecule is 3, not 2.5 as stated in the question. --- To determine the bond order of the carbon monoxide (CO) molecule, we can follow these steps: Step 1: Determine the total number of electrons in CO - Carbon (C) has 6 electrons, and Oxygen (O) has 8 electrons. - Therefore, the total number of electrons in CO = 6 (from C) + 8 (from O) = 14 electrons. Step 2: Write the molecular orbital configuration for CO - The molecular orbital (MO) configuration for CO can be represented as: - σ1s, σ1s, σ2s, σ2s, π2p_x, π2p_y, σ2p_z - The filling of the molecular orbitals with the 14 electrons is as follows: - σ1s: 2 electrons - σ1s: 2 electrons - σ2s: 2 electrons - σ2s: 2 electrons - π2p_x: 2 electrons - π2p_y: 2 electrons - σ2p_z: 2 electrons Step 3: Count the number of electrons in bonding and antibonding orbitals - Bonding orbitals: - σ1s (2) + σ2s (2) + π2p_x (2) + π2p_y (2) + σ2p_z (2) = 10 electrons - Antibonding orbitals: - σ1s (2) + σ2s (2) = 4 electrons Step 4: Calculate the bond order - The bond order formula is given by: Bond Order=12×(Number of electrons in bonding orbitals−Number of electrons in antibonding orbitals) - Plugging in the numbers: Bond Order=12×(10−4)=12×6=3 Conclusion - The bond order of the CO molecule is 3, not 2.5 as stated in the question. Topper's Solved these Questions Explore 10 Videos Explore 10 Videos Explore 10 Videos Explore 13 Videos Similar Questions The bond order of NO molecule is Bond order of Be2 molecule is Knowledge Check Assertion(A) - The bond order of NO molecule is 2.5. Reason( R )-NO molecule is paramagnetic in nature. The bond order of H2 molecule is The bond order of CO is- Bond order of Li2 molecule is The bond order of a molecule is given by __________. Bond order of helium molecule is The bond order of a molecule is given by The bond order of a molecule is given by MODERN PUBLICATION-CHEMICAL BONDING AND MOLECULAR STRUCTURE-Revision Exercises (Objective Questions)(True or False Questions) Ionic compounds are bad conductors of electricity in the solid state. The shape of SF(6) molecule is octahedral whereas that of IF(7) is squ... The bond order of CO molecule is 2.5. As N-F bond is more polar than N-H bond, NF(3) molecule has higher dip... The bond angle follows the order: NH(4)^(+) gt NH(3) gt NH(2)^(-) The dipole moment of cis C(2)H(2)Cl(2) isomer is more than that of tra... N(2)^(+) has greater bond dissociation enthalpy than N(2) molecule. Out of MgO and CaO, MgO is harder. The d - orbital involved in dsp^(2) hybridisation si d(x^(2)-y^(2)). The dipole moment of CH(3)F is greater than that of CH(3)Cl. Exams Free Textbook Solutions Free Ncert Solutions English Medium Free Ncert Solutions Hindi Medium Boards Resources Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams. Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation Contact Us
8292	https://www.facebook.com/groups/PharmacologyMCQs/posts/4976581759105027/	Pharmacology And Medicine MCQs \| Q- Which of the following treatments should be used in cases of methanol or ethylene glycol poisoning \| Facebook Log In Log In Forgot Account? Which treatment for methanol or ethylene glycol poisoning? Summarized by AI from the post below Pharmacology And Medicine MCQs Ray Makar · Admin · April 29, 2022 · Q- Which of the following treatments should be used in cases of methanol or ethylene glycol poisoning?#usmle#pharmacology#medicine A-Ethanol B-Glucagon C-Pyridoxine D-Sodium bicarbonate Answer: All reactions: 10 2 comments 1 share Like Comment Share Most relevant Mollo Alesa Ochieng A 3y Hassan Isiyaku Ramos A 3y See more on Facebook See more on Facebook Email or phone number Password Log In Forgot password? or Create new account
8293	https://online.stat.psu.edu/stat414/lesson/3/3.2	3.2 - Permutations \| STAT 414 Skip to main content ENROLL Search Search STAT 414 Introduction to Probability Theory User Preferences Font size Font family A A Mode Cards Reset Content Preview Arcu felis bibendum ut tristique et egestas quis: Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris Duis aute irure dolor in reprehenderit in voluptate Excepteur sint occaecat cupidatat non proident Lorem ipsum dolor sit amet, consectetur adipisicing elit. Odit molestiae mollitia laudantium assumenda nam eaque, excepturi, soluta, perspiciatis cupiditate sapiente, adipisci quaerat odio voluptates consectetur nulla eveniet iure vitae quibusdam? Excepturi aliquam in iure, repellat, fugiat illum voluptate repellendus blanditiis veritatis ducimus ad ipsa quisquam, commodi vel necessitatibus, harum quos a dignissimos. Close Save changes Keyboard Shortcuts Help F1 or ? Previous Page ← + CTRL (Windows) ← + ⌘ (Mac) Next Page → + CTRL (Windows) → + ⌘ (Mac) Search Site CTRL + SHIFT + F (Windows) ⌘ + ⇧ + F (Mac) Close Message ESC Breadcrumb Home 3 3.2 3.2 - Permutations Example 3-4 Section How many ways can four people fill four executive positions? Answer For the sake of concreteness, let's name the four people Tom, Rick, Harry, and Mary, and the four executive positions President, Vice President, Treasurer and Secretary. I think you'll agree that the Multiplication Principle yields a straightforward solution to this problem. If we fill the President position first, there are 4 possible people (Tom, Rick, Harry, and Mary). Let's suppose Mary is named the President. Then, since Mary can't fill more than one position at a time, when we fill the Vice President position, there are only 3 possible people (Tom, Rick, and Harry). If Tom is named the Vice President, when we fill the Treasurer position, there are only 2 possible people (Rick and Harry). Finally, if Rick is named Treasurer, when we fill the Secretary position, there is only 1 possible person (Harry). Putting all of this together, the Multiplication Principle tells us that there are: $4\times 3\times 2\times 1$ or 24 possible ways to fill the four positions. Alright, alright now... enough of these kinds of examples, eh?! The main point of this example is not to see yet another application of the Multiplication Principle, but rather to introduce the counting of the number of permutations as a generalization of the Multiplication Principle. A Generalization of the Multiplication Principle Section Suppose there are $n$ positions to be filled with $n$ different objects, in which there are: $n$ choices for the 1st position $n-1$ choices for the 2nd position $n-2$ choices for the 3rd position ... and ... 1 choice for the last position The Multiplication Principle tells us there are then in general: $n\times (n-1)\times (n-2)\times \ldots \times 1=n!$ ways of filling the $n$ positions. The symbol $n!$ is read as "$n$ -factorial," and by definition 0! equals 1. Permutation of $n$ objects an ordered arrangement of the $n$ objects We often call such a permutation a “permutation of $n$ objects taken $n$ at a time,” and denote it as $_nP_n$. That is: $_nP_n=n\times (n-1)\times (n-2) \times \ldots \times 1=n!$ Not that it really matters in this situation (since they are the same), but the first subscripted $n$ represents the number of objects you are wanting to arrange, while the second subscripted $n$ represents the number of positions you have for the objects to fill. Example 3-5 Section The draft lottery of 1969 for military service ranked all 366 days (Jan 1, Jan 2, ..., Feb 29, ..., Dec 31) of the year. The men who were eligible for service whose birthday was selected first were the first to be drafted. Those whose birthday was selected second were the second to be drafted. And so on. How many possible ways can the 366 days be ranked? Answer Well, we have 366 objects (days) and 366 positions (1st spot, 2nd spot, ... , 366th spot) to arrange them. Therefore, there are 366! ("366 factorial" ) ways of ranking the 366 possible birthdays of the eligible men. What is the probability that your birthday would be ranked first? Answer The answer, which is 1/366 = 0.0027, can be determined in (at least) two ways. The simplest way is to recognize that there is only one birthday (yours!) in the event of interest, and that there are 366 birthdays in the sample space. Therefore, the classical approach to assigning probability tells us the probability is 1 divided by 366. A second way is to recognize that there are 366! ways of ranking the 366 birthdays, and that there are 365! ways of ranking the 366 birthdays to ensure that your birthday is ranked first. Again, the classical approach to assigning probability tells us the probability is 365! divided by 366!, which after simplification is 1/366. Example 3-6 Section In how many ways can 7 different books be arranged on a shelf? Answer We could use the Multiplication Principle to solve this problem. We have seven positions that we can fill with seven books. There are 7 possible books for the first position, 6 possible books for the second position, five possible books for the third position, and so on. The Multiplication Principle tells us therefore that the books can be arranged in: $7\times 6\times 5\times 4\times 3\times 2\times 1$ or 5,040 ways. Alternatively, we can use the simple rule for counting permutations. That is, the number of ways to arrange 7 distinct objects is simply $_7P_7=7!=5040$. Example 3-7 Section With 6 names in a bag, randomly select a name. How many ways can the 6 names be assigned to 6 job assignments? If we assume that each person can only be assigned to one job, then we must select (or sample) the names without replacement. That is, once we select a name, it is set aside and not returned to the bag. Sampling without replacement occurs when an object is not replaced after it has been selected Answer If we sample without replacement, the problem reduces to simply determining the number of ways the 6 names can be arranged. We have 6 objects taken 6 at a time, and hence the number of ways is 6! = 720 possible job assignments. In this case, each person is assigned to one and only one job. What if the 6 names were sampled with replacement? That is, once we select a name, it is returned to the bag. Sampling with replacement occurs when an object is selected and then replaced before the next object has been selected Answer If we sample with replacement, we have 6 choices for each of the 6 jobs. Applying the Multiplication Principle, there are: $6\times 6\times 6\times 6\times 6\times 6=46656$ possible job assignments. In this case, each person is allowed to perform more than one job. There's even the possibility that one (rather unlucky) person gets assigned to all six jobs! The take-home message from this example is that you'll always want to ask yourself whether or not the problem involves sampling with or without replacement. Incidentally, it's not all that different from asking yourself whether or not replication is allowed. Right? Example 3-8 Section Okay, let's throw a (small) wrench into our work. How many ways can 4 people fill 3 chairs? Answer Again, for the sake of concreteness, let's name the four people Tom, Rick, Harry, and Mary and the chairs Left, Middle, and Right. If we fill the Left chair first, there are 4 possible people (Tom, Rick, Harry, and Mary). Let's suppose Tom is selected for the Left chair. Then, since Tom can't sit in more than one chair at a time when we fill the Middle chair, there are only 3 possible people (Rick, Harry, and Mary). If Rick is selected for the Middle chair, when we fill the Right chair, there are only 2 possible people (Harry and Mary). Putting all of this together, the Multiplication Principle tells us that there are: $4\times 3\times 2$ or 24 possible ways to fill the three chairs. Okay, okay! The main distinction between this example and the first example on this page is that the first example involves arranging all 4 people, whereas this example involves leaving one person out and arranging just 3 of the 4 people. This example allows us to introduce another generalization of the Multiplication Principle, namely the counting of the number of permutations of $n$ objects taken $r$ at a time, where $r\le n$. Another Generalization of the Multiplication Principle Section Suppose there are $r$ positions to be filled with $n$ different objects, in which there are: $n$ choices for the 1st position $n-1$ choices for the 2nd position $n-2$ choices for the 3rd position ... and ... $n-(r-1)$ choices for the last position The Multiplication Principle tells us there are in general: $n\times (n-1)\times (n-2)\times \ldots \times [n-(r-1)]$ ways of filling the $r$ positions. We can easily show that, in general, this quantity equals: $\dfrac{n!}{(n-r)!}$ Here's how it works: And, formally: Permutation of $n$ objects taken $r$ at a time ordered arrangement of $n$ different objects in $r$ positions. The number of such permutations is:$_nP_r=\dfrac{n!}{(n-r)!}$ The subscripted $n$ represents the number of objects you are wanting to arrange, while the subscripted $r$ represents the number of positions you have for the objects to fill. Example 3-9 Section An artist has 9 paintings. How many ways can he hang 4 paintings side-by-side on a gallery wall? « Previous3.1 - The Multiplication Principle Next3.3 - Combinations » Lesson Welcome to STAT 414! Section 1: Introduction to Probability Lesson 1: The Big Picture 1.1 - Some Research Questions 1.2 - Populations and Random Samples 1.3 - Sample Spaces 1.4 - Types of data 1.5 - Summarizing Quantitative Data Graphically Lesson 2: Properties of Probability 2.1 - Why Probability? 2.2 - Events 2.3 - What is Probability (Informally)? 2.4 - How to Assign Probability to Events 2.5 - What is Probability (Formally)? 2.6 - Five Theorems 2.7 - Some Examples Lesson 3: Counting Techniques 3.1 - The Multiplication Principle 3.2 - Permutations 3.3 - Combinations 3.4 - Distinguishable Permutations 3.5 - More Examples Lesson 4: Conditional Probability 4.1 - The Motivation 4.2 - What is Conditional Probability? 4.3 - Multiplication Rule 4.4 - More Examples Lesson 5: Independent Events 5.1 - Two Definitions 5.2 - Three Theorems 5.3 - Mutual Independence 5.4 - A Closing Example Lesson 6: Bayes' Theorem 6.1 - An Example 6.2 - A Generalization 6.3 - Another Example 6.4 - More Examples Section 2: Discrete Distributions Lesson 7: Discrete Random Variables 7.1 - Discrete Random Variables 7.2 - Probability Mass Functions 7.3 - The Cumulative Distribution Function (CDF) 7.4 - Hypergeometric Distribution 7.5 - More Examples Lesson 8: Mathematical Expectation 8.1 - A Definition 8.2 - Properties of Expectation 8.3 - Mean of X 8.4 - Variance of X 8.5 - Sample Means and Variances Lesson 9: Moment Generating Functions 9.1 - What is an MGF? 9.2 - Finding Moments 9.3 - Finding Distributions 9.4 - Moment Generating Functions Lesson 10: The Binomial Distribution 10.1 - The Probability Mass Function 10.2 - Is X Binomial? 10.3 - Cumulative Binomial Probabilities 10.4 - Effect of n and p on Shape 10.5 - The Mean and Variance Lesson 11: Geometric and Negative Binomial Distributions 11.1 - Geometric Distributions 11.2 - Key Properties of a Geometric Random Variable 11.3 - Geometric Examples 11.4 - Negative Binomial Distributions 11.5 - Key Properties of a Negative Binomial Random Variable 11.6 - Negative Binomial Examples Lesson 12: The Poisson Distribution 12.1 - Poisson Distributions 12.2 - Finding Poisson Probabilities 12.3 - Poisson Properties 12.4 - Approximating the Binomial Distribution Section 3: Continuous Distributions Lesson 13: Exploring Continuous Data 13.1 - Histograms 13.2 - Stem-and-Leaf Plots 13.3 - Order Statistics and Sample Percentiles 13.4 - Box Plots 13.5 - Shapes of distributions Lesson 14: Continuous Random Variables 14.1 - Probability Density Functions 14.2 - Cumulative Distribution Functions 14.3 - Finding Percentiles 14.4 - Special Expectations 14.5 - Piece-wise Distributions and other Examples 14.6 - Uniform Distributions 14.7 - Uniform Properties 14.8 - Uniform Applications Lesson 15: Exponential, Gamma and Chi-Square Distributions 15.1 - Exponential Distributions 15.2 - Exponential Properties 15.3 - Exponential Examples 15.4 - Gamma Distributions 15.5 - The Gamma Function 15.6 - Gamma Properties 15.7 - A Gamma Example 15.8 - Chi-Square Distributions 15.9 - The Chi-Square Table 15.10 - Trick To Avoid Integration Lesson 16: Normal Distributions 16.1 - The Distribution and Its Characteristics 16.2 - Finding Normal Probabilities 16.3 - Using Normal Probabilities to Find X 16.4 - Normal Properties 16.5 - The Standard Normal and The Chi-Square 16.6 - Some Applications Section 4: Bivariate Distributions Lesson 17: Distributions of Two Discrete Random Variables 17.1 - Two Discrete Random Variables 17.2 - A Triangular Support 17.3 - The Trinomial Distribution Lesson 18: The Correlation Coefficient 18.1 - Covariance of X and Y 18.2 - Correlation Coefficient of X and Y 18.3 - Understanding Rho 18.4 - More on Understanding Rho Lesson 19: Conditional Distributions 19.1 - What is a Conditional Distribution? 19.2 - Definitions 19.3 - Conditional Means and Variances Lesson 20: Distributions of Two Continuous Random Variables 20.1 - Two Continuous Random Variables 20.2 - Conditional Distributions for Continuous Random Variables Lesson 21: Bivariate Normal Distributions 21.1 - Conditional Distribution of Y Given X 21.2 - Joint P.D.F. of X and Y Section 5: Distributions of Functions of Random Variables Lesson 22: Functions of One Random Variable 22.1 - Distribution Function Technique 22.2 - Change-of-Variable Technique 22.3 - Two-to-One Functions 22.4 - Simulating Observations Lesson 23: Transformations of Two Random Variables 23.1 - Change-of-Variables Technique 23.2 - Beta Distribution 23.3 - F Distribution Lesson 24: Several Independent Random Variables 24.1 - Some Motivation 24.2 - Expectations of Functions of Independent Random Variables 24.3 - Mean and Variance of Linear Combinations 24.4 - Mean and Variance of Sample Mean 24.5 - More Examples Lesson 25: The Moment-Generating Function Technique 25.1 - Uniqueness Property of M.G.F.s 25.2 - M.G.F.s of Linear Combinations 25.3 - Sums of Chi-Square Random Variables Lesson 26: Random Functions Associated with Normal Distributions 26.1 - Sums of Independent Normal Random Variables 26.2 - Sampling Distribution of Sample Mean 26.3 - Sampling Distribution of Sample Variance 26.4 - Student's t Distribution Lesson 27: The Central Limit Theorem 27.1 - The Theorem 27.2 - Implications in Practice 27.3 - Applications in Practice Lesson 28: Approximations for Discrete Distributions 28.1 - Normal Approximation to Binomial 28.2 - Normal Approximation to Poisson × Save changes Close OPEN.ED@PSUExcept where otherwise noted, content on this site is licensed under a CC BY-NC 4.0 license. 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8294	https://chem-textbook.ucalgary.ca/version2/chapter-x-gases/the-boltzmann-distribution-and-gases/temperature-kinetic-energy-and-distribution-of-speeds/	Temperature, Kinetic Energy, and Distribution of Speeds - UCalgary Chemistry Textbook Skip to the content UCalgary Chemistry Textbook UCalgary Chemistry Textbook ☰ Menu Chemistry Open Textbook (Version 2) Review of Background Topics Chapter 1: Electronic Structure & Properties of Elements Chapter 2: Molecular Structures in 2D Chapter 3: Molecular Structures in 3D Chapter 4: Advanced Bonding Theories Chapter 5: Collections of Chemical Species Chapter 6: Gases Chapter 7: Chemical Kinetics Chapter 8: Chemical Equilibrium Chapter 9: Acid-Base Equilibria Chapter 10: Solubility Chapter 11: Energy and Thermodynamics Chapter 12: Electrochemistry Search for: Home → 6.3 The Boltzmann Distribution and Gases → Temperature, Kinetic Energy, and Distribution of Speeds Temperature, Kinetic Energy, and Distribution of Speeds [x] As we saw on the previous page, the particles in a gas sample will be moving at a distribution of speeds, with one most-probable (or most common) speed. To describe the effect of temperature on the particles’ motion, we need to know that The Kelvin temperature of a system is directly proportional to the average kinetic energy of that system. We can see this directly in the relation: T=2 3¯K k B T=2 3 K¯k B where: T T is the temperature in Kelvin ¯K K¯ is the average kinetic energy of a particle in the system and k B k B is the Boltzmann constant. Remembering that kinetic energy is defined as K=1 2 m v 2 K=1 2 m v 2, we can temperature to the particle speed: T=m¯v 2 3 k B T=m v¯2 3 k B where ¯v v¯ is the average speed of the particles and m m is the mass (molecular or atomic mass: we are assuming that only one type of particle is present, so there is only one mass). From here, we can see that the temperature is directly related to the square of the average speed: as temperature increases, the average speed of our particles also increases. However, since the collection of particles aren’t identical – remember, they are moving with a distribution of speeds – that doesn’t necessarily mean that every particle moves more quickly. Below, we can see the plots of F(u) – the fraction of molecules at a given speed – at various temperatures for a sample of nitrogen gas (N 2). The Boltzmann-Maxwell distributions for a sample of nitrogen gas at 100, 200, 500, and 1000 K. Investigating this figure, we can notice two important things: First, as the temperature increases, the most-probable (and average) speed increases – in other words, we see the peak (the speed with the biggest fraction of molecules) shift towards higher speeds at higher temperatures. This agrees with what we predicted from the mathematical relations above. Second, as the temperature increases, the distribution doesn’t just shift, it also gets broader. This isn’t obvious from the math above (it comes from the Boltzmann-Maxwell equation on the previous page). But, it is an important observation! At 100K, most of our molecules have a fairly similar set of speeds – they’re all gathered together in a few high-frequency zones, and the “tail” off to higher speeds is relatively small. At 1000 K, we still have some molecules moving at a lower speed, but also more at the higher speeds, so the distribution gets spread out (and “shorter” – spreading the same number of molecules out across more speeds means that our largest fraction isn’t as large as it was before). Knowing how the motion of the particles in a gas sample changes with temperature will help understand how temperature changes reaction rates and equilibria in coming chapters. This text contains content from OpenStax Chemsitry 2e. Chemistry 2e by OpenStax is licensed under Creative Commons Attribution License v4.0. Download for free here. This adaptation has been modified and added to by Drs. Erin Sullivan, Amanda Musgrove (UCalgary) & Erika Merschrod (MUN) along with many student team members. Find a typo or issue with this edition of the textbook? We are continuously editing and updating the site: please click hereto give us your feedback. UCalgary Chem Textbook Team - CC-BY-4.0
8295	https://www.youtube.com/watch?v=MkpbtCRwcCE	Converting percents to decimals \| Decimals \| Pre-Algebra \| Khan Academy Khan Academy 9090000 subscribers 329 likes Description 455143 views Posted: 4 Feb 2012 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: Remember that "per cent" means "per hundred." Knowing that, converting a percent to a decimal is done by dividing by a 100. We'll demonstrate. Practice this lesson yourself on KhanAcademy.org right now: Watch the next lesson: Missed the previous lesson? Pre-Algebra on Khan Academy: No way, this isn't your run of the mill arithmetic. This is Pre-algebra. You're about to play with the professionals. Think of pre-algebra as a runway. You're the airplane and algebra is your sunny vacation destination. Without the runway you're not going anywhere. Seriously, the foundation for all higher mathematics is laid with many of the concepts that we will introduce to you here: negative numbers, absolute value, factors, multiples, decimals, and fractions to name a few. So buckle up and move your seat into the upright position. We're about to take off! About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to KhanAcademy’s Pre-Algebra channel:: Subscribe to KhanAcademy: 31 comments Transcript: Let's see if we can write 59.2%, if we can write this as a decimal. Well, 59.2%, this literally means 59.2 per cent, which literally means 59.2 per 100, which literally means 59.2 per 100, which is the same thing as 59.2 divided by 100. And If we were to divide 59.2 by 100, what would we get? So this is the same thing is 59.2 divided by 100. So if we were to take 59.2, if we wanted to divide it by 10, we would move the decimal over to the left once. We would get 5.92. But we want to divide it by 100, so we move the decimal over to the left again. We're dividing by 10 once, and then we're dividing by 10 again. Divide by 10 twice, you're essentially dividing by 100. So you move the decimal to the left twice, you get 0.592. And we'll just put a 0 out here just to clarify things. And we're done. We've written 59.2% as a decimal. We essentially just divided this by 100, which is dividing by 10 twice, moving the decimal once to the left, and then another time. Or since we're dividing by 100, moving the decimal twice to the left. And we get 0.592.
8296	https://home.iitk.ac.in/~psraj/mth101/lecture_notes/Lecture30.pdf	1 Lecture 30 : Maxima, Minima, Second Derivative Test In calculus of single variable we applied the Bolzano-Weierstrass theorem to prove the existence of maxima and minima of a continuous function on a closed bounded interval. Moreover, we developed ﬁrst and second derivative tests for local maxima and minima. In this lecture we will see a similar theory for functions of several variables. Deﬁnition : A non-empty subset D of Rn is said to be closed if a sequence in D converges then its limit point lies in D. For example, the sets B1 = {X0 ∈R2 : ∥X ∥≤1} and H = {(x, y) ∈R2 : x ≥0, y ≥0} are closed subsets of R2. However, the sets S1 = {X ∈R2 : ∥X ∥< 1} and H+ = {(x, y) ∈R2 : x > 0, y > 0} are not closed. Deﬁnition : Let D ⊆Rn and X0 ∈D. We say that X0 is an interior point of D if there exists r > 0 such that the neighborhood Nr(X0) = {X ∈Rn : ∥X0 −X ∥< r} is contained in D. For example, all the points of S1 are interior points of B1. Similarly, all the points of H+ are interior points of H. The notions of maxima, minima, local maxima and local minima are similar to the ones deﬁned for the functions of one variable. The proof of the following theorem is similar to the proof of the existence of maximum and minimum of a continuous function on a closed bounded interval. Theorem 30.1(Existence of Maxima and Minima): Let D be a closed and bounded subset of R2 and f : D →R be continuous. Then f has a maximum and a minimum in D. Theorem 30.2(Necessary Condition for Local Maximum and Minimum): Suppose D ⊆ R2, f : D →R and (x0, y0) is an interior point of D. Let fx and fy exist at the point (x0, y0). If f has a local maximum or local minimum at (x0, y0) then fx(x0, y0) = fy(x0, y0) = 0. Proof : Note that (the one variable) functions f(x, y0) and f(x0, y) have local maximum or minimum at x0 and y0 respectively. Therefore, the derivatives of these functions are zero at x0 and y0 respectively. That is, fx(x0, y0) = fy(x0, y0) = 0. □ Note that the conditions given in the previous results are not suﬃcient. For example, consider the function f : R2 →R deﬁned by f(x, y) = xy. Note that fx(0, 0) = fy(0, 0) = 0 but (0, 0) is neither a local minimum nor a local maximum for f. Second Derivative Test for Local Maximum and Local Minimum : Suppose D ⊆R2 and f : D →R. Suppose fx and fy are continuous and they have continuous partial derivatives on D. With these assumptions we prove the following result. Theorem 30.3: Let (x0, y0) be an interior point of D and fx(x0, y0) = fy(x0, y0) = 0. Suppose (fxxfyy −f2 xy)(x0, y0) > 0. Then (i) if fxx(x0, y0) > 0 then f has a local minimum at (x0, y0). (ii) if fxx(x0, y0) < 0 then f has a local maximum at (x0, y0). Proof (): We prove (i) and the proof of (ii) is similar. Suppose fxx(x0, y0) > 0 and (fxxfyy − f2 xy)(x0, y0) > 0. Then there exists a neighborhood N of (x0, y0), such that fxx(x, y) > 0 and (fxxfyy −f2 xy)(x, y) > 0 for all (x, y) ∈N. Let (x0 + h, y0 + k) ∈N. Them by the Extended MVT (applying over N, which is possible), there 2 exists some C lying in the line joining (x0 + h, y0 + k) and (x0, y0) such that f(x0 + h, y0 + k) −f(x0, y0) = Q(C) = 1 2(h2fxx + 2hkfxy + k2fyy)(C). Note that 2fxx(C)Q(C) = {(hfxx + kfxy)(C)}2 + k2(fxxfyy −f2 xy)(C) > 0. Since fxx(C) > 0 we have Q(C) > 0 and hence f(x0 + h, y0 + k) > f(x0, y0). Therefore, f has a local minimum at (x0, y0). □ Remarks : 1. If (x0, y0) is an interior point of D, fx(x0, y0) = fy(x0, y0) = 0 and (fxxfyy − f2 xy)(x0, y0) < 0, then one can show that in every neighborhood of (x0, y0) we can ﬁnd two points (x1, y1) and (x2, y2) such that f(x1, y1) > f(x0, y0) and f(x2, y2) < f(x0, y0), that is (x0, y0) is a saddle point. 2. The above test is inconclusive when fx(x0, y0) = fy(x0, y0) = (fxxfyy −f2 xy)(x0, y0) = 0. Examples : 1. The functions f1(x, y) = −(x4 + y4) and f2(x, y) = x4 + y4 satisfy the above equation for (x0, y0) = (0, 0) but f1 has a local maximum at (0, 0) and f2 has a local minimum at (0, 0). 2. Consider the function f(x, y) = (x + y)2 −x4. This function satisﬁes the above equation for (x0, y0) = (0, 0) but it has neither a local maximum nor a local minimum at (0, 0). In fact, (0, 0) is a saddle point. This can be veriﬁed as follows. Note that for 0 < x < 1, f(x, x) > 0 and f(x, −x) < 0. 3. Let f(x, y) = x sin y. Here fx(x0, y0) = fy(x0, y0) = 0 for (x0, y0) = (0, nπ), n ∈N. Note that (fxxfyy −f2 xy)(x0, y0) < 0. Therefore, the points (0, nπ), n ∈N are saddle points. Problem 1: Let f(x, y) = 3x4 −4x2y +y2. Show that f has a local minimum at (0, 0) along every line through (0, 0). Does f have a minimum at (0, 0)? Is (0, 0) a saddle point for f? Solution : Let f(x, y) = 3x4 −4x2y + y2. Along, the x-axis, the local minimum of the function is at (0, 0). Let x = r cos θ and y = r sin θ, for a ﬁxed θ ̸= 0, π (or let y = mx). Then, f(r cos θ, r sin θ) = 3r4 sin4 θ −4r3 cos2 θ sin θ + r2 sin2 θ which is a function of one variable. By the second derivative test (for functions of one variable), we see that (0, 0) is a local minima. Since, f(x, y) = (3x2 −y)(x2 −y), we see that in the region between the parabolas 3x2 = y and y = x2, the function takes negative values and is positive everywhere else. Thus, (0, 0) is a saddle point for f. Problem 2: Let D = [−2, 2] × [−2, 2] and f : D →R be deﬁned as f(x, y) = 4xy −2x2 −y4. Find absolute maxima and absolute minima of f in D. Solution (Hints) : Note that fx(x0, y0) = fy(x0, y0) = 0 for (x0, y0) = (0, 0), (1, 1) or (−1, −1). Since these points lie in the interior of D, these are the candidates for maxima and minima for f on the set of interiors of D. Now we have to check the behavior of the function over the boundary of D. Note that (x, y) ∈D is a boundary point if and only if x = ±2 or y = ±2. So we have to consider the functions f(2, y), f(−2, y), f(x, 2) and f(x, −2) over the interval [−2, 2]. For example, f(2, y) = 8y−8−y4, y ∈ [−2, 2], has absolute maximum at y = 3 √ 2 and absolute minimum at y = −2. So, (2, 3 √ 2) and (2, −2) are the candidates for maxima and minima on the boundary line {(2, y) : y ∈[−2, 2]}. Find all possible candidates for maxima and minima and choose the maxima and minima from these candidates. The absolute maximum value of f on D is 1 which is obtained at (1, 1) and (−1, −1). The absolute minimum value of f on D is -40 and which is obtained at (2, −2) and (−2, 2).
8297	https://en.wikipedia.org/wiki/Triangle_center	Contents Triangle center In geometry, a triangle center or triangle centre is a point in the triangle's plane that is in some sense in the middle of the triangle. For example, the centroid, circumcenter, incenter and orthocenter were familiar to the ancient Greeks, and can be obtained by simple constructions. Each of these classical centers has the property that it is invariant (more precisely equivariant) under similarity transformations. In other words, for any triangle and any similarity transformation (such as a rotation, reflection, dilation, or translation), the center of the transformed triangle is the same point as the transformed center of the original triangle. This invariance is the defining property of a triangle center. It rules out other well-known points such as the Brocard points which are not invariant under reflection and so fail to qualify as triangle centers. For an equilateral triangle, all triangle centers coincide at its centroid. However the triangle centers generally take different positions from each other on all other triangles. The definitions and properties of thousands of triangle centers have been collected in the Encyclopedia of Triangle Centers. History Even though the ancient Greeks discovered the classic centers of a triangle, they had not formulated any definition of a triangle center. After the ancient Greeks, several special points associated with a triangle like the Fermat point, nine-point center, Lemoine point, Gergonne point, and Feuerbach point were discovered. During the revival of interest in triangle geometry in the 1980s it was noticed that these special points share some general properties that now form the basis for a formal definition of triangle center. Clark Kimberling's Encyclopedia of Triangle Centers contains an annotated list of over 50,000 triangle centers. Every entry in the Encyclopedia of Triangle Centers is denoted by X ( n ) {\displaystyle X(n)} or X n {\displaystyle X_{n}} where n {\displaystyle n} is the positional index of the entry. For example, the centroid of a triangle is the second entry and is denoted by X ( 2 ) {\displaystyle X(2)} or X 2 {\displaystyle X_{2}} . Formal definition A real-valued function f of three real variables a, b, c may have the following properties: If a non-zero f has both these properties it is called a triangle center function. If f is a triangle center function and a, b, c are the side-lengths of a reference triangle then the point whose trilinear coordinates are f ( a , b , c ) : f ( b , c , a ) : f ( c , a , b ) {\displaystyle f(a,b,c):f(b,c,a):f(c,a,b)} is called a triangle center. This definition ensures that triangle centers of similar triangles meet the invariance criteria specified above. By convention only the first of the three trilinear coordinates of a triangle center is quoted since the other two are obtained by cyclic permutation of a, b, c. This process is known as cyclicity. Every triangle center function corresponds to a unique triangle center. This correspondence is not bijective. Different functions may define the same triangle center. For example, the functions f 1 ( a , b , c ) = 1 a {\displaystyle f_{1}(a,b,c)={\tfrac {1}{a}}} and f 2 ( a , b , c ) = b c {\displaystyle f_{2}(a,b,c)=bc} both correspond to the centroid. Two triangle center functions define the same triangle center if and only if their ratio is a function symmetric in a, b, c. Even if a triangle center function is well-defined everywhere the same cannot always be said for its associated triangle center. For example, let f ( a , b , c ) {\displaystyle f(a,b,c)} be 0 if ⁠ a b {\displaystyle {\tfrac {a}{b}}} ⁠ and ⁠ a c {\displaystyle {\tfrac {a}{c}}} ⁠ are both rational and 1 otherwise. Then for any triangle with integer sides the associated triangle center evaluates to 0:0:0 which is undefined. Default domain In some cases these functions are not defined on the whole of ⁠ R 3 . {\displaystyle \mathbb {R} ^{3}.} ⁠ For example, the trilinears of X365 which is the 365th entry in the Encyclopedia of Triangle Centers, are a 1 / 2 : b 1 / 2 : c 1 / 2 {\displaystyle a^{1/2}:b^{1/2}:c^{1/2}} so a, b, c cannot be negative. Furthermore, in order to represent the sides of a triangle they must satisfy the triangle inequality. So, in practice, every function's domain is restricted to the region of ⁠ R 3 {\displaystyle \mathbb {R} ^{3}} ⁠ where a ≤ b + c , b ≤ c + a , c ≤ a + b . {\displaystyle a\leq b+c,\quad b\leq c+a,\quad c\leq a+b.} This region T is the domain of all triangles, and it is the default domain for all triangle-based functions. Other useful domains There are various instances where it may be desirable to restrict the analysis to a smaller domain than T. For example: a 2 b 2 + b c + c 2 ; b 2 c 2 + c a + a 2 ; c 2 a 2 + a b + b 2 . {\displaystyle a^{2}>b^{2}+bc+c^{2};\quad b^{2}>c^{2}+ca+a^{2};\quad c^{2}>a^{2}+ab+b^{2}.} Domain symmetry Not every subset D ⊆ T is a viable domain. In order to support the bisymmetry test D must be symmetric about the planes b = c, c = a, a = b. To support cyclicity it must also be invariant under 2π/3 rotations about the line a = b = c. The simplest domain of all is the line (t, t, t) which corresponds to the set of all equilateral triangles. Examples Circumcenter The point of concurrence of the perpendicular bisectors of the sides of triangle △ABC is the circumcenter. The trilinear coordinates of the circumcenter are a ( b 2 + c 2 − a 2 ) : b ( c 2 + a 2 − b 2 ) : c ( a 2 + b 2 − c 2 ) . {\displaystyle a(b^{2}+c^{2}-a^{2}):b(c^{2}+a^{2}-b^{2}):c(a^{2}+b^{2}-c^{2}).} Let f ( a , b , c ) = a ( b 2 + c 2 − a 2 ) {\displaystyle f\left(a,b,c\right)=a\left(b^{2}+c^{2}-a^{2}\right)} It can be shown that f is homogeneous: f ( t a , t b , t c ) = t a [ ( t b ) 2 + ( t c ) 2 − ( t a ) 2 ] = t 3 [ a ( b 2 + c 2 − a 2 ) ] = t 3 f ( a , b , c ) {\displaystyle {\begin{aligned}f(ta,tb,tc)&=ta{\Bigl [}(tb)^{2}+(tc)^{2}-(ta)^{2}{\Bigr ]}\[2pt]&=t^{3}{\Bigl [}a(b^{2}+c^{2}-a^{2}){\Bigr ]}\[2pt]&=t^{3}f(a,b,c)\end{aligned}}} as well as bisymmetric: f ( a , c , b ) = a ( c 2 + b 2 − a 2 ) = a ( b 2 + c 2 − a 2 ) = f ( a , b , c ) {\displaystyle {\begin{aligned}f(a,c,b)&=a(c^{2}+b^{2}-a^{2})\[2pt]&=a(b^{2}+c^{2}-a^{2})\[2pt]&=f(a,b,c)\end{aligned}}} so f is a triangle center function. Since the corresponding triangle center has the same trilinears as the circumcenter, it follows that the circumcenter is a triangle center. 1st isogonic center Let △A'BC be the equilateral triangle having base BC and vertex A' on the negative side of BC and let △AB'C and △ABC' be similarly constructed equilateral triangles based on the other two sides of triangle △ABC. Then the lines AA', BB', CC' are concurrent and the point of concurrence is the 1st isogonal center. Its trilinear coordinates are csc ⁡ ( A + π 3 ) : csc ⁡ ( B + π 3 ) : csc ⁡ ( C + π 3 ) . {\displaystyle \csc \left(A+{\frac {\pi }{3}}\right):\csc \left(B+{\frac {\pi }{3}}\right):\csc \left(C+{\frac {\pi }{3}}\right).} Expressing these coordinates in terms of a, b, c, one can verify that they indeed satisfy the defining properties of the coordinates of a triangle center. Hence the 1st isogonic center is also a triangle center. Fermat point Let Then f is bisymmetric and homogeneous so it is a triangle center function. Moreover, the corresponding triangle center coincides with the obtuse angled vertex whenever any vertex angle exceeds 2π/3, and with the 1st isogonic center otherwise. Therefore, this triangle center is none other than the Fermat point. Non-examples Brocard points The trilinear coordinates of the first Brocard point are: c b : a c : b a {\displaystyle {\frac {c}{b}}\ :\ {\frac {a}{c}}\ :\ {\frac {b}{a}}} These coordinates satisfy the properties of homogeneity and cyclicity but not bisymmetry. So the first Brocard point is not (in general) a triangle center. The second Brocard point has trilinear coordinates: b c : c a : a b {\displaystyle {\frac {b}{c}}\ :\ {\frac {c}{a}}\ :\ {\frac {a}{b}}} and similar remarks apply. The first and second Brocard points are one of many bicentric pairs of points, pairs of points defined from a triangle with the property that the pair (but not each individual point) is preserved under similarities of the triangle. Several binary operations, such as midpoint and trilinear product, when applied to the two Brocard points, as well as other bicentric pairs, produce triangle centers. Some well-known triangle centers Classical triangle centers ETC reference;Name; Symbol \| Trilinear coordinates \| Description X1 \| Incenter \| I \| 1:1:1{\displaystyle 1:1:1} \| Intersection of theangle bisectors. Center of the triangle'sinscribed circle. X2 \| Centroid \| G \| bc:ca:ab{\displaystyle bc:ca:ab} \| Intersection of themedians.Center of massof a uniform triangularlamina. X3 \| Circumcenter \| O \| cos⁡A:cos⁡B:cos⁡C{\displaystyle \cos A:\cos B:\cos C} \| Intersection of theperpendicular bisectorsof the sides. Center of the triangle'scircumscribed circle. X4 \| Orthocenter \| H \| sec⁡A:sec⁡B:sec⁡C{\displaystyle \sec A:\sec B:\sec C} \| Intersection of thealtitudes. X5 \| Nine-point center \| N \| cos⁡(B−C):cos⁡(C−A):cos⁡(A−B){\displaystyle \cos(B-C):\cos(C-A):\cos(A-B)} \| Center of the circle passing through the midpoint of each side, the foot of each altitude, and the midpoint between the orthocenter and each vertex. X6 \| Symmedian point \| K \| a:b:c{\displaystyle a:b:c} \| Intersection of the symmedians – the reflection of each median about the corresponding angle bisector. X7 \| Gergonne point \| Ge \| bcb+c−a:cac+a−b:aba+b−c{\displaystyle {\frac {bc}{b+c-a}}:{\frac {ca}{c+a-b}}:{\frac {ab}{a+b-c}}} \| Intersection of the lines connecting each vertex to the point where the incircle touches the opposite side. X8 \| Nagel point \| Na \| b+c−aa:c+a−bb:a+b−cc{\displaystyle {\frac {b+c-a}{a}}:{\frac {c+a-b}{b}}:{\frac {a+b-c}{c}}} \| Intersection of the lines connecting each vertex to the point where an excircle touches the opposite side. X9 \| Mittenpunkt \| M \| (b+c−a):(c+a−b):(a+b−c){\displaystyle (b+c-a):(c+a-b):(a+b-c)} \| Symmedianpoint of theexcentral triangle(and various equivalent definitions). X10 \| Spieker center \| Sp \| bc(b+c):ca(c+a):ab(a+b){\displaystyle bc(b+c):ca(c+a):ab(a+b)} \| Incenter of the medial triangle. Center of mass of a uniform triangular wireframe. X11 \| Feuerbach point \| F \| 1−cos⁡(B−C):1−cos⁡(C−A):1−cos⁡(A−B){\displaystyle 1-\cos(B-C):1-\cos(C-A):1-\cos(A-B)} \| Point at which the nine-point circle is tangent to the incircle. X13 \| Fermat point \| X \| csc⁡(A+π3):csc⁡(B+π3):csc⁡(C+π3).{\displaystyle \csc(A+{\tfrac {\pi }{3}}):\csc(B+{\tfrac {\pi }{3}}):\csc(C+{\tfrac {\pi }{3}}).}[a] \| Point that is the smallest possible sum of distances from the vertices. X15X16 \| Isodynamic points \| SS′ \| sin⁡(A+π3):sin⁡(B+π3):sin⁡(C+π3)sin⁡(A−π3):sin⁡(B−π3):sin⁡(C−π3){\displaystyle {\begin{aligned}\sin(A+{\tfrac {\pi }{3}}):\sin(B+{\tfrac {\pi }{3}}):\sin(C+{\tfrac {\pi }{3}})\\sin(A-{\tfrac {\pi }{3}}):\sin(B-{\tfrac {\pi }{3}}):\sin(C-{\tfrac {\pi }{3}})\end{aligned}}} \| Centers ofinversionthat transform the triangle into an equilateral triangle. X17X18 \| Napoleon points \| NN′ \| sec⁡(A−π3):sec⁡(B−π3):sec⁡(C−π3)sec⁡(A+π3):sec⁡(B+π3):sec⁡(C+π3){\displaystyle {\begin{aligned}\sec(A-{\tfrac {\pi }{3}}):\sec(B-{\tfrac {\pi }{3}}):\sec(C-{\tfrac {\pi }{3}})\\sec(A+{\tfrac {\pi }{3}}):\sec(B+{\tfrac {\pi }{3}}):\sec(C+{\tfrac {\pi }{3}})\end{aligned}}} \| Intersection of the lines connecting each vertex to the center of an equilateral triangle pointed outwards (first Napoleon point) or inwards (second Napoleon point), mounted on the opposite side. X99 \| Steiner point \| S \| bcb2−c2:cac2−a2:aba2−b2{\displaystyle {\frac {bc}{b^{2}-c^{2}}}:{\frac {ca}{c^{2}-a^{2}}}:{\frac {ab}{a^{2}-b^{2}}}} \| Various equivalent definitions. Recent triangle centers In the following table of more recent triangle centers, no specific notations are mentioned for the various points. Also for each center only the first trilinear coordinate f(a,b,c) is specified. The other coordinates can be easily derived using the cyclicity property of trilinear coordinates. ETC reference; Name \| Center functionf(a,b,c){\displaystyle f(a,b,c)} \| Year described X21 \| Schiffler point \| 1cos⁡B+cos⁡C{\displaystyle {\frac {1}{\cos B+\cos C}}} \| 1985 X22 \| Exeter point \| a(b4+c4−a4){\displaystyle a(b^{4}+c^{4}-a^{4})} \| 1986 X111 \| Parry point \| a2a2−b2−c2{\displaystyle {\frac {a}{2a^{2}-b^{2}-c^{2}}}} \| early 1990s X173 \| Congruent isoscelizers point \| tan⁡A2+sec⁡A2{\displaystyle \tan {\tfrac {A}{2}}+\sec {\tfrac {A}{2}}} \| 1989 X174 \| Yff center of congruence \| sec⁡A2{\displaystyle \sec {\tfrac {A}{2}}} \| 1987 X175 \| Isoperimetric point \| sec⁡A2cos⁡B2cos⁡C2−1{\displaystyle \sec {\tfrac {A}{2}}\cos {\tfrac {B}{2}}\cos {\tfrac {C}{2}}-1} \| 1985 X179 \| First Ajima-Malfatti point \| sec4⁡A4{\displaystyle \sec ^{4}{\tfrac {A}{4}}} \| X181 \| Apollonius point \| a(b+c)2b+c−a{\displaystyle {\frac {a(b+c)^{2}}{b+c-a}}} \| 1987 X192 \| Equal parallelians point \| bc(ca+ab−bc){\displaystyle bc(ca+ab-bc)} \| 1961 X356 \| Morley center \| cos⁡A3+2cos⁡B3cos⁡C3{\displaystyle \cos {\tfrac {A}{3}}+2\cos {\tfrac {B}{3}}\cos {\tfrac {C}{3}}} \| 1978 X360 \| Hofstadter zero point \| Aa{\displaystyle {\frac {A}{a}}} \| 1992 General classes of triangle centers Kimberling center In honor of Clark Kimberling who created the online encyclopedia of more than 32,000 triangle centers, the triangle centers listed in the encyclopedia are collectively called Kimberling centers. Polynomial triangle center A triangle center P is called a polynomial triangle center if the trilinear coordinates of P can be expressed as polynomials in a, b, c. Regular triangle center A triangle center P is called a regular triangle point if the trilinear coordinates of P can be expressed as polynomials in △, a, b, c, where △ is the area of the triangle. Major triangle center A triangle center P is said to be a major triangle center if the trilinear coordinates of P can be expressed in the form f ( A ) : f ( B ) : f ( C ) {\displaystyle f(A):f(B):f(C)} where ⁠ f ( X ) {\displaystyle f(X)} ⁠ is a function of the angle X alone and does not depend on the other angles or on the side lengths. Transcendental triangle center A triangle center P is called a transcendental triangle center if P has no trilinear representation using only algebraic functions of a, b, c. Miscellaneous Isosceles and equilateral triangles Let f be a triangle center function. If two sides of a triangle are equal (say a = b) then f ( a , b , c ) = f ( b , a , c ) ( since a b ) = f ( b , c , a ) (by bisymmetry) {\displaystyle {\begin{aligned}f(a,b,c)&=f(b,a,c)&({\text{since }}a=b)\&=f(b,c,a)&{\text{(by bisymmetry)}}\end{aligned}}} so two components of the associated triangle center are always equal. Therefore, all triangle centers of an isosceles triangle must lie on its line of symmetry. For an equilateral triangle all three components are equal so all centers coincide with the centroid. So, like a circle, an equilateral triangle has a unique center. Excenters Let f ( a , b , c ) = { − 1 if a ≥ b and a ≥ c , 1 otherwise . {\displaystyle f(a,b,c)={\begin{cases}-1&\quad {\text{if }}a\geq b{\text{ and }}a\geq c,\\;\;\;1&\quad {\text{otherwise}}.\end{cases}}} This is readily seen to be a triangle center function and (provided the triangle is scalene) the corresponding triangle center is the excenter opposite to the largest vertex angle. The other two excenters can be picked out by similar functions. However, as indicated above only one of the excenters of an isosceles triangle and none of the excenters of an equilateral triangle can ever be a triangle center. Biantisymmetric functions A function f is biantisymmetric if f ( a , b , c ) = − f ( a , c , b ) for all a , b , c . {\displaystyle f(a,b,c)=-f(a,c,b)\quad {\text{for all}}\quad a,b,c.} If such a function is also non-zero and homogeneous it is easily seen that the mapping ( a , b , c ) → f ( a , b , c ) 2 f ( b , c , a ) f ( c , a , b ) {\displaystyle (a,b,c)\to f(a,b,c)^{2}\,f(b,c,a)\,f(c,a,b)} is a triangle center function. The corresponding triangle center is f ( a , b , c ) : f ( b , c , a ) : f ( c , a , b ) . {\displaystyle f(a,b,c):f(b,c,a):f(c,a,b).} On account of this the definition of triangle center function is sometimes taken to include non-zero homogeneous biantisymmetric functions. New centers from old Any triangle center function f can be normalized by multiplying it by a symmetric function of a, b, c so that n = 0. A normalized triangle center function has the same triangle center as the original, and also the stronger property that f ( t a , t b , t c ) = f ( a , b , c ) for all t 0 , ( a , b , c ) . {\displaystyle f(ta,tb,tc)=f(a,b,c)\quad {\text{for all}}\quad t>0,\ (a,b,c).} Together with the zero function, normalized triangle center functions form an algebra under addition, subtraction, and multiplication. This gives an easy way to create new triangle centers. However distinct normalized triangle center functions will often define the same triangle center, for example f and ( a b c ) − 1 ( a + b + c ) 3 f . {\displaystyle (abc)^{-1}(a+b+c)^{3}f.} Uninteresting centers Assume a, b, c are real variables and let α, β, γ be any three real constants. Let f ( a , b , c ) = { α if a < b and a < c ( a is least ) , γ if a b and a c ( a is greatest ) , β otherwise ( a is in the middle ) . {\displaystyle f(a,b,c)={\begin{cases}\alpha &\quad {\text{if }}ab{\text{ and }}a>c&(a{\text{ is greatest}}),\[2pt]\beta &\quad {\text{otherwise}}&(a{\text{ is in the middle}}).\end{cases}}} Then f is a triangle center function and α : β : γ is the corresponding triangle center whenever the sides of the reference triangle are labelled so that a < b < c. Thus every point is potentially a triangle center. However the vast majority of triangle centers are of little interest, just as most continuous functions are of little interest. Barycentric coordinates If f is a triangle center function then so is af and the corresponding triangle center is a f ( a , b , c ) : b f ( b , c , a ) : c f ( c , a , b ) . {\displaystyle a\,f(a,b,c):b\,f(b,c,a):c\,f(c,a,b).} Since these are precisely the barycentric coordinates of the triangle center corresponding to f it follows that triangle centers could equally well have been defined in terms of barycentrics instead of trilinears. In practice it isn't difficult to switch from one coordinate system to the other. Binary systems There are other center pairs besides the Fermat point and the 1st isogonic center. Another system is formed by X3 and the incenter of the tangential triangle. Consider the triangle center function given by: f ( a , b , c ) = { cos ⁡ A if △ is acute , cos ⁡ A + sec ⁡ B sec ⁡ C if ∡ A is obtuse , cos ⁡ A − sec ⁡ A if either ∡ B or ∡ C is obtuse . {\displaystyle f(a,b,c)={\begin{cases}\cos A&{\text{if }}\triangle {\text{ is acute}},\[2pt]\cos A+\sec B\sec C&{\text{if }}\measuredangle A{\text{ is obtuse}},\[2pt]\cos A-\sec A&{\text{if either}}\measuredangle B{\text{ or }}\measuredangle C{\text{ is obtuse}}.\end{cases}}} For the corresponding triangle center there are four distinct possibilities: if reference △ is acute: cos ⁡ A : cos ⁡ B : cos ⁡ C if ∡ A is obtuse: cos ⁡ A + sec ⁡ B sec ⁡ C : cos ⁡ B − sec ⁡ B : cos ⁡ C − sec ⁡ C if ∡ B is obtuse: cos ⁡ A − sec ⁡ A : cos ⁡ B + sec ⁡ C sec ⁡ A : cos ⁡ C − sec ⁡ C if ∡ C is obtuse: cos ⁡ A − sec ⁡ A : cos ⁡ B − sec ⁡ B : cos ⁡ C + sec ⁡ A sec ⁡ B {\displaystyle {\begin{aligned}&{\text{if reference }}\triangle {\text{ is acute:}}\quad \cos A\ :\,\cos B\ :\,\cos C\[6pt]&{\begin{array}{rcccc}{\text{if }}\measuredangle A{\text{ is obtuse:}}&\cos A+\sec B\sec C&:&\cos B-\sec B&:&\cos C-\sec C\[4pt]{\text{if }}\measuredangle B{\text{ is obtuse:}}&\cos A-\sec A&:&\cos B+\sec C\sec A&:&\cos C-\sec C\[4pt]{\text{if }}\measuredangle C{\text{ is obtuse:}}&\cos A-\sec A&:&\cos B-\sec B&:&\cos C+\sec A\sec B\end{array}}\end{aligned}}} Note that the first is also the circumcenter. Routine calculation shows that in every case these trilinears represent the incenter of the tangential triangle. So this point is a triangle center that is a close companion of the circumcenter. Bisymmetry and invariance Reflecting a triangle reverses the order of its sides. In the image the coordinates refer to the (c, b, a) triangle and (using "\|" as the separator) the reflection of an arbitrary point γ : β : α {\displaystyle \gamma :\beta :\alpha } is γ \| β \| α . {\displaystyle \gamma \ \|\ \beta \ \|\ \alpha .} If f is a triangle center function the reflection of its triangle center is f ( c , a , b ) \| f ( b , c , a ) \| f ( a , b , c ) , {\displaystyle f(c,a,b)\ \|\ f(b,c,a)\ \|\ f(a,b,c),} which, by bisymmetry, is the same as f ( c , b , a ) \| f ( b , a , c ) \| f ( a , c , b ) . {\displaystyle f(c,b,a)\ \|\ f(b,a,c)\ \|\ f(a,c,b).} As this is also the triangle center corresponding to f relative to the (c, b, a) triangle, bisymmetry ensures that all triangle centers are invariant under reflection. Since rotations and translations may be regarded as double reflections they too must preserve triangle centers. These invariance properties provide justification for the definition. Alternative terminology Some other names for dilation are uniform scaling, isotropic scaling, homothety, and homothecy. Non-Euclidean and other geometries The study of triangle centers traditionally is concerned with Euclidean geometry, but triangle centers can also be studied in non-Euclidean geometry. Triangle centers that have the same form for both Euclidean and hyperbolic geometry can be expressed using gyrotrigonometry. In non-Euclidean geometry, the assumption that the interior angles of the triangle sum to 180 degrees must be discarded. Centers of tetrahedra or higher-dimensional simplices can also be defined, by analogy with 2-dimensional triangles. Some centers can be extended to polygons with more than three sides. The centroid, for instance, can be found for any polygon. Some research has been done on the centers of polygons with more than three sides. See also Notes External links
8298	https://byjus.com/chemistry/heat-capacity-cp-cv-relation/	What is Heat Capacity? When heat is absorbed by a body, the temperature of the body increases and when heat is lost, the temperature decreases. The temperature of an object is the measure of the total kinetic energy of the particles that make up that object. So when heat is absorbed by an object this heat gets translated into the kinetic energy of the particles and as a result the temperature increases. Thus, the change in temperature is proportional to the heat transfer. The formula q = n C ∆T represents the heat q required to bring about a ∆T difference in temperature of one mole of any matter. The constant C here is called the molar heat capacity of the body. Thus, the molar heat capacity of any substance is defined as the amount of heat energy required to change the temperature of 1 mole of that substance by 1 unit. It depends on the nature, size, and composition of the system. In this article, we will discuss two types of molar heat capacity – CP and CV and derive a relationship between Cp and Cv. What are Heat Capacity C, CP, and CV? The molar heat capacity C, at constant pressure, is represented by CP. At constant volume, the molar heat capacity C is represented by CV. In the following section, we will find how CP and CV are related, to an ideal gas. The relationship between CP and CV for an Ideal Gas From the equation q = n C ∆T, we can say: At constant pressure P, we have qP = n CP∆T This value is equal to the change in enthalpy, that is, qP = n CP∆T = ∆H Similarly, at constant volume V, we have qV = n CV∆T This value is equal to the change in internal energy, that is, qV = n CV∆T = ∆U We know that for one mole (n=1) of an ideal gas, ∆H = ∆U + ∆(pV )= ∆U + ∆(RT) = ∆U + R ∆T Therefore, ∆H = ∆U + R ∆T Substituting the values of ∆H and ∆U from above in the former equation, CP∆T = CV∆T + R ∆T CP = CV + R CP – CV = R To learn more about thermodynamics and heat transfer, download BYJU’S – The Learning App. Take up a quiz on Heat Capacity - Relationship between Cp and Cv Q5 Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Congrats! Visit BYJU’S for all Chemistry related queries and study materials Your result is as below 0 out of 0 arewrong 0 out of 0 are correct 0 out of 0 are Unattempted Login To View Results Did not receive OTP? Request OTP on Login To View Results Comments Leave a Comment Cancel reply Stephen Adjei October 26, 2019 at 9:20 pm very helpful Reply tera baaap hu March 25, 2021 at 10:28 pm helpful Reply Vivek July 23, 2022 at 2:31 pm Very helpful thanks a lot Reply Vivek July 23, 2022 at 2:31 pm Very useful for me thanks Reply Register with BYJU'S & Download Free PDFs
8299	https://www.sciencedirect.com/science/article/pii/S1574013724000789	Non-square grids: A new trend in imaging and modeling? - ScienceDirect Typesetting math: 100% Skip to main contentSkip to article Journals & Books ViewPDF Download full issue Search ScienceDirect Outline Highlights Abstract Graphical abstract Keywords 1. Introduction 2. Grids tessellating the plane 3. Grids for images 4. Grids in geographic information systems 5. Summary and future perspectives Declaration of competing interest Data availability References Show full outline Cited by (1) Figures (19) Show 13 more figures Tables (2) Table 1 Table 2 Computer Science Review Volume 56, May 2025, 100695 Review article Non-square grids: A new trend in imaging and modeling? Author links open overlay panelPaola Magillo Show more Outline Add to Mendeley Share Cite rights and content Under a Creative Commons license Open access Highlights •We provide a survey on the use of non-square grids for practical applications. •Such grids, especially hexagonal ones, are gaining interest and diffusion. •In imaging, the theory is well established but lacks supporting hardware. •In geography, hexagonal discrete global grids are successfully used. Abstract The raster format of images and data is commonly intended as a synonymous of a square grid. Indeed, the square is not the only shape that can tessellate the plane. Other grids are well-known, and recently they have moved out of the fields of art and mathematics, and have started being of interest for technological applications. After introducing the main types of non-square grids, this paper presents experiences of practical uses of non-square grids, especially the hexagonal one, in various fields, including digital imaging, geographic systems, and their applications in sciences like medicine, environmental monitoring, etc. We conclude with considerations on the state of the art and perspectives for the future. In our opinion, the research is mature enough to prefigure a broader diffusion of some non-square grids, especially the hexagonal one. Graphical abstract 1. Download: Download high-res image (238KB) 2. Download: Download full-size image Previous article in issue Next article in issue Keywords Non-square grids Hexagonal grid Images Discrete global grid systems Applications 1. Introduction Grids with square elements have been used by humans for discretizing the plane since ancient times (e.g., weaving and carpets). In the computer age, they have become the base for raster imaging and raster modeling of 2D scalar and vector fields. Nature, however, features other grids as well, which can be found, for instance, in honeycombs and crystals. Different grids have been used by humans for decorations (e.g., floor tilings and arabesques) and have been classified by mathematicians. In spite of this, the digital age is dominated by the square grid. Since the 1980’s, non-square grids have deserved the attention of computer scientists, who noticed the good properties of some of them (especially of the hexagonal grid). Slowly, such grids have started to be used in applications, with a recent positive trend in some fields. This is an interesting emerging phenomenon, which may lead to relevant changes in the future. Digital images based on a hexagonal grid show better connectivity structure and isotropy with respect to the square grid. Recently, hexagonal filters and other image processing operations, including the use of neural networks applied to hexagonal images, have shown good results in medical applications and feature recognition. In the field of geographic systems, in the new millennium Discrete Global Grid Systems (DGGS) have emerged as a powerful alternative to traditional cartography. This approach subdivides the entire 3D surface of the Earth into a hierarchy of grids (composed of square, rhombic, triangular, or hexagonal elements) at increasingly finer resolutions. With a DGGS, not only geographically referenced data can be rendered both in three-dimensional view and on the plane, but also calculations can be performed without the distortion introduced by projections, and at a selected level of resolution. This paper provides a survey of experiences in the use of non-square grids (especially the hexagonal one) in several and diverse disciplines. It completes, from an applicative point of view, the recent paper by Nagy , who covered the state of the art on non-square grids from the side of basic mathematical tools (coordinate systems, digital distances, geometric transformations). This paper is not aimed at presenting the current advances and challenges in the considered application fields. Our attention is restricted to the use of non-square grids in them, to show that non-square grids are used in reality, with promising performances and positive trend. This paper is organized as follows. In Section 2, we present non-square grids, their properties, and practical issues for using them in applications. In Sections 3 Grids for images, 4 Grids in geographic information systems, we provide an overview of the practical use of such grids, especially the hexagonal one, in imaging and in geographic applications, respectively. Finally, in Section 5 we try to identify promises and problems for the future. 2. Grids tessellating the plane Grids subdivide the plane and allow passing from a continuous world with real coordinates to a discrete one with integer coordinates. Other terms used to denote grids are tilings, tessellations or mosaics. In a grid, one polygonal shape, also called pixel or tile, is replicated to cover the entire plane with no overlaps and no gaps. Fig. 1 summarizes the requirements for a grid, as described in the following. Usually, edge-to-edge adjacency is required for a grid, i.e., so-called T-junctions are not allowed. This idea is formalized through the concept of cell complex. In R n, a k-dimensional cell (with 0≤k≤n) is a set of points of R n homeomorphic to the closed unit k-dimensional ball {X=(x 1,…,x k) : ‖X‖≤1}. A 2-dimensional cell complex Q is a collection of cells of dimensions 0,1,2, called vertices, edges, faces, respectively, with the following properties: 1.the boundary of each edge and of each face is the union of (lower-dimensional) cells of Q, 2.the intersection of any two cells is either empty, or the union of (lower-dimensional) cells of Q. For a 2-dimensional cell complex to be a grid, the union of its cells (i.e., its tiles) must cover the entire plane. In addition, for supporting a discrete coordinate system, it must be periodical, i.e., the spatial arrangement of cells must repeat throughout the plane. More formally, there must be a class of rigid transformations that map the grid onto itself. A necessary condition for periodicity is that all the faces have the same shape (or few different ones), and all the vertices have the same pattern (or few different patterns) of faces in their neighborhood. Another practical condition is that the tiles are polygons (i.e., edges are straight-line segments), of simple shapes, and convex. Grids with non-convex (e.g., star-shaped) tiles are used for decorations, but cause algorithmic, numerical, and approximation problems in computer processing. In summary, the tiles of a grid must be convex polygons of simple shape and congruent to each other modulo rigid transformations. Topological relations link pairs of cells in a cell complex. Incidence relations link cells of different dimensions: a cell γ 1 is incident into another cell γ 2 if γ 1 is on the boundary of γ 2 (boundary incidence) or vice-versa (co-boundary incidence). Adjacency relations link cells of the same dimension: two cells γ 1 and γ 2 are adjacent if some cell γ 3 exists, which is on the boundary of both. For two faces, we distinguish between edge-adjacency (γ 1 and γ 2 share an incident edge and its endpoint vertices) and vertex-adjacency (γ 1 and γ 2 share a vertex, but not necessarily an edge). 1. Download: Download high-res image (151KB) 2. Download: Download full-size image Fig. 1. None of these patterns satisfies all the conditions for a grid: the left one has not edge-to-edge adjacency and therefore is not a cell complex; the middle one is not periodic, the right one has non-convex tiles. The concept of duality formalizes the idea of another cell complex, which describes the topological structure of a given one. Given a 2-dimensional cell complex Q, its dual Q̄ is another 2-dimensional cell complex, where: 1.every face (edge, vertex) γ of Q corresponds to a vertex (edge, face) γ̄ of Q̄, and vice-versa; 2.if a cell γ 1 is on the boundary of a cell γ 2 in Q, the cell γ̄2 is on the boundary of γ̄1 in Q̄, and vice-versa. The dual grid of a given one can be constructed by placing a vertex in the center of each tile and joining the centers of every two edge-adjacent tiles with an edge. Fig. 2 shows two mutually dual grids. 1. Download: Download high-res image (159KB) 2. Download: Download full-size image Fig. 2. A cell complex with regular octagonal and square faces (left); its dual cell complex composed of right rectangular triangles (middle); and superimposition of the cell complex with its dual (right). 2.1. Regular grids, semi-regular grids, and their duals In regular grids, all tiles are regular polygons with the same edge length. Three regular grids exist, where tiles are squares, hexagons, and triangles, respectively. The square grid is self-dual. The hexagonal grid and the triangular grid are dual to each other. In the square and in the hexagonal grid, any tile is congruent to any other tile modulo translation. In the triangular grid, the tiles have two opposite orientations, and they are congruent to each other modulo translation and rotation of 180 degrees. In the three regular grids, not only all tiles are regular polygons with the same number of edges, but also all vertices have the same configuration of incident polygons (i.e., the dual grid has the same property). If we relax one of these two conditions, we obtain more grids. Such grids have been used in art since ancient times, and were studied by Kepler in the 17th century. In semi-regular grids, all tiles are regular polygons, possibly with different number of edges, and all vertices have the same configuration. Eleven semi-regular grids exist , , including the three regular grids. Note that the semi-regular grids are not the kind of grids we focus on, because their tiles have not the same shape; we present them only for introducing their duals. In the duals of the semi-regular grids, all tiles are congruent to each other (but they are not regular polygons), and not all vertices have the same configuration of incident polygons. Fig. 3 shows the semi-regular grids and their duals. A notation for the semi-regular grids considers the polygons incident to a vertex in radial order, and takes the number of edges of each polygon. The result is a list of numbers modulo cyclic permutation. As in , we enclose such numbers within T(...). The notation for the duals of the semi-regular grids encloses the same numbers within D(...), i.e., the dual of T(a 1…a h) is D(a 1…a h). The notations for the eleven semi-regular grids and their duals are shown in Fig. 3. For more grid types and their properties, see . 1. Download: Download high-res image (723KB) 2. Download: Download full-size image Fig. 3. The eight semi-regular grids that are not regular (in black), the dual grids (in red). Their notations T(...) and D(...), respectively, are shown. 1. Download: Download high-res image (98KB) 2. Download: Download full-size image Fig. 4. Two ways to represent a scalar field based on a grid, a square grid in this example: values at tile centers (left), and values at tile vertices (right). 2.2. Useful properties of grids for applications A grid is used to represent some information, naturally defined pointwise in the plane, in a discrete way. Formally, the grid models a (scalar or vector) field defined over the plane. Traditionally, two approaches are used: 1.Discrete approximation. A value is associated with each tile γ, summarizing the field over the area covered by γ. The field is thus approximated by a piecewise constant function, with discontinuity at tile boundaries (see Fig. 4, left). Examples are the color in an image, population density in cartography, mean temperature in physics (which are visualized as images in pseudocolors). 2.Continuous approximation. A value is associated with each vertex (i.e., each tile of the dual grid) and the field value inside a tile is approximated by interpolating the values at the tile’s vertices (see Fig. 4, right). In this way, a continuous approximation can be provided (note that, for this, the absence of T-junctions is crucial). Triangular and square tiles allow for linear and bi-linear interpolation, respectively. The grid D(4,8,8), shown in Fig. 2(middle), supports linear interpolation because it has triangular tiles, and is applicable to a square sampling pattern because it refines a square grid. A more complete version of the discrete approach associates a value with every cell of the cell complex, i.e., with each tile, edge, and vertex of the grid. This also enables considering simultaneously the grid and its dual (for instance a hexagonal grid and its dual triangular grid), and provides better accuracy in computations , . A grid where all pixels have equal area enables to weight each pixel equally during a simulation (e.g., heat propagation). All dual grids of semi-regular ones have this property, implied by the fact that the tiles have one shape. Among the regular grids, the hexagonal one is the most densely packed, i.e., the one minimizing the tile area for a fixed distance between the centers of two adjacent tiles. Moreover, it minimizes the maximum distance of a point of the tile from the tile center and the maximum distance between two points belonging to the tile (see Table 1). Therefore, the hexagonal grid is likely to minimize the approximation error when the field value sampled at the tile center is assigned to the whole tile, or when the field value at a point inside the tile is interpolated based on the values sampled at its vertices. Uniform connectivity is the property of a grid, where two tiles sharing a vertex always share an entire edge. A related property is isotropy, i.e., equal distance of a tile from its adjacent tiles. The grids whose dual has triangular tiles have uniform connectivity: among the regular and dual semiregular grids (i.e., the grids with one tile shape), only the hexagonal grid has this property. Table 1. Left: two tiles of the triangular, square, and hexagonal grid with unit distance between their centers. Right: dimensions of a regular tile when the distance between the centers of two edge-adjacent tiles is equal to 1. \| Empty Cell \| Empty Cell \| Area \| Edge \| Peri- \| Max \| Max \| --- --- \| Empty Cell \| Empty Cell \| Empty Cell \| Empty Cell \| meter \| point- \| point- \| \| Empty Cell \| Empty Cell \| Empty Cell \| Empty Cell \| Empty Cell \| center \| point \| \| Empty Cell \| Empty Cell \| Empty Cell \| Empty Cell \| Empty Cell \| dist. \| dist. \| \| \| Triangle \| 3 3/4 \| 3 \| 3 3 \| 1 \| 3 \| \| \| ≃ \| ≃ \| ≃ \| \| ≃ \| \| \| 1.3 \| 1.73 \| 5.2 \| \| 1.73 \| \| Square \| 1 \| 1 \| 4 \| 2/2 \| 2 \| \| \| \| \| \| ≃ \| ≃ \| \| \| \| \| \| 0.72 \| 1.44 \| \| Hexagon \| 3/2 \| 3/3 \| 2 3 \| 3/3 \| 2 3/3 \| \| \| ≃ \| ≃ \| ≃ \| ≃ \| ≃ \| \| \| 0.87 \| 0.58 \| 3.46 \| 0.58 \| 1.15 \| Conversely, grids (including the square one) which do not have uniform connectivity pose the question whether a tile γ must be considered as connected with all its vertex-adjacent tiles, or just to the edge-adjacent ones. Algorithms navigating the grid or applying masks to it (e.g., for image filtering), must weight adjacent pixels depending on the direction. Two connectivity types also lead to different answers about the number of connected components and holes of an object represented in a digital image. Other properties are concerned with the degrees of symmetry of a grid. When a rotation or mirroring operation does not map the grid onto itself, the tile values of the rotated image must be interpolated, with loss of information. The regular triangular and hexagonal tile have six symmetry axes, w.r.t. four axes in the square tile. Avkan et al. studied rotations with arbitrary angles in the three regular grids and showed that the triangular grid is the one where bijectivity is achieved with the largest number of rotation angles. 2.3. Grid hierarchies When a grid is used to discretize an image or to model a 2D field, the tile size defines the granularity or resolution, which is one of the parameters determining the accuracy or precision of the model. An interesting operation is reducing or enhancing resolution, i.e., rescaling the grid by changing the tile size. The square and the triangular grid can be easily downsampled by merging four (in general, a power of two) tiles into one (see Fig. 5, left and middle). The color (or field value) of the unique tile is obtained by averaging those of the original four tiles (many ways exist to do this, and they are not in the scope of this paper). By repeating this operation, we obtain a hierarchy of grids having the containment property: each (smaller) tile at higher resolution is completely contained into a (bigger) tile at lower resolution. Another hierarchy having the containment property is based on D(4,8,8), whose tiles are rectangular triangles. The grid is downsampled by merging two adjacent tiles or a group of four tiles (see Fig. 5, right). A containment hierarchy is easily described by a tree, where each tile is a node, and the children of a tile (node) γ are the tiles at the finer resolution level, whose union gives γ. The square hierarchy is the basis for the large family of quaternary trees, called quadtrees, used for representing images and scalar fields. Similarly, the hierarchy of regular triangles gives triangle quadtrees . The hierarchy of right triangles gives a binary or a quaternary tree . The hexagonal grid does not allow for a containment hierarchy. A hexagonal grid is downsampled by defining a larger tile, centered in one of the original tiles γ, which partially overlaps the six neighbors of γ. There are three ways to do this, which differ in their aperture, i.e., the ratio between the area of tiles belonging to the coarser and to the finer resolution , . In the aperture 4 hierarchy (see Fig. 6, left), the bigger hexagon has the same orientation as γ, and overlaps half of each neighbor of γ. In the aperture 3 hierarchy (see Fig. 6, middle), the bigger hexagon is rotated 90 degrees w.r.t. γ and covers one third of each neighbor of γ. The third way for defining a hierarchy of hexagons is grouping seven tiles (one tile and its six edge-adjacent ones) into one pseudo-hexagon , (see Fig. 6, right). The resulting grid has the same topological structure as a hexagonal grid. This operation can be repeated on it, thus giving a hierarchy. The tiles are roughly hexagonal in shape, and in each level of the hierarchy they are slightly rotated w.r.t. the previous level. Regarding the hexagonal apertures, Golay was the first one who noticed that a hexagonal sampling pattern can be divided into 3, 4, or 7 subgroups, such that each subgroup has still hexagonal topology . 1. Download: Download high-res image (123KB) 2. Download: Download full-size image Fig. 5. Downsampling the square grid (left), the regular triangular grid (middle), and the grid of right triangles (right). The latter one can be downsampled by merging two or four triangles. 1. Download: Download high-res image (121KB) 2. Download: Download full-size image Fig. 6. Downsampling the hexagonal grid: hierarchies with aperture 4 (left), aperture 3 (middle), aperture 7 (right). 1. Download: Download high-res image (78KB) 2. Download: Download full-size image Fig. 7. Alternative ways to downsample the hexagonal with an aperture 4 hierarchy. Other variants of aperture 4 hexagonal hierarchies are more rarely used: in the bigger tile covers three whole smaller tiles, and one third of three other ones (see Fig. 7, left); in four tiles incident in the two endpoints of an edge (roughly covering a diamond) are replaced with one bigger tile (see Fig. 7, right). Applications are sometimes interested in composing a variable-resolution grid, with tiles from different resolution levels. This happens when some parts of the domain need more details than other parts. For instance, in lossy image compression, we can keep the finest resolution only in the foreground, and use lower resolution for the background, as in Fig. 8. In a containment hierarchy, a variable-resolution grid is obtained in this way: starting from the grid at the coarsest resolution, the tiles whose resolution is sufficient are kept, while the other tiles are recursively replaced with their children. Variable-resolution grids from a quadtree (or from a triangle quadtree) are not cell complexes, as they do not have edge-to-edge adjacency (see Fig. 9, left). This problem is avoided in the hierarchy of right triangles based on the grid D(4,8,8) with aperture 2. In building a grid at variable resolution, any two right-triangles, sharing their longest edge, are replaced simultaneously with their four children. This ensures edge-to-edge adjacency in the resulting grids (see Fig. 9, right). Variable-resolution hexagonal grids require ad-hoc solutions at the junctions between two resolutions, as tiles of different size do not match geometrically. For this purpose, the hexagons are decomposed into triangles or quadrilaterals. 1. Download: Download high-res image (325KB) 2. Download: Download full-size image Fig. 8. An image, based on a quadtree, where the finest resolution is only used in the foreground. 1. Download: Download high-res image (59KB) 2. Download: Download full-size image Fig. 9. A grid at variable resolution obtained by combining tiles from different resolution levels of (left) a quadtree, and (right) a hierarchy based on right triangles. In the latter case, the result is a cell complex. 2.4. Coordinate systems for non-square grids The basic tool, necessary to deal algorithmically with a grid, is addressing its tiles with coordinates. The (discrete) coordinates are a tuple of integers which uniquely identify a tile of the grid. Moreover, it is possible to move from one tile γ to its adjacent tiles by just performing arithmetics on coordinates. The traditional Cartesian coordinates are well suited for the square grid. They are less suitable for the triangular and the hexagonal grid, because they do not reflect the three symmetry axes of such grids. The offset coordinates for the hexagonal grid are basically an ordinary Cartesian system, where the x-coordinate of each odd row is shifted horizontally by half the width of a hexagon (see Fig. 10, left). Such coordinates are used, for example, in , . A drawback is that the coordinates of the six edge-adjacent tiles of a given one (x,y) need different formulas, depending on whether y is even or odd. The array set addressing (ASA) adds an extra coordinate with value 0 or 1 depending on the even or odd row where each tile lies. This allows obtaining the six edge-adjacent tiles of any given one with the same formulas. The hexagonal grid can be encoded in two standard rectangular arrays, for tiles with 0 and 1 values, respectively. 1. Download: Download high-res image (181KB) 2. Download: Download full-size image Fig. 10. Illustration of the offset coordinate system (left), of the coordinate system with two non-orthogonal axes (middle), and of the one with three symmetric axes (right). Another class of coordinate systems for the hexagonal grid defines a pair of non orthogonal axes, forming an angle equal to either 120 or 60 degrees , , (see Fig. 10, middle). A mathematically elegant method uses complex numbers to address all the cells (tiles, vertices, edges) of a hexagonal grid and its dual triangular grid , . A Cartesian pair (x,y) is mapped to the complex number x+i y. The versor forming a 60 degree angle with the x-axis becomes the complex number ω=1/2+3/2. Each multiplication by ω performs a rotation of 60 degrees. Her , defined a coordinate system with three reference axes, forming 120 degrees with each other (see Fig. 10, right). The three coordinates of a tile are not independent and their sum is zero. This coordinate system reflects the three axes of symmetry of the hexagonal grid and simplifies many operations. Given a hexagon (x,y,z), the six adjacent ones are obtained by adding 1 to one coordinate and −1 to another one (leaving one coordinate unchanged). Affine transformations are obtained through matrix multiplications . Nagy extended this coordinate system to address not only the hexagonal tiles, but also the edges and vertices with a triplet of integer coordinates. The same coordinate system naturally extends from the hexagonal to the triangular grid, by exploiting duality . Another coordinate system for the hexagonal grid uses only one coordinate , , . This method, known as one-dimensional or spiral addressing, is indeed a path code for the aperture 7 hierarchy, and will be described in Section 2.5. Nagy and Saadat defined coordinate systems for a number of semi-regular dual grids, including D(3,3,3,3,6), D(4,3,4,6), D(3,6,3,6), and D(4,8,8). All coordinate systems are instances of a general framework, called Block Coordinate System. 2.5. Addressing tiles in grid hierarchies In a hierarchy of grids, it is necessary to identify each tile γ in each level L of the hierarchy. There are two approaches to do this, illustrated in Fig. 11. The first approach uses the level L together with the coordinates of the tile in the grid at resolution level L. To enable navigation across resolution levels, it is important to obtain the coordinates of a tile from the coordinates of the one(s) overlapping it at finer/coarser resolution. On a quadtree with N+1 levels (built over a grid with 2 N×2 N pixels), given the coordinates (x,y) of a tile γ at the finest resolution, the coordinates of the tile at level L, containing γ, can be obtained from (x,y) by keeping only the first L bits, which means dividing x,y by 2 N−L. A method for a triangle quadtree, covering a hexagonal domain, was proposed in . For the hierarchy of right triangles with aperture 4, Nagy developed a system of half-integer coordinates, where the coordinates of a bigger triangle are the average of those of the four smaller triangles covered by it. The second approach, suitable for addressing tiles in a containment hierarchy described by a tree, encodes the path from the root of the tree to γ. The path code is a variable-length string on an alphabet of N symbols (e.g., the integers from 0 to N−1) for a N-ary tree, and each digit of the string identifies the child of the current node, where the tree descent must continue. The complete geometry of a tile can be obtained from its path. Quadtrees and triangle quadtrees, as well as hierarchies of right triangles, are commonly indexed through paths, encoded as strings on an alphabet of four or two symbols (requiring 2 or 1 bit). The aperture 7 hexagonal hierarchy is described by a tree disregarding the small mismatch between the boundary of a larger hexagon and the boundaries of the union of the seven smaller hexagons refining it at the next (finer) resolution level. Therefore, each tile can be addressed by a string on the alphabet {0,1,…,6}. This approach gives the so-called one-dimensional or spiral addressing , , , also used to address tiles in a single-resolution hexagonal grid. 1. Download: Download high-res image (130KB) 2. Download: Download full-size image Fig. 11. Two ways to address a tile, shown on a quadtree. The dark pixel γ on the finest level of the hierarchy is identified either with level 3 plus its coordinates (x,y)=(5,3), or with the path code 1,2,3 telling which subpixel of the current one shall be taken to descend into the next finer resolution (the numbering of the four subpixels is shown top-right). On a quadtree, the coordinates of γ in the coarser levels can be obtained by iterative (integer) division by 2, i.e., (5,3),(2,1),(1,0),(0,0). The aperture 3 and aperture 4 hexagonal hierarchies are not represented by a tree, but rather by a directed acyclic graph where each node is a tile, and its children are those tiles, at the next (finer) resolution level, which overlap γ. Paths from the root to a tile are not unique, due to joining paths in the graph, and thus they are not suitable for addressing the tiles. A method to obtain unique paths in the aperture 3 hierarchy was proposed by Sahr . The hexagons of each grid are classifies as type A or B. Given a tile γ in the grid at level L, seven tiles intersect γ at the next (finer) level L+1: the central one (completely contained into γ) is of type B, while the others are of type A (see Fig. 12, left). In the graph, the links from a node of type A to its children of the same type are not considered for defining paths. The resulting purged graph is a tree, where nodes of type A have only one child, and nodes of type B have seven children. Other ways to reduce a hexagonal hierarchy to a tree were also proposed. In a big hexagon is considered as the parent of four hexagons, only two of them contained in it (see Fig. 12, middle); in a standard aperture 4 scheme is used, but a big hexagon is considered as the parent of the co-centered hexagon and of three of its adjacent hexagons in an alternated way (see Fig. 12, right). In both cases, the result is a quaternary tree. 1. Download: Download high-res image (275KB) 2. Download: Download full-size image Fig. 12. Left: the central hexagon (of type B) has seven children, the other six hexagons (of type A) have one child (parent and children are filled with the same color) . Middle: each hexagon has four children (marked with the same number and filled with the same color); each cluster of children covers roughly a diamond . Right: the four hexagons numbered 0–3 are children of one hexagon; each hexagon has four children (filled with the same color) . (For interpretation of the references to color in this figure legend, the reader is referred to the web version of this article.) 2.6. Grid conversion Since square and hexagonal grids are the most interesting ones, in this section we consider the problem of converting data between these two formats. As common acquisition devices produce square raster images, hexagonal images must be obtained from square ones through conversion algorithms. The used approaches differ in the following properties: •whether they provide regular hexagons or non-regular ones; •the degree of alignment between the input square and the output hexagonal grid (since the aspect ratio of squares and hexagons is different, regular hexagons cannot be aligned with squares in both axes); •whether they keep the resolution of the original square image, or not; •the way used to give a color to each hexagon (e.g., the kind of interpolation, if any). Pixel suppression , , is the most naive method: it deletes each odd row, and each second pixel on each even row of the given square grid. The remaining pixels are considered as hexagons of a grid having 1/4 the resolution of the original one (see Fig. 13). The hexagons are not regular, and no interpolation is applied. In the input (square) image is not changed, but the values of the suppressed pixels are not considered in processing. A similar approach , , , , retains the original resolution by first expanding each square pixel to 2 × 2 pixels with the same color. Then, it aggregates clusters of 2 × 2 squares in a brickwall pattern to simulate a hexagonal topology (see Fig. 14). The color of the bricks lying on odd rows is interpolated. The hexagons are not regular, but the original resolution is preserved. One could simply superimpose a regular hexagonal grid to the given square grid, and interpolate the color of each hexagon from the colors of the squares intersected by it. The hexagons may have any size (equal or bigger than the given squares) and any orientation (see Fig. 15(a)); their color is obtained by interpolating the ones of all intersected square pixels. Usually, some kind of alignment with the original square grid is considered: two sides of the hexagons are aligned with one Cartesian axis (e.g., the y axis) and either the edge or the (horizontal) width of the hexagon is set equal to the square edge (see Fig. 15, right). This idea is usually implemented by applying geometric shear transformations mapping orthogonal axes to axes forming 60 or 120 degrees , , , . Condat et al. apply three steps each consisting of a shear transformation (which only changes one coordinate) and one-dimensional filtering. The overall transformation is reversible. To have the same interpolation formulas independently of the position of the target hexagon, many authors use nearly-regular hexagons, where both dimensions are multiple of the square edge length. Each hexagon is inscribed into a rectangle of 9 × 8 squares (but other numbers are sometimes used). In this way, the two grids are aligned, but the hexagonal resolution is lower (see Fig. 16, left). To keep roughly the same resolution as the given square grid, each square is first expanded into 7 × 7 squares , , , , , . The interpolation may consider precisely all intersecting squares , or simply the 56 pixels covered by the hexagon for more than their half (see Fig. 16, left). The opposite conversion is needed to display hexagonal images on square-based raster screens, as hexagonal devices are not available. Two main approaches are used: 1. Download: Download high-res image (67KB) 2. Download: Download full-size image Fig. 13. Pixel suppression: the topology of a hexagonal grid is obtained from a square one by keeping only one every four pixels. 1. Download: Download high-res image (70KB) 2. Download: Download full-size image Fig. 14. Brickwall pattern: in order to keep the same resolution, pixels are first enlarged 2 × 2 times and then clustered to simulate hexagonal topology. 1. Download: Download high-res image (290KB) 2. Download: Download full-size image Fig. 15. Left: hexagonal grid arbitrarily oriented w.r.t. a square one. Right: a hexagonal grid partially aligned with a square one. 1. Download: Download high-res image (133KB) 2. Download: Download full-size image Fig. 16. Forming hexagonal hyperpixels from a square grid. Each hyperpixel is composed of 56 squares. In order to keep the same resolution, each square pixel is first enlarged 7 × 7 times. •hexagons are rendered with clusters of 2 × 2 squares arranged in a brickwall pattern , , , , , or •each hexagon is rendered with a so-called hyperpixel composed of many squares covering a near-hexagonal shape , , , , , . The two approaches correspond to the ones used for square-to- hexagonal grid conversion, and are shown in Fig. 14, Fig. 16, respectively (right image). In both cases, the hexagonal image needs a higher resolution display, w.r.t. the image itself, to be rendered. 3. Grids for images The standard image formats and graphic hardware are based on the square grid. In spite of this, the hexagonal grid would be the most appropriate one to represent images , , . Its properties like dense packing, uniform connectivity, good isotropy and symmetry (see Section 2.2) make it preferable with respect to the square grid. In addition, in the human vision system, retinal cells are arranged in a very similar way to a hexagonal grid . To the human eye, squares show more artifacts because we are especially sensitive to horizontal and vertical lines. Already in 1969, Golay addressed hexagonal image processing and presented an architecture with logical circuits to realize the basic operations. Since the last quarter of 20th century, authors have proposed methods for hexagonal image processing, noting that this grid allows for a more accurate representation and simpler algorithms than the traditional square grid , , . He and Jia gave a survey of hexagonal image processing algorithms. 3.1. Image processing The core of many image operations is filtering, or convolution. This operation consists of scanning the image with a moving window, the mask or kernel, which at each step is centered at a pixel γ. The mask comprises a small set of pixels around γ and gives a weight to each of them. A new value is assigned to γ, depending on the current values of the pixels covered by the mask and on the weights. On a square grid, masks usually consist of 3 × 3 pixels (including the immediate neighbors of the central pixel) or 5 × 5 pixels (including also the second-level neighbors), etc. Filtering is used in image processing for different tasks, such as denoising, gradient estimation, edge and feature detection, and thinning. Such tasks can be performed on images in any grid by means of appropriate filters. Hartman and Tanimoto defined convolution masks on a triangular grid, with application to edge detection. However, most proposals are for the hexagonal grid. Filtering a hexagonal grid allows for smaller masks w.r.t. a square grid (seven pixels for the immediate neighbors, 19 pixels for the second-level ones, etc.) and many authors show that, thanks to the uniform connectivity, it yields more accurate results, reducing aliasing and artifacts. Different hexagonal filters were proposed in , with their mathematical foundations. Coleman et al. compared various hexagonal filters, considering also different neighborhood sizes, on synthetic and real images. Hexagonal filters are smaller than their square counterpart (e.g., six immediate neighbors versus eight), therefore they save memory and computation time. Filtering in the hexagonal grid is more effective for edge and feature detection thanks to isotropy, while the square filters tend to privilege vertical and horizontal directions , , , , , . Snyder et al. described how to compute image gradient of a hexagonal image and applied it in a heat diffusion process. The consistent connectivity of the hexagonal grid makes it especially suitable for thinning , , . The purpose of thinning is finding the skeleton, i.e, the medial axis of the object represented in the image, consisting of a connected line of one-pixel width, and this can be obtained by iterative filtering. Kardos and Palagyi gave conditions for changing the color of a pixel without changing the topology of the image, in the square, triangular, and hexagonal grids , and used them in two thinning algorithms for the hexagonal grid, providing a 1- or 2-pixel wide skeleton, respectively . In , hexagonal thinning is applied to the recognition of handwritten characters. Splines approximate an unknown function, defined on a 2D domain and sampled at a set of points, by means of a mosaic of functions with finite domain, each having a fixed form (linear, bi-linear, etc.). Each function is defined locally with coefficients that depend only on samples lying in a given neighborhood (called control points because they determine the shape of that patch of the mosaic). Splines are usually defined with control points located on a square grid; hexagonal splines were defined in , , . 3.2. Frequency decomposition, multiresolution and compression The (discrete) Fourier transform, with its efficient implementation known as Fast Fourier Transform (FFT), allows passing from spatial to frequency domain, so that it is later possible to separate the various frequency bands of a 2D signal. This has many applications, e.g., smoothing or denoising, and image compression. The discrete cosine transform is also used for similar purposes. These two transformations were addressed on the hexagonal grid in , , , , . In the hexagonal FFT, in conjunction with thinning, was successfully used for fingerprint classification. In summary, almost all known operations on square images have been developed for hexagonal images as well. The main problem is that the hexagonal grid lacks standard encoding and tile addressing. The proposed solutions are often tailored for a specific architecture and therefore not immediately portable to another one. For instance, Middledon and Sivaswamy developed the FFT for a hexagonal grid with spiral addressing, while Birdsong and Rummelt considered a hexagonal grid in ASA coordinates, encoded as two rectangular arrays, and used library routines for the rectangular case. Image compression is aimed at reducing the size of an image file, and can be loss-less or lossy. Similarly to the (discrete) Fourier transform, the (discrete) wavelet transform allows separating the frequency bands of an image, but does it through a hierarchical decomposition. The image can be then reconstructed in an incremental way by adding back the wavelet coefficients in order from the coarsest level to the finest levels, thus providing a multiresolution representation. Deleting the wavelet coefficients of the finest levels in the hierarchy corresponds to removing details (or noise). The foundations of hexagonal wavelets were presented by Condat et al. . Standard wavelets are separable in two orthogonal one-dimensional components, which simplifies the computations but privileges the horizontal and vertical directions. Non-separable wavelets are defined in the hexagonal grid. presented wavelet-based multiresolution analysis of hexagonal images in spatial and frequency domain, by passing from the former to the latter and back through the Fourier transform and its inverse. In , hexagonal wavelets were used for image compression and denoising. Hexagonal image compression was addressed in with different methods (discrete wavelet transform, discrete cosine transform, and singular value decomposition). Comparison with the corresponding compression methods on the square grid shows that the hexagonal version gives better compression rates. Rashid and Alim also compared hexagonal wavelet with the Cartesian ones, showing better reconstruction quality with about half the size. 3.3. Hexagonal image acquisition With some acquisition methods used in medical imaging, such as Computer Tomography (CT) or Magnetic Resonance Imaging (MRI), the 2D image has to be reconstructed from a 1D signal through a process of filtered back projection. The patient’s body (or other object) is acquired slice by slice through a rotating device which emits a ray and measures the total quantity of matter traversed by it. Each angular direction of the ray provides a value, and the overall input for the reconstruction of the slice consists of sine signals with different amplitude and phase, called sinogram. In this context, there is a degree of freedom in choosing the target grid in which the slice will be reconstructed, therefore a hexagonal grid can be used. The isotropy of the hexagonal grid makes it more suitable than the square one, because data show rotational symmetry. By just replacing the coefficients in the standard back projection, Knaup et al. reconstructed CT images on the hexagonal grid, obtaining the same image quality with less memory and time. Abbas and Sivaswamy reconstructed hexagonal PET images, and then applied denoising with machine learning techniques. Engel et al. used the hexagonal grid inside the acquisition process of MRI images. In this way, each shot covers more space, therefore fewer shots are sufficient, leading to time saving (for the same pixel area, the hexagonal inter-center distance is larger than the square one). In the aim is finding defects in carbon fiber materials from ultrasound CT images. Reconstructed medical images are affected by noise and have low resolution, thus they need to undergo denoising and enhancement, which can be successfully performed in the hexagonal grid , . 3.4. Hexagonal graphics In common raster screens, each square pixel is made up of three subpixels, corresponding to the base colors red, green and blue. Nagy and Saadat suggested a future screen using a hexagonal raster, where the hexagonal pixel is divided into three rhombuses that will provide the three base colors. Such screen would be a D(3,6,3,6) grid. Rendering vector graphics on a (still imaginary) hexagonal display would require rasterization algorithms. Algorithms were designed to rasterize lines and circles, and other curves on a hexagonal grid , , , . Kovalevsky provided a survey of algorithms for tracing of curves and surfaces, recognizing digital straight line segments, 2D region filling, etc., on the hexagonal grid. Today, hexagonal rasterization has little practical relevance for display, but is interesting for integration of vector data into hexagonal discrete global grids (see Section 4.2). 3.5. Machine learning with hexagonal images Machine Learning (ML) approaches are used in image processing. Convolution Neural Networks (CNN) are commonly used, where the connected cells across two consecutive network levels correspond to the neighbors of a pixel. Since human retinal cells have a roughly hexagonal layout, hexagonal neural networks are especially interesting, and several scientists used them to mimic the biologic vision process. Huang and Li constructed a computer simulation of human fovea by means of a hexagonal cellular neural network. With different filters, they simulate the behavior of the different retinal cells. The system was applied for video enhancement. Other authors ported state-of-the-art Machine Learning methods from square to hexagonal images , , , . The hexagonal version provides similar or better results with fewer training parameters. Tang et al. showed that hexagonal CNN are more robust to rotations, thanks to the isotropy of the grid. In , convolution Neural Networks (CNN) were formalized for a hexagonal grid lying either on the plane or on a spherical surface (like the surface of the Earth), and were applied to the classification of remote sensing images. Some authors adopted a mixed approach, where standard square-based CNNs are used to prepare the image, which is then converted to hexagonal format and processed with hexagonal CNNs. In the purpose is identifying lesions in images of the retinal fundus, and the hexagonal grid was used due to the round shape of lesions. In the purpose is reconstructing a clean image of skin lesions. In both cases, the square image is preprocessed to remove disturbing elements (blood vessels and hair, respectively), then the image converted to hexagonal format to carry on the specific task: find the lesions with a region-growing process, and use inpainting techniques to complete the image where hair was removed. The branch of image processing which deals with human faces is an application field where ML techniques are extensively used. Hexagonal images were preferred w.r.t. square ones because facial features are not orthogonal in shape. The hexagonal grid showed to be effective for face recognition and recognition of facial expressions . A popular approach to face recognition extracts key points with the Scale-Invariant Feature Transform (SIFT) and then uses them for classification. The hexagonal SIFT was defined and successfully used in , . Cevik et al. considered the application to face recognition of some descriptors used in texture classification, comparing their square and hexagonal versions . 4. Grids in geographic information systems Geography and cartography consider information attached to locations on the Earth, i.e., geo-referenced. Traditionally, this means that latitude and longitude coordinates are attached to each datum. Geo-referenced data come from different sources and in different formats: vector (points, lines, regions, networks) or raster (fields defined pointwise on a grid). Examples of vector data are cadastrial parcels, river networks, etc. Examples of fields are temperature, population density, vegetation, land use (representing each use type with a number), etc. A Digital Elevation Model (DEM) represents elevation over the sea level. The different information kinds, referring to a given geographic area, are called layers. Each data layer is affected by a geometric (geo-location) error, depending on the acquisition device and operating conditions. A relevant issue is data integration, i.e., fusing data of different kind (for example river network in vector format and land use in raster format) and with different accuracy. Geographic data may be used for presentation to the user (e.g., assembling a map to be visualized) or further processed to compute new information (e.g., classification, statistics, simulations, etc.). Traditionally, the considered geographic domain is discretized with a square grid, but the hexagonal grid has many advantages, mainly due to its uniform connectivity, isotropy and packing density , also considering the fact that natural Earth features hardly follow orthogonal patterns. 4.1. Discrete global grids Geographic information refers to entities on the Earth surface, i.e., a sphere (more precisely, the Geoid) and it is ported onto a plane by means of cartographic projections. Traditional cartographic projections (e.g., cylindrical or conic ones) introduce distortion, and therefore can only be applied locally to a certain portion of the Earth. Latitude and longitude coordinates provide a representation at global scale, consisting of a square (rectangular) grid, but the tiles are affected by distortion, especially near the poles. At the turn between the 20th and 21th centuries, discrete Global Grid Systems have emerged, which follow a totally different approach. A Discrete Global Grid System (DGGS) subdivides the entire surface of the Earth into a grid of (almost) identical cells. The grid can then be unfolded on a plane for visualization, but calculations are performed on the 3D surface. Moreover, the system supports multiple levels of resolution by means of a hierarchy of increasingly finer grids. The basic idea behind DGGSs is the following. The first (coarsest) level of the hierarchy is one of the platonic solids inscribed in the sphere. The faces of the initial polyhedron are mapped onto the spherical surface and thus they are curved polygons. In order to obtain the next (finer) resolution, the faces (tiles of the grid) are subdivided according to fixed patterns. Since 2015, with periodical revisions, the Open Geospatial Consortium (OGC) has been defining the requirements of a DGGS . In particular, it must cover the entire Earth surface, provide multiple levels of resolution, have cells of equal area at each level, provide a unique identifier for each cell. Surveys of current DGGSs can be found in and more recently in , , , . The different DGGSs are characterized by the initial polyhedron, the tile shape and the type of hierarchy, the projection type, and the method used for indexing the cells . Table 2 summarizes the main proposals. •The pioneer systems , start from an octahedron and use a triangle quadtree on each face to generate the successive levels. •The HEALPix system , designed for astrophysical data, starts from a rhombic dodecahedron and uses a quadtree-like refinement on each face. •The rHEALPix system starts from a cube; the equatorial faces are refined into a square hierarchy by dividing each tile into nine smaller squares, while the top and bottom faces (corresponding to the poles) are subdivided with a triangular scheme. In this way the faces are always subdivided along lines parallel to the latitude and longitude. •In the initial polyhedron is a triancontahedron (a solid with 30 rhombic faces) and a quadtree-like refinement is applied on each face. •Many DGGSs are based on hexagons. As no regular polyhedron with all regular hexagonal faces exists, the hierarchy starts from a truncated regular polyhedron. The most commonly used one is the truncated icosahedron, i.e., a soccer ball polyhedron (12 pentagonal faces and 20 hexagonal ones). Any finer resolution level contains 12 pentagons and an increasing number of hexagons. The solid is oriented in such a way to place the pentagons in oceans or areas of low interest, and pentagons are treated as degenerate hexagons. For projecting from the hexagonal faces to the sphere, most DGGSs use the Icosahedral Snyder’s Equal Area (ISEA) projection . The hierarchy follows a scheme based on aperture 3 , aperture 4 , , or aperture 7 (see Fig. 17). Few authors start from other polyhedra: a truncated octahedron (8 square faces and six hexagonal ones) , or a triancontahedron by covering each rhombic face with 16 hexagons. Mahdavi et al. provided a list of applications that would benefit from a DGGS (including environmental monitoring, disaster prediction, monitoring of endangered species, health, urban planning and education). 1. Download: Download high-res image (340KB) 2. Download: Download full-size image Fig. 17. Hexagonal hierarchy starting from the truncated icosahedron. From left to right: aperture 3, aperture 4 (first three levels), and aperture 7 (first two levels). Table 2. Summary of the main discrete global grid systems , , , , , , , , , , . Li and Stefanakis compared the functionalities of a DGGS with those of a traditional geographic information system, highlighting superiority as well as aspects requiring further research. Currently open problems are listed in . 4.2. Data integration Currently, vector and raster geographic data are expressed in the latitude–longitude system, and must be converted to cell indexes in the target DGGS. The use of a DGGS adds the benefit of supporting multiple resolutions. Geographic data usually come from different sources (on ground and aerial surveys, remote sensing images obtained with different technologies). Each of these sources has a different granularity and is affected by an approximation error. A multiresolution grid allows storing each piece of information at the appropriate resolution. In vector features are ported to a triangular DGGS and the appropriate resolution is found by starting from the coarsest level and continuing the descent into the finer levels of the triangle quadtree, until all features cover disjoint sets of tiles. This approach can be extended to DGGS with cells of any shape. Raster data integration into a (hexagonal) DGGS was considered in for terrain elevation and an aperture 3 hierarchy, and in for aperture 4. A database architecture built over a hexagonal DGGS, dealing with several data formats, was described in . In data integration was addressed with specific interest for marine data and shorelines. The problem of mapping location names, present in textual documents, to DGGS cells was considered in . In general, the problem of mapping vector data to a hexagonal DGGS is similar to rasterization in hexagonal graphics (see Section 3.4), and the problem of mapping (square) raster data is similar to square-hexagonal conversion (see Section 2.6). Geographic data integration is more complex, because additional issues must be addressed, such as correct projection for data that are actually located on the Earth surface, choice of the appropriate resolution level based on the size of the squares (which depends on latitude), data extending across different faces of the base polyhedron, etc. A deeper collaboration with researchers working in hexagonal imaging would help researchers working in geographic systems, who could concentrate on the specific issues of data integration for DGGS, rather than developing the whole pipeline of square-to-hexagonal and vector-to-hexagonal conversion. Conversely, the possibility of applying their findings could motivate more research in hexagonal images. With the diffusion of DGGS, data integration across different DGGS will also become an issue. Recently, considered data mapping between different DGGS with triangular cells (which can be extended to rhombic ones) and considered mapping between hexagonal DGGS using the icosahedron or the rhombic triancontahedron as base polyhedron. 4.3. Hydrography on the hexagonal grid A large branch of geographic processing is concerned with hydrography, where the considered information is terrain elevation and the aim is calculating slope, flow direction, stream networks, water accumulation, catchment basins and several related indices . Algorithms were proposed to compute slope, gradient, hydrological properties including flow direction and accumulation, flood simulations, delineation of watershed boundaries, and valley lines on a hexagonal grid, by extending the classic square-based counterpart , , , . Since the input is a traditional (square-based) Digital Elevation Model (DEM), this must be first converted to hexagonal format. The computation of hydrographic information at different levels of detail was considered in , on a hexagonal DGGS with aperture 3. All authors remark that the hexagonal grid provides more accurate results than the standard square grid. In fact, natural terrain features usually exhibit curved rather than orthogonal shapes, and thus are better captured by hexagonal tiles. Remote sensing instruments to acquire elevation data from aerial or satellite surveys still produce square DEMs. Acquiring elevation value on a hexagonal grid would be interesting not only for the good properties of the grid, but also because it can easily model a DEM with both approaches listed in Section 2.2: by extending the value at the tile center to the entire tile (constant approximation within each tile with discontinuity at tile boundary), or by defining a piecewise continuous approximation, through a linear interpolation on the dual triangle mesh (see Fig. 18). 1. Download: Download high-res image (164KB) 2. Download: Download full-size image Fig. 18. Hexagonally sampled DEM with constant values associated to the whole tile (left) or with interpolated values by exploiting the dual triangle mesh (right). 4.4. Simulation of time-varying phenomena Many geographic processing tasks are concerned with the representation and simulation of natural or social phenomena, such as pollution or other hazards, diffusion of vegetable species, etc. A general framework for defining cellular automata on the hexagonal grid was developed in . A cellular automaton models a system where each cell changes its state based on the input received from its neighboring cells, by applying a simple transition rule. The local behavior of all cells together determines a system trend at global scale. Provided applications in are demography, urban land use, and deforestation. Kiester and Sahr developed transition rules on a DGGS taking into account not only the neighboring cells at the same resolution level, but also at the immediately higher and lower ones. This is supported by a general framework, encompassing both containment hierarchies and hexagonal hierarchies, which distinguishes the children of a given tile γ into contained and intersecting ones, and provides rules for moving from a tile to another one, both within a single resolution level and across resolutions . The hexagonal grid was used for predicting climate changes and crime rates , for simulating fire propagation , , and for earthquake response planning . Simulations were also performed by taking advantage of the multiresolution features of a hexagonal DGGS. Stough et al. studied the diffusion of CO 2 in the atmosphere, and in data related with maritime events were integrated in a DGGS for application to risk monitoring. 4.5. Other applications in geographic domain Path planning is concerned with the search for an optimal path according to some cost function. The classic setting of path planning is defined in a road network, modeled as a graph. A different problem is path planning in a planar domain where an agent can move freely, and some field (e.g., terrain slope), discretely modeled on a grid, defines pointwise the cost of the path. The hexagonal grid was used in , to discretize the domain for the computation of minimum cost paths, and in it was used to support ride services in a big city. Convolutional Neural Networks for DGGS systems were defined and used for different applications. CNNs for hexagonal DGGSs with various apertures were defined in and applied for the classification of remote sensing images. In CNNs for the HEALPix with rhombic faces were considered for the classification of astronomical measurements. 5. Summary and future perspectives We see a trend of increasing attention in non-square grid for practical, technological uses, with the hexagonal grid playing the major role. We completely agree with the final recommendation formulated in , who encouraged “the members of the image processing communities to consider also other non-traditional grids which may show better performance from various points of view”. We think that the times are now ready for an engagement of the industrial community, with the development of standard formats and hardware support. Hexagonal imaging has been studied since the last two decades of the 20th century. Many researchers developed algorithms and data structures, both for hexagonal rendering (vector-to-raster conversion) and for hexagonal image processing, showing the superiority of this grid. Passing from squares to hexagons would improve accuracy and consistency of computations and simulations, as demonstrated by experiences in the field of medical imaging, face recognition, etc., including the use of hexagonal filters and descriptors for machine learning. However, graphical devices for acquisition and display are till now based on square grids, so the engagement of the industrial and commercial world is urgently needed. Some hexagon-based acquisition devices are being developed in the field of medical imaging and astronomy. Curiously, some authors transformed the acquired hexagonal image into a square one for processing , , instead of using algorithms developed for the hexagonal grid. Hexagonal grids also need standard and universally accepted coordinate systems. The use of independently developed and different coordinate systems makes it cumbersome to exchange algorithms and software tools. The international community should examine all the approaches and agree on a unique standard format. The field of geographic systems started one decade later, but seems more advanced in the practical use of different grids as the basis for defining Discrete Global Grid Systems (DGGS). The Open Geospatial Consortium (OGC) has defined the requirements of a DGGS, which include the support for multiple levels of resolution by means of a hierarchy of increasingly finer grids, but do not impose constraints on the shape of the cells. Hexagonal-based DGGS are receiving a lot of attention in the academic community. Proposals of geographic applications developed on top of a DGGS, are being published at fast rate, considering many diverse applications. However, many of them still perform their task at a selected resolution level, and do not fully exploit multiresolution power of the DGGS. The next necessary steps for enabling a practical use of DGGSs by the wide public are many and of various nature, including technical issues (e.g., computational efficiency), theoretical issues (e.g., a data model exploiting the multiresolution nature of global grids and incorporating spatio-temporal dimensions), and cultural issues (e.g., letting end-users understand the relationship between multiresolution and data uncertainty) . Only those issues regarding the use of grids are within the scope of this paper. While traditional geographic information systems can easily interoperate because they all use the latitude–longitude grid, each DGGS has its own architecture consisting of base polyhedron, cell shape, aperture, tile addressing, etc. Even assuming that the hexagonal shape of tiles may emerge among the other options, the choice of a specific aperture appears more difficult. The aperture 3 guarantees a smooth transition between consecutive resolution levels, but has the disadvantage of rotating hexagons at each level change. The aperture 7 allows for an easy indexing of cells, being roughly a containment hierarchy, but it gives an abrupt transition between resolutions, and hexagons get slightly rotated at each new resolution change. The aperture 4, consistent with the classic aperture of quadtrees and triangles quadtrees, and providing hexagons with equal orientation at all resolutions, may be a good compromise. A mixed aperture hierarchy was explored in . Among the large variety of DGGS proposals, one or some (few) reference DGGS should be established, with standard and efficient procedures for data conversion between them (pioneer works exist , ). More cross fertilization among research communities (e.g., in imaging and in geographic systems) is also necessary, in order to coordinate the effort and avoid duplicated work. Many tools, such as data conversion and hexagonal neural networks, find applications in diverse fields and the research on them could be unified. Although hexagonal imaging has not perspectives of a rapid development in the immediate future (at least as long as hexagonal hardware does not exist), it can even now provide solutions useful in other applications, such as hexagonal DGGSs. Declaration of competing interest The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper. Recommended articles Data availability No data was used for the research described in the article. References B. Nagy Non-traditional 2D grids in combinatorial imaging - advances and challenges IWCIA 2022: Combinatorial Image Analysis, Lecture Notes in Computer Science Book Series (LNCS), vol. 13348 (2022), pp. 3-27, 10.1007/978-3-031-23612-9_1 View in ScopusGoogle Scholar A.R.H.M. 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