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1901	https://or.stackexchange.com/questions/732/how-to-select-a-constraint-programming-solver	How to select a Constraint Programming Solver - Operations Research Stack Exchange Join Operations Research By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How to select a Constraint Programming Solver Ask Question Asked 6 years, 3 months ago Modified5 years, 7 months ago Viewed 16k times This question shows research effort; it is useful and clear 41 Save this question. Show activity on this post. I would like to clarify that I am absolutely not an expert when it comes to constraint programming (I happen to use it when it seems to be the right tool for the right job). It is my understanding that the possibility to use global constraints can have a tremendous impact on how efficiently a problem can be solved. However, not all global constraint are implemented in all constraint programming solvers. For example if I use the global constraint catalog to search existing global constraints, I will find out that the AllDifferent constraint (which represents the fact that no two variables of a given subset of the variables of the problem should take the same value) is implemented, under different names, in quite some constraint programming solvers. On the other hand only two solvers are referenced as implementing the increasing constraint is referenced as only being implemented by three solvers. Here comes my question: When using constraint programming, do you first write the model, trying to use global constraints when possible, and later on select which constraint programming solver you are going to use based on the availability of those global constraint? Or do you have a set constraint programming solver that you always use, no matter which global constraint you actually need (implementing the needed global constraint yourself if needed)? solver constraint-programming Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Oct 15, 2019 at 21:08 Rodrigo de Azevedo 309 2 2 silver badges 11 11 bronze badges asked Jun 25, 2019 at 15:33 Renaud M.Renaud M. 2,436 1 1 gold badge 15 15 silver badges 28 28 bronze badges 2 2 I would not be surprised to see that this constraint-cover approach is rarely pursued. More important things are probably: do i know the solver well enough (to model it in a good way; to work around it's weak-points), does the solver allow my features (supporting branch-and-bound is probably on a higher level than supporting constraint x; how well is search tunable, which is very very important in CP imho), will the solver fit to my architecture (C++ vs. Java for example can be a criterium). Then there are more things like: "learning" vs. no-learning and different kinds of global-prop-impls.sascha –sascha 2019-06-25 18:58:53 +00:00 Commented Jun 25, 2019 at 18:58 @sascha interesting points (especially about the learning aspect and how much can the search be customised). Might be worth pointing that out in an answer to show that my view on the problem was indeed probably too narrow.Renaud M. –Renaud M. 2019-06-25 19:01:35 +00:00 Commented Jun 25, 2019 at 19:01 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 40 Save this answer. Show activity on this post. @Rob wrote a great and extensive answer, but I would like to add two systems. MiniZinc is a high level CP system that is great for learning CP, prototyping problems as well as testing different solvers. MiniZinc first flatten a MiniZinc (.mzn) model to FlagZinc format (.fzn) and there are quite a few CP solvers that supports the FlatZinc format, e.g. Gecode, Chuffed, OR-tools, SICStus Prolog, JaCoP, Picat (CP/SAT), etc (they are listed in the MiniZinc main page). There is also an MiniZinc IDE where some of these solvers are included. An example of the high level syntax of MiniZinc is the element constraint x[y] = z where x in an array of decision variables, y and z are decision variables. The constraint constrain z to be the yth value in x. Most other CP languages write this constraint as something like element(x,y,z). Thus this syntax in MiniZinc makes it a little easier to write and understand the constraint. MiniZinc support quite a few global constraints, though if a FlatZinc solver supports a specific global constraint, it will use its own instead. There is also an yearly MiniZinc Challenge where FlatZinc solvers are solving a number of different MiniZinc models. Last year, OR-Tools dominated the challenge completely. Most MiniZinc models and instances of former challenges are collected at GitHub, One drawback of MiniZinc is that it is not a - Turing compatible - programming language, but it is quite easy to spawn a MiniZinc process to solve a problem, and some systems have this integrated (e.g. SICStus Prolog, ECLiPse CLP). And if you are into C++ it is not that complicated to integrate it even more. And here's my MiniZinc page with a couple of examples, small and large. To summarize, MiniZinc is great for learning the concept of Constraint Programming and for prototyping. If one need to use another programming language, it's often quite easy to port the MiniZinc model to another CP system. Picat Well, I have to add it to the list since I am in the Picat team and I really like Picat as a CP system. :-) Picat is a logic-based multi-paradigm programming language inspired by Prolog. The creator of Picat - Neng-Fa Zhou - is also the creator of B-Prolog (which is used in Picat's engine). This Prolog inspiration is seen for example with the support of non-determinism, but Picat also supports for loops, while loops, re-assignments, indexing of lists/arrays etc. Picat supports a couple of constraint solving modules: MIP (GLPK and Gurobi), SAT, and SMT (z3 and cvc4); all these solvers has support for the same syntax/constraints (with exception of MIP solver which only supports linear constraints). The PicatSAT FlatZinc solver has done quite well the last MiniZinc Challenges. One feature of Picat that I - mostly - like is that the order of the constraints are important when using the CP module. In most CP systems the order of the constraints does not matter, but in Picat (for the CP module) the order might make a difference and this is one other way to make a model more efficient. (Picat also has a planner module for traditional planning problems, but this is a bit out of scope of the question.) We wrote a book Constraint Solving and Planning with Picat about how to use Picat for Constraint Programming problems (as well as planning problems). The freely available PDF. I hope that the two chapters about CP might be useful as an introduction to CP in general. Also, My Picat page has quite a few examples of Picat models. For harder CP problems, it's no uncommon have one have to tweak the model to make it fast enough. Most CP systems has a different ways of doing that: selecting the order of variables to test (variable selection) when testing a specific variable, selecting the order of the values to test (value selecting). Unfortunately, selecting these is - as of now - and art and one have to test different variants. Also, there are some "tricks" that often speed things up, apart from finding the best variable/value strategies, e.g. symmetry breaking and adding redundant constraints to prune the search tree. A side note: @Rob mentioned that I don't blog anymore which is correct. Instead I write - occasionally - at - Facebook - Twitter - StackOverflow, mostly answering questions about MiniZinc and Constraint Programming - And publish stuff on GitHub Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jun 29, 2019 at 8:14 hakankhakank 954 6 6 silver badges 15 15 bronze badges 2 3 Thanks for that Håkan.Rob –Rob 2019-06-29 09:14:40 +00:00 Commented Jun 29, 2019 at 9:14 Is there some resources to learn about those tricks?Joffrey L. –Joffrey L. 2020-08-13 11:30:10 +00:00 Commented Aug 13, 2020 at 11:30 Add a comment\| This answer is useful 30 Save this answer. Show activity on this post. ... Do you have a set constraint programming solver that you always use, no matter which global constraint you actually need (implementing the needed global constraint yourself if needed)? Find one that is well supported, that you understand to some extent, and that is powerful enough (speed and expressiveness) to suit your future needs. Invest your time in choosing a good tool and learning to use it well, then implementing constraints will be easier than struggling with a tool that is inadequate for the job. There are online tools for linear optimization, constraint solving, even an extension for Microsoft Excel called OpenSolver which extends Excel’s built-in Solver with more powerful solvers. Constraint Logic Solvers are available which run on many different languages: Java-based constraint solvers C-based constraint solvers C++ -based constraint solvers C, C++, Python and Tcl based constraint solvers C++/Java/Python/Scheme and Smalltalk-based constraint solvers Python-based constraint solvers PROLOG-based constraint solvers VB.NET-based constraint solvers Mozart-Oz-based constraint solvers Other constraint solvers Which to choose is partly a function of many things. Different Contraint Programming Solvers perform better than others in benchmarks, speed and ability is one aspect to consider. Focus on how easy is it to learn a system, and the modelling aspect. Consider the syntax, element constraint, reification and defining predicates (decompositions). Look for community support and the number of examples, documentation, etc. You'll notice that Google's OR-Tools has a Stack Overflow tag for OR-Tools and is free. In Java, there's OptaPlanner (StackOverflow) and Choco (StackOverflow), both open source too. OR-Tools is placed well in the benchmarks: "OR-Tools is an open source software suite for optimization, tuned for tackling the world's toughest problems in vehicle routing, flows, integer and linear programming, and constraint programming. After modeling your problem in the programming language of your choice, you can use any of a half dozen solvers to solve it: commercial solvers such as Gurobi or CPLEX, or open-source solvers such as SCIP, GLPK, or Google's GLOP and award-winning CP-SAT.". Further afield are proof solvers such as ACL2 (A Computational Logic for Applicative Common Lisp), an extension to Lisp which is itself an extensible theory in a first-order logic, and an automated theorem prover. It represents a difficult uphill climb, but it produces programs capable of proving themselves to be correct. If you visit the GCC you'll notice that they describe the global contraints in Prolog (and XML) so if you want to rely on that reference it's helpful if your Constraint Logic Solver is written in (or accepts) Prolog (or that you are fluent in that language). "Constraints Constraint logic programming extends Prolog to include concepts from constraint satisfaction. A constraint logic program allows constraints in the body of clauses, such as: A(X,Y) :- X+Y>0. It is suited to large-scale combinatorial optimisation problems and is thus useful for applications in industrial settings, such as automated time-tabling and production scheduling. Most Prolog systems ship with at least one constraint solver for finite domains, and often also with solvers for other domains like rational numbers.". There are global constraints in SICStus Prolog, and five libraries: Constraint Handling Rules, Constraint Logic Programming over Booleans (unsupported since release 4.0.7), Constraint Logic Programming over Finite Domains, Constraint Logic Programming over Rationals or Reals (unsupported), and a Finite Domain Constraint Debugger - it's quite a full featured version of Prolog, but it's not free. B-Prolog is a fast alternative and while it does have a tag on SO there's barely a dozen uses of it. But if you look at Wikipedia's comparison of Prolog implementations you'll notice that it has a feature set comparable to SICStus. So you should probably favor a Prolog implementation, but there's no reason why you couldn't use Haskell. Which underlying language is used to run whichever solver is very much a personal preference, but choosing one language and a solver framework with a large community and numerous libraries for different solvers (or the ability to output for a long list of separate solver executables) ensures that you'll always be able to progress forward rather than getting stuck implementing missing features or asking for assistance. Additional Info: One way to try different solvers is to visit the NEOS Server, a free internet-based service for solving numerical optimization problems. Hosted by the Wisconsin Institute for Discovery at the University of Wisconsin in Madison, the NEOS Server provides access to more than 60 state-of-the-art solvers in more than a dozen optimization categories. There are 3rd party submission tools that allow you to write a simple program at home and submit your job to the server. See the NEOS FAQ for details. A means to get your problem solutions reviewed is at OPTIL.io, an on-line judge system that receives algorithmic solutions of optimization problems in a form of source code from the crowd of developers, compiles it, executes in a homogeneous run-time environment and objectively evaluates using the set of test cases. Problems that are solved can be provided by external companies or scientists. Solutions can be submitted in almost any programming language. A good tool to compare some of the popular solvers is the Apache Software Foundation's Constraint Programming Solvers comparison tool. For example comparing Choco, Picat, and OR-Tools shows that OR-Tools requires quite a lengthy specification compared to Picat, while Choco is certainly more mature and supports more Global Constraints. Hakan Kjellerstrand's webpage and blog on Constraint Programming while not updated in the past year has a decade of archives and (amongst numerous other pages) a webpage about Global Constraints, also one on Operations Research. Wikipedia's has webpages on: Constraint Satisfaction and Hill Climbing (but nothing for Late Acceptance Hill Climbing (LAHC)). LAHC is one of the techniques employed by OptaPlanner, an AI constraint solver. Google returns over 70M results for "Late Acceptance Hill Climbing" - lesson: don't let one site's miss (due to chosen search terms) make you miss out on other results, check multiple search terms and search engines. Another search term to use is "Late Acceptance Anytime Algorithm" (related to LAHC) which returns another 56M results. Some of the newest directions have the least information available, it's important to do your own research and determine which direction you wish to go. Breakthroughs in AI and Neuromorphic Computing are frequent enough that best laid plans go awry a few years later; but as long as the legacy tools you have available are sufficiently capable and fast enough for your needs you shouldn't worry about bleeding edge too much, nor completely discount it either. The website arXiv has 3761 results for "Constraint Programming" and is an excellent source for researching the latest techniques. I would like to clarify that I am absolutely not an expert when it comes to constraint programming (I happen to use it when it seems to be the right tool for the right job). A description of the bottom-up and top-down proof procedures used for searching the problem space gives some insight into choosing your tool and approach. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Jun 18, 2020 at 8:36 CommunityBot 1 answered Jun 26, 2019 at 20:47 RobRob 2,151 1 1 gold badge 15 15 silver badges 35 35 bronze badges 2 1 I took a look at the solver lists at www.constraintsolving.com. There's one (dated?) reference to ILOG solver, and I couldn't find any mention of CP Optimizer. They do list some commercial products, but for some reason not that one. It might be relevant, as (a) I think it's pretty good and (b) it's free for academic use. Also, the language sorting seems to be based on the language in which the solver is coded, which need not be the language from which you call it.prubin –prubin♦ 2019-06-27 20:56:35 +00:00 Commented Jun 27, 2019 at 20:56 2 Thanks Rob. As always, the level of obsolescence of these catalog sites makes me a bit sad. It cannot be avoided I guess.Laurent Perron –Laurent Perron 2019-07-26 18:01:31 +00:00 Commented Jul 26, 2019 at 18:01 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. If you're looking for objective metrics, such as GitHub stars, commit rates, contributor counts, StackOverflow question count and other usage statistics, libhunt covers constraint solvers at least for Java: Java constraint solvers Warning: objective metrics don't tell the entire story. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Feb 11, 2020 at 12:18 Geoffrey De SmetGeoffrey De Smet 5,229 12 12 silver badges 35 35 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Operations Research Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. 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1902	https://www.youtube.com/watch?v=VDwacB2YjMY	How to solve exponential equations \| K-method \|Grade 11-12 \| Examples \| Part 3 MathGuru 2210 subscribers 68 likes Description 3726 views Posted: 4 May 2023 This video teaches you how to solve #exponentialequations Whether you're just starting out, or need a quick refresher, this is the video for you! Skills needed: Laws of exponent Algebra, Quadratic equations , Factorization Subscribe to our channel: Contact: olamamasblog@gmail.com 4 comments Transcript: hi everyone this is your math Guru in this video I'm going to show you how to solve the following exponential equations using the key method please don't forget to click on the like And subscribe button below for more easy to understand math concepts in the first example we have 3 to the power of 2x plus 6 times 3 to the power of x minus 27 equals to zero not that the law of exponents states that if I have a base of a to the power of X all raised to above Y is the same as a to the power of we're going to multiply X and Y that's X Y so I'm going to split up my exponent in each term because I'm trying to look for a common Factor so I have 3 to the power of X all raised to the power of 2 plus 6 times 3 to the power of x minus 27 equals to zero if you check each term You observe that I have 3 to the power of X comma to two of the terms so therefore using the K method I'm going to say let's my k equals 3 to the power of X so every occurrence of 3 to the power of X and the equation I'm going to replace it with K so therefore my equation becomes K squared plus 6K minus 27 equals to 0. now I have a new quadratic equation which I'm going to factorize that's K squared plus 6K minus 27 if I factorize I'm going to have K plus 9 and K minus 3 equals to zero we can have eight of the contents of the bracket equals to zero so I have K plus 9 equals to zero or k minus 3 equals to 0 for the first solution if I take plus 9 to the other side I'm going to have k equals to minus 9 remember a k is 3 to the power of X so replacing K I have 3 to the power of x equals to minus 9. however this is a no solution because I have 3 to the power of 8 must be greater than 0 for all values of x so therefore I can have 3 to the power of x equals to minus 9. for the second part of the solution I have K minus 3 equals to zero taking minus 3 to the other side I have k equals to positive three replacing my K I have 3 to the power of x equals to three there's an invisible one here and because the base are the same I'm going to equate the powers so therefore my x equals to 1. let's look at the second example in the second example I have to the power of X plus one plus two to the 12x minus 1 equals to 5. applying the laws of exponents I have a to the power of X Plus Y is the same as a to the power of x times a to the power of Y because once the B's are the same and you're multiplying you add the powers so I'm going to split out my exponents and the question given I have 2 to the power of x times so to the power of 1 plus 2 to the power of x times two to the power minus 1 equals to 5. If You observe I have 2 to the power x common so I'm going to let my k equals to the power of X so anyway you find the power of X I'm going to replace it with K so my new equation becomes k times 2 plus K times 2 to the power of minus 1 equals to 5. if I simplify for that I have 2K Plus lots of exponents again if you have a negative exponent to make it positive I'm going to find the inverse that's a to the power of positive one so therefore my K times 2 to the power minus 1 becomes k over 2 equals to 5 my LCD or my lowest common denominator a is two this is over one over one so I'm going to multiply each of my Times by two so multiplying each time by two or the LCD I'm going to have 4K oh let me do this that's times two whatever are you doing your left times two you're doing it right my new equation becomes 4K Plus k equals to 5 times 2 10. 5k equals to 10. if I divide both sides by 5 I have five here one k equals 10 divided by 5 that's 2 remember our K is 2 to the power of X so I'm going to replace my K which is above x equals to 2 there's an invisible one here and because the base are the same I'm going to equate the powers so my X is equal to 1. let's look at some more examples in the next example I have two to the power of 2x minus 6 times 2 to the power of x minus 1 plus 2 equals to zero like I did earlier I'm going to split up all my terms I have two to the power of X all raised to the power of 2 minus 6 times 2 to the power of x times 2 to the power minus 1 plus 2 equals to zero if you check each term I have 2 to the power of X common so therefore I'm going to say let my k equals 2 to the power of X so going back to my equation new equation becomes s squared minus six times K times that's a negative exponent if it becomes positive that's one over two plus two equals to zero two here one two here three so therefore my equation becomes K squared minus three K plus 2 equals to zero if I the right I'm going to have K minus 2 and K minus one either of the brackets can be equal to zero so I have K minus two because zero or k minus 1 equals to zero for the first solution I have K taking minus 2 so the other side it becomes k equals to positive two and remember the value of a k is 2 to the power of X so I have 2 to the power of x equals to two there's an invisible one here and because the base are the same I'm going to equate the powers so therefore my X equals to 1. for the other solution I have K minus 1 equals to zero taken minus one to the other side I have k equals to positive one like I did earlier I'm going to replace my K which suits to the power of x equals 1 which is the same as so to the power of zero law of exponents states that any variable or any number raised to the power of 0 is equal to one and because the base are now the same therefore my X equals to zero in my last example I have 9 to the power of x minus 3 to the power of x minus 2 equals to zero because I want to put all my terms in the base of 3 I'm going to change my 9 so 3 squared so I have 3 squared or raised to the power of x minus I still have my 3 to the power of x minus 2 equals to zero if I rearrange as the same as 3 to the power of X all to the power of 2 the same as 2x minus breaking my second term I have 3 to the power of x times three to the power of minus two equals to zero if you check each term I have 3 to the power of X common so therefore I'm going to Let's my k equals 3 to the power of X so every occurrence of three to the power of X I replace it with k that's K squared minus K making that a positive exponent I have over 9 or over 3 squared equals to zero that's K squared minus K over 3 squared is the same as 9 equals to zero my LCD or my lowest common denominator is 9 so I'm going to multiply both sides of the equation by 9 therefore expand on the brackets I have my K squared minus k equals to zero if I factorize K is common I have K and bracket like k minus 1 equals zero therefore k equals zero or my K minus 1 equals to zero note that the value of K is 3 to the power of X so I have for my left hand side 3 to the power of x equals to zero however this is in no solution because the value of 3x cannot be equal to zero on my right hand side if I take minus 1 to the other side I have my k equals to positive 1 divided by both sides by nine I have whatever nine not Stat one over nine is the same as 1 over 3 squared and making it a negative exponent the same as 3 to the power minus two so I'm just going to substitute that back I have my K which is 3 to the power of x equals 3 to the power of minus two and because the base are the same that's three therefore my x equals 2 minus two thanks for watching bye
1903	https://www.doubtnut.com/qna/642549393	In acute angled △ABC prove that tan2A+tan2B+tan2C≥9. More from this Exercise To prove that in an acute-angled triangle ABC, the inequality tan2A+tan2B+tan2C≥9 holds, we can follow these steps: Step 1: Use the Angle Sum Property In triangle ABC, we know that the sum of the angles is: A+B+C=180∘ Step 2: Apply the Tangent Addition Formula Using the tangent addition formula, we can express tan(A+B+C): tan(A+B+C)=tan180∘=0 This implies: tanA+tanB+tanC−tanAtanBtanC=0 Rearranging gives us: tanA+tanB+tanC=tanAtanBtanC Step 3: Use the AM-GM Inequality Now, we apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality: tanA+tanB+tanC3≥√(3)tanAtanBtanC Let x=tanA, y=tanB, and z=tanC. Then: x+y+z3≥√(3)xyz Step 4: Substitute Back to the Original Equation From our earlier result, we have: x+y+z=xyz Substituting this into the AM-GM inequality gives: xyz3≥√(3)xyz Cubing both sides results in: (xyz3)3≥xyz This simplifies to: (xyz)327≥xyz Assuming xyz≠0 (which is true since A, B, and C are angles in an acute triangle), we can divide both sides by xyz: (xyz)227≥1 Thus: (xyz)2≥27 Taking the square root gives: xyz≥3√3 Step 5: Apply AM-GM Again for Squares Now, we apply AM-GM again for the squares: tan2A+tan2B+tan2C3≥√(3)tan2Atan2Btan2C This leads to: tan2A+tan2B+tan2C≥3√(3)tan2Atan2Btan2C Since xyz≥3√3, we have: tan2Atan2Btan2C≥(3√3)2=27 Thus: √(3)tan2Atan2Btan2C≥3 Finally, substituting back gives: tan2A+tan2B+tan2C≥3×3=9 Conclusion Therefore, we have shown that: tan2A+tan2B+tan2C≥9 This completes the proof. To prove that in an acute-angled triangle ABC, the inequality tan2A+tan2B+tan2C≥9 holds, we can follow these steps: Step 1: Use the Angle Sum Property In triangle ABC, we know that the sum of the angles is: A+B+C=180∘ Step 2: Apply the Tangent Addition Formula Using the tangent addition formula, we can express tan(A+B+C): tan(A+B+C)=tan180∘=0 This implies: tanA+tanB+tanC−tanAtanBtanC=0 Rearranging gives us: tanA+tanB+tanC=tanAtanBtanC Step 3: Use the AM-GM Inequality Now, we apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality: tanA+tanB+tanC3≥√(3)tanAtanBtanC Let x=tanA, y=tanB, and z=tanC. Then: x+y+z3≥√(3)xyz Step 4: Substitute Back to the Original Equation From our earlier result, we have: x+y+z=xyz Substituting this into the AM-GM inequality gives: xyz3≥√(3)xyz Cubing both sides results in: (xyz3)3≥xyz This simplifies to: (xyz)327≥xyz Assuming xyz≠0 (which is true since A, B, and C are angles in an acute triangle), we can divide both sides by xyz: (xyz)227≥1 Thus: (xyz)2≥27 Taking the square root gives: xyz≥3√3 Step 5: Apply AM-GM Again for Squares Now, we apply AM-GM again for the squares: tan2A+tan2B+tan2C3≥√(3)tan2Atan2Btan2C This leads to: tan2A+tan2B+tan2C≥3√(3)tan2Atan2Btan2C Since xyz≥3√3, we have: tan2Atan2Btan2C≥(3√3)2=27 Thus: √(3)tan2Atan2Btan2C≥3 Finally, substituting back gives: tan2A+tan2B+tan2C≥3×3=9 Conclusion Therefore, we have shown that: tan2A+tan2B+tan2C≥9 This completes the proof. Topper's Solved these Questions Explore 100 Videos Explore 22 Videos Explore 5 Videos Explore 38 Videos Explore 6 Videos Similar Questions If A+B+C=π, prove that tan2A2+tan2B2+tan2C2≥1. If A+B+C=π, prove that tan2A2+tan2B2+tan2C2≥1. If A, B, C be an acute angled triangle, then the minimum value of tan4A+tan4B+tan4C will be In an actue-angled triangle ABC Statement-1: tan2A2+tan2B2+tan2C2≥1 Statement-2: tanAtanBtanC≥3√3 If A,B,C are angles of a triangle, prove that tan(B+C)+tan(C+A)+tan(A+B)tan(π−A)+tan(2π−B)+tan(3π−C)=1. In any triangle ABC , prove that following : ca−b=tan(A2)+tan(B2)tan(A2)−tan(B2) In any triangle ABC , prove that following : ca+b=1−tan(A2)tan(B2)1+tan(A2)tan(B2) In any triangle ABC , prove that following : ca+b=1−tan(A2)tan(B2)1+tan(A2)tan(B2) If A,B,C are angles of a triangle, prove that tan B+C2=cotA2. If A, B, C are the interior angles of a triangle ABC , prove that tan(C+A2)=cotB2 (ii) sin(B+C2)=cosA2 CENGAGE ENGLISH-TRIGONOMETRIC RATIOS AND TRANSFORMATION FORMULAS-Concept App. 3.10 In A B C Prove that cos^2A/2+cos^2B/2+cos^2C/2lt=9/4dot If cos^2A/2... Let alpha,beta,gamma>0a n dalpha+beta+gamma=pi/2dot Then prove that sq... In acute angled triangle ABC prove that tan ^(2)A+tan^(2)B+tan^(2)Cge9... In a Delta ABC sin Asin B sin C <= (3sqrt3)/8 In triangle ABC, prove that sin""(A)/(2)sin""(B)/(2)sin""(C)/(2)le(1)/... Exams Free Textbook Solutions Free Ncert Solutions English Medium Free Ncert Solutions Hindi Medium Boards Resources Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. 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1904	https://dspace.mit.edu/bitstream/handle/1721.1/56566/2-61Spring2004/OcwWeb/Mechanical-Engineering/2-61Spring2004/LectureNotes/index.htm	MIT OpenCourseWare \| Mechanical Engineering \| 2.61 Internal Combustion Engines, Spring 2004 \| Lecture Notes Home Courses Donate About OCW Advanced Search Help Contact Us Sub-navigation links VIEW ALL COURSES Course Home Syllabus Calendar Readings Labs Assignments Exams Projects Related Resources Lecture Notes Download Course Materials Home>Courses>Mechanical Engineering>Internal Combustion Engines>Lecture Notes Lecture Notes These materials are used to provide an organizing framework, and data, for the lectures. \| SES# \| TOPICS \| LECTURE NOTES \| --- \| 1 \| Overview Basics of Engine Operation \| (PDF) \| \| 2 \| Engine Geometry Performance Parameters \| (PDF) \| \| 3 \| Engine Disassembly In Lab \| \| \| 4 \| Ideal Cycle Analysis Combustion Stoichiometry \| (PDF 1) (PDF 2) (PDF 3) \| \| 5 \| Fuel-air Cycle Model Fuel-air Cycle Results \| \| \| 6 \| Gas Exchange: 4-Stroke Gas Exchange: 2-Stroke \| (PDF) \| \| 7 \| Spark-ignition Engine Combustion Flame Propagation and Structure \| (PDF) \| \| 8 \| Recitation \| \| \| 9 \| Knock \| (PDF) \| \| 10 \| SI Engine Emissions \| (PDF) \| \| 11 \| Emission Control Technology Review \| (PDF) \| \| 12 \| Quiz (Duration: 1-1/2 hours) Open Book \| \| \| 13 \| Mixture Preparation Engine Friction \| (PDF) \| \| 14 \| Engine Heat Transfer 2-Stroke SIE Performance \| (PDF) \| \| 15 \| 4-Stroke SIE Performance \| (PDF) \| \| 16 \| Variable Valve Control Gasoline DI Engines \| (PDF) \| \| 17 \| Diesel Overview Diesel Combustion \| (PDF 1) (PDF 2) (PDF 3) \| \| 18 \| Diesel Emissions Emission Control \| \| \| 19 \| Turbocharging Diesel Performance \| \| \| 20 \| 1-hour Lab Preparation \| \| \| 21 \| 2-hour Lab Session \| \| \| 22 \| Engine Noise Engine Dynamics \| \| \| 23 \| Future Engine Technology Discussion \| (PDF 1) (PDF 2) \| \| 24 \| Final Exam (Duration: 3 hours) Open Book \| \| RSS Feeds Privacy and Terms of Use Site Map Cite OCW Content Your use of the MIT OpenCourseWare site and course materials is subject to our Creative Commons License and other terms of use.
1905	https://www.quora.com/Why-is-acetic-acid-used-to-precipitate-barium-chromate	Why is acetic acid used to precipitate barium chromate? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Chemistry Barium Acetate Precipitation Reaction Acetic Acid Inorganic Chemistry Labor... Chemical Reactions Organic and Inorganic Che... General Inorganic Chemist... Precipitation 5 Why is acetic acid used to precipitate barium chromate? 2 Answers Sort Recommended Kumaraswamy Sathiavasan MSc in Chemistry & IAS officer(retd.) · Author has 9.2K answers and 14.4M answer views ·4y Barium chromate is insoluble in acetic acid but both calcium chromate and strontium chromate are soluble in the same acid. So, when a solution containing the bivalent calcium, strontium and barium ions is mixed with potassium chromate in presence of acetic acid, only barium will be precipitated as barium chromate (yellow solid). Ba(2+) (aq) + K2CrO4 (aq) = BaCrO4 (s) + 2K(+) (aq) Calcium and strontium ions remain in solution and thus the use of acetic acid helps separation of barium from the other two alkaline earth metals. But in the absence of acetic acid, apart from barium, calcium and stronti Continue Reading Barium chromate is insoluble in acetic acid but both calcium chromate and strontium chromate are soluble in the same acid. So, when a solution containing the bivalent calcium, strontium and barium ions is mixed with potassium chromate in presence of acetic acid, only barium will be precipitated as barium chromate (yellow solid). Ba(2+) (aq) + K2CrO4 (aq) = BaCrO4 (s) + 2K(+) (aq) Calcium and strontium ions remain in solution and thus the use of acetic acid helps separation of barium from the other two alkaline earth metals. But in the absence of acetic acid, apart from barium, calcium and strontium will also be precipitated as their respective chromates. Upvote · 9 3 Related questions More answers below Why does barium form precipitates with the addition of a buffer solution (acetic acid and ammonium acetate)? Can you use acetic acid for cleaning? What is the reaction of barium acetate, ethanol, and acetic acid? Is glacial acetic acid cancerous? Can acetic acid be used to fix the colour of the cloth? While washing, some cloths discharge colour. But with the use of acetic acid, can it be stopped? Assistant Bot · 1y Acetic acid is used to precipitate barium chromate (BaCrO₄) because it helps create the right conditions for the precipitation reaction to occur. Here’s a breakdown of the process: Formation of Barium Chromate: Barium chromate is formed when barium ions (Ba²⁺) react with chromate ions (CrO₄²⁻). The overall reaction can be represented as: Ba 2+(a q)+CrO 2−4(a q)→BaCrO 4(s)Ba 2+(a q)+CrO 4 2−(a q)→BaCrO 4(s) 2. Role of Acetic Acid: Acetic acid (CH₃COOH) is a weak acid that can lower the pH of the solution without completely neutralizing it. By adding acetic acid, the acidity of the solution Continue Reading Acetic acid is used to precipitate barium chromate (BaCrO₄) because it helps create the right conditions for the precipitation reaction to occur. Here’s a breakdown of the process: Formation of Barium Chromate: Barium chromate is formed when barium ions (Ba²⁺) react with chromate ions (CrO₄²⁻). The overall reaction can be represented as: Ba 2+(a q)+CrO 2−4(a q)→BaCrO 4(s)Ba 2+(a q)+CrO 4 2−(a q)→BaCrO 4(s) 2. Role of Acetic Acid: Acetic acid (CH₃COOH) is a weak acid that can lower the pH of the solution without completely neutralizing it. By adding acetic acid, the acidity of the solution increases, which can influence the solubility of barium chromate. 3. Precipitation Conditions: The precipitation of barium chromate is favored in an acidic environment because it helps shift the equilibrium of the chromate ions towards the formation of the less soluble barium chromate. In highly alkaline conditions, chromate ions can convert to dichromate ions (Cr₂O₇²⁻), which would not precipitate with barium. 4. Minimizing Solubility: The presence of acetic acid minimizes the solubility of barium chromate, promoting its precipitation as a solid. This is important for applications such as gravimetric analysis, where the formation of a solid precipitate is necessary for accurate measurements. In summary, acetic acid is used because it helps to create the optimal acidic conditions necessary for the effective precipitation of barium chromate from a solution, enhancing the formation of the desired solid product. Upvote · Related questions Why does barium form precipitates with the addition of a buffer solution (acetic acid and ammonium acetate)? Can you use acetic acid for cleaning? What is the reaction of barium acetate, ethanol, and acetic acid? Is glacial acetic acid cancerous? Can acetic acid be used to fix the colour of the cloth? While washing, some cloths discharge colour. But with the use of acetic acid, can it be stopped? What causes precipitate to form when barium chloride and potassium chromate react? How is acetic acid made? What is it used for? What fruits and vegetables contain acetic acid? Is 45% acetic acid safe to use? If methyl orange, which changes colour at about pH 3.7, had been used instead of phenolphthalein in an acetic acid and NaOH titration . Would you expect the calculated percentage of acetic acid to be too high or too low? Why? What are the uses of barium acetate? What is the % of acetic acid in glaciel acetic acid, acetic acid and vinegar? Is acetic acid used for injera preservation? Why does an acetic prevent precipitation of copper chromate? Why don't we use methyl orange in the titration of acetic acid? Related questions Why does barium form precipitates with the addition of a buffer solution (acetic acid and ammonium acetate)? Can you use acetic acid for cleaning? What is the reaction of barium acetate, ethanol, and acetic acid? Is glacial acetic acid cancerous? Can acetic acid be used to fix the colour of the cloth? While washing, some cloths discharge colour. But with the use of acetic acid, can it be stopped? What causes precipitate to form when barium chloride and potassium chromate react? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
1906	https://www.scribd.com/doc/165286999/Adamson-and-Gast-Physical-Chemistry-of-Surfaces	Adamson and Gast - Physical Chemistry of Surfaces Adamson and Gast - Physical Chemistry of Surfaces Uploaded by AI-enhanced description Adamson and Gast - Physical Chemistry of Surfaces Fem-Iike shapes are dendritic crystals of the protein streptavidin bound to a lipid monolayer. The island pattem is from a marbled paper, for which we thank Mrs. Phoebe Easton. No part of this publication may be reproduced, stored in a retrieval system or transmitted. Adamson and Gast - Physical Chemistry of Surfaces Adamson and Gast - Physical Chemistry of Surfaces Uploaded by AI-enhanced description Share this document Footer menu About Support Legal Social Get our free apps About Legal Support Social Get our free apps
1907	https://fiveable.me/key-terms/calculus-iv/clairauts-theorem	Clairaut's Theorem - (Calculus IV) - Vocab, Definition, Explanations \| Fiveable \| Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintablespricing my subjectsupgrade All Key Terms Calculus IV Clairaut's Theorem ∞calculus iv review key term - Clairaut's Theorem Citation: MLA Definition Clairaut's Theorem states that if a function has continuous second partial derivatives, then the order of differentiation does not matter; that is, the mixed partial derivatives are equal. This theorem is important in understanding how to compute higher-order derivatives and ensures that we can interchange the order of differentiation for functions with certain smoothness conditions, linking it to both the definition of partial derivatives and higher-order partial derivatives. 5 Must Know Facts For Your Next Test Clairaut's Theorem applies specifically when the second partial derivatives of a function are continuous at a point or over a region. The theorem can be expressed mathematically as $$ rac{{\partial^2 f}}{{\partial x \partial y}} = rac{{\partial^2 f}}{{\partial y \partial x}}$$ for a function $$f$$ of two variables. The significance of Clairaut's Theorem is that it simplifies calculations in multivariable calculus by allowing the interchange of the order of differentiation without worrying about the results being different. While Clairaut's Theorem assumes continuity of second partial derivatives, failure of this condition may lead to mixed partial derivatives being unequal. Understanding Clairaut's Theorem helps in analyzing the properties of functions that arise in physics and engineering, where multivariable functions frequently occur. Review Questions How does Clairaut's Theorem help in simplifying the computation of mixed partial derivatives? Clairaut's Theorem states that if a function has continuous second partial derivatives, you can differentiate in any order without affecting the outcome. This means that when computing mixed partial derivatives, you can choose whether to differentiate with respect to one variable first or another and still arrive at the same result. This property streamlines calculations significantly, especially in problems involving complex multivariable functions. What conditions must be satisfied for Clairaut's Theorem to hold true, and why are these conditions important? For Clairaut's Theorem to be valid, the second partial derivatives of the function must be continuous in a neighborhood around the point of interest. This condition is crucial because if the second partials are not continuous, it is possible for the mixed partial derivatives to differ. Ensuring continuity allows us to confidently interchange the order of differentiation, which is essential for accurate computations in multivariable calculus. Evaluate a function $$f(x,y) = x^2y + y^3$$ at point (1, 2) and demonstrate how Clairaut's Theorem applies when finding the mixed partial derivatives. To evaluate the mixed partial derivatives for $$f(x,y) = x^2y + y^3$$ at point (1, 2), we first compute the first-order partial derivatives: $$f_x = 2xy$$ and $$f_y = x^2 + 3y^2$$. Then we find the second-order mixed derivatives: $$f_{xy} = 2y$$ and $$f_{yx} = 2y$$ as well. Since both mixed partials are equal at (1, 2), Clairaut's Theorem holds here, confirming that we can interchange differentiation without loss of accuracy due to the continuity of the second derivatives. Related terms Partial Derivative: A derivative of a function with respect to one variable while holding the other variables constant. Mixed Partial Derivative:A derivative that involves differentiating with respect to two or more variables in succession. Continuous Function: A function that does not have any jumps, breaks, or discontinuities in its graph. 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1908	https://www.khanacademy.org/science/ap-college-physics-1/xf557a762645cccc5:kinematics/xf557a762645cccc5:mathematical-models-of-motion/v/developing-kinematic-equations-from-data	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
1909	https://kapdec.com/blog/understanding-interval-notations-simplified-guide/	The Spark Where Knowledge Finds its Voice Education is evolving — and The Spark is leading the charge with voices that matter and ideas that move. 11/01/2025 • 4 min read The Ultimate Guide to Interval Notation in Math Reading Time: 3 minutesLearn interval notation with our simplified guide. Understand how to represent ranges of numbers efficiently for algebra and calculus. by Kapdec Reading Time: 3 minutes Understanding Interval Notations \| Simplified Guide Mathematics often presents us with different ways of expressing sets of numbers, and one of the most compact and widely used methods is interval notation. Instead of writing long inequalities, interval notation provides a neat and standardized way to represent ranges of values. Understanding this notation not only helps in solving problems efficiently but also strengthens the foundation for advanced mathematical topics such as calculus, functions, and real analysis(Refer). What is Interval Notation? Interval notation is a shorthand system used to describe subsets of the real number line. Instead of using inequality symbols like <, >, ≤, or ≥, we use brackets and parentheses to represent whether a number is included in or excluded from the set. This system helps eliminate ambiguity and provides a universally accepted format, making mathematical communication smoother and clearer. Types of Intervals Intervals can be expressed in different forms depending on whether their endpoints are included or not: Open Interval (a, b): Represents all numbers strictly between a and b. The endpoints a and b are excluded. Closed Interval [a, b]: Represents all numbers between a and b, including both endpoints. Half-Open Intervals [a, b) or (a, b]: One endpoint is included while the other is excluded. Unbounded Intervals: Sometimes intervals extend infinitely. For example, (a, ∞) means all numbers greater than a, and (-∞, b] means all numbers less than or equal to b. The choice of brackets and parentheses communicates precisely whether the boundaries are part of the set. Why is Interval Notation Important? Interval notation is more than just a shorthand—it is a key to building mathematical precision. Its significance lies in: Clarity: It avoids the confusion that may arise with overlapping inequalities. Efficiency: It shortens long expressions into a concise format. Universality: It is used across textbooks, research papers, and standardized tests. Applicability: It plays a major role in higher-level mathematics like calculus, where intervals define domains, ranges, and regions of convergence. Common Uses in Mathematics Defining Domains of Functions: Interval notation helps express where a function is valid. Solving Inequalities: The solution set of an inequality is often written in interval form. Integration Bounds in Calculus: Interval notation is used to indicate the region of integration. Describing Continuous Data: It can be applied in statistics or data science to specify ranges of values. 6 FAQs About Interval Notation What do brackets and parentheses mean in interval notation? Brackets [ ] mean that the endpoint is included in the set, while parentheses ( ) mean that the endpoint is excluded. This small difference changes the set of numbers drastically. Can infinity be included in interval notation? No. Infinity ∞ and negative infinity -∞ are not actual numbers; they represent boundlessness. Therefore, they can never be included, which is why infinity is always written with parentheses ( ), not brackets [ ]. What’s the difference between open and closed intervals? An open interval excludes its boundary points, while a closed interval includes them. This difference is critical in analysis, as open intervals often appear in the study of limits and continuity, while closed intervals are used when working with compact sets. Why is interval notation better than inequalities? While inequalities like x > 2 or x ≤ 5 are intuitive, interval notation provides a cleaner, single expression to represent multiple ranges without repetition. It reduces clutter when working with complex solution sets. How does interval notation connect to the number line? Each interval corresponds to a continuous segment of the number line. Parentheses indicate that the boundary point is not shaded, while brackets show that the endpoint is shaded or included. This visual relationship makes it easier to interpret solutions graphically. Where will I use interval notation in advanced studies? Interval notation is essential in calculus, linear algebra, and real analysis. For example, when evaluating domains of logarithmic or rational functions, describing intervals of convergence in series, or setting boundaries for optimization problems, interval notation becomes indispensable. Final Thought Interval notation simplifies the way we express ranges of numbers. By mastering the use of brackets, parentheses, and infinity symbols, students and professionals alike gain a powerful tool for expressing mathematical ideas clearly and efficiently. It is not just a format but a stepping stone toward deeper mathematical understanding, ensuring accuracy in everything from algebra to advanced calculus from Kapdec. 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1910	https://brainly.com/question/24100697	[FREE] Select all conditions that would increase the effect of ion-pairing and decrease the observed van 't Hoff - brainly.com 7 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +53,3k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +17,8k Ace exams faster, with practice that adapts to you Practice Worksheets +6,8k Guided help for every grade, topic or textbook Complete See more / Chemistry Textbook & Expert-Verified Textbook & Expert-Verified Select all conditions that would increase the effect of ion-pairing and decrease the observed van 't Hoff factor. A. Lower charge on the ions B. High concentration C. Low concentration D. Higher charge on the ions 2 See answers Explain with Learning Companion NEW Asked by trevon2168 • 06/21/2021 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 41055918 people 41M 5.0 2 Upload your school material for a more relevant answer The conditions that would increase the effect of ion-pairing and decrease the observed van 't Hoff factor are a high concentration of solute and a higher charge on the ions. Explanation A number of particles are formed by dissolving a unit molecule of a solute in a solvent. The higher concentration of solution lowers the value of the van 't Hoff factor as ionic compounds do not get completely dissociated in their aqueous solution. Also, some ions form ion pairs (pairing of opposite charges in the aqueous state) and exist as a single particle which decreases the value of the van 't Hoff factor. The ions with higher charges have a greater tendency to form ion-pair due to their high electrostatic interaction. So, from this, we can conclude that conditions that would increase the effect of ion-pairing and decrease the observed van 't Hoff factor are a high concentration of solute and a higher charge on the ions. Learn more about the van 't Hoff factor here: brainly.com/question/2907004?referrer=searchResults brainly.com/question/13862200?referrer=searchResults Answered by Tringa0 •3.2K answers•41.1M people helped Thanks 2 5.0 (1 vote) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 41055918 people 41M 5.0 2 Topics in Thermodynamics of Solutions and Liquid Mixtures - Michael J Blandamer & Joao Carlos R Reis Beginning Chemistry - Ball Physical Chemistry for the Biosciences - Chang Upload your school material for a more relevant answer The conditions that would increase the effect of ion-pairing and decrease the observed van 't Hoff factor are high concentration and higher charge on the ions. So, the correct options are B and D. Lower charge on the ions and low concentration would not increase ion-pairing. Explanation To determine the conditions that would increase the effect of ion-pairing and decrease the observed van 't Hoff factor, we need to understand a few concepts about ionic solutions. Concept of Ion-Pairing Ion-pairing occurs when oppositely charged ions in solution come close enough to interact, forming a pair and behaving as a single particle rather than two separate ions. This phenomenon typically happens in concentrated solutions, leading to the observation that the actual number of particles in solution is less than what we would expect based on the initial concentration of dissolved ions. Van 't Hoff Factor (i) The van 't Hoff factor (i) is a measure of the number of particles a substance produces in solution. For example, the ideal van 't Hoff factor for sodium chloride (NaCl) is 2, since it dissociates into two ions: Na⁺ and Cl⁻. However, if ion-pairing occurs, i will be less than 2 because some of the ions remain together as pairs. Analyzing the Options Let’s examine each option provided: A. Lower charge on the ions - Lower charge typically means less strong electrostatic attraction between ions, which decreases the tendency to form ion pairs. Therefore, this condition does not promote ion-pairing. B. High concentration - High concentration increases the likelihood of ions encountering each other, thus enhancing ion-pairing. This leads to a decrease in the van 't Hoff factor due to fewer effective particles in solution. Hence, this condition increases ion-pairing and decreases the observed van 't Hoff factor. C. Low concentration - Low concentration decreases the likelihood of interactions between ions, reducing the effect of ion-pairing. Therefore, this condition does not support increased ion-pairing. D. Higher charge on the ions - Higher charges increase the electrostatic forces of attraction between the ions, which promotes the formation of ion pairs. Therefore, this condition also increases ion-pairing and decreases the van 't Hoff factor. Conclusion Thus, the two conditions that would increase the effect of ion-pairing and decrease the observed van 't Hoff factor are: B. High concentration D. Higher charge on the ions. Examples & Evidence For example, in a concentrated solution of magnesium sulfate (MgSO₄), the high charge of Mg²⁺ increases the likelihood of ion pairs forming, thus lowering the van 't Hoff factor compared to dilute solutions. Similarly, in very concentrated sodium chloride solutions, we observe a lowering of the van 't Hoff factor due to similar ion-pairing effects. Evidence from physical chemistry shows that increased concentration leads to more frequent ion interactions, supporting the idea of ion-pair formation, thereby affecting the calculated van 't Hoff factor. Thanks 2 5.0 (1 vote) Advertisement Community Answer This answer helped 191685381 people 191M 0.0 1 Answer: High concentration Higher charge on the ions Explanation: Ideally, the Van't Hoff factor is defined as the ratio between the actual concentration of particles produced when the substance is dissolved in a solution and the concentration of a substance as calculated from its mass(Wikipedia). The Van't Hoff factor is influenced by the concentration of ions in solution as well as the magnitude of charge on the ions in solution. Highly charged ions tend to remain paired in solution thereby leading to a lower value of Van't Hoff factor. Also, in highly concentrated solutions, the Van't Hoff factor is lower than in dilute solutions due to the pairing of ions Answered by pstnonsonjoku •27.6K answers•191.7M people helped Thanks 1 0.0 (0 votes) Advertisement ### Free Chemistry solutions and answers Community Answer Explain how electrostatic attraction affects the van’t Hoff factor. Why is the effect always to lower the van’t Hoff factor from the simple assumption that dissolved ionic substances always form whole number values of ions? Community Answer 4.2 19 A drink that contains 4 1/2 ounces of a proof liquor… approximately how many drinks does this beverage contain? Community Answer 5.0 7 Chemical contamination is more likely to occur under which of the following situations? When cleaning products are not stored properly When dishes are sanitized with a chlorine solution When raw poultry is stored above a ready-to-eat food When vegetables are prepared on a cutting board that has not been sanitized Community Answer 4.3 189 1. Holding 100mL of water (ebkare)__2. Measuring 27 mL of liquid(daudgtear ldnreiyc)____3. Measuring exactly 43mL of an acid (rtube)____4. Massing out120 g of sodium chloride (acbnela)____5. Suspending glassware over the Bunsen burner (rwei zeagu)____6. Used to pour liquids into containers with small openings or to hold filter paper (unfenl)____7. Mixing a small amount of chemicals together (lewl letpa)____8. Heating contents in a test tube (estt ubet smalcp)____9. Holding many test tubes filled with chemicals (estt ubet karc) ____10. Used to clean the inside of test tubes or graduated cylinders (iwer srbuh)____11. Keeping liquid contents in a beaker from splattering (tahcw sgasl)____12. A narrow-mouthed container used to transport, heat or store substances, often used when a stopper is required (ymerereel kslaf)____13. Heating contents in the lab (nuesnb bneurr)____14. Transport a hot beaker (gntos)____15. Protects the eyes from flying objects or chemical splashes(ggloges)____16. Used to grind chemicals to powder (tmraor nda stlepe) __ Community Answer Food waste, like a feather or a bone, fall into food, causing contamination. Physical Chemical Pest Cross-conta Community Answer 8 If the temperature of a reversible reaction in dynamic equilibrium increases, how will the equilibrium change? A. It will shift towards the products. B. It will shift towards the endothermic reaction. C. It will not change. D. It will shift towards the exothermic reaction. Community Answer 4.8 52 Which statements are TRUE about energy and matter in stars? Select the three correct answers. Al energy is converted into matter in stars Only matter is conserved within stars. Only energy is conserved within stars. Some matter is converted into energy within stars. Energy and matter are both conserved in stars Energy in stars causes the fusion of light elements Community Answer 4.5 153 The pH of a solution is 2.0. Which statement is correct? Useful formulas include StartBracket upper H subscript 3 upper O superscript plus EndBracket equals 10 superscript negative p H., StartBracket upper O upper H superscript minus EndBracket equals 10 superscript negative p O H., p H plus P O H equals 14., and StartBracket upper H subscript 3 upper O superscript plus EndBracket StartBracket upper O upper H superscript minus EndBracket equals 10 to the negative 14 power. Community Answer 5 Dimensional Analysis 1. I have 470 milligrams of table salt, which is the chemical compound NaCl. How many liters of NaCl solution can I make if I want the solution to be 0.90% NaCl? (9 grams of salt per 1000 grams of solution). The density of the NaCl solution is 1.0 g solution/mL solution. Community Answer 8 For which pair of functions is the exponential consistently growing at a faster rate than the quadratic over the interval mc015-1. Jpg? mc015-2. Jpg mc015-3. Jpg mc015-4. Jpg mc015-5. Jpg. New questions in Chemistry Match each discovery about atomic structure to the scientist who made the discovery: 1. Used cathode ray tubes to discover electrons. 2. Used the gold foil experiment to discover the atomic nucleus. 3. Suggested that electrons orbit the nucleus in fixed energy levels. 4. Hypothesized that elements are made of indivisible units called atoms. Which best explains why a crystal is incompressible? A. The molecules of a crystal behave like those of a gas. B. There is little, if any, space left between its adjacent molecules. C. Its molecules switch positions when they are compressed. D. Its molecules remain in position without vibrating. Which ion was formed by providing the second ionization energy to remove an electron? A. C a 2+ B. N 3− C. F e 3+ D. S 2− 2 N 2O(g)→2 N 2(g)+O 2(g) Predict the rate of change in the concentration of N 2O over this time interval. In other words, what is Δ[N 2O]/Δ t ? Express your answer in moles per liter per second to two significant figures. Consider the following reaction: 2 N 2O(g)→2 N 2(g)+O 2(g) Predict the rate of change in the concentration of N 2O over this time interval. In other words, what is Δ[N 2O]/Δ t ? Express your answer in moles per liter per second to two significant figures. Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
1911	https://www.khanacademy.org/math/cc-fourth-grade-math/imp-measurement-and-data-2/imp-money-word-problems/v/change-from-buying-apples	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
1912	https://devdude.me/blog/wavesiiiEnergy	blog Blog Waves III: Energy Propagated by a String Posted by Youssef Moawad on: 04/02/2018, in Physics Wave Energy This post carries on from the discussion in the previous two: Waves I and Waves II. It's recommended to have a look at those if you haven't already. Any wave on a string (and any in general) will transport energy. As in all physical situations, we can consider the two different forms of energy carried by the wave: kinetic and potential energy. The sum of these two parts gives the total energy transported by the wave. To do this, we will again consider the string to be formed of an infinite number of simple harmonic oscillators represented by string elements of infinitesimal length, $\delta x$, and infinitesimal mass, $\delta m$, as described in Waves I. The kinetic energy of the string element arises from the fact that the string element moves transversely, or up and down, with a velocity $v_y$ (but not horizontally; remember that the string elements themselves do not move through the string). The potential energy results because the string element is stretched from its natural length by the wave. This is elastic potential energy. Side note: Kinetic and Elastic Potential Energy Remember that the kinetic energy of an object with mass $m$ and moving with velocity $v$ is given by: $$K = \frac{1}{2}mv^2$$ The elastic energy stored in a string element is a result of the work done by the tension on the string element. For a string element of infinitesimal length, under tension $T$ that is stretched by $\Delta x$, this is given by: $$V = T\Delta x$$ Side note: Energy density It is helpful to be familiar with the concept of energy density. An energy density represents how the energy is distributed across some medium. In our case, the medium is a string and we will be interested in how the energy is distributed across its length. This would be the energy per unit length of the string. For instance, the kinetic energy density of the string is $\frac{dK}{dx}$. If we have an expression for this, we can integrate with respect to $x$ and obtain an expression for the total kinetic energy of the string. We can also do a similar treatment to potential energy density, $\frac{dV}{dx}$, and we can indeed add the energy densities up to obtain the total energy density of the string, $\frac{dE}{dx}$. Kinetic energy of a wave on a string Let's start my considering the kinetic energy of a string element. As mentioned above, the string element is under a tension, $T$, and has an infinitesimal length of $\delta x$ and mass $\delta m$. The entire string has a linear density of $\rho$ so we can say that: $$\rho = \frac{\delta m}{\delta x} \Rightarrow \delta m = \rho \delta x$$ The string element is moving transversely with velocity $v_y = \frac{\partial y}{\partial t}$ and does not move horizontally, so the kinetic energy of the string element, $\delta K$ is given by: $$\delta K = \frac{1}{2}\delta m v_y^2 = \frac{1}{2}\rho\delta x (\frac{\partial y}{\partial t})^2$$ Now let's divide by $\delta x$ and take the limit as $\delta x \rightarrow 0$ to get $\frac{dK}{dx}$. $$\Rightarrow \frac{\delta K}{\delta x} = \frac{1}{2}\rho (\frac{\partial y}{\partial t})^2$$ $$\Rightarrow \frac{dK}{dx} = \frac{1}{2}\rho (\frac{\partial y}{\partial t})^2$$ This is the kinetic energy density of the string. We can integrate this to find the total kinetic energy on the string: $$K = \int \frac{1}{2}\rho (\frac{\partial y}{\partial t})^2 dx = \frac{1}{2}\rho \int (\frac{\partial y}{\partial t})^2 dx$$ This expression gives the total kinetic energy of the waves propagating on the string. We will now consider the potential energy, which we will be able to add to the kinetic energy to get the total energy of the string. Potential energy of a wave on a string As mentioned above, the potential energy arises because the string element is stretched from its natural length by the wave. To apply the formula mentioned above, we need to determine the extension of the string element. This will be the difference between its stretched length (call this $(\delta x)'$) and its natural length, $\delta x$. Let's visualise the stretched string element to understand this more clearly: The stretched length of the string element, $(\delta x)'$ is given by $\sqrt{(\delta x)^2 + (\delta y)^2}$, by applying Pythagoras' theroem. From the diagram above, we can say that the stretched length of the string element is given by: $$(\delta x)' = \sqrt{(\delta x)^2 + (\delta y)^2} = \delta x \sqrt{1 + (\frac{\delta y}{\delta x})^2}$$ We can apply the binomial theorem here and, by ignoring terms of degree higher than 2, approximate this to: $$\Rightarrow (\delta x)' \approx \delta x (1 + \frac{1}{2}(\frac{\delta y}{\delta x})^2)$$ This is an important step so if you're not familiar with using the binomial theorem to approximate a square root term, make sure to research that and understand it well. The extension, $\Delta x$, is given by this expression minus the natural length of the element: $$\Delta x = (\delta x)' - \delta x = \delta x (1 + \frac{1}{2}(\frac{\delta y}{\delta x})^2) - \delta x $$ $$\Rightarrow \Delta x = \delta x(\frac{1}{2}(\frac{\delta y}{\delta x})^2) = \frac{1}{2} \delta x (\frac{\delta y}{\delta x})^2$$ We can now use the formula above to find the elastic potential energy of the string element: $$\delta V = T\Delta x$$ $$\Rightarrow \delta V = \frac{1}{2} T \delta x (\frac{\delta y}{\delta x})^2$$ Again dividing by $\delta x$ and taking the limit as it approaches 0: $$\Rightarrow \frac{\delta V}{\delta x} = \frac{1}{2} T (\frac{\delta y}{\delta x})^2$$ $$\Rightarrow \frac{dV}{dx} = \frac{1}{2} T (\frac{\partial y}{\partial x})^2$$ This is the potential energy density in the string. And we can again integrate with respect to $x$ to obtain the total potential energy in the string: $$V = \int \frac{1}{2} T (\frac{\partial y}{\partial x})^2 dx = \frac{1}{2} T \int (\frac{\partial y}{\partial x})^2 dx$$ Total energy in the string We can now add the kinetic and potential energies of the string to obtain an expression for the total energy propagating through the string, $E$: $$E = K + V$$ $$\Rightarrow E = \frac{1}{2}\rho \int (\frac{\partial y}{\partial t})^2 dx + \frac{1}{2} T \int (\frac{\partial y}{\partial x})^2 dx$$ Recall from Waves II, that the speed of the waves on the string is given by: $$c = \sqrt{\frac{T}{\rho}} \Rightarrow T = \rho c^2$$ Substituting T in the expression for total energy: $$E = \frac{1}{2}\rho \int (\frac{\partial y}{\partial t})^2 dx + \frac{1}{2} \rho c^2 \int (\frac{\partial y}{\partial x})^2 dx$$ $$\Rightarrow E = \frac{1}{2}\rho \int((\frac{\partial y}{\partial t})^2 + c^2(\frac{\partial y}{\partial x})^2)dx$$ Again from Waves II, that for a constant-shape travelling wave, we have $y(x,t) = y(x\pm ct)$. We had also found the relation between the first partial derivatives of $y$ with respect to $t$ and $x$ to be: $$\frac{\partial y}{\partial t} = \pm c\frac{\partial y}{\partial x}$$ Squaring both sides, we get a result that we can use in our expression for energy: $$(\frac{\partial y}{\partial t})^2 = c^2 (\frac{\partial y}{\partial x})^2$$ Substituting $(\frac{\partial y}{\partial t})^2$ back into our energy expression: $$E = \rho c^2 \int (\frac{\partial y}{\partial x})^2 dx$$ We could also instead substitute $(\frac{\partial y}{\partial x})^2 = \frac{1}{c^2} (\frac{\partial y}{\partial t})^2$ (don't get confused; this is not the wave equation!) to arrive at an equally valid result: $$E = \rho \int (\frac{\partial y}{\partial t})^2 dx$$ The next step now is to consider the power being transported by the wave, which we will do in the next post. Closing Remarks Thanks for reading! Hopefully you found this post informative and easy to follow. If there's anything you would like to be clarified or if you have any question or want to discuss something, please feel free to contact me as usual either through the comments below or privately through the website. Happy Physics-ing! Update (05/02/2018): Waves IV is now up, here. Related You may also be interested in: Waves II: Wave Speed for Waves on a String Waves I: The Wave Equation for Waves on a String Symmetries: The Beauty in Physics
1913	https://www.sanfoundry.com/1000-heat-transfer-questions-answers/	Skip to content Heat Transfer MCQ (Multiple Choice Questions) Here are 1000 MCQs on Heat Transfer (Chapterwise). What is heat transfer?a) Flow of thermal energy from low-temperature reservoir to high-temperature reservoirb) Flow of energy in the form of heat from high-temperature reservoir to low-temperature reservoirc) Flow of thermal energy irrespective of reservoir temperatured) None of the aboveView Answer Answer: bExplanation: Heat transfer is a branch of thermal engineering which deals with the study of transfer of energy from a high-temperature reservoir to low-temperature reservoir. Which of the following is a method of heat transfer?a) Convectionb) Radiationc) Conductiond) All of the mentionedView Answer Answer: dExplanation: Methods of heat transfer are:i) Conduction: Heat is transferred by solid materials.ii) Convection: Heat is transferred by gases.iii) Radiation: Heat is transferred by electromagnetic waves. Heat transfer takes place according to which of the following law?a) Newton’s second law of motionb) First law of thermodynamicsc) Newton’s law of coolingd) Second law of thermodynamicsView Answer Answer: dExplanation: The total entropy of an isolated system (the thermal energy per unit temperature that is unavailable for doing useful work) can never decrease, according to the second law of thermodynamics. Which of the following is the rate of heat transfer unit?a) Wattb) Pascalc) Jouled) NewtonView Answer Answer: aExplanation: The joule is the unit of heat transfer, while the rate of heat transfer is measured in joules per second, i.e., watts. Which of the following is an example of steady-state heat transfer?a) Electric bulb cools down by the surrounding atmosphereb) Chilling effect of cold wind on a warm bodyc) Boilers and turbinesd) Cooling of I.C engineView Answer Answer: aExplanation: The system is a perfect black body. Which way is heat transfer believed to take place in a long, hollow cylinder that is kept at consistent but varied temperatures on its inner and outer surfaces?a) Unpredictableb) Radial onlyc) No heat transfer takes placed) Axial onlyView Answer Answer: bExplanation: On the periphery of the cylinder, the ambient temperature is uniform, and the temperature is uniform. As a result, it only occurs in the radial direction. Which of the following is correct regarding one dimensional heat transfer?a) Steady – f (x, y, t), Unsteady – f (x)b) Steady – f (y, z), Unsteady – f (y)c) Steady – f (x, t), Unsteady – f (x)d) Steady – f (x), Unsteady – f (x, t)View Answer Answer: dExplanation: In a one-dimensional heat flow, steady state is determined solely by the x coordinate, whereas unstable state is determined solely by the x coordinate and time. A person prefers to sit by a fire during the cold winter months. Which of the following heat transfer types gives him with the most heat?a) Convection and radiation togetherb) Radiation will provide quick warmthc) If it is near the fire, convection sounds goodd) Conduction from the fireView Answer Answer: bExplanation: Even when separated by a medium that is colder than both of them, heat transmission by radiation can occur between two bodies. On which of the following, convective heat transfer coefficient doesn’t depend on?a) Orientation of solid surfaceb) Timec) Surface aread) SpaceView Answer Answer: cExplanation: As we know, Q = h x A x (temperature difference), the total heat transfer is affected by the surface area, but the coefficient itself is not. Coefficient is independent of the surface area. The coefficient is denoted by h and is dependent on space, time, geometry, orientation of solid surface. In liquids and gases, heat transmission is primarily caused bya) Convectionb) Radiationc) Conductiond) Conduction as well as convectionView Answer Answer: aExplanation: Convection is the transfer of thermal energy between a solid and a fluid flowing through it. For conduction heat transfer, the heat energy propagation will be minimal for __________a) Copperb) Airc) Waterd) LeadView Answer Answer: bExplanation: It’s because air has the lowest heat conductivity of all the possibilities. A 25 mm diameter egg roll (k = 1 W/m degree) is roasted with the help of microwave heating. For good quality roasting, it is desired that temperature at the center of roll is maintained at 100 degree Celsius when the surrounding temperature is 25 degree Celsius. What should be the heating capacity in W/m 3 of the microwave if the heat transfer coefficient on the surface of egg roll is 20 W/m2 degree? a) 613.31 k W/m3 b) 93.31 k W/m3 c) 6713.31 k W/m3 d) 213.31 k W/m3 View Answer Answer: dExplanation: t max = t a + q g R/2h + q g R2/4k. The rate of heat transfer for a plane wall of homogenous material with constant thermal conductivity is given by which of the following equation?a) Q = 2k/δ xb) Q = 2kAδxc) Q = kA (t1-t2)/δ d) Q = 2kAx/ δ View Answer Answer: cExplanation: The value of the temperature gradient can be substituted into the general equation to compute heat flow. The heat flow is independent of x in some way. The appropriate rate equation for convective heat transfer between a surface and adjacent fluid is prescribed by which law?a) Newton’s law of coolingb) Kirchhoff’s lawc) Newton’s first lawd) Wein’s displacement lawView Answer Answer: aExplanation: When a solid surface is cooled by a fluid, the rate equation used to describe the mechanism of convection is called Newton’s law of cooling. A wire of radius 3 mm and 1.25 m length is to be maintained at 60 degree Celsius by insulating it by a material of thermal conductivity 0.175 W/m K. The temperature of surrounding is 20 degree Celsius with heat transfer coefficient 8.5 W/ m2 K. Find percentage increase in heat loss due to insulation? a) 124.23 % b) 100.00 % c) 12.55 % d) 134.46 % View Answer Answer: dExplanation: Q = 8.5 (2 π 0.003 1.25) (60 – 20) = 8.01 W. % increase = (18.78 – 8.01/8.01) (100) = 134.46 %. Why fins are provided on a heat transfer surface?a) Pressure drop of the fluid should be minimizedb) Increase turbulence in flow for enhancing heat transferc) Surface area is maximum to promote the rate of heat transferd) Increase temperature gradient so as to enhance heat transferView Answer Answer: cExplanation: Fins are added to the surface of a heat exchanger to improve heat transmission by increasing the surface area exposed to the environment. A heating unit is made in the form of a vertical tube of 50 mm outside diameter and 1.2 m height. The tube is fitted with 20 steel fins of rectangular section with height 40 mm and thickness 2.5 mm. The temperature at the base of fin is 75 degree Celsius, the surrounding air temperature is 20 degree Celsius and the heat transfer coefficient between the fin as well as the tube surface and the surrounding air is 9.5 W/m2 K. If thermal conductivity of the fin material is 55 W/m K, find the amount of heat transferred from the tube without fin a) 118.44 W b) 98.44 W c) 8.44 W d) 908.44 W View Answer Answer: bExplanation: Q = h A d t = h (π d 0 H) (t 0 – t INFINITY). What is the rate of heat transfer from the fin in case of fin insulated at the tip?a) (h P k)1/2 (t0 – t a) tan h ml b) (h P A)1/2 (t0 – t a) tan h ml c) (h P k A)1/2 (t0 – t a) tan h ml d) (h k A)1/2 (t0 – t a) tan h ml View Answer Answer: cExplanation: It should contain all the terms i.e. h, A, P, k. Consider the following statements pertaining to heat transfer through fins(i) They must be arranged at right angles to the direction of flow of working fluid(ii) The temperature along the fin is variable and accordingly heat transfer rate varies along the fin elements(iii) Fins are equally effective irrespective whether they are on the hot side or cold side of the fluid(iv) Fins are made of materials that have thermal conductivity higher than that of wallIdentify the correct statementsa) i and iib) iii and ivc) i and ivd) ii and iiiView Answer Answer: aExplanation: The statements made in serial number 3 and 4 are incorrect. Fins are seen on the side of the body when the convective coefficient is low. advertisement In which of the following cases provision of fins on a given heat transfer surface will be more effective?a) Fewer but thin finsb) Large number of thin finsc) Large number of thick finsd) Fewer but thick finsView Answer Answer: bExplanation: Increase in ratio of perimeter P to be cross-sectional area A C brings about improvement in the effectiveness of fins. What are the relevant boundary conditions in case of heat transfer from a bar connected to two heat sources at different temperatures?a) α = α 1 at x = 1 and α = α 2 at x = 2L b) α = α 1 at x = 0 and α = α 2 at x = infinity c) α = α 1 at x = 0 and α = α 2 at x = L d) α = α 1 at x = infinity and α = α 2 at x = 1 View Answer Answer: cExplanation: It should be at x = o and x = L. Radiation heat transfer is characterized bya) Movement of discrete packets of energy as electromagnetic wavesb) Due to bulk fluid motion, there is a transport of energyc) There is the circulation of fluid by buoyancy effectsd) Thermal energy transfer as vibrational energy in the lattice structure of the materialView Answer Answer: aExplanation: EM waves is characterized by radiant heat transfer. A thin metal plate of 4 cm diameter is suspended in atmospheric air whose temperature is 290 K. This plate attains a temperature of 295 K when one of its faces receives radiant energy from a heat source at the rate of 2 W. If heat transfer coefficient on both surfaces of the plate is stated to be 87.5 W/m 2 K, workout the reflectivity of the plate. a) 0.15 b) 0.55 c) 0.25 d) 0.45 View Answer Answer: dExplanation: Heat loss by convection from both sides of the plates = 2 h A d t = 1.1 W. Energy lost by reflection = 2.0 – 1.1 = 0.9 W. Which of the following is the unit of coefficient of radiant heat transfer?a) W/m2 b) W/m K c) W/m2 K d) W/K View Answer Answer: cExplanation: Its value can be calculated from the heat flux equation for any configuration. The radiant heat transfer from a plate of 2.5 cm2 area at 1250 K to a very cold enclosure is 5.0 W. Determine the emissivity of the plate at this temperature a) 0.544 b) 0.144 c) 0.044 d) 0.244 View Answer Answer: bExplanation: Emissivity = E/σ A T4 = 0.144. The calculation of heat transmission by convection is based on the determination of a Nusselt number or convective film coefficient. To that goal, which of the following strategies have been proposed(i) Non-dimensional analysis and experimental correlations(ii) Hydrodynamic concept of velocity boundary layer(iii) Reynolds similarity between the mechanism of fluid friction in the boundary layer and the transfer of heat by convectionIdentify the correct onea) ii and iiib) i and iiic) i, ii, and iiid) i and iiView Answer Answer: aExplanation: It should be a dimensional analysis and experimental correlations. The non-dimensional parameter known as Stanton number is used in which of the following heat transfer?a) Natural convection heat transferb) Unsteady state heat transferc) Condensation heat transferd) Forced convection heat transferView Answer Answer: dExplanation: It’s the ratio of the heat transfer coefficient to the heat flow per unit temperature rise caused by fluid velocity. It can only be used for forced convection heat transfer. The convective heat transfer coefficient in the laminar flow over a flat plate __________a) Decreases with increase in free stream velocityb) Increases if a denser fluid is usedc) Increases with distanced) Increases if a higher viscosity fluid is usedView Answer Answer: bExplanation: It mostly increases if a denser fluid is used. Maximum heat transfer rate in a modern boiler is abouta) 4 10 5 W/m2 b) 5 10 5 W/m2 c) 3 10 5 W/m2 d) 2 10 5 W/m2 View Answer Answer: dExplanation: Greater importance has recently been given to the boiling heat transfer. In spite of the large heat transfer coefficient in boiling liquids, fins are used advantageously when the entire surface is exposed toa) Nucleate boilingb) Film boilingc) Transition boilingd) All modes of boilingView Answer Answer: dExplanation: Here all modes of boiling i.e. film, transition and nucleate are of great importance. The condensate film runs down a vertical flat surface where saturated steam is allowed to condense. Condensation’s local coefficient of heat transfera) Increases with increasing thickness of filmb) Increases with increasing temperature differential between the surface and vapourc) Decreases with increasing distance from the top of the surfaced) Remains constant at all heights of the surfaceView Answer Answer: cExplanation: It decreases with increasing thickness of condensate film. A steam condenser is transferring 250 k W of thermal energy at a condensing temperature of 65 degree Celsius. The cooling water enters the condenser at 20 degree Celsius with a flow rate of 7500 kg/hr. If overall heat transfer coefficient for the condenser surface is 1250 W/m2 K, what surface area is required to handle this load? a) 7.08 m2 b) 5.08 m2 c) 4.08 m2 d) 1.08 m2 View Answer Answer: aExplanation: Q = U A α m. So, A = 7.08 m2. What is the value of overall heat transfer coefficient ammonia condensers?a) 200-750 W/m2 K b) 250-2500 W/m2 K c) 1500-1750 W/m2 K d) 800-1400 W/m2 K View Answer Answer: dExplanation: It is an electrical device that stores a vast amount of energy. Which of the following is having highest value of overall heat transfer coefficient?a) Steamb) Steam condensersc) Feed water heatersd) Alcohol condensersView Answer Answer: cExplanation: Feed water heaters have an overall heat transfer coefficient of 8500 W/m2K, whereas steam, alcohol condensers, and ammonia condensers have heat transfer coefficients of 5000 W/m2 K, 630 W/m2 K and 1400 W/m2 K. For laminar film condensation on a vertical plate, the local heat transfer coefficient at the lower edge of the plate is given bya) [k 3 p 2 g h f g /4 δ l (t sat – t s)] 0.125 b) [k 3 p 2 g h f g /4 δ l (t sat – t s)] 0.1 c) [k 3 p 2 g h f g /4 δ l (t sat – t s)] 0.25 d) [k 3 p 2 g h f g /4 δ l (t sat – t s)] 0.5 View Answer Answer: cExplanation: The rate of condensation heat transfer is higher at the upper end of the plate than at the lower end. The outer surface of a vertical tube is 1.25 m long and outer diameter is 50 mm is exposed to saturated steam at atmospheric pressure. If the tube surface is maintained at 80 degree Celsius by the flow of cooling water through it, determine the rate of heat transfer to the coolanta) 32397 Wb) 19879 Wc) 48476 Wd) 25234 WView Answer Answer: bExplanation: h v = 0.943 [k 3 p 2 g h f g/δ l (t sat – t s)] 0.25 = 5046.8 W/m 2 k and heat flow rate = 19879 W. What is the value of the overall heat transfer coefficient for air to heavy tars and liquid?a) As low as 25 W/m2 K b) As low as 15 W/m2 K c) As low as 45 W/m2 K d) As low as 55 W/m2 K View Answer Answer: cExplanation: It is an incompressible fluid which has constant volume independent of pressure. Exhaust gases (c P = 1.12 k J/kg K) flowing through a tubular heat exchanger at the rate of 1200 kg/hr are cooled from 400 degree Celsius to 120 degree Celsius. This cooling is affected by water (c P = 4.18 k J/kg K) that enters the system at 10 degree Celsius at the rate of 1500 kg/hr. If the overall heat transfer coefficient is 500 k J/m2 hr degree, what heat exchanger area is required to handle the load for counter flow arrangement? a) 1.758 m2 b) 6.758 m2 c) 8.758 m2 d) 3.758 m2 View Answer Answer: dExplanation: m h c h (t h 1 – t h 2) = m c c c (t c 2 – t c 1). What is the value of the overall heat transfer coefficient for air condensers?a) 250-900 W/m2 K b) 200-1950 W/m2 K c) 350-780 W/m2 K d) 200-350 W/m2 K View Answer Answer: cExplanation: It consists of an air coil which is used to remove heat from fluids. Heat transfer coefficients for natural convection across the surface of a vertical pipe and a vertical flat plate at the same height. What could be the cause(s) of this?(i) Same height(ii) Both vertical(iii) Same fluid(iv) Same fluid flow patternSelect the correct answera) iii and ivb) ivc) i and iid) iView Answer Answer: aExplanation: The fluids must be same so their flow pattern. Which of the following is having the lowest value of overall heat transfer coefficient?a) Feed water heatersb) Air condensersc) Air to low viscosity liquidsd) Steam condensersView Answer Answer: cExplanation: Overall heat transfer coefficient for air condensers is 780 W/m2 K while that of steam condensers, air to low viscosity liquids and feed water heaters are 5000 W/m2 K, 600 W/m2 K and 8500 W/m2 K. A cross-flow type air heater has an area of 50 cm2. The overall heat transfer coefficient is 100 W/m2 K and the heat capacity of both hot and cold streams is 1000 W/m K. The value of NTU is a) 0.2 b) 6 c) 1000 d) 5 View Answer Answer: dExplanation: NTU = A U/C MIN = 5. An oil cooler in a high-performance engine has an outside surface area 0.12 m2 and a surface temperature of 65 degree Celsius. At any intermediate time air moves over the surface of the cooler at a temperature of 30 degree Celsius and gives rise to a surface coefficient equal to 45.4 W/ m 2 K. Find out the heat transfer rate? a) 564.98 W b) 324.67 W c) 190.68 W d) 768.43 W View Answer Answer: cExplanation: Q = (T2 – T1) A h = 0.12 (65-30) 45.4 = 190.68 W. Which of the following statement is incorrect according to heat transfer?a) Heat flow doesn’t depend on temperatureb) A material medium is not necessary for heat transmissionc) The process of heat transfer is an irreversible processd) For heat exchange, a temperature gradient must existView Answer Answer: aExplanation: Heat flows from higher to lower temperature. Consider a convective heat flow to water at 75 degree Celsius from a cylindrical nuclear reactor fuel rod of 50 mm diameter. The rate of heat generatioN is 50000000 W/m3 and convective heat transfer coefficient is I kW/m2 K. The outer surface temperature of the fuel element would be a) 400 degree Celsius b) 625 degree Celsius c) 700 degree Celsius d) 550 degree Celsius View Answer Answer: cExplanation: t w = t a + q g R/ 2h. For a cylindrical rod with uniformly distributed heat sources, the thermal gradient at half the radius location will bea) One halfb) One fourthc) Four timesd) TwiceView Answer Answer: aExplanation: t = t w + q g (R2 – r2)/4k. (d t /d r) r = R/2 = 1/2(d t/d r) r = R. Water (specific heat = 4 k J/kg K) enters a cross flow exchanger (both fluids unmixed) at 15 degree Celsius and flows at the rate of 7.5 kg/s. It cools air (C P = 1 k J/kg K) flowing at the rate of 10 kg/s from an inlet temperature of 120 degree Celsius. For an overall heat transfer coefficient of 780 k J/m2 hr degree and the surface area is 240 m2, determine the NTU a) 1.2 b) 8.2 c) 6.2 d) 5.2 View Answer Answer: dExplanation: NTU = U A/C MIN = 5.2. A heat exchanger to preheat oil for a furnace was designed without considering the possibility of scale formation, and the overall heat transfer coefficient based on the fuel oil side was 3200 k J/m2 hr degree. What would be the corrected coefficient of heat transfer if a fouling factor of 0.00025 m2 hr degree/k J for the fuel oil is taken into account? a) 1222.78 k J/m2 hr degree b) 1555.78 k J/m2 hr degree c) 1777.78 k J/m2 hr degree d) 1233.78 k J/m2 hr degree View Answer Answer: cExplanation: U = 1/ (R + 1/h S). Which of the following is the value of fouling factor for engine exhaust?a) 0.001 m2 K/W b) 0.002 m2 K/W c) 0.003 m2 K/W d) 0.004 m2 K/W View Answer Answer: bExplanation: Its unit is m2 hr K/kcal and it represents the reciprocal of the scale coefficient i.e. heat transfer. The value of fouling factor for industrial liquids is ______a) 0.0002 m2 K/W b) 0.0001 m2 K/W c) 0.0003 m2 K/W d) 0.0004 m2 K/W View Answer Answer: aExplanation: Its unit is m2 hr K/kcal and it represents the reciprocal of the scale coefficient i.e. heat transfer. Which of the following is the value of overall heat transfer coefficient for steam condensers?a) 2000-9500 W/m2 K b) 1500-5000 W/m2 K c) 200-9000 W/m2 K d) 3000-5500 W/m2 K View Answer Answer: bExplanation: It’s a rotating mechanism that works in the same way as a motor to control power flow in electric power transmission. The value of overall heat transfer coefficient for air to low viscosity liquid _______a) As high as 700 W/m2 K b) As high as 900 W/m2 K c) As high as 600 W/m2 K d) As high as 800 W/m2 K View Answer Answer: cExplanation: It is made up of tiny vibrating particles of matter which are held together by intermolecular bonding. Mark the system where heat transfer is given by forced convectiona) Heat flow from hot pavement to the surrounding atmosphereb) Heat exchange on the outside of cold and warm pipesc) Chilling effect of cold wind on a warm bodyd) Fluid passing through the tubes of a condenser and other heat exchange equipmentView Answer Answer: dExplanation: Forced convection occurs when the fluid motion involved in the process is produced by some external means. Chapterwise Multiple Choice Questions on Heat Transfer Our 1000+ MCQs focus on all topics of the Heat Transfer subject, covering 100+ topics. This will help you to prepare for exams, contests, online tests, quizzes, viva-voce, interviews, and certifications. You can practice these MCQs chapter by chapter starting from the 1st chapter or you can jump to any chapter of your choice. Definitions and Basic Concepts Fourier Equation and Thermal Conductivity Steady State Conduction Conduction with Heat Generation Heat Transfer from Extended Surfaces Transient (Unsteady State Heat Conduction) Radiation : Processes and Properties Radiation : Exchange between Surfaces Convection : Processes and Properties Dimensional Analysis Empirical Correlations for Free and Forced convection Condensation and Boiling Hydrodynamic and Thermal Boundary Layers Heat Exchangers and Mass Transfer 1. MCQ on Definitions and Basic Concepts Heat Transfer The section contains multiple choice questions and answers on steady and unsteady types of heat transfer and various heat transfer modes. \| \| \| --- \| \| - Modes of Heat Transfer \| - Steady and Unsteady Heat Transfer \| 2. Heat Transfer Multiple Choice Questions on Fourier Equation and Thermal Conductivity The section contains questions and answers on various properties of fourier equation and thermal conductivity which include fourier equation, heat conduction equation and thermal conductivity of different materials. \| \| \| --- \| \| - Fourier Equation - Thermal Conductivity of Materials \| - General Heat Conduction Equation - Thermal Conductivity of Different Materials \| 3. Heat Transfer MCQ on Steady State Conduction The section contains MCQs on conduction through various mechanisms which include plane wall, composite wall, cylindrical wall and a sphere. It also contains questions and answers on shape factor, variable conductivity effect and critical thickness of insulation. \| \| \| --- \| \| - Conduction Through a Plane Wall - Conduction Through a Composite Wall - Conduction Through a Cylindrical Wall - Conduction Through a Sphere \| - Shape Factor - Effect of Variable Conductivity - Critical Thickness of Insulation \| 4. Heat Transfer Multiple Choice Questions on Conduction with Heat Generation The section contains multiple choice questions and answers on heat generations through various mechanisms which include plane wall, cylinder and a sphere. Additionally, the section also contains questions on dielectric effect. \| \| \| --- \| \| - Heat Generation Through Plane Wall - Dielectric Heating \| - Heat Generation Through Cylinder - Heat Generation Through Sphere \| 5. Multiple Choice Questions on Heat Transfer from Extended Surfaces The section contains questions and answers on concepts of fins, steady heat flow along a rod, performance of fin, design consideration for fins, thermometric well properties and heat dissipation from a long fin and a fin which is insulated at the tip. \| \| \| --- \| \| - Fins - Steady Flow of Heat Along a Rod - Heat Dissipation from an Infinitely Long Fin - Heat Dissipation from a Fin Insulated at the Tip \| - Fin Performance - Design Considerations for Fins - Heat Flow Through Triangular and Parabolic Fins - Thermometric Well \| 6. Heat Transfer MCQ on Transient (Unsteady State Heat Conduction) The section contains MCQs on thermocouple response, time constant, heat conduction in solids using various mechanisms like infinite thermal conductivity, finite conduction and infinite thick solids. The section also contains questions and answers on biot number and periodic variation. \| \| \| --- \| \| - Time Constant - Response of a Thermocouple - Transient Heat with Infinite Conductivity - Transient Heat with Finite Conductivity \| - Biot Number - Lump System Analysis - Transient Heat Conduction in Infinite Thick Solids - Periodic Variation \| 7. Heat Transfer MCQ on Radiation : Processes and Properties The section contains multiple choice questions and answers on various concepts which include transmissivity, absorptivity, reflectivity, black body concepts, planck’s law, stefan-boltzman’s law, wein’s law, kirchoff’s law, radiation intensity and solar radiations. \| \| \| --- \| \| - Transmissivity - Reflectivity - Absorptivity - Black Body - Spectral And Spatial Energy Distribution - Planck’s Law - Stefan- Boltzman Law \| - Wein’s Displacement Law - Kirchoff’s Law - Gray Body and Selective Emitters - Intensity of Radiations - Solar Radiations - Solar Absorptivity \| 8. Heat Transfer Multiple Choice Questions on Radiation : Exchange between Surfaces The section contains questions and answers on Black body heat exchange, shape factor concepts, Non black body heat exchange, radiation shield, gaseous radiation, fouling factor, physical and geometrical property symbols, adiabatic surface, reradiating surface and other network approaches. \| \| \| --- \| \| - Heat Exchange Between Black Bodies - Shape Factor - Heat Exchange Between Non Black Bodies - Network Approach for Heat Exchange - Radiations Shields \| - Adiabatic and Reradiating Surfaces - Gaseous Radiations - Fouling Factor - Geomertical and Thermo Physical Properties \| 9. Heat Transfer MCQ on Convection : Processes and Properties The section contains MCQs on flow types, convection types, nusselt number, reynolds number and newton rikhman law. \| \| \| --- \| \| - Types of Flow - Reynolds Number - Types of Convection \| - Nusselt Number - Newton- Rikhman Law \| 10. Heat Transfer Multiple Choice Questions on Dimensional Analysis The section contains multiple choice questions and answers on Dimensional System and Homogeneity, rayleigh and buckhingam methods, model simulation and studies, dimensionless group importance and advantages and disadvantages of dimensional analysis. \| \| \| --- \| \| - System of dimensions - Dimensional Homogeneity - Rayleigh’s Method - Buckingham’s Pi- Method \| - Model Studies and Similitude - Dimensional Analysis – Pros and Cons - Significance of Dimensionless Groups \| 11. Heat Transfer MCQ on Empirical Correlations for Free and Forced convection The section contains questions and answers on laminar flow, turbulent flow and various correlations for forced convection. \| \| \| --- \| \| - Correlations for Forced Convection - Laminar Flow \| - Turbulent Flow \| 12. Heat Transfer Multiple Choice Questions on Condensation and Boiling The section contains MCQs on boiling, nucleate boiling, free convection boiling, condensation concepts and bubble growth. \| \| \| --- \| \| - Boiling - Bubble Growth - Nucleate Boiling - Free Convection Boiling \| - Condensation - Laminar Film Condensation - Turbulent Film Condensation \| 13. Heat Transfer MCQ on Hydrodynamic and Thermal Boundary Layers The section contains multiple choice questions and answers on reynolds analogy, pohlhausen equation, thermal boundary layer, hydrodynamic boundary layer and von-karmal equation. \| \| \| --- \| \| - Thermal Boundary Layer - Pohlhausen Equation - Reynolds Analogy \| - Hydrodynamic Boundary Layer - Von- Karmal Equation \| 14. Multiple Choice Questions on Heat Exchangers and Mass Transfer The section contains questions and answers on heat exchanger classification, mean temperature difference, discrete time system implementation, specific heat, heat transfer coefficient, transfer units number and heat exchanger effectiveness. \| \| \| --- \| \| - Classification of Heat Exchanger - Mean Value of Capacity Ratio - Mean Temperature Difference - Specific Heat \| - Heat Transfer Coefficient - Heat Flux Through a Cylindrical Wall and Plate - Heat Exchanger Effectiveness - Number of Transfer Units \| If you would like to learn "Heat Transfer" thoroughly, you should attempt to work on the complete set of 1000+ MCQs - multiple choice questions and answers mentioned above. It will immensely help anyone trying to crack an exam or an interview.Wish you the best in your endeavor to learn and master Heat Transfer! advertisement Additional Resources: Heat Transfer Operations MCQ Questions Mass Transfer MCQ Questions Thermal Engineering MCQ Questions Food Processing Unit Operations MCQ Questions Engineering Thermodynamics MCQ Questions Popular Pages: Fluidization Engineering MCQ Questions Chemical Engineering MCQ Questions Basic Chemical Engineering MCQ Questions Engineering Physics MCQ Questions Mechanical Engineering MCQ Questions I’m Manish - Founder and CTO at Sanfoundry. I’ve been working in tech for over 25 years, with deep focus on Linux kernel, SAN technologies, Advanced C, Full Stack and Scalable website designs. You can connect with me on LinkedIn, watch my Youtube Masterclasses, or join my Telegram tech discussions. If you’re in your 40s–60s and exploring new directions in your career, I also offer mentoring. Learn more here. Scroll back to top
1914	https://www.khanacademy.org/math/ap-calculus-bc/bc-advanced-functions-new/bc-9-1/v/parametric-equations-1	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
1915	https://www.northamericantutors.com/blogs/ap-calculus-formula-sheet	Tutors Pricing Blogs Resources Case Studies Hire a Tutor Tutors Pricing Blogs Resources Case Studies Hire a Tutor Free One-Page AP Calculus Formula Sheet \| Printable PDF Aug 22, 2025 Aug 22, 2025 Written by North American Tutors, reviewed by Ivy-League tutors Written by North American Tutors, reviewed by Ivy-League tutors Written by North American Tutors, reviewed by Ivy-League tutors Download Your Free One-Page AP Calculus Formula Sheet (PDF) No signup needed. Just click to download Free One-Page Calculus Formula Sheet \| Printable PDF AP Calculus requires you to know and apply a wide range of formulas, and it’s easy to lose track while studying. Having the key rules organized in one place makes practice more effective and review less stressful. TLDR: AP Calculus Formula Sheet Covers derivatives, integrals, limits, and series formulas Includes AP reference sheet + must-know rules One-page printable PDF for quick review Reviewed by Ivy-League calculus tutors Download link above Why Formulas Matter for AP Calculus Success AP Calculus exams test not only your understanding of concepts but also your ability to quickly apply the right formulas under timed conditions. Having a clear, well-organized formula sheet makes practice and test day much smoother. This guide brings together the most important AP Calculus formulas, tips from our tutors, and a free downloadable PDF for study and revision. Check out our AP Calculus Resource Hub for more AP tips, or explore our SAT Math Formula Sheet and ACT Math Formula Sheet. Key AP Calculus Formulas 1. Differentiation Formulas Power Rule: d/dx [xⁿ] = n·xⁿ⁻¹ Product Rule: (fg)’ = f’g + fg’ Quotient Rule: (f/g)’ = (f’g – fg’)/g² Chain Rule: d/dx [f(g(x))] = f’(g(x))·g’(x) Trig Derivatives: d/dx [sin x] = cos x d/dx [cos x] = –sin x d/dx [tan x] = sec²x Exponential & Logarithmic Derivatives: d/dx [e^x] = e^x d/dx [ln x] = 1/x Tutor Tip💡: Our MIT tutor reminds students that the chain rule is the most frequently tested derivative rule on the AP exam. Make it second nature. 2. Integration Formulas Basic Rules: ∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C, n ≠ –1 ∫ e^x dx = e^x + C ∫ 1/x dx = ln\|x\| + C u-Substitution: ∫ f(g(x))·g’(x) dx → ∫ f(u) du Integration by Parts: ∫ u dv = uv – ∫ v du Trig Integrals (examples): ∫ sin x dx = –cos x + C ∫ cos x dx = sin x + C ∫ sec²x dx = tan x + C 3. Limits & Continuity Standard Limit Laws: lim (f(x) + g(x)) = lim f(x) + lim g(x) L’Hôpital’s Rule: If lim f(x)/g(x) is 0/0 or ∞/∞, then lim f/g = lim f’/g’ Continuity: f(x) is continuous at c if lim x→c f(x) = f(c) 4. Sequences & Series (Quick Formulas) Geometric Series: Σ arⁿ = a/(1–r), \|r\| < 1 Taylor/Maclaurin Expansion: f(x) ≈ f(a) + f’(a)(x–a) + f’’(a)(x–a)²/2! + … Tutor Insights: How to Use These Formulas Effectively Our tutors recommend: Practice by category: Don’t just memorize. Work through practice questions using the formulas. Know when to apply: For example, chain rule is almost always tested with composite functions. Watch for traps: In integration, always double-check if u-substitution or parts is more efficient. Timed drills: Simulate exam conditions and practice applying formulas quickly. Next Steps: Get Expert Support Formulas are the foundation, but true AP Calculus success comes from applying them to challenging problems. Our Ivy League and 5+ scorers specialize in making calculus strategies simple and effective. 👉 Book a free strategy session with North American Tutors today and get personalized AP Calculus prep. 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1916	https://www.youtube.com/watch?v=j7VXw-FsnnQ	Converse, Inverse, and Contrapositive Mometrix Academy 120000 subscribers 38 likes Description 2718 views Posted: 19 Dec 2022 Hi, and welcome to this video on mathematical statements! Today, we’ll be exploring the logic that appears in the language of math. Specifically, we will learn how to interpret a math statement to create what are known as converse, inverse, and contrapositive statements. These, along with some reasoning skills, allow us to make sense of problems presented in math. Let’s get started! Check out our online test prep courses! For more resources on this topic, go to: Watch our math review playlist! Mometrix Study Guides: Mometrix Flashcards: More Test Prep Resources: Follow Mometrix Academy on Pinterest: 6 comments Transcript: Hi, and welcome to this video on mathematical statements! Today, we’ll be exploring the logic that appears in the language of math. Specifically, we will learn how to interpret a math statement to create what are known as converse, inverse, and contra-positive statements. These, along with some reasoning skills, allow us to make sense of problems presented in math. Let’s get started! Let’s first take a look at a basic statement, which can be either true or false, but never both. For example, a declarative statement pronounces a fact, like “the sun is hot.” We know that this is a statement because the sun cannot be both hot and not hot at the same time. This declarative statement could also be referred to as a proposition. Two independent statements can be related to each other in a logic structure called a conditional statement. The first statement is presented with “if”, and is referred to as the hypothesis. The second statement is linked with “then”, and is known as the conclusion. The notation associated with conditional statements typically uses the variable p for the hypothesis statement, and q for the conclusion. In words, this would be read as “if p, then q.” When the hypothesis and conclusion are identified in a statement, three other statements can be derived: 1) The converse statement is notated as "if q, then p." The original statements switch positions in the original “if-then” statement. 2) The inverse statement assumes the opposite of each of the original statements and is notated "if not p, then not q." and 3) The contra-positive statement is a combination of the previous two. The positions of p and q of the original statement are switched, and then the opposite of each is considered: "If not q, then not p." An example will help to make sense of this new terminology and notation. Let’s start with a conditional statement and turn it into our three other statements. “If it is raining, then the grass is wet.” The first step is to identify the hypothesis and conclusion statements. Conditional statements make this pretty easy, as the hypothesis follows “if” and the conclusion follows “then.” The hypothesis is “it is raining” and the conclusion is “grass is wet”. Hypothesis, is labeled as p: “it is raining” The conclusion is labeled as q: “grass is wet” Now we can use the definitions that we introduced earlier to create the other three statements. Our converse statement would be “If the grass is wet, then it is raining.” Our inverse statement would be “If it is NOT raining, then the grass is NOT wet.” And our contra-positive statement would be “If the grass is not wet, then it is not raining.” You may be wondering why we would want to go through the trouble of rearranging and considering the “opposite” of the hypothesis and conclusion statements. How is this helpful? The key is in the relationships between the statements. If we know that a statement is true (or false), then we can assume that another is also true (or false). The statements that are related in this way are considered “logically equivalent.” For example, consider the statement, “If it is raining, then the grass is wet” to be TRUE. Then you can assume that the contra-positive statement, “If the grass is NOT wet, then it is NOT raining” is also TRUE. Likewise, the converse statement, “If the grass is wet, then it is raining” is logically equivalent to the inverse statement, “If it is NOT raining, then the grass is NOT wet.” These relationships are particularly helpful in math courses when you are asked to prove theorems based on definitions that are already known. Much of that work is beyond the scope of this video, but the following examples will help to illustrate the relationships of logically equivalent statements. Here is a typical example of a TRUE statement that would be made in a geometry class based on the definition of congruent angles: “Two angles with equal measure are congruent.” As you can see, this is not a conditional statement, but we can rewrite it in the “if-then” structure to identify the hypothesis and conclusion statements as follows: “If two angles have the same measure, then the two angles are congruent” Now we have a hypothesis and a conclusion: Because the conditional statement and the contra-positive are logically equivalent, we can assume the following to be TRUE: “If the two angles are NOT congruent, then the two angles do NOT have the same measure” It follows that the converse statement, “If two angles are congruent, then the two angles have the same measure.” is logically equivalent to the inverse statement, “If two angles do NOT have the same measure, then they are NOT congruent.” Here is another example of a TRUE statement: “A square is a rectangle.” The conditional statement would be “If a figure is a square, then it is a rectangle,” which gives us our hypothesis and conclusion Because the contra-positive statement is logically equivalent, we can assume that, “If the figure is NOT a rectangle, then the figure is NOT a square” is also a TRUE statement. However, the converse statement can be disproved: “If the figure is a rectangle, then it is a square” As seen in this diagram, squares are a type of rectangle and a rectangle is a type of polygon. However, a square is a special type of rectangle that has four sides of equal length. Not all rectangles have four equal sides like a square, so our converse statement is FALSE. Accordingly, the inverse statement is also FALSE because they are logically equivalent: “If the figure is NOT a square, then it is NOT a rectangle.” In summary, the original statement is logically equivalent to the contra-positive, and the converse statement is logically equivalent to the inverse. That's a lot to take in! Let’s end this video with an example for you to process how to analyze a statement to write the converse, inverse, and contra-positive statements. For this exercise, don’t worry about whether the statements are true or false! The statement is: "All four-sided plane figures are rectangles." Now, pause the video and see if you can figure out the converse, inverse, and contra-positive statements. Remember, it helps to first turn our original statement into a conditional statement so you know the hypothesis and conclusion. Okay, let’s see if you figured it out! The conditional statement would be “If all figures are four-sided planes, then figures are rectangles.” This gave us our hypothesis and conclusion. Here are the converse, inverse, and contra-positive statements based on the hypothesis and conclusion: Converse: “If figures are rectangles, then figures are all four-sided planes” Inverse: “If figures are NOT all four-sided planes, then they are NOT rectangles” Contra-positive: “If figures are NOT rectangles, then the figures are NOT all four-sided planes” That’s all for this review! Thanks for watching, and happy studying!
1917	https://idioms.thefreedictionary.com/being+behind+bars	Being behind bars - Idioms by The Free Dictionary Being behind bars - Idioms by The Free Dictionary Printer Friendly Dictionary, Encyclopedia and Thesaurus - The Free Dictionary13,943,312,300 visits served Search / Page tools TheFreeDictionary Google ? Keyboard Word / Article Starts with Ends with Text A A A A Language: Share on FacebookTwitter Get our app Flashcards? My bookmarks? +Add current page to bookmarks 9 RegisterLog in Sign up with one click: Facebook Twitter Google Share on Facebook Get our app Dictionary Thesaurus Medical Dictionary Legal Dictionary Financial Dictionary Acronyms Idioms Encyclopedia Wikipedia Encyclopedia Tools A A A A Language: Mobile Apps: apple android For surfers: Free toolbar & extensions Word of the Day Help For webmasters: Free content Linking Lookup box Close be behind bars (redirected from being behind bars) be behind bars To be in jail or prison.I heard that Mike's uncle was behind bars for nearly 15 years. I wonder what he did.I won't feel safe until the monster who did this is behind bars!Sir,I'm afraid officers caught your son driving while intoxicated.That's why he's behind bars. See also:bar, behind Farlex Dictionary of Idioms. © 2024 Farlex,Inc,all rights reserved. See also: put (one) behind bars behind bars be put behind bars crooked on (someone or something) uncle Uncle! Mother Monster take the pledge be on the danger list be on/off the danger list Want to thank TFD for its existence? Tell a friend about us, add a link to this page, or visit the webmaster's page for free fun content. Link to this page: Facebook Twitter Feedback Flashcards & Bookmarks? Please log in or register to use Flashcards and Bookmarks. You can also log in with Facebook Twitter Google Flashcards? My bookmarks? +Add current page to bookmarks Idioms browser? ▲ behind schedule behind seven locks behind somebody's back behind someone's back behind the bit behind the counter behind the curtain behind the curve behind the eight ball behind the scenes behind the times behind the wheel behind time Behind you! behold beholden to (someone) beholder behoof behoove behoove (one) to (do something) behoove one to do beige bein be-in being being behind bars being that bejeebers bejeer bejesus belabor the point belay my last belch belch out belch up belcher belemnite battlefield belfry Belgravian Belieber belief belies believe believe (one's) ears believe (one's) eyes believe (something) of (someone) believe ears believe eyes believe in believe in (someone or something) believe in the tooth fairy ▼ Full browser? ▲ being basic being batting a thousand being be good enough being be good enough being be good enough being be good enough being be good enough being beaten being beating the bushes being becoming on her being becoming on him being becoming on me being becoming on one being becoming on somebody being becoming on someone being becoming on them being becoming on us being becoming on you being before her being before him being before me being before one being before somebody being before someone being before them being before us being before you being beforehand being beforehand being behind being behind bars being behind schedule being below full strength being below her being below him being below it being below me being below one being below par being below somebody being below someone being below something being below them being below us being below you being beneath her being beneath him being beneath it being beneath me being beneath one being beneath somebody being beneath someone being beneath something being beneath them being beneath us being beneath you being beset being better off being better placed being better than a kick in the pants being beyond ▼ Facebook Share Twitter CITE Site:Follow: Facebook Twitter Rss Mail Share: Facebook Twitter LinkedIn Mail Open / Close More fromIdioms and phrases Mobile Apps Apple Android Kindle Windows Windows Phone Free Tools For surfers: Free toolbar & extensions Word of the Day Word Finder Help For webmasters: Free content Linking Lookup box Terms of Use Privacy policy Feedback Advertise with Us Copyright © 2003-2025 Farlex, Inc Disclaimer All content on this website, including dictionary, thesaurus, literature, geography, and other reference data is for informational purposes only. This information should not be considered complete, up to date, and is not intended to be used in place of a visit, consultation, or advice of a legal, medical, or any other professional.
1918	https://www.youtube.com/watch?v=QccqtHESo4M	Proof of sin x sin y = (cos(x - y) - cos(x + y))/2 Ravi Ranjan Kumar Singh 22000 subscribers 45 likes Description 3143 views Posted: 14 Aug 2020 In this video, we will learn to prove the trigonometric identity sin x sin y = (cos(x - y) - cos(x + y))/2. Other topics of the video are: sin x sin y = (cos(x - y) - cos(x + y))/2 To prove sin x sin y = (cos(x - y) - cos(x + y))/2 What is the proof of sin x sin y = (cos(x - y) - cos(x + y))/2? How to prove sin x sin y = (cos(x - y) - cos(x + y))/2? Trigonometric identity sin x sin y = (cos(x - y) - cos(x + y))/2 To prove trigonometric identity sin x sin y = (cos(x - y) - cos(x + y))/2 What is the proof of trigonometric identity sin x sin y = (cos(x - y) - cos(x + y))/2? How to prove trigonometric identity sin x sin y = (cos(x - y) - cos(x + y))/2? I, Ravi Ranjan Kumar Singh, have produced this video. All credits of this video belong to me. You can email me at raviranjans@gmail.com You can follow me on Facebook. My Facebook account is 6 comments Transcript: In this video, we will learn to prove the trigonometric identity sine of x times sine of y is equal to half of cosine of x minus y minus cosine of x plus y. We know cosine of x minus y is equal to cosine of x times cosine of y plus sine of x times sine of y. And we also know cosine of x plus y is equal to cosine of x times cosine of y minus sine of x times sine of y. Subtracting second equation from first equation gives us, cosine of x minus y minus cosine of x plus y is equal to cosine of x times cosine of y plus sine of x times sine of y minus of cosine of x times cosine of y minus sine of x times sine of y. Here I have removed the brackets. Straight addition gives us, cosine of x minus y minus cosine of x plus y is equal to 2 times sine of x times sine of y. Taking the factor 2 to left of the equation gives us half of cosine of x minus y minus cosine of x plus y is equal to sine of x times sine of y. Thus we have proved the trigonometric identity. Thank you for being with me. I hope you like this video. Please click the like button and subscribe button. Do not forget to press the notification bell so that, you get the notifications of my new videos. Share the video and do write in the comments. I am eager to listen to your thoughts. If you have any question related to video or if you think I should make a video on any other topics in which you are interested please write in the comments. Thank you.
1919	https://www.scribd.com/document/352651362/Chebyshev-Polynomials-Wikipedia	Chebyshev Polynomials - Wikipedia \| PDF \| Polynomial \| Discrete Mathematics Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 605 views 21 pages Chebyshev Polynomials - Wikipedia Brief description about Chebyshev polynomials Full description Uploaded by Anonymous c1USJCV Go to previous items Go to next items Download Save Save Chebyshev Polynomials - Wikipedia For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Chebyshev Polynomials - Wikipedia For Later You are on page 1/ 21 Search Fullscreen Chebyshev polynomials From Wikipedia, the free encyclopedia In mathematics the Chebyshev polynomials , named after Pafnuty Chebyshev, are a sequence of orthogonal polynomials which are related to de Moivre's formula and which can be defined recursively. One usually distinguishes between Chebyshev polynomials of the first kind which are denoted T n and Chebyshev polynomials of the second kind which are denoted U n . The letter T is used because of the alternative transliterations of the name Chebyshev as Tchebycheff , Tchebyshev (French) or Tschebyschow (German).The Chebyshev polynomials T n or U n are polynomials of degree n and the sequence of Chebyshev polynomials of either kind composes a polynomial sequ ence.The Chebyshev polynomials T n are polynomials with the largest possible leading coefficient, but subject to the condition that their absolute value on the interval [−1,1] is bounded by 1. They are also the extremal polynomials for many other properties. Chebyshev polynomials are important in approximation theory because the roots of the Chebyshev polynomials of the first kind, which are also called Chebyshev nodes, are used as nodes in polynomial interpolation. The resulting interpolation polynomial minimizes the problem of Runge's phenomenon and provides an approximation that is close to the polynomial of best approximation to a continuous function under the maximum norm. This approximation leads directly to the method of Clenshaw–Curtis quadrature.In the study of differential equations they arise as the solution to the Chebyshev differential equations and for the polynomials of the first and second kind, respectively. These equations are special cases of the Sturm–Liouville differential equation. Contents 1 De fi ni ti on 1.1 T rigo nom etri c defi nit ion 1.2 Pel l equa tio n defi nit ion 1.3 Prod uct s of Che bysh ev poly nom ial s 2 Relat ions betw een Cheby shev polyn omial s of the first an d second ki nds 3 Exp lic it expr ess ion s 4 P ro pe rt ie s 4.1 Sy mm et ry 4.2 Ro ot s and e xt rem a 4.3 Diff erentia tion and integ ration C h e b y s h e v p o l y n o m i a l s - W i k i p e d i a h t t p s://e n.w i k i p e d i a.o r g/w i k i/C h e b y s h e v _ p o l y n o m i a l s 1 o f 2 1 6/2 8/1 7, 5:0 9 A M adDownload to read ad-free Plot of the first five Chebyshev T polynomials Plot of the first five Chebyshev U polynomials 4.4 Or th ogo na li ty 4.5 Mi ni mal ∞-n orm 4.6 Ot he r pro per ti es 4.7 General ized Chebyshe v polynom ials 5 E x a m p l e s 5.1 Fi rs t k in d 5.2 Se co nd ki nd 6 As a b as is se t 6.1 Ex am pl e 1 6.2 Ex am pl e 2 6.3 Par ti al sum s 6.4 Pol yno mia l in Ch ebys hev fo rm 7 Sp rea d pol yn om ial s 8 Shi fted Ch ebys hev pol ynom ial s 9 S ee a l s o 1 0 N o t e s 1 1 Re fe re nc es 12 Ex te rn al li nk s Definition The Chebyshev polynomials of the first kind are defined by the recurrence relation The ordinary generating function for T n is the exponential generating function is C h e b y s h e v p o l y n o m i a l s - W i k i p e d i a h t t p s://e n.w i k i p e d i a.o r g/w i k i/C h e b y s h e v _ p o l y n o m i a l s 2 o f 2 1 6/2 8/1 7, 5:0 9 A M adDownload to read ad-free The generating function relevant for 2-dimensional potential theory and multipole expansion is The Chebyshev polynomials of the second kind are defined by the recurrence relation The ordinary generating function for U n is the exponential generating function is Trigonometric definition The Chebyshev polynomials of the first kind can be defined as the unique polynomials satisfying or, in other words, as the unique polynomials satisfying for n = 0, 1, 2, 3, ... which is a variant (equivalent transpose) of Schröder's equation, viz. T n ( x ) is functionally conjugate to nx , codified in the nesting property below. Further compare to the spread polynomials,in the section below.The polynomials of the second kind satisfy:or C h e b y s h e v p o l y n o m i a l s - W i k i p e d i a h t t p s://e n.w i k i p e d i a.o r g/w i k i/C h e b y s h e v _ p o l y n o m i a l s 3 o f 2 1 6/2 8/1 7, 5:0 9 A M adDownload to read ad-free which is structurally quite similar to the Dirichlet kernel D n ( x ) :That cos nx is an n th-degree polynomial in cos x can be seen by observing that cos nx is the real part of one side of de Moivre's formula. The real part of the other side is a polynomial in cos x and sin x , in which all powers of sin x are even and thus replaceable through the identity cos 2 x sin 2 x = 1 . By the same reasoning, sin nx is the imaginary part of the polynomial, in which all powers of sin x are odd and thus, if one is factored out, the remaining can be replaced to create a (n-1) th-degree polynomial in cos x .The identity is quite useful in conjunction with the recursive generating formula, inasmuch as it enables one to calculate the cosine of any integral multiple of an angle solely in terms of the cosine of the base angle.Evaluating the first two Chebyshev polynomials,and one can straightforwardly determine that and so forth.T wo immediate corollaries are the composition identity (or nesting property specifying a semigroup)and the expression of complex exponentiation in terms of Chebyshev polynomials: given z = a bi , Pell equation definition The Chebyshev polynomials can also be defined as the solutions to the Pell equation C h e b y s h e v p o l y n o m i a l s - W i k i p e d i a h t t p s://e n.w i k i p e d i a.o r g/w i k i/C h e b y s h e v _ p o l y n o m i a l s 4 o f 2 1 6/2 8/1 7, 5:0 9 A M adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like J.T.schwartz - 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1920	https://online-learning-college.com/knowledge-hub/gcses/gcse-chemistry-help/esters/?srsltid=AfmBOoo1qF240i8h-scWWQrJ4Kcy8kPTULt__q9MnwnmKTXa_fhXZBjA	Understanding Esters: Properties And Naming Conventions Skip to content 0 Menu Menu Campus login Online Campus Login Courses A Level Courses A Level Business A Level Economics A Level Physics A Level Psychology A Level Sociology All courses Teaching Assistant NCFE CACHE Level 2 NCFE CACHE Level 3 NCFE CACHE Level 4 TQUK Level 4 HLTA All courses Counselling and Psychology TQUK Level 2 Certificate NCFE CACHE Level 3 Diploma Counselling Advanced Level 4 Counselling Children Level 3 All courses GCSE Courses IGCSE English Language GCSE Mathematics GCSE Business GCSE Psychology All courses Childcare & Early Years NCFE CACHE EYE Diploma Level 3 NCFE CACHE Childcare and Education Level 3 NCFE CACHE Childminder Level 3 NCFE CACHE Level 3 SENCOs NCFE CACHE Level 4 Advanced Practitioner All courses Animal Care Courses Animal Care Level 3 Dog Groomer Level 3 Equine Studies Level 3 Veterinary Support Assistant Level 3 All courses Teaching Level 3 Education and Training (PTLLS) TQUK Level 2 Autism TQUK Level 3 Autism NCFE CACHE Level 3 Learning Disabilities NCFE CACHE Level 3 Certificate in Preparing to Work in Adult Social Care All courses Health and Social Care NCFE CACHE Level 2 Extended Diploma NCFE CACHE Level 2 Health Sector NCFE CACHE Level 3 Health and Social Care NCFE CACHE Level 3 Awareness of Dementia All courses Fitness and Nutrition Personal Fitness Trainer Level 3 Sports Psychology Level 3 Sports Massage Level 3 All courses About Studying Online Tutor Support Awarding Organisations Online Campus Payment options Frequently Asked Questions Studying Online Staff training Contact 0 0 Campus login MENU MENU All posts A Levels Animal Care Business Care Careers Early Years Education Mental Health GCSEs GCSE Maths Help GCSE English Help GCSE Business Help GCSE Psychology Help GCSE Physics Help GCSE Biology Help GCSE Chemistry Help Career Guides Esters August 3, 2022 Home » GCSEs » GCSE Chemistry » Esters In this post Physical properties Making esters Naming esters Esters are organic compounds which contain the functional group -COO: Physical properties Esters are colourless, volatile compounds and therefore easily vaporise to turn from colourless liquids to colourless gases Esters are also flammable Esters are not very soluble in water and are less dense so will form a layer on top of aqueous solutions Esters have very distinctive, often sweet-smelling, scents. Some esters have a characteristic smell of pear drops whilst others smell of rum. This property makes them very useful in food flavourings and perfumes Making esters Esters are produced during esterification reactions between carboxylic acids, or compounds derived from them, and alcohols. For the purpose of this specification we will just focus on the esterification reaction between carboxylic acids and alcohols. Alcohols react with carboxylic acids, in the presence of an acid catalyst, in an esterification reaction to produce an ester and water. For example, to produce the ester ethyl ethanoate, a reaction mixture containing ethanol and ethanoic acid and a catalyst of concentrated sulphuric acid, is heated up under reflux. The ester produced can be observed as a layer formed at the top of the solution, as esters are not soluble in water. The ester layer can be removed from the reaction mixture using the process of distillation. The water is produced due to the loss of a hydrogen ion from the carboxylic acid molecule and a hydroxide ion from the alcohol molecule. The hydrogen ion and hydroxide ion bond together to form a water molecule. For example, ethanol and ethanoic acid will react in the presence of a sulphuric acid catalyst, to produce the ester ethyl ethanoate and water. The structural and displayed formulae for ethyl ethanoate are shown in the following diagram: Naming esters The names of the esters are based on the names of both the alcohol and the carboxylic acid. The first part of the ester’s name comes from the alcohol, with the suffix -anol being changed to -yl. The second part of the ester’s name comes from the carboxylic acid, with the suffix -oic acid replaced by -oate. The general structure of esters is shown in the diagram below where the carboxylic used is found on the left and the alcohol used is found on the right attached to the oxygen atom of the single C-O bond in the ester link: Where R 1 is carboxylic acid and R 2 is an alcohol. The structural and displayed formulae for any ester can be written from its name using this general structure. For example, any ester named ethyl propanoate must have been made using ethanol and propanoic acid. The propanoic acid part of the ester would be drawn on the left on the ester group and the ethanol would be attached to the right of the ester. The structural and displayed formulae for ethyl propanoate would be: The ester’s name can be determined from its structural and displayed formulae. For example, to name the ester in the diagram below we must firstly count the number of carbon atoms in each of the carboxylic acid and alcohol chains: There are three carbon atoms on the left of the ester meaning that the carboxylic acid contained three carbon atoms. The carboxylic acid must therefore be propanoic acid. There is one carbon atom on the right of the ester meaning that the alcohol only contained one carbon atom. The alcohol used must therefore have been methanol. The name of this ester is therefore methyl propanoate. Looking to get a GCSE? We offer a wide range of GCSE courses. Learn more Start your learning journey Speak to a friendly course advisor to learn more about the best qualification for you. 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1921	https://biologyjunction.com/lab-4-plant-pigments/	BIOLOGY JUNCTION Test And Quizzes for Biology, Pre-AP, Or AP Biology For Teachers And Students Lab 4 Plant Pigments Lab 4 Plant Pigments & Photosynthesis Introduction: The purpose of this lab experiment was to separate plant pigments using paper chromatography, and to measure the rate of photosynthesis in isolated chloroplasts. Because of capillary action the solvent moves up the paper causing the pigments to become visible at certain distances. The substances visible on the paper are called pigments. Chlorophyll a is the main pigment that makes up about 75% of the pigmentation in plants. Chlorophyll b makes up about 25% of the pigmentation. And carotenes and xanthophylls are accessory pigments that make up the rest of the pigmentation. Carotene is the most soluble of the pigments and as a result will be carried the farthest by the solvent. The paper will display a spectrum of the pigments found in the spinach leaves. Using the formula Rf one can determine the relationship between the distance the solvent traveled to the distance the pigment traveled. Rf=distance l2igment migrated (mm) distance solvent front migrated Light is a part of a continuum of radiation or energy waves. The energy from visible light is used in the photosynthetic process. Light is absorbed in the leaf pigments, electrons within each photosystem are boosted to a higher energy level to produce ATP and to reduce NADP and NADPH. The ATP is then used in carbon fixation. This is the incorporation of CO2 into organic molecules. To measure light transmittance in chloroplasts a spectrophotometer will be used. The reason behind measuring the light transmittance is to calculate the rate of photosynthesis in the chloroplasts. A solution called DPIP will be used in place of NADP to judge the color change of the chloroplast solutions. This technique is known as dye- reduction and it tests the hypothesis that light and chloroplasts are required for light reactions to occur. Hypothesis: In this experiment it is hypothesized that the cuvette with boiled chloroplasts and the cuvette kept in the dark containing unboiled chloroplasts will have very slight changes in light transmittance, whereas the cuvette containing unboiled chloroplasts that have been exposed to light will have an increasingly higher % transmittance over the course of time. Materials and Methods: Lab 4A: The materials used in this section of the lab were: filter paper, glass vial, small amount of solvent, a quarter, and spinach leaves. The first step was to cut a point on one end of the filter paper and draw a pencil line 1.5cm from the tip of this point. Next a spinach leave was placed on the strip of paper and rolled over with a quarter on top of the pencil line. This gives a green line across the paper, which contains the pigments of the leave. Then the strip of paper was placed into the vial with the point down in the bottom. When the pigment reached the point 1 cm from the top of the vial then it was removed. The solvent front was then quickly marked with a pencil and then each pigment front was marked as well. From the distance the pigment traveled and the distance the solvent traveled the Rf value was calculated. Lab 4B: The materials used in this lab were: a spectrophotometer, 4 cuvettes, phosphate buffer, distilled water, boiled chloroplasts, unboiled chloroplasts, and DPIP .First the cuvettes were labeled 1-4 and cleaned with lens paper because even the oil from your hands can affect the transmittance of light through the cuvette. Cuvette 2 was then wrapped with foil to keep the contents in the dark. Next 1 ml of phosphate buffer was added to all four cuvettes, 4ml of distilled water was added to cuvette 1, 3rn1 of distilled water was added to cuvettes 2,3, and 4, and Iml of DPIP was added to cuvettes 2,3, and 4. Then 3 drops of unboiled chloroplasts were added to cuvette 1, it was covered with parafilm, placed into the spectrophotometer, and set to 100% transmittance. This cuvette was used to recalibrate between readings as well. Three drops of unboiled chloroplasts were placed in cuvette 2 and 3, and three drops of boiled chloroplasts were placed in cuvette 4. The cuvettes were then covered with parafilm. Each was placed in the spectrophotometer and the % transmittance of each, every five minutes for 15 minutes, was recorded. Data:Table 4.1 Questions: 1. Which pigment migrated the furthest and why? Carotene, it was the most soluble and didn’t form bonds with the filter paper. 2. Which of the 2 types of chlorophyll is more soluble? Chlorophyll a 3. Why do leaves change color in autumn? The chlorophyll production in the leaves slows down. 4. What is the function of the chlorophyll in photosynthesis? They absorb red and blue light rays. 5 .What are the accessory pigments and what are their functions? Carotene and xanthophylls both absorb different wavelengths of light than chlorophyll does. 6. What are some other ways chromatography is used to separate plant pigments? There are three types: Column, Paper, and Thin Layer chromatography. 7 .What does the & value represent? The distance the pigment traveled and the distance the solvent traveled expressed as one value. 8. What factors involved in the separation of the pigments? In this test it was solubility. 9 .Would you expect the Rf value of a pigment to be same if a different solvent were used? Explain. No, for a different solute there would be a different solvent rate. 10. What kind of chlorophyll does the reaction center contain? What are the roles of the other pigments? Chlorophyll a” the other pigments catch different wavelengths of light. Error Analysis: Fingerprints on either the filter paper or the cuvettes may have affected the experiment because the oil from your hands can get on these things and affect the results. The spectrophotometer may have not been calibrated correctly because this was the first time this particular one had been used. Other than these there were few places for error in this lab. Conclusion: From lab 4a we discovered that the many pigments found in chloroplasts are all involved in gathering energy from sunlight. The spectrum of color displayed on the filter paper showed the pigments and the solubility of each. In lab 4b the spectrophotometer measured the light transmittance through the various cuvettes and the chloroplast solutions in each. The actual purpose of this was to observe the DPIP go from a blue color to a clear color. This indicated that photosynthesis was occurring and at what rate it was occurring. The cuvette with the unboiled chloroplasts that had been exposed to light showed the biggest change in % transmittance, which indicates that the amount of light available has a very big effect on the rate at which the light reactions of photosynthesis occur . BACK Post navigation As an Amazon Associate, this site earns from qualifying purchases. We may also earn small commissions on purchases from other retail websites. Prepping Students for College And University
1922	https://collegedunia.com/exams/hypobromous-acid-formula-structure-properties-chemistry-articleid-5629	Hypobromous Acid Formula: Structure & Properties Select Goal & City Select Goal Search for Colleges, Exams, Courses and More.. Write a Review Get Upto ₹300 Explore Explore More Study Abroad Get upto 50% discount on Visa Fees Top Universities & Colleges Abroad Exams Top Courses Exams Read College Reviews News Admission Alerts 2025 Education Loan Institute (Counselling, Coaching and More) Ask a Question College Predictor Test Series Practice Questions Course Finder Scholarship All Courses B.Tech MBA M.Tech MBBS B.Com B.Sc B.Sc (Nursing) BA BBA BCA Course Finder All Courses B.Tech MBA M.Tech MBBS B.Com B.Sc B.Sc (Nursing) BA BBA BCA B.Arch B.Ed B.Pharm B.Sc (Agriculture) BAMS LLB LLM M.Pharm M.Sc MCA Bachelor of Physiotherapy B.Des M.Planning B.Planning Agriculture Arts Commerce Computer Applications Design Engineering Law Management Medical Paramedical Pharmacy Science Architecture Aviation Dental Education Hotel Management Mass Communications Veterinary Sciences Animation Exams Chemistry Hypobromous Acid Formula: Structure & Properties Hypobromous Acid Formula: Structure & Properties Collegedunia TeamContent Curator Content Curator Hypobromous Acid, also known as Bromanol or hydroxide bromine, is a weak and unstable acid that is in +1 oxidation state. Hypobromous acid has physical and chemical properties similar to hypohalites. Hypobromous acids have disinfectant properties i.e., the capacity to kill bacteria and pathogens and are hence used as Disinfectant, Oxidiser, Deodorant, and Bleach. It is called a weak acid because bromine is lightly associated with other elements in the compound and can therefore react quickly. Table of Content Hypobromous Acid Formula and Structure of Hypobromous Acid Synthesis of Hypobromous Acid Chemical Properties of Hypobromous Acid Things to Remember Sample Questions Key Takeaways: Hypobromous Acid, Hypobromous Acid Formula, Hypobromite, HBrO formula, Formula Hypobromous Acid [Click Here for Sample Questions] Hypobromous acid, with the chemical formula HOBr and the bromine atom in the +1 oxidation state, is a weak, unstable acid. It can only be found in solution and has many of the same chemical and physical properties as hypochlorous acid (HClO). Hypobromous acid only partially dissociates into the hypobromite anion BrO and the cation H+ in aqueous solution. Hypobromite are the salts of Hypobromous acid. These salts, like the acid, are unstable and undergo a disproportionation reaction when evaporated or boiled to dryness, releasing the respective bromate and bromide salts. Also Read: Metals and Non MetalsCarbon & its Compounds Hypobromous Acid Formula [Click Here for Sample Questions] When bromine reacts with water Hypobromous acid is obtained as a result or yield product: Br 2+ H 2 O⇔ HOBr + HBr Hypobromous Acid Structure Hypobromous Acid is HOBr having one electron of Hydrogen, has six electrons of Oxygen and has seven electrons of Bromine. One electron of hydrogen forms a pair with one electron of Oxygen and other one electron of Oxygen forms a pair with one of the electrons of bromide. After the pairing oxygen has four electrons unpaired and bromine has six electrons unpaired. Hypobromous Acid Structure Also Read: Difference Between Acid and BaseAcid Test Synthesis of Hypobromous Acid [Click Here for Sample Questions] When Water is mixed with bromine, Hypobromous acid is obtained by the following reaction. Br 2+ H 2 O⇔ HOBr + HBr In Nature, bromoperoxidase, which are the enzymes that catalyse the oxidation of bromide along with hydrogen peroxide, leads to the formation of Hydrobromic acid. The reaction for the same is Br- + H 2 O 2⇔ HOBr + OH- With a pKa of 8.65, Hypobromous acid is only partially dissociated in water at pH 7. Like acids, Hypobromite salts are unstable and they would undergo a gradual disproportionation reaction to produce bromate and bromide salts. 3BrO − (aq) → 2Br − (aq) + BrO − 3(aq) Also Read: Weak BaseAcids Bases and Salt Chemical Properties of Hypobromous Acid [Click Here for Sample Questions] Below table highlights the chemical properties of Hypobromous Acid. Chemical formulaHOBr or HOBr Molecular weight96.911 g/mol Density2.470 g/cm 3 Chemical namesOxalyl bromide, bromic(I) acid, bromanol, hydroxide bromine Acidity (pKa)8.65 Boiling point20–25 °C Also Read: Decomposition ReactionDisplacement Reaction Points to Remember Following are some important points: Hypobromous acid is weak as well as unstable acid. Hypobromous acid has a molecular weight of 96.911 g/mol. Hypobromous acid is also known as Bromanolor hydroxide bromine. Hypobromous acid is in +1 oxidation state. Hypobromous acid is used as disinfectants, Deodorants, Bleach, Oxidizing Agent, Germicidal agent in spas and hot tubs. Sample Questions Ques:Write the chemical formula of Hypobromous acid?(2 Marks) Ans: HOBr is the chemical formula for Hypobromous Acid because hydrogen has one electron; Oxygen has six electrons and Bromine has seven electrons. One electron of hydrogen forms a pair with one electron of Oxygen and other one electron of Oxygen forms a pair with one of the electrons of bromide. After the pairing oxygen has four electrons unpaired and bromine has six electrons unpaired. Ques:What is the acid strength of an acid and what is the value of strength of Hypobromous acid?(2 Marks) Ans: The acid strength of an acid is defined as the ability of an acid to lose the H+ ions and an anion and is given by HA. Where H denotes the H+ ion and A denotes the Anions. The acid strength of Hypobromous acid is 8.65. Ques: What is the chemical reaction of formation of Hypobromous Acid?(2 Marks) Ans:Hypobromous acid is formed when Bromine reacts with water and the reaction of the same is as following: Br 2+ H 2 O⇔ HOBr + HBr Ques: Why does Hypobromous acid have +1 Oxidation State?(2 Marks) Ans:At 20 °C, bromine gas reacts quickly with an aqueous hydroxide (such as sodium or potassium hydroxide) by the following reaction: Br 2(aq)+2OH−(aq)⇒Br−(aq)+BrO−(aq)+H 2 O(l) Bromine disproportionates (goes through both reduction and oxidation) from oxidation state 0 (Br2) to oxidation states –1 (Br) and +1 (BrO) in this reaction. Ques: What are the chemical properties of Hypobromous Acid?(2 Marks) Ans:The chemical properties of Hypobromous Acid is: Hypobromous acid is a chemical substance with a molecular weight of 96.911 g/mol. Furthermore, the density of Hypobromous acid (HBrO), a chemical substance, is 2.470 g/cm. It has a pKa of 8.65, which indicates how acidic it is. Its boiling point is between 20 and 25 degrees Celsius. Ques. What are the uses of Hypobromous Acid? (1 Mark) Ans:Hypobromous acid is used as disinfectants, Deodorants, Bleach, Oxidizing agent, Germicidal agent in spas and hot tubs. Hypobromous Acid has the ability to kill pathogens and is hence used in Deodorants, as a Bleach, as an Oxidizer, and as a Disinfectant. It is also used as a germicidal agent in hot tubs and Spas Ques: Why is Hypobromous Acid weak? (2 Marks) Ans:Hypobromous acid is weak because the bromine molecule is loosely attached, Hypobromous acid is unstable and reacts quickly. Hypobromous acid, with the chemical formula HOBr and the bromine atom in the +1 oxidation state, is a weak, unstable acid. "Bromanol" or "hydroxide bromine" are other names for it. Ques: Why is it called Hypobromite? (2 Marks) Ans:BrO is the hypobromite ion, often known as alkaline bromine water. Bromine has an oxidation state of +1. Hypobromite is a bromine chemical similar to hypochlorites, which can be found in household bleaches and immunological cells. In many aspects, hypobromite is similar to hypochlorite, and it's likewise employed as a germicide and antiparasitic in both industrial and immune system applications. Ques: What can be the colour of Hypobromous Acid? (1 Mark) Ans.If a tank filled with 3,500 to 4,000 ppm of Hypobromous acid is left unattended for 3 days, Hypobromous acid will show orange colour. Ques: Is HOBr soluble?(1 Mark) Ans:Hypobromite is soluble in water and with a temperature range of 201K to 252K, Hypobromous acid is soluble in Sulfuric Acid. When Water is mixed with bromine, Hypobromous acid is obtained by the following reaction. 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1923	https://www.ecicaps.com/wp-content/uploads/Capacitor-Discharge-Current-Theory.pdf	1 Capacitor Discharge Current Theory Tyler Cona Electronic Concepts, Inc. Eatontown, United States of America tcona@ecicaps.com Abstract—This paper is a detailed explanation of how the current waveform behaves when a capacitor is discharged through a resistor and an inductor creating a series RLC circuit. There are several natural response cases that can occur depending on the values of the parameters in the circuit such as overdamped, underdamped and critically damped response. What this paper will focus on is a way of determining the peak discharge current achieved in the circuit. Traditionally we believe that we can use Ohm’s Law to find the peak current but this is not true in every case. I. INTRODUCTION The concept behind this paper relates back to perhaps one of the simplest circuits in electronics engineering consisting of the three most common passive components; one resistor, one inductor and one capacitor creating a series connected RLC circuit. We are interested in studying how the current behaves when the capacitor is charged to an initial voltage (𝑉 𝑜) and switched to discharge through the resistor and uncharged inductor (𝐼𝑜= 0). Depending on the values of these components one of three types of response can occur. The damping factor will be discussed shortly, and this parameter determines if the response will be underdamped or oscillating, critically damped or overdamped. The purpose of this paper is to study what happens in the transient state of the discharge cycle and how to approximate the maximum current value achieved by means of mathematical modeling and comparison of experimental results. The peak discharge current is said to be approximated by using Ohm’s Law which does not work in every case. In most overdamped cases this does show useful but as resistance gets smaller and/or inductance gets larger this concept becomes less acceptable. Fig. 1: Series RLC Circuit. R = Capacitor ESR + Discharge Circuit R L = Capacitor ESL + Discharge Circuit L C = Capacitance Vc = Initial charge voltage II. MATHEMATICAL MODELING OF THE CIRCUIT The circuit pictured in Figure 1 can be modeled using Kirchhoff’s Voltage Law summing the voltages of the components and equating to zero. Manipulating the equation using common relationships it can then be put into terms of current and solved as a second order differential equation. Applying Kirchhoff’s Voltage Law: 𝑉 𝐿+ 𝑉 𝑅+ 𝑉 𝐶= 0 Relating each expression to current: 𝑉 𝐿= 𝐿𝑑𝑖 𝑑𝑡 𝑉 𝑅= 𝑅𝑖 𝑉 𝐶= 1 𝐶∫𝑖(𝑡) 𝑑𝑡 𝑡 0 𝐿𝑑𝑖 𝑑𝑡+ 𝑅𝑖+ 1 𝐶∫𝑖(𝑡)𝑑𝑡 𝑡 0 = 0 Deriving each equation with respect to t: 𝐿𝑑2𝑖 𝑑𝑡2 + 𝑅𝑑𝑖 𝑑𝑡+ 1 𝐶𝑖(𝑡) = 0 Dividing the equation by L to put it in standard ODE form: 𝑑2𝑖 𝑑𝑡2 + 𝑅 𝐿 𝑑𝑖 𝑑𝑡+ 1 𝐿𝐶𝑖(𝑡) = 0 (1) We now have the second order differential equation in terms of current that will be solved. Transforming into Laplace Domain to solve the ODE: 𝑠2 + 𝑅 𝐿𝑠+ 1 𝐿𝐶= 0 We denote the Neper Frequency as being 𝛼= 𝑅 2𝐿 which is a measure of how fast energy is lost in an oscillating system and has a unit of measure of Nepers/second. One other important parameter is the resonant angular frequency which is defined as 𝜔𝑜= 1 √𝐿𝐶. The equation in the Laplace domain will be rewritten in terms of these two variables: 𝑠2 + 2𝛼𝑠+ 𝜔𝑜 2 = 0 2 The quadratic formula can solve for the s-variable: 𝑠1,2 = −𝛼± √𝛼2 −𝜔𝑜 2 (2) The damping variable was mentioned in the introduction and defined as the determining factor of whether the response type will be underdamped, critically damped or overdamped. We define this variable as: 𝐷𝑎𝑚𝑝𝑖𝑛𝑔 𝐹𝑎𝑐𝑡𝑜𝑟= 𝜁= 𝛼 𝜔𝑜 Under the following criteria we can distinguish which type of response to expect from our discharge circuit. 𝛼> 𝜔𝑜− 𝑜𝑣𝑒𝑟 𝑑𝑎𝑚𝑝𝑒𝑑 𝛼= 𝜔𝑜− 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙𝑙𝑦 𝑑𝑎𝑚𝑝𝑒𝑑 𝛼< 𝜔𝑜− 𝑢𝑛𝑑𝑒𝑟 𝑑𝑎𝑚𝑝𝑒𝑑 The following image has several overlaying example graphs describing what the current waveform looks like for each case. 𝜁= 1 shows critical damping, 𝜁< 1 shows the oscillatory behavior of the wave in the underdamped region and 𝜁> 1 shows the lower peak amplitude overdamped region. Fig. 2: Current behavior as zeta varies. III. OVERDAMPED RESPONSE The overdamped response is what happens when the system returns to equilibrium with no oscillations. This is defined as when 𝜁> 1 or 𝛼> 𝜔𝑂. Solving equation (2) results in both 𝑠1 and 𝑠2 to be both real and unique roots. When solving a second order differential equation with real, unique roots the general solution is described as: 𝑖(𝑡) = 𝐾 1𝑒𝑠1𝑡+ 𝐾2𝑒𝑠2𝑡 Using initial conditions the variables 𝐾 1 and 𝐾2 can be solved for. 𝑖(0) = 0 = 𝐾 1 + 𝐾2 𝑑𝑖(𝑡) 𝑑𝑡 = 𝑠1𝐾 1𝑒𝑠1𝑡+ 𝑠2𝐾2𝑒𝑠2𝑡 𝑑𝑖(0) 𝑑𝑡 = 𝑉 𝐿= 𝑠1𝐾 1 + 𝑠2𝐾2 𝐾 1 = 𝑉 𝐿(𝑠1 −𝑠2) 𝐾2 = −𝑉 𝐿(𝑠1 −𝑠2) 𝒊(𝒕) = 𝑽 𝑳(𝒔𝟏−𝒔𝟐) 𝒆𝒔𝟏𝒕+ −𝑽 𝑳(𝒔𝟏−𝒔𝟐) 𝒆𝒔𝟐𝒕 (3) IV. CRITICALLY DAMPED REPONSE The critically damped response is the borderline between the response oscillating and not oscillating. This is a special case because this means the system is balanced perfectly on the cusp of an oscillatory response. This response is defined as when 𝜁= 1 or 𝛼= 𝜔𝑂. Going back to equation (2) this solution results in two real, non-unique answers. The general solution to the differential equation with two repeating roots is described as: 𝑖(𝑡) = 𝐷1𝑡𝑒−𝛼𝑡+ 𝐷2𝑒−𝛼𝑡 Similar to the previous case the initial conditions can be used to solve for the variables 𝐷1 and 𝐷2. 𝑖(0) = 0 = 𝐷2 𝑑𝑖(𝑡) 𝑑𝑡 = 𝑉 𝐿= 𝐷1(−𝛼𝑡𝑒−𝛼𝑡+ 𝑒−𝛼𝑡) 𝑑𝑖(0) 𝑑𝑡 = 𝑉 𝐿= 𝐷1 𝒊(𝒕) = 𝑽 𝑳𝒕𝒆−𝜶𝒕 (4) V. UNDERDAMPED RESPONSE The last of the three possible response for the system is the underdamped response which results in an oscillation and continues to oscillate more and take longer to return to equilibrium as 𝜁 gets smaller. When solving equation (2) the roots are now complex as the discriminant is now negative because 𝛼< 𝜔𝑂. The general solution to the differential equation with complex roots is described as: 𝑖(𝑡) = 𝐵1𝑒−𝛼𝑡cos(𝜔𝑑𝑡) + 𝐵2𝑒−𝛼𝑡sin (𝜔𝑑𝑡) The underdamped response is a sinusoidal signal that is oscillating at a frequency dependent on the angular resonant 3 frequency and the Neper Frequency called the Damped Resonant Frequency. This variable 𝜔𝑑 is defined as: 𝜔𝑑= √𝑤𝑜 2 −𝛼2 = 𝜔𝑜√1 −𝜁2 Using initial conditions the variables 𝐵1 and 𝐵2 are solved for: 𝑖(0) = 0 = 𝐵1 𝑑𝑖(𝑡) 𝑑𝑡 = 𝑉 𝐿= 𝐵2(𝜔𝑑𝑒−𝛼𝑡cos(𝜔𝑑𝑡)) 𝑑𝑖(0) 𝑑𝑡 = 𝑉 𝐿= 𝐵2𝜔𝑑→ 𝐵2 = 𝑉 𝐿𝜔𝑑 = 𝑉 𝐿√𝜔𝑜2 −𝛼2 𝒊(𝒕) = 𝑽 𝑳√𝒘𝒐 𝟐−𝜶𝟐𝒆−𝜶𝒕𝐬𝐢𝐧 (√𝒘𝒐 𝟐−𝜶𝟐𝒕) (5) VI. APPROXIMATING PEAK CURRENT When the peak discharge current is desired, a quick way to find it in most discharge cases is using Ohm’s Law which is calculated using 𝑉= 𝐼𝑅. This is only correct in a special case where the Neper frequency 𝛼 is much greater than 𝜔0. In general this is considered an overdamped response since 𝛼> 𝜔𝑂. Equation (3) will be studied further as the inductance value becomes more insignificant to increase 𝜁. This will derive the equation that is acceptable to use when the inductance in a discharge circuit is negligible compared to the resistance value. In theory, if the Neper frequency is very large it is acceptable to use an abbreviated model for the overdamped response. To begin, equation (2) will be solved for as 𝛼≫𝜔𝑂. 𝑠1,2 = −𝛼± √𝛼2 −𝜔𝑜 2 = −𝛼 ± 𝛼 𝑠1,2 = 0, −2𝛼 Now, plugging in 𝑠1,2 values into current equation (3) and solving the amplitude first: 𝒊(𝒕) = 𝑽 𝑳(𝒔𝟏−𝒔𝟐) 𝒆𝒔𝟏𝒕+ −𝑽 𝑳(𝒔𝟏−𝒔𝟐) 𝒆𝒔𝟐𝒕 𝑉 𝐿(0 −(−2𝛼) = 𝑉 𝐿(2 ∗𝑅 2𝐿) = 𝑉 𝑅 The leading coefficient is found to be V/R. Now the extreme 𝑠1,2 values will be evaluated: 𝑖(𝑡) = 𝑉 𝑅(𝑒𝑠1𝑡−𝑒𝑠2𝑡) lim 𝐿→0 𝑒𝑠2𝑡= 0 lim 𝐿→0 𝑠1 = lim 𝐿→0 −𝛼+ √𝛼2 −𝜔02 = lim 𝐿→0 −𝑅𝐶+ √𝐶(𝑅2𝐶−4𝐿) 2𝐿𝐶 Using L’Hospital’s Rule to evaluate the limit: lim 𝐿→0 −2𝐶 2𝐶√𝐶(𝑅2𝐶−4𝐿) = −1 𝑅𝐶 ∴lim𝐿→0 𝑖(𝑡) = 𝑉 𝑅𝑒−𝑡 𝑅𝐶 (6) In conclusion, Ohm’s Law has been derived from the general solution equation (3) and proven to be an acceptable approximation for the leading coefficient of the discharge waveform if and only if the inductance of the system approaches zero, or becomes more insignificant. VII. WHY OHM’S LAW DOESN’T ALWAYS WORK Section VI discusses and concludes how the circuit will behave as the inductance of a circuit becomes negligible, which we can assume when we are discharging a capacitor through a resistor in the range of ohms and/or an inductor with a very small inductance value. This will make the Neper frequency become much larger than the angular resonant frequency which allows Ohm’s Law to be used to approximate the peak discharge current. Going more in depth in the behavior of the current waveform, unlike voltage the current waveform does not start at its peak and needs to rise from zero amps. There will always be some delay between t=0 and the instance in which it reaches its peak. It is important to note that from the instant the capacitor starts discharging, it is losing charge and therefore losing voltage since the potential across the capacitor is proportional to the charge stored in it. If the capacitor loses too much charge in the initial ramp up time it will cause the voltage to be significantly lower than the initial value, invalidating Ohm’s Law calculations using the initial charge value. An amended version of the Ohm’s Law model can be derived to give the peak discharge current with inductance and loss of charge in mind. We can calculate how long it takes the current to ramp to its peak, how much charge was lost in that time, and finally determine the voltage across the capacitor when current reaches its peak. First, evaluate how long it takes for the 4 current to reach its peak by finding the first derivative of equation (3) equating it to zero and finding t. 𝑖(𝑡) = 𝑉 𝐿(𝑠1 −𝑠2) 𝑒𝑠1𝑡+ −𝑉 𝐿(𝑠1 −𝑠2) 𝑒𝑠2𝑡 𝑑𝑖 𝑑𝑡= 0 = 𝑉𝑠1 𝐿(𝑠1 −𝑠2) 𝑒𝑠1𝑡+ −𝑉𝑠2 𝐿(𝑠1 −𝑠2) 𝑒𝑠2𝑡 𝑠1𝑒𝑠1𝑡= 𝑠2𝑒𝑠2𝑡 𝑠1 𝑠2 = 𝑒(𝑠2−𝑠1)𝑡 𝑡𝑝𝑒𝑎𝑘= ln(𝑠1 𝑠2) 𝑠2−𝑠1 (7) At this point in time, the capacitor has reached its maximum current value. Now using the total electric charge equation, the amount of charge lost during the ramp up time can be found. 𝑇𝑜𝑡𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒 𝑄= 𝐶𝑉= ∫𝑖(𝑡) 𝑑𝑡 ∞ 0 𝑄𝑙𝑜𝑠𝑡= ∫ 𝑉 𝐿(𝑠1 −𝑠2) 𝑒𝑠1𝑡+ −𝑉 𝐿(𝑠1 −𝑠2) 𝑒𝑠2𝑡 𝑑𝑡 𝑡𝑝𝑒𝑎𝑘 0 Evaluating the integral, 𝑄𝑙𝑜𝑠𝑡= 𝑉[𝑠2𝑒𝑠1𝑡𝑝𝑒𝑎𝑘−𝑠1𝑒𝑠2𝑡𝑝𝑒𝑎𝑘+ 𝑠1 −𝑠2] 𝐿𝑠1𝑠2(𝑠1 −𝑠2) By manipulating the electric charge equation, lost voltage can be found by dividing the lost charge by capacitance. Subtracting the lost voltage from the initial voltage will yield the remaining voltage across the capacitor at the time of peak current. It is at this point the resulting voltage can be divided by resistance to find the peak current value. This results in the general solution for the peak current calculation in overdamped cases when inductance is not neglected. 𝑉 𝑟𝑒𝑑𝑢𝑐𝑒𝑑= 𝑉 0 −𝑄𝑙𝑜𝑠𝑡 𝐶 𝐼𝑝𝑒𝑎𝑘= 𝑉𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑅 (8) amended Ohm’s Law equation In conclusion, when there is significant inductance present in a discharge system this will limit the peak current produced by the system. To further explain what happens when inductance approaches zero, the inductor voltage equation can be manipulated and solved for the rate of change of current rise. 𝑑𝑖 𝑑𝑡= 𝑉 𝐿 Using this equation and the concept of lost charge during the rising period proposed earlier, the theory in equation (6) can be confirmed by letting inductance approach zero. If inductance approaches zero, the slope of the rising current will approach infinity therefore making the rise time infinitely small. With an infinitely small rise time the amount of charge lost will approach zero. Finally, this will result in the voltage at peak current equal to the initial voltage. lim 𝐿→0 𝑑𝑖 𝑑𝑡= ∞ lim 𝑑𝑖 𝑑𝑡→∞ 𝑡𝑝𝑒𝑎𝑘→0 lim 𝑡𝑝𝑒𝑎𝑘→0 𝑄𝑙𝑜𝑠𝑡→0 lim 𝑄𝑙𝑜𝑠𝑡→0 𝑉 𝑟𝑒𝑑𝑢𝑐𝑒𝑑→𝑉 𝑜 VIII. COMPARING CALCULATIONS WITH EXPERIMENTS Fig. 3: 100us/Div - 500mV/Div @ 1mV/A - 2020A peak 5 Fig. 4: 100us/Div – 500A/Div – 2020A peak Figure 3 is a photo of experimental results on an oscilloscope when running capacitor discharge pulsing. This image is the results captured with R=0.126Ω L=9.3uH C=130uF and V=750V. The maximum current was found to be 2020A. Figure 4 is a computer simulated model of what the current waveform is predicted to look like when the real-world parameters are set in equation (5). It was found that the maximum current was also 2020A and the waves have the same resonant frequency as expected. This is an example of an underdamped waveform. R = 0.126Ω L = 0.5uH C = 130uF V = 750V Fig. 5: 40us/Div – 1V/Div @ 1mV/A – 4540A peak Fig. 6: 20us/Div – 500A/Div – 4356A peak Figure 5 is a photo of experimental results on an oscilloscope, showing the real findings of discharge current achieved with values of R=0.126Ω L=0.5uH C=130uF and V=750V. Figure 6 is the computer simulated waveform of the same parameters in figure 5, calculated with the values as inputs to equation (3) since they are overdamped. The only difference in figures 5,6 compared to in figures 3,4 is the inductance is lower. This will increase the 𝛼 value thus making this an overdamped case. It is important to notice the new peak current that was achieved by changing the inductance, and the peak time. Figures 3,4 reach peak current of 2020A in 56us while figures 5,6 reach peak current of 4356A in 8us. When the wave requires less time to reach the peak, it results in a higher peak current. R = 1Ω L = 6uH C = 40uF V=45V Fig. 7: 40us/Div – 100mV/Div @ 10mV/A – 36A peak Voltage Current Current Voltage Current Voltage Current 6 Fig. 8: 20us/Div – 5A/Div – 35.7A peak Figure 7 is another example of an overdamped case with values of R=1, L=6uH, C=40uF and V=45V. In this experiment a voltage probe was connected across the capacitor to follow the voltage waveform. In figure 8 the computer model used the input parameters in equation (3). It was found the experimental results and the computer model look remarkably similar and both show a peak current of 35.7A with a rise time of 14us. IX. REAL WORLD APPLICATION OF CALCULATIONS Now, figure 8 will be focused on and broken down to further analyze the waveform behavior. The R, L, C, and V values are put in equation (3). Fig. 9: Annotated graph of overdamped case example 𝑠1 = −30628.7 𝑠2 = −136038 𝑖(𝑡) = 71.15(𝑒−30628.7𝑡−𝑒−136038𝑡) 𝐴 𝑣(𝑡) = 45(1 −9.486 ∗10−6(−136038𝑒−30628.7𝑡+ 30628.7𝑒−136038𝑡 −30628.7 + 136038) 𝑉 𝛼= 83333 𝜔𝑜= 64550 This is an overdamped case so there is no oscillation and inductance is not neglected. 𝛼 is not much greater than 𝜔𝑜 so from the discussion in section 6 equation (6) cannot be used to approximate the peak current. A black line named cursor 1 is placed at the time instant of 𝑡= 𝑡𝑝𝑒𝑎𝑘 to focus on what the voltage (orange) and current (blue) waveforms look like at and before this time. An important piece of the graph is to identify where the waves are at 𝑡= 0 and 𝑡= 𝑡𝑝𝑒𝑎𝑘. Voltage starts at initial charge value of 45V and starts to decline as soon as discharging begins (𝑡= 0+). The current waveform starts at 0A at 𝑡= 0 and ramps up to its peak value of 35.7A at 𝑡= 𝑡𝑝𝑒𝑎𝑘. It can be seen that in the time range of 0 ≤𝑡≤𝑡𝑝𝑒𝑎𝑘 the voltage graph had declined by 9.3V. Since the voltage has dropped so much before 𝑡𝑝𝑒𝑎𝑘 this invalidates using Ohm’s Law to approximate the peak current using initial charge voltage. Peak current must be calculated using the mathematics in section 7 and finally using equation (8). Using equation (7) the time it takes for the current to reach its maximum point can be found: 𝑡𝑝𝑒𝑎𝑘= 14.14 𝜇𝑠 Integrating the model for the current waveform in the range of 0 ≤𝑡≤𝑡𝑝𝑒𝑎𝑘. 𝑄𝑡𝑜𝑡𝑎𝑙= 𝐶𝑉= 1.8 𝑚𝐶 𝑄𝑙𝑜𝑠𝑡= ∫ 71.15(𝑒−30628.7𝑡−𝑒−136038𝑡) 𝑑𝑡 14.14∗10−6 0 𝑄𝑙𝑜𝑠𝑡= 370 𝜇𝐶 The calculations show that 370 𝜇𝐶 was lost during the current rise period. Finally finding the lost voltage: 𝑉 𝑙𝑜𝑠𝑡= 370 𝜇𝐶 40 𝜇𝐹= 9.3 𝑉 𝑉 𝑟𝑒𝑑𝑢𝑐𝑒𝑑= 45𝑉−9.3𝑉= 35.7 𝑉 𝐼𝑝𝑒𝑎𝑘= 35.7 𝑉 1 𝛺 = 35.7 𝐴𝑝𝑒𝑎𝑘 The derived mathematical models found in this report matched the waveforms found experimentally with an oscilloscope and current probe. In addition to this, it has been proven that when inductance is not neglected and is of significance in the system, Ohm’s Law cannot be used to approximate the peak discharge current, in other words we cannot assume that in the time range 0 ≤𝑡≤𝑡𝑝𝑒𝑎𝑘 there is no charge lost. Because of the inductance impeding the rise of the discharge current there may be significant charge lost in the ramp time causing the voltage across the capacitor to be lower than expected by time the current reaches its Cursor 1: t_peak Voltage Current 7 maximum, as seen in figure 7. It is only when inductance can be neglected (𝛼≫𝜔𝑜) that it can be assumed negligible charge was lost during the rise time and Ohm’s Law holds. To emphasize this theory, one last experiment is done where a larger resistor was added to the discharge circuit thus making 𝛼≫𝜔𝑜 and allowing Ohm’s Law to be a valid calculation for peak current. The new parameters are R = 57.6Ω L = 55uH and C = 40uF. 𝑠1 = −434.19 𝑠2 = −1046874.9 𝑖(𝑡) = 0.782(𝑒−434.19𝑡−𝑒−1046874.9𝑡) 𝐴 𝑣(𝑡) = 45(1 −9.556 ∗10−7(−1046874.9𝑒−434.19𝑡+ 434.19𝑒−1046874.9𝑡 −434.19 + 1045874.9) 𝑉 𝛼= 523655 𝜔𝑜= 21320 Fig. 10: 1ms/Div – 20mV/Div @ 100mV/A – 0.78A peak Fig. 11: 1ms/Div – 0.1A/Div – 0.78A peak Figure 10 is the oscilloscope output of the experiment. Prior to discharge, voltage is constant at 45 volts and current is at 0 amps and at the time of discharge, the current is very quickly ramped up to peak. Figure 11 shows the computer simulation of this circuit using equation (3) to model the current. After observation it is shown that this simulation is a very close approximation to the real graph. Similar to the last example the charge lost in the time interval 0 ≤𝑡≤ 𝑡𝑝𝑒𝑎𝑘 will be calculated and the voltage loss will be evaluated. 𝑡𝑝𝑒𝑎𝑘= 7.44 𝜇𝑠 𝑄𝑡𝑜𝑡𝑎𝑙= 𝐶𝑉= 1.8 𝑚𝐶 𝑄𝑙𝑜𝑠𝑡= ∫ 0.782(𝑒−434.19𝑡−𝑒−1046874.9𝑡) 𝑑𝑡 7.44∗10−6 0 𝑄𝑙𝑜𝑠𝑡= 5.06 𝜇𝐶 The calculations show that only 5.06 𝜇𝐶 was lost during the current ramp up time. Compared to the last example this is only 1.3% of the charge lost previously. Finally finding the lost voltage: 𝑉 𝑙𝑜𝑠𝑡= 5.06 𝜇𝐶 40 𝜇𝐹= 0.127 𝑉 𝑉 𝑟𝑒𝑑𝑢𝑐𝑒𝑑= 45𝑉−0.127𝑉= 44.873 𝑉 This results in a negligible charge loss during the ramp up period. This concludes that if 𝛼≫𝜔𝑜 there will be a negligible voltage loss in the ramp up time and therefore it is only under this condition that Ohm’s Law can be used to calculate peak current. 𝐼𝑝𝑒𝑎𝑘= 𝑉 𝑜 𝑅= 45 𝑉 57.6 𝛺= 0.78 𝐴 X. CONCLUSIONS The objective of this paper was to outline the possibilities of discharge current waveforms and what is happening in the transient state as soon as discharge begins. The three cases that are possible for the response are underdamped (oscillatory), critically damped and overdamped and the response type is determined by the ratio between the Neper frequency and the angular resonant frequency of the system. In the underdamped case the inductance is heavily involved in the system causing the signal to oscillate at a specific frequency called the damped resonant frequency in which the current begins lagging the voltage approaching a shift of 90 degrees. Voltage Current 8 A major conclusion that have been determined is what is happening to the voltage and current waveforms in the overdamped case between 𝑡= 0 and the time when the current reaches its peak value, 𝑡= 𝑡𝑝𝑒𝑎𝑘. Since the current equations have been derived, this value can be calculated. During the time interval 0 ≤𝑡≤𝑡𝑝𝑒𝑎𝑘 the current starts at zero and is ramping up to its peak value while the voltage starts at initial charge voltage 𝑉 𝑜 and begins declining. The example on page _ shows the calculations carried out to analytically find the peak current, which was proven to be correct comparing it to real world experiment in figure (7) and computer simulation of the waveforms in figure (8). A significant conclusion is the current is not at its peak when voltage is at its peak. From the moment discharging begins, current is flowing out of the capacitor therefore it is losing charge. Since voltage across a capacitor is proportional to the charge stored, voltage potential across the capacitor will instantly begin declining as the current is ramping up. Figure 9 shows a clear demonstration of how voltage can be reduced by time the current reaches its peak value. This invalidates the ability to calculate peak discharge current using Ohm’s Law using the initial charge voltage value. Ohm’s Law is a method that may be used to approximate the peak current in a special case where the Neper Frequency is much greater than the angular resonant frequency (𝛼≫𝜔𝑜) which essentially means the ratio of resistance to inductance is very large resulting in a severely overdamped case. On page 3 it is proven that Ohm’s Law does work in this extreme case, but a more generalized approach must be considered when analyzing a circuit that isn’t necessarily severely overdamped. On page 4 it is proven that Ohm’s Law may be used when a reduction in voltage potential is considered. In conclusion the calculations shown in equations (7) and (8) are components of the general solution that can solve for the peak current in any overdamped case.
1924	https://www.studocu.com/en-us/messages/question/3979702/calculate-c10-4-using-the-combination-formula-n-kn-k-where-denotes-factorial	[Solved] calculate C10 4 using the combination formula n knk where denotes - Special Topics in Actuarial Science (STAT 495) - Studocu Skip to main content Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes Home My Library AI Notes Ask AI AI Quiz New Recent calculate C(10, 4) using the combination formula n! / [k!(n-k)!], where ! denotes factorial. Special Topics in Actuarial Science (STAT 495) My Library Courses You don't have any courses yet. Add Courses Studylists You don't have any Studylists yet. Create a Studylist Northern Illinois University Special Topics in Actuarial Science Question calculate C10 4 using the combination formula n knk where denotes Northern Illinois University Special Topics in Actuarial Science Question Anonymous Student 1 year ago calculate C(10, 4) using the combination formula n! / [k!(n-k)!], where ! denotes factorial. 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Ask a new question Discover more from: Special Topics in Actuarial Science STAT 495 Northern Illinois University 15 Documents Go to course 2 Unit 2 Special Topics in Actuarial Science Special Topics in Actuarial Science 100%(6) 6 Function-Based Treatments for Escape-Maintained Problem Behavior: A Guide Special Topics in Actuarial Science None 2 Special Topics in Actuarial Science 3-4 Special Topics in Actuarial Science None 2 Special Topics in Actuarial Science 1-2 Special Topics in Actuarial Science None Discover more from: Special Topics in Actuarial Science STAT 495Northern Illinois University15 Documents Go to course 2 Unit 2 Special Topics in Actuarial Science Special Topics in Actuarial Science 100%(6) 6 Function-Based Treatments for Escape-Maintained Problem Behavior: A Guide Special Topics in Actuarial Science None 2 Special Topics in Actuarial Science 3-4 Special Topics in Actuarial Science None 2 Special Topics in Actuarial Science 1-2 Special Topics in Actuarial Science None 1 Special Topics in Actuarial Science 5-6 Special Topics in Actuarial Science None 1 Special Topics in Actuarial Science 7-8 Special Topics in Actuarial Science None Related Answered Questions 1 year ago Check more Quichua (Inca) and aymara instruments Pinkillo Sikus (antara or zampoña) Didgeridoo Moseño 10 points QUESTION 34 Check the Quichua (Inca) and Aymara percussion instruments Bombo Wankara Chajchas Tun 10 points QUESTION 35 Check the 3 Andean instruments with Spanish influence Ocarina Charango Chirimia Harp 10 points QUESTION 36 Check the names of the Quichua emperor (Inca) and the Aztec emperor (Tlatoani) at the time of the Conquest Atahualpa Moctezuma Netzahualcóyotl Manco Capac 10 points QUESTION 37 Check the Aztec percussion instruments Huehuetl (bass drum) Bombo Tehuehuetls (turtle drums) Teponazli (log drum) 10 points QUESTION 38 Check the Aztec wind instruments Atecocolli, coyoahulli (conch shell) Huey Acátl (like the Hompak) Tzitzilli (clay flutes) Tarka 10 points QUESTION 39 Who is Carlos Chávez? Former President of Venezuela Mexican classical composer Peruvian classical composer 10 points QUESTION 40 Who composed the pieces Xochipilli? Hugo Chavez César Chavez Carlos Chávez Ricardo Chávez Special Topics in Actuarial Science (STAT 495) 1 year ago 2. Suppose a mutation occurs in a cell such that normal Okazaki fragments were created during DNA replication but were not linked together into a continuous strand. The gene for which enzyme would you predict was altered by this mutation? 3. Which of the following is false regarding the packing of DNA? A. During cell division, chromosomes loosen from a condensed state as cells enter mitosis. B. Sister chromatid are two identical halves of a replicated chromosome. C. Sister chromatids are attached at the centromere. D. Chromatin is made up of DNA wrapped around histone proteins. 4. Today's task in your research lab is to catch DNA under an electron microscope. Your PI (leading researcher) specifically wants you to capture a scan of DNA's backbone. To do this you take following three pigments and use each to highlight the different components of a nucleotide. Pigment A: nitrogenous basePigment B: phosphatePigment C: deoxyribose sugarWhich dyes would highlight DNA's backbone? A. Dye A, Dye B & Dye C B. Dye A & Dye C C. Dye A & Dye B D. Dye B & Dye C Special Topics in Actuarial Science (STAT 495) 1 year ago A company institutes an exercise break for its workers to see if this will improve job satisfaction asmeasured by a questionnaire. Scores for 25 randomly selected workers before the program wasimplemented was compared with 27 different randomly selected workers who participated in the exerciseprogram. Perform a hypothesis test at the significance level of 6% to see if the exercise break leads tohigher job satisfaction index. (Assume that all conditions are satisfied, and that each population is normallydistributed)If the test resulted in a t-value of 2.152 with 40 degrees of freedom. Compute the P-value. Detail how you got your number. Special Topics in Actuarial Science (STAT 495) 1 year ago A set of weightlifters tested a new training method. They randomly used their current and the new training method in two separate months and recorded their mass gain in pounds.The standard deviation is 0.487 and the mean is 0.733. The t value is 4.515 p-value is 0.0020.Based on your P-value, what decision do you make at a 5% significance level: Reject H0 or do not reject H0?Subject123456789New method gain1.63.32.71.61.02.02.32.22.9old method gain1.22.82.30.91.10.91.01.31.5difference (new-old)0.40.50.40.7-0.11.11.30.91.4 Special Topics in Actuarial Science (STAT 495) 1 year ago A set of weightlifters tested a new training method. They randomly used their current and the new training method in two separate months and recorded their mass gain in pounds.The standard deviation is 0.487 and the mean is 0.733. The t value is 4.515What is the p-value? round your answer to four decimal places.Subject123456789New method gain1.63.32.71.61.02.02.32.22.9old method gain1.22.82.30.91.10.91.01.31.5difference (new-old)0.40.50.40.7-0.11.11.30.91.4 Special Topics in Actuarial Science (STAT 495) 1 year ago A company institutes an exercise break for its workers to see if this will improve job satisfaction as measured by a questionnaire. Scores for 25 randomly selected workers before the program was implemented was compared with 27 different randomly selected workers who participated in the exercise program. Perform a hypothesis test at the significance level of 6% to see if the exercise break leads to higher job satisfaction index. (Assume that all conditions are satisfied, and that each population is normally distributed) what is the null and alternative hypothesis. Special Topics in Actuarial Science (STAT 495) Ask AI Home My Library Discovery Discovery Universities High Schools High School Levels Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI English (US) United States Company About us Studocu Premium Academic Integrity Jobs Blog Dutch Website Study Tools All Tools Ask AI AI Notes AI Quiz Generator Notes to Quiz Videos Notes to Audio Infographic Generator Contact & Help F.A.Q. Contact Newsroom Legal Terms Privacy policy Cookie Settings Cookie Statement Copyright & DSA English (US) United States Studocu is not affiliated to or endorsed by any school, college or university. Copyright © 2025 StudeerSnel B.V., Keizersgracht 424-sous, 1016 GC Amsterdam, KVK: 56829787, BTW: NL852321363B01 Cookies give you a personalised experience We’re not talking about the crunchy, tasty kind. 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1925	https://www.ernestpackaging.com/buzz/sustainability/excessive-packaging-problems-solutions/	Home Blog Sustainability The Problem with Overpackaging (and How to Solve It) The Problem with Overpackaging (and How to Solve It) January 13, 2025 We’ve all been there. You buy something new, in store or online, and can’t wait to tear it open—only to find it’s buried in layer after layer of packaging. Not only is overpackaging frustrating for customers, it’s terrible for the environment. Plus, it’s a waste of money, which can drive up costs for buyers and eat into margins for sellers. In this article, we’ll look at what overpackaging means, common examples and which industries are most at fault. We’ll also look at the impacts for retailers, consumers and the environment, and how you can overcome this common issue by using packaging solutions that are smart, sustainable and minimal. What is Overpackaging? Overpackaging refers to the use of excessive packaging methods and materials—meaning more than what’s needed to contain, protect, transport and sell what’s inside. Common examples include items individually wrapped in plastic, covered in multiple layers of padding or even paper, and boxes within boxes. While overpackaging is widespread these days, some industries stand out as the worst culprits. We often see excess packaging with food, including consumer packaged goods (CPG) as well as fresh goods, beauty products and cosmetics, and retail clothing. Technology and medical products also tend to excess, and online shopping is a key contributor. The Excessive Packaging Problem The environmental impact of excess packaging is significant, which is why there’s been considerable backlash in recent years. Plastics are frequently used—and rarely recycled, more often ending up in landfills or polluting our oceans, where they can do serious damage to marine wildlife as well as contribute to global warming and loss of biodiversity. While paper or cardboard are often seen as a more sustainable option, a recent Harvard Business Review article suggests that many brands are using too much paper packaging in an effort to look more eco-friendly, without actually reducing waste. Related reading: Is Zero Waste Packaging Possible? No matter the material, excessive use depletes natural resources and contributes to the global climate crisis. Plus, you might be creating a negative customer experience, because overpackaging is hard to get into and feels wasteful for green-minded shoppers. Excessive Packaging Examples So, what are the most common examples of overpackaging? Here are a few that we see all the time—and you probably do, too. Food Packaging Underfilled Packaging eCommerce Packaging Food Packaging Overpackaging is common with food products, and not just the processed goods that we expect to be packaged. Our convenience-crazed culture has also led to excessive packaging of fresh foods. For example, you might purchase hard boiled eggs, peeled and individually wrapped inside a plastic container. This makes them ready to eat, but leads to considerable waste. Readymade meals are becoming more and more popular, leading to more and more packaging waste. Individually wrapped or bagged fresh produce items, from apples to zucchini, are also becoming common in grocery stores. While some packaging is necessary to protect food and, in some cases, extend its shelf life, this kind of excess use of materials with no real benefits is problematic Under filled Packaging Have you ever purchased something that’s packaged in a much bigger-than-needed box or container? This is called underfilled or slack fill packaging, meaning the product is intentionally put in too-big packaging, creating space around the contents that serves no purpose. Sometimes this space is filled by additional packaging materials, such as packing peanuts or air pillows. Underfilled packaging is not only wasteful, it can also be deceptive. Think about when you buy a bag of chips, a bottle of laundry detergent or a container of lip gloss that’s only half full when you open it. Underfilling can put your business at risk of violating the Fair Packaging and Labeling Act, as well as damaging customer trust. e-Commerce Packaging Online shopping has led to a significant increase in both the volume of retail packages being express shipped and the amount materials being used per package. Whereas traditional retail products are typically packed into larger containers for shipment, e-commerce goods are usually packaged individually before being shipped out by the seller or fulfillment center. In many cases, they’re over packaged. For example, a box within a box, or a retail package wrapped up in excessive bubble-wrap or void-fill. Fragile products, such as consumer electronics or medical devices, are often packaged using additional cushioning materials to protect them during shipment, on top of what’s already been added to the interior packaging solution. While safe delivery is important, excessive cushioning can be unnecessary and wasteful. According to Statista, the global e-commerce industry used 2.1 billion pounds of plastic packaging in 2019, with that number expected to reach 4.5 billion by 2025. The vast majority of this material is non-biodegradable, so it often ends up in landfills. Waste from e-commerce plastic packaging was estimated to be more than one billion kilograms globally in 2019. With analysts predicting the e-commerce market coverage will grow by 25% by 2026, this is a continued source of environmental stress that must be addressed. How Does Overpackaging Impact Retailers and Consumers? Another reason to re-think your packaging is the economic fallout. Excessive packaging increases costs for retailers, especially if you’re using higher-end materials such as rigid boxes, heavy paper products and foam inserts. Extra materials usage can lead to inefficiencies across the entire supply chain, from manufacturing to storage to transportation costs. In addition to eating into margins, high shipping costs can drive up prices for consumers too—and in a tight market, that can take away your competitive advantage. Over time, wasteful packaging contributes to high energy prices, less availability of resources and other environmental factors that impact the global economy as well. Excessive Packaging Solutions Now let’s look at some of the ways you can reduce waste and increase sustainability by re-thinking your packaging approach. First, consider what materials you’re using. Are you opting for eco-friendly options, such as recyclable or reusable paper/cardboard over plastics? If you need padding inside your packaging solution, choose biodegradable materials like paper cushioning over foam. Working with a local packaging vendor will also help to reduce your carbon footprint by minimizing transportation for packaging products. Next, consider the amount of packaging you’re using. Do you really need it all? Would a ships-in-own-container (SIOC) solution, which uses the product box as the shipping box, work instead? Think minimalist. Taking any unnecessary materials out of the equation will help you reduce shipping weight and overall costs. This boosts your margins and allows you to pass the savings on to your customers. Finally, remember that reducing packaging is also a marketing opportunity. You can advertise the fact that you’re using minimalist packaging on your website or retail store. Because today’s environmentally conscious shoppers will be happy to buy from a company that cares about Mother Earth, too. How Ernest Can Help Resolve Your Overpackaging Problems If excessive packaging is a problem for your business, Ernest has solutions. Many of our products are made from reused or recyclable materials, including both corrugate and plastics. Our expert design team takes a less-is-more approach, keeping materials to a minimum and helping to lighten shipping weight and costs. We can even help you get eco-friendly about cold chain shipping, with our proprietary TempEndure solutions. And with 15 locations nationwide, we’ve always got trucks on the road locally—we often use them to help clients get product shipped. Just one more way we can help reduce your carbon footprint. If you’re ready to say goodbye to your overpackaging problems and hello to sustainable solutions, contact Ernest today. We love helping clients go green. Blazing Tales: GT Ventures BLAZING TALES The comfiest solutions for shipping beanbags. Packaging Design Packaging Design Packaging Automation Packaging Supply Chain Management Packaging Process Packaging Technology E-Team Cardboard Chaos Blazing Tales Ernest Challenge Not Hiring Ernest History Our Founder Sustainability Ernest Culture World of Packaging Like what you see? Sign up for news and the newest stuff Need a partner for the impossible? 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1926	https://math.stackexchange.com/questions/2927327/probability-of-drawing-objects-combinations-vs-permutations	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. current community your communities more stack exchange communities Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Probability of drawing objects: Combinations vs Permutations A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement the probability of getting exactly one red ball is Attempt 1 $$ P(\text{exactly 1 red ball})=\frac{^5C_1 \cdot ^3C_2}{^8C_3}=\frac{5 \cdot 3}{\frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1}}=\frac{5 \cdot 3 \cdot 3 \cdot 2}{8 \cdot 7 \cdot 6}=\frac{15}{56} $$ Attempt 2 \begin{align} P(\text{exactly 1 red ball}) & = P(RBB)+P(BRB)+P(BBR)\ & =\frac{5 \cdot 3 \cdot 2}{8 \cdot 7 \cdot 6}+\frac{3 \cdot 5 \cdot 2}{8 \cdot 7 \cdot 6}+\frac{3 \cdot 2 \cdot 5}{8 \cdot 7 \cdot 6}\ & =3 \cdot \frac{5 \cdot 3 \cdot 2}{8 \cdot 7 \cdot 6}\ & =\frac{15}{56} \end{align} Though I get same result in both methods, I think first method uses combinations(order does not matter) and second uses permutation(order matters), how can I differentiate the two ? 1 Answer 1 In Attempt 1 the underlying assumption is that you have a random mechanism selecting a $3$-element subset from a given $8$-element set, whereby all $3$-element subsets are equiprobable. You then count the number of favorable $3$-element subsets. In Attempt 2 the underlying assumption is that you have a random mechanism selecting an element from a given finite set, whereby all elements are equiprobable, and you apply this mechanism three times. Finally you compute the probability of the desired outcome. The reason that you obtain the same result in both cases is the following: Both mechanisms select (in one, resp., in three steps) a $3$-element subset of the given $8$-element set, uniformly over all ${8\choose3}$ such subsets.The first mechanism does so by its specification, and the second "by symmetry". You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Explore related questions See similar questions with these tags. Related Hot Network Questions Subscribe to RSS To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Mathematics Company Stack Exchange Network Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA . rev 2025.9.26.34547
1927	https://pynative.com/python-get-time-difference/	Python Programming Calculate Time Difference in Python In this article, you’ll learn how to find the time difference between two-time in Python. For example, we want to measure the execution time of a program. Here, we need to calculate the time difference between the program’s start and end times. Another example is we may need to capture the time change between the two events. We need to subtract one time/timestamp from another to calculate the time difference between two-time. After reading this Python article, you’ll learn: How to calculate the time interval between the two-time strings. get time difference in seconds, minutes, and hours Get the time difference between two timestamps in Python Also, see how to get the difference between two dates in days in Python. Table of contents How to get the time difference between two-time in Python Example: Calculate Time Difference in Python Get time difference in hours and minutes Time Difference between two timestamps in Python How to get the time difference between two-time in Python For example, you have a start time and stop time in the format of “10:33:26” (HH:MM:SS). The below steps show how to calculate the time difference in hours, minutes, and seconds between two-time in Python. Import datetime module Python datetime module provides various functions to create and manipulate the date and time. Use the from datetime import datetime statement to import a datetime class from a datetime module. 2. Convert time strings to a datetime object We have time data in a string format. Before calculating the difference, we need to convert both time strings to a datetime object. Use the strptime(time_str, format) function to convert a time string into a datetime object as per the corresponding format. The format codes are standard directives for mentioning the string format for parsing. For example, the %H:%M:%S format codes are for hours, minutes, and seconds. 3. Subtract the end time from the start time To get the difference between two-time, subtract time1 from time2. A result is a timedelta object. The timedelta represents a duration which is the difference between two-time to the microsecond resolution. 4. Get time difference in seconds To get a time difference in seconds, use the timedelta.total_seconds() methods. 5. Get time difference in milliseconds Multiply the total seconds by 1000 to get the time difference in milliseconds. 6. Get time difference in minutes Divide the seconds by 60 to get the difference in minutes. 7. Get time difference in hours Divide the seconds by 3600 to get the final result in hours. Example: Calculate Time Difference in Python In this example, we will see how to calculate the seconds and milliseconds between two-time. ``` from datetime import datetime start time start_time = "2:13:57" end_time = "11:46:38" convert time string to datetime t1 = datetime.strptime(start_time, "%H:%M:%S") print('Start time:', t1.time()) t2 = datetime.strptime(end_time, "%H:%M:%S") print('End time:', t2.time()) get difference delta = t2 - t1 time difference in seconds print(f"Time difference is {delta.total_seconds()} seconds") time difference in milliseconds ms = delta.total_seconds() 1000 print(f"Time difference is {ms} milliseconds") Code language: Python (python) ``` Output: Start time: 02:13:57 End time: 11:46:38 Time difference is 34361.0 seconds Time difference is 34361000.0 milliseconds Get time difference in hours and minutes Note: To calculate the time difference in minutes and hours use the below formulas. First, use the above example to get the time difference in seconds between two-time. Next, divide the seconds by 60 to get the get time difference in minutes. In the end, divide the seconds by 3600 (6060) to get the time difference in hours. Example: To get the time difference in hours and minutes between two-time. ``` from datetime import datetime start time and end time start_time = datetime.strptime("2:13:57", "%H:%M:%S") end_time = datetime.strptime("11:46:38", "%H:%M:%S") get difference delta = end_time - start_time sec = delta.total_seconds() print('difference in seconds:', sec) min = sec / 60 print('difference in minutes:', min) get difference in hours hours = sec / (60 60) print('difference in hours:', hours)Code language: Python (python) ``` Output: difference in seconds: 34361.0 difference in minutes: 572.6833333333333 difference in hours: 9.544722222222223 Time Difference between two timestamps in Python A timestamp is encoded information generally used in UNIX, which indicates the date and time at which a particular event has occurred. Now, let’s see how to find the difference between the two timestamps in hours, minutes, and seconds. First, store the start timestamp in the ‘start’ variable and the end timestamp in the ‘end’ variable. Next, use the fromtimestamp() method to convert both start and end timestamps to datetime objects. We convert these timestamps to datetime because we want to subtract one timestamp from another. Next, subtract datetime2 from datetime1 to get the difference between two timestamps in the form of timedelta Next, use the total_seconds() method to get the difference in seconds. Example 1: Difference between two timestamps in HH:MM:SS format. ``` from datetime import datetime start and end timestamps start_ts = 1652426243.907874 end_ts = 1652436243.907874 convert timestamps to datetime object dt1 = datetime.fromtimestamp(start_ts) print('Datetime Start:', dt1) dt2 = datetime.fromtimestamp(end_ts) print('Datetime End:', dt2) Difference between two timestamps in hours:minutes:seconds format delta = dt2 - dt1 print('Difference is:', delta)Code language: Python (python) ``` Output: Datetime Start: 2022-05-13 12:47:23.907874 Datetime End: 2022-05-13 15:34:03.907874 Difference is: 2:46:40 The output shows the difference between the two timestamps is 2 hours, 46 minutes, and 40 seconds. Difference between two timestamps in seconds Use the below example to get the total seconds between two timestamps. Example: ``` from datetime import datetime convert timestamps to datetime object start_ts_ = datetime.fromtimestamp(1652426243.907874) print('Datetime Start:', start_ts_) end_ts = datetime.fromtimestamp(1652436243.907874) print('Datetime End:', end_ts) Difference between two timestamps in seconds delta = end_ts - start_ts_ print('Difference is seconds:', delta.total_seconds())Code language: Python (python) ``` Output: Datetime Start: 2022-05-13 12:47:23.907874 Datetime End: 2022-05-13 15:34:03.907874 Difference is seconds: 10000.0 Did you find this page helpful? Let others know about it. Sharing helps me continue to create free Python resources. TweetF sharein shareP Pin About Vishal I’m Vishal Hule, the Founder of PYnative.com. As a Python developer, I enjoy assisting students, developers, and learners. Follow me on Twitter. 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1928	https://www.jacc.org/doi/abs/10.1016/j.jaccas.2023.102061	Acute Chest Pain and Electrical Alternans \| JACC: Case Reports Skip to main content Skip main navigation Journals JACC JACC: Advances JACC: Asia JACC: Basic to Translational Science JACC: CardioOncology JACC: Cardiovascular Imaging JACC: Cardiovascular Interventions JACC: Case Reports JACC: Clinical Electrophysiology JACC: Heart Failure Topics Topics by Journal Collections All Topics Guidelines Events Author Center Multimedia Search Search Quick Search anywhere Enter Search Terms Search 0 Sign In JACC: Case Reports Current Issue Just Accepted Archives Submit Sign In Search Quick Search anywhere Enter Search Terms Search Guidelines Author Center Topics Topics by Journal Collections All Topics JACC family of Journals JACC JACC: Advances JACC: Asia JACC Basic Transational Science JACC: CardioOncology JACC Cardiovascular Imaging JACC Cardiovascular Interventions JACC: Case Reports JACC: Clinical Electrophysiology JACC: Heart Failure Conferences Simultaneous Publications CME/MOC Multimedia About Journal Subscribe Newsletter/Email FAQs Contact Us ACC.org JACC Journals › JACC: Case Reports › Archives › Vol. 26 Open access ECG Challenge 11 October 2023 Share on Acute Chest Pain and Electrical Alternans To the Editor: Respirophasic Changes in the QRS Axis Authors: Maria Inês Barradasminesbarradas@gmail.com, Fabiana Duarte, and Inês Coutinho dos SantosAuthors Info & Affiliations Publication: JACC: Case Reports Volume 26 2,805 Metrics Total Downloads 2,805 PDFOther Formats Contents Abstract References Information & Authors Metrics & Citations View Options References Figures Tables Media Share Central Illustration Open in Viewer Download figure Download PowerPoint Abstract A young man presented with acute stabbing chest pain. A 12-lead electrocardiogram revealed electrical alternans with phasic variation of the QRS amplitude in all leads. Lung auscultation revealed absent left hemithorax breath sounds. Chest radiography confirmed a left-sided tension pneumothorax. Tension pneumothorax is a very rare cause for electrical alternans. (Level of Difficulty: Intermediate.) Formats available You can view the full content in the following formats: View PDFView Full Text References 1. Ingram D., Strecker-McGraw M.K. Electrical Alternans. StatPearls, 2022. Google Scholar 2. Schmidt D.C., Andersson C., Schultz H.H. ECG with alternating electric axis in relation to left-sided tension pneumothorax: a case report and review of the literature. Eur Clin Respir J. 2018;5:1: 1495982. Crossref PubMed Google Scholar 3. Kuritzky P., Goldfarb A.L. Unusual electrocardiographic changes in spontaneous pneumothorax. Chest. 1976;70:4: 535-537. Crossref PubMed Google Scholar Information & Authors Information Authors Information Published In JACC: Case Reports Volume 26 • 15 November 2023 Copyright Copyright © 2023, The Authors. Published by Elsevier on Behalf of The American College of Cardiology Foundation. This is an Open Access Article under the CC BY-NC-ND License ( History Received: 3 September 2023 Revised: 15 September 2023 Accepted: 19 September 2023 Published online: 11 October 2023 Keywords acute chest pain electrical alternans pneumothorax Authors Affiliations Expand All Maria Inês Barradas∗minesbarradas@gmail.com Hospital do Divino Espírito Santo de Ponta Delgada, Avenida D. Manuel I, Ponta Delgada, São Miguel, Portugal View all articles by this author Fabiana Duarte Hospital do Divino Espírito Santo de Ponta Delgada, Avenida D. Manuel I, Ponta Delgada, São Miguel, Portugal View all articles by this author Inês Coutinho dos Santos Hospital do Divino Espírito Santo de Ponta Delgada, Avenida D. Manuel I, Ponta Delgada, São Miguel, Portugal View all articles by this author Notes ∗ Address for correspondence: Dr Maria Inês Barradas da Silva, Avenida D. Manuel I, 9500-370 Ponta Delgada, Portugal. E-mail: minesbarradas@gmail.com. Twitter: @minesbarradas. Metrics & Citations Metrics Citations Metrics Metrics Downloads Citations No data available. 2,717 0 Total 6 Months 12 Months Total number of downloads See more details X (5) Facebook (1) Mendeley (5) Citations Download article citation data for: Acute Chest Pain and Electrical Alternans: Maria Inês Barradas, MD Fabiana Duarte, MD Inês Coutinho dos Santos, MD How to cite this article: Barradas, M, Duarte, F, Coutinho dos Santos, I. Acute Chest Pain and Electrical Alternans. J Am Coll Cardiol Case Rep. 2023 Nov, 26 (null) . Download article citation: If you have the appropriate software installed, you can download article citation data to the citation manager of your choice. Simply select your manager software from the list below and click Download. Format [x] Direct import View Options View options PDF View PDF Full Text View Full Text Figures Open all in viewer Download figure Download PowerPoint Go to FigureOpen in Viewer Tables Media Share Share Copy the content Link Copy Link Copied! Copying failed. Share on social media FacebookX (formerly Twitter)LinkedInemailWeChatBluesky References References 1. Ingram D., Strecker-McGraw M.K. Electrical Alternans. StatPearls, 2022. Google Scholar 2. Schmidt D.C., Andersson C., Schultz H.H. ECG with alternating electric axis in relation to left-sided tension pneumothorax: a case report and review of the literature. Eur Clin Respir J. 2018;5:1: 1495982. Crossref PubMed Google Scholar 3. Kuritzky P., Goldfarb A.L. Unusual electrocardiographic changes in spontaneous pneumothorax. Chest. 1976;70:4: 535-537. 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1929	https://irp.cdn-website.com/f5d3b9dd/files/uploaded/9808.pdf	Fundamentals of heat and mass transfer Fundamentals of heat and mass transfer 7th edition solutions pdf. Fundamentals of heat and mass transfer solution. Fundamentals of heat and mass transfer sachdeva pdf. Fundamentals of heat and mass transfer 8th edition hardcover. Fundamentals of heat and mass transfer incropera and dewitt. Fundamentals of heat and mass transfer thirumaleshwar. Fundamentals of heat and mass transfer 8th edition. Fundamentals of heat and mass transfer 8th edition solutions. Fundamentals of heat and mass transfer 4th edition. Fundamentals of heat and mass transfer 5th edition. Fundamentals of heat and mass transfer bergman. Fundamentals of heat and mass transfer chegg. Fundamentals of heat and mass transfer 9th edition. Fundamentals of heat and mass transfer 6th edition solutions. Fundamentals of heat and mass transfer eighth edition. Heat Transfer Fundamentals: A Comprehensive Guide This book provides an in-depth exploration of heat transfer, covering topics such as conduction, convection, and radiation. The introduction chapter discusses the basics of heat transfer, including the relationship to thermodynamics and units and dimensions. The second chapter delves into the analysis of heat transfer problems, highlighting methodology and relevance. It also includes a summary and problem sets. Chapters 3-5 focus on one-dimensional steady-state conduction, two-dimensional steady-state conduction, and transient conduction, respectively. These chapters cover topics such as thermal resistance, thermal conductivity, and boundary conditions. Chapter 6 explores heat transfer from extended surfaces, including fins of uniform cross-sectional area and nonuniform cross-sectional area. It also discusses overall surface efficiency. The book concludes with a summary, references, and problem sets. The text appears to be a table of contents or outline for a book on convection, specifically in the context of heat transfer and mass transfer. Chapter 5 deals with semi-infinite solids and boundary conditions, including constant temperature and heat flux conditions. Approximate solutions are also discussed, as well as periodic heating. The chapter concludes with a summary and problems to solve. Chapters 6-8 focus on convection in external flows, including the velocity and thermal boundary layers, mass transfer, and the boundary layer equations. Boundary layer similarity is also discussed, along with physical interpretation of dimensionless parameters and analogies. Chapter 7 explores external flow situations, such as flat plates, spheres, and packed beds, with discussions of methodology for convection calculations, hydrodynamic and geometric considerations, and heat and mass transfer. Chapter 8 delves into internal flows, including flow conditions, mean velocity, and pressure gradient. Thermal considerations are also covered, along with fully developed flow regions and entry regions. The chapter concludes with convection correlations for turbulent flow in circular tubes and noncircular tubes. Finally, Chapter 9 discusses free convection, covering physical considerations, governing equations for laminar boundary layers, and more. Convection and Heat Transfer The following chapters cover various aspects of convection and heat transfer. Laminar Free Convection: This topic is discussed on page 5689, followed by an explanation of the effects of turbulence (page 572). Empirical Correlations: External free convection flows are analyzed in detail, with sections dedicated to vertical plates (page 575), inclined and horizontal plates (page 578), long horizontal cylinders (page 583), spheres (page 587), and channels (page 588). Boiling and Condensation Dimensionless Parameters: These parameters are introduced on page 619. Boiling Modes: The different modes of boiling, including nucleate pool boiling (page 626) and film pool boiling (page 629), are discussed in detail. Forced Convection Boiling: This topic is covered on pages 635-636. Condensation: The physical mechanisms behind condensation are explained on page 639, followed by discussions on laminar film condensation (page 641) and turbulent film condensation (page 645). Heat Exchangers Types of Heat Exchangers: These are introduced on page 669. Overall Heat Transfer Coefficient: This topic is discussed in detail on pages 672-675. Heat Exchanger Analysis: The use of the log mean temperature difference is explained, with sections dedicated to parallel-flow and counterflow heat exchangers (pages 676-679). Radiation Fundamental Concepts: These are introduced on page 725. Radiation Heat Fluxes: The mathematical definitions and relations behind radiation heat fluxes are discussed in detail (page 729). Blackbody Radiation: The Planck distribution, Stefan-Boltzmann law, and band emission are explained on pages 741-743. Summary and Conclusion The chapters conclude with summaries of the key concepts covered (pages 599, 711, and 784). Chapter 13 covers radiation exchange and multimode heat transfer, including volumetric absorption and gaseous emission and absorption. Chapter 14 focuses on diffusion mass transfer, detailing physical origins, mixture composition, Fick's Law of Diffusion, and mass diffusivity. It also explores absolute and diffusive species fluxes, evaporation in a column, conservation of species for a stationary medium, boundary conditions, discontinuous concentrations at interfaces, solubility of gases in liquids and solids, catalytic surface reactions, and transient diffusion. Chapter A provides thermophysical properties of matter, covering metallic solids, nonmetallic solids, common materials, gases, saturated fluids, liquid metals, binary diffusion coefficients, and the solubility of selected gases and solids. Additionally, it includes total emissivity for selected surfaces and solar radiative properties for various materials. Chapters B through F offer mathematical relations and functions, such as hyperbolic functions, Gaussian error function, Bessel functions, modified Bessel functions, thermal conditions associated with uniform energy generation, the Gauss-Seidel method, convection transfer equations, Newton's Second Law of Motion, conservation of energy, and species, boundary layer equations for turbulent flow. The document also provides conversion factors, physical constants, a student companion website, and supplemental material on thermophysical properties of matter. Method of Constructing a Flux Plot....Page 9944S.1.2 Determination of the Heat Transfer Rate....Page 9954S.1.3 The Conduction Shape Factor....Page 9964S.2 The Gauss-Seidel Method: Example of Usage....Page 998Problems....Page 10035S.1 Graphical Representation of One Dimensional, Transient Conduction in the Plane Wall, Long Cylinder, and Sphere....Page 10055S.2 Analytical Solution of Multidimensional Effects....Page 1009Problems....Page 10156S.1.1 Conservation of Mass....Page 10186S.1.2 Newton’s Second Law of Motion....Page 10196S.1.3 Conservation of Energy....Page 10226S.1.4 Conservation of Species....Page 1025Problems....Page 102911S.1 Log Mean Temperature Difference Method for Multipass and Cross-Flow Heat Exchangers....Page 103311S.2 Compact Heat Exchangers....Page 1037Problems....Page 1042Index....Page 1044
1930	https://en.wikipedia.org/wiki/Balmer_series	Jump to content Balmer series العربية বাংলা Беларуская Català Čeština Deutsch Español فارسی Français Gaeilge Galego 한국어 हिन्दी Hrvatski Bahasa Indonesia Italiano Lietuvių Bahasa Melayu Nederlands 日本語 Norsk bokmål Oʻzbekcha / ўзбекча Polski Português Русский Slovenčina Slovenščina Srpskohrvatski / српскохрватски Suomi Svenska தமிழ் ไทย Türkçe Українська 中文 Edit links From Wikipedia, the free encyclopedia Sorry to interrupt, but our fundraiser won't last long. This Tuesday, we ask you to join the 2% of readers who give. If everyone reading this right now gave just $2.75, we'd hit our goal quickly. $2.75 is all we ask. September 16: Knowledge is human. We're sorry we've asked you a few times recently, but it's Tuesday, September 16—please don't wait until tomorrow to help. We're happy you consult Wikipedia often. If just 2% of our most loyal readers gave $2.75 today, we'd reach our goal quickly. Most readers donate because Wikipedia is useful, others because they realize knowledge needs humans. If you agree, please give. Any contribution helps, whether it's $2.75 one time or monthly. 25 years ago Wikipedia was a dream. A dream built piece by piece by people, not machines. Now, with 65 million articles and 260,000 volunteers across the world, Wikipedia is proof that knowledge is human—a place of free, collaborative, and accessible knowledge. Your donation isn't just supporting a website; it's investing in the world's largest collaborative project of human intelligence—crafted by humans, for humans. Please join the 2% of readers who give what they can to help keep Wikipedia strong and growing. Thank you. Hydrogen spectral series The Balmer series, or Balmer lines in atomic physics, is one of a set of six named series describing the spectral line emissions of the hydrogen atom. The Balmer series is calculated using the Balmer formula, an empirical equation discovered by Johann Balmer in 1885. The visible spectrum of light from hydrogen displays four wavelengths, 410 nm, 434 nm, 486 nm, and 656 nm, that correspond to emissions of photons by electrons in excited states transitioning to the quantum level described by the principal quantum number n equals 2. There are several prominent ultraviolet Balmer lines with wavelengths shorter than 400 nm. The series continues with an infinite number of lines whose wavelengths asymptotically approach the limit of 364.5 nm in the ultraviolet. After Balmer's discovery, five other hydrogen spectral series were discovered, corresponding to electrons transitioning to values of n other than two. Overview [edit] The Balmer series is characterized by the electron transitioning from n ≥ 3 to n = 2, where n refers to the radial quantum number or principal quantum number of the electron. The transitions are named sequentially by Greek letter: n = 3 to n = 2 is called H-α, 4 to 2 is H-β, 5 to 2 is H-γ, and 6 to 2 is H-δ. As the first spectral lines associated with this series are located in the visible part of the electromagnetic spectrum, these lines are historically referred to as "H-alpha", "H-beta", "H-gamma", and so on, where H is the element hydrogen. : \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| Transition of n \| 3→2 \| 4→2 \| 5→2 \| 6→2 \| 7→2 \| 8→2 \| 9→2 \| ∞→2 \| \| Name \| H-α / Ba-α \| H-β / Ba-β \| H-γ / Ba-γ \| H-δ / Ba-δ \| H-ε / Ba-ε \| H-ζ / Ba-ζ \| H-η / Ba-η \| Balmer break \| \| Wavelength (nm, air) \| 656.279 \| 486.135 \| 434.0472 \| 410.1734 \| 397.0075 \| 388.9064 \| 383.5397 \| 364.5 \| \| Energy difference (eV) \| 1.89 \| 2.55 \| 2.86 \| 3.03 \| 3.13 \| 3.19 \| 3.23 \| 3.40 \| \| Color \| Red \| Cyan \| Blue \| Violet \| (Ultraviolet) \| (Ultraviolet) \| (Ultraviolet) \| (Ultraviolet) \| Although physicists were aware of atomic emissions before 1885, they lacked a tool to accurately predict where the spectral lines should appear. The Balmer equation predicts the four visible spectral lines of hydrogen with high accuracy. Balmer's equation inspired the Rydberg equation as a generalization of it, and this in turn led physicists to find the Lyman, Paschen, and Brackett series, which predicted other spectral lines of hydrogen found outside the visible spectrum. The red H-alpha spectral line of the Balmer series of atomic hydrogen, which is the transition from the shell n = 3 to the shell n = 2, is one of the conspicuous colours of the universe. It contributes a bright red line to the spectra of emission or ionisation nebula, like the Orion Nebula, which are often H II regions found in star forming regions. In true-colour pictures, these nebula have a reddish-pink colour from the combination of visible Balmer lines that hydrogen emits. Later, it was discovered that when the Balmer series lines of the hydrogen spectrum were examined at very high resolution, they were closely spaced doublets. This splitting is called fine structure. It was also found that excited electrons from shells with n greater than 6 could jump to the n = 2 shell, emitting shades of ultraviolet when doing so. Balmer's formula [edit] Balmer noticed that a single wavelength had a relation to every line in the hydrogen spectrum that was in the visible light region. That wavelength was 364.50682 nm. When any integer higher than 2 was squared and then divided by itself squared minus 4, then that number multiplied by 364.50682 nm (see equation below) gave the wavelength of another line in the hydrogen spectrum. By this formula, he was able to show that some measurements of lines made in his time by spectroscopy were slightly inaccurate, and his formula also predicted lines that had not yet been observed but were found later. His number also proved to be the limit of the series. The Balmer equation could be used to find the wavelength of the absorption/emission lines and was originally presented as follows (save for a notation change to give Balmer's constant as B): Where λ is the wavelength. B is a constant with the value of 3.6450682×10−7 m or 364.50682 nm. m is the initial state n is the final state In 1888 the physicist Johannes Rydberg generalized the Balmer equation for all transitions of hydrogen. The equation commonly used to calculate the Balmer series is a specific example of the Rydberg formula and follows as a simple reciprocal mathematical rearrangement of the formula above (conventionally using a notation of m for n as the single integral constant needed): where λ is the wavelength of the absorbed/emitted light and RH is the Rydberg constant for hydrogen. The Rydberg constant is seen to be equal to ⁠4/B⁠ in Balmer's formula, and this value, for an infinitely heavy nucleus, is ⁠4/3.6450682×10−7 m⁠ = 10973731.57 m−1. Role in astronomy [edit] The Balmer series is particularly useful in astronomy because the Balmer lines appear in numerous stellar objects due to the abundance of hydrogen in the universe, and therefore are commonly seen and relatively strong compared to lines from other elements. The first two Balmer lines correspond to the Fraunhofer lines C and F. The spectral classification of stars, which is primarily a determination of surface temperature, is based on the relative strength of spectral lines, and the Balmer series in particular is very important. Other characteristics of a star that can be determined by close analysis of its spectrum include surface gravity (related to physical size) and composition. Because the Balmer lines are commonly seen in the spectra of various objects, they are often used to determine radial velocities due to doppler shifting of the Balmer lines. This has important uses all over astronomy, from detecting binary stars, exoplanets, compact objects such as neutron stars and black holes (by the motion of hydrogen in accretion disks around them), identifying groups of objects with similar motions and presumably origins (moving groups, star clusters, galaxy clusters, and debris from collisions), determining distances (actually redshifts) of galaxies or quasars, and identifying unfamiliar objects by analysis of their spectrum. Balmer lines can appear as absorption or emission lines in a spectrum, depending on the nature of the object observed. In stars, the Balmer lines are usually seen in absorption, and they are "strongest" in stars with a surface temperature of about 10,000 kelvins (spectral type A). In the spectra of most spiral and irregular galaxies, active galactic nuclei, H II regions and planetary nebulae, the Balmer lines are emission lines. In stellar spectra, the H-epsilon line (transition 7→2, 397.007 nm) is often mixed in with another absorption line caused by ionized calcium known as "H" (the original designation given by Joseph von Fraunhofer). H-epsilon is separated by 0.16 nm from Ca II H at 396.847 nm, and cannot be resolved in low-resolution spectra. The H-zeta line (transition 8→2) is similarly mixed in with a neutral helium line seen in hot stars. See also [edit] Astronomical spectroscopy Bohr model Hydrogen spectral series Lyman series Rydberg formula Balmer jump Stellar classification References [edit] ^ Nave, C. R. (2006). "Hydrogen Spectrum". HyperPhysics. Georgia State University. Retrieved March 1, 2008. ^ Jump up to: a b c d e f g Kramida, A., Ralchenko, Yu., Reader, J., and NIST ASD Team (2019). NIST Atomic Spectra Database (ver. 5.7.1), [Online]. Available: [2020, April 11]. National Institute of Standards and Technology, Gaithersburg, MD. DOI: ^ "CODATA Recommended Values of the Fundamental Physical Constants: 2006" (PDF). Committee on Data for Science and Technology (CODATA). NIST. \| Hydrogen spectral series \| \| --- \| \| Lyman series Balmer series Paschen series Brackett series Pfund series Humphreys series Other series \| \| Retrieved from " Categories: Emission spectroscopy Hydrogen physics Hidden categories: Articles with short description Short description matches Wikidata Use mdy dates from September 2021 Use American English from March 2019 All Wikipedia articles written in American English
1931	https://journals.lww.com/inpc/fulltext/2021/08020/evaluation_and_comparison_of_the_porosities_in_the.6.aspx	International Journal of Preventive and Clinical Dental Research Log in or Register Get new issue alerts Get alerts;;) Submit a Manuscript Subscribe to eTOC;;) ### Secondary Logo Enter your Email address: Privacy Policy ### Journal Logo Articles Advanced Search Toggle navigation RegisterLogin Browsing History Home Current Issue Previous Issues For Authors Information for Authors Submit a Manuscript Published Ahead-of-Print Journal Info About the Journal Editorial Board Affiliated Society Advertising Subscriptions Reprints Rights and Permissions Articles Advanced Search Apr-Jun 2021 - Volume 8 - Issue 2 Previous Article Next Article Outline INTRODUCTION MATERIALS AND METHODS Porosity analysis RESULTS DISCUSSION CONCLUSION Financial support and sponsorship Conflicts of interest REFERENCES Images Slideshow Gallery Export PowerPoint file Download PDF EPUB Cite Copy Export to RIS Export to EndNote Share Email Facebook X LinkedIn Favorites Permissions More Cite Permissions Image Gallery Article as EPUB Export All Images to PowerPoint FileAdd to My Favorites Email to Colleague Colleague's E-mail is Invalid Your Name: Colleague's Email: Separate multiple e-mails with a (;). Message: Your message has been successfully sent to your colleague. Some error has occurred while processing your request. Please try after some time. Export to End Note Procite Reference Manager [x] Save my selection Original Article Evaluation and comparison of the porosities in the acrylic mandibular denture bases processed by two different curing cycles, using two different brands of commercially available denture base resins – An in vitro study Kumbham, Kiran Babu 1,; Prasad, B. Veerendra 2; Chakravarthy, A. Kalyan 3; Avinash, R.4; Amulya, Kukutla Sai 1; Pramod Kumar, Gattu Balram 1 Author Information 1 Postgraduate Student, Department of Oral and Maxillofacial Prosthodontics and Implantology, Meghna Institute of Dental Sciences, Nizamabad, Telangana, India 2 Reader, Department of Oral and Maxillofacial Prosthodontics and Implantology, Meghna Institute of Dental Sciences, Nizamabad, Telangana, India 3 Professor and HOD, Department of Oral and Maxillofacial Prosthodontics and Implantology, Meghna Institute of Dental Sciences, Nizamabad, Telangana, India 4 Senior Lecturer, Department of Oral and Maxillofacial Prosthodontics and Implantology, Meghna Institute of Dental Sciences, Nizamabad, Telangana, India Address for correspondence: Dr. Kiran Babu Kumbham, Postgraduate Student, Department of Prosthodontics, Meghna Institute of Dental Sciences, Nizamabad - 503 001, Telangana, India. E-mail: kumbham.kiran@gmail.com Received May 06, 2021 Accepted May 10, 2021 International Journal of Preventive and Clinical Dental Research 8(2):p 52-56, Apr–Jun 2021. \| DOI: 10.4103/ijpcdr.ijpcdr_16_21 Open Abstract Purpose: This study evaluated and compared the amount of porosities in two different acrylic materials processed by two different curing cycles. Materials and Methods: A total of 200 mandibular edentulous master casts were fabricated by using a standard silicone mold. These casts were divided into two groups with 100 samples for each group and each group was divided into two subgroups with 50 samples depending on the curing cycles. On the master cast, a wax pattern was fabricated with 2 mm of thickness and processed by conventional flasking method to obtain the mandibular master specimen. This master specimen along with the master cast was flasked and mold space was created for the placement of test specimens. All these samples were subjected to curing cycles along with the test specimens followed by porosity analysis of the specimens. Results: ANOVA analysis inferred that there is a statistically significant difference between the mean porosity values between the groups. Conclusion: Analysis of mean porosity of all groups reveals that heat-polymerized specimens of Dental Products of India (DPI) processed with a long curing cycle have lesser mean percent porosity values. INTRODUCTION Porosity in acrylic resin is a complex phenomenon of multifactorial origin which depends on various factors that include: air entrapped during mixing, monomer contraction during polymerization, monomer vaporization associated with exothermic reaction and the presence of residual monomer, insufficient mixing of polymer and monomer, processing temperatures higher than 74°C, packing of the mold, and inadequate compression on the flask.[1,2,3,4] Porosity is one of the most undesirable characteristics of acrylic denture bases. High internal stress develops due to severe porosity leading to distortion and warpage. A porous surface promotes colonization of the material by oral microorganisms such as Candida albicans and facilitates the deposition of calculus resulting in staining and impaired esthetics. In addition to maintaining minimal levels of residual monomer content, dentures must be free of porosity for which an appropriate curing cycle is to be followed for polymerization to prevent high exothermic reactions. This in vitro study aims at evaluation and comparison of porosities in acrylic mandibular denture bases processed by two different curing cycles using two different brands of commercially available denture base resins. MATERIALS AND METHODS A total of 200 mandibular edentulous master casts were fabricated by using standard silicone mold [Figures 1 and 2]. The study proposal was submitted for approval and clearance was obtained from the ethical committee of our institution. Figure 1: Standard mandibular edentulous silicone mold Figure 2: Master cast obtained from the silicone mold These casts were divided into two groups with 100 samples for each group and each group was divided into two subgroups with 50 samples depending on the curing cycles. On the master cast, a wax pattern was fabricated with 2 mm of thickness and processed by conventional flasking method to obtain the mandibular master specimen [Figure 3]. Figure 3: Wax pattern over master cast This master specimen along with the master cast was flasked in a brass flask for conventional water bath curing by two-pour technique. Once the plaster sets, lower and upper members of flasks were separated and mold space was created for the placement of test specimens. All these samples were subjected to curing cycles along with the test specimens. The samples were processed under two curing cycles which include a short curing cycle involving processing the resin at 74°C for approximately 2 h and increasing the temperature of the water bath to 100°C and processing for 1 h and long curing cycle involving processing in a 74°C water bath for 8 h and then increasing the temperature to 100°C for 1 h. After processing of the samples [Figure 4] with respective processing cycles, porosity analysis of the specimens was done. Figure 4: Acrylized samples Porosity analysis Digital analytical balance is a device that is used to weigh each specimen in air and then in water [Figure 5]. The absolute density of acrylic resin was used to calculate the percent mean porosity by use of the following equation: Figure 5: Digital analytical balance W a = g (d r − d a) (v sp − v ip)---1 W w = g (d r − d w) (v sp − v ip) + g (d a − d w) v ip---2 % of porosity = V ip/V sp × 100---3 Where W a = Specimen weight in air W w = Specimen weight in water g = Gravitational constant d r = Density of acrylic resin d a = Density of the air d w = Density of water V sp = Specimen volume V ip = Internal porosity volume In the first equation, specimen volume minus volume of internal porosity was determined using the following known values. Substituting the weight of the specimen in air (W a) calculated by digital analytical balance [Figure 6] and constant values of density of resin (d r), density of air (d a), and gravitational constant (g) in equation 1, i.e. W a = g (d r − d a) (v sp − v ip)---1, the values of (V sp − V ip) can be calculated. Figure 6: Weight of the specimen in air The values are as follows: d r = 1.198 6 ± 0.01 g/ml, d a = 1.23 kg/m 3, d w = 1000 kg/m 3, and g = 9.8066 m/sec 2. Having solved the first equation for specimen volume minus volume of internal porosity, this value was used in the second equation, along with Ww measured to determine the volume of internal porosity. The weight of the specimen in water that is calculated in digital analytical balance W w [Figure 7], values of d r, d w, and g, and obtained (v sp − v ip) value from equation 1, V ip value is calculated. W w = g (d r − d w) (v sp − v ip) + g (d a − d w) v ip---2 Figure 7: Weight of the specimen in water From equation 2, v ip value is obtained. With known values for specimen volume minus volume of internal porosity, the specimen volume was calculated. As the specimen volume and internal porosity volume were known, total percent porosity was calculated using the third equation. i.e. % of porosity = V ip/V sp × 100---3 From equation 1, (v sp − v ip) is calculated and from equation 2, v ip value is calculated, substituting the v ip value in the result of the first equation gives the V sp value. Thereafter, % of porosity can be calculated by the derived values of V ip and v sp. RESULTS Analysis of mean porosity of all groups reveals that heat-polymerized specimens of DPI processed with a long curing cycle have lesser mean percent porosity values [Table 1]. Table 1: Mean percent porosity values of denture bases of all groups ANOVA test was performed and the results show that the significance value (P) was 0.001 which was <0.05 (5% level of significance). Since P< 0.05, it is inferred that there was a statistically significant difference between the mean porosity values between the groups. Post hoc test was used in conjunction with ANOVA to determine which specific group was statistically different from the other group [Table 2]. Table 2: The mean porosity percentage values of the groups are summarized DISCUSSION Polymethyl methacrylate has a high vapor pressure and processing temperatures above 100.8°C would result in vaporization of monomer leading to porosity in the processed acrylic denture bases. The greater the amount of liquid in the mass, the greater is the tendency toward porosity in the finished product when a fixed curing practice is followed. While packing the resin in the mold space if too small amount of resin is present, there will be extreme porosity in the structure distributed well throughout the mass of resin where the deficiency existed.[4,6] Taylor stated that porosity resulted when an initial processing temperature greater than 74°C was used due to a temperature rise within the specimens to above the boiling point of monomer. Incomplete packing of the mold and inadequate compression on the flask were also reported by Taylor as a cause of porosity. Al Doori, Smith et al. stated that when dentures are polymerized using rapid curing systems, the possibility of porosity in thick sections is always a major consideration. The amount of residual monomer was one of the principal factors affecting the properties of denture base materials produced under different curing cycles. Harman concluded that the degree of polymerization attained is a function of time and the temperature within the resin, it is shown that an effective curing cycle must take into account the mass and form of the sample, the ambient temperature, and the rate of temperature rise within the sample. The results of this study suggest that the best combination of material and polymerization methods remains conventional resin (DPI) polymerized in a water bath with a long curing cycle. The null hypothesis of this study was that there was no statistically significant difference in the mean percent porosity values of heat-polymerized acrylic resin of two different brands processed under two curing cycles. The results obtained from the study rejected the null hypothesis. CONCLUSION The highest mean porosity values were observed in the Acralyn H samples processed under a short curing cycle. There was no statistical difference in mean percent porosity values of specimens fabricated with DPI processed under short curing cycle and specimens of Acralyn H processed under short curing cycle. Significant statistical difference was observed with the specimens fabricated from DPI with long curing cycle and DPI with short curing cycle, DPI with long curing cycle and Acralyn H with short curing cycle, Acralyn H with long curing cycle and Acralyn H with short curing cycle, and Acralyn H with long curing cycle and DPI with short curing cycle. Within the limitations of this study, specimens fabricated with DPI under a long curing cycle were proven to have the least porosity percentage values. Financial support and sponsorship Nil. Conflicts of interest There are no conflicts of interest. REFERENCES Yannikakis S, Zissis A, Polyzois G, Andreopoulos A. 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Effects of time and temperature on polymerization of a methacrylate resin denture base J Am Dent Assoc. 1949;38:188–203 Cited Here \| PubMed \| CrossRef \| Google Scholar View full references list Keywords: Denture bases; digital analytical balance; porosity © 2021 International Journal of Preventive and Clinical Dental Research \| Published by Wolters Kluwer – Medknow View full article text Source Evaluation and comparison of the porosities in the acrylic mandibular denture bases processed by two different curing cycles, using two different brands of commercially available denture... International Journal of Preventive and Clinical Dental Research8(2):52-56, Apr-Jun 2021. Full-Size Email Favorites Export View in Gallery Email to Colleague Colleague's E-mail is Invalid Your Name: Colleague's Email: Separate multiple e-mails with a (;). Message: Your message has been successfully sent to your colleague. Some error has occurred while processing your request. Please try after some time. 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1932	https://byjus.com/maths/angle-between-line-plane/	Let us say that a line is inclined on a plane. A normal to the plane is drawn from the point where the line touches the plane. This normal forms an angle with the line. In analytic geometry, the angle between the line and the plane is equivalent to the complement of the angle between the line and the normal. In this section, we will discuss this concept in detail. Angle between a Line and a Plane A line is inclined at Φ to a plane. The vector equation of the line is given by and the vector equation of the plane can be given by Let θ be the angle between the line and the normal to the plane. Its value can be given by the following equation: Φ is the angle between the line and the plane which is the complement of θ or 90 – θ. We know that cos θ is equal to sin (90 – θ). So Φ can be given by: sin (90 – θ) = cos θ or Example Let us take up an example to understand the equations better. Problem: A line has an equation [x/6] = [(y+32)/2 ] = [(z-2)/3] . The equation of a plane is 3x + 4y – 12z = 7. Find the angle between them. Solution: Let θ be the angle between the line and the normal to the plane. In the vector form, the equations can be written as: The equation of the plane in the vector form can be given by: Finding the value of the Φ between the line and the plane: The value of Φ can be found by To solve more examples and to watch video lectures on this topic, download BYJU’S The Learning App. Quiz on Angle between line and plane Q5 Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Congrats! Visit BYJU’S for all Maths related queries and study materials Your result is as below 0 out of 0 arewrong 0 out of 0 are correct 0 out of 0 are Unattempted Login To View Results Did not receive OTP? Request OTP on Login To View Results Comments Leave a Comment Cancel reply Register with BYJU'S & Download Free PDFs
1933	https://calcworkshop.com/sequences-series/nth-term-test/	Nth Term Test for Divergence 3 Helpful Examples! // Last Updated: - Watch Video // In our previous lesson, Intro To Sequences and Series, we learned important terms such as convergence, divergence, and sequence and series. We also learned how to: Generate a sequence Simplify factorials Determine convergence for an infinite sequence We also learned our Important Acronym and Mnemonic Device for the Nine Infinite Series Tests for determining convergence/divergence for any infinite series. In this lecture we’ll explore the first of the 9 infinite series tests – The Nth Term Test, which is also called the Divergence Test. Test for Divergence This test, according to Wikipedia, is one of the easiest tests to apply; hence it is the first “test” we check when trying to determine whether a series converges or diverges. What is important to point out is that there is an nth-term test for sequences and an nth-term test for series. The steps are identical, but the outcomes are different! So it is imperative that you make sure that you are aware of whether or not you are dealing with a sequence or a series, because your conclusion of convergence or divergence depends on it. But don’t worry; it’s easy to sort out, and easy to apply! In fact, this is probably everyone’s favorite test because it’s so easy to use. Nth Term Test Video Volume 90% Closed Captions Playback Rates Rewind 10 Seconds Next Up Live 00:00 00:00 00:00 Chromecast Closed Captions Settings Fullscreen Nth Term Test Overview Example 1 Example 2 Example 3 Get more examples and over 450 HD videos with your subscription Monthly and Yearly Plans Available Get My Subscription Now Still wondering if CalcWorkshop is right for you? Take a Tour and find out how a membership can take the struggle out of learning math. 4.9 / 5 1,194 Reviews slide 5 to 8 of 50 Perry R.04-15-25 - GA, United States Jenn thoroughly explains every move that she makes! 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Lindsay06-03-25 - FL, United States I have only been using this for about a week, and I have already greatly benefited from this resource. The videos are clear and straightforward, with just the right amount of background information and explanation. I love that I am able to access any course that I need in order to fill in any gaps in skill. I am not just granted access to one particular course. I also love that when I contacted Jenn about a question that I had, she reached out with clarity and specific videos that would help my situation. This was an amazing help!! I would not have known where to find them on my own, and they really helped me to understand concepts that I needed.I highly recommend this resource:) Joann L.06-03-25 - TX, United States I am very satisfied with format Calcworkshop has to offer. I highly recommend this site. After six months of reviewing sites and texts I decided to try Calcworkshop.I stopped searching because it met all my needs.I will take my grandchildren from PreAgebra to Calculus 1 through Calcworkshop. Joann from Salado, TX Andrea A.05-30-25 - United States Thanks to all of the support offered on your website, I as able to save a lot of tears. Truly one of the best sources for calculus aid. I ended up finished Calc 1-3 with A's. I thank you and your team for the amazing website you have put together! A Reviewer05-28-25 - United States This website has helped me understand the concepts better of step by step. Grace05-24-25 - MD, United States As a college student and math major, CalcWorkshop was the perfect resource for me as it cleared up what I didn't understand in lectures or even taught me the whole topic! Jenn really does go through every step and understands where students may be confused during lessons. I totally would recommend CalcWorkshop to ANYONE, no matter what math class their taking. Jenna05-24-25 - KS, United States I abolsutely love Jenn! She made calc easier to understand and I genuinely enjoyed her videos. Without Jenn, I never would have made it through calc II. I wish I had her for Calc I! CJ05-21-25 - NC, United States I was way out of my element in Calc II in college, and my professor was no help, and I was unsuccessfully trying to teach myself calculus from a textbook. I had failed my first test and was looking for tutors, but all of the ones I tried just gave me help on homework and did not teach the content to me. I fully credit CalcWorkshop for being able to pass Calc II and not having to retake it. The way things are explained is so clear and simple, and can even help someone who is self-proclaimed "bad at math". I would recommend this site to anyone struggling in calc!! Maya05-19-25 - United States I just passed the Calc 2 course at my local collage with an A as a junior in high school and couldn't have done it without Calcworkshop! Jenn gives tons of great examples and explains things so clearly and calmly without skipping over the "little unimportant steps" that would continuously trip me up while trying to learn the new material. The website is also very well designed and makes it very easy to find exactly what you are looking for. 10/10 RetiredMD05-15-25 - FL, United States My overall experience has been excellent. I am a retired physician now teaching. Mathematics has always been important to me. I have needed review from time to time and to have this as a source was extremely helpful. Jenn shows you the thinking done in solving problems and not just the mechanics. I most likely will renew in the future but I am pleased with the quality. WeezyMathGirl05-15-25 - GA, United States These videos are amazing. She explains everything so well and this was a lifesaver for Calculus BC for my daughter! Brenda05-12-25 - KY, United States absolutely amazing! Jose C.05-12-25 - FL, United States I am very grateful to have found this site and above all I am grateful for Jen's work. It is notorious the passion for teaching and doing good to the community that continues with the hope and dream of dedicating themselves to mathematics despite the fact that the educational system or teachers that we may encounter along the way are not the most cooperative. As a personal experience, I had a terrible year academically in terms of mathematics, actually no one in my class understood the teacher, I thought that this was no longer my thing, but this online course made me regain confidence in me and knowing that from now on I will have a tool that can accompany me if I have any doubt is a great relief. John05-10-25 - NM, United States Being in my thirties and returning to school to pursue a degree in Mechanical Engineering and hadn't taken a math class in over 25 years, Calc Workshop was the only reason I survived pre calc/trig, calc 1, and calc 2. Jenn breaks down every step of every type of problem into a logical and easy to remember process. This subscription has been worth every penny and I have recommended it to everyone I know taking math. I will be re-subscribing for calc 3 and differential equations next year. Thank you Jenn for such an amazing tool and helping me achieve success in these challenging courses, I couldn't have done it without you! Deathbyintegrals05-09-25 - United States Overall it served my intended purpose which was to get me through double, triple, and line integrals. I like how the site is organized and I absolutely appreciate the clear and concise video lessons. Should I need a math refresher or assistance in a future course I will absolutely be back. Dillan05-06-25 - UT, United States Her videos are very well made, you can tell she put a lot of work into how she introduces and walks you through all the material. I used her videos to help me with calculus 3. I just canceled my subscription because I wasn't in need of the information anymore and 30$ is a lot to pay if your not gonna watch the videos. I will probably re subscribe for my future math classes. Her videos are super good!Her website is well organized and easy to get around. I did not use any of her other features on her website I just used her posted videos from the website and I found that plenty sufficient for me to learn the material. Overall, I felt like 30$ a month was a little expensive but this is such a great product that I would 100% subscribe again! Amanda M.05-02-25 - AZ, United States Calcworkshop is an outstanding resource for anyone looking to strengthen their understanding of calculus. Jen's clear and approachable teaching style made even the most complex topics feel manageable. As a math teacher, I found her step-by-step explanations incredibly helpful in breaking down difficult concepts for my own students. The scaffolding she provides is thoughtfully designed, building confidence and deepening comprehension at every stage. Whether you're looking for a refresher or a fresh perspective on solving problems, I highly recommend Calcworkshop as a go-to tool for mastering calculus. Perry R.04-15-25 - GA, United States Jenn thoroughly explains every move that she makes! She' s easy to understand and responds when emailed with questions! Jenn is saving my Differential Equations grade! William M.04-15-25 - GA, United States Calcworkshop was/is an excellent resource for anyone interested in or currently learning math. I started my Math degree at 26, spending absolutely 0 time doing math between 18 and then. I did not even know how to FOIL. The subscription got me up to speed pretty quick and then I used it for a few other courses. Couldn't recommend it enough Joey M.04-14-25 - AZ, United States They make the complex simple. They provide a systematic approach to ensure your success. Patrick04-09-25 - NJ, United States I went through the discrete math course prior to taking it at a university. Jen made difficult concepts easy to understand, leaving me well prepared for my Discrete college course. NB08-18-25 - ID, United States Very good courses with detailed videos and practice problems. It was helpful to have extra resources in calculus. Sabrina08-18-25 - CA, United States This course help prepare me for Calculus 2! Daily08-18-25 - OR, United States Calcworkshop.com for the win. I used this as a supplement for college level Pre-Calculus II, Calculus I, and Calculus II courses. The Calculus courses were during an accelerated 4-week summer term, which made them that much more difficult. Jenn's teaching ability is amazing! I was regularly able to impress my professors with my grasp of the content. They regularly commented on how much I was improving. For every topic that was covered in each class, Calcworkshop.com has the same content in an easy to digest video. U-Sub, Trig Sub, Diff-Equations, the list goes on. Best money I've spent in a very long time. Three classes supplemented with Calcworkshop.com so far, and three A letter grades! Austin08-08-25 - AZ, United States I was lost in an AP Calculus class and Jenn's explanations of things like limits and dedicates helped me out tremendously! The break downs of every lesson made it seem like I was learning basic algebra again, and allowed me to do so much better in Calculus. See more reviews on Shopper Approved
1934	https://stats.stackexchange.com/questions/37870/probability-of-visiting-all-other-states-before-return	self study - Probability of visiting all other states before return - Cross Validated Join Cross Validated By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Probability of visiting all other states before return Ask Question Asked 13 years ago Modified13 years ago Viewed 3k times This question shows research effort; it is useful and clear 6 Save this question. Show activity on this post. Question (a) Random walk on a clock. Consider the numbers 1,2,…,12 1,2,…,12 written around a clock. Consider a Markov chain that jumps with equal probability to one of the two adjacent numbers each step. What is the expected number of steps that X n X n will will take to return to its starting position? (My Work) From a result in class, we know that a doubly stochastic transition matrix p p for a Markov Chain with 12 12 states has the uniform distribution π(x)=1/12 π(x)=1/12 for all x x as a stationary distribution. We also know that if the chain is irreducible and there exists a stationary distribution (both hypotheses are satisfied) π(y)=1 E y T y π(y)=1 E y T y, so the expected time of first return (E y T y E y T y) is 12. Question (b) What is the probability that X n X n will visit all of the other states before returning to its starting position? My Question I am not sure how to compute this probability. My first intuition was to consider P(T y>12)P(T y>12), but further considering the problem, this seems incorrect because the chain does not have to visit all states before move 12. self-study stochastic-processes markov-process Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications edited Sep 24, 2012 at 0:50 StasK 32.8k 2 2 gold badges 104 104 silver badges 198 198 bronze badges asked Sep 24, 2012 at 0:08 ModeratModerat 859 1 1 gold badge 7 7 silver badges 20 20 bronze badges 6 1 I don't have a full solution, but my guess is that solving this would involving inverting the problem. Call P∗n P∗n the probability that it will visit the same state in exactly n n stops. P∗2 P∗2 = 1/12 P∗3 P∗3 = 11/122/12 = 22/144 = 11/72 Peter Flom –Peter Flom 2012-09-24 00:27:41 +00:00 Commented Sep 24, 2012 at 0:27 Oh, and then, sum from P∗2 P∗2 to P∗11 P∗11 and subtract that from 1 to get your answer.Peter Flom –Peter Flom 2012-09-24 00:35:36 +00:00 Commented Sep 24, 2012 at 0:35 This looks equivalent to computing 1−P(T y>12)1−P(T y>12) to me? How is it different?Moderat –Moderat 2012-09-24 00:56:42 +00:00 Commented Sep 24, 2012 at 0:56 Hmm. Maybe it isn't different. But then why does it assume that the chain has to visit all states before move 12? I don't see that.Peter Flom –Peter Flom 2012-09-24 01:03:00 +00:00 Commented Sep 24, 2012 at 1:03 I think that is what you assume when you "sum from 2 to 11"? Why P∗n P∗n for n≥12 n≥12 not accounted for? These are the probabilities, now I think I may need to take expectations...Moderat –Moderat 2012-09-24 01:05:04 +00:00 Commented Sep 24, 2012 at 1:05 \|Show 1 more comment 1 Answer 1 Sorted by: Reset to default This answer is useful 6 Save this answer. Show activity on this post. This looks like homework so I'm trying to give a hint, not a solution. For part (b), you definitely want to use the structure of the graph. Without loss of generality suppose you start at 12 12 and your first step is to 1 1. Can you say what the probability is that you hit 11 11 before you hit 12 12? Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Sep 24, 2012 at 1:22 Douglas ZareDouglas Zare 10.8k 2 2 gold badges 43 43 silver badges 47 47 bronze badges 6 Would this be the probability of going from 11 to 12, times the probability of being at 11? I am unsure how to rule out being at 12...Moderat –Moderat 2012-09-24 01:51:22 +00:00 Commented Sep 24, 2012 at 1:51 I'm not sure what you mean about the probability of being at 11 11. If you are at 1 1 after the first step, then with probability 1 1 you will get to 11 11 without first returning 12 12, or else you will get to 12 12 without first visiting 11 11. If you go from 1 1 to 11 11, a priori that could be two net counterclockwise steps or ten net clockwise steps, but if you know that you don't visit 12 12 on the way, you can rule out one of these.Douglas Zare –Douglas Zare 2012-09-24 03:46:00 +00:00 Commented Sep 24, 2012 at 3:46 So we went over Exit Distributions today, and for this problem, would I consider that the probability of going from 1 1 to 11 11 before 12 12 equal to g(x)=1+∑y p(x,y)g(y)g(x)=1+∑y p(x,y)g(y) where g(x)g(x) is the expected time to complete the circuit when you are at x x?Moderat –Moderat 2012-09-26 02:15:26 +00:00 Commented Sep 26, 2012 at 2:15 @jmi4: I'm not sure what you mean by the time to complete the circuit.Douglas Zare –Douglas Zare 2012-09-26 02:50:02 +00:00 Commented Sep 26, 2012 at 2:50 1 Once you get to 1 1, think of 11 11 and 12 12 as absorbing states.Douglas Zare –Douglas Zare 2012-09-26 19:51:53 +00:00 Commented Sep 26, 2012 at 19:51 \|Show 1 more comment Your Answer Thanks for contributing an answer to Cross Validated! 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1935	https://www.youtube.com/watch?v=h_-i2iTNY_s	How to Use the Angle Bisector Theorem and Stewarts' Theorem with an Example \| Minute Math #geometry Minute Math 84900 subscribers 13 likes Description 771 views Posted: 15 Aug 2022 How do you use the Angle Bisector Theorem? How do you use Stewarts' Theorem? Here is a video example of the Angle Bisector Theorem and an example of Stewarts' Theorem. We use the Angle Bisector Theorem and Stewarts' Theorem to solve for x. #geometry minutemath #mathhelp #onlinemathvideos #math #mathvideos #freemathvideos #lightboard #maths #freemathvideos #anglebisectortheorem #anglebisector #stewartstheorem Visit our website Math Help and Math Merch: Time Stamps: 00:00 - The Problem 01:05 - The Steps 04:34 - The Answer Follow us for... Tweets: Instagram: TikTok: Facebook: Personal Instagram: Business Instagram: Amazon Store: 2 comments Transcript: The Problem [Music] hi i'm sean gann and this is minute math and we're going to use the angle bisector and stewart's theorem here within this problem to solve for x okay our problem is this we have this triangle okay we have side lengths of 8 10 and 12 for the triangle we have a length cd which we have is x and we need to figure that out now c d is an angle bisector meaning these two angles a c d and d c b are congruent c d cuts the big angle directly in half with this information using these two theorems we're going to show you how well to find x here and good explanation of the angle bisector and stewart's theorem here all right but like always here we encourage you to pause this video to see if you can solve for x first look up these theorems and see how it can help you but we're going to go with an explanation of them here and solve for x okay The Steps so the first thing i'll do here with this problem here i'm going to set a value here from a d and d to b okay we know the 10 and 8 and the overall here is 12. but these individual ones we know down here that a d plus db equals a 12. so if i call the length of a d to b well w and then the length from db would have to be the length of 12 minus that length of w whatever that length is here all right now here's we're going to apply the angle bisector theorem i'm going to kind of break this down into four segments this one a and then b here then c and d with the angle bisector theorem tells us that a over b is equal to c over d in this order these two side lengths here notice it's broken down by our angle bisector it works right there okay so now with this information we can set up a proportion a over b which is 10 over 8 is equal to c over d w over well 12 minus a w now from here it'd be nice if we can figure out well what w is and then also then get what d b is with it all right so from here let's go do that we cross multiply and so we cross multiply up there and here and we get a 1 20 minus 10 w on the left is equal to an 8 w on the right we add a 10 w to both sides and we have 120 equals an 18 w divide both sides by 18. and we got a w value i'll put it right here w is equal to and i wrote down as a fraction here is a 20 over a three so we have length of here of 20 over three okay and then if we want to find well db that length we have 12 minus 820 over 3 which is the length of db and i did that work for you right here and that is a 16 over 3 when it's simplified so here is a 16 over 3. so great that gives the information here we have all side lengths and now we can apply what's called stewart's theorem okay and stewart's theorem is this with the same a b c d value and i'm using x here for this angle bisecting line here stewart's theorem says that x squared equals an a times a b minus a c times a d all right so now we have all the information that we need that's what we're going for here we can plug it in so again x squared we don't know what the x is but we know the other values a was 10 times b is 8 minus a c 20 over 3 times a d which is 16 over 3. from all that that simplifies x squared equals that's me 80 minus this number here i did it all for you or 80 here minus a 320 over 9. look at my notes here but x squared to minus one fraction is 400 over a nine so we square root both sides The Answer we know it's a plus or minus but you can assume the negative number here can't be applied so we're only deal with positive so x equals numerator square root we have 20 over a three square root of nine is a three and so our x value here x equals a 20 over a three and that was the length of c d well hopefully you learned the angle bisector theorem as well as stewart's theorem and applying both of this problem again like most theorems more theorems you know more tools in your arsenal to solve a problem like this so 20 over 3 was the amp answer here and i hoped you liked this video and if you did please consider subscribing to this youtube channel and sharing this video uh this helps us make more of these fun math videos for you and for everyone else so as always here thanks for watching [Music]
1936	https://www.scribd.com/document/475140596/Sample-Lesson-Plan-in-Linear-Programming	Sample Lesson Plan in Linear Programming \| PDF \| Mathematical Optimization \| Linear Programming Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 50%(2)50% found this document useful (2 votes) 2K views 3 pages Sample Lesson Plan in Linear Programming The lesson plan aims to teach students about linear programming through graphing systems of linear inequalities, solving optimization problems, and real-world applications. Students will wor… Full description Uploaded by otok jacildo AI-enhanced description Go to previous items Go to next items Download Save Save Sample Lesson Plan in Linear Programming For Later Share 50%50% found this document useful, undefined 50%, undefined Print Embed Translate Ask AI Report Download Save Sample Lesson Plan in Linear Programming For Later You are on page 1/ 3 Search Fullscreen Lesson Plan in Linear Algebra (Linear Programming) I. Objectves At the end of the lesson, must be able to;1. graph systems of linear inequalies,2. solve real life problems involving maximizaon, and 3. appreciate how linear programming being used in dierent elds such as manufacturing, business operaon, etc. II. Subjec Maer A. Topic: Linear Programming B. Mahematcal Conceps: Graphing s ystems of linear inequa lies 2. Maximizaon C. Skills: Crical Thinking, Problem solving, Creavity and Logical reasoning D. Maerials: laptop, projector, white board and white board markers E. Value Focused: Cooperaon and Teamwork F. Reference: Brilliant.org. 2020. Linear Progr amming \| Bri lliant Math &Science Wiki . [online] Available at: [Accessed 8 May 2020]. Faculty.washington.edu. 2020. [online] Available at: [Accessed 8 May 2020]. III. Procedure A. Preliminary Actvites 1. Inroducton The teacher will select 1 student to lead the prayer and then greet the students a good morning. Review/Motvaton The teacher will give a movaon called “Equal or Not?”. Then, the teacher will review the previous topic. B. Lesson Proper 1. Actviy: Graph me if you can (Graphing systems of linear inequalies) a. The class will be divided into 4 groups. b. Each will be given a marker, cartolina and problem set (to graph the systems of linear inequalies). c. Each group must select their reporter to report their answers. d. The acvity is good for 10 minutes. e. The acvity will be then checked by the class. Analysis: The teacher will introduce the new topic Linear Programming. a. Give the denion of Li near Programming and its uses. Linear Programming is a mathemacal technique for nding opmal soluons to problems that can be expressed using linear equaons and inequalies. adDownload to read ad-free b. Discuss the important terms ( Consrains, feasible region, and objectve functon ), give their denions and give examples. Consrains is an inequality that denes how the values of the variables in a problem are limited. Feasible region – is the region that is bound by the system of constraints Objectve functon is a funcon that denes some quanty that should be minimized or maximized. c. Introduce the maximizaon, its concepts and how to solve problems involving maximizaon. d. Give examples on real li fe problems involving maximizaon. Absracton: The teacher will discuss further the given examples to come up with a generalizaon. At this moment, the students will be able to generalize the idea of linear programming with the help of guide quesons given by the teacher. Applicaton: The teacher will give more real-life problems for the students to solve and to apply their understanding of the topic. C. Evaluaton: The teacher will test the understanding of the students. This will be then checked and recorded.Consider a chocolate manufacturing company that produces only two types of chocolate – A and B. Both the chocolates require Milk and Choco on ly. To manufacture each unit of A and B, the following quanes are required:  Each unit of A requires 1 unit of Milk and 3 units of Choco  Each unit of B requires 1 unit of Milk and 2 units of Choco The company kitchen has a total of 5 units of Milk and 12 units of Choco. On each sale, the company makes a prot of  Rs 6 per unit A sold  Rs 5 per unit B sold.Now, the company wishes to maximize its prot. How many units of A and B should it produce respecvely? D. Assignmen: A store sells two types of toys, A and B. The store owner pays $4 and $7 for each one unit of toy A and B respecvely. One unit of toys A yields a prot of $2 while a unit of toys B yields a prot of $3. The store owner esmates that no more than 20 toys will be sold every month and he does not plan to invest more than $80 in inventory of these toys. How many units of each type of toys should be stocked in order to maximize his monthly total prot?Prepared by: Lenard J. Jacildo adDownload to read ad-free Appendix 1. (Review/Motvaton)Equal or no? There are 40 feet but only 15 heads (Chicken or Pig) in the farm. The farmer owns 10 Chickens and 5 Pigs. Equal or not?2. There are 35 wheels but only 13 vehicles (Tricycle or Motorcycle) in the shop. The seller has 8 Tricycles and 5 Motorcycles. Equal or not?3. There are 23 heads but only 38 horns (Bull or Indian Rhino) in the safari. In the wild lives 15 Bulls and 8 Indian Rhinos. Equal or not?4. There are 70 legs but only 10 insects (Cockroach or Spider) in the house. There are 6 Spiders and 4 Cockroaches in the house. Equal or not? Appendix 2. (Actviy)Graph me if you can Group 1 19 x + 20 y ≥ 17 12 x + 19 y ≤ 15 10 x + 16 y 14 Group 2 13 x + 9 y < 1 11 x + 20 y≥ 16 6 x + 14 y ≥ 1 Group 3 6 x + 14 y ≥ 1 6 x + 20 y≤ 16 6 x + 4 y≥ 13 Group 4 6 x + 4 y≤ 13 8 x + 8 y 13 13 x + 12 y < 8 Read this document in other languages język polski Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Duda Problemsolutions 100% (6) Duda Problemsolutions 446 pages List of NBFC-Micro Finance Institutions (NBFC-MFIs) 0% (1) List of NBFC-Micro Finance Institutions (NBFC-MFIs) 7 pages Grade-8 PPT Squares and Square Roots No ratings yet Grade-8 PPT Squares and Square Roots 35 pages On Story - Screenwriters and Their Craft (PDFDrive) 100% (1) On Story - Screenwriters and Their Craft (PDFDrive) 197 pages EC7 Pile Design Seminar Overview 100% (2) EC7 Pile Design Seminar Overview 66 pages 04 Truss - 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1937	https://www.youtube.com/watch?v=m7N6yykWirY	NCERT Short Notes \| Class 11 Chapter 15 \| Waves Physics Galaxy 1330000 subscribers 1035 likes Description 39782 views Posted: 5 Jan 2023 Wave motion or waves is chapter 15 of NCERT text book of class 11 which consists of complete explanation and understanding of string and sound waves. This chapter is focused on mechanical waves and applications of simple harmonic motion which causes energy and disturbance propagate in a medium. Ashish Arora sir has explained the whole chapter of waves in this one shot video for making of short notes of the NCERT book quickly. Timestamps of the topics covered in this video are listed below: 00:00 Introduction to NCERT Short Notes 01:35 Introduction to Waves 04:24 Wave Pulse 07:45 Simple Harmonic Wave 10:55 Amplitude & Phase 12:45 Wavelength & Angular Wave Number 13:38 Time period, Frequency & Angular Frequency 14:23 Speed of a Travelling Wave 19:14 Principle of Superposition of Waves 27:21 Reflection of Waves 32:46 Standing Waves 36:19 Standing Waves in a Clamped String 40:36 Normal Modes in Standing Wave 43:13 Resonance 44:31 Normal Modes in Air Column 50:24 Beats 54:52 Doppler's Effect waves #wavemotion #resonance #dopplereffect #stationarywave #standingwave #wavepropagation #ncertshortnotes #physicsgalaxy #ashisharoraofficial #ashisharora Cover other chapters for NCERT Short notes: NCERT Short Notes \| Class 11 Chapter 3 \| Motion in Straight Line - NCERT Short Notes \| Class 11 Chapter 5 \| Laws of Motion - NCERT Short Notes \| Class 11 Chapter 6 \| Work, Energy & Power - NCERT Short Notes \| Class 11 Chapter 7 \| Rigid Body Dynamics - NCERT Short Notes \| Class 11 Chapter 8 \| Gravitation - NCERT Short Notes \| Class 11 Chapter 9 \| Mechanical Properties of Solid - NCERT Short Notes \| Class 11 Chapter 10 \| Mechanical Properties of Fluids - NCERT Short Notes \| Class 11 Chapter 11 \| Thermal Properties of Matter - NCERT Short Notes \| Class 11 Chapter 12 \| Thermodynamics - NCERT Short Notes \| Class 11 Chapter 13 \| Kinetic Theory of Gases - NCERT Short Notes \| Class 11 Chapter 14 \| Simple Harmonic Motion - NCERT Short Notes \| Class 11 Chapter 15 \| Waves - Download Academic resources for JEE Main, JEE Advanced & NEET, Join PG Telegram : Class 11 Revision Checklist for JEE Main & NEET : Class 12 Revision Checklist for JEE Main & NEET: Booster checklist playlist JEE Advanced Revision : Link to Playlist of all NCERT Short Notes Physics videos : Buy Physics Galaxy series of books : For Best Online Courses for JEE & NEET. Download Physics Galaxy Mobile app : Boost your JEE Advanced preparation with #1 Bestseller book of 700+ advance illustrations. Check at - 1 Bestseller book for NEET PYQs with comprehensive solutions. Check at- 1 Bestseller book for JEE Main PYQs with comprehensive solutions. Extremely useful book for NEET aspirants. Check at : 70 comments Transcript:
1938	https://www.khanacademy.org/science/hs-chemistry/x2613d8165d88df5e:chemical-bonding/x2613d8165d88df5e:ionic-nomenclature/e/apply-polyatomic-ions	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
1939	https://personalpages.manchester.ac.uk/staff/gabor.megyesi/teaching/math32062/algebra.pdf	Concepts from algebra All rings are understood to be commutative. Deﬁnition. Let R be a ring and let S be a subset of R. The ideal generated by S is the set ⟨S⟩= { n X i=1 risi \| n ≥0, ri ∈R, si ∈S, 1 ≤i ≤n}. Deﬁnition. An ideal in a ring is called ﬁnitely generated if and only if it can be generated by a ﬁnite set. An ideal is called principal if and only if it can be generated by a single element. Deﬁnition. A ring R is Noetherian if and only if all of its ideals are ﬁnitely gener-ated. (Equivalently, R is Noetherian if and only if for any increasing sequence of ideals I1 ⊆I2 ⊆I3 ⊆. . . ⊆In ⊆In+1 . . . there exists an N such that In = IN for all n ≥N.) Theorem (Hilbert Basis Theorem) If K is a ﬁeld, then the polynomial ring K[x1, x2, . . . , xn] is Noetherian for any n ≥0. Deﬁnition. Let I1, I2, . . . , In be ideals in a ring R. The product of I1, I2, . . . , In, denoted by I1I2 · · · In, is the ideal generated by all the products i1i2 · · · in, where ij ∈Ij, 1 ≤j ≤n. (Warning: The products themselves usually do not form an ideal.) Instead of all the products, it is suﬃcient to choose a generating set for each ideal. The products of the generators generate the product ideal. Deﬁnition. Let Iα, (α ∈A), be ideals in a ring R. The sum of the ideals Iα, (α ∈A), is X α∈A Iα = { X α∈A fα \| fα ∈Iα, fα = 0 for all but ﬁnitely many α}. P α∈A Iα can also be deﬁned as the ideal generated by S α∈A Iα. (Warning: The union itself is usually not an ideal.) Instead of the whole ideals, it is suﬃcient to take a generating set for each ideal, the union of these generating sets also generates P α∈A Iα. Deﬁnition. A ﬁeld K is algebraically closed if and only if every polynomial of degree at least 1 in K[x] has a root in K. For any ﬁeld K, there exists a minimal algebraically closed ﬁeld containing it, which is unique up to isomorphism, this ﬁeld is called the algebraic closure of K. 1 Deﬁnition. Let R be a ring. Let I ◁R be an ideal. The radical of I, denoted by √ I or rad I, is √ I = {x ∈R \| ∃n such that xn ∈I}. I is called a radical ideal if and only if √ I = I. Proposition. √ I is also an ideal. Deﬁnition. An ideal I in a ring R is maximal if and only if I ̸= R and I is not contained in any other proper ideal of R. Theorem. I ◁R is maximal if and only if R/I is a ﬁeld. Deﬁnition. An ideal I in a ring R is a prime ideal if and only if for all a, b ∈R, ab ∈I implies a ∈I or b ∈I. Deﬁnition. A ring R is called an integral domain if and only if for all a, b ∈R, ab ∈0 implies a = 0 or b = 0. Theorem. I ◁R is prime if and only if R/I is an integral domain. Deﬁnition. An integral domain is a unique factorisation domain (UFD) if and only if any non-zero element can be factorised into the product of primes, and any two factorisations only diﬀer by unit factors and in the order of the factors. Theorem. (Gauß’ lemma) If R is a UFD, then so is R[x]. Theorem. If K is ﬁeld, then the polynomial ring K[x] has a Euclidean algorithm, and therefore it is a UFD. Corollary. K[x1, x2, . . . , xn] is a UFD for any ﬁeld K and for any n ∈N. Theorem. A principal ideal ⟨f⟩(f ̸= 0) in a UFD is prime if and only if f is irreducible. Deﬁnition. An element r in a ring is called nilpotent if and only if there exists a positive integer n such that rn = 0. Theorem. An ideal I in a ring R is radical if and only if R/I has no nilpotent elements other than 0.
1940	https://www.labxchange.org/library/items/lb:LabXchange:74fda389:video:1	GLOBE Weather: Convection - LabXchange Skip to main contentCurriculum NewLibrary Search 33,200+ resources Educators Explore content Create a class Create content Learners Create content Join a class Log in Sign up Library catalog Video GLOBE Weather: Convection GLOBE Weather: Convection About this video Share Favorite This video on convection is a supplemental teacher resource for the GLOBE Weather Curriculum ( In this lesson, students watch a… more Published February 1, 2025\|Updated February 13, 2025 Video Player is loading. Play Video Play Loaded: 0.00% 0:00 Current Time 0:00 / Duration 4:19 1x Playback Rate 2x 1.75x 1.5x 1.25x 1x, selected 0.75x 0.5x Subtitles subtitles off, selected Captions captions settings, opens captions settings dialog captions off, selected English Fullscreen Mute This is a modal window. Beginning of dialog window. Escape will cancel and close the window. Text Color Transparency Background Color Transparency Window Color Transparency Font Size Text Edge Style Font Family Reset restore all settings to the default values Done Close Modal Dialog End of dialog window. Transcript Created with support from Tags Mylar balloon convection heat weather atmosphere Earth science climate climate change sun Report UCAR Center for Science Education This learning resource was made by UCAR Center for Science Education which is funded primarily through the National Science Foundation. WebsiteProfile \| \| Content Type \| Video \| \| \| Language \| English \| \| \| Subject \| Atmosphere and Weather, Oceans and Water Systems, Climate Change and Global Warming, Physics \| \| \| Background Knowledge \| Some \| \| \| Favorites \| 0 \| \| \| Views \| 8 \| \| \| Remixes \| 0 \| \| \| License \| LabXchange Standard License \| Suggested More by this source Organization: UCAR Center for Science Education Interactive Asset type is Interactive Sort It Out: El Niño or La Niña ------------------------------- Sort pictures of El Niño and La Niña conditions based on sea-surface temperatures. Organization: The SDG Academy Video Asset type is Video Overview of the Climate System: Global Circulation Systems ---------------------------------------------------------- This video explains the global circulation systems, including wind and ocean current systems, and what their driving factors are, as well as how they change with warmer and cooler conditions and seasons. Organization: Science Mom Cluster Asset type is Cluster Science Mom and Earth Science ----------------------------- From studying the composition of our atmosphere to modeling climate patterns and understanding geologic formations, Earth Science encompasses all the processes that have formed our planet and that will continue to shape it. Join Science Mom and Math Dad as you are introduced to the natural processes that make Earth habitable and fascinating. Organization: SubjectToClimate Teaching Guide Asset type is Teaching Guide Water Cycle Lesson: Transpiration --------------------------------- In this lesson, your students will explore transpiration’s role in cooling cities, understand its connection to climate change, and develop actionable solutions to address urban heat islands. Quick links Library My classes My content CareerXplorer My careers Messages Community People About us About Blog News Collaborators Terms of service Privacy notice Privacy & Security Your privacy choices Help Help Center Tutorials Get started Teacher resources Accessibility Connect Contact us Share your story Events Discussion Sign up for our mailing list English © 2025 The President and Fellows of Harvard College We Value Your Privacy We use cookies to improve your experience, analyze usage, and support marketing. Click 'Accept All' to consent, or adjust settings anytime. Accept All Settings Privacy Notice This video on convection is a supplemental teacher resource for the GLOBE Weather Curriculum ( In this lesson, students watch a… This learning resource was made byUCAR Center for Science Educationwhich is funded primarily through theNational Science Foundation.
1941	https://phet.colorado.edu/en/simulations/concentration	Skip to Main Content Concentration Topics Solutions Concentration Saturation Molarity Moles Volume Solubility Sample Learning Goals Describe the relationships between volume and amount of solute to solution concentration. Explain how solution color and concentration are related. Predict how solution concentration will change for any action (or combination of actions) that adds or removes water, solute, or solution, and explain why. Design a procedure for creating a solution of a given concentration. Design and justify a procedure for changing a solution from one concentration to another. Identify when a solution is saturated and predict how concentration will change for any action or combination of actions where water or solute change. Inclusive Features Alternative Input Pan and Zoom Interactive Highlights We are adding features to make our simulations more inclusive to learners with diverse needs and within diverse environments. Find all sims with inclusive features. System Requirements HTML5 simulations can run on iPads, Chromebooks, PC, Mac, and Linux systems. See full HTML5 system requirements Inclusive features are added to HTML5 sims only, and some features have platform limitations. See Accessible Sims for more details on system requirements and tested platforms. Version 1.8.2 Related Sims Beer's Law Lab Molarity Salts & Solubility Explore More Tips for Using PhET K-12 Activity Design Quick Tips PhET is a non-profit committed to providing high quality STEM resources for every classroom. Every donation counts. Donate now The PhET website does not support your browser. We recommend using the latest version of Chrome, Firefox, Safari, or Edge.
1942	https://artofproblemsolving.com/wiki/index.php/Divisibility_rules?srsltid=AfmBOooaZTYKsPUTiLNHEkBPm-GWT7P9ziIQK4q-qNeONuL4FBxRRPqQ	Art of Problem Solving Divisibility rules - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Divisibility rules Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Divisibility rules These divisibility rules help determine when positive integers are divisible by particular other integers. All of these rules apply for base-10only -- other bases have their own, different versions of these rules. Contents [hide] 1 Divisibility Videos 2 Basics 2.1 Divisibility Rule for 2 and Powers of 2 2.2 Divisibility Rule for 3 and 9 2.3 Divisibility Rule for 5 and Powers of 5 2.4 Divisibility Rule for 7 2.5 Divisibility Rule for 10 and Powers of 10 2.6 Divisibility Rule for 11 2.7 General Rule for Composites 2.7.1 Example 3 Advanced 3.1 General Rule for Primes 3.2 Divisibility Rule for 13 3.3 Divisibility Rule for 17 3.4 Divisibility Rule for 19 3.5 Divisibility Rule for 29 3.6 Divisibility Rule for 49 4 Special 4.1 Mod-preserving tests 4.1.1 Mod-preserving for 7 4.1.2 Mod-preserving for 13 4.2 Block tests 4.2.1 Small blocks -- 101 and 1001 4.2.2 Bigger blocks -- 10001 and 10000001 4.2.3 Type 2 blocks -- 111 and 11111 5 Problems 6 Resources 6.1 Books 6.2 Classes 7 See also Divisibility Videos Basics Divisibility Rule for 2 and Powers of 2 A number is divisible by if and only if the last digits of the number are divisible by . Thus, in particular, a number is divisible by 2 if and only if its units digit is divisible by 2, i.e. if the number ends in 0, 2, 4, 6 or 8. Proof Divisibility Rule for 3 and 9 A number is divisible by 3 or 9 if and only if the sum of its digits is divisible by 3 or 9, respectively. Note that this does not work for higher powers of 3. For instance, the sum of the digits of 1899 is divisible by 27, but 1899 is not itself divisible by 27. Proof Divisibility Rule for 5 and Powers of 5 A number is divisible by if and only if the last digits are divisible by that power of 5. Proof Divisibility Rule for 7 Rule 1: Partition into 3 digit numbers from the right (). The alternating sum () is divisible by 7 if and only if is divisible by 7. Proof Rule 2: Truncate the last digit of , double that digit, and subtract it from the rest of the number (or vice-versa). is divisible by 7 if and only if the result is divisible by 7. Proof Rule 3: "Tail-End divisibility." Note. This only tells you if it is divisible and NOT the remainder. Take a number say 12345. Look at the last digit and add or subtract a multiple of 7 to make it zero. In this case we get 12380 or 12310 (both are acceptable; I am using the former). Lop off the ending 0's and repeat. 1238 - 28 ==> 1210 ==> 121 - 21 ==> 100 ==> 1 NOPE. Works in general with numbers that are relatively prime to the base (and works GREAT in binary). Here's one that works. 12348 - 28 ==> 12320 ==> 1232 +28 ==> 1260 ==> 126 + 14 ==> 14 YAY! Tiny extension to tell you the remainder: Count how many zeroes you lop off and mod 6. Multiply mod 7 with the corresponding number Zeroes (mod 6) Number to multiply by 0 1 1 3 2 2 3 6 4 4 5 5 And that's the remainder. Divisibility Rule for 10 and Powers of 10 If a number is power of 10, define it as a power of 10. The exponent is the number of zeros that should be at the end of a number for it to be divisible by that power of 10. Example: A number needs to have 6 zeroes at the end of it to be divisible by 1,000,000 because . Divisibility Rule for 11 A number is divisible by 11 if the alternating sum of the digits is divisible by 11. Proof General Rule for Composites A number is divisible by , where the prime factorization of is , if the number is divisible by each of . Example For the example, we will check if 55682168544 is divisible by 36. The prime factorization of 36 to be . Thus we must check for divisibility by 4 and 9 to see if it's divisible by 36. Since the last two digits, 44, of the number is divisible by 4, so is the entire number. To check for divisibility by 9, we look to see if the sum of the digits is divisible by 9. The sum of the digits is 54 which is divisible by 9. Thus, the number is divisible by both 4 and 9 and must be divisible by 36. Advanced General Rule for Primes For every prime number other than 2 and 5, there exists a rule similar to rule 2 for divisibility by 7. For a general prime , there exists some number such that an integer is divisible by if and only if truncating the last digit, multiplying it by and subtracting it from the remaining number gives us a result divisible by . Divisibility rule 2 for 7 says that for , . The divisibility rule for 11 is equivalent to choosing . The divisibility rule for 3 is equivalent to choosing . These rules can also be found under the appropriate conditions in number bases other than 10. Also note that these rules exist in two forms: if is replaced by then subtraction may be replaced with addition. We see one instance of this in the divisibility rule for 13: we could multiply by 9 and subtract rather than multiplying by 4 and adding. is any number so that Divisibility Rule for 13 Rule 1: Truncate the last digit, multiply it by 4 and add it to the rest of the number. The result is divisible by 13 if and only if the original number was divisble by 13. This process can be repeated for large numbers, as with the second divisibility rule for 7. Proof Rule 2: Partition into 3 digit numbers from the right (). The alternating sum () is divisible by 13 if and only if is divisible by 13. Proof Rule 3: Works for . Let . If is odd add 39 to . Round up to the nearest multiple of 80, call the result . Find . Check: Is . Proof Divisibility Rule for 17 Truncate the last digit, multiply it by 5 and subtract from the remaining leading number. The number is divisible if and only if the result is divisible. The process can be repeated for any number. Proof Divisibility Rule for 19 Truncate the last digit, multiply it by 2 and add to the remaining leading number. The number is divisible if and only if the result is divisible. This can also be repeated for large numbers. Proof Divisibility Rule for 29 Truncate the last digit, multiply it by 3 and add to the remaining leading number. The number is divisible if and only if the result is divisible. This can also be repeated for large numbers. Proof Divisibility Rule for 49 Why 49? For taking pesky out of a root. Useful below 4900. Round up to a multiple of 50, call it , and subtract the original number, call this . If , it is divisible by 49. Examples: Round up: . Difference: . ? Yes! Round up: . Difference: . ? No! Round up: . Difference: . ? Yes! Extension to work for all numbers. Floor divide by 4950, multiply by 50, and add to before calculating Proof Special Mod-preserving tests These tests allow you take the modulo operation easily. Mod-preserving for 7 Multiply the first digit by 3 and add it to the rest. Mod-preserving for 13 Multiply the first digit by 3 and subtract it from the rest Block tests As a bonus, these are also mod-preserving Small blocks -- 101 and 1001 The divisibility for 101 test is simple: Alternate adding and subtracting blocks of two digits starting from the end two, which are added. Ex. 1102314 by 101 01 + 10 - 23 + 14 ← last block is always two digits and positive =0 so 1102314 is divisible by 101 The divisibility for 1001 is the same, but with blocks of three. (Starting with the end three, this time) The 1001 test also works for all it's divisors. The most useful are 7, 11, and 13. Bigger blocks -- 10001 and 10000001 10001 has block size length 4, and factors nicely into 73137. 1000001 has block size 6, and factors into 175882353. 5882353 isn't much use, but 17 is, when we're testing a large number. Type 2 blocks -- 111 and 11111 A different type of test can be yielded from adding all the blocks, but again starting with the end. 111 has a block length of three, and factors into 37 and 3. 11111 has a length of five, and factors to 41 and 271. 1111111, with a length of seven, can provide a test for 239 and 4649, if you ever need it. Problems Practice Problems on Alcumus Divisibility (Prealgebra) 2000 AMC 8 Problems/Problem 11 2006 AMC 10B Problems/Problem 25 Resources Books The AoPS Introduction to Number Theory by Mathew Crawford. The Art of Problem Solving by Sandor Lehoczky and Richard Rusczyk. Classes AoPS Introduction to Number Theory Course See also Number theory Modular arithmetic Math books Mathematics competitions Retrieved from " Category: Divisibility Rules Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
1943	https://www.snakeestate.com/venomous/black-mamba/	Unveiling the Mysteries of the Black Mamba The Black Mamba (Dendroaspis polylepis) belongs to the Elapidae snake family, which is known for its highly venomous species, slender bodies, and fixed front fangs, making it one of the most dangerous snakes in the world. Scientific Name: Dendroaspis polylepis Snake Family: Elapidae Table of Contents Introduction Habitat Diet and Feeding Behavior and Temperament Health and Lifespan Reproduction Handling and Care Breeders Snake Species Share This Page Image Credits \| CC BY 4.0 \| CC0 An overview of the species' lifestyle and environment. \| \| \| --- \| \| 🐍 Common Name \| Black Mamba \| \| 🔬 Scientific Name \| Dendroaspis polylepis \| \| 🌍 Habitat \| Savannas, rocky hills, and woodlands in sub-Saharan Africa \| \| 🥚 Diet \| Small mammals and birds \| \| 🧠 Behavior \| Extremely fast, highly venomous, and aggressive when threatened \| \| ⏳ Lifespan \| 10-20 years \| \| 📏 Size \| 8-14 ft \| \| 🛡️ Conservation Status \| Least Concern \| Think of taxonomy like a branching tree. All snakes climb the same trunk—Animalia, Chordata, Reptilia, and Serpentes. But from there, the branches split: one leads to boas, another to pythons, others to vipers and colubrids. This table shows where this particular snake lives on that tree. \| \| \| --- \| \| 👑 Kingdom \| Animalia \| \| 🦴 Phylum \| Chordata \| \| 🦎 Subphylum \| Vertebrata \| \| 🏫 Class \| Reptilia \| \| 📚 Order \| Squamata \| \| 📖 Suborder \| Serpentes \| \| 🧬 Family \| Elapidae \| \| 🧬 Subfamily \| Elapinae \| \| 🌿 Genus \| Dendroaspis \| \| 🌱 Species \| Dendroaspis polylepis \| \| 🌾 Subspecies \| None \| \| 📘 Wikipedia \| Read on Wikipedia \| An Overview of the Black Mamba The Black Mamba (Dendroaspis polylepis) is one of the most feared and respected snakes in the world. Renowned for its incredible speed and potent venom, it is native to sub-Saharan Africa and plays a vital role in its ecosystem. Known as the longest venomous snake in Africa, it is both a predator and a creature to be approached with caution, captivating the interest of herpetologists and snake enthusiasts alike. The Black Mamba's Diverse Habitat The Black Mamba thrives in a variety of environments across sub-Saharan Africa, showcasing its adaptability. From savannas and open woodlands to rocky outcrops and lowland forests, this snake is versatile in its habitat preferences. Below are key insights into where the Black Mamba lives: Geographic Range: Found across countries such as South Africa, Kenya, Ethiopia, and Senegal. Preferred Terrain: Savannas, rocky hills, and light forests with plenty of cover for hiding. Climate: Warm to hot climates, with temperatures ranging from 20°C to 40°C. In addition to these environments, the Black Mamba often occupies abandoned termite mounds or hollow tree trunks as its lair, providing safety and concealment. \| Habitat Feature \| Description \| --- \| \| Altitude Range \| Sea level to 1,800 meters \| \| Preferred Terrain \| Savannas, rocky outcrops, forests \| \| Temperature Tolerance \| 20°C to 40°C \| What Does the Black Mamba Eat? The Black Mamba is a carnivorous predator that relies on its lethal venom to subdue prey. Its diet is diverse, consisting mainly of small mammals and birds. Below are the main features of its feeding habits: Primary Diet: Rodents such as rats and hyraxes. Occasional Prey: Birds and other small reptiles. Hunting Strategy: The Black Mamba is an ambush predator, striking quickly and delivering venom that immobilizes its prey. This snake typically consumes its prey whole and can survive for weeks between meals due to its efficient metabolism. Juvenile Black Mambas often hunt smaller prey, such as lizards, before graduating to larger mammals. The Black Mamba's Behavior and Temperament Despite its reputation, the Black Mamba is not inherently aggressive and prefers to avoid conflict. However, when threatened, it exhibits several defensive behaviors. Key characteristics include: Temperament: Generally shy but highly defensive when cornered. Speed: One of the fastest snakes, capable of reaching speeds of up to 20 km/h. Defensive Display: Raises the front third of its body, flattens its neck into a hood-like shape, and hisses loudly. The Black Mamba's behavior varies with environmental factors, and it is most active during the day, using its keen senses to detect threats and prey. Health and Lifespan of the Black Mamba The Black Mamba's lifespan varies significantly between wild and captive environments. In the wild, they typically live up to 11 years, but this can extend to over 20 years in captivity with proper care. Key health considerations include: Lifespan: 11 years in the wild; up to 20 years in captivity. Predators: Birds of prey, mongoose, and larger reptiles pose threats, especially to juveniles. Common Ailments: Dehydration and parasitic infections during dry seasons. Regular veterinary care in captivity can mitigate many health issues, but Black Mambas are incredibly resilient creatures in their natural environment. The Black Mamba's Reproductive Habits The reproductive cycle of the Black Mamba is as fascinating as its other characteristics. Mating usually occurs during the spring, with males engaging in combat rituals to secure mates. Key reproductive traits include: Mating Season: Spring to early summer. Egg Clutch Size: 6-25 eggs, depending on the female's size and age. Incubation Period: Eggs hatch after 80-90 days in warm, humid conditions. Juvenile Behavior: Hatchlings are fully independent and venomous from birth. The Black Mamba's reproductive success is vital for maintaining its population, especially given its role as a top predator in its ecosystem. Safety and Handling Guidelines for the Black Mamba Handling the Black Mamba is a task reserved for trained professionals due to its speed, agility, and venomous nature. Safety tips include: Always use appropriate tools, such as snake hooks and tongs. Wear protective gear, including snake-proof gloves and boots. Ensure the enclosure is escape-proof and secure. Have access to antivenom and emergency medical support. Education and preparation are critical for anyone working with this snake to minimize risks and ensure safe interactions. Disclaimer The content in this article is intended for informational purposes only. SnakeEstate strongly advises against handling or interacting with venomous snakes, such as the Black Mamba, unless you are a trained professional. These snakes are highly dangerous, and improper handling can result in serious injury or death. Always adhere to local regulations and prioritize safety when dealing with wildlife. Other Snakes In This Species Asp Viper Black Mamba Boomslang Brazilian Coral Snake Bushmaster Coastal Taipan Common Death Adder Common European Adder Copperhead Coral Snake Cottonmouth Eastern Brown Snake Eastern Diamondback Rattlesnake Egyptian Cobra Fer-de-lance Forest Cobra Indian Cobra Inland Taipan Jameson’s Mamba King Cobra Lancehead Lataste’s Viper Malayan Krait Mojave Rattlesnake New Guinea Small-eyed Snake Nose-horned Viper Puff Adder Russell’s Viper Saw-scaled Viper Solomon Island Coral Snake South American Rattlesnake Tiger Snake Western Diamondback Rattlesnake Snake Species Curiosity bites and each of these species has a story to tell. Boa ConstrictorsCorn SnakesGarter SnakesGopher SnakesKingsnakesMilk SnakesPine SnakesPythonsRat SnakesVenomous Snakes Your Items No products in the cart. Boa Constrictors Corn Snakes Garter Snakes Gopher Snakes Milk Snakes Rat Snakes Venomous Snakes
1944	https://www.tutorela.com/math/negative-exponents/examples-exercises	Negative Exponents Practice Problems - Master Powers with Negative Exponents \| Tutorela TUTORELA Learn math Learn Math by Age Practice Math Practice Math by Grade BlogAbout Features FAQTerms of ServicePrivacy Policy Login Sign up Negative Exponents Practice Problems - Master Powers with Negative Exponents Master negative exponents with step-by-step practice problems. Learn to convert negative powers to fractions, simplify expressions, and solve complex algebraic equations. Questions 90 Videos 90 Quizzes 18 Start Learning Now 📚Master Negative Exponents Through Interactive Practice Convert expressions with negative exponents to positive fraction form using a^(-n) = 1/a^n Simplify complex fractions containing bases with negative exponents Apply negative exponent rules to algebraic expressions with variables and coefficients Solve multi-step problems involving division of powers with negative exponents Master fraction inversion when entire fractions have negative exponents Combine negative exponent rules with other power laws for complex expressions Home Learn math Exponents and Roots - Basic Exponents and Exponent rules Exponents - Special Cases Negative Exponents Negative Exponents Practice Problems - Master Powers with Negative Exponents Understanding Negative Exponents Complete explanation with examples When we see any number(positive or negative) raised to a negative power we can convert the expression into a fraction and we will do it as follows: the numerator will be 1 1 1, the denominator will be the base of the power as seen in the original exercise, but now, with a positive exponent. That is to say, in the denominator we will invert the exponent to positive. Pay attention, we will not modify the sign of the base of the potentiation even if it is negative. Property formula:a−n=1 a n a^{-n}=\frac {1}{a^n} a−n=a n 1 This property also applies to algebraic expressions. Detailed explanation Practice Negative Exponents Test your knowledge with 18 quizzes 4−1=? 4^{-1}=\text{?} 4−1=? a −1 4 -\frac{1}{4} −4 1 b 1 4 \frac{1}{4} 4 1 c 1 0.25 \frac{1}{0.25} 0.25 1 d −1 4 \frac{-1}{4} 4−1 Incorrect Correct Answer: 1 4 \frac{1}{4} 4 1 Watch Solution Continue Examples with solutions for Negative Exponents Step-by-step solutions included Exercise #1 Insert the corresponding expression: 1 2 0 2= \frac{1}{20^2}= 2 0 2 1= Step-by-Step Solution To solve this problem, we will use the properties of exponents. Specifically, we will convert the expression 1 2 0 2 \frac{1}{20^2} 2 0 2 1 into a form that uses a negative exponent. The general relationship is that 1 a n=a−n \frac{1}{a^n} = a^{-n} a n 1=a−n. Applying this rule to the given expression: Step 1: Identify the current form, which is 1 2 0 2 \frac{1}{20^2} 2 0 2 1. Step 2: Apply the negative exponent rule: 1 2 0 2=2 0−2 \frac{1}{20^2} = 20^{-2} 2 0 2 1=2 0−2. Step 3: This expression, 2 0−2 20^{-2} 2 0−2, represents 1 2 0 2 \frac{1}{20^2} 2 0 2 1 using a negative exponent. Therefore, the expression 1 2 0 2 \frac{1}{20^2} 2 0 2 1 can be expressed as 2 0−2 20^{-2} 2 0−2, which aligns with choice 1. Answer: 2 0−2 20^{-2} 2 0−2 Video Solution Exercise #2 Insert the corresponding expression: (1 20)−7= \left(\frac{1}{20}\right)^{-7}= (20 1)−7= Step-by-Step Solution To simplify the expression (1 20)−7 \left(\frac{1}{20}\right)^{-7} (20 1)−7, we will apply the rule for negative exponents. The key idea is that a negative exponent indicates taking the reciprocal and converting the exponent to a positive: Start with the expression: (1 20)−7 \left(\frac{1}{20}\right)^{-7} (20 1)−7. Apply the negative exponent rule: (1 a)−n=a n \left(\frac{1}{a}\right)^{-n} = a^n (a 1)−n=a n. For our expression: (1 20)−7 \left(\frac{1}{20}\right)^{-7} (20 1)−7 becomes 2 0 7 20^7 2 0 7. Therefore, (1 20)−7 \left(\frac{1}{20}\right)^{-7} (20 1)−7 simplifies to 2 0 7 20^7 2 0 7. Thus, the correct answer is 2 0 7 20^7 2 0 7. Answer: 2 0 7 20^7 2 0 7 Video Solution Exercise #3 Insert the corresponding expression: 1 3 2= \frac{1}{3^2}= 3 2 1= Step-by-Step Solution To solve this problem, we'll use the rule of negative exponents: Step 1: Identify that the given expression is 1 3 2\frac{1}{3^2}3 2 1. Step 2: Recognize that 1 3 2\frac{1}{3^2}3 2 1 can be rewritten using the negative exponent rule. Step 3: Apply the formula 1 a n=a−n\frac{1}{a^n} = a^{-n}a n 1=a−n to the expression 1 3 2\frac{1}{3^2}3 2 1. Now, let's work through these steps: Step 1: We have 1 3 2\frac{1}{3^2}3 2 1 where 3 is the base and 2 is the exponent. Step 2: Using the formula, convert the denominator 3 2 3^2 3 2 to 3−2 3^{-2}3−2. Step 3: Thus, 1 3 2=3−2\frac{1}{3^2} = 3^{-2}3 2 1=3−2. Therefore, the solution to the problem is 3−2 3^{-2}3−2. Answer: 3−2 3^{-2} 3−2 Video Solution Exercise #4 Insert the corresponding expression: (1 3)−4= \left(\frac{1}{3}\right)^{-4}= (3 1)−4= Step-by-Step Solution To solve this problem, we'll follow these steps: Step 1: Identify the given expression with a negative exponent. Step 2: Apply the rule for negative exponents, which allows us to convert the expression into a positive exponent form. Step 3: Perform the calculation of the new expression. Now, let's work through each step: Step 1: The expression given is (1 3)−4 \left(\frac{1}{3}\right)^{-4} (3 1)−4, which involves a negative exponent. Step 2: According to the exponent rule (a b)−n=(b a)n \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n (b a)−n=(a b)n, we can rewrite the expression with a positive exponent by inverting the fraction: (1 3)−4=(3 1)4=3 4 \left(\frac{1}{3}\right)^{-4} = \left(\frac{3}{1}\right)^4 = 3^4 (3 1)−4=(1 3)4=3 4. Step 3: Calculate 3 4 3^4 3 4. The calculation 3 4 3^4 3 4 is as follows: 3 4=3×3×3×3=81 3^4 = 3 \times 3 \times 3 \times 3 = 81 3 4=3×3×3×3=81. However, since the problem specifically asks for the corresponding expression before calculation to numerical form, the answer remains 3 4 3^4 3 4. Therefore, the answer to the problem, in terms of an equivalent expression, is 3 4 3^4 3 4. Answer: 3 4 3^4 3 4 Video Solution Exercise #5 Insert the corresponding expression: 1 4 2= \frac{1}{4^2}= 4 2 1= Step-by-Step Solution To solve the problem of expressing 1 4 2\frac{1}{4^2}4 2 1 using powers with negative exponents: Identify the base in the denominator: 4 raised to the power 2. Apply the rule for negative exponents that states 1 a n=a−n\frac{1}{a^n} = a^{-n}a n 1=a−n. Express 1 4 2\frac{1}{4^2}4 2 1 as 4−2 4^{-2}4−2. Thus, the expression 1 4 2\frac{1}{4^2}4 2 1 can be rewritten as 4−2 4^{-2} 4−2. Answer: 4−2 4^{-2} 4−2 Video Solution Show More Frequently Asked Questions Everything you need to know about Negative Exponents What is the rule for negative exponents? + The negative exponent rule states that a^(-n) = 1/a^n. When you see a negative exponent, you can rewrite it as a fraction with 1 in the numerator and the base with a positive exponent in the denominator. How do you solve problems with negative exponents step by step? + Follow these steps: 1) Identify terms with negative exponents, 2) Apply the rule a^(-n) = 1/a^n to convert to fractions, 3) Simplify the resulting fractions, 4) Combine like terms if applicable. Always keep the base unchanged, only flip the exponent sign. Does a negative exponent make the answer negative? + No, a negative exponent does not make the final answer negative. For example, 3^(-1) = 1/3, which is positive. The negative sign only affects the exponent's position, converting the expression to a fraction form. How do you handle fractions with negative exponents? + When an entire fraction has a negative exponent like (a/b)^(-n), you invert the fraction and make the exponent positive: (a/b)^(-n) = (b/a)^n. The numerator and denominator switch places. What are common mistakes when working with negative exponents? + Common errors include: • Thinking negative exponents create negative results • Changing the sign of the base instead of just the exponent • Forgetting to apply the rule to each term separately • Not simplifying fractions after conversion How do negative exponents work with variables and coefficients? + Separate the coefficient from the variable first. For example, 3·5x^(-7) = 3·5·x^(-7) = 3·5·(1/x^7) = 15/x^7. Only the variable with the negative exponent gets converted to fraction form. Can you use negative exponent rules with algebraic expressions? + Yes, negative exponent rules apply to all algebraic expressions. Whether you have numbers, variables, or complex expressions as the base, the rule a^(-n) = 1/a^n always applies consistently. What's the difference between (-3)^(-2) and -3^(-2)? + (-3)^(-2) means the entire negative number is raised to the power: (-3)^(-2) = 1/(-3)^2 = 1/9. However, -3^(-2) means only 3 has the negative exponent: -3^(-2) = -(1/3^2) = -1/9. More Negative Exponents Questions Click on any question to see the complete solution with step-by-step explanations Negative Exponents Simplify 5^-2: Negative Exponent Calculation Step-by-StepSolve: [(1/7)^-1]^4 Using Laws of Negative ExponentsSimplify the Complex Fraction: 1/(X^7/X^6) Step by StepCalculate 2^(-5): Solving Negative Exponent ExpressionSimplify x^-a: Understanding Negative Exponent Notation Continue Your Math Journey Master Negative Exponents first, then advance to these related topics that build on your fraction skills Suggested Topics to Practice in Advance Exponents - Special CasesExponents of Negative NumbersZero Exponent Rule Practice by Question Type Choose your preferred learning style and practice format for fraction problems Applying the formulaCalculating powers with negative exponentsComplete the equationconverting Negative Exponents to Positive ExponentsIdentify the greater valueMultiplying Exponents with the same basePresenting powers in the denominator as powers with negative exponentsPresenting powers with negative exponents as fractionsPresenting powers with negative exponents as fractionsUsing laws of exponents with parametersUsing the laws of exponentsUsing variables Pages Home About Learn math FAQ Terms of Service Privacy Policy Notice to Tutor Notice to Student Contact Us: support@tutorela.com
1945	https://www.quora.com/How-do-I-find-the-3rd-side-of-an-isosceles-triangle-if-two-sides-are-given	How to find the 3rd side of an isosceles triangle if two sides are given - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Isometric Triangle Geometric Properties Triangles in Geometry Classical Mathematics Solution of Triangles Congruency of Triangles Isosceles Right Triangle Triangles 5 How do I find the 3rd side of an isosceles triangle if two sides are given? All related (87) Sort Recommended Awnon Bhowmik Studied at University of Dhaka · Author has 3.7K answers and 11.2M answer views ·8y Originally Answered: How can we find the third side of an isosceles triangle if 2 equal sides are given and area of triangle is given? · For an isosceles triangle, the median from [math]A[/math] to [math]BC[/math] is also the perpendicular distance from [math]A[/math] to [math]BC[/math] Method 1: Using Appolonius Theorem, [math]c^2+b^2=2h^2+2\left(\dfrac{a}{2}\right)^2 \ \implies 2b^2=2h^2+\dfrac{a^2}{2}\\implies a^2=4b^2-4h^2\\implies a=2\sqrt{b^2-h^2}[/math] This method is not applicable unless we’re given the height, since there’s no way to find the height easily. Method 2: The cosine rule. Notice that the cosine rule is closely linked to the Appolonius Theorem. [math]\cos A=\dfrac{b^2+c^2-a^2}{2bc}\\implies \cos A=\dfrac{2b^2-a^2}{2b^2}\\implies \sin A=\sqrt{1-\left(\dfrac{2b^2-a^2}{2b^2}\right)^[/math] Continue Reading For an isosceles triangle, the median from [math]A[/math] to [math]BC[/math] is also the perpendicular distance from [math]A[/math] to [math]BC[/math] Method 1: Using Appolonius Theorem, [math]c^2+b^2=2h^2+2\left(\dfrac{a}{2}\right)^2 \ \implies 2b^2=2h^2+\dfrac{a^2}{2}\\implies a^2=4b^2-4h^2\\implies a=2\sqrt{b^2-h^2}[/math] This method is not applicable unless we’re given the height, since there’s no way to find the height easily. Method 2: The cosine rule. Notice that the cosine rule is closely linked to the Appolonius Theorem. [math]\cos A=\dfrac{b^2+c^2-a^2}{2bc}\\implies \cos A=\dfrac{2b^2-a^2}{2b^2}\\implies \sin A=\sqrt{1-\left(\dfrac{2b^2-a^2}{2b^2}\right)^2}\\implies \sin A=\dfrac{\sqrt{(2b^2)^2-(2b^2-a^2)^2}}{2b^2}\\implies \sin A=\dfrac{\sqrt{a^2(4b^2-a^2)}}{2b^2}\\implies \sin A=\dfrac{a\sqrt{4b^2-a^2}}{2b^2}[/math] [math]\Delta=\dfrac{1}{2}bc\sin A\\implies \Delta=\dfrac{1}{2}b^2\cdot \dfrac{a}{2b^2}\sqrt{4b^2-a^2}\\implies \Delta=\dfrac{1}{4}a\sqrt{4b^2-a^2}\\implies a^2(4b^2-a^2)=16\Delta^2\\implies 4a^2b^2-a^4=16\Delta^2\\implies a^4-4a^2b^2+16\Delta^2=0\\implies a^2=\dfrac{4b^2\pm\sqrt{(4b^2)^2-4(16\Delta^2)}}{2}\\implies a^2=\dfrac{4b^2\pm 4\sqrt{b^4-4\Delta^2}}{2}\\implies a^2=2b^2\pm 2\sqrt{b^4-4\Delta^2}\\implies a=\sqrt{2b^2\pm 2\sqrt{b^4-4\Delta^2}}\\text{The solution will be the value of }b\text{ for which }b^4-4\Delta^2\ge 0[/math] Upvote · 99 22 9 1 Sponsored by Grammarly Stuck on the blinking cursor? Move your great ideas to polished drafts without the guesswork. Try Grammarly today! Download 99 34 Related questions More answers below How do I find the 3rd side of an isosceles triangle if two sides and angles are given? How can I find the 3rd side of an isosceles triangle if equal sides are given? How can we find the third side of an isosceles triangle if 2 equal sides are given and area of triangle is given? How can I find the third side of an isosceles triangle if 2 equal sides and all the angles are known? What is the third side of the isosceles triangle if the equal sides are 8cm? Richard Vega Bibliophile, lifelong learner. · Author has 232 answers and 295.5K answer views ·Updated 4y Originally Answered: How do you find the third side of a triangle given two sides? · If the triangle is a right triangle, meaning it has one angle equal to 90 degrees, use the Pythagorean Theorem: [math]a^2+b^2=h^2[/math] where h is the longest side (aka the hypotenuse) If you know sides a and b, then the third side, h, is: [math]h=\sqrt{a^2+b^2}[/math] If you know sides a and h, then the third side, b, is: [math]b=\sqrt{h^2-a^2}[/math] If you know the sides b and h, then the third side, a, is: [math]a=\sqrt{h^2-b^2}[/math] You can also use the Law of Sines: [math]\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}[/math] Or, the Law of Cosines: [math]c=\sqrt{a^2+b^2–2ab\cos{\gamma}}[/math] [math]\text{where}\,\gamma\,\text{(Greek letter Gamma) is the angle opposite sid[/math] Continue Reading If the triangle is a right triangle, meaning it has one angle equal to 90 degrees, use the Pythagorean Theorem: [math]a^2+b^2=h^2[/math] where h is the longest side (aka the hypotenuse) If you know sides a and b, then the third side, h, is: [math]h=\sqrt{a^2+b^2}[/math] If you know sides a and h, then the third side, b, is: [math]b=\sqrt{h^2-a^2}[/math] If you know the sides b and h, then the third side, a, is: [math]a=\sqrt{h^2-b^2}[/math] You can also use the Law of Sines: [math]\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}[/math] Or, the Law of Cosines: [math]c=\sqrt{a^2+b^2–2ab\cos{\gamma}}[/math] [math]\text{where}\,\gamma\,\text{(Greek letter Gamma) is the angle opposite side c}[/math] [math]b=\sqrt{a^2+c^2–2ac\cos{\beta}}[/math] [math]\text{where}\,\beta\,\text{(Greek letter Beta) is the angle opposite side b}[/math] [math]a=\sqrt{b^2+c^2–2bc\cos{\alpha}}[/math] [math]\text{where}\,\alpha\,\text{(Greek letter alpha) is the angle opposite side a}[/math] Upvote · 99 11 9 1 9 1 Dean Rubine Been doing high school math since high school, circa 1975 · Upvoted by Michael Jørgensen , PhD in mathematics and Klaus Ole Kristiansen , M.Sc. Mathematics, University of Copenhagen (1992) · Author has 10.6K answers and 23.7M answer views ·7y How do I find the 3rd side of an isosceles triangle if two sides are given? If the two lengths given are unequal we know the remaining length is the same as one of those. We can’t really tell which unless one is at least twice as long as the other; in that case the unknown side has to be the long side. If we had duplicated the short side our lengths wouldn’t form a real triangle. If the two sides given are equal, all we can say about the remaining length is that it’s between zero and twice the given length. Upvote · 99 29 9 6 9 2 Robert Nichols Author has 5K answers and 15.6M answer views ·9y You can't find the third side given only 2 sides. You can however give a range of possible answers, for example if you know that the two sides are both 8 cm, then you know the third side is greater than zero cm but less than 16 cm. Or if you know that two sides are 6 cm and 11 cm then you know the third side is either 6 cm or 11 cm. Upvote · 99 10 9 2 Sponsored by JetBrains Write better C++ code with less effort. Boost your efficiency with refactorings, code analysis, unit test support, and an integrated debugger. Download 999 897 Related questions How do I find the 3rd side of an isosceles triangle if two sides and angles are given? How can I find the 3rd side of an isosceles triangle if equal sides are given? How can we find the third side of an isosceles triangle if 2 equal sides are given and area of triangle is given? How can I find the third side of an isosceles triangle if 2 equal sides and all the angles are known? What is the third side of the isosceles triangle if the equal sides are 8cm? How do I find the base of an isosceles triangle given two sides? How do you find the side of a two-sided isosceles triangle? How do I calculate the third side of an isosceles triangle when the 2 equal sides (5cm and 5cm) are given and no angles are provided (no options provided either)? How do you find the missing side of an isosceles triangle? How do you find the base of an isosceles triangle? How do you find the third side of a triangle given two sides? If we have two sides & one angle how we can find another side of a triangle? Given an isosceles triangle, two sides are 13 inches, their angle is 45 degrees. What is the length of the 3rd side? How do you find the angle of a two-sided isosceles triangle? How do I find the third side of a right-angled triangle with two sides given? Related questions How do I find the 3rd side of an isosceles triangle if two sides and angles are given? How can I find the 3rd side of an isosceles triangle if equal sides are given? How can we find the third side of an isosceles triangle if 2 equal sides are given and area of triangle is given? How can I find the third side of an isosceles triangle if 2 equal sides and all the angles are known? What is the third side of the isosceles triangle if the equal sides are 8cm? How do I find the base of an isosceles triangle given two sides? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
1946	https://www.wheelessonline.com/bones/double-major-curves/	Double Major Curves : Wheeless' Textbook of Orthopaedics Skip to content Search for: Home Advertising Information Editorial Board Alphabetical Site Index Contact Us My Account Browse Categories Arthritis Arthroscopy Bones Cranio Facial Femur Foot Hand Humerus Pelvic Radius and Ulna Spine Tibia and Fibula DVT Disaster Preparedness Joints Ankle Elbow Hip Knee Shoulder Wrist Lumbar Spine Textbook Muscles Tendons Nerves Perioperative Pain Management TAR Trauma Fractures Ortho Related Journals Lab Tests Medical Topics Medications Orthobiologics Home » Bones » Spine » Double Major Curves Double Major Curves See: Scoliosis Menu: - Discussion: frequently double curves balance out & do not produce cosmetic deformity; they do not tend to progress unless > 60 deg at skeletal maturity; operative indications: marked progression of curves; disadvantages of surgery: instrumentation must be applied to almost the entire spine; long spinal fusions may lead to: degenerative changes below level of fusion leading to chronic pain; pseudarthrosis loss of lordosis; kyphotic deformity: develops secondary to degeneration of the discs; esp common in females approaching menopause; because kyphosis may narrow of spinal canal, secondary spinal stenosis may also develop as well Categories Bones, SpineTags Spine Double Bundle ACL Doxycycline / Vibramycin Orthopaedics and the US Military National Military Family Association Wounded Warrior Project Text Author Clifford R. Wheeless, III, M.D. Orthopaedic Specialists of North Carolina Dr. Wheeless enjoys and performs all types of orthopaedic surgery but is renowned for his expertise in total joint arthroplasty (Hip and Knee replacement) as well as complex joint infections. He founded Orthopaedic Specialists of North Carolina in 2001 and practices at Franklin Regional Medical Center and Duke Raleigh Hospital. » More about Dr. Wheeless. Data Trace Internet Publishing Data Trace is the publisher of Wheeless' Textbook of Orthopaedics Data Trace specializes in Legal and Medical Publishing, Risk Management Programs, Continuing Education and Association Management. Data Trace Publishing Company 110 West Rd., Suite 227 Towson, MD 21204 Telephone: 410.494.4994 About This Site Disclaimer Editorial Board Advertising Information © 2025 Duke University Medical Center - All rights reserved.
1947	https://brainly.com/question/35624991	[FREE] In how many different ways can we arrange 3 men and 3 women such that the men are not beside each other and - brainly.com 3 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +46,3k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +35,2k Ace exams faster, with practice that adapts to you Practice Worksheets +5,3k Guided help for every grade, topic or textbook Complete See more / Mathematics Expert-Verified Expert-Verified In how many different ways can we arrange 3 men and 3 women such that the men are not beside each other and the women are not beside each other? 1 See answer Explain with Learning Companion NEW Asked by Cravens6618 • 08/10/2023 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 743156 people 743K 0.0 0 Upload your school material for a more relevant answer The total number of different ways we can arrange 3 men and 3 women, with the men not beside each other and the women not beside each other, is 864 ways. Mathematics: Permutations To find the number of different ways we can arrange 3 men and 3 women, with the men not beside each other and the women not beside each other, we can use the concept of permutations. First, we arrange the 3 men in a line. There are 3! = 3 x 2 x 1 = 6 ways to do this. Next, we arrange the 3 women in a line. Again, there are 3! = 6 ways to do this. Now, we need to consider the restrictions. Since the men cannot be beside each other, we need to insert the women between the men. There are 4 possible positions to insert the women, represented by the underscores: M_M_M. We can arrange the women in these 4 positions in 4! = 4 x 3 x 2 x 1 = 24 ways. Therefore, the total number of different ways we can arrange the 3 men and 3 women, with the men not beside each other and the women not beside each other, is 6 x 6 x 24 = 864 ways. Learn more about the topic of number here: brainly.com/question/23283166 SPJ11 Answered by triharsh1007 •8.7K answers•743.2K people helped Thanks 0 0.0 (0 votes) Expert-Verified⬈(opens in a new tab) This answer helped 743156 people 743K 0.0 0 Upload your school material for a more relevant answer The total number of ways to arrange 3 men and 3 women such that no two men are beside each other and no two women are beside each other is 144. This is achieved by first arranging the men, creating gaps for the women, and then arranging them. The arrangements are calculated using permutations and combinations considering the restrictions. Explanation To find the number of ways to arrange 3 men and 3 women such that no two men are beside each other and no two women are beside each other, we will approach this problem step-by-step. Identify the Elements: We have 3 men (M1, M2, M3) and 3 women (W1, W2, W3). Basic Arrangement Without Restrictions: If we were to arrange 6 people without any restrictions, the total arrangements would be given by the permutation of 6 distinct items, which is calculated as: 6!=720 Applying Restrictions: Since we need the men and women to not be beside each other, we can visualize the arrangement as alternating positions. We can start by arranging the men first. If we arrange the 3 men, they can be placed in the following pattern: M _ M _ M (where _ represents a space where a woman can be placed). This gives us 4 spaces (slots) to place the women: _ M _ M _ M _ Arranging the Men: The number of ways to arrange the 3 men is: 3!=6 Choosing and Arranging the Women: We have 4 gaps available (the underscores) to place 3 women, ensuring at least one gap between each man. We can choose 3 out of these 4 slots to place the women. The number of ways to select 3 slots from 4 is given by: (3 4)=4 Since the 3 women can also be arranged among themselves in: 3!=6 Therefore, the number of ways to arrange the women is: 4×6=24 Calculating Total Arrangements: Finally, multiply the arrangements for men and women together: 6×24=144 Therefore, the total number of ways to arrange the 3 men and 3 women, ensuring no two men or women are beside each other, is 144 ways. Examples & Evidence For instance, if we arrange the men first as M1, M2, M3, we can visualize the potential placements of women in the gaps created. This can look like: M1 _ M2 _ M3 _, and we can fill the gaps with the women (for example, W1, W2, W3) in the available slots, ensuring they don't sit beside each other. The calculations are based on the principles of permutations and combinations in mathematics, where the arrangement of distinct objects is determined. Thanks 0 0.0 (0 votes) Advertisement Cravens6618 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics Using the discriminant, how many solutions will this quadratic have? Explain and show all work. −3 x 2−12 x+11=0 What is $\cos \left(6^{\circ}\right) ? A. 0.10 B. 0.60 C. 0.99 D. 0.06 The power g 2 is equivalent to 81. What is the value of g−2? A. −81 B. −9 C. 81 1 D. 9 1 (7×1 0 2)(3.5×1 0 3) What is the value of (−4 3)−4 ? A. −81 256 B. −256 81 C. 256 81 D. 81 256 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
1948	https://teachmeanatomy.info/abdomen/bones/lumbar-spine/	The Lumbar Spine Written by Vicky Theakston Last updated June 17, 2020 • 36 Revisions • By Anatomography [CC-BY-SA-2.1-jp], via Wikimedia Commons Fig 1.0 Overview of the location of the lumbar vertebrae The lumbar spine is the third region of the vertebral column, located in the lower back between the thoracic and sacral vertebral segments. It is made up of five distinct vertebrae, which are the largest of the vertebral column. This supports the lumbar spine in its main function as a weight bearing structure. This article will look at the osteology of the lumbar vertebrae, examining their characteristic features, joints and their clinical correlations. Pro Feature - 3D Model You've Discovered a Pro Feature Access our 3D Model Library Explore, cut, dissect, annotate and manipulate our 3D models to visualise anatomy in a dynamic, interactive way. Learn More Characteristic Features Although the lumbar vertebrae lack some of the more distinctive features of other vertebrae, there are several characteristics that help to distinguish them. The vertebral bodies are large and kidney-shaped. They are deeper anteriorly than posteriorly, producing the lumbosacral angle (the angle between the long axis of the lumbar region and that of the sacrum). The vertebral foramen is triangular in shape. Other features of a typical lumbar vertebrae: Transverse processes are long and slender. Articular processes have nearly vertical facets. Spinous processes are short and broad. Accessory processes can be found on the posterior aspect of the base of each transverse process. They act as sites of attachment for deep back muscles. Mammillary processes can be found on the posterior surface of each superior articular process. They act as sites of attachment for deep back muscles. The fifth lumbar vertebrae, L5, has some distinctive characteristics of its own. It has a notably large vertebral body and transverse processes as it carries the weight of the entire upper body. By TeachMeSeries Ltd (2025) Fig 1.1 Superior view of a lumbar vertebrae, showing its characteristic features. Joints There are two types of joint in the lumbar spine. Both of these articulations are not unique to the lumbar vertebrae, and are present throughout the vertebral column. Between vertebral bodies – adjacent vertebral bodies are joined by intervertebral discs, made of fibrocartilage. This is a type of cartilaginous joint, known as a symphysis. Between vertebral arches – formed by the articulation of superior and inferior articular processes from adjacent vertebrae. It is a synovial type joint. Ligaments The joints of the lumbar vertebrae are supported by several ligaments. They can be divided into two groups; those present throughout the vertebral column, and those unique to the lumbar spine. Present throughout Vertebral Column Anterior and posterior longitudinal ligaments: Long ligaments that run the length of the vertebral column, covering the vertebral bodies and intervertebral discs. Ligamentum flavum:Connects the laminae of adjacent vertebrae. Interspinous ligament: Connects the spinous processes of adjacent vertebrae. Supraspinous ligament: Connects the tips of adjacent spinous processes. (Note: In the cervical spine, the interspinous and supraspinous ligaments thicken and combine to form the nuchal ligament). Unique to Lumbar Spine The lumbosacral joint (between L5 and S1 vertebrae) is strengthened by the iliolumbar ligaments. These are fan-like ligaments radiating from the transverse processes of the L5 vertebra to the ilia of the pelvis By TeachMeSeries Ltd (2025) Fig 1.2 Ligaments of the lumbar vertebrae Anatomical Relationships Throughout the vertebral column, the spinal cord travels through the vertebral canal (made up by the foramina of all vertebrae). At around the level of L1, the spinal cord terminates and the cauda equina begins. This is a bundle of lumbar, sacral and coccygeal nerve roots. Spinal nerves exit the vertebral canal through the intervertebral foramina. Clinical Relevance Abnormalities of the Lumbar Spine Lumbar Spinal Stenosis Thought to be hereditary, lumbar spinal stenosis results in a stenotic (narrow) vertebral foramen in one or several lumbar vertebrae. This can cause compression of the spinal cord and exiting nerves. By TeachMeSeries Ltd (2025) Fig 1.3 Excessive lumbar lordosis. This condition can be worsened by age-related degenerative changes, such as bulging of the intervertebral discs. Lumbar spinal stenosis can sometimes be treated surgically with a decompressive laminectomy. Excessive Lumbar Lordosis This is an abnormal anterior curvature of the vertebral column in the lumbar region, characterised by anterior tilting of the pelvis. Women develop temporary excessive lumbar lordosis in late pregnancy, but this resolves after childbirth. In either sex, obesity can similarly cause lordosis due to the increased weight of the abdomen. Both examples can cause back pain, and occur as a result of an altered line of gravity. Do you think you’re ready? Take the quiz below Pro Feature - Quiz The Lumbar Spine Question 1 of 3 How many lumbar vertebrae are there? Submitting... Skip Next Rate question: You scored 0% Skipped: 0/3 1800 More Questions Available Upgrade to TeachMeAnatomy Pro Challenge yourself with over 1800 multiple-choice questions to reinforce learning Learn More Rate This Article Recommended Reading
1949	https://www.youtube.com/watch?v=qxv3jAm7IQE	prove cos(x - pi/2) = sinx MSolved Tutoring 66100 subscribers 394 likes Description 41364 views Posted: 11 Dec 2018 prove cos(x - pi/2) = sinx 26 comments Transcript: we want to verify the following identity so the cosine of X minus PI over 2 is we'll use this property right here so we're gonna have cosine of X cosine of PI over 2 plus sine of X sine of PI over 2 so cosine of PI over 2 is the same thing as 0 plus sine of X and then the sine of PI over 2 is equal to 1 so this goes to 0 and we have sine of X times 1 which is just sine of X and that's it all right I hope this helped you out thanks for watching have a great
1950	https://www.sciencedirect.com/topics/mathematics/vector-addition	Skip to Main content My account Sign in Vector Addition In subject area:Mathematics Vector addition is defined as the process of combining two or more vectors to obtain a resultant vector, which represents the total effect of the individual vectors. This operation often involves summing the components of the vectors in each direction to determine the resultant velocity or force. AI generated definition based on: Elementary Linear Algebra (Fourth Edition), 2010 How useful is this definition? Add to Mendeley Also in subject area: Computer Science Discover other topics Chapters and Articles You might find these chapters and articles relevant to this topic. Mathematical Preliminaries 2013, Mathematical Methods for Physicists (Seventh Edition)George B. Arfken, ... Frank E. Harris Basic Properties We define a vector in a way that makes it correspond to an arrow from a starting point to another point in two-dimensional (2-D) or 3-D space, with vector addition identified as the result of placing the tail (starting point) of a second vector at the head (endpoint) of the first vector, as shown in Fig. 1.7. As seen in the figure, the result of addition is the same if the vectors are added in either order; vector addition is a commutative operation. Vector addition is also associative; if we add three vectors, the result is independent of the order in which the additions take place. Formally, this means It is also useful to define an operation in which a vector A is multiplied by an ordinary number k (a scalar). The result will be a vector that is still in the original direction, but with its length multiplied by k. If k is negative, the vector's length is multiplied by but its direction is reversed. This means we can interpret subtraction as illustrated here: and we can form polynomials such as . Up to this point we are describing our vectors as quantities that do not depend on any coordinate system that we may wish to use, and we are focusing on their geometric properties. For example, consider the principle of mechanics that an object will remain in static equilibrium if the vector sum of the forces on it is zero. The net force at the point O of Fig. 1.8 will be the vector sum of the forces labeled F1, F2, and F3. The sum of the forces at static equilibrium is illustrated in the right-hand panel of the figure. It is also important to develop an algebraic description for vectors. We can do so by placing a vector A so that its tail is at the origin of a Cartesian coordinate system and by noting the coordinates of its head. Giving these coordinates (in 3-D space) the names Ax, Ay, Az, we have a component description of A. From these components we can use the Pythagorean theorem to compute the length or magnitude of A, denoted A or , as (1.99) The components are also useful for computing the result when vectors are added or multiplied by scalars. From the geometry in Cartesian coordinates, it is obvious that if , then C will have components At this stage it is convenient to introduce vectors of unit length (called unit vectors) in the directions of the coordinate axes. Letting be a unit vector in the x direction, we can now identify as a vector of signed magnitude Ax in the x direction, and we see that A can be represented as the vector sum (1.100) If A is itself the displacement from the origin to the point (x,y,z), we denote it by the special symbol r (sometimes called the radius vector), and Eq. (1.100) becomes (1.101) The unit vectors are said to span the space in which our vectors reside, or to form a basis for the space. Either of these statements means that any vector in the space can be constructed as a linear combination of the basis vectors. Since a vector A has specific values of Ax, Ay, and Az, this linear combination will be unique. Sometimes a vector will be specified by its magnitude A and by the angles it makes with the Cartesian coordinate axes. Letting α, β, γ be the respective angles our vector makes with the x, y, and z axes, the components of A are given by (1.102) The quantities , , (see Fig. 1.9) are known as the direction cosines of A. Since we already know that , we see that the direction cosines are not entirely independent, but must satisfy the relation (1.103) While the formalism of Eq. (1.100) could be developed with complex values for the components Ax, Ay, Az, the geometric situation being described makes it natural to restrict these coefficients to real values; the space with all possible real values of two coordinates is denoted by mathematicians (and occasionally by us) IR2; the complete 3-D space is named IR3. View chapterExplore book Read full chapter URL: Book2013, Mathematical Methods for Physicists (Seventh Edition)George B. Arfken, ... Frank E. Harris Chapter Vector Analysis 2013, Mathematical Methods for Physicists (Seventh Edition)George B. Arfken, ... Frank E. Harris 3.1Review of Basic Properties In Section 1.7 we established the following properties of vectors: 1 : Vectors satisfy an addition law that corresponds to successive displacements that can be represented by arrows in the underlying space. Vector addition is commutative and associative: A + B = B + A and (A + B) + C = A + (B + C). 2 : A vector A can be multiplied by a scalar k; if k > 0 the result will be a vector in the direction of A but with its length multiplied by k; if k < 0 the result will be in the direction opposite to A but with its length mutiplied by \|k\|. 3 : The vector A − B is interpreted as A + (−1)B, so vector polynomials, e.g., A − 2B + 3C, are well-defined. 4 : A vector of unit length in the coordinate direction xi is denoted . An arbitrary vector A can be written as a sum of vectors along the coordinate directions, as The Ai are called the components of A, and the operations in Properties 1 to 3 correspond to the component formulas 5 : The magnitude or length of a vector A, denoted \|A\| or A, is given in terms of its components as 6 : The dot product of two vectors is given by the formula consequences arewhere θ is the angle between A and B. 7 : If two vectors are perpendicular to each other, their dot product vanishes and they are termed orthogonal. The unit vectors of a Cartesian coordinate system are orthogonal: (3.1) where δi j is the Kronecker delta, Eq. (1.164). 8 : The projection of a vector in any direction has an algebraic magnitude given by its dot product with a unit vector in that direction. In particular, the projection of A on the direction is , with 9 : The components of A in are related to its direction cosines (cosines of the angles that A makes with the coordinate axes) by the formulas and cos2 α + cos2 β + cos2 γ = 1. In Section 2.2 we noted that matrices consisting of a single column could be used to represent vectors. In particular, we found, illustrating for the 3-D space , the following properties. 10 : A vector A can be represented by a single-column matrix a whose elements are the components of A, as in The rows (i.e., individual elements Ai) of a are the coefficients of the individual members of the basis used to represent A, so the element Ai is associated with the basis unit vector . 11 : The vector operations of addition and multiplication by a scalar correspond exactly to the operations of the same names applied to the single-column matrices representing the vectors, as illustrated here: It is therefore appropriate to call these single-column matrices column vectors. 12 : The transpose of the matrix representing a vector A is a single-row matrix, called a row vector: The operations illustrated in Property 11 also apply to row vectors. 13 : The dot product A ⋅ B can be evaluated as aTb, or alternatively, because a and b are real, as a†b. Moreover, aTb = bTa. View chapterExplore book Read full chapter URL: Book2013, Mathematical Methods for Physicists (Seventh Edition)George B. Arfken, ... Frank E. Harris Chapter P 2015, The Linear Algebra Survival GuideFred E. Szabo PhD Illustration ■ : Vector addition in ℝ[t,4] ℝ[t,4] is the set of real polynomials of degree less than or equal to 4, in the variable t. p = 3 + t − 7 t2 + t3; q = 5 t + t4; p + q 3 + 6 t − 7 t2 + t3 + t4 ■ : Scalar multiplication in ℝ[t,6] r = 3 + 6 x - 7x2 + x3 + x4 + 3x6; Expand[− 2 r] − 6 − 12 x + 14 x2 − 2 x3 − 2 x4 − 6 x6 View chapterExplore book Read full chapter URL: Book2015, The Linear Algebra Survival GuideFred E. Szabo PhD Chapter VECTOR ALGEBRA 1984, International Edition University PhysicsGeorge B. Arfken, ... Joseph Priest 2.2Addition and Subtraction of Vectors Vectors can be added, subtracted, and multiplied. In this section we introduce vector addition, vector subtraction, and the multiplication of a vector by a scalar. The multiplication of two vectors is presented in Sections 2.4 and 2.5. Addition of Vectors In order to add vectors, we first represent them with arrows, as described in the previous section. Consider the two vectors, A and B, represented by arrows in Figure 2.3. We add these two vectors by first placing the tail end of vector B at the tip of vector A. This shift of vector B parallel to itself does not affect the vector, because neither the magnitude nor the direction is changed. We then draw an arrow from the tail end of A to the tip of B. This arrow represents vector C, the vector sum of A and B: (2.1) Example 3 Vector Addition A salesman leaves Dayton, Ohio, and drives 10 mi north on 1–75 and then 5 mi west on 1–70. Where is he relative to his starting point? The 10-mi trip north and the 5-mi trip west are vectors, since they each have both magnitude and direction. The problem is therefore one of vector addition. We will call the 10-mi trip north vector r1 and the 5-mi trip west vector r2, and then represent these vectors with arrows (Figure 2.4). In Figure 2.4, 1 cm corresponds to 2 mi. Traveling 10 mi north (from point A to point B) and then 5 mi west (from point B to point C) is equivalent to adding the two vectors r1 and r2. The resultant change of position (from A to C) is the vector sum Using a protractor, we find that the direction of r is about 26° west of north. When we measure r with a ruler we find that it is 5.7 cm long, which corresponds to 11.4 mi. The salesman, then, is 11.4 mi and 26° northwest of Dayton. We can also find the length of r algebraically by using the Pythagorean theorem. In this case, we find that the length of r is 11.18 mi. The graphical solutions, done carefully, perhaps on a larger scale, may have adequate precision. It does have the advantage of giving you a picture of vectors. The vector addition of Figure 2.3 is carried out with a triangle (ABC). If we complete the parallelogram (Figure 2.5) we see that (2.2) The order in which we add the two quantities does not affect the result. Vector addition is therefore commutative, or independent of order. By adding a third vector, D, to A + B, as in Figure 2.6, we see that the way we group the vectors does not affect the final result. (2.3) With the tail of B at the tip of A and the tail of D at the tip of B, the vector sum A + B + D runs from the tail of A to the tip of D regardless of the order of addition of the vectors. Vector addition is therefore associative, or independent of the grouping of vectors. The vectors r1 and r2 of Example 3 occur in sequence, one after the other. Vector quantities can also act simultaneously, as we see in Example 4. Example 4 Vector Addition of Forces Two forces act on a common point on the bumper of a car. One force of 20.0 newtons, acts horizontally. A second force, also 20.0 newtons, acts vertically. What is the single force equivalent to these two? First, we must draw a vector diagram (Figure 2.7) to show the two forces. We shall let 1 cm of length in the diagram correspond to 4 newtons of force. Because F1 and F2 in this example are equal, the parallelogram defined by the vectors F1 and F2 is a square. The equivalent vector F, the vector sum of F1 and F2, is given by the diagonal. Because vectors F1 and F2 form a right angle in this case, the magnitude of F is given by the Pythagorean theorem: ThusF = 28.3 newtons, 45° above the horizontal F1 In Example 4 we have superposed two forces. We have assumed that the result of the two forces acting together is equivalent to that of a single force vector given by the vector sum. This principle of superposition, or justification for treating forces as vectors, ultimately depends on experiment. Remember, we specified that the two forces F1 and F2 act on the same point. By adding the forces we can form the triangle shown in Figure 2.8. To form this triangle, F2 is displaced sideways, parallel to the original given F2. This displacement is a mathematical device for adding the two vectors, it does not imply that F2 acts on point P. The resultant F = F1 + F2, whether found as the hypotenuse of a triangle or the diagonal of a parallelogram, acts on the original point, O. Example 5 Vector Addition of Velocities A sailboat is moving due east with a velocity of 8 knots (4.12 m/s) relative to the water. Because of the tide, the water is moving northwest (45° west of north) with a velocity of 2 knots relative to the shore. What is the velocity of the sailboat relative to the shore? In this problem we are asked for the vector sum of two velocities: 8 knots east and 2 knots northwest. The graphical addition of the two velocities is shown in Figure 2.9. The magnitude and direction of the resultant velocity v may be obtained from the diagram. Measurement with a ruler and a protractor yields 6.72 knots 12.3° north of due east. Diagrams or graphs help us to picture what is going on and are also useful for checking calculations. However, diagrams and graphs are awkward to draw when we are dealing with three dimensions. Therefore, in Section 2.3, we will develop an algebraic method of vector addition, which is usually more accurate. Subtraction of Vectors If B is a vector shown as an arrow in Figure 2.10, then −B is defined to be the same vector, but in the opposite direction. The sum of a vector and its negative is zero. Therefore the subtraction of a positive vector is defined as the addition of a negative vector. For example, we write A − B as (2.4) and carry out the subtraction by adding −B. Figure 2.11 illustrates this method for the two vectors A and −B. Thus everything we have learned about vector addition may be applied to vector subtraction. Multiplication by a Scalar A vector A may be multiplied by a number or a scalar, a, to give aA. If a is greater than 1, then this multiplication corresponds to an expansion of A. If a is less than 1, then the multiplication corresponds to a contraction of A. When a is positive, the direction of A is not affected. If a is negative, the direction of A is reversed. For instance −2A has double the magnitude and is in the opposite direction from A. If the scalar has dimensions, multiplication will yield a completely new vector. For example, in mechanics the vector velocity v is often multiplied by a scalar mass m. The product mv = p, called linear momentum, has the dimensions mass × length/time. The new vector p is not at all the same as the original vector v with dimensions of length/time. Therefore, it would be nonsense dimensionally to try to add v + p. At this point in our vector algebra, these important properties of ordinary algebra apply: (2.5) (2.6) (2.7) We can combine the last two of these properties (distribution and association) and write (2.8) To a mathematician, Eq. 2.8 identifies vector algebra as a linear mathematical system. Since almost all of introductory physics is linear, vector algebra is used frequently. Questions 1. : The magnitudes of vectors A and B are different. Is it possible to have A + B = 0? 2. : Is it possible to add three vectors of equal magnitude but different directions to get zero? If you think so, illustrate with a diagram. 3. : The magnitudes of three vectors are 2 m, 4 m, and 8 m, respectively. The directions are at your disposal. Can these three vectors be added to yield zero? With a diagram, show how it is possible or why it is impossible. 4. : The vector sum of A, B, and C equals zero. if newtons and = 8 newtons, what is the maximum possible value of ? What is the minimum possible value of ? 5. : If C is the vector sum of A and B, does C have to lie in the plane determined by A and B? View chapterExplore book Read full chapter URL: Book1984, International Edition University PhysicsGeorge B. Arfken, ... Joseph Priest Chapter Mathematical Preliminaries 2013, Mathematical Methods for Physicists (Seventh Edition)George B. Arfken, ... Frank E. Harris 1.7Vectors In science and engineering we frequently encounter quantities that have algebraic magnitude only (i.e., magnitude and possibly a sign): mass, time, and temperature. These we label scalar quantities, which remain the same no matter what coordinates we may use. In contrast, many interesting physical quantities have magnitude and, in addition, an associated direction. This second group includes displacement, velocity, acceleration, force, momentum, and angular momentum. Quantities with magnitude and direction are labeled vector quantities. To distinguish vectors from scalars, we usually identify vector quantities with boldface type, as in V or x. This section deals only with properties of vectors that are not specific to three-dimensional (3-D) space (thereby excluding the notion of the vector cross product and the use of vectors to describe rotational motion). We also restrict the present discussion to vectors that describe a physical quantity at a single point, in contrast to the situation where a vector is defined over an extended region, with its magnitude and/or direction a function of the position with which it is associated. Vectors defined over a region are called vector fields; a familiar example is the electric field, which describes the direction and magnitude of the electrical force on a test charge throughout a region of space. We return to these important topics in a later chapter. The key items of the present discussion are (1) geometric and algebraic descriptions of vectors; (2) linear combinations of vectors; and (3) the dot product of two vectors and its use in determining the angle between their directions and the decomposition of a vector into contributions in the coordinate directions. Basic Properties We define a vector in a way that makes it correspond to an arrow from a starting point to another point in two-dimensional (2-D) or 3-D space, with vector addition identified as the result of placing the tail (starting point) of a second vector at the head (endpoint) of the first vector, as shown in Fig. 1.7. As seen in the figure, the result of addition is the same if the vectors are added in either order; vector addition is a commutative operation. Vector addition is also associative; if we add three vectors, the result is independent of the order in which the additions take place. Formally, this means It is also useful to define an operation in which a vector A is multiplied by an ordinary number k (a scalar). The result will be a vector that is still in the original direction, but with its length multiplied by k. If k is negative, the vector's length is multiplied by but its direction is reversed. This means we can interpret subtraction as illustrated here: and we can form polynomials such as . Up to this point we are describing our vectors as quantities that do not depend on any coordinate system that we may wish to use, and we are focusing on their geometric properties. For example, consider the principle of mechanics that an object will remain in static equilibrium if the vector sum of the forces on it is zero. The net force at the point O of Fig. 1.8 will be the vector sum of the forces labeled F1, F2, and F3. The sum of the forces at static equilibrium is illustrated in the right-hand panel of the figure. It is also important to develop an algebraic description for vectors. We can do so by placing a vector A so that its tail is at the origin of a Cartesian coordinate system and by noting the coordinates of its head. Giving these coordinates (in 3-D space) the names Ax, Ay, Az, we have a component description of A. From these components we can use the Pythagorean theorem to compute the length or magnitude of A, denoted A or , as (1.99) The components are also useful for computing the result when vectors are added or multiplied by scalars. From the geometry in Cartesian coordinates, it is obvious that if , then C will have components At this stage it is convenient to introduce vectors of unit length (called unit vectors) in the directions of the coordinate axes. Letting be a unit vector in the x direction, we can now identify as a vector of signed magnitude Ax in the x direction, and we see that A can be represented as the vector sum (1.100) If A is itself the displacement from the origin to the point (x,y,z), we denote it by the special symbol r (sometimes called the radius vector), and Eq. (1.100) becomes (1.101) The unit vectors are said to span the space in which our vectors reside, or to form a basis for the space. Either of these statements means that any vector in the space can be constructed as a linear combination of the basis vectors. Since a vector A has specific values of Ax, Ay, and Az, this linear combination will be unique. Sometimes a vector will be specified by its magnitude A and by the angles it makes with the Cartesian coordinate axes. Letting α, β, γ be the respective angles our vector makes with the x, y, and z axes, the components of A are given by (1.102) The quantities , , (see Fig. 1.9) are known as the direction cosines of A. Since we already know that , we see that the direction cosines are not entirely independent, but must satisfy the relation (1.103) While the formalism of Eq. (1.100) could be developed with complex values for the components Ax, Ay, Az, the geometric situation being described makes it natural to restrict these coefficients to real values; the space with all possible real values of two coordinates is denoted by mathematicians (and occasionally by us) IR2; the complete 3-D space is named IR3. Dot (Scalar) Product When we write a vector in terms of its component vectors in the coordinate directions, as in we can think of as its projection in the x direction. Stated another way, it is the portion of A that is in the subspace spanned by alone. The term projection corresponds to the idea that it is the result of collapsing (projecting) a vector onto one of the coordinate axes. See Fig. 1.10. It is useful to define a quantity known as the dot product, with the property that it produces the coefficients, e.g., Ax, in projections onto the coordinate axes according to (1.104) where , , are the direction cosines of A. We want to generalize the notion of the dot product so that it will apply to arbitrary vectors A and B, requiring that it, like projections, be linear and obey the distributive and associative laws (1.105) (1.106) with k a scalar. Now we can use the decomposition of B into Cartesian components as in Eq. (1.100), , to construct the dot product of the vectors A and B as (1.107) This leads to the general formula (1.108) which is also applicable when the number of dimensions in the space is other than three. Note that the dot product is commutative, with . An important property of the dot product is that is the square of the magnitude of A: (1.109) Applying this observation to , we have which can be rearranged to (1.110) From the geometry of the vector sum , as shown in Fig. 1.11, and recalling the law of cosines and its similarity to Eq. (1.110), we obtain the well-known formula (1.111) where θ is the angle between the directions of A and B. In contrast with the algebraic formula Eq. (1.108), Eq. (1.111) is a geometric formula for the dot product, and shows clearly that it depends only on the relative directions of A and B and is therefore independent of the coordinate system. For that reason the dot product is sometimes also identified as a scalar product. Equation (1.111) also permits an interpretation in terms of the projection of a vector A in the direction of B or the reverse. If is a unit vector in the direction of B, the projection of A in that direction is given by (1.112) where θ is the angle between A and B. Moreover, the dot product can then be identified as times the magnitude of the projection of A in the B direction, so . Equivalently, is equal to times the magnitude of the projection of B in the A direction, so we also have . Finally, we observe that since , Eq. (1.111) leads to the inequality (1.113) The equality in Eq. (1.113) holds only if A and B are collinear (in either the same or opposite directions). This is the specialization to physical space of the Schwarz inequality, which we will later develop in a more general context. Orthogonality Equation (1.111) shows that becomes zero when , which occurs at (i.e., at ). These values of θ correspond to A and B being perpendicular, the technical term for which is orthogonal. Thus, Checking this result for two dimensions, we note that A and B are perpendicular if the slope of B, , is the negative of the reciprocal of , or This result expands to , the condition that A and B be orthogonal. In terms of projections, means that the projection of A in the B direction vanishes (and vice versa). That is of course just another way of saying that A and B are orthogonal. The fact that the Cartesian unit vectors are mutually orthogonal makes it possible to simplify many dot product computations. Because (1.114) we can evaluate as See Chapter 3: Vector Analysis, Section 3.2: Vectors in 3-D Space for an introduction of the cross product of vectors, needed early in Chapter 2. Exercises 1.7.1 : The vector A whose magnitude is 1.732 units makes equal angles with the coordinate axes. Find , and Az. 1.7.2 : A triangle is defined by the vertices of three vectors A, B and C that extend from the origin. In terms of A, B, and C show that the vector sum of the successive sides of the triangle (AB + BC + CA) is zero, where the side AB is from A to B, etc. 1.7.3 : A sphere of radius a is centered at a point r1. (a) : Write out the algebraic equation for the sphere. (b) : Write out a vector equation for the sphere. ANS. (a) (b) r = r1 + a, where a takes on all directions but has a fixed magnitude a. 1.7.4 : Hubble's law. Hubble found that distant galaxies are receding with a velocity proportional to their distance from where we are on Earth. For the i th galaxy, with us at the origin. Show that this recession of the galaxies from us does not imply that we are at the center of the universe. Specifically, take the galaxy at r1 as a new origin and show that Hubble's law is still obeyed. 1.7.5 : Find the diagonal vectors of a unit cube with one corner at the origin and its three sides lying along Cartesian coordinates axes. Show that there are four diagonals with length Representing these as vectors, what are their components? Show that the diagonals of the cube's faces have length and determine their components. 1.7.6 : The vector r, starting at the origin, terminates at and specifies the point in space . Find the surface swept out by the tip of r if (a) : Characterize a geometrically. (b) : Describe the geometric role of a. The vector a is constant (in magnitude and direction). 1.7.7 : A pipe comes diagonally down the south wall of a building, making an angle of 45° with the horizontal. Coming into a corner, the pipe turns and continues diagonally down a west-facing wall, still making an angle of 45° with the horizontal. What is the angle between the south-wall and west-wall sections of the pipe? ANS. 120°. 1.7.8 : Find the shortest distance of an observer at the point (2,1,3) from a rocket in free flight with velocity (1,2,3) km/s. The rocket was launched at time t = 0 from (1,1,1). Lengths are in kilometers. 1.7.9 : Show that the medians of a triangle intersect in the center which is of the median's length from each vertex. Construct a numerical example and plot it. 1.7.10 : Prove the law of cosines starting from . 1.7.11 : Given the three vectors, find two that are perpendicular and two that are parallel or antiparallel. View chapterExplore book Read full chapter URL: Book2013, Mathematical Methods for Physicists (Seventh Edition)George B. Arfken, ... Frank E. Harris Chapter MOTION 1981, Physics in the Modern World (Second Edition)JERRY B. MARION Vector Addition and Subtraction The manipulation of numbers requires only the basic operations of arithmetic. But how do we handle vectors? Vectors are more complicated than numbers because they combine the essential property of numbers (namely, magnitude) with the additional property of direction. Even so, we can define in a simple way the addition and subtraction of vector quantities. We need only a few basic rules: (a) : A boat is capable of moving with a speed of 4 km/h in still water. If this boat travels downstream in a river that has a current of 3 km/h, the net speed of the boat relative to the land will be 4 km/h + 3 km/h = 7 km/h. Figure 2-10a shows the velocity vector diagram for this situation. The vector vnet is the vector sum of vboat and vriver; that is, vboat + vriver = vnet. If the same boat travels upstream, running against the current, the net speed of the boat will be reduced to 1 km/h. Figure 2-10b shows the way in which the velocity vectors are combined in this case. The vector vnet is again the sum of vboat and vriver, but now these two vectors have opposite directions, so that the magnitude of the sum vriver + vboat is only 1 km/h. Notice how we obtained the sum vector in these two diagrams. In each case we started with the vector vboat; then, we placed the origin of the vector vriver at the head (the arrow end) of vboat. The sum vector vnet was obtained by connecting the origin of vboat with the head of vriver. We follow exactly this same procedure if we wish to find the vector sum, C = A + B, of two vectors, A and B, that do not lie along the same straight line. Figure 2-11 shows this general case of vector addition. (b) : The negative of a vector A is another vector, – A, that has the same magnitude as A but has the opposite direction (Fig. 2-12). If A = 30 km/h northeast, then – A = 30 km/h southwest. (c) : How do we subtract one vector from another? If we are given the vectors A and B, how do we calculate C = A – B? This operation is carried out by using the procedures in (a) and (b). First, knowing B, we can find the vector –B; then we write (just as we can with numbers), That is, to subtract B from A, we add – B to A (Fig. 2-13). An important point regarding vector addition (or subtraction) can be seen in Fig. 2-11. The magnitude of the vector A (that is, A) plus the magnitude of the vector B (that is, B) is greater than the magnitude of the vector C (that is, C). Thus, even though C = A + B, the magnitudes are not equal: C ≠ A + B. View chapterExplore book Read full chapter URL: Book1981, Physics in the Modern World (Second Edition)JERRY B. MARION Chapter Finite Dimensional Vector Spaces 2010, Elementary Linear Algebra (Fourth Edition)Stephen Andrilli, David Hecker Highlights ■ : Vector spaces have two specified operations: vector addition (+) and scalar multiplication (·). A vector space is closed under these operations and possesses eight additional fundamental properties (as stated in the definition). ■ : The smallest possible vector space is the trivial vector space. ■ : Familiar vector spaces (under natural operations) include , , , , a line through the origin, a plane through the origin, all real-valued functions. ■ : Any scalar multiple of the zero vector equals the zero vector. ■ : The scalar 0 times any vector equals the zero vector. ■ : The scalar −1 times any vector gives the additive inverse of the vector. ■ : If a scalar multiple of a vector equals the zero vector, then either the scalar is zero or the vector is zero. View chapterExplore book Read full chapter URL: Book2010, Elementary Linear Algebra (Fourth Edition)Stephen Andrilli, David Hecker Chapter Finite Dimensional Vector Spaces 2016, Elementary Linear Algebra (Fifth Edition)Stephen Andrilli, David Hecker Highlights ▪ : Vector spaces have two specified operations: vector addition and scalar multiplication. A set with such operations is a vector space if and only if is closed under these operations and possesses the 8 additional fundamental properties stated in the definition. ▪ : The smallest possible vector space is the trivial vector space {0}. ▪ : Familiar vector spaces (under the normal operations) include , , , , a line through the origin, a plane through the origin, and all real-valued functions. ▪ : Any scalar multiple a 0 of the zero vector 0 is also equal to 0. ▪ : The scalar 0 times any vector v (that is, 0v) equals the zero vector 0. ▪ : The scalar − 1 times any vector v is equal to the additive inverse −v of the vector. ▪ : If a scalar multiple a v of a vector v is equal to the zero vector 0, then either a = 0 or v = 0. View chapterExplore book Read full chapter URL: Book2016, Elementary Linear Algebra (Fifth Edition)Stephen Andrilli, David Hecker Chapter Vector Spaces, Hilbert Spaces, and the Space 2005, Real Analysis with an Introduction to Wavelets and ApplicationsDon Hong, ... Robert Gardner EXAMPLE 5.1.3 Examples of vector spaces include: (a) : ℚn = 〈V, ℚ〉 where V = {(q1, q2, …, qn) \| q1, ∈ ℚ for 1 ≤ i ≤ n}, and scalar multiplication and vector addition are defined componentwise. (b) : ℝ = 〈V, ℝ〉 where V = {(r1, r2, …, rn) \| ri ∈ ℝ for 1 ≤ i ≤ n}, and scalar multiplication and vector addition are defined componentwise. (c) : ℂn = 〈V, ℝ〉 where V = {(r1, r2, …, cn) \| qi ∈ ℚ for 1 ≤ i ≤ n}, and scalar multiplication and vector addition are defined componentwise. (d) : Fn = 〈V, F〉 where V = {(r1, r2, · · · fn) fi ∈ F for 1 ≤ i ≤ n}, and scalar multiplication and vector addition are defined componentwise. (e) : l2(ℝ) = 〈V, ℝ〉 where and scalar multiplication and vector addition are defined componentwise. (f) : l2(ℂ) = 〈V, ℂ〉 where and scalar multiplication and vector addition are defined componentwise. The reader is probably familiar with the vector spaces of Example 5.1.3(b) and (c). However, the vector spaces of (e) and (f) may be new to you. It may not even be clear that the sets V in these examples are closed under vector addition. We will explore these examples in much more detail in the following sections, and find that they play a role as fundamental as the other examples. We would like to classify vector spaces and see what they “look like.” In that direction, we introduce several definitions. View chapterExplore book Read full chapter URL: Book2005, Real Analysis with an Introduction to Wavelets and ApplicationsDon Hong, ... Robert Gardner Chapter The Gram–Schmidt Process 2021, Programming Mathematics Using MATLAB®Lisa A. Oberbroeckling 12.1General vector spaces and subspaces 12.1.1Vector spaces Definition 12.1.1 A vector space is a set V of elements called vectors that have operations called vector addition and scalar multiplication defined so that the following conditions hold for any and scalars a, and b. • : Closure properties \| \| \| c-1. u + v ∈ V \| \| c-2. au ∈ V \| • : Addition properties \| \| \| --- \| \| a-1. u + v = v + u \| (commutativity) \| \| a-2. u + (v + w)=(u + v)+w \| (associativity) \| \| a-3. ∃ zero vector 0 ∈ V such that \| (additive identity) \| \| \| \| a-4. ∀ u ∈ V, ∃ −u ∈ V such that \| (additive inverse) \| \| \| • : (Scalar) multiplication properties \| \| \| --- \| \| m-1. a(bu)=(ab)u \| (associativity) \| \| m-2. a(u + v)=au + av \| (distributive) \| \| m-3. (a + b)u = au + bu \| (distributive) \| \| m-4. 1u = u ∀ u ∈ V \| Scalars are commonly the set of real numbers (called a real vector space) or the set of complex numbers (called a complex vector space). Examples of vector spaces 1. : : Euclidean vector space of 2. : : the set of matrices. 3. : The set of all functions with domain : • : Addition is defined pointwise: is the function . • : af is defined as the function . Note: you could define a different domain instead of . You could also change it to be all continuous functions on domain D, all differentiable functions with domain D, etc. 4. : : the set of all real-polynomials of degree ≤n. 5. : : the set of all functions whose nth derivatives exist and are continuous on (thus the function, first derivative, second derivative, … , nth derivative are all continuous on ). 6. : The set of all functions that satisfy the differential equation . Theorem 12.1.1 Properties of vectors Let V be a vector space, , and 0 the zero scalar. Then 1. : , and for any scalar c. 2. : . 3. : If , then either or . 12.1.2Subspaces Definition 12.1.2 Let V be a vector space and U be a nonempty subset of V. If U is a vector space under the same addition and scalar multiplication, then U is called a subspace of V. U is a subspace if it is closed under addition and scalar multiplication. All other vector space properties in the definition are inherited from V. Example 12.1.1 Which of the following are subspaces of ? (a) : The subset of all symmetric matrices. (b) : The subset of all matrices that are not symmetric. (c) : The subset of invertible matrices. (d) : All diagonal matrices. Example 12.1.2 Which subsets W are subspaces of the vector space V? (a) : , . (b) : , . (c) : V= set of all functions with domain being , . (d) : V= set of all functions with domain being , . (e) : V= set of all functions with domain being , } for some fixed . View chapterExplore book Read full chapter URL: Book2021, Programming Mathematics Using MATLAB®Lisa A. Oberbroeckling Related terms: Real Number Polynomial Scalar Complex Vector Space Linear Combination Real Vector Space Scalar Multiplication Vector Space Zero Vector Real Valued Function View all Topics
1951	https://www.cod.edu/academics/tasc/mathematics-assistance/pdf/diy_graphs_trig_functions.pdf	_______________ Stop by or call (630) 942-3339 To review basic Trigonometric concepts, watch the following set of YouTube videos introducing concept of basic unit circle, values of special angles, graphs of various trigonometric ratios and the inverse trigonometric functions. They are followed by several practice problems for you to try, covering all the basic concepts covered in the videos, with answers and detailed solutions. Some additional resources are included for more practice at the end. 1. Unit Circle :- 2. Values of sine and cosine of special angles on the unit circle 3. Graph of sine and cosine functions 4. Graphing sine and cosine functions with different amplitudes, periods and phase shifts 5. Graph of tan function 6. Graph of tan with transformations: 7. Graph of sec and csc function 8. Writing Equations for Trig Graphs 9. Finding a Formula for a Trigonometric Graph, Ex 2: 10. Graphing a Secant Function, EX 1: 11. Graph of inverse trig functions Practice problems: The following problems use the techniques demonstrated in the above videos. 1) Evaluate of the following without using calculator: 2) Draw the basic curves of the following trig functions over two periods: a) sin 120° = b) cos 45° = c) tan 2π = d) sec (- 60°) = e) cot ( 𝜋 4) = f) csc (− 𝜋 2) = g) sin(135°) = h) cos( 3𝜋 4 ) = i) sec (-330°) a) y = sin θ b) y = cos θ c) y = tan θ d) y = sec θ e) y = csc θ f) y = cot θ DIY: Trigonometry-Graphs of Trig Functions ______________ Stop by or call (630) 942-3339 3) Fill in the missing values of the following unit circle (_, √3 2 ) ( , _) (√3 2 , _) (1 , _) (− 1 2 , ) (−√2 2 , √2 2 ) (- √3 2 , _ ) (, 0) (−√3 2 , − 1 2) (__ , −√2 2 ) (− 1 2 , _) (0 , 1) (__, __ ) ( 1 2 , _ ) (√2 2 , ) (, − 1 2) ° 0° 𝜋 3 ° 0° 30° 0° 120° 0° _° 0° ° 0° _° 0° 225° 0° 240° 0° 300° 0° ° 0° 330° 0° 0° 0° 90° 0° 180° 0° _° 0° 𝜋 4 _ 0 _ _ _ 5𝜋 6 _ _ _ ___ 3𝜋 2 _ 7𝜋 4 11𝜋 6 ______________ Stop by or call (630) 942-3339 4) Identify the following graphs as sine, cosine, tangent or cotangent graphs a) b) c) d) _______________ Stop by or call (630) 942-3339 5) Draw the graphs of the following trigonometric functions over one period. a) y = 1+sin x b) y = cos (θ+π) c) z = tan 2β d) y = 2 sec α e) s = csc(2t) + 3 f) f(x) = cot(3x) g) y = 2 sin(2x) h) y = 5 cos( 1 2 𝜃− 𝜋 4 ) i) y = 1 2 + tan 𝜃 2 j) y = cos (2x+π) − 1 4 k) y = 1 + 1 tan𝑥 l) y = 2- sin(4θ+π) 6) Find the amplitude (if applicable),period, phase shift and vertical translation of the following functions: a) 6y = 3 + cos(24x -72π) b) 23 – y = 24 +sin (2θ) c) 10 - 3tan(x-2π) = 1-3y d) y = sin(πθ) +23 e) 5y= tan ( 𝑥 2 + 4𝜋) f) y = - 1 4 cos( 3 4 𝑥+ 𝜋 8) 7) Match the following graphs to the corresponding function: A. B. C. D. ______________ Stop by or call (630) 942-3339 E. F. a) y =-2+sec(x + 𝜋 2) b) y = − 1 2 + cos (2𝜃+ 2𝜋) c) y = 1 3 + sin ( 𝜃 2 + 2𝜋) d) y = 2sin(2θ) e) y = - csc(3θ + 𝜋 2) f) y = tan(θ − 𝜋 2) 8) Write the equation and then draw a sine function with the following characteristics: a) Amplitude of 2 units b) A period of 4π. c) Has a phase shift of 𝜋 2 to the right d) A Vertical Translation of 1 2 down 9) A rotating beacon is located at point A next to a wall as shown in the figure below. The beacon 4m from the wall. a) The distance d is given by 𝑑= 4tan (2𝜋𝑡), where t is time measured in seconds since the beacon started rotating. When t = 0, the beacon is aimed at point R. When the beacon is aimed to the right of R, the value d is positive; d is negative if the beacon is aimed to the left of R. Find d for each of the following time: (Round all answers to one decimal place) i. t=0 ii. t =0.1 iii. t= 0.2 iv. t= 0.8 v. Explain why t-values between 0.25 and 0.75 are meaningless. _______________ Stop by or call (630) 942-3339 b) The distance a is given by a = 4 \| sec2πt \| Find a for each of the following times: (Round all answers to one decimal place) i. t =0 ii. t =0.86 iii. t =1.24 10) The distance of a weight attached to a spring above its equilibrium position is s(t) = - 2 cos(20t) inches after t seconds. (s<0 means the weight is below its equilibrium position.) a) What is the maximum height that the weight rises above the equilibrium position? b) What are the frequency and period?(Hint: frequency is reciprocal of period) c) When does the weight first reach its maximum height? 11) Draw the graph of the following inverse functions: a) y = sin−1(x) b) y = cos−1(x) c) y = tan−1(x) 12) Evaluate the following: (Use calculator wherever necessary) a) sin−1(sin x), −𝜋/2 ≤𝑥≤𝜋/2 b) arccos(cos 2π) c) cos−1(0.289) d) tan(arctan(0)) e) arccot(75) Answers 1) a) √3 2 b) √2 2 c) 0 d) 2 e) 1 f) -1 g) √2 2 h) −√2 2 i) 2√3 3 2) a) b) c) d) e) f) ______________ Stop by or call (630) 942-3339 3) 4) a. 𝑦= cos(𝑥) b. 𝑦= cot (𝑥) c. 𝑦= sin (𝑥) d. 𝑦= tan (𝑥) ( 1 2 , √3 2 ) (√𝟐 𝟐 , √𝟐 𝟐) (√3 2 , 𝟏 𝟐) (1 , 0) (− 1 2 , √𝟑 𝟐) (−√2 2 , √2 2 ) (- √3 2 , 1 2 ) (−𝟏 , 0) (−√3 2 , − 1 2) (−√𝟐 𝟐 , −√2 2 ) (− 1 2 , − √𝟑 𝟐) (0 , 1) (0 , −𝟏) ( 1 2 , − √𝟑 𝟐) (√2 2 , −√𝟐 𝟐) (√𝟑 𝟐, − 1 2) 60° 0° 𝜋 3 45° 0° 30° 0° 120° 0° 135° 0° 150° 0° 210° 0° 225° 0° 240° 0° 300° 0° 315° 0° 330° 0° 0° 0° 90° 0° 180° 0° 270° 0° 𝜋 4 𝜋 6 0 𝝅 𝟐 𝟐𝝅 𝟑 𝟑𝝅 𝟒 _ 5𝜋 6 𝝅 𝟕𝝅 𝟔 𝟓𝝅 𝟒 _ 𝟒𝝅 𝟑 3𝜋 2 𝟓𝝅 𝟑 7𝜋 4 11𝜋 6 ______________ Stop by or call (630) 942-3339 5) a) y = 1+sin x b) y = cos (θ+π) c) z = tan 2β d) y = 2 sec α e) s = csc(2t) + 3 f) f(x) = cot(3x) g) y = 2 sin(2x) h) y = 5 cos( 1 2 𝜃− 𝜋 4 ) i) y = 1 2 + tan 𝜃 2 j) y = cos (2x+π) − 1 4 k) y = 1 + 1 tan𝑥 l) y = 2- sin(4θ+π) _______________ Stop by or call (630) 942-3339 6) a) Period = π 12 Phase Shift = 3π to the right Amplitude = 1 6 Vertical Translation = 1 2 unit up b) Period = π Phase Shift = 0 Amplitude =1 Vertical Translation = 1 unit down c) Period =π Phase Shift = 2π right Amplitude = not applicable Vertical Translation =3 units down d) Period = 2 Phase Shift = none Amplitude = 1 Vertical Translation = 23 units up e) Period = 2π Phase Shift = 8π left Amplitude = not applicable Vertical Translation = none f) Period = 𝟖𝝅 𝟑 Phase Shift = 𝝅 𝟔 left Amplitude = 1 4 with reflection about x-axis Vertical Translation = none 7) A. d B. c C. b D. e E. f F. a 8) y = − 1 2 + 2 sin 1 2 (𝑥− 𝜋 2) or y = − 1 2 + 2 sin( 𝑥 2 − 𝜋 4) 9) a) i. 0 ii. 2.9m iii. 12.3 m iv. -12.3 m v. During that time, the beacon is pointing away from the wall. b) i. 4m ii. 6.3 m iii. 63.70m 10) a) 2 inches b) Frequency = 10 𝜋 cycles per second, Period = 𝜋 10 seconds c) 𝑡= 𝜋 20 seconds ______________ Stop by or call (630) 942-3339 11) a) y = sin−1(x) b) y = cos−1(x) c) y = tan−1(x) 12) a) x b) 0 c) 73.2° or 1.28 radians d) 0 e) 0.764° or 0.013 radians _______________ Stop by or call (630) 942-3339 _______________ Stop by or call (630) 942-3339 ______________ Stop by or call (630) 942-3339 _______________ Stop by or call (630) 942-3339 ______________ Stop by or call (630) 942-3339 _______________ Stop by or call (630) 942-3339 ______________ Stop by or call (630) 942-3339 _______________ Stop by or call (630) 942-3339 ______________ Stop by or call (630) 942-3339 _______________ Stop by or call (630) 942-3339 ______________ Stop by or call (630) 942-3339 7. Graph A: By its shape, this graph could be either a sine or cosine function, narrowing the possible correct functions to b, c, or d. If using sine, there would be no phase shift. If using cosine, there would be a phase shift of π/2 to the right or -3π/2 to the left. Choice b does not have the correct phase shift so can be eliminated. Choice c has a vertical shift, so cannot be the correct function for graph A. This leaves choice d. 𝑦= 2𝑠𝑖𝑛2𝜃 would have an amplitude of 2 and a period of π. This is the same as shown on graph A, so the correct function is d. Graphs B and C: These graphs are also sine or cosine functions. B has a positive vertical shift (upward) while C has a negative vertical shift (downward ½ unit). Therefore, it would appear that function c corresponds to graph B and function b corresponds to graph C. Checking further: although no vertical scale is shown on graph B, we can see that the function appears to have a period of 4π (looking at the points x = -2π and 2π). This would mean the coefficient of the argument of the sine function must be 1/2. 𝑦= 1 3 + sin (𝜃 2 + 2𝜋) 𝑜𝑟 𝑦= 1 3 + 𝑠𝑖𝑛1 2 (𝜃+ 4𝜋) The phase shift for function c is -4π, which is exactly one period, so the graph would appear the same as if it had no phase shift. This also fits graph B. Graph B is function c. For graph C with a cosine function, the period appears to be π, so the coefficient of the argument would be 2π/b = π so b = 2. Function b can be written as 𝑦= − 1 2 + cos 2(𝜃+ 𝜋). The amplitudes of the graphed function C and function b are both 1, and the phase shift of function b is π, which is again exactly one period, so the function would appear the same as if it had no phase shift. Graph C is function b. Graphs D and F: These graphs are secant or cosecant function. Graph D appears to have no vertical shift and a period of 2π/3, so graph D must be function e. Graph F has a vertical translation of -2 (downward) and a period of 2π, so graph F is function a. Graph E: This graph is a tangent function hence graph E is function f _______________ Stop by or call (630) 942-3339 ______________ Stop by or call (630) 942-3339 _______________ Stop by or call (630) 942-3339 _______________ Stop by or call (630) 942-3339 __________________ Stop by or call (630) 942-3339 Additional Resources Click on the links below to download worksheets for more practice: 1. Graphing trig functions 2. Inverse trig functions Alternatively; 1. Go To 2. Under “Trigonometry” click on:  Graphing trig functions  Inverse trig functions 3. You can print out the worksheets and work on them. The solutions are provided at the end of the worksheets. 4. For help please contact the Math Assistance Area. References: Lial, Margaret L., Hornsby, John, Schneider,David I. and Daniels, Callie J. Trigonometry. Pearson, 2009. Print
1952	https://www.quora.com/Why-is-this-proof-that-1-0-incorrect	Why is this proof that 1=0 incorrect? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Logical Fallacies Mathematics Homework Ques... Proofs (mathematics) Mathematical Misconceptio... Basic Algebra Mathematical Sciences Algebra Mathematical Proof 5 Why is this proof that 1=0 incorrect? All related (48) Sort Recommended Jack Fraser-Govil Doctor of Physics, Writer of Code, Player of Games · Author has 2.6K answers and 51.5M answer views ·8y I mean, what you essentially have written here is the exact same as: 1×∞=∞1×∞=∞ 2×∞=∞2×∞=∞ Therefore: 1×∞=2×∞1×∞=2×∞ 1=2 1=2 (From which you also have 1=0 1=0, or you can set it up to prove that any two numbers are equal) To counter this, you need to remember Bustany's Rule of Infinity: Intuition and Infinity do not mix You are applying Real-Number algebra rules, to something (∞)∞) that is not a Real Number. Sure, sometimes types of infinity can be a number (the infinite ordinals are members of the Surreal Numbers, for example), Whenever you see infinity actually Continue Reading I mean, what you essentially have written here is the exact same as: 1×∞=∞1×∞=∞ 2×∞=∞2×∞=∞ Therefore: 1×∞=2×∞1×∞=2×∞ 1=2 1=2 (From which you also have 1=0 1=0, or you can set it up to prove that any two numbers are equal) To counter this, you need to remember Bustany's Rule of Infinity: Intuition and Infinity do not mix You are applying Real-Number algebra rules, to something (∞)∞) that is not a Real Number. Sure, sometimes types of infinity can be a number (the infinite ordinals are members of the Surreal Numbers, for example), Whenever you see infinity actually involved in a sum, you should always take a step back and say “hang on there” - especially when multiplying or dividing by it! “Infinity” is a trap for the unwary — you should never do algebra with it, unless you are a professional and you have a big “Don't Try This At Home” warning at the beginning! In particular, the inference: x×a=y×a⇒x=y x×a=y×a⇒x=y Does not hold for a=0,∞a=0,∞ You also seem to assert that 0×∞=0 0×∞=0 - an assertion I wouldn't be so confident with! As I have said, you need to be ridiculously careful when doing mathematics with infinities — in the same way that you need to be careful when dividing by zero (it can sort-of be done - i.e. lim x→0 sin(x)x=1 lim x→0 sin⁡(x)x=1) Unfortunately, you have not heeded this warning, and have strayed into the forbidden lands of “treating infinity like any other number”. Upvote · 999 150 9 3 9 2 Sponsored by Grammarly 92% of professionals who use Grammarly say it has saved them time Work faster with AI, while ensuring your writing always makes the right impression. Download 999 209 Related questions More answers below Can someone do a proof of 1=0 and show where it goes wrong? Is there anything wrong with this proof that 0=1? 0=1; 0^0=1^0 Is there any valid proof for 1 = 0? Is this a valid proof that 0 is not equal to 1? What is wrong with this proof that 1=-1? Alan Bustany Trinity Wrangler, Pure Mathematician · Author has 9.8K answers and 58.5M answer views ·8y Which infinity? If you want to use transfinite numbers and normal arithmetic you had better use a Field of numbers like the Surreal numbers. Infinities do not exist in the Real numbers, and normal arithmetic does not work with infinity in the Extended Real numbers. There are many transfinite Surreal numbers and every one of them has a multiplicative inverse that is not zero. For example the simplest transfinite Surreal is ω={0,1,2,3…∣{}}ω={0,1,2,3…∣{}}, which has a multiplicative inverse ε={0∣1 2,1 3,1 4,…}≠0 ε={0∣1 2,1 3,1 4,…}≠0. Hence ω×ε=1 ω×ε=1. Choosing (infinit Continue Reading Which infinity? If you want to use transfinite numbers and normal arithmetic you had better use a Field of numbers like the Surreal numbers. Infinities do not exist in the Real numbers, and normal arithmetic does not work with infinity in the Extended Real numbers. There are many transfinite Surreal numbers and every one of them has a multiplicative inverse that is not zero. For example the simplest transfinite Surreal is ω={0,1,2,3…∣{}}ω={0,1,2,3…∣{}}, which has a multiplicative inverse ε={0∣1 2,1 3,1 4,…}≠0 ε={0∣1 2,1 3,1 4,…}≠0. Hence ω×ε=1 ω×ε=1. Choosing (infinity) to be the Surreal ω ω the flaw in your proof is in claiming 1(infinity)=0 1(infinity)=0. It is not since 1 ω=ε≠0 1 ω=ε≠0. Upvote · 99 10 Assistant Bot · 1y The proof that 1=0 is incorrect because it contains a logical fallacy. The purported proof typically proceeds as follows: 1) Assume 1=0 for the sake of deriving a contradiction. 2) Divide both sides by 0, yielding 1/0 = 0/0. 3) Since 0/0 is an indeterminate form, conclude that 1/0 = 0/0 = 1. 4) Therefore, 1 = 0, which contradicts the initial assumption. However, this argument is flawed because division by zero is mathematically undefined. Dividing by zero does not yield a valid, meaningful result. Attempting to perform algebraic manipulations with division by zero introduces an invalid step that Continue Reading The proof that 1=0 is incorrect because it contains a logical fallacy. The purported proof typically proceeds as follows: 1) Assume 1=0 for the sake of deriving a contradiction. 2) Divide both sides by 0, yielding 1/0 = 0/0. 3) Since 0/0 is an indeterminate form, conclude that 1/0 = 0/0 = 1. 4) Therefore, 1 = 0, which contradicts the initial assumption. However, this argument is flawed because division by zero is mathematically undefined. Dividing by zero does not yield a valid, meaningful result. Attempting to perform algebraic manipulations with division by zero introduces an invalid step that invalidates the entire proof. Division by zero is not a legitimate mathematical operation, so any proof that relies on it contains a logical flaw and cannot be considered valid. Upvote · Ivan Adrian Koswara writes hand-crafted puzzles ·8y Related Is there anything wrong with this proof that 0=1? 0=1; 0^0=1^0 I see too many people address the part about 0 0 0 0 being undefined instead of the actual fallacy in the proof. Here, I’ll assume 0 0=1 0 0=1; I’ll show there’s still something else that’s wrong, far more problematic than that. First, writing just that is not a clear proof. Earlier lines imply later lines, so what you’re writing is actually “0=1 0=1, so 0 0=1 0 0 0=1 0”. But we can interpret that in a few other ways. One way is that you’re making an assumption: what you mean is “assume 0=1 0=1, then 0 0=1 0 0 0=1 0”. Another way is that you’re just not used to write proper proofs, and you accidentally switched the tw Continue Reading I see too many people address the part about 0 0 0 0 being undefined instead of the actual fallacy in the proof. Here, I’ll assume 0 0=1 0 0=1; I’ll show there’s still something else that’s wrong, far more problematic than that. First, writing just that is not a clear proof. Earlier lines imply later lines, so what you’re writing is actually “0=1 0=1, so 0 0=1 0 0 0=1 0”. But we can interpret that in a few other ways. One way is that you’re making an assumption: what you mean is “assume 0=1 0=1, then 0 0=1 0 0 0=1 0”. Another way is that you’re just not used to write proper proofs, and you accidentally switched the two lines around, so that you mean “0 0=1 0 0 0=1 0, so 0=1 0=1”. In the first case, we have a false statement 0=1 0=1, so anything that depends on this statement cannot be said to be true or false. It happens that 0 0=1 0 0 0=1 0 is true, but that’s of no concern. In the second case, at least we make an assumption 0=1 0=1; even if this turns out to be false, it’s still fine since we make an assumption. (For example, consider proofs by exhaustion, where you make several cases; for example, n n is even and n n is odd. In effect, in each case, you’re making an assumption about n n. Clearly for any particular n n only one of the cases will work; the other has a false assumption. But that’s fine, because we just discard the case that is wrong.) Either way, we make an assumption 0=1 0=1. How do you derive 0 0=1 0 0 0=1 0? Most likely you’re raising both sides to the 0-th power. This is fine; if a=b a=b, you can indeed prove that if either of a c,b c a c,b c is defined, then both are defined and a c=b c a c=b c. Since we accept 0 0=1 0 0=1, indeed 0 0 0 0 is defined, and thus both 0 0,1 0 0 0,1 0 are defined and they have the equal value. This statement is correct, and the proof is correct… …but now what? You’re not leading to any conclusion. In a proof, you want to show that you can reach your conclusion by following several implications. Here, the conclusion would be 0 0=1 0 0 0=1 0 instead of 0=1 0=1; you assumed 0=1 0=1 and used that to derive the conclusion 0 0=1 0 0 0=1 0. Moreover, in a proof, it just guarantees that if the premises are correct, the conclusion is correct. If the premises are false, there’s no guarantee about the conclusion at all; it can be true or false. So this “proof”, while correct, is not sound; your assumptions are wrong so you can’t derive anything from here. Finally, there’s the third case, where you meant “0 0=1 0 0 0=1 0 implies 0=1 0=1”. This time, since we assume 0 0=1 0 0=1 and we know 1 0=1 1 0=1 (that one can be proven), our premise is true. But is the conclusion true? How do you get the line 0=1 0=1? Most likely it’s by canceling exponents. However, this is not allowed in general; if you can cancel exponents, you can conclude things like 2 2=(−2)2 2 2=(−2)2 implies 2=−2 2=−2. (Or do you now think that 2=−2 2=−2, too?) Or 0×3=0×5 0×3=0×5 implies 3=5 3=5. So your step of canceling the exponents is not justified; it’s not true that you can cancel exponents like that. So 0 0=1 0 0 0=1 0 doesn’t actually imply 0=1 0=1. The conclusion 0=1 0=1 thus doesn’t follow from anywhere, and so your proof is now incorrect. So in any reasonable reading, either your proof is correct but meaningless (it doesn’t even say anything about 0=1 0=1, it just assumes it), or your proof is incorrect (canceling exponents). P.S. The consensus right now seems to be that 0 0=1 0 0=1 in certain cases, most often when the exponent is guaranteed to be an integer (e.g. in discrete math, working with polynomials or power series, etc), and 0 0 0 0 is undefined in other cases (e.g. when it appears as f(x,y)=x y f(x,y)=x y where x,y x,y are both reals). Upvote · 99 14 9 2 Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 624 Martin Jansche works in any number system, as long as it's base 10 · Upvoted by Nathan Hannon , Ph. D. Mathematics, University of California, Davis (2021) · Author has 3.7K answers and 3.7M answer views ·1y Originally Answered: Why is 1 not divisible by zero? · It depends on the type of algebraic structure we're working in and how far you’re willing to go with abuse of notation. If what you really really want is for 1 to be divisible by 0, you can work in a group. We can write the multiplicative cyclic group of order 4 with the following multiplication table: × │ 0 1 2 3 ──┼──────── 0 │ 0 1 2 3 1 │ 1 2 3 0 2 │ 2 3 0 1 3 │ 3 0 1 2 Note that the identity element is 0 0. If we define division as a/b=a b−1 a/b=a b−1, this gives us the following division table: / │ 0 1 2 3 ──┼──────── 0 │ 0 3 2 1 1 │ 1 0 3 2 2 │ 2 1 0 3 3 │ 3 2 1 0 You might try to object that this is nothing bu Continue Reading It depends on the type of algebraic structure we're working in and how far you’re willing to go with abuse of notation. If what you really really want is for 1 to be divisible by 0, you can work in a group. We can write the multiplicative cyclic group of order 4 with the following multiplication table: × │ 0 1 2 3 ──┼──────── 0 │ 0 1 2 3 1 │ 1 2 3 0 2 │ 2 3 0 1 3 │ 3 0 1 2 Note that the identity element is 0 0. If we define division as a/b=a b−1 a/b=a b−1, this gives us the following division table: / │ 0 1 2 3 ──┼──────── 0 │ 0 3 2 1 1 │ 1 0 3 2 2 │ 2 1 0 3 3 │ 3 2 1 0 You might try to object that this is nothing but modular addition and subtraction in disguise. And while that’s technically correct, there is nothing that stops us from presenting things this way, because we don’t have any further constraints on 0 0 and 1 1. It may be unusual to write the multiplicative identity as 0 0, but addition does not exist in a multiplicative group, so we don’t need to say what the additive identity might be, whether it is an annihilator for multiplication, etc. In other words, if you are willing to shoehorn things into this representation, you can have what you really really want. If you want divisibility by zero, you have got to give. Taking things like familiar properties of 0 0 and 1 1 for granted is too easy, but that’s the way it is. Upvote · 9 2 9 1 Related questions More answers below How can I prove that 1 = 0? What is the proof that 1>0? Can you prove 0 = -1 mathematically? What's the proof that 1+1 is not equal to zero? Why it is so hard to prove 1-1=0? David Joyce Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979) · Upvoted by Robby Goetschalckx , Computer scientist for 11+ years and passionate about math since childhood. and Tom McFarlane , M.S. Mathematics, University of Washington (1994) · Author has 9.9K answers and 68.4M answer views ·8y Related Is there anything wrong with this proof that 0=1? 0=1; 0^0=1^0 You have noticed that there is a particular function, namely, f(x)=x 0 f(x)=x 0, such that f(a)=f(b)f(a)=f(b) for particular values of a a and b b, and concluded that a=b a=b. This only works if the function is what is called a one-to-one function, also called an injective function. This function, f(x)=x 0 f(x)=x 0, is not a one-to-one function. Another example of a function that is not a one-to-one function is the squaring function. Note that (−3)2=3 2(−3)2=3 2, but −3≠3−3≠3. When you don’t have a one-to-one function, it is not the case that f(a)=f(b)f(a)=f(b) implies a=b a=b. There are many important one-to-one functions including linear functions li Continue Reading You have noticed that there is a particular function, namely, f(x)=x 0 f(x)=x 0, such that f(a)=f(b)f(a)=f(b) for particular values of a a and b b, and concluded that a=b a=b. This only works if the function is what is called a one-to-one function, also called an injective function. This function, f(x)=x 0 f(x)=x 0, is not a one-to-one function. Another example of a function that is not a one-to-one function is the squaring function. Note that (−3)2=3 2(−3)2=3 2, but −3≠3−3≠3. When you don’t have a one-to-one function, it is not the case that f(a)=f(b)f(a)=f(b) implies a=b a=b. There are many important one-to-one functions including linear functions like 3 x+5 3 x+5, exponential functions like 2 x 2 x, and logarithmic functions. One of the first things to notice about a function is whether or not it is a one-to-one function. Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · 99 52 Promoted by JH Simon JH Simon Author of 'How To Kill A Narcissist' ·Updated Fri How do you overcome narcissistic mental enslavement? I grew up with a narc mother and dated a narc that I just cannot get out of my head. Reclaim your True Self. Narcissistic mental enslavement begins with but goes well beyond the mind. Every narcissist mentally overwhelms their target, but what gives the narcissist’s attacks teeth is the emotional reaction. When a narcissist puts down their target, shame arises. When a narcissist threatens their target, fear arises. When a narcissist questions their target’s morality, guilt arises. These attacks bind together over time and produce a corresponding complex, making the target easier to trigger as a ripple becomes an emotional tsunami. A psychological cage is built with layer upon l Continue Reading Reclaim your True Self. Narcissistic mental enslavement begins with but goes well beyond the mind. Every narcissist mentally overwhelms their target, but what gives the narcissist’s attacks teeth is the emotional reaction. When a narcissist puts down their target, shame arises. When a narcissist threatens their target, fear arises. When a narcissist questions their target’s morality, guilt arises. These attacks bind together over time and produce a corresponding complex, making the target easier to trigger as a ripple becomes an emotional tsunami. A psychological cage is built with layer upon layer of emotional pain, which in time splits from the target’s consciousness and reduces the target to a helpless and compliant object. For example, when guilt becomes a complex, a person acts, thinks and believes as though they are guilty, regardless of the situation. Any judgement, whether justified or not, triggers a guilt attack and cripples the person’s capacity for independent thought and action. Reaction becomes the only option. Overcoming mental enslavement involves facing all emotions and integrating them into consciousness. This means having the courage to fall below the realm of thought. Through mindful exploration, you come to notice your triggers, and therefore welcome the emotional intensity into awareness. Rather than reacting to or dissociating from your emotions, you create space and allow them to rise up, and most importantly, to release. Meditation, therapy and bodywork are all potent tools to help you weave together a connection between your Higher Self and your True Self. In doing so, you create a state of wholeness and power which eclipses the mental enslavement you experience after narcissistic abuse. If you have just started your narcissistic abuse recovery journey, check out How To Kill A Narcissist. Or if you wish to immunise yourself against narcissists and move on for good, take a look at How To Bury A Narcissist. Upvote · 999 197 9 6 9 6 Falling Autumn Studied Pure Mathematics at University of Bremen · Author has 2.9K answers and 2.5M answer views ·8y Related Is there anything wrong with this proof that 0=1? 0=1; 0^0=1^0 Everything. First you can never start with something that you want to prove. Out of false claims everything can follow including correct ones. E.g. Out of “Both on the moon and the earth are rain forests” follows “there are rain forests on earth” Surely that doesn’t mean that we can import wood from the moon now. I also think that it’s wrong that people here focus on the indeterminate form. That is really not the problem. We could easily ‘fix’ it by 0⋅0=0⋅1 0⋅0=0⋅1 Or we can look at it more generally We have 2 elements a,b∈S a,b∈S a a function f:S→M f:S→M so that f(a)=f(b)f(a)=f(b) Can we now conclude that a=b a=b ? No, Continue Reading Everything. First you can never start with something that you want to prove. Out of false claims everything can follow including correct ones. E.g. Out of “Both on the moon and the earth are rain forests” follows “there are rain forests on earth” Surely that doesn’t mean that we can import wood from the moon now. I also think that it’s wrong that people here focus on the indeterminate form. That is really not the problem. We could easily ‘fix’ it by 0⋅0=0⋅1 0⋅0=0⋅1 Or we can look at it more generally We have 2 elements a,b∈S a,b∈S a a function f:S→M f:S→M so that f(a)=f(b)f(a)=f(b) Can we now conclude that a=b a=b ? No, we can absolutely not, because functions allow us to map different elements to the same element. The attribute that we can conclude that a=b a=b is called injectivity, which this is clearly not. Upvote · 9 6 9 1 Ashvika Amalan Studied at Elvebakken VGS ·4y Related What is proof for 0 being less than 1? Proof that 0<1. Lemma. For any real number x we have x^2 ≥ 0. Proof. We will consider two cases: x≥0 and x<0. In the first case x≥0 we have x^2 = x x ≥ 00 = 0 In the second case x < 0 we have −x ≥ 0 x^2 = xx = (−x)(−x) = ≥00 = 0 Hence for any real number x we have x^2 ≥ 0 Claim. We have 0 < 1. Proof. First we will show 0 ≤ 1. To see this, note that 1=11 ≥ 0 by our Lemma above. Now it suffices to show that x≠0. Indeed, suppose for a contradiction that 0 = 1. Choose any real number x≠0 that is nonzero. Note that we have x = x1 = x0 since we assumed for a contradiction that 0 = 1 = 0 This contradicts Continue Reading Proof that 0<1. Lemma. For any real number x we have x^2 ≥ 0. Proof. We will consider two cases: x≥0 and x<0. In the first case x≥0 we have x^2 = x x ≥ 00 = 0 In the second case x < 0 we have −x ≥ 0 x^2 = xx = (−x)(−x) = ≥00 = 0 Hence for any real number x we have x^2 ≥ 0 Claim. We have 0 < 1. Proof. First we will show 0 ≤ 1. To see this, note that 1=11 ≥ 0 by our Lemma above. Now it suffices to show that x≠0. Indeed, suppose for a contradiction that 0 = 1. Choose any real number x≠0 that is nonzero. Note that we have x = x1 = x0 since we assumed for a contradiction that 0 = 1 = 0 This contradicts the fact that we chose x≠0, and hence it must be the case that 0≠1. We have shown 0 ≤ 1 and 0≠1, and together these imply 0<1. Upvote · 99 43 Scott Brickner I regret not majoring in math, so I do it as a hobby · Upvoted by Robby Goetschalckx , Computer scientist for 11+ years and passionate about math since childhood. · Author has 19.1K answers and 6.5M answer views ·7y Originally Answered: Is this a valid proof that 0 is not equal to 1? · No. If you don’t know that 0≠1 0≠1, then you don’t “know enough” about numbers to talk about multiplication. When you’re doing foundational-level proofs like this, you have to state what your axioms are. Using something like a=b a=b implies a c=b c a c=b c is only valid if it’s either one of your axioms or it’s a theorem derived from your axioms. Elementary proofs about numbers usually start with Peano’s axioms. In your case, the number 1 is defined to be S(0)S(0) (where the successor function, S S, and the natural number 0 0 are defined in the axioms). The eighth axiom states that for every natural number n n, S S Continue Reading Footnotes Peano axioms - Wikipedia No. If you don’t know that 0≠1 0≠1, then you don’t “know enough” about numbers to talk about multiplication. When you’re doing foundational-level proofs like this, you have to state what your axioms are. Using something like a=b a=b implies a c=b c a c=b c is only valid if it’s either one of your axioms or it’s a theorem derived from your axioms. Elementary proofs about numbers usually start with Peano’s axioms. In your case, the number 1 is defined to be S(0)S(0) (where the successor function, S S, and the natural number 0 0 are defined in the axioms). The eighth axiom states that for every natural number n n, S(n)≠0 S(n)≠0, from which you can immediately conclude that S(0)≠0 S(0)≠0. The claim the 0≠1 0≠1 is so foundational that any proof is either going to be using axioms that make it almost trivial (as with Peano’s axioms), or else you’ll have to spell out what it is you mean by zero and one—you’d need what’s called a “model” of your numbers. Footnotes Peano axioms - Wikipedia Upvote · 9 3 Steve Powell MA(1st) in Mathematics, New College, University of Oxford (Graduated 1975) · Upvoted by David Sbabo , Master degree Mathematics, Université De Mons (2011) and Maurice Peugh , M.S. Mathematics & Physics, Stephen F. Austin State University (1995) · Author has 253 answers and 411.2K answer views ·Updated 7y Related What is wrong with this proof that 2=1? Others have pointed out that the cancelling by a−b a−b is the error here. The normal way of objecting to this step is to say “you cannot divide by zero”. The problem is, this is a negative and it isn’t easy to see how to progress when “up against a negative”. Long ago, when I was tutoring young children for elementary maths exams (as requested by nervous parents), I found that saying “you cannot do that” was rather demotivating. The issue is that you are making a series of deductions, as you were, like: \begin{align}a&=&b\ab&=&b^2\a^2-ab&=&a^2-b^2\a(a-b)&=&(a+b)(a-b)\begin{align}a&=&b\ab&=&b^2\a^2-ab&=&a^2-b^2\a(a-b)&=&(a+b)(a-b) Continue Reading Others have pointed out that the cancelling by a−b a−b is the error here. The normal way of objecting to this step is to say “you cannot divide by zero”. The problem is, this is a negative and it isn’t easy to see how to progress when “up against a negative”. Long ago, when I was tutoring young children for elementary maths exams (as requested by nervous parents), I found that saying “you cannot do that” was rather demotivating. The issue is that you are making a series of deductions, as you were, like: a=b a b=b 2 a 2−a b=a 2−b 2 a(a−b)=(a+b)(a−b)a=b a b=b 2 a 2−a b=a 2−b 2 a(a−b)=(a+b)(a−b) and each line follows (directly) from the line before (and is perfectly correct). But now we “cannot” proceed. We are not allowed to deduce: a=a+b b y c a n c e l l a t i o n a=a+b b y c a n c e l l a t i o n and most of my students were then stumped. So I decided to adopt another approach and asked “what can you do?” Or, to put it another way what (exactly) can we deduce from this equation? And the answer is: E i t h e r a=a+b o r a−b=0.E i t h e r a=a+b o r a−b=0. And that’s it! The rule is: whenever we cancel we have two alternatives: either the (cancelled) equation is true or the cancellation term is zero. One (or both) of these could be true. We can treat them entirely separately. The derivation then proceeds along two distinct possibilities, in this case leading to: E i t h e r 1=2 o r a=b.E i t h e r 1=2 o r a=b. Which deduction is perfectly correct. We therefore end with my tutor’s favourite observation “True, but not useful.” I found by consistently teaching the cancellation rule in this way, my students’ “flow” was not spoiled and they never forgot to check for zero. Upvote · 99 49 9 9 Robby Goetschalckx Computer scientist for 11+ years and passionate about math since childhood. · Author has 6.4K answers and 9.2M answer views ·5y Originally Answered: Is this a valid proof that 0 is not equal to 1? · The clue is in the last line. The word “obviously” is what we call a “weasel word”. If it is obvious, show it. Work it out. The problem here is that you did not state what set of axioms you start from. If you take the group axioms, or ring axioms, there is no problem creating a model where 0=1 0=1. And indeed, in that model, all elements are equal to 0 0. It's a group with only a single element, often referred to as “the trivial group”. If you use the Peano axioms you indeed have an axiom that prevents the trivial group of being a model, as explained in Scott Brickner's answer to Is this a valid proof Continue Reading The clue is in the last line. The word “obviously” is what we call a “weasel word”. If it is obvious, show it. Work it out. The problem here is that you did not state what set of axioms you start from. If you take the group axioms, or ring axioms, there is no problem creating a model where 0=1 0=1. And indeed, in that model, all elements are equal to 0 0. It's a group with only a single element, often referred to as “the trivial group”. If you use the Peano axioms you indeed have an axiom that prevents the trivial group of being a model, as explained in Scott Brickner's answer to Is this a valid proof that 0 is not equal to 1? Upvote · PK 9y Related Is there any valid proof for 1 = 0? No, there is no valid proof for 1=0. As it is not possible in any ways. Yaa, you must have gone through some of the ways proving 1=0. But they are not correct. The statement is clearly false but if you are so curious to prove it you have to use some undefined terms in mathematics. Using these undefined terms you can easily prove some false statements true. Undefined terms in Mathematics are:- 0/0, ∞-∞, ∞∞, ∞/∞, ∞+∞, 0/∞, 0∞, ∞/0, 1/0, 1/∞. (Note:- These are the undefined terms up-to my knowledge.) In the methods given by User-12954534289856328559, Ronak Agarwal and User-9985196108028025700, they all have Continue Reading No, there is no valid proof for 1=0. As it is not possible in any ways. Yaa, you must have gone through some of the ways proving 1=0. But they are not correct. The statement is clearly false but if you are so curious to prove it you have to use some undefined terms in mathematics. Using these undefined terms you can easily prove some false statements true. Undefined terms in Mathematics are:- 0/0, ∞-∞, ∞∞, ∞/∞, ∞+∞, 0/∞, 0∞, ∞/0, 1/0, 1/∞. (Note:- These are the undefined terms up-to my knowledge.) In the methods given by User-12954534289856328559, Ronak Agarwal and User-9985196108028025700, they all have used the undefined term 0/0 to prove 1=0. Whenever you use undefined terms, you always get a strange result. There are a numerous strange result and one of them is 1=0. Upvote · 9 4 9 2 Danya Rose I'm an Aussie feller, I've been one all my life. It may make me seem funny... · Author has 2K answers and 7M answer views ·8y Related Is there anything wrong with this proof that 0=1? 0=1; 0^0=1^0 Very wrong. The expression 0 0 0 0 is an indeterminate form. Its value is not determined uniquely the way the value of x 0 x 0 is for x≠0 x≠0, for example, or the way 0 x 0 x is (also for x≠0 x≠0). That being the case, you can’t just use it by applying an operation to some equality - especially one that starts with what you want to prove! The value of 0 0 0 0 is often chosen to be 1 1 for reasons of convenience in a particular field or situation, but it can be chosen to be otherwise for other reasons. Perhaps you can find a way to make it come to any value you like, depending on the limit of a path that approaches th Continue Reading Very wrong. The expression 0 0 0 0 is an indeterminate form. Its value is not determined uniquely the way the value of x 0 x 0 is for x≠0 x≠0, for example, or the way 0 x 0 x is (also for x≠0 x≠0). That being the case, you can’t just use it by applying an operation to some equality - especially one that starts with what you want to prove! The value of 0 0 0 0 is often chosen to be 1 1 for reasons of convenience in a particular field or situation, but it can be chosen to be otherwise for other reasons. Perhaps you can find a way to make it come to any value you like, depending on the limit of a path that approaches that expression, the way you can with 0 0 0 0. You might like to think that the fact that (−1)2=(1)2=1(−1)2=(1)2=1 gives a proof that −1=1−1=1, but this is not the case either. The function f:R→R f:R→R, such that f(x)=x 2 f(x)=x 2, is not monotonic, so it is not injective. This non-injectiveness means it has no inverse - at least, not without restricting the domain! Thus “undoing” the squaring naïvely is not valid either (and that is a far more subtle mistake than assuming what you want to prove, which is known as begging the question). Upvote · 99 14 9 1 Related questions Can someone do a proof of 1=0 and show where it goes wrong? Is there anything wrong with this proof that 0=1? 0=1; 0^0=1^0 Is there any valid proof for 1 = 0? Is this a valid proof that 0 is not equal to 1? What is wrong with this proof that 1=-1? How can I prove that 1 = 0? What is the proof that 1>0? Can you prove 0 = -1 mathematically? What's the proof that 1+1 is not equal to zero? Why it is so hard to prove 1-1=0? Is the proof of 1=2 mathematically right? What is proof that L(1) = 0? What is the logic behind the proof that 0 = 0? Why is 0/1=0 give mathematical proof? Why is the mathematical proof for 1+1=2 so long? Related questions Can someone do a proof of 1=0 and show where it goes wrong? Is there anything wrong with this proof that 0=1? 0=1; 0^0=1^0 Is there any valid proof for 1 = 0? Is this a valid proof that 0 is not equal to 1? What is wrong with this proof that 1=-1? How can I prove that 1 = 0? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
1953	https://people.math.harvard.edu/~ctm/papers/home/text/papers/gb/gb.pdf	The Gauss–Bonnet theorem for cone manifolds and volumes of moduli spaces Curtis T. McMullen∗ 17 June 2013 Abstract This paper generalizes the Gauss–Bonnet formula to a class of strat-iﬁed spaces called Riemannian cone manifolds. As an application, we compute the volumes of the moduli spaces M0,n with respect to the complex hyperbolic metrics introduced by Picard, Deligne–Mostow and Thurston. Contents 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2 Polyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 3 Spherical complexes . . . . . . . . . . . . . . . . . . . . . . . 8 4 Riemannian complexes . . . . . . . . . . . . . . . . . . . . . . 12 5 Cone manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . 15 6 Euler characteristics of strata . . . . . . . . . . . . . . . . . . 19 7 Gauss–Bonnet for cone manifolds . . . . . . . . . . . . . . . . 23 8 Volumes of moduli spaces . . . . . . . . . . . . . . . . . . . . 24 9 Hyperbolic twists and non-arithmetic groups . . . . . . . . . 28 1 Introduction The Gauss–Bonnet formula for a closed Riemannian manifold states that the Euler characteristic χ(M) is given by a curvature integral, R M Ψ(x) dv(x). Here we generalize this formula to compact Riemannian cone manifolds. By deﬁnition, an n-dimensional cone manifold M is locally isometric to a cone over an (n −1)-dimensional spherical cone manifold K (§5). The cone can be built using any Riemannian metric on the unit ball Bn that is ∗Research supported in part by the NSF. 2000 Mathematics Subject Classiﬁcation: 53C20, 32G15. Revised 24 April 2015. 1 invariant under the holonomy of K; while K itself can be assembled from spherical simplices by gluing their faces together in pairs. A cone manifold is naturally partitioned into connected strata Mσ, each of which is a totally geodesic Riemannian manifold. The solid angle of M at x, deﬁned by Θ(x) = lim r→0 voln(B(x, r)) voln(Bn)rn , is a constant along each stratum; its value on Mσ will be denoted by Θσ. Let M[n] denote the union of top–dimensional strata of M. In §7 we will show: Theorem 1.1 A compact cone manifold of dimension n satisﬁes Z M[n] Ψ(x) dv(x) = X σ χ(Mσ)Θσ. For a smooth manifold the right-hand side reduces to χ(M) and we obtain the usual Gauss–Bonnet formula. For orbifolds the right-hand terms have rational weights of the form Θσ = 1/\|Hσ\|, and we obtain Satake’s formula [Sat]. In a cone manifold Θσ can assume any positive real value, and the right-hand side provides a natural generalization of the orbifold Euler characteristic of M. Moduli spaces. Let M0,n denote the moduli space of conﬁgurations of n ≥3 ordered points on the Riemann sphere. Generalizing work of Pi-card and Deligne–Mostow, Thurston showed that for any set of real weights (µ1, . . . , µn) with 0 < µi < 1 and P µi = 2, we have a natural complex hyperbolic metric gµ on M0,n, and its metric completion P(µ) is a cone manifold [Th]. (The notation P(µ) is a reminder that the n points are ordered, so their monodromy produces pure braids.) Applying Theorem 1.1, in §8 we will show: Theorem 1.2 The complex hyperbolic volume of moduli space satisﬁes vol(M0,n, gµ) = Cn−3 X P (−1)\|P\|+1(\|P\| −3)! Y B∈P max 0, 1 − X i∈B µi !\|B\|−1 . Here P ranges over all partitions of {1, . . . , n} into blocks B, and Ck = (−4π)k/(k + 1)! is the value of Ψ(x) on a complex hyperbolic k-manifold. Previously the volume was known only in special cases [Yo], [Sau], [Par]. Orbifolds and volume ratios. There are 94 values of µ with n ≥5 such that a natural ﬁnite quotient M(µ) of P(µ) is a complex hyperbolic 2 orbifold. Their Euler characteristics are tabulated in §8, and their volumes can be obtained by multiplying by Cn−3. Deligne and Mostow observed that 16 of these orbifolds yield non-arithmetic lattices Γ(µ) in PU(1, n). In an early version of [KM], Kappes and M¨ oller showed that the Lyapunov exponents attached to these lattices and their Ga-lois conjugates are useful commensurability invariants. Motivated by their striking result, we proposed the volume ratio ρ(µ, ν) = vol(P(ν)) vol(P(µ)) = χ(P(ν)) χ(P(µ)) between P(µ) and its hyperbolic twists P(ν) as another invariant, which is easily computed using Theorem 1.2. The values of this ratio turn out to distinguish the commensurability classes of all but one of the non-arithmetic Deligne–Mostow examples (§9). Polyhedra and cell complexes. We give a self-contained account of the theory of cone manifolds, relying on a ‘unique factorization theorem’ in spherical geometry, in §5. Thurston’s (X, G) cone manifolds are a special case of those considered here. Our proof of Theorem 1.1 is based on the Gauss–Bonnet formula for Riemannian polyhedra (§2), proved in the 1940s by Allendoerfer and Weil. To apply this formula to cone manifolds, two main challenges must be ad-dressed. The ﬁrst is that the formula for polyhedra is given in terms of outer angles, instead of inner angles. We relate the two types of angles in The-orem 3.1, which is obtained by comparing a spherical complex M with its suspension S0 ∗M. As a corollary we obtain a Gauss–Bonnet formula for cell complexes with totally geodesic faces (Theorem 4.1), which states that X \|α\|=dim(M) Z Mα Ψ(x) dv(x) = X α (−1)\|α\|Θα. The second is that, when one subdivides a cone manifold into polyhedra, it may be necessary to fracture its totally geodesic strata into pieces with curved faces. These faces introduce complicated expressions involving the second fundamental form, which must be shown to cancel. This cancellation is derived from invariance of Weyl’s tube formula and properties of the spherical dual of a suspension (§6). Notes and references. Cone manifolds and related spaces appear fre-quently in diﬀerential geometry, especially in the special case of singularities 3 along a codimension two submanifold; see e.g. [Ba2], [Cg], [HMM], [FST], [Ko1], [CHK], [Br], [Tr] and [AB]. A general account of the theory of strat-iﬁed spaces is given in [Pf]. The ﬁrst proof of the Gauss–Bonnet theorem for general manifolds was given by Allendoerfer and Weil [AW], using Weyl’s tube formula [We]. Chern’s intrinsic proof [Cn] appeared shortly thereafter. The tube formula and its applications in diﬀerential geometry are treated in [Gr]. Theorem 1.1 can be extended to cone manifolds with curved strata by using the full strength of the formula for polyhedra. The volume of M0,n with respect to its natural symplectic structure is computed in [Zo]. The Euler characteristic of the Deligne-Mumford com-pactiﬁcation M0,n can be computed using inversion of powers series; see [Ge], [Mc3]. The compactiﬁcations M(µ) resulting from a choice of weights are usual diﬀerent from M0,n, and depend on µ; they are discussed from the perspective of algebraic geometry in [Ha]. Volumes of arithmetic lattices in SU(n, 1) are discussed in the recent paper [ES]. Acknowledgements. I would like to thank Kappes and M¨ oller for useful correspondence. 2 Polyhedra The results from [AW] we will need are summarized in this section and in Proposition 3.2 of the next. We begin with the Gauss–Bonnet formula for polyhedra, which may be stated as follows. Theorem 2.1 (Allendoerfer–Weil) The Euler characteristic of a com-pact Riemannian polyhedron M of dimension n satisﬁes (−1)nχ′(M) = Z M[n] Ψ(x) dv(x) + n−1 X r=0 Z M[r] dv(x) Z N(x)∗Ψ(x, ξ) dξ. Here χ′(M) = χ(M) −χ(∂M). In this section we explain the formula above and the basic properties of its integrand. Spheres and duality. Let Sn denote the unit sphere in Rn+1, and let Bn+1 denote the unit ball. We let ωn = 2 · (4π)n/2(n/2)! n! , 4 denote the volume of Sn, with the convention that (n/2)! = Γ( n 2 + 1) if n is odd. A convex spherical polyhedron is a set K ⊂Sn obtained by intersecting ﬁnitely many hemispheres. Its dual is given by K∗= {x ∈Sn : ⟨x, y⟩≥0 ∀y ∈K}. When the ambient sphere needs to be emphasized, we will write K∗= (K, Sn)∗. Manifolds. Let A be a smooth Riemannian manifold of dimension n. When n is even, the intrinsic Gauss–Bonnet integrand on A is given in terms of the Riemann curvature tensor by Ψ(x) = 2 ωn · 1 2n/2n! X i,j∈Sn ϵ(i)ϵ(j) g Ri1i2j1j2 · · · Rin−1injn−1jn. Here ϵ(i) and ϵ(j) denote the signs of the permutations i and j, and g is the determinant of the metric. When n is odd, we set Ψ(x) = 0. The Gauss–Bonnet theorem states that for any closed manifold A we have χ(A) = Z A Ψ(x) dv(x). Submanifolds. Now let A be an r-dimensional submanifold of a Rieman-nian manifold B of dimension n. Let Rijkl denote the restriction of the Riemann curvature tensor on B to A, and let Λij(ξ) denote the second fun-damental form — a symmetric tensor on A depending linearly on a normal vector ξ. In local coordinates where A ⊂B is modeled on Rr ⊂Rn, we have Λij(ξ) = ⟨∇eiej, ξ⟩. The extrinsic Gauss–Bonnet integrand is the function on the unit normal bundle to A deﬁned by Ψ(x, ξ) = X 0≤2f≤r Ψr,f(x, ξ), where Ψr,f(x, ξ) = 2 ω2fωn−2f−1 · 1 2f(2f)!(r −2f)! · X i,j∈Sr ϵ(i)ϵ(j) γ × Ri1i2j1j2 · · · Ri2f−1i2fj2f−1j2f Λi2f+1j2f+1(ξ) · · · Λirjr(ξ). Here γ is the determinant of the induced metric on A. 5 The intrinsic and extrinsic integrands on A are related by Ψ(x) = Z S(x) Ψ(x, ξ) dξ, (2.1) where S(x) is the unit sphere (of dimension n −r −1) in the normal bundle to A at x, and R S(x) dξ = ωn−r−1. In particular, if A is a closed submanifold of B then we have χ(A) = Z A dv(x) Z S(x) Ψ(x, ξ) dξ. Example. Suppose r is even, n > r and A = S2f × Sr−2f ⊂S2f × Rn−2f = B. Then Ψ(x, ξ) = Ψr,f(x, ξ). In a standard orthonormal frame there are 2f(2f)!(r −2f)! nonzero terms in the formula for Ψr,f(x, ξ), since only per-mutations in Sr which preserve the indices (1, 2, . . . , 2f) contribute, and we have Rabcd = gacgbd −gadgbc on the unit sphere. Since the second funda-mental form Λij(ξ) depends only on the projection of ξ to a 1-dimensional space, the integral of each nonzero term is given by volr(A) Z Sn−r−1 \|ξ1\|r−2f dξ = ω2f Z Sn−r−1 ωr−2f\|ξ1\|r−2f dξ = 2ω2fωn−2f−1. The product of this constant with the number of nonzero terms cancels with the leading constant factor in the formula for Ψr,f(x, ξ), yielding the expected result χ(A) = 4. Basic relation. More generally, consider a pair of smoothly embedded submanifolds A ⊂B ⊂C. Let SB(x) and SC(x) denote the ﬁber of the unit normal bundle to A at x, where A is considered as a submanifold of B and C respectively. A convex set K ⊂SB(x) ⊂SC(x) then has two duals, namely K∗ B = (K, SB(x))∗and K∗ C = (K, SC(x))∗. Similarly, the manifold A has two extrinsic Gauss–Bonnet integrands, one for B and one for C. We then have the basic relation: Z K∗ B ΨB(x, ξ) dξ = Z K∗ C ΨC(x, ξ) dξ. (2.2) Equation (2.1) can be regarded as a special case of (2.2), with A = B and K = ∅. For proofs of these relations, see [AW, (6), (7), p.121]. 6 Polyhedra. A convex Euclidean polyhedron is a set K ⊂Rn obtained by intersecting ﬁnitely many half-spaces. A smooth polyhedron is a connected Hausdorﬀspace M equipped with local charts taking values in convex Eu-clidean polyhedra, such that the transition maps between charts are smooth. When M is endowed with a smooth metric, it becomes a Riemannian polyhedron [AW]. The inward pointing normals N(x). Given an n-dimensional Rie-mannian polyhedron and r < n, we let M[r] denote the union of the r-dimensional faces of ∂M, and let M[n] = M −∂M. Each face of M can be regarded as a submanifold of a slight thickening of M[n], so its normal bun-dle and its extrinsic Gauss–Bonnet integrand are deﬁned. Given x ∈M[r], r < n, we let N(x) ⊂S(x) denote the convex set of unit vectors normal to M[r] that point into M[n]. Sketch of the proof of Gauss–Bonnet. These deﬁnitions complete the statement of the Gauss–Bonnet formula for polyhedra (Theorem 2.1). We conclude with a sketch of the proof; for details, see [AW, Theorem II]. First suppose M is a simplex. Choose an isometric embedding M , → RN+1 for some large N. Let T ⊂RN+1 be the boundary of a small tube around the image, i.e. the set of points at distance δ > 0 from M. Let γ : T →SN denote the Gauss map. For δ suﬃciently small, we also have a nearest-point projection π : T →M. M[n] T[r] M[r] Figure 1. The tube around a polyhedron in RN+1. Let T[r] = π−1(M[r]) be the subset of T closest to M[r]. Then for any t ∈T[r] we have a vector ξ = γ(t) ∈SN and a point x = π(t) ∈M[r]. The pair (x, ξ) satisﬁes ⟨ξ, v⟩= 0 for all v ∈TxM[r], and ⟨ξ, v⟩≤0 ∀v ∈N(x) ⊂TxRN+1. Conversely, if (x, ξ) satisﬁes the conditions above, then there is unique t ∈T with π(t) = x and γ(t) = v. Using this fact and Weyl’s tube formula [We], 7 one ﬁnds that the contributions to the degree of the Gauss map are given by G(M[n]) = 1 ωN Z T[n] γ∗(dξ) = Z M[n] Ψ(x) dv(x) and for r < n, G(M[r]) = 1 ωN Z T[r] γ∗(dξ) = Z M[r] dv(x) Z N(x)∗Ψ(x, ξ) dξ. (Remarkably, these quantities are independent of the choice of isometric embedding M , →RN+1.) Since M is a simplex, the Gauss map has degree one, and hence Pn r=0 G(M[r]) = 1 = (−1)nχ′(M) as desired. For the general case Allendoerfer and Weil rely on a local isometric em-bedding theorem for analytic metrics due to Cartan, since the Nash embed-ding theorem was not available at the time. They ﬁrst establish Theorem 2.1 for the simplices in a ﬁne subdivision of M, and then use an addition formula based on (2.1) and (2.2) to deduce the Gauss–Bonnet theorem for M itself. Relations (2.1) and (2.2) are proved using invariance of the tube formula under diﬀerent isometric embeddings. 3 Spherical complexes In this section we deﬁne spherical complexes, which provide models for the unit tangent bundle in the setting of singular spaces. We then discuss the inner and outer angles θα and Ωα of their cells Mα, and show: Theorem 3.1 The outer angle of a compact spherical complex M is given by Ω(M) = X α θα(1 −Ωα). Spherical complexes. A cell complex is a Hausdorﬀtopological space M equipped with a partition into disjoint cells Mα such that the following conditions hold: 1. The closure of each n-cell is homeomorphic to a closed ball in Rn, with the cell itself corresponding to the interior of the ball. 2. Whenever Mα meets Mβ we have Mα ⊂Mβ. 8 A 0-cell is simply a point. We denote the dimension of Mα by \|α\|, and we write α ≤β if Mα ⊂Mβ. A spherical complex is a cell complex built out of spherical polyhedra, in such a way that the inclusions Mα ⊂Mβ are isometries. We also require that each face of Mβ is a cell; equivalently, that for r < \|β\| we have Mβ[r] = [ {Mα : α ≤β, \|α\| = r}. (3.1) Outer angles of polyhedra. The normalized outer angle of a convex polyhedron K ⊂Sn is given by ω(K) = vol(K∗) vol(Sn) · It is easily veriﬁed that this quantity is invariant under the inclusion Sn ⊂ Sn+1; thus it depends only on K, not on the ambient sphere. The outer angle measures the proportion of Sn that is closer to the vertex of the cone R+ · K than to any other point of R+ · K; thus it measures the size of T[r] in Figure 1, explaining its relation to the tube formula. Outer angles of complexes. The outer angle of a compact spherical complex is deﬁned by: Ω(M) = X α (−1)\|α\|ω(Mα). If two spherical complexes M and M′ have isomorphic subdivisions, then Ω(M) = Ω(M′). More precisely, we have: Proposition 3.2 (Allendoerfer–Weil) Let K ⊂Sn be an open convex polyhedron, subdivided into convex cells (Kα). Then (−1)nK∗= X (−1)\|α\|K∗ α as n-chains on Sn. See [AW, Theorem 3]. The equality of chains means (−1)n R K∗f dξ = P α(−1)\|α\| R K∗ α f dξ for any f ∈C(Sn). Taking f = 1, we obtain (−1)nω(K) = P(−1)\|α\|ω(Kα), which implies that Ω(M) is invariant under subdivision. Random Euler characteristic. We note that if M is isometrically embed-ded in Sn then its outer angle is related to the expected Euler characteristic of its intersection with a random spherical hyperplane H ∼ = Sn−1: we have 2Ω(M) = χ(M) + E(χ(H ∩M)). (3.2) 9 The unit normal complex. Given x ∈Mα and β > α, we let Sαβ(x) = {v ∈TxMβ : v is a unit normal to Mα} denote the unit tangent sphere (of dimension \|β\| −\|α\| −1) normal to Mα. The directions pointing into Mβ determine an open, convex polyhedron Nαβ(x) ⊂Sαβ(x), called the inner angle. These spherical polyhedra can be assembled to form a spherical complex Nα(x) = [ β>α Nαβ(x) called the unit normal complex for Mα at x. Angles. The normalized volume of the cell Mα and the normalized inner and outer angles of the cells Nαβ(x) will be denoted by θα = vol(Mα)/ vol(S\|α\|), θαβ(x) = vol(Nαβ(x))/ vol(Sαβ(x)) and ωαβ(x) = vol(Nαβ(x)∗)/ vol(Sαβ(x)). Parallel transport along a geodesic provides an isometric map from Nαβ(x) to Nαβ(y) for any x, y ∈Mα, so in fact the angles above are independent of x. They will therefore be denoted simply by θαβ and ωαβ. Because they are constant, and dim Nαβ(x) = \|β\| −\|α\| −1, the outer angle Ω(Nα(x)) is given by Ωα = (−1)\|α\|+1 X β>α (−1)\|β\|ωαβ (3.3) for all x ∈Mα. In particular, Ωα = 0 if \|α\| = dim(M). An index α is even if \|α\| ≡0 mod 2. Theorem 3.3 The Euler characteristic of a compact spherical complex sat-isﬁes χ(M) = 2 X even α θα(1 −Ωα). Proof. Clearly χ(M) = P β(−1)\|β\| = P β χ′(Mβ). By the Gauss–Bonnet formula from Theorem 2.1, we have (−1)\|β\|χ′(Mβ) = Z Mβ Ψ(x) dv(x) + X α<β Z Mα dv(x) Z Nαβ(x)∗Ψ(x, ξ) dξ. 10 Here we have used equation (3.1) to convert the integral over M[r] into an integral over cells. Since Mβ is a spherical polyhedron of dimension n = \|β\|, we have Ψ(x) = (1/ωn)(1 + (−1)n). Since the spherical polyhedron Mα is totally geodesic, its second fundamental form vanishes and the curvature tensors on Mβ and Mα agree. Thus Ψ(x, ξ) = (1/ωr)(1/ωn−r−1)(1+(−1)r), where r = \|α\|. Integrating, we obtain χ′(Mβ) = θβ(1 + (−1)\|β\|) + (−1)\|β\| X α<β θα(1 + (−1)\|α\|) ωαβ. Summing over β and then rearranging terms, we obtain χ(M) = 2 X even α θα  1 + X β>α (−1)\|β\| ωαβ  . The stated formula now follows from equation (3.3). P e’ e Figure 2. The suspension S0 ∗P of the cell P. Suspensions. Let P be a convex polyhedron in Sn, considered as the equator of Sn+1, and let {e, e′} ∼ = S0 be the corresponding pair of poles. The suspension S0 ∗P is deﬁned by taking the convex hull of P ∪{e, e′}, as in Figure 2; it is swept out by the geodesic arcs joining the poles that pass through P, and satisﬁes vol(S0 ∗P) vol(Sn+1) = vol(P) vol(Sn)· Proof of Theorem 3.1. Let S0 ∗M denote the compact spherical complex assembled from the cells S0 ∗Mα, together with the two additional points e and e′. By Theorem 3.3, we have χ(M) = 2 X even α θα(1 −Ωα). 11 Suspension of Mα does not change the value of θα or of Ωα, but it does shift its dimension by one. To construct S0∗M we must also add two new points, e and e′, each with outer angle Ω(M). By Theorem 3.3 again, we then have χ(S0 ∗M) = 2 X odd α θα(1 −Ωα) + 2(1 −Ω(M)). Adding these two equation together and using the fact that χ(S0 ∗M) = 2 −χ(M) gives the desired formula for Ω(M). Examples and remarks. 1. We have Ω(Sn) = 1 and Ω(p) = 1/2 when p is a single point. 2. A convex spherical polygon has outer angle Ω(M) = 1/2 + \|∂M\|/4π. 3. If M is a spherical cone manifold homeomorphic to S2, then Ω(M) = 1 no matter what the cone angles are. More general, any closed spherical cone manifold of dimension two has outer angle Ω(M) = χ(M)/2. Conceptually, this follows from (3.2) and the fact that a random slice of M is a closed 1-manifold of Euler characteristic zero. (Cone manifolds will be discussed systematically in §5.) 4. The 3-sphere can be given the structure of a spherical cone manifold M in such a way that its singular locus is the Hopf link, with components of length 2πα and 2πβ. This cone manifold satisﬁes vol(M) = αβ and Ω(M) = α + β −αβ. 5. The formula in Theorem 3.1 can be iterated to yield: Ω(M) = X θα1(−θα1α2) · · · (−θαs−1αs), where the sum is over all multi-indices (α1 < α2 < · · · < αs), s ≥1. 4 Riemannian complexes In this section we deﬁne Riemannian complexes and establish the following Gauss–Bonnet formula. Theorem 4.1 Let M be a Riemannian complex with totally geodesic cells. Then X \|α\|=dim(M) Z Mα Ψ(x) dv(x) = X α (−1)\|α\|Θα. 12 This formula applies in particular to spherical, Euclidean and hyperbolic complexes. It relates the volume of M to its Euler characteristic, deﬁned as a weighted sum of simplices. The weight Θα, discussed below, is the normalized solid angle of M along Mα. Smooth complexes. A Riemannian complex M is a cell complex built out of Riemannian polyhedra, in the same way that a spherical complex is built out of spherical polyhedra (§3). We require that faces of cells are also cells, as in (3.1). The faces in a Riemannian complex, however, need not be totally geodesic. The deﬁnitions of the polyhedron Nαβ(x), the normal complex Nα(x), the inner angle θαβ(x) and the outer angle Ωα(x) generalize immediately to this setting. Gauss–Bonnet. By rearranging terms as in the proof of Theorem 3.3, we obtain: Lemma 4.2 The Euler characteristic of any compact Riemannian complex satisﬁes χ(M) = P α(−1)\|α\| R Mα Φα(x) dv(x), where Φα(x) = Ψ(x) + X β>α (−1)\|β\|−\|α\| Z Nαβ(x)∗Ψ(x, ξ) dξ. This formula has the advantage that it shifts the focus from the boundary of Mα (coming from the lower dimension cells) to its normal bundle (coming from higher dimensional cells). Solid angles. Let n = dim(M). The solid angle of M at x ∈Mα is deﬁned by Θα(x) = 1 if \|α\| = n and by Θα(x) = X β>α,\|β\|=n θαβ(x) otherwise. It can be expressed more geometrically as Θα(x) = lim r→0 voln(B(x, r)) voln(Bn) rn . To see this, write the numerator as P β voln(B(x, r) ∩Mβ) and then inter-change limit and summation. Totally geodesic cells. The Gauss–Bonnet integral of a cell will be de-noted by Vα = Z Mα Ψ(x) dv(x). 13 A cell Mα is totally geodesic if its second fundamental form vanishes under every inclusion Mα ⊂Mβ. For example, every cell in a spherical complex is totally geodesic, as is every top–dimensional cell. Whenever A and B are totally geodesic submanifolds of a Riemannian manifold C, parallel transport along a path in A∩B preserves both TA and TB. Thus if all cells in M are totally geodesic, the angles θαβ(x), Ωα(x) and Θα(x) are independent of x. As in the spherical case, we then drop the dependence on x from the notation. Moreover, we ﬁnd Φα(x) = Ψ(x)(1 −Ωα) as in the proof of Theorem 3.3. Since Ψ(x) = 0 unless \|α\| is even, Lemma 4.2 implies: Theorem 4.3 A compact Riemannian complex with totally geodesic cells satisﬁes χ(M) = X α Vα(1 −Ωα). (Note that Vα = 2θα in the spherical case, provided α is even.) Proof of Theorem 4.1. It will be useful to treat the quantities Vα, Θα, χα and χα as vectors, where χα = (−1)\|α\| and χα = χ(Mα). We will also treat θαβ, Ωαβ and Lαβ as matrices, where Ωαβ = Ω(Nαβ) = Ω  [ α<γ≤β Nαγ   and where Lαβ = ( 1 if α ≤β, and 0 otherwise. (We set θαβ = Ωαβ = 0 unless α < β.) With this notation in place, we claim that χ = χL, Ω= θ(L −Ω) and χ = V (L −Ω). The ﬁrst equation simply expresses the fact that χ(Mβ) = X α≤β (−1)\|α\|. 14 The second follows from Theorem 3.1, which gives Ωαβ = X α<γ≤β θαγ(1 −Ωγβ), and the last follows from Theorem 4.3, which gives χ(Mβ) = X α≤β Vα(1 −Ωαβ). Now the equation Ω= θ(L −Ω) implies, by adding L −Ωto both sides, that L = (I + θ)(L −Ω). Thus: V = χ(L −Ω)−1 = χL(L −Ω)−1 = χ(I + θ), or in other words Vβ = χβ + X α<β χαθαβ. Summing over all β with \|β\| = dim M, and using the deﬁnition of Θα, we obtain the formula X \|β\|=n Vβ = X α χαΘα, which is equivalent to the statement of the Theorem. 5 Cone manifolds In this section we discuss Riemannian cone manifolds. Our treatment of their strata relies on the concept of the join A∗B of two spherical complexes, deﬁned below, and the following factorization theorem. Theorem 5.1 A compact spherical cone manifold M can be canonically presented as the join M ∼ = Sa ∗B of a sphere Sa and a prime cone manifold B. We allow the limiting cases M ∼ = Sa and M ∼ = B. Here prime means there is no C such that B ∼ = S0 ∗C. Geometric cone manifolds. Let G be a group acting isometrically on a complete, connected Riemannian manifold X of dimension n. (We do not require that G acts transitively.) Let Gp ⊂G denote the stabilizer of p ∈X, 15 acting on the unit tangent sphere Xp ⊂TpX. An (X, G)–manifold is a space M equipped with charts into X whose transition functions lie in G. Following [Th], we deﬁne an (X, G) cone manifold inductively as follows. For dim(X) = 1, an (X, G) cone manifold is just an (X, G) manifold. For dim(X) > 1, an (X, G) cone manifold is a space M such that each x ∈M has a neighborhood modeled, using the geometry of X, on the cone over a compact, connected (Xp, Gp) cone manifold Sx(M), called the unit tangent cone to M at x. Spherical cone manifolds. A spherical cone manifold is an (Sn, O(n+1)) cone manifold. If a collection of n-dimensional spherical polyhedra are glued together by identifying their (n −1)-dimensional faces in pairs, the result is a spherical cone manifold. Every spherical cone manifold can be obtained in this way, as one can verify by induction on n. Outer angles of cone manifolds. In particular, a compact spherical cone manifold M can be given the structure of spherical cell complex, so its outer angle Ω(M) can be deﬁned as in §3. The outer angle is independent of the choice of subdivision of M into spherical cells, by Proposition 3.2. The unit tangent cone Sx(M) of an (X, G) manifold is a special case of a spherical cone manifold, since (Xp, Gp) ∼ = (Sn−1, H) for some H ⊂O(n). Of course when H ̸= O(n), the space Sx(M) also carries a more rigid structure. Riemannian cone manifolds. A metric space M is a Riemannian cone manifold if each x ∈M has a neighborhood U which is isometric to an open subset of an (X, G) cone manifold, for some (X, G). The pair (X, G) can vary from point to point. A local model for M can be constructed as follows. First, let g be a metric on the unit ball Bn ⊂Rn, such that every radial arc [0, 1] · x is a geodesic. Let H = O(n) ∩Isom(Bn, g). Let K be a connected (Sn−1, H) cone manifold. Then K can be described as a spherical complex (Kα), with isometric charts φα : Kα →Sn−1 that diﬀer on their overlaps by elements of the isometry group H. Since H preserves g, the isometry type of the linear cone Cg(Kα) = [0, 1] · φα(Kα) ⊂Bn is independent of the choice of chart. These polyhedral cones can therefore be assembled to form a Riemannian cell complex Cg(K), homeomorphic to the topological cone on K. 16 Every point x in a Riemannian cone manifold has a neighborhood (U, x) isometric to (Cg(K), 0), where K ∼ = Sx(M), Bn is a small ball in Tp(X), and g is the pullback of the metric on X under the exponential map. When g is the usual Euclidean metric on Bn, we refer to EK = Cg(K) as the Euclidean cone over K. Spherical joins. Let A, B be a pair of compact spherical cone manifolds. Assume A is connected unless it is isomorphic to S0, and similarly for B. The join A ∗B is the spherical cone manifold obtained from the disjoint union of A and B by adding a spherical arc [a, b] of length π/2 between every pair of points a ∈A and b ∈B. More formally, A ∗B = S(0,0)(EA × EB), where EA × EB is given the product metric. For example, Sn ∗Sm ∼ = Sn+m+1. The unit tangent cones of the join are easily computed; we have Sx(A ∗B) ∼ =      (SxA) ∗B if x ∈A, A ∗(SxB) if x ∈B, and S0 ∗(SaA) ∗(SbB) if x is in the open arc (a, b). (5.1) It is convenient to adopt the convention that S−1 = ∅and A ∗∅= A. We say A is prime if it cannot be expressed as a join A ∼ = S0 ∗B. Proof of Theorem 5.1. If M is prime we are done. Otherwise, there is least one way of writing M ∼ = A ∗B with A ∼ = Sk, k ≥0. Fix one such isometry, with k as large as possible. We can then regard A and B as subsets of M. Clearly B is prime. It remains only to show that this decomposition is unique. Let us say x ∈M is a basepoint if M ∼ = S0 ∗C with x ∈S0. Then C ∼ = Sx(M) by the computation (5.1) above. We claim that A coincides with the set of all basepoints of M. Clearly every x ∈A is a basepoint. On the other hand, if x is a basepoint and x ̸∈A, then either x ∈B or x ∈M −(A ∪B). In the ﬁrst case, we have M ∼ = S0 ∗Sx(M) ∼ = S0 ∗A ∗SxB ∼ = Sk+1 ∗SxB, by equation (5.1). But this contradicts the fact that k is maximal. We obtain a similar contradiction if x is in the interior of an arc [a, b]. Consequently A coincides with the set of basepoints of M. Similarly B coincides with the set of points in M at distance π/2 from A. Thus the factors A and B in the join are canonically determined subspaces of M, and there is a unique compatible isometry M ∼ = A ∗B. 17 Strata. Let M be an n-dimensional Riemannian cone manifold. Its a-dimensional strata are the connected components of the locus M[a] = {x ∈M : Sx(M) ∼ = Sa−1 ∗B, with B prime}. Clearly M is the disjoint union of the loci M[a] for 0 ≤a ≤n. We refer to the prime factor B above as the normal cone to M[a] at x; it will be denoted by Nx(M). Thus Sx(M) ∼ = Sa−1 ∗Nx(M) for all x ∈M[a]. Local models. Using the local model U ∼ = Cg(Sx(M)) for a neighborhood of x ∈M[a], it is easy to verify that U ∩M[a] ∼ = Cg(Sa−1); indeed, M[a] ∩U is isomorphic to (Ba, g) in suitable coordinates on Bn. Thus M[a] carries the structure of a Riemannian manifold of dimension a. The local model also shows that the geodesic rays from x to nearby points y ∈M[a] are all contained in M[a]; thus M[a] is totally geodesic in M (equivalently, Ba is totally geodesic in (Bn, g)). All other points of U belong to higher-dimensional strata, by equation (5.1). More precisely, any point y ∈U −{x} projects to a point z ∈Sx(M), and by parallel transport we have Sy(M) ∼ = S0 ∗Sz(Sx(M)). The S0 factor comes from the geodesic connecting x to y. This shows: 1. The unit tangent cones Sy(M) and Sx(M) are isomorphic whenever x and y belong to the same component of M[a]; 2. For b > a, the locus M[b]∩U is just the cone over the locus Sx(M)[b−1], with x omitted; and 3. For y ∈M[b] ∩U, we have Sy(M) ∼ = Sa ∗SwNx(M), where w is obtained by ﬁrst projecting y to Sa−1 and then to Nx(M). The top-dimensional manifold M[n] is open and dense in M. Our def-initions imply that two-dimensional cone manifolds have only isolated sin-gularities; more generally, M[n −1] is empty, and thus M is the metric completion of M[n]. 18 Cell decompositions. Every Riemannian cone manifold M admits a smooth triangulation compatible with its stratiﬁcation. More precisely, M can be given the structure of a Riemannian cell complex, M = S Mα, in such a way that its strata are unions of cells, and each cell is a simplex. The proof is by induction on n = dim M, the case n = 1 being immediate. For the inductive step, suppose x ∈M is modeled on the cone over N, dim N = n−1. Then the cone over a triangulation of N gives a triangulation of a neighborhood of x in M. By putting these local triangulations into general position within their strata, and intersecting their cells, we obtain a decomposition of M into Riemannian polyhedra. These polyhedra can then be further subdivided, by standard methods, to obtain the desired Riemannian triangulation of M (see e.g. [Ca]). Two examples. To conclude we remark that not every Riemannian cone Cg(K) is a Riemannian cone manifold, and not every Riemannian cone manifold M is an (X, G) cone manifold, To see the ﬁrst statement, let K be the 2-fold cover of S3 branched over S1. Then K is an (S3, H) cone manifold for the trivial group H. Consider a metrized ball (B4, g) that contains no totally geodesic surfaces. Then M = Cg(K) cannot be a cone manifold; if it were, then the cone over the branch locus Cg(S1) would be stratum of M, and hence totally geodesic. For the second, consider a surface M with a single irrational cone point p (meaning Θ(p) ̸∈Q), whose metric is inhomogeneous near another point q. If M were an (X, G)–manifold, then Isom(X) would contain a copy of S1 coming from the closure of the group generated by holonomy around p; but then M would admit a local Killing vector ﬁeld near q. 6 Euler characteristics of strata Let M be a compact cone manifold. We say F ⊂M is a stratiﬁed set if F is a closed union of strata of M. In this section we establish the following analogue of Theorem 4.3, whose notation is explained below. Theorem 6.1 The Euler characteristic of any stratiﬁed set F ⊂M is given by χ(F) = X Mσ⊂F V σ(1 −Ωσ(F)). Corollary 6.2 The strata of M satisfy χ(Mτ) = P σ≤τ V σ(1 −Ωστ) . 19 Strata and angles. The strata Mσ of M, deﬁned as the components of M[r] for all r ≤dim(M), give a partition of M into connected manifolds. We let \|σ\| = dim(Mσ), and we write σ ≤τ if Mσ meets Mτ; equivalently, if Mσ ⊂Mτ. Note that we use superscripts to index strata, just as we have used subscripts to index cells. For x ∈Mσ the spherical cone manifold Nx(M) coincides with the space of unit vectors normal to Mσ at x. For τ > σ, the vectors normal to Mσ that point into Mτ form a stratiﬁed subset of Nx(M), which will be denoted by Nστ(x). This stratiﬁed set can be given the structure of a spherical complex, so its outer angle is well–deﬁned, just as for a spherical cone manifold (see §3). Note that Nστ(x) might not be connected – e.g. the stratum Mτ might tend towards Mσ from two diﬀerent directions. We denote the inner and outer angles of Nστ(x) by θστ(x) = vol(Nστ(x))/ vol(S\|τ\|−\|σ\|−1) and Ωστ(x) = Ω(Nστ(x)). Using parallel transport, we ﬁnd that Sx(M) ∼ = Sy(M), Nx(M) ∼ = Ny(M), and Nστ(x) ∼ = Nστ(y) for all x, y ∈Mσ. Thus the angles above are inde-pendent of x; they will henceforth be denoted simply by θστ and Ωστ. The solid angle of M along Mσ is given by Θσ = θστ, for the unique τ ≥σ with \|τ\| = dim(M). (We adopt that convention that Θσ = 1 if \|σ\| = dim(M).) We let V σ = Z Mσ Ψ(x) dv(x) denote the intrinsic Gauss–Bonnet integral over Mσ. Cells. Let F ⊂M be a stratiﬁed set. The normal cone to Mσ ⊂F at x ∈Mσ is deﬁned by Nσ(F, x) = [ τ>σ,Mτ⊂F Nστ(x). It is a stratiﬁed subset of the spherical cone manifold Nx(M). The outer angle Ωσ(F, x) = Ω(Nσ(F, x)) 20 is again independent of x, so it will be denoted simply by Ωσ(F). As discussed in §5, a cone manifold can be decomposed into Riemannian polyhedra in a manner compatible with its stratiﬁcation. In particular, F can be given the structure of a cell complex (Fα) in such a way that Mσ is a subcomplex whenever Mσ ⊂F. Then by Lemma 4.2 we have χ(F) = X (−1)\|α\| Z Fα Φα(x) dv(x), (6.1) where Φα(x) = Ψ(x) − X β>α (−1)\|β\|−\|α\|−1 Z Nαβ(x)∗Ψ(x, ξ) dξ. (6.2) Fracture and repair. The decomposition (Fα) of F generally includes curved cells, which fracture the totally geodesic strata in F. The main step in the proof of Theorem 6.1 is to show that these curved cells make no con-tribution to the sum (6.1) giving the value of χ(F). This type of argument is essential for applications to complex hyperbolic cone manifolds, since they have no odd–dimensional totally geodesic submanifolds (of dimension > 1). Lemma 6.3 If x ∈Fα ⊂Mσ with \|α\| < \|σ\|, then Φα(x) = 0. Proof. Choose a metric g on Bn such that there is a neighborhood U of x in M which is modeled on the cone Cg(Sx(M)). Since x ∈Mσ, we have Sx(M) ∼ = S\|σ\|−1 ∗Nx(M). Since Fα ⊂Mσ has codimension \|σ\| −\|α\|, and Mσ ⊂F, we have K = Nα(x) ∼ = Sd ∗Nσ(F, x) (6.3) where d = \|σ\|−\|α\|−1 ≥0. We will use this presentation of K as a nontrivial suspension to show that Φα(x) = 0. Let Kβ = Nαβ(x) denote the spherical cells making up K, and let C = Bn. Choose submanifolds with 0 ∈A ⊂B ⊂C passing through 0 ∈C which provide local models for the cells Fα and Fβ. Let SB ⊂SC denote the ﬁbers of the unit normal bundle to A at 0, considered as a submanifold of B and C respectively. Then Kβ can be identiﬁed with a convex polyhedron inside SB and SC. Thus it has two duals, which we will denote by K′ β = (Kβ, SB)∗ and K∗ β = (Kβ, SC)∗. By the basic relation (2.2), we have Z Nαβ(x)∗Ψ(x, ξ) dξ = Z K′ β ΨB(0, ξ) dξ = Z K∗ β ΨC(0, ξ) dξ. (6.4) 21 Note that the holonomy group H of the spherical manifold Sx(M) acts isometrically on (Bn, g) and trivially on Sx(Mσ), so it also acts trivially on S0(A), and hence it preserves SC. Thus the integrand ΨC(0, ξ) is H-invariant. Since the embedding Kβ ⊂SC is well-deﬁned up to the action of H, the integral of ΨC(0, ξ) over K∗ β above is also well-deﬁned. By Proposition 3.2, for any other subdivision of K into spherical poly-hedra Lγ of dimension \|γ\|, we have X (−1)\|β\|−\|α\|−1K∗ β = X (−1)\|γ\|L∗ γ as chains in SC, modulo terms of the form C −h(C) with h ∈H (which account for diﬀerent lifts of cells to SC). Since Ψ(0, ξ) is H-invariant, it vanishes when integrated over C −h(C) and we ﬁnd X β (−1)\|β\| Z K∗ β ΨC(0, ξ) dξ = X γ (−1)\|γ\| Z L∗ γ ΨC(0, ξ) dξ. (6.5) Now by (6.3), we can write K = S0 ∗T for some spherical complex T. Let (Tγ) be a spherical cell decomposition of T. Then K can be decomposed into cells of the form Lγ ∼ = S0 ∗(Tγ) −S0, together with two 0-dimensional cells Lϵ and L′ ϵ that account for S0 itself. Each cell of the form Lγ contains an open geodesic joining the points of S0; thus the dual has L∗ γ measure zero, since it lies in the subsphere of SC orthogonal to S0. On the other hand, the two remaining cells satisfy L∗ ϵ ∪L∗ ϵ′ = SC. It follows that X γ (−1)\|γ\| Z L∗ γ ΨC(x, ξ) dξ = Z SC ΨC(x, ξ) dξ = Ψ(x) by the basic relation (2.1). Combining this equation with equations (6.2), (6.4) and (6.5) gives Φα(x) = 0. Lemma 6.4 If x ∈Fα ⊂Mσ with \|α\| = \|σ\|, then Φα(x) = Ψ(x)(1−Ωσ(F)) . Proof. In this case Fα is totally geodesic, and Nα(x) ∼ = Nσ(F, x). 22 Proof of Theorem 6.1. By the preceding two Lemmas and equation (6.1), we have χ(F) = X Mσ⊂F X Fα⊂Mσ (−1)\|α\| Z Fα Φα(x) dv(x) = X Mσ⊂F (1 −Ωσ(F)) Z Mσ Ψ(x) dv(x) = X V σ(1 −Ωσ(F)). Here we have used the fact that Φα(x) = 0 if \|α\| is odd, so we can ignore the factor (−1)\|α\|. 7 Gauss–Bonnet for cone manifolds We can now complete the proof of our main result: Theorem 7.1 For any compact cone manifold M of dimension n, we have Z M[n] Ψ(x) dv(x) = X τ χ(Mτ)Θτ. The proof will rely on: Lemma 7.2 Let M be a compact spherical cone manifold. Then the outer angle of any stratiﬁed set F ⊂M is given by Ω(F) = X Mσ⊂F θσ(1 −Ωσ(F)), where θσ = vol(Mσ)/ vol(S\|σ\|). Corollary 7.3 The outer angles of the strata in any compact cone manifold M satisfy Ωστ = X σ<ξ≤τ θσξ(1 −Ωξτ). Proof. Choose x ∈Mσ. Then F = Nστ(x) is a stratiﬁed subset of the spherical cone manifold Nx(M). For each ξ with σ < ξ ≤τ , the locus Nσξ(x) in F is a union of strata with outer angle Ωξτ and total volume θσξ. The formula then follows from the preceding result. 23 Proofs of Lemma 7.2 and Theorem 7.1. The proofs for cone manifolds closely follow the proofs for Riemannian complexes. Lemma 7.2 follows from Theorem 6.1 by comparing the formulas it gives for χ(M) to χ(S0 ∗M), just as in the proof of Theorem 3.1. As for Theorem 7.1, Corollaries 6.2 and 7.3 give the matrix equations χ = V (L −Ω) and Ω= θ(L −Ω) for strata, exactly as in the proof of Theorem 4.1. The equation χ = χL also holds for strata, provided we let χσ = χ(Mσ) −χ(∂Mσ). The same linear algebra used in the proof of Theorem 4.1 then yields Theorem 7.1. 8 Volumes of moduli spaces The traditional Gauss–Bonnet theorem implies that a closed, complex hy-perbolic manifold M ∼ = CHn/Γ satisﬁes vol(M) = Cnχ(M), where Cn = (−4π)n/(n+1)!. Here the metric is normalized so that CH1 has curvature −1. Note that M is a (G, X)–manifold for (G, X) = (PU(n, 1), CHn). Let M0,n denote the moduli space of n ordered points on the Riemann sphere. For n > 4, the universal cover of M0,n is far from being a symmetric space; in fact its automorphism group, as a complex manifold, is discrete [Roy]. Nevertheless, Thurston showed that M0,n carries a natural complex hyperbolic metric gµ for every sequence of real weights µ = (µ1, . . . , µn) with 0 < µi < 1 and P µi = 2. Moreover the metric completion P(µ) of (M0,n, gµ) is a complex hyperbolic cone manifold. The (G, X) cone manifold P(µ) both rigidiﬁes and extends the original complex manifold structure on M0,n. In this section we will use the Gauss–Bonnet theorem for cone manifolds to show: Theorem 8.1 The complex hyperbolic volume of moduli space satisﬁes vol(M0,n, gµ) Cn−3 = X P (−1)\|P\|+1(\|P\| −3)! Y B∈P max 0, 1 − X i∈B µi !\|B\|−1 . Here P ranges over all partitions of the indices (1, . . . , n) into blocks B. We regard the quantity on the right as a formula for the cone manifold Euler characteristic χ(P(µ)). 24 Shapes of polyhedra. As background for the proof, we recall that any point conﬁguration (b1, . . . , bn) determines a ﬂat metric \|ω\| = \|dz\| Q \|z −bi\|µi on the Riemann sphere with cone angles 2π(1 −µi) at the points bi. The space (b C, \|ω\|) can be isometrically embedded as a convex polyhedron in R3; conversely, a polyhedron with n marked vertices with the given cone angles determines a point conﬁguration (bi) ∈M0,n. The complex hyperbolic metric gµ on M0,n can be expressed in terms of the complexiﬁed distances between the marked points (bi) (i.e. the periods R bj bi ω), which provide local coordinates on M0,n with values in CHn−3. We also recall that the Euler characteristic of moduli space is given by χ(M0,n) = (−1)n−3(n−3)!, as is easily proved using the ﬁbration M0,n+1 → M0,n. Proof of Theorem 8.1. As shown in [Th], the metric completion M of (M0,n, gµ) is a complex hyperbolic cone manifold whose strata MP are in-dexed by partitions P of n such that P i∈B µi < 1 for all B ∈P. These strata result from collisions between the vertices of the associated polyhe-dron which keep it locally convex. The new polyhedron has \|P\| vertices, so we have MP ∼ = M0,\|P\| and thus χ(MP) = (−1)\|P\|+1(\|P\| −3)!. When the points in a block B of P break up, the nearby conﬁguration is described by \|B\| points on a ﬂat 2-dimensional cone with vertex angle 2π(1 −P i∈B µi). Thus the solid angle of MP in M is given by ΘP = Y B∈P 1 − X i∈B µi !\|B\|−1 ; cf. [Th, Prop 3.6(c)]. Provided M is compact, the Gauss–Bonnet formula for cone manifolds then gives vol(M) = Cn−3 X P χ(MP)ΘP, which yields the formula above. The case where M is ﬁnite–volume but non-compact, which occurs only when P i∈B µi = 1 for some collection of indices B ∈P, can be treated by continuity in (µi) or by using the description of cusps in [Th, §5]; see also [Le]. 25 q (pi) χ(P (µ)) χ(M(µ)) 3 1 1 1 1 1 1 -4/9 -1/1620 3 2 1 1 1 1 1/3 1/72 4 1 1 1 1 1 1 1 1 -15/64 -1/172032 4 2 1 1 1 1 1 1 25/128 5/18432 4 3 1 1 1 1 1 -1/16 -1/1920 4 2 2 1 1 1 1 -1/4 -1/192 4 3 2 1 1 1 3/16 1/32 4 2 2 2 1 1 3/8 1/32 5 2 2 2 2 2 3/5 1/200 6 1 1 1 1 1 1 1 1 1 1 1 1 -28315/419904 -809/5746705367040 6 2 1 1 1 1 1 1 1 1 1 1 5663/93312 809/48372940800 6 3 1 1 1 1 1 1 1 1 1 -119/3888 -17/201553920 6 2 2 1 1 1 1 1 1 1 1 -287/4374 -41/50388480 6 4 1 1 1 1 1 1 1 1 49/5832 7/33592320 6 3 2 1 1 1 1 1 1 1 2107/46656 301/33592320 6 5 1 1 1 1 1 1 1 -1/1296 -1/6531840 6 2 2 2 1 1 1 1 1 1 637/7776 637/33592320 6 4 2 1 1 1 1 1 1 -13/648 -13/466560 6 3 3 1 1 1 1 1 1 -11/216 -11/311040 6 3 2 2 1 1 1 1 1 -91/1296 -91/311040 6 5 2 1 1 1 1 1 5/1296 1/31104 6 4 3 1 1 1 1 1 55/1296 11/31104 6 2 2 2 2 1 1 1 1 -13/108 -13/62208 6 4 2 2 1 1 1 1 5/108 5/5184 6 3 3 2 1 1 1 1 55/648 55/31104 6 5 3 1 1 1 1 -1/54 -1/1296 6 4 4 1 1 1 1 -2/27 -1/648 6 3 2 2 2 1 1 1 55/432 55/15552 6 5 2 2 1 1 1 -1/54 -1/648 6 4 3 2 1 1 1 -5/54 -5/324 6 3 3 3 1 1 1 -1/9 -1/324 6 5 4 1 1 1 1/12 1/72 6 2 2 2 2 2 1 1 5/24 1/1152 6 4 2 2 2 1 1 -1/9 -1/108 6 3 3 2 2 1 1 -5/27 -5/216 6 5 3 2 1 1 1/12 1/24 6 4 4 2 1 1 1/6 1/24 6 4 3 3 1 1 1/6 1/24 6 3 2 2 2 2 1 -5/18 -5/432 6 5 2 2 2 1 1/12 1/72 6 4 3 2 2 1 1/4 1/8 6 3 3 3 2 1 1/3 1/18 6 3 3 2 2 2 1/2 1/24 8 3 3 3 3 3 1 -33/128 -11/5120 8 6 3 3 3 1 9/64 3/128 8 5 5 2 2 2 9/32 3/128 8 4 3 3 3 3 9/16 3/128 Table 3. Euler characteristics of the 94 orbifolds M(µ) and their cone manifold covers P(µ), with (µi) = (pi/q). 26 q (pi) χ(P (µ)) χ(M(µ)) 9 4 4 4 4 2 13/27 13/648 10 7 4 4 4 1 3/20 1/40 10 3 3 3 3 3 3 2 293/1000 293/720000 10 6 3 3 3 3 2 -26/125 -13/1500 10 9 3 3 3 2 3/100 1/200 10 6 6 3 3 2 3/10 3/40 10 5 3 3 3 3 3 -17/50 -17/6000 10 8 3 3 3 3 3/25 1/200 10 6 5 3 3 3 39/100 13/200 12 8 5 5 5 1 7/48 7/288 12 7 7 2 2 2 2 2 575/10368 115/497664 12 9 7 2 2 2 2 -23/432 -23/10368 12 7 7 4 2 2 2 -23/216 -23/2592 12 11 7 2 2 2 1/48 1/288 12 9 9 2 2 2 1/8 1/96 12 9 7 4 2 2 7/48 7/96 12 7 7 6 2 2 1/6 1/24 12 7 7 4 4 2 7/24 7/96 12 7 5 3 3 3 3 -31/144 -31/3456 12 5 5 5 3 3 3 -23/72 -23/2592 12 10 5 3 3 3 1/12 1/72 12 8 7 3 3 3 13/48 13/288 12 8 5 5 3 3 7/24 7/96 12 7 6 5 3 3 17/48 17/96 12 6 5 5 5 3 1/2 1/12 12 7 5 4 4 4 11/24 11/144 12 6 5 5 4 4 13/24 13/96 12 5 5 5 5 4 7/12 7/288 14 11 5 5 5 2 6/49 1/49 14 8 5 5 5 5 24/49 1/49 15 8 6 6 6 4 37/75 37/450 18 11 8 8 8 1 13/108 13/648 18 13 7 7 7 2 4/27 2/81 18 10 10 7 7 2 13/54 13/216 18 14 13 3 3 3 13/108 13/648 18 10 7 7 7 5 13/27 13/162 18 8 7 7 7 7 16/27 2/81 20 14 11 5 5 5 99/400 33/800 20 13 9 6 6 6 69/200 23/400 20 10 9 9 6 6 99/200 99/800 24 19 17 4 4 4 11/96 11/576 24 14 9 9 9 7 11/24 11/144 30 26 19 5 5 5 4/75 2/225 30 23 22 5 5 5 37/300 37/1800 30 22 11 9 9 9 16/75 8/225 42 34 29 7 7 7 61/588 61/3528 42 26 15 15 15 13 61/147 61/882 Table 3. (continued) 27 Volumes of 94 orbifolds. The pure moduli space P(µ) has a natural quotient M(µ) = P(µ)/Sn(µ), where Sn(µ) ⊂Sn is the group of permutations satisfying µσi = µi. Deligne and Mostow showed there are exactly 94 values of µ with n ≥5 such that M(µ) is a complex hyperbolic orbifold [DM1], [Mos], [Th]. The volumes of these orbifolds are given by vol(M(µ)) = Cn−3χ(M(µ)), (8.1) and the values of χ(P(µ)) and χ(M(µ)), computed using Theorem 8.1, are listed in Table 3. 9 Hyperbolic twists and non-arithmetic groups We conclude by discussing an invariant of non-arithmetic groups. Let T0,n denote the Teichm¨ uller space of n points on the sphere, regarded as the universal cover of M0,n. Fix real weights with 0 < µi < 1 such that Pn 1 µi = 2. Let Mod0,n(µ) denote the subgroup of ﬁnite index in the mapping-class group, analogous to Sn(µ), which preserves the weights µi. Then the natural inclusion M0,n/Sn(µ) , →M(µ) is an isometry for the metric gµ. Passing to the universal cover, we obtain a locally isometric developing map δ(µ) : T0,n →CHn−3 and a representation ρ(µ) : Mod0,n(µ) ↠Γ(µ) ⊂Isom CHn−3 satisfying δ(µ) ◦g = ρ(g) ◦δ(µ). (9.1) Now suppose M(µ) is an orbifold, deﬁned by rational weights (µi) = (pi/q) with 1 = gcd(p1, . . . , pn, q). Let ν = (kpi/q mod 1), where 1 < k < q is relatively prime to q. We say ν is a hyperbolic twist of µ if P νi = 2. 28 Theorem 9.1 If ν is a hyperbolic twist of the orbifold data µ, then there is a unique holomorphic map F making the diagram T0,n δ(µ) δ(ν) % K K K K K K K K K CHn−3 F / CHn−3 commute, and there is a unique isomorphism α : Γ(µ) →Γ(ν) such that F ◦g = α(g) ◦F for all g ∈Γ(µ). Proof. The normal subgroup Ker ρ(µ) ⊂Mod0,n(µ) is generated by rela-tions coming from the complex codimension one strata of M(µ). Namely, if τ ∈Mod0,n(µ) is a Dehn twist around a simple loop enclosing just bi and bj, and µi + µj < 1 with µi ̸= µj, then we must include the relation τ mij = 1, where mij = q/ gcd(pi + pj, q), to account for the cone angle of 2π/mij along the stratum where bi and bj collide. If µi = µj, then we must add the relation (τ 1/2)mij = 1, where τ 1/2 is a half Dehn twist exchanging bi and bj. Since gcd(k, q) = 1, the value of mij is the same for (νi) = (kpi/q) as for (µi). Thus H = Ker ρ(µ) = Ker ρ(ν). It follows that the developing map δ(ν) : T0,n →CHn−3 descends to an open holomorphic map F : X →CHn−3, where X = T0,n/H. Similarly, since M(µ) is an orbifold, δ(µ) descends to provide an embedding of X into CHn−3. Indeed, X ⊂CHn−3 is nothing more than the preimage of M0,n/Sn(µ) inside the universal cover of M(µ). The locus CHn−3 −X is a locally ﬁnite conﬁguration of complex hyperplanes, covering the lower-dimensional strata of M(µ). By the removability of singularities for bounded analytic functions, F extends to all of CHn−3. Since the kernels of ρ(µ) and ρ(ν) agree, there is a unique isomorphism between their images such that α ◦ρ(µ) = ρ(ν), and (9.1) implies that F ◦g = α(g) ◦F. We remark that Γ(ν) is essentially the Galois conjugate of Γ(µ) coming from the automorphism ζq 7→ζk q of Q(ζq)/Q, where ζq = exp(2πi/q). (The matrix entries in these groups may lie in a larger ﬁeld.) Volume ratios. The volume ratio of the hyperbolic twist is deﬁned by ρ(µ, ν) = vol M(ν) vol M(µ) = χ(M(ν)) χ(M(µ)) = χ(P(ν)) χ(P(µ)). Note that F is a contraction, by the Schwarz lemma, so ρ(µ, ν) ≤1. In the cases at hand, it can be shown that Γ(µ) is arithmetic iﬀit has no hyperbolic twist. 29 q (pi) {ρ(µ, ν)} 12 7 5 3 3 3 3 1/93 12 8 7 3 3 3 1/13 12 6 5 5 4 4 1/13 12 7 6 5 3 3 1/17 18 13 7 7 7 2 1/16 18 8 7 7 7 7 1/16 20 14 11 5 5 5 1/33, 4/33 20 10 9 9 6 6 1/33, 4/33 20 13 9 6 6 6 1/46 12 7 5 4 4 4 1/22 24 19 17 4 4 4 1/22 24 14 9 9 9 7 1/22 15 8 6 6 6 4 1/37, 4/37 30 23 22 5 5 5 1/37, 4/37 42 34 29 7 7 7 1/61, 4/61 42 26 15 15 15 13 1/61, 4/61 Table 4. Nonarithmetic groups Γ(µ), with (µi) = (pi/q). These 16 groups fall into 10 commensurability classes, which are almost all distinguished by the volume ratios given in the right column. (The group Γ(µ) for µ = (7, 5, 4, 4, 4)/12 is in a class by itself.) 30 Commensurability. For 16 of the 94 orbifolds listed in Table 3, the group Γ(µ) is non-arithmetic. The volume ratios for these groups, computed using Theorem 8.1, are listed in Table 4. As shown in [KM, §5], the set of volume ratios is a commensurability invariant for the group Γ(µ). This comes from the fact that the Euler characteristic multiplies by the degree when one passes to a ﬁnite cover, and that the boundary values of F are uniquely determined by the isomorphism Γ(µ) ∼ = Γ(ν). Thus the 16 groups listed fall into at least 9 commensurability classes. In fact, using previous work of Deligne–Mostow and Sauter one ﬁnds that the number of classes is exactly 10. The group Γ(µ) with µ = (7, 5, 4, 4, 4)/12 is not commensurable to the two other groups ρ(µ, ν) = 1/22; its trace ﬁeld is Q(cos π/6), while the other two groups have trace ﬁeld Q(cos π/12), For more details, see [Sau], [DM2] and [KM, §§6–7]. 31 Example: the octagon map. The orbifold weights µ = (3, 3, 5, 5)/8 have a hyperbolic twist ν = (1, 1, 7, 7)/8, with volume ratio ρ(µ, ν) = χ(P(ν)) χ(P(µ) = −1/4 −3/4 = 1 3· In fact P(µ) is the (4, ∞, ∞) orbifold, while P(ν) is a cone manifold of type (4/3, ∞, ∞); it has two cusps and a single cone point with angle 3π/2. The map F : H →H intertwining Γ(µ) and Γ(ν) is shown in Figure 5. It can be constructed by ﬁrst mapping a regular ideal hyperbolic octagon to a regular ideal star, and then developing by Schwarz reﬂection. The map F has a critical point of order 2 at the center of the picture, which expands the angle of a fundamental triangle from π/4 to 3π/4. The volume ratio ρ(µ, ν) gives the average of ∥DF∥2 over the octagon, with the norm measured in the hyperbolic metric. The value ρ(µ, ν) = 1/3 can also be recognized as the second Lya-punov exponent of the Teichm¨ uller curve W8 ⊂M2 associated to billiards in the regular Euclidean octagon. Indeed, W8 is isomorphic to P(µ), and F gives the natural embedding of W8 into the Hilbert modular surface (H × H)/ SL2 Z[ √ 2]. The Lyapunov exponent has a dynamical meaning: along any inﬁnite billiard path in the regular octagon, the number of bounces oﬀa given side by time T satisﬁes N(T) = aT + bT 1/3 + O(1). For a related discussion of billiards in the regular pentagon, see [Mc1, §10]. Bainbridge has shown that, more generally, any SL2(R)-invariant ergodic measure on the moduli space of 1-forms Ω1Mg(2) has Lyapunov exponent 1/3 [Ba1] (for background, see [Ko2]). In view of these examples and Theorem 9.1, the non-arithmetic orbifolds M(µ) with n > 4 can be regarded as certain higher–dimensional versions of Teichm¨ uller curves. For more on this perspective, see [Mc2, §10] and [KM]. Example: the braid group B4. The weights µ = (7, 7, 7, 7, 8)/18 are particular interesting, since in this case Sn(µ) = S4 and the map ρ(µ) : Mod0,5(µ) →Isom CH2 provides a unitary representation of the braid group B4. (More general representations of the braid group via periods are discussed in [Mc2].) In this case the stratum P(µ) has 10 components of two diﬀerent types: four with the weights (7, 7, 7, 15)/18 after 2 vertices have collided, and 6 32 F Figure 5. The hyperbolic octagon and an 8-pointed star. with the weights (7, 7, 8, 14)/18. Similarly P(µ) consists of 15 points of two diﬀerent types; 12 from the weights (7, 14, 15) and 3 from (8, 14, 14). Using Theorem 8.1, we ﬁnd χ(P(µ)) = 2 − 4 × 1 6 + 6 × 2 9 + 12 × 1 27 + 3 × 4 81 = 16 27· These weights have a unique hyperbolic twist, namely (νi) = (11µi mod 1) = (5, 5, 5, 5, 16)/20, and the corresponding volume ratio is given by ρ(µ, ν) = vol χ(P(ν))χ(P(µ)) = 1/27 16/27 = 1 16. The volume of the orbifold itself is given by vol(M(µ)) = C2 χ(P(µ)) 4! = 16π2 243 · References [AW] C. B. Allendoerfer and A. Weil. The Gauss-Bonnet theorem for Riemannian polyhedra. Trans. Amer. Math. Soc. 53(1943), 101– 129. [AB] M Atiyah and C. LeBrun. Curvature, cones and characteristic num-bers. Math. Proc. Cambridge Philos. Soc. 155(2013), 13–37. 33 [Ba1] M. Bainbridge. Euler characteristics of Teichm¨ uller curves in genus two. Geom. Topol. 11(2007), 1887–2073. [Ba2] T. Banchoﬀ. Critical points and curvature for embedded polyhedra. J. Diﬀerential Geometry 1(1967), 245–256. [Br] K. Bromberg. Hyperbolic cone-manifolds, short geodesics, and Schwarzian derivatives. J. Amer. Math. Soc. 17(2004), 783–826. [Ca] S. S. Cairns. A simple triangulation method for smooth manifolds. Bull. Amer. Math. Soc. 67(1961), 389–390. [Cg] J. Cheeger. Spectral geometry of singular Riemannian spaces. J. Diﬀ. Geom. 18(1987), 575–657. [Cn] S. S. Chern. A simple intrinsic proof of the Gauss–Bonnet formula for closed Riemannian manifold. Ann. Math. 45(1944), 747–782. [CHK] D. Cooper, C. D. Hodgson, and S. P. Kerckhoﬀ. Three-Dimensional Orbifolds and Cone-Manifolds. Mathematical Society of Japan, 2000. [DM1] P. Deligne and G. D. Mostow. Monodromy of hypergeometric functions and non-lattice integral monodromy. Publ. Math. IHES 63(1986), 5–89. [DM2] P. Deligne and G. D. Mostow. Commensurabilities among Lattices in PU(1, n). Annals of Math. Studies. Princeton Univ. Press, 1993. [ES] V. Emery and M. Stover. Covolumes of nonuniform lattices in PU(n, 1). Amer. J. Math. 136(2014), 143–164. [FST] B. Fedosov, B-W. Schulze, and N. Tarkhanov. The index of elliptic operators on manifolds with conical points. Selecta Math. 5(1999), 467–506. [Ge] E. Getzler. Operads and moduli spaces of genus 0 Riemann surfaces. In The Moduli Space of Curves, pages 199–230. Birkh¨ auser Boston, 1995. [Gr] A. Gray. Tubes. Birkh¨ auser, 2004. [HMM] A. Hassell, R. Mazzeo, and R. B. Melrose. A signature formula for manifolds with corners of codimension two. Topology 36(1997), 1055–1075. 34 [Ha] B. Hassett. Moduli spaces of weighted pointed stable curves. Adv. Math. 173(2003), 316–352. [KM] A. Kappes and M. M¨ oller. Lyapunov spectrum of ball quotients with applications to commensurability questions. Preprint, 2012. [Ko1] S. Kojima. Deformations of hyperbolic 3-cone-manifolds. J. Diﬀer-ential Geom. 49(1998), 469–516. [Ko2] M. Kontsevich. Lyapunov exponents and Hodge theory. In The Mathematical Beauty of Physics (Saclay, 1996), pages 318–332. World Sci. Publishing, 1997. [Le] E. Leuzinger. On the Gauss-Bonnet formula for locally symmetric spaces of noncompact type. Enseign. Math. 42(1996), 201–214. [Mc1] C. McMullen. Billiards and Teichm¨ uller curves on Hilbert modular surfaces. J. Amer. Math. Soc. 16(2003), 857–885. [Mc2] C. McMullen. Braid groups and Hodge theory. Math. Ann., 355(2013), 893–946. [Mc3] C. McMullen. Moduli spaces in genus zero and inversion of power series. Enseign. Math. 60(2014), 25–30. [Mos] G. D. Mostow. Generalized Picard lattices arising from half-integral conditions. Publ. Math. IHES 63(1986), 91–106. [Par] J. R. Parker. Complex hyperbolic lattices. In Discrete Groups and Geometric Structures, pages 1–42. Amer. Math. Soc., 2009. [Pf] M. J. Pﬂaum. Analytic and geometric study of stratiﬁed spaces, volume 1768 of Lecture Notes in Mathematics. Springer-Verlag, 2001. [Roy] H. L. Royden. Automorphisms and isometries of Teichm¨ uller space. In Advances in the Theory of Riemann Surfaces, pages 369–384. Princeton University Press, 1971. [Sat] I. Satake. The Gauss-Bonnet theorem for V -manifolds. J. Math. Soc. Japan 9(1957), 464–492. [Sau] J. K. Sauter. Isomorphisms among monodromy groups and applica-tions to lattices in PU(1, 2). Paciﬁc J. Math. 146(1990), 331–384. 35 [Th] W. P. Thurston. Shapes of polyhedra and triangulations of the sphere. In The Epstein Birthday Schrift, volume 1 of Geom. Topol. Monogr., pages 511–549. Geom. Topol. Publ., 1998. [Tr] M. Troyanov. On the moduli space of singular euclidean surfaces. In A. Papadopoulos, editor, Handbook of Teichm¨ uller Theory, vol-ume I, pages 507–540. Eur. Math. Soc., 2007. [We] H. Weyl. On the volumes of tubes. Amer. J. Math. 61(1939), 461– 472. [Yo] M. Yoshida. Volume formula for certain discrete reﬂection groups in PU(2, 1). Mem. Fac. Sci. Kyushu Univ. Ser. A 36(1982), 1–11. [Zo] P. Zograf. The Weil–Petersson volume of the moduli space of punctured spheres. In Mapping class groups and moduli spaces of Riemann surfaces, volume 150 of Contemp. Math., pages 367–372. Amer. Math. Soc., 1993. Mathematics Department Harvard University 1 Oxford St Cambridge, MA 02138-2901 36
1954	https://www.youtube.com/watch?v=ElXOmYyL8C8	Calculating d-spacings for different crystal structures Taylor Sparks 44800 subscribers 134 likes Description 18486 views Posted: 27 Mar 2018 Tutorial example problem solutions for determining d-spacings (interplanar spacings) for different (hkl) planes in several different crystal systems (cubic, hexagonal, orthorhombic). I show how these d-spacings exist in actual structures with VESTA software showing the planes 11 comments Transcript: okay this question says determine the inter planar spacing of the following and it gives four different crystal structures and each one of these has a different age KL plane that we'd like to forget that in place in story so inter planar spacing what does that even refer to well let's pull up the beste diagram and show what these spacings actually refer to so I'm going to pull the simple one this is body centered cubic chromium what I've done here is investor I've told it to draw the same plane located different D spacings away from the origin right so our origins give me in this back corner this add-on and then we've done D spacings which are inter planar spacings the same thing where we solved for this question away from that so again if you go up here to edit and you drop your lattice planes this is the plane that we're interested in the O 1/3 plane and I've done 1d spacing on this one I've done 2d space and then this one done 3d spaces so what we're seeing here is that if you were to look at that right that on there's a separation distance between these two planes and it's the same as the separation between these in fact there's infinite many planes that exist here and that's important for things like x-ray diffraction is because there's diffraction that can occur between these two different planes so in order to calculate that that fraction we need to figure out what is the distance between these geometrical sort of planes in the material and we have some great formula so let's do that so shown here is in a table this table shows you how to calculate the inter planar spacing for all seven different crystal structure systems you might have right whether it's cubic or hexagonal tetragonal orthorhombic those are the four simple ones or even for the more hard ones right like monoclinic triclinic or trigonal right these ones get more complicated is you have to take into account the angles between the axes in something like this this is a cubic crystal structure all the angles between these three principal axes it is 90 degrees it gets harder in the other ones because they're not so in these for example that they want us to solve for they all have 90 degrees between their angles except for the hexagonal which has a special case a 10 or 20 degrees so let's go ahead and dive into this do the first one so for question a for chromium we're given the lattice parameter and we're given the hko back to the plane so how do we figure it out we're going to use this equation right there we know that over d squared and that D is going to correspond to these specific HKS so we can even write 0 1 3 if we wanted to that's going to correspond to H squared which is going to be 0 squared plus K squared 1 squared plus l squared which is 3 squared all of that divided by just a squared which in this case is 2 point 8 8 4 9 angstroms squared so when I go ahead and plug that in I find that that is equal to let's see what we can rearrange for D D for the 0 1 3 plane if we rearrange this it's going to be equal to the square root of 2 point 8 8 4 9 angstroms squared all divided by 10 since that's what 1 squared plus 3 squared is equal to when I punch them in my calculator I get that that's equal to a zero point 9 1 2 2 angstroms that's our D spacing for that so let's go ahead and double check that with our beste software again that was for chromium so let's go ahead pull up chromium and see what that came out to be chromium with 1 D spacing that came out to be point 9 1 to 2 so it's the same value let's move on to our next one with copper copper it's now the 1 1 1 plane and we have a different lattice parameter but it's still cubic copper is still cubic so it's to be the exact same approach so we can go ahead and write this we're going to say that D for the 1 1 1 and copper is going to be equal to the square root of again it's gonna be our lattice parameter squared 3 point 6 1 5 angstroms squared now it's divided by 1 squared plus 1 squared plus 1 squared which is just 3 we punch that into a calculator I get that it is equal to two point zero eight seven angstroms so let's double check that one up over here two Vesta pull up FCC where I've already input the planes and the last planes sure enough for one D spacing it's two point zero eight seven angstroms so pretty good so the next one this is now magnesium magnesium is hexagonal close-packed so it's a hexagonal crystal structure which means we have both a and C lattice parameters so that's going to change these just a little bit the formula is now this one it's a little more complicated for example now 1 over D and this is going to be for the 1 3 2 that's going to be equal to 4/3 multiplied by H squared plus K squared plus 8 times K all of this divided by a squared meanwhile it's going to be plus l squared over C squared so let's write what that would be equal to 4/3 x okay H K L in this case that's going to be equal to 1 squared plus 3 squared plus 3 times 1 all that divided by the a lattice parameter which is 3 point 2 3 4 angstroms squared plus RL L in this case is 2 so it's gonna be 2 squared l squared is 2 squared over C squared which is 5 point 2 5 2 5 point 2 5 2 angstroms and then we have to square that so punching all that in sorry this is 1 over d squared that's equal to 1 point 8 0 2 so now solving for D D is going to be equal to the square root of 1 over 1 point 8 0 2 angstroms which I get as a value of zero point seven four four eight zero point seven four four eight angstroms okay so let's double check that over here in our pigs and crystal structure if we pull up our lattice planes let's see sure enough 0.74 496 yeah pretty close what we're getting let's do our last one this one is sodium silver molybdenum oxide this is actually a distorted spinel spinel is cubic that since this one is distorted it's now orthorhombic right so it has a B and C lattice parameters that are all different for one another and so the two zero one we can go and do this one orthorhombic is going to be a relatively simple formula it's going to be 1 over d h k l squared is equal to the h squared over a squared plus K squared over B squared plus C squared excuse me l squared over C squared and they're backwards so we go ahead and punch these in since it's the negative 2 0 1 the zero means that that second term is going to just be going away right so we have negative 2 squared that's just going to be 4 over a squared which is ten point three eight four angstroms squared plus zero squared over B squared doesn't matter it is because I'm going to be nothing and then l squared was 1 1 squared so it's just 1 over C squared and C is five point five nine three three five point five nine three three that's angstroms squared when I punch all of that in I find that it's equal to zero point zero six nine zero point zero six nine zero six angstroms right so well that's per angstrom excuse me so when we solve for that D will equal 1 over 0.06 906 ångstroms would actually be over here punching that in we get that that's equal to three point eight oh five angstroms okay let's double check that last one on Vesta so again I would have downloaded the suit file for that here's the crystal structure I go to edit those planes we've already put that in and sure enough three point eight zero five is one D spacing line so this would just be acceptor level and that's how you solve inter planar spacing you could do so with any other crystal structure if you want they just a little bit more complicated in the math
1955	https://physics.weber.edu/schroeder/statmech.pdf	A diﬀerent approach to introducing statistical mechanics Thomas A. Moore Department of Physics and Astronomy Pomona College, Claremont, CA 91711 tmoore@pomona.edu Daniel V. Schroeder Department of Physics Weber State University, Ogden, UT 84408–2508 dschroeder@cc.weber.edu The basic notions of statistical mechanics (microstates, multiplicities) are quite simple, but understanding how the second law arises from these ideas requires working with cumbersomely large numbers. To avoid getting bogged down in mathematics, one can compute multiplicities numerically for a simple model system such as an Einstein solid—a collection of identical quantum harmonic oscillators. A computer spreadsheet program or comparable software can compute the required combinatoric functions for systems containing a few hundred oscillators and units of energy. When two such systems can exchange energy, one immediately sees that some conﬁgurations are overwhelmingly more probable than others. Graphs of entropy vs. energy for the two systems can be used to motivate the theoretical deﬁnition of temperature, T = (∂S/∂U)−1, thus bridging the gap between the classical and statistical approaches to entropy. Further spreadsheet exercises can be used to compute the heat capacity of an Einstein solid, study the Boltzmann distribution, and explore the properties of a two-state paramagnetic system. c ⃝1997 American Association of Physics Teachers. Published in American Journal of Physics 65 (1), 26–36 (January, 1997). 1 I. INTRODUCTION Entropy is a crucial concept in thermal physics. A solid understanding of what entropy is and how it works is the key to understanding a broad range of physical phenomena, and enhances one’s understanding of an even wider range of phenomena. The idea of entropy and its (nearly) inevitable increase are thus core concepts that we should hope physics students at all levels would study carefully and learn to apply successfully.1 Unfortunately, the approach to teaching entropy found in virtually all introductory and many upper-level textbooks is more abstract than it needs to be. Such texts2 introduce entropy in the context of macroscopic thermodynamics, using heat engines and Carnot cycles to motivate the discussion, ultimately deﬁning entropy to be the quantity that in-creases by Q/T when energy Q is transferred by heating to a system at temperature T. In spite of the way that this approach mirrors the historical development of the idea of entropy, our students have usually found the intricate logic and abstraction involved in getting to this deﬁnition too subtle to give them any intuitive sense of what entropy is and what it measures; the concept ultimately remains for them simply a formula to memorize.3 Most introductory texts attempt to deal with this problem by including a section about how entropy quantiﬁes “disorder”. The accuracy and utility of this idea, however, depends on exactly how students visualize “disorder”, and in many situations this idea is unhelp-ful or even actively misleading.4 The macroscopic approach therefore generally does not help students develop the kind of accurate and useful intuitive understanding of this very important concept that we would like them to have. The statistical interpretation of entropy (as the logarithm of the the number of quantum microstates consistent with a system’s macrostate) is, comparatively, quite concrete and straightforward. The problem here is that it is not trivial to show from this deﬁnition that the entropy of a macroscopic system of interacting objects will inevitably increase.5 This paper describes our attempt to link the statistical deﬁnition of entropy to the second law of thermodynamics in a way that is as accessible to students as possible.6 Our primary motivation was to create a more contemporary approach to entropy to serve as the heart of one of the six modules for Six Ideas That Shaped Physics, a new calculus-based introductory course developed with the support and assistance of the Introductory University Physics Project (IUPP). However, we have found this approach to be equally useful in upper-level undergraduate courses. The key to our approach is the use of a computer to count the numbers of microstates associated with various macrostates of a simple model system. While the processes involved in the calculation are simple and easy to understand, the computer makes it practical to do the calculation for systems large enough to display the irreversible behavior of macroscopic objects. This vividly displays the link between the statistical deﬁnition of entropy and the second law without using any sophisticated mathematics. Our approach has proved to be very helpful to upper-level students and is the basis of one of the most successful sections of the Six Ideas introductory course.7 Depending on the level of the course and the time available, the computer-based approach can be used either as the primary tool or as a springboard to the more powerful analytic methods presented in statistical mechanics texts. 2 A number of choices are available for the speciﬁc computational environment: a stan-dard spreadsheet program such as Microsoft Excel or Borland QuattroPro; a stand-alone program specially written for the purpose; or even a graphing calculator such as the TI-82. We will use spreadsheets in our examples in the main body of this paper, and brieﬂy describe the other alternatives in Section VI. After some preliminary deﬁnitions in Section II, we present the core of our approach, the consideration of two interacting “Einstein solids”, in Section III. In the remaining sections we derive the relation between entropy, heat, and temperature, then go on to describe a number of other applications of spreadsheet calculations in statistical mechanics. II. PHYSICS BACKGROUND A. Basic Concepts of Statistical Physics The aim of statistical mechanics is to explain the behavior of systems comprised of a very large number of particles. We can characterize the physical state of such a system in either of two ways, depending on whether we focus on its characteristics at the macroscopic or molecular level. The macrostate of a system is its state as viewed at a macroscopic level. For instance, to describe the macrostate of a tank of helium gas, we could specify the number of molecules N it contains, the total volume V that they occupy, and their total internal energy U. As long as the gas is in internal thermal equilibrium, these three parameters suﬃce to determine its macrostate. (Other thermodynamic variables, such as the temperature T and the pressure P of the gas, can then be calculated using various “equations of state” such as the ideal gas law PV = NkT and the relation U = 3 2NkT, where k is Boltzmann’s constant.) In general, the macrostate of a system is determined by the various conservation laws and physical constraints that it is subject to. The microstate of a system is its state as viewed at the molecular level. To describe the microstate of our tank of helium gas, we would have to specify the exact position and velocity (or quantum mechanical wavefunction) of each individual molecule. Note that if we know a system’s microstate, we also know its macrostate: The total energy U, for instance, is just the sum of the energies of all the molecules. On the other hand, knowing a system’s macrostate does not mean we’re anywhere close to knowing its microstate. The number of microstates that are consistent with a given macrostate is called the multiplicity of that macrostate, denoted Ω. Equivalently, Ωis the number of accessible microstates, limited only by the constraints that determine the macrostate. For our tank of helium gas, the multiplicity would be a function of three macroscopic variables, for instance, Ω(U, V, N). In most cases the multiplicity will be a very large number, since there will be an enormous number of diﬀerent ways to distribute a given amount of energy among the system’s molecules (and, for ﬂuids, to distribute the molecules within the available volume). The fundamental assumption of statistical mechanics is that an isolated system in a given macrostate is equally likely to be in any of its accessible microstates. This means 3 that the probability of ﬁnding the system in any given microstate is 1/Ω, since there are Ωmicrostates, all equally probable. In the next section we will see how this assumption leads to the conclusion that macroscopic objects must exhibit irreversible behavior, such as the spontaneous ﬂow of energy from a hot object to a cold one. In order to make this argument concrete, though, we will focus our attention on a speciﬁc model for a thermodynamic system, since the exact way that a macrostate’s multiplicity depends on N, U, and V depends on what physical system we are considering. Once we understand how the argument works for our speciﬁc model, we will be able to understand (at least qualitatively) how it would apply to other physical systems. B. The Einstein Solid Model There are only a few realistic systems whose multiplicities can be calculated using ele-mentary methods. Even an ideal gas is too complicated in this regard—a direct derivation of its multiplicity function requires some fairly sophisticated tools.8 A much simpler system would be one whose particles are all ﬁxed in place (so the volume is irrelevant), storing en-ergy in units that are all the same size. This idealized model is treated in many statistical mechanics texts.9 Physically, it corresponds to a collection of identical quantum harmonic oscillators. In 1907, Einstein proposed such a model to describe the thermal properties of a simple crystalline solid, treating each atom as an independent three-dimensional harmonic oscillator. We will therefore call this model an Einstein solid. We will use N to denote the number of one-dimensional oscillators in an Einstein solid. (So the number of atoms is N/3.) If we neglect the static zero-point energy (which plays no role in the thermal behavior), then the energy of any oscillator can be written E = ϵn, (1) where n is a nonnegative integer and ϵ is a constant (ϵ = hf, where f is the oscillator’s natural frequency). In other words, each oscillator can store any number of units of energy, all of the same size. The total energy of the entire solid can thus be written U = ϵq, (2) where q is another nonnegative integer. The multiplicity of an Einstein solid is some function of N and q: the number of ways of arranging q units of energy among N oscillators. For a very small Einstein solid one can simply count all the ways of arranging the energy among the oscillators. For example, if N = q = 3, the following 10 distinct distributions are possible (where the ﬁrst number in each triplet speciﬁes the number of energy units in the ﬁrst oscillator, the second the number of energy units in the second oscillator, etc.): 300, 030, 003, 210, 120, 201, 102, 021, 012, 111. Thus the multiplicity of this macrostate is Ω= 10. By directly counting microstates in this manner, one can not only vividly illustrate the concepts of macrostates and microstates to ﬁrst-year students, but also show how the multiplicity increases very sharply with both N and q, without using any diﬃcult mathematics. The general formula for the multiplicity of an Einstein solid is Ω(N, q) = q + N −1 q = (q + N −1)! q!(N −1)! = (U/ϵ + N −1)! (U/ϵ)!(N −1)! . (3) 4 An elegant derivation of this formula can be found in the textbook of Callen.10 In brief, the trick is to represent any microstate as a sequence of dots (representing energy units) and lines (representing partitions between one oscillator and the next). A total of q dots and N −1 lines are required, for a total of q + N −1 symbols. The number of possible arrangements is just the number of ways of choosing q of these symbols to be dots. Whether or not one derives formula (3) in class, students have little trouble believing it after they have veriﬁed a few simple cases by brute-force counting. In an introductory course, this is all we usually have them do. III. INTERACTING SYSTEMS To understand irreversible processes and the second law, we now move on to consider a system of two Einstein solids, A and B, which are free to exchange energy with each other but isolated from the rest of the universe. First we should be clear about what is meant by the “macrostate” of such a composite system. For simplicity, we assume that our two solids are weakly coupled, so that the exchange of energy between them is much slower than the exchange of energy among atoms within each solid. Then the individual energies of the two solids, UA and UB, will change only slowly; over suﬃciently short time scales they are essentially ﬁxed. We will use the word “macrostate” to refer to the state of the combined system, as speciﬁed by the (temporarily) constrained values of UA and UB. For any such macrostate we can compute the multiplicity, as we shall soon see. However, on longer time scales the values of UA and UB will change, so we will also talk about the total multiplicity of the combined system, considering all microstates to be “accessible” with only Utotal = UA + UB held ﬁxed. In any case, we must also specify the number of oscillators in each solid, NA and NB, in order to fully describe the macrostate of the system. Let us start with a very small system, in which each of the “solids” contains only three oscillators and the solids share a total of six units of energy:11 NA = NB = 3; qtotal = qA + qB = 6. (4) Given these parameters, we must still specify the individual value of qA or qB to describe the macrostate of the system. There are seven possible macrostates, with qA = 0, 1, . . . 6. For each of these macrostates, we can compute the multiplicity of the combined system as the product of the multiplicities of the individual systems: Ωtotal = ΩAΩB. (5) This works because the systems are independent of each other: For each of the ΩA mi-crostates accessible to solid A, there are ΩB microstates accessible to solid B. A table of the various possible values of ΩA, ΩB, and Ωtotal is shown in Fig. 1. Students can easily 5 Two Einstein Solids NA= 3 NB= 3 q_total= 6 q A OmegaA qB OmegaB OmegaTotal 0 1 6 2 8 2 8 1 3 5 2 1 6 3 2 6 4 1 5 9 0 3 1 0 3 1 0 1 0 0 4 1 5 2 6 9 0 5 2 1 1 3 6 3 6 2 8 0 1 2 8 4 6 2 q_A Omega_Total 0 2 0 4 0 6 0 8 0 1 0 0 0 1 2 3 4 5 6 Fig. 1. Computation of the multiplicies for two Einstein solids, each containing three oscillators, which share a total of six units of energy. The individual multiplicities (ΩA and ΩB) can be computed by simply counting the various microstates, or by applying Eq. (3). The total multiplicity Ωtotal is the product of the individual multiplicities, because there are ΩB microstates of solid B for each of the ΩA microstates of solid A. The total number of microstates (462) is computed as the sum of the last column, but can be checked by applying Eq. (3) to the combined system of six oscillators and six energy units. The ﬁnal column is plotted in the graph at right. construct this table for themselves, computing the multiplicities by brute-force counting if necessary. From the table we see that the fourth macrostate, in which each solid has three units of energy, has the highest multiplicity (100). Now let us invoke the fundamental assumption: On long time scales, all of the 462 microstates are accessible to the system, and therefore, by assumption, are equally probable. The probability of ﬁnding the system in any given macrostate is therefore proportional to that macrostate’s multiplicity. For instance, in this example the fourth macrostate is the most probable, and we would expect to ﬁnd the system in this macrostate 100/462 of the time. The macrostates in which one solid or the other has all six units of energy are less probable, but not overwhelmingly so. Another way to construct the table of Fig. 1 is to use a computer spreadsheet program. Both Microsoft Excel and Borland QuattroPro include built-in functions for computing binomial coeﬃcients: q + N −1 q = COMBIN(q + N −1, q) in Excel; MULTINOMIAL(q, N −1) in QuattroPro. (6) The formulas from the Excel spreadsheet that produced Fig. 1 are shown in Fig. 2. The advantage in using a computer is that it is now easy to change the parameters to consider larger systems. Changing the numbers of oscillators in the two solids can be accomplished by merely typing new numbers in cells B2 and D2. Changing the number of energy units requires also adding more lines to the spreadsheet. The spreadsheet becomes rather cumbersome with much more than a few hundred energy units, while overﬂow errors can occur if there are more than a few thousand oscillators in each solid. Fortunately, most of the important qualitative features of thermally interacting systems can be observed long before these limits are reached. 6 1 2 3 4 5 6 7 8 9 1 0 1 1 A B C D E F Two Einstein Solids NA= 3 NB= 3 q_total= 6 q A OmegaA qB OmegaB OmegaTotal 0 =COMBIN($B$2-1+A4,A4) =$F$2-A4 =COMBIN($D$2-1+C4,C4) =B4D4 =A4+1 =COMBIN($B$2-1+A5,A5) =$F$2-A5 =COMBIN($D$2-1+C5,C5) =B5D5 =A5+1 =COMBIN($B$2-1+A6,A6) =$F$2-A6 =COMBIN($D$2-1+C6,C6) =B6D6 =A6+1 =COMBIN($B$2-1+A7,A7) =$F$2-A7 =COMBIN($D$2-1+C7,C7) =B7D7 =A7+1 =COMBIN($B$2-1+A8,A8) =$F$2-A8 =COMBIN($D$2-1+C8,C8) =B8D8 =A8+1 =COMBIN($B$2-1+A9,A9) =$F$2-A9 =COMBIN($D$2-1+C9,C9) =B9D9 =A9+1 =COMBIN($B$2-1+A10,A10=$F$2-A10 =COMBIN($D$2-1+C10,C10=B10D10 =SUM(E4:E10 Two Einstein Solids NA= 3 0 0 NB= 2 0 0 q_total= 1 0 0 q A OmegaA qB OmegaB OmegaTotal 0 1 1 0 0 3E+81 2.77E+81 1 3 0 0 9 9 9E+80 2.78E+83 2 45150 9 8 3E+80 1.39E+85 3 5E+06 9 7 1E+80 4.62E+86 4 3E+08 9 6 3E+79 1.15E+88 5 2E+10 9 5 1E+79 2.27E+89 6 1E+12 9 4 4E+78 3.73E+90 7 5E+13 9 3 1E+78 5.23E+91 8 2E+15 9 2 4E+77 6.39E+92 9 6E+16 9 1 1E+77 6.91E+93 1 0 2E+18 9 0 4E+76 6.7E+94 1 1 5E+19 8 9 1E+76 5.88E+95 1 2 1E+21 8 8 3E+75 4.71E+96 1 3 3E+22 8 7 1E+75 3.47E+97 1 4 7E+23 8 6 3E+74 2.36E+98 1 5 2E+25 8 5 1E+74 1.5E+99 q_A Omega_Total 0 1E+114 2E+114 3E+114 4E+114 5E+114 6E+114 7E+114 0 2 0 4 0 6 0 8 0 1 0 0 Fig. 2. Formulas from the Excel spreadsheet that produced Fig. 1. (Most of the formulas can be generated automatically by copying from one row to the next. Dollar signs indicate “absolute” references, which remain unchanged when a formula is copied.) Fig. 3. Spreadsheet calculation of the multiplicity of each macrostate, for two larger Einstein solids sharing 100 energy units. Here solids A and B contain 300 and 200 energy units, respectively. Only the ﬁrst few lines of the table are shown; the table continues down to qA = 100. The ﬁnal column is plotted in the graph at right. Notice from the graph that the multiplicity at qA = 60 is more than 1033 times larger than at qA = 0. Figure 3 shows a spreadsheet-generated table, and a plot of Ωtotal vs. qA, for the parameters NA = 300, NB = 200, qtotal = 100. (7) We see immediately that the multiplicities are now quite large. But more importantly, the multiplicities of the various macrostates now diﬀer by factors that can be very large—more than 1033 in the most extreme case. If we again make the fundamental assumption, this result implies that the most likely macrostates are 1033 times more probable than the least likely macrostates. If this system is initially in one of the less probable macrostates, and we then put the two solids in thermal contact and wait a while, we are almost certain to ﬁnd it fairly near the most probable macrostate of qA = 60 when we check again later. In other 7 words, this system will exhibit irreversible behavior, with net energy ﬂowing spontaneously from one solid to the other, until it is near the most likely macrostate. This conclusion is just a statement of the second law of thermodynamics. There is one important limitation in working with systems of only a few hundred particles. According to the graph of Fig. 3, there are actually quite a few macrostates with multiplicities nearly as large as that of the most likely macrostate. In relative terms, however, the width of this peak will decrease as the numbers of oscillators and energy units increase. This is already apparent in comparing Figs. 1 and 3. If the numbers were increased further, we would eventually ﬁnd that the relative width of the peak (compared to the entire horizontal scale of the graph) is completely negligible. Unfortunately, one cannot verify this statement directly with the spreadsheets. Students in an introductory class have little trouble believing this result, when reminded of the fact that macroscopic systems contain on order of 1023 particles and energy units. In an upper-division course, one can prove using Stirling’s approximation that the relative width of the peak is of order 1/ √ N or 1/√q, whichever is larger. Thus, statistical ﬂuctuations away from the most likely macrostate will be only about one part in 1011 for macroscopic systems.12 Using this result, we can state the second law more strongly: Aside from ﬂuctua-tions that are normally much too small to measure, any isolated macroscopic system will inevitably evolve toward its most likely accessible macrostate. Although we have proved this result only for a system of two weakly coupled Einstein solids, similar results apply to all other macroscopic systems. The details of the counting become more complex when the energy units can come in diﬀerent sizes, but the essential properties of the combinatorics of large numbers do not depend on these details. If the subsystems are not weakly coupled, the division of the whole system into subsystems becomes a mere mental exercise with no real physical basis; nevertheless, the same counting arguments apply, and the system will still exhibit irreversible behavior if it is initially in a macrostate that is in some way “inhomogeneous”. IV. ENTROPY AND TEMPERATURE Instead of always working with multiplicities, it is convenient to deﬁne the entropy of a system as S = k ln Ω. (8) The factor of Boltzmann’s constant is purely conventional, but the logarithm is important: It means that, for any given macrostate, the total entropy of two weakly interacting13 systems is the sum of their individual entropies: Stotal = k ln(ΩAΩB) = k ln ΩA + k ln ΩB = SA + SB. (9) Since entropy is a strictly increasing function of multiplicity, we can restate the second law as follows: Aside from ﬂuctuations that are normally much too small to measure, any 8 q_A Entropy 0 5 0 1 0 0 1 5 0 2 0 0 2 5 0 3 0 0 0 2 0 4 0 6 0 8 0 1 0 0 S_total S_B S_A equilibrium isolated macroscopic system will inevitably evolve toward whatever (accessible) macrostate has the largest entropy. It is a simple matter to insert columns for SA, SB, and Stotal into the spreadsheet in Fig. 3. A graph of these three quantities is shown in Fig. 4. Notice that Stotal is not a strongly peaked function. Nevertheless, the analysis of the previous section tells us that, when this system is in equilibrium, it will almost certainly be found in a macrostate near qA = 60. Fig. 4. A spreadsheet-generated plot of SA, SB, and Stotal for the system of two Einstein solids represented in Fig. 3. At equilibrium, the slopes of the individual entropy graphs are equal in magnitude. Away from equilibrium, the solid whose slope dS/dq is steeper tends to gain energy, while the solid whose slope dS/dq is shallower tends to lose energy. Now let us ask how entropy is related to temperature. The most fundamental way to deﬁne temperature is in terms of energy ﬂow and thermal equilibrium: Two objects in thermal contact are said to be at the same temperature if they are in thermal equilibrium, that is, if there is no spontaneous net ﬂow of energy between them. If energy does ﬂow spontaneously from one to the other, then we say the one that loses energy has the higher temperature while the one that gains energy has the lower temperature. Referring again to Fig. 4, it is not hard to identify a quantity that is the same for solids A and B when they are in equilibrium. Equilibrium occurs when Stotal reaches a maximum, that is, when the slope dStotal/dqA = 0. But this slope is the sum of the slopes of the individual entropy graphs, so their slopes must be equal in magnitude: dSA dqA = −dSB dqA at equilbrium. (10) If we were to plot SB vs. qB instead of qA, its slope at the equilbrium point would be equal, in sign and magnitude, to that of SA(qA). So the quantity that is the same for both solids 9 when they are in equilibrium is the slope dS/dq. Temperature must be some function of this quantity.14 To identify the precise relation between temperature and the slope of the entropy vs. energy graph, look now to a point in Fig. 4 where qA is larger than its equilibrium value. Here the entropy graph for solid B is steeper than that for solid A, meaning that if a bit of energy were to pass from A to B, solid B would gain more entropy than solid A loses. Since the total entropy would increase, the second law tells us that this process will happen spontaneously. In general, the steeper an object’s entropy vs. energy graph, the more it “wants” to gain energy (in order to obey the second law), while the shallower an object’s entropy vs. energy graph, the less it “minds” losing a bit of energy. We therefore conclude that temperature is inversely related to the slope dS/dU. In fact, the reciprocal (dS/dU)−1 has precisely the units of temperature, so we might guess simply 1 T = dS dU . (11) The factor of Boltzmann’s constant in the deﬁnition (8) of entropy eliminates the need for any further constants in this formula. To conﬁrm that this relation gives temperature in ordinary Kelvin units one must check a particular example, as we will do in the following section. Our logical development of the concepts of entropy and temperature from statistical considerations is now complete. In the following section, we will apply these ideas to the calculation of heat capacities and other thermal properties of some simple systems. However, at this point in the development, many instructors will wish to ﬁrst generalize the discussion to include systems (such as gases) whose volume can change. The derivatives in Eqs. (10) and (11) then become partial derivatives at ﬁxed volume, and one cannot simply rearrange Eq. (11) to obtain dS = dU/T. There are at least two ways to get to the correct relation between entropy and temperature, and we now outline these brieﬂy. The standard approach in statistical mechanics texts15 is to ﬁrst consider two interact-ing systems separated by a movable partition, invoking the second law to identify pressure by the relation P = T(∂S/∂V )U. Combining this relation with Eq. (11) one obtains the thermodynamic identity dU = TdS + (−PdV ). For quasistatic processes (in which the change in volume is suﬃciently gradual), the second term is the (inﬁnitesimal) work W done on the system.16 But the ﬁrst law says that for any process, dU = Q + W, where Q is the (inﬁnitesimal) energy added by heating. Therefore, for quasistatic processes only, we can identify TdS = Q or dS = Q T (quasistatic processes only). (12) From this relation one can go on to treat a variety of applications, most notably engines and refrigerators. In an introductory course, however, there may not be time for the general discussion outlined in the previous paragraph. Fortunately, if one does not need the thermodynamic 10 identity, some of the steps can be bypassed (at some cost in rigor). We have already seen that, when the volume of a system does not change, its entropy changes by dS = dU T (constant volume). (13) Now consider a system whose volume does change, but that is otherwise isolated from its environment. If the change in volume is quasistatic (i.e., gradual), then the entropy of the system cannot change. Why? Theoretically, because while the energies of the system’s quantum states (and of the particles in those states) will be shifted, the process is not violent enough to knock a particle from one quantum state to another; thus the multiplicity of the system remains ﬁxed. Alternatively, we know from experiment that by reversing the direction of an adiabatic, quasistatic volume change we can return the system to a state arbitrarily close to its initial state; if its entropy had increased, this would not be possible even in the quasistatic limit. By either argument, the work done on a gas as it is compressed should not be included in the change in energy dU that contributes to the change in entropy. The natural generalization of Eq. (13) to non-constant-volume quasistatic processes is therefore to replace dU with dU −W = Q, the energy added to the system by heating alone. Thus we again arrive at Eq. (12). V. FURTHER SPREADSHEET CALCULATIONS So far in this paper we have used spreadsheet calculations only to gain a qualitative understanding of the second law and the relation between entropy and temperature. Now that we have developed these concepts, however, we can use the same tools to make quantitative predictions about the thermal behavior of various systems. A. Heat Capacity of an Einstein Solid Figure 5 shows a spreadsheet calculation of the entropy of a single Einstein solid of N = 50 oscillators with anywhere from 0 to 100 units of energy. Then, in the fourth column, we have calculated the temperature T = ∆U/∆S using a pair of closely spaced entropy and energy values. We have actually used a centered-diﬀerence approximation, so the cell D4, for example, contains the formula 2/(C5 −C3). Since we have omitted the constants k and ϵ from the formulas, the temperatures are expressed in units of ϵ/k. We now know the temperature for every energy, or vice versa. In practice, however, one usually compares such predictions to experimental measurements of the heat capacity, deﬁned (for systems at constant volume) as C = dU dT (constant volume). (14) This quantity, normalized to the number of oscillators, is computed in the ﬁfth column of the table, again using a centered-diﬀerence approximation (for instance, cell E5 contains the formula (2/(D6 −D4))/$E$1). Since we are still neglecting dimensionful constants, the entries in the column are actually for the quantity C/(Nk). 11 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 A B C D E F G H Einstein Solid # of oscillators = 5 0 q Omega S/k kT/eps C/Nk 0 1 0.00 1 5 0 3.91 0.28 2 1 2 7 5 7.15 0.33 0.45 3 22100 10.00 0.37 0.54 4 292825 12.59 0.40 0.59 5 3162510 14.97 0.44 0.64 6 28989675 17.18 0.47 0.67 7 2.32E+08 19.26 0.49 0.70 8 1.65E+09 21.23 0.52 0.73 9 1.06E+10 23.09 0.55 0.75 1 0 6.28E+10 24.86 0.58 0.77 1 1 3.43E+11 26.56 0.60 0.78 1 2 1.74E+12 28.19 0.63 0.80 1 3 8.31E+12 29.75 0.65 0.81 1 4 3.74E+13 31.25 0.68 0.82 Temperature Heat Capacity 0.00 0.20 0.40 0.60 0.80 1.00 0.00 1.00 2.00 3.00 Fig. 5. Spreadsheet calculation of the entropy, temperature, and heat capacity of an Einstein solid. (The table continues down to q = 100.) The graph at right shows the heat capacity as a function of temperature. To see the behavior at lower temperatures one can simply increase the number of oscillators in cell E1. To the right of the table is a spreadsheet-generated graph of the heat capacity vs. tem-perature for this system. This theoretical prediction can now be compared to experimental data, or to the graphs often given in textbooks.17 The quantitative agreement with exper-iment veriﬁes that the formula 1/T = dS/dU is probably correct.18 At high temperatures we ﬁnd the constant value C = Nk, as expected from the equipartition theorem (with each oscillator contributing two degrees of freedom). At temperatures below ϵ/k, however, the heat capacity drops oﬀsharply. This is just the qualitative behavior that Einstein was trying to explain with his model in 1907. To investigate the behavior of this system at lower temperatures, one can simply in-crease the number of oscillators in cell E1. With 45000 oscillators the temperature goes down to about 0.1ϵ/k, corresponding to a heat capacity per oscillator of about .005k. At this point it is apparent that the curve becomes extremely ﬂat at very low temperatures. Experimental measurements on real solids conﬁrm that the heat capacity goes to zero at low temperature, but do not conﬁrm the detailed shape of this curve. A more accurate treatment of the heat capacities of solids requires more sophisticated tools, but is given in standard textbooks.19 A collection of identical harmonic oscillators is actually a more accurate model of a completely diﬀerent system: the vibrational degrees of freedom of an ideal diatomic gas. Here again, experiments conﬁrm that the equipartition theorem is satisﬁed at high temperatures but the vibration (and its associated heat capacity) “freezes out” when kT ≪ ϵ. Graphs showing this behavior are included in many textbooks.20 B. The Boltzmann Distribution Perhaps the most useful tool of statistical mechanics is the Boltzmann factor, e−E/kT, which gives the relative probability of a system being in a microstate with energy E, when 12 Single Oscillator and Reservoir NA= 1 NB= 1 0 0 0 q_total= 1 0 0 q A OmegaA qB OmegaB OmegaTotal ln Omega 0 1 1 0 0 1E+144 1.3E+144 331.8 1 1 9 9 1E+143 1.2E+143 329.4 2 1 9 8 1E+142 1.1E+142 327.0 3 1 9 7 9E+140 9.5E+140 324.6 4 1 9 6 8E+139 8.4E+139 322.2 5 1 9 5 7E+138 7.4E+138 319.8 6 1 9 4 6E+137 6.4E+137 317.3 7 1 9 3 5E+136 5.5E+136 314.9 8 1 9 2 5E+135 4.7E+135 312.4 9 1 9 1 4E+134 3.9E+134 309.9 1 0 1 9 0 3E+133 3.3E+133 307.4 1 1 1 8 9 3E+132 2.7E+132 304.9 1 2 1 8 8 2E+131 2.2E+131 302.4 1 3 1 8 7 2E+130 1.8E+130 299.9 1 4 1 8 6 1E+129 1.4E+129 297.4 1 5 1 8 5 1E+128 1.1E+128 294.9 1 6 1 8 4 9E+126 9.0E+126 292.3 1 7 1 8 3 7E+125 7.0E+125 289.8 1 8 1 8 2 5E+124 5.3E+124 287.2 1 9 1 8 1 4E+123 4.1E+123 284.6 2 0 1 8 0 3E+122 3.0E+122 282.0 q_A ln(Omega_Total) 280.0 285.0 290.0 295.0 300.0 305.0 310.0 315.0 320.0 325.0 330.0 335.0 0 5 1 0 1 5 2 0 it is in thermal contact with a much larger system (or “reservoir”) at temperature T. A fairly straightforward derivation of this result can be found in most textbooks on statistical mechanics.21 A less general derivation, based on the exponentially decreasing density of an isothermal atmosphere, is given in Serway’s introductory text.22 Although it is no substitute for an analytic treatment, a spreadsheet calculation can easily display this exponential dependence for the case of a single one-dimensional oscillator in thermal contact with a large Einstein solid.23 Figure 6 shows such a calculation, for the parameters NA = 1, NB = 1000, and qtotal = 100. Here solid A is the “system”, while solid B, which is much larger, is the “reservoir”. The graph shows the natural logarithm of the total multiplicity, plotted as a function of qA. According to the fundamental assumption, the total multiplicity is proportional to the probability of ﬁnding the small system in each given microstate. Note that the graph is a straight line with a negative slope: This means that the probability does indeed decay exponentially (at least approximately) as the energy E = ϵqA of the small system increases. Fig. 6. States of a single oscillator in thermal contact with a large reservoir of identical oscillators. A graph of the logarithm of the multiplicity vs. the oscillator’s energy has a constant, negative slope, in accord with the Boltzmann distribution. According to the Boltzmann factor, the slope of the graph of multiplicity (or probabil-ity) vs. qA = E/ϵ should equal −ϵ/kT. Since the small system always has unit multiplicity and zero entropy, the entropy of the combined system is the same as the entropy of the reservoir: Stotal = SB. Once they know this, students can check either numerically or analytically that the slope of the graph is ∂ ∂qA ln Ωtotal = ϵ k ∂SB ∂UA = −ϵ k ∂SB ∂UB = −ϵ kTB , (15) 13 where the derivatives and the temperature TB are all evaluated at UA = 0. C. Thermal Behavior of a Two-State Paramagnet We have used the Einstein solid model in our examples for several reasons: it is fairly realistic, its behavior is intuitive, and, most importantly, it is mathematically simple. Other systems generally lack at least one of these qualities. To carry out our spreadsheet calculations, however, all we require is mathematical simplicity. The simplest possible system would consist of “atoms” that have only two allowed energy levels. A common example is a paramagnetic material, with one unpaired electron per atom, immersed in a uniform magnetic ﬁeld B. Let us say that B points in the +z direction, and quantize the spin (hence the magnetic moment) of the atom along this axis. When the magnetic moment points up, the atom is in the ground state, while when the magnetic moment points down, the atom is in the excited state. The energy levels of these two states are up : E↑= −µB, down : E↓= +µB, (16) where µ is the magnitude of the atom’s magnetic moment. In a collection of N such atoms, the total energy will be U = N↑E↑+ N↓E↓= µB(N↓−N↑), (17) where N↑and N↓are the numbers of atoms in the up and down states. These numbers must add up to N: N↑+ N↓= N. (18) The total magnetization of the sample is given by M = µN↑−µN↓= µ(N↑−N↓) = −E/B. (19) To describe a macrostate of this system we would have to specify the values of N and U, or equivalently, N and N↑(or N and N↓). The multiplicity of any such macrostate is the number of ways of choosing N↑atoms from a collection of N: Ω(N, N↑) = N N↑ = N! N↑!(N −N↑)! = N! N↑!N↓!. (20) From this simple formula we can calculate the entropy, temperature, and heat capacity. Figure 7 shows a spreadsheet calculation of these quantities for a two-state paramagnet of 100 “atoms”. The rows of the table are given in order of increasing energy, which means decreasing N↑and decreasing magnetization. Alongside the table is a graph of entropy vs. energy. Since the multiplicity of this system is largest when N↑and N↓are equal, the entropy also reaches a maximum at this point (U = 0) and then falls oﬀas more energy is added. This behavior is unusual and counter-intuitive, but is a straightforward consequence of the fact that each “atom” can store only a limited amount of energy. 14 Paramagnet # of spins = 1 0 0 N_up U/uB Omega S/k kT/uB C/Nk M/Nu 1 0 0 - 1 0 0 1 0.0 0 1.00 9 9 - 9 8 1 0 0 4.6 0.47 0.07 0.98 9 8 - 9 6 4 9 5 0 8.5 0.54 0.31 0.96 9 7 - 9 4 2E+05 12.0 0.60 0.36 0.94 9 6 - 9 2 4E+06 15.2 0.65 0.40 0.92 9 5 - 9 0 8E+07 18.1 0.70 0.42 0.90 9 4 - 8 8 1E+09 20.9 0.75 0.43 0.88 9 3 - 8 6 2E+10 23.5 0.79 0.44 0.86 9 2 - 8 4 2E+11 25.9 0.84 0.44 0.84 9 1 - 8 2 2E+12 28.3 0.88 0.44 0.82 9 0 - 8 0 2E+13 30.5 0.93 0.44 0.80 8 9 - 7 8 1E+14 32.6 0.97 0.43 0.78 8 8 - 7 6 1E+15 34.6 1.02 0.42 0.76 8 7 - 7 4 7E+15 36.5 1.07 0.41 0.74 8 6 - 7 2 4E+16 38.3 1.12 0.40 0.72 8 5 - 7 0 3E+17 40.1 1.17 0.38 0.70 8 4 - 6 8 1E+18 41.7 1.22 0.37 0.68 8 3 - 6 6 7E+18 43.3 1.28 0.35 0.66 Energy Entropy 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 - 1 0 0 - 5 0 0 5 0 1 0 0 Fig. 7. Spreadsheet calculation of the entropy, temperature, heat capacity, and magneti-zation of a two-state paramagnet. (The symbol u represents µ, the magnetic moment.) At right is a graph of entropy vs. energy, showing that as energy is added, the temperature (given by the reciprocal of the slope) goes to inﬁnity and then becomes negative. To appreciate the consequences of this behavior, imagine that the system starts out with U < 0 and that we gradually add more energy. At ﬁrst it behaves like an Einstein solid: The slope of its entropy vs. energy graph decreases, so its tendency to suck in energy (in order to maximize its entropy in accord with the second law) decreases. In other words, the system becomes hotter. As U approaches zero, the system’s desire to acquire more energy disappears. Its temperature is now inﬁnite, meaning that it will willingly give up a bit of energy to any other system whose temperature is ﬁnite (since it loses no entropy in the process). When U increases further and becomes positive, our system actually wants to give up energy. Intuitively, we would say that its temperature is higher than inﬁnity. But this is mathematical nonsense, so it is better to think about the reciprocal of the temperature, 1/T, which is now lower than zero, in other words, negative. In these terms, our intuition even agrees with the precise relation (11). The paramagnetic system can be an invaluable pedagogical tool, since it forces students to abandon the naive notion that temperature is merely a measure of the average energy in a system, and to embrace instead the more general relation (11), with all its connotations. One can even imagine using the paramagnet in place of the Einstein solid as the main paradigm for the lessons of Sections III and IV.24 In particular, it is not hard to set up a spreadsheet model of two interacting paramagnets, to look at how energy can be shared between them. However, this system exhibits behavior not commonly seen in other systems or in students’ prior experience. We therefore prefer not to use it as a ﬁrst example, but rather to save it for an upper-level course in which students have plenty of time to retrain their intuition. Putting these fundamental issues aside, we can go on to calculate both the heat capacity 15 Heat Capacity Temperature 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.00 5.00 10.00 Magnetization Temperature - 1 . 0 0 - 0 . 5 0 0.00 0.50 1.00 - 1 0 . 0 0 - 5 . 0 0 0.00 5.00 10.00 Fig. 8. Graphs of the heat capacity and magnetization as a function of temperature for a two-state paramagnet, generated from the last three columns of the spreadsheet shown in Fig. 7. and the magnetization of our two-state paramagnet as a function of temperature. Plots of these quantities are shown in Fig. 8. Since the heat capacity goes to zero at high temperature, the equipartition theorem fails utterly to describe this system (as it must, since it applies only to degrees of freedom for which the energy is a quadratic function of a coordinate or momentum). The magnetization is complete at T just greater than zero, but drops oﬀas the temperature is increased. The behavior in the high-temperature limit is known as Curie’s law, as described in many textbooks.25 VI. IMPLEMENTATION DETAILS The ideas presented in this paper have been tested, in various forms, in calculus-based introductory courses at Pomona College, Grinnell College, Smith College, Amherst College, and St. Lawrence University, and in an upper-division thermodynamics course at Weber State University. We have found that students at either level are quite able to understand the fundamental concepts of statistical mechanics that we are trying to teach. In particular, the fact that the statistical approach to the second law is more “modern” does not, in our opinion, make it unsuitable for ﬁrst-year students. On the contrary, we have had far more success in teaching the statistical approach than in teaching the concept of entropy from the older classical viewpoint used in most texts. The use of computers in our approach does not seem to pose a barrier to today’s students. Many of them have had previous experience with spreadsheet programs, and the rest are able to learn very quickly with the help of a one-page instruction sheet that we have prepared.26 16 A. Sample Exercises We generally present all of the material of Section III in class, and hand out reproduc-tions of the ﬁgures from that section. For homework, we ask students to reproduce Fig. 1, then change the parameters to look at some larger systems. Here is a typical homework assignment, written for students who will be using Excel: 1. List all the possible microstates for an Einstein solid containing three oscillators and ﬁve units of energy. Verify that Ω= 5+3−1 5 . 2. Set up an Excel spreadsheet identical (initially) to the one shown on the handouts: two Einstein solids, each containing three harmonic oscillators, with a total of six units of energy. You’ll need to type in all the labels and formulas down to row 4, as well as cell A5. After that, you can use the “Fill Down” command to ﬁll in the rest (except for cell E11). Use the “chart wizard tool” (the second icon from the right) to create the graph. Check that your spreadsheet yields all the same results as on the handout. Then modify your spreadsheet to show the case where one Einstein solid contains six harmonic oscillators and the other contains four harmonic oscillators (with the total number of energy units still equal to six). Turn in a printout of the modiﬁed spreadsheet. Assuming that all microstates are equally likely, what is the most probable macrostate, and what is its probability? What is the least probable macrostate, and what is its probability? 3. Modify your spreadsheet from the previous problem to show the case where solid A contains 100 oscillators, solid B contains 200 oscillators, and there are 100 units of energy in total. It’s easiest to delete the old graph and make a new one, and this time it’s best to make it an “XY (scatter)” graph instead of a column graph. Turn in a printout of the ﬁrst several rows of the spreadsheet as well as the graph. What is the most probable macrostate, and what is its probability? What is the least probable macrostate, and what is its probability? Calculate the entropies of these two macrostates (expressed as unitless numbers, neglecting the factor of Boltzmann’s constant). 4. Starting with your spreadsheet from the previous problem, add columns for the entropy of solid A, the entropy of solid B, and the total entropy. Print a graph showing all three entropies, plotted vs. the energy of solid A. Use the formula T = ∆U/∆S to calculate the temperatures of both solids (in units of ϵ/k) at the equilbrium point and also at the point qA = 60. Do your results make sense? Explain. We have also given computer-based homework assignments on the material of Sec-tion V. In upper-division courses we have asked students to generate Fig. 5 before seeing it in class. The treatment of this material can vary greatly depending on how the class is taught, however, so we will not reproduce our homework exercises here. B. Alternative Computational Environments Spreadsheet programs provide an almost ideal computational environment for the cal-culations described in this paper: they are versatile, easy to learn, and readily available on most campuses. Other computing options may be more suitable in some contexts, however. 17 Before we discovered spreadsheets, one of us (T. A. M.) wrote a stand-alone program for the Macintosh to produce tables and graphs such as those shown in Figs. 1 and 3. The stand-alone program is even easier to use than a spreadsheet, and can conveniently handle somewhat larger systems and values of q. It also allows one to completely omit formula (3) from the course if desired—students can simply be asked to verify the computer’s computations in a few simple cases. This program is suﬃcient to teach the core of our approach, as outlined in Section III. A copy of the program is available on request.27 A much cheaper, but less convenient, option is to perform the calculations on a modern graphing calculator. We do not particularly recommend this; in our experience, graphing calculators are much harder to use than spreadsheet programs. However, a motivated student who already knows how to use the calculator should be able to reproduce all the major results in this paper without too much diﬃculty. As an example, we describe here an appropriate set of instructions for the TI-82.28 To set up tables like those shown in Section III, one can use the “table” feature of the TI-82, which displays seven rows at a time, calculating each row as it is displayed. The ﬁrst column of the table, labeled X, should be used for qA. The following formulas in the “Y=” screen will then generate the remaining four columns: Y1 = (X + A −1) nCr X Y2 = Q −X Y3 = (Y2 + B −1) nCr Y2 Y4 = Y1 ∗Y3 Here we are using variables A, B, and Q to hold the constants NA, NB, and qtotal, respectively. Before plotting a graph, it is best to set Xmin to zero and Xmax to 94; this insures that each pixel corresponds to an integer value of X. To calculate temperatures and heat capacities on the TI-82, one must use “lists” instead of “tables”. The following sequence of instructions will reproduce most of Fig. 5, storing q, Ω, S/k, kT/ϵ, and C/Nk in lists L1 through L5: seq(X, X, 0, 50, 1) →L1 seq((X + A −1) nCrX, X, 0, 50, 1) →L2 ln(L2) →L3 seq(2/(L3(X + 2) −L3(X)), X, 1, 49, 1) →L4 seq(2/(L4(X + 2) −L4(X))/A, X, 1, 47, 1) →L5 Again we assume that the number of oscillators is stored in variable A. Lists 4 and 5 will be oﬀset upward by one and two places respectively, but this can be ﬁxed by entering “stat-edit” mode and inserting zeros. One should also add zeros at the end to make all lists the same length. It is then a simple matter to plot the heat capacity vs. the temperature, as shown in Fig. 5. Unfortunately, the lists will not update automatically when the value of A is changed. 18 Notes 1An overview of various ways of thinking about entropy has been given by Keith Andrew, “Entropy,” Am. J. Phys. 52(6), 492–496 (1984). 2For instance, David Halliday, Robert Resnick, and Jearl Walker, Fundamentals of Physics (Wiley, New York, 1993), 4th ed.; Raymond A. Serway, Physics for Scientists and Engineers (Saunders College Publishing, Philadelphia, 1996), 4th ed.; Francis W. Sears and Gerhard L. Salinger, Thermodynamics, Kinetic Theory, and Statistical Thermodynamics (Addison-Wesley, Reading, MA, 1975), 3rd ed. 3Rudolf Clausius himself remarked that “the second law of thermodynamics is much harder for the mind to grasp than the ﬁrst,” as quoted in Abraham Pais, Subtle is the Lord (Oxford University Press, Oxford, 1982), p. 60. 4For example, a deep understanding of entropy and the behavior of self-organizing sys-tems makes it clear that biological life is fully consistent with (and is indeed an expression of) the increase of entropy, even though life apparently involves an increase of order in the universe. Similarly, it is not obvious that a disorderly pile of ice chips has less entropy than a simple and seemingly orderly cup of the same amount of water. 5Boltzmann himself had a great deal of trouble convincing the scientiﬁc community of the validity of this statistical approach to entropy. See Pais, op. cit., pp. 60–68, 82– 83. Arthur Beiser comments on page 315 in his Concepts of Modern Physics text (4th ed., McGraw-Hill, New York, 1987) that “battles with nonbelievers left him severely depressed” and contributed to his eventual suicide in 1906, just as experimental evidence in favor of the atomic hypothesis and Boltzmann’s approach to thermal physics began to become overwhelming. 6For another attempt to address the same issue, see Ralph Baierlein, “Entropy and the second law: A pedagogical alternative,” Am. J. Phys. 62(1), 15–26 (1994). Baierlein’s approach is less quantitative than ours, but full of physical insight. 7A complete text for a three-week introductory-level unit on thermal physics based on this approach is available. Please contact Thomas Moore for more information. 8A partial derivation can be found in Keith Stowe, Introduction to Statistical Mechan-ics and Thermodynamics (Wiley, New York, 1984), pp. 191–192. A very nice indirect approach to the multiplicity of an ideal gas has been given by David K. Nartonis, “Using the ideal gas to illustrate the connection between the Clausius and Boltzmann deﬁnitions of entropy,” Am. J. Phys. 44(10), 1008–1009 (1976). 9See, for example, Henry A. Bent, The Second Law (Oxford University Press, New York, 1965), pp. 144–162; Herbert B. Callen, Thermodynamics and an Introduction to Thermostatistics (Wiley, New York, 1985), 2nd ed., p. 334; W. G. V. Rosser, An Introduc-tion to Statistical Physics (Ellis Horwood Limited, Chichester, 1982), pp. 31–41; Charles A. Whitney, Random Processes in Physical Systems (Wiley, New York, 1990), pp. 97–139. An even simpler model is a two-state system, as treated, for example, in Charles Kittel and Herbert Kroemer, Thermal Physics (Freeman, San Francisco, 1980), 2nd ed., pp. 10–21. 19 We treat this model in Section V.C below, and also discuss there why we don’t use it as our primary example. 10Callen, op. cit., p. 334. 11This same example has been used independently by Whitney, op. cit., p. 126. We are grateful to one of our reviewers for pointing out this reference to us. 12A nice calculation of the width of the peak for a simpler system has been given by Ralph Baierlein, “Teaching the approach to thermodynamic equilibrium: Some pictures that help,” Am. J. Phys. 46(10), 1042–1045 (1978). 13The deﬁnition of entropy, like that of macrostate or multiplicity, depends on the time scale under consideration. Here we are interested in time scales that are long compared to the relaxation time of each individual system, but short compared to the relaxation time of the composite system. For a strongly coupled system this clean separation of time scales would not be possible. Note that on very long time scales, even our weakly coupled system would have a slightly larger entropy, equal to k ln Ωwhere Ωis the total multiplicity for all allowed values of UA and UB. For macroscopic systems this number is practically indistinguishable from the quantity k ln(ΩAΩB) computed for the most likely macrostate. A careful discussion of the role of time scales in deﬁning entropy is given in L. D. Landau and E. M. Lifshitz, Statistical Physics, tr. J. B. Sykes and M. J. Kearsley (Pergamon, Oxford, 1980), 3rd edition, Part 1, p. 27. 14Notice that the preceding argument depends crucially on the additivity of S = k ln Ω for the composite system. Thus the logarithm in the deﬁnition of S is essential, if S and T are to be related in a simple way. 15For instance, F. Mandl, Statistical Physics (Wiley, Chichester, 1988), 2nd ed., pp. 47, 83–87. 16The standard notation is d ¯W, but it is too tempting to read this incorrectly as the “change in work” (a meaningless phrase), rather than correctly as an inﬁnitesimal “amount of work”. We therefore call it simply W, with the understanding that in this context it is an inﬁnitesimal quantity. Similarly, we use Q instead of d ¯Q. Note also that we take W to be positive when work is done on the system. 17See, for instance, Halliday, Resnick, and Walker, op. cit., Fig. 20-2. 18Most textbooks deﬁne the temperature scale operationally in terms of a gas ther-mometer. The most direct veriﬁcation of 1/T = dS/dU would therefore be to apply this formula directly to an ideal gas. But the gas thermometer can be used to measure the heat capacity of a solid, and this result, in turn, can be compared to our theoretical pre-diction. Note, in particular, that the factor of Boltzmann’s constant in the deﬁnition of S conveniently makes the temperature come out in ordinary Kelvin units, with no further constants needed. 19For instance, Mandl, op. cit., chapter 6. 20For instance, Halliday, Resnick, and Walker, op. cit., Fig. 21-13. 21For instance, Callen, op. cit., p. 350; Mandl, op. cit., pp. 52–56; Stowe, op. cit., pp. 284–287. 20 22Serway, op. cit., pp. 599–601. 23Similar calculations can be done by hand, albeit for a somewhat smaller “large” system. See Whitney, op. cit., p. 128. 24The two-state model system is used to introduce statistical concepts in the popular text of Kittel and Kroemer, op. cit.. This text includes a nice calculation of the sharpness of the multiplicity function for a system of two interacting paramagnets. 25Halliday, Resnick, and Walker, op. cit., p. 928. 26This instruction sheet, which is written for users of Excel on the Macintosh, is avail-able on request from Daniel Schroeder. 27Please send requests to Thomas Moore. Note that the program runs only on the Macintosh; no version for DOS or Windows is available. 28Since the TI-82 overﬂows at 10100, one is limited to somewhat smaller systems than we have used in our earlier examples. In practice this is not much of a limitation. 21
1956	https://www.quora.com/How-much-blood-loss-is-possible-for-a-broken-femur-bone-operation	How much blood loss is possible for a broken femur bone operation? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Medicine and Healthcare Femur Fracture Complications Surgical Procedures Blood Loss Orthopedics Bleeding Leg Fractures Femur 5 How much blood loss is possible for a broken femur bone operation? All related (38) Sort Recommended Edward Leahy Lives in Earth (1961–present) · Upvoted by Liang-Hai Sie , Retired general internist, former intensive care physician. · Author has 14.1K answers and 10M answer views ·7y Just breaking a femur can cause a 2 unit blood loss into the thigh. Further blood loss will occur during surgery and depends on where the fracture is and the actual procedure selected. Some procedures like per cutaneous IM nailing usually don't cause a lot of bleeding. Bigger open procedures can cause more blood loss. Obviously an injury to the femoral artery can cause rapid exsanguination, though that would likely be a fatal injury. Its been a long time since I worked in a trauma center so most of the people with femur fractures that I see are old people who fell and broke their femur, often a Continue Reading Just breaking a femur can cause a 2 unit blood loss into the thigh. Further blood loss will occur during surgery and depends on where the fracture is and the actual procedure selected. Some procedures like per cutaneous IM nailing usually don't cause a lot of bleeding. Bigger open procedures can cause more blood loss. Obviously an injury to the femoral artery can cause rapid exsanguination, though that would likely be a fatal injury. Its been a long time since I worked in a trauma center so most of the people with femur fractures that I see are old people who fell and broke their femur, often at the femoral neck. Most of the repairs I see are IM nails or hemiarthroplasties so the blood loss isn't usually too bad. Upvote · 9 6 Sponsored by Grammarly 92% of professionals who use Grammarly say it has saved them time Work faster with AI, while ensuring your writing always makes the right impression. Download 999 207 Related questions More answers below What will happen to a broken femur bone in which the rod is broken due to heavy weight? How long does it take to recover from a broken femur? How long will it take to walk normal after femur surgery? What if a fractured femur is left untreated? My grandmother's femur fractured in November and surgery is not possible because of high blood pressure. She is bed-bound and it’s February. What can be done to recover her? Can I remove a femur fracture rod 4 years later? Amelia Andiamo Studied Orthopedics&Oncology (Graduated 2000) · Author has 1K answers and 4.4M answer views ·7y As a general rule I ask for 4 units depending on the seriousness of the issue. Serious trauma might require up to 6 units. My record is 7 on a soldier who was almost hit with a 50 cal. The thinking was we would have to amputate however I decided to try a repair. We had nothing to lose, if we failed we could amputate. But back to your question; lots. Lots of blood will come out. Upvote · 9 6 9 1 Christopher Fox Orthopedic Surgeon, spine surgeon · Author has 11.7K answers and 45.4M answer views ·7y It depends. Where is the femur broken? How much energy is involved with the fracture? What operation was done? When was it done compared to the fracture? What ancillary anatomical structures were injured? These are all important questions. Upvote · 9 2 Michael L. Jirka Former EMT, Navy Corpsman, Army Medic, at Clinical Laboratory Scientist (1969–2009) · Upvoted by Liang-Hai Sie , Retired general internist, former intensive care physician. · Author has 16K answers and 24.6M answer views ·7y I would think 0–2 units. Usually, femur breaks are done by young adults doing stupid shit … they can lose a lot of blood because they are so healthy. Older people who break the femur at the ball will usually get 1–2 units if their hemoglobin is marginal (they were low to start with). Upvote · 9 2 Sponsored by Book Geists If you're a Kiwi, this could be the best day of your life! Available to Kiwis only. Read today. Learn More 1.3K 1.3K Related questions More answers below What not to do after a broken femur? Can I walk ninety days after a distal femur bone fracture? What should I expect during the recovery of a distal femur fracture? Why do so many seniors die after breaking their femur bone? Why does a leg shrink or become thin after a femur fracture and post surgery? Assistant Bot · Sep 6 Blood loss during surgery to fix a fractured femur varies widely depending on fracture type, surgical approach, fixation method, patient factors, and whether the injury is open or associated with other trauma. Typical ranges and key factors: Typical estimated blood loss (EBL) Intramedullary nailing (closed femoral shaft fracture, single limb, isolated injury): commonly 100–500 mL. Many straightforward cases are toward the lower end (100–250 mL); complex reductions or reaming can push toward 300–500 mL. Plate-and-screw fixation (open reduction internal fixation, displaced peri‑trochanteric or subt Continue Reading Blood loss during surgery to fix a fractured femur varies widely depending on fracture type, surgical approach, fixation method, patient factors, and whether the injury is open or associated with other trauma. Typical ranges and key factors: Typical estimated blood loss (EBL) Intramedullary nailing (closed femoral shaft fracture, single limb, isolated injury): commonly 100–500 mL. Many straightforward cases are toward the lower end (100–250 mL); complex reductions or reaming can push toward 300–500 mL. Plate-and-screw fixation (open reduction internal fixation, displaced peri‑trochanteric or subtrochanteric fractures): commonly 200–800 mL, often higher than closed nailing because of more soft‑tissue dissection. Hemiarthroplasty or total hip arthroplasty for femoral neck fractures: typically 300–1,200 mL depending on approach (posterior vs. lateral), cement use, and patient factors; intracapsular vs. displaced fractures change bleeding risk. Severe open fractures, multiple fractures, polytrauma, or vascular injury: blood loss can exceed 1,000–2,000+ mL and may require transfusion and massive‑bleeding protocols. Perioperative and hidden blood loss Hidden (postoperative) loss into tissues and the femoral canal can add 20–50% more than intraoperative measured loss, particularly after intramedullary nailing or arthroplasty. Cumulative total blood loss = measured intraoperative EBL + postoperative drain output + calculated hidden loss (often estimated from hemoglobin drop and blood volume). Patient and operative modifiers that increase bleeding Anticoagulation or antiplatelet drugs not fully reversed Osteoporotic bone with comminution or multiple fragments Reamed intramedullary nailing, prolonged operative time, or revision surgery Obesity, difficult exposure, or extensive soft‑tissue injury Preexisting anemia or coagulopathy Transfusion thresholds and preparation Many centers transfuse if hemoglobin <7–8 g/dL in otherwise stable patients; higher thresholds (8–10 g/dL) for symptomatic patients, cardiovascular disease, or ongoing bleeding. Preop optimization (correct anemia, stop/reverse anticoagulants when appropriate), intraoperative blood-conservation (controlled hypotension, tranexamic acid, cell salvage), and postoperative monitoring reduce transfusion need. Practical guidance (for clinicians and patients) Expect modest blood loss (100–500 mL) for routine closed intramedullary nailing of an isolated femoral shaft fracture; expect higher volumes for open procedures, arthroplasty, or complex/open fractures. Plan preoperative labs (CBC, coagulation), crossmatch blood if risk factors present, and use tranexamic acid unless contraindicated. Monitor hemoglobin postoperatively and account for hidden loss when assessing need for transfusion. Summary Blood loss for femur fracture surgery typically ranges from about 100 mL for straightforward closed nailing to several hundred milliliters for open fixation or arthroplasty, with severe or complex cases exceeding 1,000 mL. Hidden postoperative loss and patient factors can substantially increase total blood loss, so perioperative planning and blood‑conservation measures are important. (Information reflects literature and practice patterns through May 2024.) Upvote · Liang-Hai Sie Retired general internist, former intensive care physician. · Author has 63.1K answers and 215.9M answer views ·7y The femur is full of blood vessels, so according to Medscape Femur Injuries and Fractures a broken femur in 40% of the cases will cause a blood loss of 2 to 3 units (pints) of blood. Don’t think the surgery if needed would cause a lot more blood loss on top of that. Upvote · 9 2 Chelsea Smith RN, BSN, Certified Emergency Nurse · Author has 22.4K answers and 48M answer views ·7y All of it. Truly. It's possible to exsanguinate during surgery, especially one involving repairing a fracture in a major (and highly vascular) bone so close to the femoral artery. Your particular surgical risk would be determined by the surgeon and no one can possibly speculate on your particular case. Upvote · 9 1 Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 620 Roger Skebelsky Orthopedic and Emergency Medicine Physician Assistant · Author has 741 answers and 4.8M answer views ·7y This is all dependent upon the complexity of the fracture and time it take to complete the operation from open to close. Usually a patient is typed and crossed for 2 units of blood pre operatively. However, on average for this type of surgery it can be from 250cc to 500cc. You want to limit as much blood loss as possible to avoid post operative complications and risking transfusion reactions. Upvote · 9 1 Wilk Dedwylder Former Retired emergency physician · Upvoted by Richard Smith , formerly pharmacist and ENT surgeon · Author has 19.3K answers and 31.7M answer views ·1y Related How much blood is lost from bone fractures? I need the answer as gar as I can to get the 7 degrees from the doctor. How to answer this? Break a carpal navicular and it may take an MRI to see anything at all, with no appreciable blood loss, though the blood supply to the proximal part of the bone may be compromised. In a typical Colles fracture, the local hematoma is normally quite visible, and it isn’t at all difficult to estimate (though hardly worth doing). On the other end of the spectrum, if you want to hear stories of massive transfusion, talk to a doctor about the more complicated pelvic fractures. The personal record I ever saw involved transfusion of over sixty units of packed red cells (and other repl Continue Reading How to answer this? Break a carpal navicular and it may take an MRI to see anything at all, with no appreciable blood loss, though the blood supply to the proximal part of the bone may be compromised. In a typical Colles fracture, the local hematoma is normally quite visible, and it isn’t at all difficult to estimate (though hardly worth doing). On the other end of the spectrum, if you want to hear stories of massive transfusion, talk to a doctor about the more complicated pelvic fractures. The personal record I ever saw involved transfusion of over sixty units of packed red cells (and other replacements, of course). Upvote · 9 6 9 1 Sponsored by State Bank of India Stay Informed!! Stay Protected!! Our Contact Centre calls only from numbers beginning with 1600 or 140 series. Learn More 99 45 Sue Johnson Studied EMT-B; EMT-I; Vet Assistant · Author has 1.2K answers and 1.8M answer views ·Updated 1y Related How long does it take to walk again after femur fracture surgery? As someone who broke their hip/femur this past August, this is what I’ve read: It depends by and large on the method used to fix it. Bed rest and no fixation, quite a long time. I’m not sure on plates and screws. I had internal fixation called pinning. It involved a long rod placed the length of my femur, and a shorter one through it heading up into the head of the femur. This: Yeah, ouch. But! It’s capable of full weight bearing immediately. It’s pretty much dictated by how much pain you can handle to stand, and then walk. I was up two days after, sent to an inpatient acute rehab at four days, and Continue Reading As someone who broke their hip/femur this past August, this is what I’ve read: It depends by and large on the method used to fix it. Bed rest and no fixation, quite a long time. I’m not sure on plates and screws. I had internal fixation called pinning. It involved a long rod placed the length of my femur, and a shorter one through it heading up into the head of the femur. This: Yeah, ouch. But! It’s capable of full weight bearing immediately. It’s pretty much dictated by how much pain you can handle to stand, and then walk. I was up two days after, sent to an inpatient acute rehab at four days, and released from there another seven days later, able to walk around with a walker or crutches. I’m 11 weeks out. I was released from outpatient physical therapy last week. I’m only using a cane when I can’t walk without a pain limp (that leg is now shorter, so there’s still definitely a hitch in the gait). And I’ve started going to dog agility lessons with my dog again. Happy days! Though that’s slow and we’re concentrating on learning distance handling so I don’t have to run, yet. But there’s still a ton of work to do. In no small part because I also “cheese grated” the ligaments around it when it displaced. Technique, time, and commitment. That’s what determines your recovery. So ask your ortho how it was pieced back together, then spend a bit of time with your favorite search engine reading about the technique and its prognosis. Best of luck. Editing with two updates from this. First, the hardware from the original fixation broke at five months and had to be entirely replaced. Second, at two and a half years out, the pain was still persisting and they finally agreed to new imaging. CT showed I had a nonunion, meaning it NEVER actually healed! I had my hip replaced in May 2023. It’s soooo much better. But still some ways to go, as they say all such surgeries take at least a year to be considered fully recovered. “At least” is an individual thing. So it’ll take as long as it takes. Upvote · 99 20 9 2 David Grubbs 25+ years in the medical field: RN, Paramedic, Firefighter · Author has 141 answers and 439.5K answer views ·Updated 4y Related Can you die from a broken femur? Yes, especially if you don't get treatment quickly. The average blood loss from a closed femur fracture is 1–1.5 LITERS!!! If the fracture is open, or especially if the bone lacerated an artery, it can be even greater. I have watched someone literally bleed to death right in front of my eyes from an open femur fracture while we were trying to cut him out of his car… so I guarantee you it can happe Continue Reading Yes, especially if you don't get treatment quickly. The average blood loss from a closed femur fracture is 1–1.5 LITERS!!! If the fracture is open, or especially if the bone lacerated an artery, it can be even greater. I have watched someone literally bleed to death right in front of my eyes from an open femur fracture while we were trying to cut him out of his car… so I guarantee you it can happen!!! That said, at 17 I was in a car wreck. I broke my femur and had a tibial plateau fracture along with various other injuries. The responders were able to cut me free, get a traction splint on me, and get me to a hospital in time to save my life. After multiple surgeries, about 6 months of rehab, and about a year of life…. I joined a volunteer fire department and eventually earned my FFI, FFII (what you need to be a paid firefighter in most states), extrication technician, HAZMAT technician, fire officer II, fire instructor II, fire inspector I, pump operator, aerial operator, a fire science degree, my EMT, Paramedic, my Emergency medical services degree, and eventually my nursing degree as a RN. I spent 18 years in the fire service and became the Department Chief of a combination department at 28 yrs old. No one in my family had ever been in the f... Upvote · 99 19 9 8 Shane Miller Former CTO (Chief Technology Officer) ·2y Related How do I treat a broken femur? Part 1 of 5:Treating a Broken Femur With Surgery There are three phases of surgical treatment – acute, recovery and maintenance. The first priority when it comes to surgery is to rule out any other life threatening injuries and to stabilize the affected area. Acute Phase – surgical stabilization is necessary at this point as well as the restoration of limb alignment. Traction is administered for initial stabilization to take place. Inter-medullary nailing is the treatment of choice. Recovery Phase – the goal here is to encourage immediate weight bearing tolerance, improvement of hip and knee rang Continue Reading Part 1 of 5:Treating a Broken Femur With Surgery There are three phases of surgical treatment – acute, recovery and maintenance. The first priority when it comes to surgery is to rule out any other life threatening injuries and to stabilize the affected area. Acute Phase – surgical stabilization is necessary at this point as well as the restoration of limb alignment. Traction is administered for initial stabilization to take place. Inter-medullary nailing is the treatment of choice. Recovery Phase – the goal here is to encourage immediate weight bearing tolerance, improvement of hip and knee range of motion, and strengthening of the affected area. Gait training and crutch use is done in this phase. Low extremity aerobic training such as biking, swimming, and the use of an elliptical trainer is necessary for as long as it is tolerated. Maintenance Phase - in this phase, weight bearing is permitted once complete healing has been achieved. Activities that can be done here include 30 to 45 minutes of bike riding in a flat surface. Walking for 3 to 5 miles (4.8 to 8.0 km) per week is also encouraged. Running can be done after 3 weeks then gradual return to usual training regimen. After surgery, physical therapy, discussed below, is needed in order to gradually restore function to the thighs. Usually, range of motion exercises and stretches are done to give strength to the broken femur. Gradual increase in physical activity is also considered. Be aware of when surgery will occur. If you are suffering from closed femoral fracture, the surgeon will wait until you become stable before conducting the surgery. However, for open fracture, an emergency surgery is done immediately to prevent infection. Surgery for a broken femur can last up to three or four hours. Internal fixation is the choice surgery done to treat a broken femur. In this type of surgery, a cut is made to open the fracture, then internal fixator or special metal devices are used to hold the bone in place. Afterwards, a rod or a large nail is placed in the center of the femur to support the bone until it heals. Sometimes a plate is also placed next to the bone which is attached using screws in order to frame the outside of the leg. Metal plates and screws will reduce the risk of further complications from occurring due to displacement. Investigate external fixation to promote healing. For external fixation, the surgeon will cut an opening to surgically place metal pins and screws into the bone above and below the fracture site. The pins and screws are attached to a bar outside the skin. This bar holds the bones together, when the skin and muscles have been injured, to promote healing. Pins are generally used when joints are not involved. Consider intra-medullary nailing to stabilize the area. During intramedullary nailing, metal rods are inserted into the marrow canal of the bones through a small incision. They are then screwed to the bone at both ends. This procedure keeps the bone in its correct position during healing. This surgery provides strong, stable and full-length fixation. Part 2 of 5:Managing a Broken Femur With Medication NSAIDS contain analgesic, anti-inflammatory as well as antipyretic properties. These drugs act upon certain body chemicals to prevent the progress of inflammation during injuries. Health care practitioners do not encourage intake of this drug for the first 48 hours of injury, to let the body follow its natural course of healing. The most commonly prescribed drugs to manage pain and swelling includes Celecoxib (Celebrex), Ibuprofen (Advil), and Naproxen (Aleve). Try analgesics as an alternate to NSAIDS. These medications ensure comfort, and manage pain and swelling. They also contain sedating properties that can ease the stress of dealing with pain. Acetaminophen (Tylenol) is usually the drug of choice. Request prescription painkillers if the pain of your broken leg is intolerable. In the case of a broken femur, pain can be severe and NSAIDS and other light pain medications are often insufficient. During consultation, healthcare providers generally address pain and inflammation. Prescription drugs, such as stronger painkillers, can be given to ease the sensation of pain and lessen the degree of inflammation. Examples of prescription painkillers are codeine and tramadol. These painkillers are only prescribed if simple painkillers do not reach the desired level of pain reduction. Stronger types of painkillers must be taken at their proper dosage to avoid addiction and dependence. Consider using antibiotics. Antibiotics may be necessary to avoid infection. Especially in the case of open femoral fracture and directly after surgery, antibiotics may be used to prevent infection of the site. The physician will first obtain a wound culture by getting a sample of wound discharge and getting it analyzed. If an infection is found, the physician will prescribe the right antibiotic. Open femoral fractures may occur from trauma cause by skin opening in the wound area. Part 3 of 5:Maintaining the Healing Process While at Home Take good care while showering. When showering, make sure to cover your brace or immobilizer with a plastic bag to keep it dry. Also, be careful to not fall or further injure yourself. If an immobilizer is not present, carefully wash the incision site with soap and. Pat it dry gently afterwards. Remember not to rub the incision site or to put any kind of cream or lotion over it. Avoid falling by sitting on a stool while showering. Never soak in a tub or pool unless otherwise advised by the doctor. Install hand rails in the shower, near the tub or next to the toilet. Put a slip proof mat on the bathroom floor inside the shower and even in the tub. Change the bandage incision daily to monitor it and refresh it. Change the bandage on a daily basis in order to expose the incision site to air and allow it to heal. Make sure that the wound looks healthy, and is not suffering from infection. Watch out for the presence of redness, excessive drainage and unnecessary opening of the wound. Inform your health care team of your injury. Let your primary care physician and your dentist know that there are rods and pins in your leg. This will allow your doctor and dentist to take any possible metal in your system into consideration when they are treating you for other ailments. Note that antibiotics should be taken before any kind of dental work is done, to reduce the risk of infection. Antibiotics are a prophylactic therapy done for as long as pins and rods are inside your body. Lower your bed to allow for easy access. The bed where you sleep in should be low enough to allow your feet to touch the floor. This will not only allow you to get in and out of bed without straining your femur, it also helps to prevent falls and other accidents from occurring. Set up your living area to accommodate you. Make sure that your house and frequented areas are set up to give you easy access and prevent falls. Also, make sure hazards are clear from trafficked areas in order to avoid accidents. Remove loose cords, wires, and rugs from the areas where you usually walk. Do not keep small animals because you might accidentally step on them. Repair uneven flooring. Provide good lighting. Protect yourself from extreme heat to prevent bleeding. Individuals are advised to avoid contact with high temperatures such as hot baths, saunas and heat packs, because these can dilate blood vessels and increase incidents of bleeding. Although heat may feel good at first, it's potentially damaging effects need to be avoided. Immobilize the limb. Stability of the femur must be maintained to prevent additional damage to the area and its surrounding tissues. Immobilization can be achieved by putting a splint and bandage on the affected area. Extreme activities and the application of too much pressure in the injured area should also be avoided to prevent further damage to the affected area. Avoid massage and other manipulation of the affected area until healing is complete. Apply a cold compress to reduce the risk of bleeding and swelling. Cold compress can usually be placed over the injured area to promote constriction of blood vessels that reduces the risk of bleeding and swelling. Compression also aids in stimulating the flow of lymph fluid, which carries important nutrients to the damaged tissues surrounding the injury. Lymph fluid also removes waste from cells and body tissues which are an important part of the tissue regeneration process. Elevate the affected limb to aid in circulation. The injured area should be kept elevated above the heart level. This action aids in proper blood circulation by promoting good venous blood return. Because the action of the veins is to return blood to the heart, elevating the injured area will allow gravity to aid venous blood return. Part 4 of 5:Rehabbing a Broken Femur With Physical Therapy Do hip and knee bends to strengthen the muscles. Doing hip and knee bends will increase the flexibility of your quadriceps and hamstrings, helping to restore mobility and function. Lie on a flat surface with the head elevated 30 to 45 degrees on a pillow. As far as is possible and comfortable, bend the affected knee towards the chest. Return to your starting position. Repeat this 10 to 20 times provided that there is no pain increase. Try static quadriceps contraction. This is a low-stress exercise that is good as a starting point for strengthening. Lie on a flat surface with a towel under the affected leg. Start by tightening the front thigh muscles (quadriceps) by pushing the knee in a downward motion into the towel. Hold for 5 seconds. Place the middle and pointer finger on the inner quadriceps in order to feel the tightening of muscles. This will make sure that proper tightening is accomplished. Repeat tightening the front thigh muscles and holding them for 5 seconds for 10 repetitions. Exercise and muscle movement promotes good blood circulation and oxygen distribution to the affected area. This will speed up the repair of the damaged tissue around the femur. This is because oxygen is essential for maintaining cellular integrity, function, and repair when tissues are injured. Most femoral fractures take four to six months to completely heal. Even with proper treatment, the length of recovery will depend on the severity of the fracture and the presence of other injuries to the body. Eat foods that are rich in calcium such as green leafy vegetables and make it a habit to regularly drink milk. Consume vitamin D supplements to help strengthen your bones. This will ensure that you do not re-break your femur. Wear protective equipment. Especially if you are training for a big event, wear protective equipment to ensure that you do not break a bone. Also, make sure that you take short breaks in between your training period to help your bones recover from the pressure, strain and exhaustion. Consider cross training. Do not overtrain because it can cause too much strain on the femur and your other bones as well. Alternate exercises so that your bones have a period to rest. You can alternate running with non-impact activities such as cycling and swimming. Upvote · 9 2 9 1 Stu Lewine Former CVICU Staff RN · Author has 1.2K answers and 1.4M answer views ·2y Related How long does it take to heal from a fractured femur? I have one and a metal rod in my femur to fix it. My son fractured his femur in half 30 some years ago and had an intramedulary titanium rod placed. He was and still is a super athletic self motivated individual. He rehabbed his leg incredibly well. He participated in 2 Iron Man Triathlons, he snowboards, does mountain biking and skis, and is a firefighter, nurse, paramedic (do I sound proud?) He passed on the option for rod removal because he wa Continue Reading My son fractured his femur in half 30 some years ago and had an intramedulary titanium rod placed. He was and still is a super athletic self motivated individual. He rehabbed his leg incredibly well. He participated in 2 Iron Man Triathlons, he snowboards, does mountain biking and skis, and is a firefighter, nurse, paramedic (do I sound proud?) He passed on the option for rod removal because he was and still is doing so well at 53 years of age. He lives in Truckee California and doesn’t complain about the cold snowy weather bothering his leg. Rod removal is not an easy procedure and from what I remember was told the surgery and rehab is no walk in the park. He doesn’t complain, he has excellent mobility and range of motion, probabl... Upvote · 9 7 Related questions What will happen to a broken femur bone in which the rod is broken due to heavy weight? How long does it take to recover from a broken femur? How long will it take to walk normal after femur surgery? What if a fractured femur is left untreated? My grandmother's femur fractured in November and surgery is not possible because of high blood pressure. She is bed-bound and it’s February. What can be done to recover her? Can I remove a femur fracture rod 4 years later? What not to do after a broken femur? Can I walk ninety days after a distal femur bone fracture? What should I expect during the recovery of a distal femur fracture? Why do so many seniors die after breaking their femur bone? Why does a leg shrink or become thin after a femur fracture and post surgery? Is the metal rod in a femur fracture permanent? After femur fracture can we come back to a normal walk as before? What happens if the rod is never taken out after a femur surgery? How much time will it take to cure a femoral neck fracture? Can I walk normal after left leg femur and right tibia surgery with rods and nails? Related questions What will happen to a broken femur bone in which the rod is broken due to heavy weight? How long does it take to recover from a broken femur? How long will it take to walk normal after femur surgery? What if a fractured femur is left untreated? My grandmother's femur fractured in November and surgery is not possible because of high blood pressure. She is bed-bound and it’s February. What can be done to recover her? Can I remove a femur fracture rod 4 years later? What not to do after a broken femur? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
1957	https://chem.libretexts.org/Courses/Heartland_Community_College/HCC%3A_Chem_162/24%3A_Chemistry_of_Coordination_Compounds/24.3%3A_Equilibrium_of_Metal_Complexes	Skip to main content 24.3: Equilibrium of Metal Complexes Last updated : May 19, 2019 Save as PDF 24.2: Ligands 24.4: Nomenclature of Complexes Page ID : 55129 ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives To be introduced to complex ions, including ligands. Previously, you learned that metal ions in aqueous solution are hydrated—that is, surrounded by a shell of usually four or six water molecules. A hydrated ion is one kind of a complex ion (or, simply, complex), a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons, such as the [Al(H2O)6]3+ ion. A complex ion forms from a metal ion and a ligand because of a Lewis acid–base interaction. The positively charged metal ion acts as a Lewis acid, and the ligand, with one or more lone pairs of electrons, acts as a Lewis base. Small, highly charged metal ions, such as Cu2+ or Ru3+, have the greatest tendency to act as Lewis acids, and consequently, they have the greatest tendency to form complex ions. As an example of the formation of complex ions, consider the addition of ammonia to an aqueous solution of the hydrated Cu2+ ion {[Cu(H2O)6]2+}. Because it is a stronger base than H2O, ammonia replaces the water molecules in the hydrated ion to form the [Cu(NH3)4(H2O)2]2+ ion. Formation of the [Cu(NH3)4(H2O)2]2+ complex is accompanied by a dramatic color change, as shown in Figure . The solution changes from the light blue of [Cu(H2O)6]2+ to the blue-violet characteristic of the [Cu(NH3)4(H2O)2]2+ ion. The Formation Constant The replacement of water molecules from by ammonia occurs in sequential steps. Omitting the water molecules bound to for simplicity, we can write the equilibrium reactions as follows: with with with with The sum of the stepwise reactions is the overall equation for the formation of the complex ion: The hydrated Cu2+ ion contains six H2O ligands, but the complex ion that is produced contains only four ligands, not six. The equilibrium constant for the formation of the complex ion from the hydrated ion is called the formation constant (). The equilibrium constant expression for Kf has the same general form as any other equilibrium constant expression. In this case, the expression is as follows: The formation constant () has the same general form as any other equilibrium constant expression. Water, a pure liquid, does not appear explicitly in the equilibrium constant expression, and the hydrated Cu2+(aq) ion is represented as Cu2+ for simplicity. As for any equilibrium, the larger the value of the equilibrium constant (in this case, Kf), the more stable the product. With Kf = 2.1 × 1013, the [Cu(NH3)4(H2O)2]2+ complex ion is very stable. The formation constants for some common complex ions are listed in Table . Table : Formation Constants for Selected Complex Ions in Aqueous Solution \| \| Complex Ion \| Equilibrium Equation \| \| \| Reported values are overall formation constants. Source: Data from Lange’s Handbook of Chemistry, 15th ed. (1999). \| \| \| \| \| Ammonia Complexes \| [Ag(NH3)2]+ \| Ag+ + 2NH3 ⇌ [Ag(NH3)2]+ \| 1.1 × 107 \| \| [Cu(NH3)4]2+ \| Cu2+ + 4NH3 ⇌ [Cu(NH3)4]2+ \| 2.1 × 1013 \| \| [Ni(NH3)6]2+ \| Ni2+ + 6NH3 ⇌ [Ni(NH3)6]2+ \| 5.5 × 108 \| \| Cyanide Complexes \| [Ag(CN)2]− \| Ag+ + 2CN− ⇌ [Ag(CN)2]− \| 1.1 × 1018 \| \| [Ni(CN)4]2− \| Ni2+ + 4CN− ⇌ [Ni(CN)4]2− \| 2.2 × 1031 \| \| [Fe(CN)6]3− \| Fe3+ + 6CN− ⇌ [Fe(CN)6]3− \| 1 × 1042 \| \| Hydroxide Complexes \| [Zn(OH)4]2− \| Zn2+ + 4OH− ⇌ [Zn(OH)4]2− \| 4.6 × 1017 \| \| [Cr(OH)4]− \| Cr3+ + 4OH− ⇌ [Cr(OH)4]− \| 8.0 × 1029 \| \| Halide Complexes \| [HgCl4]2− \| Hg2+ + 4Cl− ⇌ [HgCl4]2− \| 1.2 × 1015 \| \| [CdI4]2− \| Cd2+ + 4I ⇌ [CdI4]2− \| 2.6 × 105 \| \| [AlF6]3− \| Al3+ + 6F− ⇌ [AlF6]3− \| 6.9 × 1019 \| \| Other Complexes \| [Ag(S2O3)2]3− \| Ag+ + 2S2O32− ⇌ [Ag(S2O3)2]3− \| 2.9 × 1013 \| \| [Fe(C2O4)3]3− \| Fe3+ + 3C2O42− ⇌ [Fe(C2O4)3]3− \| 2.0 × 1020 \| Example If 12.5 g of is added to 500 mL of 1.00 M aqueous ammonia, what is the equilibrium concentration of ? Given: mass of Cu2+ salt and volume and concentration of ammonia solution Asked for: equilibrium concentration of Cu2+(aq) Strategy: Calculate the initial concentration of due to the addition of copper(II) nitrate hexahydrate. Use the stoichiometry of the reaction shown in Equation to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations of all species in solution. Substitute the final concentrations into the expression for the formation constant () to calculate the equilibrium concentration of . Solution Adding an ionic compound that contains to an aqueous ammonia solution will result in the formation of [Cu(NH3)4]2+(aq), as shown in Equation We assume that the volume change caused by adding solid copper(II) nitrate to aqueous ammonia is negligible. A The initial concentration of from the amount of added copper nitrate prior to any reaction is as follows: Because the stoichiometry of the reaction is four NH3 to one Cu2+, the amount of NH3 required to react completely with the Cu2+ is 4(0.0846) = 0.338 M. The concentration of ammonia after complete reaction is 1.00 M − 0.338 M = 0.66 M. These results are summarized in the first two lines of the following table. Because the equilibrium constant for the reaction is large (2.1 × 1013), the equilibrium will lie far to the right. Thus we will assume that the formation of [Cu(NH3)4]2+ in the first step is complete and allow some of it to dissociate into Cu2+ and NH3 until equilibrium has been reached. If we define x as the amount of Cu2+ produced by the dissociation reaction, then the stoichiometry of the reaction tells us that the change in the concentration of [Cu(NH3)4]2+ is −x, and the change in the concentration of ammonia is +4x, as indicated in the table. The final concentrations of all species (in the bottom row of the table) are the sums of the concentrations after complete reaction and the changes in concentrations. \| ICE \| [Cu2+] \| [NH3] \| [[Cu(NH3)4]2+] \| --- --- \| \| initial \| 0.0846 \| 1.00 \| 0 \| \| after complete reaction \| 0 \| 0.66 \| 0.0846 \| \| change \| +x \| +4x \| −x \| \| final \| x \| 0.66 + 4x \| 0.0846 − x \| B Substituting the final concentrations into the expression for the formation constant () and assuming that , which allows us to remove from the sum and difference, The value of x indicates that our assumption was justified. The equilibrium concentration of Cu2+(aq) in a 1.00 M ammonia solution is therefore 2.1 × 10−14 M. Exercise The ferrocyanide ion () is very stable, with a of . Calculate the concentration of cyanide ion in equilibrium with a 0.65 M solution of . Answer : 2 × 10−6 M The Effect of the Formation of Complex Ions on Solubility What happens to the solubility of a sparingly soluble salt if a ligand that forms a stable complex ion is added to the solution? One such example occurs in conventional black-and-white photography. Recall that black-and-white photographic film contains light-sensitive microcrystals of AgBr, or mixtures of AgBr and other silver halides. AgBr is a sparingly soluble salt, with a Ksp of 5.35 × 10−13 at 25°C. When the shutter of the camera opens, the light from the object being photographed strikes some of the crystals on the film and initiates a photochemical reaction that converts AgBr to black Ag metal. Well-formed, stable negative images appear in tones of gray, corresponding to the number of grains of AgBr converted, with the areas exposed to the most light being darkest. To fix the image and prevent more AgBr crystals from being converted to Ag metal during processing of the film, the unreacted AgBr on the film is removed using a complexation reaction to dissolve the sparingly soluble salt. The reaction for the dissolution of silver bromide is as follows: with The equilibrium lies far to the left, and the equilibrium concentrations of Ag+ and Br− ions are very low (7.31 × 10−7 M). As a result, removing unreacted AgBr from even a single roll of film using pure water would require tens of thousands of liters of water and a great deal of time. Le Chatelier’s principle tells us, however, that we can drive the reaction to the right by removing one of the products, which will cause more AgBr to dissolve. Bromide ion is difficult to remove chemically, but silver ion forms a variety of stable two-coordinate complexes with neutral ligands, such as ammonia, or with anionic ligands, such as cyanide or thiosulfate (S2O32−). In photographic processing, excess AgBr is dissolved using a concentrated solution of sodium thiosulfate. The reaction of Ag+ with thiosulfate is as follows: with The magnitude of the equilibrium constant indicates that almost all Ag+ ions in solution will be immediately complexed by thiosulfate to form [Ag(S2O3)2]3−. We can see the effect of thiosulfate on the solubility of by writing the appropriate reactions and adding them together: Comparing with shows that the formation of the complex ion increases the solubility of by approximately 3 × 1013. The dramatic increase in solubility combined with the low cost and the low toxicity explains why sodium thiosulfate is almost universally used for developing black-and-white film. If desired, the silver can be recovered from the thiosulfate solution using any of several methods and recycled. If a complex ion has a large Kf, the formation of a complex ion can dramatically increase the solubility of sparingly soluble salts. Example : Common Ion Effect in Complexation Due to the common ion effect, we might expect a salt such as AgCl to be much less soluble in a concentrated solution of KCl than in water. Such an assumption would be incorrect, however, because it ignores the fact that silver ion tends to form a two-coordinate complex with chloride ions (AgCl2−). Calculate the solubility of AgCl in each situation: in pure water in 1.0 M KCl solution, ignoring the formation of any complex ions the same solution as in part (b) except taking the formation of complex ions into account, assuming that AgCl2− is the only Ag+ complex that forms in significant concentrations At 25°C, Ksp = 1.77 × 10−10 for AgCl and Kf = 1.1 × 105 for AgCl2−. Given: Ksp of AgCl, Kf of AgCl2−, and KCl concentration Asked for: solubility of AgCl in water and in KCl solution with and without the formation of complex ions Strategy: Write the solubility product expression for AgCl and calculate the concentration of Ag+ and Cl− in water. Calculate the concentration of Ag+ in the KCl solution. Write balanced chemical equations for the dissolution of AgCl and for the formation of the AgCl2− complex. Add the two equations and calculate the equilibrium constant for the overall equilibrium. Write the equilibrium constant expression for the overall reaction. Solve for the concentration of the complex ion. Solution If we let x equal the solubility of AgCl, then at equilibrium Substituting this value into the solubility product expression, Thus the solubility of in pure water at 25°C is 1.33 × 10−5 M. If x equals the solubility of AgCl in the KCl solution, then at equilibrium [Ag+] = x M and [Cl−] = (1.0 + x) M. Substituting these values into the solubility product expression and assuming that : If the common ion effect were the only important factor, we would predict that is approximately five orders of magnitude less soluble in a 1.0 M KCl solution than in water. To account for the effects of the formation of complex ions, we must first write the equilibrium equations for both the dissolution and the formation of complex ions. Adding the equations corresponding to Ksp and Kf gives us an equation that describes the dissolution of AgCl in a KCl solution. The equilibrium constant for the reaction is therefore the product of Ksp and Kf: If we let equal the solubility of AgCl in the KCl solution, then at equilibrium [AgCl2−] = x and [Cl−] = 1.0 − x. Substituting these quantities into the equilibrium constant expression for the net reaction and assuming that , That is, AgCl dissolves in 1.0 M KCl to produce a 1.9 × 10−5 M solution of the AgCl2− complex ion. Thus we predict that AgCl has approximately the same solubility in a 1.0 M KCl solution as it does in pure water, which is 105 times greater than that predicted based on the common ion effect. (In fact, the measured solubility of AgCl in 1.0 M KCl is almost a factor of 10 greater than that in pure water, largely due to the formation of other chloride-containing complexes.) Exercise Calculate the solubility of mercury(II) iodide () in each situation: pure water a 3.0 M solution of NaI, assuming [HgI4]2− is the only Hg-containing species present in significant amounts Ksp = 2.9 × 10−29 for HgI2 and Kf = 6.8 × 1029 for [HgI4]2−. Answer a : 1.9 × 10−10 M Answer b : 1.4 M Complexing agents, molecules or ions that increase the solubility of metal salts by forming soluble metal complexes, are common components of laundry detergents. Long-chain carboxylic acids, the major components of soaps, form insoluble salts with Ca2+ and Mg2+, which are present in high concentrations in “hard” water. The precipitation of these salts produces a bathtub ring and gives a gray tinge to clothing. Adding a complexing agent such as pyrophosphate (O3POPO34−, or P2O74−) or triphosphate (P3O105−) to detergents prevents the magnesium and calcium salts from precipitating because the equilibrium constant for complex-ion formation is large: with However, phosphates can cause environmental damage by promoting eutrophication, the growth of excessive amounts of algae in a body of water, which can eventually lead to large decreases in levels of dissolved oxygen that kill fish and other aquatic organisms. Consequently, many states in the United States have banned the use of phosphate-containing detergents, and France has banned their use beginning in 2007. “Phosphate-free” detergents contain different kinds of complexing agents, such as derivatives of acetic acid or other carboxylic acids. The development of phosphate substitutes is an area of intense research. Commercial water softeners also use a complexing agent to treat hard water by passing the water over ion-exchange resins, which are complex sodium salts. When water flows over the resin, sodium ion is dissolved, and insoluble salts precipitate onto the resin surface. Water treated in this way has a saltier taste due to the presence of Na+, but it contains fewer dissolved minerals. Complexing Agents in MRIs Another application of complexing agents is found in medicine. Unlike x-rays, magnetic resonance imaging (MRI) can give relatively good images of soft tissues such as internal organs. MRI is based on the magnetic properties of the 1H nucleus of hydrogen atoms in water, which is a major component of soft tissues. Because the properties of water do not depend very much on whether it is inside a cell or in the blood, it is hard to get detailed images of these tissues that have good contrast. To solve this problem, scientists have developed a class of metal complexes known as “MRI contrast agents.” Injecting an MRI contrast agent into a patient selectively affects the magnetic properties of water in cells of normal tissues, in tumors, or in blood vessels and allows doctors to “see” each of these separately (Figure ). One of the most important metal ions for this application is Gd3+, which with seven unpaired electrons is highly paramagnetic. Because Gd3+(aq) is quite toxic, it must be administered as a very stable complex that does not dissociate in the body and can be excreted intact by the kidneys. The complexing agents used for gadolinium are ligands such as DTPA5− (diethylene triamine pentaacetic acid), whose fully protonated form is shown in Figure . Summary The formation of complex ions can substantially increase the solubility of sparingly soluble salts if the complex ion has a large Kf. A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. The equilibrium constant for the formation of the complex ion is the formation constant (Kf). The formation of a complex ion by adding a complexing agent increases the solubility of a compound. 24.2: Ligands 24.4: Nomenclature of Complexes
1958	https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/21%3A_Organic_Chemistry/21.04%3A_Aldehydes_Ketones_Carboxylic_Acids_and_Esters	Published Time: 2022-10-18T17:27:11Z 21.4: Aldehydes, Ketones, Carboxylic Acids, and Esters - Chemistry LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 21: Organic Chemistry Chemistry - 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Chemistry - Atoms First 2e (OpenStax) 5. 21: Organic Chemistry 6. 21.4: Aldehydes, Ketones, Carboxylic Acids, and Esters Expand/collapse global location Chemistry - Atoms First 2e (OpenStax) Front Matter 1: Essential Ideas of Chemistry 2: Atoms, Molecules, and Ions 3: Electronic Structure and Periodic Properties 4: Chemical Bonding and Molecular Geometry 5: Advanced Theories of Covalent Bonding 6: Composition of Substances and Solutions 7: Stoichiometry of Chemical Reactions 8: Gases 9: Thermochemistry 10: Liquids and Solids 11: Solutions and Colloids 12: Thermodynamics 13: Fundamental Equilibrium Concepts 14: Acid-Base Equilibria 15: Equilibria of Other Reaction Classes 16: Electrochemistry 17: Kinetics 18: Representative Metals, Metalloids, and Nonmetals 19: Transition Metals and Coordination Chemistry 20: Nuclear Chemistry 21: Organic Chemistry 22: Appendices Back Matter 21.4: Aldehydes, Ketones, Carboxylic Acids, and Esters Last updated Aug 13, 2023 Save as PDF 21.3: Alcohols and Ethers 21.5: Amines and Amides Page ID 415559 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. Aldehydes and Ketones 1. Example 21.4.1 21.4.1: Oxidation and Reduction in Organic Chemistry 1. Solution 2. Exercise 21.4.1 21.4.1/21:_Organic_Chemistry/21.04:_Aldehydes_Ketones_Carboxylic_Acids_and_Esters#Exercise_.5C(.5CPageIndex.7B1.7D.5C)) Carboxylic Acids and Esters Learning Objectives By the end of this section, you will be able to: Describe the structure and properties of aldehydes, ketones, carboxylic acids and esters Another class of organic molecules contains a carbon atom connected to an oxygen atom by a double bond, commonly called a carbonyl group. The trigonal planar carbon in the carbonyl group can attach to two other substituents leading to several subfamilies (aldehydes, ketones, carboxylic acids and esters) described in this section. Aldehydes and Ketones Both aldehydes and ketones contain a carbonyl group, a functional group with a carbon-oxygen double bond. The names for aldehyde and ketone compounds are derived using similar nomenclature rules as for alkanes and alcohols, and include the class-identifying suffixes -al and -one, respectively: In an aldehyde, the carbonyl group is bonded to at least one hydrogen atom. In a ketone, the carbonyl group is bonded to two carbon atoms: As text, an aldehyde group is represented as –CHO; a ketone is represented as –C(O)– or –CO–. In both aldehydes and ketones, the geometry around the carbon atom in the carbonyl group is trigonal planar; the carbon atom exhibits sp 2 hybridization. Two of the sp 2 orbitals on the carbon atom in the carbonyl group are used to form σ bonds to the other carbon or hydrogen atoms in a molecule. The remaining sp 2 hybrid orbital forms a σ bond to the oxygen atom. The unhybridized p orbital on the carbon atom in the carbonyl group overlaps a p orbital on the oxygen atom to form the π bond in the double bond. Like the Figure 21.4.1 21.4.1: The carbonyl group is polar, and the geometry of the bonds around the central carbon is trigonal planar. The importance of molecular structure in the reactivity of organic compounds is illustrated by the reactions that produce aldehydes and ketones. We can prepare a carbonyl group by oxidation of an alcohol—for organic molecules, oxidation of a carbon atom is said to occur when a carbon-hydrogen bond is replaced by a carbon-oxygen bond. The reverse reaction—replacing a carbon-oxygen bond by a carbon-hydrogen bond—is a reduction of that carbon atom. Recall that oxygen is generally assigned a –2 oxidation number unless it is elemental or attached to a fluorine. Hydrogen is generally assigned an oxidation number of +1 unless it is attached to a metal. Since carbon does not have a specific rule, its oxidation number is determined algebraically by factoring the atoms it is attached to and the overall charge of the molecule or ion. In general, a carbon atom attached to an oxygen atom will have a more positive oxidation number and a carbon atom attached to a hydrogen atom will have a more negative oxidation number. This should fit nicely with your understanding of the polarity of C–O and C–H bonds. The other reagents and possible products of these reactions are beyond the scope of this chapter, so we will focus only on the changes to the carbon atoms: Example 21.4.1 21.4.1: Oxidation and Reduction in Organic Chemistry Methane represents the completely reduced form of an organic molecule that contains one carbon atom. Sequentially replacing each of the carbon-hydrogen bonds with a carbon-oxygen bond would lead to an alcohol, then an aldehyde, then a carboxylic acid (discussed later), and, finally, carbon dioxide: What are the oxidation numbers for the carbon atoms in the molecules shown here? Solution In this example, we can calculate the oxidation number (review the chapter on oxidation-reduction reactions if necessary) for the carbon atom in each case (note how this would become difficult for larger molecules with additional carbon atoms and hydrogen atoms, which is why organic chemists use the definition dealing with replacing C–H bonds with C–O bonds described). For CH 4, the carbon atom carries a –4 oxidation number (the hydrogen atoms are assigned oxidation numbers of +1 and the carbon atom balances that by having an oxidation number of –4). For the alcohol (in this case, methanol), the carbon atom has an oxidation number of –2 (the oxygen atom is assigned –2, the four hydrogen atoms each are assigned +1, and the carbon atom balances the sum by having an oxidation number of –2; note that compared to the carbon atom in CH 4, this carbon atom has lost two electrons so it was oxidized); for the aldehyde, the carbon atom’s oxidation number is 0 (–2 for the oxygen atom and +1 for each hydrogen atom already balances to 0, so the oxidation number for the carbon atom is 0); for the carboxylic acid, the carbon atom’s oxidation number is +2 (two oxygen atoms each at –2 and two hydrogen atoms at +1); and for carbon dioxide, the carbon atom’s oxidation number is +4 (here, the carbon atom needs to balance the –4 sum from the two oxygen atoms). Exercise 21.4.1 21.4.1 Indicate whether the marked carbon atoms in the three molecules here are oxidized or reduced relative to the marked carbon atom in ethanol: There is no need to calculate oxidation states in this case; instead, just compare the types of atoms bonded to the marked carbon atoms: Answer (a) reduced (bond to oxygen atom replaced by bond to hydrogen atom); (b) oxidized (one bond to hydrogen atom replaced by one bond to oxygen atom); (c) oxidized (2 bonds to hydrogen atoms have been replaced by bonds to an oxygen atom) Aldehydes are commonly prepared by the oxidation of alcohols whose –OH functional group is located on the carbon atom at the end of the chain of carbon atoms in the alcohol: Alcohols that have their –OH groups in the middle of the chain are necessary to synthesize a ketone, which requires the carbonyl group to be bonded to two other carbon atoms: An alcohol with its –OH group bonded to a carbon atom that is bonded to no or one other carbon atom will form an aldehyde. An alcohol with its –OH group attached to two other carbon atoms will form a ketone. If three carbons are attached to the carbon bonded to the –OH, the molecule will not have a C–H bond to be replaced, so it will not be susceptible to oxidation. Formaldehyde, an aldehyde with the formula HCHO, is a colorless gas with a pungent and irritating odor. It is sold in an aqueous solution called formalin, which contains about 37% formaldehyde by weight. Formaldehyde causes coagulation of proteins, so it kills bacteria (and any other living organism) and stops many of the biological processes that cause tissue to decay. Thus, formaldehyde is used for preserving tissue specimens and embalming bodies. It is also used to sterilize soil or other materials. Formaldehyde is used in the manufacture of Bakelite, a hard plastic having high chemical and electrical resistance. Dimethyl ketone, CH 3 COCH 3, commonly called acetone, is the simplest ketone. It is made commercially by fermenting corn or molasses, or by oxidation of 2-propanol. Acetone is a colorless liquid. Among its many uses are as a solvent for lacquer (including fingernail polish), cellulose acetate, cellulose nitrate, acetylene, plastics, and varnishes; as a paint and varnish remover; and as a solvent in the manufacture of pharmaceuticals and chemicals. Carboxylic Acids and Esters The odor of vinegar is caused by the presence of acetic acid, a carboxylic acid, in the vinegar. The odor of ripe bananas and many other fruits is due to the presence of esters, compounds that can be prepared by the reaction of a carboxylic acid with an alcohol. Because esters do not have hydrogen bonds between molecules, they have lower vapor pressures than the alcohols and carboxylic acids from which they are derived (see Figure 21.4.2 21.4.2). Figure 21.4.2 21.4.2: Esters are responsible for the odors associated with various plants and their fruits. Both carboxylic acids and esters contain a carbonyl group with a second oxygen atom bonded to the carbon atom in the carbonyl group by a single bond. In a carboxylic acid, the second oxygen atom also bonds to a hydrogen atom. In an ester, the second oxygen atom bonds to another carbon atom. The names for carboxylic acids and esters include prefixes that denote the lengths of the carbon chains in the molecules and are derived following nomenclature rules similar to those for inorganic acids and salts (see these examples): The functional groups for an acid and for an ester are shown in red in these formulas. The hydrogen atom in the functional group of a carboxylic acid will react with a base to form an ionic salt: Carboxylic acids are weak acids (see the chapter on acids and bases), meaning they are not 100% ionized in water. Generally only about 1% of the molecules of a carboxylic acid dissolved in water are ionized at any given time. The remaining molecules are undissociated in solution. We prepare carboxylic acids by the oxidation of aldehydes or alcohols whose –OH functional group is located on the carbon atom at the end of the chain of carbon atoms in the alcohol: Esters are produced by the reaction of acids with alcohols. For example, the ester ethyl acetate, CH 3 CO 2 CH 2 CH 3, is formed when acetic acid reacts with ethanol: The simplest carboxylic acid is formic acid, HCO 2 H, known since 1670. Its name comes from the Latin word formicus, which means “ant”; it was first isolated by the distillation of red ants. It is partially responsible for the pain and irritation of ant and wasp stings, and is responsible for a characteristic odor of ants that can be sometimes detected in their nests. Acetic acid, CH 3 CO 2 H, constitutes 3–6% vinegar. Cider vinegar is produced by allowing apple juice to ferment without oxygen present. Yeast cells present in the juice carry out the fermentation reactions. The fermentation reactions change the sugar present in the juice to ethanol, then to acetic acid. Pure acetic acid has a penetrating odor and produces painful burns. It is an excellent solvent for many organic and some inorganic compounds, and it is essential in the production of cellulose acetate, a component of many synthetic fibers such as rayon. The distinctive and attractive odors and flavors of many flowers, perfumes, and ripe fruits are due to the presence of one or more esters (Figure 21.4.3 21.4.3). Among the most important of the natural esters are fats (such as lard, tallow, and butter) and oils (such as linseed, cottonseed, and olive oils), which are esters of the trihydroxyl alcohol glycerine, C 3 H 5(OH)3, with large carboxylic acids, such as palmitic acid, CH 3(CH 2)14 CO 2 H, stearic acid, CH 3(CH 2)16 CO 2 H, and oleic acid, Oleic acid is an unsaturated acid; it contains a double bond. Palmitic and stearic acids are saturated acids that contain no double or triple bonds. Figure 21.4.3 21.4.3: Over 350 different volatile molecules (many members of the ester family) have been identified in strawberries. (credit: Rebecca Siegel) This page titled 21.4: Aldehydes, Ketones, Carboxylic Acids, and Esters is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax. Back to top 21.3: Alcohols and Ethers 21.5: Amines and Amides Was this article helpful? Yes No Recommended articles 21: Organic ChemistryOrganic chemistry involving the scientific study of the structure, properties, and reactions of organic compounds and organic materials, i.e., matter ... Article typeSection or PageAuthorOpenStaxLicenseCC BYLicense Version4.0OER program or PublisherOpenStaxShow Page TOCno on page Tags Author tag:OpenStax © Copyright 2025 Chemistry LibreTexts Powered by CXone Expert ® ? 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1959	https://pubchem.ncbi.nlm.nih.gov/compound/Aluminum-fluoride	Aluminum fluoride \| AlF3 \| CID 2124 - PubChem An official website of the United States government Here is how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. NIH National Library of Medicine NCBI PubChem About Docs Submit Contact Search PubChem compound Summary Aluminum fluoride PubChem CID 2124 Structure Primary Hazards Laboratory Chemical Safety Summary (LCSS) Datasheet Molecular Formula AlF 3 Synonyms aluminum fluoride Aluminum trifluoride trifluoroalumane trifluoridoaluminium aluminium trifluoride View More... Molecular Weight 83.976748 g/mol Computed by PubChem 2.2 (PubChem release 2025.04.14) Dates Create: 2004-09-16 Modify: 2025-08-23 Description Aluminum fluoride appears as odorless white powder or granules. Denser than water. Solubility in water at 25 °C equals 0.559 g / 100 mL. CAMEO Chemicals Aluminium trifluoride is an aluminium coordination entity. ChEBI 1 Structures 1.1 2D Structure Structure Search Get Image Download Coordinates Chemical Structure Depiction Full screen Zoom in Zoom out PubChem 1.2 3D Status Conformer generation is disallowed since MMFF94s unsupported element PubChem 2 Names and Identifiers 2.1 Computed Descriptors 2.1.1 IUPAC Name trifluoroalumane Computed by Lexichem TK 2.7.0 (PubChem release 2025.04.14) PubChem 2.1.2 InChI InChI=1S/Al.3FH/h;31H/q+3;;;/p-3 Computed by InChI 1.07.2 (PubChem release 2025.04.14) PubChem 2.1.3 InChIKey KLZUFWVZNOTSEM-UHFFFAOYSA-K Computed by InChI 1.07.2 (PubChem release 2025.04.14) PubChem 2.1.4 SMILES FAlF Computed by OEChem 2.3.0 (PubChem release 2025.04.14) PubChem 2.2 Molecular Formula AlF 3 Computed by PubChem 2.2 (PubChem release 2025.04.14) Australian Industrial Chemicals Introduction Scheme (AICIS); CAMEO Chemicals; PubChem AlF 3 ILO-WHO International Chemical Safety Cards (ICSCs) 2.3 Other Identifiers 2.3.1 CAS 7784-18-1 Australian Industrial Chemicals Introduction Scheme (AICIS); CAMEO Chemicals; CAS Common Chemistry; ChemIDplus; EPA Chemical Data Reporting (CDR); EPA Chemicals under the TSCA; EPA DSSTox; European Chemicals Agency (ECHA); Hazardous Substances Data Bank (HSDB); ILO-WHO International Chemical Safety Cards (ICSCs); New Zealand Environmental Protection Authority (EPA); NJDOH RTK Hazardous Substance List 2.3.2 Deprecated CAS 1202643-05-7, 1328946-66-2, 856859-79-5 ChemIDplus 1202643-05-7, 1328946-66-2, 2816047-54-6, 856859-79-5 EPA Chemicals under the TSCA 1328946-66-2, 856859-79-5 EPA DSSTox 2.3.3 European Community (EC) Number 232-051-1 European Chemicals Agency (ECHA) 2.3.4 ChEBI ID CHEBI:49464 ChEBI 2.3.5 DSSTox Substance ID DTXSID8030712 EPA DSSTox 2.3.6 ICSC Number 1324 ILO-WHO International Chemical Safety Cards (ICSCs) 2.3.7 Wikidata Q409057 Wikidata 2.3.8 Wikipedia Aluminium trifluoride Wikipedia Aluminium fluoride Wikipedia 2.4 Synonyms 2.4.1 MeSH Entry Terms aluminum fluoride aluminum trifluoride Medical Subject Headings (MeSH) 2.4.2 Depositor-Supplied Synonyms aluminum fluoride Aluminum trifluoride trifluoroalumane trifluoridoaluminium aluminium trifluoride Fluorid hlinity Aluminum fluoride (AlF3) Aluminium fluorure Z77H3IKW94 DTXSID8030712 CHEBI:49464 Aluminium Fluoratum DTXCID6010712 232-051-1 7784-18-1 Aluminium fluoride RefChem:5537 AlF3 Aluminum fluoride, anhydrous MFCD00003426 Trifluoroaluminum Aluminum fluoride anhydrous aluminium(III) fluoride Fluorid hlinity [Czech] Aluminium fluorure [French] CCRIS 2282 HSDB 600 EINECS 232-051-1 aluminium(3+) fluoride Aluminum fluoride, 97% [AlF3] UNII-Z77H3IKW94 CBiol_002057 SCHEMBL79383 Aluminum fluoride, reagent grade Bio1_000343 Bio1_000832 Bio1_001321 AKOS015902486 SY061137 CS-0031741 EC 232-051-1 Q409057 Aluminum fluoride, anhydrous, >=99.99% trace metals basis Aluminum fluoride, anhydrous, powder, >=99.9% trace metals basis PubChem 3 Chemical and Physical Properties 3.1 Computed Properties Property Name Property Value Reference Property Name Molecular Weight Property Value 83.976748 g/mol Reference Computed by PubChem 2.2 (PubChem release 2025.04.14) Property Name Hydrogen Bond Donor Count Property Value 0 Reference Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14) Property Name Hydrogen Bond Acceptor Count Property Value 3 Reference Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14) Property Name Rotatable Bond Count Property Value 0 Reference Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14) Property Name Exact Mass Property Value 83.9767479 Da Reference Computed by PubChem 2.2 (PubChem release 2025.04.14) Property Name Monoisotopic Mass Property Value 83.9767479 Da Reference Computed by PubChem 2.2 (PubChem release 2025.04.14) Property Name Topological Polar Surface Area Property Value 0 Å² Reference Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14) Property Name Heavy Atom Count Property Value 4 Reference Computed by PubChem Property Name Formal Charge Property Value 0 Reference Computed by PubChem Property Name Complexity Property Value 8 Reference Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14) Property Name Isotope Atom Count Property Value 0 Reference Computed by PubChem Property Name Defined Atom Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Undefined Atom Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Defined Bond Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Undefined Bond Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Covalently-Bonded Unit Count Property Value 1 Reference Computed by PubChem Property Name Compound Is Canonicalized Property Value Yes Reference Computed by PubChem (release 2025.04.14) PubChem 3.2 Experimental Properties 3.2.1 Physical Description Aluminum fluoride appears as odorless white powder or granules. Denser than water. Solubility in water at 25 °C equals 0.559 g / 100 mL. CAMEO Chemicals Pellets or Large Crystals; Dry Powder EPA Chemical Data Reporting (CDR) White solid; [Hawley] Hygroscopic; [ICSC] Odorless powder; [MSDSonline] Haz-Map, Information on Hazardous Chemicals and Occupational Diseases HYGROSCOPIC WHITE OR COLOURLESS CRYSTALS. ILO-WHO International Chemical Safety Cards (ICSCs) 3.2.2 Color / Form White, hexagonal crystals Lide, DR (ed.). CRC Handbook of Chemistry and Physics. 81st Edition. CRC Press LLC, Boca Raton: FL 2000, p. 4-39 Hazardous Substances Data Bank (HSDB) 3.2.3 Melting Point 1291 °C Lewis, R.J. Sax's Dangerous Properties of Industrial Materials. 10th ed. Volumes 1-3 New York, NY: John Wiley & Sons Inc., 1999., p. V2: 131 Hazardous Substances Data Bank (HSDB) 3.2.4 Solubility Loses water at 100 °C, more at 200 °C. ... slight solubility. /Trihydrate/ O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 62 Hazardous Substances Data Bank (HSDB) In water, 0.559 g/100 mL @ 25 °C O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 62 Hazardous Substances Data Bank (HSDB) Sparingly soluble in acids and alkalies, even hot concentrated H2SO4 has little effect. O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 62 Hazardous Substances Data Bank (HSDB) INSOL IN ALC & ACETONE Weast, R.C. (ed.) Handbook of Chemistry and Physics. 69th ed. Boca Raton, FL: CRC Press Inc., 1988-1989., p. B-68 Hazardous Substances Data Bank (HSDB) Solubility in water, g/100ml at 20Â °C: 0.5 ILO-WHO International Chemical Safety Cards (ICSCs) 3.2.5 Density 2.88 at 77 °F (USCG, 1999) - Denser than water; will sink U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office. CAMEO Chemicals 3.10 Lide, DR (ed.). CRC Handbook of Chemistry and Physics. 81st Edition. CRC Press LLC, Boca Raton: FL 2000, p. 4-39 Hazardous Substances Data Bank (HSDB) 2.9 g/cmÂ³ ILO-WHO International Chemical Safety Cards (ICSCs) 3.2.6 Vapor Pressure 1 mm Hg @ 1238 °C Lewis, R.J. Sax's Dangerous Properties of Industrial Materials. 10th ed. Volumes 1-3 New York, NY: John Wiley & Sons Inc., 1999., p. V2 131 Hazardous Substances Data Bank (HSDB) Vapor pressure, Pa at 1238Â °C: 133 ILO-WHO International Chemical Safety Cards (ICSCs) 3.2.7 Stability / Shelf Life Low melting solids or colorless, volatile liquids. /Alkylaluminum halides/ O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 60 Hazardous Substances Data Bank (HSDB) Less sensitive than trialkylaminums to oxidation upon exposure to air. /Alkylaluminum halides/ O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 60 Hazardous Substances Data Bank (HSDB) 3.2.8 Autoignition Temperature Not flammable (USCG, 1999) U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office. CAMEO Chemicals 3.2.9 Decomposition When heated to decomposition it emits highly toxic fumes of /hydrogen fluoride/. Lewis, R.J. Sax's Dangerous Properties of Industrial Materials. 9th ed. Volumes 1-3. New York, NY: Van Nostrand Reinhold, 1996., p. 116 Hazardous Substances Data Bank (HSDB) 3.2.10 Other Experimental Properties Sublimes at 1272 °C, 760 mmHg The Merck Index. 10th ed. Rahway, New Jersey: Merck Co., Inc., 1983., p. 51 Hazardous Substances Data Bank (HSDB) Heat of formation: -1510.4 (crystalline); -1204.6 (gas) kJ/mol @ 298.15 K; Gibbs energy of formation: -1431.1 (crystalline); -1188.2 (gas) kJ/mol @ 298.15 K; standard molar entropy: 66.5 (crystaline); 277.1 (gas) J/mol K @ 298.15 K; molar heat capacity at constant pressure: 75.1 (crystalline); 62.6 (gas) J/mol K @ 298.15 K Lide, DR (ed.). CRC Handbook of Chemistry and Physics. 81st Edition. CRC Press LLC, Boca Raton: FL 2000, p. 5-5 Hazardous Substances Data Bank (HSDB) Enthalpy of fusion: 98 kJ/mol @ triple point Lide, DR (ed.). CRC Handbook of Chemistry and Physics. 81st Edition. CRC Press LLC, Boca Raton: FL 2000, p. 6-121 O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 62 Hazardous Substances Data Bank (HSDB) Hydrolyzed by superheated steam at 300-400 °C O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 62 Hazardous Substances Data Bank (HSDB) 3.3 Chemical Classes Metals -> Metals, Inorganic Compounds Haz-Map, Information on Hazardous Chemicals and Occupational Diseases Corrosives NJDOH RTK Hazardous Substance List 3.3.1 Cosmetics Oral care; Antiplaque S13 \| EUCOSMETICS \| Combined Inventory of Ingredients Employed in Cosmetic Products (2000) and Revised Inventory (2006) \| DOI:10.5281/zenodo.2624118 NORMAN Suspect List Exchange 4 Spectral Information 4.1 1D NMR Spectra 4.1.1 19F NMR Spectra Instrument Name Varian A56/60A Copyright Copyright © 2002-2025 Wiley-VCH GmbH. All Rights Reserved. Thumbnail SpectraBase 4.2 IR Spectra 4.2.1 FTIR Spectra Technique MULLED IN MINERAL OIL Source of Sample The Matheson Company, Inc., Cincinnati, Ohio Copyright Copyright © 1980, 1981-2025 John Wiley & Sons, Inc. All Rights Reserved. Thumbnail SpectraBase 5 Related Records 5.1 Related Compounds with Annotation Follow these links to do a live 2D search or do a live 3D search for this compound, sorted by annotation score. This section is deprecated (see the neighbor discontinuation help page for details), but these live search links provide equivalent functionality to the table that was previously shown here. PubChem 5.2 Related Compounds Same Connectivity Count 2 Mixtures, Components, and Neutralized Forms Count 138 Similar Compounds (2D) View in PubChem Search Similar Conformers (3D) View in PubChem Search PubChem 5.3 Substances 5.3.1 PubChem Reference Collection SID 481106877 PubChem 5.3.2 Related Substances All Count 395 Same Count 167 Mixture Count 228 PubChem 5.3.3 Substances by Category PubChem 5.4 Entrez Crosslinks PubMed Count 9 Protein Structures Count 98 Taxonomy Count 4 Gene Count 13 PubChem 5.5 Associated Chemicals Aluminum ion (3+);22537-23-1 Hazardous Substances Data Bank (HSDB) 6 Chemical Vendors PubChem 7 Drug and Medication Information 7.1 FDA National Drug Code Directory National Drug Code (NDC) Directory 7.2 Drug Labels Homeopathic product and label DailyMed 7.3 Drug Warnings Food and Environmental Agents: Effect on Breast-Feeding: Reported Sign or Symptom in Infant or Effect on Lactation: Fluorides: None. /from Table 7/ Report of the American Academy of Pediatrics Committee on Drugs in Pediatrics 93 (1): 142 (1994) Hazardous Substances Data Bank (HSDB) 8 Pharmacology and Biochemistry 8.1 Absorption, Distribution and Excretion Fluorides are absorbed from GI tract, lung, and skin. Intestine is the major site of absorption. The relatively soluble compounds, such as sodium fluoride, are almost completely absorbed. ... Fluoride has been detected in all organs and tissues examined. ... There is no evidence that it is concentrated in any tissues except bone, thyroid, aorta and perhaps kidney. Fluoride is preponderantly deposited in the skeleton and teeth, the degree of skeletal storage is related to intake and age. ... Major route of ... excretion is via the kidneys; however, small amounts of fluoride appear in sweat, milk, and intestinal secretions. ... About 90% of fluoride ion filtered by glomerulus is reabsorbed by renal tubules. /Fluoride/ Hardman, J.G., L.E. Limbird, P.B., A.G. Gilman. Goodman and Gilman's The Pharmacological Basis of Therapeutics. 10th ed. New York, NY: McGraw-Hill, 2001., p. 1735 Hazardous Substances Data Bank (HSDB) Virtually all fluoride in plasma ... is ultrafilterable. Renal excretion of radiofluoride depends on glomerular filtration and variable tubular reabsorption. Probably, reabsorption is largely passive ... . Fluoride excretion increases when plasma concentration is increased. Procedures that increase urinary flow rate (eg, administration of osmotic diuretics, hypertonic saline, or diuretic drugs) increase the clearance of fluoride. /Fluoride/ National Research Council. Drinking Water & Health Volume 1. Washington, DC: National Academy Press, 1977., p. 376 Hazardous Substances Data Bank (HSDB) Rats given (18)-fluoride ion as a radiotracer by continuous iv infusion of sodium fluoride for 3 hr showed at sublethal dose rates, blood fluoride concentration nears steady state proportional to fluoride infusion rate. Blood, kidney, and lung had highest concentration at doses up to 3 mg fluoride/kg/hr, but at 6 mg/kg/hr the fluoride of the liver, spleen and hollow organs increased sharply. Amount above this was not well processed by excretory mechanism. Rats infused 3 hr with 6 mg fluoride/kg/hr: during infusion fluoride concentration of bone and other tissues was high, bone the highest. Of soft tissues, lung had the highest, brain, testes, and fat pads the least concentration. During depletion phase, tissue fluoride concentration decreased, bone fluoride remained constant, and substantial amount remained in the lung. /Sodium fluoride/ KNAUS RM ET AL; TOX APPL PHARM 38 (2): 335-43 (1976) Hazardous Substances Data Bank (HSDB) Following ingestion, soluble fluorides are rapidly absorbed from the gastrointestinal tract at least to the extent of 97%. Absorbed fluoride is distributed throughout the tissues of the body by the blood. Fluoride concentrations in soft tissues fall to pre-exposure levels within a few hours of exposure. Fluoride exchange with hydroxyl radicals of hydroxyapatite (the inorganic constituent of bone) to form fluorohydroxyapatite. Fluoride that is not retained is excreted rapidly in urine. In adults under steady state intake conditions, the urinary concentration of fluoride tends to approximate the concentration of fluoride in the drinking water. This reflects the decreasing retention of fluoride (primarily in bone) with increasing age. Under certain conditions perspiration may be an important route of fluoride excretion. The concentration of fluoride retained in bones and teeth is a function of both the concentration of fluoride intake and the duration of exposure. Periods of excessive fluoride exposure will result in increased retention in the bone. However, when the excessive exposure is eliminated, the bone fluoride concentration will decrease to a concentration that is again reflective of intake. /Fluoride/ USEPA; Drinking Water Criteria Document for Fluoride p.III-19 (1985) EPA Contract No. 68-03-3279 Hazardous Substances Data Bank (HSDB) For more Absorption, Distribution and Excretion (Complete) data for ALUMINUM FLUORIDE (8 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 8.2 Mechanism of Action Fluoride is very reactive and capable of inhibiting a number of enzymes, including preglycolytic enzymes, phosphatases, and cholinesterase. The result is inhibition of cellular glucose phosphorylation (hence subsequent glycolysis) and respiration and increased sensitivity of cholinergic mechanisms to acetylcholinesterase. /Sodium fluoride/ Booth, N.H., L.E. McDonald (eds.). Veterinary Pharmacology and Therapeutics. 5th ed. Ames, Iowa: Iowa State University Press, 1982., p. 1014 Hazardous Substances Data Bank (HSDB) Inhibition of one or more enzymes controlling cellular glycolysis (and perhaps respiration) may result in a critical lesion. ... Binding or precipitation of calcium as calcium fluoride ... Suggested as mechanism underlying many diverse signs & symptoms in fluoride poisoning, particularly if death is delayed. ... At least in some species fluoride interferes with both contractile power of heart and the mechanism of beat in a way that cannot be ascribed to hypocalcemia. /Fluoride/ Gosselin, R.E., R.P. Smith, H.C. Hodge. Clinical Toxicology of Commercial Products. 5th ed. Baltimore: Williams and Wilkins, 1984., p. II-112 Hazardous Substances Data Bank (HSDB) Aluminum fluoride (AlF4-) inhibited guanine nucleotide-activated phospholipase D (PLD) in rat submandibular gland cell-free lysates in a concentration-dependent response. This effect was consistent in permeabilized cells with endogenous phospholipid PLD substrates. Inhibition was not caused by either fluoride or aluminum alone and was reversed by aluminum chelation. Inhibition of PLD by aluminum fluoride was not mediated by cAMP, phosphatases 1, 2A or 2B, or phosphatidate phosphohydrolase. AlF4- had a similar inhibitory effect on rArf-stimulated PLD, but did not block the translocation of Arf from cytosol to membranes, indicating a post-GTP-binding-protein site of action. Oleate-sensitive PLD, which is not guanine nucleotide-dependent, was also inhibited by AlF4-, supporting a G protein-independent mechanism of action. A submandibular Golgi-enriched membrane preparation had high PLD activity which was also potently inhibited by AlF4-, leading to speculation that the known fluoride inhibition of Golgi vesicle transport may be PLD-mediated. PMID:10570952 Li L, Fleming N; FEBS Lett 458 (3): 419-423 (1999) Hazardous Substances Data Bank (HSDB) Aluminum fluoride (AlF(4)(-)) inhibited phospholipase D (PLD) purified from cabbage in both PIP(2)-dependent and PIP(2)-independent assays. ... Enzyme kinetic studies confirmed that PLD followed Hill kinetics, characteristic for allosteric enzymes, with an apparent Hill coefficient (n(app)) of 3.8, indicating positive cooperativity among multiple substrate-binding sites and suggesting possible functional oligomerization of the enzyme. AlF(4)(-) modification of PLD kinetics was consistent with a competitive mode of enzyme inhibition. PMID:10561485 Li L, Fleming N; FEBS Lett 461 (1-2): 1-5 (1999) Hazardous Substances Data Bank (HSDB) For more Mechanism of Action (Complete) data for ALUMINUM FLUORIDE (6 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 9 Use and Manufacturing 9.1 Uses EPA CPDat Chemical and Product Categories The Chemical and Products Database, a resource for exposure-relevant data on chemicals in consumer products, Scientific Data, volume 5, Article number: 180125 (2018), DOI:10.1038/sdata.2018.125 EPA Chemical and Products Database (CPDat) Sources/Uses Used in ceramics; as flux in metallurgy, ceramic glazes, and enamels; in aluminum manufacture (to lower the melting point and increase the conductivity of the electrolyte); as inhibitor of fermentation; as catalyst in organic reactions; in the manufacture of aluminum silicate; [HSDB] Haz-Map, Information on Hazardous Chemicals and Occupational Diseases Industrial Processes with risk of exposure Aluminum Producing [Category: Industry] Metal Preparation and Pouring [Category: Foundry] Haz-Map, Information on Hazardous Chemicals and Occupational Diseases Activities with risk of exposure Ceramics making [Category: Hobbies] Enameling [Category: Hobbies] Haz-Map, Information on Hazardous Chemicals and Occupational Diseases In ceramics, as flux in metallurgy, in aluminum manufacture, as inhibitor of fermentation, as catalyst in organic reactions. O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 62 Hazardous Substances Data Bank (HSDB) Production of aluminum to lower the melting point & increase the conductivity of the electrolyte, flux in ceramic glazes & enamels, manufacture of aluminum silicate, catalyst. Lewis, R.J., Sr (Ed.). Hawley's Condensed Chemical Dictionary. 13th ed. New York, NY: John Wiley & Sons, Inc. 1997., p. 43 Hazardous Substances Data Bank (HSDB) Ceramics (production of white enamel). /Aluminum fluoride, hydrate/ Lewis, R.J., Sr (Ed.). Hawley's Condensed Chemical Dictionary. 13th ed. New York, NY: John Wiley & Sons, Inc. 1997., p. 43 Hazardous Substances Data Bank (HSDB) 9.1.1 Use Classification Cosmetics -> Oral care; Antiplaque S13 \| EUCOSMETICS \| Combined Inventory of Ingredients Employed in Cosmetic Products (2000) and Revised Inventory (2006) \| DOI:10.5281/zenodo.2624118 NORMAN Suspect List Exchange 9.1.2 Industry Uses Intermediates Pigments Other Processing aids not otherwise specified Process regulators Filler Processing aids, not otherwise listed Catalyst EPA Chemical Data Reporting (CDR) 9.1.3 Consumer Uses Pigment EPA Chemical Data Reporting (CDR) 9.2 Methods of Manufacturing ...By heating ammonium hexafluoroaluminate to red heat in a stream of nitrogen O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 62 Hazardous Substances Data Bank (HSDB) Obtained by the action of fluorine or hydrogen fluoride on the metal at high temperatures, or by fusing cryolite or sodium fluoride with aluminum sulfate. HASZELDINE RN, SHARPE AG; FLUORINE AND ITS COMPOUNDS (1951) p.33 Hazardous Substances Data Bank (HSDB) Action of hydrogen fluoride gas on alumina trihydrate; reaction of hydrogen fluoride on a suspension of aluminum trihydrate, followed by calcining the hydrate formed; reaction of fluosilicic acid on aluminum hydrate. Lewis, R.J., Sr (Ed.). Hawley's Condensed Chemical Dictionary. 13th ed. New York, NY: John Wiley & Sons, Inc. 1997., p. 43 Hazardous Substances Data Bank (HSDB) 9.3 Impurities 0.5 to 2.0 ppm of Silicon, Magnesium, Iron, Nickel, Chromium Aldrich; Catalog Hdbk Fine Chem p.40 (1984) Hazardous Substances Data Bank (HSDB) 9.4 Formulations / Preparations Grade: Technical Lewis, R.J., Sr (Ed.). Hawley's Condensed Chemical Dictionary. 13th ed. New York, NY: John Wiley & Sons, Inc. 1997., p. 43 Hazardous Substances Data Bank (HSDB) Aluminum fluoride grade: 87% CHEMCYCLOPEDIA 1985 p.168 Hazardous Substances Data Bank (HSDB) Grades or purity: 90.7% WEISS, Hdbk HAZARD CHEM p.69 Hazardous Substances Data Bank (HSDB) Grade: Technical, CP /Aluminum fluoride, hydrate/ Lewis, R.J., Sr (Ed.). Hawley's Condensed Chemical Dictionary. 13th ed. New York, NY: John Wiley & Sons, Inc. 1997., p. 43 Hazardous Substances Data Bank (HSDB) 9.5 U.S. Production Aggregated Product Volume 2019: 50,000,000 - <100,000,000 lb 2018: 50,000,000 - <100,000,000 lb 2017: 10,000,000 - <50,000,000 lb 2016: 10,000,000 - <50,000,000 lb EPA Chemical Data Reporting (CDR) (1972) 1.26X10+11 GRAMS SRI Hazardous Substances Data Bank (HSDB) (1975) 1.18X10+11 GRAMS SRI Hazardous Substances Data Bank (HSDB) (1982) 3.82X10+7 g/TECHNICAL/ BUREAU OF THE CENSUS. CURRENT INDUSTRIAL REPORTS: INORGANICS p.6 (1983) Hazardous Substances Data Bank (HSDB) 9.6 General Manufacturing Information Industry Processing Sectors Furniture and Related Product Manufacturing Printing and Related Support Activities Custom Compounding of Purchased Resins Primary Metal Manufacturing Construction Plastics Product Manufacturing Fabricated Metal Product Manufacturing Paper Manufacturing Transportation Equipment Manufacturing Paint and Coating Manufacturing Adhesive Manufacturing EPA Chemical Data Reporting (CDR) EPA TSCA Commercial Activity Status Aluminum fluoride (AlF3): ACTIVE EPA Chemicals under the TSCA Aluminum fluoride and cryolite are the two aluminum-fluorine compounds of industrial importance. Their main application is in baths for electrolytic aluminum production, accounting for over 95% of their use. Ullmann's Encyclopedia of Industrial Chemistry. 6th ed.Vol 1: Federal Republic of Germany: Wiley-VCH Verlag GmbH & Co. 2003 to Present, p. V14 409 (2003) Hazardous Substances Data Bank (HSDB) Loses water at 100 °C, more at 200 °C. It does not seem possible to obtain the anhydrous compound free from oxides by dehydration of the hydrates. /Trihydrate/ O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 62 Hazardous Substances Data Bank (HSDB) 10 Safety and Hazards 10.1 Hazards Identification 10.1.1 GHS Classification 1 of 3 items View All Note This chemical does not meet GHS hazard criteria for 95.2% (360 of 378) of all reports. GHS Hazard Statements Not Classified Reported as not meeting GHS hazard criteria by 360 of 378 companies (only 4.8% companies provided GHS information). For more detailed information, please visit ECHA C&L website. ECHA C&L Notifications Summary Aggregated GHS information provided per 378 reports by companies from 11 notifications to the ECHA C&L Inventory. Reported as not meeting GHS hazard criteria per 360 of 378 reports by companies. There are 9 notifications provided by 18 of 378 reports by companies with hazard statement code(s). Information may vary between notifications depending on impurities, additives, and other factors. The percentage value in parenthesis indicates the notified classification ratio from companies that provide hazard codes. Only hazard codes with percentage values above 10% are shown. For more detailed information, please visit ECHA C&L website. European Chemicals Agency (ECHA) 10.1.2 Hazard Classes and Categories Not Classified European Chemicals Agency (ECHA) View More... 10.1.3 Health Hazards ACUTE: respiratory irritation; possible nose bleeding or vomiting; CHRONIC: aggravates bronchitis/asthma; increased bone density. (USCG, 1999) U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office. CAMEO Chemicals 10.1.4 Fire Hazards Special Hazards of Combustion Products: When heated to sublimation condition, emits toxic fumes of fluoride (USCG, 1999) U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office. CAMEO Chemicals Not combustible. Gives off irritating or toxic fumes (or gases) in a fire. ILO-WHO International Chemical Safety Cards (ICSCs) 10.1.5 Hazards Summary A severe eye irritant; A respiratory tract irritant; [HSDB] A skin, eye, and respiratory tract irritant; Repeated or prolonged inhalation exposure may cause asthma. The substance may have effects on the bone, nervous system, resulting in bone alterations (fluorosis), and nervous system impairment. [ICSC] A corrosive substance that can cause injury to the skin, eyes, and respiratory tract; Inhalation may cause chemical pneumonitis; [MSDSonline] See FLUORIDES. Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 10.1.6 Fire Potential Certain polymerization catalysts, such as aluminum alkyls, react & burn violently on contact with water. /Aluminum alkyls/ National Research Council. Prudent Practices for Handling Hazardous Chemicals in Laboratories. Washington, DC: National Academy Press, 1981., p. 221 Hazardous Substances Data Bank (HSDB) Aluminum alkyls are organic aluminum compounds that are highly reactive and dangerous because of spontaneous burning in air. /Aluminum alkyls/ Grant, W.M. Toxicology of the Eye. 3rd ed. Springfield, IL: Charles C. Thomas Publisher, 1986., p. 73 Hazardous Substances Data Bank (HSDB) Prophoric material in flammable solvent. Vapors are heavier than air & may travel to a source of ignition & flash back. /Aluminum alkyls/ Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 49-16 Hazardous Substances Data Bank (HSDB) 10.1.7 Skin, Eye, and Respiratory Irritations A severe eye irritant. Lewis, R.J. Sax's Dangerous Properties of Industrial Materials. 9th ed. Volumes 1-3. New York, NY: Van Nostrand Reinhold, 1996., p. 116 Hazardous Substances Data Bank (HSDB) Irritating to nose and throat. U.S. Coast Guard, Department of Transportation. CHRIS - Hazardous Chemical Data. Volume II. Washington, D.C.: U.S. Government Printing Office, 1984-5. Hazardous Substances Data Bank (HSDB) Irritation of the eyes has been noted in patients who have been exposed /to aluminum alkyls/. /Aluminum alkyls/ Grant, W.M. Toxicology of the Eye. 3rd ed. Springfield, IL: Charles C. Thomas Publisher, 1986., p. 73 Hazardous Substances Data Bank (HSDB) May cause minor irritation to lungs or eyes. /Aluminum (dust or powder)/ Fire Protection Guide to Hazardous Materials. 12 ed. Quincy, MA: National Fire Protection Association, 1997., p. 49-17 Hazardous Substances Data Bank (HSDB) 10.2 Safety and Hazard Properties 10.2.1 Explosive Limits and Potential ...Violently explosive when they come into contact with water. /Aluminum alkyls/ International Labour Office. Encyclopedia of Occupational Health and Safety. Vols. I&II. Geneva, Switzerland: International Labour Office, 1983., p. 422 Hazardous Substances Data Bank (HSDB) 10.2.2 OSHA Standards Permissible Exposure Limit: Table Z-1 8-hr Time Weighted Avg: 2.5 mg/cu m. /Fluorides, as F/ 29 CFR 1910.1000; U.S. National Archives and Records Administration's Electronic Code of Federal Regulations. Available from, as of July 1, 2004: Hazardous Substances Data Bank (HSDB) Vacated 1989 OSHA PEL TWA 2 mg/cu m is still enforced in some states. /Aluminum (soluble salts & alkyls, as Al)/ NIOSH. NIOSH Pocket Guide to Chemical Hazards. DHHS (NIOSH) Publication No. 97-140. Washington, D.C. U.S. Government Printing Office, 1997., p. 359 Hazardous Substances Data Bank (HSDB) 10.2.3 NIOSH Recommendations Recommended Exposure Limit: 10-Hr Time-Weighted Avg: 10 mg/cu m (total). /Aluminum/ NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety& Health. DHHS (NIOSH) Publication No. 2004-103 (2003). Hazardous Substances Data Bank (HSDB) Recommended Exposure Limit: 10 Hr Time-Weighted Avg: 5 mg/cu m (resp). /Aluminum/ NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety& Health. DHHS (NIOSH) Publication No. 2004-103 (2003). Hazardous Substances Data Bank (HSDB) Recommended Exposure Limit: 10 Hr Time-Weighted Avg: 2 mg/cu m. /Aluminum (soluble salts and alkyls, as Al)/ NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety& Health. DHHS (NIOSH) Publication No. 2004-103 (2003). Hazardous Substances Data Bank (HSDB) Recommended Exposure Limit: 10 Hr Time-Weighted Avg: 5 mg/cu m. /Aluminum (pyro powders and welding fumes, as Al)/ NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety& Health. DHHS (NIOSH) Publication No. 2004-103 (2003). Hazardous Substances Data Bank (HSDB) 10.3 First Aid Measures Inhalation First Aid Fresh air, rest. Refer for medical attention. ILO-WHO International Chemical Safety Cards (ICSCs) Skin First Aid Rinse skin with plenty of water or shower. ILO-WHO International Chemical Safety Cards (ICSCs) Eye First Aid First rinse with plenty of water for several minutes (remove contact lenses if easily possible), then refer for medical attention. ILO-WHO International Chemical Safety Cards (ICSCs) Ingestion First Aid Rinse mouth. Give one or two glasses of water to drink. Refer for medical attention . ILO-WHO International Chemical Safety Cards (ICSCs) 10.3.1 First Aid For acute poisoning, oral administration of limewater, intravenous infusion of glucose, and intravenous injections of calcium gluconates. (USCG, 1999) U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office. CAMEO Chemicals 10.4 Fire Fighting In case of fire in the surroundings, use appropriate extinguishing media. ILO-WHO International Chemical Safety Cards (ICSCs) 10.4.1 Fire Fighting Procedures Stop flow of liquid before extinguishing fire. Use dry chemical or carbon dioxide. DO NOT use water as straight stream directly on spilled material. Water fog can be used to control fire. DO NOT use halogenated extinguishing agents on spilled material. Violent reaction may result. Use water spray to keep fire-exposed containers cool. Fight fire from protected location or maximum possible distance. /Aluminum Alkyls/ Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 49-17 Hazardous Substances Data Bank (HSDB) 10.5 Accidental Release Measures 10.5.1 Spillage Disposal Personal protection: particulate filter respirator adapted to the airborne concentration of the substance. Sweep spilled substance into covered containers. If appropriate, moisten first to prevent dusting. ILO-WHO International Chemical Safety Cards (ICSCs) 10.5.2 Disposal Methods Aluminum compounds are treated under anhydrous conditions to prevent violent reactions, recover solvent, and form Al compounds suitable for landfill by reaction with anhydrous hydrolysis agent, eg calcium hydroxide. /Aluminum compounds/ Lee TE; Disposal of Aluminum-Containing Waste Materials, US Patent Number 4018867 4/19/77 (ETHYL CORP) Hazardous Substances Data Bank (HSDB) SRP: The most favorable course of action is to use an alternative chemical product with less inherent propensity for occupational exposure or environmental contamination. Recycle any unused portion of the material for its approved use or return it to the manufacturer or supplier. Ultimate disposal of the chemical must consider: the material's impact on air quality; potential migration in soil or water; effects on animal, aquatic, and plant life; and conformance with environmental and public health regulations. Hazardous Substances Data Bank (HSDB) 10.5.3 Preventive Measures SRP: The scientific literature for the use of contact lenses in industry is conflicting. The benefit or detrimental effects of wearing contact lenses depend not only upon the substance, but also on factors including the form of the substance, characteristics and duration of the exposure, the uses of other eye protection equipment, and the hygiene of the lenses. However, there may be individual substances whose irritating or corrosive properties are such that the wearing of contact lenses would be harmful to the eye. In those specific cases, contact lenses should not be worn. In any event, the usual eye protection equipment should be worn even when contact lenses are in place. Hazardous Substances Data Bank (HSDB) Work clothing that becomes wet or significantly contaminated should be removed and replaced. /Aluminum (soluble salts and alkyls, as Al)/ NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety& Health. DHHS (NIOSH) Publication No. 2004-103 (2003). Hazardous Substances Data Bank (HSDB) Workers whose clothing may have become contaminated should change into uncontaminated clothing before leaving the work premises. /Aluminum (soluble salts and alkyls, as Al)/ NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety& Health. DHHS (NIOSH) Publication No. 2004-103 (2003). Hazardous Substances Data Bank (HSDB) The worker should immediately wash the skin when it becomes contaminated. /Aluminum (soluble salts and alkyls, as Al)/ NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety& Health. DHHS (NIOSH) Publication No. 2004-103 (2003). Hazardous Substances Data Bank (HSDB) SRP: Contaminated protective clothing should be segregated in such a manner so that there is no direct personal contact by personnel who handle, dispose, or clean the clothing. Quality assurance to ascertain the completeness of the cleaning procedures should be implemented before the decontaminated protective clothing is returned for reuse by the workers. Contaminated clothing should not be taken home at end of shift, but should remain at employee's place of work for cleaning. Hazardous Substances Data Bank (HSDB) 10.6 Handling and Storage 10.6.1 Safe Storage Dry. Well closed. ILO-WHO International Chemical Safety Cards (ICSCs) 10.6.2 Storage Conditions Extremely reactive with air, moisture and compounds containing active hydrogen and therefore must be kept under a blanket of inert gas. /Aluminum alkyls/ International Labour Office. Encyclopaedia of Occupational Health and Safety. 4th edition, Volumes 1-4 1998. Geneva, Switzerland: International Labour Office, 1998., p. 62.3 Hazardous Substances Data Bank (HSDB) In general, materials ... toxic as stored or which can decompose into toxic components ... Should be stored in cool ... ventilated place, out of ... sun, away from ... fire hazard ... be periodically inspected and monitored. Incompatible materials should be isolated ... Sax, N.I. Dangerous Properties of Industrial Materials. 4th ed. New York: Van Nostrand Reinhold, 1975., p. 726 Hazardous Substances Data Bank (HSDB) All possibility of contact with water must be avoided. Solution containing not more than 20% of these compounds in non-reactive solvents, however, can be handled without risk of spontaneous ignition. /Aluminum alkyls/ International Labour Office. Encyclopedia of Occupational Health and Safety. Vols. I&II. Geneva, Switzerland: International Labour Office, 1983., p. 422 Hazardous Substances Data Bank (HSDB) Must be stored in an inert atmosphere; sensitive to oxidation and hydrolysis in air. /Trialkylaluminums/ O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 60 Hazardous Substances Data Bank (HSDB) Separate from air, water, halocarbons, alcohols. Store in a cool, dry, well-ventilated location. Outside or detached storage is preferred. Inside storage should be in a standard flammable liquid storage warehouse, room, or cabinet. /Aluminum alkyls/ Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 49-17 Hazardous Substances Data Bank (HSDB) 10.7 Exposure Control and Personal Protection Exposure Summary Biological Exposure Indices (BEI) [ACGIH] - Fluorides in urine = 2 mg/L prior to shift or 3 mg/L at end of shift; (Repeated measurements recommended.) ACGIH - Documentation of the TLVs and BEIs, 7th Ed. Cincinnati: ACGIH Worldwide, 2020. Haz-Map, Information on Hazardous Chemicals and Occupational Diseases Maximum Allowable Concentration (MAK) 1.0 [mg/m3], inhalable fraction, as F[German Research Foundation (DFG)] Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 10.7.1 Permissible Exposure Limit (PEL) 2.5 [mg/m3], as F Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 10.7.2 Threshold Limit Values (TLV) 2.5 [mg/m3], as F Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 8 hr Time Weighted Avg (TWA): 2.5 mg/cu m. /Fluorides, as F/ American Conference of Governmental Industrial Hygienists TLVs and BEIs. Threshold Limit Values for Chemical Substances and Physical Agents and Biological Exposure Indices. Cincinnati, OH, 2008, p. 31 Hazardous Substances Data Bank (HSDB) Biological Exposure Index (BEI): Determinant: fluorides in urine; Sampling Time: prior to shift; BEI: 3 mg/g creatinine. Determinant: fluorides in urine; Sampling Time: end of shift; BEI: 10 mg/g creatinine. The determinant may be present in biological specimens collected from subjects who have not been occupationally exposed, at a concentration which could affect interpretation of the result. Such background concentrations are incorporated in the BEI value. The determinant is nonspecific, since it is also observed after exposure to other chemicals. /Fluorides/ American Conference of Governmental Industrial Hygienists TLVs and BEIs. Threshold Limit Values for Chemical Substances and Physical Agents and Biological Exposure Indices. Cincinnati, OH, 2008, p. 105 Hazardous Substances Data Bank (HSDB) A4; Not classifiable as a human carcinogen. /Fluorides, as F/ American Conference of Governmental Industrial Hygienists TLVs and BEIs. Threshold Limit Values for Chemical Substances and Physical Agents and Biological Exposure Indices. Cincinnati, OH, 2008, p. 31 Hazardous Substances Data Bank (HSDB) Excursion Limit Recommendation: Excursions in worker exposure levels may exceed 3 times the TLV-TWA for no more than a total of 30 minutes during a work day, and under no circumstances should they exceed 5 times the TLV-TWA, provided that the TLV-TWA is not exceeded. /Fluorides, as F/ American Conference of Governmental Industrial Hygienists TLVs and BEIs. Threshold Limit Values for Chemical Substances and Physical Agents and Biological Exposure Indices. Cincinnati, OH, 2008, p. 5 Hazardous Substances Data Bank (HSDB) For more Threshold Limit Values (TLV) (Complete) data for ALUMINUM FLUORIDE (7 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) (as F): 2.5 mg/m ILO-WHO International Chemical Safety Cards (ICSCs) 10.7.3 Occupational Exposure Limits (OEL) EU-OEL (as F): 2,5 mg/m ILO-WHO International Chemical Safety Cards (ICSCs) MAK (Maximale Arbeitsplatz Konzentration) (as F, inhalable fraction): 1 mg/m ILO-WHO International Chemical Safety Cards (ICSCs) 10.7.4 Other Standards Regulations and Guidelines USSR: 1 mg/cu m /Fluorides/ International Labour Office. Encyclopedia of Occupational Health and Safety. Vols. I&II. Geneva, Switzerland: International Labour Office, 1983., p. 892 Hazardous Substances Data Bank (HSDB) 10.7.5 Inhalation Risk Evaporation at 20Â °C is negligible; a harmful concentration of airborne particles can, however, be reached quickly. ILO-WHO International Chemical Safety Cards (ICSCs) 10.7.6 Effects of Short Term Exposure The aerosol is irritating to the eyes, skin and respiratory tract. ILO-WHO International Chemical Safety Cards (ICSCs) 10.7.7 Effects of Long Term Exposure Repeated or prolonged inhalation may cause asthma. The substance may have effects on the bone and central nervous system. This may result in bone alterations (fluorosis) and nervous system impairment. ILO-WHO International Chemical Safety Cards (ICSCs) 10.7.8 Acceptable Daily Intakes 4 mg/L (Adjusted Acceptable Daily Intake) /Fluoride/ USEPA Drinking Water Criteria Documents; Fluoride p.1X-22 (1985) USEPA PB86-118163 Hazardous Substances Data Bank (HSDB) 10.7.9 Personal Protective Equipment (PPE) Goggles to protect against airborne particles and possibly respirator for intermittent heavy dust exposures. (USCG, 1999) U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office. CAMEO Chemicals Wear appropriate personal protective clothing to prevent skin contact. /Aluminum (soluble salts and alkyls, as Al/ NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety& Health. DHHS (NIOSH) Publication No. 2004-103 (2003). Hazardous Substances Data Bank (HSDB) Wear appropriate eye protection to prevent eye contact. /Aluminum (soluble salts and alkyls, as Al)/ NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety& Health. DHHS (NIOSH) Publication No. 2004-103 (2003). Hazardous Substances Data Bank (HSDB) Protective clothing and a high standard of training in the necessary precautionary measures are essential for the handling of the materials. /Aluminum alkyls/ International Labour Office. Encyclopedia of Occupational Health and Safety. Volumes I and II. New York: McGraw-Hill Book Co., 1971., p. 271 Hazardous Substances Data Bank (HSDB) 10.7.10 Exposure Prevention PREVENT DISPERSION OF DUST! ILO-WHO International Chemical Safety Cards (ICSCs) 10.7.11 Inhalation Prevention Use local exhaust or breathing protection. ILO-WHO International Chemical Safety Cards (ICSCs) 10.7.12 Skin Prevention Protective gloves. ILO-WHO International Chemical Safety Cards (ICSCs) 10.7.13 Eye Prevention Wear safety goggles. ILO-WHO International Chemical Safety Cards (ICSCs) 10.7.14 Ingestion Prevention Do not eat, drink, or smoke during work. Wash hands before eating. ILO-WHO International Chemical Safety Cards (ICSCs) 10.8 Stability and Reactivity 10.8.1 Air and Water Reactions Slightly soluble in water CAMEO Chemicals 10.8.2 Reactive Group Salts, Basic CAMEO Chemicals 10.8.3 Reactivity Profile ALUMINUM FLUORIDE when heated to sublimation condition, emits toxic fumes of fluoride (USCG, 1999). U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office. CAMEO Chemicals 10.8.4 Hazardous Reactivities and Incompatibilities Violently impact-sensitive when in contact with sodium and potassium. Lewis, R.J. Sax's Dangerous Properties of Industrial Materials. 9th ed. Volumes 1-3. New York, NY: Van Nostrand Reinhold, 1996., p. 116 Hazardous Substances Data Bank (HSDB) ... On contact with acid or acid fumes, they emit highly toxic fumes. /Fluorides/ Sax, N.I. Dangerous Properties of Industrial Materials. 6th ed. New York, NY: Van Nostrand Reinhold, 1984., p. 176 Hazardous Substances Data Bank (HSDB) Extremely reactive with air, moisture, and compounds containing active hydrogen... /Alkyl aluminum compounds/ International Labour Office. Encyclopaedia of Occupational Health and Safety. 4th edition, Volumes 1-4 1998. Geneva, Switzerland: International Labour Office, 1998., p. 63.2 Hazardous Substances Data Bank (HSDB) Reacts violently with a broad range of materials including air &water. /Aluminum alkyls/ Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 49-17 Hazardous Substances Data Bank (HSDB) 10.9 Transport Information 10.9.1 Shipment Methods and Regulations The International Maritime Dangerous Goods Code lays down basic principles for transporting hazardous chemicals. Detailed recommendations for individual substances and a number of recommendations for good practice are included in the classes dealing with such substances. A general index of technical names has also been compiled. This index should always be consulted when attempting to locate the appropriate procedures to be used when shipping any substance or article. IMDG; International Maritime Dangerous Goods Code; International Maritime Organization p.4221 (1998) Hazardous Substances Data Bank (HSDB) The International Air Transport Association (IATA) Dangerous Goods Regulations are published by the IATA Dangerous Goods Board pursuant to IATA Resolutions 618 and 619 and constitute a manual of industry carrier regulations to be followed by all IATA Member airlines when transporting hazardous materials. IATA. Dangerous Goods Regulations. 45 th Ed. Montreal, Canada and Geneva, Switzerland. International Air Transport Association. Dangerous Goods Regulations, 2004., p. 93 Hazardous Substances Data Bank (HSDB) No person may /transport,/ offer or accept a hazardous material for transportation in commerce unless that person is registered in conformance ... and the hazardous material is properly classed, described, packaged, marked, labeled, and in condition for shipment as required or authorized by ... /the hazardous materials regulations (49 CFR 171-177)./ 49 CFR 171.2; U.S. National Archives and Records Administration's Electronic Code of Federal Regulations. Available from, as of February 5, 2004: Hazardous Substances Data Bank (HSDB) 10.10 Regulatory Information The Australian Inventory of Industrial Chemicals Chemical: Aluminium fluoride (AlF3) Australian Industrial Chemicals Introduction Scheme (AICIS) REACH Registered Substance Status: Active Update: 05-10-2022 European Chemicals Agency (ECHA) New Zealand EPA Inventory of Chemical Status Aluminium fluoride: Does not have an individual approval but may be used under an appropriate group standard New Zealand Environmental Protection Authority (EPA) New Jersey Worker and Community Right to Know Act The New Jersey Worker and Community Right to Know Act requires public and private employers to provide information about hazardous substances at their workplaces. (N.J.S.A. 34:5A-1 et. seq.) NJDOH RTK Hazardous Substance List 10.10.1 Federal Drinking Water Guidelines EPA 50-200 ug/l /Aluminum/ USEPA/Office of Water; Federal-State Toxicology and Risk Analysis Committee (FSTRAC). Summary of State and Federal Drinking Water Standards and Guidelines (11/93) To Present Hazardous Substances Data Bank (HSDB) 10.10.2 State Drinking Water Standards (CA) CALIFORNIA 1000 ug/l /Aluminum/ USEPA/Office of Water; Federal-State Toxicology and Risk Analysis Committee (FSTRAC). Summary of State and Federal Drinking Water Standards and Guidelines (11/93) To Present Hazardous Substances Data Bank (HSDB) 10.10.3 State Drinking Water Guidelines (AZ) ARIZONA 73 ug/l /Aluminum/ USEPA/Office of Water; Federal-State Toxicology and Risk Analysis Committee (FSTRAC). Summary of State and Federal Drinking Water Standards and Guidelines (11/93) To Present Hazardous Substances Data Bank (HSDB) (CA) CALIFORNIA 200 ug/l /Aluminum/ USEPA/Office of Water; Federal-State Toxicology and Risk Analysis Committee (FSTRAC). Summary of State and Federal Drinking Water Standards and Guidelines (11/93) To Present Hazardous Substances Data Bank (HSDB) (ME) MAINE 1430 ug/l /Aluminum/ USEPA/Office of Water; Federal-State Toxicology and Risk Analysis Committee (FSTRAC). Summary of State and Federal Drinking Water Standards and Guidelines (11/93) To Present Hazardous Substances Data Bank (HSDB) 10.11 Other Safety Information Chemical Assessment IMAP assessments - Aluminium fluoride (AlF3): Human health tier I assessment Australian Industrial Chemicals Introduction Scheme (AICIS) 10.11.1 Special Reports Anon; Information profiles on Potential Occupaional Hazards: Aluminum& Compounds, 2nd Draft (Revised); GRA & I Issue 21 NTIS/PB89-216238 (1989)> TD3: Information profiles are working papers used by the National Institute for Occupational Safety and Health to assist the Institute in establishing priorities. The proile summarizes data on organic and inorganic substances containing aluminum as the only metal. Each summary presents data on known and suspected health effects, the extent of worker exposure, physical and chemical properties and the industrial importance of the following aluminum compounds: aluminum metal, aluminum ammonium sulfate, aluminum chlorhydrate, aluminum chloride anhydrous, aluminum chloride hydrous, aluminum distearate, aluminum ethoxide, aluminum fluoride, aluminum hydride, aluminum hydroxide, aluminum nitrate, aluminum oxide, aluminum ortho-phosphate, aluminum potassium sulfate, aluminum silicate, aluminum sodium sulfate, aluminum sulfate, calcium aluminum silicate, diethylaluminum chloride, sodium aluminate, tri-n-butylaluminum, triethylaluminum, tri-n-hexylaluminum, triisobutylaluminum, tri-n-octylaluminum. Detailed literature searches are conducted to identify information to be used in the profile sumaries. Sponsored by National Inst. for Occupational Safety and Health, Rockville, MD. Hazardous Substances Data Bank (HSDB) 11 Toxicity 11.1 Toxicological Information 11.1.1 Evidence for Carcinogenicity A4: Not classifiable as a human carcinogen. /Aluminum metal and insouble compound/ American Conference of Governmental Industrial Hygienists TLVs and BEIs. Threshold Limit Values for Chemical Substances and Physical Agents and Biological Exposure Indices. Cincinnati, OH, 2008, p. 11 Hazardous Substances Data Bank (HSDB) A4; Not classifiable as a human carcinogen. /Fluorides as F/ American Conference of Governmental Industrial Hygienists TLVs and BEIs. Threshold Limit Values for Chemical Substances and Physical Agents and Biological Exposure Indices. Cincinnati, OH, 2008, p. 31 Hazardous Substances Data Bank (HSDB) 11.1.2 Exposure Routes The substance can be absorbed into the body by inhalation of dust and by ingestion. ILO-WHO International Chemical Safety Cards (ICSCs) 11.1.3 Signs and Symptoms Inhalation Exposure Cough. Shortness of breath. Sore throat. ILO-WHO International Chemical Safety Cards (ICSCs) Skin Exposure Pain. Redness. ILO-WHO International Chemical Safety Cards (ICSCs) Eye Exposure Redness. Pain. ILO-WHO International Chemical Safety Cards (ICSCs) 11.1.4 Adverse Effects Dermatotoxin - Skin burns. Asthma - Reversible bronchoconstriction (narrowing of bronchioles) initiated by the inhalation of irritating or allergenic agents. Toxic Pneumonitis - Inflammation of the lungs induced by inhalation of metal fumes or toxic gases and vapors. Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 11.1.5 Acute Effects ChemIDplus 11.1.6 Interactions In the gut, antacids form insoluble compounds with numerous substances. Interactions are especially prominent with aluminum-containing antacids. Aluminum forms insoluble aluminum phosphate, which decreases the bioavailability of phosphate. Although aluminum fluoride is somewhat soluble, aluminum-containing antacids interfere with the absorption of fluoride, probably through the formation of more complex compounds, such as fluoroapatite. Aluminum-containing antacids adsorb bile acids, lysolecithin, and various proteins. /Aluminum-Containing antacids/ Gilman, A.G., L.S.Goodman, and A. Gilman. (eds.). Goodman and Gilman's The Pharmacological Basis of Therapeutics. 7th ed. New York: Macmillan Publishing Co., Inc., 1985., p. 981 Hazardous Substances Data Bank (HSDB) ... Pretreatment of rats with fluoride increased their sensitivity to succinylcholine, demeton and parathion. /Fluoride/ Gosselin, R.E., R.P. Smith, H.C. Hodge. Clinical Toxicology of Commercial Products. 5th ed. Baltimore: Williams and Wilkins, 1984., p. II-112 Hazardous Substances Data Bank (HSDB) 11.1.7 Antidote and Emergency Treatment /Exposure to aluminum fluoride/ dust: If inhaled, irritating to nose and throat. Move to fresh air. U.S. Coast Guard, Department of Transportation. CHRIS - Hazardous Chemical Data. Volume II. Washington, D.C.: U.S. Government Printing Office, 1984-5. Hazardous Substances Data Bank (HSDB) 11.1.8 Medical Surveillance Fluoride levels in urine should be checked periodically and all workers should be subjected to periodical skeletal X-ray exam particularly of the pelvis. /Fluoride and cmpd/ International Labour Office. Encyclopedia of Occupational Health and Safety. Vols. I&II. Geneva, Switzerland: International Labour Office, 1983., p. 894 Hazardous Substances Data Bank (HSDB) 11.1.9 Human Toxicity Excerpts /SIGNS AND SYMPTOMS/ /Anhydrous aluminum fluoride is a/ strong irritant to tissue. Lewis, R.J., Sr (Ed.). Hawley's Condensed Chemical Dictionary. 13th ed. New York, NY: John Wiley & Sons, Inc. 1997., p. 43 Hazardous Substances Data Bank (HSDB) /SIGNS AND SYMPTOMS/ Symptoms Following Exposure /to aluminum fluoride/: Acute: respiratory irritation, possible nose bleeding or vomiting: Chronic: aggravates bronchitis/ asthma; increased bone density. U.S. Coast Guard, Department of Transportation. CHRIS - Hazardous Chemical Data. Volume II. Washington, D.C.: U.S. Government Printing Office, 1984-5. Hazardous Substances Data Bank (HSDB) /SIGNS AND SYMPTOMS/ Fluorosis. Inhalation of fluoride poses a potential hazard in workers in primary aluminium production, but the majority of workers are clinically unaffected. Clinical fluorosis, which is rare, commences with stiffness in the lower back followed by pain and then limitation of rotation of the trunk. Later, the spine becomes rigid and stiff, with restriction of chest movement and of the large joints, particularly the hip, that is accompanied by osteosclerosis. /Aluminum fluorides/ IARC. Monographs on the Evaluation of the Carcinogenic Risk of Chemicals to Humans. Geneva: World Health Organization, International Agency for Research on Cancer, 1972-PRESENT. (Multivolume work). Available at: p. 34(53) 1984 Hazardous Substances Data Bank (HSDB) /SIGNS AND SYMPTOMS/ Variable degrees of bony fluorosis have been seen. The first stage of this condition consists simply of an increase in bone density, particularly marked in the vertebral bodies and pelvis. As fluoride is further absorbed into bone, calcification of the ligaments of the pelvis may be seen ... In the event of extreme and protracted exposure to fluoride, calcification of the paraspinal and other ligamentous structures as well as about joints are noted. /Aluminum fluorides/ International Labour Office. Encyclopedia of Occupational Health and Safety. Vols. I&II. Geneva, Switzerland: International Labour Office, 1983., p. 133 Hazardous Substances Data Bank (HSDB) For more Human Toxicity Excerpts (Complete) data for ALUMINUM FLUORIDE (12 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 11.1.10 Non-Human Toxicity Excerpts /LABORATORY ANIMALS: Acute Exposure/ /Acute poisoning/ if sufficient fluoride is absorbed ... Fluoride ion increases capillary permeability and also produces a coagulation defect. These actions lead to hemorrhagic gastroenteritis and hemorrhages, congestion, and edema in various organs including the brain. Clinical manifestations include excitability, muscle tremors, weakness, urination, defecation, salivation, emesis, sudden collapse, clonic convulsions, coma, and death due to respiratory and cardiac failure. Cyanosis and early rigor mortis ... /Fluoride/ Booth, N.H., L.E. McDonald (eds.). Veterinary Pharmacology and Therapeutics. 5th ed. Ames, Iowa: Iowa State University Press, 1982., p. 1014 Hazardous Substances Data Bank (HSDB) /LABORATORY ANIMALS: Acute Exposure/ After intraperitoneal administration of a single large dose of fluoride (sodium fluoride, 35 mg/kg body weight), the calcium contents of the renal cortex and medulla of fluoride intoxicated rats were increased by 33 and 10 times, respectively. /Sodium Fluoride/ Suketa Y et al; Toxicol Appl Pharmacol 39: 313-19 (1977) as cited in WHO; Environ Health Criteria: Fluorine and Fluorides p.54 (1984) Hazardous Substances Data Bank (HSDB) /LABORATORY ANIMALS: Acute Exposure/ The ionic fluoride levels in plasma following intraperitoneal administration of 15, 20, or 25 mg of fluoride per kg body weight to 200 g rats /was studied/. In animals given 25 mg/kg, the mean ionic fluoride level in plasma was 38 mg/liter after 10 min and the animals invariably died within 1 hr. All animals receiving 15 or 20 mg/kg survived, despite mean ionic fluoride levels in plasma of 22.9 and 29.2 mg/l, respectively. /Sodium Fluoride/ Singer L et al; Proc Soc Exp Biol Med 157: 363-68 (1978) as cited in WHO; Environ Health Criteria: Fluorine and Fluorides p.53 (1984) Hazardous Substances Data Bank (HSDB) /LABORATORY ANIMALS: Acute Exposure/ The acute and subacute physiological and pathological effects of fluoride (as sodium fluoride) administered intravenously and orally to male and female dogs /were described/. When fluoride was infused intravenously in four dogs at the rate of 5.4 mg Fluoride/min, the mean acute lethal dose was 36.0 + or - 0.5 mg Fluoride/kg with death occurring after 59 to 64 minutes of infusion. The principal effects observed were a progressive decline in blood pressure, heart rate, central nervous system activity (pupil size, response to light, tendon reflexes) with vomiting and defecation. /Sodium Fluoride/ Leone NC et al; Public Health Rep 71: 459-67 (1956) as cited in USEPA; Drinking Water Criteria Document for Fluoride p.III-9 (1985) EPA Contract No. 68-03-3279 Hazardous Substances Data Bank (HSDB) For more Non-Human Toxicity Excerpts (Complete) data for ALUMINUM FLUORIDE (14 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 11.1.11 Non-Human Toxicity Values LD50 Mouse oral 103 mg/kg Lewis, R.J. Sax's Dangerous Properties of Industrial Materials. 9th ed. Volumes 1-3. New York, NY: Van Nostrand Reinhold, 1996., p. 116 Hazardous Substances Data Bank (HSDB) 11.1.12 Populations at Special Risk Populations that appear to be at increased risk from the effects of fluoride are individuals that suffer from diabetes insipidus or some forms of renal impairment. /Fluoride/ USEPA; Drinking Water Criteria Document for Fluoride p.I-5 (1985) EPA Contract No. 68-03-3279 Hazardous Substances Data Bank (HSDB) 11.2 Ecological Information 11.2.1 Ecotoxicity Values LC50 Chlorella vulgaris (Green algae) 100 uM/3 days; static /technical product/ Rai LC et al; Aquatic Toxicology 42 (1): 67-84 (1998) Hazardous Substances Data Bank (HSDB) LC50 Chlorella vulgaris (Green algae) 180 uM/15 days; static /technical product/ Rai LC et al; Aquatic Toxicology 42 (1): 67-84 (1998) Hazardous Substances Data Bank (HSDB) 11.2.2 Probable Routes of Human Exposure Chronic fluorosis generally develops after prolonged (10-20 years) exposure to industrial dusts, insecticides, or water where fluorides exceed 3 to 4 ppm. This is especially true in workers involved in the production of aluminum, steel, or glass. /Aluminum fluorides/ Ellenhorn, M.J. and D.G. Barceloux. Medical Toxicology - Diagnosis and Treatment of Human Poisoning. New York, NY: Elsevier Science Publishing Co., Inc. 1988., p. 533 Hazardous Substances Data Bank (HSDB) Production of aluminum sulfate and aluminum fluoride is associated with exposure to several irritant substances. Such production has been connected with reversible bronchial obstruction or asthma. In an aluminum plant in Sweden, 6 cases of asthma occurred in 1975, and 7 in 1976. The number of exposed workers was 35-40. The levels of aluminum fluoride (personal sampling) were measured during these two years, and the mean concentrations were 3-6 mg/cu m. In 1977, improvements were made at this plant thereby reducing the mean levels of aluminum fluoride to 0.4-1.0 mg/cu m. During the years 1978-1980, only two cases of asthma occurred. /Aluminum fluorides/ Friberg, L., Nordberg, G.F., Kessler, E. and Vouk, V.B. (eds). Handbook of the Toxicology of Metals. 2nd ed. Vols I, II.: Amsterdam: Elsevier Science Publishers B.V., 1986., p. 15 Hazardous Substances Data Bank (HSDB) Electrolytic production of aluminum can lead to a substantial exposure to fluorides and carcinogenic tar oils, including polyaromatic hydrocarbons. /Aluminum fluorides/ Friberg, L., Nordberg, G.F., Kessler, E. and Vouk, V.B. (eds). Handbook of the Toxicology of Metals. 2nd ed. Vols I, II.: Amsterdam: Elsevier Science Publishers B.V., 1986., p. 1 Hazardous Substances Data Bank (HSDB) Primary aluminum production plants are located in about 40 countries. The two main methods used for aluminum production are Soderberg and prebake, which encompass a number of processes and job categories. Substantial exposures to airborne polynuclear aromatic compounds have been measured in certain occupational settings in this industry. Exposures have been higher in potrooms of plants using the Soderberg process than in those using the prebake process; some workers may have exposed to both process. Exposures to fluorides and a variety of other contaminants also occur in potrooms. /Aluminum fluorides/ IARC. Monographs on the Evaluation of the Carcinogenic Risk of Chemicals to Humans. Geneva: World Health Organization, International Agency for Research on Cancer, 1972-PRESENT. (Multivolume work). Available at: p. 34(57)1984 Hazardous Substances Data Bank (HSDB) 12 Associated Disorders and Diseases Associated Occupational Diseases with Exposure to the Compound Asthma, occupational [Category: Airway Disease] Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 13 Literature 13.1 Consolidated References PubChem 13.2 NLM Curated PubMed Citations Medical Subject Headings (MeSH) 13.3 Springer Nature References Springer Nature 13.4 Thieme References Thieme Chemistry 13.5 Chemical Co-Occurrences in Literature PubChem 13.6 Chemical-Gene Co-Occurrences in Literature PubChem 13.7 Chemical-Disease Co-Occurrences in Literature PubChem 13.8 Chemical-Organism Co-Occurrences in Literature PubChem 14 Patents 14.1 Depositor-Supplied Patent Identifiers PubChem Link to all deposited patent identifiers PubChem 14.2 WIPO PATENTSCOPE Patents are available for this chemical structure: PATENTSCOPE (WIPO) 14.3 Chemical Co-Occurrences in Patents PubChem 14.4 Chemical-Disease Co-Occurrences in Patents PubChem 14.5 Chemical-Gene Co-Occurrences in Patents PubChem 14.6 Chemical-Organism Co-Occurrences in Patents PubChem 15 Interactions and Pathways 15.1 Protein Bound 3D Structures RCSB Protein Data Bank (RCSB PDB) View 98 proteins in NCBI Structure PubChem 15.1.1 Ligands from Protein Bound 3D Structures PDBe Ligand Code AF3 PDBe Structure Code 5UKD PDBe Conformer Protein Data Bank in Europe (PDBe) 15.2 Chemical-Target Interactions Comparative Toxicogenomics Database (CTD) 16 Biological Test Results 16.1 BioAssay Results PubChem 17 Classification 17.1 MeSH Tree Medical Subject Headings (MeSH) 17.2 ChEBI Ontology ChEBI 17.3 ChemIDplus ChemIDplus 17.4 CAMEO Chemicals CAMEO Chemicals 17.5 UN GHS Classification GHS Classification (UNECE) 17.6 EPA CPDat Classification EPA Chemical and Products Database (CPDat) 17.7 NORMAN Suspect List Exchange Classification NORMAN Suspect List Exchange 17.8 EPA DSSTox Classification EPA DSSTox 17.9 EPA TSCA and CDR Classification EPA Chemicals under the TSCA 17.10 FDA Drug Type and Pharmacologic Classification National Drug Code (NDC) Directory 17.11 PFAS and Fluorinated Organic Compounds in PubChem PubChem 17.12 EPA Substance Registry Services Tree EPA Substance Registry Services 17.13 MolGenie Organic Chemistry Ontology MolGenie 17.14 Chemicals in PubChem from Regulatory Sources PubChem 18 Information Sources Filter by Source Australian Industrial Chemicals Introduction Scheme (AICIS)LICENSE Aluminium fluoride (AlF3) Aluminium fluoride (AlF3) CAMEO ChemicalsLICENSE CAMEO Chemicals and all other CAMEO products are available at no charge to those organizations and individuals (recipients) responsible for the safe handling of chemicals. 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1960	https://rhetoric.byu.edu/Primary%20Texts/Quintilian.htm	\| \| \| Quintilian Institutio Oratoria (95 C.E.) \| \| \| \| \| --- \| \| Rhetoric Timeline \| Primary Source Synopses \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| Quintilian was the celebrated orator and rhetorician from the first century who brought forward rhetorical theory from ancient Greece and from the heyday of Roman rhetoric in the prior century. This theory he compiled in his Institutio Oratoria, an exhaustive and pedagogically oriented treatement of rhetoric in twelve books. Many later rhetoricians, especially from the Renaissance, derived their rhetorical theory directly from this text. Compact and detailed outlines follow below. The complete text online is now available (John Selby Watson translation, 1856), courtesy of Lee Honeycutt of Iowa State University. Compact Outline: \| \| \| --- \| \| Book I \| Elementary Education (Prior to Rhetoric) \| \| Book II \| The Nature and Rudiments of Rhetoric \| \| Book III \| Invention: Kinds of Oratory \| \| Book IV \| Arrangement (The Parts of a Speech) \| \| Book V \| Arrangement and Proofs \| \| Book VI \| Arrangement, Pathos, Judgment \| \| Book VII \| Arrangement, Laws, Reasoning \| \| Book VIII \| Eloquence: Style, Words, Tropes \| \| Book IX \| Eloquence: Figures of Thought and Speech \| \| Book X \| Training Eloquence: Reading, Composition, Speaking \| \| Book XI \| Kairos, Memory, and Delivery \| \| Book XII \| The Character of an Orator: Duties, Studies, etc. \| Detailed Outline: \| \| \| Book One \| \| 1.1 \| Elementary Education \| \| 1.2 \| Public vs. Private Education \| \| 1.3 \| Capacity and Treatment of Students \| \| 1.4 \| Grammar \| \| 1.5 \| Purity of Language and Vices of Diction \| \| 1.6 \| Origin and Usage of Words \| \| 1.7 \| Spelling \| \| 1.8 \| Reading, Authors to be Read \| \| 1.9 \| Composition \| \| 1.10 \| Studies Adjunct to Rhetoric \| \| 1.11 \| Lessons from the Theatre: Delivery, Gesture, Recitation, Gymnastics \| \| 1.12 \| Studying Multiple Subjects Simultaneously \| \| \| \| \| \| Book Two \| \| 2.1 \| Rhetoric vis-a-vis Grammar \| \| 2.2 \| Choice of Teacher; The Teacher - Pupil Relationship \| \| 2.3 \| Inferior Teachers \| \| 2.4 \| Progymnasmata (Elementary Rhetorical Exercises) \| \| 2.5 \| Rhetorical Analysis of Literary Authors \| \| 2.6 \| Declamation - Amount of Guidance for; Imitation \| \| 2.7 \| Declamation and Memorization \| \| 2.8 \| Customizing Methods to Pupil's Abilities \| \| 2.9 \| Students to Treat Teachers as Parents \| \| 2.10 \| Themes for Declamation to have Verisimilitude \| \| 2.11 \| Necessity of Rhetorical Instruction \| \| 2.12 \| Untrained Speakers - Pro and Con \| \| 2.13 \| Need for Adapting of Rules \| \| 2.14 \| "Rhetoric" and "Oratory" Considered \| \| 2.15 \| Definitions of Oratory \| \| 2.16 \| The Value of Oratory Attacked and Defended \| \| 2.17 \| The Art, Morality, and Truth of Oratory \| \| 2.18 \| Rhetoric as a Practical Art \| \| 2.19 \| Nature and Art \| \| 2.20 \| Is Rhetoric a Virtue? \| \| 2.21 \| The Subject of Rhetoric. Oratory vs. Philosophy; An Orator's Broad Knowledge. \| \| \| \| \| \| Book Three \| \| 3.1 \| Greek and Roman Writers on Rhetoric \| \| 3.2 \| Origin of Oratory \| \| 3.3 \| Divisions and Order of the Art of Rhetoric \| \| 3.4 \| Views on the Number of Kinds of Oratory \| \| 3.5 \| Things vs. Words; Questions; Definitions of a Cause \| \| 3.6 \| Stasis Theory (The Status of a Case) \| \| 3.7 \| Panegyric \| \| 3.8 \| Deliberative Oratory \| \| 3.9 \| Forensic Oratory; Parts of a Forensic Speech \| \| 3.10 \| The Nature of the Cause \| \| 3.11 \| The Question, Mode of Defence, Point for Decision, Foundation of the Case, etc. \| \| \| \| \| \| Book Four \| \| 4.1 \| Parts of a Speech: The Introduction (Exordium) \| \| 4.2 \| Parts of a Speech: The Statement of Facts (Narratio) \| \| 4.3 \| Digressions \| \| 4.4 \| Propositions Preparatory to Proof \| \| 4.5 \| Parts of a Speech: Partition (Partitio) \| \| \| \| \| \| Book Five \| \| 5.1 \| Parts of a Speech: Artificial and Unartificial Proofs \| \| 5.2 \| Previous Decisions \| \| 5.3 \| Public Opinion \| \| 5.4 \| Evidence from Torture \| \| 5.5 \| Refutation of Documents \| \| 5.6 \| Taking Oaths: Pro and Con \| \| 5.7 \| Evidence: Documentary, Oral, Witnesses, Supernatural \| \| 5.8 \| Aritifical Proofs \| \| 5.9 \| Signs, Circumstantial Evidence, Prognostics \| \| 5.10 \| Arguments \| \| 5.11 \| Examples and Instances \| \| 5.12 \| Arguments, cont'd \| \| 5.13 \| Parts of a Speech: Refutation and Proof \| \| 5.14 \| Enthymeme, Epicheireme, and Syllogism \| \| \| \| \| \| Book Six \| \| 6.1 \| Parts of a Speech: Conclusion (Peroration) \| \| 6.2 \| The Judge's Temperament; Pathos, Ethos \| \| 6.3 \| Wit and Humor \| \| 6.4 \| Altercatio or Debate \| \| 6.5 \| Judgment and Sagacity \| \| \| \| \| \| Book Seven \| \| 7.1 \| Arrangement \| \| 7.2 \| Conjecture \| \| 7.3 \| Definition \| \| 7.4 \| Quality \| \| 7.5 \| Points of Law \| \| 7.6 \| Letter of the Law / Intention \| \| 7.7 \| Contradictory Laws \| \| 7.8 \| Syllogism \| \| 7.9 \| Ambiguity \| \| 7.10 \| Status of a Case; Limits of Rules \| \| \| \| \| \| Book Eight \| \| 8.1 \| Style \| \| 8.2 \| Clarity (perspicuitas) \| \| 8.3 \| Ornament: Merits and Faults \| \| 8.4 \| Amplification and Diminution \| \| 8.5 \| The Value of General Reflections in Oratory \| \| 8.6 \| Tropes \| \| \| \| \| \| Book Nine \| \| 9.1 \| Figures of Thought and Speech \| \| 9.2 \| Figures of Thought (Detail) \| \| 9.3 \| Figures of Speech (Detail) \| \| 9.4 \| Apt Use of Structure, Rhythm, Metrical Feet \| \| \| \| \| \| Book Ten \| \| 10.1 \| Reading Curriculum \| \| 10.2 \| Imitation \| \| 10.3 \| Writing \| \| 10.4 \| Correction \| \| 10.5 \| Composition Exercises: Translation, Paraphrase, Theses, Commonplaces, Declamations \| \| 10.6 \| Thought and Premeditation \| \| 10.7 \| Extemporaneous Speaking \| \| \| \| \| \| Book Eleven \| \| 11.1 \| Respecting kairos When Speaking \| \| 11.2 \| Memory \| \| 11.3 \| Delivery, Gesture, Dress \| \| \| \| \| \| Book Twelve \| \| 12.1 \| The Great Orator as Good Man \| \| 12.2 \| Strengthening Character; Philosophical Study \| \| 12.3 \| The Study of Civil Law \| \| 12.4 \| Orator Prepared with Examples and Precedents \| \| 12.5 \| Firmness; Presence of Mind; Cultivating Natural Advantages \| \| 12.6 \| The Orator's Age \| \| 12.7 \| Causes to be Undertaken; Remuneration \| \| 12.8 \| Careful Study of the Case \| \| 12.9 \| Not Aiming for Applause; Restraining Invective; Preparation Through Writing and Extemporaneous Speaking \| \| 12.10 \| Styles of Oratory \| \| 12.11 \| Retirement from Speaking; Successful Training of an Orator \| \| \| \| \| This work is licensed under a Creative Commons Attribution 3.0 License. Gideon O. Burton, Brigham Young University Please cite "Silva Rhetoricae" (rhetoric.byu.edu) Trees \|SILVA RHETORICAE \| Flowers \|
1961	https://lweb.cfa.harvard.edu/~dfabricant/huchra/ay145/constants.html	Physical & Astronomical Constants Here is a list of useful constants: 1 inch = 2.54 cm 1 mile = 1.6 km 1 km = 1000 m 1 m = 100 cm = 1000 mm = 109 nm Planck Legnth = 1.616252x10-35 m = sqrt (hbar G/c3) Classical Electron Radius = 2.81794x10-15 m = e2/ me c2 Proton Radius = 0.83x10-15 m = 0.83 femtometer = 0.83x10-13 cm 1 Angstrom = 10-10 m = 10-8 cm 1 Earth Radius = 6.37814x108 cm = 6.37814x106 m (Equatorial) 1 Jupiter Radius = 7.1492x109 cm = 7.1492x107 m (Equatorial) 1 Solar Radius = 6.959910 cm = 6.9599x108 m (Equatorial) 1 Moon's Mean Orbital Radius = 384,400 km = 3.8440x108 m 1 AU = 1.4960x1011 m = 1.4960x108 km (Astronomical Unit) 1 ly = 9.4605x1015 m = 9.4605x1012 km (light year) 1 pc = 3.0857x1016 m = 3.0857x1013 km = 3.261633 ly = 206264.806 AU (parsec) = 3.0857x1018 cm 1 Mpc = 106 pc (Megaparsec) = 3.0857x1024 cm DH = c/H0 = 3000 h-1 Mpc = 9.26x1027 h-1 cm (The Hubble Scale) Angles: 1° = 1 degree = 60' = 60 arcminutes 1' = 60" = 60 arcseconds 1 radian = 360° /2 pi = 57.2957795131° = 206264.806" Area of a Sphere = 41252.96124 square degrees = 4 pi Steradians Mass: 1 amu = 1.6605402 x10-24 gram = 1.6605402 x10-27 kg 1 amu c2 = 931.49432 MeV 1 Hydrogen Atom Mass = 1.007825 amu = 1.673534 x10-24 gram 1 Helium 4 Atom Mass = 4.00260325415 amu 1 Carbon 12 Atom Mass = 12.0000000 amu 1 Proton Mass = 1.6726231 x10-24 gram 1 Neutron Mass = 1.674920 x10-24 gram 1 Electron Mass = 9.1093897 x10-28 gram 1 kg = 2.20462 lb 1 Solar Mass = 1.989x1033 gram = 1.989x1030 kg 1 Jupiter Mass = 1.899x1030 gram = 1.899x1027 kg 1 Earth Mass = 5.9736x1027 gram = 5.9736x1024 kg 1 Lunar Mass = 7.3477x1025 gram = 7.3477x1022 kg (Proton Mass)/(Electron Mass) = 1836.15 Earth's Mean Density = 5515.3 kg/m3 Moon's Mean Density = 3346.4 kg/m3 Time: 1 Planck Time = 5.39124x10-44 s = sqrt (hbar G/c5) 1 Sidereal Day = 23h 56m 04.09054s 1 Solar Day = 24h = 86400 s 1 Sidereal Year = 3.155815x107 s 1 Tropical Year = 3.155693x107 s Energy: 1 Joule = 2.39x10-1 calorie 1 Joule = 107 ergs 1 eV = 1.602177x10-12 erg = 1.602177x10-19 Joule 1 Solar Luminosity = 3.826x1033 ergs/s = 3.826x1026 Joules/s = 3.826x1026 Watts Sun's Absolute Magnitude V = 4.83, B = 5.48, K = 3.28 Vega's Absolute Magnitude V = 0.58, B = 0.58, K = 0.58 Temperature: Solar Surface Effective Temperature = 5770 K Physical Constants: Avogadro's Number = NA = 6.0221367x1023 /mole Gravitational Constant = G = 6.67259x10-8 cm3 gram-1 s-2 = 4.301x10-9 km2 Mpc MSun-1 s-2 Planck's Constant = h = 6.6260755x10-27 erg s Speed of Light = c = 2.99792458x1010 cm s-1 = 2.99792458x108 m s-1 Boltzmann's Constant = k = 1.380658x10-16 erg K-1 Stefan-Boltzmann Constant = &sigma = 5.67051x10-5 erg cm-2 K-4 s-1 Radiation Density Constant = a = 7.56591x10-15 erg cm-3 K-4 Rydberg = RH = 1.09677585x10 5 cm-1 Electron charge = e = 4.8032x10-10 esu = 1.6022x10-19 Coulomb 1 Coulomb = 6.24151x1018 e
1962	https://www.cliffsnotes.com/study-notes/16610472	Formal Lab Report •SCH3UP-22 (docx) - CliffsNotes Lit NotesStudy GuidesDocumentsQ&AAsk AI Chat PDF Log InSign Up Literature NotesStudy GuidesDocumentsHomework QuestionsChat PDFLog InSign Up Formal Lab Report •SCH3UP-22 .docx School Northview Heights Secondary SchoolWe aren't endorsed by this school Course CHEMISTRY 3U Subject Chemistry Date Jul 31, 2024 Pages 19 Uploaded by JudgeIbis4460 Download Helpful Unhelpful Download Helpful Unhelpful Home/ Chemistry SCH3UP-22 Sodium bicarbonate, NaHCO 3, reacted with acetic acid, CH 3 COOH, to produce a percent yield of carbon dioxide gas, CO 2 , closest to 100%, collected over water. Ava Bird Marcellus Chen Patrick Duong Sophie Low Ryleigh Sabaratnam Alyssa Strelet 11 U Chemistry Mrs. Hutchison Wednesday June 12th, 2024 1 Abstract This experiment was aimed to achieve a percent yield of carbon dioxide gas, CO 2 , closest to 100% when 5 mL of 5% (w/v) acetic acid, CH 3 COOH, and 0.25 g of sodium bicarbonate, NaHCO 3 , are reacted. The gas produced in the reaction is collected over water using a buchner flask, a hose, a gas collection tray, and a graduated cylinder. This would, suggestively, operate as an efficient form of gas collection. The percent yield of carbon dioxide is expected to be close to 100% since the lab has few sources of error. Potential sources of error include impurities and experimental errors. For instance, sodium bicarbonate may absorb water or gases from the atmosphere, mixing with carbon dioxide gas in the buchner flask. As well as not all of the carbon dioxide gas produced from the reaction being collected in the graduated cylinder. Four trials were performed under the same conditions resulting in an average actual yield of carbon dioxide gas of 59 mL, or 0.002267 mol. Since the theoretical yield of carbon dioxide gas is 0.002976 mol, the percent yield of carbon dioxide is 76.18%. Introduction This lab aims to achieve a yield of carbon dioxide gas, CO 2 , closest to 100% when 5 mL of 5% (w/v) acetic acid, CH 3 COOH, and 0.25g of sodium bicarbonate, NaHCO 3 , are reacted. A percent yield close to 100% of a product indicates an efficient reaction. An efficient reaction will cut down the costs of reactions when the product is required. There will be lower product loss resulting in fewer reactions having to be done, thus saving materials. A percent yield of a gas close to 100% is useful when collecting a harmful gas. In these situations, the greatest amount of gas must be collected so that harmful gases are not released into the atmosphere to harm the environment or people. An example of a gas that would be collected to avoid harming others is carbon monoxide gas, CO, since it is deadly if a high concentration is present in the atmosphere. Various sources produce carbon monoxide gas, CO, since it is the result of incomplete combustion. Collecting close to 100% of the carbon monoxide gas in an incomplete combustion reaction would reduce the concentration of carbon monoxide gas, CO, in the atmosphere, protecting many people and maintaining the quality and health of the environment. 2 Hypothesis If 5 mL of 5% (w/v) acetic acid is mixed with 0.25g of sodium bicarbonate, then the percent yield of carbon dioxide gas produced will be less than 100% because of impurities. This results in a lower genuine mass of reactant compounds, reducing their ability to fully react to a complete yield. Gas will escape from the graduated cylinder if it is not completely flat or a hole is present. CO 2 could also be lost in the air because it is gaseous, resulting in an inaccurate percent yield. Lastly, not all of the CO 2 gas may make it into the graduated cylinder because it may still be present in the Buchner flask and possibly the hose following the reaction. Sodium bicarbonate, NaHCO 3 , has hygroscopic properties which means it absorbs moisture from the air. The moisture absorbed, primarily water vapour, alters the mass of sodium bicarbonate. This means the mass of sodium bicarbonate weighed will be greater than the mass of pure sodium bicarbonate. This changes the yield since less sodium bicarbonate, NaHCO 3 , is reacted than accounted for. Physical and Chemical Properties of carbon dioxide Gas, CO 2 The physical properties of carbon dioxide gas, CO 2 , are colourless, odourless, and soluble in water, ethanol, and acetone. The melting point is -56.5°C and the boiling point is -78.5°C (National Library of Medicine, 2024) . This is directly related to carbon dioxide gas, CO 2 , being a covalent molecule with non-polar covalent bonds. Carbon dioxide is also known as an acidic oxide that reacts with water to create carbonic acid. Lastly, carbon dioxide forms with alkali metals to create carbonates and bicarbonates (Vedantu, 2024) as seen in Table 1. Table 1 Reactions between carbon dioxide and Basic Substances carbon dioxide reacts with water to create carbonic acid carbon dioxide reacts with alkalis to give carbonates and bicarbonate CO 2+ H 2 O → H 2 CO 3 CO 2+ NaOH → NaHCO 3 Electronegativity of Covalent Bonds Between carbon and oxygen Atoms Electronegativity is the tendency of an atom to attract a shared pair of electrons toward itself. The higher the electronegativity of a molecule, the more likely it is to attract a pair of electrons. Carbon dioxide has an electronegativity difference of 0.8, thus further proving that it is a non- polar covalent bond. 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The model is based on the idea that the reactant passes through a transition state within the enzymesubstrate The International school of Choueifat BIO 101 Gizmo+Balancing+Chem+Equations.pdf Name: _ Date: _ Student Exploration: Balancing Chemical Equations Vocabulary: coefficient, combustion, compound, decomposition, double replacement, element, molecule, product, reactant, single replacement, subscript, synthesis Prior Knowledge Questions (D Lake Zurich High School CHEM 433 2024 HKDSE CHEM.pdf 2024-DSE CHEM PAPER 1B HONG KONG EXAMINATIONS AND ASSESSMENT AUTHORITY HONG KONG DIPLOMA OF SECONDARY EDUCATION EXAMINATION 2024 CHEMISTRY SECTION Candidate Number PAPER1 B: Question-Answer Book B This paper must be answered In English INSTRUCTIONS FOR SE The University of Hong Kong MATH 2011 AP Chem Unit 1 FRQ.pdf UNIT 1 FREE-RESPONSE QUESTION Atomic Structure and Properties Directions: This question is a long free-response question that requires about 23 minutes to answer and is worth 10 points. 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Mother baby , d : cost i . hand , not ethical e . - w/ arguing parents over + regulated def in Ch same beliefs insurance ensure Breath rea Referra Carteret Community College NUR 111 PS6 chem 436.pdf Chemistry 436, Spring 2017 Problem Set 6 Due: Friday April 14th 2017 Name: UIN: To receive credit, answers must be clearly written in the appropriate spaces (you must turn in all answers on this sheet), also each intermediate of every mechanism and synthe University of Illinois, Urbana Champaign CHEM 436 MDMA 2.docx write a short proposal in 3 paragraph on the following job description is given : Freelancer will be asked to use a text, reference book, and/or research paper (will be provided by the client, no external sources can be used) to write 10 well-articulated National University of Sciences & Technology, Islamabad CHE 325 Lecture Notes Ch 7 Blank.pdf CHEM 1465 59 Chapter 7. Introduction to Chemical Bonding 7.1 Introduction to Ionic and Covalent Bonding 1. Three types of chemical bonding: a) Ionic bonding involves attractions between charged particles with a complete b) of . Covalent bonding is typical University of Texas, Arlington CHEM 1465 G. C. I-5 Limiting Reagent Study 37.pdf )é/ 1 5 \J Z/ G.C. \|-5 Limiting Reagent Study i /)& G.C. \|-5: LABORATORY Name MS(Z(\% REPORT Parks Lab partners' names: A\|€X(s TA's Name: Poyscoy S . 37 Date {O/S/72™> Brancstiotiey YOU MUST SHOW ALL YOUR CALCULATIONS TO RECEIVE CREDIT FOR THEM! Proper Florida Atlantic University CHM 2045L CliffsNotes study guides are written by real teachers and professors, so no matter what you're studying, CliffsNotes can ease your homework headaches and help you score high on exams. 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1963	https://askfilo.com/user-question-answers-smart-solutions/theater-seating-arithmetic-sequence-there-are-28-seats-in-3335343635323135	Question asked by Filo student Theater Seating Arithmetic Sequence There are 28 seats in the front row of a theater. Each successive row contains two more seats than the previous row. If there are 24 rows, how many seats are in the last row of the theater? Views: 5,247 students Updated on: Jun 16, 2025 Text SolutionText solutionverified iconVerified Concepts arithmetic sequences, nth term of sequence Explanation We're asked for the number of seats in the last (24th) row. The number of seats increases by 2 with each row. This forms an arithmetic sequence: The nth term of an arithmetic sequence is: an=a1+(n−1)d Step-By-Step Solution Step 1 Identify the variables: Step 2 Plug values into the nth term formula: an=a1+(n−1)d a24=28+(24−1)×2 Step 3 Calculate 24−1: 24−1=23 Step 4 Multiply the difference by the increment: 23×2=46 Step 5 Add result to first term: 28+46=74 Final Answer There are 74 seats in the last row of the theater. Students who ask this question also asked Views: 5,744 Topic: Smart Solutions View solution Views: 5,528 Topic: Smart Solutions View solution Views: 5,705 Topic: Smart Solutions View solution Views: 5,554 Topic: Smart Solutions View solution Stuck on the question or explanation? Connect with our tutors online and get step by step solution of this question. \| \| \| --- \| \| Question Text \| Theater Seating Arithmetic Sequence There are 28 seats in the front row of a theater. Each successive row contains two more seats than the previous row. If there are 24 rows, how many seats are in the last row of the theater? \| \| Updated On \| Jun 16, 2025 \| \| Topic \| All topics \| \| Subject \| Smart Solutions \| \| Class \| Class 10 \| \| Answer Type \| Text solution:1 \| Are you ready to take control of your learning? Download Filo and start learning with your favorite tutors right away! Questions from top courses Explore Tutors by Cities Blog Knowledge © Copyright Filo EdTech INC. 2025
1964	https://education.nationalgeographic.org/resource/condensation/	ENCYCLOPEDIC ENTRY ENCYCLOPEDIC ENTRY Condensation Condensation is the process where water vapor becomes liquid Grades 5 - 8 Subjects Earth Science, Meteorology, Physics ‌ ‌‌‌ ‌‌‌‌‌‌‌‌‌‌ Loading ... Powered by Condensation is the process where water vapor becomes liquid. It is the reverse action of evaporation, where liquid water becomes a vapor.Condensation happens one of two ways: Either the air is cooled to its dew point or it becomes so saturated with water vapor that it cannot hold any more water.Dew PointDew point is the temperature at which condensation happens. (Dew is simply condensed water in the atmosphere.) Air temperatures can reach or fall below the dew point naturally, as they often do at night. That's why the ground, structures, and objects left outside are often coated with water droplets in the morning.Condensation can also produce water droplets on the outside of soda cans or glasses of cold water. When warm air hits the cold surface, it reaches its dew point and condenses. This leaves droplets of water on the glass or can.When a pocket of air becomes full of water vapor, clouds form. The point at which condensation starts can be easily viewed in cumulus clouds, which have flat bottoms. Those flat bottoms are where vapor begins to condense into water droplets.SaturationClouds are simply masses of water droplets in the atmosphere. Molecules in water vapor are far apart from one another. As more water vapor collects in clouds, they can become saturated with water vapor. Saturated clouds cannot hold any more water vapor. When clouds are saturated with water vapor, the density, or closeness, of the molecules increases. The vapor condenses and becomes rain.Cold air holds less water vapor than warm air. This is why warm climates are often more humid than cold ones: Water vapor remains in the air instead of condensing into rain. Cold climates are more likely to have rain, because water vapor condenses more easily there. Fast Fact Making RainCloud condensation nuclei (CCN) are microscopic bits of clay, salt, or solid pollutants such as ash from smoke. Water in clouds condenses around these condensation nuclei to form raindrops. Reference National Weather Service: Clouds and ContrailsNational Geographic Science: Clouds, explainedNOAA: Cloud Classification Credits Media Credits The audio, illustrations, photos, and videos are credited beneath the media asset, except for promotional images, which generally link to another page that contains the media credit. The Rights Holder for media is the person or group credited. Writers Kim Rutledge Melissa McDaniel Santani Teng Hilary Hall Tara Ramroop Erin Sprout Jeff Hunt Diane Boudreau Hilary Costa Illustrators Mary Crooks, National Geographic Society Tim Gunther Editors Jeannie Evers, Emdash Editing, Emdash Editing Kara West Educator Reviewer Nancy Wynne Producer National Geographic Society other Last Updated October 19, 2023 User Permissions For information on user permissions, please read our Terms of Service. If you have questions about how to cite anything on our website in your project or classroom presentation, please contact your teacher. They will best know the preferred format. When you reach out to them, you will need the page title, URL, and the date you accessed the resource. Media If a media asset is downloadable, a download button appears in the corner of the media viewer. If no button appears, you cannot download or save the media. Text Text on this page is printable and can be used according to our Terms of Service. Interactives Any interactives on this page can only be played while you are visiting our website. You cannot download interactives. Related Resources
1965	https://physics.stackexchange.com/questions/757508/writing-the-fourth-kinematic-formula-v2-v-02-2a-delta-x-in-a-differen	Writing the fourth kinematic formula ($v^2 = v_0^2 + 2a \Delta x$) in a different form - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Writing the fourth kinematic formula (v 2=v 2 0+2 a Δ x v 2=v 0 2+2 a Δ x) in a different form Ask Question Asked 2 years, 6 months ago Modified2 years, 6 months ago Viewed 1k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I have noticed that if we play with the formula a bit, we get v 2(v+v 0)2⋅(v−v 0)v¯¯¯⋅Δ v=v 2 0+2 a Δ x=a Δ x=a Δ x v 2=v 0 2+2 a Δ x(v+v 0)2⋅(v−v 0)=a Δ x v¯⋅Δ v=a Δ x Meaning the average velocity v¯¯¯v¯ times the change in velocity Δ v Δ v is equal to the (constant) acceleration times the change in place Is this correct? If so, is there a "physical meaning" behind this form of the formula? Meaning, why does multiplying the average velocity with the difference yields acceleration times displacement? kinematics Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Mar 30, 2023 at 19:30 Kyle Kanos 29.2k 41 41 gold badges 70 70 silver badges 138 138 bronze badges asked Mar 30, 2023 at 18:28 Ak2399Ak2399 133 4 4 bronze badges 3 "why does multiplying the average velocity with the difference yields acceleration times displacement?" You have shown why in your post BioPhysicist –BioPhysicist 2023-03-30 20:15:19 +00:00 Commented Mar 30, 2023 at 20:15 Why was this closed as "homework-like"? It gives an interesting result and asks whether it has a nice physical interpretation.Michael Seifert –Michael Seifert 2023-03-30 21:03:09 +00:00 Commented Mar 30, 2023 at 21:03 @MichaelSeifert FWIW, my vote was not for the homework reason. Although I can see the take of "check my work" given the explicit "is this correct?" (Many people often see "homework" and overlook the latter part of the closure reason)BioPhysicist –BioPhysicist 2023-03-30 22:34:38 +00:00 Commented Mar 30, 2023 at 22:34 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. The formula is valid and has a geometric and a physical interpretation. The hand side corresponds to the kinetic energy of a unit mass, and the right-hand side the work done by a force on the same unit mass. Multiply both sides with m m the mass and get m v¯¯¯Δ v=m a Δ x(1)(1)m v¯Δ v=m a Δ x Use F=m a F=m a and the definition of work W=F Δ x W=F Δ x to get W=m v¯¯¯Δ v(2)(2)W=m v¯Δ v but you can show that the RHS is the change in kinetic energy as W=m(v+v 0)2⋅(v−v 0)=1 2 m v 2−1 2 m v 2 0=Δ E(3)(3)W=m(v+v 0)2⋅(v−v 0)=1 2 m v 2−1 2 m v 0 2=Δ E which goes down to the fundamental principle of work = change in energy. But there is a geometric interpretation also. Take the velocity plot vs. time, which under constant acceleration, it is a straight line between v 0 v 0 and v v The area under the curve is the distance traveled. From the triangle area formula that is Δ x=1 2 Δ t(v−v 0)(4)(4)Δ x=1 2 Δ t(v−v 0) but the distance traveled can also be found by the average speed v¯¯¯=v 0+v 2 v¯=v 0+v 2 which you can see above as the area under the rectangle defined by the dashed line. Δ x=v¯¯¯Δ t(5)(5)Δ x=v¯Δ t Combine this with the kinematic equation Δ v=a Δ t Δ v=a Δ t which is solve for time Δ t=Δ v a Δ t=Δ v a to get Δ x=v¯¯¯Δ v a(6)(6)Δ x=v¯Δ v a or a Δ x=v¯¯¯Δ v(7)(7)a Δ x=v¯Δ v Note that if/when you learn calculus, you will deal with infinitesimal changes and the equation a d x=v d v a d x=v d v is solve by integration (summation of all the small parts) as ∫a d x=∫v d v=1 2 v 2−1 2 v 2 0(8)(8)∫a d x=∫v d v=1 2 v 2−1 2 v 0 2 which is the solution for non-constant a a and matches with equation (3) above as it is interpreted as work = change in kinetic energy. An example is when you have a spring with F=−k x F=−k x and thus a=−1 m k x a=−1 m k x, you use it in (8) to get −∫1 m k x x=−1 2 k m x 2=1 2 v 2−1 2 v 2 0−∫1 m k x x=−1 2 k m x 2=1 2 v 2−1 2 v 0 2 which is equivalent to the energy balance between the potential energy of the spring and the kinetic energy of the mass at initial conditions (rhs) and at current conditions (lhs). 1 2 k x 2+1 2 m v 2=1 2 m v 2 0 1 2 k x 2+1 2 m v 2=1 2 m v 0 2 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Mar 30, 2023 at 20:20 answered Mar 30, 2023 at 20:09 John AlexiouJohn Alexiou 40.5k 6 6 gold badges 74 74 silver badges 187 187 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. The average acceleration is defined as a¯=Δ v Δ t a¯=Δ v Δ t, and the average velocity is v¯=Δ x Δ t v¯=Δ x Δ t. For a constant acceleration a=a¯a=a¯, so the right hand side of your expression could be rewritten: a Δ x=Δ v Δ t Δ x=Δ v Δ x Δ t=Δ v v¯.a Δ x=Δ v Δ t Δ x=Δ v Δ x Δ t=Δ v v¯. This shows that your expression is correct, but I can't find any deep physical significance in it. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Mar 30, 2023 at 19:32 Paul T.Paul T. 7,544 23 23 silver badges 43 43 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Average velocity by basic definition is the average distance covered by an object divided by the average time required to do so. Hence, v avg = s/t Also, acceleration is the rate of change of velocity, or simply change in velocity divided by the time interval. Hence, the LHS is the displacement of the body times its acceleration Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Mar 30, 2023 at 19:27 Vedaansh AgarwalVedaansh Agarwal 101 2 2 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Physics Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. 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1966	https://math.stackexchange.com/questions/25791/definition-of-an-ordered-pair	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Definition of an Ordered Pair Ask Question Asked Modified 11 years, 9 months ago Viewed 7k times 16 $\begingroup$ "The ordered pair $(a,b)$ is defined to be the set ${{a},{a,b}}$." ~ Hungerford's Algebra (p.6) I think this is the first time that i've seen this definition. I've read the wiki page. Is it defined this way, as opposed to a definition relating to functions as in a Cartesian product, because this definition is considered more elementary (or foundational) being that it is related directly to sets? Also, the definition of an ordered $n$-tuple, according to the wiki page seems vague (perhaps i'm misunderstanding it). For an ordered triple it gives the example: $$(1,2,3) = {{(1,2)},{(1,2),3}}$$ but how do we know this is not the ordered pair $((1,2),3)$? Or is the difference between $(1,2,3)$ and $((1,2),3)$ considered trivial? Thirdly, and perhaps unrelated, what does it mean for a natural number to be defined $$2_{\mathbb{N}} = {\emptyset,{\emptyset} },$$ and is this also done so that we can define $\mathbb{N}$ in terms of sets? elementary-set-theory definition Share edited Apr 13, 2017 at 12:21 CommunityBot 1 asked Mar 8, 2011 at 17:27 dandielliedandiellie 30022 silver badges99 bronze badges $\endgroup$ 6 6 $\begingroup$ You want to define an arbitrary product of sets as some kind of function. You want to define a function as a special kind of relation. You want to define a relation as a subset of the Cartesian product of two sets. You want to define the Cartesian product of two sets as the collection of ordered pairs. So you need to define 'ordered pair' as some kind of set (the only thing there is to begin with). $\endgroup$ wildildildlife – wildildildlife 2011-03-08 17:36:07 +00:00 Commented Mar 8, 2011 at 17:36 $\begingroup$ @wildildildlife Thanks, i wasn't even thinking about how to define a function. It certainly makes no sense to define an ordered pair in terms of functions and turn around and define functions in terms of things which are defined in terms of ordered pairs. $\endgroup$ dandiellie – dandiellie 2011-03-08 19:09:09 +00:00 Commented Mar 8, 2011 at 19:09 $\begingroup$ A similar question arose on MathOverflow at mathoverflow.net/questions/32181/ambiguity-in-ordered-tuples $\endgroup$ JDH – JDH 2011-03-08 20:32:48 +00:00 Commented Mar 8, 2011 at 20:32 2 $\begingroup$ If you defined an ordered pair naively as ${a, b}$ you wouldn't be able to distinguish between $(a, b)$ and $(b, a)$ as different ordered pairs, since order isn't important in sets. The definition give above distinguishes $a$ as the first element of the ordered pair, and $b$ as the second element of the ordered pair, as the intersection and symmetric difference between the two sets, respectively. Is that what you're asking? $\endgroup$ Uticensis – Uticensis 2011-03-09 04:47:00 +00:00 Commented Mar 9, 2011 at 4:47 1 $\begingroup$ No, certainly not. I was actually thinking that we could define an ordered pair as a function. For example, $(a,b)$ would be defined as the function $f:{1,2}\rightarrow{a,b}$ such that $f(1)=a$ and $f(2)=b$. This is the way i was taught to define a Cartesian product, as the set of functions from an indexing set to the union of the sets being considered. $\endgroup$ dandiellie – dandiellie 2011-03-09 05:03:18 +00:00 Commented Mar 9, 2011 at 5:03 \| Show 1 more comment 6 Answers 6 Reset to default 11 $\begingroup$ You may be interested in reading Kuratowski's "Set Theory". Here's what I remember from it: First one defines pairs $\langle a,b\rangle = {{a},{a,b}}$, with this definition one can define $A\times B$ as the set of all pairs $(a,b)$ with $a\in A$ and $b\in B$. However, that's not a good way to proceed, because of the problems you note. With this definition one defines $\prod_{i\in I} X_i$ as the set of functions $f\colon I \to X_i$ such that $f(i)\in X_i$, here ${X_i \mid i\in I}$ is a collection of sets (in other words a function $I\to \mathcal P(\cup X_i)$). In particular $A^2$ is the set of functions $2\to A$, where $2 = {\varnothing, {\varnothing}}$, and $A\times B$ is the set of functions $f\colon 2 \to A\cup B$ where $f(\varnothing)\in A$ and $f({\varnothing})\in B$. Now we forget about that first definition, and proceed with the latter. (Even though we use the former definition to state the latter!) The practical advantage is that now in fact $A\times B\times C$ is actually well-defined, just like any other product, no matter how large the index-set $I$. It is still not true that $(A\times B)\times C = A\times (B\times C)$, and in fact both are still different from $A\times B\times C$. However, there are bijections between these three sets that are so obvious that for all practical purposes one may consider them to be equal. Share answered Mar 8, 2011 at 18:01 MyselfMyself 8,9973232 silver badges4747 bronze badges $\endgroup$ 4 $\begingroup$ So is this what you're saying? We define an ordered pair in terms of sets so that we have a solid, foundational definition and so that we avoid getting in trouble down the road, but it's not actually the working definition? The more useful definition is the one using functions. $\endgroup$ dandiellie – dandiellie 2011-03-09 05:31:01 +00:00 Commented Mar 9, 2011 at 5:31 $\begingroup$ @dandiellie Yes, that's about how it works if you want to do everything "by the book". For finite products you could in fact go without the definition as a function, but it's just not as smooth. When doing real mathematics both are similar enough so it wouldn't really matter much anyway. $\endgroup$ Myself – Myself 2011-03-09 05:39:05 +00:00 Commented Mar 9, 2011 at 5:39 $\begingroup$ Ok, thanks. I haven't gotten to real mathematics yet, so imma have to stick with "the book" for now. :p But i understand what you mean. $\endgroup$ dandiellie – dandiellie 2011-03-09 05:44:19 +00:00 Commented Mar 9, 2011 at 5:44 $\begingroup$ Dear@myself,similarly with fifth point,it's not true that $R[x_1...,x_n]=(R[x_1,...,x_{n-1}])[x_n]$ thay're isomorphism but many people think they're equal. $\endgroup$ R Salimi – R Salimi 2013-12-20 13:54:19 +00:00 Commented Dec 20, 2013 at 13:54 Add a comment \| 3 $\begingroup$ We want to define the most with as little as possible. That way we only define what sets are, and by that we define ordered pairs, and so on. The usual way is to define an ordered pair $\langle a,b\rangle = {{a},{a,b}}$. This is just because it's easy to work with. You can define an ordered pair as the image of a function from the domain which is the power set of the power set of the empty set, the first element is the image of $\emptyset$ and the second is the image of ${\emptyset}$. (Yes, functions are usually defined as collections of ordered pairs. I'm talking about existence of a formula with two free variables.) Again, these are just conventions and we work with that we find comfortable and as clear as possible. As for the second issue, we only define pairs, but there is a natural identification between $\langle a,\langle b,c\rangle\rangle$ and $\langle a,b,c\rangle$ and of course $\langle\langle a,b\rangle, c\rangle$. So once again we only define as little as possible and somewhat abuse our own notation because we know that the formal backbone exists and is strong. And lastly, as I said before, we want to define the most with as little as possible. In the world of set theory it's nice to have only sets. So we define $0=\emptyset$, and inductively we can define the natural numbers in terms of sets, so $n={0,\ldots,n-1}$. Share edited Mar 8, 2011 at 18:56 dandiellie 30022 silver badges99 bronze badges answered Mar 8, 2011 at 17:35 Asaf Karagila♦Asaf Karagila 407k4848 gold badges646646 silver badges1.1k1.1k bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ For your first question, you are correct. An ordered pair defined as "syntactic sugar" for the set construction is more fundamental. For your second question, when we think about a $k$-tuple, the fundamental property is that we have "projections" from the $k$-tuple to a particular element of the tuple. In other words, all that we essentially require is that we can somehow discern one element from another, and there is an ordering. With the definition given, $((1,2),3)$ is indeed equivalent to $(1,2,3)$. This is almost like Lisp where (1 2 3) is equivalent to (1 . (2 . (3 . ()))). The principle is the same; we are just making a sort of binary tree. As you might imagine, $(1,2,3,4)$ would be equivalent to $(((1,2),3),4)$ by the definition given. This is really an unfortunate problem with notation, and not so much the definition of tuples themselves. For your last question, basically we can give a formal construction to the natural numbers by using more primitive pieces of mathematics. You can think of it as just a particular encoding for natural numbers --- or even a canonical encoding if you want. Another encoding might be a natural number defined as a composition of "successor" functions, etc. But of course they're all isomorphic. Share answered Mar 8, 2011 at 17:41 QuadrescenceQuadrescence 22011 silver badge77 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ As regards your first question, yes, your intuition that this way of defining things is more set-theory-like and foundational is correct. Since in a set no ordering is defined (i.e., the set ${a,b}$ is the same as the set ${ b, a }$ ), if you want to express everything just in terms of sets you have to fix a convention to do so. So, in a sense, writing ${{a},{a,b}}$ is a bit like saying, "we are interested in the set ${a, b}$, but with $a$ in a distinguished role". Share answered Mar 8, 2011 at 17:42 DaGDaG 48522 silver badges1212 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ The need for an explicit definition of ordered pair comes from the simplicity of he concept of set: there is no order defined in the set axioms - an object is either a member of a set or not. But it is a convenient mechanism to be able to have the first of a series (and the second also). And the definition of 'ordered pair' allows that. Of course you have to show that, whatever your definition of 'ordered pair' is, you can show that if $(a, b) = (c, d)$ then $a = c$ and $b=d$, which you can do with the given definition (the notation ${b,a,d,c}$ is the same set as ${a,b,c,d}$, but $(b,a,d,c)$ is different from $(a,b,c,d)$ To get n-tuples, it should be straightforward that (though not equivalent in underlying set notation) one can say $(a,b,c)$ can be implemented using ordered pairs as $((a,b),c)$ or $(a, (b,c))$ where you can extract the first element easily, but also to get the second element of the triple you get the first element of the $(b,c)$ which is the second element of the pair $(a,(b,c))$. Share answered Mar 8, 2011 at 20:13 MitchMitch 8,83733 gold badges4040 silver badges7272 bronze badges $\endgroup$ Add a comment \| -1 $\begingroup$ (a,b,c) = {{a},{a,b},{a,b,c}}? [(a,b,c) = (d,e,f)] <==> [[a=d & b=e] & c=f] [ [a=d & b=e] & c=f ] ==> [ {{a},{a,b},{a,b,c}} = {{d},{d,e},{d,e,f}} ] Now suppose instead that [[a=d & b=f] & c=e]: {{a},{a,b},{a,b,c}} = {{d},{d,f},{d,e,f}} The set {{d},{d,f},{d,e,f}} is not the same as the set {{d},{d,e},{d,e,f}} because the set {d,f} (which is in {{d},{d,f},{d,e,f}}) is not the same as the set {d,e} (which is in {{d},{d,e},{d,e,f}}). For example: {a,b} = {d,e} if a=d & b=e OR if b=d and a=e. However, there is no order ({d,e} or {e,d}, which are the same) in which {d,e} = {d,f}. From this, {{d},{d,e},{d,e,f}} is not equivalent to {{d},{d,f},{d,e,f}}. If {{a},{a,b,},{a,b,c}} = {{d},{d,f},{d,e,f}}, then {{a},{a,b},{a,b,c}} is not equivalent to {{d},{d,e},{d,e,f}}. So: [ {{a},{a,b},{a,b,c}} = {{d},{d,e},{d,e,f}} ] <===> [[a=d & b=e] & c=f] Which means if... x = {a} y = {a,b} z = {a,b,c} then... (x,y,z) = {x,y,z} {x,y,z} = {{a},{a,b},{a,b,c}} {{a},{a,b},{a,b,c}} = {{d},{d,e},{d,e,f}} if and only if a=d, b=e and c=f. Or something like that, I think. I've only just started learning so forgive me, I'm still mainly thinking in terms of predicate logic. Also forgive the presentation. Share edited Dec 20, 2013 at 10:44 answered Dec 20, 2013 at 10:38 dionysusdionysus 111 bronze badge $\endgroup$ 3 $\begingroup$ Please, use Tex when posting on this site. $\endgroup$ user38355 – user38355 2013-12-20 11:01:28 +00:00 Commented Dec 20, 2013 at 11:01 $\begingroup$ I have no idea how to, though. As I indicated, I'm really only just beginning with set theory, from which it can be more or less easily deduced that I probably haven't been posting on this site for very long and probably don't know how to do so properly. This is also why I mentioned the poor presentation of my post at the end of it, so I'm not sure as to the intentions of your comment. $\endgroup$ dionysus – dionysus 2013-12-20 21:45:57 +00:00 Commented Dec 20, 2013 at 21:45 $\begingroup$ My comment was intended as a courteous reminder to please learn the standard way, Tex/MathJax, of posting math equations on this site. $\endgroup$ user38355 – user38355 2013-12-21 12:13:27 +00:00 Commented Dec 21, 2013 at 12:13 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions elementary-set-theory definition See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked How can an ordered pair be expressed as a set? 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1967	https://eprint.iacr.org/2020/1355.pdf	Modular Lagrange Interpolation of the Mod Function for Bootstrapping of Approximate HE Charanjit S. Jutla IBM T. J. Watson Research Center Nathan Manohar UCLA Abstract We introduce a novel variant of Lagrange interpolation called modular Lagrange inter-polation that allows us to obtain and prove error bounds for explicit low-degree polynomial approximations of a function on a union of equally-spaced small intervals even if the func-tion overall is not continuous. We apply our technique to the mod function and obtain explicit low-degree polynomial approximations with small error. In particular, for every k and N >> k, we construct low-degree polynomials that approximate f(x) = x mod N, for \|f(x)\| ≤1 and \|x\| ≤kN, to within O(1/N) additive approximation. For k = O(log N), the result is generalized to give O(d)-degree polynomial approximations to within O(N −d) error for any d ≥1. Literature in approximation theory allows for arbitrary precision poly-nomial approximation of only smooth functions, whereas the mod function is only piecewise linear. These polynomials can be used in bootstrapping for approximate homomorphic encryp-tion, which requires computing the mod function near multiples of the modulus. Our work bypasses the fundamental error of approximation in prior works caused by ﬁrst approxi-mating the mod function by a scaled sine function. We implement the bootstrapping of HEAAN using our polynomials and proﬁle various parameter settings. For example, we demonstrate bootstrapping that can achieve 67 bit message precision, larger than the pre-cision of a double variable, whereas the most advanced prior work was only capable of up to 40 bit message precision. 1 Introduction Polynomial interpolation is a fundamental problem in mathematics. It is well known that given n + 1 distinct points x0, x1, . . . , xn and their corresponding values y0, y1, . . . , yn, there is a unique polynomial p(x) of degree at most n such that p(xi) = yi for all i ∈{0, 1, . . . , n}. Using Lagrange interpolation, it is possible to easily determine this unique polynomial p(x). Polynomial interpolation’s utility goes beyond ﬁnding polynomials that pass through points in 1 a dataset. Suppose we are given some arbitrary smooth function f(x) on a contiguous interval. Then, using Lagrange interpolation, we can hope to ﬁnd a low-degree polynomial that is a good approximation of f on this interval. This immediately raises two questions: 1) How many points should we use for interpolation? and 2) How should these points be chosen? It is well-known that Lagrange interpolation of even a smooth function (on a contiguous interval) where the interpolation points are chosen as equally spaced points leads to bad ap-proximations (see e.g. ). This is known as the “Runge phenomenon”, and a more technical explanation can be found in Section 3. However, it is also known that such smooth functions can be asymptotically approximated with high precision if the density of the points in the contiguous interval, say [−1, 1], is proportional to (1 −x2)−1/2 (that is, the density of points is higher near the edges of the interval, see e.g. [11, 5]). A common choice of points are the Chebyshev nodes or roots of the nth Chebyshev polynomial Tn(x), which asymptotically sat-isfy this density function, which we will refer to as the Chebyshev density function. These lie in the interval [−1, 1] and are deﬁned by the formula xj = cos π2(j + 1) 2n for j = 0, 1, . . . , n −1. Figure 1 shows the function sin(πx) on the interval [−1, 1] with points chosen according to the above formula for n = 25 and 50. For an arbitrary interval [a, b], one can simply scale the distribution of points on [−1, 1] to this interval. -1.0 -0.5 0.5 1.0 -1.0 -0.5 0.5 1.0 (a) 25 Chebyshev points -1.0 -0.5 0.5 1.0 -1.0 -0.5 0.5 1.0 (b) 50 Chebyshev points Figure 1: Plots of sin(πx) on [−1, 1] with Chebyshev points. In fact, for a polynomial f(x) that is n + 1 times diﬀerentiable on an interval [a, b], the polynomial pn of degree ≤n obtained via Lagrange interpolation at n + 1 distinct points 2 t0, t1, . . . , tn ∈[a, b] will have error at any point x ∈[a, b] given by the formula f(x) −pn(x) = f(n+1)(ψx) (n + 1)! · n Y i=0 (x −ti), where ψx is some point in [a, b] depending on x. Since determining ψx is diﬃcult, error analysis typically proceeds by upper bounding f(n+1)(·) on [a, b] and then bounding \| Qn i=0(x −ti)\|. It is known that setting the ti’s as the Chebyshev nodes scaled to [a, b] minimizes this latter quantity. Thus, for a single contiguous interval, it is known how to obtain a good low-degree polynomial approximation and prove a good bound on the error. Algorithmic search methods that determine the minimax polynomial approximation of a continuous function on a closed interval are also well-studied, e.g. the Remez Algorithm . However, in this work we will focus on constructive methods as that leads to better understanding of the tradeoﬀs in target applications. This is discussed in more detail in Section 9. In this work, we ask a related question, motivated by its applications to cryptography. Suppose we have many far apart, equally spaced, (contiguous) intervals and a piecewise smooth function f on these intervals, such that f is not continuous on a single closed interval containing these set of intervals. Is it possible to obtain a low-degree polynomial that approximates f well, restricted to the set of small intervals? We show how to achieve the above and prove good error bounds for bounded piecewise linear (or low-degree) functions using a new technique that we call modular Lagrange inter-polation. If we simply tried to apply standard Lagrange interpolation, we would immediately run into problems trying to prove an error bound, since the Lagrange interpolation theorem requires the function to be n + 1 times diﬀerentiable, but the piece-wise smooth function is not continuous (or even necessarily deﬁned) on a closed interval that contains these intervals. A common approach in previous works has been to approximate the piecewise linear function with a continuous function such as the sine function. But, this limits the approximation pos-sible due to the error inherent in the continuous approximation. However, as discussed above, on a single interval, we know how to use Lagrange interpolation to ﬁnd a low-degree poly-nomial approximation. Modular Lagrange interpolation refers to the method of “combining” these low-degree polynomials that work well on one speciﬁc interval into a single low-degree polynomial that simultaneously approximates the function well on all intervals. Crucial to the modular Lagrange interpolation technique is that the construction allows one to prove good er-ror bounds. We remark that it is not equivalent to Lagrange interpolation on judiciously picked points in the intervals, e.g. those approximating the Chebyshev density function. However, our “combiners” will mimic the Chebyshev density function. The combiners behave similarly to Lagrange basis polynomials, which have the “delta function” property meaning that they evaluate to one at a speciﬁc point and to zero at all other points. However, the combiners must 3 still approximate this delta function by a low-degree polynomial over intervals that are small but not negligible. To keep the degree small, we leverage the trade-oﬀbetween the number of unit-sized intervals k and the size N separating these intervals, and a strategy similar to the Chebyshev density function works well to avoid the Runge phenomenon. While this strategy of making the “combiners” similar to Chebyshev-weighted Lagrange basis polynomials works well for simple approximations, to get approximations of O(N−c) (for c > 1) we must use a more advanced application of Lagrange interpolation, which approximates the function in each interval by at least (c + 1) points, and the combiners now depend on each such point rather than just the whole interval. A more technical introduction to modular Lagrange interpolation is given in Section 3. 1.1 Application to Cryptography The problem of ﬁnding a single low-degree polynomial approximation to a piece-wise function over many intervals is a natural math question and interesting in its own right. However, it turns out that this problem also has a major application to cryptography with regards to FHE bootstrapping, which we will now describe. The work of [9, 8] presented a new homomorphic encryption (HE) scheme for approximate arithmetic (called the CKKS HE scheme) over real/complex numbers. The CKKS HE scheme was considerably more eﬃcient than other schemes for evaluating arithmetic circuits and lever-aged properties of approximate arithmetic to achieve these eﬃciency gains. It has found many applications, among them privacy-preserving machine learning and secure genome analysis (see [15, 20, 4, 17, 23, 16] for some examples). However, the initial CKKS HE scheme lacked a bootstrapping procedure, and, thus, it was not a fully homomorphic encryption (FHE) scheme. This was remedied when introduced the ﬁrst bootstrapping procedure for the CKKS HE scheme, which followed the general template introduced by Gentry of evaluating the de-cryption circuit homomorphically. The challenge here is that the decryption procedure for CKKS requires computing the mod function, which is not easily representable via an arith-metic circuit. In fact, the mod function modulo q on the interval [−Kq, Kq] for some integer K is not even a continuous function. However, made the clever observation that in the CKKS HE scheme, we have an upper bound m on the size of the message, which can be made much smaller than q. In this situation, we actually only need to be able to compute the mod function on points in [−Kq, Kq] that are a distance at most m from a multiple of q. In this case, the mod function is periodic with period q and is linear on each of the small intervals around a multiple of q. Figure 2 shows the mod function along with the small intervals for approximation. The work of observed that the mod function [t]q on these intervals can be approximated 4 -20 -10 10 20 -4 -2 2 4 Figure 2: The mod function with modulus q = 10. The solid red lines represent the small intervals on which we need to approximate. via a scaled sine function S(t) = q 2π sin 2πt q . This approximation introduces an inherent error that depends on the message upper bound m. Let ϵ denote the ratio m q . Then, it can be shown that \|[t]q −S(t)\| ≤2π2 3 qϵ3. If ϵ is small enough, then this error can be suﬃciently small for use in bootstrapping provided that S(t) can be well-approximated by a low degree polynomial. The work of along with several followup works [6, 14, 18] proceeded to provide methods of approximating this scaled sine function (or scaled cosine function in the case of and scaled sine/cosine and inverse sine in the case of ) by a low-degree polynomial, which can then be plugged into the bootstrapping procedure of . However, due to the inherent error between the mod function [t]q and the scaled sine function S(t), this approach has a “fundamental error” that will occur regardless of how S(t) is approximated. One of the problems with this is that in order for the error to be O(1) (and, therefore, not destroy the message), m must be O(q2/3). This means that we must begin bootstrapping while the size of the encrypted message is considerably smaller than q, which is a source of ineﬃciency in the bootstrapping procedure, particularly in applications that require high precision. Compounding this problem is the fact that for large q, the size of the coeﬃcients of the approximation of S(t) grow. This requires the basis polynomials to be computed to higher precision and aﬀects the stability of the computation, where the errors introduced by the approximate arithmetic are ampliﬁed due to the large 5 coeﬃcients. As a consequence, arbitrarily high-precision bootstrapping for CKKS is unknown. The reason obtaining high-precision bootstrapping for CKKS is important is that one of the main applications for CKKS is privacy-preserving machine learning. However, many ML algorithms require high precision computation in order to converge. This may be especially true during the learning phase of neural networks, which involves back propagation and integer division by private integers. Additional nonlinear steps involve pooling functions, threshold functions, etc. Moreover, due to their high depth, computing these ML algorithms homomor-phically without bootstrapping is infeasible. Thus, for privacy-preserving ML applications, high-precision bootstrapping is required. Finding polynomial approximations of the mod function with error smaller than the “funda-mental error” (by directly approximating the mod function instead of ﬁrst approximating it via a scaled sine function) is one of the main ways one can hope to obtain high-precision bootstrap-ping for approximate homomorphic encryption. Finding such a method was explicitly posed as an open problem in , which stated that they believe “ﬁnding another approximation of [·]q operation ... can be a new breakthrough of improving the bootstrapping.” 1.2 This Work In this work, we introduce a method called modular Lagrange interpolation that can be used to obtain explicit low-degree polynomials that are provably good approximations of the mod function on small intervals around multiples of q, exactly what is required for bootstrapping for approximate homomorphic encryption. Moreover, by avoiding approximating the mod function by a scaled sine function, our approximation does not inherently have a ﬂat error associated with this approximation, and we are able to obtain more accurate approximations. Furthermore, we can explicitly upper bound the size of the coeﬃcients of the polynomial in the Chebyshev polynomial basis, which is the preferred basis as it leads to a more stable calculation. The size of the coeﬃcients in the Chebyshev polynomial basis directly aﬀects the precision required for computing the approximating polynomial itself. This is the ﬁrst work to give a comprehensive and robust solution to approximating the mod function by even bounding the size of the Chebyshev polynomial coeﬃcients In particular, we give a degree sixty-three polynomial that approximates the mod function to O(N−1) on intervals of size one, modulo N. The constants hide the dependence on the number of intervals k, but we show that these polynomials work well for k = 12, a common parameter setting of current HE schemes, with N = 106. We generalize this result to obtain O(N−d) error-approximation for any d ≥1, with O(d)-degree polynomials. For d = 4, we get a polynomial of degree 127 with an approximation error of 2−72, which improves upon the fundamental error of 2−57 associated with the sine approximation found in previous approaches. 6 A degree 127 polynomial can be computed by a circuit with multiplicative depth 7 and having low-depth is crucial, as it aﬀects the HE-levels consumed during bootstrapping. For k = 12, polynomials that work well for smaller N, i.e. N ≥128, are shown with degree sixty-nine. Similarly, the result generalizes to O(N−d) error-approximation for any d ≥1. Once again, we ﬁnd that setting d = 4 is suﬃcient to obtain an approximation error of 2−22, which improves upon the fundamental error of 2−18 associated with the sine approximation. This degree 159 polynomial can be computed by a circuit with multiplicative depth 8. Even better approximations can be found in Tables 1 through 4 in Section 9. We implement and report CKKS bootstrapping using our polynomials and demonstrate that they enable high-precision bootstrapping for a variety of parameter settings. For example, our bootstrapping achieves 67 bit message precision, larger than the precision of a double variable, whereas prior work was only capable of up to 40 bit message precision. Our full bootstrapping results can be found in Table 6. 1.3 Organization In Section 2, we provide some background on Lagrange interpolation and the Chebyshev basis polynomials and show lemmas that will be useful in proving error bounds. In Section 3, we introduce our modular Lagrange interpolation for the mod function by considering the simpler setting of only 3 intervals. In Section 4, we increase the number of intervals to be parameterized by k and show an O(1/N)-approximation for large N. In Section 5, we show a more general method that leads to O(N−d) approximations for each d ≥1. In Section 6, we show how to obtain O(N−d) approximations for small N (≈100). In Section 7, we provide an upper bound on the magnitude of coeﬃcients of our polynomials in the Chebyshev basis. In Section 8, we show that using standard error analysis for Lagrange interpolation cannot lead to as good of error bounds as modular Lagrange interpolation. In Section 9, we describe the application to bootstrapping for approximate homomorphic encryption in more detail and evaluate the performance of our polynomials for typical parameter settings. In Section 10, we give our implementation results for CKKS bootstrapping using our polynomials for a variety of parameter settings and show that they enable high-precision bootstrapping. In Section 11, we discuss how to apply modular Lagrange interpolation to functions other than the mod function. 2 Preliminaries For integers a, b, we use the notation [a..b] to represent the set {a, a + 1, . . . , b}. For reals a, b, the notation [a, b] will represent the usual closed interval. 7 We recall some background on Lagrange interpolation and relevant lemmas that will be useful when proving error bounds. 2.1 Lagrange Interpolation Given m + 1 distinct points t0, ..., tm, the jth Lagrange basis polynomial ℓ(m) j (x) is given by ℓ(m) j;t0,...,tm(x) △ = Y i∈[0..m]{j} x −ti tj −ti (1) When m is clear from context, we drop the superscript. Similarly, when t0, ..., tm is clear from context, we drop these from the subscript. When t0, ..., tm are collectively referred to as a set S, we also abuse notation and write, for any z ∈S, ℓ(m) z;S (x) △ = Y t∈S{z} x −t z −t (2) The Lagrange interpolation theorem states that for any polynomial f(x) (over a unique factorization domain) of degree m f(x) = X j∈[0..m] f(tj) ∗ℓ(m) j (x). Using the Lagrange interpolation theorem, we can prove the following lemma 2, which will be useful later in proving our error bounds. We start by proving a simpler version of the lemma. Lemma 1 Let R be a unique-factorization domain. If m ≥1, then for any distinct t0, ..., tm in R, P j∈0..m ∗ℓ(m) j;t0,...,tm(x) is identically zero. Proof: This follows from the fact that X j∈[0..m] (x −tj) ∗ℓ(m) j;t0,...,tm(x) = x ∗ X j∈[0..m] ℓ(m) j (x) − X j∈[0..m] tj ∗ℓ(m) j (x) = x ∗1 −x = 0, where we use the Lagrange interpolation theorem twice, as both 1 and x are polynomials of degree ≤m. □ The above lemma generalizes as follows. 8 Lemma 2 Let R be a unique-factorization domain. Let F(x, y) be a bivariate polynomial (over R) of degree m′ in y and with a factor (x −y). In other words F(x, y) can be written as F(x, y) = (x −y) ∗ X k∈[0..m′−1] ak(x) yk where ak(x) are polynomials (over R) in x. If m′ ≤m, then for any distinct t0, ..., tm in R, P j∈[0..m] F(x, tj) ∗ℓ(m) j;t0,...,tm(x) is identically zero. Proof: We have X j∈[0..m] F(x, tj) ∗ℓ(m) j;t0,...,tm(x) = X j∈[0..m] (x −tj) ∗ℓ(m) j (x) ∗ X k∈[0..m′−1] ak(x) tk j = X k∈[0..m′−1] ak(x) ∗ X j∈[0..m] (x −tj) ∗ℓ(m) j (x) ∗tk j = X k∈[0..m′−1] ak(x) ∗  x ∗ X j∈[0..m] ℓ(m) j (x) ∗tk j − X j∈[0..m] ℓ(m) j (x) ∗tk+1 j   = X k∈[0..m′−1] ak(x) ∗[x ∗xk −xk+1] = 0 where the second last equality follows by Lagrange interpolation theorem, recalling that m′ ≤ m. □ The following is a well-known bound on the error of polynomial interpolation on an interval [a, b]. Theorem 3 (Polynomial Interpolation) Let f be an n + 1 times diﬀerentiable function on [a, b] and pn be a polynomial of degree ≤n that interpolates f at n + 1 distinct points t0, t1, . . . , tn ∈[a, b], meaning pn(ti) = f(ti) for all 0 ≤i ≤n. Then, for each t ∈[a, b], there exists a point ψt ∈[a, b] such that f(t) −pn(t) = f(n+1)(ψt) (n + 1)! · n Y i=0 (t −ti). 9 2.2 Chebyshev Basis Polynomials and Nodes Instead of the standard polynomial basis {1, x, x2, . . .}, one can instead work with the Cheby-shev basis {T0(x), T1(x), T2(x), . . .}. The Chebyshev basis polynomials mimic the double-angle cosine formula. We have eix = cos x + i sin x, and hence e2ix = cos 2x + i sin 2x = (eix)2 = cos2 x −sin2 x + 2i sin x cos x. The real part of this is the same as 2 cos2 x −1. In general, Re(enix) = cos(nx) and Im(enix) = sin(nx). The n-th Chebyshev basis polynomial Tn(x) is deﬁned to be cos(n arccos(x)). Thus, T0(x) = 1, T1(x) = x, T2(x) = 2x2 −1, and so on. Induc-tively we have, T2n(x) = 2Tn(x)2 −1 and T2n+1(x) = 2xT2n(x) −T2n−1(x). The Chebyshev basis polynomials form an orthogonal basis under the following inner product. ⟨Ti, Tj⟩= d−1 X k=0 Ti(xk)Tj(xk) =    0 if i ̸= j and i, j < d d if i = j = 0 d 2 if i = j ̸= 0 and i < d where xk are Chebyshev nodes (roots) of Td(x), i.e. xk = cos π 2k + 1 2d for k = 0, 1, ..., d −1. The orthogonality relation is proved using trigonometric identities (see e.g. or see for an alternate proof). Polynomials expressed in the Chebyshev basis also have the nice property that the Cheby-shev basis polynomials always evaluate to within −1 and 1 on the interval [−1, 1] (just like the standard polynomial basis). For numerical stability of evaluation, it is well known that using the Chebyshev basis polynomials leads to smaller coeﬃcients, which determines required precision of evaluation of the basis polynomials. 3 Good Polynomial Approximation of the Mod Function In this section, we describe modular Lagrange interpolation and show how to obtain a good polynomial approximation of the mod function on small intervals in the simpler case where we only care about 3 intervals. Let the three intervals be the intervals of size one centered at −N, 0, and N, with N >> 11. More precisely, the three intervals are −N +[−1/2, 1/2], [−1/2, 1/2], and N + [−1/2, 1/2]. For any d ≥1, consider d + 1 (possibly equally-spaced) points in each of the three intervals. For each ℓ∈{−1, 0, 1}, deﬁne Sℓto be the set containing the d + 1 points in the ℓ-th interval. We will let S stand for the union of these sets, i.e. S−1 ∪S0 ∪S1. 1Earlier works consider intervals of size 2ϵ with mod taken with respect to 1. Thus, N can be viewed as 1/2ϵ. 10 Let fN be the mod function modulo N. In other words, fN(z) = z −ℓN for z in the ℓ-th interval. For any c ≥1, deﬁne the polynomial gc,d (of degree 3d + 2c) as gc,d(x) = X ℓ∈{−1,0,1} X z∈Sℓ fN(z) ∗   Y w∈Sℓ{z} x −w z −w  ∗ Y ℓ′∈{−1,0,1} ℓ′̸=ℓ x −ℓ′N z −ℓ′N d+c (3) Observe that if c is set to one, then gd,1(x) is almost the degree 3d + 2 polynomial that is the Lagrange interpolation of fN(x) at the 3(d + 1) points S. This follows since gd,1(x) can then be written as gd,1(x) = X ℓ∈{−1,0,1} X z∈Sℓ fN(z) ∗   Y w∈Sℓ{z} x −w z −w  ∗ Y ℓ′∈{−1,0,1} ℓ′̸=ℓ x −ℓ′N z −ℓ′N d+1 = X ℓ∈{−1,0,1} X z∈Sℓ fN(z) ∗   Y w∈Sℓ{z} x −w z −w  ∗ Y ℓ′∈{−1,0,1} ℓ′̸=ℓ Y w∈Sℓ′ x −ℓ′N z −ℓ′N ≈ X ℓ∈{−1,0,1} X z∈Sℓ fN(z) ∗   Y w∈Sℓ{z} x −w z −w  ∗ Y ℓ′∈{−1,0,1} ℓ′̸=ℓ Y w∈Sℓ′ x −w z −w (4) = X z∈S fN(z) ∗   Y w∈S{z} x −w z −w  . (5) Note, the last expression (5) is the degree 3d+2 polynomial that is the Lagrange interpolation of fN(x) at the 3(d + 1) points S. Observe that the diﬀerence from gd,1 is that when adding in the term for Lagrange interpolation with respect to some z, for points w that are not in the same interval as z, we consider the ratio x−ℓ′N z−ℓ′N instead of x−w z−w (that is, we replace w with the central point ℓ′N of the interval that w is in). The reason for this alteration is that, as we will soon discuss, c will not be a constant, and will be required to depend on ℓ′. In particular, c can be much larger than d, and hence more than the number of points chosen in each interval. Experiments we have conducted show that low-degree Lagrange interpolation on points chosen in the intervals, even mimicking an overall Chebyshev density, does not lead to good approximation, and especially so when we are seeking approximations of error O(N−c), for 11 c > 1. Moreover, the standard error analysis also does not produce good bounds. See Section 8 for a lower bound on the standard error analysis for low-degree Lagrange interpolation. We will refer to the above interpolation gd,c(x), as well as its slight variations that we will consider later, as the modular Lagrange interpolation of the 3(d + 1) points. It is easy to see that for x ∈Sℓ(ℓ∈{−1, 0, 1}), the summand in the deﬁnition of gd,1(x) corresponding to ℓis still equal to fN(x) (as in Lagrange interpolation). However, the other summands are no longer zero. Nevertheless, we will show that the other summands are close to zero. Thus, in this simple case, modular-Lagrange interpolation is almost the polynomial interpolation of the 3(d + 1) points. We now discuss, why keeping c constant does not lead to a good approximation, especially when the number of intervals is increased to more than three. To this end, we ﬁrst discuss why Lagrange interpolation of a smooth function is a bad approximation if the points are chosen equally-spaced. The contributions of the denominator in the Lagrange interpolation (5) are known as the barycentric weights, i.e. for each z ∈S, the barycentric weight is wz = 1 Q w∈S\z(z −w). When interpolating a smooth function over a contiguous interval, if the interpolation points, e.g. S above, are chosen equally spaced then the barycentric weights are lopsided. In particular, they are larger near the center (when z ∈S is near the center of the contiguous interval) compared to at the edges (when z ∈S is near the edge of the contiguous interval) by a factor of 2d, where d is the number of points. Now, the evaluation of the Lagrange interpolation (5) at x near the edge, and in particular near a point z′ ∈S near the edge, is supposed to approximate f(x) (which for a smooth function should be close to f(z′)). Indeed, the summand in (5) corresponding to (z =) z′ is close to f(z′). Additional error in the approximation comes from the other summands, i.e. for z ̸= z′. This error, for any z ̸= z′, is proportional to the product of wz and Q w∈S,w̸=z(x −w), the latter being more or less independent of z. Also, this latter product is large when x is near the edge, as compared to when it is near the center. Since the barycentric weights are large when z is near the center, this then implies that the summand corresponding to z near the center, for x near the edge, has a signiﬁcantly larger error. This is in contrast to the error for x near the center, and any z, whether near the center or near the edge. The reader may wonder if the diﬀerent summands may have opposing signs and some cancellation of error may ensue. While there is some such cancellation, the overall lopsided error accumulation continues to hold. This is known as the “Runge phenomenon”. To get a good approximation to the smooth function, the points are therefore chosen using the Chebyshev density function. In the modular-Lagrange interpolation above, the barycentric weights can be deﬁned as (for 12 -10 -5 5 10 1.×10-18 2.×10-18 3.×10-18 4.×10-18 Figure 3: The barycentric weights wz for the point set [−12..12]. z ∈Sℓ), wz = 1 Q w∈Sℓ\z(z −w) ∗Q ℓ′∈{−1,0,1} ℓ′̸=ℓ (z −ℓ′N)d+1 With N >> 1, these are almost the same as the original barycentric weights, and hence the lopsided nature of these weights continues to hold, especially when there are many more intervals than the simple case of three intervals we have considered so far. Thus, instead of keeping c constant as in deﬁnition (3), we will make it vary with the interval number, with larger values for intervals at the edges. The idea is similar to choosing points for Lagrange interpolation with density proportional to (1−x2)−1/2; however, the rate at which c increases, especially at the edges, will be less drastic as we also capitalize on both the intervals being tiny compared to N as well as the number of intervals being small. Thus, we will be able to keep the degree of the polynomial reasonably low. 4 O(1/N)-approximation for Large N If one is satisﬁed with O(1/N)-error approximation, then there is an even simpler polynomial than gd,c(x) given above. Moreover, the analysis is also considerably simpler. So, in this sec-tion, we ﬁrst deﬁne this simpler polynomial, but generalized to 2k + 1 intervals, and prove approximation bounds for it. In later sections, we will deﬁne polynomials that are generaliza-tions of deﬁnition (3) and give better approximations with error O(N−d), for any d > 1. For the simpler polynomial, it suﬃces to take d = c = 1 and also remove the dependence on z in the multiplicative term with power (d + c) in deﬁnition (3). 13 Let the (2k + 1) intervals be −kN + [−1/2, 1/2] to kN + [−1/2, 1/2]. For each integer ℓ∈{−k, −k+1, . . . , k}, consider (d+1 =)2 distinct points Sℓ= {zℓ 0, zℓ 1} in each of the (2k+1) intervals. In particular, zℓ i = ℓ∗N +i/2 for i ∈[0..1]. In other words, Sℓ= {ℓ∗N, ℓ∗N +1/2}. We will now deﬁne a polynomial of degree (4k+1)+ν, that approximates the mod function on these intervals. The value ν is deﬁned below. Recall, the mod function f is deﬁned as fN(x) = x −ℓN for all x ∈ℓN + [−1/2, 1/2]. Let L = {−k, −k + 1, . . . , k}. Deﬁne ˆ g(x) = X ℓ∈L X z∈Sℓ fN(z) ∗ Y w∈Sℓ{z} x −w z −w ∗ Y ℓ′∈L ℓ′̸=ℓ x −ℓ′N (ℓ−ℓ′)N 2+p(ℓ′) (6) where p(ℓ′) is deﬁned as p(ℓ′) =    0 for \|ℓ′\| ≤⌊k/2⌋ 1 for ⌊(11/12) ∗k⌋≥\|ℓ′\| > ⌊k/2⌋ 2 for \|ℓ′\| > ⌊(11/12) ∗k⌋ With Lagrange interpolation on each Sℓ, the above deﬁnition (6) simpliﬁes to ˆ g(x) = X ℓ∈L (x −ℓN) ∗ Y ℓ′∈L ℓ′̸=ℓ x −ℓ′N (ℓ−ℓ′)N 2+p(ℓ′) (7) The value ν is deﬁned to be the sum of p(ℓ′) over all ℓ′ ∈L. With the above deﬁnition of p(ℓ′), ν = 4∗k−2∗⌊k/2⌋−2∗⌊11/12∗k⌋. If k is a multiple of 12, then the degree of ˆ g is 31 6 k+1 (= 4k+1+ν). It is not diﬃcult to check that ˆ g is an odd polynomial, i.e. ˆ g(−x) = −ˆ g(x). We make k a multiple of 12 since k = 12 is a typical parameter setting that results in bootstrapping for approximate homomorphic encryption. By deﬁning p(ℓ′) with 12 in the denominator and setting k to be a multiple of 12, we make the analysis simpler by ensuring that the values in the deﬁnition of p(ℓ′) are all integers. This can easily be generalized to arbitrary values of k. We will show that ˆ g(x) is a good approximation to the mod function via two lemmas. First, Lemma 4 shows that the error between the summand corresponding to the interval that x is in and the mod function is small. Then, Lemma 5 shows that the summand corresponding to the interval that x is not in is close to 0. Combining Lemmas 4 and 5, we immediately arrive at Theorem 6, which shows that ˆ g(x) approximates the mod function well on all the intervals. 14 Lemma 4 For N > k, for every ℓ∈[−k..k], if x is in the ℓ-th interval, then fN(x) −fN(x) ∗ Y ℓ′∈L ℓ′̸=ℓ x −ℓ′N (ℓ−ℓ′)N 2+p(ℓ′) ≤2 ∗ln(e ∗k) N + 2 ∗k2 N2 Lemma 5 For N > k, and k a multiple of 12, for every t ∈[−k..k] and x in the t-th interval, X ℓ∈L ℓ̸=t (x −ℓN) ∗ Y ℓ′∈L ℓ′̸=ℓ x −ℓ′N ℓN −ℓ′N 2+p(ℓ′) ≤ 1 2 ∗ln(e ∗k) ∗max {0.924 ∗k 1 2 ∗1.1221k N , 0.098 ∗k 3 2 ∗3.81k N2 , 0.014 ∗k2 ∗9.813k N3 } A brief proof sketch of Lemma 5 follows below in section 4.1. The detailed proofs of Lemmas 4 and 5 can be found in Supplementary Material B and C, respectively. Combining the above lemmas gives the following theorem. Theorem 6 For N > k, and for k a multiple of 12, for the mod function fN(·), for any x such that \|x −t ∗N\| ≤1/2 with t ∈{−k, −k + 1, . . . , k}, we have \|ˆ g(x) −fN(x)\| < N−1 ∗ 2 ∗ln(e ∗k) + 2 ∗k2 N + 0.462 ∗ln(e ∗k) ∗ √ k ∗1.1221k ∗max {1, 0.106 ∗k ∗3.4k N , 0.015 ∗k 3 2 ∗8.745k N2 } Focusing on the typical parameter setting k = 12, we obtain the following corollary. Corollary 7 For the mod function fN(·), for k = 12, N > 12, for any x such that \|x−t∗N\| ≤ 1/2 with t ∈{−k, −k + 1, . . . , k}, we have that ˆ g is an odd polynomial with deg(ˆ g) = 63, and \|ˆ g(x) −fN(x)\| < N−1 ∗ 7 + 300 N + 22.3 ∗max{1, 3.03 ∗106 N , 1.25 ∗1011 N2 } With N ≥106, we get the following corollary. Corollary 8 For the mod function fN(·), for k = 12, N ≥106, for any x such that \|x−t∗N\| ≤ 1/2 with t ∈{−k, −k + 1, . . . , k}, we have that ˆ g is an odd polynomial with deg(ˆ g) = 63, and \|ˆ g(x) −fN(x)\| < 75/N 15 Remark. If we want good bounds for N ≈100 with the degree of ˆ g only marginally higher than 63, then we need a more complicated deﬁnition of ˆ g (in particular, p(ℓ′)), which we describe in a later section (Section 6). 4.1 Proof Sketch of Lemma 5 While the detailed proof of Lemma 5 can be found in Supplementary Material C, we give a brief sketch of the proof here. We start by analyzing, for every t, ℓ∈[−k..k], t ̸= ℓ, and x in the t-th interval (writing x = x′ + tN), the quantity (x −ℓN) ∗ Y ℓ′∈L ℓ′̸=ℓ x −ℓ′N ℓN −ℓ′N 2 = (x′ + (t −ℓ)N) ∗ Y ℓ′∈L ℓ′̸=ℓ x′ + (t −ℓ′)N (ℓ−ℓ′)N 2 i.e. when p(ℓ′) is removed from the exponent. Since \|x′\| ≤1/2 and N >> 1, the absolute value of the above is approximately \|(t −ℓ)\|N ∗ 1/2 \|ℓ−t\|N 2 ∗ Y ℓ′∈L ℓ′̸=ℓ ℓ′̸=t \|t −ℓ′\| \|ℓ−ℓ′\| 2 = 1/4 \|ℓ−t\|N ∗ Y ℓ′∈L ℓ′̸=ℓ ℓ′̸=t \|t −ℓ′\| \|ℓ−ℓ′\| 2 . When t = k and ℓ= 0, the denominator in the big product is close to (k!k!)2, whereas the numerator is close to (2k!)2, making the product about (2k!/k!k!)2, which is well approximated by 24k. Even with N ≈220, this then leads to an error of 248−20, for k = 12. Note however, if t = 1 and ℓ= 0, then the product is close to 1, and we get an error of about 1/N. This is indeed the lopsided barycentric weight problem, also known as the Runge phenomenon, described earlier. So, to correct this phenomenon, we add more weights in the exponent for larger \|ℓ′\|, and this is the function p(ℓ′). For example, since p(k) = 2, then the above expression for t = k will now be close to \|(k −ℓ)\|N ∗ 1/2 \|ℓ−k\|N 2+2 ∗ Y ℓ′∈L ℓ′̸=ℓ ℓ′̸=t \|k −ℓ′\| \|ℓ−ℓ′\| 2+p(ℓ′) If the big product at ℓ= 0 remained as before, i.e. (2k!/k!k!)2 ≈24k, then the above quantity is now approximately 24k/N 3. With k = 12 and N = 220, this quantity is less than 1. However, the big product is not exactly (2k!/k!k!)2 anymore, because of the p(ℓ′) in the exponent. However, using good upper and lower bounds on the factorial function, we can upper bound 16 this quantity by about 10k, which is even better than the previous 24k = 16k. It is possible that this product changes for the worse for other values of t and ℓ. However, we can show that this quantity behaves in a fairly smooth fashion, with only one local minimum and one local maximum as t varies from 0 to k (while ℓis kept ﬁxed at 0). The quantity is also lower bounded by 1, and, thus, the upper bound also holds for any ℓ. When k is ﬁxed, such as k = 12, one can also explicitly calculate all possible 12∗25 values of the above product, say, conveniently using a computer. However, since we can prove a general bound for arbitrary k, we state the result in general terms, since other or future applications may require larger k, say for enhanced security. However, we do report this computer-assisted bound in Appendix C.2, which shows an improvement of a factor of ten from the more general analysis. We also take this computer-assisted approach in Section 6 where more complicated p(·) functions are considered. 5 General Case: Obtaining an O(N −d)-error Upper Bound In this section, we give a modular Lagrange interpolation that leads to an approximation of the mod function with error O(N−d) for each d ≥1. As opposed to the interpolation in Section 4, speciﬁcally deﬁnition (6), where the second product was independent of z, to get the improved bound we must have the second product depend on z, similar to that in (3). Let L = {−k, −k + 1, . . . , k}. Let the 2k + 1 intervals be ℓN + [−1/2, +1/2] for ℓ∈L. For each ℓ∈L, consider d + 1 distinct points Sℓ= {zℓ 0, ..., zℓ d} in each of the (2k + 1) intervals (chosen symmetrically around ℓN). With the aim of obtaining a polynomial approximation of the mod function with error O(N−d) for every d ≥1, we now deﬁne polynomials of degree O(kd). The mod function fN is deﬁned as fN(x) = x −ℓN for all x ∈[ℓN −1/2, ℓN + 1/2]. For any b, d ≥1 and c ≥0, let gb,c,d(x) = X ℓ∈L X z∈Sℓ fN(z) ∗ Y w∈Sℓ{z} x −w z −w ∗ Y ℓ′∈L ℓ′̸=ℓ x −ℓ′N z −ℓ′N d+c+b∗p(ℓ′) (8) where p(ℓ′) is deﬁned as before in Section 4. When b = ⌈d/2⌉and d + c = 2b, we refer to the polynomial simply as gd(x). Since the points Sℓare chosen symmetrically around ℓN, and the intervals are also symmetric around zero, and further p(ℓ′) = p(\|ℓ′\|), it is an easy exercise to check that gb,c,d(x) is an odd function, i.e. gb,c,d(−x) = −gb,c,d(x). Note that the degree of the polynomial gd(x) is d+b∗(4k+ν), where the value ν is deﬁned to be the sum of p(ℓ′) over all ℓ′ ∈L. Let m = b ∗(4k + ν). 17 Theorem 9 For the mod function fN(·), for any d > 0, N > 0.22 ∗k ∗3.4k and N > (m + d + 1)/(d + 1), k ≥12 and k a multiple of twelve, for any x such that \|x −t ∗N\| ≤1/2 with t ∈{−k, −k + 1, . . . , k}, we have \|g2d(x) −fN(x)\| < N−2d ∗22d ∗ 1 + em d 2d + N−2d ∗ 4k ∗N + (4d + 6) ∗(33 ∗k2)d ∗ √ k ∗1.1221k/2 d and \|g2d+1(x) −fN(x)\| < N−2d−1 ∗22d+1 ∗ 1 + 2em 2d + 1 2d+1 + N−2d−1 ∗ 4k + (4d + 10) ∗(33 ∗k2)d+1/N ∗ √ k ∗1.1221k/2 d+1 The proof of this theorem is similar to the proof of Theorem 6, but now it uses the more advanced Lagrange interpolation lemma 2. The detailed proof can be found in Supplementary Material D. Note that a small calculation shows that, for k ≥12, (17k2+1/21.1221k) < (.22∗k∗3.4k)1/4, hence the above is upper bounded by N−2d(N ∗Nd/4 + (4d + 10)Nd/4), which in turn is upper bounded by (4d + 10) ∗N−7/4d+1. Thus, if N is exponential in 2k + 1, more precisely N > e2k+1 > 0.22 ∗k ∗3.4k, then the density of the points need not grow as drastically as (1 −x2)−1/2, and the error can still approach zero, asymptotically as N−(7/4)∗d+1, with degree at most 3(d+1)k. Note that the above is referring to the weights of p(ℓ′) for modular Lagrange interpolation, and not the number of points per interval as in standard Lagrange interpolation. 6 Enhanced Interpolation for Small N While the modular Lagrange interpolation deﬁned in Section 4 works well for large N, and in particular N ≥106, to get a good approximation that works for smaller N, e.g. N ≈100, we need to make the p(ℓ′) function rise faster with ℓ′. Thus, we can deﬁne a new modular Lagrange interpolation with the following p(ℓ′) function to get a better approximation. p(ℓ′) =        0 for \|ℓ′\| ≤⌊k/2⌋ 1 for ⌊(10/12) ∗k⌋≥\|ℓ′\| > ⌊k/2⌋ 2 for ⌊(11/12) ∗k⌋≥\|ℓ′\| > ⌊(10/12) ∗k⌋ 4 for \|ℓ′\| > ⌊11/12 ∗k⌋ 18 A theorem bounding the approximation-error similar to Theorem 6 can be proven, but the case analysis is proportionately more complicated. Instead, we will restrict ourselves to k = 12, and then a proof can be given by explicitly calculating (say, using a computer) the ratios Pt/Pℓfor all t, ℓ∈[−12..12]. Theorem 10 For the mod function fN(·), for k = 12, N > 12, for any x such that \|x−t∗N\| ≤ 1/2 with t ∈{−k, −k + 1, . . . , k}, we have that ˆ g deﬁned in (6) using above p(ℓ′) is an odd polynomial with deg(ˆ g) = 69, and \|ˆ g(x) −fN(x)\| < 1 N ∗ 1 4 + 3.5 N + 144 N2 ∗ 2 + max {1, 60 N , 2919 N2 , 1.23 ∗107 N4 } The proof along with pseudocode can be found in Supplementary Material E. To get approximation error of O(N−d), we generalize the above deﬁnition of ˆ g as in Sec-tion 5. We deﬁne a polynomial of degree (25 ∗d + 24 ∗c + 20 ∗b), for b, d ≥1 and c ≥0, as in (8). When b = ⌈d/2⌉and d + c = 2b, we just write gb,c,d(x) as gd(x). We have the following upper bound on the approximation-error of gd(x). Theorem 11 For the mod function fN(·), for any d > 0, b = ⌈d/2⌉, k = 12, and N > 60, for any x such that \|x −t ∗N\| ≤1/2 with t ∈{−k, −k + 1, . . . , k}, we have for gd deﬁned in (8) using above p(ℓ′), \|gd(x) −fN(x)\| ≤N−d ∗(744)d + 24 ∗N + 1 + (2d + 3) ∗(2310)d ∗N−2b The proof of Theorem 11 in Appendix E.1 is a generalization of the proof of Theorem 10 and is analogous to the proof of Theorem 9 generalizing Theorem 6. While the theorem as stated would only give useful bounds for N > 2310, the number 2310 above was obtained by a very loose analysis. As experiments described in Section 9 show, for N = 1024, the polynomials described have maximum error close to N−2/3∗d for d = 3..9. 7 Upper Bounding the Coeﬃcients in the Chebyshev Basis The modular Lagrange interpolation as described previously is in the usual polynomial basis {1, x, x2, . . .}. However, for the stability of evaluation of this polynomial at points in the var-ious intervals, it is better to ﬁrst transform the polynomial into the Chebyshev basis. The transformed polynomial will be referred to as the Chebyshev transform. Recall that polyno-mials expressed in the Chebyshev basis also have the nice property that the Chebyshev basis 19 polynomials always evaluate to within −1 and 1 on the interval [−1, 1], and hence the pre-cision to which these basis values need to be calculated is dictated by the magnitude of the coeﬃcients of the Chebyshev transform. The Chebyshev transform of f(x) of degree d is then given by {aℓ}ℓ∈[0..d] such that f(x) = Pd ℓ=0 aℓ∗Tℓ(x). Using the orthogonality relations of the Chebyshev basis polynomials (see Section 2), we can obtain aℓ= ⟨Tℓ, f⟩ ⟨Tℓ, Tℓ⟩. Moreover, by Cauchy-Schwartz inequality, for each ℓ, \|aℓ\|2 ≤⟨f, f⟩/(d/2). Thus, the coeﬃ-cients of the Chebyshev transform (of f) are upper bounded by a fraction of the L2-norm of f evaluated at the Chebyshev nodes, which then is at most two times the max-norm of f. While Theorem 6 and other similar theorems bound ˆ g (more precisely, the diﬀerence from the mod function) on the speciﬁed small intervals, here we need the max-norm of ˆ g at the Cheby-shev nodes, or more conservatively, at any point in [−kN −1/2, kN + 1/2] (this range being normalized to be [−1, 1] before obtaining the Chebyshev transform). So, we now estimate the maximum value of ˆ g(x) for any x ∈[−kN −1/2, kN + 1/2]. In the proof of Lemma 5, while bounding (13) by (15), now instead of x′ being small, i.e. \|x′\| ≤1/2, we now have that \|x′\| < N/2. Thus, in (15), the factor x′2+p(t) ∗N−1−p(t) now becomes N. Thus, we get the following upper bound on absolute value of each coeﬃcient aℓ(for all ℓ∈{0, . . . , d}) of Chebyshev transform of (6): (31 6 k + 1) ∗0.5 ∗ln(e ∗k)∗ max {0.924 ∗k 1 2 ∗1.1221k ∗N, 0.098 ∗k 3 2 ∗3.81k ∗N, 0.014 ∗k2 ∗9.813k ∗N}, which is easily seen to be less than 0.007 ∗( 31 6 k + 1) ∗ln(e ∗k) ∗k2 ∗9.813k ∗N. A simple calculation shows that for k = 12, N = 106, this value is 1.8 ∗1020. Similarly, we get the following upper bound on absolute value of each coeﬃcient aℓ(for all ℓ∈{0, . . . , d}) of Chebyshev transform of polynomials of Section 6: 69 ∗N ∗1.23 ∗107, which for N = 210 (and k set to 12), is about 1012. The size of the coeﬃcients of the Chebyshev transform is important as one can round these coeﬃcients to be integers (or to the nearest decimal upto which approximation-error is sought), and then the Chebyshev basis polynomials (which take values in [−1, 1]) need to be computed to a precision which is same as the precision of the rounded coeﬃcients. 20 8 Lower Bound on Standard Error Analysis for Lagrange In-terpolation In this section, we provide an argument that using standard Lagrange interpolation as in to approximate the mod function on multiple small intervals cannot be improved to obtain as good of error bounds as our modular Lagrange interpolation when using the standard error analysis for Lagrange interpolation. They had the clever idea of using Lagrange interpolation, but instead of picking the points for interpolation via the Chebyshev method (which would end up picking points over the entire space), they only picked points inside the small intervals on which we wish to approximate the mod function. The points in each interval were picked according to the Chebyshev method, and the number of points to pick inside each interval was determined via a greedy algorithm. We will show that it is not possible to allocate numbers of points to these small interval and then pick points inside these small intervals and obtain error bounds that match our modular Lagrange interpolation using the standard error analysis. We note that standard Lagrange interpolation can only be applied to a continuous function. Thus, in , they ﬁrst approximate the mod function [t]q by a scaled sine function2. We will ignore the inherent error associated with approximating the mod function by the scaled sine function and, instead, focus on the error that occurs when the scaled sine function is approximated by a degree n polynomial. We show that even this error cannot be as good as our bounds using standard Lagrange interpolation using the standard error analysis. We defer this section to Supplementary Material F. 9 Application to Bootstrapping for Approximate HE In Section 1.1, we explained that approximating the mod function on small intervals around the modulus is a necessary step in bootstrapping for approximate homomorphic encryption (CKKS). In this section, we will brieﬂy overview the bootstrapping procedure for the CKKS-FHE scheme introduced in and explain how our modular Lagrange interpolation leads to explicit polynomials that enable high-precision bootstrapping. We evaluate the performance of our polynomials for various typical settings of parameters and ﬁnd that they bypass the fundamental error introduced by approximating the mod function with a scaled sine function even for low-degree polynomials computable in depth 7 or 8. 2By shifting the center of the intervals, approximates the mod function by a scaled cosine function. We will perform our analysis on the scaled sine function and not shift the center of the intervals 21 Notation and Necessary Preliminaries: Let M be a power of 2 and ΦM(X) = XN + 1 be the Mth cyclotomic polynomial of degree N = M/2. Let R = Z[X]/ΦM(X). For an integer q, let Rq = Zq[X]/ΦM(X). Using the canonical embedding σ, it is possible to map an element m(X) ∈R into CN by evaluating m(X) at the Mth primitive roots of unity. Using the same canonical embedding, it is also possible to deﬁne an isometric ring isomorphism between S = R[X]/ΦM(X) and CN/2, where for an element m(X) ∈S, it has the canonical embedding norm \|\|m\|\|can ∞= \|\|σ(m)\|\|∞. Overview of the CKKS-FHE Scheme: The CKKS-FHE scheme is an FHE scheme for approximate arithmetic over real/complex numbers. Its security is based on the ring-LWE (RLWE) assumption. The message space of the scheme is polynomials m(X) in R with \|\|m\|\|can ∞< q/2 for a prime q. Using the canonical embedding and appropriate scaling, one can map a vector in CN/2 of ﬁxed precision into R. The fact that canonical embedding induces an isometric ring isomorphism between S and CN/2 implies that operations on the message space R map to the same operations performed coordinate-wise on CN/2. Thus, the CKKS-FHE scheme supports packing N/2 complex numbers into a single plaintext and operating on them in single instruction multiple data (SIMD) manner. Please refer to for more details on this encoding procedure. We will refer to m(X) ∈R as the plaintext/message and the corresponding vector in CN/2 as the plaintext “slots.” A ciphertext ct encrypting a message m ∈R is an element of R2 qℓfor some ℓ∈{0, . . . , L}. ℓrefers to the “level” of the ciphertext. In , qℓ= pℓ∗q for primes p and q. However, qℓcan be set in other ways (such as via an RNS basis ). The decryption structure is ⟨ct, sk⟩mod qℓ= m + e for some small error e ∈R. Observe that there is no way to remove e and some of the least signiﬁcant bits of m are unrecoverable. A fresh ciphertext is generated at the highest level L. Homomorphic operations increase the magnitude of the error and the message and one must apply a rescaling procedure or modular reduction to bring a ciphertext to a lower level to continue homomorphic computation. Eventually, a ciphertext is at the lowest level (an element of R2 q), and no further operations can be performed. Bootstrapping Procedure for CKKS-FHE: introduced the ﬁrst bootstrapping pro-cedure for the CKKS-FHE scheme. Subsequent works [6, 13, 14] improved various aspects of bootstrapping, but the overall procedure remains the same. The goal is to take a ciphertext at the lowest level and bring it up to a higher level so that homomorphic computation can continue. Thus, given a ciphertext ct at the lowest level, we want to obtain another ciphertext ct′ such that ⟨ct, sk⟩mod q ≈⟨ct′, sk⟩mod qℓ 22 for some ℓ> 1. For simplicity in the following, we will include the starting decryption error in the message m. That is, we will assume that ⟨ct, sk⟩mod q = m. Bootstrapping is done via the following sequence of steps: 1. Modulus Raising: By simply considering ct as a ciphertext at the highest level, it follows that ⟨ct, sk⟩mod qL = qI + m for some I ∈R. 2. Coeﬃcients to Slots: We need to perform the modular reduction on the polynomial coeﬃcients of t = qI + m. However, recall that homomorphic computations evaluate coordinate-wise on the plaintext “slots,” not the polynomial coeﬃcients. Thus, we need to transform our ciphertext so that the polynomial coeﬃcients are in the “slots.” This can be done by evaluating a linear transformation homomorphically. 3. Compute the Mod Function: We need a procedure to compute/approximate the mod function homomorphically. This is a signiﬁcant challenge since we can only com-pute arithmetic operations homomorphically. 4. Slots to Coeﬃcients: Finally, we need to undo the coeﬃcients to slots step. This can be done by homomorphically evaluating the inverse of the previous linear transform. Observe that if we can approximate the mod function, then the above procedure will give us a ct′ at some higher level ℓthat decrypts to m + e for some small error e. Since we are dealing with approximate arithmetic, this error from bootstrapping can be absorbed into the other errors that occur during approximate arithmetic and homomorphic evaluation. Prior Approaches to Approximating the Mod Function: We can upper bound \|I\| < K for some integer K (a typical value is K = 12) so that we only need to approximate the mod function on the interval [−Kq −m, Kq +m], where we have overloaded notation to make m an upper bound on the size of the message for consistency of notation with prior works. However, ﬁnding a good polynomial approximation for the mod function on this interval is diﬃcult since it is not even a continuous function. As described in the introduction, observed that if m is suﬃciently small, then the mod function [t]q can be approximated by the scaled sine function S(t) = q 2π sin 2πt q . This approximation introduces a “fundamental error” of 2π2 3 qϵ3, where ϵ = m/q. Thus, to obtain 23 O(1) error, we require m = O(q2/3), meaning that we must begin bootstrapping prior to m becoming too large. The work then proceeded by approximating S(t) using a Taylor expansion to degree O(Kq) so that the error of approximation with S(t) is about the same as the error between S(t) and [t]q. Since they are approximating a scaled sine function, they are able to use double-angle formulas for sine to reduce the computational cost of evaluating the approximation polynomial by ﬁrst approximating a scaled-down version sin 2πt 2r∗q to a degree d0 = O(1) and then using this approximation to approximate S(t). The required setting of r is O(log Kq) and so the multiplicative depth (alternatively, the ciphertext levels consumed) remains the same. The work improved upon this method by instead using Chebyshev interpolation to approximate S(t), which lowered the error of approximation and the required degree. In Chebyshev interpolation, instead of working with the polynomial basis {1, x, x2, . . .}, one works with the Chebyshev basis {T0(x), T1(x), T2(x), . . .} and uses the Chebyshev nodes as points for interpolation. Approximating S(t) via Chebyshev interpolation pn(t) of degree ≤n gives an error of \|S(t) −pn(t)\| ≤qKn+1 πn (n + 1)!. Observe that the above error does not depend on ϵ (that is, it is a good approximation on the entire space [−Kq, Kq] and does not utilize the fact that we only need a good approximation close to multiples of q). However, for a typical parameter setting K = 12, this error bound only improves on the trivial bound of q for degree n ≥98. The work improved on the approximation of the scaled sine function by leveraging the fact that we only care that our approximation is good near multiples of q. To do this, uses Chebyshev interpolation on the union of these small intervals instead of the entire space [−Kq, Kq]. Implicit in this, they consider the ratio between the maximum size of a message and q. This procedure allows them to reduce the degree of the polynomial required for approximation and allows the error of approximation to depend on the ratio ϵ = m/q. For approximating the scaled sine function on 2K + 1 intervals near multiples of q (near −Kq, . . . , Kq), the error of approximation in any particular interval is O ϵd , where d is the number of points chosen for Chebyshev interpolation in that interval. However, due to the constants hidden in the big-O notation (which can depend exponentially on K), choosing the same number of points for Chebyshev interpolation in all intervals does not give the best approximation, and the authors choose d for each interval via a greedy algorithm. The above approaches all require ﬁrst approximating [t]q via a scaled sine function, and, therefore, will always at least have error 2π2 3 qϵ3. If we want to have a smaller error, it is 24 necessary to use a diﬀerent method that avoids the scaled sine function. A pair of recent works by the same authors [19, 18] attempt to avoid the scaled sine function by instead trying to ﬁnd the optimal minimax polynomial of a ﬁxed degree that approximates the mod function via algorithmic search. uses L2-norm minimization and uses a variant of the Remez algorithm to obtain an approximation to the optimal minimax polynomial of a given degree that approximates the modular reduction function on the union of intervals containing points close to multiples of q. However, in both of these works, the polynomial is found via algorithmic search. Moreover, the degree of the polynomial is ﬁxed a priori before any approximation is computed. Without any bounds showing trade-oﬀs between the polynomial degree, size of the coeﬃcients, and the error of approximation, it is hard to develop strategies for picking the degree. Unfortunately, as observed by , the size of the coeﬃcients of these polynomials are too large to enable high-precision bootstrapping. By using a composition of sine/cosine and the inverse sine function, are able to improve on , but their bootstrapping is only capable of up to 40 bit message precision. Our Approach to Approximating the Mod Function: In contrast to the above ap-proaches, using modular Lagrange interpolation, we give explicit low-degree polynomials that directly approximate the mod function on intervals around multiples of the modulus to ar-bitrarily small error. We are able to formally prove error bounds for our polynomials and bounds on the magnitudes of the coeﬃcients in the Chebyshev basis. Our approach avoids the “fundamental error” associated with using the scaled sine function as an intermediate approx-imation, and, therefore, the restriction that m = O(q2/3) is removed. Moreover, the size of the coeﬃcients of our polynomials are suﬃciently small, so the Chebyshev basis polynomials are capable of being evaluated homomorphically to the required precision during bootstrap-ping. As we demonstrate in Section 10, this enables high-precision bootstrapping, which was previously unknown. In previous sections, the p(ℓ′) function weighting the various intervals was chosen with the setting K = 12 in mind. However, our approach is general, and for a diﬀerent value of K, one could deﬁne an appropriate p(ℓ′) function to obtain good approximations. Our polynomials are deﬁned in terms of a modulus N and intervals of length 1. Thus, to evaluate [t]q for bootstrapping, where t = qI + m′ for some \|m′\| < m and \|I\| < K, one would ﬁrst compute t 2m = q 2mI + m′ 2m. Setting N = q 2m and evaluating the appropriate polynomial gives an approximation to m′ 2m, which can then be multiplied by 2m to obtain an approximation of m′ as desired. Tables 1, 2, 3, and 4 show experimental results of implementations of polynomials obtained using Macaulay2 for various settings of m/q = ϵ. For all polynomials, we set c = 0 since experimental results show that this performs best. Consistent with prior works, we set K = 12 25 Table 1: Polynomials for m/q = 2−7 Degree Points/Interval (d + 1) b c Depth Precision† Error†† 69 3 1 0 7 30 bits 2−11 89 3 2 0 7 20 bits 2−12 115 4 2 0 7 37 bits 2−15 139 5 2 0 8 39 bits 2−21 159 5 3 0 8 40 bits 2−22 185 6 3 0 8 60 bits 2−26 229 7 4 0 8 63 bits 2−30 255 8 4 0 8 83 bits 2−34 299 9 5 0 9 83 bits 2−37 325 10 5 0 9 107 bits 2−37 † This is the precision at which the Chebyshev polynomials need to be evaluated. †† The error is relative to q in line with earlier conventions. Table 2: Polynomials for m/q = 2−10 Degree Points/Interval (d + 1) b c Depth Precision Error 69 3 1 0 7 37 bits 2−17 115 4 2 0 7 40 bits 2−24 139 5 2 0 8 67 bits 2−34 185 6 3 0 8 73 bits 2−40 209 7 3 0 8 97 bits 2−49 255 8 4 0 8 103 bits 2−57 279 9 4 0 9 127 bits 2−65 325 10 5 0 9 133 bits 2−72 349 11 5 0 9 155 bits 2−90 26 Table 3: Polynomials for m/q = 2−15 Degree Points/Interval (d + 1) b c Depth Precision Error 63 3 1 0 6 50 bits 2−22 77 3 2 0 7 50 bits 2−28 103 4 2 0 7 83 bits 2−40 141 5 3 0 8 103 bits 2−56 167 6 3 0 8 130 bits 2−65 205 7 4 0 8 153 bits 2−81 231 8 4 0 8 177 bits 2−86 269 9 5 0 9 203 bits 2−105 295 10 5 0 9 220 bits 2−107 Table 4: Polynomials for m/q = 2−20 Degree Points/Interval (d + 1) b c Depth Precision Error 63 3 1 0 6 67 bits 2−36 103 4 2 0 7 93 bits 2−55 127 5 2 0 7 127 bits 2−72 167 6 3 0 8 153 bits 2−92 191 7 3 0 8 187 bits 2−106 231 8 4 0 8 213 bits 2−129 255 9 4 0 8 247 bits 2−140 295 10 5 0 9 277 bits 2−166 Table 5: Fundamental Error Between the Scaled Sine and Mod Functions m/q Error 2−7 2−18 2−10 2−27 2−15 2−42 2−20 2−57 27 and evaluate the scaled down mod function that takes an input in a small interval [I −ϵ, I + ϵ] for I ∈{−K, . . . , K}. The precision refers to the number of bits of precision we need to evaluate the Chebyshev basis polynomials in order to obtain the best error. The polynomials in Tables 1 and 2 were obtained using the enhanced interpolation from Section 6, while the polynomials in Tables 3 and 4 are from Section 5 (Section 4 for the degree 63 polynomials)3. Table 5 shows the fundamental error between the scaled sine function and the mod function for various settings of m/q. We observe that once we take d + 1 = 5 points per interval (and set b to 2 or 3), our polynomials have smaller error than this fundamental error by many orders of magnitude. This gain is only increased as we increase the number of points per interval. These polynomials are computable in depth 7 for m/q = 2−20 and in depth 8 for m/q = 2−7, 2−10, 2−15. If one is willing to use a depth 8 circuit for m/q = 2−20 or a depth 9 circuit for m/q = 2−7, 2−10, 2−15, then the error decreases even further. 10 Implementation To demonstrate the applicability of our polynomials to high precision bootstrapping for ap-proximate homomorphic encryption, we updated the bootstrapping procedure of the HEAAN library to utilize our polynomials during the “Compute the Mod Function” step (see Sec-tion 9). Additionally, we updated HEAAN to use the quadmath library, since we wanted to achieve bootstrapping error smaller than the precision of a double. We ran our implementation using a PC with an AMD Ryzen 5 3600 3.6 GHz Six-Core CPU. Table 6 gives our bootstrapping results for various settings of parameters. As before, ϵ represents the ratio m/q, where m is an upper bound on the size of the message (including any errors associated from the approximate arithmetic and prior homomorphic operations) and q is the size of the modulus prior to bootstrapping. qL denotes the modulus of the largest level, which is the modulus of a fresh ciphertext prior to any homomorphic operations. N denotes the ring dimension, which we increase as qL increases to maintain 128-bit security [2, 3]. All of our results were obtained using 8 slots. We observe that we could have utilized N/4 slots (out of a possible N/2) without aﬀecting the running time of the “Compute the Mod Function” step that we updated in the bootstrapping procedure since the (up to) N/2 coeﬃcients of the message polynomial could ﬁt into the slots of a single ciphertext. However, we opted to use a ﬁxed number of slots since the runtimes of the “Coeﬃcients To Slots” and the “Slots to Coeﬃcients” steps in the bootstrapping procedure (which we did not modify) scale with the number of slots and would perform poorly if the number is too large. qℓ′ denotes the 3The polynomials in these Tables 3 and 4 used the enhanced strategy as described in Section 6 where p(ℓ′) is forced to zero if \|ℓ\| ≥\|ℓ′\|. 28 Table 6: High-Precision Bootstrapping Results Gap Input Modulus Ring Dim. Degree Depth Modulus Error Output Runtime†† Precision† (Fresh) of poly of poly (After) (Bootstrap) Precision (secs) −log2 ϵ log2 m log2 qL N log2 qℓ′ βbs 7 21 1200 216 139 8 519 2−14 13 75 10 30 1600 216 139 8 665 2−24 23 114 10 43 2400 217 209 8 1148 2−37 36 613 10 63 2400 217 279 9 603 2−55 54 693 10 50 3000 217 349 9 1274 2−46 45 1244 10 70 3000 217 349 9 1002 2−58 57 1206 10 90 3000 217 349 9 720 2−68 67 1071 15 30 1600 216 103 7 627 2−25 24 85 15 58 2400 217 167 8 708 2−53 52 494 20 20 1200 216 63 6 393 2−15 14 40 20 20 1600 216 63 6 793 2−15 14 63 20 20 3600 218 63 6 2791 2−15 14 949 20 56 2400 217 127 7 806 2−51 50 411 20 76 3000 217 167 8 752 2−64 63 640 20 76 3600 218 167 8 1351 2−64 63 1893 † The modulus qℓof the ciphertext prior to bootstrapping is m/ϵ. The number of bits of qℓis the sum of the ﬁrst two entries in a row in the table. †† Includes runtime of “Coeﬃcients to Slots” and “Slots to Coeﬃcients” steps. Number of slots ﬁxed to be 8 so that the “Compute the Mod Function” step dominates runtime. Results reported are from a 3.6 GHz six-core AMD CPU using quadmath, NTL and GMP software libraries. 29 modulus of the ciphertext after bootstrapping. The reported error is the decryption error after performing bootstrapping4. We see that our polynomials are capable of achieving high precision bootstrapping, with the resulting message precision as large as 67 bits, larger even than the precision of a double variable. Prior to our work, the highest precision bootstrapping of CKKS was the recent work of which could achieve a resulting message precision of up to 40 bits. Thus, we view our result as a substantial improvement for bootstrapping in settings where high precision is required, such as the inference step of a convolution neural network or even the learning stage of the neural network. As mentioned earlier, since CKKS is for approximate arithmetic, it is only possible to have unlimited computation for stable computations that do not lose precision. However, even such stable computations lose precision in early stages prior to convergence. Thus, it is important to begin such computations with high precision and, later, one can switch to smaller precision during the stable regime. 11 Modular Lagrange Interpolation Beyond the Mod Function The techniques developed in earlier sections also apply to other bounded piecewise smooth functions, especially of low-degree. However, many times, a more judicious algebraic manip-ulation can lead to a smaller degree or better approximation interpolation. For example, for x ∈ℓN + [−1/2, 1/2], ℓ∈[−k..k], deﬁne I(x) to be ℓ. Then the mod function fN(x) on these intervals can be written as x −I(x) ∗N. and I(x) = (x −fN(x))/N. Thus, an O(N−1)-approximation g(x) of fN(x) gives us a O(N−2) approximation of I(x), as (x−g(x))/N. This is not surprising, as a linear function already approximates I(x) to O(N−1) and, thus, the modu-lar interpolation is likely to give a much better approximation, i.e. an O(N−2)-approximation, whereas the bounds in Theorem 6 only give O(N−1). We have the following generalization of Theorem 9. Let {fℓ}ℓ∈[−k..k] be a set of functions each of degree s, with \|fℓ(x)\| bounded by B when x is in the ℓ-th interval ℓN + [−1/2, 1/2]. Deﬁne, for b, d ≥1, and c ≥0, gb,c,d(x) = X ℓ∈L X z∈Sℓ fℓ(z) ∗ Y w∈Sℓ{z} x −w z −w ∗ Y ℓ′∈L ℓ′̸=ℓ x −ℓ′N z −ℓ′N d+c+b∗p(ℓ′) (9) where p(ℓ′) is deﬁned as before in Section 4. When b = ⌈d/2⌉and d + c = 2b, we refer to the polynomial simply as gd(x). 4 In other words, if the decryption before bootstrapping would have resulted in message slot value m, then the decryption after bootstrapping would result in a message slot value m′ such that \|m′ −m\| ≤βbs\|m\|. Thus, the resulting precision of m′ is −log2 βbs −1 bits. 30 Theorem 12 For any k, d > 0, N > 0.22 ∗k ∗3.4k and N > (m + d + 1)/(d + 1), k ≥12, and k a multiple of twelve, for any x such that \|x −ℓN\| ≤1/2 with ℓ∈{−k, −k + 1, . . . , k}, we have \|g2d(x) −fℓ(x)\| < N−2d+s ∗B ∗22d ∗ 1 + em d 2d + N−2d+s ∗ 4k ∗N + (4d + 6) ∗(33 ∗k2)d ∗ √ k ∗1.1221k/2 d and \|g2d+1(x) −fℓ(x)\| < N−2d−1+s ∗B ∗22d+1 ∗ 1 + 2em 2d + 1 2d+1 + N−2d−1+s ∗ 4k + (4d + 10) ∗(33 ∗k2)d+1/N ∗ √ k ∗1.1221k/2 d+1 12 Conclusion In this work, we introduced a novel technique called modular Lagrange interpolation that al-lows one to ﬁnd low-degree polynomial approximations of a function on the union of small intervals even if the function is not continuous. By using modular Lagrange interpolation, we are able to prove explicit error bounds on these polynomials. We demonstrated the eﬃcacy of our technique by constructing explicit low-degree polynomials that approximate the mod function well on intervals around the modulus. These polynomials have major applications to bootstrapping for approximate homomorphic encryption and allow us to bypass the funda-mental error of approximation inherent in prior works which ﬁrst used a scaled sine function to approximate the mod function on these intervals. We implemented the bootstrapping of HEAAN using our polynomials and gave performance results for various parameter settings. We found that our polynomials enable high-precision bootstrapping and, as an example, achieved bootstrapping with 67 bit message precision. Further improvements to the ﬂoating point pre-cision of the CKKS embedding in implementations can lead to even higher precision with our techniques. References Heaan, 10 Albrecht, M.R.: On dual lattice attacks against small-secret lwe and parameter choices in helib and seal. In: Coron, J.S., Nielsen, J.B. (eds.) Advances in Cryptology – EURO-CRYPT 2017. pp. 103–129. 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SIAM Review 41, No. 1, 135–147 (2004) 2.2 Supplementary Material A Upper Bound on Lagrange Basis Polynomial In this section, we will determine an upper bound on the absolute value of the Lagrange basis polynomials ℓ(m) j;t0,...,tm(x) when t0, . . . , tm are evenly spaced points in the closed interval [−1/2, 1/2] and x ∈[−1/2, 1/2]. We can write these points as ti = i m −1 2 for i ∈{0, . . . , m}. 33 First, we will consider the denominator of ℓ(m) j;t0,...,tm(x), which is independent of x. For the jth Lagrange basis polynomial, the denominator is given by Y i∈{0,...,m}{j} tj −ti. The term tj −ti = j−i m . Thus, the absolute value of the denominator can be written as j!(m −j)! mm . Now, we will consider the numerator. For the jth Lagrange basis polynomial, the numerator can be written as Y i∈{0,...,m}{j} x −ti = Y i∈{0,...,m}{j} x − i m −1 2 for x ∈[−1/2, 1/2]. Setting y = x + 1 2, we have Y i∈{0,...,m}{j} y −i m for y ∈[0, 1]. Consider the m + 1 points 0, 1 m, 2 m, . . . , 1. The maximum possible value of the numerator occurs when y is set to maximize the product of the distances between y and all but one these points. First, we observe that for any ﬁxed y, the value is maximized when we exclude the point closest to y. Having done this, we observe that the maximum occurs when y = 1 and we remove the point 1. This follows from the fact that the nearest point is 1 m away, which is the maximum possible. The second nearest point is 2 m away, which, again, is the maximum possible and so forth. Thus, the maximum absolute value of the numerator can be bounded by m! mm . Dividing the numerator and denominator, we arrive at the bound m! j!(m −j)! = m j for the jth Lagrange basis polynomial. 34 B Proof of Lemma 4 For any ℓ∈[−k..k], and x ∈[−1/2 + ℓN, +1/2 + ℓN], X z∈Sℓ fN(z) ∗   Y w∈Sℓ{z} x −w z −w  ∗ Y ℓ′∈L ℓ′̸=ℓ x −ℓ′N (ℓ−ℓ′)N 2+p(ℓ′) = Y ℓ′∈L ℓ′̸=ℓ x −ℓ′N (ℓ−ℓ′)N 2+p(ℓ′) ∗ X z∈Sℓ fN(z) ∗   Y w∈Sℓ{z} x −w z −w   = (x −ℓN) ∗ Y ℓ′∈L ℓ′̸=ℓ x −ℓ′N (ℓ−ℓ′)N 2+p(ℓ′) = (x −ℓN) ∗ Y ℓ′∈L ℓ′̸=ℓ 1 −(x −ℓN)/((ℓ′ −ℓ)N) 2+p(ℓ′) = x′ ∗ Y ℓ′∈L ℓ′̸=ℓ 1 −x′/((ℓ′ −ℓ)N) 2+p(ℓ′) (10) where we let x′ stand for (x −ℓN). If ℓis zero, then the big product above can be simpliﬁed, by noting that p(ℓ′) = p(−ℓ′), to Y ℓ′∈[1..k] 1 −(x)2/(ℓ′N)22+p(ℓ′) When ℓ̸= 0, we can still try to pairoﬀ(ℓ′ −ℓ) around ℓas much as possible, but we must address the slight complication due to p(ℓ′). So, deﬁne L(1) to be subset of L where p(ℓ′) ≥1, and similarly L(2) to be subset of L where p(ℓ′) ≥2. Then, we have Y ℓ′∈L ℓ′̸=ℓ 1 −x′/((ℓ′ −ℓ)N) 2+p(ℓ′) = Y ℓ′∈L ℓ′̸=ℓ 1 − x′ (ℓ′ −ℓ)N 2 ∗ Y ℓ′∈L(1) ℓ′̸=ℓ 1 − x′ (ℓ′ −ℓ)N ∗ Y ℓ′∈L(2) ℓ′̸=ℓ 1 − x′ (ℓ′ −ℓ)N 35 Since, the above is one plus or minus other terms involving x′/N, it follows that the quan-tity (10) diﬀers from fN(x) = x′ (with x in the ℓ-th interval) in absolute value by at most 1 2 ∗    −1 + Y ℓ′∈L ℓ′̸=ℓ 1 + \|x′\| \|ℓ′ −ℓ\| ∗N 2 ∗ Y ℓ′∈L(1) ℓ′̸=ℓ 1 + \|x′\| \|ℓ′ −ℓ\| ∗N ∗ Y ℓ′∈L(2) ℓ′̸=ℓ 1 + \|x′\| \|ℓ′ −ℓ\| ∗N     (11) Focusing just on one of these products we have, and assuming N > k, Y ℓ′∈L ℓ′̸=ℓ 1 + \|x′\| \|ℓ′ −ℓ\| ∗N ≤ Y i∈[1..k] 1 + \|x′\| i ∗N 2 ≤  1 + X i∈[1..k] \|x′\| i ∗N + X j∈[2..k] k j \|x′\| N j   2 ≤  1 + ln(ek) ∗\|x′\| N + X j∈[2..k] kj ∗ \|x′\| N j   2 ≤  1 + ln(ek) ∗\|x′\| N + X j∈[2..k] \|x′\| ∗k N j   2 ≤ 1 + ln(ek) ∗\|x′\| N + \|x′\| ∗k N 2 ∗ 1 1 −\|x′\| ∗k/N !2 ≤ 1 + 2 ∗ln(ek) ∗\|x′\| N + 4 ∗ \|x′\| ∗k N 2! ≤ 1 + 2 ∗ln(ek) ∗\|x′\| N + k2 N2 (12) 36 Thus, using (11), it follows that the quantity (10) diﬀers from fN(x) = x′ (with x in the ℓ-th interval) in absolute value by at most 2 ∗ln(ek) N + 2k2 N2 C Proof of Lemma 5 Proof: With \|x −tN\| ≤1/2 for t ∈{−k, −k + 1, . . . , k}, we now analyze the other summands (i.e. ℓ̸= t) of (7). Let x = x′ + tN, with \|x′\| ≤1/2. We have (recalling ℓ̸= t), X z∈Sℓ fN(z) ∗ Y w∈Sℓ{z} x′ + t ∗N −w z −w ∗ Y ℓ′∈L ℓ′̸=ℓ x′ + t ∗N −ℓ′N ℓN −ℓ′N 2+p(ℓ′) = (x′ −(ℓ−t)N) ∗ Y ℓ′∈L ℓ′̸=ℓ x′ −(ℓ′ −t)N (ℓ−ℓ′)N 2+p(ℓ′) (13) For N >> 1, the absolute value of the above is upper bounded by approximately (x′)2+p(t) ∗N−1−p(t) ∗\|t −ℓ\|−1−p(t) ∗ Y ℓ′∈L ℓ′̸=ℓ ℓ′̸=t \|ℓ′ −t\| \|ℓ′ −ℓ\| 2+p(ℓ′) (14) We can be more precise and use (10), (11) and (12) to get that expression 13 is upper bounded by (x′)2+p(t) ∗(1 + 4 ∗ln(ek) N + 4k2 N2 ) ∗N−1−p(t) ∗\|t −ℓ\|−1−p(t) ∗ Y ℓ′∈L ℓ′̸=ℓ ℓ′̸=t \|ℓ′ −t\| \|ℓ′ −ℓ\| 2+p(ℓ′) (15) We now upper bound this quantity for diﬀerent values of t and ℓ. Along the way, we will show that the denominator above is smallest when ℓis zero. We will split the analysis according to the three cases depending on the value of p(t). So, we ﬁrst consider t ≤k/2, where p(t) = 0. In this case the above quantity has factor N−1, i.e. has no additional negative powers of N. 37 Thus, we would like to prove that, for ℓ= 0 and t ≤k/2, the following quantity is small: \|t −ℓ\|−1−p(t) ∗ Y ℓ′∈L ℓ′̸=ℓ ℓ′̸=t \|ℓ′ −t\| \|ℓ′ −ℓ\| 2+p(ℓ′) = \|t −ℓ\|−1−p(t) ∗\|t −ℓ\|p(t)−p(ℓ) ∗Pt Pℓ = \|t −ℓ\|−1−p(ℓ) ∗Pt Pℓ (16) where Pi = Q ℓ′∈L,ℓ′̸=i \|ℓ′ −i\|2+p(ℓ′). Consider the following sequence of integers: a0 = 0, a1 = ⌊k/2⌋, a2 = ⌊(11/12) ∗k⌋, and a3 = k. Since, we assume that k is a multiple of 12, a1 = k/2 and a2 = (11/12) ∗k. Then, p(ℓ′) = k for integers ℓ′ in the range [ak + 1..ak+1], for all k ∈[0..2]. Also, p(a0) = 0. Thus, for i ≤a1, Pi = ((a3 −i)!(a3 + i)!)4 ∗((a1 −i)!(a1 + i)!)−1 ∗((a2 −i)!(a2 + i)!)−1 , and for a1 < i ≤a2, Pi = ((a3 −i)!(a3 + i)!)4 ∗((a1 + i)!/(i −a1)!)−1 ∗((a2 −i)!(a2 + i)!)−1 , and for a2 < i ≤a3 = k, Pi = ((a3 −i)!(a3 + i)!)4 ∗((a1 + i)!/(i −a1)!)−1 ∗((a2 + i)!/(i −a2)!)−1 . In Lemma 13 below we show that for all \|ℓ\| ≤k and all \|t\| ≤k/2, we have Pt/Pℓ< 0.924 ∗ √ k ∗(1.1221)k, and, for all \|ℓ\| ≤k and all \|t\| ≤11/12 ∗k, we have Pt/Pℓ< 0.098 ∗ √ k3 ∗(3.81)k, and, for all \|ℓ\| ≤k and all \|t\| ≤k, we have Pt/Pℓ< e ∗0.0051 ∗k2 ∗(9.813)k ≤0.014 ∗k2 ∗9.813k. Thus, in deﬁnition 6 of ˆ g(x), using 15, and well-known bound on harmonic series, and assuming N > max{k, 80∗ln(ek)}, we have for all t ∈L, and for all x ∈[−1/2+t∗N, 1/2+t∗N], X ℓ∈L ℓ̸=t X z∈Sℓ fN(z) ∗ Y w∈Sℓ{z} x −w z −w ∗ Y ℓ′∈L ℓ′̸=ℓ x −ℓ′N z −ℓ′N 2d+p(ℓ′) (17) ≤(1/4) ∗1.1 ∗2 ∗ln(ek) ∗max{0.9 ∗k 1 2 ∗1.1221k N , 0.24 ∗k 3 2 ∗3.81k N2 , 0.014 ∗k2 ∗9.813k N3 } 38 □ Lemma 13 For k a multiple of twelve, and k ≥12, for all \|ℓ\| ≤k and all \|t\| ≤k/2, we have Pt/Pℓ< 0.924 ∗ √ k ∗(1.1221)k, and, for all \|ℓ\| ≤k and all \|t\| ≤11/12 ∗k, we have Pt/Pℓ< 0.098 ∗ √ k3 ∗(3.81)k, and, for all \|ℓ\| ≤k and all \|t\| ≤k, we have Pt/Pℓ< e ∗0.0051 ∗k2 ∗(9.813)k ≤0.014 ∗k2 ∗9.813k. Proof: We will prove the lemma by upper bounding Pt/P0 in each of the three regions. Along the way, we will also prove that Pt/P0 is lower bounded by one, for all t, 0 ≤t ≤k. We ﬁrst focus on t ≤a1 = k/2. Then, Pt P0 = (a3 −t)!(a3 + t)! a3!a3! 4 ∗ (a1 −t)!(a1 + t)! a1!a1! −1 ∗ (a2 −t)!(a2 + t)! a2!a2! −1 = (a3 −t)!(a3 + t)! a3!a3! 4 ∗ a1!a1! (a1 −t)!(a1 + t)! ∗ a2!a2! (a2 −t)!(a2 + t)!, To upper bound this quantity, we will use Robbins inequalities , which are explicit bound versions of Sterling’s approximation of the factorial function. √ 2π nn+ 1 2 e−ne 1 12n+1 < n! < √ 2π nn+ 1 2 e−ne 1 12n . (18) Using the fact that t ≤k/2, we get Pt P0 ≤e 12 12·1 ∗ ((k + t)(k −t))4 k8 ∗ a2 1 (a1 −t)(a1 + t) ∗ a2 2 (a2 −t)(a2 + t) 1/2 ∗ (19) ((k −t)k−t(k + t)k+t)4 (a1 −t)a1−t(a1 + t)a1+t ∗(a2 −t)a2−t(a2 + t)a2+t ∗ k8k a2a1 1 ∗a2a2 2 !−1 (20) Writing the second factor in (19) as Ct and the quantity (20) as Qt, the above inequality can be written as Pt P0 ≤Ct ∗Qt (21) 39 The same inequalities (18) also yield Pt P0 ≥e 12 12·1+1 ∗Ct ∗Qt (22) For t = a1 = k/2 In section C.1, we show that 1 ≤Ct ≤0.34 √ k. As for Qt, we have Qt = ((k −t)k−t(k + t)k+t)4 ∗(a1)2a1(a2)2a2 (a1 −t)a1−t(a1 + t)a1+t ∗(a2 −t)a2−t(a2 + t)a2+t ∗k8k = ((1 −τ)k−t(1 + τ)k+t)4 (α1 −τ)a1−t(α1 + τ)a1+t ∗(α2 −τ)a2−t(α2 + τ)a2+t ∗(α1)2a1 ∗(α2)2a2 =   ((1 −τ 2)( 1+τ 1−τ )τ)4 (α2 1 −τ 2)α1 α1+τ α1−τ τ ∗(α2 2 −τ 2)α2 α2+τ α2−τ τ ∗(α1)2α1 ∗(α2)2α2   k , (23) where α1 = a1/k = 1/2 and α2 = a2/k = 11/12. In order to ﬁnd the value of τ, 0 < τ ≤1/2, where the above quantity is maximized, we take the derivative of the logarithm of the above quantity (23) w.r.t. τ, using the fact that d dt a ∗log(a2 −t2) + t ∗log a + t a −t = log a + t a −t . The required derivative (w.r.t. τ) is then k ∗ 4 ∗log 1 + τ 1 −τ −log α1 + τ α1 −τ −log α2 + τ α2 −τ Thus, the derivative is zero at roots of (1 + τ)4(α1 −τ)(α2 −τ) −(1 −τ)4(α1 + τ)(α2 + τ) = τ ∗(124 ∗τ 4 −128 ∗τ 2 + 20) Thus, the roots are 0, ± q (16 ± 10 √ 1.01)/31. Other than zero, the only root in the region [0, 1/2] is q (16 −10 √ 1.01)/31 which is approximately 0.438109. It is not diﬃcult to check that this value of τ makes a local maxima of (23). Using this value of τ in (23), we get that for t ≤k/2, Qt < (1.1221)k 40 Since, the above had a single maxima at τ is 0.438109, one can also calculate this expression at τ = 1/2. In that case, the expression has declined to (1.067)k. Thus, Qt is lower bounded in this region by 1 (i.e. its value at t = 0). Hence, for t > 0, 1 ≤Pt/P0 ≤e ∗0.34 √ k ∗(1.1221)k Thus, for all \|ℓ\| ≤k/2 and all \|t\| ≤k/2, we have Pt/Pℓ< 0.34 ∗ √ k ∗e ∗(1.1221)k In the region a1 ≤\|t\| ≤a2, we have Pt P0 ≤e ∗ ((k + t)(k −t))4 k8 ∗a2 1(t −a1) (a1 + t) ∗ a2 2 (a2 −t)(a2 + t) 1/2 ∗ (24) ((k −t)k−t(k + t)k+t)4 (t −a1)a1−t(a1 + t)a1+t ∗(a2 −t)a2−t(a2 + t)a2+t ∗ k8k a2a1 1 ∗a2a2 2 !−1 (25) Writing the second factor in (24) as C(1) t and the quantity in (25) as Q(1) t , we can write the above as (in the region a1 ≤t ≤a2) C(1) t ∗Q(1) t ≤Pt P0 ≤e ∗C(1) t ∗Q(1) t (26) Using the same techniques as above, one can show that in the region a1 ≤\|t\| ≤a2, the above expression (23) (but, with (1/2 −τ) now replaced by (τ −1/2)) has a (single) local minima at τ = 0.5335 (where the value is (1.0285)k, which is more than Q0 (= 1)), and attains its maximum value at \|t\| = (0.916644) ∗k of (3.81)k . Moreover, at τ = 8/12, its value is 1.256. As for C(1) t , we show in Section C.1 that for a1 ≤t ≤8/12, 0.26 ∗ √ k ≤C(1) t ≤0.036 ∗ √ k3, and for 8/12 ≤t ≤11/12, 0.011 ∗k ≤C(1) t ≤0.036 ∗ √ k3. Thus, in this region (a1 ≤t ≤a2), the upper bound is given by Pt/P0 ≤e ∗0.036 √ k3 ∗3.81k. As for the lower bound, we have for a1 ≤t ≤8/12, 0.26 ∗ √ k ∗1.0285k ≤Pt/P0, and for 8/12 ≤t ≤11/12, 0.011 ∗k ∗1.256k ≤Pt/P0. 41 Now, for k ≥12, both these lower bounds are more than 1, and hence in this entire region Pt/P0 > 1. Thus, for all \|ℓ\| ≤11/12 ∗k and all \|t\| ≤k/2, we have Pt/Pℓ< e ∗0.34 ∗ √ k ∗(1.1221)k ≤0.924 ∗ √ k ∗(1.1221)k, and, for all \|ℓ\| ≤11/12 ∗k and all \|t\| ≤11/12 ∗k, we have Pt/Pℓ< 0.098 ∗ √ k3 ∗(3.81)k. Next, in the region a2 ≤\|t\| ≤a3, we have Pt P0 ≤e ∗ ((k + t)(k −t))4 k8 ∗a2 1(t −a1) (a1 + t) ∗a2 2(t −a2) (a2 + t) 1/2 ∗ (27) ((k −t)k−t(k + t)k+t)4 (t −a1)a1−t(a1 + t)a1+t ∗(t −a2)a2−t(a2 + t)a2+t ∗ k8k a2a1 1 ∗a2a2 2 !−1 (28) Writing the second factor in (27) as C(2) t and the quantity in (28) as Q(2) t , we can write the above as (in the region a2 ≤t ≤a3) e 8 12k+1 + 4 13 ∗C(2) t ∗Q(2) t ≤Pt P0 ≤e ∗C(2) t ∗Q(2) t (29) Using the same techniques as above, the above expression (23) (but, now additionally (11/12− τ) replaced by (τ −11/12)) is strictly increasing, attaining its maximum value at \|t\| = k of (9.813)k. Moreover, in Section C.1 we show that in this region, 0.7/ √ k ≤C(2) t ≤.0051 ∗k2. Thus, in this region Pt/P0 > e 8 12k+1 + 4 13 ∗0.7/ √ k ∗3.81k which is greater than 1 (for k ≥12). Thus, for all \|ℓ\| ≤k and all \|t\| ≤k/2, we have Pt/Pℓ< 0.924 ∗ √ k ∗(1.1221)k, and, for all \|ℓ\| ≤k and all \|t\| ≤11/12 ∗k, we have Pt/Pℓ< 0.098 ∗ √ k3 ∗(3.81)k, and, for all \|ℓ\| ≤k and all \|t\| ≤k, we have Pt/Pℓ< e ∗0.0051 ∗k2 ∗(9.813)k ≤0.014 ∗k2 ∗(9.813)k. □ 42 C.1 Analysis of Ct, C(1) t and C(2) t Ct: Recall, for 0 ≤t ≤a1 = k/2, we deﬁned Ct = ((k + t)(k −t))4 k8 ∗ a2 1 (a1 −t)(a1 + t) ∗ a2 2 (a2 −t)(a2 + t) 1/2 Robbin’s bounds (18) only apply for n ≥1, and for n = 0, we have n! = 1. Thus, Ct is more correctly deﬁned as follows5. Ct = ((k + t)(k −t))4 k8 ∗ a2 1 µ(a1 −t) ∗(a1 + t) ∗ a2 2 (a2 −t)(a2 + t) 1/2 where µ(x) = x for x ≥1, and µ(x) = 1 for x < 1. Now, at the boundaries we have C0 = 1 and Ca1 = Ck/2 = ((3/4)4 ∗(1/4)k ∗1/(1 −(6/11)2))1/2 = 0.34 ∗ √ k. For the intermediate region, one can calculate that the roots of the derivative of Ct lie outside this intermediate region. Thus, for k ≥12, the minimum value of Ct is 1 and the maximum value is 0.34 ∗ √ k. C(1) t : We have, for t such that k/2 = a1 ≤t ≤a2 = 11/12 ∗k, C(1) t = ((k + t)(k −t))4 k8 ∗a2 1 ∗µ(t −a1) (a1 + t) ∗ a2 2 µ(a2 −t) ∗(a2 + t) 1/2 . Again, at the boundaries we have, C(1) k/2 = 0.34 ∗ √ k, and C(1) 11/12∗k = (1 −(11/12)2)4 ∗(1/4) ∗11/12 −1/2 11/12 + 1/2 ∗k3 k ∗ (11/12)2 ∗k 11/12 + 11/12 1/2 = 0.005 ∗k3/2 In the intermediate region, instead of computing the roots of the derivative of C(1) t , we compute the extremum of both the numerator and the denominator to bound C(1) t . The derivative of the numerator has a root at t = 0.623 ∗k, whereas the derivative of the denominator has no root in the region. At t = 0.623k, the numerator is (0.00357k13)1/2, which is a local maxima. For k ≥12, the denominator has minimum value at t = a2, 5 No such correction is required in the deﬁnition of Qt, as it is well known that limit of nn, as n tends to 0, is one. 43 which is ((a2 + 1/2) ∗(a2 + a2)k10)1/2. Thus, C(1) t is upper bounded in the intermediate region by 0.036 ∗k3/2. Note that for k ≥12, 0.036 ∗k > 0.34. Thus, the maximum value in the whole region is upper bounded by 0.036 ∗k3/2. The minimum value is lower bounded by splitting the region into two parts: (a) from a1 to 10/12∗k, and (b) from 8/12∗k to a2. In each region, the minimum is lower bounded by taking the ratio of the minimum of the numerator and the maximum of the denominator. Note, that the numerator has a single local extremum (maxima) at 0.623 ∗k, and hence the minimum of the numerator in both regions is at the boundaries. – At t = 11/12 ∗k, the numerator is (5.7 ∗10−5 ∗k13)1/2. – At t = 8/12 ∗k, the numerator is (0.0033 ∗k13)1/2. Thus, the minimum of the numerator in region (b) is at t = 11/12 ∗k. – At t = k/2, the numerator is ((0.0665/k) ∗k13)1/2. Since, for k ≥12, 0.0033 ∗k = 0.04, in region (a) the numerator is lower bounded by ((0.04/k) ∗k13)1/2 Again, since the denominator’s derivative has no zeroes in the entire region, it is a strictly decreasing function for k ≥12. At t = 8/12 ∗k, the denominator is (0.462 ∗k11)1/2. At t = k/2, the denominator is (0.59 ∗k11)1/2. Thus, the minimum is lower bounded in region (a) by 0.26 ∗ √ k. And the minimum is lower bounded in region (b) by 0.011 ∗k. To summarize, for a1 ≤t ≤8/12, 0.011 ∗k ≤C(1) t ≤0.036 ∗ √ k3, and for 8/12 ≤t ≤a2, 0.26 ∗ √ k ≤C(1) t ≤0.036 ∗ √ k3. C(2) t : We have, for t such that 11/12 ∗k = a2 ≤t ≤a3 = k, C(2) t = ((k + t)µ(k −t))4 k8 ∗a2 1 ∗µ(t −a1) (a1 + t) ∗a2 2 ∗µ(t −a2) (a2 + t) 1/2 . Again, at the boundaries we have, C(2) 11/12∗k = 0.005 ∗k3/2, and C(2) k = ((2k)4/k8 ∗ k2/12 ∗(11/12)2k2 ∗(1/23))1/2 = 0.221. Since, k ≥12, the numerator is upper bounded in the region by ((2k)4 ∗(k/12)4 ∗ k3/8 ∗(11/12)2k2 ∗(k/12))1/2. The denominator is lower bounded by (k8 ∗(17/12) ∗k ∗ 44 (22/12)∗k)1/2. Thus, C(2) t is upper bounded by .0051∗k2. Similarly, it is lower bounded by (((1 + 11/12)k)4 ∗k3/8 ∗(11/12)2 ∗k2)1/2 divided by (k8 ∗(3/2) ∗k ∗(23/12 ∗k))1/2, which is 0.7 ∗k−1/2. Since, this is less than 0.221 for k ≥12, we have, for a2 ≤t ≤a3, 0.7/ √ k ≤C(2) t ≤.0051 ∗k2 C.2 Alternate Computer Assisted Proof for k = 12 When k is ﬁxed, for example k = 12, we can compute all possible 12 ∗25 possibilities in (15). The computer-assited proof leads to the following upper bound. For N > k, and k = 12, for every t ∈[−k..k] and x in the t-th interval, X ℓ∈L ℓ̸=t (x −ℓN) ∗ Y ℓ′∈L ℓ′̸=ℓ x −ℓ′N ℓN −ℓ′N 2+p(ℓ′) ≤ N−1 ∗1 2 ∗ln(e ∗k) ∗1.622 ∗max {1, 4.5 ∗105 N , 1.68 ∗109 N2 } D Proof of General Case Theorem Proof: (of Theorem 9) In this section we prove the counterpart of Lemma 4 for the general case. In the next subsection we prove the counterpart of Lemma 5 for the general case. Let’s analyze the ℓ-th summand in gb,c,d(x) deﬁned in (8), assuming \|x −ℓN\| ≤1/2 (and noting z ∈Sℓimplies \|x −z\| ≤1, and writing x′ for (x −ℓN) and z′ for (z −ℓN)) X z∈Sℓ fN(z) ∗   Y w∈Sℓ{z} x −w z −w  ∗ Y ℓ′∈L ℓ′̸=ℓ 1 + x′ −z′ z′ + (ℓ−ℓ′)N d+c+b∗pℓ′) (30) 45 Focusing on the second product we have Y ℓ′∈L ℓ′̸=ℓ 1 + x′ −z′ z′ + (ℓ−ℓ′)N d+c+b∗pℓ′) Y ℓ′∈L ℓ′̸=ℓ 1 + z′ −x′ (ℓ′ −ℓ)N(1 −z′/((ℓ−ℓ′)N) d+c+b∗pℓ′) Y ℓ′∈L ℓ′̸=ℓ 1 + (z′ −x′) (ℓ′ −ℓ)N ∗(1 −z′/((ℓ−ℓ′)N))−1 d+c+b∗pℓ′) (31) Since fN(z) = 0, for z′ = z −ℓN = 0, we can exclude such z from the sum in (30), and then the power series expansion of (1 −z′/((ℓ−ℓ′)N))−1 is convergent. So, suppose the above (31) is 1 + X i>0 X j≥0 ai,j(z′ −x′)i ∗(z′)j ∗N−i−j Then, (30) can be written as X z∈Sℓ fN(z) ∗   Y w∈Sℓ{z} x −w z −w  ∗  1 + X i>0 X j≥0 ai,j(z′ −x′)i ∗(z′)j ∗N−i−j   = X z∈Sℓ fN(z) ∗   Y w∈Sℓ{z} x −w z −w  ∗  1 + X i>0 X j≥0,i+j≥d ai,j(z′ −x′)i ∗(z′)j ∗N−i−j   = (x −ℓN) + X z∈Sℓ fN(z) ∗   Y w∈Sℓ{z} x −w z −w  ∗ X i>0 X j≥0,i+j≥d ai,j(z′ −x′)i ∗(z′)j ∗N−i−j, where the ﬁrst equality follows from Lemma 2, and the second by Lagrange interpolation over 46 Sℓ. Now, consider the following alternate expression (instead of (31)) Y ℓ′∈L ℓ′̸=ℓ 1 + z′ −x′ N ∗(1 −z′/N)−1 d+c+b∗pℓ′) = 1 + z′ −x′ N ∗(1 −z′/N)−1 (d+c)∗2k+b∗P ℓ′̸=ℓp(ℓ′) = 1 + X i>0 X j≥0 bi,j(z′ −x′)i ∗(z′)j ∗N−i−j for some bi,j. Since \|(ℓ−ℓ′)\| ≥1, it follows that for all i, j, \|bi,j\| ≥\|ai,j\|. Then (30) diﬀers from (x −ℓN) in absolute value by at most X z∈Sℓ \|fN(z)\| ∗ Y w∈Sℓ{z} x −w z −w ∗ X i>0 X j≥0,i+j≥d \|ai,j\| ∗\|z′ −x′\|i ∗\|z′\|j ∗N−i−j ≤ X z∈Sℓ \|fN(z)\| ∗ Y w∈Sℓ{z} x −w z −w ∗ X i>0 X j≥0,i+j≥d \|bi,j\| ∗\|z′ −x′\|i ∗\|z′\|j ∗N−i−j (32) Now, \|bi,j\| is itself upper bounded by (let m = (d + c) ∗2k + b ∗ν ≥d) m i i −1 + j j 47 Thus, we have the following inequalities X i>0 X j≥0,i+j≥d \|bi,j\| ∗\|z′ −x′\|i ∗\|z′\|j ∗N−i−j ≤ X m≥i>0 X j≥0,i+j≥d m i i −1 + j j ∗2−j ∗N−i−j = X j≥0 X m≥i>0,i+j≥d m i i −1 + j j ∗2−j ∗N−i−j = X j≥0 X m+j≥i+j>j,i+j≥d m i i −1 + j j ∗2−j ∗N−i−j = X j≥0 X m+j≥i>j,i≥d m i −j i −1 j ∗2−j ∗N−i = X i≥d N−i ∗ X m+j≥i>j≥0 m i −j i −1 j ∗2−j ≤ X i≥d N−i ∗ X m+j≥i>j≥0 (m)i−j (i −j)! i j ∗2−j ≤ X i≥d N−i ∗ X i≥j≥0 (m)i−j (i −j)! ∗2i−j ≤ X i≥d N−i ∗ X i≥j≥0 j! i! ∗ i j ∗(2m)i−j ≤ X i≥d N−i ∗ X i≥j≥0 i j ∗ e i i−j ∗(2m)i−j ≤ X i≥d N−i ∗ 1 + 2em i i ≤ X i≥d N−i ∗ 1 + 2em d i ≤N−d ∗ 1 + 2em d d X i≥0 N−i ∗ 1 + 2em d i ≤2 ∗N−d ∗ 1 + 2em d d Thus, using (32), and the upper bound on Lagrange basis polynomials (see Section A), we 48 have that (30) diﬀers from (x −ℓN) in absolute value by at most (d + 1) ∗2d ∗N−d ∗ 1 + 2em d d , (33) where m = (d + c) ∗2k + b ∗ν. This proves the counterpart of Lemma 4 for the general case. D.1 General Upper Bound on x in another interval In this subsection we prove the counterpart of Lemma 5 for the general case. For this part of the prove, we will assume that d + c = 2b, as that makes the proof simpler. Let x be in the t-th interval, \|x −t ∗N\| < 1/2, and let ℓ̸= t, for t, ℓ∈L. Let Rℓ(x) △ = Y ℓ′∈L ℓ′̸=ℓ x −ℓ′N (ℓ−ℓ′)N d+c+b∗p(ℓ′) , and Lℓ,z(x) △ = fN(z) ∗ Y w∈Sℓ{z} x −w z −w . Then, for z ∈Sℓ, writing z = z′ + ℓN, X z∈Sℓ fN(z) ∗ Y w∈Sℓ{z} x −w z −w ∗ Y ℓ′∈L ℓ′̸=ℓ x −ℓ′N z′ + (ℓ−ℓ′)N d+c+b∗p(ℓ′) = Rℓ(x) ∗ X z∈Sℓ Lℓ,z(x) ∗ Y ℓ′∈L ℓ′̸=ℓ 1 1 + z′/((ℓ−ℓ′)N) d+c+b∗p(ℓ′) = Rℓ(x) ∗    (x −ℓN) + X z∈Sℓ Lℓ,z(x) ∗    −1 + Y ℓ′∈L ℓ′̸=ℓ 1 1 + z′/((ℓ−ℓ′)N) d+c+b∗p(ℓ′)         (34) where the last equality follows by Lagrange interpolation of (x −ℓN). Now consider the second summand inside the big parenthesis: X z∈Sℓ Lℓ,z(x) ∗    −1 + Y ℓ′∈L ℓ′̸=ℓ 1 + z′/((ℓ−ℓ′)N) −(d+c+b∗p(ℓ′))     49 Since fN(ℓN) = 0, in the above sum we can exclude z such that z = ℓN, i.e. z′ = 0. Moreover, \|z′\|/N < 1, and hence the power series of the big product above is convergent. Suppose the power series of the above product is 1 + P i>0 ai ∗(z′/N)i, then the absolute value of above can be written as X z∈Sℓ Lℓ,z(x) ∗ X i>0 ai ∗(z′/N)i = X d>i>0 ai ∗(x −ℓN)i ∗N−i + X z∈Sℓ Lℓ,z(x) ∗ X i≥d ai ∗(z′/N)i ≤ X d>i>0 \|ai\| ∗\|x −ℓN\|i ∗N−i + X z∈Sℓ \|Lℓ,z(x)\| ∗ X i≥d \|ai\| ∗\|z′/N\|i (35) Now, consider an alternate expression, where m = (d + c) ∗2k + b ∗ν ≥d, X z∈Sℓ Lℓ,z(x) ∗    −1 + Y ℓ′∈L ℓ′̸=ℓ 1 −z′/N −(d+c+b∗p(ℓ′))    = X z∈Sℓ Lℓ,z(x) ∗ −1 + 1 −z′/N −m and let the convergent power series of the second factor above be P i>0 bi ∗(z′/N)i. Note for every i, bi ≥0. Indeed, bi = m+i−1 m . Also, for each i, bi = \|bi\| ≥\|ai\|, as \|ℓ−ℓ′\| ≥1. 50 Thus, (35) is upper bounded as follows, for N > m+d+1 (d+1) , X d>i>0 \|ai\| ∗\|x −ℓN\|i ∗N−i + X z∈Sℓ \|Lℓ,z(x)\| ∗ X i≥d \|ai\| ∗\|z′/N\|i ≤ X d>i>0 bi ∗\|x −ℓN\|i ∗N−i + X z∈Sℓ \|Lℓ,z(x)\| ∗ X i≥d bi ∗\|z′/N\|i ≤ X d>i>0 bi ∗\|1 + (t −ℓ)N\|i ∗N−i + 2d ∗\|1 + (t −ℓ)N\|d ∗ X i≥d bi ∗\|z′/N\|i ≤ X d>i>0 m + i −1 m ∗\|N−1 + (t −ℓ)\|i + (d + 1) ∗2d ∗\|1 + (t −ℓ)N\|d ∗ X i≥d N−i ∗ m + i −1 m ∗2−i ≤ X d>i>0 m + i −1 i −1 ∗\|N−1 + (t −ℓ)\|i + (d + 1) ∗\|N−1 + (t −ℓ)\|d ∗ X i≥0 N−i ∗ m + d + i −1 m ∗2−i ≤ m + d −2 d −2 ∗ X d>i>0 \|N−1 + (t −ℓ)\|i + (d + 1) ∗\|N−1 + (t −ℓ)\|d ∗ X i≥0 (2N)−i ∗ m + d m ∗ m + d + 1 d + 1 i−1 ≤ m + d −2 d −2 ∗\|N−1 + (t −ℓ)\|d + (d + 1) ∗\|N−1 + (t −ℓ)\|d ∗ m + d m ∗ 1 1 −m+d+1 (d+1)∗2N ≤(2d + 3) ∗\|N−1 + (t −ℓ)\|d ∗ m + d m ≤(2d + 3) ∗(2k + 1)d ∗ m + d d where we used bounds on Lagrange basis polynomials from Section A and the following in-equality (which is easily proved by induction on i ≥0), for m ≥s, m + i s ≤ m + 1 m + 1 −s i ∗ m s . Thus, 34 is upper bounded by \|Rℓ(x)\| ∗ (x −ℓN) + (2d + 3) ∗(2k + 1)d ∗ m + d d ≤\|Rℓ(x)\| ∗ 2k ∗N + 1 + (2d + 3) ∗((2k + 1) ∗e ∗(m + d)/d)d (36) 51 Now, Rℓ(x) is same as 13 in proof of Lemma 5. So, it is bounded similarly as follows, by writing x = x′ + tN, and recalling t ̸= ℓ, Rℓ(x) = Y ℓ′∈L ℓ′̸=ℓ x −ℓ′N (ℓ−ℓ′)N d+c+b∗p(ℓ′) = x′ (ℓ−t)N d+c+b∗p(t) ∗ Y ℓ′∈L ℓ′̸=ℓ ℓ′̸=t x −ℓ′N (ℓ−ℓ′)N d+c+b∗p(ℓ′) = x′ (ℓ−t)N d+c+b∗p(t) ∗ Y ℓ′∈L ℓ′̸=ℓ ℓ′̸=t (t −ℓ′)N (ℓ−ℓ′)N d+c+b∗p(ℓ′) ∗ Y ℓ′∈L ℓ′̸=ℓ ℓ′̸=t x′ + (t −ℓ′)N (t −ℓ′)N d+c+b∗p(ℓ′) (37) Now, we upper bound the third product as in Lemma 4, and the second product as in Lemma 5. Here we will assume that d + c = 2b, as that makes the proof simpler. Starting with the third product, just as in the proof of Lemma 4 (even though one factor is missing now, namely ℓ′ = t) Y ℓ′∈L ℓ′̸=ℓ ℓ′̸=t x′ + (t −ℓ′)N (t −ℓ′)N b(2+∗p(ℓ′)) ≤ 1 + 4 ∗ln(ek) N + 4k2 N2 b (38) For the ﬁrst two products above, following proof of Lemma 5, and in particular (16), we have x′ (ℓ−t)N 2d+d∗p(t) ∗ Y ℓ′∈L ℓ′̸=ℓ ℓ′̸=t (t −ℓ′)N (ℓ−ℓ′)N 2b+b∗p(ℓ′) ≤(1/2)2b+b∗p(t) 1 \|ℓ−t\|N 2b+b∗p(ℓ) ∗ Pt Pℓ b ≤(2\|l −t\|N)−2b ∗ Pt Np(ℓ)Pℓ b ≤(2\|l −t\|N)−2b ∗ max{0.9 ∗k 1 2 ∗1.1221k, 0.24 ∗k 3 2 ∗3.81k N , 0.014 ∗k2 ∗9.813k N2 } b 52 Thus, Rℓ(x) is upper bounded by 1 + 4∗ln(ek) N + 4k2 N2 4\|l −t\|2N2 !b ∗ max{0.9 ∗k 1 2 ∗1.1221k, 0.24 ∗k 3 2 ∗3.81k N , 0.014 ∗k2 ∗9.813k N2 } b Thus, using (36) we have that (34) is upper bounded by 2k ∗N + 1 + (2d + 3) ∗((2k + 1) ∗e ∗(m + d)/d)d ∗ 1 + 4∗ln(ek) N + 4k2 N2 4\|l −t\|2N2 !b ∗ max{0.9 ∗k 1 2 ∗1.1221k, 0.24 ∗k 3 2 ∗3.81k N , 0.014 ∗k2 ∗9.813k N2 } b This completes the proof of the counterpart of Lemma 5 for the general case. □ To simplify the presentation of the above result, note that max{0.9 ∗k 1 2 ∗1.1221k, 0.24 ∗k 3 2 ∗3.81k N , 0.014 ∗k2 ∗9.813k N2 = 0.9 ∗ √ k ∗1.1221k ∗max {1, 0.22 ∗k ∗3.4k N , 0.015 ∗k 3 2 ∗8.745k N2 } Thus, if we assume N > 0.22∗k∗3.4k, k ≥12, b = ⌈d/2⌉, and also noting that m = b∗(4k+ν), the above upper bound simpliﬁes to 2k ∗N + (2d + 3) ∗(33 ∗k2)d ∗ √ k ∗1.1221k 2\|l −t\|2N2 !b Combining with the upper bound (33), we have that for the mod function fN(·), for any d > 0, N > 0.22 ∗k ∗3.4k, k ≥12, k a multiple of twelve, for any x such that \|x −t ∗N\| ≤1/2 with t ∈{−k, −k + 1, . . . , k}, we have \|g2d(x) −fN(x)\| < 22d ∗N−2d ∗ 1 + 2em 2d 2d + N−2d ∗ 4k ∗N + (4d + 6) ∗(33 ∗k2)d ∗ √ k ∗1.1221k/2 d and \|g2d+1(x) −fN(x)\| < 22d+1 ∗N−2d−1 ∗ 1 + 2em 2d + 1 2d+1 + N−2d−1 ∗ 4k + (4d + 10) ∗(33 ∗k2)d+1/N ∗ √ k ∗1.1221k/2 d+1 53 E Proof of Theorem 10 for Small N Proof: (of Theorem 10) The proof that X z∈Sℓ fN(z) ∗   Y w∈Sℓ{z} x −w z −w  ∗ Y ℓ′∈L ℓ′̸=ℓ x −ℓ′N (ℓ−ℓ′)N 2+p(ℓ′) (39) diﬀers from fN(x) = x′ (with x in the ℓ-th interval) in absolute value by at most 2∗ln(ek) N + 2k2 N2 (now with k = 12), remains the same as the proof of Lemma 4. The proof of a lemma similar to Lemma 5 is also similar except for the bounds on Pt/Pℓ, for t, ℓ∈[−12..12]. Just as in the proof of Lemma 5, we have similar to (15) (recalling, \|x −tN\| ≤1/2 for t ∈{−k..k}, t ̸= ℓ, and x = x′ + tN, and now k = 12) X z∈Sℓ fN(z) ∗ Y w∈Sℓ{z} x′ + t ∗N −w z −w ∗ Y ℓ′∈L ℓ′̸=ℓ x′ + t ∗N −ℓ′N ℓN −ℓ′N 2+p(ℓ′) ≤1 4 ∗(1 + 4 ∗ln(ek) N + 4k2 N2 ) ∗N−1−p(ℓ,t) ∗\|t −ℓ\|−1−p(t) ∗ Y ℓ′∈L ℓ′̸=ℓ ℓ′̸=t \|ℓ′ −t\| \|ℓ′ −ℓ\| 2+p(ℓ′) (40) For each of m = 0..4, for all t such that p(t) = m, we can calculate the upper bound on Γm △ = \|t −ℓ\|−1−p(t) ∗ Y ℓ′∈L ℓ′̸=ℓ ℓ′̸=t \|ℓ′ −t\| \|ℓ′ −ℓ\| 2+p(ℓ′) A simple computer program (given at the end of this section) shows that Γ0 ≤1, Γ1 ≤60, Γ2 ≤2919 and Γ4 ≤1.23 ∗107. Note Γ3 is undeﬁned. Thus, X z∈Sℓ fN(z) ∗ Y w∈Sℓ{z} x′ + t ∗N −w z −w ∗ Y ℓ′∈L ℓ′̸=ℓ x′ + t ∗N −ℓ′N ℓN −ℓ′N 2+p(ℓ′) ≤N−1 ∗(1 4 + 3.5 N + 144 N2 ) ∗max {1, 60 N , 2919 N2 , 1.23 ∗107 N4 } (41) 54 Combining (39) and (41) we have For the mod function fN(·), for k = 12, N ≥320, for any x such that \|x −t ∗N\| ≤1/2 with t ∈{−k, −k + 1, . . . , k}, we have that ˆ g deﬁned in (6) is an odd polynomial with deg(ˆ g) = 69, and \|ˆ g(x) −fN(x)\| < 1 N ∗ 1 4 + 3.5 N + 144 N2 ∗ 2 + max {1, 60 N , 2919 N2 , 1.23 ∗107 N4 } □ Below we provide the computer program written in the language Macaulay 2 to upper bound the various Γ. k=12; for i from 0 to 4 do gamma_(i) =0; for t from 0 to k do ( plt = 0; if (t > 6) then plt = plt +1; if (t > 10) then plt = plt +1; if (t > 11) then plt = plt +2; p_t = plt; ); for ell from -k to k do ( for t from 0 to k do ( f = 1; for ellp from -1k to k do ( absl = abs(ellp); if ((ellp !=ell) and ( ellp !=t)) then ( f = f (abs(ellp - t)/abs(ellp-ell))^(2+p_(absl)); ); ); f = f/(abs(t -ell))^(1 + p_t); if (f > gamma_(p_t) ) then gamma_(p_t) = f; ); ); for i from 0 to 4 do << "max Gamma at " << i << " is " << toRR(32,gamma_(i)) << "\n"; 55 E.1 General Case Following the proof in section D, we also have in the general case that the mod function fN(·), for any d > 0, b = ⌈d/2⌉, N > 60, k = 12, for any x such that \|x −t ∗N\| ≤1/2 with t ∈{−k, −k + 1, . . . , k}, we have \|gd(x) −fN(x)\| ≤2d ∗N−d ∗ 1 + 2em d d + 2k ∗N + 1 + (2d + 3) ∗((2k + 1) ∗e ∗(m + d)/d)d ∗ 1 + 4∗ln(ek) N + 4k2 N2 4\|l −t\|2N2 !b ∗ max {1, 60 N , 2919 N2 , 1.23 ∗107 N4 } b ≤2d ∗N−d ∗(372b/d)d + 24 ∗N + 1 + (2d + 3) ∗(2310)d ∗N−2b F Lower Bound on the Standard Error Analysis for Lagrange Interpolation (cont’d) Recall the following well-known theorem regarding the error of polynomial interpolation from Section 2. Theorem 14 (Polynomial Interpolation) Let f be an n + 1 times diﬀerentiable function on [a, b] and pn be a polynomial of degree ≤n that interpolates f at n + 1 distinct points t0, t1, . . . , tn ∈[a, b], meaning pn(ti) = f(ti) for all 0 ≤i ≤n. Then, for each t ∈[a, b], there exists a point ψt ∈[a, b] such that f(t) −pn(t) = f(n+1)(ψt) (n + 1)! · n Y i=0 (t −ti). Error analysis for Lagrange interpolation proceeds by upper bounding f(n+1)(·) on [a, b] and then bounding Qn i=0(t −ti). We are interested in approximating S(t) = sin(2πkt) on the 2k + 1 intervals of width 2ϵ/k centered at (1/k) ∗[−k, . . . , k]. Observe that since sin x and cos x lie between −1 and 1, we can bound \|S(n+1)(ψt)\| ≤(2πk)n+1 56 for any ψt. So the error analysis will be able to bound the error between S(t) and the polynomial pn(t) by \|S(t) −pn(t)\| ≤(2πk)n+1‘ (n + 1)! · n Y i=0 (t −ti) . We will now show that no matter how the ti’s are chosen in these small intervals6, the resulting error bound will be worse than that obtained via modular Lagrange interpolation. Let di be the number of points for interpolation chosen in the interval centered at i/k for i ∈{−k, . . . , k}. Then, the points in the ith interval can be written as ti,j = i/k + δi,j for δi,j ∈[−ϵ/k, ϵ/k] and 1 ≤j ≤di. It follows that n + 1 = Pk −k di. Consider an arbitrary point tℓ= ℓ/k + δ in the interval centered around ℓ/k. Then, k Y i=−k di Y j=1 (tℓ−ti,j) = k Y i=−k di Y j=1 \|(ℓ/k −i/k + δ −δi,j)\| ≥   dℓ Y j=1 \|(δ −δℓ,j)\|     k Y i=−k,i̸=ℓ di Y j=1 (1/k) ∗(\|ℓ−i\| −2ϵ)   ≥ ϵ k ∗dℓ dℓ   k Y i=−k,i̸=ℓ di Y j=1 (1/k) ∗(\|ℓ−i\| −2ϵ)   for the worst-case δ ∈[−ϵ/k, ϵ/k] when dℓ≥1. Considering only the highest order in ϵ error term, we will approximate (\|ℓ−i\| −2ϵ) with \|ℓ−i\|. To obtain an O(ϵ2) approximation (consistent with our O(1/N) approximation), we must set di ≥2 for all i. However, if we set di ≥3, then we exceed the degree of our O(1/N) approximation in Section 4. Let w(i) be a function that represents the number of points beyond 2 that are in the ith interval. That is, w(i) = di −2. Let ν = Pk i=−k w(i). Thus, n + 1 = 2 ∗(2k + 1) + ν. Then, for a worst-case interval ℓ(one where dℓ= 2), the above error approximation gives (2π)n+1 (n + 1)! ∗ ϵ 2 2 ∗   k Y i=−k,i̸=ℓ (\|ℓ−i\|)2  ∗   k Y i=−k,i̸=ℓ (\|ℓ−i\|)w(i)  . 6One could consider taking points not inside these intervals, but this can be seen to be suboptimal if the number of points is even. Since approximations to the mod function are odd polynomials, the number of points chosen for interpolation will be even. 57 Since Qk i=−k,i̸=ℓ(\|ℓ−i\|)2 is minimized when ℓ= 0 and increases as \|ℓ\| increases, a reasonable assumption to minimize the error across all intervals is that w(i) is an increasing function with \|i\| and is symmetric (w(i) = w(−i)). Let α denote the largest \|i\| for which w(i) = 0. Then, the error in the 0th interval can be bounded by (2π)4k+2+ν (4k + 2 + ν)! ∗ ϵ 2 2 ∗(k!)4 ∗ k! α! 2 ∗(α + 1)ν−2(k−α) ≥ (2π)4k+2+ν (4k + 2 + ν)2+ν ∗ ϵ 2 2 ∗(k!)4 (4k)! ∗ k! α! 2 ∗(α + 1)ν−2(k−α) ≈ (2π)4k+2+ν (4k + 2 + ν)2+ν ∗ ϵ 2 2 ∗1 44k ∗ k! α! 2 ∗(α + 1)ν−2(k−α) = ϵ2 ∗ π4k+2+ν 24k−ν ∗(4k + 2 + ν)2+ν ∗ k! α! 2 ∗(α + 1)ν−2(k−α) When k is a multiple of 12, we give polynomials in Section 4 with ν = 14(k/12). Thus, the minimum possible value of α is k −7(k/12) = 5(k/12). The lower bound then becomes ϵ2 ∗ π(31k/6)+2 2(17k/6) ∗((31k/6) + 2)2+7(k/6) ∗ k! (5k/12)! 2 . For k = 12, this gives a lower bound of 769137ϵ2. For comparison, we obtained a bound of 75ϵ2. Note that this lower bound only focused on the error for the 0th interval, and this error is already larger than the error of our polynomials. G Explicit Polynomials In this section, we give some explicit polynomials for approximating the mod function obtained via modular Lagrange interpolation. The following (odd) polynomials are listed with the constant term ﬁrst and the highest degree term last. Suppose q = N ∗M, where M is the maximum size of the message m, and the input x is of the form I ∗q + m, for I ∈[−12..12]. Then, the input x should be ﬁrst normalized by dividing by 12q + 1, so as to be evaluated by Chebyshev polynomials. Call the normalized input x′. Next, the various (odd) Chebyshev basis polynomials T2j+1, with 1 ≤2j + 1 ≤d, should be evaluated on x′ upto the prescribed precision. Here, d is the degree of the polynomial. The result, which approximates m, is P(d−1)/2 j=0 a2j+1 ∗T2j+1. 58 G.1 N = 1024 Polynomial for N = 1024, d = 7, c = 0, b = 3, max-error = 2−49 (additive to m/q), degree d = 209, precision required for Cheby polys: 109 bits, for message space of precision 43 bits (i.e. log M). (the following odd polynomial is listed with the constant term ﬁrst and the highest degree term last): 0, 1168086454234386721728786093, 0, 857343731807213129276527371, 0, 321577746951483724417855166, 0, −292231639690436254023249313, 0, −817558357077948111433871545, 0, −1115078287892665095295847460, 0, −1110902683997716285124711721, 0, −815466383301985986929060172, 0, −318118048203314366189979744, 0, 240336163193778344064042767, 0, 708546663412296489590609529, 0, 966802652802823190347418980, 0, 958410990971409284025257086, 0, 701859806234615738681506707, 0, 281315417920828815838724026, 0, −180619100645381391833095478, 59 0, −559074895353411112211879488, 0, −761193870352871464575970113, 0, −748961547193141244456643106, 0, −545019098411792243865207141, 0, −221182605886666986588184985, 0, 125631214744033173329756970, 0, 402176605497046862164911192, 0, 543822384945637454963785227, 0, 529041122166502663418034057, 0, 380454231175936390963182110, 0, 153583874550410836778346038, 0, −82158926416441517047319102, 0, −264016877090445503291895215, 0, −352017418757438912124391674, 0, −336726247305874507782050670, 0, −237542040580981854329586741, 0, −93285861002735563557789426, 0, 51203641438492931279476695, 0, 158138557109275412094214802, 0, 205902506436761190012091838, 0, 192272901361176496521724271, 0, 131771332034736267678084566, 0, 48836718997887305615420090, 0, −30567669476018418201737430, 0, −86288586492669414092635654, 0, −108382466605506825823721887, 0, −97870349385353469283431522, 0, −64308240709636355867793447, 0, −21461468941244398795157840, 0, 17326592921743672239788373, 0, 42700061643388303938158218, 0, 50996981335404100973959652, 0, 43999901398812509501638598, 0, 27218306023500888628165559, 0, 7511351204331858459027720, 0, −9124894335827475847590487, 0, −19011264986878466920308735, 0, −21240027011005272177515640, 0, −17235488182308094169059958, 0, −9762047684096432167581061, 0, −1801405931639871012067336, 0, 4343202668899288644999705, 0, 7513256624328692928829408, 0, 7707337571415601635186575, 0, 5756436611224478681862020, 0, 2848796879491757763177843, 0, 85869737980867178544936, 0, −1808098547283706804642313, 0, −2580773528142545311733665, 0, −2379644644727334869970039, 0, −1584099449739462886296632, 0, −620801233570437909456226, 0, 177002015939962120138178, 0, 637322616908069820961416, 0, 746708447295636591621208, 0, 600320533243730473229437, 0, 335551277339171148995831, 0, 75303469471198484934870, 0, −104484715159419839814631, 0, −181085471349235784144282, 0, −172186740047558875429158, 0, −114989546962237138479380, 0, −46933405878291066774262, 0, 6278911197203248699968, 0, 34210824196182898522253, 0, 38872665499010244728055, 0, 28960040671209758802896, 0, 14214605200094274586116, 0, 1734974669062679927928, 0, −5301450813792947532094, 0, −7037641220180290308568, 0, −5423997599405783363784, 0, −2702130219676810140085, 0, −440520134223430783698, 0, 745852086848961038498, 0, 975349130781356798319, 0, 687152308562330003311, 0, 294829853406611648652, 0, 23932836640143372608, 0, −83474940988996328415, 0, −83385665165129747914, 0, −46136189137266827627, 0, −14190050009111049552, 0, 932967800085450602, 0, 4059717598672962384, 0, 2677340276445438938, 0, 1049770524170609837, 0, 256678428266378497, 0, 31549863120212428. 60 G.2 N = 220 Polynomial for N = 220, d = 3, c = 0, b = 1, max-error = 2−36 (additive to m/q), degree d = 63, precision required for Cheby polys: 145 bits, for message space of precision 20 bits (i.e. log M). (the following odd polynomial is listed with the constant term ﬁrst and the highest degree term last): 0, −1018138773718856523, 0, −792255707788299537, 0, −398036457380237928, 0, 65464815475745761, 0, 485190908761807690, 0, 765021604222736345, 0, 850929892529953988, 0, 742813330658976390, 0, 489822309534211986, 0, 171645897353125137, 0, −127102284142356358, 0, −340499429680023172, 0, −435817908378521076, 0, −416018785796205163, 0, −311821255067144991, 0, −167840365987534680, 0, −27972299265682149, 0, 75830917899224394, 0, 129111891480935373, 0, 134457712479775307, 0, 106377864700523206, 0, 63925757517659405, 0, 23779346922819549, 0, −4084115434202789, 0, −17254849720858790, 0, −18883394301970980, 0, −14376926743580044, 0, −8506558748904986, 0, −3962809189252832, 0, −1406881258274829, 0, −347428595975786, 0, −46018339611917. 61
1968	https://pressbooks.bccampus.ca/introductorygeneralphysics2phys1207/chapter/22-11-more-applications-of-magnetism/	Skip to content Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. Chapter 7 Magnetic field produced by moving electric charges 7.2 More Applications of Magnetism – Mass spectrometry and MRI Summary Describe some applications of magnetism. Mass Spectrometry The curved paths followed by charged particles in magnetic fields can be put to use. A charged particle moving perpendicular to a magnetic field travels in a circular path having a radius r It was noted that this relationship could be used to measure the mass of charged particles such as ions. A mass spectrometer is a device that measures such masses. Most mass spectrometers use magnetic fields for this purpose, although some of them have extremely sophisticated designs. Since there are five variables in the relationship, there are many possibilities. However, if v, q, and Bcan be fixed, then the radius of the path r is simply proportional to the mass mof the charged particle. Let us examine one such mass spectrometer that has a relatively simple design. (See Figure 1.) The process begins with an ion source, a device like an electron gun. The ion source gives ions their charge, accelerates them to some velocity v, and directs a beam of them into the next stage of the spectrometer. This next region is a velocity selector that only allows particles with a particular value of v to get through. The velocity selector has both an electric field and a magnetic field, perpendicular to one another, producing forces in opposite directions on the ions. Only those ions for which the forces balance travel in a straight line into the next region. If the forces balance, then the electric force F = qE equals the magnetic force F=qvB, so that qe=qvB. Noting that q cancels, we see that is the velocity particles must have to make it through the velocity selector, and further, that v can be selected by varying E and B. In the final region, there is only a uniform magnetic field, and so the charged particles move in circular arcs with radii proportional to particle mass. The paths also depend on charge q, but since qis in multiples of electron charges, it is easy to determine and to discriminate between ions in different charge states. Mass spectrometry today is used extensively in chemistry and biology laboratories to identify chemical and biological substances according to their mass-to-charge ratios. In medicine, mass spectrometers are used to measure the concentration of isotopes used as tracers. Usually, biological molecules such as proteins are very large, so they are broken down into smaller fragments before analyzing. Recently, large virus particles have been analyzed as a whole on mass spectrometers. Sometimes a gas chromatograph or high-performance liquid chromatograph provides an initial separation of the large molecules, which are then input into the mass spectrometer. Cathode Ray Tubes—CRTs—and the Like What do non-flat-screen TVs, old computer monitors, x-ray machines, and the 2-mile-long Stanford Linear Accelerator have in common? All of them accelerate electrons, making them different versions of the electron gun. Many of these devices use magnetic fields to steer the accelerated electrons. Figure 2 shows the construction of the type of cathode ray tube (CRT) found in some TVs, oscilloscopes, and old computer monitors. Two pairs of coils are used to steer the electrons, one vertically and the other horizontally, to their desired destination. Figure 2. The cathode ray tube (CRT) is so named because rays of electrons originate at the cathode in the electron gun. Magnetic coils are used to steer the beam in many CRTs. In this case, the beam is moved down. Another pair of horizontal coils would steer the beam horizontally. Magnetic Resonance Imaging Magnetic resonance imaging (MRI) is one of the most useful and rapidly growing medical imaging tools. It non-invasively produces two-dimensional and three-dimensional images of the body that provide important medical information with none of the hazards of x-rays. MRI is based on an effect called nuclear magnetic resonance (NMR) in which an externally applied magnetic field interacts with the nuclei of certain atoms, particularly those of hydrogen (protons). These nuclei possess their own small magnetic fields, similar to those of electrons and the current loops discussed earlier in this chapter. When placed in an external magnetic field, such nuclei experience a torque that pushes or aligns the nuclei into one of two new energy states—depending on the orientation of its spin (analogous to the N pole and S pole in a bar magnet). Transitions from the lower to higher energy state can be achieved by using an external radio frequency signal to “flip” the orientation of the small magnets. (This is actually a quantum mechanical process. The direction of the nuclear magnetic field is quantized as is energy in the radio waves. We will return to these topics in later chapters.) The specific frequency of the radio waves that are absorbed and reemitted depends sensitively on the type of nucleus, the chemical environment, and the external magnetic field strength. Therefore, this is a resonance phenomenon in which nuclei in a magnetic field act like resonators (analogous to those discussed in the treatment of sound in Chapter 16 Oscillatory Motion and Waves) that absorb and reemit only certain frequencies. Hence, the phenomenon is named nuclear magnetic resonance (NMR). NMR has been used for more than 50 years as an analytical tool. It was formulated in 1946 by F. Bloch and E. Purcell, with the 1952 Nobel Prize in Physics going to them for their work. Over the past two decades, NMR has been developed to produce detailed images in a process now called magnetic resonance imaging (MRI), a name coined to avoid the use of the word “nuclear” and the concomitant implication that nuclear radiation is involved. (It is not.) The 2003 Nobel Prize in Medicine went to P. Lauterbur and P. Mansfield for their work with MRI applications. The largest part of the MRI unit is a superconducting magnet that creates a magnetic field, typically between 1 and 2 T in strength, over a relatively large volume. MRI images can be both highly detailed and informative about structures and organ functions. It is helpful that normal and non-normal tissues respond differently for slight changes in the magnetic field. In most medical images, the protons that are hydrogen nuclei are imaged. (About 2/3 of the atoms in the body are hydrogen.) Their location and density give a variety of medically useful information, such as organ function, the condition of tissue (as in the brain), and the shape of structures, such as vertebral disks and knee-joint surfaces. MRI can also be used to follow the movement of certain ions across membranes, yielding information on active transport, osmosis, dialysis, and other phenomena. With excellent spatial resolution, MRI can provide information about tumours, strokes, shoulder injuries, infections, etc. An image requires position information as well as the density of a nuclear type (usually protons). By varying the magnetic field slightly over the volume to be imaged, the resonant frequency of the protons is made to vary with position. Broadcast radio frequencies are swept over an appropriate range and nuclei absorb and reemit them only if the nuclei are in a magnetic field with the correct strength. The imaging receiver gathers information through the body almost point by point, building up a tissue map. The reception of reemitted radio waves as a function of frequency thus gives position information. These “slices” or cross sections through the body are only several mm thick. The intensity of the reemitted radio waves is proportional to the concentration of the nuclear type being flipped, as well as information on the chemical environment in that area of the body. Various techniques are available for enhancing contrast in images and for obtaining more information. Scans called T1, T2, or proton density scans rely on different relaxation mechanisms of nuclei. Relaxation refers to the time it takes for the protons to return to equilibrium after the external field is turned off. This time depends upon tissue type and status (such as inflammation). While MRI images are superior to x rays for certain types of tissue and have none of the hazards of x rays, they do not completely supplant x-ray images. MRI is less effective than x rays for detecting breaks in bone, for example, and in imaging breast tissue, so the two diagnostic tools complement each other. MRI images are also expensive compared to simple x-ray images and tend to be used most often where they supply information not readily obtained from x rays. Another disadvantage of MRI is that the patient is totally enclosed with detectors close to the body for about 30 minutes or more, leading to claustrophobia. It is also difficult for the obese patient to be in the magnet tunnel. New “open-MRI” machines are now available in which the magnet does not completely surround the patient. Over the last decade, the development of much faster scans, called “functional MRI” (fMRI), has allowed us to map the functioning of various regions in the brain responsible for thought and motor control. This technique measures the change in blood flow for activities (thought, experiences, action) in the brain. The nerve cells increase their consumption of oxygen when active. Blood hemoglobin releases oxygen to active nerve cells and has somewhat different magnetic properties when oxygenated than when deoxygenated. With MRI, we can measure this and detect a blood oxygen-dependent signal. Most of the brain scans today use fMRI. Other Medical Uses of Magnetic Fields Currents in nerve cells and the heart create magnetic fields like any other currents. These can be measured but with some difficulty since their strengths are about 10-6to 10-8 less than the Earth’s magnetic field. Recording of the heart’s magnetic field as it beats is called a magnetocardiogram (MCG), while measurements of the brain’s magnetic field is called a magnetoencephalogram (MEG). Both give information that differs from that obtained by measuring the electric fields of these organs (ECGs and EEGs), but they are not yet of sufficient importance to make these difficult measurements common. In both of these techniques, the sensors do not touch the body. MCG can be used in fetal studies, and is probably more sensitive than echocardiography. MCG also looks at the heart’s electrical activity whose voltage output is too small to be recorded by surface electrodes as in EKG. It has the potential of being a rapid scan for early diagnosis of cardiac ischemia (obstruction of blood flow to the heart) or problems with the fetus. MEG can be used to identify abnormal electrical discharges in the brain that produce weak magnetic signals. Therefore, it looks at brain activity, not just brain structure. It has been used for studies of Alzheimer’s disease and epilepsy. Advances in instrumentation to measure very small magnetic fields have allowed these two techniques to be used more in recent years. What is used is a sensor called a SQUID, for superconducting quantum interference device. This operates at liquid helium temperatures and can measure magnetic fields thousands of times smaller than the Earth’s. Finally, there is a burgeoning market for magnetic cures in which magnets are applied in a variety of ways to the body, from magnetic bracelets to magnetic mattresses. The best that can be said for such practices is that they are apparently harmless, unless the magnets get close to the patient’s computer or magnetic storage disks. Claims are made for a broad spectrum of benefits from cleansing the blood to giving the patient more energy, but clinical studies have not verified these claims, nor is there an identifiable mechanism by which such benefits might occur. PhET Explorations: Magnet and Compass Ever wonder how a compass worked to point you to the Arctic? Explore the interactions between a compass and bar magnet, and then add the Earth and find the surprising answer! Vary the magnet’s strength, and see how things change both inside and outside. Use the field meter to measure how the magnetic field changes. Figure 3. Magnet and Compass Section Summary Crossed (perpendicular) electric and magnetic fields act as a velocity filter, giving equal and opposite forces on any charge with velocity perpendicular to the fields and of magnitude . Conceptual Questions 1: Measurements of the weak and fluctuating magnetic fields associated with brain activity are called magnetoencephalograms (MEGs). Do the brain’s magnetic fields imply coordinated or uncoordinated nerve impulses? Explain. 2: Discuss the possibility that a Hall voltage would be generated on the moving heart of a patient during MRI imaging. Also discuss the same effect on the wires of a pacemaker. (The fact that patients with pacemakers are not given MRIs is significant.) 3: A patient in an MRI unit turns his head quickly to one side and experiences momentary dizziness and a strange taste in his mouth. Discuss the possible causes. 4: You are told that in a certain region there is either a uniform electric or magnetic field. What measurement or observation could you make to determine the type? (Ignore the Earth’s magnetic field.) 5: An example of magnetohydrodynamics (MHD) comes from the flow of a river (salty water). This fluid interacts with the Earth’s magnetic field to produce a potential difference between the two river banks. How would you go about calculating the potential difference? 6: Draw gravitational field lines between 2 masses, electric field lines between a positive and a negative charge, electric field lines between 2 positive charges and magnetic field lines around a magnet. Qualitatively describe the differences between the fields and the entities responsible for the field lines. Problems & Exercises 1: Indicate whether the magnetic field created in each of the three situations shown in Figure 4 is into or out of the page on the left and right of the current. Figure 4. 2: What are the directions of the fields in the center of the loop and coils shown in Figure 5? Figure 5. 3: What are the directions of the currents in the loop and coils shown in Figure 6? Figure 6. 4: To see why an MRI utilizes iron to increase the magnetic field created by a coil, calculate the current needed in a 400-loop-per-meter circular coil 0.660 m in radius to create a 1.20-T field (typical of an MRI instrument) at its centre with no iron present. The magnetic field of a proton is approximately like that of a circular current loop 0.650 x 10-15 m in radius carrying 1.05 x 104 A. What is the field at the centre of such a loop? 5: Inside a motor, 30.0 A passes through a 250-turn circular loop that is 10.0 cm in radius. What is the magnetic field strength created at its centre? 6: Nonnuclear submarines use batteries for power when submerged. (a) Find the magnetic field 50.0 cm from a straight wire carrying 1200 A from the batteries to the drive mechanism of a submarine. (b) What is the field if the wires to and from the drive mechanism are side by side? (c) Discuss the effects this could have for a compass on the submarine that is not shielded. 7: How strong is the magnetic field inside a solenoid with 10,000 turns per meter that carries 20.0 A? 8: What current is needed in a solenoid with 3000 turns per metre to produce a magnetic field 104 times the Earth’s magnetic field of 5.00 x 10-5 T. 9: How far from the starter cable of a car, carrying 150 A, must you be to experience a field less than the Earth’s 5.00 x 10-5 T? Assume a long straight wire carries the current. (In practice, the body of your car shields the dashboard compass.) 10: Measurements affect the system being measured, such as the current loop in Figure 8. (a) Estimate the field the loop creates by calculating the field at the centre of a circular loop 20.0 cm in diameter carrying 5.00 A. (b) What is the smallest field strength this loop can be used to measure, if its field must alter the measured field by less than 0.0100%? 11: Figure 7 shows a long straight wire just touching a loop carrying a current I1. Both lie in the same plane. (a) What direction must the current I2 in the straight wire have to create a field at the centre of the loop in the direction opposite to that created by the loop? (b) What is the ratio of I1/ I2 that gives zero field strength at the centre of the loop? (c) What is the direction of the field directly above the loop under this circumstance? Figure 7. 12: Find the magnitude and direction of the magnetic field at the point equidistant from the wires in Chapter 22.10 Figure 5(a), using the rules of vector addition to sum the contributions from each wire. 13: Find the magnitude and direction of the magnetic field at the point equidistant from the wires in Chapter 22.10 Figure 5(b), using the rules of vector addition to sum the contributions from each wire. 14: What current is needed in the top wire in Chapter 22.10 Figure 5(a) to produce a field of zero at the point equidistant from the wires, if the currents in the bottom two wires are both 10.0 A into the page? 15: Calculate the size of the magnetic field 20 m below a high voltage power line. The line carries 450 MW at a voltage of 300,000 V. 16: Integrated Concepts (a) A pendulum is set up so that its bob (a thin copper disk) swings between the poles of a permanent magnet as shown in Figure 8. What is the magnitude and direction of the magnetic force on the bob at the lowest point in its path, if it has a positive 0.250 μC charge and is released from a height of 30.0 cm above its lowest point? The magnetic field strength is 1.50 T. (b) What is the acceleration of the bob at the bottom of its swing if its mass is 30.0 grams and it is hung from a flexible string? Be certain to include a free-body diagram as part of your analysis. Figure 8. 17: Integrated Concepts (a) What voltage will accelerate electrons to a speed of 6.00 x 10-7 m/s? (b) Find the radius of curvature of the path of a proton accelerated through this potential in a 0.500-T field and compare this with the radius of curvature of an electron accelerated through the same potential. 18: Integrated Concepts Find the radius of curvature of the path of a 25.0-MeV proton moving perpendicularly to the 1.20-T field of a cyclotron. 19: Integrated Concepts To construct a non-mechanical water meter, a 0.500-T magnetic field is placed across the supply water pipe to a home and the Hall voltage is recorded. (a) Find the flow rate in liters per second through a 3.00-cm-diameter pipe if the Hall voltage is 60.0 mV. (b) What would the Hall voltage be for the same flow rate through a 10.0-cm-diameter pipe with the same field applied? 20: Integrated Concepts (a) Using the values given for an MHD drive in Exercise 2, and assuming the force is uniformly applied to the fluid, calculate the pressure created in N/m2. (b) Is this a significant fraction of an atmosphere? 21: Integrated Concepts (a) Calculate the maximum torque on a 50-turn, 1.50 cm radius circular current loop carrying 50 μA in a 0.500-T field. (b) If this coil is to be used in a galvanometer that reads 50 μA full scale, what force constant spring must be used, if it is attached 1.00 cm from the axis of rotation and is stretched by the 60.0 degree arc moved? 22: Integrated Concepts A current balance used to define the ampere is designed so that the current through it is constant, as is the distance between wires. Even so, if the wires change length with temperature, the force between them will change. What percent change in force per degree will occur if the wires are copper? 23: Integrated Concepts (a) Show that the period of the circular orbit of a charged particle moving perpendicularly to a uniform magnetic field is T = 2πm/(qB) . (b) What is the frequency f? (c) What is the angular velocity ω? Note that these results are independent of the velocity and radius of the orbit and, hence, of the energy of the particle. (Figure 9.) Figure 9. Cyclotrons accelerate charged particles orbiting in a magnetic field by placing an AC voltage on the metal Dees, between which the particles move, so that energy is added twice each orbit. The frequency is constant, since it is independent of the particle energy—the radius of the orbit simply increases with energy until the particles approach the edge and are extracted for various experiments and applications. 24: Integrated Concepts A cyclotron accelerates charged particles as shown in Figure 9. Using the results of the previous problem, calculate the frequency of the accelerating voltage needed for a proton in a 1.20-T field. 25: Integrated Concepts (a) A 0.140-kg baseball, pitched at 40.0 m/s horizontally and perpendicular to the Earth’s horizontal 5.00 x 10-5 T field, has a 100-nC charge on it. What distance is it deflected from its path by the magnetic force, after traveling 30.0 m horizontally? (b) Would you suggest this as a secret technique for a pitcher to throw curve balls? 26: Integrated Concepts (a) What is the direction of the force on a wire carrying a current due east in a location where the Earth’s field is due north? Both are parallel to the ground. (b) Calculate the force per meter if the wire carries 20.0 A and the field strength is 3.00x 10-5 T. 27: Integrated Concepts One long straight wire is to be held directly above another by repulsion between their currents. The lower wire carries 100 A and the wire 7.50 cm above it is 10-gauge (2.588 mm diameter) copper wire. (a) What current must flow in the upper wire, neglecting the Earth’s field? (b) What is the smallest current if the Earth’s 3.00x 10-5 T. field is parallel to the ground and is not neglected? (c) Is the supported wire in a stable or unstable equilibrium if displaced vertically? If displaced horizontally? 28: Unreasonable Results (a) Find the charge on a baseball, thrown at 35.0 m/s perpendicular to the Earth’s 3.00x 10-5 T field, that experiences a 1.00-N magnetic force. (b) What is unreasonable about this result? (c) Which assumption or premise is responsible? 29: Unreasonable Results A charged particle having mass 6.64 x 10-27 kg (that of a helium atom) moving at 8.70 x 105 m/s perpendicular to a 1.50-T magnetic field travels in a circular path of radius 16.0 mm. (a) What is the charge of the particle? (b) What is unreasonable about this result? (c) Which assumptions are responsible? 30: Unreasonable Results An inventor wants to generate 120-V power by moving a 1.00-m-long wire perpendicular to the Earth’s 5.00 x 10-5 T field. (a) Find the speed with which the wire must move. (b) What is unreasonable about this result? (c) Which assumption is responsible? 31: Unreasonable Results Frustrated by the small Hall voltage obtained in blood flow measurements, a medical physicist decides to increase the applied magnetic field strength to get a 0.500-V output for blood moving at 30.0 cm/s in a 1.50-cm-diameter vessel. (a) What magnetic field strength is needed? (b) What is unreasonable about this result? (c) Which premise is responsible? 32: Unreasonable Results A surveyor 100 m from a long straight 200-kV DC power line suspects that its magnetic field may equal that of the Earth and affect compass readings. (a) Calculate the current in the wire needed to create a 5.00 x 10-5 T field at this distance. (b) What is unreasonable about this result? (c) Which assumption or premise is responsible? 33: Construct Your Own Problem Consider a mass separator that applies a magnetic field perpendicular to the velocity of ions and separates the ions based on the radius of curvature of their paths in the field. Construct a problem in which you calculate the magnetic field strength needed to separate two ions that differ in mass, but not charge, and have the same initial velocity. Among the things to consider are the types of ions, the velocities they can be given before entering the magnetic field, and a reasonable value for the radius of curvature of the paths they follow. In addition, calculate the separation distance between the ions at the point where they are detected. 34: Construct Your Own Problem Consider using the torque on a current-carrying coil in a magnetic field to detect relatively small magnetic fields (less than the field of the Earth, for example). Construct a problem in which you calculate the maximum torque on a current-carrying loop in a magnetic field. Among the things to be considered are the size of the coil, the number of loops it has, the current you pass through the coil, and the size of the field you wish to detect. Discuss whether the torque produced is large enough to be effectively measured. Your instructor may also wish for you to consider the effects, if any, of the field produced by the coil on the surroundings that could affect detection of the small field. Glossary magnetic resonance imaging (MRI) : a medical imaging technique that uses magnetic fields create detailed images of internal tissues and organs nuclear magnetic resonance (NMR) : a phenomenon in which an externally applied magnetic field interacts with the nuclei of certain atoms magnetocardiogram (MCG) : a recording of the heart’s magnetic field as it beats magnetoencephalogram (MEG) : a measurement of the brain’s magnetic field Solutions Problems & Exercises 1: (a) right-into page, left-out of page (b) right-out of page, left-into page (c) right-out of page, left-into page 3: (a) clockwise (b) clockwise as seen from the left (c) clockwise as seen from the right 4: 1.01 x 1013 T 6: (a) 4.80 x 10-4 T (b) Zero (c) If the wires are not paired, the field is about 10 times stronger than Earth’s magnetic field and so could severely disrupt the use of a compass. 8: 133 A 10: (a) 3.14 x 10-5 T (b) 0.314 T 12: 7.55 x 10-5 T , 23.4 degrees 14: 10.0 A 16: (a) 9.09 x 10-7 N, upward . (b) 3.03 x 10-5 m/s2 18: 60.2 cm 20: (a) 1.02 x 103 N/m2(b) Not a significant fraction of an atmosphere which is 101 kPa. 22: 17.0 x 10-4 oC 24: 18.3 MHz 26: (a) Straight up (b) 6.00 x 10-4 N/m (c) 94.1 μm (d) 2.47 Ωm, 49.4 V/m 28: (a) 571 C (b) Impossible to have such a large separated charge on such a small object. (c) The 1.00-N force is much too great to be realistic in the Earth’s field. 30: (a) 2.40 x106 m/s (b) The speed is too high to be practical ≤ 1% speed of light (c) The assumption that you could reasonably generate such a voltage with a single wire in the Earth’s field is unreasonable 32: (a) 25.0 kA (b) This current is unreasonably high. It implies a total power delivery in the line of 50.0×10^9 W, which is much too high for standard transmission lines. (c) 100 meters is a long distance to obtain the required field strength. Also coaxial cables are used for transmission lines so that there is virtually no field for DC power lines, because of cancellation from opposing currents. The surveyor’s concerns are not a problem for his magnetic field measurements. License Douglas College Physics 1207 Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.
1969	https://math.stackexchange.com/questions/2540453/finding-invariant-measures-to-solve-recurrences	Skip to main content Finding invariant measures (to solve recurrences) Ask Question Asked Modified 7 years, 8 months ago Viewed 87 times This question shows research effort; it is useful and clear 7 Save this question. Show activity on this post. INTRO: If you don't feel like reading the justification for my question, just skip down to the question. Some recurrence relations can be solved by finding invariants, or quantities that remain unchanged. For example, consider the sequences defined by an+1=anbnan+bn+1 bn+1=an+bn a0=b0=−1 If one defines the function μ as μ(x,y)=x+xy+y then it is easily proven that μ(x,y)=μ(xyx+y+1,x+y) From this, it follows that μ(an+1,bn+1)=μ(an,bn) and μ(an,bn)=μ(a0,b0) μ(an,bn)=−1 an+anbn+bn=−1 an=−1−bn1+bn an=−1 and so bn+1=−1+bn bn+1=bn−1 ...yielding explicit formulas for an and bn: an=−1, bn=−1−n This was perhaps a trivial (and manufactured) example, but it demonstrates how invariants are used to solve recursions. QUESTION: I would like to know the following: what are some methods used to find invariants? That is, what methods might one use to find a particular nontrivial solution μ to the functional equation μ(x,y)=μ(f(x,y),g(x,y)) where f,g are given? sequences-and-series recurrence-relations invariance Share CC BY-SA 3.0 Follow this question to receive notifications edited Nov 28, 2017 at 23:57 Franklin Pezzuti Dyer asked Nov 28, 2017 at 0:28 Franklin Pezzuti DyerFranklin Pezzuti Dyer 41k99 gold badges8181 silver badges175175 bronze badges 3 Nice question. Experience and "educated guesses" provide reliable tools, but in general it might be a non-trivial problem. Related: math.stackexchange.com/questions/2459643/… – Jack D'Aurizio Commented Nov 28, 2017 at 1:00 The problem you want is even a little harder than you have written, since you want f and g to be functions of both x and y, as per your example, not just single variable like you wrote. – Will Fisher Commented Nov 28, 2017 at 1:35 @WillFisher Ooh, good observation. I should probably clarify in my question. – Franklin Pezzuti Dyer Commented Nov 28, 2017 at 23:57 Add a comment \| 0 Reset to default You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions sequences-and-series recurrence-relations invariance See similar questions with these tags. Featured on Meta Upcoming initiatives on Stack Overflow and across the Stack Exchange network... 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1970	https://publish.obsidian.md/pkc/Hub/Theory/Category+Theory/Logic/empty+set	empty set - PKC - Obsidian Publish PKC Excalidraw Hub ExcaliBrain Excalidraw Operations Tech Theory Category Theory Cognitive Science Combinatorics Logic Proof Type Theory abacus Abstract² Interpretation AC AdaBoost algebraic logic All analytic philosophy Analytical Philosophy Antecedent anti-symmetric anti-symmetry antisymmetric appearance Aristotelian logic Aristotelian Syllogistic Aristotle’s logic Atomic essay autoepistemic logic Autonomous Syntax Axiom of Choice Axiomatic Design Banach space Banach-Tarski Paradox Bell inequalities Beyond Metrics BHK Interpretation Bible Binary logic Bottom bound variable British Empiricism movement Brouwer-Heyting-Kolmogorov interpretation butterfly effect Calculus of Constructions cardinality Cartesian Linguistics Categorical Logic Categories in Aristotelian Logic Causal Diagraming 1 Causal Diagraming Causal Propagation cell Charles Sanders Peirce and the Foundations 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In other words, it is a set that contains nothing. The notation for the empty set is {}, or sometimes Ø. It is important to note that the empty set is not the same as a set containing nothing. See empty set > Unit Type and Empty set. It is a distinct concept representing the absence of any elements. The empty set has several interesting properties in mathematics: Unique: There is only one empty set. Any two empty sets are considered equal. Subset of Every Set: The empty set is a subset of every other set because it does not contain any elements. For example, if A and B are sets, and A is the empty set, then A ⊆ B. Union: The union of the empty set with any other set results in the same set. For example, if A is any non-empty set, then A ∪ {} = A. Intersection: The intersection of the empty set with any other set results in the empty set itself. For example, if A is any non-empty set, then A ∩ {} = {}. Complement: The complement of the empty set in any universal or superset U is equal to U itself. In other words, U - {} = U. Unit Type and Empty set The logical interpretation of a unit set is that it contains a single element. It is represented by enclosing the element within curly braces, such as {x}. For example, if x represents the number 3, then {x} would be the unit set containing only 3. On the other hand, an empty set (also known as a null set) is a set that contains no elements. It is denoted by the symbol Ø or {} with no elements inside. The empty set is considered a subset of every set because it does not have any elements that are not in another set. In terms of their relation, the empty set can be seen as the absence of any elements, whereas a unit set represents the presence of one specific element. So in a way, they are related in that both represent specific conditions regarding the existence or non-existence of elements within a set. However, they are fundamentally different as one contains an element while the other does not. For a more detailed discussion, see @ThreeOneLogic2022 > Contractability. Conclusion The concept of the empty set has applications in various branches of mathematics and logic. It helps define concepts such as subsets, intersections, and complements more precisely and consistently. References dataview Table title as Title, authors as Authors where contains(subject, "empty set") or contains(subject, "Unit Type") empty set Not found This page does not exist Interactive graph On this page Powered by Obsidian Publish
1971	https://pdhstar.com/wp-content/uploads/2019/06/CE-089-Manning-Equation-for-Open-Channels.pdf?srsltid=AfmBOoqxNjvlIiZmMFGWbvmmQ-gLMTyRQfT8TFGOGGIuRpO2Az-CFL-5	PDH Star \| T / F: (833) PDH‐STAR (734‐7827) \| E: info@pdhstar.com CE‐089 Manning Equation for Open Channels Instructor: Harlan Bengtson, PhD, P.E. Course ID: CE‐089 PDH Hours: 4 PDH 1 Manning Equation - Open Channel Flow Calculations Harlan H. Bengtson, PhD, P.E. COURSE CONTENT 1. Introduction The Manning equation is a widely used empirical equation for uniform open channel flow of water. It provides a relationship among several open channel flow parameters of interest: flow rate or average velocity, bottom slope of the channel, cross-sectional area of flow, wetted perimeter, and Manning roughness coefficient for the channel. Open channel flow takes place in natural channels like rivers and streams, as well as in manmade channels like those used to transport wastewater and in circular sewers flowing partially full. The main topic of this course is uniform open channel flow, in which the channel slope, water velocity and water depth remain constant. This includes a variety of example calculations with the Manning equation and the use of Excel spreadsheets for those calculations. Figure 1. Bighorn River in Montana – a Natural Open Channel 2Image Source: National Park Service, Bighorn Canyon National Recreational Area website at: Figure 2. Irrigation Canal Branch in Sinai – A man-made open channel Image Source: Egypt-Finland Agric. Res Proj 2. Learning Objectives At the conclusion of this course, the student will  Know the difference between laminar & turbulent open channel flow.  Know the difference between steady state & unsteady state open channel flow.  Know the difference between uniform & non-uniform open channel flow.  Be able to calculate the hydraulic radius for flow of a specified depth in an open channel with specified cross-sectional shape and size. 3  Be able to calculate the Reynolds Number for a specified open channel flow and determine whether the flow will be laminar or turbulent flow.  Be able to use tables such as the examples given in this course to determine a value for Manning roughness coefficient for flow in a manmade channel.  Be able to use the Manning Equation to calculate volumetric flow rate, average velocity, Manning roughness coefficient, or channel bottom slope, if given adequate information about a reach of open channel flow  Be able to use the Manning Equation, with an iterative procedure, to calculate normal depth for specified volumetric flow rate, channel bottom slope, channel shape & size, and Manning roughness coefficient for a reach of open channel flow  Be able to make Manning Equation calculations in either U.S. units or S.I. units  Be able to calculate the Manning roughness coefficient for a natural channel based on descriptive information about the channel.  Be able to use the Manning Equation to make calculations for the flow of water in a circular pipe either flowing full or flowing half full.  Be able to carry out a variety of calculations for full or partially full flow of water under gravity in a circular pipe site . 3. Topics Covered in this Course I. Open Channel Flow vs Pipe Flow II. Classifications of Open Channel Flow A. Laminar or Turbulent Flow 4B. Steady State or Unsteady State Flow C. Supercritical, Subcritical or Critical Flow D. Uniform or Nonuniform flow III. Manning Equation/Uniform Open Channel Flow Basics A. The Manning Equation B. Manning Roughness Coefficient C. Reynolds Number D. Hydraulic Radius E. The Manning Equation in S.I. Units F. The Manning Equation in Terms of V Instead of Q IV. Manning Equation Calculations for Manmade Channels A. The Easy Parameters to Calculate with the Manning Equation B. The Hard Parameter to Calculate - Determination of Normal Depth C. Circular Pipes Flowing Full D. Circular Pipes Flowing Partially Full V. Uniform Flow Calculations for Natural Channels A. The Manning Roughness Coefficient for Natural Channels B. Manning Equation Calculations VI. Summary VII. References and Websites 4. Open Channel Flow vs Pipe Flow The term “open channel flow” is used to refer to flow with a free surface at atmospheric pressure, in which the driving force for flow is gravity. Pipe flow, on the other hand is used to refer to flow in a closed conduit under pressure, in which the primary driving force is typically pressure. Open channel flow occurs in natural channels, such as rivers and streams and in manmade channels, as for storm water, waste water and irrigation water. This course is about open channel flow, and in particular, about uniform open channel flow. The next 5section covers several different classifications of types of open channel flow, including clarification of the difference between uniform and nonuniform open channel flow. 5. Classifications of Open Channel Flow A. Turbulent and Laminar Flow: Description of a given flow as being either laminar or turbulent is used for several fluid flow applications (like pipe flow and flow past a flat plate) as well as for open channel flow. In each of these fluid flow applications a Reynolds number is used for the criterion to determine whether a given flow will be laminar or turbulent. For open channel flow a Reynolds number below 500 is typically used as the criterion for laminar flow, while the flow will typically be turbulent for a Reynolds number greater than 12,500. For a flow with Reynolds number between 500 and 12,500, other conditions, like the upstream channel conditions and the roughness of the channel walls will determine whether the flow is laminar or turbulent. Background on Laminar and Turbulent Flow : Osborne Reynolds reported in the late 1800s on experiments that he performed observing the difference between laminar and turbulent flow in pipes and quantifying the conditions for which each would occur. In his classic experiments, he injected dye into a transparent pipe containing a flowing fluid. He observed that the dye flowed in a streamline and didn’t mix with the rest of the fluid under some conditions, which he called laminar flow. Under other conditions, however, he observed that the net velocity of the fluid was in the direction of flow, but there were eddy currents in all directions that caused mixing of the fluid. Under these turbulent flow conditions, the entire fluid became colored with the dye. The figure below illustrates laminar and turbulent open channel flow. 6 Figure 3. Dye injection into laminar & turbulent open channel flow Laminar flow, sometimes also called streamline flow, occurs for flows with high fluid viscosity and/or low velocity. Turbulent flow takes place for flows with low fluid viscosity and/or high velocity. More discussion of the Reynolds number and its calculation for open channel flow are given in Section 6 of this course. For most practical cases of water transport in either manmade or natural open channels, the Reynolds number is greater than 12,500, and thus the flow is turbulent. One notable exception is flow of a thin liquid layer on a large flat surface, such as rainfall runoff from a parking lot, highway, or airport runway. This type of flow, often called sheet flow, is typically laminar. B. Unsteady State and Steady State Flow: The concepts of steady state and unsteady state flow are used for a variety of fluid flow applications, including open channel flow. Steady state flow is taking place whenever there are no changes in velocity pattern or magnitude with time at a given channel cross section. When unsteady state flow is present, however, there are changes of velocity with time at any given cross section in the flow. Steady state open channel flow takes place when a constant flow rate of liquid is passing through the channel. Unsteady state open channel flow takes place when there is a changing flow rate, as for example in a river after a rainstorm. Steady state or nearly steady state conditions are present for many practical open channel flow situations. The equations and calculations presented in this course are all for steady state flow. 7 C. Critical, Subcritical, and Supercritical Flow: Any open channel flow must be one of these three classifications: supercritical, subcritical or critical flow. The interpretation of these three classifications of open channel flow, and the differences among them, aren’t as obvious or intuitive as the interpretation and the differences for the other classifications (steady and unsteady state, laminar and turbulent, and uniform and non-uniform flow). Some of the behaviors for subcritical and supercritical flow and the transitions between them may not be what you would intuitively expect. Supercritical flow takes place when there is a relatively high liquid velocity and relatively shallow depth of flow. Subcritical flow, as one might expect, takes place when there is a relatively low liquid velocity and relatively deep flow. The Froude number (Fr = V/(gl) 1/2 ) provides information about whether a given flow is supercritical, subcritical or critical. For subcritical flow, Fr is less than one; for supercritical flow, it is greater than one; and for critical flow it is equal to one. Further details about subcritical, supercritical and critical flow are beyond the scope of this course. D. Non-Uniform and Uniform Flow: Uniform flow will occur in a reach of open channel whenever there is a constant flow rate of liquid passing through the channel, the bottom slope is constant, the channel surface roughness is constant, and the cross-section shape & size are constant. Under these conditions, the depth of flow and the average velocity of the flowing liquid will remain constant in that reach of channel. Non-uniform flow will be present for reaches of channel where there are changes in the bottom slope, channel surface roughness, cross-section shape, and/or cross-section size. Whenever the bottom slope, surface roughness, and channel cross-section shape and size become constant in a downstream reach of channel, another set of uniform flow conditions will occur there. This is illustrated in Figure 4. 8 Figure 4. Non-uniform and Uniform Open Channel Flow 6. Manning Equation/Uniform Open Channel Flow Basics As just described above, uniform open channel flow takes place in a channel reach that has constant channel cross-section size and shape, constant surface roughness, and constant bottom slope, with a constant volumetric flow rate of liquid passing through the channel. These conditions lead to flow at a constant depth of flow and constant liquid velocity, as illustrated in Figure 2. A. The Manning Equation is an empirical equation that was developed by the French engineer, Philippe Gauckler in 1867. It was redeveloped by the Irish engineer, Robert Manning, in 1890. Although this equation is also known as the Gauckler-Manning equation, it is much more commonly known simply as the Manning equation or Manning formula in the United States. This formula gives the relationship among several parameters of interest for uniform flow of water in an open channel. Not only is the Manning equation empirical, it is also a dimensional equation. This means that the units to be used for each of the parameters must be specified for a given constant in the equation. For commonly used U.S. units the Manning Equation and the units for its parameters are as follows: 9 Q = (1.49/n)A(R h2/3 )S 1/2 (1) Where:  Q = the volumetric flow rate of water passing through the channel reach in ft 3/sec.  n = the Manning roughness coefficient for the channel surface ( a dimensionless, empirical constant).  A = the cross-sectional area of water normal to the flow direction in ft 2.  Rh = the hydraulic radius (R h = A/P). (A is the cross-sectional area as defined just above in ft 2 and P is the wetted perimeter of the cross-sectional area of flowing water in ft.  S = the bottom slope of the channel in ft/ft (dimensionless ). S is actually the slope of the energy grade line. For uniform flow, however, the depth of flow is constant and the velocity head is constant, so the slope of the energy grade line is the same as that of the hydraulic grade line and is the same as the slope of the water surface, which is the same as the channel bottom slope. For convenience, the channel bottom slope is typically used for S in the Manning Equation. B. Manning Roughness Coefficient, n, is a dimensionless, empirical constant, as just described above. Its value depends on the nature of the channel and its surfaces. There are tables with values of n for various channel types and surfaces in many handbooks and textbooks, as well as at several online sources. Table 1 below is a typical table of this type. This table gives n values for several manmade open channel surfaces. Values of n for natural channels will be addressed in Section 8. 10 Table 1. Manning Roughness Coefficient, n, for Selected Surfaces Source for n values in Table 1: 11 C. The Reynolds number for open channel flow is defined as Re = VR h/, where Rh is the hydraulic radius, as defined above, V is the liquid velocity (= Q/A), and  and  are the density and viscosity respectively of the flowing fluid. Since the Reynolds number is dimensionless, any consistent set of units can be used for Rh V,  and . If done properly, all of the units will cancel out, leaving Re dimensionless. The flow must be in the turbulent regime in order to use the Manning equation for uniform open channel flow. It is fortunate that nearly all practical instances of water transport through an open channel have Re greater than 12,500, in which case the flow is turbulent,and the Manning equation can be used. The Manning equation is specifically for the flow of water, and no water properties are required in the equation. In order to calculate the Reynolds number to check on whether the flow is turbulent, however, values of density and viscosity for the water in question are needed. Tables of density and viscosity values for water as a function of temperature are available in many textbooks, handbooks, and websites. Table 2 below gives values for density and viscosity of water from 32 oF to 70 oF. Table 2. Density and Viscosity of Water 12 Example #1: Water at 60 oF is flowing 1.2 feet deep in a 3 foot wide rectangular open channel, as shown in the diagram below. The channel is made of concrete (made with wooden forms) and has a bottom slope of 0.0008. Determine whether the flow is laminar or turbulent. Solution: Based on the problem statement, this will be uniform flow. The flow is probably turbulent, however the velocity is needed in order to calculate the Reynolds number. Hence we will assume that the flow is turbulent, use the Manning equation to calculate Q and V. Then Re can be calculated to check on whether the flow is indeed turbulent. The parameters needed for the right side of the Manning equation are as follows: From Table 1, for concrete made with wooden forms: n = 0.015 A = (1.2)(3) = 3.6 ft 2P = 3 + (2)(1.2) = 5.4 ft Rh = A/P = 3.6/5.4 = 0.6667 ft S = 0.0008 (given in problem statement) Substituting into the Manning equation ( Q = (1.49/n)A(R h2/3 )S 1/2 ) : 13 Q = 1.49/0.015)(3.6)(0.6667 2/3 )(0.0008 1/2 ) = 7.72 cfs Now the average velocity, V, can be calculated: V = Q/A = 7.72/3.6 = 2.14 ft/sec From Table 2, for 60 oF:  = 1.938 slugs/ft 3 and  = 2.334 x 10 -5 lb-sec/ft 2Substituting into Re = VR h/ : Re = (1.938)(2.14)(0.6667)/2.334 x 10 -5 = 118,470 Since Re > 12,500, this open channel flow is turbulent D. Hydraulic Radius is a parameter that must be calculated for various channel shapes in order to use the Manning Equation. Some common cross-sectional shapes used for open channel flow are rectangular, trapezoidal triangular, circular, and semicircular . Formulas for the hydraulic radius for each of these channel shapes will now be presented. A rectangular channel allows easy calculation of the hydraulic radius. The bottom width will be represented by b and the depth of flow will be represented by y, as shown in the Figure 5 below. Figure 5. Rectangular Open Channel Cross-Section 14 The area and wetted perimeter will be as follows: A = by P = 2y + b Then R h = A/P, or: For a rectangular channel: R h = by/(2y + b) (2) A trapezoidal cross-section is used for some manmade open channels and can be used as an approximation of the cross-sectional shape for some natural channels. Figure 6 shows the parameters typically used to describe the size and shape of a trapezoidal channel. Figure 6. Trapezoidal Open Channel Cross-section The channel bottom width and the water depth are represented by b and y, the same as with the rectangular channel. Additional parameters for the trapezoidal channel shape are:  B, the water surface width;  l, the wetted length of the sloped side;  , the angle of the sloped side of the channel from the vertical; and 15  z, the channel side slope expressed as horiz :vert = z :1. The size and shape of a trapezoidal channel are often specified with the bottom width, b, and the side slope, z. The hydraulic radius for flow in a trapezoidal open channel can be expressed in terms of y, b, and z, as follows: i) The cross-sectional area of flow, A, is the area of the trapezoid in Figure 4: A = y(b + B)/2 = (y/2)(b + B) From Figure 6, it can be seen that the surface width, B, is greater than the bottom width, b, by zy at each end, or: B = b + 2zy Substituting for B into the equation for A gives: A = (y/2)(b + b + 2zy) = (y/2)(2b + 2zy) Simplifying gives: A = by + zy 2 As shown in Figure 6, the wetted perimeter of the cross-sectional area of flow is P = b + 2 l By Pythagoras’ Theorem for the triangle at each end of the trapezoid: l2 = y 2 + (yz) 2 or l = [y 2 + (yz) 2 ]1/2 Substituting for l into the equation for P and simplifying gives: P = b + 2y(1 + z 2) 1/2 Substituting for A and P in R h = A/P gives: For a trapezoidal open channel: R h = (by + zy 2 )/[b + 2y(1 + z 2 ) 1/2 ] (3) 16 Example #2: Determine the hydraulic radius of water flowing 1.5 ft deep in a trapezoidal open channel with a bottom width of 2 ft and side slope of horiz :vert = 3 :1. Solution: From the problem statement, y = 1.5 ft, b = 2 ft, and z = 3. Substituting these values into the expression for hydraulic radius gives: Rh = (21.5 + 31.5 2)/[2 + 21.5(1 + 3 2) 1/2 ] = 0.849 ft This type of calculation can conveniently be done using an Excel spreadsheet like the simple one shown in the screenshot in Figure 7 below. This particular spreadsheet is set up to allow user entry of the channel bottom width, the depth of flow, and the side slope expressed as z. The spreadsheet then calculates the cross-sectional area of flow, A, the wetted perimeter, P, and the hydraulic radius, Rh, for the trapezoidal channel. The equations shown at the bottom of the worksheet are the same as those presented and discussed in this course. 17 Figure 7. Hydraulic Radius Calculator Spreadsheet The triangular open channel cross-sectional shape is the third one that we’ll be considering. Figure 8 below shows the parameters typically used to specify the size and shape of a triangular channel. They are:  B, the surface width of the water in the channel  y, the water depth in the channel, measured from the triangle vertex 18  l, the wetted length of the sloped side; and  z, the channel side slope expressed as horiz :vert = z :1. Figure 8. Triangular Open Channel Cross-Section For a triangular open channel, it‘s convenient to have the hydraulic radius expressed in terms of y and z, which can be done as follows: The area of the triangle in Figure 8, which represents the area of flow is: A = (1/2)By, but as shown in the figure, B = 2zy. Substituting for B in the equation for A and simplifying gives: A = y 2 z Also from Figure 8, it can be seen that the wetted perimeter is: P = 2 l.Substituting l = [y 2 + (yz) 2]1/2 (as shown above for the trapezoid), and simplifying gives: P = 2[y 2(1 + z 2)] 1/2 Substituting for A and P in R h = A/P gives: For a trianglular open channel: R h = y 2z/{2[y 2(1 + z 2 )] 1/2 } (4) 19 Circular pipes are used for open channel (gravity) flow for applications like storm sewers, sanitary sewers, and circular culverts. These pipe typically flow only partially full most of the time, but the full flow scenario is often used for hydraulic design. Hydraulic radius expressions for the full flow and half full cross-sections will be developed here. There will be additional discussion of partially full pipe flow in Section 7. Figure 9 shows a diagram for a pipe flowing full and a pipe flowing half full. The only parameters needed for either of these cases are the diameter and the radius of the pipe. Figure 9. Circular and semicircular Open Channel Cross-Sections The hydraulic radius for a circular pipe of diameter D, flowing full, can be calculated as follows: The cross-sectional area of flow is: A = D2/4 The wetted perimeter is: P = DThe hydraulic radius is: R h = A/P = ( D2/4)/( D) Or simply (for a pipe flowing full): R h = D/4 (5) 20 For the semicircular shape of a pipe flowing exactly half full, the area, A, and the wetted perimeter, P, will each be half of the values for the full pipe flow. Thus the hydraulic radius will remain the same, so (for a pipe flowing half full): R h = D/4 (6) E. The Manning Equation in SI Units is the same as that for U.S. units except that the constant is 1.00 instead of 1.49: Q = (1.00/n)A(R h2/3 )S 1/2 (7) Where:  Q = the volumetric flow rate of water passing through the channel reach in m 3/s.  n = the Manning roughness coefficient for the channel surface ( a dimensionless, empirical constant).  A = the cross-sectional area of water normal to the flow direction in m 2.  Rh = the hydraulic radius in m (R h = A/P). (A is the cross-sectional area as defined just above in m 2 and P is the wetted perimeter of the cross-sectional area of flowing water in m.  S = the bottom slope of the channel in m/m (dimensionless ). F. The Manning Equation in terms of Average Velocity: For some calculations, it is better to have the Manning Equation expressed in terms of average velocity, V, instead of in terms of volumetric flow rate. The definition of average velocity is V = Q/A , where Q and A are as previously defined. Substituting Q = VA into the Manning equation as given in Equation (1), and solving for V gives the following form of the Manning equation. For U.S. units: V = (1.49/n)(R h2/3 )S 1/2 (8) 21 For S.I. units, the constant is 1.00 instead of 1.49, giving: For S.I. units: V = (1.00/n)(R h2/3 )S 1/2 (9) 7. Manning Equation Calculations for Manmade Channels A. The Easy Parameters to Calculate with the Manning Equation: Several different parameters can be the “unknown” to be calculated with the Manning equation, based on known values for enough other parameters. If Q and V, S, or n is the unknown parameter to be calculated, and enough information is known to calculate the hydraulic radius, then the solution involves simply substituting values into the Manning equation and solving for the desired unknown parameter. These four parameters are thus the “easy parameters to calculate.” This type of Manning equation calculation is illustrated with several examples here. Then in the next section, we’ll take a look at the hard parameter to calculate, normal depth. Example #3: Use the Manning equation to determine the volumetric flow rate and average velocity of water flowing 0.9 m deep in a trapezoidal open channel with bottom width equal to 1.2 m and side slope of horiz :vert = 2 :1. The channel is concrete poured with steel forms and its bottom slope is 0.0003. Solution: The hydraulic radius can be calculated from the specified information (y = 0.9 m, b = 1.2 m, & z = 2) using the formula for a trapezoidal channel as follows: Rh = (by + zy 2)/[b + 2y(1 + z 2) 1/2 ] = (1.20.9 + 20.9 2)/[1.2 + 20.9(1 + 2 2) 1/2 ]22 Rh = 0.517 m (also: A = by + zy 2 = 1.20.9 + 20.9 2 = 2.70 m 2Substituting R h and A into Equation (1) along with S = 0.0003 (given) and n = 0.011 (from Table 1) gives: Q = (1.00/n)A(R h2/3 )S 1/2 = (1.00/0.011)(2.70)(0.517 2/3 )(0.0003 1/2 ) Q = 2.74 m 3/s Now the average velocity, V, can be calculated from V = Q/A = 2.74/2.70 m/s V = 1.01 m/s This type of calculation is also easy to make with an Excel spreadsheet, like the one shown in the Figure 8 screenshot on the next page. Example #4: What would be the required slope for a 15 inch diameter circular storm sewer made of centrifugally spun concrete, if it needs to have an average velocity of at least 3.0 ft/sec when it’s flowing full? Solution: For the 15” diameter sewer, R h = D/4 = (15/12)/4 = 0.3125 ft. From Table 1, for centrifugally spun concrete, n = 0.013. Substituting these values for R h and n, along with the given value of V = 3.0 ft/sec, into Equation (8) and solving for S gives: S = {(0.013)(3.0)/[1.49(0.3125) 2/3 ]} 2 = 0.003231 = S 23 Figure 10. A Spreadsheet for Q & V in a Trapezoidal Channel Determination of the required Manning roughness coefficient, n, for a specified flow rate or velocity, bottom slope, and adequate information to calculate the hydraulic radius, would be a less common calculation, but would proceed in a manner very similar to Example #3 and Example #4. B. The Hard Parameter to Calculate - Determination of Normal Depth: When the depth of flow, y, is the unknown parameter to be determined using the Manning equation, an iterative calculation procedure is often required. This is because an equation with y as the only unknown can typically be obtained, but 24 the equation usually can’t be solved explicitly for y, making this “the hard parameter to calculate.” The depth of flow for a given flow rate through a channel reach of known shape size & material and known bottom slope is called the normal depth, and is sometimes represented by the symbol, y o. The typical situation requiring determination of the normal depth, y o, will have specified values for the flow rate, Q, the Manning roughness coefficient, n, and channel bottom slope, S, along with adequate channel size and shape information to allow A and R h to be expressed as functions of y o. The approach for calculating the normal depth, y o, for a situation as described above, is to rearrange the Manning equation to: AR h2/3 = Qn/(1.49S 1/2 ) (10) The right side of this equation will be a constant and the left side will be an expression with y o as the only unkown. The next couple of examples illustrate calculation of y o using an iterative calculation with Equation (10). Example #5: Determine the normal depth for a water flow rate of 20 ft 3/sec, through a rectangular channel with a bottom slope of 0.00025, bottom width of 4 ft, and Manning roughness coefficient of 0.012. Solution: Substituting the expressions for A and R h for a rectangular channel into the left hand of Equation (10) and substituting the given values for Q, n, and S into the right side, gives: 4y o(4y o/(4 + 2y o)) 2/3 = (20)(0.012)/[1.49 (0.00025 1/2 )] = 10.187 This equation has y o as the only unknown. The equation can’t be solved explicitly for y o, but it can be solved by an iterative (trial and error) process as illustrated in the Excel spreadsheet screenshot in Figure 11 on the next page. The spreadsheet screenshot shows the solution to be: y o = 2.40 ft , accurate to 3 significant figures. Note that this type of iterative calculation can also be accomplished with Excel's Goal Seek or Solver tool. 25 Figure 11. A Spreadsheet for Normal Depth in a Rectangular Channel Example #6: Determine the normal depth for a water flow rate of 20 ft 3/sec, through a trapezoidal channel with a bottom slope of 0.00025, bottom width of 4 ft, side slope of horiz :vert = 2 :1, and Manning roughness coefficient of 0.012. Solution: The values of Q, n, & S are the same as for Example #5, so the right hand side of Equation (10) will remain the same at 10.187. The left hand side 26 will be somewhat more complicated with the expression for R h as a function of y o for a trapezoid. Equation (10) for this calculation is: (4y o + 2y o2){(4y o + 2y o2)/[4 + 2y o(1 + 2 2) 1/2 ]} 2/3 = 10.187 The iterative calculations leading to y o = 1.49 ft are shown below. The solution is y o = 1.49 ft, because 10.228 is closer to the target value of 10.187, than the value of 10.094 for y o = 1.48 or 10.363 for y o = 1.50. C. Circular Pipes Flowing Full : Because of the simple form of the equations for hydraulic radius and cross-sectional area as functions of the diameter for a circular pipe flowing full ( R h = D/4 and A = D2/4 ), the Manning equation can be conveniently used to calculate Q and V, D, S, or n if the other parameters are known. Several useful forms of the Manning equation for a circular pipe flowing full under gravity are: Q = (1.49/n)( D2/4)((D/4) 2/3 )S 1/2 (11) V = (1.49/n)((D/4) 2/3 )S 1/2 (12) D = 4[Vn/(1.49S 1/2 )] 3/2 (13) D = 1.33Qn/S 1/2 (14) Note that these four equations are for the U.S. units previously specified. For S.I. units, the 1.49 should be replaced with 1.00 in the first three equations. In Equation (14), 1.33 should be replaced with 0.893. 27 Hydraulic design of storm sewers is typically based on full pipe flow using equations (11) through (14). Example #7: What would be the flow rate and velocity in a 30 inch diameter storm sewer that has n = 0.011 and slope = 0.00095, when it is flowing full under gravity? Solution: Substituting the given values of n, D, and S into Equation (12) gives: V = (1.49/0.011)((30/12)/4) 2/3 = 3.052 ft/sec = V Then Q can be calculated from Q = VA = V( D2/4) Q = (3.05 ft/sec)[ (2.5 2)/4 ft 2] = 15.0 cfs = Q D. Circular Pipes Flowing Partially Full : Although hydraulic design of storm sewers is typically done on the basis of the circular pipe flowing full, a storm sewer will often flow partially full due to a storm of intensity less than the design storm. Thus, there is sometimes interest in calculations for partially full pipe flow, such as the flow rate or velocity at a given depth of flow or the depth of flow for a given velocity or flow rate. Figure 12 shows the depth of flow, y, and the diameter, D, as used for partially full pipe flow calculations. 28 Figure 12. Depth of flow, y, and Diameter, D, for Partially Full Pipe Flow Graphical Solution: One common way of handling partially full pipe flow calculations is through the use of a graph that correlates V/V full and Q/Q full to y/D, as shown in Figure 13. Figure 13. Flow Rate and Velocity Ratios in Pipes Flowing Partially Full 29 If values of D, V full and Q full are known or can be calculated, then the velocity, V, and flow rate, Q, can be calculated for any depth of flow, y, in that pipe through the use of figure 13. Example #8: What would be the velocity and flow rate in the storm sewer of Example #7 (D = 30”, n = 0.011, S = 0.00095) when it is flowing at a depth of 12 inches? Solution: From the solution to Example #7: V full = 3.052 ft/sec and Q full = 15.0 cfs. From the given y and D values: y/D = 12/30 = 0.40. From Figure 13, for y/D = 0.40, V/V full = 0.70 and Q/Q full = 0.25. V = (V/V full )V full = (0.70)(3.052) ft/sec = 2.14 ft/sec = V Q = (Q/Q full )Q full = (0.25)(15.0) cfs = Q Background on Equations for Partially Full Pipe Flow: There are equations available to calculate the A and P for any depth of flow in a circular pipe (as presented below). These equations allow calculation of the hydraulic radius for partially full pipe flow. If the hydraulic radius calculated by this method is used with the Manning equation, using the full pipe value for n, the calculated flow rate and velocity don’t agree well with experimental measurements. This was observed by T.R Camp in 1946 (reference #3). Camp developed a method that uses Manning roughness, n, to be variable as a function of y/D, which makes calculated results agree with experimental measurements. Camp is the original source for a diagram like Figure 13, which gives V/V full , Q/Q full , and n/n full as functions of y/D. The graphs in Figure 13 were created using values read from a similar graph in Steel & McGhee (reference # 2). Equations for less than half full pipe flow: The diagram and equations below summarize the calculation of A, P, & R h for a pipe flowing less than half full 30 Figure 14. Diagram and Equations for Less Than Half Full Pipe Flow For known pipe diameter, D, and depth of flow, y, the equations above allow calculation of cross-sectional area of flow, A, and wetted perimeter, P. Then the hydraulic radius can be calculated from R h = A/P, for a pipe flowing less than half full. Equations for more than half full pipe flow: The diagram and equations below summarize the calculation of A, P, & R h for a pipe flowing more than half full: 31 Figure 15. Diagram and Equations for More Than Half Full Pipe Flow Similarly, this set of equations allow calculation of A, P, and R h if the pipe diameter and depth of flow are known for more than half full pipe flow. Equation for n/n full : As discussed above, in addition to the value of the hydraulic radius, the value of Mannings roughness coefficient is needed at the given y/D value in order to proceed with a Manning equation calculation. The following equation can be used to calculate n/n full as a function of y/D :32 The source for this equation is: Goswami, I., Civil Engineering All-in-One PE Exam Guide Breadth and Depth, 2nd Ed, McGraw-Hill, NY, NY, 2012, Equation 303.32 These equations provide all of the tools necessary to make Manning equation calculations for partially full pipe flow. With all of the equations and all of the steps required, Excel spreadsheets are ideal for this type of calculation. For a more detailed discussion of Manning equation calculations for partially full pipe flow, and the use of spreadsheets for those calculations, see reference #6 at the end of this course. 8. Uniform Flow Calculations for Natural Open Channels The Manning equation is used a lot for natural channel flow calculations, as well as with the manmade channel examples we’ve already considered. One of the primary differences in using the Manning equation for natural channel flow is the lack of precision in estimation of Manning roughness coefficient values. A. The Manning Roughness Coefficient: There are several approaches available for determining the Manning roughness coefficient, n, for flow in a natural open channel, including i) experimental determination of n; ii) use of a table or tables that give maximum, minimum and average n values for a variety of channel descriptions; and iii) a method devised by Cowan (reference #4) that uses a base n value determined by the general type of channel and modifies that base n value based on various descriptors of the channel. A bit more about each of these methods follows. 33 i) Experimental determination of the Manning roughness coefficient can be accomplished by measuring the depth of flowing water, the size and shape of the channel cross-section, and the volumetric flow rate for a given reach of channel. These measured values can be used to calculate an empirical value for n that can then be used for subsequent Manning equation calculations for that reach of channel. Example #9: Calculate the Manning roughness coefficient, n, for a reach of river channel that has a bottom slope of 0.00028, with a cross-section that can be approximated as a trapezoid with bottom width equal to 8 ft and side slopes of horiz:vert = 4:1, if the flow rate has been estimated to be 75 cfs in that reach when the depth of water is 3 ¼ ft. Solution: The Manning equation can be solved for n to give: n = (1.49/Q)A(R h2/3 )S 1/2 The area and hydraulic radius of the trapezoidal cross-section are calculated as follows: A = by + zy 2 = 83.25 + 43.25 2 = 68.25 ft 2P = b + 2y(1 + z 2) 1/2 = 8 + 23.25(1 + 4 2) 1/2 = 34.80 ft Rh = A/P = 68.25/34.80 = 1.96 ft Substituting values into the above equation for n gives: n = (1.49/75)(68.25)(1.96 2/3 )(0.00028 1/2 ) = 0.0355 = n ii) There are Tables of n values in many textbooks and handbooks, as well as on websites. The table on the next two pages is an example from the Indiana Department of Transportation Design Manual (website ref # 1 ). Similar tables of n values are available on many state agency websites. The table below from the Indiana DOT Design Manual gives minimum, maximum, and normal values of the Manning roughness coefficient for a range of excavated or dredged and natural stream channels. 34 35 Table 3. Manning Roughness Coefficient, n, for Natural Channels 36 Example #10: What are the minimum, maximum, and normal values of the Manning roughness coefficient, n, for a minor mountain stream with no vegetation in the channel, banks usually steep, trees and brush along banks submerged at high stages, and cobbles with large boulders on the bottom, based on Table 3 below, from the Indiana DOT Design Manual? Solution: From Table 3, the values of n for the described natural channel are: nmin = 0.040, n max = 0.07, n normal = 0.050 iii) The Cowan procedure was first presented in reference #4. There is also a good description of this method in McCuen (reference #5). This procedure uses a base n value with several terms added to it based on characteristics of the channel, as described below. 1. The Base Roughness Coefficient, n 1, is selected from the following based on the character of the channel:  Channels in earth: n 1 = 0.02  Channels cut into rock: n 1 = 0.025  Channels in fine gravel: n 1 = 0.024  Channels in coarse gravel: n 1 = 0.028 2. The value for the Irregularity Modifier, n 2, is selected from the following based on the degree of irregularity:  Smooth (surface comparable to the best attainable for the materials involved) n 2 = 0.000  Minor (good dredged channels; slightly eroded or scoured side slopes of canals) n 2 = 0.005 37  Moderate (fair to poor deredged channels; moderately sloughed or eroded canal side slopes) n 2 = 0.010  Severe (badly sloughed banks of natural streams; badly eroded or sloughed sides of canals or drainage channels; unshaped, jagged and irregular surfaces of channels excavated in rock n 2 = 0.020 3. The value for the Cross Section Modifier, n 3, is selected from the following based on the character of variations in size & shape of cross section:  Change in size or shape occurs gradually n 3 = 0.000  Large & small sections alternate occasionally or shape changes cause occasional shifting of main flow from side to side n 3 = 0.005  Large & small sections alternate frequently or shape changes cause frequent shifting of main flow from side to side n 3 = 0.010 – 0.020 4. The value for the Obstruction Modifier, n 4, is selected from the following based on the relative effect of obstructions:  Negligible n 4 = 0.000  Minor n 4 = 0.010 – 0.015  Appreciable n 4 = 0.020 – 0.030  Severe n 4 = 0.040 – 0.060 38 5. The value for the Vegetation Modifier, n 5, is selected from the following based on the degree of vegetation effect on n:  Low n 5 = 0.005 – 0.010  Medium n 5 = 0.010 – 0.020  Appreciable n 5 = 0.020 – 0.050  Very High n 5 = 0.050 – 0.100 6. The value for the Menadering Modifier, n 6, is selected from the following based on the degree of meander:  Minor ( meander length :straight length = 1.0 – 1.2 ) n 6 = 0.000  Appreciable ( meander length :straight length = 1.2 – 1.5 ) n 6 = 0.15n s  Severe ( meander length :straight length > 1.5 ) n 6 = 0.30n sWhere n s = n 1 + n 2 + n 3 + n 4 + n 57. The value of the Manning Roughness Coefficient is calculated from: n = n 1 + n 2 + n 3 + n 4 + n 5 + n 6 Example #11: Estimate the value for the Manning roughness coefficient for a channel in earth with minor irregularity, only gradual changes in size or shape, minor obstructions, medium effect of vegetation, and minor meander. 39 Solution: From the lists above: n 1 = 0.02, n 2 = 0.005, n 3 = 0.000, n 4 =0.010 – 0.015, n 5 = 0.010 – 0.020, and n 6 = 0.000. Choosing the midpoint of the ranges given for n 4 and n 5 (n 4 = 0.0125 and n 5 = 0.015) gives the following equation for n: n = 0.02 + 0.005 + 0.000 + 0.0125 + 0.015 + 0.000 n = 0.0525 B. Manning Equation Calculations for natural channels are the same as for manmade channels except for less precision in estimating the Manning roughness coefficient and greater difficulty in determining the hydraulic radius if the channel cross-section isn’t a simple shape. Example #12: A reach of channel of a minor stream on a plain is described as clean, winding, with some pools. The bottom slope is fairly constant at 0.00031 for this reach of channel. Its cross-section over this reach can be approximated as a trapezoid with a bottom width of 6 feet and side slopes of horiz :vert = 3:1. Find the range of flow rates that could be expected for this reach of channel for a 3.75 ft depth of flow, based on the maximum and minimum values of the Manning roughness coefficient from Table 3. Solution: From the information given in the problem statement, b = 6 ft, y = 3.75 ft, z = 3, and S = 0.00031. From Table 3 (for the channel description given in the problem statement), n min = 0.033 and n max = 0.045. From the equations for A, P, & R h for a trapezoidal open channel: Rh = [(6)(3.75) + 3(3.75 2)]/[6 + (2)(3.75)(1 + 3 2) 1/2 ] = 2.177 ft A = (6)(3.75) + 3(3.75 2) = 64.69 ft 2Substituting values into the Manning Equation [Q = (1.49/n)A(R h2/3 )S 1/2 ] gives the following results: 40 For n min (0.033): Q max = (1.49/0.033)(64.69)(2.177 2/3 )(0.00031) 1/2 Qmax = 86.4 ft 3/sec For n max (0.045): Q min = (1.49/0.045)(64.69)(2.177 2/3 )(0.00031) 1/2 Qmin = 63.3 ft 3/sec 9. Summary Open channel flow, which has a free liquid surface at atmospheric pressure, occurs in a variety of natural and man-made settings. Open channel flow may be classified as i) laminar or turbulent, ii) steady state or unsteady state, iii) critical, subcritical, or supercritical, and iv) uniform or nonuniform flow. Many practical cases of open channel flow can be treated as turbulent, steady state, uniform flow. Several open channel flow parameters are related through the empirical Manning Equation, for turbulent, uniform open channel flow ( Q = (1.49/n)A(R h2/3 )S 1/2 ). The use of the Manning equation for uniform open channel flow calculations and for the calculation of parameters in the equation, such as cross-sectional area and hydraulic radius, are illustrated in this course through worked examples for manmade channels and for natural channels. 10. References and Websites Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4 th Ed., New York: John Wiley and Sons, Inc, 2002. 2. Steel, E.W. & McGhee, T.J., Water Supply and Sewerage, 5 th Ed., New York, McGraw-Hill Book Company, 1979. 41 3. Camp, T.R., “Design of Sewers to Facilitate Flow,” Sewage Works Journal, 18 (3), 1946. 4. Cowan, W.L., "Estimating Hydraulic Roughness Coefficients," Agricultural Engineering, Vol. 37: 473-475, 1956. 5. McCuen, R.H., Hydrologic Analysis and Design , 2nd Ed., Prentice Hall, Upper Saddle River, NJ, 1998. Bengtson, H.H., “Partially Full Pipe flow Calculations with Excel Spreadsheets,” an online article at www.engineeringexcelspreadsheets.com. 7. Bengtson, H.H., “The Manning Equation for Open Channel Flow Calculations”, available as a paperback and as an Amazon Kindle e-book. Websites: Indiana Department of Transportation Design Manual, available on the internet at: 2. Illinois Department of Transportation Drainage Manual, available on the internet at: 3. Low Cost Easy to Use Excel Spreadsheets for Engineering Calculations, at www.engineeringexceltemplates.com.
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1973	https://math.montana.edu/jobo/st446/documents/ho2a.pdf	2 SIMPLE PROBABILITY SAMPLES 2.1 Types of Probability Samples • A simple random sample (SRS) of size n is a sample of n units selected from a population in such a way that every sample of size n has an equal chance of being selected. – The Board of Regents at a university is deciding whether or not every one of its students must have two semesters of statistics coursework in order to fulﬁll their degree requirements. The student senate selected 200 students at random from the entire student body and asked their opinion on this issue. It was found that 1% of the students sampled supported a requirement of two semesters of statistics coursework. – A sample of twenty MSU instructors is selected based on the following scheme: An alphabetical list of all the instructors is prepared, then a unique number is assigned in sequence (1, 2, 3, . . .) to each instructor, and ﬁnally, using a random number generator, twenty instructors are randomly chosen. • A stratiﬁed random sample is a sample selected by ﬁrst dividing the population into non-overlapping groups called strata and then taking a simple random sample within each stratum. Dividing the population into strata should be based on some criterion so that units are similar within a stratum but are diﬀerent between strata. – The Career Services staﬀwants to know if they are adequately meeting the needs of the students they serve. They design a survey which addresses this question, and they send the survey to 50 students randomly chosen from each of the university’s colleges (50 agriculture students, 50 arts and science students, 50 engineering students, etc.). – A biologist is interested in estimating a deer population total in a small geographic region. The region contains two habitat types which are known to inﬂuence deer abundance. From each habitat type, 10 plots are randomly selected to be surveyed. – Note: Stratum sample sizes do not have to be equal. • A systematic sample is a sample in which units are selected in a “systematic” pattern in the population of interest. To take a systematic sample you will need to divide the sampling frame into groups of units, randomly choose a set of starting points in the ﬁrst group, and then sample from every group using the same positions of the starting points. – You are a quality engineer at Intel and are testing the quality of newly-produced computer chips. You need to take a sample of chips and test their quality. As the chips roll oﬀthe production line, you decided to test every 50th chip produced starting with the third chip (i.e., sample chips 3, 53, 103, 153, ...). • Suppose the observation units in a population are grouped into non-overlapping sets called clusters. A cluster sample is a SRS of clusters. – You work for the Department of Agriculture and wish to determine the percentage of farmers in the United States who use organic farming techniques. It would be diﬃcult and costly to collect a SRS or a systematic sample because both of these sampling designs would require visiting many individual farms that are located far from each other. A convenience sample of farmers from a single county would be 12 biased because farming practices vary from region to region. You decide to select several dozen counties from across the United States and then survey every farmer in each of these selected counties. Each county contains a cluster of farmers and data is collected on every farm within the randomly selected counties (clusters). • A multistage sample is a sample acquired by successively selecting smaller groups within the population in stages. The selection process at any stage may employ any sampling design (such as a SRS or a stratiﬁed sample). – A city oﬃcial is investigating rumors about the landlord of a large apartment building 13 complex. To get an idea of the tenants’ opinions about their landlord, the oﬃcial takes a SRS of buildings in the complex followed by a SRS of apartments from each selected building. From each chosen apartment a resident is interviewed. – A U.S. national opinion poll was taken as follows: First, the U.S. is stratiﬁed into 4 regions. Then a random sample of counties was selected from each region followed by a random sample of townships within each of these counties. Finally, a random sample of households (clusters) within each township is taken. 2.2 Probability Sampling Designs • Suppose N is the population size. That is, there are N units in the universe or ﬁnite population of interest. • The N units in the universe are denoted by an index set of labels: U = { 1, 2, 3, . . . N } Note: Some texts will denote U = {u1, u2, u3, . . . uN}. • From this universe (or population) a sample of n units is to be taken. Let S represent a sample of n units from U. • Associated with each of the N units is a measurable value related to the population characteristic of interest. Let yi be the value associated with unit i, and the population of y-values is {y1, y2, . . . , yN}. • A point of clariﬁcation is needed here. On page 28, Lohr states that S is “a subset consisting of n of the units in U. – S will only be a ‘subset’ in a strict mathematical sense if sampling is done without replacement. That is, a unit can only appear in the sample at most once. – However, S will not be a ‘subset’ if sampling is done with replacement. That is, a unit can appear in the sample multiple times. • Sampling designs that are based on planned randomness are called probability samples, and a probability P(S) is assigned to every possible sample S. • The probability that unit i will be included in a sample is denoted πi and is called the inclusion probability for unit i. • In many sampling procedures, diﬀerent units in the population have diﬀerent probabilities of being included in a sample (i.e., diﬀerent inclusion probabilities) that depend on either (i) the type of sampling procedure or (ii) the probabilities may be imposed by the researcher to obtain better estimates by including “more important” units with higher probability. • Example: Suppose unit selections are drawn with probability proportional to unit size and that sampling is done with replacement of units. – The population total = 16. There are N = 5 sampling units. The ﬁgure shows the units, labeled 1 to 5, and the ﬁve yi values. 14 – Sampling plan: You are to select two units. The ﬁrst unit ui is selected with proba-bility pi proportional to its size and the yi is recorded. The unit is then put back. A second unit uj is selected using the same method for selecting the ﬁrst unit, and its yj value is recorded. – Note that the same unit can be sampled twice. This is an example of sampling with replacement. The following table describes the population. 1 2 3 4 5 yi − → pi − → .4 .3 .1 .1 .1 – The table below shows all 15 possible pairs of sampled y-values (but not ordered). E.g., (1,2) means you selected unit 1 then unit 2 or you selected unit 2 then unit 1. In either case, you end up with the same sample of size 2. Note 15 X i=1 P(Si) = 1. Sample Units y-values P(Sj) Calculation S1 {1, 2} 7 , 4 0.24 (.4)(.3) + (.3)(.4) S2 {1, 3} 7 , 0 0.08 (.4)(.1) + (.1)(.4) S3 {1, 4} 7 , 2 0.08 (.4)(.1) + (.1)(.4) S4 {1, 5} 7 , 3 0.08 (.4)(.1) + (.1)(.4) S5 {2, 3} 4 , 0 0.06 (.3)(.1) + (.1)(.3) S6 {2, 4} 4 , 2 0.06 (.3)(.1) + (.1)(.3) S7 {2, 5} 4 , 3 0.06 (.3)(.1) + (.1)(.3) S8 {3, 4} 0 , 2 0.02 (.1)(.1) + (.1)(.1) S9 {3, 5} 0 , 3 0.02 (.1)(.1) + (.1)(.1) S10 {4, 5} 2 , 3 0.02 (.1)(.1) + (.1)(.1) S11 {1, 1} 7 , 7 0.16 (.4)(.4) S12 {2, 2} 4 , 4 0.09 (.3)(.3) S13 {3, 3} 0 , 0 0.01 (.1)(.1) S14 {4, 4} 2 , 2 0.01 (.1)(.1) S15 {5, 5} 3 , 3 0.01 (.1)(.1) The inclusion probability πi is found by summing P(Si) over all samples containing unit i. Thus, the inclusion probabilities when sampling with replacement are π1 = = .24 + .08 + .08 + .08 + .16 π2 = = .24 + .06 + .06 + .06 + .09 π3 = = .08 + .06 + .02 + .02 + .01 π4 = = .08 + .06 + .02 + .02 + .01 π5 = = .08 + .06 + .02 + .02 + .01 • Example Suppose unit selections are drawn with probability proportional to unit size, and that sampling is done without replacement of units. – The population is the same as the previous example. 15 1 2 3 4 5 yi − → 7 4 0 2 3 pi − → .4 .3 .1 .1 .1 – Sampling plan: You are to select two units. The ﬁrst unit ui is selected with prob-ability pi proportional to its size and the yi is recorded. The unit is not put back. A second unit uj is selected using sampling proportional to size for the remaining 4 units, and its yj value is recorded. – Note that the same unit cannot be sampled twice. This is an example of sampling without replacement. The following table shows all 10 possible pairs of sampled y-values (but not ordered). Note: 10 X i=1 P(Si) = 315/315 = 1. – Once that once the ﬁrst unit is selected, the probabilities for selecting the second unit are no longer pi. The probabilities are proportional to sizes of the remaining 4 units. – For example, to ﬁnd the probability of selecting units 1 and 2, we need to calculate the probabilities of (i) selecting unit 1 ﬁrst, then unit 2 and (ii) selecting unit 2 ﬁrst, then unit 1. ∗The probability of selecting unit 1 ﬁrst is .4. Now only units 2, 3, 4, and 5 remain with unit 2 accounting for 3/6 (=.3/(.3+.1+.1+.1) = .3/.6) of the remaining sizes. Thus, the probability of selecting unit 1 then unit 2 = (.4)(.3/.6). ∗The probability of selecting unit 2 ﬁrst is .3. Now only units 1, 3, 4, and 5 remain with unit 1 accounting for 4/7 (=.4/(.4+.1+.1+.1) = .4/.7) of the remaining sizes. Thus, the probability of selecting unit 2 then unit 1 = (.3)(.4/.7). ∗Thus, the probability of sampling units 1 and 2 is the sum of these two proba-bilities: (.4)(.3/.6) + (.3)(.4/.7). Sample Units y-values P(Sj) Calculation S1 {1, 2} 7 , 4 13/35 = 117/315 ≈.371 (.4)(.3/.6) + (.3)(.4/.7) S2 {1, 3} 7 , 0 1/9 = 35/315 ≈.111 (.4)(.1/.6) + (.1)(.4/.9) S3 {1, 4} 7 , 2 1/9 = 35/315 ≈.111 (.4)(.1/.6) + (.1)(.4/.9) S4 {1, 5} 7 , 3 1/9 = 35/315 ≈.111 (.4)(.1/.6) + (.1)(.4/.9) S5 {2, 3} 4 , 0 8/105 = 24/315 ≈.076 (.3)(.1/.7) + (.1)(.3/.9) S6 {2, 4} 4 , 2 8/105 = 24/315 ≈.076 (.3)(.1/.7) + (.1)(.3/.9) S7 {2, 5} 4 , 3 8/105 = 24/315 ≈.076 (.3)(.1/.7) + (.1)(.3/.9) S8 {3, 4} 0 , 2 1/45 = 7/315 ≈.022 (.1)(.1/.9) + (.1)(.1/.9) S9 {3, 5} 0 , 3 1/45 = 7/315 ≈.022 (.1)(.1/.9) + (.1)(.1/.9) S10 {4, 5} 2 , 3 1/45 = 7/315 ≈.022 (.1)(.1/.9) + (.1)(.1/.9) Thus, the inclusion probabilities when sampling without replacement are π1 = 74/105 = (117 + 35 + 35 + 35)/315 ≈ π2 = 63/105 = (117 + 24 + 24 + 24)/315 ≈ π3 = 73/315 = (35 + 24 + 7 + 7)/315 ≈ π4 = 73/315 = (35 + 24 + 7 + 7)/315 ≈ π5 = 73/315 = (35 + 24 + 7 + 7)/315 ≈ 16 2.2.1 Parameters, Statistics, Expectations, and Estimation Bias • One goal of sampling is to draw conclusions about a population of interest based on the data collected. This process of drawing conclusions is called statistical inference. • A parameter is a value which describes some characteristic of a population (or possibly describes the entire population). • A statistic is a value that can be computed from the observed (sample) data without making use of any unknown parameters. • Unless the statistic and parameter are explicitly stated, we will use b θ and θ to represent an unspeciﬁed statistic and parameter of interest, respectively. • In general, the value of a population parameter is unknown. Statistics computed from sampling data can provide information about the unknown parameter. • Common statistics of interest: Let y1, y2, . . . , yn be a sample of y-values. – The sample mean is y = 1 n n X i=1 yi. – The sample variance is s2 = 1 n −1 (y1 −y)2 + (y2 −y)2 + · · · + (yn −y)2 = 1 n −1 n X i−1 (yi −y)2 = 1 n −1 X y2 i −1 n X yi 2 – The sample standard deviation s is √ s2. • Common parameters of interest: – Notation: Let parameter t be the population total and parameter yU be the pop-ulation mean from a ﬁnite population of size N. Thus, t = N X i=1 yi yU = 1 N N X i=1 yi = (1) – The population variance parameter S2 is deﬁned as: S2 = 1 N −1 N X i=1 (yi −yU)2 (2) = 1 N −1 N X i=1 y2 i −t2 N ! = 1 N −1 N X i=1 y2 i −Ny2 U ! – The population standard deviation parameter S is deﬁned as S = √ S2. – In other texts, τ, µ, σ2 and σ are used to represent the population total t, mean yU, variance S2, and standard deviation S. 17 • Because only a part of the population is sampled in any sampling plan, the value of a statistic b θ will vary in repeated random sampling. The inherent variability of b θ associated with sampling is called sampling variability. • The sampling distribution of a statistic b θ is the probability distribution of the values that can be observed for the statistic over all possible samples for a given sampling scheme. • The expected value of b θ, denoted E[b θ] is the mean of the sampling distribution of b θ: E[b θ] = X S b θSP(S) (3) = X k kP(b θ = k) (4) – In (3): b θS is the value of b θ calculated for sample S and the summation is taken over all possible samples (S). Thus, E[b θ] is the weighted average of b θ calculated over all possible samples with weights P(S). – In (4): The summation is taken over k = the set of possible values that can be observed for b θ. Thus, E[b θ] is the weighted average of the possible values of b θ with weights P(b θ = k) = the probability of observing b θ = k. – These represent two approaches for calculating E[b θ]. • The (estimation) bias of the estimator b θ for estimating a parameter θ is the numerical diﬀerence between E[b θ] and the parameter value θ. That is, Bias[b θ] = E[b θ] −θ. • An estimator b θ is unbiased to estimate a parameter θ if Bias[b θ] = 0. • Estimation example: Consider the small population of N = 4 values: 0, 3, 6, 12. For this population, the population mean yU = 21/4 = and the population variance S2 = 1 N −1 N X i=1 (yi −yU)2 = 1 3 5 X i=1 (yi −5.25)2 = 1 3 (0 −5.25)2 + (3 −5.25)2 + (6 −5.25)2 + (12 −5.25)2 = 1 3 (−5.25)2 + (−2.25)2 + (0.75)2 + (6.75)2 = 1 3(27.5625 + 5.0625 + 0.5625 + 45.5625) = 78.75 3 = . Thus, S = √ 26.25 ≈ . Now, consider the following two sampling schemes, and assume the probabilities for se-lecting a sampling unit are all equal within each stage. Scheme I: Take a sample of size n = 2 with replacement. Because there are 4×4 = 16 ordered sampling sequences, each one has probability = 1/16. See Table 2.1A. Scheme II: Take a sample of size n = 2 without replacement. Because there are 4 × 3 = 12 ordered sampling sequences, each one has probability = 1/12. See Table 2.1B. 18 Table 2.1A: Sample means, variances, and standard deviations for all possible samples selected with replacement and n = 2 S y-values P(S) yS s2 S sS ySP(S) s2 SP(S) sSP(S) S1 0 , 3 P(S1) = 2/16 1.5 4.5 2.1213 3/16 9/16 0.2652 S2 0 , 6 P(S2) = 2/16 3.0 18.0 4.2426 6/16 36/16 0.5303 S3 0 , 12 P(S3) = 2/16 6.0 72.0 8.4853 12/16 144/16 1.0606 S4 3 , 6 P(S4) = 2/16 4.5 4.5 2.1213 9/16 9/16 0.2652 S5 3 , 12 P(S5) = 2/16 7.5 40.5 6.3640 15/16 81/16 0.7955 S6 6 , 12 P(S6) = 2/16 9.0 18.0 4.2426 18/16 36/16 0.5303 S7 0 , 0 P(S7) = 1/16 0.0 0.0 0 0/16 0/16 0 S8 3 , 3 P(S8) = 1/16 3.0 0.0 0 5/16 0/16 0 S9 6 , 6 P(S9) = 1/16 6.0 0.0 0 6/16 0/16 0 S10 12 , 12 P(S10) = 1/16 12.0 0.0 0 12/16 0/16 0 84/16 315/16 ≈3.4471 Table 2.1B: Sample means, variances, and standard deviations for all possible samples selected without replacement and n = 2 S y-values P(S) yS s2 S sS ySP(S) s2 SP(S) sSP(S) S1 0 , 3 P(S1) = 2/12 1.5 4.5 2.1213 3/12 9/12 0.3536 S2 0 , 6 P(S2) = 2/12 3.0 18.0 4.2426 6/12 36/12 0.7071 S3 0 , 12 P(S3) = 2/12 6.0 72.0 8.4853 12/12 144/12 1.4142 S4 3 , 6 P(S4) = 2/12 4.5 4.5 2.1213 9/12 9/12 0.3536 S5 3 , 12 P(S5) = 2/12 7.5 40.5 6.3640 15/12 81/12 1.0607 S6 6 , 12 P(S6) = 2/12 9.0 18.0 4.2426 18/12 36/12 0.7071 63/12 315/12 ≈4.9497 Estimation using equation (3): • Recall: the parameter values are yU = 5.25, S2 = 26.25, and S ≈5.1235. • Using equation (3) and Table 2.1A, we get E[y] = X S ySP(S) = 84/16 = 5.25 − →Bias[y] = 5.25 −5.25 = (5) E[s2] = X S s2 SP(S) = 315/16 = 19.6875 − →Bias[s2] = 19.6875 −26.25 = E[s] = X S sSP(S) ≈3.4471 − →Bias[s] ≈3.4471 −5.1235 = Therefore, y is an unbiased estimator for yU, but s2 and s are biased estimators for S2 for S for Scheme I. • Using equation (3) and Table 2.1B, we get E[y] = X S ySP(S) = 63/12 = 5.25 − →Bias[y] = 5.25 −5.25 = (6) E[s2] = X S S2 SP(S) = 315/12 = 26.25 − →Bias[s2] = 315/12 −26.25 = E[s] = X S sSP(S) ≈4.9497 − →Bias[s] ≈4.9497 −5.1235 = Therefore, y and s2 are an unbiased estimators for yU and S2, respectively, but s is a biased estimator for S for Scheme II. 19 Estimation using equation (4): • To ﬁnd the sampling distributions of y, s2, and s, you determine all possible values that can be observed for each statistic and the associated probabilities for observing each of these values. • Let k represent a possible value of a statistic. • Tables 2.2A and 2.2B show the sampling distributions of y, s2, and s for Scheme I (sampling with replacement) and II (sampling without replacement), respectively. A third (product) column is included for each statistic to calculate expectations. • Note that we will get the same results concerning expected values and biases as those summarized in (5) and (6). Table 2.2A: Sampling distribution of y, s2, and s for Scheme I and expected value calculations for E(y), E(s2), and E(s) k P(y = k) kP(y = k) k P(s2 = k) kP(s2 = k) k P(s = k) kP(s = k) 0.0 1/16 0/16 0.0 4/16 0/16 √ 0.0 4/16 0 1.5 2/16 3/16 4.5 4/16 18/16 √ 4.5 4/16 ≈0.5303 3.0 3/16 9/16 18.0 4/16 72/16 √ 18.0 4/16 ≈1.0607 4.5 2/16 9/16 40.5 2/16 81/16 √ 40.5 2/16 ≈0.7955 6.0 3/16 18/16 72.0 2/16 144/16 √ 72.0 2/16 ≈1.0607 7.5 2/16 15/16 9.0 2/16 18/16 12.0 1/16 12/16 E[y] = 84/16 E[S2] = 315/16 E[S] ≈ Table 2.2B: Sampling distribution of y, s2, and s for Scheme II and expected value calculations for E(y), E(s2), and E(s) k P(y = k) kP(y = k) k P(s2 = k) kP(s2 = k) k P(s = k) kP(s = k) 1.5 2/12 3/12 4.5 4/12 18/12 √ 4.5 4/12 ≈0.7071 3.0 2/12 6/12 18.0 4/12 72/12 √ 18.0 4/12 ≈1.4142 4.5 2/12 9/12 40.5 2/12 81/12 √ 40.5 2/12 ≈1.0607 6.0 2/12 12/12 72.0 2/12 144/12 √ 72.0 2/12 ≈1.4142 7.5 2/12 15/12 9.0 2/12 18/12 E[y] = 63/12 E[S2] = 315/12 E[S] ≈ • Important: You cannot make general statements about bias (such as “statistic b θ is an unbiased (biased) estimator of parameter θ”. You must also know the probability sampling plan to determine whether or not a statistic is biased or unbiased. • For example, it will not always be true that y will be an unbiased estimator of yU. That is, there will be sampling plans for which E(y) ̸= yU. 20 2.2.2 The Variance and Mean Squared Error of a Statistic • So far the focus has been on the expected value of a statistic to check for bias. Another natural concern is the variability (spread) of the statistic. It is certainly possible for an unbiased statistic to have large variability. • We will consider two measures of variability: the variance and the mean squared error. • The variance of the sampling distribution of b θ (or simply, V (b θ)) is deﬁned to be V (b θ) = E h (b θS −E(b θ))2i = X S P(S) h (b θS −E(b θ) i2 (7) where b θS is the value of b θ calculated for sample S. • The mean squared error is deﬁned to be MSE[b θ] = E h (b θ −θ)2i = X S P(S) h (b θS −θ) i2 (8) • The MSE, however, can be rewritten: MSE[b θ] = E h (b θ −θ)2i = E (b θ −E[b θ]) + (E[b θ] −θ) 2 = E h (b θ −E[b θ])2 + (E[b θ] −θ)2 + 2(b θ −E[b θ])(E[b θ] −θ) i = E h (b θ −E[b θ])2i + E h (E[b θ] −θ)2i + 2E h (b θ −E[b θ]) i (E[b θ] −θ) = E h (b θ −E[b θ])2i + E h (E[b θ] −θ)2i + 2 (0) (E[b θ] −θ) = (9) We used (A + B)2 = A2 + B2 + 2AB with A = b θ −E[b θ] and B = E[b θ] −θ. • The relationship between variance and mean squared error: – The variance is the average of the squared deviations of the statistic values about the mean (expected value) of the statistic. – The mean squared error is the average of the squared deviations of the statistic values about the parameter. – Thus, for an unbiased statistic, the variance and the mean squared error are identical (i.e. V (b θ) = MSE(b θ)). 21
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Reject All Save My Preferences Accept All Skip to ContentGo to accessibility pageKeyboard shortcuts menu Log in Concepts of Biology Introduction Concepts of BiologyIntroduction Contents Contents Highlights Table of contents Preface The Cellular Foundation of Life Cell Division and Genetics Molecular Biology and Biotechnology Evolution and the Diversity of Life Animal Structure and Function Ecology 19 Population and Community Ecology 20 Ecosystems and the Biosphere Introduction 20.1 Energy Flow through Ecosystems 20.2 Biogeochemical Cycles 20.3 Terrestrial Biomes 20.4 Aquatic and Marine Biomes Key Terms Chapter Summary Visual Connection Questions Review Questions Critical Thinking Questions 21 Conservation and Biodiversity A \| The Periodic Table of Elements B \| Geological Time C \| Measurements and the Metric System Index Search for key terms or text. Close Figure 20.1 The (a) Karner blue butterfly and (b) wild lupine live in oak-pine barren habitats in North America. (credit a: modification of work by John & Karen Hollingsworth, USFWS) Chapter Outline 20.1 Energy Flow through Ecosystems 20.2 Biogeochemical Cycles 20.3 Terrestrial Biomes 20.4 Aquatic and Marine Biomes Ecosystem ecology is an extension of organismal, population, and community ecology. The ecosystem comprises all the biotic components (living things) and abiotic components (non-living things) in a particular geographic area. Some of the abiotic components include air, water, soil, and climate. Ecosystem biologists study how nutrients and energy are stored and moved among organisms and the surrounding atmosphere, soil, and water. Wild lupine and Karner blue butterflies live in an oak-pine barren habitat in portions of Indiana, Michigan, Minnesota, Wisconsin, and New York (Figure 20.1). This habitat is characterized by natural disturbance in the form of fire and nutrient-poor soils that are low in nitrogen—important factors in the distribution of the plants that live in this habitat. Researchers interested in ecosystem ecology study the importance of limited resources in this ecosystem and the movement of resources (such as nutrients) through the biotic and abiotic portions of the ecosystem. Researchers also examine how organisms have adapted to their ecosystem. PreviousNext Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. 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1975	https://www.youtube.com/watch?v=p_NuJoAO-cQ	Definite Integral of 1/sqrt(x) from 1 to 4 The Math Sorcerer 1220000 subscribers 98 likes Description 8418 views Posted: 1 Sep 2022 Definite Integral of 1/sqrt(x) from 1 to 4 If you enjoyed this video please consider liking, sharing, and subscribing. Udemy Courses Via My Website: Free Homework Help : My FaceBook Page: There are several ways that you can help support my channel:) Consider becoming a member of the channel: My GoFundMe Page: My Patreon Page: Donate via PayPal: Udemy Courses(Please Use These Links If You Sign Up!) Abstract Algebra Course Advanced Calculus Course Calculus 1 Course Calculus 2 Course Calculus 3 Course Calculus 1 Lectures with Assignments and a Final Exam Calculus Integration Insanity Differential Equations Course Differential Equations Lectures Course (Includes Assignments + Final Exam) College Algebra Course How to Write Proofs with Sets Course How to Write Proofs with Functions Course Trigonometry 1 Course Trigonometry 2 Course Statistics with StatCrunch Course Math Graduate Programs, Applying, Advice, Motivation Daily Devotionals for Motivation with The Math Sorcerer Thank you:) 3 comments Transcript: hi in this problem we're going to work out this definite integral so we're integrating from 1 to 4 and we have 1 over the square root of x so to do this problem we first have to write the integrand in a way that allows us to integrate it so because we have a square root of x here what we can do is we can write it as x to the one-half so i'll do it in steps the first step will be to rewrite the integral so we're going from one to four and it's one over x to the one-half and then here we still have the dx and again we're not in a place yet where we can integrate this so what we're going to do now is we're going to bring this upstairs so this is one to four x to the negative one-half and here we have the dx okay so now we're in a good place now we have x to a power so we can integrate recall the formula for integration says basically that you add one to this and you divide by the result so negative one-half plus one is really just one-half right because you take a minus half from a whole so you get a half so you add that to that right one plus negative one-half is one-half and then you divide by one-half right the formula says you add one to that and divide by the result and we're going from one to four so you draw a bracket and you go from one to four okay um because we're dividing by one half we're really multiplying by two over one so i'm just going to basically flip it and put it here this is two and then x to the one half is really the square root of x and we're going from one to four all right very good and the rule now says that first you plug in the 4 then you subtract then you plug in the 1. so this is 2. plug in the 4 subtract 2 plug in the 1. so this is equal to 2 times the square root of 4 is 2 so you get 2 times 2 minus and the square root of 1 is 1 so you get 2 times 1. 2 times 2 is 4 and then minus 2 and so you get 2. and so that would be the answer in this problem that would be the value 2 is the answer to this problem here good luck
1976	https://eptumed.com/skillLab3/modelShowOrgan.php?catalog=NDQmMTIyNA==	:: Skill Lab Learning :: Digestive System Admin Main Model Stomach Stomach The stomach is the expanded part of the digestive tract between the esophagus and small intestine. It is specialized for the accumulation of ingested food, which it chemically and mechanically prepares for digestion and passage into the duodenum. The stomach acts as a food blender and reservoir; its chief function is enzymatic digestion. The gastric juice gradually converts a mass of food into a semiliquid mixture, chyme (Greek juice), which passes fairly quickly into the duodenum. An empty stomach is only of slightly larger caliber than the large intestine; however, it is capable of considerable expansion and can hold 2-3 L of food. Position, Parts, and Surface Anatomy of Stomach The size, shape, and position of the stomach can vary markedly in persons of different body types (bodily habitus) and may change even in the same individual as a result of diaphragmatic movements during respiration, the stomach's contents (empty vs. after a heavy meal), and the position of the person. In the supine position, the stomach commonly lies in the right and left upper quadrants, or epigastric, umbilical, and left hypochondrium and flank regions. In the erect position, the stomach moves inferiorly. In asthenic (thin, weak) individuals, the body of the stomach may extend into the pelvis. The stomach has four parts: •Cardia: the part surrounding the cardial orifice (opening), the superior opening or inlet of the stomach. In the supine position, the cardial orifice usually lies posterior to the 6 th left costal cartilage, 2-4 cm from the median plane at the level of the T 11 vertebra. • Fundus: the dilated superior part that is related to the left dome of the diaphragm, limited inferiorly by the horizontal plane of the cardial orifice. The cardial notch is between the esophagus and the fundus. The fundus may be dilated by gas, fluid, food, or any combination of these. In the supine position, the fundus usually lies posterior to the left 6 th rib in the plane of the MCL. • Body: the major part of the stomach between the fundus and pyloric antrum. • Pyloric part: the funnel-shaped outflow region of the stomach; its wider part, the pyloric antrum, leads into the pyloric canal, its narrower part. The pylorus (Greek, gatekeeper) is the distal, sphincteric region of the pyloric part. It is a marked thickening of the circular layer of smooth muscle that controls discharge of the stomach contents through the pyloric orifice (inferior opening or outlet of the stomach) into the duodenum. Intermittent emptying of the stomach occurs when intragastric pressure overcomes the resistance of the pylorus. The pylorus is normally tonically contracted so that the pyloric orifice is reduced, except when emitting chyme (semifluid mass). At irregular intervals, gastric peristalsis pushes the chyme through the pyloric canal and orifice into the small intestine for further mixing, digestion, and absorption. In the supine position, the pyloric part of the stomach lies at the level of the transpyloric plane, midway between the jugular notch superiorly and the pubic crest inferiorly. The plane transects the 8 th costal cartilages and the L 1 vertebra. When erect its location varies from the L 2 through L 4 vertebra. The pyloric orifice is approximately 1.25 cm right of the midline. The stomach also features two curvatures: • Lesser curvature: forms the shorter concave right border of the stomach. The angular incisure (notch), the most inferior part of the curvature, indicates the junction of the body and pyloric part of the stomach. The angular incisure lies just to the left of the midline. • Greater curvature: forms the longer convex left border of the stomach. It passes inferiorly to the left from the junction of the 5 th intercostal space and MCL, then curves to the right, passing deep to the 9 th or 10 th left cartilage as it continues medially to reach the pyloric antrum. Because of the unequal lengths of the lesser curvature on the right and the greater curvature on the left, in most people the shape of the stomach resembles the letter J. Interior of Stomach The smooth surface of the gastric mucosa is reddish brown during life, except in the pyloric part, where it is pink. In life, it is covered by a continuous mucous layer that protects its surface from the gastric acid the stomach's glands secrete. When contracted, the gastric mucosa is thrown into longitudinal ridges or wrinkles called gastric folds (gastric rugae); they are most marked toward the pyloric part and along the greater curvature. During swallowing, a temporary groove or furrow-like gastric canal forms between the longitudinal gastric folds along the lesser curvature. It can be observed radiographically and endoscopically. The gastric canal forms because of the firm attachment of the gastric mucosa to the muscular layer, which does not have an oblique layer at this site. Saliva and small quantities of masticated food and other fluids drain along the gastric canal to the pyloric canal when the stomach is mostly empty. The gastric folds diminish and obliterate as the stomach is distended (fills). Relations of Stomach The stomach is covered by visceral peritoneum, except where blood vessels run along its curvatures and in a small area posterior to the cardial orifice. The two layers of the lesser omentum extend around the stomach and leave its greater curvature as the greater omentum. Anteriorly, the stomach is related to the diaphragm, left lobe of liver, and anterior abdominal wall. Posteriorly, the stomach is related to the omental bursa and pancreas; the posterior surface of the stomach forms most of the anterior wall of the omental bursa. The transverse colon is related inferiorly and laterally to the stomach as it courses along the greater curvature of the stomach to the left colic flexure. The bed of the stomach, on which the stomach rests in the supine position, is formed by the structures forming the posterior wall of the omental bursa. From superior to inferior, the bed of the stomach is formed by the left dome of the diaphragm, spleen, left kidney and suprarenal gland, splenic artery, pancreas, and transverse mesocolon. Vessels and Nerves of Stomach The rich arterial supply of the stomach arises from the celiac trunk and its branches. Most blood is supplied by anastomoses formed along the lesser curvature by the right and left gastric arteries, and along the greater curvature by the right and left gastro-omental (gastro-epiploic) arteries. The fundus and upper body receive blood from the short and posterior gastric arteries. The veins of the stomach parallel the arteries in position and course. The right and left gastric veins drain into the hepatic portal vein; the short gastric veins and left gastro-omental veins drain into the splenic vein, which joins the superior mesenteric vein (SMV) to form the hepatic portal vein. The right gastro-omental vein empties in the SMV. A prepyloric vein ascends over the pylorus to the right gastric vein. Because this vein is obvious in living persons, surgeons use it for identifying the pylorus. The gastric lymphatic vessels accompany the arteries along the greater and lesser curvatures of the stomach. They drain lymph from its anterior and posterior surfaces toward its curvatures, where the gastric and gastro-omental lymph nodes are located. The efferent vessels from these nodes accompany the large arteries to the celiac lymph nodes. The following is a summary of the lymphatic drainage of the stomach: • Lymph from the superior two thirds of the stomach drains along the right and left gastric vessels to the gastric lymph nodes; lymph from the fundus and superior part of the body of the stomach also drains along the short gastric arteries and left gastro-omental vessels to the pancreaticosplenic lymph nodes. • Lymph from the right two thirds of the inferior third of the stomach drains along the right gastro-omental vessels to the pyloric lymph nodes. • Lymph from the left one third of the greater curvature drains to the pancreaticoduodenal lymph nodes, which are located along the short gastric and splenic vessels. The parasympathetic nerve supply of the stomach is from the anterior and posterior vagal trunks and their branches, which enter the abdomen through the esophageal hiatus. The anterior vagal trunk, derived mainly from the left vagus nerve (CN X), usually enters the abdomen as a single branch that lies on the anterior surface of the esophagus. It runs toward the lesser curvature of the stomach, where it gives off hepatic and duodenal branches, which leave the stomach in the hepatoduodenal ligament. The rest of the anterior vagal trunk continues along the lesser curvature, giving rise to anterior gastric branches. The larger posterior vagal trunk, derived mainly from the right vagus nerve, enters the abdomen on the posterior surface of the esophagus and passes toward the lesser curvature of the stomach. The posterior vagal trunk supplies branches to the anterior and posterior surfaces of the stomach. It gives off a celiac branch, which passes to the celiac plexus, and then continues along the lesser curvature, giving rise to posterior gastric branches. The sympathetic nerve supply of the stomach, from the T 6 through T 9 segments of the spinal cord, passes to the celiac plexus through the greater splanchnic nerve and is distributed through the plexuses around the gastric and gastro-omental arteries. Tooth Tongue Hard palate Soft palate Palatine uvula Pharynx Esophagus Stomach Cardia Fundus of stomach Lesser curvature Greater curvature Pylorus Duodenum Pancreas Spleen Gallbladder Right lobe of liver Left lobe of liver Quadrate lobe Cystic duct Common hepatic duct Hepatic portal vein Proper hepatic artery Jejunum Ileum Ileocecal valve (Bauhin's valve) Ascending colon Transverse colon Descending colon Caecum Appendix Rectum Copyright 2018, Chulabhorn International College of Medicine www.cicm.tu.ac.th
1977	https://reshka.feniks.help/ekonomika/zadachi-na-procenty-i-pribyl	Задачи на проценты и прибыль: задачи, примеры, решения +7 495 727-22-67 Стать автором Войти Время — это деньги! Не нашли решение вашей задачи? Теперь Решка решает все задачи по любому предмету за 30 секунд Получить решение Задачи на проценты и прибыль Главная Экономика Задачи на проценты и прибыль Найти прибыль каждого предприятия за каждый месяц (июнь, июль, август) на основании количества произведенной продукции и данных о прибыли на единицу продукции для каждого вида фрукта Расчет чистого дисконтированного дохода, индекса прибыльности, срока окупаемости Найти трудоемкость продукции Виды работ: Автореферат Аннотация Аспирантский реферат Аттестационная работа Бакалаврская работа Бизнес-план Выпускная квалификационная работа Диплом MBA Дипломная работа Дипломная работа колледжа Диссертация Дистанционный экзамен, on-line тест Дневник практики Доклад Докторская диссертация Другое Исправление и доработка готовой работы Кандидатская диссертация Кейс Контрольная работа Копирайтинг Курсовая работа Лабораторная работа Литературный обзор, интерпретация текста Магистерская диссертация Маркетинговое исследование Монография Набор текста Научная статья Научно-исследовательская работа Онлайн-помощь Ответы на экзаменационные вопросы Отчет по практике Перевод текста Повышение уникальности Практическая работа Презентация Проверка выполненной работы Рабочая тетрадь, конспект РГР Реферат Рецензирование работ Речь, выступление, раздаточный материал Решение задач Самостоятельная работа Семестровая работа Статья ВАК Статья, рецензия Творческая работа Тезисный план Технико-экономическое обоснование Чертеж Шпаргалка Эссе Контакты: +7 495 727-22-67 info@feniks.help 09.00-21.00 МСК без выходных Услуги Решка Feniks.Help Партнерская программа ВУЗы Лента заказов Работа для студентов Информация для клиентов Информация для авторов Пользовательское соглашение Контакты © 2010 - 2025 Feniks.Help. Напишем БЕСПЛАТНО любую работу за 30 минут
1978	https://www.cuemath.com/geometry/triangle-inequality-theorem/	LearnPracticeDownload Triangle Inequality Theorem The triangle inequality theorem states that, in a triangle, the sum of lengths of any two sides is greater than the length of the third side. Suppose a, b and c are the lengths of the sides of a triangle, then, the sum of lengths of a and b is greater than the length c. Similarly, b + c > a, and a+ c > b. If, in any case, the given side lengths are not able to satisfy these conditions, it means it is not possible to draw a triangle with those measurements. \| \| \| --- \| \| 1. \| What Is Triangle Inequality Theorem? \| \| 2. \| Triangle Inequality Theorem Formula \| \| 3. \| FAQs on Triangle Inequality Theorem \| What is Triangle Inequality Theorem? The triangle inequality theorem states, "The sum of any two sides of a triangle is greater than its third side." This theorem helps us to identify whether it is possible to draw a triangle with the given measurements or not without actually doing the construction. Let's understand this with the help of an example. Triangle ABC has side lengths of 6 units, 8 units, and 12 units. Here, AB = 6 units, BC = 8 units and CA = 12 units. The sum of sides AB and BC is 6 + 8 = 14 units and 14 is greater than side CA (12 units). The sum of sides BC and CA is 8 + 12 = 20 units and 20 is greater than side AB (6 units). The sum of sides CA and AB that is 12 + 6 = 18 units and 18 is greater than side BC (8 units). Thus, lengths of all the sides satisfy the triangle inequality theorem. In this, not only one, but all 3 cases should satisfy the triangle inequality theorem. Let's take another example. Let's check whether a triangle with sides lengths 5 units, 3 units, and 10 units satisfy the triangle inequality theorem or not. Here, 5 + 3 = 8 which is less than 10 3 + 10 = 13 which is greater than 5 10 + 5 = 15 which is greater than 3 We can see that two cases are satisfying the triangle inequality theorem but one case is not satisfying. This means the triangle with these side lengths does not exist. All three sides should satisfy the triangle inequality theorem. Triangle Inequality Theorem Formula Before understanding the formula, first, we need to understand the proof of the triangle inequality theorem. Consider a triangle ABC as shown below. Let us extend side AB to the point D such that AC = AD and △BDC will form a right angled triangle at angle C. Applying angle sum property in △BDC, we get, ∠BDC + ∠CBD + ∠BCD = 180° ∠BDC + ∠CBD + 90° = 180° ∠BDC + ∠CBD = 90° This implies, ∠BCD > ∠BDC. As the side opposite to the greater angle is longer, we have BD > BC. This implies: BD > BC AB + AD > BC AB + AC > BC Hence proved. Similarly, we can prove that AC + BC > AB and AB + BC > AC. So, the triangle inequality theorem formula is, AB + AC > BC AC + BC > AB AB + BC > AC Related Articles on Triangle Inequality Theorem Check the following articles to learn more about the triangle inequality theorem. Triangle Triangle Inequality Types of Triangles Area of Triangle Triangle Inequality Theorem Examples Example 1: Suzie has three sticks of lengths 4 units, 8 units, and 2 units. Using the triangle inequality theorem, find out whether Suzie can form a triangle using these sticks or not? Solution: The triangle formed by the given sticks must satisfy the triangle inequality theorem. Let's check if the sum of the two sides is greater than the third side. 4 + 8 > 2 ⟹ 12 > 2 ⟹ TRUE 2 + 8 > 4 ⟹ 10 > 4 ⟹ TRUE 4 + 2 > 8 ⟹ 6 > 8 ⟹ FALSE So, the lengths of the sticks do not satisfy the triangle inequality theorem. Thus, Suzie can't form a triangle using the sticks of the given lengths. 2. Example 2: Ron wants to decorate his triangular flag with a ribbon. The two sides of the flag are 8 units and 2 units. Using the triangle inequality theorem, find out how much ribbon is required for the third side? Solution: By using the triangle inequality theorem, we can say that the length of the third side must be less than the sum of the other two sides. So, the third side is less than 8 units + 2 units = 10 units. Also, the third side cannot be less than the difference between the other two sides. So, the third side is greater than 8 units - 2 units= 6 units. Thus, the length of the ribbon can be 7, 8, or 9 units. View More > go to slidego to slide Have questions on basic mathematical concepts? Become a problem-solving champ using logic, not rules. Learn the why behind math with our certified experts Book a Free Trial Class Practice Questions on Triangle Inequality Theorem Check Answer > go to slidego to slide FAQs on Triangle Inequality Theorem What is Meant by Triangle Inequality Theorem? The triangle inequality theorem states that the sum of any two sides of a triangle is greater than the third side, and if the sum of any two sides of a triangle is not greater than the third side it means the triangle does not exist. Why is Triangle Inequality Theorem Important? The triangle inequality theorem is important to find out whether the triangle with the given three measurements exists or not. As the theorem states that sum of any two sides should be greater than the measurement of the third side. For example, the triangle with sides 3 units, 4 units, and 9 units does not exist as it does not satisfy the triangle inequality theorem. 3 + 4 > 9 ⟹ 7 > 9 ⟹ False 4 + 9 > 3 ⟹ 13 > 3 ⟹ True 9 + 3 > 4 ⟹ 12 > 4 ⟹ True Thus, by using the triangle inequality theorem we can say that the given measurements do not form a triangle. How is the Triangle Inequality Theorem used in Real Life? One example of the application of the triangle inequality theorem in real life is by Engineers. Civil engineers use the triangle inequality theorem in real life. Since their work is related to surveying, transportation, and urban planning. With the help of the triangle inequality theorem, they calculate the unknown lengths and estimate the remaining dimension. Does the Triangle Inequality Theorem Apply to all Triangles? Yes, the triangle inequality theorem applies to all triangles. Any side of a triangle must be shorter than the sum of the other two sides. If a side is greater than or equal to the sum of the other two sides, then it is not a triangle. What is an example of the Triangle Inequality Theorem? Following is the example of the triangle inequality theorem. Triangle with side lengths 5, 7, and 9 units exists, as lengths of all sides satisfy the theorem. 5 + 7 > 9 ⟹ 12 > 9 ⟹ TRUE 7 + 9 > 5 ⟹ 16 > 5 ⟹ TRUE 9 + 5 > 7 ⟹ 14 > 7 ⟹ TRUE How do you write a Triangle Inequality Theorem? Suppose ABC is a triangle. We will write the triangle inequality theorem in this form: AB + AC > BC AC + BC > AB AB + BC > AC Download FREE Study Materials Related Worksheet Math worksheets andvisual curriculum FOLLOW CUEMATH Facebook Youtube Instagram Twitter LinkedIn Tiktok MATH PROGRAM Online math classes Online Math Courses online math tutoring Online Math Program After School Tutoring Private math tutor Summer Math Programs Math Tutors Near Me Math Tuition Homeschool Math Online Solve Math Online Curriculum NEW OFFERINGS Coding SAT Science English MATH ONLINE CLASSES 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math ABOUT US Our Mission Our Journey Our Team QUICK LINKS Maths Games Maths Puzzles Our Pricing Math Questions Blogs Events FAQs MATH TOPICS Algebra 1 Algebra 2 Geometry Calculus math Pre-calculus math Math olympiad MATH TEST Math Kangaroo AMC 8 MATH CURRICULUM 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math FOLLOW CUEMATH Facebook Youtube Instagram Twitter LinkedIn Tiktok MATH PROGRAM Online math classes Online Math Courses online math tutoring Online Math Program After School Tutoring Private math tutor Summer Math Programs Math Tutors Near Me Math Tuition Homeschool Math Online Solve Math Online Curriculum NEW OFFERINGS Coding SAT Science English MATH CURRICULUM 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math MATH TEST CAASPP CogAT STAAR NJSLA SBAC Math Kangaroo AMC 8 ABOUT US Our Mission Our Journey Our Team MATH TOPICS Algebra 1 Algebra 2 Geometry Calculus math Pre-calculus math Math olympiad Numbers Measurement QUICK LINKS Maths Games Maths Puzzles Our Pricing Math Questions Blogs Events FAQs MATH ONLINE CLASSES 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math Terms and ConditionsPrivacy Policy
1979	https://www.youtube.com/watch?v=uCHgfa5q3WY	sin, cos and tan of 0°, 90°, 180°, 270°, ... Guru Tutor 4050 subscribers 88 likes Description 10437 views Posted: 13 May 2022 Calculating sin, cos and tan of angles bordering the quadrants. Link to Trigonometry playlist (Algebra 2): 2 comments Transcript: [Music] we're now moving on to something called quadrantal angles now i never heard that word until i used this textbook quadrantal angles what does that mean quadrantal angles are the angles on the borders of the quadrant so in other words 0 degrees plus and minus 90 degrees plus minus 180 degrees plus minus 270 degrees and so on and so forth okay they are the angles that are on like the x-axis and the y-axis so we have a quadrantal angle here we're going to do example one let's say we're talking about the quadrantal angle 180 degrees right it borders between two quadrants that's what a quadrantal angle is we're going to evaluate the six trigonometric ratios for 180 degrees so in other words what we're finding is sine 180 what is cos 180 10 100 and what is cosec sec cot so how do we do it we think back to here but what i suggest you do is redraw it each time okay it's got to come from up here you can't have a little cheat sheet anyway so just do a circle draw the theta theta is theta theta and what did that mean again uh you've got to remember that sine is y and cos is x but what if you can't remember those let's quickly do that diagram x y what was sine sine theta is opposite over hypotenuse sine is y and cos theta is x all right so what is sine of 180 well now we think if theta is 180 where is it theta is 180 we're talking about this point theta is 180 here's the point here and what are the coordinates of that point coordinates so that's minus one zero so sine of 180 is always going to be the y coordinate so here we've got sine of 180 what is the y coordinate of that point zero that's it cos 180 so cos is always the x coordinate and you look at the coordinates of that points remember we're just thinking like this because 180 is the x coordinate that's gonna be minus one tan 180 tan is always y over x so what is the y divided by the x so it's zero divided by minus one which is zero okay so they are the three values there the sign causing tan for theta here what about cosec 180 so cosec is one over sine so it's one over zero that's undefined yeah and by the way you can use your uh gdc here to check to test these if you do sine of 180 degrees these should all agree sec 180 remember sec is 1 over cos 180 what was cos 180 negative one one over negative one oh that's defined minus one and finally cot 180 cot is one over ten and what was tan 180 zero one over zero again we've got something undefined here okay so for each value of theta figure out where it is okay remember that the the x coordinate represents cos of that angle and the y coordinate represents sine of that angle okay and you work from there all right can you do that b we're going to do an example in radians so theta is negative 5 pi on 2. so it's a multiple of pi over 2 which is a multiple of 90 degrees so it's going to be one of those bordering angles bordering between two quadrants so let's draw it now we're going to think if theta is negative 5 pi 2 where's that so we're starting from zero and we're going in multiples of pi over two which is multiples of 90. and we're going in the negative direction five times so 1 2 3 4 5. theta is negative 5 pi 2 is down there so where is our point on the unit circle there it is what are the coordinates of that that is 0 negative 1. and remember what this means is that the sine is always the y coordinate so that is sine negative 5 pi over 2. and this is cos okay so it's always the x y is always the cause of the angle and the sign of the angle so what is sine negative 5 pi 2 sine is always the y coordinate which is negative 1. what's cos of theta here so theta is down here cos is always the x coordinate zero and what about tan tan is always y over x so it's minus one over zero okay remember tan theta is always y over x coming over here and that is undefined okay so if you did tan of negative 5 pi and 2 on your gdc it should give you some sort of an error message cosec so this is going to be cosec is 1 over sine so i might just save a bit of time here i won't write 1 over sine negative 5.2 cosec is 1 over sine so this is 1 over negative 1 which is negative 1. sec is 1 over cos so it's 1 over 0 undefined okay remember 0 on the bottom we can't do zero on the top is okay and cot is one of a tan so it's now this is a bit tricky can't do one over undefined can but what we can do is think tan theta is y over x then cot theta is x over y so think of it that way x is zero and y is minus one so that equals zero that is defined okay so there we are the six trigonometric ratios for theta is negative five pi on two
1980	https://puzzling.stackexchange.com/questions/129769/minimum-number-of-squares-to-color	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Minimum Number of Squares to Color Ask Question Asked Modified 8 months ago Viewed 1k times 4 $\begingroup$ This is the December 23rd puzzle from Matthew Scroggs' annual puzzle advent calendar. It has me stumped, so I'm looking for a hint on how to proceed. The text of the puzzle is below: In a grid of squares, each square is friendly with itself and friendly with every square that is horizontally, vertically, or diagonally adjacent to it (and is not friendly with any other squares). In a $5\times5$ grid, it is possible to colour $8$ squares so that every square is friendly with at least two coloured squares: It is not possible to do this by colouring fewer than $8$ squares. What is the fewest number of squares that need to be coloured in a $23\times23$ grid so that every square is friendly with at least two coloured squares? My usual plan for puzzles like this is to start small and look for a pattern. In a $1\times1$ grid it is impossible for the lone square to be "friendly" with $2$ squares. In a $2\times2$ grid, coloring any $2$ squares will do the job. In a $3\times3$ grid, coloring the middle row or column of $3$ squares works and seems to be the minimal solution. In a $4\times4$ grid, I think the minimal solution is $8$ squares (like the $5\times5$ grid). That's because no two corners have any common neighbors and each needs two colored neighbors. Coloring the $8$ squares along the two diagonals works, for example. So here's where I am stuck, because I haven't been able to "scale up". The one theory I have (which is likely wrong): I note that all the minimal solutions so far have been Fibonacci Numbers, which has been a hallmark of this advent calendar in past years. Since $4\times4$ and $5\times5$ have the same solution, it may be that $(2n)\times(2n)$ and $(2n+1)\times(2n+1)$ continue to have the same solution? Following the Fibonacci pattern, this would lead to a final answer of $610$ colored squares in the $23\times23$ grid. This is in line with the contest rules that each day must have a $3$ digit answer. I'm not necessarily looking for a full answer to this. Just hints on how to identify the pattern and move up to larger grids. mathematics visual pattern optimization Share Improve this question edited Dec 24, 2024 at 0:14 RobPratt 17.7k11 gold badge4040 silver badges7373 bronze badges asked Dec 23, 2024 at 21:49 DreiCleanerDreiCleaner 32722 silver badges99 bronze badges $\endgroup$ 4 $\begingroup$ My conjectured answer can't be right because $23\times23$ grid only has $529$ total squares. Oops! $\endgroup$ DreiCleaner – DreiCleaner 2024-12-23 22:02:57 +00:00 Commented Dec 23, 2024 at 22:02 $\begingroup$ I also don't think the Fibonacci pattern holds. The $6 \times 6$ grid can be done with 10 coloured squares, the second and fifth columns almost completely coloured, with the fourth cells left out. $\endgroup$ Avastar – Avastar 2024-12-23 22:21:33 +00:00 Commented Dec 23, 2024 at 22:21 1 $\begingroup$ This is part of an ongoing contest. Part of the rules are: Ten randomly selected people who solve all the puzzles, open the warehouse, and fill in the entry form behind the door on the 25th will win prizes! ¦ The winners will be randomly chosen from all those who submit the entry form before the end of 2024. Each day's puzzle (and the entry form on Christmas Day) will be available from 5:00am GMT. But as the winners will be selected randomly, there's no need to get up at 5am on Christmas Day to enter! $\endgroup$ Will.Octagon.Gibson – Will.Octagon.Gibson 2024-12-24 04:59:09 +00:00 Commented Dec 24, 2024 at 4:59 $\begingroup$ Now that the contest has ended, Matthew Scroggs has posted the answers to all the problems. In particular he made a lengthy blog post about this one: mscroggs.co.uk/blog/114 $\endgroup$ DreiCleaner – DreiCleaner 2025-01-02 13:00:30 +00:00 Commented Jan 2 at 13:00 Add a comment \| 5 Answers 5 Reset to default 5 $\begingroup$ I would begin from the observation that in the 3x3 block centred on any square (which, if the square is at the edge, may have fewer than 9 actual squares in it) there must be at least two coloured squares. (This is deliberately not a complete answer, because OP specifically says "I'm not necessarily looking for a full answer to this" and after all this is an ongoing contest of sorts. I am pretty sure Matthew Scroggs will not be upset by this amount of online discussion. But if you, yes you, are reading this and happen to be Matthew Scroggs, do please comment and tell me if I'm wrong!) Share Improve this answer answered Dec 23, 2024 at 22:17 Gareth McCaughan♦Gareth McCaughan 124k77 gold badges330330 silver badges474474 bronze badges $\endgroup$ 4 $\begingroup$ I'm not sure how to proceed from this observation. There is so much overlap among all the $3\times3$ blocks. Two different blocks could share either 0, 1, 2, 3, 4, or 6 squares (not even considering blocks at the edge of the grid). $\endgroup$ DreiCleaner – DreiCleaner 2024-12-23 23:55:23 +00:00 Commented Dec 23, 2024 at 23:55 2 $\begingroup$ As you realized - overlaps are not good. So, just don't overlap. I'm afraid that further hint will ruin the fun for you. $\endgroup$ WhatsUp – WhatsUp 2024-12-24 00:57:21 +00:00 Commented Dec 24, 2024 at 0:57 1 $\begingroup$ I'm very much not upset by this amount of discussion (and very happy that other people also found this puzzle interesting). I found this discussions because someone linked to it from the OEIS sequence I made (oeis.org/A379726). I've started working out terms for the sequence with 3 and 4 squares friendly to each square if anyone wants an additional challenge. $\endgroup$ mscroggs – mscroggs 2025-01-02 20:04:09 +00:00 Commented Jan 2 at 20:04 $\begingroup$ Thanks, @mscroggs! Your conjecture about the multiple-of-3 case is correct; check your email :-). $\endgroup$ Gareth McCaughan – Gareth McCaughan ♦ 2025-01-03 04:41:56 +00:00 Commented Jan 3 at 4:41 Add a comment \| 3 $\begingroup$ For a fixed $n\times n$ grid, you can solve the problem via integer linear programming as follows. Let binary decision variable $x_{ij}$ indicate whether cell $(i,j)$ is colored. The problem is to minimize $\sum_{i=1}^n \sum_{j=1}^n x_{ij}$ subject to \begin{align} \sum_{r=i-1}^{i+1} \sum_{c=j-1}^{j+1} x_{rc} &\ge 2 &&\text{for all $(i,j)\in {1,\dots,n} \times {1,\dots,n}$} \end{align} Here are the resulting minimum values for $n \in {2,\dots,23}$: 2 3 8 8 10 18 18 21 32 32 36 50 50 55 72 72 78 98 98 105 128 128 In fact, the linear programming relaxation yields these same minimum values, and the dual variables provide a short certificate of optimality. It appears that $(3n+1)\times(3n+1)$ and $(3n+2)\times(3n+2)$ yield the same minimum, as you observed for $4\times4$ and $5\times5$. Here are two hints for $23\times23$: You can obtain the minimum value by never having more than two adjacent ones in a connected component. $2m^2$ Share Improve this answer edited Dec 24, 2024 at 3:34 answered Dec 24, 2024 at 2:41 RobPrattRobPratt 17.7k11 gold badge4040 silver badges7373 bronze badges $\endgroup$ Add a comment \| 2 $\begingroup$ I had another answer, but after thinking more about it, I figured that holes could be poked in it. So here's another: Since it's a $23\times23$ square, it'll take at least $ceil(23\times2/3)=16$ painted squares to cover an entire length: Repeat this $7$ times, and a $23\times21$ area is covered. The remaining area also takes $16$, so $16\times8=128$. Share Improve this answer edited Jan 2 at 7:41 answered Jan 1 at 10:57 NautilusNautilus 8,3371414 silver badges3636 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ Clearly not a full answer, just a lower bound: It's obvious that for large n, an nxn grid requires at least $\frac{2n^2}{9}$ coloured squares - each square has 9 neighbours, and each requires 2 neighbouring coloured squares. We can make a slight improvement to this by noticing that at least 4 coloured squares must be on the edge (one per corner), and there are 3 "missing" squares for each of those, so that brings us to a lower bound of $\frac{2n^2+12}{9}$. Share Improve this answer answered Dec 25, 2024 at 6:26 SpitemasterSpitemaster 1,3531111 silver badges1515 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ This is too late for the contest, but I want to try and give a visually intuitive solution for this puzzle, both the specific problem as well as a somewhat general case. This answer builds on ideas from @Gareth McCaughan and @RobPratt's answers, so thanks to them! We begin with the fact that each $9 \times 9$ block centred on one of the squares in our grid must contain at least two coloured squares within it. If our centre square happens to be a grid edge, our block contains only 6 squares, two of which we want to colour. It is easy to see that the only way to satisfy the friendliness requirements for each of these 6 squares is if the central two squares are coloured. What we can do now is use this $3\times2$ block to tile the top edge of a grid (as long as it is sufficiently large). Each of the squares in these tiles then will be friendly with exactly two coloured squares. This tiling produces a curious property. If in our block we had coloured a different pair of squares, we would not have had the friendliness condition be satisfied for all 6 squares. However, this tiling procedure erases this problem, as those squares in one block that do not have enough friendly coloured squares from their own tile, simply get the ones they need (and no more) from the next tile. Of course, the rightmost squares above have only one friendly coloured square, but this can be solved either by placing another tile to the right, or by some other means. Regardless, this property of our tiles will be used later. Consider now a square of side length $3n + 2$, where $n$ is some positive integer. We can begin lining the top edge of this with our original tiles, placing $n$ of them, thus using $2n$ coloured squares and giving $6n$ squares the friendly neighbours they want. Unfortunately, this leaves us with the last two columns not being able to fit another block. Notice though that the remaining length of this edge is the same as the breadth of each of our blocks. What if we rotate our blocks by $90$ degrees and start placing them vertically, repeating our process for the side edge? Happily, this suggests a process where we repeat the above steps for each of the four edges of our $(3n+2) \times (3n+2)$ grid, covering the 'first' $3n$ squares along each edge with tiles having their long side parallel to the grid edge, with the remaining 2 squares of each edge taken care of by the tiles of the adjacent edge. But before we get to that point, there's a problem. The first two squares in the third column from the right have three friendly coloured squares. Since we would like to minimise the total number of coloured squares we use, it is in our advantage to not have these overlaps in the effective range of our tiles, i.e. give every square only two friendly coloured squares. Focusing our efforts on tweaking the rotated tile we just added, let's try and ensure that every square in the first two rows is given exactly the two friendly coloured squares it needs, and no more. The only way to do this is by changing the rotated tile to be as below. Now, using the property of these tiles we noted above, (almost) the entirety of the last two columns can also be taken care of now, by using the changed vertical tile. Working similarly for the bottom-right and bottom-left grid corners, we end up with a tiling of the grid as below, where I have only shown the corner tiles. It is easy to see now that every square in the two outermost 'rings' of the grid now has their friendliness requirements exactly satisfied. We've used up $n$ tiles with $2n$ coloured squares per edge, for a total of $8n$ coloured blocks. What of the squares inside? The third ring from the outside is a bit of an odd one, as some of the squares in it already have one friendly coloured square, while the others have none. Let's ignore this ring for now and focus on the grid inside it, where no squares have any friendly coloured squares. The side length of this smaller grid is $(3n + 2) - 3 - 3 = 3(n-2) + 2$, placing it in the very family of grids we are studying. Effectively then, we can treat it in exactly the same way as we've treated the larger grid, having the same tiles laid out in the same arrangement. Since this smaller grid will have another even smaller grid inside it satisfying the same side length condition, we have a recursive relation here. Coloured Squares of the $3n+2$ grid $= 8n\: + $ Coloured Squares of the $3(n-2)+2$ grid $= 8n + 8(n-2)\: + $ Coloured Squares of the $3(n-4)+2$ grid and so on. Before we jump though to calculating the number of coloured squares we need, let's make sure that this recursive process leaves our third ring (for each grid) with exactly 2 friendly coloured squares. It is visually clear that the third ring too has now been taken care of, with no squares having more than 2 friendly coloured squares. So, finally, our method of tiling the grid is complete, and we can calculate how many coloured squares we've used. Since we're reducing $n$ by $2$ in each step, there are two cases, $n$ being odd or even. If $n$ is odd, the smallest grid we reach is when $n=1$, i.e. a $5\times5$ grid. As can be seen from the image in the question (by tweaking it slightly), this too can be tiled the way we've described above. So, the total number of coloured squares can be easily calculated as $S = 8(n + (n-2) + \ldots + 3 + 1) = \frac{8}{2}(n + 1)(\frac{n+1}{2}) = 2(n+1)^2$ If $n$ is even, the smallest grid we reach is when $n=0$, i.e. a $2\times2$ grid. This doesn't follow the tiling process, so we'll add its 2 coloured squares separately. The total number of coloured squares can then again be easily calculated as $S = 8(n + (n-2) + \ldots + 4 + 2) + 2 = \frac{8}{2}(n+2)(\frac{n}{2}) + 2 = 2((n+1)^2 - 1) + 2 = 2(n+1)^2$ Both cases then give exactly the same result (as well as being what @RobPratt hints at in his answer). Using this then to solve our specific puzzle, since $23 = 3(7) + 2$, our $n=7$, meaning that the number of coloured squares we must have so that every square has two friendly coloured squares is $2(n+1)^2 = 2(7+1)^2 = \mathbf{128}$. I'm not very sure if this is rigorous enough to be a proper proof that this will be the minimum number of coloured squares we need, but since my answer matches that of both @RobPratt and @Nautilus (with the squares themselves being arranged in almost the same form as that given by the latter), I am inclined to believe that this is so. Share Improve this answer edited Jan 1 at 17:18 answered Jan 1 at 16:56 AvastarAvastar 2,89311 gold badge1818 silver badges3636 bronze badges $\endgroup$ Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions mathematics visual pattern optimization See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure The USAMTS attracts a lot of cheating attempts Related Four-by-four square 11 Turn on all squares Coloring the Chess Board Red and White Squares Minimum number of lines to draw 111 squares Minimum number of moves to win this number game 2 Minimum moves to lose grid game 3 Lowest possible ratio of number of borders to countries annexable 13 Can you color the 8x8 grid red and blue? 11 Can you make all the squares the same color? 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1981	https://www.scribd.com/document/858540200/The-CDC-Field-Epidemiology-Manual-1st-Edition-Full-Book-Access	The CDC Field Epidemiology Manual - 1st Edition Full Book Access \| PDF \| Centers For Disease Control And Prevention \| Epidemiology Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 100%(10)100% found this document useful (10 votes) 426 views 16 pages The CDC Field Epidemiology Manual - 1st Edition Full Book Access The CDC Field Epidemiology Manual, edited by Sonja A. Rasmussen and Richard A. Goodman, serves as a comprehensive guide for public health professionals conducting field investigations. 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1982	https://solventlearning.com/calculating-unit-rates-and-unit-prices/	Solvent Learning Calculating Unit Rates and Unit Prices Table of Contents Video Lessons > Calculating Unit Rates and Unit Prices Introduction and Video In this video lesson, we will learn strategies for calculating unit rates and unit prices. We'll discuss the definition of unit rate and unit price and look at examples of calculating unit rates and unit prices. Before beginning this lesson, it's important to understand rates and how to write them. Lesson Notes for Unit Rates and Unit Prices Let's look at the details of the video lesson, including definitions and examples of calculating unit rates and prices. Definition of Unit Rate A unit rate is a specific type of rate where the second value, also known as the denominator when expressed as a fraction, is the number one. This concept has many practical applications. For instance, the expression "60 miles per 1 hour" or, in fraction form, "60 miles/1 hour" is an example of a unit rate. With unit rates, it's common to omit the 1 in the second value. So instead of saying "60 miles per 1 hour," we can simply say "60 miles per hour." Example of Calculating Unit Rate Ian can type 170 words in five minutes and we want to find his typing speed as a unit rate. Definition of Unit Price Unit price is a particular kind of unit rate where the numerator is a measure of money. Similar to unit rates, the second value or denominator is one. For instance, if a box of cookies costs $3, the unit price can be expressed as "$3 per one box," "$3 over one box" in fraction form, or using the front slash symbol "$3/box." Example of Calculating Unit Price Shayla bought three boxes of cookies for $12.75 and we want to find the unit price per box of cookies. Summary and Practice In this video lesson, we learned the definition of unit rate and unit price and how to calculate them. Remember, when finding a unit rate or unit price, it's important to first set up the problem based on the provided information, ideally in fraction form. If one of the values involves money, it's placed as the numerator. If there is a unit of time, that value is placed in the denominator. Division is then used to determine the unit rate or unit price. And finally, be sure to include the units in your final answer. Try this practice activity to see what you learned. Now we're going to learn about unit rates and unit prices. A unit rate is a rate where the second value, or denominator if it's written as a fraction, is the number one. So for our first example, we have… after 1 hour of driving, the car has traveled 60 miles. We can write this as 60 miles per 1 hour, or as a fraction, 60 miles over 1 hour. And we can recognize that it's a unit rate because the second value, or the denominator, is the number one. However, it's more common for us to not even write the number one. So if you ever see a rate written this way with no number with the second unit of measurement, then it's implied that it's just a one. We can also write it a third way using the front slash symbol instead of the word “per.” So we would read this as 60 miles per hour, but we use the front slash instead of the word “per.” Let's look at our first example. Ian can type 170 words in five minutes. Find his typing speed as a unit rate. So another way we can read this problem is… find how many words Ian can type per minute. Now let's look at the information that we have. We know that he can type 170 words in five minutes. We can also represent this ratio as a fraction as 170 words over five minutes. And remember that whenever one of your values represents time, you want to write that as the second value, or if it's a fraction, write that value in the denominator. So that's why we put five minutes in the denominator of our fraction. But keep in mind here, this is not a unit rate because we have five minutes as our second value instead of one. So we're going to convert this to a unit rate. So we'll start with our fraction. And remember that fractions represent division. So by simplifying this fraction with division, we're actually going to be able to convert this to a unit rate. So when we're dividing, we'll take the numerator and divide by the denominator. So it becomes 170 divided by five, which gives us 34. And that's how many words Ian can type in 1 minute. So we can write this as 34 words per minute, or if we want to use the forward slash instead of the word “per,” we can write it this way, but we would still read it as 34 words per minute. Next, we have unit price. So unit price is a type of unit rate where the first value, or the numerator, is a measurement of money. But since it is still a unit rate, the second value will be a one. So for example, a box of cookies costs $3. Let's write this as a unit price. We could say that it costs $3 per one box, or as a fraction, $3 over one box. And most commonly, we don't even bother writing the number one for the second value. And then the third way is to use the front slash instead of the word “per.” So any of these three variations represent the unit price for this box of cookies. Shayla bought three boxes of cookies for $12.75. Find the unit price for a box of cookies. We can also think of this problem as asking us to find the cost of one box of cookies, because that's the same thing as a unit price for a box of cookies. So let's set this problem up based on the information they give us. We have $12.75 for three boxes, or as a fraction, $12.75 over three boxes. And notice these are not unit prices, because even though our first value does represent a dollar amount, our second value is not the number one. So we'll need to convert that fraction into a value that has a one as the second value. And we'll use division - numerator divided by denominator, which gives us $12.75 divided by three, and that's $4.25. So when we write it out with our units, we have $4.25 per box. Or we can use the front slash symbol and still read it as $4.25 per box. So keep in mind, when you are setting up a problem that asks you to find the unit price or the unit rate, we need to first set it up based on the information they give us, ideally as a fraction. And if it does involve some type of money, we place that as the numerator of the fraction. And then we use division to figure out the value of the unit price or unit rate. And remember to keep the units with the values. Related Standard: Common Core 6.RP.A.2
1983	https://math.stackexchange.com/questions/3284880/find-the-sum-of-the-even-degree-coefficients	algebra precalculus - Find the sum of the even-degree coefficients - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Find the sum of the even-degree coefficients [closed] Ask Question Asked 6 years, 2 months ago Modified6 years, 2 months ago Viewed 71 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Closed. This question does not meet Mathematics Stack Exchange guidelines. It is not currently accepting answers. Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc. Closed 6 years ago. Improve this question I am doing practice problems from the AMC-10 math competition. I do not understand how to go about solving the problem Let (1+x+x 2)n=a 0+a 1 x+a 2 x 2+...+a 2 n x 2 n(1+x+x 2)n=a 0+a 1 x+a 2 x 2+...+a 2 n x 2 n be an identity of x x. If we let s=a 0+a 2+a 4+...+a 2 n s=a 0+a 2+a 4+...+a 2 n, find s s. The choices are A,2 n A,2 n, B,(2 n)+1 B,(2 n)+1, C,3 n−1 2 C,3 n−1 2 , D,3 n 2 D,3 n 2 , and E,3 n+1 2 E,3 n+1 2 Thanks in advance algebra-precalculus contest-math Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jul 6, 2019 at 11:49 hmakholm left over Monica 292k 25 25 gold badges 441 441 silver badges 706 706 bronze badges asked Jul 6, 2019 at 11:43 JamlandiaJamlandia 81 4 4 bronze badges 3 10 Set x=−1,1 x=−1,1 and add lab bhattacharjee –lab bhattacharjee 2019-07-06 11:45:14 +00:00 Commented Jul 6, 2019 at 11:45 2 Just a note: a 0 a 0 can be typed as $a_0$. Here is also a MathJax page that I find useful.Toby Mak –Toby Mak 2019-07-06 11:47:40 +00:00 Commented Jul 6, 2019 at 11:47 1 ...or, if you are lazy and if its a multiple choice test, just try for n=1,2 n=1,2 dcolazin –dcolazin 2019-07-06 11:51:10 +00:00 Commented Jul 6, 2019 at 11:51 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. As noted by lab in a comment, for x=1,−1 x=1,−1 you get 3 n=a o+a 1+…+a 2 n 1=a o−a 1+…+a 2 n 3 n=a o+a 1+…+a 2 n 1=a o−a 1+…+a 2 n so 3 n+1=2 a 0+2 a 2+…+2 a 2 n 3 n+1=2 a 0+2 a 2+…+2 a 2 n. Another (lazy) idea if you already have some candidates: for n=1,a 0+a 2=2 n=1,a 0+a 2=2, for n=2,(1+x+x 2)2=1+2 x+3 x 2+2 x 3+x 4 n=2,(1+x+x 2)2=1+2 x+3 x 2+2 x 3+x 4, so a 0+a 2+a 4=5 a 0+a 2+a 4=5. Then the only candidate possible is 3 n+1 2 3 n+1 2. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jul 6, 2019 at 12:02 dcolazindcolazin 2,262 9 9 silver badges 15 15 bronze badges Add a comment\| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algebra-precalculus contest-math See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 1Let f(x)=a 2 0 x n+a 1 x n−1+a 2 x n−2+...........+a n,f(x)=a 0 2 x n+a 1 x n−1+a 2 x n−2+...........+a n, where a 0,a 1,....,a n∈R a 0,a 1,....,a n∈R 3Coefficient Problem (polynomial expansion) 4A summation series of binomial coefficients 7Find a polynomial with integer coefficients 1Regarding algebric manipulation in order to find a 2 0−a 2 1+a 2 2+...a 2 2 n a 0 2−a 1 2+a 2 2+...a 2 n 2 7Will high school competition math materials help me prepare for the Putnam? 0Is there always a point (h,k)(h,k) in a polynomial function P(x)P(x) for which P P becomes odd or even by translating (h,k)(h,k) to (0,0)(0,0)? 0What's the smallest value of the greatest number in the sum? 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1984	https://pt.scribd.com/doc/78536359/Thermodynamic-Versus-Kinetic-Reaction-Control	Organic Reaction Control Explained \| PDF \| Physical Chemistry \| Organic Chemistry Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 2K views 15 pages Organic Reaction Control Explained The document discusses the concepts of kinetic vs thermodynamic control in organic reactions. It gives the examples of addition of HBr to 1,3-butadiene and sulfonation of naphthalene. For th… Full description Uploaded by Udasi Raqs Kerti Hai AI-enhanced title and description Go to previous items Go to next items Download Save Save Thermodynamic Versus Kinetic Reaction Control For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Thermodynamic Versus Kinetic Reaction Control For Later You are on page 1/ 15 Search Fullscreen Ki ne ti c vs.Thermodynamic Control in Organic Reactions John Cullen August, 1997 The concept of kinetic vs. thermodynamic control of reactions is an important one in organic chemistry. There are a number of reactions known in which there are two (or more) possible reaction products of a r eaction, and one product (kinetic product) predominates when the reaction is done at low temperature. The other (thermodynamic product) predominates when the reaction is done at a higher temperature. This concept is investigated using the CAChe TM molecular modeling software. I. Addition of HBr to 1,3-Butadiene HBr adds to 1,3-butadiene to form a mixture of 3-bromo-1-butene (1,2-addition) and 1-bromo-2-butene (1,4-addition). 1 At -80 o C , 3-bro mo-1-but ene is the major pro duct,and at 40 o C 1-b romo-2- butene is the major product.Procedure In the CACh e Editor , dra w the struc ture s of the all ylic carb oca tion (eit her reso nan ce form), 3-bromo- 1-butene and 1-bromo-2-butene (cis and trans isomers). Be certain to have the correct hybridization for the carbon atoms and a +1 charge on the carbocation. You need not draw hydrogen atoms. From the Beautify menu adDownload to read ad-free select Comprehensive to obtain a drawing of the molecule with hydrogens added and approximately correct bond angles.Minimizing the energy of structures is done by selecting Applications and then MOPAC.Select Optimize Geometry under calculation type and AM1 parameters. Run the program and obtain the value of the heat of formation found in the dialog box.In order to view the charge density of the carbocation, go to Applications and select Visualizer. From the View menu select Atom Shape. Set the parameters in the dialog box to display Element and Charge as Spheres, shaded and radius for partial charge. The partial charge on each atom will be displayed color-coded with red representing partial positive charge and yellow representing partial negative charge. The size of sphere is proportional to the amount of partial charge on the atom. Alternatively numerical values of the partial charge on each atom and heats of formation for all species can be obtained using Project Leader TM .Results Results of these calculations show that in the allylic carbocation intermediate there is more partial positive charge on C-3 (+0.29) than on C-1 (+0.12). Thus under kinetic conditions nucleophilic attack by bromide occurs at the secondary allylic carbon rather than the primary allylic carbon. The calculated heats of formation verify that 1-bromo-2-butene (thermodynamic product) is more stable than 3-bromo-1-butene. Compound Heat of Formation(kcal/mole) 3-b r o m o-1-b u t e n e 6.4 0 t r a n s-1-b r o m o-2-b u t e n e 0.8 9 c i s-1-b r o m o-2-b u t e n e 1.9 5 II. Sulfonation of Naphthalene Another reaction that illustrates the concept of kinetic vs. thermodynamic ontrol is the sulfonation of naphthalene with SO 3 in H 2 SO 4 . 2 When the reaction is run at 80 o C , 1- naphthalenesulfonic acid is the major product; at 160 o C , 2- Representation of allylic carbocation color coded to show partial charges adDownload to read ad-free naphthalenesulfonic acid is the major product. The intermediates in the sulfonation are arenium ions. The relative energies of the arenium ions should approximate the energies of the transition states leading to the formation of the sulfonic acids (Hammond Postulate). 2-Naphthalenesulfonic acid is more stable than 1- naphthalenesulfonic acid because of an unfavorable steric interaction between the sulfonic acid group in the 1-position and the hydrogen in the 8-position. The arenium ion leading to 1- naphthalenesulfonic acid is more stable than the arenium ion leading to the 2-isomer because of better resonance stabilization. The reaction should be able to be modeled by calculating the heats of formation of the isomeric naphthalenesulfonic acids and the arenium ion intermediates leading to each. 3 Procedure In the CAChe Editor , construc t structure s for 1-naphth alenesu lfonic aci d, 2-naphthalenesulfonic and the 1-arenium ion and 2-arenium ions shown above. Then use MOPAC to minimize the energy of each and calculate the heat of formation of each.Results adDownload to read ad-free Calculation of the heats of formations gave results that are in agreement with expectation, as shown in the table below. Structure Heat of Formation(kcal/mole) 1-N a p h t h a l e n e s u l f o n i c a c i d-7 5.3 6 2-N a p h t h a l e n e s u l f o n i c a c i d-7 7.3 2 1-Arenium ion(precurso r to 1-naphthalen esulfonic acid)111.68 2-Arenium ion(precurso r to 2-naphthalen esulfonic acid)112.75 ......................Thermodynamic versus kinetic reaction control From Wikipedia, the free encyclopedia Thermodynamic reaction control or kinetic reaction control in achemicalreactioncan decide the composition in a reaction product mixture when competing pathways lead to different products and the reaction conditions influence theselectivity.The distinction is relevant when product A forms faster (which is called the kinetically controlled product) than product B because theactivation energyfor product A is lower adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Medan LPG Terminal Overview 100% (1) Medan LPG Terminal Overview 38 pages (Hooker and Monas, 2008) Shoestring Venture - 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1985	https://www.aimspress.com/article/doi/10.3934/math.2021168?viewType=HTML	About AIMS Press Open Access Policy Contact us Copyright © AIMS Press AIMS Mathematics 2021, Volume 6, Issue 3: 2756-2763. doi: 10.3934/math.2021168 Previous Article Next Article Research article Asymptotic expansion of a finite sum involving harmonic numbers Ling Zhu , Department of Mathematics, Zhejiang Gongshang University, Hangzhou, Zhejiang, China Received: 06 September 2020 Accepted: 30 December 2020 Published: 05 January 2021 MSC : 11M06, 33B15, 41A60 Abstract Full Text(HTML) Download PDF Download PDF In the paper, we obtain asymptotic expansion of the finite sum of some sequences Sn=∑nk=1(n2+k)−1 by using the Euler's standard one of the harmonic numbers. Keywords:+ asymptotic expansion, + finite sum of some sequences, + harmonic numbers Citation: Ling Zhu. Asymptotic expansion of a finite sum involving harmonic numbers[J]. 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AIMS Mathematics, 2024, 9(12): 33591-33609. doi: 10.3934/math.20241603 \| ### Abstract In the paper, we obtain asymptotic expansion of the finite sum of some sequences Sn=∑nk=1(n2+k)−1 by using the Euler's standard one of the harmonic numbers. ### 1. Introduction In [1, p.458], Graham, Knuth and Patashnik proposed a problem: Find the asymptotic value of the sum \| \| \| --- \| \| Sn=n∑i=11n2+i=1n2+1+1n2+2+⋯+1n2+n \| (1.1) \| with an absolute error of O(n−7). The following answer \| \| \| --- \| \| Sn=1n−12n2−16n3+14n4−215n5+112n6+O(1n7) \| (1.2) \| appeared in [1, p.459] and modestly be interpreted as probably correct. In this paper we obtained an asymptotic expression of Sn with absolute error O(1/n4l+1) (l∈N) by Euler's standard one of the harmonic numbers, and solved the above problem as a special case. ### 2. Main result and its proof Theorem 2.1. Let l a given natural number and B2w be an even-indexed Bernoulli number. Then as n→∞, \| \| \| --- \| \| Sn=n∑i=11n2+i=1n2+1+1n2+2+⋯+1n2+n=1n−12n2+4l−1∑k=3k≠4mj=[k/4]Aknk+∑k=4i1≤i≤l−1j=i−1Aknk+O(1n4l+1), \| (2.1) \| where \| \| \| --- \| \| Ak=(−1)k−1k+(−1)k−22+j∑w=1B2w2w(−1)k−(2w+1)(2w−1)!(k−(2w+1))!(k−4w)! \| (2.2) \| for k≥3. Proof. Let γ be the Euler's constant and P2l+1(x) be a periodic Bernoulli polynomial. Then from Equation (7), Inequalities (8) and (9) in : \| \| \| Hn=11+12+⋯+1n=lnn+γ+12n−l∑k=1B2k2k1n2k+∫∞nP2l+1(x)x2l+1dx, \| \| \| \| \|P2l+1(x)\|≤2(2l+1)!(2π)2l+1∞∑r=11r2l+1, \| \| \| \| \|∫∞nP2l+1(x)x2l+1dx\|≤4n√lπ(lnπe)2l, \| one can obtain the Euler's standard asymptotic expansion of the harmonic numbers: as n→∞, \| \| \| --- \| \| Hn=lnn+γ+12n−l∑k=1B2k2k1n2k+O(1n2l+1). \| (2.3) \| By (2.3) we obtain \| \| \| --- \| \| Sn=Hn2+n−Hn2=ln(n2+n)−lnn2+12(n2+n)−l∑i=1B2i2i1(n2+n)2i−∞∑i=l+1B2i2i1(n2+n)2i−(12n2−l∑i=1B2i2i1n4i)+∞∑i=l+1B2i2i1n4i+O(1n4l+2)=ln(1+1n)+12(n2+n)−l∑i=1B2i2i1(n2+n)2i−∞∑i=l+1B2i2i1(n2+n)2i−12n2+l∑i=1B2i2i1n4i+∞∑i=l+1B2i2i1n4i+O(1n4l+2). \| (2.4) \| Since \| \| \| --- \| \| (a+t)−n=(−1)n−1(a+t)−1(n−1)!, a∈R, \| (2.5) \| and for −1<t<1, \| \| \| [ln(1+t)]′=(1+t)−1=∞∑k=0(−1)ktk,ln(1+t)=∞∑k=0(−1)kk+1tk+1=∞∑k=1(−1)k−1ktk, \| we have \| \| \| 1n2+n=1n211+1n=∞∑k=0(−1)k1nk+2,1(n2+n)2i=1n4i1(1+1n)2i=−1(2i−1)!∞∑k=2i−1(−1)kk!(k−(2i−1))!1nk+2i+1, \| and \| \| \| Sn=∞∑k=1(−1)k−1k1nk+12∞∑k=0(−1)k1nk+2+l∑i=1B2i(2i)!∞∑k=2i−1(−1)kk!(k−(2i−1))!1nk+2i+1+∞∑i=l+1B2i(2i)!∞∑k=2i−1(−1)kk!(k−(2i−1))!1nk+2i+1−12n2+l∑i=1B2i2i1n4i+∞∑i=l+1B2i2i1n4i=∞∑k=1(−1)k−1k1nk+12∞∑k=0(−1)knk+2+[l−1∑i=1B2i(2i)!∞∑k=2i−1(−1)kk!(k−(2i−1))!1nk+2i+1−B2l2l1n4l]−12n2+[l−1∑i=1B2i2i1n4i+B2l2l1n4l] \| \| \| \| =∞∑k=1(−1)k−1k1nk+12∞∑k=0(−1)knk+2+l−1∑i=1B2i(2i)!∞∑k=2i−1(−1)kk!(k−(2i−1))!1nk+2i+1−12n2+l−1∑i=1B2i2i1n4i=∞∑k=1(−1)k−1k1nk+12∞∑k=0(−1)knk+2+B22!∞∑k=1(−1)kk!(k−1)!1nk+3+B44!∞∑k=3(−1)kk!(k−3)!1nk+5+B66!∞∑k=5(−1)kk!(k−5)!1nk+7+⋯+B2l−2(2l−2)!∞∑k=2l−3(−1)kk!(k−(2l−3))!1nk+2l−1−12n2+B221n4+B441n8+B661n12+⋯+B2l−22l−21n4l−4 \| \| \| \| =(−1)011n+(−1)121n2+∞∑k=3(−1)k−1k1nk+12(−1)0n2+12∞∑k=1(−1)knk+2+B22!∞∑k=1(−1)kk!(k−1)!1nk+3+B44!∞∑k=3(−1)kk!(k−3)!1nk+5+B66!∞∑k=5(−1)kk!(k−5)!1nk+7+⋯+B2l−2(2l−2)!∞∑k=2l−3(−1)kk!(k−(2l−3))!1nk+2l−1−12n2+B221n4+B441n8+B661n12+⋯+B2l−22l−21n4l−4 \| \| \| \| =(−1)011n+(−1)121n2+12(−1)0n2−12n2+∞∑k=3(−1)k−1k1nk+12∞∑k=1(−1)knk+2+B22!∞∑k=1(−1)kk!(k−1)!1nk+3+B44!∞∑k=3(−1)kk!(k−3)!1nk+5+B66!∞∑k=5(−1)kk!(k−5)!1nk+7+⋅⋅⋅+B2l−2(2l−2)!∞∑k=2l−3(−1)kk!(k−(2l−3))!1nk+2l−1+B221n4+B441n8+B661n12+⋅⋅⋅+B2l−22l−21n4l−4 \| \| \| \| =1n−121n2+[∞∑k=3(−1)k−1k1nk+12∞∑k=1(−1)k1nk+2]+B22!∞∑k=1(−1)kk!(k−1)!1nk+3+B44!∞∑k=3(−1)kk!(k−3)!1nk+5+B66!∞∑k=5(−1)kk!(k−5)!1nk+7+⋯+B2l−2(2l−2)!∞∑k=2l−3(−1)kk!(k−(2l−3))!1nk+2l−1+B221n4+B441n8+B661n12+⋯+B2l−22l−21n4l−4. \| Similarly, from the brackets above we can extract the cubic term of 1/n: \| \| \| Sn=1n−121n2−16n3+[∞∑k=4(−1)k−1k1nk+12∞∑k=2(−1)knk+2+B22!∞∑k=1(−1)kk!(k−1)!1nk+3]+B44!∞∑k=3(−1)kk!(k−3)!1nk+5+B66!∞∑k=5(−1)kk!(k−5)!1nk+7+⋯+B2l−2(2l−2)!∞∑k=2l−3(−1)kk!(k−(2l−3))!1nk+2l−1+B221n4+B441n8+B661n12+⋯+B2l−22l−21n4l−4. \| Carrying out this idea and according to k is a multiple of 4 or not, we can classify the coefficients of (1/nk) as follows: \| \| \| Sn=1n−12n2+4l−1∑k=3k≠4mj=[k/4][(−1)k−1k+(−1)k−22+j∑w=1B2w2w(−1)k−(2w+1)(2w−1)!(k−(2w+1))!(k−4w)!]1nk+∑k=4i1≤i≤l−1j=i−1[(−1)k−1k+(−1)k−22+2j−2∑w=1B2w2w(−1)k−(2w+1)(2w−1)!(k−(2w+1))!(k−4w)!]1nk+O(1n4l+1), \| which is equivalent to (2.1). ### 3. Calculation of Ak and error analysis By (2.1) and (2.2) we can obtain many concrete conclusions. Letting k=3,4,5,⋯,13,⋯ in (2.2) gives (1) when k=3, we have \| \| \| A3=(−1)23+12(−1)=−16; \| (2) when k=4, Then i=1 and we have \| \| \| A4=(−1)34+(−1)22=14; \| (3) when k=5, then j=1 and \| \| \| A5=(−1)45+(−1)32+162(−1)21!2!1!=−215; \| (4) when k=6, then j=1 and \| \| \| A6=(−1)56+(−1)42+162(−1)31!3!2!=112; \| (5) when k=7, then j=1 and \| \| \| A7=(−1)67+(−1)52+162(−1)41!4!3!=−142; \| (6) when k=8, then i=2, j=1 and \| \| \| A8=(−1)78+(−1)62+162(−1)51!5!4!=−124; \| (7) when k=9, then j=2 and \| \| \| A9=(−1)89+(−1)72+162(−1)61!6!5!+(−130)4(−1)43!4!1!=790; \| (8) when k=10, then j=2 and \| \| \| A10=(−1)910+(−1)82+162(−1)71!7!6!+(−130)4(−1)53!5!2!=−110; \| (9) when k=11, then j=2 and \| \| \| A11=(−1)1011+(−1)92+162(−1)81!8!7!+(−130)4(−1)63!6!3!=111; \| (10) when k=12, then i=3, j=2 and \| \| \| A12=(−1)1112+(−1)102+162(−1)91!9!8!+(−130)4(−1)73!7!4!=−124; \| (11) when k=13, then i=3, j=2 and \| \| \| A13=(−1)1213+(−1)112+162(−1)101!10!9!+(−130)4(−1)83!8!5!++1426(−1)65!6!1!=−892730; \| and so on. From this we can get the following corollary. Corollary 3.1. Let \| \| \| Sn=n∑i=11n2+i=1n2+1+1n2+2+⋯+1n2+n. \| Then as n→∞, \| \| \| --- \| \| Sn=1n−12n2−16n3+14n4−215n5+112n6−142n7−124n8+790n9 −110n10+111n11−124n12−892730n13+O(1n14). \| (3.1) \| As we can see, (3.1) affirms (1.2). Noting down \| \| \| Tn,k=k∑i=1Aini, \| from Theorem 2.1 we have that as n→∞, \| \| \| Sn=n∑i=11n2+i=1n2+1+1n2+2+⋯+1n2+n=Tn,k+O(1nk+1), k≥3. \| The following is the error analysis table of Sn and Tn,k: \| \| \| \| \| \| \| --- --- --- \| \| n \| 5 \| 10 \| 20 \| 30 \| , \| \| Sn−Tn,6 \| −3.8017×10−7 \| −2.7290×10−9 \| −2.0086×10−11 \| −1.1484×10−12 \| \| Sn−Tn,8 \| 3.1262×10−8 \| 6.8643×10−11 \| 1.4258×10−13 \| 3.7872×10−15 \| \| Sn−Tn,10 \| 1.6799×10−9 \| 8.6526×10−13 \| 4.3339×10−16 \| 5.0517×10−18 \| \| Sn−Tn,13 \| 1.5455×10−11 \| 1.096×10−15 \| 7.2370×10−20 \| 2.5462×10−22 \| \| Show Table DownLoad: CSV which shows that Tn,k has a good approximation to Sn. ### 4. Remarks One of the reviewers pointed out that the key formula (2.4) of this paper can be obtained by the other two methods. The following two notes benefit from the reviewer's reminder. I hope the three methods provided in this paper will be helpful for further research in related fields. Remark 4.1. Without referring to Euler-Maclaurin summation formula for the harmonic numbers Hn, we can obtain (2.4) when takinging f(t)=1/(n2+t) into the Euler-Maclaurin summation formula. In fact, by using the Euler-Maclaurin summation formula (see [3, p.45]) \| \| \| n∑k=0f(k)=∫n0f(t)dt+12[f(0)+f(n)]+m∑k=1B2k(2k)![f(2k−1)(n)−f(2k−1)(0)]+Rm, \| where \| \| \| Rm=nB2m+2(2m+2)!f(2m+2)(θn), 0<θ<1, n,m∈N \| and (2.5), the key formula (2.4) follows from the relations \| \| \| ∫n0f(t)dt=ln(n2+n)−lnn2=ln(1+1n),12[f(0)+f(n)]−1n2=12(n2+n)−12n2 \| and \| \| \| Sn=n∑k=0f(k)−1n2. \| Remark 4.2. The sum Sn can be written in the form \| \| \| Sn=ψ(n2+n+1)−ψ(n2+1), ψ(z)=ddzlnΓ(z), \| where the gamma function Γ(z) is denoted by \| \| \| Γ(z)=∫∞0tz−1e−tdt. (see [3,p.221]) \| From known asymptotic expansion of the psi function ψ(z): \| \| \| ψ(z)∼lnz−12z−∞∑k=1B2k(2k)z2k, (see [3,p.224]) \| and a property of ψ(z): \| \| \| ψ(z+1)=ψ(z)+1z, (see [3,p.224]) \| the formula (2.4) follows. ### Acknowledgments The author is thankful to reviewers for reviewers' careful corrections and valuable comments on the original version of this paper. This paper is supported by the Natural Science Foundation of China grants No.61772025. ### Conflict of interest The author declares no conflict of interest in this paper. ### References \| \| \| --- \| \| \| R. L. Graham, D. E. Knuth, O. Patashnik, Concrete mathematics: A foundation for computer science, 2 Eds., Amsterdam: Addison-Wesley Publishing Company, 1994. \| \| \| D. E. Knuth, Euler's constant to 1271 places, Math. Comput., 16 (1962), 275–280. \| \| \| A. Jeffrey, Handbook of mathematical formulas and integrals, 3 Eds., Elsevier Academic Press, 2004. \| This article has been cited by: \| \| \| \| --- \| 1. \| Yong-Guo Shi, Xiaoyu Luo, Zhi-jie Jiang, Asymptotics on a heriditary recursion, 2024, 9, 2473-6988, 30443, 10.3934/math.20241469 \| \| ##### Reader Comments © 2021 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License ( 通讯作者: 陈斌, bchen63@163.com 1. 沈阳化工大学材料科学与工程学院沈阳 110142 本站搜索百度学术搜索万方数据库搜索 CNKI搜索 AIMS Mathematics 1.8 3.1 Metrics Article views(2661) PDF downloads(43) Cited by(1) Preview PDF Download XML Export Citation Article outline Abstract Introduction Main result and its proof Calculation of Ak and error analysis Remarks Acknowledgments Conflict of interest References Other Articles By Authors On This Site Ling Zhu On Google Scholar Ling Zhu Related pages on Google Scholar on PubMed Tools Email to a friend About AIMS Press Open Access Policy Contact us Copyright © AIMS Press DownLoad: Full-Size Img PowerPoint Return \| \| \| \| \| \| \| --- --- --- \| \| n \| 5 \| 10 \| 20 \| 30 \| , \| \| Sn−Tn,6 \| −3.8017×10−7 \| −2.7290×10−9 \| −2.0086×10−11 \| −1.1484×10−12 \| \| Sn−Tn,8 \| 3.1262×10−8 \| 6.8643×10−11 \| 1.4258×10−13 \| 3.7872×10−15 \| \| Sn−Tn,10 \| 1.6799×10−9 \| 8.6526×10−13 \| 4.3339×10−16 \| 5.0517×10−18 \| \| Sn−Tn,13 \| 1.5455×10−11 \| 1.096×10−15 \| 7.2370×10−20 \| 2.5462×10−22 \| \| Show Table DownLoad: CSV \| \| \| \| \| \| \| --- --- --- \| \| n \| 5 \| 10 \| 20 \| 30 \| , \| \| Sn−Tn,6 \| −3.8017×10−7 \| −2.7290×10−9 \| −2.0086×10−11 \| −1.1484×10−12 \| \| Sn−Tn,8 \| 3.1262×10−8 \| 6.8643×10−11 \| 1.4258×10−13 \| 3.7872×10−15 \| \| Sn−Tn,10 \| 1.6799×10−9 \| 8.6526×10−13 \| 4.3339×10−16 \| 5.0517×10−18 \| \| Sn−Tn,13 \| 1.5455×10−11 \| 1.096×10−15 \| 7.2370×10−20 \| 2.5462×10−22 \| Catalog Abstract 1. Introduction 2. Main result and its proof 3. Calculation of Ak and error analysis 4. Remarks Acknowledgments Conflict of interest References
1986	https://brainly.com/question/48281283	[FREE] Calculate the net price factor and net price (in dollars). For convenience, round the net price factor to - brainly.com 7 Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Try advanced features for free Join for free Log in / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Calculate the net price factor and net price (in dollars). For convenience, round the net price factor to five decimal places and the net price to the nearest cent, when necessary. List Price: $495.80 Trade Discount Rate: 26/15/5 Net Price Factor: Net Price: 1 See answer See Expert answer Asked by mgbdub2004 • 02/23/2024 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 4394068 people 4M 4.0 0 Upload your school material for a more relevant answer The net price factor is calculated by multiplying the complements of the trade discount series of 26/15/5, which gives 0.59515 after rounding to five decimal places. To find the net price, the list price of $495.80 is multiplied by the net price factor, resulting in $294.88 when rounded to the nearest cent. Explanation To calculate the net price factor and net price for the given list price with a trade discount series of 26/15/5, we need to apply the discount rates successively. The trade discount series means the item is discounted by 26%, then 15% of the remaining price, and finally 5% of that remaining price. First, we convert the percentage discounts into decimal form to find the complement (1 - discount rate) for each discount: Initial price complement: 1 - 0.26 = 0.74 Second price complement: 1 - 0.15 = 0.85 Third price complement: 1 - 0.05 = 0.95 Next, we multiply these complements to find the net price factor: Net price factor = 0.74 × 0.85 × 0.95 = 0.59515 (rounded to five decimal places) We then multiply the net price factor by the list price to find the net price: Net price = $495.80 × 0.59515 = $294.88 (rounded to the nearest cent) Answered by TaylorSmith778 •26K answers•4.4M people helped Thanks 0 4.0 (1 vote) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 4394068 people 4M 4.0 0 Economics Economics - Theory Through Applications - LibreTexts Principles of Economics Upload your school material for a more relevant answer The net price factor is 0.59515, and the net price calculated from the list price after applying the trade discounts is $294.88. Explanation To find the net price factor and net price, we first need to calculate the discounts based on the trade discount series of 26/15/5 applied to the list price of $495.80. Convert the Trade Discounts to Decimal Form: The first discount is 26%, so the complement is: 1−0.26=0.74 The second discount is 15%, giving us: 1−0.15=0.85 The third discount is 5%, resulting in: 1−0.05=0.95 Calculate the Net Price Factor: Now, we will multiply these complements to find the net price factor: Net Price Factor=0.74×0.85×0.95 =0.59515 We round this to five decimal places. Calculate the Net Price: Next, we multiply the list price by the net price factor to find the net price: Net Price=495.80×0.59515 =294.88 This is rounded to the nearest cent. Thus, the calculations yield: Net Price Factor: 0.59515 Net Price: $294.88 Examples & Evidence For example, if a product's list price is $100 and has a trade discount of 20/10/5, you would first calculate the net price factor by finding the complements of each discount and multiplying them, then apply that factor to find the net price. The calculations shown are based on standard methods for applying trade discounts and calculating net price factors in business and mathematics. Thanks 0 4.0 (1 vote) Advertisement mgbdub2004 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer calculate the net price factor and net price (in dollars). for convenience, round the net price factor to five decimal places and the net price to the nearest cent, when necessary. list price trade discount rate net price factor net price $0.49 20/9/4 0.83333 incorrect: your answer is incorrect. $ 0.41 incorrect: your answer is incorrect. Community Answer Calculate the net price factor (as a %) and net price (in $) by using the complement method. If the list price is $1244.25 and the trade discount rate is 45.5%. Round your answer to the nearest cent. Community Answer Calculate the net price factor (as a %) and net price (in $) by using the complement method. Round your answer to the nearest cent. a) Net Price Factor: 92% b) Net Price: $125.78 c) Net Price Factor: 87% d) Net Price: $138.42 Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? New questions in Mathematics Consider function f. f(x)=7 x−21 Place the steps for finding f−1(x) in the correct order. [\begin{array}{c} x^2+21=7 y \ \frac{1}{7}\left(x^2-21\right)=f^{-1}(x), \text { where } x \geq 0 \ (x+21)^2=7 y \ x=\sqrt{7 y-21} \ x^2=7 y-21 \ y=\sqrt{7 x-21} \ \frac{1}{7} x^2+3=f^{-1}(x), \text { where } x \geq 0 \ \frac{1}{7} x^2+3=y \end{array}] The function g is defined below. g(x)=x 2−8 x−9 x 2+4 x−12 Find all values of x that are NOT in the domain of g. If there is more than one value, separate them with commas. x= The table and corresponding frequency polygon show information about the heights of members of a tennis club. What fraction of members were between 120 and 130 cm tall? \| Height (x cm) \| Midpoint \| ---\| \| $110 \| $120 \| $130 \| $140 \| $150 The base of a triangle is 21 inches and the height is 12 inches. Which of these expressions correctly shows how to calculate the area of a triangle? (21+12)×2(21×12)÷2(21+12)÷2(21×12)×2 Match each expression on the right with a similar term in the column on the left. $\begin{array}{c} 16 a \ -16 m n \ 4 a^3 \ 11 a^2 b \end{array}$ Previous questionNext question What’s your question? 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1987	https://congruentmath.com/blog/5-fun-ways-to-teach-and-practice-integers-opposites-and-absolute-values/	5 Fun Ways to Teach and Practice Integers, Opposites, and Absolute Values by Ping · September 13, 2023 Pin This In my 6th grade math classroom, we're all about practicing math concepts with a creative and practical twist. In this blog post, I'm going to give you the lowdown on 5 awesome activities for teaching integers, opposites, and absolute values. First, why do I love teaching integers, opposites, and absolute values? Because it’s easy to connect it to real life. Positive and negative integers are perfect for describing all kinds of situations, from elevations (are you above or below sea level?) to money (did you make a profit or loss?), and even to temperatures (are you above or below freezing?). And it’s a very visual topic. You can plot positive and negative integers on number lines as you practice comparing and ordering. When solving for absolute value, you can visually represent the distance away from zero on a number line. Now, onto the good stuff. Here are 5 fun ways to teach integers, opposites, and absolute values. On this page 1) Get artsy with it — with visual guided notes with doodles and coloring!-Get-artsy-with-it-with-visual-guided-notes-with-doodles-and-coloring!) Integers, Opposites & Absolute Value Guided Notes Compare & Order Integers & Absolute Values Guided Notes 2) Try some digital self-checking pixel art.-Try-some-digital-self-checking-pixel-art.) 3) Mix things up with Doodle Math - a twist on color by number!-Mix-things-up-with-Doodle-Math-a-twist-on-color-by-number!) 4) Get competitive with Kahoot, Blooket, or Quizziz.-Get-competitive-with-Kahoot-Blooket-or-Quizziz.) 5) Give your students some choice.-Give-your-students-some-choice.) Integers, opposites, and absolute value activities Want free resources? Get my free resource library with digital & print activities—plus tips over email. Join for free resources → 1) Get artsy with it — with visual guided notes with doodles and coloring! Who says math lessons can't be colorful and creative? Integers, Opposites & Absolute Value Guided Notes Pin This These guided notes include steps for plotting integers on a number line, identifying opposites, calculating absolute values of rational numbers, and much more. Plus, there's plenty of opportunity for coloring and doodling, so your students won't get bored. Each guided note also includes mazes and color-by-number activities! It covers the CCSS standards: CCSS 6.NS.C.5, CCSS 6.NS.C.6.a, CCSS 6.NS.C.6, CCSS 6.NS.C.6.c, CCSS 6.NS.C.7.a, CCSS 6.NS.C.7 Compare & Order Integers & Absolute Values Guided Notes Pin This Want to practice comparing and ordering integers and rational numbers instead? I got you! This resource was created after I realized students need a way to visually compare integers using number lines. It covers the CCSS standards: CCSS 6.NS.C.7, CCSS 6.NS.C.7.a, CCSS 6.NS.C.7.b, CCSS 6.NS.C.7.c, CCSS 6.NS.C.7.d Also, if it’s the first time you’re teaching this topic, click here for a formal lesson plan on how to teach Integers, Opposites, and Absolute Values. 2) Try some digital self-checking pixel art. Let your students work at their own pace with digital self-checking pixel art. As they type in the correct answers in the Google Sheets, magical pixels are unlocked to reveal parts of mystery images. It's a fun, interactive way to practice math, and you can even put your students in small groups to help each other out. Pin This There’s a version with integers, opposites, and absolute values. Another version focuses on comparing and ordering rational numbers. 3) Mix things up with Doodle Math - a twist on color by number! If you have a chatty class, Doodle Math is perfect for getting them to practice math while giving them some peace and quiet. As students solve math problems, they unlock doodle patterns to fill in the corresponding sections of the image. It's a great way to get students engaged and focused. Pin This Doodle Math activities are great for sub plans or homework as well since they are self-checking. As a previous buyer said, “This activity is fantastic. The students really liked the doodling as a change of pace to a traditional worksheet.” 4) Get competitive with Kahoot, Blooket, or Quizziz. Who doesn't love a little friendly competition? These game-based platforms are perfect for getting your students excited about math and practicing their skills in a fun way. Kahoot has activities for definitions and drilling absolute value, Blooket has activities for comparing integers, and Quizziz has quizzes for real-life applications of integers. 5) Give your students some choice. Pin This Let your students choose between 2-3 different activities to work on for the day. It's a great way to promote independence and responsibility, and it gives your students some control over their learning. Plus, you can even offer incentives like no homework for the day if the class stays on task and earns enough positive marks. Students go wild (in a good way) for this type of incentive! Get all the resources by grabbing the Integers, Opposites, Absolute Value Digital and Print Bundle. Integers, opposites, and absolute value activities So there you have it, 5 awesome activities for teaching integers, opposites, and absolute values. Don't forget to check out my Integers, Opposites, Absolute Value Digital and Print Bundle to take your lessons to the next level. Hi, I'm Ping! I spent 7 years in the classroom working to make math fun and relevant in middle school, by integrating math, art, and technology. I started Congruent Math to share this all with you. About Congruent Math More to read #### 7 Ways to Help Students Practice Adding & Subtracting Rational Numbers September 4, 2023 #### Engaging Warm Ups for Middle School Math July 8, 2024 #### 5 Ways to Integrate Real-Life Applications In Every Math Lesson October 8, 2023 Want more ideas and freebies? Get my free resource library with digital & print activities—plus tips over email. Join for Free Resources →
1988	https://plato.stanford.edu/archives/sum2005/entries/probability-interpret/index.html	Interpretations of Probability (Stanford Encyclopedia of Philosophy/Summer 2005 Edition) This is a file in the archives of the Stanford Encyclopedia of Philosophy. version history HOW TO CITE THIS ENTRY#### Stanford Encyclopedia of Philosophy A \| B \| C \| D \| E \| F \| G \| H \| I \| J \| K \| L \| M \| N \| O \| P \| Q \| R \| S \| T \| U \| V \| W \| X \| Y \| Z This document uses XHTML/Unicode to format the display. If you think special symbols are not displaying correctly, see our guide Displaying Special Characters.last substantive content change MAY 7 2003 Interpretations of Probability ‘Interpreting probability’ is a commonly used but misleading name for a worthy enterprise. The so-called ‘interpretations of probability’ would be better called ‘analyses of various concepts of probability’, and ‘interpreting probability’ is the task of providing such analyses. Normally, we speak of interpreting a formal system, that is, attaching familiar meanings to the primitive terms in its axioms and theorems, usually with an eye to turning them into true statements about some subject of interest. However, there is no single formal system that is ‘probability’, but rather a host of such systems. To be sure, Kolmogorov's axiomatization, which we will present shortly, has achieved the status of orthodoxy, and it is typically what philosophers have in mind when they think of ‘probability theory’. Nevertheless, several of the leading ‘interpretations of probability’ fail to satisfy all of Kolmogorov's axioms, yet they have not lost their title for that. Moreover, various other quantities that have nothing to do with probability do satisfy Kolmogorov's axioms, and thus are interpretations of it in a strict sense: normalized mass, length, area, volume, and indeed anything that falls under the scope of measure theory, the abstract mathematical theory that generalizes such quantities. Nobody seriously considers these to be ‘interpretations of probability’, however, because they do not play the right role in our conceptual apparatus. Instead, we will be concerned here with various probability-like concepts that purportedly do. Be all that as it may, we will follow common usage and drop the cringing scare quotes in our survey of what philosophers have taken to be the chief interpretations of probability. Whatever we call it, the project of finding such interpretations is an important one. Probability is virtually ubiquitous. It plays a role in almost all the sciences. It underpins much of the social sciences -- witness, for example, the prevalence of the use of statistical testing, confidence intervals, regression methods, and so on. It finds its way, moreover, into much of philosophy. In epistemology, the philosophy of mind, and cognitive science, we see states of opinion being modeled by subjective probability functions, and learning being modeled by the updating of such functions. Since probability theory is central to decision theory and game theory, it has ramifications for ethics and political philosophy. It figures prominently in such staples of metaphysics as causation and laws of nature. It appears again in the philosophy of science in the analysis of confirmation of theories, scientific explanation, and in the philosophy of specific scientific theories, such as quantum mechanics, statistical mechanics, and genetics. It can even take center stage in the philosophy of logic, the philosophy of language, and the philosophy of religion. Thus, problems in the foundations of probability bear at least indirectly, and sometimes directly, upon central scientific, social scientific, and philosophical concerns. The interpretation of probability is one of the most important such foundational problems. 1. Kolmogorov's Probability Calculus 2. Criteria of adequacy for the interpretations of probability 3. The Main Interpretations 3.1 Classical Probability 3.2 Logical Probability 3.3 Frequency Interpretations 3.4 Propensity Interpretations 3.5 Subjective Probability Probability as degree of belief The betting interpretation and the Dutch Book argument Probabilities and utilities Orthodox Bayesianism, and further constraints on rational credences 4. Conclusion: Future Prospects? Bibliography Other Internet Resources Related Entries 1. Kolmogorov's Probability Calculus Probability theory was inspired by games of chance in 17 th century France and inaugurated by the Fermat-Pascal correspondence. However, its axiomatization had to wait until Kolmogorov's classic Foundations of the Theory of Probability (1933). Let Ω be a non-empty set (‘the universal set’). A field (or algebra) on Ω is a set F of subsets of Ω that has Ω as a member, and that is closed under complementation (with respect to Ω) and union. Let P be a function from F to the real numbers obeying: (Non-negativity) P(A) ≥ 0, for all A ∈ F. (Normalization) P(Ω) = 1. (Finite additivity) P(A ∪ B) = P(A) + P(B) for all A, B ∈ F such that A ∩ B = Ø. Call P a probability function, and (Ω, F, P) a probability space. The assumption that P is defined on a field guarantees that these axioms are non-vacuously instantiated, as are the various theorems that follow from them. The non-negativity and normalization axioms are largely matters of convention, although it is non-trivial that probability functions take at least the two values 0 and 1, and that they have a maximal value (unlike various other measures, such as length, volume, and so on, which are unbounded). We will return to finite additivity at a number of points below. We may now apply the theory to various familiar cases. For example, we may represent the results of tossing a single die once by the set Ω={1, 2, 3, 4, 5, 6}, and we could let F be the set of all subsets of Ω. Under the natural assignment of probabilities to members of F, we obtain such welcome results as P({1}) = 1/6, P(even) = P({2} ∪ {4} ∪ {6}) = 3/6, P(odd or less than 4) = P(odd) + P(less than 4) − P(odd ∩ less than 4) = 1/2 + 1/2 − 2/6 = 4/6, and so on. We could instead attach probabilities to members of a collection S of sentences of a formal language, closed under (countable) truth-functional combinations, with the following counterpart axiomatization: P(A) ≥ 0 for all A ∈ S. If T is a logical truth (in classical logic), then P(T) = 1. P(AB) = P(A) + P(B) for all A ∈ S and B ∈ S such that A and B are logically incompatible. Now let us strengthen our closure assumptions regarding F, requiring it to be closed under complementation and countable union; it is then called a sigma field (or sigma algebra) on Ω. It is controversial whether we should strengthen finite additivity, as Kolmogorov does: 3′. (Countable additivity) If {A i} is a countably infinite collection of (pairwise) disjoint sets, each of which is an element of F, then Kolmogorov comments that infinite probability spaces are idealized models of real random processes, and that he limits himself arbitrarily to only those models that satisfy countable additivity. This axiom is the cornerstone of the assimilation of probability theory to measure theory. The conditional probability of A given B is then given by the ratio of unconditional probabilities: P(A \| B)=P(A ∩ B) P(B), provided P(B) > 0. This is often taken to be the definition of conditional probability, although it should be emphasized that this is a technical usage of the term that may not align perfectly with a pretheoretical concept that we might have (see Hájek, forthcoming). Indeed, some authors take conditional probability to be the primitive notion, and axiomatize it directly (e.g. Popper 1959b, Renyi 1970, van Fraassen 1976, Spohn 1986 and Roeper and Leblanc 1999). There are other axiomatizations that give up normalization; that give up countable additivity, and even additivity; that allow probabilities to take infinitesimal values (positive, but smaller than every positive real number); that allow probabilities to be vague (interval-valued, or more generally sets of numerical values). For now, however, when we speak of ‘the probability calculus’, we will mean Kolmogorov's approach, as is standard. Given certain probabilities as inputs, the axioms and theorems allow us to compute various further probabilities. However, apart from the assignment of 1 to the universal set and 0 to the empty set, they are silent regarding the initial assignment of probabilities. For guidance with that, we need to turn to the interpretations of probability. First, however, let us list some criteria of adequacy for such interpretations. 2. Criteria of adequacy for the interpretations of probability What criteria are appropriate for assessing the cogency of a proposed interpretation of probability? Of course, an interpretation should be precise, unambiguous, and use well-understood primitives. But those are really prescriptions for good philosophizing generally; what do we want from our interpretations of probability, specifically? We begin by following Salmon (1966, 64), although we will raise some questions about his criteria, and propose some others. He writes: Admissibility._We say that an interpretation of a formal system is admissible if the meanings assigned to the primitive terms in the interpretation transform the formal axioms, and consequently all the theorems, into true statements. A fundamental requirement for probability concepts is to satisfy the mathematical relations specified by the calculus of probability… _Ascertainability. This criterion requires that there be some method by which, in principle at least, we can ascertain values of probabilities. It merely expresses the fact that a concept of probability will be useless if it is impossible in principle to find out what the probabilities are… Applicability. The force of this criterion is best expressed in Bishop Butler's famous aphorism, “Probability is the very guide of life.”… It might seem that the criterion of admissibility goes without saying: ‘interpretations’ of the probability calculus that assigned to P the interpretation ‘the number of hairs on the head of’ or ‘the political persuasion of’ would obviously not even be in the running, because they would render the axioms and theorems so obviously false. The word ‘interpretation’ is often used in such a way that ‘admissible interpretation’ is a pleonasm. Yet it turns out that the criterion is non-trivial, and indeed if taken seriously would rule out several of the leading interpretations of probability! As we will see, some of them fail to satisfy countable additivity; for others (certain propensity interpretations) the status of at least some of the axioms is unclear. Nevertheless, we regard them as genuine candidates. It should be remembered, moreover, that Kolmogorov's is just one of many possible axiomatizations, and there is not universal agreement on which is ‘best’ (whatever that might mean). Indeed, Salmon's preferred axiomatization differs from Kolmogorov's. Thus, there is no such thing as admissibility tout court, but rather admissibility with respect to this or that axiomatization. It would be unfortunate if, perhaps out of an overdeveloped regard for history, one felt obliged to reject any interpretation that did not obey the letter of Kolmogorov's laws and that was thus ‘inadmissible’. In any case, if we found an inadmissible interpretation that did a wonderful job of meeting the criteria of ascertainability and applicability, then we should surely embrace it. So let us turn to those criteria. It is a little unclear in the ascertainability criterion just what “in principle” amounts to, though perhaps some latitude here is all to the good. Understood charitably, and to avoid trivializing it, it presumably excludes omniscience. On the other hand, understanding it in a way acceptable to a strict empiricist or a verificationist may be too restrictive. ‘Probability’ is apparently, among other things, a modal concept, plausibly outrunning that which actually occurs, let alone that which is actually observed. Most of the work will be done by the applicability criterion. We must say more (as Salmon indeed does) about what sort of a guide to life probability is supposed to be. Mass, length, area and volume are all useful concepts, and they are ‘guides to life’ in various ways (think how critical distance judgments can be to survival); moreover, they are admissible and ascertainable, so presumably it is the applicability criterion that will rule them out. Perhaps it is best to think of applicability as a cluster of criteria, each of which is supposed to capture something of probability's distinctive conceptual roles; moreover, we should not require that all of them be met by a given interpretation. They include: Non-triviality:_an interpretation should make non-extreme probabilities at least a conceptual possibility. For example, suppose that we interpret ‘_P’ as the truth_function: it assigns the value 1 to all true sentences, and 0 to all false sentences. Then trivially, all the axioms come out true, so this interpretation is admissible. We would hardly count it as an adequate _interpretation of__probability, however, and so we need to exclude it. It is essential to probability that, at least in principle, it can take intermediate_values. All of the interpretations that we will present meet this criterion, so we will discuss it no more. _Applicability to frequencies: an interpretation should render perspicuous the relationship between probabilities and (long-run) frequencies. Among other things, it should make clear why, by and large, more probable events occur more frequently than less probable events. Applicability to rational belief: an interpretation should clarify the role that probabilities play in constraining the degrees of belief, or credences, of rational agents. Among other things, knowing that one event is more probable than another, a rational agent will be more confident about the occurrence of the former event. Applicability to ampliative inference: an interpretation will score bonus points if it illuminates the distinction between ‘good’ and ‘bad’ ampliative inferences, while explicating why both fall short of deductive inferences. The next criterion may be redundant, given our list so far, but including it will do no harm: _Applicability to science:_an interpretation should illuminate paradigmatic uses of probability in science (for example, in quantum mechanics and statistical mechanics). Perhaps there are further metaphysical desiderata that we might impose on the interpretations. For example, there appear to be connections between probability and modality. Events with positive probability can happen, even if they don't. Some authors also insist on the converse condition that only events with positive probability can happen, although this is more controversial -- see our discussion of ‘regularity’ in Section 4. (Indeed, in uncountable probability spaces this condition will require the employment of infinitesimals, and will thus take us beyond the standard Kolmogorov theory -- ‘standard’ both in the sense of being the orthodoxy, and in its employment of standard, as opposed to ‘non-standard’ real numbers. See Skyrms 1980.) In any case, our list is already long enough to help in our assessment of the leading interpretations on the market. 3. The Main Interpretations 3.1 Classical Probability The classical interpretation owes its name to its early and august pedigree. Championed by Laplace, and found even in the works of Pascal, Bernoulli, Huygens, and Leibniz, it assigns probabilities in the absence of any evidence, or in the presence of symmetrically balanced evidence. The guiding idea is that in such circumstances, probability is shared equally among all the possible outcomes, so that the classical probability of an event is simply the fraction of the total number of possibilities in which the event occurs. It seems especially well suited to those games of chance that by their very design create such circumstances -- for example, the classical probability of a fair die landing with an even number showing up is 3/6. It is often presupposed (usually tacitly) in textbook probability puzzles. Here is a classic statement by Laplace: The theory of chance consists in reducing all the events of the same kind to a certain number of cases equally possible, that is to say, to such as we may be equally undecided about in regard to their existence, and in determining the number of cases favorable to the event whose probability is sought. The ratio of this number to that of all the cases possible is the measure of this probability, which is thus simply a fraction whose numerator is the number of favorable cases and whose denominator is the number of all the cases possible. (1814, 1951 6-7) There are numerous questions to be asked about this formulation. When are events of the same kind? Intuitively, ‘heads’ and ‘tails’ are equally likely outcomes of tossing a fair coin; but if their kind is ‘ways the coin could land’, then ‘edge’ should presumably be counted alongside them. The “certain number of cases” and “that of all the cases possible” are presumably finite numbers. What, then, of probabilities in infinite spaces? Apparently, irrational-valued probabilities such as 1/√2 are automatically eliminated, and thus theories such as quantum mechanics that posit them cannot be accommodated. (We will shortly see, however, that Laplace's theory has been refined to handle infinite spaces.) Who are “we”, who “may be equally undecided”? Different people may be equally undecided about different things, which suggests that Laplace is offering a subjectivist interpretation in which probabilities vary from person to person depending on contingent differences in their evidence. This is not his intention. He means to characterize the objective probability assignment of a rational agent in an epistemically neutral position with respect to a set of “equally possible” cases. But then the proposal risks sounding empty: for what is it for an agent to be “equally undecided” about a set of cases, other than assigning them equal probability? This brings us to one of the key objections to Laplace's account. The notion of “equally possible” cases faces the charge of either being a category mistake (for ‘possibility’ does not come in degrees), or circular (for what is meant is really ‘equally probable’). The notion is finessed by the so-called ‘principle of indifference’, a coinage due to Keynes. It states that whenever there is no evidence favoring one possibility over another, they have the same probability. Thus, it is claimed, there is no circularity in the classical definition after all. However, this move may only postpone the problem, for there is still a threat of circularity, albeit at a lower level. We have two cases here: outcomes for which we have no evidence at all, and outcomes for which we have symmetrically balanced evidence. There is no circularity in the first case unless the notion of ‘evidence’ is itself probabilistic; but artificial examples aside, it is doubtful that the case ever arises. For example, we have a considerable fund of evidence on coin tossing from the results of our own experiments, the testimony of others, our knowledge of some of the relevant physics, and so on. In the second case, the threat of circularity is more apparent, for it seems that some sort of weighing of the evidence in favor of each outcome is required, and it is not obvious that this can be done without reference to probability. Indeed, the most obvious characterization of symmetrically balanced evidence is in terms of equality of conditional probabilities: given evidence E and possible outcomes O 1, O 2, …, O n, the evidence is symmetrically balanced iff P(O 1 \| E) = P(O 2 \| E) = … = P(O n \| E). Then it seems that probabilities reside at the base of the interpretation after all. Still, it would be an achievement if all probabilities could be reduced to cases of equal probability. As we have seen, Laplace's classical theory is restricted to finite sample spaces. When the spaces are countably infinite, the spirit of the classical theory may be upheld by appealing to the information-theoretic principle of maximum entropy, a generalization of the principle of indifference. Entropy is a measure of the lack of ‘informativeness’ of a probability distribution. The more concentrated is the distribution, the less is its entropy; the more diffuse it is, the greater is its entropy. For a discrete distribution P = (p 1, p 2, …), the entropy of P is defined as: The principle of maximum entropy enjoins us to select from the family of all distributions consistent with our background knowledge the distribution that maximizes this quantity. In the special case of choosing the most uninformative prior over a finite set of possible outcomes, this is just the familiar ‘flat’ classical distribution discussed previously. Things get more complicated in the infinite case, since there cannot be a flat distribution over denumerably many outcomes, on pain of violating the standard probability calculus (with countable additivity). Rather, the best we can have are sequences of progressively flatter distributions, none of which is truly flat. We must then impose some further constraint that narrows the field to a smaller family in which there is a distribution of maximum entropy. This constraint has to be imposed from outside as background knowledge, but there is no general theory of which external constraint should be applied when. Let us turn now to uncountably infinite spaces. It is easy -- all too easy -- to assign equal probabilities to the points in such a space: each gets probability 0. Non-trivial probabilities arise when uncountably many of the points are clumped together in larger sets. If there are finitely many clumps, Laplace's classical theory may be appealed to again: if the evidence bears symmetrically on these clumps, each gets the same share of probability. Enter Bertrand's paradoxes. They all arise in uncountable spaces and turn on alternative parametrizations of a given problem that are non-linearly related to each other. Some presentations are needlessly arcane; length and area suffice to make the point. The following example (adapted from van Fraassen 1989) nicely illustrates how Bertrand-style paradoxes work. A factory produces cubes with side-length between 0 and 1 foot; what is the probability that a randomly chosen cube has side-length between 0 and 1/2 a foot? The tempting answer is 1/2, as we imagine a process of production that is uniformly distributed over side-length. But the question could have been given an equivalent restatement: A factory produces cubes with face-area between 0 and 1 square-feet; what is the probability that a randomly chosen cube has face-area between 0 and 1/4 square-feet? Now the tempting answer is 1/4, as we imagine a process of production that is uniformly distributed over face-area. This is already disastrous, as we cannot allow the same event to have two different probabilities (especially if this interpretation is to be admissible!). But there is worse to come, for the problem could have been restated equivalently again: A factory produces cubes with volume between 0 and 1 cubic feet; what is the probability that a randomly chosen cube has volume between 0 and 1/8 cubic-feet? Now the tempting answer is 1/8, as we imagine a process of production that is uniformly distributed over volume. And so on for all of the infinitely many equivalent reformulations of the problem (in terms of the fourth, fifth, … power of the length, and indeed in terms of every non-zero real-valued exponent of the length). What, then, is the probability of the event in question? The paradox arises because the principle of indifference can be used in incompatible ways. We have no evidence that favors the side-length lying in the interval [0, 1/2] over its lying in [1/2, 1], or vice versa, so the principle requires us to give probability 1/2 to each. Unfortunately, we also have no evidence that favors the face-area lying in any of the four intervals [0, 1/4], [1/4, 1/2], [1/2, 3/4], and [3/4, 1] over any of the others, so we must give probability 1/4 to each. The event ‘the side-length lies in [0, 1/2]’, receives a different probability when merely redescribed. And so it goes, for all the other reformulations of the problem. We cannot meet any pair of these constraints simultaneously, let alone all of them. How does the classical theory of probability fare with respect to our criteria of adequacy? Let us begin with admissibility. It is claimed that (Laplacean) classical probabilities are only finitely additive (see, e.g., de Finetti 1974). It would be more correct to say that classical probabilities are countably additive, but trivially so. As we have seen, classical probabilities are only defined on finite sample spaces. The statement 3′ of countable additivity, recall, is a conditional; its antecedent, “{A i} is a countably infinite collection of (pairwise) disjoint sets,” is never satisfied in such spaces. Thus, the conditional is vacuously true. Classical probabilities are ascertainable, assuming that the space of possibilities can be determined in principle. They bear a relationship to the credences of rational agents; the concern, as we saw above, is that the relationship is vacuous, and that rather than constraining the credences of a rational agent in an epistemically neutral position, they merely record them. Without supplementation, the classical theory makes no contact with frequency information. However the coin happens to land in a sequence of trials, the possible outcomes remain the same. Indeed, even if we have strong empirical evidence that the coin is biased towards heads with probability, say, 0.6, it is hard to see how the unadorned classical theory can accommodate this fact -- for what now are the ten possibilities, six of which are favorable to heads? Laplace does supplement the theory with his Rule of Succession: "Thus we find that an event having occurred successively any number of times, the probability that it will happen again the next time is equal to this number increased by unity divided by the same number, increased by two units." (1951, 19) That is: Pr(success on N+1st trial\|N consecutive successes)=N+1 N+2 Thus, inductive learning is possible. We must ask, however, whether such learning can be captured once and for all by such a simple formula, the same for all domains and events. We will return to this question when we discuss the logical interpretation below. Science apparently invokes at various points probabilities that look classical. Bose-Einstein statistics, Fermi-Dirac statistics, and Maxwell-Boltzmann statistics each arise by considering the ways in which particles can be assigned to states, and then applying the principle of indifference to different subdivisions of the set of alternatives, Bertrand-style. The trouble is that Bose-Einstein statistics apply to some particles (e.g. photons) and not to others, Fermi-Dirac statistics apply to different particles (e.g. electrons), and Maxwell-Boltzmann statistics do not apply to any known particles. None of this can be determined a priori, as the classical interpretation would have it. Moreover, the classical theory purports to yield probability assignments in the face of ignorance. But as Fine (1973) writes: If we are truly ignorant about a set of alternatives, then we are also ignorant about combinations of alternatives and about subdivisions of alternatives. However, the principle of indifference when applied to alternatives, or their combinations, or their subdivisions, yields different probability assignments (170). This brings us to one of the chief points of controversy regarding the classical interpretation. Critics accuse the principle of indifference of extracting information from ignorance. Proponents reply that it rather codifies the way in which such ignorance should be epistemically managed -- for anything other than an equal assignment of probabilities would represent the possession of some knowledge. Critics counter-reply that in a state of complete ignorance, it is better to assign vague probabilities (perhaps vague over the entire [0, 1] interval), or to eschew the assignment of probabilities altogether. 3.2 Logical probability Logical theories of probability retain the classical interpretation's idea that probabilities can be determined a priori by an examination of the space of possibilities. However, they generalize it in two important ways: the possibilities may be assigned unequal weights, and probabilities can be computed whatever the evidence may be, symmetrically balanced or not. Indeed, the logical interpretation, in its various guises, seeks to encapsulate in full generality the degree of support or confirmation that a piece of evidence E confers upon a given hypothesis H, which we may write as c(H, E). In doing so, it can be regarded also as generalizing deductive logic and its notion of implication, to a complete theory of inference equipped with the notion of ‘degree of implication’ that relates E to H. It is often called the theory of ‘inductive logic’, although this is a misnomer: there is no requirement that E be in any sense ‘inductive’ evidence for H. ‘Non-deductive logic’ would be a better name, yet even this overlooks the fact that deductive logic's relations of implication and incompatibility are also accommodated as extreme cases in which the confirmation function takes the values 1 and 0 respectively. Nevertheless, what is significant is that the logical interpretation provides a framework for induction. Early proponents of logical probability include Johnson (1921), Keynes (1921), and Jeffreys (1939). However, by far the most systematic study of logical probability was by Carnap. His formulation of logical probability begins with the construction of a formal language. In (1950) he considers a class of very simple languages consisting of a finite number of logically independent monadic predicates (naming properties) applied to countably many individual constants (naming individuals) or variables, and the usual logical connectives. The strongest (consistent) statements that can be made in a given language describe all of the individuals in as much detail as the expressive power of the language allows. They are conjunctions of complete descriptions of each individual, each description itself a conjunction containing exactly one occurrence (negated or unnegated) of each predicate of the language. Call these strongest statements state descriptions. Any probability measure m(−) over the state descriptions automatically extends to a measure over all sentences, since each sentence equivalent to a disjunction of state descriptions; m in turn induces a confirmation function c(−, −): c(h, e)=m(h&e) m(e) There are obviously infinitely many candidates for m, and hence c, even for very simple languages. Carnap argues for his favored measure “m” by insisting that the only thing that significantly distinguishes individuals from one another is some qualitative difference, not just a difference in labeling. Call a structure description a maximal set of state descriptions, each of which can be obtained from another by some permutation of the individual names. m assigns each structure description equal measure, which in turn is divided equally among their constituent state descriptions. It gives greater weight to homogenous state descriptions than to heterogeneous ones, thus ‘rewarding’ uniformity among the individuals in accordance with putatively reasonable inductive practice. The induced c allows inductive learning from experience. Consider, for example, a language that has three names, a, b and c, for individuals, and one predicate F. For this language, the state descriptions are: Fa&Fb&Fc ¬Fa&Fb&Fc Fa& ¬Fb&Fc Fa&Fb& ¬Fc ¬Fa& ¬Fb&Fc ¬Fa&Fb& ¬Fc Fa& ¬Fb& ¬Fc ¬Fa& ¬Fb& ¬Fc There are four structure descriptions: {1}, “Everything is F”; {2, 3, 4}, “Two F s, one ¬F”; {5, 6, 7}, “One F, two ¬F s”; and {8}, “Everything is ¬F”. The measure m assigns numbers to the state descriptions as follows: first, every structure description is assigned an equal weight, 1/4; then, each state description belonging to a given structure description is assigned an equal part of the weight assigned to the structure description: State description Structure description Weight _m_ 1. Fa.Fb.Fc I. Everything is F 1/4 1/4 ¬Fa.Fb.Fc 1/12 Fa.¬Fb.Fc II. Two F s, one ¬F 1/4 1/12 Fa.Fb.¬Fc 1/12 ¬Fa.¬Fb.Fc 1/12 ¬Fa.Fb.¬Fc III. One F, two ¬F s 1/4 1/12 Fa.¬Fb.¬Fc 1/12 ¬Fa.¬Fb.¬Fc IV. Everything is ¬F 1/4 1/4 Notice that m gives greater weight to homogenous state descriptions than to heterogeneous ones, thus ‘rewarding’ uniformity among the individuals, in accordance with putatively reasonable inductive practice. This will manifest itself in the inductive support that hypotheses can gain from appropriate evidence statements. Consider the hypothesis statement h = Fc, true in 4 of the 8 state descriptions, with a priori probability m(h) = 1/2. Suppose we examine individual “a” and find it has property F -- call this evidence e. Intuitively, e is favorable (albeit weak) inductive evidence for h. We have: m(h&e) = 1/3, m(e) = 1/2, and hence c(h, e)=m(h&e) m(e)=2/3. This is greater than the a priori probability m(h) = 1/2, so the hypothesis has been confirmed. It can be shown that in general m yields a degree of confirmation c that allows learning from experience. Note, however, that infinitely many confirmation functions, defined by suitable choices of the initial measure, allow learning from experience. We do not have yet a reason to think that c is the right choice. Carnap claims nevertheless that c stands out for being simple and natural. He later generalizes his confirmation function to a continuum of functions c λ. Define a family of predicates to be a set of predicates such that, for each individual, exactly one member of the set applies, and consider first-order languages containing a finite number of families. Carnap (1963) focuses on the special case of a language containing only one-place predicates. He lays down a host of axioms concerning the confirmation function c, including those induced by the probability calculus itself, various axioms of symmetry (for example, that c(h, e) remains unchanged under permutations of individuals, and of predicates of any family), and axioms that guarantee undogmatic inductive learning, and long-run convergence to relative frequencies. They imply that, for a family {P n}, n = 1, …, k (k> 2): c λ(individual s + 1 is P j, s j of the first s individuals are P j)=(s j + λ/k) s + λ, where λ is a positive real number. The higher the value of λ, the less impact evidence has: induction from what is observed becomes progressively more swamped by a classical-style equal assignment to each of the k possibilities regarding individual s + 1. The problem remains: what is the correct setting of λ, or said another way, how ‘inductive’ should the confirmation function be? Also, it turns out that for any such setting, a universal statement in an infinite universe always receives zero confirmation, no matter what the (finite) evidence. Many find this counterintuitive, since laws of nature with infinitely many instances can apparently be confirmed. Earman (1992) discusses the prospects for avoiding the unwelcome result. Significantly, Carnap's various axioms of symmetry are hardly logical truths. More seriously, we cannot impose further symmetry constraints that are seemingly just as plausible as Carnap's, on pain of inconsistency -- see Fine (1973, 202). Goodman taught us: that the future will resemble the past in some respect is trivial; that it will resemble the past in all respects is contradictory. And we may continue: that a probability assignment can be made to respect some symmetry is trivial; that one can be made to respect all symmetries is contradictory. This threatens the whole program of logical probability. Another Goodmanian lesson is that inductive logic must be sensitive to the meanings of predicates, strongly suggesting that a purely syntactic approach such as Carnap's is doomed. Scott and Krauss (1966) use model theory in their formulation of logical probability for richer and more realistic languages than Carnap's. Still, finding a canonical language seems to many to be a pipe dream, at least if we want to analyze the “logical probability” of any argument of real interest -- either in science, or in everyday life. Logical probabilities are admissible. Countable additivity is vacuously satisfied in any finite language, since there cannot be an infinite sequence of pairwise incompatible sentences. Given a choice of language, the values of a given confirmation function are ascertainable; thus, if this language is rich enough for a given application, the relevant probabilities are ascertainable. The whole point of the theory of logical probability is to explicate ampliative inference, although given the apparent arbitrariness in the choice of language and in the setting of λ -- thus, in the choice of confirmation function -- one may wonder how well it achieves this. The problem of arbitrariness of the confirmation function also hampers the extent to which the logical interpretation can truly illuminate the connection between probabilities and frequencies. The arbitrariness problem, moreover, stymies any compelling connection between logical probabilities and rational credences. And a further problem remains even after the confirmation function has been chosen: if one's credences are to be based on logical probabilities, they must be relativized to an evidence statement, e. But which is to be? Carnap's recommendation is that e should be one's total evidence, that is, the maximally specific information at one's disposal, the strongest proposition of which one is certain. However, when we go beyond toy examples, it is not clear that this is well-defined. Suppose I have just watched a coin toss, and thus learned that the coin landed heads. Perhaps ‘the coin landed heads’ is my total evidence? But I also learned a host of other things: as it might be, that the coin landed at a certain time, bouncing in a certain way, making a certain noise as it did so … Call this long conjunction of facts X. I also learned a potentially infinite set of de se propositions: ‘I learned that X’, ‘I learned that I learned that X’ and so on. Perhaps, then, my total evidence is the infinite intersection of all these propositions, although this is still not obvious -- and it is not something that can be represented by a sentence in one of Carnap's languages, which is finite in length. More significantly, the total evidence criterion goes hand in hand with positivism and a foundationalist epistemology according to which there are such determinate, ultimate deliverances of experience. But perhaps learning does not come in the form of such ‘bedrock’ propositions, as Jeffrey (1992) has argued -- maybe it rather involves a shift in one's subjective probabilities across a partition, without any cell of the partition becoming certain. Then it may be the case that the strongest proposition of which one is certain is expressed by a tautology T -- hardly an interesting notion of ‘total evidence’. In connection with the ‘applicability to science’ criterion, a point due to Lakatos is telling. By Carnap's lights, the degree of confirmation of a hypothesis depends on the language in which the hypothesis is stated and over which the confirmation function is defined. But scientific progress often brings with it a change in scientific language (for example, the addition of new predicates and the deletion of old ones), and such a change will bring with it a change in the corresponding c-values. Thus, the growth of science may overthrow any particular confirmation theory. There is something of the snake eating its own tail here, since logical probability was supposed to explicate the confirmation of scientific theories. 3.3 Frequency Interpretations Gamblers, actuaries and scientists have long understood that relative frequencies bear an intimate relationship to probabilities. Frequency interpretations posit the most intimate relationship of all: identity. Thus, we might identify the probability of ‘heads’ on a certain coin with the frequency of heads in a suitable sequence of tosses of the coin, divided by the total number of tosses. A simple version of frequentism, which we will call finite frequentism, attaches probabilities to events or attributes in a finite reference class in such a straightforward manner: the probability of an attribute A in a finite reference class B is the relative frequency of actual occurrences of A within B. Thus, finite frequentism bears certain structural similarities to the classical interpretation, insofar as it gives equal weight to each member of a set of events, simply counting how many of them are ‘favorable’ as a proportion of the total. The crucial difference, however, is that where the classical interpretation counted all the possible outcomes of a given experiment, finite frequentism counts actual outcomes. It is thus congenial to those with empiricist scruples. It was developed by Venn (1876), who in his discussion of the proportion of births of males and females, concludes: “probability is nothing but that proportion” (p. 84, his emphasis). Finite frequentism gives an operational definition of probability, and its problems begin there. For example, just as we want to allow that our thermometers could be ill-calibrated, and could thus give misleading measurements of temperature, so we want to allow that our ‘measurements’ of probabilities via frequencies could be misleading, as when a fair coin lands heads 9 out of 10 times. More than that, it seems to be built into the very notion of probability that such misleading results can arise. Indeed, in many cases, misleading results are guaranteed. Starting with a degenerate case: according to the finite frequentist, a coin that is never tossed, and that thus yields no actual outcomes whatsoever, lacks a probability for heads altogether; yet a coin that is never measured does not thereby lack a diameter. Perhaps even more troubling, a coin that is tossed exactly once yields a relative frequency of heads of either 0 or 1, whatever its bias. Famous enough to merit a name of its own, this is the so-called ‘problem of the single case’. In fact, many events are most naturally regarded as not merely unrepeated, but in a strong sense unrepeatable -- the 2000 presidential election, the final game of the 2001 NBA play-offs, the Civil War, Kennedy's assassination, certain events in the very early history of the universe. Nonetheless, it seems natural to think of non-extreme probabilities attaching to some, and perhaps all, of them. Worse still, some cosmologists regard it as a genuinely chancy matter whether our universe is open or closed (apparently certain quantum fluctuations could, in principle, tip it one way or the other), yet whatever it is, it is ‘single-case’ in the strongest possible sense. The problem of the single case is particularly striking, but we really have a sequence of related problems: ‘the problem of the double case’, ‘the problem of the triple case’ … Every coin that is tossed exactly twice can yield only the relative frequencies 0, 1/2 and 1, whatever its bias… A finite reference class of size n, however large n is, can only produce relative frequencies at a certain level of ‘grain’, namely 1/n. Among other things, this rules out irrational probabilities; yet our best physical theories say otherwise. Furthermore, there is a sense in which any of these problems can be transformed into the problem of the single case. Suppose that we toss a coin a thousand times. We can regard this as a single trial of a thousand-tosses-of-the-coin experiment. Yet we do not want to be committed to saying that that experiment yields its actual result with probability 1. The problem of the single case is that the finite frequentist fails to see intermediate probabilities in various places where others do. There is also the converse problem: the frequentist sees intermediate probabilities in various places where others do not. Our world has myriad different entities, with myriad different attributes. We can group them into still more sets of objects, and then ask with which relative frequencies various attributes occur in these sets. Many such relative frequencies will be intermediate; the finite frequentist automatically identifies them with intermediate probabilities. But it would seem that whether or not they are genuine probabilities, as opposed to mere tallies, depends on the case at hand. Bare ratios of attributes among sets of disparate objects may lack the sort of modal force that one might expect from probabilities. I belong to the reference class consisting of myself, the Eiffel Tower, the southernmost sandcastle on Santa Monica Beach, and Mt Everest. Two of these four objects are less than 7 ft. tall, a relative frequency of 1/2; moreover, we could easily extend this class, preserving this relative frequency (or, equally easily, not). Yet it would be odd to say that my probability of being less than 7 ft. tall, relative to this reference class, is 1/2, even though it is perfectly acceptable (if uninteresting) to say that 1/2 of the objects in the reference class are less than 7 ft. tall. Some frequentists (notably Venn 1876, Reichenbach 1949, and von Mises 1957 among others), partly in response to some of the problems above, have gone on to consider infinite reference classes, identifying probabilities with limiting relative frequencies of events or attributes therein. Thus, we require an infinite sequence of trials in order to define such probabilities. But what if the actual world does not provide an infinite sequence of trials of a given experiment? Indeed, that appears to be the norm, and perhaps even the rule. In that case, we are to identify probability with a hypothetical or counterfactual limiting relative frequency. We are to imagine hypothetical infinite extensions of an actual sequence of trials; probabilities are then what the limiting relative frequencies would be if the sequence were so extended. Note that at this point we have left empiricism behind. A modal element has been injected into frequentism with this invocation of a counterfactual; moreover, the counterfactual may involve a radical departure from the way things actually are, one that may even require the breaking of laws of nature. (Think what it would take for the coin in my pocket, which has only been tossed once, to be tossed infinitely many times -- never wearing out, and never running short of people willing to toss it!) One may wonder, moreover, whether there is always -- or ever -- a fact of the matter of what such counterfactual relative frequencies are. Limiting relative frequencies, we have seen, must be relativized to a sequence of trials. Herein lies another difficulty. Consider an infinite sequence of the results of tossing a coin, as it might be H, T, H, H, H, T, H, T, T, … Suppose for definiteness that the corresponding relative frequency sequence for heads, which begins 1/1, 1/2, 2/3, 3/4, 4/5, 4/6, 5/7, 5/8, 5/9, …, converges to 1/2. By suitably reordering these results, we can make the sequence converge to any value in [0, 1] that we like. (If this is not obvious, consider how the relative frequency of even numbers among positive integers, which intuitively ‘should’ converge to 1/2, can instead be made to converge to 1/4 by reordering the integers with the even numbers in every fourth place, as follows: 1, 3, 5, 2, 7, 9, 11, 4, 13, 15, 17, 6, …) To be sure, there may be something natural about the ordering of the tosses as given -- for example, it may be their temporal ordering. But there may be more than one natural ordering. Imagine the tosses taking place on a train that shunts backwards and forwards on tracks that are oriented west-east. Then the spatial ordering of the results from west to east could look very different. Why should one ordering be privileged over others? A well-known objection to any version of frequentism is that relative frequencies must be relativised to a reference class. Consider a probability concerning myself that I care about -- say, my probability of living to age 80. I belong to the class of males, the class of non-smokers, the class of philosophy professors who have two vowels in their surname, … Presumably the relative frequency of those who live to age 80 varies across (most of) these reference classes. What, then, is my probability of living to age 80? It seems that there is no single frequentist answer. Instead, there is my probability-qua-male, my probability-qua-non-smoker, my probability-qua-male-non-smoker, and so on. This is an example of the so-called reference class problem for frequentism (although it can be argued that analogues of the problem arise for the other interpretations as well). And as we have seen in the previous paragraph, the problem is only compounded for limiting relative frequencies: probabilities must be relativized not merely to a reference class, but to a sequence within the reference class. We might call this the reference sequence problem. The beginnings of a solution to this problem would be to restrict our attention to sequences of a certain kind, those with certain desirable properties. For example, there are sequences for which the limiting relative frequency of a given attribute does not exist; Reichenbach thus excludes such sequences. Von Mises (1957) gives us a more thoroughgoing restriction to what he calls collectives -- hypothetical infinite sequences of attributes (possible outcomes) of specified experiments that meet certain requirements. Call a place-selection an effectively specifiable method of selecting indices of members of the sequence, such that the selection or not of the index i depends at most on the first i − 1 attributes. The axioms are: Axiom of Convergence:_the limiting relative frequency of any attribute exists. _Axiom of Randomness: the limiting relative frequency of each attribute in a collective ω is the same in any infinite subsequence of ω which is determined by a place selection. The probability of an attribute A, relative to a collective ω, is then defined as the limiting relative frequency of A in ω. Note that a constant sequence such as H, H, H, …, in which the limiting relative frequency is the same in any infinite subsequence, trivially satisfies the axiom of randomness. This puts some strain on the terminology -- offhand, such sequences appear to be as non-random as they come -- although to be sure it is desirable that probabilities be assigned even in such sequences. Be that as it may, there is a parallel between the role of the axiom of randomness in von Mises' theory and the principle of maximum entropy in the classical theory: both attempt to capture a certain notion of disorder. Collectives are abstract mathematical objects that are not empirically instantiated, but that are nonetheless posited by von Mises to explain the stabilities of relative frequencies in the behavior of actual sequences of outcomes of a repeatable random experiment. Church (1940) renders precise the notion of a place selection as a recursive function. Nevertheless, the reference sequence problem remains: probabilities must always be relativized to a collective, and for a given attribute such as ‘heads’ there are infinitely many. Von Mises embraces this consequence, insisting that the notion of probability only makes sense relative to a collective. In particular, he regards single case probabilities as nonsense: “We can say nothing about the probability of death of an individual even if we know his condition of life and health in detail. The phrase ‘probability of death’, when it refers to a single person, has no meaning at all for us” (11). Some critics believe that rather than solving the problem of the single case, this merely ignores it. And note that von Mises understates the commitments of his theory: by his lights, the phrase ‘probability of death’ also has no meaning at all when it refers to a million people, or a billion, or any finite number. Let us see how the frequentist interpretations fare according to our criteria of adequacy. Finite relative frequencies of course satisfy finite additivity. In a finite reference class, only finitely many events can occur, so only finitely many events can have positive relative frequency. In that case, countable additivity is satisfied somewhat trivially: all but finitely many terms in the infinite sum will be 0. Limiting relative frequencies violate countable additivity (de Finetti 1972, §5.22). Indeed, the domain of definition of limiting relative frequency is not even a field, let alone a sigma field (de Finetti 1972, §5.8). So such relative frequencies do not provide an admissible interpretation of Kolmogorov's axioms. Finite frequentism has no trouble meeting the ascertainability criterion, as finite relative frequencies are in principle easily determined. The same cannot be said of limiting relative frequencies. On the contrary, any finite sequence of trials (which, after all, is all we ever see) puts literally no constraint on the limit of an infinite sequence; still less does an actual finite sequence put any constraint on the limit of an infinite hypothetical sequence, however fast and loose we play with the notion of ‘in principle’ in the ascertainability criterion. It might seem that the frequentist interpretations resoundingly meet the applicability to frequencies criterion. Finite frequentism meets it all too well, while limiting relative frequentism meets it in the wrong way. If anything, finite frequentism makes the connection between probabilities and frequencies too tight, as we have already observed. A fair coin that is tossed a million times is very unlikely to land heads exactly half the time; one that is tossed a million and one times is even less likely to do so! Facts about finite relative frequencies should serve as evidence, but not conclusive evidence, for the relevant probability assignments. Limiting relative frequentism fails to connect probabilities with finite frequencies. It connects them with limiting relative frequencies, of course, but again too tightly: for even in infinite sequences, the two can come apart. (A fair coin could land heads forever, even if it is highly unlikely to do so.) To be sure, science has much interest in finite frequencies, and indeed working with them is much of the business of statistics. Whether it has any interest in highly idealized, hypothetical extensions of actual sequences, and relative frequencies therein, is another matter. The applicability to rational opinion goes much the same way: it is clear that such opinion is guided by finite frequency information, unclear that it is guided by information about limits of hypothetical frequencies. 3.4 Propensity Interpretations Like the frequency interpretations, propensity interpretations locate probability ‘in the world’ rather than in our heads or in logical abstractions. Probability is thought of as a physical propensity, or disposition, or tendency of a given type of physical situation to yield an outcome of a certain kind, or to yield a long run relative frequency of such an outcome. This view was motivated by the desire to make sense of single-case probability attributions such as ‘the probability that this radium atom decays in 1600 years is 1/2’. Indeed, Popper (1957) advances his propensity theory as an account of such quantum mechanical probabilities. Popper develops the theory further in (1959a). For him, a probability p of an outcome of a certain type is a propensity of a repeatable experiment to produce outcomes of that type with limiting relative frequency p. For instance, when we say that a coin has probability 1/2 of landing heads when tossed, we mean that we have a repeatable experimental set-up -- the tossing set-up -- that has a propensity to produce a sequence of outcomes in which the limiting relative frequency of heads is 1/2. With its heavy reliance on limiting relative frequency, this position risks collapsing into von Mises-style frequentism according to some critics. Giere (1973), on the other hand, explicitly allows single-case propensities, with no mention of frequencies: probability is just a propensity of a repeatable experimental set-up to produce sequences of outcomes. This, however, creates the opposite problem to Popper's: how, then, do we get the desired connection between probabilities and frequencies? It is thus useful to follow Gillies (2000) in distinguishing long-run propensity theories and single-case propensity theories: A long-run propensity theory is one in which propensities are associated with repeatable conditions, and are regarded as propensities to produce in a long series of repetitions of these conditions frequencies which are approximately equal to the probabilities. A single-case propensity theory is one in which propensities are regarded as propensities to produce a particular result on a specific occasion (822). Hacking (1965) and Gillies offer long-run (though not infinitely long-run) propensity theories; Fetzer (1982, 1983) and Miller (1994) offer single-case propensity theories. Note that ‘propensities’ are categorically different things depending on which sort of theory we are considering. According to the long-run theories, propensities are tendencies to produce relative frequencies with particular values, but the propensities are not the probability values themselves; according to the single-case theories, the propensities are the probability values. According to Popper, for example, a fair die has a propensity -- an extremely strong tendency -- to land ‘3’ with long-run relative frequency 1/6. The small value of 1/6 does not measure this tendency. According to Giere, on the other hand, the die has a weak tendency to land ‘3’. The value of 1/6 does measure this tendency. It seems that those theories that tie propensities to frequencies do not provide an admissible interpretation of the (full) probability calculus, for the same reasons that relative frequencies do not. It is prima facie unclear whether single-case propensity theories obey the probability calculus or not. To be sure, one can stipulate that they do so, perhaps using that stipulation as part of the implicit definition of propensities. Still, it remains to be shown that there really are such things -- stipulating what a witch is does not suffice to show that witches exist. Indeed, to claim, as Popper does, that an experimental arrangement has a tendency to produce a given limiting relative frequency of a particular outcome, presupposes a kind of stability or uniformity in the workings of that arrangement (for the limit would not exist in a suitably unstable arrangement). But this is the sort of ‘uniformity of nature’ presupposition that Hume argued could not be known either a priori, or empirically. Now, appeals can be made to limit theorems -- so called ‘laws of large numbers’ -- whose content is roughly that under suitable conditions, such limiting relative frequencies almost certainly exist, and equal the single case propensities. Still, these theorems make assumptions (e.g., that the trials are independent and identically distributed) whose truth again cannot be known, and must merely be postulated. Part of the problem here, say critics, is that we do not know enough about what propensities are to adjudicate these issues. There is some property of this coin tossing arrangement such that this coin would land heads with a certain long-run frequency, say. But as Hitchcock (2002) points out, “calling this property a ‘propensity’ of a certain strength does little to indicate just what this property is.” Said another way, propensity accounts are accused of giving empty accounts of probability, à la Molière's ‘dormative virtue’ (Sober 2000, 64). Similarly, Gillies objects to single-case propensities on the grounds that statements about them are untestable, and that they are “metaphysical rather than scientific” (825). Some might level the same charge even against long-run propensities, which are supposedly distinct from the testable relative frequencies. This suggests that the propensity account has difficulty meeting the applicability to science criterion. Some propensity theorists (e.g., Giere) liken propensities to physical magnitudes such as electrical charge that are the province of science. But Hitchcock observes that the analogy is misleading. We can only determine the general properties of charge -- that it comes in two varieties, that like charges repel, and so on -- by empirical investigation. What investigation, however, could tell us whether or not propensities are non-negative, normalized and additive? More promising, perhaps, is the idea that propensities are to play certain theoretical roles, and that these place constraints on the way they must behave, and hence what they could be (in the style of the Ramsey/Lewis/‘Canberra plan’ approach to theoretical terms -- see Lewis 1970 or Jackson 2000). The trouble here is that these roles may pull in opposite directions, overconstraining the problem. The first role, according to some, constrains them to obey the probability calculus (with finite additivity); the second role, according to others, constrains them to violate it. On the one hand, propensities are said to constrain the degrees of belief, or credences, of a rational agent. We will have more to say in the next section about what credences are and what makes them rational, but for now recall the ‘applicability to rational belief’ criterion: an interpretation should clarify the role that probabilities play in constraining the credences of rational agents. One such putative role for propensities is codified by Lewis' ‘Principal Principle’ (1980). Roughly, the principle is that rational credences strive to ‘track’ propensities -- sometimes called “chances” -- so that if a rational agent knows the propensity of a given outcome, her degree of belief will be the same. More generally, where ‘P’ is the subjective probability function of a rational agent, and ‘ch’ is the propensity (chance) function, P(A\| ch(A) = x) = x, for all A_and for all _x_such that _P(ch(A) = x) >0. For example, my degree of belief that this coin toss lands heads, given that its propensity of landing heads is 3/4, is 3/4. The Principal Principle underpins an argument (Lewis 1980) that whatever they are, propensities must obey the usual probability calculus (with finite additivity). After all, it is argued, rational credences, which are guided by them, do -- see the next section. On the other hand, Humphreys (1985) gives an influential argument that propensities do not obey the probability calculus. The idea is that the probability calculus implies Bayes' theorem, which allows us to ‘invert’ a conditional probability: P(A \| B)=P(B \| A).P(A) P(B) Yet propensities seem to be measures of ‘causal tendencies’, and much as the causal relation is asymmetric, so these propensities supposedly do not invert. Suppose that we have a test for an illness that occasionally gives false positives and false negatives. A given sick patient may have a (non-trivial) propensity to give a positive test result, but it apparently makes no sense to say that a given positive test result has a (non-trivial) propensity to have come from a sick patient. ‘Humphreys' paradox’, as it is known, has prompted Fetzer and Nute (in Fetzer 1981) to offer a “probabilistic causal calculus” which looks quite different from Kolmogorov's calculus. Thus, we have an argument that whatever they are, propensities must not obey the usual probability calculus. Perhaps all this shows that the notion of ‘propensity’ bifurcates: on the one hand, there are propensities that bear an intimate connection to relative frequencies and rational credences, and that obey the probability calculus (with finite additivity); on the other hand, there are causal propensities that behave rather differently. In that case, there would be still more interpretations of probability than have previously been recognized. 3.5 Subjective probability 3.5.1 Probability as degree of belief We may characterize subjectivism (also known as personalism and subjective Bayesianism) with the slogan: ‘Probability is degree of belief’. We identify probabilities with degrees of confidence, or credences, or “partial” beliefs of suitable agents. Thus, we really have many interpretations of probability here, as many as there are doxastic states of suitable agents: we have Aaron's degrees of belief, Abel's degrees of belief, Abigail's degrees of belief,…, or better still, Aaron's degrees of belief-at-time-t 1, Aaron's degrees of belief-at-time-t 2, Abel's degrees of belief-at-time-t 1,…. Of course, we must ask what makes an agent ‘suitable’. What we might call unconstrained subjectivism places no constraints on the agents -- anyone goes, and hence anything goes. Various studies by psychologists (see, e.g., several articles in Kahneman et al. 1982) show that people commonly violate the usual probability calculus in spectacular ways. We clearly do not have here an admissible interpretation (with respect to any probability calculus), since there is no limit to what agents might assign. Unconstrained subjectivism is not a serious proposal. More interesting, however, is the claim that the suitable agents must be, in a strong sense, rational. Beginning with Ramsey (1926), various subjectivists have wanted to assimilate probability to logic by portraying probability as the logic of partial belief. A rational agent is required to be logically consistent, now taken in a broad sense. These subjectivists argue that this implies that the agent obeys the axioms of probability (although perhaps with only finite additivity), and that subjectivism is thus (to this extent) admissible. Before we can present this argument, we must say more about what degrees of belief are. 3.5.2 The betting interpretation and the Dutch Book argument Subjective probabilities are traditionally analyzed in terms of betting behavior. Here is a classic statement by de Finetti (1980): Let us suppose that an individual is obliged to evaluate the rate p_at which he would be ready to exchange the possession of an arbitrary sum _S(positive or negative) dependent on the occurrence of a given event E, for the possession of the sum pS; we will say by definition that this number p_is the measure of the degree of probability attributed by the individual considered to the event _E, or, more simply, that p_is the probability of _E(according to the individual considered; this specification can be implicit if there is no ambiguity). (62) This boils down to the following analysis: Your degree of belief in E_is _p_iff _p_units of utility is the price at which you would buy or sell a bet that pays 1 unit of utility if _E, 0 if not E. The analysis presupposes that, for any E, there is exactly one such price -- let's call this the agent's fair price for the bet on E. This presupposition may fail. There may be no such price -- you may refuse to bet on E at all (perhaps unless coerced, in which case your genuine opinion about E may not be revealed), or your selling price may differ from your buying price, as may occur if your probability for E is vague. There may be more than one fair price -- you may find a range of such prices acceptable, as may also occur if your probability for E is vague. For now, however, let us waive these concerns, and turn to an argument that uses the betting interpretation purportedly to show that rational degrees of belief must conform to the probability calculus (with at least finite additivity). A Dutch book (against an agent) is a series of bets, each acceptable to the agent, but which collectively guarantee her loss, however the world turns out. Ramsey notes, and it can be easily proven (e.g., Skyrms 1984), that if your subjective probabilities violate the probability calculus, then you are susceptible to a Dutch book. For example, suppose that you violate the additivity axiom by assigning P(A ∪ B) <P(A) + P(B), where A and B are mutually exclusive. Then a cunning bettor could buy from you a bet on A ∪ B for P(A ∪ B) units, and sell you bets on A and B individually for P(A) and P(B) units respectively. He pockets an initial profit of P(A) + P(B) − P(A ∪ B), and retains it whatever happens. Ramsey offers the following influential gloss: “If anyone's mental condition violated these laws [of the probability calculus], his choice would depend on the precise form in which the options were offered him, which would be absurd.” (1980, 41) Equally important, and often neglected, is the converse theorem that establishes how you can avoid such a predicament. If your subjective probabilities conform to the probability calculus, then no Dutch book can be made against you (Kemeny 1955); your probability assignments are then said to be coherent. In a nutshell, conformity to the probability calculus is necessary and sufficient for coherence. But let us return to the betting analysis of credences. It is an attempt to make good on Ramsey's idea that probability “is a measurement of belief qua basis of action” (34). While he regards the method of measuring an agent's credences by her betting behavior as “fundamentally sound” (34), he recognizes that it has its limitations. The betting analysis gives an operational definition of subjective probability, and indeed it inherits some of the difficulties of operationalism in general, and of behaviorism in particular. For example, you may have reason to misrepresent your true opinion, or to feign having opinions that in fact you lack, by making the relevant bets (perhaps to exploit an incoherence in someone else's betting prices). Moreover, as Ramsey points out, placing the very bet may alter your state of opinion. Trivially, it does so regarding matters involving the bet itself (e.g., you suddenly increase your probability that you have just placed a bet). Less trivially, placing the bet may change the world, and hence your opinions, in other ways (betting at high stakes on the proposition ‘I will sleep well tonight’ may suddenly turn you into an insomniac). And then the bet may concern an event such that, were it to occur, you would no longer value the pay-off the same way. (During the August 11, 1999 solar eclipse in the UK, a man placed a bet that would have paid a million pounds if the world came to an end.) These problems stem largely from taking literally the notion of entering into a bet on E, with its corresponding payoffs. The problems may be avoided by identifying your degree of belief in a proposition with the betting price you regard as fair, whether or not you enter into such a bet; it corresponds to the betting odds that you believe confer no advantage or disadvantage to either side of the bet (Howson and Urbach 1993). There is something of the Rawlsian ‘veil of ignorance’ reasoning here: imagine that you are to set the price for the bet, but you do not yet know which side of the bet you are to take. At your fair price, you should be indifferent between taking either side. De Finetti speaks of “an arbitrary sum” as the prize of the bet on E. The sum had better be potentially infinitely divisible, or else probability measurements will be precise only up to the level of ‘grain’ of the potential prizes. For example, a sum that can be divided into only 100 parts will leave probability measurements imprecise beyond the second decimal place, conflating probabilities that should be distinguished (e.g., those of a logical contradiction and of ‘a fair coin lands heads 8 times in a row’). More significantly, if utility is not a linear function of such sums, then the size of the prize will make a difference to the putative probability: winning a dollar means more to a pauper more than it does to Bill Gates, and this may be reflected in their betting behaviors in ways that have nothing to do with their genuine probability assignments. De Finetti responds to this problem by suggesting that the prizes be kept small; that, however, only creates the opposite problem that agents may be reluctant to bother about trifles, as Ramsey points out. Better, then, to let the prizes be measured in utilities: after all, utility is infinitely divisible, and utility is a linear function of utility. 3.5.3 Probabilities and utilities Utilities (desirabilities) of outcomes, their probabilities, and rational preferences are all intimately linked. The Port Royal Logic (Arnauld, 1662) showed how utilities and probabilities together determine rational preferences; de Finetti's betting interpretation derives probabilities from utilities and rational preferences; von Neumann and Morgenstern (1944) derive utilities from probabilities and rational preferences. And most remarkably, Ramsey (1926) (and later, Savage 1954 and Jeffrey 1966) derives both probabilities and utilities from rational preferences alone. First, he defines a proposition to be ethically neutral -- relative to an agent and an outcome -- if the agent is indifferent between having that outcome when the proposition is true and when it is false. The idea is that the agent doesn't care about the ethically neutral proposition as such -- it is a means to a end that he might care about, but it has no intrinsic value. Now, there is a simple test for determining whether, for a given agent, an ethically neutral proposition N has probability 1/2. Suppose that the agent prefers A to B. Then N has probability 1/2 iff the agent is indifferent between the gambles: A_if _N, B_if not _B if N, A if not. Ramsey assumes that it does not matter what the candidates for A and B are. We may assign arbitrarily to A and B any two real numbers u(A) and u(B) such that u(A) >u(B), thought of as the desirabilities of A and B respectively. Having done this for the one arbitrarily chosen pair A and B, the utilities of all other propositions are determined. Given various assumptions about the richness of the preference space, and certain ‘consistency assumptions’, he can define a real-valued utility function of the outcomes A, B, etc -- in fact, various such functions will represent the agent's preferences. He is then able to define equality of differences in utility for any outcomes over which the agent has preferences. It turns out that ratios of utility-differences are invariant -- the same whichever representative utility function we choose. This fact allows Ramsey to define degrees of belief as ratios of such differences. For example, suppose the agent is indifferent between A, and the gamble “B if X, C otherwise.” Then it follows from considerations of expected utility that her degree of belief in X, P(X), is given by: P(X)=u(A) - u(C) u(B) - u(C) Ramsey shows that degrees of belief so derived obey the probability calculus (with finite additivity). He calls what results “the logic of partial belief,” and indeed he opens his essay with the words “In this essay the Theory of Probability is taken as a branch of logic….” Ramsey avoids some of the objections to the betting interpretation, but not all of them. Notably, the essential appeal to gambles again raises the concern that the wrong quantities are being measured. And his account has new difficulties. It is unclear what facts about agents fix their preference rankings. It is also dubious that consistency requires one to have a set of preferences as rich as Ramsey requires, or that one can find ethically neutral propositions of probability 1/2. This in turn casts some doubt on Ramsey's claim to assimilate probability theory to logic. Savage (1954) likewise derives probabilities and utilities from preferences among options that are constrained by certain putative 'consistency' principles. For a given set of such preferences, he generates a class of utility functions, each a positive linear transformation of the other (i.e. of the form U 1 = aU 2 + b, where a> 0), and a unique probability function. Together these are said to ‘represent’ the agent's preferences. Jeffrey (1966) refines the method further. The result is theory of decision according to which rational choice maximizes ‘expected utility’, a certain probability-weighted average of utilities. Some of the difficulties with the behavioristic betting analysis of degrees of belief can now be resolved by moving to an analysis of degrees of belief that is functionalist in spirit. According to Lewis (1986a, 1994a), an agent's degrees of belief are represented by the probability function belonging to a utility function/probability function pair that best rationalizes her behavioral dispositions, rationality being given a decision-theoretic analysis. There is a deep issue that underlies all of these accounts of subjective probability. They all presuppose the existence of necessary connections between desire-like states and belief-like states, rendered explicit in the connections between preferences and probabilities. In response, one might insist that such connections are at best contingent, and indeed can be imagined to be absent. Think of an idealized Zen Buddhist monk, devoid of any preferences, who dispassionately surveys the world before him, forming beliefs but no desires. It could be replied that such an agent is not so easily imagined after all -- even if the monk does not value worldly goods, he will still prefer some things to others (e.g., truth to falsehood). Once desires enter the picture, they may also have unwanted consequences. For example, how does one separate an agent's enjoyment or disdain for gambling from the value she places on the gamble itself? Ironically, a remark that Ramsey makes in his critique of the betting interpretation seems apposite here: “The difficulty is like that of separating two different co-operating forces” (1980, 35). The betting interpretation makes subjective probabilities ascertainable to the extent that an agent's betting dispositions are ascertainable. The derivation of them from preferences makes them ascertainable to the extent that his or her preferences are known. However, it is unclear that an agent's full set of preferences is ascertainable even to himself or herself. Here a lot of weight may need to be placed on the ‘in principle’ qualification in the ascertainability criterion. The expected utility representation makes it virtually analytic that an agent should be guided by probabilities -- after all, the probabilities are her own, and they are fed into the formula for expected utility in order to determine what it is rational for her to do. 4.5.4 Orthodox Bayesianism, and further constraints on rational credences But do they function as a good guide? Here it is useful to distinguish different versions of subjectivism. Orthodox Bayesians in the style of de Finetti recognize no rational constraints on subjective probabilities beyond: conformity to the probability calculus, and a rule for updating probabilities in the face of new evidence, known as conditioning. An agent with probability function P 1, who becomes certain of a piece of evidence E, should shift to a new probability function P 2 related to P 1 by: (Conditioning) P_2(_X) = P_1(_X\| E) (provided P_1(_E) >0). This is a permissive epistemology, licensing doxastic states that we would normally call crazy. Thus, you could assign probability 1 to this sentence ruling the universe, while upholding such extreme subjectivism -- provided, of course, that you assign probability 0 to this sentence not ruling the universe, and that your other probability assignments all conform to the probability calculus. Some otherwise extreme subjectivists impose the further rationality requirement of regularity: only a priori falsehoods get assigned probability 0. This is sometimes also called ‘strict coherence’, and it is advocated by authors such as Kemeny (1955), Jeffreys (1961), Edwards et al. (1963), Shimony (1970), and Stalnaker (1970). It is meant to capture a form of open-mindedness and responsiveness to evidence. But then, perhaps unintuitively, someone who assigns probability 0.999 to this sentence ruling the universe can be judged rational, while someone who assigns it probability 0 is judged irrational. Note also that the requirement of regularity seems to afford a new argument for the non-existence of God as traditionally conceived: an omniscient agent, who gives probability 1 to all truths, would be convicted of irrationality. Thus regularity seems to require ignorance, or false modesty. See, e.g., Levi (1978) for further opposition to regularity. Probabilistic coherence plays much the same role for degrees of belief that consistency plays for ordinary, all-or-nothing beliefs. What an extreme subjectivist, even one who demands regularity, lacks is an analogue of truth, some yardstick for distinguishing the ‘veridical’ probability assignments from the rest (such as the 0.999 one above), some way in which probability assignments are answerable to the world. It seems, then, that the subjectivist needs something more. And various subjectivists offer more. Having isolated the “logic” of partial belief as conformity to the probability calculus, Ramsey goes on to discuss what makes a degree of belief in a proposition reasonable. After canvassing several possible answers, he settles upon one that focuses on habits of opinion formation -- “e.g. the habit of proceeding from the opinion that a toadstool is yellow to the opinion that it is unwholesome” (50). He then asks, for a person with this habit, what probability it would be best for him to have that a given yellow toadstool is unwholesome, and he answers that “it will in general be equal to the proportion of yellow toadstools which are in fact unwholesome” (50). This resonates with more recent proposals (e.g., van Fraassen 1984, Shimony 1988) for evaluating degrees of belief according to how closely they match the corresponding relative frequencies -- in the jargon, how well calibrated they are. Since relative frequencies obey the axioms of probability (up to finite additivity), it is thought that rational credences, which strive to track them, should do so also. However, rational credences may strive to track various things. For example, we are often guided by the opinions of experts. We consult our doctors on medical matters, our weather forecasters on meteorological matters, and so on. Gaifman (1988) coins the terms “expert assignment” and “expert probability” for a probability assignment that a given agent strives to track: “The mere knowledge of the [expert] assignment will make the agent adopt it as his subjective probability” (193). This idea may be codified as follows: (Expert) P(A\| pr(A) = x) = x, for all x_such that _P(pr(A) = x) >0 where ‘P’ is the agent's subjective probability function, and ‘pr(A)’ is the assignment that the agent regards as expert. For example, if you regard the local weather forecaster as an expert on your local weather, and she assigns probability 0.1 to it raining tomorrow, then you may well follow suit: P(rain\| pr(rain) = 0.1) = 0.1 More generally, we might speak of an entire probability function as being such a guide for an agent over a specified set of propositions. Van Fraassen (1989, 198) gives us this definition: “If P is my personal probability function, then q is an expert function for me concerning family F of propositions exactly if P(A \| q(A) = x) = x for all propositions A in family F.” Let us define a universal expert function for a given rational agent as one that would guide all of that agent's probability assignments in this way: an expert function for the agent concerning all propositions. Van Fraassen (1984, 1995a), following Goldstein (1983), argues that an agent's future probability functions are universal expert functions for that agent. He enshrines this idea in his Reflection Principle, where P t is the agent's probability function at time t, and P t+Δ is her function at a later time t+Δ: P__t(A\| P__t+Δ(A) = x) = x, for all A_and for all _x_such that _P__t(P__t+Δ(A) = x) >0. The principle encapsulates a certain demand for ‘diachronic coherence’ imposed by rationality. Van Fraassen defends it with a ‘diachronic’ Dutch Book argument (one that considers bets placed at different times), and by analogizing violations of it to the sort of pragmatic inconsistency that one finds in Moore's paradox. We may go still further. There may be universal expert functions for all rational agents. Let us call such a function a universal expert function, without any relativization to an agent. The Principle of Direct Probability regards the relative frequency function as a universal expert function; we have already seen the importance that proponents of calibration place on it. Let A be an event-type, and let relfreq(A) be the relative frequency of A (in some suitable reference class). Then for any rational agent with probability function P, we have P(A\| relfreq(A) = x) = x, for all A_and for all _x_such that _P(relfreq(A) = x) >0. (Cf. Hacking 1965.) Lewis, as we have seen, posits a similar universal expert role for the objective chance function, ch, in his Principal Principle: P(A\| ch(A) = x) = x, for all A_and for all _x_such that _P(ch(A) = x) >0. A frequentist who thinks that chances just are relative frequencies would presumably think that the Principal Principle just is the Principle of Direct Probability; but Lewis' principle may well appeal to those who have a very different view about chances -- e.g., propensity theorists. The argument that we saw in the previous section, using the Principal Principle to show that propensities (chances) must obey the probability calculus, can now be turned on its head: assuming that they do obey it, rational degrees of belief, which aim to track these propensities, must do so too. The ultimate expert, presumably, is the truth function -- the function that assigns 1 to all the true propositions and 0 to all the false ones. Knowledge of its values should surely trump knowledge of the values assigned by human experts (including one's future selves), frequencies, or chances. Note that for any putative expert q, P(A\| q(A) = x∩ A) = 1, for all A_and for all _x_such that _P(q(A) = x ∩ A) >0 -- the truth of A overrides anything the expert might say. So all of the proposed expert probabilities above should really be regarded as defeasible. Joyce (1998) portrays the rational agent as estimating truth values, seeking to minimize a measure of distance between them and her probability assignments. He argues that for any measure of distance that satisfies certain intuitive properties, any agent who violates the probability axioms could serve this epistemic goal better by obeying them instead, however the world turns out. There are some unifying themes in these approaches to subjective probability. An agent's degrees of belief determine her estimates of certain quantities: the values of bets, or the desirabilities of gambles more generally, or the probability assignments of various ‘experts’ -- humans, relative frequencies, objective chances, or truth values. The laws of probability then are claimed to be constraints on these estimates: putative necessary conditions for minimizing her ‘losses’ in a broad sense, be they monetary, or measured by distances from the assignments of these experts. 4. Conclusion: Future Prospects? It should be clear from the foregoing that there is still much work to be done regarding the interpretation of probability. Each interpretation that we have canvassed seems to capture some crucial insight into it, yet falls short of doing complete justice to it. Perhaps the full story about probability is something of a patchwork, with partially overlapping pieces. In that sense, the above interpretations might be regarded as complementary, although to be sure each may need some further refinement. My bet, for what it is worth, is that we will retain at least three distinct notions of probability: one quasi-logical, one objective, and one subjective. There are already signs of the rehabilitation of classical and logical probability, and in particular the principle of indifference and the principle of maximum entropy, by authors such as Stove (1986), Bartha and Johns (2001), Festa (1993), Paris and Vencovská (1997), and Maher (2000, 2001). Relevant here may also be advances in information theory and complexity theory (see Fine 1973, Li and Vitanyi 1997). These theories have already proved to be fruitful in the study of randomness (Kolmogorov 1965, Martin-Löf 1966), which obviously is intimately related to the notion of probability. Refinements of our understanding of randomness, in turn, should have a bearing on the frequency interpretations (recall von Mises' appeal to randomness in his definition of ‘collective’), and on propensity accounts (especially those that make explicit ties to frequencies). Given the apparent connection between propensities and causation adumbrated in Section 3, powerful causal modeling techniques by authors such as Spirtes, Glymour and Scheines (1993) and Pearl (2000), and recent work on causation more generally (e.g., Hall 2003, Woodward forthcoming) should also prove fruitful here. An outgrowth of frequentism is Lewis' (1986b, 1994b) account of chance. It runs roughly as follows. The laws of nature are those regularities that are theorems of the best theory: the true theory of the universe that best balances simplicity, strength, and likelihood (that is, the probability of the actual course of history, given the theory). If any of the laws are probabilistic, then the chances are whatever these laws say they are. Now, it is somewhat unclear exactly what ‘simplicity’ and ‘strength’ consist in, and exactly how they are to be balanced. Perhaps insights from statistics and computer science may be helpful here: approaches to statistical model selection, and in particular the ‘curve-fitting’ problem, that attempt to characterize simplicity, and its trade-off with strength -- e.g., the Akaike Information Criterion (see Forster and Sober 1994), the Bayesian Information Criterion (see Kieseppä 2001), Minimum Description Length theory (see Rissanen 1999) and Minimum Message Length theory (see Wallace and Dowe 1999). State-of-the-art contributions to the subjectivist theory of probability include Schervish, Seidenfeld and Kadane's (2000) research on degrees of incoherence (measuring the extent of departures from obedience to the probability calculus) and on the aggregation of the opinions of multiple agents (Seidenfeld et al. 1989; see also Hild forthcoming). These promise to be fertile areas of future research. We may expect that further criteria of adequacy for subjective probabilities will be developed -- perhaps refinements of ‘scoring rules’ (Winkler 1996), and more generally, candidates for playing a role for subjective probability analogous to the role that truth plays for belief. Here we may come full circle. For belief is answerable both to logic and to objective facts. A refined account of degrees-of-belief may be answerable both to a refined quasi-logical and a refined objective notion of probability. Well may we say that probability is a guide to life; but the task of understanding exactly how and why it is has still to be completed, and will surely prove to be a guide to future theorizing about it. Suggested Further Reading Kyburg (1970) contains a vast bibliography of the literature on probability and induction pre-1970. Also useful for references before 1967 is the bibliography for “Probability” in the Macmillan Encyclopedia of Philosophy. Earman (1992) and Howson and Urbach (1993) have more recent bibliographies, and give detailed presentations of the Bayesian program. Skyrms (2000) is an excellent introduction to the philosophy of probability. Von Plato (1994) is more technically demanding and more historically oriented, with another extensive bibliography that has references to many landmarks in the development of probability theory in the last century. Fine (1973) is still a highly sophisticated survey of and contribution to various foundational issues in probability, with an emphasis on interpretations. Billingsley (1995) and Feller (1968) are classic textbooks on the mathematical theory of probability. Bibliography Arnauld, A. Logic, or, The Art of Thinking ("The Port Royal Logic"), 1662, tr. J. Dickoff and P. James, Indianapolis: Bobbs-Merrill, 1964 Bartha, P. and Johns, R., 2001, “Probability and Symmetry”, Philosophy of Science 68 (Proceedings), S109-S122 Billingsley, P., 1995, Probability and Measure, 3rd ed., New York: John Wiley & Sons Carnap, R., 1950, Logical Foundations of Probability, Chicago: University of Chicago Press -----, 1952, The Continuum of Inductive Methods, Chicago: University of Chicago Press -----, 1963, “Replies and Systematic Expositions” in The Philosophy of Rudolf Carnap, P. A. Schilpp, (ed.), Open Court, Illinois: La Salle Church, A., 1940, “On the Concept of a Random Sequence”, Bulletin of the American Mathematical Society 46: 130-135 De Finetti, B., 1937, “La Prévision: Ses Lois Logiques, Ses Sources Subjectives”, Annales de l”Institut Henri Poincaré, 7: 1-68; translated as “Foresight. Its Logical Laws, Its Subjective Sources”, in Studies in Subjective Probability, H. E. Kyburg, Jr. and H. E. Smokler (eds.), Robert E. Krieger Publishing Company, 1980 -----, 1972, Probability, Induction and Statistics, New York: Wiley -----, 1990, (originally published 1974), Theory of Probability, Vol. 1, Wiley Classics Library, John Wiley & Sons Earman, J., 1992, Bayes or Bust, Cambridge: MIT Press Edwards, W., Lindman, H., and Savage, L. 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B., 2000, “How sets of coherent probabilities may serve as models for degrees of incoherence”, Journal of Uncertainty, Fuzziness, and Knowledge-based Systems 8, No. 3 (June), 347-356 Scott D., and Krauss P.,1966, “Assigning Probabilities to Logical Formulas”, in Aspects of Inductive Logic, J. Hintikka and P. Suppes, (eds.), Amsterdam: North-Holland Seidenfeld, T., Kadane, J. and Schervish, M. 1989, “On the Shared Preferences of Two Bayesian Decision Makers”, Journal of Philosophy, 86: 225-244 Shimony, A., 1970, “Scientific Inference”, in The Nature and Function of Scientific Theories, R. Colodny (ed.), Pittsburgh: University of Pittsburgh Press -----, 1988, “An Adamite Derivation of the Calculus of Probability”, Probability and Causality, in J.H. Fetzer (ed.), Dordrecht: D. Reidel Skyrms, B., 1980, Causal Necessity, New Haven: Yale University Press -----, 1984, Pragmatics and Empiricism, New Haven: Yale University Press -----, 2000, Choice and Chance, 4 th ed, Wadsworth, Inc. Sober, E., 2000, Philosophy of Biology, 2 nd ed, Westview Press Spirtes, P., Glymour, C. and Scheines, R., 1993, Causation, Prediction, and Search, New York: Springer-Verlag Spohn, W., 1986, “The Representation of Popper Measures”, Topoi 5 Stalnaker, R., 1970, “Probabilities and Conditionals”, Philosophy of Science 37: 64-80 Stove, D. C., 1986, The Rationality of Induction, Oxford: Oxford University Press van Fraassen, B., 1984, “Belief and the Will”, Journal of Philosophy 81: 235-256 -----, 1989, Laws and Symmetry, Oxford: Clarendon Press -----, 1995a, “Belief and the Problem of Ulysses and the Sirens”, Philosophical Studies 77: 7-37 -----, 1995b, “Fine-grained Opinion, Conditional Probability, and the Logic of Belief”, Journal of Philosophical Logic 24: 349-377 Venn, J., 1876, The Logic of Chance, 2 nd ed., Macmillan and co; reprinted, New York, 1962. von Mises R., 1957, Probability, Statistics and Truth, revised English edition, New York: Macmillan von Neumann, J. and Morgenstern, O., 1944, Theory of Games and Economic Behavior, Princeton: Princeton University Press; New York: John Wiley and Sons, 1964. von Plato J., 1994, Creating Modern Probability, Cambridge: Cambridge University Press Wallace, C. S. and Dowe, D. L., 1999, “Minimum Message Length and Kolmogorov Complexity”, Computer Journal (special issue on Kolmogorov complexity), 42 (4) 270-283 Winkler, R. L., 1996, “Scoring Rules and the Evaluation of Probabilities”, Test, 5 (1) 1-60 Woodward, J., forthcoming, A Theory of Explanation: Causation, Invariance and Intervention, Oxford: Oxford University Press Other Internet Resources "Probability", short entry by Michael Cohen (U. Wales) in The Oxford Companion to Philosophy, hosted at xrefer.com. "Difficulties of the standard interpretations of probability", by Lazlo E. Szabo (History and Philosophy of Science, Eötvös University, Budapest) "Probability, (in PDF), lectures by Paul Bartha (Philosophy, University of British Columbia). "Interpretations of Probability, (in PDF), Chapter 4 of a dissertation by Janneke Hendriek van Lith (University of Utrecht). Related Entries Carnap, Rudolf \| causation: probabilistic \| epistemology: Bayesian \| LaPlace, Pierre Simon \| Popper, Karl \| Ramsey, Frank \| Reichenbach, Hans Acknowledgements I thank Branden Fitelson, Matthias Hild, Christopher Hitchcock, Ralph Miles, Teddy Seidenfeld, Elliott Sober, and Jim Woodward for their many helpful comments, and especially Jim Joyce, who gave me very detailed and incisive feedback. Copyright © 2003 Alan Hájek ahajek@hss.caltech.edu A \| B \| C \| D \| E \| F \| G \| H \| I \| J \| K \| L \| M \| N \| O \| P \| Q \| R \| S \| T \| U \| V \| W \| X \| Y \| Z Stanford Encyclopedia of Philosophy
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1990	https://www.math.lsu.edu/~bogdan/7410/handout.pdf	Math 7410 Graph Theory Bogdan Oporowski Department of Mathematics Louisiana State University April 14, 2021 Deﬁnition of a graph Deﬁnition 1.1 A graph G is a triple (V, E, I) where ◮V (or V (G)) is a ﬁnite set whose elements are called vertices; ◮E (or E(G)) is a ﬁnite set disjoint from V whose elements are called edges; and ◮I, called the incidence relation, is a subset of V × E in which each edge is in relation with exactly one or two vertices. Example 1.2 ◮V = {v1, v2, v3, v4} ◮E = {e1, e2, e3, e4, e5, e6, e7} ◮I = {(v1, e1), (v1, e4), (v1, e5), (v1, e6), (v2, e1), (v2, e2), (v3, e2), (v3, e3), (v3, e5), (v3, e6), (v4, e3), (v4, e4), (v4, e7)} v3 v2 v4 v1 e1 e2 e3 e4 e5 e6 e7 Simple graphs Deﬁnition 1.3 ◮Edges incident with just one vertex are loops. ◮Edges incident with the same pair of vertices are parallel. ◮Graphs with no parallel edges and no loops are called simple. v3 v2 v4 v1 e1 e2 e3 e4 e5 e6 e7 v3 v2 v4 v1 e1 e2 e3 e4 e5 e6 e7 Edges of a simple graph can be described as two-element subsets of the vertex set. Example 1.4 E = {{v1, v2}, {v2, v3}, {v3, v4}, {v1, v4}, {v1, v3}}. Note 1.5 Graph Terminology Deﬁnition 1.6 ◮The graph G is empty if V = ∅, and is trivial if E = ∅. ◮The cardinality of the vertex-set of a graph G is called the order of G and denoted \|G\|. ◮The cardinality of the edge-set of a graph G is called the size of G and denoted ∥G∥. ◮Two vertices incident with the same edge are adjacent or neighbors. ◮Similarly, two edges incident with the same vertex are adjacent. ◮The number of edges incident with a vertex v of G, with loops counted twice, is the degree of v and is written as d(v) or dG(v). ◮The set of neighbors of a vertex v of G, other than v itself, is denoted by N(v) or by NG(v). ◮Similarly, if U is a subset of the vertex set of G, then N(U) is the set of those vertices that are not in U, but are adjacent to a vertex in U. Isomorphism Deﬁnition 1.7 The graphs G1 = (V1, E1, I1) and G2 = (V2, E2, I2) are isomorphic, written G1 ∼ = G2, if there are bijections ϕ : V1 →V2 and ψ : E1 →E2 such that (v, e) ∈I1 if and only if (ϕ(v), ψ(e)) ∈I2. Such a pair of bijections is an isomorphism. Note 1.8 ◮If G1 and G2 are simple, then an isomorphism may be deﬁned as a bijection ϕ : V1 →V2 such that u and v are adjacent in G1 if and only if ϕ(u) and ϕ(v) are adjacent in G2. ◮Isomorphic graphs are usually considered “the same”. Theorem 1.9 (Babai, 2015–2016) Graph isomorphism problem can be solved in quasi-polynomial time. There is a constant c and an algorithm that can decide whether two graphs on n vertices are isomorphic or not in at most 2O((log n)c) steps. Isomorphism Example Example 1.10 Which of the following graphs are isomorphic? 1 2 3 4 5 6 1 3 5 2 4 6 G1 G2 G3 G ∼G ≇G Automorphism Deﬁnition 1.11 An automorphism of a graph is an isomorphism from the graph to itself. Note 1.12 ◮The automorphisms of a graph form a group. ◮Computer software for ﬁnding automorphism groups of graphs is a part of the Sage system, available at Theorem 1.13 (Frucht, 1938) For every ﬁnite group X there is a graph whose automorphism group is X. Problem 1 For every positive integer n, construct a simple graph with exactly n automorphisms. Subgraphs Deﬁnition 1.14 A graph G1 = (V1, E1, I1) is a subgraph of a graph G2 = (V2, E2, I2), written G1 ⩽s G2, if ◮V1 ⊆V2, ◮E1 ⊆E2, and ◮I1 is induced by I2. Alternately, we may think of G1 as obtained from G2 by ◮Deleting vertices (denoted G −v or G −U), and ◮Deleting edges (denoted G \ e or G \ F). Deﬁnition 1.15 G1 is an induced subgraph of G2 if E1 consists of all those elements of E2 whose incident vertices lie in V1. Altrenately, we may think of G1 as obtained from G2 by deleting only vertices. Subgraph Example Example 1.16 G1 G2 G3 ◮G2 is a subgraph of G1, but it is not an induced subgraph. ◮G3 is an induced subgraph of G1. Reconstruction Conjectures The deck of a graph G is the collection of graphs G −v over all v ∈V (G). A graph is reconstructible if no other graph (up to isomorphism) has the same deck. Conjecture 1.17 (Reconstruction Conjecture) Every simple graph on at least three vertices is reconstructible. The edge-deck of a graph G is the collection of graphs G \ e over all e ∈E(G). A graph is edge-reconstructible is no other graph has the same edge-deck. Conjecture 1.18 (Edge-Reconstruction Conjecture) Every simple graph on at least four edges is edge-reconstructible. Walks, Trails, Paths, and Cycles Deﬁnition 1.19 ◮A walk is a sequence v0, e1, v1, e2, v2, . . . , en, vn, where each edge ei is incident with vertices vi−1 and vi. ◮The length of a walk is the number of edges in it. ◮A walk is closed if its ﬁrst and last vertices coincide. ◮A trail is a walk in which no edge is repeated. ◮A path is a trail with no repeated vertices. ◮A cycle is a trail with no vertices repeated except that the ﬁrst vertex is the same as the last. ◮For a path or a cycle, we will often blur the distinction between the sequence of vertices and edges, and the graph it forms. ◮The graph that is a path on n vertices (which has length n −1) will be denoted as Pn. ◮The graph that is a cycle on n vertices (which has length n) will be denoted as Cn. ◮A graph is connected if each pair of its vertices can be connected by a walk (equivalently, a trail or a path). Complete Graphs and Complements Deﬁnition 1.20 ◮A complete graph on n vertices, denoted by Kn, is a simple graph in which every two of its n vertices are connected by an edge. ◮If G is a simple graph, then the complement of G, denoted by G, is the simple graph on the same vertex set as G, and in which two vertices are adjacent if and only if they are not adjacent in G. ◮A simple graph is self-complementary if it is isomorphic to its own complement. Problem 2 Suppose that G is a non-trivial simple graph such that both G and G are connected. Prove that G has P4 as an induced subgraph. Problem 3 (a) Show that the order of a self-complementary graph is congruent to 0 or 1 modulo 4. (b) Construct a self-complementary graph of order n for every positive integer n congruent to 0 or 1 modulo 4. Hand-Shaking Lemma Theorem 1.21 (Hand-Shaking Lemma) X v∈V (G) d(v) = 2∥G∥ Corollary 1.22 The number of vertices of odd degree is even. Trees Deﬁnition 2.1 ◮A graph having no cycles is acyclic or a forest. ◮A connected forest is a tree. ◮A leaf or a pendant vertex is a vertex of degree one. ◮A subgraph of G is spanning if it has all the vertices of G. ◮The distance between vertices u and v of G, written d(u, v) or dG(u, v), is the length of the shortest path in G that contains both u and v. (Such a path is called a uv-path and u and v are its ends.) If a uv-path does not exist, then d(u, v) = ∞. ◮The distance between sets U and W of vertices of G, written d(U, W ), is the length of a shortest uw-path where u ∈U and w ∈W , or inﬁnity if no such path exists. Induction for Trees Theorem 2.2 Every tree with at least two vertices has at least two leaves. Deleting a leaf from a tree of order n produces a tree of order n −1. Proof. In an acyclic graph, the ends of a maximal non-trivial path have degree one. Let v be a leaf of a tree T and let T ′ = T −v. Then T ′ is acyclic. Suppose u and w are vertices of T ′. Then, in T there is a uw-path P. But P cannot contain v as dT (v) = 1, and so it also lies in T ′. Characterization of Trees Theorem 2.3 For a simple graph G of order n the following are equivalent: (A) G is connected and acyclic; (B) G is connected and has size n −1; (C) G is acyclic and has size n −1; and (D) For every two vertices u and v, the graph G contains exactly one uv-path. A ⇒B. Induction on n. Trivial for n = 1. For the inductive step, let v be a leaf, which exists by 2.2, and consider G −v. By the induction hypothesis, G −v has size n −2, so G has size n −1. B ⇒C. Delete edges from G, one by one, until the graph has no cycles, and call the resulting connected and acyclic graph G′. Then G′ satisﬁes (A), and so also satisﬁes (B), and so has size n −1. This implies that G′ = G. The remainder of the proof is left as an exercise. Edge Exchange Theorem 2.4 If T and T ′ are two spanning trees of a connected graph G and e ∈E(T ) \ E(T ′), then there is an edge e′ ∈E(T ′) \ E(T ) such that T \ e ∪e′ is a spanning tree of G. Proof. Consider T \ e: it is disconnected with exactly two connected components (maximal connected subgraphs) S and S′. Since T ′ is connected, it must have an edge e′ with one endpoint in each S and S′. Clearly, T \ e ∪e′ is a spanning tree of G. Minimum Cost Spanning Tree Suppose G is a graph and c : E(G) →N is a cost function. The cost of a subgraph H of G is P e∈E(H) c(e). We want to ﬁnd a minimum-cost spanning tree T of G. Algorithm 2.5 (Kruskal) ◮Start with V (T ) = V (G) and E(T ) = ∅. ◮Order the edges of G so that their costs are non-decreasing. ◮Proceed with each edge of G, one by one, in the above order: if its joins two components of T , add it to T ; otherwise do nothing. Example 2.6 b b b b b b b 6 8 5 4 9 11 12 1 2 3 4 7 10 10 b b b b b b b Proof of Kruskal’s Theorem Theorem 2.7 (Kruskal) In a connected graph, Kruskal’s Algorithm produces a minimum-cost spanning tree. Proof. It is clear that the algorithm produces a spanning tree. Let T be the resulting graph, and suppose T ′ is a spanning tree of minimum cost. If T ′ = T , then there is nothing to prove. If T ̸= T ′, let e be the ﬁrst edge chosen for T that is not in T ′. Adding e to T ′ creates a cycle C, but since T does not have cycles, T ′ has an edge e′ / ∈E(T ). Consider the spanning tree T ′ \ e′ ∪e. Since T ′ contains e′ and all edges of T chosen before e, both e and e′ are available when the algorithm chooses e, and hence c(e) ≤c(e′). Thus T ′ \ e′ ∪e is a spanning tree with cost at most T ′ that agrees with T for a longer initial list of edges than T ′ does. Repeating this argument yields a minimum-cost spanning tree that equals T , proving that the costs of T and T ′ are the same. Enumerating Labeled Trees We would like to know how many diﬀerent (and here we really mean diﬀerent rather than non-isomorphic) trees with the vertex set {1, 2, . . . , n} are there? Theorem 2.8 (Cayley’s Formula due to Borchardt (1860)) There are nn−2 trees with vertex set {1, 2, . . . , n}. Proof. There are nn−2 sequences of length n −2 with entries from {1, 2, . . . , n}. We will establish a bijection between such sequences and trees on the vertex set {1, 2, . . . , n}. Pr¨ ufer Sequences To ﬁnd a Pr¨ ufer sequence f(T ) of a labeled tree T , ◮delete the leaf with the smallest label, and ◮append the label of its neighbor to the sequence until one edge remains. Example 2.9 b 3 b 4 b 5 b 2 b 6 b 9 b 10 b 8 b 1 b 7 b 11 Pr¨ ufer sequence: 6 , 2 , 2 , 6 , 1 , 8 , 8 , 1 , 7 Trees from Sequences Now we describe how to produce a tree from a Pr¨ ufer sequence. ◮Begin with a forest having n isolated vertices labeled 1, 2, . . . , n. ◮Proceed with all n −2 elements of the sequence, and, at the ith step, ◮let x be the label in position i. ◮let y be the smallest label that does not appear at the ith or later position and has not yet been marked as “ﬁnished”. ◮add the edge xy, and ◮mark y as ﬁnished. ◮Join the two remaining unﬁnished vertices with an edge. Example 2.10 Sequence: 6, 2, 2, 6, 1, 8, 8, 1, 7 Finished: Sequence: 6, 2, 2, 6, 1, 8, 8, 1, 7 Finished: 3 Sequence: 6, 2, 2, 6, 1, 8, 8, 1, 7 Finished: 3, 4 Sequence: 6, 2, 2, 6, 1, 8, 8, 1, 7 Finished: 3, 4, 5 Sequence: 6, 2, 2, 6, 1, 8, 8, 1, 7 Finished: 3, 4, 5, 2 Sequence: 6, 2, 2, 6, 1, 8, 8, 1, 7 Finished: 3, 4, 5, 2, 6 Sequence: 6, 2, 2, 6, 1, 8, 8, 1, 7 Finished: 3, 4, 5, 2, 6, 9 Sequence: 6, 2, 2, 6, 1, 8, 8, 1, 7 Finished: 3, 4, 5, 2, 6, 9, 10 Sequence: 6, 2, 2, 6, 1, 8, 8, 1, 7 Finished: 3, 4, 5, 2, 6, 9, 10, 8 Sequence: 6, 2, 2, 6, 1, 8, 8, 1, 7 Finished: 3, 4, 5, 2, 6, 9, 10, 8, 1 b 3 b 6 b 9 10 b 8 b 1 7 Proof of Cayley’s Formula Now we show that the two operations described previously are inverses of each other. First, we show that when we start with a sequence, we indeed produce a tree. Note that we start of the ith step with n −i + 1 unﬁnished vertices and n −i −1 remaining vertices in the sequence. Therefore y can be chosen as described, and the algorithm produces a graph of order n and size n −1. Each step joins two unﬁnished vertices and marks one of them as ﬁnished. Thus after i steps the graph has n −i components, each containing exactly one unﬁnished vertex. The ﬁnal step connects the graph thereby creating a tree. Now we need to show that the obtained tree is the same as the one that created the sequence. In each step of computing the sequence, we can mark the deleted leaf as “ﬁnished”. The labels that do not yet appear in the remainder of the sequence we generate are the unﬁnished vertices that are not leaves. Because the next leaf deleted is the least, the edge deleted in each stage of computing the sequence is precisely the edge added when constructing the graph. Therefore the correspondence between the sequences and labeled trees is a bijection. Counting Trees with Prescribed Degrees Corollary 2.11 The number of trees with vertex set {1, 2, . . . , n} in which vertices 1, 2, . . . , n have respective degrees d1, d2, . . . , dn is (n −2)! Q(di −1)!. Proof. When we delete vertex x from T when constructing the Pr¨ ufer sequence, all neighbors of x except for one have already been deleted. We record x in the sequence once for each deleted neighbor and x does not appear in the sequence again. Hence x appears in the sequence d(x) −1 times. Therefore we count the trees by counting sequences of length n −2 having di −1 copies of i, for each i. If we distinguish between various copies of i, then there are (n −2)! such sequences. Since we really cannot distinguish between the copies, we have over-counted by a factor of (di −1)! for each i. Minors Deﬁnition 2.12 ◮If e is an edge of G incident with two distinct vertices u and v, then the contraction of e is the operation of deleting e and identifying u and v. ◮Contracting a loop is the same as deleting it. ◮The graph obtained from G by contracting e is denoted G/e (extended to G/F if F ⊆E(G)). ◮A graph H is a minor of G if it can be obtained from G by a sequence of operation each of which is one of the following: ◮deleting an edge; ◮deleting an isolated vertex; and ◮contracting an edge. ◮We write H ⩽m G to indicate that H is isomorphic to a minor of G. Note 2.13 The order of operations of deleting and contracting to get a minor of a graph is irrelevant. Counting Spanning Trees Theorem 2.14 Let τ(G) denote the number of distinct spanning trees of a (labeled) graph G. If e is a non-loop edge of G, then τ(G) = τ(G \ e) + τ(G/e). Example 2.15 b b b b e G b b b b G \ e b b b G/e τ(G) = τ(G \ e) + τ(G/e) = 4 + 4 = 8 Proof of Spanning Tree Formula ◮The spanning trees of G \ e are precisely the spanning trees of G that avoid e. ◮The spanning trees of G/e correspond to the spanning trees of G using e. (If T is a spanning tree of G/e, then E(T ) ∪e form the edge-set of a spanning tree of G.) ◮The formula follows. Using the deletion-contraction formula for calculating the number of spanning trees is ineﬃcient. A much more eﬃcient method is to construct a special matrix, called the Laplacian of the graph, and to compute its determinant. Bipartite Graphs Deﬁnition 3.1 A graph G is bipartite if the vertex set of G can be partitioned into sets X and Y such that every edge of G joins a vertex in X to a vertex in Y . The simple bipartite graph in which each of the m vertices of X is joined to each of the n vertices of Y is called complete bipartite and denoted Km,n. Example 3.2 X Y K3,3 b b b b Characterization of Bipartite Graphs Theorem 3.3 A graph is bipartite if and only if it has no cycles of odd length. Proof. Necessity is clear: every cycle of G must alternate between a vertex in X and a vertex in Y , and so it must be of even length. For suﬃciency, we may assume that G is connected. Now, pick a vertex x of G, and let X be the set of vertices whose distance from x is even, and let Y be the set of vertices whose distance from x is odd. Clearly, {X, Y } is a partition of V (G). Suppose now that some two vertices of X or some two vertices of Y , say x1 and x2, are adjacent. Let P1 be a shortest path from x to x1 and let P2 be a shortest path from x to x2. Let u be vertex on P1 ∩P2 that the the farthest from x, and let P ′ 1 and P ′ 2 be the subpaths of, respectively, P1 and P2, from u to x1 and from u to x2. Then P ′ 1 and P ′ 2 have the same length, and so the cycle P ′ 1 ∪P ′ 2 ∪x1x2 has odd length. This proves that x1 and x2 cannot be adjacent. Matching Deﬁnition 3.4 ◮A matching is a set of pairwise non-adjacent edges. ◮A matching is perfect (is a 1-factor) if it meets every vertex of the graph. ◮A matching saturates the set X of vertices if each vertex in X is incident with an edge in the matching. Example 3.5 b b b b b b b b b b b b b b b b b b b b Does G have a matching that saturates all vertices on the left side? No! Look at S, which has 3 elements, and N(S), which has only 2 elements. Hall’s Marriage Theorem Theorem 3.6 (Hall’s Marriage Theorem, 1935) Suppose G is a bipartite graph with bipartition {X, Y }. The graph G has a matching saturating X if and only if \|N(S)\| ≥\|S\| for every subset S of X. Deﬁnition 3.7 ◮Given a matching M, an M-alternating path is a path that alternates between edges in M and edges not in M. ◮A non-trivial M-alternating path P that begins and ends at M-unsaturated vertices is an M-augmenting path. ◮Replacing M ∩E(P) by E(P) \ M produces a new matching M ′ that has one more edge than M. Example 3.8 b b b b b b b b b b M unsaturated b b b b b b b b M ′ Augmenting Paths Theorem 3.9 (Berge 1957) A matching M in a bipartite graph G is a maximum matching in G if and only if G has no M-augmenting path. Proof. It is clear that if G has an M-augmenting path, then M is not maximum. Suppose now that G has a matching M ′ that is larger than M and let F be the subgraph of G induced by the symmetric diﬀerence of M and M ′, that is, by all those edges that are in exactly one of M and M ′. The the maximum degree of F is at most 2, each component of F is a path or a cycle. Every path and every cycle in F alternates between edges in M and edges in M ′. Thus each cycle in F has the same number of edges from M and from M ′. Since \|M ′\| > \|M\|, there must be a component of F that is a path with more edges from M ′ than from M—an M-augmenting path. Proof of Hall’s Theorem Recall the Hall’s Condition: \|N(S)\| ≥\|S\| for every S ⊆X. Necessity is clear. To prove suﬃciency, suppose the Hall’s condition holds, let M be a maximum matching, and suppose u ∈X is unsaturated. Let S and T be subsets of X and Y , respectively, that are reachable from u by M-alternating paths. These paths reach Y from u along edges not in M, and reach X along edges in M. Hence every vertex in S −u is reached along an edge in M from a vertex in T . Since there are no augmenting paths, every vertex in T is saturated. Hence the edges of M establish a bijection between T and S −u. Note that an edge between S and y ∈Y −T would be an edge not in M, and thus create an M-augmenting path to y, which contradicts y / ∈T . Hence T = N(S), and \|N(S)\| = \|T \| = \|S\| −1 < \|S\|; a contradiction. Regular Graphs Deﬁnition 3.10 ◮A graph is regular if all vertices have the same degree. ◮If every vertex degree has the same value k, then the graph is k-regular. ◮A 3-regular graph is sometimes called cubic. Corollary 3.11 If G is a k-regular bipartite graph for some k > 0, then G has a perfect matching. Proof. Counting the edges by endpoints in X and by endpoints in Y , we conclude that k\|X\| = k\|Y \|, and so \|X\| = \|Y \|, and so every matching saturating X is perfect. Consider S ⊆X, and suppose that there are m edges between S and N(S). Since G is k-regular, we have m = k\|S\|. Since these m edges are incident to N(S), we have m ≤k\|N(S)\|. Hence k\|S\| ≤k\|N(S)\| and the Hall’s condition holds. Vertex Covers Deﬁnition 3.12 ◮A vertex cover of G is a set S of vertices such that every edge of G is incident with at least one element of S. ◮The vertices in S cover the edges of G. Theorem 3.13 (K¨ onig-Egerv´ ary 1931) If G is a bipartite graph, then the maximum size of a matching in G equals the minimum size of a vertex cover in G. Easy Direction. Since distinct vertices must be used to cover the edges of a matching, we have \|U\| ≥\|M\| whenever U is a vertex cover and M is a matching. Proof of K¨ onig-Egerv´ ary Theorem, Continued Given a minimum vertex cover U, we construct a matching of size \|U\|. Suppose G has bipartition {X, Y }. Let R = U ∩X and T = U ∩Y . Let H and H′ be the subgraphs of G induced by R ∪(Y −T ) and T ∪(X −R), respectively. We use 3.6 to show H has a matching saturating R, and H′ has a matching saturating T . Suppose S ⊆R and consider NH(S) ⊆Y −T . If \|NH(S)\| < \|S\|, then we can substitute NH(S) for S in U to obtain a smaller vertex cover, which is impossible. Hence H satisﬁes the Hall’s condition and so has a matching of size \|R\|. Likewise, H′ has a matching of size \|T \|. The union of these two matchings is a matching of G of size \|U\|. b b b b b b X Y b b b b U b b b b Matchings in Non-Bipartite Graphs Note 3.14 ◮Hall’s Marriage Theorem 3.6 does not make sense for non-bipartite graphs. ◮K¨ onig-Egerv´ ary Theorem 3.13 fails, in general, for non-bipartite graphs. Does the graph below have a perfect matching? b b b b No, since removing the two vertices in the middle leaves more than two components of odd order. Tutte’s 1-Factor Theorem Deﬁnition 3.15 ◮A graph (or component) is odd (even) if it has odd (even) order. ◮The number of odd components in a graph G will be denoted by q(G). Theorem 3.16 (Tutte 1-Factor) A graph G has a perfect matching if and only if q(G −S) ≤\|S\| for every S ⊆V (G). Necessity. If Q is an odd component of G −S, then a perfect matching must contain at least one edge between Q and S. Since edges in a matching are non-adjacent, the condition follows. Proof of Tutte’s 1-Factor Theorem Recall the Tutte Condition: q(G −S) ≤\|S\| for every S ⊆V (G). We assume that the condition holds and produce a perfect matching. We proceed by induction on the order of G. The claim is trivial if \|G\| ≤2. Now suppose that the Tutte Condition holds for G, which has order n > 2, and that the theorem holds for all graphs of smaller order. First note that q(G −v) = 1 = \|{v}\|, and so we may pick S0 to be a maximal subset of V (G) such that q(G −S0) = \|S0\|. Let Q1, Q2, . . . , Qm be the odd components of G −S0, and let D1, D2, . . . , Dk be the even components of G −S0. We will show that: (1) each Dj has a perfect matching; (2) if v ∈V (Qi), then Qi −v has a perfect matching; and (3) G contains a set s1v1, s2v2, . . . , smvm of edges such that S0 = {s1, s2, . . . , sm} and vi ∈V (Qi) for all i. Note that after (1)–(3) are established, the proof is complete. Proof of Tutte’s 1-Factor Theorem, Continued To prove (1), which says that every Dj has a perfect matching, we want to apply the induction hypothesis, and so we need to verify that every Dj satisﬁes the Tutte Condition. Suppose S ⊆V (Dj). Since the Tutte Condition holds for G, we have q(G −(S ∪S0)) ≤\|S ∪S0\| = \|S\| + \|S0\|. To count the odd components of G −(S ∪S0) = (G −S0) −S, note that when S is deleted from G −S0, none of the Qi’s is aﬀected, and so q(G −(S ∪S0)) = q(G −S0) + q(Dj −S) = \|S0\| + q(Dj −S). Combining the previous inequality with the last equation, we get \|S0\| + q(Dj −S) ≤\|S\| + \|S0\|, and so q(Dj −S) ≤\|S\|, which means that Tutte Condition holds for Dj, as required. Proof of Tutte’s 1-Factor Theorem, Continued Now, we prove (2), which states that each Qi −v has a perfect matching. Let v ∈V (Qi) and suppose that the Tutte Conditions fails for Qi −v, that is, there is a set S ⊆Qi −v such that q(Qi −v −S) > \|S\|. Now, \|V (Qi)\| = \|S ∪{v}\| + X even components Bt of Qi−v−S \|V (Bt)\| + X odd components Rs of Qi−v−S \|V (Rs)\|. Reducing this equation modulo 2, gives 1 ≡\|S\| + 1 + q(Qi −v −S) (mod 2), and thus q(Qi −v −S) ≡\|S\| (mod 2), and so q(Qi −v −S) ≥\|S\| + 2. Now notice that upon deleting {v} ∪S from G −S0 the only component of G −S0 that is aﬀected is Qi, which is lost, and the number of new odd components formed is q(Qi −v −S). Hence q(G −S0 −v −S) = q(G −S0) −1 + q(Qi −v −S). Now, since G satisﬁes the Tutte Condition for S0 ∪{v} ∪S, we have \|S0\| + 1 + \|S\| ≥q(G −S0 −v −S) = q(G −S0) −1 + q(Qi −v −S) ≥\|S0\| −1 + \|S\| + 2. But that implies that q(G −S0 −v −S) = \|S0 ∪{v} ∪S\|, which contradicts the maximality if S0, and so (2) follows. Proof of Tutte’s 1-Factor Theorem, Continued Now we turn to (3), which states that G contains a set s1v1, s2v2, . . . , smvm of edges such that S0 = {s1, s2, . . . , sm} and vi ∈V (Qi) for all i. For that, we form a bipartite graph H with X = {Q1, Q2, . . . , Qm} and Y = S0, in which Qi is joined to a vertex sj in S0 if and only if G has an edge from sj to Qi. To prove (3), we need to show that H has a perfect matching. We need to check that H satisﬁes the Hall Condition. Let A ⊆X. But NH(A) is also a set of vertices of G, so G satisﬁes the Tutte Condition for NH(A), that is, q(G −NH(A)) ≤\|NH(A)\|. But every odd component Q of G −S0 that is in A is also a component of G −NH(A). Thus q(G −NH(A)) ≥\|A\|, and so \|NH(A)\| ≥\|A\|, as required. Hence H has a perfect matching, and hence (3) is proved, and so is Tutte’s 1-Factor Theorem. Homework Set 2 Problem 4 Derive the suﬃciency (the non-obvious direction) of the Hall’s Marriage Theorem from the Tutte’s 1-Factor Theorem. Problem 5 Prove that a 3-regular simple graph has a 1-factor if and only if it decomposes into copies of P4. Problem 6 Prove that a tree T has a perfect matching if and only if q(T −v) = 1 for every v ∈V (T ). Do not invoke Tutte’s 1-Factor Theorem. Petersen’s Theorem Deﬁnition 3.17 A an edge e of G is a cut-edge if G \ e has more connected components than G. Corollary 3.18 (Petersen 1891) Every simple 3-regular graph with no cut-edge has a perfect matching. Proof. We prove that G satisﬁes the Tutte Condition. Let S ⊆V (G), and count the edges between S and the odd components of G −S. Since G is 3-regular, every vertex in S is incident to at most three such edges. If each odd component H of G −S is incident to at least three such edges, then 3q(G −S) ≤3\|S\|, and the Tutte Condition holds. Let m be the number of edges from S to H. The sum of vertex degrees in H is 3\|H\| −m, which must be even. Since \|H\| is odd, m must be also odd, but it cannot be 1 since G would have a cut-edge. Thus m must be at least 3 and the Tutte Condition holds. Deﬁnition 4.1 ◮A closed trail that uses every edge of the graph is called an Euler tour. ◮A graph is Eulerian if it has an Euler tour. Example 4.2 b b b b b b 1 2 3 4 5 6 7 8 9 10 11 12 Not Eulerian! Characterization of Eulerian Graphs Theorem 4.3 (Euler 1873) A graph is Eulerian if and only if all its vertices have even degrees and all of its edges belong to a single component. Lemma 4.4 Non-trivial maximal trails in graphs with all degrees even are closed. Proof. Let T be a maximal non-trivial trail in some graph G with all degrees even. Since T is maximal, it includes all edges of G incident with its ﬁnal vertex v. If T is not closed, then the degree of v must be odd, which is impossible. Proof of Euler’s Theorem Necessity is clear. For suﬃciency, suppose that G is non-trivial with all degrees even and all edges in same component. Let T be a trail of maximum length. By Lemma 4.4, T is closed. Let G′ = G \ E(T ) and suppose G′ is non-trivial. Since the degree of every vertex in G and in T is even, so it is in G′. Since all edges of G lie in the same component, there is an edge e of G′ adjacent to an edge in T . Let T ′ be a maximal trail in G′ with e as its ﬁrst edge. Again by Lemma 4.4, T ′ is closed. Hence we may detour T along T ′ to produce a longer trail; a contradiction. Vertex Connectivity Deﬁnition 5.1 ◮A separating set or a vertex cut of a graph G is a set S ⊆V (G) such that G −S has more than one component. ◮Vertex connectivity or connectivity κ(G) of a graph G is deﬁned as follows: ◮κ(G) = 0 if G is disconnected; ◮κ(G) = \|G\| −1 if G is connected, but has no pair of distinct non-adjacent vertices. ◮κ(G) = j if G is connected, but has a pair of non-adjacent vertices, and j is the smallest integer such that G has a j-element vertex cut. ◮If k is a positive integer, then G is k-connected or k-vertex-connected if k ≤κ(G). Note 5.2 ◮Vertex connectivity is not aﬀected by adding or deleting loops and parallel edges. ◮K1 is connected although κ(K1) = 0. Connectivity Examples Example 5.3 ◮κ(Kn) = n −1 for n ≥2; ◮κ(Km,n) = min(m, n); ◮If T is a non-trivial tree, then κ(T ) = 1. ◮κ(Cn) = 2 for all n ≥3. ◮An n-wheel Wn is obtained from Cn by adding a new vertex and joining it to all vertices of Cn. If n ≥3, then κ(Wn) = 3. Edge Connectivity Deﬁnition 5.4 ◮A disconnecting set of edges of a graph G with \|G\| > 1 is a set F ⊆E(G) such that G \ F has more than one component. ◮A graph is k-edge-connected if every disconnecting set has at least k edges. ◮The edge connectivity of G, written κ′(G) is the maximum k such that G is k-edge-connected. ◮Given S, T ⊆V (G), we write [S, T ] for the set of edges with one endpoint in S and the other in T . ◮An edge cut is a set of edges of the form [S, S] where S is a non-empty proper subset of V (G). ◮A bond is a minimal non-empty edge cut. Example 5.5 disconnecting set, but not an edge cut edge cut, but not a bond bond b b b b b b b b b b Whitney’s Theorem Note 5.6 The edge connectivity of a graph is unaﬀected by adding or deleting loops, but is aﬀected by adding and deleting edges in parallel. Theorem 5.7 (Whitney 1932) If G is graph with \|G\| > 1, then κ(G) ≤κ′(G) ≤δ(G). Proof. The edges incident to a vertex form a disconnecting set, so κ′ ≤δ. Clearly, κ(G) ≤\|G\| −1. Suppose [S, S] is a minimum edge cut of size k′ = κ′(G). If every vertex in S is adjacent to every vertex in S, then k′ = \|S\|\|S\| ≥\|G\| −1, and the inequality follows. Hence we may assume that there are vertices s ∈S and s ∈S that are non-adjacent. Let T be the vertex set consisting of all neighbors of s in S and all vertices in S −s that have neighbors in S. Then T is a vertex cut consisting of one endpoint of each edge in [S, S]. Hence κ ≤κ′. Connectivity Example Example 5.8 κ = 1 < κ′ = 2 < δ = 3 Theorem 5.9 If G is a connected graph and S is a non-empty proper subset of V (G), then F = [S, S] is a bond if and only if G \ F has two components. Equivalently, if and only if the subgraphs of G induced by each of S and S are connected. Proof. If G \ F has two components, then F is a bond, since G \ F ′ is connected for every proper subset F ′ of F. If G \ F has more than two components, then we may assume S = A ∪B with no edges between A and B. Then [A, A] is an edge cut which is a proper subset of F; a contradiction. Tutte Connectivity Deﬁnition 5.10 ◮A k-separation of a graph G is a pair of subgraphs {A, B} of G such that each of A and B has size at least k, A ̸= G, B ̸= G, A ∪B = G, and A ∩B is trivial of order at most k. ◮If G has a k-separation for some k, then Tutte connectivity of G is min{j : G has a j separation}, and ∞if no k-separation exists. Example 5.11 1-separations: b b 2-separations: b b b b ∞-te Tutte connectivity: ∅ b b Tutte Connectivity vs. Vertex Connectivity Theorem 5.12 If G is a graph on at least 3 vertices and G ≇K3, then the Tutte connectivity of G is min(κ(G), g(G)), where g(G) is the girth of G, that is, the length of a shortest cycle in G. Proof: Exercise. Deﬁnition 5.13 ◮A component of a graph G is a maximal subgraph of G that has Tutte connectivity at least 1. ◮A block of a graph G is a maximal subgraph of G that has Tutte connectivity at least 2. Note 5.14 A block of a non-empty graph is an isolated vertex, a loop-graph, a graph on two vertices with a positive number of edges between those vertices, or is vertex-2-connected. Block Tree Note 5.15 Two distinct blocks in a graph share at most one vertex since otherwise their union would be Tutte-2-connected. Deﬁnition 5.16 The block-tree of a connected graph G is a tree T whose vertex set is the disjoint union of the blocks of G and those vertices of G that belong to more than one block. The only edges in T are those that join vertices of G to bloks that contain them. Example 5.17 b b b b b b b b b b b b b Whitney’s Characterization of 2-Connected Graphs Deﬁnition 5.18 Two paths are internally-disjoint if neither contains a non-endpoint of the other. Theorem 5.19 (Whitney) A graph with at least three vertices is 2-connected if and only if each pair u and v of vertices is connected by a pair internally-disjoint uv-paths. Proof. If G has two internally-disjoint uv-paths, then deletion of one vertex cannot separate u from v. Hence G has no one-element vertex-cuts and so is 2-connected. Proof of Suﬃciency For the converse, suppose that G is 2-connected. We prove by induction on d(u, v) that G has two internally-disjoint uv-paths. When d(u, v) = 1, the graph G \ uv is connected since κ′(G) ≥κ(G) ≥2. A uv-path in G \ uv is internally disjoint from the uv-path consisting of the edge uv only. For the induction step, consider d(u, v) = k > 1 and assume that G has internally-disjoint xy-paths whenever 1 ≤d(x, y) < k. Let w be the vertex just before v on a shortest uv-path. Then d(u, w) = k −1 and, by the induction hypothesis, G has two internally-disjoint uw-paths P and Q. Since G −w is connected, it has a uv-path R. If R meets P and Q only in u, then the conclusion follows. Let z be the last vertex on R that belongs to P ∪Q. By symmetry, we may assume that z ∈V (P). We combine the uz-subpath of P with the zv-subpath of R to obtain a uv-path internally-disjoint from Q ∪wv. Expansion Lemma Lemma 5.20 (Expansion Lemma) If G is a k-connected graph and G′ is obtained from G by adding a new vertex y adjacent to at least k vertices of G, then G′ is also k-connected. Proof. Suppose S is a separating set of G′. If y ∈S, then S −y separates G, so \|S\| ≥k + 1. If y / ∈S and N(y) ⊆S, then \|S\| ≥k. Otherwise, S must separate G, and again \|S\| ≥k. Characterization of 2-Connected Graphs Theorem 5.21 If G is simple and \|G\| ≥3, then the following are equivalent (and characterize simple 2-connected graphs): (A) G is connected and has no cut-vertex; (B) For every two vertices x and y of G, there are two internally-disjoint xy-paths; (C) For every two vertices x and y of G, there is a cycle through x and y. (D) δ ≥1 and every pair of edges of G lies on a common cycle. Proof. Whitney’s Theorem 5.19 establishes the equivalence of (A) and (B). Clearly, (B) and (C) are equivalent. To see that (D) implies (C), apply (D) to edges incident to the desired x and y. Proof, Continued (A) G is connected and has no cut-vertex; (C) For every two vertices x and y of G, there is a cycle through x and y. (D) δ ≥1 and every pair of edges of G lies on a common cycle. We prove that (A) and (C) imply (D). Suppose G is 2-connected and uv and xy are edges of G. Add to G vertices w and z, and connect w with u and v, and connect z to x and y. By The Expansion Lemma 5.20, the resulting graph G′ is also 2-connected. Hence w and z lie on a common cycle C′ of G′. Replace the paths uwv and xyz by uv and xz, respectively, to obtain the desired cycle of G. Subdivisions Deﬁnition 5.22 ◮Subdividing an edge uv of a graph G is the operation of deleting uv and adding a path uwv through a new vertex w. ◮A graph G is a subdivision of a graph H if G can be obtained from H by successively subdividing (zero or more) edges. ◮A graph H is a topological minor of G, written H ⩽t G, if a subgraph of G is a subdivision of H. ◮A graph is a topological minor of G if it can be obtained from G by a sequence of operations each of which is one of the following: ◮deleting an edge; ◮deleting a vertex; and ◮contracting an edge incident with a vertex of degree two (un-subdividing an edge). Subdivisions and 2-Connectedness Corollary 5.23 A subdivision of a 2-connected graph is also 2-connected. Proof. Suppose G′ is formed by subdividing an edge uv of G with a new vertex w. By Theorem 5.21, it suﬃces to ﬁnd a cycle through two arbitrary edges e and f of G′. If e, f ∈E(G), then we can use the cycle of G, unless it uses uv, in which case we reroute the cycle through w. When e ∈E(G) and f ∈{uw, wv}, we modify a cycle passing through e and uv. When {e, f} = {uw, wv}, we modify a cycle through uv. Ears Deﬁnition 5.24 ◮A path addition to G is the addition to G of a path of length ℓ≥1 between two vertices of G, introducing ℓ−1 new vertices. ◮The added path is an ear. ◮An ear decomposition is a partition of E(G) into sets R0, R1, . . . , Rk so that C = R0 is a cycle, and Ri, for i > 0, is a path addition to the graph R0 ∪R1 ∪. . . ∪Ri−1. Example 5.25 b b b b b b Whitney’s Ear Decomposition Theorem 5.26 (Whitney’s Ear Decomposition) A simple graph is 2-connected if and only if it has an ear decomposition. Furthermore, every cycle in a 2-connected graph is the initial cycle of some ear decomposition. Proof. First we prove that graph with an ear decomposition is 2-connected. Since cycles in simple graphs are 2-connected, it suﬃces to show that path addition preserves 2-connectedness. Let u and v be the endpoints of an ear P to be added to a 2-connected graph G. Adding an edge e joining u and v (if u and v are non-adjacent) cannot reduce vertex-connectivity, so G ∪e is 2-connected. A succession of subdivions covers G ∪e into G ∪P. By Corollary 5.23, each subdivision preserves 2-connectedness. Proof of Suﬃciency Now, given a 2-connected graph G, we build an ear decomposition of G from a cycle C of G. Let G0 = C. Suppose we have built up a subgraph Gi by adding ears. If Gi ̸= G, then we may choose an edge uv of G \ E(Gi) and an edge xy ∈E(Gi). Because G is 2-connected, uv and xy lie on a common cycle C′. Let P be the path of C that contains uv and exactly two vertices of Gi, one at each end of P. Now P is an ear that can be added to Gi to obtain a larger subgraph Gi+1 of G. The process ends when all edges of G have been absorbed. Closed-Ear Decomposition Deﬁnition 5.27 A closed-ear decomposition of a graph G is a partition of E(G) into sets R0, R1, . . . , Rk such that R0 is a cycle and Ri for i > 0 is either a path addition or a cycle with exactly one vertex in R0 ∪R1 ∪. . . Ri−1 (closed ear). Theorem 5.28 A simple graph is 2-edge-connected if and only if it has a closed-ear decomposition. Moreover, every cycle in a 2-edge-connected graph is the initial cycle in some closed-ear decomposition. Proof omitted. The Menger Theorem Theorem 5.29 (Menger 1927) If x and y are non-adjacent distinct vertices of a graph G, then the minimum size of a vertex-cut separating x from y equals the maximum number of pairwise internally-disjoint xy-paths. Proof. Let κ(x, y) denote the minimum size of a vertex-cut separating x from y. Let λ(x, y) denote the maximum number of pairwise internally-disjoint xy-paths. An vertex-cut separating x from y must contain an internal vertex from every xy-path, and so κ(x, y) ≥λ(x, y). To prove the opposite inequality, we use induction on \|G\|. If \|G\| = 2, then κ(x, y) = λ(x, y) = 0. For the induction step, suppose \|G\| > 2 and let k = κ(x, y); we construct k pairwise internally-disjoint xy-paths. Proof of the Menger Theorem, Case 1 Case 1: G has a minimum xy-vertex-cut S not containing N(x) and not containing N(y). Let V1 be the set of vertices on xS-paths, and let V2 be the set of vertices on Sy-paths. Clearly, V1 ∩V2 = S. Form H1 from the subgraph of G induced by V1 by adding a vertex y′ and connecting it to all edges in S. Similarly, form H2 from the subgraph of G induced by V2 by adding a vertex x′ and connecting it to all edges in S. Every xy-path in G starts with an xS-path (which is contained in H1), so every xy′-cut in H1 is an xy-cut in G. Hence κH1(x, y′) = k. Hence κH2(x′, y) = k. Since V1 omits y ∪N(y) −S and V2 omits x ∪N(x) −S, each of H1 and H2 has fewer vertices than G. Hence the induction hypothesis gives λH1(x, y′) = k = λH2(x′, y). Combining k xS-paths in H1 that meet only at x with k Sy-paths in H2 that meet only in y, we get k pairwise internally-disjoint xy-paths in G. Proof of the Menger Theorem, Case 2 Case 2: Every minimum xy-vertex-cut contains N(x) or N(y). Since N(x) and N(y) are themselves xy-vertex-cuts, the condition implies that that there are no minimum xy-vertex-cuts other than N(x) and N(y). If G has a vertex v outside {x, y} ∪N(x) ∪N(y), then v is in no minimum xy-vertex-cut; hence κG−v(x, y) = k and applying the induction hypothesis to G −v yields the desired xy-paths in G. If v ∈N(x) ∩N(y), then v appears in every xy-vertex-cut, and so kG−v(x, y) = k −1. Now, applying the induction hypothesis to G −v yields k −1 xy-paths, which combine with the path xvy to get the desired k paths. We may thus assume that N(x) and N(y) are disjoint and together with {x, y} form the entire V (G). Let G′ be the bipartite graph with bipartition N(x) and N(y) and the edge set [N(x), N(y)]. Every xy-path in G uses some edge of G′, and so the xy-vertex-cuts are precisely the vertex covers of G′. By the K¨ onig-Egerv´ ary Theorem 3.13, G has a matching of size k. The edges of the matching together with the edges incident with x and y form the desired k paths. The Edge Version of Menger’s Theorem Theorem 5.30 (Edge Version of Menger’s Theorem) If x and y are distinct vertices of a graph, then the minimum size κ′(x, y) of the set of edges that separate x from y equals the maximum number λ′(x, y) of pairwise edge-disjoint xy-paths. Deﬁnition 5.31 The line graph of a graph G, written L(G), is a simple graph whose vertex set is E(G) with two vertices adjacent if the corresponding edges are adjacent in G. Example 5.32 e f g h b b b b e f g h Proof of Edge Version of Menger’s Theorem Modify G to obtain G′ by adding two vertices s and t and two new edges sx and yt. This operation does not aﬀect κ′(x, y) and λ′(x, y). A set of edges disconnects x from y in G if and only if the corresponding vertices of L(G′) form a set of edges separating sx from yt. Similarly, edge-disjoint xy-paths in G become internally-disjoint paths from sx to yt in L(G′), and vice versa. Applying the vertex version of the Menger Theorem 5.29 yields κ′ G(x, y) = κL(G′)(sx, yt) = λL(G′)(sx, yt) = λ′ G(x, y). Theorem 5.33 The connectivity of G equals the maximum k such that λ(x, y) ≥k for all vertices x and y of G. The edge connectivity of G equals the maximum k such that λ′(x, y) ≥k for all vertices x and y of G. Proof. The edge version follows immediately from Theorem 5.30 since κ′(G) = minx,y∈V (G) κ′(x, y). For connectivity, we get κ(x, y) = λ(x, y) if x and y are non-adjacent, and κ(G) is the minimum of these values. If x and y are adjacent, we get λG(x, y) = 1 + λG\xy(x, y) = 1 + κG\xy(x, y) ≥1 + κ(G \ xy) ≥κ(G). Tutte’s Wheel Theorem Theorem 5.34 (Tutte’s Wheel Theorem) If G is a Tutte-3-connected graph on at least four vertices that is not a wheel, then there is an edge e of G such that at least one of G/e and G \ e is also Tutte-3-connected. Lemma 5.35 (Thomassen 1980) Every 3-connected graph G on at least ﬁve vertices has an edge e such that G/e is 3-connected. Proof of the Lemma Assume that for each edge e the graph G/e is not 3-connected, and so has a 2-element vertex cut. Since G is 3-connected, one of the elements of this vertex cut must come from contracting e = xy. Let z be the other element of this vertex cut. Then {x, y, z} is a vertex cut in G. Choose e = xy and the corresponding z so that the graph G −{x, y, z} has a component H with the largest possible order. Let H′ be another component of G −{x, y, z}. Since {x, y, z} is a minimal vertex cut, each of x, y, and z has a neighbor in ech of H and H′. Let u be a neighbor of z in H′, and let v be a vertex such that {z, u, v} disconnects G. The subgraph of G induced by V (H) ∪{x, y} is connected. Deleting v from this subgraph (if it occurs there) cannot disconnect it, since then {z, v} would disconnect G. Therefore all elements of V (H) ∪{x, y} −v belong to the same component of G −{z, u, v}, which has more vertices than H; a contradiction. Proof of Tutte’s Wheel Theorem Every T3C graph G on at least 4 vertices has an edge e such that G/e or G \ e is T3C, unless G is a wheel. Suppose G is T3C and has at least 4 vertices, but has no edge whose deletion or contraction results in a T3C graph. If \|G\| = 4, then G ∼ = K4, and the conclusion holds; so we may assume that G has at least 5 vertices. Then Lemma 5.35 implies that G has an edge whose contraction results in a 3-connected graph. Let F (a k-fan) be a subgraph of G such that: ◮F consists of a path on vertices (listed in order) v0, v1, . . . , vk, and a vertex c adjacent to all vertices of the path; ◮G/vk−1vk is 3-connected; ◮each of v1, v2, . . . , vk−1 has degree 3 in G; and ◮k is maximal, subject to the conditions above. Lemma 5.35 implies that k ≥1. The graph G \ cvk is not 3-connected, so it has a vertex cut of size 2. One of the cut vertices is vk−1; let’s name the other one u. Vertices c and vk are in distinct components A and B, respectively, of G \ cvk −{vk−1, u}. Consider G/vk−1vk. It is 3-connected, which implies that vk is the only vertex of B, and so vk also has degree 3. Note that each of v1, v2, . . . , vk−2 has degree 3 and is not adjacent to vk, which implies that u is not among those vertices. Clearly, u ̸= vk and u ̸= c. Proof, continued Case 1: u ̸= v0 Look at G/uvk. If this graph is not 3-connected, then G has a vertex-cut of size 3 containing u and vk. But we concluded that the degree of vk is 3, which means that the vertex cut containing u and vk separates vk−1 from c, which are adjacent; a contradiction. This means that G/uvk is not simple. One possibility of this happening is that u, c, and vk form a triangle, but that would imply that G has a larger k-fan; a contradiction. Otherwise, u, vk−1, and vk form a triangle. But then, if k −1 > 0, the vertex vk−1 would be adjacent to vk−2, vk, c, and u, which is impossible, the degree of vk−1 is 3. So k −1 = 0. In that case, however, v0, v1, c, and u form a 2-fan; again a contradiction. Case 2: u = v0 Note that v1, v2, . . . , vk have degree 3. So if G contained another vertex, say z, it would be disconnected from v1, v2, . . . , vk, by deleting v0 and c; which is impossible. It follows that G is a (k + 1)-wheel, which completes the proof. Clique Sums Deﬁnition 5.36 ◮A clique-sum of two graphs G and H is obtained from the disjoint union of G and H by identifying a complete subgraph of G with a complete subgraph (of the same order) of H, and then deleting the edges of the identiﬁed subgraph. ◮If the identiﬁed complete subgraph has order k, then the clique-sum is called k-sum and written G ⊕k H. Example 5.37 ◮A 0-sum is a disjoint union. ◮A 1-sum consists of two subgraphs that share exactly one vertex. ◮Every graph can be obtained by repeatedly 0-summing graphs, starting with connected graphs. ◮Every connected graph can be obtained by repeatedly 1-summing graphs, starting with blocks. Decomposition of 2-Connected Graphs Deﬁnition 5.38 A 3-block is a cycle of length at least 3, a loopless graph on two vertices with at least 3 edges between them (co-cycles), or Tutte-3-connected graph. Theorem 5.39 Every Tutte-2-connected graph of size at least 3 can be obtained by repeatedly 2-summing graphs, starting with 3-blocks. Moreover, in this process, no two cycles are 2-summed together, and two co-cycles are 2-summed together. The decomposition is unique. b b Homework Set 3 Problem 7 Suppose G is a graph that is non-trivial, connected, and such that every edge e is in some two cycles that meet only at e. What is the highest edge-connectivity of G that can be inferred from these properties? Problem 8 Find all non-negative integers k for which the following statement is true: For every simple k-regular graph G on at least two vertices, κ(G) = κ′(G). Problem 9 Suppose G is a simple r-connected graph of even order with no K1,r+1 as an induced subgraph for a positive integer r. Prove that G has a perfect matching. Polygonal Paths Note 6.1 A graph may be viewed as a topological space when ◮the vertices are points; ◮the edges are homeomorphic images of the unit interval; and ◮the incidences are reﬂected by the vertices (points) being in the closure of the edges. Deﬁnition 6.2 A polygonal curve in the plane is the union of ﬁnitely many line segments such that each segment starts at the end of the previous one and no point lies in more than one segment, except the end of one segment and the beginning of the next one coincide. A simple open polygonal curve is one homeomorphic to a closed interval. A simple closed polygonal curve is one homeomorphic to a unit circle. Plane Graphs Deﬁnition 6.3 ◮A drawing of a graph G is a function that maps each vertex v ∈V (G) to a point f(v) in the plane, and each uv-edge to a simple polygonal f(u)f(v)-curve in the plane. ◮A point f(e) ∩f(e′) other than the a common endpoint is a crossing. ◮A graph is planar if it has a drawing without crossings. Such a drawing is a planar embedding of G. ◮A plane graph is a particular drawing of a a graph in the plane with no crossings. Note 6.4 A plane embedding corresponds to an embedding of the graph in the sphere through a stereographic projection. Faces Theorem 6.5 (Jordan Curve Theorem) If C is a simple closed polygonal curve in the plane, then the complement of C in the plane consists of two connected components each with C as the boundary. Deﬁnition 6.6 ◮The connected components of the complement of a plane graph are the faces of the embedding. ◮The length of a face is the number of edges in the boundary of the face, with cut-edges counted twice. ◮The dual graph G∗of a non-empty plane graph G is the graph such that ◮the vertices of G∗are the faces of G; ◮the edges of G∗are the edges of G; ◮a vertex and an edge of G∗are incident if and only if the edge is the boundary of the corresponding face of G. Example of a Dual Graph Example 6.7 b b b b b b Properties of Dual Graphs Note 6.8 Diﬀerent graphs may have the same dual. (G∗)∗= G if and only if G is connected. Theorem 6.9 Edges in a plane graph form a cycle if and only if the edges in the dual graph form a bond. Proof. Suppose D is a set of edges of G that contains a cycle. By Jordan Curve Theorem 6.5, some face u∗of G lies inside this cycle, and some other v∗lies outside. Then every u∗v∗-path in G∗must contain an edge of D∗. Conversely, suppose D contains no cycle. Then it is possible to reach every face of G∗from every other without crossing D∗. Hence G∗\ D∗is connected so D∗contains no bond. Properties of Dual Graphs, Continued Theorem 6.10 If a plane graph G is connected, then the Tutte connectivity of G is the same as the Tutte connectivity of G∗. Proof: Exercise. Theorem 6.11 The following are equivalent for a plane graph G: (A) G is bipartite; (B) every face of G has even length; (C) G∗is Eulerian. Proof. To see that (A) implies (B), note that the boundary of every face of G is the union of closed walks, and if the total length is odd, then one of the walks must be of odd length, and so contain an odd-length cycle. Proof, Continued Conversely, suppose that G has an odd cycle C. Since G has no crossings, C is laid out as a simple closed polygonal curve. Let F be the region enclosed by C. Every face of G is completely within F, or completely outside of F. Summing up the face lengths for the faces inside F gives an even number since every face is even. This sum counts each edge of C once, and every edge inside F twice. Hence C is even; a contradiction. The equivalence of (B) and (C) follows from the fact the the dual graph is connected and its vertex degrees are the face lengths of G. Note 6.12 ◮Deleting an edge or a vertex from a plane graph results in a plane graph. ◮Contracting an edge in a plane graph can be visualized as sliding the two endvertices towards each other until they meet, pulling all incident edges along. ◮Thus the class of planar graphs is minor-closed, that is, all minors of planar graphs are also planar. Euler’s Formula Theorem 6.13 (Euler’s Formula) If a connected non-empty plane graph has v vertices, e edges, and f faces, then v −e + f = 2. Proof. We proceed by induction on v. If v = 1, then G has only loops, each a closed curve in the embedding. If e = 0, then f = 1, and the formula holds. Each added loop adds one more edge and one more face, and so the formula holds when v = 1. Suppose v > 1. Since G is connected, it has a non-loop edge. Contract such an edge to obtain a plane graph with v′ = v −1 vertices, e′ = e −1 edges, and f ′ = f faces. Applying the inductive hypothesis, we get v′ −e′ + f ′ = 2, and so (v −1) −(e −1) + (f) = v −e + f = 2, as desired. Corollaries of Euler’s Formula Note 6.14 ◮Euler’s Formula implies that all plane embeddings of connected graphs have the same number of faces. ◮Contracting a non-loop edge of G has the eﬀect of deleting the corresponding edge in G∗. Similarly, deleting a non-cut edge of G has the eﬀect of contracting the corresponding edge in G∗. ◮Euler’s Formula (as stated) fails for disconnected graphs. Corollary 6.15 If G is a planar graph whose order v is at least 3, whose size is e, and whose girth g is at least 3 but ﬁnite, then e ≤(v −2)g g −2 . If G is simple, then e ≤3v −6. Proof Without loss of generality, we may assume that G is plane and connected. Let fi denote the number of faces of G of length i. Since every edge appears in two faces or in the same face twice, we have 2e = P ifi ≥gf. Substituting this into Euler’s Formula gives v −e + 2e/g ≥2. e(2 g −1) ≥2 −v e ≤v −2 2 g −g g = (v −2)g g −2 Note that when G is simple, g ≥3 and so e ≤3v −6. Example 6.16 Is K5 planar? No, since e = 10 > 3v −6 = 9. Is K3,3 planar? No, since e = 9 > (v −2)g g −2 = 8. Platonic Solids We want to ﬁnd all graphs that are ◮planar, ◮simple, ◮3-connected, ◮k-regular (k ≥3), ◮l-co-regular (that is, each face has same length l ≥3). We have kv = 2e = lf, and so the Euler’s Formula 6.13 gives us 2e/k −e + 2e/l = 2. Thus e(2/k −1 + 2/l) = 2 and e = 2kl 2k + 2l −kl . Then −kl + 2l + 2k > 0, and so −kl + 2l + 2k −4 > −4, And so (k −2)(l −2) < 4, and so k, l ≥3 and k, l ≤5. The Zoo of Platonic Solids e = 2kl 2k + 2l −kl v = 2e k f = 2e l k l (k −2)(l −2) e v f name 3 3 1 6 4 4 tetrahedron 3 4 2 12 8 6 cube 4 3 2 12 6 8 octahedron 3 5 3 30 20 12 dodecahedron 5 3 3 30 12 20 icosahedron Statement of the Kuratowski Theorem Theorem 6.17 (Kuratowski 1930) A graph is planar if and only if it has neither K5 nor K3,3 as a topological minor. Theorem 6.18 (Wagner) A graph is planar if and only if it has neither K5 nor K3,3 as a minor. Lemma 6.19 If F is the edge-set of the boundary of a face of a plane graph G, then G has an plane embedding in which F is the boundary of the inﬁnite face. Proof. Apply stereographic projection twice. Minimally No-Planar Graphs Deﬁnition 6.20 A graph is minimally non-planar if it is non-planar, but every proper subgraph of it is planar. Lemma 6.21 Every minimal non-planar graph is 2-connected. Proof. If G is disconnected, we can embed one component of G inside one face of the rest of G. Similarly, if G has a cut-vertex v, let G1, G2, . . . , Gk be the subgraphs of G induced by v together with the components of G −v. By the minimality of G, these subgraphs are planar. It is easy to see that the plane embeddings of these subgraphs can be put together to form a plane embedding of G. Kuratowski Graphs and Topological Minors Lemma 6.22 Suppose G = H1 ⊕2 H2 is non-planar. Then at least one of H1 and H2 is non-planar. Proof. Let e be the common edge of H1 and H2. Suppose both H1 and H2 are planar. By Lemma 6.19, each of H1 and H2 can be embedded in the plane with e in the boundary of the inﬁnite face. It is now easy to put together the embeddings of H1 and H2 into a plane embedding of G. Deﬁnition 6.23 ◮A Kuratowski subgraph is a subgraph isomorphic to a subdivision of K5 or of K3,3. ◮A vertex of a graph G is a branch vertex of a Kuratowski subgraph H of G, if its degree in H exceeds two. proof Lemma 6.24 If G/e has a Kuratowski subgraph, then so does G. Proof. Let H be a Kuratowski subgraph of G′ = G/e, and let z be the vertex of G′ obtained by contracting e = xy. If z is not a branch vertex of H, then G also has a Kuratowski subgraph obtained from H by lengthening a path through z if necessary. If z is a branch vertex of H and at most one of the edges incident to z in H is incident to x in G, then z can be expanded into xy to lengthen that path, and y becomes the corresponding branch vertex of a Kuratowski subgraph of G. The only remaining case to consider is when H is a subdivision of K5, z is a branch vertex of H, and each of x and y is incident in G to two of the four edges incident to z in H. Let u1, u2 be the branch vertices of H that are at the other ends of paths leaving z on the edges incident with x, and let v1, v2 be the other branch vertices of H. By deleting the edges of the u1u2-path and the v1v2-path, we obtain a subdivision of K3,3. Tutte’s Version of Kuratowski’s Theorem Deﬁnition 6.25 A plane embedding is convex if every face except the inﬁnite one is a convex polygon. Theorem 6.26 (Tutte 1960–63) If G is a simple 3-connected graph with neither K5 nor K3,3 as the topological minor, then G has a convex embedding in the plane with no three vertices on a line. Proof. We proceed by induction on \|G\|. The only 3-connected simple graph on at most 4 vertices is K4 and it has such an embedding. Let G be a graph on n ≥5 vertices and suppose the theorem holds for all graphs on fewer than n vertices. Proof, Continued By Lemma 5.35, G has an edge e = xy such that G/e is also 3-connected; let z denote the vertex resulting from contracting e. By Lemma 6.24, G/e has no Kuratowski subgraph, and so by the induction hypothesis, H = G/e has a convex embedding with no 3 vertices on a line. Let C denote the cycle that is the boundary of the face H −z that is not a face of H. Since we started with a convex embedding, the face bounded by C contains straight line segments from z to all of its neighbors. Let x1, x2, . . . , xk be the neighbors of x in the cyclic order on C. If all neighbors of y lie on the portion of C from xi to xi+1, then we obtain a convex embedding of G. If this does not occur, then either: 1. y shares three neighbors with x, in which case we obtain a subdivision of K5; or 2. y has neighbors u, v in C that are in diﬀerent components of C −{xixi+1} for some i, in which case we obtain a subdivision of K3,3. Proof of Kuratowski Theorem Recall: G is planar if and only if neither K5 nor K3,3 is a topological minor of G. Proof. Without loss of generality, we may asume that G is simple. We showed in Example 6.16 that K5 and K3,3 are both non-planar. Therefore any subdivision of K3 or of K3,3 is also non-planar, as is any supergraph of such a subdivision. Suppose the converse implication fails, and G is a counter-example of the possible smallest order, that is, G is non-planar but has no Kuratowski subgraph. Then G is minimally non-planar, and, by Lemma 6.21, G is 2-connected. Now, Lemma 6.22 implies that G is 3-connected. But then we get a contradiction with the Tutte Theorem 6.26, which states that a 3-connected graph with no Kuratowski subgraph is planar. Problem Set 4 Problem 10 Prove that every n-vertex plane graph isomorphic to its dual has 2n −2 edges. For each n ≥4, construct a simple n-vertex plane graph isomorphic to its dual. Problem 11 Let G be a connected plane graph such that every vertex of G is incident with two faces of length four, one face of length six, and no other faces. Use Euler’s Formula to determine the number of vertices, edges, and faces of G. Draw G. Problem 12 A plane graph is outerplane if it has a face incident with all the vertices. A graph is outerplanar if it isomorphic to an outerplane graph. Prove that a graph is outerplanar if and only if it has neither K4 nor K2,3 as a topological minor. Chromatic Number Deﬁnition 7.1 ◮A k-coloring of a graph G is a labeling f : V (G) →{1, 2, . . . , k}. ◮The labels are colors. ◮The vertices with color i are a color class. ◮A k-coloring f is proper if f(x) ̸= f(y) whenever x and y are adjacent. ◮The chromatic number χ(G) is the minimum k such that G is k-colorable. ◮If χ(G) = k, then G is k-chromatic. ◮If χ(G) = k, but χ(H) < k for every proper subgraph H of G, then G is k-color-critical or k-critical. ◮Let ω(G) denote the clique number of G, that is, the order of a largest complete subgraph of G. ◮Let α(G) denote the independence number of G, that is, the largest number of vertices of G no two of which are adjacent. Facts About Chromatic Number ◮For vertex coloring, all graphs will be considered simple. ◮χ(G) ≥ω(G), and χ(G) = ω(G) when G is complete. ◮χ(G) may exceed ω(G), for example, consider C2r+1 ∨Ks, that is, the graph formed from the disjoint union of C2r+1 and Ks by joining each vertex of C2r+1 to each vertex of Ks. ◮There are triangle-free graphs of arbitrarily high chromatic number (Mycielski) ◮Let M2 = K2. ◮Construct Mn+1 from Mn by: ◮Start with a copy M of Mn; ◮for each vertex v in M add a new vertex u and connect it to the neighbors of v in M; ◮add a new vertex w and connect it to all vertices not in M. ◮χ(G) ≥\|G\|/α(G) ◮χ(G) ≤∆(G) + 1 Proof. Color greedily: Order the vertices arbitrarily as v1, v2, . . . , vn. Starting with k = 1, color each vertex vk with the smallest color not used among the vertices v1, v2, . . . , vk−1 that are neighbors of vk. Brooks’ Theorem Theorem 7.2 (Brooks 1941) If G is a connected simple graph other than a clique and an odd cycle, then χ(G) ≤∆(G). Proof. Suppose G is connected, not a clique, not an odd cycle, and let k = ∆(G). If G is not not regular, then choose vn so that d(vn) < k. Since G is connected, we can grow a spanning tree of G from vn, assigning indices in the decreasing order as we reach vertices. Then each vertex vi, other than vn, has at least one neighbor among vi+1, vi+2, . . . , vn, and so each vertex vi (including vn) has at most k −1 neighbors among v1, v2, . . . , vi−1. Hence the greedy coloring requires at most k colors. In the remaining case, G is k-regular. If G has a cut-vertex x, then let G′ be the subgraph of G induced by a component of G −x together with x. Then the degree of x in G′ is less than k, and so G′ has a proper k-coloring. After permuting the colors, if necessary, it is easy to combine the k-colorings of such subgraphs into a k-coloring of G. Proof of Brooks’ Theorem, Continued Thus we may assume that G is 2-connected. Suppose that G has an induced 3-vertex path with vertices we label v1, vn, v2, in order, such that G −{v1, v2} is connected. We could then index the vertices of a spanning tree of G −{v1, v2} as before, so that every vi other than vn has at least one neighbor among vi+1, vi+2, . . . , vn. Then every vertex vi other than vn would have at most k −1 neighbors among v1, v2, . . . , vi−1, while two of the neighbors of vn, namely v1 and v2 would both receive color 1 by the greedy coloring. Hence, it suﬃces to show that every 2-connected k-regular graph G with k ≥3, not a clique, not an odd cycle, has 3-vertex induced path v1, vn, v2 such that G −{v1, v2} is connected. Choose a vertex x. If κ(G −x) ≥2, then let v1 = x, and let v2 be a vertex of distance two from x, which exists, since G is connected, regular and not a clique. If κ(G −x) = 1, then x has a neighbor in every block of G −x that is a leaf in the block tree of G −x, since G has no cut-vertex. Neighbors v1 and v2 of x in two such blocks are non-adjacent. Furthermore, G −{x, v1, v2} is connected, since blocks have no cut-vertices. Now, G −{v1, v2} is also connected, since k ≥3 and so x has a neighbor other than v1 and v2. Let vn = x to complete the proof. 5-Color Theorem Theorem 7.3 (Heawood 1890) Every loopless planar graph has a proper 5-coloring. Proof. Suppose G is a plane graph that is a minimal counter-example. Then G is simple, and so ∥G∥≤3\|G\| −6 by Corollary 6.15. It follows that G has a vertex v of degree 5 or less. Then G −v has a 5-coloring f by the minimality of G. Since G is not 5-colorable, each color appears at one of the neighbors of v (and so d(v) = 5). We may assume that the colors on the neighbors of v appear as 1, 2, 3, 4, and 5 as they are inspected clockwise. Let Gi,j denote the subgraph of G −v induced by the colors i and j. Note that we an exchange the two colors on any component of Gi,j to obtain another proper coloring of G −v. If some two neighbors of v were in diﬀerent components of Gi,j, then switching colors on one such component would result in two neighbors of G being colored the same, Thus allowing to extend the coloring to v. Thus we may assume that every two neighbors of v are in the same component of Gi,j. Proof, Continued Let Pi,j be a path in Gi,j joining the neighbors of v colored i and j. Consider the cycle C of G induced by P1,3 together with v, which separates the neighbor of v colored 2 from the one colored 4. Hence P2,4 must cross C, which is impossible. Theorem 7.4 (4-Color Theorem, Appel and Haken 1977) Every loopless planar graph has a proper 4-coloring. Proof omitted. Discharging Method Theorem 7.5 Suppose G is a plane triangulation with δ(G) ≥5. Then G contains two adjacent vertices one of which has degree 5, and the other has degree 5 or 6. Proof. Suppose G is as described, but the conclusion fails. Let v, e, and f be, respectively, the number of vertices, edges, and faces of G. Since G is a plane triangulation, 3f = 2e, and the Euler Formula implies e = 3v −6. To each vertex u, assign a charge of 6 −d(u). Note that the sum of all the charges in G is X u∈V (G) (6 −d(u)) = 6v −2e = 6v −2(3v −6) = 12. Now, set the discharging rule that each degree-5 vertex sends 1/5 to each of its neighbors. The ﬁnal charge of vertices of degree 5 or 6 is zero. If d(u) ≥7, then the ﬁnal charge of u is at most 6 −d(u) + d(u)/5, and so it can be positive only if d(u) = 7. But a degree-7 vertex with a positive ﬁnal charge would have to have six neighbors of degree 5, which implies that two of such neighbors must be adjacent since G is a triangulation; a contradiction. Chromatic Polynomial Let PG(x) denote the number of ways to properly color a (labeled) graph G with x colors. If G has loops, then PG(x) = 0. If G is edgeless of order n, then PG(x) = xn. If e is a non-loop non-multiple edge of G incident with u and v, then the proper colorings of G \ e with x colors can be partitioned into two sets: A, in which u and v receive the same color, and B, in which they do not. Then A corresponds to proper colorings of G/e with x colors, and B corresponds to proper colorings of G. Hence PG(x) = PG\e(x) −PG/e(x). PG(x) is called the chromatic polynomial of G. Theorem 7.6 (Four-Color Theorem, restated) If G is a planar loopless graph, then PG(4) > 0. Perfect Graphs Deﬁnition 7.7 A graph G is perfect if χ(H) = ω(H) for every induced subgraph H of G. Theorem 7.8 (Perfect Graph Theorem, Lov´ asz 1972) A graph is perfect if and only if its complement is perfect. Theorem 7.9 (Strong Graph Theorem (formerly Berge’s Strong Graph Conjecture), Chudnovsky, Robertson, Seymour, Thomas 2002) A graph is perfect if and only if it has no induced subgraph that is an odd cycle of length at least ﬁve or its complement. Edge Colorings Deﬁnition 7.10 ◮A k-edge-coloring of a graph G is a labeling f : E(G) →{1, 2, . . . , k}. ◮The labels are colors and the edge-set with one color is a color class. ◮A k-edge-coloring is proper if adjacent edges have diﬀerent colors, or equivalently, if every color class is a matching. ◮A graph is k-edge-colorable if it has a proper k-edge-coloring. ◮The chromatic index or edge chromatic number χ′(G) of a loopless graph G is the least k such that G is k-edge-colorable. Note 7.11 ∆(G) ≤χ′(G). Edge-Coloring of Bipartite Graphs Theorem 7.12 (K¨ onig 1916) If G is bipartite, then χ′(G) = ∆(G). Proof. We showed in Corollary 3.11 that every non-trivial regular bipartite graph H has a perfect matching. By induction on ∆(H), this yields a proper ∆-edge-coloring of H. Thus it suﬃces to prove that every bipartite graph G of maximum degree k is a subgraph of a k-regular bipartite graph H. Add the vertices to the smaller side to equalize the sides, if necessary. If the resulting graph is not k-regular, then each side has a vertex with degree less than k. Add an edge joining such a pair of vertices. Continue adding edges until the graph becomes regular (although not necessarily simple). Vizing’s Theorem Theorem 7.13 (Vizing 1964–65, Gupta 1966) If G is simple, then χ′(G) ≤∆(G) + 1. Proof. Suppose uv is an edge left uncolored by a proper (∆(G) + 1)-edge-coloring of a proper subgraph G′ of G. After possibly re-coloring some edges, we extend this coloring to include uv; call this an augmentation. After an appropriate number of augmentations, we obtain a proper (∆(G) + 1)-edge-coloring of G. Since the number of colors exceeds ∆(G), every vertex has some color not appearing on its incident edges. Let a0 be a color missing at u and let a1 be a color missing at v. We may assume that a1 appears at u, or it could be used for the edge uv. Let v1 be the neighbor of u along the edge colored a1. At v1 some color a2 is missing. We may assume that a2 appears at u, or we could re-color uv1 from a1 to a2, and then use a1 on uv to augment the coloring. Proof of Vizing’s Theorem, Continued For i ≥2, we continue this process: Having selected a new color ai that appears at u, let vi be the neighbor of u along the edge colored ai. If ai+1 is missing at u, then we shift color aj from uvj to uvj−1 for 1 ≤j ≤i (where v0 = v) to complete the augmentation (downshifting from i). We are ﬁnished unless ai+1 appears at u, in which case the process continues. Since we have ∆(G) + 1 to choose from, this iterative process of selecting ai+1 eventually repeats a color. (Note that we do not need the fact that each vertex has at most ∆neighbors to augment the coloring by one edge; we just need that there is a color missing at each vertex.) Let l be the smallest index such that the color al+1 missing at vl is in the list a1, a2, . . . , al; suppose al+1 = ak. Note that if a0 is missing at vl, then we downshift from vl and use color a0 on uvl to complete the augmentation. Hence we may assume that a0 appears at vl, but ak does not. Let P be the (unique) maximal path of edges colored a0 or ak that begins at vl. Switching on P means interchanging the colors a0 and ak on the edges of P. Proof of Vizing’s Theorem, Continued If P reaches vk, then it does so along an edge colored a0, continues along the edge colored ak, and stops at u. In this case, we downshift from k and switch on P. Similarly, if P reaches vk−1, then it does so along an edge colored a0, and stops there. In that case, we downshift from k −1, give color a0 to uvk−1, and switch on P. Finally, suppose that P reaches neither vk nor vk−1, and so it ends outside {u, vl, vk, vk−1}. In that case, we downshift from l, give color a0 to uvl, and switch on P. List Colorings Suppose G is a graph with the vertex set V , and L = (Lv)v∈V associates with each vertex v a list Lv of colors available to color v. We say that G admits an L-coloring if there is a proper coloring of G such that, for every vertex v, the color of v is in the list Lv. The graph G is k-list-colorable or k-choosable if G admits an L-coloring for every L = (Lv)v∈V with \|Lv\| = k for every vertex v. The smallest k such that G is k-choosable is called the list-chromatic number of G and is denoted by ch(G). List colorings of edges are deﬁned analogously, as is the list-chromatic index ch′(G). Note that if L = (Lv)v∈V is such that all Lv’s are identical and of cardinality k, then G admitting an L-coloring is equvalent to G being k-colorable. An analogous statement holds for edge-colorings. Thus ch(G) ≥χ(G) and ch′(G) ≥χ′(G). But there are graphs for which ch(G) ̸= χ(G). Consider K3,3 where each side of the bipartition has lists {1, 2}, {1, 3}, and {2, 3}. The list-chromatic number of this graph is 3, while the chromatic number is 2. Every Planar Graph Is 5-Choosable Theorem 7.14 (Thomassen 1994) Every planar graph is 5-choosable. In fact, we will prove a somewhat stronger statement: Suppose that G is a plane graph such that each internal face is a triangle, and the external face is bounded by a cycle C with vertices v1, v2, . . . vk (in this order). Let L = (Lv)v∈V (G) be the set of lists such that Lv1 = {1}, Lv2 = {2}, \|Lvi\| ≤3 for all i ∈{3, 4, . . . , k}, and \|Lw\| = 5 for all vertices w not on C. Then G admits an L-coloring. We prove this by induction. The claim is obvious for the smallest graph for which it makes sense, that is, a triangle. Suppose the claim is true for every graph on fewer than n vertices, and suppose that G is like described above, and \|G\| = n. Case 1 Suppose that C has a chord vw. By re-indexing the vertices if necessary, we may assume that v2 = w. Consider the two cycles C1 and C2 contained in C ∪vw with v1 lying on C1 but not on C2, and the graphs G1 and G2 bounded by C1 and C2, respectively. Create a list L1 by restricting L to V (G1). Applying the inductive hypothesis to G1 we get an L1 coloring of G1. Let c be the color assigned by the coloring of G1 to v, which must be diﬀerent from the color of w. Construct a list L2 by restricting L to V (G2 −v) and assigning {c} as the list for v. Now, applying the inductive hypothesis to G2, we get an L2-coloring of G2, which can be combined with the coloring of G1 to get an L-coloring of G. Case 2 Suppose now that C has no chord. Consider the neighbors of vk that are v1, u1, u2, . . . , um, vk−1, in this order. Let j, and l be two colors in Lvk that are diﬀerent from 1, and remove j and l (if present) from Lui for all i ∈{1, 2, . . . , m} to create a list L′. Applying the inductive hypothesis to G −vk results in an L′-coloring of G −vk. Extend the coloring to an L-coloring of G by assigning to vk the color from {j, l} that is diﬀerent from the color of vk−1. List Colorings Theorem 7.15 There are simple planar graphs that are not 4-choosable. Conjecture 7.16 ch′(G) = χ′(G). Flows and Circulations A ﬂow is an assignment of “values” to directed edges of a graph G so that for every vertex x ∈V (G) the net ﬂow into x is zero. Deﬁnition 7.17 ◮Let ⃗ E = {(e, x, y) : e ∈E, x ∈V, y ∈V, e = xy}. Thus an edge e = xy with x ̸= y has two directions (e, x, y) and (e, y, x). A loop e = xx has only one direction. ◮Let H be an abelian group written additively with neutral element 0 (usually H = Z or H = Zk, that is, integers modulo k). ◮Given X, Y ⊆V (G) and ⃗ F ⊆⃗ E, deﬁne ⃗ F(X, Y ) = {(e, x, y) ∈⃗ F : x ∈X, y ∈Y, x ̸= y}. ◮Given X, Y ⊆V (G) and f : ⃗ E →H, we write f(X, Y ) = X ⃗ e∈⃗ E(X,Y ) f(⃗ e). Flows and Circulations Deﬁnition 7.18 ◮A function f : ⃗ E →H is a circulation or H-circulation if (F1) f(e, x, y) = −f(e, y, x) for all (e, x, y) ∈⃗ E; and (F2) f(x, V (G)) = 0 for all x ∈V (G) (Kirchhoﬀ’s Law) ◮A function f : ⃗ E →H is nowhere-zero if f(⃗ e) ̸= 0 for all ⃗ e ∈⃗ E. ◮A nowhere-zero H-circulation is called an H-ﬂow. Note 7.19 ◮If f satisﬁes (F1), then f(X, X) = 0 for all X ⊆V . ◮If f satisﬁes (F2), then f(X, V ) = 0. ◮If f is a circulation, then f(X, X) = 0 for every X ⊆V . ◮If f is a circulation and e = xy is a cut-edge, then f(e, x, y) = 0. Flow polynomial Suppose \|H\| = x and let G be a graph. We want to ﬁnd the number FG(x) of H-ﬂows in G. If G has a cut-edge, then FG(x) = 0. If G has n loops and no other edges, then FG(x) = (x −1)n. Let e be a non-loop edge with endpoints u and v of G. Count the number of H-ﬂows in G/e. Those ﬂows can be partitioned into two sets: A, those that induce a ﬂow in G \ e, and B, those that do not. Those ﬂows in A cannot be extended to a ﬂow on G, whereas those in B can. So FG(x) = FG/e(x) −FG\e(x). Clearly, FG(x) is a polynomial, and is called the ﬂow polynomial of G. It follows: Corollary 7.20 If H and H′ are two ﬁnite abelian groups of equal order, then G has an H-ﬂow if and only if it has an H′-ﬂow. Deﬁnition 7.21 ◮A Z-ﬂow f such that 0 < \|f(⃗ e)\| < k for all ⃗ e ∈⃗ E is a k-ﬂow. ◮The ﬂow number of a graph G, denoted by ϕ(G), is the smallest k such that G has a k-ﬂow, or inﬁnite if no k-ﬂow exists. Theorem 7.22 (Tutte 1950) A graph admits a k-ﬂow if and only if it admits a Zk-ﬂow. Proof of ⇒only. Use the natural map i 7→i from Z to Zk. k-Flows for Small k Theorem 7.23 A graph has a 2-ﬂow if and only if all vertices have even degree. Proof. By Corollary 7.22, a graph has a 2-ﬂow if and only if it has a Z2-ﬂow, that is, the constant map ⃗ E →Z2 with value 1 satisﬁes (F2). This is the case if and only if every vertex degree is even. Theorem 7.24 (Tutte 1949) A cubic graph has a 3-ﬂow if and only if it is bipartite. Proof. Let G be a cubic graph. Suppose ﬁrst that G has a 3-ﬂow, and thus a Z3-ﬂow. We show that every cycle C = x0x1 . . . xlx0 has even length. Consider two consecutive edges of C: ei−1 = xi−1xi and ei = x1xi+1. If f(ei−1, xi−1, xi) = f(ei, xi, xi+1), then f could not satisfy (F2) at xi due to a non-zero value of the third edge at xi. Therefore f assigns 1 and 2 to the edges of C alternately, and so C must be even. Conversely, let G be bipartite with bipartition (X, Y ). Since G is cubic, the map ⃗ E →Z3 deﬁned by f(e, x, y) = 1 and (e, y, x) = 2 for all edges xy with x ∈X and y ∈Y is a Z3-ﬂow. Flow Number of Cliques Theorem 7.25 ϕ(Kn) =      2 if n is odd; 4 if n = 4; and 3 if n is even and exceeds 4. Proof. The case for n odd follows from Theorem 7.23, and n = 4 can be checked directly. We handle the remaining cases by induction. Note that K6 is the edge-disjoint union of G1, G2, and G3 where G1 ∼ = G2 ∼ = K3 and G3 ∼ = K3,3. Each of G1 and G2 has a 2-ﬂow, while K3,3 has a 3-ﬂow by Theorem 7.24. The union of these ﬂows is a 3-ﬂow on G. Proof, Continued Now let n be even and greater than 6, and assume that the assertion holds for n −2. Now G can be written as edge-disjoint union of Kn−2 and G′ = Kn−2 ∨K2. The Kn−2 has a 3-ﬂow by induction hypothesis. Therefore it suﬃces to ﬁnd a Z3-ﬂow on G′. Let x and y be the vertices of K2. Then each triangle xyz has a constant Z3-ﬂow. Adding all of those ﬂows produces a circulation on G′ that is non-zero, except possibly on xy. If that is the case, the multiply exactly one of the ﬂows by 2 before adding them all up. The result follows. 4-Flows Theorem 7.26 (i) A graph has a 4-ﬂow if and only if it is the union of two subgraphs whose vertices have all degrees even. (ii) A cubic graph has a 4-ﬂow if and only if it is 3-edge-colorable. Proof. Let H = Z2 × Z2. By Theorems 7.20 and 7.22, a graph has a 4-ﬂow if and only if it as an H-ﬂow. Now (i) follows immediately from Theorem 7.23. Assume a cubic graph G has an H-ﬂow f. It is easy to check that f gives a 3-edge-coloring. Conversely, since the non-zero elements of H sum up to 0, every proper 3-edge-coloring of G using colors H \ 0 deﬁnes an H-ﬂow on G. Flow-Coloring Duality Theorem 7.27 (Tait 1878) A simple 2-edge-connected 3-regular plane graph G is 3-edge-colorable if and only if its dual is 4-colorable. Proof. Suppose G is 4-face-colored with elements of Z2 × Z2. For each edge, assign to it the color that is the sum of the colors of the two incident faces. Then it is easy to check that this results in proper 3-edge-coloring. Conversely, suppose the edges of G can be colored with colors from Z2 × Z2 \ {(0, 0)}. Let H1 be the subgraph induced by the edges colored (1, 0) or (1, 1), and let H2 be the subgraph induced by the edges colored (0, 1) or (1, 1). Note that each of H1 and H2 is the disjoint union of cycles. To each face of G, assign the color (p1, p2) where pi is the parity (0 for even, 1 for odd) of the number of cycles that contain it inside. Problem Set 5 Problem 13 Prove that every loopless planar graph on fewer than thirteen vertices admits a proper 4-coloring. In other words, prove the Four-Color Theorem for graphs on at most twelve vertices. Problem 14 Show that every graph without a cut-edge admits a ﬂow. Problem 15 Show that if a graph has a spanning cycle, then it admits a 4-ﬂow. Tutte’s Flow Conjectures Conjecture 7.28 (Tutte) ◮(5-Flow Conjecture, 1954) Every graph with no cut-edge has a 5-ﬂow. ◮(4-Flow Conjecture, 1966) Every graph with no cut-edge and no Petersen graph minor has a 4-ﬂow. ◮(3-Flow Conjecture, 1972) Every graph with no edge-cuts of size 1 and 3 has a 3-ﬂow. Theorem 7.29 (Seymour 1981) Every graph with no cut-edge has a 6-ﬂow. Theorem 7.30 (Robertson, Sanders, Seymour, Thomas 2000) Every cubic graph with no cut-edge and no Petersen graph minor has a 4-ﬂow. Theorem 7.31 (Gr¨ otzsch 1959) Every planar graph with no edge-cuts of size 1 and 3 has a 3-ﬂow. Miscellany of Coloring Results Theorem 7.32 (Tutte 1954) χ(G) = ϕ(G∗) Proof. Conjecture 7.33 (Hadwiger 1943) If n is an integer exceeding 1, and G has no Kn-minor, then χ(G) < n. Note 7.34 ◮Trivial for n = 2. ◮Obvious for n = 3. ◮Easy for n = 4. ◮Equivalent to Four-Color Theorem for n = 5. ◮Proved by Robertson, Seymour, and Thomas for n = 6. ◮Unknown for n ≥7. Hamilton Cycles Deﬁnition 8.1 ◮A spanning subgraph that is a cycle or a path is called a Hamilton cycle or a Hamilton path. ◮A graph is Hamiltonian if it has a Hamilton cycle. Suﬃcient Condition for Hamiltonicity Theorem 8.2 (Dirac 1952) Every graph of order n ≥3 and δ ≥n/2 is Hamiltonian. Proof. Let G be a graph as described. Note that G is connected; otherwise a vertex in a smallest component would have degree less than n/2. Let P = x0x1 . . . xk be a longest path in G. By the maximality of P, all neighbors of x0 and all neighbors of xk lie on P. Hence at least n/2 of the vertices x0, x1, . . . , xk−1 are adjacent to xk, and at least n/2 of the same k < n vertices xi are such that x0xi+1 ∈E(G). By the Pigeon-Hole Principle, there is a vertex xi that has both properties, that is, x0xi+1 ∈E(G) and xixk ∈E(G). Let C be the cycle obtained from P by deleting the edge xixi+1 and adding edges x0xi+1 and xixk. If C is not Hamilton, then, since G is connected, C would have a neighbor in G −C, which would yield a path longer than P; a contradiction. Note on Dirac’s Theorem Note 8.3 Note that n/2 in Dirac’s Theorem 8.2 is the best possible. We cannot replace it with ⌊n/2⌋if n is odd, since then G which is a 1-sum of two copies of K⌈n/2⌉would have δ = ⌊n/2⌋, but no Hamilton cycle. Another Suﬃcient Condition Theorem 8.4 Every graph G with \|G\| ≥3 and κ(G) ≥α(G) is Hamiltonian. Proof. Let k = κ(G) and let C be a longest cycle in G. Enumerate the vertices of C cyclically so that V (C) = {vi : i ∈Zn} with vivi+1 ∈E(C) for all i ∈Zn. If C is not a Hamiltonian cycle, pick a vertex v not in C. Let F = {Pi : i ∈I} be a maximum-cardinality collection of vC-paths that pairwise meet only in v and so that Pi contains vi. Then vvj / ∈E(G) for every j / ∈I, and \|I\| ≥min{k, \|C\|} by Menger’s Theorem 5.29. For every i ∈I, we have i + 1 / ∈I, otherwise (C ∪Pi ∪Pi+1) \ vivi+1 would be a cycle longer than C. Thus \|I\| < \|C\| and hence \|I\| = \|F\| ≥k. Furthermore, vi+1vj+1 / ∈E(G) for all i, j ∈I, as otherwise (C ∪Pi ∪Pj ∪vi+1vj+1) \ vivi+1 \ vjvj+1 would be a cycle longer than C. Hence {vi+1 : i ∈I} ∪{v} is a set of at least k + 1 independent vertices in G, contradicting the assumption that α(G) ≤k. A Necessary Condition Theorem 8.5 If G is a Hamiltonian graph, then for every set ∅̸= S ⊆V (G), the graph G −S has at most S components. Proof. When leaving a component of G −S, a Hamilton cycle can go only to S and the arrivals in S must occur at diﬀerent vertices of S. Hence S must have at least as many vertices as G −S has components. Note that if we managed to prove that every 3-connected cubic plane graph is Hamiltonian, then we woud have proved that every such graph has a 4-ﬂow, and so is 3-edge-colorable, and so is 4-face-colorable. Unfortunately, there are 3-connected cubic plane graphs that are not Hamiltonian. Grinberg’s Theorem Theorem 8.6 (Grinberg 1968) If G is a loopless plane graph with a Hamilton cycle C, and G has f ′ i faces of length i inside C and f ′′ i faces of length i outside C, then P i(i −2)(f ′ i −f ′′ i ) = 0. Proof. Want to show that P i(i −2)f ′ i = P i(i −2)f ′′ i . It suﬃces to show that P i(i −2)f ′ i remains invariant as we add edges inside a cycle C of length n. If there are no edges inside C, then the sum is n −2. Suppose P i(i −2)f ′ i = n −2 for any graph with k edges inside C. We can obtain any graph with k + 1 edges inside C by adding an edge to such graph. The edge addition cuts a face of length r into faces of lengths s and t. We have s + t = r + 2, and so (s −2) + (t −2) = r −2 and so the total contribution remains the same. Corollary 8.7 The Tutte graph is not Hamiltonian. Theorem 8.8 (Tutte 1956) Every 4-connected planar graph is Hamiltonian. Theorem 8.9 (Thomas, Yu 1994) Every 4-connected projective graph is Hamiltonian. Theorem 8.10 (Thomas, Yu 1997) Every 5-connected toroidal graph is Hamiltonian. Suppose we have a ﬁnite family F of pairwise disjoint convex polygons (with interiors) in the plane with all sides of length 1. Suppose that all these polygons together have m sides σ1, σ2, . . . , σm where m is even. Orient arbitrarily each of the sides σi by choosing one of its endpoints as the initial point, and choose an arbitrary partition of the sides into pairs. From the disjoint union of polygons F, form a topological space S by identifying sides in each pair of our partition so that the orientations agree. If S is connected (which will be assumed from now on), we get a surface S, that is, a compact Hausdorﬀspace that is locally homeomorphic to the unit disk in the plane. The identiﬁed edges of the polygons can be viewed as edges of a graph, the endpoints of those edges as vertices, and the interiors of the polygons are faces. What results is an embedding of a connected graph into some surface, which will be called 2-cell embedding. Theorem 9.1 Every surface is homeomorphic to a triangulated surface. Proof omitted. Consider now two disjoint triangles T1 and T2 (such that all sides have same length) in a face F of a surface S with a 2-cell embedded graph G. We form a new surface S′ by deleting from F the interiors of T1 and T2, and identifying T1 with T2 such that their clockwise orientations (as deﬁned by F) disagree. We say that the surface S′ is obtained from S by adding a handle. If we identify T1 and T2 so that their orientations agree to obtain a surface S′′, then S′′ is obtained from S by adding a twisted handle. Finally, let T be a square in F. Let S′′′ be obtained from S by deleting the interior of T and identifying the diametrically opposite points of the square. Then S′′′ is obtained from S by adding a crosscap. It is easy to show that S′, S′′, and S′′′ are independent (up to homeomorphism) of where in F the triangles T1 and T2 and the square T come from, or whether they come from distinct faces. Consider the sphere S0 (tetrahedron). If we add h handles to S0, then we obtain Sh, which we call the orientable surface of genus h. If we add h crosscaps to S0, we obtain Nh, the non-orientable surface of genus h. The surfaces S1, S2, N1, and N2 are called, respectively torus, double torus, projective plane, and Klein bottle. A twisted handle can be always replaced with two crosscaps, and, as long as there is a crosscap, a handle can also be replaced by two crosscaps. Theorem 9.2 Let S be the surface obtained from the sphere by adding h handles, t twisted handles, and c crosscaps. If t = c = 0, then S = Sh. Otherwise, S = N2h+2t+c. Theorem 9.3 Let S be a surface and let G be a graph that is 2-cell embedded in S with v vertices, e edges and f faces. Then S is homeomorphic to either Sh or Nk, where v −e + f = 2 −2h = 2 −k. Euler’s Characteristic Deﬁnition 9.4 The Euler characteristic χ(S) of a surface S is deﬁned as χ(S) = ( 2 −2h, if S = Sh; 2 −k, if S = Nk. π-Walks Suppose G is a connected non-trivial graph. Suppose that for each v ∈V (G) we have a cyclic permutation πv of edges incident with v. Let’s consider a closed walk W = v1e1v2e2v3 . . . vkekv1, which is determined by the ﬁrst edge e1 = v1v2 and the requirement that for each i we have πvi(ei) = ei+1 where ek+1 = e1 and k is minimal with this property. Note, however, that some edges might occur in W twice, traversed in opposite directions. We will not distinguish W from its cyclic shifts. If π = {πv : v ∈V (G)} (the rotation system), then W is a π-walk. For each π-walk, take a polygon with as many sides as the length of the walk, disjoint from other polygons—call it a π-polygon. Now take all π-polygons. Each edge appears exactly twice in the π-walks, and this determines their orientation. By identifying each side with its mate we obtain a 2-cell embedding of graph isomorphic to G in some orientable surface. Theorem 9.5 Every cellular embedding (an embedding where each face is homeomorphic to an open disk) into an orientable surface is determined by its rotation system. Embedding Schemes An embedding scheme is a pair Π = (π, λ) where π is a rotation system, and λ : E(G) →{−1, 1} is a signature. Two embedding schemes are equivalent if one can be obtained from the other by a sequence of operations, each involving a change of clockwise to counter-clockwise orientation at a vertex v and the corresponding change of signatures of edges incident with v. Now we deﬁne a surface embedding using π-walks and π-polygons as before, with the following modiﬁcation. Whenever we traverse an edge of negative signature, we switch the permutation at a vertex from π(v) to π−1(v). Traversing a π-walk stops whenever we are about to traverse the same edge in the same direction and we are in the same mode (clockwise or counter-clockwise). The resulting surface is non-orientable if and only if G contains a cycle with an odd number of edges of negative signature. Theorem 9.6 Every cellular embedding of a graph in some surface is uniquely determined, up to homeomorphism, by its embedding scheme. Genus of a Graph Deﬁnition 9.7 The genus γ(G) and the non-orientable genus ˜ γ(G) of a graph G are the minimum h and the minimum k, respectively, such that G has an embedding into the surface Sh, respectively into Nk. An embedding into such surface of minimum genus is minimum genus, respectively minimum non-orientable genus, embedding. Theorem 9.8 Every minimum (orientable) genus embedding of a connected graph is cellular. Theorem 9.9 Let G be a connected graph. If ˜ γ(G) < 2γ(G) + 1, then every non-orientable minimum genus embedding of G is cellular. If ˜ γ(G) = 2γ(G) + 1 and G is not a tree, then G has both a cellular and a non-cellular embedding in N˜ γ(G). Cycle Double-Cover Conjecture Conjecture 9.10 (Cycle Double-Cover Conjecture) Every 2-edge-connected graph G can be expressed as a union of cycles so that every edge of G appears in exactly two cycles. Conjecture 9.11 Every 2-edge-connected graph has an embedding in some surface so that every face with the boundary is homeomorphic to the closed unit disk. Holds for 4-connected graphs. Bounds on Genus Theorem 9.12 Let G be a simple connected graph with v vertices (v ≥3) and e edges. Then γ ≥ l e 6 −v 2 + 1 m and ˜ γ ≥ l e 3 −v + 2 m . Proof. Let f denote the number of facial walks. The sum of lengths of those walks is 2e. Since the graph is simple, 2e ≥3f. By Euler’s Formula, 3χ = 3v −3e + 3f ≤3v −e, and so χ ≤v −e/3. Note that γ = 1 −χ/2 and ˜ γ = 2 −χ. In the orientable case, we have γ ≥1 −v −e/3 2 = e 6 −v 2 + 1. Similarly, in the nonorientable case, we have ˜ γ ≥2 −(v −e 3) = e 3 −v + 2. The theorem follows now from the fact that both γ and ˜ γ are integres. Genera of Complete Graphs Corollary 9.13 γ(Kn) ≥ (n −3)(n −4) 12 and ˜ γ(Kn) ≥ (n −3)(n −4) 6 Heawood’s Theorem Theorem 9.14 (Ringel, Youngs) If n ≥3 and n ̸= 7, then γ(Kn) = (n −3)(n −4) 12 and ˜ γ(Kn) = (n −3)(n −4) 6 Prof of Heawood’s Formula Theorem 9.15 (Heawood’s Formula) Let S be a surface with Euler genus g = 2 −(v −e + f) > 0 and let G be a loopless graph embedded in S. Then χ(G) ≤ 7 + √1 + 24g 2 . Proof. Suppose the theorem fails and G is a minimal counter-example. Let c = χ(G), and note that since g > 0 and G is a counter-example, we have c ≥7. From Theorem 9.12, we have e ≤3v −6 + 3g, and so 2e ≤6v −12 + 6g. The minimality of G implies that δ(G) ≥c −1. Clearly, δv ≤2e, and since c −1 ≤δ, as implied by the minimality of G, we have (c −1)v ≤2e. Combining this with an earlier inequality, we get (c −1)v ≤6v −12 + 6g, and so (c −7)v + 12 −6g ≤0. Since (c −7) ≥0 and c ≤v, we have c2 −7c + 12 −6g ≤0. Using the quadratic formula ﬁnishes the proof. Theorem 9.16 (Ringel–Youngs) The bound in Heawood’s Formula is the best possible, except that maximum chromatic number of graphs embedded in the Klein bottle is 6. Homework Problem 16 Prove that for every number n there is a bipartite graph whose choosability number is greater than n. Problem 17 Find the (orientable) genus of the Petersen graph. Problem 18 Does K5 have cellular embeddings into two diﬀerent orientable surfaces? Into two diﬀerent non-orientable surfaces? Well-Quasi-Ordering Deﬁnition 10.1 ◮A relation is a quasi-ordering if it is reﬂexive and transitive. ◮A quasi-ordering ≼on X is a well-quasi-ordering, or a wqo, if for every inﬁnite sequence x0, x1, . . . in X there are indices i < j such that xi ≼xj. ◮Then (xi, xj) is a good pair for the sequence. ◮An inﬁnite sequence containing a good pair is good; otherwise it is bad. Antichains and Decreasing Sequences Theorem 10.2 A quasi ordering ≼on X is a wqo if and only if X contains neither an inﬁnite antichain nor an inﬁnite strictly descending chain x0 ≻x1 ≻. . . . Proof. The forward implication is obvious. Conversely, let x = x0, x1, . . . be a bad sequence in X. We deﬁne inﬁnite subsequences y, x0, x1 . . . of x, and two sets A and B as follows. We start with y as the empty sequence, A and B as empty sets, and x0 = x. Suppose that for some i = 0, 1, . . . , the ﬁrst i elements of y and the sequence xi have been deﬁned, and each of the ﬁrst i elements of y has been placed in exactly one of A and B. Consider the ﬁrst element x of xi. ◮If inﬁnitely many elements of xi are incomparable with x, make xi+1 be an inﬁnite subsequence of xi consisting of the elements incomparable with x, and put x into A. ◮If this doesn’t happen, there are inﬁnitely many elements of xi smaller than x. In that case, let xi+1 be an inﬁnite subsequence of xi consisting of the elements x′ such that x ≻x′ and put x into B. Proof, Continued After this inductive construction, at least one of the sets A and B is inﬁnite. If A is inﬁnite, its elements form an antichain. If B is inﬁnite, the sequence y restricted to B is an inﬁnite strictly descending sequence. Theorem 10.3 If X is a wqo, then every inﬁnite sequence in X has an inﬁnite increasing subsequence. Proof: Exercise. Ordering of Sequences Deﬁnition 10.4 ◮The set of all ﬁnite sequences of elements of a set X will be denoted by X<ω. ◮The set of all ﬁnite subsets of elements of a set X will be denoted by [X]<ω. ◮The relation ≼on X is extended to X<ω as follows: If x = (x1, x2, . . . , xm) and y = (y1, y2, . . . , yn), we write x ≼y whenever there is a strictly increasing function f : {1, 2, . . . , m} →{1, 2, . . . , n} such that xi ≼yf(i) for all i ∈{1, 2, . . . , m}. ◮The relation ≼on X is extended to [X]<ω by saying that two elements x and y of [X]<ω satisfy x ≼y whenever there is a injection f from x to y such that x ≼f(x) for all x ∈x. ◮A quasi-order X is well-founded if it has no inﬁnite strictly descending chains. Higman’s Theorem Theorem 10.5 (Higman) If X is well-quasi-ordered by ≼, then so is X<ω. Proof. Suppose X<ω is not a wqo. Observe that X<ω is well-founded. We construct a minimal bad sequence (xn)n∈N in X<ω. Given n ∈N, assume inductively that xi has been deﬁned for every i < n and that there is a bad sequence in X<ω starting with x0, x1, . . . , xn−1. This is clearly true for n = 0. Choose xn ∈X<ω so that some bad sequence starts with x0, x1, . . . , xn and xn is minimal with this property (exists as X<ω is well-founded). Proof of Higman’s Theorem, Continued Clearly, (xn)n∈N is a bad sequence. For each n, let yn be xn with the last element xn deleted. By Theorem 10.3, the sequence (xn)n∈N has an inﬁnite increasing subsequence (xni)i∈N. By the minimality of xn0, the sequence x0, x1, . . . , xn0−1, yn0, yn1, . . . is good, so it has a good pair. This good pair must be of the form (yi, yj). Extending the injection yi 7→yj by xi 7→xj, we get a good pair (xi, xj); a contradiction. Corollary 10.6 If X is well-quasi-ordered by ≼, then so is [X]<ω. Note 10.7 (Rado) Higman’s Theorem does not extend to inﬁnite sequences. Ordering Trees Consider two trees T and T ′ with roots, respectively r and r′. Note that the root r induces a natural partial order on the vertices of the tree T . Speciﬁcally, u ⩽T v if u lies on the rv-path. We write T ≼T ′ if there is an isomorphism ϕ from some subdivision S of T to a subtree of S′ of T ′ that preserves the tree order, that is, u ⩽S v if and only if ϕ(u) ⩽S′ ϕ(v). Note that if T ≼T ′, then T is a topological minor of T ′. Theorem 10.8 (Kruskal 1960) Trees are well-quasi-ordered by the topological minors relation. Proof of Kruskal’s Theorem We show that rooted trees are well-quasi-ordered by ≼. Suppose not. Let T0, T1, . . . be a minimal bad sequence, with ri being the root of Ti. For each n, let An denote the set of components of Tn −rn, made into rooted trees by choosing the neighbors of rn as the roots. First, we show that A = S An is a wqo. Let (Sk)k∈N be a sequence of elements of A. For each k, let n = n(k) denote the An that contains Sk. Pick a k with the smallest n(k). Then T0, T1, . . . , Tn(k)−1, Sk, Sk+1, . . . is a good sequence, by the minimality of (Tn). Clearly, a good pair of that sequence must be of the form (Si, Sj). Proof of Kruskal’s Theorem, Continued By Corollary 10.6, the sequence (An)n∈N has a good pair (Ai, Aj). Let f : Ai →Aj be an injection such that T ≼f(T ) for all T ∈Ai. We extend the union of those embeddings to a map ϕ from V (Ti) to V (Tj) by letting ϕ(ri) = rj. The map ϕ is an embedding that preserves the tree order, proving that (Ti, Tj) is a good pair; a contradiction. Tree-Decomposition Let G be a graph, T be a tree, and let V = {Vt}t∈V (T ) be a family of vertex sets Vt ⊆V (G) (called bags). Deﬁnition 10.9 The pair (T, V) is called a tree-decomposition of G if (T1) V (G) = S t∈V (T ) Vt; (T2) For every edge e of G, there is a t ∈V (T ) such that both endpoints of e are in Vt; and (T3) Vr ∩Vt ⊆Vs whenever s lies between r and t in T . (T3) (Alternate version) For every v ∈V (G), the subgraph Tv induced by those t for which v ∈Vt is connected. The width of the decomposition (T, V) is the maximum of \|Vt\| −1 taken over all v ∈V (T ). The tree-width of G, denoted by tw(G) is the minimum width over all possible tree-decomposisions. Properties of Tree-Decompositions Theorem 10.10 If H is a subgraph of G, and (T, {Vt}t∈V (T )) is a tree-decomposition of G, then (T, {Vt ∩V (H)}t∈V (T )) is a tree-decomposition of H. Proof is very easy. Lemma 10.11 Let t1t2 be an edge of T , and let T1 and T2 be the components of T \ t1t2, with t1 ∈V (T1) and t2 ∈V (T2). Then Vt1 ∩Vt2 separates U1 = S t∈V (T1) Vt from U2 = S t∈V (T2) Vt. Proof. Both t1 and t2 lie on every s1s2-path in T with s1 ∈V (T1) and s2 ∈V (T2). Therefore U1 ∩U2 ⊆Vt1 ∩Vt2 by (T3). What is left to show is that G has no edge u1u2 with u1 ∈U1 −U2 and u2 ∈U2 −U1. If G has such an edge u1u2, then (T2) implies that there is a t ∈V (T ) such that Vt contains both u1 and u2. But t can be in neither T1 nor in T2; a contradiction. Lemma 10.12 For every W ⊆V (G), one of the following holds: (i) there is a t ∈V (T ) such that W ⊆Vt; (ii) there are vertices w1 and w2 in W and an edge t1t2 of T such that w1 and w2 lie outside of Vt1 ∩Vt2 and are separated by it in G. Proof. We orient the edges of T as follows. For each edge t1t2 of T , deﬁne U1 and U2 as in Lemma 10.11; then Vt1 ∩Vt2 separates U1 from U2. If Vt1 ∩Vt2 does not separate any two vertices of W , then W ⊆Ui for some i ∈{1, 2}; we orient t1t2 towards that ti. Let t be the last vertex of a maximal directed path in T ; we claim that W ⊆Vt. Suppose w ∈W and let t′ ∈V (T ) be such that w ∈Vt′. If t′ ̸= t, then the edge e at t that separates t from t′ is directed towards t, so w also lies in Vt′′ for some t′′ in the component of T \ e containing t. Therefore w ∈Vt by (T3). Corollary 10.13 If H is a complete subgraph of G, and (T, {Vt}t∈V (T )) is a tree-decomposition of G, then there is a bag Vt that contains all vertices of H. Theorem 10.14 If G is a minor of H, then tw(G) ≤tw(H). Proof. If G is obtained from H by deleting an edge, then a tree-decomposition of H is also a tree-decomposition of G. If G is obtained from H by deleting a vertex, then a tree-decomposition of H may be modiﬁed by removing the vertex from all bags to form a tree-decomposition of G. If G is obtained from H by ontracting an edge uv to a new vertex w, then a tree-decomposition of H may be modiﬁed by replacing each occurence by u and v by w to form a tree-decomposition of G. Corollary 10.15 For every integer k, the class of graphs of tree-width at most k is closed under the taking of minors. Theorem 10.16 (Robertson-Seymour 1990) For every positive integer k, the graphs of tree-width less than k are well-quasi-ordered by the minor relation. Theorem 10.17 ◮tw(G) < 2 if and only if K3 is not a minor of G. ◮tw(G) < 3 if and only if K4 is not a minor of G. Brambles Deﬁnition 10.18 ◮Two subsets U and W of V (G) touch if a vertex of U is in W or is a neighbor of a vertex in W . ◮A set of mutually touching connected vertex sets in G is a bramble. ◮A subset of V (G) covers a bramble if it meets each of its elements. ◮The smallest number of vertices that cover a bramble is the order of the bramble. Lemma 10.19 Any set of vertices separating two covers of a bramble also covers that bramble. Proof. Since each set in a bramble is connected and meets both of the covers, it also meets any set separating these covers. Tree-width vs. Bramble Theorem 10.20 (Seymour-Thomas 1993) Let k be a non-negative integer. tw(G) ≥k if and only if G contains a bramble of order greater than k. Let B be a bramble in G. We show that every tree-decomposition (T, (Vt)t∈V (T )) of G has a bag that covers B. Orient the edges t1t2 as in the proof of Lemma 10.12. If X = Vt1 ∩Vt2 covers B, the conclusion holds. If not, then for each B ∈B disjoint from X there is an i ∈{1, 2} such that B ⊆Ui \ X. This i is the same for all such B, because they touch. Orient the edge t1t2 towards ti. Then if t is the last vertex of a maximal directed path in T , then Vt covers B. Theorem 10.21 The tree-width of an n × n grid (n > 1) is n. Theorem 10.22 (Robertson-Seymour 1986) For every integer r there is an integer k such that every graph of tree-width at least k has an r × r grid minor. Theorem 10.23 Every planar graph is a minor of a suﬃciently large grid. Theorem 10.24 (Robertson-Seymour) Planar graphs are well-quasi-ordered by the minor relation. Representativity Deﬁnition 10.25 Suppose G is a graph embedded in a surface S. The representativity of G is the smallest number of points that a homotopically non-trivial cruve in S intersects the graph. The S-representativity of an abstract graph H is the smallest representativity of all embbedings of H in S, or zero if no embedding exists. Theorem 10.26 (Robertson-Seymour) Every graph embeddable on a surface S is a minor of a graph of suﬃciently high S-representativity. Theorem 10.27 For every surface S (orientable or not), the graphs embeddable in S are well-quasi-ordered by the minors relation. Graphs Almost Embedded on Surfaces Let r, s, t, and u be non-negative integers. Let H(r, s, t, u) is the class of graphs G such that after deleting at most r vortices each of depth at most s from G, and after deleting at most t vertices, the resulting graph has Euler genus at most u. Let G(r, s, t, u) be the class of graphs that are obtained by repeated clique-summing graphs from H(r, s, t, u). Theorem 10.28 (Robertson-Seymour) The class G(r, s, t, u) is well-quasi-ordered by the minor relation. Graph Minors Theorem Theorem 10.29 (Robertson-Seymour) For every integer k there are integers r, s, t, and u such that every graph without Kk-minor belongs to G(r, s, t, u). Corollary 10.30 Every minor-closed class of graphs other than the class of all graphs is a subclass of some G(r, s, t, u). Corollary 10.31 The class of all (ﬁnite) graphs is well-quasi-ordered by the minor relation. Problem 19 For each integer n exceeding one, ﬁnd a bramble of order n + 1 in the n × n grid. Problem 20 A tree T is a caterpillar if T contains a path P such that every vertex of T either lies on P or is adjacent to a vertex of P. A caterpillar forest is a disjoint union of caterpillars. Find the minor-minimal graphs that are not caterpillar forests. Problem 21 What is the tree-width of the graph obtained from the Petersen graph by deleting one edge? The Tur´ an Graph Question: Given a graph H, what is the greatest possible number of edges in a simple graph of order n that does not have H as a subgraph? We will answer this question when H is a complete graph. Deﬁnition 11.1 The unique complete r-partite graph on n ≥r vertices whose partition sets diﬀer by at most 1 is called the Tur´ an graph T r(n). The size of T r(n) will be denoted by tr(n). Theorem 11.2 (Tur´ an 1941) Given integers r and n exceeding 1, the unique simple graph of order n without Kr as a subgraph of maximum possible size is T r−1(n). Proof of Tur´ an’s Theorem First, observe that among all simple k-partite (k < r) graphs on n vertices, T r−1(n) has the largest size. Observe also that T r−1(n) does not have Kr as a subgraph, but T r(n) does. Thus it suﬃces to show that a simple graph G of order n, the maximum size without Kr as a subgraph is complete multipartite. By duplicating a vertex v, we mean adding a new vertex v′ and joining it to all neighbors of v (but not v itself). If not, then non-adjacency is not an equivalence relation on V (G), that is, there are vertices y1, x, and y2 such that y1x and xy2 do not form edges of G, but y1y2 does. If d(y1) > d(x), then deleting x and duplicating y1 yields another Kr-free graph with more edges than G. So d(y1) ≤d(x) and d(y2) ≤d(x). But then deleting y1 and y2 and duplicating x twice gives a Kr-free graph with more edges than G; a contradiction. Erd˝ os-Stone Theorem Corollary 11.3 If G is a simple graph of order n and size more than tr−1(n), then G contains Kr as a subgraph. Theorem 11.4 (Erd˝ os-Stone, 1946) For all integers r ≥2 and s ≥1, and every ǫ > 0, there is an integer n0 such that every simple graph of order n ≥n0 and size at least tr−1(n) + ǫn2 contains the complete r-partite graph with each part of cardinality s. Proof omitted. Deﬁnition 11.5 Given a simple graph H and an integer n, let hn(H) denote the maximum edge density that a simple H-free graph of order n can have; that is, the maximum number of edges that a simple H-free graph of order n can have divided by n 2 . Lemma 11.6 lim n→∞hn(Kr) = r −2 r −1. Corollary of Erd˝ os-Stone Theorem Corollary 11.7 For every simple, non-trivial graph H, lim n→∞hn(H) = χ(H) −2 χ(H) −1. Proof. Let r = χ(H). Then H is not a subgraph of T r−1(n) for all n, and so hn(Kr) ≤hn(H). On the other hand, if Ks r denotes the complete r-partite graph on rs vertices with every part of cardinality s, then hn(H) ≤hn(Ks r) for all suﬃciently large s. Moreover, Erd˝ os-Stone Theorem 11.4 implies that for every ǫ > 0 and n large enough hn(Ks r) < hn(Kr) + ǫn2 n 2 . Hence for large n, we have hn(Kr) ≤hn(H) ≤hn(Ks r) < hn(Kr)+ǫn2 n 2 = hn(Kr)+ 2ǫ 1 −1/n ≤hn(Kr)+4ǫ. Then Lemma 11.6 ﬁnishes the proof. Ramsey Theorem Theorem 11.8 (Ramsey 1930) For every natural number r there is a natural number n such that every simple graph of order at least n contains either Kr or Kr as an induced subgraph. Proof. Trivial for r = 1; assume r ≥2. Let n = 22r−3, and let G be a simple graph of order at least n. We will deﬁne a sequence V1, V2, . . . , V2r−2 of sets and choose vertices vi ∈Vi so that the following hold: (i) \|Vi\| = 22r−2−i for all i ∈{1, 2, . . . , 2r −2}; (ii) Vi ⊆Vi−1 \ {vi−1} for all i ∈{1, 2, . . . , 2r −2}; and (iii) vi−1 is adjacent either to all vertices in Vi or to no vertex in Vi for all i ∈{1, 2, . . . , 2r −2}. Let V1 be any set of 22r−3 vertices of and pick v1 ∈V1 arbitrarily. Then (i)–(iii) hold trivially. Inductively, \|Vi−1 \ {vi−1}\| = 22r−1−i −1, and Vi−1 \ {vi−1} contains a subset Vi satisfying (i)–(iii); pick vi arbitrarily in Vi. Among the vertices v1, v2 . . . , v2r−3, at least r −1 show the same behavior described in (iii). Those r −1 vertices together with v2r−2 induce either Kr or Kr. Inﬁnite version of Ramsey’s Theorem Recall that [X]k denotes the set of k-elements subsets of a set X. Given a c-coloring, that is, partitioning into c classes, of elements of [X]k, we call a set Y ⊆X monochromatic if all elements of [Y ]k receive the same color. Then Ramsey’s Theorem can be re-stated as: For every r there is an n such that if X is an n-element set and [X]2 is 2-colored, then X has a monochromatic subset of cardinality r. Theorem 11.9 Let k and c be positive integers, and let X be an inﬁnite set. If [X]k is c-colored, then X has an inﬁnite monochromatic subset. Proof We proceed by induction on k. If k = 1, then the claim clearly holds. Let k > 0 and assume that the theorem holds for all smaller values of k. Let [X]k be colored with c colors. We will construct an inﬁnite sequence X0, X1, . . . of inﬁnite subsets of X and choose elements xi ∈Xi such that (for all i): (i) Xi+1 ⊆Xi \ {xi}; and (ii) all k-element sets of the form {xi} ∪Z where Z ⊆[X]k−1 have the same color, which we associate with xi. Start with X0 = X and pick x0 ∈X0 arbitrarily. Having chosen Xi and xi ∈Xi, we c-color [Xi \ {xi}]k−1 by giving each set Z the color of {xi} ∪Z in our c-coloring of [X]k. By the induction hypothesis, Xi \ {xi} has an inﬁnite monochromatic subset, which we choose as Xk+1. Pick xk+1 ∈Xk+1 arbitrarily. Since c is ﬁnite, one of the colors is associated with inﬁnitely many xi—they form an inﬁnite monochromatic subset of X. Theorem 11.10 (K¨ onig Inﬁnity Lemma) Let V0, V1, . . . be an inﬁnite sequence of disjoint non-empty ﬁnite sets, and let G be an inﬁnite graph on their union. Assume that every vertex v in Vn, for n ≥1, has a neighbor f(v) in Vn−1. Then G contains a ray, that is a one-way-inﬁnite path, v0v1 . . . with vn ∈Vn for all n. Proof. Let P be the set of all ﬁnite paths of the form vf(v)f(f(v)) . . . . Since V0 is ﬁnite, but P is inﬁnite, inﬁnitely many of the paths in P begin at the same vertex v0. On these inﬁnitely many paths, inﬁnitely many agree on v1 ∈V1, because V1 is ﬁnite. This gives rise to the inductive deﬁnition of vn for every n so that v0v1 . . . form a ray. Theorem 11.11 For all positive integers k, c, and r there is an integer n ≥k such that every n-element set X has a monochromatic r-element subset with respect to any c-coloring of [X]k. Proof: To simplify notation, we will also use n denote the set {0, 1, . . . , n −1}. Suppose the theorem fails for some k, c, and r. Then for every n ≥k there is a c-coloring of [n]k such that n contains no monochromatic r-element subset. We will call such colorings bad. For every n ≥k, let Vn be the (nonempty) set of bad colorings of [n]k. For n > k, the restriction f(g) of any g ∈Vn to [n −1]k is still bad, and so lies in Vn−1. By K¨ onig Inﬁnity Lemma 11.10, there is an inﬁnite sequence gk, gk+1, . . . of bad colorings gn ∈Vn such that f(gn) = gn−1 for all n > k. For every m ≥k, all colorings gn with n ≥m agree on [m]k, so for each Y ∈[N]k the value gn(Y ) coincides for all n > max Y . We deﬁne g(Y ) as this common value gn(Y ). Then g is a bad coloring of [N] since every r-element subset S of N is contained in some suﬃciently large [n], and so S cannot be monochromatic since g coincides on [n]k with the bad coloring gn. This contradicts 11.9. Ramsey numbers Deﬁnition 11.12 The least integer n associated with k, c, and r as in Theorem 11.11 is the Ramsey number for k, c and r, and is denoted by R(k, c, r). We will also use the notation R(H1, H2) to denote the least order n such that H1 is a subgraph of G or H2 is a subgraph of G for every graph G of order n. If H1 = H2, then R(H1, H2) may be written as R(H1). We proved before that R(2, 2, 3) = 6, and that R(K3, K3) = 6. In most cases the exact Ramsey numbers are not known. Most known values and bounds are listed at Theorem 11.13 Let s and t be positive integers, and let T be a tree of order t. Then R(T, Ks) = (s −1)(t −1) + 1. Proof. The disjoint union of s −1 copies of Kt−1 contains no copy of T , while the complement of this graph, the complete s −1-partite graph Kt−1 s−1, does not contain Ks. Thus R(T, Ks) ≥(s −1)(t −1) + 1. Conversely, suppose that G is any graph of order n = (s −1)(t −1) + 1 whose complement contains no copy of Ks. Then s > 1 and in any proper vertex coloring of G, at most s −1 vertices can get the same color. Hence χ(G) ≥⌈n/(s −1)⌉= t, and so G has a subgraph H with δ(H) ≥t −1 (greedy coloring). Then H contains T as a subgraph. Theorem 11.14 (Chvat´ al, R¨ odl, Szemer´ edi and Trotter 1983) For every positive integer ∆there is a constant c such that R(H) ≤c\|H\| for all graphs H with ∆(H) ≤∆. Proof omitted—uses the Regularity Lemma. Ramsey Graphs Ramsey’s Theorem can be restated as follows: For every graph H = Kr there is a graph G such that every 2-coloring of the edges of G gives an induced monochromatic subgraph isomorphic to H. (In fact, a suﬃciently large complete graph will work for G.) Question: Given an arbitrary graph H, is there a graph G such that every 2-coloring of G gives an induced monochromatic subgraph isomorphic to H? Theorem 11.15 (Deuber; Erd˝ os, Hajnal, P´ osa; R¨ odl 1973) Every graph has a Ramsey graph. For every graph H there is a graph G such that, for every partition {E1, E2} of E(G), has an induced subgraph H with E(H) ⊆E1 or E(H) ⊆E2. Proof Given two graphs G = (V, E) and H, and U ⊆V , we write G[U →H] to denote the graph obtained from G by replacing each vertex u in U by a copy H(u) of H, joining H(u) completely to H(u′) whenever uu′ ∈E, and joining each H(u) to v whenever uv ∈E and v ∈V \ U. We will prove the following strengthening of the theorem () For any two graphs H1 and H2, there is a graph G = G(H1, H2) such that every edge-coloring of G with colors 1 and 2 yields either an induced H1 ⊆G with all edges colored 1, or an induced H2 ⊆G with all edges colored 2. We proceed by induction on \|H1\| + \|H2\|. If either H1 or H2 has no edges, in particular, \|H1\| + \|H2\| ≤1, then () holds with G = Kn for suﬃciently large n. For the induction step, assume that both H1 and H2 have at least one edge, and that () holds for all pairs (H′ 1, H′ 2) with smaller \|H′ 1\| + \|H′ 2\|. For each i ∈{1, 2}, pick a vertex xi ∈Hi that is incident with an edge, let H′ i = Hi −xi, and let H′′ i be the subgraph of Hi induced by the neighbors of xi. We will construct a sequence G0, G1, . . . , Gn of disjoint graphs with Gn being the desired Ramsey graph G(H1, H2). We will also deﬁne subsets Vi ⊆V (Gi) and a map f : V 1 ∪V 2 ∪· · · ∪V n →V 0 ∪V 1 ∪· · · ∪V n−1 such that f(V i) = V i−1 for all i ≥1. We will also write f i for composing f with itself i times, with f 0 understood as the identity map on V (G0). Note that f i(v) ∈V 0 for all v ∈V i; we call f i(v) the origin of v. Vertices in V i with diﬀerent origins are adjacent in Gi if and only if their origins are adjacent in G0. By the induction hypothesis, there are Ramsey graphs G1 = G(H1, H′ 2) and G2 = G(H′ 1, H2). Let G0 be a copy of G1, and let V 0 = V (G0). Let W ′ 0, W ′ 1, . . . , W ′ n−1 be the subsets of V 0 spanning an H′ 2 in G0. Thus n is deﬁned as the number of induced copies of H′ 2 in G0. For i ∈{0, 1, . . . , n −1}, let W ′′ i be the image of V (H′′ 2 ) under some isomorphism H′ 2 →G0[W ′ i]. Assume now that G0, G1, . . . , Gi−1 and V 0, V 1, . . . , V i−1 have been deﬁned for i ≥1 and that f has been deﬁned on V 1 ∪. . . ∪V i−1 as described above. We construct Gi from Gi−1 in two steps. For the ﬁrst step, consider the set U i of all the vertices v ∈V i−1 whose origin f i−1(v) lies in W ′′ i−1. (U 0 = W ′′ 0 ) Expand Gi−1 to a new graph ˜ Gi−1 by replacing every vertex u ∈U i−1 with a copy of G2(u) of G2, that is, let ˜ Gi−1 = Gi−1[U i−1 →G2]. Set f(u′) = u for all u ∈U i−1 and u′ ∈G2(u), and f(v′) = v for all v′ = (v, ∅) with v ∈V i−1 \ U i−1. ((v, ∅) is the unexpanded copy of a vertex v ∈Gi−1 in ˜ Gi−1) For the second step, let F denote the set of all families F of the form F = (H′(u)\|u ∈U i−1), where each H′ 1(u) is an induced subgraph of G2(u). For each F in F, add new vertex x(F) to ˜ Gi−1 and join it, for every u ∈U i−1, to all the vertices in the image H′′ 1 (u) ⊆H′ 1(u) under some isomorphism from H′ 1 o the H′ 1(u) ⊆G2(u) selected by F. Denote the resulting graph by Gi. Now we show that Gn satisﬁes (). We prove the following: () For every edge coloring with colors 1 and 2, the graph Gi contains either and induced H1 colored 1, or an induced H2 colored 2, or an induced graph colored 2 such that V (H) ⊆V i and the restriction of f i to V (H) is an isomorphism between H and G0[W ′ k] for some k ∈{i, i + 1, . . . , n −1}. For i = 0, () follows from the choice of G0 as a copy of G1 = G(H1, H′ 2). Now let 1 ≤i ≤n, and assume that () holds for all smaller values of i. Let an edge coloring of Gi be given. For each u ∈U i−1 there is a copy of G2 in Gi: Gi ⊇G2(u) ∼ = G(H′ 1, H2). If some G2(u) contains an induced H2 colored 2, then the conclusion holds. If not, then every G2(u) has an induced subgraph H′ 1(u) ∼ = H′ 1 colored 1. Let F be the family of these graphs H′ 1(u), one for each u ∈U i−1 and let x = x(F). If, for some u ∈U i−1, all the x −H′′ 1 (u) edges in Gi are also colored 1, then we have an induced copy of H1 in Gi and again the conclusion holds. So we may assume that each H′′ 1 (u) has a vertex yu for which the edge xyu is colored 2. The restriction yu 7→u of f to ˆ U i−1 = {yu\|u ∈U i−1} ⊆V i extends by (v, ∅) 7→v to an isomorphism from ˆ Gi−1 = Gi[ ˆ U i−1 ∪{(v, ∅)\|v ∈V (Gi−1) \ U i−1}] to Gi−1, and so our edge-coloring of Gi induces an edge-coloring of Gi−1. If this coloring gives an induced H1 ⊆Gi−1 colored 1, or an induced H2 ⊆Gi−1 colored 2, we have these also in ˆ Gi−1 ⊆Gi and again the conclusion holds. By () for i −1 we may then assume that Gi−1 has an induced H′ colored 2 with V (H′) ⊆V i−1 and such that the restriction of f i−1 to V (H′) is an isomorphism from H′ to G0[W ′ k] ∼ = H′ 2 for some k ∈{i −1, . . . , n −1}. Let ˆ H′ be the corresponding induced subgraph of ˆ Gi−1 ⊆Gi (also colored 2). then V ( ˆ H′) ⊆Vi, f i(V ( ˆ H′)) = f i−1(V (H′)) = W ′ k, and F i : ˆ H′ →G0[W ′ k] is an isomorphism. If k ≥i, then the proof of () is complete with H = ˆ H′. We thus assume that k < i, and so k = i −1. By deﬁnition of U i−1 and ˆ Gi−1, the inverse image of W ′′ i−1 under isomorphism f i : ˆ H′ →G0[W ′ i−1] is a subset of U i−1. Since x is adjacent to those vertices that lie in ˆ U i−1 and all those edges are colored 2, the graph ˆ H′ and x together induce in Gi a copy of H2 colored 2. Problem 22 Prove that for every positive integer k there is an integer N such that if G is a 2-connected graph of order at least N, then G has a subdivision of Ck or K2,k. Find an upper bound on N it terms of k. Problem 23 Find a Ramsey graph for C4, that is, ﬁnd a graph G such that if the edges of G are partitioned into {E1, E2}, then G has a induced subgraph isomorphic to C4 all of whose edges belong to one of E1 or E2. Ramsey Theorem for Connected Graphs Theorem 11.16 For every positive integer k there is an integer N such that if G is a connected graph of order at least N, then G contains Pk or K1,k as a subgraph. Theorem 11.17 For every positive integer k there is an integer N such that if G is a 2-connected graph of order at least N, then G contains Ck or K2,k as topological minors. Theorem 11.18 For every positive integer k there is an integer N such that if G is a 3-connected graph of order at least N, then G contains a topological minor Wk (wheel) or Vk (wheel with center replaced by a path) or K3,k. Theorem 11.19 For every positive integer k there is an integer N such that if G is an 4-connected graph of order at least N, then G contains as a topological minor DWk (double wheel) or Zk (zig-zag ladder) or Mk (M¨ obius ladder) or K4,k. Parallel Minors Recall that a graph is a topological minor of another if it can be obtained by ◮deleting edges or isolated vertices ◮contracting edges in series that are in series with another edge Topological minors are also called series minors A graph is a parallel minor of another if it can be obtained by ◮contracting edges ◮deleting edges that are in parallel with other edges (simplifying) Unavoidable Parallel Minors Theorem 11.20 For every positive integer k there is an integer N such that if G is a connected graph of order at least N, then G contains as a parallel minor Pk or K1,k or Kk or Ck. Theorem 11.21 For every positive integer k there is an integer N such that if G is a 2-connected graph of order at least N, then G contains as a parallel minor K′ 2,k (K2,k plus an edge) or Fk (fan) or Kk or Ck. Theorem 11.22 For every positive integer k there is an integer N such that if G is a 3-connected graph of order at least N, then G contains as a parallel minor K′ 3,k (K2,k plus a triangle) or DFk (double fan) or Kk or Wk. Theorem 11.23 For every positive integer k there is an integer N such that if G is a 4-connected graph of order at least N, then G contains as a parallel minor K′ 4,k or DWk or DW ′ k (double wheel with an axel) or T Fk (triple fan) or Zk or Mk or Kk. Partitioning Planar Graphs Conjecture 11.24 (Chartrand, Geller, Hedetniemi) Every planar graph is a union of two outerplanar graphs. “Proof”: “Proof”: ◮WLOG G is a triangulation ◮G is 4-connected ⇒Hamilton cycle C ⇒(C+ edges inside C) ∪(C+ edges outside C) ◮min counter-example G not 4-connected ⇒G is 0-, 1-, 2-, or 3-sum of A and B ⇒decompose each A and B and make parts “ﬁt together” FAIL! ◮later, proved by true by Gon¸ calves Theorem 11.25 (Ding, O., Sanders, Vertigan; Kedlaya) Every planar graph is a union of two series-parallel graphs. Theorem 11.26 (Ding, O., Sanders, Vertigan; Kedlaya) Every planar has an edge-partition into two graphs of tree-width ≤2. Partitioning Graphs on Surfaces Theorem 11.27 (Ding, O., Sanders, Vertigan) Every projective graph has a vertex-partition into two graphs of tw ≤2. Theorem 11.28 (DOSV) Every graph of non-negative Euler characteristic has a vertex-partition and an edge-partition into two graphs of tw ≤3. Note 11.29 This is best possible for toroidal graphs. Theorem 11.30 (DOSV) Every graph G has ◮vertex-partition into two graphs of tw ≤6 −2χ(G) ◮edge-partition into two graphs of tw ≤9 −3χ(G) Proofs ◮Set v ∈V (G) and Vk = set of vertices distance k from v. ◮Vertex-partitions: graphs induced by [ k even Vk and [ k odd Vk ◮Edge-partitions: let Hk = induced by edges [Vk, Vk] and [Vk, Vk+1] [ k even Hk and [ k odd Hk Minor-Closed Classes Conjecture 11.31 (Thomas) For every G there is an integer k such that every graph with no G-minor has a vertex-partition and edge-partition into two graphs of tw ≤k. Theorem 11.32 (DOSV, DeVos, Reed, Seymour) For every minor-closed class of graphs other than the class of all graphs there is a number k such that every member of the class has a vertex-partition and edge-partition into two graphs of tw ≤k. Theorem 11.33 (Robertson and Seymour) All members of any minor-closed class of graphs other than the class of all graphs are clique-sums of graphs that can “almost” be embedded on surfaces of bounded genus. A graph is a minor of another if it can be obtained by ◮deleting edges and/or vertices ◮contracting edges Minor-closed class of graphs: if G is in the class, then so are all of its minors. Theorem 11.34 (R&S) Every minor-closed class of graphs can be characterized by excluding ﬁnitely many graphs as minors. Partitions and Contractions Question 11.35 (Oxley) Can every co-graphic matroid be partitioned into two series-parallel matroids? Can the edges of every graph be partitioned into E1 and E2 such that each of G/E1 and G/E2 is series parallel? ◮yes if planar: dualize our theorem ◮yes if 4-connected: 2 edge-disjoint spanning trees by Nash-Williams Theorem 11.36 (Morgan, O.) The edges of every projective graph can be partitioned into E1 and E2 such that each of G/E1 and G/E2 has tw ≤3. Theorem 11.37 (MO) The edges of every toroidal graph can be partitioned into E1 and E2 such that of tw(G/E1) ≤3 and tw(G/E2) ≤4. Theorem 11.38 (Demaine, Hajiaghayi, Mohar) The edges of a graph of genus g can be partitioned into E1 and E2 such that each of G/E1 and G/E2 has tw ≤O(g2). Deﬁnition of T(k, l, r) ◮Start with Kk, and assign all of its vertices level 0 ◮Inductively, for each Kk subgraph H of level n −1, add r new vertices, join each of them to all vertices of H and declare all newly created vertices and Kk subgraphs to have level n. ◮Stop after having created all level-l subgraphs. T (2, 4, 1) b b level 0 b level 1 b b level 2 b b b b level 3 b b b b b b b b level 4 Partitioning k-Trees Deﬁnition 11.39 k-tree: T (k, l, r) where l is arbitrary, and r can very arbitrarily at every stage. Theorem 11.40 (DOSV) Every (k1 + k2 + 1)-tree has a vertex-partition into a k1-tree and a k2-tree. Theorem 11.41 (DOSV) Every (k1 + k2)-tree has an edge-partition into a k1-tree and a k2-tree. Ramsey-Type Results Theorem 11.42 (DOSV) For every k1, k2, l, and r there is L such that for every vertex-partition {G1, G2} of T (k1 + k2, L, r): T (k1, l, r) ⊆G1 or T (k2, l, r) ⊆G2. Corollary 11.43 For every k, l, and r there is L such that for every vertex-partition {G1, . . . , Gk} of T (k, L, r): at least one Gi contains T (1, l, r). Conjecture 11.44 (DOSV) For every k, l, and r there is L such that for every edge-partition {G1, . . . , Gk} of T (k, L, r): at least one Gi contains a subdivision of T (1, l, r). Large k-Trees and Edge-Partitions T (2, 4, 1) b b level 0 b level 1 b b level 2 b b b b level 3 b b b b b b b b level 4 b b b b b b b b b b b b b b b b b Theorem 11.45 (DOSV) For every l and r there are L and R such that if T (2, L, R) has its edges colored red and blue, then it contains a red T (1, l, r) or a blue subdivision of T (1, l, r). Partitioning Into Graphs With Only Small Components Theorem 11.46 (Alon, Ding, O, Vertigan) If k is tree-width and ∆is maximum degree of G, then ◮there is a vertex-partition of G into 2 graphs with components on only ≤24k∆vertices; ◮there is an edge-partition of G into 2 graphs with components on only ≤24k∆(∆+ 1) vertices. Q: Is it enough to bound just the tree-width? A: No, consider ◮large star for edge-partitions, ◮and a large fan for vertex-partitions. Q: Is it enough to bound just the vertex degree? A: No, there are 4-regular graphs of arbitrarily large girth (Erd˝ os, Sachs) ◮one part of an edge-partition will contain a cycle ◮for vertex-partitions, consider line graphs of those graphs ∆≤4 and Vertex-Partiions Theorem 11.47 (ADOV) If ∆(G) ≤4, then G has a vertex-partition into two graphs on components with at most 57 vertices. Theorem 11.48 (Haxell, Szab´ o, Tardos) 57 can be reduced to 6. Note: 5 is a lower bound. Theorem 11.49 (Haxell, Szab´ o, Tardos) If ∆(G) ≤5, then G has a vertex-partition into two graphs on components with at most 6,053,628,175 vertices. Three-Color Theorem for Planar Graphs? Question 11.50 Is there a number c such that the every planar graph can have its vertices colored with 3-colors so that each monochromatic component has at most c vertices? Answer: No! For a positive integer n, take n disjoint copies of a fan on n2 + n + 1 vertices. Then add one more vertex v0 joining it to all vertices of all the fans; name the graph Un. Theorem 11.51 In every vertex 3-coloring of Un, there is a monochromatic component on more than n vertices. Proof. Without loss of generality, the color of v0 is red. If each fan has a vertex colored red, then the conclusion follows. So suppose that one of the fans F has its vertices colored with only two colors, and suppose the tip v of F is blue. If F has n other vertices colored blue, then the conclusion follows. In the remaining case, the blue vertices cut the path F −v into at most n monochromatic segments, and so at least one of those segments must have more than n vertices. Four-Color Theorem for Minor-Closed Classes of Graphs Theorem 11.52 (ADOV+S) Let G be a minor-closed class of graphs other than the class of all graphs, and pick ∆. There is a number c(G, ∆) such that every member of G whose max degree is ≤∆can be vertex 4-colored so that all monochromatic components have at most c vertices. Proof. ◮Every graph G in G has a vertex-partition into two graphs of tw ≤w(G). ◮If ∆(G) ≤∆, then each of those can be 2-colored with components on at most c vertices. ◮This gives a 4-coloring of G with components on at most c vertices.
1991	https://www.metmuseum.org/met-publications/heilbrunn-timeline-of-art-history	Heilbrunn Timeline of Art History The Metropolitan Museum of Art 2000 Webby Awards, Best Art Website (2017) Museums and the Web, Best Research Site (2005) American Association of Museums Media and Technology Committee MUSE Awards, Education & Interpretation (2005) The Metropolitan Museum of Art's Heilbrunn Timeline of Art History, originally launched in 2000, presents the Met's collection via a chronological, geographical, and thematic exploration of global art history. Targeted at students and scholars of art history, it is an invaluable reference, research, and teaching tool. Authored by The Met's experts—curators, conservators, scientists, and educators—the Timeline comprises 300 timelines, more than 1,000 essays, more than 7,000 objects, and is regularly updated and enriched to provide new scholarship and insights on the collection. You May Also Like A slider containing 5 items. Press the down key to skip to the last item. Art = Discovering Infinite Connections in Art History 2020 The Arts of Korea: A Resource for Educators 2001 Free to download The Metropolitan Museum of Art: Masterpiece Paintings 2016 Greek Art From Prehistoric to Classical: A Resource for Educators 2000 Free to download The Care and Handling of Art Objects: Practices in The Metropolitan Museum of Art 2019 Free to download
1992	https://catdir.loc.gov/catdir/toc/ecip0618/2006025826.html	Table of contents for Fluid mechanics Table of contents for Fluid mechanics / Frank M. White. Bibliographic record and links to related information available from the Library of Congress catalog. Note: Contents data are machine generated based on pre-publication provided by the publisher. Contents may have variations from the printed book or be incomplete or contain other coding. Preface xi Chapter 1 Introduction 3 1.1 Preliminary Remarks 3 1.2 The Concept of a Fluid 4 1.3 The Fluid as a Continuum 6 1.4 Dimensions and Units 7 1.5 Properties of the Velocity Field 14 1.6 Thermodynamic Properties of a Fluid 16 1.7 Viscosity and Other Secondary Properties 23 1.8 Basic Flow Analysis Techniques 38 1.9 Flow Patterns: Streamlines, Streaklines, and Pathlines 39 1.10 The Engineering Equation Solver 44 1.11 Uncertainty of Experimental Data 45 1.12 The Fundamentals of Engineering (FE) Examination 46 1.13 Problem-Solving Techniques 47 1.14 History and Scope of Fluid Mechanics 47 Problems 49 Fundamentals of Engineering Exam Problems 57 Comprehensive Problems 57 References 59 Chapter 2 Pressure Distribution in a Fluid 63 2.1 Pressure and Pressure Gradient 63 2.2 Equilibrium of a Fluid Element 65 2.3 Hydrostatic Pressure Distributions 67 2.4 Application to Manometry 74 2.5 Hydrostatic Forces on Plane Surfaces 78 2.6 Hydrostatic Forces on Curved Surfaces 84 2.7 Hydrostatic Forces in Layered Fluids 86 2.8 Buoyancy and Stability 89 2.9 Pressure Distribution in Rigid-Body Motion 94 2.10 Pressure Measurement 102 Summary 106 Problems 106 Word Problems 128 Fundamentals of Engineering Exam Problems 129 Comprehensive Problems 130 Design Projects 131 References 132 Chapter 3 Integral Relations for a Control Volume 135 3.1 Basic Physical Laws of Fluid Mechanics 135 3.2 The Reynolds Transport Theorem 139 3.3 Conservation of Mass 147 3.4 The Linear Momentum Equation 153 3.5 The Angular Momentum Theorem 166 3.6 The Energy Equation 172 3.7 Frictionless Flow: The Bernoulli Equation 182 Summary 191 Problems 192 Word Problems 220 Fundamentals of Engineering Exam Problems 221 Comprehensive Problems 222 Design Project 223 References 223 Chapter 4 Differential Relations for Fluid Flow 225 4.1 The Acceleration Field of a Fluid 225 4.2 The Differential Equation of Mass Conservation 227 4.3 The Differential Equation of Linear Momentum 234 4.4 The Differential Equation of Angular Momentum 240 4.5 The Differential Equation of Energy 242 4.6 Boundary Conditions for the Basic Equations 244 4.7 The Stream Function 249 4.8 Vorticity and Irrotationality 257 4.9 Frictionless Irrotational Flows 259 4.10 Some Illustrative Plane Potential Flows 264 4.11 Some Illustrative Incompressible Viscous Flows 269 Summary 278 Problems 278 Word Problems 288 Fundamentals of Engineering Exam Problems 289 Comprehensive Problems 289 References 290 Chapter 5 Dimensional Analysis and Similarity 293 5.1 Introduction 293 5.2 The Principle of Dimensional Homogeneity 296 5.3 The Pi Theorem 302 5.4 Nondimensionalization of the Basic Equations 309 5.5 Modeling and Its Pitfalls 318 Summary 328 Problems 329 Word Problems 337 Fundamentals of Engineering Exam Problems 337 Comprehensive Problems 338 Design Projects 339 References 340 Chapter 6 Viscous Flow in Ducts 343 6.1 Reynolds Number Regimes 343 6.2 Internal versus External Viscous Flows 348 6.3 Head Loss--The Friction Factor 351 6.4 Laminar Fully Developed Pipe Flow 353 6.5 Turbulence Modeling 355 6.6 Turbulent Pipe Flow 361 6.7 Three Types of Pipe Flow Problems 369 6.8 Flow in Noncircular Ducts 375 6.9 Minor Losses in Pipe Systems 384 6.10 Multiple-Pipe Systems 393 6.11 Experimental Duct Flows: Diffuser Performance 399 6.12 Fluid Meters 404 Summary 425 Problems 426 Word Problems 443 Fundamentals of Engineering Exam Problems 444 Comprehensive Problems 445 Design Projects 447 References 447 Chapter 7 Flow Past Immersed Bodies 451 7.1 Reynolds Number and Geometry Effects 451 7.2 Momentum Integral Estimates 455 7.3 The Boundary Layer Equations 458 7.4 The Flat-Plate Boundary Layer 461 7.5 Boundary Layers with Pressure Gradient 470 7.6 Experimental External Flows 476 Summary 503 Problems 503 Word Problems 516 Fundamentals of Engineering Exam Problems 517 Comprehensive Problems 517 Design Project 518 References 519 Chapter 8 Potential Flow and Computational Fluid Dynamics 523 8.1 Introduction and Review 523 8.2 Elementary Plane Flow Solutions 526 8.3 Superposition of Plane Flow Solutions 528 8.4 Plane Flow past Closed-Body Shapes 535 8.5 Other Plane Potential Flows 544 8.6 Images 549 8.7 Airfoil Theory 551 8.8 Axisymmetric Potential Flow 563 8.9 Numerical Analysis 568 Summary 583 Problems 583 Word Problems 594 Comprehensive Problems 594 Design Projects 596 References 596 Chapter 9 Compressible Flow 599 9.1 Introduction 599 9.2 The Speed of Sound 604 9.3 Adiabatic and Isentropic Steady Flow 606 9.4 Isentropic Flow with Area Changes 612 9.5 The Normal Shock Wave 619 9.6 Operation of Converging and Diverging Nozzles 627 9.7 Compressible Duct Flow with Friction 632 9.8 Frictionless Duct Flow with Heat Transfer 644 9.9 Two-Dimensional Supersonic Flow 649 9.10 Prandtl-Meyer Expansion Waves 659 Summary 671 Problems 672 Word Problems 685 Fundamentals of Engineering Exam Problems 685 Comprehensive Problems 686 Design Projects 687 References 688 Chapter 10 Open-Channel Flow 691 10.1 Introduction 691 10.2 Uniform Flow; The Chazy Formula 697 10.3 Efficient Uniform-Flow Channels 702 10.4 Specific Energy; Critical Depth 704 10.5 The Hydraulic Jump 711 10.6 Gradually Varied Flow 716 10.7 Flow Measurement and Control by Weirs 724 Summary 731 Problems 731 Word Problems 742 Fundamentals of Engineering Exam Problems 743 Comprehensive Problems 743 Design Projects 744 References 745 Chapter 11 Turbomachinery 747 11.1 Introduction and Classification 747 11.2 The Centrifugal Pump 750 11.3 Pump Performance Curves and Similarity Rules 756 11.4 Mixed- and Axial-Flow Pumps: The Specific Speed 766 11.5 Matching Pumps to System Characteristics 772 11.6 Turbines 779 Summary 792 Problems 793 Word Problems 804 Comprehensive Problems 804 Design Project 806 References 806 Appendix A Physical Properties of Fluids 808 Appendix B Compressible Flow Tables 813 Appendix C Conversion Factors 830 Appendix D Equations of Motion in Cylindrical Coordinates 832 Appendix E Introduction to EES 834 Answers to Selected Problems 846 Index 853 Library of Congress Subject Headings for this publication: Fluid mechanics.
1993	https://books.google.com/books/about/Solution_Manual_for_Engineering_Economic.html?id=f_z6BreeUx8C	Solution Manual for Engineering Economic Analysis - Donald G. Newnan, Jerome P. Lavelle, Ted G. Eschenbach - Google Books Sign in Hidden fields Try the new Google Books Books Add to my library Try the new Google Books Check out the new look and enjoy easier access to your favorite features Try it now No thanks Try the new Google Books My library Help Advanced Book Search Get print book No eBook available Amazon.com Barnes&Noble.com Books-A-Million IndieBound Find in a library All sellers» ### Get Textbooks on Google Play Rent and save from the world's largest eBookstore. Read, highlight, and take notes, across web, tablet, and phone. Go to Google Play Now » My library My History Solution Manual for Engineering Economic Analysis Donald G. Newnan, Jerome P. Lavelle, Ted G. Eschenbach Oxford University Press, Incorporated, 2000 - Technology & Engineering - 432 pages Bibliographic information Title Solution Manual for Engineering Economic Analysis AuthorsDonald G. Newnan, Jerome P. Lavelle, Ted G. Eschenbach Edition illustrated Publisher Oxford University Press, Incorporated, 2000 ISBN 157645052X, 9781576450529 Length 432 pages SubjectsTechnology & Engineering › Engineering (General) Technology & Engineering / Engineering (General) Export CitationBiBTeXEndNoteRefMan About Google Books - Privacy Policy - Terms of Service - Information for Publishers - Report an issue - Help - Google Home
1994	https://jingyan.baidu.com/article/aa6a2c148dccd74c4c19c4c9.html	空间点到平面的距离公式推导是什么-百度经验分享到一键分享 QQ空间新浪微博百度云收藏人人网腾讯微博百度相册开心网腾讯朋友百度贴吧豆瓣网搜狐微博百度新首页 QQ好友和讯微博更多... 百度分享新闻网页贴吧知道经验音乐图片视频地图百科文库发布经验百度首页登录写经验领红包首页分类美食/营养游戏/数码手工/爱好生活/家居健康/养生运动/户外职场/理财情感/交际母婴/教育时尚/美容认证悬赏令回享商城视频经验知道百度经验>生活/家居>生活常识空间点到平面的距离公式推导是什么浏览：945 \| 更新：2022-12-06 15:19 空间点到平面的距离公式推导：设直线的方向向量是s，Q是这直线上任意一点，则空间点P转这直线的距离：d=\|QP×s\|/\|s\|，这里QP表示以Q为起点、P为终点的向量。距离d是以向量QP、向量s为邻边的平行四边形s边上的高，所以 d=\|QP\|sin=/\|s\|=\|QP×s\|/\|s\|。两平行线之间的距离公式：设两条直线方程为。 Ax+By+C1=0。 Ax+By+C2=0。则其距离公式为｜C1-C2\|/√（A²+B²)。相关推导两平行直线间的距离就是从一条直线上任一点到另一条直线的距离，设点P(a，b)在直线Ax+By+C1=0上，则满足Aa+Bb+C1=0，即Aa+Bb=-C1，由点到直线距离公式，P到直线Ax+By+C2=0距离为。 d=\|Aa+Bb+C2\|/√（A²+B²)。 =\|-C1+C2\|/√（A²+B²)。 =\|C1-C2\|/√（A²+B²)。经验内容仅供参考，如果您需解决具体问题(尤其法律、医学等领域)，建议您详细咨询相关领域专业人士。展开阅读全部换一批相关经验如何在英语考试临近的时候快速学习？2018.10.05 PPT中滑板车转弯动画设置方法2021.10.27 就业压力来自哪些方面?2017.11.06 如何查询开封最新限号情况2020.12.26 如何清洁Ugg拖鞋2020.06.11 今日支出元写经验有钱赚 >> 狄念容作者的经验钱币符号 WINDOWS系统与IOS系统有什么区别 CAD制图时，为什么我用虚线画的线条... 关于月亮的传说有哪些引发偏激共振现象的原因如要投诉，请到百度经验投诉中心，如要提出意见、建议，请到百度经验管理吧反馈。此内容有帮助? 2 我的财富值去登录我的现金去登录帮助意见反馈投诉举报 ©2025Baidu使用百度前必读百度经验协议作者创作作品协议企业推广京ICP证030173号-1 京网文【2023】1034-029号分享收藏返回顶部新浪微博QQ 空间 ◆ 请扫描分享到朋友圈辅助模式
1995	https://www.frontiersin.org/journals/immunology/articles/10.3389/fimmu.2017.00992/full	Frontiers \| Dermatomyositis and Immune-Mediated Necrotizing Myopathies: A Window on Autoimmunity and Cancer Frontiers in Immunology About us About us Who we are Mission and values History Leadership Awards Impact and progress Frontiers' impact Our annual reports Publishing model How we publish Open access Peer review Research integrity Research Topics FAIR² Data Management Fee policy Services Societies National consortia Institutional partnerships Collaborators More from Frontiers Frontiers Forum Frontiers Planet Prize Press office Sustainability Career opportunities Contact us All journalsAll articlesSubmit your researchSearchLogin Frontiers in Immunology Sections Sections Alloimmunity and Transplantation Antigen Presenting Cell Biology Autoimmune and Autoinflammatory Disorders : Autoimmune Disorders Autoimmune and Autoinflammatory Disorders: Autoinflammatory Disorders B Cell Biology Cancer Immunity and Immunotherapy Comparative Immunology Cytokines and Soluble Mediators in Immunity Immunological Memory Immunological Tolerance and Regulation Inflammation Microbial Immunology Molecular Innate Immunity Mucosal Immunity Multiple Sclerosis and Neuroimmunology NK and Innate Lymphoid Cell Biology Nutritional Immunology Parasite Immunology Primary Immunodeficiencies Systems Immunology T Cell Biology Vaccines and Molecular Therapeutics Viral Immunology ArticlesResearch TopicsEditorial board About journal About journal Scope Field chief editors Mission & scope Facts Journal sections Open access statement Copyright statement Quality For authors Why submit? Article types Author guidelines Editor guidelines Publishing fees Submission checklist Contact editorial office About us About us Who we are Mission and values History Leadership Awards Impact and progress Frontiers' impact Our annual reports Publishing model How we publish Open access Peer review Research integrity Research Topics FAIR² Data Management Fee policy Services Societies National consortia Institutional partnerships Collaborators More from Frontiers Frontiers Forum Frontiers Planet Prize Press office Sustainability Career opportunities Contact us All journalsAll articlesSubmit your research Frontiers in Immunology Sections Sections Alloimmunity and Transplantation Antigen Presenting Cell Biology Autoimmune and Autoinflammatory Disorders : Autoimmune Disorders Autoimmune and Autoinflammatory Disorders: Autoinflammatory Disorders B Cell Biology Cancer Immunity and Immunotherapy Comparative Immunology Cytokines and Soluble Mediators in Immunity Immunological Memory Immunological Tolerance and Regulation Inflammation Microbial Immunology Molecular Innate Immunity Mucosal Immunity Multiple Sclerosis and Neuroimmunology NK and Innate Lymphoid Cell Biology Nutritional Immunology Parasite Immunology Primary Immunodeficiencies Systems Immunology T Cell Biology Vaccines and Molecular Therapeutics Viral Immunology ArticlesResearch TopicsEditorial board About journal About journal Scope Field chief editors Mission & scope Facts Journal sections Open access statement Copyright statement Quality For authors Why submit? Article types Author guidelines Editor guidelines Publishing fees Submission checklist Contact editorial office Frontiers in Immunology Sections Sections Alloimmunity and Transplantation Antigen Presenting Cell Biology Autoimmune and Autoinflammatory Disorders : Autoimmune Disorders Autoimmune and Autoinflammatory Disorders: Autoinflammatory Disorders B Cell Biology Cancer Immunity and Immunotherapy Comparative Immunology Cytokines and Soluble Mediators in Immunity Immunological Memory Immunological Tolerance and Regulation Inflammation Microbial Immunology Molecular Innate Immunity Mucosal Immunity Multiple Sclerosis and Neuroimmunology NK and Innate Lymphoid Cell Biology Nutritional Immunology Parasite Immunology Primary Immunodeficiencies Systems Immunology T Cell Biology Vaccines and Molecular Therapeutics Viral Immunology ArticlesResearch TopicsEditorial board About journal About journal Scope Field chief editors Mission & scope Facts Journal sections Open access statement Copyright statement Quality For authors Why submit? Article types Author guidelines Editor guidelines Publishing fees Submission checklist Contact editorial office Submit your researchSearchLogin Your new experience awaits. Try the new design now and help us make it even better Switch to the new experience REVIEW article Front. Immunol., 21 August 2017 Sec. Inflammation Volume 8 - 2017 \| Dermatomyositis and Immune-Mediated Necrotizing Myopathies: A Window on Autoimmunity and Cancer Audrey Aussy 1Olivier Boyer1Nadège Cordel1,2 1 Normandie University, UNIROUEN, INSERM, U1234, Rouen University Hospital, Department of Immunology, Rouen, France 2 Unit of Dermatology and Internal Medicine, Pointe-à-Pitre University Hospital, University of the French West Indies, Fouillole, Pointe-à-Pitre, Guadeloupe Autoimmune myopathies (myositides) are strongly associated with malignancy. The link between myositis and cancer, originally noticed by Bohan and Peter in their classification in 1975 (1), has been evidenced by large population-based cohort studies and a recent meta-analysis. The numerous reports of cases in which the clinical course of myositis reflects that of cancer and the short delay between myositis and cancer onset support the notion that myositis may be an authentic paraneoplastic disorder. Thus, cancer-associated myositis raises the question of cancer as a cause rather than a consequence of autoimmunity. Among myositides, dermatomyositis and more recently, although to a lesser extent, immune-mediated necrotizing myopathies are the most documented forms associated with cancer. Interestingly, the current diagnostic approach for myositis is based on the identification of specific antibodies where each antibody determines specific clinical features and outcomes. Recent findings have shown that the autoantibodies anti-TIF1γ, anti-NXP2 and anti-HMGCR are associated with cancers in the course of myositis. Herein, we highlight the fact that the targets of these three autoantibodies involve cellular pathways that intervene in tumor promotion and we discuss the role of cancer mutations as autoimmunity triggers in adult myositis. Introduction The link between autoimmunity and cancer has become a topic of unceasing interest over the past decade. Although it is increasingly evident that the risk of cancer is augmented in patients affected by several types of autoimmune diseases (AID), the nature of the interplay between autoimmunity and cancer remains elusive (2–4). One important question here recalls the old egg-and-chicken dilemma: is the autoimmune background in AID a seedbed for cancer development or, alternatively, may cancer cause autoimmunity? Rheumatic AID such as systemic lupus erythematosus, rheumatoid arthritis, or Sjögren syndrome promote cancer development after several years of chronic inflammation and also exposure to immunosuppressive drugs (2, 5, 6). In the case of lupus for instance, the broadness of cancer type spectrum is striking, including hematological—mostly virus induced—malignancies but also numerous kinds of solid tumors such as vulva, lung, thyroid, and liver cancer (7). Here, the risk of cancer cannot only be ascribed to the sole autoimmune status but also presumably to iatrogenic immunosuppression. This view is counterbalanced by the examples of paraneoplastic neurological syndromes, systemic sclerosis, and myositis. Paraneoplastic neurological syndromes with anti-neuronal autoantibodies (aAbs) include a most illustrative example, i.e., anti-Hu encephalitis associated with small cell lung carcinoma (SCLC) (8). Anti-Hu aAbs recognize the HuD autoantigen normally restricted to neurons but ectopically expressed on SCLC cells (9). Expression of the immunogenic HuD molecule by SCLC elicits the production of anti-Hu aAb and CD8+ cytotoxic T cells, explaining the parallel clinical course of neurological symptoms and SCLC evolution and demonstrating the direct link between cancer and tissue-specific AID (10). Cancer-induced breakage of tolerance can also be caused by tumoral somatic mutations, as recently highlighted by the case of systemic sclerosis associated with anti-polymerase III (POL3) aAbs. The immunogenic peptides generated by POLR3 gene mutations induce a POL3-specific CD4+ T cell response with production of specific antibodies that secondarily target wild-type POL3 by epitope spreading (11). This present review focuses on the forms of myositides, i.e., dermatomyositis (DM) and immune-mediated necrotizing myopathies (IMNMs) which have been identified as associated with cancer and represent a paradigm of cancer-associated AID. DM, Risk of Cancer, and Diagnostic Contribution of Autoantibodies Autoimmune myopathies or myositides constitute a heterogeneous group of severe acquired myopathies. They are characterized clinically by symmetrical proximal muscle weakness, associated or not with systemic features, and histologically by various levels of myofiber necrosis/regeneration and interstitial mononuclear infiltrates. Clinical and histopathological patterns define different diseases: polymyositis, DM, overlap myositis, sporadic inclusion-body myositis, and IMNM (12–17). Dermatomyositis affects both adults and children among all ethnic groups with an unbalanced 1/2-sex-ratio in favor of women. Its annual incidence varies from 1.9 to 7.7 cases per million inhabitants according to data in the literature with a peak of frequency in 40–60-year-old adults and in 5–14-year-old children (13, 18). The appearance of specific cutaneous manifestations is typical of DM and is among its diagnostic criteria. Cutaneous manifestations typically consist of erythematous scaly papules over the metacarpophalangeal knuckles (Gottron’s papules) (Figure 1A); a symmetrical reddish-violet periorbital edema that predominates on the upper eyelids (heliotrope erythema) but may affect the rest of the face; lupus-like erythema which involves low neck (V sign) (Figure 1B), shoulders (shawl sign), extensor surfaces of the limbs, dorsal side of hands and fingers and scalp; poikiloderma of the upper trunk (Figure 1C); and centripetal flagellate erythema affecting the trunk and or proximal extremities (Figure 1D). Cutaneous manifestations of DM also include non-specific lesions such as (i) vascular lesions, i.e., periungual erythema with telangiectatic capillary loops, nail fold dilated capillaries visible to the naked eye, cuticular hypertrophy (Figures 1E,F), vasculitis, cutaneous necrosis, or Raynaud’s phenomenon, which are more prevalent in the course of juvenile dermatomyositis and (ii) several other dermatological features such as pruritus (present in 30% of DM), photosensitivity, mucinosis, and calcifications, which are more frequent in children rather than in adults, i.e., 30–70 versus 10%. Dermatological particularities of DM have been reported in several ethnic groups. In Afro-Caribbeans, edema of the face is usually predominant whereas in Eurasians, the Wong-type DM which mimicks a pytiriasis rubra pilaris seems to be more frequent (19). FIGURE 1 Figure 1. Clinical and histological features of dermatomyositis (DM). (A) Gottron’s sign: erythematous scaly papules over the metacarpophalangeal joints. (B) «V sign» in a white European male patient with DM. (C) Poikiloderma (i.e., erythema, atrophy, variable pigmentary changes) on the upper trunk of an African Caribbean female patient with DM. (D) Typical centripetal flagellate erythema affecting the upper trunk of a male patient with DM. (E,F) Periungual erythema and telangiectatic capillary loops in patients with DM. (G–I) Histological feature of a Gottron’s papule. (G) Slight hyperkeratosis, basal cell vacuolar degeneration, upper dermal edema, and perivascular inflammatory cell infiltrate with enlarged capillaries (HES staining, ×20). (H) DM interface dermatitis with vacuolar changes of the basal cell layer, perivascular inflammatory cell infiltrate with capillary dilatation, endothelial cell turgescence, and pigmentary incontinence (HES staining, ×40). (I) Positive alcian-blue staining attesting dermal mucin deposits (×20). Specific lesions of DM are histologically characterized by an interface dermatitis with basal layer vacuolar changes that are associated in various degrees with hyperkeratosis, epidermal atrophy, basement membrane thickening, upper dermal edema, pigmentary incontinence, mucine deposits, and light perivascular CD4+ T lymphocyte infiltrate of the superficial dermis (Figures 1G–I) (20). Dermatological features usually precede muscle weakness by 3–6 months but may appear several years before. However, muscular manifestations may be absent, defining the amyopathic form of DM with an associated cancer rate, which is theoretically the same as classic DM (21, 22). Diagnosis of DM is supported by serum creatine kinase (CK) elevation, which mirrors muscle lysis and by electromyographic data, but diagnosis is confirmed by muscle biopsy except for amyopathic DM. Histological examination of muscle tissue (Figure 2) typically shows perifascicular atrophy, necrotic and regenerative muscle fibers, septal and/or perivascular inflammatory cell infiltrate, and endomysial microangiopathy with membranolytic attack complex C5b-9 capillary staining that may also be present in cutaneous lesions. Although these capillary injuries have long been considered as indirect evidence of an initial endothelial target in DM, recent scientific findings demonstrate that capillary lesions might be non-specific in relationship to ischemia-reperfusion injury of perimysial arcade arteries (23). The role for interferons in DM pathogenesis is increasingly evident, since a typical interferon type 1 pathway signature was found in both muscle (24) and skin (25), with a correlation with disease activity (26). FIGURE 2 Figure 2. Muscle biopsy of dermatomyositis. (A) Perifascicular atrophy. (B) Area of contiguous necrotic myofibers (arrow) corresponding to a microinfarct. (C) Punch-out vacuoles within myofibers (arrows) assessing focal myosinolysis. (D) Ubiquitous myofiber reexpression of MHC-class I with perifascicular reinforcement. (E) Neural cell adhesion molecule (NCAM) immunostaining showing large areas of positive myofibers indicating muscle ischemia. (F) Platelet endothelial cell adhesion molecule (PECAM) immunostaining for endothelial cells showing marked endomysial capillary drop out. (G) Complement activation assessed by the presence of membrane attack complex deposits at the level of endomysial capillaries (arrows). Frozen sections, light microscopy; hematoxylin-eosin (A–C), immunoperoxydase technique (D–G), HLA-ABC (D), CD56/NCAM (E), CD31/PECAM (F), and C5b-9 (G). The outcome of DM is variable with benign forms highly responsive to treatment and severe forms associated with cardiac or lung involvement and/or malignancies. In juvenile DM, cancer association is uncommon. Nonetheless, a poor prognosis may be due to the intensity of vasculitis and vascular damage that involves the skin and the digestive system. Calcifications on the areas around joints may also lead to severe functional impairment in children. Interestingly, several dermatological manifestations of DM such as cutaneous necrosis, skin vasculitis or pruritus, or histological patterns such as leucocytoclastic vasculitis are reported to be associated with cancer whatever the ethnic group (27, 28). Conversely, several features seem to protect against cancer such as Afro-Caribbean ethnicity (29). Globally, myositides are rare diseases and epidemiologic data remain scarce. Studies are limited by their retrospective character and small population size. Yet, since its first descriptions (30, 31), the association between cancer and myositis has been confirmed by several studies published between 1975 and 2012, reporting a global malignancy rate from 6.7 to 32% (32–35). Importantly, several studies also point to cancer as the main cause of death in cancer-associated myositis patients (34, 36, 37). Yet, it may be difficult to distinguish between cancer on the one hand, and myositis exacerbation and its complications on the other hand, as the actual cause of death. Most observations highlighted the short delay between the onset of myositis and the discovery of cancer. In some cases, cancer even preceded myositis, reinforcing the view that myositis might be the consequence rather than the cause of cancer and even leading to individualizing “cancer-associated myositis” in one classification in 2005 (14). Cancer risk is particularly established in DM. A recent meta-analysis confirmed adult (but not juvenile) DM as a risk-factor of cancer, with a standardized incident ratio (SIR) for occurrence of cancer of 5.5 [4.3–6.7], mostly peaking at 1 year around DM diagnosis (38). A limitation to previous studies, including those reviewed in this meta-analysis, is that the classification of myositis was essentially based on the widely used Bohan and Peter classification which tends to classify IMNM, overlap myositis, and polymyositis in a single category (1). Whereas this classification remains popular for its practical value in DM, progress in the definition of clinical, pathological, and serological patterns has led to newer classifications that are useful for the diagnosis of inclusion-body myositis (12), overlap myositis (14), or IMNM (13). In the light of these different existing classifications for myositis, a new unifying classification would provide a much-awaited tool. The detection of myositis-specific aAbs (MSA) has proved most helpful in classifying the different forms and predicting the outcome of myositis. Indeed, in clinical practice, MSA define particular clinical ± histological patterns (39). In DM, anti-Mi-2 are classically associated with absence of cancer, sensitivity to treatment and, overall, good prognosis (40–42). Anti-melanoma differentiation antigen 5 (MDA-5) identifies a subgroup of DM patients with interstitial lung disease, necrotizing cutaneous lesions, skin ulcerations, and tender palmar papules while muscle signs are minimal (43–45). Anti-SAE (small ubiquitin-like modifier activating enzyme) has a low prevalence (1–4%) and patients have classic cutaneous signs of DM. Interestingly, a significant but low rate of cancer has been reported in this group (46–48). Anti-TIF1γ (transcriptional intermediary factor 1 gamma) is the leader of cancer-associated aAbs, with a rate of associated malignancy ranging from 60 to 80% of patients (49, 50). Anti-NXP2 (nuclear matrix protein 2) is another biomarker of risk of malignancy in adult patients, with a 30% rate of cancer among DM patients (51, 52). Besides DM, IMNM is another form of myositis recently suggested to be associated with malignancy (53–55). IMNM may be subdivided in three groups: seronegative, anti-HMGCR (3-hydroxy-3-methylglutaryl-coenzyme-A-reductase), and anti-SRP (signal recognition particle). In contrast to anti-SRP positive patients, one study has recently suggested that, among IMNM, seronegative and anti-HMGCR positive patients have a significantly higher risk of cancer (56). Cancer-Associated Myopathies According to Autoantibodies DM with Anti-TIF1γ, the Leader in Cancer-Associated Myositis Anti-TIF1γ was first described in 2006 as an antibody directed against a 155 kDa protein, especially in patients with DM (49, 57). This protein was rapidly identified as TIF1γ. Anti-TIF1γ aAbs scored positive in 20–30% of adult DM and 30–40% of juvenile DM with some differences according to geographical origin (39, 58). HLA DQA10301 was associated with anti-TIF1γ DM (49). The high prevalence of cancer was largely confirmed with a rate from 18 up to 80% of adult patients especially in the 2 years surrounding DM (22, 52, 59). A meta-analysis performed by Trallero-Araguás and colleagues estimated a 78% sensitivity and 89% specificity of these antibodies for diagnosing an associated cancer (50). Since 2001, age has been recognized as a risk factor for cancer among DM patients (33). Recently, two publications strengthened the role of advancing age in the increasing risk of cancer among adult anti-TIF1γ positive DM patients (60, 61). No study found any predominance of one type of cancer; those occurring in adults with DM were generally comparable to those in the general population, stratified by age and sex (breast, lung, colorectal, bladder) as well as to some more rare cancers such as gastric or thymus cancer (59). Anti-TIF1γ DM associates classic but severe cutaneous signs with moderate muscular symptoms, frequent dysphagia but decreased systemic features compared to other DM (60). The specific clinical and histopathological features of anti-TIF1γ DM are summarized in Table 1. Regarding juvenile DM, no increased risk of cancer is observed (62). The median age of DM is 6.8 years of age. Anti-TIF1γ juvenile DM more often presents chronic or polycyclic courses associated with more severe prognosis and profuse cutaneous involvement (63). TABLE 1 Table 1. Characteristics of anti-TIF1γ, anti-NXP2, and anti-HMGCR myositis. DM with Anti-NXP2, the Second Actor in Cancer-Associated Myositis A novel aAb directed against a 140 kDa protein was found in a 1997 cohort of juvenile DM and named anti-MJ (64). The 140 kDa protein was next identified as NXP2, also known as MORC3 (84). Anti-NXP2 aAbs are present in 22–25% of juvenile DM patients and in 1–17% of adults with DM, depending on the method of detection (61, 70). Forms of myositides other than DM may occasionally be associated with anti-NXP2 (67, 84). Cancer was detected in 24–37.5% of adults who scored positive for anti-NXP2 in several retrospective series (60, 61, 67). As for anti-TIF1γ DM, no specific type of tumor was mentioned. Clinical manifestations are partially distinct from anti-TIF1γ DM (Table 1). Indeed, classic cutaneous features are less severe but there is a higher prevalence of calcinosis (70). In addition, muscular involvement is constant and more severe. Similarities between anti-TIF1γ and anti-NXP2 include dysphagia and prevalence of peripheral edema (61, 70). Interstitial lung disease has only been reported in one cohort (85) and Raynaud’s phenomenon is found in 20% of cases (64, 85). In children, the median age at onset has been calculated at 5.8 years of age (58). Two series found an increased risk of calcinosis in children and a high prevalence of severe muscular involvement with functional disabilities, muscle cramps, and dysphagia but no cancer (Table 1) (39, 51). IMNMs, New Players in Cancer-Associated Myositis? Immune-mediated necrotizing myopathies are a recently described entity, based on specific histological pattern with poor inflammatory infiltrate and presence of significant necrotizing and regenerative fibers (13, 72). As mentioned above, three IMNM subsets have been identified according to serologic status: anti-SRP, anti-HMGCR, or negative serology. Anti-HMGCR and seronegative IMNMs seem to be associated with a higher risk of cancer, with a SIR score of 2.79 and 8.35, respectively (56). Between 13 and 36% of anti-HMGCR positive patients have an associated cancer (56, 68, 69). A genetic study found that HLA-DRB111:01 is associated with a higher risk of anti-HMGCR IMNM in both white American and African American adult populations, whereas HLA-DRB107:01 seem to be associated with a risk of anti-HMGCR myositis in small series of children (86, 87). Anti-HMGCR aAbs were discovered in 2010 in a group of patients who developed myositis after exposure to statins, without resolution by stopping statins (65, 66). Histological features corresponded to IMNM (88–90). While the association with statins was confirmed in recent studies, it involved a wide range between 37 and 94% of patients (56, 91). Thus, statins are not necessarily required to develop anti-HMGR myositis. Clinically, patients present severe and acute muscle weakness, dramatically elevated CK level, but extra-muscular disorders are uncommon (65, 66, 71). Anti-HMGCR aAbs have also been identified in the sera of juvenile myositis, without exposure either to statins or cancer. Clinical features in children may wrongly shift toward muscular dystrophy (92). Targets of Cancer-Associated Autoimmune Response in Myositis Intriguingly, all three aAb targets in cancer-associated myositis are involved to some extent in cancer pathogenesis. TIF1γ, Encoded by the TRIM33 Gene TIF1γ, also known as ecto, RETfused7, or TRIM33, was discovered in 1999 and identified as the third member of the TIF1 protein (after TIF1α and TIF1β) (73). These three proteins belong to the TRIM (tripartite motif) protein family defined by a particular RING-finger domain (93). TIF1 proteins are a subfamily characterized by several domains from 3′ to 5′, including the RING-finger domain, 1 or 2B-boxes, a coiled-coil domain, a plant homeodomain (PHD), and a bromodomain (94). They are involved in multiple critical biological processes. TIF1γ is particularly known for being involved in embryonic development, hematopoiesis, mitosis and cycle regulation, DNA repair, innate and adaptive immunity, osteoblast differentiation, viral transcription, and oncogenesis in case of dysregulation. TIF1γ can exert its role as an E3-ubiquitin ligase, as a histone-binding protein or by sumoylating proteins. The main interacting pathways of TIF1γ are the TGF-β canonical (Smad4-dependent) and non-canonical (Smad4-independent) pathways (95–97). In the canonical pathway, TIF1γ acts either as a repressor (via ubiquitination), a competitor or a partner of Smad4, depending on the cellular context (96). In the canonical TGF-β pathway, the activation of the TGF-β receptor leads to the recruitment of RSmad (Smad 2 or 3) and phosphorylation allows the formation of a RSmad/Smad4 complex, which next enters the nucleus to activate the transcription of targeted genes. During mammalian embryonic development, TIF1γ plays a major role in patterning and polarizing embryonic cells before gastrulation by inhibiting Nodal/Smad4 signaling, which promotes endodermic proliferation (98). Later in embryonic development, Smad4 and TIF1γ cooperate or act redundantly to promote both proliferation and differentiation of neural stem cells and palate development (99, 100). It has been shown that TIF1γ participates in the differentiation of stem cells in collaboration with Smad4, by direct interaction with histone via PHD–bromodomain leading to the assembling of Smad4/Smad2–3 complex on targeted genes (101). In adult tissues, TIF1γ promotes the terminal differentiation of mammary gland and lactation by antagonization of Smad4 (102), supports both osteoblast proliferation and differentiation under stimulation of bone morphogenetic proteins (BMP) via the activation of a particular RSmad complex (Smad1/5) (103), regulates granulopoiesis in mice (104), and participates in the development of iNKT cells (105). During erythropoiesis, Smad4 and TIF1γ competitively bind to phosphorylated SMad2/3 (RSmad) in response to TGF-β to promote the proliferation and maturation of erythroblasts (106, 107). Otherwise, TIF1γ is involved in the balance between lymphoid and myeloid lineage and protects hematopoietic stem cells from aging (108), through TAL1 and PU1 DNA-binding protein, whose transcriptional activity depend on TGF-β (109, 110). TIF1γ has many other functions in cells, mediated by different pathways and functions. Regarding the innate immune system, TIF1γ directly represses the transcription of the interferon-b gene (ifnb) at late phase of macrophage activation (74). Also, it binds multiple chromatin sites in monocyte to promote production of macrophage, binds other chromatin sites in mature macrophage to regulate the responses after toll-like receptor (TLR) activation by lipo-polysaccharide (LPS) (111), and is directly involved in proteasome activation via the ubiquitination of DHX33 (112). TIF1γ also has many roles in cell homeostasis; TIF1γ is strongly involved in antiproliferative cellular effect by (i) mediating ubiquitination and then the degradation of LIM-domain-binding protein which is involved in the transcription of cycle activator genes (113) and (ii) interaction with APC/C (anaphase-promoting complex/cyclosome) to promote the alignment and stability of chromosomes during mitosis and to prevent abnormal metaphase–anaphase transition (75, 76). Moreover, TIF1γ is largely involved in DNA repair by recruiting different proteins promoting chromatin relaxation and repair (114, 115). Next, TIF1γ is a strong tumor suppressor by preventing β-catenin degradation (116), epithelial-to-mesenchymal transition (117, 118), and by regulation of the chromatin (119, 120). The role of TIF1γ as a tumor suppressor has been directly shown in chronic myelomonocytic leukemia (121), pancreatic tumor (122–124), hepatocellular carcinoma (125), renal cell carcinoma (126), and non-small cell lung cancer (127), where TIF1γ decreased expression or inactivation promotes proliferation and probably epithelial-to-mesenchymal transition (97–104). Paradoxically, overexpression of TIF1γ is involved in oncogenesis notably in breast cancer, where TIF1γ interferes with TGF-β to promote poorer prognosis (128). Moreover, hyper-expression of TIF1γ has been shown in a significant proportion of colorectal adenocarcinomas (129). NXP2, Encoded by the MORC3 Gene MORC3 (microrchidia3) is a nuclear protein which belongs to a highly conserved nuclear matrix protein family which has recently been identified (77). Four MORC members have been identified and are characterized by three common structural domains (ATPase domain, Zinc finger domain, and coiled-coil domain) involved in chromatin remodeling and epigenetic regulation (77). MORC3 includes two other domains, a nuclear matrix binding site and an RNA binding site, essential for the regulation of transcription (78). Its mRNA expression is relatively ubiquitous but the expression of MORC3 protein is particularly elevated in immune cells (78, 79). MORC3 plays several critical roles in cell regulation, illustrated by an early death at birth or day 1 in MORC3 knockout mice (130). MORC3 promotes the activation of p53, a tumor suppressor inducing cellular senescence in response to oncogenic factors and supports the architecture of the nucleus (78, 130). The ATPase domain of MORC3 interacts with the coiled-coil domain to form homodimers that can bind DNA (131). Zinc finger is implicated in the nucleus localization and binds MORC3 to histone (132). These two mechanisms seem to be involved in DNA repair and epigenetic regulation (77, 80). MORC3 interacts with ROR1, a tyrosine kinase involved in pre-B cell receptor signaling pathway promoting cell proliferation (133). Finally, MORC3 plays a role in the transduction of calcium homeostasis regulator and is involved in bone remodeling (134). HMGCR, Encoded by the HMGCR Gene HMGCR is a limiting enzyme from the cholesterol biosynthetic chain, catalyzing the reduction reaction of HMGCoA to mevalonate (81). This enzyme has a large transmembrane domain in the endoplasmic reticular membrane and the harbored cytosolic N and C terminal domain. The catalytic domain is in the C-terminal domain (82, 83). This catalytic domain is the target of both statins and specific autoantibodies. Interestingly, it has been recently shown that its functions were indirectly crucial in many pathways. Indeed, HMGCR positively regulates the growth and migration of glioblastoma cells and could play a role in the metastatic capacities of tumoral cells, as well as other enzymes involved in lipidic metabolism (135). The role of HMGCR and the mevalonate pathway in oncogenesis has been suspected for two decades and cases of tumors have been reported to exhibit higher level and activity of HMGCR (136, 137). In 2010, Clendening et al. (138) confirmed this hypothesis by showing that the dysregulation of HMGCR promotes transformation of normal breast epithelial cells and growth of transformed cells (138). Moreover, different levels of HMGCR are associated with different response to radiotherapy in bladder cancer (139) and upregulation of HMGCR is also known to promote proliferation and migration of malignant cells in both glioblastoma and gastric cancer (135, 140). In vitro inhibition of HMGCR performed by specific miRNA limits proliferation, invasion, and metastasis process of breast cancer cells (141). TIF1γ, NXP2, and HMGCR: Targets of an Antitumoral Response? In the model proposed by Joseph and colleagues in systemic sclerosis, the immune response directed against POL3 was initially an antitumor immune response. Oncogenesis results from random and additive mutations on several genes some of which are involved in mitotic checkpoint, DNA repair, or differentiation, under the effect of viruses, oncogenes, or radiation. Later, tumoral cells can over-express, ectopically express, or express mutated forms of distinct proteins. The immune system may recognize these newly synthetized forms as neoantigens, which may ultimately lead to T cell and B cell responses (142, 143). In many human cancers, TIF1γ is considered as a tumor suppressor by inhibiting the TGF-β pathway. HMGCR also seems to be involved to some extent in oncogenesis through its role in metabolic pathways. Today, NXP2 is not directly known to be involved in cancer but it interacts with the well documented p53 tumor suppressor. Therefore, the targets of these three MSA appear to be proteins involved in cellular pathways that intervene in tumor promotion. Thus, it is reasonable to hypothesize that somatic mutations of TIF1γ, NXP2, or HMGCR genes in tumors may provoke a specific antitumoral immune response which may secondarily extend to the target organs of myositis (muscle, skin) by cross-reactivity and/or a process of epitope spreading (Figure 3). It is tempting to speculate that absence of cancer in some myositis patients may result from an efficacious antitumoral immune response: in this view, myositis might be the immunological price to pay for tumor eradication. FIGURE 3 Figure 3. Hypothetical model of specific antitumoral response as a trigger of dermatomyositis through the example of TIF1γ. According to this model, TIF1γ (or NXP2, or HMGCR) is modified in the tumor (gene mutation, overexpression, ectopic expression, posttranslational modification), leading to the development of a TIF1γ- (or NXP2-, or HMGCR-) specific T and B cell antitumor response. Secondarily, breakage of tolerance results from cross-reactivity and/or epitope spreading, promoting a response against muscle and skin. The hypothesis developed herein implies that the abovementioned antitumoral response provokes/sustains myositis. How this response could promote myositis is unclear. Indeed, the three TIF1γ, NXP-2, and HMGCR proteins are ubiquitously expressed, raising the question of how they could be muscle/skin-specific. In this regard, it should be remembered that most if not all MSAs, associated with cancer or not, are specific for ubiquitously expressed, intracellular proteins such as t-RNA synthetases, MDA-5, and signal recognition particles. Some elements support the direct pathogenicity of aAbs in some forms of myositis. For instance, aAb level correlates with clinical evolution in IMNM (72, 144–146). The pathogenic effect of aAbs is particularly real for anti-HMGCR aAb. Indeed, Arouche-Delaperche and colleagues recently demonstrated that anti-HMGCR aAbs promoted muscle atrophy and impaired regeneration and expression of inflammatory cytokines (147). A similar effect could be expected for anti-TIF1γ and NXP2 aAbs. TIF1γ has been shown to be over-expressed in regenerating muscle (148), which could explain the recognition of TIF1γ in muscle by aAb. The Therapeutic Dilemma of Cancer-Associated Myositis Most myositis clinical trials exclude patients with cancer, limiting the spectrum of our therapeutic knowledge. While beneficial for treating myositis, immunosuppression is of difficult use in the presence of cancer. Reciprocally, myositis alters the patient’s condition and complicates the therapeutic strategy, i.e., administering antitumoral drugs or performing surgery. Guidelines will be required to help manage patients in this context. The hypothesis of myositis triggered by cancer, rather than the opposite, supports a therapeutic strategy of performing an active antitumoral treatment compatible with the patient’s myositis status. Among new cancer therapies, immune checkpoint inhibitors may expose the patient to an exacerbation of autoimmunity that may yield a myositis flare. Since aAbs may be directly involved in myositis pathogenesis, therapeutic trials evaluating plasma exchanges are welcome. Conclusion Taken together, available data point to a unicist view of cancer-associated myositis, in which AID may result from an antitumoral response. This response may be triggered by mutations, overexpression, or posttranslational modification of the autoantigen in tumor. Elucidating these mechanisms will provide strong clues to better understand the potential role of cancer as a cause of autoimmunity. Author Contributions The first version of this review was written by AA and NC. OB revised it critically for important intellectual content. All authors approved the version to be published. Conflict of Interest Statement The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest. Acknowledgments We are grateful to Nikki Sabourin-Gibbs (Rouen University Hospital) for her help in editing the manuscript. 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Expression of the dermatomyositis autoantigen transcription intermediary factor 1γ in regenerating muscle. Arthritis Rheumatol (2015) 67(1):266–72. doi:10.1002/art.38863 CrossRef Full Text \| Google Scholar Keywords: cancer, myositis, autoimmunity, TIF1gamma, autoantibody Citation: Aussy A, Boyer O and Cordel N (2017) Dermatomyositis and Immune-Mediated Necrotizing Myopathies: A Window on Autoimmunity and Cancer. Front. Immunol. 8:992. doi: 10.3389/fimmu.2017.00992 Received: 23 June 2017; Accepted: 03 August 2017; Published: 21 August 2017 Edited by: Massimo Gadina, National Institute of Arthritis and Musculoskeletal and Skin Diseases, United States Reviewed by: Silvia Brunelli, University of Milano-Bicocca, Italy Juerg Hamacher, Lindenhof Hospital, Switzerland Copyright: © 2017 Aussy, Boyer and Cordel. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) or licensor are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms. Correspondence: Olivier Boyer, olivier.boyer@chu-rouen.fr Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher. Frontiers' impact Articles published with Frontiers have received 12 million total citations Your research is the real superpower - learn how we maximise its impact through our leading community journals Explore our impact metrics Download article Download PDF ReadCube EPUB XML Share on Export citation EndNote Reference Manager Simple Text file BibTex 12,9K Total views 3,7K Downloads 92 Citations Citation numbers are available from Dimensions View article impact View altmetric score Share on Edited by Massimo Gadina National Institute of Arthritis and Musculoskeletal and Skin Diseases (NIH), United States Reviewed by Silvia Brunelli University of Milano-Bicocca, Italy Juerg Hamacher Lindenhofspital, Switzerland Table of contents Abstract Introduction DM, Risk of Cancer, and Diagnostic Contribution of Autoantibodies Cancer-Associated Myopathies According to Autoantibodies IMNMs, New Players in Cancer-Associated Myositis? Targets of Cancer-Associated Autoimmune Response in Myositis TIF1γ, NXP2, and HMGCR: Targets of an Antitumoral Response? The Therapeutic Dilemma of Cancer-Associated Myositis Conclusion Author Contributions Conflict of Interest Statement Acknowledgments References Export citation EndNote Reference Manager Simple Text file BibTex Check for updates People also looked at Tissue-Resident Memory CD8+ T Cells: From Phenotype to Function David J. Topham and Emma C. 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1996	https://research.engineering.nyu.edu/~rlevicky/Files/Other/Handout14_6333.pdf	CBE 6333, R. Levicky 1 Potential Flow Part I. Theoretical Background. Potential Flow. Potential flow is irrotational flow. Irrotational flows are often characterized by negligible viscosity effects. Viscous effects become negligible, for example, for flows at high Reynolds number that are dominated by convective transport of momentum. Thus potential flow is often useful for analyzing external flows over solid surfaces or objects at high Re, provided the flows still remain laminar. Moreover, when the flow over a surface is rapid (high Re), the viscous boundary layer region (within which potential flow would be a bad assumption) that forms next to the solid body is very thin. Then, to a very good approximation, the presence of the boundary layer can be neglected when analyzing the potential flow region. That is, the potential flow can be assumed to follow the contours of the solid surface, as if the boundary layer was not present. When the thickness of the boundary layer is small compared to the dimensions of the object over which the potential flow is occurring, we can proceed as follows to analyze the total (potential flow + boundary layer flow) problem: i). First, determine the velocities and pressure distribution in the potential flow region, assuming that the potential flow extends all the way to any solid surfaces present (ie. neglecting the presence of the boundary layer). ii). Solve the flow inside the boundary layer using the pressure distribution obtained from the potential flow solution (i) as input. In other words, the potential flow imposes the pressure on the boundary layer (see the earlier discussion of boundary layers). At the edge of the boundary layer, the velocities are matched with those obtained from the potential flow solution (i) through the use of appropriate boundary conditions. The Velocity Potential. In potential flow the velocity field v is irrotational. This means that vorticity =  =   v = 0 (1) When   v = 0 the rate of rotation of an infinitesimal element of fluid is zero. From vector calculus (see the first handout on vector analysis) we know that if a velocity field is irrotational then it can be expressed as the gradient of a "scalar potential" : v = - (irrotational flow) (2) In equation (2),  is the "velocity potential." Using the definition of the  operator, in cartesian coordinates: v1 = -1 x   v2 = -2 x   v3 = -3 x   (2a) In cylindrical coordinates: vr = -r   v = -  Φ r 1 vz = -z Φ   (2b) CBE 6333, R. Levicky 2 In spherical coordinates: vr = -r Φ   v = -  Φ r 1 v = -    sin 1 r (2c) If, in addition, the flow is incompressible, then •v = 0  • = 0 what is equivalent to 2 = 0 (irrotational, incompressible flow) (3) The Stream Function. The stream function  can be defined for any two-dimensional flow, whether the flow is irrotational or not, compressible or incompressible. Two-dimensional means that at least one of the velocity components is zero (in other words, at most two of the velocity components are nonzero). Some flow types for which the stream function is useful, and the accompanying definitions of the stream function, are: Flow in Cartesian coordinates, with v3 = 0: v1 =     x2 v2 =    x1 (4a) Flow in cylindrical coordinates with vZ = 0: vr = -     r 1 v = r   (4b) Flow in cylindrical coordinates with v = 0: vr = z r   1 vZ = - r r   1 (4c) Flow in spherical coordinates with v = 0: vr = -     sin 1 2 r v = r r    sin 1 (4d) The definitions (4a) through (4d) are joined by the requirement that the stream function automatically satisfy the equation of continuity for an incompressible fluid, •v = 0. The divergence •v will automatically equal zero because of the equivalence of the mixed second derivatives of the stream function. For instance, for incompressible flow in Cartesian coordinates with v3 = 0 •v = 0 1 2 2 2 1 2 2 2 1 1                x x x x x v x v   (5) After the second equal sign in equation (5), the definitions (4a) of the stream function were used. The equality to zero comes about because the mixed second derivatives of  are equal. CBE 6333, R. Levicky 3 The stream function is constant along a streamline. The change d in the stream function due to an infinitesimal displacement ds along a streamline can be written as d =  • ds =  • vdt = t v x t v x d d 2 2 1 1        (6) In equation (6), the displacement ds along a streamline was expressed as v dt (t is time). Because the streamline is everywhere tangent to the velocity field v, ds and v must point in the same direction. Both vectors are infinitesimally small, and it is then justified to take ds as equal to vdt. Inserting equations (4a) into (6), d = (v2v1 - v1v2) dt = 0 (7) Since the change d in the stream function arising from a displacement along a streamline is zero, the stream function must be constant along a streamline. The change in the stream function between a pair of streamlines equals the volumetric flowrate between those two streamlines. Consider the flow in Figure 1. We will define flowrate from right to left as positive. We wish to calculate the volumetric flowrate Q' (per unit width into the page) that occurs between streamlines 1 and 2. To calculate Q', we will integrate the rate at which fluid flows across the curve that connects points 1 and 2 in the figure. The flowrate Q' is given by Q' = -   2 1 dl n v = -     2 1 2 2 1 1 dl n v n v (8) where dl is an infinitesimal displacement along the curve and n is a unit normal to the curve pointing opposite to the fluid flow (see Figure 1). n1 is the component of n along the x1 direction, v1 is the velocity component along the x1 direction, etc. Figure 1 From figure 1, n1dl = n cos dl = n dx2 = dx2 (9) CBE 6333, R. Levicky 4 In equation (9), we first set n1 = n cos, where n is the magnitude of n and  is as indicated in Figure 1. Second, we observed that cos dl equals the displacement dx2 that occurs in the x2 direction when position along the curve changes by dl; thus, we substituted cos dl = dx2. Finally, the magnitude n of n was set equal to 1 since n is a unit vector. Similarly, it can be shown that n2dl = n sin dl = - n dx1 = - dx1 (10) In equation (10), we used dx1 = -sin dl. The minus sign must be included because the displacement dx1 that occurs in the x1 direction when position along the curve shifts by dl is negative, although sin dl evaluates to a positive number. Inserting equations (9) and (10) into (8), Q' =           2 1 1 1 2 2 1 2 2 1 2 1 ) d d ( ) d d ( x x x x x v x v   Q' = 1 2 2 1 d       (11) In arriving at equation (11), we used the fact that d = (/x1)dx1 + (/x2)dx2. Equation (11) states that the volumetric flowrate Q', per unit width into the page, between streamlines 1 and 2 equals the difference in the stream function. Note that Q' will be positive for flow from right to left, and negative for flow from left to right. Lines of constant  (streamlines) are perpendicular to lines of constant  (velocity potential lines). On a streamline,  is constant, so that d =  • ds = 0 d d 2 2 1 1       x x x x   (12) The stream function does not change (ie. d = 0) because the displacement ds = 1dx1 + 2dx2 is taken along a streamline. Making use of equations (4a), equation (12) becomes v2dx1 - v1dx2 = 0 or, after rearrangement, dx2 / dx1 = v2 / v1 (13) Equation (13) gives the slope of the streamline, as depicted in Figure 2. Figure 2. On a line of constant velocity potential,  is constant. Therefore, CBE 6333, R. Levicky 5 d =  • ds = 0 d d 2 2 1 1       x x Φ x x Φ (14) In equation (14), the displacement ds occurs along a line of constant velocity potential. Inserting equations (2a) into equation (14), - v1dx1 - v2dx2 = 0 or, after rearrangement, dx2 / dx1 = - v1 / v2 (15) Equation (15) gives the slope of a line of constant velocity potential. Since the slopes of a streamline (equation (13)) and of a line of constant velocity potential (equation (15)) are negative reciprocals of one another, streamlines and velocity potential lines must be mutually perpendicular. For irrotational, incompressible, two-dimensional flows, the stream function and the velocity potential obey the same differential equation. From equation (3) for irrotational, incompressible flow, 2 = 0 (irrotational, incompressible flow) (3) The condition of irrotationality means that   v = ( 1 2 x v   - 2 1 x v   ) 3 = 0 (16) In equation (16), for convenience the two-dimensional flow was assumed to take place in the x1x2 plane, so that v3 = 0. Equation (16) requires that 1 2 x v   - 2 1 x v   = 0 (16b) Inserting the definitions of the velocities in terms of the stream function (equations (4a)) into equation (16b) 0 2 2 2 2 1 2       x x   which is equivalent to 2 = 0 (2D, irrotational flow) (17) Equations (3) and (17) show that for a two-dimensional, irrotational, incompressible flow, the velocity potential and the stream function both obey "Laplace's equation": 2 = 0 and 2 = 0 (18) CBE 6333, R. Levicky 6 Functions that obey Laplace's equation are called "harmonic" functions. Therefore, under these conditions both the stream function and the velocity potential are harmonic. Important: Since  and  obey the same differential equation (for 2D, irrotational, incompressible flow), a solution to one potential flow problem can be directly used to generate a solution to a second potential flow by interchanging  and . Specifically, if 1 and 1 represent potential flow 1, then the interchange 2 = 1 and 2 = -1 will represent some other potential flow 2. The minus sign in 2 = -1 is needed to ensure consistency in the sign of the velocity components derived from  and  (see examples below). Different potential flows can be added together to generate new potential flows (the Principle of Superposition). Laplace's equation (equation (18)) is linear. The linear property means that if the stream function and velocity potential are known for two different flows, say flows 1 and 2, then the sum of flows 1 and 2 will also be a solution to Laplace's equation. By "solution to Laplace's equation" we mean that, if 21 = 0 and 22 = 0 then 2(1 + 2) = 21 + 22 = 0 (19) Here, 1 is the stream function for flow 1 and 2 is the stream function for flow 2. Equation (19) shows that the sum of the stream functions for flows 1 and 2 is also a solution to Laplace's equation. Same comments apply to the sum of the velocity potentials, 1 + 2, for flows 1 and 2. Now, if  and  of a potential flow equal the sum of the stream functions and velocity potentials of two other flows 1 and 2, so that  = 1 + 2 and  = 1 + 2, then the velocity field of that flow will equal the sum of the velocity fields of flows 1 and 2. This statement can be easily verified. For example, in terms of the stream functions (equivalently, we could have used the velocity potentials), and using a single prime to indicate velocities for flow 1 and double prime those for flow 2, Flow 1 (stream function 1): v1' = 2 1 x     v2' = 1 1 x   (20) Flow 2 (stream function 2): v1'' = 2 2 x     v2'' = 1 2 x   (21) Flow 1 + Flow 2 (stream function  = 1 + 2): v1 = ' ' 1 ' 1 2 2 2 1 2 ) 2 1 ( 2 v v x x x x                       (22) v2 = ' ' 2 ' 2 1 2 1 1 1 ) 2 1 ( 1 v v x x x x                    (23) CBE 6333, R. Levicky 7 Equations (22) and (23) show that if the stream functions from flows 1 and 2 are added to create a new flow, then the velocity field of the new flow will equal the sum of the velocity fields of the two constituent flows. We thus see that the addition of stream functions or velocity potentials physically results in the superposition (addition) of the flows represented by those stream functions and velocity potentials. This superposition principle is a powerful way to generate solutions to potential flow problems. We will look at an example of superposition in the next section. Part II. Applications. Uniform Potential Flow. Figure 3 depicts the case of uniform flow in the x1 direction. For this uniform flow, v1 = Vo v2 = 0 Therefore, -   x Vo 1    = - Vox1 + f(x2) (24) The derivative of  with respect to x2 is zero,  / x2 = - v2 = 0. Therefore, the function of integration f(x2) in equation (24) can at most be a constant. Since a constant of integration would not influence the velocities obtained from , the value of the constant is arbitrary. For simplicity, we set the constant equal to 0 so that equation (24) becomes  = -Vox1 (25a) Figure 3. Figure 4. By using equations (4a), it is straightforward to show that the stream function is  = -Vox2 (25b) CBE 6333, R. Levicky 8 Figure 4 plots the lines of constant  and constant . As discussed earlier, since lines of constant  are streamlines, they point along the flow direction. Also, as pointed out earlier, the streamlines and lines of constant  are mutually orthogonal. If we choose the streamlines for x2 = 1 and x2 = 3 in Figure 4, according to equation (11) the volumetric flowrate between the streamlines is, Q' = 3 - 1 = -3Vo - (-Vo) = -2 Vo (26) Recall that Q' is the volumetric flowrate per unit width into the page. Because the flow occurs from left to right, according to our earlier convention Q' comes out to a negative number. If  and  are interchanged, then  = Vox1 (27a)  = -Vox2 (27b) Note that, to preserve consistency of sign in the definition of the flow fields derived from  and , a minus sign was inserted in (27a). Equations (27) again describe a uniform potential flow, but now the flow field is (you can easily verify it) v1 = 0 and v2 = Vo The interchange of  and  thus resulted in a uniform flow in the x2 direction. In the context of Figure 4, the streamlines and the lines of constant velocity potential were interchanged. Source and Sink Flows. In two dimensions, a source is a line (into the page) from which fluid flows outward, and a sink is a line at which fluid flows inward and is removed (Figure 5). For these flows, v = 0 and vr = Q'/(2r) (28) Figure 5. In equation (28), Q' is the total volumetric flowrate outward from the source, per unit depth into the page. Q' > 0 for a source, Q' < 0 for a sink. Equation (28) for vr ensures that the total radial volumetric flowrate is constant. In other words, at each radial distance r from the source or sink the volumetric flowrate per unit depth, given by 2r vr, equals the constant Q'. This volumetric flowrate is the product of the area per unit depth into the page (ie. 2rW would be the area of flow if the extent of the source into the page is W), times the radial flow velocity vr. Using equations (2b) and proceeding as for the case of uniform flow, the velocity potential is found to be CBE 6333, R. Levicky 9 -r  = vr = Q'/(2r)   = - Q'/(2) lnr + f()   = - Q'/(2) lnr (29) In equation (29), the function of integration f() was set to zero since v = -  Φ r 1 = 0 implies that at most f can be an arbitrary constant. Similarly, the stream function can be found using equations (4b) -     r 1 = vr = Q'/(2r)   = - Q'/(2)  + f(r)   = - Q'/(2)  (30) where similar arguments can be made to set f(r) = 0. From equation (29), lines of constant  are equivalent to lines of constant r, while from equation (30) we see that streamlines (lines of constant ) coincide with lines of constant . The velocity potential lines and streamlines are illustrated in Figure 6. Again, we see that the curves of  and  are orthogonal, as required. Figure 6. Potential Vortex Flow. If the velocity potential and stream function for the source/sink flow are interchanged, we get a new flow for which  = Q'/(2) lnr and  = - Q'/(2)  (31) The velocities for this new flow can be derived from  and  by using equations (4b) or (2b).The result is vr = 0 and v = Q'/(2r) (32) Equations (32) describe so-called "potential vortex" flow, a useful model for phenomena such as tornadoes and whirlpools. In a potential vortex, fluid moves in concentric circles with a velocity v that decreases as 1/r, where r is the radial distance from the center of the vortex. The diagram of streamlines and velocity potential lines for a potential vortex looks exactly like Figure 6, except that the streamlines and velocity potential lines are interchanged. When referring to potential vortex flow, it is customary to CBE 6333, R. Levicky 10 use the symbol  instead of Q'.  / 2 is known as the "strength of the vortex."  is positive for counterclockwise vortex flow, and negative for clockwise vortex flow. From equation (32) it is clear that the higher  / 2 is, the greater the velocity v. In terms of , the potential vortex equations are  = /(2) lnr  = - /(2)  v = /(2r) (33) Superposition of Potential Flows. As an example of the superposition of two potential flows, we will superpose (add) a uniform and a potential vortex flow. The functions  and  become,  = -Vox2 + /(2) lnr = -Vox2 + /(4) ln(x1 2 + x2 2) (34)  = -Vox1 - /(2)  = -Vox1 - /(2) tan-1(x2 / x1) (35) In equations (34) and (35), the first term on the right comes from the uniform flow and the second from the potential vortex flow.  and  have been expressed in cartesian coordinates, and so can be used to calculate the CCS velocity components v1 and v2 directly. For instance, equations (4a) can be applied to the stream function to yield v1 = Vo -  x2 /{2 (x1 2 + x2 2)} (36) v2 =  x1 /{2 (x1 2 + x2 2)} (37) Setting Vo = 1 m/s and  / 2 = 2 m2/s, the resultant flow is depicted in Figure 7. At each point an arrow shows the direction of the local velocity field. The length of the arrows is fixed, and so does not represent the magnitude of the local velocity. There is a "stagnation point" on the top side of the vortex at the position (0,2). At a stagnation point, all velocity components are zero. CBE 6333, R. Levicky 11 -2 0 2 0 2 4 stagnation point x2 (m) x1 (m) Figure 7. The Method of Images. At a solid boundary, the magnitude of potential flow perpendicular to the boundary must be zero (i.e. no fluid flow can occur across the solid surface). The previous statement is approximate in that the potential flow only extends to the outer edge of a boundary layer, where a small, perpendicular velocity component is typically present (e.g. as could be obtained from the numerical Blasius solution for the laminar boundary layer over a flat surface). However, if the rate of boundary layer growth along the surface is very slow, as will be true if the flow over the surface is sufficiently rapid, then the perpendicular velocity component will be small and so will the error introduced by neglecting it. Since no flow occurs across a streamline, a streamline meets the same stipulation as a solid boundary. Indeed, any streamline of a potential flow can be considered as a solid surface, with no resultant change to the flow pattern. Since fluid flows along a streamline, thinking of a streamline as a solid surface implies flow (i.e. slip) along that surface. This implication does not pose a physical paradox however since in reality potential flow does not extend all the way to a surface (it only extends to the outer edge of the boundary layer) and therefore is not subject to a no-slip boundary condition. The method of images is a technique by which streamlines representing solid surfaces can be generated in a systematic fashion. This approach takes advantage of the fact that there can be no flow across lines of symmetry separating identical flows. For instance, the left side of Figure 8 shows two identical sources; due to symmetry, it must be true that on the line of symmetry at x2 = 0 the velocity in the x2 direction is zero. The line x2 = 0 is a streamline that is coincident with the x1 axis. According to the previous paragraph, this streamline can be thought of as a solid surface. Thus the flow resulting from the superposition of the two identical sources in Fig. 8 left will be a solution to the problem of a single source next to a solid wall, as illustrated in Fig. 8 right. CBE 6333, R. Levicky 12 Figure 8. Left: A pair of identical sources, both of strength Q', located at x2 = a and x2 = -a. The sources are "images" of one another. Right: A single source next to a solid wall. From equation (29), the velocity potential for a single source centered at the origin is:  = - Q'/(2) lnr = - Q'/(4) ln(x1 2 + x2 2) (38) If the source is centered at x2 = a, then  = - Q'/(4) ln(x1 2 + (x2 - a)2) (39) As indicated above, the velocity potential t for the case of a single source a distance a from a solid wall can be found from the superposition of two identical source flows, one located at x2 = a and the other at x2 = -a,  t = - Q'/(4) ln(x1 2 + (x2 - a)2) - Q'/(4) ln(x1 2 + (x2 + a)2) (40) Equation (40) is the velocity potential for both of the flows depicted in Figure 8. Of course, for the single source next to a solid wall only the region x2 > 0 is relevant. The stream function can be obtained by identical procedures, and the velocities can be calculated from either the velocity potential or the stream function. The method of images works by superposing identical flows in order to create lines of symmetry across which there is no flow. This approach can be helpful in devising solutions for flows near solid surfaces when the location of the surfaces can be made coincident with the lines of symmetry. To repeat, since there is no flow across a streamline, any streamline for any potential flow can be viewed as a solid surface. This principle can be very powerful in analyzing potential flows even for rather complex geometries. CBE 6333, R. Levicky 13 The Use of Bernoulli Equation in Potential Flow. By definition, potential flows are irrotational. If in addition the flow is incompressible, frictionless, and steady state, then the Extended Bernoulli Equation (EBE) becomes (for details, refer to the earlier handout on Bernoulli equation) V V p p g z z ws 2 2 1 2 2 2 1 2 1        ( ) (41) Since potential flow is irrotational, the Bernoulli equation holds between any two points 1 and 2, so that the points do not have to lie on the same streamline. We have already come across this fact during our previous derivation of the Bernoulli's equation. In contrast, if the vorticity is not zero, then points 1 and 2 must be located on the same streamline. In potential flow, Bernoulli's equation is most often used to compute the pressure distribution. Typically, the shaft work term will be zero. Then once the velocity field is calculated (perhaps using one of the methods described above), the pressure distribution p2 - p1 is obtained directly from equation (41).
1997	https://www.youtube.com/watch?v=ABPm9gYctQU	Geometry Construction - Perpendicular to Line through Point turksvids 29700 subscribers 8 likes Description 2668 views Posted: 18 Jan 2016 Using GeoGebra for Compass & Straight Edge Constructions. In this video we construct a line perpendicular to a given line through a given point. CCSS.MATH.CONTENT.HSG.CO.D.12 Transcript: okay in this video we're going to do another geometric construction um in this case uh we're going to construct perpendicular line from uh when we're given a point and a line so we want the perpendicular through a point um to a particular line so we have basically this situation um and the way we do this uh if you know how to construct a midpoint uh it's really really similar so what we'll do is I'm going to start off I'm going to put an extra point on uh this just so that I can fix my compass so what I'm going to do is I'm going to go from C to D make a circle uh with the compass so from C to D well actually I'm going to go from D to C pops up and then I'm going to click again to fix it there so have that uh now what I want to do is I want to find this intersection point right here okay so I have that and now what I'm basically going to do is uh the same deal that I did when I found the mid point of a segment so I want to find the midpoint of um de and if I can find the midpoint of De that'll allow me to draw a line segment through that um through the midpoint and c and that line will be perpendicular to AB and then I'll be done so let's try to do that so uh first I'm going to go from E to D and click and I'm going to go from D to e and click did I miss it and click okay so first thing I did was I started at Point C and I went from C to D made a circle and then I found the intersection the additional intersection at e and then I used uh D to e to make one circle e to D to make another Circle and then what I'm actually going to do is I'm just going to find an intersection any intersection of this circle and this circle will allow me to draw the perpendicular so I'm going to use the one that's down here because it's the most obvious I guess that um now if I use this tool to create a line well obviously I could just use a perpendicular line tool geogebra um so I click that and so my claim is that uh this line will be you can see the way that I'm doing these constructions sometimes they disappear so if I go whoops because the things I use to construct it are gone won't work but if I find this point uh right here here and all I got to do is measure this angle it should be 90° and it should stay 90° forever and there you go 90 degrees so if I move this stays 90 degrees so there you go um turns out you use like tons of circles when you're doing constructions which makes sense because you're using a straight edge and a compass and a compass really is just a tool that creates circles um in the real world you don't make the entire circle you just swing out tiny little arcs um but I don't have that option because I'm using geogebra but anyway there's your construction of a line perpendicular to a given line through a given point hope you found this helpful and uh good luck
1998	https://fermi.jhuapl.edu/denscalc.html	Sea Water Equation of State Calculator HomeStaffResearchLinksOutreachSoftware DensCalc A Sea Water Equation of State Calculator ======================================== The JavaScript calculator below will allow you to compute the UNESCO International Equation of State (IES 80) as described in Fofonoff (1985). The calculator is quite flexible. To use the calculator: 1. You must enter either the depth or the pressure. 2. Check any 2 of the bottom 4 checkboxes to indicate what parameters will be entered. 3. Enter changes to the checked parameters and the other parameters will be automatically recalculated. The depth and pressure are related according to a simplified approximation for a standard ocean (by convention an ocean at 0 deg C and 35 psu). The depth label includes an approximation symbol as a reminder. The quantities below the horizontal rule (Sound speed,...) are output only. - [x] ~Depth:m - [x] Pressure:dbar - [x] Temperature:deg C - [x] Conductivity:S/m - [x] Salinity:PSU - [x] Density:kg/m^3 Sound speed:m/s Specific heat:J/(kg °C) Freezing point:°C Adiabatic lapse rate:mdeg/dbar Potential temperature:°C You must have JavaScript enabled in your browser to work with this calculator. Click here for an alternative version of this calculator that computes temperature, salinity and the speed of sound in seawater. This could be useful for estimating salinity when only temperature and sound speed measurements are available. ### DensCalc - A Palm OS Density Calculator: An improved version of this calculator, called DensCalc, is now available for the Palm OS. Click on the image of the Palm to read more about this free program or to download DensCalc. ### DISCUSSION: The equation of state of water is a complicated curve fit to very precise measurements. The relationship between the Practical Salinity Scale and older measures of salinity (typically measured in parts per thousand) is nearly, but not quite 1:1. In addition the conversion from absolute conductivity to salinity is discussed in some detail in Fofonoff. The interested reader is urged to read that article for a more complete discussion of these issues. The general relationship between temperature, salinity and density is illustrated in the figure below. The isopycnals (lines of constant density) are labelled in units of sigma-t, where sigma-t = (density-1000)(kg/m^3). This is a standard oceanographic notation. Although the actual equation of state is quite complex there is a simple rule of thumb to remember that can be quite helpful: density increases by roughly 1 part per 1000 when the temperature decreases by 5 deg C, the salinity increases by 1 psu or the pressure increases by 200 dbar (about 200 m depth). #### Range of Validity: The underlying equations are valid for temperatures from -2 to 35 deg C, pressures from 0 to 10,000 dbar, and practical salinity from 2 to 42. #### Numerics: The Javascript functions contained within this page provide a direct calculation of salinity from temperature, conductivity and pressure, and density from temperature, salinity and pressure. All of the other calculations are done by numerical inversion. The error criterion for convergence is set to 0.001% of the iterated parameter, but the reader is cautioned that the author cannot certify the numerical accuracy and precision of Javascript on your system. While the numerics are reasonably accurate for "reasonable" values, I have not incorporated extensive error checking. Thus it is easy to enter nonsensical values and get nonsensical results. The underlying equations are valid for temperatures from -2 to 35 deg C, pressures from 0 to 10,000 dbar, and practical salinity from 2 to 42. The actual Javascript code that performs the calculations is easily accessible. With this page open in your broswer, select "Source" under the "View" menu. The functions within the code can be used in other Javascript programs or are easily ported to C or Java. C-language versions of the routines are also available by downloading the source code to the Palm OS version of this calculator. (You don't need a Palm to look at the source code.) #### Relationship of Depth and Pressure: This calculator relates depth and pressure according to the quadratic relationship proposed by Saunders (1981) evaluated at a latitude of 30 deg. This is a simplified approximation to a more complicated form describing the depth/pressure relationship for a standard ocean (by convention this is an ocean at 0 deg C and 35 psu). The latitude enters in through the variation in gravity, effecting the depth of the 5000 dBar surface at a rate of 40 cm per degree of latitude at 30 deg. The net result is that the depth is at best an approximation. At 30 deg latitude it should be good to a few meters, but the errors will grow to as much as 15 m near the poles. #### Units: The pressures in this calculator are not absolute pressures, but gauge pressures. The difference is that gauge pressure is measured relative to 1 standard atmosphere, so the gauge pressure at the surface is 0, not 1000 mbar. I've also used decibar's throughout. This reflects common oceanographic usage, but is not accepted as an SI unit. The internationally preferred units are Pascals, where 1 dbar = 1e4 Pascal. #### Auxiliary Parameters: The calculator computes the values for several additional properties of seawater based on temperature, salinity and pressure. These include the following properties: Speed of sound (units m/s) Specific heat: the energy in Joules required to to raise the temperature of one kg of seawater one °C at constant pressure (units Joules/(kg °C)) Freezing point: valid for practical salinity ranging from 4 to 40 (units °C) Adiabatic lapse rate: the change of temperature per unit pressure for an adiabatic change of pressure of an element of seawater (units millideg/decibar) Potential temperature: temperature an element of seawater would have if raised adiabatically with no change in salinity to atmospheric pressure (units °C) The speed of sound calculation is based on Chen and Millero (1977). The actual Javascript was a port of Fortran code provided to me by Pascal Vernin (pvernin@cea.fr). The values for the other four parameters are based on algorithms described in Fofonoff and Millard (1983). #### Other Formulations: Other formulations of the equation of state exist and may be superior for some applications. Feistel (2003) describes a Gibbs potential formulation of seawater thermodynamics that is claimed to be more accurate than the IES 80 and is valid over a slightly wider range of temperatures. Feistel (2005) describes the numerical implementation of these functions. Wright (1977) produced an equation of state formulation for efficient use in numerical models. McDougall, et al. (2005) likewise modified the 2003 Gibbs potential functions for more efficient computation. Anati (1999) discussed the problems in the measurement of salinity in highly saline brines and showed how the density becomes strongly dependent on the ionic composition of the brine. #### References: Anati, D. A., "The salinity of hypersalinme brines: Concepts and misconceptions," International Journal of Salt Lake Research, Vol. 8, 55-70, 1999. Chen and Millero, "Speed of sound in seawater at high pressures," Journal of the Acoustical Society of America, Vol. 62, No. 5, 1129-1135, Nov 1977. Feistel, R. "A new extended Gibbs thermodynamic potential of seawater," Progress in Oceanography, 58, 43-115, 2003, Corrigendum, 61, 99, 2004. Feistell, R., "Numerical implementation and oceanographic application of the Gibbs thermodynamic potential of seawater," Ocean Science, 1, 9-16, 2005. Fofonoff, N. P., "Physical Properties of Seawater: A New Salinity Scale and Equation of State of Seawater," Journal of Geophysical Research, Vol 90 No. C2, pp 3332-3342, March 20, 1985. Fofonoff and Millard, "Algorithms for computation of fundamental properties of seawater," UNESCO Technical papers in marine science No. 44, 1983. McDougall, T. J., Jackett, D. R., Wright, D. G., and Feistel, R., "Accurate and Computationally Efficient Algorithms for Potential Temperature and Density of Seawater," Journal of Atmospheric and Oceanic Technology, 20, 730–741, 2003. Saunders, Journal of Physical Oceanography, Vol. 11, pp. 573-574, April 1981. UNESCO, "Background papers and supporting data on the International Equation of State of Seawater 1980," UNESCO Technical papers in marine science No. 38, 1981. Wright, D. G., "An equation of state for use in ocean models: Eckart's formula revisited," Journal of Atmospheric and Oceanic Technology, 14, 735– 740, 1997. #### Revision History: Text was updated on November 24, 2006 to include reference list and links to UNESCO papers. The calculator was updated on January 22, 2006. A description of how to access the Javascript code was added to address a frequenctly asked question. Additional comments were also added to the Javascript code to make it slightly easier to read. This calculator was updated on June 17, 1998 and again on November 8, 1999 to fix bugs in the interpretation of pressure units (dbar vs bar). The updated version has been checked against additional test points and seems to be accurate. The calculator now also updates automatically if the pressure or depth units are changed. The depth and pressure calculations were updated on Feb 23, 2000, based on the suggestion of Pascal Vernin. Rick Chapman © The Johns Hopkins University Applied Physics Laboratory Update:Nov 26 17:17:54 2006
1999	https://en.wikipedia.org/wiki/IQ_classification	Jump to content IQ classification العربية Català Español فارسی Français 한국어 Հայերեն हिन्दी 中文 Edit links From Wikipedia, the free encyclopedia Categorisation of people's intelligence based on IQ This article is about the classifications of IQ. For the history and applications of IQ, see Intelligence quotient. IQ classification is the practice of categorizing human intelligence, as measured by intelligence quotient (IQ) tests, into categories such as "superior" and "average". With the usual IQ scoring methods, an IQ score of 100 means that the test-taker's performance on the test is of average performance in the sample of test-takers of about the same age as was used to norm the test. An IQ score of 115 means performance one standard deviation above the mean, while a score of 85 means performance one standard deviation below the mean, and so on. This "deviation IQ" method is used for standard scoring of all IQ tests in large part because they allow a consistent definition of IQ for both children and adults. By the existing "deviation IQ" definition of IQ test standard scores, about two-thirds of all test-takers obtain scores from 85 to 115, and about 5 percent of the population scores above 125 (i.e. normal distribution). When IQ testing was first created, Lewis Terman and other early developers of IQ tests noticed that most child IQ scores come out to approximately the same number regardless of testing procedure. Variability in scores can occur when the same individual takes the same test more than once. Further, a minor divergence in scores can be observed when an individual takes tests provided by different publishers at the same age. There is no standard naming or definition scheme employed universally by all test publishers for IQ score classifications. Even before IQ tests were invented, there were attempts to classify people into intelligence categories by observing their behavior in daily life. Those other forms of behavioral observation were historically important for validating classifications based primarily on IQ test scores. Some early intelligence classifications by IQ testing depended on the definition of "intelligence" used in a particular case. Contemporary IQ test publishers take into account reliability and error of estimation in the classification procedure. Differences in individual IQ classification [edit] IQ scores can differ to some degree for the same person on different IQ tests, so a person does not always belong to the same IQ score range each time the person is tested (IQ score table data and pupil pseudonyms adapted from description of KABC-II norming study cited in Kaufman 2009). \| Pupil \| KABC-II \| WISC-III \| WJ-III \| \| Asher \| 90 \| 95 \| 111 \| \| Brianna \| 125 \| 110 \| 105 \| \| Colin \| 100 \| 93 \| 101 \| \| Danica \| 116 \| 127 \| 118 \| \| Elpha \| 93 \| 105 \| 93 \| \| Fritz \| 106 \| 105 \| 105 \| \| Georgi \| 95 \| 100 \| 90 \| \| Hector \| 112 \| 113 \| 103 \| \| Imelda \| 104 \| 96 \| 97 \| \| Jose \| 101 \| 99 \| 86 \| \| Keoku \| 81 \| 78 \| 75 \| \| Leo \| 116 \| 124 \| 102 \| IQ tests generally are reliable enough that most people 10 years of age and older have similar IQ scores throughout life. Still, some individuals score very differently when taking the same test at different times or when taking more than one kind of IQ test at the same age. About 42% of children change their score by 5 or more points when re-tested. For example, many children in the famous longitudinal Genetic Studies of Genius begun in 1921 by Lewis Terman showed declines in IQ as they grew up. Terman recruited school pupils based on referrals from teachers, and gave them his Stanford–Binet IQ test. Children with an IQ above 140 by that test were included in the study. There were 643 children in the main study group. When the students who could be contacted again (503 students) were retested at high school age, they were found to have dropped 9 IQ points on average in Stanford–Binet IQ. Some children dropped by 15 IQ points or by 25 points or more. Yet parents of those children thought that the children were still as bright as ever, or even brighter. Because all IQ tests have error of measurement in the test-taker's IQ score, a test-giver should always inform the test-taker of the confidence interval around the score obtained on a given occasion of taking each test. IQ scores are ordinal scores and are not expressed in an interval measurement unit. Besides the reported error interval around IQ test scores, an IQ score could be misleading if a test-giver failed to follow standardized administration and scoring procedures. In cases of test-giver mistakes, the usual result is that tests are scored too leniently, giving the test-taker a higher IQ score than the test-taker's performance justifies. On the other hand, some test-givers err by showing a "halo effect", with low-IQ individuals receiving IQ scores even lower than if standardized procedures were followed, while high-IQ individuals receive inflated IQ scores. The categories of IQ vary between IQ test publishers as the category labels for IQ score ranges are specific to each brand of test. The test publishers do not have a uniform practice of labeling IQ score ranges, nor do they have a consistent practice of dividing up IQ score ranges into categories of the same size or with the same boundary scores. Thus psychologists should specify which test was given when reporting a test-taker's IQ category if not reporting the raw IQ score. Psychologists and IQ test authors recommend that psychologists adopt the terminology of each test publisher when reporting IQ score ranges. IQ classifications from IQ testing are not the last word on how a test-taker will do in life, nor are they the only information to be considered for placement in school or job-training programs. There is still a dearth of information about how behavior differs between people with differing IQ scores. For placement in school programs, for medical diagnosis, and for career advising, factors other than IQ can be part of an individual assessment as well. The lesson here is that classification systems are necessarily arbitrary and change at the whim of test authors, government bodies, or professional organizations. They are statistical concepts and do not correspond in any real sense to the specific capabilities of any particular person with a given IQ. The classification systems provide descriptive labels that may be useful for communication purposes in a case report or conference, and nothing more. — Alan S. Kaufman and Elizabeth O. Lichtenberger, Assessing Adolescent and Adult Intelligence (2006) IQ classification tables [edit] There are a variety of individually administered IQ tests in use. Not all report test results as "IQ", but most report a standard score with a mean score level of 100. When a test-taker scores higher or lower than the median score, the score is indicated as 15 standard score points higher or lower for each standard deviation difference higher or lower in the test-taker's performance on the test item content. Wechsler Intelligence Scales [edit] Main article: Wechsler Adult Intelligence Scale See also: Wechsler Intelligence Scale for Children and Wechsler Preschool and Primary Scale of Intelligence The Wechsler intelligence scales were developed from earlier intelligence scales by David Wechsler. David Wechsler, using the clinical and statistical skills he gained under Charles Spearman and as a World War I psychology examiner, crafted a series of intelligence tests. These eventually surpassed other such measures, becoming the most widely used and popular intelligence assessment tools for many years. The first Wechsler test published was the Wechsler–Bellevue Scale in 1939. The Wechsler IQ tests for children and for adults are the most frequently used individual IQ tests in the English-speaking world and in their translated versions are perhaps the most widely used IQ tests worldwide. The Wechsler tests have long been regarded as the "gold standard" in IQ testing. The Wechsler Adult Intelligence Scale—Fourth Edition (WAIS–IV) was published in 2008 by The Psychological Corporation. The Wechsler Intelligence Scale for Children—Fifth Edition (WISC–V) was published in 2014 by The Psychological Corporation, and the Wechsler Preschool and Primary Scale of Intelligence—Fourth Edition (WPPSI–IV) was published in 2012 by The Psychological Corporation. Like all contemporary IQ tests, the Wechsler tests report a "deviation IQ" as the standard score for the full-scale IQ, with the norming sample mean raw score defined as IQ 100 and a score one standard deviation higher defined as IQ 115 (and one deviation lower defined as IQ 85). During the First World War in 1917, adult intelligence testing gained prominence as an instrument for assessing drafted soldiers in the United States. Robert Yerkes, an American psychologist, was assigned to devise psychometric tools to allocate recruits to different levels of military service, leading to the development of the Army Alpha and Army Beta group-based tests. The collective efforts of Binet, Simon, Terman, and Yerkes laid the groundwork for modern intelligence test series. Wechsler (WAIS–IV, WPPSI–IV) IQ classification \| IQ Range ("deviation IQ") \| IQ Classification \| \| 130 and above \| Very Superior \| \| 120–129 \| Superior \| \| 110–119 \| High Average \| \| 90–109 \| Average \| \| 80–89 \| Low Average \| \| 70–79 \| Borderline \| \| 69 and below \| Extremely Low \| Wechsler Intelligence Scale for Children–Fifth Edition (WISC-V) IQ classification \| IQ Range ("deviation IQ") \| IQ Classification \| \| 130 and above \| Extremely High \| \| 120–129 \| Very High \| \| 110–119 \| High Average \| \| 90–109 \| Average \| \| 80–89 \| Low Average \| \| 70–79 \| Very Low \| \| 69 and below \| Extremely Low \| Psychologists have proposed alternative language for Wechsler IQ classifications. The term "borderline", which implies being very close to being intellectually disabled (defined as IQ under 70), is replaced in the alternative system by a term that doesn't imply a medical diagnosis. Alternate Wechsler IQ Classifications (after Groth-Marnat 2009) \| Corresponding IQ Range \| Classifications \| More value-neutral terms \| \| 130 and above \| Very superior \| Upper extreme \| \| 120–129 \| Superior \| Well above average \| \| 110–119 \| High average \| High average \| \| 90–109 \| Average \| Average \| \| 80–89 \| Below average \| Below average \| \| 70–79 \| Borderline \| Well below average \| \| 69 and below \| Extremely low \| Lower extreme \| Stanford–Binet Intelligence Scale Fifth Edition [edit] Main article: Stanford–Binet Intelligence Scales The fifth edition of the Stanford–Binet scales (SB5) was developed by Gale H. Roid and published in 2003 by Riverside Publishing. Unlike scoring on previous versions of the Stanford–Binet test, SB5 IQ scoring is deviation scoring in which each standard deviation up or down from the norming sample median score is 15 points from the median score, IQ 100, just like the standard scoring on the Wechsler tests. Stanford–Binet Fifth Edition (SB5) classification \| IQ Range ("deviation IQ") \| IQ Classification \| \| 140 and above \| Very gifted or highly advanced \| \| 130–139 \| Gifted or very advanced \| \| 120–129 \| Superior \| \| 110–119 \| High average \| \| 90–109 \| Average \| \| 80–89 \| Low average \| \| 70–79 \| Borderline impaired or delayed \| \| 55–69 \| Mildly impaired or delayed \| \| 40–54 \| Moderately impaired or delayed \| \| 19-39 \| Profound mental damage \| Woodcock–Johnson Test of Cognitive Abilities [edit] Main article: Woodcock–Johnson Tests of Cognitive Abilities The Woodcock–Johnson a III NU Tests of Cognitive Abilities (WJ III NU) was developed by Richard W. Woodcock, Kevin S. McGrew and Nancy Mather and published in 2007 by Riverside. The WJ III classification terms are not applied. Woodcock–Johnson R \| IQ Score \| WJ III Classification \| \| 131 and above \| Very superior \| \| 121 to 130 \| Superior \| \| 111 to 120 \| High Average \| \| 90 to 110 \| Average \| \| 80 to 89 \| Low Average \| \| 70 to 79 \| Low \| \| 69 and below \| Very Low \| Kaufman Tests [edit] The Kaufman Adolescent and Adult Intelligence Test was developed by Alan S. Kaufman and Nadeen L. Kaufman and published in 1993 by American Guidance Service. Kaufman test scores "are classified in a symmetrical, nonevaluative fashion", in other words the score ranges for classification are just as wide above the mean as below the mean, and the classification labels do not purport to assess individuals. KAIT 1993 IQ classification \| 130 and above \| Upper Extreme \| \| 120–129 \| Well Above Average \| \| 110–119 \| Above average \| \| 90–109 \| Average \| \| 80–89 \| Below Average \| \| 70–79 \| Well Below Average \| \| 69 and below \| Lower Extreme \| Main article: Kaufman Assessment Battery for Children The Kaufman Assessment Battery for Children, Second Edition was developed by Alan S. Kaufman and Nadeen L. Kaufman and published in 2004 by American Guidance Service. KABC-II 2004 Descriptive Categories \| Range of Standard Scores \| Name of Category \| \| 131–160 \| Upper Extreme \| \| 116–130 \| Above Average \| \| 85–115 \| Average Range \| \| 70–84 \| Below Average \| \| 40–69 \| Lower Extreme \| Cognitive Assessment System [edit] Main article: Cognitive Assessment System The Das-Naglieri Cognitive Assessment System test was developed by Jack Naglieri and J. P. Das and published in 1997 by Riverside. Cognitive Assessment System 1997 full scale score classification \| Standard Scores \| Classification \| \| 130 and above \| Very Superior \| \| 120–129 \| Superior \| \| 110–119 \| High Average \| \| 90–109 \| Average \| \| 80–89 \| Low Average \| \| 70–79 \| Below Average \| \| 69 and below \| Well Below Average \| Differential Ability Scales [edit] Main article: Differential Ability Scales The Differential Ability Scales Second Edition (DAS–II) was developed by Colin D. Elliott and published in 2007 by Psychological Corporation. The DAS-II is a test battery given individually to children, normed for children from ages two years and six months through seventeen years and eleven months. It was normed on 3,480 noninstitutionalized, English-speaking children in that age range. The DAS-II yields a General Conceptual Ability (GCA) score scaled like an IQ score with the mean standard score set at 100 and 15 standard score points for each standard deviation up or down from the mean. The lowest possible GCA score on DAS–II is 30, and the highest is 170. DAS-II 2007 GCA classification \| GCA \| General Conceptual Ability Classification \| \| ≥ 130 \| Very high \| \| 120–129 \| High \| \| 110–119 \| Above average \| \| 90–109 \| Average \| \| 80–89 \| Below average \| \| 70–79 \| Low \| \| ≤ 69 \| Very low \| Reynolds Intellectual Ability Scales [edit] Reynolds Intellectual Ability Scales (RIAS) were developed by Cecil Reynolds and Randy Kamphaus. The RIAS was published in 2003 by Psychological Assessment Resources. RIAS 2003 Scheme of Verbal Descriptors of Intelligence Test Performance \| Intelligence test score range \| Verbal descriptor \| \| ≥ 130 \| Significantly above average \| \| 120–129 \| Moderately above average \| \| 110–119 \| Above average \| \| 90–109 \| Average \| \| 80–89 \| Below average \| \| 70–79 \| Moderately below average \| \| ≤ 69 \| Significantly below average \| Historical IQ classification tables [edit] Lewis Terman, developer of the Stanford–Binet Intelligence Scales, based his English-language Stanford–Binet IQ test on the French-language Binet–Simon test developed by Alfred Binet. Terman believed his test measured the "general intelligence" construct advocated by Charles Spearman (1904). Terman differed from Binet in reporting scores on his test in the form of intelligence quotient ("mental age" divided by chronological age) scores after the 1912 suggestion of German psychologist William Stern. Terman chose the category names for score levels on the Stanford–Binet test. When he first chose classification for score levels, he relied partly on the usage of earlier authors who wrote, before the existence of IQ tests, on topics such as individuals unable to care for themselves in independent adult life. Terman's first version of the Stanford–Binet was based on norming samples that included only white, American-born subjects, mostly from California, Nevada, and Oregon. Terman's Stanford–Binet original (1916) classification \| IQ Range ("ratio IQ") \| IQ Classification \| \| Above 140 \| "Near" genius or genius \| \| 120–140 \| Very superior intelligence \| \| 110–120 \| Superior intelligence \| \| 90–110 \| Normal, or average, intelligence \| \| 80–90 \| Dullness, rarely classifiable as feeble-mindedness \| \| 70–80 \| Borderline deficiency, sometimes classifiable as dullness, often as feeble-mindedness \| \| 69 and below \| Definite feeble-mindedness \| Rudolph Pintner proposed a set of classification terms in his 1923 book Intelligence Testing: Methods and Results. Pintner commented that psychologists of his era, including Terman, went about "the measurement of an individual's general ability without waiting for an adequate psychological definition." Pintner retained these terms in the 1931 second edition of his book. Pintner 1923 IQ classification \| IQ Range ("ratio IQ") \| IQ Classification \| \| 130 and above \| Very Superior \| \| 120–129 \| Very Bright \| \| 110–119 \| Bright \| \| 90–109 \| Normal \| \| 80–89 \| Backward \| \| 70–79 \| Borderline \| Albert Julius Levine and Louis Marks proposed a broader set of categories in their 1928 book Testing Intelligence and Achievement. Some of the entries came from contemporary terms for people with intellectual disability. Levine and Marks 1928 IQ classification \| IQ Range ("ratio IQ") \| IQ Classification \| \| 175 and over \| Precocious \| \| 150–174 \| Very superior \| \| 125–149 \| Superior \| \| 115–124 \| Very bright \| \| 105–114 \| Bright \| \| 95–104 \| Average \| \| 85–94 \| Dull \| \| 75–84 \| Borderline \| \| 50–74 \| Morons \| \| 25–49 \| Imbeciles \| \| 0–24 \| Idiots \| The second revision (1937) of the Stanford–Binet test retained "quotient IQ" scoring, despite earlier criticism of that method of reporting IQ test standard scores. The term "genius" was no longer used for any IQ score range. The second revision was normed only on children and adolescents (no adults), and only "American-born white children". Terman's Stanford–Binet Second Revision (1937) classification \| IQ Range ("ratio IQ") \| IQ Classification \| \| 140 and above \| Very superior \| \| 120–139 \| Superior \| \| 110–119 \| High average \| \| 90–109 \| Normal or average \| \| 80–89 \| Low average \| \| 70–79 \| Borderline defective \| \| 69 and below \| Mentally defective \| A data table published later as part of the manual for the 1960 Third Revision (Form L-M) of the Stanford–Binet test reported score distributions from the 1937 second revision standardization group. Score Distribution of Stanford–Binet 1937 Standardization Group \| IQ Range ("ratio IQ") \| Percent of Group \| \| 160–169 \| 0.03 \| \| 150–159 \| 0.2 \| \| 140–149 \| 1.1 \| \| 130–139 \| 3.1 \| \| 120–129 \| 8.2 \| \| 110–119 \| 18.1 \| \| 100–109 \| 23.5 \| \| 90–99 \| 23.0 \| \| 80–89 \| 14.5 \| \| 70–79 \| 5.6 \| \| 60–69 \| 2.0 \| \| 50–59 \| 0.4 \| \| 40–49 \| 0.2 \| \| 30–39 \| 0.03 \| David Wechsler, developer of the Wechsler–Bellevue Scale of 1939 (which was later developed into the Wechsler Adult Intelligence Scale) popularized the use of "deviation IQs" as standard scores of IQ tests rather than the "quotient IQs" ("mental age" divided by "chronological age") then used for the Stanford–Binet test. He devoted a whole chapter in his book The Measurement of Adult Intelligence to the topic of IQ classification and proposed different category names from those used by Lewis Terman. Wechsler also criticized the practice of earlier authors who published IQ classification tables without specifying which IQ test was used to obtain the scores reported in the tables. Wechsler–Bellevue 1939 IQ classification \| IQ Range ("deviation IQ") \| IQ Classification \| Percent Included \| \| 128 and over \| Very Superior \| 2.2 \| \| 120–127 \| Superior \| 6.7 \| \| 111–119 \| Bright Normal \| 16.1 \| \| 91–110 \| Average \| 50.0 \| \| 80–90 \| Dull normal \| 16.1 \| \| 66–79 \| Borderline \| 6.7 \| \| 65 and below \| Defective \| 2.2 \| In 1958, Wechsler published another edition of his book Measurement and Appraisal of Adult Intelligence. He revised his chapter on the topic of IQ classification and commented that "mental age" scores were not a more valid way to score intelligence tests than IQ scores. He continued to use the same classification terms. Wechsler Adult Intelligence Scales 1958 Classification \| IQ Range ("deviation IQ") \| IQ Classification \| (Theoretical) Percent Included \| \| 128 and over \| Very Superior \| 2.2 \| \| 120–127 \| Superior \| 6.7 \| \| 111–119 \| Bright Normal \| 16.1 \| \| 91–110 \| Average \| 50.0 \| \| 80–90 \| Dull normal \| 16.1 \| \| 66–79 \| Borderline \| 6.7 \| \| 65 and below \| Defective \| 2.2 \| The third revision (Form L-M) in 1960 of the Stanford–Binet IQ test used the deviation scoring pioneered by David Wechsler. For rough comparability of scores between the second and third revision of the Stanford–Binet test, scoring table author Samuel Pinneau set 100 for the median standard score level and 16 standard score points for each standard deviation above or below that level. The highest score obtainable by direct look-up from the standard scoring tables (based on norms from the 1930s) was IQ 171 at various chronological ages from three years six months (with a test raw score "mental age" of six years and two months) up to age six years and three months (with a test raw score "mental age" of ten years and three months). The classification for Stanford–Binet L-M scores does not include terms such as "exceptionally gifted" and "profoundly gifted" in the test manual itself. David Freides, reviewing the Stanford–Binet Third Revision in 1970 for the Buros Seventh Mental Measurements Yearbook (published in 1972), commented that the test was obsolete by that year. Terman's Stanford–Binet Third Revision (Form L-M) classification \| IQ Range ("deviation IQ") \| IQ Classification \| \| 140 and above \| Very superior \| \| 120–139 \| Superior \| \| 110–119 \| High average \| \| 90–109 \| Normal or average \| \| 80–89 \| Low average \| \| 70–79 \| Borderline defective \| \| 69 and below \| Mentally defective \| The first edition of the Woodcock–Johnson Tests of Cognitive Abilities was published by Riverside in 1977. The classifications used by the WJ-R Cog were "modern in that they describe levels of performance as opposed to offering a diagnosis." Woodcock–Johnson R \| IQ Score \| WJ-R Cog 1977 Classification \| \| 131 and above \| Very superior \| \| 121 to 130 \| Superior \| \| 111 to 120 \| High Average \| \| 90 to 110 \| Average \| \| 80 to 89 \| Low Average \| \| 70 to 79 \| Low \| \| 69 and below \| Very Low \| The revised version of the Wechsler Adult Intelligence Scale (the WAIS-R) was developed by David Wechsler and published by Psychological Corporation in 1981. Wechsler changed a few of the boundaries for classification categories and a few of their names compared to the 1958 version of the test. The test's manual included information about how the actual percentage of people in the norming sample scoring at various levels compared to theoretical expectations. Wechsler Adult Intelligence Scales 1981 Classification \| IQ Range ("deviation IQ") \| IQ Classification \| Actual Percent Included \| Theoretical Percent Included \| \| 130 and above \| Very Superior \| 2.6 \| 2.2 \| \| 120–129 \| Superior \| 6.9 \| 6.7 \| \| 110–119 \| High Average \| 16.6 \| 16.1 \| \| 90–109 \| Average \| 49.1 \| 50.0 \| \| 80–89 \| Low Average \| 16.1 \| 16.1 \| \| 70–79 \| Borderline \| 6.4 \| 6.7 \| \| 69 and below \| Mentally Retarded \| 2.3 \| 2.2 \| The Kaufman Assessment Battery for Children (K-ABC) was developed by Alan S. Kaufman and Nadeen L. Kaufman and published in 1983 by American Guidance Service. K-ABC 1983 Ability Classifications \| Range of Standard Scores \| Name of Category \| Percent of Norm Sample \| Theoretical Percent Included \| \| 130 and above \| Upper Extreme \| 2.3 \| 2.2 \| \| 120–129 \| Well Above Average \| 7.4 \| 6.7 \| \| 110–119 \| Above Average \| 16.7 \| 16.1 \| \| 90–109 \| Average \| 49.5 \| 50.0 \| \| 80–89 \| Below Average \| 16.1 \| 16.1 \| \| 70–79 \| Well Below Average \| 6.1 \| 6.7 \| \| 69 and below \| Lower Extreme \| 2.1 \| 2.2 \| The fourth revision of the Stanford–Binet scales (S-B IV) was developed by Thorndike, Hagen, and Sattler and published by Riverside Publishing in 1986. It retained the deviation scoring of the third revision with each standard deviation from the mean being defined as a 16 IQ point difference. The S-B IV adopted new classification terminology. After this test was published, psychologist Nathan Brody lamented that IQ tests had still not caught up with advances in research on human intelligence during the twentieth century. Stanford–Binet Intelligence Scale, Fourth Edition (S-B IV) 1986 classification \| IQ Range ("deviation IQ") \| IQ Classification \| \| 132 and above \| Very superior \| \| 121–131 \| Superior \| \| 111–120 \| High average \| \| 89–110 \| Average \| \| 79–88 \| Low average \| \| 68–78 \| Slow learner \| \| 67 or below \| Mentally retarded \| The third edition of the Wechsler Adult Intelligence Scale (WAIS-III) used different classification terminology from the earliest versions of Wechsler tests. Wechsler (WAIS–III) 1997 IQ test classification \| IQ Range ("deviation IQ") \| IQ Classification \| \| 130 and above \| Very superior \| \| 120–129 \| Superior \| \| 110–119 \| High average \| \| 90–109 \| Average \| \| 80–89 \| Low average \| \| 70–79 \| Borderline \| \| 69 and below \| Extremely low \| Classification of low IQ [edit] Main article: Intellectual disability See also: Borderline intellectual functioning The earliest terms for classifying individuals of low intelligence were medical or legal terms that preceded the development of IQ testing. The legal system recognized a concept of some individuals being so cognitively impaired that they were not responsible for criminal behavior. Medical doctors sometimes encountered adult patients who could not live independently, being unable to take care of their own daily living needs. Various terms were used to attempt to classify individuals with varying degrees of intellectual disability. Many of the earliest terms are now considered extremely offensive. In modern medical diagnosis, IQ scores alone are not conclusive for a finding of intellectual disability. Recently adopted diagnostic standards place the major emphasis on the adaptive behavior of each individual, with IQ score a factor in diagnosis in addition to adaptive behavior scales. Some advocate for no category of intellectual disability to be defined primarily by IQ scores. Psychologists point out that evidence from IQ testing should always be used with other assessment evidence in mind: "In the end, any and all interpretations of test performance gain diagnostic meaning when they are corroborated by other data sources and when they are empirically or logically related to the area or areas of difficulty specified in the referral." In the United States, the Supreme Court ruled in the case Atkins v. Virginia, 536 U.S. 304 (2002) that states could not impose capital punishment on people with "mental retardation", defined in subsequent cases as people with IQ scores below 70.[citation needed] This legal standard continues to be actively litigated in capital cases. Historical [edit] Historically, terms for intellectual disability eventually became perceived as an insult, in a process commonly known as the euphemism treadmill. The terms mental retardation and mentally retarded became popular in the middle of the 20th century to replace the previous set of terms, which included "imbecile", "idiot", "feeble-minded", and "moron", among others. By the end of the 20th century, retardation and retard became widely seen as disparaging and politically incorrect, although they are still used in some clinical contexts. The long-defunct American Association for the Study of the Feeble-minded divided adults with intellectual deficits into three categories in 1916: Idiot indicated the greatest degree of intellectual disability in which a person's mental age is below three years. Imbecile indicated an intellectual disability less severe than idiocy and a mental age between three and seven years. Moron was defined as someone a mental age between eight and twelve. Alternative definitions of these terms based on IQ were also used.[citation needed] Mongolism and Mongoloid idiot were terms used to identify someone with Down syndrome, as the doctor who first described the syndrome, John Langdon Down, believed that children with Down syndrome shared facial similarities with the now-obsolete category of "Mongolian race". The Mongolian People's Republic requested that the medical community cease the use of the term; in 1960, the World Health Organization agreed the term should cease being used. Retarded comes from the Latin retardare, 'to make slow, delay, keep back, or hinder', so mental retardation meant the same as mentally delayed. The first record of retarded in relation to being mentally slow was in 1895. The term mentally retarded was used to replace terms like idiot, moron, and imbecile because retarded was not then a derogatory term. By the 1960s, however, the term had taken on a partially derogatory meaning. The noun retard is particularly seen as pejorative; a BBC survey in 2003 ranked it as the most offensive disability-related word. The terms mentally retarded and mental retardation are still fairly common, but organizations such as the Special Olympics and Best Buddies are striving to eliminate their use and often refer to retard and its variants as the "r-word". These efforts resulted in U.S. federal legislation, known as Rosa's Law, which replaced the term mentally retarded with the term intellectual disability in federal law. Classification of high IQ [edit] Genius [edit] Main article: Genius Francis Galton (1822–1911) was a pioneer in investigating both eminent human achievement and mental testing. In his book Hereditary Genius, written before the development of IQ testing, he proposed that hereditary influences on eminent achievement are strong, and that eminence is rare in the general population. Lewis Terman chose "'near' genius or genius" as the classification label for the highest classification on his 1916 version of the Stanford–Binet test. By 1926, Terman began publishing about a longitudinal study of California schoolchildren who were referred for IQ testing by their schoolteachers, called Genetic Studies of Genius, which he conducted for the rest of his life. Catherine M. Cox, a colleague of Terman's, wrote a whole book, The Early Mental Traits of 300 Geniuses, published as volume 2 of The Genetic Studies of Genius book series, in which she analyzed biographical data about historic geniuses. Although her estimates of childhood IQ scores of historical figures who never took IQ tests have been criticized on methodological grounds, Cox's study was thorough in finding out what else matters besides IQ in becoming a genius. By the 1937 second revision of the Stanford–Binet test, Terman no longer used the term "genius" as an IQ classification, nor has any subsequent IQ test. In 1939, Wechsler wrote "we are rather hesitant about calling a person a genius on the basis of a single intelligence test score." The Terman longitudinal study in California eventually provided historical evidence on how genius is related to IQ scores. Many California pupils were recommended for the study by schoolteachers. Two pupils who were tested but rejected for inclusion in the study because their IQ scores were too low, grew up to be Nobel Prize winners in physics: William Shockley and Luis Walter Alvarez. Based on the historical findings of the Terman study and on biographical examples such as Richard Feynman, who had an IQ of 125 and went on to win the Nobel Prize in physics and become widely known as a genius, the view of psychologists and other scholars[who?] of genius is that a minimum IQ, about 125, is strictly necessary for genius,[citation needed] but that IQ is sufficient for the development of genius only when combined with the other influences identified by Cox's biographical study: an opportunity for talent development along with the characteristics of drive and persistence. Charles Spearman, bearing in mind the influential theory that he originated—that intelligence comprises both a "general factor" and "special factors" more specific to particular mental tasks—wrote in 1927, "Every normal man, woman, and child is, then, a genius at something, as well as an idiot at something." Giftedness [edit] Main article: Intellectual giftedness A major point of consensus among all scholars of intellectual giftedness is that there is no generally agreed upon definition of giftedness. Although there is no scholarly agreement about identifying gifted learners, there is a de facto reliance on IQ scores for identifying participants in school gifted education programs. In practice, many school districts in the United States use an IQ score of 130, including roughly the upper 2 to 3 percent of the national population as a cut-off score for inclusion in school gifted programs. Five levels of giftedness have been suggested to differentiate the vast difference in abilities that exists between children on varying ends of the gifted spectrum. Although there is no strong consensus on the validity of these quantifiers, they are accepted by many experts of gifted children. Levels of Giftedness (M.U. Gross) \| Classification \| IQ Range \| σ \| Prevalence \| \| Mildly gifted \| 115–129 \| +1.00–+1.99 \| 1:6–1:44 \| \| Moderately gifted \| 130–144 \| +2.00–+2.99 \| 1:44–1:1,000 \| \| Highly gifted \| 145–159 \| +3.00–+3.99 \| 1:1,000–1:10,000 \| \| Exceptionally gifted \| 160–179 \| +4.00–+5.33 \| 1:10,000–1:1,000,000 \| \| Profoundly gifted \| 180– \| +5.33– \| < 1:1,000,000 \| As long ago as 1937, Lewis Terman pointed out that error of estimation in IQ scoring increases as IQ score increases, so that there is less and less certainty about assigning a test-taker to one band of scores or another as one looks at higher bands. Modern IQ tests also have large error bands for high IQ scores. As an underlying reality, such distinctions as those between "exceptionally gifted" and "profoundly gifted" have never been well established. All longitudinal studies of IQ have shown that test-takers can bounce up and down in score, and thus switch up and down in rank order as compared to one another, over the course of childhood. IQ classification categories such as "profoundly gifted" are those based on the obsolete Stanford–Binet Third Revision (Form L-M) test. The highest reported standard score for most IQ tests is IQ 160, approximately the 99.997th percentile. IQ scores above this level have wider error ranges as there are fewer normative cases at this level of intelligence. Moreover, there has never been any validation of the Stanford–Binet L-M on adult populations, and there is no trace of such terminology in the writings of Lewis Terman. Although two contemporary tests attempt to provide "extended norms" that allow for classification of different levels of giftedness, those norms are not based on well validated data. See also [edit] Cattell–Horn–Carroll theory Creativity and intelligence Dyscalculia Dysgraphia Educational psychology Evolution of human intelligence Fluid and crystallized intelligence Heritability of IQ High-IQ society Learning disability Psychometrics Savant syndrome Spearman's hypothesis Standardized test References [edit] ^ Wechsler 1958, Chapter 3: The Classification of Intelligence ^ Matarazzo 1972, Chapter 5: The Classification of Intelligence ^ Gregory 1995, entry "Classification of Intelligence" ^ Jump up to: a b c Kamphaus 2005, pp. 518–20 section "Score Classification Schemes" ^ Gottfredson 2009, pp. 31–32 ^ Hunt 2011, p. 5 "As mental testing expanded to the evaluation of adolescents and adults, however, there was a need for a measure of intelligence that did not depend upon mental age. Accordingly the intelligence quotient (IQ) was developed. ... The narrow definition of IQ is a score on an intelligence test ... where 'average' intelligence, that is the median level of performance on an intelligence test, receives a score of 100, and other scores are assigned so that the scores are distributed normally about 100, with a standard deviation of 15. Some of the implications are that: 1. Approximately two-thirds of all scores lie between 85 and 115. 2. Five percent (1/20) of all scores are above 125, and one percent (1/100) are above 135. Similarly, five percent are below 75 and one percent below 65." ^ Aiken 1979, p. 139 ^ Anastasi & Urbina 1997, p. 326 "Correlation studies of test scores provide actuarial data, applicable to group predictions. ... Studies of individuals, on the other hand, may reveal large upward or downward shifts in test scores." ^ Kaufman 2009, pp. 151–153 "Thus, even for tests that measure similar CHC constructs and that represent the most sophisticated, high–quality IQ tests ever available at any point in time, IQs differ." ^ Jump up to: a b Terman 1916, p. 79 "What do the above IQ's imply in such terms as feeble-mindedness, border-line intelligence, dullness, normality, superior intelligence, genius, etc.? When we use these terms two facts must be borne in mind: (1) That the boundary lines between such groups are absolutely arbitrary, a matter of definition only; and (2) that the individuals comprising one of the groups do not make up a homogeneous type." ^ Jump up to: a b Wechsler 1939, p. 37 "The earliest classifications of intelligence were very rough ones. To a large extent they were practical attempts to define various patterns of behavior in medical-legal terms." ^ Kaufman 2009, Figure 5.1 IQs earned by preadolescents (ages 12–13) who were given three different IQ tests in the early 2000s ^ Kaufman 2013, Figure 3.1 "Source: A. S. Kaufman. IQ Testing 101 (New York: Springer, 2009). Adapted with permission." ^ Mackintosh 2011, p. 169 "after the age of 8–10, IQ scores remain relatively stable: the correlation between IQ scores from age 8 to 18 and IQ at age 40 is over 0.70." ^ Uzieblo et al. 2012, p. 34 "Despite the increasing disparity between total test scores across intelligence batteries—as the expanding factor structures cover an increasing amount of cognitive abilities (Flanagan, et al., 2010)—Floyd et al. (2008) noted that still 25% of assessed individuals will obtain a 10-point IQ score difference with another IQ battery. Even though not all studies indicate significant discrepancies between intelligence batteries at the group level (e.g., Thompson et al., 1997), the absence of differences at the individual level cannot be automatically assumed." ^ Ryan, Joseph J.; Glass, Laura A.; Bartels, Jared M. (2010-02-10). "Stability of the WISC-IV in a Sample of Elementary and Middle School Children". Applied Neuropsychology. 17 (1): 68–72. doi:10.1080/09084280903297933. ISSN 0908-4282. PMID 20146124. S2CID 205615200. ^ Shurkin 1992, pp. 89–90 (citing Burks, Jensen & Terman, The Promise of Youth: Follow–up Studies of a Thousand Gifted Children 1930) "Twelve even dropped below the minimum for the Terman study, and one girl fell below 104, barely above average for the general population. ... Interestingly, while his tests measured decreases in test scores, the parents of the children noted no changes at all. Of all the parents who filled out the home questionnaire, 45 percent perceived no change in their children, 54 percent thought their children were getting brighter, including the children whose scores actually dropped." ^ Sattler 2008, p. 121 "Whenever you report an overall standard score (e.g., a Full Scale IQ or a similar standard score), accompany it with a confidence interval (see Chapter 4). The confidence interval is a function of both the standard error of measurement and the confidence level: the greater the confidence level (e.g., 99% > 95% > 90% > 85% > 68%) or the lower the reliablility of the test (rxx = .80 < rxx = .85 < rxx = .90), the wider the confidence interval. Psychologists usually use a confidence interval of 95%." ^ Matarazzo 1972, p. 121 "The psychologist's effort at classifying intelligence utilizes, at present, an ordinal scale, and is akin to what a layman does when he tries to distinguish colors of the rainbow." (emphasis in original) ^ Gottfredson 2009, pp. 32–33 "We cannot be sure that IQ tests provide interval–level measurement rather than just ordinal–level (i.e., rank–order) measurement. ... we really do not know whether a 10–point difference measures the same intellectual difference at all ranges of IQ." ^ Mackintosh 2011, pp. 33–34 "Although many psychometricians have argued otherwise (e.g., Jensen 1980), it is not immediately obvious that IQ is even an interval scale, that is, one where, say, the ten–point difference between IQ scores of 110 and 100 is the same as the ten–point difference between IQs of 160 and 150. The most conservative view would be that IQ is simply an ordinal scale: to say that someone has an IQ of 130 is simply to say that their test score lies within the top 2.5% of a representative sample of people the same age." ^ Jensen 2011, p. 172 "The problem with IQ tests and virtually all other scales of mental ability in popular use is that the scores they yield are only ordinal (i.e., rank-order) scales; they lack properties of true ratio scales, which are essential to the interpretation of the obtained measures." ^ Flynn 2012, p. 160 (quoting Jensen, 2011) ^ Kaufman & Lichtenberger 2006, pp. 198–202 (section "Scoring Errors") "Bias errors were in the direction of leniency for all subtests, with Comprehension producing the strongest halo effect." ^ Reynolds & Horton 2012, Table 4.1 Descriptions for Standard Score Performances Across Selected Pediatric Neuropsychology Tests ^ Aiken 1979, p. 158 ^ Sattler 1988, p. 736 ^ Sattler 2001, p. 698 "Tests usually provide some system by which to classify scores. Follow the specified classification system strictly, labeling scores according to what is recommended in the test manual. If you believe that a classification does not accurately reflect the examinee's status, state your concern in the report when you discuss the reliability and validity of the findings." ^ Gottfredson 2009, p. 32 "One searches in vain, for instance, for a good accounting of the capabilities that 10-year-olds, 15-year-olds, or adults of 110 usually possess but similarly aged individuals of IQ 90 do not ... IQ tests are not intended to isolate and measure highly specific skills and knowledge. This is the job of suitably designed achievement tests." ^ Kaufman & Lichtenberger 2006, p. 89 ^ Jump up to: a b c d e f g h i Urbina 2011, Table 2.1 Major Examples of Current Intelligence Tests ^ Flanagan & Harrison 2012, chapters 8-13, 15-16 (discussing Wechsler, Stanford–Binet, Kaufman, Woodcock–Johnson, DAS, CAS, and RIAS tests) ^ Mackintosh 2011, p. 32 "The most widely used individual IQ tests today are the Wechsler tests, first published in 1939 as the Wechsler–Bellevue Scale." ^ Saklofske et al. 2003, p. 3 "To this day, the Wechsler tests remain the most often used individually administered, standardized measures for assessing intelligence in children and adults" (citing Camara, Nathan & Puente, 2000; Prifitera, Weiss & Saklofske, 1998) ^ Georgas et al. 2003, p. xxv "The Wechsler tests are perhaps the most widely used intelligence tests in the world" ^ Meyer & Weaver 2005, p. 219 Campbell 2006, p. 66 Strauss, Sherman & Spreen 2006, p. 283 Foote 2007, p. 468 Kaufman & Lichtenberger 2006, p. 7 Hunt 2011, p. 12 ^ Carducci, Bernardo J.; Nave, Christopher S.; Fabio, Annamaria; Saklofske, Donald H.; Stough, Con, eds. (2020-09-18). The Wiley Encyclopedia of Personality and Individual Differences. doi:10.1002/9781119547174. ISBN 9781119057536. ^ Weiss et al. 2006, Table 5 Qualitative Descriptions of Composite Scores ^ Jump up to: a b c Sattler 2008, inside back cover ^ Kaufman, Alan S.; Engi Raiford, Susan; Coalson, Diane L. (2016). Intelligent Testing With the WISC-V. Hoboken, New Jersey: John Wiley & Sons. p. 237. ISBN 978-1-118-58923-6. ^ Kamphaus 2005, p. 519 "Although the Wechsler classification system for intelligence test scores is by far the most popular, it may not be the most appropriate (Reynolds & Kaufman 1990)." ^ Groth-Marnat 2009, p. 136 ^ Groth-Marnat 2009, Table 5.5 ^ Jump up to: a b Kaufman 2009, p. 112 ^ Jump up to: a b c Kamphaus 2005, p. 337 ^ Kamphaus 2005, pp. 367–68 ^ Kaufman et al. 2005, Table 3.1 Descriptive Category System ^ Gallagher & Sullivan 2011, p. 347 ^ Naglieri 1999, Table 4.1 Descriptive Categories of PASS and Full Scale Standard Scores ^ Dumont, Willis & Elliot 2009, p. 11 ^ Dumont, Willis & Elliot 2009, p. 20 ^ Dumont & Willis 2013, "Range of DAS Subtest Scaled Scores" (Web resource) ^ Dumont, Willis & Elliot 2009, Table Rapid Reference 5.1 DAS-II Classification Schema ^ Reynolds & Kamphaus 2003, p. 30 (Table 3.2 RIAS Scheme of Verbal Descriptors of Intelligence Test Performance) ^ Spearman 1904 ^ Wasserman 2012, pp. 19–20 "The scale does not pretend to measure the entire mentality of the subject, but only general intelligence. (citing Terman, 1916, p. 48, emphasis in original) ^ Wasserman 2012, p. 19 "No foreign-born or minority children were included. ... The overall sample was predominantly white, urban, and middle-class" ^ Jump up to: a b Terman 1916, p. 79 ^ Kaufman 2009, p. 110 ^ Naglieri 1999, p. 7 "The concept of general intelligence was assumed to exist, and psychologists went about 'the measurement of an individual's general ability without waiting for an adequate psychological definition.' (Pintner, 1923, p. 52)." ^ Pintner 1931, p. 117 ^ Jump up to: a b Levine & Marks 1928, p. 131 ^ Jump up to: a b Kamphaus et al. 2012, pp. 57–58 (citing Levine and Marks, page 131) ^ Wasserman 2012, p. 35 "Inexplicably, Terman and Merrill made the mistake of retaining a ratio IQ (i.e., mental age/chronological age) on the 1937 Stanford–Binet, even though the method had long been recognized as producing distorted IQ estimates for adolescents and adults (e.g., Otis, 1917). Terman and Merrill (1937, pp. 27–28) justified their decision on the dubious ground that it would have been too difficult to reeducate teachers and other test users familiar with ratio IQ." ^ Jump up to: a b c d Terman & Merrill 1960, p. 18 ^ Terman & Merrill 1937, p. 20 ^ Wasserman 2012, p. 35 "The 1939 test battery (and all subsequent Wechsler intelligence scales) also offered a deviation IQ, the index of intelligence based on statistical difference from the normative mean in standardized units, as Arthur Otis (1917) had proposed. Wechsler deserves credit for popularizing the deviation IQ, although the Otis Self-Administering Tests and the Otis Group Intelligence Scale had already used similar deviation-based composite scores in the 1920s." ^ Wechsler 1939, pp. 39–40 "We have seen equivalent Binet I.Q. ratings reported for nearly every intelligence test now in use. In most cases the reporters proceeded to interpret the I.Q.'s obtained as if the tests measured the same thing as the Binet, and the indices calculated were equivalent to those obtained on the Stanford–Binet. ... The examiners were seemingly unaware of the fact that identical I.Q.'s on the different tests might well represent very different orders of intelligence." ^ Wechsler 1958, pp. 42–43 "In brief, mental age is no more an absolute measure of intelligence than any other test score." ^ Wechsler 1958, p. 42 Table 3 Intelligence classification of WAIS IQ's ^ Terman & Merrill 1960, pp. 276–296 (scoring tables for 1960 Stanford–Binet) ^ Freides 1972, pp. 772–773 "My comments in 1970 [published in 1972] are not very different from those made by F. L. Wells 32 years ago in The 1938 Mental Measurements Yearbook. The Binet scales have been around for a long time and their faults are well known." ^ Jump up to: a b Gregory 1995, Table 4 Ability classifications, IQ ranges, and percent of norm sample for contemporary tests ^ Naglieri 1999, p. 7 "In fact, the stagnation of intelligence tests is apparent in Brody's (1992) statement: 'I do not believe that our intellectual progress has had a major impact on the development of tests of intelligence' (p. 355)." ^ Sattler 1988, Table BC-2 Classification Ratings on Stanford–Binet: Fourth Edition, Wechsler Scales, and McCarthy Scales ^ Kaufman 2009, p. 122 ^ American Psychiatric Association 2013, pp. 33–37 Intellectual Disability (Intellectual Development Disorder): Specifiers "The various levels of severity are defined on the basis of adaptive functioning, and not IQ scores, because it is adaptive functioning that determines the level of supports required. Moreover, IQ measures are less valid in the lower end of the IQ range." ^ Flanagan & Kaufman 2009, p. 134 (emphasis in original) ^ Flynn 2012, Chapter 4: Death, Memory, and Politics ^ Shaw, Steven R.; Anna M.; Jankowska (2018). Pediatric Intellectual Disabilities at School. Brooklyn, New York: Springer. p. 5. ISBN 978-3-030-02990-6. ^ Gernsbacher, Morton Ann; Raimond, Adam R.; Balinghasay, M. Theresa; Boston, Jilana S. (2016-12-19). ""Special needs" is an ineffective euphemism". Cognitive Research: Principles and Implications. 1 (1): 29. doi:10.1186/s41235-016-0025-4. ISSN 2365-7464. PMC 5256467. PMID 28133625. ^ Nash, Chris; Hawkins, Ann; Kawchuk, Janet; Shea, Sarah E (February 2012). "What's in a name? Attitudes surrounding the use of the term 'mental retardation'". Paediatrics & Child Health. 17 (2): 71–74. doi:10.1093/pch/17.2.71. ISSN 1205-7088. PMC 3299349. PMID 23372396. ^ Rafter, Nicole Hahn (1998). Creating Born Criminals. University of Illinois Press, ISBN 978-0-252-06741-9 ^ Cummings NA, Wright RH (2005). "Chapter 1, Psychology's surrender to political correctness". Destructive trends in mental health: the well-intentioned path to harm. New York: Routledge. ISBN 978-0-415-95086-2. ^ Treadway, Walter L. (1916). "The Feeble-Minded: Their Prevalence and Needs in the School Population of Arkansas". Public Health Reports. 31 (47): 3231–3247. doi:10.2307/4574285. hdl:2027/loc.ark:/13960/t4hm5zr5h. ISSN 0094-6214. JSTOR 4574285. S2CID 68261373. ^ Howard-Jones N (January 1979). "On the diagnostic term "Down's disease"". Medical History. 23 (1): 102–4. doi:10.1017/s0025727300051048. PMC 1082401. PMID 153994. ^ "Worst Word Vote". Ouch. BBC. 2003. Archived from the original on 2007-03-20. Retrieved 2007-08-17. ^ Rosa's Law, Pub. L. 111-256, 124 Stat. 2643 (2010). ^ "SpecialOlympics.org". SpecialOlympics.org. Archived from the original on 2010-07-30. Retrieved 2010-06-29. ^ Pintner 1931, pp. 356–357 "From a study of these boyhood records, estimates of the probable I.Q.s of these men in childhood have been made. ... It is of course obvious that much error may creep into an experiment of this sort, and the I.Q. assigned to any one individual is merely a rough estimate, depending to some extent upon how much information about his boyhood years has come down to us." ^ Shurkin 1992, pp. 70–71 "She, of course, was not measuring IQ, she was measuring the length of biographies in a book. Generally, the more information, the higher the IQ. Subjects were dragged down if there was little information about their early lives." ^ Eysenck 1995, p. 59 "Cox might well have been advised to reject a few of her geniuses for lack of evidence." Eysenck 1998, p. 126 "Cox found that the more was known about a person's youthful accomplishments, that is, what he had done before he was engaged in doing the things that made him known as a genius, the higher was his IQ ... So she proceeded to make a statistical correction in each case for lack of knowledge; this bumped up the figure considerably for the geniuses about whom little was in fact known. ... I am rather doubtful about the justification for making the correction." ^ Cox 1926, pp. 215–219, 218 (Chapter XIII: Conclusions) "3. That all equally intelligent children do not as adults achieve equal eminence is in part accounted for by our last conclusion: youths who achieve eminence are characterized not only by high intellectual traits, but also by persistence of motive and effort, confidence in their abilities, and great strength or force of character." (emphasis in original) ^ Kaufman 2009, p. 117 "Terman (1916), as I indicated, used near genius or genius for IQs above 140, but mostly very superior has been the label of choice" (emphasis in original) ^ Wechsler 1939, p. 45 ^ Eysenck 1998, pp. 127–128 "Terman, who originated those 'Genetic Studies of Genius', as he called them, selected ... children on the basis of their high IQs, the mean was 151 for both sexes. Seventy–seven who were tested with the newly translated and standardized Binet test had IQs of 170 or higher–well at or above the level of Cox's geniuses. What happened to these potential geniuses–did they revolutionize society? ... The answer in brief is that they did very well in terms of achievement, but none reached the Nobel Prize level, let alone that of genius. ... It seems clear that these data powerfully confirm the suspicion that intelligence is not a sufficient trait for truly creative achievement of the highest grade." ^ Simonton 1999, p. 4 "When Terman first used the IQ test to select a sample of child geniuses, he unknowingly excluded a special child whose IQ did not make the grade. Yet a few decades later that talent received the Nobel Prize in physics: William Shockley, the cocreator of the transistor. Ironically, not one of the more than 1,500 children who qualified according to his IQ criterion received so high an honor as adults." ^ Shurkin 2006, p. 13 (See also "The Truth About the 'Termites'"; Kaufman, S. B. 2009) ^ Leslie 2000. "We also know that two children who were tested but didn't make the cut -- William Shockley and Luis Alvarez -- went on to win the Nobel Prize in Physics. According to Hastorf, none of the Terman kids ever won a Nobel or Pulitzer." ^ Park, Lubinski & Benbow 2010. "There were two young boys, Luis Alvarez and William Shockley, who were among the many who took Terman's tests but missed the cutoff score. Despite their exclusion from a study of young 'geniuses,' both went on to study physics, earn PhDs, and win the Nobel prize." ^ Gleick 2011, p. 32 "Still, his score on the school IQ test was a merely respectable 125." ^ Robinson 2011, p. 47 "After all, the American physicist Richard Feynman is generally considered an almost archetypal late 20th-century genius, not just in the United States but wherever physics is studied. Yet, Feynman's school-measured IQ, reported by him as 125, was not especially high" ^ Spearman 1927, p. 221 ^ Sternberg, Jarvin & Grigorenko 2010, Chapter 2: Theories of Giftedness ^ McIntosh, Dixon & Pierson 2012, pp. 636–637 ^ Jump up to: a b Gross 2000, pp. 3–9 ^ Terman & Merrill 1937, p. 44 "The reader should not lose sight of the fact that a test with even a high reliability yields scores which have an appreciable probable error. The probable error in terms of mental age is of course larger with older than with young children because of the increasing spread of mental age as we go from younger to older groups. For this reason it has been customary to express the P.E. [probable error] of a Binet score in terms of I.Q., since the spread of Binet I.Q.'s is fairly constant from age to age. However, when our correlation arrays [between Form L and Form M] were plotted for separate age groups they were all discovered to be distinctly fan-shaped. Figure 3 is typical of the arrays at every age level. From Figure 3 it becomes clear that the probable error of an I.Q. score is not a constant amount, but a variable which increases as I.Q. increases. It has frequently been noted in the literature that gifted subjects show greater I.Q. fluctuation than do clinical cases with low I.Q.'s ... we now see that this trend is inherent in the I.Q. technique itself, and might have been predicted on logical grounds." ^ Lohman & Foley Nicpon 2012, Section "Conditional SEMs" "The concerns associated with SEMs [standard errors of measurement] are actually substantially worse for scores at the extremes of the distribution, especially when scores approach the maximum possible on a test ... when students answer most of the items correctly. In these cases, errors of measurement for scale scores will increase substantially at the extremes of the distribution. Commonly the SEM is from two to four times larger for very high scores than for scores near the mean (Lord, 1980)." ^ Lohman & Foley Nicpon 2012, Section "Scaling Issues" "The spreading out of scores for young children at the extremes of the ratio IQ scale is viewed as a positive attribute of the SB-LM by clinicians who want to distinguish among the highly and profoundly gifted (Silverman, 2009). Although spreading out the test scores in this way may be helpful, the corresponding normative scores (i.e., IQs) cannot be trusted both because they are based on out-of-date norms and because the spread of IQ scores is a necessary consequence of the way ratio IQs are constructed, not a fact of nature." ^ Hunt 2011, p. 8 ^ Perleth, Schatz & Mönks 2000, p. 301 "Norm tables that provide you with such extreme values are constructed on the basis of random extrapolation and smoothing but not on the basis of empirical data of representative samples." ^ Urbina 2011, Chapter 2: Tests of Intelligence. "[Curve-fitting] is just one of the reasons to be suspicious of reported IQ scores much higher than 160" ^ Lohman & Foley Nicpon 2012, Section "Scaling Issues" "Modern tests do not produce such high scores, in spite of heroic efforts to provide extended norms for both the Stanford Binet, Fifth Edition (SB-5) and the WISC-IV (Roid, 2003; Zhu, Clayton, Weiss, & Gabel, 2008)." Bibliography [edit] Aiken, Lewis (1979). Psychological Testing and Assessment (Third ed.). Boston: Allyn and Bacon. ISBN 978-0-205-06613-1. American Psychiatric Association (2013). Diagnostic and Statistical Manual of Mental Disorders (Fifth ed.). Arlington, VA: American Psychiatric Publishing. ISBN 978-0-89042-555-8. Anastasi, Anne; Urbina, Susana (1997). Psychological Testing (Seventh ed.). Upper Saddle River (NJ): Prentice Hall. ISBN 978-0-02-303085-7. Campbell, Jonathan M. (2006). "Chapter 3: Mental Retardation/Intellectual Disability". In Campbell, Jonathan M.; Kamphaus, Randy W. (eds.). 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Assessment of Children (Third ed.). San Diego (CA): Jerome M. Sattler, Publisher. ISBN 978-0-9618209-0-9. Sattler, Jerome M. (2001). Assessment of Children: Cognitive Applications (Fourth ed.). San Diego (CA): Jerome M. Sattler, Publisher. ISBN 978-0-9618209-7-8. Sattler, Jerome M. (2008). Assessment of Children: Cognitive Foundations. La Mesa (CA): Jerome M. Sattler, Publisher. ISBN 978-0-9702671-4-6. Shurkin, Joel (1992). Terman's Kids: The Groundbreaking Study of How the Gifted Grow Up. Boston (MA): Little, Brown. ISBN 978-0-316-78890-8. Shurkin, Joel (2006). Broken Genius: The Rise and Fall of William Shockley, Creator of the Electronic Age. Macmillan. ISBN 978-1-4039-8815-7. Simonton, Dean Keith (1999). Origins of genius: Darwinian perspectives on creativity. Oxford: Oxford University Press. ISBN 978-0-19-512879-6. Spearman, C. (April 1904). ""General Intelligence," Objectively Determined and Measured" (PDF). American Journal of Psychology. 15 (2): 201–292. doi:10.2307/1412107. JSTOR 1412107. Archived from the original (PDF) on 7 April 2014. Retrieved 31 May 2013. Spearman, Charles (1927). The Abilities of Man: Their Nature and Measurement. New York (NY): Macmillan. Sternberg, Robert J.; Jarvin, Linda; Grigorenko, Elena L. (2010). Explorations in Giftedness. Cambridge: Cambridge University Press. ISBN 978-0-521-74009-8. Strauss, Esther; Sherman, Elizabeth M.; Spreen, Otfried (2006). A Compendium of Neuropsychological Tests: Administration, Norms, and Commentary (Third ed.). Cambridge: Oxford University Press. ISBN 978-0-19-515957-8. Terman, Lewis M. (1916). The Measurement of Intelligence: An Explanation of and a Complete Guide to the Use of the Stanford Revision and Extension of the Binet–Simon Intelligence Scale. Riverside Textbooks in Education. Ellwood P. Cubberley (Editor's Introduction). Boston: Houghton Mifflin. Retrieved 26 June 2010. Terman, Lewis M.; Merrill, Maude (1937). Measuring Intelligence: A Guide to the Administration of the New Revised Stanford–Binet Tests of Intelligence. Boston: Houghton Mifflin. Terman, Lewis Madison; Merrill, Maude A. (1960). Stanford–Binet Intelligence Scale: Manual for the Third Revision Form L-M with Revised IQ Tables by Samuel R. Pinneau. Boston (MA): Houghton Mifflin. Urbina, Susana (2011). "Chapter 2: Tests of Intelligence". In Sternberg, Robert J.; Kaufman, Scott Barry (eds.). The Cambridge Handbook of Intelligence. Cambridge: Cambridge University Press. pp. 20–38. ISBN 978-0-521-73911-5. Uzieblo, Katarzyna; Winter, Jan; Vanderfaeillie, Johan; Rossi, Gina; Magez, Walter (2012). "Intelligent Diagnosing of Intellectual Disabilities in Offenders: Food for Thought" (PDF). Behavioral Sciences & the Law. 30 (1): 28–48. doi:10.1002/bsl.1990. PMID 22241548. Retrieved 15 July 2013. Wasserman, John D. (2012). "Chapter 1: A History of Intelligence Assessment". In Flanagan, Dawn P.; Harrison, Patti L. (eds.). Contemporary Intellectual Assessment: Theories, tests, and issues (Third ed.). New York (NY): Guilford Press. pp. 3–55. ISBN 978-1-60918-995-2. Wechsler, David (1939). The Measurement of Adult Intelligence (first ed.). Baltimore (MD): Williams & Witkins. LCCN 39014016. Wechsler, David (1958). The Measurement and Appraisal of Adult Intelligence (fourth ed.). Baltimore (MD): Williams & Witkins. Retrieved 4 June 2013. Weiss, Lawrence G.; Saklofske, Donald H.; Prifitera, Aurelio; Holdnack, James A., eds. (2006). WISC-IV Advanced Clinical Interpretation. Practical Resources for the Mental Health Professional. Burlington (MA): Academic Press. ISBN 978-0-12-088763-7. This practitioner's handbook includes chapters by L.G. Weiss, J.G. Harris, A. Prifitera, T. Courville, E. Rolfhus, D.H. Saklofske, J.A. Holdnack, D. Coalson, S.E. Raiford, D.M. Schwartz, P. Entwistle, V. L. Schwean, and T. Oakland. Further reading [edit] Gordon, Robert A. (1997). "Everyday life as an intelligence test: Effects of intelligence and intelligence context." Intelligence 24(1): 203-320. doi:10.1016/S0160-2896(97)90017-9 Gottfredson, Linda S. (1997). "Why g matters: The complexity of everyday life." Intelligence 24, 79–132. doi:10.1016/S0160-2896(97)90014-3 Cattell, Raymond (1987). Intelligence: Its Structure, Growth and Action. New York: North-Holland. 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