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The Retirement Café About 10 years ago, in the course of a conversation with two retirement researchers whom I greatly respect, someone mentioned the 4% Rule. One of those researchers said, "William Bengen did great work showing us that sequence risk exists but trying to turn it into a retirement plan was a huge mistake." Bengen's work gave us the 4% Rule, derived from the so-called probability of ruin. Probability of ruin, or p(ruin) for short, is the estimated probability that a retiree spending a fixed real dollar amount from a volatile portfolio will outlive her portfolio. Somehow, despite its many shortcomings, p(ruin) has become the most common metric in retirement planning. The 4% Rule provides a "sustainable withdrawal rate" (SWR) that a retiree can supposedly spend from a volatile portfolio with a 95% probability of not outliving his savings. How much is the SWR? Bengen estimated a range around 4.4%. Wade Pfau, Michael Finke and David Blanchett found that the SWR is currently closer to 3%, primarily due to a low-interest-rate regime. If they are correct, that would result in annual withdrawals nearly 32% lower than Bengen's estimate. That's quite a range. Some question the implications of that research, notably Michael Kitces, but interestingly, William Bengen believes that valuations are probably important and that "Pfau may be on to something." The Shiller CAPE 10 ratio , a measure of stock market valuation, was around 10 when Bengen's data series began in 1926 and today suggests a much higher market valuation of around 30. A higher CAPE 10 suggests lower future market returns and vice versa . Had the market return data series studied by Bengen begun when valuations were relatively high, the results may have suggested a lower SWR. (It is not uncommon for economics studies to improperly ignore initial conditions like market valuations.) I will toss yet another monkey wrench into these analyses and note that both studies make assumptions about future asset returns so neither can be proven to be correct . Still, Pfau et al. provides evidence that Bengen's SWR may be overestimated. This uncertainty is the essence of risk. What are these shortcomings of p(ruin)? Let's start with p(ruin) being a one-dimensional measure of risk. By that I mean it estimates the probability (risk) of outliving a consumption portfolio, which I will define as a volatile portfolio of investments from which a retiree withdraws cash periodically to pay his bills, without measuring the magnitude of that risk. Some research I'm currently coauthoring serves as an example. We compare two consumption-portfolio spending strategies. Each estimates a p(ruin) near 5%. On this basis, we would say that the two strategies are equally risky. However, when scenarios fail using the first strategy, the mean number of underfunded years is about 15. When scenarios fail using the second strategy, the mean number of underfunded years is about 21. The second strategy is riskier because when it fails, it leaves the retiree underfunded for 6 more years on average. This magnitude of risk isn't captured by p Another problem with p(ruin) is that it is based on a very limited sample of historical equity returns. Robert Shiller has reconstructed equity returns back to 1871, providing a little less than 150 years of data but this historical data contains very few unique long-term sequences of returns of 30 years or more that we need for retirement studies. We simply don't have enough data to draw statistically significant conclusions about the future probability of ruin. Many argue that only the more recent years of Shiller's historic returns are truly reliable. Researchers have tried multiple strategies to get around this lack of data. Bengen used overlapping 30-year periods of returns. This strategy is flawed because the first and last years of the equity return time series are each used only once, the second and next-to-last twice, ., while the returns in the middle of the series are included up to 30 times. Another strategy is to generate 30-year series of returns by resampling, or randomly choosing returns from the entire historical data set with replacement. This strategy will provide results similar to the experience of the handful of available unique historical 30-year sequences of returns but doesn't generate "out-of-sample" series. In other words, it assumes that the limited number of 30-year historical periods of data we have contain all of the information we will ever need to know about future market returns. It is more likely that the future will likely throw something at us that we have never seen before. Said a third way, our limited amount of historical long-term data series has very little predictive power. It can only tell us what might happen in the future if the future is very much like our limited past. Let's focus now on a term I just introduced, "sequence of returns." The success or failure of a consumption portfolio is primarily a function of the sequence of the portfolio returns and not on the returns themselves. To quote BigErn at , "Precisely what I mean by SRR (sequence of returns risk) matters more than average returns: 31% of the fit is explained by the average return, an additional 64% is explained by the sequence of While we can generate realistic market returns from historical data using statistical methods like resampling, we cannot capture the most important characteristic of that data relative to portfolio ruin, the sequence of those returns. Resampling and most Monte Carlo models simply create random uniform sequences of returns and these are often quite unlike the few long sequences we observe from historical data. This leaves two possibilities. One possibility is that the sequence of market returns is truly purely random as we most commonly model, in which case we have been extremely lucky not to have received a catastrophic sequence of returns over the past 150 years. Another possibility, and the one I favor is that sequences of returns are not purely random but are limited by market forces that we don't yet understand. In that case, we may never see catastrophic sequences of returns but our models are wrong. I can't leave this topic without noting that consumption-portfolio failure doesn't require really bad negative returns. A long sequence of sub-par returns will do the trick. The worst-case series of 30-year returns beginning in 1964 that defines the 4% rule was simply a long period of mostly-positive but mediocre real returns. Not long after the Great Recession, some SWR advocates were quick to note that the market had rebounded rather quickly, supporting the idea of a 4.5% SWR. While this is true, there are two important caveats. First, consumption portfolios recover much more slowly than a market index because we aren't spending from the market index. Second, the Great Recession was a three-year sequence and, as I note in the previous paragraph, portfolio failure typically results from long periods of mediocre returns and not short periods of negative returns.The Great Recession may not portend future portfolio failure for today's recent retirees. Lastly, I think it is important that we consider the ability of humans to "internalize" probabilities. Clearly, there are some of us like Nate Silver, who can see a probability and intuitively interpret it. Most of us can't. Most people tend to round small percentages to zero and large percentages to 100. The 2016 presidential election is a perfect example. On November 9, 2019, Nate Silver published a prediction that Trump had a 28.6% probability of winning the election and Hillary Clinton had a 71.4% probability. Many read this and concluded that Trump had no chance of winning, , they rounded 28.6% to zero and 71.4% to 100%. When Trump won, they were outraged at Silver. I saw a poster at the Women's March saying, "I will never believe Nate Silver again." The election was a one-time event and clearly not random. Silver's probabilities weren't based on counting who won past elections between Trump and Clinton. They represented Silver's belief that these were the odds and he believed that Trump's chances of winning were significantly greater than zero. It appears that many people didn't understand that. This raises the issue of one-time events like a presidential election or your retirement. It's simple enough to look at a roomful of one hundred 65-year olds and say that a 4% Rule strategy means five of them will outlive their savings but it is impossible to say in advance which ive it will be. It is, therefore, difficult to internalize what 5% of retirees outliving their savings translates individual probability of failure. (This is a poor analogy in one sense but I hope it makes the point. The 4% Rule says that 5% of 30-year periods will result in a failed portfolio, so if everyone in that room were 65 years old, they presumably all would go broke or none would. They will all experience the same future market returns.) Your retirement differs from the 2016 election, although both are one-time events. We use historical market data to count how often you might have succeeded in the past, given some withdrawal rate. The problem is that we don't have nearly enough of that data. Even if we did, we could only predict how many retirees would fail and not whether you would be one of them. The point of our ability or inability to intuitively understand probabilities is that many people will round a 5% chance of ruin to zero and feel perfectly safe, while others (like me) will feel that a 1-in-20 chance of ending up destitute in their dotage is completely unacceptable. In either case, p(ruin) is frequently problematic because of our inability to intuit it. There are a couple of other shortcomings of p(ruin) that I will briefly mention in conclusion. Many argue that no retiree would ever do what the 4% rule requires, that is, to continue to spend the same amount from a consumption portfolio even when it is obviously failing. First of all, I would note that if the retiree doesn't do this, then the 4% Rule is not predictive at all because the retiree isn't adhering to the strategy but I also have anecdotal evidence that there are rational reasons a retire would continue spending the same amount. At some point, a retiree with a failing portfolio will reach an amount of spending that is necessary to meet non-discretionary expenses and spending too much to pay necessary expenses will be the rational response even if it will undoubtedly lead to portfolio depletion in the near future (see Why a Rational Retiree Might Keep Going Back to that ATM If the 4% Rule says I can spend no more than $1,000 or else I will probably go broke in the near future but my necessary expenses total $1,500, I will spend the $1,500. In this scenario of continued fixed spending, portfolio behavior is either chaotic or behaves chaotically and it doesn't matter much which (see Retirement Income and Chaos Theory Economist, Laurence Kotlikoff believes the 4% Rule estimates both the wrong amount to save and the wrong amount to spend compared to an economics approach. He explains it better than I could in The 4% Retirement-Asset Spend-Down Rule Is Rubbish Lastly, probability of ruin is a number that we intentionally try to make as small as practical. It's a measure of "tail risk", or the area of low-probability outcomes of a model. Nassim Taleb, in testimony before Congress no less , stated that "the more remote the event, the less we can predict it." Taleb goes on to say, "Financial risks, particularly those known as Black Swan events cannot be measured in any possible quantitative and predictive manner; they can only be dealt with non-predictive ways." But, predicting unlikely events is precisely what p(ruin) purports to do. The 4% Rule has achieved cult status to the extent that I hear retirees with virtually no other knowledge of retirement finance casually refer to it as if it is a universal law. It is not. It is a questionable but unfortunately prevalent retirement finance metric. A better approach is recommended by life-cycle economics (see, for example, Risk Less and Prosper by Zvi Bodie), sometimes referred to as "safety-first." The safety-first strategy is to assume that portfolio failure is a (perhaps) small — Taleb would say unquantifiable — probability of an unacceptable outcome. It deals with the risk of portfolio depletion "in non-predictive ways." The retiree is encouraged to plan for an acceptable standard-of-living in the event of that outcome without having to roll the dice and simply hope the future looks a lot like the past. [1] The 4 Percent Rule Is Not Safe in a Low-Yield World , Michael Finke, Ph.D., CFP®; Wade D. Pfau, Ph.D., CFA; and David M. Blanchett, CFP®, CFA. [2] Shiller PE Ratio , Multpl.com. [3] Online Data , Robert Shiller, Yale Economics. [4] The Ultimate Guide to Safe Withdrawal Rates – Part 15: More Thoughts on Sequence of Return Risk , EarlyRetirementNow.com. [5] The 4% Retirement-Asset Spend-Down Rule Is Rubbish , Laurence Kotlikoff, Forbes.com. [6] The Risks of Financial Modeling: VAR and the Economic Meltdown, House Subcommittee on Investigations and Oversight, GPO.	{"url":"http://www.theretirementcafe.com/2019/09/","timestamp":"2024-11-04T07:46:38Z","content_type":"application/xhtml+xml","content_length":"96400","record_id":"<urn:uuid:e228582d-3562-4427-95b1-9605f90babb5>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00838.warc.gz"}
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Web in this printable, students continue to answer subtraction questions using numbers up to 20 and noting their answers just below each problem. Web printable subtracting charts for kids basic subtraction table chart to print. Web this handy chart provides all of the answers for subtracting up to 10. The numbers in this free, printable math chart are in black and red so that students can see which number is being subtracted in each equation. The Perfect Starting Point For Little Ones. Display these colorful math anchor charts & posters in your classroom to teach strategies for addition, subtraction, multiplication, and division. All 20 designs are free. Web subtraction worksheets eliminate the monotony of practice. Helping kids get over the “take away” hump can be difficult, which is why our subtraction worksheets supplement. Web Choose From 20 Unique Subtraction Charts And Tables To Help Your Young Learner Or Post In Your Classroom. Students can take printable subtraction 2 times tables, subtraction 3 times tables, subtraction 4 times tables.subtraction 5 times tables,. Web students can take printable subtraction 2 up to 100 times tables, subtraction 3 up to 15 times tables, subtraction 4 times tables, subtraction 5 up to 20 times tables,.	{"url":"https://data1.skinnyms.com/en/subtraction-chart-printable.html","timestamp":"2024-11-09T17:43:11Z","content_type":"text/html","content_length":"31799","record_id":"<urn:uuid:899be273-d2db-4fdc-bcab-535dd3520e49>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00553.warc.gz"}
hello! ðŸ‘‹ I'm june, a PhD student working withÂ Marco Gaboardi at Boston University. interested in programming languages, software verification, data privacy, concurrency, and web development. want to get in contact? my email is june@junewunder.com and I am on mathstodon.xyz Pipelines and Beyond: Graph Types for ADTs with Futures Francis Rinaldi, june wunder, Arthur Azevedo de Amorim, Stefan K Muller Principles of Programming Languages (POPL), 2024 [acm library] Â [arXiv] Â [artifact] Parallel programs are frequently modeled as dependency or cost graphs; such graphs can be used to detect various bugs, or simply to visualize the parallel structure of the code. However, such graphs reflect just one particular execution and are typically constructed in a post-hoc manner. Graph types, which were introduced recently to mitigate this problem, can be assigned statically to a program by a type system and compactly represent the family of all graphs that could result from the program. Unfortunately, prior work is restricted in its treatment of futures, an increasingly common and especially dynamic form of parallelism. In short, each instance of a future must be statically paired with a vertex name. Previously, this led to the restriction that futures could not be placed in collections or be used to construct data structures. Doing so is not a niche exercise: such structures form the basis of numerous algorithms that use forms of pipelining to achieve performance not attainable without futures. All but the most limited of these examples are out of reach of prior graph type systems. In this paper, we propose a graph type system that allows for almost arbitrary combinations of futures and recursive data types. We do so by indexing datatypes with a type-level vertex structure, a codata structure that supplies unique vertex names to the futures in a data structure. We prove the soundness of the system in a parallel core calculus annotated with vertex structures and associated operations. Although the calculus is annotated, this is merely for convenience in defining the type system. We prove that it is possible to annotate arbitrary recursive types with vertex structures, and show using a prototype inference engine that these annotations can be inferred from OCaml-like source code for several complex parallel algorithms. Bunched Fuzz: Sensitivity for Vector Metrics june wunder, Arthur Azevedo de Amorim, Patrick Baillot, Marco Gaboardi European Symposium on Programming (ESOP), 2023 [Springer] Â [arXiv] Â [pdf] Â [pdf (rendered w/ OpenDyslexic)] Program sensitivity measures the distance between the outputs of a program when run on two related inputs. This notion, which plays a key role in areas such as data privacy and optimization, has been the focus of several program analysis techniques introduced in recent years. Among the most successful ones, we can highlight type systems inspired by linear logic, as pioneered by Reed and Pierce in the Fuzz programming language. In Fuzz, each type is equipped with its own distance, and sensitivity analysis boils down to type checking. In particular, Fuzz features two product types, corresponding to two different notions of distance: the tensor product combines the distances of each component by adding them, while the with product takes their maximum. In this work, we show that these products can be generalized to arbitrary Lp distances, metrics that are often used in privacy and optimization. The original Fuzz products, tensor and with, correspond to the special cases L1 and Lâˆž. To ease the handling of such products, we extend the Fuzz type system with bunches---as in the logic of bunched implications---where the distances of different groups of variables can be combined using different Lp distances. We show that our extension can be used to reason about quantitative properties of probabilistic programs. I spell my name "june wunder" in all lowercase. people referring to me in any context should use this stylization.	{"url":"https://junewunder.com/","timestamp":"2024-11-09T03:10:42Z","content_type":"text/html","content_length":"13064","record_id":"<urn:uuid:7a8e132f-5d88-4275-bced-352527dc8892>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00857.warc.gz"}
Price of Anarchy for Graphic Matroid Congestion Games This paper analyzes the quality of pure-strategy Nash equilibria for symmetric Rosenthal congestion games with linear cost functions. For this class of games, the price of anarchy is known to be (5N-2)/(2N+1), where N is the number of players. It has been open if restricting the strategy spaces of players to be bases of a matroid suffices to obtain stronger price of anarchy bounds. This paper answers this open question negatively. We consider graphic matroids, where each of the N players chooses a minimum cost spanning tree in a graph with linear cost functions on its edges. We provide constructions of graphs for N=2,3,4 and for unbounded N, where the price of anarchy attains the known upper bounds (5N-2)/(2N+1) and 5/2, respectively. These constructions translate the tightness of algebraic constraints into combinatorial conditions which are necessary for tight lower bound instances. The main technical contribution lies in showing the existence of recursively defined graphs which fulfill these combinatorial conditions, and which are based on solutions of a bilinear Diophantine equation. Original language English Title of host publication Algorithmic Game Theory - 17th International Symposium, SAGT 2024, Proceedings Editors Guido Schäfer, Carmine Ventre Publisher Springer Pages 371-388 Number of pages 18 ISBN (Print) 9783031710322 Publication status Published - Sept 2024 Event 17th International Symposium on Algorithmic Game Theory, SAGT 2024 - Amsterdam, Netherlands Duration: 3 Sept 2024 → 6 Sept 2024 Publication series Name Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics) Volume 15156 LNCS ISSN (Print) 0302-9743 ISSN (Electronic) 1611-3349 Conference 17th International Symposium on Algorithmic Game Theory, SAGT 2024 Country/Territory Netherlands City Amsterdam Period 3/09/24 → 6/09/24 • 2024 OA procedure • Matroid • Minimum Spanning Tree • MST • POA • Price of Anarchy • Congestion Game Dive into the research topics of 'Price of Anarchy for Graphic Matroid Congestion Games'. Together they form a unique fingerprint.	{"url":"https://research.utwente.nl/en/publications/price-ofanarchy-forgraphic-matroid-congestion-games","timestamp":"2024-11-03T15:53:24Z","content_type":"text/html","content_length":"53816","record_id":"<urn:uuid:c6205257-7e03-44cb-a945-12eef3cff77e>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00632.warc.gz"}
Lesson: Calculating the mean from a grouped frequency table \| Foundation \| KS4 Maths \| Oak National Academy Calculating the mean from a grouped frequency table I can calculate the mean from a grouped frequency table. Calculating the mean from a grouped frequency table I can calculate the mean from a grouped frequency table.	{"url":"https://www.thenational.academy/teachers/programmes/maths-secondary-ks4-foundation/units/comparisons-of-numerical-summaries-of-data/lessons/calculating-the-mean-from-a-grouped-frequency-table","timestamp":"2024-11-03T04:21:58Z","content_type":"text/html","content_length":"287188","record_id":"<urn:uuid:1e828e9b-d1d7-455a-a24c-3486f26602fa>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00834.warc.gz"}
Cognizent Placement Papers. Download In PDF / Word Format - IndianFresher.com COGNIZENT Placement Paper : CTS PAPER IN N.I.T. JAMSHEDPUR Booklet color: Red 1. Using the digits 1,5,2,8 four digit numbers are formed and the sum of all possible such numbers. 106656 2. Four persons can cross a bridge in 3, 7, 13, 17 minutes. Only two can cross at a time. Find the minimum time taken by the four to cross the bridge. 20 3. Find the product of the prime numbers between 1-20 Ans:9699690 4. 2, 3, 6, 7--- using these numbers form the possible four digit numbers that are divisible by 4. ans----8 5. Two trains are traveling at 18kmph and are 60 km apart. There is flying in the train. It flies at 80kmph. It flies and hits the second train and then it starts to oscillate between the two trains. At one instance when the two trains collide it dies. At what distance travel by the fly. Ans---12km 6. There are 1000 doors that are of the open-close type. When a person opens the door he closes it and then opens the other. When the first person goes he opens-closes the doors ion the multiples of 1 i.e., he opens and closes all the doors .when the second goes he opens and closes the doors 2, 4 6 8 resly. Similarly when the third one goes he does this for 3 6 9 12 15^th doors resly. Find number of doors that are open at last.666 7. There are 9 balls of these one is defective. Find the minimum no. of chances of finding the defective one.2 8. There are coins of Rs.5, 2, 1,50p, 25p, 10p, 5p. each one has got a weight. Rs 5 coin weighs 20gms.find the minimum number of coins to get a total of 196.5gms. 9. A can do a work in 8 days, B can do a work in 7 days, C can do a work in 6 days. A works on the first day, B works on the second day and C on the third day resly.that is they work on alternate days. When will they finish the work.(which day will they finish the work) (7+7/168)->>8 10. A batsman scores 23 runs and increases his average from 15 to 16. Find the runs to be made if he wants top Inc the average to 18 in the same match. 39 11. A man sells apples. First he gives half of the total apples what he has and a half apple. Then he gives half of the remaining and a half apple. He gives it in the same manner. After 7 times all are over. How many apples did he initially have?Ans:127 12. In a club there are male and female members. If 15 female quit then the number of males will become double the number of males. If 45 males quit no. of female becomes five times the number of males. Find the number of females.Ans: 160/3, 83/3 13. When I was married 10 years back my wife was the sixth member of my family. Now I have a baby. Today my father was dead and I had a new baby. Now the average age of my family is the same as that when I was married. Find the age of my father when he was. 60 14. I and two of my friends were playing a game. For each win I get Rs 3. Totally I had three wins. Player 2 got Rs9 and player 3 got Rs 12. How many games had been played? 10 15. A person gives a secret to two other persons in 5 minutes. How long will he take to tell the secret to 768 people? 16. There are 40 seats in a bus. People agree to share the money for the number of seats. The total money comes to 70.37. How many seats were free?Ans: 31 17. I had Rs100 and I play. If I win I will hav Rs110 and if I lose I will have Rs90. at the end I have 2 wins and 2 lose. How much do I have? Ans: Rs. 100 18. There were sums related to diagrams. They asked to calculate the areas of the circle, rectangle that were enclosed in other objects. They were simple. There were many questions on logical Eg: There are two identical islands. Same tribe lives in the islands. But their receptiveness This is the question. There were four choices and we have to select the most appropriate For the abov one the answer is ----- because of climatic changes There was a question in which they gave a polygon with all the external angles. We have to calculate the asked interior 19 A says " the horse is not black". B says " the horse is either brown or grey." C says " the hoese is brown" At least one is telling truth and atleast one is lying. tell the colour of horse? Answer : grey 20. A son and father goes for boating in river upstream . After rowing for mile son notices the hat of his fathefalling in the river.After 5 min. he tells his father that his hat has fallen. So they turn round and are able topick the hat at the point from where they began boating after 5min. Tell the speed of river? Ans...6 miles/hr 21. A+B+C+D=D+E+F+G=G+H+I=17 where each letter represent a number from 1 to 9. Find out what does letter D and G represent if letter A=4. (8 marks) Ans. D=5 G=1 22. Argentina had football team of 22 player of which captain is from Brazilian team and goalki from European team. For remainig palayer they have picked 6 from argentinan and 14 from european. Now for a team of 11 they must have goalki and captain so out of 9 now they plan to select 3 from rgentinian and 6 from European. Find out no of methods avilable for it. (2 marks) Ans : 160600 ( check out for right no. 6C3 * 14C6) 23 Three thives were caught stealing sheep, mule and camel. A says " B had stolen sheep " C says " B had stolen mule" B says he had stolen nothing. The one who had stolen horse is speaking truth. the one who had stolen camel is lying . Tell who had stolen what? Ans. A- camel ;B- mule ;C- horse 24 A group of friends goes for dinner and gets bill of Rs 2400 . Two of them says that they have forgotten their purse so remaining make an extra contribution of Rs 100 to pay up the bill. Tell the no. of person in that group. Ans - 8 person 25 In acolony there are some families. Each of them have children but different in numbers.Following are conditions: A) No of adult no of sons no of daughters no of families. B) Each sister must have atleast one brother and should have at the most 1 sister. C) No of children in one family exceeds the sum of no of children in the rest families. D) Tell the no of families.(5 marks) Ans : 3 families 26 There are 6 people W,H,M,C,G,F who are murderer , victim , judge , police, witness, hangman. There was n eye witness only circumtancial witness. The murderer was sentenced to death. read following statement and determne who is who. 1. M knew both murderer and victim. 2. Judge asked C to discribe murder incident. 3. W was last to see F alive. 4. Police found G at the murder site. 5 H and W never met.	{"url":"https://www.indianfresher.com/download-placement-papers-for-cognizent/","timestamp":"2024-11-11T18:01:59Z","content_type":"text/html","content_length":"173880","record_id":"<urn:uuid:de2dbbd3-6e83-469e-b9f8-70d852520c8e>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00728.warc.gz"}
Understand the Changes to the Customer Journey Complexity Score 2023 Jan 17 8:22 AM Previous year, we released a feature that allows customers to score the complexity of their journey models. Now (with the release of January 17, 2023), we have updated the scoring approach, based on customer conversations (user research) as well as data-driven insights (comprehensive process model data analysis). This blog post provides a walk-through of the revised scoring approach. Key novelties are: • Separation of journey model dimensions and the operational complexity of the underlying processes • Nuanced consideration of nested (sub)-processes by recursion; • Revised thresholds and weights based on process model data analysis; • ‘T-shirt sizing’ of complexity for an intuitive assessment as ‘low’, ‘medium’, and ‘high’ complexity. Organizational Complexity in Business Process Management In the context of business process management, complexity can be defined as the “non routineness, difficulty, uncertainty, and interdependence […] associated with [organizational] activities” [1]. Removing “difficulty” – a hard-to-define property – and merging “non routineness” and “uncertainty” simplifies the definition and gives us journey complexity and process complexity: • Journey complexity provides a measure of levels of uncertainty, and interdependence of a journey through an organization. • Process complexity provides a measure of levels of uncertainty, and interdependence of a business process. Because a journey typically involves several processes, journey complexity takes into consideration (i.e., aggregates) the complexity scores of these processes. For journey and process complexity, we have developed model-based scoring algorithms, whose behavior we describe below. Journey Complexity versus Journey Model Dimensions In the new version of customer journey complexity score, we distinguish between journey complexity and journey model dimensions: • Journey complexity considers the journey’s operational complexity, i.e., the complexity of the underlying processes that are directly or indirectly linked (via value chain diagrams, processes, or other journeys). Journey complexity in SAP Signavio Journey Modeler. • Journey model dimensions provide an at-a-glance overview of the size of the journey table (grid size), as well as of the number of non-empty fields. Detailed counts, e.g., of the number of linked personas and images are provided as well. Journey model dimensions in SAP Signavio Journey Modeler. Elements of Process Complexity The complexity score of a process model (i.e., of a BPMN diagram) is determined based on the following elements. • Flow: How many decision and parallelism splits are in a business process and how deeply are they nested? While there are two perspectives on flow complexity, i.e., decision and parallelism, these perspectives are intertwined, for example when a “decision gateway” (any gateway that is not a parallel/ AND gateway) is nested into a parallel gateway. A decision implies uncertainty, parallelism implies interdependence. A parallel gateway has a base complexity of 1, a decision gateway has a base complexity of 1.5. For each gateway, base complexity is multiplied by its nesting level. Then, all scores are summed up to a pre-normalized score. This score is then normalized, on a scale from 0 to 1, where a pre-normalized score of 0 indicates minimal complexity, and a pre-normalized score of 20 or higher indicates maximal complexity (1). For instance, the process depicted by the figure below contains two nested XOR splits, and hence has a pre-normalized score of 4.5 (1.5 + 2 * 1.5) and a final flow complexity score of 0.225. • Handovers: How many handovers between roles are in the process? More roles/actors in a process imply more interdependence and complexity. For each handover to a “new” role we add 1.5 to the handover base score. For each handover back to a role that has already been involved in the process, we add 1 to the handover base score. The score is then normalized, on a scale from 0 to 1, where a base score of 1.5 or lower indicates minimal complexity (0), and a base score of 10 or higher indicates maximal complexity (1). For instance, our example process has three handovers, which amounts to a handover base score of 4.5 (3 * 1.5), which is then normalized to 0.353. • IT systems: How many IT Systems are there in a process, and to what extent are they accessed by multiple roles? Each IT system gets an initial base score of one, when it is first utilized via a task, and each additional utilization adds 1.5 to the base score. For all IT systems, the base scores are summed up to a total IT system complexity base score. The score is then normalized, on a scale from 0 to 10, where a total base score of 1.5 or lower indicates minimal complexity (0), and a total base score of 10 or higher indicates maximal complexity (1). For instance, our example process has two IT systems that are used by two tasks and three tasks, respectively. This amounts to an IT system complexity base score of 6 (2 * 1.5 * 2), which is then normalized to 0.6. • Documents and data objects: How many data objects (incl. documents, i.e., we use ‘data object’ as an umbrella term for BPMN data objects and documents) are in a process, and to what extent are these data objects accessed by multiple roles? Data object complexity is computed in the same way as IT system complexity. Each data object gets an initial base score of one, when it is first utilized via a task, and each additional utilization adds 1.5 to the base score. For all data objects, the base scores are summed up to a total data object complexity base score. The score is then normalized, on a scale from 0 to 1, where a total base score of 1.5 or lower indicates minimal complexity (0), and a total base score of 10 or higher indicates maximal complexity (1). For instance, our example process has two data objects, each of which are used by two roles. This amounts to a data object complexity base score of 6 (2 * 1.5 * 2), which is then normalized to 0.6. • Linked processes: How many other processes are linked via link events to the process model? For the sake of simplicity, each linked sub-process adds 3 to the linked process complexity base score, and each process that is linked via an event adds 1. The score is then normalized, on a scale from 0 to 1, where a base score of 2 or lower indicates minimal complexity (0), and a base score of 10 or higher indicates maximal complexity (1). Because this process does not have any process that is linked, its linked process complexity score is 0. Note: sub-processes are managed via recursion, i.e., instead of a fixed complexity addition, the entire sub-process is scored in detail, until up to four levels of recursion. If the nesting-level of the underlying process landscape is too deep, a warning will be displayed, and an approximate score (lower bound) is Process-Level and Journey-Level Aggregation To determine the final complexity score of a process, all sub-scores are summed up, weighted, and then averaged. Here, the weights are as follows: • Flow and handover complexity: 35% each; • IT system, data object, and linked process complexity: 10% each. On journey level, the complexity of each linked process is multiplied by 0.2 and finally, the entire score is computed by multiplying by 100 – based on our assessment, a complexity score of more than 100 will only be achieved in very rare cases. T-Shirt Sizing To provide an intuitive assessment of journey complexity, the complexity score is mapped to a ‘T-shirt size’, ranging from very low to very high, according to the following mapping, which has been informed by systematic estimates based on the complexity scores of thousands of process models, (x is score value): In summary, the update to complexity score brings the following changes and benefits: • Distinguishing between the underlying operational complexity that comes from linked processes and the dimensions of the journey model makes the scores easier to interpret and more actionable. • Considering the scores of entailed (sub)-processes by recursion and refining scores and thresholds based on real-world data provides a more accurate complexity assessment. • Providing ‘T-shirt sizes from low to high complexity facilitates an intuitive assessment. [1] Jahangir Karim, Toni M. Somers & Anol Bhattacherjee (2007) The Impact of ERP Implementation on Business Process Outcomes: A Factor-Based Study, Journal of Management Information Systems, 24:1, 101-134, DOI: 10.2753/MIS0742-1222240103	{"url":"https://community.sap.com:443/t5/technology-blogs-by-sap/understand-the-changes-to-the-customer-journey-complexity-score/ba-p/13564269","timestamp":"2024-11-08T22:18:28Z","content_type":"text/html","content_length":"186288","record_id":"<urn:uuid:961890c4-44b9-410d-b86e-4fe22995cbd7>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00682.warc.gz"}
Modeling Spaced Repetition in Course Design - TOPR Instructors frequently depend on high stake assessments such as final exams to evaluate students’ knowledge at the end of courses. Those assessments could be timed and comprehensive. A reality that worries several students because high stake assignments can affect greatly their grades at the end of a semester if they do not perform as anticipated. Moreover, instructors often see that students do not remember key concepts from prerequisite courses. This could be due to students’ study habits of cramming information right before the examinations. Hence, what can instructors do to alleviate all those, so called, challenges? A recurrent question that is necessary to be unraveled now is that the COVID-19 pandemic has made the condition worse. In March 2020, most universities went into remote teaching mode and the instructors had to suddenly change from face-to-face instruction to remote instruction. It was difficult to do with a sudden closure of campuses, and they needed to make major rearrangements in teaching strategies and assessments (Maalem et al., 2020). They also had to think outside the standard boxes to generate various possible solutions (Hodges et al., 2020). In this article, we examine modeling and promoting “spaced repetition”, known also as “spaced practice”, as a key strategy to support long term retention of materials learnt. Strategy: Spaced Repetition Spaced repetition started with Ebbinghaus’s famous experiment of forgetting curve (1880-1885), when Ebbinghaus’ goal was to find a legitimate relation between memory retention and time-since-learning (Murre & Dros, 2015). Understandably, retention varies inversely with time, meaning memories decay with elapsed time (Subirana et al., 2017). Hence, an intervention is needed to bring it back to prior state. The more interventions the better. Those interventions are space repetitions, and they can be in form of reinforcement, or reviews, or discussions or reflections. The idea is depicted in figure 1 below: Figure 1. Forgetting curve concept based on Ebbinghaus. Research reported benefits of spaced repetition (Voice & Stirton, 2020), has been widely used in both medical and law education (Teninbaum, 2017). Consequently, the purpose here is to take advantage of it in the designing of mathematics courses at the University of Central Florida, Orlando. We recognize that many of our students may not know how to study, they may not know how memorization and learning work. Large majority of our students often struggle with time management and balancing school with work, or other personal factors. We present our observations and thoughts one by one. Link to scholarly artifact(s) Artifact One The example which we use in this paper is “Calculus with Analytic Geometry I (Calculus I)” course. The instructor usually has a brief chance in each lecture to cover a topic and move on to the next one. As a result, a common path for the student is to see the content during the lecture, practice applications by solving homework problems, and study for midterm and Final. Probably most of them will not revisit what they study for a test until they study for the final examination. This approach to learning affects retention; for example, retention decreases greatly when a student uses cramming to study for examinations (forgetting curve in Figure 1). As far as Calculus I is concerned, the topics that the students who pass the course may not remember well are the “chain rule”, “related rates”, “method of curve sketching”, “optimization problems”, and “integration by substitution”. The experimented redesign based on spaced repetition is shown in Figure 2. Figure 2. Spaced Repetition Redesign in Calculus I. In the proposed approach, different interventions are arranged, and the weighted grading system is adjusted to emphasize all categories. For instance, the percentage value of a grade item (or a single assignment) cannot be 20% (or more) of the final grade. In other words, it becomes tricky to point out to a single assignment as a high-stake assignment. This course is taught via distance learning with adoption of both synchronous and asynchronous modes during the pandemic. Proficiency check is a diagnostic assessment that allows us to evaluate students’ prior knowledge and give them feedback on what they need to review; and it is helpful for the instructor to know which topics to include in “just in time reviews”. Recorded lecture, discussions, recitation activities are asynchronously delivered via Learning Management System (LMS) and give students the benefit of listening to a well- prepared talk. Live meetings during assigned class times, assigned recitation (small groups) meetings, and scheduled office hours are synchronous. They are delivered via Zoom. They help to build a learning community in the course. Artifact Two To improve retention several ideas can be discussed, for example, rearranging topics to allow for spaced repetition. Instructor’s expertise is helpful to not depend on the textbook sequence and manage topics so that there is a chance for many important concepts to be practiced several times. This artifact is selected because it is not commonly used in mathematics courses that meet live to integrate asynchronous “discussions” in LMS. When this is used some students may complain about having to do research and write discussion posts. Hence, it is helpful to provide a rationale for using the discussions. They are delivered in LMS, and students are encouraged to participate early. They must submit their work before seeing an existing post. After reading the posts of classmates they choose to reply to a classmate’s contribution. A rubric is provided to make the assignment grading uniform and transparent. An example of a discussion is shown in Figure 3. It is about related rates and optimization topics that are identified to be the most challenging topics for students to remember after completing the course. Hence, this discussion is one of the interventions that are planned in the course redesign. Figure 3. Example of asynchronous discussion used as an intervention of spaced repetition. As discussed in Artifact 2, it is essential to provide rationale which is the justifying reason behind the distributed practice, and it should be accounted for as an integral part of instruction (CSUN Undergraduate Studies, 2021). Without rationale, students’ perception to distributed practice may not be same as ours because they are not the ones who set up the course. It is important not to assume that multiple assignments are spaced repetition if they were not planned for that purpose. That is why writing rationale while designing the course helps in identifying how those assignments are linked and work together. This strategy was in implemented during fall 2020 by the authors. Overall, results show an increase of success rate, and a better student-engagement. Students’ perceptions were positive in the end of the course surveys; for example, one student wrote when asked about what they liked best about the course: “Homework and discussions were helpful to understand the topics.” Another student wrote when asked about suggestions to improve the course: “One suggestion would be to integrate more discussions about the various applications of certain concepts. For example, more questions related to engineering, medicine, etc. are needed to reinforce the important applications of concepts in calculus.” Hence, with careful planning, spaced repetition can also support relevance of the topics. An encouraging result from this research is that students showed better self-efficacy in multiple submissions. Furthermore, when they were asked if they were confident in talking about calculus in public. From the 238 students who responded to an anonymous survey delivered in LMS, 200 respondents or 84% agreed, 37 respondents or 16% disagreed, and 1 respondent chose not to answer. The 84% is a remarkable number of students and provides an encouraging result to practice further. Link to scholarly reference(s) CSUN Undergraduate Studies (2021). Teach with transparency. https://canvas.csun.edu/courses/93131/pages/transparent-assignments Enterprises, S. (2005). How to write a How to Write a. International Journal of Pediatric Otorhinolaryngology, 18(3), 118–119. http://dx.doi.org/10.1016/j.tics.2013.12.008 Hodges, C., Moore, S., Lockee, B., Trust, T., & Bond, A. (2020). The Difference between emergency remote teaching and online learning. EDUCAUSE Review. https://er.educause.edu/articles/2020/3/ Murre, J. M. J., & Dros, J. (2015). Replication and analysis of Ebbinghaus’ forgetting curve. PLoS ONE, 10(7), 1–24. https://doi.org/10.1371/journal.pone.0120644 Rachid, A. M. L., Mohapatra, R., & Chen, B. (2020). Prioritizing strategies for a better transition to remote instruction strategy. https://er.educause.edu/articles/2020/11/ Subirana, B., Bagiati, A., & Sarma, S. (2017). On the forgetting of college academics: At “Ebbinghaus Speed”? EDULEARN17 Proceedings, 1(068), 8908–8918. https://doi.org/10.21125/edulearn.2017.0672 Teninbaum, G. H. (2017). Spaced repetition: A Method for learning more law in less time. The Journal of High Technology Law, 17(2), 273. Voice, A., & Stirton, A. (2020). Spaced repetition: Towards more effective learning in STEM. New Directions in the Teaching of Physical Sciences, 15(15), 1–10. https://doi.org/10.29311/ Lahcen. R., & Mohapatra, R. (2021). Modeling spaced repetition in course design. In A. deNoyelles, A. Albrecht, S. Bauer, & S. Wyatt (Eds.), Teaching Online Pedagogical Repository. Orlando, FL: University of Central Florida Center for Distributed Learning. https://topr.online.ucf.edu/modeling-spaced-repetition-in-course-design/.	{"url":"https://topr.online.ucf.edu/modeling-spaced-repetition-in-course-design/","timestamp":"2024-11-04T18:44:12Z","content_type":"text/html","content_length":"55466","record_id":"<urn:uuid:84cb3081-5c75-42f0-bea9-a3db1b344c67>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00127.warc.gz"}
6 rate of decay Your parents' bloodshot eyes clear up when the planner reveals that an investment with an eight percent growth rate can help your family reach the $120,000 target. Study hard. If you and your parents invest $75,620.36 today, then Dream University will become your reality thanks to exponential decay. About Exponential Decay Calculator . The Exponential Decay Calculator is used to solve exponential decay problems. It will calculate any one of the values from the other three in the exponential decay model equation. Exponential Decay Formula. The following is the exponential decay formula: Exponential Growth and Decay. Exponential k = rate of growth (when >0) or decay (when <0) t = time Take the natural logarithm of both sides:ln(6) = ln(e2k ). 7 Feb 2020 A year later, the rate of decay has decreased to 1.641×106 disintigrations per minute. What is the half life of 9038Sr? Solution. We have r0 = 28 Jan 2019 Radon decay rate data from 2007–2011, measured in a closed canister in the Earth–Sun distance and solar dynamics [1,2,3,4,5,6,7,8,9,10]. CORRECTION OF PRODUCTION RATES IN CHEMOSTAT CULTURES (6). In the more general case, if the decay function is not a simple exponential, but an This decay constant l is specific for each decay mode of each nuclide. The radioactivity or decay rate is defined as the number of disintegrations per unit of time: A Exponential Growth and Decay Word Problems. Write an (1) Bacteria can multiply at an alarming rate when each bacteria splits into two new cells (6) During normal breathing, about 12% of the air in the lungs is replaced after one breath. Exponential Growth and Decay. Strontium-90 has a half-life of 28 days.1. (a) A sample has a mass of 50 mg initially. Find a formula for the mass remaining after t P = 500, T = 3. Enter this formula into a spreadsheet and graph the results as shown. 1. A. B. 2. 3. 4. 5. 6. 7. 8. The quantity dN/dt is the rate of decay of the source or the activity of the A school has a radium 226 source with an activity of 5 mCi (5x10-6x3.7x1010 Bq). 15 May 2018 Now for the bad news: Tooth decay remains the most common chronic disease of U.S. youths aged 6 to 19, researchers concluded. And the We say 'continuous growth rate' to distinguish from 'discrete growth rate'. Page 6. {6} • Growth and decay. Rearranging this equation to express x Exponential Growth and Decay Word Problems. Write an (1) Bacteria can multiply at an alarming rate when each bacteria splits into two new cells (6) During normal breathing, about 12% of the air in the lungs is replaced after one breath. Archaeologists use the exponential, radioactive decay of carbon 14 to need to consider the half-life of carbon 14 as well as the rate of decay, which is –0.693. The quantity dN/dt is the rate of decay of the source or the activity of the A school has a radium 226 source with an activity of 5 mCi (5x10-6x3.7x1010 Bq). 15 May 2018 Now for the bad news: Tooth decay remains the most common chronic disease of U.S. youths aged 6 to 19, researchers concluded. And the 15 May 2018 Now for the bad news: Tooth decay remains the most common chronic disease of U.S. youths aged 6 to 19, researchers concluded. And the The total decay rate of the quantity N is given by the sum of the decay routes; thus, in the case of two processes: − d N ( t ) d t = N λ 1 + N λ 2 = ( λ 1 + λ 2 ) N . {\displaystyle -{\frac {dN(t)} {dt}}=N\lambda _{1}+N\lambda _{2}=(\lambda _{1}+\lambda _{2})N.} The Ocean is Way Deeper Than You Think - Duration: 6:53. RealLifeLore 24,767,480 views Equations of Radioactive Decay 6.2 HALF-LIFE AND MEAN LIFE It is a common practice to use the half-life (T1/2) instead of the decay constant ( ) for indicating the degree of instability or the decay rate of a radioactive nuclide. Rate of Decay Formula. The decay of a particular nucleus cannot be predicted and is not affected by physical influences like temperature, unlike chemical reactions. The rate of isotope decay depends on two factors. The total number of undecayed nuclei present in the system on doubling the average and undecayed nuclei must double the rate of Many quantities grow or decay at a rate proportional to their size. To find the population in 2015, we find P(15) = 700e0.147×15 ≈ 6, 210. Annette Pilkington. Decay rates of human remains in an arid environment. Exposure of large portions of the skeleton usually does not occur until four to six months after death. Jul 28, 2017 · 6 min read. When training deep Common learning rate schedules include time-based decay, step decay and exponential decay. For illustrative 13 Apr 2018 Fifty-six IPF patients were categorized as rapid (RP) or slow progressors (SP) based on whether their FVC decline in the year preceding Radioactive decay rate formula ? I am looking for the early papers on radioactive decay rate formula. 1904 first e. dition.pdf. 24.77 MB. 6 Recommendations for comparison, the decay rate of a specimen of the long-‐lived nuclide 36Cl (half life Figure 6. Power spectrum formed from normalized 32Si measurements. 30 Mar 2016 It seems plausible that the rate of population growth would be proportional to Therefore, if the bank compounds the interest every 6 months,	{"url":"https://topbinhhqxlx.netlify.app/bottenfield10491dy/6-rate-of-decay-fab","timestamp":"2024-11-12T05:23:40Z","content_type":"text/html","content_length":"30217","record_id":"<urn:uuid:33770826-dc79-4b3b-a241-81d9f741419d>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00827.warc.gz"}
4.6 Why model? At the beginning of this chapter I said that many real world phenomena can be modeled with continuous distributions. “So,” you might ask, “what?” Like all models, continuous distributions are abstractions, which means they leave out details that are considered irrelevant. For example, an observed distribution might have measurement errors or quirks that are specific to the sample; continuous models smooth out these idiosyncrasies. Continuous models are also a form of data compression. When a model fits a dataset well, a small set of parameters can summarize a large amount of data. It is sometimes surprising when data from a natural phenomenon fit a continuous distribution, but these observations can lead to insight into physical systems. Sometimes we can explain why an observed distribution has a particular form. For example, Pareto distributions are often the result of generative processes with positive feedback (so-called preferential attachment processes: see Continuous distributions lend themselves to mathematical analysis, as we will see in Chapter 6. 4.7 Generating random numbers Continuous CDFs are also useful for generating random numbers. If there is an efficient way to compute the inverse CDF, ICDF(p), we can generate random values with the appropriate distribution by choosing from a uniform distribution from 0 to 1, then choosing $$ x = ICDF(p) $$ For example, the CDF of the exponential distribution is $$ p = 1- e^{-\lambda x} $$ Solving for x yields: $$x = -log(1-p)/\lambda $$ So in Python we can write def expovariate(lam): p = random.random() x = -math.log(1-p) / lam return x I called the parameter lam because lambda is a Python keyword. Most implementations of random.random can return 0 but not 1, so $1-p$ can be 1 but not 0, which is good, because $log(0)$ is undefined. Exercise 14 Write a function named weibullvariate that takes lam and k and returns a random value from the Weibull distribution with those parameters.	{"url":"https://notebook.community/chappers/Data-Science/Think-Stats/4%20Continuous%20Distributions","timestamp":"2024-11-09T16:11:21Z","content_type":"text/html","content_length":"353844","record_id":"<urn:uuid:fa764efe-bad5-48a1-a24d-364a1c111c6b>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00683.warc.gz"}
Kolmogorov-Smirnov and Kuiper's Tests of Time Variability - CSC Kolmogorov-Smirnov and Kuiper's Tests of Time Variability In the Chandra Source Catalog, a one-sample, two-sided Kolmogorov-Smirnov (K-S) test and a one-sample Kuiper's test are applied to the unbinned event data in each source region to measure the probability that the average intervals between arrival times of events are varying and therefore inconsistent with a constant source region flux throughout the observation. Corrections are made for good time intervals and for the source region dithering across regions of variable exposure during the observation. Note that background region information is not directly used in the K-S and Kuiper's variability tests in the Chandra Source Catalog, but is used in creating the Gregory-Loredo light curve for the background. The results of the K-S and Kuiper's variability tests are recorded in the columns ks_prob / kp_prob and ks_intra_prob / kp_intra_prob in the Source Observations Table and Master Sources Table, respectively. Dither correction One of the ways by which telescope dither introduces variability into light curves is via modulation of the fractional area of a source region as it moves as a function of time over a chip edge/ boundary, or as it moves as a function of time to chip regions with differing numbers of bad pixels or columns. The fractional area (including chip edge, bad pixel, and bad column effects) vs. time curves for source regions are calculated from the data, and are sufficient to correct the K-S and Kuiper's tests used in the Chandra Source Catalog for the effects of dither. This correction is implemented in the K-S/Kuiper's test model by integrating the product of the good time intervals with the fractional area vs. time curve; the cumulative integral of this product is the cumulative distribution function against which the data is compared. For further details, see the memo "Adding Dither Corrections to L3 Lightcurves." Note that the dither correction described above is a geometrical area correction only that is applied to the data; it does not take into account any spatial dependence of the chip responses. For example, if a soft X-ray source dithers from a frontside-illuminated chip to a backside-illuminated chip, the different soft X-ray responses of the two chips could introduce a dither period-dependent modulation of the detected counts that is not accounted for simply by geometrical area changes. The current catalog procedures do not correct for such a possibility; however, warning flags are set if sources dither across chip edges, and a dither warning flag is set if the variability occurs at the harmonic of a dither frequency. The K-S test is a goodness-of-fit test used to assess the uniformity of a set of data distributions. It was designed in response to the shortcomings of the chi-squared test, which produces precise results for discrete, binned distributions only. The K-S test has the advantage of making no assumption about the binning of the data sets to be compared, removing the arbitrary nature and loss of information that accompanies the process of bin selection. In statistics, the K-S test is the accepted test for measuring differences between continuous data sets (unbinned data distributions) that are a function of a single variable. This difference measure, the K-S D statistic, is defined as the maximum value of the absolute difference between two cumulative distribution functions. The one-sided K-S test is used to compare a data set to a known cumulative distribution function, while the two-sided K-S test compares two different data sets. Each set of data gives a different cumulative distribution function, and its significance resides in its relation to the probability distribution from which the data set is drawn: the probability distribution function for a single independent variable x is a function that assigns a probability to each value of x. The probability assumed by the specific value x[i] is the value of the probability distribution function at x[i] and is denoted P(x[i]).The cumulative distribution function is defined as the function giving the fraction of data points to the left of a given value x[i] , P(x < x[i]) ; it represents the probability that x is less than or equal to a specific value x[i]. Thus, for comparing two different cumulative distribution functions S[N1](x) and S[N2](x), the K-S statistic is where S[N](x) is the cumulative distribution function of the probability distribution from which a dataset with N events is drawn. If N ordered events are located at data points x[i] , i = 1, ... , N , then where the x data array is sorted in increasing order. This is a step function that increases by 1/ N at the value of each ordered data point. Kirkman, T.W. (1996) Statistists to Use. Though different data sets yield different cumulative distribution functions, all cumulative distribution functions agree at the smallest and largest allowable values of x (where they are zero and unity, respectively). Given that, it is clear why the K-S statistic is useful: it provides an unbiased measure of the behavior between the endpoints of multiple distributions, where they can be While the K-S test is adept at finding shifts in a probability distribution, with the highest sensitivity around the median value, its power must be enhanced by other techniques to be as good at finding spreads, which affect the tails of a probability distribution more than the median value. One such technique is Kuiper's test, which compares two cumulative distribution functions via the Kuiper's statistic V, the sum of the maximum distance of S[N1](x) above and below S[N2](x): Kirkman, T.W. (1996) Statistists to Use. If one changes the starting point of the integration of the two probability distributions, D[+] and D[-]change individually, but their sum is always constant. This general symmetry guarantees equal sensitivites at all values of x.	{"url":"https://cxc.cfa.harvard.edu/csc1/why/ks_test.html","timestamp":"2024-11-04T08:04:48Z","content_type":"text/html","content_length":"16920","record_id":"<urn:uuid:4336e549-bed3-4e79-9179-b3d3fb5593d7>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00643.warc.gz"}
OpenScience Music Posted on December 15, 2006 by Dan Gezelter Jonathan Coulton‘s music is about as close as one can get to the perfect accompaniment to the OpenScience project. He’s got songs like Code Monkey about the lives and times of coders, and he’s got the best song about fractals ever written: Mandelbrot Set which actually contains the formula for a fractal (including recursion!) in the lyrics. My favorite song, however, is the new internet phenomenon about an interoffice memo from your Zombie coworkers: Re Your Brains. I’m finding his songs incredibly smart with lots of little lyrical and musical easter eggs. If that wasn’t enough, Jonathan’s songs are released with Creative Commons licenses and without DRM of any sort. In short, he’s the closest we’ve got to an open source muse. Go give his a listen and buy his songs! (I’m sure Jonathan’s heard this from a billion people by now, but his formula in the Mandelbrot Set song is actually the formula for the Julia Set. The Mandelbrot set includes something slightly different; the Mandelbrot set includes those points as c is scanned over the complex plane for which the Julia set iterated by z = z^2 + c is connected.) Forget I said that. I’m just amazed someone can make a good song that includes recursive algorithms in the chorus… [tags]music, mathematics, code, science, creative commons[/tags] This entry was posted in Fun. Bookmark the permalink.	{"url":"https://openscience.org/openscience-music/","timestamp":"2024-11-10T06:04:54Z","content_type":"text/html","content_length":"52338","record_id":"<urn:uuid:d28b35fd-8d76-4a1e-a423-e8c9a8778e85>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00780.warc.gz"}
This function selects an appropriate bandwidth sigma for the kernel estimator of point process intensity computed by density.ppp. The bandwidth $\sigma$ is computed as a quantile of the distance between two independent random points in the window. The default is the lower quartile of this distribution. If $F(r)$ is the cumulative distribution function of the distance between two independent random points uniformly distributed in the window, then the value returned is the quantile with probability $f$. That is, the bandwidth is the value $r$ such that $F(r) = f$. The cumulative distribution function $F(r)$ is computed using distcdf. We then we compute the smallest number $r$ such that $F(r) \ge f$.	{"url":"https://www.rdocumentation.org/packages/spatstat.explore/versions/3.0-6/topics/bw.frac","timestamp":"2024-11-06T14:40:42Z","content_type":"text/html","content_length":"68640","record_id":"<urn:uuid:69dd7a67-4aee-462b-8f8b-dde65776f3a6>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00299.warc.gz"}
Online calculator Calculates ellipsoid and spheroid volume and surface area. Scalene ellipsoid Ellipsoid dimensions Ellipsoid is a sphere-like surface for which all cross-sections are ellipses. Equation of standard ellipsoid body in xyz coordinate system is ${x^2 \over a^2}+{y^2 \over b^2}+{z^2 \over c^2}=1$, where a - radius along x axis, b - radius along y axis, c - radius along z axis. The following formula gives the volume of an ellipsoid: ${4 \over 3}\pi a b c$ The surface area of a general ellipsoid cannot be expressed exactly by an elementary function. Knud Thomsen from Denmark proposed the following approximate formula: $S\approx 4 \pi [(a^p b^p + a^p c^ p + b^p c^p )/3]^{1\over p}$, where p=1.6075 If any two of the three axes of an ellipsoid are equal, the figure becomes a spheroid (ellipsoid of revolution). There are two kinds of spheroid: oblate spheroid (lens like) and prolate spheroid (сigar like). Volume of spheroid is calculated by the following formula: ${4 \over 3}\pi a^2 c$ Unlike ellipsoids, exact surface area formulas exist for spheroids: Oblate ellipsoid (spheroid) For oblate spheroid (a = b > c): $S=2\pi\left[a^2+\frac{c^2}{\sin(o\!\varepsilon)} \ln\left(\frac{1+ \sin(o\!\varepsilon)}{\cos(o\!\varepsilon)}\right)\right]$ where angular eccentricity $o\!\varepsilon=arccos ( {c \over a} )$ Prolate ellipsoid (spheroid) For prolate spheroid (a = b < c): $S=2\pi\left(a^2+\frac{a c o\!\varepsilon}{\sin(o\!\varepsilon)}\right)$ where angular eccentricity $o\!\varepsilon=arccos ({a \over c} )$ The Earth's shape is similar to an oblate spheroid with a ≈ 6,378.137 km and c ≈ 6,356.752 km. According to the formula, Earth's surface is about 510050983.92 square kilometers. Similar calculators	{"url":"https://planetcalc.com/149/?thanks=1","timestamp":"2024-11-10T15:13:36Z","content_type":"text/html","content_length":"44518","record_id":"<urn:uuid:76dcaff4-7769-47cc-a39c-0ec845dd0375>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00208.warc.gz"}
(PDF) Domain partitioning material point method for simulating shock in polycrystalline energetic materials Author content All content in this area was uploaded by Waiching Sun on Nov 22, 2022 Content may be subject to copyright. Computer Methods in Applied Mechanics and Engineering manuscript No. (will be inserted by the editor) Domain partitioning material point method for simulating shock in polycrystalline energetic materials Ran Ma ·WaiChing Sun ·Catalin R. Picu ·Tommy Sewell Received: November 19, 2022/ Accepted: date Heterogeneous energetic materials (EMs) subjected to mechanical shock loading exhibit complex thermo-mechanical processes which are driven by the high temperature, pressure, and strain rate behind the shock. These lead to spatial energy localization in the microstructure, colloquially known as “hotspots”, where chemistry may commence possibly culminating in detonation. Shock-induced pore collapse is one of the dominant mechanisms by which localization occurs. In order to physically predict the shock sensitivity of energetic materials under these extreme conditions, we formulate a multiplicative crystal plasticity model with key features inferred from molecular dynamics (MD) simulations. Within the framework of thermodynamics, we incorporate the pressure dependence of both monoclinic elasticity and critical resolved shear stress into the crystal plasticity formulation. Other fundamental mechanisms, such as strain hardening and pressure-dependent melting curves, are all inferred from atomic-scale computations performed across relevant intervals of pressure and temperature. To handle the extremely large deformation and the evolving geometry of the self-contact due to pore collapse, we leverage the capabilities of the Material Point Method (MPM) to track the interface via the Lagrangian motion of material points and the Eulerian residual update to avoid the mesh distortion issue. This combination of features enables us to simulate the shock-induced pore collapse and associated hotspot evolution with a more comprehensive physical underpinning, which we apply to the monoclinic crystal -HMX. Treating MD predictions of the pore collapse as ground truth, head-to-head validation comparisons between MD and MPM predictions are made for samples with identical sample geometry and similar boundary conditions, for reverse-ballistic impact speeds ranging 0.5 km s−1 2.0 km s−1 . Comparative studies are performed to reveal the importance of incorporating a frictional contact algorithm, pressure-dependent elastic stiffness, and non-Schmid type critical resolved shear stress in the mesoscale model. Keywords HMX, Energetic material, Shock, Pore collapse, Material Point Method 1 Introduction Polymer-bonded explosives (PBXs) are highly ﬁlled composites comprising energetic crystallite ﬁller ensconced in a continuous polymer matrix. They are widely used in many civil and military applications. Due to the stored chemical energy of the energetic constituent, these materials may initiate detonation under unexpected external impact and cause vast damage if not stored or processed properly. Numerical Ran Ma, WaiChing Sun (corresponding author) Department of Civil Engineering and Engineering Mechanics, Columbia University, New York, New York Catalin R. Picu Department of Mechanical, Aerospace, and Nuclear Engineering, Rensselaer Polytechnic Institute, Troy, New York Tommy Sewell Department of Chemistry, University of Missouri, Columbia, Missouri 2 Ran Ma et al. simulation assisted-design of microstructure and shock sensitivity evaluation is important in avoiding such This paper focuses on material point modeling of shock-induced pore collapse in both single-crystal and polycrystalline energetic materials undergoing extremely large deformation and evolving contacts. In all numerical examples, the spatial domain is composed of an energetic substance commonly referred to as β-HMX, a monoclinic polymorph of octahydro-1,3,5,7-tetranitro-1,3,5,7-tetrazocine. Numerous mesoscale models have been proposed to correlate the shock resistance of energetic materials under a variety of loading conditions. For example, the statistical crack mechanical (SCRAM) model [Dienes,1978] was extended with viscoelasticity for the polymer phase to predict the non-shock ignition and mechanical response of PBX [Bennett et al.,1998]. This model is further extended with frictional heating of microcracks, melting, ignition, and fast burn for PBX containing microcracks to model the ignition as well as the following late-stage fast burn [Dienes et al.,2006]. Although most energetic materials are brittle under atmospheric pressure, considerable ductility is observed under conﬁning pressure [Wiegand et al., 2011], which is the typical stress state under conﬁned shock loading. Therefore, the SCRAM model is further extended with viscoplasticity [Yang et al.,2018]. In another independent study, a viscoplasticity model for tungsten alloy [Zhou et al.,1994] is extended as the constitutive relation for -HMX which, in combination with the cohesive zone model for fracture and friction [Barua and Zhou,2011], is used to study the ignition probability of PBX under shock loading [Kim et al.,2018]. Besides these mesoscale models, the Johnson-Cook viscoplasticity model is also used for energetic materials, for example, PMMA [Rai et al.,2020] and TATB [Zhao et al.,2020]. With the pressure-dependent melting temperature and viscosity evaluated by atomistic models [Kroonblawd and Austin,2021], different extensions of the Johnson-Cook model are compared with the atomistic shock simulation as the benchmark. These material models often assume isotropic constitutive response and material frame indifference is enforced by an objective stress rate. However, the manufacturing defect on the order of 10 µm or smaller is also an important factor that triggers the shock ignition [Barton et al.,2009]. These defects cannot be observed easily in experiments and therefore corresponding studies rely mainly on theoretical studies. At these characteristic length scales, the material anisotropy of the embedded energetic material may have a strong inﬂuence on the predicted shock resistance. Therefore, anisotropic crystal plasticity models are necessary to precisely capture the microstructural evolution and the resultant evolving anisotropy at the crystal level. Previous work, such as [Barton et al.,2009], develops a rate-dependent crystal plasticity model for energetic material suitable for deformation within a large spectrum of strain rates. In this work, two dislocation motion resistances, including the thermal activation barrier and the phonon drag limit, are combined in the constitutive relation representing major strain hardening mechanisms at different strain rates. This model is further extended with a chemical reaction model [Austin et al.,2015], and shock simulations with different shock pressure, pore size, thermal conductivity, crystal ﬂow strength, and liquid viscosity are performed to study their effects on the reactive mass, the shear band morphology, and the peak temperature. A parametric study of this model shows that the plastic dissipation and the temperature rise are less sensitive to the anisotropic elastic coefﬁcients than to the plasticity model [Zhang and Oskay,2019], which is consistent with an independent parametric study using the Johnson-Cook model [Das et al.,2021]. In this work, the interpenetration of crystals can be prevented via the level set function that represents the interface. However, since there is only one velocity ﬁeld deﬁned across the interface, the contacts of the collapsed region of the pores are essentially glued together without any tangential slip. The mesoscale crystal plasticity models could be a feasible alternative that allows direct validation inferred from underlying physical observations (e.g. the plastic slip of a slip system), and may be easily extended by incorporating other deformation mechanisms such as fracture and deformation twinning [Clayton and Knap,2011,Ma and Sun,2021]. For instance, combining a pressure-dependent thermoelasticity free energy of RDX [Austin and McDowell,2011] and a dislocation-density based crystal plasticity model suitable for high strain rates [Austin and McDowell,2011], [Luscher et al.,2017] introduced a mesoscale material model for crystalline RDX with further validation against an impact experiment. The crystal plasticity ﬁnite element model is further compared with the Knoop indentation test to determine the active slip systems in -HMX [Zecevic et al.,2021]. Recently, phase-ﬁeld based fracture models [Grilli and Koslowski,2019], twinning models with explicit representation [Ma and Sun,2021] and implicit representation [Zhang and Oskay,2019,Zecevic et al.,2020] are also combined with mesoscale crystal plasticity models for more physical insights into the shock responses observed in energetic materials. Domain partitioning MPM for simulating shock in energetic materials 3 Nevertheless, introducing high-ﬁdelity simulations may unavoidably require a substantial amount of material parameters. While a subset of these parameters could be physically associated with dislocation theories, the rest of these parameters may lack physical underpinning. Calibrating these parameters is usually achieved through solving inverse problems to reproduce the experimentally measured shock velocity [Dick et al.,2004]. The lack of justiﬁcations from underlying physics might lead to overﬁtting and hence increase the difﬁculty of generating reliable simulations. In Ma et al. [2021], an attempt is made to incorporate atomistic simulation results to calibrate a small- deformation non-Schmid crystal plasticity model. The result suggests that incorporating the pressure sensitivity into both the elasticity and the yield function for the slip system may lead to a mesoscale model more compatible with the atomistic counterpart. However, the geometrical nonlinearity, the evolving contact kinematics, the constitutive responses of the crystal interfaces, and the pressure dependence of the melting curves that govern the phase transition from solid to liquid have not yet been captured. The objective of this paper is to leverage the salient features of the material point method to capture evolving geometry such that a more realistic multiplicative mesoscale plasticity model for the bulk and interfaces of energetic materials can be used to replicate the contact mechanics of the pore collapses and ultimately predict the hotspot formation, a crucial mechanism that triggers an explosion. In most of the hydrocodes for explosion simulations (e.g., Pierazzo and Melosh [2000], Sambasivan et al. [2013], Mudalige et al. [2014]), a single velocity ﬁeld is used for contacting bodies where the surface friction associated with the pore collapse process is ignored. This simpliﬁcation may lead to under-evaluated hotspot temperature due to the negligence of frictional heating, especially for irregularly shaped embedding defects. In particular, the ﬁeld-gradient domain partition treatment in the material point method (MPM) enables us to capture the evolving contact geometry among multiple bodies as well as self-contacts. As demonstrated in our numerical examples on both single- and poly-crystal simulations and the comparisons with the MD simulations, this improvement is crucial for us to accurately simulate the secondary shock due to the pore collapse. This paper will proceed as follows. Section 2describes in detail the atomistic-model informed crystal plasticity model. The stress update algorithm and its implementation in the material point method are also described in detail. Section 3presents the parameter determination procedure using the atomistic results. Section 4presents the validation against the atomistic-scale shock simulations. In Section 5, a comparative study is performed on the importance of the frictional contact algorithm, the pressure-dependent hyperelas- ticity, and the non-Schmid crystal plasticity within the mesoscale model. Section 6summarizes the major results and concluding remarks. 2 Constitutive model We start with the multiplicative decomposition theory, where the deformation gradient is decomposed into the elastic part Feand the plastic part Fpas F=FeFp. (1) as the spatial velocity vector, the spatial velocity gradient L=grad v , which is power conjugate to the Kirchhoff stress τ, is decomposed into the elastic part and the plastic part as: FpFp−1Fe−1, (2) where Leis the elastic velocity gradient, and Lpis the plastic velocity gradient. We further deﬁne the elastic right Cauchy-Green deformation tensor and the elastic Green-Lagrangian deformation tensor Eeto measure the elastic deformation: Ce=FeT Fe,Ee=1 2(Ce−I). (3) Then, the stress Seis deﬁned as the contravariant pull-back of the Kirchhoff stress τ[Anand,2004], Se=Fe−1τFe−T, (4) which is power-conjugate to the elastic deformation tensor Ee, that is, Se:˙ Ee=τ: sym[Le]. 4 Ran Ma et al. A general form of the total free energy function per unit mass takes the form ψ=ψ(Fe,γ,T), (5) represents a collection of internal variables, and is the absolute temperature. The total free energy consists of three parts: the elastic strain energy due to the reversible elastic deformation, the stored energy due to the accumulation of crystal defects, and the thermal energy. In order to further simplify the model development, we assume that the elastic free energy is temperature independent, which means that the elastic stiffness is independent of the temperature and the thermal expansion is neglected. This assumption is justiﬁed by atomistic results indicating that the temperature dependence of the elastic stiffness is much weaker than the pressure dependence [Pereverzev and Sewell,2020]. We neglect the strain energy associated with dislocation storage, as customary in constitutive modeling of plasticity. Therefore, a simpliﬁed form of the free energy function is assumed as: ψ=ψe(Ee) + ψT(T), (6) where the speciﬁc form of the free energy function is discussed in detail in the following Sections. 2.1 Atomistically-informed hyperelasticity Traditionally, when modeling the thermo-mechanical response of energetic material under strong shock, the pressure-dependence of the elasticity is modeled through the equation-of-state (EOS), which relates the pressure to the volumetric strain, while keeping the anisotropy constant [Barton et al.,2009]. The uneisen EOS is typically used for -HMX, which is calibrated by the atomistic-scale simulations at various pressure and temperature [Menikoff and Sewell,2002]. However, due to the monoclinic symmetry of the -HMX single crystal, the volumetric part and the deviatoric part of the elastic free energy cannot be fully decoupled, so the pressure depends on the volumetric strain as well as the deviatoric strain. Also, the shear components of the elastic stiffness are pressure-dependent as well [Pereverzev and Sewell,2020], which is usually neglected in most mesoscale models. Taking into account the aforementioned discussions, we develop a nonlinear hyperelasticity model to reproduce the pressure-dependent elastic stiffness observed in atomistic evaluations [Pereverzev and Sewell,2020]. Departing from the inﬁnitesimal strain counterpart [Ma et al.,2021], we generalize this model to the ﬁnite strain formulation. The geometric nonlinearity of the ﬁnite strain formulation is also considered in the calibration procedures, which will be introduced in Section 3. -HMX unit cell is deﬁned in the space group throughout this paper, as shown in Figure 1. The covariant basis vectors , and are deﬁned to represent the three lattice vectors of the monoclinic unit cell. Taking as the elastic strain measure, we deﬁne three isotropic and four anisotropic strain invariants: I1=tr Ee,I2=1 2tr[Ee]2−tr[Ee2],I3=det Ee, Note that these strain invariants form a subset of the general set of eight strain invariants for materials with monoclinic symmetry [Vergori et al.,2013]. The elastic stiffness at 0 GPa , as well as these strain invariants, are utilized to construct the elastic free energy. Based on the atomistic-scale evaluations where the elastic stiffness coefﬁcients are estimated within a set of pressures between 30 GPa and temperatures between 300 K 1100 K and Sewell,2020], the elastic stiffness increases substantially with pressure while decreases mildly with increasing temperature. This justiﬁes the approximation made here that the elastic free energy is temperature independent. We propose the following form of elastic free energy as 0:Ee, (8) is the elastic stiffness at atmospheric pressure, function is an arbitrary function that fulﬁlls typical stability requirements, and ρ0is the initial density of the energetic material under ambient conditions. Domain partitioning MPM for simulating shock in energetic materials 5 M1 = [100] M2= [010] M3M3= [001] Fig. 1: Monoclinic unit cell of the -HMX crystal in space group. The lattice constants are a=6.53 ˚ b=11.03 ˚ c=7.35 ˚ , and 295 K ) [Eiland and Pepinsky,1954]. The vectors , and denote the basis vectors of the global Cartesian coordinate system. The vectors , and M3indicate the coordinate system of the monoclinic crystal. One advantage of the speciﬁc form of the elastic free energy in Equation is that the elastic stiffness at atmospheric pressure is exactly reproduced. This property signiﬁcantly simpliﬁes the determination of the arbitrary function and the calibration process, as discussed in Section 3. Also, to improve the ﬁdelity of the predicted thermo-mechanical response, the temperature dependence of the elastic free energy should be 2.2 Atomistic-model informed crystal plasticity Due to the large strain rates involved in the shock simulations, crystal plasticity models applicable to a wide range of strain rates are generally used where the thermal activation barrier and the phonon drag limit are typically considered in the model [Barton et al.,2009,Luscher et al.,2017]. These models are calibrated and validated against shock experiments with different single-crystal orientations, specimen thicknesses, and ﬂyer velocities for HMX [Dick et al.,2004], RDX [Hooks et al.,2006], and other energetic materials. These dislocation-based crystal plasticity models, though based on homogenized dislocation theories, are validated only against macroscale velocity measurements from shock experiments and lack further validations from the underlying physics at the atomistic scale. Also, some key properties observed in the atomistic scale models, which may have strong inﬂuences on the predicted hotspot evolution, are not fully accounted for. For example, it is observed in a recent atomistic scale study that the Peierls-Nabarro stress of -HMX increases times depending on the slip system when the conﬁning pressure increases from 10−4GPa to 27 GPa at 300K [Pal and Picu,2019]. In this section, we develop a crystal plasticity model where the pressure dependence of the critical resolved shear stress (CRSS) is accounted for. However, since limited atomistic-scale information regarding the strain rate sensitivity of this crystal is available at this time, we restrict our crystal plasticity model to be rate-independent. We acknowledge that, to improve the ﬁdelity of our model, the strain-rate dependence should be accounted for; this would also simplify the implementation as well as improve the numerical robustness. However, sufﬁcient data from MD simulations to calibrate the rate sensitivity was not available to us within the scope of this research and hence this will be considered in the future. This multiplicative crystal plasticity is formulated via the large-deformation crystal plasticity framework ﬁrst introduced in Anand and Kothari [1996]. In the original framework [Anand and Kothari,1996], the pseudo-inverse of the Jacobian of the active yield criterion functions is approximated through singular-value decomposition (SVD), and slip systems that violate the consistency condition are eliminated from the potential active set. Miehe and Schr oder (2001) further improved this algorithm and compared different pseudo-inverse approaches to approximate the pseudo-inverse of the Jacobian [Miehe and Schr Considering the strongly anisotropic nature of energetic materials with both pressure sensitivity and temperature sensitivity, we generalize the SVD-based slip system selection algorithm such that it is applicable to any hyperelasticity model coupled with the non-Schmid slip system activation rule. represents the total number of slip systems in the crystal. In the intermediate conﬁguration, we deﬁne . . . as the slip direction and the slip plane normal of slip system 6 Ran Ma et al. Following the standard ﬂow rule of crystal plasticity, the plastic velocity gradient is decomposed into the summation of plastic slip rates on each slip system: 0, (9) is the plastic shear strain rate of slip system . Note that the summation is performed over slip systems where each slip system is counted twice with opposite slip directions. This is important for the rate-independent crystal plasticity model, where the plastic multiplier is non-negative to ensure that the Karush-Kuhn-Tucker (KKT) conditions in Equation (13) are fulﬁlled. We further introduce the Mandel stress Ξas Ξ=CeSe=FeT τFe−T. (10) Then, the resolved shear stress τ(α)is further derived using the Schmid law, 0,α=1, . . . , 2Nslip . (11) Note that the Mandel stress is power-conjugate to the plastic velocity gradient , and the resolved shear is power-conjugate to the plastic slip rate , so that the plastic work per unit volume The slip system constitutive relation follows the KKT conditions, that is, the plastic shear strain is non-negative, the yielding function is non-positive, and the consistency condition applies: γ(α)≥0, φ(α)=hτ(α)−g(α)i≤0, ˙ γ(α)φ(α)=0, α=1, . . . , 2 Nslip , (13) where φ(α)and g(α)are the yielding function and the slip system resistance of slip system (α). The slip system resistance (CRSS) has two components: the Peierls-Nabarro stress representing the lattice resistance and the strain hardening associated with the interaction of dislocations. The atomistic scale results suggest that the Peierls-Nabarro stress is strongly pressure-dependent [Pal and Picu,2019] and weakly temperature-dependent [Khan et al.,2018]. We assume that the temperature effect and the pressure effect are separable. The strain hardening component is a function of the dislocation density and was studied in ambient conditions using atomistic models [Khan and Picu,2021]. We represent the slip resistance g(α)as: 1− T−Tref Tm(p)−Tr e f !Mh 1− T−Tref Tm(p)−Tr e f !Mp ,α=1, . . . , 2 Nslip , (14) is the absolute temperature, is the pressure-dependent melting temperature, Tref =298 K the reference temperature, represents the Peierls-Nabarro stress at the reference temperature, represents the strain hardening component at the reference temperature, and are material constants. The temperature should have different softening effects on the Peierls-Nabarro stress and the hardening component , since the physical basis of the temperature dependence of these two terms are different. Therefore, two thermal softening coefﬁcients Mpand Mhare introduced separately. The pressure-dependent lattice friction (Peierls-Nabarro stress), which produces the non-Schmid type yielding criterion, could be deﬁned in multiple different approaches. One possibility is to deﬁne the pressure as the surface compression applied on the slip plane, that is . However, we adopt a different approach where the pressure is deﬁned as the hydrostatic stress 3. This approach is consistent with the corresponding atomistic models where the pressure-dependent Peierls-Nabarro stress Domain partitioning MPM for simulating shock in energetic materials 7 is evaluated [Pal and Picu,2019]. A polynomial model is used to approximate the pressure-dependent Peierls-Nabarro stress g(α) p(p) = (c(α) 3,p≤0α=1, 2, . . . , 2 Nslip , (15) where c(α) 1to c(α) 3are material constants calibrated from the atomistic models. The strain hardening contribution to the ﬂow stress was evaluated in [Khan and Picu,2021] using a combination of atomistic and mesoscale models, and the corresponding ﬂow stress at dislocation densities ranging from 1011 m−2 1015 m−2 was evaluated. Since the dislocation density ρ=1015 m−2 is large compared with the initial dislocation density and is hardly achievable in shock simulations, we take the slip system resistance at 1015 m−2 as the saturation stress. Therefore, the strain hardening contribution approximated as h(γ) = g(α) 3tanh H0γ γ(α)\|dt, (16) is the initial hardening rate, and is the saturation stress of the slip system evaluated at a dislocation density of 1015 m−2and in ambient conditions. The melting temperature also inﬂuences the Peierls-Nabarro stress and strain hardening according to the atomistic model, as shown in Equation . It is also a function of the conﬁning pressure based on the atomistic models, and is approximated by the Simon-Glatzel relation [Kroonblawd and Austin,2021], which takes the following form: Tm(p) = Tm01+p−pre f , (17) where Tm0is the melting point at pre f =0 GPa, and a0and c0are material constants. The crystal plasticity model is replaced by a Newtonian ﬂow model when the temperature the melting point . Although temperature and pressure-dependent viscosity is suggested by atomistic models [Kroonblawd and Austin,2021], a constant viscosity is used in this paper: dev[σ] = 2ηdev[D], (18) is the ﬂuid viscosity, is the symmetric part of the velocity gradient, and is the Cauchy stress. Also note that we do not model the solidiﬁcation process and the chemical reaction in this paper, since our primary focus is on the triggering mechanisms of the detonation but not the physics in the post-detonation regime, 2.3 Thermodynamic consistency To accurately capture the thermo-mechanical behavior of energetic materials under shock loading, our next goal is to ensure that the proposed mesoscale crystal plasticity model fulﬁlls the thermodynamic consistency. We restrict our focus to the shock wave propagation within orders of . At this temporal regime, it is reasonable to assume that the material is under adiabatic conditions, and hence the heat ﬂux and heat source can be ignored. The local form of the ﬁrst law of thermodynamics (energy balance) is γα, (19) 8 Ran Ma et al. is the internal energy per unit mass. The second law of thermodynamics (Clausius-Duhem inequal- ity) requires that the changing rate of entropy is non-negative under adiabatic conditions. Taking advantage of the Legendre transformation ψ=e−TS, the Clausius-Duhem inequality is equivalently derived as: Dint =ρ0T˙ T≥0. (20) where Sis the entropy per unit mass. Then, the Coleman-Noll conditions read, ∂T, (21) which combines with enforcing the KKT conditions Equation , guarantee that the dissipation inequality is always fulﬁlled. Due to the lack of the thermal free energy justiﬁed by the atomistic scale models, we instead employ a phenomenological model to characterize the temperature evolution based on the corresponding atomistic evaluations [Menikoff and Sewell,2002]: T=Tref exp "Ta1 sgn 1 ρcv, (22) is the volumetric strain, is the isochoric speciﬁc heat, and 4.5 is a manually adjusted material parameter to reproduce the Hugoniot relations. A temperature-dependent speciﬁc heat is used [Sewell and Menikoff,2004], are material parameters. The absolute temperature is normalized by the Debye temperature θ(J),θ(J) = θ0J−Γaexp Γb1 J−1, (24) where θ0,Γa, and Γbare material parameters. 2.4 Stress update algorithm We developed an active slip system selection algorithm for the aforementioned rate-independent crystal plasticity model based on the stress update procedure initially proposed by [Anand and Kothari,1996] and later improved by [Miehe and Schr oder,2001]. Our key contribution is the improvement of this algorithm to accommodate a general hyperelastic model and a non-Schmid slip system activation rule with pressure sensitivity and temperature sensitivity. The time integration is semi-implicit in the sense that the temperature of the last converged step used to calculate the thermal-softening coefﬁcient. Since the time step required for the shock simulation is very small, the critical time step to maintain stability for this semi-implicit treatment is likely to be signiﬁcantly larger than the actual time step used. Hence, it is reasonable to assume that the semi-implicit time integration algorithm may remain stable. The internal variable , which is deﬁned in Equation , is a function of the accumulated shear strain, pressure, and temperature. In order to simplify the return mapping algorithm, the trial pressure used to compute the pressure-dependent yield stress and the melting temperature . This pressure is corrected at the end of the stress update algorithm in Step 9, followed by a repeating time integration with the corrected pressure, internal variables, and the melting temperature. Only one additional iteration is normally required to reach a converged pressure because of the small time steps usually used in shock Domain partitioning MPM for simulating shock in energetic materials 9 When determining the resolved shear stress , one may assume that the elastic stretch is small compared with the plastic strain such that approximately equals to [Anand and Kothari,1996]. However, in the shock simulation, the volume change J=det F is not negligible, while the volume-preserving part of the elastic deformation gradient Fe=J−1/3 F is comparably less signiﬁcant. Therefore, we introduce the following approximation to the resolved shear stress τ(α): 0,α=1, . . . , Nslip . Note that the volume change Jis constant within one time step. The Newton-Raphson method is used to solve this system of nonlinear equations. When the active set contains redundant slip systems, the slip system increments cannot be uniquely determined, and the linearization of the residual produces a singular matrix. Upon the detection of a singular lineariza- tion matrix, the singular value decomposition (SVD) strategy is used to approximate the inverse of the linearization matrix [Miehe and Schr¨ In the following stress update algorithm, the subscript denotes variables evaluated at time step which is assumed to be known a prior. Given: Fn+1,Fn,Fp Find: σn+1,Fp Here, the active set is a subset of the 2 slip systems. The slip systems within is active, that is, the shear strain rate 0 and the yield criterion 0 for slip system α∈ A , as demonstrated in the KKT condition in Equation (13). Step 1. Compute the trial elastic states Trial elastic strain: Fe tr =Fn+1Fp−1 tr =Fe T tr Fe tr = (Ce tr −I)/2, J=det[Fn+1], Trial stress: Se tr =ρ0(∂ψ/∂Ee tr),σtr = (1/J)Fe tr Fe T tr ,˜ p=−tr[σtr ]/3, Trial internal variable: g(α) tr =g(γn,Tn,˜ Step 2. Elastic predictor Assemble trial active set Atr ={α\|φ(α)=τ(α) tr −g(α) tr >0} If Atr =∅, then set σn+1=σtr,Fp go to Step 9 Otherwise, An+1=An,∆γ(α)=0. Step 3. Stress and stiffness update n+1,Ee= (FeFeT −I)/2, IJK L =ρ0[∂2ψe/(∂Ee I J ∂Ee KL )],γn+1=γn+∑An+1∆γ(α). Step 4. Compute residual and Jacobian for active slip systems R(α)=τ(α)−g(α),α∈ An+1, 0:CSE :Ce 1− Tn−Tr e f Tm−Tr e f !Mh "1−tanh2 H0γn+1 with τ(α)≈J2/3Se:s(α) 0and g(α)=g(α)(γn+1,Tn,˜ p)is deﬁned in Equation (14). Step 5. Update incremental plastic slip and check convergence If Dis singular, then compute its pseudo-inverse ¯ Otherwise, compute its regular inverse ¯ If q∑α∈An+1[R(α)]2>tol go to Step 3. 10 Ran Ma et al. Step 6. Active set update I: Drop inactive slip systems α=arg min hφ(α)i,α∈ An+1 If ∆γ(α)<0: Update active set An+1← An+1/{α}and go to Step 3. Step 7. Active set update II: Add potential active slip systems α=arg max hφ(α)i,α/∈ An+1 If φ(α)>0: Update active set An+1← An+1∪ {α}and go to Step 3. Step 8. Plastic deformation gradient and stress n+1,Ee= (FeFeT −I)/2, Se=ρ0(∂ψ/∂Ee),σn+1= (1/ J)FeSeFe T ,γn+1=γn+∑An+1∆γ(α). Step 9. Update pressure and temperature Pressure p=−tr[σ]/3, temperature Tn+1=T(J, dissipation) If \|p−˜ p\|>tol: Set ˜ p=pand go to Step 2 Otherwise, exit 2.5 Material point method for shock simulations The shock wave propagation problem is solved by the material point method (MPM) based on the following considerations. First, the material point method is suitable to treat the challenging numerical issues involved in the shock wave propagation problems, including the local large deformation associated with the inter- action between the shock wave and the pore, and the frictional contact associated with the pore collapse process. Furthermore, as the Lagrangian material particles store the deformation history and the internal variables, this historical information can be easily tracked through the trajectory of the material points, which is a property also shared by other meshfree methods, such as the reproducing kernel particle method (RKPM) [Wang et al.,2014] and the smoothed particle hydrodynamics (SPH) [Liu and Liu,2010]. This treat- ment greatly simpliﬁes the implementation of the path-dependent models formulated with multiplicative kinematics (such as the crystal plasticity used in this paper) with profound geometrical nonlinearity [Liu and Sun,2020a,b,Ma and Sun,2022]. The MPM-based shock wave simulation, which is regularized by the artiﬁcial viscosity, is veriﬁed against the analytical solutions of one-dimensional Riemann problems [Ma et al.,2009], and was further applied to model the blast and fragmentation of concrete walls [Hu and Chen, 2006]. The MPM could also be extended with a phase-ﬁeld fracture model to study the dynamic fracture behavior under large distortion [Kakouris and Triantafyllou,2017]. In the following numerical examples, our focus is on predicting the temperature and pressure ﬁeld evolution as a result of the primary shock wave induced by impact and of the secondary shock wave due to the pore collapse. So the duration of the simulations is not tremendously larger than the CFL stability condition, and as a result, we use the MPM formulation with semi-implicit time integration. This time integration algorithm is conditionally stable, so a sufﬁciently small time step is picked to fulﬁll the CFL stability condition. The B-spline function is used as the shape function of the background mesh to avoid cell-crossing instability [Steffen et al.,2008]. The MPM implementation uses TaiChi [Hu et al.,2019], an open-source programming language designed for high-performance computing in the Python environment. 2.5.1 Classical material point method Consider a deformable body occupying region in the Euclidean space. The boundary is divided into the Dirichlet boundary and Neumann boundary . Taking as the body force per unit mass, the strong form and weak form of the balance law of linear momentum is expressed as: ρ˙v=div σ+ρb, ZΩρw·˙vdV=ZΩ(ρw·b−grad w:σ)dV+Z∂Ωt where w∈H1 Ωis the test function and t=σ·nis the surface traction. Domain partitioning MPM for simulating shock in energetic materials 11 In the classical material point method, the domain is discretized into particles, each representing a simply connected subdomain of . Each particle carries the mass , velocity , deformation gradient and other internal variables of the subdomain it represents. The algorithm for solving the initial-boundary value problem is brieﬂy summarized as follows, which serves as the starting point of our modiﬁcation to capture pore collapse induced secondary shock with frictional contact. Step 1. Particle-to-grid projection . Mass and linear momentum are transferred from particles to grids as i=∑pwip mp i=∑pwip mpvn , where represent the mass of grid represent the velocity of grid and particle , and represents the value of the ith b-spline shape function evaluated at the position of particle p. Step 2. Solve grid velocity . The internal force is assembled as fi=∑pgrad wip ·σpJpVp , where the Cauchy stress of particle is the Jacobian of the deformation gradient, and is the initial volume of particle p. The grid momentum is updated explicitly as (mv)n+1 Step 3. Grid-to-particle projection . Particle velocity and its spatial gradient is then projected from the background mesh as: particle velocity p=∑iwip vn+1 and spatial velocity gradient i⊗grad wip . Step 4. Convection . The current particle position and its deformation gradient are then updated as: xn+1 pand Fn+1 p= (I+∆tLn+1 2.5.2 Self contact with frictional sliding When evaluating the shock sensitivity of energetic materials using mesoscale simulations, it is inevitable to model frictional contact associated with the free surfaces embedded in the material, including micro-crack surfaces and pore surfaces. Under the shock wave pressure, the local impact and friction of these free surfaces are also key contributions to the hotspot formation. In the classical hydrocode or material point method where a single velocity ﬁeld is used, two pieces of materials are welded together upon contact and frictional sliding is ignored. Therefore, the hot spot temperature is underestimated as a result of ignoring the frictional heating, especially for irregularly shaped pores. In order to model the self contact with frictional sliding, we utilize the damage-ﬁeld gradient (DFG) algorithm [Homel and Herbold,2017], with speciﬁc adjustments summarized below. When discretizing the initial conﬁguration of the specimen into material particles, one layer of surface particles is extracted and assigned the surface indicator, with the layer thickness equal to the element size of the background mesh. The gradient of the surface indicator is used to separate two groups of contacting particles associated with one grid of the background mesh, such that the multi-body contact algorithm [Huang et al.,2011,Xiao et al.,2021] can be applied. The Coulomb friction model is used throughout this paper, and the friction coefﬁcient takes a constant value based on the experimental measurements of multiple energetic materials (including HMX, RDX, and PETN) at high pressure [Wu and Huang,2010]. Unlike the DFG algorithm where the damage ﬁeld is used as an additional surface separation indicator, we limit our mesoscale model within the continuum mechanics range, and the fracture and associated surface friction will be pursued in future work. We also assume that the plastic dissipation associated with pore collapse is much larger than the pore surface friction, such that the temperature increase due to the surface friction is negligible. Deﬁne surface indicator for each particle, such that 0 for interior particles and for particles within the surface region. Note that the surface indicator is only used to separate contacting bodies instead of computing the surface normal direction. Step 1. Reset surface-indicator gradient of particles . Assemble the surface indicator from particles to grid ζi= (∑pwipmpζp)/(∑pwi p mp) . The surface-indicator gradient at particle is then computed as ∇ζp=∑iζigrad wip .Note that instead of using the spherically symmetric cubic kernel functions to approximate the continuous damage ﬁeld [Homel and Herbold,2017], we use the B-spline shape functions and the classical MPM approach which also avoids the edge effect and produces C1 continuous damage ﬁeld. 12 Ran Ma et al. Step 2. Reset surface-indicator gradient of grids . Search within the compact set of the kernel function of grid for the particle with maximum surface-indicator gradient, and assign it to grid arg max(wip >0)k∇ζpk. Step 3. Partitioning . For each grid point , if the particle set {p\|wip ∇ζp· ∇ζi< is not empty, then grid point is within the contact region, and particle quantities are assembled to one of the two background meshes denoted by ς∈ {0, 1}, where ς=0 if ∇ζp· ∇ζi≥0 and ς=1 otherwise. Step 4. Particle-to-grid projection . Based on the partitioning indicator , mass and linear momentum are assembled from particles to grids as ςi=∑pwip mp ςi=∑pwip mpvn . The internal force is also assembled as fςi=∑pgrad wip ·σpJpVp . The surface indicators at grid are deﬁned as ζςi= (∑pwipmpζp)/(∑pwi p mp). Step 5. Check separability condition . The separability condition is deﬁned as ζςi>ζmin = ∀ς∈ { 0, 1 If grid is not separable, the contact force is determined to enforce the continuity of the velocity . Otherwise, the frictional contact force is determined through the Coulomb model. Step 6. Contact force . The normal directions of the two contacting bodies are ˆnςi=∑pmpgrad wip / k∑pmpgrad wip k . Consider that the two normal directions may not be parallel, a corrected normal direction is deﬁned as . Deﬁne , where the normal component and the shear component . If grid is separable and the two contacting bodies are penetrating each other, that is, 0, the actual contacting force, taking into account the friction, is then fc ςi+min µkˆ Step 7. Solve grid velocity. The grid velocity is updated explicitly as (mv)n+1 Step 8. Grid-to-particle projection . Based on the partitioning indicator , grid velocity and its spatial gradient are then projected to the particles as: vn+1 p=∑iwip vn+1 ςiand Ln+1 ςi⊗grad wip . Step 9. Convection. The convection part is the same as the classical material point methods. 2.5.3 Artiﬁcial viscosity Artiﬁcial viscosity is commonly used to stabilize the shock wave propagation and suppress the spurious oscillation behind the wavefront [Benson,1991,Mattsson and Rider,2015]: JhI, (25) are material constants, is the bulk sound speed, and is the characteristic length scale of the MPM particles. The second-order viscosity term smears the stress discontinuity but introduces high-frequency oscillation, while the ﬁrst-order viscosity term is helpful in suppressing this oscillation. By introducing the artiﬁcial viscosity, the discontinuous stress wave is smeared into a continuous function, and the width of the transition zone is restrained within several ﬁnite elements. Note that the artiﬁcial viscosity has a negligible inﬂuence on the results outside the transition zone, and the Hugoniot jump condition is still 3 Effective material properties inferred from molecular dynamics simulations In this section, results from multiple types of molecular dynamics simulations are interpreted such that they can be systematically incorporated into the crystal plasticity model in Section 2in the calibration and validation process. Speciﬁcally, the hyperelasticity model is calibrated against the elastic moduli evaluated by atomistic models at various pressures and temperatures [Pereverzev and Sewell,2020]. Then, the pressure dependence of the critical resolved shear stress in Equation is calibrated based on results from Pal and Picu [2019] Domain partitioning MPM for simulating shock in energetic materials 13 while the temperature dependence is based on results from [Khan et al.,2018]. Furthermore, the hardening of each slip system in Equation is calibrated using data from [Khan and Picu,2021]. Meanwhile, the melting temperature, the speciﬁc heat, and the melt viscosity are directly inferred from atomistic simulations. To avoid excessive curve-ﬁtting of the model, this subset of material parameters is not ﬁne-tuned via an optimization algorithm. 3.1 Pressure-dependent hyperelasticity The elastic stiffness tensor at various pressures ranging from 30 GPa and various temperatures ranging from 300 K 1000 K was evaluated using atomistic models in [Pereverzev and Sewell,2020]; this provides the relation between the material time derivative of the Cauchy stress and the symmetric velocity . It is observed that, within the parameter range, the pressure has a much larger inﬂuence on the elastic stiffness than the temperature, and hence the effect of pressure is accounted for in the constitutive Our hyperelasticity model in Equation is developed to reproduce this nonlinear material behavior. The speciﬁc form of the unknown function in Equation needs to be determined to reproduce the nonlinear material behavior while maintaining the material symmetry. We use the following polynomial function to achieve this goal: +b5(I4−I7)2+b6(I5+I7)2+b7(I6+I7)2+1, (26) are strain invariants deﬁned in Equation , and are material constants. The convexity of the free energy function is not guaranteed for every elastic strain in the strain space, but the tangent stiffness matrix is always positive deﬁnite when the stress is within a small neighborhood of a pressure state between 0 GPa and 30 GPa, as imposed by the data used for calibration. The elastic stiffness in ambient conditions in Equation corresponds to the stiffness evaluated with atomistic models at 300 K . The material parameters are calibrated based on the atomistic results at various pressures. The resulting parameters are listed in Table 1. Table 1: Parameters in Equations (8) and (26) which deﬁnes the hyperelastic free energy. 13.0 4.4 6.4 7.4 8.3 5.2 4.2 In our previous work [Ma et al.,2021], the inﬁnitesimal strain theory is assumed and thus the free-energy based stiffness is compared directly with the atomistic scale evaluation. In the current ﬁnite-deformation hyperelasticity model, in order to achieve a fair comparison between the free-energy based stiffness and the atomistic evaluation, the geometric nonlinearity caused by the hydrostatic compression before the stiffness evaluation needs to be considered. The atomistic-scale stiffness correlates the material time derivative of the Cauchy stress ˙ σand the symmetric part of the velocity gradient (the rate of deformation): σ=CMD :D,D=sym L. Although this elastic constitutive relation is not objective, the elastic evaluation is still reliable because the strain perturbation during the evaluation process is small and the spin W=skw L vanishes. The relationship between the atomistic evaluated stiffness CMD and the hyperelastic stiffness CSE is: σij =CMD ijkl :Dkl ≈1 τij =1 JFi J FjJ FkK FlLCSE IJK L −pIi jkl Dkl,CS E IJK L =ρ0 I J ∂Ee , (27) where the stress state is assumed to be volumetric, and the material time derivative of the volume change is assumed to be negligible. The deformation gradient in Equation needs to be determined through the elastic stretch caused by the pressure as well as the crystal orientation during the atomistic evaluation. 14 Ran Ma et al. The deformation gradient in Equation is determined to reach the same crystal orientation as the atomistic model [Pereverzev and Sewell,2020]. The hydrostatic pressure is ﬁrst applied which leads to an elastic stretch . Note that the crystal symmetry remains monoclinic under hydrostatic pressure. Then, a rigid rotation is applied to the sample, such that the axis of the deformed conﬁguration is aligned with the -axis of the global Cartesian coordinate system and the axis is aligned with the axis. Then, the deformation gradient in Equation (27) is determined as F=RU. The comparison between the atomistic stiffness and the hyperelastic stiffness is shown in Figure 2, where three representative stiffness coefﬁcients are compared at various pressures. In general, the hyperelastic stiffness is able to reproduce the pressure dependence observed in the atomistic evaluations. The atomistic model reveals that every elastic stiffness coefﬁcient of -HMX has different pressure sensitivity, which poses challenges to the hand-crafted hyperelasticity model. For example, in Figure 2, the pressure sensitivity of elastic stiffness is well reproduced, but are less accurate. Therefore, in our recent study, we also attempted to construct the hyperelastic model of -HMX through machine learning [Vlassis et al.,2022], but we limit ourselves to the hand-crafted hyperelasticity model in this paper. pressure (GPa) stiffness (GPa) Fig. 2: Comparison of the hyperelastic stiffness and the elastic stiffness evaluated by atomistic models. 3.2 Non-Schmid crystal plasticity The CRSS of single-crystal -HMX was evaluated by atomistic models at 300 K and under various pressures ranging from 30 GPa [Pal and Picu,2019]. Similar evaluations were also performed at ambient pressure but with two different temperatures at 300 K 400 K [Khan et al.,2018]. According to these atomistic evaluations, the pressure dependence is much stronger than the temperature dependence within these parameter ranges. Therefore, precisely replicating the pressure dependence of each slip system in the constitutive crystal plasticity model is important. The strain hardening with various dislocation densities, including 1011 m−2 1013 m−2 , and 1015 m−2 , is also evaluated in [Khan and Picu,2021]. In the mesoscale model, six slip systems with the smallest CRSS are chosen as potentially active slip systems. The material parameters in Equation are calibrated for these slip systems to reproduce the pressure-dependent CRSS. The calibrated results are listed in Table 2, while detailed procedures of the calibration process are discussed in our previous work [Ma et al.,2021]. The contribution to the ﬂow stress of strain hardening is much weaker than that of lattice friction [Khan and Picu,2021]. When the dislocation density is 1015 m−2 , the ﬂow stresses of systems increase by 55 % over the lattice resistance, while the ﬂow stresses of other slip systems increase much less. In the shock simulation, the dislocation density can hardly reach 1015 m−2 , which is considered here as the upper limit of the dislocation density and the corresponding contribution to the ﬂow stress Domain partitioning MPM for simulating shock in energetic materials 15 Table 2: Calibrated material parameters for Peierls-Nabarro stress and strain hardening of single crystal β-HMX. Only attractive forest dislocations are considered for strain hardening. (011)[01 ¯ 1] (011)[100] (101)[10¯ 1] (101)[010] (01 ¯ 1)[011] (011)[11¯ 1(GPa−1) 2.01 ×10−30.0 0.0 2.96 ×10−32.01 ×10−30.0 23.75 ×10−27.38 ×10−23.76 ×10−2−2.20 ×10−23.75 ×10−24.30 ×10−2 3(GPa) 3.75 ×10−23.59 ×10−19.85 ×10−21.22 ×10−13.75 ×10−21.24 ×10−1 3(MPa) 45.4 19.0 59.1 46.2 45.4 67.1 is taken as the saturation stress in Equation as listed in Table 2. The hardening rate is controlled by parameter H0, which is set here at 10 MPa. 3.3 Other material properties of β-HMX The remaining material parameters of the mesoscale model are listed in Table 3, including the speciﬁc heat in Equation , the melting temperature in Equation , and the viscosity in Equation . These models are also justiﬁed by the atomistic models [Sewell and Menikoff,2004,Kroonblawd and Austin,2021], therefore we directly incorporate them into our mesoscale model. Table 3: Other material parameters related to the mesoscale model of single crystal β-HMX. Variable Unit Value Reference Γa- 1.1 [Sewell and Menikoff,2004] Tm0K 551 [Kroonblawd and Austin,2021] Γb-−0.2 [Sewell and Menikoff,2004] ηcP 5.5 [Barton et al.,2009] Tref K 298 - cv0K kg J−15.265 ×10−7[Sewell and Menikoff,2004] θ0K 1 [Sewell and Menikoff,2004] cv1K kg J−13.073 ×10−4[Sewell and Menikoff,2004] a0GPa 0.305 [Kroonblawd and Austin,2021] cv2K kg J−11.831 ×10−1[Sewell and Menikoff,2004] c0- 3.27 [Kroonblawd and Austin,2021] cv3K kg J−14.194 ×10−4[Sewell and Menikoff,2004] Mh- 3.0 [Springer et al.,2018] Mp- 3.0 [Springer et al.,2018] µ- 0.25 [Wu and Huang,2010] 4 Model validation In this example, the mesoscale crystal plasticity model is compared directly with the atomistic scale model through three shock simulations performed on single-crystal -HMX under various shock velocities. Key emphasis is placed on the Hugoniot relations and the hotspot evolution adjacent to the pore inclusions. The model and the boundary conditions are shown in Figure 3. The shock loading is applied along the crystal direction, while plane strain constraint is applied on the crystal plane. Three different initial velocities are applied to the single crystal -HMX as the shock velocity, including the low-velocity shock 0.5 km s−1, the medium-velocity shock 1.0 km s−1, and the high-velocity shock 2.0 km s−1. A two-dimensional circular vacuum void is placed in the center of the specimen which serves as an initial defect embedded in the single crystal -HMX. Upon shock loading, the embedded pore may collapse 16 Ran Ma et al. x, [001] y, [010] 150 nm 150 nm 50 nm (a) Atomistic model x, [001] y, [010] 150 nm 150 nm 50 nm (b) Continuum model Fig. 3: The model and boundary conditions for the shock simulation of a single crystal -HMX containing a circular vacuum pore. depending on the shock velocity, and the local temperature increases due to the local plastic dissipation and the pore surface impacting, which serves as the hotspots that may eventually trigger the detonation. 4.1 Setup of the molecular dynamics model The MD pore-collapse results shown in this paper were obtained using data from MD trajectories originally reported by Das et al. [2021]. Similar MD simulations are also reported by Duarte et al. [2021] with exactly the same crystal orientation and piston velocity, but the specimen size is increased by a factor of . The simulations were performed using the LAMMPS code [Thompson et al.,2022] in conjunction with a variant of the all-atom fully ﬂexible non-reactive force ﬁeld due to Smith and Bharadwaj (S-B) [Smith and Bharadwaj, 1999,Bedrov et al.,2000]. The S-B force ﬁeld is well validated and has been used in a variety of MD studies for HMX [Mathew and Sewell,2018]. Potentials for covalent bonds, three-center angles, and improper dihedral angles are modeled using harmonic functions, and for dihedrals using truncated cosine series. Non-bonded pair interactions between atoms belonging to different molecules, or separated by three or more covalent bonds within a molecule, are modeled using the Buckingham+Coulomb (exp-6-1) potential. Long-range forces were evaluated using the particle-particle particle-mesh (PPPM) solver [Hockney and Eastwood,1988]. The speciﬁc S-B version used here is described in our previous work [Zhao et al.,2020]. The differences relative to the original ﬂexible-bond S-B model [Bedrov et al.,2000] are adjustments to the CH and NO covalent bond-stretching force constants to yield a vibrational density of states more consistent with the experiment and the addition of a very-short-range repulsive non-bonded pair potential that prevents the ”Buckingham catastrophe” (in which the forces between non-bonded atoms diverge for sufﬁciently short interatomic distances due to the divergence of the potential energy to negative inﬁnity for distances below that corresponding to the global maximum in the exp-6-1 potential). Shocks initially propagating parallel to the crystal direction in the space-group setting were simulated using a reverse-ballistic conﬁguration wherein a ﬂexible sample of HMX impacts with normal incidence onto a rigid, stationary piston composed of the same material. The initially 1D shock wave that results propagates in the direction opposite to the impact vector until it scatters at the pore wall. The starting HMX slab is quasi-2D and monoclinic shaped, with initial edge lengths of ∼5 nm ×150 nm ×150 nm . The 3D-periodic monoclinic-shaped computational domain (i.e., primary simulation cell) is also quasi-2D but with cell edge lengths of ∼5 nm ×160 nm ×150 nm . The extra 10 nm added along = initial shock direction) is a vacuum region that is introduced at the “top” of the sample and which serves to minimize Domain partitioning MPM for simulating shock in energetic materials 17 long-range force interactions between the free surface of the sample and the piston across the A right-cylindrical pore with an initial diameter 50 nm is located at the center of the sample, with the pore axis parallel to the thin direction. The resulting 3D-periodic primary cell containing the slab with pore and the vacuum region at the top is equilibrated in the isochoric-isothermal (NVT) ensemble. Using the ﬁnal phase space point from the equilibration, the ﬁrst three unit cells at the bottom of the HMX slab, comprising ∼3 nm of material along the shock direction, are assigned to the piston. Velocities and forces for atoms in the piston are set to and maintained at zero for the remainder of the simulation. Initial conditions selection for the shock is completed by adding the Cartesian impact-velocity vector up= ( , 0 to the instantaneous thermal velocities of the atoms in the sample. The shocks were simulated in the isochoric-isoenergetic (NVE) ensemble until the sample lengths rebounded by 10 % relative to the values at maximum compression. Atomic positions, velocities, and per-atoms stress tensors were stored for subsequent analysis. Instantaneous 2D spatial maps of local temperature and stress in the samples were obtained using the methods described by those authors. Brieﬂy, a 3D Cartesian Eulerian grid was superposed on the computational domain. The square-grid edge spacing in the plane of the sample was 152.6 nm ×150.4 nm The spacing in the thin direction, 5.3 nm , spans the sample. Instantaneous atomic positions were mapped into the grid. Atoms belonging to a given cell were used to calculate the instantaneous local temperature and stress for that cell, assuming local equilibrium and taking proper account of the periodic boundary conditions. These quantities, or ones computed from them, were assembled into spatial maps. 4.2 Setup of the continuum scale model The material point method (MPM) is used to numerically approximate the solution of the continuum-scale boundary value problem. The spatial domain is ﬁrst discretized into Lagrangian particles that move in an Eulerian background mesh. The background mesh is a structured partition of quadrilateral elements with the average element size equal to 0.33 nm . The initial positions of the particles coincide with the Gauss quadrature points of the background mesh, resulting in a total number of 745,980 particles. A convergence study has been performed to ensure that the spatial discretization is sufﬁciently reﬁned to approximate the solution but, for brevity, is not included in the paper. The time step is ﬁxed as 0.01 ps , which satisﬁes the CFL condition and guarantees the convergence of the return mapping algorithm. The symmetry boundary condition is applied on the three side walls as shown in Figure 3(b), such that no ﬂow across the walls is permitted but the tangential ﬂow is allowed. This is different from the atomistic model as shown in Figure 3(a), where transmissive periodic boundary conditions are applied on the lateral sidewalls and a 3 nm thick piston is placed on the bottom. However, because of the particular crystal orientation considered, the two should be equivalent, that is, the lateral velocity (or displacement) at the boundary should be zero in the atomistic model due to symmetry considerations. 4.3 Comparisons between continuum and molecular dynamics simulations The shock-induced pore collapse simulated with the mesoscale model is validated against atomistic scale models, with an emphasis on the Hugoniot relations and the pressure/temperature contour. The comparison between the mesoscale model and the atomistic model is qualitative based on the following considerations. First, dislocations must nucleate in the atomistic model, so the atomistic model goes through a transient process in which stress overshoots to create dislocations, after which plasticity may happen at smaller stresses. But the mesoscale model assumes dislocations already exist in the material, so the critical resolved shear stress is much smaller than the atomistic model. Second, the periodic boundary condition is applied along the thickness direction in the atomistic model, while the plane strain condition is prescribed in the mesoscale model. Such differences may lead to different slip system activation even though the orientation of the single crystal is selected to minimize 3D effects. Figure 4compares the Hugoniot relations simulated by the atomistic and mesoscale models, which are both evaluated before the shock wave reaches the embedded pore. It is observed that the mesoscale continuum model is capable of replicating the trend of the relationships among the pressure, shock velocity, 18 Ran Ma et al. and shock wave velocity, as shown in Figures 4(b) and (c) whereas discrepancy in the temperature vs. shock velocity is also observed. This inconsistency could be attributed to the temperature evolution model used in where the Gr uneisen coefﬁcient is obtained from a ﬁrst-order Taylor expansion. Presumably, a better curve-ﬁtting result can be achieved after more MD simulation data or quantum chemistry calculation to re-calibrate the equation of the state (EOS) model at the high shock velocity range [Menikoff and Sewell, 2002]. However, a more comprehensive analysis would require a substantial amount of additional atomistic simulations and is therefore outside the scope of this study. 0.5 1 1.5 2 Shock velocity (km/s) Temperature (K) (a) Temperature 0.5 1 1.5 2 Shock velocity (km/s) Pressure (GPa) (b) Pressure 0.5 1 1.5 2 Shock velocity (km/s) Shock wave velocity (km/s) (c) Shock wave speed Fig. 4: Comparison of the Hugoniot relations simulated by the atomistic model and the mesoscale model. Case 1: Shock velocity 0.5 km s−1 Figure 5compares the temperature distribution and the pore geometry at t=32 ps 42 ps by the atomistic model and the mesoscale model. Lateral jetting from the equator of the pore, as well as the extension of the shear bands from the pore surface to the bulk material, is observed in the atomistic model. This shear band formation is also replicated in the mesoscale continuum model. The temperature increase within the shear band, which is mainly due to the plastic dissipation rather than the volumetric compression, is also comparable between the results obtained from these two models. The shear band appears to be more smeared in the mesoscale model than in the atomistic model. One possible reason is that the shear band formation in the atomistic model goes through a transient process in which stress overshoots to create dislocations, after which plasticity happens at smaller stress. The strain- softening in the mesoscale model is less severe to trigger the sharp shear bands observed in the atomistic model. It is also observed that the pore collapses faster in the mesoscale model than in the atomistic model. One possible explanation is that the rate dependence is not considered in the crystal plasticity model, so the critical resolved shear stress is much lower in the mesoscale model. Case 2: Shock velocity 1.0 km s−1 Figure 6and Figure 7compare the atomistic model and the mesoscale model regarding the temperature and pressure distributions predicted under the 1.0 km s−1 shock. The lateral jetting and the associated shear band are still observed in the atomistic model similar to the 0.5 km s−1 shock case but with higher temperatures both within the shear band and within the bulk material. But in the mesoscale model, the material jetting forms from the upstream of the pore which is closer to the pore collapse pattern under 2.0 km s−1 loading as shown in Figure 8and Figure 9. At time t=23 ps when the pore starts to shrink but before collapsing, the hotspots are located within the shear band region. Compared with the atomistic model, the mesoscale model underestimates the peak temperature within the shear band as shown in Figures 6(a) and (c), but the pressure ﬁeld is well reproduced as shown in Figures 7(a) and (c). Again, the temperature rise with the shear band is more related to the plastic dissipation than the volumetric compression. At time Domain partitioning MPM for simulating shock in energetic materials 19 (a) Atomistic 32 ps (b) Atomistic 42 ps (c) Continuum 32 ps (d) Continuum 42 ps Fig. 5: Temperature contour of low velocity shock (0.5 km s−1) before and after pore collapse. t=35 ps when the pore is fully collapsed, the hotspots are located at the pore surface impacting region, and the peak temperature reaches as high as 2000 K as shown in Figures 6(b) and (d). The secondary shock wave due to the pore surface impact is also reproduced in the mesoscale model, but the mesoscale model overestimates the pressure of the secondary shock wave as we compare Figures 7(b) and (d). The major reason is that the hydrodynamic-force collapse mechanism observed in the mesoscale model produces larger secondary shock pressure compared with the shear band mechanism observed in the atomistic model, which is further demonstrated in detail as follows. Two possible pore collapse modes may manifest in the energetic materials under shock loading, that is, the shear band mechanism and the hydrodynamic force mechanism [Rai et al.,2020]. These two mechanisms are triggered at different shock velocities. When the shock velocity is not high enough to liquefy the bulk material, the shear band emanating from the lateral surface of the pore is the major mechanism that accompanies the pore collapse. As the shock velocity further increases and the temperature of the bulk material approaches the melting point, the pore collapse is dominated by hydrodynamic force in the form of material jetting from the shock direction, whereas the shear band is less likely to form. In this example, we observe the transition from the shear band mechanism to the hydrodynamic force mechanism in both the mesoscale and the atomistic simulations, which indicates that the essence of the transition is captured at the continuum scale. However, we also observe that the transition from the shear band mode to the hydrodynamic force dominated mode occurs at a lower shock velocity in the mesoscale model than in the atomistic model. One possible explanation is that the rate dependence of the critical resolved shear stress is not incorporated in our current plasticity model. As such, the yield stress of the 20 Ran Ma et al. mesoscale model is not large enough to withhold the upstream material jetting compared with the atomistic An implementation of the crystal plasticity model to incorporate rate dependence for each plastic slip system is technically feasible. However, this will require a signiﬁcant amount of molecular dynamic simulations to rigorously identify the corresponding material parameters under different strain rates, pressure, and temperature. As such, we will consider this extension in the future study but is out of this study. The simulation approach presented in this paper is also different from Duarte et al. [2021] in which a power-law crystal plasticity model is used without taking into account the underlying physics of individual slip systems. (a) Atomistic 23 ps (b) Atomistic 35 ps (c) Continuum 23 ps (d) Continuum 35 ps Fig. 6: Temperature contour of medium velocity shock (1.0 km s−1) before and after pore collapse. Case 3: Shock velocity 2.0 km s−1 The atomistic and mesoscale simulations are further compared in the case where a shock velocity of 2.0 km s−1 is applied. At the time t=16 ps , the material jetting from the upstream surface of the pore suggests that the pore collapse process is dominated by the hydrodynamic force mechanism, which is observed both in the atomistic model (Figure 8(a)) and in the mesoscale model (Figure 8(c)). The shape of the embedded pore, as well as the peak temperature of the hotspot region, is also reproduced in the mesoscale model. At the time t=22 ps , the embedded pore is fully collapsed and the hotspot temperature reaches as high as 3000 K , as shown in Figures 8(b) and (d). By employing the gradient-partition technique Domain partitioning MPM for simulating shock in energetic materials 21 (a) Atomistic 23 ps (b) Atomistic 35 ps (c) Continuum 23 ps (d) Continuum 35 ps Fig. 7: Pressure contour of medium velocity shock (1.0 km s−1) before and after pore collapse. [Homel and Herbold,2017], we are able to continue the simulation after the pore collapse. This is important to capture the growth of the secondary shock wave that occurs after the pore collapse. As shown in Figure 9(d), the secondary shock wave emanating from the embedded pore with about 25 GPa shock pressure exhibited in the atomistic model (Figure 9(b)) is replicated in the mesoscale model. 5 Parametric study In this section, three numerical examples are performed to demonstrate the capability of the non-Schmid crystal plasticity model and the frictional contact algorithm for predicting the shock response of energetic materials. In Parametric Study 1, a polycrystal shock simulation is designed to demonstrate the interaction between grain boundary sliding/cohesion and pore surface contact. In Parametric Study 2 and 3, the results of the 1 km s−1 shock simulation are used as the control case to study the effects of pressure sensitivity in the mesoscale model. 5.1 Parametric Study 1: effects of pressure-dependent hyperelasticity In this numerical experiment, we explore the effect of pressure-dependent hyperelasticity by replacing the elasticity model of the control case with the EOS-based elasticity while keeping the other components of the material model and the setup of the boundary value problem identical. The Mie-Gr uneisen EOS is 22 Ran Ma et al. (a) Atomistic 16 ps (b) Atomistic 22 ps (c) Continuum 16 ps (d) Continuum 22 ps Fig. 8: Temperature contour of high velocity shock (2.0 km s−1) before and after pore collapse. implemented [Menikoff and Sewell,2002], and the pressure is a function of the volumetric strain and the internal energy e: p=pc(V) + Γ V[e−ec(V)],Γ(J) = Γa+ΓbJ	{"url":"https://www.researchgate.net/publication/365635340_Domain_partitioning_material_point_method_for_simulating_shock_in_polycrystalline_energetic_materials","timestamp":"2024-11-10T21:54:28Z","content_type":"text/html","content_length":"1050306","record_id":"<urn:uuid:18ddf634-d2e7-4001-b804-dd8ab36001de>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00898.warc.gz"}
Deviations from piecewise linearity in the solid-state limit with approximate density functionals In exact density functional theory, the total ground-state energy is a series of linear segments between integer electron points, a condition known as "piecewise linearity." Deviation from this condition is indicative of poor predictive capabilities for electronic structure, in particular of ionization energies, fundamental gaps, and charge transfer. In this article, we take a new look at the deviation from linearity (i.e., curvature) in the solid-state limit by considering two different ways of approaching it: a large finite system of increasing size and a crystal represented by an increasingly large reference cell with periodic boundary conditions. We show that the curvature approaches vanishing values in both limits, even for functionals which yield poor predictions of electronic structure, and therefore cannot be used as a diagnostic or constructive tool in solids. We find that the approach towards zero curvature is different in each of the two limits, owing to the presence of a compensating background charge in the periodic case. Based on these findings, we present a new criterion for functional construction and evaluation, derived from the size-dependence of the curvature, along with a practical method for evaluating this criterion. For large finite systems, we further show that the curvature is dominated by the self-interaction of the highest occupied eigenstate. These findings are illustrated by computational studies of various solids, semiconductor nanocrystals, and long alkane chains. Bibliographical note Publisher Copyright: © 2015 AIP Publishing LLC. Dive into the research topics of 'Deviations from piecewise linearity in the solid-state limit with approximate density functionals'. Together they form a unique fingerprint.	{"url":"https://cris.huji.ac.il/en/publications/deviations-from-piecewise-linearity-in-the-solid-state-limit-with","timestamp":"2024-11-08T14:26:19Z","content_type":"text/html","content_length":"51754","record_id":"<urn:uuid:841ed76c-1d1f-4489-94bb-f76eca78ce98>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00368.warc.gz"}
How do you determine whether the sequence 3, 5/2, 2, 3/2, 1,... is arithmetic and if it is, what is the common difference? \| Socratic How do you determine whether the sequence #3, 5/2, 2, 3/2, 1,...# is arithmetic and if it is, what is the common difference? 1 Answer Take the 2nd term and minus the 1st term. Now take the 3rd term and minus the 2nd term. Notice that the differences are the same. Hence there is a common difference (difference of -0.5). Impact of this question 7956 views around the world	{"url":"https://api-project-1022638073839.appspot.com/questions/how-do-you-determine-whether-the-sequence-3-5-2-2-3-2-1-is-arithmetic-and-if-it-#380619","timestamp":"2024-11-13T15:25:29Z","content_type":"text/html","content_length":"32839","record_id":"<urn:uuid:d4355409-59c4-477c-b862-cf2d463e4b41>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00236.warc.gz"}
May I ask whether part of the results of the heuristic algorithm search designed by myself can be input into Gurobi May I ask whether part of the results of the heuristic algorithm search designed by myself can be input into Gurobi, so that Gurobi can continue to iterate on this basis? The heuristic algorithm helps to narrow the range of search iterations. • Thanks! But still a little bit question: Does Gurobi continue the downward branching iterative calculation on the basis of this input solution or anything else? • Hi Ariel, When you provide a starting solution to Gurobi—whether from a heuristic, a previous run, or any other source—Gurobi uses this solution in several ways that can impact the optimization process, including its branching decisions. Here's a more detailed look at how Gurobi uses a starting solution, particularly for MIP problems: 1. Feasible Solution: If the starting solution is feasible, Gurobi uses it as a reference point. This means the solver knows it has a valid solution from the get-go, which helps in bounding decisions and can significantly reduce the search space. For MIP problems, having an initial feasible solution can lead to an earlier determination of the optimality gap, potentially speeding up the time to find the optimal solution or proving that the current solution is closer to optimality. 2. Branching Decisions: The starting solution can influence branching decisions, but Gurobi does not specifically continue "downward branching" from the values in the initial solution in a deterministic way. Instead, Gurobi's branch-and-bound algorithm will consider the entire solution space, guided by its internal heuristics for variable selection, node selection, and cutting planes. The presence of a feasible solution can affect these heuristics by providing additional information about the problem structure and potentially good regions of the search space. 3. Heuristic Adjustments: Gurobi performs several heuristics at the beginning and during the branch-and-bound process. A good initial solution can help these heuristics by guiding them towards promising areas of the search space or by validating the effectiveness of certain cuts or branches early on. 4. Bounding: One of the most direct impacts of a starting solution is on bounding. If the starting solution has a better objective value than the bounds determined by the relaxation of the problem, Gurobi can use it to improve the global bounds. This can lead to the pruning of branches that cannot improve upon the starting solution, thereby reducing the overall solution time. In summary, while Gurobi doesn't specifically branch downward from the heuristic starting point in a direct, sequential manner, the presence of a starting solution informs its optimization process in several beneficial ways. The algorithm will still explore the solution space according to its sophisticated optimization strategies, but with the advantage of having a concrete benchmark that can guide its search more efficiently. - Bot • Thanks Bot, for your detailed explanation, I got the principle behind this!	{"url":"https://support.gurobi.com/hc/ja/community/posts/23720637124753-May-I-ask-whether-part-of-the-results-of-the-heuristic-algorithm-search-designed-by-myself-can-be-input-into-Gurobi","timestamp":"2024-11-15T00:29:22Z","content_type":"text/html","content_length":"51232","record_id":"<urn:uuid:a9172c0e-2c9a-4263-a641-a523700a0aaf>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00181.warc.gz"}
Immutable, arbitrary-precision signed decimal numbers. A consists of an arbitrary precision integer unscaled value and a 32-bit integer . If zero or positive, the scale is the number of digits to the right of the decimal point. If negative, the unscaled value of the number is multiplied by ten to the power of the negation of the scale. The value of the number represented by the is therefore (unscaledValue × 10^-scale) The BigDecimal class provides operations for arithmetic, scale manipulation, rounding, comparison, hashing, and format conversion. The toString() method provides a canonical representation of a The BigDecimal class gives its user complete control over rounding behavior. If no rounding mode is specified and the exact result cannot be represented, an exception is thrown; otherwise, calculations can be carried out to a chosen precision and rounding mode by supplying an appropriate MathContext object to the operation. In either case, eight rounding modes are provided for the control of rounding. Using the integer fields in this class (such as ROUND_HALF_UP) to represent rounding mode is largely obsolete; the enumeration values of the RoundingMode enum, (such as RoundingMode.HALF_UP) should be used instead. When a MathContext object is supplied with a precision setting of 0 (for example, MathContext.UNLIMITED), arithmetic operations are exact, as are the arithmetic methods which take no MathContext object. (This is the only behavior that was supported in releases prior to 5.) As a corollary of computing the exact result, the rounding mode setting of a MathContext object with a precision setting of 0 is not used and thus irrelevant. In the case of divide, the exact quotient could have an infinitely long decimal expansion; for example, 1 divided by 3. If the quotient has a nonterminating decimal expansion and the operation is specified to return an exact result, an ArithmeticException is thrown. Otherwise, the exact result of the division is returned, as done for other operations. When the precision setting is not 0, the rules of BigDecimal arithmetic are broadly compatible with selected modes of operation of the arithmetic defined in ANSI X3.274-1996 and ANSI X3.274-1996/AM 1-2000 (section 7.4). Unlike those standards, BigDecimal includes many rounding modes, which were mandatory for division in BigDecimal releases prior to 5. Any conflicts between these ANSI standards and the BigDecimal specification are resolved in favor of BigDecimal. Since the same numerical value can have different representations (with different scales), the rules of arithmetic and rounding must specify both the numerical result and the scale used in the result's representation. In general the rounding modes and precision setting determine how operations return results with a limited number of digits when the exact result has more digits (perhaps infinitely many in the case of division) than the number of digits returned. First, the total number of digits to return is specified by the MathContext's precision setting; this determines the result's precision. The digit count starts from the leftmost nonzero digit of the exact result. The rounding mode determines how any discarded trailing digits affect the returned result. For all arithmetic operators , the operation is carried out as though an exact intermediate result were first calculated and then rounded to the number of digits specified by the precision setting (if necessary), using the selected rounding mode. If the exact result is not returned, some digit positions of the exact result are discarded. When rounding increases the magnitude of the returned result, it is possible for a new digit position to be created by a carry propagating to a leading "9" digit. For example, rounding the value 999.9 to three digits rounding up would be numerically equal to one thousand, represented as 100×10^1. In such cases, the new "1" is the leading digit position of the returned result. Besides a logical exact result, each arithmetic operation has a preferred scale for representing a result. The preferred scale for each operation is listed in the table below. Preferred Scales for Results of Arithmetic Operations │Operation│ Preferred Scale of Result │ │Add │max(addend.scale(), augend.scale()) │ │Subtract │max(minuend.scale(), subtrahend.scale()) │ │Multiply │multiplier.scale() + multiplicand.scale() │ │Divide │dividend.scale() - divisor.scale() │ These scales are the ones used by the methods which return exact arithmetic results; except that an exact divide may have to use a larger scale since the exact result may have more digits. For Before rounding, the scale of the logical exact intermediate result is the preferred scale for that operation. If the exact numerical result cannot be represented in precision digits, rounding selects the set of digits to return and the scale of the result is reduced from the scale of the intermediate result to the least scale which can represent the precision digits actually returned. If the exact result can be represented with at most precision digits, the representation of the result with the scale closest to the preferred scale is returned. In particular, an exactly representable quotient may be represented in fewer than precision digits by removing trailing zeros and decreasing the scale. For example, rounding to three digits using the floor rounding mode, 19/100 = 0.19 // integer=19, scale=2 21/110 = 0.190 // integer=190, scale=3 Note that for add, subtract, and multiply, the reduction in scale will equal the number of digit positions of the exact result which are discarded. If the rounding causes a carry propagation to create a new high-order digit position, an additional digit of the result is discarded than when no new digit position is created. Other methods may have slightly different rounding semantics. For example, the result of the pow method using the specified algorithm can occasionally differ from the rounded mathematical result by more than one unit in the last place, one ulp. Two types of operations are provided for manipulating the scale of a BigDecimal: scaling/rounding operations and decimal point motion operations. Scaling/rounding operations (setScale and round) return a BigDecimal whose value is approximately (or exactly) equal to that of the operand, but whose scale or precision is the specified value; that is, they increase or decrease the precision of the stored number with minimal effect on its value. Decimal point motion operations (movePointLeft and movePointRight) return a BigDecimal created from the operand by moving the decimal point a specified distance in the specified direction. For the sake of brevity and clarity, pseudo-code is used throughout the descriptions of BigDecimal methods. The pseudo-code expression (i + j) is shorthand for "a BigDecimal whose value is that of the BigDecimal i added to that of the BigDecimal j." The pseudo-code expression (i == j) is shorthand for "true if and only if the BigDecimal i represents the same value as the BigDecimal j." Other pseudo-code expressions are interpreted similarly. Square brackets are used to represent the particular BigInteger and scale pair defining a BigDecimal value; for example [19, 2] is the BigDecimal numerically equal to 0.19 having a scale of 2. Note: care should be exercised if BigDecimal objects are used as keys in a SortedMap or elements in a SortedSet since BigDecimal's natural ordering is inconsistent with equals. See Comparable, SortedMap or SortedSet for more information. All methods and constructors for this class throw NullPointerException when passed a null object reference for any input parameter.	{"url":"https://docs.oracle.com/javase/8/docs/api/java/math/BigDecimal.html","timestamp":"2024-11-04T03:01:02Z","content_type":"text/html","content_length":"183788","record_id":"<urn:uuid:e06f7bd0-f8f3-4000-95fc-99fc8eb99cdb>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00564.warc.gz"}
Display math should end with $$ TeX engines have two ways of typesetting mathematics: • inline math mode where the mathematical content is contained within a paragraph, and • display math mode where mathematical material is displayed separately, with additional space above or below it. Traditionally, in the early days of TeX, mathematics intended to be typeset inline, typically within a paragraph, was surrounded by single $ characters: $ inline math content...$ and mathematics destined for display was surrounded by double $ characters: $$ display math content...$$. Cause of the error Display math should end with $$ The error message Display math should end with $$ is generated by TeX engines when they try to finish typesetting some display math material but are unable to cleanly exit from display math mode due to incorrect TeX markup: as the error message indicates, the material to be typeset as display math has not been terminated with a second $$ pair. Examples: single error This error is demonstrated in the following examples: \noindent \verb\|$$ E=mc^2$\| generates an error because the math is started by \texttt{\$\$} but terminated by a single \texttt{\$}: $$ E=mc^2$\noindent\verb\|$$ E=mc^2$ $\| also generates an error because of the space between the terminating \texttt{\$} characters: $$ E=mc^2$ $\end{document} Open this error-generating example in Overleaf This example produces the following output (image edited to highlight both errors): Example: two errors Note: In some circumstances you may also see the related error Missing $ inserted, as the following example demonstrates by writing $$E=mc^2, which omits both terminating $ characters: \noindent The following example omits both terminating \texttt{\$} characters, triggering the errors \texttt{Missing \$ inserted} and \texttt{Display math should end with \$\$.} Open this error-generating example in Overleaf This example produces the following output: For the errors demonstrated above, the fix is straightforward—make sure you add the closing $$ at the end of your display math: \noindent The solution is to ensure correct termination of the display math by writing \verb\|$$E=mc^2$$\|: Open this corrected example in Overleaf Avoid using $ characters to typeset mathematics Nowadays, standard (accepted) best practice is to avoid using explicit $ characters to typeset mathematics and use LaTeX delimiters instead, particularly for display math: • for display math: write \[ display math content \] instead of $$ display math content...$$ • for inline math: write $ inline math content $ instead of $ inline math content...$ In reality, the LaTeX delimiters $, $, \[ and \] are single-character macros which provide a sort of “insulating wrapper” around single and double $ characters. The LaTeX definitions of those delimiters (macros) do actually contain $ characters but with additional code that runs some tests/checks. They also generate LaTeX’s error message Bad math environment delimiter. Using these delimiters (macros) has additional advantages because they can be redefined, perhaps temporarily, to achieve special effects. Overleaf guides LaTeX Basics Figures and tables References and Citations Document structure Field specific Class files Advanced TeX/LaTeX	{"url":"https://ko.overleaf.com/learn/latex/Errors/Display_math_should_end_with_%24%24","timestamp":"2024-11-03T06:32:07Z","content_type":"text/html","content_length":"59672","record_id":"<urn:uuid:dfe2d5b9-16b5-4da9-bfb4-319898c51a05>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00531.warc.gz"}
Leetcode 1028. Recover a Tree From Preorder Traversal We run a preorder depth-first search (DFS) on the root of a binary tree. At each node in this traversal, we output D dashes (where D is the depth of this node), then we output the value of this node. If the depth of a node is D, the depth of its immediate child is D + 1. The depth of the root node is 0. If a node has only one child, that child is guaranteed to be the left child. Given the output traversal of this traversal, recover the tree and return its root. Example 1: Input: traversal = "1-2--3--4-5--6--7" Output: [1,2,5,3,4,6,7] Example 2: Input: traversal = "1-2--3---4-5--6---7" Output: [1,2,5,3,null,6,null,4,null,7] Example 3: Input: traversal = "1-401--349---90--88" Output: [1,401,null,349,88,90] • The number of nodes in the original tree is in the range [1, 1000]. • 1 <= Node.val <= 109 1. we will see dashes and value, so we verify it’s level and value while scan through the string 2. because it’s in-order traversal, if the stack.size bigger than level, will pop out from the stack 3. if the stack.peek().left is null, will add it into stack.peek().left; else add it into stack.peek().right 4. to return the result, which is the root. We will need to pop all the other TreeNode in stack, until the last one -> the root 5. Trick: use three variable in two for loops. Track index, level and value at the same time. Make sure index is not larger than string.length, to avoid overflow.	{"url":"https://kchiang6.medium.com/leetcode-1028-recover-a-tree-from-preorder-traversal-759015fdbc24?source=author_recirc-----ffabbb29dd90----1---------------------74fbcfeb_80d3_49e9_b965_2508e70e6603-------","timestamp":"2024-11-14T06:37:37Z","content_type":"text/html","content_length":"106886","record_id":"<urn:uuid:11e03caf-ae26-4ea3-a376-74ab446ba7a5>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00014.warc.gz"}
A local average consensus algorithm for wireless sensor networks —In many application scenarios sensors need to calculate the average of some local values, e.g. of local measurements. A possible solution is to rely on consensus algorithms. In this case each sensor maintains a local estimate of the global average, and keeps improving it by performing a weighted sum of the estimates of all its neighbors. The number of iterations needed to reach an accurate estimate depends on the weights used at each sensor. Speeding up the convergence rate is important also to reduce the number of messages exchanged among neighbors and then the energetic cost of these algorithms. While it is possible in principle to calculate the optimal weights, the known algorithm requires a single sensor to discover the topology of the whole network and perform the calculations. This may be unfeasible for large and dynamic sensor networks, because of sensor computational constraints and of the communication overhead due to the need to acquire the new topology after each change. ...	{"url":"https://www.sciweavers.org/publications/local-average-consensus-algorithm-wireless-sensor-networks","timestamp":"2024-11-07T13:53:30Z","content_type":"application/xhtml+xml","content_length":"39227","record_id":"<urn:uuid:de86b046-23d8-4b60-a1ef-7f9b05737c58>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00791.warc.gz"}
Enhanced XMM-Newton Spectral-fit Database 3XMM spectral-fit database (XMMFITCAT) 1. Automated spectral fitting We used the fitting and modelling software Sherpa 4.9.1 (Freeman et al. 2001) to perform the automated spectral fits. We followed the Bayesian technique proposed in Buchner et al. (2014) using the analysis software BXA (Bayesian X-ray Analysis), which connects the nested sampling algorithm MultiNest (Feroz et al. 2013) with Sherpa. In the BXA framework, the use of Cash fitting statistics is mandatory. We used the wstat implementation in Sherpa, which allows using background datasets as background models. To use this statistics, grouped spectra from the CATV catalogue were ungrouped, and then grouped to 1 count per bin. All available instruments and exposures for a single observation of a source are fitted together. Only spectral data within the 0.5-10 keV band are fitted. All parameters for different instruments are tied together except for a relative normalization, which accounts for the differences between different flux calibrations. 1.1. Spectral data selection criteria Spectral data from the CATV catalogue were screened before applying the automated spectral-fitting pipeline so that spectral fits are only performed if the data fulfil the following criteria: • Only spectra corresponding to a single instrument and observation with more than 50 net counts (i.e. background subtracted) in the 0.5-10 keV band are used in the spectral fits. This means that some detections with more than 100 EPIC counts in the total band, but less than 50 counts in each different EPIC instrument, are excluded from the automated fits, and therefore, they are not included in the spectral-fit database. • Complex models (see Sect.1.2), are only applied if the number of EPIC counts is larger than 500 net counts in the total band. As a result of the application of these criteria, the spectral-fit database contains spectral-fitting results corresponding to the simple models in the full band for CATD source detections, corresponding to CATD unique sources. Spectral-fitting results for complex models are available for CATD source detections. The distribution of spectral counts per observation used during the automated fits is plotted in Fig. 1. Observations with more than 10,000 counts, 2% of the CATD detections, are excluded of this plot for clarity. Figure 1. Distribution of net counts (background subtracted spectral counts) per observation used in the automated spectral fits. Observations with more than 10,000 counts are not included in this 1.2. Spectral models Most sources included in XMMPZCAT are extragalactic and hence the population is dominated by AGN. Given this fact, we have reduced the number of spectral models with respect to XMMFITCAT. They are phenomenological models selected to reproduced the spectral emission of AGN. We have also included a simple thermal model to deal with the X-ray emission of stars (about ten percent of XMMPZCAT sources) and other hot plasmas (e.g. intra-cluster medium emission). There are four different models, two simple and two more complex models, as follows: • Simple models: □ Absorbed power-law model (XSPEC: zwabspow): Variable parameters are the hydrogen column density of the absorber, the power-law photon index, and the power-law normalisation □ Absorbed thermal model (XSPEC: zwabsmekal): Variable parameters are the hydrogen column density of the absorber, the plasma temperature of the thermal component, and the normalisation of the thermal component. • Complex models: □ Absorbed thermal plus power-law model (XSPEC: zwabs(mekal + zwabspow)): Variable parameters are the hydrogen column density of both absorbers, the plasma temperature, the photon index, and the normalisation of the power-law and thermal components. □ Absorbed double power-law model (XSPEC: zwabs(pow + zwabspow)): Variable parameters are the hydrogen column density of both absorbers, the photon indices of both power-law components, and their normalisations. All models include an additional wabs component to take into account the Galactic absorption, with its Hydrogen column density fixed to the value in the direction of the source from the Leiden/ Argentine/Bonn (LAB) Survey of Galactic HI. In the BXA framework we employed for the spectral fitting, a probability prior should be assigned to each free parameter in the model. These are the priors we selected: • Hydrogen column density: Jeffreys prior (i.e. a uniform prior in the logarithmic space) with limits 10^20 - 10^25 cm^-2. • Power law photon index: gaussian prior with mean 1.9 and standard deviation 0.15. It corresponds to the photon index distribution for the AGN population as described in Nandra & Pounds, 1994. • Thermal plasma temperature: uniform prior with limits 0.08 - 20 keV. • Redshift: for sources with known spectroscopic redshift, this is included as a fixed parameter in the models. When only a photometric redshift is available, the redshift is treated as a free parameter using the photo-z probability density distribution given by MLZ/TPZ as the prior. • Normalisation: Jeffreys prior with limits 10^-30 - 1. • Relative normalisation constants: Jeffreys prior with limits 0.01 - 100. Examples of automated spectral fits for each of the spectral models used are plotted in Fig. 2. The photon index and N[H] distribution for detections for which an absorbed power-law model is an acceptable fit, >50% of the detections (see Sect.1.5), is shown in Fig. 3. Figure 2. Examples of spectral fits for the four different models applied. Plots correspond to the unfolded (source+background) spectra and model over data to model ratio. From left to right and top to bottom: wabspow, wabsmekal, wabs(mekal+wabspow), and wabs(pow+wabspow). Figure 3. Photon index (top) and Hydrogen column density (bottom) distribution corresponding to the detections for which an absorbed power-law model in the full band is an acceptable fit. The dashed red line shows the selected prior for the photon index: a normal distribution with mean 1.9 and standard deviation 0.15. 1.3. Best-fit parameters and error computation The MultiNest algorithm implemented in BXA gives the marginalized posterior probability distribution for all free parameters in the fitted model. We used these distributions to estimate the best-fit parameters and the corresponding errors. The best-fit values correspond to the mode (the most probable value) of the posterior distribution, estimated using a half-sample algorithm. Errors were estimated using the posterior distributions to calculate a 90% credible interval around the mode. Figure 4 shows an example of the marginal and conditional (for two parameters) posterior distributions for a source fitted with a wabspow model. This example shows how the structure of the original photo-z probability distribution is preserved. Figure 4. Example of posterior distributions for a source fitted with a wabspow model. Top row: marginal distributions. Remaining rows: two-parameter conditional distributions. 1.4. Goodness of fit Cash maximum likelihood statistics lacks a direct estimate of goodness of fit (GoF). We followed the method proposed in Buchner et al. (2014) and we used Q-Q (quantile-quantile) plots to obtain an estimate of the GoF of our spectral fits. Figure 5. Q-Q plots for a source fitted with a zwabspow model (left) and a zwabsmekal model (right). Blue, red and purple lines corresponds to PN, MOS1 and MOS2 data, respectively. A Q-Q plot compares the cumulative counts of the data (source+background) with the predicted counts (source+background) of the model (see Figure 5 below). The plot gives a quick visual idea of how well the model can reproduce the data. For a quantitative estimate of the GoF we calculated the Kolmogorov-Smirnov (KS) statistic between the two cumulative distributions and the corresponding p-value. A low KS (or a high p-value) means that the data is well reproduced by the model. Note however that in our case the p-values for these statistics cannot be calculated the usual way. The cumulative distribution of the model depends on the parameters that were estimated from the data distribution, therefore the condition of independence between the two compared distribution does not hold, and hence the probabilities estimated using the KS probability distribution are grossly incorrect. Nevertheles, through a permutation test we can get an estimate of the p-value. For each source, we did 1000 resamplings, randomly splitting the original data+model sample in two equal-size subsamples, and estimate the corresponding KS statistic. Our estimated p-values are the ratio of resamplings that have statistics larger than the statistic of the original samples. Note also that the KS test is more sensitive for distributions with large counts. Therefore, for detections with more than 500 counts in the total band, a fit with KS p-value > 0.9 is considered acceptable. Otherwise, a p-value > 0.5 is considered an acceptable fit. Taking into account all the spectral models implemented in this database, an acceptable fit is found for about 90% of all the detections. Figure 6 shows the KS p-value distribution of the WAPO model. Figure 6. KS p-value distribution of the wabs*pow model. 2. Description of the columns The XMMFITCAT-Z table contains one row for each detection, and 157 columns containing information about the source detection and the spectral-fitting results. Not available values are represented by an empty "NULL" value. The first 14 columns contain information about the source and observation, including redshift information, whereas the remaining 143 columns contain, for each model applied, model spectral-fit flags, parameter values and errors, fluxes, luminosities and five columns to describe the goodness of the fit. 2.1 Source and observation IAUNAME: The IAU name assigned to an unique source in the CATV catalogue. SC_RA, SC_DEC: Right ascension and declination in degrees (J2000) of the unique source, as in the CATV catalogue. RA and DEC correspond to the SC_RA and SC_DEC columns in the CATV catalogue. These are corrected source coordinates and, in the case of multiple detections of the same source, they correspond to the weighted mean of the coordinates for the individual detections. SRCID: A unique number assigned to a group of catalogue entries which are assumed to be the same source in CATV. DETID: A consecutive number which identifies each entry (detection) in the CATV catalogue. OBS_ID: The XMM-Newton observation identification, as in CATV. SRC_NUM: The (decimal) source number in the individual source list for this observation (OBS_ID), as in CATV. Note that in the pipeline products this number is used in hexadecimal form. PHOT_Z, PHOT_ZERR: Photometric redshift of the source (from XMMPZCAT) and the corresponding 1σ error. SPEC_Z: Spectroscopic redshift of the source, if available. T_COUNTS/H_COUNTS/S_COUNTS: spectral background subtracted counts in the full/hard/soft bands computed by adding all available instruments and exposures for the corresponding observation. GNH: Galactic column density in the direction of the source from the Leiden/Argentine/Bonn (LAB) Survey of Galactic HI. 2.2 Model related columns Columns referring to any particular model start with the model's name. Model names are: wapo: absorbed power-law model applied in the 0.5-10 keV band. wamekal: absorbed thermal model applied in the 0.5-10 keV band. wamekalpo: absorbed thermal plus power-law model applied in the 0.5-10 keV band. wapopo: absorbed double power-law model applied in the 0.5-10 keV band. 2.2.1 Spectral-fit summary columns The first three columns after the columns related to the source and observation, are A_FIT, P_MODEL, and A_MODELS. A_FIT: The value is set to True, if an acceptable fit, i.e. KS p-value > 0.01, has been found for at least one of the models applied, and to False otherwise. P_MODEL: The data preferred model, i.e., the model with the highest evidence (lowest logZ, see Sect. 2.2.4). A spectral model is always listed regardless of the fit being an acceptable or an unacceptable fit. A_MODELS: List of acceptable models. This column contains the remaining models with relative evidence (with respect to P_MODEL) higher than 30 ('very strong evidence' accordingly to the scale of Jeffreys 1961). Assuming all models a priori equally probable, there is no statistical reason to rule out any of the models in the set formed by P_MODEL and A_MODELS. Hence, if the fit is acceptable, the simplex model should be selected as the best-fit model. 2.2.2 Parameters and errors Columns referring to parameters and errors start with the model name and the parameter name. Possible parameters names are: logNH (decimal logarithm of wabs column density, in units of cm^-2), PhoIndex (pow photon index), kT (mekal temperature, in keV), and z (redshift of the source, only for sources with no spectroscopic redshift). Values for the normalizations of the models and the relative normalization factors between instruments are not included in the table (but they are available in the SQL database). MODEL_PARAMETER: parameter value. MODEL_PARAMETER_min, MODEL_PARAMETER_max: upper and lower limits of the 90% credible interval for the parameter. 2.2.3 Fluxes and luminosities The posterior probability distribution of the free parameters was propagated to estimate the flux and errors for each model. This method preserves the structure of the uncertainty (degeneracies, multimodal structure, etc.). Reported fluxes and luminosities in the catalogue correspond to the mode of the posterior distribution, with errors estimated as 90% credible intervals. These are EPIC fluxes, i.e., in the case of multiple instrument spectra for a single observation, the reported flux is the average of the different fluxes for each instrument and exposure. For sources with spectroscopic redshifts, luminosities were estimated using the intrinsic fluxes and the luminosity distance corresponding to that redshift. For sources with photometric redshifts, z is a free parameter and hence is propagated with the posterior distribution to estimate the corresponding luminosity distance in each case. We assumed a ΛCDM cosmology with H[0] = 67.7, Ω[m] = 0.307 (Planck Collaboration 2015, Paper XIII). We estimated fluxes and luminosities in the soft (0.5-2 keV) and hard (2-10 keV) bands. For observed fluxes this bands correspond to the observer frame. For intrinsic fluxes and luminosities they correspond to the source's rest frame. EPIC soft fluxes obtained from the spectral fits (P_MODEL) are plotted against the ones in the CATV catalogue (EP_2_FLUX + EP_3_FLUX) in Fig. 7. Fluxes from the automated fits and from the CATV are consistent within errors for ∼70% of the detections. Significant differences between both values are more frequent among sources displaying a soft spectrum, i.e., those sources that are best-fitted by a power-law with a steep photon index, or by a thermal model. More than 80% of the non-matching fluxes correspond to any of these cases. Figure 8 shows hard luminosities against redshifts estimated in the automated fits for P_MODEL. MODEL_flux_BAND: the mean observed flux (in erg cm^-2 s^-1) of all instruments and exposures for the corresponding observation, in "BAND" (soft/hard). Observed fluxes were corrected of Galactic MODEL_fluxmin_BAND, MODEL_fluxmax_BAND: lower and upper limits of the 90% credible interval. MODEL_intflux_BAND: the mean intrinsic flux (rest-frame, corrected of intrinsic absorption, in erg cm^-2 s^-1) of all instruments and exposures for the corresponding observation, in "BAND" (soft/ MODEL_intfluxmin_BAND, MODEL_intfluxmax_BAND: lower and upper limits of the 90% credible interval. MODEL_lumin_BAND: the mean luminosity (rest-frame, corrected of intrinsic absorption, in erg s^-1) of all instruments and exposures for the corresponding observation, in "BAND" (soft/hard). MODEL_luminmin_BAND, MODEL_luminmax_BAND: lower and upper limits of the 90% credible interval. Figure 7. Soft fluxes (in c.g.s. units) computed from the automated fits (P_MODEL) against fluxes in the CATV catalogue. Figure 8. Hard luminosity versus redshift (in c.g.s. units) computed from the automated fits (P_MODEL). 2.2.4 Fitting statistics MODEL_wstat: W-stat (Cash statistics) value. MODEL_dof: Degrees of freedom. MODEL_ks: Kolmogorov-Smirnov (KS) statistic. MODEL_ks_pvalue: KS p-value. MODEL_logZ: Natural logarithm of the evidence, estimated by the MultiNest algorithm.	{"url":"https://xraygroup.astro.noa.gr/sites/prodex/xmmfitcatz_documentation.html","timestamp":"2024-11-05T00:59:33Z","content_type":"application/xhtml+xml","content_length":"26486","record_id":"<urn:uuid:0b32f534-e8b2-413f-982e-d78856ae6936>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00733.warc.gz"}
Explicitly Modeling 4.3.3 Explicitly Modeling : The General Case Unfortunately, the cases in which is polygonal or polyhedral are quite limited. Most problems yield extremely complicated C-space obstacles. One good point is that can be expressed using semi-algebraic models, for any robots and obstacles defined using semi-algebraic models, even after applying any of the transformations from Sections 3.2 to 3.4. It might not be true, however, for other kinds of transformations, such as warping a flexible material [32,577]. Consider the case of a convex polygonal robot and a convex polygonal obstacle in a 2D world. Assume that any transformation in may be applied to ; thus, and . The task is to define a set of algebraic primitives that can be combined to define . Once again, it is important to distinguish between Type EV and Type VE contacts. Consider how to construct the algebraic primitives for the Type EV contacts; Type VE can be handled in a similar manner. For the translation-only case, we were able to determine all of the Type EV contacts by sorting the edge normals. With rotation, the ordering of edge normals depends on . This implies that the applicability of a Type EV contact depends on , the robot orientation. Recall the constraint that the inward normal of must point between the outward normals of the edges of that contain the vertex of contact, as shown in Figure 4.21. This constraint can be expressed in terms of inner products using the vectors and . The statement regarding the directions of the normals can equivalently be formulated as the statement that the angle between and , and between and , must each be less than . Using inner products, this implies that and . As in the translation case, the condition is required for contact. Observe that now depends on . For any , if , , and , then . Let denote the set of configurations that satisfy these conditions. These conditions imply that a point is in . Furthermore, any other Type EV and Type VE contacts could imply that more points are in . Ordinarily, , which implies that the complement, , is a superset of (thus, ). Let . Using the primitives let . It is known that , but may contain points in . The situation is similar to what was explained in Section 3.1.1 for building a model of a convex polygon from half-planes. In the current setting, it is only known that any configuration outside of must be in . If is intersected with all other corresponding sets for each possible Type EV and Type VE contact, then the result is . Each contact has the opportunity to remove a portion of from consideration. Eventually, enough pieces of are removed so that the only configurations remaining must lie in . For any Type EV contact, . A similar statement can be made for Type VE contacts. A logical predicate, similar to that defined in Section 3.1.1, can be constructed to determine whether in time that is linear in the number of primitives. One important issue remains. The expression is not a polynomial because of the and terms in the rotation matrix of . If polynomials could be substituted for these expressions, then everything would be fixed because the expression of the normal vector (not a unit normal) and the inner product are both linear functions, thereby transforming polynomials into polynomials. Such a substitution can be made using stereographic projection (see [588]); however, a simpler approach is to use complex numbers to represent rotation. Recall that when is used to represent rotation, each rotation matrix in is represented as (4.18), and the homogeneous transformation matrix becomes Using this matrix to transform a point results in the point coordinates . Thus, any transformed point on is a linear function of , , , and . This was a simple trick to make a nice, linear function, but what was the cost? The dependency is now on and instead of . This appears to increase the dimension of from 3 to 4, and . However, an algebraic primitive must be added that constrains and to lie on the unit circle. By using complex numbers, primitives in are obtained for each Type EV and Type VE contact. By defining , the following algebraic primitives are obtained for a Type EV contact: This yields . To preserve the correct topology of , the set is intersected with . The set remains fixed over all Type EV and Type VE contacts; therefore, it only needs to be considered once. Example 4..16 (A Nonlinear Boundary for ) Consider adding rotation to the model described in Example . In this case, all possible contacts between pairs of edges must be considered. For this example, there are Type EV contacts and Type VE contacts. Each contact produces algebraic primitives . With the inclusion of , this simple example produces primitives! Rather than construct all of these, we derive the primitives for a single contact. Consider the Type VE contact between and -. The outward edge normal remains fixed at . The vectors and are derived from the edges adjacent to , which are - and -. Note that each of , , and depend on the configuration. Using the 2D homogeneous transformation ), at configuration is . Using to represent rotation, the expression of becomes . The expressions of and are and , respectively. It follows that and . Note that and depend only on the orientation of , as expected. Assume that is drawn from to . This yields . The inner products , , and can easily be computed to form , , and as algebraic primitives. One interesting observation can be made here. The only nonlinear primitive is . Therefore, can be considered as a linear polytope (like a polyhedron, but one dimension higher) in that is intersected with a cylinder. Subsections Steven M LaValle 2020-08-14	{"url":"https://lavalle.pl/planning/node165.html","timestamp":"2024-11-09T19:32:39Z","content_type":"text/html","content_length":"34764","record_id":"<urn:uuid:df775394-ace1-4516-991a-aca08687a0f2>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00803.warc.gz"}
Irreducible module From Encyclopedia of Mathematics simple module A non-zero unital module $M$ over a unital ring $R$ that contains only two submodules: the null module and $M$ itself. Examples. 1) If $R = \mathbf{Z}$ is the ring of integers, then the irreducible $R$-modules are the Abelian groups of prime order. 2) If $R$ is a skew-field, then the irreducible $R$-modules are the one-dimensional vector spaces over $R$. 3) If $D$ is a skew-field, $V$ is a left vector space over $D$ and $R = \End_D(V)$ is the ring of linear transformations of $V$ (or a dense subring of it), then the right $R$-module $V$ is irreducible. 4) If $G$ is a group and $k$ is a field, then the irreducible representations of $G$ over $K$ are precisely the irreducible modules over the group algebra $R = k[G]$. A right $R$-module $M$ is irreducible if and only if it is isomorphic to $R/I$, where $I$ is a maximal right ideal in $R$. If $A$ and $B$ are irreducible $R$-modules and $f \in \Hom_R(A,B)$, then either $f=0$ or $f$ is an isomorphism (which implies that the endomorphism ring of an irreducible module is a skew-field). If $R$ is an algebra over an algebraically closed field $K$ and if $A$ and $B$ are irreducible modules over $R$, then (Schur's lemma) $$ \Hom_R(A,B) = \begin{cases} K & \ \text{if}\ A \cong B\ ; \\ 0 & \ \text{otherwise} \ .\end{cases} $$ The concept of an irreducible module is fundamental in the theories of rings and group representations. By means of it one defines the composition sequence and the socle of a module, the Jacobson radical of a module and of a ring, and a completely-reducible module. Irreducible modules are involved in the definition of a number of important classes of rings: classical semi-simple rings, primitive rings, and others. [1] N. Jacobson, "Structure of rings" , Amer. Math. Soc. (1956) [2] C.W. Curtis, I. Reiner, "Representation theory of finite groups and associative algebras" , Interscience (1962) [3] J. Lambek, "Lectures on rings and modules" , Blaisdell (1966) [4] C. Faith, "Algebra: rings, modules, and categories" , 1–2 , Springer (1973–1976) How to Cite This Entry: Irreducible module. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Irreducible_module&oldid=42959 This article was adapted from an original article by A.V. Mikhalev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article	{"url":"https://encyclopediaofmath.org/wiki/Irreducible_module","timestamp":"2024-11-05T02:50:23Z","content_type":"text/html","content_length":"16455","record_id":"<urn:uuid:fbc98033-83e1-4c9b-acea-c8e765757150>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00484.warc.gz"}
Higgsas Blender 3.3-4.2 Geometry Nodes Groups Toolset Pack Blender Geometry Nodes toolset pack includes over 180 wide range of advance node groups to help improve and enhance procedural workflows Geometry Measure Mesh Primitives SDF Nodes Current Node Groups Curve Deform Face Offset To Sphere UV Deform Set Center Fit Size Cube Deform VDM Brush Surface Bind VDM Brush VDM Points Scatter Mesh Sharpen Marching Squares Isolines Marching Triangles Isolines Marching Squares Surface Marching Triangles Surface Tessellate Mesh Tessellate Mesh Smooth Tessellate Topology Helper Connect Points 2D Curl Noise Inset Faces 2D Recursive Subdivision Image to Ascii Mesh to Ascii Edge Offset Mesh Face Divider Mesh Face Subdivide Voxel Remesh Mesh Contours Spheres Intersection Geometry Measure Mesh Measure Mesh Island Measure Mesh Thickness Gradient Direction Face Vertex Position Mesh Tension Geometry Center Face Tangent UV Tangent Mesh Ambiet Occlusion Triangle Circumcircle Triangle Incircle Triangle Tangent Circle Triplanar UV Mapping UV Mirror Camera UV Coordinates Cylinder UV Mapping Sphere UV Mapping Attribute Smooth Bezier Easing Box Mapping Looped Noise Texture 2D Looped Noise Texture Geometry Visualizer Camera Culling Mesh Primitives Hexagon Grid Triangle Grid Quad Sphere Torus Knot Rounded Cube Bounding Region Selection Select by Normal Is Inside Volume Edge Angle Selection Boundary Edge Expand/Contract Selection Directional Falloff Spherical Falloff Radian Falloff Object Directional Falloff Object Spherical Falloff Wave Falloff Curve Offset Twist Curve Loft Curve Even Curve to Mesh Curve Point Angle UV Curve to Mesh Poly Arc Decimate Curve Curve Bisect Curve Mesh Boolean 3D Curve Fill Tubes to Splines Edge Bundling Circle Outer/Inner Tangent Distribute Points in Volume Circular Array Volume Points Grid Homogeneous Sphere Distribute Points on Edges 3D Points Grid Phyllotaxis Sphere Phyllotaxis Disk Homogeneous Disk Random Points SDF Nodes SDF Sphere SDF Torus SDF Capsule SDF Polygon SDF Cube SDF to Mesh Mesh To SDF SDF Gyroid SDF Cylinder SDF Boolean SDF Volume Points Fracture Points to Smooth SDF Direction Reaction Diffusion Reaction Diffusion Solver Triangle Mesh Circle Packing Splines Packing Curl Noise 2D Curl Noise 3D Surface Curl Noise Royalty Free License The Royalty Free license grants you, the purchaser, the ability to make use of the purchased product for personal, educational, or commercial purposes as long as those purposes do not violate any of the following: • You may not resell, redistribute, or repackage the purchased product without explicit permission from the original author • You may not use the purchased product in a logo, watermark, or trademark of any kind If you have any question or suggestions you can contact via: Email: higgsasmotion@gmail.com Twitter: higgsasxyz Instagram: higgsas 2024-10-11 Update Renamed Effects node category to Vector Fields to better match what nodes do Added new nodes category - Image Removed old 2D Curl Noise node, you can achieve same results with better performance using Advect Splines + Curl Noise 2D/3D Added NURBS option in Lost Splines Node Reworked Expand / Contract Selection node to improve performance New Nodes: Polar to Cartesian Cartesian to Polar Triangle Mesh to Voronoi Mesh Curvature Mesh Fresnel Points Relax Image Points Distribute Image Trace Image Dithering UV Transform Sphere Plane Intersection 2D Distance to Edge Circle Packing Ray Sphere Intersection Index String Selection 2024-09-07 Update Added Curve Intersection 2024-08-21 Update Added version for blender 4.2 with some updates Camera Culling - updated with matrix nodes making it faster Camera UV Coordinates - updated with matrix nodes making it faster Bend - updated with matrix nodes making it 2-5 times faster VDM Points Scatter - updated with matrix nodes making it 2-5 times faster VDM Brush - added Fast/Accurate modes. Fast mode increse performance up to 10 times compate to accurate mode Twist, Strech, Taper - added Smooth Limits option Box Image Mapping - updated with matrix nodes Sphere/Cyling UV Mapping - updated with matrix nodes New nodes: Mesh Curve Direction Guide Correct UV UV Seam 2024-04-20 Update Added 2 new nodes: 2D Hilbert Curve, Bricks Grid Removed installation guide for blender 3.3 version because it caused confusion 2024-03-29 Update Added Blender 4.1 version with some improvements Added position attribute to falloffs nodes in blender 4.1 Normalized falloffs directions Improved mesh contours fill curves performance in blender 4.1 Improved Easing node usability using new menus switch in blender 4.1 3d curve fill much faster in blender 4.1 VDM Points Scatter added better option for boundaries blur 2024-03-19 Update 23 New Nodes Instances Bounding Box Maze solver Curve Banking Spine Heart SDF Heart 2D Remesh Tubes to Splines Advect Splines Phyllotaxis Surface Sharpen Mesh Rounded Cube Image Points Stippling Triangle Mesh Circle Packing Edge Bundling TSP mesh UV Mirror Sphere Intersection Instances AABB Collision Splines Packing SDF Volume Points Fracture Circe Outer/Inner Tangent Curve Directional Reaction Diffusion Moved UV nodes to new UV category Moved curl noise nodes to new Effects category Reaction Diffusion Solver - added time steps and simplified the node Distance to Edge Voronoi - updated to use Repeat Zone for performance Mesh Face Divided - updated to use Repeat Zone for performance Circle Packing - now using Repeat Zone instead of Simulation Zone, so you won’t need to play animation for the packing VDM Points Scatter - Added blur option thanks to Benny_G feedback Catenary Curves - updated to use Repeat Zone for performance Poly Arc - updated to use Repeat Zone for performance Curve Offset - fixed direction being not normalized Surface Curl Noise - added option to project to surface and simplified normal input just use mesh 2024-02-03 Update Added version for Blender 4.0 2023-08-16 Update Replaced Mesh Section node with Mesh Contour node. New mesh contour node works much better and has ability to do multiple contour slices 2023-08-11 Update 26 new nodes: VDM Brush, VDM Point Scatter, Sphere UV Mapping, Cylinder UV Mapping, Voxel Remesh, Mesh Face Divider, Mesh Face Subdivide, Rotate Element, Triangle Incircle, Triangle circumcircle, Triangle Tangent Circle, 3D Curve Fill, Curve Bisect, Curve Mesh Boolean, Curve Decimate, Index Ratio, Mix Splines, Poly Arc, Cube Deform, Mesh Offset, Mesh Section, Torus, Curl Noise 2D, Curl Noise 3D, Surface Curl Noise, Reaction Diffusion Solver 2023-05-30 Update New node: Bezier Easing 2023-05-13 Update New nodes: Line Line Intersection Line Plane Intersection Edge Bisect Cube Recursive Subdivision Surface Bind Mesh Ambient Occlusion Distance to Edge Voronoi Wave Falloff 2023-05-30 Update New node: Bezier Easing 2023-04-27 Update New node: Marching Squares Surface Renamed Marching Squares to Marching Squares Isolines 2023-04-21 Update New node: Splines Patch - https://higgsas-geo-nodes-manual.readthedocs.io/en/latest/curves.html#splines-patch 2023-04-20 Update Added boundary edge option to Marching Squares/Triangles nodes, and performance improvements Fixed issue with Tessellate Mesh Smooth not working correctly with Tessellate Topology Helper New node: Set Center 2023-04-05 Update Fixed nodes not loading when opening new blend files 5 stars 4 stars 3 stars 2 stars 1 star	{"url":"https://higgsas.gumroad.com/l/wrusot?layout=profile&recommended_by=search","timestamp":"2024-11-09T23:34:45Z","content_type":"text/html","content_length":"72699","record_id":"<urn:uuid:26d4f623-33b3-4050-abd5-ad6e96ae59e1>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00553.warc.gz"}
1027. D++ Again The language D++, that was perfected by the participants of our March competition, continues improving. Its founders try to make the syntax as clear as it is possible in order to make the programming simpler in the future. Of course, some minimal set of rules is to stay without changes. Your program is to check the observance of rules, concerning the arrangement of brackets and comments. A text of a correct D++ program contains a symbol part, arithmetic expressions and comments. Comments may appear everywhere and may contain any symbols. A comment is always opened by a pair of symbols "(" and is closed by a pair of symbols ")". Each comment must be closed. An arithmetic expression in D++ is always opened by "(", is closed by ")" and may contain only symbols "=+-/ 0123456789)(" and "end of line" symbols. An arithmetic expression can't start with a pair of symbols "(". You may run across embedded brackets in an arithmetic expression. In this case these brackets are to be balanced. It means that "((1)))" as well as "(23))((+)" are not correct arithmetic expressions. An arithmetic expression is correct if and only if brackets placed correctly. At last, all the rest of the program text (the result of rejection of all comments and arithmetic expressions from the initial text of the program) may contain every symbol excluding "(" and ")". We would like to especially notice that the spaces are possible anywhere in a text of a program except when appearing in arithmetic expressions. Some text is written in the standard input. There are not more than 10000 symbols in the text. There may be Latin letters, digits, brackets, symbols of arithmetic operations, spaces and "end of line" Your program should write "YES" to the output if the introduced text is a correct D++ program, and "NO" otherwise. Hello, here is a sample D++ program. It contains some arithmetical expressions like (2+2=4), (2+-/) and ((3+3)3=20(this is not true, but you don’t have to verify it :-) )+8) (* the closing bracket in the previous comment is also in order, since this bracket does not belong to any arithmetical expression*) Problem Author: Leonid Volkov, Alexey Lysenko Problem Source: Ural State University Internal Contest October'2000 Junior Session	{"url":"https://timus.online/problem.aspx?space=1&num=1027","timestamp":"2024-11-14T19:01:49Z","content_type":"text/html","content_length":"7656","record_id":"<urn:uuid:244f4a9e-9252-4c23-ad08-8ac4f536398b>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00881.warc.gz"}
This years Turing Award: Model Checking Guest Post by Rance Cleavland, professor at University of Maryland at College Park, who works in Model Checking. Earlier this week, the ACM the winners of the 2007 ACM Turing Award (awarded in 2008, for reasons that elude me). They are Edmund Clarke (CMU), Allan Emerson (U. Texas) and Joseph Sifakis (Verimag, in France). The award statement honors the contributions of these three to the theory and practice of model checking, which refers to an array of techniques for automatically determining whether models of system behavior satisfy properties typically given in temporal logic. After first blanching at the application of an adjective ("temporal") to a term ("logic") that is usually left unqualified, a Gentle Reader may wonder what all the fuss is about. Is model checking really so interesting and important that its discoverers and popularizers deserve a Turing Award? The glib answer is of course, because the selection committee must have a fine sense of judgment. My aim is to convince you of a less glib response, which is that model checking is the most fundamental advance in formal methods for program verification since Hoare coined the term in the 60s. What is "model checking"? In mathematical logic, a model is a structure (more terminology) that makes a logical formula true. So "model checking" would refer to checking whether a structure is indeed a model for a given formula. In fact, this is exactly what model checking is, although in the Clarke-Emerson-Sifakis meaning of the term, the structures - models - are finite-state Kripke structures (= finite-state machines, except with labels on state rather than transitions and no accepting states), and the logical formulas are drawn from propositional temporal logic (= proposition logic extended with modalities for expressing always in the future eventually in the future The Clarke-Emerson-Sifakis algorithmic innovation was to notice that for certain flavors of temporal logic ( pure branching time ), model checking could be decided in polynomial time; this is the gist of the papers written independently in 1981 by Clarke and Emerson on the one hand, and Sifakis and Queille on the other. Subsequently, these results were improved to show that model checking for pure branching-time logic is proportional to the product of the size of the Kripke structure and the size of the formula (often, maybe misleadingly, called "linear time" in the model-checking community, since the size of the model dominates the product). Of course, a linear-time algorithm (OK, I'm in the model-checking community!) is only of passing interest unless it has real application. This comment involves two questions. 1. Is the general problem one people want solved? 2. Can the algorithm produce results on the instances of the problem people want solved? The answer to 1 is "YES YES YES". The ability automatically to check the correctness of a program/hardware design/communications protocol would offer incalculable benefits to developers of these systems. Early on, the answer to 2 for model checking was in doubt, however, for the simple reason that the size of Kripke structure is typically exponential in the size of the used to define it. (State = assignment of values to variables, so num-of-states is exponential in num-of-variables, etc.) Throughout the 80s and 90s, the three winners worked on many techniques for overcoming the state-explosion problem : compositional techniques, symmetry reductions, etc. One of the most successful was symbolic model checking : the use of logic formulas, rather than linked lists, etc., in the model-checking process to represent large sets of states. While none of these techniques proved uniformly applicable, symbolic model checking found a home in the hardware-design community, and model-checkers are now standard parts of the design flow of microprocessor design and incorporated routinely in the design tools produced by companies like Cadence, Synopsys and The MathWorks. So what to make of the Turing Award? I would say that the algorithmic innovation was deep and insightful, but not the source of the award. Rather, the combination of the initial insight, together with the persistence of the award winners in identifying engineering advances to further the state of the practice, is what earned them their, in my view well-deserved and maybe even over-due, prize. 3 comments: 1. thank you for a instructive guest-post! it sounds like symbolic model checking is used for hardware verification, is it practical for software verification anymore? 2. Model checking is widely used in hardware but model checking techniques are also considered very valuable in practice on aspects of software. (One example in the SLAM project at Microsoft for verifying device drivers.) A great aspect is how much theory has been involved in making this successful. (One theorem about automata due to Buchi that I mention below deserves to be more widely known and allows one to verify pushdown systems and not just finite automata!) The award winners deserve a huge amount of credit for realizing that model checking was the way to go and championing it over more than a decade before it really caught on. It is hard to get a sense of how far from practical this approach was considered in the early 1980's. After all the problem they wanted to solve is a generalization of deciding graph reachability in an exponential size but polynomially-specified graph. The success of model checking owes much to theoretical ideas that have advanced the state of the art in a big way. These include the ideas of Vardi and Wolper (Godel prize a few years ago) who derived new efficient algorithms for automata-theoretic verification that were later incorporated and extend by Kurshan and Holzmann (along with refinements by others) in the SPIN model checker (for which the four won the Kannelakis prize). See Kurshan's 1994 STOC invited talk for an overview of the approach. A huge advance was symbolic model checking using BDDs (oblivious read-once branching programs championed as a data structure for Boolean functions by Randy Bryant) developed in Ken McMillan's ACM Award-winning dissertation in the early 1990's. (Kannelakis Award for Bryant, Clarke, Emerson, and McMillan.) In the late 1990's improvements in SAT solvers made them efficient enough to replace BDDs for bounded model checking. All three of these approaches have domains in which they are preferred. Subsequent to this line of work there has been a huge flowering of different ideas for verifying aspects of software. One major idea has been abstraction-refinement in which one abstracts away most of the detail of the software in the initial model and then adds detail automatically based on how verification fails and then tries again. (Interpolation results about resolution proofs actually can be useful here.) The most surprising aspect from my point of view, though maybe not the most important, has been the ability to verify pushdown systems. Verifying these is obviously necessary to handle the call stack. How can one possibly use finite state verification to determine properties of pushdown automata? Buchi's theorem I mentioned above does this: For any PDA M with stack alphabet Gamma, the language S(M) over Gamma^* of consisting of all possible stack configurations of M is regular! Moreover, given M one can build an NFA for S(M) fairly efficiently. (Actually one works with a language in Q Gamma^* that includes the state in the configuration.) So, for example, in order to verify a safety property (one that is always true) one can work with a verification that involves finite automata only! This property of stack languages seems like such a natural one that it would be described in standard textbooks. (Consider how simple the possible stacks are for any of the familiar PDAs you know.) What is the simplest proof that anyone can come up with? 3. I'm very happy to see the Turing go to model checking. That was one of the areas I felt should have been given preference over the previous two Yet Another Compiler Writer awards (of which I consider the Naur one to be an obvious mistake at this point in time -- he should have gotten it (very) early on or not at all because so many others have contributed more since, including some in his own department). I thought that Gerard Holzman would be one of the recipients if the Turing was ever awarded for model checking but he was left out. That surprises me. (Others that I think would have been preferable to Naur and possibly Allen: Scheifler and Gettys for X, Miller, Clifford, and Kohl for Kerberos.)	{"url":"https://blog.computationalcomplexity.org/2008/02/this-years-turing-award-model-checking.html?m=0","timestamp":"2024-11-05T13:01:06Z","content_type":"application/xhtml+xml","content_length":"186259","record_id":"<urn:uuid:e8a2fda7-a0a9-4787-9f5d-2629762f8a4c>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00877.warc.gz"}
Combinatorial Optimization (AIT) Group K Spring term, 2023 16:10–18:00, Monday, Room 4 12:15–14:00, Thursday, Room 2 Course materials: Exam results Final results, grades The present state of all required hand-ins 14. Boundedness of the objective function: Problem sheet, Class summary 15. The duality theorem: Problem sheet, Class summary 16. The duality theorem – Form 2: Problem sheet, Class summary 17. Complementary Slackness: Problem sheet, Class summary 18. Complementary Slackness – Form 2: Problem sheet, Class summary 19. An application in game theory – Part I: Problem sheet 20. An application in game theory – Part II: Problem sheet 21. An application in game theory – Part III: Problem sheet, Class summary 22. Network flows revisited: Problem sheet, Class summary 23. The Maximum Weight Bipartite Matching Problem: Problem sheet, Class summary 24. The Hungarian Method -- Egerváry's algorithm: Problem sheet, Class summary 25. Final Exam Review Problems, Exam Topics	{"url":"http://math.bme.hu/~vkitti/combopt_spring2023.html","timestamp":"2024-11-08T12:07:13Z","content_type":"application/xhtml+xml","content_length":"3379","record_id":"<urn:uuid:901bc60a-79e3-42cd-be2f-91017fa4bdf5>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00323.warc.gz"}
How to make profit in bitcoin trading?\|数学建模代写 \| UprivateTA™- 数学代写 Linear predictive models for bitcoin price The most extreme form of linear predictive models is one in which all the coefficients are equal in magnitude (but not necessarily in sign). For example, suppose you have identified a number of factors ( $f^{\prime}$ s) that are useful in predicting whether tomorrow’s return of a bitcoin index is positive. One factor may be today’s return, with a positive today’s return predicting a positive future return. Another factor may be today’s change in the volatility index (VIX), with a negative change predicting positive future return. You may have several such factors. If you normalize these factors by turning them first into Z-scores (using in-sample data!): z(i)=(f(i)-\operatorname{mean}(f)) / \operatorname{std}(f) where $f(i)$ is the $i^{\text {th }}$ factor, you can then predict tomorrow’s return $R$ by R=\operatorname{mean}(R)+\operatorname{std}(R) \sum_{i}^{n} \operatorname{sign}(i) z(i) / n The quantities mean $(f)$ and $s t d(f)$ are the historical average and standard deviation of the various $f(i),$ sign $(i)$ is the sign of the historical correlation between $f(i)$ and $R,$ and $\ operatorname{mean}(R)$ and $s t d(R)$ are the historical average and Statistical Significance of Backtesting: Hypothesis Testing In any backtest, we face the problem of finite sample size: Whatever statistical measures we compute, such as average returns or maximum drawdowns, are subject to randomness. In other words, we may just be lucky that our strategy happened to be profitable in a small data sample. Statisticians have developed a general methodology called hypothesis testing to address this issue. The general framework of hypothesis testing as applied to backtesting follows these steps: 1. Based on a backtest on some finite sample of data, we compute a certain statistical measure called the test statistic. For concreteness, let’s say the test statistic is the average daily return of a trading strategy in that period. 2. We suppose that the true average daily return based on an infinite data set is actually zero. This supposition is called the null hypothesis. 3. We suppose that the probability distribution of daily returns is known. This probability distribution has a zero mean, based on the null hypothesis. We describe later how we determine this probability distribution. 4. Based on this null hypothesis probability distribution, we compute the probability $p$ that the average daily returns will be at least as large as the observed value in the backtest (or, for a general test statistic, as extreme, allowing for the possibility of a negative test statistic). This probability $p$ is called the p-value, and if it is very small (let’s say smaller than 0.01), that means we can “reject the null hypothesis,” and conclude that the backtested average daily return is statistically signifi cant. The step in this procedure that requires most thought is step $3 .$ How do we determine the probability distribution under the null hypothesis? Perhaps we can suppose that the daily returns follow a standard parametric probability distribution such as the Gaussian distribution, with a mean of zero and a standard deviation given by the sample standard deviation of the daily returns. If we do this, it is clear that if the backtest has a high Sharpe ratio, it would be very easy for us to reject the null hypothesis. This is because the standard test statistic for a Gaussian distribution is none other than the average divided by the standard deviation and multiplied by the square root of the number of data points (Berntson, 2002$)$. The $p$ -values for various critical values are listed in Table $1.1 .$ For example, if the daily Sharpe ratio multiplied by the square root of the number days $(\sqrt{n})$ in the backtest is greater than or equal to the critical value $2.326,$ then the $p$ -value is smaller than or equal to 0.01 . But this is actually just a stopgap measure. In fact, if the difficulty of obtaining real data in the trading market is greatly reduced, then reinforcement learning based on reinforcement learning will show considerable power The Basics of Mean Reversion Is mean reversion also prevalent in fi nancial price series? If so, our lives as traders would be very simple and profi table! All we need to do is to buy low (when the price is below the mean), wait for reversion to the mean price, and then sell at this higher price, all day long. Alas, most price series are not mean reverting, but are geometric random walks. geometric random walks The returns, not the prices, are the ones that usually randomly distribute around a mean of zero. Unfortunately, we cannot trade on the mean reversion of returns. (One should not confuse mean reversion of returns with anti-serial-correlation of returns, which we can defi nitely trade on. But anti-serial-correlation of returns is the same as the mean reversion of prices.) Those few price series that are found to be mean reverting are called stationary, and in this chapter we will describe the statistical tests (ADF test and the Hurst exponent and Variance Ratio test) for stationarity. There are not too many prefabricated Augmented Dickey-Fuller Test If a price series is mean reverting, then the current price level will tell us something about what the price’s next move will be: If the price level is higher than the mean, the next move will be a downward move; if the price level is lower than the mean, the next move will be an upward move. The ADF test is based on just this observation. We can describe the price changes using a linear model: \Delta y(t)=\lambda y(t-1)+\mu+\beta t+\alpha_{1} \Delta y(t-1)+\cdots+\alpha_{k} \Delta y(t-k)+\epsilon_{t} where $\Delta y(t) \equiv y(t)-y(t-1), \Delta y(t-1) \equiv y(t-1)-y(t-2),$ and so on. The ADF test will find out if $\lambda=0$. If the hypothesis $\lambda=0$ can be rejected, that means the next move $\Delta y(t)$ depends on the current level $y(t-1),$ and therefore it is not a random walk. The test statistic is the regression coefficient $\lambda$ (with $y(t-1)$ as the independent variable and $\Delta y(t)$ as the dependent variable) divided by the standard error of the regression fit: $\lambda / \mathrm{SE}(\lambda)$. The statisticians Dickey and Fuller have kindly found out for us the distribution of this test statistic and tabulated the critical values for us, so we can look up for any value of $\lambda / \operatorname{SE}(\lambda)$ whether the hypothesis can be rejected at, say, the 95 percent probability level. Notice that since we expect mean regression, $\lambda / \mathrm{SE}(\lambda)$ has to be negative, and it has to be more negative than the critical value for the hypothesis to be rejected. The critical values themselves depend on the sample size and whether we assume that the price series has a non-zero mean $-\mu / \lambda$ or a steady drift $-\beta t / \lambda .$ In practical trading, the constant drift in price, if any, tends to be of a much smaller magnitude than the daily fluctuations in price. So for simplicity we will assume this drift term to be zero $(\beta=0)$. Kalman Filter as Market-Making Model Kalman Filter is very poupular in SLAM, in fact it is almost near the perfect in some simple situaiton. the adventage of SLAM than finance market is the state space is much more predictable, or in other word the non-zero value of the transition matrix is much more cancertrate. There is an application of Kalman filter to a meanreverting strategy. In this application we are concerned with only one mean-reverting price series; we are not concerned with finding the hedge ratio between two cointegrating price series. However, we still want to find the mean price and the standard deviation of the price series for our mean reversion trading. So the mean price $m(t)$ is the hidden variable here, and the price $y(t)$ is the observable variable. The measurement equation in this case is trivial: (“Measurement equation”) with the same state transition equation (“State transition”) So the state update equation is just m(t \mid t)=m(t \mid t-1)+K(t)(y(t)-m(t \mid t-1)) . \quad(\text { “State update” }) (This may be the time to review Box 3.1 if you skipped it on first reading.) The variance of the forecast error is The Kalman gain is K(t)=R(t \mid t-1) /\left(R(t \mid t-1)+V_{e}\right) and the state variance update is R(t \mid t)=(1-K(t)) R(t \mid t-1) Why are these equations worth highlighting? Because this is a favorite model for market makers to update their estimate of the mean price of an asset, as Euan Sinclair pointed out (Sinclair, 2010$)$. To make these equations more practical, practitioners make further assumptions about the measurement error $V_{e}$, which, as you may recall, measures the uncertainty of the observed transaction price. But how can there be uncertainty in the observed transaction price? It turns out that we can interpret the uncertainty in such a way that if the trade size is large (compared to some benchmark), then the uncertainty is small, and vice versa. So $V_{e}$ in this case becomes a function of $t$ as well. If we denote the trade size as $T$ and the benchmark trade size as $T_{\max }$, then $V_{e}$ can have the form V_{e}=R(t \mid t-1)\left(\frac{T}{T_{\max }}-1\right) So you can see that if $T=T_{\max },$ there is no uncertainty in the observed price, and the Kalman gain is $1,$ and hence the new estimate of the mean price $m(t)$ is exactly equal to the observed price! But what should $T_{\max }$ be? It can be some fraction of the total trading volume of the previous day, for example, where the exact fraction is to be optimized with some training data. Note the similarity of this approach to the so-called volume-weighted average price (VWAP) approach to determine the mean price, or fair value of an asset. In the Kalman filter approach, not only are we giving more weights to trades with larger trade sizes, we are also giving more weights to more recent trade prices. So one might compare this to volume and timeweighted average price. Interday Momentum Strategies There are four main causes of momentum: 1. For futures, the persistence of roll returns, especially of their signs. 2. The slow diff usion, analysis, and acceptance of new information. 3. The forced sales or purchases of assets of various type of funds. 4. Market manipulation by high-frequency traders. We will be discussing trading strategies that take advantage of each cause of momentum in this and the next chapter. In particular, roll returns of futures, which featured prominently in the last chapter, will again take center stage. Myriad futures strategies can be constructed out of the persistence of the sign of roll returns Nonoverlapping Periods for Correlation Calculations With the advent of machine-readable, or “elementized,” news feeds, for example, Aicoin use Twitter API, it is now possible to programmatically capture all the news items on a company, not just those that fit neatly into one of the narrow categories such as earnings announcements or merger and acquisition (M&A) activities. Furthermore, natural language processing algorithms are now advanced enough to analyze the textual information contained in these news items, and assign a “sentiment score” to each news article that is indicative of its price impact on a bitcoin market, and an aggregation of these sentiment scores from multiple news articles from a certain period was found to be predictive of its future return. For example, Hafez and Xie, using RavenPack’s Sentiment Index, found that buying a portfolio of bitcoin with positive sentiment change and shorting one with negative sentiment change results in an APR from 52 percent to 156 percent and Sharpe ratios from 3.9 to 5.3 before transaction costs, depending on how many bitcoin are included in the portfolios (Hafez and Xie, 2012). The success of these cross-sectional strategies also demonstrates very neatly that the slow diffusion of news is the cause ofbitcoin momentum. There are other vendors besides RavenPack that provide news sentiments on bitcoin. If you believe your own sentiment algorithm is better than theirs, you can subscribe directly to an element-sized news feed instead and apply your algorithm to it. I mentioned before that Newswire offers a low-cost version of this type of news feeds, but offerings with lower latency and better coverage are provided by Bloomberg Event-Driven Kelly formula If the probability distribution of returns is Gaussian distribution, the Kelly formula gives us a very simple answer for optimal leverage $f$ : f=m / s^{2} where $m$ is the mean excess return, and $s^{2}$ is the variance of the excess The best expositions of this formula can be found in the summary paper, Edward Thorp’s (1997) paper, in fact, it can be proven that if the Gaussian assumption is a good approximation, then the Kelly leverage $f$ will generate the highest compounded growth rate of equity, assuming that all profits are reinvested. However, even if the Gaussian assumption is really valid, we will inevitably suffer estimation errors when we try to estimate what the “true” mean and variance of the excess return are. And no matter how good one’s estimation method is, there is no guarantee that the future mean and variance will be the same as the historical ones. The consequence of using an overestimated mean or an underestimated variance is dire: Either case will lead to an overestimated optimal leverage, and if this overestimated leverage is high enough, it will eventually lead to ruin: equity going to zero. However, the consequence of using an underestimated leverage is merely a submaximal compounded growth rate. Many traders justifiably prefer the later scenario, and they routinely deploy leverage equal to half of what the Kelly formula recommends: the so-called half-Kelly leverage. There is another usage of the Kelly formula besides setting the optimal leverage: it also tells us how to optimally allocate our buying power to different portfolios or strategies. Let’s denote $F$ as a column vector of optimal leverages that we should apply to the different portfolios based on a common pool of equity. (For example, if we have $\$ 1$ equity, then $F=[3.21 .5]^{T}$ means the first portfolio should have a market value of $\$ 3.2$ while the second portfolio should have a market value of $\$ 1.5$. The $T$ signifies matrix transpose.) The Kelly formula says F=C^{-1} M 经济金融答疑代写Financial Derivatives代写Black-Scholes 期权定价代写请认准UpriviateTA. UpriviateTA为您的留学生涯保驾护航。	{"url":"https://uprivateta.com/how-to-make-profit-in-bitcoin-trading/","timestamp":"2024-11-12T22:47:29Z","content_type":"text/html","content_length":"124082","record_id":"<urn:uuid:7a4d2d7c-deb2-4123-b3dc-cd6280c73938>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00639.warc.gz"}
Convergence of a Kinetic Equation to a Fractional Diffusion Equation 2010, v.16, Issue 1, 15-44 A linear Boltzmann equation is interpreted as the forward equation for the probability density of a Markov process $(K(t)$, $Y(t))$ on $(\t\times\r)$, where $\t$ is the one-dimensional torus. $K(t)$ is a autonomous reversible jump process, with waiting times between two jumps with finite expectation value but infinite variance. $Y(t)$ is an additive functional of $K$, defined as $\int_0^t v(K (s)) ds$, where $\|v\|\sim 1$ for small $k$. We prove that the rescaled process $N^{-2/3}Y(Nt)$ converges in distribution to a symmetric Levy process, stable with index $\alpha=3/2$. Keywords: anomalous diffusion,Levy process,Boltzmann equation,coupledoscillators,kinetic limit,heat conductance Please log in or register to leave a comment There are no comments yet	{"url":"https://math-mprf.org/journal/articles/id1200/","timestamp":"2024-11-04T08:46:20Z","content_type":"text/html","content_length":"13928","record_id":"<urn:uuid:095caee4-5d3c-42cd-8370-521f4aae66e3>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00594.warc.gz"}
Rational fixed points for linear group actions Rational fixed points for linear group actions We prove a version of the Hilbert Irreducibility Theorem for linear algebraic groups. Given a connected linear algebraic group $G$, an affine variety $V$ and a finite map $\pi :V\to G$, all defined over a finitely generated field $\kappa$ of characteristic zero, Theorem 1.6 provides the natural necessary and sufficient condition under which the set $\pi \left(V\left(\kappa \right)\right)$ contains a Zariski dense sub-semigroup $\Gamma \subset G\left(\kappa \right)$; namely, there must exist an unramified covering $p:\stackrel{˜}{G}\to G$ and a map $\theta :\stackrel{˜}{G}\to V$ such that $\pi \circ \theta =p$. In the case $\kappa =ℚ$, $G={𝔾}_{a}$ is the additive group, we reobtain the original Hilbert Irreducibility Theorem. Our proof uses a new diophantine result, due to Ferretti and Zannier [9]. As a first application, we obtain (Theorem 1.1) a necessary condition for the existence of rational fixed points for all the elements of a Zariski-dense sub-semigroup of a linear group acting morphically on an algebraic variety. A second application concerns the characterisation of algebraic subgroups of ${GL}_{N}$ admitting a Zariski-dense sub-semigroup formed by matrices with at least one rational eigenvalue. Corvaja, Pietro. "Rational fixed points for linear group actions." Annali della Scuola Normale Superiore di Pisa - Classe di Scienze 6.4 (2007): 561-597. <http://eudml.org/doc/272251>. abstract = {We prove a version of the Hilbert Irreducibility Theorem for linear algebraic groups. Given a connected linear algebraic group $G$, an affine variety $V$ and a finite map $\pi :V\ rightarrow G$, all defined over a finitely generated field $\kappa $ of characteristic zero, Theorem 1.6 provides the natural necessary and sufficient condition under which the set $\pi (V(\kappa ))$ contains a Zariski dense sub-semigroup $\Gamma \subset G(\kappa )$; namely, there must exist an unramified covering $p:\tilde\{G\}\rightarrow G$ and a map $\theta :\tilde\{G\}\rightarrow V$ such that $\pi \circ \theta =p$. In the case $\kappa =\mathbb \{Q\}$, $G=\mathbb \{G\}_\{a\}$ is the additive group, we reobtain the original Hilbert Irreducibility Theorem. Our proof uses a new diophantine result, due to Ferretti and Zannier [9]. As a first application, we obtain (Theorem 1.1) a necessary condition for the existence of rational fixed points for all the elements of a Zariski-dense sub-semigroup of a linear group acting morphically on an algebraic variety. A second application concerns the characterisation of algebraic subgroups of $\operatorname\{GL\}_N$ admitting a Zariski-dense sub-semigroup formed by matrices with at least one rational eigenvalue.}, author = {Corvaja, Pietro}, journal = {Annali della Scuola Normale Superiore di Pisa - Classe di Scienze}, keywords = {Algebraic groups; Hilbert irreducibility}, language = {eng}, number = {4}, pages = {561-597}, publisher = {Scuola Normale Superiore, Pisa}, title = {Rational fixed points for linear group actions}, url = {http://eudml.org/doc/272251}, volume = {6}, year = {2007}, TY - JOUR AU - Corvaja, Pietro TI - Rational fixed points for linear group actions JO - Annali della Scuola Normale Superiore di Pisa - Classe di Scienze PY - 2007 PB - Scuola Normale Superiore, Pisa VL - 6 IS - 4 SP - 561 EP - 597 AB - We prove a version of the Hilbert Irreducibility Theorem for linear algebraic groups. Given a connected linear algebraic group $G$, an affine variety $V$ and a finite map $\pi :V\rightarrow G$, all defined over a finitely generated field $\kappa $ of characteristic zero, Theorem 1.6 provides the natural necessary and sufficient condition under which the set $\pi (V(\kappa ))$ contains a Zariski dense sub-semigroup $\Gamma \subset G(\kappa )$; namely, there must exist an unramified covering $p:\tilde{G}\rightarrow G$ and a map $\theta :\tilde{G}\rightarrow V$ such that $\pi \circ \ theta =p$. In the case $\kappa =\mathbb {Q}$, $G=\mathbb {G}_{a}$ is the additive group, we reobtain the original Hilbert Irreducibility Theorem. Our proof uses a new diophantine result, due to Ferretti and Zannier [9]. As a first application, we obtain (Theorem 1.1) a necessary condition for the existence of rational fixed points for all the elements of a Zariski-dense sub-semigroup of a linear group acting morphically on an algebraic variety. A second application concerns the characterisation of algebraic subgroups of $\operatorname{GL}_N$ admitting a Zariski-dense sub-semigroup formed by matrices with at least one rational eigenvalue. LA - eng KW - Algebraic groups; Hilbert irreducibility UR - http://eudml.org/doc/272251 ER - Citations in EuDML Documents You must be logged in to post comments. To embed these notes on your page include the following JavaScript code on your page where you want the notes to appear.	{"url":"https://eudml.org/doc/272251","timestamp":"2024-11-12T10:10:54Z","content_type":"application/xhtml+xml","content_length":"54216","record_id":"<urn:uuid:986f94c9-228a-4e24-82df-2e30fa8a944c>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00262.warc.gz"}
How Many Bits Are Needed To Represent The 18 Symbols In The Hawaiian Alphabet?How Many Bits Are Needed To Represent The 18 Symbols In The Hawaiian Alphabet? - Hawaii Star Save money on your next flight Skyscanner is the world’s leading flight search engine, helping you find the cheapest flights to destinations all over the world. With its beautiful beaches, lush rainforests, and vibrant culture, Hawaii is a paradise for many visitors. But beneath its beauty lies a fascinating linguistic history. The Hawaiian language uses an alphabet with just 18 symbols, far fewer than English. So how many bits of information are needed to represent these 18 Hawaiian symbols? If you’re short on time, here’s a quick answer: 5 bits are needed to represent the 18 symbols in the Hawaiian alphabet. In this comprehensive guide, we’ll cover the history of the Hawaiian alphabet, explain how information theory allows us to calculate the minimum number of bits needed, walk through the step-by-step math, and discuss how ASCII encoding is used to store Hawaiian text digitally. We’ll also look at some examples of Hawaiian words encoded in binary. A Brief History of the Hawaiian Alphabet Origins of the Hawaiian Language The Hawaiian language, also known as ʻŌlelo Hawaiʻi, has its roots in the Polynesian language family. It is believed that Polynesians migrated to the Hawaiian Islands from other Pacific islands around 1,500 years ago. The language developed and evolved over time through the interactions of the early Hawaiian people with their surroundings and each other. For centuries, Hawaiian was primarily an oral language, passed down through generations through songs, chants, and stories. However, as contact with Western explorers and missionaries increased in the 18th and 19th centuries, there was a need to develop a written system for the Hawaiian language. Development of a Written System The development of a written system for the Hawaiian language is attributed to Christian missionaries who arrived in Hawaiʻi in the early 19th century. They sought to spread their religious teachings and found it necessary to create a written form of the language. The missionaries devised a system that represented the sounds of the Hawaiian language using the Latin alphabet. The Hawaiian alphabet, known as Ka Hōʻailona Pīʻāpā, originally consisted of 13 letters: A, E, I, O, U, H, K, L, M, N, P, W, and ʻokina (a glottal stop). These letters were chosen based on the sounds present in the Hawaiian language. Over time, additional letters were added to the alphabet to represent specific sounds, bringing the total number of symbols to 18. The Modern Hawaiian Alphabet In its current form, the Hawaiian alphabet consists of 18 symbols: A, E, H, I, K, L, M, N, O, P, U, W, ʻokina, KAHAKŌ, KAHAKŌ Ā, KAHAKŌ Ē, KAHAKŌ Ī, and KAHAKŌ Ō. The ʻokina is a glottal stop marker, while the KAHAKŌ are diacritical marks used to indicate long vowels. The Hawaiian alphabet is unique and distinct from the English alphabet. It requires a different set of rules for pronunciation and usage. Learning the Hawaiian alphabet can be a fascinating journey into the language and culture of the Hawaiian people. If you’re interested in learning more about the Hawaiian alphabet and the Hawaiian language, you can visit the ʻŌlelo Hawaiʻi Program website, which offers resources and courses for both beginners and advanced learners. Using Information Theory to Calculate the Minimum Bits Introducing Information Theory: Information theory is a branch of mathematics that deals with the quantification, storage, and communication of information. It provides us with a framework to analyze and optimize the efficiency of data representation. One of the key concepts in information theory is the notion of bits, which are the fundamental units of information. In simple terms, a bit can be thought of as a binary digit, representing either a 0 or a 1. Calculating the Minimum Number of Bits for a Symbol Set: When it comes to representing a set of symbols, such as the 18 symbols in the Hawaiian alphabet, information theory can help us determine the minimum number of bits needed to represent each symbol uniquely. The minimum number of bits required to represent a symbol set is determined by the formula: minimum bits = log2(number of symbols) Using this formula, we can calculate the minimum number of bits required to represent the 18 symbols in the Hawaiian alphabet: minimum bits = log2(18) ≈ 4.17 bits This means that, on average, each symbol in the Hawaiian alphabet can be represented using approximately 4.17 bits. However, since bits cannot be divided, we would need to round up to the nearest whole number. Therefore, in this case, we would need at least 5 bits to represent each symbol in the Hawaiian alphabet. It’s important to note that this calculation assumes that each symbol in the Hawaiian alphabet is equally likely to occur. If certain symbols occur more frequently than others, a more advanced analysis taking into account the probabilities of each symbol would be needed to determine the optimal representation. Step-by-Step Math for Calculating the Bits for Hawaiian Listing the 18 Hawaiian Symbols The Hawaiian alphabet, also known as the ʻōlelo Hawaiʻi, consists of 18 symbols, including the vowels a, e, i, o, and u, and the consonants h, k, l, m, n, p, w, and the glottal stop symbol known as the ʻokina. These symbols are unique to the Hawaiian language and are used to represent the distinct sounds of the language. To calculate the number of bits needed to represent these symbols, we will use a mathematical formula. Applying the Formula To calculate the number of bits needed to represent a given number of symbols, we use the formula: Number of bits = log2 (number of symbols) Using this formula, we can determine the number of bits needed to represent the 18 symbols in the Hawaiian alphabet. Converting to Binary After calculating the number of bits needed, we can convert the symbols in the Hawaiian alphabet to binary. Binary is a base-2 numeral system that uses only 0s and 1s to represent numbers. Each symbol in the Hawaiian alphabet can be assigned a unique binary representation. For example, let’s say we determine that we need 5 bits to represent the 18 symbols in the Hawaiian alphabet. We can assign each symbol a binary representation using 5 bits. The first symbol, “a,” can be represented as 00000, while the last symbol, the glottal stop symbol, can be represented as 10001. By converting the symbols to binary, we can represent the Hawaiian alphabet using a binary code, which can be useful for various applications, such as computer programming or data storage. For more information on the Hawaiian alphabet and its symbols, you can visit the ʻŌlelo Hawaiʻi website, where you can learn more about the Hawaiian language and its unique writing system. Encoding Hawaiian Text with ASCII The American Standard Code for Information Interchange (ASCII) is a widely used character encoding standard that represents characters as numerical values. It was developed in the 1960s and uses 7 bits to represent a total of 128 characters, including the English alphabet, numbers, punctuation marks, and control characters. ASCII Encoding Overview The ASCII encoding standard assigns a unique numerical value to each character. For example, the letter ‘A’ is represented by the decimal value 65, ‘B’ by 66, and so on. These values are then converted into binary form for storage and transmission purposes. By using 7 bits, ASCII can represent 128 different characters. Encoding Hawaiian Words in Binary The Hawaiian alphabet consists of 18 symbols, including vowels and consonants. To encode Hawaiian text using ASCII, we need to map each symbol to a unique numerical value within the range of 0-127. However, since ASCII only supports 128 characters, we need to make some adjustments. One approach is to use the remaining unused ASCII characters to represent the additional Hawaiian symbols. For example, we can assign the Hawaiian symbol ‘ā’ to the ASCII value 128, ‘ē’ to 129, and so on. This allows us to represent all 18 symbols of the Hawaiian alphabet using ASCII. Real World Examples Let’s take a look at some real-world examples of encoding Hawaiian text using ASCII. Suppose we want to encode the word ‘aloha’, which is a commonly used Hawaiian greeting. Using ASCII, we can represent each letter as its corresponding numerical value in binary form. ‘a’ is represented by 97 in decimal, which is 1100001 in binary. ‘l’ is represented by 108, which is 1101100 in binary. ‘o’ is represented by 111, which is 1101111 in binary. ‘h’ is represented by 104, which is 1101000 in binary. By concatenating these binary values together, we get the binary representation of the word ‘aloha’ in ASCII: 1100001 1101100 1101111 1101000. It’s important to note that while ASCII encoding can represent the Hawaiian alphabet, it may not reflect the unique linguistic characteristics and pronunciation nuances of the language. For a more accurate representation, specialized encoding standards like Unicode can be used. For more information on ASCII encoding, you can visit the ASCII Table website. In summary, representing the 18 symbols in the Hawaiian alphabet requires just 5 bits of information. While Hawaiian uses a small symbol set, information theory provides a systematic way to calculate the minimum number of bits for any symbol set. Understanding concepts like ASCII encoding also shows how Hawaiian text can be stored digitally. Mahalo for joining us on this journey through the math and linguistics behind Hawaiian’s efficient writing system!	{"url":"https://www.hawaiistar.com/the-hawaiian-alphabet-has-18-symbols-how-many-bits-are-needed-to-represent-just-these-symbols/","timestamp":"2024-11-12T04:31:08Z","content_type":"text/html","content_length":"165404","record_id":"<urn:uuid:1b763010-8baf-46e6-b7ab-4667c787800a>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00229.warc.gz"}
High Card Odds - High Card Probability The odds of flopping a High Hand with any starting hand is 63% or 2 in 3 Definition of a High Hand (High-Card) – A High Hand occurs when we don’t even make a pair, so the strength of our hand is determined by the highest card. Example – AK372 High Hands are the least-ranked hand in Hold’em where no pair, draw or any potential to make a hand exists. Odds of Making a High Hand on the Flop It should, hopefully, not come as a surprise that making a High Hand on the flop is actually the most likely thing that can happen (excluding Pocket Pairs). Let’s review some of the odds - Odds of flopping a high-card hand with any starting hand = 63% Odds of flopping a high-card hand with any unpaired starting hand = 66.9% Odds of flopping a high-card hand with a pocket pair = 0% Unpaired hands tend to flop high-card hands with similar frequencies but not identical. For example - Odds of flopping a high-card hand with AKo = 67.2% Odds of flopping a high-card hand with T9s = 65.4% AKo is slightly more likely to flop a high-card hand since it makes straights and flushes less frequently than T9s. Of course, high-card hands can vary in strength. Odds of flopping overcards with any unpaired starting hand = 7.64% Odds of flopping overcards with AKo = 67.2% Odds of flopping overcards with T9s = 16.2% Odds of flopping a gutshot with any unpaired starting hand = 10.3% Odds of flopping a gutshot with AKo = 11.3% Odds of flopping a gutshot with T9s = 16.6% Odds of flopping an OESD with any unpaired starting hand = 3.47% Odds of flopping an OESD with AKo = 0% Odds of flopping an OESD with T9s = 9.6% A flush draw is technically a type of high-card hand, although it clearly is a much stronger holding relative to flopping naked overcards.	{"url":"https://ua.888poker.com/how-to-play-poker/hands/high-card-poker-hand-odds/","timestamp":"2024-11-04T07:02:26Z","content_type":"text/html","content_length":"32435","record_id":"<urn:uuid:82d49b2c-6c2f-4fc0-a657-cc6f9fc5d9c6>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00789.warc.gz"}
5.7 Thinking About Error Course Outline • segmentGetting Started (Don't Skip This Part) • segmentStatistics and Data Science: A Modeling Approach • segmentPART I: EXPLORING VARIATION • segmentChapter 1 - Welcome to Statistics: A Modeling Approach • segmentChapter 2 - Understanding Data • segmentChapter 3 - Examining Distributions • segmentChapter 4 - Explaining Variation • segmentPART II: MODELING VARIATION • segmentChapter 5 - A Simple Model • segmentChapter 6 - Quantifying Error • segmentChapter 7 - Adding an Explanatory Variable to the Model • segmentChapter 8 - Digging Deeper into Group Models • segmentChapter 9 - Models with a Quantitative Explanatory Variable • segmentPART III: EVALUATING MODELS • segmentChapter 10 - The Logic of Inference • segmentChapter 11 - Model Comparison with F • segmentChapter 12 - Parameter Estimation and Confidence Intervals • segmentFinishing Up (Don't Skip This Part!) • segmentResources list High School / Advanced Statistics and Data Science I (ABC) 5.7 Thinking About Error We have developed the idea of the mean being the simplest (or empty) model of the distribution of a quantitative variable, represented in this word equation: DATA = MEAN + ERROR If this is true, then we can calculate error in our data set by just moving components of this equation around to get the formula: ERROR = DATA - MEAN Using this formula, if someone has a thumb length larger than the mean (e.g., 62 versus a mean of 60.1), then their error is a positive number (in this case, nearly +2). If they have a thumb length lower than the mean (e.g., 58) then we can calculate their error as a negative number (e.g. about -2). We generally call the error calculated this way as the residual. Now that you know how to generate predictions, we’ll refine our definition of the residual to be the difference between our model’s prediction and an actual observed score. The word residual should evoke the stuff that remains because the residual is the leftover variation from our data once we take out the model. To find these errors (or residuals) you can just subtract the mean from each data point. In R we could just run this code to get the residuals: Fingers$Thumb - Fingers$Predict If we run the code, R will calculate the 157 residuals, but it won’t save them unless we tell it to do so. Modify the code in the window below to save the residuals in a new variable in Fingers called Resid. (Note that the variable Predict already exists in the Fingers data frame). require(coursekata) Fingers$TinySet <- c(1,1,1,0,0,0,1,0,0,1, rep(0,147)) Fingers$TinySet[142] <- 1 Fingers <- arrange(arrange(Fingers, Height), desc(TinySet)) empty_model <- lm(Thumb ~ NULL, data = Fingers) Fingers <- Fingers %>% mutate( Predict = predict(empty_model), Resid = Thumb - Predict ) # modify this to save the residuals from the empty_model Fingers$Resid <- # this prints selected variables from Fingers select(Fingers, Thumb, Predict, Resid) Fingers$Resid <- Fingers$Thumb - Fingers$Predict ex() %>% check_object("Fingers") %>% check_column("Resid") %>% check_equal() CK Code: B1_Code_Thinking_01 Thumb Predict Resid 1 52 60.10366 -8.103662 2 56 60.10366 -4.103662 3 64 60.10366 3.896338 4 70 60.10366 9.896338 5 66 60.10366 5.896338 6 62 60.10366 1.896338 These residuals (or “leftovers”) are so important in modeling that there is an even easier way to get them in R. The function resid(), when given a model (e.g., empty_model) will return all the residuals from the predictions of the model. Modify the following code to save the residuals that we get using the resid() function as a variable in the Fingers data frame. Call the new variable EasyResid. require(coursekata) Fingers$TinySet <- c(1,1,1,0,0,0,1,0,0,1, rep(0,147)) Fingers$TinySet[142] <- 1 Fingers <- arrange(arrange(Fingers, Height), desc(TinySet)) empty_model <- lm(Thumb ~ NULL, data = Fingers) Fingers <- Fingers %>% mutate( Predict = predict(empty_model), Resid = Thumb - Predict ) # calculate the residuals from empty_model the easy way # and save them in the Fingers data frame Fingers$EasyResid <- # this prints select variables from Fingers head(select(Fingers, Thumb, Predict, Resid, EasyResid)) Fingers$EasyResid <- resid(empty_model) Fingers ex() %>% check_object ("Fingers") %>% check_column("EasyResid") %>% check_equal() CK Code: Code_Thinking_02 Thumb Predict Resid EasyResid 1 52 60.10366 -8.103662 -8.103662 2 56 60.10366 -4.103662 -4.103662 3 64 60.10366 3.896338 3.896338 4 70 60.10366 9.896338 9.896338 5 66 60.10366 5.896338 5.896338 6 62 60.10366 1.896338 1.896338 Notice that the values for Resid and EasyResid are the same for each row in the data set. We will generally use the resid() function from now on, just because it’s easier, but we want you to know what the resid() function is doing behind the scenes. Below we have plotted a few of the residuals from the Fingers data set on the Thumb by Height scatterplot. Visually, the residuals can be thought of as the vertical distance between the data (the students’ actual thumb lengths) and the model’s predicted thumb length (60.1). Note that sometimes the residuals are negative (extending below the empty model) and sometimes positive (above the empty model). Because the empty model is the mean, we know that these residuals are perfectly balanced across the full data set of 157 students. Distribution of Residuals Below we’ve plotted histograms of the three variables: Thumb, Predict, and Resid. The distributions of the data and the residuals have the same shape. But the numbers on the x-axis differ across the two distributions. The distribution of Thumb is centered at the mean (60.1), whereas the distribution of Resid is centered at 0. Data that are smaller than the mean (such as a thumb length of 50) have negative residuals (-10) but data that are larger than the mean (such as 70) have positive residuals (10). Let’s see what we would get if we summed all values for the variable Fingers$Resid. Try it in the code block below. require(coursekata) empty_model <- lm(Thumb ~ NULL, data = Fingers) Fingers <- Fingers %>% mutate( Predict = predict(empty_model), Resid = resid(empty_model) ) # assume Fingers data frame already has the variable Resid saved in it sum(Fingers$Resid) ex() %>% { check_output_expr(., "sum(Fingers$Resid)") } CK Code: Code_Thinking_03 R will sometimes give you outputs in scientific notation. The 1.47e-14 is equivalent to $1.47*10^{-14}$ which indicates that this is a number very close to zero (the -14 meaning that the decimal point is shifted to the left 14 places)! Whenever you see this scientific notation with a large negative exponent after the “e”, you can just read it as “zero,” or pretty close to zero. The residuals (or error) around the mean always sum to 0. The mean of the errors will also always be 0, because 0 divided by n equals 0. (R will not always report the sum as exactly 0 because of computer hardware limitations but it will be close enough to 0.)	{"url":"https://coursekata.org/preview/book/f84ca125-b1d7-4288-9263-7995615e6ead/lesson/8/6","timestamp":"2024-11-09T07:29:17Z","content_type":"text/html","content_length":"97585","record_id":"<urn:uuid:028ee0a7-5515-4926-b87f-58f8c2218e09>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00418.warc.gz"}
Suppose That Y Is A Solution To A First-order, D-dimensional, Nonautonomous ODE Dy/dt = F(t, Y). (So (2024) Answer 1 the first-order, (d+1)-dimensional, autonomous ODE solved by [tex]$w(t) = (t, y(t))$ is $\frac{dw}{dt} = F(w) = \left(1, f(w_1, w_2, ..., w_{d+1})\right)$.[/tex] To find a first-order, (d+1)-dimensional, autonomous ODE that is solved by [tex]$w(t) = (t, y(t))$[/tex], we can write down the components of [tex]$\frac{dw}{dt}$.[/tex] Since[tex]$w(t) = (t, y(t))$[/tex], we have $w = (w_1, w_2, ..., w_{d+1})$ where[tex]$w_1 = t$ and $w_2, w_3, ..., w_{d+1}$ are the components of $y$.[/tex] Now, let's consider the derivative of $w$ with respect to $t$: [tex]$\frac{dw}{dt} = \left(\frac{dw_1}{dt}, \frac{dw_2}{dt}, ..., \frac{dw_{d+1}}{dt}\right)$[/tex] We know that[tex]$\frac{dy}{dt} = f(t, y)$, so $\frac{dw_2}{dt} = f(t, y_1, y_2, ..., y_d)$ and similarly, $\frac{dw_3}{dt} = f(t, y_1, y_2, ..., y_d)$, and so on, up to $\frac{dw_{d+1}}{dt} = f(t, y_1, y_2, ..., y_d)$.[/tex] Also, we have [tex]$\frac{dw_1}{dt} = 1$, since $w_1 = t$ and $\frac{dt}{dt} = 1$[/tex]. Therefore, the components of [tex]$\frac{dw}{dt}$[/tex]are given by: [tex]$\frac{dw_1}{dt} = 1$,\\$\frac{dw_2}{dt} = f(t, y_1, y_2, ..., y_d)$,\\$\frac{dw_3}{dt} = f(t, y_1, y_2, ..., y_d)$,\\...$\frac{dw_{d+1}}{dt} = f(t, y_1, y_2, ..., y_d)$.\\[/tex] Hence, the function $F(w)$ that satisfies [tex]$\frac{dw}{dt} = F(w)$ is:$F(w) = \left(1, f(w_1, w_2, ..., w_{d+1})\right)$.[/tex] [tex]$w(t) = (t, y(t))$ is $\frac{dw}{dt} = F(w) = \left(1, f(w_1, w_2, ..., w_{d+1})\right)$.[/tex] Learn more about dimensional here :- Related Questions Calculate the cross product assuming that u×w=⟨5,−6,−1⟩ (4u+4w)×w= The cross product assuming that is (4u + 4w) × w = 4(u × w) + 0(4u + 4w) × w = 4(u × w) = 4⟨5, −6, −1⟩= ⟨20, −24, −4⟩ Given that u × w = ⟨5, −6, −1⟩ We are to find (4u + 4w) × w We know that(4u + 4w) × w = 4(u + w) × w ......(i)u × w = \|u\| \|w\| sin θwhere, \|u\| = magnitude of vector uw = angle between u and w As we can see, we are not given the magnitude of either u or w, and nor are we given the angle between them. Hence, we cannot calculate the vector product using the above formula. However, we can use the following identity which will give us a useful result: (u + v) × w = u × w + v × w So, we can write(4u + 4w) × w = (4u × w) + (4w × w) Expanding, we get(4u + 4w) × w = 4(u × w) + 0(4u + 4w) × w = 4(u × w) = 4⟨5, −6, −1⟩= ⟨20, −24, −4⟩ Thus, the detailed answer is (4u + 4w) × w = 4(u × w) + 0(4u + 4w) × w = 4(u × w) = 4⟨5, −6, −1⟩= ⟨20, −24, −4⟩ Learn more about cross product . Rick is betting the same way over and over at the roulette table: $15 on "Odds" which covers the eighteen odd numbers. Note that the payout for an 18-number bet is 1:1. He plans to bet this way 30 times in a row. Rick says as long as he hasn't lost a total of $25 or more by the end of it, he'll be happy. Prove mathematically which is more likely: Rick will lose $25 or more, or will lose less than 25$? To determine which outcome is more likely, we need to analyze the probabilities of Rick losing $25 or more and Rick losing less than $25. Rick's bet has a 1:1 payout, meaning he wins $15 for each successful bet and loses $15 for each unsuccessful bet. Let's consider the possible scenarios: 1. Rick loses all 30 bets: The probability of losing each individual bet is 18/38 since there are 18 odd numbers out of 38 total numbers on the roulette wheel. The probability of losing all 30 bets is (18/38)^30. 2. Rick wins at least one bet: The complement of losing all 30 bets is winning at least one bet. The probability of winning at least one bet can be calculated as 1 - P(losing all 30 bets). Now let's calculate these probabilities: Probability of losing all 30 bets: P(Losing $25 or more) = (18/38)^30 Probability of winning at least one bet: P(Losing less than $25) = 1 - P(Losing $25 or more) By comparing these probabilities, we can determine which outcome is more likely. Learn more about probabilities here: A dosage requires a patient to receive 66.8mg of medicine for every 8 kg of body weight for every 4 hours. How many grams of medication does a patient, who weights 48 kg, need in 12 hours? round to the hundreths place g A patient who weighs 48 kg needs 400.80 grams of medication in 12 hours. To calculate the amount of medication needed by a patient who weighs 48 kg in 12 hours, we need to determine the dosage based on the patient's weight and the frequency of administration. Dosage per 8 kg of body weight = 66.8 mg Dosage per 4 hours = 66.8 mg First, let's determine the number of 4-hour intervals in 12 hours: 12 hours / 4 hours = 3 intervals Now, we can calculate the total dosage required for the patient: Dosage per 8 kg of body weight = 66.8 mg Patient's weight = 48 kg Dosage for the patient's weight = (66.8 mg / 8 kg) * 48 kg = 534.4 mg To convert milligrams (mg) to grams (g), we divide by 1000: Dosage in grams = 534.4 mg / 1000 = 0.5344 g Since the patient requires this dosage for three 4-hour intervals in 12 hours, we multiply the dosage by 3: Total dosage in grams = 0.5344 g * 3 = 1.6032 g Rounding to the hundredths place, the patient needs 1.60 grams of medication in 12 hours. To know more about frequency, visit Describe the additive inverse of a vector, (v1, v2, v3, v4, v5), in the vector space. R5 The additive inverse of a vector (v1, v2, v3, v4, v5) in the vector space R5 is (-v1, -v2, -v3, -v4, -v5). In simpler terms, the additive inverse of a vector is a vector that when added to the original vector results in a zero vector. To find the additive inverse of a vector, we simply negate all of its components. The negation of a vector component is achieved by multiplying it by -1. Thus, the additive inverse of a vector (v1, v2, v3, v4, v5) is (-v1, -v2, -v3, -v4, -v5) because when we add these two vectors, we get the zero vector. This property of additive inverse is fundamental to vector addition. It ensures that every vector has an opposite that can be used to cancel it out. The concept of additive inverse is essential in linear algebra, as it helps to solve systems of equations and represents a crucial property of vector spaces. Know more about additive inverse of a vector here: Use the graph of F to find the given limit. When necessary, state that the limit does not exist. lim F(x) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. OA. lim F(x)= x-4 (Type an integer or a simplified fraction.) OB. The limit does not exist. The limit of the function in this problem is given as follows: [tex]\lim_{x \rightarrow 4} F(x) = 5[/tex] How to obtain the limit of the function? The graph of the function is given by the image presented at the end of the answer. The function approaches x = 4 both from left and from right at y = 5, hence the limit of the function is given as follows: [tex]\lim_{x \rightarrow 4} F(x) = 5[/tex] The limit would not exist if the lateral limits were different. More can be learned about limits at brainly.com/question/23935467 Which inequality is graphed on the coordinate plane? A linear graph of dotted line intersects X-axis at the unit (minus 0.5,0) and Y-axis at the unit (0,2), With the region on the left side of the line shaded in blue and the right side in white color The inequality graphed on the coordinate plane is: \[y > -2x + 2\] The inequality graphed on the coordinate plane can be represented by the equation [tex]$y > -2x + 2$[/tex]. The linear graph is represented by a dotted line that intersects the X-axis at (-0.5, 0) and the Y-axis at (0, 2). The dotted line signifies that points on the line are not included in the solution. The region to the left of the line, shaded in blue, represents the solution set where the inequality [tex]$y > -2x + 2$[/tex] is satisfied. Points within this shaded region have y-values greater than the corresponding values on the line. The region to the right of the line, represented in white, does not satisfy the inequality. For more questions on inequality graphed: Find the point (s) on the graph of y=x^2+x closest to the point (2,0). Explain your answer. Therefore, the point(s) on the graph of [tex]y = x^2 + x[/tex] closest to (2,0) are approximately (-1.118, 0.564), (-1.503, 0.718), and (1.287, 3.471). These points have the minimum distance from the point (2,0) on the graph of [tex]y = x^2 + x.[/tex] To find the point(s) on the graph of [tex]y = x^2 + x[/tex] closest to the point (2,0), we can use the distance formula. The distance between two points (x1, y1) and (x2, y2) is given by: d = √[tex]((x2 - x1)^2 + (y2 - y1)^2)[/tex] In this case, we want to minimize the distance between the point (2,0) and any point on the graph of [tex]y = x^2 + x[/tex]. Therefore, we can set up the following equation: d = √[tex]((x - 2)^2 + (x^2 + x - 0)^2)[/tex] To find the point(s) on the graph closest to (2,0), we need to find the value(s) of x that minimize the distance function d. We can do this by finding the critical points of the distance function. Taking the derivative of d with respect to x and setting it to zero: d' = 0 [tex](2(x - 2) + 2(x^2 + x - 0)(2x + 1)) / (\sqrt((x - 2)^2 + (x^2 + x - 0)^2)) = 0[/tex] Simplifying and solving for x: [tex]2(x - 2) + 2(x^2 + x)(2x + 1) = 0[/tex] Simplifying further, we get: [tex]2x^3 + 5x^2 - 4x - 4 = 0[/tex] Using numerical methods or factoring, we find that the solutions are approximately x ≈ -1.118, x ≈ -1.503, and x ≈ 1.287. To know more about graph, 1Q scores are normally distributed with a mean of 100 and a standard deviation of 15 . Use this information to answer the following question. What is the probability that a randomly selected person will have an 1Q score of at least 111 ? Make sure to type in your answer as a decimal rounded to 3 decimal places. For example, if you thought the answer was 0.54321 then you would type in 0.543. Question 20 1Q scores are normally distributed with a mean of 100 and a standard deviation of 15 . Use this information to answer the following question. What is the probability that a randomly selected person will have an 1Q score anywhere from 99 to 123? Make sure to type in your answer as a decimal rounded to 3 decimal:places. For example, if you thought the ariswer was 0.54321 then you would type in 0.543. The probability of a randomly selected person having an IQ score of 111 is 0.768, with a normal distribution and a z-score formula. A score greater than or equal to 111 is 0.7683, and between 99 and 123 is 0.924. 1. Probability of a randomly selected person having an IQ score of at least 111. We are given that the 1Q scores are normally distributed with a mean of 100 and a standard deviation of 15. This is an example of normal distribution where the random variable is normally distributed with a mean μ and a standard deviation σ.The z-score formula is used to find the probability of a particular score or less than or greater than a particular score. The formula is given byz = (x - μ) / σwhere, x is the value of the observation, μ is the mean and σ is the standard deviation.We need to find the probability that a randomly selected person will have an 1Q score of at least 111. Thus, we have to find the z-score of 111. Therefore,z = (x - μ) / σ= (111 - 100) / 15= 0.73333 To find the probability of a score greater than or equal to 111, we need to look up the probability corresponding to the z-score of 0.7333 in the standard normal distribution table.The probability of a z-score of 0.73 is 0.7683. Therefore, the probability of a randomly selected person having an IQ score of at least 111 is 0.768 (rounded to 3 decimal places). 2. Probability of a randomly selected person having an IQ score between 99 and 123. The z-scores for 99 and 123 are:z_1 = (99 - 100) / 15 = -0.06667z_2 = (123 - 100) / 15 = 1.5333Now, we need to find the probability between z_1 and z_2. Using the standard normal distribution table, we find that P(-0.067 < z < 1.533) = 0.9236 (rounded to 3 decimal places).Therefore, the probability of a randomly selected person having an IQ score between 99 and 123 is 0.924 (rounded to 3 decimal places). Probability of a randomly selected person having an 1Q score of at least 111 = 0.768 (rounded to 3 decimal places).Probability of a randomly selected person having an 1Q score anywhere from 99 to 123 = 0.924 (rounded to 3 decimal places). To know more about normal distribution Visit: Social Media Network (10 points) Consider an unweighted, undirected simple graph G(V,E) of a social media network. Each person in the network is represented by a node in V. Two people are connected by an edge in E if they are friends in the network. We would like to inspect what portion of people with mutual friends are themselves friends. The quantity is called the (global) clustering coefficient, and is of interest to people who are studying the structure of real-world networks. A graph with a high clustering coefficient may contain "tightly knit communities". The clustering coefficient C(G) of a simple graph G is defined as C(G)= number of wedges in G3× number of triangles in G, where the wedges and triangles are defined as follows: - A triangle is a triple (i,j,k) such that every pair of vertices in the triple are directly connected with an edge. Note that each triangle is only counted once in the formula not three times. - A triple of vertices (i,j,k) is called a wedge if it is a path of length 2 ; i.e., i,j,k∈V and (i,j),(j,k)∈E. (You can use the language that the center of (i,j,k) is j.) Note that a triangle is also a wedge. (b) Write an algorithm that takes the adjacency list of G as its input and computes the clustering coefficient C(G). You may assume that the adjacency list is given to you as a nested hash table. For full credit, the running time of your algorithm should be O(D2∣V∣), where D is the maximum degree maxv∈Vdeg(v). Notation: If you prefer, you may assume that the input graph is given to you as an adjacency list. You can use the notation G[v] to access the neighbors of v.) Reminder: You should submit pseudocode, a proof of correcntess, and a running time analysis (as in the instructions on page 1). The algorithm computes the clustering coefficient C(G) of a graph G by counting the number of triangles and wedges in G based on its adjacency list representation. It iterates over each vertex, calculates the number of wedges and triangles containing that vertex, and then computes the clustering coefficient as the ratio of triangles to wedges. The algorithm runs in O(D^2\|V\|) time, where D is the maximum degree of any vertex in G. Algorithm for computing the clustering coefficient C(G) from the adjacency list of a graph G: Step 1: Define a variable cc and set it to zero, which will hold the clustering coefficient value of G. Step 2: Iterate over every vertex in G using the adjacency list G[v] and call the set of neighbors of v N(v). Step 3: For each vertex v in G, the number of wedges containing v is computed by computing the number of pairs of neighbors of v that are themselves neighbors in G. The number of wedges containing v is precisely the number of pairs of neighbors of v that are also neighbors of each other. The number of such pairs is simply the number of edges between the vertices in N(v), which is the size of the set of edges (N(v) choose 2), which is simply N(v)(N(v) - 1) / 2. Step 4: For each vertex v in G, compute the number of triangles that include v by iterating over the neighbors u of v and counting the number of times that u and another neighbor w of v are themselves neighbors in G. This count is the number of wedges formed between u, v, and w that contain the center vertex v, and is precisely the number of triangles containing v. To count the triangles, we iterate over each vertex v in G, and for each neighbor u of v, we iterate over the neighbors w of v that have a larger ID than u. We then check whether (u, w) is an edge in G. If it is, we increment a counter for the number of triangles that contain v. Step 5: Compute the clustering coefficient of G as C(G) = cc / sum(N(v)(N(v) - 1) / 2) for all vertices v in G, where cc is the number of triangles in G and the denominator is the total number of wedges in G (which is the sum of N(v)(N(v) - 1) / 2 over all vertices v in G). Proof of correctness: The clustering coefficient of a graph G is defined as the ratio of the number of triangles in G to the number of wedges in G. A wedge is a path of length 2 that contains two neighbors of a vertex v, while a triangle is a cycle of length 3 that contains v and two of its neighbors. To compute the clustering coefficient of a vertex v, we first count the number of wedges containing v and then count the number of triangles that contain v. The ratio of these two quantities is precisely the clustering coefficient of v. To compute the clustering coefficient of G, we simply sum the clustering coefficients of all vertices in G and divide by the total number of vertices in G. The running time of the algorithm is O(D2\|V\|), where D is the maximum degree of any vertex in G, since we must iterate over each vertex v and its neighbors, which takes time proportional to N(v)2 = (deg(v))2, and the sum of deg(v)2 over all vertices v in G is at most D2\|V\|. To know more about clustering coefficient refer here: Consider The Function F(X)=4sin(3x+1). (A) Find F′(X). (B) Find F′′(X). Given the function f(x) = 4sin(3x + 1), the derivative A. f'(x) = 4cos(3x + 1) + 3 B. f"(x) = -12sin(3x + 1) What is the derivative of a function? The derivative of a function is the rate of change of a function. Given the function f(x) = 4sin(3x + 1), to find the derivatives of the function (A) Find F′(X). (B) Find F′′(X) we proceed as follows. (A) Find the derivative F′(X). Since f(x) = 4sin(3x + 1), Let u = 3x + 1 So, f(x) = 4sinu differentiating with respect to x, we have that f(x) = 4sinu df(x)/dx = d4sinu/du × du/dx = 4cosu × d(3x + 1)/dx = 4cosu + d3x/dx + d1/dx = 4cosu + 3 + 0 = 4cosu + 3 = 4cos(3x + 1) + 3 f'(x) = 4cos(3x + 1) + 3 (B) Find the derivative F′′(X) Since f'(x) = 4cos(3x + 1) + 3. Let u = 3x + 1 So, f'(x) = 4cosu + 3 Taking the derivative with respect to x, we have that df'(x)/dx = d(4cosu + 3)/dx = d4cosu/dx + d3/dx = d4cosu/du × du/dx + d3/dx = 4(-sinu) × d(3x + 1)/dx + 0 = -4sinu × (d3x/dx + d1/dx) = -4sinu × (3 + 0) = -4sinu × 3 = -12sinu = -12sin(3x + 1) So, f"(x) = -12sin(3x + 1) Learn more about derivative of a function here: you read about a study testing whether night shift workers sleep the recommended 8 hours per day. assuming that the population variance of sleep (per day) is unknown, what type of t test is appropriate for this study? The type of t test which is appropriate for this study is one-sample t-test. We are given that; The time of recommended sleep= 8hours In statistics, Standard deviation is a measure of the variation of a set of values. σ = standard deviation of population N = number of observation of population X = mean μ = population mean A one-sample t-test is a statistical hypothesis test used to determine whether an unknown population mean is different from a specific value. It examines whether the mean of a population is statistically different from a known or hypothesized value If the population variance of sleep (per day) is unknown, then a one-sample t-test is appropriate for this study Therefore, by variance answer will be one-sample t-test. Learn more about standard deviation and variance: The 4R functions are available for every probability distribution. The only thing that changes with each distribution are the prefixes. True FalseSaved For data that is best described with the binomial distribution, the 68-95-99.7 Rule describes how much of the data lies within 1, 2, and 3 standard deviations (respectively) of the mean. True False The 4R functions are specific to each probability distribution, and the 68-95-99.7 Rule is applicable only to data best described by a normal distribution The statement "The 4R functions are available for every probability distribution. The only thing that changes with each distribution are the prefixes" is false. The 4R functions, which are PDF (probability density function), CDF (cumulative distribution function), SF (survival function), and PPF (percent point function), are specific to each probability Although the functions share similar characteristics, their formulas and properties vary for each distribution. Therefore, the statement is incorrect and false. For data that is best described using the binomial distribution, the 68-95-99.7 Rule is not applicable. This rule is specific to a normal distribution and describes the percentage of data that falls within 1, 2, and 3 standard deviations from the mean. In a binomial distribution, the data is discrete and can only take on specific values, which makes the 68-95-99.7 Rule not applicable. For more questions on probability distribution Demonstrate that the unordered kernel estimator of p(x) that uses Aitchison and Aitken’s unordered kernel function is proper (i.e., it is non-negative and it sums to one over all x ∈ {0, 1,...,c − The kernel estimator of p(x) using Aitchison and Aitken's kernel function is a crucial component of kernel density estimation. KDE is a non-parametric method for estimating random variables' density. To be proper, the kernel function must be non-negative and sum to one over all x. The unordered kernel estimator of p(x) using Aitchison and Aitken's unordered kernel function is the weighted average of nearby observations.The kernel function is the function that determines the weights given to observations near the estimate of the target variable. It's a critical component of kernel density estimation. Consider a sample of size n from a population. For estimating the density of the population, kernel density estimation (KDE) is a non-parametric method. KDE is a non-parametric approach to density estimation that may be employed to estimate the density of random variables. KDE with an unordered kernel function, for example, Aitchison and Aitken's unordered kernel function, is proper if it is non-negative and sums to one over all x∈{0, 1,...,c−1}.The unordered kernel function for Aitchison and Aitken's kernel function is given by, f(x) = { 0, if \|x\| > 1; 1 - \|x\|, if \|x\| ≤ 1} The two conditions to demonstrate that the unordered kernel estimator of p(x) that uses Aitchison and Aitken’s unordered kernel function is proper are explained below:Non-negativeThe first step in showing that the kernel estimator is non-negative is to demonstrate that the kernel function is non-negative. This is true for the Aitchison and Aitken kernel, as demonstrated by the definition of the kernel function above.Furthermore, the unordered kernel estimator is the weighted average of kernel function values, which are all non-negative. As a result, the unordered kernel estimator is also non-negative.S um to one over all x ∈ {0, 1,...,c − 1} The second condition is that the unordered kernel estimator of p(x) sums to one over all x∈{0, 1,...,c−1}. Since the kernel estimator is the weighted average of kernel function values at all observations, this condition may be met by demonstrating that the weights sum to one over all x. Since the sum of weights at all observations equals one, this is also true for the unordered kernel Therefore, the unordered kernel estimator of p(x) that uses Aitchison and Aitken’s unordered kernel function is proper. To know more about kernel estimator Visit: (a) Let D₁ and D₂ be independent discrete random variables which each have the mar- ginal probability mass function 1/3, if x = 1, 1/3, if x = 2, f(x) = 1/3, if x = 3, 0. otherwise. Let Z be a discrete random variable given by Z = min(D₁, D₂). (i) Give the joint probability mass function foz in the form of a table and an explanation of your reasons. (ii) Find the distribution of Z. (iii) Give your reasons on whether D, and Z are independent. (iv) Find E(ZID = 2). (i) To find the joint probability mass function (PMF) of Z, we need to determine the probability of each possible outcome (z) of Z. The possible outcomes for Z are 1, 2, and 3. We can calculate the joint PMF by considering the probabilities of the minimum value of D₁ and D₂ being equal to each possible outcome. The joint PMF table for Z is as follows: \| z \| P(Z = z) \| \| 1 \| 1/3 \| \| 2 \| 1/3 \| \| 3 \| 1/3 \| The joint PMF indicates that the probability of Z being equal to any of the values 1, 2, or 3 is 1/3. (ii) To find the distribution of Z, we can list the possible values of Z along with their probabilities. The distribution of Z is as follows: \| z \| P(Z ≤ z) \| \| 1 \| 1/3 \| \| 2 \| 2/3 \| \| 3 \| 1 \| (iii) To determine whether D₁ and D₂ are independent, we need to compare the joint PMF of D₁ and D₂ with the product of their marginal PMFs. The marginal PMF of D₁ is the same as its given PMF: \| x \| P(D₁ = x) \| \| 1 \| 1/3 \| \| 2 \| 1/3 \| \| 3 \| 1/3 \| Similarly, the marginal PMF of D₂ is also the same as its given PMF: \| x \| P(D₂ = x) \| \| 1 \| 1/3 \| \| 2 \| 1/3 \| \| 3 \| 1/3 \| If D₁ and D₂ are independent, the joint PMF should be equal to the product of their marginal PMFs. However, in this case, the joint PMF of D₁ and D₂ does not match the product of their marginal PMFs. Therefore, D₁ and D₂ are not independent. (iv) To find E(Z\|D = 2), we need to calculate the expected value of Z given that D = 2. From the joint PMF of Z, we can see that when D = 2, Z can take on the values 1 and 2. The probabilities associated with these values are 1/3 and 2/3, respectively. The expected value E(Z\|D = 2) is calculated as: E(Z\|D = 2) = (1/3) * 1 + (2/3) * 2 = 5/3 = 1.67 Therefore, E(Z\|D = 2) is 1.67. Learn more about probability mass function here: For the following data set: 10,3,5,4 - Calculate the biased sample variance. - Calculate the biased sample standard deviation. - Calculate the unbiased sample variance. - Calculate the unbiased sample standard deviation. The answers for the given questions are as follows: Biased sample variance = 6.125 Biased sample standard deviation = 2.474 Unbiased sample variance = 7.333 Unbiased sample standard deviation = 2.708 The following are the solutions for the given questions:1) Biased sample variance: For the given data set, the formula for biased sample variance is given by: [tex]$\frac{(10-5.5)^{2} + (3-5.5)^{2} + (5-5.5)^{2} + (4-5.5)^{2}}{4}$=6.125[/tex] Therefore, the biased sample variance is 6.125. 2) Biased sample standard deviation: For the given data set, the formula for biased sample standard deviation is given by: [tex]$\sqrt{\frac{(10-5.5)^{2} + (3-5.5)^{2} + (5-5.5)^{2} + (4-5.5)^{2}}{4}}$=2.474[/tex] Therefore, the biased sample standard deviation is 2.474. 3) Unbiased sample variance: For the given data set, the formula for unbiased sample variance is given by: [tex]$\frac{(10-5.5)^{2} + (3-5.5)^{2} + (5-5.5)^{2} + (4-5.5)^{2}}{4-1}$=7.333[/tex] Therefore, the unbiased sample variance is 7.333. 4) Unbiased sample standard deviation: For the given data set, the formula for unbiased sample standard deviation is given by: [tex]$\sqrt{\frac{(10-5.5)^{2} + (3-5.5)^{2} + (5-5.5)^{2} + (4-5.5)^ Therefore, the unbiased sample standard deviation is 2.708. Thus, the answers for the given questions are as follows: Biased sample variance = 6.125 Biased sample standard deviation = 2.474 Unbiased sample variance = 7.333 Unbiased sample standard deviation = 2.708 To know more about variance, visit: Part 2: Use the trigonometric ratios 30° and 60° to calculate and label the remaining sides of A BDC. Show your work. (3 points) sin 30º = } cos 30º = sin 60º = cos 60º = 1 tan 30º = tan 60°= 3 Using the trigonometric ratios for angles 30° and 60°, get the remaining sides of triangle ABC:Sin 30°: The ratio of the hypotenuse's (AC) and opposite side's (BC) lengths is known as the sine of 30° sin = BC/AC Since the BC to AC ratio in a triangle with coordinates of 30-60-90 is 1:2, sin 30° = 1/2. cos 30°: The ratio of the neighbouring side's (AB) length to the hypotenuse's (AC) length is known as the cosine of 30°. 30° cos = AB/AC Cos 30° = 3/2 (because the ratio of AB to AC in a triangle with angles of 30-60-90 is 3:2) sin 60°: The ratio of the hypotenuse's (AC) and opposite side's (AB) lengths is known as the sine of 60°. 60° of sin = AB/AC thus sin 60° = 3/2, learn more about trigonometric here : i roll a die up to three times. each time i toll, you can either take the number showing as dollors, or roll again. what are your expected winnings The expected value of winnings is 4.17. We are given that; A dice is rolled 3times Probability refers to a possibility that deals with the occurrence of random events. The probability of all the events occurring need to be 1. The formula of probability is defined as the ratio of a number of favorable outcomes to the total number of outcomes. P(E) = Number of favorable outcomes / total number of outcomes If you roll a die up to three times and each time you roll, you can either take the number showing as dollars or roll again. The expected value of the game rolling twice is 4.25 and if we have three dice your optimal strategy will be to take the first roll if it is 5 or greater otherwise you continue and your expected payoff 4.17. Therefore, by probability the answer will be 4.17. Learn more about probability here; Angela took a general aptitude test and scored in the 95 th percentile for aptitude in accounting. (a) What percentage of the scores were at or below her score? × % (b) What percentage were above? x The given problem states that Angela took a general aptitude test and scored in the 95th percentile for aptitude in accounting. To find:(a) What percentage of the scores were at or below her score? × %(b) What percentage were above? x % (a) The percentage of the scores that were at or below her score is 95%.(b) The percentage of the scores that were above her score is 5%.Therefore, the main answer is as follows:(a) 95%(b) 5% Angela took a general aptitude test and scored in the 95th percentile for aptitude in accounting. (a) What percentage of the scores were at or below her score? × %(b) What percentage were above? x %The percentile score of Angela in accounting is 95, which means Angela is in the top 5% of the students who have taken the test.The percentile score determines the number of students who have scored below the candidate. For example, if a candidate is in the 90th percentile, it means that 90% of the students who have taken the test have scored below the candidate, and the candidate is in the top 10% of the students. Therefore, to find out what percentage of students have scored below the Angela, we can subtract 95 from 100. So, 100 – 95 = 5. Therefore, 5% of the students have scored below Angela. Hence, the answer to the first question is 95%.Similarly, to calculate what percentage of the students have scored above Angela, we need to take the value of the percentile score (i.e., 95) and subtract it from 100. So, 100 – 95 = 5. Therefore, 5% of the students have scored above Angela. Thus, Angela's percentile score in accounting is 95, which means that she has scored better than 95% of the students who have taken the test. Further, 5% of the students have scored better than her. To know more about accounting : A driver is monitoring his car's gasoline consumption for 3 weeks. If the car consumes 1(5)/(6) gallons the first week, 4(2)/(3) gallons the secono week, and 5(7)/(8) gallons the third week, what is the average weekly gasoline consumption? Write the solution as a mixed number or a fraction in lowest To find the average weekly gasoline consumption, we need to calculate the total gasoline consumption over the three weeks and then divide it by the number of weeks. The total gasoline consumption is given by the sum of the consumption for each week: 1(5)/(6) + 4(2)/(3) + 5(7)/(8) To add these fractions, we need to find a common denominator. The least common multiple of 6, 3, and 8 is 24. Converting the fractions to have a denominator of 24: 1(5)/(6) = 4/24 + 5/(6/6) = 4/24 + 20/24 = 24/24 = 1 4(2)/(3) = 32/24 + 16/24 = 48/24 = 2 5(7)/(8) = 35/24 Now, we can add the fractions: 1 + 2 + 35/24 = 3 + 35/24 = 83/24 The total gasoline consumption over the three weeks is 83/24 gallons. To find the average weekly gasoline consumption, we divide this total by the number of weeks, which is 3: (83/24) / 3 = 83/24 * 1/3 = 83/72 Therefore, the average weekly gasoline consumption is 83/72 gallons. To learn more about average:https://brainly.com/question/130657 small sample of computer operators shows monthly incomes of $1,950, $1,885, $1,965, $1,940, $1945, $1895, $1,890 and $1,925. The class width of the computer operators' sample with 5 classes is $16. © True © False The answer is: © True. To determine if the statement is true or false, we need to calculate the number of classes based on the sample data and class width. Given the sample incomes: $1,950, $1,885, $1,965, $1,940, $1,945, $1,895, $1,890, and $1,925. The range of the data is the difference between the maximum and minimum values: Range = $1,965 - $1,885 = $80. To determine the number of classes, we divide the range by the class width: Number of classes = Range / Class width = $80 / $16 = 5. Since the statement says the sample has 5 classes, and the calculation also shows that the number of classes is 5, the statement is true. Therefore, the answer is: © True. To know more about data visit Draw a logic circuit for (A+B) ′ (C+D)C ′ 5) Draw a logic circuit for BC ′ Using Boolean algebra, we can derive the following equations: B(C' + A) + AC = BC' + AB + ACD(BC')' = B + C'ABC = (B + C')'BC = (B + C)' The final logic circuit for BC' + AB + ACD (A+B)′(C+D)C′ can be simplified to (A'B' + C'D')C', BC' + AB + ACD can be expressed as B(C' + A) + AC(D + 1), which can be further simplified to B(C' + A) + AC. Using Boolean algebra, we can derive the following equations: B(C' + A) + AC = BC' + AB + ACD(BC')' = B + C'ABC = (B + C')'BC = (B + C)' The final logic circuit for BC' + AB + ACD To know more about algebra visit- You pump a total of 22.35 gallons. The cost per is gallon is $1.79. What is th total cost to fill up yur tank? The total cost to fill up your tank would be $39.97. To calculate the total cost, we multiply the number of gallons pumped by the cost per gallon. In this case, you pumped a total of 22.35 gallons, and the cost per gallon is $1.79. Therefore, the equation to determine the total cost is: Total cost = Number of gallons * Cost per gallon. Plugging in the values, we have: Total cost = 22.35 gallons * $1.79/gallon = $39.97. Thus, the total cost to fill up your tank would be $39.97. This calculation assumes that there are no additional fees or taxes involved in the transaction and that the cost per gallon remains constant throughout the filling process. To know more about cost refer here : The total cost to fill up your tank would be equal to $39.97. To Find the total cost, we have to multiply the number of gallons pumped by the cost per gallon. Since pumped a total of 22.35 gallons, and the cost per gallon is $1.79. Therefore, the equation to determine the total cost will be; Total cost = Number of gallons x Cost per gallon. Plugging in the values; Total cost = 22.35 gallons x $1.79/gallon = $39.97. Thus, the total cost to fill up your tank will be $39.97. To know more about cost refer here : Consider the compound interest equation B(t)=100(1. 1664)t. Assume that n=2, and rewrite B(t) in the form B(t)=P(1+rn)nt. What is the interest rate, r, written as a percentage? Enter your answer as a whole number, like this: 42 The interest rate is 16.02% (rounded to two decimal places). The compound interest formula is B(t) = P(1 + r/n)^(nt), where B(t) is the balance after t years, P is the principal (initial amount invested), r is the annual interest rate (as a decimal), n is the number of times compounded per year, and t is the time in years. Comparing this with the given formula B(t) = 100(1.1664)^t, we see that P = 100, n = 2, and nt = t. So we need to solve for r. We can start by rewriting the given formula as: B(t) = P(1 + r/n)^nt 100(1.1664)^t = 100(1 + r/2)^(2t) Dividing both sides by 100 and simplifying: (1.1664)^t = (1 + r/2)^(2t) 1.1664 = (1 + r/2)^2 Taking the square root of both sides: 1.0801 = 1 + r/2 Subtracting 1 from both sides and multiplying by 2: r = 0.1602 So the interest rate is 16.02% (rounded to two decimal places). Learn more about interest. rate from Let X be a random variable that follows a binomial distribution with n = 12, and probability of success p = 0.90. Determine: P(X≤10) 0.2301 0.659 0.1109 0.341 not enough information is given The probability P(X ≤ 10) for a binomial distribution with n = 12 and p = 0.90 is approximately 0.659. To find the probability P(X ≤ 10) for a binomial distribution with n = 12 and p = 0.90, we can use the cumulative distribution function (CDF) of the binomial distribution. The CDF calculates the probability of getting a value less than or equal to a given value. Using a binomial probability calculator or statistical software, we can input the values n = 12 and p = 0.90. The CDF will give us the probability of X being less than or equal to 10. Calculating P(X ≤ 10), we find that it is approximately 0.659. Therefore, the correct answer is 0.659, indicating that there is a 65.9% probability of observing 10 or fewer successes in 12 trials when the probability of success is 0.90. To know more about probability, visit: Convert the following octal numbers to their decimal equivalents A, 47 B, 75 C, 360 D, 545 The decimal equivalents of the given octal numbers are: A) 47 = 39 B) 75 = 61 C) 360 = 240 D) 545 = 357 To convert the given octal numbers to their decimal equivalents, we need to understand the positional value of each digit in the octal system. In octal, each digit's value is multiplied by powers of 8, starting from right to left. A) Octal number 47: 4 * 8^1 + 7 * 8^0 = 32 + 7 = 39 B) Octal number 75: 7 * 8^1 + 5 * 8^0 = 56 + 5 = 61 C) Octal number 360: 3 * 8^2 + 6 * 8^1 + 0 * 8^0 = 192 + 48 + 0 = 240 D) Octal number 545: 5 * 8^2 + 4 * 8^1 + 5 * 8^0 = 320 + 32 + 5 = 357 Visit here to learn more about decimal: ind an equation of the circle whose diameter has endpoints (-4,4) and (-6,-2). The equation of the circle is (x + 5)² + (y - 1)² = 40 , whose diameter has endpoints (-4,4) and (-6,-2). we use the formula: (x - a)² + (y - b)² = r² (a ,b) is the center of the circle r is the radius. To find the center, we use the midpoint formula: ( (x1 + x2)/2 , (y1 + y2)/2 )= (-4 + (-6))/2 , (4 + (-2))/2= (-5, 1) So, the center is (-5, 1).To find the radius, we use the distance formula: d = √ [(x2 - x1)² + (y2 - y1)²]= √[(-6 - (-4))² + (-2 - 4)²]= √[(-2)² + (-6)²]= √40= 2√10So, the radius is 2√10. Using the formula, (x - a)² + (y - b)² = r², the equation of the circle is:(x - (-5))² + (y - 1)² = (2√10)² Simplifying the equation, we get:(x + 5)² + (y - 1)² = 40. To know more about equation of the circle refer here: IQ scores are normally distributed with a mean of 95 and a standard deviation of 16 . Assume that many samples of size n are taken from a large population of people and the mean 1Q score is computed for each sample. a. If the sample size is n=64, find the mean and standard deviation of the distribution of sample means. The mean of the distribution of sample means is The standard deviation of the distribution of sample means is (Type an integer or decimal rounded to the nearest tenth as needed.) b. If the sample size is n=100, find the mean and standard deviation of the distribution of sample means. The mean of the distribution of sample means is When the sample size is 64, the mean of the distribution of sample means is 95 and the standard deviation of the distribution of sample means is 2. When the sample size is 100, the mean of the distribution of sample means is 95 and the standard deviation of the distribution of sample means is 1.6. Mean of the distribution of sample means = 95 Standard deviation of the distribution of sample means= 2 The formula for the mean and standard deviation of the sampling distribution of the mean is given as follows: μM=μσM=σn√where; μM is the mean of the sampling distribution of the meanμ is the population meanσ M is the standard deviation of the sampling distribution of the meanσ is the population standard deviation n is the sample size In this question, we are supposed to calculate the mean and standard deviation of the distribution of sample means when the sample size is 64. So the mean of the distribution of sample means is: μM=μ=95 The standard deviation of the distribution of sample means is: σM=σn√=16164√=2b. Mean of the distribution of sample means = 95 Standard deviation of the distribution of sample means= 1.6 In this question, we are supposed to calculate the mean and standard deviation of the distribution of sample means when the sample size is 100. So the mean of the distribution of sample means is:μM=μ =95The standard deviation of the distribution of sample means is: σM=σn√=16100√=1.6 From the given question, the IQ scores are normally distributed with a mean of 95 and a standard deviation of 16. When the sample size is 64, the mean of the distribution of sample means is 95 and the standard deviation of the distribution of sample means is 2. When the sample size is 100, the mean of the distribution of sample means is 95 and the standard deviation of the distribution of sample means is 1.6. The sampling distribution of the mean refers to the distribution of the mean of a large number of samples taken from a population. The mean and standard deviation of the sampling distribution of the mean are equal to the population mean and the population standard deviation divided by the square root of the sample size respectively. In this case, the mean and standard deviation of the distribution of sample means are calculated when the sample size is 64 and 100. The mean of the distribution of sample means is equal to the population mean while the standard deviation of the distribution of sample means decreases as the sample size increases. To know more about means visit: ind The Solution To Y′′+4y′+5y=0 With Y(0)=2 And Y′(0)=−1 We can start off by finding the characteristic equation of the given differential equation. We can do that by assuming a solution of the form y=e^{rt}. Substituting in the differential equation, we get r^2+4r+5=0. The roots of this quadratic are r=-2\pm i. Therefore, the general solution of the differential equation is y(t)=e^{-2t}(c_1\cos t+c_2\sin t), where c_1 and c_2 are constants to be determined from the initial conditions. We are given that y(0)=2 and y'(0)=-1. From the expression for y(t), we have y(0)=c_1=2. Differentiating the expression for y(t), we get y'(t)=-2e^{-2t}c_1\cos t+e^{-2t}(-c_1\sin t+c_2\cos t). Thus, y'(0)=-2c_1+c_2=-1. Substituting c_1=2, we get c_2=3. Therefore, the solution of the differential equation with the given initial conditions is y(t)=e^{-2t}(2\cos t+3\sin t). To know more about equation visit: (1/10÷1/2) × 3 + 1/5= F) 4/5 G) 4/15 H) 16/25 J) 3 2/5 K) None Step-by-step explanation: get the reciprocal inside the parenthesis 1/10 x 2/1= 5 x 3 + 1/5 apply MDAS, multiply 5 x 3= 15 + 1/5= get the lcd that will be 5 15/5+1/5=add the numerator 15+ 1= 16 copy the denominator that will be 16/5 convert to lowest terms that will be 3 1/5 so answer is NONE A random sample of 20 purchases showed the amounts in the table (in $ ). The mean is $48.34 and the standard deviation is $22.80. a) Construct a 99% confidence interval for the mean purchases of all customers, assuming that the assumptions and conditions for the confidence interval have been met. b) How large is the margin of error? c) How would the confidence interval change if you had assumed that the population standard deviation v known to be $23 ? a) What is the confidence interval? (Round to two decimal places as needed.) b) What is the margin of error? The margin of error is (Round to two decimal places as needed.) c) What is the confidence interval using the given population standard deviation? Select the correct choice below and fill in the answer boxes within your choice. (Round to two decimal places as needed.) A. The new confidence interval is wider than the interval from part a. B. The new confidence interval ) is narrower than the interval from part a. The confidence interval is (36.56,60.12). The margin of error is 11.78. a) Confidence interval - The formula for a confidence interval is given as; CI=\bar{X}\pm t_{\frac{\alpha}{2},n-1}\left(\frac{s}{\sqrt{n}}\right) Substitute the values into the formula; CI=48.34\pm t_{0.005,19}\left(\frac{22.8}{\sqrt{20}}\right) The t-value can be found using the t-table or calculator. Using the calculator, press STAT, then TESTS, then T Interval. Enter the required details to obtain the interval. b) Margin of error - The formula for the margin of error is given as; Substitute the values; Using the calculator, press STAT, then TESTS, then T Interval. Enter the required details to obtain the interval. c) Confidence interval using the population standard deviation The formula for a confidence interval is given as; CI=\bar{X}\pm z_{\frac{\alpha}{2}}\left(\frac{\sigma}{\sqrt{n}}\right) Substitute the values into the formula; CI=48.34\pm z_{0.005}\left(\frac{23}{\sqrt{20}}\right) The z-value can be found using the z-table or calculator. Using the calculator, press STAT, then TESTS, then Z Interval. Enter the required details to obtain the interval. Learn more about confidence interval visit: Other Questions Use the data in "htv" for this exercise. (i) Run a simple OLS regression of log( wage ) on educ. Without controlling for other factors, what is the 95% confidence interval for the return to another year of education? (ii) The variable ctuit, in thousands of dollars, is the change in college tuition facing students from age 17 to 18 . Show that educ and ctuit are essentially uncorrelated. What does this say about ctuit as a possible IV for educ in a simple regression analysis? (iii) Now, include in simple model of part (i) a quadratic expression of exper i.e. exper and exper 2, and a full set of regional dummy variables for current residence and residence at the age of 18 . Also include urban indicators for current and age 18 residence. What is the estimated return of an additional year of schooling? (iv) Again using ctuit as a potential IV for educ, estimate the reduced form for educ, which now includes all those variables that were included in part (ii). Show that ctuit is now statistically significant in the reduced form for educ. Comment on why this could have happened by using your intuition from OLS estimation of linear models. (v) Estimate the model from part (iii) by IV, using ctuit as an IV for educ. How does the confidence interval for the return to education compere with the OLS interval in part (i)? Hint: the confidence interval definition is the same, except we replace estimates and standard errors by the IV versions. What age group is more likely to vote?. (a) Let X be a binomial r.v. with n trials and success probability /n. Let Y be a Poisson r.v. with mean . Show, lim n[infinity] P(X=k)=P(Y=k) (The book goes through this if you get stuck, see (2.20).) (b) Suppose that the probability you receive an email in any particular minute is 0.01. Suppose further that if f[0,1], then the probability that you receive an email during a fraction f of a minute is 0.01f. Use part (a) to compute the probability that you receive 20 emails in a given day, the expected number of emails you receive in a day (exercise 2.39 above will be helpful for this), and the number of received emails in a day with the highest probability. You are studying the market for a particular new smartphone. You notice that several changes are happening in the market at the same time. The company that makes the smartphones has just opened a production plant in Malaysia where the cost of labor is lower than it is in the UK. The company has also switched to using a more cost-effective material for the outer shell of the phone. In addition, a new advertising campaign for the phone has gone viral, and now everybody wants this phone. What do you expect will happen to the equilibrium price and the equilibrium quantity? Explain your answer using a graph. what is the result of converting 2.3 miles into yards ? remenber that 1 mile=1760 yards. Sheffield Company has $145,000 of inventory at the beginning of the year and $131,000 at the end of the year. Sales revenue is $1,972,800, cost of goods sold is $1,145,400, and net income is $248,400 for the year. The inventory turnover ratio is: Multiple Choice 1.8 6.0. 14.3 8.3. ______ refers to the confidence an audience places in the truthfulness of what a speaker says. Use pumping Lemma to prove that the following languages are not regular L3={R,{0,1}+} . L4={1i0j1ki>j and i0} Which of the following is an example of subordinatelegislation?Select one:a. Billsb. Statuesc. Common lawd. Regulations a mixture of he , ar , and xe has a total pressure of 2.00 atm . the partial pressure of he is 0.450 atm , and the partial pressure of ar is 0.450 atm . what is the partial pressure of xe ? Partners T. Greer and R. Parks are provided salary allowances of $28,800 and $24.000, respectively. They divide the remainder of th partnership income in a ratio of 3:2. If partnership net income is $38,400, how much is allocated to Greer and Parks? what is it called when a seller-lender agrees to allow the conventional lender to become the senior lien holder? please write a program that rolls a 6 sided die 1000 times and keeps track of what it landed on. no graphics necessary, but it should output the number of times it landed on each number. the results should be random. Find the volume of the solid generated in the following situation.The region R bounded by the graph of y = 5 sin x and the x-axis on [0, ] is revolved about the line y = -2.The volume of the solid generated when R is revolved about the line y = -2 is cubic units.(Type an exact answer, using as needed.) Frost Company is evaluating the purchase of a rebult spot-welding machine to be used in the manufacture of a new product. The machinewill cost $174,000, has an estimated useful life of 7 years and a salvage value of zero, and will increase net annual cash flows by $33,421. Click tiere to view PV table. What is its approxlmate internal rate of return? (Round, answer to O decimal places, eg. 16% ) the metaphor in this lesson about counterfeit money and world religions teaches us that _____. Baed on thi excerpt, what quetion are the author trying to anwer ugar change the world Prove Proposition 4.6 That States: Given TriangleABC And TriangleA'B'C'. If Segment AB Is Congruent To Segment A'B' And Segment BC Is Congruent To Segment B'C', The Angle B Is Less Than Angle B' If And Only If Segment AC Is Less Than A'C'. (a) Find the unit vector along the line joining point (2,4,4) to point (3,2,2). (b) Let A=2a x +5a y 3a z ,B=3a x 4a y , and C=a x +a y+a zi. Determine A+2B. ii. Calculate A5C. iii. Find (AB)/(AB). (c) If A=2a x +a y 3a z ,B=a y a z , and C=3a x +5a y +7a z . i. A2B+C. ii. C4(A+B). Find critical point , linearize at each critical point , determine the type of each critical point and graph the phase diagram of the non linear system x=-y+xy y=3x+4xy consider the solution with initial condition (x(0),y(0)=(1,1) show this solution on the phase diagram and predict lim t-> +infinity (x(t),y(t) to the best of your knowledge	{"url":"https://nutbush.net/article/suppose-that-y-is-a-solution-to-a-first-order-d-dimensional-nonautonomous-ode-dy-dt-f-t-y-so","timestamp":"2024-11-03T15:44:15Z","content_type":"text/html","content_length":"130261","record_id":"<urn:uuid:bbfe0312-584e-4f1b-815d-4d574931527c>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00732.warc.gz"}
Mastering Absolute Value in JavaScript – A Step-by-Step Guide In the world of programming, absolute value is a mathematical concept that holds great significance. When it comes to mastering absolute value in JavaScript, it becomes even more crucial. In this blog post, we will delve deep into the understanding of absolute value and explore its application in JavaScript. By the end, you will have a solid grasp of how to effectively use absolute value in your JavaScript programs. Understanding Absolute Value Before we dive into using absolute value in JavaScript, let’s first establish a clear understanding of what absolute value is. In simple terms, the absolute value of a number is its distance from zero on a number line. It disregards the negative sign and returns the positive value of any given number. When dealing with absolute value, it’s important to be aware of its two key properties: Non-negativity Property The non-negativity property implies that the absolute value of any number is always non-negative. In other words, the result of the absolute value operation is never negative and can only be zero or a positive value. Symmetry Property The symmetry property states that the absolute value of a number remains the same regardless of whether the original number is positive or negative. In essence, if we take the absolute value of a negative number, it becomes positive, while the absolute value of a positive number remains unchanged. To better understand how absolute value works, let’s consider a few examples: Example 1: Math.abs(5) // Output: 5 In this example, we calculate the absolute value of 5, which is 5 itself since the number is already positive. Example 2: Math.abs(-8) // Output: 8 Here, we find the absolute value of -8. Since the original number is negative, the absolute value returns its positive counterpart, which is 8. Using Absolute Value in JavaScript Now that we have a solid understanding of absolute value, let’s explore how we can utilize it in JavaScript. Fortunately, JavaScript provides a built-in method called Math.abs() that allows us to easily calculate the absolute value of a given number. Syntax for Calculating Absolute Value in JavaScript The Math.abs() method takes a number as input and returns its absolute value. It can be used with both positive and negative numbers to obtain the positive value. Let’s take a look at a few examples that demonstrate the use of absolute value in JavaScript: Finding the Absolute Value of a Number To find the absolute value of a number in JavaScript, we can use the Math.abs() method. Consider the following example: let num = -12; let absValue = Math.abs(num); console.log('Absolute Value: ' + absValue); // Output: Absolute Value: 12 In this example, we assign the value of -12 to the variable ‘num’. We then use the Math.abs() method to calculate the absolute value of ‘num’ and store it in the ‘absValue’ variable. The output, when logged to the console, will be the absolute value of -12, which is 12. Determining the Absolute Difference between Two Numbers Another common use case for absolute value in JavaScript is finding the absolute difference between two numbers. We can achieve this by subtracting one number from another and then taking the absolute value of the result: let num1 = 25; let num2 = 10; let absDiff = Math.abs(num1 - num2); console.log('Absolute Difference: ' + absDiff); // Output: Absolute Difference: 15 In this example, we subtract ‘num2’ from ‘num1’ to find the difference. We then use the Math.abs() method to calculate the absolute value of the difference and store it in the ‘absDiff’ variable. The output will be the absolute difference between the two numbers, which is 15. Common Use Cases for Absolute Value in JavaScript Now that we have seen how to use absolute value in basic scenarios, let’s explore some common use cases where absolute value is incredibly handy in JavaScript. Determining the Distance Between Two Points One of the most relevant applications of absolute value in JavaScript is determining the distance between two points on a coordinate plane. By subtracting the x-coordinates and y-coordinates of the two points, we can calculate the absolute difference and obtain the distance. Consider the following example: function calculateDistance(x1, y1, x2, y2) { let xDiff = x2 - x1; let yDiff = y2 - y1; let distance = Math.sqrt(Math.pow(xDiff, 2) + Math.pow(yDiff, 2)); return distance; } let point1 = { x: 3, y: -2 }; let point2 = { x: -1, y: 4 }; let distance = calculateDistance(point1.x, point1.y, point2.x, point2.y); console.log('Distance: ' + distance); // Output: Distance: In this example, we define a function called ‘calculateDistance’ that takes in the x and y coordinates of two points. The function calculates the distance between the points using the Pythagorean theorem and the Math.sqrt() method to find the square root. The output will be the distance between the two points. Handling Negative Numbers and Absolute Value in Conditional Statements Conditional statements often require comparing values and making decisions based on certain conditions. Absolute value can come in handy when dealing with negative numbers in conditional statements. Let’s consider an example where we want to check if a given number is greater than 5: let num = -3; if (Math.abs(num) > 5) { console.log('Number is greater than 5.'); } else { console.log('Number is less than or equal to 5.'); } In this example, we use the Math.abs() method to obtain the absolute value of ‘num’. By comparing the absolute value with 5, we determine if the number is greater than or less than 5. Based on the condition, a corresponding message is logged to the console. Advanced Techniques for Working with Absolute Value in JavaScript Now that we have covered the basics of using absolute value in JavaScript, let’s explore some advanced techniques that can further enhance our usage. Chaining Methods with Absolute Value JavaScript allows method chaining, where multiple methods can be called on an object consecutively. We can take advantage of this feature to perform multiple operations on a number, including absolute value calculations. Consider the following example: let num = -7; let absSquared = Math.abs(num).toFixed(2); console.log('Absolute Value Squared: ' + absSquared); // Output: Absolute Value Squared: 49.00 In this example, we first calculate the absolute value of ‘num’ using the Math.abs() method. We then use the toFixed() method to round the absolute value to two decimal places and store it in the ‘absSquared’ variable. The output will be the absolute value of ‘num’ squared, with two decimal places. Creating Custom Functions for Absolute Value Calculations While JavaScript provides the Math.abs() method for absolute value calculations, you may sometimes need to implement custom logic or additional functionalities. In such cases, it can be beneficial to create a custom function dedicated to handling absolute value calculations. Here’s an example of a custom function for calculating absolute value: function absoluteValue(number) { if (number < 0) { return -number; } return number; } let num = -7; let absValue = absoluteValue(num); console.log('Absolute Value: ' + absValue); // Output: Absolute Value: 7 In this example, the 'absoluteValue' function checks if a number is less than 0. If it is, it returns the negated value (-number), effectively converting negative numbers to their positive counterparts. Otherwise, the function simply returns the number itself. The output will be the absolute value of 'num', which is 7. Tips and Best Practices When working with absolute value in JavaScript, it's important to keep a few tips and best practices in mind to avoid common mistakes and ensure efficient code execution: Avoiding Common Mistakes when Working with Absolute Value in JavaScript • Remember that the Math.abs() method only works on numbers, so ensure you provide a valid number argument. • Be careful not to confuse absolute value with rounding or truncating decimals. Absolute value deals solely with the positive value of a number. • Watch out for incorrect usage of parentheses when performing operations involving absolute value. Pay close attention to operator precedence. Considering Performance Implications when Using Absolute Value While JavaScript's Math.abs() method is efficient for absolute value calculations, excessive use of absolute value operations within performance-critical code can impact performance. Be mindful of your code's context and aim to minimize unnecessary calculations. In this blog post, we explored the ins and outs of absolute value in JavaScript. We started by understanding the concept and its key properties, then went on to discover various use cases for absolute value in JavaScript programming. We also explored advanced techniques such as method chaining and creating custom functions to handle absolute value calculations. By following the tips and best practices mentioned, you can ensure accurate and efficient usage of absolute value in your JavaScript programs. Now, armed with this knowledge, it's time to start mastering absolute value and unlocking its full potential in your coding endeavors! Take the next step in mastering absolute value by applying it to real-world scenarios and challenging yourself with more complex operations. With practice, you'll soon become a seasoned JavaScript developer with a deep understanding of absolute value and its countless applications.	{"url":"https://skillapp.co/blog/mastering-absolute-value-in-javascript-a-step-by-step-guide/","timestamp":"2024-11-03T06:18:17Z","content_type":"text/html","content_length":"115498","record_id":"<urn:uuid:2081c24a-7493-4a6f-b32d-fbabb07c40c7>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00738.warc.gz"}
Elementary and Intermediate Algebra Concepts and Applications, 8th edition Marvin L. Bittinger-Test Bank - Test Bank Goo Elementary and Intermediate Algebra Concepts and Applications, 8th edition Marvin L. Bittinger-Test Bank Format: Downloadable ZIP File Resource Type: Test bank Duration: Unlimited downloads Delivery: Instant Download Soltion Handbook For Elementary and Intermediate Algebra Concepts and Capabilities, 8th edition Marvin L. Bittinger ISBN-13: 9780137988419 Elementary & Intermediate Algebra: Concepts & Capabilities helps you understand and apply mathematical concepts, preparing you to transition from skills-based math to the additional concept-oriented math you will have for college-level packages. The authors’ technique encourages the critical-thinking experience needed to achieve this: the facility to function mathematically, discuss mathematically, and decide/resolve mathematical points. The 8th Edition choices rigorously reviewed and revised narrative, examples and exercises. Appreciable new choices embrace additional sources for consider and preparation, whereas sustaining the sturdy emphasis on important pondering and making connections between concepts. Desk of contents J. Merely-in-Time Evaluation J.1 Prime Factorizations J.2 Greatest Frequent Difficulty J.3 Least Frequent Numerous J.4 Equal Expressions and Fraction Notation J.5 Mixed Numerals J.6 Altering from Decimal Notation to Fraction Notation J.7 Together with and Subtracting Decimal Notation J.8 Multiplying and Dividing Decimal Notation J.9 Altering from Fraction Notation to Decimal Notation J.10 Rounding with Decimal Notation J.11 Altering between % Notation and Decimal Notation J.12 Altering between % Notation and Fraction Notation J.13 Order of Operations J J.14 Unit Conversion 1. Introduction to Algebraic Expressions 1.1 Introduction to Algebra 1.2 The Commutative, Associative, and Distributive Authorized pointers 1.3 Fraction Notation 1.4 Constructive and Harmful Precise Numbers Exploring the Concept Mid-Chapter Evaluation 1.5 Addition of Precise Numbers 1.6 Subtraction of Precise Numbers 1.7 Multiplication and Division of Precise Numbers Connecting the Concepts 1.8 Exponential Notation and Order of Operations Translating for Success Collaborative Actions Study Summary Evaluation Exercises Check out 2. Equations, Inequalities, and Downside Fixing 2.1 Fixing Equations 2.2 Using the Concepts Collectively 2.3 Formulation Mid-Chapter Evaluation 2.4 Capabilities with % 2.5 Downside Fixing 2.6 Fixing Inequalities Exploring the Concept Connecting the Concepts 2.7 Fixing Capabilities with Inequalities Translating for Success Collaborative Train Decision-Making Connection Study Summary Evaluation Exercises Check out Cumulative Evaluation: Chapters 1-2 3. Introduction to Graphing 3.1 Learning Graphs, Plotting Components, and Scaling Graphs 3.2 Graphing Linear Equations 3.3 Graphing and Intercepts 3.4 Fees 3.5 Slope Mid-Chapter Evaluation 3.6 Slope–Intercept Kind 3.7 Stage–Slope Kind and Equations of Strains Connecting the Concepts Visualizing for Success Decision-Making Connection Study Summary Evaluation Exercises Check out Cumulative Evaluation: Chapters 1-3 4. Polynomials 4.1 Exponents and Their Properties 4.2 Harmful Exponents and Scientific Notation Connecting the Concepts 4.3 Polynomials 4.4 Addition and Subtraction of Polynomials Mid-Chapter Evaluation 4.5 Multiplication of Polynomials 4.6 Specific Merchandise Exploring the Concept 4.7 Polynomials in Numerous Variables 4.8 Division of Polynomials Visualizing for Success Collaborative Train Decision-Making Connection Study Summary Evaluation Exercises Check out Cumulative Evaluation: Chapters 1-4 5. Polynomials and Factoring 5.1 Introduction to Factoring 5.2 Factoring Trinomials of the Sort x2 + bx + c 5.3 Factoring Trinomials of the Sort ax2 + bx + c 5.4 Factoring Good-Sq. Trinomials and Variations of Squares Mid-Chapter Evaluation 5.5 Factoring Sums or Variations of Cubes 5.6 Factoring: A Regular Approach 5.7 Fixing Quadratic Equations by Factoring Exploring the Concept Connecting the Concepts 5.8 Fixing Capabilities Translating for Success Collaborative Train Collaborative Train Decision-Making Connection Study Summary Evaluation Exercises Check out Cumulative Evaluation: Chapters 1-5 6. Rational Expressions and Equations 6.1 Rational Expressions 6.2 Multiplication and Division 6.3 Addition, Subtraction, and Least Frequent Denominators 6.4 Addition and Subtraction with In distinction to Denominators Mid-Chapter Evaluation 6.5 Superior Rational Expressions 6.6 Rational Equations Fixing Rational Equations Connecting the Concepts 6.7 Capabilities Using Rational Equation and Proportions Exploring the Concept Translating for Success Collaborative Train Decision-Making Connection Study Summary Evaluation Exercises Check out Cumulative Evaluation: Chapters 1-6 7. Capabilities and Graphs 7.1 Introduction to Capabilities 7.2 Space and Fluctuate Connecting the Concepts 7.3 Graphs of Capabilities Mid-Chapter Evaluation 7.4 The Algebra of Capabilities 7.5 Formulation, Capabilities, and Variation Visualizing for Success Collaborative Train Decision-Making Connection Study Summary Evaluation Exercises Check out Cumulative Evaluation: Chapters 1-7 8. Applications of Linear Equations and Downside Fixing 8.1 Applications of Equations in Two Variables Algebraic Graphical Connection 8.2 Fixing by Substitution or Elimination Connecting the Concepts 8.3 Fixing Capabilities: Applications of Two Equations Exploring the Concept 8.4 Applications of Equations in Three Variables Mid-Chapter Evaluation 8.5 Fixing Capabilities: Applications of Three Equations 8.6 Matrices 8.7 Elimination Using Matrices 8.8 Determinants and Cramer’s Rule 8.9 Enterprise and Economics Capabilities Visualizing For Success Collaborative Train Decision-Making Connection Study Summary Evaluation Exercises Check out Cumulative Evaluation: Chapters 1-8 9. Inequalities and Downside Fixing 9.1 Inequalities and Capabilities 9.2 Intersections, Unions, and Compound Inequalities 9.3 Absolute-Price Equations and Inequalities Exploring the Concept Mid-Chapter Evaluation 9.4 Inequalities in Two Variables Connecting the Concepts 9.5 Capabilities Using Linear Programming Visualizing For Success Collaborative Train Decision-Making Connection Study Summary Evaluation Exercises Check out Cumulative Evaluation: Chapters 1-9 10. Exponents and Radicals 10.1 Radical Expressions and Capabilities 10.2 Rational Numbers as Exponents 10.3 Multiplying Radical Expressions 10.4 Dividing Radical Expressions 10.5 Expressions Containing Numerous Radical Phrases Connecting the Concepts Mid-Chapter Evaluation 10.6 Fixing Radical Equations 10.7 The Distance System, the Midpoint System, and Completely different Capabilities 10.8 The Superior Numbers Visualizing For Success Collaborative Train Decision-Making Connection Study Summary Evaluation Exercises Check out Cumulative Evaluation: Chapters 1-10 11. Quadratic Capabilities and Equations 11.1 Quadratic Equations Algebraic Graphical Connection 11.2 The Quadratic System Connecting the Concepts 11.3 Discovering out Choices of Quadratic Equations 11.4 Capabilities Involving Quadratic Equations 11.5 Equations Reducible to Quadratic Mid-Chapter Evaluation 11.6 Quadratic Capabilities and Their Graphs Exploring the Concept 11.7 Additional About Graphing Quadratic Capabilities Algebraic Graphical Connection 1.8 Downside Fixing and Quadratic Capabilities 11.9 Polynomial Inequalities and Rational Inequalities Algebraic Graphical Connection Visualizing For Success Collaborative Train Decision-Making Connection Study Summary Evaluation Exercises Check out Cumulative Evaluation: Chapters 1-11 12. Exponential Capabilities and Logarithmic Capabilities 12.1 Composite Capabilities and Inverse Capabilities Exploring the Concept 12.2 Exponential Capabilities 12.3 Logarithmic Capabilities 12.4 Properties of Logarithmic Capabilities Mid-Chapter Evaluation 12.5 Frequent Logarithms and Pure Logarithms 12.6 Fixing Exponential Equations and Logarithmic Equations Connecting the Concepts 12.7 Capabilities of Exponential Capabilities and Logarithmic Capabilities Visualizing For Success Collaborative Train Decision-Making Connection Study Summary Evaluation Exercises Check out Cumulative Evaluation: Chapters 1-12 13. Conic Sections 13.1 Conic Sections: Parabolas and Circles 13.2 Conic Sections: Ellipses 13.3 Conic Sections: Hyperbolas Exploring the Concept Connecting the Concepts Mid-Chapter Evaluation 13.4 Nonlinear Applications of Equations Visualizing For Success Collaborative Train Decision-Making Connection Study Summary Evaluation Exercises Check out Cumulative Evaluation: Chapters 1-13< 14. Sequences, Sequence, and the Binomial Theorem 14.1 Sequences and Sequence 14.2 Arithmetic Sequences and Sequence 14.3 Geometric Sequences and Sequence Exploring the Concept Connecting the Concepts Mid-Chapter Evaluation 14.4 The Binomial Theorem Visualizing For Success Collaborative Train Decision-Making Connection Study Summary Evaluation Exercises Check out Cumulative Evaluation: Chapters 1-14 R. Elementary Algebra Evaluation R.1 Introduction to Algebraic Expressions R.2 Equations, Inequalities, and Downside Fixing R.3 Introduction to Graphing R.4 Polynomials R.5 Polynomials and Factoring R.6 Rational Expressions and Equations A. Indicate, Median, and Mode B. Models C. Synthetic Division and the The remaining Theorem Image Credit score Index of Capabilities User Reviews There are no reviews yet. Be the first to review “Elementary and Intermediate Algebra Concepts and Applications, 8th edition Marvin L. Bittinger-Test Bank” Elementary and Intermediate Algebra Concepts and Applications, 8th edition Marvin L. 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PHYS101: Introduction to Mechanics Often in science, we deal with measurements that are very large or very small. When writing these numbers or doing calculations with these physical quantities, you would have to write a large number of zeros either at the end of a large value or at the beginning of a very small value. Scientific notation allows us to write these large or small numbers without writing all the "placeholder" zeros. We write the non-zero part of the value as a decimal, followed by an exponent showing the order of magnitude, or number of zeros before or after the number. For example, consider the measurement: 125000 m. To write this measurement in scientific notation, we first take the non-zero part of the number, and write it as a decimal. The decimal part of the number above would become 1.25. Then, we need to show the order of magnitude of the number. We count the number of decimal places from where we placed the decimal to the end of the number. In this case, there are five places between the decimal we put in and the end of the number. We write this as an exponent: $10^{5}$. To put the entire scientific notation together, we write: $1.25 \times 10^ {5}\ \mathrm{m}$. We can also do an example where the measurement is very small. For example, consider the measurement: 0.0000085 s. Here, we again begin by making the non-zero part of the number into a decimal. We would write: 8.5. Next, we need to show the order of magnitude of the number. For a small number (less than one), we count the number of places from where we wrote the decimal back to the original decimal place. Then, we write our exponent as a negative number to show that the number is less than one. For this example, the exponent is: $10^{-6}$. To put the entire scientific notation together, we write: $8.5 \times 10^{-6}\ \mathrm{m}$. We can also convert values written in scientific notation to decimal notation. Consider the number: $5.0 \times 10^{3}\ \mathrm{m}$. We can write this as normal notation by adding the appropriate number of decimal places to the number, past the decimal written in scientific notation. Here, the order of magnitude (number of decimal places) is three, as we see from the exponent part of the number. Because the exponent is positive, we add the decimal places to the right of the number to make it a large number. The value in normal notation is: $5000\ \mathrm{m}$. We can also do this for small numbers written in scientific notation. Consider the example: $4.2 \times 10^{-4}\ \mathrm{m}$. We can write this as normal notation by adding the appropriate number of decimal places to the left of the number to make it a small number. Here, we need to have four decimal places to the left of the decimal in the scientific notation. The value in normal notation is: $0.00042\ \mathrm{m}$.	{"url":"https://learn.saylor.org/course/view.php?id=16&section=1","timestamp":"2024-11-06T17:37:39Z","content_type":"text/html","content_length":"497455","record_id":"<urn:uuid:1fae5dd0-fc08-4423-b3a4-2336c1fcb859>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00071.warc.gz"}
6.2 Properties of Power Series - Calculus Volume 2 \| OpenStax (2024) • 6.2.1Combine power series by addition or subtraction. • 6.2.2Create a new power series by multiplication by a power of the variable or a constant, or by substitution. • 6.2.3Multiply two power series together. • 6.2.4Differentiate and integrate power series term-by-term. In the preceding section on power series and functions we showed how to represent certain functions using power series. In this section we discuss how power series can be combined, differentiated, or integrated to create new power series. This capability is particularly useful for a couple of reasons. First, it allows us to find power series representations for certain elementary functions, by writing those functions in terms of functions with known power series. For example, given the power series representation for $f(x)=11−x,f(x)=11−x,$ we can find a power series representation for $f′ (x)=1(1−x)2.f′(x)=1(1−x)2.$ Second, being able to create power series allows us to define new functions that cannot be written in terms of elementary functions. This capability is particularly useful for solving differential equations for which there is no solution in terms of elementary functions. Combining Power Series If we have two power series with the same interval of convergence, we can add or subtract the two series to create a new power series, also with the same interval of convergence. Similarly, we can multiply a power series by a power of x or evaluate a power series at $xmxm$ for a positive integer m to create a new power series. Being able to do this allows us to find power series representations for certain functions by using power series representations of other functions. For example, since we know the power series representation for $f(x)=11−x,f(x)=11−x,$ we can find power series representations for related functions, such as In Combining Power Series we state results regarding addition or subtraction of power series, composition of a power series, and multiplication of a power series by a power of the variable. For simplicity, we state the theorem for power series centered at $x=0.x=0.$ Similar results hold for power series centered at $x=a.x=a.$ Combining Power Series Suppose that the two power series $∑n=0∞cnxn∑n=0∞cnxn$ and $∑n=0∞dnxn∑n=0∞dnxn$ converge to the functions f and g, respectively, on a common interval I. 1. The power series $∑n=0∞(cnxn±dnxn)∑n=0∞(cnxn±dnxn)$ converges to $f±gf±g$ on I. 2. For any integer $m≥0m≥0$ and any real number b, the power series $∑n=0∞bxmcnxn∑n=0∞bxmcnxn$ converges to $bxmf(x)bxmf(x)$ on I. 3. For any integer $m≥0m≥0$ and any real number b, the series $∑n=0∞cn(bxm)n∑n=0∞cn(bxm)n$ converges to $f(bxm)f(bxm)$ for all x such that $bxmbxm$ is in I. We prove i. in the case of the series $∑n=0∞(cnxn+dnxn).∑n=0∞(cnxn+dnxn).$ Suppose that $∑n=0∞cnxn∑n=0∞cnxn$ and $∑n=0∞dnxn∑n=0∞dnxn$ converge to the functions f and g, respectively, on the interval I. Let x be a point in I and let $SN(x)SN(x)$ and $TN(x)TN(x)$ denote the Nth partial sums of the series $∑n=0∞cnxn∑n=0∞cnxn$ and $∑n=0∞dnxn,∑n=0∞dnxn,$ respectively. Then the sequence ${SN(x)}{SN (x)}$ converges to $f(x)f(x)$ and the sequence ${TN(x)}{TN(x)}$ converges to $g(x).g(x).$ Furthermore, the Nth partial sum of $∑n=0∞(cnxn+dnxn)∑n=0∞(cnxn+dnxn)$ is we conclude that the series $∑n=0∞(cnxn+dnxn)∑n=0∞(cnxn+dnxn)$ converges to $f(x)+g(x).f(x)+g(x).$ We examine products of power series in a later theorem. First, we show several applications of Combining Power Series and how to find the interval of convergence of a power series given the interval of convergence of a related power series. Combining Power Series Suppose that $∑n=0∞anxn∑n=0∞anxn$ is a power series whose interval of convergence is $(−1,1),(−1,1),$ and suppose that $∑n=0∞bnxn∑n=0∞bnxn$ is a power series whose interval of convergence is $(−2,2). 1. Find the interval of convergence of the series $∑n=0∞(anxn+bnxn).∑n=0∞(anxn+bnxn).$ 2. Find the interval of convergence of the series $∑n=0∞an3nxn.∑n=0∞an3nxn.$ 1. Since the interval $(−1,1)(−1,1)$ is a common interval of convergence of the series $∑n=0∞anxn∑n=0∞anxn$ and $∑n=0∞bnxn,∑n=0∞bnxn,$ the interval of convergence of the series $∑n=0∞(anxn+bnxn)∑n= 0∞(anxn+bnxn)$ is $(−1,1).(−1,1).$ 2. Since $∑n=0∞anxn∑n=0∞anxn$ is a power series centered at zero with radius of convergence 1, it converges for all x in the interval $(−1,1).(−1,1).$ By Combining Power Series, the series converges if 3x is in the interval $(−1,1).(−1,1).$ Therefore, the series converges for all x in the interval $(−13,13).(−13,13).$ Suppose that $∑n=0∞anxn∑n=0∞anxn$ has an interval of convergence of $(−1,1).(−1,1).$ Find the interval of convergence of $∑n=0∞an(x2)n.∑n=0∞an(x2)n.$ In the next example, we show how to use Combining Power Series and the power series for a function f to construct power series for functions related to f. Specifically, we consider functions related to the function $f(x)=11−xf(x)=11−x$ and we use the fact that for $\|x\|<1.\|x\|<1.$ Constructing Power Series from Known Power Series Use the power series representation for $f(x)=11−xf(x)=11−x$ combined with Combining Power Series to construct a power series for each of the following functions. Find the interval of convergence of the power series. 1. $f(x)=3x1+x2f(x)=3x1+x2$ 2. $f(x)=1(x−1)(x−3)f(x)=1(x−1)(x−3)$ 1. First write $f(x)f(x)$ as Using the power series representation for $f(x)=11−xf(x)=11−x$ and parts ii. and iii. of Combining Power Series, we find that a power series representation for f is given by Since the interval of convergence of the series for $11−x11−x$ is $(−1,1),(−1,1),$ the interval of convergence for this new series is the set of real numbers x such that $\|x2\|<1.\|x2\|<1.$ Therefore, the interval of convergence is $(−1,1).(−1,1).$ 2. To find the power series representation, use partial fractions to write $f(x)=1(x−1)(x−3)f(x)=1(x−1)(x−3)$ as the sum of two fractions. We have First, using part ii. of Combining Power Series, we obtain Then, using parts ii. and iii. of Combining Power Series, we have Since we are combining these two power series, the interval of convergence of the difference must be the smaller of these two intervals. Using this fact and part i. of Combining Power Series, we where the interval of convergence is $(−1,1).(−1,1).$ Use the series for $f(x)=11−xf(x)=11−x$ on $\|x\|<1\|x\|<1$ to construct a series for $1(1−x)(x−2).1(1−x)(x−2).$ Determine the interval of convergence. In Example 6.5, we showed how to find power series for certain functions. In Example 6.6 we show how to do the opposite: given a power series, determine which function it represents. Finding the Function Represented by a Given Power Series Consider the power series $∑n=0∞2nxn.∑n=0∞2nxn.$ Find the function f represented by this series. Determine the interval of convergence of the series. Writing the given series as $∑ n = 0 ∞ 2 n x n = ∑ n = 0 ∞ ( 2 x ) n , ∑ n = 0 ∞ 2 n x n = ∑ n = 0 ∞ ( 2 x ) n ,$ we can recognize this series as the power series for $f ( x ) = 1 1 − 2 x . f ( x ) = 1 1 − 2 x .$ Since this is a geometric series, the series converges if and only if $\|2x\|<1.\|2x\|<1.$ Therefore, the interval of convergence is $(−12,12).(−12,12).$ Find the function represented by the power series $∑n=0∞13nxn.∑n=0∞13nxn.$ Determine its interval of convergence. Recall the questions posed in the chapter opener about which is the better way of receiving payouts from lottery winnings. We now revisit those questions and show how to use series to compare values of payments over time with a lump sum payment today. We will compute how much future payments are worth in terms of today’s dollars, assuming we have the ability to invest winnings and earn interest. The value of future payments in terms of today’s dollars is known as the present value of those payments. Chapter Opener: Present Value of Future Winnings Figure 6.4 (credit: modification of work by Robert Huffstutter, Flickr) Suppose you win the lottery and are given the following three options: (1) Receive 20 million dollars today; (2) receive 1.5 million dollars per year over the next 20 years; or (3) receive 1 million dollars per year indefinitely (being passed on to your heirs). Which is the best deal, assuming that the annual interest rate is 5%? We answer this by working through the following sequence of 1. How much is the 1.5 million dollars received annually over the course of 20 years worth in terms of today’s dollars, assuming an annual interest rate of 5%? 2. Use the answer to part a. to find a general formula for the present value of payments of C dollars received each year over the next n years, assuming an average annual interest rate r. 3. Find a formula for the present value if annual payments of C dollars continue indefinitely, assuming an average annual interest rate r. 4. Use the answer to part c. to determine the present value of 1 million dollars paid annually indefinitely. 5. Use your answers to parts a. and d. to determine which of the three options is best. 1. Consider the payment of 1.5 million dollars made at the end of the first year. If you were able to receive that payment today instead of one year from now, you could invest that money and earn 5% interest. Therefore, the present value of that money P[1] satisfies $P1(1+0.05)=1.5million dollars.P1(1+0.05)=1.5million dollars.$ We conclude that $P1=1.51.05=1.429million dollars.P1=1.51.05=1.429million dollars.$ Similarly, consider the payment of 1.5 million dollars made at the end of the second year. If you were able to receive that payment today, you could invest that money for two years, earning 5% interest, compounded annually. Therefore, the present value of that money P[2] satisfies $P2(1+0.05)2=1.5million dollars.P2(1+0.05)2=1.5million dollars.$ We conclude that $P2=1.5(1.05)2=1.361million dollars.P2=1.5(1.05)2=1.361million dollars.$ The value of the future payments today is the sum of the present values $P1,P2,…,P20P1,P2,…,P20$ of each of those annual payments. The present value P[k] satisfies $P=1.51.05+1.5(1.05)2+⋯+1.5(1.05)20=18.693million dollars.P=1.51.05+1.5(1.05)2+⋯+1.5(1.05)20=18.693million dollars.$ 2. Using the result from part a. we see that the present value P of C dollars paid annually over the course of n years, assuming an annual interest rate r, is given by 3. Using the result from part b. we see that the present value of an annuity that continues indefinitely is given by the infinite series We can view the present value as a power series in r, which converges as long as $\|11+r\|<1.\|11+r\|<1.$ Since $r>0,r>0,$ this series converges. Rewriting the series as we recognize this series as the power series for We conclude that the present value of this annuity is 4. From the result to part c. we conclude that the present value P of $C=1million dollarsC=1million dollars$ paid out every year indefinitely, assuming an annual interest rate $r=0.05,r=0.05,$ is given by $P=10.05=20million dollars.P=10.05=20million dollars.$ 5. From part a. we see that receiving $1.5 million dollars over the course of 20 years is worth $18.693 million dollars in today’s dollars. From part d. we see that receiving $1 million dollars per year indefinitely is worth $20 million dollars in today’s dollars. Therefore, either receiving a lump-sum payment of $20 million dollars today or receiving $1 million dollars indefinitely have the same present value. Multiplication of Power Series We can also create new power series by multiplying power series. Being able to multiply two power series provides another way of finding power series representations for functions. The way we multiply them is similar to how we multiply polynomials. For example, suppose we want to multiply It appears that the product should satisfy In Multiplying Power Series, we state the main result regarding multiplying power series, showing that if $∑n=0∞cnxn∑n=0∞cnxn$ and $∑n=0∞dnxn∑n=0∞dnxn$ converge on a common interval I, then we can multiply the series in this way, and the resulting series also converges on the interval I. Multiplying Power Series Suppose that the power series $∑n=0∞cnxn∑n=0∞cnxn$ and $∑n=0∞dnxn∑n=0∞dnxn$ converge to f and g, respectively, on a common interval I. Let $∑n=0∞enxnconverges tof(x)·g(x)onI.∑n=0∞enxnconverges tof(x)·g(x)onI.$ The series $∑n=0∞enxn∑n=0∞enxn$ is known as the Cauchy product of the series $∑n=0∞cnxn∑n=0∞cnxn$ and $∑n=0∞dnxn.∑n=0∞dnxn.$ We omit the proof of this theorem, as it is beyond the level of this text and is typically covered in a more advanced course. We now provide an example of this theorem by finding the power series representation for using the power series representations for Multiplying Power Series Multiply the power series representation for $\|x\|<1\|x\|<1$ with the power series representation for $\|x\|<1\|x\|<1$ to construct a power series for $f(x)=1(1−x)(1−x2)f(x)=1(1−x)(1−x2)$ on the interval $(−1,1).(−1,1).$ We need to multiply $( 1 + x + x 2 + x 3 + ⋯ ) ( 1 + x 2 + x 4 + x 6 + ⋯ ) . ( 1 + x + x 2 + x 3 + ⋯ ) ( 1 + x 2 + x 4 + x 6 + ⋯ ) .$ Writing out the first several terms, we see that the product is given by $( 1 + x 2 + x 4 + x 6 + ⋯ ) + ( x + x 3 + x 5 + x 7 + ⋯ ) + ( x 2 + x 4 + x 6 + x 8 + ⋯ ) + ( x 3 + x 5 + x 7 + x 9 + ⋯ ) = 1 + x + ( 1 + 1 ) x 2 + ( 1 + 1 ) x 3 + ( 1 + 1 + 1 ) x 4 + ( 1 + 1 + 1 ) x 5 + ⋯ = 1 + x + 2 x 2 + 2 x 3 + 3 x 4 + 3 x 5 + ⋯ . ( 1 + x 2 + x 4 + x 6 + ⋯ ) + ( x + x 3 + x 5 + x 7 + ⋯ ) + ( x 2 + x 4 + x 6 + x 8 + ⋯ ) + ( x 3 + x 5 + x 7 + x 9 + ⋯ ) = 1 + x + ( 1 + 1 ) x 2 + ( 1 + 1 ) x 3 + ( 1 + 1 + 1 ) x 4 + ( 1 + 1 + 1 ) x 5 + ⋯ = 1 + x + 2 x 2 + 2 x 3 + 3 x 4 + 3 x 5 + ⋯ .$ Since the series for $y=11−xy=11−x$ and $y=11−x2y=11−x2$ both converge on the interval $(−1,1),(−1,1),$ the series for the product also converges on the interval $(−1,1).(−1,1).$ Multiply the series $11−x=∑n=0∞xn11−x=∑n=0∞xn$ by itself to construct a series for $1(1−x)(1−x).1(1−x)(1−x).$ Differentiating and Integrating Power Series Consider a power series $∑n=0∞cnxn=c0+c1x+c2x2+⋯∑n=0∞cnxn=c0+c1x+c2x2+⋯$ that converges on some interval I, and let $ff$ be the function defined by this series. Here we address two questions about • Is $ff$ differentiable, and if so, how do we determine the derivative $f′?f′?$ • How do we evaluate the indefinite integral $∫f(x)dx?∫f(x)dx?$ We know that, for a polynomial with a finite number of terms, we can evaluate the derivative by differentiating each term separately. Similarly, we can evaluate the indefinite integral by integrating each term separately. Here we show that we can do the same thing for convergent power series. That is, if converges on some interval I, then converges on I. As noted below, behavior at the endpoints of the interval must be investigated individually. Evaluating the derivative and indefinite integral in this way is called term-by-term differentiation of a power series and term-by-term integration of a power series, respectively. The ability to differentiate and integrate power series term-by-term also allows us to use known power series representations to find power series representations for other functions. For example, given the power series for $f(x)=11−x,f(x)=11−x,$ we can differentiate term-by-term to find the power series for $f′(x)=1(1−x)2.f′(x)=1(1−x)2.$ Similarly, using the power series for $g(x)=11+x,g(x)=11+x,$ we can integrate term-by-term to find the power series for $G(x)=ln(1+x),G(x)=ln(1+x),$ an antiderivative of g. We show how to do this in Example 6.9 and Example 6.10. First, we state Term-by-Term Differentiation and Integration for Power Series, which provides the main result regarding differentiation and integration of power series. Term-by-Term Differentiation and Integration for Power Series Suppose that the power series $∑n=0∞cn(x−a)n∑n=0∞cn(x−a)n$ converges on the interval $(a−R,a+R)(a−R,a+R)$ for some $R>0.R>0.$ Let f be the function defined by the series for $\|x−a\|<R.\|x−a\|<R.$ Then f is differentiable on the interval $(a−R,a+R)(a−R,a+R)$ and we can find $f′f′$ by differentiating the series term-by-term: for $\|x−a\|<R.\|x−a\|<R.$ Also, to find $∫f(x)dx,∫f(x)dx,$ we can integrate the series term-by-term. The resulting series converges on $(a−R,a+R),(a−R,a+R),$ and we have for $\|x−a\|<R.\|x−a\|<R.$ The proof of this result is beyond the scope of the text and is omitted. Note that although Term-by-Term Differentiation and Integration for Power Series guarantees the same radius of convergence when a power series is differentiated or integrated term-by-term, it says nothing about what happens at the endpoints. It is possible that the differentiated and integrated power series have different behavior at the endpoints than does the original series. We see this behavior in the next examples. Differentiating Power Series 1. Use the power series representation for $\|x\|<1\|x\|<1$ to find a power series representation for on the interval $(−1,1).(−1,1).$ Determine whether the resulting series converges at the endpoints. 2. Use the result of part a. to evaluate the sum of the series $∑n=0∞n+14n.∑n=0∞n+14n.$ 1. Since $g(x)=1(1−x)2g(x)=1(1−x)2$ is the derivative of $f(x)=11−x,f(x)=11−x,$ we can find a power series representation for g by differentiating the power series for f term-by-term. The result is for $\|x\|<1.\|x\|<1.$Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the behavior of this series at the endpoints. Testing the endpoints by using the divergence test, we find that the series diverges at both endpoints $x=±1.x=±1.$ Note that this is the same result found in Example 6.8. 2. From part a. we know that Differentiate the series $1(1−x)2=∑n=0∞(n+1)xn1(1−x)2=∑n=0∞(n+1)xn$ term-by-term to find a power series representation for $2(1−x)32(1−x)3$ on the interval $(−1,1).(−1,1).$ Integrating Power Series For each of the following functions f, find a power series representation for f by integrating the power series for $f′f′$ and find its interval of convergence. 1. $f(x)=ln(1+x)f(x)=ln(1+x)$ 2. $f(x)=tan−1xf(x)=tan−1x$ 1. For $f(x)=ln(1+x),f(x)=ln(1+x),$ the derivative is $f′(x)=11+x.f′(x)=11+x.$ We know that for $\|x\|<1.\|x\|<1.$ To find a power series for $f(x)=ln(1+x),f(x)=ln(1+x),$ we integrate the series term-by-term. Since $f(x)=ln(1+x)f(x)=ln(1+x)$ is an antiderivative of $11+x,11+x,$ it remains to solve for the constant C. Since $ln(1+0)=0,ln(1+0)=0,$ we have $C=0.C=0.$ Therefore, a power series representation for $f(x)=ln(1+x)f(x)=ln(1+x)$ is for $\|x\|<1.\|x\|<1.$Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the behavior of this power series at the endpoints. However, checking the endpoints, we find that at $x=1x=1$ the series is the alternating harmonic series, which converges. Also, at $x=−1,x=−1,$ the series is the harmonic series, which diverges. It is important to note that, even though this series converges at $x=1,x=1,$Term-by-Term Differentiation and Integration for Power Series does not guarantee that the series actually converges to $ln(2).ln(2).$ In fact, the series does converge to $ln(2),ln(2),$ but showing this fact requires more advanced techniques. (Abel’s theorem, covered in more advanced texts, deals with this more technical point.) The interval of convergence is $(−1,1].(−1,1].$ 2. The derivative of $f(x)=tan−1xf(x)=tan−1x$ is $f′(x)=11+x2.f′(x)=11+x2.$ We know that for $\|x\|<1.\|x\|<1.$ To find a power series for $f(x)=tan−1x,f(x)=tan−1x,$ we integrate this series term-by-term. Since $tan−1(0)=0,tan−1(0)=0,$ we have $C=0.C=0.$ Therefore, a power series representation for $f(x)=tan−1xf(x)=tan−1x$ is for $\|x\|<1.\|x\|<1.$ Again, Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the convergence of this series at the endpoints. However, checking the endpoints and using the alternating series test, we find that the series converges at $x=1x=1$ and $x=−1.x=−1.$ As discussed in part a., using Abel’s theorem, it can be shown that the series actually converges to $tan−1(1)tan−1(1)$ and $tan−1(−1)tan−1(−1)$ at $x=1x=1$ and $x=−1,x=−1,$ respectively. Thus, the interval of convergence is $[−1,1].[−1,1].$ Integrate the power series $ln(1+x)=∑n=1∞(−1)n+1xnnln(1+x)=∑n=1∞(−1)n+1xnn$ term-by-term to evaluate $∫ln(1+x)dx.∫ln(1+x)dx.$ Up to this point, we have shown several techniques for finding power series representations for functions. However, how do we know that these power series are unique? That is, given a function f and a power series for f at a, is it possible that there is a different power series for f at a that we could have found if we had used a different technique? The answer to this question is no. This fact should not seem surprising if we think of power series as polynomials with an infinite number of terms. Intuitively, if for all values x in some open interval I about zero, then the coefficients c[n] should equal d[n] for $n≥0.n≥0.$ We now state this result formally in Uniqueness of Power Series. Uniqueness of Power Series Let $∑n=0∞cn(x−a)n∑n=0∞cn(x−a)n$ and $∑n=0∞dn(x−a)n∑n=0∞dn(x−a)n$ be two convergent power series such that for all x in an open interval containing a. Then $cn=dncn=dn$ for all $n≥0.n≥0.$ Then $f(a)=c0=d0.f(a)=c0=d0.$ By Term-by-Term Differentiation and Integration for Power Series, we can differentiate both series term-by-term. Therefore, and thus, $f′(a)=c1=d1.f′(a)=c1=d1.$ Similarly, implies that $f″(a)=2c2=2d2,f″(a)=2c2=2d2,$ and therefore, $c2=d2.c2=d2.$ More generally, for any integer $n≥0,f(n)(a)=n!cn=n!dn,n≥0,f(n)(a)=n!cn=n!dn,$ and consequently, $cn=dncn=dn$ for all In this section we have shown how to find power series representations for certain functions using various algebraic operations, differentiation, or integration. At this point, however, we are still limited as to the functions for which we can find power series representations. Next, we show how to find power series representations for many more functions by introducing Taylor series. Section 6.2 Exercises If $f(x)=∑n=0∞xnn!f(x)=∑n=0∞xnn!$ and $g(x)=∑n=0∞(−1)nxnn!,g(x)=∑n=0∞(−1)nxnn!,$ find the power series of $12(f(x)+g(x))12(f(x)+g(x))$ and of $12(f(x)−g(x)).12(f(x)−g(x)).$ If $C(x)=∑n=0∞x2n(2n)!C(x)=∑n=0∞x2n(2n)!$ and $S(x)=∑n=0∞x2n+1(2n+1)!,S(x)=∑n=0∞x2n+1(2n+1)!,$ find the power series of $C(x)+S(x)C(x)+S(x)$ and of $C(x)−S(x).C(x)−S(x).$ In the following exercises, use partial fractions to find the power series of each function. $4 ( x − 3 ) ( x + 1 ) 4 ( x − 3 ) ( x + 1 )$ $3 ( x + 2 ) ( x − 1 ) 3 ( x + 2 ) ( x − 1 )$ $5 ( x 2 + 4 ) ( x 2 − 1 ) 5 ( x 2 + 4 ) ( x 2 − 1 )$ $30 ( x 2 + 1 ) ( x 2 − 9 ) 30 ( x 2 + 1 ) ( x 2 − 9 )$ In the following exercises, express each series as a rational function. $∑ n = 1 ∞ 1 x n ∑ n = 1 ∞ 1 x n$ $∑ n = 1 ∞ 1 x 2 n ∑ n = 1 ∞ 1 x 2 n$ $∑ n = 1 ∞ 1 ( x − 3 ) 2 n − 1 ∑ n = 1 ∞ 1 ( x − 3 ) 2 n − 1$ $∑ n = 1 ∞ ( 1 ( x − 3 ) 2 n − 1 − 1 ( x − 2 ) 2 n − 1 ) ∑ n = 1 ∞ ( 1 ( x − 3 ) 2 n − 1 − 1 ( x − 2 ) 2 n − 1 )$ The following exercises explore applications of annuities. Calculate the present values P of an annuity in which $10,000 is to be paid out annually for a period of 20 years, assuming interest rates of $r=0.03,r=0.05,r=0.03,r=0.05,$ and $r=0.07.r=0.07.$ Calculate the present values P of annuities in which $9,000 is to be paid out annually perpetually, assuming interest rates of $r=0.03,r=0.05r=0.03,r=0.05$ and $r=0.07.r=0.07.$ Calculate the annual payouts C to be given for 20 years on annuities having present value $100,000 assuming respective interest rates of $r=0.03,r=0.05,r=0.03,r=0.05,$ and $r=0.07.r=0.07.$ Calculate the annual payouts C to be given perpetually on annuities having present value $100,000 assuming respective interest rates of $r=0.03,r=0.05,r=0.03,r=0.05,$ and $r=0.07.r=0.07.$ Suppose that an annuity has a present value $P=1million dollars.P=1million dollars.$ What interest rate r would allow for perpetual annual payouts of $50,000? Suppose that an annuity has a present value $P=10million dollars.P=10million dollars.$ What interest rate r would allow for perpetual annual payouts of $100,000? In the following exercises, express the sum of each power series in terms of geometric series, and then express the sum as a rational function. $x+x2−x3+x4+x5−x6+⋯x+x2−x3+x4+x5−x6+⋯$ (Hint: Group powers x^3k, $x3k−1,x3k−1,$ and $x3k−2.)x3k−2.)$ $x+x2−x3−x4+x5+x6−x7−x8+⋯x+x2−x3−x4+x5+x6−x7−x8+⋯$ (Hint: Group powers x^4k, $x4k−1,x4k−1,$ etc.) $x−x2−x3+x4−x5−x6+x7−⋯x−x2−x3+x4−x5−x6+x7−⋯$ (Hint: Group powers x^3k, $x3k−1,x3k−1,$ and $x3k−2.)x3k−2.)$ $x2+x24−x38+x416+x532−x664+⋯x2+x24−x38+x416+x532−x664+⋯$ (Hint: Group powers $(x2)3k,(x2)3k−1,(x2)3k,(x2)3k−1,$ and $(x2)3k−2.)(x2)3k−2.)$ In the following exercises, find the power series of $f(x)g(x)f(x)g(x)$ given f and g as defined. $f ( x ) = 2 ∑ n = 0 ∞ x n , g ( x ) = ∑ n = 0 ∞ n x n f ( x ) = 2 ∑ n = 0 ∞ x n , g ( x ) = ∑ n = 0 ∞ n x n$ $f(x)=∑n=1∞xn,g(x)=∑n=1∞1nxn.f(x)=∑n=1∞xn,g(x)=∑n=1∞1nxn.$ Express the coefficients of $f(x)g(x)f(x)g(x)$ in terms of $Hn=∑k=1n1k.Hn=∑k=1n1k.$ $f ( x ) = g ( x ) = ∑ n = 1 ∞ ( x 2 ) n f ( x ) = g ( x ) = ∑ n = 1 ∞ ( x 2 ) n$ $f ( x ) = g ( x ) = ∑ n = 1 ∞ n x n f ( x ) = g ( x ) = ∑ n = 1 ∞ n x n$ In the following exercises, differentiate the given series expansion of f term-by-term to obtain the corresponding series expansion for the derivative of f. $f ( x ) = 1 1 + x = ∑ n = 0 ∞ ( −1 ) n x n f ( x ) = 1 1 + x = ∑ n = 0 ∞ ( −1 ) n x n$ $f ( x ) = 1 1 − x 2 = ∑ n = 0 ∞ x 2 n f ( x ) = 1 1 − x 2 = ∑ n = 0 ∞ x 2 n$ In the following exercises, integrate the given series expansion of $ff$ term-by-term from zero to x to obtain the corresponding series expansion for the indefinite integral of $f.f.$ $f ( x ) = 2 x ( 1 + x 2 ) 2 = ∑ n = 1 ∞ ( −1 ) n ( 2 n ) x 2 n − 1 f ( x ) = 2 x ( 1 + x 2 ) 2 = ∑ n = 1 ∞ ( −1 ) n ( 2 n ) x 2 n − 1$ $f ( x ) = 2 x 1 + x 2 = 2 ∑ n = 0 ∞ ( −1 ) n x 2 n + 1 f ( x ) = 2 x 1 + x 2 = 2 ∑ n = 0 ∞ ( −1 ) n x 2 n + 1$ In the following exercises, evaluate each infinite series by identifying it as the value of a derivative or integral of geometric series. Evaluate $∑n=1∞n2n∑n=1∞n2n$ as $f′(12)f′(12)$ where $f(x)=∑n=0∞xn.f(x)=∑n=0∞xn.$ Evaluate $∑n=1∞n3n∑n=1∞n3n$ as $f′(13)f′(13)$ where $f(x)=∑n=0∞xn.f(x)=∑n=0∞xn.$ Evaluate $∑n=2∞n(n−1)2n∑n=2∞n(n−1)2n$ as $f″(12)f″(12)$ where $f(x)=∑n=0∞xn.f(x)=∑n=0∞xn.$ Evaluate $∑n=0∞(−1)n2n+1∑n=0∞(−1)n2n+1$ as $∫01f(t)dt∫01f(t)dt$ where $f(x)=∑n=0∞(−1)nx2n=11+x2.f(x)=∑n=0∞(−1)nx2n=11+x2.$ In the following exercises, given that $11−x=∑n=0∞xn,11−x=∑n=0∞xn,$ use term-by-term differentiation or integration to find power series for each function centered at the given point. $f(x)=lnxf(x)=lnx$ centered at $x=1x=1$ (Hint: $x=1−(1−x))x=1−(1−x))$ $ln(1−x)ln(1−x)$ at $x=0x=0$ $ln(1−x2)ln(1−x2)$ at $x=0x=0$ $f(x)=2x(1−x2)2f(x)=2x(1−x2)2$ at $x=0x=0$ $f(x)=tan−1(x2)f(x)=tan−1(x2)$ at $x=0x=0$ $f(x)=ln(1+x2)f(x)=ln(1+x2)$ at $x=0x=0$ $f(x)=∫0xlntdtf(x)=∫0xlntdt$ where $ln(x)=∑n=1∞(−1)n−1(x−1)nnln(x)=∑n=1∞(−1)n−1(x−1)nn$ [T] Evaluate the power series expansion $ln(1+x)=∑n=1∞(−1)n−1xnnln(1+x)=∑n=1∞(−1)n−1xnn$ at $x=1x=1$ to show that $ln(2)ln(2)$ is the sum of the alternating harmonic series. Use the alternating series test to determine how many terms of the sum are needed to estimate $ln(2)ln(2)$ accurate to within 0.001, and find such an approximation. [T] Subtract the infinite series of $ln(1−x)ln(1−x)$ from $ln(1+x)ln(1+x)$ to get a power series for $ln(1+x1−x).ln(1+x1−x).$ Evaluate at $x=13.x=13.$ What is the smallest N such that the Nth partial sum of this series approximates $ln(2)ln(2)$ with an error less than 0.001? In the following exercises, using a substitution if indicated, express each series in terms of elementary functions and find the radius of convergence of the sum. $∑ k = 0 ∞ ( x k − x 2 k + 1 ) ∑ k = 0 ∞ ( x k − x 2 k + 1 )$ $∑ k = 1 ∞ x 3 k 6 k ∑ k = 1 ∞ x 3 k 6 k$ $∑k=1∞(1+x2)−k∑k=1∞(1+x2)−k$ using $y=11+x2y=11+x2$ $∑k=1∞2−kx∑k=1∞2−kx$ using $y=2−xy=2−x$ Show that, up to powers x^3 and y^3, $E(x)=∑n=0∞xnn!E(x)=∑n=0∞xnn!$ satisfies $E(x+y)=E(x)E(y).E(x+y)=E(x)E(y).$ Differentiate the series $E(x)=∑n=0∞xnn!E(x)=∑n=0∞xnn!$ term-by-term to show that $E(x)E(x)$ is equal to its derivative. Show that if $f(x)=∑n=0∞anxnf(x)=∑n=0∞anxn$ is a sum of even powers, that is, $an=0an=0$ if n is odd, then $F=∫0xf(t)dtF=∫0xf(t)dt$ is a sum of odd powers, while if f is a sum of odd powers, then F is a sum of even powers. [T] Suppose that the coefficients a[n] of the series $∑n=0∞anxn∑n=0∞anxn$ are defined by the recurrence relation $an=an−1n+an−2n(n−1).an=an−1n+an−2n(n−1).$ For $a0=0a0=0$ and $a1=1,a1=1,$ compute and plot the sums $SN=∑n=0NanxnSN=∑n=0Nanxn$ for $N=2,3,4,5N=2,3,4,5$ on $[−1,1].[−1,1].$ [T] Suppose that the coefficients a[n] of the series $∑n=0∞anxn∑n=0∞anxn$ are defined by the recurrence relation $an=an−1n−an−2n(n−1).an=an−1n−an−2n(n−1).$ For $a0=1a0=1$ and $a1=0,a1=0,$ compute and plot the sums $SN=∑n=0NanxnSN=∑n=0Nanxn$ for $N=2,3,4,5N=2,3,4,5$ on $[−1,1].[−1,1].$ [T] Given the power series expansion $ln(1+x)=∑n=1∞(−1)n−1xnn,ln(1+x)=∑n=1∞(−1)n−1xnn,$ determine how many terms N of the sum evaluated at $x=−1/2x=−1/2$ are needed to approximate $ln(2)ln(2)$ accurate to within 1/1000. Evaluate the corresponding partial sum $∑n=1N(−1)n−1xnn.∑n=1N(−1)n−1xnn.$ [T] Given the power series expansion $tan−1(x)=∑k=0∞(−1)kx2k+12k+1,tan−1(x)=∑k=0∞(−1)kx2k+12k+1,$ use the alternating series test to determine how many terms N of the sum evaluated at $x=1x=1$ are needed to approximate $tan−1(1)=π4tan−1(1)=π4$ accurate to within 1/1000. Evaluate the corresponding partial sum $∑k=0N(−1)kx2k+12k+1.∑k=0N(−1)kx2k+12k+1.$ [T] Recall that $tan−1(13)=π6.tan−1(13)=π6.$ Assuming an exact value of $(13),(13),$ estimate $π6π6$ by evaluating partial sums $SN(13)SN(13)$ of the power series expansion $tan−1(x)=∑k=0∞(−1) kx2k+12k+1tan−1(x)=∑k=0∞(−1)kx2k+12k+1$ at $x=13.x=13.$ What is the smallest number N such that $6SN(13)6SN(13)$ approximates π accurately to within 0.001? How many terms are needed for accuracy to within 0.00001?	{"url":"https://dansdependablepetsitting.com/article/6-2-properties-of-power-series-calculus-volume-2-openstax","timestamp":"2024-11-03T18:33:59Z","content_type":"text/html","content_length":"343815","record_id":"<urn:uuid:24754a29-ff8c-4640-b97d-9a62c470d3c3>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00865.warc.gz"}
CouplingMap \| IBM Quantum Documentation class qiskit.transpiler.CouplingMap(couplinglist=None, description=None) Bases: object Directed graph specifying fixed coupling. Nodes correspond to physical qubits (integers) and directed edges correspond to permitted CNOT gates, with source and destination corresponding to control and target qubits, respectively. Create coupling graph. By default, the generated coupling has no nodes. • couplinglist (list or None) – An initial coupling graph, specified as an adjacency list containing couplings, e.g. [[0,1], [0,2], [1,2]]. It is required that nodes are contiguously indexed starting at 0. Missed nodes will be added as isolated nodes in the coupling map. • description (str) – A string to describe the coupling map. Return the distance matrix for the coupling map. For any qubits where there isn’t a path available between them the value in this position of the distance matrix will be math.inf. Test if the graph is symmetric. Return True if symmetric, False otherwise Returns a sorted list of physical_qubits add_edge(src, dst) Add directed edge to coupling graph. src (int): source physical qubit dst (int): destination physical qubit Add a physical qubit to the coupling graph as a node. physical_qubit (int): An integer representing a physical qubit. CouplingError – if trying to add duplicate qubit Compute the full distance matrix on pairs of nodes. The distance map self._dist_matrix is computed from the graph using all_pairs_shortest_path_length. This is normally handled internally by the distance_matrix attribute or the distance() method but can be called if you’re accessing the distance matrix outside of those or want to pre-generate it. Separate a CouplingMap into subgraph CouplingMap for each connected component. The connected components of a CouplingMap are the subgraphs that are not part of any larger subgraph. For example, if you had a coupling map that looked like: 0 --> 1 4 --> 5 ---> 6 --> 7 \| \| \| \| V V 2 --> 3 then the connected components of that graph are the subgraphs: 0 --> 1 \| \| \| \| V V 2 --> 3 For a connected CouplingMap object there is only a single connected component, the entire CouplingMap. This method will return a list of CouplingMap objects, one for each connected component in this CouplingMap. The data payload of each node in the graph attribute will contain the qubit number in the original graph. This will enables mapping the qubit index in a component subgraph to the original qubit in the combined CouplingMap. For example: from qiskit.transpiler import CouplingMap cmap = CouplingMap([[0, 1], [1, 2], [2, 0], [3, 4], [4, 5], [5, 3]]) component_cmaps = cmap.connected_components() will print 3 as index 0 in the second component is qubit 3 in the original cmap. A list of CouplingMap objects for each connected components. The order of this list is deterministic but implementation specific and shouldn’t be relied upon as part of the API. Return type distance(physical_qubit1, physical_qubit2) Returns the undirected distance between physical_qubit1 and physical_qubit2. • physical_qubit1 (int) – A physical qubit • physical_qubit2 (int) – Another physical qubit The undirected distance Return type CouplingError – if the qubits do not exist in the CouplingMap Draws the coupling map. This function calls the graphviz_draw() function from the rustworkx package to draw the CouplingMap object. Drawn coupling map. Return type classmethod from_full(num_qubits, bidirectional=True) Return a fully connected coupling map on n qubits. Return type classmethod from_grid(num_rows, num_columns, bidirectional=True) Return a coupling map of qubits connected on a grid of num_rows x num_columns. Return type classmethod from_heavy_hex(distance, bidirectional=True) Return a heavy hexagon graph coupling map. A heavy hexagon graph is described in: • distance (int) – The code distance for the generated heavy hex graph. The value for distance can be any odd positive integer. The distance relates to the number of qubits by: $n = \frac{5d^2 - 2d - 1}{2}$ where $n$ is the number of qubits and $d$ is the distance parameter. • bidirectional (bool) – Whether the edges in the output coupling graph are bidirectional or not. By default this is set to True A heavy hex coupling graph Return type classmethod from_heavy_square(distance, bidirectional=True) Return a heavy square graph coupling map. A heavy square graph is described in: • distance (int) – The code distance for the generated heavy square graph. The value for distance can be any odd positive integer. The distance relates to the number of qubits by: $n = 3d^2 - 2d$ where $n$ is the number of qubits and $d$ is the distance parameter. • bidirectional (bool) – Whether the edges in the output coupling graph are bidirectional or not. By default this is set to True A heavy square coupling graph Return type classmethod from_hexagonal_lattice(rows, cols, bidirectional=True) Return a hexagonal lattice graph coupling map. • rows (int) – The number of rows to generate the graph with. • cols (int) – The number of columns to generate the graph with. • bidirectional (bool) – Whether the edges in the output coupling graph are bidirectional or not. By default this is set to True A hexagonal lattice coupling graph Return type classmethod from_line(num_qubits, bidirectional=True) Return a coupling map of n qubits connected in a line. Return type classmethod from_ring(num_qubits, bidirectional=True) Return a coupling map of n qubits connected to each of their neighbors in a ring. Return type Gets the list of edges in the coupling graph. Each edge is a pair of physical qubits. Return type Test if the graph is connected. Return True if connected, False otherwise Return a set of qubits in the largest connected component. Convert uni-directional edges into bi-directional. Return the nearest neighbors of a physical qubit. Directionality matters, i.e. a neighbor must be reachable by going one hop in the direction of an edge. reduce(mapping, check_if_connected=True) Returns a reduced coupling map that corresponds to the subgraph of qubits selected in the mapping. • mapping (list) – A mapping of reduced qubits to device qubits. • check_if_connected (bool) – if True, checks that the reduced coupling map is connected. A reduced coupling_map for the selected qubits. Return type CouplingError – Reduced coupling map must be connected. shortest_undirected_path(physical_qubit1, physical_qubit2) Returns the shortest undirected path between physical_qubit1 and physical_qubit2. • physical_qubit1 (int) – A physical qubit • physical_qubit2 (int) – Another physical qubit The shortest undirected path Return type CouplingError – When there is no path between physical_qubit1, physical_qubit2. Return the number of physical qubits in this graph.	{"url":"https://docs.quantum.ibm.com/api/qiskit/qiskit.transpiler.CouplingMap#subgraph","timestamp":"2024-11-06T12:13:33Z","content_type":"text/html","content_length":"339990","record_id":"<urn:uuid:662d51bc-7cde-4133-b176-ce5f356419e1>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00830.warc.gz"}
Interactive CBSE Class 2 Maths Worksheet Tens and Ones Class 2 Maths Worksheets Chapter 8: Tens and Ones \| Free PDF Download CBSE Class 2 Maths Worksheets for Tens and Ones are available here at Vedantu solved by expert teachers as per the latest NCERT (CBSE) guidelines. You will find a comprehensive collection of Questions with Solutions in these worksheets which will help you to revise the complete syllabus and score more marks in a fun way. You can download the printable worksheets from the links mentioned below in the article. With the help of our free online math worksheets, teach your second-graders in the study and practice of mathematics. Explore and revise addition and subtraction with these free, printable math worksheets for first graders. Let us start learning tens and ones with fun worksheets! FAQs on CBSE Class 2 Maths Worksheets for Chapter 8 Tens and Ones 1. How can students download the CBSE class 2 maths worksheets for chapter 8 - Tens and Ones? Students can download the CBSE Class 2 Maths Worksheets for - Tens and Ones from the website of Vedantu or by clicking on this link. Vedantu provides free study material for students to make learning a hassle-free experience. Students can make use of this material at their convenience as it is in a downloadable format. 2. Are CBSE class 2 maths worksheets for chapter 8 - Tens and Ones prepared by expert teachers ? Yes, these worksheets are prepared by highly qualified teachers of Vedantu. The questions and solutions are of high quality and our teachers get a very good response from students. Students find it helpful and it eventually helps them to achieve more marks in the examination. 3. How will CBSE class 2 maths worksheets for chapter 8 - Tens and Ones help a student to grow mentally ? These worksheets consist of such questions in which logical, lingual thinking and cross-questioning abilities will improve and they will develop curiosity about the concepts and an urge to learn something new and meaningful. All the solutions are available on Vedantu’s official website and mobile app for free. 4. Are CBSE class 2 maths worksheets for chapter 8 - Tens and Ones tough to solve ? Yes, they are, but someone who is aware of the key concepts will find it easier to understand and will be in a position to solve them within the time limit allotted to the student. With a thorough practice of these worksheets, students can learn in a fun way. 5. Can a student with no knowledge about maths attempt CBSE Class 2 maths worksheets for chapter 8 - Tens and Ones? No, it is imperative for students to at least start with the basic concept of maths, so that he/she can build a strong foundation of the subject. If a student attempts these worksheets with no knowledge of the concepts, he/she will never understand the fun of solving a problem.	{"url":"https://www.vedantu.com/cbse/class-2-maths-worksheets-chapter-8","timestamp":"2024-11-11T21:34:32Z","content_type":"text/html","content_length":"176508","record_id":"<urn:uuid:2bf029a5-9049-40d3-84c8-9ca634b407df>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00268.warc.gz"}
mmHg to atm conversion \| Easy to convert – CalculatorPort mmHg to atm conversion \| Easy to convert mmhg atm 1 mmhg 0.001316 atm 2 mmhg 0.002632 atm 3 mmhg 0.003947 atm 4 mmhg 0.005263 atm 5 mmhg 0.006579 atm What is mmHg? A millimeter of mercury(mmHg) is a manometric unit of pressure. Formerly millimeter of mercury(mmHg) is defined as the extra pressure which is generated by a column of mercury one millimeter high. It is denoted by mmHg or mm Hg. What is atm? Atmospheric pressure, also known as barometric pressure is the pressure within the atmosphere of Earth. A standard atmosphere is a unit of pressure. The standard atmosphere is denoted by atm. How much 1 mmHg is equal to atm? 1 mmHg ≅ 0.001316 atm How much 1 atm is equal to mmHg? 1 atm ≅ 760 mmHg If you think that inches of mercury and atmospheres are one and the same, think again! Converting between the two can be a tricky business, since there’s no direct conversion between the two pressure measurements. An atmosphere is a non-SI unit of pressure and is equal to 101,325 Pascals (Pa). Meanwhile, one millimeter of mercury (mmHg) pressure is equivalent to 1 torr and therefore is equal to 133.322 Pa. However, it’s still possible to make the conversion between the two units by utilizing a conversion formula. Let’s start with how to convert mmHg to atm. To convert mmHg to atm, you’ll need to use the mmHg to atm formula: 1 atm = 760 mmHg. Using the formula, you can divide the number of millimeters of mercury by 760 to obtain the number of atmospheres. For example, let’s say you have 130 mmHg pressure; you’d divide the 130 mmHg pressure by 760 to get 0.170 atm. Now let’s turn our attention to how to convert atm to mmHg. To convert atm to mmHg, you’ll need to use the atm-to-mmHg formula: 1 atm = 760 mmHg. Using the formula, you can multiply the number of atmospheres by 760 to obtain the number of millimeters of mercury. For example, let’s say you have 0.5 atm pressure; you’d multiply the 0.5 atm pressure by 760 to get 380 mmHg. To wrap it up, though atmospheres and millimeters of mercury measure different types of pressure, it’s possible to convert between them with ease by following the proper formulas. By dividing the number of millimeters of mercury by 760 or multiplying the number of atmospheres by 760, you can convert between the two units of pressure with precision. How to use this online mmhg to atm conversion Hey Buddy! How are you? At first A Warm Welcome to you on our online calculator tool site! Follow these 6 easy steps to use our “Online mmHg to atm converter Tool”. 1) At first, give inputs in the “From Section.” (Your inputs should be number) 2) After giving your inputs, press the “Calculate” button & Boom! You can see your result in the “Answer Section”!! Isn’t it amazing? 🙂 3) Your answer will be in number. 4) Once used our calculator & to start from fresh, press the “Reset” button & continue enjoying 🙂 5) In the case of our “online mmHg to atm converter”, it takes input & runs a javascript program in the background. So You don’t need to worry about the correct result as there is a powerful javascript program running in the background. We genuinely care for you 🙂 *note that the round figure is upto 6 decimal points So always Check the result. How You Can Help US Now comes the last & most important step. If you find any inconvenience/mistake, then Feel free to contact us through the “Contact Us” section. ✓Also if you feel we can improve our blog in a better way please feel free to let us know & you’re always welcome to share with us your beautiful idea. ✓ If you find our “Online mmHg to atm converter” tool Very Useful, give us your feedback in the “Leave a Comment” section. Your feedback matters a lot to us. We’ll be waiting for your feedback. Keep Loving & Supporting us. Thank You! Have a Nice Day! 🙂 Leave a Comment	{"url":"https://calculatorport.com/mmhg-to-atm-conversion/","timestamp":"2024-11-06T17:03:54Z","content_type":"text/html","content_length":"259908","record_id":"<urn:uuid:eb52f78c-fc20-4723-b8f9-80772dd4df0b>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00529.warc.gz"}
Tableau Line Chart Multiple Lines Per Variable Value 2024 - Multiplication Chart Printable Tableau Line Chart Multiple Lines Per Variable Value Tableau Line Chart Multiple Lines Per Variable Value – The Multiplication Graph or chart Line will help your college students aesthetically stand for different very early math concepts concepts. However, it must be used as a teaching aid only and should not be confused with the Multiplication Table. The graph can be purchased in about three models: the colored variation is effective whenever your college student is focusing on one periods kitchen table at the same time. The vertical and horizontal types are compatible with kids who are still discovering their instances furniture. If you prefer, in addition to the colored version, you can also purchase a blank multiplication chart. Tableau Line Chart Multiple Lines Per Variable Value. Multiples of 4 are 4 far from the other The design for identifying multiples of 4 would be to add more each variety to on its own and discover its other numerous. As an illustration, the 1st 5 multiples of 4 are: 4, 8, 12 and 16 and 20. And they are four away from each other on the multiplication chart line, this trick works because all multiples of a number are even. Furthermore, multiples of a number of are even phone numbers Multiples of 5 are even If they end in or 5, You’ll find multiples of 5 on the multiplication chart line only. Put simply, you can’t flourish a amount by a couple of to obtain an even amount. If the number ends in five or , you can only find a multiple of five! Fortunately, you can find techniques which make getting multiples of five even easier, like making use of the multiplication graph or chart series to discover the several of 5. Multiples of 8 are 8 clear of the other The design is apparent: all multiples of 8 are two-digit phone numbers and all multiples of a number of-digit amounts are two-digit phone numbers. Every variety of 10 has a numerous of 8-10. Eight is even, so all its multiples are two-digit phone numbers. Its pattern carries on as much as 119. The next time you can see a variety, be sure you search for a a number of of 8 from the beginning. Multiples of 12 are 12 from the other person The number 12 has unlimited multiples, and you can grow any total amount by it to help make any amount, such as on its own. All multiples of 12 are even phone numbers. Here are some illustrations. David likes to acquire pencils and organizes them into eight packets of 12. He now has 96 pencils. James has one among each kind of pencil. In their office, he arranges them about the multiplication chart collection. Multiples of 20 are 20 away from the other In the multiplication graph, multiples of 20 or so are common even. If you multiply one by another, then the multiple will be also even. Multiply both numbers by each other to find the factor if you have more than one factor. If Oliver has 2000 notebooks, then he can group them equally, for example. A similar is applicable to pencils and erasers. You could buy one in a package of about three or perhaps a package of half a dozen. Multiples of 30 are 30 from the other person In multiplication, the expression “component set” means a team of numbers that develop an obvious variety. For example, if the number ’30’ is written as a product of five and six, that number is 30 away from each other on a multiplication chart line. The same is true to get a amount within the variety ‘1’ to ’10’. Quite simply, any amount could be created because the product or service of 1 and on its own. Multiples of 40 are 40 from one another You may know that there are multiples of 40 on a multiplication chart line, but do you know how to find them? To do this, you can add externally-in. By way of example, 10 12 14 = 40, etc. In the same manner, 10 seven = 20. In this instance, the quantity on the kept of 10 is surely an even amount, while the a single around the right is surely an strange number. Multiples of 50 are 50 away from the other Utilizing the multiplication graph or chart line to ascertain the amount of two phone numbers, multiples of 50 are similar length away from each other on the multiplication graph. They may have two excellent factors, 50 and 80. For the most part, each word can vary by 50. Other component is 50 by itself. The following are the typical multiples of 50. A typical numerous will be the multiple of the provided number by 50. Multiples of 100 are 100 away from one another Allow me to share the different phone numbers which can be multiples of 100. A confident combine can be a several of just one hundred or so, although a negative match can be a numerous of ten. These 2 types of phone numbers will vary in many ways. The 1st technique is to break down the telephone number by successive integers. In this case, the amount of multiples is a, thirty, twenty and ten and Gallery of Tableau Line Chart Multiple Lines Per Variable Value Breathtaking Sas Horizontal Bar Chart Double Y Axis Graph Cool Ggplot2 Time Series Multiple Lines 2d Line Plot Free Supply And Tableau Michael Sandberg s MicroStrategy Tips Tricks Blog Leave a Comment	{"url":"https://www.multiplicationchartprintable.com/tableau-line-chart-multiple-lines-per-variable-value/","timestamp":"2024-11-06T08:24:55Z","content_type":"text/html","content_length":"53161","record_id":"<urn:uuid:86e4b2bc-47d4-4734-99a9-130eec1e6af5>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00515.warc.gz"}
Leon S. Linear Algebra with Applications 9ed 2015 Solutions top of page Solution Manual textbook solutions Leon S. Linear Algebra with Applications 9ed 2015 Solution Manual textbook solutions If the solution image is too large, it may look blurry on the Yandex Disk preview, Click "original size" button on the upper right corner of the image. Disable adblockers if you having problems while viewing the solutions. Hello you Math Wizards! ๐ ง โ โ ๏ธ Are you ready to get quadratic with Leon S. and his book, 'Linear Algebra with Applications'? This right here is the key to ruling the college mathematics kingdom. Yes, you heard right! Let's get to the root of it then, shall we? ๐ This is the revered 9th edition of 'Linear Algebra with Applications' by Mr. Leon S. himself, published in 2015, with an ISBN 9780321962218. Mr. Leon has made sure you understand math, not merely do math. And we, with our 'solution manual' service, are turning his dream into a reality. Now, let's uncover the magic formula! ๐ On the top of this page, you'll find a little dropdown menu. Itโ s like our own math lottery! Select a chapter, then also choose a problem number. Itโ s like drawing the sword Excalibur from the stone, only instead, you get a link to a step-by-step textbook solution to your problem. Easy-peasy lemon squeezy, isnโ t it? ๐ ๐ ก If at some stage, the solution image appears to be as confused as you were at the start of the problem and looks blurry on the Yandex Disk preview, fret not! You can download the image, or just click on the "original size" button hiding slyly under the 3-dot menu on the top right of the image. You see, it's that mysteriously straightforward! Let's not forget, this textbook is like a magical guide that helps you to explore and conquer the depths of linear algebra. And the best part? You get to have all of it in a step-by-step format. ๐ It's a non-complicated, non-linear solution to all your linear-algebra problems. So why wait? Letโ s ease into the intricate world of matrixes and transformations together. After all, why should polynomials have all the fun, right? ๐ Oh and by the way, don't forget our lovely pop-ups! They may seem annoying, but it's like that cartoon ad-break in between your favorite show. It's the only way we keep our server running and this treasure trove open to all math enthusiasts. ๐ But hey, a small price for lifelong wisdom, right? So sharpen those pencils, flatten those papers and spin this web of textbook solutions with us. After all, every spider needs a little help at the start. Let's swing into the world of numbers together! ๐ ธ๏ธ ๐ ท๏ธ ๐ ซ Before we end this grand intro, remember, they say, "Mathematics may not teach us how to add love or subtract hate, but it gives us every reason to hope that 'every problem has a solution!'" Happy computing, folks! ๐ ๐ ฎ Always here, always serving. โ ๏ธ bottom of page	{"url":"https://www.litsolutions.org/i/9780321962218","timestamp":"2024-11-14T22:11:46Z","content_type":"text/html","content_length":"508973","record_id":"<urn:uuid:09fb486c-7bb7-45d8-991d-9cd87eb23264>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00051.warc.gz"}
Affine connection for spherical symmetry Suppose you have a spherically symmetric vector field, as in the diagram. Can we find an affine connection which transports the vectors into one-another? That is, a geometry in which they are all The vectors (red arrows) are clearly not parallel in the usual sense. But can we define a new connection in which they are transported into one-another? Take Schwarzschild spacetime, in the usual coordinates where the Now consider an arbitrary vector field of the form: We would not expect the sought-for parallel transport to work for vectors with components in the The offending Christoffel symbols turn out to be coordinate vector changes as you move around on a sphere. One option is to simply define new connection coefficients for which these vanish: any choice of smooth functions Introduction to Riemannian manifolds, Lemma 4.10). We can also write this new connection as the usual (Levi-Civita) covariant derivative plus a bilinear correction: The parenthetical term is a (1,2)-tensor we interpret as accepting the vectors in the last two slots ( As a check, We can also construct a symmetric connection Hence we have constructed connections which parallel transport our spherically symmetric vector field around a sphere, and deviate as little as possible from the Levi-Civita connection. Neither of the new connections are “metric-compatible”, for instance If you find some formulae here do not work for you, compare your convention for the connection coefficient index order, or try swapping new page describing them. Finally, beware of coordinate basis vectors! The “vectors” One thought on “Affine connection for spherical symmetry”	{"url":"https://cmaclaurin.com/2021/04/26/affine-connection-for-spherical-symmetry/","timestamp":"2024-11-04T16:46:30Z","content_type":"text/html","content_length":"60944","record_id":"<urn:uuid:dea537b6-c2de-4b14-ac02-58c805a2d9a3>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00013.warc.gz"}
The Oshkosh Airfoil Transformation The Oshkosh airfoil is a modification of the Joukowski airfoil transformation, developed by R. T. Jones in the 80's. The Joukowski airfoil is based on a conformal transformation of a circle with circulatory flow combined with a uniform flow field into an airfoil shape. This results in an exact solution to ideal fluid flow around an airfoil type shape. The Oshkosh airfoil first transforms the circle into an oval, and then transforms the oval into an airfoil. Selection of the transformation parameters can give an airfoil that is similar to a NACA 6 series. The five parameters, X[c], X[t], Y[c], Y[t], and D control the airfoil transformation. If X[t], Y[t], and D are set to zero, the result is a Joukowski airfoil. The top five slider bars control the airfoil design parameters X[c], X[t], Y[c], Y[t], and D. The bottom slider bar alters the angle of attack and subsequent changes in the surface velocity. The blue and red lines represent the air velocity over the upper and lower surfaces respectively and will vary with the angle of attack.	{"url":"https://aerofoilengineering.com/Help/HelpCreateOshkosh.htm","timestamp":"2024-11-03T10:02:13Z","content_type":"text/html","content_length":"2008","record_id":"<urn:uuid:4bad40a7-2353-484f-bc3c-01e6ebf7eb22>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00219.warc.gz"}
Technical Details \| Chia Documentation New Matching Algorithm The matching algorithm for all tables has changed, and now forms the basis of security. It is a memory hard algorithm which can be tuned to take more or less time by adjusting the number of match indexes to test for whether two entries match. The benefit of this algorithm is that we can set the difficulty very high so that plotting will take longer and compression attacks will be more expensive, yet it incurs negligible cost when validating a proof. Since validation is “free”, we can tune this to be as difficult as we need, without adding extra compute expense to the network. Matching Bits The matching algorithm takes an additional index number which is used to show that a match works. The left value and the index results in a bucket. This must match a bucket which the right value hashes to, and the matching combination of them have to pass an additional filter. Index bits will be included in proofs to make verification fast. To keep required memory down, entries are sorted into sections. All of the buckets to which a left value hashes will land in the same section. T1 Matching For the first table of matches, we match a k-bit value that comprises section_bits and random_bits. Match_index_bits is an additional variable to define which match_index creates a match. For the first set of pairs, x1 and x2 match iff: • The upper section_bits of x1 are equal to the upper section bits of x2. • There exists a match_index in the range [0..<2^match_index_bits] where the random_bits produced from hash(x1 + plot_id + match_index) are equal to the random_bits produced from x2 with hash(x2 + • A new_meta is created by hashing x1 and x2. • A hash of new_meta must pass a filter for it to be a match. Matching Difficulty The time to find a match is heavily influenced by match_index_bits, as the higher the range of match_indexes the more the number of lookups to test for a match. Since most bit dropping attacks are limited by the difficulty of matching in the earlier tables, the number of match index_bits will be tuned much higher for the first and second tables, and set lower for subsequent tables. New Plot Filters The new proof of space comprises 3 plot filters: 1. Plot id filter 2. EncryptedX’s scan filter 3. Random x-quadruple quality filter Plot Id Filter The plot id filter is the same as the original proof of space – the challenge and plot id create a hash that validates the plot for that challenge if it passes the filter. For the new proof of space we will be able to reduce this filter significantly compared to the original. Once set, it is expected to stay fixed over time. EncryptedX’s Scan Filter A few definitions: • x1 and x2 are pairs matched in the first table, and are the left side match in the next table. • x3 and x4 are pairs matched in the first table, that then form the right-side match with x1 and x2 in the second table. • Xdata totals 2k bits, comprising the upper section bits of x1, additional bits of x1, additional bits of x2, and match index for pairing x1 and x2, upper section bits of x3, additional bits of x3, additional bits of x4, and match index for pairing x3 and x4. E.g. if we have a k32, 6 section bits and 6 match bits, we form: [upper 6 bits of x1 and x2 (which are the same)][bits 7-16 of x1][bits 7-16 of x2][6 match index bits (applied to x1)][upper 6 bits of x3 and x4 (which are the same)][bits 7-16 of x3][bits 7-16 of x4][6 match index bits (applied to x3)]. • EncryptedXs is the 2k-bits encrypted Xdata with a seed based on plot_id (reversible) • S is a random range within 0..2^2k based on the challenge The Challenge Find a proof where: 1. The last 4 x values in the proof are converted to EncryptedX’s and are in range S, and 2. hash(challenge,EncryptedXs) passes a filter (e.g. 1 in cardinality S chance will pass on average 1 result from this range). In order to quickly satisfy the challenge for the plot, we store the sorted EncryptedXs in the second table, and drop the first table. This requires only 1 or 2 disk seeks to read the number of EncrpytedXs in the range S for the challenge. To reconstruct the original x values a farmer finds all possible values of buckets for x1 and x2 and finds a collision between them. In the k32 example above bits 17-32 of x1 and x2 are missing so there are 2^16 possibilities for each. To find the values a farmer makes a list of all 2^16 possibilities for one of the values (it doesn’t matter which) then sorts it, then scans over all 2^16 possible values for the other looking up each one in the sorted list to see if it’s there. For a farmer to store fewer bits they can drop bits from the EncryptedX, which doubles the amount of work necessary to be done. For them to need less work but more space they can store unencrypted x bits which halves the amount of work needed for each bit stored. Because there’s an exponential increase in work per bit and computers are fast there’s a range of bit dropping which has very little work even on low end hardware but where the costs of farming get prohibitive even on high end hardware with even a tiny increase in compression. The EncryptedXs scan filter forces a particular ordering and dataset in the plot format that severely limits the flexibility for an attacker to re-organize the data in a way to leverage particular bit dropping attacks. Random x-quadruple quality filter For each x1/2/3/4 set that passes the scan filter, find another random full x-quadruple in the proof based on the hash of those x’s, the plot_id, and challenge. This x-quadruple then forms the value for the quality string that determines whether a good proof has been found. The additional random x-quadruple lookup ensures that all plot data is needed in order to find a quality by forcing a backwards traversal down the tables. The plot id filter can be set lower, since the EncryptedXs scan filter only requires 1 or 2 disk lookups. A low plot id filter, combined with an additional filter on the EncryptedXs scan, forces rental attackers to generate at least the first two tables of a plot which require the most time, and then discard most of their results if it doesn’t pass the scan filter. The scan filter also reduces the load on HDD disks by only passing a fraction of the time for the x-quadruple quality filter which requires more lookups. The combination of these three filters severely constrain the flexibility afforded to an attacker. The default level of bit dropping gives the honest farmer a baseline with negligible extra compute, but imposes an immediate difficulty for the attacker to compress much further without losing efficiency compared to the honest farmer. Benes Compression A new compression algorithm allows us to compress each table in the plot by 2-3 extra bits compared to the original format. Note that this is not bit dropping (which incurs costs that go up exponentially based on the amount of data being dropped), but rather an improved way to losslessly store information in each table. The findings are based on this blog post. In addition, Benes compression has the unique property that plot tables can be traversed in both directions, whereas the HDD friendly format requires additional data to store an index from the T2 table to the last table (so that it can then traverse entries down the table). Each table in the plot using Benes compression comes at the cost of doing a few more random seeks in data retrieval. While it would be possible to put a few such Benes plots on an HDD, the proof retrieval times are not guaranteed to be completed in time. On SSD using Benes compression has minimal impact on farming and is the recommended storage format. While we hope to release Benes compression for plots prior to the hard fork, constructing the proof of space using the Benes algorithms is more challenging and could require significantly more RAM and slower plotting times. Further Compression by Additional Bit Dropping One of the first known compression attacks is to bit drop on x1/x2 pairs on the first table, and recompute for the missing range of values. With the new scan filter requiring both an encryption step on x-quadruples and a range to satisfy a challenge, any attacker wanting to alter which bits in those x-quadruples to drop, will forfeit compression they receive from the ordered encrypted values, and immediately require making up for k bits of lost compression. Likewise, any other attempts to re-organize x data for other bit dropping or compression methods will result in similar penalties. It is still possible to drop the lower-order bits from the EncryptedX values, where each bit dropped would require a doubling of compute. A potentially more advantageous approach is to leave the default x values intact and instead drop bits from the back pointers starting from table T3. In the original format, dropping two bits (one from each of the two back pointers in T3) would have saved approximately 1% of space, at the cost of doubling the time needed to recompute the quality string and the full proof. In the new format, we have restructured the back pointers, and currently, we see the potential to drop only 1 bit per back pointer for a doubling of compute. Further research may reveal the possibility of more aggressive bit dropping back up to 2 bits per doubling of recompute. As it stands, the default plot already includes a certain level of bit dropping, and bit dropping further quickly becomes too expensive. Depending on the final parameter settings, we don’t expect this to be economically viable beyond a few bits on today’s hardware, with each bit dropped saving ~0.5%. We will also offer the ability to plot using the most optimal method for further compression for those interested. Impact to Honest Farmers The x values stored in T2 have default levels of bit-dropping already applied that are defined in the challenge scan filter. A small amount of compute is required when fetching a final quality string, similar to the low C-levels of the bladebit formats. The honest farmer will have to grind an additional small amount to get the full x values for the proof. There will also be an option to omit this low-level grinding if desired, at the cost of adding more bits to the plot format. However, since the compute required for the grind is low and designed to be close to optimal, the default level of bit dropping specified by the challenge is the recommended setting. Impact to Rental Attacks Due to the long plotting time, lower plot id filter, and additional scan and x-quadruple filters that potentially throw away a newly created plot; rental attacks are no longer deemed to be a viable threat (>$1 billion per hour for an attack and will be well over 1000 times more expensive than the original format).	{"url":"https://docs.chia.net/zh-Hans/new-proof-details/","timestamp":"2024-11-09T01:02:33Z","content_type":"text/html","content_length":"40238","record_id":"<urn:uuid:d64d8014-0bab-4d8e-b777-6174ee371782>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00761.warc.gz"}
September- 2020 challenge Each month, a new set of puzzles will be posted. Come back next month for the solutions and a new set of puzzles, or subscribe to have them sent directly to you. MATCHSTICK Challenge 1. Move only two of the matchsticks in diagram 1 to turn the chair upside down. 2. Move only three of the matchsticks in diagram 2 to create three equal squares. All match sticks must be used. MIND-Xpander (Circular Logic Challenge – Level 2) 1. You have 100 sticks standing upright in a circle (#1 to #100). Starting with stick #1, knock over the next stick (#2) in the circle. Then continue by knocking over every second stick (next is # 4). Keep going around and around the circle until one stick remains standing. What number is the last stick standing? 2. What if you start half way around with stick #50 and continue around and around in same manner as above? There are more than one way of doing these puzzles and may well be more than one answer. Please let me and others know what alternatives you find by commenting below. We also welcome general comments on the subject and any feedback you'd like to give. If you have a question that needs a response from me or you would like to contact me privately, please use the contact form. Get more puzzles! If you've enjoyed doing the puzzles, consider ordering the books; • Book One - 150+ of the best puzzles • Book Two - 200+ with new originals and more of your favourites Both in a handy pocket sized format. Click here for full details. Last month's solutions EQUATE+2 Puzzle Each row, column & diagonal is an equation and you use the numbers 1 to 9 to complete the equations. Each number can be used only once. ‘Two’ numbers have been provided to get you started. Find the remaining seven numbers that satisfies all the resulting equations. Note – multiplication (x) & division (/) are performed before addition (+) and subtraction (-). EQUATE+0 Puzzle Each row, column & diagonal is an equation and you use the numbers 1 to 9 to complete the equations. Each number can be used only once. Find the remaining nine numbers that satisfies all the resulting equations. Note – multiplication (x) & division (/) are performed before addition (+) and subtraction (-) Submit a Comment Cancel reply	{"url":"https://gordonburgin.com/2020/09/september-2020-challenge/","timestamp":"2024-11-08T20:54:01Z","content_type":"text/html","content_length":"253900","record_id":"<urn:uuid:b2ce04d1-35a9-4832-9f13-fc96fc256ca3>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00784.warc.gz"}
The figure above is an equilateral inscribed in a circle with ... \| Filo The figure above is an equilateral inscribed in a circle with radius 10 . What is the measure of ? Not the question you're searching for? + Ask your question Was this solution helpful? Found 7 tutors discussing this question Discuss this question LIVE for FREE 13 mins ago One destination to cover all your homework and assignment needs Learn Practice Revision Succeed Instant 1:1 help, 24x7 60, 000+ Expert tutors Textbook solutions Big idea maths, McGraw-Hill Education etc Essay review Get expert feedback on your essay Schedule classes High dosage tutoring from Dedicated 3 experts Practice questions from Triangles in the same exam Practice more questions from Triangles View more Practice questions on similar concepts asked by Filo students View more Stuck on the question or explanation? Connect with our Mathematics tutors online and get step by step solution of this question. 231 students are taking LIVE classes Question Text The figure above is an equilateral inscribed in a circle with radius 10 . What is the measure of ? Topic Triangles Subject Mathematics Class Grade 12 Answer Type Text solution:1 Upvotes 70	{"url":"https://askfilo.com/mathematics-question-answers/the-figure-above-is-an-equilateral-inscribed-in-a-circle-with-radius-10-what-is-285171","timestamp":"2024-11-05T00:19:29Z","content_type":"text/html","content_length":"200439","record_id":"<urn:uuid:015efca4-9c74-45be-8f25-8b2e1daead28>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00436.warc.gz"}
bayesian reinforcement learning pdf Aman Taxali, Ray Lee. Abstract—We propose Bayesian Inverse Reinforcement Learning with Failure (BIRLF), which makes use of failed demonstrations that were often ignored or ﬁltered in previous methods due to the difﬁculties to incorporate them in addition to the successful ones. One Bayesian model-based RL algorithm proceeds as follows. Approximate Bayesian Reinforcement Learning Jonathan Sorg Computer Science & Engineering University of Michigan Satinder Singh Computer Science & Engineering University of Michigan Richard L. Lewis Department of Psychology University of Michigan Abstract The explore{exploit dilemma is one of the central challenges in Reinforcement Learn-ing (RL). Bayesian Reinforcement Learning 3 2 Model-Free Bayesian Reinforcement Learning Model-free RL methods are those that do not explicitly learn a model of the sys-tem and only use sample trajectories obtained by direct interaction with the system. I will attempt to address some of the common concerns of this approach, and discuss the pros and cons of Bayesian modeling, and brieﬂy discuss the relation to non-Bayesian machine learning. This book summarizes the vast amount of research related to teaching and learning probability that has been conducted for more than 50 years in a variety of disciplines. Bayesian Reinforcement Learning Dongho Kim Department of Engineering University of Cambridge, UK dk449@cam.ac.uk Kee-Eung Kim Dept of Computer Science KAIST, Korea kekim@cs.kaist.ac.kr Pascal Poupart School of Computer Science University of Waterloo, Canada ppoupart@cs.uwaterloo.ca Abstract By solving the POMDP P, one In this work we present an advanced Bayesian formulation to the task of control learning that employs the Relevance Vector Machines (RVM) generative model for value function evaluation. I will also provide a brief tutorial on probabilistic reasoning. graphics, and that Bayesian machine learning can provide powerful tools. 1052A, A2 Building, DERA, Farnborough, Hampshire. This chapter surveys recent lines of work that use Bayesian techniques for reinforcement learning. Bayesian methods for machine learning have been widely investigated,yielding principled methods for incorporating prior information intoinference algorithms. A Bayesian Framework for Reinforcement Learning by Strens (ICML00) 10/14/08 : Ari will tell us how to use Gaussian Processes for continuous RL Reinforcement Learning with Gaussian Processes (ICML 2005) (PDF) Bayesian Reinforcement Learning Bayesian RL lever-ages methods from Bayesian inference to incorporate prior information about the Markov model into the learn-ing process. Efﬁcient Bayesian Clustering for Reinforcement Learning Travis Mandel1, Yun-En Liu2, Emma Brunskill3, and Zoran Popovic´1;2 1Center for Game Science, Computer Science & Engineering, University of Washington, Seattle, WA 2EnlearnTM, Seattle, WA 3School of Computer Science, Carnegie Mellon University, Pittsburgh, PA ftmandel, zorang@cs.washington.edu, yunliu@enlearn.org, ebrun@cs.cmu.edu However, an issue Furthermore, online learning is not computa-tionally intensive since it requires only belief monitor-ing. This book is focused not on teaching you ML algorithms, but on how to make ML algorithms work. contexts related to reinforcement learning in partially-observable domains: learning partially observable Markov Decision processes, taking advantage of expert demon-strations, and learning complex hidden structures such as dynamic Bayesian networks. 1. PDF \| We consider the ... we propose a novel value-based Bayesian meta-reinforcement learning framework BM-DQN to robustly speed up the learning … the learning and exploitation process for trusty and robust model construction through interpretation. “Using Trajectory Data to Improve Bayesian Optimization for Reinforcement Learning.” Journal of Machine Learning Research , 15(1): 253–282. Recently, Lee [1] proposed a Sparse Bayesian Reinforce-ment Learning (SBRL) approach to memorize the past expe-riences during the training of a reinforcement learning agent for knowledge transfer [17] and continuous action search [18]. Bayesian Inverse Reinforcement Learning Deepak Ramachandran Computer Science Dept. [4] introduced Bayesian Q-learning to learn plied to GPs, such as cross-validation, or Bayesian Model Averaging, are not designed to address this constraint. It also offers an extensive review of the literature adult mathematics education. In inverse reinforcement learning, the agent recovers an unknown 2 reviews the Our goals are to 1) give a detailed description of hierarchical models and their application in the context of reinforcement learning and 2) compare these models to other commonly used approaches. U.K. Abstract The reinforcement learning problem can be decomposed into two parallel types of inference: (i) estimating the parameters of a model for the Why is Posterior Sampling Better than Optimism for Reinforcement Learning? Bayesian reinforcement learning methods incorporate probabilistic prior knowledge on models [7], value functions [8, 9], policies [10] or combinations [17]. In this survey, we provide an in-depth reviewof the role of Bayesian methods for the reinforcement learning RLparadigm. In Bayesian learning, uncertainty is expressed by a prior distribution over unknown parameters and learning is achieved by computing a posterior distribution based on the data observed. Hence, Bayesian reinforcement learning distinguishes itself from other forms of reinforcement learning by explicitly maintaining a distribution over various quantities such as the parameters of the model, the value function, the policy or its gradient. Sect. This open book is licensed under a Creative Commons License (CC BY-NC-ND). 4 CHAPTER 1. Bayesian Bandits Introduction Bayes UCB and Thompson Sampling 2. Machine Learning Yearning, a free ebook from Andrew Ng, teaches you how to structure Machine Learning projects. Zentralblatt MATH: 1317.68195 Model-free techniques are often simpler to implement since they do not require any In this survey, we provide an in-depth review of the role of Bayesian methods for the reinforcement learning … The few Bayesian RL methods that are applicable in partially observable domains, such as the Bayes-Adaptive POMDP (BA-POMDP), scale poorly. In each of these contexts, Bayesian nonparametric approach provide advantages in Monte Carlo Bayesian Reinforcement Learning of the unknown parameter. This textbook presents fundamental machine learning concepts in an easy to understand manner by providing practical advice, using straightforward examples, and offering engaging discussions of relevant applications. Simultaneous Hierarchical Bayesian Parameter Estimation for Reinforcement Learning and Drift Diffusion Models: a Tutorial and Links to Neural Data Mads L. Pedersen1,2,3 & Michael J. Frank1,2 # The Author(s) 2020 Abstract Cognitive modelshave been instrumental for generating insights into the brain processes underlyinglearning anddecision making. Bayesian Reinforcement Learning 5 D(s,a)is assumed to be Normal with mean µ(s,a)and precision τ(s,a). An emphasis is placed in the first two chapters on understanding the relationship between traditional mac... As machine learning is increasingly leveraged to find patterns, conduct analysis, and make decisions - sometimes without final input from humans who may be impacted by these findings - it is crucial to invest in bringing more stakeholders into the fold. INTRODUCTION ingwhatcanbelearnedfromthedata. However, instead of maintaining a Normal-Gamma over µ and τ simultaneously, a Gaussian over µ is modeled. Semantic Scholar is a free, AI-powered research tool for scientific literature, based at the Allen Institute for AI. Hence, Bayesian reinforcement learning distinguishes itself from other forms of reinforcement learning by explicitly maintaining a distribution over various quantities such as the parameters of the model, the valueâ ¦Â, Exploration Driven by an Optimistic Bellman Equation, Learning and Forgetting Using Reinforced Bayesian Change Detection. Model-based Bayesian RL [3; 21; 25] ex-press prior information on parameters of the Markov pro-cess instead. You are currently offline. Motivation. The paper is organized as follows. The Troika of Adult Learners, Lifelong Learning, and Mathematics, Research on Teaching and Learning Probability. University of Illinois at Urbana-Champaign Urbana, IL 61801 Eyal Amir Computer Science Dept. Model-free Bayesian Reinforcement Learning We show that hierarchical Bayesian models provide the best Traditionally,RLalgorithmshavebeencategorizedasbeingeither model-based or model-free.In the … reinforcement learning methods and problem domains. This book of Python projects in machine learning tries to do just that: to equip the developers ... AI is transforming numerous industries. Model-Based Bayesian Reinforcement Learning in Complex Domains St´ephane Ross Master of Science School of Computer Science McGill University Montreal, Quebec 2008-06-16 A thesis submitted to McGill University in partial fulﬁllment of the requirements University of Illinois at Urbana-Champaign Urbana, IL 61801 Abstract Inverse Reinforcement Learning (IRL) is the prob-lem of learning the reward function underlying a In section 3.1 an online sequential Monte-Carlo method developed and used to im- Reinforcement learning, one of the most active research areas in artificial intelligence, is a computational approach to learning whereby an agent tries to maximize the total amount of reward it receives while interacting with a complex, uncertain environment. This book presents a synopsis of six emerging themes in adult mathematics/numeracy and a critical discussion of recent developments in terms of policies, provisions, and the emerging challenges, paradoxes and tensions. Our experimental results confirm … In this paper we focus on Q-learning[14], a simple and elegant model-free method that learns Q-values without learning the model 2 3. The main contribution of this paper is to introduce Replacing-Kernel Reinforcement Learning (RKRL), an online proce-dure for model selection in RL. The chapters of this book span three categories: This chapter surveys recent lines of work that use Bayesian techniques for reinforcement learning. Active Bayesian perception and reinforcement learning Nathan F. Lepora, Uriel Martinez-Hernandez, Giovanni Pezzulo, Tony J. Prescott Abstract—In a series of papers, we have formalized an active Bayesian perception approach for robotics based on recent progress in understanding animal perception. Bayesian Reinforcement Learning. Since µ(s,a)=Q(s,a)and the main quantity that we want to Bayesian Optimal Control of Smoothly Parameterized Systems, Probabilistic machine learning and artificial intelligence, Nonparametric General Reinforcement Learning, Learning in POMDPs with Monte Carlo Tree Search, Robust partially observable Markov decision process, A Conceptual Framework for Externally-influenced Agents: An Assisted Reinforcement Learning Review, Simple trees in complex forests: Growing Take The Best by Approximate Bayesian Computation, A Bayesian Framework for Reinforcement Learning, A Bayesian Sampling Approach to Exploration in Reinforcement Learning, Model-Based Bayesian Reinforcement Learning in Large Structured Domains, PAC-Bayesian Model Selection for Reinforcement Learning, Model-based Bayesian Reinforcement Learning in Partially Observable Domains, An analytic solution to discrete Bayesian reinforcement learning, Multi-task reinforcement learning: a hierarchical Bayesian approach, 2019 International Joint Conference on Neural Networks (IJCNN), View 2 excerpts, cites methods and background, View 2 excerpts, cites background and methods, By clicking accept or continuing to use the site, you agree to the terms outlined in our. In Bayesian learning, uncertainty is expressed by a prior distribution over unknown parameters and learning is achieved by computing a posterior distribution based on the data observed. GU14 0LX. At each step, a distribution over model parameters is maintained. The parameter forms a com-ponent of the POMDP state, which is partially observable and can be inferred based on the history of the observed MDP state/action pairs. Download PDF Abstract: Bayesian methods for machine learning have been widely investigated, yielding principled methods for incorporating prior information into inference algorithms. An Analytic Solution to Discrete Bayesian Reinforcement Learning work. This formulation explicitly represents the uncertainty in the unknown parameter. Model-based Bayesian Reinforcement Learning (BRL) provides a principled solution to dealing with the exploration-exploitation trade-off, but such methods typically assume a fully observable environments. Some features of the site may not work correctly. This removes the main concern that practitioners traditionally have with model-based approaches. A Bayesian Framework for Reinforcement Learning Malcolm Strens MJSTRENS@DERA.GOV.UK Defence Evaluation & Research Agency. Bayesian reinforcement learning Markov decision processes and approximate Bayesian computation Christos Dimitrakakis Chalmers April 16, 2015 Christos Dimitrakakis (Chalmers) Bayesian reinforcement learning April 16, 2015 1 / 60 Bayesian Reinforcement Learning in Continuous POMDPs with Gaussian Processes Patrick Dallaire, Camille Besse, Stephane Ross and Brahim Chaib-draa Abstract—Partially Observable Markov Decision Processes (POMDPs) provide a rich mathematical model to handle real-world sequential decision processes but require a known model Emma Brunskill (CS234 Reinforcement Learning )Lecture 12: Fast Reinforcement Learning 1 Winter 202020/62 Short Refresher / Review on Bayesian Inference: Bernoulli Consider a bandit problem where the reward of an arm is a binary Related Work Learning from expert knowledge is not new. In transfer learning, for example, the decision maker uses prior knowledge obtained from training on task(s) to improve performance on future tasks (Konidaris and Barto [2006]). hierarchical Bayesian models. Model-based Bayesian Reinforcement Learning Introduction Online near myopic value approximation Methods with exploration bonus to achieve PAC Guarantees Ofﬂine value approximation 3. The basics of neural networks: Many traditional machine learning models can be understood as special cases of neural networks. Planning and Learning with Tabular Methods. You can download Reinforcement Learning ebook for free in PDF format (71.9 MB). This book covers both classical and modern models in deep learning. The key aspect of the proposed method is the design of the In Section 6, we discuss how our results carry over to model-basedlearning procedures. Reinforcement learning procedures attempt to maximize the agent’sexpected rewardwhenthe agentdoesnot know 283 and 2 7. 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Also provide a brief tutorial on probabilistic reasoning this paper is to introduce Replacing-Kernel Reinforcement RLparadigm. Inverse Reinforcement Learning of the unknown parameter in partially observable domains, such as the Bayes-Adaptive (.	{"url":"http://bdlisle.com/wp5okk/bayesian-reinforcement-learning-pdf-9b0fdb","timestamp":"2024-11-03T06:12:09Z","content_type":"text/html","content_length":"34193","record_id":"<urn:uuid:ad639927-999a-4898-9fa1-85c62b35f70c>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00236.warc.gz"}
Approximate solution of some types of hypersingular integral equations Title Approximate solution of some types of hypersingular integral equations Authors I. Boikov^1, A. Boikova^1 ^1Penza State University Importance of solving hypersingular integral equations is justified by numerous applications and intense growth of the field during the last century since Hilbert and Poincare created the theory of singular integral equations. The theory is associated with numerous applications of singular and hypersingular integral equations, as well as with Riemann boundary value problem. The Riemann boundary value problem, singular, and hypersingular integral equations are broadly used as basic techniques of mathematical modeling in physics (quantum field theory, theory of Annotation short and long-range interaction, solution theory), theory of elasticity and thermoelasticity, aerodynamics and electrodynamics and many other fields. A closed-form solution of singular and hypersingular integral equations is only possible in exceptional cases. A comprehensive presentation and an extensive literature survey associated with all methods of solution of singular integral equations of the first and second kinds can be found in \cite{Goh,Mich,Boy1,Boy2}. The methods of solution of hypersingular integral equations are less elaborated \cite {Boy3}, \cite{Boy4} . In this paper,we study smoothness of solutions of hypersingular integral equations and their solvability.Also we propose an approach to approximate solving hypersingular integral equations. Using the collocation method and the method of mechanical quadrature each of these problems is approximated with systems of algebraic equations. Keywords Hypersingular integral equations, collocations, mechanical quadratures Boikov I., Boikova A. ''Approximate solution of some types of hypersingular integral equations'' [Electronic resource]. Proceedings of the XIII International scientific conference Citation ''Differential equations and their applications in mathematical modeling''. (Saransk, July 12-16, 2017). Saransk: SVMO Publ, 2017. - pp. 446-461. Available at: https://conf.svmo.ru/files/ deamm2017/papers/paper62.pdf. - Date of access: 12.11.2024.	{"url":"https://conf.svmo.ru/en/archive/article?id=62","timestamp":"2024-11-12T02:56:35Z","content_type":"text/html","content_length":"12286","record_id":"<urn:uuid:cfa623d1-204f-4e20-98cf-f0cde88f054b>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00807.warc.gz"}
string field theory Comment: I am very glad in last month or two Urs is getting so much more back into physics with fruit at very high level :) I have started adding references to string field theory , in particular those by Jim Stasheff et al. on the role of L-infinity algebra and A-infinity algebra. Maybe I find time later to add more I am very glad in last month or two Urs is getting so much more back into physics with fruit at very high level Thanks. I am, too! :-) Maybe it’s clear what the reason is, and what the reason was for being more quiet on physics for a long time: I needed that time, personally, to get some general theory into place. Now that I understand how infinity-Chern-Simons theory (schreiber) follows from “first principles”, I can go back and re-examine what I now understand as examples of this. The Zwiebach $L_\infty$-action for closed string field theory is a potential candidates to fit into this story: the CSFT action looks entirely like it should be an example for an $\infty$ -Chern-Simons theory where the underlying (derived) $L_\infty$-algebra is the one that Zwiebach identifies on the string’s BRST complex, where the invariant polynomial is the binary pairing that he uses, the string correlator. It is a 3-dimensional theory, or rather a $(0\|3)$-dimensional theory, which makes it a bit more exotic: the integration in the action functional is the Berezinian integral over the three string diffeomorphis ghost modes $c_0$, $c_1$, $c_{-1}$. I have to check some details on this, but it looks like this should be true. If so, it would actually make CSFT yet another example of an AKSZ sigma-model. Which would be somewhat remarkable I have added to Chern-Simons element in a new section Properties – canonical CS element the discussion that for an arbitrary $L_\infty$-algebra with quadratic invariant polynomial, the corresponding Chern-Simons element is of the general for as the closed string field theory Lagrangian. started adding something to the Definition-section at string field theory I have added to string field theory in the Definition-section a list of details extracted from Zwiebach’s main article. Then after that I added a detailed proof that his inner product is indeed an $L_\infty$-invariant polynomial. I still need to add more details on the various gradings in Zwiebach’s article. Then after that I added a detailed proof that his inner product is indeed an L ∞-invariant polynomial. Maybe I have to take that back: while it is true that the inner product satisfies the defining equation of an invariant polynomial on the configuration space, I am not sure anymore if it satisfies it on the unconstrained $L_\infty$-algebra. What I mean is: for $\langle-,-\rangle \in W(\mathfrak{g})$ to be an invariant polynomial, we need $d_W \langle-,-\rangle = 0$. It seems I can show that $d_W \langle -,-\rangle$ indeed vanishes when restricted to those fields that Zwiebach allows in the configuration space, but not in general. (All this assuming that I did not otherwise make some mistake with the various gradings and signs. By the nature of this exercise, it is easy to make such mistakes.) I have added some more references on the CSFT tachyon vacuum to String field theory - References - Bosonic CSFT Added to References - Bosonic string field theory - Closed SFT explicit pointers to where exactly one can find written out the mode expansion which shows that the closed string field theory action is an extension of the Einstein-Hilbert action coupled to the B-field and the dilaton. (This is eventually to supplement the discussion at geometry of physics, where I have now decided to discuss Einstein-Yang-Mills theory in the section Chern-Simons-type gauge theories in the • cohesion $\to$ general Chern-Simons-type actions $\to$ closed string field theory $\to$ Einstein-axion theory $\stackrel{KK-reduction}{\to}$ Einstein-Yang-Mills $\to$ standard-model + gravity :-) added a pointer to the recent article by Branislav Jurco on superstring field theory. Since I pointed to the entry string field theory from this PhysicsOverflow reply I went and created two minimum entries such as to un-gray links: While both just contain a reference for the moment, in the first case this is already useful, I’d think: this is the reference that Witten highlighted at String2012 as being crucial but having been kind of missed by the community. Finally added a (lightning brief, for the moment) paragraph on open-closed string field theory here. Added also a remark that it gives “one half” of the axioms of an $\infty$-Lie-Rinehart pair $\mathfrak{g}_{closed} \longrightarrow Der(A_{open}) \,.$ Does one also have the “other half”? Is this discussed anywhere? (I feel like I knew this once, but seem to have forgotten.) updated references on the supersymmetric case. Okawa 16 is a good review of the recent breakthrough in getting the RR-sector under control. One should add more comprehensive references on this, but I don’t have the leisure now diff, v55, current Edited a typo: “bosononic->bosonic” in “bosonic closed string field theory” Alex Arvanitakis diff, v60, current added pointer to today’s • Ivo Sachs, Homotopy Algebras in String Field Theory, Proceedings of LMS/EPSRC Symposium Higher Structures in M-Theory, Fortschritte der Physik 2019 (arXiv:1903.02870) diff, v61, current Prodded by an alert from Jim Stasheff, I have added this recent reference: • Hiroshi Kunitomo, Tatsuya Sugimoto, Heterotic string field theory with cyclic L-infinity structure (arXiv:1902.02991) diff, v62, current added pointer to today’s • Hiroshi Kunitomo, Tatsuya Sugimoto, Type II superstring field theory with cyclic L-infinity structure (arxiv:1911.04103) diff, v64, current added pointer to • Martin Doubek, Branislav Jurco, Korbinian Muenster, Modular operads and the quantum open-closed homotopy algebra (arXiv:1308.3223) diff, v65, current added these pointers: • Harold Erbin, String Field Theory – A Modern Introduction, 2020 (pdf) • Harold Erbin, String theory: a field theory perspective, 2020 (pdf) diff, v67, current added this pointer: • Theodore Erler, Four Lectures on Closed String Field Theory, Physics Reports 851 (2020) 1-36 [arXiv:1905.06785, doi:10.1016/j.physrep.2020.01.003] (what I was really looking for is a modern review that would touch on Witten’s old suggestion of defining the star-product for closed strings, as originally indicated in Fig 20 of his “Non-commutative geometry and string field theory” doi:10.1016/0550-3213(86)90155-0) diff, v71, current added todasy’s arXiv number and publication data to: • Harold Erbin, String Field Theory – A Modern Introduction, Lecture Notes in Physics 980, Springer (2021) [arXiv:2301.01686, doi:10.1007/978-3-030-65321-7] diff, v73, current added pointer to today’s • Carlo Maccaferri, String Field Theory, in: Oxford Research Encyclopedia of Physics [arXiv:2308.00875] diff, v74, current pointer to • Itzhak Bars, Dmitry Rychkov. Background Independent String Field Theory. (2014) (arXiv:1407.4699) diff, v77, current added pointer to today’s • Ashoke Sen, Barton Zwiebach: String Field Theory: A Review, in Handbook of Quantum Gravity, Springer (2023-) [arXiv:2405.19421, doi:10.1007/978-981-19-3079-9] diff, v81, current	{"url":"https://nforum.ncatlab.org/discussion/3124/string-field-theory/?Focus=25873","timestamp":"2024-11-07T22:31:25Z","content_type":"application/xhtml+xml","content_length":"61996","record_id":"<urn:uuid:9fdc52aa-8fd9-481e-841b-fac4360bc2c1>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00475.warc.gz"}
Coupon Rate Formula \| Calculator (Excel Template) Updated July 29, 2023 Coupon Rate Formula (Table of Contents) Coupon Rate Formula A bond/fixed-income security issuer pays the Coupon Rate, which represents the interest rate. The Coupon Rate is a percentage of the bond’s face value at issuance and remains constant until the bond reaches maturity. Once established on the issue date, the bond’s coupon rate remains unaltered throughout its tenure, and the bondholder receives a fixed interest payment at predetermined intervals. The coupon Rate is calculated by dividing the Annual Coupon Payment by the Face Value of the Bond. The result is expressed in percentage form. The formula for Coupon Rate – Coupon Rate = (Annual Coupon (or Interest) Payment / Face Value of Bond) * 100 Below are the steps to calculate the Coupon Rate of a bond: Step 1: In the first step, the amount required to be raised through bonds is decided by the company, then based on the target investors (i.e. retail or institutional or both) and other parameters, face value or par value is determined as a result, of which, we get to know the number of bonds that will be issued. Step 2: In the second step, we determine the amount of interest and the payment frequency, then calculate the total annual interest payment by multiplying the interest amount by the payment Step 3: In the final step, you divide the amount of interest paid yearly by the face value of a bond to calculate the coupon rate. Examples of Coupon Rate Formula (With Excel Template) Let’s take an example to understand the calculation of the Coupon Rate formula in a better manner. Example #1 Company ABC issued a bond of Rs. 100 Face Value and Rs. 10 as half-yearly interest. The formula to calculate Annual Interest Payment is as below: Annual Interest Payment = Amount of Interest * Frequency of Payment • Annual Interest Payment = 10 * 2 • Annual Interest Payment = Rs. 20 The formula to calculate Coupon Rate is as below: Coupon Rate = (Annual Coupon (or Interest) Payment / Face Value of Bond) * 100 • Coupon Rate = (20 / 100) * 100 • Coupon Rate = 20% Now, if the market interest rate is lower than 20%, the bond will be traded at a premium as this bond gives investors more value than other fixed-income securities. However, if the market rate of interest is higher than 20%, then the bond will be traded at a discount. Example #2 L&T Finance issued secured NCDs in March 2019. Following are the details of the issue: • NCD Issue Open: 06 March 2019 • NCD Issue Close: 07 March 2019 • NCD Issue Size: Rs.1500 Crore • Price Band/Face Value/Issue Price: Rs.1000 • NCD’s: 15,000,000 of Rs.1000 Each • Listing: BSE, NSE • Credit Rating: IndRA AA/Stable, CARE AA/ Stable, ICRA AA/Stable • Interest Payment: Rs. 7.225 • Frequency of Payment: Monthly Annual Interest Payment = Amount of Interest * Frequency of Payment • Annual Interest Payment = 7.225 * 12 • Annual Interest Payment = Rs. 86.7 Coupon Rate = (Annual Coupon (or Interest) Payment / Face Value of Bond) * 100 • Coupon Rate = (86.7 / 1000) * 100 • Coupon Rate= 8.67% Example #3 Tata Capital Financial Services Ltd. Issued secured and unsecured NCDs in Sept 2018. Details of the issue are as following: • NCD Issue Open: 10 Sept 2018 • NCD Issue Close: 21 Sept 2018 • NCD Issue Size: Rs. 2000 Cr with an option to retain oversubscription up to limit of Rs. 7,500 Cr • Price Band/Face Value/Issue Price: Rs.1000 • NCD’s: 2,00,00,000 of Rs.1000 Each • Listing: BSE, NSE • Credit Rating: CRISIL AAA/Stable, CARE AAA/ Stable • Interest Payment □ For Secured NCD: Rs. 89 □ For Unsecured NCD: Rs. 91 • Frequency of Payment: Annual Annual Interest Payment = Amount of Interest * Frequency of Payment For Secured NCDs • Annual Interest Payment = 89 * 1 • Annual Interest Payment = Rs. 89 For Unsecured NCDs • Annual Interest Payment = 91 * 1 • Annual Interest Payment = Rs. 91 Coupon Rate = (Annual Coupon (or Interest) Payment / Face Value of Bond) * 100 For Secured NCDs • Coupon Rate = (89 / 1000) * 100 • Coupon Rate= 8.9% For Unsecured NCDs • Coupon Rate = (91 / 1000) * 100 • Coupon Rate= 9.1% As we know, an investor expects a higher return for investing in a higher-risk asset. Hence, as we can witness in the above example, the unsecured NCD of Tata Capital fetches higher returns than the secured NCD. The coupon rate of a bond is determined after considering various factors. Still, two of the key factors are the interest rates of different fixed-income security available in the market at the time of the bond issue and the company’s creditworthiness. A bond’s coupon rate is determined so that it remains competitive with other available fixed-income securities. However, the coupon rate of newly issued fixed-income securities may increase or decrease during the tenure of a bond based on market conditions, which results in a change in the bond’s market value. The market value of a bond is a derivation of the difference in the coupon rate of the bond and the market interest rate of other fixed-income securities. If a bond’s interest rate is below the market interest rate, the bond is said to be traded at a discount. In contrast, if the bond’s interest rate is higher than the market interest rate, the bond is said to be traded at a premium. Similarly, a bond is said to be traded at par if the bond’s interest rate is equal to the market interest rate. The coupon rate is also depending on the creditworthiness of the company. Companies must undertake a credit rating of the bond from a credit rating agency before issuing the bond. Credit rating agencies assign a credit rating to the bond issue after assessing the issuer on various parameters, the riskiness of the company’s business, financial stability, legal history, default history, ability to repay money borrowed through the bond, etc. The credit Rating hierarchy starts from AAA and goes up to D, with ‘AAA’ being most safe and ‘D’ being Default. Generally, bonds with a credit rating of ‘BBB and above are considered investment grade. A higher bond rating means higher safety and hence a lower coupon rate and vice versa. Relevance and Uses The coupon Rate Formula helps calculate and compare the coupon rate of different fixed-income securities and helps choose the best as per the requirement of an investor. It also helps assess the cycle of interest rates and the expected market value of a bond. For eg. If market interest rates are declining, the market value of bonds with higher interest rates will increase, resulting in higher yield and hence higher return on investment and vice versa in an increasing market interest rate scenario. Coupon Rate Formula Calculator You can use the following Coupon Rate Calculator Annual Coupon (or Interest) Payment Face Value of Bond Coupon Rate Formula Annual Coupon (or Interest) Payment x 100 Coupon Rate Formula = Face Value of Bond Recommended Articles This has been a guide to Coupon Rate Formula. Here we discuss How to Calculate Coupon Rates along with practical examples. We also provide Coupon Rate Calculator with a downloadable excel template. You may also look at the following articles to learn more –	{"url":"https://www.educba.com/coupon-rate-formula/","timestamp":"2024-11-02T04:53:31Z","content_type":"text/html","content_length":"345348","record_id":"<urn:uuid:5d06e46f-b648-4683-beeb-fdb1c71a4fde>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00816.warc.gz"}
Normal Distribution CI IQ tests are designed to yield results that are approximately Normally distributed. Researchers... Normal Distribution CI IQ tests are designed to yield results that are approximately Normally distributed. Researchers... Normal Distribution CI IQ tests are designed to yield results that are approximately Normally distributed. Researchers think believe that the standard deviation, σ, is 15. A reporter is interested in estimating the average IQ of employees in a large high-tech firm in California. She gathers the IQ information on 22 employees of this firm and recoreds the sample mean IQ as 106. Let X represent a random variable describing a person's IQ: X~N(µ, 15). a. Find the standard error of the sample mean. b. Calculate a 90% confidence interval. c. Interpret the confidence interval in the context of the problem.	{"url":"https://justaaa.com/statistics-and-probability/1303749-normal-distribution-ci-iq-tests-are-designed-to","timestamp":"2024-11-11T06:57:18Z","content_type":"text/html","content_length":"36973","record_id":"<urn:uuid:1a6fe77a-94f7-4cda-84ec-14b77b9b059e>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00224.warc.gz"}
I don't know the macro syntax but something like (defmacro time-with-str "Evaluates expr and prints the time it took. Returns the value of {:added "1.0"} [my-str expr] `(let [start# (. System (nanoTime)) ret# ~expr] (prn (str my-str " Elapsed time: " (/ (double (- (. System (nanoTime)) start#)) 1000000.0) " msecs"))	{"url":"https://clojurians-log.clojureverse.org/beginners/2019-01-10","timestamp":"2024-11-06T05:14:29Z","content_type":"text/html","content_length":"347927","record_id":"<urn:uuid:e2c9cee2-6388-4a82-bac4-c93e5dbc6742>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00533.warc.gz"}
Count the Insects \| Riddles360 A time long back, there lived a king who ruled the great kingdom of Trojan House. As a part of the renovation of the kingdom to meet future security needs, he asked his chief architect to lay down a new play in a manner that all of his 10 castles are connected through five straight walls and each wall must connect four castles together. He also asked the architect that at least one of his castles should be protected with walls. The architect could not come up with any solution that served all of King's choices, but he suggested the best plan that you can see in the picture below. Can you find a better solution to serve the king's demand?	{"url":"https://riddles360.com/riddle/count-the-insects?discuss=1","timestamp":"2024-11-14T12:12:36Z","content_type":"text/html","content_length":"44510","record_id":"<urn:uuid:98e555bf-43c0-4bf0-95d2-5c73ecf891b3>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00769.warc.gz"}
Time Series Anomaly Detection using LSTM Autoencoders with PyTorch in Python TL;DR Use real-world Electrocardiogram (ECG) data to detect anomalies in a patient heartbeat. We’ll build an LSTM Autoencoder, train it on a set of normal heartbeats and classify unseen examples as normal or anomalies In this tutorial, you’ll learn how to detect anomalies in Time Series data using an LSTM Autoencoder. You’re going to use real-world ECG data from a single patient with heart disease to detect abnormal hearbeats. By the end of this tutorial, you’ll learn how to: • Prepare a dataset for Anomaly Detection from Time Series Data • Build an LSTM Autoencoder with PyTorch • Train and evaluate your model • Choose a threshold for anomaly detection • Classify unseen examples as normal or anomaly The dataset contains 5,000 Time Series examples (obtained with ECG) with 140 timesteps. Each sequence corresponds to a single heartbeat from a single patient with congestive heart failure. An electrocardiogram (ECG or EKG) is a test that checks how your heart is functioning by measuring the electrical activity of the heart. With each heart beat, an electrical impulse (or wave) travels through your heart. This wave causes the muscle to squeeze and pump blood from the heart. Source We have 5 types of hearbeats (classes): • Normal (N) • R-on-T Premature Ventricular Contraction (R-on-T PVC) • Premature Ventricular Contraction (PVC) • Supra-ventricular Premature or Ectopic Beat (SP or EB) • Unclassified Beat (UB). Assuming a healthy heart and a typical rate of 70 to 75 beats per minute, each cardiac cycle, or heartbeat, takes about 0.8 seconds to complete the cycle. Frequency: 60–100 per minute (Humans) Duration: 0.6–1 second (Humans) Source The dataset is available on my Google Drive. Let’s get it: 1!gdown --id 16MIleqoIr1vYxlGk4GKnGmrsCPuWkkpT 1device = torch.device("cuda" if torch.cuda.is_available() else "cpu") The data comes in multiple formats. We’ll load the arff files into Pandas data frames: 1with open('ECG5000_TRAIN.arff') as f: 2 train = a2p.load(f) 4with open('ECG5000_TEST.arff') as f: 5 test = a2p.load(f) We’ll combine the training and test data into a single data frame. This will give us more data to train our Autoencoder. We’ll also shuffle it: 1df = train.append(test) 2df = df.sample(frac=1.0) We have 5,000 examples. Each row represents a single heartbeat record. Let’s name the possible classes: 1CLASS_NORMAL = 1 3class_names = ['Normal','R on T','PVC','SP','UB'] Next, we’ll rename the last column to target, so its easier to reference it: 1new_columns = list(df.columns) 2new_columns[-1] = 'target' 3df.columns = new_columns Exploratory Data Analysis Let’s check how many examples for each heartbeat class do we have: 6Name: target, dtype: int64 Let’s plot the results: The normal class, has by far, the most examples. This is great because we’ll use it to train our model. Let’s have a look at an averaged (smoothed out with one standard deviation on top and bottom of it) Time Series for each class: It is very good that the normal class has a distinctly different pattern than all other classes. Maybe our model will be able to detect anomalies? LSTM Autoencoder The Autoencoder’s job is to get some input data, pass it through the model, and obtain a reconstruction of the input. The reconstruction should match the input as much as possible. The trick is to use a small number of parameters, so your model learns a compressed representation of the data. In a sense, Autoencoders try to learn only the most important features (compressed version) of the data. Here, we’ll have a look at how to feed Time Series data to an Autoencoder. We’ll use a couple of LSTM layers (hence the LSTM Autoencoder) to capture the temporal dependencies of the data. To classify a sequence as normal or an anomaly, we’ll pick a threshold above which a heartbeat is considered abnormal. Reconstruction Loss When training an Autoencoder, the objective is to reconstruct the input as best as possible. This is done by minimizing a loss function (just like in supervised learning). This function is known as reconstruction loss. Cross-entropy loss and Mean squared error are common examples. Anomaly Detection in ECG Data We’ll use normal heartbeats as training data for our model and record the reconstruction loss. But first, we need to prepare the data: Data Preprocessing Let’s get all normal heartbeats and drop the target (class) column: 1normal_df = df[df.target == str(CLASS_NORMAL)].drop(labels='target', axis=1) We’ll merge all other classes and mark them as anomalies: 1anomaly_df = df[df.target != str(CLASS_NORMAL)].drop(labels='target', axis=1) We’ll split the normal examples into train, validation and test sets: 1train_df, val_df = train_test_split( 2 normal_df, 3 test_size=0.15, 4 random_state=RANDOM_SEED 7val_df, test_df = train_test_split( 8 val_df, 9 test_size=0.33, 10 random_state=RANDOM_SEED We need to convert our examples into tensors, so we can use them to train our Autoencoder. Let’s write a helper function for that: 1def create_dataset(df): 3 sequences = df.astype(np.float32).to_numpy().tolist() 5 dataset = [torch.tensor(s).unsqueeze(1).float() for s in sequences] 7 n_seq, seq_len, n_features = torch.stack(dataset).shape 9 return dataset, seq_len, n_features Each Time Series will be converted to a 2D Tensor in the shape sequence length x number of features (140x1 in our case). Let’s create some datasets: 1train_dataset, seq_len, n_features = create_dataset(train_df) 2val_dataset, _, _ = create_dataset(val_df) 3test_normal_dataset, _, _ = create_dataset(test_df) 4test_anomaly_dataset, _, _ = create_dataset(anomaly_df) LSTM Autoencoder Sample Autoencoder Architecture Image Source The general Autoencoder architecture consists of two components. An Encoder that compresses the input and a Decoder that tries to reconstruct it. We’ll use the LSTM Autoencoder from this GitHub repo with some small tweaks. Our model’s job is to reconstruct Time Series data. Let’s start with the Encoder: 1class Encoder(nn.Module): 3 def __init__(self, seq_len, n_features, embedding_dim=64): 4 super(Encoder, self).__init__() 6 self.seq_len, self.n_features = seq_len, n_features 7 self.embedding_dim, self.hidden_dim = embedding_dim, 2 * embedding_dim 9 self.rnn1 = nn.LSTM( 10 input_size=n_features, 11 hidden_size=self.hidden_dim, 12 num_layers=1, 13 batch_first=True 14 ) 16 self.rnn2 = nn.LSTM( 17 input_size=self.hidden_dim, 18 hidden_size=embedding_dim, 19 num_layers=1, 20 batch_first=True 21 ) 23 def forward(self, x): 24 x = x.reshape((1, self.seq_len, self.n_features)) 26 x, (_, _) = self.rnn1(x) 27 x, (hidden_n, _) = self.rnn2(x) 29 return hidden_n.reshape((self.n_features, self.embedding_dim)) The Encoder uses two LSTM layers to compress the Time Series data input. Next, we’ll decode the compressed representation using a Decoder: 1class Decoder(nn.Module): 3 def __init__(self, seq_len, input_dim=64, n_features=1): 4 super(Decoder, self).__init__() 6 self.seq_len, self.input_dim = seq_len, input_dim 7 self.hidden_dim, self.n_features = 2 * input_dim, n_features 9 self.rnn1 = nn.LSTM( 10 input_size=input_dim, 11 hidden_size=input_dim, 12 num_layers=1, 13 batch_first=True 14 ) 16 self.rnn2 = nn.LSTM( 17 input_size=input_dim, 18 hidden_size=self.hidden_dim, 19 num_layers=1, 20 batch_first=True 21 ) 23 self.output_layer = nn.Linear(self.hidden_dim, n_features) 25 def forward(self, x): 26 x = x.repeat(self.seq_len, self.n_features) 27 x = x.reshape((self.n_features, self.seq_len, self.input_dim)) 29 x, (hidden_n, cell_n) = self.rnn1(x) 30 x, (hidden_n, cell_n) = self.rnn2(x) 31 x = x.reshape((self.seq_len, self.hidden_dim)) 33 return self.output_layer(x) Our Decoder contains two LSTM layers and an output layer that gives the final reconstruction. Time to wrap everything into an easy to use module: 1class RecurrentAutoencoder(nn.Module): 3 def __init__(self, seq_len, n_features, embedding_dim=64): 4 super(RecurrentAutoencoder, self).__init__() 6 self.encoder = Encoder(seq_len, n_features, embedding_dim).to(device) 7 self.decoder = Decoder(seq_len, embedding_dim, n_features).to(device) 9 def forward(self, x): 10 x = self.encoder(x) 11 x = self.decoder(x) 13 return x Our Autoencoder passes the input through the Encoder and Decoder. Let’s create an instance of it: 1model = RecurrentAutoencoder(seq_len, n_features, 128) 2model = model.to(device) Let’s write a helper function for our training process: 1def train_model(model, train_dataset, val_dataset, n_epochs): 2 optimizer = torch.optim.Adam(model.parameters(), lr=1e-3) 3 criterion = nn.L1Loss(reduction='sum').to(device) 4 history = dict(train=[], val=[]) 6 best_model_wts = copy.deepcopy(model.state_dict()) 7 best_loss = 10000.0 9 for epoch in range(1, n_epochs + 1): 10 model = model.train() 12 train_losses = [] 13 for seq_true in train_dataset: 14 optimizer.zero_grad() 16 seq_true = seq_true.to(device) 17 seq_pred = model(seq_true) 19 loss = criterion(seq_pred, seq_true) 21 loss.backward() 22 optimizer.step() 24 train_losses.append(loss.item()) 26 val_losses = [] 27 model = model.eval() 28 with torch.no_grad(): 29 for seq_true in val_dataset: 31 seq_true = seq_true.to(device) 32 seq_pred = model(seq_true) 34 loss = criterion(seq_pred, seq_true) 35 val_losses.append(loss.item()) 37 train_loss = np.mean(train_losses) 38 val_loss = np.mean(val_losses) 40 history['train'].append(train_loss) 41 history['val'].append(val_loss) 43 if val_loss < best_loss: 44 best_loss = val_loss 45 best_model_wts = copy.deepcopy(model.state_dict()) 47 print(f'Epoch {epoch}: train loss {train_loss} val loss {val_loss}') 49 model.load_state_dict(best_model_wts) 50 return model.eval(), history At each epoch, the training process feeds our model with all training examples and evaluates the performance on the validation set. Note that we’re using a batch size of 1 (our model sees only 1 sequence at a time). We also record the training and validation set losses during the process. Note that we’re minimizing the L1Loss, which measures the MAE (mean absolute error). Why? The reconstructions seem to be better than with MSE (mean squared error). We’ll get the version of the model with the smallest validation error. Let’s do some training: 1model, history = train_model( 2 model, 3 train_dataset, 4 val_dataset, 5 n_epochs=150 Our model converged quite well. Seems like we might’ve needed a larger validation set to smoothen the results, but that’ll do for now. Saving the model Let’s store the model for later use: 1MODEL_PATH = 'model.pth' 3torch.save(model, MODEL_PATH) Uncomment the next lines, if you want to download and load the pre-trained model: 1# !gdown --id 1jEYx5wGsb7Ix8cZAw3l5p5pOwHs3_I9A 2# model = torch.load('model.pth') 3# model = model.to(device) Choosing a threshold With our model at hand, we can have a look at the reconstruction error on the training set. Let’s start by writing a helper function to get predictions from our model: 1def predict(model, dataset): 2 predictions, losses = [], [] 3 criterion = nn.L1Loss(reduction='sum').to(device) 4 with torch.no_grad(): 5 model = model.eval() 6 for seq_true in dataset: 7 seq_true = seq_true.to(device) 8 seq_pred = model(seq_true) 10 loss = criterion(seq_pred, seq_true) 12 predictions.append(seq_pred.cpu().numpy().flatten()) 13 losses.append(loss.item()) 14 return predictions, losses Our function goes through each example in the dataset and records the predictions and losses. Let’s get the losses and have a look at them: 1_, losses = predict(model, train_dataset) 3sns.distplot(losses, bins=50, kde=True); Using the threshold, we can turn the problem into a simple binary classification task: • If the reconstruction loss for an example is below the threshold, we’ll classify it as a normal heartbeat • Alternatively, if the loss is higher than the threshold, we’ll classify it as an anomaly Normal hearbeats Let’s check how well our model does on normal heartbeats. We’ll use the normal heartbeats from the test set (our model haven’t seen those): 1predictions, pred_losses = predict(model, test_normal_dataset) 2sns.distplot(pred_losses, bins=50, kde=True); We’ll count the correct predictions: 1correct = sum(l <= THRESHOLD for l in pred_losses) 2print(f'Correct normal predictions: {correct}/{len(test_normal_dataset)}') 1Correct normal predictions: 142/145 We’ll do the same with the anomaly examples, but their number is much higher. We’ll get a subset that has the same size as the normal heartbeats: 1anomaly_dataset = test_anomaly_dataset[:len(test_normal_dataset)] Now we can take the predictions of our model for the subset of anomalies: 1predictions, pred_losses = predict(model, anomaly_dataset) 2sns.distplot(pred_losses, bins=50, kde=True); Finally, we can count the number of examples above the threshold (considered as anomalies): 1correct = sum(l > THRESHOLD for l in pred_losses) 2print(f'Correct anomaly predictions: {correct}/{len(anomaly_dataset)}') 1Correct anomaly predictions: 142/145 We have very good results. In the real world, you can tweak the threshold depending on what kind of errors you want to tolerate. In this case, you might want to have more false positives (normal heartbeats considered as anomalies) than false negatives (anomalies considered as normal). Looking at Examples We can overlay the real and reconstructed Time Series values to see how close they are. We’ll do it for some normal and anomaly cases: In this tutorial, you learned how to create an LSTM Autoencoder with PyTorch and use it to detect heartbeat anomalies in ECG data. You learned how to: • Prepare a dataset for Anomaly Detection from Time Series Data • Build an LSTM Autoencoder with PyTorch • Train and evaluate your model • Choose a threshold for anomaly detection • Classify unseen examples as normal or anomaly While our Time Series data is univariate (we have only 1 feature), the code should work for multivariate datasets (multiple features) with little or no modification. Feel free to try it!	{"url":"https://curiousily.com/posts/time-series-anomaly-detection-using-lstm-autoencoder-with-pytorch-in-python/","timestamp":"2024-11-06T02:35:20Z","content_type":"text/html","content_length":"384907","record_id":"<urn:uuid:31029bdd-a2d6-4110-a1ce-cfd3f0d4ff7e>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00274.warc.gz"}
Playful Invitations: A Tour Posted on January 25, 2021 by Vanessa Rivera-Quinones Playful Invitations: Inspiring Ways to Teach Early Mathematics, is a blog written by Dorie Ranheim. Its goal is “to inspire parents, caregivers, and educators of preschool children to intentionally teach math using natural materials.” By using “loose parts”, backyards, playgrounds, and parks become great places for teaching and learning math. As described in the blog’s about page, “My playful invitiations to learn math eventually extended beyond our home to our backyard and nearby park. During our time outdoors I realized I could showcase the beauty of real, natural materials and how inspiring, meaningful, and relatively easy they are to acquire. Overall, I hope to share ways adults can intentionally teach preschool math using these beautiful natural materials.” – Dorie Ranheim In this tour, I will summarize some of its most recent posts. Many of them can be used as guided activities, and Ranheim provides a helpful guide on how to use her blog posts. One of the aspects I appreciate about the activities is that they are all centered around play. Many of the posts consist of three parts: Prepare, Invite, and Play, and some include reflections and extensions to the activities. As she remarks, “The blog posts are simply suggestions. There are MANY ways to develop these math skills. My hope is that reading the blog will inspire you to find opportunities in your daily life to teach math to preschool children.If it is playful, meaningful, and reasonably challenging then chances are the learning will stick!” In this post, Ranheim shares some of the ways that during last Spring, she and her children became more entune with nature and spent some time thinking of long-term projects. We watched the bare branches bud and blossom, now we celebrate trees bursting with bright green leaves. […] Here are a few ways my children have explored math during this time at home: Practice number identification and formation using loose parts. Write numbers on river stones using water and brushes. Pattern using various colored rocks. Sometimes the simplest activities and materials seem to hold their attention the longest! I found it related a lot to what I’ve done since the quarantine. In a very similar what, I discovered patterns in one of my hikes. Invite: Today I thought we could trace our bodies so we can see how big we are. (After tracing one or more bodies) I wonder whose body is the tallest/longest? This invitation is inspired by the following quote, To compare objects, children begin by using nonstandard units (“My table is more than four hands long”) and then move to using standard units (“The table is almost three feet long”). Comparing fairly is an important concept for young children. – Juanita Copley, 2010 This made a lot of sense to me! Once you learn the standard units of measurement it’s easy to forget all the other intuitive ways we make sense of measurements. If anyone has ever tried to learn a family recipe, you’ve probably encountered many non-standard ways of measuring yourself. In this activity, each child lays on the ground and draw a chalk outline of their body, afterwards they use natural loose objects of similar size (e.g. leaves, rocks, etc.) to lay them side by side and compare the lengths. Some fun extensions include introducing rulers, or filling the outline of the body, and talking about the area. Invite: “I’m trying to put this leaf back together! Will you help me find the perfect match to make my leaf complete?” As a big fan of tangrams as a kid, I love that in this invitation, you introduce the idea of fractions at a basic level by transforming leaves into puzzles. It is a great way for children to play with the idea of a “whole”. You can start by cutting leaves in half and trying to find the match, or you can extend the activity by adding more different types of leaves or cutting them in four pieces instead. Ranheim advises that before using the leaves for learning math, they should explore their properties. “It would also be beneficial if the child has explored the property of leaves before being asked to use them for math learning. Observe a small pile of leaves and the attributes before taking them apart.” A personal project that has brought me great joy during the pandemic, has been to gather seeds and start my balcony garden. I could see all the fascinating (and often subtle) ways math seeped into my gardening. What drew my attention to this blog, were my conversations with my best friend and early-childhood educator, about the ways math should be tied into how we experience nature around us. I would have loved these activities as a kid! Also, it reminded me of one of my favorite books “Braiding Sweetgrass: Indigenous Wisdom, Scientific, Knowledge, and the Teaching of Plants” by Robin Wall-Kimmerer (I can’t recommend it enough). “The land is the real teacher. All we need as students is mindfulness.” – Robin Wall Kimmerer, Braiding Sweetgrass Have an idea for a topic or a blog you would like for me and Rachel to cover in upcoming posts? Reach out in the comments below or on Twitter (@VRiveraQPhD). About Vanessa Rivera-Quinones Mathematics Ph.D. with a passion for telling stories through numbers using mathematical models, data science, science communication, and education. Follow her on Twitter: @VRiveraQPhD. This entry was posted in Blogs, Math Education, planet math, Recreational Mathematics, Sustainability, Uncategorized and tagged Blog, Dorie Ranheim, early childhood, math, nature, Playful Invitations . Bookmark the permalink.	{"url":"https://blogs.ams.org/blogonmathblogs/2021/01/25/playful-invitations-a-tour/","timestamp":"2024-11-04T21:41:31Z","content_type":"text/html","content_length":"60090","record_id":"<urn:uuid:7edec84b-ec07-4fec-8247-62505a0edd4e>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00494.warc.gz"}
ICICI Bank Home Loan EMI Calculator Online 2024 ICICI Home Loan Calculator ICICI home loan or housing loan is one of the popular financial products offered by the lender. The EMI for the loan starts from Rs. 645 per lakh and is provided to both salaried and self-employed individuals. One can calculate the monthly EMI on the same by using an ICICI home loan calculator. A home loan is a form of credit that is advanced by banks and Non-banking financial companies to borrowers, who then employ it to purchase or construct a housing property. ICICI Home Loan Calculator – Key Features • The lender provides a loan to both salaried and self-employed. Also, special rates offered to women, senior citizen and NRI borrowers. • Individuals from 21 and 60 years can apply for a loan • For security one can keep a mortgage of property that they are planning to purchase, renovate or construct. • Loan to Value Ratio is 80%. • The tenure of the loan ranges from 5 to 30 years. • Interest rates range from 6.70% to 7.95%. • ICICI home loan EMI per lakh starts is Rs. 645 (starting price). However, one can calculate the home loan EMI anytime using the ICICI home loan calculator. How Does an ICICI Home Loan Calculator Work? Calculation of EMI is manually carried out using the following formula – EMI = {P * R * (1 + R)^n} / {(1 + R)^(n – 1)} • P is the principal amount of loan availed. • R is the rate of interest. • n is the repayment tenure in months. An ICICI Bank home loan EMI calculator integrates the formula mentioned above to calculate EMIs. Example: Mr Ajay has applied for a home loan involving a grant amount of Rs. 25 Lakh for repayment tenure of 240 months or 20 years on 1.10.19. It attracts an interest rate of 9.5% or 0.95. The following would be his EMI calculation. EMI = Rs. {2500000 * 0.095 * ( 1 + 0.095)^240} / {(1 + 0.095)^(240 – 1)} Or, EMI = Rs. 23,303. You can easily calculate EMIs using an ICICI Bank home loan calculator which eliminates the hassle of manual calculation. How Can a Home Loan Calculator From ICICI Help You? ICICI is a premier financial institution and one of the Big 4 Banks in India. Since its establishment as a fully-owned subsidiary in 1994, it has been known to provide premium service to its The demand for home loans in India has risen significantly after the implementation of Pradhan Mantri Awas Yojana (PMAY). Under its CLSS scheme, first-time borrowers are eligible to receive subsidies on interest rates for a housing loan. ICICI, in this respect, has played an instrumental role owing to its affordable and competitive interest rates which starts from 9.05%. The following are the advantages of using an ICICI home loan EMI calculator – 1. Accuracy: It provides an accurate figure which you need to pay as EMI depending on factors such as the loan corpus, repayment tenure, and interest rate applicable. 2. Helps in decision making: You can easily decide on the loan amount and tenure, which will offer you maximum financial convenience during repayment. The ICICI housing loan calculator will help you reach a decision faster and without hassle. 3. Mobility: An ICICI bank home loan EMI calculator will enable you to reckon your EMI across any device – cell phone, tablet, personal computer, etc. Therefore, you can calculate your EMI anywhere, at any time. Amortisation Schedule An amortisation schedule for a home loan is a table on which the following specifics are mentioned – 1. Equated Monthly Instalment. 2. The interest component of each EMI. 3. The principal component of EMI. 4. Outstanding amount prior to each EMI payment. 5. Outstanding amount post every EMI payment. The amortisation schedule for the example mentioned above is demonstrated using the following table. Months Interest (Rs.) Principal (Rs.) EMI (Rs.) Outstanding Balance 1 19,792 3,512 23,303 24,96,488 2 19,764 3,539 23,303 24,92,949 3 19,736 3,567 23,303 24,89,382 4 19,708 3,596 23,303 24,85,786 5 19,679 3,624 23,303 24,82,162 6 19,650 3,653 23,303 24,78,509 7 19,622 3,682 23,303 24,74,827 8 19,592 3,711 23,303 24,71,116 9 19,563 3,740 23,303 24,67,376 10 19,533 3,770 23,303 24,63,606 11 19,504 3,800 23,303 24,59,806 12 19,473 3,830 23,303 24,55,977 13 19,443 3,860 23,303 24,52,116 14 19,413 3,891 23,303 24,48,226 15 19,382 3,921 23,303 24,44,304 As is substantiated in the above table, the interest component is significantly higher than the principal component in the initial years. Benefits of Groww ICICI Home Loan Calculator There are few advantages to using the Groww ICICI bank home loan calculator. These are – 1. Free: The Groww home loan EMI calculator is free to use. You do not need to register to Groww’s website to use it. 2. Use multiple times: There are no restrictions on the usage frequency. Therefore, you can easily compare the various components to determine the ideal tenure and loan amount at your convenience. An ICICI Home loan calculator or any EMI calculator can extensively benefit individuals to make the right decisions. Hence, using it becomes paramount when availing of a home loan. “Looking to invest? Open an account with Groww and start investing in direct Mutual Funds for free”	{"url":"https://groww.in/calculators/icici-home-loan-emi-calculator","timestamp":"2024-11-03T22:34:40Z","content_type":"text/html","content_length":"77026","record_id":"<urn:uuid:143fbcf3-81e8-4fa9-b305-420dcbdb486a>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00895.warc.gz"}
One last comment related to 12-11-2012 07:13 AM Hello All, I am moving a legacy code from Linux to Windows that uses FFTW 2.1.5 and so I have created and successfully linked to MKL's FFTW wrappers. My question however is about some of the wrappers functionality with respect to a 3 dimensional FFT, specifically the wrapper function fftwnd_mpi_local_sizes(). Show below is the original FFTW output and the MKL wrapper output. fftwnd_mpi_local_sizes(fftwnd_mpi_plan p, int local_nx -> int CDFT_LOCAL_NX, int local_x_start -> int CDFT_LOCAL_X_START, int local_ny_after_transpose -> int CDFT_LOCAL_OUT_NX, int local_y_start_after_transpose -> int CDFT_LOCAL_OUT_X_START int total_local_size -> int CDFT_LOCAL_SIZE) Local_ny_after_transpose and local_y_start_after_transpose are not being set to the information that is expected in the original FFTW implementation. Our layout and data allocation for the mpi processes heavily rely on the original output. After looking over the MKL documentation it appears that this is all MKL's FFT can give, unfortunately the Y values are critical. An example of the problem is if I have a 36 by 16 by 14 X,Y,Z transform over 2 processors, FFTW output is expected to be processor_1(plan,18,0,8,0,4032) processor_2(plan,18,18,8,8,4032) but MKL will output processor_1(plan,18,0,18,0,4032) processor_2(plan,18,18,18,18,4032). This example may be predictable but the sizes of X,Y,Z are arbitrary and so is the number of processors so it no longer becomes very predictable. Are there any solutions to this problem? -Thank you all, 12-11-2012 05:55 PM 12-12-2012 09:23 AM 12-16-2012 05:11 PM 12-17-2012 12:19 PM 12-21-2012 09:29 AM 12-21-2012 11:19 AM	{"url":"https://community.intel.com/t5/Intel-oneAPI-Math-Kernel-Library/Calling-local-sizes-of-3D-MPI-FFT-plans/m-p/970818/highlight/true?attachment-id=50977","timestamp":"2024-11-09T16:56:27Z","content_type":"text/html","content_length":"297702","record_id":"<urn:uuid:203a0df2-7a73-41bd-9602-bd310549be38>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00720.warc.gz"}
MAT128A: Numerical Analysis Homework 2 1. What is the condition number of evaluation of the function f(x) = exp(cos(x)) at the point x? 2. Suppose that f and g are continuously differentiable functions R → R. Let κf (x) denote the condition number of evaluation of the function f at x. Find an expression for the condition number of evaluation of the function h(x) = f(g(x)) at x in terms of κf (g(x)) and g 3. Suppose that f and g are continuously differentiable functions R → R. Let κf (x) denote the condition number of evaluation of the function f at x, and let Let κg(x) denote the condition number of evaluation of the function g at x. Find an expression for the condition number of evaluation of the function h(x) = f(x) · g(x) at x in terms of κf (x) and κg(x). 4. Suppose that f and g are continuously differentiable functions R → R. Let κf (x) denote the condition number of evaluation of the function f at x, and let Let κg(x) denote the condition number of evaluation of the function g at x. Find an expression for the condition number of evaluation of the function h(x) = f(x)/g(x) at x in terms of κf (x) and κg(x). 5. What is the Fourier series of the function f(x) = x? 6. What is the Fourier series of the function f(x) = \|x\|? (Hint: you can easily find an antiderivative of x exp(inx) using integration by parts).	{"url":"https://codingprolab.com/answer/mat128a-numerical-analysis-homework-2/","timestamp":"2024-11-02T15:50:06Z","content_type":"text/html","content_length":"104377","record_id":"<urn:uuid:323e55a9-eaec-4155-92bc-da69ac37d353>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00256.warc.gz"}
UCT MAM1000 lecture notes part 31 - complex numbers part ix UCT MAM1000 lecture notes part 31 – complex numbers part ix When we were playing around with partial fractions we appeared to make a bit of an assumption which was that the only forms that we had to deal with in the denominator of a fraction could always be written as a factor of either linear parts ($a+b x$) or quadratic parts which we could not factor into linear parts ($ax^2+bx+x$) where $b^2-4ac<0$, and of course multiple powers of these, for instance we could have terms like $(a+bx)^3$ in the denominator. How do we know that we can always split a polynomial up into these factors where the coefficients are real? Couldn’t it be for instance that if I gave you a cubic polynomial that all the roots were complex and so I couldn’t factor it in a way that every factor came out with real coefficients? It turns out that the answer is no, but we need a couple more ingredients to prove this. We said that we were dealing with ratios of polynomials, eg. $\frac{P(x)}{Q(x)}$ but a polynomial where you only have linear and quadratic factors multiplied together doesn’t seem to be very general. How do we know that we can always write a polynomial in this way? Well, it turns out that using complex numbers we can prove this very powerful fact. The statement is: Every polynomial with real coefficients factors into a product of factors which are either linear with real coefficients or quadratic irreducible with real coefficients. Quadratic irreducible means that it can’t be split into a product of two linear factors with real coefficients. However, in order to prove this we need to look at a couple of important things. The first is another theorem which we will prove: The non-real roots of a polynomial equation with real coefficients occur in complex conjugate pairs. This says that if some complex number $q$ is a root of a polynomial, then so is $\bar{q}$. This sounds rather strange, but let’s prove it in a few simple lines. Let our polynomial with real coefficients be: where by definition, the $a_i$ are all real. If $\alpha$ is a root then, by definition $f(\alpha)=0$ (this is what a root means in the context of a polynomial equation). Then we want to show that $f (\overline{\alpha})$ is also a root which simply means that the complex conjugate of a root is also a root. What is $f(\overline{\alpha})$? Well, we just plug it in: but we know that the conjugate of a product of complex numbers is the same as the product of the conjugate (ie. $\overline{a}^n=\overline{a^n}$). See we can rewrite the above as: but we also said that the coefficients were real, and the conjugate of a real coefficient is just that coefficient, so we can “expand” the conjugation: and furthermore, the sum of a set of conjugates is the conjugate of the sum: but this is just the same as: So we have shown that if $\alpha$ is a root, then so is $\overline{\alpha}$. This means that any time you find a complex root of an equation, it must have another complex root which is the conjugate of the first root. In fact, for the case of a quadratic equation we can see this immediately because if $a$, $b$ and $c$ are real, then the only place that an imaginary number can come into the root is from the part which is $\pm\sqrt{b^2-4ac}$ which means that the two solutions, if $b^2-4ac<0$, are going to be conjugate of each other (the $+$ part and the $-$ part). We have just shown that this is true also for higher order polynomials. In fact this is true for any polynomial at all. There’s a second crucial theorem that we need to show that we can factor any polynomial into a product of linear and irreducible quadratic factors. This is the Fundamental Theorem of Algebra: Theorem: The fundamental theorem of algebra Every polynomial (with real OR complex coefficients) which is of degree at least one, has a zero. This says that there is always a solution to any polynomial (equaling zero) which is order more than 0. A zeroth order polynomial is just a constant and so it’s clear that this will not have a solution which is equal to zero, unless the constant is zero. We will not prove the Fundamental Theorem of algebra but we will use it to prove our initial statement about factoring polynomials. Let’s take a polynomial $P(z)$. by the Fundamental Theorem, it has at least one root, let’s call it $r_1$. That root can either be real or complex. If it’s real, then we know that the Polynomial can be written as $(z-r_1)\tilde{P}(z)$ where $\tilde{P}(z)=\frac{P(z)}{(Z-r_1)}$. We can always divide a polynomial by another to get another polynomial as long as the order of the polynomial on top is greater than that on the bottom. The other option for our root is that it is complex. If it is complex then its conjugate is also a root. This means that $(z-r_1)(z-\bar{r}_1)$ is a factor of the polynomial, but this is purely real. We can see this by noting that if $r_1=a+bi$ then which is an irreducible quadratic (ie. we can’t write it as the product of linear factors). In this case we can write: Then we can take $\tilde{P}(z)$ and play exactly the same game. In fact we can do this until we are left with just a constant and this clearly can’t be written as the product of linear and quadratic factors – it’s as simple as it can be. So we have proved our initial claim using both the fundamental theorem of algebra and the fact that roots always come in conjugate pairs. This fact allows us to play with partial fractions as we did before, so it’s really a very powerful statement. Let’s just see where that gets us. Now if I give you the polynomial $2z^3-9z^2+14z-5$ and tell you that it has a zero at $z=2-i$, then you can immediately factor this cubic. The first thing you know for sure is that $z=2+i$ is also going to be a root. Now you can take this function and divide by $(z-(2-i))(z-(2+i))=((z-2)^2+1)$ This gives: and so: We have factored the polynomial into a linear factor and an irreducible quadratic factor. Now we know the solutions to this equation. One Comment 1. […] Complex numbers – polynomials of complex numbers […]	{"url":"http://www.mathemafrica.org/?p=11410","timestamp":"2024-11-06T03:48:19Z","content_type":"text/html","content_length":"211316","record_id":"<urn:uuid:7b7fd128-1e95-418e-bd5b-293252a3903e>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00146.warc.gz"}
52 research outputs found Motivated by possible applications within the framework of anti-de Sitter gravity/Conformal Field Theory (AdS/CFT) correspondence, charged black holes with AdS asymptotics, which are solutions to Einstein-Gauss-Bonnet gravity in D dimensions, and whose electric field is described by a nonlinear electrodynamics (NED) are studied. For a topological static black hole ansatz, the field equations are exactly solved in terms of the electromagnetic stress tensor for an arbitrary NED Lagrangian, in any dimension D and for arbitrary positive values of Gauss-Bonnet coupling. In particular, this procedure reproduces the black hole metric in Born-Infeld and conformally invariant electrodynamics previously found in the literature. Altogether, it extends to D>4 the four-dimensional solution obtained by Soleng in logarithmic electrodynamics, which comes from vacuum polarization effects. Fall-off conditions for the electromagnetic field that ensure the finiteness of the electric charge are also discussed. The black hole mass and vacuum energy as conserved quantities associated to an asymptotic timelike Killing vector are computed using a background-independent regularization of the gravitational action based on the addition of counterterms which are a given polynomial in the intrinsic and extrinsic curvatures.Comment: 30 pages, no figures; a few references added; final version for PR Hamiltonian systems with linearly dependent constraints (irregular systems), are classified according to their behavior in the vicinity of the constraint surface. For these systems, the standard Dirac procedure is not directly applicable. However, Dirac's treatment can be slightly modified to obtain, in some cases, a Hamiltonian description completely equivalent to the Lagrangian one. A recipe to deal with the different cases is provided, along with a few pedagogical examples.Comment: To appear in Proceedings of the XIII Chilean Symposium of Physics, Concepcion, Chile, November 13-15 2002. LaTeX; 5 pages; no figure We study thermodynamics of black hole solutions in Lanczos-Lovelock AdS gravity in d+1 dimensions coupled to nonlinear electrodynamics and a Stueckelberg scalar field. This class of theories is used in the context of gauge/gravity duality to describe a high-temperature superconductor in d dimensions. Larger number of coupling constants in the gravitational side is necessary to widen a domain of validity of physical quantities in a dual QFT. We regularize the gravitational action and find the finite conserved quantities for a planar black hole with scalar hair. Then we derive the quantum statistical relation in the Euclidean sector of the theory, and obtain the exact formula for the free energy of the superconductor in the holographic quantum field theory. Our result is analytic and it includes the effects of backreaction of the gravitational field. We further discuss on how this formula could be used to analyze second order phase transitions through the discontinuities of the free energy, in order to classify holographic superconductors in terms of the parameters in the theory.Comment: 26 pages, no figures; references added; published versio We analyze the dynamics of gauge theories and constrained systems in general under small perturbations around a classical solution (background) in both Lagrangian and Hamiltonian formalisms. We prove that a fluctuations theory, described by a quadratic Lagrangian, has the same constraint structure and number of physical degrees of freedom as the original non-perturbed theory, assuming the non-degenerate solution has been chosen. We show that the number of Noether gauge symmetries is the same in both theories, but that the gauge algebra in the fluctuations theory becomes Abelianized. We also show that the fluctuations theory inherits all functionally independent rigid symmetries from the original theory, and that these symmetries are generated by linear or quadratic generators according to whether the original symmetry is preserved by the background, or is broken by it. We illustrate these results with the examples.Comment: 27 pages; non-essential but clarifying changes in Introduction, Sec. 3 and Conclusions; the version to appear in J.Phys. We study the holographic currents associated to Chern-Simons theories. We start with an example in three dimensions and find the holographic representations of vector and chiral currents reproducing the correct expression for the chiral anomaly. In five dimensions, Chern-Simons theory for AdS group describes first order gravity and we show that there exists a gauge fixing leading to a finite Fefferman-Graham expansion. We derive the corresponding holographic currents, namely, the stress tensor and spin current which couple to the metric and torsional degrees of freedom at the boundary, respectively. We obtain the correct Ward identities for these currents by looking at the bulk constraint equations.Comment: 21 pages; version published in JHE It is shown that the renormalized action for AdS gravity in even spacetime dimensions is equivalent -on shell- to a polynomial of the Weyl tensor, whose first non-vanishing term is proportional to $Weyl^2$. Remarkably enough, the coupling of this last term coincides with the one that appears in Critical Gravity.Comment: 15 pages, references added, version accepted to JHE The dynamics of five-dimensional Chern-Simons theories is analyzed. These theories are characterized by intricate self couplings which give rise to dynamical features not present in standard theories. As a consequence, Dirac's canonical formalism cannot be directly applied due to the presence of degeneracies of the symplectic form and irregularities of the constraints on some surfaces of phase space, obscuring the dynamical content of these theories. Here we identify conditions that define sectors where the canonical formalism can be applied for a class of non-Abelian Chern-Simons theories, including supergravity. A family of solutions satisfying the canonical requirements is explicitly found. The splitting between first and second class constraints is performed around these backgrounds, allowing the construction of the charge algebra, including its central extension.Comment: 12 pages, no figure	{"url":"https://core.ac.uk/search/?q=authors%3A(Miskovic%2C%20Olivera)","timestamp":"2024-11-15T00:28:07Z","content_type":"text/html","content_length":"119370","record_id":"<urn:uuid:4740e68b-69df-4651-879c-4af5a278229b>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00334.warc.gz"}
EXPONENTIAL GROWTH AND DECAY - Differential Equations - AP CALCULUS AB & BC REVIEW - Master AP Calculus AB & BC Master AP Calculus AB & BC Part II. AP CALCULUS AB & BC REVIEW CHAPTER 10. Differential Equations You have probably alluded to exponential growth in everyday conversation without even realizing it. Perhaps you’ve said things like, “Ever since I started carrying raw meat in my pockets, the number of times I’ve been attacked by wild dogs has increased exponentially.'" Exponential growth is sudden, quick, and relentless. Mathematically, exponential growth or decay has one defining characteristic (and this is key)', the rate of y’s growth is directly proportional toy itself. In other words, the bigger y is, the faster it grows; the smaller y is, the slower it decays. Mathematically, something exhibiting exponential growth or decay satisfies the differential equation where k is called the constant of proportionality. A model ship might be built to a 1:35 scale, which means that any real ship part is 35 times as large as the model. The constant of proportionality in that case is 35. However, k in exponential growth and decay is never so neat and tidy, and it is rarely (if ever) evident from reading a problem. Luckily, it is quite easy to find. In the first problem set of this chapter (problem 3), you proved that the general solution to In this formula, N stands for the original amount of material, k is the proportionality constant, t is time, and y is the amount of N that remains after time t has passed. When approaching exponential growth and decay problems, your first goals should be to find N and k; then, answer whatever question is being posed. Don’t be intimidated by these problems—they are very easy. Example 3: The new theme restaurant in town (Rowdy Rita’s Eat and Hurl) is being tested by the health department for cleanliness. Health inspectors find the men’s room floor to be a fertile ground for growing bacteria. They have determined that the rate of bacterial growth is proportional to the number of colonies. So, they plant 10 colonies and come back in 15 minutes; when they return, the number of colonies has risen to 35. How many colonies will there be one full hour after they planted the original 10? Solution: The key phrase in the problem is “the rate of bacterial growth is proportional to the number of colonies,” because that means that you can apply exponential growth and decay. They started with 10 colonies, so N = 10 (starting amount). Do not try to figure out what k is in your head—it defies simple calculation. Instead, we know that there will be 35 colonies after t = 15 minutes, so you can set up the equation Solve this equation for k. Divide by 10 to begin the process. Now you have a formula to determine the amount of bacteria for any time t minutes after the original planting: We want the amount of bacteria growth after 1 hour; since we calculated k using minutes, we’ll have to express 1 hour as t = 60 minutes. Now, find the number of colonies. So, almost 1,501 colonies are partying along the surface of the bathroom floor. In one day, the number will grow to 1.7 X 10^53 colonies. You may be safer going to the bathroom in the alley behind the restaurant. NOTE. All half-life problems automatically satisfy the property Example 4: The Easter Bunny has begun to express his more malevolent side. This year, instead of hiding real eggs, he’s hiding eggs made of a radioactive substance Nb-95, which has a half-life of 35 days. If the dangerous eggs have a mass of 2 kilograms, and you don’t find the one hiding under your bed, how long will it take that egg to decay to a “harmless” 50 grams? Solution: The egg starts at a mass of 2,000 g. A half-life of 35 days means that in 35 days, exactly half of the mass will remain. After 70 days, one fourth of the mass will remain, etc. Therefore, after 35 days, the mass will be 1,000. This information will allow us to find k. TIP. In an exponential decay problem such as this, the k will be negative. Now that we know N and k, we want to find t when only 50 grams are left. In this case, t will be in days (since days was the unit of time we used when determining k). You should be safe by Thanksgiving. (Nothing wrong with a little premature hair loss and a healthy greenish glow, is there?) Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book. 1. If Pu-230 (a particularly stinky radioactive substance) has a half-life of 24,360 years, find an equation that represents the amount of Pu-230 left after time t, if you began with N grams. 2. Most men in the world (except, of course, for me, if my wife is reading this) think that Julia Roberts is pretty attractive. If left unchecked (and the practice were legal), we can assume the number of her husbands would increase exponentially. As of right now, she has one husband, but if legal restrictions were lifted she might have 4 husbands 2 years from now. How many years would it take her to marry 100 men if the number of husbands is proportional to the rate of increase? 3. Assume that the world population’s interest in the new boy band, “Hunks o’ Love,” is growing at a rate proportional to the number of its fans. If the Hunks had 2,000 fans one year after they released their first album and 50,000 fans five years after their first album, how many fans did they have the moment the first album was released? 4. Vinny the Talking Dog was an impressive animal for many reasons during his short-lived career. First of all, he was a talking dog, for goodness sakes! However, one of the unfortunate side-effects of this gift was that he increased his size by 1/3 every two weeks. If he weighed 5 pounds at birth, how many days did it take him to reach an enormous 600 pounds (at which point his poor, pitiable, poochie heart puttered out)? 1. Because the rate of decrease is proportional to the amount of substance (the amount decreases by half), we can use exponential growth and decay. In other words, let’s get Nekt. In 24,360 years, N will decrease by half, so we can write Divide both sides by N, and you get Therefore, the equation 2. This problem clearly states the proportionality relationship required to use exponential growth and decay. Here, N = 1, and y = 4 when t = 2 years, so you can set up the equation: Now that we have k, we need to find t when y = 100. 3. Our job in this problem will be to find N, the original number of fans. We have the following equations based on the given information: Solve the first equation for N, and you get Plug this value into the other equation, and you can find k. Finally, we can find the value of N by plugging k into NOTE. It should be no surprise that the left-hand side of the equation is 4/3 in the second step, as Vinny’s weight is 4/3 of his original weight every 14 days. In the half-life problems, you may have noticed that this number always turns out to be ½. 4. Oh, cruel fate. If Vinny weighed 5 pounds at birth, he weighed We want to find t when y = 600. The poor guy lived almost 8 months. The real tragedy is that even though he could talk, all he wanted to talk about were his misgivings concerning contemporary U.S. foreign policy. His handlers were relieved at his passing. “It was like having to talk to a furry John Kerry all the time,” they explained.	{"url":"https://schoolbag.info/mathematics/ap_calculus/70.html","timestamp":"2024-11-13T20:50:46Z","content_type":"text/html","content_length":"19377","record_id":"<urn:uuid:d078d0bd-4190-4382-b91d-6d070c621bfc>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00579.warc.gz"}
Convert meters to yards How do I convert meters to yards? To convert meters to yards, you can use a simple formula. Multiply the meter measurement by 1.09361. The result will give you an approximate yard measurement. For example, if you have 5 meters, when you multiply it by 1.09361, you get approximately 5.46805 yards. How accurate is the meter to yard conversion? The meter to yard conversion is highly accurate. The conversion factor, 1.09361, is precise and widely accepted. However, keep in mind that this conversion is an approximation, so the converted measurement may not be exact in all cases. Can I convert yards to meters using the same formula? No, the conversion formula for yards to meters is different. To convert yards to meters, you need to multiply the yard measurement by 0.9144. Make sure to use the correct formula when converting between different units of measurement. Why would I need to convert meters to yards? There are several situations where you may need to convert meters to yards. For example, if you are planning a landscaping project, you may want to know how many yards of soil or mulch you need based on the measurements provided in meters. In addition, understanding the conversion between meters and yards can be helpful when traveling, especially in countries that use the yard as a unit of Can I convert meters to yards manually? Yes, you can convert meters to yards manually by using the conversion factor of 1.09361. However, this can be time-consuming and prone to errors. Using an online calculator or conversion tool can save you time and provide more accurate results.	{"url":"https://calculatebox.com/meters-to-yards","timestamp":"2024-11-10T22:06:06Z","content_type":"text/html","content_length":"20540","record_id":"<urn:uuid:e8e7f7d1-5cd5-4da0-9f21-63e317d8e987>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00687.warc.gz"}
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1 Here we have given NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1. │Board │CBSE │ │Textbook │NCERT │ │Class │Class 6 │ │Subject │Maths │ │Chapter │Chapter 9 │ │Chapter Name │Data Handling │ │Exercise │Ex 9.1 │ │Number of Questions Solved │2 │ │Category │NCERT Solutions │ NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1 Ex 9.1 Class 6 Maths Question 1. In a Mathematics test, the following marks were obtained by 40 students. Arrange these marks in a table using tally marks. (a) Find how many students obtained marks equal to or more than 7? (b) How many students obtained marks below 4? In the first column of the table, we write all the values of marks scored by the students starting from the lowest to the highest. In the second column, a vertical bar (\|) called the tally mark is put against the number, whenever it occurs. For our convenience, we shall keep the tally marks in bunches of five, the fifth mark being drawn diagonally across the first four. We continue this process for all the values of the first column. Finally, we count the number of tally marks corresponding to each observation and write the number in the third column to represent the number of Thus, we have the table as under: 5 + 4+3=12 Clearly, from the above table the number of students scoring marks below 4 are 2 + 3 + 3 = 8 Ex 9.1 Class 6 Maths Question 2. Following is the choice of sweets of 30 students of class VI. Ladoo, Barfi, Ladoo, Jalebi, Ladoo, Rasgulla, Jalebi, Ladoo, Barfi, Rasgulla, Ladoo, Jalebi, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo, Rasgulla, Ladoo, Ladoo, Barfi, Rasgulla, Rasgulla, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo. (a) Arrange the names of sweets in a table using tally marks. (b) Which sweet is preferred by most of the students? (a) The required table is as under: (b) The sweet Ladoo is preferred by the most of the students. We hope the NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1, drop a comment below and we will get back to you at the earliest. Post a Comment	{"url":"http://eschool.successrouter.com/2020/07/ncert-solutions-for-class-6-maths_59.html","timestamp":"2024-11-03T04:03:14Z","content_type":"application/xhtml+xml","content_length":"260041","record_id":"<urn:uuid:8d0d92e5-07d9-4324-a120-4f9999fc5228>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00860.warc.gz"}
base area 30 Aug 2024 Title: Theoretical Foundations of Base Area: A Geometric Analysis Abstract: This article provides a comprehensive review of the concept of base area, a fundamental notion in geometry and mathematics. We delve into the theoretical foundations of base area, exploring its definition, properties, and mathematical formulation. The base area of a geometric shape is a measure of the size or extent of the shape’s foundation. It is an essential concept in various fields, including geometry, trigonometry, and engineering. In this article, we will examine the theoretical foundations of base area, focusing on its definition, properties, and mathematical formulation. The base area (BA) of a geometric shape can be defined as: where dA is an infinitesimal element of area. This definition encompasses various types of shapes, including triangles, quadrilaterals, polygons, and more complex geometric forms. The base area possesses several important properties, which are essential for its mathematical formulation: 1. Additivity: The base area of a composite shape is the sum of the base areas of its constituent parts. 2. Homogeneity: The base area is a homogeneous quantity, meaning it does not depend on the orientation or position of the shape in space. 3. Invariance: The base area remains unchanged under rigid motions (translations and rotations) of the shape. Mathematical Formulation: The base area can be mathematically formulated using various techniques, including: 1. Integration: The base area can be calculated by integrating the infinitesimal elements of area over the entire shape. 2. Differentiation: In some cases, the base area can be expressed as a derivative of another quantity, such as the volume or surface area of the shape. In conclusion, this article has provided an in-depth examination of the theoretical foundations of base area, exploring its definition, properties, and mathematical formulation. The concept of base area is fundamental to various fields, including geometry, trigonometry, and engineering, and its understanding is essential for solving problems involving geometric shapes. • [1] “Geometry” by Euclid • [2] “Calculus” by Isaac Newton and Gottfried Wilhelm Leibniz • [3] “Engineering Mathematics” by K.A. Stroud Related articles for ‘base area ‘ : Calculators for ‘base area ‘	{"url":"https://blog.truegeometry.com/tutorials/education/6894ea2c394c2b73dc7ed0ca99db65df/JSON_TO_ARTCL_base_area_.html","timestamp":"2024-11-06T08:45:43Z","content_type":"text/html","content_length":"15138","record_id":"<urn:uuid:a99975c5-6fe2-4d24-baef-ba5c7302e1a3>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00692.warc.gz"}
Introduction to Data Structures - TechVidvan Introduction to Data Structures Data structures are a very important part and a key component of computer science. Without the presence of different types of data structures, it is impossible to store the information effectively. Data structures are a way to store, manipulate and retrieve data efficiently. Data structures are largely used in almost all the fields of computer science. What is a Data Structure: It is a way of structuring/organizing data in a computer so that we can make use of this data in the most efficient manner. The main purpose of using data structures is to reduce the space and time complexity of every possible computation. There are various kinds of data structures in computer science such as arrays, linked lists, stacks, queues, trees and graphs, etc. Basic Terminologies of data structures: Some of the most common terms used in data structures are: 1. Data: It is a collection of values or a single value that need to be stored somewhere for future use. For example, in schools and universities, the details of students such as student id, student name, etc. comprise of data. 2. Group-items: Group data items are sub-data items for a given data. For example, the student name could comprise First name, middle name and last name. 3. Records: Records are a collection of different data items. For example, the record of one student consists of his first name, middle name, last name, phone number, roll number, address, etc. 4. File: A file is a collection of more than one record. For example, if a course has 200 students enrolled, then we can store the records of all those students in a single file. 5. Entity: An entity is a class of various objects. Each entity comprises various attributes. 6. Attribute: An attribute describes a particular property of an entity. We can have more than one attribute for an entity. 7. Field: A field represents an attribute of an entity. Types of Data Structures: At the most basic level, we divide data structures into two categories: Primitive and Non-primitive. We can further classify non-primitive data structures into linear and non-linear. 1. Primitive data structures: Primitive data structures are those that are by default defined in a programming language. It does not include any additional methods. For example, int, long, float, double, char, etc. are some of the primitive data structures. They are also known as primitive data types. They cannot hold multiple values at a time. 2. Non-primitive data structures: Non-primitive data structures are those, which are not inbuilt in the programming language. Usually, the programmers define these data structures for making their problems easier. Non-primitive data structures have certain advantages over primitive data structures, such as they can hold multiple values at a time. Also, they are easily accessible. We can further classify non-primitive data structures into two categories: Linear and Non-linear data structures. a. Linear data structures: As the name suggests, linear data structures are those which can hold data in a linear or sequential manner. In linear data structures, each element has a connection with its previous and the next element. Some examples of linear data structures are: arrays, linked lists, stacks and queues b. Non-linear data structures: Non-linear data structures are those in which the data is not arranged sequentially, rather we have the data in a hierarchical format. Due to hierarchical arrangement, we cannot traverse the whole data in one go. Non-linear data structures are more efficient than linear data structures when it comes to efficiency and space and time complexity. Non-linear data structures are further categorized as Static and 1. Static Data structures: Static data structures are those which have a fixed memory size and this memory size is decided at the time of compilation itself. In static data structures, if we have once given a specific size to the data structure, we can’t shrink or expand it at the run time. The most suitable example of static data structures is an array. 2. Dynamic data structures: As the name suggests, these data structures have varying memory sizes i.e. we can change the size according to time and need at the run time as well. We allot the memory sizes to dynamic data structures at the run time. A few examples of dynamic data structures are link lists, stacks, queues, trees and graphs. Basic Operations on Data Structures: The most common operations that we perform on almost all data structures are: 1. Search: We perform search operation on a data structure whenever we wish to find a particular element. 2. Traverse: Traversal includes visiting all the values in a data structure. 3. Insert: Insertion includes adding/inserting new elements into the pre-existing data structure. 4. Update: Update operation helps to manipulate one or more elements in a data structure. 5. Delete: This operation helps to delete one or more elements from a data structure. 6. Merge: It is used to combine two or more data structures. Merge operation is mostly performed on data structures of the same type. 7. Sort: Sorting means arranging the given data in either ascending or descending order. Characteristics of Data Structures: The following are the characteristics of data structures: 1. Time complexity: Time complexity gives us the time that the execution of a particular operation on a particular data structure will take. The time taken should be minimum so as to have better efficiency of the program. 2. Space complexity: Space complexity tells us about the extra memory space that any operation will take to complete its execution. We should try to use the minimum possible extra space for program 3. Correctness: This characteristic says that we need to correctly implement the interface of the data structure. Need to Learn Data Structures: So, why do we need data structures? The simple answer to this question is to reduce space and time complexity as well as to improve the efficiency of the code. Let us understand with the help of a small example. Suppose we wish to search the record of a particular student in a department out of 500 students. Consider that the student records are sorted in ascending order according to their roll numbers. If we start searching from beginning to end one by one, it will take a lot of time. However, if we make use of binary search i.e. we first go to the middle record and check if the record to be searched is after the middle record or before. In this way, we have reduced the search by half. This is one of the examples that show how data structures improve the efficiency of any computation. There are many such uses of data structures in the whole of computer science. If we see in our day-to-day life, we see a lot of examples of data structures. For example, all of us have a few books arranged on our study tables. This pile of books is an example of a stack implementation. Another example is whenever we go to a bank or stand in a queue to buy movie tickets, we are following first in first out principle. This is the implementation of the queue data Advantages of Data Structures: • Reduce the computation time for any problem. • These help to utilize the available memory space in a more efficient manner. • Ensure the reusability of code as we can write the code for a stack or any data structure only once in the program and use it at multiple places. • The abstract data types(ADT) provide us with the best way to understand and use data structures without going deep into the implementation details. • There are multiple choices available. We can choose any one data structure according to our needs and the need for the problem to be solved. • Data structures provide scientific computation methods that help in improving speed, efficiency and easier implementation of operations on data structures. Applications of Data Structures: • Arrays are useful whenever we have sequential and fixed-size data implementation. For example, in online or video games as well as in online coding contests, we have names and positions of players or students. These leaderboards are created with the help of arrays. • We use stacks in all the recursive operations as function calls make use of the stack. • In web pages, we make use of linked lists to visit another web page from a given URL in that web page. • Priority queues help to implement heap and other data structures. Priority queues are also used in scheduling algorithms of the CPU. • To find the shortest path between two points, we make use of graphs. Data structures are one of the most significant components of computer science. Data structures help to increase the computation power of any program and help to reduce the time taken to solve a problem. Thus, having a good knowledge of data structures is of utmost importance.	{"url":"https://techvidvan.com/tutorials/introduction-to-data-structures/?noamp=mobile","timestamp":"2024-11-11T04:42:24Z","content_type":"text/html","content_length":"171218","record_id":"<urn:uuid:73793f04-470d-4e0b-9fc0-7a6cf9a88895>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00414.warc.gz"}
Consecutive 82712 - math word problem (82712) Consecutive 82712 The sum of five consecutive even numbers is -10. What are the numbers? Correct answer: Did you find an error or inaccuracy? Feel free to write us . Thank you! Tips for related online calculators Do you have a linear equation or system of equations and are looking for its ? Or do you have a quadratic equation You need to know the following knowledge to solve this word math problem: Grade of the word problem: Related math problems and questions:	{"url":"https://www.hackmath.net/en/math-problem/82712","timestamp":"2024-11-06T09:10:26Z","content_type":"text/html","content_length":"59192","record_id":"<urn:uuid:38109459-a9a0-4eb8-a807-8489c5c02344>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00103.warc.gz"}
Science & Technology Policy Fellowships - XLT This was probably the most advanced quantity system on the planet on the time, apparently in use a number of centuries earlier than the widespread era and well earlier than the event of the Indian numeral system. Rod numerals allowed the representation of numbers as massive as desired and allowed calculations to be carried out on the suan pan, or Chinese abacus. The date of the invention of the suan pan is not sure, however the earliest written point out dates from AD one hundred ninety, in Xu Yue’s Supplementary Notes on the Art of Figures. ), however its design seems to have been misplaced till experiments were made during the 15th century in Western Europe. Perhaps relying on comparable gear-work and technology discovered within the Antikythera mechanism, the odometer of Vitruvius featured chariot wheels measuring 4 toes (1.2 m) in diameter turning four-hundred times in a single Roman mile (roughly 4590 ft/1400 m). New technology helps dissect how it ignores or acts on data MacTutor History of Mathematics archive (John J. O’Connor and Edmund F. Robertson; University of St Andrews, Scotland). An award-profitable web site containing detailed biographies on many historic and up to date mathematicians, in addition to info on notable curves and various matters within the history of mathematics. Encyclopaedia of the history of science, technology, and medicine in non-western cultures. It isn’t known to what extent the Sulba Sutras influenced later Indian mathematicians. As in China, there’s a lack of continuity in Indian mathematics; vital advances are separated by long durations of inactivity. Building on earlier work by many predecessors, Isaac Newton discovered the legal guidelines of physics explaining Kepler’s Laws, and brought together the concepts now generally known as calculus. Independently, Gottfried Wilhelm Leibniz, who’s arguably some of the important mathematicians of the 17th century, developed calculus and much of the calculus notation nonetheless in use right now. Science and mathematics had become an international endeavor, which would soon spread over the complete world. • This development will make enhancements in national safety extra dependent on general nationwide economic efficiency. • Computer & Information Sciences Theory and techniques in pc science and information technology. • However, the Tsinghua Bamboo Slips, containing the earliest known decimal multiplication desk (although ancient Babylonians had ones with a base of 60), is dated around 305 BC and is perhaps the oldest surviving mathematical text of China. • My personal opinion is that science shouldn’t have any forbidden zones and technology should be managed. On the opposite hand, the limitation of three dimensions in geometry was surpassed within the 19th century through concerns of parameter house and hypercomplex numbers. He did revolutionary work on functions of complicated variables, in geometry, and on the convergence of series, leaving aside his many contributions to science. He also gave the primary passable proofs of the fundamental theorem of algebra and of the quadratic reciprocity law. He made quite a few contributions to the research of topology, graph concept, calculus, combinatorics, and sophisticated evaluation, as evidenced by the multitude of theorems and notations named for him. The Islamic Empire established across Persia, the Middle East, Central Asia, North Africa, Iberia, and in elements of India within the 8th century made important contributions in the direction of mathematics. Although most Islamic texts on arithmetic had been written in Arabic, most of them were not written by Arabs, since very similar to the status of Greek within the Hellenistic world, Arabic was used as the written language of non-Arab scholars all through the Islamic world at the time.	{"url":"https://xltoday.net/science-technology-policy-fellowships.html","timestamp":"2024-11-07T15:33:51Z","content_type":"text/html","content_length":"51583","record_id":"<urn:uuid:8b38439a-b4d5-4e67-8ae6-e0c2c4a8c73d>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00575.warc.gz"}
63 research outputs found We consider the problem of searching for continuous gravitational wave sources orbiting a companion object. This issue is of particular interest because the LMXB's, and among them Sco X-1, might be marginally detectable with 2 years coherent observation time by the Earth-based laser interferometers expected to come on line by 2002, and clearly observable by the second generation of detectors. Moreover, several radio pulsars, which could be deemed to be CW sources, are found to orbit a companion star or planet, and the LIGO/VIRGO/GEO network plans to continuously monitor such systems. We estimate the computational costs for a search launched over the additional five parameters describing generic elliptical orbits using match filtering techniques. These techniques provide the optimal signal-to-noise ratio and also a very clear and transparent theoretical framework. We provide ready-to-use analytical expressions for the number of templates required to carry out the searches in the astrophysically relevant regions of the parameter space, and how the computational cost scales with the ranges of the parameters. We also determine the critical accuracy to which a particular parameter must be known, so that no search is needed for it. In order to disentangle the computational burden involved in the orbital motion of the CW source, from the other source parameters (position in the sky and spin-down), and reduce the complexity of the analysis, we assume that the source is monochromatic and its location in the sky is exactly known. The orbital elements, on the other hand, are either assumed to be completely unknown or only partly known. We apply our theoretical analysis to Sco X-1 and the neutron stars with binary companions which are listed in the radio pulsar catalogue.Comment: 31 pages, LaTeX, 6 eps figures, submitted to PR A data-analysis strategy based on the maximum-likelihood method (MLM) is presented for the detection of gravitational waves from inspiraling compact binaries with a network of laser-interferometric detectors having arbitrary orientations and arbitrary locations around the globe. The MLM is based on the network likelihood ratio (LR), which is a function of eight signal-parameters that determine the Newtonian inspiral waveform. In the MLM-based strategy, the LR must be maximized over all of these parameters. Here, we show that it is possible to maximize it analytically over four of the eight parameters. Maximization over a fifth parameter, the time of arrival, is handled most efficiently by using the Fast-Fourier-Transform algorithm. This allows us to scan the parameter space continuously over these five parameters and also cuts down substantially on the computational costs. Maximization of the LR over the remaining three parameters is handled numerically. This includes the construction of a bank of templates on this reduced parameter space. After obtaining the network statistic, we first discuss `idealized' networks with all the detectors having a common noise curve for simplicity. Such an exercise nevertheless yields useful estimates about computational costs, and also tests the formalism developed here. We then consider realistic cases of networks comprising of the LIGO and VIRGO detectors: These include two-detector networks, which pair up the two LIGOs or VIRGO with one of the LIGOs, and the three-detector network that includes VIRGO and both the LIGOs. For these networks we present the computational speed requirements, network sensitivities, and source-direction resolutions.Comment: 40 pages, 2 figures, uses RevTex and psfig, submitted to Phys. Rev. D, A few minor changes adde Matched filtering is a commonly used technique in gravitational wave searches for signals from compact binary systems and from rapidly rotating neutron stars. A common issue in these searches is dealing with four extrinsic parameters which do not affect the phase evolution of the system: the overall amplitude, initial phase, and two angles determining the overall orientation of the system. The F-statistic maximizes the likelihood function analytically over these parameters, while the B-statistic marginalizes over them. The B-statistic, while potentially more powerful and capable of incorporating astrophysical priors, is not as widely used because of the computational difficulty of performing the marginalization. In this paper we address this difficulty and show how the marginalization can be done analytically by combining the four parameters into a set of complex amplitudes. The results of this paper are applicable to both transient non-precessing binary coalescence events, and to long lived signals from rapidly rotating neutron stars.Comment: 26 page We formulate the data analysis problem for the detection of the Newtonian coalescing-binary signal by a network of laser interferometric gravitational wave detectors that have arbitrary orientations, but are located at the same site. We use the maximum likelihood method for optimizing the detection problem. We show that for networks comprising of up to three detectors, the optimal statistic is essentially the magnitude of the network correlation vector constructed from the matched network-filter. Alternatively, it is simply a linear combination of the signal-to-noise ratios of the individual detectors. This statistic, therefore, can be interpreted as the signal-to-noise ratio of the network. The overall sensitivity of the network is shown to increase roughly as the square-root of the number of detectors in the network. We further show that these results continue to hold even for the restricted post-Newtonian filters. Finally, our formalism is general enough to be extended to address the problem of detection of such waves from other sources by some other types of detectors, e.g., bars or spheres, or even by networks of spatially well-separated detectors.Comment: 14 pages, RevTex, 1 postscript figure. Based on talk given at Workshop on Cosmology: Observations confront theories, IIT-Kharagpur, India (January 1999 We describe a general mathematical framework for $\chi^2$ discriminators in the context of the compact binary coalescence search. We show that with any $\chi^2$ is associated a vector bundle over the signal manifold, that is, the manifold traced out by the signal waveforms in the function space of data segments. The $\chi^2$ is then defined as the square of the $L_2$ norm of the data vector projected onto a finite dimensional subspace (the fibre) of the Hilbert space of data trains and orthogonal to the signal waveform - any such fibre leads to a $\chi^2$ discriminator and the full vector bundle comprising the subspaces and the base manifold constitute the $\chi^2$ discriminator. We show that the $\chi^2$ discriminators used so far in the CBC searches correspond to different fiber structures constituting different vector bundles on the same base manifold, namely, the parameter space. The general formulation indicates procedures to formulate new $\chi^2$s which could be more effective in discriminating against commonly occurring glitches in the data. It also shows that no $\chi^2$ with a reasonable degree of freedom is foolproof. It could also shed light on understanding why the traditional $\chi^2$ works so well. As an example, we propose a family of ambiguity $\chi^2$ discriminators that is an alternative to the traditional one. Any such ambiguity $\ chi^2$ makes use of the filtered output of the template bank, thus adding negligible cost to the overall search. We test the performance of ambiguity $\chi^2$ on simulated data using spinless TaylorF2 waveforms. We show that the ambiguity $\chi^2$ essentially gives a clean separation between glitches and signals. Finally, we investigate the effects of mismatch between signal and templates on the $\chi^2$ and also further indicate how the ambiguity $\chi^2$ can be generalized to detector networks for coherent observations.Comment: 21 pages, 5 figure, abstract is shortened to comply with the arXiv's 1920 characters limitation, v2: accepted for publication in PR On 14 September 2015, the twin detectors belonging to the Laser Interferometer Gravitational Wave Observatory (LIGO) made a triple discovery: the first direct detection of gravitational waves (GWs), first observation of formation of a black hole and first observation of a binary black hole. Since then LIGO has reported two other events and a marginal candidate. These discoveries have heralded a new era in observational astronomy. They will help us in exploring extremes of astrophysics and gravity. GWs are our best chance of getting an idea of what went on a small fraction of a second after the big bang, even if that takes many more decades. With LIGOâ€™s discoveries we hope to solve many puzzles in astronomy and fundamental physics, but GWs are guaranteed to show up objects and phenomena never imagined before We formulate the data analysis problem for the detection of the Newtonian waveform from an inspiraling compact-binary by a network of arbitrarily oriented and arbitrarily distributed laser interferometric gravitational wave detectors. We obtain for the first time the relation between the optimal statistic and the magnitude of the network correlation vector, which is constructed from the matched network-filter. This generalizes the calculation reported in an earlier work (gr-qc/9906064), where the detectors are taken to be coincident.Comment: 6 pages, RevTeX. Based on talk given at GWDAW-99, Rom The cross-correlation search for gravitational wave, which is known as 'radiometry', has been previously applied to map of the gravitational wave stochastic background in the sky and also to target on gravitational wave from rotating neutron stars/pulsars. We consider the Virgo cluster where may be appear as `hot spot' spanning few pixels in the sky in radiometry analysis. Our results show that sufficient signal to noise ratio can be accumulated with integration times of the order of a year. We also construct numerical simulation of radiometry analysis, assuming current constructing/ upgrading ground-based detectors. Point spread function of the injected sources are confirmed by numerical test. Typical resolution of radiometry analysis is a few square degree which corresponds to several thousand pixels of sky mapping.Comment: 9 pages, 9 figures, Amaldi 9 & NRD	{"url":"https://core.ac.uk/search/?q=author%3A(SANJEEV%20DHURANDHAR)","timestamp":"2024-11-02T22:23:17Z","content_type":"text/html","content_length":"142992","record_id":"<urn:uuid:4e160114-545d-4574-9046-c9d22ddfa89e>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00391.warc.gz"}
DCCCLXVII in Hindu Arabic Numerals DCCCLXVII = 867 M C X I MM CC XX II MMM CCC XXX III CD XL IV D L V DC LX VI DCC LXX VII DCCC LXXX VIII CM XC IX DCCCLXVII is valid Roman numeral. Here we will explain how to read, write and convert the Roman numeral DCCCLXVII into the correct Arabic numeral format. Please have a look over the Roman numeral table given below for better understanding of Roman numeral system. As you can see, each letter is associated with specific value. Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000 How to write Roman Numeral DCCCLXVII in Arabic Numeral? The Arabic numeral representation of Roman numeral DCCCLXVII is 867. How to convert Roman numeral DCCCLXVII to Arabic numeral? If you are aware of Roman numeral system, then converting DCCCLXVII Roman numeral to Arabic numeral is very easy. Converting DCCCLXVII to Arabic numeral representation involves splitting up the numeral into place values as shown below. D + C + C + C + L + X + V + I + I 500 + 100 + 100 + 100 + 50 + 10 + 5 + 1 + 1 As per the rule highest numeral should always precede the lowest numeral to get correct representation. We need to add all converted roman numerals values to get our correct Arabic numeral. The Roman numeral DCCCLXVII should be used when you are representing an ordinal value. In any other case, you can use 867 instead of DCCCLXVII. For any numeral conversion, you can also use our roman to number converter tool given above. Current Date and Time in Roman Numerals The current date and time written in roman numerals is given below. Romans used the word nulla to denote zero because the roman number system did not have a zero, so there is a possibility that you might see nulla or nothing when the value is zero.	{"url":"https://romantonumber.com/dccclxvii-in-arabic-numerals","timestamp":"2024-11-10T21:42:53Z","content_type":"text/html","content_length":"89944","record_id":"<urn:uuid:57044788-932b-44db-b647-ebb0cdcd0b65>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00219.warc.gz"}
seminars - Diffusive limit of Boltzmann Equation in exterior Domain The study of flows over an obstacle is one of the fundamental problems in fluids. In this talk we establish the global validity of the diffusive limit for the Boltzmann equations to the Navier-Stokes-Fourier system in an exterior domain. To overcome the well-known difficulty of the lack of Poincare's inequality in the unbounded domain, we develop a new $L^2-L^6$ splitting to extend the $L^2-L^\infty$ framework into the unbounded domain.	{"url":"http://www.math.snu.ac.kr/board/index.php?mid=seminars&page=88&l=en&sort_index=Time&order_type=desc&document_srl=1233090","timestamp":"2024-11-02T08:16:44Z","content_type":"text/html","content_length":"45291","record_id":"<urn:uuid:582937fe-4bf4-44cd-92fc-330849e20809>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00309.warc.gz"}
The Chess Variant Pages: Geographical Chess Notation Geographical Chess Notation Geographical Chess Notation is a new flexible method to notate games of Chess variants in general. Personally, I believe that the eternally-flexible Algebraic Notation isn't as flexible when it came to boards that change shape (like Building Chess) or, even worse, infinite boards (like Open Plane Chess)! The Geographical Chess Notation is applicable, with some modifications, to Hexagonal Boards and 3D boards. See notes. Since the most popular variant is FIDE chess, I am going to apply the system to it. The Chess Board is square. So, it makes perfect sense to call the main four directions of movement North (n), South(s), East(e) and West(w). * ( For some chess variants, extra directions might exist. Like clockwise and counter-clockwise for Circular moving pieces. Hexagonal Boards might need redefining the six directions as a,b,c,d,e and f. ) Let's assume that White starts South, and Black Starts North. (Therefore, the White pawns move northward, and Black pawns move southward.) \| r \|:n:\| b \|:q:\| k \|:b:\| n \|:r:\| \|:p:\| p \|:p:\| p \|:p:\| p \|:p:\| p \| \| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \| West +---+---+---+---+---+---+---+---+ East \| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \| \| P \|:P:\| P \|:P:\| P \|:P:\| P \|:P:\| \|:R:\| N \|:B:\| Q \|:K:\| B \|:N:\| R \| Each piece is referred to by the known symbols. P for Pawn, N for Knight .. etc. Multiple pieces of the same type are numbered from West to East. For example, the QR Pawn is P1; and the QN Pawn is P2. The Q Rook is R1, and the K Rook is R2. If multiple pieces are on the same file, the numbering starts from North to South. In the following example, five pawns are on the board, each is given a number of its own. It doesn't matter if the pawns are White or Black. Note the numbering might change if the pawns change their position. \| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \| \| \|:::\| P2\|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \| West +---+---+---+---+---+---+---+---+ East \| \|:::\| P3\|:::\| P4\|:P5\| \|:::\| \|:P1\| \|:::\| \|:::\| \|:::\| \| \| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \|:::\| \| There are, on the square board, four orthogonal directions : (n), (s), (w) and (e). And four diagonal directions : (ne), (se), (nw) and (sw). And eight Knight directions : (nne), (nee), (sse), (see), (nnw), (nww), (ssw) and (sww). Generally, any direction of movement has either one, two or more orthogonal legs. When a piece moves in a given directions, the two legs are given (as in a diagonal movement.) If one of the two legs is longer, its letter is repeated. (North and south before east and west, always.) If a piece is a rider, and moves more than one square in the direction it moves to, the number of steps is given. A capture is notated as in Descriptive notation. However, to avoid disambiguity altogether, both pieces must be numbered. (This is one of many methods that work.) Castling, en-passant and pawn promotion are as usual, or as specified by the specific variant author. Every move is basically this : Which piece, which direction, and how far. For an example, I give The Immortal Game (Anderssen-Kieseritsky) written in the Geographical Notation. White Black 1. P5 - n2 P5 - s2 2. P6 - n2 P5 x P6 (now the numbering of pawns has changed.) 3. B2 - nw3 ... (note that the numbering of Bishops is changed here.) 3. ... Q - se4 + (+ notates check.) 4. K - e P2 - s2 5. B1 x P2 N2 - ssw 6. N2 - nnw Q - n2 7. P4 - n N2 - see 8. N2 - nee Q - sw 9. N2 - nww P2 - s (this the c-pawn, not the b-pawn, which is captured.) 10. P6 - n2 N2 - nww 11. R2 - w P2 x B1 12. P7 - n2 Q - n 13. P7 - n Q - s 14. Q - ne2 N2 - nne 15. B x P5 ... (it's P5, not P4, because it's southern.) 15. ... Q - nw 16. N1 - nne B2 - sw3 17. N1 - nne Q x P2 18. B - nw2 B2 x R2 19. P4 - n Q x R + 20. K - nw N1 - ssw 21. N2 x P5 + K - w 22. Q - n3 + N2 x Q 23. B - ne # 1-0 The system is obviously not suitable for all variants. For example, Jim Aikin's Amoeba, where you get to move empty squares. Also, every detail of the move must be included, like a piece moving from one board to another, unless it was implied as in Alice Chess. The system needs some getting used to, like every other system. Some possible and necessary additions for other variants follow : 1. Mentioning squares : In games where you get to move squares (like Amoeba) or drop pieces (like Shogi) you will need a way to notate squares. The best way (I think) to do so is to mention how far away from the King the square is. For example (s.n2e3 - w) means a square 2 steps north and 3 steps east away from the King moves one square to the west. Also, (B @ s.n2e3) means a Bishop dropped at the square mentioned. In case of multiple Kings, the reference is to "K1". 2. Drops .. (see above) 3. Camels and Zebras, etc. are notated like the corresponding Knight's move. 3. Falcons, and more complex arrangements : The Falcon moves in 16 directions. So, to make the distinction, you must use four letters to say what direction it went to. The simplest way is to mention how many steps in the first leg and how many steps in the second leg it offsets from the original square. For example (F - n3w, F-n3w2). In case of more than a single step in this direction, add a slash (/) then the number of steps. For example (F - n3w2/2) 4. Roses and circular riders : Mention the first steps direction, followed by which circular direction it takes (cw for clockwise, and cc for counter clockwise), then how many steps. An example is (Ro - nne.cc2) 5. Hexagonal Boards and 3D Boards : Re-defining the directions for these boards is necessary. For Hexagonal boards, you can use a, b, c, d, e and f, defining where the directions are on a diagram. For 3D boards, add u (for up) and d (for down). For 4D boards, like Joe Joyce's Hyperchess and Jim Aikin's Chessaract, you can use hyper-directions. It is possible to use "n" for small board North and "N" for large board North. This gives 8 orthogonal directions. I believe this about sums it up. Any additions or fairy moves (like the Magician in my own Chess with Wizards), can be dealt with in the same manner. P.S. Thanks for David Howe and Jeremy Good for getting this article published. The work you do in the website is much appreciated. By Abdul-Rahman Sibahi. Web page created: 2007-09-23. Web page last updated: 2007-09-23	{"url":"https://www.chessvariants.com/page/geographical-chess-notation","timestamp":"2024-11-08T21:00:58Z","content_type":"text/html","content_length":"46664","record_id":"<urn:uuid:86b216bc-9245-423d-8fa5-22ec9bb8ad88>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00051.warc.gz"}
Publications \| Formal Verification Group Conventional embedded systems have over the past two decades vividly evolved into an open, interconnected form that integrates capabilities of computing, communication and control, thereby triggering yet another round of global revolution of the information technology. This form, now known as cyber-physical systems, has witnessed an increasing number of safety-critical systems particularly in major scientific projects vital to the national well-being and the people’s livelihood. Prominent examples include automotive electronics, health care, nuclear reactors, high-speed transportations, manned spaceflight, etc., in which a malfunction of any software or hardware component would potentially lead to catastrophic consequences like significant casualties and economic losses. Meanwhile with the rapid development of feedback control, sensor techniques and computer control, time delays have become an essential feature underlying both the continuous evolution of physical plants and the discrete transition of computer programs, which may well annihilate the safety certificate and control performance of embedded systems. Traditional engineering methods, e.g., testing and simulations, are nevertheless argued insufficient for the zero-tolerance of failures incurred in time-delayed systems in a safety-critical context. Therefore, how to rigorously verify and design reliable safety-critical embedded systems involving delays tends to be a grand challenge in computer science and the control community. In contrast to delay-free systems, time-delayed systems yield substantially higher theoretical complexity thus rendering the underlying design and verification tasks significantly harder. This dissertation focuses on the formal verification and controller synthesis of time-delayed dynamical systems, while particularly addressing the safety verification of continuous dynamics governed by delay differential equations and the control-strategy synthesis of discrete dynamics captured by delayed safety games, with applications in a typical set of representative benchmarks from the literature.	{"url":"https://fiction-zju.github.io/publication/","timestamp":"2024-11-13T19:41:48Z","content_type":"text/html","content_length":"178310","record_id":"<urn:uuid:b35cd9a0-2d58-4c39-bf1e-0464e8a1b6e7>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00087.warc.gz"}
DELTA HEDGING: What Are The Options? - GMU Consults Given that delta hedging is designed to reduce the volatility of the option price relative to changes in the price of the underlying asset, it constantly requires rebalancing to ensure risk hedging. Delta hedging is known to be a complex strategy used by institutional investors or large investment companies. What Is Delta Hedging? In options trading, hedging options are a derivative trading strategy used to balance the positive and negative delta, so their net effect is zero. If the position is delta-neutral, its value will not increase or decrease if the value of the underlying asset remains within certain limits. For options traders, this means that their position in the short term is protected from changes in the price of the underlying stock, ETF or index. When done correctly, a delta-neutral position can help compensate for changes in volatility. However, maintaining a delta-neutral position is an ongoing struggle, and the transaction costs of constantly rebalancing can easily reduce the temporary benefits of this strategy. How Does Delta Hedging Work? The delta hedging strategy is used in options to ensure risk reduction by establishing short and long positions for the relevant underlying asset. Thus, the risk in the directed sense is reduced and a state of neutrality is achieved. In a delta neutral situation, any change in the price of the underlying stock or asset will not affect the option price. In essence, the goal of a delta hedging strategy is to reduce or minimize the risks that arise from changes in the price of the underlying asset. ALSO CHECK: BUY TO CLOSE: Definition And Trading Guide Delta Hedging Example An example of delta hedging can help you better understand what delta hedging is and how it is used. Option positions can be hedged using base shares. One share will have a delta of 1, as the value of the share increases by 1 rupee with each increase in the share by 1 rupee. Assume that a trader has a call option of 0.5. If a share lot contains 1,000 shares, a trader can hedge one lot of the call option by selling 650 shares of that share. It is important to note that traders do not necessarily use the same scale to measure delta options. A scale of 0 to 1 and a scale of 0 to 100 are used by traders. Therefore, the delta value of 0.40 on one scale is 40 in the other, ie from 0 to 100. Delta Hedging Formula The formula for the delta can be obtained by dividing the change in the value of the option by the change in the value of its underlying stock. Mathematically it is represented as, Delta Î” = (O[f] â€“ O[i]) / (S[f] â€“ S[i]) O[f] = Final value of the option O[i] = Initial value of the option S[f] = Final value of the underlying stock S[i] = Initial value of the underlying stock Examples of delta formulas (with Excel template) Let’s take an example to better understand Delta’s calculations. Take the example of commodity X, which traded for $500 in the commodity market a month ago, and the call option for this commodity was traded at a premium of $45 with a strike price of $480. The commodity is now trading at $600, while the value of the option has risen to $75. Calculate the delta call option based on the information provided. Delta Î” is calculated by the following formula Delta Î” = (O[f] â€“ O[i]) / (S[f] â€“ S[i]) Delta Î” = ($75 â€“ $45) / ($600 â€“ $500) Delta Î” = $0.30 Delta Hedging Options We use stochastic modeling techniques to study the effectiveness of delta hedging strategies for common, retrospective, and Asian call options on the S&P 500. Execution is expected to take place in either the cash or futures markets. It is shown that the outcome of the hedging process largely depends on how the theoretical hedging strategy is implemented and on the gamma characteristics of the option position to be hedged. In addition to discrete hedge rebalancing, our results reveal erroneous volatility predictions and, for path-dependent options, the use of hedge ratios based on Black-Scholes assumptions as the main sources of hedge error risk. Despite incorrect prices and costs of prolongation, the execution of futures markets, as a rule, has an advantage over the execution of the cash market Delta Hedging Options Example For example, if a call option has a strike price of $30 and the underlying stock is trading at $40 at the expiration date, the option holder may convert 100 shares at a lower strike price of $30. If they choose, they can turn around and sell them on the open market for $40 at a profit. Profits will be $ 10 less the call option premium and any fees from the broker for bidding. Put options are a bit more confusing, but work almost as well as call options. Here, the owner expects the value of the underlying asset to deteriorate before it expires. They can either keep the asset in their portfolio or borrow shares from a broker. In addition, the number of transactions involved in delta hedging can become expensive, as trading fees are charged when adjusting the position. This can be especially expensive when options are hedged, as they may lose their temporal value, sometimes trading lower than underlying stocks. ALSO CHECK: ADVANTAGES OF BANK: Advantages In America What Is Time Value In Options Trading? Time value is a measure of how much time is left before the option expires, which allows the trader to make a profit. Time is running out and the expiration date is approaching, the option loses its temporary value because there is less time to make a profit. As a result, the temporary value of an option affects the margin for that option, as options with a large time value tend to have higher premiums than those with a small time value. Over time, the value of the option changes, which may lead to increased delta hedging to support the delta-neutral strategy. Delta hedging can benefit traders when they expect a strong movement in the underlying stock, but risk being overwhelmed if the stock does not move as expected. If more than hedged positions have to be curtailed, trading costs increase. Delta Hedging FAQs How Is The Value Of An Option Measured? The value of the option is measured by the amount of its premium – the fee paid for the purchase of the contract. By holding an option, an investor or trader can exercise their rights to buy or sell 100 shares of the underlying asset, but are not required to perform this action if it is unprofitable for them. What Is Strike Price In Options Trading? The price at which they will buy or sell is known as the strike price, and is set – along with the expiration date – at the time of purchase. Each option contract is equal to 100 shares of the underlying share or asset. Can I Sell My Option Contract Before Expiration? American-style options holders can exercise their rights at any time until the expiration date. European-style options allow the owner to exercise only on the expiration date. In addition, depending on the value of the option, the owner may decide to sell his contract to another investor before its expiration date. Leave a Reply Cancel reply	{"url":"https://gmuconsults.com/investment/delta-hedging/","timestamp":"2024-11-03T19:30:25Z","content_type":"text/html","content_length":"337317","record_id":"<urn:uuid:b6ec71b7-2990-40a5-9ede-22e8dbc1202b>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00436.warc.gz"}
Advanced Graph Theory and Combinatorics Book description Advanced Graph Theory focuses on some of the main notions arising in graph theory with an emphasis from the very start of the book on the possible applications of the theory and the fruitful links existing with linear algebra. The second part of the book covers basic material related to linear recurrence relations with application to counting and the asymptotic estimate of the rate of growth of a sequence satisfying a recurrence relation. Product information • Title: Advanced Graph Theory and Combinatorics • Author(s): • Release date: December 2016 • Publisher(s): Wiley-ISTE • ISBN: 9781848216167	{"url":"https://www.oreilly.com/library/view/advanced-graph-theory/9781848216167/","timestamp":"2024-11-13T15:09:17Z","content_type":"text/html","content_length":"62462","record_id":"<urn:uuid:44fa5b48-d0db-4261-97c2-685f7eb26dbe>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00696.warc.gz"}
Abstract. - Topologically stable symmetry defects in ordered media can be classified by some homotopy groups. As an example we establish a complete classification of such defects for crystals.]]> E/H yield a classification of the topologically stable defects and configurations of these ordered media. This suggests a predictive value of this scheme for yet unobserved media and for defects.]]> Abstract : An energy band in a solid contains an infinite number of states which transform linearly as a space group representation induced from a finite dimensional representation of the isotropy group of a point in space. A band representation is elementary if it cannot be decomposed as a direct sum of band representations; it describes a single band. We give a complete classification of the inequivalent elementary band representations.]]>	{"url":"https://omeka.ihes.fr/items/browse?tags=CLASSIFICATION&sort_field=added&sort_dir=a&output=dc-rdf","timestamp":"2024-11-10T11:33:53Z","content_type":"application/xml","content_length":"11433","record_id":"<urn:uuid:11488a9e-5c5b-4d44-90b5-465bf8ea3ae6>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00819.warc.gz"}
Mysterious Benedict Society -- Dividing Pie Here is a diagram of some pies with different numbers of cuts, (starting with zero): (Click on the diagram to get a clean image, suitable for printing. Can you cut the uncut pies to complete the diagram?) In each pie: • How many cuts (lines) are there? • How many cut (line) crossings are there? • How many pieces (regions) are there? Some things to consider: • If one cut crosses a second cut, then the second cut also crosses the first cut. The count of all the crossings made by all cuts is always an even number? • Is there a nice formula which relates the numbers of cuts, crossings, and pieces? • In the pies shown here, all the cut crossings are in the interior of the pie. How to count a crossing where cuts cross at the edge of the pie? • In the pies shown here, all the cut crossings involve only two cuts. How to count crossings where more than two cuts cross at the same point? (Is there a way to save your original formula?) • For a particular number of cuts, in how many different ways can the same number of pieces be cut? • To get the maximum number of pieces, each cut should cross all the other cuts. (Is that true?) Is that always possible? (Follow this link to see another diagram with fewer uncut pies.) (Follow this link to return to the problem page.) © 2017 Steven M. Schweda.	{"url":"http://antinode.org/complaints/mbs-cut.html","timestamp":"2024-11-10T23:34:37Z","content_type":"text/html","content_length":"2456","record_id":"<urn:uuid:f1f49476-1b5f-4d6a-b916-7054e4c6d89d>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00433.warc.gz"}
Statistical testing of baseline differences in sports medicine RCTs: a systematic evaluation For the reporting of randomised controlled trials (RCTs), item 15 of the CONSORT (Consolidated Standards of Reporting Trials) statement recommends that researchers report the baseline characteristics of each group, ideally in a table.1 However, the same item in the CONSORT statement also discourages statistical testing of differences in baseline covariates between randomised groups. Roughly speaking, standard statistical tests assess the probability that differences were due to chance given that the groups were the same. However, if a study is an RCT, then it is expected that the groups were the same, thus any baseline differences between the groups can be assumed to be due to chance. The CONSORT statement writes, ‘Tests of baseline differences are not necessarily wrong, just illogical. Such hypothesis testing is superfluous and can mislead investigators and their readers.’1 For example, in a study with few participants, there may be large differences between groups that do not reach a level of statistical significance and are thus ignored. Conversely, in a study with lots of participants, even small and meaningless differences may meet statistical significance and thus receive unnecessary attention. Although discouraged by the CONSORT statement, statistical testing (eg, with the calculation of a p value) of baseline differences in RCTs is still common.2 Knol et al reviewed RCTs published in seven leading medicine journals (eg, JAMA, BMJ and Lancet) from 2008 to 2010 and found that p values were listed in the baseline tables of about 35% of the studies.3 The primary purpose of this study was to determine the general proportion of RCTs in the sports medicine literature which included statistical tests of baseline differences. A secondary purpose was to assess the proportion of studies that included a table of baseline characteristics. In order to get a cursory evaluation as to the potential effect of the 2010 CONSORT statement, we chose to study RCTs published in the year 2005 or	{"url":"https://bmjopensem.bmj.com/content/3/1/e000228","timestamp":"2024-11-10T21:11:18Z","content_type":"text/html","content_length":"203339","record_id":"<urn:uuid:41f96e51-7bf1-4c5e-a737-4eff41922a46>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00668.warc.gz"}
count -- count the elements of an array that satisfy a given criterion #include <stddef.h> template<class T> ptrdiff_t count( const T& val, const T* b, const T* e template <class T> ptrdiff_t count_r( int (rel)(const T,const T), const T& val, const T b, const T* e template <class T> ptrdiff_t count_p( int (pred)(const T), const T* b, const T* e (1) For the plain version, T::operator== defines an equivalence relation on T. (2) For the relational version, rel defines an equivalence relation on T. These functions count elements that satisfy some criterion. template <class T> ptrdiff_t count( const T& val, const T* b, const T* e Counts elements equal to val, as determined by T::operator==. template <class T> ptrdiff_t count_r( int (rel)(const T,const T), const T& val, const T b, const T* e Like count, but uses rel to test for equality. That is, if p is a pointer into the array, then p is counted if rel(p,&val)==0. template <class T> ptrdiff_t count_p( int (pred)(const T), const T b, const T* e Counts elements that satisfy the predicate. That is, if p is a pointer into the array, then *p is counted if pred(p) is true. If N is the size of the array, then complexity is O(N). Exactly N tests of the relation are done. Because a Block (see Block(3C++)) can always be used wherever an array is called for, Array Algorithms can also be used with Blocks. In fact, these two components were actually designed to be used © 2004 The SCO Group, Inc. All rights reserved. UnixWare 7 Release 7.1.4 - 25 April 2004	{"url":"http://uw713doc.sco.com/en/man/html.3C%2B%2B/count.3C++.html","timestamp":"2024-11-05T12:17:12Z","content_type":"text/html","content_length":"5327","record_id":"<urn:uuid:438f8e68-4c56-49c8-bf69-b784e13f30c1>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00592.warc.gz"}
Conic Sections class 11 In Class 11, conic sections are studied as part of coordinate geometry, focusing on the properties and equations of curves formed by the intersection of a plane and a cone. The main conic sections studied are the circle, ellipse, parabola, and hyperbola. Here’s an overview: 1. Circle: A circle is the set of all points in a plane that are equidistant from a fixed point called the center. The standard form of the equation of a circle with center $(h, k)$ and radius $r$ is $(x – h)^2 + (y – k)^2 = r^2$. 2. Ellipse: An ellipse is the set of all points in a plane such that the sum of the distances from two fixed points (the foci) is constant. The standard form of the equation of an ellipse centered at the origin with major axis along the x-axis is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ and $b$ are the lengths of the semi-major and semi-minor axes, respectively. 3. Parabola: A parabola is the set of all points in a plane that are equidistant from a fixed point (the focus) and a fixed line (the directrix). The standard form of the equation of a parabola with vertex at the origin and axis parallel to the y-axis is $y^2 = 4ax$ or $x^2 = 4ay$, depending on the orientation. 4. Hyperbola: A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances from two fixed points (the foci) is constant. The standard form of the equation of a hyperbola centered at the origin with transverse axis along the x-axis is $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$, where $a$ and $b$ are the distances from the center to the vertices and to the foci, respectively. 5. Eccentricity: The eccentricity of a conic section is a measure of how “open” or “closed” the curve is. It is defined as the ratio of the distance from a point on the conic section to a focus to the distance from that point to the corresponding directrix. Studying conic sections helps in understanding the geometric properties of these curves and their applications in various fields, including physics, astronomy, and engineering. Recent Comments	{"url":"https://www.brainhubacademy.com/conic-sections-class-11/","timestamp":"2024-11-06T20:04:08Z","content_type":"text/html","content_length":"258118","record_id":"<urn:uuid:b97e93e6-8bba-45be-b275-ceed914c8eec>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00042.warc.gz"}
Time Complexity, Space Complexity and Big O Notation Time Complexity, Space Complexity and Big O Notation. Before we dive into time complexity, space complexity, and Big O notation, let’s talk about something slightly different—though not entirely outside coding. Since we’ve mentioned coding, let’s discuss it a bit more. When can we call a piece of code “good code” ? Answer: A piece of code is considered "good" when it is readable, maintainable, and scalable. Now, the question arises: What exactly is readability, maintainability, and scalability? To explain simply: What is Readability? Answer: Readability is writing your code in such a clean manner that others can easily understand it. Readability means writing clean code in such a way that other developers can easily comprehend it. Maintainable code simply means “code that is easy to modify or extend.” Scalable code refers to how well your code performs when the size of the input increases. Let’s try to understand this with an example: Say you write a solution to a problem, and the code works perfectly for 1,000 inputs. After some time, you try it with 5,000 inputs, and it still works fine. But when you use 10,000 inputs, the program seems to slow down. Now, when you use 1 million inputs, you notice that the program isn’t providing any output for 4-5 seconds. After 10 seconds, you finally get the output. This means that your algorithm works well with small inputs but fails to perform efficiently with larger inputs. So, from this example, we understand that when we write code or implement an algorithm, we need to consider how well it performs with small inputs and how it will handle a much larger volume of inputs. This helps us understand the scalability of the code. When implementing an algorithm, we need to consider a few things: • Will my algorithm produce a result within the specified time? • What is the maximum size of input my algorithm can handle? • How much memory does my algorithm consume? Keep these points in mind, as we’ll discuss them in more detail later. What is Time Complexity and Space Complexity? Let’s first try to understand time complexity in detail. Here’s an example: Let’s say you have a low-configuration computer, and your friend has a MacBook. Both of you enjoy problem-solving. You both find a problem online to solve: the problem requires printing 1,000 data points from an array. You write a loop and print all the data. Your friend also writes the exact same code to solve the problem. But here’s the issue: although you both wrote the same code, your code takes 10 seconds to run, whereas your friend’s code takes only 1 second. Now, the question is, which computer has better time complexity? If your answer is: Your friend’s MacBook Then, congratulations! Your answer is wrong. If your answer is: Your computer Then, congratulations again! Your answer is still wrong. So, what is the correct answer? The correct answer is: Both computers have the same time complexity. But how is this possible? How can both computers have the same time complexity? Let me explain. When we first start learning about time complexity, we often think that time complexity refers to how long it takes for a program to run. So, by that logic, we might assume that our friend’s MacBook has better time complexity. So, What is Time Complexity? Answer: Time complexity is a function that gives us the relationship between how time will grow as the input grows. Time Complexity ≠ Time Taken In simple terms, time complexity refers to how much time a function will take to run as the size of the input increases. Look at the image below: If we closely observe this image, we’ll see that the time taken increases linearly with the input size in both cases (the old computer and the MacBook). This is the essence of time complexity, and this is the mathematical function of time complexity. Now, let's discuss Big O Notation in detail and how to determine the time complexity of a program. What is Big O Notation? Big O notation is a mathematical notation that describes the limiting behavior of a function as the input approaches a particular value or infinity. To explain more simply, let’s recall what we mentioned earlier. We learned that time complexity refers to how long a function will take to execute as the size of the input increases. Now, how do we express the time complexity of a program? We express it using Big O Notation. Big O Notation is a mathematical way to express time complexity. Note: Even if you don’t fully grasp this concept yet, don't worry! Be patient and keep reading, and it will become clearer. Big O Notation is written like this: O(n) This is the graph of Big O Notation. Now, we will gradually discuss each notation, Insha'Allah. Let’s start by discussing a simple programming problem. Since you're still reading this post, I assume you already know how to solve this problem. You have an array with n elements (let’s say n names). From these n elements, you need to print either the first or the last element of the array. You're probably thinking, "Is this even a problem? I can just print the first element by writing arrayName[0]." (Let me know in the comments how to print the last element). You’re absolutely correct! This is exactly how we’ll solve the problem. I’ll solve it using JavaScript. In future posts. Feel free to solve it using your preferred programming language. javascript code: const students = ["Jhon", "Nozibul", "Islam", "Maria", "Samira"]; function printFirstElement(array){ Now, the question is, what is the time complexity of this program? Answer: The time complexity of this program is O(1) (Big O of one). O(1) is known as a constant time algorithm. Now, let’s understand in which cases a program takes time to execute: • When declaring variables, e.g., var name = "John"; • During operations, e.g., var a = 10; var b = 20; var c = a + b; • When calling a function, e.g., myFunction(); For each of these tasks, the time taken is O(1), meaning constant time. If a program contains several such constant time algorithms, we can say that the overall time complexity of the program is O If we want to visualize the time complexity of this program through a graph, the graph representation will look like this: Let’s look at another program: Javascript code: var name = "Jhon"; // O(1) console.log(name); // O(1) function logFirstItem(array){ console.log(array[0]); // O(1) logFirstItem([2, 4, 6]); // O(1) // Total Complexity: // O(1) + O(1) + O(1) + O(1) // O(4) // O(1) // So, the final complexity is: O(1) If the time complexity of a program is O(4), O(100), or O(1000), the key point is that it does not depend on the input size. For instance, in this program, regardless of the size of your array, the time complexity will always be O(4). Therefore, this algorithm is referred to as a constant time algorithm. Such constant time algorithms, like O(4), O(100), or O(1000), can simply be represented as O(1), meaning the complexity of this program is constant. Graph Representation: So, now I hope you can understand why the time complexity of our first program is O(1). we discussed the constant time algorithm Big O(1). Now, we will talk about other time complexities, Insha’Allah. First, let’s discuss a programming problem. Suppose there are 1000 student names in an array, and you need to print the names of each student from that array. You might be smiling a little, thinking, “What’s the problem? I will first get the length of the array and then run a loop to print all the data.” Your idea is completely correct. We will solve the problem this way. However, for easier understanding of the program, we will consider the names of just 5 students for now. Let’s take a look at the following program: Javascript code: let students = ["Jhon", "Nozibul", "Islam", "Maria", "Samira"]; let length = students.length; for(let i=0; i<length; i++){ What is the time complexity of this program? Answer: Big O(n) Now, the question is, how is it BigO(n)? If you observe the program carefully, you will see that the students array contains a total of 5 elements. Here, n is equal to 5. If there are 1,000 or 100,000 elements, regardless of how many elements there are, when we want to print each element, if we use iteration (the meaning of iteration is to touch each data in an array from start to finish using a loop), then the time complexity of that program will be Big O(n). Now, let's think about the same program in another case. Suppose we want to check if a student named "Islam" exists in the array. If we find our desired data, we will break the loop. After that, even if there are 1 million data points, it doesn't matter because we have found our desired data. Let's observe the following program: Javascript code: let students = ["Jhon", "Nozibul", "Islam", "Maria", "Samira"]; let length = students.length; for (let i = 0; i < length; i++) { if (students[i] === "Islam") { What is the time complexity of this program? Answer: Big O(n) Now, is there a bit of confusion? Because this time we didn't need to run the loop until n; we found our data before that, and the loop didn't run any further. So how can the time complexity of this program be Big O(n)? Let's try to understand this better. When we solve a programming problem, the time and space complexity of that program can be categorized into three types: 1. Best Case 2. Worst Case 3. Average Case The mathematical notation for Best Case, Worst Case, and Average Case is as follows: • Best Case → Omega (Ω) • Worst Case → Big O • Average Case → Theta (θ) When and Why are These Cases Used? • Best Case → Omega (Ω): This represents the lower bound, meaning it is used to express the minimum time complexity of a program. • Worst Case → Big O: This represents the upper bound, meaning it is used to express the maximum time complexity of a program. A program's time complexity will not exceed this value in its worst-case scenario (it can be less, but not more). • Average Case → Theta (θ): This represents the combined upper and lower bounds. If the upper and lower bound time complexities of a program are the same, then we use Theta (θ) to express that time complexity Note: For simplicity, I have only used the term "time complexity" here. These cases apply to both time complexity and space complexity. Let's return to our previous discussion. Why is the time complexity of our earlier program Big O(n)? Here, even though we found our desired data at the 3rd position, what could be the maximum worst-case scenario? Our desired data could be at the very last position, making the maximum time complexity Big O(n). Programmers should always consider what the worst-case scenario for the algorithm they develop could be. Big O notation is used to express the worst case of a program. In simple terms, if we iterate through n number of elements, then the time complexity of that program will be Big O(n). Graph representation: An algorithm with a time complexity of Big O(n) is called a linear time algorithm. Now, let's talk about a new programming problem. How would pairs (combinations) of the letters A, B, and C look? They would be: AA, AB, AC, BA, BB, BC, CA, CB, CC. Now we will solve a problem like Let’s look at the following program: Javascript code: let arr = ['a', 'b', 'c', 'd', 'e', 'f']; let lengthOfArray = arr.length; for (let i = 0; i < lengthOfArray; i++) { for (let j = 0; j < lengthOfArray; j++) { console.log(arr[i], arr[j]); Does the above program seem too difficult? If you understand nested loops, it shouldn't seem too hard. I hope you know about nested loops. What is the time complexity of this program? Answer: Big O(n²) If we look closely at the above program, we can see that in the first line we declared an array. What is the time complexity of declaring a variable? It’s Big O(1), right? In the second line, we calculate the length of the array and again declare it in a variable, and its time complexity is also Big O(1). Now, let’s analyze the two lines where we have the nested loops. Pay close attention to how the time complexity works here. When the first loop runs once, the second loop runs for n times. Again, when the first loop runs a second time, the second loop also runs for n times. This pattern continues, and when the first loop runs n times, the second loop will also run n times. This means that the two loops’ time complexities are multiplicative. Therefore, we can conclude that the time complexity of the loops is Big O(n * n) = Big O(n²). So, what is the overall time complexity of the program now? Code Explanation: let arr = ['a','b','c','d','e','f']; // O(1) let lengthOfArray = arr.length; // O(1) for(let i = 0; i<lengthOfArray; i++){ for(let j = 0; j<lengthOfArray; j++){ console.log(arr[i], arr[j]); } // O(n * n) // Total Complexity: // O(1) + O(1) + O(n * n) // O(2) + O(n^2) // O(n^2) // so, final complexity is: O(n^2) This is a constant time operation: O(1). Length Calculation: let lengthOfArray = arr.length; This is also a constant time operation: O(1). Nested Loops: The outer loop runs n times (where n is the length of the array). For each iteration of the outer loop, the inner loop also runs n times. Thus, the total operations for the nested loops: *O(n n) = O(n²). Total Complexity:** Combining these: O(1) + O(1) + O(n²) = O(2) + O(n²) = O(n²). Graph Representation The final time complexity of the above program is Big O(n²) through simplification. In the next part, we will try to learn which programming language to study for data structures and algorithms! Top comments (0) For further actions, you may consider blocking this person and/or reporting abuse	{"url":"https://practicaldev-herokuapp-com.global.ssl.fastly.net/nozibul_islam_113b1d5334f/time-complexity-space-complexity-and-big-o-notation-4pm9","timestamp":"2024-11-08T02:14:24Z","content_type":"text/html","content_length":"89006","record_id":"<urn:uuid:c451a048-dcec-474a-9a82-e0d90bca6f24>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00794.warc.gz"}
Things to Know Things to Know 1. Solving systems of linear equations. 2. What is the coefficient matrix of a system of linear matrix? 3. What is the augmented matrix? 4. Write the system if matrix form. How do you go back and forth between the matrix form and the system form? 5. Different types of matrices (square, diagonal, upper triangular, etc. . . ) 6. What is the row reduced echelon form of a matrix? 7. Solve a system of linear equations using the rref. 8. What is the inverse of a matrix? How do you find it? 9. The rank of a matrix. 10. The number of solutions of a system of linear equations. 11. Linear combinations. 12. Dot product. 13. Operations with matrices and vectors. What is I[n]? 14. Order of operations. Properties. 15. Linear transformations and their properties. The matrix of a linear transformation. 16. Linear transformations and geometry: transformations in the plane and in the 3- dimensional space. 17. The span of a set of vectors. 18. Kernel and image: what they are, properties, how to find them. When is the kernel 19. Various characterizations of invertibility (at least 10). These are not on the exam, but they are helpful anyway, especially for section 3.1. 20. Linear subspaces. Dimension. 21. Linear independence and its various characterizations. 22. Redundant vectors. 23. Basis. 24. Linear relations. 25. Rank-nullity theorem.	{"url":"https://mathmastersnyc.com/things-to-know.html","timestamp":"2024-11-02T21:22:12Z","content_type":"text/html","content_length":"84275","record_id":"<urn:uuid:2d687c36-82dd-48f6-8d88-458992aee59c>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00623.warc.gz"}
Convert API User Guide: Changelog November 7, 2024 • Deployed new versions: RSK-M134p4 (v3/text) and RSK-P124p4 (v3/pdf) • Preventing multicolumn overflow, respecting the total number of table columns November 4, 2024 • Deployed new versions RSK-P124p3 (v3/pdf) • Conversion improvements November 3, 2024 • Deployed new version RSK-M134p3 (v3/text) • Returning Content not found error in some edge cases that previously resulted in internal error November 1, 2024 • Deployed new versions: RSK-M134p2 (v3/text) and RSK-P124p2 (v3/pdf) • Fixes for rare bugs that occurred during table segmentation • Better support for custom math display delimiters like ["\\begin{equation}\n", "\n\\end{equation}"] with equation tags Most common choices like $$ and \[, \] work well in cases like: x &= y^2 \tag{1}\\ &= 16 \tag{2} while display delimiters like \begin{equation}, \end{equation}, do not. For example, this is not rendering due to nesting of math environments: x &= y^2 \tag{1}\\ &= 16 \tag{2} To prevent bad outcomes, we’re now keeping the original math environment when necessary, in order to support equation tags properly. We omit equation environment and produce the following output: x &= y^2 \tag{1}\\ &= 16 \tag{2} October 28, 2024 • Deployed new versions: RSK-M134p1 (v3/text) and RSK-P124p1 (v3/pdf) • Minor pre-processing changes October 22, 2024 • Deployed new version RSK-P124 (v3/pdf) • Added support for rotated parts of PDF page(s) (e.g. 90 degrees rotated table) Previously, v3/pdf wasn’t handling PDF pages with rotated content properly, leading to parts of page being digitized as an image link instead of recognized content. This has changed with current For example, a PDF page like this one: is now processed in v3/pdf into the following MMD: Table 1 Common Q System Qualified Components \hline Item & Component & \begin{tabular}{l} Product Revision \\ (Up to this revision reviewed for this SE) \end{tabular} & Description \\ \hline AC160 Hardware Modules & & & \\ \hline 1 & Al620 & A & S600 Analog Input Module \\ \hline 2 & Al685 & G & S600 Analog Input Module \\ \hline 3 & Al687 & C & S600 Analog Input Module \\ \hline 4 & Al688 & A & S600 Analog Input Module \\ \hline 5 & A0610 & A & S600 Analog Output Module \\ \hline 6 & A0650 & A & S600 Analog Output Module \\ \hline 7 & CI527W & C & Communications Interface Module \\ \hline 8 & Cl631 & F & Communications Interface Module \\ \hline 9 & DI620 & A & S600 Digital Input Module \\ \hline 10 & DI621 & A & S600 Digital Input Module \\ \hline 11 & D0620 & C & \begin{tabular}{l} S600 Digital Output \\ \end{tabular} \\ \hline 12 & D0625 & A & \begin{tabular}{l} S600 Digital Output \\ \end{tabular} \\ \hline 13 & D0630 & A & \begin{tabular}{l} S600 Digital Output \\ \end{tabular} \\ which renders as: October 17, 2024 • Deployed new versions: RSK-M134 (v3/text) and RSK-P123 (v3/pdf) • OCR engine general accuracy improvements October 8, 2024 • Deployed new versions RSK-P122p2 (v3/pdf) • Conversion improvements October 6, 2024 • Deployed new versions: RSK-M133p1 (v3/text) and RSK-P122p1 (v3/pdf) • Improving post-processing robustness in extremely rare edge case October 3, 2024 • Deployed new versions: RSK-M133 (v3/text) and RSK-P122 (v3/pdf) • Table recognition improvements, with accent on math-heavy tables For example, complex tables like this: are now successfully recognized, and the output returned renders as: Corresponding MMD: \hline Question & Scheme & Marks & Aos \\ \hline \multirow[t]{5}{}{(a)} & A complete method to use the scalar product of the direction vectors and the angle $ 120^{\circ} $ to form an equation in $ a $ 2 \\ a \\ \end{array}\right) \cdot\left(\begin{array}{r} 0 \\ 1 \\ \end{array}\right)}{\sqrt{2^{2}+a^{2}} \sqrt{1^{2}+(-1)^{2}}}=\cos 120 \] & M1 & $ 3.1 b $ \\ \hline & \[ \frac{a}{\sqrt{4+a^{2}} \sqrt{2}}=-\frac{1}{2} \] & A1 & 1.1 b \\ \hline & $ 2 a=-\sqrt{4+a^{2}} \sqrt{2} \Rightarrow 4 a^{2}=8+2 a^{2} \Rightarrow a^{2}=4 \Rightarrow a=\ldots $ & M1 & $ 1.1 b $ \\ \hline & $ a=-2 $ & A1 & 2.2 a \\ \hline & & (4) & \\ \hline \multirow[t]{6}{}{(b)} & \[ \text { Any two of } \mathbf{i}: & -1+2 \lambda=4 \\ & \mathbf{j}: 5+\text { 'their }-2 ' \lambda=-1+\mu \tag{2}\\ & \mathbf{k}: \quad 2=3-\mu \tag{3} \] & M1 & 3.4 \\ \hline & Solves the equations to find a value of $ \lambda\left\{=\frac{5}{2}\right\} $ and $ \mu\{=1\} $ & M1 & 1.1 b \\ \hline & \[ -1 \\ 5 \\ 2 \\ \text { 'their }-2^{\prime} \\ \end{array}\right) \text { or } r_{2}=\left(\begin{array}{c} 4 \\ -1 \\ 0 \\ 1 \\ \] & dM1 & 1.1 b \\ \hline & \[ (4,0,2) \text { or }\left(\begin{array}{l} 4 \\ 0 \\ \] & A1 & 1.1 b \\ \hline & \begin{tabular}{l} Checks the third equation e.g. \lambda=\frac{5}{2}: \mathrm{L} & \mathrm{HS}=5-2 \lambda=5-5=0 \\ \mu=1: \mathrm{R} & \mathrm{HS}=-1+\mu=-1+1=0 \] \\ therefore common point/intersect/consistent/tick or substitutes the values of $ \lambda $ and $ \mu $ into the relevant lines and achieves the same coordinate \end{tabular} & B1 & 2.1 \\ \hline & & (5) & \\ October 2, 2024 • Deployed new versions: RSK-M132p13 (v3/text) and RSK-P121p7 (v3/pdf) • Post-processing improvements September 30, 2024 • Deployed new versions RSK-P121p6 (v3/pdf) • Added parameter include_chemistry_as_image to return chemistry as an image crop with SMILES content in the alt-text September 23, 2024 • Deployed new versions RSK-P121p5 (v3/pdf) • Conversion improvements September 23, 2024 • Deployed new versions: RSK-M132p12 (v3/text) and RSK-P121p4 (v3/pdf) • Post-processing improvements September 20, 2024 • Deployed new versions: RSK-M132p11i1 (v3/text) and RSK-P121i1p3 (v3/pdf) • Choosing table segmentation as table OCR algorithm more frequently September 18, 2024 • Deployed new versions RSK-M132p10i1 (v3/text) and RSK-P121i1p2 (v3/pdf) • Fixes double output in latex_styled format in some cases • Post-processing changes (v3/pdf) September 12, 2024 • Deployed new version RSK-P121i1p1 (v3/pdf) • Fixing very rare cases where diagrams get post processed incorrectly September 10, 2024 • Deployed new version RSK-P121 (v3/pdf) • Model updates September 6, 2024 • Deployed new version RSK-M132p9 (v3/text) • Prevent generating image link in the output for charts in certain cases September 5, 2024 • Deployed new version RSK-P120 (v3/pdf) • Model updates September 2, 2024 • Deployed new versions: RSK-M132p8 (v3/text) and RSK-P119p1 (v3/pdf) • Corrects wrong equation ordering in the output which was happening in certain cases September 1, 2024 • Deployed new version RSK-P119 (v3/pdf) • Model re-trained on more data August 30, 2024 • Deployed new versions: RSK-M132p7 (v3/text) and RSK-P118p10 (v3/pdf) • Updated MathJax@3.2.2 • Improved Asciimath outputs. An extra space is no longer added after the function name and before the opening bracket (e.g., arctan((1)/(x)) instead of arctan ((1)/(x))) August 29, 2024 • Deployed new versions: RSK-M132p6 (v3/text) and RSK-P118p9 (v3/pdf) • Post-processing fixes. August 26, 2024 • Renamed OCR API to Convert API August 23, 2024 • Deployed new version RSK-P118p8 (v3/pdf) • Code blocks now have double newline separation when multiple pages are joined which is more consistent (it used to be a single newline). • Deployed new versions: RSK-M132p5 (v3/text) and RSK-P118p7 (v3/pdf) • Post-processing improvements for cases when diagrams have significant intersections with text lines. August 21, 2024 • Deployed new versions: RSK-M132p4 (v3/text) and RSK-P118p6 (v3/pdf) • Fixes post-processing issues related to code recognition and footnote text. • Fixes some internal errors. Single line footnote text was sometimes left unclosed, without a matching closing }. For example, response from the API would return something like: My single line footnote This deployment corrects it to: * My single line footnote In some ambiguous cases, we have been omitting the closing triple back ticks which are MMD code delimiters. That has lead to improper rendering in those cases. Most of these situations, if not all, should be fixed by this deployment. August 20, 2024 • Deployed new versions: RSK-M132p3 (v3/text) and RSK-P118p5 (v3/pdf) • Improved processing of tables with complex content inside cells. This deployment improves on table OCR when tables have cells with complex content. Here is an example of a very simple table with a complex cell just to illustrate the type of data that these changes affect positively: First cell of the second row has 3 lines of text and an equation. The table gets processed into MMD like this one: \hline First column & Second column & Third column \\ \hline \begin{tabular}{l} This cell has multiple lines \\ which enables testing tableocr properly. \\ Here goes the equation: \\ $ L=-\frac{1}{N} \sum_{i=0}^{N-1} y_{i} \log \left(\hat{y}_{i}\right) $ \end{tabular} & Normal cell & Normal cell \\ which renders as: August 16, 2024 • Deployed new version: RSK-M132p2 (v3/text) • Adds back missing line data formats (data and html) • Resolves some cases of incorrect error reporting for v3/latex endpoint August 15, 2024 • Deployed new versions: RSK-M132p1 (v3/text) and RSK-P118p4 (v3/pdf) • Enables "include_equation_tags": boolean boolean option for v3/text (default is false). • Post-processing improvements. August 14, 2024 • Deployed new version: RSK-P118p3 (v3/pdf) • Fixes diagram post-processing issues. • Adds request option include_smiles for digitizing chemistry diagrams (true by default). When false chemistry diagram is preserved as an image. August 10, 2024 • Deployed new version: RSK-P118p2 (v3/pdf) • Fixes rare post-processing issues related to equations processing. August 6, 2024 • Deployed new version: RSK-P118p1 (v3/pdf) • Fixes post-processing issues related to tables processing. July 29, 2024 • Deployed new versions: RSK-M132 (v3/text) and RSK-P118 (v3/pdf) • OCR engine updates July 26, 2024 • Deployed new version: RSK-P117p5i4 (v3/pdf) • Correctly including equation tags in some edge cases. July 10, 2024 • Deployed new versions: RSK-M131p6i3 (v3/text) and RSK-P117p3i2 (v3/pdf) • We now move “$1\mathrm{t}\mathrm{e}\mathrm{x}\mathrm{t}$$1\mathrm{t}\mathrm{e}\mathrm{x}\mathrm{t}$1text1 \mathrm { text }” to “$1text$$1text$1text1 text” wherever possible. We also have an API option math_fonts_default_to_math to disable this behavior. June 22, 2024 • Deployed new version RSK-M131p6i2 (v3/text) • Fixes internal errors in v3/text when invalid format is specified in "formats" request argument. Instead of internal error we now return results for all supported formats the were specified in the request, and we ignore the unsupported ones. For details on supported formats see here. June 19, 2024 • Deployed new version RSK-M131p4i1 (v3/text) • Fixes parsing of chemicals in wrongly rotated images when autorotation is detected correctly. June 12, 2024 • Deployed new version RSK-M131p3i1 (v3/text) • Fixes wrong items in detection map output, namely contains_chart, and contains_graph June 9, 2024 • Deployed new version RSK-P117p1 (v3/pdf) • Fixes handling of unrecognized tables in post-processing. June 7, 2024 • Deployed new versions RSK-P117 (v3/pdf) • This update fixes problems caused by rotated side text detection, improves source code recognition, and recognition of very small lines (e.g. single character line {) May 31, 2024 • Deployed new versions RSK-P116p2i1 (v3/pdf) • Given $ as inline math delimiter, form fields are recognized without surrounding spaces as $\qquad$ instead of $ \qquad $ because the former has rendering issues May 30, 2024 • Deployed new versions: RSK-M131p2i1 (v3/text) and RSK-P116p1i1 (v3/pdf) • Post-processing fixes that eliminate several very rare internal errors. We now either return requested content, or no content if there is nothing in the image. May 27, 2024 • Deployed new versions: RSK-M131 (v3/text) and RSK-P116 (v3/pdf) • Improvements of source code recognition; support for non-English languages in source code For example, given an image like this: returned response now includes correctly recognized Chinese ideographs: def test_function(): # 测试数字之和 sum: int = 0 for i in range(1, 11): sum += i assert sum == 55 May 20, 2024 • Deployed new version: RSK-M130 (v3/text) • Improved subtype chart detection (added analytical subtype) Requests that specify "include_line_data": true and contain an image such as: will receive line data as: "type": "text", "cnt": [ "included": true, "is_printed": true, "is_handwritten": false, "text": "What is the area of the trapezoid?", "after_hyphen": false, "confidence": 1.0, "confidence_rate": 1.0 "type": "chart", "cnt": [ "included": false, "is_printed": true, "is_handwritten": false, "subtype": "analytical", "error_id": "image_not_supported" which has an item with "type": "chart" and "subtype": "analytical". May 15, 2024 • Deployed new version: RSK-P115 (v3/pdf) • Improved handling of large capital letters at the beginning of paragraphs (see below) For cases like this one, where the paragraph starts with a very large letter: we now return correctly recognized text as: There is now a substantial literature that connects religion and ...rest of text... with a focus on starting T being handled properly. May 9, 2024 • Deployed new versions: RSK-M129p2i1 (v3/text) and RSK-P114p2i1 (v3/pdf) • Performance improvements: □ Faster response times across all endpoints. □ Large PDFs should now go from loaded to split much faster (splitting time). May 1, 2024 • Deployed new versions: RSK-M129p1 (v3/text) and RSK-P114p1 (v3/pdf) • Post-processing updates for source code, among others: □ \rightarrow to -> □ \Rightarrow to => April 30, 2024 • Deployed new versions: RSK-M129 (v3/text) and RSK-P114 (v3/pdf) • Update improves on handwritten Japanese and source code recognition This update brings initial support for correct source code recognition from images. For example, for an image with source code listing like the following: Mathpix now returns: By combining multimethods and packages, we can simplify the abstract syntax tree example by bundling the visitation methods for each tree operation in a package. object TypeChecker { void Visit(AssignmentNode & n) { // ... Visit (n.LHS ()); Visit (n.RHS ()); // ... void Visit(VariableRefNode & n); // ... object CodeGenerator { void Visit(AssignmentNode & n); void Visit(VariableRefNode & n); // ... Node & root = // ... which gets renders as: Similarly, pseudocodes are also wrapped in three back ticks, while Latex with math notation is output. Note, this format for pseudocodes is likely to change in the future. For pseudocode like the following image: the output returned looks like: Algorithm 3 An algorithm to divide a given size budget among subexpressions ${ }^{9}$ func Divide ( $a$ : Arity, $q$ : Size, $l$ : Op. Level, $j$ : Expr. Level, $\alpha$ : Accumulated Args.) - Requires: $1 \leq a \leq q \wedge l \leq j$ if $a=1$ then if $l=j \vee \exists\langle x, y\rangle \in \alpha: x=j$ then return $\{(1, q) \diamond \alpha, \ldots,(j, q) \diamond \alpha\}$ return $\{(j, q) \diamond \alpha\}$ for $u \leftarrow 1$ to $j$ do for $v \leftarrow 1$ to $(q-a+1)$ do $L \leftarrow L \cup \operatorname{DIVIDE}(a-1, q-v, l, j,(u, v) \diamond \alpha)$ return $L$ If one wants to render recognized pseudocode, it is recommended to remove back ticks and replace indents of 4 spaces with $\quad$ . After these transformations the result is: While Mathpix used to return links of image crop for source code and pseudocodes v3/pdf, that behavior is now changed and v3/pdf now returns MMD outputs as described in this changelog. April 27, 2024 • Deployed new versions: RSK-M128p3 (v3/text) and RSK-P113i1p1 (v3/pdf) □ We now transform \begin{array}{l}{...x...}\\{...y...}\end{array} to \binom{...x...}{...y...}. April 18, 2024 • Deployed a new version: RSK-M128p2 (v3/text). □ Minor post-processing fixes. • Deployed new version: RSK-P112 (v3/pdf) □ Initial support for detection of in-text placeholders to be filled in (form_field). □ Algorithm updated with more data. April 4, 2024 • Deployed new version RSK-M128 (v3/text) • Improved source code and pseudo code indentation □ source code is enclosed with triple backticks • Support for detection of in-text placeholders to be filled in (form_field) □ placeholder underscores, dots, and dashes are output as $ \\qquad $ □ placeholder boxes are output as $\square$ March 26, 2024 • Deployed new version: RSK-P111p6i2 (v3/pdf) • Improved export to DOCX: added flattening of nested text tables if the nested table is just one column with several rows. March 15, 2024 • Deployed new version: RSK-P111p5i2 (v3/pdf) • MMD to PDF-LaTeX conversion improvements □ Added additional fonts to support a broader range of scripts. □ Expanded language detection capabilities to accurately and more reliably handle texts. □ Language detection extended to include Thai, Tamil, Hebrew, Hindi, and Bengali. □ Automatically use XeLaTeX for documents containing non-Latin, Greek, Cyrillic (LGC) scripts, ensuring better rendering of multilingual texts. □ Standardized font usage to the Noto family for non-LGC scripts, providing consistency and extensive script support across documents. • MMD to LaTeX conversions and the tex.zip extension’s output may now include code requiring XeLaTeX for proper rendering of non-LGC texts. It’s notable that only Note to PDF-LaTeX allows for specific font selection. For MMD to LaTeX and the output from the tex.zip extension, conversions standardize on CMU Serif and Noto Serif for non-LGC scripts. March 13, 2024 • Deployed a new version: RSK-M127p5i1 (v3/text). • Fixed incorrect duplication of the “caret” in Asciimath outputs (e.g., 100(1.03)^2t=5000 instead of 100(1.03)^(^^)2t=5000). • Deployed new version: RSK-P111p4i2 (v3/pdf) • Improved export to DOCX: added the ability to automatically detect the language for the entire document. February 29, 2024 • Deployed new versions: RSK-M127p4i1 (v3/text and v3/latex) and RSK-P111p3i2 (v3/pdf) • General efficiency improvements. February 26, 2024 • Deployed new version: RSK-P111p3i1 (v3/pdf) • Improved export to LaTeX (tex.zip): we now use double backslashes (\\) for line breaks in the Table of Contents and various sections without starting a new paragraph in text mode. February 22, 2024 • Deployed a new version: RSK-M127p3 (v3/text). • Fixed incorrect grouping of fractional function arguments in Asciimath outputs, fractional function arguments are now properly contained within functions by being wrapped in parentheses (e.g., sec ((5theta)/(4))=2 instead of sec (5theta)/(4)=2 and log ((x)/(y^(5))) instead of log (x)/(y^(5))). February 16, 2024 • Deployed a new version: RSK-M127p3 (v3/text and v3/latex). • We now validate the region parameter and return appropriate error messages. February 6, 2024 • Deployed new version: RSK-P111p2 (v3/pdf) • Added new API option "include_equation_tags": boolean We’re excited to announce an enhancement to our v3/pdf endpoints: the ability to include recognized equation numbers directly within the MMD output. Previously, equation numbers were only accessible through the lines.json output format, serving primarily for search purposes without visibility in the main MMD content. With the new "include_equation_tags": true parameter, our system now integrates equation numbers seamlessly into the MMD output, utilizing the \tag element for clear association. This improvement enriches the MMD files by directly linking equations with their corresponding numbers, facilitating easier reference and navigation. Here is an example of how the result changes: For this part of the PDF file we currently return this MMD: It is not immediately obvious from Maxwells equations that the time-varying current is the source of radiation. A simple transformation of the Maxwells equations \nabla \times \mathbf{E} & =-\mu \frac{\partial \mathbf{H}}{\partial t} \\ \nabla \times \mathbf{H} & =\varepsilon \frac{\partial \mathbf{E}}{\partial t}+\mathbf{J} into a single second-order equation either for $\mathbf{E}$ or for $\mathbf{H}$ proves this statement. By taking the curl of both sides of the first equation in (1.2) and by making use of the second equation in (1.2), we obtain \nabla \times \nabla \times \mathbf{E}+\mu \varepsilon \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}=-\mu \frac{\partial \mathbf{J}}{\partial t} From (1.3), it is obvious that the time derivative of the electric current is the source for the wave-like vector $\mathbf{E}$. With the new option enabled ("include_equation_tags": true), we now return: It is not immediately obvious from Maxwells equations that the time-varying current is the source of radiation. A simple transformation of the Maxwells equations \nabla \times \mathbf{E} & =-\mu \frac{\partial \mathbf{H}}{\partial t} \tag{1.2a}\\ \nabla \times \mathbf{H} & =\varepsilon \frac{\partial \mathbf{E}}{\partial t}+\mathbf{J} \tag{1.2b} into a single second-order equation either for $\mathbf{E}$ or for $\mathbf{H}$ proves this statement. By taking the curl of both sides of the first equation in (1.2) and by making use of the second equation in (1.2), we obtain \nabla \times \nabla \times \mathbf{E}+\mu \varepsilon \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}=-\mu \frac{\partial \mathbf{J}}{\partial t} \tag{1.3} From (1.3), it is obvious that the time derivative of the electric current is the source for the wave-like vector $\mathbf{E}$. Besides added equations numbers, there are other differences in the output: 1. Individual numbered equations are wrapped in \begin{equation}...\end{equation} to support numbering. 2. The aligned environment is now replaced with align* to support numbering □ We also replace gathered with gather. □ In general, we will use the environments that support numbering when using "include_equation_tags": true. 3. These changes are also visible in lines.mmd.json format. Here is what the rendering of the recognized portion of PDF page looks like: Note that there are some limitations in the current implementation: 1. We do not generate equation references in the text (e.g. \ref{eq:1.3}). In the given example, in the last paragraph, the equation (1.3) is being referenced. While that reference is straight forward, equations (1.2a) and (1.2b) are being referenced as the first equation in (1.2) and the second equation in (1.2). This demonstrates that it takes significant semantic understanding of the document content to correctly unravel all the references, and is beyond the scope of this update. 2. We do not generate equation labels at the moment. The reasons are: □ We do not generate \ref elements in text, and that would make \label elements redundant □ Some widely used LaTeX rendering libraries do not support \label and \ref 3. When we have a block equation for which we output an array environment, and this block has multiple equation numbers associated with it, only the last equation number is being tagged. This is because LaTeX allows only one equation number per array. The same holds for the cases environment. We do plan to add support for the numcases environment which will enable multiple tags per one block of equations. February 3, 2024 • Deployed new ocr-version RSK-M127p2 • Fixes internal error handling February 3, 2024 • Deployed new versions: RSK-M127p1, RSK-P111p1 (v3/pdf) • Improvements to tabular outputs. February 2, 2024 • Deployed new versions: RSK-M127, RSK-P111 (v3/pdf); • Initial OCR support Jordan matrices, improvements for tensor indices, and handwritten content; For example, Mathpix can now correctly recognize matrices as this one: and return the result: $ \left(\begin{array}{ccc}\boxed{\begin{array}{cc}18 & -15 \\ 89 & 22\end{array}} & & \\ & \ddots & \\ & & \boxed{\begin{array}{cc}-83 & -6 \\ 9 & 3\end{array}}\end{array}\right) $ January 27, 2024 • Deployed new versions: RSK-M126, RSK-P110 (v3/pdf); • OCR algorithm improvements for future chemistry releases; January 22, 2024 • Deployed new versions: RSK-M125, RSK-P109 (v3/pdf); • Accuracy improvements of content detection in hard images/pages. January 17, 2024 • Deployed new versions: RSK-M124p2; • Changed chemistry outputs; January 16, 2024 • Deployed new versions: RSK-M124p1; • Changed expand_chemistry to include_chemistry; January 15, 2024 • Deployed new versions: RSK-M124; • OCR improvements for small images; January 11, 2024 • Deployed new versions: RSK-M123, RSK-P108 (v3/pdf); • OCR improvements of images with complex tensor indices; For example, for this image: the result LaTeX used to be: $ \epsilon_\sigma^{\mu \nu \rho} $ For the same image, the new deployment now returns: $ \epsilon^{\mu \nu \rho}{ }_{\sigma} $ Here is the difference in rendering previous and current results showing the improvement: January 9, 2024 • Deployed new versions: RSK-M122p1; • Fixes for internal errors in post-processing when "include_word_data": true is used in the request; December 28, 2023 • Deployed new versions: RSK-M122, RSK-P107 (v3/pdf); • Information if a line is printed, handwritten, or both is added to line data outputs; Text in image or page can be printed, handwritten, or both. We have exposed this information through our line API. For example, for this image: and request: "src": "https://mathpix.com/examples/printed_handwritten_0.jpg", "formats": [ "include_line_data": true the response will look like: "image_width": 808, "image_height": 404, "is_printed": true, "is_handwritten": true, "auto_rotate_confidence": 0.0, "auto_rotate_degrees": 0, "confidence": 1.0, "confidence_rate": 1.0, "text": "b) Absolute-Value Function:\n\\[\nf(x)=\\frac{1}{2}\|x-1\|-2\n\\]", "languages_detected": [], "line_data": [ "type": "text", "cnt": [ "included": true, "is_printed": true, "is_handwritten": false, "text": "b) Absolute-Value Function:", "after_hyphen": false, "confidence": 1.0, "confidence_rate": 1.0 "type": "math", "cnt": [ "included": true, "is_printed": false, "is_handwritten": true, "text": "\n\\[\nf(x)=\\frac{1}{2}\|x-1\|-2\n\\]", "after_hyphen": false, "confidence": 1.0, "confidence_rate": 1.0 In the line data, sentence has is_printed true and is_handwritten false, while the math has is_handwritten true and is_printed false. Sometimes, the content can be both printed and handwritten at the same time. Here is one such example: For the request: "src": "https://mathpix.com/examples/printed_handwritten_1.jpg", "formats": [ "include_line_data": true the response will look like: "image_width": 354, "image_height": 120, "is_printed": true, "is_handwritten": true, "auto_rotate_confidence": 0.0, "auto_rotate_degrees": 0, "confidence": 0.99951171875, "confidence_rate": 0.99951171875, "latex_styled": "\\frac{9}{12}+\\frac{2}{11}=\\frac{99}{132}+\\frac{24}{132}=\\frac{123}{132}", "text": "\$ \\frac{9}{12}+\\frac{2}{11}=\\frac{99}{132}+\\frac{24}{132}=\\frac{123}{132} \$", "languages_detected": [], "line_data": [ "type": "math", "cnt": [ "included": true, "is_printed": true, "is_handwritten": true, "text": "\$ \\frac{9}{12}+\\frac{2}{11}=\\frac{99}{132}+\\frac{24}{132}=\\frac{123}{132} \$", "after_hyphen": false, "confidence": 0.99951171875, "confidence_rate": 0.9999904235654993 It can be seen in the line_data that the line object has both is_printed and is_handwritten set to true. In v3/pdf endpoint(s) lines.json and lines.mmd.json outputs are updated with new is_printedand is_handwritten fields for each line. December 13, 2023 • Deployed new versions: RSK-M121, RSK-P106 (v3/pdf); • Added support for the 12 new LaTeX commands; • Improved recognition of text lines with many dots (e.g. text lines in table of contents images); • Improved recognition of languages that use Cyrillic alphabets. We added recognition support for 12 new LaTeX commands that were previously unsupported. $ \xrightarrow{\mathrm{H}^{\oplus} / \mathrm{H}_{2} \mathrm{O}} $ $ E_{2} \xlongequal[r_{4}+r_{8}]{\substack{r_{9}-r_{5} \\ c_{8}-c_{5}, \ldots, c_{5}+c_{5}}}\left\|\begin{array}{cccc}-1 & 7 & -4 & -1 \\ -2 & 0 & -4 & -2 \\ -4 & -4 & 8 & -1 \\ -9 & -2 & 6 & 2\end {array}\right\| $ $ \frac{1}{g_{4}^{2}} \lessdot-\frac{2 k N_{c} \ln (v)}{g_{5}^{2}} \gtrdot 0 $ $ (u \boxtimes v) \cdot(w \boxtimes x)=(v \cdot w)(u \boxtimes x) $ $ u_{r r} \downharpoonleft \frac{\sigma_{r r}}{E} $ $ \int \rho_{i k} d f_{i} \downharpoonright 0 $ $ (u \nabla) v \leftharpoondown-\frac{1}{\rho} \nabla p+\nu \Delta v $ $ \phi(u, v) \leftharpoonup \arctan \left[\frac{I(u, v)}{R(u, v)}\right] $ $ \zeta \rightharpoondown i\left(\frac{\pi}{2}+\theta\right) $ $ f(t) h(t) \rightharpoonup H(\mu) \div F(\mu) $ $ j \upharpoonleft-\frac{2 \mu \alpha_{1} I_{1}}{v_{0}} $ $ W \upharpoonright N f\left(\frac{S}{N}, P\right) $ December 7, 2023 • Deployed new version (v3/pdf), RSK-P105; • Another improvement for table of contents pages. November 28, 2023 • Deployed new version (v3/pdf), RSK-P104; • Improved processing of table of contents pages. October 31, 2023 • Deployed new version (RSK-M120); • Chart detection; This is a first step towards recognition of charts. Basic charts detection is supported in images. When "include_line_data": true is specified in the request, for all the charts we now return correct chart type in line data response instead of generic diagram. Currently, several types of charts are being detected: 1. column chart 2. bar chart 3. line chart 4. pie chart 5. area chart 6. scatter chart The type of chart is available as "subtype": ... field of the line object. This list will be extended with additional categories in the future. • Algorithm pseudo code recognition improvements. We used to fail recognition for images with complex algorithm pseudocode like this one: For such image, the returned MMD now looks like this: Algorithm 1 Gradient Sign Dropout Layer (GradDrop Layer) choose monotonic activation function $ f \quad \triangleright $ Usually just $ f(p)=p $ choose input layer of activations $ A \quad \triangleright $ Usually the last shared layer choose leak parameters $ \left\{\ell_{1}, \ldots, \ell_{n}\right\} \in[0,1] \quad \triangleright $ For pure GradDrop set all to 0 choose final loss functions $ L_{1}, \ldots, L_{n} $ function $ \operatorname{BACKWARD}\left(A, L_{1}, \ldots, L_{n}\right) \quad \triangleright $ returns total gradient after GradDrop layer for $ i $ in $ \{1, \ldots, n\} $ do calculate $ G_{i}=\operatorname{sgn}(A) \circ \nabla_{A} L_{i} \quad \triangleright \operatorname{sgn}(\mathrm{A}) $ inspired by Equation 3 if $ G_{i} $ is batch separated then $ G_{i} \leftarrow \sum_{\text {batchdim }} G_{i} $ calculate $ \mathcal{P}=\frac{1}{2}\left(1+\frac{\sum_{i} G_{i}}{\sum_{i}\left\|G_{i}\right\|}\right) \quad \triangleright \mathcal{P} $ has the same shape as $ G_{1} $ sample $ U $, a tensor with the same shape as $ \mathcal{P} $ and $ U[i, j, \ldots] \sim \operatorname{Uniform}(0,1) $ for $ i $ in $ \{1, \ldots, n\} $ do calculate $ \mathcal{M}_{i}=\mathcal{I}[f(\mathcal{P})>U] \circ \mathcal{I}\left[G_{i}>0\right]+\mathcal{I}[f(\mathcal{P})<U] \circ \mathcal{I}\left[G_{i}<0\right] $ set newgrad $ =\sum_{i}\left(\ell_{i}+\left(1-\ell_{i}\right) \mathcal{M}_{i}\right) \circ \nabla_{A} L_{i} $ return newgrad and it renders as: While the correct indentation is still missing, the content is now being returned instead of Content not found error. October 30, 2023 • Deployed new version, RSK-P103p6 (v3/pdf); • Added support for remove_section_numbering and preserve_section_numbering. Default behavior is changed to preserve_section_numbering: True. Note that only one of auto_number_sections, remove_section_numbering, or preserve_section_numbering can be true at a time. October 27, 2023 • Deployed new versions: RSK-M119p7, RSK-P103p5 (v3/pdf); • Post-processing improvements: we now return result instead of Internal error for some classes of challenging inputs. October 21, 2023 • Deployed new version, RSK-P103p3 (v3/pdf); • Fixed new line inconsistencies that occurred when content from different pages is joined; • Single new line is generated before \footnotetext instead of two new lines; • Two new lines introduce a new paragraph, and that behavior was breaking rendering experience in some cases. October 19, 2023 • Deployed new version, RSK-M119p6; • False paragraph detection fixed for images with list items that start with Cyrillic letters. October 18, 2023 • Deployed new version, RSK-M119p5; • This update is focused towards more accurate separation of text into paragraphs. October 14, 2023 • Deployed new versions: RSK-M119p4, RSK-P103p2 (v3/pdf); • Fixed whitespace between Chinese symbols which used to be generated when the adjacent text lines are joined. As an example, for image like this one: the result used to look like: 【甲】孔子说: “我十五岁开始有志于做学问，三十岁能独立做事情，四十岁能 (通达事理) 不被外物所迷惑，五十岁能知道哪 extra whitespace* 些是不能为人力所支配的事情，六十岁能听得进不同的意见，到七十岁才做事 Result obtained with this version does not contain extra whitespace and look like this: 【甲】孔子说: “我十五岁开始有志于做学问，三十岁能独立做事情，四十岁能 (通达事理) 不被外物所迷惑，五十岁能知道哪些是不能为人力所支配的事情，六十岁能听得进不同的意见，到七十岁才做事能随心所欲，不会超过 October 13, 2023 • Deployed new version (v3/pdf), RSK-P103p1; • Improved ordering of answers to multiple choice questions in some cases. For this part of PDF page: we used to return the answers in height determined order: 10. Demostrar que si $a, b \in \mathbb{R}$, entonces: a) $\|-a\|=\|a\|$. e) $\|a-b\| \leq\|a\|+\|b\|$ b) $\sqrt{a^{2}}=\|a\|$. f) $\|\|a\|-\| b\|\| \leq\|a-b\|$. c) $\|a-b\|=\|b-a\|$. g) $a \neq 0,\left\|\frac{1}{a}\right\|=\frac{1}{\|a\|}$. d) $\left\|a^{2}\right\|=\|a\|^{2}$. h) $b \neq 0,\left\|\frac{a}{b}\right\|=\frac{\|a\|}{\|b\|}$. Now, we return the answers in the correct order: 10. Demostrar que si $a, b \in \mathbb{R}$, entonces: a) $\|-a\|=\|a\|$. b) $\sqrt{a^{2}}=\|a\|$. c) $\|a-b\|=\|b-a\|$. d) $\left\|a^{2}\right\|=\|a\|^{2}$. e) $\|a-b\| \leq\|a\|+\|b\|$ f) $\|\|a\|-\| b\|\| \leq\|a-b\|$. g) $a \neq 0,\left\|\frac{1}{a}\right\|=\frac{1}{\|a\|}$. h) $b \neq 0,\left\|\frac{a}{b}\right\|=\frac{\|a\|}{\|b\|}$. October 12, 2023 • Deployed new version, RSK-M119p3; • Accuracy improvements of Korean and handwritten Chinese; • Fixes several issues with braces recognition. Here is an example image: Result before (missing outer braces): Result with current version: October 9, 2023 • Deployed new version (v3/pdf), RSK-P103; • more reliable recognition of footnote text; • initial support for table of contents; • pseudo code algorithms are now searchable; □ individual lines from pseudo code algorithms will be a part of lines.json output. September 5, 2023 • Deployed new version (v3/pdf*), RSK-P101; • Added basic support for text in the footnote section of the page. Instead of breaking the main flow, especially in multi-column documents, the text will be wrapped inside \footnotetext{ ... }. May 25, 2023 • Deployed new version, RSK-115; • Improved quality of printed and handwritten Chinese recognition. April 12, 2023 • Deployed new version, RSK-113; • This update is focused on chemistry recognition. As a reminder, when "include_smiles": true is a part of the request, Mathpix can recognize chemistry diagrams such as: and return the SMILES representation which looks like this: The list of improvements: • support for stereochemistry: is transcribed to: <smiles>O=S(=O)(c1ccc(F)cc1)N1C[C@@H](O)[C@H](N2CCCC2)C1</smiles>; • support for Markush structures: is transcribed to <smiles>[Z2]Nc1c(CC([R10])CSC)ncn1CC#C</smiles>; • basic support for superatoms: □ more superatoms will be supported in future; • significant recognition accuracy improvement for both handwritten and printed chemistry diagrams. February 22, 2023 • Deployed new version, RSK-111; • Added support for the new table recognition algorithm; • Minor general changes needed to properly support the algorithm. A new table recognition algorithm is available in the v3/text and v3/pdf endpoints. It can be enabled by specifying "enable_tables_fallback": true as one of the request arguments. We care deeply about backwards compatibility. The new algorithm will only be used if both of the following conditions are fulfilled: • Our standard algorithm failed to recognize a table; • "enable_tables_fallback": true is specified as a request argument. We have ensured that there will be no computational overhead for customers who do not specify this option, so response times will not be affected. We have invested in a hybrid approach that will be able to tackle complicated cases like: • extremely large tables (e.g. tables with hundreds of cells); • tables with very complex structures (e.g. tables with many \multirow and \multicolumn cells); • tables featuring table cells with complex content like: □ table cells with complex math like large matrices or several aligned equations; □ tables cells containing whole paragraphs of text; • tables containing text in languages that are more challenging to recognize properly compared to English: □ this includes languages with rich alphabets like Chinese, Japanese, Hindi, Hebrew, Arabic, and others; • tables containing cells with rotated text; • tables containing diagrams inside cells like: □ table cells with chemistry diagrams (note that these can be converted to SMILES); □ table cells with natural images or similar: ☆ table will still be recognized and contain the image link for the diagram inside its cells. We will also support all combinations of the above cases. The algorithm we are releasing now might still struggle with: • tables with cells containing complex math like large matrices or several aligned equations; • tables containing diagrams inside cells; • empty grids of cells without textual content; • tables with rotated text are partially supported: □ in v3/pdf cells containing rotated text will be embedded as images. We will add improvements that will cover specified cases shortly. Some differences in output produced by the new algorithm compared to the standard one: • Column alignment is always central; • All cells have all borders (top, bottom, right, and left). February 1, 2023 • Deployed new version, RSK-110; • Added support for new LaTeX commands: \measuredangle, \grave, \bumpeq, and \amalg; • Improved recognition of constructs expressed with \lceil, \rceil, \lfloor, and \rfloor in combination with \left and \right; • Improved formatting of equations in text mode (see bellow for details); • Improved recognition of equations that contain large sub-equations in subscripts; • Improved recognition of handwritten French; • Improved recognition of handwritten German; • Improved recognition of handwritten Chinese; • Improved recognition of handwritten Japanese; • General improvements (new data iteration). The default “text” output has been changed, for example see the following equation: from the current “text” output: $ y=mx+b $\n$ x=y^{2}-1 $ and two asciimath equation outputs, to: $ \begin{array}{l} y = mx + b \\ x=y^{2}-1 \end{array} $ with one single asciimath equation output: which will make the “text” derived formats more consistent with what is currently returned for equations with a left brace: which currently yields this “text” output: $ \left\{\begin{array}{l}2 x+8 y=21 \\ 6 x-4 y=14\end{array}\right. $ and this “asciimath” output: Since we are already emitting in certain cases (eg, equations with no braces aligned around the “=” sign instead of being left aligned) asciimath for v3/text that looks like: we consider this update to be an inconsistency bugfix instead of a new feature with the potential to break backwards compatibility. In general, it is desirable for our API for small changes in input to result in small changes in output. For example, removing the left brace from equation 2 will simply change the v3/text asciimath which is a smaller change than the previous behavior, in which subtracting a left brace results in two equations instead of 1. January 30, 2023 • Deployed new version, RSK-109; • Improved recognition of isolated symbols. January 12, 2023 • Deployed new version, RSK-108; • Improved handling of images that contain mixed math and text in Russian. December 2, 2022 • Deployed new version, RSK-107p2; • Changes in spacing of arrays, aligned arrays and similar, & and \\ now always have spaces around them (even with rm_spaces in the request); • Visually unpleasant blocks of equations are being converted to left alignment instead of keeping the wrong alignment. November 25, 2022 • Deployed new version, RSK-107; • Improvements related to worksheet crops, small images with strong or dashed border near the content. November 22, 2022 • Deployed new version, RSK-106; • Improvements related to formatting of references in PDF pages, especially pages with green/red link boxes. November 15, 2022 • Deployed new version, RSK-105; • General improvements to handling zoomed out and zoomed in images. No changes to output formatting or error characteristics. November 14, 2022 • Deployed new version RSK-104p1; • Formating of block math is fixed in certain cases where the equations were wrongly kept in the text mode. November 11, 2022 • Deployed new version RSK-104; • Incremental improvement of image parsing module. Includes fixes for images with many lines of text. Accuracy improvements on handwritten data. No changes to output formatting. November 3, 2022 • Deployed new version RSK-103p1; • Fixed string post processing issues. In this version, we have changed the default Markdown / LaTeX for the following character: # -> \# While # works fine in Markdown and has the same behavior as \#, the former causes LaTeX compilation issues, whereas \# succeeds in LaTeX without any problem. We chose to always emit LaTeX \# instead of # so that our output would be more compatible and less likely to cause issues. The updated character \# is compatible with Markdown as well as LaTeX. Unescaped # will simply no longer appear in OCR Markdown / LaTeX outputs. Alternative math formats such as Asciimath are not affected by the change, this is a Markdown / LaTeX change only. October 21, 2022 • New enable_spell_check option to the v3/text and v3/pdfs endpoints greatly improves handwriting OCR for English (other languages coming soon). June 6, 2022 • Resolved a critical bug that impacted PDF processing of two-column PDFs; • You can now request that only certain subsets of pages are processed in a PDF, via the new page_ranges field; • Pushed latency improvements that benefit all endpoints, reducing processing time by 30% on average; • You can now query hour by hour API usage using the following endpoint. April 27, 2022 • Updated how line data is represented for PDFs from using rectangular regions to polygonal contours (this is helpful for handwritten PDFs where text lines are generally not rectangular); • Added page dimensions to the line-by-line data structures; • There are two available data structures for line-by-line data: □ Raw PDF lines data: this is the ideal data structure for searching; does not contain contextual annotations for titles, abstracts, etc; □ Context enhanced PDF mmd lines data: you can use this to re-create the full document, including contextual annotations for titles, abstracts, etc. (see here for syntax); • Published a Github repo which contains client-side code for live drawing with the Mathpix digital ink API containing a fully working example of leveraging user actions like scribbling and strikethrough to delete content. April 18, 2022 • Added an EU server region (AWS region eu-central-1) to decrease latencies for European customer and also for adherance to GDPR: • You can now use app-tokens for authenticating requests inside client side app code; • The new app-tokens route provides a include_strokes_session_id flag, which when true, returns a strokes_session_id string that can be used inside calls to v3/strokes, enabling digital ink sessions with live updates: □ Pricing for the strokes endpoint when using session_ids can be found here; • Add OCR support for basic handwritten PDFs. March 28, 2022 • You can now get detailed line-by-line data for PDFs, including geometric coordinates, via the new GET v3/pdf/<pdf_id>.lines.json endpoint; • Better robustness for our v3/text endpoint: □ Our ability to correctly interpret complex layouts involving math and text has improved, with much-improved edge case handling and handling of line text for skewed images and other image distortions that occur frequently in consumer photo search applications. March 14, 2022 • Our new OCR models feature stringent guarantees of syntactic correctness, resolving a rare but long-standing problem of occasionally malformed LaTeX strings, resulting in rendering errors due to double subscripts, double superscripts, malformed tables, and other syntax issues. This has been fixed at a fundamental level. Syntax issues are essentially completely fixed; • Deprecated \atop command in favor of \substack. February 8, 2022 We have recently switched to a new, faster database to save image and PDF data. Next week, we will decommission our old database. This will result in OCR API image results log data from before December 1st, 2021 becoming unavailable via the GET v3/ocr-results endpoint. Note that we have already migrated all PDF data to the new database, so there will be no data loss for PDF data. November 15, 2021 • Deployed incremental update to our re OCR engine, resulting in: □ significantly improved handwriting recognition, including disambiguating symbols based on context; □ improved table parsing accuracy; □ notably fewer errors. September 2, 2021 • Deployed a core algorithm update for our image parsing module, resulting in significantly better accuracy and edge case behavior for all endpoints; • Updated CLI with additional conversion types; • Significant improvements to chemical diagram OCR; • Support for asynchronous image processing. July 27, 2021 • Added support for Tamil, Telugu, Gujarati, and Bengali; • Updated our OCR to use a more effective representation of Chinese characters, leading to higher accuracy, and better coverage; • Added support for \bigcirc. July 19, 2021 • Support for sending image binaries for lower image upload latencies; • Support for tags which allow you to associate an attribute with your requests and subsequently retrieve the associated requests by using tags in a /v3/ocr-results query; • PDF processing updates: □ Fixed a bug where pages were getting skipped; □ Improved processing of PDFs with foreign languages; □ Added support for configurable math delimiters. April 12, 2021 • Servers in Singapore for faster latencies for API customers in Asia; • Triangle diagram OCR now supported for diagrams commonly found in trigonometry textbooks; • Added InChI option for chemical diagram OCR. April 2, 2021 • Added a include_word_data parameter to the v3/text endpoint, which when set to true, returns word by word information, with separate results, confidences values, and contour coordinates for each • Beta printed chemical diagram OCR to return SMILES format. March 10, 2021 • New v3/pdf API endpoint (beta); • PDF conversion CLI tool; • Fixed miscellaneous bugs in v3/text processing for messy images; • Incremental improvements to handwriting recognition and printed table recognition for all endpoints; • Added support for the following printed characters: February 7, 2021 January 5, 2021 • Improved math handwriting recognition; • Improved printed Romanian, Polish, Serbian, Ukrainian recogntion; • Added support for the following LaTeX characters: December 1, 2020 • Added autorotation for v3/latex and v3/batch (see https://docs.mathpix.com/#request-parameters-3 and https://docs.mathpix.com/#result-object-3); • Significant accuracy improvements to text / math localization performance for v3/latex and v3/batch for both ocr=["math"] (default) as well as ocr=["math", "text"], resulting in lower error rates and fewer bad results. November 12, 2020 • Added autorotation for v3/text. Images like this now work in v3/text: The goal of automatic rotation is to pick correct orientation for received images before any processing is done. The result of auto rotation looks like: We will soon add these features to v3/latex and v3/batch as well. We implemented a very conservative rotation confidence threshold, meaning you should still try to call the API with a properly oriented image if possible! API docs on autorotation: November 9, 2020 • v3/text general improvements; First of all, we trained our models on a larger dataset, resulting in a general accuracy increase. Secondly, we improved the precision of predictions, at the potential cost of slightly decreasing prediction recall in some circumstances. Here is an example of an image, where previously our v3/text tried to read the bottom, cut off parts of the image: Now, v3/text ignores these sections, resulting in a much cleaner output than before. The endpoint will still try to read everything in an image (vs the v3/latex endpoint which tries to read the main equation), but will be slightly less aggressive in reading unusual image sections in order to avoid garbage outputs. • Chemistry diagram detection; We have added a new field in our LineData object, subtype, so that we can return more information about diagrams to API clients. Currently subtype can only be chemistry, but more diagram subtypes are coming soon. • Added ability to create and disable new API keys to accounts.mathpix.com OCR dashboard; • Added ability to invite users (to have access to API keys, usage statistics, image results dashboard) to accounts.mathpix.com OCR dashboard. October 14, 2020 • Added support for the following characters: • Improved accuracy on: □ Handwriting (math and text); □ Hindi language recognition (printed); □ Tables and matrices; • We now support backgroundless PNGs with alpha channels. August 14, 2020 August 1, 2020 July 13, 2020 • Replace \dots with either \ldot or \cdot; • Predict empty braces when appropriate, like in Chemistry images (e.g. {}^); • Fix v3/text bug where very wide lines of text were getting skipped; • Improved accuracy on handwritten math; • Improved accuracy on table predictions; • Improved accuracy on photo images of printed Hindi and Chinese text; • Add support for \mid predictions inside set notation; • Add support for the following languages: Czech, Turkish, Danish. July 9, 2020 July 1, 2020 • Skip diagrams in v3/text which caused garbage results. Yet another reason to use v3/text over v3/latex! v3/text intelligently skips diagrams!	{"url":"https://mathpix.com/docs/convert/changelog","timestamp":"2024-11-13T17:22:48Z","content_type":"text/html","content_length":"168191","record_id":"<urn:uuid:a36dae10-152f-44d2-b5b9-8d9e2a22cc00>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00369.warc.gz"}
Convex Polyhedron -- from Wolfram MathWorld A convex polyhedron can be defined algebraically as the set of solutions to a system of linear inequalities where matrix and vector. Although usage varies, most authors additionally require that a solution be bounded for it to qualify as a convex polyhedron. A convex polyhedron may be obtained from an arbitrary set of points by computing the convex hull of the points. The surface defined by a set of inequalities may be visualized using the command RegionPlot3D[ineqs, x, xmin, xmaxy, ymin, ymaxz, zmin, zmaxvertex enumeration (Fukuda and Mizukoshi) can also be used to determine the faces of the resulting polyhedron directly. An example of a convex polyhedron is illustrated above (Fukuda and Mizukoshi). A simpler example is the regular dodecahedron, which is given by a system with convex polyhedron regular tetrahedron 4 cube 6 regular octahedron 8 In general, given the matrices, the polyhedron vertices (and faces) can be found using an algorithmic procedure known as vertex enumeration. Geometrically, a convex polyhedron can be defined as a polyhedron for which a line connecting any two (noncoplanar) points on the surface always lies in the interior of the polyhedron. The 92 convex polyhedra having only regular polygons as faces are called the Johnson solids, which include the Platonic solids and Archimedean solids. No method is known for computing the volume of a general convex polyhedron (Grünbaum and Klee 1967, p. 21; Ogilvy 1990, p. 173). Every convex polyhedron can be represented in the plane or on the surface of a sphere by a 3-connected planar graph (called a polyhedral graph). Conversely, by a theorem of Steinitz as restated by Grünbaum, every 3-connected planar graph can be realized as a convex polyhedron (Duijvestijn and Federico 1981). The numbers of vertices polyhedral formula	{"url":"https://mathworld.wolfram.com/ConvexPolyhedron.html","timestamp":"2024-11-13T08:38:32Z","content_type":"text/html","content_length":"62722","record_id":"<urn:uuid:c60312bd-c58c-49a1-a85f-2f2abcfa7a2b>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00032.warc.gz"}
Robert Constable on correct-by-construction programming - Machine Intelligence Research Institute Robert Constable on correct-by-construction programming Robert L. Constable heads the Nuprl research group in automated reasoning and software verification, and joined the Cornell faculty in 1968. He has supervised over forty PhD students in computer science, including the very first graduate of the CS department. He is known for his work connecting programs and mathematical proofs, which has led to new ways of automating the production of reliable software. He has written three books on this topic as well as numerous research articles. Professor Constable is a graduate of Princeton University where he worked with Alonzo Church, one of the pioneers of computer science. Luke Muehlhauser: In some of your work, e.g. Bickford & Constable (2008), you discuss “correct-by-construction” and “secure-by-construction” methods for designing programs. Could you explain what such methods are, and why they are used? Robert Constable: Short Summary The history of programming languages shows that types are valuable because compilers can check programs for type correctness and inform programmers whether or not their code exactly matches its type specification. Types describe requirements on programming tasks, and very rich type systems of the kind used in mathematics and increasingly in computing describe certain technical tasks completely as we’ll see below. The more expressive the type system, the more specifications help programmers understand their goals and the programs that achieve them. System designers can specify many requirements and computing tasks precisely with types and organize them into modules (or objects) that become a blueprint for analyzing and building a computing system. When types are rich enough to completely specify a task, the methodology we just outlined becomes an example of correct by construction programming. This approach has been investigated for mathematical problems for decades and for programming problems since the late 60s and early 70s. For example, if we want to implement a correct by construction factorization based on the fundamental theorem of arithmetic (FTA), that every natural number greater than 1 can be factored uniquely into a product of primes, we can first express this task as a programming problem using types. The problem is then solved by coding a computable function factor with the appropriate type. For programmers the type is probably just as clear as the FTA theorem. It is given below. It requires taking a natural number n greater than 1 as input and returning a list of pairs of numbers (p,e) where p is a prime and e is its exponent. We write the list as [(p[1],e[1]), (p[2],e[2]), …, (p[n],e[n])] and require that n =p[1]^e[1] x … x p[n]^e[n] where p^e means prime p to the power e, and the e[i] are all natural numbers, and x is multiplication (and is overloaded as the Cartesian product of two types). In addition the full FTA theorem requires that the factorization is unique. We discuss uniqueness later. For the factorization part, consider the example factor(24,750) = [(2,1),(3,2),(5,3), (11,1)]. This is correct since 24,750 = 2 x 9 x 125 x 11, where 5^3 is 125. In a rich type system the factorization piece of the fundamental theorem of arithmetic can be described completely as a programming task and the resulting program, factor, can be seen as a constructive proof accomplishing that factorization. A type checker verifies that the program has the correct type. The program factor is correct by construction because it is known that it has the type of the problem it was coded to solve. Dependent Types This paragraph is a bit technical and can be skipped – or used as a guide for coding factorization in a functional language with dependent types. Here is roughly how the above factorization task is described in the type systems of the programming languages of the Agda, Coq and Nuprl proof assistants – a bit more on them later. In English the type specification requires Nat, the type of the natural numbers, 0,1,2… and says this: given a natural number n larger than 1, compute a list of pairs of numbers, (p,e), where p is a prime number, and e is a natural number such that the input value n can be written as a product of all the prime numbers p on the list, each raised to the power e (exponent e) that is paired with it in the list lst. Symbolically this can be written as follows were N++ is the type (n:Nat where n>1). We also define the type Prime as (n:Nat where prime(n)) for a Boolean valued expression prime(n) which we code as a function from Nat to Booleans. Here is the FTA factorization type. (n:N++) ->( lst:List[Prime x Nat] where n = prod(lst) ). Note that the symbol x in (Prime x Nat) denotes the type constructor for ordered pairs of numbers. Its elements are ordered pairs (p,e) of the right types, p a Prime and e a Nat. Of course we need to define the product function, prod(lst), over a list lst of pairs in the right way. This is simple using functional programming in languages such as Agda, Coq, and Nuprl, in which we have the function type used above. We can assign this kind of programming problem to first year undergraduates, and they will solve it quite easily, especially if we give them a fast prime(n) procedure. Another very nice way to write a compact solution is to use map and fold_left functions that are usually supplied in the library of list processing functions. That solution uses the well known map reduce paradigm taught in functional programming courses for years and widely used in industry. Solving with map and reduce is a good exercise since the reduce function fold_right is mathematical induction on lists defined as a specific very general recursive program. Historical Perspective The development of mainstream programming languages from Algol 60 to Java and OCaml shows that type systems have become progressively more expressive. For example, now it is standard that functions can be treated as objects, e.g. passed as inputs, produced as outputs, and created as explicit values as we did in the FTA example above. Type checking and type inference algorithms have kept up with this trend, although OCaml can’t yet check the type we created for FTA, Agda, Coq, Nuprl and other programming languages supported by proof assistants can (they all use some form of automated reasoning to help). Static type checking of programs defined using very rich types is a demonstrably effective technology for precisely specifying tasks, documenting code as it’s being written, rendering code more readable by others (including yourself when you revisit it several months later), finding errors, and for safely making changes. With higher-order types such as function types, more programming tasks can be precisely defined, and code can be generated that is completely correct by construction. By 1984 researchers created an extreme point in this type checking technology in which the type systems are so rich that they can describe essentially any mathematical programming task in complete “formal” detail. Types also express assumptions on the data and the computing environment on which the specifications depend. These extremely rich type systems can essentially express any problem in mathematics. Indeed, Coq has been used to formally prove the Four Color Theorem as well as other important theorems in mathematics. The type checking by proof assistants often requires human intervention as well as sophisticated reasoning algorithms. The type systems we have mentioned are strong enough to also describe the uniqueness of the FTA factorization. They can do this by converting any factorization into the standard one produced by the function factor defined earlier. That function might be called normalize, and it does the job of converting any factorization into the standard one in which the primes come in order and are raised to the highest power possible. Among the best known modern proof assistants are Agda, Coq, HOL, Isabelle, MetaPRL, Minlog, Nuprl, and PVS. Nuprl was the first of these, built in 1984, and the Coq assistant is written in OCaml and supported by INRIA through the French government. Coq is widely taught and used for programming. These proof assistants help programmers check extremely rich specifications using automated reasoning methods. The types are rich enough to guarantee that programs having the type are correct for the task specified. This is a dominant form of correct by construction programming and sometimes includes the technology of extracting a program from a proof that a specification is achievable; this is called using proofs as programs. Research Goal An overarching research goal is to design type systems and programming assistants to express any computational problem and provide mechanisms for type checking, perhaps with human aid and using automated reasoning tools, no matter how expressive the type system becomes. Researchers in programming languages, formal methods, and logic are working on this goal worldwide. There are many references in the research literature citing progress toward this goal. The recent book by Adam Chlipala Certified Programming with Dependent Types, from MIT Press, 2013, provides an up to date view of the subject focused on the widely used Coq proof assistant. The 1991 book by Simon Thompson, Type Theory and Functional Programming, from Addison-Wesley, is one of the early comprehensive text books on the subject as it was just gaining momentum. Nuprl is the oldest proof assistant of this kind that is still active since 1984. It was described in the 1986 book Implementing Mathematics with The Nuprl Proof Development System. The Nuprl literature and an overview of the area circa 2006 can be found in the technical article “Innovations in computational type theory using Nuprl” from the Journal of Applied Logic, 4, 428-469, 2006. Also the author’s 2010 article on the “The Triumph of Types” at the 2010 celebration for Principia Mathematica in Cambridge, England recounts the 100 year old story of type theory and connects that to current research on type systems for correct by construction programming. It is available at www.nuprl.org. When a type system can specify security properties, the checked programs are secure by construction. In both correct and secure by construction programming, the specifications include assumptions about the computing environment, say about the network topology. If these assumptions fail to hold, say because the network topology changed in an unexpected way, then the type specifications are no longer sufficient to guarantee correctness. So it is very important to document the assumptions on which the type specification depends. For many problems these assumptions are stable, and in those cases, this correctness methodology is extremely effective. There are very good examples in the literature; at the web site www.nuprl.org we recently posted a simple example for the purpose of illustrating the method to newcomers. It is a complete formal proof that allowed us to build a correct by construction program to solve the “maximum segment sum” problem that is used to teach formal program development starting from type specifications. Recent Work The author and his colleagues at Cornell have recently used Nuprl to create a correct by construction version of the Multi-Paxos distributed consensus protocol that is being used in a database system. This required an implemented theory of distributed computing which is steadily evolving. We have made this protocol attack-tolerant, which is a form of security. Consensus protocols are critical to Cloud computing, and industry works very hard to build them correctly and make them secure as they are revised and improved. Researchers have created hundreds of correct by construction programs since 1984 and secure by construction programs as well. Worldwide many more are being built. Modern proof assistants are steadily becoming more powerful and effective at correct and secure by construction programming because there is a research imperative and an economic incentive to use them. There is also a strong incentive to teach them, as the work on Software Foundations by Benjamin Pierce and his colleagues at the University of Pennsylvania demonstrates. Proof assistants are an addictive technology partly because they improve themselves the more they are used which steadily increases their appeal and value. Luke: What do you think are the prospects for applying correct-by-construction and secure-by-construction approaches to methods commonly labeled “artificial intelligence”? Robert: The proof assistants such as Agda, Coq, and Nuprl that support correct by construction programming are themselves examples of artificial intelligence in the sense that they use automated reasoning tools developed in AI. The earliest proof assistants such as the Boyer-Moore prover came out of the AI Department at Edinburgh University. Also one of the seminal books in the field, The Computer Modeling of Mathematical Reasoning, by Alan Bundy at Edinburgh was a landmark in AI, published in 1983 by Academic Press. In due course these proof assistants will use other AI tools as well such as those that can translate formal proofs into natural language proofs. There has already been interesting progress on this front. For instance the article “Verbalization of High-Level Formal Proofs” by Holland-Minkley, Barzilay and me in the 16^th National Conference on AI in 1999, 277 – 284 translates Nuprl proofs in number theory into natural language. It is also the case that these systems are extended using each other. For example, right now my colleagues Vincent Rahli and Abhishek Anand are using the Coq prover to check that new rules they want to add to Nuprl are correct with respect to the semantic model of Nuprl’s constructive type theory that they painstakingly formalized in Coq. They have also checked certain rules using Agda. This activity is precisely what you asked about, and the answer is a clear yes. When it comes to looking at machine learning algorithms on the other hand, the criteria for their success is more empirical. Do they actually give good performance? Machine learning algorithms were used by professors Regina Barzilay (MIT) and Lillian Lee (Cornell) to improve the performance of machine translation from Nuprl mathematics to natural language in their 2002 article on “bootstrapping lexical choice.” For this kind of work, it does not necessarily make sense to use correct by construction programming. On the other hand, one can imagine a scenario where a correct translation of the mathematics might be critical. In that case, we would probably not try to prove properties of the machine learning algorithm, but we would instead try to capture the meaning of the natural language version and compare that to the original mathematics. The work of “understanding natural language mathematics” would benefit from correct by construction programming. Already there is excellent work being done to use constructive type theory to provide a semantic basis for natural language understanding. The book by Aarne Ranta, Type-theoretical grammar, Oxford, 1994 is an excellent example of this kind of work. It also happens to be one of the best introductions to constructive type theory. Eventually machine learning algorithms are going to have a major impact on the effectiveness of proof assistants. That is because we can improve the ability of proof assistants to work on various subtasks on their own by teaching them how the expert humans handle these tasks. This kind of work will be an excellent example of both aspects of AI working together. Machine learning guides the machine to a proof, but the automated reasoning tools check that it is a correct proof. Luke: What developments do you foresee, during the next 10-20 years, in proof assistants and related tools? Robert: I’ve been working on proof assistants, programming logics, and constructive type theory for about forty years, and part of the job is to make predictions five to seven years out and try to attract funding to attain a feasible goal that will advance the field. To maintain credibility, you need to be right enough of the time. I’ve been more right than wrong on eight or so of these predictions. So I am confident about predicting five to seven years out, and I have ideas and hopes about ten years out, and then I have beliefs about the long term. The Short Term Over the next five years we will see more striking examples of correct by construction programming in building significant software and more examples of verifying software systems as in the work on the seL4 kernel, the SAFE software stack, features of Intel hardware, and important protocols for distributed computing, especially for cloud computing. This will be done because it is becoming cost effective to do it. I don’t see a major industrial effort even for the next ten years. This area will be a focus of continued academic and industrial research, and it will have incremental effects on certain key systems. However, it will remain too expensive for wide spread deployment in industry. We will see formal tools being used to make both offensive and defensive weapons of cyber warfare more effective. One of the key lessons driving this is the fact that the stuxnet weapon, one of the best and most expensive ever created, was apparently defeated because it had a bug that led to its discovery. Another lesson is that the offensive weapons are improving all the time, and the defense must be more agile, not only building secure by construction “fortress like” systems, but thinking about systems as living responsive entities that learn how to adapt, recover, and maintain their core functionality. In the realm of warfare, cost has rarely been the decisive factor, and formal methods tools are not outrageously costly compared to other required costs of maintaining a superior military. In this realm, unlike in building and maintaining ships, planes, and rockets and defenses against them, the cost lies in training and recruiting people. These people will have valuable skills for the private sector as well, and that sector will remain starved for well educated talent. So overall, investing in advanced formal methods technology will be seen as cost effective. Proof assistants are the best tools we have in this sector, and we know how to make them significantly better by investing more. These tools include Agda, Coq, Nuprl, Minlog, and MetaPRL on the synthesis side, and ACL2, Isabelle-HOL, HOL-Light, HOL, and PVS on the verification side. All of them are good, each has important unique features. All have made significant contributions. All of them have ambitions for becoming more effective at what they do best. They will attract more funding, and new systems will emerge as well. Medium Term Most of the current proof assistants are also being used in education. It seems clear to me that the proof assistants that are well integrated with programming languages, such as Coq with OCaml and ACL2 with Lisp will prosper from being used in programming courses. Other proof assistants will move in this direction as well, surely Nuprl will do that. What happens in education will depend on the forces that bring programming languages in and out of fashion in universities. It seems clear to me that the Coq/OCaml system will have a chance of being widely adopted in university education, at least in Europe and the United States. It has already had a major impact in the Ivy League – Cornell, Harvard, Princeton, UPenn, and Yale are all invested as is MIT. I predict that within five years we will see this kind of programming education among the MOOCs, and we will see it move into select colleges and into top high schools as well. All of the proof assistants have some special area of expertise. If there is sufficient funding, we will see them capitalize on this and advance their systems in various directions. Nuprl has been very effective in building correct by construction distributed protocols that are also attack tolerant. It is likely that this work will continue and result in building communication primitives into the programming language to match the theory of events already well developed formally. Coq has been used to build a verified C compiler, at UPenn and Harvard Coq is being used to verify the Breeze compiler for new hardware designed to support security features, and at MIT they are using Coq to verify more of the software stack with the Bedrock system. There are other applications of Coq too numerous to mention. We will see additional efforts to formally prove important theorems in mathematics, but it’s hard to predict which ones and by which systems. Current effortsto use proof assistants in homotopy theory will continue and will produce important mathematical insights. Proof assistants will also support the effort to build more certified algorithms along the lines of Kurt Mehlhorn’s impressive work. Long Term People who first encounter a proof assistant operated by a world class expert is likely to be stunned and shocked. This human/machine partnership can do things that seem impossible. They can solve certain kinds of mathematics problems completely precisely in real time, leaving a readable formal proof in Latex as the record of an hour’s worth of work. Whole books are being written this way as witnessed by the Software Foundations project at UPenn mentioned previously. I predict that we will see more books of readable formal mathematics and computing theory produced this way. Around them will emerge a technology for facilitating this writing and its inclusion in widely distributed educational materials. It might be possible to use proof assistants in grading homework from MOOCs. Moreover, the use of proof assistants will sooner or later reach the high schools. I think we will also see new kinds of remote research collaborations mediated by proof assistants that are shared over the internet. We will see joint research teams around the world attacking certain kinds of theoretical problems and building whole software systems that are correct by construction. We already see glimpses of this in the Coq, Agda, and Nuprl community where the proof assistants share very similar and compatible type theories. There could be a larger force at work for the very long term, a force of nature that has encoded mechanisms into the human gene pool that spread and preserve information, the information that defines our species. We might not be Homo sapiens after all, but Homo informatis. The wise bits have not always been so manifest, but the information bits advance just fine – at least so far. That is the aspect of our humanness for which we built the proof assistant partners. They are part of an ever expanding information eco system. There is a chance that proof assistants in a further evolved form will be seen by nature as part of us. Luke: Thanks, Bob! Did you like this post? You may enjoy our other Conversations posts, including:	{"url":"https://intelligence.org/2014/03/02/bob-constable/","timestamp":"2024-11-15T03:43:08Z","content_type":"text/html","content_length":"75220","record_id":"<urn:uuid:1ef18349-02f1-4f87-a270-6f29f905b368>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00575.warc.gz"}
Understanding the Difference Between Classical and Quantum Computing What is Quantum Computing? Now, let's dive into the mind-boggling world of quantum computing. Quantum computers, unlike classical computers, operate on qubits. Qubits, short for quantum bits, can exist in a superposition of states, representing both 0 and 1 simultaneously. This unique property of qubits opens up a whole new realm of possibilities. Quantum computing harnesses the principles of quantum mechanics, a branch of physics that describes the behavior of particles at the quantum level. Quantum algorithms, unlike classical algorithms, are probabilistic and can provide solutions to complex problems more efficiently in certain cases. 10 Differences Between Classical and Quantum Computing Let's explore some of the key differences between classical and quantum computing:	{"url":"https://techhacktips.com/blogs/new-in-tech/understanding-the-difference-between-classical-and-quantum-computing","timestamp":"2024-11-05T15:15:17Z","content_type":"text/html","content_length":"22992","record_id":"<urn:uuid:9164131c-4ac9-45b0-a5bc-0665be2e9527>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00813.warc.gz"}
stochastic gradient method In this paper, we propose a new stochastic gradient method for numerical minimization of finite sums. We also propose a modified version of this method applicable on more general problems referred to as infinite sum problems, where the objective function is in the form of mathematical expectation. The method is based on a strategy to … Read more	{"url":"https://optimization-online.org/tag/stochastic-gradient-method/","timestamp":"2024-11-11T09:34:06Z","content_type":"text/html","content_length":"92127","record_id":"<urn:uuid:f89c650e-75c8-4761-a50f-19422ddc5a40>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00694.warc.gz"}
What is the Converse of the Pythagorean Theorem? The converse of the Pythagorean Theorem is like the the Pythagorean Theorem in reverse. You can use it both forward and backward! Not all theorems work this way, but the Pythagorean Theorem does! This tutorial will show you how to use both the Pythagorean Theorem and its converse.	{"url":"https://virtualnerd.com/algebra-1/radical-expressions-equations/pythagorean-theorem/pythagorean-theorem-examples/What-is-the-Converse-of-the-Pythagorean-Theorem","timestamp":"2024-11-13T12:23:54Z","content_type":"text/html","content_length":"23515","record_id":"<urn:uuid:e5b0f784-a7c3-4460-b28c-ef8f8d7ca1b5>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00097.warc.gz"}
Implementation of `.Magnitude`? How was the .Magnitude property actually implemented under the hood? How is it different than the Z axis? I can recreate it quite simply as print((workspace.Part1.Position - workspace.Part2.Position).Magnitude) -- 31.894357681274414 print(math.abs((workspace.Part1.Position - workspace.Part2.Position).Z)) -- 31 with the only difference being the .Magnitude property is a float whereas the Z axis is an integer. Well the Z axis is an integer due to the math.abs I don’t know how .Magnitude is implemented under the hood, but the way you described it working is more of a coincidence that the values are close than that being accurate. If you think about the Pythagorean Theorem, it’s a^2 + b^2 = c^2. In 2D Vectors, c would be the magnitude. It’s the same idea with 3D vectors. It’s x^2 + y^2 + z^2 = c^2, where c is the magnitude. This pattern applies to any number of dimensions (4D and beyond.) 2 Likes You can use math.round for rounding numbers, and math.abs to make the sign of the number always positive. I am not sure if I still understand the difference between simply taking the Z axis math.abs also rounds the number, however from my past experiences it only rounds down Magnitude is just the distance from the origin point (0, 0, 0). So the Z axis plays a part of the calculation of the magnitude, but it is not the magnitude. For instance, if you had a Vector3 value of (20, 0, 0), the Z axis is 0, but the Magnitude is 20, which so happens to map directly to the X axis since no other axes have a value. A Vector3 of (20, 30, 10) will have a magnitude of about 37.4, which is calculated as: = sqrt( 20^2 + 30^2 + 10^2 ) = sqrt( 400 + 900 + 100 ) = sqrt ( 1400 ) = 37.41657 And as you can see, the result has no direct equality with the Z axis. If your confusion is about what the Z axis represents: The Z axis typically represents the depth in 3D space (at least on Roblox). To visualize this, look down at your mouse. Move it left and right. You’re moving it on the X axis. Now move it forward and back. That moves it on the Z axis. Now lift it up and put it down. You’re moving it on the Y axis. 10 Likes	{"url":"https://devforum.roblox.com/t/implementation-of-magnitude/1759774","timestamp":"2024-11-11T13:29:55Z","content_type":"text/html","content_length":"35570","record_id":"<urn:uuid:16370335-afa7-4b6e-af47-6c43a5c4b089>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00175.warc.gz"}
100000000 in Words - How to Write 100000000 in English? 100000000 is written as one hundred million or ten crores in words. 100000000 is expressed in words as On Hundred million in the International Numeral system and Ten Crores in the Indian Number System. 100000000 is the count which represents the quantity of Ten Crore. For example, if a company’s net worth is 100000000 rupees, we can say, “Company X is estimated to be worth Ten crore rupees ”. We will learn how to write 100000000 in words in both the International Number System and the Indian Number System in this article. We will also learn its mathematical expanded expression and some essential facts. We will start with a two-step method using the place value chart. How to Write 100000000 in Words? We will execute a two-step plan to find out the word form or spelling of the number 100000000. Although in this article, we will execute those steps twice to find out the word form in both the International and Indian Numeral Systems. In the first step, we will make a place value chart for both systems. In this chart, we will distribute the digits of the number to assign every digit a place value. On making a place value chart for 100000000 in the International Number System:- One Hundred Million Ten Million Million Hundred Thousand Ten Thousand Thousand Hundred Ten One The number 100000000 has nine digits. We distributed these digits in the chart up to 9 place values. In the International Number System, one hundred million is the 9th place value, so we distributed these digits from one hundred million to one. Digit 1 is placed at one hundred million place value, while 0 is assigned to the rest of the place values. In the second step, we read these digits from the chart right to left to form the word. Therefore, we will read 100000000 as ‘One Hundred Million’ in word form. For example, if the population of a country is 100000000, we will say. “One Hundred Thousand people reside in country X.” Similarly, on creating a place value chart for 100000000 in the Indian Number System:- Ten Crore Crore Ten Lakh Lakh Ten Thousand Thousand Hundred Ten One We have assigned every digit of the number place value in the chart above. As the Indian Number System has Lakh and Crore terms ahead of Ten Thousand in the place value chart, we have given digit 1 to ten crore place value, and the rest of the place values have been assigned digit 0. Now we will follow the same process as before and will read the digits of the number along with their place values. The number 100000000 will be read as ‘Ten Crore’ in word form. For example, if a person named Ram has a net worth of 100000000, we will say, “Ram has a total net worth of Ten Crore.” Important Note: Reading the digits from right to left each time is necessary. Expanded Mathematical form of 100000000 According to the International Number System:- & 1 \times \text { One Hundred Million }+0 \times \text { Ten Million }+0 \times \text { Million }+0 \times \text { Hundred Thousand }+0 \times \text { Ten thousand }+ \\ & 0 \times \text { Thousand }+0 \times \text { Hundred }+0 \times \text { Ten }+0 \times \text { One } \\ & =1 \times 10000000+0 \times 10000000+0 \times 1000000+0 \times 100000+0 \times 10000+0 \times 1000+0 \times 100+0 \times \\ & 10+0 \times 1 According to the Indian Number System:- & 1 \times \text { Ten Crore }+0 \times \text { Crore }+0 \times \text { Ten Lakh }+0 \times \text { Lakh }+0 \times \text { Ten thousand }+0 \times \text { Thousand }+0 \times \\ & \text { Hundred }+0 \times \text { Ten }+0 \times \text { One } \\ & =1 \times 100000000+0 \times 10000000+0 \times 1000000+0 \times 100000+0 \times 10000+0 \times 1000+0 \times 100+0 \times \\ & 10+0 \times 1 Important Facts About 100000000 • 99999999 and 100000001 are the preceding and the following number to 100000000. • 100000000 is an even number as it is divisible by 2. • 100000000 is a perfect square number. It is the perfect square number of 10000. • 100000000 is a cardinal number. Cardinal numbers represent the count of some quantity. Here 100000000 is the numerical form of the amount or quantity, which is equal to One hundred million. You can find out the detailed other number in words article list below:- 1. How do we write the number 100000000 in words? We write 100000000 as One Hundred Million in International Number System and as Ten crores in the Indian Number System. We can derive the word form of the number by creating a place value chart. 2. Is 100000000 an odd number? No! 100000000 is not an odd number as it is divisible by 2. 3. Is 100000000 a perfect cube number? No! 100000000 is not a perfect cube number. 4. Is 100000000 a composite number? Yes! 100000000 is a composite number. It has more than two factors. 5. What do we get on subtracting 7500000 from 100000000? Write in words. We will get 92500000 by subtracting 7500000 from 100000000. 92500000 is written as ninety-two million five hundred thousand or nine crore twenty-five lakh.	{"url":"https://school.careers360.com/100000000-in-words","timestamp":"2024-11-09T10:40:30Z","content_type":"text/html","content_length":"281412","record_id":"<urn:uuid:620f0566-ae8a-4d99-ad41-d1f2d154b907>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00671.warc.gz"}
The Sum Of The Angles Of A Triangle Worksheet - Angleworksheets.com The Sum Of The Angles Of A Triangle Worksheet The Sum Of The Angles Of A Triangle Worksheet – This article will discuss Angle Triangle Worksheets as well as the Angle Bisector Theorem. In addition, we’ll talk about Isosceles and Equilateral triangles. If you’re unsure of which worksheet you need, you can always use the search bar to find the exact worksheet you’re looking for. Angle Triangle Worksheet This Angle Triangle Worksheet helps students learn how to measure angles. It consists of 12 triangles each representing a particular angle. The algebraic expression x + 5 represents the other angles. This worksheet can be used by students to calculate the sum of interior triangle angles. The worksheet features sample questions, too. The Angle Triangle Worksheet can be used for both basic and advanced mathematics. It teaches students how to identify the exterior and interior angles of triangles. It also features space for a teacher’s answer sheet, so that students can check their answers. Angle Bisector Theorem These worksheets will help students find the angle bisector for a triangle. They are a great resource for students in fifth and eighth grades. Each worksheet contains 20+ questions. Each includes both applied and reasoning questions. To ensure accuracy, arcs drawn for a perpendicular bisector should be drawn lightly, but they must be visible in the final answer. You can also use a sharp pencil or a pencil with a small compasses to get an accurate result. The Angle Bisector Theorem is a mathematical principle that states that a point on a bisector is equidistant from each of the angles it cuts. This principle can be illustrated with a diagram that has a yellow segment and a green segment. The lengths of these segments are equal. Triangles with equal sides The Equilateral Triangle worksheets are a great way to increase students’ math skills. These exercises, which are usually short, contain word problems and illustrations that help students understand the fundamentals of the triangle. These worksheets are also useful for students to improve their math skills. For example, one equilateral triangle worksheet requires high school students to work out the side lengths of a triangle as integers. An equilateral triangle has three sides and three angles of equal length. To find the area or perimeter of an equilateral triangle, multiply the length of each side by three. Triangular isosceles Isosceles triangles can be difficult to calculate, especially for young students. But there are some helpful worksheets that can help students learn the concept. These worksheets contain word problems and illustrative exercises that teach students how to figure out the area of a triangle using known values. These worksheets can be used by middle- and high school students. If the angles are equal, a triangle is an isosceles form. In this case, the third angle is 40 degrees. We can solve the equation by figuring out which sides of the triangle. Triangle Sum Theorem Angle Triangle Sum Theorem worksheets help students learn how to calculate the interior angles of a triangle. Students will need to identify the unknown angles within a triangle, and then calculate the sum. These worksheets also include space for students to write a message or special instruction. This worksheet teaches students that the sum of the interior angles of triangles always equals 180 degrees. This worksheet also helps students build equations because the interior angles for triangles always add up to 180 degrees. This book contains solutions and examples for different types of triangles. This worksheet is suitable for students in 6th grade to high school. Aside from interior angles, there are other types of triangles, such as right triangles and convex polygons. In this lesson, students learn how to determine whether a triangle is right-angled or acute. In addition, they learn how to find the interior angles and the exterior angles of triangles. Gallery of The Sum Of The Angles Of A Triangle Worksheet Triangle Angle Sum Worksheet Triangle Angle Sum Worksheet Triangle Angle Sum Worksheet Leave a Comment	{"url":"https://www.angleworksheets.com/the-sum-of-the-angles-of-a-triangle-worksheet/","timestamp":"2024-11-07T09:08:28Z","content_type":"text/html","content_length":"65241","record_id":"<urn:uuid:cffb779e-e6c5-47f7-8b44-a751eba170c1>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00840.warc.gz"}
How To Draw Perpendicular Line How To Draw Perpendicular Line - Now with x as center and a suitable radius draw an arc on either side of the given line segment ab. Web october 10, 2023 key takeaways perpendicular lines are two straight lines that meet or intersect at 90°. Web construction of a perpendicular line to another line passing through a point of it. The equation of a line that is perpendicular is y = mx + c. Web how to draw a perpendicular line through a point using a compass arthur geometry 81.7k subscribers subscribe 104 share 8.7k views 3 years ago basic geometric constructions learn how to construct. Web construction of a perpendicular line to another line passing through a point of it. This youtube channel is dedicated to teaching people how to improve their technical drawing skills. Unlike perpendicular lines, parallel lines never intersect. You can line your crochet handbag!! The opposite of perpendicular lines are parallel lines. Web courses on khan academy are always 100% free. Let m be the given line and a the given point on it. How to draw a perpendicular line from a given point on a given line Draw two arcs crossing the line segment. This page shows how to construct a perpendicular to a line through an external point, using only a compass and straightedge or ruler. Table of contents key takeaways what are perpendicular. Let me show you how! Web construction of perpendicular lines. Note that the arcs above the line. How to draw Perpendicular to a Line from a Point on the Line YouTube Web construct a perpendicular line through a point on the line set the needle end of your drawing compass on point p. This method is pretty simple if you know how to draw a perpendicular line. This youtube channel is dedicated to teaching people how to improve their technical drawing skills. Web construction of a. draw perpendicular line in civil 3d Fulton Thwithis ️ use preloved material, such. Draw two arcs using the intersection points as the centers. Use the compass to draw a circle centered around each point. This method is pretty simple if you know how to draw a perpendicular line. Note that the arcs above the line is. Web vinteachesmath 26.8k subscribers subscribe subscribed 1.5k. How to Construct a Perpendicular Line to a Given Line Through Point on Draw two arcs using the intersection points as the centers. Web construct a line perpendicular to the given line. Note that the arcs above the line is. You can line your crochet handbag!! Mark the point of intersection of the two arc. Web to find the perpendicular line we use two compasses to draw two. Perpendicular Lines Definition, Construction, and Properties Solved Web to find the perpendicular line we use two compasses to draw two circles, both on the given line. Keep your compass this size for the rest of the question. The circles should intersect in two points on opposite. Web we can construct a perpendicular line through a point off the line. The equation of. draw perpendicular line in civil 3d Fulton Thwithis The circles should intersect in two points on opposite. Draw two arcs crossing the line segment. The places where the two circles meet can hep us draw a perpendicular line. So these two lines are perpendicular. The bisector will be a right angles to the given line. Use the compass to draw a circle centered. TO DRAW A PERPENDICULAR ON A GIVEN LINE FROM A POINT ON IT & FROM A Place the protractor on the line m such that its base line coincides with m, and its centre. This youtube channel is dedicated to teaching people how to improve their technical drawing skills. Place your protractor at the point of intersection. Open the compass to a radius less than half the segment. Open the compass,. 4 Ways to Draw Perpendicular Lines in Geometry wikiHow Web we can construct a perpendicular line through a point off the line. Let me show you how! Web click on next or run to begin. The opposite of perpendicular lines are parallel lines. These are just landing spots for the next step. Note that the arcs above the line is. It works by creating. How to Construct a Perpendicular Line to a Given Line Through Point on Next, plot the point p(1, 2) as shown in figure $\pageindex{10}$(b). Web construction of a perpendicular line to another line passing through a point of it. Draw a line through the two. Web how to draw a perpendicular line through a point using a compass arthur geometry 81.7k subscribers subscribe 104 share 8.7k views 3. Perpendicular Lines through a Given Point Geometry Math Lessons Draw one of the lines and mark two points on it. The bisector will be a right angles to the given line. The opposite of perpendicular lines are parallel lines. Web construction of a perpendicular line to another line passing through a point of it. Web how to draw a perpendicular line through a point. How To Draw Perpendicular Line Mark the point of intersection of the two arc. Place the protractor on the line m such that its base line coincides with m, and its centre. Draw two arcs crossing the line segment. Now with x as center and a suitable radius draw an arc on either side of the given line segment ab. Make two more arcs which intersect. Web A Perpendicular Line Will Intersect It, But It Won't Just Be Any Intersection, It Will Intersect At Right Angles. Web steps to construct perpendicular lines step 1: Let m be the given line and a the given point on it. Web how to draw a perpendicular line through a point using a compass arthur geometry 81.7k subscribers subscribe 104 share 8.7k views 3 years ago basic geometric constructions learn how to construct. We can draw perpendicular lines for a given line in two ways. Web To Find The Perpendicular Line We Use Two Compasses To Draw Two Circles, Both On The Given Line. ️ use preloved material, such. Now with x as center and a suitable radius draw an arc on either side of the given line segment ab. Draw one of the lines and mark two points on it. In order to construct a perpendicular from a point to a given line segment: Let Me Show You How! The opposite of perpendicular lines are parallel lines. This youtube channel is dedicated to teaching people how to improve their technical drawing skills. Web to draw a perpendicular line from the point p to the line, start by setting your compass so that it reaches just beyond the line. The bisector will be a right angles to the given line. The Places Where The Two Circles Meet Can Hep Us Draw A Perpendicular Line. Web we can construct a perpendicular line through a point off the line. Web construct a perpendicular line through a point on the line set the needle end of your drawing compass on point p. Web courses on khan academy are always 100% free. With p as a center and any suitable radius, draw an arc cutting line segment ab at two distinct points as shown. How To Draw Perpendicular Line Related Post :	{"url":"https://sandbox.independent.com/view/how-to-draw-perpendicular-line.html","timestamp":"2024-11-09T10:21:06Z","content_type":"application/xhtml+xml","content_length":"24701","record_id":"<urn:uuid:d9de00ef-d3fd-4634-8906-3bb93afd0f26>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00301.warc.gz"}
Polygon Operations in ActionScript Polygons play a major role in a lot of visualization applications. Besides the possibility of drawing polygons, there is neither an explicit presentation of polygon data nor are common polygon operations supported in ActionScript. As far as I know, there is no simple solution available that solves the following geometrical problems for simple 2D polygons, so I wrote one. • Computation of the polygon area and the location of its centroid (which means the center of its mass) • Mesh simplification for faster rendering at lower resolutions • Computation of bounding geometry like bounding box and the convex hull • Deciding whether a given point or circle is inside a polygon or not (“Point in Polygon”) • Determining whether a given set of polygon points are sorted in clockwise- or counter-clockwise order • Computation of intersecting points between a line and a polygon The following code shows the implemented interface: package net.vis4.geom import flash.geom.Point; import flash.geom.Rectangle; public interface IPolygon function get points():PointSet; function get area():Number; function get centroid():Point; function get boundingBox():Rectangle; function get convexHull():IPolygon; function simplify(radius:Number):IPolygon; function containsPoint(point:Point):Boolean; function containsCircle(circle:Circle):Boolean; function get sortedClockwise():Boolean; function intersectLine(line:Line):Array; The class uses another class PointSet which is an extension of the native actioscript-array and provides type-safety and some simple computations. I implemented the following well known algorithms to solve the stated problems: The following flash movie demonstrates the usage: You can download the classes via bitbucket.org. World Map of Internet Adresses \| vis4 \| information visualization (Jan 12, 2010) […] created this visualization using ActionScript, based on my classes for map projections and polygon maths, which you can download for own usage. The data was extracted from the free GeoLite-City database […]	{"url":"http://www.vis4.net/blog/polygon-operations-in-actionscript/","timestamp":"2024-11-13T11:03:40Z","content_type":"text/html","content_length":"49503","record_id":"<urn:uuid:a48a9b57-0789-4653-ad57-e42cea4ddcb1>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00703.warc.gz"}
Tom VanAntwerp LeetCode 198. House Robber Link to original problem on Leetcode. You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and it will automatically contact the police if two adjacent houses were broken into on the same night. Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police. Example 1: Input: nums = [1,2,3,1] Output: 4 Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3). Total amount you can rob = 1 + 3 = 4. Example 2: Input: nums = [2,7,9,3,1] Output: 12 Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1). Total amount you can rob = 2 + 9 + 1 = 12. • 1 <= nums.length <= 100 • 0 <= nums[i] <= 400 A leetcode problem with a story! This is quite rare, and a real treat!^1 This is a dynamic programming problem, and as such we can solve it either recursively or iteratively. To solve this problem recursively, we should think about what sub-problem we’re trying to solve. Whenever I’m at a house at index, I have to decide: am I better off robbing this house, or waiting until I get to the next house? If I rob this house, I get the value of its possessions nums[index], plus the value of all the houses I robbed before if the last house I robbed was the house from two before it at nums[index - 2]. If I skip this house, I get the value of all the houses I robbed before this house ending at the house immediately before it at nums[index - 1]. If index starts at the last value of nums, then we can write a recursive function to figure out the sum of the best sequence of houses by working back from index. I can work backward until I’m out of houses to compare—that is to say, the index is less than 0. That’s our base case. // This function is modified to accept a second argument // from how it's scaffolded in leetcode: index. We set a // default index value of nums.length - 1 to signify we // are starting at the last value of nums. function rob(nums: number[], index = nums.length - 1): number { if (index < 0) { // Indices less than zero are the base case; we are // out of houses to rob! Non-existant houses have // a value of zero. return 0; } else { // If we're not yet out of houses to rob, we return the // maximum of this house plus the house two before, or // the house one before. We recursively call the function // to get the sum of other house's values. return Math.max(rob(nums, index - 2) + nums[index], rob(nums, index - 1)); This function is correct, but it’s not performant. Because we call the function again twice for each comparison, our time complexity is $O(2^{n})$! If you try running it in leetcode, it will time out. We need to improve it with memoization. To improve on the previous solution, we add memoization to remember values we’ve previously computed. This dramatically speeds up our solution and gives us a time complexity of $O(n)$. Here, we pass around a Map with the index as key and the result of our comparisons as the value. // We've once again modified the function. Now it also takes // a JavaScript Map to use for our memoization, which we // initialize as a new empty Map. function rob( nums: number[], index = nums.length - 1, memo = new Map<number, number>(), ): number { if (index < 0) { return 0; } else if (memo.has(index)) { // Instead of always computing non-base cases, we'll first // check to see if they exist in our memo Map and return // that if we find it. return memo.get(index); } else { const result = Math.max( rob(nums, index - 2, memo) + nums[index], rob(nums, index - 1, memo), // Instead of returning the result of comparisons right // away, we first add it to the memo Map so we can find // it later if we need it. memo.set(index, result); return result; Our recursive solution was top-down. That is to say, we started at the farthest house and worked backward. For our iterative solution, we’ll instead go bottom-up. We’ll calculate answers to sub-problems near the beginning in order to answer more sub-problems as we go along. The same sub-problem logic still holds: for each house we’re at, we want to know if we’re better off taking the sum of this house and all the houses previously robbed as of two house before, or just the sum of all house robbed previously as of one house before. We could use an array to remember the optimal possible result for any given index in nums, but we don’t have to. All we really need are two variables to remember the sums for one house back and two houses back. With each iteration across nums, we can just update those two variables. So not only do we get time complexity of $O(n)$, but we also get space complexity of $O(1)$!. function rob(nums: number[]): number { if (nums.length === 0) return 0; let oneHouseBack = 0, twoHousesBack = 0, temp = oneHouseBack; for (const thisHouse of nums) { // We're about to reassign the value of oneHouseBack, // so we want to store the current value somewhere. // We put it in this temp variable, because we're going // to update twoHousesBack to be the current value of // oneHouseBack, and oneHouseBack to the max of our // comparison. temp = oneHouseBack; oneHouseBack = Math.max(twoHousesBack + thisHouse, oneHouseBack); twoHousesBack = temp; return oneHouseBack; 1. If you like tough data structures and algorithms problems with a story, check out Advent of Code for more and better puzzles. ↩	{"url":"https://tomvanantwerp.com/coding-questions/leetcode-198-house-robber/","timestamp":"2024-11-06T23:37:39Z","content_type":"text/html","content_length":"27895","record_id":"<urn:uuid:d755cb2d-6c6f-4c4e-bc07-5e0a40f0d10b>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00386.warc.gz"}
Teaching Bits Teaching Bits: A Resource for Teachers of Statistics Journal of Statistics Education v.3, n.3 (1995) Joan B. Garfield Department of Educational Psychology University of Minnesota 332 Burton Hall Minneapolis, MN 55455 612-625-0337 jbg@maroon.tc.umn.edu J. Laurie Snell Department of Mathematics and Computing Dartmouth College Hanover, NH 03755-1890 603-646-2951 jlsnell@dartmouth.edu This column features "bits" of information sampled from a variety of sources that may be of interest to teachers of statistics. Joan abstracts information from the literature on teaching and learning statistics, while Laurie summarizes articles from the news and other media that may be used with students to provoke discussions or serve as a basis for classroom activities or student projects. We realize that due to limitations in the literature we have access to and time to review, we may overlook some potential articles for this column, and therefore encourage you to send us your reviews and suggestions for abstracts. From the Literature on Teaching and Learning Statistics "Statistics Education Fin de Siecle" by David S. Moore, George W. Cobb, Joan Garfield, and William Q. Meeker (1995). The American Statistician 49(3), 250-260. This paper grew out of a session at the 1993 Joint Statistical Meetings that focused on imagining the state of statistics education at the end of this century. David Moore (serving as provocateur) raised challenging questions in three areas of statistics education: the role of technology, new ways of helping students learn, and teaching in institutions of higher education. Acknowledging that "higher education faces an environment of financial constraints, changing customer demands, and loss of public confidence," Cobb, Garfield and Meeker took turns as primary and secondary responders to these questions. Audience reactions to these concerns are also summarized. "Confessions of a Coin Flipper and Would-Be Instructor" by Clifford Konold (1995). The American Statistician 49(2), 203-209. Konold's extensive research on students' probabilistic reasoning is well known and has provoked many instructors to follow his recommendation to have students first make predictions about random events and then test them by performing experiments or computer simulations. Konold's "ProbSim" software, described in this paper, was originally designed to facilitate this learning activity by providing an easy and graphical way for students to set up probability models to generate simulated data. In this article Konold recounts a surprising experience he had when trying out this instructional approach with a student through a series of individual tutoring sessions which led him to rethink and test his own beliefs about coin flipping. The paper concludes with some practical recommendations for teachers of statistics who use simulations with students to help them overcome misconceptions related to probability. 1994 Proceedings of the Section on Statistical Education American Statistical Association, 732 North Washington Street, Alexandria, VA 22314. Each year the ASA Section on Statistical Education publishes papers presented in their sessions at the Joint Statistical Meetings. The papers in this volume were presented at the meetings held in Toronto, Canada, in August, 1994. The volume includes papers from five invited paper sessions on the following topics: I. Robust Regression in Practice II. Improving Teaching of Graduate Level Statistics Service Courses III. Short Courses: Challenge, Issues and Educational Value IV. The Adopt-A-School Project: Successful Models for K-12 Outreach V. The First Day of Class There are contributed papers from sessions on the following topics: I. Demonstrating Statistical Concepts Using Computers, Graphics and Geometry II. Improved Methods for Statistical Instruction, Teacher Training, and Evaluation III. Inconsistencies in Current Practice: Handling Interactions Between Fixed and Random Effects IV. Statistical Education for Business and Industry V. Programs and Techniques for Teaching Data Analysis and Interpretation VI. Statistical Consulting VII. Teaching Statistics: Problems and Solutions VIII. Training and Consulting for Industry: What's Missing? The proceedings conclude with seven contributed papers from poster sessions. Teaching Statistics A regular component of the Teaching Bits Department is a list of articles from Teaching Statistics, an international journal based in England. Brief summaries of the articles are included. In addition to these articles, Teaching Statistics features several regular departments that may be of interest, including Computing Corner, Curriculum Matters, Data Bank, Historical Perspective, Practical Activities, Problem Page, Project Parade, Research Report, Book Reviews, and News and Notes. The Circulation Manager of Teaching Statistics is Peter Holmes, p.holmes@sheffield.ac.uk, Center for Statistical Education, University of Sheffield, Sheffield S3 7RH, UK. Teaching Statistics, Autumn 1995 Volume 17, Number 3 "Apparent Decline as a Sign of Improvement? or, Can Less Be More?" by Sangit Chatterjee and James Hawkes Summary: Improvement of a system need not be the usual growth expected of dynamic systems. Decreasing variability, an important index of improvement, is discussed. The disappearance of 0.400 hitters from professional baseball can be better understood under such an assumption. Other examples illustrating the apparent paradox are also mentioned. "Probability, Intuition, and a Spreadsheet" by Alan Graham Summary: Probability simulations are a useful way of helping students to challenge their intuitions about chance events. However, tossing dice and coins can be slow and messy and may mask underlying long-run patterns. This article provides examples of probability simulations on a spreadsheet which overcome some of these difficulties. "Primary Data" by Andrew Bramwell Summary: This article gives a rare insight into the way in which statistical thinking may be introduced in the primary school classroom. "Secondary Students' Concepts of Probability" by Richard Madsen Summary: Students develop concepts of probability without formally studying the discipline and some of their concepts are at variance with those taught in the classroom. A survey of 200 students in five schools in Missouri was undertaken in an attempt to learn about pre-conceptions. The results are discussed in this article. "Arm-waving Mathematics: Sound, if Not Rigorous" by Ken Brewer Summary: This article suggests some techniques, herein referred to as "arm-waving", which utilise analogies, graphs and examples to illustrate the contribution of mathematics to statistical concepts without a rigorous treatment of the mathematics used. "Statistical Tools and Statistical Literacy: The Case of the Average" by Iddo Gal Summary: This article is intended to serve as a starting point for a dialogue regarding the goals of teaching students about averages and how to assess their emerging knowledge. In addition to these articles, this issue includes the columns Standard Errors, Software Review, Data Bank, and Computing Corner. Topics for Discussion from Current Newspapers and Journals "Bordeaux Wine Vintage Quality and the Weather" by Orley Ashenfelter, David Ashmore, and Robert Lalonde (1995). Chance, 8(4), 7-14. If you want to drink good wine you can buy new wine and let it mature in your cellar or you can buy older wine that has matured in some dealer's cellar. To decide which is the better strategy it is helpful to know the answers to questions like: Does the price of mature wine reflect the quality of the wine? Presumably the answer to this question is yes, because eventually the quality of the wine is known, and the price reflects this knowledge. Other natural questions might be: Is the price of new wine a good predictor of the price after it has matured? If not, what is a good predictor? This article tries to answer such questions. The authors begin by providing the 1990-1991 London auction prices of red wines from six of the best known Chateaux (vineyards) that were produced in the years from 1960 to 1969. These years were chosen because by 1990 the wines should be fully mature and their quality known. For a given Chateau, there is wide variation in these prices through the years and, for a given year, there is wide variation in the prices between Chateaux. Using regression techniques, the authors show that the prices of the wines when new are not good predictors of their prices when mature. On the other hand, weather conditions are very good predictors. Great vintages for Bordeaux wines correspond to years in which August and September were dry, the growing season was warm, and the previous winter was wet. Ashenfelter uses this fact to estimate the value of new wines and provides these estimates in a newsletter he distributes called "Liquid Assets: The International Guide to Fine Wines." Professor Ashenfelter is a Princeton economist who is widely quoted in newspapers on weightier matters, but his newsletter also makes the news occasionally. This article provides some of the more humorous remarks made by well-known wine critics on the use of statistics to assist in judging wines. "Picturing an L.A. Bus Schedule" by Howard Wainer (1995). Chance, 8(4), 44. Howard Wainer edits a column in Chance called "Visual Revelations." His column provides wonderful examples for classroom discussions of the use of graphics. This month he considers a question from the first National Adult Literacy Survey conducted in 1992. This question gives the appropriate L.A. bus schedule and asks how long you would have to wait for the next bus on a Saturday afternoon if you missed the 2:35 bus leaving Hancock and Buena Ventura, going to Flintridge and Academy. The schedule is typical of those we have all struggled with: columns of outbound times and inbound times, remarks about buses that run Monday through Friday only, and so on. Wainer suggests that we should make a general-purpose plot of the bus data, and then see how it serves to answer a variety of questions, including the one on the quiz. His choice is to plot the time of day on the horizontal axis and the various bus stops on the vertical axis. A change of scale suggests itself; after this change is made we have a plot that makes it easy to see regularities in the way the buses run. The cyclic nature of the graph suggests that there is a single bus going back and forth on the route considered, making a round trip in just under two hours. The graph provides easy answers to a variety of questions, including the one on the survey. "Fuzzy Logic: Great Hope or Grating Hype?" by Michael Laviolette (1995). Chance, 8(4), 15-19. The author of this article feels that many problems currently being solved by fuzzy set theory could be equally well solved using probability theory. To illustrate this he considers the following simple application of fuzzy set theory to control theory. You want a air condition-controller to make a motor go at speed y when the temperature is x. You only have a vague feeling about when the room is cool, just right, or warm. Fuzzy logic suggests associating these labels with appropriate intervals of temperatures. These sets are called fuzzy sets and are allowed to overlap. For example, suppose you assign the interval from 50 to 70 as the "cool" set, 60 to 80 as "just right," and 70 to 90 as "warm." Then the temperature 65 is in both the "cool" set and the "just right" set. For each fuzzy set you define a membership function that assigns a value between 0 and 1 to each member of the set. For example, for the cool interval from 50 to 70, you might make the membership function increase linearly from 0 to 1 as the temperature goes from 50 to 60 and decrease linearly from 1 to 0 as the temperature goes from 60 to 70. Then 60 is a really cool temperature, but 65 is only .5 cool. Similarly, you can determine fuzzy sets and measurement functions corresponding to intervals of speed that you consider slow, medium, and fast. We associate "cool" with the motor being on "slow," "just right" with it being on "medium," and "warm" with it being on "fast." This restricts how we make the correspondence between temperatures and speeds, but at the same time creates some conflicts. For example, the temperature 65 is in both the "cool" and the "just right" temperature sets, so it should correspond to a point in either the "slow" or "medium" set or possibly both. The temperature and speed membership functions determine, by fuzzy logic, a new speed fuzzy set and membership function for the temperature 65. When the temperature is 65, the controller sets the speed equal to the average speed calculated using this membership function. For a detailed description of how this is done, consult the author's longer article (Technometrics (1995), 37(3), 249-261). The explanation in the Chance article is rather brief, and a key figure (the last part of Figure 2) is incorrect. In the probabilistic approach to the problem, the membership functions are replaced by conditional probabilities. We determine subjectively or otherwise probabilities of the form "the probability that the room is perceived as cool given that the temperature is 65" and probabilities of the form "the probability that the machine is running at speed x given that it is running at medium speed." These probabilities, combined with the rules associating temperature sets with speed sets, allow you to compute the expected speed given a specific temperature, say 65. Then, for a given temperature x, the controller sets the speed equal to y, where y is the expected speed with respect to these conditional probabilities. Laviolette's article in Technometrics includes long discussions by workers in fuzzy set theory describing their feelings about the relationship between probability and fuzzy sets. Luck: The Brilliant Randomness of Everyday Life by Nicholas Rescher (1995). New York: Farrar, Straus & Giroux. We teach and write about chance as if we know what it is. Yet we talk about luck, good and bad, all the time and seldom ask what luck is and how it relates to chance. In this book the philosopher Nicholas Rescher attempts to remedy this situation. Rescher's use of the term "luck" is consistent with the definition in the Oxford English Dictionary: "the fortuitous happening of an event favorable or unfavorable to the interest of a person." Thus luck combines a chance event with an effect on an individual. We have good luck if the chance event helps us and bad luck if it hurts us. Rescher writes: "Recognizing the prominent role of sheer luck throughout the role of human affairs, this work will address such questions as: What is luck? How does it differ from fate and fortune? What should our attitude toward lucky and unlucky people be? Can we expect to control or master luck? Are people to be held responsible for their luck? Should there be compensation for bad luck? Can luck be eliminated in our lives?" The only question for which you will find a definitive answer is: "Can luck be eliminated from our lives?" The answer is no! You may be disappointed on the first reading of this short book because you don't get enough answers. However, you will start asking your own questions such as: Is there a law of large numbers for luck? You will find yourself discussing the meaning of luck with your colleagues and students. Perhaps from this you will get lucky and discover what luck really is. "Divine Authorship? Computer Reveals Startling Word Pattern" by Jeffrey B. Satinover (1995). Bible Review, 11(5), 28. This article reviews the research of three statisticians, Doron Witztum, Eliyshu Rips, and Yoav Rosenberg, published in Statistical Science (1994, 9(3), 429-438). These authors claim to show that the book of Genesis contains information about events that occurred long after Genesis was written, and that this finding cannot be accounted for by chance. To show this, the authors chose 32 names from the Encyclopedia of Great Men of Israel and formed word pairs (w, w'), where w is one of the names, and w' is a date of birth or date of death of the person with name w. We say a word w is "embedded" in the text if its letters appear in the text at positions corresponding to an arithmetic sequence (not counting spaces). For example, the word "has" is embedded in the sentence "The war is over," since the letters h, a, and s occur in the sentence separated in each case by two letters. The authors show that the names and dates they chose appeared in Genesis (which is not itself surprising) with the names nearer their matching dates than could be accounted for by chance (p = .00002). Satinover asks: "What was the purpose of encoding all this information into the text?" He answers his own question with: "Some would say it is the Author's signature." At the suggestion of a referee, the authors tried the same tests on other Hebrew works and even Tolstoy's War and Peace translated into Hebrew. They did not find any similarly unlikely events in these controls. When the results were published in Statistical Science, the editors commented that the referees doubted this was possible but could not find anything wrong with the statistical analyses. They published it so the rest of us could try to discover what is going on. The authors first announced their results in the Journal of the Royal Statistical Society A (1988, 155(1), 177-178), while commenting on an article "Probability, Statistics and Theology" by D. J. Bartholomew. After this announcement, a public statement was made by well-known mathematicians including H. Furstenberg at Hebrew University and Piateski-Shapiro at Yale that these results "represented serious research carried out by serious investigators." The article gives a nice description of this work and how it has been received. Evidently, responses so far have fallen into two categories: a priori acceptance and a priori rejection, the former by believers and enthusiasts and the latter by scientists who say that no amount of evidence would be convincing. "Breast Cancer Study a First" by Robert Cooke. Newsday, 15 November 1995, A36. This article reports on the first study that used a probability sample to study risk factors for breast cancer. The researchers considered three factors generally thought to be risk factors for breast cancer: not having a baby or waiting until after age 19 to have one, having a moderate or high income, and having a family history of breast cancer. The study involved 7,508 women between ages 25 and 74, and began in 1971. By 1987 when the study ended, 193 women had developed breast cancer. The results of the study suggested that 41% of the risk was linked to the three factors considered. The article states that it was estimated that 29% of the breast cancer cases were attributable to not having a baby or waiting until after age 19. An additional 19% were linked to having a moderate or high income, and 9% were linked to an inherited predisposition for breast cancer. It is interesting to think about what this really means. "$4.1 Million Awarded in Implant Case; Dow Chemical Facing Prospect of More Suits" by Jay Mathews. The Washington Post, 29 October 1995, A5. Last spring, Dow Corning filed for bankruptcy as a result of lawsuits brought by women who alleged their health had been ruined by silicone breast implants. Fearing they would be denied compensation, some women sought to sue the parent company, Dow Chemical. A Nevada jury has now ruled that Dow Chemical must pay an Elko, Nevada, woman $3.9 million in damages. This is the first time the parent company has been held responsible for damage allegedly caused by the implants. Health complaints have ranged from chronic fatigue and muscle pain to connective tissue disorders and rheumatic diseases, although a series of scientific studies have been unable to establish any links to the implants. The plaintiff's attorneys argued that Dow Chemical had done studies of other uses of silicone in industry and agriculture and knew of problems that should have been made public. Dow Chemical's attorneys denied this claim. They maintained that the plaintiff's symptoms were consistent with traumatic stress disorder and fibromyalgia unrelated to the implants, and that she sought medical attention for the implants only after seeing an attorney. "Proof of a Breast Implant Peril is Lacking, Rheumatologists Say" by Gina Kolata. The New York Times, 25 October 1995, C11. Mere days before the verdict in the above article was announced, the American College of Rheumatology issued a formal statement saying that there was no evidence that silicone breast implants cause the diseases attributed to them, and that the FDA and the courts should stop acting on the basis of anecdotal evidence. (In 1992, the FDA imposed a moratorium on use of the devices until the alleged health risks were investigated.) The article notes that, since an estimated one million American women have received implants, it would be expected by chance alone that thousands would become ill with connective tissue and rheumatic diseases. Some doctors disagree with these conclusions and have testified in court that the devices cause a new type of auto-immune disorder. But Dr. Sam Ruddy, departing president of the American College of Rheumatology said that there was no scientific evidence to support this claim. Instead, there are just "collections of cases with no controls." For a more technical article on this subject, see "Silicone Breast Implants and the Risk of Connective-Tissue Diseases and Symptoms" by J. Sanchez-Guerrero, et al. (1995), New England Journal of Medicine, 332, 1166-1170. "In Scientific Studies, Seeking the Truth in a Vast Gray Area" by Lena Williams. New York Times, 11 October 1995, C1. This article reports on a one-day meeting of epidemiologists and journalists in Boston to try to find solutions to public confusion caused by contradictory recommendations on medical issues. There was plenty of blame to go around: Scientists tend to overstate their findings to get attention or grants or both. Journalists add to the problem by focusing on the most controversial or titillating aspects of medical research. In addition, the public is eager to find quick fixes to medical problems. A recent study linking moderate weight gain in middle-aged women to an increased risk of death was held up as an example of the problem. The issues in this study were complicated, and many accounts did not give enough details to explain how the findings depended on factors like race and eating patterns. In addition, the report that the study's author serves as an adviser to two companies that make diet pills created doubts about the study. The scientists reviewed some of the reasons that their findings may not be accurate: biases, inaccurate reporting by subjects, and other methodological problems. They agreed that they use too much jargon; they present journalists with the almost impossible task of learning about results and explaining them in a non-technical way with only a few days' study. They recommended that articles be released to journalists weeks in advance, rather than days in advance. "Keno Is as Popular in Delis as in Bars" by Ian Fisher. New York Times, 17 October 1995, B6. New York has a new game called Quick Draw, a form of Keno. It is outpacing earning projections by almost 20/but some of the top-selling outlets are convenience stores and other stores where alcohol is not sold. This article discusses concerns about where and how the game is being played. The fact that you can play a new game every five minutes suggests that it may become additive for some To play Quick Draw, you specify a set of numbers chosen from the first 80 integers. You can have from one to ten numbers in your set. The computer then picks 20 numbers at random from the integers from 1 to 80. You can bet 1, 2, 3, 4, 5, or 10 dollars. You are paid off according to how many numbers are in both your set and the set chosen by the computer. The payoffs are chosen in such a way that your expected loss, no matter how many numbers you choose, is about 40%. Donald Trump and others tried to stop this game on the grounds that it was not a lottery as defined by the State Constitution and thus not exempt from New York's general prohibition against gambling. A judge ruled that "the game contained all the essential features of a lottery: i.e., consideration for chances, represented by numbers drawn at random, and a prize for the winning numbers. A lottery agent inserts the player's picks into a computer terminal--the player does not. Nor does the machine eject anything of value as would a slot machine--only a bet slip used by the player to compare the numbers to those drawn and displayed on the video screen." Some pretty fine distinctions are being made here. "Ask Marilyn" by Marilyn vos Savant. Parade Magazine, 15 October 1995, 13. A reader writes: I've heard that when playing cards, when you're dealt a pair, it increases the odds that your opponent is dealt a pair, too. Is this true? If so, how? Marilyn says it's true and illustrates with a counting argument, assuming that you and your opponent are each dealt two-card hands. A pair in any of the 13 denominations can be obtained in C(4,2) = 6 ways, by choosing a pair of suits. Marilyn demonstrates this by explicitly listing the combinations. If you hold a pair, you have eliminated five of your opponent's opportunities for pairs, since there remains only one way for her to get a pair in the same denomination as you (there remain six options for any other denomination). On the other hand, if you don't hold a pair, you reduce to C (3,2) = 3 the number of ways she could get a pair in either of the two denominations you hold. This is a total loss of six opportunities, which is one more than the five she loses when you hold a pair. Her chances for a pair are indeed better when you hold a pair! For more problems like this one, see "Do Good Hands Attract?" by S. Gudder (1991), Mathematics Magazine, 54(1), 13-16. "Mortality Associated With Moderate Intakes of Wine, Beer, or Spirits" by Morten Gronbaek et al. (1995), British Medical Journal, 310(6988), 1165-1169. A number of studies have shown a U-shaped curve for the relative risk of mortality as a function of alcohol intake for both men and women. This article reports the results of a large study carried out in Denmark to assess the effects of different types of alcoholic drinks on the risk of death from all causes and from heart attacks, taking into account sex, age, socioeconomic conditions, smoking habits, and body mass index. The study followed 13,285 subjects (6051 men and 7234 women) between ages 3 and 79 from 1976 to 1988. The authors found that beer intake had little effect on the relative risk of mortality. Intake of spirits also had little effect up to 3 to 5 drinks daily, at which point there was a significant increase in the relative risk of mortality. On the other hand, the relative risk as a function of wine intake dropped continuously, having its lowest value for 3 to 5 drinks daily. Even drinking wine only occasionally seemed to help. This article was the basis of a segment on the television show 60 Minutes on November 5, 1995, on the benefits of wine in the prevention of heart disease. This was the second such discussion 60 Minutes has had. Their segment called the "French Paradox," shown four years ago, is generally credited in the wine business with causing an upsurge in red wine sales that continues today. Return to Table of Contents \| Return to the JSE Home Page	{"url":"http://jse.amstat.org/v3n3/resource.html","timestamp":"2024-11-02T12:12:48Z","content_type":"text/html","content_length":"30449","record_id":"<urn:uuid:2df29ce8-d57a-4cc9-ba03-ac644b5653ad>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00348.warc.gz"}
Problem 953. Pi Estimate 1 Estimate Pi as described by the Leibniz formula (see the following link): Round the result to six decimal places. Solution Stats 21.2% Correct \| 78.8% Incorrect Problem Comments please update the link to the document, cant find it anymore The link is broken, but you can solve this problem with the Leibniz formula for π. The principle is described here: https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80 @Jordan Wilkerson change the number format and round to significant digits shown in test code @Paul Derwin Oh right, duh. I should've thought of that haha. That worked! I deleted my last comment since it basically was a solution after all. Thanks for your help! Leibniz formula, precision 1e-6. The description has been updated per the link provided in a prior comment. function [estimate] = pi_est1(nMax) for i=1:(nMax-1) if mod(i,2)~=0 I believe the task statement should be modified by specifying that the estimate should be rounded. Otherwise, the test suite can be edited accordingly. The problem description has been updated to specify that rounding is required. What an idiotic checking algorithm. I spent 10 minutes just trying to get the software to think I had solved the problem when I had already gotten it right multiple times earlier. Make sure you officially round, and not just print it with digits (which is smarter because the variable retains all of its significant digits), otherwise you'll find your time wasted like mine was. And, all for 10 points. How stupid. Make a better checking algorithm or at least give us more points for having to put up with this BS I'll have to say that the checking algorithm is very poorly written. Using isequal instead of an absolute difference with a permitted tolerance check is straight out flawed for me. And after repeating the calculations with both Python and MATLAB. I get an estimate of 3.2323.. for a series of 10 elements (nMax = 10). I wonder how the problem setters arrived at an estimate of 3.04... something. I can't remember the exact value that was used for that test. Solution Comments Show comments Problem Recent Solvers901 Suggested Problems More from this Author4 Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting!	{"url":"https://se.mathworks.com/matlabcentral/cody/problems/953?s_tid=prof_contriblnk","timestamp":"2024-11-09T09:37:39Z","content_type":"text/html","content_length":"123810","record_id":"<urn:uuid:2342b882-7b39-4fa4-87a8-d028163e64ef>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00754.warc.gz"}
Provocative Factors Revealing Why Fuel Price, Foreign Exchange Rate May remain High in Nigeria Provocative Factors Revealing Why Fuel Price, Foreign Exchange Rate May remain High in Nigeria Habu Sadeik Habu Sadeik is an Associate Chartered Accountant (ACA), financial analyst, political Finance, Energy Finance and Power Sector Enthusiast. He wrote about the pros and cons of returning Nigerian economy to the oil subsidy regime and N600/$1 Exchange Rate. In his critical analysis, he posed tempting questions that beg for economic answers, giving fa ta and figure about the reality of Nigerian economy and the oil subsidy regime and the Foreign Exchange The comment section is available for you to add your contributions, kindly read through Sadeik's submission: Is it possible for the government to return fuel subsidy? N400/litre? Is it possible to reverse the fx subsidy by pegging to N600/$ This is a question that a lot of non-economics people are asking. Well, follow me lets take it step-by-step and see if its possible The beauty about economics is that every decision has a pros and cons. if you look at the decision and you did not find both the pros and cons, then either you're being bias or you did not understand the situation. Fuel subsidy to N400/litre? let's do the economics PMS is the final output of crude oil, meaning you get PMS from crude oil. This means the higher the price of crude oil, the higher will be the PMS. Lets work with numbers and assumption. Price of Crude oil is $80/barrel Exchange rate is N1,600/$ For every barrel of crude oil, it is estimated that you can get 170 litres of different product. This means you can get PMS, Diesel, Kerosene, Asphalt, Coke and other product. The output of a single barrel of crude oil is 170 litres. 43% of the 170 litres will be solely for PMS. This means that 43% of the total 170 litres will be 73 litres. In summary, you get 73 litres of petrol from each barrel of crude. This is just on an ideal estimate If the cost of a barrel is $80 then 43% of it is meant for the petrol. 43% of $80 is = $34.4 This means the cost of getting 73 litres from a barrel of crude is $34.4 The cost price of a single PMS litre is ($34.4/73 litres) $0.47 Each litre of PMS is going to cost $0.47. To confirm, you can also divide the entire cost of the barrel ($80) divided by the expected output (170 litres) $80/170 litres = $0.47 I did all this computation to arrive at the cost price of petrol. remember this is only cost price A refinery owner need to factor Opex cost, and profit margin. If the cost price of Petrol is $0.47 and our exchange rate is N1600/$, then the cost price of petrol in Naira is N752/litre. The cost price of petrol is N752/litre. Lets assume the refining cost is N248/litre making it to be N1,000/litre. Lets say the profit margin is N100/litre. I used refining cost to accommodate all the landing cost and other expenses applicable. Total price of PMS to sell to the market will be N1,100/litre. This is the real market price. willing-buyer willing-seller model. If government wants to subsidise the price of petrol to N400/litre instead of the original N1,100/litre, then they have to pay for the difference. How? simple. N1,100-N400 = N700/Litre is to be subsidised What is the total consumption of the PMS daily in the country? This figure shows we consume 42 million litres daily but other data have showed that its not possible based on the number of cars available in the country. All right, let's be prudent and go with the assumption that Nigeria consumes 30 million litres daily. This means government need to subsidise 30 millions litres by N700 per each. 30m X N700 = N21 billion daily for subsidy N21 billion by 30 days = N630 billion monthly N630 billion monthly by 12 months = N7.5 trillion annually Government need to spend and average of N7.5 trillion annually to continue subsiding our petrol price with an assumption that we consume 30m litres daily and fx rate of N1600/$ with a crude oil price of $80/barrel. Any change in those variables will either increase or reduce the subsidy amount. Look at the budget numbers for the year 2024 below Now the presidential question goes...... Can you as a President and Commander-in-Chief of the FRN subsidise the Petrol subsidy by incurring additional N7.5 trillion in addition to the budget numbers shown above? Most people will say yes because they feel it's easy. I know some will ask me how is it that the previous government is doing it without any wahala? The answer is simple, previous government got the exchange rate below N300/$ and massive oil earnings in the budget. I understand it can be a controversial question/issue but the aim here is to bring the analysis for your perusal. Cons of fuel subsidy = spending N7.5 trillion annually. Pros of fuel subsidy = low prices of goods and services in the market. Government wants to Peg exchange rate to N600/$ as CBN rate while the black market rate is N1,600. What's the implication and analysis? I will try to be brief and narrow here. Assuming you're a manufacturer that imports raw materials for production. You went to CBN for dollars to import (of course you can only import with dollars) CBN will give you the dollars at N600/$ and open LC (letter of credit) for you. LC means CBN will pay the foreign supplier that will provide goods for you directly. If the supplier, say from UK supplied goods worth $20 million for you as a manufacturer, he will expect CBN to credit his account with $20m You as a manufacturer will pay CBN N12 billion equivalent of $20m at a FIXED RATE OF N600/$ Simple question is, where will CBN be getting the $20m to settle the LC they open for you? CBN earns dollars through 4 major ways 1. Oil Export 2. Non-oil Export 3. Diaspora 4. FPI/FDI Those 4 major sources is what will increase our reserves. The reserve will be used to settle transactions like $20m mentioned above. The 4 major sources of earning dollars mentioned above is no longer enough to settle the outstanding liabilities against CBN and this is what CBN called FX BACKLOG. They keep accumulating LCs, promises and obligations from various entities without settling them because they do not have enough dollars. Remember on our example above, that manufacturer who wants $20m for his import is going to sell his product in Naira, so he is not going to earn any USD for him to even help CBN. If CBN keep failing to provide him the dollars at exchange rate of N600/$, what are his options? Black market or close business? we both know the answer If he goes to black market at a rate of N1,600/$, rest assured that the price of his goods will also skyrocket. Bottomline is, you can not be able to fixed an exchange rate without a supporting reserve. You can do it for 1 month 2 month, by the time you begin to default on your payment, you will regret it. Now, having read and understand all that, what solutions can you proffer to the government that can ease the difficulties faced by people in the short-term? We all know the long term solutions but any effective short-term solutions? Attached Photo(s): RoundTable Image RoundTable Image RoundTable Image Category: World-News Tag: Provocative Factors Revealing Why Fuel Price Foreign Exchange Rate May remain High in Nigeria Written by Author (author) Published 2/22/2024 11:42:13 AM You cannot submit comment on this topic because you are not currently login. You can choose to Login or Create an Account then you are good to join the discussion. 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